where a 0 , a 1 , , a n are the coefficients of the polynomial U n (z), and the a –k are the coefficients of the expansion of the function Ψ(z), which are given by the obvious formula a –k = – 1 2πi  L κ  j=1 (τ – β j ) p j H(τ)τ k–1 X + (τ) dτ. The solvability conditions acquire the form a n = a n–1 = ···= a n–p+ν+2 =0. If a solution must satisfy the additional condition Φ – (∞) = 0, then, for ν – p > 0, in formulas (53) we must take the polynomial P ν–p–1 (z), and for ν – p <0,p – ν conditions must be satisfied. 12.3-10. The Riemann Problem for a Multiply Connected Domain Let L = L 0 + L 1 + ···+ L m be a collection of m + 1 disjoint contours, and let the interior of the contour L 0 contain the other contours. By Ω + we denote the (m + 1)-connected domain interior for L 0 and exterior for L 1 , , L m .ByΩ – we denote the complement of Ω + + L in the entire complex plane. To be definite, we assume that the origin lies in Ω + . The positive direction of the contour L is that for which the domain Ω + remains to the left, i.e., the contour L 0 must be traversed counterclockwise and the contours L 1 , , L m , clockwise. We first note that the jump problem Φ + (t) – Φ – (t)=H(t) is solved by the same formula Φ(z)= 1 2πi  L H(τ) dτ τ – z as in the case of a simply connected domain. This follows from the Sokhotski–Plemelj formulas, which have the same form for a multiply connected domain as for a simply connected domain. The Riemann problem (homogeneous and nonhomogeneous) can be posed in the same way as for a simply connected domain. We write ν k = 1 2π [arg D(t)] L k (all contours are passed in the positive direction). By the index of the problem we mean the number ν = m  k=0 ν k . (54) If ν k (k =1, , m) are zero for the inner contours, then the solution of the problem has just the same form as for a simply connected domain. To reduce the general case to the simplest one, we introduce the function m  k=1 (t – z k ) ν k , where the z k are some points inside the contours L k (k =1, , m). Taking into account the fact that [arg(t – z k )] L j = 0 for k ≠ j and [arg(t – z j )] L j = –2π, we obtain 1 2π  arg m  k=1 (t – z k ) ν k  L j = 1 2π  arg(t – z j ) ν j  L j = –ν j , j =1, , m. Page 621 © 1998 by CRC Press LLC © 1998 by CRC Press LLC Hence,  arg  D(t) m  k=1 (t – z k ) ν k  L j =0, j =1, , m. Let us calculate the increment of the argument of the function D(t) m  k=1 (t – z k ) ν k with respect to the contour L 0 : 1 2π  arg  D(t) m  k=1 (t – z k ) ν k  L 0 = 1 2π  arg D(t)  L 0 + 1 2π m  k=1 [ν k arg(t – z k )] L 0 = ν 0 + m  k=1 ν k = ν. Since the origin belongs to the domain Ω + , it follows that [arg t] L k =0, k =1, , m, [arg t] L 0 =2π. Therefore,  arg  t –ν m  k=1 (t – z k ) ν k D(t)  L = 0. (55) 1 ◦ . The Homogeneous Problem. Let us rewrite the boundary condition Φ + (t)=D(t)Φ – (t) (56) in the form Φ + (t)= t ν m  k=1 (t – z k ) ν k  t –ν m  k=1 (t – z k ) ν k D(t)  Φ – (t). (57) The function t –ν m  k=1 (t – z k ) ν k D(t) has zero index on each of the contours L k (k =1, , m), and hence it can be expressed as the ratio t –ν m  k=1 (t – z k ) ν k D(t)= e G + (t) e G – (t) , (58) where G(z)= 1 2πi  L ln  τ –ν m  k=1 (τ – z k ) ν k D(τ)  dτ τ – z . (59) The canonical function of the problem is given by the formulas X + (z)= m  k=1 (z – z k ) –ν k e G + (z) , X – (z)=z –ν e G – (z) . (60) Now the boundary condition (57) can be rewritten in the form Φ + (t) X + (t) = Φ – (t) X – (t) . Page 622 © 1998 by CRC Press LLC © 1998 by CRC Press LLC As usual, by applying the theorem on analytic continuation and the generalized Liouville theorem (see Subsection 12.3-1), we obtain Φ + (z)= m  k=1 (z – z k ) –ν k e G + (z) P ν (z), Φ – (z)=z –ν e G – (z) P ν (z). (61) We can see that this solution differs from the above solution of the problem for a simply connected domain only in that the function Φ + (z) has the factor m  k=1 (z – z k ) –ν k . Under the additional condition Φ – (∞) = 0, in formulas (61) we must take the polynomial P ν–1 (z). Applying the Sokhotski–Plemelj formulas, we obtain G ± (t)=± 1 2 ln[t –ν Π(t)D(t)] + G(t), where G(t) is the Cauchy principal value of the integral (59) and Π(t)= m  k=1 (t – z k ) ν k . On passing to the limit as z → t in formulas (60) we obtain X + (t)=  D(t) t ν Π(t) e G(t) , X – (t)= 1 √ t ν Π(t)D(t) e G(t) . (62) The sign of the root is determined by the (arbitrary) choice of a branch of the function ln[t –ν Π(t)D(t)]. 2 ◦ . The Nonhomogeneous Problem. By the same reasoning as above, we represent the boundary condition Φ + (t)=D(t)Φ – (t)+H(t) (63) in the form Φ + (t) X + (t) – Ψ + (t)= Φ – (t) X – (t) – Ψ – (t), where Ψ(z)isdefined by the formula Ψ(z)= 1 2πi  L H(τ) X + (τ) dτ τ – z . This gives the general solution Φ(z)=X(z)[Ψ(z)+P ν (z)] (64) or Φ(z)=X(z)[Ψ(z)+P ν–1 (z)], (65) if the solution satisfies the condition Φ – (∞)=0. For ν < 0, the nonhomogeneous problem is solvable if and only if the following conditions are satisfied:  L H(t) X + (t) t k–1 dt = 0, (66) where k ranges from 1 to –ν – 1 if we seek solutions bounded at infinity and from 1 to –ν if we assume that Φ – (∞)=0. Under conditions (66), the solution can also be found from formulas (64) or (65) by setting P ν ≡ 0. If the external contour L 0 is absent and the domain Ω + is the plane with holes, then the main difference from the preceding case is that here the zero index with respect to all contours L k (k =1, , m) is attained by the function m  k=1 (t – z k ) ν k D(t) that does not involve the factor t –ν . Therefore, to obtain a solution to the problem, it suffices to repeat the above reasoning on omitting this factor. Page 623 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 12.3-11. The Cases of Discontinuous Coefficients and Nonclosed Contours Assume that the functions D(t) and H(t) in the boundary condition of the Riemann problem (63) satisfy the H ¨ older condition everywhere on L except for points t 1 , , t m at which these functions have jumps, and assume that L is a closed curve. None of the limit values vanishes, and the boundary condition holds everywhere except for the discontinuity points at which it makes no sense. A solution to the problem is sought in the class of functions that are integrable on the contour. Therefore, a solution is everywhere continuous, in the sense of the H ¨ older condition, possibly except for the points t k . For these points, there are different possibilities. 1 ◦ . We can assume boundedness at all discontinuity points, and thus seek a solution that is every- where bounded. 2 ◦ . We can assume that a solution is bounded at some discontinuity points and admit an integrable singularity at the other discontinuity points. 3 ◦ . We can admit integrable singularity at all points which are admitted by the conditions of the problem. The first class of solutions is the narrowest, the second class is broader, and the third class is the largest. The number of solutions depends on the class in which it is sought, and it can turn out that a problem that is solvable in a broader class is unsolvable in a narrower class. We make a few remarks on the Riemann problem for nonclosed contours. Assume that a contour L consists of a collection of m simple closed disjoint curves L 1 , , L m whose endpoints are a k and b k (the positive direction is from a k to b k ). Assume that D(t) and H(t) are functions given on L and