Karlstad University

Division for Engineering Science, Physics and Mathematics

Yury V. Shestopalov and Yury G. Smirnov

Integral Equations

A compendium

Karlstad 2002

Contents

1 Preface 4

2 Notion and examples of integral equations (IEs). Fredholm IEs of the firstand second kind 52.1 Primitive function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3 Examples of solution to integral equations and ordinary differential equations 6

4 Reduction of ODEs to the Volterra IE and proof of the unique solvabilityusing the contraction mapping 84.1 Reduction of ODEs to the Volterra IE . . . . . . . . . . . . . . . . . . . . . . . 84.2 Principle of contraction mappings . . . . . . . . . . . . . . . . . . . . . . . . . . 10

5 Unique solvability of the Fredholm IE of the 2nd kind using the contractionmapping. Neumann series 125.1 Linear Fredholm IE of the 2nd kind . . . . . . . . . . . . . . . . . . . . . . . . . 125.2 Nonlinear Fredholm IE of the 2nd kind . . . . . . . . . . . . . . . . . . . . . . . 135.3 Linear Volterra IE of the 2nd kind . . . . . . . . . . . . . . . . . . . . . . . . . . 145.4 Neumann series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145.5 Resolvent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

6 IEs as linear operator equations. Fundamental properties of completely con-tinuous operators 176.1 Banach spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176.2 IEs and completely continuous linear operators . . . . . . . . . . . . . . . . . . . 186.3 Properties of completely continuous operators . . . . . . . . . . . . . . . . . . . 20

7 Elements of the spectral theory of linear operators. Spectrum of a completelycontinuous operator 21

8 Linear operator equations: the Fredholm theory 238.1 The Fredholm theory in Banach space . . . . . . . . . . . . . . . . . . . . . . . 238.2 Fredholm theorems for linear integral operators and the Fredholm resolvent . . . 26

9 IEs with degenerate and separable kernels 279.1 Solution of IEs with degenerate and separable kernels . . . . . . . . . . . . . . . 279.2 IEs with degenerate kernels and approximate solution of IEs . . . . . . . . . . . 319.3 Fredholm’s resolvent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

10 Hilbert spaces. Self-adjoint operators. Linear operator equations with com-pletely continuous operators in Hilbert spaces 3410.1 Hilbert space. Selfadjoint operators . . . . . . . . . . . . . . . . . . . . . . . . . 3410.2 Completely continuous integral operators in Hilbert space . . . . . . . . . . . . . 3510.3 Selfadjoint operators in Hilbert space . . . . . . . . . . . . . . . . . . . . . . . . 3710.4 The Fredholm theory in Hilbert space . . . . . . . . . . . . . . . . . . . . . . . . 37

2

10.5 The Hilbert–Schmidt theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

11 IEs with symmetric kernels. Hilbert–Schmidt theory 4211.1 Selfadjoint integral operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4211.2 Hilbert–Schmidt theorem for integral operators . . . . . . . . . . . . . . . . . . 45

12 Harmonic functions and Green’s formulas 4712.1 Green’s formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4712.2 Properties of harmonic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

13 Boundary value problems 50

14 Potentials with logarithmic kernels 5114.1 Properties of potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5214.2 Generalized potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

15 Reduction of boundary value problems to integral equations 56

16 Functional spaces and Chebyshev polynomials 5916.1 Chebyshev polynomials of the first and second kind . . . . . . . . . . . . . . . . 5916.2 Fourier–Chebyshev series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

17 Solution to integral equations with a logarithmic singulatity of the kernel 6317.1 Integral equations with a logarithmic singulatity of the kernel . . . . . . . . . . 6317.2 Solution via Fourier–Chebyshev series . . . . . . . . . . . . . . . . . . . . . . . . 64

18 Solution to singular integral equations 6618.1 Singular integral equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6618.2 Solution via Fourier–Chebyshev series . . . . . . . . . . . . . . . . . . . . . . . . 66

19 Matrix representation 7119.1 Matrix representation of an operator in the Hilbert space . . . . . . . . . . . . . 7119.2 Summation operators in the spaces of sequences . . . . . . . . . . . . . . . . . . 72

20 Matrix representation of logarithmic integral operators 7420.1 Solution to integral equations with a logarithmic singularity of the kernel . . . . 76

21 Galerkin methods and basis of Chebyshev polynomials 7921.1 Numerical solution of logarithmic integral equation by the Galerkin method . . . 81

3

1 Preface

This compendium focuses on fundamental notions and statements of the theory of linear oper-ators and integral operators. The Fredholm theory is set forth both in the Banach and Hilbertspaces. The elements of the theory of harmonic functions are presented, including the reductionof boundary value problems to integral equations and potential theory, as well as some methodsof solution to integral equations. Some methods of numerical solution are also considered.

The compendium contains many examples that illustrate most important notions and thesolution techniques.

The material is largely based on the books Elements of the Theory of Functions and Func-tional Analysis by A. Kolmogorov and S. Fomin (Dover, New York, 1999), Linear IntegralEquations by R. Kress (2nd edition, Springer, Berlin, 1999), and Logarithmic Integral Equa-tions in Electromagnetics by Y. Shestopalov, Y. Smirnov, and E. Chernokozhin (VSP, Utrecht,2000).

4

2 Notion and examples of integral equations (IEs). Fred-

holm IEs of the first and second kind

Consider an integral equation (IE)

f(x) = φ(x) + λ∫ b

aK(x, y)φ(y)dy. (1)

This is an IE of the 2nd kind. Here K(x, y) is a given function of two variables (the kernel)defined in the square

Π = (x, y) : a ≤ x ≤ b, a ≤ y ≤ b,f(x) is a given function, φ(x) is a sought-for (unknown) function, and λ is a parameter.

We will often assume that f(x) is a continuous function.The IE

f(x) =∫ b

aK(x, y)φ(y)dy (2)

constitutes an example of an IE of the 1st kind.We will say that equation (1) (resp. (2)) is the Fredholm IE of the 2nd kind (resp. 1st kind)

if the the kernel K(x, y) satisfies one of the following conditions:(i) K(x, y) is continuous as a function of variables (x, y) in the square Π;(ii) K(x, y) is maybe discontinuous for some (x, y) but the double integral

∫ b

a

∫ b

a|K2(x, y)|dxdy < ∞, (3)

i.e., takes a finite value (converges in the square Π in the sense of the definition of convergentRiemann integrals).

An IEf(x) = φ(x) + λ

∫ x

0K(x, y)φ(y)dy, (4)

is called the Volterra IE of the 2nd kind. Here the kernel K(x, y) may be continuous in thesquare Π = (x, y) : a ≤ x ≤ b, a ≤ y ≤ b for a certain b > a or discontinuous and satisfyingcondition (3); f(x) is a given function; and λ is a parameter,

Example 1 Consider a Fredholm IE of the 2nd kind

x2 = φ(x)−∫ 1

0(x2 + y2)φ(y)dy, (5)

where the kernel K(x, y) = (x2 + y2) is defined and continuous in the square Π1 = (x, y) : 0 ≤x ≤ 1, 0 ≤ y ≤ 1, f(x) = x2 is a given function, and the parameter λ = −1.

Example 2 Consider an IE of the 2nd kind

f(x) = φ(x)−∫ 1

0ln |x− y|φ(y)dy, (6)

5

were the kernel K(x, y) = ln |x − y| is discontinuous (unbounded) in the square Π1 along theline x = y, f(x) is a given function, and the parameter λ = −1. Equation (6) a Fredholm IEof the 2nd kind because the double integral

∫ 1

0

∫ 1

0ln2 |x− y|dxdy < ∞, (7)

i.e., takes a finite value—converges in the square Π1.

2.1 Primitive function

Below we will use some properties of the primitive function (or (general) antiderivative) F (x)defined, for a function f(x) defined on an interval I, by

F ′(x) = f(x). (8)

The primitive function F (x) for a function f(x) (its antiderivative) is denoted using theintegral sign to give the notion of the indefinite integral of f(x) on I,

F (x) =∫

f(x)dx. (9)

Theorem 1 If f(x) is continuous on [a, b] then the function

F (x) =∫ x

af(x)dx. (10)

(the antiderivative) is differentiable on [a, b] and F ′(x) = f(x) for each x ∈ [a, b] .

Example 3 F0(x) = arctan√

x is a primitive function for the function

f(x) =1

2√

x(1 + x)

because

F ′0(x) =

dF0

dx=

1

2√

x(1 + x)= f(x). (11)

3 Examples of solution to integral equations and ordi-

nary differential equations

Example 4 Consider a Volterra IE of the 2nd kind

f(x) = φ(x)− λ∫ x

0ex−yφ(y)dy, (12)

where the kernel K(x, y) = ex−y, f(x) is a given continuously differentiable function, and λ isa parameter. Differentiating (12) we obtain

f(x) = φ(x)− λ∫ x

0ex−yφ(y)dy

f ′(x) = φ′(x)− λ[φ(x) +∫ x

0ex−yφ(y)dy].

6

Subtracting termwise we obtain an ordinary differential equatin (ODE) of the first order

φ′(x)− (λ + 1)φ(x) = f ′(x)− f(x) (13)

with respect to the (same) unknown function φ(x). Setting x = 0 in (12) we obtain the initialcondition for the unknown function φ(x)

φ(0) = f(0). (14)

Thus we have obtained the initial value problem for φ(x)

φ′(x) − (λ + 1)φ(x) = F (x), F (x) = f ′(x)− f(x), (15)

φ(0) = f(0). (16)

Integrating (15) subject to the initial condition (16) we obtain the Volterra IE (12), whichis therefore equivalent to the initial value problem (15) and (16).

Solve (12) by the method of successive approximations. Set

K(x, y) = ex−y 0 ≤ y ≤ x ≤ 1,

K(x, y) = 0 0 ≤ x ≤ y ≤ 1.

Write two first successive approximations:

φ0(x) = f(x), φ1(x) = f(x) + λ∫ 1

0K(x, y)φ0(x)dy, (17)

φ2(x) = f(x) + λ∫ 1

0K(x, y)φ1(y)dy = (18)

f(x) + λ∫ 1

0K(x, y)f(y)dy + λ2

∫ 1

0K(x, t)dt

∫ 1

0K(t, y)f(y)dy. (19)

Set

K2(x, y) =∫ 1

0K(x, t)K(t, y)dt =

∫ x

yex−tet−ydt = (x− y)ex−y (20)

to obtain (by changing the order of integration)

φ2(x) = f(x) + λ∫ 1

0K(x, y)f(y)dy + λ2

∫ 1

0K2(x, y)f(y)dy. (21)

The third approximation

φ3(x) = f(x) + λ∫ 1

0K(x, y)f(y)dy +

+ λ2∫ 1

0K2(x, y)f(y)dy + λ3

∫ 1

0K3(x, y)f(y)dy, (22)

where

K3(x, y) =∫ 1

0K(x, t)K2(t, y)dt =

(x− y)2

2!ex−y. (23)

7

The general relationship has the form

φn(x) = f(x) +n∑

m=1

λm∫ 1

0Km(x, y)f(y)dy, (24)

where

K1(x, y) = K(x, y), Km(x, y) =∫ 1

0K(x, t)Km−1(t, y)dt =

(x− y)m−1

(m− 1)!ex−y. (25)

Assuming that successive approximations (24) converge we obtain the (unique) solution to IE(12) by passing to the limit in (24)

φ(x) = f(x) +∞∑

m=1

λm∫ 1

0Km(x, y)f(y)dy. (26)

Introducing the notation for the resolvent

Γ(x, y, λ) =∞∑

m=1

λm−1Km(x, y). (27)

and changing the order of summation and integration in (26) we obtain the solution in thecompact form

φ(x) = f(x) + λ∫ 1

0Γ(x, y, λ)f(y)dy. (28)

Calculate the resolvent explicitly using (25):

Γ(x, y, λ) = ex−y∞∑

m=1

λm−1 (x− y)m−1

(m− 1)!= ex−yeλ(x−y) = e(λ+1)(x−y), (29)

so that (28) gives the solution

φ(x) = f(x) + λ∫ 1

0e(λ+1)(x−y)f(y)dy. (30)

Note that according to (17)

Γ(x, y, λ) = e(λ+1)(x−y), 0 ≤ y ≤ x ≤ 1,

Γ(x, y, λ) = 0, 0 ≤ x ≤ y ≤ 1.

and the series (27) converges for every λ.

4 Reduction of ODEs to the Volterra IE and proof of

the unique solvability using the contraction mapping

4.1 Reduction of ODEs to the Volterra IE

Consider an ordinary differential equatin (ODE) of the first order

dy

dx= f(x, y) (31)

8

with the initial conditiony(x0) = y0, (32)

where f(x, y) is defined and continuous in a two-dimensional domain G which contains thepoint (x0, y0).

Integrating (31) subject to (32) we obtain

φ(x) = y0 +∫ x

x0

f(t, φ(t))dt, (33)

which is called the Volterra integral equation of the second kind with respect to the unknownfunction φ(t). This equation is equivalent to the initial value problem (31) and (32). Note thatthis is generally a nonlinear integral equation with respect to φ(t).

Consider now a linear ODE of the second order with variable coefficients

y′′ + A(x)y′ + B(x)y = g(x) (34)

with the initial conditiony(x0) = y0, y′(x0) = y1, (35)

where A(x) and B(x) are given functions continuous in an interval G which contains the pointx0. Integrating y′′ in (34) we obtain

y′(x) = −∫ x

x0

A(t)y′(x)dx−∫ x

x0

B(x)y(x)dx +∫ x

x0

g(x)dx + y1. (36)

Integrating the first integral on the right-hand side in (36) by parts yields

y′(x) = −A(x)y(x)−∫ x

x0

(B(x)− A′(x))y(x)dx +∫ x

x0

g(x)dx + A(x0)y0 + y1. (37)

Integrating a second time we obtain

y(x) = −∫ x

x0

A(x)y(x)dx−∫ x

x0

∫ x

x0

(B(t)−A′(t))y(t)dtdx+∫ x

x0

∫ x

x0

g(t)dtdx+[A(x0)y0+y1](x−x0)+y0.

(38)Using the relationship ∫ x

x0

∫ x

x0

f(t)dtdx =∫ x

x0

(x− t)f(t)dt, (39)

we transform (38) to obtain

y(x) = −∫ x

x0

A(t)+(x− t)[(B(t)−A′(t))]y(t)dt+∫ x

x0

(x− t)g(t)dt+[A(x0)y0 +y1](x−x0)+y0.

(40)Separate the known functions in (40) and introduce the notation for the kernel function

K(x, t) = −A(t) + (t− x)[(B(t)− A′(t))], (41)

f(x) =∫ x

x0

(x− t)g(t)dt + [A(x0)y0 + y1](x− x0) + y0. (42)

Then (40) becomes

y(x) = f(x) +∫ x

x0

K(x, t)y(t))dt, (43)

which is the Volterra IE of the second kind with respect to the unknown function φ(t). Thisequation is equivalent to the initial value problem (34) and (35). Note that here we obtain alinear integral equation with respect to y(x).

9

Example 5 Consider a homogeneous linear ODE of the second order with constant coefficients

y′′ + ω2y = 0 (44)

and the initial conditions (at x0 = 0)

y(0) = 0, y′(0) = 1. (45)

We see that here

A(x) ≡ 0, B(x) ≡ ω2, g(x) ≡ 0, y0 = 0, y1 = 1,

if we use the same notation as in (34) and (35).Substituting into (40) and calculating (41) and (42),

K(x, t) = ω2(t− x),

f(x) = x,

we find that the IE (43), equivalent to the initial value problem (44) and (45), takes the form

y(x) = x +∫ x

0(t− x)y(t)dt. (46)

4.2 Principle of contraction mappings

A metric space is a pair of a set X consisting of elements (points) and a distance which is anonnegative function satisfying

ρ(x, y) = 0 if and only if x = y,

ρ(x, y) = ρ(y, x) (Axiom of symmetry), (47)

ρ(x, y) + ρ(y, z) ≥ ρ(x, z) (Triangle axiom)

A metric space will be denoted by R = (X, ρ).A sequence of points xn in a metric space R is called a fundamental sequence if it satisfies

the Cauchy criterion: for arbitrary ε > 0 there exists a number N such that

ρ(xn, xm) < ε for all n, m ≥ N. (48)

Definition 1 If every fundamental sequence in the metric space R converges to an element inR, this metric space is said to be complete.

Example 6 The (Euclidian) n-dimensional space Rn of vectors a = (a1, . . . , an) with the Eu-clidian metric

ρ(a, b) = ||a− b||2 =

√√√√n∑

i=1

(ai − bi)2

is complete.

10

Example 7 The space C[a, b] of continuous functions defined in a closed interval a ≤ x ≤ bwith the metric

ρ(φ1, φ2) = maxa≤x≤b

|φ1(x)− φ2(x)|

is complete. Indeed, every fundamental (convergent) functional sequence φn(x) of continuousfunctions converges to a continuous function in the C[a, b]-metric.

Example 8 The space C(n)[a, b] of componentwise continuous n-dimensional vector-functionsf = (f1, . . . , fn) defined in a closed interval a ≤ x ≤ b with the metric

ρ(f1, f

2) = || max

a≤x≤b|f 1

i (x)− f 2i (x)|||2

is complete. Indeed, every componentwise-fundamental (componentwise-convergent) functionalsequence φn(x) of continuous vector-functions converges to a continuous function in theabove-defined metric.

Definition 2 Let R be an arbitrary metric space. A mapping A of the space B into itself issaid to be a contraction if there exists a number α < 1 such that

ρ(Ax,Ay) ≤ αρ(x, y) for any two points x, y ∈ R. (49)

Every contraction mapping is continuous:

xn → x yields Axn → Ax (50)

Theorem 2 Every contraction mapping defined in a complete metric space R has one and onlyone fixed point; that is the equation Ax = x has only one solution.

We will use the principle of contraction mappings to prove Picard’s theorem.

Theorem 3 Assume that a function f(x, y) satisfies the Lipschitz condition

|f(x, y1)− f(x, y2)| ≤ M |y1 − y2| in G. (51)

Then on a some interval x0 − d < x < x0 + d there exists a unique solution y = φ(x) to theinitial value problem

dy

dx= f(x, y), y(x0) = y0. (52)

Proof. Since function f(x, y) is continuous we have |f(x, y)| ≤ k in a region G′ ⊂ G whichcontains (x0, y0). Now choose a number d such that

(x, y) ∈ G′ if |x0 − x| ≤ d, |y0 − y| ≤ kd

Md < 1.

Consider the mapping ψ = Aφ defined by

ψ(x) = y0 +∫ x

x0

f(t, φ(t))dt, |x0 − x| ≤ d. (53)

11

Let us prove that this is a contraction mapping of the complete set C∗ = C[x0 − d, x0 + d] ofcontinuous functions defined in a closed interval x0− d ≤ x ≤ x0 + d (see Example 7). We have

|ψ(x)− y0| =∣∣∣∣∫ x

x0

f(t, φ(t))dt

∣∣∣∣ ≤∫ x

x0

|f(t, φ(t))| dt ≤ k∫ x

x0

dt = k(x− x0) ≤ kd. (54)

|ψ1(x)− ψ2(x)| ≤∫ x

x0

|f(t, φ1(t))− f(t, φ2(t))| dt ≤ Md maxx0−d≤x≤x0+d

|φ1(x)| − φ2(x). (55)

Since Md < 1, the mapping ψ = Aφ is a contraction.From this it follows that the operator equation φ = Aφ and consequently the integral

equation (33) has one and only one solution.This result can be generalized to systems of ODEs of the first order.To this end, denote by f = (f1, . . . , fn) an n-dimensional vector-function and write the

initial value problem for a system of ODEs of the first order using this vector notation

dy

dx= f(x, y), y(x0) = y0. (56)

where the vector-function f is defined and continuous in a region G of the n + 1-dimensionalspace Rn+1 such that G contains the ’n+1-dimensional’ point x0, y0. We assume that f satisfiesthe the Lipschitz condition (51) termwise in variables y0. the initial value problem (56) isequivalent to a system of IEs

φ(x) = y0 +∫ x

x0

f(t, φ(t))dt. (57)

Since function f(t, y) is continuous we have ||f(x, y)|| ≤ K in a region G′ ⊂ G which contains(x0, y0). Now choose a number d such that

(x, y) ∈ G′ if |x0 − x| ≤ d, ||y0 − y|| ≤ Kd

Md < 1.

