intro to method of applied math

8/19/2019 Intro to method of applied math

1/187

Methods of Applied Mathematics I

Math 583, Fall 2015

Jens Lorenz

December 16, 2015

Department of Mathematics and Statistics,UNM, Albuquerque, NM 87131

Contents

1 A Simple Boundary Value Problem 51.1 Existence, Uniqueness, Green’s Function . . . . . . . . . . . . . . 51.2 Formalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 Perturbed Equations and the Contraction Mapping Theorem 112.1 An Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 The Contraction Mapping Theorem . . . . . . . . . . . . . . . . 112.3 Example: extended . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.4 The Mean Value Theorem in Integral Form . . . . . . . . . . . . 152.5 Completeness of (C (Ω), | · |∞) . . . . . . . . . . . . . . . . . . . . 162.6 Completeness of (C 2[0, l], | · |∞,2) . . . . . . . . . . . . . . . . . . 18

3 Fredholm’s Alternative and Green’s Function 193.1 Linear Second–Order BVPs . . . . . . . . . . . . . . . . . . . . . 193.2 The Green’s Function . . . . . . . . . . . . . . . . . . . . . . . . 213.3 Estimates for the Integral Operator G . . . . . . . . . . . . . . . 233.4 The Formal Adjoint . . . . . . . . . . . . . . . . . . . . . . . . . 25

4 Introduction to Distribution Theory 284.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.2 Testfunctions and Distributions . . . . . . . . . . . . . . . . . . . 304.3 Examples of Regular and Singular Distributions . . . . . . . . . . 314.4 Differentiation of Distributions . . . . . . . . . . . . . . . . . . . 324.5 The Simplest Differential Equations for Distributions . . . . . . . 334.6 A Simple Equation Involving Distributions . . . . . . . . . . . . 374.7 The Formal Adjoint . . . . . . . . . . . . . . . . . . . . . . . . . 384.8 Further Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 404.9 The Green’s Function Revisited . . . . . . . . . . . . . . . . . . . 42

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5 Three Types of Linear Problems 465.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

5.1.1 Inhomogeneous Problems . . . . . . . . . . . . . . . . . . 46

5.1.2 Spectral Problems . . . . . . . . . . . . . . . . . . . . . . 465.1.3 Problems of Evolution; Semi–Group Theory . . . . . . . . 47

5.2 BVPs as Operator Equations . . . . . . . . . . . . . . . . . . . . 475.3 Inhomogeneous Problems . . . . . . . . . . . . . . . . . . . . . . 495.4 Spectral Problems . . . . . . . . . . . . . . . . . . . . . . . . . . 495.5 Problems of Evolution: Semigroup Theory . . . . . . . . . . . . . 49

6 Linear Operators: Riesz Index, Projectors, Resolvent 506.1 Nullspaces and Ranges . . . . . . . . . . . . . . . . . . . . . . . . 506.2 The Neumann Series . . . . . . . . . . . . . . . . . . . . . . . . . 566.3 Spectral Theory of Matrices . . . . . . . . . . . . . . . . . . . . . 58

6.3.1 Eigenvalues and Resolvent . . . . . . . . . . . . . . . . . . 586.3.2 Jordan Normal Form and Riesz Index . . . . . . . . . . . 596.3.3 Bases for N r and Rr . . . . . . . . . . . . . . . . . . . . . 606.3.4 Pro jectors . . . . . . . . . . . . . . . . . . . . . . . . . . . 616.3.5 Integral Representation of the Projector P . . . . . . . . . 626.3.6 Direct Sum Decompositions for Different Eigenvalues . . . 63

7 Riesz–Schauder Theory of Linear Compact Operators 667.1 Auxiliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . 667.2 Null Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 697.3 Range Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 717.4 The Spectrum of Linear Compact Operators . . . . . . . . . . . . 76

8 Hilbert–Schmidt Kernels and Compact Operators 798.1 Integral Operators With Continuous Kernels . . . . . . . . . . . 798.2 Limits of Compact Operators . . . . . . . . . . . . . . . . . . . . 808.3 Hilbert–Schmidt Kernels . . . . . . . . . . . . . . . . . . . . . . . 81

9 Best Approximations in Inner Product Spaces 849.1 The Gram-Schmidt Process . . . . . . . . . . . . . . . . . . . . . 849.2 Best Approximations . . . . . . . . . . . . . . . . . . . . . . . . . 849.3 The Decomposition H = V ⊕ V ⊥ . . . . . . . . . . . . . . . . . . 869.4 Riesz Representation Theorem . . . . . . . . . . . . . . . . . . . 87

10 Spectral Theory of Compact Hermitian Operators 8910.1 Motivation: The Matrix Case . . . . . . . . . . . . . . . . . . . . 8910.2 Use of the Quadratic Form (u, Au) . . . . . . . . . . . . . . . . . 9010.3 Existence of an Orthonormal Sequence of Eigenvectors . . . . . . 9510.4 Density of C ∞0 in L2 . . . . . . . . . . . . . . . . . . . . . . . . . 9910.5 Application to Ordinary BVPs . . . . . . . . . . . . . . . . . . . 10210.6 The Sign of Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . 10510.7 Lower Bound for Eigenvalues . . . . . . . . . . . . . . . . . . . . 10710.8 Expansion of the Green’s function . . . . . . . . . . . . . . . . . 110

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10.9 Two Formulas for Green’s Functions: An Example . . . . . . . . 11210.9.1 First Formula for the Green’s Function: Eigenfunction

Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

10.9.2 Second Formula for the Green’s Function: Via Delta–Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

11 Initial–Boundary Value Problems on a Finite x–Interval 11711.1 The Formal Solution Process for an IBVP . . . . . . . . . . . . . 11711.2 A Particular Problem . . . . . . . . . . . . . . . . . . . . . . . . 11911.3 An IBVP on 0 ≤ x ≤ l . . . . . . . . . . . . . . . . . . . . . . . . 122

11.3.1 Solution Via Separation of Variables . . . . . . . . . . . . 12211.3.2 Solution Via Laplace Transform in Time . . . . . . . . . . 122

12 A Problem on the Half–Line 0 ≤ x < ∞ 12412.1 An ODE for 0 ≤ x < ∞ . . . . . . . . . . . . . . . . . . . . . . . 12412.2 Transformation to a First–Order System . . . . . . . . . . . . . . 12412.3 The Green’s Function . . . . . . . . . . . . . . . . . . . . . . . . 12812.4 The Case 0 < λ < ∞ . . . . . . . . . . . . . . . . . . . . . . . . . 12912.5 Estimate of u by u and u . . . . . . . . . . . . . . . . . . 13412.6 S ummary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

13 An Initial–Boundary Value Problem 13713.1 Application of Laplace Transform . . . . . . . . . . . . . . . . . . 13713.2 Fourier and Laplace Transform . . . . . . . . . . . . . . . . . . . 13813.3 Inverse Laplace Transform: The Simplest Example . . . . . . . . 13913.4 Inverse Laplace Transform: The Case a = 0 in Theorem 13.1 . . 143

13.5 Inverse Laplace Transform: Proof of Theorem 13.1 . . . . . . . . 14513.6 Solution Formula and Example . . . . . . . . . . . . . . . . . . . 14713.7 The Strip Problem Reconsidered . . . . . . . . . . . . . . . . . . 149

14 Fourier Transformation: Application to the Equation u − λu =f 15214.1 Fourier Transformation . . . . . . . . . . . . . . . . . . . . . . . . 15214.2 The Equation u(x) − λu(x) = f (x) for x ∈ R . . . . . . . . . . . 153

14.2.1 The Case λ /∈ iR . . . . . . . . . . . . . . . . . . . . . . . 15414.2.2 The Case λ ∈ iR . . . . . . . . . . . . . . . . . . . . . . . 155

14.3 Extension of a Bounded Linear Operator . . . . . . . . . . . . . . 156

15 The Cauchy Problem for the Heat Equation 15915.1 Fourier Transform in Space . . . . . . . . . . . . . . . . . . . . . 15915.2 Laplace Transform in Time . . . . . . . . . . . . . . . . . . . . . 16115.3 Transform in Space and Time . . . . . . . . . . . . . . . . . . . . 16215.4 Remarks on the Heat Kernel and Non–Uniqueness . . . . . . . . 165

16 The Cauchy Problem for the Wave Equation 16916.1 T he 1D Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

16.1.1 The 1D Case with g ≡ 0 . . . . . . . . . . . . . . . . . . . 16916.1.2 The 1D Case with f ≡ 0 . . . . . . . . . . . . . . . . . . . 171

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16.2 T he 3D Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17216.2.1 The 3D Case with f ≡ 0 . . . . . . . . . . . . . . . . . . . 17216.2.2 The 3D Case with g

≡0 . . . . . . . . . . . . . . . . . . . 175

16.2.3 Light Cones . . . . . . . . . . . . . . . . . . . . . . . . . . 17616.3 The 2D Wave Equation: Method of Descent . . . . . . . . . . . . 177

17 Elliptic Equations with Constant Coefficients 179

18 Remarks on Hyperbolic Equations 185

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1 A Simple Boundary Value Problem

1.1 Existence, Uniqueness, Green’s Function

Let f ∈ C [0, l] denote a given function. Try to find a function u ∈ C 2[0, l]satisfying

−u(x) = f (x) for 0 ≤ x ≤ l, u(0) = u(l) = 0 . (1.1)Here the differential equation −u = f is supplemented by two homogeneousDirichlet boundary conditions:

u(0) = u(l) = 0 .

Uniqueness: Suppose that u1 and u2 solve the BVP (1.1). We set u =

u1 − u2 and obtain thatu(x) = 0 for 0 ≤ x ≤ l, u(0) = u(l) = 0 .

It follows that u(x) = A + Bx and the boundary conditions yield that A = B =0. Therefore, the BVP has at most one solution.

Existence: By integration, we first obtain a special solution of the differ-ential equation,

usp(x) = − x0

t0

f (s) ds .

