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 Fredholms 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 HilbertSchmidt theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

11 IEs with symmetric kernels. HilbertSchmidt theory 4211.1 Selfadjoint integral operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4211.2 HilbertSchmidt theorem for integral operators . . . . . . . . . . . . . . . . . . 45

12 Harmonic functions and Greens formulas 4712.1 Greens 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 FourierChebyshev 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 FourierChebyshev series . . . . . . . . . . . . . . . . . . . . . . . . 64

18 Solution to singular integral equations 6618.1 Singular integral equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6618.2 Solution via FourierChebyshev 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) = (x) + baK(x, y)(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

= {(x, y) : a x b, a y b},f(x) is a given function, (x) is a sought-for (unknown) function, and is a parameter.

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

f(x) = baK(x, y)(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 ;(ii) K(x, y) is maybe discontinuous for some (x, y) but the double integral b

a

ba|K2(x, y)|dxdy a or discontinuous and satisfyingcondition (3); f(x) is a given function; and is a parameter,

Example 1 Consider a Fredholm IE of the 2nd kind

x2 = (x) 10(x2 + y2)(y)dy, (5)

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

Example 2 Consider an IE of the 2nd kind

f(x) = (x) 10ln |x y|(y)dy, (6)

5

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

0

10ln2 |x y|dxdy

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

(x) (+ 1)(x) = f (x) f(x) (13)

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

(0) = f(0). (14)

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

(x) (+ 1)(x) = F (x), F (x) = f (x) f(x), (15)(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) = exy 0 y x 1,K(x, y) = 0 0 x y 1.

Write two first successive approximations:

0(x) = f(x), 1(x) = f(x) + 10K(x, y)0(x)dy, (17)

2(x) = f(x) + 10K(x, y)1(y)dy = (18)

f(x) + 10K(x, y)f(y)dy + 2

10K(x, t)dt

10K(t, y)f(y)dy. (19)

Set

K2(x, y) = 10K(x, t)K(t, y)dt =

xyextetydt = (x y)exy (20)

to obtain (by changing the order of integration)

2(x) = f(x) + 10K(x, y)f(y)dy + 2

10K2(x, y)f(y)dy. (21)

The third approximation

3(x) = f(x) + 10K(x, y)f(y)dy +

+ 2 10K2(x, y)f(y)dy +

3 10K3(x, y)f(y)dy, (22)

where

K3(x, y) = 10K(x, t)K2(t, y)dt =

(x y)22!

exy. (23)

7

The general relationship has the form

n(x) = f(x) +n

m=1

m 10Km(x, y)f(y)dy, (24)

where

K1(x, y) = K(x, y), Km(x, y) = 10K(x, t)Km1(t, y)dt =

(x y)m1(m 1)! e

xy. (25)

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

(x) = f(x) +m=1

m 10Km(x, y)f(y)dy. (26)

Introducing the notation for the resolvent

(x, y, ) =m=1

m1Km(x, y). (27)

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

(x) = f(x) + 10(x, y, )f(y)dy. (28)

Calculate the resolvent explicitly using (25):

(x, y, ) = exym=1

m1(x y)m1(m 1)! = e

xye(xy) = e(+1)(xy), (29)

so that (28) gives the solution

(x) = f(x) + 10e(+1)(xy)f(y)dy. (30)

Note that according to (17)

(x, y, ) = e(+1)(xy), 0 y x 1,(x, y, ) = 0, 0 x y 1.

and the series (27) converges for every .

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

(x) = y0 + xx0f(t, (t))dt, (33)

which is called the Volterra integral equation of the second kind with respect to the unknownfunction (t). This equation is equivalent to the initial value problem (31) and (32). Note thatthis is generally a nonlinear integral equation with respect to (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) = xx0A(t)y(x)dx

xx0B(x)y(x)dx+

xx0g(x)dx+ y1. (36)

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

y(x) = A(x)y(x) xx0(B(x) A(x))y(x)dx+

xx0g(x)dx+ A(x0)y0 + y1. (37)

Integrating a second time we obtain

y(x) = xx0A(x)y(x)dx

xx0

xx0(B(t)A(t))y(t)dtdx+

xx0

xx0g(t)dtdx+[A(x0)y0+y1](xx0)+y0.

(38)Using the relationship x

x0

xx0f(t)dtdx =

xx0(x t)f(t)dt, (39)

we transform (38) to obtain

y(x) = xx0{A(t)+(x t)[(B(t)A(t))]}y(t)dt+

xx0(x t)g(t)dt+[A(x0)y0+y1](xx0)+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) =

xx0(x t)g(t)dt+ [A(x0)y0 + y1](x x0) + y0. (42)

Then (40) becomes

y(x) = f(x) + xx0K(x, t)y(t))dt, (43)

which is the Volterra IE of the second kind with respect to the unknown function (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 + 2y = 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) 2, 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) = 2(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+ x0(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

(x, y) = 0 if and only if x = y,

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

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

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

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

(xn, xm) < 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

(a, b) = ||a b||2 = n

i=1

(ai bi)2

is complete.

