Аннотация

Аннотация

An exact real solution of linear Volterra - Fredholm and Volterra loaded integro-differential equation Bx = f is presented.

Полный текст

Introduction. Volterra - Fredholm integro-differential equations (IDEs) appear in various fields of science such as physics, biology, and engineering. Such equations are the mathematical models in the problems of mechanics, electromagnetic theory, population dynamics, fluid dynamics, pharmacokinetic studies and many others [1-5]. Usually it is difficult to solve Volterra - Fredholm IDE using analytical methods. By Adomian decomposition method in [6-7] the solution of such IDEs is given in an infinite series of components that can be recurrently determined. Spline collocation method was used to solve Volterra integro-differential equations in [8]. The numerical solution of linear Fredholm - Volterra integro-differential equations was studied in [9]. In [10] was developed an existence and uniqueness theorem of the solution to an initial value problems for a class of second-order impulsive integro-differential equations of Volterra type in a real Banach space by using the generalized Banach fixed point theorem. Existence and Uniqueness Theorems are also illustrated for the systems of Volterra IDEs of first order in [2], [11-14] and for Volterra IDEs of the first and second order in [6-7, 20, 22]. Exact solutions to Volterra - Fredholm integro-differential equations were obtained in [4, 19]. We investigate the problem (1) where an operator B contains a linear differential operator , the linear Volterra integral operator K with kernel of convolution type and the inner product of vectors g(x)Φ(u) with linear bounded functionals Φi, i = 1, ..., m. Similar problem for arbitrary Banach spaces X, Y and Fredholm integro-differential (loaded differential) equation , , where is a correct operator and are bounded integral functionals (loaded part of differential equations), was investigated in [20] and for with nonlocal boundary conditions in [21]. We prove in this paper that the operator is continuously invertible and, using Oinarov extensions of linear operators in Banach space [22], we obtain exact real solutions in closed form of Problem (1). The technique which we present, is simple to use and can be easily incorporated to any Computer Algebra System (CAS). Terminology and notation. Let Х be a complex Banach space and Х* its adjoint space, i.e. the set of all complex-valued linear and bounded functionals on X. We denote by f(x) 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 operator, if kerB = {0}. An operator A is called invertible if there exists the inverse operator A-1. An operator A is called continuously invertible if it is invertible and the inverse operator A-1 is continuous. Remind that if a linear operator is injective, then it is invertible. A bounded operator B: X → Y is said to be compact if the image {Bun} of any bounded sequence {un} of X contains a Cauchy subsequence. A function f(x) is said to be of exponential order α if |f(x)| ≤ Meαx, 0 ≤ x < ∞, where M and α are constants. By F(s) = L[f(x)] we denote Laplace transform of function f(x) and by L-1[F(s)] = f(x) the inverse Laplace transform of F(s). An equation is called loaded equation if it contains the solution function on a manifold with dimension less than the dimension of domain of this function [23]. For example an ordinary loaded differential equation is represented by where xj are fixed points. If Φi ∈ X∗, i = 1, ..., m, then we denote by Φ = = col(Φ1, ..., Φm) and Φ(x) = col(Φ1(x), ..., Φm(x)). Let g = (g1, ..., gm) be a vector of Xm. We will denote by Φ(g) the m × m matrix whose i,j-th entry is the value of functional Φi on element gj and by Im the identity m × m matrix, respectively. Note that Φ(gC) = Φ(g)C, where C is a m×m constant matrix. By c and 0 we will denote a vector and zero