MA 201: Partial Differential EquationsLecture - 7

MA201(2018):PDE

Second Order Linear Equations

• Consider the second order linear equation in two independentvariables

(Auxx + Buxy + Cuyy + Dux + Euy + Fu)(x , y) = −G(x , y). (1)

• In operator notation

(T (u))(x , y) = −G(x , y) = f (x , y) (Say), (2)

withT (u) = Auxx + Buxy + Cuyy + Dux + Euy + Fu.

• Since T is linear (why?), we have

T (u1 + u2) = T (u1) + T (u2) & T (cu) = cT (u) ∀c ∈ R. (3)

Remark:

• Equation (2) is called homogeneous, if f = 0, otherwise it is callednon-homogeneous.

MA201(2018):PDE

The Principle of Superposition

Theorem

Suppose u1 solves linear PDE T (u) = f1 and u2 solves T (u) = f2, thenu = c1u1 + c2u2 solves T (u) = c1f1 + c2f2. In particular, if u1 and u2both solve the same homogeneous linear PDE T (u) = 0, so doesu = c1u1 + c2u2.

Remark:

• Any linear combination of solutions of linear homogeneous PDE isalso a solution.

• A solution u = u(x , y) to a homogeneous equation T (u) = 0 iscalled general solution if it contains two arbitrary functions.

• If u is a general solution to homogeneous equation T (u) = 0 and upis a particular solution to non-homogeneous equation T (w) = f ,then u + up is also a solution to the non-homogeneous equation andit is called general solution to T (w) = f .

MA201(2018):PDE

Linear Equations with Constant Coefficients

With the notations D = ∂/∂x and D ′ = ∂/∂y , a PDE with constantcoefficients can be written as

F (D,D ′)u = f . (4)

We further classify the PDE (4) into two main types:

• Reducible: Equation (4) is called reducible if it can be written asthe product of linear factors of the form aD + bD ′ + c , withconstants a, b, c . For example, the equation

uxx − uyy = 0.

In this case

F (D,D ′) = D2 − (D ′)2 = (D + D ′)(D − D ′).

• Irreducible: Equation (4) is called irreducible if it not reducible. Forexample when F (D,D ′) = D2 − D ′.

MA201(2018):PDE

Linear Equations with Constant Coefficients: Reducible Equation

An n-th order reducible PDE can be written as

F (D,D ′)u =(

n∏

r=1

(arD + brD′ + cr )

)

u = f . (5)

Theorem

If (arD + brD′ + cr ) is a factor of F (D,D ′), ar 6= 0, then

ur = exp

{

−crx

ar

}

φr (brx − ary)

is a solution of the equation F (D,D ′)u = 0. Here, φr is an arbitrary realvalued single variable function.

Theorem

If (brD′ + cr ) is a factor of F (D,D ′) and φr is an arbitrary real valued

single variable function, then

ur = exp

{

−cry

br

}

φr (brx)

is a solution of the equation F (D,D ′)u = 0.MA201(2018):PDE

Linear Equations with Constant Coefficients: Reducible Equation

Theorem

If (aD + bD ′ + c)m (m ≤ n, a 6= 0) is a factor of F (D,D ′) andφ1, φ2, . . . , φm are arbitrary real valued single variable functions, then

exp{

−cx

a

}

m∑

i=1

x i−1φi (bx − ay)

is a solution of the equation F (D,D ′)u = 0.

Theorem

If (bD ′ + c)m (m ≤ n) is a factor of F (D,D ′) and φ1, φ2, . . . , φm arereal valued single variable functions, then

exp{

−cy

b

}

m∑

i=1

x i−1φi (bx)

is a solution of the equation F (D,D ′)u = 0.

MA201(2018):PDE

Reducible Equations: Examples

Example

• General solution ofuxx − uyy = 0

is given byu = φ1(x − y) + φ2(x + y),

φ1 and φ2 are arbitrary real valued single variable functions.

• General solution of

∂4u

∂x4+∂4u

∂y4= 2

∂4u

∂x2∂y2

is given by

u = xφ1(x − y) + φ2(x − y) + xψ1(x + y) + ψ2(x + y).

