HANDBOOK OFINTEGRAL EQUATIONS phần 8 docx

Ifλis a regular value, then both the Fredholm integral equation of the second kind and the transposed equation are solvable for any right-hand side, and both the equations have unique so

Example 1 Let us solve the integral equation

y(x) – λ

 1 0

xty(t) dt = f (x), 0≤ x ≤ 1,

by the method of successive approximations Here we haveK(x, t) = xt, a = 0, and b = 1 We successively define

K1(x, t) = xt, K2(x, t) =

 1 0

(xz)(zt) dz = xt

3, K3(x, t) =

1 3

 1 0

(xz)(zt) dz = xt

3 2 , , K n(x, t) = xt

3n–1 According to formula (5) for the resolvent, we obtain

For some Fredholm equations, the Neumann series (5) for the resolvent is convergent for all values

ofλ Let us establish this fact.

Assume that two kernelsK(x, t) and L(x, t) are given These kernels are said to be orthogonal

if the following two conditions hold:

 b

a K(x, z)L(z, t) dz = 0,

 b

a

for all admissible values ofx and t.

There exist kernels that are orthogonal to themselves For these kernels we haveK2(x, t)≡ 0,

whereK2(x, t) is the second iterated kernel It is clear that in this case all the subsequent iterated

kernels also vanish, and the resolvent coincides with the kernelK(x, t).

Example 2 Let us find the resolvent of the kernelK(x, t) = sin(x – 2t), 0 ≤ x ≤ 2π, 0 ≤ t ≤ 2π.

z=0 = 0.

Thus, in this case the resolvent of the kernel is equal to the kernel itself:

R(x, t; λ) ≡ sin(x – 2t),

so that the Neumann series (6) consists of a single term and clearly converges for anyλ.

Remark 2 If the kernelsM(1)(x, t), , M(n)(x, t) are pairwise orthogonal, then the resolvent

corresponding to the sum

is equal to the sum of the resolvents corresponding to each of the summands

• References for Section 11.3: S G Mikhlin (1960), M L Krasnov, A I Kiselev, and G I Makarenko (1971),

J A Cochran (1972), V I Smirnov (1974), A J Jerry (1985).

11.4 Method of Fredholm Determinants

11.4-1 A Formula for the Resolvent

A solution of the Fredholm equation of the second kind

y(x) – λ

 b a

The functionD(x, t; λ) is called the Fredholm minor and D(λ) the Fredholm determinant The

series (4) converge for all values ofλ and hence define entire analytic functions of λ The resolvent

R(x, t; λ) is an analytic function of λ everywhere except for the values of λ that are roots of D(λ).

These roots coincide with the characteristic values of the equation and are poles of the resolvent

R(x, t; λ).

Example 1 Consider the integral equation

y(x) – λ

 1 0

xe t y(t) dt = f (x), 0≤ x ≤ 1, λ ≠ 1.

We have

A0(x, t) = xe t, A1(x, t) =

 1 0

 1 0

Let us find the coefficientsB n:

B1=

 1 0

K(t1,t1)dt1=

 1 0

t1e t1dt1= 1, B2=

 1 0

 1 0

It is clear thatB n= 0 for all subsequent coefficients as well.

According to formulas (4), we have

In practice, the calculation of the coefficients A n(x, t) and Bn of the series (4) by means of

formulas (5) and (6) is seldom possible However, formulas (5) and (6) imply the following

recurrent relations:

A n(x, t) = Bn K(x, t) – n

 b a

(–s) ds = –1 Formula (7) implies the relation

A1(x, t) = – x – 2t

2 –

 1 0

(x – 2s)(s – 2t) ds = –x – t + 2xt +2 Furthermore, we have

B2=

 1 0

 –2s + 2s2+23

ds = 13,

A2(x, t) = x – 2t

3 – 2

 1 0

(x – 2s) –s – t + 2st +2 

ds = 0, B3=B4=· · · = 0, A3(x, t) = A4(x, t) = · · · = 0.

Hence,

D(λ) = 1 + 1λ +1λ2 ; D(x, t; λ) = x – 2t + λ

x + t – 2xt – 2 

The resolvent has the form

11.5 Fredholm Theorems and the Fredholm Alternative

11.5-1 Fredholm Theorems

THEOREM1 Ifλis a regular value, then both the Fredholm integral equation of the second kind

and the transposed equation are solvable for any right-hand side, and both the equations have unique

solutions The corresponding homogeneous equations have only the trivial solutions

THEOREM2 For the nonhomogeneous integral equation to be solvable, it is necessary and

sufficient that the right-hand sidef (x)satisfies the conditions

 b a

f (x)ψ k(x) dx = 0, k = 1, , n,

whereψ k(x)is a complete set of linearly independent solutions of the corresponding transposed

homogeneous equation

THEOREM3 Ifλis a characteristic value, then both the homogeneous integral equation and the

transposed homogeneous equation have nontrivial solutions The number of linearly independent

solutions of the homogeneous integral equation is finite and is equal to the number of linearly

independent solutions of the transposed homogeneous equation

THEOREM4 A Fredholm equation of the second kind has at most countably many characteristic

values, whose only possible accumulation point is the point at infinity

11.5-2 The Fredholm Alternative

The Fredholm theorems imply the so-called Fredholm alternative, which is most frequently used in

the investigation of integral equations

THEFREDHOLM ALTERNATIVE Either the nonhomogeneous equation is solvable for any

right-hand side or the corresponding homogeneous equation has nontrivial solutions

The first part of the alternative holds if the given value of the parameter is regular and the second

if it is characteristic

Remark The Fredholm theory is also valid for integral equations of the second kind with weak

singularity

• References for Section 11.5: S G Mikhlin (1960), M L Krasnov, A I Kiselev, and G I Makarenko (1971),

J A Cochran (1972), V I Smirnov (1974), A J Jerry (1985), D Porter and D S G Stirling (1990), C Corduneanu (1991),

J Kondo (1991), W Hackbusch (1995), R P Kanwal (1997).

11.6 Fredholm Integral Equations of the Second Kind

With Symmetric Kernel

11.6-1 Characteristic Values and Eigenfunctions

Integral equations whose kernels are symmetric, that is, satisfy the condition K(x, t) = K(t, x), are

called symmetric integral equations.

Each symmetric kernel that is not identically zero has at least one characteristic value

For anyn, the set of characteristic values of the nth iterated kernel coincides with the set of nth

powers of the characteristic values of the first kernel

The eigenfunctions of a symmetric kernel corresponding to distinct characteristic values are

orthogonal, i.e., if

ϕ1(x) = λ1

 b

a K(x, t)ϕ1(t) dt, ϕ2(x) = λ2

 b

a K(x, t)ϕ2(t) dt, λ1≠ λ2,

(ϕ1,ϕ2) = 0, (ϕ, ψ)≡

 b a ϕ(x)ψ(x) dx.

The characteristic values of a symmetric kernel are real

The eigenfunctions can be normalized; namely, we can divide each characteristic function by its

norm If several linearly independent eigenfunctions correspond to the same characteristic value, say,

ϕ1(x), , ϕn(x), then each linear combination of these functions is an eigenfunction as well, and

these linear combinations can be chosen so that the corresponding eigenfunctions are orthonormal

Indeed, the function

ψ1(x) = ϕ1(x)

ϕ1, ϕ1 =(ϕ1,ϕ1),has the norm equal to one, i.e.,ψ1 = 1 Let us form a linear combination αψ1+ϕ2and chooseα

so that

(αψ1+ϕ2,ψ1) = 0,i.e.,

is orthogonal toψ1(x) and has the unit norm Next, we choose a linear combination αψ1+βψ2+ϕ3,

where the constantsα and β can be found from the orthogonality relations

(αψ1+βϕ2+ϕ3,ψ1) = 0, (αψ1+βψ2+ϕ3,ψ2) = 0

For the coefficientsα and β thus defined, the function

ψ3= αψ1+βψ2+ϕ2

αψ1+βϕ2+ϕ3

is orthogonal toψ1andψ2and has the unit norm, and so on

As was noted above, the eigenfunctions corresponding to distinct characteristic values are

orthogonal Hence, the sequence of eigenfunctions of a symmetric kernel can be made orthonormal

In what follows we assume that the sequence of eigenfunctions of a symmetric kernel is

or-thonormal

We also assume that the characteristic values are always numbered in the increasing order of

their absolute values Thus, if

|λ1| ≤ |λ2| ≤ · · · ≤ |λ n | ≤ · · · (5)

If there are infinitely many characteristic values, then it follows from the fourth Fredholm

theorem that their only accumulation point is the point at infinity, and henceλ n → ∞ as n → ∞.

The set of all characteristic values and the corresponding normalized eigenfunctions of a

sym-metric kernel is called the system of characteristic values and eigenfunctions of the kernel The

system of eigenfunctions is said to be incomplete if there exists a nonzero square integrable function

that is orthogonal to all functions of the system Otherwise, the system of eigenfunctions is said to

be complete.

11.6-2 Bilinear Series

Assume that a kernelK(x, t) admits an expansion in a uniformly convergent series with respect to

the orthonormal system of its eigenfunctions:

If a symmetric kernel K(x, t) has finitely many characteristic values, then it is degenerate,

because in this case we have

and the above quadratic functional vanishes forϕ(x) = 0 only Such a kernel has positive characteristic

values only A negative definite kernel is defined similarly.

Each symmetric positive definite (or negative definite) continuous kernel can be decomposed in

a bilinear series in eigenfunctions that is absolutely and uniformly convergent with respect to the

variablesx, t.

The assertion remains valid if we assume that the kernel has finitely many negative (positive,

respectively) characteristic values

If a kernelK(x, t) is symmetric, continuous on the square S = {a ≤ x ≤ b, a ≤ t ≤ b}, and has

uniformly bounded partial derivatives on this square, then this kernel can be expanded in a uniformly

convergent bilinear series in eigenfunctions

11.6-3 The Hilbert–Schmidt Theorem

If a functionf (x) can be represented in the form

f (x) =

 b a

where the symmetric kernelK(x, t) is square integrable and g(t) is a square integrable function,

then f (x) can be represented by its Fourier series with respect to the orthonormal system of

eigenfunctions of the kernelK(x, t):

f (x)ϕ k(x) dx, k = 1, 2,

a

then the series (12) is absolutely and uniformly convergent for any functionf (x) of the form (11).

Remark 1 In the Hilbert–Schmidt theorem, the completeness of the system of eigenfunctions

is not assumed

11.6-4 Bilinear Series of Iterated Kernels

By the definition of the iterated kernels, we have

K m(x, t) =

 b

a

The Fourier coefficientsa k(t) of the kernel Km(x, t), regarded as a function of the variable x, with

respect to the orthonormal system of eigenfunctions of the kernelK(x, t) are equal to

In formula (16), the sum of the series is understood as the limit in mean-square If in addition to the

above assumptions, inequality (13) is satisfied, then the series in (16) is uniformly convergent

11.6-5 Solution of the Nonhomogeneous Equation

Let us represent an integral equation

y(x) – λ

 b a

 b

a

f (x)ϕ k(x) dx = yk–f k.Taking into account the expansion (8), we obtain

λ

 b a K(x, t)y(t) dt = λ

then, fork ≠ p, p + 1, , q, the terms (20) preserve their form For k = p, p + 1, , q, formula (19)

implies the relationf k =A k(λ – λk)/λ, and by (21) we obtain fp=f p+1=· · · = f q = 0 The last

a

f (x)ϕ k(x) dx = 0fork = p, p+1, , q, i.e., the right-hand side of the equation must be orthogonal to the eigenfunctions

that correspond to the characteristic valueλ.

In this case, the solutions of Eqs (17) have the form

where the terms in the first of the sums (22) with indicesk = p, p + 1, , q must be omitted (for

these indices,f k andλ – λ kvanish in this sum simultaneously) The coefficientsC kin the second

sum are arbitrary constants

Remark 2 On the basis of the bilinear expansion (8) and the Hilbert–Schmidt theorem, the

solution of the symmetric Fredholm integral equation of the first kind

 b a K(x, t)y(t) dt = f (x), a ≤ x ≤ b,

can be constructed in a similar way in the form

and the necessary and sufficient condition for the existence and uniqueness of such a solution

inL2(a, b) is the completeness of the system of the eigenfunctions ϕk(x) of the kernel K(x, t) together

with the convergence of the series∞

k=1

f2

k λ2

k, where theλ kare the corresponding characteristic values

It should be noted that the verification of the last condition for specific equations is quite

complicated In the solution of Fredholm equations of the first kind, the methods presented in

Chapter 10 are usually applied

11.6-6 The Fredholm Alternative for Symmetric Equations

The above results can be unified in the following alternative form

A symmetric integral equation

y(x) – λ

 b

a

for a given λ, either has a unique square integrable solution for an arbitrarily given function

f (x) ∈ L2(a, b), in particular, y = 0 for f = 0, or the corresponding homogeneous equation has

finitely many linearly independent solutionsY1(x), , Yr(x), r > 0

For the second case, the nonhomogeneous equation has a solution if and only if the right-hand

sidef (x) is orthogonal to all the functions Y1(x), , Yr(x) on the interval [a, b] Here the solution

is defined only up to an arbitrary additive linear combinationA1Y1(x) +· · · + A r Y r(x)

11.6-7 The Resolvent of a Symmetric Kernel

The solution of a Fredholm equation of the second kind (23) can be written in the form

Here the collectionsϕ k(x) and λk form the system of eigenfunctions and characteristic values of

Eqs (23) It follows from formula (25) that the resolvent of a symmetric kernel has only simple

poles

11.6-8 Extremal Properties of Characteristic Values and Eigenfunctions

Let us introduce the notation

(u, w) =

 b a u(x)w(x) dx, u2= (u, u),(Ku, u) =

 b a

 b a K(x, t)u(x)u(t) dx dt,

where (u, w) is the inner product of functions u(x) and w(x),u is the norm of a function u(x),

and (Ku, u) is the quadratic form generated by the kernel K(x, t)

Letλ1be the characteristic value of the symmetric kernelK(x, t) with minimum absolute value

and lety1(x) be the eigenfunction corresponding to this value Then

1

|λ1| = maxy/≡0

|(Ky, y)|

in particular, the maximum is attained, andy = y1is a maximum point

Letλ1, , λ nbe the firstn characteristic values of a symmetric kernel K(x, t) (in the ascending

order of their absolute values) and lety1(x), , yn(x) be orthonormal eigenfunctions corresponding

toλ1, , λ n, respectively Then the formula

1

|λ n+1| = max

|(Ky, y)|

is valid for the characteristic value λ n+1 followingλ n The maximum is taken over the set of

functionsy which are orthogonal to all y1, , y nand are not identically zero, that is,y≠ 0

in particular, the maximum in (27) is attained, andy = y n+1is a maximum point, wherey n+1is any

eigenfunction corresponding to the characteristic valueλ n+1which is orthogonal toy1, , y n

Remark 3 For a positive definite kernelK(x, t), the symbol of modulus on the right-hand sides

of (27) and (28) can be omitted

11.6-9 Integral Equations Reducible to Symmetric Equations

An equation of the form

y(x) – λ

 b

a

whereK(s, t) is a symmetric kernel and ρ(t) > 0 is a continuous function on [a, b], can be reduced to

a symmetric equation Indeed, on multiplying Eq (29) by√

ρ(x) and introducing the new unknown

ρ(x), L(x, t) = K(x, t)

whereL(x, t) is a symmetric kernel.

