THÔNG TIN TÀI LIỆU
3.1-3. Kernels Containing Integer Powers of x and t or Rational Functions 11. b a x – t 3 y(t) dt = f(x). Let us remove the modulus in the integrand: x a (x – t) 3 y(t) dt + b x (t – x) 3 y(t) dt = f (x). (1) Differentiating (1) twice yields 6 x a (x – t)y(t) dt +6 b x (t – x)y(t) dt = f xx (x). This equation can be rewritten in the form 3.1.2: b a |x – t| y(t) dt = 1 6 f xx (x). (2) Therefore the solution of the integral equation is given by y(x)= 1 12 y xxxx (x). (3) The right-hand side f(x) of the equation must satisfy certain conditions. To obtain these conditions, one must substitute solution (3) into (1) with x = a and x = b and into (2) with x = a and x = b, and then integrate the four resulting relations by parts. 12. b a x 3 – t 3 y(t) dt = f(x). This is a special case of equation 3.8.3 with g(x)=x 3 . 13. b a xt 2 – t 3 y(t) dt = f(x)0≤ a < b < ∞. The substitution w(t)=t 2 y(t) leads to an equation of the form 3.1.2: b a |x – t|w(t) dt = f(x). 14. b a x 2 t – t 3 y(t) dt = f(x). The substitution w(t)=|t| y(t) leads to an equation of the form 3.1.8: b a x 2 – t 2 w(t) dt = f(x). 15. a 0 x 3 – βt 3 y(t) dt = f(x), β >0. This is a special case of equation 3.8.4 with g(x)=x 3 and β = λ 3 . Page 200 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 16. b a x – t 2n+1 y(t) dt = f(x), n =0,1,2, Solution: y(x)= 1 2(2n + 1)! f (2n+2) x (x). (1) The right-hand side f (x) of the equation must satisfy certain conditions. To obtain these conditions, one must substitute solution (1) into the relations b a (t – a) 2n+1 y(t) dt = f (a), b a (t – a) 2n–k y(t) dt = (–1) k+1 A k f (k+1) x (a), A k =(2n + 1)(2n) (2n +1– k); k =0,1, ,2n, and then integrate the resulting equations by parts. 17. ∞ 0 y(t) dt x + t = f(x). The left-hand side of this equation is the Stieltjes transform. 1 ◦ . By setting x = e z , t = e τ , y(t)=e –τ/2 w(τ), f (x)=e –z/2 g(z), we obtain an integral equation with difference kernel of the form 3.8.15: ∞ –∞ w(τ) dτ 2 cosh 1 2 (z – τ) = g(z), whose solution is given by w(z)= 1 √ 2π 3 ∞ –∞ cosh(πu) ˜g(u)e iux du, ˜g(u)= 1 √ 2π ∞ –∞ g(z)e –iuz dz. • Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975). 2 ◦ . Under some assumptions, the solution of the original equation can be represented in the form y(x) = lim n→∞ (–1) n (n – 1)! (n + 1)! x 2n+1 f (n) x (x) (n+1) x , (1) which is the real inversion of the Stieltjes transform. An alternative form of the solution is y(x) = lim n→∞ (–1) n 2π e n 2n x 2n f (n) x (x) (n) x . (2) To obtain an approximate solution of the integral equation, one restricts oneself to a specific value of n in (1) or (2) instead of taking the limit. • Reference: I. I. Hirschman and D. V. Widder (1955). Page 201 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 3.1-4. Kernels Containing Square Roots 18. a 0 √ x – √ t y(t) dt = f (x), 0 < a < ∞. This is a special case of equation 3.8.3 with g(x)= √ x. Solution: y(x)= d dx √ xf x (x) . The right-hand side f(x) of the equation must satisfy certain conditions. The general form of the right-hand side is f(x)=F (x)+Ax + B, A = –F x (a), B = 1 2 aF x (a) – F (a) – F (0) , where F (x) is an arbitrary bounded twice differentiable function (with bounded first deriva- tive). 