MA201(2018):PDE

One Dimensional Wave Equation

MA201(2018):PDE

Vibrating string and the wave equation

• Consider a stretched string of length π with the ends fastened to the ends x = 0and x = π.

• Suppose that the string is set to vibrate by displacing it from its equilibriumposition.

• Let u(x , t) denote the transverse displacement at time t ≥ 0 of the point on thestring at position x . That is, we assume that each point of the string moves onlyin the vertical direction.

• In particular, u(x , 0) denotes the initial shape of the string and ut(x , 0) denotesthe initial velocity.

• That is, the string is set to vibrate by supplying an initial velocity ut(x , 0) to itfrom its equilibrium position u(x , 0) and then releasing it.

MA201(2018):PDE

Vibrating string and the wave equation (Contd.)• Under some physical assumptions, we arrive at the following equation known asthe one-dimensional wave equation:

utt(x , t) = c2uxx(x , t), (6)

which governs the entire process.

• Here c represents a physical quantity.

Note:

• The above equation is derived without considering any external force acting onthe string.

• Suppose that we do wish to include such an external force on the string (due toits weight or other impressed external forces (like gravity or pressure)), referredto as a load, given by

f (x , t) = vertical force per unit length at point x , at time t.

MA201(2018):PDE

Vibrating string and the wave equation (Contd.)• Then, the result will be the nonhomogeneous wave equation

utt − c2uxx = F (x , t), (7)

where F (x , t) = 1ρf (x , t), with ρ as the mass per unit length of the string.

Two special cases of (7) particularly generate interest:

• When the external force is due to the gravitational acceleration g only, theequation becomes

utt = c2uxx − g . (8)

• When the external force is due to the resistance of the medium (a stringvibrating in a fluid), the equation becomes

utt − c2uxx = −kut , k is a positive constant, (9)

known as damped wave equation.

MA201(2018):PDE

Wave EquationIt is to be noted that the use of string problem to demonstrate the waveequation is a matter of convenience. There are more applications inPhysics and Engineering.

For instance,u(x , t) = sin(x ± ct)

represents sinusoidal waves traveling with speed c in the positive andnegative directions, respectively, without change of shape.

Figure : Water Wave

MA201(2018):PDE

Wave Equation

Figure : Sound Wave

MA201(2018):PDE

Wave Equation (Contd.)

• Consider one-dimensional homogeneous wave equation

utt = c2uxx , (10)

• The solution is given by

u(x , t) = f (x − ct) + g(x + ct) (11)

where f and g are arbitrary real valued single variable functions.

• The physical interpretations of the functions f and g are quiteinteresting.

MA201(2018):PDE

Wave Equation (Contd.)To see this let us first consider the solution u = f (x − ct).

At t = 0, it defines the curve u = f (x) = u0(x)(say ), and after timet = t1, it defines the curve u = f (x − ct1) = u1(x)(say ).

But these curves u0 and u1 are identical in the sense that

u1(x) = f (x − ct1) = u0(X ),

that is, the origin is shifted by a distance ct1 towards right.

Figure : A Progressive Wave

MA201(2018):PDE

Wave equation (Contd.)For instance, consider

y (x , t) = sin(x − ct)

• A particle at the location (0, 0) at time t = 0 moves to location(ct1, 0) at time t = t1.

MA201(2018):PDE

Wave Equation (Contd.)Thus the entire configuration moves along the positive direction of thex-axis a distance of ct1 in time t1.

The velocity with which the wave is propagated is, therefore,

v =ct1

t1= c

Similarly the function g(x + ct) defines a wave progressing in thenegative direction of the x-axis with constant velocity c.

The total solution is, therefore, the algebraic sum of these two travelingwaves.

Solution (11) is a very convenient representation for progressive waveswhich travel large distances through a uniform medium.

MA201(2018):PDE

The D’Alembert solution of the wave equationLet us consider the following initial value problem

utt − c2uxx = 0, −∞ < x <∞, (12)

Displacement: u(x , 0) = φ(x), (13)

Velocity: ut(x , 0) = ψ(x), (14)

that is, we consider the vibration of a thin string of infinite length withsome initial displacement and initial velocity.

From solution (11), we find that

f (x) + g(x) = φ(x), (15)

−cf ′(x) + c g ′(x) = ψ(x), (16)

for all values of x .

