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APPLIED

MATHEMATICS

ATH. KEHAGIAS

THESSALONIKI 2019

DIFFERENTIAL EQUATIONS

AND INTEGRAL TRANSFORMS

Ath.

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Copyright 2019 Ath. Kehagias

Sofistis Publications

Free Usage

1st Edition, October 2019

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9/30/2019 The metamorphosis of Escher

https://escher.ntr.nl/en/eindeloos/p488bclEIDUBIcOUBNYw 1/1

Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . i

Notation and Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii

I Ordinary Differential Equations

1 First Order ODEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Linear ODEs: Solution Properties . . . . . . . . . . . . . . . . . . . . . . . . . 3

3 Linear ODEs: Solution Methods . . . . . . . . . . . . . . . . . . . . . . . . . . 11

4 Systems of Linear ODEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

5 Series Solutions of ODEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

II Laplace

6 Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

7 Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

8 Convolution and Integral Equations . . . . . . . . . . . . . . . . . . . . . 69

9 Dirac Delta and Generalized Functions . . . . . . . . . . . . . . . . . . 75

10 Difference Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

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III Fourier

11 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

12 Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

IV Partial Differential Equations

13 PDEs for Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

14 PDEs for Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

15 PDEs for Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

16 Bessel Functions and Applications . . . . . . . . . . . . . . . . . . . . . . 189

17 Vector Spaces of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

18 Sturm-Liouville Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

V Appendices

A Definitions of the Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231

B Distribution Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239

C Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241

D Numerical Solution of PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243

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Preface

I hear it, I forget it.I see it, I remember it.

I do it, I learn it.

In these notes we mainly study ordinary differential equations (ODEs) i.e.,equations involving derivatives with respect to a single variable, such as

dydx

+ y = et ;

and partial differential equations (PDEs) i.e., equations involving derivatives withrespect to more than one variable, such as

∂ 2u∂x2 +

∂ 2u∂y2 = 0.

In addition we study integral transforms, such as the Laplace and Fourier trans-forms, from a general point of view as well as in connection to the solution of ODEsand PDEs.

The notes are in an unfinished state and are known to contain some errors.

Ath. KehagiasThessaloniki, September 2019

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Notation and Preliminaries

1. We use the symbol := to denote a definition (of a function, set etc.).2. N is the set of positive integers 1,2,3, ... .3. N0 is the set of nonnegative integers 0,1,2,3, ... .4. Z is the set of integers ...,−2,−1,0,1,2, ... .5. R is the set of real numbers (−∞,+∞).6. R∗ is the extended set of real numbers [−∞,+∞].7. R+

0 is the set of nonnegative real numbers [0,+∞).8. R−0 is the set of nonpositive real numbers (−∞,0].9. C is the set of complex numbers (x+ iy) : x,y ∈ R.

10. Q[0,1] the set of rational numbers in [0,1].11. SL is the solution set y : L(y) = 0, where L is a linear differential operator.12. FL is the set of functions which satisfy the Dirichlet conditions in [−L,L].13. The notation (an)n∈A for a sequence (where A ∈ N) but also for a set. So we

write (with an abuse of notation) (an)n∈A ⊆ R. Similarly when ( fn)n∈A is asequence of functions, we will write (for example) ( fn)n∈A ⊆ CN .

14. 0(x) is the zero function, i.e.,

∀x : 0(x) := 0.

15. 1A (x) is the characteristic function of set A, i.e.,

1A (x) :=

1 x ∈ A0 x /∈ A .

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1 First Order ODEs . . . . . . . . . . . . . . . . . . 3

2 Linear ODEs: Solution Properties 3

3 Linear ODEs: Solution Methods . 11

4 Systems of Linear ODEs . . . . . . . . . 25

5 Series Solutions of ODEs . . . . . . . . 35

Ordinary DifferentialEquations

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1. First Order ODEs

Here we study differential equations of the form

dydx

= F (x,y) .

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1.1 Theory and Examples 1

1.1 Theory and Examples

1.2 Solved Problems

1.3 Unsolved Problems

1.4 Advanced Problems

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2. Linear ODEs: Solution Properties

Here we study differential equations of the form

dnydxn +an−1 (x)

dn−1ydxn−1 + ...+a1 (x)

dydx

+a0 (x)y = g(x) .

We focus on theoretical properties of the solutions y(x).

2.1 Theory and Examples2.1.1 Definition. An n-th order linear differential equation has the form

dnydxn +an−1 (x)

dn−1ydxn−1 + ...+a1 (x)

dydx

+a0 (x)y = g(x) . (2.1)

Sometimes we attach to (2.1) initial conditions

y(x0) = c0, y′ (x0) = c1, ..., y(n−1) (x0) = cn−1. (2.2)

If g(x) is identically zero, the DE is called homogeneous, otherwise nonhomoge-neous.

2.1.2 Example. This is a 2-nd order homogeneous linear differential equation

d2ydx2 +(1− x)

dydx

+ x2y = 0.

This is a 3-rd order nonhomogeneous linear differential equation

d3ydx3 +3

d2ydx2 +3

dydx

+ y = x.

2.1.3 Example. This is a 2-nd order nonhomogeneous linear differential equationwith initial conditions

d2ydx2 +3

dydx

+2y = 0

y(0) = 1y′ (0) = 1

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4 Chapter 2. Linear ODEs: Solution Properties

2.1.4 Theorem. If a0 (x), ..., an−1 (x) are continuous in some interval (A,B), (2.1)-(2.2) has a unique solution in (A,B), i.e., there exists a unique function y(x) whichsatisfies (2.1)-(2.2) for every x ∈ (A,B).Proof. We will only give a part of the proof. In particular, we will show that: whenn = 2, if (2.1)-(2.2) has a solution, it is unique1. Suppose u(x) and v(x) are twosolutions of (2.1)-(2.2) and let w(x) = u(x)− v(x). Then

w′′ (x)+a1 (x)w′ (x)+a0 (x)w(x) = 0, w(x) = w′ (x) = 0.

Also define z(x) = (w′ (x))2 +(w(x))2, then z(x0) = 0. Now choose any x1,x2 suchthat

x0 ∈ [x1,x2]⊂ (A,B) .

Then, for any x ∈ [x1,x2]

z′ (x) = 2w′ (x)w′′ (x)+2w(x)w′ (x)

=−2w′ (x)[a1 (x)w′ (x)+a0 (x)w(x)

]+2w(x)w′ (x)

=−2a1 (x)(w′ (x)

)2+2w(x)w′ (x)(1−a0 (x)) .

By continuity, there exists some M such that

∀x ∈ [x1,x2] : |a0 (x)|< M, |a1 (x)|< M.

Hence ∣∣z′ (x)∣∣≤ 2 |a1 (x)|(w′ (x)

)2+2 |w(x)|

∣∣w′ (x)∣∣ |1−a0 (x)|

≤ 2M(w′ (x)

)2+2(1+M) |w(x)|

∣∣w′ (x)∣∣≤ 2M

((w′ (x)

)2+(w(x))2

)+(1+M)

((w′ (x)

)2+(w(x))2

)= (1+3M)

((w′ (x)

)2+(w(x))2

)= (1+3M)z(x) .

Hence−(1+3M)z(x)≤ z′ (x)≤ (1+3M)z(x) .

Letting K = 1+3M we have:

z′ (x)−Kz(x)≤ 0⇒(z′ (x)−Kz(x)

)e−Kx ≤ 0⇒

(z(x)e−Kx)′ ≤ 0.

Hence z(x)e−Kx is non increasing, which means that(∀x ∈ [x0,x2] : 0≤ z(x)e−Kx ≤ z(x0)e−Kx = 0

)⇒ (∀x ∈ [x0,x2] : z(x) = 0) .

Similarly we have

z′ (x)+Kz(x)≥ 0⇒(z′ (x)+Kz(x)

)eKx ≥ 0⇒

(z(x)eKx)′ ≥ 0.

1This proof was given by B. Travis in “Uniqueness of Initial Value Problems”, DivulgacionesMatematicas vol. 5, No. 1/2 (1997), pp. 39–41. The full proof of the theorem (existence anduniqueness for any n ∈ N) is omitted because it requires more advanced concepts.

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Hence z(x)eKx is non decreasing, which means that(∀x ∈ [x1,x0] : 0≤ z(x)eKx ≤ z(x0)eKx = 0

)⇒ (∀x ∈ [x1,x0] : z(x) = 0) .

In short:

∀x ∈ [x1,x2]⊂ (A,B) :(w′ (x)

)2+(w(x))2 = 0⇒

∀x ∈ (A,B) : (u(x)− v(x))2 = 0⇒∀x ∈ (A,B) : u(x) = v(x) .

2.1.5 Definition. For any an−1 (x) , ...,a1 (x) ,a0 (x) we can define the n-th orderdifferential operator of L() as follows:

L() =dn

dxn +an−1 (x)dn−1

dxn−1 + ...+a1 (x)ddx

+a0 (x) .

2.1.6 What this means is that L() is a function which has as domain and rangefunction sets; it takes as input a function y(x) and produces as output anotherfunction

L(y) =dnydxn +an−1 (x)

dn−1ydxn−1 + ...+a1 (x)

dydx

+a0 (x)y.

Hence (2.1) can be rewritten as

L(y) = g(x) .

2.1.7 Example. Defining

L() =d2

dx2 +3ddx

+2

we can write the 2-nd order nonhomogeneous linear differential equation

d2ydx2 +3

dydx

+2y = 0

asL(y) = 0.

2.1.8 Theorem. The n-th order differential operator L() is linear, i.e.,

∀κ,λ ∈ C,∀u,v : L(κu+λv) = κL(u)+λL(v) .

Proof. This is obvious:

L(κu+λv) =dn

dxn (κu+λv)+ ...+a1 (x)ddx

(κu+λv)+a0 (x)(κu+λv)

= κ

(dnudxn + ...+a1 (x)

dudx

+a0 (x)u)+λ

(dnvdxn + ...+a1 (x)

dvdx

+a0 (x)v)

= κL(u)+λL(v) .

2.1.9 Definition. The set of functions y1 (x) , ...,yK (x) is called linearly indepen-dent on X (where X ⊆ R) iff

(∀x ∈ X : c1y1 (x)+ ...+ cKyK (x) = 0)⇒ (∀k : ck = 0) . (2.3)

The set is called linearly dependent on X iff (2.3) does not hold.

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2.1.10 Example. The set

1,x,x2 is lin. ind. on R because(∀x ∈ X : c01+ c1x+ c2x2 = 0

)⇒ c0 + c10+ c202 = 0

c0 + c11+ c212 = 0c0 + c12+ c222 = 0

⇒ c0 = c1 = c2 = 0

since

∣∣∣∣∣∣1 0 01 1 11 2 4

∣∣∣∣∣∣= 2 6= 0.

2.1.11 Example. The set 1,x,1+ x is lin. ind. on R because

∀x ∈ X : c11+ c2x+ c3 (1+ x) = 0

can be satrisfied with c1 = 1,c2 = 1,c3 =−1.

2.1.12 Definition. The Wronskian W (x|y1, ...,yK) of a function set y1 (x) , ...,yK (x)(we also write simply W (x)) is defined to be the determinant

W (x) =

∣∣∣∣∣∣∣∣y1 y2 ... yKy′1 y′2 ... y′K... ... ... ...

y(K−1)1 y(K−1)

2 ... y(K−1)K

∣∣∣∣∣∣∣∣ .2.1.13 Example. The Wronskian of 1,x,1+ x is

W (x) =

∣∣∣∣∣∣1 x 1+ x0 1 10 0 0

∣∣∣∣∣∣= 0.

2.1.14 Example. The Wronskian of

1,x,x2 is

W (x) =

∣∣∣∣∣∣1 x x2

0 1 2x0 0 2

∣∣∣∣∣∣= 2.

2.1.15 Example. The Wronskian of

cosx,sinx,eix is

W (x) =

∣∣∣∣∣∣cosx sinx eix

−sinx cosx ieix

−cosx −sinx −eix

∣∣∣∣∣∣= 0.

2.1.16 Theorem. Given [a,b] and L(y) = 0.1. If y1,y2, ...,yK are solutions of L(y) = 0 on [a,b] and

∀x ∈ [a,b] : W (x|y1, ...,yK) = 0

then the set y1,y2, ...,yK is lin. dep. on [a,b].

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2. If∃x ∈ [a,b] : W (x|y1, ...,yK) 6= 0

then the set y1,y2, ...,yK is lin. ind. on [a,b].

2.1.17 The terms linear operator”, “linearly independent” etc. remind us of LinearAlgebra. This is not accidental, there is a connection between linear DE’s andLinear Algebra, as will now become obvious.

2.1.18 Definition. Given the linear differential operator

L() =dn

dxn +an−1 (x)dn−1

dxn−1 + ...+a1 (x)ddx

+a0 (x)

we define the solution set of L by

SL := L(y) = 0 .

In other words, y ∈SL is equivalent to

dnydxn +an−1 (x)

dn−1ydxn−1 + ...+a1 (x)

dydx

+a0 (x)y = 0. (2.4)

2.1.19 Theorem. The set of solutions SL is a vector space.Proof. We need to prove:

∀κ,λ ∈ C,∀u,v ∈SL : κu+λv ∈SL.

Indeed:(u ∈SL⇒ L(u) = 0⇒ κL(u) = 0v ∈SL⇒ L(v) = 0⇒ λL(v) = 0

)⇒ κL(u)+λL(v)=L(κu+λv)= 0⇒ κu+λv∈SL.

2.1.20 We will next show that: if L is an n-th order differential operator, thenthe solution space SL has dimension n. To this end we need the following twostandard theorems of Linear Algebra2.

2.1.21 Theorem. If T is a linear transformation from V to W , then the followingstatements are equivalent.

1. T is one-to-one on V .2. T is invertible and T−1 is linear.3. ∀x ∈V : T (x) = 0⇒ x = 0.

2.1.22 Theorem. If T is a linear transformation from V to W and dim(V ) = n < ∞,then the following statements are equivalent.

1. T is one-to-one on V .2. If e1, ...,en is a linearly independent set, then T (e1) , ...,T (en) is a linearly

independent set.3. dim(T (V )) = n.4. If e1, ...,en is a basis of V , then T (e1) , ...,T (en) is a basis of T (V ).2For the proofs see, e.g., T. Apostol, Calculus, vol.2, Wiley, 1969.

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2.1.23 Theorem. Let L be an n-th order linear differential operator. The vectorspace SL = y : L(y) = 0 has dimension n.Proof. Let T be a linear transformation which maps SL to Cn by (for some fixedx0)

T (y) =[y(x0) ,y′ (x0) , ...,y(n−1) (x0)

].

(Why is it linear?) Then:1. by the uniqueness Theorem, T (y) = 0⇒ y(x) = 0(x);2. by Theorem 2.1.21 T is one-to-one on SL;3. by Theorem 2.1.22 dim(SL) = dim(Cn) = n.

2.1.24 Corollary. Let L be an n-th order linear differential operator. If y1, ...,ynis a linearly independent set of solutions of L(y) = 0, then it is a basis of SL =y : L(y) = 0. In other words, every solution y of L(y) = 0 can be written as

y = c1y1 + ...+ cnyn.

2.1.25 Hence, if we find n linearly independent solutions y1, ...,yn of

dnydxn +an−1 (x)

dn−1ydxn−1 + ...+a1 (x)

dydx

+a0 (x)y = 0. (2.5)

then every solution y of L(y) = 0 can be written as

y = c1y1 + ...+ cnyn. (2.6)

We call (2.6) the general solution of (2.5).

2.1.26 Example. The differential equation

d2ydx2 + y = 0

has two solutions y1 (x) = cosx,y2 (x) = sinx (check it!) which are form a lin.ind. seton R, since(

c1 cos0+ c2 sin0 = 0c1 cos π

2 + c2 sin π

2 = 0

)⇒(

c11+ c20 = 0c10+ c21 = 0

)⇒(

c1 = 0c2 = 0

).

Hence cosx,sinx is a basis of SL (where L = d2

dx2 + 1). Any other solution canbe written as a linear combination of cosx and sinx and, conversely, any linearcombination of cosx and sinx is a solution (why?).

The theorem tells us that the DE does not have a bigger and lin.ind. set ofsolutions. It does have more solutions, e.g., eix, e−ix are also solutions (check it!).In fact

eix,e−ix is another basis of SL (and so are

cosx,eix,

cosx,e−ix etc.).

But the set cosx,sinx,cosx+ sinx is a lin.dep. set (obviously) and the same istrue of

cosx,sinx,eix since

∀x ∈ R : cosx+ isinx− eix = 0.

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2.1.27 Example. Let SL be the set of solutions of

d2ydx2 +5

dydx

+6y = 0.

A basis of SL is

e−2x,e−3x. The general solution is

y(x) = c1e−2x + c2e−3x.

2.1.28 Definition. We say that y(x) is the general solution of L(y) = g(x) iff everysolution can be written in the form of y(x).

2.1.29 Theorem. Consider the nonhomogeneous DE

dnydxn +an−1 (x)

dn−1ydxn−1 + ...+a1 (x)

dydx

+a0 (x)y = g(x) (2.7)

which can also be written asL(y) = g(x) .

It has general solutiony(x) = yh (x)+ yp (x)

where1. yh (x) is the general solution of L(y) = 0,2. yp (x) is some solution of L(y) = g(x).

Proof. By assumption we have

L(yh) = 0,L(yp) = g(x) .

Adding these two together we get

L(yh)+L(yp) = g(x)⇒ L(yh + yp) = g(x) .

This shows that yh + yp is a solution L(y) = g(x). Why is it the general solution?In other words, how do you prove that every solution of L(y) = g(x) can be writtenin the form yh + yp?

2.1.30 There is an analogous theorem of Linear Algebra about the solutions ofAx = b. What does it say?

2.1.31 Example. The general solution of d2ydx2 + y = 1 is

y(x) = c1 cosx+ c2 sinx+1.

Here yh (x) = c1 cosx+ c2 sinx is the general solution of d2ydx2 + y = 0 and yp (x) = 1 is

a solution of d2ydx2 + y = 0.

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2.2 Solved Problems

2.3 Unsolved Problems1. Find the Wronskian of 3x,2x.

Ap.∣∣∣∣ 3x 2x

3 2

∣∣∣∣= 0.

2. Find the Wronskian of 1,2x.

Ap.∣∣∣∣ 1 2x

0 2

∣∣∣∣= 2.

3. Find the Wronskian of

1+ x+ x2,1−2x,x2.

Ap.

∣∣∣∣∣∣1+ x+ x2 1−2x x2

1+2x −2 2x2 0 2

∣∣∣∣∣∣=−6.

4. Find the Wronskian of ex,e−x,xex.

Ap.

∣∣∣∣∣∣ex e−x xex

ex −e−x ex (x+1)ex e−x ex (x+2)

∣∣∣∣∣∣=−4e2xe−x.

5. Is 3x,2x lin. ind.?Ap. No.

6. Is 1,2x lin. ind.?Ap. Yes.

7. Is

1+ x+ x2,1−2x,x2 lin. ind.?Ap. Yes.

8. Is ex,e−x,xex lin. ind.?Ap. Yes.

2.4 Advanced Problems1. Show that dx

dt = F (ax+by) can be converted to a separable DE by the changeof variable v = ax+bt.

2. Show that yF (xy)+ xG(xy)dy = 0 has the solution lnx =∫ G(z)

z(G(z)−F(z))dz+ c.

3. Show that dydx = F

( yx

)has the solution lnx =

∫ dzF(z)−z + c.

4. Show that every solution of dxdt = f (t,x), x(0) = c is also a solution of the

integral equation x(t) = c+∫ t

0 f (s,x(s))ds.5. The problem dx

dt = f (t,x), x(0) = c has solution x(t). Show that the functionsequence obtained from the iteration

x0 (t) = c, ∀n : xn+1 (t) = c+∫ t

0f (s,xn (s))ds

satisfies∀t : lim

n→∞xn (t) = x(t) .

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3. Linear ODEs: Solution Methods

Here we present methods which compute solutions of differential equations suchas

dnydxn +an−1 (x)

dn−1ydxn−1 + ...+a1 (x)

dydx

+a0 (x)y = g(x) ,

dnydxn +an−1

dn−1ydxn−1 + ...+a1

dydx

+a0y = g(x) .

We focus on theoretical properties of the solutions y(x).

3.1 Theory and Examples for Constant Coefficient DEs

3.1.1 Theorem. The 2nd order homogeneous linear differential equation withconstant coefficients

d2ydx2 +a1

dydx

+a0y = 0 (3.1)

has characteristic equation

r2 +a1r+a0 = 0

with roots r1,r2. The general solution of (3.1) has the form1. y(x) = c1er1x + c2er2x when r1 6= r2.2. y(x) = c1er1x + c2xer1x when r1 = r2.

Proof. First we check that ernx is a solution (for n ∈ 1,2)

d2

dx2 ernx +a1ddx

ernx +a0ernx = r2nernx +a1rnernx +a0ernx

=(r2

n +a1rn +a0)

ernx = 0.

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12 Chapter 3. Linear ODEs: Solution Methods

Then also the linear combination c1er1x + c2er2x is a solution

d2

dx2 (c1er1x + c2er2x)+a1ddx

(c1er1x + c2er2x)+a0 (c1er1x + c2er2x)

= c1

[d2

dx2 er1x +a1ddx

er1x +a01er1x]+ c2

[d2

dx2 er2x +a1ddx

er2x +a01er2x]

= c10+ c20 = 0.

Now, when r1 6= r2, it is easy to check that er1x,er2x is a lin.ind. set. But this isnot the case when r1 = r2. However, in that case xer1x is also a solution:

d2

dx2 (xer1x)+a1ddx

(xer1x)+a0xer1xernx

= a1exr1 (xr1 +1)+ r1exr1 (xr1 +2)+ xa0exr1

= a1exr1 (1)+ r1exr1 (2)

=(r2

1 +a1r1 +a0)

xer1x +(a1 +2r1)er1x = 0

since r1 = r2 implies that (a) a21−4a0 = 0 and (b) r1 = r2 =−a1

2 . Hence er1x,xer1xis a lin.ind. set of solutions of (3.1).

3.1.2 Example. The DEd2ydx2 +3

dydx

+2y = 0

has characteristic equationr2 +3r+2 = 0

with roots r1 = −1,r2 = −2. Hence two solutions are e−x,e−2x;

e−x,e−2x is abasis of the solution set; and every solution can be written in the form y(x) =c1e−x + c2e−2x.

3.1.3 Example. The DEd2ydx2 +2

dydx

+ y = 0

has characteristic equationr2 +2r+1 = 0

with roots r1 = r2 =−1. Hence two solutions are e−x,xe−x; e−x,xe−x is a basis ofthe solution set; and every solution can be written in the form y(x) = c1e−x+c2xe−x.

3.1.4 Example. The DEd2ydx2 +

dydx

+ y = 0

has characteristic equationr2 + r+1 = 0

with roots r1 =12 i√

3− 12 ,r2 =−1

2 i√

3− 12 . Hence two solutions are ex−1+i

√3

2 ,ex−1−i√

32

and every solution can be written in the form y(x) = c1ex−1+i√

32 +c2ex−1−i

√3

2 . However,note that

ex−1+i√

32 = e−

x2 ei

√3

2 x = e−x2

(cos

(√3

2x

)+ isin

(√3

2x

))= e−

x2 cos

(√3

2x

)+ie−

x2 sin

(√3

2x

).

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3.1 Theory and Examples for Constant Coefficient DEs 13

Similarly

ex−1−i√

32 = e−

x2 ei

√3

2 x = e−x2

(cos

(√3

2x

)− isin

(√3

2x

))= e−

x2 cos

(√3

2x

)−ie−

x2 sin

(√3

2x

).

Hence the general solution can be written as

y(x) = c1ex−1+i√

32 + c2ex−1−i

√3

2

= c1ex−1+i√

32 + c2ex−1−i

√3

2

= c1

(e−

x2 cos

(√3

2x

)+ ie−

x2 sin

(√3

2x

))+ c2

(e−

x2 cos

(√3

2x

)− ie−

x2 sin

(√3

2x

))

= (c1 + c2)e−x2 cos

(√3

2x

)+ i(c1− c2)e−

x2 sin

(√3

2x

)

= p1e−x2 cos

(√3

2x

)+ p2e−

x2 sin

(√3

2x

).

This (cosine/sine) form is the one we will prefer to use.

3.1.5 The above theorem generalizes to n-th order linear homogeneous DEs withconstant coefficients.

3.1.6 Theorem. The homogeneous linear differential equation with constantcoefficients

dnydxn +an−1

dn−1ydxn−1 + ...+a1

dydx

+a0y = 0 (3.2)


rn +an−1rn−1 + ...+a1r+a0 = 0. (3.3)

Suppose that (3.3) has roots r1 with multiplicity m1, ..., rK with multiplicity mK (wehave m1 +m2 + ...+mK = n). The general solution of (3.2) is

y(x) =K

∑k=1

mk

∑m=1

(ckmtm−1erkt) .

Proof. Left to the reader.

3.1.7 Example. The DE

d3ydx3 −6

d2ydx2 +11

dydx−6y = 0


r3−6r2 +11r−6 = 0

with roots, r1 = 1,r2 = 3,r3 = 2. Hence its general solution is

y(x) = c1ex + c2e2x + c3e3x

(check it!).

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3.1.8 Example. The DE

d3ydx3 +3

d2ydx2 +3

dydx

= y = 0


r3 +3r2 +3r+1 = 0

with roots, r1 = r2 = r3 =−1. Hence its general solution is

y(x) = c1e−x + c2xe−x + c3x2e−x.

(check it!).

3.1.9 We now turn to nonhomogeneous equations. There are two methods tosolve these:

1. The method of undetermined coefficients. This is simpler; we will explainit with examples.

2. The method of variation of parameters. This is less simple but more general,we will present it a little later.

3.1.10 Example. To solved2ydx2 −

dydx−2y = 1 (3.4)

we will exploit Theorem ??. From this we know that the solution is y(x) =yh (x)+ yp (x), where yh (x) is the general solution of

d2ydx2 −

dydx−2y = 0 (3.5)

and yp (x) is some solution of (3.4).The char. eq. of (3.5) is r2− r− 2 = 0 with roots r1 = −1,r2 = 2. Hence the

general solution isyh (x) = c1e−x + c2e2x.

To find some solution of (3.4) we take a guess. A reasonable guess is yp (x) = A(why?). Then we must have

d2Adx2 −

dAdx−2A = 1⇒−2A = 1⇒ A =−1

2⇒ yp (x) =−

12.

Hence the general solution of (3.4) is

y(x) = c1e−x + c2e2x− 12.


dydx−2y = x2 (3.6)

we take y(x) = yh (x)+ yp (x), where yh (x) is the general solution of

d2ydx2 −

dydx−2y = 0; (3.7)

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this was already found to be

yh (x) = c1e−x + c2e2x.

For the particular solution we guess is yp (x) = Ax2 +Bx+ c (why?). Then we musthave

d2 (Ax2 +Bx+C)

dx2 −d(Ax2 +Bx+C

)dx

−2(Ax2 +Bx+C

)= x2⇒

2A− (2Ax+B)−2(Ax2 +Bx+C

)= x2⇒

−2Ax2 +(−2A−2B)x+(2A−B−2C) = x2.

Then we must have

−2A = 1−2A−2B = 0

2A−B−2C = 0

Henceyp (x) =−

12

x2 +12

x− 34

andy(x) = c1e−x + c2e2x− 1

2x2 +

12

x− 34.


dydx−2y = ex (3.8)


d2ydx2 −

dydx−2y = 0; (3.9)


yh (x) = c1e−x + c2e2x.

For the particular solution we guess yp (x) = Aex (why?). Then we must have

d2 (Aex)

dx2 − d (Aex)

dx−2(Aex) = ex⇒ Aex−Aex−2Aex = ex⇒ A =−1

2⇒ yp (x) =−

12

ex.


y(x) = c1e−x + c2e2x− 12

ex.


dydx−2y = e−x (3.10)

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d2ydx2 −

dydx−2y = 0; (3.11)


yh (x) = c1e−x + c2e2x.

For the particular solution we should not guess yp (x) = Ae−x, because e−x alreadyappears in yp (x). So we try yp (x) = Axe−x and we have

d2 (Axe−x)

dx2 − d (Axe−x)

dx−2(Axe−x)= e−x⇒

Ae−x (x−1)−2Axe−x +Ae−x (x−2) = e−x⇒(A−2A+A)xe−x +(−A−2A)e−x = e−x⇒

A =−13⇒ yp (x) =−

13

xe−x.


y(x) = c1e−x + c2e2x− 13

e−x.

3.1.14 We can summarize various “reasonable guesses” for particular solutionsof L(y) = g(x) in the following table.

For g(x) guess yp (x)A0 +A1x+ ...+Amxm B0 +B1x+ ...+Bmxm

AePx BeQx

C1 cos(wx)+C2 sin(wx) D1 cos(wx)+D2 sin(wx)ePx (A0 +A1x+ ...+Amxm) eQx (B0 +B1x+ ...+Bmxm)(C1 cos(wx)+C2 sin(wx))(A0 + ...+Amxm) (D1 cos(wx)+D2 sin(wx))(B0 + ...+Bmxm)

3.1.15 We now turn to the method of variation of parameters, which is alsopresented by examples.

3.1.16 Example. Let us first solve

d2ydx2 −2

dydx

+ y = ex

(which we could also solve by undetermined coefficients) just to illustrate themethod of variation of parameters. We note that the homogeneous equation

d2ydx2 −2

dydx

+ y = 0

has general solution c1ex + c2xex and we now consider a particular solution of theform

v1 (x)ex + v2 (x)xex.

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Now we solve the system

v′1 (x)(ex)+ v′2 (x)(xex) = 0

v′1 (x)(ex)′+ v′2 (x)(xex)′ = ex

which becomes

v′1 (x)(ex)+ v′2 (x)(xex) = 0

v′1 (x)(ex)+ v′2 (x)(e

x + xex) = ex

with solution

v′1 (x) =

∣∣∣∣ 0 xex

ex ex + xex

∣∣∣∣∣∣∣∣ ex xex

ex ex + xex

∣∣∣∣ =−x⇒ v1 (x) =−x2

2,

v′2 (x) =

∣∣∣∣ ex 0ex ex

∣∣∣∣∣∣∣∣ ex xex

ex ex + xex

∣∣∣∣ = 1⇒ v2 (x) = x.

Finally then

yp (x) =−x2

2ex + x2ex =

x2

2ex

and

y(x) = yh (x)+ yp (x) = c1ex + c2xex +x2

2ex.

3.1.17 Example. To solved2ydx2 −2

dydx

+ y =ex

xwe note that the homogeneous equation

d2ydx2 −2

dydx

+ y = 0

has general solution c1ex + c2xex and we now consider a particular solution of theform

v1 (x)ex + v2 (x)xex.


v′1 (x)(ex)+ v′2 (x)(xex) = 0

v′1 (x)(ex)′+ v′2 (x)(xex)′ =

ex

x

which becomes

v′1 (x)(ex)+ v′2 (x)(xex) = 0

v′1 (x)(ex)+ v′2 (x)(e

x + xex) =ex

x

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with solution

v′1 (x) =

∣∣∣∣ 0 xex

ex

x ex + xex

∣∣∣∣∣∣∣∣ ex xex

ex ex + xex

∣∣∣∣ =−1⇒ v1 (x) = x,

v′2 (x) =

∣∣∣∣ ex 0ex ex

x

∣∣∣∣∣∣∣∣ ex xex

ex ex + xex

∣∣∣∣ =1x⇒ v2 (x) = ln |x| .

Finally thenyp (x) =−xex + ln |x|xex

and

y(x) = yh (x)+ yp (x) = c1ex + c2xex− xex + ln |x|xex = c1ex + c3xex + ln |x|xex.

3.1.18 Example. To solved2ydx2 +9y = 3tan3x

we note that the homogeneous equation

d2ydx2 +9y = 0

has general solution c1 cos3x+ c2 sin3x and we now consider a particular solutionof the form

v1 (x)cos3x+ v2 (x)sin3x.


v′1 (x)cos3x+ v′2 (x)sin3x = 0−v′1 (x)3sin3x′+ v′2 (x)3cos3x = 3tan3x

with solution

v′1 (x) =

∣∣∣∣ 0 sin3x3tan3x 3cos3x

∣∣∣∣∣∣∣∣ cos3x sin3x−3sin3x 3cos3x

∣∣∣∣ =−(sin3x) tan3x =−sin2 3xcos3x

,

v′2 (x) =

∣∣∣∣ cos3x 0−3sin3x 3tan3x

∣∣∣∣∣∣∣∣ cos3x sin3x−3sin3x 3cos3x

∣∣∣∣ = cos3x tan3x = sin3x.

Then

v1 =−∫ sin2 3x

cos3xdx =

16

ln∣∣∣∣2−2sin3x2+2sin3x

∣∣∣∣+ 13

sin3x

v2 =∫

sin3xdx =−13

cos3x

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3.2 Theory and Examples for Variable Coefficient DEs 19

and finally

y(x) = c1 cos3x+ c2 sin3x+(

16


∣∣∣∣+ 13

sin3x)

cos3x− 13

cos3xsin3x

= c1 cos3x+ c2 sin3x+16


∣∣∣∣cos3x.

3.2 Theory and Examples for Variable Coefficient DEs3.2.1 Theorem. The general solution of the 1st order linear differential equation

dydx

+a0 (x)y = f (x)

isy(x) = e−

∫a0(x)dx

(c+

∫f (x)e

∫a0(x)dxdx

).

3.2.2 Since we have studied 1st order linear differential equations in a previouscourse, we will not deal with them in these notes.

3.2.3 For the n-th order linear differential equation

dnydxn +an−1 (x)

dn−1ydxn−1 + ...+

dydx

+a0 (x)y = f (x)

we can again use variation of parameters, as shown in the following examples.Note that this depends on having found the general solution of

dnydxn +an−1 (x)

dn−1ydxn−1 + ...+

dydx

+a0 (x)y = 0

which is not always easy.

3.2.4 Example. We solve

x2 d2ydx2 −2x

dydx

+2y = x lnx

given that two lin. ind. solutions of

x2 d2ydx2 −2x

dydx

+2y = 0

are x and x2. In this case, we first divide by x2 to get

d2ydx2 −

2x

dydx

+2x2 y =

lnxx

v′1x+ v′2x2 = 0

v′1 + v′22x =lnxx

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and, solving, we get

v′1 =−lnxx⇒ v1 =−

12(lnx)2

v′2 =−lnxx2 ⇒ v1 =−

1+ lnxx

.

Then the general solution is

y(x) = c1x+ c2x2− x2(lnx)2− x(1+ lnx) = c3x+ c2x2− x lnx− x

2(lnx)2 .

3.2.5 Example. We solve

(x2−1

) d2ydx2 −2x

dydx

+2y =(x2−1

)2

given that two lin. ind. solutions of

x2 d2ydx2 −2x

dydx

+2y = 0

are x and x2 +1. In this case, we first divide by x2−1 to get

d2ydx2 −

2x2−1

dydx

+2

x2−1y =

(x2−1

)v′1x+ v′2

(x2 +1

)= 0

v′1 + v′22x = x2−1

and, solving, we get

v′1 =−x2−1⇒ v1 =−x3

3− x

v′2 = x⇒ v2 =x2

2


y(x) = c1x+ c2(x2 +1

)+ x(−x3

3− x)+(x2 +1

) x2

2= c1x+ c2x2 +

16

x2 (x2−3).

3.3 Solved Problems

3.4 Unsolved Problems1. Solve d2y

dx2 −3dydx +2y = ex.

Ans. C1e2x− ex +C2ex− xex.2. Solve d2y

dx2 −4dydx +2y = ex.

Ans. C1ex(√

2+2)− ex +C2e−x(√

2−2).

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3.4 Unsolved Problems 21

3. Solve d2ydx2 −4dy

dx +4y = ex.Ans.ex +C1e2x +C2xe2x.

4. Solve d2ydx2 +

dydx + y = ex,.

Ans. 13ex +C44

(cos 1

2

√3x)

e−12 x−C45

(sin 1

2

√3x)

e−12 x.


dx +2y = x2−1.Ans. 3

2x+C1e2x +C2ex + 12x2 + 5

4 .


dx +2y = x2−1.

Ans. 2x+C1ex(√

2+2) +C2e−x(√

2−2) + 12x2 +3.


dx +4y = x2−1.Ans.1

2x+C1e2x + 14x2 +C2xe2x + 1

8 .

8. Solve d2ydx2 +

dydx + y = x.

Ans. x+C1(cos 1

2

√3x)

e−12 x−C2

(sin 1

2

√3x)

e−12 x−1.


dx +2y = sinx.Ans. 3

10 cosx+ 110 sinx+C1e2x +C2ex.

10. Solve d2ydx2 +5dy

dx +6y = cosx.Ans. 1

10 cosx+ 110 sinx+C1e−2x +C2e−3x.


dx +4y = sinx+ cosx.Ans. 7

25 cosx− 125 sinx+C1e2x +C2xe2x.

12. Solve d2ydx2 +

dydx + y = sinx.

Ans. C1(cos 1

2

√3x)

e−12 x− cosx−C2

(sin 1

2

√3x)

e−12 x.


dx +2y = ex = ex.Ans. C1e2x− ex +C2ex− xex.

14. Solve d2ydx2 +5dy

dx +6y = e−2x.Ans. C1e−2x +C2e−3x + 1

e2x (x−1).


dx +4y = e2x.Ans. C1e2x + 1

2x2e2x +C2xe2x.

16. Solve d2ydx2 +

dydx + y = e−

x2 sin

√3x2 .

Ans. (sin 12

√3x−√

3xcos 12

√3x)

3e12 x

+C1(cos 1

2

√3x)

e−12 x +C2

(sin 1

2

√3x)

e−12 x.


dx + y = ex

x5 .Ans. C1ex + xC2ex + 1

12x3 ex.

18. Solve d2ydx2 + y = 1

cosx .Ans. C1 cosx−C2 sinx+ xsinx+ cosx ln(cosx).


dx +3y = ex

1+ex .Ans. c1et + c2e3t + et

2 ln(1+ et).

20. Solve d2ydx2 + y = 1

cosx .Ans. C1 cosx−C2 sinx+ xsinx+ cosx ln(cosx).

21. Solve x2 d2ydx2 + xdy

dx − y = 2x2 + 2, given that two solutions of the associated

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homogeneous DE are x and 1x .

Ans. C1x+ C2x + 4x3−12x

6x .

22. Solve xd2ydx2 +(2−2x) dy

dx +(x−2)y = e2x, given that two solutions of the associ-ated homogeneous DE are ex and ex

x .Ans. 1

x e2x +C1ex +C2ex

x .

23. Solve x2 d2ydx2 − 4xdy

dx + 6y = x5/2, given that two solutions of the associatedhomogeneous DE are x2 and x3.Ans. c1x2 + c2x3−4x

52 .

24. Solve(x2 + x

) d2ydx2 +

(2− x2) dy

dx − (2+ x)y = x(x+1)2, given that two solutionsof the associated homogeneous DE are ex and 1

x .Ans. C1

x +C2ex− x− 13x2−1.

3.5 Advanced Problems1. Suppose x1 (t) ,x2 (t) are solutions of p(t) d2x

dt2 +q(t) dxdt + r (t)x = 0 with Wron-

skian W (x1 (t) ,x2 (t)). Show that p(t) dWdt + q(t)W = 0.

2. Suppose x1 (t) ,x2 (t) are solutions of d2xdt2 + p(t) dx

dt + r (t)x = 0, with p(t) ,q(t)continuous in [t1, t2]. Show that, for all t, t0 ∈ (t1, t2) we have

W (x1,x2) =W (x1,x2) |t=t0e−∫ t

t0p(s)ds

.

3. Show that the set of solutions of dxdt =

√|x| is not a vector space.

4. Suppose x1 (t) , ...,xN (t) are continuous in [t1, t2] and define

ai j =∫ t2

t1xi (t)x j (t)dt.

Prove that x1 (t) , ...,xN (t) is linearly independent in [t1, t2] iff∣∣∣∣∣∣∣∣a11 a12 ... a1na21 a22 ... a2n... ... ... ...an1 an2 ... ann

∣∣∣∣∣∣∣∣ 6= 0.

5. Show that d2

dt2 x(t)+a ddt x(t)+bx(t) = F (t), with characteristic equation roots

z1 6= z2, has the following solution.

x(t) = c1ez1t + c2ez2t +ez1t

z1− z2

∫e−z1tF (t)dt +

ez2t

z2− z1

∫e−z2tF (t)dt.

6. An Euler equation is a DE of the form

tn dnxdtn +an−1tn−1 dn−1x

dtn−1 + ...+a1tdxdt

+a0x = 0

Show that every Euler equation can be converted to a linear DE by thesubstitution t = eu.

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3.5 Advanced Problems 23

7. Find a transformation which converts

(pt +q)2 d2xdt2 +a(pt +q)

dxdt

+a0x = 0

to a constant coefficient DE and solve it. What conditions must a,b satisfy?8. Form constant coefficient homeogeneous linear DEs with the smallest possible

order and having as solutions: (i) xe−3x, (ii) x2 sinx, (iii) cosx+ e−2x.9. Show that a particular solution of

dnxdtn +an−1

dn−1xdtn−1 + ...+a1

dxdt

+a0x = Aect

isx(t) =

Aect

cn +an−1cn−1 + ...+a1c+a0

under the condition cn +an−1cn−1 + ...+a1c+a0 6= 0.10. Suppose limt→∞ f (t) = 0. Prove that all solutions of d2x

dt2 + f (t)x(t) = 0 arebounded.

11. Suppose f (t) is continuously differentiable, limt→∞ f (t) = 0 and∫

∞

0 | f (t)|dt <∞. Prove that all solutions of d2x

dt2 +(1+ f (t))x(t) = 0 are bounded.

12. Suppose x(t) is a solution of t d2xdt2 +

dxdt + tx(t) = 0 in (0,∞). Show that g(x) =

x1/2x(t) is bounded in (0,∞).

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4. Systems of Linear ODEs

We study properties and solution methods for systems of ordinary differentialequations, such as

dx1

dt= a11x1 (t)+a12x2 (t)+ ...+a1NxN (t)+ f1 (t) ,

...

dxN

dt= aN1x1 (t)+aN2x2 (t)+ ...+aNNxN (t)+ fN (t) .

4.1 Theory and Examples4.1.1 Notation. We will limit ourselves to the study of ODE systems with constantcoefficients, such as

dxdt

= a11x(t)+a12y(t)+ f (t) , x(0) = x0,

dydt

= a21x(t)+a22y(t)+g(t) , y(0) = y0.

This can be rewritten, using the notation of Linear Algebra, asdzdt

= Az+u(t) , z(0) = z0

where

z(t) =[

x(t)y(t)

], z0 =

[x0y0

], u(t) =

[f (t)g(t)

], A =

[a11 a12a21 a22

]This notation can be extended to any N×N matrix A and N×1 vector z(t).4.1.2 Theorem. Given any N×N matrix A and N×1 vectors z(t), u(t), z0, the

systemdzdt

= Az+u(t) , z(0) = z0

has solutionz(t) = eAtz0 +

∫ t

0eA(t−τ)u(τ)dτ.

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26 Chapter 4. Systems of Linear ODEs

4.1.3 The above theorem is esthetically pleasing, as it is a generalization of thefact that the scalar system

dxdt

= ax+u, x(0) = x0

has solutionx(t) = eatx0 +

∫ t

0ea(t−τ)u(τ)dτ.

However, application of the theorem requires the computation of

eAt = I+At +12!

A2t2 + ... (4.1)

4.1.4 We can compute eAt by rewriting (4.1) as a finite sum, using the Cayley-Hamilton Theorem. However, we will omit this approach and present an allternativesolution method, which uses matrix diagonalization to decouple the ODEs of thesystem. We present the method by examples.

4.1.5 Example. Let us solve

dxdt

= 2x+ y, x(0) = 1,

dydt

= 3x+4y, y(0) = 1.

This can be rewritten as

ddt

[x(t)y(t)

]=

[2 13 4

][x(t)y(t)

],

[x(0)y(0)

]=

[11

].

Now, defining

z(t)=[

x(t)y(t)

], z0=

[11

], A=

[2 13 4

].

we havedzdt

= Az, z(0) = z0 (4.2)

We will now diagonalize A using eigenvalues and eigenvectors. Note that A haseigenvectors and eigenvalues:

1↔[−11

], 5↔

[ 131

]Defining

U =

[−1 1

31 1

], U−1 =

[−3

414

34

34

], Λ =

[1 00 5

]we have

AU = ΛU ⇒U−1AU = Λ.

or, more specifically,[−1 1

31 1

]−1[ 2 13 4

][−1 1

31 1

]=

[1 00 5

].

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Now, defining z=U−1z, A=U−1AU and multiplying from the left we have

U−1 dzdt

=U−1Az⇒ ddt

(U−1z

)=U−1AUU−1z⇒ dz

dt=Λz⇒

[ dxdtdydt

]=

[1 00 5

][xy

].

Similarly

z0 =U−1z0 =

[−1 1

31 1

]−1[ 11

]=

[−1

232

]This breaks down very nicely to

dxdt

= x, x(0) =−12

dydt

= 5y, y(0) =32

which can be solved separately to get

x(t) =−12

et , y(t) =32

e5t

To find x(t) ,y(t) we use[x(t)y(t)

]=

[−1 1

31 1

][−1

2et

32e5t

]=

[ 12et + 1

2e5t

32e5t− 1

2et

]4.1.6 Example. We now solve

dxdt

=−x+ y, x(0) = 1

dydt

= x− y+1, y(0) = 2

In matrix form we have

ddt

[xy

]=

[−1 11 −1

][xy

]+

[01

],

[x(0)y(0)

]=

[12

]We find eigenvectors of the matrix

A =

[−1 11 −1

]They are [

11

]↔ 0,

[−11

]↔−2.

Hence we have

U =

[1 −11 1

],U−1 =

[1 −11 1

]−1

=

[ 12

12

−12

12

]and

U−1AU =

[ 12

12

−12

12

][−1 11 −1

][1 −11 1

]=

[0 00 −2

]= Λ.

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Furthermore,

u =

[ 12

12

−12

12

][01

]=

[ 1212

], z0 =

[ 12

12

−12

12

][12

]=

[ 3212

].

Hence we have the system[ dxdtdydt

]=

[0 00 −2

][xy

]+

[ 1212

],

[x(0)y(0)

]=

[ 3212

]or (

dxdt

=12, x(0) =

32

)⇒ x(t) =

t2+

32(

dydt

=−2y+12, y(0) =

12

)⇒ y(t) =

14

e−2t +14

Then [x(t)y(t)

]=U

[x(t)y(t)

]=

[1 −11 1

][ t2 +

32

14e−2t + 1

4

]=

[ 12t− 1

4e−2t + 54

12t + 1

4e−2t + 74

].

4.1.7 The method can be generalized and applied to the system

dzdt

= Az+u(t) , z(0) = z0 (4.3)

for any N×N matrix A, provided that A is diagonalizable, i.e., it can be transformedto a diagonal matrix Λ; this is true iff A has N linearly independent eigenvectors.

4.1.8 Not every matrix is diagonalizable. If A is not diagonalizable, the methodcannot be applied. However, every matrix can be transformed to Jordan canonicalform, and a variation of the method can then be applied (but the resulting ODEswill not be fully decoupled).


dxdt

= x+2y, x(0) = 1

dydt

= y, y(0) = 2

In matrix form we have

ddt

[xy

]=

[1 20 1

][xy

],

[x(0)y(0)

]=

[12

].

The matrix is not diagonalizable, but is already in Jordan canonicl form. We firstsolve

dydt

= y, y(0) = 2

and gety(t) = 2et .

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Then we havedxdt

= x+4et , x(0) = 1

which has solution solution is:

x(t)et +4tet .

4.1.10 Now we will present another solution method which reduces the system offirst order DE to a single N-th order DE. Again we present the method by examples.


dxdt

= 2x+ y, x(0) = 1,

dydt

= 3x+4y, y(0) = 1.

by transforming the system to companion form. We define

1 =[

1 0], S =

[1

1A

], A = SAS−1, z = Sz

and, multiplying from the left, we get

dzdt

= Az⇒ dSzdt

= SAS−1Sz⇒ dzdt

= Az.

So we have

1A =[

1 0][ 2 1

3 4

]=[

2 1], S =

[1 02 1

], S−1 =

[1 0−2 1

]and

A =

[1 02 1

][2 13 4

][1 0−2 1

]=

[0 1−5 6

],

[x0y0

]=

[1 02 1

][11

]=

[13

].

Hence we have

ddt

[xy

]=

[0 1−5 6

][xy

],

[x(0)y(0)

]=

[13

]This implies that

dxdt

= y, x(0) = 1,dydt

=−5x+6y, y(0) = 3,d2xdt2 =

dydt

Combining these we get

d2xdt2 =−5x+6

dxdt

, x(0) = 1, x′ (0) = 3

which has solution

x(t) =12

et +12

e5t

y(t) =ddt

x(t) =12

et +52

e5t

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then [x(t)y(t)

]=

[ 12et + 1

2e5t

12et + 5

2e5t

]and finally [

x(t)y(t)

]=

[1 0−2 1

][ 12et + 1

2e5t

12et + 5

2e5t

]=

[ 12et + 1

2e5t

32e5t− 1

2et

]4.1.12 Note how nicely we got a single second order ODE from the system of twofirst order ODEs; also note that one variable is the derivative of the other.

4.1.13 Example. Let us now solve

dxdt

= 2x+ y+1, x(0) = 1,

dydt

= 3x+4y, y(0) = 1.

We have

1A =[

1 0][ 2 1

3 4

]=[

2 1], S =

[1 02 1

], S−1 =

[1 0−2 1

]and

A =

[1 02 1

][2 13 4

][1 0−2 1

]=

[0 1−5 6

],[

x0y0

]=

[1 02 1

][11

]=

[13

], u =

[1 02 1

][10

]=

[12

].

Hence we haveddt

[xy

]=

[0 1−5 6

][xy

]+

[12

],

[x(0)y(0)

]=

[13

]This implies that

dxdt

= y+1, x(0) = 1,dydt

=−5x+6y+2, y(0) = 3,d2xdt2 =

dydt

Combining these we get

d2xdt2 =−5x+6

dxdt−6+2, x(0) = 1, x′ (0) = y′ (0)+1 = 4

which has solution

x(t) =54

et +1120

e5t− 45

y(t) =ddt

x(t)−1 =54

et +114

e5t−1

then [x(t)y(t)

]=

[ 54et + 11

20e5t− 45

54et + 11

4 e5t−1

]and finally[

x(t)y(t)

]=

[1 0−2 1

][ 54et + 11

20e5t− 45

54et + 11

4 e5t−1

]=

[ 54et + 11

20e5t− 45

3320e5t− 5

4et + 35

]

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4.2 Solved Problems 31

4.1.14 This method can also be generalized to any system

dzdt

= Az+u(t) , z(0) = z0 (4.4)

with N×N matrix A. Again there is a condition: the so-called observability matrix

S =

1

1A...

1An−1

must be invertible. When S is not invertible (we say the system is unobservable)the method cannot be applied.

4.1.15 There is a "reverse” of the above metod, by which we can write any N-thorder linear ODE as a single first order vector ODE. We illustrate this by anexample. Consider

d3xdt3 −3

d2xdt2 +3

dxdt− x = u(t) (4.5)

and define

z1 = x, z2 =dxdt

, z3 =d2xdt2 .

Then (4.5) is equivalent to

ddt

z1z2z3

=

0 1 00 0 13 −3 1

z1z2z3

+ 0

0u(t)

.4.1.16 There is a dual method of reducing the system (4.2) to a single ODE, whichmakes use of the controllability matrix

T =[

1′ A1′ ... AN−11].

How do you think it works?

4.2 Solved Problems


1. Solve the system of DEs:

dxdt =−x+ ydydt = x− yx(0) = 4y(0) = 1

.

Ans. x(t) = 32e−2t + 5

2 ,y(t) =52 −

32e−2t .


dxdt =−x+ y+ et

dydt = x− y− et

x(0) = 0y(0) = 1

.

Ans. y(t) = 56e−2t− 1

3et + 12 ,x(t) =

13et− 5

6e−2t + 12 .

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dxdt =−x+ y+ et

dydt = x− y− et

x(0) = 4y(0) = 1

.

Ans. y(t) = 52 −

76e−2t− 1

3et ,x(t) = 13et + 7

6e−2t + 52 .


dxdt =−x+ y+ cos t

dydt = x− yx(0) = 0y(0) = 0

.

Ans. y(t) = 25 sin t− 1

5 cos t + 15e2t ,x(t) = 1

5 cos t + 35 sin t− 1

5e2t .


dxdt =−4x+6ydydt =−3x+5y

x(0) = 1y(0) = 2

.

Ans. x(t) = 3e2t−2e−t ,y(t) = 3e2t− e−t .


dxdt =−4x+6y+1dydt =−3x+5y+2

x(0) = 0y(0) = 1

.

Ans. y(t) = 72e2t− 5

2 ,x(t) =72e2t− 7

2 .


dxdt =−4x+6y+ et

dydt =−3x+5y

x(0) = 1y(0) = 0

.

Ans. y(t) = 32et + 1

2e−t−2e2t ,x(t) = 2et + e−t−2e2t .


dxdt =−4x+6y

dydt =−3x+5y+ e−t

x(0) = 1y(0) = 2

.

Ans. y(t) = 113 e2t− 5

3e−t− te−t ,x(t) = 113 e2t− 8

3e−t−2te−t .


dxdt = 3x+4ydydt =−x+7y

x(0) = 1y(0) = 0

.

Ans. x(t) = e5t−2te5t ,y(t) =−te5t .


dxdt = 3x+4ydydt =−x+7y

x(0) = 1y(0) = 1

.

Ans. x(t) = e5t +2te5t ,y(t) = e5t + te5t .


dxdt = 3x+4y

dydt =−x+7y+1

x(0) = 0y(0) = 1

.