Since the general solution of the homogeneous differential equation −u = 0 isuhom(x) = A + Bx

we obtain that

u(x) = usp(x) + A + Bx

is the general solution of the equation −u = f . We now impose the boundaryconditions to obtain A and B :

0 = u(0) = A, 0 = u(l) = usp(l) + Bl .

This yields the following solution of the BVP:

u(x) = usp(x) − usp(l) xl

.

Motivation for a Green’s Function: We want to find a continuousfunction

g : [0, l] × [0, l] → Rwith the following property: For every f ∈ C [0, l] the solution u(x) of the BVP(1.1) is given by

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u(x) =

l0

g(x, y)f (y) dy . (1.2)

If this holds then we have

0 = u(0) =

l0

g(0, y)f (y) dy for all f ∈ C [0, l] ,

which implies that

g(0, y) ≡ 0 .Similarly,

g(l, y) ≡ 0 .If we differentiate (1.2) twice and formally put the derivatives under the integralsign, then we obtain

f (x) = −u(x) = − l0

gxx(x, y)f (y) dy .

Since, formally,

l0

δ (x − y)f (y) dy = f (x) ,

we are lead to require

−gxx(x, y) = δ (x − y) .This motivates the following construction.

Construction of a Solution Via a Green’s Function: Fix at point0 < x0 < l and try to find a function g(x) with

−g(x) = δ (x − x0), g(0) = g(l) = 0 .Here we first take an intuitive view of the δ –function and think of the right–handside of the above equation as a function satisfying

l0 δ (x − x0) dx = 1, δ (x − x0) = 0 for x = x0 .

Thus, the conditions for g(x) are the boundary conditions

g(0) = g(l) = 0

and

g(x) = 0 for x = x0 and 1 = − l0

g(x) dx = g (0) − g(l)

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In addition, we require that g ∈ C [0, l].The differential equation g (x) = 0 for x = x0 yields

g(x) =

Ax + B for 0 ≤ x < x0Cx + D for x0 < x ≤ l

The boundary conditions imply

B = 0 and D = −Cl .Therefore,

g(x) =

Ax for 0 ≤ x < x0

C (x − l) for x0 < x ≤ lThe continuity condition g ∈ C [0, l] yields that

Ax0 = C (x0 − l) ,thus

C = Ax0x0 − l .

Obtain:

g(x) =

Ax for 0 ≤ x ≤ x0Ax0x0−l (x − l) for x0 ≤ x ≤ l

and

g(x) =

A for 0 ≤ x < x0Ax0x0−l for x0 < x ≤ l

Therefore,

g(0) − g(l) = A(1 − x0x0 − l ) = A

−lx0 − l .

The condition g(0) − g(l) = 1 yields that

A = l − x0

l .

We have obtained:

g(x) = 1

l

x(l − x0) for 0 ≤ x ≤ x0x0(l − x) for x0 ≤ x ≤ l

Clearly, the function depends on x as well as x0.With a change of notation, let

g(x, y) = 1

l

x(l − y) for 0 ≤ x ≤ yy(l − x) for y ≤ x ≤ l

We then have that g ∈ C ([0, l] × [0, l]) and

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−gxx(x, y) = δ (x − y), g(0, y) = g(l, y) = 0 .

The function g(x, y) is called the Green’s function for the BVP (1.1). It istrivial, but important, to note that the function g(x, y) does not depend on theinhomogeneous term f (x) in the differential equation.

Note that

gx(x, y) = 1

l

l − y for 0 ≤ x < y

−y for y < x ≤ l

Theorem 1.1 For any f ∈ C [0, l] the solution of the BVP (1.1) is

u(x) =

l0

g(x, y)f (y) dy . (1.3)

It is easy to check that the function u(x) defined in (1.3) satisfies u(0) =u(l) = 0. Thus, it remains to show that −u = f .

Formal Argument: If we formally differentiate under the integral sign,we obtain

−u(x) = l0

−gxx(x, y)f (y) dy

=

l0

δ (x − y)f (y) dy

= f (x)

Rigorous Argument: From

u(x) =

l0

g(x, y)f (y) dy

obtain that

ux(x) =

l0

gx(x, y)f (y) dy

= x0 −

y

l f (y) dy +

1

l lx (l − y)f (y) dy

and

uxx(x) = −xl

f (x) − 1l

(l − x)f (x) = f (x) .

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1.2 Formalization

Let

V = C [0, l]

denote the space of right–hand sides f of the differential equation −u = f inBVP.

Let

U = {u ∈ C 2[0, l] : u(0) = u(l) = 0}denote the solution space. Note that the boundary conditions are taken care of by imposing them onto the functions in the solution space.

We have shown that the differential operator

L :

U → V u → −u

is 1–1 and onto. The inverse of L is the integral operator

G :

V → U

f → l0 g(x, y)f (y) dy = (Gf )(x)On V = C [0, l] define the norm

|f |∞ = max0≤x≤l

|f (x| for f ∈ V .

On C 2

[0, l] define the norm

|u|∞,2 = |u||∞ + |u|∞ + |u|∞ for u ∈ C 2[0, l] .Since U ⊂ C 2[0, l] the norm | · |∞,2 is also a norm on U .

Theorem 1.2 1) For all f ∈ V = C [0, l] the estimate

|Gf |∞ ≤ C 0|f |∞ (1.4)holds with

C 0 = max0≤x≤l l

0 g(x, y) dy .

2) Let f ∈ V and set u = Gf . There is a constant C > 0, independent of f , so that the estimate

|u|∞ + |u|∞ + |u|∞ ≤ C |f |∞ (1.5)holds.

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Proof: 1) Let u = Gf . We have for all x ∈ [0, l]:

|u(x)| ≤ l0

|g(x, y)||f (y)| dy

≤ |f |∞ l0

|g(x, y)| dy

Taking the maximum over 0 ≤ x ≤ l and noting that g(x, y) ≥ 0, the firstestimate follows.

2) By 1) we have |u|∞ ≤ C 0|f |∞. Also, since −u = f we have |u|∞ = |f |∞.It remains to estimate |u|∞. Since

u(0) = u(l) = 0

there exists 0 ≤ ξ ≤ l with u(ξ ) = 0. Then we have

u(x) = u(x) − u(ξ ) = xξ

u(s) ds ,

thus

|u(x)| ≤ |x − ξ ||u|∞ ≤ l|u|∞ = l|f |∞ .Therefore,

|u|∞,2 ≤ (1 + l + C 0)|f |∞ .

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2 Perturbed Equations and the Contraction Map-

ping Theorem

2.1 An Example

Consider the nonlinear boundary value problem

−u(x) = f (x) + ε cos u(x) for 0 ≤ x ≤ l, u(0) = u(l) = 0 , (2.1)where f ∈ V := C [0, l] is a given function.

Applying the operator G introduced in Section 1.2 yields the fixed pointequation

u = Gf + εG cos u .

For fixed f ∈ V , define the following map

Φ :

V → U ⊂ V

u → Φ(u) = Gf + εG cos uUsing Theorem 1.2 we obtain for all u1, u2 ∈ V :

|Φ(u1) − Φ(u2)|∞ ≤ |ε|C 0 | cos u1 − cos u2|∞≤ |ε|C 0 |u1 − u2|∞

The second estimate follows from the mean value theorem and the bound

| sin ξ | ≤ 1 for all real ξ .Therefore, if |ε|C 0 < 1, the contraction mapping theorem implies that there

exists a unique u ∈ V with

u = Φ(u) .

It then follows that u ∈ U , and u is the unique solution of the BVP (2.1).

2.2 The Contraction Mapping Theorem

Metric Spaces. Let X denote any nonempty set. A function

d :

X × X → [0, ∞)(u, v) → d(u, v)is called a metric on X if the following holds for all u, v, w ∈ X :

1) d(u, v) = d(v, u)2) d(u, w) ≤ d(u, v) + d(v, w)3) d(u, v) = 0 if and only if u = v.

A pair (X, d) where X is a nonempty set and d is a metric on X is called ametric space.

A sequence un ∈ X is called a Cauchy sequence if for all ε > 0 there existsan integer N (ε) so that

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d(um, un) < ε for all m, n ≥ N (ε) .

A metric space (X, d) is called complete if every Cauchy sequence in X hasa limit in X .If

U, ·

is a normed space then

d(u, v) = u − v for u, v ∈ U defines a metric on U . If the corresponding metric space is complete, then

U, ·

is called a Banach space.

Example: Let Ω ⊂ RN denote a compact set and let C (Ω) denote thevector space of all continuous functions u : Ω → R. Define the maximum norm

|u|∞ = maxx∈Ω |u(x)|, u ∈ C (Ω) .One can show that

C (Ω), | · |∞

is a Banach space.

Definition: Let (X, d) denote a metric space. A mapping Φ : X → X iscalled a contraction if there exists 0 ≤ q


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d(un+1, un) ≤ q nd(u1, u0) .

Therefore, if n > m:

d(un, um) ≤ d(un, un−1) + . . . + d(um+1, dm)≤

q n−1 + q n−2 + . . . + q m

d(u1, u0)

≤ q m(1 + q + . . .)d(u1, u0)=

q m

1 − q d(u1, u0)

Since q m → 0 as m → ∞ it follows that un is a Cauchy sequence in X , thusun → u∗ for some u∗ ∈ X .

In the equation

un+1 = Φ(un)

we let n → ∞. Using the continuity of Φ we obtain that

u∗ = Φ(u∗) ,

i.e., u∗ is a fixed point. Furthermore, in the estimate

d(un, um) ≤ q m

1 − q d(u1, u0)

one can let n go to infinity. Then un →

u∗ and the error estimate (2.2) follows.

2.3 Example: extended

Let

φ : [0, l] ×R3 → Rdenote a C 1 function and consider B V P ε:

−u(x) = f (x) + εφ(x, u(x), u(x), u(x)), u(0) = u(l) = 0 , (2.3)where f ∈ V = C [0, l] is a given function. Define the operator Q : C 2[0, l] →C [0, l] by

Q(u)(x) = φ(x, u(x), u(x), u(x)) for 0 ≤ x ≤ l .As above, we apply the operator G to the differential equation (2.3) and BV P εbecomes the fixed point equation

u = Gf + εGQ(u), u ∈ C 2[0, l] .If ε = 0 then BV P 0 has the unique solution

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u0 := Gf .

