Application of the Fourier Transform to the Solution of Singular Integral ...

Mathematical Notes, vol. 74, no. 5, 2003, pp. 695–700. Translated from Matematicheskie Zametki, vol. 74, no. 5, 2003, pp. 739–744. c Original Russian Text Copyright
2003 by I. A. Soldatenkov.

Application of the Fourier Transform to the Solution of Singular Integral Equations I. A. Soldatenkov Received December 19, 2001; in final form, January 14, 2002

Abstract—A method for solving a certain system of singular integral equations with constant coefficients is proposed. It is based on a procedure for reducing singular equations to equations with continuous difference kernel; the solution of the latter is constructed by using the classical Fourier transform in the class of absolutely integrable functions. Explicit expressions for the solution of the singular integral equations under consideration are obtained. Key words: Fourier transform, singular integral equation, continuous difference kernel.

1. INTRODUCTION This paper is concerned with the construction in closed form of the solution of the following system of linear singular equations with constant coefficients:


a

Aq(x) + B −a

q(ξ) dξ = f (x), ξ−x

x ∈ (−a, a),

(1)

where A and B are constant N × N matrices, N = 1, 2, . . . , f (x) and q(x) are the given and unknown vector functions of dimension N , respectively, and x is a real variable. The classical approach to the solution of systems of the form (1) is based on their reduction to the linear conjugation problem for several functions of a complex variable [1 2]. Using this approach, one can construct solutions of more general (than (1)) systems of singular integral equations involving, in particular, x-dependent matrices A and B as well as more complicated contours of integration. However, the corresponding algorithms for constructing solutions include cumbersome procedures for determining the canonical solutions of the conjugation problem; as a result, the construction in closed form of the solution of Eq. (1) on the basis of the classical approach gives rise to difficulties. 2. MAIN RELATIONS The notation ϕ(x) ∈ H ∗ [−a, a] means that ϕ(x) = ϕ∗ (x)(a + x)−ε1 (a − x)−ε2 ,

ϕ∗ (x) ∈ H[−a, a],

εk < 1,

(2)

where H[−a, a] is the H¨ older class [1]. Unless stated otherwise, we assume that the components of the vector functions f (x) and q(x) are contained in H ∗ [−a, a] . Without loss of generality, we assume that the components of f , q , A , and B are real; moreover, det A = 0 . Let us transform system (1). We multiply it by (x − t)−1 for some t ∈ (−a, a) and integrate over x ∈ [−a, a] . Next, setting det A = 0 , we multiply the resulting relation by BA−1 and again 0001-4346/2003/7456-0695 $25.00

c
2003 Plenum Publishing Corporation

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I. A. SOLDATENKOV

using (1), we replace the integral B equation

a

q(x)/(x − t) dx in it by f (t) − Aq(t) . Then we obtain the

−a




a

dt t−x

Aq(x) + n −a




a

q(ξ) dξ = d(x), ξ−t

(3)

f (ξ) dξ ∈ H ∗ [−a, a]. ξ−x

(4)

−a

in which n = −BA−1 B and −1

d(x) = f (x) − BA




a −a

Proposition 1. The solution set of Eq. (3) contains all solutions of the original equation (1). In what follows, we shall also use the following expression equivalent to (3), which can be obtained by interchanging the order of integration in (3) on the basis of the Poincar´e–Bertrand formula [1]:
a

k(ξ , x)q(ξ) dξ = d(x),

mq(x) + n

(5)

−a

where

  a + x  . L(x) = ln a − x

L(ξ) − L(x) , k(ξ , x) = ξ−x

2

m = A − π n,

∗ 3. SOLUTION OF EQ. (3) OF CLASS H1/2 ∗ The notation ϕ(x) ∈ H1/2 [−a, a] means that the function ϕ(x) can be expressed in the form (2) ∗ for εk < 1/2 . Here we assume that f (x) and q(x) belong to H1/2 [−a, a] ; in addition, we assume ∗ that the function d(x) is also contained in H1/2 [−a, a] [1]: ∗ [−a, a]. {f (x), q(x), d(x)} ∈ H1/2

(6)

We introduce a new variable s and the corresponding functions 2s = L(x),

x = a tanh s,

Q(s) = q(a tanh s),

D(s) = d(a tanh s).

(7)

Then Eq. (5) can be rewritten as


∞

mΦ(s) + 2n −∞

(s − η) Φ(η) dη = R(s), sinh(s − η)

(8)

where Φ(s) = aQ(s)/ cosh s , R(s) = aD(s)/ cosh s ; by (6), we have {Φ(s), R(s)} ∈ H(−∞, ∞) ∩ L(−∞, ∞).

