1396 INTEGRAL EQUATIONS 1 ◦ . To solve this integral equation, direct and inverse Laplace transforms are used. The solution can be represented in the form y(x)=f(x)– x a R(x – t)f(t) dt.(1) Here the resolvent R(x) is expressed via the kernel K(x) of the original equation as follows: R(x)= 1 2πi c+i∞ c–i∞ R(p)e px dp, R(p)= K(p) 1 + K(p) , K(p)= ∞ 0 K(x)e –px dx. 2 ◦ .Letw = w(x) be the solution of the simpler auxiliary equation with a = 0 and f ≡ 1: w(x)+ x 0 K(x – t)w(t) dt = 1.(2) Then the solution of the original integral equation with arbitrary f = f (x) is expressed via the solution of the auxiliary equation (2) as y(x)= d dx x a w(x – t)f(t) dt = f (a)w(x – a)+ x a w(x – t)f t (t) dt. T11.3. Linear Equations of the First Kind with Constant Limits of Integration 1. b a |x – t| y(t) dt = f(x), 0 ≤ a < b < ∞. Solution: y(x)= 1 2 f xx (x). The right-hand side f (x) of the integral equation must satisfy certain relations. The general form of f (x)isasfollows: f(x)=F (x)+Ax + B, A =– 1 2 F x (a)+F x (b) , B = 1 2 aF x (a)+bF x (b)–F (a)–F (b) , where F(x) is an arbitrary bounded twice differentiable function (with bounded first deriva- tive). 2. a 0 y(t) √ |x – t| dt = f(x), 0<a ≤ ∞. Solution: y(x)=– A x 1/4 d dx a x dt (t – x) 1/4 t 0 f(s) ds s 1/4 (t – s) 1/4 , A = 1 √ 8π Γ 2 (3/4) . T11.3. LINEAR EQUATIONS OF THE FIRST KIND WITH CONSTANT LIMITS OF INTEGRATION 1397 3. b a y(t) |x – t| k dt = f(x), 0<k <1. It is assumed that |a| + |b| < ∞. Solution: y(x)= 1 2π cot( 1 2 πk) d dx x a f(t) dt (x – t) 1–k – 1 π 2 cos 2 ( 1 2 πk) x a Z(t)F (t) (x – t) 1–k dt, where Z(t)=(t – a) 1+k 2 (b – t) 1–k 2 , F (t)= d dt t a dτ (t – τ) k b τ f(s) ds Z(s)(s – τ) 1–k . 4. b 0 y(t) |x λ – t λ | k dt = f(x), 0<k <1, λ >0. Solution: y(x)=–Ax λ(k–1) 2 d dx b x t λ(3–2k)–2 2 dt (t λ – x λ ) 1–k 2 t 0 s λ(k+1)–2 2 f(s) ds (t λ – s λ ) 1–k 2 , A = λ 2 2π cos πk 2 Γ(k) Γ 1 + k 2 –2 , where Γ(k) is the gamma function. 5. ∞ –∞ y(t) |x – t| 1–λ dt = f(x), 0<λ <1. Solution: y(x)= λ 2π tan πλ 2 ∞ –∞ f(x)–f(t) |x – t| 1+λ dt. It assumed that the condition ∞ –∞ |f(x)| p dx < ∞ is satisfied for some p, 1 < p < 1/λ. 6. ∞ –∞ sign(x – t) |x – t| 1–λ y(t) dt = f(x), 0<λ <1. Solution: y(x)= λ 2π cot πλ 2 ∞ –∞ f(x)–f(t) |x – t| 1+λ sign(x – t) dt. 7. ∞ –∞ a + b sign(x – t) |x – t| 1–λ y(t) dt = f(x), 0<λ <1. Solution: y(x)= λ sin(πλ) 4π a 2 cos 2 1 2 πλ + b 2 sin 2 1 2 πλ ∞ –∞ a + b sign(x – t) |x – t| 1+λ f(x)–f(t) dt. 1398 INTEGRAL EQUATIONS 8. ∞ 0 y(x + t) – y(x – t) t dt = f(x). Solution: y(x)=– 1 π 2 ∞ 0 f(x + t)–f(x – t) t dt. In equations T11.3.9 and T11.3.10 and their solutions, all singular integrals are under- stood in the sense of the Cauchy principal value. 9. ∞ –∞ y(t) dt t – x = f(x). Solution: y(x)=– 1 π 2 ∞ –∞ f(t) dt t – x . The integral equation and its solution form a Hilbert transform pair (in the asymmetric form). 10. b a y(t) dt t – x = f(x). This equation is encountered in hydrodynamics in solving the problem on the flow of an ideal inviscid fluid around a thin profile (a ≤ x ≤ b). It is assumed that |a| + |b| < ∞. 1 ◦ . The solution bounded at the endpoints is y(x)=– 1 π 2 (x – a)(b – x) b a f(t) √ (t – a)(b – t) dt t – x , provided that b a f(t) dt √ (t – a)(b – t) = 0. 