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Ans. y(t) = 2825e5t + 12

5 te5t− 325 ,x(t) =

245 te5t− 4

25e5t + 425 .


dxdt = 3x+4y

dydt =−x+7y+ e−t

x(0) = 1y(0) = 2

.

Ans. y(t) = 199 e5t− 1

9e−t + 103 te5t ,x(t) = 1

9e−t + 89e5t + 20

3 te5t .


dxdt =−y

dydt = x−2yx(0) = 1y(0) = 2

.

Ans. x(t) = e−t− te−t ,y(t) = 2e−t− te−t .


dxdt =−y+1dydt = x−2yx(0) = 1y(0) = 2

.

Ans. y(t) = e−t−2te−t +1,x(t) = 2−2te−t− e−t .


dxdt =−y+ sin t

dydt = x−2yx(0) = 1y(0) = 2

.

Ans. y(t) = 52et − 1

2tet − 1

2 cos t,x(t) = 12 sin t− cos t− 1

2tet +

2et .


dxdt =−y+1

dydt = x−2y+1

x(0) = 1y(0) = 0

.

Ans. y(t) = te−t− e−t +1,x(t) = te−t +1.


dxdt = x+2y

dydt = 2x+ y+1

x(0) = 1y(0) = 0

.

Ans. y(t) = 23e3t− e−t + 1

3 ,x(t) = e−t + 23e3t− 2

3 .


dxdt = x+2ydydt = 2x+ yx(0) = 1y(0) = 1

.

Ans. x(t) = e3t ,y(t) = e3t .


dxdt = x+2y+ et

dydt = 2x+ y+1

x(0) = 1y(0) = 0

.

Ans. y(t) = 1112e3t− 3

4e−t− 12et + 1

3 ,x(t) =34e−t + 11

12e3t− 23 .

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dxdt = x+2y+ e3t

dydt = 2x+ y+1

x(0) = 1y(0) = 0

.

Ans. y(t) = 1324e3t− 7

8e−t + 12te3t + 1

3 ,x(t) =78e−t + 19

24e3t + 12te3t− 2

3 .


dxdt = x+4ydydt = x+ yx(0) = 1y(0) = 0

.

Ans. x(t) = 12e−t + 1

2e3t ,y(t) = 14e3t− 1

4e−t .


dxdt = x+4y− et

dydt = x+ yx(0) = 1y(0) = 1

.

Ans. y(t) = 14et + 1

8e−t + 58e3t ,x(t) = 5

4e3t− 14e−t .


dxdt = x+4y− e−t

dydt = x+ yx(0) = 1y(0) = 0

.

Ans. y(t) = 316e3t− 3

16e−t + 14te−t ,x(t) = 5

8e−t + 38e3t− 1

2te−t .


dxdt = x+4y+1dydt = x+ y−1

x(0) = 1y(0) = 2

.

Ans. y(t) = 32e−t + 7

6e3t− 23 ,x(t) =

73e3t−3e−t + 5

3 .


dxdt =−x−4y+1dydt =−x− y−1

x(0) = 1y(0) = 2

.

Ans. y(t) = 43e−3t + 2

3 ,x(t) =83e−3t− 5

3 .


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5. Series Solutions of ODEs

This chapter is devoted to the solution of ODEs by expansion into infinite series.

5.1 Theory and Examples5.1.1 In this chapter we present methods to solve the linear ODEs

P0 (x)d2ydx2 +P1 (x)

dydx

+P2 (x)y = 0 (5.1)

P0 (x)d2ydx2 +P1 (x)

dydx

+P2 (x)y = f (x) (5.2)

using infinite series expansions. In what follows we always assume P0 (x), P1 (x),P2 (x) have no common factors. Note that we can always rewrite the above as

d2ydx2 +

P1 (x)P0 (x)

dydx

+P2 (x)P0 (x)

y = 0 (5.3)

d2ydx2 +

P1 (x)P0 (x)

dydx

+P2 (x)P0 (x)

y = f (x) (5.4)

5.1.2 Definition. We call x0 an ordinary point of (5.1) iff P0 (x0) 6= 0 and a singularpoint iff P0 (x0) = 0.

5.1.3 Theorem. Suppose that1. P0 (x) ,P1 (x) ,P2 (x) are polynomials with no common factor,2. x0 is an ordinary point of (5.1) and3. ρ is the distance between x0 and the closest zero of P0 (x) (with ρ = ∞ if P0 (x)

has no zeros).Then every solution of (5.1) can be written as a power series

∞

∑n=0

an (x− x0)n

which converges in (−ρ,ρ).

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36 Chapter 5. Series Solutions of ODEs

5.1.4 Example. We start with a simple example. Let us solve

d2ydx2 + y = 0.

Of course we know that the general solution is

c1 cosx+ c2 sinx,

but let us see what the power series approach yields. We assume that

y = a0 +a1x+a2x2 +a3x3 +a4x4 + ...

Then

dydx

= a1 +2a2x+3a3x2 +4a4x3 +5a5x4...

d2ydx2 = 2a2 +6a3x+12a4x2 +20a5x3...

So we have

0 =d2ydx2 + y = (a0 +2a2)+(a1 +6a3)x+(a2 +12a4)x2 +(a3 +20a5)x3 + ...

and we get

a2 =−12

a0,

a3 =−16

a1

a4 =−1

12a2 =

124

a0

a5 =−1

20a3 =

1120

a1

...

and so

y(x) = a0

(1− 1

2x2 +

124

x4− ...

)+a1

(x− 1

6x3 +

1120

x5− ...

).

More generally:

(n+2)(n+1)an+2 = an⇒ an+2 =1

(n+1)(n+2)an

and so

y(x) = a0

(∑

n∈0,2,4,...

(−1)n+1

n!xn

)+a1

(∑

n∈1,3,5,...

(−1)n+1

n!xn

)

which, as expected, equalsa0 cosx+a1 sinx.

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5.1.5 Example. Now let us solve

d2ydx2 + y = x

y(0) = 0y′ (0) = 1

d2ydx2 + y = x, y(0) = 0, y′ (0) = 1.

Assumingy = a0 +a1x+a2x2 +a3x3 +a4x4 + ...

we get, as before, that

x =d2ydx2 + y = (a0 +2a2)+(a1 +6a3)x+(a2 +12a4)x2 +(a3 +20a5)x3 + ... .

Also, from y(0) = 0 we

a0 = 0

a2 =12

a0 = 0

a4 =1

12a2 = 0

...

From y′ (0) = 1 we get

a1 = 1a1 +6a3 = 1⇒ a3 = 0

a3 +20a5 = 0⇒ a5 = 0...

Hence we havey = x.

5.1.6 Example. Now we will solve

d2ydx2 − x

dydx

+2y = 0

Assumingy = a0 +a1x+a2x2 +a3x3 + ...

we get

2y = 2a0 +2a1x+2a2x2 +2a3x3 +2a4x4 + ...

−xdydx

=−a1x−2a2x2−3a3x3−4a4x4− ...

d2ydx2 = 2a2 +6a3x+12a4x2 + ...

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So we have

0 =d2ydx2 − x

dydx

+2y

= (2a0 +2a2)+(a1 +6a3)x+12a4x2 + ...

and we get

a2 =−a0,

a3 =−1

2 ·3a1,

a4 = 0,

a5 =1

4 ·5a3 =−

12 ·3 ·4 ·5

a1,

a6 = 0,

a7 =6

6 ·7a5 =−

1 ·32 ·3 ·4 ·5 ·6 ·7

a1,

a8 = 0...

Finally:

y(x) = a0(1− x2)+a1

(x− 1

6x3− 1

120x5− ...

).

5.1.7 Example. To solve

d2ydx2 − (x−2)

dydx

+2y = 0

in powers of (x−2) we first perform the change of variable u = x−2. Since dydu = dy

dx

and d2ydu2 =

d2ydx2 (why?) the DE becomes

d2ydu2 −u

dydu

+2y = 0.

This is the DE we solved in the previous example, where we got

y(u) = a0(1−u2)+a1

(u− 1

6u3− 1

120u5− ...

)or(

1− (x−2)2)=

y(x) = a0

(1− (x−2)2

)+a1

((x−2)− 1

6(x−2)3− 1

120(x−2)5− ...

)= a0

(−x2 +4x−3

)++a1

((x−2)− 1

6(x−2)3− 1

120(x−2)5− ...

).

5.1.8 Definition. We call x0 an regular singular point of (5.1) iff

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1. it can be written in the form

(x− x0)2 A(x)

d2ydx2 +(x− x0)B(x)

dydx

+C (x)y = 0

where A(x), B(x), C (x) are polynomials and2. A(x0) 6= 0, B(x0) 6= 0, C (x0) 6= 0.

Otherwise we call x0 an irregular singular point of (5.1).

5.1.9 Theorem (Frobenius). Suppose that x0 = 0 is regular singular point of

x2A(x)d2ydx2 + xB(x)

dydx

+C (x)y = 0, (5.5)

where A(x) ,B(x) ,C (x) are polynomials with no common factor. Then there existconstants ρ > 0 and λ such that (5.5) has at least one solution of the form

y(x) = xλ∞

∑n=0

anxn

valid in the interval (0,ρ).


8x2y′′+10xy′+(x−1)y = 0 (5.6)

which can be rewritten as

y′′+54x

y′+x−18x2 y = 0. (5.7)

Clearly x0 = 0 is a singular point of (5.7) but, since

xP(x) = x54x

=54

analytic

x2Q(x) = x2 x−18x2 =

x−18

analytic

it is a regular singular point. So we will try to find a solution of the form

y(x) = xλ∞

∑n=0

anxn =∞

∑n=0

anxn+λ (5.8)

Taking derivatives of (5.8) in (5.6) we get

0 =xλ [8λ (λ −1)+10λ −1]a0+

xλ+1 ([8(λ +1)λ +10(λ +1)−1]a1−a0)+

xλ+2 ([8(λ +3)(λ +1)+10(λ +3)−1]a3−a1)+ ...

or

0 =[8λ (λ −1)+10λ −1]a0+

x([8(λ +1)λ +10(λ +1)−1]a1−a0)+

x2 ([8(λ +3)(λ +1)+10(λ +3)−1]a3−a1)+ ... .

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This can be satisfied if the indicial equation

[8λ (λ −1)+10λ −1] = 0

holds (and then we specify the an’s by a recursion, as previously). Now, the indicialequation can be rewritten as

8λ2 +2λ −1 = 0,

which has solutions: λ1 =14 , λ2 =−1

2 . Furthermore, we get the requrrence relation

an =−1

[4(λ +n)−1] [2(λ +n)+1]an−1.

So from the two λ values we get

for λ1 =14

: an =−1

2n(4n+3)an−1,

for λ2 =−12

: an =−1

2n(4n−3)an−1.

Consequently we get two linearly independent solutions

y1 (x) = c1x14

(1− 1

14x+

1616

x2− ...

)y2 (x) = c2x−

12

(1− 1

2x+

140

x2− ...

)and the general solution is

y(x) = c1y1 (x)+ c2y2 (x) .

5.1.11 In the previous example the indicial equation had two distinct roots whichdid not differ by an integer. Consider the next examples.

5.1.12 Example. Let us solve the ODE

x2 d2ydx2 +2x

dydx

+ x2y = 0. (5.9)

Assuming, as usual,

y = xλ∞

∑n=0

anxn (5.10)

after some algebra we get

∞

∑n=0

(n+λ )(n+λ +1)anxn +∞

∑n=2

an−2xn = 0⇒

λ (λ +1)a0 +(λ +1)(λ +2)a1x+∞

∑n=2

((n+λ )(n+λ +1)an +an−2)xn = 0.

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Then the indicial equation isλ (λ +1) = 0

with roots λ1 = 0,λ2 =−1. With λ1 = 0 we get

a1 = 0, an =−an−2

n(n+1)

and soa1 = 0, a2 =−

13!

a0, a3 = 0, a4 =15!

a0, ...

and (taking for simplicity a0 = 1):

y1 (x) =(

1− x2

3!+

x4

5!− ...

)=

(1− x3

3! +x5

5! − ...)

x=

sinxx

.

Taking now λ2 =−1 we similarly get

0a1 = 0, an =−1

n(n−1)an−2.

Since we are looking for some solution, let us take a1 = 0. Then we get

a2 =−12!

a0, a4 =14!

a0, ...

and, taking for simplicity a0 = 1,

y2 (x) = x−1(

1− x2

2!+

x4

4!− ...

)=

cosxx

.

So the general solution is

y(x) = c1sinx

x+ c2

cosxx

.

5.1.13 Example. Let us solve the ODE (it is the Bessel equation, which we willstudy in greater detail in Chapter 17):

x2 d2ydx2 + x

dydx

+(x2−n2)y = 0 (5.11)

assuming n ∈ 1,2,3, ... . Again we assume

y(x) = xλ∞

∑k=0

ckxk =∞

∑k=0

ckxk+λ

So (x2−n2)y = xλ

∞

∑k=0

ckxk+2−n2xλ∞

∑k=0

ckxk = xλ∞

∑k=2

ck−2xk−n2xλ∞

∑k=0

ckxk,

xdydx

= xλ∞

∑k=0

(k+λ )ckxk,

x2 d2ydx2 = xλ

∞

∑k=0

(k+λ )(k+λ −1)ckxk.

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Adding the above we get

xλ∞

∑k=0

[(k+λ )(k+λ −1)ck +(k+λ )ck + ck−2−n2ck

]xk = 0

and so [(k+λ )(k+λ −1)ck +(k+λ )ck + ck−2−n2ck

]= 0⇒[

(k+λ )2−n2]

ck =−ck−2

Letting k = 0 (and since c−2 = 0 and we do not want c0 = 0) we get the indicialequation

λ2 = n2

When λ = n, then((k+n)2−n2

)ck =−ck−2⇒(

k2 +2kn)

ck =−ck−2⇒ ck =−1

k (k+2n)ck−2

So one solution is

y1 (x) = xn(

1− x2

2(2+2n)+

x4

2 ·4 · (2+2n) · (4+2n)+ ...

). (5.12)

When λ =−n, we get a second solution

y2 (x) = x−n(

1− x2

2(2−2n)+

x4

2 ·4 · (2−2n) · (4−2n)+ ...

)(5.13)

but, since we assumed n ∈ 1,2,3, ..., the second series does not exist. Still (forn∈ 1,2,3, ...) we have found one solution of (5.11), namely y1 (x). This is as statedby Frobenius’ Theorem.

5.1.14 Example. Finally, let us consider a case in which the indicial equationhas a double root. We solve

x2 d2ydx2 −2x2 dy

dx+

(x2 +

14

)y = 0.

By the usual methods we get the indicial equation

λ2−λ +

14= 0

with roots λ1 = λ2 =12 . Assuming

y(x) = xλ∞

∑k=0

ckxk =∞

∑k=0

ckxk+λ

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we get, after the usual manipulations,

0=(2λ −1)2+((2λ +1)2 a1−8λa0

)x+

∞

∑n=2

((2n+2λ −1)2 an−8(n+λ −1)an−1 +4an−2

)xn

Setting λ = 12 we get

a1 = a0, a2 =12!

a0, a3 =13!

a0, ... .

Hence we get a solution

y1 = x1/2(

1+ x+x2

2!+ ...

)= x1/2ex.

The Frobenius method cannot get another solution! (We can get another linearlyindependent solution, by other methods. It turns out to be y2 (x) = x1/2ex lnx.)

5.2 Solved Problems

5.3 Unsolved Problems1. Find as a power series around x0 = 0 the general solution of

(1+ x2) d2y

dx2 +

6xdydx +6y = 0.

Ans. y(0)+ xy′ (0)−3x2y(0)−2x3y′ (0)+5x4y(0)+O(x5).

2. Find as a power series around x0 = 0 the general solution of(1+ x2) d2y

dx2 +

5xdydx +3y = 0.

Ans. y(0)+ xy′ (0)− 32x2y(0)− 4

3x3y′ (0)+ 158 x4y(0)+O

(x5).

3. Find as a power series around x0 = 0 the general solution of(1+3x2) d2y

dx2 −6xdy

dx +10y = 0.Ans. y(0)+ xy′ (0)−5x2y(0)− 2

3x3y′ (0)+ 53x4y(0)+O

(x5).

4. Find as a power series around x0 = 0 the general solution of(1− x2) d2y

dx2 +

6xdydx +6y = 0.

Ans. y(0)+ xy′ (0)−3x2y(0)−2x3y′ (0)+4x4y(0)+O(x5).


dx2 −5xdy

dx +6y = 0.Ans. .

6. Find as a power series around x0 = 0 the general solution of d2ydx2 − xy = 0.

Ans. y(0)+ xy′ (0)+ 16x3y(0)+ 1

12x4y′ (0)+O(x5).

7. Find as a power series around x0 = 0 the general solution of d2ydx2 + xy = 0.

Ans. y(0)+ xy′ (0)− 16x3y(0)− 1

12x4y′ (0)+O(x5).


dx2 +

6x2 dydx +6xy = 0.

Ans. y(0)+ xy′ (0)− x3y(0)− x4y′ (0)+O(x5).

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dx2 +

6xdydx +6y = 0.

Ans. y(0)+ xy′ (0)−3x2y(0)−2x3y′ (0)+ 92x4y(0)+O

(x5).


dx2 +

6xdydx +6y = 0.

Ans. .11. Find as a power series around x0 = 0 the general solution of

(1− x3) d2y

dx2 +

x3 dydx + x2y = 0.

Ans. y(0)+ xy′ (0)− 112x4y(0)+O

(x5).

12. Find as a power series around x0 = 0 the solution of

(1− x2) d2y

dx2 + xdydx + y = 0

y(0) = 1y′ (0) = 1

.

Ans. 1+ x− 12x2− 1

3x3 + 124x4 +O

(x5).


(1− x2) d2y

dx2 + xdydx + y = 0

y(0) = 0y′ (0) = 2

.

Ans. 2x− 23x3 +O

(x6).


(1+ x2) d2y

dx2 + xdydx + y = 0

y(0) = 1y′ (0) = 1

.

Ans. 1+ x− 12x2− 1

3x3 + 524x4 +O

(x5).

15. Find as a power series around x0 = 0 the solution of(1− x) d2y

dx2 +(1− x)2 dydx + y = 0

y(0) = 1y′ (0) = 1

.

Ans. 1+ x− x2 + 16x3− 7

24x4 +O(x5).


(1− x4) d2y

dx2 +2xdydx + x2y = 0

y(0) = 1y′ (0) = 0

.

Ans. 1− 112x4 +O

(x5).

17. Find power series solutions (around x0 = 0) of 2x2 d2ydx2 +x(3+2x) dy

dx−(1− x)y=0.Ans. 1

x − 1 + 12x− 1

6x2 + 124x3 + O

(x4), √x− 2

5x32 + 4

35x52 − 8

315x72 + 16

3465x92 +

O(

x112

).

18. Find power series solutions (around x0 = 0) of 2x2 d2ydx2 +x(5+ x) dy

dx−(2−3x)y=0.Ans. 1

x2 +13x +

13 −

13x+ 1

9x2 +O(x3), √x− 1

2x32 + 1

8x52 − 1

48x72 + 1

384x92 +O

(x

112

).

.19. Find power series solutions (around x0 = 0) of 3x2 d2y

dx2 + x(1+ x) dydx − y = 0.

Ans. 13√x− 1

3x23 + 1

18x53 − 1

162x83 + 1

1944x113 + O

(x

143

), x− 1

7x2 + 170x3 − 1

910x4 +1

14560x5 +O(x6).

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20. Find power series solutions (around x0 = 0) of x2 (8+ x) d2ydx2 + x(2+3x) dy

dx +(1+ x)y = 0.Ans.

√x− 9

40x32 + 5

128x52 − 245

39936x72 + 6615

7241728x92 +O

(x

112

), 4√

x− 2596x

54 + 675

14336x94 −

380255046272x

134 + 732615

645922816x174 +O

(x

214

).

21. Find power series solutions (around x0 = 0) of 8x2 d2ydx2 + x

(2+ x2) dy

dx + y = 0.

Ans. 4√

x− 1112x

94 + 3

17920x174 +O

(x

214

),√

x− 172x

52 + 5

19584x92 +O

(x

112

).

22. Find power series solutions (around x0 = 0) of x(1+ x) d2ydx2 +(1− x) dy

dx + y = 0.Ans. lnx− x(lnx−4)+O

(x5), 1− x+O

(x5).

23. Find power series solutions (around x0 = 0) of x2 d2ydx2 −x(1− x) dy

dx +(1− x2)y =

0.Ans. x lnx−x2 (lnx−1)+x3 (3

4 lnx−1)− ..., x−x2+ 3

4x3− 1336x4+ 79

576x5+O(x6)

.24. Find power series solutions (around x0 = 0) of 4x2 d2y

dx2 +(1+4x)y = 0.

Ans.√

x lnx− x32 (lnx−2)+ x

52(1

4 lnx− 34

)− x

72( 1

36 lnx− 11108

),√

x− x32 + 1

4x52 −

136x

72 + 1

576x92 +O

(x

112

).

25. Find power series solutions (around x0 = 0) of xd2ydx2 −5dy

dx + xy = 0.Ans. −86400−10800x2−1350x4 +O

(x5), x6− 1

16x8 + 1640x10 +O

(x11).

26. Find power series solutions (around x0 = 0) of x(1+ x) d2ydx2 −4dy

dx −2y = 0.Ans. x5−3x6 +6x7−10x8 +15x9 +O

(x10), 2880−1440x+480x2 +O

(x5).

27. Find power series solutions (around x0 = 0) of x2 d2ydx2 −3xdy

dx +(3+4x)y = 0.Ans. x3− 4

3x4+ 23x5− 8

45x6+ 4135x7+O

(x8),−2x−8x2+16x3 lnx−x4 (64

3 lnx− 2569

)+

x5 (323 lnx− 200

9

)+O

(x6).

28. Find power series solutions (around x0 = 0) of xd2ydx2 + y = 0.

Ans. x− 12x2 + 1

12x3− 1144x4 + ..., 1− x lnx+ x2 (1

2 lnx− 34

)− x3 ( 1

12 lnx− 736

)+ ...

.29. Find power series solutions (around x0 = 0) of xd2y

dx2 + xdydx + y = 0.

Ans. 1− x(lnx+1) + x2 lnx− x3 (12 lnx− 1

4

)+ ..., x− x2 + 1

2x3− 16x4 + 1

24x5 +

O(x6).

30. Find power series solutions (around x0 = 0) of 6x2 (1+2x2) d2ydx2 +x

(1+50x2) dy

dx +(1+30x2)y = 0.

Ans.√

x−2x52 +4x

92 +O

(x

112

), 3√

x−2x73 +4x

133 +O

(x

163

).

31. Find power series solutions (around x0 = 0) of 2x2 (1+ x) d2ydx2−x(1−3x) dy

dx +y=0.Ans. x− x2 + x3− x4 + x5 +O

(x6), √x− x

32 + x

52 − x

72 + x

92 +O

(x

112

).

32. Find power series solutions (around x0 = 0) of x2 (8+ x) d2ydx2 + x(2+3x) dy

dx +(1+ x)y = 0.Ans.

√x− 9

40x32 + 5

128x52 − 245

39936x72 + ...,

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4√

x− 2596x

54 + 675

14336x94 − 38025

5046272x134 + ... .


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6 Laplace Transform . . . . . . . . . . . . . . . 49

7 Differential Equations . . . . . . . . . . . . 59

8 Convolution and Integral Equations69

9 Dirac Delta and Generalized Func-tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

10 Difference Equations . . . . . . . . . . . . . 87

Laplace

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6. Laplace Transform

The Laplace transform can be seen as a generalization of the Taylor series expansionto continuous powers s ∈ R+

0 rather than discrete powers n ∈ N0 .

6.1 Theory and Examples6.1.1 Definition. Let f (t) be defined on [0,∞). The Laplace transform of f (t) isdefined to be

F (s) = L ( f (t)) :=∫

∞

0−e−st f (t)dt (6.1)

provided the integral is well defined.

6.1.2 For the time being we assume s ∈ R. Furthermore, for the time being wecan assume the lower limit of integration is 0, i.e., we can use the definition

F (s) = L ( f (t)) :=∫

∞

0e−st f (t)dt. (6.2)

This will change in Chapter 5.

6.1.3 Example. To find the Laplace transform of f (t) = 1, we have

L ( f (t)) =∫

∞

0e−st f (t)dt

=

(−1

se−st

)∞

t=0=−1

se−s∞ +

1s

e−s0 ==−1s

e−∞ +1s

e−0

=1s.

6.1.4 Example. To find the Laplace transform of f (t) = t, we have

L ( f (t)) =∫

∞

0e−st f (t)dt =

∫∞

0te−stdt =

(− 1

s2 e−st (st +1))∞

t=0

=

(− 1

s2 e−s∞ (s∞+1))−(− 1

s2 e−s0 (s0+1))=− 1

s2 e−∞ (∞+1)+1s2 e−0 (0+1)

=1s2 .

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50 Chapter 6. Laplace Transform

6.1.5 Example. Similarly we find

L (tn) =∫

∞

0e−sttndt =

n!sn+1 .

6.1.6 Definition. We say that f (t) is piecewise continuous in [a,b] iff [a,b] canbe partitioned as

[a,b] = [a0,a1]∪ [a1,a2]∪ ...∪ [aK−1,aK]

so that1. for each k ∈ 0,1, ...,K−1 : f (t) is continous in [ak,ak+1],2. limt→a−0

f (t) and limt→a+Kf (t) exist,

3. for each k ∈ 1, ...,K−1 : limt→a−kf (t) and limt→a+k

f (t) exist.

6.1.7 Definition. We say that f (t) is of exponential order γ after N iff there existM > 0 such that

∀t > N : | f (t)|< Meγt .

Often we use the simpler: “ f (t) is of exponential order γ”, or “ f (t) is of exponentialorder”.

6.1.8 Theorem. If for every N: f (t) is of exponential order γ after N and piecewisecontinuous in [0,N], then L ( f (t)) exists for all s > γ.

6.1.9 Example. To find the Laplace transform of f (t) = eat we have

L ( f (t)) =∫

∞

0e−steatdt =

(e(a−s)t

a− s

)t=∞

t=0

=

(e(a−s)∞

a− s

)+

(e(a−s)0

a− s

)=

1s−a

, defined for all s > a.

6.1.10 Example. To find the Laplace transforms of f (t) = cos t and g(t) = sin t weuse

L (cos t)+ iL (sin t) = L ((cos t + isin t))

=∫

∞

0e−st (cos t + isin t)dt =

∫∞

0e−steitdt =

1s− i

=s+ i

s2 +1

=s

s2 +1+ i

1s2 +1

.

HenceL (cos t) =

ss2 +1

, L (sin t) =1

s2 +1.

6.1.11 Example. Trying to find the Laplace transform of f (t) = et2we note that

for all s ∈ R we haveL ( f (t)) =

∫∞

0e−stet2

dt = ∞.

So the Laplace transform of et2does not exist.

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6.1.12 Theorem (Linearity). L (c1 f1 (t)+ c2 f2 (t)) = c1L ( f1 (t))+ c2L ( f2 (t)) .Proof. Obvious, from the linearity of the integral operator.

6.1.13 Example. L(5t +3e−2t)= 5

s2 +3

s+2 .

6.1.14 Theorem (Shift). If L ( f (t)) = F (s), then L (eat f (t)) = F (s−a) .Proof.

L(eat f (t)

)=∫

∞

0eat f (t)e−stdt =

∫∞

0f (t)e−(s−a)tdt = F (s−a) .

6.1.15 Example. L(e−2t cos t

)= s+2

(s+2)2+1.

6.1.16 Theorem. If L ( f (t)) = F (s) and

g(t) =

f (t− t0) t > t00 t < t0

then L (g(t)) = e−t0sF (s) .Proof.

L (g(t)) =∫

∞

0g(t)e−stdt =

∫∞

t0f (t)e−stdt

=∫

∞

0f (t− t0)e−s(t−t0)e−st0d (t− t0) = e−st0

∫∞

0f (t− t0)e−s(t−t0)dt

= e−st0F (s) .

6.1.17 Theorem (Scaling). If a> 0 and L ( f (t)) =F (s), then L ( f (at)) = 1aF( s

a

).

Proof. Obvious.

6.1.18 Example. L (cos3t) = s/3

( s3)

2+1

= 3s2+9 .

6.1.19 Theorem (Transform of Derivative). If for every N: (i) f (t) is continuousand of exponential order γ after N, (ii) f ′ (t) is piecewise continuous in [0,N] and(iii) L ( f (t)) = F (s), then

L(

f ′ (t))= sF (s)− f (0) .

Proof.

L(

f ′ (t))=∫

∞

0e−st d f

dtdt =

∫∞

0e−std f

= e−st f (t) |∞t=0−∫

∞

0f (t)de−st = e−st f (t) |∞t=0 +

∫∞

0f (t)se−stdt

= e−s∞ f (∞)− e−s0 f (0)+ sL( f (t)) = sF (s)− f (0) .

6.1.20 Theorem. If f (t) L ( f (t)) = F (s) and under conditions similar to thoseof the previous theorem, we have

L(

f ′′ (t))= s2F (s)− s f (0)− f ′ (0) ,

L(

f (n) (t))= snF (s)− sn−1 f (0)− sn−2 f ′ (0)− ....− f (n−1) (0) .

Proof. Similar to the previous one.

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6.1.21 Example. L (t) = 1s2 , then

L (1) = L((t)′)= sL (t)− (t)t=0 = s

1s2 −0 =

1s

6.1.22 Theorem (Transform of Integral). If L ( f (t)) = F (s), then

L

(∫ t

0f (u)du

)=

1s

F (s) .

Proof. Easy.

6.1.23 Example. L (t) = 1s2 , then

L(t2)= 2

∫ t

0tdt = 2

1s

1s2 =

2s3 .

6.1.24 Example. Using mathematical induction: L(t0) = L (1) = 1

s ; assumeL (tn) = n!

sn+1 , then

sL(tn+1)− (t)0 = L ((n+1) tn)⇒ sL

(tn+1)= (n+1)

n!sn+1 ⇒

L(tn+1)= (n+1)!

sn+2 .

6.1.25 Apparently, multiplying by s gives differentiation in the time domain;multiplying by 1

s gives integration in the time domain.

6.1.26 Example. Suppose we want to solve

d fdt

+ f = 0, f (0) = 3.

Then we have

sF (s)− f (0)+F (s) = 0⇒ (s+1)F (s) = f (0)⇒

F (s) =3

s+1⇒ f (t) = 3e−t .

Note we have converted the DE to an algebraic equation, solved that and foundf (t) by inverting F (s).

6.1.27 Example. Suppose we want to solve

d2 fdt2 +3

d fdt

+2 f = 1, f (0) = 0, f ′ (0) = 0.

Then we have

s2F− s f (0)− f ′ (0)+3(sF− f (0))+2F =1s⇒(

s2 +3s+2)

F =1s⇒

F =1

s(s2 +3s+2).

To continue the solution we must decompose 1s(s2+3s+2)

into partial fractions; we

will show how this is done in Chapter 5.

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6.1.28 Theorem (Inverse Transform of Derivative). If L ( f (t)) = F (s), thenL (tn f (t)) = (−1)n dnF

dsn .Proof. Easy by mathematical induction.

6.1.29 Example. L(te2t)= (−1)1 (L (

e2t))′ =−( 1s−2

)′= 1

(s−2)2 .

6.1.30 Theorem (Inverse Transform of Integral). If L ( f (t)) = F (s), thenL(1

t f (t))=∫

∞

u F (u)du provided limt→0f (t)

t exists.Proof. Easy.

6.1.31 Example. L( sin t

t

)=∫

∞

udu

u2+1 = arctan 1s .

6.1.32 Apparently multiplying (dividing) by t in the time domain yields differentia-tion (integration) in the Laplace domain.

6.1.33 Theorem. If f (t) is periodic with period T (i.e., f (t +T ) = f (t)) then

L ( f (t)) =∫ T

0 e−st f (t)dt1− e−sT .

Proof. Easy.

6.1.34 Theorem (First Limit Theorem). If L ( f (t)) = F (s) then lims→∞ F (s) = 0.Proof. lims→∞ F (s) = lims→∞ L ( f (t)) = lims→∞

∫e−st f (t)dt =

∫lims→∞ e−st f (t)dt =

0.

6.1.35 Theorem (Second Limit Theorem). If L ( f (t)) = F (s) and the limitsexist, then

limt→0

f (t) = lims→∞

sF (s) , limt→∞

f (t) = lims→0

sF (s) .

Proof. SincesF (s)− f (0) = L

(f ′ (t)

)=∫

∞

0e−st f ′ (t)dt.

we have

lims→∞

(sF (s)− f (0)) = lims→∞

∫∞

0e−st f ′ (t)dt =

∫∞

0lims→∞

e−st f ′ (t)dt = 0.

Hencelims→∞

sF (s) = f (0) = limt→0

f (t) .

Similarly

lims→0

(sF (s)− f (0))= lims→0

∫∞

0e−st f ′ (t)dt− f (0)=

∫∞

0f ′ (t)dt− f (0)= lim

t→∞f (t)− f (0) .

Hencelims→0

sF (s)− f (0) = limt→∞

f (t)− f (0)⇒ lims→0

sF (s) = limt→∞

f (t) .

6.1.36 Definition. If F (s) = L ( f (t)), then we also write f (t) = L −1 (F (s)) andwe say that F (s) is the inverse Laplace transform of F (s).

6.1.37 Theorem (Lerch). If, for every N, f (t) is sectionally continuous in [0,N]and of exponential order γ after N and F (s) = L ( f (t)), then L −1 (F (s)) is uniqueand equal to f (t).

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6.1.38 Example. L −1 ( 1s+3

)= e−3t (since L

(e−3t)= 1

s+3 ).

6.1.39 Example. L −1(

1s2

)= t.

6.1.40 Theorem (Linearity). L −1 (c1F1 + c2F2) = c1L−1 (F1)+ c2L

−1 (F2) .Proof. Easy.

6.1.41 Example.

L −1(

1s2−a2

)= L −1

(1

2a(s−a)− 1

2a(s+a)

)=

12a

(eat− e−at)= sinh(at)

a.

6.1.42 Theorem (Shift). L −1 (F (s−a)) = eat f (t) .Proof. Easy.

6.1.43 Example.

L −1(

1s2 +2s+5

)= L −1

(1

(s+1)2 +22

)=

12

e−t sin(2t) .

6.1.44 Theorem (Shift).

L −1 (e−t0sF (s))=

(f (t− t0) when t ≥ t0

0 when t < t0

).

Proof. Easy.

6.1.45 Theorem (Scaling). L −1 (F (ks)) = 1k f( t

k

)(with k > 0).

Proof. Easy.

6.1.46 Example.

L −1(

s4s2 +64

)= L −1

(12

2s

(2s)2 +82

)=

12

(12

cos(

8t2

))=

14

cos(4t) .

6.1.47 Theorem (Inverse Transform of Derivative). L −1 (F ′ (s)) =−t f (t).Proof. Easy.

6.1.48 Example. L −1(

s(s2+1)

2

). We note that

(1

s2+1

)′=− 2s

(s2+1)2 . Then

L −1

(s

(s2 +1)2

)=−1

2L −1

((1

s2 +1

)′)=

(−1

2

)(−1) tL −1

(1

s2 +1

)=

12

t sin t.

6.1.49 Theorem (Inverse Transform of Integral). L −1 (∫

∞

s F (u)du) = f (t)t .

Proof. Easy.

6.1.50 Example. L −1 (ln( s+1s

)). We note that∫

∞

s

(1u− 1

u+1

)du = (lnu− ln(u+1))∞

s=0 = ln(

s+1s

).

Hence

L −1(

ln(

s+1s

))=

1tL −1

(1s− 1

s+1

)=

1− e−t

t.

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6.1.51 Theorem. If f (0) = 0, then L −1 (sF (s)) = f ′ (t).Proof. Easy.

6.1.52 Example. L (sin t) = 1s2+1 , sin0 = 0. Then L −1

(s

s2+1

)= (sin t)′ = cos t.

6.1.53 Example. How about L (cos t) = ss2+1 , what is then L −1

(s2

s2+1

)? It turns

out to be L −1(

s2

s2+1

)=−sin t +δ (t). Now −sin t = (cos t)′, but where did the δ (t)

come from? Note that in this case we have s2

s2+1 , what is different in this fractionfrom what we have seen previously?

6.1.54 Question. What would the L −1 (sF (s)) be if f (0) 6= 0? Note that then wewould have an abrupt change of f (t) at t = 0. Could it be

L −1 (sF (s)) = f ′ (t)+L −1 ( f (0))

What would L −1 ( f (0)) be? L −1 (1) ?

6.1.55 Theorem (Inverse Transform of Integral). If L −1(

F(s)s

)=∫ t

0 f (u)du.Proof. Easy.

6.1.56 Example. L (cos t) = ss2+1 . Then

L −1(

1s

ss2 +1

)= L −1

(1

s2 +1

)=∫ t

0cos tdt = sin t.

6.1.57 Example. L −1(

1s2+5s+6

). Since

1s2 +5s+6

=1

s+2− 1

s+3,

we haveL −1

(1

s2 +5s+6

)= L −1

(1

s+2− 1

s+3

)= e−2t− e−3t

6.1.58 Example. L −1(

3s+7s2−2s−3

). Since

3s+7s2−2s−3

=4

s−3− 1

s+1,

we haveL −1

(3s+7

s2−2s−3

)= L −1

(4

s−3− 1

s+1

)= 4e3t− e−t

6.1.59 Example. L −1(

2s2−4s3+6s2+11s+6

). Since

2s2−4s3 +6s2 +11s+6

=7

s+3− 4

s+2− 1

s+1,

we have

L −1(

2s2−4s3 +6s2 +11s+6

)= L −1

(7

s+3− 4

s+2− 1

s+1

)= 7e−3t−4e−2t− e−t

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6.1.60 Example. L −1(

3s+1s3−s2+s−1

). Since

3s+1s3− s2 + s−1

=2

s−1− 2s−1

s2 +1,

we have

L −1(

3s+1s3− s2 + s−1

)= L −1

(2

s−1− 2s−1

s2 +1

)= 2et−2cos t + sin t

6.1.61 We summarize basic pairs in the following table.

f (t) F (s)1 1

st 1

s2

tn n!sn+1

eat 1s−a

cosat ss2+a2

sinat as2+a2

6.1.62 We summarize basic properties in the following table

f (t) F (s)c1 f1 + c2 f2 c1F1 + c2F2a f (at) F

( sa

)eat f (t) F (s−a)

f (t− t1) t > t10 t < t1

e−st1F (s)

f ′ (t) sF (s)− f (0)f ′′ (t) s2F (s)− s f (0)− f ′ (0)−t f (t) F ′ (s)t2 f (t) F ′′ (s)∫ t

0 f (u)du 1s F (s)

f (t)t

∫∞

s F (s)ds

6.2 Solved Problems

6.3 Unsolved Problems1. Find the Laplace transform of f (t) = sinh t.

Ans. Laplace transform is: 1s2−1 .

2. Find the Laplace transform of f (t) =

2 when t < 50 when t > 5 .

Ans. 21−e−5s

s .3. Find the Laplace transform of f (t) = t2e4t .

Ans. Laplace transform is: 2(s−4)3 .

4. Find the Laplace transform of f (t) = e−4t sin2t.Ans. Laplace transform is: 2

(s+4)2+4.

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5. Find the Laplace transform of f (t) = cosh5t.Ans. Laplace transform is: s

s2−25 .6. Find the Laplace transform of f (t) = 2cos5t−3sin5t.

Ans. Laplace transform is: 2 ss2+25 −

15s2+25 .

7. Find the Laplace transform of f (t) = t sinat.Ans. Laplace transform is: 2a s

(a2+s2)2 .

8. Find the Laplace transform of f (t) = t2 cos t.Ans. Laplace transform is: 8 s3

(s2+1)3 −6 s

(s2+1)2 .

9. Evaluate∫

∞

0 e−2t cos tdt.Ans. 2

5 .10. Evaluate

∫∞

0 te−2t cos tdt.Ans. 3

25 .11. Find the inverse Laplace transform of 3

4s2+16 .Ans. 3

8 sin2t.12. Find the inverse Laplace transform of 4s

s2−25 .Ans. 4cosh5t.

13. Find the inverse Laplace transform of 3s−5s2+5s+6 .

Ans. 3e−52 t (cosh 1

2t− 253 sinh 1

2t).

14. Find the inverse Laplace transform of 3s−5s2−5s+6 .

Ans. 3e52 t (cosh 1

2t + 53 sinh 1

2t).

15. Find the inverse Laplace transform of 4s+12s2+8s+12 .

Ans. 4e−4t (cosh2t− 12 sinh2t

).

16. Find the inverse Laplace transform of 3s+7s2+2s+13 .

Ans. 3e−t (cos2√

3t + 29

√3sin2

√3t).

17. Find the inverse Laplace transform of e−πs(s+1)s2+s+ 1

4.

Ans. e12 π− 1

2 t Heaviside(t−π)(1

2t− 12π +1

).

18. Find the inverse Laplace transform of s(s2+4)

2 .

Ans. 14t sin2t.

19. Find the inverse Laplace transform of s2

(s2+4)2 .

Ans. 14 sin2t + 1

2t cos2t.20. Find the inverse Laplace transform of 1

s2(s2+1).

Ans. t− sin t.21. Find the inverse Laplace transform of 3s+7

s2−2s−3 .Ans. 3et (cosh2t + 5

3 sinh2t).

22. Find the inverse Laplace transform of 2s2−1(s+1)(s+2)(s+3) .

Ans. 12e−t−7e−2t + 17

2 e−3t .23. Find the inverse Laplace transform of 2s2−1

s3+6s2+5s−12 .Ans. 1

20et− 174 e−3t + 31

5 e−4t .24. Find the inverse Laplace transform of 3s+1

s3−s2+s−1 .

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Ans. 2et−2cos t + sin t.25. Find the inverse Laplace transform of 5−2s

6s2+7s+2 .

Ans. −13e−

712 t (cosh 1

12t−37sinh 112t).

6.4 Advanced Problems1. Prove that L (h(t)+h(t−T )+h(t−2T )+ ...) = 1

1−eT s .2. Suppose f (t) has period T . Prove that

L ( f (t)) =1

1− eT s

∫ T

0e−sT f (t)dt.

3. Find a function which is not of exponential order.

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7. Differential Equations

We present the application of the Laplace transform to the solution of ODEs. Thebasic idea is that the Laplace transform can convert an ODE to an algebraicequation, which is easier to solve.


7.1.1 To solve a differential equation with unknown y(t) using the Laplace trans-form, we first convert to an algebraic equation with unknown Y (s), then we solvethe algebraic equation to obtain Y (s) and finally we invert to obtain the desiredy(t).


d2ydt2 + y = t, y(0) = 1, y′ (0) =−2

we take the Laplace transform and we have

s2Y − sy(0)− y′ (0)+Y =1s2 ⇒ s2Y − s+2+Y =

1s2 ⇒(

s2 +1)

Y = s−2+1s2 ⇒ Y =

s−2s2 +1

+1

s2 (s2 +1)⇒

Y (s) =s

s2 +1− 2

s2 +1+

1s2 −

1s2 +1

⇒

y(t) = cos t−3sin t + t.


d2ydt2 +2

dydt

+5y = e−t sin t, y(0) = 0, y′ (0) = 1.

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60 Chapter 7. Differential Equations


s2Y − sy(0)− y′ (0)+2(sY − y(0))+5Y =1

(s+1)2 +1⇒

(s2 +2s+5

)Y −1 =

1

(s+1)2 +1⇒

Y (s) =1

s2 +2s+5+

1(s2 +2s+2)(s2 +2s+5)

=1

s2 +2s+5+

13(s2 +2s+2)

− 13(s2 +2s+5)

y(t) =13

e−t sin t +13

e−t sin2t.


d2ydt2 +9y = cos2t, y(0) = 1, y

(π

2

)=−1.

we let y′ (0) = a and, taking Laplace transform, we have

s2Y − sy(0)− y′ (0)+9Y =s

s2 +4⇒(

s2 +9)

Y − s−a =s

s2 +4⇒

Y (s) =s+as2 +9

+s

(s2 +4)(s2 +9)⇒

Y (s) =s

s2 +9+

as2 +9

+s

5(s2 +4)− s

5(s2 +9)⇒

y(t) =45

cos3t +a3

sin3t +15

cos2t.

Now

−1 = y(

π

2

)=

45

cos3π

2+

a3

sin3π

2+

15

cos2π

2⇒

−1 =−a3− 1

5⇒ a =

125

And soy(t) =

45

cos3t +45

sin3t +15

cos2t

7.1.5 Example. We define

h(t) =

0 when t < 01 whent > 0 .

Clearly L (h(t)) = 1.Now, to solve

dydt

+2y = h(t−4) , y(0) = 3

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sY − y(0−)+2Y =

e−4s

s⇒

(s+2)Y = 3+e−4s

s⇒

Y (s) =3

s+2+

1s(s+2)

e−4s⇒

Y (s) =3

s+2+

12

e−4s(

1− 1s+2

)⇒

y(t) = 3e−2t +12

h(t−4)(

1− e−2(t−4)).

7.1.6 The Laplace transform method can also be applied to the solution of systemsof ODEs, as seen in the following examples.


dxdt

= 2x+ y+1, x(0) = 0

dydt

= 3x+4y, y(0) = 0

we have

sX− x(0) = 2X +Y +1s

sY − y(0) = 3X +4Y

(s−2)X−Y =1s

−3X +(s−4)Y = 0

Solution is:

X =s−4

5s−6s2 + s3 =s−4

s(s−1)(s−5)=

34(s−1)

+1

20(s−5)− 4

5s

Y =3

5s−6s2 + s3 =3

s(s−1)(s−5)=

320(s−5)

− 34(s−1)

+35s

and so

x =1

20e5t +

34

et− 45

y =3

20e5t− 3

4et +

35

7.2 Solved Problems7.2.1 Problem. Solve

d2ydt2 −3

dydt

+2y = 4e2t , y(0) =−3, y′ (0) = 5.

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Solution. Taking Laplace transform we get

s2Y − sy(0)− y′ (0)−3(sY − y(0))+2Y =4

s−2⇒(

s2−3s+2)

Y +3s−5−9 =4

s−2⇒

=− 3s−14(s−1)(s−2)

+4

(s−2)2 (s−1)

=4

s−2− 7

s−1+

4

(s−2)2

y(t) = 4e2t−7et +4te2t .

7.2.2 Problem. Solve

d3ydt3 −3

d2ydt2 +3

dydt− y = t2et , y(0) = 1, y′ (0) = 0, y′′ (0) =−2.

Solution. Taking Laplace transform we get(s3−3s2 +3s−1

)Y − s2 +3s−1 =

2

(s−1)3 ⇒(s3−3s2 +3s−1

)Y =

2

(s−1)3 + s2−3s+1⇒

Y =2

(s−1)6 +s2−3s+1

(s−1)3

=2

(s−1)6 +s2−2s+1− s

(s−1)3

=2

(s−1)6 +(s−1)2

(s−1)3 −s−1

(s−1)3 +1

(s−1)3

y(t) =2ett5

5!+ et− tet +

t2et

2!=

ett5

60+ et− tet +

t2et

2.

7.2.3 Problem. Solved3ydt3 −3

d2ydt2 +3

dydt− y = t2et .

Solution. We set y(0) = a, y′ (0) = b, y′′ (0) = c.(s3Y −as2−bs− c

)−3(s2Y −as−b

)+3(sY −a)−Y =

2

(s−1)3 ⇒(s3−3s2 +3s−1

)Y −as2− (b−3a)s− (c−3b+3a) =

2

(s−1)3 ⇒(s3−3s2 +3s−1

)Y =

2

(s−1)3 +as2 +(b−3a)s+(3a−3b+ c)

(s−1)6 ⇒

Y =2

(s−1)6 +c3

(s−1)3 +c2

(s−1)2 +c1

s−1

y(t) =ett5

60+ c1et + c2tet + c3t2et .

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d2ydt2 + y = h(t−2)−h(t−4) , y(0) = 1, y′ (0) = 0


s2Y − s+Y =e−2s

s− e−4s

s⇒(

s2 +1)

Y = s+e−2s

s− e−4s

s⇒

Y =s

s2 +1+

e−2s

s(s2 +1)− e−4s

s(s2 +1)

We have

Y1 =s

s2 +1

Y2 =1s− s

s2 +1and hence

y1 (t) = cos ty2 (t) = 1− cos t

and

y(t) = y1 (t)+h(t−2)y2 (t−2)−h(t−4)y2 (t−4)= cos t +h(t−2)(1− cos(t−2))−h(t−4)(1− cos(t−4))

7.2.5 Problem Solve

d2ydt2 +

dydt

+54

y = h(t)−h(t−5) , y(0) = 0, y′ (0) = 0


s2Y + sY +54

Y =1− e−5s

s⇒

Y =(

1− e−5s) 1

s(s2 + s+ 5

4

)We will first compute z(t) when

Z (s) =1

s(s2 + s+ 5

4

)Since s2 + s+ 5

4 =(s+ 1

2

)2+1 we have

1s(s2 + s+ 5

4

) = 45s−

45s+ 4

5

s2 + s+ 54

=45

(1s− s

s2 + s+ 54

− 1s2 + s+ 5

4

)

=45

(1s− s(

s+ 12

)2+ 1

4

− 12

2(s+ 1

2

)2+ 1

4

)

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and

z(t) =45

(1− e−2t cos(t)− e−2t

2sin(t)

)Then

y(t) = z(t)− z(t−5)h(t−5)

7.2.6 Problem Solve

d2ydt2 + y = f (t) , y(0) = 0, y′ (0) = 0

wheref (t) = (−1)m when t ∈ [mπ,(m+1)π)


f (t) = 1+2∞

∑n=1

(−1)n h(t−nπ)

F (s) =1s

(1+2

∞

∑n=1

(−1)n e−nπs

)Then

Y (s) =1

s(s2 +1)

(1+2

∞

∑n=1

(−1)n e−nπs

)Now

L −1(

1s(s2 +1)

)= L −1

(1s− s

s2 +1

)= 1− cos t

L −1(

e−nπs

s(s2 +1)

)= h(t−nπ)(1− cos(t−nπ)) = h(t−nπ)(1− (−1)n cos(t))

Hence

y(t) = 1− cos t +2∞

∑n=1

(−1)n h(t−nπ)(1− (−1)n cos(t))

Note that when t ∈ [mπ,(m+1)π) we get

y(t) = 1− cos t +2m

∑n=1

(−1)n h(t−nπ)(1− (−1)n cos(t))

= (−1)m− (2m+1)cos t


dxdt

= 2x−3y, x(0) = 8

dydt

=−2x+ y, y(0) = 3

Solution. We have

sX− x(0) = 2X−3YsY − y(0) =−2X +Y ⇒ (s−2)X +3Y = 8

2X +(s−1)Y = 3 ⇒

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X =

∣∣∣∣ 8 33 s−1

∣∣∣∣∣∣∣∣ s−2 32 s−1

∣∣∣∣ =8s−17

s2−3s−4=

5s+1

+3

s−4⇒ x = 5e−t +3e4t

Y =

∣∣∣∣ s−2 82 3

∣∣∣∣∣∣∣∣ s−2 32 s−1

∣∣∣∣ =3s−22

s2−3s−4=

5s+1

− 2s−4

⇒ y = 5e−t−2e4t

7.2.8 Problem. Solvedxdt

= y+1, x(0) = 1

dydt

=−x+ t, y(0) = 0

Solution. We have

sX− x(0) = Y + 1s

sY − y(0) =−X + 1s2⇒ sX−Y = 1+ 1

sX + sY = 1

s2⇒(

X =s2 + s3 +1

s2 + s4 ,Y =− 1s2 +1

)Then

s2 + s3 +1s2 + s4 =

s2

s2 + s4 +s3

s2 + s4 +1

s2 (1+ s2)

=1

1+ s2 +s

1+ s2 +1s2 −

1s2 +1

=s

1+ s2 +1s2

andx = L −1

(s

1+ s2 +1s2

)= cos t + t

andy = L −1

(− 1

s2 +1

)=−sin t.


1. Given( dx

dt +2x = 1x(0) = 3

), find X (s) = L (x(t)) and x(t).

Ans. x(t) = 12 +5e−2t , X (s) = 5

s+2 +12s .

2. Given

d2xdt2 +3dx

dt +2x = 0x(0) = 3x′ (0) = 1

, find X (s) = L (x(t)) and x(t).

Ans. x(t) = 7e−t−4e−2t , X (s) = 7s+1 −

4s+2 .

3. Given

d2xdt2 +4dx

dt +4x = 0x(0) = 3x′ (0) = 1


Ans. x(t) = 3e−2t +7te−2t , X (s) = 3s+2 +

7(s+2)2 .

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4. Given

d2xdt2 +4dx

dt +4x = e−t

x(0) = 3x′ (0) = 1


Ans. x(t) = 2e−2t +6te−2t + 1et , X (s) = 2

s+2 +6

(s+2)2 +1

s+1 .

5. Given

d2xdt2 +4dx

dt +4x = e−2t

x(0) = 1x′ (0) = 1


Ans. x(t) = e−2t +3te−2t + 12t2e−2t , X (s) = 1

s+2 +3

(s+2)2 +1

(s+2)3 .

6. Given

d2xdt2 +7dx

dt +10x = 0x(0) = 1

x′ (0) =−1


Ans. x(t) = 43e−2t− 1

3e−5t , X (s) = 43(s+2) −

13(s+5) .

7. Given

d2xdt2 +10dx

dt +25x = 1x(0) = 1x′ (0) = 1


Ans. x(t) = 2425e−5t + 29

5 te−5t + 125 , X (s) = 24

25(s+5) +29

5(s+5)2 +1

25s .

8. Given

d2xdt2 +

dxdt +

14x = e−t/2

x(0) = 1x′ (0) = 0


Ans. x(t) = e−12 t + 1

2te−12 t + 1

2t2e−t/2, X (s) = 1s+ 1

2+ 1

2(s+ 12)

2 +1

(s+ 12)

3 .