satisfy the H ¨ older condition, and D(t) ≠ 0 everywhere. It is required to find a function Φ(z) that is analytic on the entire plane except for the points of the contour L, and whose boundary values Φ + (t) and Φ – (t), when tending to L from the left and from the right, are integrable functions satisfying the boundary condition (63). As can be seen from the setting, the Riemann problem for a nonclosed contour principally differs from the problem for a closed contour in that the entire plane with the cut along the curve L forms a single domain, and instead of two independent analytic functions Φ + (z) and Φ – (z), we must find a single analytic function Φ(z) for which the contour L is the line of jumps. The problem posed above can be reduced to that for a closed contour with discontinuous coefficients. The details on the Riemann boundary value problem with discontinuous coefficients and non- closed contours can be found in the references cited below. 12.3-12. The Hilbert Boundary Value Problem Let a simple smooth closed contour L and real H ¨ older functions a(s), b(s), and c(s) of the arc length s on the contour be given. By the Hilbert boundary value problem we mean the following problem. Find a function f(z)=u(x, y)+iv(x, y) that is analytic on the domain Ω + and continuous on the contour for which the limit values of the real and the imaginary part on the contour satisfy the linear relation a(s)u(s)+b(s)v(s )=c(s ). (67) For c(s) ≡ 0 we obtain the homogeneous problem and, for nonzero c(s), a nonhomogeneous. The Hilbert boundary value problem can be reduced to the Riemann boundary value problem. The methods of this reduction can be found in the references cited at the end of the section. • References for Section 12.3: F. D. Gakhov (1977), N. I. Muskhelishvili (1992). Page 624 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 12.4. Singular Integral Equations of the First Kind 12.4-1. The Simplest Equation With Cauchy Kernel Consider the singular integral equation of the first kind 1 πi  L ϕ(τ) τ – t dτ = f(t), (1) where L is a closed contour. Let us construct the solution. In this relation we replace the variable t by τ 1 , multiply by 1 πi dτ 1 τ 1 – t , integrate along the contour L, and change the order of integration according to the Poincar ´ e–Bertrand formula (see Subsection 12.2-6). Then we obtain 1 πi  L f(τ 1 ) τ 1 – t dτ 1 = ϕ(t)+ 1 πi  L ϕ(τ) dτ 1 πi  L dτ 1 (τ 1 – t)(τ – τ 1 ) . (2) Let us calculate the second integral on the right-hand side of (2):  L dτ 1 (τ 1 – t)(τ – τ 1 ) = 1 τ – t   L dτ 1 τ 1 – t –  L dτ 1 τ 1 – τ  = 1 τ – t (iπ – iπ)=0. Thus, ϕ(t)= 1 πi  L f(τ) τ – t dτ. (3) The last formula gives the solution of the singular integral equation of the first kind (1) for a closed contour L. 12.4-2. An Equation With Cauchy Kernel on the Real Axis Consider the following singular integral equation of the first kind on the real axis: 1 πi  ∞ –∞ ϕ(t) t – x dt = f (x), –∞ < x < ∞. (4) Equation (4) is a special case of the characteristic integral equation on the real axis (see Subsec- tion 13.2-4). In the class of functions vanishing at infinity, Eq. (4) has the solution ϕ(x)= 1 πi  ∞ –∞ f(t) t – x dt, –∞ < x < ∞. (5) Denoting f(x)=F (x)i –1 , we rewrite Eqs. (4) and (5) in the form 1 π  ∞ –∞ ϕ(t) t – x dt = F (x), ϕ(x)=– 1 π  ∞ –∞ F (t) t – x dt. – ∞ < x < ∞. (6) The two formulas (6) are called the Hilbert transform pair (see Subsection 7.6-3). Page 625 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 12.4-3. An Equation of the First Kind on a Finite Interval Consider the singular integral equation of the first kind 1 π  b a ϕ(t) t – x dt = f (x), a ≤ x ≤ b, (7) on a finite interval. Its solutions can be constructed by using the theory of the Riemann boundary value problem for a nonclosed contour (see Subsection 12.3-11). Let us present the final results. 