11.6-10 Skew-Symmetric Integral Equations

By a skew-symmetric integral equation we mean an equation whose kernel is skew-symmetric, i.e.,

an equation of the form

Equation (31) with the skew-symmetric kernel (32) has at least one characteristic value, and all

its characteristic values are purely imaginary

• References for Section 11.6: E Goursat (1923), R Courant and D Hilbert (1931), S G Mikhlin (1960), M L Krasnov,

A I Kiselev, and G I Makarenko (1971), J A Cochran (1972), V I Smirnov (1974), A J Jerry (1985), F G Tricomi

(1985), D Porter and D S G Stirling (1990), C Corduneanu (1991), J Kondo (1991), W Hackbusch (1995), R P Kanwal

(1997).

11.7 An Operator Method for Solving Integral Equations

of the Second Kind

11.7-1 The Simplest Scheme

Consider a linear equation of the second kind of the special form

where L is a linear (integral) operator such that L2=k, k = const.

Let us apply the operator L to Eq (1) We obtain

Remark In Section 9.4, various generalizations of the above method are described

11.7-2 Solution of Equations of the Second Kind on the Semiaxis

1◦ Consider the equation

We obtain the solution by formula (3) taking into account Eq (5):

and acts by the rule L2= 1 (see Subsection 7.6-1)

We obtain the solution by formula (3), fork = 1, taking into account Eq (8):

• Reference for Section 11.7: A D Polyanin and A V Manzhirov (1998).

11.8 Methods of Integral Transforms and Model

Solutions

11.8-1 Equation With Difference Kernel on the Entire Axis

Consider an integral equation of convolution type of the second kind with one kernel

y(x) + √1

2π

 ∞

–∞ K(x – t)y(t) dt = f (x), –∞ < x < ∞, (1)

wheref (x) and K(x) are the known right-hand side and the kernel of the integral equation and y(x)

is the unknown function Let us apply the (alternative) Fourier transform to Eq (1) In this case,

taking into account the convolution theorem (see Subsection 7.4-4), we obtain

Thus, on applying the Fourier transform we reduce the solution of the original integral equation (1)

to the solution of the algebraic equation (2) for the transform of the unknown function The solution

of Eq (2) has the form

Formula (3) gives the transform of the solution of the original integral equation in terms of the

transforms of the known functions, namely, the kernel and the right-hand side of the equation The

solution itself can be obtained by applying the Fourier inversion formula:

1 +K(u) e–iux du. (4)

In fact, formula (4) solves the problem; however, sometimes it is not convenient because it

requires the calculation of the transformF (u) for each right-hand side f (x) In many cases, the

representation of the solution of the nonhomogeneous integral equation via the resolvent of the

original equation is more convenient To obtain the desired representation, we note that formula (3)

can be transformed to the expression

Thus, to determine the solution of the original integral equation (1), it suffices to find the

func-tionR(x) by formula (7).

The functionR(x) is a solution of Eq (1) for a special form of the function f (x) Indeed, it

follows from formulas (3) and (5) that forY(u) = R(u) the function F(u) is equal to K(u) This

means that, forf (x) ≡ K(x), the function y(x) ≡ R(x) is a solution of Eq (1), i.e., the resolvent of

Eq (1) satisfies the integral equation

R(x) + √1

2π

 ∞

Note that to calculate direct and inverse Fourier transforms, one can use the corresponding tables

from Supplements 6 and 7 and the books by H Bateman and A Erd´elyi (1954) and by V A Ditkin

Assume thatλ < 12α In this case the integral (13) makes sense and can be calculated by means of the theory of residues on

applying the Jordan lemma (see Subsections 7.1-4 and 7.1-5) After some algebraic manipulations, we obtain

R(x) = – √

2π √ αλ

α2 – 2αλexp

 –|x| √ α2 – 2αλ

f (t) dt, –∞ < x < ∞. (15)

11.8-2 An Equation With the Kernel K(x, t) = t Q(x/t) on the Semiaxis

Here we consider the following equation on the semiaxis:

wheres = σ + iτ is a complex variable (σ1<σ < σ2) and ˆf (s) is the transform of the function f (x).

In what follows, we briefly denote the Mellin transform by M{f(x)} ≡ M{f(x), s}.

For known ˆf (s), the original function can be found by means of the Mellin inversion formula

f (x) = M–1{ ˆf(s)} ≡ 1

2πi

 c+i ∞ c–i∞

ˆ

f (s)x–s ds, σ1<c < σ2, (18)where the integration path is parallel to the imaginary axis of the complex planes and the integral

is understood in the sense of the Cauchy principal value

On applying the Mellin transform to Eq (16) and taking into account the fact that the integral

with such a kernel is transformed into the product by the rule (see Subsection 7.3-2)

Under the application of this analytical method of solution, the following technical difficulties

can occur: (a) in the calculation of the transform for a given kernelK(x) and (b) in the calculation

of the solution for the known transform ˆy(s) To find the corresponding integrals, tables of direct

and inverse Mellin transforms are applied (e.g., see Supplements 8 and 9) In many cases, the

relationship between the Mellin transform and the Fourier and Laplace transforms is first used:

11.8-3 Equation With the Kernel K(x, t) = t Q(xt) on the Semiaxis

Consider the following equation on the semiaxis:

y(x) –

 ∞

0

To solve this equation, we apply the Mellin transform On multiplying Eq (25) by x s–1 and

integrating with respect tox from zero to infinity, we obtain

11.8-4 The Method of Model Solutions for Equations on the Entire Axis

Let us illustrate the capability of a generalized modification of the method of model solutions (see

Subsection 9.6) by an example of the equation

The right-hand side of (35) can be regarded as a functional equation for the kernele pxof the inverse

Laplace transform To solve it, we replacep by –p – β in Eq (33) We finally obtain

Since here –∞ < x < ∞, one must set p = iu and use the formulas from Subsection 9.6-3 Then the

solution of Eq (32) for an arbitrary functionf (x) can be represented in the form

• References for Section 11.8: M L Krasnov, A I Kiselev, and G I Makarenko (1971), V I Smirnov (1974),

P P Zabreyko, A I Koshelev, et al (1975), F D Gakhov and Yu I Cherskii (1978), A D Polyanin and A V Manzhirov

(1997, 1998).

11.9 The Carleman Method for Integral Equations of

Convolution Type of the Second Kind

11.9-1 The Wiener–Hopf Equation of the Second Kind

Equations of convolution type of the second kind of the form*

frequently occur in applications Here the domain of the kernelK(x) is the entire real axis.

Let us extend the equation domain to the negative semiaxis by introducing one-sided functions,

which coincides with (1) forx > 0.

The auxiliary functiony–(x) is introduced to compensate for the left-hand side of Eq (2) for

x < 0 Note that y–(x) is unknown for x < 0 and is to be found in solving the problem

Let us pass to the Fourier integrals in Eq (2) (see Subsections 7.4-3, 10.4-1, and 10.4-2) We

obtain a Riemann problem in the form

then we rewrite the Riemann problem in the usual form

Y+(u) =D(u)Y–(u) +H(u), –∞ < u < ∞, (4)where

1 +K(u), H(u) =

F(u)

The Riemann problem (4) is equivalent to Eq (1); in particular, these equations are

simulta-neously solvable or unsolvable and have an equal number of arbitrary constants in their general

solutions If the index ν of the Riemann problem, which is given by the relation

(which is also sometimes called the index of the Wiener–Hopf equation of the second kind), is

positive, then the homogeneous equation (1) (f (x)≡ 0) has exactly ν linearly independent solutions,

* Prior to reading this section looking through Sections 10.4 and 10.5 is recommended.