9. Given

d2xdt2 +4dx

dt +4x = sin2tx(0) = 1x′ (0) = 0


Ans. x(t) = 94

te2t − 1

8 cos2t + 98e−2t , X (s) = 9

8(s+2) −18

ss2+4 +

94

1s+2 .

10. Given

d2xdt2 +3dx

dt +2x = t sin tx(0) = 0x′ (0) = 0


Ans. x(t) = 1750 cos t + 3

25 sin t− 310t cos t + 1

10t sin t− 12et +

425e−2t , X (s) = 17

50s

s2+1 +15

s(s2+1)

2 +4

25(s+2) −35

s2

(s2+1)2 +

2150(s2+1)

− 12

1s+1 .

11. Given

dxdt = 2x−4y+1

dydt =−x− y

x(0) = 1x′ (0) = 0

, find X (s)=L (x(t)) , Y (s)=L (y(t)) and x(t) ,y(t).

Ans. y(t) = 710e−2t − 7

60e3t + 16 , x(t) = 7

10e−2t + 715e3t − 1

6 , X (s) = 710(s+2) +

715(s−3) −

16s , Y (s) = 7

10(s+2) −7

60(s−3) +16s .

12. Given

dxdt = 2x−4y

dydt =−x− y+ e−2t

x(0) = 1x′ (0) = 0

, find X (s) = L (x(t)) , Y (s) = L (y(t)) and

x(t) ,y(t).

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Ans. y(t) = 1425e−2t − 3

50e3t + 45te−2t , x(t) = 19

25e−2t + 625e3t + 4

5te−2t , X (s) =19

25(s+2) +4

5(s+2)2 +6

25(s−3) , Y (s) = 1425(s+2) +

45(s+2)2 − 3

50(s−3) .

13. Given

dxdt = 2x−4y+1dydt =−x− y+1

x(0) = 1x′ (0) = 0


Ans. y(t) = 310e−2t− 1

20e3t + 12 , x(t) = 3

10e−2t + 15e3t + 1

2 , X (s) = 310(s+2)+

15(s−3)+

12s , Y (s) = 3

10(s+2) −1

20(s−3) +12s .

14. Given

dxdt = 2x−4y+ sin t

dydt =−x− y

x(0) = 0x′ (0) = 0


x(t) ,y(t).Ans. y(t) = 7

50 sin t− 150 cos t+ 1

25e2t − 150e3t , x(t) = 1

25e2t − 425 sin t− 3

25 cos t+ 225e3t ,

X (s) = 225(s−3) −

325

ss2+1 −

425(s2+1)

+ 125

1s−2 , Y (s) = 7

50(s2+1)− 1

50(s−3) −150

ss2+1 +

125

1s−2 .

15. Given

dxdt = x+2y

dydt = 4x+3y

x(0) = 2x′ (0) =−1

, find X (s) =L (x(t)) , Y (s) =L (y(t)) and x(t) ,y(t).

Ans. x(t) = 116 e−t + 1

6e5t , y(t) = 13e5t − 11

6 e−t , X (s) = 116(s+1) +

16(s−5) , Y (s) =

13(s−5) −

116(s+1) .

16. Given

dxdt = x+2y+ et

dydt = 4x+3y+1

x(0) = 1x′ (0) = 0


Ans. y(t) = 310e5t − e−t − 1

2et + 15 , x(t) = 1

4et + e−t + 320e5t − 2

5 , X (s) = 14(s−1) +

1s+1 +

320(s−5) −

25s , Y (s) = 3

10(s−5) −1

s+1 −1

2(s−1) +15s .

17. Given

dxdt = x+2y

dydt = 4x+3y+ e−t

x(0) = 1x′ (0) = 0


x(t) ,y(t).Ans. y(t) = 4

9e5t − 1718e−t + 1

3te−t , x(t) = 79e−t + 2

9e5t − 13te−t , X (s) = 7

9(s+1) −1

3(s+1)2 +2

9(s−5) , Y (s) = 13(s+1)2 − 17

18(s+1) +4

9(s−5) .

18. Given

dxdt = x+2y+ e−t

dydt = 4x+3y+1

x(0) = 1x′ (0) = 0


x(t) ,y(t).Ans. y(t) = 11

45e5t − 139 e−t − 2

3te−t + 15 , x(t) = 23

18e−t + 1190e5t + 2

3te−t − 25 , X (s) =

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2318(s+1) +

23(s+1)2 +

1190(s−5) −

25s , Y (s) = 11

45(s−5) −2

3(s+1)2 − 139(s+1) +

15s .

7.4 Advanced Problems1. Suppose f (t) is continuous in [0,∞) and of exponential order. Prove that

every solution of dxdt +ax(t) = f (t) is of exponential order.

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8. Convolution and Integral Equations

We present the convolution of two functions, a concept basic in understanding thenature of solutions of differential equations. Convolution also generates integralequations which can be converted, by the Laplace transform, to algebraic equations.

8.1 Theory and Examples8.1.1 Definition. The convolution of f (t), g(t) is

( f ∗g)(t) =∫ t

0f (τ)g(t− τ)dτ

8.1.2 Example.

et ∗ e−3t =∫ t

0eτe−3(t−τ)dτ =

∫ t

0e−3te4τdτ

= e−3t∫ t

0e4τdτ = e−3t

(14

e4τ

)t

τ=0=

14

e−3t (e4t−1)=

et− e−3t

4

8.1.3 Theorem (Convolution). Let F (s) = L ( f (t)), G(s) = L (g(t)). Then

F (s)G(s) = L

(∫ t

0f (τ)g(t− τ)dτ

)= L ( f ∗g)

Proof. Changing integration order we have

L

(∫ t

τ=0f (τ)g(t− τ)dτ

)=∫

∞

t=0e−st

(∫ t

τ=0f (τ)g(t− τ)dτ

)dt

=∫

∞

t=0

∫ t

τ=0e−st f (τ)g(t− τ)dτdt =

∫∞

τ=0

∫∞

t=τ

e−st f (τ)g(t− τ)dtdτ

Now letting v = t− τ,dv = dt (for τ constant) so

=∫

∞

τ=0

∫∞

v=0e−s(v+τ) f (τ)g(v)dvdτ =

(∫∞

τ=0e−sτ f (τ)dτ

)(∫∞

v=0e−svg(v)dv

)=F (s)G(s) .

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70 Chapter 8. Convolution and Integral Equations

8.1.4 Example. From the previous example we know

L(et ∗ e−3t)= L

(et− e−3t

4

)=

14(s−1)

− 14(s+3)

.

From the Convolution Theorem we get

L(et ∗ e−3t)= L

(et)L (

e−3t)= 1(s−1)(s+3)

=1

4(s−1)− 1

4(s+3).

8.1.5 Example. Similarly to the previous:

(t)∗(te−t)= ∫ t

0τe−τ (t− τ)dτ =

∫ t

0e−τ(tτ− τ

2)dτ

=(−e−τ

(tτ− τ

2))τ

τ=0−∫ t

0e−τ (t−2τ)dτ

=(−e−τ

(tτ− τ

2)− (t−2τ)e−τ +(−2)(−e−τ

))tτ=0

= te−t +2e−t + t−2.

But alsoL (t) =

1s2 , L

(te−t)= 1

(s+1)2

andL (t)L

(te−t)= 1

(s+1)2 s2=

2s+1

+1

(s+1)2 −2s+

1s2 .

Hence

L −1

(1

(s+1)2 s2

)= L −1

(1s2 +

2s+1

+1

(s+1)2 −2s

)= t +2e−t + te−t−2.

8.1.6 Theorem. The following hold when the respective convolutions are welldefined.

1. f ∗g = g∗ f ,2. ( f ∗g)∗ p = f ∗ (g∗ p),3. ( f +g)∗ p = f ∗ p+g∗ p.4. With o(t) being the zero function (∀t : o(t) = 0): f ∗ o = o (so o is the zero

element).Proof. Immediate.

8.1.7 Question. What is the unit element of convolution? I.e., what is u :f ∗u = f ? It must be such that L (u) = 1. Have you seen a function u(t) such thatL (u) is a constant?

8.1.8 To obtain an intuitive interpretation of the convolution, consider the solutionof the ODE

d fdt

+a0 f = g(t)

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which, as you recall, is

f (t) = e−a0t(∫

ea0tg(t)dt + c)= ce−a0t + e−a0t

∫ea0tg(t)dt.

To also satisfy f (0) = f0 we c = f0 and so we have

f (t) = f0e−a0t + e−a0t∫ t

0ea0τg(τ)dτ = f0e−a0t +

∫ t

0e−a0(t−τ)g(τ)dτ

= f0e−a0t +g∗ e−a0t .

So f (t) is the superposition of two components:1. f (0)e−a0t is the initial condition transmitted in time with a modulation

(attenuation) e−a0t

2. g∗e−a0t =∫ t

0 e−a0(t−τ)g(τ)dτ can be interpreted as: the sum of all inputs g(τ)(i.e., at each time τ ∈ (0, t]) modulated by e−a0(t−τ) (i.e., attenuated by a timet− τ, which is the length of time from when g(τ) it was applied until thecurrent time. .

8.1.9 With the Laplace transform, we can generalize the above to the problem

dn fdtn +an−1

dn−1 fdtn−1 + ...+a0 f = g(t) , f (0) = f0, f ′ (0) = f1, ..., f (n−1) (0) = fn−1

For simplicity we take f0 = f1 = ...= fn−1 = 0 (but similar results can be obtainedfor nonzero initial conditions). In this case we have(

sn +an−1sn−1 + ...+a1s+a0)

F = G

orF =

1sn +an−1sn−1 + ...+a1s+a0

G = PG

and, with

p(t) = L −1 (P) = L −1(

1sn +an−1sn−1 + ...+a1s+a0

),

we get

f (t) = g∗ p =∫ t

0g(τ) p(t− τ)dτ.

In other words: f (t) (at time t) is the sum of the inputs g(τ) (at all times τ ∈ (0, t])modulated by p for a length of time t− τ (from τ until current time t). If

dn fdtn +an−1

dn−1 fdtn−1 + ...+a0 f = g

describes the output of a (physical or other) system when the input is g(t), then wecan picture the operation of the system as a box with input g(t) and output f (t)(and g(t) can be changeable). Then we have F = PG and we call P the transferfunction of the system. It gives the output F (s) (in Laplace space) for every imputG(s) (we can also say that P = F

G , where the function P(s) is fixed for every (G,F)pair and describes the system behavior). We call p(t) the impulse function of thesystem (why?).

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8.1.10 The situation is similar when the system has equation

dn fdtn +an−1

dn−1 fdtn−1 + ...+a0 f = bm

dmgdtn +bm−1

dm−1gdtn−1 + ...+b0g

(so the input is also processed by the system) in which case the transfer function is

P(s) =bmsm + ...+b1s+b0

sn +an−1sn−1 + ...+a1s+a0.


d2ydt2 +w2y = f (t) , y(0) = 1, y′ (0) =−2

we take the Laplace transform and get

s2Y − sy(0)− y′ (0)+w2Y = F (s)⇒(s2 +w2)Y − s+2 = F (s)⇒ Y =

s−2s2 +w2 +

F (s)s2 +w2

y(t) = coswt− 2w

sinwt + f (t)∗ sinwtw

= coswt− 2w

sinwt +1w

∫ t

0f (τ)sin(w(t− τ))dτ.

8.1.12 The Laplace transform can also be used to solve integral equations (usuallyof the convolution type, but other types can also be handled).


x(t) = t2 +∫ t

0x(u)sin(t−u)du

we have

X =2s3 +X

1s2 +1

⇒ X(

1− 1s2 +1

)=

2s3 ⇒

X(

s2

s2 +1

)=

2s3 ⇒ X =

2(s2 +1

)s5 =

2s3 +

2s5 ⇒

x = t2 +1

12t4.


x(t) =t3

6−∫ t

0x(u)(t−u)du

we have

X =1s4 −X

1s2 ⇒ X

(1+

1s2

)=

1s4 ⇒

X(s2 +1

)s2 =

1s4 ⇒ X =

1s2 (s2 +1)

=1s2 −

1s2 +1

⇒

x = t− sin t

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8.1.15 Example. To solve∫ t

0x(u)(t−u)du = 2x(t)+ t−2

we have(X (s))2 = 2X (s)+

1s2 −

2s⇒ X2−2X =

1s2 −

2s

and we note immediately that that X = 1s satisfies the equation. Hence x(t) = 1.

8.1.16 To solvex(t) = t−

∫ t

0x(u)(t−u)du

we have

X =1s2 −X

1s2 ⇒ X

(1+

1s2

)=

1s2 ⇒

X(s2 +1

)s2 =

1s2 ⇒ X =

1s2 +1

⇒

x = sin t

8.2 Solved Problems

8.3 Unsolved Problems1. Compute the convolution (1)∗ (1).

Ans. t.2. Compute the convolution (t)∗ (cos t).

Ans. 1− cos t.3. Compute the convolution (sin t)∗ (cos t).

Ans. 12t sin t.

4. Compute the convolution (et)∗ (cos t).Ans. 1

2et− 12 cos t + 1

2 sin t.5. Compute the convolution 1∗1∗ ...∗1

n times.

Ans. tn−1

(n−1)! .

6. Solve the integral equation∫ t

0 x(τ)e−(t−τ)dτ = t2−2e−t−2t +2.Ans. t2.


0 x(τ)e−(t−τ)dτ = 12 cos t + 1

2 sin t− 12e−t .

Ans. cos t.8. Solve the integral equation

∫ t0 x(τ)e−(t−τ)dτ = 1

2t cos t− 12 sin t + 1

2t sin t.Ans. t cos t.


0 x( τ)e−(t−τ)dτ = 12et

(e2t−1

).

Ans. et .10. Solve the integral equation

∫ t0 x(τ)e−(t−τ)dτ = 1

4e−t

(e2(−t)+2t−1

).

Ans. tet .11. Solve the integral equation

∫ t0 x(τ)e−(t−τ)dτ = t2−3e−t−2t +3.

Ans. t2 +1.

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0 x(τ)e−(t−τ)dτ = t2− e−t− t +1.Ans. t + t2.


0 x(τ)e−(t−τ)dτ = 25e−t− 2

5 cos2t + 15 sin2t.

Ans. sin(2t).14. Solve the integral equation

∫ t0 x(τ)e−(t−τ)dτ = 1

10 cos3t− 110e−t + 3

10 sin3t.Ans. cos(3t).


0 x(τ)(t− τ)dτ = t3

6 .Ans. t.


0 x(τ)(t− τ)dτ = 1− cos t.Ans. cos t.


0 x(τ)(t− τ)dτ = et− t−1.Ans. et .


0 x(τ)cos(t− τ)dτ = 12et− 1

2 cos t + 12 sin t.

Ans. et .19. Solve the integral equation

∫ t0 x(τ)cos(t− τ)dτ = 1

2 sin t + 12t cos t.

Ans. cos t.20. Solve the integral equation

∫ t0 x(τ)cos(t− τ)dτ = 1

2t sin t.Ans. sin t.


0 x(τ)e−(t−τ)dτ = 2t.Ans. 2+2t.


0 x(τ)(2+ t− τ)dτ = t.Ans. 1

2e−t/2.23. Solve the integral equation

∫ t0 x(τ)e−2(t−τ)dτ = sin t.

Ans. cos t +2sin t.24. Solve the integral equation x(t)+

∫ t0 x(τ)sin(t− τ)dτ = cos t + 1

2t sin t.Ans. cos(t).

25. Solve the integral equation x(t)+∫ t

0 x(τ)e−2(t−τ)dτ = e−2t +(ln(et))e−2t .Ans. e−2t .


0 x(τ)e−(t−τ)dτ = 2t + e−t−1.Ans. t.


0 x(τ)et−τdτ = t.Ans. 2t−t2

2 .28. Solve the integral equation x(t)+

∫ t0 x(τ)(t− τ)dτ = t.

Ans. sin t.29. Solve the integral equation x(t)+

∫ t0 x(τ)(t− τ)dτ = sin t.

Ans. 12 sin t + 1

2t cos t.30. Solve the integral equation x(t)+

∫ t0 x(τ)et−τdτ = et+sin t

2 .Ans. sin t+cos t

2 .

8.4 Advanced Problems1. Suppose f (t) ,g(t) are piecewise continuous in [0,∞) and of exponential order.

Show that f (t)∗g(t) is of exponential order.

Ath.

Keha

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9. Dirac Delta and Generalized Functions

The Dirac delta function δ (x) is very useful in the solution of ODEs, especially onesrelated to engineering problems. However, despite its name, it is not a functionin the usual sense. Rather, it is a generalized function. Our main task in thischapter to clarify this concept.

9.1 Theory and Examples9.1.1 We return to the full definition of the Laplace transform.

L (x(t)) :=∫

∞

0−x(t)e−stdt. (9.1)

As mentioned, when applied to functions continuous at 0 the use of 0− is notimportant. But now, dealing with discontinuous functions, its significance willbecome clear.

9.1.2 All Laplace transform properties hold under (9.1), but we must modify thederivative formulas as follows

L(x′ (t)

)= sX− x

(0−)

L(x′′ (t)

)= s2X− sx

(0−)− x′

(0−)

etc.

Here x(0−) means limt→0− x(t) and similarly for x′ (0−) etc.

9.1.3 In many applications we study systems with discontinuous input. Forexample in a linear circuit we turn a switch on and voltage jumps from zero to one.We model this by the Heaviside step function, previously denoted by h(t); fromnow on we will also use the notation Heaviside(t):

Heaviside(t) :=

1 t < 00 t > 0 , Heaviside(t− t0) :=

1 t < t00 t > t0

.

As will be seen, the actual value at the point of discontinuity (0 or t0) is notimportant.

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76 Chapter 9. Dirac Delta and Generalized Functions

9.1.4 The Laplace transform of Heaviside(t) is

L (Heaviside(t)) =∫

∞

0−Heaviside(t)e−stdt = lim

ε→0+lim

M→∞

∫ M

−ε

Heaviside(t)e−stdt

= limε→0+

limM→∞

(∫ 0

−ε

0e−stdt +∫ M

01e−stdt

)= lim

M→∞

∫ M

01e−stdt = lim

M→∞

(−1

se−sM +

1s

e−s0)=−0+

1s.

(Note that it is the same as the Laplace transform of x(t) = 1; Heaviside(t) and 1are the same on (0,∞)). The Laplace transform of Heaviside(t− t0) is, as usual

L (Heaviside(t− t0)) = e−st0L (Heaviside(t)) =e−st0

s.

9.1.5 In other applications the input can be a switch which is turned on at t0 andoff at t1. For such situations we define the functions

Heaviside(t) −Heaviside(t− t1) , Heaviside(t− t0) −Heaviside(t− t1− t0) etc.

So

Heaviside(t− t0) −Heaviside(t− t1− t0) =

1 t ∈ (t0, t1)0 t /∈ (t0, t1)

.

Their Laplace transforms are

L (Heaviside(t) −Heaviside(t− t1)) =1− e−t1s

s,

L (Heaviside(t− t0) −Heaviside(t− t1− t0)) =1− e−t1s

se−t0s

9.1.6 Finally the input can be an impulse, i.e., it has short duration and highvalue. This can be modeled by

δε (t) =Heaviside(t) −Heaviside(t− ε)

ε

(and similarly for δε (t− t0)). Note that

δε (t) := 1

εt ∈ (0,ε)

0 t /∈ (0,ε).

So∀ε > 0 :

∫∞

0δε (t)dt = 1;

a constant amount of ‘energy’ or ‘momentum’ is delivered. Now we can take thelimit as ε → 0+ and define

δ (t) := limε→0+

δε (t) =

∞ t = 00 t 6= 0

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The Laplace transform of δε (t) is

L (δε (t)) = L (δε (t)) = L

(Heaviside(t) −Heaviside(t− ε)

ε

)=

1− e−εs

εs

We can compute L(

δ (t))

in two different ways.

L(

δ (t))=∫

∞

0−δ (t)e−stdt = 0

L(

δ (t))= lim

ε→0+L (δε (t)) = lim

ε→0+

1− e−εs

εs= 1

The first answer is correct but we do not like it because it suggests δ (t) is thesame as the zero function. We like the second answer better but it is possiblywrong, because we exchanged limit with integration (and this, generally, is notacceptable).

9.1.7 We also like that L(

δ (t))= 1 because it gives us

L(Heaviside′ (t)

)= sH (s)−Heaviside

(0−)= s

1s−0 = 1 = L

(δ (t)

). (9.2)

But the above is also strictly wrong, because the derivative Heaviside′ (t) does notexist. However (9.2) suggests that Heaviside′ (t), the derivative of the step function,is δ (t) which looks “kind of right”: it is zero everywhere except at t = 0, where it isinfinite (corresponding to an infinite rate of increase).

9.1.8 We will try to fix these issues, by defining the generalized derivativeHeaviside′ (t) as a new entity: Dirac(t).

9.1.9 We will define the generalized derivatives

Dirac(t) := Heaviside′ (t) , Dirac(t− t0) := Heaviside′ (t− t0) .

or, for simplicity of notation

δ (t) := Heaviside′ (t) , δ (t− t0) := Heaviside′ (t− t0) .

Here δ (t) , δ (t− t0) are generalized functions.

9.1.10 The above “definitions” are not rigorous, but can be made so. This requiresmore advanced mathematics; one can use either

1. the Stieltjes integral (which generalizes the standard Riemann integral) or2. the theory of distributions.In both of these approaches one can establish that Heaviside′ (t) is a certain “en-

tity” which has the derivative interpretation and many useful derivative properties.(Actually it is even better to take Heaviside′ (t)dt "together" and write

Heaviside′ (t)dt = d Heaviside;

then ∫ b

ax(t)Heaviside′ (t)dt =

∫ b

ax(t)d Heaviside,

the Stieltjes integral of x(t) (with respect to Heaviside(t)). For more detils, seeAppendix A.)

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9.1.11 In particular (as we will take for granted) the following hold∫Dirac(t)dt =

∫Heaviside′ (t)dt = Heaviside(t)+ c, (9.3)∫

Dirac(t− t0)dt =∫

Heaviside(t− t0)dt = Heaviside(t− t0)+ c, (9.4)∫ b

aDirac(t)dt =

∫ b

aHeaviside′ (t)dt = Heaviside(b)−Heaviside(a) , (9.5)∫ b

aDirac(t− t0)dt =

∫ b

aHeaviside′ (t− t0)dt = Heaviside(b)−Heaviside(a) (9.6)

The most useful ones are (9.5)-(9.6) which are written more precisely as follows

∀a,b ∈ R∗ :∫ b

aDirac(t)dt =

∫ b

aHeaviside′ (t)dt =

1 0 ∈ (a,b)0 0 /∈ [a,b] (9.7)

∀a,b ∈ R∗ :∫ b

aDirac(t− t0)dt =

∫ b

aHeaviside′ (t− t0)dt =

1 t0 ∈ (a,b)0 t0 /∈ [a,b] (9.8)

Note that we do not deal with the case where 0 or t0 ∈ a,b. In other words, it isacceptable to have a discontinuity inside the integration interval, but not on itsboundaries.

9.1.12 It can also be proved that (under the Stieltjes integral) integration by partsholds for generalized derivatives, i.e., for all a,b ∈ R∗ and all differentiable x(t) wehave ∫ b

ax(t)Heaviside′ (t)dt = (x(t)Heaviside(t))t=b

t=a−∫ b

ax′ (t)Heaviside(t)dt.

Then we also have the following results..

9.1.13 Theorem. The following hold:

∀a,b ∈ R∗ :∫ b


x(0) if 0 ∈ (a,b)

0 if 0 /∈ [a,b] (9.9)

∀a,b ∈ R∗ :∫ b


x(t0) if t0 ∈ (a,b)

0 if t0 /∈ [a,b] (9.10)

Proof. Using (9.5) and integration by parts we have∫ b

ax(t)Heaviside′ (t)dt = (x(t)Heaviside(t))t=b

t=a−∫ b

ax′ (t)Heaviside(t)dt

and distinguish three cases

a < b < 0 :∫ b

ax(t)Heaviside′ (t)dt = x(b)0− x(a)0−0 = 0,

a < 0 < b :∫ b

ax(t)Heaviside′ (t)dt = x(b)1− x(a)0− x(b)+ x(0) = x(0) ,

0 < a < b :∫ b

ax(t)Heaviside′ (t)dt = x(b)1− x(a)1− x(b)+ x(a) = 0.

This completes the proof.

Ath.

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9.1.14 Using the above we can, for example, write the following∫ 0−

−∞

Dirac(t)dt = 0,∫ 0+

−∞

Dirac(t)dt = (t)dt = 1,∫ 0+

0−Dirac(t)dt = 1 (9.11)

Note that these integrals involve double limits which can, in this case, be inter-changed. E.g.∫ 0−

−∞

δ (t)dt = limδ→0+

(lim

M→∞

∫ −δ

−MDirac(t)

)dt = lim

δ→0+

(lim

M→∞

∫ −δ

−MDirac(t)dt

)= lim

M→∞

∫ −a

−MDirac(t)dt + lim

δ→0−

∫ −δ

−aDirac(t)dt = 0+0 = 0.

9.1.15 Now we can resolve the previous contradiction on the relationship betweenDirac(t) and Heaviside(t):

L (Dirac(t)) =∫

∞

0−δ (t)e−stdt = e−s0 = 1,

L (Dirac(t)) = sL (Heaviside(t))−Heaviside(0−)= s

1s−0 = 1.

So, in this sense too, Dirac(t) is the (generalized) derivative of Heaviside(t).

9.1.16 Theorem. For every x(t) which has a Laplace transform we have

∀t > 0 : x(t)∗Dirac(t) =∫ t

0−x(τ)Dirac(t− τ)dτ = x(t) ,

∀t > t0 : x(t)∗Dirac(t− t0) =∫ t

0−x(τ)Dirac(t− t0− τ)dτ = Heaviside(t− t0)x(t− t0) .

Proof. We only prove the first one, by the previous theorem. First define f (t) =x(−t). Now

x(t)∗Dirac(t) = Dirac(t)∗ x(t) =∫ t

0−Dirac(τ)x(t− τ)dτ =

=∫ t

0−Dirac(τ) f (τ− t)dτ = f (0− t) = f (−t) = x(t)

since 0 ∈ (0−, t) .We can also prove it by Laplace transformation:

L (x(t)∗Dirac(t)) = X (s)L (Dirac(t)) = X (s)1 = X (s) = L (x(t))

and, similarly,

L (x(t)∗Dirac(t− t0))=X (s)L (Dirac(t))e−t0s =X (s)e−t0s =L (Heaviside(t− t0)x(t− t0)) .

9.1.17 Summarizing, the basic properties of Dirac(t) are:

when t0 < t :∫ t

0−Dirac(t− t0)x(t)dt = x(t0) ,

when t0 < t : x(t)∗Dirac(t− t0) = x(t− t0) .

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9.1.18 Now we are ready to solve differential equations which involve generalizedfunctions.


dydt

+2y = Dirac(t−1) , y(0−)= 1

we have

sY −1+2Y = e−s⇒ Y =1

s+2+

1s+2

e−s

⇒ y(t) = e−2t +Heaviside(t−1)e−2(t−1).


dydt

+2y = Dirac(t) , y(0−)= 1

we have

sY −1+2Y = 1⇒ Y =2

s+2⇒ y(t) = 2e−2t .

But note that y(0) = 2! Is this a contradiction to the given initial condition?No, it is not! The initial condition requires y(0−) = 1. The solution should more

accurately be written as

y(t) = e−2t +Heaviside(t)e−2t

(why?) and then it is clear that

y(0−)= 1, y

(0+)= 2.


d2ydt2 +2

dydt

+2y = Dirac(t−1) , y(0−)= 0, y′

(0−)= 0

we have

s2Y +2sY +2Y = e−s⇒Y =e−s

s2 +2s+2=

e−s

(s+1)2 +1⇒ y(t)=Heaviside(t−1)e−(t−1) sin(t−1)

(since 1(s+1)2+1

is Laplace transform of e−t sin t).


d2ydt2 +2

dydt−15y = 6Dirac(t−9) , y

(0−)=−5, y′

(0−)= 7

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we have, taking Laplace transformation,

s2Y +5s−7+2sY +10−15Y = 6e−9s⇒(s2 +2s−15

)Y =−(5s+3)+6e−9s⇒

Y =− 5s+3s2 +2s−15

+6e−9s

(s2 +2s−15)

Now

Y1 =5s+3

s2 +2s−15=

94(s−3)

+11

4(s+5)

Y2 =6

s2 +2s−15=

68(s−3)

− 68(s+5)

so

y1 (t) =94

e3t +114

e−5t

y2 (t) =68

e3t− 68

e−5t

andy(t) =−y1 (t)+Heaviside(t−9)y2 (t−9) .


d2ydt2 +5

dydt

+6y = e−t Dirac(t−2) , y(0−)= 2, y

(0−)=−5

we have

s2Y −2s+5+5sY −10+6Y = e−2(s+1)⇒(s2 +5s+6

)Y = 2s+5+ e−2(s+1)⇒

Y =2s+5

s2 +5s+6+ e−2 e−2s

s2 +5s+6.

Setting

Y1 =2s+5

s2 +5s+6=

1s+2

+1

s+3

Y2 =1

s2 +5s+6=

1s+2

− 1s+3

we get

y1 (t) = e−2t + e−3t

y2 (t) = e−2t− e−3t

and finally

y(t) = e−2t + e−3t + e−2 Heaviside(t−2)(

e−2(t−2)− e−3(t−2))

= e−2t + e−3t +Heaviside(t−2)(e2e−2t− e4e−3t) .

Ath.

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d2ydt2 + y =

∞

∑k=1

Dirac(t− kπ) , y(0−)= 0, y′

(0−)= 0

we have(s2 +1

)Y =

∞

∑k=1

e−kπs⇒ Y =1

s2 +1

∞

∑k=1

e−kπs⇒

y(t) =∞

∑k=1

Heaviside(t− kπ)sin(t− kπ) =

(∞

∑k=1

Heaviside(t− kπ)(−1)k

)sin t

or

y(t) =

0 t ∈ [2mπ,(2m+1)π)−sin t t ∈ [2(m+1)π,(2m+2)π)

.


d2ydt2 − y = δ (t−3) , y

(0−)= 1, y

(0−)= 0

Solution. We have (s2−1

)Y − s = e−3s⇒ Y =

ss2−1

+e−3s

s2−1Setting

Y1 =s

s2−1=

12(s−1)

+1

2(s+1)

Y2 =1

s2−1we get

y1 (t) =12

et +12

e−t =12

cosh t

y2 (t) =12

1s−1

+12

1s+1

=12

sinh t

we then get

y(t) =12

cosh t +12

h(t−3)sinh(t−3)


d2ydt2 +2

dydt

+ y = δ (t−1) , y(0−)= 0, y′

(0−)= 0

Solution. We have

s2Y +2sY +Y = 1⇒ Y =1

s2 +2s+1=

1(s+1)

⇒

y(t) = h(t−1)e−(t−1) (t−1)

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d2ydt2 +4y = δ (t−π)−δ (t−2π) , y

(0−)= 0, y

(0−)= 0

Solution. We have(s2 +4

)Y = e−πs− e−2πs⇒ Y =

e−πs

s2 +4+

e−2πs

s2 +4Setting

Y1 =1

s2 +4, y1 (t) =

12

sin2t

we get

y(t) =12

h(t−π)sin(2t−2π)−h(t−2π)sin(2t−4π)

y(t) =12(h(t−π)−h(t−2π))sin(2t)


d2ydt2 +2

dydt

+2y = δ (t−π) , y(0−)= 1, y′

(0−)= 1

Solution. We have

s2Y − s−1+2sY −2+2Y = e−πs⇒

Y =e−πs

(s+1)2 +1+

s+3

(s+1)2 +1⇒

Y =e−πs

(s+1)2 +1+

s+1

(s+1)2 +1+

s

(s+1)2 +1⇒

Setting

Y1 =1

(s+1)2 +1

Y2 =s+1

(s+1)2 +1

Y3 =2

(s+1)2 +1we get

y1 (t) = e−t sin t

y2 (t) = e−t cos t

y3 (t) = 2e−t sin t

and so

y(t) = h(t−π)y1 (t−π)+ y2 (t)+ y3 (t)

= h(t−π)e−(t−π) sin(t−π)+ e−t cos t +2e−t sin t

=−h(t−π)e−(t−π) sin t + e−t cos t +2e−t sin t

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d2ydt2 −2

dydt−3y = 2δ (t−1)−δ (t−3) , y

(0−)= 2, y′

(0−)= 2

Solution. We have

s2Y −2s−2−2sY +4−3sY = 2e−s− e−3s⇒

Y =2s−2

s2−2s−3+

2e−s

s2−2s−3− e−3s

s2−2s−3⇒

Y =2s−2

s2−2s−3+

2e−s

s2−2s−3− e−3s

s2−2s−3

Setting

Y1 =2s−2

s2−2s−3=

1s+1

+1

s−3

Y2 =1

s2−2s−3=

14(s−3)

− 14(s+1)

we get

y1 (t) = e−t + e3t

y2 (t) =14

e3t− 14

e−t

and so

y(t) = y1 (t−π)+2h(t−1)y2 (t−1)−h(t−3)y2 (t−1)

= e−t + e3t +12

h(t−1)(

e3(t−1)− e−(t−1))− 1

4h(t−3)

(e3(t−3)− e−(t−3)

)9.3 Unsolved Problems

1. Solve( dx

dt + x = Dirac(t−1)x(0−) = 0

).

Ans. Heaviside(t−1)e−(t−1).

2. Solve( dx

dt + x = Dirac(t−5)x(0−) = 1

).

Ans. e−t +Heaviside(t−5)e−(t−5).

3. Solve

d2xdt2 +3dx

dt +2x = Dirac(t−1)x(0−) = 0x′ (0−) = 0

.

Ans. −Heaviside(t−1)e−2(t−2)+Heaviside(t−1)e−(t−1).

4. Solve

d2xdt2 +2dx

dt + x = Dirac(t−1)x(0−) = 1x′ (0−) = 1

.

Ans. e−t +2te−t−(

Heaviside(t−1)e−(t−1)− t Heaviside(t−1)e−(t−1))

.

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Keha

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5. Solve

d2xdt2 +3dx

dt +2x = Dirac(t−1)x(0−) = 1x′ (0−) = 2

.

Ans. 4e−t−3e−2t−Heaviside(t−1)e−2(t−1)+Heaviside(t−1)e−(t−1).

6. Solve

d2xdt2 +4dx

dt +4x = Dirac(t−1)x(0−) = 0x′ (0−) = 0

.

Ans. −Heaviside(t−1)e−2(t−1)+ t Heaviside(t−1)e−2(t−1).

7. Solve

d2xdt2 +4dx

dt +4x = Dirac(t−1)x(0−) = 1x′ (0−) = 2

.

Ans. e−2t +4te−2t−Heaviside(t−1)e−2(t−1)+ t Heaviside(t−1)e−2(t−1).

8. Solve( dx

dt = x+2y+Dirac(t−1) , x(0−) = 0dydt = 4x+3y, y(0−) = 0

).

Ans. y(t)= 23e5t−5 Heaviside(t−1)− 2

3e1−t Heaviside(t−1), x(t)= 23e1−t Heaviside(t−1)+

13e5t−5 Heaviside(t−1).

9. Solve( dx

dt = x+2y+Dirac(t−1) , x(0−) = 1dydt = 4x+3y, y(0−) = 1

).

Ans. y(t)= 43e5t− 1

3e−t− 23e1−t Heaviside(t−1)+ 2

3e5t−5 Heaviside(t−1), x(t)=13e−t + 2

3e5t + 23e1−t Heaviside(t−1)+ 1


10. Solve( dx

dt = x+2y+Dirac(t−1) , x(0−) = 1dydt = 4x+3y+Dirac(t−1) , y(0−) = 0

).

Ans. y(t)= 23e5t− 2

3e−t− 13e1−t Heaviside(t−1)+ 4

3e5t−5 Heaviside(t−1), x(t)=23e−t + 1

3e5t + 13e1−t Heaviside(t−1)+ 2


11. Solve( dx

dt + x = Dirac(t)x(0−) = 0

).

Ans. Heaviside(t)e−t .

12. Solve( dx

dt + x = Dirac(t−5)x(0−) = 1

).

Ans. e−t +Heaviside(t−5)e−(t−5).

13. Solve

d2xdt2 +2dx

dt + x = Dirac(t)+Heaviside(t−1)x(0−) = 1x′ (0−) = 1

.

Ans. e−t + 32te−t + te−t Heaviside(t)−t Heaviside(t−1)e−(t−1)+Heaviside(t−1).

14. Solve

d2xdt2 +3dx

dt +2x = Dirac(t)x(0−) = 0x′ (0−) = 0

.

Ans. 12e−2t− 1

2e−t− 1e2t (Heaviside(t)− et Heaviside(t)).

15. Solve

d2xdt2 +3dx

dt +2x = Dirac(t)x(0−) = 1x′ (0−) = 2

.

Ans. 72e−t− 5

2e−2t− 1e2t (Heaviside(t)− et Heaviside(t)).

Ath.

Keha

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16. Solve

d2xdt2 +4dx

dt +4x = Dirac(t)x(0−) = 0x′ (0−) = 0

.

Ans. te2t Heaviside(t)− 1

2te−2t .

17. Solve

d2xdt2 +4dx

dt +4x = Dirac(t)x(0−) = 1x′ (0−) = 2

.

Ans. e−2t + 72te−2t + t

e2t Heaviside(t) .

18. Solve

d2xdt2 +3dx

dt +2x = 1+Dirac(t)x(0−) = 0x′ (0−) = 0

.

Ans. Heaviside(t−1)e−(t−1)-Heaviside((t−1)e−2(t−1)+1/2+1/2e−2t− e−t .

19. Solve

dxdt = x+2y+Dirac(t)

x(0−) = 0dydt = 4x+3yy(0−) = 0

.

Ans. y(t)= 13e−t− 1

3e5t− 23e−t Heaviside(t)+ 2

3e5t Heaviside(t), x(t)= 23e−t Heaviside(t)−

16e5t− 1

3e−t + 13e5t Heaviside(t).

20. Solve

dxdt = x+2y+Dirac(t)

x(0−) = 1dydt = 4x+3yy(0−) = 1

.

Ans. y(t)= e5t− 23e−t Heaviside(t)+ 2

3e5t Heaviside(t), x(t)= 12e5t + 2

3e−t Heaviside(t)+13e5t Heaviside(t).


Ath.

Keha

gias



10. Difference Equations

In this chapter we present difference equations, i.e., sequences which are definedby a recursive equation.


10.1.1 We are interested in discrete time functions (i.e., with domain N0). Toemphasize this, we will write xn instead of x(t). In particular, we are interested insolving difference equations, e.g.,

xn+2 +3xn+1 +2xn = 0, x0 = 1, x1 = 1.

10.1.2 We will start our analysis studying corresponding continuous time func-tions and their Laplace transforms; later we will turn to a completely discrete-timepoint of view.

10.1.3 Notation. We will use the functions

u(t) = h(t) −h(t−1) , u(t−n) = h(t−n) −h(t−n−1) .

(Here and in what follows, h(t) is the Heaviside(t) function.)

10.1.4 Given a discrete time function xt , we define a function

x(t) = xn when n≤ t < n+1;

this can also be written as

x(t) =∞

∑n=0

xnu (t−n) =∞

∑n=0

xn [h(t−n) −h(t−n−1) ] .

Ath.

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88 Chapter 10. Difference Equations

To find L (x(t)) we have

L (x(t)) =∫ 1

0x(t)e−stdt

=∫ 1

0x0e−stdt +

∫ 2

1x1e−stdt +

∫ 3

2x2e−stdt + ...

= x01− e−s

s+ x1

e−s− e−2s

s+ x2

e−2s− e−3s

s+ ...

= x01− e−s

s+ x1e−s 1− e−s

s+ x2e−2s 1− e−s

s+ ...

=1− e−s

s

(x0 + x1e−s + x1e−2s + ...

)=

1− e−s

s

∞

∑n=0

xne−ns.

In short

X (s) = L (x(t)) =1− e−s

s

∞

∑n=0

xne−ns.

10.1.5 Example. If , with |r|< 1, we have

∀n ∈ N0 : xn = rn

x(t) =∞

∑n=0

rnu (t−n)

then

X (s) = L (x(t)) =1− e−s

s

(1+ re−s + r2e−2s + ...

)=

1− e−s

s1

1− re−s =1− e−s

s(1− re−s)

In the last step we have used, with z−1 = re−s (and assuming∣∣z−1

∣∣< 1), the factthat

1+ re−s + r2e−2s + ...= 1+ z−1 + z−2 + ...=1

1− z−1 =1

1− re−s .

10.1.6 Given x(t) = ∑∞n=0 xnu (t−n), let us also compute

L (x(t +1)) =∫

∞

0e−stx(t +1)dt =

∫∞

1e−s(τ−1)x(τ)dτ = es

∫∞

1e−sτx(τ)dτ

= es∫

∞

0e−sτx(τ)dτ− es

∫ 1

0e−sτx(τ)dτ

= esX (s)− es∫ 1

0e−sτx(τ)dτ = esX (s)− es

∫ 1

0e−sτx0dτ

= esX (s)− es (1− e−s)

sx0.

In this manner we get

L (x(t +1)) = esX (s)− es (1− e−s)

sx0,

L (x(t +2)) = e2sX (s)− es (1− e−s)

s(x0es + x1) ,

...

Do you see the similarity with the Laplace transform of derivatives?

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10.1.7 Example. To solve the difference equation

x0 = 0, x1 = 1, ∀n ∈ N0 : xn+2−5xn+1 +6xn = 0.

we let

x(t) =∞

∑n=0

xnu (t−n)

and we get the equivalent equation

x(0) = 0, x(1) = 1, ∀t ≥ 0 : x(t +2)−5x(t +1)+6x(t) .

Transforming we get

e2sX (s)− es (1− e−s)

s−5esX (s)+6X (s) = 0⇒

(e2s−5es +6

)X (s) =

es (1− e−s)

s⇒

X (s) =es (1− e−s)

s(es−3)(es−2)

=(1− e−s)

s

(es

es−3− es

es−2

)=

1− e−s

s

(1

1−3e−s −1

1−2e−s

)=

1− e−s

s(1−3e−s)− 1− e−s

s(1−2e−s)⇒

L (x(t)) =1− e−s

s(1−3e−s)− 1− e−s

s(1−2e−s).

Since1− e−s

s(1− re−s)= L

(∞

∑n=0

rnu (t−n)

),

we get

x(t) =∞

∑n=0

3nu (t−n)−∞

∑n=0

2nu (t−n)

and∀n ∈ N0 : xn = 3n−2n.

10.1.8 Obviously the above method can be used for any nonhomogeneous lineardifference equation

∀n ∈ N0 : xn+K + p1xn+K−1 + ...+ pK−1xn+1 + pKxn = un+K.

10.1.9 We should not have to go through continuous time; the problem is es-sentially one of discrete time and should be solvable entirely in the discrete timedomain. To achieve this we introduce a new transform.

10.1.10 Definition. The Z transform of xn is:

X (z) = Z (x) :=∞

∑n=0

xnz−n

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To ensure the sum is well defined we assume

∀n :∣∣∣∣xn

zn

∣∣∣∣< 1,

i.e., we assume that |z| is sufficiently large.

10.1.11 Since X (z) is a MacLaurin series in z−1, if two functions xn and yn haveX (z) = Y (z), then also xn = yn. In other words, the Z transform is invertible.

10.1.12 Example. The Z transform of xn = rn is

Z (xn) = 1+ rz−1 + r2z−2 + ...= limn

1−( r

z

)n

1− rz

=1

1− rz−1 =z

z− r

Recalling that the continuous-time function x(t) = (∑∞n=0 rnu (t−n)) has Laplace

transform

L (x(t)) =1− e−s

s

(1+ re−s + r2e−2s + ...

)=

1− e−s

s1

1− re−s

we recognize that z plays the role of es but, by turning to discrete time, we havegotten rid of the cumbersome 1−e−s

s factor.

10.1.13 So we can think of the Z transform as a special case of the Laplacetransform, but we prefer to think of it as a completely independent transform. It isreasonable then to expect that, if we obtain Z properties analogous to the Laplaceproperties, we can also perform similar operations. In particular, we should beable to solve difference equations with Z -transform, similarly to solving differentialequations with Laplace transform.

10.1.14 Example. The Z transform of hn = 1 is

Z (hn) =∞

∑n=0

1z−n =1

1− z−1 =z

z−1

10.1.15 Example. The Z transform of

δn :=

1 n = 00 n 6= 0

is

∆(z) = Z (δn) =∞

∑n=0

δnz−n = 1+0z−1 +0z−2 + ...= 1

10.1.16 Example. The Z transform of xn = n is

X (z) = Z (xn) = 0+1z−1 +2z−2 +3z−3...

= u(1+2u+3u2 + ...

)= u

ddu

(u+u2 +u3 + ...

)= u

ddu

(u

1−u

)=

u

(1−u)2 =z−1

(1− z−1)2 =

z2z−1

(z−1)2

=z

(z−1)2 .

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10.1.17 Theorem. Z (κxn +λyn) = κZ (xn)+λZ (yn) .Proof. Easy.

10.1.18 Theorem. For m≥ 0: Z (xn−mhn−m) = z−mZ (xn).Proof. Because

Z (xn−mhn−m) =∞

∑n=0

xn−mhn−mz−n

=∞

∑n=m

xn−mz−n = z−m∞

∑n=m

xn−mz−(n−m)

= z−m∞

∑k=0

xkz−k = z−mZ (xn) .

10.1.19 Theorem. For m≥ 0: Z (xn+m) = zm (Z (xn)−∑m−1k=0 xkz−k).

Proof. Because

Z (xn+m) =∞

∑n=0

xn+mz−n = zm∞

∑n=0

xn+mz−(n+m)

= zm

(∞

∑n=0

xnz−n−m−1

∑n=0

xnz−n

).

10.1.20 Example. So for example

Z (xn+1) = zX (z)− zx0

Z (xn+2) = z2X (z)− z2x0− zx1

10.1.21 Theorem. The following final and initial value properties hold:

x0 = limz→∞

X (z) , limn→∞

xn = limz→1

z−1z

X (z) .

Proof. Easy.

10.1.22 Theorem. If X (z) = Z (xn), then X( z

a

)= Z (anxn).

Proof. Easy.

10.1.23 Theorem. If Z (xn) = X (z), then Z (nxn) =−zdXdz .

Proof. Because

Z (nxn) =∞

∑n=0

nxnz−n = z∞

∑n=0

nxnz−n−1

=−z∞

∑n=0

nxn1n

(z−n)′ =−z

(∞

∑n=0

xnz−n

)′=−z

dXdz

10.1.24 Theorem (Convolution). If Z (xn) = X (z), Z (yn) = Y (z), then

Z

(n

∑m=0

xmxn−m

)= X (z)Y (z) .

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Proof. Because

Z

(n

∑m=0

xmyn−m

)=

∞

∑n=0

(n

∑m=0

xmyn−m

)z−n

=∞

∑n=0

n

∑m=0

xmyn−mz−(n−m)z−m

=∞

∑m=0

∞

∑n=m

xmyn−mz−(n−m)z−m

=∞

∑m=0

∞

∑k=0

xmykz−kz−m

=∞

∑m=0

xmz−m

(∞

∑k=0

ykz−k

)

=

(∞

∑m=0

xmz−m

)(∞

∑k=0

ykz−k

)= X (z)Y (z)

10.1.25 Theorem. The following pairs hold:

an zz−a

n z(z−1)2

n2 z(z+1)(z−1)3

cos(wn) z(z−cosw)z2−2zcosw+1

sin(wn) zsinwz2−2zcosw+1

Proof. Left to the reader.

10.1.26 Inversion methods are similar to those of Laplace transform, as will beseen in the following examles of solving difference equations.


3xn+1−2xn = (−1)n , x0 = 2

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we take the Z -transforms:

3(zX− zx0)−2X =z

z+1⇒

3zX−6z−2X =z

z+1⇒

(3z−2)X = 6z+z

z+1⇒

X =6z

(3z−2)+

z(z+1)(3z−2)

⇒

Xz=

6(3z−2)

+1

(z+1)(3z−2)⇒

X (z) =11z

5(z− 2

3

) − z5(z+1)

⇒

xn =115

(23

)n

− 15(−1)n .


xn+2−32

xn+1 +12

x [n] =13n , x0 = 4, x1 = 0

we take the Z transforms:

z2X− z2x0− zx1−32(zX− zx0)+

12

X =z

z− 13

⇒

z2X−4z2− 32

zX +6z+12

X =z

z− 13

⇒(z2− 3

2z+

12

)X = 4z2−6z+

zz− 1

3

⇒

Xz=

4z−6(z− 1

2

)(z−1)

+1(

z− 13

)(z− 1

2

)(z−1)

⇒

X (z) =9z

z− 13

− 4zz− 1

2

− zz−1

⇒

xn = 9(

13

)n

−4(

12

)n

−1


xn+2−7xn+1 +10xn = 16n, x0 = 6, x1 = 2

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z2X− z2x0− zx1−7(zX− zx0)+10X =16z

(z−1)2 ⇒

z2X−6z2−2z−7(zX−6z)+10X =16z

(z−1)2 ⇒(z2−7z+10

)X = 6z2−40z+

16z

(z−1)2 ⇒

Xz=

5z−1

+4

(z−1)2 +4

z−2− 3

z−5⇒

X [z] =5z

z−1+

4z

(z−1)2 +4z

z−2− 3z

z−5⇒

xn = 5+4n+4 ·2n−3 ·5n.


xn+2−2xn+1 + xn = 3n, x0 = 0, x1 = 2


z2X− z2x0− zx1−2(zX− zx0)+X =3z

(z−1)2 ⇒

z2X−2z−2zX +X =3z

(z−1)2 ⇒(z2−2z+1

)X = 2z+

3z

(z−1)2 ⇒

X (z) =3z

(z−1)4 +2z

(z−1)2 ⇒

xn =12

n(n−1)(n−2)+2n

10.1.31 To introduce the Z transform we did not really need the Laplace transform,we could have started from scratch. (In fact many people have done this, in separatedomains, for instance, generating functions in Probability Theory exploit the sameidea). But it is useful to have seen the connection of L to Z .

10.1.32 Recall the analogy

Z (x) =∞

∑n=0

xn(z−1)n

=∞

∑n=0

xnun,

L (x) =∫

∞

0x(t)

(e−s)t dt =

∫∞

0x(t)vtdt.

So the Laplace transform is like a "continuous" Taylor series.

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10.2 Solved Problems


1. Find xn and X (z) = Z (x(n)) given that(

xn+1 +12xn = 0

x0 = 1

).

Ans. xn =((−1

2

))n.


xn+1 +12xn = 1

x0 = 0

).

Ans. xn =23 −

23

((−1

2

))n.


xn+1 +12xn = 1

x0 = 2

).

Ans. xn =43

((−1

2

))n+ 2

3 .


xn+1 +12xn =

(13

)n

x0 = 1

).

Ans. –15

(−1

2

)n+ 6

5

(13

)n.


xn+1 +12xn = n

x0 = 0

).

Ans. 139

(−1

2

)n− 49 +

23n.


xn+1 +12xn =

(12

)n

x0 = 1

).

Ans. − 12n .


xn+1 +12xn =

(12

)n

x0 = 3

).

Ans. 2(−1

2

)n+(1

2

)n.


xn+1 +12xn =

(−1

2

)n

x0 = 3

).

Ans. 5(−1

2

)n− (2n+2)(−1

2

)n.


xn+1 + xn = 1x0 = 2

).

Ans. 32 (−1)n + 1

2 .


xn+1− xn = 1x0 = 2

).

Ans. 2+n.

11. Find xn and X (z) = Z (x(n)) given that

xn+2 +56xn+1 +

16xn = 0

x0 = 1x1 = 1

.

Ans. 9(−1

3

)n−8(−1

2

)n.


xn+2 +56xn+1 +

16xn = 0

x0 = 1x1 = 1

.

Ans. 9(−1

3

)n−8(−1

2

)n.


xn+2 + xn+1 +14xn = 0

x0 = 1x1 = 1

.

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Ans. 4(−1

2

)n− (3n+3)(−1

2

)n.


xn+2 +56xn+1 +

16xn = (−1)n

x0 = 0x1 = 0

.

Ans. 9(−1

3

)n−12(−1

2

)n+3(−1)n.


xn+2 + xn+1 +14xn = (−1)n

x0 = 1x1 = 1

.

Ans. 4(−1

2

)n− (7n+7)(−1

2

)n+4(−1)n.


xn+2 + xn+1 +14xn =

(12

)n

x0 = 0x1 = 1

.

Ans. − 12n ((−1)n−1).


xn+2 + xn+1 +14xn = 0

x0 = 1x1 = 1

.

Ans. − 12n (n(−1)n−1).


xn+2 +3xn+1 +2xn = 1x0 = 1x1 = 1

.

Ans. 52 (−1)n− 5

3 (−2)n + 16 .


xn+2 +3xn+1 +2xn = nx0 = 0x1 = 1

.

Ans. 54 (−1)n− 10

9 (−2)n + n6 −

536 .


xn+2 +3xn+1 +2xn = 4n

x0 = 0x1 = 1

.

Ans. 56 (−1)n− 6

7 (−2)n + 1425n.

21. Find xn,yn and X (z) = Z (x(n)) ,Y (z) = Z (y(n)) given that

xn+1 =12

xn + yn, x0 = 0,

yn+1 =13

yn, y0 = 1

Ans. x(n) = 6(1

2

)n−6(1

3

)n, y(n) =(1

3

)n.22. Find xn,yn and X (z) = Z (x(n)) ,Y (z) = Z (y(n)) given that

xn+1 =12

xn + yn, x0 = 0,

yn+1 =13

yn +1, y0 = 0

Ans. x(n) = 3+9(1

3

)n−12(1

2

)n, y(n) = 32 −

32

(13

)n.