19. a 0 √ x – β √ t y(t) dt = f (x), β >0. This is a special case of equation 3.8.4 with g(x)= √ x and β = √ λ. 20. a 0 √ x – t y(t) dt = f (x). This is a special case of equation 3.8.5 with g(x)= √ x (see item 3 ◦ of 3.8.5). 21. a 0 x – √ t y(t) dt = f (x). This is a special case of equation 3.8.6 with g(t)= √ t (see item 3 ◦ of 3.8.6). 22. a 0 y(t) √ |x – t| dt = f (x), 0 < a ≤ ∞. This is a special case of equation 3.1.29 with k = 1 2 . 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) . 23. ∞ –∞ y(t) √ |x – t| dt = f (x). This is a special case of equation 3.1.34 with λ = 1 2 . Solution: y(x)= 1 4π ∞ –∞ f(x) – f(t) |x – t| 3/2 dt. Page 202 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 3.1-5. Kernels Containing Arbitrary Powers 24. a 0 |x k – t k | y(t) dt = f (x), 0 < k <1, 0<a < ∞. 1 ◦ . Let us remove the modulus in the integrand: x 0 (x k – t k )y(t) dt + a x (t k – x k )y(t) dt = f (x). (1) Differentiating (1) with respect to x yields kx k–1 x 0 y(t) dt – kx k–1 a x y(t) dt = f x (x). (2) Let us divide both sides of (2) by kx k–1 and differentiate the resulting equation. As a result, we obtain the solution y(x)= 1 2k d dx x 1–k f x (x) . (3) 2 ◦ . Let us demonstrate that the right-hand side f(x) of the integral equation must satisfy certain relations. By setting x=0 and x=a, in (1), we obtain two corollaries a 0 t k y(t) dt= f(0) and a 0 (a k – t k )y(t) dt = f (a), which can be rewritten in the form a 0 t k y(t) dt = f (0), a k a 0 y(t) dt = f (0) + f(a). (4) Substitute y(x) of (3) into (4). Integrating by parts yields the relations af x (a)=kf(a)+kf(0) and af x (a)=2kf(a)+2kf(0). Hence, the desired constraints for f (x) have the form f(0) + f(a)=0, f x (a)=0. (5) Conditions (5) make it possible to find the admissible general form of the right-hand side of the integral equation: f(x)=F (x)+Ax + B, A = –F x (a), B = 1 2 aF x (a) – F (a) – F (0) , where F (x) is an arbitrary bounded twice differentiable function with bounded first derivative. The first derivative may be unbounded at x = 0, in which case the conditions x 1–k F x x=0 =0 must hold. 25. a 0 |x k – βt k | y(t) dt = f (x), 0 < k <1, β >0. This is a special case of equation 3.8.4 with g(x)=x k and β = λ k . 26. a 0 |x k t m – t k+m | y(t) dt = f (x), 0 < k <1, 0<a < ∞. The substitution w(t)=t m y(t) leads to an equation of the form 3.1.24: a 0 |x k – t k |w(t) dt = f(x). Page 203 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 27. 1 0 |x k – t m | y(t) dt = f (x), k >0, m >0. The transformation z = x k , τ = t m , w(τ)=τ 1–m m y(t) leads to an equation of the form 3.1.1: 1 0 |z – τ|w(τ) dτ = F (z), F (z)=mf(z 1/k ). 28. b a |x – t| 1+λ y(t) dt = f (x), 0 ≤ λ <1. For λ = 0, see equation 3.1.2. Assume that 0 < λ <1. 