and the nonhomogeneous equation is unconditionally solvable and its solution depends onν arbitrary

complex constants

In the caseν ≤ 0, the homogeneous equation has no nonzero solutions For ν = 0, the

nonhomo-geneous equation is unconditionally solvable, and the solution is unique If the indexν is negative,

then the conditions

 ∞

–∞

F(u) du

X+(u)[1 +K(u)](u + i) k = 0, k = 1, 2, , –ν, (7)are necessary and sufficient for the solvability of the nonhomogeneous equation (see Subsec-

whereY+(u) is the solution of the Riemann problem (4) and (5) that is constructed by the scheme of

Subsection 10.4-4 (see Fig 3) The last formula shows that the solution does not depend on Y–(u),

i.e., is independent of the choice of the extension of the equation to the negative semiaxis

2◦ Now let us study the exceptional case of the integral equation (1) in which the normality

condition for the Riemann problem (3) (see Subsections 10.4-6 and 10.4-7) is violated In this case,

the coefficientD(u) = [1 + K(u)]–1 has no zeros, and its order at infinity isη = 0 The general

solution to the boundary value problem (3) can be obtained by formulas (63) of Subsection 10.4-7

forα i = 0 The solution of the original integral equation (1) can be determined from the solution of

the boundary value problem on applying formula (8)

Figure 4 depicts a scheme of solving the Wiener–Hopf equations (see also Subsection 10.5-1)

Example Consider the equation

1 +K(u) = P (u)

(u2 + 1) 2 , P (z) = z4+ 2(a – b + 1)z2+ 2a + 2b + 1.

On the basis of the normality condition, we assume that the constantsa and b are such that the polynomial P (z) has no real

roots Letα + iβ be a root of the biquadratic equation P (z) = 0 such that α > 0 and β > 0 Since the coefficients of the

equation are real, it is clear that (α – iβ), (–α + iβ), and (–α – iβ) are the other three roots Since the function 1 + K(u) is

real as well, it follows that it has zero index, and hence Eq (9) is uniquely solvable.

On factorizing, we obtain the relation 1 +K(u) = X– (u)/ X+ (u), where

X+ (u) = (u + i)

2

(u + α + iβ)(u – α + iβ), X– (u) = (u – α – iβ)(u + α – iβ)

(u – i)2 Applying this result, we represent the boundary condition (4), (5) in the form

It follows from the theorem on the analytic continuation and the generalized Liouville theorem (see Subsection 10.4-3) that

both sides of the above relation are equal to

C1

u – α – iβ +

C2

u + α – iβ,

Fig 4 Scheme of solving the Wiener–Hopf integral equations Forβ = 0, we have the equation of the

first kind, and forβ = 1, we have the equation of the second kind.

where the constantsC1andC2must be defined Hence,

Since the problem is more or less cumbersome, we pass from the transform (11) to the corresponding original function

in two stages We first find the inverse transform of the summand

We can calculate the integrals

(u + iβ – α)(u + iβ + α)(u + α – iβ)

by means of the residue theory (see Subsections 7.1-4 and 7.1-5) and substitute the values (12) into the constantsC1andC2.

11.9-2 An Integral Equation of the Second Kind With Two Kernels

Consider an integral equation of convolution type of the second kind with two kernels of the form

desired function as the difference of one-sided functions,

Applying the Fourier integral transform (see Subsection 7.4-3), we obtain

[1 +K1(u)]Y+(u) – [1 +K2(u)]Y–(u) =F(u). (19)This implies the relation

Y+(u) = 1 +K2(u)

1 +K1(u)Y–(u) + F(u)

HereK1(u),K2(u), andF(u) stand for the Fourier integrals of known functions The unknown

transformsY+(u) andY–(u) are the boundary values of functions that are analytic on the upper and

lower half-planes, respectively Thus, we have obtained a Riemann boundary value problem

1◦ Assume that the normality conditions are satisfied, i.e.,

1 +K1(u)≠ 0, 1 + K2(u)≠ 0,then we can rewrite the Riemann problem in the usual form (see Subsection 10.4-4):

Y+(u) =D(u)Y–(u) +H(u), –∞ < u < ∞, (21)where

D(u) = 1 +K2(u)

1 +K1(u), H(u) = F(u)

The Riemann problem (21), (22) is equivalent to Eq (16): these problems are solvable or

unsolvable simultaneously, and have the same number of arbitrary constants in their general solutions

If the index

ν = Ind1 +K2(u)

is positive, then the homogeneous equation (16) (f (x) ≡ 0) has precisely ν linearly independent

solutions, and the nonhomogeneous equation is unconditionally solvable; moreover, the solution of

this equation depends onν arbitrary complex constants.

In the caseν ≤ 0, the homogeneous equation has no nonzero solutions The nonhomogeneous

equation is unconditionally solvable forν = 0, and the solution is unique For the case in which the

indexν is negative, the conditions

 ∞

–∞

F(u) du

X+(u)[1 +K1(u)](u + i)k = 0, k = 1, 2, , –ν, (24)are necessary and sufficient for the solvability of the nonhomogeneous equation

In all cases for which the solution of Eq (16) exists, this solution can be found by the formula

the scheme of Subsection 10.4-4 (see Fig 3)

Thus, the solution of Eq (16) is equivalent to the solution of a Riemann boundary value problem

and is reduced to the calculation of finitely many Fourier integrals

2◦ Now let us study the exceptional case of an integral equation of the form (16) Assume that

the functions 1 +K1(u) and 1 +K2(u) can have zeros, and these zeros can be both different and

coinciding points of the contour Let us write out the expansion of these functions on selecting the

Herea i ≠ b j, but it is possible that some pointsd k(k = 1, , p) coincide with either aiorb j This

corresponds to the case in which the functions 1 +K1(u) and 1 +K2(u) have a common zero of

different multiplicity We do not select these points especially because their presence does not affect

the solvability conditions and the number of solutions of the problem

It follows from Eq (19) and from the condition that a solution must be finite on the contour that,

for the solvability of the problem, and all the more for the solvability of Eq (16), it is necessary that

the functionF(u) have zero of order γ kat any pointd k, i.e.,F(u) must have the form

Since the functionsK1(u) andK2(u) vanish at infinity, it follows that the point at infinity is a

regular point ofD(u).

Assume that conditions (28) are satisfied In this case the Riemann boundary value problem (20)

can be rewritten in the form (see Subsections 10.4-6 and 10.4-7)

On finding its general solution in the exceptional case under consideration, we obtain the general

solution of the original equation by means of formula (25)

Let us state the conclusions on the solvability conditions and on the number of solutions of

Eq (16) For the solvability of Eq (16), it is necessary that the Fourier transform of the right-hand

side of the equation satisfiesl conditions of the form (27) If these conditions are satisfied, then,

forν – n > 0, problem (20) and the integral equation (16) have exactly ν – n linearly independent

solutions Forν – n ≤ 0, we must take the polynomial P ν–n–1(z) to be identically zero, and, for

the case in whichν – n < 0, the right-hand side must satisfy another n – ν conditions If the latter

conditions are satisfied, then the integral equation has a unique solution

Example Consider Eq (16) for which

whereα and β are real constants In this case, K1(x – t) = 0 for x < t and K2(x – t) = 0 for x < t Hence, the equation

under consideration has the form

The solution of the Riemann problem depends on the signs ofα and β.

1◦ Letα > 0 and β > 0 In this case we have ν = Ind D(u) = 0 The left-hand side and the right-hand side of the boundary

condition contain functions that have analytic continuations to the upper and the lower half-plane, respectively On applying

the theorem on the analytic continuation directly and the generalized Liouville theorem (Subsection 10.4-3), we see that

2π e x(1 + 2x) forx < 0.