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xn+1 = xn +2yn, x0 = 1,yn+1 = 4xn +3yn, y0 = 0

Ans. x(n) = 135n +2(−1)n, y(n) = 2

35n− 23 (−1)n.


xn+1 = xn +2yn +1, x0 = 0,yn+1 = 4xn +3yn, y0 = 0

Ans. x(n) = 1125n− 1

3 (−1)n + 14 , y(n) = 1

65n + 13 (−1)n− 1

2 .25. Find xn,yn and X (z) = Z (x(n)) ,Y (z) = Z (y(n)) given that

xn+1 = xn +2yn +2n, x0 = 0,yn+1 = 4xn +3yn, y0 = 0

Ans. x(n) = 192n + 1

95n− 29 (−1)n + 1

4 , y(n) = 295n + 2

9 (−1)n− 492n.


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11 Fourier Series . . . . . . . . . . . . . . . . . . . . 101

12 Fourier Transform . . . . . . . . . . . . . . . 115

Fourier

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11. Fourier Series

A Fourier series expansion of a periodic function f (t) is a representation of f (t) asa sum of sines and cosines (or, as a sum of complex exponentials).


11.1.1 Example. Consider the following problem

dydt

+ay =∞

∑n=0

(−1)n Heaviside(t−nπ) , y(0) = 0. (11.1)

In the plot we see the (periodic) input and the output of (11.1).

The problem can be solved in the standard manner by Laplace transform:

sY +aY =1s

∞

∑n=0

(−1)n e−nπs⇒ Y =1

s(s+a)

∞

∑n=0

e−nπs⇒

Y =1a

(1s− 1

s+a

)∞

∑n=0

(−1)n e−nπs

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102 Chapter 11. Fourier Series

Letting

Yn (s) =(−1)n

a

(1s− 1

s+a

)e−nπs

yn (t) =(−1)n

aHeaviside(t−nπ)

(1− eat)

we get

Y (s) =∞

∑n=0

Yn (s)

y(t) =∞

∑n=0

(−1)n

aHeaviside(t−nπ)

(1− eat) .

11.1.2 In the plot we see clearly the almost-periodicity of y(t). This is reasonable:we apply a periodic input and we get a periodic output. However, periodicity isnot obvious from the form of y(t). We want another solution method, which willshow the periodicity clearly (actually we want to separate the transient part of thesolution from the steady-state, periodic one).

11.1.3 Example. Before proceeding to this end, we will also solve the followingeasy problems. First we consider

dydt

+ay = 1, y(0) = 0 (11.2)

which has solution:

(s+a)Y =1s⇒ Y =

1a

(1s− 1

s+a

)⇒ y(t) =

1a

(1− e−at) .

Second we considerdydt

+ay = sin(nt) , y(0) = 0 (11.3)

which has solution:

sY +aY =n

s2 +n2 ⇒ Y =n

(s2 +n2)(s+a)⇒

Y =n

a2 +n2

(1

s+a+

an2 + s2 −

sn2 + s2

)⇒

y(t) =n

a2 +n2

(e−at +asin(nt)− cos(nt)

).

In the plot we see again the almost-periodicity of y(t); this is also reflected in thesteady-state component of the solution:

na2 +n2 (asin(nt)− cos(nt)) .

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11.1.4 Note that, by linearity, if

dydt

+ay = u(t) , x(0) = 0. (11.4)

with input u(t) = un (t) has solution yn (t) then with input

u(t) = ∑n

κnun (t)

it has solutiony(t) = ∑

nκnyn (t) .

11.1.5 Now we will do two things:1. present a method by which by which periodic functions are written as sums

of sines and cosines;2. use this method to write the solution of our Example 11.1.1 as the sum of a

transient part and a periodic (sum of sines and cosines) part.

11.1.6 Definition: We call f (t) periodic, iff there exists some T such that

∀t : f (t +T ) = f (t) .

The smallest such T is called the period of f (t).

11.1.7 Definition: We call f (t) piecewise continuous in [t1, t2] iff1. There exist t1 = τ0 < τ1 < τ2 < ... < τn−1 < τn = t2 such that

[t1, t2] = [τ0,τ1]∪ [τ1,τ2]∪ ...∪ [τn−1,τn] ,

2. f (t) is continuous inside each [τi−1,τi] and3. for each τi (i ∈ 1, ...,n) the side limits limt→τ

−i

f (t), limt→τ+i

f (t) exist (exceptfor limt→t−1

f (t) and limt→t+2f (t) ).

11.1.8 Theorem: Let f (t) satisfy the following (Dirichlet) conditions:1. f (t) is defined in (0,2L),2. f (t) and f ′ (t) are piecewise continuous in (0,2L),3. f (t) is periodic with period 2L.

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Then, using the constants

an =1L

∫ 2L

0f (t)cos

nπtL

dt, bn =1L

∫ 2L

0f (t)sin

nπtL

dt. (11.5)

we have:1. at every point of continuity t:

f (t) =a0

2+

∞

∑n=1

an cosnπtL

+∞

∑n=1

bn sinnπtL

, (11.6)

2. at every point of discontinuity t:

12

(lim

τ→t−f (τ)+ lim

τ→t+f (τ)

)=

a0

2+

∞

∑n=1

an cosnπtL

+∞

∑n=1

bn sinnπtL

, (11.7)

The right side of (11.6), when an,bn are given by (11.5), is called the trigonometricFourier series of f (t).Proof. This is a partial proof. Assuming that f (t) can be written as

f (t) =a0

2+

∞

∑n=1

an cosnπtL

+∞

∑n=1

bn sinnπtL

, (11.8)

we will show that an, bn are given by (11.5).Indeed, since

∀n ∈ N :∫ 2L

0cos

nπtL

dt =∫ 2L

0sin

nπtL

dt = 0.

we have:∫ 2L

0f (t)dt =

∫ 2L

0

(a0

2+

∞

∑n=1

an cosnπtL

+∞

∑n=1

bn sinnπtL

)dt

=∫ 2L

0

a0

2dt +

∞

∑n=1

(∫ 2L

0an cos

nπtL

dt)+

∞

∑n=1

(∫ 2L

0bn sin

nπtL

dt)

=∫ 2L

0

a0

2dt +0

In short ∫ 2L

0f (t)dt =

a0

22L⇒ a0 =

1L

∫ 2L

0f (t)dt =

1L

∫ 2L

0f (t)cos

0πtL

dt.

This proves the a0 formula.Taking any m ∈ N, we have∫ 2L

0f (t)cos

mπtL

dt =∫ 2L

0

(a0

2+

∞

∑n=1

an cosnπtL

+∞

∑n=1

bn sinnπtL

)cos

mπtL

dt

=∫ 2L

0

a0

2cos

mπtL

dt +∞

∑n=1

(∫ L

−Lan cos

nπtL

cosmπt

Ldt)

+∞

∑n=1

(∫ 2L

0bn sin

nπtL

cosmπt

Ldtdt

). (11.9)

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Since

∀n ∈ N0 :∫ 2L

0sin

nπtL

cosmπt

Ldt = 0,

∀n ∈ N0,n 6= m :∫ 2L

0cos

nπtL

cosmπt

Ldt = 0

and ∫ 2L

0cos2 mπt

Ldt = L

replacing the above in (11.9) we get∫ 2L

0f (t)cos

mπtL

dt = amL⇒ am =1L

∫ 2L

0f (t)cos

mπtL

dt

and we have proved the an formula for n ∈ N. The bn formula (for n ∈ N) is provedsimilarly.

11.1.9 For a full proof we would need to define the finite sum

fN (t) =a0

2+

N

∑n=1

an cosnπtL

+N

∑n=1

bn sinnπtL

and show that (at every continuity point t) we have

limN→∞

fN (t) = f (t) .

11.1.10 Remark: The Dirichlet conditions are sufficient but not necessary for theexistence of the trigonometric Fourier series of f (t).

11.1.11 Example: Let f (t) be

f (t) =

1 when t ∈ (2kπ,(2k+1)π) , k ∈ N00 when t ∈ ((2k+1)π,(2k+2)π) , k ∈ N0

Obviously f (t) is periodic with T = 2π. We compute its Fourier series as follows.

a0 =1π

∫ 2π

0f (t)dt =

1π

∫π

01dt +

1π

∫ 2π

π

0dt = 1.

For any n≥ 1 we have

an =1π

∫ 2π

0f (t)cos(nt)dt =

1π

∫π

01cos(nt)dt = 0.

Also

bn =1π

∫ 2π

0f (t)dt =

1π

∫π

0sin(nt)dt =− 1

π

(1n

cos(nt))π

0

=− 1nπ

(cos(nπ)− cos(0)) = 2

nπwhen n = 2k+1

0 when n = 2k.

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Hence

f (t) =12+

2π

(sin(t)

1+

sin(3t)3

+sin(5t)

5+ ...

)=

12+

2π

∑n∈1,3,5,...

sin(nt)n

11.1.12 Example. To solve Example 11.1.1, we first take u0 (t) = 12 and get that

y0 (t) =1

2a

(1− e−at) .

Then, with un (t) = 2nπ

sin(nt) we get (from the second problem) that

2(a2 +n2)π

e−at− 2(a2 +n2)πn

(ncos(nt)−asin(nt)) .

And finally, with

u(t) =12+

2π

∑n∈1,3,5,...

sin(nt)n

the solution is the sum of transient and steady-state periodic parts:

y(t) = y(t)+ y(t)

with

y(t) =−e−at

2a+ ∑

n∈1,3,5,...

2(a2 +n2)π

e−at

y(t) =1

2a− ∑

n∈1,3,5,...

2(ncos(nt)−asin(nt))(a2 +n2)πn

.

11.1.13 Theorem: Let f (t) satisfy the Dirichlet conditions. Then:1. f (t) is even iff its trigonometric Fourier series has only cosine terms;2. f (t) is odd iff its trigonometric Fourier series has only sine terms.

Proof. Assume that the f (t) Fourier series has only cosine terms:

f (t) =a0

2+

∞

∑n=1

an cosnπtL

.

Then f (t) is even, since it is a sum of even functions.Conversely, since

bn =1L

∫ L

−Lf (t)sin

nπtL

dt

if f (t) is even then f (t)sin nπtL is odd and bn = 0, i.e., f (t) has only cosine terms.

We have proved the first part of the theorem; the second part is proved similarly.

11.1.14 If f (t) is defined only on a finite interval [t1, t2], we can extend it on R bydefining

∀t ∈ [t1, t2] ,n ∈ N : f (t +n · (t2− t1)) = f (t) .

The extended f (t) is periodic with T = t2− t1 and the formulas (11.6), (11.7) hold.

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11.1.15 Example: Define f : [−5,5]→ R as follows:

f (t) =

0 when −5 < t < 03 when 0≤ t < 5 .

To find the Fourier series of f (t), we extend f (t) to all of R. Then f (t) satisfiesthe Dirichlet conditions and has half period L = 5, hence

an =15

∫ 5

−5f (t)cos

nπt5

dt =15

∫ 0

−50cos

nπt5

dt +15

∫ 5

03cos

nπt5

dt

=35

(5

nπsin

nπt5

)t=5

t=0=

3 when n = 00 when n 6= 0 .

Similarly

bn =15

∫ 5

−5f (t)sin

nπt5

dt =15

∫ 0

−50sin

nπt5

dt +15

∫ 5

03sin

nπt5

dt

=35

(− 5

nπcos

nπt5

)t=5

t=0=

3(1− cosnπ)

nπ.

Finally

f (t) =32+

6π

(sin

πt5+

13

sin3πt5

+15

sin5πt5

+ ...

).

Nota Bene:

∀n ∈ Z : f (5n) = limt→5n−

f (t)+ limt→5n+

f (t) =32.

11.1.16 Theorem. Let f (t) satisfy the Dirichlet conditions. Then at every conti-nuity point t we have

f (t) =∞

∑n=−∞

cneinπt/L, (11.10)

where

cn =1

2L

∫ L

−Lf (t)e−inπt/Ldt. (11.11)

At every discontinuity point t, the left side of (11.10) is replaced by

12

(lim

t→τ−f (t)+ lim

t→τ+f (t)

).

The right side of (11.10) is called the exponential Fourier series of f (t).Proof. By Theorem 11.1.8, since f (t) satisfies the Dirichlet conditions, at everycontinuity point t we have

f (t) =a0

2+

∞

∑n=1

an cosnπtL

+∞

∑n=1

bn sinnπtL

.

Ath.

Keha

gias


Replacing cos nπtL , sin nπt

L we get

f (t) =a0

2+

∞

∑n=1

ane

inπtL + e−

inπtL

2− i

∞

∑n=1

bne

inπtL − e−

inπtL

2

=a0

2+

∞

∑n=1

an− ibn

2e

inπtL +

∞

∑n=1

an + ibn

2e−

inπtL

=∞

∑n=−∞

cneinπt

L

where, for n = 0,

c0 =a0

2=

12L

∫ L

−Lf (t)dt =

12L

∫ L

−Lf (t)ei 0πt

L dt,

for n ∈ Z+,

cn =an− ibn

2=

12L

(∫ L

−Lf (t)cos

nπtL

dt− i∫ L

−Lf (t)sin

nπtL

dt)

=1

2L

∫ L

−Lf (t)

(cos

nπtL− isin

nπtL

)dt

=1

2L

∫ L

−Lf (t)e−i nπt

L dt

and for n ∈ Z−,

cn =a−n + ib−n

2=

12L

(∫ L

−Lf (t)cos

−nπtL

dt + i∫ L

−Lf (t)sin

−nπtL

dt)

=1

2L

∫ L

−Lf (t)

(cos

nπtL− isin

nπtL

)dt

=1

2L

∫ L

−Lf (t)e−i nπt

L dt.

In short, for each n ∈ Z :

cn =1

2L

∫ L

−Lf (t)e−i nπt

L dt.

11.1.17 Example. Continuing Example 11.1.12, let us get the exponential Fourierseries of f (t). We have

cn =1

2π

∫ 2π

0f (t)e−intdt =

12π

(∫π

01e−intdt +

∫ 2π

π

0e−intdt).

Since the second integral is zero, we get: for n = 0

c0 =1

2π

∫π

0dt =

12

and for n 6= 0:

cn =1

2π

∫π

0e−intdt =− 1

2inπe−int |π0 =− 1

2inπ

(e−inπ −1

)=

1inπ

=− inπ

when n = 2k+10 when n = 2k.

Ath.

Keha

gias


and

f (t) =12+ ∑

n∈...,−3,−1,1,3,...

1inπ

eint .

The exponential series is equivalent to the trigonometric:

f (t) =12+ ∑

n∈...,−3,−1,1,3,...

1inπ

eint =12+

2π

eit− e−it

2i+

23π

ei3t− e−i3t

2i+ ....

=12+

2π

sin(t)+2

3πsin(3t)+ ....=

12+

2π

∑n∈1,3,5,...

sin(nt)n

.

11.1.18 Theorem: Let f (t) be periodic with period T and f (t) , d fdt be piecewise

continuous. Then, at the continuity points of f (t) and d fdt , the derivative (resp. inte-

gral) of f (t) equals the term-by-term derivative (resp. integral) of the corresponding(trig / exponential) Fourier series.

11.1.19 Theorem (Parseval): Let f (t), g(t) satisfy the Dirichlet conditions withhalf-period L, and Fourier series

f (t) =a0

2+

∞

∑n=1

an cosnπtL

+∞

∑n=1

bn sinnπtL

,

g(t) =p0

2+

∞

∑n=1

pn cosnπtL

+∞

∑n=1

qn sinnπtL

.

Then we have

1L

∫ L

−Lf (t)g(t)dt =

a0 p0

2+

∞

∑n=1

(an pn +bnqn) , (11.12)

1L

∫ L

−L( f (t))2 dt =

a20

2+

∞

∑n=1

(a2

n +b2n). (11.13)

Also, if f (t) ,g(t) have exponential Fourier series

f (t) =∞

∑n=−∞

cneinπt/L, g(t) =∞

∑n=−∞

rneinπt/L,

then

12L

∫ L

−Lf (t)g(t)dt =

∞

∑n=−∞

cnrn, (11.14)

12L

∫ L

−L| f (t)|2 dt =

∞

∑n=−∞

|cn|2 . (11.15)

Ath.

Keha

gias


Proof. We only prove the exponential forms(11.14)-(11.15). We have

∫ L

−Lf (t)g(t)dt =

∫ L

−L

(∞

∑n=−∞

cnei nπtL

)(∞

∑m=−∞

rmei mπtL

)dt

=∫ L

−L

(∞

∑n=−∞

cnei nπtL

)(∞

∑m=−∞

rme−i mπtL

)dt

=∫ L

−L

(∞

∑n=−∞

∞

∑m=−∞

cnei nπtL rme−i mπt

L

)dt

=∞

∑n=−∞

∞

∑m=−∞

cnrm

(∫ L

−Lei nπt

L e−i mπtL dt

). (11.16)

We see that ∫ L

−Lei nπt

L e−i mπtL dt =

2L when n = m

0 when n 6= m . (11.17)

Hence (11.16) yields ∫ L

−Lf (t)g(t)dt = 2L

∞

∑n=−∞

cnrn

which is equivalent to (11.14). For (11.15) let f (t) = g(t), then f (t)g(t) = | f (t)|2,cnrn = |cn|2.

11.1.20 The proof of Theorem 11.1.19 reminds us of inner products and vectorspaces. Let us show the connection between these and Fourier analysis.

11.1.21 Notation. We denote by FL the set of functions which satisfy the Dirichletconditions (for a fixed L).

11.1.22 Theorem: FL is a vector space.Proof. Let f (t) ,g(t) ∈ FL and κ,λ ∈ C. Define h(t) = κ f (t)+λg(t).

1. Since f (t), g(t) are defined in (−L,L), the same is true of h(t).2. Sicne f (t) ,g(t) and f ′ (t) ,g′ (t) are piecewise continuous in (−L,L), the same

is true for h(t), h′ (t).3. Since f (t) ,g(t) are periodic with period 2L, the same is true of h(t).

Hence h(t) = κ f (t)+λg(t) ∈ FL, i.e., FL is a vector space.

11.1.23 Definition. We say ..., f−1 (t) , f0 (t) , f1 (t) , ... ⊆ FL is linearly indepen-dent iff (

∀t :∞

∑n=−∞

κn fn (t) = 0

)⇒ (∀n ∈ Z : κn = 0) .

11.1.24 Definition. We say ..., f−1 (t) , f0 (t) , f1 (t) , ... ⊆ FL is a basis of FL iff (i)..., f−1 (t), f0 (t), f1 (t), ... is linearly independent and (ii) every f (t) ∈ FL can bewritten as linear combination of the fn (t)’s:

f (t) =∞

∑n=−∞

κn fn (t) .

Ath.

Keha

gias


11.1.25 Definition. We say that ..., f−1 (t) , f0 (t) , f1 (t) , ... ⊆ FL is orthogonal iff

∀m 6= n :∫ L

−Lfm (t)gm (t) = 0.

11.1.26 Theorem. The set

ei nπtL

∞

n=−∞

is an orthogonal basis of FL.Proof. With the above definitions we see the following.

1.

ei nπtL

∞

n=−∞

is a subset of FL and is orthogonal; we showed that in (11.17)of Theorem 11.1.19.

2. Since

ei nπtL

∞

n=−∞

is orthogonal it is also linearly independent.

3. From Theorem 11.1.19 every f (t) ∈ FL can be written as

f (t) =∞

∑n=−∞

cnei nπtL ,

a linear combination of

ei nπtL

∞

n=−∞

elements; hence

ei nπtL

∞

n=−∞

is a basisof FL .

11.1.27 Theorem. The set cosnt∞

n=0∪sinnt∞

n=1 is an orthogonal basis of FL.Proof. Similar to that of Theorem 11.1.26.


11.3 Unsolved Problems1. Compute the trigonometric Fourier series of

f (x) =

2 gia −π < x < 01 gia 0 < x < π

.

Ans. f (x) = 32 −

2π ∑

∞n=0

12n+1 sin [(2n+1)x].

2. Compute the trigonometric Fourier series of

f (x) =

0 gia −5 < x < 03 gia 0 < x < 5 .

Ans. f (x) = 32 +∑

∞n=0

6(2n+1)π sin

((2n+1)πx

5

).

3. Compute the trigonometric Fourier series of f (x) = cosx with period 2π.Ans. f (x) = cosx.

4. Compute the trigonometric Fourier series of f (x) = |x| (−1 < x < 1).Ans. f (x) = 1

2 −4

π2 ∑∞n=0

cos[(2n+1)πx](2n+1)2 .

5. Compute the trigonometric Fourier series of

f (x) =

1 gia 0 < x < 2−1 gia 2 < x < 4 .

Ans. 4π ∑

∞n=0

sin (2n+1)πx2

2n+1 .

Ath.

Keha

gias


6. Compute the trigonometric Fourier series of f (x) = 4x (0 < x < 10).

Ans. 20− 40π ∑

∞n=1

sin( nπx5 )

n .7. Compute the trigonometric Fourier series of f (x) = cos2 x.

Ans. 12 +

12 cos2x.

8. Compute the exponential Fourier series of f (x) = 1 (0 < x < 2π).Ans. c0 = 1, cn = 0 for n 6= 0.

9. Compute the exponential Fourier series of f (x) = x2 (0 < x < 2π ).Ans. cn =

2n2 + i2

n , n = 0,±1,±2, ... .10. Compute the exponential Fourier series of

f (x) =

1 gia 0 < x < 2−1 gia 2 < x < 4 .

Ans. cn =2n2 + i2π

n , n = 0,±1,±2, ... .11. Compute the exponential Fourier series of f (x) = cosx (0 < x < 2π ).

Ans. c1 = c−1 =12 , all the other coefficients are 0.

12. Compute the trigonometric and exponential Fourier series of f (t) = t (on[−π,π]).Ans.

f (t) = 2 sin(t)− sin(2 t)+2/3 sin(3 t)−1/2 sin(4 t)+2/5 sin(5 t)+ ...,

f (t) =−ieit + ie−it + i/2e2 it− i/2e−2 it− i/3e3 it + i/3e−3 it + i/4e4 it− i/4e−4 it + ...

13. Compute the trigonometric and exponential Fourier series of f (t) = t2 (on[−π,π]).Ans.

f (t) = 1/3π2−4 cos(t)+ cos(2 t)−4/9 cos(3 t)+1/4 cos(4 t)− ...

f (t) = 1/3π2−2eit−2e−it +1/2e2 it +1/2e−2 it−2/9e3 it−2/9e−3 it + ...

14. Compute the trigonometric and exponential Fourier series of f (t) = t2 + t (on[−π,π]).Ans.

f (t) =13

π2−4 cos(t)+2 sin(t)+ cos(2 t)− sin(2 t)− 4

9cos(3 t)+

23

sin(3 t)+ ...

f (t) =13

π2− (2+ i)eit− (2− i)e−it +

(12+

i2

)e2 it +

(12− i

2

)e−2 it− ...

15. Compute the trigonometric and exponential Fourier series of f (t)=

0 −π < t < 03 0 < t < π

.

Ap.

f (t) =32+6

sin(t)π

+2sin(3 t)

π+

65

sin(5 t)π

+ ...

f (t) = 3/2− 3 ieit

π+

3 ie−it

π− ie3 it

π+

ie−3 it

π− 3/5 ie5 it

π+

3/5 ie−5 it

π+ ...

Ath.

Keha

gias



0 −1 < t < 01 0 < t < 1 .

Ans.

f (t) =12+2

sin(π t)π

+23

sin(3π t)π

+25

sin(5π t)π

+ ...

f (t) =12− ieiπ t

π+

ie−iπ t

π− ie3 iπ t

3π+

ie−3 iπ t

3π− ie5 iπ t

5π+

ie−5 iπ t

5π+ ...


0 −5 < t < 03 0 < t < 5 .

Ans.

f (t) = 3/2+6sin(1/5π t)

π+2

sin(3/5π t)π

+6/5sin(π t)

π+ ...

f (t) = 3/2− 3 iei/5π t

π+

3 ie−i/5π t

π− ie3/5 iπ t

π+

ie−3/5 iπ t

π− 3/5 ieiπ t

π+

3/5 ie−iπ t

π+ ...

18. Compute the trigonometric and exponential Fourier series of f (t) = (t−1)2

(−π < t < π )Ans.

f (t) =103

3−400

cos(1/10π t)π2 −40

sin(1/10π t)π

+100cos(1/5π t)

π2 +20sin(1/5π t)

π− ...

f (t) =1033

+

(−200π

−2 +20 iπ

)ei/10π t +

(−200π

−2− 20 iπ

)e−i/10π t + ...

19. Compute the trigonometric and exponential Fourier series of f (t) = |sin t|(−π < t < π )Ans.

f (t) = 2π−1−4/3

cos(2 t)π

− 4 cos(4 t)15π

+ ...

f (t) = 2π−1−2/3

e2 it

π−2/3

e−2 it

π−2/15

e4 it

π−2/15

e−4 it

π+ ...

20. Compute the trigonometric and exponential Fourier series of f (t) = e|t| (−1 <t < 1)Ans.

f (t) =2eπ −2

2π+

(−eπ −1)cos(t)π

+25(eπ −1)cos(2 t)

π+

15(−eπ −1)cos(3 t)

π+ ...

f (t) =12

2eπ −2π

+12(−eπ −1)eit

π+

12(−eπ −1)e−it

π+

15(eπ −1)e2 it

π+

15(eπ −1)e−2 it

π+ ...


1. Find necessary and sufficient conditions so that the exponential Fourierseries of f (t) has only real (only imaginary) terms.

Ath.

Keha

gias


2. Show that

f (t) =a0

2+

∞

∑n=1

an cosnπtT

+∞

∑n=1

bn sinnπtT

.

can also be written as

f (t) =∞

∑n=0

pn cos(nπt

T+φn

).

3. Prove that ∑∞n=1

sin(na)sin(nb)n2 = a(π−b)

2 .


1(2n+1)4 =

π4

96 .


1n4 =

π4

90 .6. Compute ∑

∞n=1

1n2 .

7. Compute ∑∞n=1

14n2−1 .

8. Suppose f (t) = ∑∞n=−∞ anein2πt , g(t) = ∑

∞n=−∞ bnein2πt and

p(t) =∫ 2π

0f (t− τ)g(τ)dτ =

∞

∑n=−∞

cnein2πt .

9. If g(t) = ∑∞n=−∞ anein2πt , what is the exponential Fourier series of f (t) =

tg(t)?What is the relatrionship between (an)∞

n=−∞,(bn)

∞

n=−∞,(cn)

∞

n=−∞?

10. Prove that: at every x ∈ R, f (t) = ∑∞n=1

sin(n2t)n2 is continuous but not differen-

tiable.

Ath.

Keha

gias



12. Fourier Transform

The Fourier transform is the extension of the Fourier series when the finite interval[−L,L] is replaced by (−∞,∞).

12.1 Theory and Examples12.1.1 For periodic functions, with period T , we have

f (t) =∞

∑n=−∞

cnei n2π

T t , cn =1T

∫ T/2

−T/2f (t)e−i n2π

T tdt. (12.1)

What happens when T → ∞? Intuitively we see the following.1. f (t) is not periodic.2. 2π

T → 0.3. The sum of (12.1) tends to an integral.4. For (12.1) to be meaningful, the integral must be well defined; this is not

guaranteed when the integration limits tend to ±∞.

12.1.2 Theorem. Suppose that f (t)1. is absolutely integrable, i.e.,

∫∞

−∞| f (t)|dt < ∞;

2. satisfies, for every T , the Dirichlet conditions in every finite interval[−T

2 ,T2

]⊆

(−∞,∞).

Then1. at every continuity point t of f (t) we have

f (t) =1

2π

∫∞

−∞

F (ω)eiωtdω, (12.2)

whereF (ω) =

∫∞

−∞

f (t)e−iωtdt. (12.3)

2. at every discontinuity point t of f (t) we have:

12

(lim


τ→t+f (τ)

)=

12π

∫∞

−∞

F (ω)eiωtdω. (12.4)

Ath.

Keha

gias

116 Chapter 12. Fourier Transform

Proof. The following “proof” is not rigorous but captures the basic ideas. For agiven f (t) choose some T and for each t ∈

(−T

2 ,T2

)define fT (t) = f (t). Then

∀t ∈(−T

2,T2

): fT (t) =

∞

∑n=−∞

c(n)ei 2πnT t , (12.5)

where

c(n) =1T

∫ T/2

−T/2fT (t)e−i 2πn

T tdt. (12.6)

Letting δω = 2π

T and ω = nδω = 2πnT we get from (12.5) that

fT (t) =∞

∑n=−∞

c(n)einδωt

=∞

∑n=−∞

(1T

∫ T/2

−T/2fT (t)e−inδωtdt

)einδωt

=∞

∑n=−∞

(1

2π

∫ T/2

−T/2fT (t)e−iωtdt

)eiωt

δω

=1

2π∑

ω∈...,− 2π

T ,0, 2π

T ,...FT (ω)eiωt

δω

where

FT (ω) =∫ T/2

−T/2fT (t)e−iωtdt.

Taking limits (and assuming they exist)

f (t) = limT→∞

fT (t) , (12.7)

F (ω) = limT→∞

FT (ω) =∫

∞

−∞

f (t)e−iωtdt. (12.8)

Substituting (12.7)-(12.8) in (12.5) and assuming that, as T →∞, the sum of (12.5)tends to an integral, we get

f (t) =1

2π

∫∞

−∞

F (ω)eiωtdt.

12.1.3 Definition. We define F (ω), the Fourier transform of f (t) by

F (ω) :=∫

∞

−∞

f (t)e−iωtdt

12.1.4 Roughly speaking (and ignoring discontinuity points) we have

F (ω) =∫

∞

−∞

f (t)e−iωtdt⇔ f (t) =1

2π

∫∞

−∞

F (ω)eiωtdω (12.9)

From (12.9) we see that the Fourier transform is invertible.

Ath.

Keha

gias


12.1.5 Notation. We write

F ( f (t)) = F (ω) and F−1 (F (ω)) = f (t)

12.1.6 Theorem. Suppose that f (t)1. is absolutely integrable, i.e.,

∫∞

−∞| f (t)|dt < ∞;

2. satisfies the Dirichlet conditions in every finite interval[−T

2 ,T2

]⊆ (−∞,∞).

Then1. at every continuity point t of f (t) we have

f (t) =∫

∞

0(a(ω)cosωt +b(ω)cosωt)dω, (12.10)

where

a(ω) =1π

∫∞

−∞

f (t)cosωtdt, b(ω) =1π

∫∞

−∞

f (t)sinωtdt; (12.11)

2. at every discontinuity point t of f (t) we have:

12

(lim


τ→t+f (τ)

)=∫

∞

0(a(ω)cosωt +b(ω)cosωt)dω. (12.12)

Proof. It follows immediately from Theorem 12.1.8.

12.1.7 Definition. We define FC (ω), the Fourier cosine transform of f (t), by

FC (ω) :=∫

∞

−∞

f (t)cos(ωt)dt

and FS (ω), the Fourier sine transform of f (t), by

FS (ω) :=∫

∞

−∞

f (t)sin(ωt)dt

We writeFC ( f (t)) = FC (ω) , F−1

C (F (ω)) = f (t)

andFS ( f (t)) = FS (ω) , F−1

S (F (ω)) = f (t) .

12.1.8 Intuitively, (12.10) is the “limit” of the Fourier series and (12.11) is the“limit” of the formula for the Fourier coefficients (in both cases, when T → ∞).

12.1.9 Note that (12.11) can also be written as

f (t) =1

2π

∫∞

−∞

∫∞

−∞

f (τ)cos(w(t− τ))dτdw.

12.1.10 Example: To compute the Fourier transform of

f (t) =

1 when |t|< a0 when |t|> a

we have

F (ω) =∫

∞

−∞

f (t)e−iωtdt =∫ a

−ae−iωtdt =

e−iωt

−iω|t=at=−a

=e−iωa− eiωa

−iω= 2

sin(ωa)ω

= 2sinc(wa) .

Ath.

Keha

gias


12.1.11 Example: To compute the Fourier transform of Dirac(t− t0) we have

F (δ (t− t0)) =∫

∞

−∞

Dirac(t− t0)e−iωtdt = e−iωt0

12.1.12 Example. To compute the Fourier transform of f (t) = e−|t|:

F(

e−|t|)=∫

∞

−∞

e−|t|e−iωtdt

=∫

∞

0e−te−iωtdt +

∫ 0

−∞

ete−iωtdt.

Next (with a = 1+ iω )∫∞

0e−te−iωtdt =

∫∞

0e−atdt =−1

ae−at |t=∞

t=0 =−1a=

11+ iω

(since e−a∞ = e−(1+i)∞ = 0). Similarly∫ 0

−∞

ete−iωtdt =1

1− iω

henceF(

e−|t|)=

11− iω

+1

1+ iω=

2ω2 +1

.

12.1.13 Theorem (Linearity). Suppose f (t), g(t) satisfy the conditions of 12.1.2.Then

F (κ f +λg) = κF ( f )+λF (g)

Proof. Obvious.

12.1.14 Theorem (Parseval). Suppose f (t) ,g(t) satisfy the conditions of 12.1.2and have Fourier transforms F (ω) ,G(ω). Then∫

∞

−∞

f (t)g(t)dt =1

2π

∫∞

−∞

F (ω)G(ω)dω, (12.13)∫∞

−∞

| f (t)|2 dt =1

2π

∫∞

−∞

|F (ω)|2 dω. (12.14)

Proof. We have∫∞

−∞

f (t)g(t)dt =∫

∞

−∞

f (t)(

12π

∫∞

−∞

G(ω)eiωtdω

)dt

=1

2π

∫∞

−∞

f (t)(∫

∞

−∞

G(ω)e−iωtdω

)dt

=1

2π

∫∞

−∞

G(ω)

(∫∞

−∞

f (t)e−iωtdt)

dω

=1

2π

∫∞

−∞

G(ω)F (ω)dω.

This proves (12.13); then (12.14) follows, taking g(t) = f (t).

Ath.

Keha

gias


12.1.15 Example. To compute the integral∫

∞

−∞dt

(1+t2)2 we have F

(1

1+t2

)= πe−|ω|

and, letting f (t) = 11+t2 , we get∫

∞

−∞

dt

(1+ t2)2 =

∫∞

−∞

| f (t)|2 dt =1

2π

∫∞

−∞

∣∣∣πe−|ω|∣∣∣2 dω = π

∫∞

0e−2ωdω =

π

2.

12.1.16 Theorem (Duality). Suppose f (t) satisfies the conditions of 12.1.2 andhas Fourier transform F (ω). Then

F (F (t)) = 2π f (−ω) .

Proof. We have

F ( f (t)) = F (ω) =∫

∞

−∞

f (t)e−iωtdt, (12.15)

F−1 (F (ω)) = f (t) =1

2π

∫∞

−∞

F (ω)eiωtdω. (12.16)

Substituting in (12.16) t with −t we get

2π f (−t) =∫

∞

−∞

F (ω)e−iωtdω.

Exchanging t and ω we get

2π f (−ω) =∫

∞

−∞

F (t)e−iωtdt = F (F (t)) .

12.1.17 Example: To compute the Fourier transform of eiω0t , we know thatF (Dirac(t− t0)) = e−iωt0 and so F (Dirac(t + t0)) = eiωt0. By the duality theorem

F(eiω0t)= 2π Dirac(−w+w0) = 2π Dirac(w−w0) .

Or, we can take

F−1 (2πδ (w−w0)) =1

2π2π

∫∞

−∞

Dirac(w−w0)eiwtdw = eiw0t .

12.1.18 Nota Bene! From this follows that

F (1) = Dirac(t) .

What is F (Heaviside(t))? We will answer this a little later.

12.1.19 Example: To compute the Fourier transform of cos(w0t), sin(w0t), we use

F (cos(w0t)) = F

(eiω0t + e−iω0t

2

)= π (Dirac(w+w0)+Dirac(w−w0)) .

Similarly we get

F (sin(w0t)) = F

(eiω0t− e−iω0t

2i

)= πi(Dirac(w+w0)−Dirac(w−w0)) .

Ath.

Keha

gias


12.1.20 These results actually make intuitive sense, because they show thepresence of pure frequencies.

12.1.21 Example. To compute the Fourier transform of f (t) = 11+t2 , we use

F

(12

e−|t|)=

11+ω2

Then, by Theorem 12.1.16, we have

F

(1

1+ t2

)= 2π f (−ω) = πe−|ω|.

12.1.22 Theorem (Scaling). Suppose f (t) satisfies the conditions of 12.1.2 andhas Fourier transform F (ω). Then. for each a 6= 0:

F ( f (at)) =1|a|

F(

ω

a

).

Proof. Let a > 0. Then

F ( f (at)) =1a

∫∞

−∞

f (at)e−i ω

a atd (at)

=1a

∫∞

−∞

f (s)e−i ω

a sds =1a

F(

ω

a

).

The proof is similar when a < 0.

12.1.23 Example. To compute the Fourier transform of f (t) = 14+9t2 , we have

F

(1

4+9t2

)=

14F

(1

1+ 94t2

)=

14F

(1

1+(3

2t)2

).

Hence taking a = 23 in Theorem 12.1.22 we get

F

(1

4+9t2

)=

14

132

πe−|2ω/3| =π

6e−|2ω/3|.

12.1.24 Theorem (Shifting). Suppose f (t) satisfies the conditions of 12.1.2 andhas Fourier transform F (ω). Then

F ( f (t− t0)) = F (ω)e−iωt0, F(

f (t)eiω0t)= F (ω−ω0) .

Proof. We have

F ( f (t− t0)) =∫

∞

−∞

f (t− t0)e−iωtdt

=∫

∞

−∞

f (t− t0)e−iω(t−t0)e−iωt0d (t− t0) = F (ω)e−iωt0.

Also

F(

f (t)eiω0t)= ∫ ∞

−∞

f (t)e−iωteiω0tdt

=∫

∞

−∞

f (t)e−i(ω−ω0)tdt = F (ω−ω0) .

Ath.

Keha

gias


12.1.25 Example. To compute the Fourier transform of f (t) = 1t2−4t+5 we note

thatt2−4t +5 = 1+(t−2)2 .

Hence

F

(1

1+ t2

)= πe−|ω|,

F

(1

t2−4t +5

)= F

(1+(t−2)2

)= πe−|ω|e−i2ω .

12.1.26 Theorem (Differentiation). Suppose f (t) satisfies the conditions of12.1.2 and has Fourier transform F (ω). Then

F

(d fdt

)= iωF (ω) , F

(∫f dt)=

1iω

F (ω) , F (t f (t)) = idFdω

.

Proof. We have

F

(d fdt

)=∫

∞

−∞

d fdt

e−iωtdt =∫

∞

−∞

e−iωtd f

= f (t)e−iωt |t=∞t=−∞−

∫∞

−∞

f (t)d(e−iωt)

= f (t)e−iωt |t=∞t=−∞−

∫∞

−∞

f (t)e−iωt (−iω)dt.

From∫

∞

−∞| f (t)|dt < ∞ we conclude that limt→−∞ f (t) = limt→∞ f (t) = 0; hence

F

(d fdt

)= iω

∫∞

−∞

f (t)e−iωtdt = iωF ( f ) = iωF (ω) .

For the second part of the theorem let g(t) =∫

f (t)dt; then dgdt = f (t) and

F (∫

f dt) = F (g(t)) = G(ω). Now

F (ω) = F ( f ) = F

(dgdt

)= iωG(ω)⇒ G(ω) =

1iω

F (ω) .

The third part is proved similarly to the first.

12.1.27 Example. To compute the Fourier transform of f (t) = t(t2+1)

2 , we use

ddt

(1

1+ t2

)=− 2t

(t2 +1)2 .

Hence

F

(− 2t

(t2 +1)2

)= iωF

(1

1+ t2

)= iωπe−|ω|.

And so

F

(t

(t2 +1)2

)=−1

2iωF

(1

1+ t2

)=− iωπ

2e−|ω|.

Ath.

Keha

gias


12.1.28 Example. To compute the Fourier transform of f (t) = tt2+1 we use

F

(1

1+ t2

)= πe−|ω|.

Hence

F

(t

1+ t2

)=

1−i

F

(−it

1+ t2

)= i

ddω

(πe−|ω|

).

For ω > 0 we have F (ω) = F(

t1+t2

)= −πie−ω ; for ω < 0 we have F (ω) = πieω .

What happens at ω = 0?

12.1.29 Example: To compute the Fourier transform of Heaviside(t) and Heaviside(t− t0),we have

d Heavisidedt

= Dirac(t)⇒ iwH (w) = 1⇒ Heaviside(w) =1iw

.

This cannot be right, because H (w) is purely imaginary and then Heaviside(t)should be odd. What we are missing is a constant of integration, which should beapplied to Heaviside(t) to get an odd function. Indeed, we can define

s(t) = 2Heaviside(t)−1.

Now we also have

dsdt

= 2Dirac(t)⇒ iwS (w) = 2⇒ S (w) =2iw

.

This respects oddness. Now

Heaviside(t) =s(t)+1

2⇒ H (w) =

S (w)+12

=1iw

+π Dirac(w) .

Of course this is not a rigorous proof. To verify we must compute

F−1(

1iw

+π Dirac(w))=

12π

∫∞

−∞

(1iw

+π Dirac(w))

dw

=1

2π

∫∞

−∞

1iw

e−iwtdw+1

2π

∫∞

−∞

πδ (w)e−iwtdw.

=12

s(t)+12= Heaviside(t)

(where1

2π

∫∞

−∞

1iw

e−iwtdw =12

s(t)

requires complex integration). Finally

F (Heaviside(t− t0)) =(

1iw

+π Dirac(w))

e−iwt0

Ath.

Keha

gias


12.1.30 Definition. The convolution of f (t) and g(t) is defined by

f ∗g =∫

∞

−∞

f (τ)g(t− τ)dτ.

Note the difference from the definition of convolution in the context of Laplacetransforms.

12.1.31 Example. Given

g(t) =

1 otan |t|< 10 otan |t|> 1

we compute

(g∗g)(t) =∫

∞

−∞

g(τ)g(t− τ)dτ.

by taking cases.

1. t <−2. When |τ|< 1 we have

t− τ ≤−2+1 =−1⇒ g(τ)g(t− τ) = 0,

when |τ|> 1 we haveg(τ)g(t− τ) = 0 = 0.

Hence∀t ∈ (−∞,−2) : (g∗g)(t) = 0.

2. t ∈ (−2,0). When |τ|< 1 we have in (−1, t +1):

g(τ)g(t− τ) = 1

when |τ|> 1 we haveg(τ)g(t− τ) = 0.

Hence

∀t ∈ (−2,0) : (g∗g)(t) =∫

∞

−∞

g(τ)g(t− τ)dτ =∫ t+1

−1dτ = t +2.

3. t ∈ (0,2) and (2,∞) are treated similarly.4. Finally

(g∗g)(t) =

0 otan |t|> 22+ t otan t ∈ (−2,0)2− t otan t ∈ (0,2)

.

12.1.32 Theorem (Convolution): Let f (t), g(t) satisfy the conditions of Theorem12.1.2. Then

F ( f ∗g) = F ( f ) ·F (g) .

Ath.

Keha

gias


Proof. We have

F ( f ∗g) =∫

∞

−∞

(∫∞

−∞

f (τ)g(t− τ)dτ

)e−iωtdt

=∫

∞

−∞

(∫∞

−∞

f (τ)e−iωτg(t− τ)e−iω(t−τ)dτ

)dt

=∫

∞

−∞

(∫∞

−∞

f (τ)e−iωτg(t− τ)e−iω(t−τ)dt)

dτ

=∫

∞

−∞

f (τ)e−iωτ

(∫∞

−∞

g(t− τ)e−iω(t−τ)d (t− τ)

)dτ

=∫

∞

−∞

f (τ)e−iωτG(ω)dτ = G(ω)∫

∞

−∞

f (τ)e−iωτdτ = F (ω)G(ω) .

12.1.33 Example. Let us verify Theorem 12.1.32 for g∗g, where

g(t) =

1 otan |t|< 10 otan |t|> 1 .

We have computed (g∗g)(t) in Example 12.1.31. Its Fourier transform is

F (g∗g) =∫

∞

−∞

(g∗g)(t)eiωtdt

=∫ 0

−2(2+ t)eiωtdt +

∫ 2

0(2− t)eiωtdt

... after many calculations ...

=2(1− cos(2ω))

ω2 =4sin2 (ω)

ω2 .

But, since F (g) = 2sin(ω)ω

, we get immediately

F (g∗g) = G(ω)2 =4sin2 (ω)

ω2

which verifies Theorem 12.1.32. The second way to compute F (g∗g) is mucheasier.

12.1.34 Theorem. When the indicated convolutions exist, the following hold:

f ∗g = g∗ ff ∗ (g∗h) = ( f ∗g)∗h

f ∗ (g+h) = f ∗g+ f ∗h.

Proof. All the properties follow from Theorem 12.1.32.

12.1.35 Consider an arbitrary function f (t) and define

φ (t) := Heaviside(t)e−σt f (t) =

e−σt f (t) t > 00 t < 0

Ath.

Keha

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Then the Fourier transform of φ (t) is

F (φ (t)) =∫

∞

−∞

φ (t)e−iwtdt =∫

∞

0f (t)e−σte−iwtdt =

∫∞

0f (t)e−stdt = L ( f (t))

with s = σ + iw. So the Laplace transform of f (t) is the Fourier transform of afunction φ (t) which is the restriction of f (t) on (0,∞) with an e−σt attenuation.

1. Adding the attenuation e−σt increases the class of functions ( f (t)e−σt )which are integrable (e.g., when f (t) = Heaviside(t) is not integrable bute−σt Heaviside(t) is).

2. We want to restrict f (t) to t > 0 because we have a beginning of time (whenthe system started working).

3. This is also why the Laplace transform of a derivative needs the initialconditions; while in Fourier transform, the beginning of time is at t = −∞

and it is assumed that at t = 0 the effect of the "initial” conditions at t =−∞

has worn away.

12.1.36 Let us now examine the application of the Fourier transform to thesolution of differential equations. We will work with examples.


dxdt

+ x = Heaviside(t)−Heaviside(t−1) , x(0−)= 1

first with Laplace, then with Fourier transform.With Laplace we have

(s+1)X = 1+1− e−st

s⇒ X =

1s+1

+(1− e−st) 1

s(s+1)

⇒ X =1

s+1+(1− e−st)(1

s− 1

s+1

)⇒ xL (t) = e−t +Heaviside(t)

(1− e−t)−Heaviside(t−1)

(1− e−(t−1)

)With Fourier things are quite (but not totally) similar; we have

F (Heaviside(t)−Heaviside(t−1)) =∫ 1

0e−iwtdt = i

1− e−iw

w=

e−iw−1iw

and then

(iw+1)X =e−iw−1

iw⇒ X =

(e−iw−1

) 1iw(iw+1)

⇒ X =(e−iw−1

)( 1iw− 1

iw+1

)⇒ xF (t) = Heaviside(t)

(1− e−t)−Heaviside(t−1)

(1− e−(t−1)

).

We see thatxL (t) = xF (t)+ e−t .

Ath.

Keha

gias


So the two solutions are not identical. But this is not surprising. The differencecan be traced back to the dx

dt formulas:

L

(dxdt

)= X (s)− sx

(0−), F

(dxdt

)= X (w) .

And indeed the difference between xL (t) and xF (t) is the term x(0−)e−t , whichthe intial condition propagated in time; the Fourier transform assumes no initialconditions or, better, that the "initial" conditions were given at t =−∞ and theirinfluence has worn out.

12.1.38 Example. Now we solve

dxdt

+ x = Dirac(t−1) , x(0−)= 0

first with Laplace, then with Fourier transform.With Laplace we have

(s+1)X = e−s⇒ X =e−s

s+1⇒ xL (t) = Heaviside(t−1)e−(t−1)

With Fourier we have (with exactly analogous calculations and with F (Dirac(t−1))=e−iw):

(iw+1)X = e−iw⇒ X =e−iw

iw+1⇒ xF (t) = Heaviside(t−1)e−(t−1)

We see thatxL (t) = xF (t) .

12.1.39 Example. Now we solve

d2xdt2 +2

dxdt

+ x = Heaviside(t)−Heaviside(t−1) , x(0−)= 0, x′

(0−)= 0

first with Laplace, then with Fourier transform.With Laplace we have(

s2 +2s+1)

X =1− e−st

s⇒ X =+

(1− e−st) 1

s(s2 +2s+1)⇒ xL (t) = Heaviside(t) x(t)−Heaviside(t−1) x(t−1)

wherex(t) = 1− te−t− e−t

and with Fourier(−w2 +2iw+1

)X =

1− e−iwt

iw⇒ X =

(1− e−iwt) 1

iw(−w2 +2iw+1)

⇒ X =(1− e−iwt)( 1

iw− 1

(iw+1)2 −1

iw+1

)⇒ xF (t) = Heaviside(t) x(t)−Heaviside(t−1) x(t−1)

We see thatxL (t) = xF (t) .

Ath.

Keha

gias


12.1.40 Example. Here is one final example. We solve

d2xdt2 +2

dxdt

+ x = sin t, x(0−)= 0, x′

(0−)= 0

first with Laplace, then with Fourier transform.With Laplace we have(

s2 +2s+1)

X =1

s2 +1⇒ X =

1(s2 +1)(s2 +2s+1)

⇒ X =1

2(s+1)− 1

2s

s2 +1+

1

2(s+1)2

⇒ xL (t) =12

e−t +12

te−t− 12

cos t.

With Fourier we note that

F−1 (eiw)= Dirac(t−1) and F−1 (e−iw)= Dirac(t +1)

hence

F−1 (sinw) = F−1(

eiw− e−iw

2i

)=

12i(Dirac(t−1)−Dirac(t +1) )

Then, by duality,

F (sin t) = πi(Dirac(w+1)−Dirac(w−1) )

and (−w2 +2iw+1

)X = πi(Dirac(w+1)−Dirac(w−1) )

⇒ X = πiDirac(w+1)−Dirac(w−1)

(iw+1)2

Since

12π

∫∞

−∞

iπDirac(w+1)

(iw+1)2 eiwtdw =i

2(i(−1)+1)2 e−iwt =−14

e−it

12π

∫∞

−∞

iπDirac(w−1)

(iw+1)2 eiwtdw =i

2(i1+1)2 eit =14

πeit

we get

xF (t) =−12

e−it + eit

2=−1

2cos t

We see thatxL (t) = xF (t)+

12

e−t +12

te−t .

Here the cause of the discrepancy is different. Namely the part 12e−t + 1

2te−t is notsquare integrable in (−∞,∞) and so Fourier transform canot capture it.

Ath.

Keha

gias


12.1.41 Notation.

Heaviside(t) =

1 for t ≥ 0,0 for t < 0.

Dirac(t) = d Heavisidedt

sgn(t) =

1 for t > 0,−1 for t < 0.

Π(t) =

1 for |t| ≤ 1,0 for |t|> 1.

Λ(t) =

1−|t| for |t| ≤ 1,0 for |t|> 1.

sinc(t) = sin tt

12.1.42 The basic properties of the Fourier transform are as follows.

f (t) F (ω) = F ( f (t))κ f1 (t)+λ f2 (t) κF1 (ω)+λF2 (ω)

f (at) 1|a|F

(ω

a

)f (−t) F (−ω)f (t− t0) F (ω)e−iωt0

f (t)eiω0t F (ω−ω0)

f (t)cos(ω0t) 12F (ω−ω0)+

12F (ω +ω0)

f (t)sin(ω0t) 12iF (ω−ω0)− 1

2iF (ω +ω0)F (t) 2π f (−ω)d fdt iωF (ω)∫

x(τ)dτ1

iω F (ω)

−it f (t) ddω

F (ω)∫∞

−∞| f (t)|2 dt

∫∞

−∞|F (ω)|2 dω∫

∞

−∞f (t)g(t)dt

∫∞

−∞F (w)G(ω)dω

f (t)∗g(t) F (ω)G(ω)

12.1.43 Some basic Fourier transform pairs are the following.

Ath.

Keha

gias


f (t) F (ω) = F ( f (t))Heaviside(t− t0)

( 1iw +π Dirac(w)

)e−iwt0

Dirac(t− t0) e−iwt0

eiw0t Dirac(w−w0)cos(w0t) π (Dirac(w−w0)+Dirac(w+w0))sin(w0t) πi(−Dirac(w−w0)+Dirac(w+w0))

e−at Heaviside(t) 1a+iω

e−a|t| 2aa2+ω2

e−at2√

π

a e−ω2/4a

Π(t) 2sinc(ω)1π

sinc(t) Π(ω)

te−at Heaviside(t) 1(a+iω)2

1a2+t2

π

a e−a|ω|

1t πi(1−2Heaviside(ω))



1. Compute the Fourier transform of f (t) =

1− t2 an |t| ≤ 10 an |t|> 1.

Ans. 4 sinω−ω cosω

ω3 .2. Compute the Fourier transform of f (t) = 3

t2+4Ans. 3/2π

(e−2w Heaviside(w)+ e2w Heaviside(−w)

)3. Compute the Fourier transform of f (t) = 1

2+t .Ans. ie2 iwπ (1−2 Heaviside(w))

4. Compute the Fourier transform of f (t) = e−3 t Heaviside(t).Ans. (3+ iw)−1

5. Compute the Fourier transform of f (t) = te−3 t Heaviside(t).Ans. (3+ iw)−2

6. Compute the Fourier transform of f (t) = e−4 t2.

Ans. 1/2e−1/16w2√π

7. Compute the inverse Fourier transform of F(ω) =

1−ω2 an |ω| ≤ 10 an |ω|> 1.

Ans. f (t) = 2πt3 (sin t− t cos t)

8. Compute the inverse Fourier transform of F(ω) = sin(ωt0)ωt0

.

Ans. f (t) = 1

2t0an |t|< |t0|

0 an |t|> |t0|.)

9. Compute the inverse Fourier transform of F(ω) = 1√2π

e−ω22 .

Ans. f (t) = 12π

e−12 t2

.