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

(1, 2) = maxaxb

|1(x) 2(x)|is complete. Indeed, every fundamental (convergent) functional sequence {n(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

(f1, f

2) = || max

axb|f 1i (x) f 2i (x)|||2

is complete. Indeed, every componentwise-fundamental (componentwise-convergent) functionalsequence {n(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 < 1 such that

(Ax,Ay) (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 Picards 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 = (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| kdMd < 1.

Consider the mapping = A defined by

(x) = y0 + xx0f(t, (t))dt, |x0 x| d. (53)

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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

|(x) y0| = xx0f(t, (t))dt

xx0|f(t, (t))| dt k

xx0dt = k(x x0) kd. (54)

|1(x) 2(x)| xx0|f(t, 1(t)) f(t, 2(t))| dt Md max

x0dxx0+d|1(x)| 2(x). (55)

Since Md < 1, the mapping = A is a contraction.From this it follows that the operator equation = A 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

(x) = y0 + xx0f(t, (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|| KdMd < 1.

Then we consider the mapping = A defined by

(x) = y0 + xx0f(t, (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) = (x) + baK(x, y)(y)dy, (59)

12

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

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

g(x) = (x) + baK(x, y)(y)dy, (60)

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

g1(x) = baK(x, y)f1(y)dy + (x),

g2(x) = baK(x, y)f2(y)dy + (x),

(g1, g2) = maxaxb

|g1(x) g2(x)|

|| ba|K(x, y)||f1(y) f2(y)|dy ||M(b a) max

ayb|f1(y)| f2(y)| (61)

< (f1, f2) if || < 1M(b a) (62)

Consequently, the mapping A is a contraction if

|| < 1M(b a) , (63)

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

fn(x) = (x) + baK(x, y)fn1(y)dy. (64)

5.2 Nonlinear Fredholm IE of the 2nd kind

The method is applicable to nonlinear IEs

f(x) = (x) + baK(x, y, f(y))dy, (65)

where the kernel K and are continuous functions, and

|K(x, y, z1)K(x, y, z2)| M |z1 z2|. (66)If satisfies (63) we can prove in the same manner that the mapping g = Af defined by

g(x) = (x) + baK(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.

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5.3 Linear Volterra IE of the 2nd kind

Consider a Volterra IE of the 2nd kind

f(x) = (x) + xaK(x, y)(y)dy, (68)

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

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 = limkAknx, then Ax0 = x0. In fact Ax0 = limkAknAx. Sincethe mapping An is a contraction, we have

(AknAx,Aknx) (A(k1)nAx,A(k1)nx) . . . k(Ax, x).Consequently,

limk

(AknAx,Aknx) = 0,

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

g(x) = (x) + xaK(x, y)(y)dy, (69)

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

5.4 Neumann series

Consider an IE

f(x) = (x) baK(x, y)(y)dy, (70)

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

(x) = f(x) + baK(x, y)(y)dy, (71)

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

0(x) = f(x). (72)

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

1(x) = f(x) + baK(x, y)0(y)dy, (73)

14

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

n+1(x) = f(x) + baK(x, y)n(y)dy. (74)

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

|K(x, y)| M, (x, y) .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 n of successive approxima-tions (74) converges uniformly for all satisfying

1B, B =

ba

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

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

Proof. Write two first successive approximations (74):

1(x) = f(x) + baK(x, y)f(y)dy, (77)

2(x) = f(x) + baK(x, y)1(y)dy = (78)

f(x) + baK(x, y)f(y)dy + 2

baK(x, t)dt

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

Set

K2(x, y) = baK(x, t)K(t, y)dt (80)

to obtain (by changing the order of integration)

2(x) = f(x) + baK(x, y)f(y)dy + 2

baK2(x, y)f(y)dy. (81)

In the same manner, we obtain the third approximation

3(x) = f(x) + baK(x, y)f(y)dy +

+ 2 baK2(x, y)f(y)dy +

3 baK3(x, y)f(y)dy, (82)

where

K3(x, y) = baK(x, t)K2(t, y)dt. (83)

15

The general relationship has the form

n(x) = f(x) +n

m=1

m baKm(x, y)f(y)dy, (84)

where

K1(x, y) = K(x, y), Km(x, y) = baK(x, t)Km1(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) = baKr(x, t)Kmr(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)

(x) = f(x) +m=1

m baKm(x, y)f(y)dy. (87)

In order to prove the convergence of this series, write ba|Km(x, y)|2dy Cm x [a, b], (88)

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

Km(x, y) = baKm1(x, t)K(t, y)dt, (m = 2, 3, . . .). (89)

Applying to (89) the Schwartz inequality we obtain

|Km(x, y)|2 = ba|Km1(x, t)|2

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

Integrating (90) with respect to y yields ba|Km(x, y)|2dy B2

ba|Km1(x, t)|2 B2Cm1 (B =

ba

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

which givesCm B2Cm1, (92)

and finally the required estimateCm B2m2C1. (93)

Denoting

D =

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

and applying the Schwartz inequality we obtain baKm(x, y)f(y)dy

2

ba|Km(x, y)|2dy

ba|f(y)|2dy D2C1B2m2. (95)

16

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

DC1||mBm1, (96)

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

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

DC1||n+1Bn1 ||B . (97)

5.5 Resolvent

Introduce the notation for the resolvent

(x, y, ) =m=1

m1 baKm(x, y) (98)

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

(x) = f(x) + ba(x, y, )f(y)dy. (99)

One can show that the resolvent satisfies the IE

(x, y, ) = K(x, y) + baK(x, t)(t, y, )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 = maxaxb

|f(x)|

[and the metric(1, 2) = ||1 2||C = max

axb|1(x)| 2(x)|]

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

Definition 4 A setM 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 {} defined on the closed interval [a, b] is said to be uni-formly bounded if there exists a number M > 0 such that

|(x)| < Mfor all x and for all in the family.