2◦ Letα < 0 and β < 0 Here we again have ν = 0, X+ (z) = (z – iβ)(z – iα)–1 , andX– (z) = 1 On grouping the

terms containing the boundary values of functions that are analytic in each of the half-planes and then applying the analytic

continuation theorem and the generalized Liouville theorem (Subsection 10.4-3), we see that

Y+ (z)

X+ (z)+

β + 1 i(β – 1)

whereC is an arbitrary constant Now, by means of the Fourier inversion formula, we obtain the general solution of the

integral equation in the form

It can be seen from the expression forY– (z) that the singularity of the function Y– (z) at the point iβ disappears if we set

β = –1 The last condition is exactly the solvability condition of the Riemann problem In this case we have the unique

Remark 1 Some equations whose kernels contain not the difference but certain other

combina-tions of arguments, namely, the product or, more frequently, the ratio, can be reduced to equacombina-tions

considered in Subsection 11.9-2 For instance, the equation

Y (ξ) +

 1 0

1

τ N1



ξ τ



Y (τ ) dτ = g(ξ), ξ > 0, (31)

becomes a usual equation with two kernels after the following changes of the functions and their

arguments: ξ = e x,τ = e t,N1(ξ) = K1(x), N2(ξ) = K2(x), g(ξ) = f (x), and Y (ξ) = y(x)

11.9-3 Equations of Convolution Type With Variable Integration Limit

1◦ Consider the Volterra integral equation of the second kind

where the interval [0,T ) can be either finite or infinite In contrast with Eq (1), where the kernel is

defined on the entire real axis, here the kernel is defined on the positive semiaxis

Equation (32) can be regarded as a special case of the one-sided equation (1) of Subsection 11.9-1

To see this, we can rewrite Eq (32) in the form

the upper half-plane, possibly except for finitely many poles that are zeros of the function 1 +K+(z)

(we assume that 1 +K+(z)≠ 0 on the real axis) Therefore, the index ν of the problem is always

nonpositive,ν ≤ 0 On rewriting the problem in the form [1 + K+(u)]Y+(u) =Y–(u) +F+(u), we

see thatY–(u)≡ 0, which implies

Y+(u) = F+(u)

Consider the following cases

1.1 The function 1 +K+(z) has no zeros on the upper half-plane (this means that ν = 0) In this

case, Eq (32) has a unique solution for an arbitrary right-hand sidef (x), and this solution can be

expressed via the resolvent:

1.2 The function 1 +K+(z) has zeros at the points z = a1, , a mof the upper half-plane (in

this case we haveν < 0, and ν is equal to the minus total order of the zeros) The following two

possibilities can occur

(a) The functionF+(z) vanishes at the points a1, , a m, and the orders of these zeros are not

less than the orders of the corresponding zeros of the function 1 +K+(z) In this case, the function

F+(z)[1 +K+(z)]–1has no poles again, and thus the equation has the unique solution (34)

The assumption d k F+(aj)/dzk = 0 on the zeros of the function F+(z) is equivalent to the

conditions  ∞

–∞

f (t)e–ia j t t k dt = 0, k = 0, , η j– 1, j = 1, , m, (35)

whereη j is the multiplicity of the zero of the function 1 +K+(z) at the point aj In this case,

conditions (35) are imposed directly on the right-hand side of the equation

(b) The functionF+(z) does not vanish at the points a1, , a m(or vanishes with less

multi-plicity than 1 +K+(z)) In this case, the functionF+(z)[1 +K+(z)]–1has poles, and therefore the

function (33) does not belong to the class under consideration Equation (32) has no solutions in the

chosen class of functions In this case, conditions (35) fail

The last result does not contradict the well-known fact that a Volterra equation always has a

unique solution Equation (32) belongs to the class of Volterra type equations, and therefore is also

solvable in case (b), but in a broader space of functions with exponential growth

2◦ Another simple special case of Eq (1) in Subsection 11.9-1 is the following equation with

variable lower limit:

This corresponds to the case in which the functionK(x) in Eq (1) is left one-sided: K(x) = K–(x)

Under the assumption 1 +K–(u)≠ 0, the Riemann problem becomes

Y+(u) = Y–(u)

1 +K–(u)+

F+(u)

2.1 The function 1 +K–(z) has no zeros on the lower half-plane This means that the inverse

transform of the functionY–(u)[1 +K–(u)]–1is left one-sided, and such a function does not influence

the relation between the inverse transforms of (37) forx > 0 Thus, if we introduce the function

R–(u) = – K–(u)

1 +K–(u)

(for convenience of the final formula), then by applying the Fourier inversion formula to Eq (37)

and by settingx > 0 we obtain the unique solution to Eq (36),

2.2 The function 1 +K–(z) has zeros in the lower half-plane Since this function is nonzero

both on the entire real axis and at infinity, it follows that the number of zeros is finite The Riemann

problem (37) has a positive index which is just equal to the number of zeros in the lower half-plane

(the zeros are counted according to their multiplicities):

1 +K–(u) = – Ind[1 +K–(u)] = η1+· · · + η n> 0

Hereη kare the multiplicities of the zerosz kof the function 1 +K–(z), k = 1, , n

Let

C1k

z – z k

+ C2k(z – zk)2 +· · · + C η k k

(z – zk)η k

be the principal part of the Laurent series expansion of the functionY–(z)[1 +K–(z)]–1in powers of

(z – zk),k = 1, , n In this case, Eq (37) becomes

where the dots denote a function whose inverse transform vanishes forx > 0 Under the passage to

the inverse transforms in Eq (38), forx > 0 we obtain

Here the P k(x) are polynomials of degree ηk – 1 We can verify that the function (39) is a

solution of Eq (36) for arbitrary coefficients of the polynomials Since the number of linearly

independent solutions of the homogeneous equation (36) is equal to the index, it follows that the

above solution (39) is the general solution of the nonhomogeneous equation

11.9-4 Dual Equation of Convolution Type of the Second Kind

Consider the dual integral equation of the second kind

y(x) + √1

2π

 ∞

–∞ K1(x – t)y(t) dt = f (x), 0 <x < ∞, y(x) + √1

2π

 ∞

–∞ K2(x – t)y(t) dt = f (x), –∞ < x < 0,

(40)

in which the functiony(x) is to be found.

In order to apply the Fourier transform technique (see Subsections 7.4-3, 10.4-1, and 10.4-2),

we extend the domain of both conditions in Eq (40) by formally rewriting them for all real values

ofx This can be achieved by introducing new unknown functions into the right-hand sides These

functions must be chosen so that the conditions given on the semiaxis are not violated Hence, the

first condition in (40) must be complemented by a summand that vanishes on the positive semiaxis

and the second by a summand that vanishes on the negative semiaxis Thus, the dual equation can

be written in the form

y(x) + √1

2π

 ∞

–∞ K1(x – t)y(t) dt = f (x) + ξ–(x),y(x) + √1

2π

 ∞

–∞ K2(x – t)y(t) dt = f (x) + ξ+(x),

–∞ < x < ∞, (41)

where theξ ±(x) are some right and left one-sided functions so far unknown

On applying the Fourier integral transform, we arrive at the relations

[1 +K1(u)]Y(u) = F(u) + Ξ–(u), [1 +K2(u)]Y(u) = F(u) + Ξ+(u) (42)

Here the three functionsY(u), Ξ+(u), and Ξ–(u) are unknown

Now on the basis of (42) we can find

Y(u) = F(u) + Ξ–(u)

1 +K1(u) =

F(u) + Ξ+(u)

and eliminate the function Y(u) from relations (42) by applying formula (43) We obtain the

Riemann boundary value problem in the form

then we can rewrite the Riemann problem (44) in the usual form (see Subsection 10.4-4)

Ξ+(u) =D(u)Ξ–(u) +H(u), –∞ < u < ∞, (45)where

D(u) = 1 +K2(u)

1 +K1(u), H(u) = K2(u) –K1(u)

The Riemann problem (45), (46) is equivalent to Eq (40); in particular, they are solvable and

unsolvable simultaneously and have the same number of arbitrary constants in the general solutions

If the index

ν = Ind1 +K2(u)

is positive, then the homogeneous equation (40) (f (x) ≡ 0) has exactly ν linearly independent

solutions, and the nonhomogeneous equation is unconditionally solvable and the solution depends

onν arbitrary complex constants.