2 ◦ . The solution bounded at the endpoint x = a and unbounded at the endpoint x = b is y(x)=– 1 π 2 x – a b – x b a b – t t – a f(t) t – x dt. 3 ◦ . The solution unbounded at the endpoints is y(x)=– 1 π 2 √ (x – a)(b – x) b a √ (t – a)(b – t) t – x f(t) dt + C , where C is an arbitrary constant. The formula b a y(t) dt = C π holds. 11. b a e λ|x–t| y(t) dt = f(x), –∞ < a < b < ∞. Solution: y(x)= 1 2λ f xx (x)–λ 2 f(x) . The right-hand side f (x) of the integral equation must satisfy certain relations. The general form of the right-hand side is given by f(x)=F (x)+Ax + B, A = 1 bλ – aλ – 2 F x (a)+F x (b)+λF (a)–λF (b) , B =– 1 λ F x (a)+λF (a)+Aaλ + A , where F (x) is an arbitrary bounded, twice differentiable function. T11.3. LINEAR EQUATIONS OF THE FIRST KIND WITH CONSTANT LIMITS OF INTEGRATION 1399 12. b a ln |x – t| y(t) dt = f(x). Carleman’s equation. 1 ◦ . Solution for b – a ≠ 4: y(x)= 1 π 2 √ (x – a)(b – x) b a √ (t – a)(b – t) f t (t) dt t – x + 1 ln 1 4 (b – a) b a f(t) dt √ (t – a)(b – t) . 2 ◦ .Ifb – a = 4, then for the equation to be solvable, the condition b a f(t)(t – a) –1/2 (b – t) –1/2 dt = 0 must be satisfied. In this case, the solution has the form y(x)= 1 π 2 √ (x – a)(b – x) b a √ (t – a)(b – t) f t (t) dt t – x + C , where C is an arbitrary constant. 13. b a ln |x – t| + β y(t) dt = f(x). By setting x = e –β z, t = e –β τ, y(t)=Y (τ ), f (x)=e –β g(z), we arrive at an equation of the form T11.3.12: B A ln |z – τ | Y (τ) dτ = g(z), A = ae β , B = be β . 14. a –a ln A |x – t| y(t) dt = f(x), –a ≤ x ≤ a. Solution for 0 < a < 2A: y(x)= 1 2M (a) d da a –a w(t, a)f(t) dt w(x, a) – 1 2 a |x| w(x, ξ) d dξ 1 M (ξ) d dξ ξ –ξ w(t, ξ)f(t) dt dξ – 1 2 d dx a |x| w(x, ξ) M (ξ) ξ –ξ w(t, ξ) df (t) dξ, where the prime stands for the derivative and M(ξ)= ln 2A ξ –1 , w(x, ξ)= M(ξ) π ξ 2 – x 2 . 1400 INTEGRAL EQUATIONS 15. a 0 ln x + t x – t y(t) dt = f(x). Solution: y(x)=– 2 π 2 d dx a x F (t) dt √ t 2 – x 2 , F(t)= d dt t 0 sf(s) ds √ t 2 – s 2 . 16. ∞ 0 cos(xt)y(t) dt = f(x). Solution: y(x)= 2 π ∞ 0 cos(xt)f (t) dt. Up to constant factors, the function f (x) and the solution y(t) are the Fourier cosine transform pair. 17. ∞ 0 sin(xt)y(t) dt = f (x). Solution: y(x)= 2 π ∞ 0 sin(xt)f(t) dt. Up to constant factors, the function f (x) and the solution y(t) are the Fourier sine transform pair. 18. π/2 0 y(ξ) dt = f(x), ξ = x sin t. Schl ¨ omilch equation. Solution: y(x)= 2 π f(0)+x π/2 0 f ξ (ξ) dt , ξ = x sin t. 19. 2π 0 cot t – x 2 y(t) dt = f(x), 0 ≤ x ≤ 2π. Here the integral is understood in the sense of the Cauchy principal value and the right-hand side is assumed to satisfy the condition 2π 0 f(t) dt = 0. Solution: y(x)=– 1 4π 2 2π 0 cot t – x 2 f(t) dt + C, where C is an arbitrary constant. It follows from the solution that 2π 0 y(t) dt = 2πC. The equation and its solution form a Hilbert transform pair (in the asymmetric form). 20. ∞ 0 tJ ν (xt)y(t) dt = f(x), ν >– 1 2 . Here, J ν (z) is the Bessel function of the first kind. Solution: y(x)= ∞ 0 tJ ν (xt)f(t) dt. The function f (x) and the solution y(t) are the Hankel transform pair. T11.4. LINEAR EQUATIONS OF THE SECOND KIND WITH CONSTANT LIMITS OF INTEGRATION 1401 21. ∞ –∞ K 0 |x – t| y(t) dt = f(x). Here, K 0 (z) is the modified Bessel function of the second kind. Solution: y(x)=– 1 π 2 d 2 dx 2 – 1 ∞ –∞ K 0 |x – t| f(t) dt. 22. ∞ –∞ K(x – t)y(t) dt = f(x). The Fourier transform is used to solve this equation. 