Existence and Uniqueness of Solution for a Fractional Order Integro-Differential Equation with Non-Local and Global Boundary Conditions

In this paper, we prove an important existence and uniqueness theorem for a fractional order Fredholm – Volterra integro-differential equation with non-local and global boundary conditions by converting it to the corresponding well known Fredholm integral equation of second kind. The considered in this paper has been solved already numerically in [1].


Introduction
Let's consider a problem under boundary condition containing non-local and global terms for a fractional order integro-differential equation q a a D y x f x K x t y t t K x t y t t q m m x a b

Existence and Uniqueness of Solution
Theorem.Let the functions   ,m are real constant, the boundary conditions (2) are linearly independent and condition (15) is satisfied.Then the boundary value problem (1)-( 2) has unique solution.
Proof: Acting in Equation ( 1) by fractional order derivative operator m q D  [2], we get since then we get the equation where Now, we write Equation (3) in the general from and accept that is known, then the fundamental solution (see [3]) is in the form where is Heaviside's unit function.Now, we try to get some basic relations.The first of these relations is Lagrange's formula.We multiply both ides of Equation (3) by fundamental solution (5) and integrate the obtained expression on (see [4,5]) to get where, 1) integrating by parts on the left hand side of expression (7) and taking into account that ( 5) is a fundamental solution of (3.1), give the first basic relation in the form Hence, the first expressions for the necessary conditions are obtained in the form It is easy to see that the second expression in (9) turns into an identity.Indeed, as it is seen from ( 5)-( 6 , therefore, the second one of necessary conditions (9) turns into identity.Now, we construct the second basic expression to get the second group of necessary conditions.For that, we multiply both sides of (3) by the derivative of ( 5) and integrate on   , a b [6,7]: Integrating by parts on the left side of the obtained expression and taking into account ( 5) and ( 6), we get the second basic relation as follows: and so the second group of the necessary conditions are obtained as Similar to the second expression of (9), we can show that the second expression of (11) turns into identity.If we continue this process, in order to get the m-th basic relation, we multiply (3) the   -th order derivative of ( 5) and integrate on to get: Here, once integrating by parts on the left side of the obtained expression gives Thus, if we take into account that ( 5) is the fundamental solution, the last relation (m-th) will be as follows: Therefore, the last group of necessary conditions will be in the form: here, as above, the second necessary condition turns into identity.Now, we join to the given m linearly independent boundary condition (7), the necessary conditions in (9), (11) and etc. (13) that are not identities, and write the system of 2m linear algebraic equations obtained with respect to the boundary values of the unknown function in the following way.
( 1) Copyright © 2011 SciRes.AM For solving the system (14) by the Cramer's rule, it is necessary that its basic determinant differ from zero.
Accept that the following condition is satisfied Then, from system (14), we get where ( , )   p q  denotes the cofactor of the elements at the intersection of p-th row and q-th column of the determi-nant  .Calculate the following expression: Then, we get: and so, Finally, coming back to (8), we take into account ( 16) and ( 17) and write the second kind Fredholm type inte-gral equation [8] for which the boundary value problem (1)-( 2) is reduced to: where By the hypothesis of theorem on the functions   constants, and boundary conditions (2) are linearly independent.
), the integral at the right side of the second condition contains the value of the function  , x a the the summation in the second expression contains the Heaviside function which is zero for = Equation (19)  has unique solution and so in all conducted operations we can come back and we conclude that the solution of (19) is the unique solution of boundary value problem (1)-(2).