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

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

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

Theorem 6 (Arzelas 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) = (x) + baK(x, y)(y)dy, (101)

which can be written in the operator form as

f = + A. (102)

Here the linear (integral) operator A defined by

f(x) = baK(x, y)(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

= {(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 = k(x), k = 1, 2, . . . n, (104)

where k(x) are continuous functions.

18

Proof. To prove the theorem, it is sufficient to show (according to Arzelas 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]. SetM = sup

|K(x, y)|,

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

|y k(x)| < 12Mn

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

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

|K(x, y)K(x, y)| < (b a)13 (106)

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

|f(x) f(x)| ba|K(x, y)K(x, y)||(y)|dy (107)

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

|y k(x)| < 12Mn

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

|y k(x)| < 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

3M. Therefore

A|K(x, y)K(x, y)||(y)|dy 2

3||||. (108)

The integral over the complement B can be obviously estimated byB|K(x, y)K(x, y)||(y)|dy

3||||. (109)

Therefore|f(x) f(x)| ||||. (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 (x) with bounded norm ||||. Theuniform boundedness of f(x) (according to Definition (5)) corresponding to the functions (x)with bounded norms follows from the inequality

|(x)| B|K(x, y)||(y)|dy M(b a)||||. (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 elements{A3x(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|| < 2c and N such that

||Akx(n)n Akx(m)m || N and m > N (which is possible because the sequence ||Akx(n)n ||converges). As a result,

||Ax(n)n Ax(m)m || < , (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 = AA1 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 is said to be a characteristic value of the operator A if there existsa nonzero vector x (characteristic vector) such that Ax = x.

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

Det |A I| = 0.The totality of those values of for which the inverse of the operator AI does not exist

is called the spectrum of the operator A.The values of for which the operator A I has the inverse are called regular values of

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

R = (A I)1is called the resolvent of the operator A.

In an infinite-dimensional space, the spectrum of the operator A necessarily contains all thecharacteristic values and maybe some other numbers.

In this connection, the set of characteristic values is called the point spectrum of the operatorA. 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 for which A I 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 for which A I has a (bounded) inverse whose domain isnot dense in E, this part is called the residual spectrum.

If the point is regular, i.e., if the inverse of the operator AI exists, then for sufficientlysmall the operator A (+ )I also has the inverse, i.e., the point + 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 I exists for arbitrary for which || > ||A||.

Proof. Since

A I = (I 1

A),

the resolvent

R = (A I)1 = 1

(I A

)1= 1

k=0

Ak

k.

This operator series converges for ||A/|| < 1, i.e., the operator A I 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 arbitrary > 0 only a finite number of linearly independent characteristic vectors which correspond tothe characteristic values whose absolute values are greater than .

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

Axn = nxn, |n| > > 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 = 1x1 + . . .+ kxk, (119)

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

Ax = 11x1 + . . .+ kkxk, (120)

and, consequently,A1x = (1/1)x1 + . . .+ (k/k)xk (121)

and

||A1x|| < 1||x||.

Therefore, the operator A1 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 = xn1,

. . . .

The element xn belongs to subspace En for each n but does not belong to subspace En1. Infact,

T nxn = Tn1xn1 = . . . = Tx1 = 0,

but

T n1xn = T n2xn1 = . . . = 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 (yn+1, En) 12,

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

(yn+1, En) = inf {||yn+1 x||, x En}.

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

||Ayp Ayq|| = ||yp (yq + Typ Tyq)|| 12,

since yq + Typ Tyq Ep1. 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 Af 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 Af is solvable for all h, then the equation f Af = 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 = xAx be solvable isthat f(y) = 0 for all f for which f Af = 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) Af(x)

i.e., f(y) = 0 for all f for which f Af = 0 .2. Now assume that f(y) = 0 for all f for which f Af = 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 xAx. Thus, itis necessary to prove that an element y1 which cannot be represented in the form xAx 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 Af1 = 0.

24

These conditions are equivalent to the following:

f1(y1) 6= 0, f1(x Ax) = 0 x.

In fact,(f1 Af1)(x) = f1(x) Af1(x) = f1(x) f1(Ax) = f1(x Ax).

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

f1(z + y1) = .

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

Corollary If the equation f Af = 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 Af = h be solvableis that h(x) = 0 for all x for which x Ax = 0.

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

h(x) = f(x) Af(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 xAx = 0. We shall show that the equation

f Af = h is solvable. Consider the set F of all elements of the form y = xAx. 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 Af = h. This completes the proof of thetheorem.