1 ◦ . A solution that is unbounded at both endpoints: ϕ(x)=– 1 π 1 √ (x – a)(b – x)   b a √ (t – a)(b – t) t – x f(t) dt + C  , (8) where C is an arbitrary constant and  b a ϕ(t) dt = C. (9) 2 ◦ . A solution bounded at the endpoint a and unbounded at the endpoint b: ϕ(x)=– 1 π  x – a b – x  b a  b – t t – a f(t) t – x dt. (10) 3 ◦ . A solution bounded at both endpoints: ϕ(x)=– 1 π  (x – a)(b – x)  b a f(t) √ (t – a)(b – t) dt t – x , (11) under the condition that  b a f(t) dt √ (t – a)(b – t) = 0. (12) Solutions that have a singularity point s inside the interval [a, b] can also be constructed. These solutions have the following form: 4 ◦ . A singular solution that is unbounded at both endpoints: ϕ(x)=– 1 π 1 √ (x – a)(b – x)   b a √ (t – a)(b – t) t – x f(t) dt + C 1 + C 2 x – s  , (13) where C 1 and C 2 are arbitrary constants. 5 ◦ . A singular solution bounded at one endpoint: ϕ(x)=– 1 π  (x – a)(b – x)   b a  b – t t – a f(t) t – x dt + C x – s  , (14) where C is an arbitrary constants. 6 ◦ . A singular solution bounded at both endpoints: ϕ(x)=– 1 π  (x–a)(b– x)   b a f(t) √ (t–a)(b– t) dt t–x + A x–s  , A =  1 –1 f(t) dt √ (t–a)(b– t) . (15) Page 626 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 12.4-4. The General Equation of the First Kind With Cauchy Kernel Consider the general equation of the first kind with Cauchy kernel 1 πi  L M(t, τ) τ – t ϕ(τ) dτ = f (t), (16) where the integral is understood in the sense of the Cauchy principal value and is taken over a closed or nonclosed contour L. As usual, the functions a(t), f(t), and M (t, τ)onL are assumed to satisfy the H ¨ older condition, where the last function satisfies this condition with respect to both variables. We perform the following manipulation with the kernel: M(t, τ) τ – t = M(t, τ) – M(t, t) τ – t + M(t, t) τ – t and write M(t, t)=b(t), 1 πi M(t, τ) – M(t, t) τ – t = K(t, τ). (17) We can rewrite Eq. (16) in the form b(t) πi  L ϕ(τ) τ – t dτ +  L K(t, τ)ϕ(τ) dτ = f(t). (18) It follows from formulas (17) that the function b(t) satisfies the H ¨ older condition on the entire contour L and K(t, τ) satisfies this condition everywhere except for the points with τ = t at which this function satisfies the estimate |K(t, τ)| < A |τ – t| λ ,0≤ λ <1. The general singular integral equation of the first kind with Cauchy kernel is frequently written in the form (18). The general singular integral equation of the first kind is a special case of the complete singular integral equation whose theory is treated in Chapter 13. In general, it cannot be solved in a closed form. However, there are some cases in which such a solution is possible. Let the function M(t, τ ) in Eq. (16), which satisfies the H ¨ older condition with respect to both variables on the smooth closed contour L by assumption, have an analytic continuation to the domain Ω + with respect to each of the variables. If M(t, t) ≡ 1, then the solution of Eq. (16) can be obtained by means of the Poincar ´ e–Bertrand formula (see Subsection 12.2-6). This solution is given by the relation ϕ(t)= 1 πi  L M(t, τ) τ – t f(τ) dτ. (19) Eq. (16) can be solved without the assumption that the function M(t, τ ) satisfies the condition M(t, t) ≡ 1. Namely, assume that the function M (t, τ ) has the analytic continuation to Ω + with respect to each of the variables and that M(z, z) ≠ 0 for z ∈ Ω + . In this case, the solution of Eq. (16) has the form ϕ(t)= 1 πi 1 M(t, t)  L M(t, τ) M(τ, τ) f(τ) τ – t dτ. (20) In Section 12.5, a numerical method for solving a special case