Then we consider the mapping ψ = Aφ defined by

ψ(x) = y0 +∫ x

x0

f(t, φ(t))dt, |x0 − x| ≤ d. (58)

and prove that this is a contraction mapping of the complete space C∗ = C[x0 − d, x0 + d] ofcontinuous (componentwise) vector-functions into itself. The proof is a termwise repetitiion ofthe reasoning in the scalar case above.

5 Unique solvability of the Fredholm IE of the 2nd kind

using the contraction mapping. Neumann series

5.1 Linear Fredholm IE of the 2nd kind

Consider a Fredholm IE of the 2nd kind

f(x) = φ(x) + λ∫ b

aK(x, y)φ(y)dy, (59)

12

where the kernel K(x, y) is a continuous function in the square

Π = (x, y) : a ≤ x ≤ b, a ≤ y ≤ b,so that |K(x, y)| ≤ M for (x, y) ∈ Π. Consider the mapping g = Af defined by

g(x) = φ(x) + λ∫ b

aK(x, y)φ(y)dy, (60)

of the complete space C[a, b] into itself. We have

g1(x) = λ∫ b

aK(x, y)f1(y)dy + φ(x),

g2(x) = λ∫ b

aK(x, y)f2(y)dy + φ(x),

ρ(g1, g2) = maxa≤x≤b

|g1(x)− g2(x)| ≤

≤ |λ|∫ b

a|K(x, y)||f1(y)− f2(y)|dy ≤ |λ|M(b− a) max

a≤y≤b|f1(y)| − f2(y)| (61)

< ρ(f1, f2) if |λ| < 1

M(b− a)(62)

Consequently, the mapping A is a contraction if

|λ| < 1

M(b− a), (63)

and (59) has a unique solution for sufficiently small |λ| satisfying (63).The successive approximations to this solution have the form

fn(x) = φ(x) + λ∫ b

aK(x, y)fn−1(y)dy. (64)

5.2 Nonlinear Fredholm IE of the 2nd kind

The method is applicable to nonlinear IEs

f(x) = φ(x) + λ∫ b

aK(x, y, f(y))dy, (65)

where the kernel K and φ are continuous functions, and

|K(x, y, z1)−K(x, y, z2)| ≤ M |z1 − z2|. (66)

If λ satisfies (63) we can prove in the same manner that the mapping g = Af defined by

g(x) = φ(x) + λ∫ b

aK(x, y, f(y))dy, (67)

of the complete space C[a, b] into itself is a contraction because for this mapping, the inequality(61) holds.

13

5.3 Linear Volterra IE of the 2nd kind

Consider a Volterra IE of the 2nd kind

f(x) = φ(x) + λ∫ x

aK(x, y)φ(y)dy, (68)

where the kernel K(x, y) is a continuous function in the square Π for some b > a, so that|K(x, y)| ≤ M for (x, y) ∈ Π.

Formulate a generalization of the principle of contraction mappings.

Theorem 4 If A is a continuous mapping of a complete metric space R into itself such thatthe mapping An is a contraction for some n, then the equation Ax = x has one and only onesolution.

In fact if we take an arbitrary point x ∈ R and consider the sequence Aknx, k = 0, 1, 2, . . ., therepetition of the proof of the classical principle of contraction mappings yields the convergenceof this sequence. Let x0 = limk→∞ Aknx, then Ax0 = x0. In fact Ax0 = limk→∞ AknAx. Sincethe mapping An is a contraction, we have

ρ(AknAx,Aknx) ≤ αρ(A(k−1)nAx,A(k−1)nx) ≤ . . . ≤ αkρ(Ax, x).

Consequently,limk→∞

ρ(AknAx,Aknx) = 0,

i.e., Ax0 = x0.Consider the mapping g = Af defined by

g(x) = φ(x) + λ∫ x

aK(x, y)φ(y)dy, (69)

of the complete space C[a, b] into itself.

5.4 Neumann series

Consider an IE

f(x) = φ(x)− λ∫ b

aK(x, y)φ(y)dy, (70)

In order to determine the solution by the method of successive approximations and obtain theNeumann series, rewrite (70) as

φ(x) = f(x) + λ∫ b

aK(x, y)φ(y)dy, (71)

and take the right-hand side f(x) as the zero approximation, setting

φ0(x) = f(x). (72)

Substitute the zero approximation into the right-hand side of (73) to obtain the first approxi-mation

φ1(x) = f(x) + λ∫ b

aK(x, y)φ0(y)dy, (73)

14

and so on, obtaining for the (n + 1)st approximation

φn+1(x) = f(x) + λ∫ b

aK(x, y)φn(y)dy. (74)

Assume that the kernel K(x, y) is a bounded function in the square Π = (x, y) : a ≤ x ≤b, a ≤ y ≤ b, so that

|K(x, y)| ≤ M, (x, y) ∈ Π.

or even that there exists a constant C1 such that

∫ b

a|K(x, y)|2dy ≤ C1 ∀x ∈ [a, b], (75)

The the following statement is valid.

Theorem 5 Assume that condition (75) holds Then the sequence φn of successive approxima-tions (74) converges uniformly for all λ satisfying

λ ≤ 1

B, B =

∫ b

a

∫ b

a|K(x, y)|2dxdy. (76)

The limit of the sequence φn is the unique solution to IE (70).

Proof. Write two first successive approximations (74):

φ1(x) = f(x) + λ∫ b

aK(x, y)f(y)dy, (77)

φ2(x) = f(x) + λ∫ b

aK(x, y)φ1(y)dy = (78)

f(x) + λ∫ b

aK(x, y)f(y)dy + λ2

∫ b

aK(x, t)dt

∫ b

aK(t, y)f(y)dy. (79)

Set

K2(x, y) =∫ b

aK(x, t)K(t, y)dt (80)

to obtain (by changing the order of integration)

φ2(x) = f(x) + λ∫ b

aK(x, y)f(y)dy + λ2

∫ b

aK2(x, y)f(y)dy. (81)

In the same manner, we obtain the third approximation

φ3(x) = f(x) + λ∫ b

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

+ λ2∫ b

aK2(x, y)f(y)dy + λ3

∫ b

aK3(x, y)f(y)dy, (82)

where

K3(x, y) =∫ b

aK(x, t)K2(t, y)dt. (83)

15

The general relationship has the form

φn(x) = f(x) +n∑

m=1

λm∫ b

aKm(x, y)f(y)dy, (84)

where

K1(x, y) = K(x, y), Km(x, y) =∫ b

aK(x, t)Km−1(t, y)dt. (85)

and Km(x, y) is called the mth iterated kernel. One can easliy prove that the iterated kernelssatisfy a more general relationship

Km(x, y) =∫ b

aKr(x, t)Km−r(t, y)dt, r = 1, 2, . . . , m− 1 (m = 2, 3, . . .). (86)

Assuming that successive approximations (84) converge we obtain the (unique) solution to IE(70) by passing to the limit in (84)

φ(x) = f(x) +∞∑

m=1

λm∫ b

aKm(x, y)f(y)dy. (87)

In order to prove the convergence of this series, write

∫ b

a|Km(x, y)|2dy ≤ Cm ∀x ∈ [a, b], (88)

and estimate Cm. Setting r = m− 1 in (86) we have

Km(x, y) =∫ b

aKm−1(x, t)K(t, y)dt, (m = 2, 3, . . .). (89)

Applying to (89) the Schwartz inequality we obtain

|Km(x, y)|2 =∫ b

a|Km−1(x, t)|2

∫ b

a|K(t, y)|2dt. (90)

Integrating (90) with respect to y yields

∫ b

a|Km(x, y)|2dy ≤ B2

∫ b

a|Km−1(x, t)|2 ≤ B2Cm−1 (B =

∫ b

a

∫ b

a|K(x, y)|2dxdy) (91)

which givesCm ≤ B2Cm−1, (92)

and finally the required estimateCm ≤ B2m−2C1. (93)

Denoting

D =

√∫ b

a|f(y)|2dy (94)

and applying the Schwartz inequality we obtain

∣∣∣∣∣∫ b

aKm(x, y)f(y)dy

∣∣∣∣∣2

≤∫ b

a|Km(x, y)|2dy

∫ b

a|f(y)|2dy ≤ D2C1B

2m−2. (95)

16

Thus the common term of the series (87) is estimated by the quantity

D√

C1|λ|mBm−1, (96)

so that the series converges faster than the progression with the denominator |λ|B, which provesthe theorem.

If we take the first n terms in the series (87) then the resulting error will be not greaterthan

D√

C1|λ|n+1Bn

1− |λ|B . (97)

5.5 Resolvent

Introduce the notation for the resolvent

Γ(x, y, λ) =∞∑

m=1

λm−1∫ b

aKm(x, y) (98)

Changing the order of summation and integration in (87) we obtain the solution in the compactform

φ(x) = f(x) + λ∫ b

aΓ(x, y, λ)f(y)dy. (99)

One can show that the resolvent satisfies the IE

Γ(x, y, λ) = K(x, y) + λ∫ b

aK(x, t)Γ(t, y, λ)dt. (100)

6 IEs as linear operator equations. Fundamental prop-

erties of completely continuous operators

6.1 Banach spaces

Definition 3 A complete normed space is said to be a Banach space, B.

Example 9 The space C[a, b] of continuous functions defined in a closed interval a ≤ x ≤ bwith the (uniform) norm

||f ||C = maxa≤x≤b

|f(x)|

[and the metricρ(φ1, φ2) = ||φ1 − φ2||C = max

a≤x≤b|φ1(x)| − φ2(x)|]

is a normed and complete space. Indeed, every fundamental (convergent) functional sequenceφn(x) of continuous functions converges to a continuous function in the C[a, b]-metric.

Definition 4 A set M in the metric space R is said to be compact if every sequence of elementsin M contains a subsequence that converges to some x ∈ R.

17

Definition 5 A family of functions φ defined on the closed interval [a, b] is said to be uni-formly bounded if there exists a number M > 0 such that

|φ(x)| < M

for all x and for all φ in the family.

Definition 6 A family of functions φ defined on the closed interval [a, b] is said to be equicon-tinuous if for every ε > 0 there is a δ > 0 such that

|φ(x1)− φ(x2)| < ε

for all x1, x2 such that |x1)− x2| < δ and for all φ in the given family.

Formulate the most common criterion of compactness for families of functions.

Theorem 6 (Arzela’s theorem). A necessary and sufficient condition that a family of contin-uous functions defined on the closed interval [a, b] be compact is that this family be uniformlybounded and equicontinuous.

6.2 IEs and completely continuous linear operators

Definition 7 An operator A which maps a Banach space E into itself is said to be completelycontinuous if it maps and arbitrary bounded set into a compact set.

Consider an IE of the 2nd kind

f(x) = φ(x) + λ∫ b

aK(x, y)φ(y)dy, (101)

which can be written in the operator form as

f = φ + λAφ. (102)

Here the linear (integral) operator A defined by

f(x) =∫ b

aK(x, y)φ(y)dy (103)

and considered in the (Banach) space C[a, b] of continuous functions in a closed interval a ≤x ≤ b represents an extensive class of completely continuous linear operators.

Theorem 7 Formula (103) defines a completely continuous linear operator in the (Banach)space C[a, b] if the function K(x, y) is to be bounded in the square

Π = (x, y) : a ≤ x ≤ b, a ≤ y ≤ b,and all points of discontinuity of the function K(x, y) lie on the finite number of curves

y = ψk(x), k = 1, 2, . . . n, (104)

where ψk(x) are continuous functions.

18

Proof. To prove the theorem, it is sufficient to show (according to Arzela’s theorem) that thefunctions (103) defined on the closed interval [a, b] (i) are continuous and the family of thesefunctions is (ii) uniformly bounded and (iii) equicontinuous. Below we will prove statements(i), (ii), and (iii).

Under the conditions of the theorem the integral in (103) exists for arbitrary x ∈ [a, b]. Set

M = supΠ|K(x, y)|,

and denote by G the set of points (x, y) for which the condition

|y − ψk(x)| < ε

12Mn, k = 1, 2, . . . n, (105)

holds. Denote by F the complement of G with respect to Π. Since F is a closed set and K(x, y)is continuous on F , there exists a δ > 0 such that for points (x′, y) and (x′′, y) in F for which|x′ − x′′| < δ the inequality

|K(x′, y)−K(x′′, y)| < (b− a)1

3ε (106)

holds. Now let x′ and x′′ be such that |x′ − x′′| < δ holds. Then

|f(x′)− f(x′′)| ≤∫ b

a|K(x′, y)−K(x′′, y)||φ(y)|dy (107)

can be evaluated by integrating over the sum A of the intervals

|y − ψk(x′)| <

ε

12Mn, k = 1, 2, . . . n,

|y − ψk(x′′)| <

ε

12Mn, k = 1, 2, . . . n,

and over the complement B of A with respect to the closed interval [a, b]. The length of A doesnot exceed ε

3M. Therefore

∫

A|K(x′, y)−K(x′′, y)||φ(y)|dy ≤ 2ε

3||φ||. (108)

The integral over the complement B can be obviously estimated by∫

B|K(x′, y)−K(x′′, y)||φ(y)|dy ≤ ε

3||φ||. (109)

Therefore|f(x′)− f(x′′)| ≤ ε||φ||. (110)

Thus we have proved the continuity of f(x) and the equicontinuity of the functions f(x) (ac-cording to Definition (6)) corresponding to the functions φ(x) with bounded norm ||φ||. Theuniform boundedness of f(x) (according to Definition (5)) corresponding to the functions φ(x)with bounded norms follows from the inequality

|φ(x)| ≤∫

B|K(x, y)||φ(y)|dy ≤ M(b− a)||φ||. (111)

19

6.3 Properties of completely continuous operators

Theorem 8 If An is a sequence of completely continuous operators on a Banach space Ewhich converges in (operator) norm to an operator A then the operator A is completely contin-uous.

Proof. Consider an arbitrary bounded sequence xn of elements in a Banach space E suchthat ||xn|| < c. It is necessary to prove that sequence Axn contains a convergent subsequence.

The operator A1 is completely continuous and therefore we can select a convergent subse-quence from the elements A1xn. Let

x(1)1 , x

(1)2 , . . . , x(1)

n , . . . , (112)

be the inverse images of the members of this convergent subsequence. We apply operator A2 toeach members of subsequence (112)). The operator A2 is completely continuous; therefore wecan again select a convergent subsequence from the elements A2x

(1)n , denoting by

x(2)1 , x

(2)2 , . . . , x(2)

n , . . . , (113)

the inverse images of the members of this convergent subsequence. The operator A3 is also com-pletely continuous; therefore we can again select a convergent subsequence from the elementsA3x

(2)n , denoting by

x(3)1 , x

(3)2 , . . . , x(3)

n , . . . , (114)

the inverse images of the members of this convergent subsequence; and so on. Now we can forma diagonal subsequence

x(1)1 , x

(2)2 , . . . , x(n)

n , . . . . (115)

Each of the operators An transforms this diagonal subsequence into a convergent sequence.If we show that operator A also transforms (115)) into a convergent sequence, then this willestablish its complete continuity. To this end, apply the Cauchy criterion (48) and evaluate thenorm

||Ax(n)n − Ax(m)

m ||. (116)

We have the chain of inequalities

||Ax(n)n − Ax(m)

m || ≤ ||Ax(n)n − Akx

(n)n ||+

+ ||Akx(n)n − Akx

(m)m ||+ ||Akx

(m)m − Ax(m)

m || ≤≤ ||A− Ak||(||x(m)

m ||+ ||x(m)m ||) + ||Akx

(n)n − Akx

(m)m ||.

Choose k so that ||A− Ak|| < ε2c

and N such that

||Akx(n)n − Akx

(m)m || < ε

2

holds for arbitrary n > N and m > N (which is possible because the sequence ||Akx(n)n ||

converges). As a result,||Ax(n)

n − Ax(m)m || < ε, (117)

i.e., the sequence ||Ax(n)n || is fundamental.

20

Theorem 9 (Superposition principle). If A is a completely continuous operator and B is abounded operator then the operators AB and BA are completely continuous.

Proof. If M is a bounded set in a Banach space E, then the set BM is also bounded. Conse-quently, the set ABM is compact and this means that the operator AB is completely continuous.Further, if M is a bounded set in E, then AM is compact. Then, because B is a continuousoperator, the set BAM is also compact, and this completes the proof.

Corollary A completely continuous operator A cannot have a bounded inverse in a finite-dimensional space E.

Proof. In the contrary case, the identity operator I = AA−1 would be completely continuousin E which is impossible.

Theorem 10 The adjoint of a completely operator is completely continuous.

7 Elements of the spectral theory of linear operators.

Spectrum of a completely continuous operator

In this section we will consider bounded linear operators which map a Banach space E intoitself.

Definition 8 The number λ is said to be a characteristic value of the operator A if there existsa nonzero vector x (characteristic vector) such that Ax = λx.

In an n-dimensional space, this definition is equivalent to the following:the number λ is said to be a characteristic value of the operator A if it is a root of thecharacteristic equation

Det |A− λI| = 0.

The totality of those values of λ for which the inverse of the operator A−λI does not existis called the spectrum of the operator A.

The values of λ for which the operator A − λI has the inverse are called regular values ofthe operator A.

Thus the spectrum of the operator A consists of all nonregular points.The operator

Rλ = (A− λI)−1

is called the resolvent of the operator A.In an infinite-dimensional space, the spectrum of the operator A necessarily contains all the

characteristic values and maybe some other numbers.In this connection, the set of characteristic values is called the point spectrum of the operator

A. The remaining part of the spectrum is called the continuous spectrum of the operator A. Thelatter also consists, in its turn, of two parts: (1) those λ for which A − λI has an unboundedinverse with domain dense in E (and this part is called the continuous spectrum); and (2) the

21

remainder of the spectrum: those λ for which A− λI has a (bounded) inverse whose domain isnot dense in E, this part is called the residual spectrum.

If the point λ is regular, i.e., if the inverse of the operator A−λI exists, then for sufficientlysmall δ the operator A− (λ + δ)I also has the inverse, i.e., the point δ + λ is regular also. Thusthe regular points form an open set. Consequently, the spectrum, which is a complement, is aclosed set.

Theorem 11 The inverse of the operator A− λI exists for arbitrary λ for which |λ| > ||A||.

Proof. Since

A− λI = −λ(I − 1

λA

),

the resolvent

Rλ = (A− λI)−1 = −1

λ

(I − A

λ

)−1

= −1

λ

∞∑

k=0

Ak

λk.

This operator series converges for ||A/λ|| < 1, i.e., the operator A− λI has the inverse.

Corollary. The spectrum of the operator A is contained in a circle of radius ||A|| with centerat zero.

Theorem 12 Every completely continuous operator A in the Banach space E has for arbitraryρ > 0 only a finite number of linearly independent characteristic vectors which correspond tothe characteristic values whose absolute values are greater than ρ.

Proof. Assume, ad abs, that there exist an infinite number of linearly independent characteristicvectors xn, n = 1, 2, . . ., satisfying the conditions

Axn = λnxn, |λn| > ρ > 0 (n = 1, 2, . . .). (118)

Consider in E the subspace E1 generated by the vectors xn (n = 1, 2, . . .). In this subspacethe operator A has a bounded inverse. In fact, according to the definition, E1 is the totality ofelements in E which are of the form vectors

x = α1x1 + . . . + αkxk, (119)

and these vectors are everywhere dense in E1. For these vectors

Ax = α1λ1x1 + . . . + αkλkxk, (120)

and, consequently,A−1x = (α1/λ1)x1 + . . . + (αk/λk)xk (121)

and

||A−1x|| < 1

ρ||x||.