Corollary If the equation x Ax = 0 does not have nonzero solution, then the equationf Af = 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 Af = h is solvable for all h. Then, by Corollary 2to Thereom 13, the equation f Af = 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 = 1 is a characteristic value for the operator A.

All the results obtained above for the equation x Ax = y remain valid for the equationxAx = y (the adjoint equation is f Af = h). It follows that either the operator I Ahas an inverse or the number 1/ 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 Af = 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

A(x) = baK(x, y)(y)dy (123)

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

(x) baK(x, y)(y)dy = f(x) (124)

will be called regular values. The numbers for which the Fredholm resolvent (x, y, ) 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 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),

(x) = f(x) + ba(x, y, )f(y)dy. (125)

26

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

(x) baK(x, y)(y)dy = 0 (126)

has only the zero solution (x) = 0.

9 IEs with degenerate and separable kernels

9.1 Solution of IEs with degenerate and separable kernels

An IE

(x) baK(x, y)(y)dy = f(x) (127)

with a degenerate kernel

K(x, y) =ni=1

ai(x)bi(y) (128)

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

(x) ni=1

ai(x) babi(y)(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 = babi(y)(y)dy, (130)

which are unknown constants. Equation (129)) becomes

(x) = f(x) + ni=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,

ni=1

ai(x)

{ci

babi(y)

[f(y) +

nk=1

ckak(y)

]dy

}= 0. (132)

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

ci babi(y)

[f(y) +

nk=1

ckak(y)

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

27

Denoting

fi = babi(y)f(y)dy, aik =

babi(y)ak(y)dy (134)

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

ci n

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() =

1 a11 a12 . . . a1na21 1 a22 . . . a2n

. . . . . .an1 an2 . . . 1 ann

. (136)

D() is a polynomial in powers of of an order not higher than n. Note that D() is notidentically zero because D(0) = 1. Thus there exist not more than n different numbers = ksuch that D(k) = 0. When =

k, then system (135) together with IE (129) are either

not solvable or have infinitely many solutions. If 6= k, then system (135) and IE (129) areuniquely solvable.

Example 10 Consider a Fredholm IE of the 2nd kind

(x) 10(x+ y)(y)dy = f(x). (137)

Here

K(x, y) = x+ y =2i=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 is a parameter.Look for the solution to (137) in the form (131) (note that here n = 2)

(x) = f(x) + 2i=1

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

Denoting

fi = 10bi(y)f(y)dy, aik =

10bi(y)ak(y)dy, (139)

28

so that

a11 = 10b1(y)a1(y)dy =

10ydy =

1

2,

a12 = 10b1(y)a2(y)dy =

10dy = 1, (140)

a21 = 10b2(y)a1(y)dy =

10y2dy =

1

3,

a22 = 10b2(y)a2(y)dy =

10ydy =

1

2,

and

f1 = 10b1(y)f(y)dy =

10f(y)dy, (141)

f2 = 10b2(y)f(y)dy =

10yf(y)dy,

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

ci 2

k=1

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

or, explicitly,

(1 /2)c1 c2 = f1,(/3)c1 + (1 /2)c2 = f2,

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

D() =

1 /2 /3 1 /2 = 1 + 2/4 2/3 = 1 2/12 = 112(2 + 12 12)(143)

is a polynomial of order 2. (D() is not identically, and D(0) = 1). There exist not more than2 different numbers = k such that D(

k) = 0. Here there are exactly two such numbers; they

are

1 = 6 + 43, (144)

2 = 6 43.

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

system (143)

D1 =

f1 f2 1 /2 = f1(1 /2) + f2 =

= (1 /2) 10f(y)dy +

10yf(y)dy =

10[(1 /2) + y]f(y)dy, (145)

29

D1 =

f1 f2 1 /2 = f1(1 /2) + f2 =

= (1 /2) 10f(y)dy +

10yf(y)dy =

10[(1 /2) + y]f(y)dy, (146)

D2 =

(1 /2) f1/3 f2 = f2(1 /2) + f1/3 =

= (1 /2) 10yf(y)dy + (/3)

10f(y)dy =

10[y(1 /2) + /3]f(y)dy, (147)

Then, the solution of system (143) is

c1 =D1D

= 122 + 12 12

10[(1 /2) + y]f(y)dy =

=1

2 + 12 12 10[(6 12) 12y]f(y)dy, (148)

c2 =D2D

= 122 + 12 12

10[y(1 /2) + /3]f(y)dy =

=1

2 + 12 12 10[y(6 12) 4]f(y)dy (149)

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

(x) = f(x) + (c1x+ c2) =

= f(x) +

2 + 12 12 10[6( 2)(x+ y) 12xy 4]f(y)dy. (150)

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

Example 11 Consider a Fredholm IE of the 2nd kind

(x) 2pi0

sinx cos y(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 is a parameter.Rewrite IE (151) in the form (131) (here n = 1)

(x) = f(x) + c sinx, c = 2pi0

cos y(y)dy. (152)

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

c = 2pi0

f(x) cos xdx (153)

30

Therefore the solution is

(x) = f(x) + c sinx = f(x) + 2pi0

sinx cos yf(y)dy. (154)

Example 12 Consider a Fredholm IE of the 2nd kind

x2 = (x) 10(x2 + y2)(y)dy, (155)