We want to show that for |ε| sufficiently small B V P ε has a solution uε close tou0, and this solution is locally unique.

On the space U we use the norm

|u|∞,2 = |u|∞ + |u|∞ + |u|∞ .The estimate (1.5) becomes

|Gh|∞,2 ≤ C G|h|∞ for all h ∈ V = C [0, l] .On C 2[0, l] define the operator Φε by

Φε(u) = u0 + εGQ(u) .

We obtain that Φε : C 2[0, l] → U and BV P ε is equivalent to the fixed point

equation

u = Φε(u), u ∈ C 2[0, l] .If the derivatives of the function φ are unbounded, one cannot expect that theoperator Φε is globally a contraction.

We will restrict Φε to a finite ball centered at u0. In fact, let

B1 = {u ∈ U : |u − u0|∞,2 ≤ 1} .

For u ∈ B1 we have

|Φε(u) − u0|∞,2 = |ε||GQ(u)|∞,2≤ |ε|C G|Q(u)|∞

It is not difficult to prove that there is a constant M Q so that

|Q(u)|∞ ≤ M Q|u|∞,2 for all u ∈ B1 .In fact, one can choose

M Q = max{|

φ(x,α ,β ,γ )|

: 0≤

x≤

l,|α−

u0(x)|+

|β −

u0

(x)|+γ

−u0

(x)| ≤

1}

.

Then the above estimate for |Φε(u) − u0|∞,2 yields that

|Φε(u) − u0|∞,2 ≤ |ε|C GM Q for u ∈ B1 .Therefore, if

|ε|C GM Q ≤ 1 ,then the operator Φε maps the ball B1 into itself.

For u1, u2 ∈ B1 we have

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|Φε(u1)

−Φε(u2)

|∞,2

≤ |ε

|C G

|Q(u1)

−Q(u2)

|∞≤ |ε|C GC Q|u1 − u2|∞,2for a suitable constant C Q. The second estimate follows from the mean valuetheorem; see Section 2.4.

If |ε| is small enough then we obtain by the contraction mapping theoremthat BV P ε has a unique solution uε ∈ B1. (A solution exists in B1 and isunique in B1, but other solutions outside of B1 may also exist.)

We can apply the error estimate (2.2) of the contraction mapping theoremwith

u∗ = uε, m = 0, u1 = Φε(u0) = u0 + εGQ(u0) .

Since |u1 − u0|∞,2 ≤ C |ε| the bound (2.2) yields that

|uε − u0|∞,2 = O(ε) .

This must be expected since BV P ε differs from BV P 0 by an O(ε)–perturbationterm.

2.4 The Mean Value Theorem in Integral Form

A standard form of the mean value theorem of calculus is the following: If f : [a, b] → R is a C 1 function then there exists ξ ∈ [a, b] so that

f (b) − f (a) = f (ξ )(b − a) .In analysis, it is often better to express the value f (ξ ) in integral form, i.e., asan average of derivatives.

Let H : Rm → R denote a C 1 map and let P, Q ∈ Rm. The function

P + t(Q − P ), 0 ≤ t ≤ 1 ,parameterizes the straight line from P to Q.

Set

h(t) = H (P + t(Q

−P )), 0

≤t

≤1 .

By the chain rule,

h(t) = ∇H (P + t(Q − P )) · (Q − P ) ,and by the fundamental theorem of calculus,

h(1) − h(0) = 10

h(t) dt .

Since h(0) = H (P ) and h(1) = H (Q) one obtains

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H (Q)−

H (P ) = 1

0 ∇H (P + t(Q

−P ))

·(Q

−P ) dt

=

10

∇H (P + t(Q − P )) dt · (Q − P )

Here the integral is defined componentwise.If H : Rm → Rs is vector valued, one obtains in the same way:

Theorem 2.2 (Mean value theorem in integral form) Let H : Rm → Rs denote a C 1 function and let P, Q denote two points in Rm. Then the identity

H (Q) − H (P ) = 10

DH (P + t(Q − P )) dt (Q − P )

holds. Here DH (x) ∈ Rs×m denotes the Jacobian of H at x and the integral is defined elementwise.

Application: Let

φ : [0, l] ×R3 → Rdenote a C 1 function and define the operator

Q : C 2[0, l] → C [0, l]

by

Q(u)(x) = φ(x, u(x), u(x), u(x)), 0 ≤ x ≤ l .Note that the derivatives of φ(x,α,β,γ ) are bounded if (α ,β ,γ ) varies in abounded set. Therefore, for all u1, u2 ∈ B1:

|Q(u1)(x) − Q(U 2)(x)| = |φ(x, u1(x), u1(x), u1(x)) − φ(x, u2(x), u2(x), u2(x))|≤ C Q

|u1(x) − u2(x)| + |u1(x) − u2(x)| + |u1(x) − u2(x)|

≤ C Q|u1 − u2|∞,2

2.5 Completeness of (C (Ω), | · |∞)Theorem 2.3 Let Ω ⊂ RN denote a compact set and let C (Ω) denote the vector space of all continuous functions u : Ω → R. Then

|u|∞ = maxx∈Ω

|u(x)|

defines a norm on C (Ω) and the space

C (Ω), | · |∞

is complete.

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2.6 Completeness of (C 2[0, l], | · |∞,2)Let un ∈ C 2[0, l] denote a Cauchy sequence w.r.t. the norm

|u|∞,2 = |u|∞ + |u|∞ + |u|∞ .It follows that the three sequences

un, un, u

n

are Cauchy sequences in C [0, l] w.r.t. |·|∞. Therefore, there exist limits u, v, w ∈C [0, l] with

|un − u|∞ → 0, |un − v|∞ → 0, |un − w|∞ → 0 . (2.4)In the equation

un(x) = un(0) +

x0

un(s) ds for 0 ≤ x ≤ l

we let n → ∞ and obtain that

u(x) = u(0) +

x0

v(s) ds for 0 ≤ x ≤ l .

This implies that u ∈ C 1[0, l] and u = v. In the same way, obtain thatv ∈ C 1[0, l] and v = w. Therefore, u ∈ C 2[0, l] and u = w. Then (2.4) yieldsthat

|un − u|∞,2 → 0 as n → ∞ .We have proved that the normed space

C 2[0, l], | · |∞,2

is complete.

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3 Fredholm’s Alternative and Green’s Function

3.1 Linear Second–Order BVPs

Let a0, a1, a2 ∈ C [a, b] and assume that

a0(x) = 0 for all x ∈ [a, b] .Define the differential operator

Lu = a0u + a1u + a2u for u ∈ C 2[a, b] .

Also, define the boundary operator

Ru = R1uR2u

= α1u(a) + α2u(a)

β 1u(b) + β 2u(b) where (α1, α2) = (0, 0) = (β 1, β 2) .

For given f ∈ C [a, b] consider the BVP

Lu = f, Ru = 0 . (3.1)

For simplicity, we assume all functions to be real.

Auxiliary Results on Initial–Value Problems: We use the same nota-tion as above.

Lemma 3.1 1) Every IVP

Lu(x) = f (x), a ≤ x ≤ b, u(a) = α, u(a) = β ,has a unique solution u ∈ C 2[a, b].

2) The homogeneous equation

Lu = 0 in a ≤ x ≤ b ,has two linearly independent solutions, u1(x) and u2(x).

3) Let u1(x), u2(x) denote two solutions of Lu = and let

W (x) = det

u1(x) u2(x)u1(x) u

2(x)

denote their Wronskian. If u1(x) and u2(x) are linearly independent functions,then W (x) = 0 for all a ≤ x ≤ b. If u1(x) and u2(x) are linearly dependent,then W (x) ≡ 0.

Proof of 3): Let u1, u2 be linearly independent and suppose that W (a) = 0.There exists a vector

αβ

=

00

with

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u1(a) u2(a)u1(a) u

2(a)

αβ

=

00

.

The function

u(x) = αu1(x) + βu2(x)

satisfies

Lu = 0, u(a) = u(a) = 0 .

It follows that u ≡ 0. The functions u1 and u2 are linearly dependent. We will prove Fredholm’s Alternative for the BVP Lu = f,Ru = 0.

Essentially, the alternative follows from the above results for IVPs together

with Fredholm’s alternative for 2 × 2 linear matrix systems.Theorem 3.1 (Fredholm’s Alternative) The BVP (3.1) has a unique solution u ∈ C 2[a, b] if and only if the homogeneous problem

Lu = 0, Ru = 0 . (3.2)

has only the trivial solution.

Proof: Assume that the homogeneous problem has only the trivial solution.We must show that the inhomogeneous problem is solvable.

Let usp(x) denote the solution of the initial value problem

Lu = f, u(a) = u(a) = 0 ,

and let u1, u2 denote two linearly independent solutions of the homogeneousequation Lu = 0. Every function of the form

u(x) = usp(x) + c1u1(x) + c2u2(x)

satisfies Lu = f . Imposing the boundary conditions on u(x) leads to two linearequations for c1 and c2:

c1R1u1 + c2R1u2 =

−R1usp

c1R2u1 + c2R2u2 = −R2uspIf the corresponding homogeneous system has the nontrivial solution ( c∗1, c

∗2)

then the homogeneous BVP has the nontrivial solution

c∗1u1(x) + c∗2u2(x) ,

which contradicts our assumption. Therefore, the above system for c1 and c2is uniquely solvable, leading to a solution of the inhomogeneous BVP.

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3.2 The Green’s Function

See Stakgold I, p. 66.

Assume unique solvability of the BVP (3.1). We want to write the solutionin the form

u(x) =

ba

g(x, y)f (y) dy

where g (x, y) is the Green’s function corresponding to (L, R).First fix a < y < b. We want to construct the function

g(x) = g(x, y)

satisfying the following conditions:1) (Lg)(x) = 0 for a

≤x < y and for y < x

≤b.