Therefore, the operator A−1 defined for vectors of the form (119) by means of (120) can be ex-tended by continuity (and, consequently, with preservation of norm) to all of E1. The operator

22

A, being completely continuous in the Banach space E, is completely continuous in E1. Accord-ing to the corollary to Theorem (9), a completely continuous operator cannot have a boundedinverse in infinite-dimensional space. The contradiction thus obtained proves the theorem.

Corollary. Every nonzero characteristic value of a completely continuous operator A in theBanach space E has only finite multiplicity and these characteristic values form a bounded setwhich cannot have a single limit point distinct from the origin of coordinates.

We have thus obtained a characterization of the point spectrum of a completely continuousoperator A in the Banach space E.

One can also show that a completely continuous operator A in the Banach space E cannothave a continuous spectrum.

8 Linear operator equations: the Fredholm theory

8.1 The Fredholm theory in Banach space

Theorem 13 Let A be a completely continuous operator which maps a Banach space E intoitself. If the equation y = x− Ax is solvable for arbitrary y, then the equation x− Ax = 0 hasno solutions distinct from zero solution.

Proof. Assume the contrary: there exists an element x1 6= 0 such that x1 − Ax1 = Tx1 = 0Those elements x for which Tx = 0 form a linear subspace E1 of the banach space E. Denoteby En the subspace of E consisting of elements x for which the powers T nx = 0.

It is clear that subspaces En form a nondecreasing subsequence

E1 ⊆ E2 ⊆ . . . En ⊆ . . . .

We shall show that the equality sign cannot hold at any point of this chain of inclusions. Infact, since x1 6= 0 and equation y = Tx is solvable for arbitrary y, we can find a sequence ofelements x1, x2, . . . , xn, . . . distinct from zero such that

Tx2 = x1,

Tx3 = x2,

. . . ,

Txn = xn−1,

. . . .

The element xn belongs to subspace En for each n but does not belong to subspace En−1. Infact,

T nxn = T n−1xn−1 = . . . = Tx1 = 0,

but

T n−1xn = T n−2xn−1 = . . . = Tx2 = x1 6= 0.

23

All the subspaces En are linear and closed; therefore for arbitrary n there exists an elementyn+1 ∈ En+1 such that

||yn+1|| = 1 and ρ(yn+1, En) ≥ 1

2,

where ρ(yn+1, En) denotes the distance from yn+1 to the space En:

ρ(yn+1, En) = inf ||yn+1 − x||, x ∈ En.

Consider the sequence Ayk. We have (assuming p > q)

||Ayp − Ayq|| = ||yp − (yq + Typ − Tyq)|| ≥ 1

2,

since yq + Typ − Tyq ∈ Ep−1. It is clear from this that the sequence Ayk cannot containany convergent subsequence which contradicts the complete continuity of the operator A. Thiscontradiction proves the theorem

Corollary 1. If the equation y = x − Ax is solvable for arbitrary y, then this equation hasunique solution for each y, i.e., the operator I − A has an inverse in this case.

We will consider, together with the equation y = x− Ax the equation h = f − A∗f whichis adjoint to it, where A∗ is the adjoint to A and h, f are elements of the Banach space E, theconjugate of E.

Corollary 2. If the equation h = f −A∗f is solvable for all h, then the equation f −A∗f = 0has only the zero solution.

It is easy to verify this statement by recalling that the operator adjoint to a completelycontinuous operator is also completely continuous, and the space E conjugate to a Banachspace is itself a Banach space.

Theorem 14 A necessary and sufficient condition that the equation y = x−Ax be solvable isthat f(y) = 0 for all f for which f − A∗f = 0.

Proof. 1. If we assume thast the equation y = x− Ax is solvable, then

f(y) = f(x)− f(Ax) = f(y) = f(x)− A∗f(x)

i.e., f(y) = 0 for all f for which f − A∗f = 0 .2. Now assume that f(y) = 0 for all f for which f −A∗f = 0. For each of these functionals

we consider the set Lf of elements for which f takes on the value zero. Then our assertion isequivalent to the fact that the set ∩Lf consists only of the elements of the form x−Ax. Thus, itis necessary to prove that an element y1 which cannot be represented in the form x−Ax cannotbe contained in ∩Lf . To do this we shall show that for such an element y1 we can construct afunctional f1 satisfying

f1(y1) 6= 0, f1 − A∗f1 = 0.

24

These conditions are equivalent to the following:

f1(y1) 6= 0, f1(x− Ax) = 0 ∀x.

In fact,(f1 − A∗f1)(x) = f1(x)− A∗f1(x) = f1(x)− f1(Ax) = f1(x− Ax).

Let G0 be a subspace consisting of all elements of the form x− Ax. Consider the subspaceG0, y1 of the elements of the form z + αy1, z ∈ G0 and define the linear functional by setting

f1(z + αy1) = α.

Extending the linear functional to the whole space E (which is possible by virtue of the Hahn–Banach theorem) we do obtain a linear functional satisfying the required conditions. Thiscompletes the proof of the theorem.

Corollary If the equation f − A∗f = 0 does not have nonzero solution, then the equationx− Ax = y is solvable for all y.

Theorem 15 A necessary and sufficient condition that the equation f − A∗f = h be solvableis that h(x) = 0 for all x for which x− Ax = 0.

Proof. 1. If we assume that the equation f − A∗f = h is solvable, then

h(x) = f(x)− A∗f(x) = f(x− Ax)

i.e., h(x) = 0 if x− Ax = 0.2. Now assume that h(x) = 0 for all x such that x−Ax = 0. We shall show that the equation

f −A∗f = h is solvable. Consider the set F of all elements of the form y = x−Ax. Constructa functional f on F by setting

f(Tx) = h(x).

This equation defines in fact a linear functional. Its value is defined uniquely for each y becauseif Tx1 = Tx2 then h(x1) = h(x2). It is easy to verify the linearity of the functional and extendit to the whole space E. We obtain

f(Tx) = T ∗f(x) = h(x).

i.e., the functional is a solution of equation f − A∗f = h. This completes the proof of thetheorem.

Corollary If the equation x − Ax = 0 does not have nonzero solution, then the equationf − A∗f = h is solvable for all h.

The following theorem is the converse of Thereom 13.

Theorem 16 If the equation x − Ax = 0 has x = 0 for its only solution, then the equationx− Ax = y has a solution for all y.

25

Proof. If the equation x − Ax = 0 has only one solution, then by virtue of the corollary tothe preceding theorem, the equation f − A∗f = h is solvable for all h. Then, by Corollary 2to Thereom 13, the equation f − A∗f = 0 has f = 0 for its only solution. Hence Corollary toThereom 14 implies that the equation x− Ax = y has a solution for all y.

Thereoms 13 and 16 show that for the equation

x− Ax = y (122)

only the following two cases are possible:(i) Equation (122) has a unique solution for each y, i.e., the operator I − A has an inverse.(ii) The corresponding homogeneous equation x − Ax = 0 has a nonzero solution, i.e., thenumber λ = 1 is a characteristic value for the operator A.

All the results obtained above for the equation x − Ax = y remain valid for the equationx−λAx = y (the adjoint equation is f − λA∗f = h). It follows that either the operator I −λAhas an inverse or the number 1/λ is a characteristic value for the operator A. In other words,in the case of a completely continuous operator, an arbitrary number is either a regular pointor a characteristic value. Thus a completely continuous operator has only a point spectrum.

Theorem 17 The dimension n of the space N whose elements are the solutions to the equationx − Ax = 0 is equal to the dimension of the subspace N∗ whose elements are the solutions tothe equation f − A∗f = 0.

This theorem is proved below for operators in the Hilbert space.

8.2 Fredholm theorems for linear integral operators and the Fred-holm resolvent

One can formulate the Fredholm theorems for linear integral operators

Aφ(x) =∫ b

aK(x, y)φ(y)dy (123)

in terms of the Fredholm resolvent.The numbers λ for which there exists the Fredholm resolvent of a Fredholm IE

φ(x)− λ∫ b

aK(x, y)φ(y)dy = f(x) (124)

will be called regular values. The numbers λ for which the Fredholm resolvent Γ(x, y, λ) is notdefined (they belong to the point spectrum) will be characteristic values of the IE (124), andthey coincide with the poles of the resolvent. The inverse of characteristic values are eigenvaluesof the IE.

Formulate, for example, the first Fredholm theorem in terms of regular values of the resol-vent.

If λ is a regular value, then the Fredholm IE (124) has one and only one solution for arbitraryf(x). This solution is given by formula (99),

φ(x) = f(x) + λ∫ b

aΓ(x, y, λ)f(y)dy. (125)

26

In particular, if λ is a regular value, then the homogeneous Fredholm IE (124)

φ(x)− λ∫ b

aK(x, y)φ(y)dy = 0 (126)

has only the zero solution φ(x) = 0.

9 IEs with degenerate and separable kernels

9.1 Solution of IEs with degenerate and separable kernels

An IE

φ(x)− λ∫ b

aK(x, y)φ(y)dy = f(x) (127)

with a degenerate kernel

K(x, y) =n∑

i=1

ai(x)bi(y) (128)

can be represented, by changing the order of summation and integration, in the form

φ(x)− λn∑

i=1

ai(x)∫ b

abi(y)φ(y)dy = f(x). (129)

Here one may assume that functions ai(x) (and bi(y) ) are linearly independent (otherwise, thenumer of terms in (128)) can be reduced).

It is easy to solve such IEs with degenerate and separable kernels. Denote

ci =∫ b

abi(y)φ(y)dy, (130)

which are unknown constants. Equation (129)) becomes

φ(x) = f(x) + λn∑

i=1

ciai(x), (131)

and the problem reduces to the determination of unknowns ci. To this end, substitute (131)into equation (129) to obtain, after simple algebra,

n∑

i=1

ai(x)

ci −

∫ b

abi(y)

[f(y) + λ

n∑

k=1

ckak(y)

]dy

= 0. (132)

Since functions ai(x) are linearly independent, equality (132) yields

ci −∫ b

abi(y)

[f(y) + λ

n∑

k=1

ckak(y)

]dy = 0, i = 1, 2, . . . n. (133)

27

Denoting

fi =∫ b

abi(y)f(y)dy, aik =

∫ b

abi(y)ak(y)dy (134)

and changing the order of summation and integration, we rewrite (133) as

ci − λn∑

k=1

aikck = fi, i = 1, 2, . . . n, (135)

which is a linear equation system of order n with respect to unknowns ci.Note that the same system can be obtained by multiplying (131) by ak(x) (k = 1, 2, . . . n)

and integrating from a to bSystem (135) is equivalent to IE (129) (and (127), (128)) in the following sense: if (135) is

uniquely solvable, then IE (129) has one and only one solution (and vice versa); and if (135)has no solution, then IE (129) is not solvable (and vice versa).

The determinant of system (135) is

D(λ) =

∣∣∣∣∣∣∣∣∣

1− λa11 −λa12 . . . −λa1n

−λa21 1− λa22 . . . −λa2n

. . . . . .−λan1 −λan2 . . . 1− λann

∣∣∣∣∣∣∣∣∣. (136)

D(λ) is a polynomial in powers of λ of an order not higher than n. Note that D(λ) is notidentically zero because D(0) = 1. Thus there exist not more than n different numbers λ = λ∗ksuch that D(λ∗k) = 0. When λ = λ∗k, then system (135) together with IE (129) are eithernot solvable or have infinitely many solutions. If λ 6= λ∗k, then system (135) and IE (129) areuniquely solvable.

Example 10 Consider a Fredholm IE of the 2nd kind

φ(x)− λ∫ 1

0(x + y)φ(y)dy = f(x). (137)

Here

K(x, y) = x + y =2∑

i=1

ai(x)bi(y) = a1(x)b1(y) + a2(x)b2(y),

wherea1(x) = x, b1(y) = 1, a2(x) = 1, b2(y) = y,

is a degenerate kernel, f(x) is a given function, and λ is a parameter.Look for the solution to (137) in the form (131) (note that here n = 2)

φ(x) = f(x) + λ2∑

i=1

ciai(x) = f(x) + λ(c1a1(x) + c2a2(x)) = f(x) + λ(c1x + c2). (138)

Denoting

fi =∫ 1

0bi(y)f(y)dy, aik =

∫ 1

0bi(y)ak(y)dy, (139)

28

so that

a11 =∫ 1

0b1(y)a1(y)dy =

∫ 1

0ydy =

1

2,

a12 =∫ 1

0b1(y)a2(y)dy =

∫ 1

0dy = 1, (140)

a21 =∫ 1

0b2(y)a1(y)dy =

∫ 1

0y2dy =

1

3,

a22 =∫ 1

0b2(y)a2(y)dy =

∫ 1

0ydy =

1

2,

and

f1 =∫ 1

0b1(y)f(y)dy =

∫ 1

0f(y)dy, (141)

f2 =∫ 1

0b2(y)f(y)dy =

∫ 1

0yf(y)dy,

and changing the order of summation and integration, we rewrite the corresponding equation(133) as

ci − λ2∑

k=1

aikck = fi, i = 1, 2, (142)

or, explicitly,

(1− λ/2)c1 − λc2 = f1,

−(λ/3)c1 + (1− λ/2)c2 = f2,

which is a linear equation system of order 2 with respect to unknowns ci.The determinant of system (143) is

D(λ) =

∣∣∣∣∣1− λ/2 −λ−λ/3 1− λ/2

∣∣∣∣∣ = 1 + λ2/4− λ− λ2/3 = 1− λ− λ2/12 = − 1

12(λ2 + 12λ− 12)(143)

is a polynomial of order 2. (D(λ) is not identically, and D(0) = 1). There exist not more than2 different numbers λ = λ∗k such that D(λ∗k) = 0. Here there are exactly two such numbers; theyare

λ∗1 = −6 + 4√

3, (144)

λ∗2 = −6− 4√

3.

If λ 6= λ∗k, then system (143) and IE (137) are uniquely solvable.To find the solution of system (143) apply Cramer’s rule and calculate the determinants of

system (143)

D1 =

∣∣∣∣∣f1 −λf2 1− λ/2

∣∣∣∣∣ = f1(1− λ/2) + f2λ =

= (1− λ/2)∫ 1

0f(y)dy + λ

∫ 1

0yf(y)dy =

∫ 1

0[(1− λ/2) + yλ]f(y)dy, (145)

29

D1 =

∣∣∣∣∣f1 −λf2 1− λ/2

∣∣∣∣∣ = f1(1− λ/2) + f2λ =

= (1− λ/2)∫ 1

0f(y)dy + λ

∫ 1

0yf(y)dy =

∫ 1

0[(1− λ/2) + yλ]f(y)dy, (146)

D2 =

∣∣∣∣∣(1− λ/2) f1

−λ/3 f2

∣∣∣∣∣ = f2(1− λ/2) + f1λ/3 =

= (1− λ/2)∫ 1

0yf(y)dy + (λ/3)

∫ 1

0f(y)dy =

∫ 1

0[y(1− λ/2) + λ/3]f(y)dy, (147)

Then, the solution of system (143) is

c1 =D1

D= − 12

λ2 + 12λ− 12

∫ 1

0[(1− λ/2) + yλ]f(y)dy =

=1

λ2 + 12λ− 12

∫ 1

0[(6λ− 12)− 12yλ]f(y)dy, (148)

c2 =D2

D= − 12

λ2 + 12λ− 12

∫ 1

0[y(1− λ/2) + λ/3]f(y)dy =

=1

λ2 + 12λ− 12

∫ 1

0[y(6λ− 12)− 4λ]f(y)dy (149)

According to (138), the solution of the IE (137) is

φ(x) = f(x) + λ(c1x + c2) =

= f(x) +λ

λ2 + 12λ− 12

∫ 1

0[6(λ− 2)(x + y)− 12xy − 4λ]f(y)dy. (150)

When λ = λ∗k (k = 1, 2), then system (143) together with IE (137) are not solvable.

Example 11 Consider a Fredholm IE of the 2nd kind

φ(x)− λ∫ 2π

0sin x cos yφ(y)dy = f(x). (151)

HereK(x, y) = sin x cos y = a(x)b(y)

is a degenerate kernel, f(x) is a given function, and λ is a parameter.Rewrite IE (151) in the form (131) (here n = 1)

φ(x) = f(x) + λc sin x, c =∫ 2π

0cos yφ(y)dy. (152)

Multiplying the first equality in (152) by b(x) = cos x and integrating from 0 to 2π we obtain

c =∫ 2π

0f(x) cos xdx (153)

30

Therefore the solution is

φ(x) = f(x) + λc sin x = f(x) + λ∫ 2π

0sin x cos yf(y)dy. (154)

Example 12 Consider a Fredholm IE of the 2nd kind

x2 = φ(x)−∫ 1

0(x2 + y2)φ(y)dy, (155)

where K(x, y) = x2 +y2 is a degenerate kernel f(x) = x2 is a given function, and the parameterλ = 1.

9.2 IEs with degenerate kernels and approximate solution of IEs

Assume that in an IE

φ(x)− λ∫ b

aK(x, y)φ(y)dy = f(x) (156)

the kernel K(x, y and f(x) are continuous functions in the square

Π = (x, y) : a ≤ x ≤ b, a ≤ y ≤ b.

Then φ(x) will be also continuous. Approximate IE (156) with an IE having a degenerate kernel.To this end, replace the integral in (156) by a finite sum using, e.g., the rectangle rule

∫ b

aK(x, y)φ(y)dy ≈ h

n∑

k=1

K(x, yk)φ(yk), (157)

where

h =b− a

n, yk = a + kh. (158)

The resulting approximation takes the form

φ(x)− λhn∑

k=1

K(x, yk)φ(yk) = f(x) (159)

of an IE having a degenerate kernel. Now replace variable x in (159) by a finite number ofvalues x = xi, i = 1, 2, . . . , n. We obtain a linear equation system with unknowns φk = φ(xk)and right-hand side containing the known numbers fi = f(xi)

φi − λhn∑

k=1

K(xi, yk)φk = fi, i = 1, 2, . . . , n. (160)

We can find the solution φi of system (160) by Cramer’s rule,

φi =D

(n)i (λ)

D(n)(λ), i = 1, 2, . . . , n, (161)

31

calculating the determinants

D(n)(λ) =

∣∣∣∣∣∣∣∣∣

1− hλK(x1, y1) −hλK(x1, y2) . . . −hλK(x1, yn)−hλK(x2, y1) 1− hλK(x2, y2) . . . −hλK(x2, yn)

. . . . . .−hλK(xn, y1) −hλK(xn, y2) . . . 1− hλK(xn, yn)

∣∣∣∣∣∣∣∣∣. (162)

and so on.To reconstruct the approximate solution φ(n) using the determined φi one can use the IE

(159) itself:

φ(x) ≈ φ(n) = λhn∑

k=1

K(x, yk)φk + f(x). (163)

One can prove that φ(n) → φ(x) as n →∞ where φ(x) is the exact solution to IE (159).

9.3 Fredholm’s resolvent

As we have already mentioned the resolvent Γ(x, y, λ) of IE (159) satisfies the IE

Γ(x, y, λ) = K(x, y) + λ∫ b

aK(x, t)Γ(t, y, λ)dt. (164)

Solving this IE (164) by the described approximate method and passing to the limit n → ∞we obtain the resolvent as a ratio of two power series

Γ(x, y, λ) =D(x, y, λ)

D(λ), (165)

where

D(x, y, λ) =∞∑

n=0

(−1)n

n!Bn(x, y)λn, (166)

D(λ) =∞∑

n=0

(−1)n

n!cnλ

n, (167)

B0(x, y) = K(x, y),

Bn(x, y) =∫ b

a. . .

∫ b

a

∣∣∣∣∣∣∣∣∣

K(x, y) K(x, y1) . . . K(x, yn)K(y1, y) K(y1, y1) . . . K(y1, yn)

. . . . . .K(yn, y) K(yn, y1) . . . K(yn, yn)

∣∣∣∣∣∣∣∣∣dy1 . . . dyn; (168)

c0 = 1,

32

cn =∫ b

a. . .