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

9.2 IEs with degenerate kernels and approximate solution of IEs

Assume that in an IE

(x) baK(x, y)(y)dy = f(x) (156)

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

= {(x, y) : a x b, a y b}.Then (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)(y)dy h

nk=1

K(x, yk)(yk), (157)

where

h =b an

, yk = a+ kh. (158)

The resulting approximation takes the form

(x) hn

k=1

K(x, yk)(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 k = (xk)and right-hand side containing the known numbers fi = f(xi)

i hn

k=1

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

We can find the solution {i} of system (160) by Cramers rule,

i =D

(n)i ()

D(n)(), i = 1, 2, . . . , n, (161)

31

calculating the determinants

D(n)() =

1 hK(x1, y1) hK(x1, y2) . . . hK(x1, yn)hK(x2, y1) 1 hK(x2, y2) . . . hK(x2, yn)

. . . . . .hK(xn, y1) hK(xn, y2) . . . 1 hK(xn, yn)

. (162)

and so on.To reconstruct the approximate solution (n) using the determined {i} one can use the IE

(159) itself:

(x) (n) = hn

k=1

K(x, yk)k + f(x). (163)

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

9.3 Fredholms resolvent

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

(x, y, ) = K(x, y) + baK(x, t)(t, y, )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

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

D(), (165)

where

D(x, y, ) =n=0

(1)nn!

Bn(x, y)n, (166)

D() =n=0

(1)nn!

cnn, (167)

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

Bn(x, y) = ba. . .

ba

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 = ba. . .

ba

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() is called the Fredholm determinant, and D(x, y, ), the first Fredholm minor.

If the integral ba

ba|K(x, y)|2dxdy

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

D(x, y, ) =

=1

n=0

(1)nn!

Bn(x, y)n = B0(x, y) B1(x, y) =

= x+ y (1

2(x+ y) xy 1

3

), (176)

D() =

=2

n=0

(1)nn!

cnn = c0 c1+ 1

2c2

2 =

= 1 1122 = 1

12(2 + 12 12)

and

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

D()= 12x+ y

(12(x+ y) xy 1

3

))

2 + 12 12 . (177)

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

(x) = f(x) + 10(x, y, )f(y)dy,

so that we obtain, using (176),

(x) = f(x) +

2 + 12 12 10[6( 2)(x+ y) 12xy 4]f(y)dy, (178)

which coincides with formula (150).The numbers = k (k = 1, 2) determined in (144),

1 = 6 +

4 and 2 = 6

4, are

(simple, i.e., of multiplicity one) poles of the resolvent because thay are (simple) zeros of theFredholm determinant D(). When = 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,

(x, y) = (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),

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

(x, x) 0, (x, x) = 0 if and only if x = 0.Note that in a complex linear space

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

The norm in the Euclidian space is introduced by

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

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

(Ax, y) = (x,Ay), 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

(x) = f(x) + baK(x, y)(y)dy. (179)

Assume that K(x, y) is a HilbertSchmidt kernel, i.e., a square-integrable function in the square = {(x, y) : a x b, a y b}, so that b

a

ba|K(x, y)|2dxdy , (180)

and f(x) L2[a, b], i.e., ba|f(x)|2dx .

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

A(x) = baK(x, y)(y)dy. (181)

If K(x, y) is a HilbertSchmidt kernel, then operator (181) will be called a HilbertSchmidtoperator.

Rewrite IE (179) as a linear operator equation

= A(x) + f, f, 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

ba|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

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

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

|(x)|2 = baK(x, y)(y)dy

2

ba|K(x, y)|2dy

ba|(y)|2dy = (186)

= ||||2 ba|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

||A||2 = ba|(y)|2dy ||||2

ba

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

which proves (1) that (x) is a square-integrable function, i.e., ba|(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() which converges in (operator) norm to the operator A. Let {n} be an orthonormalsystem in L2[a, b]. Then {m(x)n(y)} form an orthonormal system in L2(). Consequently,

K(x, y) =

m,n=1

amnm(x)n(y). (188)

By setting

KN(x, y) =N

m,n=1

amnm(x)n(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 {m}Nm=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 ba

ba|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 be an eigenvalue of a selfadjoint operator A, i.e., Ax = x, x H, x 6= 0.Then we have

(x, x) = (x, x) = (Ax, x) = (x,Ax) = (x, x) = (x, x), (192)

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

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

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

Ax = x, x H, x 6= 0; Ay = y, y H, y 6= 0.Then we have

(x, y) = (x, y) = (Ax, y) = (x,Ay) = (x, y) = (x, y), (193)

which gives( )(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 HilbertSchmidt con-dition b

a

ba|K(x, y)|2dxdy , (195)

and write IE (179)

(x) = baK(x, y)(y)dy + f(x) (196)

37

in the operator form (182),

= A+ f, (197)

or

T = f, T = I A. (198)The corresponding homogeneous IE

T0 = 0, (199)

and the adjoint equations

T = g, T = I A, (200)and

T 0 = 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 T = f is solvable if and only if f is orthogonal toevery solution of the adjoint homogeneous equation T 0 = 0 .