10. Compute the inverse Fourier transform of F(ω) =

1−|ω| an |ω| ≤ 10 an |ω|> 1.

Ans. f (t) = 1π

1−cos tt2 .

Ath.

Keha

gias


11. Compute the inverse Fourier transform of F(ω) = 1√1−iw

.

Ans. et Heaviside(−t)√π√−t

12. Compute the inverse Fourier transform of F(ω) = 3w2+1 .

Ans. 3/2et Heaviside(−t)+3/2 Heaviside(t)e−t

13. Compute the inverse Fourier transform of F(ω) = 3 ww2+1 .

Ans. 3/2 i(−et Heaviside(−t)+Heaviside(t)e−t)

14. Compute the inverse Fourier transform of F(ω) = sin(w)w2+1 .

Ans. i(e−t+1 Heaviside(t−1)+et−1 Heaviside(−t+1)−et+1 Heaviside(−t−1)−e−t−1 Heaviside(t+1))

4

15. Compute the inverse Fourier transform of F(ω) = sin(w)w2+9 .

Ans. i(e−3 t+3 Heaviside(t−1)+e3 t−3 Heaviside(−t+1)−e−3 t−3 Heaviside(t+1)−e3 t+3 Heaviside(−t−1))

1216. Compute the Fourier transform of f (t) = e−t2

.Ans. F (ω) =

√2

2 e−14 ω2

.17. Compute the sine Fourier transform of f (t) = e−|t|.

Ans. FS (ω) =√

2π

ww2+1 .

18. Compute the cosine Fourier transform of f (t) =

1−|t| an |t| ≤ 10 an |t|> 1.

Ans. Fc (ω) =√

2π

1−cosω

ω2 .

19. Compute the cosine Fourier transform of f (t) = sin(ω0t)ω0t .

Ans. Fc (ω) =

√π

2 an |ω|< |ω0|0 an |ω|> |ω0|

.

20. Compute the cosine Fourier transform of f (t) =

t2 an 0≤ |t| ≤ 10 an |t|> 1.

Ans. Fc (ω) =√

2π

ω2 sinω−2sinω+2ω cosω

πω3 .

12.4 Advanced Problems1. Prove:

Heaviside(at +b)=Heaviside(

t +ba

)Heaviside(a)+Heaviside

(−t− b

a

)Heaviside(−a) .

2. Prove: ddt Λ =−Π

( t2

)sgn(t) .

3. Prove: ddt sinc(t) = cosπt

t −sinπtπt2 .

4. Compute∫

∞

−∞dt

(1+t2)2 .

Ans. 12π.

5. Compute∫

∞

−∞t2dt

(1+t2)2 .

Ans. 12π.

6. Compute∫

∞

−∞t4dt

(1+t2)4 .

Ans. 116π.

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7. Solve the integral equation:∫

∞

−∞

y(τ)(t−τ)2+a2 dτ = 1

t2+b2 .

Ans. y(t) = b−abπ(t2+(b−a)2)

8. Prove: Λ(t) = Π(t)∗Π(t) .9. Prove:

ddt

[ f (t)∗g(t)] =(

ddt

f (t))∗g(t)

ddt

[ f (t)∗h(t)] = f (t) .

10. Prove:Heaviside(t)∗ [ f (t)Heaviside(t)] =

∫ t

0f (τ)dτ.

11. Prove: 1πt ∗

−1πt = δ (t) .

12. Compute

Heaviside(t)∗Heaviside(t) ,Heaviside(t)∗Heaviside(t)∗Heaviside(t) ,...

limn→∞

Heaviside(t)∗Heaviside(t)∗ ...∗Heaviside(t)︸︷︷︸n times

13. Prove: if f (t) is real valued, then |F (ω)|2 is even.14. Prove: ∫

∞

−∞

t f (t)dt =−F ′ (0)−2iπ

,∫∞

−∞

t2 f (t)dt =−F ′′ (0)4π2 .

15. What is the relationship between f (t) and F (F ( f (t)))? Between f (t) andF (F (...(F ( f (t))) ...))?

16. Prove: ∫∞

−∞

(d fdt

)(d fdt

)dt = 4π

2∫

∞

−∞

(dFdω

)(dFdω

)dω.

17. Prove:| f (t)| ≤

∫∞

−∞

|F (ω)|dω.

18. Prove: ∫∞

−∞

e−πt2cos(2πωt)dt = e−πω2

.

19. Find f (t) such that f (t) = F (t) (where F (w) = F ( f (t))).20. Prove:

limN→

(1−ω

2)(1− ω2

4

)...

(1− ω2

N2

)...= A · e−ω2/B.

What are the A,B values?

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13 PDEs for Equilibrium . . . . . . . . . . . . 135

14 PDEs for Diffusion . . . . . . . . . . . . . . . 149

15 PDEs for Waves . . . . . . . . . . . . . . . . . . 173

16 Bessel Functions and Applications189

17 Vector Spaces of Functions . . . . . 201

18 Sturm-Liouville Problems . . . . . . . 215

Partial Differential Equations

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13. PDEs for Equilibrium

Equilibrium PDEs describe phenomena which, after having evolved in time, havereached a steady state. The classic equilibrium PDE is the Laplace equation

∇2u = 0,

i.e., uxx +uyy = 0, uxx +uyy +uzz = 0 etc.

13.1 Theory and Examples13.1.1 In this chapter we study the Laplace equation

uxx +uyy = 0 (13.1)

and its variants. Note that this differential equation involves a function u(x,y)of two variables and its partial derivatives. Such equations are called partialdifferential equations (PDE).

13.1.2 The function u(x,y) can be an electric potential, a temperature, a probabilitydistribution etc. Note that u(x,y) does not involve time; the Laplace equation isused to describe phenomena in equilibrium.

13.1.3 Why does (13.1) describe equilibrium? The full answer will come in laterchapters, where we will see that

u(x,y) = limt→∞

u(x,y, t)

i.e., u(x,y) is the steady state limit of another function u(x,y, t) which involvestime. However, some preliminary arguments can be given here.

13.1.4 We first present an informal argument, based on the approximation ofderivatives by finite differences. Recall that, for a (sufficiently smooth) function ofone variable, we have

f ′ (x)' f (x+∆x)− f (x)∆x

, f ′′ (x)' f (x+∆x)−2 f (x)+ f (x−∆x)∆x2

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136 Chapter 13. PDEs for Equilibrium

and the approximation becomes better as ∆x→ 0. Applying the correspondingdiscretization to (13.1) we get the discrete version of the Laplace equation:

u(x+∆x,y)−2u(x,y)+u(x−∆x,y)∆x2 +

u(x,y+∆y)−2u(x,y)+u(x,y−∆y)∆y2 = 0

Assuming ∆x = ∆y, this yields

u(x,y) =u(x+∆x,y)+u(x−∆x,y)+u(x,y+∆y)+u(x,y−∆y)

4.

In otherwords, if u satisfies the Laplace equation in the neighborhood of somepoint (x,y), then at that point u(x,y) equals the average of the value of u at thefour nearest neighbors of (x,y). This is a very rough argument, but it captures theessence of the matter.

13.1.5 A more rigorous argument depends on the theory of complex functions.Recall the following facts.

1. A function u which satisfies uxx +uyy = 0 is called harmonic.2. Given an holomorphic function f (z) = u(x,y)+ iv(x,y), its real part u(x,y)

and its imaginary part v(x,y) are harmonic functions.3. Given a function f (z) = u(x,y)+ iv(x,y), which is holomorphic inside and on

a circle centered at z0 and with radius R, we have

f (z0) =1

2π

∫ 2π

0f(

z0 +R · eiθ)

dθ ; (13.2)

in other words the value at z0 is the average of the value on the circle.It is easily seen that the (13.2) implies that the average property also holds for thereal and imaginary parts of f (z):

u(x0,y0) =1

2π

∫ 2π

0u(x0 +Rcosθ ,y0 +Rsinθ ,)dθ ; (13.3)

v(x0,y0) =1

2π

∫ 2π

0v(x0 +Rcosθ ,y0 +Rsinθ ,)dθ . (13.4)

In short: if a function satisfies the Laplace equation inside and on a circle, its valueat the center of the circle equals its average value on the circle. We will soon seemore connections of the Laplace equation to complex functions.

13.1.6 To find a specific solution of the Laplace equation (13.1) two additionalfactors must be specified.

1. The region on which the equation holds.2. The conditions on the boundary of the region (boundary conditions).

The situation is analogous to that of the “ordinary” differential equations studiedin previous chapters. Whereas in these cases a particular solution was fullydetermined by the differential equation and initial conditions, the situation is morecomplex now. Since the boundary will be a curve on the plane, the boundaryconditions consist in specifying functions (rather than numerical values) on theboundary.

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13.1.7 A similar situation appears, once again, in the theory of complex functions.Namely, Cauchy’s Integral Formula tells us that a function f (z) is analytic insideand on the boundary of a region (the boundary being a simple closed curve) thenthe value of f (z) at some point z0 inside the region is given by

f (z0) =1

2πi

∮ f (z)z− z0

dz.

Do you see the connection to solutions of the Laplace equation? Recall that, iff (z) = u(x,y)+ iv(x,y) is analytic then uxx +uyy = 0, vxx + vyy = 0.

13.1.8 In general we distinguish the following kinds of Laplace equation problems,according to the nature of the boundary conditions..

1. Dirichlet problems: boundary conditions on u(x,y).2. Neumann problems: boundary conditions on the partial derivatives (e.g.,

ux (x,y), uy (x,y)).3. Mixed Dirichlet / Neumann problems.

13.1.9 Let us now solve what is perhaps the simplest Dirichlet problem.

0 < x < L and 0 < y < M : uxx +uyy = 0 (13.5)0 < x < L : u(x,0) = f x) (13.6)0 < x < L : u(x,M) = 0 (13.7)0 < y < M : u(0,y) = 0 (13.8)0 < y < M : u(L,y) = 0. (13.9)

We will solve the above using the method of separation of variables. Assume thatu(x,y) = X (x)Y (y). Then (13.5) becomes

X ′′Y +XY ′′ = 0⇒ X ′′

X=−Y ′′

Y=−b2 (13.10)

(the choice −b2 will be explained a little later). From (13.10) we get

X ′′+b2X = 0 (13.11)

Y ′′−b2Y = 0. (13.12)

Fron (13.8), (13.9) follows that the solutions of (13.11) have the form

Xn (x) = sin(bnx) (13.13)

with bn =nπ

L (n ∈ 0,±1,±2, ...). The solutions of (13.12) have the form

Yn (y) =Cnebny +Dne−bny,

but can be written equivalently as

Yn (y) = En sinh(bn (y+Fn)) . (13.14)

From (13.7) we get Fn =−M, hence

Yn (y) = En sinh(bn (y−M)) (13.15)

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and

u(x,y) =∞

∑n=1

En sin(nπ

Lx)

sinh(nπ

L(y−M)

). (13.16)

We must also satisfy (13.6). Here is why we chose the constant −b2 (i.e., negative):now we can expand f (x) in a Fourier series:

f (x) = u(x,0) =∞

∑n=1

En sin(nπ

Lx)

sinh(−nπM

L

). (13.17)

HenceEn =

−2Lsinh

(nπML

) ∫ L

0f (x)sin

(nπ

Lx)

dx (13.18)

In short, (13.16) and (13.18) give the solution of (13.5)–(13.9).

13.1.10 Example. Let us solve (13.5)–(13.9) with L = M = π and f (x) = sin2 (x) .Then

En =−2

π sinh(nπ)

∫π

0sin2 (x)sin(nx)dx =

− 4

π sinhπncos(πn)−1n(n2−4)

n 6= 2

0 n = 2(13.19)

Hence

u(x,y) =8π

∑n=1,3,5,...

1π sinh(nπ)

· 1n(n2−4)

· sin(nx)sinh(n(y−π)) . (13.20)

13.1.11 Here is a somewhat more complex version of (13.5)–(13.9):

0 < x < L and 0 < y < M : uxx +uyy = 0 (13.21)0 < x < L : u(x,0) = f (x) (13.22)0 < x < L : u(x,M) = g(x) (13.23)0 < y < M : u(0,y) = h(x) (13.24)0 < y < M : u(L,y) = k(x) (13.25)

To find u(x,y) solving (13.21)–(13.25) we use the principle of superposition: weassume u(x,y)= u1 (x,y)+ u2 (x,y)+ u3 (x,y)+ u4 (x,y), where:

1. u1 (x,y) satisfies uxx+uyy = 0 and u(x,0) = f (x), u(x,M) = u(0,y) = u(L,y) = 0.2. u2 (x,y) satisfies uxx+uyy = 0 and u(x,L) = g(x), u(x,0) = u(0,y) = u(L,y) = 0.3. u3 (x,y) satisfies uxx+uyy = 0 and u(0,y) = h(x), u(x,0) = u(x,M) = u(L,y) = 0.4. u4 (x,y) satisfies uxx+uyy = 0 and u(L,y) = k (x), u(x,0) = u(x,M) = u(0,y) = 0.

We solve each of these subproblems in the same way that we solved (13.5)–(13.9);hence we find u1, u2, u3, u4 and then u(x,y).

13.1.12 Here is a simple Neumann problem:

0 < x < L and 0 < y < M : uxx +uyy = 0 (13.26)0 < x < L : uy (x,0) = f (x) (13.27)0 < x < L : uy(x,M) = 0 (13.28)0 < y < M : ux(0,y) = 0 (13.29)0 < y < M : ux (L,y) = 0. (13.30)

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Once again we separate variables and obtain

X ′′+b2X = 0 (13.31)

Y ′′−b2Y = 0. (13.32)

For X we have the boundary conditions X ′ (0) = X ′ (L) = 0. Hence the only accept-able solutions are

Xn (x) = cos(nπ

Lx).

For n ∈ 1,2, .. we can write the Y solutions in the form

Yn (y) = En sinh(nπ

Ly)+Fn cosh

(nπ

Ly).

Hence the general solution is

u(x,y) =∞

∑n=0

(En sinh

(nπ

Ly)+Fn cosh

(nπ

Ly))

cos(nπ

Lx)

= E0 +∞

∑n=1

(En sinh

(nπ

Ly)+Fn cosh

(nπ

Ly))

cos(nπ

Lx). (13.33)

Now (13.27), (13.28) yield

f (x) = uy (x,0) =∞

∑n=1

nπ

LFn cos

(nπ

Lx)

(13.34)

0 = uy (x,M) =∞

∑n=1

nπ

L

(En sinh

(nπ

LM)+Fn cosh

(nπ

LM))

cos(nπ

Lx)

(13.35)

From (13.34) we conclude that nπ

L Fn are the coefficients of the cosine series off (x) which must, however, have zero coefficient on cos

(0π

L x). In other words, the

problem only has a solution if the following compatibility condition holds:

∫ L

0f (x)dx = 0. (13.36)

If (13.36) holds, then F1,F2, ... are determined by

Fn =2

nπ

∫ L

0f (x)cos

(nπ

Lx)

dx (13.37)

and E1,E2, ... are determined by solving (for n ∈ 1,2, ...) equation (13.35) whichbecomes

En sinh(nπ

LM)+Fn cosh

(nπ

LM)= 0. (13.38)

E0 remains undeterined (this is reasonable: the boundary conditions only detrminethe derivatives).

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13.1.13 Here is a mixed Dirichlet-Neumann problem:

0 < x < L and 0 < y < M : uxx +uyy = 0 (13.39)0 < x < L : u(x,0) = f (x) (13.40)0 < x < L : u(x,M) = 0 (13.41)0 < y < M : ux(0,y) = 0 (13.42)0 < y < M : ux (L,y) = 0. (13.43)

We separate the variables and get

X ′′+b2X = 0

Y ′′−b2Y = 0.

For X we also have X ′ (0) = X ′ (L) = 0; hence the only acceptable solutions are

Xn (x) = cos(nπ

Lx).

For n ∈ 1,2, .. we can write the Y solutions as


L(y+Fn)

);

since we must have Yn (M) = 0 we finally get


L(y−M)

).

Especially for n = 0 we have

Yn (y) = E0M− y

M.

So, finally, the general solution is

u(x,y) = E0M− y

M+

∞

∑n=1

En sinh(nπ

L(y−M)

)cos(nπ

Lx). (13.44)

To satisfy (13.40) we must have

f (x) = u(x,0) = E0−∞

∑n=1

En sinh(nπ

LM)

cos(nπ

Lx)

(13.45)

hence

E0 =1L

∫ L

0f (x)dx (13.46)

∀n ∈ 0,1,2, ... : En =−2

Lsinh(nπ

L M) ∫ L

0f (x)cos

(nπ

Lx)

dx. (13.47)

The solution of (13.39)–(13.43) can be simplified to

u(x,y) =A0

2

(M− y

M

)+

∞

∑n=1

An

sinh(nπM

L

) sinh(nπ

L(M− y)

)cos(nπ

Lx)

∀n ∈ 0,1,2, ... : An =2L

∫ L

0f (x)cos

(nπ

Lx)

dx.

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13.1.14 Let us now solve the Laplace equation on an infinite region, namely onthe half-plane:

−∞ < x < ∞, 0 < y < ∞ : uxx +uyy = 0 (13.48)−∞ < x < ∞ : u(x,0) = f (x). (13.49)

Separating variables we get again

X ′′+b2X = 0 (13.50)

Y ′′−b2Y = 0. (13.51)

The solutions of (13.50) again have the form cos(bx) and sin(bx) but now we haveno constraint on the b values The solutions of (13.51) have the form eby, e−by but(assuminh b > 0) eby is rejected because it gives an unbounded u(x,y). Finally (bysuperposition) a solution of (13.48) is

u(x,y) =∫

∞

0e−by (A(b)cos(bx)+B(b)sin(bx))db. (13.52)

Letting y = 0 in (13.52) we get

f (x) = u(x,0) =∫

∞

0(A(b)cos(bx)+B(b)sin(bx))db. (13.53)

Hence

A(b) =1π

∫∞

−∞

f (x)cos(bx)dx, B(b) =1π

∫∞

−∞

f (x)sin(bx)dx (13.54)

Hence (13.52) and (13.54) solve (13.48)–(13.49).We can write the solution in another form, which shows clearly how the

boundary conditions determine the solution. Replacing (13.54) in (13.52) we get

u(x,y) =1π

∫∞

0

(∫∞

−∞

e−by f (z)cos(b(z− x))dz)

db (13.55)

=1π

∫∞

−∞

f (z)(∫

∞

0e−by cos(b(z− x))db

)dz. (13.56)

Then∫∞

0e−by cos(b(z− x))db =

1

y2 +(z− x)2

(−ye−by cosb(z− x)− (x− z)e−by sinb(z− x)

)∞

b=0

=y

y2 +(z− x)2 .

Henceu(x,y) =

1π

∫∞

−∞

y f (z)

y2 +(z− x)2 dz. (13.57)

This is the Poisson formula for the half-plane (known to us from the theory ofcomplex functions). The interpretation of (13.57) is this: the value of u(x,y) at(x,y) is the average of the u(z,0) = f (z) (i.e., the values on the x axis boundary),weighted by 1

y2+(z−x)2 , i.e., the inverse square of the distance between (z,0) and

(x,y).

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−∞ < x < ∞, 0 < y < ∞ : uxx +uyy = 0

−∞ < x < ∞ : u(x,0) = f (x) =

0 x < 01 x > 0

we have:

u(x,y) =1π

∫∞

−∞

y f (z)

y2 +(x− z)2 dz =1π

∫∞

0

y

y2 +(x− z)2 dz

=1π

(tan−1 z− x

y

)∞

z=0=

1π·(

π

2− tan−1

(xy

))=

12− 1

πtan−1

(xy

).

13.1.16 Let us also solve the Laplace equation on a half-strip:

0 < x < 1,0 < y < ∞ : uxx +uyy = 00 < x < 1 : u(x,0) = f (x)0 < y < ∞ : u(0,y) = u(1,y) = 0

By separation of variables we get

u(x,y) = X (x)Y (y) ,X (x) = sin(nπx) ,

Y (y) = e−nπy

(Y (y) = enπy is unbounded and hence rejected). It follows that

u(x,y) =∞

∑n=1

Ane−nπy sin(nπx)

and

An = 2∫ 1

0f (x)sin(nπx)dx.

If, e.g., f (x) = 1, then An =2π

1−cosnπ

n and

u(x,y) =4π

(e−y sin(πx)+

13

e−3y sin(3πx)+15

e−5y sin(5πx)+ ...

).

Try to show that the above is equivalent to

u(x,y) =2π

tan−1(

sin(πx)sinh(y)

).

13.1.17 In many cases we must solve the Laplace equation on a region withcurved boundary. We will consider the simplest case: the region is a disk. It isnatural to use polar coordinates. Recall that

uxx +uyy = urr +1r

ur +1r2 urr, (13.58)

hence the Laplace equation in polar coordinates becomes

urr +1r

ur +1r2 urr = 0. (13.59)

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13.1.18 For example let us solve:

0≤ r < 1, 0≤ θ ≤ 2π : urr +1r

ur +1r2 uθθ = 0 (13.60)

0≤ θ ≤ 2π : u(1,θ) = f (θ). (13.61)

Lettingu(r,θ) = R(r)Θ(θ)

we get

urr +1r

ur +1r2 uθθ = 0⇒

R′′Θ+1r

R′Θ+1r2 RΘ

′′ = 0⇒

R′′

R+

1r

R′

R+

1r2

Θ′′

Θ= 0⇒

r2 R′′

R+ r

R′

R=−Θ′′

Θ= a. (13.62)

From (13.62) we see that

Θ′′+aΘ = 0 (13.63)

r2R′′+ rR′−aR = 0. (13.64)

Note that Θ must be periodic with period 2π. Let us consider the possible valuesof a.

1. If a =−b2 < 0 then Θ(θ) =Cebθ +De−bθ which is not periodic.2. If a = 0 then Θ(θ) =Cθ +D and periodicity is satisified only for C = 0. Then

we get R(r) = A+B logr and R(0) is not well defined unless B = 0. In short,for a = 0 the only acceptable solution is the constant u(x, t) = AD.

3. If a > 0 we haveΘ(θ) =C cos

(√aθ)+Dsin

(√aθ)

(13.65)

and periodicity is satisfied if√

a = bn = n. Also

R(r) = Ar√

a +Br−√

a = Arn +Br−n (13.66)

and, for R(0) to be well defined, we must have B = 0.From the above we see that a solution of (13.60) must have the form

u(r,θ) =C0

2+

∞

∑n=1

rn · (Cn cos(nθ)+Dn sin(nθ)) . (13.67)

To also satisfy (13.61) we must set

f (θ) = u(1,θ) =C0

2+

∞

∑n=1

(Cn cos(nθ)+Dn sin(nθ)) (13.68)

and let Cn, Dn be the Fourier coefficient of f (θ):

Cn =1π

∫ 2π

0f (θ)cos(nθ)dθ , Dn =

1π

∫ 2π

0f (θ)sin(nθ)dθ (13.69)

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13.1.19 We can also express the solution u(r,θ) with the Poisson formula for theunit circle. Replacing (13.69) in (13.67) we get

u(r,θ) =1

2π

∫ 2π

0f (θ)dθ +

1π

∞

∑n=1

rn ·(∫ 2π

0f (θ)(cos(nφ)cos(nθ)+ sin(nφ)sin(nθ))dφ

)=

12π

∫ 2π

0f (φ) ·

[1+2

∞

∑n=1

rn · cos [n · (θ −φ)]

]dφ

=1

2π

∫ 2π

0f (φ) ·

[1+

∞

∑n=1

(rnein·(θ−φ)+ rne−in·(θ−φ)

)]dφ

=1

2π

∫ 2π

0f (φ) ·

[1+

∞

∑n=1

rnein·(θ−φ)+∞

∑n=1

rne−in·(θ−φ)

]dφ

=1

2π

∫ 2π

0f (φ) ·

[1+

rei·(θ−φ)

1− rei·(θ−φ)+

re−i·(θ−φ)

1− re−i·(θ−φ)

]dφ

=1

2π

∫ 2π

0f (φ) · 1− r2

1−2r cos(θ −φ)+ r2 ·dφ (13.70)

The interpretation of (13.70) is this: the value of u(r,θ) at (r,θ) is the average ofthe values u(1,φ) = f (φ) (the values on the unit circle) weighted by 1

1−2r cos(θ−φ)+r2

(i.e., by the inverse squared distance between (1,φ) and (r,θ)).

13.1.20 Example. To solve:

0≤ r < 1, 0≤ θ ≤ 2π : urr +1r

ur +1r2 uθθ = 0 (13.71)

0≤ θ ≤ 2π : u(1,θ) = f (θ) =

1 0 < θ < π

0 π < θ < 2π(13.72)

Using separation of variables, a solution of (13.71) is:

u(r,θ) =C0

2+

∞

∑n=1

rn · (Cn cos(nθ)+Dn sin(nθ)) . (13.73)

and

Cn =1π

∫ 2π

0f (θ)cos(nθ)dθ =

1π

∫π

0cos(nθ)dθ =

1 gia n = 00 gia n > 0 , (13.74)

Dn =1π

∫ 2π

0f (θ)sin(nθ)dθ =

1π

∫π

0sin(nθ)dθ =

1nπ

(1− cos(nπ)) . (13.75)

Hence

u(r,θ) =12+

1π

(r sin(θ)+

13

r3 sin(3θ)+15

r5 sin(5θ)+ ...

)(13.76)

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Using the Poisson formula:

u(r,θ) =1

2π

∫ 2π

0f (φ) · 1− r2

1−2r cos(θ −φ)+ r2 dφ

=1

2π

∫π

0

1− r2

1−2r cos(θ −φ)+ r2 dφ

=12+

1π

tan−1(

2r sin(θ)1− r2

). (13.77)

Prove that the two forms of the solution are equivalent. It suffices to show that(r sin(θ)+

13

r3 sin(3θ)+15

r5 sin(5θ)+ ...

)=

12

tan−1(

2r sin(θ)1− r2

). (13.78)

13.1.21 Here is Dirichlet problem for the exterior of the unit circle:

1 < r, 0≤ θ ≤ 2π : urr +1r

ur +1r2 uθθ = 0 (13.79)

0≤ θ ≤ 2π : u(1,θ) = f (θ). (13.80)

Omitting the details, we finally get

u(r,θ) =C0

2+

∞

∑n=1

r−n (Cn cos(nθ)+Dn sin(nθ)) . (13.81)

Cn =1π

∫ 2π


1π

∫ 2π

0f (θ)sin(nθ)dθ (13.82)

or :

u(r,θ) =1

2π

∫ 2π

0f (φ) · r2−1

1−2r cos(θ −φ)+ r2 ·dφ . (13.83)

13.1.22 Another variant is the Dirichlet problem on a ring:

r1 < r < r2, 0≤ θ ≤ 2π : urr +1r

ur +1r2 uθθ = 0 (13.84)

0≤ θ ≤ 2π : u(r1,θ) = f1(θ) (13.85)0≤ θ ≤ 2π : u(r2,θ) = f2 (θ) . (13.86)

Separating variables, we get the same families of solutions as in 13.1.18. But,because r = 0 is not included in the region, we can now accept solutions of theforms A+B ln(r), rn, r−n. So we can finally assume

u(r,θ) =12(A0 +B0 ln(r))+

∞

∑n=1

[(Anrn +Bnr−n)cos(nθ)+

(Cnrn +Dnr−n)sin(nθ)

].

(13.87)To satisfy (13.85), (13.86) we determine ta A0,B0 from

A0 +B0 ln(r1) =1π

∫ 2π

0f1 (θ)dθ (13.88)

A0 +B0 ln(r2) =1π

∫ 2π

0f2 (θ)dθ (13.89)

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and An,Bn (for n ∈ 1,2, ...) from

Anrn1 +Bnr−n

1 =1π

∫ 2π

0f1 (θ)cos(nθ)dθ (13.90)

Cnrn1 +Dnr−n

1 =1π

∫ 2π

0f1 (θ)sin(nθ)dθ (13.91)

Anrn2 +Bnr−n

2 =1π

∫ 2π

0f2 (θ)cos(nθ)dθ (13.92)

Cnrn2 +Dnr−n

2 =1π

∫ 2π

0f2 (θ)sin(nθ)dθ (13.93)

13.1.23 Here is a Neumann problem on the unit circle:

1 < r, 0≤ θ ≤ 2π : urr +1r2 uθθ +

1r

ur = 0 (13.94)

0≤ θ ≤ 2π : ur (1,θ) = f (θ). (13.95)

By separation of variables we get

u(r,θ) =C0

2+

∞

∑n=1

rn (Cn cos(nθ)+Dn sin(nθ)) ; (13.96)

differentiating with respect to r we get

ur (r,θ) =∞

∑n=1

nrn−1 (Cn cos(nθ)+Dn sin(nθ)) (13.97)

hence

f (θ) = ur (1,θ) =∞

∑n=1

n · (Cn cos(nθ)+Dn sin(nθ)) (13.98)

and so, for n ∈ 1,2, ... we have:

Cn =1

nπ

∫ 2π


1nπ

∫ 2π

0f (θ)sin(nθ)dθ . (13.99)

Once again, to be able to expand f (θ) in a Fourier series the compatibility condition∫ 2π

0f (θ)dθ = 0. (13.100)

Also note that C0 is an arbitrary constant (this is reasonable, since boundaryconditions only detrmine derivatives of the solution). Finally, omitting details, letus give a Poisson - like formula for the Neumann problem on the unit circle:

u(r,θ) =C0

2− 1

2π

∫ 2π

0f (φ) ln

[1−2r cos(θ −φ)+ r2]dφ (13.101)

with C0 an arbitrary constant.

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1. Solve uxx+uyy = 0 (0 < x < π and 0 < y < π ), u(x,0) = 0 (0 < x < π), u(x,π) = 0(0 < x < π), u(0,y) = g(y) (0 < y < π), u(π,y) = 0 (0 < y < π).Ans. u(x,y) = −4∑

∞n=1

1+2·(−1)n

n3 sinh(nπ)sinh(n(π− x))sin(ny).

2. Solve uxx + uyy = 0 (0 < x < π and 0 < y < π), u(x,0) = x2 (π− x) (0 < x < π),u(x,π) = 0 (0 < x < π), u(0,y) = 0 (0 < y < π), u(π,y) = 0 (0 < y < π).Ans. u(x,y) = ∑

∞n=1

Ansinh(nπ) sinh(n(π− x))sin(ny) me An =

2π

∫π

0 g(y)sin(ny)dy.3. Solve uxx+uyy = 0 (0 < x < π and 0 < y < π ), u(x,0) = x2 (0 < x < π), u(x,π) =

x2 (0 < x < π), u(0,y) = 0 (0 < y < π), u(π,y) = 0 (0 < y < π).Ans. u(x,y) = πx− 8

π ∑∞n=1

1(2n−1)3 cosh( 2n−1

2 π)· cosh

((2n−1)

(π

2 − y))· sin((2n−1)x).

4. Solve uxx+uyy = 0 (0< x< π and 0< y< π ), u(x,0) = x2 (0 < x < π), u(x,π) = 0(0 < x < π), ux(0,y) = 0 (0 < y < π), ux (π,y) = 0 (0 < y < π).Ans. u(x,y) = 1

3π (π− y)+ 4∑∞n=1

(−1)n

n2 sinh(nπ)sinh(n(π− y))cos(nx).

5. Solve uxx + uyy = 0 (0 < x < 1 and 0 < y < 1), u(x,0) = (1− x)2 (0 < x < 1),u(x,1) = 0 (0 < x < 1), ux(0,y) = 0 (0 < y < 1), u(1,y) = 0 (0 < y < 1).Ans. u(x,y) = 4∑

∞n=1 An sinh

[(n− 1

2

)π (1− y)

]cos[(

n− 12

)πx]

me An =1

π2(n− 12)

2 + (−1)n

π3(n− 12)

3 .

6. Solve uxx + uyy = 0 (0 < x < π and 0 < y < π), uy (x,0) = f (x) (0 < x < π),u(x,π) = 0 (0 < x < π), u(0,y) = 0 (0 < y < π), ux (π,y) = 0 (0 < y < π).Ans. u(x,y) = −∑

∞n=1 An

1(n− 1

2)cosh( 2n−12 π)· cosh((2n−1)(π− y)) · sin

(2n−12 x

)me Cn =

2π

∫π

0 g(y)sin(ny)dy.

7. Solve uxx+uyy = 0 (−∞ < x < ∞ and 0 < y < ∞), u(x,0) = f (x) =−1 x < 0

1 x > 0 .

Ans. u(x,y) = 2π

tan−1(

xy

).

8. Solve uxx+uyy = 0 (−∞< x<∞ and 0< y<∞), u(x,0)= f (x)=

0 x <−11 −1 < x < 10 1 < x

Ans. u(x,y) = 1π

tan−1(

1+xy

)+ 1

πtan−1

(1−x

y

).

9. Solve:

uxx +uyy = 0 (−∞ < x < ∞, 0 < y < 1)u(x,0) = f (x) (−∞ < x < ∞)

u(x,1) = 0 (−∞ < x < ∞)

and show

u(x,y) =1π

∫∞

b=0

∫∞

z=−∞

f (z)sinh(by)sinh(ba)

cos(bz−bx)dzdb.

10. Solve urr +1r ur +

1r2 uθθ = 0 (0 < r < 1, 0 ≤ θ ≤ 2π), u(1,θ) = 120+ 60cos2θ

(0≤ θ ≤ 2π).Ans. u(r,θ) = 120+60r2 cos(2θ).

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1r2 uθθ = 0 (0 < r < 1, 0≤ θ ≤ 2π ), u(1,θ) = sin3

θ (0≤ θ ≤ 2π ).Ans. u(r,θ) = 1

4

(3r sin(θ)− r3 sin(3θ)

).


1r2 uθθ = 0 (0 < r < 4, 0≤ θ ≤ 2π ), u(4,θ) = 256cosθ (0≤ θ ≤

2π ).Ans. u(r,θ) = 1

8

[768+64r2 cos(2θ)+ r4 cos(4θ)

].


1r2 uθθ = 0 (0 < r < 1, 0≤ θ ≤ π ), u(1,θ) = θ · (π−θ) (0≤ θ ≤

π ), u(r,0) = u(r,π) = 0 (0≤ r < 1).Ans. u(r,θ) = 8

π ∑∞k=1

r2k−1

(2k−1)3 sin [(2k−1)θ ].14. A plate has the shape of a circular sector with uinit radius and angle

θ0 : (r,θ) : 0≤ r ≤ 1,0≤ θ ≤ θ0.

If the side (1,θ): 0≤ θ ≤ θ0is kept at the temperature and the sides (r,θ):0 ≤ r ≤ 1, θ = 0, (r,θ): θ = θ0 are kept at zero temperature, find thetemperature u(r,θ) in steady state.Ans. u(r,θ)= 2

θ0∑

∞n=1 rn·

(∫ θ00 f (φ)sin

(nπ

φ

θ0

)dφ

)· sin

(nπ

θ

θ0

).


Ath.

Keha

gias



14. PDEs for Diffusion

Diffusion PDEs describe phenomena which evolve in time but tend to reach (in thelimit of infinite time) a steady state. The classic diffusion PDE is the heat equation

∂u∂ t

= c2∇

2u,

e.g, ut = c2uxx, ut = c2 (uxx +uyy) etc.


14.1.1 In this chapter we will study PDEs which describe diffusion phenomena.Our typical example will be the transmission of heat, but the same equationsdescribe the diffusion of a fluid in a porous medium, the change of the probabilityfunction of a random process etc.

14.1.2 In the context of heat transmission, consider a thin rod. The temperatureof the rod is denoted by u(x, t); this implies that different parts of the rod can be atdifferent temperatures at different times. Without justification (it can be found ina physics textbook) we postulate that u(x, t) is governed by the following diffusionequation:

ut = a2uxx. (14.1)

We will later introduce variations of (14.1) to describe variations of the basic heattransmission problem.

14.1.3 Similarly to the Laplace equation, to obtain a particular solution of thediffusion equation we must specify boundary conditions (at the ends of the rod);now we must also specify initial conditions (what is the temperature of each part ofthe rod at initial time t = 0).

14.1.4 Let us now solve what is probably the simplest version of the diffusion

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150 Chapter 14. PDEs for Diffusion

equation.

0 < x < 1 and 0 < t < ∞ : ut = a2uxx (14.2)0 < t < ∞ : u(0, t) = u(1, t) = 0 (14.3)0 < x < 1 : u(x,0) = f (x) . (14.4)

As in Chapter 11, we assume that u(x, t) can be written in the form

u(x, t) = X(x)T (t) (14.5)

Then (14.2) yields

X(x)T ′(t) = a2X ′′(x)T (t)⇒ T ′(t)a2T (t)

=X ′′(x)X(x)

. (14.6)

The left (resp. right) part is a function of t only (resp. x only). For these two partsto be equal, for every x and t, they must both be equal to to a constant, which wedenote by −b2. Then we have

T ′(t)a2T (t)

=X ′′(x)X(x)

=−b2⇒(

T ′(t) =−a2b2T (t)X ′′(x) =−b2X(x).

)(14.7)

We have used −b2 because we want −a2b2 to be nonpositive (if it was positive, thenT (t) and u(x, t) would be unbounded as t→ ∞).

We now can solve the two DEs of (14.7). We get

T (t) =Ce−a2b2t (14.8)X(t) = Asin(bx)+Bcos(bx) . (14.9)

Hence every function

u(x, t) = T (t)X(x) = e−a2b2t · (Asin(bx)+Bcos(bx))

solves (14.2). But we also must satisfy the boundary conditions (14.3) which nowbecome:

0 = u(0, t) = Asin(0)+Bcos(0) = B (14.10)0 = u(1, t) = Asin(b)+Bcos(b) = Asin(b) . (14.11)

Hence b must satisfyb = 0,±π,±2π, ... (14.12)

Consequently, for every integer n, we have un(x, t) = e−a2n2π2tAn sin(nπx), whichsatisfies (14.2) and (14.3); then this is also true of

u(x, t) =∞

∑n=1

un (x, t) =∞

∑n=1

e−a2n2π2tAn sin(nπx) . (14.13)

Finally, we must the initial condition (14.4):

0 < x < 1 : f (x) = u(x,0) =∞

∑n=1

An sin(nπx) . (14.14)

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This is a typical problem of Fourier series expansion. We define g(x) to be the oddextension of f (x) in [−1,1]; then we can find An from

An =∫ 1

−1g(x)sin(nπx)dx = 2

∫ 1

0f (x)sin(nπx)dx. (14.15)

In conclusion, the function

u(x, t) =∞

∑n=1

e−a2n2π2tAn sin(nπx) (14.16)

∀n ∈ 1,2, ... : An = 2∫ 1

0f (x)sin(nπx)dx. (14.17)

solves (14.2)–(14.4).

14.1.5 From the physics point of view, the problem (14.2)–(14.4) corresponds tothe situation in which the ends of the rod are kept at zero temperature. Note thatlimt→∞ u(x, t) = 0, i.e., eventually the rod reaches a steady-state of zero temperature.Does this make physical sense?


0 < x < 3 kai 0 < t < ∞ : ut = 2uxx (14.18)0 < t < ∞ : u(0, t) = u(3, t) = 0 (14.19)0 < x < 3 : u(x,0) = 5sin(4πx)−3sin(8πx)+2sin(10πx) (14.20)

Unlike the problem of (14.2)–(14.4), now the ends of the rod are at x = 0,x = 3(not at x = 0,x = 1). This is easy to solve with the variable change x′ = x/3; thenbn = nπ/3. Hence the solution has the form

u(x, t) =∞

∑n=1

e−2n2π2t/9An sin(nπ

3x)

(14.21)

To satisfy the initial condition we must have

5sin(4πx)−3sin(8πx)+2sin(10πx) = u(x,0) =∞

∑n=1

An sin(nπ

3x). (14.22)

We see that (14.22) is satisfied with A12 = 5, A24 =−3, A30 = 2 and the remainingAn = 0. Then the solution is

u(x, t) = 5e−288π2t/9 sin(4πx)−3e−1152π2t/9 sin(8πx)+2e−1800π2t/9 sin(10πx)(14.23)

= 5e−32π2t sin(4πx)−3e−128π2t sin(8πx)+2e−200π2t sin(10πx) . (14.24)


0 < x < 3 kai 0 < t < ∞ : ut = 2uxx (14.25)0 < t < ∞ : u(0, t) = u(3, t) = 0 (14.26)0 < x < 3 : u(x,0) = 25 (14.27)

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The solution has the form

u(x, t) =∞

∑n=1

e−2n2π2t/9An sin(nπ

3x)

(14.28)

For the initial condition we need

25 = u(x,0) =∞

∑n=1

An sin(nπ

3x). (14.29)

Hence

An =23

∫ 3

025sin

(nπx3

)= 50

1− cosnπ

nπ

and the solution is

u(x, t) = 50∞

∑n=1

1− cosnπ

nπe−2n2π2t/9 sin

(nπ

3x). (14.30)

14.1.8 We will now solve a problem with nonhomogeneous boundary conditions:

0 < x < 1 kai 0 < t < ∞ : ut = a2uxx (14.31)0 < t < ∞ : u(0, t) = k1 (14.32)0 < t < ∞ : u(1, t) = k2 (14.33)0 < x < 1 : u(x,0) = f (x) (14.34)

This problem corresponds to the situation in which the ends of the rod are kept atconstant temperatures k1 and k2. If we try to solve by separation of variables wewill find out there exists no solution of the form u(x, t) = T (t)X(x).

However we can obtain the solution by applying a preliminary transform (whichis suggested by the nature of the physical problem). Namely, it is reasonable toassume that at steady state the rod temperature is given by the function

w(x) = k1 +(k2− k1)x, (14.35)

i.e., a uniform temperature variation along the rod. Note that w(x) satisfies (14.32)“by construction”. Also w(x) satisfies (14.31) because

wt(x) = wxx(x) = 0. (14.36)

But w(x) does not satisfy (14.34).Let us now check whether there exists some v(x, t) such that

u(x, t) = w(x)+ v(x, t) (14.37)

solves the original problem. Replacing u(x, t) by w(x)+ v(x, t) in (14.31), sincewt(x) = wtt(x) = 0, we get

vt = a2vxx. (14.38)

We also get

v(0, t) = v(1, t) = 0 (14.39)v(x,0) = f (x)− k1− (k2− k1)x = h(x) (14.40)

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But (14.38)–(14.40) is of the same form as the problemn of 14.1.4, which we knowhow to solve it. It is

v(x, t) =∞

∑n=1

e−a2b2tAn sin(nπx) (14.41)

∀n ∈ 1,2, ... : An = 2∫ 1

0h(x)sin(nπx)dx. (14.42)

Hence (14.31)–(14.34) has solution

u(x, t) = w(x)+ v(x, t)

where

u(x, t) = k1 +(k2− k1)x+∞

∑n=1

Ane−a2b2t sin(nπx) (14.43)

∀n ∈ 1,2, ... : An = 2∫ 1

0( f (x)− k1− (k2− k1)x)sin(nπx)dx. (14.44)


0 < x < 3 and 0 < t < ∞ : ut = 2uxx (14.45)0 < t < ∞ : u(0, t) = 10 (14.46)0 < t < ∞ : u(3, t) = 40 (14.47)0 < x < 3 : u(x,0) = 25 (14.48)

We set u(x, t) = w(x)+ v(x, t) and

w(x) = 10+10x. (14.49)

Now we solve

0 < x < 3 and 0 < t < ∞ : vt = 2uxx (14.50)0 < t < ∞ : v(0, t) = 0 (14.51)0 < t < ∞ : v(3, t) = 0 (14.52)0 < x < 3 : v(x,0) = 15−10x (14.53)

which gives

v(x, t) =∞

∑n=1

Ane−2n2π2t/9 sin(nπ

3x)

(14.54)

with

An =23

∫ 3

0(15−10x)sin

(nπ

3x)

dx =30nπ

(cos(nπ)−1) . (14.55)

Finally the full solution is

u(x, t) = 10+10x+∞

∑n=1

30nπ

(cos(nπ)−1)e−2n2π2t/9 sin(nπ

3x)

(14.56)

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14.1.10 Let us now solve

0 < x < 1 and 0 < t < ∞ : ut = a2uxx− cu (14.57)0 < t < ∞:u(0, t) = 0 (14.58)0 < t < ∞ : u(1, t) = 0 (14.59)0 < x < 1 : u(x,0) = f (x). (14.60)

The term −cu (where we assume c ≥ 0) corresponds to heat radiation to theenvironment, i.e., the rod is not fully thermally insulated.

Mathematically, we expect −cu to generate in the solution u(x, t) a term of theform e−ct (this is the attenuation effect). Let us chekc whether this assumptionhelps in simplifying the problem. Assume that the solution has the form

u(x, t) = e−ctv(x, t). (14.61)

Indeed, we then getut =−cv+ e−ctvt , uxx = e−ctvxx (14.62)

and hence

ut = a2uxx− cu⇒ (14.63)

−cv+ e−ctvt = a2e−ctvxx− cv⇒ (14.64)

e−ctvt = e−cta2vxx. (14.65)

We also see that

0 = u(0, t) = e−ctv(0, t)⇒ v(0, t) = 0 (14.66)

0 = u(1, t) = e−ctv(1, t)⇒ v(1, t) = 0 (14.67)

andf (x) = u(x,0) = e−c·0v(x,0) = v(x,0). (14.68)

In short, v(x, t) is a solution of the problem

0 < x < 1 and 0 < t < ∞ : vt = a2vxx (14.69)0 < t < ∞ : v(0, t) = v(1, t) = 0 (14.70)0 < x < 1 : v(x,0) = f (x) (14.71)

which we have already solved; hence u(x, t) is given by

u(x, t) = e−ct∞

∑n=1

e−a2n2π2tAn sin(nπx) (14.72)

∀n ∈ 1,2, ... : An = 2∫ 1

0f (x)sin(nπx)d. (14.73)

14.1.11 A harder problem is:

0 < x < 1, 0 < t < ∞ : ut = uxx− cu (14.74)0 < t < ∞ : u(0, t) = k0 (14.75)0 < t < ∞ : u(1, t) = k1 (14.76)0 < x < 1 : u(x,0) = 0. (14.77)

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We assume thatu(x, t) = w(x)+ v(x, t)e−ct

where w(x) satisfies

0 < x < 1 : wxx− cw = 0, w(0) = k0, w(1) = k1.

Then we have

ut = 0+ vte−ct− cve−ct

uxx = wxx + vxxe−ct

−cu =−cw− cve−ct .

Hence for 0 < x < 1,0 < t < ∞ we have

ut = uxx− cu⇒vte−ct− cve−ct = wxx + vxxe−ct− cw− cve−ct ⇒

vte−ct = vxxe−ct +wxx− cw⇒vt = vxx.

And for 0 < x < 1 we have

0 = u(x,0) = w(x)+ v(x,0)e−c·0⇒v(x,0) =−w(x) .

Finally, for 0 < t < ∞ we have

k0 = u(0, t) = w(0)+ v(0, t) · e−ct ⇒k0 = k0 + v(0, t)e−ct ⇒0 = v(0, t)

and we can similarly show that for 0 < t < ∞ we have

0 = v(1, t) .

Summarizing, we will solve the subproblems

0 < x < 1 : wxx− cw = 0, w(0) = k0, w(1) = k1 (14.78)

and

0 < x < 1, 0 < t < ∞ : vt = vxx (14.79)0 < t < ∞ : v(0, t) = 0 (14.80)0 < t < 1 : v(1, t) = 0 (14.81)0 < x < 1 : v(x,0) =−w(x) . (14.82)

Let us first solve (14.78). We have

w(x) = asinh(√

cx)+bcosh

(√cx), w(0) = k0, w(1) = k1

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hence

asinh(0)+bcosh(0) = k0

asinh(√

c)+bcosh

(√c)= k1

The solution of the system is

b = k0, a =−k0 cosh(√

c)− k1

sinh(√

c)

hence

w(x) = k0 cosh(√

cx)− k0 cosh(

√c)− k1

sinh(√

c)sinh

(√cx).

Next we solve (14.79)–(14.82). With separation of variables we get v(x, t) = X (x)T (t)and

T ′

T=

X ′′

X=−b2.

Hence

T (t) = e−b2t

X (x) = Asin(bx)+Bcos(bx) .

To satisfy X (0) = X (1) = 0 we take bn = 0,±π,±2π, ... and Bn = 0 for all n. Inconclusion, a general solution is

v(x, t) =∞

∑n=1

Ane−n2π2t sin(nπx)

and we also have

−k0 cosh(√

cx)+

k0 cosh(√

c)− k1

sinh(√

c)sinh

(√cx)=−w(x) = v(x,0) =

∞

∑n=1

An sin(nπx)

Hence, we have

∀n∈1,2, ... : An = 2∫ 1

0

(−k0 cosh

(√cx)+

k0 cosh(√

c)− k1

sinh(√

c)sinh

(√cx))

sin(nπx)dx.

or

∀n∈1,2, ... : An =−2k0

∫ 1

0cosh

(√cx)

sin(nπx)dx+2k0 cosh(

√c)− k1

sinh(√

c)

∫ 1

0sinh

(√cx)

sin(nπx)dx.

Finally, (14.74)-(14.77) has solution

u(x, t)= k0 cosh(√

cx)− k0 cosh(

√c)− k1

sinh(√

c)sinh

(√cx)+

(∞

∑n=1


)·e−ct .

As t→ ∞ we get

limt→∞

u(x, t) = k0 cosh(√

cx)− k0 cosh(

√c)− k1

sinh(√

c)sinh

(√cx)+ lim

t→∞

(∞

∑n=1


)· e−ct

= k0 cosh(√

cx)− k0 cosh(

√c)− k1

sinh(√

c)sinh

(√cx)= w(x) .

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If the attenuation factor is small, c' 0, then

cosh(√

cx)=

e√

cx + e−√

cx

2' 1+

√cx+1−

√cx

2= 1

sinh(√

cx)=

e√

cx− e−√

cx

2' 1+

√cx−1+

√cx

2=√

cx

and cosh(c)' 1, sinh(√

c)'√

c. Hence

limt→∞

u(x, t) = k0 cosh(√

cx)− k0 cosh(

√c)− k1

sinh(√

c)sinh

(√cx)

' k0 ·1−k0 ·1− k1√

c√

cx

= k0− (k0− k1)x

which approximates the steady state of the already solved problem:

0 < x < 1, 0 < t < ∞ : zt = zxx (14.83)0 < t < ∞ : z(0, t) = k0 (14.84)0 < t < ∞ : z(1, t) = k1 (14.85)0 < x < 1 : z(x,0) = 0. (14.86)

14.1.12 So far we have examined problems in which the boundary conditions areon u(x, t). This is not always the case; in some physical problems we may have thecondition that one or both ends of the rod are insulated. This means that ux(x, t)is zero at the respective end (there is no heat flow across this end). For example,consider the following problem.

0 < x < 1 and 0 < t < ∞ : ut = a2uxx (14.87)0 < t < ∞ : ux(0, t) = ux(1, t) = 0 (14.88)0 < x < 1 : u(x,0) = f (x) (14.89)

Using the separation of variables assumption, after some algebra we conslude that

u(x, t) = T (t)X(x) = e−a2b2t (Asin(bx)+Bcos(bx))

solves (14.87). To satisfy the boundary conditions (14.88) we must have

0 = ux(0, t) = Acos(0)−Bsin(0) = A (14.90)0 = ux(1, t) = Acos(b)−Bsin(b) = Bsin(b) (14.91)

which impliesb ∈ 0,±π,±2π, ... . (14.92)

Hence for every integer n, the function un(x, t) = e−a2n2π2tBn cos(nπx) satisfies both(14.87) and (14.88) and the same is true for

u(x, t) =B0

2+

∞

∑n=1

e−a2n2π2tBn cos(nπx) . (14.93)

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To also satisfy the initial condition (14.89) we must have

0 < x < 1 : f (x) = u(x,0) =∞

∑n=0

Bn cos(nπx) . (14.94)

We can achieve this by defining g(x) to be the even extension of f (x) on [−1,1];then we find Bn from

Bn =∫ 1

−1g(x)cos(nπx)dx = 2

∫ 1

0f (x)cos(nπx)dx. (14.95)

In conclusion

u(x, t) =B0

2+

∞

∑n=1

Bne−a2n2π2t cos(nπx) (14.96)

∀n ∈ 0,1,2, ..Bn = 2∫ 1

0f (x)cos(nπx)dx (14.97)

solves (14.87)–(14.89).

14.1.13 Example. The solution of

0 < x < 1 and 0 < t < ∞ : ut = uxx (14.98)0 < t < ∞ : ux(0, t) = ux(1, t) = 0 (14.99)0 < x < 1 : u(x,0) = x (14.100)

is

u(x, t) =B0

2+

∞

∑n=1

e−a2n2π2tBn cos(nπx) (14.101)

B0 = 1 (14.102)

Bn = 2∫ 1

0xcos(nπx)dx = 2

cosnπ−1n2π2 (n = 1,2, ..) (14.103)

14.1.14 Example. The solution of

0 < x < 1 and 0 < t < ∞ : ut = uxx (14.104)0 < t < ∞ : ux(0, t) = ux(1, t) = 0 (14.105)

0 < x < 1 : u(x,0) = 1+ x2 (14.106)

is

u(x, t) =B0

2+

∞

∑n=1

e−a2n2π2t ·Bn cos(nπx) (14.107)

B0 =83, (14.108)

n ∈ 1,2, .. : kai Bn = 2∫ 1

0

(1+ x2)cos(nπx)dx = 4

cosnπ

n2π2 . (14.109)

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14.1.15 We now consider a problem of heat transmission in an infinite rod:

−∞ < x < ∞, 0 < t < ∞ : ut = a2uxx, (14.110)−∞ < x < ∞ : u(x,0) = f (x). (14.111)

Because x ∈ (−∞,∞) we will apply the Fourier transform with respect x. In otherwords, we define

U(w, t) = F (u(x, t)) =∫

∞

−∞

e−iwxu(x, t)dx. (14.112)

Then (14.110)-(14.111) becomes

Ut =−w2a2U, (14.113)U(w,0) = F ( f (x)) = F(w). (14.114)

Now, (14.113) has the solution

U(w, t) =C(w)e−w2a2t (14.115)

and, since U(w,0) = F(w), we get

U(w, t) = F(w)e−w2a2t . (14.116)

Note that (14.116) is a convolution. Since

e−w2a2t = F

(1

2a√

πte−

x2

4a2t

), (14.117)

the inverse of (14.116) gives

u(x, t) = f (x)∗ 12a√

πte−

x2

4a2t , (14.118)

i.e.,u(x, t) =

12a√

πt

∫∞

−∞

f (z)e−(x−z)2/4a2tdz. (14.119)

This is the solution of (14.110)-(14.111).