1 ◦ . Let us remove the modulus in the integrand: x a (x – t) 1+λ y(t) dt + b x (t – x) 1+λ y(t) dt = f (x). (1) Let us differentiate (1) with respect to x twice and then divide both the sides by λ(λ + 1). As a result, we obtain x a (x – t) λ–1 y(t) dt + b x (t – x) λ–1 y(t) dt = 1 λ(λ +1) f xx (x). (2) Rewrite equation (2) in the form b a y(t) dt |x – t| k = 1 λ(λ +1) f xx (x), k =1– λ. (3) See 3.1.29 and 3.1.30 for the solutions of equation (3) for various a and b. 2 ◦ . The right-hand side f(x) of the integral equation must satisfy certain relations. By setting x = a and x = b in (1), we obtain two corollaries b a (t – a) 1+λ y(t) dt = f (a), b a (b – t) 1+λ y(t) dt = f (b). (4) On substituting the solution y(x) of (3) into (4) and then integrating by parts, we obtain the desired constraints for f(x). 29. a 0 y(t) |x – t| k dt = f (x), 0 < k <1, 0<a ≤ ∞. 1 ◦ . Solution: y(x)=–Ax k–1 2 d dx a x t 1–2k 2 dt (t – x) 1–k 2 t 0 f(s ) ds s 1–k 2 (t – s) 1–k 2 , A = 1 2π cos πk 2 Γ(k) Γ 1+k 2 –2 , where Γ(k) is the gamma function. 2 ◦ . The transformation x = z 2 , t = ξ 2 , w(ξ)=2ξy(t) leads to an equation of the form 3.1.31: √ a 0 w(ξ) |z 2 – ξ 2 | k dξ = f z 2 . Page 204 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 30. 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 . • Reference: F. D. Gakhov (1977). 31. a 0 y(t) |x 2 – t 2 | k dt = f (x), 0 < k <1, 0<a ≤ ∞. Solution: y(x)=– 2Γ(k) cos 1 2 πk π Γ 1+k 2 2 x k–1 d dx a x t 2–2k F (t) dt (t 2 – x 2 ) 1–k 2 , F(t)= t 0 s k f(s ) ds (t 2 – s 2 ) 1–k 2 . • Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975). 32. b a y(t) |x λ – t λ | k dt = f (x), 0 < k <1, λ >0. 1 ◦ . The transformation z = x λ , τ = t λ , w(τ)=τ 1–λ λ y(t) leads to an equation of the form 3.8.30: B A w(τ) |z – τ| k dτ = F (z), where A = a λ , B = b λ , F (z)=λf(z 1/λ ). 2 ◦ . Solution with a =0: 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. 33. 1 0 y(t) |x λ – t m | k dt = f (x), 0 < k <1, λ >0, m >0. The transformation z = x λ , τ = t m , w(τ)=τ 1–m m y(t) leads to an equation of the form 3.8.30: 1 0 w(τ) |z – τ| k dτ = F (z), F (z)=mf(z 1/λ ). Page 205 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 34. ∞ –∞ 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. • Reference: S. G. Samko, A. A. Kilbas, and A. A. Marichev (1993). 35. ∞ –∞ y(t) |x 3 – t| 1–λ dt = f (x), 0 < λ <1. The substitution z = x 3 leads to an equation of the form 3.1.34: ∞ –∞ y(t) |z – t| 1–λ dt = f z 1/3 . 36. ∞ –∞ y(t) |x 3 – t 3 | 1–λ dt = f (x), 0 < λ <1. The transformation z = x 3 , τ = t 3 , w(τ)=τ –2/3 y(t) leads to an equation of the form 3.1.34: ∞ –∞ w(τ) |z – τ| 1–λ dτ = F (z), F (z)=3f z 1/3 . 37. ∞ –∞ 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. • Reference: S. G. Samko, A. A. Kilbas, and A. A. Marichev (1993). 38. ∞ –∞ 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. • Reference: S. G. Samko, A. A. Kilbas, and A. A. Marichev (1993). 