Handbook of mathematics for engineers and scienteists part 204 pps

T11.1. LINEAR EQUATIONS OF THE FIRST KIND WITH VARIABLE LIMIT OF INTEGRATION 1389 26.  x a  cos[λ(x – t)] + b  y(t) dt = f(x), f(a) =0. For b = 0, see equation T11.1.24. For b =–1, see equation T11.1.25. 1 ◦ . Solution for b(b + 1)>0: y(x)= f  x (x) b + 1 + λ 2 k(b + 1) 2  x a sin[k(x – t)]f  t (t) dt,wherek = λ  b b + 1 . 2 ◦ . Solution for b(b + 1)<0: y(x)= f  x (x) b + 1 + λ 2 k(b + 1) 2  x a sinh[k(x – t)]f  t (t) dt,wherek = λ  –b b + 1 . 27.  x a sin[λ(x – t)]y(t) dt = f(x), f(a) = f  x (a) =0. Solution: y(x)= 1 λ f  xx (x)+λf(x). 28.  x a sin  λ √ x – t  y(t) dt = f(x), f(a) =0. Solution: y(x)= 2 πλ d 2 dx 2  x a cosh  λ √ x – t  √ x – t f(t) dt. 29.  x a J 0  λ(x – t)  y(t) dt = f(x). Here, J ν (z) is the Bessel function of the first kind and f (a)=0. Solution: y(x)= 1 λ  d 2 dx 2 + λ 2  2  x a (x – t) J 1  λ(x – t)  f(t) dt. 30.  x a J 0  λ √ x – t  y(t) dt = f(x). Here, J ν (z) is the Bessel function of the first kind and f (a)=0. Solution: y(x)= d 2 dx 2  x a I 0  λ √ x – t  f(t) dt. 31.  x a I 0  λ(x – t)  y(t) dt = f(x). Here, I ν (z) is the modified Bessel function of the first kind and f(a)=0. Solution: y(x)= 1 λ  d 2 dx 2 – λ 2  2  x a (x – t) I 1  λ(x – t)  f(t) dt. 1390 INTEGRAL EQUATIONS 32.  x a I 0  λ √ x – t  y(t) dt = f(x). Here, I ν (z) is the modified Bessel function of the first kind and f(a)=0. Solution: y(x)= d 2 dx 2  x a J 0  λ √ x – t  f(t) dt. 33.  x a [g(x) – g(t)]y(t) dt = f(x). It is assumed that f(a)=f  x (a)=0 and f  x /g  x ≠ const. Solution: y(x)= d dx  f  x (x) g  x (x)  . 34.  x a [g(x) – g(t) + b]y(t) dt = f (x), f (a) =0. For b = 0, see equation T11.1.33. Solution for b ≠ 0: y(x)= 1 b f  x (x)– 1 b 2 g  x (x)  x a exp  g(t)–g(x) b  f  t (t) dt. 35.  x a [g(x) + h(t)]y(t) dt = f(x), f(a) =0. For h(t)=–g(t), see equation T11.1.33. Solution: y(x)= d dx  Φ(x) g(x)+h(x)  x a f  t (t) dt Φ(t)  , Φ(x)=exp   x a h  t (t) dt g(t)+h(t)  . 36.  x a K(x – t)y(t) dt = f(x). 1 ◦ .LetK(0)=1 and f(a)=0. Differentiating the equation with respect to x yields a Volterra equation of the second kind: y(x)+  x a K  x (x – t)y(t) dt = f  x (x). The solution of this equation can be represented in the form y(x)=f  x (x)+  x a R(x – t)f  t (t) dt.(1) Here the resolvent R(x) is related to the kernel K(x) of the original equation by R(x)=L –1  1 p  K(p) – 1  ,  K(p)=L  K(x)  , T11.2. LINEAR EQUATIONS OF THE SECOND KIND WITH VARIABLE LIMIT OF INTEGRATION 1391 where L and L –1 are the operators of the direct and inverse Laplace transforms, respectively.  K(p)=L  K(x)  =  ∞ 0 e –px K(x) dx, R(x)=L –1   R(p)  = 1 2πi  c+i∞ c–i∞ e px  R(p) dp. 2 ◦ .LetK(x) have an integrable power-law singularity at x = 0. Denote by w = w(x)the solution of the simpler auxiliary equation (compared with the original equation) with a = 0 and constant right-hand side f ≡ 1,  x 0 K(x – t)w(t) dt = 1.(2) Then the solution of the original integral equation with arbitrary right-hand side is expressed in terms of w as follows: 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. 37.  x a  g(x) – g(t) y(t) dt = f(x), f(a) =0, g  x (x) >0. Solution: y(x)= 2 π g  x (x)  1 g  x (x) d dx  2  x a f(t)g  t (t) dt √ g(x)–g(t) . 