∫ b

a

∣∣∣∣∣∣∣∣∣

K(y1, y1) K(y1, y2) . . . K(y1, yn)K(y2, y1) K(y2, y2) . . . K(y2, yn)

. . . . . .K(yn, y1) K(yn, y2) . . . K(yn, yn)

∣∣∣∣∣∣∣∣∣dy1 . . . dyn. (169)

D(λ) is called the Fredholm determinant, and D(x, y, λ), the first Fredholm minor.If the integral ∫ b

a

∫ b

a|K(x, y)|2dxdy < ∞, (170)

then seires (167) for the Fredholm determinant converges for all complex λ, so that D(λ) is anentire function of λ. Thus (164)–(167) define the resolvent on the whole complex plane λ, withan exception of zeros of D(λ) which are poles of the resolvent.

We may conclude that for all complex λ that do not coincide with the poles of the resolvent,the IE

φ(x)− λ∫ b

aK(x, y)φ(y)dy = f(x) (171)

is uniquely solvable, and its solution is given by formula (99)

φ(x) = f(x) + λ∫ b

aΓ(x, y, λ)f(y)dy. (172)

For practical computation of the resolvent, one can use the following recurrent relationships:

B0(x, y) = K(x, y), c0 = 1,

cn =∫ b

aBn−1(y, y)dy, n = 1, 2, . . . , (173)

Bn(x, y) = cnK(x, y)− n∫ b

aK(x, t)Bn−1(t, y)dt, n = 1, 2, . . . . (174)

Example 13 Find the resolvent of the Fredholm IE (137) of the 2nd kind

φ(x)− λ∫ 1

0(x + y)φ(y)dy = f(x). (175)

We will use recurrent relationships (173) and (174). Here

K(x, y) = x + y

is a degenerate kernel. We have c0 = 1, Bi = Bi(x, y) (i = 0, 1, 2), and

B0 = K(x, y) = x + y,

c1 =∫ 1

0B0(y, y)dy =

∫ 1

02ydy = 1,

B1 = c1K(x, y)−∫ 1

0K(x, t)B0(t, y)dt =

= x + y −∫ 1

0(x + t)(t + y)dt =

1

2(x + y)− xy − 1

3,

c2 =∫ 1

0B1(y, y)dy =

∫ 1

0

(1

2(y + y)− y2 − 1

3

)dy = −1

6,

B2 = c2K(x, y)−∫ 1

0K(x, t)B1(t, y)dt =

= −1

6(x + y)− 2

∫ 1

0(x + t)

(1

2(t + y)− ty − 1

3

)dt = 0.

33

Since B2(x, y) ≡ 0, all subsequent c3, c4, . . ., and B3, B4, . . ., vanish (see (173) and (174)),and we obtain

D(x, y, λ) =

=1∑

n=0

(−1)n

n!Bn(x, y)λn = B0(x, y)− λB1(x, y) =

= x + y − λ(

1

2(x + y)− xy − 1

3

), (176)

D(λ) =

=2∑

n=0

(−1)n

n!cnλ

n = c0 − c1λ +1

2c2λ

2 =

= 1− λ− 1

12λ2 = − 1

12(λ2 + 12λ− 12)

and

Γ(x, y, λ) =D(x, y, λ)

D(λ)= −12

x + y − λ(

12(x + y)− xy − 1

3

))

λ2 + 12λ− 12. (177)

The solution to (175) is given by formula (99)

φ(x) = f(x) + λ∫ 1

0Γ(x, y, λ)f(y)dy,

so that we obtain, using (176),

φ(x) = f(x) +λ

λ2 + 12λ− 12

∫ 1

0[6(λ− 2)(x + y)− 12xy − 4λ]f(y)dy, (178)

which coincides with formula (150).The numbers λ = λ∗k (k = 1, 2) determined in (144), λ∗1 = −6 +

√4 and λ∗2 = −6−√4, are

(simple, i.e., of multiplicity one) poles of the resolvent because thay are (simple) zeros of theFredholm determinant D(λ). When λ = λ∗k (k = 1, 2), then IE (175) is not solvable.

10 Hilbert spaces. Self-adjoint operators. Linear opera-

tor equations with completely continuous operators

in Hilbert spaces

10.1 Hilbert space. Selfadjoint operators

A real linear space R with the inner product satisfying

(x, y) = (y, x) ∀x, y ∈ R,

(x1 + x2, y) = (x1, y) + (x2, y), ∀x1, x2, y ∈ R,

(λx, y) = λ(x, y), ∀x, y ∈ R,

(x, x) ≥ 0, ∀x ∈ R, (x, x) = 0 if and only if x = 0.

34

is called the Euclidian space.In a complex Euclidian space the inner product satisfies

(x, y) = (y, x),

(x1 + x2, y) = (x1, y) + (x2, y),

(λx, y) = λ(x, y),

(x, x) ≥ 0, (x, x) = 0 if and only if x = 0.

Note that in a complex linear space

(x, λy) = λ(x, y).

The norm in the Euclidian space is introduced by

||x|| = √x, x.

The Hilbert space H is a complete (in the metric ρ(x, y) = ||x − y||) infinite-dimensionalEuclidian space.

In the Hilbert space H, the operator A∗ adjoint to an operator A is defined by

(Ax, y) = (x,A∗y), ∀x, y ∈ H.

The selfadjoint operator A = A∗ is defined from

(Ax, y) = (x,Ay), ∀x, y ∈ H.

10.2 Completely continuous integral operators in Hilbert space

Consider an IE of the second kind

φ(x) = f(x) +∫ b

aK(x, y)φ(y)dy. (179)

Assume that K(x, y) is a Hilbert–Schmidt kernel, i.e., a square-integrable function in the squareΠ = (x, y) : a ≤ x ≤ b, a ≤ y ≤ b, so that

∫ b

a

∫ b

a|K(x, y)|2dxdy ≤ ∞, (180)

and f(x) ∈ L2[a, b], i.e., ∫ b

a|f(x)|2dx ≤ ∞.

Define a linear Fredholm (integral) operator corresponding to IE (179)

Aφ(x) =∫ b

aK(x, y)φ(y)dy. (181)

If K(x, y) is a Hilbert–Schmidt kernel, then operator (181) will be called a Hilbert–Schmidtoperator.

Rewrite IE (179) as a linear operator equation

φ = Aφ(x) + f, f, φ ∈ L2[a, b]. (182)

35

Theorem 18 Equality (182) and condition (180) define a completely continuous linear opera-tor in the space L2[a, b]. The norm of this operator is estimated as

||A|| ≤√∫ b

a

∫ b

a|K(x, y)|2dxdy. (183)

Proof. Note first of all that we have already mentioned (see (75)) that if condition (180) holds,then there exists a constant C1 such that

∫ b

a|K(x, y)|2dy ≤ C1 almost everywhere in [a, b], (184)

i.e., K(x, y) ∈ L2[a, b] as a function of y almost for all x ∈ [a, b]. Therefore the function

ψ(x) =∫ b

aK(x, y)φ(y)dy is defined almost everywhere in [a, b] (185)

We will now show that ψ(x) ∈ L2[a, b]. Applying the Schwartz inequality we obtain

|ψ(x)|2 =

∣∣∣∣∣∫ b

aK(x, y)φ(y)dy

∣∣∣∣∣2

≤∫ b

a|K(x, y)|2dy

∫ b

a|φ(y)|2dy = (186)

= ||φ||2∫ b

a|K(x, y)|2dy.

Integrating with respect to x and replacing an iterated integral of |K(x, y)|2 by a double integral,we obtain

||Aφ||2 =∫ b

a|ψ(y)|2dy ≤ ||φ||2

∫ b

a

∫ b

a|K(x, y)|2dxdy, (187)

which proves (1) that ψ(x) is a square-integrable function, i.e.,

∫ b

a|ψ(x)|2dx ≤ ∞,

and estimates the norm of the operator A : L2[a, b] → L2[a, b].Let us show that the operator A : L2[a, b] → L2[a, b] is completely continuous. To this end,

we will use Theorem 8 and construct a sequence An of completely continuous operators on thespace L2(Π) which converges in (operator) norm to the operator A. Let ψn be an orthonormalsystem in L2[a, b]. Then ψm(x)ψn(y) form an orthonormal system in L2(Π). Consequently,

K(x, y) =∞∑

m,n=1

amnψm(x)ψn(y). (188)

By setting

KN(x, y) =N∑

m,n=1

amnψm(x)ψn(y). (189)

define an operator AN which is finite-dimensional (because it has a degenerate kernel and mapstherefore L2[a, b] into a finite-dimensional space generated by a finite system ψmN

m=1) andtherefore completely continuous.

36

Now note that KN(x, y) in (189) is a partial sum of the Fourier series for K(x, y), so that

∫ b

a

∫ b

a|K(x, y)−KN(x, y)|2dxdy → 0, N →∞. (190)

Applying estimate (183) to the operator A− AN and using (190) we obtain

||A− AN || → 0, N →∞. (191)

This completes the proof of the theorem.

10.3 Selfadjoint operators in Hilbert space

Consider first some properties of a selfadjoint operator A (A = A∗) acting in the Hilbert spaceH.

Theorem 19 All eigenvalues of a selfadjoint operator A in H are real.

Proof. Indeed, let λ be an eigenvalue of a selfadjoint operator A, i.e., Ax = λx, x ∈ H, x 6= 0.Then we have

λ(x, x) = (λx, x) = (Ax, x) = (x, Ax) = (x, λx) = λ(x, x), (192)

which yields λ = λ, i.e., λ is a real number.

Theorem 20 Eigenvectors of a selfadjoint operator A corresponding to different eigenvaluesare orthogonal.

Proof. Indeed, let λ and µ be two different eigenvalues of a selfadjoint operator A, i.e.,

Ax = λx, x ∈ H, x 6= 0; Ay = µy, y ∈ H, y 6= 0.

Then we haveλ(x, y) = (λx, y) = (Ax, y) = (x,Ay) = (x, µy) = µ(x, y), (193)

which gives(λ− µ)(x, y) = 0, (194)

i.e., (x, y) = 0.

Below we will prove all the Fredholm theorems for integral operators in the Hilbert space.

10.4 The Fredholm theory in Hilbert space

We take the integral operator (181), assume that the kernel satisfies the Hilbert–Schmidt con-dition

∫ b

a

∫ b

a|K(x, y)|2dxdy ≤ ∞, (195)

and write IE (179)

φ(x) =∫ b

aK(x, y)φ(y)dy + f(x) (196)

37

in the operator form (182),

φ = Aφ + f, (197)

or

Tφ = f, T = I − A. (198)

The corresponding homogeneous IE

Tφ0 = 0, (199)

and the adjoint equations

T ∗ψ = g, T ∗ = I − A∗, (200)

and

T ∗ψ0 = 0. (201)

We will consider equations (198)–(201) in the Hilbert space H = L2[a, b]. Note that (198)–(201) may be considered as operator equations in H with an arbitrary completely continuousoperator A.

The four Fredholm theorems that are formulated below establish relations between thesolutions to these four equations.

Theorem 21 The inhomogeneous equation Tφ = f is solvable if and only if f is orthogonal toevery solution of the adjoint homogeneous equation T ∗ψ0 = 0 .

Theorem 22 (the Fredholm alternative). One of the following situations is possible:(i) the inhomogeneous equation Tφ = f has a unique solution for every f ∈ H(ii) the homogeneous equation Tφ0 = 0 has a nonzero solution .

Theorem 23 The homogeneous equations Tφ0 = 0 and T ∗ψ0 = 0 have the same (finite)number of linearly independent solutions.

To prove the Fredholm theorems, we will use the definiton of a subspace in the Hilbert spaceH:M is a linear subspace of H if M closed and f, g ∈ M yields αf + βg ∈ M where α and β arearbitrary numbers,and the following two properties of the Hilbert space H:(i) if h is an arbitrary element in H, then the set horth of all elements f ∈ H orthogonal to h,horth = f : f ∈ H, (f, h) = 0 form a linear subspace of H which is called the orthogonalcomplement,

and(ii) if M is a (closed) linear subspace of H, then arbitrary element f ∈ H is uniquely representedas f = h + h′, where h ∈ M , and h′ ∈ M orth and M orth is the orthogonal complement to M .

Lemma 1 The manifold ImT = y ∈ H : y = Tx, x ∈ H is closed.

38

Proof. Assume that yn ∈ Im T and yn → y. To prove the lemma, one must show that y ∈ Im T ,i.e., y = Tx. We will now prove the latter statement.

According to the definition of Im T , there exist vectors xn ∈ H such that

yn = Txn = xn − Axn. (202)

We will consider the manifold Ker T = y ∈ H : Ty = 0 which is a closed linear subspaceof H. One may assume that vectors xn are orthogonal to Ker T , i.e., xn ∈Ker T orth; otherwise,one can write

xn = hn + h′n, hn ∈ KerT,

and replace xn by hn = xn − h′n with h′n ∈ KerT orth. We may assume also that the set ofnorms ||xn|| is bounded. Indeed, if we assume the contrary, then there will be an unboundedsubsequence xnk with ||xnk|| → ∞. Dividing by ||xnk|| we obtain from (202)

xnk

||xnk|| − Axnk

||xnk|| → 0.

Note that A is a completely continuous operator; therefore we may assume that

A

xnk

||xnk||

is a convergent subsequence, which converges, e.g., to a vector z ∈ H. It is clear that ||z|| = 1and Tz = 0, i.e., z ∈ KerT . We assume however that vectors xn are orthogonal to Ker T ;therefore vector z must be also orthogonal to Ker T . This contradiction means that sequence||xn|| is bounded. Therefore, sequence Axn contains a convergent subsequence; consequently,sequence xn will also contain a convergent subsequence due to (202). Denoting by x the limitof xn, we see that, again y = Tx due to (202). The lemma is proved.

Lemma 2 The space H is the direct orthogonal sum of closed subspaces KerT ∗ and ImT ∗,

KerT + ImT ∗ = H, (203)

and also

KerT ∗ + ImT = H. (204)

Proof. The closed subspaces Ker T and Im T ∗ are orthogonal because if h ∈Ker T then (h, T ∗x) =(Th, x) = (0, x) = 0 for all x ∈ H. It is left to prove there is no nonzero vector orthogonal toKer T and Im T ∗. Indeed, if a vector z is orthogonal to Im T ∗, then (Tz, x) = (z, T ∗x) = 0 forall x ∈ H, i.e., z ∈Ker T . The lemma is proved.

Lemma 2 yields the first Fredholm theorem 21. Indeed, f is orthogonal to Ker T ∗ if andonly if f ∈ Im T ∗, i.e., there exists a vector φ such that Tφ = f .

Next, set Hk = Im T k for every k, so that, in particular, H1 = Im T . It is clear that

. . . ⊂ Hk ⊂ . . . ⊂ H2 ⊂ H1 ⊂ H, (205)

all the subsets in the chain are closed, and T (Hk) = Hk+1.

39

Lemma 3 There exists a number j such that Hk+1 = Hk for all k ≥ j.

Proof. If such a j does not exist, then all Hk are different, and one can construct an orthonormalsequence xk such that xk ∈ Hk and xk is orthogonal to Hk+1. Let l > k. Then

Axl − Axk = −xk + (xl + Txk − Txl).

Consequently, ||Axl − Axk|| ≥ 1 because xl + Txk − Txl ∈ Hk+1. Therefore, sequence Axkdoes not contain a convergent subsequence, which contradicts to the fact that A is a completelycontinuous operator. The lemma is proved.

Lemma 4 If KerT = 0 then ImT = H.

Proof. If Ker T = 0 then T is a one-to-one operator. If in this case Im T 6= H then the chain(205) consists of different subspaces which contradicts Lemma 3. Therefore Im T = H.

In the same manner, we prove that Im T ∗ = H if Ker T ∗ = 0.Lemma 5 If ImT = H then KerT = 0.Proof. Since Im T = H, we have Ker T ∗ = 0 according to Lemma 3. Then Im T ∗ = Haccording to Lemma 4 and Ker T = 0 by Lemma 2.

The totality of statements of Lemmas 4 and 5 constitutes the proof of Fredholm’s alternative22.

Now let us prove the third Fredholm’s Theorem 23.To this end, assume that Ker T is an infinite-dimensional space: dimKer T = ∞. Then this

space conyains an infinite orthonormal sequence xn such that Axn = xn and ||Axl − Axk|| =√2 if k 6= l. Therefore, sequence Axn does not contain a convergent subsequence, which

contradicts to the fact that A is a completely continuous operator.Now set µ = dimKer T and ν = dimKer T ∗ and assume that µ < ν. Let φ1, . . . , φµ be an

orthonormal basis in Ker T and ψ1, . . . , ψν be an orthonormal basis in Ker T ∗. Set

Sx = Tx +µ∑

j=1

(x, φj)ψj.

Operator S is a sum of T and a finite-dimensional operator; therefore all the results provedabove for T remain valid for S.

Let us show that the homogeneous equation Sx = 0 has only a trivial solution. Indeed,assume that

Tx +µ∑

j=1

(x, φj)ψj = 0. (206)

According to Lemmas 2 vectors psij are orthogonal to all vectors of the form Tx (206) yields

Tx = 0

and

(x, φj) = 0, quad1 ≤ j ≤ µ.

40

We see that, on the one hand, vector x must be a linear combination of vectors psij, and onthe other hand, vector x must be orthogonal to these vectors. Therefore, x = 0 and equationSx = 0 has only a trivial solution. Thus according to the second Fredholm’s Theorem 22 thereexists a vector y such that

Ty +µ∑

j=1

(y, φj)ψj = ψµ+1.

Calculating the inner product of both sides of this equality and ψµ+1 we obtain

(Ty, ψµ+1) +µ∑

j=1

(y, φj)(ψj, ψµ+1) = (ψµ+1, ψµ+1).

Here (Ty, ψµ+1) = 0 because Ty ∈ Im T and Im T is orthogonal to Ker T ∗. Thus we have

(0 +µ∑

j=1

(y, φj)× 0 = 1.

This contradiction is obtained because we assumed µ < ν. Therefore µ ≥ ν. Replacing nowT by T ∗ we obtain in the same manner that µ ≤ ν. Therefore µ = ν. The third Fredholm’stheorem is proved.

To sum up, the Fredholm’s theorems show that the number λ = 1 is either a regular pointor an eigenvalue of finite multiplicity of the operator A in (197).

All the results obtained above for the equation (197) φ = Aφ+f (i.e., for the operator A−I)remain valid for the equation λφ = Aφ + f (for the operator A − λI with λ 6= 0). It followsthat in the case of a completely continuous operator, an arbitrary nonzero number is either aregular point or an eigenvalue of finite multiplicity. Thus a completely continuous operator hasonly a point spectrum. The point 0 is an exception and always belongs to the spectrum of acompletely continuous operator in an infinite-dimensional (Hilbert) space but may not be aneigenvalue.

10.5 The Hilbert–Schmidt theorem

Let us formulate the important Hilbert–Schmidt theorem.

Theorem 24 A completely continuous selfadjoint operator A in the Hilbert space H has an or-thonormal system of eigenvectors ψk corresponding to eigenvalues λk such that every elementξ ∈ H is represented as

ξ =∑

k

ckψk + ξ′, (207)

where the vector ξ′ ∈KerA, i.e., Aξ′ = 0. The representation (207) is unique, and

Aξ =∑

k

λkckψk, (208)

andλk → 0, n →∞, (209)

if ψk is an infinite system.

41

11 IEs with symmetric kernels. Hilbert–Schmidt theory

Consider first a general a Hilbert–Schmidt operator (181).

Theorem 25 Let A be a Hilbert–Schmidt operator with a (Hilbert–Schmidt) kernel K(x, y).Then the operator A∗ adjoint to A is an operator with the adjoint kernel K∗(x, y) = K(y, x).

Proof. Using the Fubini theorem and changing the order of integration, we obtain

(Af, g) =∫ b

a

∫ b

aK(x, y)f(y)dy

g(x)dx =

=∫ b

a

∫ b

aK(x, y)f(y)g(x)dydx =

=∫ b

a

∫ b

aK(x, y)g(x)dx

f(y)dy =

=∫ b

af(y)

∫ b

aK(x, y)g(x)dx

dy,

which proves the theorem.