39. ∞ 0 y(t) dt (ax + bt) k = f(x), a >0, b >0, k >0. By setting x = 1 2a e 2z , t = 1 2b e 2τ , y(t)=be (k–2)τ w(τ), f (x)=e –kz g(z). we obtain an integral equation with the difference kernel of the form 3.8.15: ∞ –∞ w(τ) dτ cosh k (z – τ) = g(z). Page 206 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 40. ∞ 0 t z–1 y(t) dt = f (z). The left-hand side of this equation is the Mellin transform of y(t)(z is treated as a complex variable). Solution: y(t)= 1 2πi c+i∞ c–i∞ t –z f(z) dz, i 2 = –1. For specific f(z), one can use tables of Mellin and Laplace integral transforms to calculate the integral. • References: H. Bateman and A. Erd ´ elyi (vol. 2, 1954), V. A. Ditkin and A. P. Prudnikov (1965). 3.1-6. Equation Containing the Unknown Function of a Complicated Argument 41. 1 0 y(xt) dt = f (x). Solution: y(x)=xf x (x)+f(x). The function f(x) is assumed to satisfy the condition xf(x) x=0 =0. 42. 1 0 t λ y(xt) dt = f (x). The substitution ξ = xt leads to equation x 0 ξ λ y(ξ) dξ = x λ+1 f(x). Differentiating with respect to x yields the solution y(x)=xf x (x)+(λ +1)f(x). The function f(x) is assumed to satisfy the condition x λ+1 f(x) x=0 =0. 43. 1 0 Ax k + Bt m )y(xt) dt = f (x). The substitution ξ = xt leads to an equation of the form 1.1.50: x 0 Ax k+m + Bξ m y(ξ) dξ = x m+1 f(x). 44. 1 0 y(xt) dt √ 1 – t = f(x). The substitution ξ = xt leads to Abel’s equation 1.1.36: x 0 y(ξ) dξ √ x – ξ = √ xf(x). 45. 1 0 y(xt) dt (1 – t) λ = f(x), 0 < λ <1. The substitution ξ = xt leads to the generalized Abel equation 1.1.46: x 0 y(ξ) dξ (x – ξ) λ = x 1–λ f(x). Page 207 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 46. 1 0 t µ y(xt) (1 – t) λ dt = f (x), 0 < λ <1. The transformation ξ = xt, w(ξ)=ξ µ y(ξ) leads to the generalized Abel equation 1.1.46: x 0 w(ξ) dξ (x – ξ) λ = x 1+µ–λ f(x). 47. ∞ 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. • Reference: V. A. Ditkin and A. P. Prudnikov (1965). 3.1-7. Singular Equations In this subsection, all singular integrals are understood in the sense of the Cauchy principal value. 48. ∞ –∞ 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). • Reference: V. A. Ditkin and A. P. Prudnikov (1965). 49. 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. Solutions that have a singularity point x = s inside the interval [a, b] can be found in Subsection 12.4-3. • Reference: F. D. Gakhov (1977). Page 208 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 3.2. Equations Whose Kernels Contain Exponential Functions 3.2-1. Kernels Containing Exponential Functions 1. b a e λ|x–t| y(t) dt = f (x), –∞ < a < b < ∞. 1 ◦ . Let us remove the modulus in the integrand: x a e λ(x–t) y(t) dt + b x e λ(t–x) y(t) dt = f (x). (1) Differentiating (1) with respect to x twice yields 2λy(x)+λ 2 x a e λ(x–t) y(t) dt + λ 2 b x e λ(t–x) y(t) dt = f xx (x). (2) By eliminating the integral terms from (1) and (2), we obtain the solution y(x)= 1 2λ f xx (x) – λ 2 f(x) . (3) 2 ◦ . The right-hand side f(x) of the integral equation must satisfy certain relations. By setting x = a and x = b in (1), we obtain two corollaries