2) g satisfies the boundary conditions, i.e., Rg = 0.3) g ∈ C [a, b].4) g(x) satisfies a jump condition at x = y so that

(Lg)(x) = δ (x − y) .Let u1(x) and u2(x) denote two linearly independent solutions of the homo-

geneous equation Lu = 0. Then g(x) has the form

g(x) =

Au1(x) + Bu2(x) for x < yCu1(x) + Du2(x) for y < x

Here the constants A,B, C, D will generally depend on y.

Lemma 3.2 Let u1, u2 denote two linearly independent solutions of Lu = 0and consider the boundary operator

R1u = α1u(a) + α2u(a) with (α1, α2) = (0, 0) .

Then

R1u1 = 0 or R1u2 = 0 .

Proof: If R1u1 = R1u2 = 0 then

u1(a) u1(a)u2(a) u

2(a)

α1α2

=

00

.

However, W (a) = 0, a contradiction. Assume first that R1u1 = 0 and set

uh1(x) = −R1u2R1u1

u1(x) + u2(x) .

Then uh1 is a nontrivial solution of Lu = 0 with

R1uh1 = 0 .

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Similarly, if R1u2 = 0, consider

uh1

(x) = u1(x)−

R1u1

R1u2u2(x) .

In both cases, we obtain a nontrivial solution uh1(x) of

Lu = 0, R1u = 0 .

Therefore, enforcing the boundary conditions on g (x) leads to

g(x) =

A1uh1(x) for x < yA2uh2(x) for y < x

Here uh1 and uh2 are two nontrivial solutions of Lu = 0 with

R1uh1 = 0, R2uh2 = 0 .The solutions uh1 and uh2 are linearly independent. Otherwise, there would bea nontrivial solution u of Lu = 0 satisfying R1u = R2u = 0.

We note that

g(x) =

A1uh1(x) for x < y

A2uh2(x) for y < x

The condition g ∈ C [a, b] leads to the requirement

A1uh1(y) − A2uh2(y) = 0 .

Integrate the condition

(Lg)(x) = δ (x − y) for a ≤ x ≤ bover the interval

y − ε ≤ x ≤ y + ε .We have that

y+εy−ε

δ (x − y) dy = 1 for all ε > 0 .

Also,

y+εy−ε

g(x) dy = g (y + ε) − g(y − ε) .

This leads to the jump condition

A2uh2(y) − A1uh1(y) =

1

a0(y) .

The system for the coefficients A1, A2 becomes

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uh1(y) −uh2(y)

−uh1(y) u

h2(y)

A1A2

=

01

a0(y)

The determinant of the matrix is

det(y) = uh1(y)uh2(y) − uh2(y)uh1(y) ,

which is the Wronskian of the functions uh1 and uh2. Since det(y) = 0 oneobtains a unique solution for A1 and A2.

Solving for A1 and A2 we obtain

A1 = A1(y) = uh2(y)

a0(y)det(y), A2 = A2(y) =

uh1(y)

a0(y)det(y) .

The Green’s function is

g(x, y) = 1

a0(y)det(y)

uh1(x)uh2(y) for a ≤ x ≤ y ≤ buh2(x)uh1(y) for a ≤ y ≤ x ≤ b .

3.3 Estimates for the Integral Operator G

Consider the BVP Lu = f, Ru = 0 as specified above. Assume that the homo-geneous equation

Lu = 0, Ru = 0

only has the trivial solution u = 0. Thus, we can construct a Green’s function

g ∈ C

[a, b] × [a, b]

so that the integral operator G defined by

(Gf )(x) =

ba

g(x, y)f (y) dy

is the solution operator for BVP. Clearly, for all f ∈ C [a, b],

|Gf |∞ ≤ M |f |∞with

M = maxx

b

a |g(x, y)

|dy .

Theorem 3.2 There exists a constant C so that

|Gf |∞,2 ≤ C |f |∞ for all f ∈ C [a, b] .

Proof: First assume that

Lu = ( pu) + qu

with p ∈ C 1[a, b], q ∈ C [a, b] and

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| p(x)| ≥ p0 > 0 for a ≤ x ≤ b .

Also, assume that

R1u = α1u(a) + α2u(a) and α2 = 0 .

In the following, M 1, M 2 etc. denote constants independent of f .If u = Gf then R1u = 0, thus

|u(a)| ≤α1

α2

|u(a)| ≤ M 1|f |∞ .From the equation

( pu) = f − quobtain that

( pu)(x) − ( pu)(a) = xa

(f − qu)(s) ds .

Therefore,

|u|∞ ≤ M 2|f |∞ .Also,

pu + pu + qu = f ,

and one obtains the bound

|u|∞ ≤ M 3|f |∞ .This proves the estimate of the theorem if Lu = ( pu) + qu and α2 = 0.

If α2 = 0 but β 2 = 0, we can argue as above and obtain the estimate. If α2 = β 2 = 0, then

u(a) = u(b) = 0 ,

thus u(ξ ) = 0 for some ξ . The estimate follows as above.Now consider the equation

Lu = a0u + a1u + a2u = f

where a j ∈ C and a0(x) = 0 for all a ≤ x ≤ b. We multiply the equation by afunction α(x) which will be determined below. We have

αa0u + αa1u + αa2u = αf (3.3)

and want the choose α so that

αa0u + αa1u = ( pu) = pu + pu .

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We obtain the conditions

p = αa0, p = αa1 ,

thus

(ln p) = p

p =

a1a0

.

Set

p(x) = exp x

a(a1/a0)(s) ds

and set

α(x) =

p(x

a0(x) .

Then the equation (3.3) takes the form

( pu) + αa2u = αf .

Our previous estimate applies, and we obtain

|u|∞,2 ≤ M 3|α|∞|f |∞ = M 4|f |∞ .This proves the theorem.

3.4 The Formal Adjoint

Assume that

a0 ∈ C 2, a1 ∈ C 1, a2 ∈ C .The formal adjoint of L is

L∗v = (a0v) − (a1v) + a2v= a0v

+ (2a0 − a1)v + (a0 − a1 + a2)v

Note that L = L∗ if and only if a 0 = a1. If this holds then L is called formally

self–adjoint and can be written as

Lu = (a0u) + a2u .

Define the inner product for (real valued) functions u, v ∈ C [a, b] by

(u, v) =

ba

u(x)v(x) dx .

For functions u, v with compact support in I = [a, b] we have that

(Lu,v) = (u, L∗v) .

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We now assume that u, v ∈ C 2(I ). Then one obtains through integration byparts

(Lu,v) − (u, L∗v) = J (u, v)ba

with

J (u, v) = a0(uv − uv) + (a1 − a0)uv .

We now want to impose boundary conditions on u and v so that the boundaryterms vanish. Consider

J a = J (u, v)(a) ,

for definiteness. Suppose we have

0 = Rau = α1u(a) + α2u(a) .

Case 1: α2 = 0.Then we have u(a) = 0. If we require that v(a) = 0 then J a = 0.

Case 2: α2 = 0.The boundary condition Rau = 0 yields that

u(a) = −α1α2

u(a) .

One obtains that

J a = u(a)

α∗1v(a) + α∗2v(a)

=: u(a)R∗av

with

α∗1 = −a0(a)α1α2

+ (a1 − a0)(a), α∗2 − −a0(a) .

Thus, one can obtain a boundary operator R∗ so that the following holds:If u, v ∈ C 2 and Ru = R∗v = 0 then

(Lu,v) = (u, L∗v) .

One calls (L∗, R∗) the adjoint of (L, R).Note: If L = L∗ then a0 = a1 and one can choose R∗ = R.

Theorem 3.3 a) Assume that

Lu = 0, Ru = 0 implies u = 0 .

Then

L∗v = 0, R∗v = 0 implies v = 0 .

b) Let g(x, y) denote the Green’s function for

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Lu = f 1, Ru = 0

and let h(x, y) denote the Green’s function for

L∗v = f 2, R∗v = 0 .

Then we have

g(x, y) = h(y, x) .

Proof: a) Let f ∈ C be arbitrary and let Lu = f, Ru = 0. Assume that

L∗v = 0, R∗v = 0 .

Then we have

(f, v) = (Lu,v)

= (u, L∗v)= 0

Since v is orthogonal to all f ∈ C it follows that v = 0.b) Let

u(x) =

ba

g(x, y)f 1(y) dy

and

v(y) =

ba

h(y, x)f 2(x) dx .

Since

(Lu,v) = (u, L∗v)

we have that

(f 1, v) = (u, f 2) .

This yields that

ba

f 1(y)

ba

h(y, x)f 2(x) dxdy =

ba

f 2(x)

ba

g(x, y)f 1(y) dydx .

Therefore,

ba

ba

h(y, x) − g(x, y)

f 1(y)f 2(x) dxdy = 0 .

The claim follows.

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4 Introduction to Distribution Theory

The English theoretical physicist Paul Dirac (1902–1984) introduced the δ –

function. The French mathematician Laurent Schwartz (1915–2002) pioneereddistribution theory. He received the fields medal in 1950.

4.1 Motivation

Intuitively, we may think of the delta–function as a function from R to R whichhas the following two properties:

1) δ (x) = 0 for x = 0;2)

∞−∞ δ (x) dx = 1.

However, such a function does not exist.One can try to replace δ (x) by a sequence of functions sn(x) defined for

x∈R.

Example 1:

sn(x) = 1

π

n

1 + n2x2, n = 1, 2, . . .

Example 2:

sn(x) = n√

π e−n

2x2 , n = 1, 2, . . .

For both sequences, the following holds:

limn→∞ sn(x) = 0 for x = 0, limn→∞ sn(0) = ∞ .

and

∞−∞

sn(x) dx = 1 for all n .

Example 3: Set

sn(x) =

0 for |x| > 1n2n for 12n < |x| ≤ 1n

−n for

|x

| ≤ 12n

In this case,

limn→∞ sn(x) = 0 for x = 0, limn→∞ sn(0) = −∞ .

and

∞−∞

sn(x) dx = 1 for all n .