of the general equation of the first kind is given, which is of independent interest from the viewpoint of applications. Remark 1. The solutions of complete singular integral equations that are constructed in Sub- section 12.4-4 can also be applied for the case in which the contour L is a collection of finitely many disjoint smooth closed contours. Page 627 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 12.4-5. Equations of the First Kind With Hilbert Kernel 1 ◦ . Consider the simplest singular integral equation of the first kind with Hilbert kernel 1 2π  2π 0 cot  ξ – x 2  ϕ(ξ) dξ = f (x), 0 ≤ x ≤ 2π, (21) under the additional assumption  2π 0 ϕ(x) dx = 0. (22) Equation (21) can have a solution only if a solvability condition is satisfied. This condition is obtained by integrating Eq. (21) with respect to x from zero to 2π and, with regard for the relation  2π 0 cot  ξ – x 2  dx =0, becomes  2π 0 f(x) dx = 0. (23) To construct a solution of Eq. (21), we apply the solution of the simplest singular integral equation of the first kind with Cauchy kernel by assuming that the contour L is the circle of unit radius centered at the origin (see Subsection 12.4-1). We rewrite the equation with Cauchy kernel and its solution in the form 1 π  L ϕ 1 (τ) τ – t dτ = f 1 (t), (24) ϕ 1 (t)=– 1 π  L f 1 (τ) τ – t dτ, (25) which is obtained by substituting the function ϕ 1 (t) instead of ϕ(t) and the function f 1 (t)i –1 instead of f(t) into the relations of 12.4-1. We set t = e ix and τ = e iξ and find the relationship between the Cauchy kernel and the Hilbert kernel: dτ τ – t = 1 2 cot  ξ – x 2  dξ + i 2 dξ. (26) On substituting relation (26) into Eq. (24) and into the solution (25), with regard to the change of variables ϕ(x)=ϕ 1 (t) and f(x)=f 1 (t) we obtain 1 2π  2π 0 cot  ξ – x 2  ϕ(ξ) dξ + i 2π  2π 0 ϕ(ξ) dξ = f (x), (27) ϕ(x)=– 1 2π  2π 0 cot  ξ – x 2  f(ξ) dξ – i 2π  2π 0 f(ξ) dξ. (28) Equation (21), under the additional assumption (22), coincides with Eq. (27), and hence its solution is given by the expression (28). Taking into account the solvability conditions (23), on the basis of (28) we rewrite a solution of Eq. (21) in the form ϕ(x)=– 1 2π  2π 0 cot  ξ – x 2  f(ξ) dξ. (29) Formulas (21) and (29), together with conditions (22) and (23), are called the Hilbert inversion formula. Page 628 © 1998 by CRC Press LLC © 1998 by CRC Press LLC Remark 2. Equation (21) is a special case of the characteristic singular integral equation with Hilbert kernel (see Subsections 13.1-2 and 13.2-5). 2 ◦ . Consider the general singular integral equation of the first kind with Hilbert kernel 1 2π  2π 0 N(x, ξ) cot  ξ – x 2  ϕ(ξ) dξ = f (x). (30) Let us represent its kernel in the form N(x, ξ) cot ξ – x 2 =  N(x, ξ) – N(x, x)  cot ξ – x 2 + N(x, x) cot ξ – x 2 . We introduce the notation N(x, x)=–b(x), 1 2π  N(x, ξ) – N(x, x)  cot ξ – x 2 = K(x, ξ), (31) and rewrite Eq. (30) as follows: – b(x) 2π  2π 0 cot  ξ – x 2  ϕ(ξ) dξ +  2π 0 K(x, ξ)ϕ(ξ) dξ = f (x), (32) It follows from formulas (31) that the function b(x) satisfies the H ¨ older condition, whereas the kernel K(x, ξ) satisfies the H ¨ older condition everywhere except possibly for the points x = ξ,at which the following estimate holds: |K(x, ξ)| < A |ξ – x| λ , A = const < ∞,0≤ λ <1. The general singular integral equation of the first kind with Hilbert kernel is frequently written in the form (32). It is a special case of the complete singular integral equation with Hilbert kernel, which is treated in Subsections 13.1-2 and 13.4-8. • References for Section 12.4: F. D. Gakhov (1977), F. D. Gakhov and Yu. I. Cherskii (1978), S. G. Mikhlin and S. Pr ¨ ossdorf (1986), N. I. Muskhelishvili (1992), I. K. Lifanov (1996). 