b a e λt y(t) dt = e λa f(a), b a e –λt y(t) dt = e –λb f(b). (4) On substituting the solution y(x) of (3) into (4) and then integrating by parts, we see that e λb f x (b) – e λa f x (a)=λe λa f(a)+λe λb f(b), e –λb f x (b) – e –λa f x (a)=λe –λa f(a)+λe –λb f(b). Hence, we obtain the desired constraints for f(x): f x (a)+λf(a)=0, f x (b) – λf(b) = 0. (5) The general form of the right-hand side satisfying conditions (5) 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. 2. b a Ae λ|x–t| + Be µ|x–t| y(t) dt = f (x), –∞ < a < b < ∞. Let us remove the modulus in the integrand and differentiate the resulting equation with respect to x twice to obtain 2(Aλ + Bµ)y(x)+ b a Aλ 2 e λ|x–t| + Bµ 2 e µ|x–t| y(t) dt = f xx (x). (1) Eliminating the integral term with e µ|x–t| from (1) with the aid of the original integral equation, we find that 2(Aλ + Bµ)y(x)+A(λ 2 – µ 2 ) b a e λ|x–t| y(t) dt = f xx (x) – µ 2 f(x). (2) For Aλ+Bµ = 0, this is an equation of the form 3.2.1, and for Aλ+ Bµ ≠ 0, this is an equation of the form 4.2.15. The right-hand side f (x) must satisfy certain relations, which can be obtained by setting x = a and x = b in the original equation (a similar procedure is used in 3.2.1). Page 209 © 1998 by CRC Press LLC © 1998 by CRC Press LLC [...]... – sinh(µt) y(t) dt = f (x), 8 β > 0, µ > 0 0 This is a special case of equation 3.8 .4 with g(x) = sinh(βx) and λ = µ/β b sinh3 λ|x – t| y(t) dt = f (x) 9 a Using the formula sinh3 β = 1 4 sinh 3β – 3 4 sinh β, we arrive at an equation of the form 3.3.6: b a b 1 4 A sinh 3λ|x – t| – 3 A sinh λ|x – t| y(t) dt = f (x) 4 n Ak sinh λk |x – t| y(t) dt = f (x), 10 a –∞ < a < b < ∞ k=1 1◦ Let us remove the... sin(βx) – sin(µt) y(t) dt = f (x), 10 β > 0, µ > 0 0 This is a special case of equation 3.8 .4 with g(x) = sin(βx) and λ = µ/β b sin3 λ|x – t| y(t) dt = f (x) 11 a Using the formula sin3 β = – 1 sin 3β + 4 3 4 sin β, we arrive at an equation of the form 3.5.8: b – 1 A sin 3λ|x – t| + 3 A sin λ|x – t| y(t) dt = f (x) 4 4 a b n Ak sin λk |x – t| y(t) dt = f (x), 12 a –∞ < a < b < ∞ k=1 1◦ Let us remove the... LLC © 1998 by CRC Press LLC Page 216 3.3 -4 Kernels Containing Hyperbolic Cotangent b coth(λx) – coth(λt) y(t) dt = f (x) 21 a This is a special case of equation 3.8.3 with g(x) = coth(λx) b cothk x – cothk t y(t) dt = f (x), 22 0 < k < 1 0 This is a special case of equation 3.8.3 with g(x) = cothk x 3 .4 Equations Whose Kernels Contain Logarithmic Functions 3 .4- 1 Kernels Containing Logarithmic Functions... exp[–g(x)t2 ]y(t) dt = f (x) 0 Assume that g(0) = ∞, g(∞) = 0, and gx < 0 1 The substitution z = leads to equation 3.2.12: 4g(x) 1 √ πz ∞ exp – 0 t2 y(t) dt = F (z), 4z 2 where the function F (z) is determined by the relations F = √ f (x) π by means of eliminating x g(x) and z = 1 4g(x) 3.3 Equations Whose Kernels Contain Hyperbolic Functions 3.3-1 Kernels Containing Hyperbolic Cosine b cosh(λx) – cosh(λt)... Ik = Ik (x) (3) 2◦ With the aid of (1), the integral equation can be rewritten in