vector, respectively. We also denote u = u(x), g = g(x), f = f(x), Ω = [a, b] × [a, b], C0[a, b] = = {u(x)∈C[a; b]: u(a) = 0}, C0n[a; b] = {u(x) ∈ Cn[a; b]: u(j-1)(a) = 0, j = 1, …, n}, and where α0, α1, ..., αn are constants, g1(x), ..., gm(x) are linearly independent in C[a,b] functions, g(x) = (g1(x), ..., gm(x)), Φ1, ..., Φm the set of bounded linearly independent on Cn-1[a,b] functionals, Φ(u) = col(Φ1(u), ..., Φm(u)) and Kj(0) = 0, j = 0,1, ..., n. Denote Kj,0(z) = Kj(z), j = 0,1, ..., n. Let Kj,i+1(z) be the antiderivative of Kj,i(z), such that Kj,i(0) = 0, j = 0,1, ..., n, i = 0,1, ..., n-1. The index i in Kj,i(z) shows the number of integrations of the function Kj(z). Also denote by (Kj,i)(z) = Kj,i(z). Theorem 1. Let α0 ≠ 0. Then: (i) Volterra integro-differential equation u (x) - Ku(x) = 0 or (2) is reduces to Volterra integral equation (3) where and has a unique zero solution. (ii) The operator is closed and continuously invertible. (iii) If the functions u(x), g(x), f(x) are of exponential order α then the nonhomogenious equation (x)-Ku(x) = f(x) for each f(x) has a unique solution (4) Where Proof. (i), (ii). First for j = 0,1, ..., n, i = 0,1, ..., n-1 we prove two formulas (5) (6) It is easy to verify that (7) Further by using Fubbini theorem and integrating by parts we obtain Now taking into account (7) we get So we proved (5). Further by using Fubbini theorem and (7) we have So we proved (6). Now we will prove that is injective. Rewrite (2) as Integrating with respect to x both sides of the above equation and using the initial conditions and (5), (6) obtain The second integration gives By the third integration we get By n-th integration obtain or (2). Since is a continuous function, the equation (3) has a unique zero solution and so (2) has also a unique zero solution. Then the operator -K is invertible. The closedness of the operator is proved in [24]. The operator K is bounded as compact operator. Remind that and So We will show that -K is closed. Let un ∈ and Then and The last relation follows from the boundedness of K. From since is closed, follows that u0 ∈ D( ) and = + Ku0. Then So the operator is closed, which implies the closedness of Now, since R( - K) = C[a,b], by Closed Graph Theorem, the operator (A - K)-1 is bounded and the operator is continuously invertible. (iii) By using Laplace transform on both sides of (x) - Ku(x) = f(x) and then by inverse Laplace transform we obtain (4). The theorem is proved. The next theorem is useful for solving of linear Volterra - Fredholm or Volterra loaded integro-differential equations with initial boundary conditions. Theorem 2. Let the operator be defined by the equality Bu(x) = (x) - Ku(x) - gΦ(u)(x) = f(x). (8) Then: (i) R(K) ⊂ C0[a,b], the operator K is compact and R( - K) = C[a,b]. (ii) The operator B is invertible if and only if detW ≠ 0, where (9) Proof. (i) Let Then z(a) = 0 and R(K)⊂C0[a, b]. Denote by Then We will show that the operators are compact. Note that for x1, x2 ∈ [a, b] hold Since Kj(x - t) is continuous, it is uniformly continuous and |Kju(x1) - Kju(x2)| can be made arbitrarily small by taking |x1 - x2| small. How small |x1 - x2| should be depends only on In other words, the Kju are equicontinuous for a bounded set of u. Sinse {Kju} is equibounded for a bounded set {u} for a similar reason, by Ascoli - Arzeia criterion follows that {Kjui} contains a uniformly convergent subsequence if {ui} is bounded. Since this means that {Kjui} contains a Cauchy subsequence in C[a, b], then Kj, j = 0, 1, ..., n is compact. From the last follows that is a compact operator as a finite sum of compact operators. Now we find R( -K). Let ( -K)u=y, y C[a, b]. This equation is equivalent to The operator is compact, because K is compact and is bounded. By second Fredholm Theorem , where ψ ∈ ker(I - K∗ ). By the first Fredholm Theorem dim ker(I - K∗ ) = dim ker But ker = ker( -K) = {0}, since -K by Theorem 1 is invertible. So ker(I - K∗ ) ={0} and R = Then R( - K) = C[a, b]. (ii) Since the operator B is linear, it is sufficiently to prove that B is injective. Let Bu = 0 and detW ≠ 0. Then Bu = u - Ku - gΦ(u) = 0. (10) Taking into account that the operator - K is invertible and R( - K) = = C[a, b] we can use the inverse operator ( - K)-1 on both sides of Equation (10) and get u = ( - K)-1gΦ(u). (11) Applying the vector Φ on (11) and using the linearity of Φ1, ..., Φm, we arrive at the equations (12) From the last equation, since detW ≠ 0, follows that Φ(u) = 0. Substituting Φ(u) = 0 into (11) we obtain u = 0. So kerB = {0} and the operator B is invertible. Conversely. Let det W = 0. Then there exists a nonzero vector c = = col(c1, ..., cm) such that Wc = 0. Consider the element u0 = ( - K)-1gc. Observe that u0 ≠ 0 since g1, ..., gm is a linearly independent set in C[a, b]. Then Bu0 = ( -K)u0-gΦ(u0) = gc-gΦ(( -K)-1g)c = g[Im-Φ(( - K)-1g)]c = = gWc = 0. Consequently, u0 ∈ ker B and B is not injective. Hence the operator B is invertible if and only if det W ≠ 0. The theorem is proved. Corollary. The equation Bu = f for each f ∈ C[a, b] has a unique solution (13) if and only if detW ≠ 0. The operator B is continuously invertible on C[a, b] and (14) Proof. Acting by the inverse operator ( - K)-1 on both sides of Equation (8) we get u = ( - K)-1f + ( - K)-1gΦ(u). (15) Applying the vector Φ on (15) and using the linearity of Φ1, ..., Φm, we arrive at the equations (16) Substituting (16) into (15) we obtain the exact solution (13) of (8). The boundedness of the operator ( - K)-1 on C[a, b] and the boundedness of the functionals Φ1, ..., Φm on Cn-1[a, b] imply the boundedness of the operator B-1 on C[a, b]. In the next examples we assume that a function u(x) is of exponential order α, 0 ≤ x < ∞. Example 1. The next linear Volterra - Fredholm integro-differential equation (17) has the unique exact solution (18) Proof. If we compare the equation (17) with (8) it is natural to take n = 2, m = 1, u(x)= (x), D( )={u(x) C2[0,π]:u(0)=0, (0)=0}, Ku(x)= , Φ(u)= , Bu= Ku - gΦ(u), D(B) = D( ), g(x )= f(x) = Note that . This means that Φ ∈C[0, π]∗. Then Φ ∈ C1[0, π]∗. We can use Theorem 2. Let be the operators of the direct and inverse Laplace transforms, respectively. Denote by [u(x)] = = U(s) and [y(x)] = Y(s). The functions sin2x are continuous on each closed interval [0, b], b < ∞ and of exponential order 0. So we can use Laplace transform. Note that every solution of (17) on [0, ∞) is also the solution of (17) on [0, π]. From by using the Laplace transform and convolution operator we get or , or Now by using the inverse Laplace transform we obtain or Then for and we get Further we compute Substituting these values into (13) we obtain (18). ▲ Example 2. The next linear Volterra loaded integro-differential equation (19) u(0) = 0, (0) = 0, u(x) ∈ [0, π], has the unique exact solution u(x) = sin2x - 2x. (20) Proof. If we compare the equation (19) with (8) it is natural to take n = = 2, m = 1, u (x) = , D( ) = ∈ C2[0, π]: u(0) = 0, , Ku(x) = - , Φ(u) = u(π), Bu = u - Ku - gΦ(u), D(B) = = D( ), g(x) = sin x, f(x) = (2π - 8)sin x. Note that |Φ(u)| = |u(π)| ≤ ||u||C. This means that Φ ∈ C[0, π]∗. We can use Theorem 2. Let be the operators of the direct and inverse Laplace transforms, respectively. Denote by [u(x)] = U(s) and [y(x)] = Y (s). The functions sin x, cos x are continuous on each closed interval [0, b], b < ∞ and of exponential order 0. So we can use Laplace transform. Note that every solution of (19) on [0, ∞) is also the solution of (19) on [0, π]. From by using the Laplace transform and convolution operator we get or or Now by using the inverse Laplace transform we obtain or Then for g(x) = sinx and f(x) = (2π - 8)sinx we get and Further we compute Substituting these values into (13) we obtain (20).