HANDBOOK OFINTEGRAL EQUATIONS phần 1 pdf

... f(x). For µ =0 ,1, 2, , see equations 1. 1 .1, 1. 1.2, 1. 1.4, 1. 1 .12 , and 1. 1.23. For 1& lt;µ < 0, see equation 1. 1.42. Set µ = n – λ, where n =1, 2, and 0 ≤ λ < 1, and f(a)=f  x (a)=···= f (n 1) x (a)=0. On ... Degenerate One 11 .13 -1. Approximation of the Kernel 11 .13 -2. The Approximate Solution 11 .14 . The Bateman Method 11 .14 -1. The General Scheme of the M...

Ngày tải lên: 23/07/2014, 16:20

... – 1  – 1 sinh 2n 1 x + n 1  k =1 ( 1) k 1 2 k (n – 1) (n – 2) (n – k) (2n – 3)(2n – 5) (2n – 2k – 1) 1 sinh 2n–2k 1 x  , n =1, 2, 31.  dx sinh 2n +1 x = cosh x 2n  – 1 sinh 2n x + n 1  k =1 ( 1) k 1 (2n 1) (2n–3) ... a) –3/2 2a 1/ 2 – 2(πp) 1/ 2 e ap erfc  √ ap  10 x 1/ 2 (x + a) 1 (π/p) 1/ 2 – πa 1/ 2 e ap erfc  √ ap  11 x 1/ 2 (x + a) 1 πa 1/...

Ngày tải lên: 23/07/2014, 16:20

... Subsections 12 .2-5, 12 .3-6, and 12 .3 -10 ): Φ + (t)=X + (t)  1 2 H(t) X + (t) +Ψ(t)– 1 2 P ν 1 (t)  , Φ – (t)=X – (t)  – 1 2 H(t) X + (t) +Ψ(t)– 1 2 P ν 1 (t)  , (9) where Ψ(t)= 1 2πi  L H(τ) X + (τ) dτ τ ... equation has the form y(x)= 1 1+λ 2 π 2  f(x)+  1 0 ξ α (1 – x) α x α (1 – ξ) α  1 ξ – x – 1 x + ξ – 2xξ  f(ξ) dξ  + C (1 – x) β x 1+ β , α = 2...

Ngày tải lên: 23/07/2014, 16:20

... B n : B 1 =  1 0 K(t 1 , t 1 ) dt 1 =  1 0 t 1 e t 1 dt 1 =1, B 2 =  1 0  1 0     t 1 e t 1 t 1 e t 2 t 2 e t 1 t 2 e t 2     dt 1 dt 2 =0. Page 536 © 19 98 by CRC Press LLC © 19 98 by ... kernel. • References for Section 11 .14 : H. Bateman (19 22), E. Goursat (19 23), L. V. Kantorovich and V. I. Krylov (19 58). 11 .15 . The Collocation Method...

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... h i 1  j =1 β j K ij y j 1 – 1 2 hK ii , i =2, , n, x i = a +(i – 1) h, n = b – a h +1, β j =  1 2 for j =1, 1 for j > ;1, where the notation coincides with that introduced in Subsection 9 .10 -1. ... rule: A 1 = A 2 = ···= A n 1 = h, A n =0, h = b – a n – 1 , x i = a + h(i – 1) (i =1, , n). (2) Trapezoidal rule: A 1 = A n = 1 2 h, A 2 = A 3 = ···= A n 1 = h, h = b...

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... 6.2. 31 with f(t)=λ, a = 0, and b =1. 38. y(x)+A  1 0 y(t)y(x + λt) dt =0. This is a special case of equation 6.2.35 with f(t) ≡ A, a = 0, and b =1. 1 ◦ . A solution: y(x)= C(λ +1) A [1 – e C(λ +1) ] e Cx , where ... y(x)+  b a f(t)y(t)y(x + λt) dt =0, λ >0. 1 ◦ . Solutions: y 1 (x)=– 1 I 0 exp(–Cx), y 2 (x)= I 2 – I 1 x I 2 1 – I 0 I 2 exp(–Cx), I m =  b a t m exp...

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... A 2 are given by A 1 = f 1 + λ(f 1 s 3 – f 2 s 2 ) (s 2 2 – s 1 s 3 )λ 2 – (s 1 – s 3 )λ +1 , A 2 = –f 2 + λ(f 2 s 1 – f 1 s 2 ) (s 2 2 – s 1 s 3 )λ 2 – (s 1 – s 3 )λ +1 , f 1 =  b a f(x)g(x) ... dx. 1 ◦ . Solution with λ ≠ λ 1, 2 : y(x)=f(x)+λ(A 1 + A 2 x), where the constants A 1 and A 2 are given by A 1 = f 1 – λ  B(f 1 h 1 + f 2 h 2 ) – 1 2 Af 2 (b...

Ngày tải lên: 23/07/2014, 16:20

... A 2 ), where A 1 = 12 f 1 +6λ ( f 1 ∆ 2 –2f 2 ∆ 1 ) λ 2 ∆ 4 1 +12 , A 2 = 12 f 2 +2λ ( 3f 2 ∆ 2 –2f 1 ∆ 3 ) λ 2 ∆ 4 1 +12 , f 1 =  b a f(x) dx, f 2 =  b a xf(x) dx, ∆ n = b n – a n . Page 247 © 19 98 by ... 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)...

Ngày tải lên: 23/07/2014, 16:20

... β k sin(β k x)  , where 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. 32. ... form y(x)=f(x)+  x a {B 1 sin[µ 1 (x – t)] + B 2 sin  µ 2 (x – t)  f(t) dt, µ i =  |z i |, where B 1 and B 2 are determined from the system B 1 µ 1 λ...

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... equation (1) is known and is determined by formula (5). (a) For ϕ(x)=x λ , y(x)=λx λ 1 L 1  x λ , 1 λx λ 1 d dx  f(x)+λ 2 x λ 1 L 2  x λ , 1 λx λ 1 d dx   x a R  x λ , t λ  t λ 1 f(t) dt. (b) ... f(x). Solution: y(x)=A  d 2 dx 2 + λ 2  n  x a (x – t) n–ν 1 J n–ν 1 [λ(x – t)] f(t) dt, A =  2 λ  n 1 Γ(ν +1) Γ(n – ν) Γ(2ν +1) Γ(2n – 2ν – 1) , where – 1 2 < ν <...

Ngày tải lên: 23/07/2014, 16:20