Consider a symmetric IE

φ(x)− λ∫ b

aK(x, y)φ(y)dy = f(x) (210)

with a symmetric kernel

K(x, y) = K(y, x) (211)

for complex-valued or

K(x, y) = K(y, x) (212)

for real-valued K(x, y).According to Theorem 25, an integral operator (181) is selfadjoint in the space L2[a, b], i.e.,

A = A∗, if and only if it has a symmetric kernel.

11.1 Selfadjoint integral operators

Let us formulate the properties (19) and (20) of eigenvalues and eigenfunctions (eigenvectors)for a selfadjoint integral operator acting in the Hilbert space H.

Theorem 26 All eigenvalues of a integral operator with a symmetric kernel are real.

Proof. Let λ0 and φ0 be an eigenvalue and eigenfunction of a integral operator with a symmetrickernel, i.e.,

φ0(x)− λ0

∫ b

aK(x, y)φ0(y)dy = 0. (213)

42

Multiply equality (213) by φ0 and integrate with respect to x between a and b to obtain

||φ0||2 − λ0(Kφ0, φ0) = 0, (214)

which yields

λ0 =||φ0||2

(Kφ0, φ0). (215)

Rewrite this equality as

λ0 =||φ0||2

µ, µ = (Kφ0, φ0), (216)

and prove that µ is a real number. For a symmetric kernel, we have

µ = (Kφ0, φ0) = (φ0, Kφ0). (217)

According to the definition of the inner product

(φ0, Kφ0) = (Kφ0, φ0).

Thus

µ = (Kφ0, φ0) = µ = (Kφ0, φ0),

so that µ = µ, i.e., µ is a real number and, consequently, λ0 in (215) is a real number.

Theorem 27 Eigenfunctions of an integral operator with a symmetric kernel corresponding todifferent eigenvalues are orthogonal.

Proof. Let λ and µ and φλ and φµ be two different eigenvalues and the corresponding eigen-functions of an integral operator with a symmetric kernel, i.e.,

φλ(x) = λ∫ b

aK(x, y)φλ(y)dy = 0, (218)

φµ(x) = λ∫ b

aK(x, y)φµ(y)dy = 0. (219)

Then we have

λ(φλ, φµ) = (λφλ, φµ) = (Aφλ, φµ) = (φλ, Aφµ) = (φλ, µφµ) = µ(φλ, φµ), (220)

which yields(λ− µ)(φλ, φµ) = 0, (221)

i.e., (φλ, φµ) = 0.

One can also reformulate the Hilbert–Schmidt theorem 24 for integral operators.

43

Theorem 28 Let λ1, λ2, . . . , and φ1, φ2, . . . , be eigenvalues and eigenfunctions (eigenvectors)of an integral operator with a symmetric kernel and let h(x) ∈ L2[a, b], i.e.,

∫ b

a|h(x)|2dx ≤ ∞.

If K(x, y) is a Hilbert–Schmidt kernel, i.e., a square-integrable function in the square Π =(x, y) : a ≤ x ≤ b, a ≤ y ≤ b, so that

∫ b

a

∫ b

a|K(x, y)|2dxdy < ∞, (222)

then the function

f(x) = Ah =∫ b

aK(x, y)h(y)dy (223)

is decomposed into an absolutely and uniformly convergent Fourier series in the orthonornalsystem φ1, φ2, . . . ,

f(x) =∞∑

n=1

fnφn(x), fn = (f, φn).

The Fourier coefficients fn of the function f(x) are coupled with the Fourier coefficients hn ofthe function h(x) by the relationships

fn =hn

λn

, hn = (h, φn),

so that

f(x) =∞∑

n=1

hn

λn

φn(x) (fn = (f, φn), hn = (h, φn)). (224)

Setting h(x) = K(x, y) in (224), we obtain

K2(x, y) =∞∑

n=1

ωn(y)

λn

φn(x), ωn(y) = (K(x, y), φn(x)), (225)

and ωn(y) are the Fourier coefficients of the kernel K(x, y). Let us calculate ωn(y). We have,using the formula for the Fourier coefficients,

ωn(y) =∫ b

aK(x, y)φn(x)dx, (226)

or, because the kernel is symmetric,

ωn(y) =∫ b

aK(x, y)φn(x)dx, (227)

Note that φn(y) satisfies the equation

φn(x) = λn

∫ b

aK(x, y)φn(y)dy. (228)

44

Replacing x by y abd vice versa, we obtain

1

λn

φn(y) =∫ b

aK(y, x)φn(x)dx, (229)

so that

ωn(y) =1

λn

φn(y).

Now we have

K2(x, y) =∞∑

n=1

φn(x)φn(y)

λ2n

. (230)

In the same manner, we obtain

K3(x, y) =∞∑

n=1

φn(x)φn(y)

λ3n

,

and, generally,

Km(x, y) =∞∑

n=1

φn(x)φn(y)

λmn

. (231)

which is bilinear series for kernel Km(x, y).For kernel K(x, y), the bilinear series is

K(x, y) =∞∑

n=1

φn(x)φn(y)

λn

. (232)

This series may diverge in the sense of C-norm (i.e., uniformly); however it alwyas convergesin L2-norm.

11.2 Hilbert–Schmidt theorem for integral operators

Consider a Hilbert–Schmidt integral operator A with a symmetric kernel K(x, y). In this case,the following conditions are assumed to be satisfied:

i Aφ(x) =∫ b

aK(x, y)φ(y)dy,

ii∫ b

a

∫ b

a|K(x, y)|2dxdy ≤ ∞, (233)

iii K(x, y) = K(y, x).

According to Theorem 18, A is a completely continuous selfadjoint operator in the space L2[a, b]and we can apply the Hilbert–Schmidt theorem 24 to prove the following statement.

45

Theorem 29 If 1 is not an eigenvalue of the operator A then the IE (210), written in theoperator form as

φ(x) = Aφ + f(x), (234)

has one and only one solution for every f . If 1 is an eigenvalue of the operator A, then IE(210) (or (234)) is solvable if and only if f is orthogonal to all eigenfunctions of the operator Acorresponding to the eigenvalue λ = 1, and in this case IE (210) has infinitely many solutions.

Proof. According to the Hilbert–Schmidt theorem, a completely continuous selfadjoint oper-ator A in the Hilbert space H = L2[a, b] has an orthonormal system of eigenvectors ψkcorresponding to eigenvalues λk such that every element ξ ∈ H is represented as

ξ =∑

k

ckψk + ξ′, (235)

where ξ′ ∈Ker A, i.e., Aξ′ = 0. Set

f =∑

k

bkψk + f ′, Af ′ = 0, (236)

and look for a solution to (234) in the form

φ =∑

k

xkψk + φ′, Aφ′ = 0. (237)

Substituting (236) and (237) into (234) we obtain

∑

k

xkψk + φ′ =∑

k

xkλkψk +∑

k

bkψk + f ′, (238)

which holds if and only if

f ′ = φ′,

xk(1− λk) = bk (k = 1, 2, . . .)

i.e., when

f ′ = φ′,

xk =bk

1− λk

, λk 6= 1

bk = 0, λk = 1. (239)

The latter gives the necessary and sufficient condition for the solvability of equation (234). Notethat the coefficients xn that correspond to those n for which λn = 1 are arbitrary numbers.The theorem is proved.

46

12 Harmonic functions and Green’s formulas

Denote by D ∈ R2 a two-dimensional domain bounded by the closed smooth contour Γ.A twice continuously differentiable real-valued function u defined on a domain D is called

harmonic if it satisfies Laplace’s equation

∆ u = 0 in D, (240)

where

∆u =∂2u

∂x2+

∂2u

∂y2(241)

is called Laplace’s operator (Laplacian), the function u = u(x), and x = (x, y) ∈ R2. We willalso use the notation y = (x0, y0).

The function

Φ(x,y) = Φ(x− y) =1

2πln

1

|x− y| (242)

is called the fundamental solution of Laplace’s equation. For a fixed y ∈ R2, y 6= x, the functionΦ(x,y) is harmonic, i.e., satisfies Laplace’s equation

∂2Φ

∂x2+

∂2Φ

∂y2= 0 in D. (243)

The proof follows by straightforward differentiation.

12.1 Green’s formulas

Let us formulate Green’s theorems which leads to the second and third Green’s formulas.Let D ∈ R2 be a (two-dimensional) domain bounded by the closed smooth contour Γ

and let ny denote the unit normal vector to the boundary Γ directed into the exterior of Γand corresponding to a point y ∈ Γ. Then for every function u which is once continuouslydifferentiable in the closed domain D = D + Γ, u ∈ C1(D), and every function v which is twicecontinuously differentiable in D, v ∈ C2(D), Green’s first theorem (Green’s first formula) isvalid ∫ ∫

D

(u∆v + gradu · grad v)dx =∫

Γ

u∂v

∂ny

dly, (244)

where · denotes the inner product of two vector-functions. For u ∈ C2(D) and v ∈ C2(D),Green’s second theorem (Green’s second formula) is valid

∫ ∫

D

(u∆v − v∆u)dx =∫

Γ

(u

∂v

∂ny

− v∂u

∂ny

)dly, (245)

Proof. Apply Gauss theorem ∫ ∫

D

divAdx =∫

Γ

ny · Adly (246)

47

to the vector function (vector field) A = (A1.A2) ∈ C1(D) (with the components A1. A2 beingonce continuously differentiable in the closed domain D = D +Γ, Ai ∈ C1(D), i = 1, 2) definedby

A = u · grad v =

[u

∂v

∂x, u

∂v

∂y

], (247)

the components are

A1 = u∂v

∂x, A2 = u

∂v

∂y. (248)

We have

divA = div (u grad v) =∂

∂x

(u

∂v

∂x

)+

∂

∂y

(u∂v

∂y

)= (249)

∂u

∂x

∂v

∂x+ u

∂2v

∂x2+

∂u

∂y

∂v

∂y+ u

∂2v

∂y2= gradu · grad v + u∆v. (250)

On the other hand, according to the definition of the normal derivative,

ny · A = ny · (ugrad v) = u(ny · grad v) = u∂v

∂ny

, (251)

so that, finally,

∫ ∫

D

divAdx =∫ ∫

D

(u∆v + gradu · grad v)dx,∫

Γ

ny · Ad ly =∫

Γ

u∂v

∂ny

dly, (252)

which proves Green’s first formula. By interchanging u and v and subtracting we obtain thedesired Green’s second formula (245).

Let a twice continuously differentiable function u ∈ C2(D) be harmonic in the domain D.Then Green’s third theorem (Green’s third formula) is valid

u(x) =∫

Γ

(Φ(x,y)

∂u

∂ny

− u(y)∂Φ(x,y)

∂ny

)dly, x ∈ D. (253)

Proof. For x ∈ D we choose a circle

Ω(x, r) = y : |x− y| =√

(x− x0)2 + (y − y0)2 = rof radius r (the boundary of a vicinity of a point x ∈ D) such that Ω(x, r) ∈ D and direct theunit normal n to Ω(x, r) into the exterior of Ω(x, r). Then we apply Green’s second formula(245) to the harmonic function u and Φ(x,y) (which is also a harmonic function) in the domainy ∈ D : |x− y| > r to obtain

0 =∫ ∫

D\Ω(x,r)

(u∆Φ(x,y)− Φ(x,y)∆u)dx = (254)

∫

Γ∪Ω(x,r)

(u(y)

∂Φ(x,y)

∂ny

− Φ(x,y)∂u(y)

∂ny

)dly. (255)

48

Since on Ω(x, r) we have

grad yΦ(x,y) =ny

2πr, (256)

a straightforward calculation using the mean-value theorem and the fact that

∫

Γ

∂v

∂ny

dly = 0 (257)

for a harmonic in D function v shows that

limr→0

∫

Ω(x,r)

(u(y)

∂Φ(x,y)

∂ny

− Φ(x,y)∂u(y)

∂ny

)dly = u(x), (258)

which yields (253).

12.2 Properties of harmonic functions

Theorem 30 Let a twice continuously differentiable function v be harmonic in a domain Dbounded by the closed smooth contour Γ. Then

∫

Γ

∂v(y)

∂ny

dly = 0. (259)

Proof follows from the first Green’s formula applied to two harmonic functions v and u = 1.

Theorem 31 (Mean Value Formula) Let a twice continuously differentiable function u beharmonic in a ball

B(x, r) = y : |x− y| =√

(x− x0)2 + (y − y0)2 ≤ rof radius r (a vicinity of a point x) with the boundary Ω(x, r) = y : |x − y| = r (a circle)and continuous in the closure B(x, r). Then

u(x) =1

πr2

∫

B(x,r)

u(y)dy =1

2πr

∫

Ω(x,r)

u(y)dly, (260)

i.e., the value of u in the center of the ball is equal to the integral mean values over both theball and its boundary.

Proof. For each 0 < ρ < r we apply (259) and Green’s third formula to obtain

u(x) =1

2πρ

∫

|x−y|=ρ

u(y)dly, (261)

whence the second mean value formula in (260) follows by passing to the limit ρ → r. Multi-plying (261) by ρ and integrating with respect to ρ from 0 to r we obtain the first mean valueformula in (260).

49

Theorem 32 (Maximum–Minimum Principle) A harmonic function on a domain cannotattain its maximum or its minimum unless it is constant.

Corollary. Let D ∈ R2 be a two-dimensional domain bounded by the closed smooth contour Γand let u be harmonic in D and continuous in D. Then u attains both its maximum and itsminimum on the boundary.

13 Boundary value problems

Formulate the interior Dirichlet problem: find a function u that is harmonic in a domain Dbounded by the closed smooth contour Γ, continuous in D = D ∪ Γ and satisfies the Dirichletboundary condition:

∆ u = 0 in D, (262)

u|Γ = −f, (263)

where f is a given continuous function.Formulate the interior Neumann problem: find a function u that is harmonic in a domain D

bounded by the closed smooth contour Γ, continuous in D = D ∪ Γ and satisfies the Neumannboundary condition

∂u

∂n

∣∣∣∣∣Γ

= −g, (264)

where g is a given continuous function.

Theorem 33 The interior Dirichlet problem has at most one solution.

Proof. Assume that the interior Dirichlet problem has two solutions, u1 and u2. Their differenceu = u1 − u2 is a harmonic function that is continuous up to the boundary and satisifes thehomogeneous boundary condition u = 0 on the boundary Γ of D. Then from the maximum–minimum principle or its corollary we obtain u = 0 in D, which proves the theorem.

Theorem 34 Two solutions of the interior Neumann problem can differ only by a constant.The exterior Neumann problem has at most one solution.

Proof. We will prove the first statement of the theorem. Assume that the interior Neumannproblem has two solutions, u1 and u2. Their difference u = u1− u2 is a harmonic function thatis continuous up to the boundary and satisifes the homogeneous boundary condition ∂u

∂ny= 0

on the boundary Γ of D. Suppose that u is not constant in D. Then there exists a closed ballB contained in D such that ∫ ∫

B

|gradu|2dx > 0.

50

From Green’s first formula applied to a pair of (harmonic) functions u, v = u and the interiorDh of a surface Γh = x− hnx : x ∈ Γ parallel to the Ds boundary Γ with sufficiently smallh > 0, ∫ ∫

Dh

(u∆u + |gradu|2)dx =∫

Γ

u∂u

∂ny

dly, (265)

we have first ∫ ∫

B

|gradu|2dx ≤∫ ∫

Dh

|gradu|2dx, (266)

(because B ∈ Dh); next we obtain

∫ ∫

Dh

|gradu|2dx ≤∫

Γ

u∂u

∂ny

dly = 0. (267)

Passing to the limit h → 0 we obtain a contradiction∫ ∫

B

|gradu|2dx ≤ 0.

Hence, the difference u = u1 − u2 must be constant in D.

14 Potentials with logarithmic kernels

In the theory of boundary value problems, the integrals

u(x) =∫

C

E(x,y)ξ(y)dly, v(x) =∫

C

∂

∂ny

E(x,y)η(y)dly (268)

are called the potentials. Here, x = (x, y), y = (x0, y0) ∈ R2; E(x,y) is the fundamental solutionof a second-order elliptic differential operator;

∂

∂ny

=∂

∂ny

is the normal derivative at the point y of the closed piecewise smooth boundary C of a domainin R2; and ξ(y) and η(y) are sufficiently smooth functions defined on C. In the case of Laplace’soperator ∆u,

g(x,y) = g(x− y) =1

2πln

1

|x− y| . (269)

In the case of the Helmholtz operator H(k) = ∆ + k2, one can take E(x,y) in the form

E(x,y) = E(x− y) =i

4H

(1)0 (k|x− y|),

where H(1)0 (z) = −2ig(z) + h(z) is the Hankel function of the first kind and zero order (one

of the so-called cylindrical functions) and1

2g(x− y) =

1

2πln

1

|x− y| is the kernel of the two-

dimensional single layer potential; the second derivative of h(z) has a logarithmic singularity.

51

14.1 Properties of potentials

Statement 1. Let D ∈ R2 be a domain bounded by the closed smooth contour Γ. Then thekernel of the double-layer potential

V (x,y) =∂Φ(x,y)

∂ny

, Φ(x,y) =1

2πln

1

|x− y| , (270)

is a continuous function on Γ for x, y ∈ Γ.

Proof. Performing differentiation we obtain

V (x,y) =1

2π

cos θx,y

|x− y| , (271)

where θx,y is the angle between the normal vector ny at the integration point y = (x0, y0) andvector x − y. Choose the origin O = (0, 0) of (Cartesian) coordinates at the point y on curveΓ so that the Ox axis goes along the tangent and the Oy axis, along the normal to Γ at thispoint. Then one can write the equation of curve Γ in a sufficiently small vicinity of the point yin the form

y = y(x).

The asumption concerning the smoothness of Γ means that y0(x0) is a differentiable functionin a vicinity of the origin O = (0, 0), and one can write a segment of the Taylor series

y = y(0) + xy′(0) +x2

2y′′(ηx) =

1

2x2y′′(ηx) (0 ≤ η < 1)

because y(0) = y′(0) = 0. Denoting r = |x − y| and taking into account that x = (x, y) andthe origin of coordinates is placed at the point y = (0, 0), we obtain

r =√

(x− 0)2 + (y − 0)2 =√

x2 + y2 =

√x2 +

1

4x4(y′′(ηx))2 = x

√1 +

1

4x2(y′′(ηx))2;

cos θx,y =y

r=

1

2

xy′′(ηx)√1 + x2(1/4)(y′′(ηx))2

,

andcos θx,y

r=

y

r2=

1

2

y′′(ηx)

1 + x2(1/4)(y′′(ηx))2.

The curvature K of a plane curve is given by

K =y′′

(1 + (y′)2)3/2,

which yields y′′(0) = K(y), and, finally,

limr→0

cos θx,y

r=

1

2K(y)

52

which proves the continuity of the kernel (270).

Statement 2 (Gauss formula) Let D ∈ R2 be a domain bounded by the closed smoothcontour Γ. For the double-layer potential with a constant density

v0(x) =∫

Γ

∂Φ(x,y)

∂ny

dly, Φ(x,y) =1

2πln

1

|x− y| , (272)

where the (exterior) unit normal vector n of Γ is directed into the exterior domain R2 \ D, wehave

v0(x) = −1, x ∈ D,

v0(x) = −1

2, x ∈ Γ, (273)

v0(x) = 0, x ∈ R2 \ D.

Proof follows for x ∈ R2 \ D from the equality

∫

Γ

∂v(y)

∂ny

dly = 0 (274)

applied to v(y) = Φ(x,y). For x ∈ D it follows from Green’s third formula

u(x) =∫

Γ

(Φ(x,y)

∂u

∂ny

− u(y)∂Φ(x,y)

∂ny

)dly, x ∈ D, (275)

applied to u(y) = 1 in D.Note also that if we set

v0(x′) = −1

2, x′ ∈ Γ,

we can also write (273) as

v0±(x′) = lim

h→±0v(x + hnx′) = v0(x′)± 1

2, x′ ∈ Γ. (276)

Corollary. Let D ∈ R2 be a domain bounded by the closed smooth contour Γ. Introducethe single-layer potential with a constant density

u0(x) =∫

Γ

Φ(x,y)dly, Φ(x,y) =1

2πln

1

|x− y| . (277)

For the normal derivative of this single-layer potential

∂u0(x)

∂nx

=∫

Γ

∂Φ(x,y)

∂nx

dly, (278)

53

where the (exterior) unit normal vector nx of Γ is directed into the exterior domain R2 \ D, wehave

∂u0(x)

∂nx

= 1, x ∈ D,

∂u0(x)

∂nx

=1

2, x ∈ Γ, (279)

∂u0(x)

∂nx

= 0, x ∈ R2 \ D.