14.1.16 We can also solve (14.110), (14.111) using the Laplace transform. Beforesolving the general case of (14.110), (14.111), let us consider a special case:

−∞ < x < ∞, 0 < t < ∞ : vt = a2vxx (14.120)−∞ < x < ∞ : v(x,0) = δ (x) (14.121)

(i.e., we assume f (x) = δ (x)). Defining

V (x,s) = L (v(x, t)) (14.122)

we get the transform of (14.120):

sV − v(x,0) = a2Vxx (14.123)

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Using (14.121) we geta2Vxx− sV =−δ (x) . (14.124)

A bounded solution of (14.124) is

V (x,s) =e−√

s|x|/a

2a√

s; (14.125)

inverting we get

v(x, t) = L −1

(e−√

s|x|/a

2a√

s

)=

e−x2/4a2t

2a√

πt. (14.126)

14.1.17 Before solving the general problem, consider the physical interpretationof (14.126). It gives the temperature u(x, t) if we place a “point heat source” atx = 0. Note that e−x2/4a2t/2a

√πt is a Gaussian bell curve, centered at 0 and with

time increasing “spread” 2at. This tells us that the initial “heat spike” is diffusingin time; this is very reasonable in the context of heat transmission. The functione−x2/4a2t/2a

√πt is called the heat kernel, with basic characteristic the smoothing

of the initial conditions.

14.1.18 Let us return to the general problem. We can represent the initialconditons (14.110)-(14.111) as a convolution u(x,0) = f (x) = f (x) ∗ δ (x). Sinceconvolution is a linear combination, the solution u(x, t) will also be the convolutionof f (x) with v(x, t):

u(x, t) = f (x)∗ v(x, t) =1

2a√

πt

∫∞

−∞

f (z)e−(x−z)2/4a2tdz. (14.127)

This gives the solution of (14.110), (14.111) by Laplace transform; it is the same as(14.119), the solution by Fourier transform.

14.1.19 Example. If the initial conditions is f (x) = e−x2, then (14.110)-(14.111)

has solution

u(x, t) =1

2a√

πt

∫∞

−∞

e−z2e−(x−z)2/4a2tdz =

e−x2/(1+4a2t)√(1+4a2t)

. (14.128)

For every x we have

limt→∞

u(x, t) = limt→∞

e−x2/(1+4a2t)√(1+4a2t)

= 0 (14.129)

which means that in steady state the rod reaches zero temperature. Does thismake physical sense?

14.1.20 Example. If the initial condition is

f (x) =

1 gia |x| ≤ 10 gia |x|> 1. (14.130)

Then the solution of (14.110)-(14.111) is

u(x, t) =1

2a√

πt

∫∞

−∞

f (z)e−(x−z)2/4a2tdz =12·(

erf(

x+12a√

t

)− erf

(x−12a√

t

)).

(14.131)

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where the error function erf(x) is

erf(x) :=2√π

∫ x

0e−z2

dz. (14.132)

Once again we see that

limt→∞

u(x, t) = limt→∞

12·(

erf(

x+12a√

t

)− erf

(x−12a√

t

))= 0. (14.133)

14.1.21 Returning to the general solution

u(x, t) =1

2a√

πt

∫∞

−∞

f (z)e−(x−z)2/4a2tdz (14.134)

we see that the temperature at (x, t) is a weighted sum of the temperatures atall points of the (infinite) rod; the weight on the influence of point z to point x isthe value of the kernel e−(x−z)2/4a2t ; hence the influence of z on x decreases withdistance, but increases with time. What physical intuition do you see in this?

14.1.22 Here is yet another approach to the infinite rod heat diffusion problem:

−∞ < x < ∞, 0 < t < ∞ : ut = a2uxx (14.135)−∞ < x < ∞ : u(x,0) = f (x). (14.136)

We will now use separation of variables. Assuming

u(x, t) = X(x)T (t) (14.137)

and applying the usual steps, we get

T ′(t) =−a2b2T (t)X ′′(x) =−b2X(x)

(14.138)

and conclude that any function of the form

(A(b)sin(bx)+B(b)cos(bx))e−a2b2t (14.139)

is a solution of (14.135) and so is the linear combination

u(x, t) =∫

∞

0(A(b)sin(bx)+B(b)cos(bx))e−a2b2t db. (14.140)

But, in the absence of boundary conditions, b can take any value in R. Hence, tohave

f (x) = u(x,0) =∫

∞

0(A(b)sin(bx)+B(b)cos(bx))db

we choose A(b), B(b) to be the trigonometric Fourier transforms:

A(b) =1π

∫∞

−∞

f (z)sin(bz)dz,

B(b) =1π

∫∞

−∞

f (z)cos(bz)dz.

Finally, the solution is

u(x, t)=1π

∫∞

0

([∫∞

−∞

f (z)sin(bz)dz]

sin(bx)+[∫

∞

−∞

f (z)cos(bz)dz]

cos(bx))

e−a2b2t db.

(14.141)

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14.1.23 To see the connection between (14.141) and solution

u(x, t) =1

2a√

πt

∫∞

−∞

f (z)e−(x−z)2/4a2tdz (14.142)

obtained by the Laplace and Fourier transforms, let us change the order ofintegration in (14.141), to get

u(x, t) =1π

∫∞

−∞

f (z)(∫

∞

0[sin(bz)sin(bx)+ cos(bz)cos(bx)]e−a2b2t db

)dz

=1π

∫∞

−∞

f (z)(∫

∞

0cos(b [z− x])e−a2b2t db

)dz. (14.143)

Note that ∫∞

0cos(b [z− x])e−a2b2t db = Re

[∫∞

0eib[z−x]e−a2b2t db

].

This integral can be computed by complex integration and turns out to be∫∞

0cos(b [z− x])e−a2b2t db =

√π

2a√

te−(x−z)2/4a2t .

Replacing in (14.143) we once again get the solution in terms of the heat kernel

u(x, t) =1

2a√

πt

∫∞

−∞

f (z)e−(x−z)2/4a2tdz. (14.144)

Hence the application of integral transforms can be seen as a limiting case ofseparation of variables.

14.1.24 Let us now return to our claim of Chapter 11: the Laplace equationdescribes phenomena in equilibrium. Essentially, we have already seen why thisis the case. Namely, in all the cases we have studied in this chapter the solutionu(x, t) tends to a limit:

limt→∞

u(x, t) = u(x)

and the limit function satisfiesuxx = 0

(why?). This means that u(x) the solution of the one-dimensional Laplace equation(satisfying the appropriate boundary conditions).

14.1.25 In short, the solution of the diffusion equation is the solution of theLaplace equation. In this chapter we have only seen examples of this involvingone spatial variable x but, as we will see later, it is also true for more than onevariables (x,y, ...).

14.1.26 Finally let us solve the Heat equation on rectangle (SchFourier 41, 2.29)

ut = κ (uxx +uyy)

u(0,y, t) = u(1,y, t) = u(x,0, t) = u(x,1, t) = 0u(x,y,0) = F (x,y)|u(x,y)|< M

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We assumeu(x,y, t) = X (x)Y (y)T (t)

and then we getXY T ′ = κ

(X ′′Y T +XY ′′T

)or, dividing by κXY T

T ′

κT=

X ′′

X+

Y ′′

YSo we can set

T ′

κT=−λ

2 =

(X ′′

X+

Y ′′

Y

)to get

T ′ =−κλ2T

−λ2 =

X ′′

X+

Y ′′

Y

The first equation has solution T = e−κλ 2t . The second equation can be written as

X ′′

X=−Y ′′

Y−λ

2

orX ′′

X=−µ

2 =−Y ′′

Y−λ

2

which gives

X ′′+µ2X = 0

Y ′′+(λ

2−µ2)Y = 0

These equations have solutions

X (x) = aµ cos(µx)+bµ sin(µx)

Y (x) = cµ,λ cos(√

λ 2−µ2x)+dµ,λ sin

(√λ 2−µ2x

)From u(0,y, t) = 0 we get bµ = 0. u(1,y, t) = 0 we get µ = mπ. From u(x,0, t) = 0 weget dµ = 0. From u(x,1, t) = 0 we get

√λ 2−µ2 = nπ. So finally the solution is

u(x,y, t) =∞

∑m=1

∞

∑n=1

Bm,n sin(mπx)sin(nπx)e−κ(m2+n2)πt

Letting t = 0 we get

F (x,y) = u(x,y,0) =∞

∑m=1

∞

∑n=1

Bm,n sin(mπx)sin(nπx)

and so

Bm,n = 4∫ 1

0

∫ 1

0F (x,y)sin(mπx)sin(nπx)dxdy

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∀t > 0,x ∈ (0,L) :∂u∂ t

= k∂ 2u∂x2 (14.145)

∀t > 0 : u(0, t) = 0, u(L, t) = 0 (14.146)

∀x ∈ (0,L) : u(x,0) = 4sin3πx

L(14.147)

Solution. Here we can simply take

B3 = 4, Bn = 0 when n 6= 3

and get the solution

u(x, t) = 4sin(

3πxL

)e−k( 3π

L )2t


∀t > 0,x ∈ (0,3) :∂u∂ t

= 2∂ 2u∂x2 (14.148)

∀x ∈ (0,3) : u(x,0) = 25 (14.149)∀t > 0 : u(0, t) = 10, u(3, t) = 40 (14.150)

Solution. To solve this we assume u(x, t) = v(x, t) +w(x). Then the originalproblem becomes

∀t > 0,x ∈ (0,3) :∂v∂ t

= 2∂ 2v∂x2 +2

d2wdx2 (14.151)

∀x ∈ (0,3) : v(x,0)+w(x) = 25 (14.152)∀t > 0 : v(0, t)+w(0) = 10, v(3, t)+w(3) = 40 (14.153)

We find w(x) to be the solution of the problem

0 =d2wdx2 , w(0) = 10, w(3) = 40; (14.154)

obviouslyw(x) = 10+10x.

Then v(x, t) solves the problem

∀t > 0,x ∈ (0,L) :∂v∂ t

= 2∂ 2v∂x2 (14.155)

∀x ∈ (0,L) : v(x,0) = 25− (10+10x) = 15−10x (14.156)∀t > 0 : v(0, t) = 0, v(3, t) = 0 (14.157)

Now we get

v(x, t) =∞

∑n=1

Bn sin(nπx

3

)e−

2n2π29 t

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and we must have

15−10x = v(x,0) =∞

∑n=1

Bn sin(nπx

3

).

From which we get

Bn =23

∫ 3

0(15−10x)sin

(nπx3

)dx =

30nπ

(1+ cos(nπ)) .

So the required solution is

u(x, t) = w(x, t)+ v(x, t)

= 10+10x+∞

∑n=1

30nπ

(1+ cos(nπ))sin(nπx

3

)e−

2n2π29 t

Note thatlimt→∞

u(x, t) = 10+10x.

This is the steady state temperature.


∀t > 0,x ∈ (0,L) :∂u∂ t

= k∂ 2u∂x2 (14.158)

∀t > 0 : u(0, t) = 0, u(L, t) = 0 (14.159)

∀x ∈ (0,L) : u(x,0) = 4sin3πx

L+7sin

8πxL

(14.160)

Solution. Here we can take

B3 = 4, B8 = 7, Bn = 0 when n /∈ 3,8

and get the solution

u(x, t) = 4sin(

3πxL

)e−k( 3π

L )2t +7sin

(8πx

L

)e−k( 8π

L )2t


∀t > 0,x ∈ (0,L) :∂u∂ t

=∂ 2u∂x2 (14.161)

∀t > 0 : u(0, t) = 0, u(1, t) = 0 (14.162)∀x ∈ (0,1) : u(x,0) = 100 (14.163)

Solution. We must have

100 = u(x,0) =∞

∑n=1

Bn sin(nπx

L

)Then we will also have∫ L

0100sin(mπx)dx =

∫ L

0

(∞

∑n=1

Bn sin(nπx)

)sin(mπx)dx

=∞

∑n=1

Bn

∫ L

0sin(nπx)sin(mπx)dx

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And since∫ L

0100sin(mπx)dx = 100

(− 1

mπcos(mπx)

)x=1

x=0=

100mπ

(1− cos(mπ))

∫ L

0sin(nπx)sin(mπx)dx =

(− 1

π

mcos(mπ)sin(nπ)−nsin(nπ)cos(mπ)m2−n2 for m 6= n

− 1π

cos(mπ)sin(mπ)−mπ

2m for m = n

)=

(0 for m 6= n12 for m = n

)we get

100mπ

(1− cos(mπ)) = Bm12⇒ Bm =

200mπ

(1− cos(mπ)) .

Finally the solution is

u(x, t) =∞

∑n=1

200nπ

(1− cos(nπ))sin(nπx)e−(nπ)2t

=400π

sin(πx)e−π2t +4003π

sin(3πx)e−(3π)2t +4005π

sin(5πx)e−(3π)2t + ...

Clearly we have expanded f (x) = 100 (actually its odd expansion) in a Fourier sineseries.


∀t > 0,x ∈ (0,25) :∂u∂ t

= 100∂ 2u∂x2 (14.164)

∀t > 0 : ux (0, t) = 0, ux (25, t) = 0 (14.165)∀x ∈ (0,25) : u(x,0) = x (14.166)

Solution. Here

A0 =1π

∫π

0xdx =

π

2

Am =225

∫ 25

0f (x)cos

mπxL

dx =50

m2π2 (−1+(−1)m)

and

u(x, t) =252+

∞

∑m=1

50m2π2 (−1+(−1)m)cos

(mπx25

)e−100(mπ

25 )2t

=252− 100

π2 cos(

πx25

)e−

100625 π2t− 100

9π2 cos(

3πx25

)e−

900625 π2t− 100

25π2 cos(

5πx25

)e−

2500625 π2t− ...

14.2.6 Problem. Solve the heat diffusion problem on a ring:

∀t > 0,x ∈ (−L,L) :∂u∂ t

= k∂ 2u∂x2 (14.167)

∀x ∈ (−L,L) : u(x,0) = f (x) (14.168)∀t > 0 : u(−L, t) = u(L, t) (14.169)∀t > 0 : ux (−L, t) = ux (L, t) (14.170)

Ath.

Keha

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Solution. Assumeu(x, t) = φ (x)g(t) .

Then1kg

dgdt

=−λ =1φ

d2φ

dx2

As usualg(t) = g(0)e−λkt

and, fromd2φ

dx2 +λφ = 0,

we get

φ (x) = Acos(√

λx)+Bsin

(√λx)

with boundary conditions

Acos(−√

λL)+Bsin

(−√

λL)= Acos

(√λL)+Bsin

(√λL)

−Asin(−√

λL)+Bcos

(−√

λL)=−Asin

(√λL)+Bcos

(√λL).

From the first one we get

Bsin(√

λL)= 0

so√

λ = nπ which also satisfies the second.So finally we get solutions of the form

u(x, t) = A0 +∞

∑n=1

An cos(nπx

L

)e−k( nπ

L )2t +

∞

∑n=1

Bn sin(nπx

L

)e−k( nπ

L )2t

We still have to satisfy (??):

∀x ∈ (0,L) : u(x,0) = f (x)

We must have

f (x) = A0 +∞

∑n=1

An cos(nπx

L

)+

∞

∑n=1

Bn sin(nπx

L

)So we get

A0 =1L

∫ L

−Lf (x)dx

Am =1

2L

∫ L

−Lf (x)cos

mπxL

dx

Bm =1

2L

∫ L

−Lf (x)sin

mπxL

dx

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Keha

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14.2.7 Problem. Solve the problem

∀t > 0,x ∈ (−π,π) :∂u∂ t

= k∂ 2u∂x2 (14.171)

∀x ∈ (−π,π) : u(x,0) = x2−π2 (14.172)

∀t > 0 : u(−π, t) = u(π, t) (14.173)∀t > 0 : ux (−π, t) = ux (π, t) (14.174)

Solution. Applying the previous results, we have

A0 =1π

∫π

0xdx =−4

3π

2

Am =1

2L

∫ L

0f (x)cosmxdx = (−1)m 4

m2

Bm = 0

and

u(x, t) =−4(

π2

3+ cos(x)e−kt− 1

4cos(2x)e−k4t + ...

)14.2.8 Problem. Solve

0 < x,0 < t : ut = kuxx

0 < x : u(x,0) = 00 < t : u(0, t) = g(t)0 < x,0 < t : |u(x, t)|< M

Soolution. Taking Laplace wrt t we get

sU (x,s)−u(x,0) = kUxx (x,s)U (0,s) = G(s)

ThenU (x,s) = c1e

√ sk x + c2e−

√ sk x

and from boundedness we get c1 = 0 and from IC

U (x,s) = G(s)e−√ s

k x

Then we geth(t) = L−1

(e−√ s

k x)=

x√2πk

t−3/2e−x2/4kt

and

u(x, t) = g(t)∗h(t)

=∫ t

0

x√2πk

u−3/2e−x2/4kug(t−u)du =∫ t

0

x√2πk

(t−u)−3/2 e−x2/4k(t−u)g(t)du

Ath.

Keha

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0 < x < L,0 < t : ut = kuxx

0 < x : u(x,0) =U0

0 < t : ux (0, t) = 00 < t : u(L, t) =U1

Solution. Taking Laplace wrt t we get

sU (x,s)−u(x,0) = kUxx (x,s)U (x,0) =U0

which becomes

Uxx (x,s)−sk

U (x,s) =−U0

k

Ux (0,s) = 0,U (L,s) =U1

s

Then

U (x,s) = c1 cosh(√

sk

x)+ c2 sinh

(√sk

x)+

U0

s

From Ux (0,s) = 0 we get c2 = 0 and from U (L,s) = U1s we get

U (x,s) =U0

s+(U1−U0)

cosh(√ s

k x)

scosh(√ s

k L)

To invert this we use tables and get

u(x, t) =U0 +4π(U1−U0)

∞

∑n=1

(−1)n

2n−1cos((2n−1)πx

2L

)e−

(2n−1)2π2kt4L2


0 < x,0 < t : ut = kuxx

0 < x : u(x,0) = 00 < t : u(0, t) =U0

0 < x,0 < t : |u(x, t)|< M

Solution. Taking Laplace wrt t we get

Uxx (x,s)−sk

U (x,s)−u(x,0) = 0

U (0,s) =U0

s

ThenU (x,s) = c1e

√ sk x + c2e−

√ sk x

Ath.

Keha

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From boundedness we get c1 = 0 and from U (0,s) = U0s we get

U (x,s) =U0

se−√ s

k x

To invert this we use tables and get

u(x, t) =U0 erfc(

x2√

kt

)=U0

(1− 2√

π

∫ x2√

kt

0e−u2

du)


0 < x,0 < t : ut = kuxx

0 < x : u(x,0) = 00 < t : u(0, t) =U0

0 < x,0 < t : |u(x, t)|< M

using Fourier transforms.Solution. Here it is important to remember the following facts

Fs

(dudx

)=−bFc (u)

Fc

(dudx

)= bFs (u)−u(0)

Fs

(d2udx2

)=−b2Fs (u)+bu(0)

Fc

(d2udx2

)=−b2Fc (u)− pu′ (0)

Because we have u(x,0) = 0, we take sine transform and we get

Ut (b, t) =−kb2U (b, t)+ kbu(0, t)

Ut (b, t)+ kb2U (b, t) = kU0b

ThenU (x,b) = c1e−kb2t +

U0

bSince we must have

U (x,0) =∫

∞

0u(x,0)sin(bx)dx = 0

we get

0 =U (x,0) = c1 +U0

b⇒ c1 =−

U0

b.

So finally

U (x,b) =U0

b

(1− e−kb2t

)and

u(x, t) =2U0

π

∫∞

0

1b

(1− e−kb2t

)sin(bx)db

Ath.

Keha

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Compare

Fourier Sol: u(x, t) =2U0

π

∫∞

0

1b

(1− e−kb2t

)sin(bx)db

Laplace Sol: u(x, t) =U0

(1− 2√

π

∫ x2√

kt

0e−u2

du)

Can you show that these two are equivalent?

14.3 Unsolved Problems1. Solve ut = uxx (0 < x < 1, 0 < t < ∞), u(0, t) = u(1, t) = 0 (0 < t < ∞), u(x,0) = 1

(0 < x < 1).Ans. u(x, t) = 4

π

(e−π2t sin(πx)+ 1

3e−9π2t sin(3πx)+ 15e−25π2t sin(5πx)+ ...

).

2. Solve ut = uxx (0 < x < 1, 0 < t < ∞), u(0, t) = u(1, t) = 0 (0 < t < ∞), u(x,0) =x− x2 (0 < x < 1).Ans. u(x, t) = 8

π3



).

3. Solve ut = 2uxx (0< x< 4, 0< t <∞), u(0, t)= u(4, t)= 0 (0< t <∞), u(x,0)= 25x(0 < x < 4).Ans. u(x, t) =−200

π ∑∞n=1

cos(nπ)n e−n2π2t/8 sin

(nπx4

).

4. Solve ut = 4uxx (0 < x < 1, 0 < t < ∞), u(0, t) = u(4, t) = 0 (0 < t < ∞), u(x,0) =x2− x3 (0 < x < 1).Ans. u(x, t) = 2∑

∞n=1

2(−1)n+1−1n3π3 e−4n2π2t sin(nπx).

5. Solve ut = uxx (0 < x < π, 0 < t < ∞), u(0, t) = u(π, t) = 0 (0 < t < ∞), u(x,0) =sin3 (x) (0 < x < π ).Ans. u(x, t) = 3

4e−t sin(x)− 14e−9t sin(3x).

6. Solve ut = uxx (0< x< 1, 0< t <∞), u(0, t)= 0, u(1, t)= 1 (0< t <∞), u(x,0)= x2

(0 < x < 1).Ans. u(x, t)= x− 8

π3



).

7. Solve ut = uxx (0 < x < 1, 0 < t < ∞), u(0, t) = 1, u(1, t) = 2 (0 < t < ∞), u(x,0) =x+1+ sin(πx) (0 < x < 1).Ans. u(x, t) = 1+ x+ e−π2t sin(πx).

8. Solve ut = uxx (0 < x < 10, 0 < t < ∞), u(0, t) = 150, u(10, t) = 100 (0 < t < ∞),u(x,0) = 150−5 · x (0 < x < 10).

9. Solve ut = uxx (0 < x < 1, 0 < t < ∞), u(0, t) = 0, u(1, t) = 2 (0 < t < ∞), u(x,0) =x2 · (1− x) (0 < x < 1).

10. Solve ut = uxx (0 < x < L, 0 < t < ∞), u(0, t) = a, u(L, t) = b (0 < t < ∞), u(x,0) =sin(

πxL

)(0 < x < L).

11. Solve ut = uxx− cu (0 < x < 1, 0 < t < ∞), u(0, t) = 0, u(1, t) = 0 (0 < t < ∞),u(x,0) = 1 (0 < x < 1).Ans. u(x, t) = 2e−ct

∑∞n=1

1−cosnπ

nπ· e−a2n2π2t · sin(nπx).

12. Solve ut = uxx− cu (0 < x < π, 0 < t < ∞), u(0, t) = 0, u(π, t) = 0 (0 < t < ∞),u(x,0) = x · (1− x) (0 < x < 1).Ans. u(x, t) = 2e−ct

∑∞n=1

2−nπ sinnπ−2cosnπ

n3π3 · e−a2n2π2t · sin(nπx).13. Solve ut = uxx− cu (0 < x < 1, 0 < t < ∞), u(0, t) = 0, u(1, t) = 0 (0 < t < ∞),

u(x,0) = x2 · (1− x) (0 < x < 1).

Ath.

Keha

gias


Ans. u(x, t) = 2e−ct∑

∞n=1

6sinnπ−4nπ cosnπ−n2π2 sinnπ−2nπ

n4π4 · e−a2n2π2t · sin(nπx).14. Solve ut = uxx− u (0 < x < 1, 0 < t < ∞), u(0, t) = 1, u(1, t) = 2 (0 < t < ∞),

u(x,0) = 0 (0 < x < 1).15. Solve ut = uxx− u (0 < x < 1, 0 < t < ∞), u(0, t) = 1, u(1, t) = 2 (0 < t < ∞),

u(x,0) = x+1 (0 < x < 1).16. Solve ut = uxx (0 < x < π, 0 < t < ∞), u(0, t) = ux(π, t) = 0 (0 < t < ∞), u(x,0) =

cos(x) (0 < x < π ).Ans. u(x, t) = e−a2t · cos(x).

17. Solve ut = uxx (0 < x < π, 0 < t < ∞), ux(0, t) = ux(π, t) = 0 (0 < t < ∞), u(x,0) =sin(x) (0 < x < π).Ans. u(x, t) = 2−∑

∞n=1

cosnπ+1n2−1 · e

−a2n2t · cos(nx).18. Solve ut = uxx (0< x< 1, 0< t <∞), ux(0, t) = ux(1, t) = 0 (0< t <∞), u(x,0) = x2

(0 < x < 1).Ans. u(x, t) = 1

3 +4∑∞n=1

cos(nπ)n2π2 · e−a2n2π2t · cos(nπx).

19. Solve ut = uxx (0 < x < 1, 0 < t < ∞), ux(0, t) = ux(1, t) = 0 (0 < t < ∞), u(x,0) =x · (1− x) (0 < x < 1).Ans. u(x, t) = 1

6 −2∑∞n=1

(cos(nπ)+1)n2π2 · e−a2n2π2t · cos(nπx).


Ath.

Keha

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15. PDEs for Waves

Wave PDEs describe oscillatory phenomena (in both space and time). The classicwave PDE is

∂ 2u∂ t2 = c2

∇2u,

e.g, utt = c2uxx, utt = c2 (uxx +uyy) etc.

15.1 Theory and Examples15.1.1 We start the study of wave PDEs with a simple example: a thin string isplaced along the x-axis with its two ends fastened at x = 0 and x = L (the lengthof the string is L). Call u(x, t) the (transverse to the x axis) displacement of thepoint x at time t. It can be shown (see a physics textbook) that u(x, t) satisfies (for0≤ x≤ L and 0≤ t < ∞) the wave equation:

utt = c2uxx. (15.1)

15.1.2 Why is (15.1) a wave equation? To understand this we will employ arough argument, using discretization of the time derivative uxx. Let us define forn ∈ 0,1,2, ...,N and with δx = L/N the functions

vn(t) = u(nδx, t). (15.2)

Then (15.1) becomes

∀n ∈ 1,2, ...,N−1 : (vn)tt =( c

δx

)2(vn+1−2vn + vn−1) (15.3)

or, with k = 2( c

δx

)2,

∀n ∈ 1,2, ...,N−1 : (vn)tt + k2vn = k2 (vn+1 + vn−1) (15.4)

We also assume that boundary conditions are given for v0 and vN . Now, (15.4)is a system of ordinary DEs. Setting w = k2 (vn+1 + vn−1) we get

(vn)tt + k2vn = w (15.5)

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174 Chapter 15. PDEs for Waves

which is the equation of a harmonic oscillator driven by w; in fact w= k2 (vn+1 + vn−1)describes the coupling of the displacment vn to vn+1,vn−1, the displacements of theneighboring elements.

15.1.3 To solve the system (15.4) we also need initial conditions on v1, ...,vN−1and their derivatives. Similarly, to solve (15.1) we need:

1. boundary conditions u(0, t), u(L, t) gia t ≥ 0:2. initial conditions u(x,0) = f (x), ut(x,0) = g(x) for x ∈ [0,L].

15.1.4 Let us now solve the basic problem

0 < x < L, 0 < t : utt = c2uxx (15.6)0 < t : u(0, t) = u(L, t) = 0 (15.7)0 < x < L : u(x,0) = f (x) (15.8)0 < x < L : ut(x,0) = g(x) (15.9)

with separation of variables. Assuming u(x, t) = X(x)T (t) we get from (15.6) that

1c2 ·

T ′′

T=

X ′′

X=−b2 (15.10)

(we will explain the −b2 choice a little later). Now(15.10) yields

T ′′+b2c2T = 0, (15.11)

X ′′+b2X = 0; (15.12)

where (15.11) has solutions sin(bct), cos(bct) and (15.12) has solutions sin(bx),cos(bx) which must equal zero at x = 0 and x = L. Consequently: (i) we reject thecosine solutions; (ii) we only accept sine solutions of the form sin(bnx), bn =

nπ

L ,n ∈ 0,±1,±2, .... Hence a general form of u(x, t) is

u(x, t) =∞

∑n=1

(An cos

(nπcL

t)+Bn sin

(nπcL

t))

sin(nπ

Lx)

(15.13)

which satisfies (15.6), (15.7). To satisfy (15.8)-(15.9) we must have

f (x) = u(x,0) =∞

∑n=1

(An cos(0)+Bn sin(0))sin(nπ

Lx)=

∞

∑n=1

An sin(nπ

Lx)

(15.14)

g(x) = ut(x,0) =nπc

L

∞

∑n=1

(−An sin(0)+Bn cos(0))sin(nπ

Lx)=

nπcL

∞

∑n=1

Bn sin(nπ

Lx).

(15.15)

We see from (15.14) that the An are the Fourier coefficients1 of the odd extensionof f (x) on [−L,L] and are given by

n ∈ 1,2, ... : An =2L

∫ L

0f (x)sin

(nπ

Lx)

(15.16)

1Here is why we chose the negative constant −b2. With a positive constant we would not be ableto obtain f (x) as a Fourier series..

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Similarly, from (15.15) we see that the Bn are the Fourier coefficients of the oddextension of g(x) on [−L,L] and are given by

n ∈ 1,2, ... : Bn =2

nπc

∫ L

0g(x)sin

(nπ

Lx)

(15.17)

15.1.5 From (15.13) we see that every point of the string performs a linearcombination of oscillations of the for

An cos(nπc

Lt)+Bn sin

(nπcL

t).

For each n we get a different harmonic frequency, all of which are multiles of thefundamental frequency πc

L . Every such oscillation(An cos

(nπcL

t)+Bn sin

(nπcL

t))

sin(nπ

Lx)

is a stationary wave.


0 < x < 1,0 < t : utt = c2uxx

0 < t : u(0, t) = u(1, t) = 00 < x < 1 : u(x,0) = 10 < x < 1 : ut (x,0) = 0.

According to the previous, u(x, t) is given by

u(x, t) =∞

∑n=1

An cos(nπct)sin(nπx)

where

An = 2∫ 1

01sin(nπx)dx = 2

1− cosnπ

nπ.

In other words

u(x, t) =4π

(cos(πct) · sin(πx)+

13

cos(3πct)sin(3πx)+15

cos(5πct)sin(5πx)+ ...

)15.1.7 Example. Let us solve

0 < x < 1,0 < t : utt = c2uxx

0 < t : u(0, t) = u(1, t) = 00 < t : u(x,0) = w(x)0 < t : ut (x,0) = 0

where

w(x) =

2x 0≤ x≤ 12

2−2x 12 < x≤ 1

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Hence u(x, t) is given by

u(x, t) =∞

∑n=1


where

An = 2∫ 1/2

02xsin(nπx)dx+2

∫ 1

1/2(2−2x)sin(nπx)dx

=−2−2sin 1

2nπ +nπ cos 12nπ

n2π2 +2−2sinnπ +nπ cos 1

2nπ +2sin 12nπ

n2π2 =8sin nπ

2n2π2 .

Finally

u(x, t) =∞

∑n=1

8sin nπ

2n2π2 cos(nπct) · sin(nπx)


0 < x < 1,0 < t : utt = c2uxx

0 < t : u(0, t) = u(1, t) = 0

0 < x < 1 : u(x,0) = x2

0 < x < 1 : ut (x,0) = 0.

In this case, we have Bn = 0 for all n and u(x, t) is given by

u(x, t) =∞

∑n=1


where

An = 2∫ 1

0x2 sin(nπx)dx =

2cosnπ−2−n2π2 cosnπ

n3π3

15.1.9 Now let us consider the infinite string with zero initial velocity:

−∞ < x < ∞, 0 < t : utt = c2uxx (15.18)−∞ < x < ∞ : u(x,0) = f (x) (15.19)−∞ < x < ∞ : ut(x,0) = 0. (15.20)

Setting u(x, t) = X(x)T (t) we get, in the usual manner, we get solutions of the form

(A(b)cos(bx)+B(b)sin(bx))cos(bct) . (15.21)

Since it is an infinite string, we have no boundary conditions and restrictions onthe value of b. Hence, by superposition, we get the solution

u(x, t) =∫

∞

0(A(b)cos(bx)+B(b)sin(bx))cos(bct)db. (15.22)

Setting t = 0 in (15.22), we get

f (x) = u(x,0) =∫

∞

0(A(b)cos(bx)+B(b)sin(bx))db. (15.23)

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For (15.23) to hold we must have

A(b) =1π

∫∞

−∞

f (z)cos(bz)dz, (15.24)

B(b) =1π

∫∞

−∞

f (z)sin(bz)dz. (15.25)

Hence (15.22)-(15.25) give the solution of (15.18) - (15.20).

15.1.10 We can solve similarly problems with other initial and boundary conditions.However we will now see an alternative way to solve certain wave PDEs.

15.1.11 Let us for the time being ignore initial and boundary conditions and tryto find solutions of utt = c2uxx only. Using can be rewritten using the differentialoperators Dx and Dt :

Dttu− c2Dxxu = 0⇒Dt (Dtu)− c2Dx (Dxu) = 0⇒

(Dt− cDx)(Dt + cDx)u = 0.

The final equation is satisfied when either Dtu− cDxu = 0 or Dtu+ cDxu = 0, i.e.,

ut− cux = 0 or ut + cux = 0. (15.26)

It is easy to check2 that rhe general solution of ut− cux = 0 is φ (x+ ct) where φ isan arbitrary (differentiable) function. Indeed:

Dt (φ (x+ ct))− cDx (φ (x+ ct)) = cφ′ (x+ ct)− cφ

′ (x+ ct) = 0.

Similarly, the general solution of ut + cux = 0 is ψ (x− ct), where ψ is an arbitraryfunction. . Hence a solution of utt− cuxx = 0 is

u(x, t) = φ (x+ ct)+ψ (x− ct) (15.27)

which can be verified by computing the second derivatives. The solution (15.27) iscalled the d’Alembert solution of the wave PDE.

15.1.12 Consider the infinite string with zero initial velocity. It is described by

−∞ < x < ∞,0 < t < ∞ : utt = c2uxx (15.28)−∞ < x < ∞ : u(x,0) = f (x) (15.29)−∞ < x < ∞ : ut(x,0) = 0. (15.30)

So we must havef (x) = u(x,0) = φ (x)+ψ (x) (15.31)

and0 = ut (x,0) = φ

′ (x)c−ψ′ (x)c⇒ K = φ (x)−ψ (x) . (15.32)

Henceφ (x) =

f (x)+K2

, ψ (x) =f (x)−K

2(15.33)

2These first order PDEs can be solved by the method of characteristics (see Chapter ??).

Ath.

Keha

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kai arau(x, t) =

12

f (x− ct)+12

f (x+ ct) . (15.34)

From the physical point of view, the solution consists of two functions f (x+ ct)and f (x− ct) which are travelling waves: they move in space maintaining theiroriginal shape; f (x−ct) moves towards +∞ and f (x+ ct) moves towards −∞; bothhave propagation speed equal to c. Note that there is no interaction between thetwo waves.

15.1.13 We have previously solved the same problem with separation of variablesblhtwn and obtained

u(x, t) =∫

∞

0(A(b)cos(bx)+B(b)sin(bx))cos(bct)db (15.35)

where

A(b) =1π

∫∞

−∞

f (z)cos(bz)dz, B(b) =1π

∫∞

−∞

f (z)sin(bz)dz. (15.36)

Let us show that (15.35)-(15.36) are equivalent to (15.34). Replacing (15.36) in(15.35), we get

u(x, t) =1π

∫∞

0

[(∫∞

−∞

f (z)cos(bz)dz)

cos(bx)+(∫

∞

−∞

f (z)sin(bz)dz)

sin(bx)]

cos(bct)db

=1π

∫∞

0

[∫∞

−∞

( f (z) [cos(bz)cos(bx)+ sin(bz)sin(bx)]cos(bct))dz]

db

=1π

∫∞

0

[∫∞

−∞

f (z)cos [b(z− x)]cos(bct)dz]

db

=1

2π

∫∞

0

[∫∞

−∞

f (z)(cos [b(z+ ct− x)]+ cos [b(z− ct− x)])dz]

db

=1

2π

∫∞

0

[∫∞

−∞

f (z)cos [b(z+ ct− x)]dz]

db+1

2π

∫∞

0

[∫∞

−∞

f (z)cos [b(z− ct− x)]dz]

db

(15.37)

Using the representation

f (x) =1π

∫∞

0

[∫∞

−∞

f (z)cos [b(z− x)]dz]

db

we see that (15.37) is exactly equal to 12 f (x+ ct)+ 1

2 f (x− ct).

15.1.14 Example. The problem

−∞ < x < ∞, 0 < t < ∞ : utt = c2uxx (15.38)

−∞ < x < ∞ : u(x,0) = e−x sin(x) (15.39)−∞ < x < ∞ : ut(x,0) = 0 (15.40)

has the solution

u(x, t) =12

e−(x−ct) sin(x− ct)+12

e−(x+ct) sin(x+ ct)

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15.1.15 A somewhat harder problem is

−∞ < x < ∞, 0 < t < ∞ : utt = c2uxx (15.41)−∞ < x < ∞ : u(x,0) = f (x) (15.42)−∞ < x < ∞ : ut(x,0) = g(x) . (15.43)

Now (15.32) becomes

g(x) = ut (x,0) = φ′ (x) · c−ψ

′ (x) · c⇒ 1c

∫ x

x0

g(z)dz+K = φ (x)−ψ (x) (15.44)

where x0 and K are arbitrary. Solving (15.31) and (15.44) with respect to φ and ψ

we get

φ (x) =12

f (x)+12c

∫ x

x0

g(z)dz+K2

(15.45)

ψ (x) =12

f (x)− 12c

∫ x

x0

g(z)dz− K2. (15.46)

Hence

u(x, t) =12

f (x− ct)+12

f (x+ ct)+12c

∫ x+ct

x0

g(z)dz− 12c

∫ x−ct

x0

g(z)dz (15.47)

and we finally get the general solution of the wave equation for an infinite string.

u(x, t) =12

f (x− ct)+12

f (x+ ct)+12c

∫ x+ct

x−ctg(z)dz. (15.48)

15.1.16 Example. The problem

−∞ < x < ∞, 0 < t < ∞ : utt = c2uxx (15.49)−∞ < x < ∞ : u(x,0) = 0 (15.50)−∞ < x < ∞ : ut(x,0) = sin(x) (15.51)

has the solution

u(x, t) =12c

∫ x+ct

x−ctsin(z)dz =

12c

(cos(x− ct)− cos(x+ ct)) .

15.1.17 In the remainder of this scetion we present some examples of solving thePDE using integral transforms.

15.1.18 To solve

x > 0, t > 0 : utt = c2uxx

x > 0 : u(x,0) = 0x > 0 : ut (x,0) = 0t > 0 : u(0, t) = Asin(wt)x > 0, t > 0 : |u(x, t)|< M

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we apply Laplace transform (with respect to t) and get

s2U (x,s)− su(x,0)−ut (x,0) = c2Uxx (x,s) =⇒

Uxx (x,s) =s2

c2U (x,s) =⇒U (x,s) = c1esxc + c2e−

sxc

andu(0, t) = Asin(wt) =⇒U (0,s) =

Aws2 +w2 .

From boundedness we reject the esxa part of the solution and get

U (x,s) = c2e−sxa .

Then

Aws2 +w2 =U (0,s) = c2e−

s0c = c2

U (x,s) =Aw

s2 +w2 e−sxc

So finally (since L −1(

Aws2+w2

)= sin(wt)) we get

u(x, t) = sin[w(

t− xc

)]Heaviside

(t− x

c

)15.1.19 To solve

0 < x < 1, t > 0 : utt = uxx + sin(πx)0 < x < 1 : u(x,0) = ut (x,0) = 0t > 0 : u(0, t) = u(1, t) = 0


s2U (x,s)− su(x,0)−ut (x,0) =Uxx (x,s)+sin(πx)

s=⇒

Uxx (x,s)− s2U (x,s) =−sin(πx)s

U (0,s) =U (L,s) = 0.

We find a particular solution for the inhomogeneous equation:

U = Acosπx+Bsinπx

Ux =−πAsinπx+Bπ cosπx

Uxx =−π2Acosπx−Bπ

2 sinπx.

So we must have

−(π

2A+ s2A)

cosπx−(π

2B+ s2B)

sinπx =−sin(πx)s

⇒ A = 0,B =1

(s2 +π2)s

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andU (x,s) =

1(s2 +π2)s

sin(πx) .

Hence the general solution is

U (x,s) = c1esx + c2e−sx +1

(s2 +π2)ssin(πx)

From the boundary conditions we get

0 =U (0,s) = A1 +A2

0 =U (L,s) = A1esL +A2e−sL

and since ∣∣∣∣ 1 1esL e−sL

∣∣∣∣= e−sL− esL 6= 0,

we get A1 = 0,A2 = 0. Hence the transformed solution is

U (x,s) =1

(s2 +π2)ssin(πx) .

Finally

L −1(

1(s2 +π2)s

)=

1π2 −

1π2 cosπt

andu(x, t) =

(1− cosπt)sin(πx)π2

15.1.20 To solve (Churchill OM p.130, Sec 42)

0 < x, t > 0 : utt = uxx−g0 < x : u(x,0) = ut (x,0) = 0 (15.52)t > 0 : u(0, t) = lim

x→∞ux (x, t) = 0


s2U (x,s)− su(x,0)−ut (x,0) =Uxx−gs=⇒

Uxx (x,s)− s2U (x,s) =gs

U (0,s) = limx→∞

Ux (x,s) = 0.

A particular solution is:U (x,s) =−gsk

so0+gsk+2 =

gs⇒ k =−3⇒ U (x,s) =− g

s3 .


U (x,s) = c1esx + c2e−sx− gs3 .

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0 =U (0,s) = c1 + c2−gs3

0 = limx→∞

Ux (x,s) = limx→∞

(c1sesx− c2se−sx) .

Then c1 = 0 and c2 =gs3 and the transformed solution is

U (x,s) =gs3

(e−sx−1

).

Finally

u(x, t) = L −1( g

s3

(e−sx−1

))=

g2

(Heaviside(t− x)(t− x)2− t2

)15.1.21 To solve (Schaum Laplace, p.225, Ex. 8.6)

0 < x < L, t > 0 : utt = c2uxx

x > 0 : u(x,0) = 0x > 0 : ut (x,0) = 0t > 0 : u(0, t) = 0t > 0 : ux (L, t) = F


s2U (x,s)− su(x,0)−ut (x,0) = c2Uxx =⇒

Uxx−s2

c2U = 0

U (0,s) = 0,Ux (L,s) =Fs.

The general solution is

U (x,s) = c1 coshsxc+ c2 sinh

sxc.


0 =U (0,s) = c1⇒ c1 = 0

andFs=Ux (L,s) = c2

sc

coshsLc⇒ c2 =

Fcs2 cosh sL

c.

Hence

U (x,s) =Fcsinh sx

c

s2 cosh sLc

To invert we use tables. We get

u(x, t) = F

[x+

8Lπ2

∞

∑n=1

(−1)n

(2n−1)2 sin((2n−1)πx

2L

)cos((2n−1)πct

2L

)]

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15.1.22 Finally let us solve the wave equation on rectangle (SchFourier 45, 2.33):

utt = c2 (uxx +uyy)

u(0,y, t) = u(1,y, t) = u(x,0, t) = u(x,1, t) = 0u(x,y,0) = F (x,y)

ut (x,y,0) = 0|u(x,y)|< M

We assumeu(x,y, t) = X (x)Y (y)T (t)

and then we getXY T ′ = c2 (X ′′Y T +XY ′′T

)or, dividing by κXY T

T ′′

c2T=

X ′′

X+

Y ′′

YSo we can set

T ′′

c2T=−λ

2 =

(X ′′

X+

Y ′′

Y

)to get

T ′′ = c2λ

2T

−λ2 =

X ′′

X+

Y ′′

Y

The first equation has solution T = pλ cos(λ t)+qλ sin(λ t). The second equationcan be written as

X ′′

X=−Y ′′

Y−λ

2

orX ′′

X=−µ

2 =−Y ′′

Y−λ

2

which gives

X ′′+µ2X = 0

Y ′′+(λ

2−µ2)Y = 0

These equations have solutions

X (x) = aµ cos(µx)+bµ sin(µx)

Y (x) = cµ,λ cos(√

λ 2−µ2x)+dµ,λ sin

(√λ 2−µ2x

)Using the initial and boundary conditions we get

1. From ut (x,y,0) = 0 we get qλ = 0.2. From u(0,y, t) = 0 we get bµ = 0. From u(1,y, t) = 0 we get µ = mπ.3. From u(x,0, t) = 0 we get dµ = 0. From u(x,1, t) = 0 we get

√λ 2−µ2 = nπ.

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So finally the solution is

u(x,y, t) =∞

∑m=1

∞

∑n=1

Bm,n sin(mπx)sin(nπx)cos(

c√

m2 +n2πt)

Letting t = 0 we get

F (x,y) = u(x,y,0) =∞

∑m=1

∞

∑n=1

Bm,n sin(mπx)sin(mπx)

and so

Bm,n = 4∫ 1

0

∫ 1

0F (x,y)sin(mπx)sin(mπx)dxdy


For 0 < x < 1,0 < t : utt = c2uxx,

For 0 < t : u(0, t) = u(1, t) = 0,For 0 < x < 1 : u(x,0) = 1,For 0 < x < 1 : ut (x,0) = 0.

Solution. We have

u(x, t) =∞

∑n=1

An cos(nπct) · sin(nπx)

where

An = 2∫ 1

01 · sin(nπx)dx = 2 · 1− cosnπ

nπ.

In other words

u(x, t) =∞

∑n=1

2 · 1− cosnπ

nπcos(nπct) · sin(nπx)

=4π

(cos(πct) · sin(πx)+

13

cos(3πct) · sin(3πx)+15

cos(5πct) · sin(5πx)+ ...

)15.2.2 Problem. Solve

∀x ∈ (0,π) , t > 0 :∂ 2u∂ t2 = 5

∂ 2u∂x2

∀t > 0 : u(0, t) = u(π, t) = 0∀x ∈ (0,π) : u(x,0) = xsinx∀x ∈ (0,π) : ut (x,0) = 0.

Solution. Applying the previous results we have 2π

∫π

0 xsin2 xdx = 12π

Am =2π

∫π

0xsin2 xdx =

π

2

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and for m > 1

Am =2π

∫π

0xsinxsin(mx)dx =

4m

(m2−1)2π

((−1)m+1−1

).

Hence the solution is

u(x, t) =π

2sinxcos

(√5t)+

∞

∑m=1

4m

(m2−1)2π

((−1)m+1−1

)sin(mx)cos

(m√

5t)

=π

2sinxcos

(√5t)− 16

9sin(2x)cos

(2√

5t)− 32

225sin(4x)cos

(4√

5t)− ...


For 0 < x < 1,0 < t : utt = c2uxx,

For 0 < t : u(0, t) = u(1, t) = 0,

For 0 < x < 1 : u(x,0) = h(x) =

2x 0≤ x≤ 12

2−2x 12 < x≤ 1

,

For 0 < x < 1 : ut (x,0) = 0.

Solution. We have

u(x, t) =∞

∑n=1

An cos(nπct) · sin(nπx)

where

An = 2∫ 1/2

02xsin(nπx)dx+2

∫ 1

1/2(2−2x)sin(nπx)dx

=−2−2sin 1

2nπ +nπ cos 12nπ

n2π2 +2−2sinnπ +nπ cos 1

2nπ +2sin 12nπ

n2π2 =8sin nπ

2n2π2 .

So finally

u(x, t) =∞

∑n=1

8sin nπ

2n2π2 cos(nπct) · sin(nπx)


For −∞ < x < ∞, 0 < t : utt = c2uxx, (15.53)For −∞ < x < ∞ : u(x,0) = Π(x), (15.54)For −∞ < x < ∞ : ut(x,0) = 0, (15.55)

Solution. We have

u(x, t) =∫

∞

0(A(b)cos(bx)+B(b)sin(bx))cos(bct)db (15.56)

where

A(b) =1π

∫∞

−∞

Π(z)cos(bz)dz =1π

∫ 1

−1cos(bz)dz =

2sin(b)πb

(15.57)

B(b) =1π

∫∞

−∞

Π(z)sin(bz)dz = 0. (15.58)

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Since

2sin(b)πb

= A(b) =1π

∫∞

−∞

Π(z)cos(bz)dz,B(b) = 0⇒

Π(x) =∫

∞

0A(b)cos(bx)db =

∫∞

0

2sin(b)πb

cos(bx)db

we have

u(x, t) =∫

∞

0

2sin(b)πb

cos(bx)cos(bct)db (15.59)

=∫

∞

0

2sin(b)πb

12(cos [b(x− ct)]+ cos [b(x+ ct)])db (15.60)

=12

∫∞

0

2sin(b)πb

cos [b(x− ct)]db+12

∫∞

0

2sin(b)πb

cos [b(x+ ct)]db (15.61)

=12

Π(x− ct)+12

Π(x+ ct) (15.62)


For −∞ < x < ∞ : 0 < t < ∞ : utt = c2uxx (15.63)

For −∞ < x < ∞ : u(x,0) = e−x2sin(x) , (15.64)

For −∞ < x < ∞ : ut(x,0) = 0 (15.65)

Solution. The problem has the solution

u(x, t) =12

e−(x−ct)2sin(x− ct)+

12

e−(x+ct)2sin(x+ ct)


For −∞ < x < ∞ : utt = c2uxx, (15.66)For −∞ < x < ∞ : u(x,0) = 0, (15.67)For −∞ < x < ∞ : ut(x,0) = sinx. (15.68)

Solution.The problem has the solution

u(x, t) =12c

∫ x+ct

x−ctsin(z)dz =

12c

(cos(x− ct)− cos(x+ ct)) .

15.2.7 Problem. Solve (Schaum Laplace, p.226, Ex. 8.5)

0 < x < L, t > 0 : utt = a2uxx

x > 0 : u(x,0) = mx(L− x)x > 0 : ut (x,0) = 0t > 0 : u(0, t) = 0t > 0 : u(L, t) = 0

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Solution. We apply Laplace transform (with respect to t) and get

s2U− su(x,0)−ut (x,0) = c2Uxx =⇒

Uxx−s2

c2U =−msx(L− x)c2

U (0,s) =U (L,s) = 0

the general solution is

U = A1 coshsxa+A2 sinh

sxa+

mx(L− x)s

− 2c2ms3 .


U (x,s) =2c2m

s3

cosh[

s(2x−L)2c

]cosh

[ sL2c

] +mx(L− x)

s− 2c2m

s3

To invert (the cosh part) we use tables. We get

u(x, t)=8mL2

π3

∞

∑n=1

1

(2n−1)3 sin((2n−1)πx

L

)cos((2n−1)πct

L

)+mx(L− x)−c2mt2


1. Solve utt = c2uxx (0< x< 1, 0< t), u(0, t)= u(1, t)= 0 (0< t), u(x,0)=

1 0≤ x≤ 12

−1 12 < x≤ 1

,

ut (x,0) = 0.Ans. u(x, t)= 4

π(sin(2πx)sin(2πt)+ 1

3 sin(6πx)sin(6πt)+ 15 sin(10πx)sin(10πt)+...

.2. Solve utt = c2uxx (0 < x < 1, 0 < t), u(0, t) = u(1, t) = 0 (0 < t), u(x,0) = x,

ut (x,0) = 0.

Ans. u(x, t) = 2π ∑

∞n=1

(−1)n+1

n sin(nπx)cos(nπct).3. Solve utt = c2uxx (0 < x < 1, 0 < t), u(0, t) = u(1, t) = 0 (0 < t), u(x,0) = 0,

ut (x,0) = 1.Ans. u(x, t) = ∑

∞n=1

2−2cos(nπ)n2π2c sin(nπx)sin(nπct).

4. Solve utt = c2uxx (0 < x < 1, 0 < t), u(0, t) = u(1, t) = 0 (0 < t), u(x,0) = 1,ut (x,0) = 1.Ans. u(x, t)=∑n=1,3,5,...

4nπ

sin(nπx)cos(nπct)+ ∑n=1,3,5,...( 2

nπ

)2sin(nπx)sin(nπct).

5. Solve utt = c2uxx (0 < x < 1, 0 < t), u(0, t) = u(1, t) = 0 (0 < t), u(x,0) = x ·(1− x),ut(x,0) = 0.Ans. u(x, t) = ∑

∞n=1

4n3π3 (1− (−1)n)sin(nπx)cos(nπct).

6. Solve utt = c2uxx (0 < x < π, 0 < t), u(0, t) = u(π, t) = 0 (0 < t), u(x,0) = 0,ut (x,0) = 8sin2 x.Ans. u(x, t) = ∑n=1,3,4,...

32πcn2(n2−4)

((−1)n−1)sin(nx)sin(nct).

7. Solve utt = c2uxx (0 < x < π, 0 < t), u(0, t) = u(π, t) = 0 (0 < t), u(x,0) = 0,ut (x,0) = x2.Ans. u(x, t) = ∑n=1,3,4,...

2πc(n2−4)

(1− (−1)n)sin(nx)sin(nct).