(9)

Here H(−∞, ∞) denotes the class of functions satisfying H¨ older’s condition in a neighborhood of any point of the real axis. Further, we use the classical Fourier transform [3]: 1 f˜(λ) = √ 2π




∞

iλs

f (s)e −∞

dλ,

1 f (s) = √ v. p. 2π




∞

f˜(λ)e−iλs dλ

−∞

and the following relations for the convolution (f ∗ g)(s): (f ∗ g)(s) ∈ C(−∞, ∞) ∩ L(−∞, ∞),


(f ∗ g)(λ) = f˜(λ)˜ g (λ)

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for {f (s), g(s)} ∈ C(−∞, ∞) ∩ L(−∞, ∞) , |f (s)| ≤ const ;
∞ 1 (f ∗ g)(s) = √ v. p. f˜(λ)˜ g (λ)e−iλs dλ 2π −∞

(11)

for {f (s), g(s), f˜(λ)} ∈ L(−∞, ∞) , f (s) ∈ H(−∞, ∞) , g(s) ∈ C(−∞, ∞) . Let us take the Fourier transform of both sides of Eq. (8), use the second relation in (10), and take (9) into account. As a result, denoting π ζ(λ) = cosh2 λ, Ka (λ) = mζ(λ) + π 2 n, 2 we obtain the relation   = ζ(λ)R(λ), Ka (λ)Φ(λ) in which, in view of (9), we have [3]  R(λ) ∈ C(−∞, ∞),

 lim R(λ) = 0,

λ→±∞

 |R(λ)| ≤ const .

 The resulting relation is an algebraic equation for the Fourier image Φ(λ) of the unknown function Φ(s) , whose condition for solvability is of the form [4] det Ka (λ) = 0,

λ ∈ (−∞, ∞).

˜ a (λ) = K −1 (λ) and In this case, there exists a matrix h a ˜ a (λ)R(λ).   Φ(λ) = ζ(λ)h

(12)

˜ a (λ) = M (λ)/ det Ka (λ) , where M (λ) is the matrix of N × N algebraic compleExpressing h ments of the elements of the matrix Ka (λ) , and setting det m = 0 , we can establish the following ˜ a (λ): properties of h ˜ (k) (λ) ∈ C(−∞, ∞) ∩ L(−∞, ∞), (1 + |λ|l )h a l ˜ (k) k, l = 0, 1, . . . , lim λ h (λ) = 0, λ→±∞

(13)

a

˜ a (λ) such that [3] which imply the existence of the Fourier preimage ha (s) of the matrix function h lim sk h(l) a (s) = 0,

h(l) a (s) ∈ C(−∞, ∞) ∩ L(−∞, ∞),

s→±∞

k, l = 0, 1, 2, . . . .

(14)

˜ a (λ)π 2 nm−1 R(λ)    −h and take the inverse Fourier Let us express (12) as Φ(λ) = m−1 R(λ)  transform of both sides, invoking the properties (discussed above) of the function R(λ) as well as taking into account (9), (11), (13), and (14). As a result, returning to the functions Q(s) = a−1 Φ(s) cosh s

and

we can find the following expression for Q(s): −1

Q(s) = m

1 D(s) − √ cosh s 2π




∞

−∞

D(s) = a−1 R(s) cosh s,

ha (s − η)π 2 nm−1 D(η)

dη . cosh η

(15)

The expression (15) satisfies assumption (6) if we strengthen the first relation in (14) and require the exponential decay at infinity of the well-known matrix function ha (s) and of its derivative: −β h(l) s), a (s) = O(cosh

β > 0,

l = 0, 1.

(16)

Below we shall examine the question of whether expression (15) can be used as the solution of the original equation (1).

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4. SOLUTION OF EQ. (3) OF CLASS H ∗ Further, instead of (6) we assume {f (x), q(x), d(x)} ∈ H ∗ [−a, a].

(17)

To construct the solution of Eq. (3) of class H ∗ [−a, a] , we carry out an identical transformation of this equation. Namely, we invert the doubly iterated integral in (3) using the well-known formula [1, 5]: 
a 
a  2
1 a − ξ 2 ψ(ξ) 1 a ϕ(ξ) dξ ϕ(x) dx − = ψ(x) ⇐⇒ ϕ(x) = √ dξ . π −a ξ − x ξ−x π a2 − x2 −a −a As a result, we obtain
 A a 2 a − ξ 2 q(ξ)k(ξ , x) dξ m a2 − x2 q(x) − 2 π −a 
a  1 1 0 1 2 2 2 2 a − ξ d(ξ)k(ξ , x) dξ − πnP + nP L(x) , = a − x d(x) − 2 π −a π where 0

P ≡




a

1




P ≡

q(x) dx, −a




a

a

dx −a

−a

q(ξ) dξ = ξ−x




(18)

a

q(x)L(x) dx.