Theorem 22 (the Fredholm alternative). One of the following situations is possible:(i) the inhomogeneous equation T = f has a unique solution for every f H(ii) the homogeneous equation T0 = 0 has a nonzero solution .

Theorem 23 The homogeneous equations T0 = 0 and T0 = 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 f + g M where and 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) ifM 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 ImT and yn y. To prove the lemma, one must show that y ImT ,i.e., y = Tx. We will now prove the latter statement.

According to the definition of ImT , there exist vectors xn H such that

yn = Txn = xn Axn. (202)

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

xn = hn + hn, hn KerT,

and replace xn by hn = xn hn with hn 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|| A

xnk||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 KerT ;therefore vector z must be also orthogonal to KerT . 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 KerT and ImT are orthogonal because if h KerT then (h, T x) =(Th, x) = (0, x) = 0 for all x H. It is left to prove there is no nonzero vector orthogonal toKerT and ImT . Indeed, if a vector z is orthogonal to ImT , then (Tz, x) = (z, T x) = 0 forall x H, i.e., z KerT . The lemma is proved.

Lemma 2 yields the first Fredholm theorem 21. Indeed, f is orthogonal to KerT if andonly if f ImT , i.e., there exists a vector such that T = f .

Next, set Hk =ImT k for every k, so that, in particular, H1 =ImT . 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 {Axk}does 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 KerT = {0} then T is a one-to-one operator. If in this case ImT 6= H then the chain(205) consists of different subspaces which contradicts Lemma 3. Therefore ImT = H.

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

The totality of statements of Lemmas 4 and 5 constitutes the proof of Fredholms alternative22.

Now let us prove the third Fredholms Theorem 23.To this end, assume that KerT is an infinite-dimensional space: dimKerT =. 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 =dimKerT and =dimKerT and assume that < . Let 1, . . . , be an

orthonormal basis in KerT and 1, . . . , be an orthonormal basis in KerT. Set

Sx = Tx+j=1

(x, j)j.

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+j=1

(x, j)j = 0. (206)

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

Tx = 0

and

(x, j) = 0, quad1 j .

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 Fredholms Theorem 22 thereexists a vector y such that

Ty +j=1

(y, j)j = +1.

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

(Ty, +1) +j=1

(y, j)(j, +1) = (+1, +1).

Here (Ty, +1) = 0 because Ty ImT and ImT is orthogonal to KerT . Thus we have

(0 +j=1

(y, j) 0 = 1.

This contradiction is obtained because we assumed < . Therefore . Replacing nowT by T we obtain in the same manner that . Therefore = . The third Fredholmstheorem is proved.

To sum up, the Fredholms theorems show that the number = 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) = A+f (i.e., for the operator AI)remain valid for the equation = A + f (for the operator A I with 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 HilbertSchmidt theorem

Let us formulate the important HilbertSchmidt theorem.

Theorem 24 A completely continuous selfadjoint operator A in the Hilbert space H has an or-thonormal system of eigenvectors {k} corresponding to eigenvalues k such that every element H is represented as

=k

ckk + , (207)

where the vector KerA, i.e., A = 0. The representation (207) is unique, andA =

k

kckk, (208)

andk 0, n, (209)

if {k} is an infinite system.

41

11 IEs with symmetric kernels. HilbertSchmidt theory

Consider first a general a HilbertSchmidt operator (181).

Theorem 25 Let A be a HilbertSchmidt operator with a (HilbertSchmidt) 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) = ba

{ baK(x, y)f(y)dy

}g(x)dx =

= ba

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

= ba

{ baK(x, y)g(x)dx

}f(y)dy =

= baf(y)

{ baK(x, y)g(x)dx

}dy,

which proves the theorem.

Consider a symmetric IE

(x) baK(x, y)(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 0 and 0 be an eigenvalue and eigenfunction of a integral operator with a symmetrickernel, i.e.,

0(x) 0 baK(x, y)0(y)dy = 0. (213)

42

Multiply equality (213) by 0 and integrate with respect to x between a and b to obtain

||0||2 0(K0, 0) = 0, (214)

which yields

0 =||0||2

(K0, 0). (215)

Rewrite this equality as

0 =||0||2

, = (K0, 0), (216)

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

= (K0, 0) = (0, K0). (217)

According to the definition of the inner product

(0, K0) = (K0, 0).

Thus

= (K0, 0) = = (K0, 0),

so that = , i.e., is a real number and, consequently, 0 in (215) is a real number.

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

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

(x) = baK(x, y)(y)dy = 0, (218)

(x) = baK(x, y)(y)dy = 0. (219)

Then we have

(, ) = (, ) = (A, ) = (, A) = (, ) = (, ), (220)

which yields( )(, ) = 0, (221)

i.e., (, ) = 0.

One can also reformulate the HilbertSchmidt theorem 24 for integral operators.

43

Theorem 28 Let 1, 2, . . . , and 1, 2, . . . , 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 HilbertSchmidt kernel, i.e., a square-integrable function in the square ={(x, y) : a x b, a y b}, so that b

a

ba|K(x, y)|2dxdy

Replacing x by y abd vice versa, we obtain

1

nn(y) =

baK(y, x)n(x)dx, (229)

so that

n(y) =1

nn(y).