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8. Solve utt = uxx (−∞ < x < ∞, 0 < t < ∞) kai u(x,0) = e−x2, ut (x,0) = 0 (−∞ <

x < ∞).Ans. u(x, t) = 1

2

[e−(x−t)2

+ e−(x+t)2].

9. Solve utt = uxx (−∞ < x < ∞, 0 < t < ∞) kai u(x,0) = 0, ut (x,0) = xe−x2(−∞ <

x < ∞).Ans. u(x, t) = 1

4

[e−(x−t)2

− e−(x+t)2]).

10. Solve utt = uxx (−∞ < x < ∞, 0 < t < ∞), u(x,0) = 0, ut (x,0) = 1 (−∞ < x < ∞).Ans. u(x, t) = t.

11. Solve utt = uxx (−∞< x<∞, 0< t <∞), u(x,0) = sinx, ut (x,0) = x2 (−∞< x<∞).Ans. u(x, t) = sin(x)cos(ct)+ x2t + 1

3c2t3.12. Solve utt = uxx (−∞ < x < ∞, 0 < t < ∞), u(x,0) = x3, ut (x,0) = x (−∞ < x < ∞).

Ans. u(x, t) = x3 +3c2xt2 + xt.13. Solve utt = uxx (−∞ < x < ∞, 0 < t < ∞), u(x,0) = cosx, ut (x,0) = 1/e (−∞ <

x < ∞).Ans. u(x, t) = cos(x)cos(ct)+ t

e .14. Solve utt = uxx (−∞ < x < ∞, 0 < t < ∞), u(x,0) = log 1

1+x2 , ut (x,0) = 2 (−∞ <x < ∞).Ans. u(x, t) = 2t+ log

(√1+ x2 +2cxt + c2t2

)+ log

(√1+ x2−2cxt + c2t2

).

15. Solve utt = uxx (0 < x < ∞, 0 < t < ∞), u(x,0) = xe−x2, ut (x,0) = 0 (−∞ < x < ∞)

u(0, t) = 0 (0 < t < ∞).


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16. Bessel Functions and Applications

Bessel functions can be seen as the analog of sines and cosines for problems incylindrical coordinates. Much like sines and cosines, they emerge as solutions ofan ODE, in this case Bessel’s equation.

16.1 Theory and Examples16.1.1 In Chapter 5 we have studied the ODE

x2 d2ydx2 + x

dydx

+(x2−ν

2)y = 0,n≥ 0. (16.1)

This is Bessel’s equation and its solutions, the Bessel functions, have manyapplications in PDE problems. In this chapter we will study Bessel functions ingreater detail.

16.1.2 As we will see, the general solution of (16.1) has the form

y = c1Jν (x)+ c2Yν (x) (16.2)

where Jν (x) (resp. Yν (x)) are called Bessel function of first (resp. second) kind andν-th order.

16.1.3 We will determine Jν (x) and Yν (x) in several steps. Let us first continuethe solution of (16.1) from where we stopped in Example 5.1.13. We have alreadyseen that, assuming solutions of the form

y(x) = xr∑ckxk

we get the indicial equationr2 = ν

2

with solutions r1 = ν, r2 = −ν. We have also seen that, with r = ν, we get thesolution

y1 (x) = xν

(1− x2

2(2+2ν)+

x4

2 ·4 · (2+2ν) · (4+2ν)+ ...

). (16.3)

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190 Chapter 16. Bessel Functions and Applications

Similarly, when r =−ν, we get the solution

y2 (x) = x−ν

(1− x2

2(2−2ν)+

x4

2 ·4 · (2−2ν) · (4−2ν)+ ...

). (16.4)

When ν = 0 the two series are identical; when ν ∈ 1,2,3, ... the second seriesdoes not exist; when ν /∈ 0,1,2,3, ... the two series are linearly independent.

16.1.4 Definition. The Bessel function of the first kind and ν-th order is definedto be

Jν (x) :=xν

2νΓ(ν +1)

(1− x2

2(2+2ν)+

x4

2 ·4 · (2+2ν) · (4+2ν)+ ...

)(16.5)

which can be rewritten as

Jν (x) =∞

∑n=0

(−1)n ( x2

)ν+2n

n!Γ(ν +n−1). (16.6)

In the following plot we see J0 (x), J1 (x), J2 (x).

PlotBessel

16.1.5 Clearly, (16.5) is a rescaled version of the solution y1 (x) of (16.3). Thefunction Γ(x) appearing in (16.5)-(16.6) is the Gamma function, the generalizationof the factorial function to the real numbers (see Appendix C); in particular:∀n ∈N : Γ(n+1) = n!. The generalization is required because n can be non-integer.

16.1.6 Definition. We extend the definition of Jν (x) to values ν ∈ −1,−2,−3, ...by replacing ν ∈ 1,2,3, ... with −ν in (16.5).

16.1.7 Theorem. The following hold

∀ν ∈ Z : J−ν (x) = (−1)ν Jν (x) , (16.7)∀ν /∈ Z : J−ν (x) and Jν (x) are linearly independent. (16.8)

In the case ∀ν /∈ Z, the general solution of (16.1) is

c1Jν (x)+ c2J−ν (x) .

Proof. For (16.7) we have

Jν (x) =∞

∑n=0

(−1)n ( x2

)ν+2n

n!Γ(ν +n−1)

J−ν (x) =∞

∑n=0

(−1)n ( x2

)−ν+2n

n!Γ(−ν +n−1)

=ν−1

∑n=0

(−1)n ( x2

)−ν+2n

n!Γ(−ν +n−1)+

∞

∑n=ν

(−1)n ( x2

)−ν+2n

n!Γ(−ν +n−1)

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It is a known property of the Gamma function that Γ(−ν +n−1) = ∞ for n =0,1, ...,ν−1. Hence the first sum is 0. In the second sum we place n = ν +k and itbecomes

J−ν (x) =∞

∑k=0

(−1)ν+k ( x2

)ν+2k

(ν + k)!Γ(k+1)= (−1)ν

∞

∑k=0

(−1)k ( x2

)ν+2k

k!Γ(ν + k+1)= (−1)ν Jν (x)

So for ν ∈ 1,2,3, ... Jν (x) and J−ν (x) are linearly dependent.We omit the proof of (16.8); it is straightforward but tedious.For the last part of the theorem, when ν /∈ Z we have that J−ν (x) and Jν (x) are

linearly independent. So we have two linearly indeendent solutions (16.1), whichis a second order linear equation. By the results of Chapter 1, J−ν (x) ,Jν (x) is abasis of the space of solutions of (16.1).

16.1.8 Definition. The Bessel function of the 2nd kind and ν-th order is definedto be

∀ν /∈ Z : Yν (x) =Jν (x)cos(νπ)− J−ν (x)

sin(νπ),

∀ν ∈ Z : Yν (x) = limµ→ν

Jµ (x)cos(µπ)− J−µ (x)sin(µπ)

.

16.1.9 Theorem. For every ν ∈ R : limx→0Yν (x) =−∞.

16.1.10 Theorem. The ODE

x2 d2ydx2 + x

dydx

+(x2−ν

2)y = 0, ν ≥ 0

has general solutiony = c1Jν (x)+ c2Yν (x)

Proof. Omitted.

16.1.11 In what follows we will concentrate on the Bessel functions of the firstkind, which are the ones most often appearing in applications. These functionshave a large number of properties (many of them useful in applications). We willonly prove only a few useful identities; the reader can find more in the Problemssections.

16.1.12 Theorem. We have

∀ν ∈ R :ddx

(xνJν (x)) = xνJν−1 (x)

Proof. We have

ddx

(xνJn (x)) =ddx

(∞

∑n=0

(−1)n x2ν+2n

22ν+2nn!Γ(ν +n−1)

)=

∞

∑n=0

(−1)n x2ν+2n−1

2ν+2n−1n!Γ(ν +n)

= xν∞

∑r=0

(−1)n x(ν−1)+2n

2(ν−1)+2nn!Γ((ν−1)+n+1)= xνJν−1 (x)

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16.1.13 Theorem. We have

J1/2 (x) =

√2

πxsinx, (16.9)

J−1/2 (x) =

√2

πxcosx. (16.10)

Proof. We only prove (16.9). We have

J1/2 (x) =∞

∑n=0

(−1)n ( x2

)2n+ 12

n!Γ(n+ 3

2

)=

( x2

) 12

Γ(3

2

) − ( x2

) 52

1!Γ(5

2

) + ( x2

) 92

2!Γ(9

2

) − ...

=

( x2

) 12

√π

2

−( x

2

) 52

1!32

√π

2

+

( x2

) 92

2!32

52

√π

2

− ...

=

( x2

) 12

√π

2

(1− x2

3!+

x4

5!− ...

)=

( x2

) 12

√π

2

sinxx

.

16.1.14 Theorem. For large x we have the asymptotic formulas

Jν (x)∼√

2πx

cos

(x−

(n+ 1

2

)π

2

),

J−1/2 (x)∼√

2πx

sin

(x−

(n+ 1

2

)π

2

).

Proof. Omitted.

16.1.15 Theorem. When λ 6= µ, we have

∫ 1

0xJν (λx)Jν (µx)dx =

µJν (λ )J′ν (µ)−λJν (µ)J′ν (λ )λ 2−µ2

Proof. It is easy to check that, by a change of variables y1 (x) = Jν (λx) is a solutionof

x2 d2y1

dx2 + xdy1

dx+(λ

2x2−ν2)y1 = 0 (16.11)

and y2 (x) = Jν (µx) is a solution of

x2 d2y2

dx2 + xdy2

dx+(µ

2x2−ν2)y2 = 0. (16.12)

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Multiplying (16.11) by y2 and (16.12) by y1 and subtracting we get

x2(

y2d2y1

dx2 − y1d2y2

dx2

)+ x(

y2dy1

dx− y1

dy2

dx

)=(µ

2−λ2)x2y1y2⇒

xddx

(y2

dy1

dx− y1

dy2

dx

)+

(y2

dy1

dx− y1

dy2

dx

)=(µ

2−λ2)xy1y2⇒

ddx

(x(

y2dy1

dx− y1

dy2

dx

))=(µ

2−λ2)xy1y2⇒

x(

y2dy1dx − y1

dy2dx

)µ2−λ 2 =

∫xy1y2dx⇒

x(λJν (µx)J′ν (λx)−µJν (λx)J′ν (µx))µ2−λ 2 =

∫xJν (λx)Jν (µx)dx⇒

(λJn (µ)J′ν (λ )−µJν (λ )J′ν (µ))µ2−λ 2 =

∫ 1

0xJν (λx)Jν (µx)dx

16.1.16 Theorem (Orthogonality). Let c1,c2 be constants such that c1c2 6= 0,and let λ ,µ be two different roots of

c1Jν (x)+ c2xJ′ν (x) = 0. (16.13)

Then we have ∫ 1

0xJν (λx)Jν (µx)dx = 0.

Hence, for any λ ,µ as above, the set √

xJν (λx) ,√

xJν (µx) is orthogonal in (0,1).Proof. This theorem follows from more general results of Chapter 17. However wecan present and independent proof here. Since λ ,µ are roots of (16.13), we have

c1Jν (λ )+ c2λJ′ν (λ ) = 0 (16.14)c1Jν (µ)+ c2µJ′ν (µ) = 0 (16.15)

For this system in c1, c2 to have a nonzero solution, we must have∣∣∣∣ Jν (λ ) λJ′ν (λ )Jν (µ) µJ′ν (µ)

∣∣∣∣= Jν (λ )µJ′ν (µ)− Jν (µ)λJ′ν (λ ) = 0.

Since we also know that∫ 1

0xJν (λx)Jν (µx)dx =

(λJν (µ)J′ν (λ )−µJν (λ )J′ν (µ))µ2−λ 2 ,

we immediately get the required result.

16.1.17 Theorem. For any ν ∈ R, Jν (x) has an infinite number of distinct roots

0 < λν ,1 < λν ,2 < λν ,3....

We also have∀n : λν−1,1 < λν ,1 < λν+1,1.

Proof. It follows from more general results which will be presented in Chapter 17.

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16.1.18 Theorem (Bessel-Fourier Series). For some n ∈ R let 0 < λν ,1 < λν ,2 <λν ,3... be the roots of Jν (x). Furthermore, let f (x) be a function which is piecewisedifferentiable on [0,1].

1. If f (x) is continuous at x ∈ [0,1] then

f (x) = limN→∞

N

∑n=1

cnJν (λν ,nx) . (16.16)

2. If f (x) is discontinuous at x ∈ [0,1] then

12

(lim

ξ→x−f (ξ )+ lim

ξ→x+f (ξ )

)= lim

N→∞

N

∑n=1

cnJν (λν ,nx) (16.17)

(with the obvious modification if x ∈ 0,1).In both (16.16) and (16.17), we have

∀n ∈ N : cn =2(

Jν+1(λν ,k

))2

∫ 1

0x f (x)Jν

(λν ,kx

)dx. (16.18)

Proof. This also follows from results presented in Chapter 5.

16.1.19 The above theorems remind us of Fourier series expansions, with thecn’s of (16.18) playing the role of Fourier coefficients. In fact, trigonometric seriesand series of Bessel functions are just two special cases of a more general theoryof ortogonal functions. As already hinted, the related results will be presented inChapter 16.

16.1.20 We now present some examples of the application of Bessel functions toPDE problems.

16.1.21 Example. Let us solve the heat equation on a disk:

0 < ρ < 1,0 < t : ut = κ

(uρρ +

1ρ

uρ

)0 < t : u(1, t) = 0,0 < ρ < 1 : u(ρ,0) = F (ρ) ,

0 < ρ < 1,0 < t : |u(ρ, t)|< M

Lettingu(ρ, t) = P(ρ)T (t)

we get

P(ρ)T ′ (t) = κ

(P′′ (ρ)T +

1ρ

P′ (ρ)T (t))⇒

T ′

κT=

P′′

P+

1ρ

PP

′=−λ

2

So we have

T ′ =−κλ2T

P′′+1ρ

P′+λ2P = 0 (16.19)

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We know the first ODE has solution T = e−κλ 2t . Now we want solutions of thesecond ODE, which can be rewritten as

ρ2P′′+ρP′+λ

2ρ

2P = 0. (16.20)

We recognize the ODE (16.20) as Bessel’s equation of order ν = 0 ; hence it hassolutions

P = Aλ J0 (λρ)+BλY0 (λρ)

where J0 and Y0 are the Bessel functions (of zero order) of the first and secondkind respectively. Since we want u(0, t) = P(0)T (t) bounded, we see that Bλ = 0(because limρ→0Y0 (λρ) = ∞). Since u(1, t) = 0 we get

0 = P(1) = Aλ J0 (λ )

so λ ∈ λ1,λ2, ..., the set of the Bessel function J0 (λ ) roots. Thus a solution of(16.19) is

AmJ0 (λmρ)e−λ 2mt

and the general solution is

u(ρ, t) =∞

∑m=1

AmJ0 (λmρ)e−λ 2mt .

Now we choose the Am so that

F (ρ) = u(ρ,0) =∞

∑m=1

AmJ0 (λmρ)

In other words, we want a Bessel series expansion of F (ρ). That such an expansionis possible follows from Theorem 16.1.18.

16.1.22 Let us solve the heat equation on a cylinder:

0 < ρ < 1,0 < z < 1,0 < t : ut = κ

(uρρ +

1ρ

uρ +uzz

)0 < ρ < 1,0 < z < 1 : u(ρ,z,0) = F (ρ,z) ,0 < ρ < 1,0 < t : u(ρ,0, t) = 0,0 < ρ < 1,0 < t : u(ρ,1, t) = 0,0 < z < 1,0 < t : u(1,z, t) = 0,0 < ρ < 1,0 < z < 1,0 < t : |u(ρ,z, t)|< M.

Lettingu(ρ, t) = P(ρ)Z (z)T (t)

we get

PZT ′ = κ

(P′′ZT +

1ρ

P′ZTt +PZ′′T)⇒

T ′

κT=

P′′

P+

1ρ

PP

′+

Z′

Z=−λ

2

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So we haveT ′ =−κλ

2T ⇒ T (t) = c1e−κλ 2t

Next we write the second as

P′′

P+

1ρ

PP

′=−λ

2− Z′

Z=−µ

2

Now we have

P′′

P+

1ρ

PP

′=−µ

2⇒ P(ρ) = c2J0 (µρ)+ c3Y0 (µρ)

andZ′

Z= µ

2−λ2 = ν

2⇒ Z (z) = c4eνz + c5e−νz

From |u(ρ,z, t)|< M we get c3 = 0. So a solution is

u(ρ,z, t) = J0 (µρ)(Aνeνz +Bνe−νz)e−κλ 2t

Now from0 = u(ρ,0, t) = J0 (µρ)(Aν +Bν)e−κλ 2t .

So Bν =−Aν andu(ρ,z, t) = AνJ0 (µρ)

(eνz− e−νz)e−κλ 2t

Next from0 = u(ρ,1, t) = AνJ0 (µρ)

(eν − e−ν

)e−κλ 2t

we get eν − e−ν = 0 which means e2ν = 1 and ν = kπi. So the solution becomes

u(ρ,z, t) = AνJ0 (µρ)sin(kπz)e−κλ 2t

From u(1,z, t) = 0 we get

0 = AνJ0 (µ)sin(kπz)e−κλ 2t

so µ ∈ µ1,µ2, ..., the set of the roots of J0 (µ) = 0. Furthermore

λ2 = µ

2−ν2 = µ

2 + k2π

2

So the solution becomes

u(ρ,z, t) = ∑k

∑m

Am,kJ0 (µmρ)sin(kπz)e−κ(µ2m+k2π2)t

Now

F (ρ,z) = u(ρ,z,0) = ∑k

(∑m

Am,kJ0 (µmρ)

)sin(kπz) = ∑

kBk (ρ)sin(kπz)

So, taking ρ as constant we get

Bk (ρ) =21

∫ 1

0F (ρ,z)sin(kπz)dz

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And then

Am,k =2

J1 (µm)2

∫ 1

0ρBk (ρ)J0 (µmρ)dρ

or

Am,k =4

J1 (µm)2

∫ 1

0

∫ 1

0ρF (ρ,z)J0 (µmρ)sin(kπz)dρdz

16.1.23 Let us solve the wave equation on a circular drum:

0 < ρ < 1,0 < φ < 2π,0 < t : utt = c2(

uρρ +1ρ

uρ +1

ρ2 uφφ

)0 < φ < 2π,0 < t : u(1,φ , t) = 0,0 < ρ < 1,0 < φ < 2π,0 < t : ut (ρ,φ ,0) = 0,0 < ρ < 1,0 < φ < 2π,0 < t : u(ρ,φ ,0) = F (ρ,φ)

Lettingu(ρ,φ , t) = P(ρ)Φ(φ)T (t)

we get

PΦT ′′ = c2(

P′′ΦT +1ρ

P′ΦT +1

ρ2 PΦφφ T)⇒

T ′′

c2T=

P′′

P+

1ρ

PP

′+

1ρ2

Φ′′

Φ=−λ

2

With the usual methods we get

T ′′+λ2c2T = 0

Φ′′+µ

2Φ = 0

ρ2R′′+ρP′+

(λ

2ρ

2−µ2)P = 0

The general solutions are

T (t) = A1 cos(λct)+B1 sin(λct)Φ(φ) = A2 cos(µφ)+B2 sin(µφ)

P(ρ) = A3Jµ (λρ)+B3Yµ (λρ)

Using the boundary conditions we get

B1 = 0µ = m

B3 = 0

So the solution becomes

u(ρ,φ , t) = Jm (λρ)(Acos(mφ)+Bsin(mφ))cos(λct)

From0 = u(1,φ , t) = Jm (λ )(Acos(mφ)+Bsin(mφ))

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we get λ ∈ λmk where for every m, λmk is the k-th root of Jm (λ ). Then the generalsolution is

u(ρ,φ , t) =∞

∑m=0

∞

∑k=1

Jm (λmkρ)(Amk cos(mφ)+Bmk sin(mφ))cos(λmkct) .

Using the same approach as before we get

Cm (ρ) =1π

∫ 2π

0F (ρ,φ)cos(mφ)dφ

Dm (ρ) =1π

∫ 2π

0F (ρ,φ)sin(mφ)dφ

and

Amk =2

Jm+1 (λmk)2

∫ 1

0ρJm (λmkρ)Cm (ρ)dρ

Bmk =2

Jm+1 (λmk)2

∫ 1

0ρJm (λmkρ)Dm (ρ)dρ


16.3 Unsolved Problems1. Prove that J′0 (x) =−J1 (x).2. Prove that J′ν (x) =

12 (Jν−1 (x)− Jν+1 (x)) .

3. Prove that J′′ν (x) =

14 (Jν−2 (x)−2Jν (x)+ Jν+2 (x)).

4. Evaluate∫

x3J2 (x)dx.Ans. x3J3 (x)+ c.

5. Evaluate∫ 1

0 x3J0 (x)dx.Ans. 2J0 (1)−3J1 (1) .

6. Evaluate∫

x2J0 (x)dx.Ans. x2J1 (x)+ xJ0 (x)−

∫J0 (x)dx.

7. Evaluate∫

J1 ( 3√

x)dx.Ans. 6 3

√xJ1 ( 3√

x).8. Evaluate

∫ J2(x)x2 dx.

Ans. −J2(x)3x −

J1(x)3 + 1

3∫

J0 (x)dx.9. Evaluate

∫J0 (x)sinxdx.

Ans. xJ0 (x)sinx− xJ1 (x)cosx = c.10. Solve: ut = k

(uρρ +

1ρ

uρ

)on an axially symmetric cylinder of infinite length

and radius R, subject to

0 < t : u(R, t) = 0,0 < ρ < R : u(ρ,0) = f (ρ) .

Ans. u(ρ, t)= 2R2 ∑

∞n=1 AnJ0 (λnρ)e−kλ 2

n t , with An =1

(J1(λnR))2

∫ 10 ρ f (ρ)J0 (λnρ)dρ.

11. Prove that J20 (x)+∑

∞k=1 J2

k (x) = 1.

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12. Solve: utt = uρρ +1ρ

uρ +1

ρ2 uφφ on the unit disk with

0 < φ < 2π,0 < t : u(1,φ , t) = 0,0 < φ < 2π,0 < ρ < 1 : u(ρ,φ ,0) = ρ cos(3φ) ,

0 < φ < 2π,0 < ρ < 1 : ut (ρ,φ ,0) = 0.

Ans. u(ρ,φ , t)=∑∞n=1 AnJ3 (λnρ)cos(3φ)cos(λnt), with An =

2((λ 2n−8)J0(λn)−6λnJ1(λn)+8)

λ 3n J2

4 (λn).

16.4 Advanced Problems1. Prove that J0 (x)+2J2 (x)+2J4 (x)+ ...= 1.2. Prove that J1 (x)− J3 (x)+ J5 (x)− J7 (x)+ ...= 1

2 sinx.3. Prove that xJ′ν (x) = νJν (x)− Jν+1 (x) .4. Prove that xJ′ν (x) = xJν−1 (x)−νJν (x) .5. Prove that 1

2Jν+1 (x) = 2vx Jν (x)− Jν−1 (x) .

6. Prove that (xνJν (x))′ = xνJν−1 (x) .

7. Prove that (x−νJν (x))′=−xν−1Jν+1 (x) .

8. Prove that J3/2 (x) =√

2πx

( sinxx − cosx

).

9. Prove that J0 (x)+2∑∞k=1 (−1)k J2k (x) = cosx.

10. Prove that ∑∞k=1 (2k−1)J2k−1 (x) = x

2 .

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17. Vector Spaces of Functions

We have already mentioned that many concepts of Linear Algebra (linear combi-nation, vector space, independence, orthogonality, ...) can be applied to sets offunctions which, under appropriate conditions, can be viewed as vector spaces. Inthis chapter we will revisit this idea and explore it in more depth.


17.1.1 Hint. Whenever a new definition or theorem is introduced, try to find thecorresponding definition or theorem concerning the finite dimensional vector spaceRN (which you have probably seen in Linear Algebra).

17.1.2 Definition. Given a set of functions F, we say that F is a vector space iff

∀ f ,g ∈ F and ∀κ,λ ∈ R : κ f +λg ∈ F.

17.1.3 Definition. Given a vector space F and a vector space G⊆ F, we say thatG is a vector subspace of F.

17.1.4 Example. Let PN be the set of polynomials of degree at most N. PN is avector space. Indeed, if f (t) ,g(t) are polynomials of degree at most N, the sameis true of h(t) = κ f (t)+λg(t). For any m,n ∈ N0, if M ≤ N then PM is a vectorsubspace of PN .

17.1.5 Example. SL, the solution set of L(y) = 0 (where L is a linear differentialoperator) is a vector space, as we have seen in Chapter 2.

17.1.6 Example. The set FL of functions which satisfy the Dirichlet conditionson [−L,L] is a vector space, as we have seen in Chapter 9.

17.1.7 Definition. Let CN (a,b) (with a,b ∈ R) be the set of functions which havecontinuous derivatives of orders 0,1, ...,N on [a,b]. C∞ (a,b) is the set of functionswhich have continuous derivatives of all orders 0,1, ... on [a,b].

17.1.8 Definition. Let CN (a,b) (with a,b ∈ R) be the set of functions which havepiecewise continuous derivatives of orders 0,1, ...,N on [a,b]. C∞ (a,b) is the set

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202 Chapter 17. Vector Spaces of Functions

of functions which have piecewise continuous derivatives of all orders 0,1, ... on[a,b].

17.1.9 Example. It is easily checked that, for any N ∈ N∗, CN (a,b) is a vectorspace and CN (a,b) is a vector subspace of CN (a,b). Also, if M ≥ N, CM (a,b) is avector subspace of CN (a,b).

17.1.10 Definition. Let F be a vector space of functions. We say that ( fn)n∈A ⊆ Fis linearly independent in F iff(

∀x : ∑n∈A

κn fn (x) = 0

)⇒ (∀n ∈ 1,2, ...,N : κn = 0) .

17.1.11 Example. Take P2 and its elements f0 (x) = 1, f1 (x) = x, f2 (x) = x2. Thenthe set

( fn)n∈0,1,2 = ( f0, f1, f2) =(1,x,x2)

is a linearly independent set in P2.

17.1.12 Definition. Let F be a vector space of functions and take ( fn)n∈A ⊆ A⊆N.The span of ( fn)n∈Ais

Span(( fn)n∈A

):=

g : g = ∑

n∈Aκn fn, ∀n : κn ∈ R

.

The definition covers the cases of both finite and countably infinite sequences offunctions.

17.1.13 Example. It is easily checked that, for the space of second orderpolynomials P2, we have

P2 = Span(1,x,x2)= Span

(1+ x,1− x,x2)= Span

(1,4,1− x,3x2) .

17.1.14 Theorem. Let F be a vector space of functions and take ( fn)n∈A ⊆ F.Then Span

(( fn)n∈A

)is a vector space.

Proof. Suppose g,h ∈ Span(( fn)n∈A

)and κ,λ ∈ C. Then

g = ∑n∈A

an fn, h = ∑n∈A

bn fn.

Then

κg+λh = κ ∑n∈A

an fn +λ ∑n∈A

bn fn = ∑n∈A

(κan +λbn) fn ∈ Span(( fn)n∈A

).

17.1.15 Definition. Given a vector space F and some strictly positive functionw(x), we define for all f ,g ∈ F the inner product of f ,g with respect to w by

( f ,g)w :=∫ b

af (x)g(x)w(x)dx.

In case w(x) = 1, we simply write ( f ,g).

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17.1.16 Theorem. For all f ,g,h ∈ F and for all κ,λ ∈ R the following hold.1. ( f ,g) = (g, f ) .2. (κ f +λg,h) = κ ( f ,h)+λ (g,h) .3. ( f , f )≥ 0.4. ( f , f ) = 0 iff f (x) = 0 for almost all x ∈ [a,b].

Proof. The proofs of 1-3 are easy. For example we have

( f ,g) =∫ b

af (x)g(x)dx =

∫ b

ag(x) f (x)dx = (g, f )

and this proves 1; we can similarly prove 2 and 3.Regarding 4, if we have

0 = ( f , f ) =∫ b

a| f (x)|2 dx,

we cannot conclude that f (x) = 0 for all x ∈ [a,b]. For example, consider thefunction

f (x) =

0 x 6= a+b2

1 x = a+b2

.

Then ∫ b

a| f (x)|2 dx = 0.

This happens because the value of the integral does not depend on the value of thefunction at single point. In fact, when the integral is understood in the Lebesguesense, the value of the integral does not change if we change the function values ata countable set of points or even at any point set of measure zero1. However, usingthe theory of Lebesgue integration, we can prove that

∫ ba | f (x)|

2 dx = 0 iff f (x) 6= 0on a set of zero measure, which is exactly the statement of 4.

17.1.17 We can conclude that

( f , f ) = 0⇔ (∀x ∈ [a,b] : f (x) = 0)

if, for example, we assume that f (x) is a continuous function. For this and relatedreasons, in what follows we will deal with the space C2 (a,b) equipped with theinner product (., .). The analysis also holds if we use the weighted inner product(., .)w; we have omitted this aspect in the interest of brevity. Also, much of thefollowing analysis also holds for many other vector spaces of functions.

17.1.18 Definition. The norm of f ∈ C2 (a,b) is defined by

‖ f‖ := ( f , f )1/2 .

17.1.19 Recall that a,b are finite (a,b ∈ R). Hence: ∀ f ∈ C2 (a,b) : ‖ f‖< ∞ (why?).

17.1.20 Theorem. For all f ∈ C2 (a,b) and κ ∈ R, the following hold.1. ‖ f‖ ≥ 0.2. ‖ f‖= 0 iff f (x) = 0(x).1For the exact meanings of these terms see Chapter B.

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3. ‖κ f‖= |κ|‖ f‖.Proof. For 1, we note that

‖ f‖= ( f , f )1/2 =

(∫ b

a| f (x)|2 dx

)≥ 0.

For 2, we have already remarked that, when f is continuous,

‖ f‖2 = ( f , f ) = 0⇒ f (x) = 0(x) .

For 3, we have

‖κ f‖=(κ f ,κ f )1/2 =

(∫ b

a|κ f (x)|2 dx

)1/2

=

(κ

2∫ b

a| f (x)|2 dx

)1/2

= |κ|( f , f )1/2 = |κ|‖ f‖ .

17.1.21 Theorem (Schwartz inequality). For all f ,g ∈ C2 (a,b) we have

|( f ,g)| ≤ ‖ f‖ · ‖g‖ .

Proof. Let∀λ ∈ R : a(λ ) := ( f +λg, f +λg) = ‖ f +λg‖2 ≥ 0.

Now

( f +λg, f +λg) = ( f , f )+λ ( f ,g)+λ (g, f )+λ2 (g,g)

= λ2 (g,g)+2( f ,g)λ +( f , f )

= λ2 ‖g‖2 +2( f ,g)λ +‖ f‖2 .

Since a(λ ) is a trinomial and for all λ ∈ R we have a(λ )≥ 0 (which also holds for‖g‖2) the discriminant of a(λ ) must be negative. Then

(2( f ,g))2−4‖g‖2 ‖ f‖2 ≤ 0⇒ |( f ,g)|2 ≤ ‖g‖2 ‖ f‖2⇒ |( f ,g)| ≤ ‖g‖‖ f‖ .

17.1.22 Theorem (Triangle inequality). For all f ,g ∈ C2 (a,b) we have

‖ f +g‖ ≤ ‖ f‖+‖g‖ .

Proof. Using the a(λ ) of the previous proof, we have

‖ f +g‖2 = a(1) = ‖g‖2 +2( f ,g)+‖ f‖2 ≤ ‖g‖2 +2‖g‖‖ f‖+‖ f‖2

hence‖ f +g‖2 ≤ (‖ f‖+‖g‖)2⇒‖ f +g‖ ≤ ‖ f‖+‖g‖ .

17.1.23 Definition. Given a function g ∈ C2 (a,b) and a sequence of functions(gn)

∞

n=0 ⊆ C2 (a,b), we say that gn converges to g in the mean iff

limn→∞‖g−gn‖= 0

or, more explicitly,

limn→∞

∫ b

a|g(t)−gn (t)|2 dt = 0.

We also use the notationl.i.m.n→∞

gn = g.

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17.1.24 Example. Define

gn (x) :=

1 x ∈[0, 1

n

]0 x ∈ (1

n ,1].

Then limn→∞ ‖0−gn‖= 0, in the sense

limn→∞

∫ 1

0|0(t)−gn (t)|2 dt = 0.

17.1.25 Definition. We say that f ,g ∈ C2 (a,b) are orthogonal (to each other) in[a,b] iff ( f ,g) = 0.

17.1.26 Definition. We say that the set ( fn)n∈A ⊆ C2 (a,b) is orthogonal in [a,b] iff

∀m 6= n : ( fm, fn) = 0.

We say that ( fn)n∈A is orthonormal in [a,b] iff (a) it is orthogonal in [a,b] and(b) ∀n ∈ A : ‖ fn‖= 1.

17.1.27 Example. Since

∀m,n ∈ N :∫ 2π

0sinmxsinnxdx =

0 m 6= nπ m = n ,

it follows that (in the interval [0,2π])1. sinx and sin2x are orthogonal to each other;2. the set (sinnx)∞

n=1 is orthogonal;3. the set

(1√π

sinnx)∞

n=1is orthonormal.

17.1.28 Theorem. If the set ( fn)n∈A is orthogonal then it is linearly independent.Proof. We have

∑n∈A

κn fn = 0⇒∀m ∈ N : fm ∑n∈A

κn fn = 0

⇒∀m ∈ N : ∑n∈A

κn fm fn = 0

⇒∀m ∈ N : ∑n∈A

κn

∫ b

afm (x) fn (x)dx =

∫ b

a0(x)dx

⇒∀m ∈ N : ∑n∈A

κn ( fm, fn) = 0

⇒∀m ∈ N : κn ‖ fm‖2 = 0⇒∀m ∈ N : κn = 0.

17.1.29 Definition. Given functions g, f ∈ C2 (a,b), the projection of g to f isdefined by

Proj(g| f ) := arg minh=κ f‖g−h‖ .

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Since ∫ 1

0x · x2dx =

14,∫ 1

0

(x2)2

dx =15

we haveProj

(x|x2)= 1/4

1/5x2 =

54

x2.

Furthermore we have(x−Proj

(x|x2) ,Proj

(x|x2))= ∫ 1

0

(x− 5

4x2)

x2dx = 0

and

‖x‖2 =∫ 1

0x2dx =

13∥∥x−Proj

(x|x2)∥∥2

=∫ 1

0

(x− 5

4x2)2

dx =148

‖Proj(x|sinx)‖2 =∫ 1

0

(54

x2)2

dx =5

16

hence‖x‖2 =

∥∥x−Proj(x|x2)∥∥2

+∥∥Proj

(x|x2)∥∥2

.

17.1.32 Example. In [0,2π], the projection of g(x) = x to f (x) = sinx is

(g, f )

‖ f‖2 f =∫ 2π

0 xsinxdx∫ 2π

0 sin2 xdxsinx.

Since ∫ 2π

0xsinxdx =−2π,

∫ 2π

0sin2 xdx = π

we haveProj(x|sinx) =

−2π

πsinx =−2sinx.

Furthermore we have

(x−Proj(x|sinx) ,Proj(x|sinx)) =∫ 2π

0(x+2sinx)sinxdx = 0

and

‖x‖2 =∫ 2π

0x2dx =

83

π3

‖x−Proj(x|sinx)‖2 =∫ 2π

0(x+2sinx)2 dx =

83

π3−4π

‖Proj(x|sinx)‖2 =∫ 2π

0(−2sinx)2 dx = 4π

hence‖x‖2 = ‖x−Proj(x|sinx)‖2 +‖Proj(x|sinx)‖ .

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17.1.33 Definition. Given a function g ∈ C2 (a,b) and a vector subspace V ⊆C2 (a,b), the projection of g to V is

Proj(g|V) := argminh∈V‖g−h‖ .

17.1.34 Theorem. Given a function g ∈ C2 (a,b) and an orthonormal set ( fn)Nn= ⊆

C2 (a,b), let V= Span(( fn)

Nn=1

). Then we have

Proj(g|V) =N

∑n=1

Proj(g| fn) fn =N

∑n=1

(g, fn) fn. (17.1)

Furthermore g−Proj(g|V) is orthogonal to Proj(g|V) :

(g−Proj(g|V) ,Proj(g|V)) = 0; (17.2)

and, in fact, to every h ∈V:

∀h ∈V : (g−Proj(g|V) ,h) = 0; (17.3)

we also have‖g−Proj(g|V)‖2 +‖Proj(g|V)‖2 = ‖g‖2 . (17.4)

Finally∑n∈A|(g, fn)|2 ≤ ‖g‖2 . (17.5)

Proof. Since V= Span(( fn)

Nn=1

), we want to minimize

J (κ1, ...κN) =

∥∥∥∥∥g−∑n∈A

κn fn

∥∥∥∥∥2

=

(g−∑

n∈Aκn fn,g−∑

n∈Aκn fn

)

= ‖g‖2 +

∥∥∥∥∥∑n∈A

κn fn

∥∥∥∥∥2

+2

(g, ∑

n∈Aκn fn

)= ‖g‖2 + ∑

n∈Aκ

2n ‖ fn‖2 +2 ∑

n∈Aκn (g, fn)

= ‖g‖2 + ∑n∈A

κ2n +2 ∑

n∈Aκn (g, fn)

(since ‖ fn‖2 = 1). Setting the partial derivatives of J (κ1, ...κN) equal to zero wehave

∀n : 2κn−2(g, fn) = 0

which implies the only critical point of J (κ1, ...κN) is

∀n : κ∗n = (g, fn)

and it is clearly a global minimum. Hence, by definition,

Proj(g|V) =N

∑n=1

(g, fn) fn

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The results (17.2)-(17.4) are proved as in the proof of Theorem 17.1.30. Then, from(17.4), we have

‖Proj(g|V)‖2 ≤ ‖g‖2

But, by orthogonality of ( fn)Nn=1, we have

‖Proj(g|V)‖2 =

∥∥∥∥∥ N

∑n=1

(g, fn) fn

∥∥∥∥∥2

=

(N

∑n=1

(g, fn) fn,N

∑n=1

(g, fn) fn

)=

N

∑n=1‖(g, fn)‖2 .

which yields (17.5).

17.1.35 Example. With f1 = sinx√π

, f2 = sin2x√π

, the set ( fn)2n=1 is orthonormal in

[−π,π] (check it). Then with g = x and V= span(

sinx√π, sin2x√

π

)we have

(g, f1) =∫

π

−π

xsinx√

πdx = 2

√π,

Proj(g| f1) = (g, f1)1 = 2sinx;

(g, f2) =∫

π

−π

xsin2x√

πdx =−

√π,

Proj(g| f2) = (g, f2)sin2x√

π=−sin2x.

Hence

Proj(g|V) = Proj(g| f1)+Proj(g| f2) = 2sinx− sin2x,g−Proj(g|V) = x−2sinx+ sin2x.

Also

‖g‖2 =∫

π

−π

x2dx =23

π3,

‖Proj(g|V)‖2 =∫

π

−π

(2sinx− sin2x)2 dx = 5π,

‖g−Proj(g|V)‖2 =∫

π

−π

(x−2sinx+ sin2x)2 dx =23

π3−5π

and so‖g−Proj(g|V)‖2 +‖Proj(g|V)‖2 = ‖g‖2 .

17.1.36 Example. With f1 = 1, f2 = 1−3x the set ( fn)2n=1 is orthonormal in [0,1]

(check it). Then with g = x2 and V= span(1,1−3x) we have

(g, f1) =

(∫ 1

0x21dx

)=

13,

Proj(g| f1) = (g, f1)1 =13

;

(g, f2) =∫ 1

0x2 (1−3x)dx =− 5

12,

Proj(g| f2) = (g, f2)(1−3x) =54

x− 512

.

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And so

Proj(g|V) = Proj(g| f1)+Proj(g| f2) =13+

54

x− 512

=54

x− 112

,

g−Proj(g|V) = x2− 54

x+1

12

and

‖g‖2 =∫ 1

0x4dx =

15=

13,

‖Proj(g|V)‖2 =∫ 1

0

(54

x− 112

)2

dx =61

144,

‖g−Proj(g|V)‖2 =∫ 1

0

(x2− 5

4x+

112

)2

dx =13

240

and‖g−Proj(g|V)‖2 +‖Proj(g|V)‖2 = ‖g‖2 .

17.1.37 The following two theorems are easy corollaries of Theorem 17.1.34.

17.1.38 Theorem (Pythagorean). Given a function g ∈ C2 (a,b) and a vectorsubspace V⊆ C2 (a,b), we can decompose g as follows:

g = g+ g

where‖g‖2 = ‖g‖2 +‖g‖2 and (g, g) = 0.

17.1.39 Theorem. Given a function g∈ C2 (a,b) and an orthonormal set ( fn)∞

n=1⊆C2 (a,b). Then

limn→∞

(g, fn) = 0.

17.1.40 Definition. Given an orthonormal set ( fn)∞

n=1 ⊆ C2 (a,b), we say that( fn)

∞

n=1 is complete iff

∀g ∈ CN (a,b) : limn→∞

∥∥∥∥∥g−N

∑n=1

(g, fn) fn

∥∥∥∥∥= 0.

17.1.41 Theorem. Given an orthonormal set ( fn)∞

n=1 ⊆ C2 (a,b), ( fn)∞

n=1 is com-plete iff

∀g ∈ C2 (a,b) :∞

∑n=1|(g, fn)|2 = ‖g‖2 . (17.6)

Proof. Let VN = Span(( fn)

Nn=1

). Then we have

‖g‖2 = ‖g−Proj(g|VN)‖2 +‖Proj(g|VN)‖2 =

∥∥∥∥∥g−N

∑n=1

(g, fn) fn

∥∥∥∥∥2

+

∥∥∥∥∥ N

∑n=1

(g, fn) fn

∥∥∥∥∥2

.

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If ( fn)∞

n=1 is complete, then

∀g ∈ C2 (a,b) : limn→∞

∥∥∥∥∥g−N

∑n=1

(g, fn) fn

∥∥∥∥∥= 0

and so

∀g ∈ C2 (a,b) : ‖g‖2 = limn→∞

∥∥∥∥∥ N

∑n=1

(g, fn) fn

∥∥∥∥∥2

=

∥∥∥∥∥ ∞

∑n=1

(g, fn) fn

∥∥∥∥∥2

=∞

∑n=1|(g, fn)|2 .

Conversely, if

∀g ∈ C2 (a,b) :∞

∑n=1|(g, fn)|2 = ‖g‖2 (17.7)

then

∀g ∈ C2 (a,b) : 0 =

∥∥∥∥∥g−N

∑n=1

(g, fn) fn

∥∥∥∥∥2

(17.8)

which means ( fn)∞

n=1 is complete.

17.1.42 Definition. Given a complete orthonormal set ( fn)∞

n=1 ⊆ C2 (a,b), thegeneralized Fourier series (with respect to ( fn)

∞

n=1) of g ∈ C2 (a,b) is

∞

∑n=1

(g, fn) fn. (17.9)

17.1.43 We give two final theorems without proof.

17.1.44 Theorem. Let F be a vector space of functions and ( fn)n∈A, (gn)n∈B twocomplete (in F) sequences. Then they have the same cardinality, i.e., |A|= |B|17.1.45 Definition. Let F be a vector space of functions and ( fn)n∈A a complete

(in F) sequence. The dimension of F is the number of elements in ( fn)n∈A, i.e.,

dim(F) = |A| .


17.3 Unsolved Problems1. Prove that 0(x) belongs to every vector space of functions.2. Prove that ‖ f −g‖2 +‖ f +g‖2 = 2‖ f‖2 +2‖g‖2.3. Prove that ( f ,g) = 1

4 ‖ f +g‖2− 14 ‖ f −g‖2.

4. Prove that ( fn)n∈A is linearly independent iff there is some n∗ ∈ A and (κn)n6=n∗

such thatfn∗ = ∑

n6=n∗κn fn.

5. Prove that: limn→∞ ‖ fn− f‖= 0⇒ limn→∞ ‖ fn‖= ‖ f‖ .The following problems are with respect to the vector space C2 (a,b).

6. Let f1 = 1, f2 = x, f3 = x2. Prove that f1, f2, f3 is linearly independent.

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7. Let f1 = 1, f2 = |x|, f3 = x2. Prove that f1, f2, f3 is linearly independent.8. Let f1 = 1, f2 = x, f3 = x2. With the usual inner product defined on [a,b] =

[−1,1], compute ( f1, f2), ( f2, f3), ( f1,2 f1 +4 f3), ‖ f1‖2, ‖ f3‖2, ‖− f1 +2 f3‖2.9. Find a,b such that the set x,a+bx is orthogonal on [a,b] = [−1,1].

10. Find a,b such that the set x,a+bx is orthogonal on [a,b] = [0,1].11. Let f1 = 1, f2 = x, f3 = x2. Find functions g1,g2,g3 such that: (a) Span( f1, f2, f3)=

Span(g1,g2,g3) and g1,g2,g3is orthogonal on [a,b] = [−1,1].12. Let f1 = 1, f2 = x, f3 = x2. Find functions g1,g2,g3 such that: (a) Span( f1, f2, f3)=

Span(g1,g2,g3) and g1,g2,g3is orthogonal on [a,b] = [0,1].13. Let f1 = 1, f2 = x, f3 = x2. Find functions g1,g2,g3 such that: (a) Span( f1, f2, f3)=

Span(g1,g2,g3) and g1,g2,g3is orthogonal on [a,b] = [0,2].14. Find necessary and sufficient conditions (on f ,g) so that ‖ f +g‖= ‖ f‖+‖g‖.15. Find necessary and sufficient conditions (on f ,g) so that |( f ,g)|= ‖ f‖‖g‖.16. With [a,b] = [−1,1] find the projections of x and x2 to Span(cos(πx) ,cos(2πx)).17. With [a,b] = [−1,1] find the projections of x and x2 to Span(sin(πx) ,sin(2πx)).18. With [a,b] = [−1,1] find the projections of x and x2 to Span(cos(πx) ,cos(2πx) ,sin(πx) ,sin(2πx)).19. With [a,b] = [−1,1] find the projection of x2 to Span(cos(πx) ,cos(2πx) ,cos(3πx)).20. With [a,b] = [0,1] find the projections of x and x2 to Span(cos(πx) ,cos(2πx)).21. With [a,b] = [0,1] find the projections of x and x2 to Span(sin(πx) ,sin(2πx)).22. With [a,b] = [0,1] find the projections of x and x2 to Span(cos(πx) ,cos(2πx) ,sin(πx) ,sin(2πx)).23. With [a,b] = [0,1] find the projection of x2 to Span(cos(πx) ,cos(2πx) ,cos(3πx)).24. With [a,b] = [−1,1] find the projection of cos2 (πx) to Span(cos(πx) ,cos(2πx) ,cos(3πx)).25. With [a,b] = [−1,1] find the projection of sin(πx) to Span(cos(πx) ,cos(2πx) ,cos(3πx)).26. With [a,b] = [−1,1] find the projection of cos(x) to Span(cos(πx) ,cos(2πx) ,cos(3πx)).27. Find conditions on a,b so that the set cosx,sinx is orthogonal on [a,b].28. Find conditions on a,b so that the set (cos(nx))∞

n=0 is orthogonal on [a,b].29. Given the sequence ( fn)

∞

n=0 defined by

f0 (x) := 1, ∀n ∈ N : fn (x) :=1

2nn!dn

dxn

((x2−1

)n)

;

prove that ( fn)∞

n=0 is orthonormal in [0,1].30. Given the sequence ( fn)

∞

n=0 defined by

∀n ∈ N0 : fn (x) := ex dn

dxn

(xne−x) ;

prove that(

e−x/2

n! fn

)∞

n=0is orthonormal in [0,∞).

17.4 Advanced ProblemsThe following problems use the set of functions

L2 (a,b) :=

f :∫ b

a| f (x)|2 dx < ∞

.

1. Prove that L2 (a,b) is a vector space of functions.2. Prove that sin(x) ∈ L2 (0,2π).

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3. Prove that sin(x) /∈ L2 (−∞,∞).4. Prove that e−|x| ∈ L2 (−∞,∞).5. Let S be a vector subspace of L2 (a,b) and define the orthogonal complement

of S byS⊥ = f : ∀g ∈S : ( f ,g) = 0 .

Is S⊥ a vector space?6. Repeat the above problem with S⊆ L2 (a,b) but not a vector subspace.7. Is

(S⊥)⊥

=S?8. Show that S⊥ =

⋂x∈Sx⊥.

9. Show that S⊥∩S⊆ 0.10. Let S be a vector subspace of L2 (a,b) and define the projection operator PS

byPS ( f ) = Proj( f |S) .

Prove the following and provide a geometric interpretation.(a) PS is a linear operator.(b) PS is idempotent: PS (PS ( f )) = PS ( f ).(c) PS is distance reducing: ‖PS ( f )−PS (g)‖ ≤ ‖ f −g‖.

11. Prove that: if Q is a linear, idempotent and distance reducing operator, then itis a projection operator into some vector subspace S of L2 (a,b) (i.e., Q = PS).

12. Prove that dim(L2 (a,b)) = ℵ0.

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18. Sturm-Liouville Problems

We now present the family of the so-called Sturm-Liouville problems. This is classof boundary value ODE problems which arise often in PDE applications andgeneralize the Fourier and Bessel problems seen in previous chapters1.

18.1 Theory and Examples18.1.1 All functions appearing in this chapter belong to C2 (a,b), the space of allfunctions which are continuous and have continuous first and second derivativeson [a,b].

18.1.2 Definition. A regular Sturm-Liouville problem is one of the form

∀x ∈ (a,b) :ddx

(p(x)

dydx

)+q(x)y+λw(x)y = 0, (18.1)

Ay(a)+By′ (a) = 0, (18.2)Cy(b)+Dy′ (b) = 0; (18.3)

with the following additional assumptios:1. a,b ∈ R;2. p(x), q(x), w(x) are real-valued functions defined on [a,b], and such that for

all x ∈ [a,b] :(a) p(x) , p′ (x) ,q(x) ,q′ (x) are continuous,(b) for all p(x) and w(x) are strictly positive;

3. A,B,C,D ∈ R and AB 6= 0, CD 6= 0.The Sturm-Liouville operator corresponding to (18.1) is defined by

L(y) :=ddx

(p(x)

dydx

)+q(x)y.

18.1.3 Note that (18.1)-(18.3) always has the trivial solution y(x) = 0.1While the results presented in this chapter are quite interesting and useful, most of the required

proofs are beyond the scope of these notes. Hence in this chapter we present theorems withouttheir proofs.

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216 Chapter 18. Sturm-Liouville Problems

18.1.4 Theorem. Consider the regular Sturm-Liouville problem

∀x ∈ (a,b) :ddx

(p(x)

dydx

)+q(x)y+λw(x)y = 0, (18.4)

Ay(a)+By′ (a) = 0, (18.5)Cy(b)+Dy′ (b) = 0. (18.6)

For a fixed λ , the solution space of (18.4)-(18.6) is a vector space and the corre-sponding Sturm-Liouville operator is linear.Proof. Immediate.

18.1.5 Definition. Suppose that for some λ ∈R there exists some nontrivial solu-tion yλ (x) of (18.1)-(18.3). Then we call λ an eigenvalue and yλ (x) a correspondingeigenfunction of (18.1)-(18.3).

18.1.6 Example. Consider the problem

∀x ∈ (0,π) :d2ydx2 +λy = 0,

y(0) = 0,y(π) = 0.

This is a regular Sturm-Liouville problem with: a = 0,b = π, p(x) = 1,q(x) =0,w(x) = 1, A = C = 1, B = D = 0. To solve the problem we first get the generalsolution in the form

y(x) = c1 cos(√

λx)+ c2 sin

(√λx)

(note that we can have λ < 0 and√

λ ∈ C). Now we apply the boundary conditionsand get

c1 cos(0)+ c2 sin(0) = 0

c1 cos(√

λπ

)+ c2 sin

(√λπ

)= 0

which has the following solutions1. (c1,c2) = (0,0) for any λ value (but this is the trivial solution) or2. (0,c2) = (0,c2) for any c2 ∈ R, provided λ = n2, n ∈ Z−0.

So the SL problem has an infinity of solutions but they all have the form

y(x) =∞

∑n=1

bn sin(nx) ;

in other words, the solution space is

Span(sin(x) ,sin(2x) ,sin(3x) , ...) .

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18.1.7 Example. Next consider the problem

∀x ∈ (0,π) :d2ydx2 +λy = 0

y′ (0) = 0y′ (π) = 0

Again, it is a regular Sturm-Liouville problem with: a = 0,b = π, p(x) = 1,q(x) =0,w(x) = 1, A =C = 0, B = D = 0. To solve, we get the general solution in the form

y(x) = c1 cos(√

λx)+ c2 sin

(√λx)

and from the boundary conditions we get

−c1 sin(0)+ c2 cos(0) = 0

−c1√

λ sin(√

λπ

)+ c2√

λ cos(√

λπ

)= 0

which, apart from the trivial solution, has solutions (c1,0) for any c1 ∈ R, providedλ = n2, n ∈ Z.

So we get an infinity of solutions, all having the form

y(x) =∞

∑n=1

bn cos(nx) ;

the solution space is

Span(

sin(

12

x),sin

(32

x),sin

(52

x), ...

).

18.1.8 Example. Now consider the problem

∀x ∈ (0,π) :d2ydx2 +λy = 0

y(0) = 0y′ (π) = 0

Again, it is a regular Sturm-Liouville problem with: a = 0,b = π, p(x) = 1,q(x) =0,w(x) = 1, A = D = 1, B =C = 0. To solve, we get the general solution in the form

y(x) = c1 cos(√

λx)+ c2 sin

(√λx)


c1 cos(0)+ c2 sin(0) = 0

−c1√

λ sin(√

λπ

)+ c2√

λ cos(√

λπ

)= 0

which, apart from the trivial solution, has solutions (0,c2) for any c2 ∈ R, provided

λ =(n+ 1

2

)2, n ∈ Z.