For the case ν ≤ 0, the homogeneous equation has no nonzero solutions For ν = 0, the

nonhomogeneous equation is unconditionally solvable, and a solution is unique If the indexν is

negative, then the conditions

For all cases in which a solution of Eq (40) exists, it can be found by the formula

whereΞ+(u), Ξ–(u) is a solution of the Riemann problem (45), (46) that is constructed by the scheme

of Subsection 10.4-4 (see Fig 3)

2◦ Let us investigate the exceptional case of the integral equation (40) Assume that the functions

1 +K1(u) and 1 +K2(u) can have zeros that can be either different or coinciding points of the contour

Take the expansions of these functions on selecting the coinciding zeros in the form of (26) and

further repeat the reasoning performed for the equations of convolution type of the second kind with

two kernels After finding the general solution of the Riemann boundary value problem (44) in this

exceptional case (see Subsection 10.4-7), we obtain the general solution of the original equation (40)

by formula (49)

The conclusions on the solvability conditions and on the number of solutions of Eq (40) are

similar to those made above for the equations with two kernels in Subsection 11.9-2

Remark 2 Equations treated in Section 11.9 are sometimes called characteristic equations of

convolution type.

• Reference for Section 11.9: F D Gakhov and Yu I Cherskii (1978).

11.10 The Wiener–Hopf Method

Assume that the parameterz that enters the transform (1) can take complex values as well Let us

study the properties of the functionY(z) regarded as a function of the complex variable z To this

end, we represent the functiony(x) in the form*

In this case the transformY(z) of the function y(x) is clearly equal to the sum of the transforms Y+(z)

andY–(z) of the functions y+(x) and y–(x), respectively Let us clarify the analytic properties of the

functionY(z) by establishing the analytic properties of the functions Y+(z) andY–(z) Consider the

functiony+(x) given by relations (3) Its transform is equal to

whereM is a constant, then the function Y+(z) given by formula (4) is an analytic function of the

complex variablez = u + iv in the domain Im z > v–, and in this domain we haveY+(z)→ 0 as

|z| → ∞ We can also show that the functions y+(x) andY+(z) are related as follows:

y+(x) = √1

2π

 ∞+iv

–∞+iv Y+(z)e–izx dz, (6)

* Do not confuse the functionsy ±(x) and Y ±(x) introduced in this section with the functions y ±(x) and Y ±(x) introduced

in Subsection 10.4-2 and used in solving the Riemann boundary value problem on the real axis.

where the integration is performed over any line Imz = v > v– in the complex planez, which is

parallel to the real axis

Forv– < 0 (i.e., for functionsy(x) with exponential decay at infinity), the real axis belongs

to the domain in which the functionY+(z) is analytic, and we can integrate over the real axis in

formula (6) However, if the only possible values ofv–are positive (for instance, if the functiony+(x)

has nontrivial growth at infinity, which does not exceed the exponential growth with linear exponent),

then the analyticity domain of the functionY+(z) is strictly above the real axis of the complex plane z

(and in this case, the integral (4) can be divergent on the real axis) Similarly, if the functiony–(x)

in relations (3) satisfies the condition

is an analytic function of the complex variablez in the domain Im z < v+ The functiony–(x) can

be expressed viaY–(z) by means of the relation

It is clear that forv–<v+, the functionY(z) defined by formula (1) is an analytic function of the

complex variablez in the strip v–< Imz < v+ In this case, the functionsy(x) and Y(z) are related

by the Fourier inversion formula

y(x) = √1

2π

 ∞+iv

where the integration is performed over an arbitrary line in the complex planez belonging to the

stripv– < Imz < v+ In particular, forv– < 0 andv+> 0, the functionY(z) is analytic in the strip

containing the real axis of the complex planez.

Example 1 Forα > 0, the function K(x) = e–α |x|has the transform

K(z) = √1

2π

2α

α2 +z2 , which is an analytic function of the complex variablez in the strip –α < Im z < α, which contains the real axis.

11.10-2 The Homogeneous Wiener–Hopf Equation of the Second Kind

Consider a homogeneous integral Wiener–Hopf equation of the second kind in the form

y(x) =

 ∞

0

whose solution can obviously be determined up to an arbitrary constant factor only Here the domain

of the functionK(x) is the entire real axis This factor can be found from additional conditions of

the problem, for instance, from normalization conditions

We assume that Eq (11) defines a functiony(x) for all values of the variable x, positive and

negative Let us introduce the functions y–(x) and y+(x) by formulas (3) Obviously, we have

y(x) = y+(x) + y–(x), and Eq (11) can be rewritten in the form

That is, the functiony+(x) can be determined by the solution of the integral equation (12) and the

functiony–(x) can be expressed via the functions y+(x) and K(x) by means of formulas (13) In

this case, we have the relation

y+(x) + y–(x) =

 ∞

–∞

which is equivalent to the original equation (11)

Let the functionK(x) satisfy the condition

is analytic in the stripv–< Imz < v+

Let us seek the solution of Eq (11) satisfying the condition

|y+(x)| < M1e µx as x → ∞, (17)whereµ < v+(such a solution exists) In this case we can readily verify that the integrals on the

right-hand sides in (12) and (13) are convergent, and the functiony–(x) satisfies the estimate

|y–(x)| < M2e v+x as x → –∞. (18)

It follows from conditions (17) and (18) that the transformsY+(z) andY–(z) of the functions

y+(x) and y–(x) are analytic functions of the complex variable z for Im z > µ and Im z < v+,

respectively

Let us pass to the solution of the integral equation (11) or of Eq (14), which is equivalent

to (11) To this end, we apply the (alternative) Fourier transform By the convolution theorem (see

Subsection 7.4-4), it follows from (14) that

Y+(z) +Y–(z) =√

2πK(z)Y+(z),or

where

Thus, by means of the Fourier transform, we succeeded in the passage from the original integral

equation to an algebraic equation for the transforms However, in this case Eq (19) involves two

unknown functions In general, a single algebraic equation cannot uniquely determine two unknown

functions The Wiener–Hopf method makes it possible to solve this problem for a certain class of

functions This method is mainly related to the study of the analyticity domains of the functions that

enter the equation and to a special representation of this equation The main idea of the Wiener–Hopf

method is as follows

Let Eq (19) be representable in the form

W+(z)Y+(z) = –W–(z)Y–(z), (21)where the left-hand side is analytic in the upper half-plane Imz > µ and the right-hand side is

analytic in the lower half-plane Imz < v+, whereµ < v+, so that there exists a common analyticity

strip of these functions:µ < Im z < v+ Since the analytic continuation is unique, it follows that there

exists a unique entire function of the complex variable that coincides with the left-hand side of (21)

in the upper half-plane and with the right-hand side of (21) in the lower half-plane, respectively

If, in addition, the functions that enter Eq (21) have at most power-law growth with respect toz

at infinity, then it follows from the generalized Liouville theorem (see Subsection 10.4-3) that the

entire function under consideration is a polynomial In particular, for the case of a function that is

bounded at infinity we obtain

W+(z)Y+(z) = –W–(z)Y–(z) = const (22)These relations uniquely determine the functionsY+(z) andY–(z)

Thus, let us apply the above scheme to the solution of Eq (19) It follows from the above

reasoning that the analyticity domains of the functionsY+(z), Y–(z), andW(z) = 1 – √2πK(z),

respectively, are the upper half-plane Imz > µ, the lower half-plane Im z < v+, and the strip

v– < Imz < v+ Therefore, this equation holds in the strip*µ < Im z < v+, which is the common

analyticity domain for all functions that enter the equation In order to transform Eq (19) to the

form (21), we assume that it is possible to decompose the functionW(z) as follows:

W(z) = W+(z)

where the functionsW+(z) andW–(z) are analytic for Im z > µ and Im z < v+, respectively Moreover,

we assume that, in the corresponding analyticity domains, these functions grow at infinity no faster

thanz n, wheren is a positive integer A representation of an analytic function W(z) in the form (23)

is often called a factorization of W(z).