12.5. Multhopp–Kalandiya Method Consider a general singular integral equation of the first kind with Cauchy kernel on the finite interval [–1, 1] of the form 1 π  1 –1 ϕ(t) dt t – x + 1 π  1 –1 K(x, t)ϕ(t) dt = f(x). (1) This equation frequently occurs in applications, especially in aerodynamics and 2D elasticity. We present here a method of approximate solution of Eq. (1) under the assumption that this equation has a solution in the classes indicated below. 12.5-1. A Solution That is Unbounded at the Endpoints of the Interval According to the general theory of singular integral equations (e.g., see N. I. Muskhelishvili (1992)), such a solution can be represented in the form ϕ(x)= ψ(x) √ 1 – x 2 , (2) Page 629 © 1998 by CRC Press LLC © 1998 by CRC Press LLC where ψ(x) is a bounded function on [–1, 1]. Let us substitute the expression (2) into Eq. (1) and introduce new variables θ and τ by the relations x = cos θ and t = cos τ,0≤ θ ≤ π,0≤ τ ≤ π.In this case, Eq. (1) becomes 1 π  π 0 ψ(cos τ) dτ cos τ – cos θ + 1 π  π 0 K(cos θ, cos τ )ψ(cos τ ) dτ = f(cos x). (3) Let us construct the Lagrange interpolation polynomial for the desired function ψ(x) with the Chebyshev nodes x m = cos θ m , θ m = 2m – 1 2n π, m =1, , n. This polynomial is known to have the form L n (ψ; cos θ)= 1 n n  l=1 (–1) l+1 ψ(cos θ l ) cos nθ sin θ l cos θ – cos θ l . (4) Note that for each l the fraction on the right-hand side in (4) is an even trigonometric polynomial of degree ≤ n – 1. We define the coefficients of this polynomial by means of the known relations 1 π  π 0 cos nτ dτ cos τ – cos θ = sin nθ sin θ ,0≤ θ ≤ π, n =0,1,2, (5) and rewrite (4) in the form L n (ψ; cos θ)= 2 n n  l=1 ψ(cos θ l ) n–1  m=0 cos mθ l cos mθ – 1 n n  l=1 ψ(cos θ l ). (6) On the basis of the above two relations we write out the following quadrature formula for the singular integral: 1 π  1 –1 ϕ(t) dt t – x = 2 n sin θ n  l=1 ψ(cos θ l ) n–1  m=1 cos mθ l sin mθ. (7) This formula is exact for the case in which ψ(t) is a polynomial of order ≤ n – 1int. To the second integral on the left-hand side of Eq. (1), we apply the formula 1 π  1 –1 P (x) dx √ 1 – x 2 = 1 n n  l=1 P (cos θ l ), (8) which holds for any polynomial P(x) of degree ≤ 2n – 1. In this case, by (8) we have 1 π  1 –1 K(x, t)ϕ(t) dt = 1 n n  l=1 K(cos θ, cos θ l )ψ(cos θ l ). (9) On substituting relations (7) and (9) into Eq. (1), we obtain 2 n sin θ n  l=1 ψ(cos θ l ) n–1  m=1 cos mθ l sin mθ + 1 n n  l=1 K(cos θ, cos θ l )ψ(cos θ l )=f(cos θ). (10) Page 630 © 1998 by CRC Press LLC © 1998 by CRC Press LLC [...]... solution of singular integral equations are discussed as well • References for Section 12.5: A I Kalandiya ( 197 3), N I Muskhelishvili ( 199 2), S M Belotserkovskii and I K Lifanov ( 199 3), and I K Lifanov ( 199 6) © 199 8 by CRC Press LLC © 199 8 by CRC Press LLC Page 632 Chapter 13 Methods for Solving Complete Singular Integral Equations 13.1 Some Definitions and Remarks 13.1-1 Integral Equations With Cauchy Kernel... integral equations are Fredholm kernels Remark 3 The complete and characteristic singular integral equations are sometimes called singular integral equations of the second kind • References for Section 13.1: F D Gakhov ( 197 7), F G Tricomi ( 198 5), S G Mikhlin and S Pr¨ ssdorf ( 198 6), o A Dzhuraev ( 199 2), N I Muskhelishvili ( 199 2), I K Lifanov ( 199 6) 13.2 The Carleman Method for Characteristic Equations. .. for Section 13.2: P P Zabreyko, A I Koshelev, et al ( 197 5), F D Gakhov ( 197 7), F G Tricomi ( 198 5), N I Muskhelishvili ( 199 2) © 199 8 by CRC Press LLC © 199 8 by CRC Press LLC Page 645 13.3 Complete Singular Integral Equations Solvable in a