Now we have

K2(x, y) =n=1

n(x)n(y)

2n. (230)

In the same manner, we obtain

K3(x, y) =n=1

n(x)n(y)

3n,

and, generally,

Km(x, y) =n=1

n(x)n(y)

mn. (231)

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

K(x, y) =n=1

n(x)n(y)

n. (232)

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

11.2 HilbertSchmidt theorem for integral operators

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

i A(x) = baK(x, y)(y)dy,

ii ba

ba|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 HilbertSchmidt 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

(x) = A+ 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 = 1, and in this case IE (210) has infinitely many solutions.

Proof. According to the HilbertSchmidt theorem, a completely continuous selfadjoint oper-ator A in the Hilbert space H = L2[a, b] has an orthonormal system of eigenvectors {k}corresponding to eigenvalues k such that every element H is represented as

=k

ckk + , (235)

where KerA, i.e., A = 0. Set

f =k

bkk + f, Af = 0, (236)

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

=k

xkk + , A = 0. (237)

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

xkk + =

k

xkkk +k

bkk + f, (238)

which holds if and only if

f = ,

xk(1 k) = bk (k = 1, 2, . . .)

i.e., when

f = ,

xk =bk

1 k , k 6= 1bk = 0, k = 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 n = 1 are arbitrary numbers.The theorem is proved.

46

12 Harmonic functions and Greens formulas

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

harmonic if it satisfies Laplaces equation

u = 0 in D, (240)

where

u =2u

x2+2u

y2(241)

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

The function

(x,y) = (x y) = 12pi

ln1

|x y| (242)

is called the fundamental solution of Laplaces equation. For a fixed y R2, y 6= x, the function(x,y) is harmonic, i.e., satisfies Laplaces equation

2

x2+2

y2= 0 in D. (243)

The proof follows by straightforward differentiation.

12.1 Greens formulas

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

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

D

(uv + gradu grad v)dx =

uv

nydly, (244)

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

D

(uv vu)dx =

(uv

ny v u

ny

)dly, (245)

Proof. Apply Gauss theorem D

divAdx =

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+, Ai C1(D), i = 1, 2) definedby

A = u grad v =[uv

x, u

v

y

], (247)

the components are

A1 = uv

x, A2 = u

v

y. (248)

We have

divA = div (u grad v) =

x

(uv

x

)+

y

(uv

y

)= (249)

u

x

v

x+ u

2v

x2+u

y

v

y+ u

2v

y2= gradu grad v + uv. (250)

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

ny A = ny (ugrad v) = u(ny grad v) = u vny

, (251)

so that, finally, D

divAdx =

D

(uv + gradu grad v)dx,

ny Ad ly =

uv

nydly, (252)

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

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

u(x) =

((x,y)

u

ny u(y)(x,y)

ny

)dly, x D. (253)

Proof. For x D we choose a circle(x, r) = {y : |x y| =

(x x0)2 + (y y0)2 = r}

of radius r (the boundary of a vicinity of a point x D) such that (x, r) D and direct theunit normal n to (x, r) into the exterior of (x, r). Then we apply Greens second formula(245) to the harmonic function u and (x,y) (which is also a harmonic function) in the domainy D : |x y| > r to obtain

0 = D\(x,r)

(u(x,y) (x,y)u)dx = (254)

(x,r)

(u(y)

(x,y)

ny (x,y)u(y)

ny

)dly. (255)

48

Since on (x, r) we have

grad y(x,y) =ny2pir

, (256)

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

v

nydly = 0 (257)

for a harmonic in D function v shows that

limr0

(x,r)

(u(y)

(x,y)

ny (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 . Then

v(y)

nydly = 0. (259)

Proof follows from the first Greens 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 r}

of radius r (a vicinity of a point x) with the boundary (x, r) = {y : |x y| = r} (a circle)and continuous in the closure B(x, r). Then

u(x) =1

pir2

B(x,r)

u(y)dy =1

2pir

(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 < < r we apply (259) and Greens third formula to obtain

u(x) =1

2pi

|xy|=

u(y)dly, (261)

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

49

Theorem 32 (MaximumMinimum 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 and 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 , continuous in D = D and satisfies the Dirichletboundary condition:

u = 0 in D, (262)

u| = f, (263)where f is a given continuous function.

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

u

n

= 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 of D. Then from the maximumminimum 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 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 Greens first formula applied to a pair of (harmonic) functions u, v = u and the interiorDh of a surface h = {x hnx : x } parallel to the Ds boundary with sufficiently smallh > 0,

Dh

(uu+ |gradu|2)dx =

uu

nydly, (265)

we have first B

|gradu|2dx Dh

|gradu|2dx, (266)

(because B Dh); next we obtain Dh

|gradu|2dx

uu

nydly = 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)(y)dly, v(x) =C

nyE(x,y)(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 (y) and (y) are sufficiently smooth functions defined on C. In the case of Laplacesoperator u,

g(x,y) = g(x y) = 12pi

ln1

|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) = i4H

(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

2piln

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 . Then thekernel of the double-layer potential

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

ny, (x,y) =

1

2piln

1

|x y| , (270)

is a continuous function on for x, y .Proof. Performing differentiation we obtain

V (x,y) =1

2pi

cos x,y|x y| , (271)

where x,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 curve so that the Ox axis goes along the tangent and the Oy axis, along the normal to at thispoint. Then one can write the equation of curve in a sufficiently small vicinity of the point yin the form

y = y(x).

