Tham khảo tài liệu 'mixed boundary value problems episode 2', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | Overview 17 Hafen10 derived Equation and Equation in 1910. Example Consider the integral equation of the form x K x2 t2 f t dt g x 0 x 0 where K is known. We can solve equations of this type by reducing them to I K T F t dT G t 0 Jo . . via the substitutions x VỆ t y T F t f ỵ T 2ỵ T G g v . Taking the Laplace transform of both sides of Equation we have by the convolution theorem L K t L F t L G t . Defining L L t 1 sL K t we have by the convolution theorem _ d r i F d I L T G t dT d Jo 0 or f x 2d dx ix I tg t L x2 t2 dt 0 To illustrate this method we choose K t cos y t which has the Laplace transform L K t ỵ ne-k2 4s y s. Then L t L-1 ek2 4s y ns cosh t n vt Therefore the integral equation rx cosh Wx2 t2 X ------ f t dt g x 0 x 0 x2 t2 has the solution 2 d f x n dx cosh feựx2 t2 Vx2 t tg t dt 10 Hafen M. 1910 Studien uber einige Probleme der Potentialtheorie. Math. Ann. 69 517-537. 2008 by Taylor Francis Group LLC 18 Mixed Boundary Value Problems In particular if g 0 0 then f x cosh feựx2 Vx2 t2 t2 . g t dt. 2x r n Jo Example In 1970 Cooke11 proved that the solution to the integral equation T x t ln Jo x t h t dt nf x 0 x 1 is - . h t 2 d f 1 aS a da 2 n dt _Jt g a2 t2 ntVl t2 where S f di. 0 V a2 i2 We will use this result several times in this book. For example at the beginning of Chapter 4 we must solve the integral equation 1 Ấ g t ln n J 0 tanh 3x tanh 3t tanh 3x tanh 3t x d h 0 x 1 where 2h3 n. How does Equation help us here If we introduce the variables tanh 3t tanh 3 r and tanh 3x tanh 3 X then Equation transforms into an integral equation of the form Equation . Substituting back into the original variables we find that h t -w d n2 dt i1 tanh 3a S a 1 da t cosh2 3a granh2 3a tanh2 3t where q ._ Ỉa di b a i . . 0 v tanh2 3a tanh2 3i Another useful result derived by Cooke is that the solution to the .