Theorem 35 Let D ∈ R2 be a domain bounded by the closed smooth contour Γ. The double-layer potential

v(x) =∫

Γ

∂Φ(x,y)

∂ny

ϕ(y)dly, Φ(x,y) =1

2πln

1

|x− y| , (280)

with a continuous density ϕ can be continuously extended from D to D and from R2 \ D toR2 \D with the limiting values on Γ

v±(x′) =∫

Γ

∂Φ(x′,y)

∂ny

ϕ(y)dly ± 1

2ϕ(x′), x′ ∈ Γ, (281)

or

v±(x′) = v(x′)± 1

2ϕ(x′), x′ ∈ Γ, (282)

wherev±(x′) = lim

h→±0v(x + hnx′). (283)

Proof.1. Introduce the function

I(x) = v(x)− v0(x) =∫

Γ

∂Φ(x,y)

∂ny

(ϕ(y)− ϕ0)dly (284)

where ϕ0 = ϕ(x′) and prove its continuity at the point x′ ∈ Γ. For every ε > 0 and every η > 0there exists a vicinity C1 ⊂ Γ of the point x′ ∈ Γ such that

|ϕ(x)− ϕ(x′)| < η, x′ ∈ C1. (285)

SetI = I1 + I2 =

∫

C1

. . . +∫

Γ\C1

. . . .

Then|I1| ≤ ηB1,

54

where B1 is a constant taken according to

∫

Γ

∣∣∣∣∣∂Φ(x,y)

∂ny

∣∣∣∣∣ dly ≤ B1 ∀x ∈ D, (286)

and condition (286) holds because kernel (270) is continuous on Γ.Taking η = ε/B1, we see that for every ε > 0 there exists a vicinity C1 ⊂ Γ of the point

x′ ∈ Γ (i.e., x′ ∈ C1) such that|I1(x)| < ε ∀x ∈ D. (287)

Inequality (287) means that I(x) is continuous at the point x′ ∈ Γ.2. Now, if we take v±(x′) for x′ ∈ Γ and consider the limits v±(x′) = limh→±0 v(x + hny) frominside and outside contour Γ using (276), we obtain

v−(x′) = I(x′) + limh→−0

v0(x′ − hnx′) = v(x′)− 1

2ϕ(x′), (288)

v+(x′) = I(x′) + limh→+0

v0(x′ + hnx′) = v(x′) +1

2ϕ(x′), (289)

which prove the theorem.

Corollary. Let D ∈ R2 be a domain bounded by the closed smooth contour Γ. Introducethe single-layer potential

u(x) =∫

Γ

Φ(x,y)ϕ(y)dly, Φ(x,y) =1

2πln

1

|x− y| . (290)

with a continuous density ϕ. The normal derivative of this single-layer potential

∂u(x)

∂nx

=∫

Γ

∂Φ(x,y)

∂nx

ϕ(y)dly (291)

can be continuously extended from D to D and from R2 \ D to R2 \D with the limiting valueson Γ

∂u(x′)∂nx ±

=∫

Γ

∂Φ(x′,y)

∂nx′ϕ(y)dly ∓ 1

2ϕ(x′), x′ ∈ Γ, (292)

or∂u(x′)∂nx ±

=∂u(x′)∂nx

∓ 1

2ϕ(x′), x′ ∈ Γ, (293)

where∂u(x′)∂nx′

= limh→±0

nx′ · grad v(x′ + hnx′). (294)

55

14.2 Generalized potentials

Let SΠ(Γ) ∈ R2 be a domain bounded by the closed piecewise smooth contour Γ. We assumethat a rectilinear interval Γ0 is a subset of Γ, so that Γ0 = x : y = 0, x ∈ [a, b].

Let us say that functions Ul(x) are the generalized single layer (SLP) (l = 1) or double layer(DLP) (l = 2) potentials if

Ul(x) =∫

Γ

Kl(x, t)ϕl(t)dt, x = (x, y) ∈ SΠ(Γ),

whereKl(x, t) = gl(x, t) + Fl(x, t) (l = 1, 2),

g1(x, t) = g(x, y0) =1

πln

1

|x− y0| , g2(x, t) =∂

∂y0

g(x,y0) [y0 = (t, 0)],

F1,2 are smooth functions, and we shall assume that for every closed domain S0Π(Γ) ⊂ SΠ(Γ),the following conditions hold

i) F1(x, t) is once continuously differentiable with respect to thevariables of x and continuous in t;

ii) F2(x, t) and

F 12 (x, t) =

∂

∂y

t∫

q

F2(x, s)ds, q ∈ R1,

are continuous.

We shall also assume that the densities of the generalized potentials ϕ1 ∈ L(1)2 (Γ) and

ϕ2 ∈ L(2)2 (Γ), where functional spaces L

(1)2 and L

(2)2 are defined in Section 3.

15 Reduction of boundary value problems to integral

equations

Green’s formulas show that each harmonic function can be represented as a combination ofsingle- and double-layer potentials. For boundary value problems we will find a solution in theform of one of these potentials.

Introduce integral operators K0 and K1 acting in the space C(Γ) of continuous functionsdefined on contour Γ

K0(x) = 2∫

Γ

∂Φ(x,y)

∂ny

ϕ(y)dly, x ∈ Γ (295)

and

K1(x) = 2∫

Γ

∂Φ(x,y)

∂nx

ψ(y)dly, x ∈ Γ. (296)

The kernels of operators K0 and K1 are continuous on Γ As seen by interchanging the orderof integration, K0 and K1 are adjoint as integral operators in the space C(Γ) of continuousfunctions defined on curve Γ.

56

Theorem 36 The operators I −K0 and I −K1 have trivial nullspaces

N(I −K0) = 0, N(I −K1) = 0, (297)

The nullspaces of operators I + K0 and I + K1 have dimension one and

N(I + K0) = span 1, N(I + K1) = span ψ0 (298)

with ∫

Γ

ψ0dly 6= 0. (299)

Proof. Let φ be a solution to the homogeneous integral equation φ − K0φ = 0 and define adouble-layer potential

v(x) =∫

Γ

∂Φ(x,y)

∂ny

ϕ(y)dly, x ∈ D (300)

with a continuous density φ. Then we have, for the limiting values on Γ,

v±(x) =∫

Γ

∂Φ(x,y)

∂ny

ϕ(y)dly ± 1

2ϕ(x) (301)

which yields2v−(x) = K0ϕ(x)− ϕ = 0. (302)

From the uniqueness of the interior Dirichlet problem (Theorem 33) it follows that v = 0 in thewhole domain D. Now we will apply the contunuity of the normal derivative of the double-layerpotential (300) over a smooth contour Γ,

∂v(y)

∂ny

∣∣∣∣∣+

− ∂v(y)

∂ny

∣∣∣∣∣−= 0, y ∈ Γ, (303)

where the internal and extrenal limiting values with respect to Γ are defined as follows

∂v(y)

∂ny

∣∣∣∣∣−= lim

y→Γ,y∈D

∂v(y)

∂ny

= limh→0

ny · grad v(y − hny) (304)

and∂v(y)

∂ny

∣∣∣∣∣+

= limy→Γ,y∈R2\D

∂v(y)

∂ny

= limh→0

ny · grad v(y + hny). (305)

Note that (303) can be written as

limh→0

ny · [grad v(y + hny)− grad v(y − hny)] = 0. (306)

From (304)–(306) it follows that

∂v(y)

∂ny

∣∣∣∣∣+

− ∂v(y)

∂ny

∣∣∣∣∣−= 0, y ∈ Γ. (307)

Using the uniqueness for the exterior Neumann problem and the fact that v(y) → 0, |y| → ∞,we find that v(y) = 0, y ∈ R2 \ D. Hence from (301) we deduce ϕ = v+ − v− = 0 on Γ. ThusN(I −K0) = 0 and, by the Fredholm alternative, N(I −K1) = 0.

57

Theorem 37 Let D ∈ R2 be a domain bounded by the closed smooth contour Γ. The double-layer potential

v(x) =∫

Γ

∂Φ(x,y)

∂ny

ϕ(y)dly, Φ(x,y) =1

2πln

1

|x− y| , x ∈ D, (308)

with a continuous density ϕ is a solution of the interior Dirichlet problem provided that ϕ is asolution of the integral equation

ϕ(x)− 2∫

Γ

∂Φx,y)

∂ny

ϕ(y)dly = −2f(x), x ∈ Γ. (309)

Proof follows from Theorem 14.1.

Theorem 38 The interior Dirichlet problem has a unique solution.

Proof The integral equation φ−Kφ = 2f of the interior Dirichlet problem has a unique solutionby Theorem 36 because N(I −K) = 0.

Theorem 39 Let D ∈ R2 be a domain bounded by the closed smooth contour Γ. The double-layer potential

u(x) =∫

Γ

∂Φ(x,y)

∂ny

ϕ(y)dly, x ∈ R2 \ D, (310)

with a continuous density ϕ is a solution of the exterior Dirichlet problem provided that ϕ is asolution of the integral equation

ϕ(x) + 2∫

Γ

∂Φ(x,y)

∂ny

ϕ(y)dly = 2f(x), x ∈ Γ. (311)

Here we assume that the origin is contained in D.

Proof follows from Theorem 14.1.

Theorem 40 The exterior Dirichlet problem has a unique solution.

Proof. The integral operator K : C(Γ) → C(Γ) defined by the right-hand side of (310) iscompact in the space C(Γ) of functions continuous on Γ because its kernel is a continuousfunction on Γ. Let ϕ be a solution to the homogeneous integral equation ϕ + Kϕ = 0 on Γand define u by (310). Then 2u = ϕ + Kϕ = 0 on Γ, and by the uniqueness for the exteriorDirichlet problem it follows that u ∈∈ R2 \ D, and

∫Γ ϕdly = 0. Therefore ϕ + Kϕ = 0 which

means according to Theorem 36 (because N(I + K) = span 1), that ϕ = const on Γ. Now∫Γ ϕdly = 0 implies that ϕ = 0 on Γ, and the existence of a unique solution to the integral

equation (311) follows from the Fredholm property of its operator.

58

Theorem 41 Let D ∈ R2 be a domain bounded by the closed smooth contour Γ. The single-layer potential

u(x) =∫

Γ

Φ(x,y)ψ(y)dly, x ∈ D, (312)

with a continuous density ψ is a solution of the interior Neumann problem provided that ψ isa solution of the integral equation

ψ(x) + 2∫

Γ

∂Φ(x,y)

∂nx

ψ(y)dly = 2g(x), x ∈ Γ. (313)

Theorem 42 The interior Neumann problem is solvable if and only if∫

Γψdly = 0 (314)

is satisfied.

16 Functional spaces and Chebyshev polynomials

16.1 Chebyshev polynomials of the first and second kind

The Chebyshev polynomials of the first and second kind, Tn and Un, are defined by the followingrecurrent relationships:

T0(x) = 1, T1(x) = x; (315)

Tn(x) = 2xTn−1(x)− Tn−2(x), n = 2, 3, ...; (316)

U0(x) = 1, U1(x) = 2x; (317)

Un(x) = 2xUn−1(x)− Un−2(x), n = 2, 3, ... . (318)

We see that formulas (316) and (318) are the same and the polynomials are different becausethey are determined using different initial conditions (315) and (317).

Write down few first Chebyshev polynomials of the first and second kind; we determinethem directly from formulas (315)–(318):

T0(x) = 1,T1(x) = x,T2(x) = 2x2 − 1,T3(x) = 4x3 − 3x,T4(x) = 8x4 − 8x2 + 1,T5(x) = 16x5 − 20x3 + 5x,T6(x) = 32x6 − 48x4 + 18x2 − 1;

(319)

U0(x) = 1,U1(x) = 2x,U2(x) = 4x2 − 1,U3(x) = 8x3 − 4x,U4(x) = 16x4 − 12x2 + 1,U5(x) = 32x5 − 32x3 + 6x,U6(x) = 64x6 − 80x4 + 24x2 − 1.

(320)

59

We see that Tn and Un are polynomials of degree n.Let us breifly summarize some important properties of the Chebyshev polynomials.If |x| ≤ 1, the following convenient representations are valid

Tn(x) = cos(n arccos x), n = 0, 1, ...; (321)

Un(x) =1√

1− x2sin ((n + 1) arccos x), n = 0, 1, ... . (322)

At x = ±1 the right-hand side of formula (322) should be replaced by the limit.Polynomials of the first and second kind are coupled by the relationships

Un(x) =1

n + 1T ′

n+1(x), n = 0, 1, ... . (323)

Polynomials Tn and Un are, respectively, even functions for even n and odd functions for oddn:

Tn(−x) = (−1)n Tn(x),Un(−x) = (−1)n Un(x).

(324)

One of the most important properties is the orthogonality of the Chebyshev polynomialsin the segment [−1, 1] with a certain weight function (a weight). This property is expressed asfollows:

1∫

−1

Tn(x)Tm(x)1√

1− x2dx =

0, n 6= mπ/2, n = m 6= 0π, n = m = 0

; (325)

1∫

−1

Un(x)Um(x)√

1− x2 dx =

0, n 6= mπ/2, n = m

. (326)

Thus in the segment [−1, 1], the Chebyshev polynomials of the first kind are orthogonalwith the weight 1/

√1− x2 and the Chebyshev polynomials of the second kind are orthogonal

with the weight√

1− x2.

16.2 Fourier–Chebyshev series

Formulas (325) and (326) enable one to decompose functions in the Fourier series in Chebyshevpolynomials (the Fourier–Chebyshev series),

f(x) =a0

2T0(x) +

∞∑

n=1

anTn(x), x ∈ [−1, 1], (327)

or

f(x) =∞∑

n=0

bnUn(x), x ∈ [−1, 1]. (328)

Coefficients an and bn are determined as follows

an =2

π

1∫

−1

f(x)Tn(x)dx√

1− x2, n = 0, 1, ...; (329)

60

bn =2

π

1∫

−1

f(x)Un(x)√

1− x2dx, n = 0, 1, ... . (330)

Note that f(x) may be a complex-valued function; then the Fourier coefficients an and bn

are complex numbers.Consider some examples of decomposing functions in the Fourier–Chebyshev series

√1− x2 =

2

π− 4

π

∞∑

n=1

1

4n2 − 1T2n(x), −1 ≤ x ≤ 1;

arcsin x =4

π

∞∑

n=0

1

(2n + 1)2T2n+1(x), −1 ≤ x ≤ 1;

ln (1 + x) = − ln 2 + 2∞∑

n=1

(−1)n−1

nTn(x), −1 < x ≤ 1.

Similar decompositions can be obtained using the Chebyshev polynomials Un of the secondkind. In the first example above, an even function f(x) =

√1− x2 is decomposed; therefore its

expansion contains only coefficients multiplying T2n(x) with even indices, while the terms withodd indices vanish.

Since |Tn(x)| ≤ 1 according to (321), the first and second series converge for all x ∈ [−1, 1]uniformly in this segment. The third series represents a different type of convergence: it con-verges for all x ∈ (−1, 1] (−1 < x ≤ 1) and diverges at the point x = −1; indeed, thedecomposed function f(x) = ln(x + 1) is not defined at x = −1.

In order to describe the convergence of the Fourier–Chebyshev series introduce the functionalspaces of functions square-integrable in the segment (−1, 1) with the weights 1/

√1− x2 and√

1− x2. We shall write f ∈ L(1)2 if the integral

1∫

−1

|f(x)|2 dx√1− x2

(331)

converges, and f ∈ L(2)2 if the integral

1∫

−1

|f(x)|2√

1− x2dx. (332)

converges. The numbers specified by (convergent) integrals (331) and (332) are the squared

norms of the function f(x) in the spaces L(1)2 and L

(2)2 ; we will denote them by ‖f‖2

1 and ‖f‖22,

respectively. Write the symbolic definitions of these spaces:

L(1)2 := f(x) : ‖f‖2

1 :=

1∫

−1

|f(x)|2 dx√1− x2

< ∞,

L(2)2 := f(x) : ‖f‖2

2 :=

1∫

−1

|f(x)|2√

1− x2dx < ∞.

61

Spaces L(1)2 and L

(2)2 are (infinite-dimensional) Hilbert spaces with the inner products

(f, g)1 =

1∫

−1

f(x)g(x)dx√

1− x2,

(f, g)2 =

1∫

−1

f(x)g(x)√

1− x2 dx.

Let us define two more functional spaces of differentiable functions defined in the segment(−1, 1) whose first derivatives are square-integrable in (−1, 1) with the weights 1/

√1− x2 and√

1− x2:W 1

2 := f(x) : f ∈ L(1)2 , f ′ ∈ L

(2)2 ,

andW 1

2 := f(x) : f ∈ L(1)2 , f ′(x)

√1− x2 ∈ L

(2)2 ;

they are also Hilbert spaces with the inner products

(f, g)W 12

=

1∫

−1

f(x)dx√

1− x2

1∫

−1

g(x)dx√

1− x2+

1∫

−1

f ′(x)g′(x)√

1− x2 dx,

(f, g)W 1

2=

1∫

−1

f(x)dx√

1− x2

1∫

−1

g(x)dx√

1− x2+

1∫

−1

f ′(x)g′(x) (1− x2)3/2 dx

and norms

‖f‖2W 1

2=

∣∣∣∣∣∣

1∫

−1

f(x)dx√

1− x2

∣∣∣∣∣∣

2

+

1∫

−1

|f ′(x)|2√

1− x2 dx,

‖f‖2W 1

2

=

∣∣∣∣∣∣

1∫

−1

f(x)dx√

1− x2

∣∣∣∣∣∣

2

+

1∫

−1

|f ′(x)|2 (1− x2)3/2 dx.

If a function f(x) is decomposed in the Fourier–Chebyshev series (327) or (328) and is

an elemet of the space L(1)2 or L

(2)2 (that is, integrals (331) and (332) converge), then the

corresponding series always converge in the root-mean-square sense, i.e.,

‖f − SN‖1 → 0 and ‖f − SN‖2 → 0 as N →∞,

where SN are partial sums of the Fourier–Chebyshev series. The ’smoother’ function f(x) inthe segment [−1, 1] the ’faster’ the convergence of its Fourier–Chebyshev series. Let us explainthis statement: if f(x) has p continuous derivatives in the segment [−1, 1], then

∣∣∣∣∣f(x)− a0

2T0(x)−

N∑

n=1

anTn(x)

∣∣∣∣∣ ≤C ln N

Np, x ∈ [−1, 1], (333)

where C is a constant that does not depend on N .

62

A similar statement is valid for the Fourier–Chebyshev series in polynomials Un: if f(x) hasp continuous derivatives in the segment [−1, 1] and p ≥ 2, then

∣∣∣∣∣f(x)−N∑

n=0

bnUn(x)

∣∣∣∣∣ ≤C

Np−3/2, x ∈ [−1, 1]. (334)

We will often use the following three relationships involving the Chebyshev polynomials:

1∫

−1

1

x− yTn(x)

dx√1− x2

= πUn−1(y), −1 < y < 1, n ≥ 1; (335)

1∫

−1

1

x− yUn−1(x)

√1− x2dx = −πTn(y), −1 < y < 1, n ≥ 1; (336)

1∫

−1

ln1

|x− y|Tn(x)dx√

1− x2=

π ln 2, n = 0;π Tn(y)/n, n ≥ 1.

(337)

The integrals in (335) and (336) are improper integrals in the sense of Cauchy (a detailedanalysis of such integrals is performed in [8, 9]). The integral in (337) is a standard convergentimproper integral.

