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Handbook of mathematics for engineers and scienteists part 120

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Handbook of mathematics for engineers and scienteists part 120. 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. | Chapter 16 Integral Equations 16.1. Linear Integral Equations of the First Kind with Variable Integration Limit 16.1.1. Volterra Equations of the First Kind 16.1.1-1. Some definitions. Function and kernel classes. A Volterra linear integral equation of the first kind has the general form K x t y t dt f x 16.1.1.1 J a where y x is the unknown function a x b K x t is the kernel of the integral equation and f x is a given function the right-hand side of equation 16.1.1.1 . The functions y x and f x are usually assumed to be continuous or square integrable on a b . The kernel K x t is usually assumed either to be continuous on the square S a x b a t b or to satisfy the condition fb fb 2 2 K2 x t dxdt B2 to 16.1.1.2 aa where B is a constant that is to be square integrable on this square. It is assumed in 16.1.1.2 that K x t 0 for t x. The kernel K x t is said to be degenerate if it can be represented in the form K x t g1 x h1 t gn x hn t . The kernel K x t of an integral equation is called difference kernel if it depends only on the difference of the arguments K x t K x -1 . Polar kernels K x t L x t x -t -3 M x t 0 3 1 16.1.1.3 and logarithmic kernels kernels with logarithmic singularity K x t L x t ln x -1 M x t 16.1.1.4 where the functions L x t and M x t are continuous on S and L x x 0 are often considered as well. Polar and logarithmic kernels form a class of kernels with weak singularity. Equations containing such kernels are called equations with weak singularity. In case the functions K x t and f x are continuous the right-hand side of equation 16.1.1.1 must satisfy the following conditions 1 . If K a a 0 then f x must be constrained by f a 0. 801 802 Integral Equations 2 . If K a a Kx a a K a 1 a a 0 0 Kæra a a œ then the right-hand side of the equation must satisfy the conditions f a f x a f n a 0. 3 . If K a a Kx a a K n- 1 a a 0 K Xa a œ then the right-hand side of the equation must satisfy the conditions f a f x a f n-1 a 0. For polar kernels of the form .

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