HYERS-ULAM STABILITY OF SOME FREDHOLM INTEGRAL EQUATION

We prove the Hyers-Ulam stability of some kinds of Fredholm integral equation. That is, if φ(t) is an approximate solution of a Fredholm integral equation, then there exists an exact solution of the differential equation near to φ(t). AMS Subject Classification: 34K20, 26D10


Introduction and Preliminaries
S.M. Ulam [15] gave a wide-ranging talk about a series of important unsolved problems in 1940.The question concerning the stability of group homomorphisms is one of them.D.H. Hyers (see [1]) proved the stability for the case of approximately additive mappings under the assumption that G 1 and G 2 are Banach spaces.After then, the Hyers-Ulam stability of function equation (see [2,11,12,13]) and differential function (see [3,4,6,7,8,9,10,14]) was investigated by several mathematicians.
Fredholm integral equation is a kind of important integral equation.In this paper, we will try to investigate the Hyers-Ulam stability to some kinds of Fredholm integral equation: The first class of convolutional Fredholm integral and the second class of Fredholm integral and where the kernel of integral k(t) is continuous and satisfies some properties, and f (t) is continuous.Firstly, we should introduce these prerequisite knowledge.Fourier transform: Inverse Fourier transform: and we know that Fourier transform has some properties: we have And the following inequality is useful to our proof to the main results.
(1)Firstly, we will prove the stability.Let φ be a solution of Ineq.(2.1) and put where It follows Fourier transform and the property(e) that Then we have It follows the inverse Fourier transform that Then, we can know from the property (a) and (e) that It follows the inverse Fourier transform and properties (c) that and if we let g(t) = 0, we get a solution of Eq. (1.1) Then, follows the Theorem 1.1 Moreover, following property (g), we know that (2) Now, we will prove the existence and uniqueness of Eq.(1.1), and that the solution is real function.
Uniqueness.If there exists another solution φ1 (t) satisfies Eq.(1.1) and the fourier transform of it exists.By using fourier transform, we have follows the property (c), we know φ1 (t) = φ 1 (t).

.10)
It follows the inverse Fourier transform that Then, we can know from the property(a) and (e) that (2.12) It follows the inverse Fourier transform and properties(c) that and if we let g(t) = 0, we get a solution of Eq.(1.2) e isx dsdx.