17 Solution to integral equations with a logarithmic sin-

gulatity of the kernel

17.1 Integral equations with a logarithmic singulatity of the kernel

Many boundary value problems of mathematical physics are reduced to the integral equationswith a logarithmic singulatity of the kernel

Lϕ :=

1∫

−1

ln1

|x− y| ϕ(x)dx√

1− x2= f(y), −1 < y < 1 (338)

and

(L + K)ϕ :=

1∫

−1

(ln1

|x− y| + K(x, y)) ϕ(x)dx√

1− x2= f(y), −1 < y < 1. (339)

In equations (338) and (339) ϕ(x) is an unknown function and the right-hand side f(y) is agiven (continuous) function. The functions ln (1/|x− y|) in (338) and ln (1/|x− y|) + K(x, y)in (339) are the kernels of the integral equations; K(x, y) is a given function which is assumedto be continuous (has no singularity) at x = y. The kernels go to infinity at x = y andtherefore referred to as functions with a (logarithmic) singulatity, and equations (338) and(339) are called integral equations with a logarithmic singulatity of the kernel. In equations(338) and (339) the weight function 1/

√1− x2 is separated explicitly. This function describes

63

the behaviour (singularty) of the solution in the vicinity of the endpoints −1 and +1. Linearintegral operators L and L + K are defined by the left-hand sides of (338) and (339).

Equation (338) is a particular case of (339) for K(x, y) ≡ 0 and can be solved explicitly.Assume that the right-hand side f(y) ∈ W 1

2 (for simplicity, one may assume that f is acontinuously differentiable function in the segment [−1, 1]) and look for the solution ϕ(x) in

the space L(1)2 . This implies that integral operators L and L + K are considered as bounded

linear operatorsL : L

(1)2 → W 1

2 and L + K : L(1)2 → W 1

2 .

Note that if K(x, y) is a sufficiently smooth function, then K is a compact (completely contin-uous) operator in these spaces.

17.2 Solution via Fourier–Chebyshev series

Let us describe the method of solution to integral equation (338). Look for an approximationϕN(x) to the solution ϕ(x) of (338) in the form of a partial sum of the Fourier–Chebyshev series

ϕN(x) =a0

2T0(x) +

N∑

n=1

anTn(x), (340)

where an are unknown coefficients. Expand the given function f(y) in the Fourier–Chebyshevseries

fN(y) =f0

2T0(y) +

N∑

n=1

fnTn(y), (341)

where fn are known coefficients determined according to (329). Substituting (340) and (341)into equation (338) and using formula (337), we obtain

a0

2ln 2T0(y) +

N∑

n=1

1

nanTn(y) =

f0

2T0(y) +

N∑

n=1

fnTn(y). (342)

Now equate coefficients multiplying Tn(y) with the same indices n:

a0

2ln 2 =

f0

2,

1

nan = fn,

which gives unknown coefficients an

a0 =f0

ln 2, an = nfn

and the approximation

ϕN(x) =f0

2 ln 2T0(x) +

N∑

n=1

fnTn(x). (343)

The exact solution to equation (338) is obtained by a transition to the limit N →∞ in (343)

ϕ(x) =f0

2 ln 2T0(x) +

∞∑

n=1

nfnTn(x). (344)

64

In some cases one can determine the sum of series (344) explicitly and obtain the solutionin a closed form. One can show that the approximation ϕN(x) of the solution to equation (338)

defined by (340) converges to the exact solution ϕ(x) as N →∞ in the norm of space L(1)2 .

Consider integral equation (339). Expand the kernel K(x, y) in the double Fourier–Chebyshevseries, take its partial sum

K(x, y) ≈ KN(x, y) =N∑

i=0

N∑

j=0

kijTi(x)Tj(y), (345)

and repeat the above procedure. A more detailed description is given below.

Example 14 Consider the integral equation

1∫

−1

(ln1

|x− y| + xy) ϕ(x)dx√

1− x2= y +

√1− y2, −1 < y < 1.

Use the method of solution described above and set

ϕN(x) =a0

2T0(x) +

N∑

n=1

anTn(x)

to obtain

y +√

1− y2 =2

πT0(y) + T1(y)− 4

π

∞∑

k=1

T2k(y)

4k2 − 1,

K(x, y) = T1(x)T1(y).

Substituting these expressions into the initial integral equation, we obtain

a0

2ln 2T0(y) +

N∑

n=1

an

nTn(y) +

π

2a1T1(y) =

2

πT0(y) + T1(y)− 4

π

[N/2]∑

k=1

T2k(y)

4k2 − 1,

here [N2] denotes the integer part of a number. Equating the coefficients that multiply Tn(y) with

the same indices n we determine the unknowns

a0 =4

π ln 2, a1 =

2

π, a2n+1 = 0, a2n = − 8

π

n

4n2 − 1, n ≥ 1.

The approximate solution has the form

ϕN(x) =4

π ln 2+

2

πx− 8

π

[N/2]∑

n=1

n

4n2 − 1T2n(x).

The exact solution is represented as the Fourier–Chebyshev series

ϕ(x) =4

π ln 2+

2

πx− 8

π

∞∑

n=1

n

4n2 − 1T2n(x).

The latter series converges to the exact solution ϕ(x) as N → ∞ in the norm of space L(1)2 ,

and its sum ϕ ∈ L(1)2 .

65

18 Solution to singular integral equations

18.1 Singular integral equations

Consider two types of the so-called singular integral equations

S1ϕ :=

1∫

−1

1

x− yϕ(x)

dx√1− x2

= f(y), −1 < y < 1; (346)

(S1 + K1)ϕ =

1∫

−1

(1

x− y+ K1(x, y)

)ϕ(x)

dx√1− x2

= f(y), −1 < y < 1; (347)

S2ϕ :=

1∫

−1

1

x− yϕ(x)

√1− x2dx = f(y), −1 < y < 1; (348)

(S2 + K2)ϕ :=

1∫

−1

(1

x− y+ K2(x, y)

)ϕ(x)

√1− x2 dx = f(y), −1 < y < 1. (349)

Equations (346) and (348) constitute particular cases of a general integral equation (347)and can be solved explicitly. Singular integral operators S1 and S1 + K1 are considered asbounded linear operators in the weighted functional spaces introduced above, S1 : L

(1)2 → L

(2)2 ,

S1 + K1 : L(1)2 → L

(2)2 . Singular integral operators S2 S2 + K2 are considered as bounded linear

operators in the weighted functional spaces S2 : L(2)2 → L

(1)2 , S2 +K2 : L

(2)2 → L

(1)2 . If the kernel

functions K1(x, y) and K2(x, y) are sufficiently smooth, K1 and K2 will be compact operatorsin these spaces.

18.2 Solution via Fourier–Chebyshev series

Look for an approximation ϕN(x) to the solution ϕ(x) of the singular integral equations in theform of partial sums of the Fourier–Chebyshev series. From formulas (335) and (336) it followsthat one should solve (346) and (347) by expanding functions ϕ in the Chebyshev polynomialsof the first kind and f in the Chebyshev polynomials of the second kind. In equations (348)and (349), ϕ should be decomposed in the Chebyshev polynomials of the second kind and f inthe Chebyshev polynomials of the first kind.

Solve equation (346). Assume that f ∈ L(2)2 and look for a solution ϕ ∈ L

(1)2 :

ϕN(x) =a0

2T0(x) +

N∑

n=1

anTn(x); (350)

fN(y) =N−1∑

k=0

fkUk(y), (351)

where coefficients fk are calculated by formulas (330). Substituting expressions (350) and (351)into equation (346) and using (335), we obtain

πN∑

n=1

anUn−1(y) =N−1∑

k=0

fkUk(y). (352)

66

Coefficient a0 vanishes because (see [4])

1∫

−1

1

x− y

dx√1− x2

= 0, −1 < y < 1. (353)

Equating, in (352), the coefficients multiplying Un(y) with equal indices, we obtain an =fn−1/π, n ≥ 1. The coefficient a0 remains undetermined. Thus the solution to equation (346)depends on an arbitrary constant C = a0/2. Write the corresponding formulas for the approx-imate solution,

ϕN(x) = C +1

π

N∑

n=1

fn−1 Tn(x); (354)

and the exact solution,

ϕ(x) = C +1

π

∞∑

n=1

fn−1Tn(x). (355)

The latter series converges in the L(1)2 -norm.

In order to obtain a unique solution to equation (346) (to determine the constant C),one should specify an additional condition for function ϕ(x), for example, in the form of anorthogonality condition

1∫

−1

ϕ(x)dx√

1− x2= 0. (356)

(orthogonality of the solution to a constant in the space L(1)2 ). Then using the orthogonality

(325) of the Cebyshev polynomial T0(x) ≡ 1 to all Tn(x), n ≥ 1, and calculating the innerproduct in (355) (multiplying both sides by 1/

√1− x2 and integrating from -1 to 1) we obtain

C =1

π

1∫

−1

ϕ(x)dx√

1− x2= 0.

Equation (347) can be solved in a similar manner; however first it is convenient to decomposethe kernel function K1(x, y) in the Fourier–Chebyshev series and take the partial sum

K1(x, y) ≈ K(1)N (x, y) =

N∑

i=0

N−1∑

j=0

k(1)ij Ti(x)Uj(y), (357)

and then proceed with the solution according to the above.If K1(x, y) is a polynomial in powers of x and y, representation (357) will be exact and can

be obtained by rearranging explicit formulas (315)–(320).

Example 15 Solve a singular integral equation

1∫

−1

(1

x− y+ x2y3

)ϕ(x)

dx√1− x2

= y, −1 < y < 1

67

subject to condition (356). Look for an approximate solution in the form

ϕN(x) =a0

2T0(x) +

N∑

n−1

anTn(x).

The right-hand side

y =1

2U1(y).

Decompose K1(x, y) in the Fourier–Chebyshev series using (320),

K1(x, y) =(

1

2T0(x) +

1

2T2(x)

) (1

4U1(y) +

1

8U3(y)

).

Substituting the latter into equation (335) and using the orthogonality conditions (325), weobtain

πN∑

n=1

anUn−1(y) +(

1

4U1(y) +

1

8U3(y)

)π(a0 + a2)

4=

1

2U1(y),

Equating the coefficients multiplying Un(y) with equal indices, we obtain

πa1 = 0, πa2 +π(a0 + a2)

16=

1

2, πa3 = 0,

πa4 +π(a0 + a2)

32= 0; πan = 0, n ≥ 5.

Finally, after some algebra, we determine the unknown coefficients

a0 = C, a2 =1

17(8

π− C), a4 =

1

68(2C − 1

π), an = 0, n 6= 0, 2, 4,

where C is an arbitrary constant. In order to determine the unique solution that satisfies (356),calculate

C = 0, a0 = 0, a2 =8

17π, a4 = − 1

68π.

The final form of the solution is

ϕ(x) =8

17πT2(x)− 1

68πT4(x),

or

ϕ(x) = − 2

17πx4 +

18

17πx2 − 33

68π.

Note that the approximate solution ϕN(x) = ϕ(x) for N ≥ 4. We see that here, one can obtainthe solution explicitly as a polynomial.

Consider equations (348) and (349). Begin with (348) and look for its solution in the form

ϕN(x) =N−1∑

n=0

bnUn(x). (358)

68

Expand the right-hand side in the Chebyshev polynomials of the first kind

fN(y) =f0

2T0(y) +

N∑

k=1

fkTk(y), (359)

where coefficients fk are calculated by (329). Substitute (358) and (359) into (348) and use(336) to obtain

−πN−1∑

n=0

bnTn+1(y) =f0

2T0(y) +

N∑

k=1

fkTk(y), (360)

where unknown coefficients bn are determined form (360),

bn = − 1

πfn+1, n ≥ 0.

Equality (360) holds only if f0 = 0. Consequently, we cannot state that the solution to (348)exists for arbitrary function f(y). The (unique) solution exists only if f(y) satisfies the orthog-onality condition

1∫

−1

f(y)dy√

1− y2= 0, (361)

which is equivalent to the requirement f0 = 0 (according to (329) taken at n = 0). Note that,when solving equation (348), we impose, unlike equation (356), the condition (361) for equation(346), on the known function f(y). As a result, we obtain

a unique approximate solution to (348) for every N ≥ 2

ϕN(x) = − 1

π

N−1∑

n=0

fn+1Un(x), (362)

anda unique exact solution to (348)

ϕ(x) = − 1

π

∞∑

n=0

fn+1Un(x). (363)

We will solve equations (348) and (349) under the assumption that the right-hand side

f ∈ L(1)2 and ϕ ∈ L

(2)2 . Series (363) converges in the norm of the space L

(2)2 .

The solution of equation (349) is similar to that of (348). The difference is that K2(x, y) isapproximately replaced by a partial sum of the Fourier–Chebyshev series

K2(x, y) ≈ K(2)N (x, y) =

N−1∑

i=0

N∑

j=0

k(2)ij Ui(x)Tj(y). (364)

Substituting (358), (359), and (364) into equation (349) we obtain a linear equation systemwith respect to unknowns bn. As well as in the case of equation (348), the solution to (349)exists not for all right-hand sides f(y).

The method of decomposing in Fourier–Chebyshev series allows one not only to solve equa-tions (348) and (349) but also verify the existence of solution. Consider an example of anequation of the form (349) that has no solution (not solvable).

69

Example 16 Solve the singular integral equation

1∫

−1

(1

x− y+

√1− y2

)ϕ(x)

√1− x2 dx = 1, −1 < y < 1; ϕ ∈ L

(2)2 .

Look for an approximate solution, as well as in the case of (348), in the form of a series

ϕN(x) =N−1∑

n=0

bnUn(x).

Replace K2(x, y) =√

1− y2 by an approximate decomposition

K(2)N (x, y) =

2

πT0(y)− 4

π

[N/2]∑

k=1

1

4k2 − 1T2k(y).

The right-hand side of this equation is 1 = T0(y). Substituting this expression into the equation,we obtain

−πN−1∑

n=0

bnTn+1(y) +

T0(y)− 2

[N/2]∑

k=1

1

4k2 − 1T2k(y)

b0 = T0(y).

Equating the coefficients that multiply T0(y) on both sides of the equation, we obtain b0 = 1;equating the coefficients that multiply T1(y), we have −πb0 = 0,which yields b0 = 0. Thiscobtradiction proves that the considered equation has no solution. Note that this circumstanceis not connected with the fact that we determine an approxiate solution ϕN(x) rather than theexact solution ϕ(x). By considering infinite series (setting N = ∞) in place of finite sums wecan obtain the same result.

Let us prove the conditions for function f(y) that provide the solvability of the equation

1∫

−1

(1

x− y+

√1− y2

)ϕ(x)

√1− x2 dx = f(y), −1 < y < 1

with the right-hand side f(y) To this end, replace the decomposition of the right-hand sidewhich was equal to one by the expansion

fN(y) =f0

2T0(y) +

N∑

n=1

fnTn(y).

Equating the coefficients that multiply equal-order polynomials, we obtain

b0 =f0

2, b0 = − 1

πf1,

b2k−1 = − 1

π

(f2k +

2

4k2 − 1b0

),

b2k = − 1

πf2k+1; k ≥ 1.

70

We see that the equation has the unique solution if and only if

f0

2= − 1

πf1.

Rewrite this condition using formula (329)

1

2

1∫

−1

f(y)dy√

1− y2= − 1

π

1∫

−1

y f(y)dy√

1− y2.

Here, one can uniquely determine coefficient bn, which provides the unique solvability of theequation.

19 Matrix representation of an operator in the Hilbert

space and summation operators

19.1 Matrix representation of an operator in the Hilbert space

Let ϕn be an orthonormal basis in the Hilbert space H; then each element f ∈ H can

be represented in the form of the generalized Fourier series f =∞∑

n=1

anϕn, where the sequence

of complex numbers an is such that the series∞∑

n=1

|an|2 < ∞ converges in H. There exists a

(linear) one-to-one correspondence between such sequences an and the elements of the Hilbertspace.

We will use this isomorphism to construct the matrix representation of a completely contin-uous operator A and a completely continuous operator-valued function A(γ) holomorphic withrespect to complex variable γ acting in H.

Assume that g = Af , f ∈ H, and

gk = gk(γ) = (A(γ)f, ϕk),

i.e., an infinite sequence (vector) of the Fourier coefficients gn = (g1, g2, ...) corresponds to

an element g ∈ H and we can write g = (g1, g2, ...). Let f =∞∑

k=1

fkϕk; then f = (f1, f2, ...). Set

fN = Pf =N∑

k=1

fkϕk and Aϕn =∞∑

k=1

aknϕk, n = 1, 2, . . .. Then,

limN→∞

AfN = limN→∞

N∑

n=1

fn

∞∑

k=1

aknϕk = Af =∞∑

k=1

( ∞∑

n=1

aknfn

)ϕk.

On the other side, Af =∞∑

k=1

gkϕk, hence, gk =∞∑

n=1

akn(Af, ϕk) = (g, ϕk). Now we set Aϕk = gk

and(gk, ϕj) = (Aϕk, ϕj) = ajk,

71

where ajk are the Fourier coefficients in the expansion of element gk over the basis ϕj.Consequently,

Aϕk = gk =∞∑

j=1

ajkϕj,

and the series ∞∑

j=1

|ajk|2 =∞∑

j=1

|(gk, ϕj)|2 < ∞; k = 1, 2, . . . .

Infinite matrix ajk∞j,k=1 defines operator A; that is, one can reconstruct A by this matrix andan orthonormal basis ϕk of the Hilbert space H: find the unique element Af for each f ∈ H.

In order to prove the latter assertion, we will assume that f =∞∑

k=1ξkϕk and fN = Pf . Then,

AfN =∞∑

k=1

ηNk ϕk, where ηN

k =N∑

j=1

akjξj. Applying the continuity of operator A, we have

Ck = (Af, ϕk) = limN→∞

(AfN , ϕk) = limN→∞

ηNk =

∞∑

j=1

akjξj.

Hence, for each f ∈ H,

Af =∞∑

k=1

Ckϕk, Ck =∞∑

j=1

akjξj,

and the latter equalities define the matrix representation of the operator A in the basis ϕkof the Hilbert space H.

19.2 Summation operators in the spaces of sequences

In this section, we will employ the approach based on the use of Chebyshev polynomials toconstruct families of summation operators associated with logarithmic integral equations.

Let us introduce the linear space hp formed by the sequences of complex numbers an suchthat ∞∑

n=1

|an|2np < ∞, p ≥ 0,

with the inner product and the norm

(a, b) =∞∑

n=1

anbnnp, ‖a‖p =( ∞∑

n=1

|an|2np) 1

2

.

One can easily verify the validity of the following statement.

Theorem 43 hp, p ≥ 0 is the Hilbert space.

Let us introduce the summation operators

L =

1 0 . . . 0 . . .0 2−1 . . . 0 . . .. . . . . . . . . . . . . . .0 . . . n−1 0 . . .. . . . . . . . . . . . . . .

72

or, in other notation,

L = lnj∞n,j=1, lnj = 0, n 6= j, lnn =1

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

and

L−1 =

1 0 . . . 0 . . .0 2 . . . 0 . . .. . . . . . . . . . . . . . .0 . . . n 0 . . .. . . . . . . . . . . . . . .

orL−1 = lnj∞n,j=1, lnj = 0, n 6= j, lnn = n, n = 1, . . . .

that play an important role in the theory of integral operators and operator-valued functionswith a logarithmic singularity of the kernel.

Theorem 44 L is a linear bounded invertible operator acting from the space hp to the space hp+2:L : hp → hp+2 (ImL = hp+2, Ker L = ∅). The inverse operator L−1 : hp+2 → hp, is bounded,p ≥ 0 .

2 Let us show that L acts from hp to hp+2 and L−1 acts from hp+2 to hp. Assume that a ∈ hp.Then, for La we obtain

∞∑

n=1

∣∣∣∣an

n

∣∣∣∣2

np+2 =∞∑

n=1

|an|2np < ∞.

If b ∈ hp+2, we obtain similar relationships for L−1b:

∞∑

n=1

|bnn|2np =∞∑

n=1

|bn|2np+2 < ∞.