For all three examples, the following is easy to check: If φ ∈ C ∞0 (R) then

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∞

−∞

sn(x)φ(x) dx → φ(0) as n → ∞ .

Proof for Example 1: First note that

∞−∞

sn(x) dx = 1

π

∞−∞

n dx

1 + n2x2 (nx = y, ndx = dy)

= 1

π

∞−∞

dy

1 + y2

= arctan y∞−∞

= 1

For any φ ∈ C ∞0 we have ∞−∞

sn(x)φ(x) dx =

∞−∞

sn(x)

φ(x) − φ(0)

dx + φ(0) .

Denote the integral on the right–hand side by J n. We must show that J n → 0as n → ∞.

Let ε > 0 be given. There exists δ > 0 with

|φ(x) − φ(0)| ≤ ε2

for |x| ≤ δ .Therefore,

|J n| ≤ δ−δ

· · · + |x|≥δ

· · ·

≤ ε2

+ 4|φ|∞ 1π

∞δ

n dx

1 + n2x2

The last integral equals

J (n, δ ) =

∞nδ

dy

1 + y2

and it is easy to see that J (n, δ ) → 0 as n → ∞.

Proof for Example 3: Proceeding as for Example 1, one obtains that

|J n| ≤ ε2

+ 4|φ|∞ ∞δ

|sn(x)| dx .

If n is large enough, then sn(x) = 0 for x ≥ δ and, therefore, |J n| ≤ ε2 forn ≥ N (ε).

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4.2 Testfunctions and Distributions

The space

D = C ∞0 (R,C)consists of all functions φ : R → C which are infinitely often differentiable andwhich have compact support.

In general, if φ : R → C is any function, then its support is the closure of the set

{x ∈ R : φ(x) = 0} .Thus, a function φ : R → C has compact support if and only if there exists afinite interval [a, b] so that

φ(x) = 0 if x /∈ [a, b] .The space D is called the space of test functions. An example of a test

function is

φ(x) =

e−1/(1−x2) for |x| 0 the function

ψ(x) = φx − x0

ε

is also a test function.In the space D one introduces a convergence concept of sequences as follows:Definition 3.1: Let φn, φ ∈ D. Then one says

φn → φ in Dif the following two conditions hold:

1) There exists a finite interval [a, b] so that φ(x) = φn(x) = 0 for all n andall x /∈ [a, b].

2) For all derivatives Dk = dk

dxk:

|Dk(φ

−φn)

|∞ →0 as n

→ ∞, k = 0, 1, 2, . . .

This convergence concept is very restrictive: For

φn → φ in Dto hold, all derivatives of φn must converge uniformly to the correspondingderivatives of φ and, outside some finite interval, the functions φn(x) and φ(x)all agree with each other.

A functional t on the vector space D is simply a map from D to the field of scalars C:

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t :

D → Cφ

→ t(φ)

A functional t : D → C is called linear if

t(aφ1 + bφ2) = at(φ1) + bt(φ2) for all a, b ∈ C and φ1, φ2 ∈ D .

If t : D → C is a linear functional, we also write

t(φ) = t, φ for φ ∈ D .

Definition 3.2: A linear functional t : D → C is called continuous if

φn → φ in D implies t, φn → t, φ ∈ C .Since the condition for convergence φn → φ in D is very strong, the condition

for continuity of a linear functional t : D → C is very mild. Typically, if youwrite down a linear functional t : D → C, it will be continuous.Definition 3.3: The linear space of all continuous linear functionals t : D → Cis called the space of distributions (in one space dimension). It is denoted byD. This space is also called the dual space to D. If tn, t ∈ D then convergence

tn → t in D

means, by definition, that

tn, φ → t, φ in C for all φ ∈ D .

4.3 Examples of Regular and Singular Distributions

If f ∈ L1loc = L1loc(R,C) then the assignment

φ → f, φ = ∞−∞

f (x)φ(x) dx, φ ∈ D ,

defines a distribution, which is identified with f . One can prove: If f, g ∈ L1locthen f and g determine the same distribution if and only if f (x) = g(x) almost

everywhere.Continuity of the linear functional

φ → ∞−∞

f (x)φ(x) dx

on D follows from Lebesgue’s Dominated Convergence Theorem (LDCT).The distributions of the form

φ → ∞−∞

f (x)φ(x) dx

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with f ∈ L1loc are called regular distributions. All other distributions are calledsingular.

For any fixed a

∈R the assignment

φ → δ a, φ = φ(a)is a singular distribution. One also writes

δ a, φ = φ(a) = ∞−∞

δ (x − a)φ(x) dx

although the integral is meaningless as a Riemann or Lebesgue integral.Example: Let

H (x) =

1 for x > 00 for x


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The operator D on distributions is a continuous operator in the followingsense:

Lemma 4.1 Let tn, t ∈ D. If tn → t in D then then Dtn → Dt in D.Proof: For all φ ∈ D:

Dtn, φ = −tn, Dφ→ − t,Dφ = Dt,φ .

4.5 The Simplest Differential Equations for Distributions

Example 1: Find all t ∈ D with

Dt = 0 .

a) Let c ∈ C denote any constant. We have for all φ ∈ D:

Dc,φ = −c,Dφ= −c

∞−∞

φ(x) dx

= 0

This shows that the distribution t = c satisfies Dc = 0.b) We will show that the equation Dt = 0 has no other solutions. Thus, let

t ∈ D and assume that Dt = 0. Fix a testfunction φ0 with ∞−∞

φ0(x) dx = 1

and set

t, φ0 =: c .If φ ∈ D is arbitrary, then we set

J =

∞

−∞

φ(x) dx

and obtain the decomposition

φ(x) = J φ0(x) +

φ(x) − Jφ0(x)

=: J φ0(x) + ψ(x)

with

∞−∞

ψ(x) dx = 0 .

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Note that

w(x) := x

−∞ψ(s) ds

defines a testfunction w with Dw = ψ. Therefore,

t, ψ = t, w = −Dt,w = 0 .The decomposition

φ = J φ0 + ψ

implies

t, φ = J t, φ0= cJ

= c

∞−∞

φ(x) dx

= c, φ

This proves that the distribution t equals the constant c = t, φ0. Example 2: Find all t ∈ D with

Dt = δ 0 .

We know that DH = δ 0 and thus

D(H + c) = δ 0

for any constant c. Thus, any distribution t of the form t = H + c satisfiesDt = δ 0. Conversely, if Dt = δ 0 then D(t− H ) = 0, thus t = H + c by Example1. This shows that t = H + c is the general solution of the differential equationDt = δ 0.

Example 3: Let f ∈ C , i.e., f : R → C is a continuous function. Find allsolutions t ∈ D of

Dt = f .

Set

F (x) =

x0

f (s) ds ,

thus F ∈ C 1(R,C). For all φ ∈ D holds:

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DF,φ

=

−F, φ

= − ∞−∞

F (x)φ(x) dx

=

∞−∞

f (x)φ(x) dx

= f, φ

This shows that t = F satisfies Dt = f . All solutions of the equation Dt = f are given by

t = c +

x0

f (s) ds .


D2t = 0 .

a) If c1 and c2 are constants then let t = c1x + c2. We have for all φ ∈ D:

D2t, φ = t, φ=

∞−∞

(c1x + c2)φ(x) dx

= ∞

−∞(c

1x + c

2)φ(x) dx

= 0

Thus, the distribution t = c1x + c2 satisfies D2t = 0.

b) Show that there are no other solutions. If D2t = 0 then D(Dt) = 0. ByExample 1 we obtain that Dt = c1. Then Example 3 yields that

t = c1x + c2 .


D

2

t = δ 0 .We know from Example 2 that

Dt = c + H .

Set

H 1(x) =

x0

H (s) ds .

We claim that DH 1 = H .

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Proof:

DH 1, φ

=

−H 1, φ

= − ∞0

xφ(x) dx

= −xφ(x)∞x=0

+

∞0

φ(x) dx

= H, φ

It follows that all solutions t of the equation D2t = δ 0 are given by

t = c1x + c2 + H 1(x) .

Continuity of the Differentiation Operator in DLet tn, t ∈ D and let

tn → t in D .This means that

tn, φ → t, φ in C for all φ ∈ D .One obtains that

Dtn, φ = −tn, φ → − t, φ = Dt,φ in C for all φ ∈ D .

Thus,

Dtn → Dt in D .The operator D is continuous in D.

Example: Consider the sequence of functions

sn(x) = 1

n sin(enx) .

Each sn defines the distribution

sn, φ

=

1

n R

sin(enx)φ(x) dx .

If the interval [a, b] contains the support of φ then

|sn, φ| ≤ b − an

|φ|∞ ,thus sn → 0 in D. The derivative of sn(x) is

sn(x) = en

n cos(enx) with |sn|∞ → ∞ as n → ∞ .

Nevertheless, sn → 0 in the sense of distributions.

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4.6 A Simple Equation Involving Distributions

Multiplication of a distribution by a C ∞–function: Let a ∈ C ∞ and lett ∈ D. One defines the product at ∈ D by

at,φ = t,aφ for all φ ∈ D .Example: Determine all t ∈ D with

xt = 0 .

If t = cδ 0, where c ∈ C is an arbitrary constant, then

xt,φ = cδ 0, xφ = 0for all φ ∈ D. Thus the distribution t = cδ 0 solves the equation xt = 0.

We want to show that the equation xt = 0 has only the solutions t = cδ 0.Assume xt = 0. Fix a function φ0 ∈ D with φ0(0) = 1 and set

t, φ0 = c .For any φ ∈ D write

φ(x) = φ(0)φ0(x) +

φ(x) − φ(0)φ0(x)

=: φ(0)φ0(x) + ψ(x) .

Here ψ ∈ D and ψ(0) = 0.We can write

ψ(x) = x

0ψ(t) dt (substitute t = xs, dt = xds)

= x

10

ψ(xs) ds

= xw(x)

with w ∈ D.This yields the decomposition

φ(x) = φ(0)φ0(x) + ψ(x) = φ(0)φ0(x) + xw(x) .