38.  x a y(t) dt  g(x) – g(t) = f(x), g  x (x) >0. Solution: y(x)= 1 π d dx  x a f(t)g  t (t) dt √ g(x)–g(t) . T11.2. Linear Equations of the Second Kind with Variable Limit of Integration 1. y(x) – λ  x a y(t) dt = f(x). Solution: y(x)=f(x)+λ  x a e λ(x–t) f(t) dt. 2. y(x) + λ  x a (x – t)y(t) dt = f (x). 1 ◦ . Solution for λ > 0: y(x)=f(x)–k  x a sin[k(x – t)]f (t) dt, k = √ λ. 2 ◦ . Solution for λ < 0: y(x)=f(x)+k  x a sinh[k(x – t)]f(t) dt, k = √ –λ. 1392 INTEGRAL EQUATIONS 3. y(x) + λ  x a (x – t) 2 y(t) dt = f(x). Solution: y(x)=f(x)–  x a R(x – t)f(t) dt, R(x)= 2 3 ke –2kx – 2 3 ke kx  cos  √ 3 kx  – √ 3 sin  √ 3 kx   , k =  1 4 λ  1/3 . 4. y(x) + λ  x a (x – t) 3 y(t) dt = f(x). Solution: y(x)=f(x)–  x a R(x – t)f(t) dt, where R(x)=  k  cosh(kx)sin(kx)–sinh(kx)cos(kx)  , k =  3 2 λ  1/4 for λ > 0, 1 2 s  sin(sx)–sinh(sx)  , s =(–6λ) 1/4 for λ < 0. 5. y(x) + A  x a (x – t) n y(t) dt = f(x), n =1, 2, 1 ◦ . Differentiating the equation n + 1 times with respect to x yields an (n+1)st-order linear ordinary differential equation with constant coefficients for y = y(x): y (n+1) x + An! y = f (n+1) x (x). This equation under the initial conditions y(a)=f(a), y  x (a)=f  x (a), , y (n) x (a)=f (n) x (a) determines the solution of the original integral equation. 2 ◦ . Solution: y(x)=f(x)+  x a R(x – t)f (t) dt, R(x)= 1 n + 1 n  k=0 exp(σ k x)  σ k cos(β k x)–β k sin(β k x)  , where the coefficients σ k and β k are given by σ k = |An!| 1 n+1 cos  2πk n + 1  , β k = |An!| 1 n+1 sin  2πk n + 1  for A < 0; σ k = |An!| 1 n+1 cos  2πk + π n + 1  , β k = |An!| 1 n+1 sin  2πk + π n + 1  for A > 0. 6. y(x) + λ  x a y(t) dt √ x – t = f(x). Abel equation of the second kind. This equation is encountered in problems of heat and mass transfer. Solution: y(x)=F (x)+πλ 2  x a exp[πλ 2 (x – t)]F (t) dt, where F (x)=f(x)–λ  x a f(t) dt √ x – t . T11.2. LINEAR EQUATIONS OF THE SECOND KIND WITH VARIABLE LIMIT OF INTEGRATION 1393 7. y(x) – λ  x 0 y(t) dt (x – t) α = f(x), 0<α <1. Generalized Abel equation of the second kind. 1 ◦ . Assume that the number α can be represented in the form α = 1 – m n ,wherem = 1, 2, , n = 2, 3, (m < n). In this case, the solution of the generalized Abel equation of the second kind can be written in closed form (in quadratures): y(x)=f(x)+  x 0 R(x – t)f (t) dt, where R(x)= n–1  ν=1 λ ν Γ ν (m/n) Γ(νm/n) x (νm/n)–1 + b m m–1  μ=0 ε μ exp  ε μ bx  + b m n–1  ν=1 λ ν Γ ν (m/n) Γ(νm/n)  m–1  μ=0 ε μ exp  ε μ bx   x 0 t (νm/n)–1 exp  –ε μ bt  dt  , b = λ n/m Γ n/m (m/n), ε μ =exp  2πμi m  , i 2 =–1, μ = 0, 1, , m – 1. 2 ◦ . Solution for any α from 0 < α < 1: y(x)=f(x)+  x 0 R(x – t)f(t) dt,whereR(x)= ∞  n=1  λΓ(1 – α)x 1–α  n xΓ  n(1 – α)  . 8. y(x) + A  x a e λ(x–t) y(t) dt = f(x). Solution: y(x)=f(x)–A  x a e (λ–A)(x–t) f(t) dt. 9. y(x) + A  x a  e λ(x–t) –1  y(t) dt = f(x). 1 ◦ . Solution for D ≡ λ(λ – 4A)>0: y(x)=f(x)– 2Aλ √ D  x a R(x – t)f(t) dt, R(x)=exp  1 2 λx  sinh  1 2 √ Dx  . 2 ◦ . Solution for D ≡ λ(λ – 4A)<0: y(x)=f(x)– 2Aλ √ |D|  x a R(x – t)f(t) dt, R(x)=exp  1 2 λx  sin  1 2  |D| x  . 