The norm ‖La‖p+2 = ‖a‖p. The element a = L−1b, a ∈ hp, is the unique solution of the equationLa = b, and ker L = ∅. According to the Banach theorem, L−1 is a bounded operator. 2

Note that ‖L‖ = ‖L−1‖ = 1 and

L−1L = LL−1 = E,

where E is the infinite unit matrix.Assume that the summation operator A acting in hp is given by its matrix anj∞n,j=1 and

the following condition holds:

∞∑

n=1

∞∑

j=1

np+2

jp|anj|2 ≤ C2 < ∞, C = const, C > 0.

Then, we can prove the following statement.

Theorem 45 The operator A : hp → hp+2 is completely continuous

73

2 Let ξ ∈ hp and Aξ = η. Then,

∞∑

n=1

np+2|ηn|2 =∞∑

n=1

np+2

∣∣∣∣∣∣

∞∑

j=1

anjξj

∣∣∣∣∣∣

≤ ‖ξ‖2p

∞∑

n=1

∞∑

j=1

np+2 |anj|2jp

≤ C2‖ξ‖2p,

hence, η ∈ hp+2. Now we will prove that A is completely continuous. In fact, operator A canbe represented as a limit with respect to the operator norm of a sequence of finite-dimensionaloperators AN , which are produced by the cut matrix A and defined by the matrices AN =anjN

n,j=1:

‖(A− AN)ξ‖2p+2 =

N−1∑

n=1

np+2

∣∣∣∣∣∣

∞∑

j=N

anjξj

∣∣∣∣∣∣

2

+∞∑

n=N

np+2

∣∣∣∣∣∣

∞∑

j=1

anjξj

∣∣∣∣∣∣

2

≤ ‖ξ‖2p

∞∑

n=N

∞∑

j=1

np+2 |anj|2jp

+ ‖ξ‖2p

N−1∑

n=1

∞∑

j=N

|anj|2jp

np+2

≤ ‖ξ‖2p

[ ∞∑

n=N

( ∞∑

j=1

np+2 |anj|2jp

)+

∞∑

j=N

( ∞∑

n=1

np+2 |anj|2jp

)],

and, consequently, the following estimates are valid with respect to the operator norm:

‖A− AN‖ ≤∞∑

n=N

( N∑

j=1

np+2 |anj|2jp

)+

∞∑

j=N

( ∞∑

n=1

np+2 |anj|2jp

)→ 0, N →∞. 2

Corollary 1 The sequence of operators AN strongly converges to operator A: ‖A−AN‖ → 0,N → ∞, and ‖A‖ < C, C = const. In order to provide the complete continuity of A, it issufficient to require the fulfilment of the estimate

∞∑

n=1

∞∑

j=1

np+2|anj|2 < C, C = const. (365)

Definition 9 Assume that condition (365) holds for the matrix elements of summation oper-ator A. Then, the summation operator F = L + A : hp → hp+2 is called the L-operator.

Theorem 46 L-operator F = L + A : hp → hp+2 is a Fredholm operator; ind (L + A) = 0.

20 Matrix Representation of Logarithmic Integral Op-

erators

Consider a class Φ of functions ϕ such that they admit in the interval (−1, 1) the representationin the form of the Fourier–Chebyshev series

ϕ(x) =1

2T0(x)ξ0 +

∞∑

n=1

ξnTn(x); −1 ≤ x ≤ 1, (366)

74

where Tn(x) = cos(n arccos x) are the Chebyshev polynomials and ξ = (ξ0, ξ1, ...ξn, ...) ∈ h2;that is,

‖ξ‖22 =

1

2|ξ0|2 +

∞∑

n=1

|ξn|2n2 < ∞.

In this section we add the coordinate ξ0 to vector ξ and modify the corresponding formulas. It iseasy to see that all results of previous sections are also valid in this case. Recall the definitionsof weighted Hilbert spaces associated with logarithmic integral operators:

L(1)2 = f(x) : ‖f‖2

1 =

1∫

−1

|f(x)|2 dx√1− x2

< ∞,

L(2)2 = f(x) : ‖f‖2

2 =

1∫

−1

|f(x)|2√

1− x2dx < ∞,

W 12 = f(x) : f ∈ L

(1)2 , f ′ ∈ L

(2)2 ,

and

W 12 = f(x) : f ∈ L

(2)2 , f ∗(x) = f(x)

√1− x2, f ∗′ ∈ L

(2)2 , f ∗(−1) = f ∗(1) = 0.

Below, we will denote the weights by p1(x) =√

1− x2 and p−11 (x).

Note that the Chebyshev polynomials specify orthogonal bases in spaces L(1)2 and L

(2)2 . In

order to obtain the matrix representation of the integral operators with a logarithmic singularityof the kernel, we will prove a similar statement for weighted space W 1

2 .

Lemma 6 W 12 and Φ are isomorphic.

2 Assume that ϕ ∈ Φ. We will show that ϕ ∈ W 12 . Function ϕ(x) is continuous, and (366) is

its Fourier–Chebyshev series with the coefficients

ξn =2

π

1∫

−1

Tn(x)ϕ(x)p−11 (x)dx.

The series∞∑

n=1

|ξn|2n2 converges and, due to the Riesz–Fischer theorem, there exists an element

f ∈ L(1)2 , such that its Fourier coefficients in the Chebyshev polynomial series are equal to nξn;

that is, setting Un(x) = sin(n arccos x), one obtains

nξn =2

π

1∫

−1

nTn(x)ϕ(x)p−11 (x)dx = − 2

π

1∫

−1

U ′n(x)ϕ(x)dx

= − 2

π

1∫

−1

U ′n(x)(p1(x)ϕ(x))

dx

p1(x)=

2

π

1∫

−1

Un(x)f(x)dx

p1(x), n ≥ 1.

75

Since f(x) is an integrable function, the integral

1∫

−1

|f(x)|p−11 (x)dx converges, and

F (x) =

x∫

−1

f(t)dt

p1(t)

is absolutely continuous (as a function in the form of an integral depending on the upper limitof integration), F ′(x) = f(x)p−1

1 (x), and

2

π

1∫

−1

Un(x)f(x)dx

p1(x)=

2

π

1∫

−1

Un(x)F ′(x)dx = − 2

π

1∫

−1

U ′n(x)F (x)dx.

The latter equalities yield

1∫

−1

U ′n(x)[ϕ(x)− F (x)]dx = −n

1∫

−1

Tn(x)[ϕ(x)− F (x)]dx

p1(x)= 0, n ≥ 1.

Hence, ϕ(x) − F (x) = C = const. The explicit form of F ′(x) yields ϕ′(x) = f(x)p−11 (x);

therefore, ϕ′(x)p1(x) ∈ L(1)2 and ϕ′(x) ∈ L

(2)2 .

Now, let us prove the inverse statement. Let ϕ ∈ W 12 and be represented by the Fourier–

Chebyshev series (366). Form the Fourier series

p1(x)ϕ′(x) =∞∑

n=1

ηnUn(x), |x| ≤ 1,

where

ηn =2

π

1∫

−1

Un(x)(p1(x)ϕ′(x))dx

p1(x)= − 2

π

1∫

−1

U ′n(x)ϕ(x)dx =

=2n

π

1∫

−1

Tn(x)ϕ(x)dx

p1(x).

Now, in order to prove that ξ ∈ h2, it is sufficient to apply the Bessel inequality to ηn = nξn,where

ξn =2

π

1∫

−1

ϕ(x)Tn(x)dx

p1(x).

2

20.1 Solution to integral equations with a logarithmic singularity ofthe kernel

Consider the integral operator with a logarithmic singularity of the kernel

Lϕ = − 1

π

1∫

−1

ln |x− s|ϕ(s)ds

p1(s), ϕ ∈ W 1

2 ,

76

and the integral equationLϕ + Nϕ = f, ϕ ∈ W 1

2 , (367)

where

ΦN(x) = Nϕ =

1∫

−1

N(q)ϕ(s)ds

πp1(x), q = (x, s),

and N(q) is an arbitrary complex-valued function satisfying the continuity conditions

N(q) ∈ C1(Π1);∂2N

∂x2(q) ∈ L2[Π1, p

−11 (x)p−1

1 (s)],

Π1 = ([−1, 1]× [−1, 1]).

Write the Fourier–Chebyshev expansions

ϕ(s) =1√2ξ0T0 +

∞∑

n=1

ξnTn(s), ξ = (ξ0, ξ1, ..., ξn, ...) ∈ h2, (368)

and

f(x) =f0

2T0(x) +

∞∑

n=1

fnTn(x),

where the coefficients

fn =2

π

1∫

−1

f(x)Tn(x)dx√

1− x2, n = 0, 1, . . . .

Function ΦN(x) is continuous and may be thus also expanded in the Fourier–Chebyshev series

ΦN(x) =1

2b0T0(x) +

∞∑

n=1

bnTn(x), (369)

where

bn = 2∫ ∫

Π1

N(q)Tn(x)ϕ(s)dxds

π2p1(x)p1(s). (370)

According to Lemma 6 and formulas (368) and (369), equation (367) is equivalent in W 12 to the

equation

ln 2ξ0√2T0(x) +

∞∑

n=1

n−1ξnTn(x) +b0

2T0(x) +

∞∑

n=1

bnTn(x)

=f0

2T0(x) +

∞∑

n=1

fnTn(x), |x| ≤ 1.(371)

Note that since ΦN(x) is a differentiable function of x in [−1, 1], the series in (371) convergeuniformly, and the equality in (371) is an identity in (−1, 1).

Now, we substitute series (368) for ϕ(x) into (370). Taking into account that Tn(x) is abasis and one can change the integration and summation and to pass to double integrals (thesestatements are verified directly), we obtain an infinite system of linear algebraic equations

√2 ln 2ξ0 + b0 = f0,

n−1ξn + bn = fn, n ≥ 1.(372)

77

In terms of the summation operators, this systems can be represented as

Lξ + Aξ = f ; ξ = (ξ0, ξ1, . . . , ξn, . . .) ∈ h2; f = (f0, f1, . . . , fn, . . .) ∈ h4.

Here, operators L and A are defined by

L = lnj∞n,j=0, lnj = 0, n 6= j; l00 = ln 2; lnn =1

n, n = 1, 2, . . . ;

A = anj∞n,j=0, anj = εnj

∫ ∫

Π1

N(q)Tn(x)Tj(s)dxds

π2p1(x)p1(s), (373)

where

εnj =

1, n = j = 0,2, n ≤ 1, j ≤ 1,√

2, for other n, j.

Lemma 7 Let N(q) satisfy the continuity conditions formulated in (367). Then

∞∑

n=1,j=0

|anj|2n4 +∞∑

j=0

|a0j|2 1

ln2 2≤ C0, (374)

where

C0 =1

ln2 2

1∫

−1

∣∣∣∣∣∣

1∫

−1

N(q)dx

πp1(x)

∣∣∣∣∣∣

2

ds

πp1(s)+

∫ ∫

Π1

p21(x) |(p1(x)N ′

x)′x|2

dxds

π2p1(x)p1(s). (375)

2 Expand a function of two variables belonging to the weighted space L2[Π1, p−11 (x)p−1

1 (s)] inthe double Fourier–Chebyshev series using the Fourier–Chebyshev system Tn(x)Tm(s):

−p1(x)(p1(x)N ′x(q))′x =

∞∑

n=0

∞∑

m=0

CnmTn(x)Tm(s),

where

Cnm = −ε2nm

π

1∫

−1

Tm(s)ds

p1(s)

1∫

−1

Tn(x)(p1(x)N ′x(q))′xdx

=ε2

nm

π

1∫

−1

Tm(s)ds

p1(s)

1∫

−1

T ′n(x)p1(x)N ′

x(q)dx

=nε2

nm

π2

1∫

−1

Tm(s)

p1(s)ds

1∫

−1

Un(x)N ′x(q)dx

=n2ε2

nm

π2

∫ ∫

Π1

N(q)Tn(x)Tm(s)

p1(x)p1(s)dxds = n2εnmanm.

The Bessel inequality yields

∞∑

n=1

∞∑

m=0

n4π2|anm|2 ≤∫ ∫

Π1

p21(x)|[p1(x)N ′

x(q)]′x|2dxds

p1(x)p1(s).

78

Let us expand the function

g(s) =1

ln 2

1∫

−1

N(q)dx

π2p1(x)

in the Fourier–Chebyshev series. Again, due to the Bessel inequality, we obtain

1

ln2 2

∞∑

m=0

|a0m|2 ≤ 1

ln2 2

1∫

−1

∣∣∣∣∣∣

1∫

−1

N(q)ds

πp1(x)

∣∣∣∣∣∣

2

ds

πp1(s). 2

21 Galerkin methods and basis of Chebyshev polyno-

mials

We describe the approximate solution of linear operator equations considering their projectionsonto finite-dimensional subspaces, which is convenient for practical calculations. Below, it isassumed that all operators are linear and bounded. Then, the general results concerning thesubstantiation and convergence the Galerkin methods will be applied to solving logarithmicintegral equations.

Definition 10 Let X and Y be the Banach spaces and let A : X → Y be an injective operator.Let Xn ⊂ X and Yn ⊂ Y be two sequences of subspaces with dim Xn = dim Yn = n andPn : Y → Yn be the projection operators. The projection method generated by Xn and Pn

approximates the equationAϕ = f

by its projectionPnAϕn = Pnf.

This projection method is called convergent for operator A, if there exists an index N such thatfor each f ∈ ImA, the approximating equation PnAϕ = Pnf has a unique solution ϕn ∈ Xn forall n ≥ N and these solutions converge, ϕn → ϕ, n →∞, to the unique solution ϕ of Aϕ = f .

In terms of operators, the convergence of the projection method means that for all n ≥N , the finite-dimensional operators An := PnA : Xn → Yn are invertible and the pointwiseconvergence A−1

n PnAϕ → ϕ, n →∞, holds for all ϕ ∈ X. In the general case, one may expectthat the convergence occurs if subspaces Xn form a dense set:

infψ∈Xn

‖ψ − ϕ‖ → 0, n →∞ (376)

for all ϕ ∈ X. This property is called the approximating property (any element ϕ ∈ X can beapproximated by elements ψ ∈ Xn with an arbitrary accuracy in the norm of X). Therefore,in the subsequent analysis, we will always assume that this condition is fulfilled. Since An =PnA is a linear operator acting in finite-dimensional spaces, the projection method reduces tosolving a finite-dimensional linear system. Below, the Galerkin method will be considered asthe projection method under the assumption that the above definition is valid in terms of theorthogonal projection.

Consider first the general convergence and error analysis.

79

Lemma 8 The projection method converges if and only if there exists an index N and a positiveconstant M such that for all n ≥ N , the finite-dimensional operators An = PnA : Xn → Yn areinvertible and operators A−1

n PnA : X → X are uniformly bounded,

‖A−1n PnA‖ ≤ M. (377)

The error estimate is valid

‖ϕn − ϕ‖ ≤ (1 + M) infψ∈Xn

‖ψ − ϕ‖. (378)

Relationship (378) is usually called the quasioptimal estimate and corresponds to the errorestimate of a projection method based on the approximation of elements from X by elementsof subspaces Xn. It shows that the error of the projection method is determined by the qualityof approximation of the exact solution by elements of subspace Xn.

Consider the equationSϕ + Kϕ = f (379)

and the projection methodsPnSϕn = Pnf (380)

andPn(S + K)ϕn = Pnf (381)

with subspaces Xn and projection operators Pn : Y → Yn.

Lemma 9 Assume that S : X → Y is a bounded operator with a bounded inverse S−1 : Y → Xand that the projection method (380) is convergent for S. Let K : X → Y be a compact boundedoperator and S + K is injective. Then, the projection method (381) converges for S + K.

Lemma 10 Assume that Yn = S(Xn) and ‖PnK − K‖ → 0, n → ∞. Then, for sufficientlylarge n, approximate equation (381) is uniquely solvable and the error estimate

‖ϕn − ϕ‖ ≤ M‖PnSϕ− Sϕ‖

is valid, where M is a positive constant depending on S and K.

For operator equations in Hilbert spaces, the projection method employing an orthogonalprojection into finite-dimensional subspaces is called the Galerkin method.

Consider a pair of Hilbert spaces X and Y and assume that A : X → Y is an injectiveoperator. Let Xn ⊂ X and Yn ⊂ Y be the subspaces with dim Xn = dim Yn = n. Then, ϕn ∈ Xn

is a solution of the equation Aϕ = f obtained by the projection method generated by Xn andoperators of orthogonal projections Pn : Y → Yn if and only if

(Aϕn, g) = (f, g) (382)

for all g ∈ Yn. Indeed, equations (382) are equivalent to Pn(Aϕn − f).Equation (382) is called the Galerkin equation.

80

Assume that Xn and Yn are the spaces of linear combinations of the basis and probe func-tions: Xn = u1, . . . , un and Yn = v1, . . . , vn. Then, we can express ϕn as a linear combination

ϕn =n∑

k=1

γnun,

and show that equation (382) is equivalent to the linear system of order n

n∑

k=1

γk(Auk, vj) = (f, vj), j = 1, . . . , n, (383)

with respect to coefficients γ1, . . . , γn.Thus, the Galerkin method may be specified by choosing subspaces Xn and projection

operators Pn (or the basis and probe functions). We note that the convergence of the Galerkinmethod do not depend on a particular choice of the basis functions in subspaces Xn and probefunctions in subspaces Yn. In order to perform a theoretical analysis of the Galerkin methods,we can choose only subspaces Xn and projection operators Pn. However, in practice, it is moreconvenient to specify first the basis and probe functions, because it is impossible to calculatethe matrix coefficients using (383).

21.1 Numerical solution of logarithmic integral equation by the Galerkinmethod

Now, let us apply the projection scheme of the Galerkin method to logarithmic integral equation

(L + K)ϕ =

1∫

−1

(ln1

|x− y| + K(x, y)) ϕ(x)dx√

1− x2= f(y), −1 < y < 1. (384)

Here, K(x, y) is a smooth function; accurate conditions concerning the smoothness of K(x, y)that guarantee the compactness of integral operator K are imposed in Section 2.

One can determine an approximate solution of equation (384) by the Galerkin methodtaking the Chebyshev polynomials of the first kind as the basis and probe functions. Accordingto the scheme of the Galerkin method, we set

ϕN(x) =a0

2T0(x) +

N−1∑

n=1

anTn(x),

XN = T0, . . . , TN−1, YN = T0, . . . , TN−1, (385)

and find the unknown coefficients from finite linear system (383). Since L : XN → YN is abijective operator, the convergence of the Galerkin method follows from Lemma 10 applied tooperator L. If operator K is compact and L+K is injective, then we can also prove the conver-gence of the Galerkin method for operator L + K using Lemma 9. Note also that quasioptimalestimate 378 in the L

(1)2 norm is valid for ϕN(x) → ϕ(x), N →∞ .

Theorem 47 If operator L + K : L(1)2 → W 1

2 is injective and operator K : L(1)2 → W 1

2 iscompact, then the Galerkin method (385) converges for operator L + K.

81

The same technique can be applied to constructing an approximate solution to the hyper-singular equation

(H + K)ϕ =d

dy

1∫

−1

(1

x− y+ K(x, y)

)ϕ(x)

√1− x2dx = f(y), −1 < y < 1. (386)

The conditions concerning the smoothness of K(x, y) that guarantee the compactness of integraloperator K are imposed in Section 2.

Choose the Chebyshev polynomials of the second kind as the basis and probe functions, i.e.,set

XN = U0, . . . , UN−1, YN = U0, . . . , UN−1. (387)

Unknown approximate solution ϕN(x) is determined in the form

ϕN(x) =N−1∑

n=0

bnUn(x). (388)

Note that hypersingular operator H maps XN onto YN and is bijective.From Lemma 10, it follows that the Galerkin method for operator H converges. By virtue

of Lemma 9, this result is also valid for operator H + K under the assumption that K iscompact and H + K is injective. Quasioptimal formula (378) estimates the rate of convergenceof approximate solution ϕN(x) to exact one ϕ(x), N →∞ in the norm of space W 1

2 . Unknowncoefficients an are found from finite linear system (383).

Theorem 48 If operator H +K : W 12 → L

(2)2 is injective and K : W 1

2 → L(2)2 is compact, then

the Galerkin method (387) converges for operator H + K.

82