the form n Ak Ik = f (x) (4) k=1 Differentiating (4) with respect to x twice and taking into account (3), we obtain n n Ak λ2 Ik = fxx (x), k σ1 y(x) + k=1 σ1 = 2 Ak λ k (5) k=1 Eliminating the integral In from (4) and (5) yields n–1 Ak (λ2 – λ2 )Ik = fxx (x) – λ2 f (x) k n n σ1 y(x) + (6) k=1 Differentiating (6) with... satisfy certain relations (see item 2◦ of equation 3.8.3) b ln |x – t| y(t) dt = f (x) 2 a Carleman’s equation 1◦ Solution with b – a ≠ 4: 1 y(x) = 2 √ π (x – a)(b – x) b √ a (t – a)(b – t) ft (t) dt + t–x π ln 1 1 4 (b – a) b √ a f (t) dt (t – a)(b – t) 2◦ If b – a = 4, then for the equation to be solvable, the condition b f (t)(t – a)–1/2 (b – t)–1/2 dt = 0 a must be satisfied In this case, the solution... t| + β y(t) dt = f (x) 3 a By setting x = e–β z, t = e–β τ , f (x) = e–β g(z), y(t) = Y (τ ), we arrive at an equation of the form 3 .4. 2: B ln |z – τ | Y (τ ) dτ = g(z), A = aeβ , B = beβ A a 4 A y(t) dt = f (x), –a ≤ x ≤ a |x – t| This is a special case of equation 3 .4. 3 with b = –a Solution with 0 < a < 2A: ln –a y(x) = a 1 d 2M (a) da – – 1 2 w(t, a)f (t) dt w(x, a) –a a w(x, ξ) |x| 1 d 2 dx a |x|... form 1 e 4n n 2n 2n 2n n d y(x) = lim D x D x D f (x), D = 2 n→∞ n 4 dx To calculate the solution approximately, one should restrict oneself to a specific value of n in this formula instead of taking the limit 0 • Reference: I I Hirschman and D V Widder (1955) b ln |xβ – tβ | y(t) dt = f (x), 13 β > 0 a The transformation z = xβ , τ = tβ , w(τ ) = t1–β y(t) leads to Carleman’s equation 3 .4. 2: B ln... dτ = F (z), A = aβ , B = bβ , A where F (z) = βf z 1/β 1 ln |xβ – tµ | y(t) dt = f (x), 14 β > 0, µ > 0 0 The transformation z = xβ , τ = tµ , w(τ ) = t1–µ y(t) leads to an equation of the form 3 .4. 2: 1 ln |z – τ |w(τ ) dτ = F (z), F (z) = µf z 1/β 0 © 1998 by CRC Press LLC © 1998 by CRC Press LLC Page 219 3 .4- 3 An Equation Containing the Unknown Function of a Complicated Argument 1 A ln t + B)y(xt)... 1998 by CRC Press LLC © 1998 by CRC Press LLC Page 2 14 where the primes denote the derivatives with respect to x By comparing formulas (1) and (2), we find the relation between Ik and Ik : Ik = 2λk y(x) + λ2 Ik , k Ik = Ik (x) (3) 2◦ With the aid of (1), the integral equation can be rewritten in the form n Ak Ik = f (x) (4) k=1 Differentiating (4) with respect to x twice and taking into account (3), . = 1 2 . 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) . 23. ∞ –∞ y(t) √ |x – t| dt = f (x). This is a special case of equation 3.1. 34 with λ = 1 2 . Solution: y(x)= 1 4 ∞ –∞ f(x). condition x λ+1 f(x) x=0 =0. 43 . 1 0 Ax k + Bt m )y(xt) dt = f (x). The substitution ξ = xt leads to an equation of the form 1.1.50: x 0 Ax k+m + Bξ m y(ξ) dξ = x m+1 f(x). 44 . 1 0 y(xt) dt √ 1. LLC 46 . 1 0 t µ y(xt) (1 – t) λ dt = f (x), 0 < λ <1. The transformation ξ = xt, w(ξ)=ξ µ y(ξ) leads to the generalized Abel equation 1.1 .46 : x 0 w(ξ) dξ (x – ξ) λ = x 1+µ–λ f(x). 47 . ∞ 0 y(x
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