(19)

−a

As in the previous section, we can introduce a new variable s and the corresponding functions Q(s) and D(s) (see (7)) as well as the functions Ψ(s) =

a2 Q(s) , cosh2 s

W (s) =

a2 D(s) . cosh2 s

Using these functions, we can express Eq. (18) in a form similar to (8); next, as in the previous section, taking its Fourier transform, we can obtain the following expression for Q(s):
∞ 1 dη 2 −1 hb (s − η)(Am−1 − E)D(η) Q(s) = m D(s) + √ cosh s cosh2 η 2π −∞ 1 − 2 cosh2 s[Γ0 (s)P 0 + Γ1 (s)P 1 ], (20) a where E is the unit matrix, hb (s) is the Fourier preimage of the matrix function ˜ b (λ) = K −1 (λ), h b

Kb (λ) = mζ(λ) − A,

Γ0 (s) and Γ1 (s) are the Fourier preimages of the matrix functions   ˜ b (λ)n, ˜ b (λ)n.  0 (λ) = π/2πa cosh(πλ/2)h  1 (λ) = i π/2a sinh(πλ/2)h Γ Γ ˜ b (λ) is guaranteed by the condition [4] The existence of a function h det Kb (λ) = 0,

λ ∈ (−∞, ∞)

and expression (18) satisfies assumption (17) if (l)

hb (s) = O(cosh−β s),

(l)

Γ0,1 (s) = O(cosh−ε s),

β > 0,

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5. SOLUTION OF THE ORIGINAL EQUATION (1) ∗ [−a, a] is of the form (15), The passage from Eq. (1) to Eq. (3), whose solution q(x) ∈ H1/2 was carried out under the assumptions det Ka (λ) = 0 , λ ∈ (−∞, ∞) , (6), and (16), while the solution q(x) ∈ H ∗ [−a, a] is defined by formula (20) under the assumptions det Kb (λ) = 0 , λ ∈ (−∞, ∞) , (17), and (21); moreover, in both cases we assumed that det A = 0 and det m = 0 . By Proposition 1, any solution of Eq. (1) satisfies (3). Let us examine the possibility of using formulas (15) and (20) for solving the original equation (1).

Remark 1. Formula (15) uniquely defines the solution of Eq. (3), while formula (20) depends on two constant vectors P 0 and P 1 whose choice is arbitrary and leads to the solution q(x) of Eq. (3) satisfying conditions (19). A similar result was obtained in [1, 2]. ∗ [−a, a] of Proposition 3. In the case det Ka (λ) = 0 , λ ∈ (−∞, ∞) , the solution q(x) ∈ H1/2 ∗ [−a, a] satisfies Eq. (1). In the case det Kb (λ) = 0 , λ ∈ (−∞, ∞) , the Eq. (3) for f (x) ∈ H1/2 solution q(x)∈H ∗ [−a, a] of Eq. (3) for f (x) ∈ H ∗ [−a, a] satisfies Eq. (1) provided that
a 0 1 f (x) dx, det B = 0. (22) AP + BP = −a

By Proposition 3, we can use formulas (15) and (20) for solving the original equation (1); the constants P 0 and P 1 in (20) turn out to be interrelated. 6. EXAMPLE Consider Eq. (1) in the form occurring, for example, in problems of elasticity theory [6, 7]. Suppose that N = 2 , and the elements Aij and Bij of the matrices A and B are defined as follows: Aij = Bij = 0,

except

A11 = A1 ,

A22 = A2 ,

B12 = B1 ,

B21 = B2 ,

where the Ai and Bi are constants. In this case, the elements of the matrices n and m are as follows: nij = mij = 0,

B1 B2 B1 B2 , n22 = − , A2 A1 = m1 ≡ A1 − π 2 n1 , m22 = m2 ≡ A2 − π 2 n2 .

n11 = −

except

m11

For the conditions det A =  0 , det m = 0 , and the last condition in (22) to hold, we put Ak = 0 , mk = 0 , Bk = 0 . The functions ha,b (s) and Γ0,1 (s) in the case under consideration are of the form ha,b (s) =

(2/π)1/2 Za,b (s) −1

m , 2 ωa,b 1 ± ωa,b Γ1 (s) =

πa cosh Ωb s Γ0 (s) =  nm−1 , 1 − ωb2 cosh s

a sinh Ωb s nm−1 , ωb cosh s

where m−1 the matrix inverse to m and A1 A2 A1 A2 −1≡ − 1, ωb2 = ≡ , m1 m2 m1 m2 2 2 Ωb = arcsin ωb , Ωa = asinh ωa , π π sin Ωa s sinh Ωb s , Zb (s) = . Za (s) = sinh s sinh s

ωa2 =

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The conditions det Ka (λ) = 0 and det Kb (λ) = 0 , where λ ∈ (−∞, ∞) , hold if 0