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Introduction The Volterra and Fredholm integro-differential equations (IDEs) appear in modeling many situations in areas such as mechanics, electromagnetic theory, population dynamics, pharmacokinetic studies, forestry and many others [1-6]. Existence and Uniqueness Theorems are illustrated for the systems of Volterra IDEs of first order in [7-10]. The finding for exact solutions to nonlinear integro-differential equations is a difficult problem, because this problem is reduced to a nonlinear system of algebraic (transcendental) equations. The solutions of nonlinear Volterra and Fredholm IDEs are obtained in the most cases by numerical methods [11, 12]. Exact solutions of Volterra and Fredholm integro-differential equations where studied in [4], [13-21]. In this paper we investigate two problems with equation . In the first problem the operator B contains a linear differential operator A, the Volterra operator K with kernel of convolution type and the inner product of vectors with nonlinear bounded functionals i.e the next problem (1) where In the second problem the operator B contains a linear differential operator A and the inner product of vectors with nonlinear bounded on functionals where denotes the nonlinear Fredholm integral. This paper is the continuation of the paper [16], and the generalization of the papers [18, 21], where is a linear correct operator, X, Z are Banach spaces and are linear bounded functionals on Z in [18], or are nonlinear bounded functionals on Z, in [21]. Note that, usually, a nonlinear Fredholm integro-differential problem is not correct because it is transformed to nonlinear algebraic (transcendental) equation which has most from one solutions. 1. Terminology and notation Let X, Y be complex Banach spaces and is an adjoint to X space, i.e. the set of all complex-valued linear and bounded functionals f on X. We denote by the value of f on x. We write D(A) and R(A) for the domain and the range of the operator A, respectively. A linear operator B is said to be an injective, if . An operator is called invertible if there exists the inverse operator . An operator is called continuously invertible if it is invertible and the inverse operator is continuous on Y. Remind that every linear injective operator is invertible. A function is said to be of exponential order α if where M, α are constants. Everywhere in this paper we use the notations Remind that a linear operator is called correct if and the inverse exists and is bounded on Y. Denote where is the antiderivative of and or the index i in shows the number of integrations of the function . From Theorems 1, 2 [17] immediately follows the next theorem. Theorem 1. Let . Then (i) Volterra integro-differentizal equation or (2) is reduces to Volterra integral equation (3) and has a unique zero solution. (ii) The operator A-K is closed and continuously invertible, i.e. and is continuous on . (iii) If the functions are of exponential order then nonhomogeneous equation for each has a unique solution (4) where where L is Laplace Transform and is Inverse Laplace Transform. We prove the next theorem. Theorem 2. Let the operators the vectors be defined as in (1) and A-K be an injective operator. Then: (i) For the exact solution to the problem (1) is given by (5) for every vector that solves nonlinear algebraic (transcendental) system of n equations (6) (ii) If (6) has no solution, then (1) also has no solution. Proof. (i), (ii). Since A - K is a linear injective operator and by Theorem 1, then from (1) we get (7) Acting by functional vector Φ on both sides of (7), we obtain (8) Denoting by b = Φ(u) and substituting in (8) we arrive to the nonlinear algebraic (transcendental) system of n equations (9) Let b∗ be a solution of this system satisfying b∗ = Φ(u). By substitution of b∗ into Equation (7) we obtain the solution (5). It is evident that if (6) has no solution, then (1) also has no solution. The theorem is proved. Remark. If the initial conditions in (1) are nonhomogeneous, i.e. then by transformation the problem (1) is reduced to the case with homogeneous initial conditions and then we can apply Theorem 2. The next theorem is useful to solve the nonlinear Fredholm IDEs. Theorem 3. Let the operators B, A: , A is a linear correct operator, (10) and the vectors nonlinear bounded functionals on Then: (i) For the exact solution of Problem (10) is given by (11) for every vector that solves nonlinear algebraic (transcendental) system of n equations (12) (ii) If (12) has no solution, then Problem (10) also has no solutions. Proof. (i), (ii). From (10), since A is a linear correct operator, we obtain , (13) (14) Acting by functional vector F on both sides of (13) we obtain (15) Denoting by and substituting in (15) we arrive to the nonlinear algebraic (transcendental) system of n equations . Let be a solution of this system satisfying . By substitution of into Equation (14) we obtain (11). It is evident that if (12) has no solution, then (10) also has no solution. The theorem is proved. Below we give two examples which show the usefulness of our results. Example 1. The next nonlinear integro-differential Volterra and Fredholm equation (16) has two exact solutions (17) Proof. If we compare the equation (16) with (1) it is natural to take We can use Theorem 2. Let be the operators of the direct and inverse Laplace transform, respectively. Denote by and ). The functions are continuous on each closed interval Furthermore is of exponential order 1 and is of exponential order 0. So, we can use Laplace transform. Note that every solution of (16) on is also the solution of (16) on . From by using the Laplace transform and convolutions operator we get or Now by using the inverse Laplace transform we obtain or Then Now by using (6) we obtain Solving this equation, we find Substituting these values into (5) we obtain (17). The next nonlinear Fredholm IDE we solve by Theorem 3. Example 2. Let the operator be defined by (18) Then Problem (18) has two exact real solutions (19) Proof. If we compare the equation (18) with (10) it is natural to take Then and (20) From equation (12) and (20) we get Using Derive program we compute (21) We remind that (22) Then substituting the values from (21) into (11) and using (22), we obtain which gives (19).

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M. M Baiburin

L.N. Gumilyov Eurasian National University

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