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y(x) =∞

∑n=1

bn sin((

n+12

)x)

;


Span(

sin(

12

x),sin

(32

x),sin

(52

x), ...

).

18.1.9 Example. Now consider the problem

∀x ∈ (0,1) : x2 d2ydx2 + x

dydx

+(x2−ν

2)y = 0

y(0) = 1y′ (1) = 0

The differential equation is the Bessel equation. The problem is equivalent to aregular Sturm-Liouville one, with: a = 0,b = 1, p(x) = x,q(x) = x,w(x) = 1

x , A =D = 1, B =C = 0. To see this, consider

ddx

(x

dydx

)+q(x)y+λw(x)y = 0

and take p(x) = x,q(x) = x,w(x) = 1x , λ =−ν2 to get

ddx

(x

dydx

)+ xy+λ

1x

y = 0⇒

xd2ydx2 +

dydx

+ xy+λ1x

y = 0⇒

xd2ydx2 +

dydx

+

(x− ν2

x

)y⇒

x2 d2ydx2 + x

dydx

+(x2−ν

2)y = 0.

So the general solution has the form

y(x) = c1Jν

(√λx)+ c2Yν

(√λx)


c1 cos(0)+ c2 sin(0) = 0

−c1√

λ sin(√

λπ

)+ c2√

λ cos(√

λπ

)= 0

which, apart from the trivial solution, has solutions (0,c2) for any c2 ∈ R, provided

λ =(n+ 1

2

)2, n ∈ Z.

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y(x) =∞

∑n=1

bn sin((

n+12

)x)

;


Span(

sin(

12

x),sin

(32

x),sin

(52

x), ...

).

18.1.10 Definition. A periodic Sturm-Liouville problem is one of the form

∀x ∈ (a,b) :ddx

(p(x)

dydx

)+q(x)y+λw(x)y = 0,

y(a) = y(b) ,y′ (a) = y′ (b) .

The assumptions on a,b, p(x) ,q(x) ,w(x) are the same as for the regular problemand, in addition, we assume p(x) ,q(x) ,w(x) are periodic with period b−a (notethere exist no A,B,C,D constants in the boundary conditions).

18.1.11 Example. Consider

∀x ∈ (0,π) :d2ydx2 +λy = 0

y(0) = y(1)y′ (0) = y′ (1)

It is a periodic Sturm-Liouville problem with: a = 0,b = π, p(x) = 1,q(x) = 0,w(x) =1. The general solution has the form

y(x) = c1 cos(√

λx)+ c2 sin

(√λx)


c1 cos(0)+ c2 sin(0) = c1 cos(√

λπ

)+ c2 sin

(√λπ

)−c1 sin

(√λ0)+ c2 cos

(√λ0)=−c1 sin

(√λπ

)+ c2 cos

(√λπ

)

c1 = c1 cos(√

λπ

)c2 = c2 cos

(√λπ

)which yields nontrivial solutions of the form (c1,c1) for any c1 ∈R, provided λ = 4n2,n ∈ Z. So we get the general solution

y(x) = c1

∞

∑n=0

(cos(2nx)+ sin(2nx)) .

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18.1.12 In what follows we will always be referring to a regular Sturm-Liouvilleproblem; hence the appearing a,b, p(x) ,q(x) ,w(x) and A,B,C,D are always as-sumed to satisfy the corresponding Sturm-Liouville regulrity conditions.

18.1.13 Theorem. Consider the regular Sturm-Liouville problem

∀x ∈ (a,b) :ddx

(p(x)

dydx

)+q(x)y+λw(x)y = 0, (18.7)

Ay(a)+By′ (a) = 0, (18.8)Cy(b)+Dy′ (b) = 0. (18.9)

Suppose that, for a fixed λ , the general solution of (18.7) can be written asc1uλ + c2vλ . Then λ is an eigenvector of (18.7)-(18.9) iff∣∣∣∣ Auλ (a)+Bu′

λ(a) Avλ (a)+Bv′

λ(a)

Cuλ (b)+Du′λ(b) Cvλ (b)+Dv′

λ(b)

∣∣∣∣= 0. (18.10)

18.1.14 Example. Recalling Example 18.1.6

∀x ∈ (0,π) :d2ydx2 +λy = 0

y(0) = 0y(π) = 0

we see that (18.10) becomes (with a = 0, b = π, A = C = 1, B = D = 0, uλ (x) =cos(√

λx)

, vλ (x) = sin(√

λx)

)

∣∣∣∣∣∣ cos(√

λ0)

sin(√

λ0)

cos(√

λπ

)sin(√

λπ

) ∣∣∣∣∣∣= sinπ√

λ = 0.

which is exactly the condition we used in the solution of Example 18.1.6.

18.1.15 Theorem (Symmetry). The Sturm-Liouville operator L() : C2 (a,b)×C2 (a,b)→ C2 (a,b) satisfies

(u,L(v)) = (L(u) ,v)

for all solutions u,v of the corresponding SL problem.

18.1.16 Example. The Sturm-Liouville operator corresponding to Example 18.1.6is L() = d2

dx2 . Take u,v solutions of the problem, then

(u,L(v)) =∫

π

0u

d2vdx2 dx =

∫π

0u

ddx

(dvdx

)dx =

=

(u

dvdx

)π

0−∫

π

0

dvdx

dudx

dx =−∫

π

0

dvdx

dudx

dx,

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since u(0) = u(π) = 0. Similarly

(L(u) ,v) =∫

π

0

d2udx2 vdx =

∫π

0

ddx

(dudx

)vdx =

=

(dudx

v)π

0−∫

π

0

dvdx

dudx

dx =−∫

π

0

dvdx

dudx

dx.

So (u,L(v)) = (L(u)), as required by the theorem. Note that we have not reallyused the special form of L for this particular problem. We could use the specialform of the solutions, to check that

∫π

0 sin(5x) d2

dx2 (sin(3x))dx = 0 =

(u,L(v)) =∫

π

0sin(mx)

d2

dx2 (sin(nx))dx = 0,

(L(u) ,v) =∫

π

0

d2

dx2 (sin(mx))sin(nx)dx = 0.

18.1.17 Theorem (Orthogonality). If λ ,µ are different eigenvalues of a regularor periodic Sturm-Liouville problem, and uλ , uµ are corresponding eigenfunctions,then uλ , uµ are orthogonal with respect to w(x), i.e.

(uλ ,uµ

)w = 0 or, explicitly:∫ b

auλ (x)uµ (x)w(x)dx = 0.

18.1.18 Example. Continuing with Example 18.1.6, taking eigenvalues λ = m2 6=n2 = µ, we have corresponding eigenvectors uλ (x) = sin(mx), uµ (x) = sin(nx) andclearly ∫

π

0uλ (x)uµ (x)dx =

∫π

0sin(mx)sin(nx) = 0.

18.1.19 Theorem (Real Eigenvalues). If λ is an eigenvalue of a regular orperiodic Sturm-Liouville problem, then λ ∈ R. Let Vλ be the span of the set of alleigenfunctions corresponding to λ . Then Vλ is a vector space and there exists anorthonormal set of real-valued eigenfunctions which is complete in Vλ .

18.1.20 Theorem (Eigenvalue Multiplicity). If λ is an eigenvalue of a regularSturm-Liouville problem and uλ ,vλ two eigenfunctions corresponding to λ , thenuλ ,vλ are linearly dependent.

18.1.21 Example. Continuing with Example 18.1.6, we have seen that theeigenvalues have the form λ = n2 ∈ R and for each such λ the correspondingeigenfunction is uλ (x) = sin(nx). Also Vλ = Span(sin(nx)), a vector space; any twoeigenfunctions belonging to Vλ will have the form c1 sin(nx), c2 sin(nx) and hencewill be linearly dependent.

18.1.22 Theorem (Eigenvalue Monotonicity). The set of all eigenvalues of aregular Sturm-Liouville problem forms a strictly increasing sequence

λ1 < λ2 < ... < λn < λn+1 < ...

which is unbounded:limn→

λn = ∞.

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18.1.23 Example. Staying with Example 18.1.6, we can enumerate the eigneval-ues as λn = n2 ; then we clearly have

λ1 < λ2 < ... and limn→∞

λn = ∞.

18.1.24 Example.

18.1.25 Theorem (Eigenvalue Monotonicity). The set of all eigenvalues ofa periodic Sturm-Liouville problem (taking multilicities into account) forms anincreasing sequence

λ1 ≤ λ2 ≤ ..≤ λn ≤ λn+1 ≤ ...

which is unbounded:limn→

λn = ∞.

18.1.26 Example. For this theorem the appropriate example is Example 18.1.11,where we have

1. For λ1 = 0 the eigenfunction f1 (x) = 1;2. with n∈ N, for λ2n = λ2n+1 = n the eigenfunctions f2n (x) = cos(nx) and

f2n+1 (x) = sin(nx) .

So we haveλ1 < λ2 = λ3 < λ4 = λ5 < ... .

18.1.27 The next two theorems are the most important ones of this Chapter.

18.1.28 Theorem. Let (un)∞

n=1 be the orthonormal set of all eigenfunctions of aregular or periodic Sturm-Liouville problem and let g ∈ C2 (a,b). Then

if f (x) is continuous at x : f (x) =∞

∑n=1

(g, fn) fn (x) , (18.11)

if f (x) is discontinuous at x :12

(lim

ξ→x−f (ξ )+ lim

ξ→x+f (ξ )

)=

∞

∑n=1

(g, fn) fn (x)

(18.12)

(with the obvious modifications for x ∈ a,b).18.1.29 Theorem. The orthonormal set (un)

∞

n=1 of all eigenfunctions of a regularor periodic Sturm-Liouville problem is complete in C2 (a,b).

18.1.30 Example. Continuing with 18.1.11, we see that the orthonormal set(un)

∞

n=1 of all eigenfunctions is

U= 1⋃( 1√

πcos(nx)

)∞

n=1

⋃( 1√π

sin(nx))∞

n=1

=

1,

1√π

cos(x) ,1√π

sin(x) ,1√π

cos(2x) ,1√π

sin(2x) , ....

Now, we already know that (18.11)-(18.12) hold; this is just the fact that anyfunction continuous in [0,1] has Fourier series representation in terms of U; so

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Theorem 18.1.28 does not tell us anything new. On the othe rhand, from Theorem18.1.29 we see that any function g ∈ C2 (0,1) also satisfies

limN→∞

∥∥∥∥∥g−N

∑n=0

(an√

πcos(nx)+

bn√π

cos(nx))∥∥∥∥∥

2

= 0,

i.e., the Fourier approximation holds in themean square sense as well.

18.1.31 Example. Looking again at Example 18.1.6, we see that we can approxi-mate any g ∈ C2 (0,1) (in both the pointwise and the mean square sense) by sinesonly. This may seem a little surprising at first, but can actually be justified in twodifferent ways:

1. by taking g(x), the odd extension of g in [−1,1]: since g will be odd, it willhave a Fourier series in sines only;

2. or, as a consequence of Theorems 18.1.28 and 18.1.29.

18.1.32 Example. Looking at Example 18.1.13, we see why any function g ∈C2 (0,1) can be approximated (in both the pointwise and the mean square sense)by a series of Bessel functions; it is a direct consequence of Theorems 18.1.28 and18.1.29.

18.1.33 Example. Approximation with Legendre ....

18.1.34 Example. Let us solve the problem

y′′+λy = 0y(0) = 0

Cy(1)+ y′ (1) = 0

with C > 0. It is clearly a regular SL problem. Let us consider possible values forλ .

1. If λ < 0 then from the first boundary condition the general solution is

y(x) = c1 sinh(√−λx

)and then the second boundary condition gives

0 < tanh(√−λ

)=−√−λ

C< 0.

Hence negative λ values are rehjected.2. If λ = 0, then the general solution is y(x) = c0 + c1x but then the boundary

conditions require

c0 = 0Cc0 + c1 = 0

which only gives the trivial solution.

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3. Finally, for every λ > 0 we get an eigenfunction which, by the first boundarycondition must have the form

uλ (x) = cλ sin(√

λx)

;

but by the second boundary condition λ must satisfy

tan√

λ =−√

λ

C. (18.13)

Now, (18.13) cannot be solved exactly. However, solving graphically

eq1823_solutions

we see that we will have an infinity of solutions (eigenvalues), which we canenumerate in ioncreasing order:

0 = λ0 < λ1 < λ2 < ... .

Hence the general solution will have the form

y(x) =∞

∑n=0

cn sin(√

λnx).

18.1.35 Example. Let us solve the problem

y′′+λy = 0y(0)+ y′ (0) = 0

y(1)+3y′ (1) = 0

It is easily seen that its ia regular SL problem. For the same reasons as in Example18.1.34, we get the following caes.

1. We cannot have λ = 0.2. If we have λ =−k2 < 0 then the solution has the form

y = c1 coshkx+ c2 sinhkx

and the boundary conditions give

c1 + kc2 = 0(coshk+3k sinhk)c1 +(sinhk+3k coshk)c2 = 0

To get a nontrivial solution we must have

0 =

∣∣∣∣ 1 k(coshk+3k sinhk) (sinhk+3k coshk)

∣∣∣∣= (1−3k2)sinhk+2k coshk

So k must satisfy

tanhk =− 2k1−3k2

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We see graphically that this has a single solution computed numerically tobe k0 = 1.122..; then λ0 =−1.2587483. Now we solve the first equation of thesystem

c1 +1.122c2 = 0⇒ c1 =−1.122c2

hence we get an eigenfunction:

y0 (x) = 1.122cosh(1.122x)− sinh(1.122x)

3. If λ = k2 > 0 then the solution is

y = c1 coskx+ c2 sinkx

and the BCs give

c1 + kc2 = 0(cosk−3k sink)c1 +(sink+3k cosk)c2 = 0

To get a nontrivial solution we must have

0 =

∣∣∣∣ 1 k(cosk−3k sink) (sink+3k cosk)

∣∣∣∣= sink+3k2 sink+2k cosk

So k must satisfy

tank =− 2k1+3k2

Again we note graphically that this has an infinity of solutions, which we cancompute numerically:

k1 = 2.9256k2 = 6.1766k3 = 9.3528...

The corresponding eigenvalues are

λ1 = 8.5596λ2 = 38.1502λ3 = 87.4953...

So we get a family of solutions:

yn (x) = kn cos(knx)− sin(knx) .

The general solutiuon has the form

y(x) =∞

∑n=0

cnyn (x) .

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1. Put the ODEd2ydx2 +b

dydx

+ cy = 0

in Sturm-Liouville form.Ans. d

dx

(ebx dy

dx

)+ cebxy. //13.2.1

2. Put the ODE (1− x2) d2y

dx2 − xdydx

+α2y = 0


dx

(√1− x2 dy

dx

)+ α2√

1−x2 y. //13.2.33. Put the ODE

x2 d2ydx2 +bx

dydx

+ cy = 0


dx

(xb dy

dx

)+ cxb−2y. //13.2.4

4. Put the ODEd2ydx2 −2xy

dydx

+2αy = 0


dx

(e−x2 dy

dx

)+2αe−x2

y. // 13.2.55. Put the ODE

xd2ydx2 +(1− x)y

dydx

+αy = 0


dx

(xe−x dy

dx

)+αe−xy. // 13.2.6

6. Put the ODE (1− x2) d2y

dx2 −2xydydx

+α (α +1)y = 0


dx

((1− x2) dy

dx

)+α (α +1)y. // 13.2.7

7. Find the eigenvalues of

d2ydx2 +λy = 0, y(0)+2y′ (0) = 0, y(2) = 0.

Ans. λ0 = 0, λ1 = 5.047, λ2 = 14.919, ... . // 13.2.118. Solve the problem

d2ydx2 +2y

dydx

+ y+λy = 0, y(0) = 0, y(1) = 0.

Ans. .13.2.9

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9. Solve the problem

d2ydx2 +2

dydx

+ y+λy = 0, y′ (0) = 0, y′ (1) = 0.

Ans. λn = n2π2, yn (x) = e−x sin(nπx). // 13.2.1010. Solve the problem

d2ydx2 +2

dydx

+ y+λy = 0, y(0) = 0, y′ (1) = 0.

Ans. λ0 = −1,y0 = 1; λn = n2π2, yn (x) = e−x (nπ cos(nπx)+ sin(nπx)). //13.2.21


x2 d2ydx2 −2x

dydx

+2y+λx2y = 0, y(1) = 0, y(2) = 0.

Ans. λn = n2π2, yn (x) = xsin(nπ (x−2)). // 13.2.2212. Solve the problem

d2ydx2 +λy = 0, y(0)+αy′ (0) = 0, y(π)+αy′ (π) = 0.

Ans. λ0 = −1/α2,y0 = e−x/α ; λn = n2, yn (x) = nα cos(nx)− sin(nx). //13.2.26


d2ydx2 +λy = 0, y(0)+αy′ (0) = 0, y(π)+αy′ (π) = 0.

Ans. λ0 = −1/α2,y0 = e−x/α ; λn = n2, yn (x) = nα cos(nx)− sin(nx). //13.2.26

18.4 Advanced Problems1. Prove that all the eigenvalues of the problem

d2ydx2 +

(λ − x2)y = 0

u′ (0) = u′ (1) = 0

are positive.2. Prove that all the eigenvalues of the problem

d2ydx2 +

(λ − x2)y = 0

u′ (0) = limx∞

u′ (x) = 0

are positive.

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3. Let u be a nontrivial solution of

∀x ∈ (a,b) :d2ydx2 +w(x)y = 0.

Suppose that∀x ∈ [a,b] : w(x)≤ 0.

Show that u(x) = 0 has at most one root (in [a,b]).4. Let u1 and u2 be nontrivial solutions of

∀x ∈ (a,b) :d2ydx2 +w1 (x)y = 0,

∀x ∈ (a,b) :d2ydx2 +w2 (x)y = 0,

respectively. Suppose that

∀x ∈ [a,b] : w1 (x)> w2 (x) .

Show that between every two roots of u2 (x) = 0 there exists one root ofu1 (x) = 0.

5. Does d2ydx2 +

(1+ sin2 x

)y = 0 have oscillating solutions in (0,∞)?

6. Does d2ydx2 − x2y = 0 have oscillating solutions in (0,∞)?

7. Does d2ydx2 +

1x y = 0 have oscillating solutions in (0,∞)?

8. If L1 (·) and L2 (·) are two Sturm-Liouville operators, let us define L12 (·) =L1 (L2 (·)) and L21 (·) = L2 (L1 (·)). Find necessary and/or sufficient condi-tions so that L12 (·) = L21 (·).

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A Definitions of the Integral . . . . . . 231

B Distribution Theory . . . . . . . . . . . . . 239

C Gamma Function . . . . . . . . . . . . . . . . 241

D Numerical Solution of PDEs . . . . 243

Appendices

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A. Definitions of the Integral

A.1 Riemann Integral

A.1.1 In previous parts of these notes we have used definite integrals of the form∫ ba f (x)dx without much discussion. In most cases what we used was the Riemann

integral which, as you recall, is defined as the limit of finite sums.

A.1.2 We will now give a rigorous definition of∫ b

a. f (x)dx in the Riemann sense.To this end we need some preliminaries.

A.1.3 Definition. A partition of a closed interval [a,b] is a finite sequence of pointsP = (x0,x1, ...,xN) such that

a = x0 < x1 < ... < xN = b.

The intervals of the partition are [x0,x1], ..., [xN−1,xN ]. The norm w(P) of thepartition is the largest interval width:

w(P) = max1≤n≤N

(xN− xN−1) .

A.1.4 Definition. A tagged partition of a closed interval [a,b] is a pair of finiteseqeuences (P,Q) where P = (x0,x1, ...,xN) is a partition of [a,b] and Q = (ξ1, ...,ξN)is a finite sequence of points such that

∀n : ξn ∈ [xn−1,xn] .

A.1.5 Definition. Let f be a function defined on [a,b]. The Riemann integral of fon [a,b] is a number I such that

∀ε > 0∃δ > 0 : ((P,Q) is a tagged partition with w(P)< δ )⇒

∣∣∣∣∣I− N

∑n=1

f (ξn)(xn− xn−1)

∣∣∣∣∣< ε.

(A.1)The number I is usually denoted by

∫ ba f (x)dx.

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232 Chapter A. Definitions of the Integral

A.1.6 What the above definition says is:∫ b

a. f (x)dx = I iff the difference between Iand any sum over a finite partition can be made as small as we want, provided thenorm of the partition is sufficiently small. A concise way to write (A.1) is∫ b

af (x)dx := lim

w(P)→∞

N

∑n=1

f (ξn)(xn− xn−1) .

A.1.7 If we fix f and a we can define a function F (x) by

F (x) :=∫ x

af (z)dz.

(when (a,x) and f are such that∫ x

a f (z)dz exists) and, as is well known, F ′ (x) =f (x).

A.1.8 Now consider the converse question: given some F (x), can we represent itas a Riemann integral

∫ xa f (z)dz? The answer is: not always. For example, if F (x)

is discontinuous at some point, it cannot be so represented since, as we know,∫ xa f (z)dz is a continuous function of x.

A.1.9 In the next section we will show how to represent a discontinuous F (x)as a a Stieltjes integral (which is a generalization of the Riemann integral). Theconnection of this problem to the representation of Heaviside(x) as the integral ofits derivative (i.e., as

∫ ba Dirac(x)dx) is obvious.

A.2 Stieltjes Integral, Heaviside and Dirac FunctionsA.2.1 Definition. Let f be a function defined on [a,b] ⊂ R and φ a monotone

function on [a,b]. The Stieltjes integral of f on [a,b] with respect to φ is a number Isuch that

∀ε > 0∃δ > 0 : (P,Q) is a tag.part. with w(P)< δ

⇒

∣∣∣∣∣I− N

∑n=1

f (ξn)(xn− xn−1)(φ (xn)−φ (xn−1))

∣∣∣∣∣< ε.

(A.2)

The number I usually denoted by∫ b

a f (x)dφ and (A.2) can be wriiten in short formas ∫ b

af (x)dφ := lim

w(P)→∞

N

∑n=1

f (ξn)(xn− xn−1)(φ (xn)−φ (xn−1)) .

A.2.2 Note that, when φ (x) = x, the Stieltjes integral reduces to the Riemannintegral. Also, it can be proved that, when φ (x) is differentiable:∫ b

af (x)dφ =

∫ b

af (x)φ

′ (x)dx.

A.2.3 Now, taking φ (x) =Heaviside(x), let us calculate the Stieltjes integral∫ b

af (x)dHeaviside.

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A.3 About Sets 233

For some tagged partition (P,Q), consider the sum

J =N

∑n=1

f (ξn)(Heaviside(xn)−Heaviside(xn−1)) .

1. If 0 /∈ [a,b], clearly J = 0.2. If 0 ∈ (xn−1,xn) (for some n), then J = f (0). As w(P)→ 0, there will always

exist some [xn−1,xn] 3 0. Hence we distinguish two subcases.(a) If f (x) is continuous at x = 0, then J →

w(P)→0f (0). This can be rewritten

as ∫ b

af (x)dHeaviside = f (0) or

∫ b

af (x)Dirac(x)dx = f (0) .

This is the relationship (9.9).(b) If f (x) is discontinuous at x = 0, small changes of some ξn can result in

different values of J, and J cannot tend to a limit; hence the Stieltjesintegral does not exist.

A.3 About Sets

A.3.1 For a really rigorous treatment of Fourier analysis neither the Riemann northe Stieltjes integral suffices. What is required is the Lebesgue integral. Beforethis can be defined, we need some preliminaries regarding sets.

A.3.2 Definition. Given sets A and B, we define:

A∩B := x : x ∈ A and x ∈ B , A∪B := x : x ∈ A or x ∈ B or both .

These operations can be expanded to an infinite collection of sets A1,A2, ... asfollows:

∞⋂n=1

An = x : x ∈ An for every n ∈ N ,∞⋃

n=1

An = x : x ∈ An for at least one n ∈ N .

A.3.3 Definition. When A⊆ R, we define the translation of A by x0 ∈ R by

A+ x0 = y : y = x+ x0, x ∈ A .

A.3.4 Definition. Given some set A, we define its cardinality by

|A| := “the number of elements of A”.

A.3.5 If a set contains a finite number n of elements, then we say that its is afinite countable set and its cardinality is, obviously, n. In other words:

|a1,a2, ...,an|= n.

Things are more interesting when A contains an infinite number of elements. Thenwe distinguish two cases.

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1. If the elements of A can be placed into a 1-to-1 correspondence with N, thenwe say A is infinitely countable or enumerable and we write |A|= ℵ0. In otherwords

|a1,a2, ...|= ℵ0.

Intuitively, when we say that a1,a2, ... is (finitely or infinitely) countablewe mean that we can enumerate its elements: there is a first element (a1), asecond element (a2) and so on.

2. If the elements of A cannot be placed into a 1-to-1 correspondence with N,then we say A is uncountable. It may not be obvious but such sets do exist.For example, it can be proved that we cannot put the elements of the interval(0,1) into a 1-to-1 correspondence with N. Intuitively, this means that therecannot be a first, second etc. element of (0,1). To put this in another way,while both N and (0,1)have an infinite number of elements, (0,1) has moreelements than N. This suggests that there are different kinds of infinity.

A.3.6 Definition. Given a set A, we define its powerset

℘(A) := “the set of all subsets of A”.

A.3.7 Example. We have

℘(1,2) = /0,1 ,2 ,1,2

and|℘(1,2)|= 4 = 22 = 2|1,2|.

More generally, for any finite set A with cardinality |A| ∈ N 0 we have |℘(A)|= 2|A|.And we also have

∀A : |A| ∈ N⇒|℘(A)|= 2|A| > |A| .It is interesting to note that we also have

∀A : |A|= ℵ0⇒|℘(A)|= 2ℵ0 > ℵ0.

However, to prove the above we need to define the arithmetic of cardinalities (so thatexpressions such as 2ℵ0 make rigorous sense). We will not need these concepts inour treatment of the Lebesgue integral.

A.4 Lebesgue IntegralA.4.1 We will now define the Lebesgue integral in several steps.

A.4.2 We first need to define the Lebesgue measure of a set; intuitively, it is ageneralization of the length of an interval.

A.4.3 In other words: the length of (a,b) is

l (a,b) := b−a;

what would be the analog for a general set A (not an interval)? Let us call thisanalog µ; note that it is a set function, i.e., it assigns real numbers to sets:

µ :℘(R)→ R.

Ideally, µ should have the following properties for all sets of real numbers..

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A.4 Lebesgue Integral 235

P1 The measure of an interval (a,b) is its length:

∀(a,b) : µ (a,b) = b−a.

P2 The measure is an increasing function:

∀A,B : A⊆ B⇒ µ (A)≤ µ (B) .

P3 The measure is translation invariant:

∀A : µ (A+ x0) = µ (A) .

P4 Measures of nonintersecting sets add up:

∀A,B : A∩B = /0 we have µ (A∪B) = µ (A)+µ (B)

and more generally

∀A1,A2, ... : j 6= k⇒ A j∩Ak = /0 we have µ (∪∞i=1Ai) =

∞

∑i=1

µ (Ai) .

A.4.4 Unfortunately, it can be proved that no function µ : ℘→ R satisfies allP1-P4. So we will try to find such a µ which works on a restricted range of sets.

A.4.5 Definition. We define the outer Lebesgue measure µ∗as follows: for anyA⊆ R

µ∗ (A) := inf

∞

∑n=1

l (In) : A⊆ ∪∞n=1In

where the inf is taken over sequences of open intervals (In)

∞

n=1 (i.e., for all n :In = (an,bn)).

A.4.6 Theorem. The outer Lebesgue measure has the following properties for allsets A,B, A1,A2, ... :

1. ∀A : µ∗ (A)≥ 0 (nonnegativity of measure).2. ∀(a,b) : µ∗ ((a,b)) = b−a (outer measure of an interval is its length).3. A⊆ B⇒ µ∗ (A)≤ µ∗ (B) (monotonicity).4. |A| ≤ℵ0⇒ µ∗ (A) = 0 (countable sets have zero measure).5. ∀x0 ∈ R : µ∗ (A+ x0) = µ∗ (A) (translation invariance).6. µ∗

(∪∞

n=1A)≤ ∑

∞n=1 µ∗ (An) (subadditivity).

A.4.7 Example. Consider Q(0,1), the set of rational numbers in (0,1). Because∣∣Q(0,1)∣∣ = ℵ0 (the rationals are countable), it follows that µ∗

(Q(0,1)

)= 0. Now

consider J(0,1), the set of irrational numbers in (0,1). We have

J(0,1) ⊆ (0,1)⇒ µ∗ (J(0,1))≤ µ

∗ ((0,1)) = 1.

Also, becauseQ(0,1)∪J(0,1) = (0,1) and Q(0,1)∩J(0,1) = /0,

we have

1 = µ∗ ((0,1)) = µ

∗ (Q(0,1)∪J(0,1))≤ µ

∗ (Q(0,1))+µ

∗ (J(0,1))= 0+µ∗ (J(0,1)) .

In short1≤ µ

∗ (J(0,1))≤ 1⇒ µ∗ (J(0,1))= 1.

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A.4.8 Definition. A set A⊆ R is called Lebesgue measurable iff

∀B⊆ R : µ∗ (A) = µ

∗ (A∩B)+µ∗ (A∩Bc) .

If A is measurable, we define its Lebesgue measure by

µ (A) := µ∗ (A) .

A.4.9 Theorem. The family of measurable sets satisfies the following.1. Both /0 and R are measurable.2. If A,B are measurable then A∩B, A∪B, Ac, A+ x0 (for any x0 ∈ R ) are

measurable.3. If for some (An)

∞

n=1 we have : ∀n : An is measurable, then ∪∞n=1An and ∩∞

n=1Anare measuable.

4. If µ∗ (A) = 0 then A is measurable.5. Every open and every closed set is measurable (so, in particular, all intervals

are measurable).6. If A,B are measurable and A⊆ B then µ (A)≤ µ (B).7. For any sequence (An)

∞

n=1 of measurable sets such that m 6= n⇒ Am∩An = /0,we have

µ

(∞⋃

n=1

An

)=

∞

∑n=1

µ (An) .

A.4.10 It can be proved that there exist sets which are not measurable.

A.4.11 We are finally ready to define the Lebesgue integral.

A.4.12 Definition. The characteristic function of a set A⊆ R is defined by

1A (x) =

1 iff x ∈ A,0 iff x /∈ A.

A.4.13 Definition. Every function of the form

φ (x) =N

∑n=1

κn1An (x)

is called a simple function, provided that:1. all the sets An are measurable,2. m 6= n⇒ Am∩An = /0,3. m 6= n⇒ κm 6= κn = /0.

A.4.14 Definition. For every simple function φ (x) = ∑Nn=1 κn1An (x), the Lebesgue

integral over a measurable set A is defined by∫A

φ (x)dx :=N

∑n=1

κnµ (An∩A) .

A.4.15 Definition. For every bounded function f which is zero outside a (mea-surable) set A with µ (A)< ∞ define

IA ( f ) = inf∫

Aφ (x)dx where φ is simple and f ≤ φ

,

IA ( f ) = sup∫

Aφ (x)dx where φ is simple and φ ≤ f

.

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A.4 Lebesgue Integral 237

If IA ( f ) = IA ( f ) we say that f is (Lebesgue) integrable on A and define the Lebesgueintegral of f on A by ∫

Af (x)dx := IA ( f ) .

The integral of f over some B different from the domain A of f is defined by∫B

f (x)dx :=∫

Af (x)1B (x)dx.

A.4.16 The meaning of the above definitions is actually simple. To compute theLebesgue integral of a function f we do the following.

1. If f is simple, we partition its range into (a finite number of) sets, where fhas constant value in each set; the integral is the sum of the measures ofthe sets, each measure weighted by the respective value of f .

2. If f is not simple, we approximate it from above and below by simple functions;the integral of f is the liniting common value of the integrals of the upperand lower bounding functions.

A.4.17 Compare the above with the Riemann integral, where we partition thedomain of the function.

A.4.18 Example.∫R 1A (x)dx = µ (A),

∫A dx = µ (A), ...

A.4.19 Example.

A.4.20 Example.

A.4.21 Example.

A.4.22 Theorem. If f is Riemann integrable, it is also Lebesgue integrable andthe the Riemann and Lebesgue integrals are equal.

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B. Distribution Theory

Here is another way to justify dHeavisidedx =Dirac(x).

B.1 Theory and Examples

B.1.1 We can think that of the definite integral∫ b

a as a functional, i.e. a functionwhich has as domain a set of functions and as range a set of numbers.

B.1.2 More rigorously, fix a,b (such that 0 ∈ (a,b)) and define C∞0 to be the set

of functions which have continuous derivatives of all orders and vanish in someneighborhood of a and in some neighborhood of b. Now chose a real valued functionf (x) and define

∀φ ∈ C∞0 : A f (φ) :=

∫ b

aφ (x) f (x)dx (B.1)

Then clearlyA f : C∞

0 → R.

B.1.3 It is in fact easy to see that A f is a linear functional, i.e.,

∀κ1,κ2 ∈ R and ∀φ1,φ2 ∈ C∞0 : A f (κ1φ1 +κ2φ2) = κ1A f (φ1)+κ2A f (φ2) .

B.1.4 We can also construct linear functionals which are not obtained from anintegral. As an example, we define the functional

ADirac : C∞0 → R

by∀φ ∈ C∞

0 : ADirac (φ) := φ (0) .

It is easy to show that ADirac is a linear functional. It is not so easy but can beshown (do it!) that ADirac cannot be written as an integral of the form (B.1) for anyfunction f .

B.1.5 Now, keeping a,b fixed, denote the set of all functionals from C∞0 to R by Q.

Furthermore denote

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240 Chapter B. Distribution Theory

1. by QR the subset of regular functionals, i.e., the ones which can be obtained(for some f ) by an integral of the form (B.1) and

2. by QS the subset of singular functionals, i.e., the ones which cannot beobtained (for some f ) by an integral of the form (B.1).

B.1.6 We can think of the elements of Q (i.e., the functionals) as (representativesof) generalized functions. The elements of QR (i.e., the regular functionals) arerepresentatives of actual functions: each A f is a representative of some functionf . This is not true of the elements of QS (i.e., the singular functionals): theycannot be obtained from (i.e., represent) some function f . Hence Q generalizes(i.e., enlarges) the set of functions.

B.1.7 Now we can ask: given some generalized function A ∈Q, which generalizedfunction B ∈Q should represent the derivative of A?

B.1.8 If A = A f ∈QR it is natural to choose B = A f ′, i.e., the regular generalizedfunction obtained (by an integral) from f ′. More explicitly, with

A f (φ) =∫ b

aφ (x) f (x)dx

A f ′ (φ) =∫ b

aφ (x) f ′ (x)dx

we call A f ′ the derivative of A f .

B.1.9 A consequence of the above is that (integrating by parts and using the factthat φ (a) = φ (b) = 0):

A f ′ (φ) =∫ b

aφ (x) f ′ (x)dx = (φ (x) f (x))x=b

x=a−∫ b

aφ′ (x) f (x)dx =−A f

(φ′) . (B.2)

B.1.10 Now we can use (B.2) to define the derivative of a generalized function(“generalized derivative”). In other words, we define(

A f (φ))′ :=−A f

(φ′) . (B.3)

The advantage of (B.3) is that it applies to all generalized functions, not only theregular ones! (Recall that φ ′ is always defined, since φ ∈ C∞

0 .)

B.1.11 Now let us take a =−∞ and b = ∞ and compute the generalized derivativeof Heaviside(x). First note that

AHeaviside(x) (φ) =∫

∞

−∞

φ (x)Heaviside(x)dx =∫

∞

0φ (x)dx.

Then

AHeaviside’(x) (φ)=−AHeaviside(x)(φ′)=−∫ ∞

0φ′ (x)dx=−φ (∞)+φ (0)= φ (0)=ADirac (φ) .

In shortAHeaviside’(x) (φ) = ADirac (φ) .

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C. Gamma Function

The Gamma function Γ(n) is the generalization of the factorial function n!.

C.1 Theory and ExamplesC.1.1 Definition. The Gamma function is defined by

∀n ∈ (0,∞) : Γ(n) :=∫

∞

0xn−1e−xdx.

C.1.2 Example. We have

Γ(1) =∫

∞

0x0e−xdx =

∫∞

0e−xdx = 1.

Also, with x = u2 we have dx = 2udu and so

Γ

(12

)=∫

∞

0x−

12 e−xdx =

∫∞

0

1u

e−u22udu = 2

∫∞

0e−u2

du =√

π.

C.1.3 Theorem. The Gamma function satisfies

∀n ∈ (0,∞) : Γ(n+1) = nΓ(n)

Proof. We have ∫xne−xdx =−xne−x−

∫ (−e−x)nxn−1dx.

Hence

Γ(n+1) =∫

∞

0xne−xdx =

(−xne−x)x=∞

x=0 −∫

∞

0

(−e−x)nxn−1dx

=−0+0+n∫

∞

0xn−1e−xdx = nΓ(n) .

C.1.4 Theorem. The Gamma function satisfies

∀n ∈ N0 : Γ(n+1) = n!

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242 Chapter C. Gamma Function

Proof. We have

Γ(1) =∫

∞

0x1−1e−xdx = 1 = 0!

Γ(2) = Γ(1+1) = 1Γ(1) = 1 = 1!,Γ(3) = Γ(2+1) = 2Γ(2) = 2 = 2!,Γ(4) = Γ(3+1) = 3Γ(3) = 3 ·2 = 3!,

...

and, by induction, we easily obtain the required result.

C.1.5 Definition. We extend the definition of Γ(n) as follows. Define

Z−0 := 0,−1,−2, ...

and define∀n ∈ (−∞,0) : Γ(n) :=

Γ(n+1)n

.

C.1.6 Example. Γ(−1

2

)=

Γ( 12)− 1

2=−2

√π.

C.1.7 Theorem. We have

∀n ∈ Z−0 :

limx→(2n)+ Γ(x) = +∞,

limx→(2n)− Γ(x) =−∞,

limx→(2n−1)+ Γ(x) =−∞,

limx→(2n−1)− Γ(x) =−∞.

C.1.8 Example. Γ(0+) = limx→0+ Γ(x) = +∞, Γ(0−) = limx→0− Γ(x) =−∞.

C.1.9 Theorem (Stirling). For large n:

n!'√

2πn(n

e

)n.

C.1.10 Example. Γ(10) = 362880,√

2π ·10(10

e

)10= 3.5987× 106. Γ(112.3) = 7.

2544×10180,√

2π ·112.3(112.3

e

)112.3= 8.1406×10182.

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D. Numerical Solution of PDEs

In this chapter we will discuss the numerical solution of PDEs by computer.

D.1 Basic Methods

D.1.1 We start with some very simple (almost primitive) methods. While thesewill not yield accurate solutions, they are useful in highlighting the basic ideasbehind numerical solution of PDEs. For a better understanding, we also presentthe MATLAB code which implements the methods.

D.1.2 Example. Our first example involves a first order PDE:

0 < t : ut +3ux = 0, (D.1)0 < x : u(x,0) = f (x). (D.2)

It is easily checked that the analytical solution is

u(x, t) = f (x−3t). (D.3)

Hence (D.1) describes the transmission of the initial condition to the left withvelocity 3. For example, if

f (x) =

0 gia x≤ 10x−10

10 gia 10≤ x≤ 201− 20−x

10 gia 20≤ x≤ 300 gia 30≤ x.

(D.4)

then u(x, t) looks like this:

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244 Chapter D. Numerical Solution of PDEs

Let us now solve the same PDE numerically. Letting

vn(t) = u(n, t)

we get

ut =dvn

dt, ux ' vn− vn−1. (D.5)

Let us take n = 0,1, ...,40 and assume v0(t) = 0 for every t. Then we must solve thesystem of ODEs

dv1

dt=−3 · (v1− v0)

...

dvn

dt=−3 · (vn− vn−1) (D.6)

...

dv40

dt=−3 · (v40− v39)

with initial conditions vn(0) = f (n) (using f (x) of (D.4)). This can be achieved bythe following MATLAB commands.

clearT=[0:0.1:10];u0=[zeros(1,10) [0.1:0.1:1] [0.9:-0.1:0.1] zeros(1,10)];[t,u]=ode23(’flux11’,T,u0);

which use the function file flux11.m:

function ut=flux(t,u)N=length(u);ut(1,1)=-3*u(1);for n=2:Nut(n,1)=-3*(u(n)-u(n-1));end

The results appear in the following figures; we can see that they are in goodagreement with the exact solution.

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D.1 Basic Methods 245

D.1.3 Example. Let us solve one more first order PDE

ut +3uux = 0, u(x,0) = f (x) (D.7)

where f (x) is again given by (D.4). The discretization of this is given by changingthe flux11.m so that it corresponds to ut =−3uux. Then we get the followingresults. Where is the nonlinear behavior?

D.1.4 Let us now solve some second order PDEs. The basic idea is to replace thetime and space derivatives with finite differences:

ut 'u(m ·δx,(n+1) ·δ t)−u(m ·δx,n ·δ t)

δ t(D.8)

ux 'u(m ·δx,n ·δ t)−u((m−1) ·δx,n ·δ t)

δx(D.9)

' u((m+1) ·δx,n ·δ t)−u(m ·δx,n ·δ t)δx

(D.10)

uxx 'u((m+1) ·δx,n ·δ t)+u((m−1) ·δx,n ·δ t)−2u(m ·δx,n ·δ t)

δx2 . (D.11)

Lettingum,n = u(m ·δx,n ·δ t) (D.12)

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the above equations become

ut 'um,n+1−um,n

δ t(D.13)

ux 'um,n−um−1,n

δx(D.14)

'um+1,n−um,n

δx(D.15)

uxx 'um+1,n +um−1−2um,n

δx2 . (D.16)

and the diffusion equation ut = a2uxx becomes

um,n+1−um,n

δ t= a2 um+1,n +um−1,n−2um,n

δx2 ⇒ (D.17)

um,n+1 = um,n +δ t ·a2

δx2 · (um+1,n +um−1,n−2um,n) . (D.18)

D.1.5 Example. We solve the diffusion PDE

0 < x < 40 : ut = a2uxx (D.19)0 < t < 100 : u(0, t) = 0, (D.20)0 < t < 100 : u(40, t) = 0, (D.21)0 < x < 40 : u(x,0) = f (x). (D.22)

We use for f (x) a triangular function with its apex at x = 20 and we use δx =1,δ t = 0.1. The MATLAB code is

clear

u(:,1)=[zeros(1,10) [0.1:0.1:1] [0.9:-0.1:0.1] zeros(1,10)]’;

M=length(u);

a2=4;

dt=0.1;

for n=1:999

u(1,n+1)=u(1,n);

for m=2:M-1

u(m,n+1)=u(m,n)+dt*a2*(u(m+1,n)+u(m-1,n)-2*u(m,n));

end

u(M,n+1)=u(M,n);

end

figure(1); surf(u); shading flat

The results are plotted below, for two values: a = 2 and a = 3. We can clearly seethe smooothing effect of the heat kernel e−x2/4a2t , which becomes faster with largerdiffusion coefficient a.

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D.1.6 Example. Next we solve the same problem but with a negative diffusioncoefficient

0 < x < 40 : ut =−a2uxx (D.23)0 < t < 100 : u(0, t) = 0, (D.24)0 < t < 100 : u(40, t) = 0, (D.25)0 < x < 40 : u(x,0) = f (x). (D.26)

This results show a behavior opposite of smoothing: local differences are accentu-ated. There is also evidence of numerical errors.

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D.1.7 Example. Our final diffusion example involves the nonlinear reaction-diffusion equation1:

0 < x < 40 : ut = a2uxx +bu(1−u) (D.27)0 < t < 100 : u(0, t) = 0, (D.28)0 < t < 100 : u(40, t) = 0, (D.29)

0 < x < 40 : u(x,0) =

0 x≤ 301 30 < x≤ 50 . (D.30)

In (D.27) the time evolution of u depends both on the diffusion term a2uxx and anonlinear “reaction” term bu(1−u). For u ∈ (0,1) the reaction term is positive,and this favors the increase of u; note also that increase rate becomes 0 at u = 0and u = 1. The results are seen in the following figures, for two sets of values:with (a,b) = (2,1) we have reaction-diffusion and with (a,b) = (2,0) we have purereaction.

1It is used to model chemical reactions.

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D.1.8 Example. Now consider the Laplace PDE

0 < x < 30, 0 < y < 40 : uxx +uyy = 0, (D.31)0 < y < 40 : u(0,y) = 1, (D.32)0 < y < 40 : u(30,y) = 0, (D.33)

0 < x < 30 : u(x,0) =(

1− x30

)2, (D.34)

0 < x < 30 : u(x,40) =(

1− x30

)1/4. (D.35)

Here the discretizations are as follows:

vm,n = u(m ·δx,n ·δy)

uxx 'vm+1,n + vm−1,n−2vm,n

δx2

uyy 'vm,n+1 + vm,n−1−2vm,n

δy2 .

These, in conjunction with the boundary conditions, yield the system (with δx =δy = 1 and m = 0,1, ...,30, n = 0,1, ...,40):

0 < m < 30, 0 < n < 40 : vm+1,n + vm−1,n + vm,n+1 + vm,n−1−4vm,n = 0 (D.36)0≤ n≤ 40 : v0,n = 1 (D.37)0≤ n≤ 40 : v30,n = 0 (D.38)

0 < m < 30 : vm,0 =(

1− m30

)2(D.39)

0 < m < 30 : um,40 =(

1− m30

)1/4. (D.40)

Equations (D.36)–(D.40) are a system of linear algebraic equations with unknownsv0,0, v0,1, ..., v0,40, v1,0, v1,1, ..., v1,40, ..., v30,40. It can be shown that the systemmatrix is invertible and can be solved to obtain u(m ·δx,n ·δy) ' vm,n (for m =0,1, ...,30 and n= 0,1, ...,40). However, we will follow a different approach. Considerthe system of difference equations (for m = 0,1, ...,30, n = 1, ...,40, t = 0,1,2, ...):

0 < m < 30, 0 < n < 40 : vtm,n =

vt−1m+1,n + vt−1

m−1,n + vt−1m,n+1 + vt−1

m,n−1 +4vt−1m,n

8(D.41)

0≤ n≤ 40 : vt0,n = 1 (D.42)

0≤ n≤ 40 : vt30,n = 0 (D.43)

0 < m < 30 : vtm,0 =

(1− m

30

)2(D.44)

0 < m < 30 : utm,40 =

(1− m

30

)1/4(D.45)

with random initial conditions v0m,n (m = 0,1, ...,30, n = 1, ...,40, t = 0,1,2, ...). While

not obvious, it can be proved that, as t→ ∞, the system (D.41)–(D.45) converges

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to the solution of the system (D.36)–(D.40). In other words, for m = 0,1, ...,30,n = 1, ...,40, we have

limt→∞

vtm,n = vm,n. (D.46)

Then, the numerical solution of (D.41)–(D.45) gives an approximate solution of(D.31)–(D.35). The algorithm is implemented by the following MATLAB code.

M=30;N=40;T=100;u=rand(M,N);u(1,:)=ones(1,N);u(M,:)=zeros(1,N);u(:,1)=(([1:-1/M:1/M]).^2)’;u(:,N)=(([1:-1/M:1/M]).^(1/4))’;for t=1:T

uold=u;for m=2:M-1

for n=2:N-1u(m,n)=(4*uold(m,n)+uold(m-1,n)+uold(m+1,n)+uold(m,n-1)+uold(m,n+1))/8;

endenddisp([t max(max(abs(u-uold)))])

end

The results are plotted below.

D.1.9 Example. The above method can be applied to more general problems. Thefollowing code solves a Laplace problem on more complicated region.

clearM=30;N=40;T=100;

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D.2 Advanced Methods 251

u=rand(M,N);u(1,:)=ones(1,N);u(M,:)=zeros(1,N);u(:,1)=(([1:-1/M:1/M]).^2)’;u(:,N)=(([1:-1/M:1/M]).^(1/4))’;

for m=11:M-10for n=11:N-25u(m,n)=1;

endendfor t=1:T

uold=u;for m=2:M-1

for n=2:N-1u(m,n)=(4*uold(m,n)+uold(m-1,n)+uold(m+1,n)+uold(m,n-1)+uold(m,n+1))/8;

endendfor m=11:M-10

for n=11:N-25u(m,n)=1;

endenddisp([t max(max(abs(u-uold)))])

end

The results are as follows.

D.2 Advanced MethodsD.2.1 Let us now turn to more sophisticated methods for the numerical solutionof PDEs. There is an extensive literature on this subject, and a vast collection ofsolution methods have been developed. We are not going to explain these methods(a full course could be devoted to this).

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D.2.2 Instead we will look at the problem for the computer user ’s perspective.A large number of computer PDE solvers is available; some of them must bepurchased and others are freely available. We will mention one example from eachcategory.

D.2.1 MATLAB Partial Differential Equation Toolbox

D.2.3 A good proprietary PDE solver is the MATLAB Partial Differential EquationToolbox. It provides functions for solving PDEs using finite element analysis. Italso provides a GUI which facilitates the definition of the problem (especially thespecification of complex regions).

D.2.4 Example. As an example of the use of the PDE toolbox, we present thecommands required for the solution of the problem

0≤ x2 + y2 < 1 : uxx +utt = δ (x,y) ,

x2 + y2 = 1 : u(x,y) = 0.

This is a Poisson equation, i.e., a Laplace equation with input. In this case, theimput is δ (x,y), which is the two dimensional Dirac delta function:∫ ∫

Ω

f (x,y)δ (x,y)dxdy =

f (0,0) if (0,0) ∈Ω,0 if (0,0) /∈Ω.

The equation must hold on the unit disk; and we have zero boundary conditionsat the unit circle. The required MATLAB commands are:

% Problem Definitionc = 1;a = 0;f = @circlef;% Create the modelnumberOfPDE = 1;pdem = createpde(numberOfPDE);

% Create geometry and append to the PDE Modelg = @circleg;geometryFromEdges(pdem,g);

% Define the boundary conditionsfigure;pdegplot(pdem, ’edgeLabels’, ’on’);axis equaltitle ’Geometry With Edge Labels Displayed’;% Solution is zero at all four outer edges of the circleapplyBoundaryCondition(pdem,’Edge’,(1:4), ’u’, 0);

% Solve the equation and plot the solution[u,p,e,t] = adaptmesh(g,pdem,c,a,f,’tripick’,’circlepick’,’maxt’,2000,’par’,1e-3);

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figure;pdeplot(p,e,t,’xydata’,u,’zdata’,u,’mesh’,’off’);

The plot is as follows.

D.2.2 FlexPDE

D.2.5 The “Lite” version of the FlexPDE solver is available for free. There is also a“Pro”, paid version, but the free version is extremely capable and comes with greatdocumentaion and many examples.

D.2.6 You can dowload it from https://www.pdesolutions.com/sdmenu7.html.Make sure that you also download the free book Fields of Physics by Finite ElementAnalysis - An Introduction, by G. Backstrom, from https://www.pdesolutions.com/bookstore.html;it explains a large number of the examples included in the FlexPDE distribution.

D.2.7 Example. Here is an example from the above mentioned book. It concernsfinding the electric potential u(x,y) in a trapezoidal plate under given boundaryconditions. In the interior of the region, the potential satisfies the Laplace equation

uxx +uyy = 0.

The boundary conditions are rather complicated to list in mathematical notation,but they can be read easily from the respective code:

TITLE cond3.pde ’Conduction in a Trapezoidal Plate’ Find the potential U in a plate under an impressed voltage. From "Fields of Physics" by Gunnar Backstrom

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SELECTerrlim= 1e-4

VARIABLESU

DEFINITIONSL1= 0.5L2= 0.25Ly= 1cond= 5.99e7Ex= -dx(U)Ey= -dy(U)E= -grad(U)Em= magnitude(E)Jx= cond*ExJy= cond*EyJ= cond*EJm= magnitude(j)eqn= div( -cond*grad(U))

EQUATIONSdiv( -cond*grad(U))= 0

BOUNDARIESregion 1

start(-L1,0) value(U)= 0line to (L1,0) natural(U)= 0line to (L2,Ly) value(U)= 1.0line to (-L2,Ly) natural(U)= 0line to close

END

Here is the plot of the results

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D.2.8 Example. Here is one more example u(x,y). It concerns finding the steadystate temperature u(x,y) in a cylindrical insulator which contains two hot watertubes. In the interior of the region, u(x,y) satisfies the Laplace equation

uxx +uyy = 0.

The boundary conditions are rather complicated to list in mathematical notation,but they can be read easily from the respective code:

TITLE heat1.pde ’Two Insulated Tubes’

Temperature field in an insulator containing two hot-water tubes. From "Fields of Physics" by Gunnar Backstrom SELECT

errlim= 1e-4VARIABLES

tempDEFINITIONS

r0= 0.1d= 0.15r1= 0.5Lx= 0.3Ly= 0.2k= 0.03 Thermal conductivity of insulation fluxd_x= -k*dx(temp)

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fluxd_y= -k*dy(temp)fluxd= -k*grad( temp)fluxdm= magnitude( fluxd)f_angle= sign(fluxd_y)* arccos(fluxd_x/fluxdm)/pi*180

EQUATIONSdiv(-k*grad(temp))= 0

BOUNDARIESregion 1

start ’outer’ (0,-r1) value(temp)= 273 Frozen soil arc( center= 0,0) angle= 360 closestart ’left’ (-d-r0,0) value(temp)= 323 Cutout for hot water tube 2 arc( center= -d,0) angle= -360 closestart ’right’ (d-r0,0) value(temp)= 353 Cutout for hot water tube 1 arc( center= d,0) angle= -360 close

END

The results are plotted below.

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D.3 Projects 257

D.3 Projects1. assa

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Bibliography

[1] Haberman, Richard. Elementary applied partial differential equations. Vol. 987.Englewood Cliffs, NJ: Prentice Hall, 1983.

kehagias - users.auth.gr · matematicas vol. 5, no. 1/2 (1997), pp. 39–41. the full proof of the...

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