We have

t, ψ = t,xw = xt,w = 0 .Therefore,

t, φ = t, φ(0)φ0= φ(0)c

= cδ 0, φWe have shown that

t = cδ 0 .

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4.7 The Formal Adjoint

A General Linear Differential Operator

Let a0, a1, . . . , an ∈ C ∞(R,C) and setLu = a0D

nu + a1Dn−1u + . . . + an−1Du + a0u for u ∈ D .

For t ∈ D and φ ∈ D we have:

Lt,φ =n

k=0

an−kDkt, φ

=n

k=0

Dkt, an−kφ

=n

k=0

(−1)kt, Dk(an−kφ)

= t, L∗φ

with

L∗φ =∞k=0

(−1)kDk(an−kφ) .

The operator L∗ is called the formal adjoint of L.

Solutions of Lt = δ 0

We now assume that a0(x) = 0 for all x ∈ R. We expect that we candetermine solutions t ∈ D of the equation Lt = δ 0 in the following way:

Let g denote a piecewise smooth function of the form

g(x) =

u(x) for x 0

(4.1)

where

u ∈ C ∞(−∞, 0] and (Lu)(x) = 0 for x ≤ 0and

v ∈ C ∞[0, ∞) and (Lv)(x) = 0 for x ≥ 0 .Furthermore, assume that g ∈ C n−2(R), i.e.,

Dku(0) = Dkv(0) for k = 0, . . . , n − 2 . (4.2)Our vague considerations also suggest that we should consider

ε−ε

a0Dng dx +

ε−ε

a1Dn−1g dx + . . . +

ε−ε

ang dx =

ε−ε

δ 0(x) dx = 1 .

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This formally leads to the requirement

Dn−1v(0)−

Dn−1u(0) = 1

a0(0) . (4.3)

Theorem 4.1 Let g : R → C denote a function of the form (4.1) satisfying (4.2) and (4.3). Then g defines a regular distribution and

Lg = δ 0 .

Proof: We must show that

Lg,φ = φ(0) for all φ ∈ D .Here

Lg,φ = g, L∗φ

=n

k=0

(−1)k ∞−∞

g(x)Dk(an−kφ)(x) dx

=

nk=0

(−1)k 0−∞

u(x)Dk(an−kφ)(x) dx +n

k=0

(−1)k ∞0

v(x)Dk(an−kφ)(x) dx

=: U + V

We now use integration by parts on smooth functions and must show that

U + V = φ(0).

Lemma 4.2 Let u ∈ C ∞(−∞, 0] and let p ∈ D. Then we have 0−∞

u(x)Dk p(x) dx = (−1)k 0−∞

(Dku(x)) p(x) dx + BT leftk

with

BT leftk =k−1 j=0

(−1) j(D ju Dk−1− j p)(0) .

Similarly, if v ∈

C ∞[0,∞

) then

∞0

v(x)Dk p(x) dx = (−1)k ∞0

(Dkv(x)) p(x) dx + BT rightk

with

BT rightk = −k−1 j=0

(−1) j(D jv Dk−1− j p)(0) .

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Proof: Through integration by parts:

0−∞

uDk p dx = (uDk−1 p)(0) − 0−∞

Du Dk−1 p dx

= (uDk−1 p)(0) − (DuDk−2 p)(0) + 0−∞

D2u Dk−2 p dx

= . . .

= (−1)k 0−∞

(Dku(x)) p(x) dx + BT leftk

We continue the proof of the theorem. Using the fact that u and v solvethe homogeneous equations, Lu = 0 and Lv = 0, one obtains that

g, L∗φ = nk=0

(−1)k

BT leftk + BT rightk

=n

k=0

(−1)kk−1 j=0

(−1) j

(D ju)Dk−1− j(an−kφ)(0) − (D jv)Dk−1− j(an−kφ)(0)

Using the equations (4.2), it follows that all terms cancel except the termsfor k = n and j = n − 1. One then obtains that

g, L∗φ = (−1)n(−1)n−1((Dn−1u)a0φ)(0) − ((Dn−1v)a0φ)(0)

= −(a0φ)(0)(Dn−1u − Dn−1v)(0)= φ(0)

The last equation follows from (4.3).

4.8 Further Examples

Example: Consider the function

f (x) =

ln x for x > 0

0 for x 00 for x


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Df,φ

=

−f, φ

= − ∞0

(ln x)φ(x) dx

= − limε→0+

∞ε

(ln x)φ(x) dx

= − limε→0+

− (ln ε)φ(ε) −

∞ε

φ(x)

x dx

= limε→0+

(ln ε)φ(ε) +

∞ε

φ(x)

x dx

Note that the existence of the limit is guaranteed since the distribution Df iswell–defined. One cannot expect, however, that the two limits

limε→0+(ln ε)φ(ε) and limε→0+

∞ε

φ(x)

x dx

exist separately. (The two limits will only exist if φ(0) = 0.)

Example: Consider the function

f (x) = ln |x| for x = 0with pointwise derivative

f (x) = 1

x for x = 0 .

Note that

f ∈ L1loc, but f /∈ L1loc .What is Df ?

We have for all φ ∈ D:

Df,φ = −f, φ= −

∞−∞

(ln |x|)φ(x) dx

= − limε→0+

−ε−∞

. . . + ∞ε

. . .

Here

−ε−∞

ln(−x)φ(x) dx = ln(ε)φ(−ε) − −ε−∞

φ(x)

x dx

and ∞ε

ln(x)φ(x) dx = − ln(ε)φ(ε) − ∞ε

φ(x)

x dx .

Noting that

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limε→0+

ln(ε)(φ(ε) − φ(−ε)) = 0

one obtains that

Df,φ = limε→0+

−ε−∞

φ(x)

x dx +

∞ε

φ(x)

x dx

= p.v.

∞−∞

φ(x)

x dx

Here p.v. stands for Cauchy principle value.We have obtained that

Df,φ

= p.v.

∞

−∞f (x)φ(x) dx

for the function

f (x) = ln |x| .

4.9 The Green’s Function Revisited

Let a0, a1, a2 ∈ C ∞(R) and assume that a0(x) = 0 for all x ∈ R. Let

Lu = a0u + a1u + a2u, u ∈ C 2(R) .

For simplicity, we ignore boundary conditions and consider the equation

Lu = f ,

where f ∈ C (R) is a given function with compact support.The Relation Lg(x) = δ (x − y).When constructing the Green’s function g(x, y) we fix y ∈ R and construct

a function g(x) = g(x, y) with

g ∈ C (R), g ∈ C ∞(−∞, y], g ∈ C ∞[y, ∞)and

Lg(x) = δ (x − y) .For simplicity, let y = 0. We obtain the conditions

Lg(x) = 0 for − ∞ < x ≤ 0Lg(x) = 0 for 0 ≤ x < ∞

with one sided derivatives at x = y. Also, we obtain the jump condition

g(0+) − g(0−) = 1a0(0)

.

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Theorem 4.2 Assume that g(x) satisfies the above conditions. Then we have

Lg = δ 0 in

D .

Proof: For all φ ∈ D we have

Lg,φ = g, L∗φ=

∞−∞

g(x)(L∗φ)(x) dx

=

0−∞

. . . +

∞0

. . .

Here, through integration by parts,

0−∞

g(a0φ) dx =

g(a0φ)

− ga0φ

(0−) + 0−∞

ga0φ dx

and, similarly,

∞0

g(a0φ) dx =

− g(a0φ) + ga0φ

(0+) +

∞0

ga0φ dx .

It follows that

∞−∞

g(a0φ) dx = (a0φ)(0)g

(0+) − g(0−) + 0−∞

ga0φ dx + ∞0

ga0φ dx .

Also,

− ∞−∞

g(a1φ) dx =

∞−∞

ga1φ dx .

One obtains that

Lg,φ = g, L∗φ= (a0φ)(0)

g(0+) − g(0−)

= φ(0)

This proves that Lg = δ 0 in the sense of distributions. Solution of the Inhomogeneous Equation Lu = f .We use the same notation as above. Assume that g(x, y) has been con-

structed by the process described above. We assume:1) g ∈ C (R×R)2) Let

T right = {(x, y) ∈ R2 : x ≥ y} and T left = {(x, y) ∈ R2 : x ≤ y} .

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Then

g

∈C ∞(T right) and g

∈C ∞(T left)

with one sided derivatives along the diagonal where x = y.3) For every fixed y:

Lxg(x, y) = 0 for x ≤ yLxg(x, y) = 0 for x ≥ y

with one–sided derivatives at x = y.4) For every fixed y the jump condition

D1g(y+, y)

−D1g(y

−, y) =

1

a0(y)

(4.4)

holds.

Theorem 4.3 Assume that g(x, y) satisfies the above conditions and let f ∈C (R) have compact support. Set

u(x) =

∞−∞

g(x, y)f (y) dy .

Then u ∈ C 2 and

Lu(x) = f (x) for x ∈ R .

Proof: We have

u(x) =

x−∞

g(x, y)f (y) dy +

∞x

g(x, y)f (y) dy .

Therefore,

u(x) = g(x, x)f (x) + x−∞

D1g(x, y)f (y) dy − g(x, x)f (x) + ∞x

D1g(x, y)f (y) dy

=

x

−∞

D1g(x, y)f (y) dy +

∞x

D1g(x, y)f (y) dy

and

u(x) = D1g(x, x−)f (x)+ x−∞

D21g(x, y)f (y) dy−D1g(x, x+)f (x)+ ∞x

D21g(x, y)f (y) dy .

It follows that

Lu(x) = a0(x)

D1g(x, x−) − D1g(x, x+)

f (x) .

The jump condition (4.4) yields

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D1g(x+, x) − D1g(x−, x) = 1a0(x)

.

By assumption, the function D1g(x, y) has a smooth extension from T right tothe diagonal. Therefore,

D1g(x+, x) = D1g(x, x−) .Similarly,

D1g(x−, x) = D1g(x, x+) .One obtains that

Lu(x) = f (x) .