Thus, as the result of factorization, the original equation is reduced to the form (21) It follows

from the above reasoning that this equation determines an entire function of the complex variablez.

Since Y ±(z)→ 0 as |z| → ∞ and the growth of the functions W ±(z) does not exceed that

of a power function z n, it follows that the entire function under consideration can be only a

polynomialP n–1(z) of degree at most n – 1

If the growth of the functionsW ±(z) at infinity is only linear with respect to the variable z, then

it follows from relations (22), by virtue of the Liouville theorem (see Subsection 10.4-3), that the

corresponding entire function is a constantC In this case we obtain the following relations for the

unknown functionsY+(z) andY–(z):

which define the transform of the solution up to a constant factor, which can be found at least from

the normalization conditions In the general case, the expressions

Y+(z) = P n–1(z)

W+(z) , Y–(z) = –P n–1(z)

define the transform of the desired solution of the integral equation (11) up to indeterminate constants,

which can be found from the additional conditions of the problem The solution itself is defined by

means of the Fourier inversion formula (6), (9), and (10)

Example 2 Consider the equation

y(x) = λ

 ∞

0

whose kernel has the formK(x) = λe–|x|.

Let us find the transform of the functionK(x):

1 – 2λ, and √

1 – 2λ ≤ µ < 1 For λ > 1

2 , the functionW+(z) is analytic

and nonzero in the domain Imz > 0 It is clear that the function W–(z) is a nonzero analytic function in the domain Im z < 1.

Therefore, for 0 <λ <1 both functions satisfy the required conditions in the domainµ < Im z < 1.

Forλ > 1 , the strip 0 < Imz < 1 is the common domain of analyticity of the functions W+ (z) and W– (z) Thus, we

have obtained the desired factorization of the function (28).

Consider the expressionsY ±(z) W ±(z) Since Y ±(z) → 0 as |z| → ∞, and, according to (29), the growth of the

functionsW ±(z) at infinity is linear with respect to z, it follows that the entire function P n–1(z) that coincides with

Y+ (z) W+ (z) for Im z > µ and with Y– (z) W– (z) for Im z < 1 can be a polynomial of zero degree only Therefore,

On closing the integration contour forx > 0 by a semicircle in the lower half-plane and estimating the integral over this

semicircle by means of the Jordan lemma (see Subsections 7.1-4 and 7.1-5), after some calculations we obtain

y+(x) = C

 cos(√

11.10-3 The General Scheme of the Method The Factorization Problem

In the general case, the problem which is solved by the Wiener–Hopf method can be reduced to the

following problem It is required to find functionsY+(z) andY–(z) of the complex variable z that

are analytic in the half-planes Imz > v–and Imz < v+, respectively (v–<v+), vanish as|z| → ∞ in

their analyticity domains, and satisfy the following functional equation in the strip (v–< Imz < v+):

HereA(z), B(z), and C(z) are given functions of the complex variable z that are analytic in the strip

v–< Imz < v+, and the functionsA(z) and B(z) are nonzero in this strip.

The main idea of the solution of this problem is based on the possibility of a factorization of the

expressionA(z)/B(z), i.e., of a representation in the form

A(z) B(z) =

W+(z)

where the functions W+(z) andW–(z) are analytic and nonzero in the half-planes Im z > v– and

Imz < v+, and the stripsv–< Imz < v+andv– < Imz < v+ have a nonempty common part In this

case Eq (34), with regard to Eq (35), can be rewritten in the form

where the functions D+(z) and D–(z) are analytic in the half-planes Im z > v– and Imz < v +,

respectively, and all three stripsv–< Imz < v+,v– < Imz < v+, andv– < Imz < v+have a nonempty

common part, for a stripv0< Imz < v0

+, then, in this common strip, the following functional equationholds:

W+(z)Y+(z) +D+(z) = –W–(z)Y–(z) –D–(z) (38)The left-hand side of Eq (38) is a function analytic in the half-planev0< Imz, and the right-hand

side is a function analytic in the domain Imz < v0

+ Since these functions coincide in the strip

v0< Imz < v0

+, it follows that there exists a unique entire function that coincides with the left-hand

side and the right-hand side of (38) in their analyticity domains, respectively If the growth at infinity

of all functions that enter the right-hand sides of Eqs (35) and (37), in their analyticity domains,

is at most that ofz n, then it follows from the limit relationY ±(z)→ 0 as |z| → ∞ that this entire

function is a polynomialP n–1(z) of degree at most n – 1 Thus, the relations

Y+(z) = P n–1(z) –D+(z)

W+(z) , Y–(z) = –P n–1(z) –D–(z)

determine the desired functions up to constants These constants can be found from the additional

conditions of the problem

The application of the Wiener–Hopf method is based on the representations (35) and (37) If a

functionG(z) is analytic in the strip v– < Imz < v+and if in this strip the functionG(z) uniformly

tends to zero as|z| → ∞, then in this strip the following representation is possible:

where the functionG+(z) is analytic in the half-plane Im z > v–, the functionG–(z) is analytic in the

The integrals (41) and (42), being regarded as integrals depending on a parameter, define analytic

functions of the complex variablez under the assumption that the point z does not belong to the

integration contour

In particular,G+(z) is an analytic function in the half-plane Im z > v–andG–(z) in the half-plane

Imz > v +

Moreover, if a functionH(z) is analytic and nonzero in the strip v–< Imz < v+and ifH(z) → 1

uniformly in this strip as|z| → ∞, then the following representation holds in the strip:

H+(z) = exp

12πi

H–(z) = exp

– 12πi

where the functionsH+(z) andH–(z) are analytic and nonzero in the half-planes Im z > v– and

Imz < v+, respectively The representation (43) is called a factorization of the function H(z).

11.10-4 The Nonhomogeneous Wiener–Hopf Equation of the Second Kind

Consider the Wiener–Hopf equation of the second kind

y(x) –

 ∞

0

Suppose that the kernelK(x) of the equation and the right-hand side f (x) satisfy conditions (15).

Let us seek the solutiony+(x) to Eq (46) for which condition (17) is satisfied

In this case, reasoning similar to that in the derivation of the functional equation (19) for a

homogeneous integral equation shows that, in the case of Eq (46), the following functional equation

must hold on the stripµ < Im z < v+:

Y+(z) +Y–(z) =√

or

whereW(z) is subjected to condition (20), as well as in the case of a homogeneous equation.

We now note that Eq (48) is a special case of Eq (34) In the strip v– < Imz < v+, the

functionW(z) is analytic and uniformly tends to 1 as |z| → ∞ because |K(z)| → 0 as |z| → ∞ In

this case, this function has the representation (see (43)–(45))

W(z) = W+(z)

where the functionW+(z) is analytic in the upper half-plane Im z > v–andW–(z) is analytic in the

lower half-plane Imz < v+, and the growth at infinity of the functionsW ±(z) does not exceed that

respectively

To establish the possibility of a representation (51), we note that the functionF+(z) is analytic

in the upper half-plane Imz > v–and uniformly tends to zero as|z| → ∞ The function W–(z) is

analytic in the lower half-plane Imz < v+, and, according to the method of its construction, we can

perform the factorization (49) so that the functionW–(z) remains bounded in the strip v–< Imz < v+

as|z| → ∞ Hence (see (40)–(42)), the functions F+(z)W–(z) in the strip v–< Imz < v+satisfy all

conditions that are sufficient for the validity of the representation (51)