1 ◦ . Solution: y(x)= 1 2π ∞ –∞ f(u) K(u) e iux du, f(u)= 1 √ 2π ∞ –∞ f(x)e –iux dx, K(u)= 1 √ 2π ∞ –∞ K(x)e –iux dx. The following statement is valid. Let f (x) L 2 (–∞, ∞)andK(x) L 1 (–∞, ∞). Then for a solution y(x) L 2 (–∞, ∞) of the integral equation to exist, it is necessary and sufficient that f(u)/ K(u) L 2 (–∞, ∞). 2 ◦ . Let the function P (s)defined by the formula 1 P (s) = ∞ –∞ e –st K(t) dt be a polynomial of degree n with real roots of the form P (s)= 1 – s a 1 1 – s a 2 1 – s a n . Then the solution of the integral equation is given by y(x)=P (D)f(x), D = d dx . 23. ∞ 0 K(x – t)y(t) dt = f(x). Wiener–Hopf equation of the first kind. This equation is discussed in the books by Gakhov and Cherskii (1978), Mikhlin and Pr ¨ ossdorf (1986), Muskhelishvili (1992), and Polyanin and Manzhirov (1998) in detail. T11.4. Linear Equations of the Second Kind with Constant Limits of Integration 1. y(x) – λ b a (x – t)y(t) dt = f (x). Solution: y(x)=f(x)+λ(A 1 x + A 2 ), 1402 INTEGRAL EQUATIONS where A 1 = 12f 1 + 6λ (f 1 Δ 2 – 2f 2 Δ 1 ) λ 2 Δ 4 1 + 12 , A 2 = –12f 2 + 2λ (3f 2 Δ 2 – 2f 1 Δ 3 ) λ 2 Δ 4 1 + 12 , f 1 = b a f(x) dx, f 2 = b a xf(x) dx, Δ n = b n – a n . 2. y(x) + A b a |x – t| y(t) dt = f(x). 1 ◦ .ForA < 0, the solution is given by y(x)=C 1 cosh(kx)+C 2 sinh(kx)+f (x)+k x a sinh[k(x – t)]f(t) dt, k = √ –2A,(1) where the constants C 1 and C 2 are determined by the conditions y x (a)+y x (b)=f x (a)+f x (b), y(a)+y(b)+(b – a)y x (a)=f(a)+f(b)+(b – a)f x (a). (2) 2 ◦ .ForA > 0, the solution is given by y(x)=C 1 cos(kx)+C 2 sin(kx)+f(x)–k x a sin[k(x – t)]f (t) dt, k = √ 2A,(3) where the constants C 1 and C 2 are determined by conditions (2). 3 ◦ . In the special case a = 0 and A > 0, the solution of the integral equation is given by formula (3) with C 1 = k I s (1 +cosλ)–I c (λ +sinλ) 2 + 2 cos λ + λ sin λ , C 2 = k I s sin λ + I c (1 +cosλ) 2 + 2 cos λ + λ sin λ , k = √ 2A, λ = bk, I s = b 0 sin[k(b – t)]f (t) dt, I c = b 0 cos[k(b – t)]f(t) dt. In equations T11.4.3 and T11.4.4 and their solutions, all singular integrals are under- stood in the sense of the Cauchy principal value. 3. Ay(x) + B π 1 –1 y(t) dt t – x = f(x), –1 < x <1. Without loss of generality we may assume that A 2 + B 2 = 1. 1 ◦ . The solution bounded at the endpoints: y(x)=Af(x)– B π 1 –1 g(x) g(t) f(t) dt t – x , g(x)=(1 + x) α (1 – x) 1–α ,(1) where α is the solution of the trigonometric equation A + B cot(πα)=0 (2) on the interval 0 < α < 1. This solution y(x) exists if and only if 1 –1 f(t) g(t) dt = 0. . equation and its solution form a Hilbert transform pair (in the asymmetric form). 10. b a y(t) dt t – x = f(x). This equation is encountered in hydrodynamics in solving the problem on the flow of. 2πC. The equation and its solution form a Hilbert transform pair (in the asymmetric form). 20. ∞ 0 tJ ν (xt)y(t) dt = f(x), ν >– 1 2 . Here, J ν (z) is the Bessel function of the first kind. Solution: y(x)= ∞ 0 tJ ν (xt)f(t). (s)defined by the formula 1 P (s) = ∞ –∞ e –st K(t) dt be a polynomial of degree n with real roots of the form P (s)= 1 – s a 1 1 – s a 2 1 – s a n . Then the solution of the integral