MA201(2018):PDE

The D’Alembert solution of the wave equation (Contd.)Integrating the second equation with respect to x

−f (x) + g(x) =1

c

∫ x

x0

ψ(τ)dτ + A. (17)

where A is an integration constant and τ is a dummy variable.

From (15) and (17)

f (x) =1

2

[

φ(x)−1

c

∫ x

x0

ψ(τ)dτ

]

− A/2, (18)

g(x) =1

2

[

φ(x) +1

c

∫ x

x0

ψ(τ)dτ

]

+ A/2, (19)

Substituting these expressions into (11)

u(x , t) =1

2[φ(x − ct) + φ(x + ct)] +

1

2c

∫ x+ct

x−ct

ψ(τ)dτ. (20)

This is D’Alembert’s (1717-1783) solution of one-dimensional wave

equation, which he actually derived around 1746.

MA201(2018):PDE

Non-Homogeneous Wave Equation: Duhamel’s PrincipleFind u(x , t) such that

utt(x , t)− c2uxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞) (21)

with ICs u(x , 0) = 0, ut(x , 0) = 0 ∀x ∈ (−∞,∞). (22)

Then the solution u is given by

u(x , t) =

∫ t

0

v(x , t − τ, τ)dτ. (23)

What is v?

MA201(2018):PDE

Duhamel’s Principle (Contd.)Here, v is the solution of the problem

vtt − αvxx = 0, (x , t) ∈ (0, L)× (0,∞) (24)

with ICs v(x , 0) = 0, vt(x , 0) = f (x , s), s > 0, (25)

for some real parameter s > 0.

Note that the solution v of the above problem depends on x , t andparameter s.

Thus, v = v (x , t, s). Accordingly, we can modify the IC’s as

v(x , 0, s) = 0 = φ (say), vt(x , 0, s) = f (x , s) = ψ(x) (say), s > 0.

Due to D’Alembert’s, v = v (x , t, s) is given by

v(x , t, s) =1

2c

∫ x+ct

x−ct

ψ(τ)dτ =1

2c

∫ x+ct

x−ct

f (τ, s)dτ.

MA201(2018):PDE

Leibniz Rule: Differentiation under the sign of integration

d

dt

(

∫ b(t)

a(t)

G (x , t)dx)

=

∫ b(t)

a(t)

Gtdx + G (b(t), t)b′(t)

−G (a(t), t)a′(t). (26)

Use Leibniz rule on

u(x , t) =

∫ t

0

v (x , t − τ, τ)dτ

to have

ut =

∫ t

0

vt(x , t − τ, τ)dτ + v(x , t − t, t)×dt

dt− v(x , t − 0, 0)× 0

=

∫ t

0

vt(x , t − τ, τ)dτ + v(x , 0, t) =

∫ t

0

vt(x , t − τ, τ)dτ.

Similarly

utt(x , t) =

∫ t

0

vtt(x , t − τ, τ)dτ + vt(x , 0, t) =

∫ t

0

vtt(x , t − τ, τ)dτ + f (x , t)

c2uxx (x , t) = c2∫ t

0

vxx(x , t − τ, τ)dτ so that utt − c2uxx = f (x , t).

MA201(2018):PDE

Completely Non-Homogeneous Wave Equation: Duhamel’s

PrincipleFind u(x , t) such that

utt(x , t)− c2uxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞) (27)

with ICs u(x , 0) = φ(x), ut(x , 0) = ψ(x) ∀x ∈ (−∞,∞). (28)

Try to break the problem as (ρ, θ) problems as: Find θ such that

θtt(x , t)− c2θxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞) (29)

with IC’s θ(x , 0) = 0, θt(x , 0) = 0 ∀x ∈ (−∞,∞). (30)

Find ρ such that

ρtt(x , t)− c2ρxx = 0, (x , t) ∈ (−∞,∞)× (0,∞) (31)

with IC’s ρ(x , 0) = φ(x), ρt(x , 0) = ψ(x) ∀x ∈ (−∞,∞). (32)

Principle of Super Position tells that ρ+ θ solves the completelynon-homogeneous problem.

MA201(2018):PDE