The above reasoning makes it possible to take into account the fact that the growth at infinity of

the functionsW ±(z) does not exceed that of zn, and thus to present the transform of the solution of

the nonhomogeneous integral equation (46) in the form

11.10-5 The Exceptional Case of a Wiener–Hopf Equation of the Second Kind

Consider the exceptional case of a Wiener–Hopf equation of the second kind in which the

func-tionW(z) = 1 – √2πK(z) has finitely many zeros N (counted according to their multiplicities) in

the stripv– < Imz < v+ In this case, the factorization is also possible To this end, it suffices to

introduce the auxiliary function

W1(z) = ln

(z2+b2)N/2 W(z)

i

(z – zi)–α i



whereα iis the multiplicity of the zeroz iand a positive constantb > {|v–|, |v+|} is chosen so that the

function in the square brackets has no additional zeros in the stripv–< Imz < v+

However, in the exceptional case, the Wiener–Hopf method gives the answer only if the number

of zeros of the functionW(z) is even This restriction is due to the fact that only for the case in

which the number of zeros is even is it possible to achieve the necessary behavior at infinity (for

the application of the Wiener–Hopf method) of the function (z2+b2)N/2 (see F D Gakhov and

Yu I Cherskii (1978)) The last restriction makes no real obstacle to the broad use of the Wiener–

Hopf method in solving applied problems in which the kernelK(x) of the corresponding integral

equation is frequently an even function, and thus the reasoning below can be applied completely

Remark 1 The Wiener–Hopf equation of the second kind for functions vanishing at infinity can

be reduced to a Riemann boundary value problem on the real axis (see Subsection 11.9-1) In this

case, the assumption that the number of zeros of the functionW(z) is even, as well as the assumption

that the kernelK(x) is even in the exceptional case, are unessential.

Remark 2 For functions with nontrivial growth at infinity, the complete solution of Wiener–

Hopf equations of the second kind is presented in the cited book by F D Gakhov and Yu I Cherskii

(1978)

of the first kind under the assumption that the kernels of these equations are even

• References for Section 11.10: B Noble (1958), A G Sveshnikov and A N Tikhonov (1970), V I Smirnov (1974),

F D Gakhov (1977), F D Gakhov and Yu I Cherskii (1978).

11.11 Krein’s Method for Wiener–Hopf Equations

11.11-1 Some Remarks The Factorization Problem

Consider the Wiener–Hopf equation of the second kind

y(x) –

 ∞

0

wheref (x), y(x) ∈ L1(0,∞) and K(x) ∈ L1(–∞, ∞) Let us use the classes of functions that

can be represented as Fourier transforms (alternative Fourier transform in the asymmetric form, see

Subsection 7.4-3), of functions fromL1(–∞, ∞), L1(0,∞), and L1(–∞, 0) For brevity, instead of

these symbols we simply writeL, L+, andL– Let functionsh(x), h1(x), and h2(x) belong to L, L+,

andL–, respectively; in this case, their transforms can be represented in the form

LetT (x) belong to L and let ˇ T (u) be its transform Assume that

1 – ˇT (u) ≠ 0, Ind[1 – ˇT (u)] = 1

This formula readily implies the relation ln[1 – ˇT (u)] → 0 as u → ±∞.

In what follows, we apply the factorization of functions ˇ M(u) of the class Q that are continuous

on the interval –∞ ≤ u ≤ ∞ Here the factorization means a representation of the function ˇM(u) in

the form of a product

whereMˇ–(z) and Mˇ+(z) are analytic functions in the corresponding half-planes Im z > 0 and

Imz < 0 continuous up to the real axis Moreover,

ˇ

M+(z)≠ 0 for Im z ≥ 0 and Mˇ–(z)≠ 0 for Im z ≤ 0. (6)

Relation (5) implies the formula

k = Ind ˇ M(u).

The factorization (5) is said to be canonical provided that k = 0.

In what follows we consider only functions of the form

Let us state the main results concerning the factorization problem

A function (7) admits a canonical factorization if and only if the following two conditions hold:

ˇ

In this case, the canonical factorization is unique Moreover, if conditions (9) hold, then there exists

a functionM (x) in the class L such that

 0 –∞

M (x)e iux dx



Hence, we have ˇM(u) ∈ Q and ˇ M ±(u)∈ Q ± The factors in the canonical factorization are also

described by the following formulas:

In the general case of the factorization, the following assertion holds A function (7) admits a

factorization (5) if and only if the following condition is satisfied:

M(u) = ˇ M–(u) ˇM+(u), –∞ < u < ∞

The last relation implies the canonical factorization for the function

11.11-2 The Solution of the Wiener–Hopf Equations of the Second Kind

THEOREM1 For Eq (1) to have a unique solution of the classL+for an arbitraryf (x) ∈ L+, it

is necessary and sufficient that the following conditions hold:

THEOREM2 If condition (15) holds, then the inequalityν > 0is necessary and sufficient for the

existence of nonzero solutions in the classL+of the homogeneous equation

y(x) –

 ∞

0

The set of these solutions has a basis formed byν functionsϕ k(x)(k = 1, , ν) that tend to zero

asx → ∞and that are related as follows:

whereCis a nonzero constant and the functionsϕ k(t)andψ(t)belong toL+

THEOREM3 If condition (15) holds and ifν > 0, then for anyf (x) ∈ L+Eq (1) has infinitely

many solutions inL+

However, ifν < 0, then, for a given f (x) ∈ L+, Eq (1) has either no solutions fromL+or a unique

solution For the latter case to hold, it is necessary and sufficient that the following conditions be

1◦ If conditions (15) and (16) hold, then there exists a unique factorization

[1 – ˇK(u)]–1= ˇM+(u) ˇM–(u), (21)and

where 0≤ x < ∞, 0 ≤ t < ∞, R+(x) = 0, and R–(x) = 0 for x < 0, so that, for f (x) from L+, the

solution of the equation is determined by the expression

y(x) = f (x) +

 ∞

0

Formula (23) can be rewritten as follows:

we have the representation (22) and formula (23) for the resolvent

Moreover, fork = 1, , ν, the following representations hold:

i k Mˇ+(u)(u – i)k =

 ∞

0

whereg k(x) is the solution of the homogeneous equation (17) The solutions ϕk(x) mentioned in

Theorem 2 can also naturally be expressed via the functionsg k(x)

3◦ Ifν = – Ind[1 – ˇ K(u)] < 0, then the transposed equation

y(x) –

 ∞

0

has the index –ν > 0 If formula (28) defines a factorization for Eq (1), then the transposed equation

admits a factorization of the form

[1 – ˇK(u)]–1= ˇM–(–u) ˇM+(–u),and ˇM–(–u) plays the role of ˇM+(u), and ˇM+(–u) plays the role of ˇM–(u)

11.11-3 The Hopf–Fock Formula

Let us give a useful formula that allows one to express the solution of Eq (1) with an arbitrary

right-hand sidef (x) via the solution to a simpler auxiliary integral equation with an exponential

On the basis of formula (34), we can obtain the solution of Eq (1) for a generalf (x) as well

(see also Section 9.6):

Remark 1 All results obtained in Section 11.11 concerning Wiener–Hopf equations of the

sec-ond kind remain valid for continuous, square integrable, and some other classes of functions, which

are discussed in detail in the paper by M G Krein (1958) and in the book by C Corduneanu (1973)

Remark 2 The solution of the Wiener–Hopf equation can be also obtained in other classes of

functions for the exceptional case in which 1 – ˇK(u) = 0 (see Subsections 11.9-1 and 11.10-5).

• References for Section 11.11: V A Fock (1942), M G Krein (1958), C Corduneanu (1973), V I Smirnov (1974),

P P Zabreyko, A I Koshelev, et al (1975).

11.12 Methods for Solving Equations With Difference

Kernels on a Finite Interval

11.12-1 Krein’s Method

Consider a method for constructing exact analytic solutions of linear integral equations with an

arbitrary right-hand side The method is based on the construction of two auxiliary solutions of

simpler equations with the right-hand side equal to 1 The auxiliary solutions are used to construct

a solution of the original equation for an arbitrary right-hand side