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Handbook of mathematics for engineers and scienteists part 129

Handbook of mathematics for engineers and scienteists part 129. 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. | 864 Integral Equations For the case in which t y is a polynomial in y i.e. t y po t pi t y Pn t yn 16.5.3.7 where p0 t . pn t are for instance continuous functions of t on the interval a b system 16.5.3.6 becomes a system of nonlinear algebraic equations for A1 . Am. The number of solutions of the integral equation 16.5.3.3 is equal to the number of solutions of system 16.5.3.6 . Each solution of system 16.5.3.6 generates a solution 16.5.3.5 of the integral equation. 2 . Consider the Urysohn equation with a degenerate kernel of the special form rb _ n y x 9k x fk t y t f dt h x . 16.5.3.8 7 a I k 1 Its solution has the form n y x h x 2 Xk9k x 16.5.3.9 where the constants Xk can be defined by solving the algebraic or transcendental system of equations Xm i fm t h t XkQk t dt 0 m 1 . n. 16.5.3.10 Ja k i To different roots of this system there are different corresponding solutions of the nonlinear integral equation. It may happen that real solutions are absent. A solution of the Urysohn equation with a degenerate kernel in the general form i b n 1 f x y x J 1229k x y x hk t y t dt 0 16.5.3.11 can be represented in the implicit form f x y x 22 Xk 9k x y x 0 16.5.3.12 k 1 where the parameters Xk are determined from the system of algebraic or transcendental equations Xk - Hk X 0 k 1 . n c b 16.5.3.13 Hk X hk t y t dt X X1 . Xn . J a Into system 16.5.3.13 we must substitute the function y x y x X which can be obtained by solving equation 16.5.3.12 . The number of solutions of the integral equation is defined by the number of solutions obtained from 16.5.3.12 and 16.5.3.13 . It can happen that there is no solution. 16.5. Nonlinear Integral Equations 865 Example 1. Let us solve the integral equation with parameter A. We write y x a i Jo xty3 t dt ty3 t dt. In this case it follows from 16.5.3.14 that y x AAx. 16.5.3.14 16.5.3.15 16.5.3.16 A On substituting y x in the form 16.5.3.16 into relation 16.5.3.15 we obtain A i tA3A3t3 dt. Jo Hence A 5 A3 A3. For A 0 equation .