3 ◦ . Solution for λ = 4A: y(x)=f(x)–4A 2  x a (x – t)exp  2A(x – t)  f(t) dt. 1394 INTEGRAL EQUATIONS 10. y(x) + A  x a (x – t)e λ(x–t) y(t) dt = f(x). 1 ◦ . Solution for A > 0: y(x)=f(x)–k  x a e λ(x–t) sin[k(x – t)]f (t) dt, k = √ A. 2 ◦ . Solution for A < 0: y(x)=f(x)+k  x a e λ(x–t) sinh[k(x – t)]f(t) dt, k = √ –A. 11. y(x) + A  x a cosh[λ(x – t)]y(t) dt = f(x). Solution: y(x)=f(x)+  x a R(x – t)f (t) dt, R(x)=exp  – 1 2 Ax   A 2 2k sinh(kx)–A cosh(kx)  , k =  λ 2 + 1 4 A 2 . 12. y(x) + A  x a sinh[λ(x – t)]y(t) dt = f(x). 1 ◦ . Solution for λ(A – λ)>0: y(x)=f(x)– Aλ k  x a sin[k(x – t)]f(t) dt,wherek =  λ(A – λ). 2 ◦ . Solution for λ(A – λ)<0: y(x)=f(x)– Aλ k  x a sinh[k(x – t)]f(t) dt,wherek =  λ(λ – A). 3 ◦ . Solution for A = λ: y(x)=f(x)–λ 2  x a (x – t)f(t) dt. 13. y(x) – λ  x 0 J 0 (x – t)y(t) dt = f(x). Here, J 0 (z) is the Bessel function of the first kind. Solution: y(x)=f(x)+  x 0 R(x – t)f (t) dt, where R(x)=λ cos  √ 1 – λ 2 x  + λ 2 √ 1 – λ 2 sin  √ 1 – λ 2 x  + λ √ 1 – λ 2  x 0 sin  √ 1 – λ 2 (x – t)  J 1 (t) t dt. T11.2. LINEAR EQUATIONS OF THE SECOND KIND WITH VARIABLE LIMIT OF INTEGRATION 1395 14. y(x) –  x a g(x)h(t)y(t) dt = f (x). Solution: y(x)=f(x)+  x a R(x, t)f(t) dt,whereR(x, t)=g(x)h(t)exp   x t g(s)h(s) ds  . 15. y(x) +  x a (x – t)g(x)y(t) dt = f(x). 1 ◦ . Solution: y(x)=f(x)+ 1 W  x a  Y 1 (x)Y 2 (t)–Y 2 (x)Y 1 (t)  g(x)f(t) dt,(1) where Y 1 = Y 1 (x)andY 2 = Y 2 (x) are two linearly independent solutions (Y 1 /Y 2 const) of the second-order linear homogeneous differential equation Y  xx + g(x)Y = 0. In this case, the Wronskian is a constant: W = Y 1 (Y 2 )  x – Y 2 (Y 1 )  x ≡ const. 2 ◦ . Given only one nontrivial solution Y 1 = Y 1 (x) of the linear homogeneous differential equation Y  xx +g(x)Y = 0, one can obtain the solution of the integral equation by formula (1) with W = 1, Y 2 (x)=Y 1 (x)  x b dξ Y 2 1 (ξ) , where b is an arbitrary number. 16. y(x) +  x a (x – t)g(t)y(t) dt = f(x). 1 ◦ . Solution: y(x)=f(x)+ 1 W  x a  Y 1 (x)Y 2 (t)–Y 2 (x)Y 1 (t)  g(t)f(t) dt,(1) where Y 1 = Y 1 (x)andY 2 = Y 2 (x) are two linearly independent solutions (Y 1 /Y 2 const) of the second-order linear homogeneous differential equation Y  xx + g(x)Y = 0. In this case, the Wronskian is a constant: W = Y 1 (Y 2 )  x – Y 2 (Y 1 )  x ≡ const. 2 ◦ . Given only one nontrivial solution Y 1 = Y 1 (x) of the linear homogeneous differential equation Y  xx +g(x)Y = 0, one can obtain the solution of the integral equation by formula (1) with W = 1, Y 2 (x)=Y 1 (x)  x b dξ Y 2 1 (ξ) , where b is an arbitrary number. 17. y(x) +  x a K(x – t)y(t) dt = f(x). Renewal equation. . 1  ,  K(p)=L  K(x)  , T11.2. LINEAR EQUATIONS OF THE SECOND KIND WITH VARIABLE LIMIT OF INTEGRATION 1391 where L and L –1 are the operators of the direct and inverse Laplace transforms, respectively.  K(p)=L  K(x)  =  ∞ 0 e –px K(x). T11.1. LINEAR EQUATIONS OF THE FIRST KIND WITH VARIABLE LIMIT OF INTEGRATION 1389 26.  x a  cos[λ(x – t)] + b  y(t) dt = f(x), f(a) =0. For b = 0, see equation T11.1.24. For b =–1, see equation. w(x)the solution of the simpler auxiliary equation (compared with the original equation) with a = 0 and constant right-hand side f ≡ 1,  x 0 K(x – t)w(t) dt = 1.(2) Then the solution of the original