Closed Form In contrast with characteristic equations and their transposed equations, complete singular integral equations cannot be solved in the closed form in general... the cases in which the contour is finite and infinite are essentially the same Therefore, the theory of singular integral equations on an infinite contour in the class of functions that vanish at infinity coincides with the theory of equations on a finite contour © 199 8 by CRC Press LLC © 199 8 by CRC Press LLC Page 641 Just as for the case of a finite contour, the characteristic integral equation a(x)ϕ(x) +... differs from the transformation (2) and (3) because the above transformation of Eq (1) does not lead to the desired decomposition For equations with discontinuous coefficients, see the cited books © 199 8 by CRC Press LLC © 199 8 by CRC Press LLC Page 634 13.1-2 Integral Equations With Hilbert Kernel A complete singular integral equation with Hilbert kernel has the form 1 2π a(x)ϕ(x) + 2π N (x, ξ) cot... KK equation Thus, we have transformed the singular integral equation ( 19) into the Fredholm integral equation (21) for the same unknown function ϕ(t) This is the first regularization method, which is called left regularization © 199 8 by CRC Press LLC © 199 8 by CRC Press LLC Page 651 2◦ Right Regularization On replacing in Eq ( 19) the desired function by the expression (20), ˜ ϕ(t) = K[ω(t)], (22) where... consideration because the index of the regularizer K∗◦ is equal to –ν ≤ 0 © 199 8 by CRC Press LLC © 199 8 by CRC Press LLC Page 652 2◦ Right Regularization Consider Eq (24) and the corresponding regularized equation ˜ KK[ω(t)] = f (t), ( 29) ˜ K[ω(t)] = ϕ(t) (30) which is obtained by substitution If ωj (t) is a solution of Eq ( 29) , then formula (30) gives the corresponding solution of the original equation... characteristic part of the operator K only and is equal to its index, i.e., n – m = ν (33) Corollary The number of linearly independent solutions of characteristic equations is minimal among all singular equations with given index ν © 199 8 by CRC Press LLC © 199 8 by CRC Press LLC Page 653 13.4-6 The Carleman–Vekua Approach to the Regularization Let us transfer the regular part of a singular equation to the right-hand... of linear algebraic equations for the approximate values ζl of the desired function ζ(x) at the nodes: n bkl ζl = fk , fk = f (cos θk ), k = 1, , n, l=1 bkl θl 1 θk θk ± θl = tan cot – 1 + 2 sin2 K(cos θk , cos θl ) n 2 2 2 (17) After solving the system (17), the corresponding approximate solution to Eq (1) can be found by formulas (13) and (14) © 199 8 by CRC Press LLC © 199 8 by CRC Press LLC Page... general case the contour L consists of m + 1 closed smooth curves L = L0 + L1 + · · · + Lm For equations with nonclosed contours, see, for example, the books by F D Gakhov ( 197 7) and N I Muskhelishvili ( 199 2) Remark 1 The above relationship between Eqs (1) and (4) that involves the properties of these equations is violated if we modify the condition and assume that in Eq (1) the function M (t, τ ) . ( 197 7), F. G. Tricomi ( 198 5), S. G. Mikhlin and S. Pr ¨ ossdorf ( 198 6), A. Dzhuraev ( 199 2), N. I. Muskhelishvili ( 199 2), I. K. Lifanov ( 199 6). 13.2. The Carleman Method for Characteristic Equations 13.2-1 singular integral equations (e.g., see N. I. Muskhelishvili ( 199 2)), such a solution can be represented in the form ϕ(x)= ψ(x) √ 1 – x 2 , (2) Page 6 29 © 199 8 by CRC Press LLC © 199 8 by CRC Press. integral equations are discussed as well. • References for Section 12.5: A. I. Kalandiya ( 197 3), N. I. Muskhelishvili ( 199 2), S. M. Belotserkovskii and I. K. Lifanov ( 199 3), and I. K. Lifanov ( 199 6). Page