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MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 16

Tham khảo tài liệu 'mathematical method in science and engineering episode 16', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | SOLUTION OF INTEGRAL EQUATIONS 553 Using f x in Equation 18.35 we can also write y x X2 - 2f x 18.38 which when substituted back into Equation 18.37 gives a differential equation to be solved for f x - x3 2xf x 18.39 the solution of which is f x Ce- 2 x2-l . Finally substituting this into Equation 18.38 gives US the solution for the integral equation as y x 1 -Cex 18.40 Because an integral equation also contains the boundary conditions constant of integration is found by substituting this solution Eq. 18.40 into the integral Equation 18.35 as c 1. We now consider the Volterra equation y x g x 4- A ex ty t dt 18.41 Jo and differentiate it with respect to X as y x g x Xy x A ex ty t dt Jo where we have used Equation 18.5 . Eliminating the integral between these two formulas we obtain y x - A l 2 z g x - g x . 18.43 The boundary condition to be imposed on this differential equation follows from integral equation 18.41 as t 0 g 0 . 18.5 SOLUTION OF INTEGRAL EQUATIONS Because the unknown function appears under an integral sign integral equations are in general more difficult to solve than differential equations. However there are also quite a few techniques that one can use in finding their solutions. In this section we introduce some of the most commonly used techniques. 554 INTEGRAL EQUATIONS 18.5.1 Method of Successive Iterations Neumann Series Consider a Fredholm equation given as f x g x X i K x t f t dt. 18.44 J a We start the Neumann sequence by taking the first term as fo x g x . 18.45 Using this as the approximate solution of Equation 18.44 we write fi x g x i K x t fo t dt. 18.46 J a We keep iterating like this to construct the Neumann sequence as fo x g x 18.47 fi x g x X i K x t fo t dt 18.48 J a fo x g x A i K x t fi t dt 18.49 J a fn ĩ x g x A i K x t fn t dt 18.50 J a This gives us the Neumann series solution as fb fb f x g x X 1 K x x g x dx X2 1 dx Jo J Ũ f dx K x x K x x g x -ị J a 18.51 If we take 6 fb 1 C a í 2 dxdt B2 J a J a B 0 18.52 SOLUTION OF .