The asumption concerning the smoothness of 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(x) =

1

2x2y(x) (0 < 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(x))2 = x

1 +

1

4x2(y(x))2;

cos x,y =y

r=

1

2

xy(x)1 + x2(1/4)(y(x))2

,

andcos x,y

r=

y

r2=

1

2

y(x)1 + x2(1/4)(y(x))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,

limr0

cos x,yr

=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 . For the double-layer potential with a constant density

v0(x) =

(x,y)

nydly, (x,y) =

1

2piln

1

|x y| , (272)

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

v0(x) = 1, x D,v0(x) = 1

2, x , (273)

v0(x) = 0, x R2 \ D.

Proof follows for x R2 \ D from the equality

v(y)

nydly = 0 (274)

applied to v(y) = (x,y). For x D it follows from Greens third formula

u(x) =

((x,y)

u

ny u(y)(x,y)

ny

)dly, x D, (275)

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

v0(x) = 12, x ,

we can also write (273) as

v0(x) = lim

h0v(x+ hnx) = v

0(x) 12, x . (276)

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

u0(x) =

(x,y)dly, (x,y) =1

2piln

1

|x y| . (277)

For the normal derivative of this single-layer potential

u0(x)

nx=

(x,y)

nxdly, (278)

53

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

have

u0(x)

nx= 1, x D,

u0(x)

nx=

1

2, x , (279)

u0(x)

nx= 0, x R2 \ D.

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

v(x) =

(x,y)

ny(y)dly, (x,y) =

1

2piln

1

|x y| , (280)

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

v(x) =

(x,y)ny

(y)dly 12(x), x , (281)

or

v(x) = v(x) 12(x), x , (282)

wherev(x) = lim

h0v(x+ hnx). (283)

Proof.1. Introduce the function

I(x) = v(x) v0(x) =

(x,y)

ny((y) 0)dly (284)

where 0 = (x) and prove its continuity at the point x . For every > 0 and every > 0

there exists a vicinity C1 of the point x such that

|(x) (x)| < , x C1. (285)

SetI = I1 + I2 =

C1

. . .+

\C1. . . .

Then|I1| B1,

54

where B1 is a constant taken according to

(x,y)ny dly B1 x D, (286)

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

x (i.e., x C1) such that|I1(x)| < x D. (287)

Inequality (287) means that I(x) is continuous at the point x .2. Now, if we take v(x) for x and consider the limits v(x) = limh0 v(x+ hny) frominside and outside contour using (276), we obtain

v(x) = I(x) + limh0

v0(x hnx) = v(x) 12(x), (288)

v+(x) = I(x) + lim

h+0v0(x + hnx) = v(x) +

1

2(x), (289)

which prove the theorem.

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

u(x) =

(x,y)(y)dly, (x,y) =1

2piln

1

|x y| . (290)

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

u(x)

nx=

(x,y)

nx(y)dly (291)

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

u(x)nx

=

(x,y)nx

(y)dly 12(x), x , (292)

oru(x)nx

=u(x)nx

12(x), x , (293)

whereu(x)nx

= limh0

nx grad v(x + hnx). (294)

55

14.2 Generalized potentials

Let S() R2 be a domain bounded by the closed piecewise smooth contour . We assumethat a rectilinear interval 0 is a subset of , so that 0 = {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) =

Kl(x, t)l(t)dt, x = (x, y) S(),

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

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

1

piln

1

|x y0| , g2(x, t) =

y0g(x,y0) [y0 = (t, 0)],

F1,2 are smooth functions, and we shall assume that for every closed domain S0() S(),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

tq

F2(x, s)ds, q R1,

are continuous.

We shall also assume that the densities of the generalized potentials 1 L(1)2 () and2 L(2)2 (), where functional spaces L(1)2 and L(2)2 are defined in Section 3.

15 Reduction of boundary value problems to integral

equations

Greens 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() of continuous functionsdefined on contour

K0(x) = 2

(x,y)

ny(y)dly, x (295)

and

K1(x) = 2

(x,y)

nx(y)dly, x . (296)

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

56

Theorem 36 The operators I K0 and I K1 have trivial nullspacesN(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 {0} (298)with

0dly 6= 0. (299)

Proof. Let be a solution to the homogeneous integral equation K0 = 0 and define adouble-layer potential

v(x) =

(x,y)

ny(y)dly, x D (300)

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

v(x) =

(x,y)

ny(y)dly 1

2(x) (301)

which yields2v(x) = K0(x) = 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 ,

v(y)

ny

+

v(y)ny

= 0, y , (303)where the internal and extrenal limiting values with respect to are defined as follows

v(y)

ny

= limy,yDv(y)

ny= lim

h0ny grad v(y hny) (304)

andv(y)

ny

+

= limy,yR2\D

v(y)

ny= lim

h0ny grad v(y + hny). (305)

Note that (303) can be written as

limh0

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 . (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 = v+ v = 0 on . ThusN(I K0) = {0} and, by the Fredholm alternative, N(I K1)