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5 Three Types of Linear Problems

5.1 Overview

Let U denote a normed space over C. Let D(A) denote a subspace of U andlet A : D(A) → U denote a linear operator.

5.1.1 Inhomogeneous Problems

Consider the equation

Au = f

where f ∈ U is given. Does Fredholm’s Alternative hold? If G = A−1 exists, isthe operator G : U → U bounded? In which norms? Is G : U → U compact?

Often, the following holds: If the equation Au = f comes from a BVP in afinite region, then Fredholm’s Alternative holds. If Au = 0 implies u = 0, then

G = A−1 : U → U is compact. For BVPs on infinite regions the situation is often more difficult.

5.1.2 Spectral Problems

Consider the equation

Au = λu .

Which λ ∈ C are eigenvalues? If λ is not an eigenvalue, how does (λI − A)−1

behave ?Assume that Fredholm’s Alternative holds for the equation Au = f and

that Au = 0 implies u = 0. Also, assume that G = A−1 : U → U is compact.For λ = 0 the equation Au = λu is equivalent to

Gu = 1

λ u .

Therefore, we will study spectral theory for compact operators G.If s ∈ C is not an eigenvalue of A then

sI − A = (sG − I )A(sI − A)−1 = G(sG − I )−1

One therefore studies (sG − I )−1 for compact operators G and obtains infor-mation about the resolvent (sI − A)−1.

The best results are obtained if G is compact and G = G∗.

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5.1.3 Problems of Evolution; Semi–Group Theory

Consider the IVP

u(t) = Au(t), u(0) = u0 .

Formally, the solution is

u(t) = eAtu0 ,

but the exponential series

eAt =

∞ j=0

1

j! (At) j

typically does not converge if A is unbounded. If

ũ(s) =

∞0

u(t)e−st dt

denotes the Laplace transform of u(t) then, formally,

sũ(s) − u0 = Aũ(s)ũ(s) = (sI − A)−1u0u(t) =

1

2πi

Γ

(sI − A)−1est ds

Note that estimates of the resolvent (sI −A)−1

become important. This relatesto spectral theory.

5.2 BVPs as Operator Equations

Boundary value problems for ODEs and PDEs can often be cast in the followingabstract setting:

Let U denote a Banach space, let D(A) denote a linear subspace of U andlet

A : D(A) → U (5.1)

denote a linear operator. For a given f ∈ U consider the equationAu = f .

Typical questions are: Does Fredholm’s Alternative hold? What are allsolutions of the equation Au = 0? If there is a unique solution u = A−1f forevery f ∈ U , how can one represent u? Can one estimate norms of u in termsof f ?

Examples of ordinary BVPs: Let I = [a, b], let a0, a1, a2 ∈ C (I ) and let

Lu = a0u + a1u + a2u .

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Also, consider boundary operators

Rau = α1u(a) + α2u(a), Rbu = β 1u(b) + β 2u(b)

where

(α1, α2) = (0, 0) and (β 1, β 2) = (0, 0) .We also set

Ru =

RauRbu

.

If f ∈ C (I ) is a given function, the BVP reads

Lu = f, Ru = 0 .

To put this into the abstract setting (5.1) one has to specify the Banach spaceU and the subspace D(A). If one wants to work with classical functions, onecan set

U = C (I ) and D(A) = {u ∈ C 2(I ) : Ru = 0} .A norm on U is

|f |∞ = maxx∈I

|f (x)| for f ∈ U

and (U, | · |∞) is a Banach space.The operator A then is

Au = Lu for u ∈ D(A) .A disadvantage of this setting is that the space C (I ) with maximum norm

is not a Hilbert space.One can also set

U = L2(a, b)

with L2–innerproduct

(f, g) = ba f (x)g(x) dx .

Then consider the Sobolev space

H 2(a, b) = {u : u,Du,D2u ∈ L2(a, b)}where u,Du, D2u are understood in the sense of distributions, i.e., as elementsof the dual of the space of test functions C ∞0 (a, b).

One can prove that

H 2(a, b) ⊂ C 1[a, b] .

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Therefore, it makes sense to apply the boundary operator R to an elementu ∈ H 2(a, b).

Define

D(A) = {u ∈ H 2(a, b) : Ru = 0} .The BVP

Lu = f, Ru = 0 ,

becomes the operator equation

Au = f for f ∈ L2(a, b), u ∈ D(A) .

5.3 Inhomogeneous Problems

Let (U, · ) denote a Banach space; let D(A) ⊂ U and let A : D(A) → U denote a linear operator. Consider the inhomogeneous problem

Au = f

where f ∈ U is given and u ∈ D(A) is unknown.Questions: 1) Under what assumptions does Fredholm’s alternative hold?

2) Suppose that the inverse operator A−1 : U → D(A) exists. Can one boundthe solution u of the equation Au = f in terms of f ? In which norms? 3) Underwhat assumptions is the operator A−1 compact?

If A−1 is compact and symmetric, then one can derive expansion theorems

for the solution u in terms of the eigenfunctions of A.

5.4 Spectral Problems

Let (U, ·) denote a complex Banach space; let D(A) ⊂ U and let A : D(A) →U denote a linear operator.

In spectral theory one studies the equation

Au = su for s ∈ C .

5.5 Problems of Evolution: Semigroup Theory

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6 Linear Operators: Riesz Index, Projectors, Resol-

vent

6.1 Nullspaces and Ranges

Let U denote a vector space and let L : U → U denote a linear operator. Forevery j = 0, 1, 2, . . . we set

N j = ker(L j), R j = range(L

j) .

It is clear that

{0} = N 0 ⊂ N 1 ⊂ N 2 ⊂ . . . ⊂ U and

U = R0 ⊃ R1 ⊃ R2 ⊃ . . . ⊃ {0} .

Example 1: Let U = C3 and

L =

0 1 00 0 1

0 0 0

, L2 =

0 0 10 0 0

0 0 0

, L3 = 0 .

We have

N 1 = a

00

: a ∈ C, N 2 =

a

b0

: a, b ∈ C, N 3 = C3 .

Therefore,

{0} = N 0 ⊂ N 1 ⊂ N 2 ⊂ N 3 = N 4 = C3 ,where the inclusions are strict.

For the ranges we have:

R1 =

a

b0

: a, b ∈C

, R2 =

a

00

: a ∈C

, R3 = {0}

and

C3 = R0 ⊃ R1 ⊃ R2 ⊃ R3 = R4 = {0} ,

where the inclusions are strict.We note that the smallest index j with N j = N j+1 is j = 3 and the smallest

index k with Rk = Rk+1 is k = 3.

Example 2: Let U = C ∞(R) and let L denote the derivative operator,

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Lu = u for u ∈ U .

Then the space N j consists of all polynomials of degree ≤ j − 1 and, therefore,the inclusion

N j ⊂ N j+1is strict for all j .

For the ranges we have R j = U for all j, thus

U = R0 = R1 = R2 = . . .

There never is a strict inclusion.

Example 3: Let U = C ∞(R) and let L denote the integration operator,

(Lu)(x) =

x0

u(s) ds for u ∈ U .

If f = Lu then f = u. Therefore, if Lu = 0 then u = 0. This shows thatN 1 = {0}. In fact, if f = L ju then D jf = 0. Therefore, if L ju = 0 then u = 0.Thus,

{0} = N 0 = N 1 = N 2 = . . .All null–spaces are trivial.

Let f = Lu ∈ R1. Clearly, f (0) = 0. Conversely, assume that f ∈ U andf (0) = 0. Then

f (x) =

x0

f (s) ds ,

thus f = Lf ∈ R1. We have shown that

R1 = {f : f ∈ U, f (0) = 0} .It is easy to generalize:

R j = {f : f ∈ U, f (0) = Df (0) = . . . = D j−1f (0) = 0 .Therefore,

U = R0 ⊃ R1 ⊃ R2 ⊃ . . .where all inclusions are strict.

Examples 2 and 3 may suggest that nothing can be said, in general, abouta number j with

N j = N j+1

and a number k with

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Rk = Rk+1 .

Nevertheless, the remarkable Theorem 6.1 (see below) holds.The following result is not surprising.

Lemma 6.1 a) If N j = N j+1 then N j+1 = N j+2.b) If R j = R j+1 then R j+1 = R j+2.

Proof: a) The inclusion N j+1 ⊂ N j+2 always holds. Let u ∈ N j+2, thusL j+1(Lu) = 0. It then follows that

Lu ∈ N j+1 = N j ,Thus L j+1u = 0, thus u ∈ N j+1. We have shown that u ∈ N j+2 implies

u ∈ N j+1, i.e., N j+2 ⊂ N j+1.b) The inclusion R j+2 ⊂ R j+1 always holds. Let u ∈ R j+1. There existsv ∈ U with u = L j+1v. We have that

L jv ∈ R j = R j+1 ,thus L jv = L j+1w for some w ∈ U . This yields that

u = L j+1v = L j+2w ∈ R j+2 .

If there exists a finite j with N j = N j+1 we set

r = min{ j ≥ 0 : N j = N j+1} . (6.1)Similarly, if there exists a finite k with Rk = Rk+1 we set

s = min{k ≥ 0 : Rk = Rk+1} . (6.2)

The next result, which only uses linearity, is remarkable.

Theorem 6.1 Assume there exist j and k with

N j = N j+1 and Rk = Rk+1

and define the indices r and s by (6.1) and (6.2). Then we have

r = s .

Furthermore,

U = N r ⊕ Rrand the spaces N r and Rr are invariant under L:

L(N r) ⊂ N r, L(Rr) = Rr .The restriction of L to Rr is a bijection from Rr onto Rr.

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Proof: a) First suppose that r > s. We then have

N r−1

⊂N r = N r+1 and N r

−1

= N r .

We also have

Rr−1 = Rr since r > s .

