tailieunhanh - Handbook of mathematics for engineers and scienteists part 205

Handbook of mathematics for engineers and scienteists part 205. Tài liệu toán học quốc tế để phục vụ cho các bạn tham khảo, tài liệu bằng tiếng anh rất hữu ích cho mọi người. | 1396 Integral Equations 1 . To solve this integral equation direct and inverse Laplace transforms are used. The solution can be represented in the form rx y x f x -J 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 1 f c i - K p f R x - R p epx dp R p ---------- K p K x e-px dx. 2rn J c-i o 1 K p Jo 2 . Let w w x be the solution of the simpler auxiliary equation with a 0 and f 1 w 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 d fx fx y x w x - t f t dt f a w x - a w x - t f t dt. dx jn jn j a j a . Linear Equations of the First Kind with Constant Limits of Integration 1. J x -t y t dt f x 0 a b œ. Solution y x 2 f x . The right-hand side f x of the integral equation must satisfy certain relations. The general form of f x is as follows f x F x Ax B A -2 fx a fx b B 2 aFx a bFx b - F a - F b where F x is an arbitrary bounded twice differentiable function with bounded first derivative . fa y t 2. y dt f x 0 a œ. Solution Adia dt if f s ds y x1 4 dx Jx t - x 1 4 Jo s1 4 t - s 1 4 1 V8n r2 3 4 . A . Linear Equations of the First Kind with Constant Limits of Integration 1397 yb y t 3- J X-tk dt f X 0 k 1 It is assumed that u fi oo. Solution x where y x cot 2nk - - Í 2n 2 dx Ja x f t dt - cos2 n2 x - t 1 k Z t F t ------- dt x - t 1-k i fe i-fe Z t t - a 2 b -1 2 F t d ff dT b f s ds dt 1a t - T k JT Z s s - T 1-k A íb y t t 4 Jo dt f x 0 k 1 A 0. Solution X k-l d y x -Ax 2 X 3-2 fc -2 b t 2 dt x tX - xX 2 0 a cosf k- W r 2n 2 X fe l -2 S 2 f s ds tX - sX i k y2 a where r k is the gamma function. í y t 5 L X - t 1-A dt f x o A 1 Solution A y x tan 2n f x - f t x dt. œ X - t 1 A It assumed that the condition I f x fdx o is satisfied for some p 1 p 1 A. 7-œ sign x -1 z x x 6. ----1 -y t dt f x 0 A 1. J- x - t 1-A Solution A nX a œ f x - f t . yw 2n - - x - t i x slgn x -1 dt a b sign x -1 zx 7 .