Let u ∈ N r be arbitrary. We have

Lr−1u ∈ Rr−1 = Rr ,thus

Lr−1u = Lrv

for some v ∈ U . Since u ∈ N r we have0 = Lru = Lr+1v ,

thus

v ∈ N r+1 = N r .This yields that

0 = Lrv = Lr−1u ,

which implies that u

∈N r

−1. Thus, we obtain that N r

⊂N r

−1. Together with

the inclusion N r−1 ⊂ N r we obtain that N r−1 = N r. This contradiction provesthat r > s is not possible.

b) Second, suppose that s > r. We then have

Rs−1 ⊃ Rs = Rs+1 and Rs−1 = Rs .We also have

N s = N s−1 since s > r .

Let v ∈ Rs−1 be arbitrary, thus v = Ls−1u for some u ∈ U . We have

Lv = Ls

u ∈ Rs = Rs+1 ,thus

Lv = Ls+1w

for some w ∈ U . From

Lsu = Lv = Ls+1w

we conclude that

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Ls(u − Lw) = 0 ,

thus

u − Lw ∈ N s = N s−1 .Therefore,

0 = Ls−1(u − Lw), Ls−1u = Lsw ,and

v = Ls−1u = Lsw ∈ Rs .We have shown that v ∈ Rs−1 implies v ∈ Rs, thus the inclusion

Rs ⊂ Rs−1cannot be strict. This contradiction proves that s > r is not possible.

For the remaining parts of the proof we will assume r ≥ 1 since the claimsare trivial for r = 0. Just note that

N 0 = {0} and R0 = U .c) We claim that

N r

∩Rr =

{0

} . (6.3)

To show this, let v ∈ N r ∩ Rr, thus

Lrv = 0 and v = Lru

for some u ∈ U . One obtains that

L2ru = 0, thus u ∈ N 2r = N r ,thus v = Lru = 0.

d) To prove the space decomposition U = N r ⊕ Rr, we must show that forevery u ∈ U there exists a unique v ∈ N r and a unique w ∈ Rr with

u = v + w .

The uniqueness of v and w follows from (6.3), and it remains to show existence.To this end, let u ∈ U be arbitrary. We then have

Lru ∈ Rr = R2r ,thus

Lru = L2rq

for some q ∈ U . We then have

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u = (u − Lrq ) + Lrq

with Lr

q ∈ Rr. Also,Lr(u − Lrq ) = Lru − L2rq = 0 ,

thus u − Lrq ∈ N r.e) Claim: L(N r) ⊂ N r. To show this, let u ∈ L(N r) be arbitrary. Then

u = Lv for some v ∈ N r. This yields that Lru = L(Lrv) = L0 = 0, thusu ∈ N r.

f) Claim: L(Rr) ⊂ Rr. To show this, let u ∈ L(Rr) be arbitrary. Then u =Lv for some v ∈ Rr; thus v = Lrw for w ∈ U . This implies that u = Lr(Lw),thus u ∈ Rr.

g) Claim: Rr

⊂L(Rr). To show this, let u

∈Rr = Rr+1 be arbitrary. There

exists v ∈ U withu = Lr+1v = L(Lrv) = Lw with w = Lrv ∈ Rr .

This shows that u ∈ L(Rr).h) Claim: The operator L is one–to–one on Rr. To show this, let u ∈ Rr

and assume that Lu = 0. Then we have u ∈ N 1 ⊂ N r and (6.3) implies thatu = 0.

This completes the proof of the theorem. Definition: If L : U → U is a linear operator as in Theorem 6.1 then the

number r ∈ {0, 1, 2, . . .} is called the Riesz index of L.Terminology: For any u ∈ U there exists a unique v ∈ N r and a unique

w ∈ Rr with

u = v + w .

The mappings

P :

U → U

u → vand

Q : U →

U

u → ware linear and are projectors, i.e., P 2 = P and Q2 = Q. The projector P iscalled the projector onto N r along Rr; the projector Q is called the projectoronto Rr along N r. Clearly,

P + Q = I .

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6.2 The Neumann Series

Theorem 6.2 Let U denote a Banach space and let A : U → U denote a bounded linear operator with

A


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u − Au = f .

c) We have shown that L = I − A : U → U is one–to–one and onto.Therefore, the inverse operator L−1 : U → U exists. We claim that L−1 is abounded operator. For every f ∈ U we have

L−1f = ∞

j=0

A jf ≤ 11 − q f .

This proves that L−1 is bounded and

L−1 ≤ 11 − q .

d) Define the operators

S n =n

j=0

A j for n = 1, 2, . . .

Then we have for every f ∈ U :

L−1f − S nf = ∞

j=n+1

A jf ≤ q n+1

1 − q f .

This proves that

L−1

− S n ≤ q n+1

1 − q .Thus, the operators S n converge to L

−1 in operator norm. Remark: We have shown that

n j=0

A j → (I − A)−1 as n → ∞ ,

where the convergence holds in operator norm. One often writes

(I − A)−1 =∞

j=0

A j

although the convergence in the above series may not be clear. A precise resultis

n

j=0

A j − (I − A)−1 → 0 as n → ∞ .

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6.3 Spectral Theory of Matrices

6.3.1 Eigenvalues and Resolvent

Let A ∈ Cn×n. The characteristic polynomial of A is

pA(λ) = det(A − λI )= (λ1 − λ)α1 · · · (λq − λ)αq

where λ1, . . . , λq are the distinct eigenvalues of A and α j is the algebraic mul-tiplicity of λ j. The set

σ(A) = {λ1, . . . , λq}of eigenvalues of A is called the spectrum of A. We have

j

α j = n .

The set

ρ(A) = C \ σ(A)is called the resolvent set of A. The matrix valued function

R(z) = (zI − A)−1, z ∈ ρ(A) ,is called the resolvent of A.

The resolvent

R : ρ(A) → Cn×n

is a meromorphic function of z in every entry rij(z). In fact, by determinantrules for the inverse of a matrix, each function rij(z) is a rational function of z .

Let z0 ∈ ρ(A) be fixed and let z ∈ C denote a point near z0 with

|z − z0||R(z0)|


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Remarks on Laplace Transform and Resolvent: Let A ∈ Cn×n, u0 ∈Cn. The IVP

u(t) = Au(t), u(0) = u0 , (6.4)

has the solution

u(t) = eAtu0

where

eAt =

∞ j=0

t j

j! A j .

If A is an unbounded operator, then one cannot use the above power series to

define the solution operator eAt

. If

û(s) =

∞0

u(t)e−st dt

denotes the Laplace transform of u(t) then the IVP (6.4) transforms to

sû(s) − u0 = Aû(s) ,which yields that

û(s) = R(s)u0 with R(s) = (sI − A)−1 for s ∈ (C \ σ(A)) .

This indicates that the solution operator eAt is the inverse Laplace transformof the resolvent R(s). Even if A is unbounded (e.g., a differential operator) onecan often use the resolvent of A to study the IVP (6.4).

6.3.2 Jordan Normal Form and Riesz Index

Let A ∈ Cn×n. Using Schur’s Theorem and the Blocking Lemma, we know thatthere exists a nonsingular matrix T ∈ Cn×n so that

T −1AT =

B1 0

. . .

0 Bq

where B j has dimension α j × α j , B j is upper triangular, and the diagonalelements of B j are all λ j.

Thus,

B j = λ jI + U j

where U j is strictly upper triangular and, therefore, nilpotent.In fact, one can transform to Jordan normal form. The matrices

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J 1 = (0), J 2 = 0 1

0 0 , J 3 =

0 1 00 0 1

0 0 0

, etc.

are called elementary Jordan matrices. It is easy to see that

J j−1 j = 0, J j j = 0 .Thus, the Riesz index of J j equals j .

A block diagonal matrix J is called a Jordan matrix (to the eigenvalue zero)if each diagonal block of J is an elementary Jordan matrix.

If J ∈ Cα×α is a Jordan matrix and the dimension of the largest elementaryJordan block of J is r × r, then

J r

−1

= 0, J r

= 0 .Therefore, the Riesz index of J equals r, the dimension of the largest Jordanblock of J .

6.3.3 Bases for N r and Rr

Let A ∈ Cn×n and let λ1 denote an eigenvalue of A. Let α denote the algebraicmultiplicity of λ1 and let β = n − α.

There is a transformation matrix T so that

T −1(A − λ1I )T =

U 00 B

=: Ã . (6.5)

Here U has dimension α × α and

U r−1 = 0, U r = 0 .The matrix B of dimension β × β is nonsingular. We have

Ãr =

0 00 Br

.

If we set

Ñ j = ker( Ã j), R̃ j = range( Ã

j)

then we have

Ñ r = {

ξ I

0

: ξ I ∈ Cα}

R̃r = {

0ξ II

: ξ II ∈ Cβ}

Split the transformation matrix T as

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T = (T I , T II )

where T

I

contains the first α columns of T .Let

N r = ker((A − λ1I )r)Rr = range((A − λ1I )r)

We have

x ∈ N r ⇔ (A − λ1I )rx = 0⇔ ÃrT −1x = 0

⇔ T −1

x = ξ I

0

for some ξ I

∈ Cα

⇔ x = T I ξ I for some ξ I ∈ Cα

Similarly,

x ∈ Rr ⇔ x = T II ξ II for some ξ II ∈ Cβ .If x = T ξ then the decomposition

x = T I ξ I + T II ξ II with T I ξ I ∈ N r, T II ξ II ∈ Rrcorresponds to the decomposition

Cn = N r ⊕ Rr .The first α columns of T form a basis for N r, the last β = n − α columns of T form a basis for Rr.

6.3.4 Projectors

In this section, let T denote a transformation matrix with (6.5) where λ1 is aneigenvalue of A. Let r denote the Riesz index of L = A − λ1I and let

N r = kerLr, Rr = rangeL

r .

Let x∈Cn and let ξ = T −1x. If P denotes the projector onto N r along Rr

then we have

P x = T I ξ I

= T

ξ I

0

= T

I 00 0

ξ

= T

I 00 0

T −1x

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Therefore,

P = T I α 0

0 0

T −1

.

Similarly, if Q = I −P denotes the projector onto Rr along N

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