Closed-Form Solutions of Linear Ordinary Differential Equations with General Boundary Conditions

This paper deals with the solution of boundary value problems for ordinary differential equations with general boundary conditions. We obtain closed-form solutions in a symbolic form of problems with the general n-th order differential operator, as well as the composition of linear operators. The method is based on the theory of the extensions of linear operators in Banach spaces.


Introduction
Differential equations model numerous phenomena and processes in sciences and engineering. Boundary value problems for elementary differential equations with classical boundary conditions have been studied exhaustively by many researchers and comprehensive material is now included in various standard texts. A more difficult and less investigated subject is the general or nonlocal boundary value problems. In many applications, the incorporation of general boundary conditions such as multipoint and integral conditions is inevitable. For example, in [1], the necessity of integral conditions in certain models of epidemics and population growth and the effects when neglecting these conditions are explained.
Ordinary differential equations with non-local boundary conditions were first studied at the beginning of the 20th century in [2][3][4], and later in [5]. Abstract non-local boundary value problems were considered in [6][7][8]. Operator methods for solving differential equations are analyzed in the books [9][10][11]. A description of the theory and the different directions of differential equations with non-local boundary conditions are given in the monograph [12]. An overview of non-local boundary value problems and their historical evolution can also be found in the survey papers [13][14][15][16]. Boundary value problems with integral constraints have been considered in [17][18][19][20][21][22][23][24][25], to mention but a few. Boundary value problems with multipoint and integral conditions have been studied in [26][27][28][29][30][31][32], and others. The present paper aims at providing a framework for symbolic computations for the solution of linear ordinary differential equations of order n with the most general multipoint and integral conditions, and boundary value problems for powers and products of differential operators.
In C[a, b], let the general n-th order linear ordinary differential operator, A = a 0 (x) d n dx n + a 1 (x) d n−1 dx n−1 + · · · + a n−1 (x) d dx + a n (x), where the coefficients a i (x), i = 0, . . . , n, are all continuous functions on the interval [a, b], a 0 (x) = 0, and D(A) = C n [a, b] and R(A) = C[a, b] are its domain and range, respectively.
We are concerned with the solution in the closed-form of boundary value problem for the differential equation and the boundary conditions where Υ = col(Υ 1 , . . . , Υ n ) is a vector of linear bounded functionals of the general form where . . , n, k = 0, . . . , n − 1, j = 0, . . . , m, are constants, and u (k) designates the k-th order derivative of u. The non-homogeneous term f (x) ∈ C[a, b], while the nonhomogeneous term b = col(β 1 , . . . , β n ) is a constant vector. The function u(x) ∈ C n [a, b] is the sought solution. We formulate the above problem in a convenient symbolic form and establish uniqueness solvability criteria and derive the solution in closed-form. Solution formulae to some special boundary value problems for composite differential operators are also obtained. The method is based on the theory of the extensions of linear operators in Banach spaces, see, for example, [33][34][35][36], and is an extension of the work [37] by the authors. The paper is organized as follows. In Section 2, we give some results needed for the analysis in later sections. Sections 3 and 4 contain the main findings of our investigation. In Section 5, the implementation of the technique is explained by solving two example problems. Finally, some conclusions are quoted in Section 6.

Preliminaries
Let X, Y be Banach spaces and P : X → Y a linear operator. The operator P is injective or one-to-one if for every u 1 , u 2 ∈ D(P), u 1 = u 2 implies Pu 1 = Pu 2 . The operator P is surjective or onto Y if R(P) = Y. If P is both injective and onto, then there exists the inverse operator P −1 : Y → X defined by P −1 f = u if and only if Pu = f for each f ∈ Y; in this case, R(P −1 ) = D(P).
The operator P is called closed if for every sequence u m in D(P) converging to u 0 with Pu m → y 0 , y 0 ∈ Y, it follows that u 0 ∈ D(P) and Pu 0 = y 0 . A closed operator P is called maximal if R(P) = Y and ker P = {0}.
The operator P is correct if it is both injective and onto, and the inverse operator P −1 is bounded on Y. The problem Pu = f is correct if the operator P is correct.
An operator P r : X → Y is a restriction of P, or P is an extension of P r , if D(P r ) ⊂ D(P) and P r u = Pu for all u ∈ D(P r ).
Let Ψ = col(Ψ 1 , . . . , Ψ n ) be a column vector of functionals Ψ i ∈ X * , i = 1, . . . , n, and v = (v 1 , . . . , v n ) a row vector of elements v i ∈ X, i = 1, . . . , n. By Ψ(v), we symbolize the n × n matrix where N is a n × m constant matrix. Proposition 1. Let X, Y be real Banach spaces, A : X on → Y a linear operator, z = (z 1 , . . . , z n ) a basis of ker A, and A the restriction of A defined by where the components of the vector Φ = col(Φ 1 , . . . , Φ n ) are linear bounded functionals on X. Then: (ii) If additionally to (i), the operator A −1 is bounded on the whole Y, then the operator A is correct.
(ii) Since A is injective and the operator A −1 is bounded on the whole Y by hypothesis, it follows that the operator A is correct. Proposition 2. Let X, Y be real Banach spaces, A : X on → Y a linear operator, and z = (z 1 , . . . , z n ) a basis of ker A. If there exists a correct restriction A of A defined by where the components of the vector Φ = col(Φ 1 , . . . , Φ n ) are linear bounded functionals on X, then A is closed and so A is a maximal operator.
Proof. Let u m ∈ D(A), u m → u and Au m → y, m → ∞. Denote Au m = y m . Since A is a linear operator and A −1 y m is a particular solution of Au m = y m , then every solution u m to this equation can be represented as where a m = col(a m1 , . . . , a mn ) ∈ R n . Acting by the vector Φ on (6), we obtain Since, by hypothesis, the operator A is correct, it follows from Proposition 1 that det Φ(z) = 0 and hence Substitution of (7) into (6) yields , we conclude that u ∈ D(A) and Au = y. So the operator A is closed and hence A is a maximal operator.

General Boundary Conditions
In this Section, we study boundary value problems for ordinary differential equations with general homogeneous and nonhomogeneous boundary conditions.
Let now For example, the operator . . , n, with x 0 being a fixed point in [a, b], known as Cauchy boundary conditions, is correct. In the particular case where A = d n dx n , the inverse A and the unique solution of the correct problem Au = f , for any f ∈ X, is given explicitly by From the above and Proposition 2, it is concluded that the n-th order linear operator A in (1) is closed and maximal.

Homogeneous Boundary Conditions
First, we consider the boundary value problem with homogeneous boundary conditions, namely where the linear operator A 0 : X → X is a restriction of the n-th order linear operator A in (1), the components of the vector Υ = col(Υ 1 , . . . , Υ n ) ∈ [X n−1 n ] * are as in (3), (4), and f ∈ X. Lemma 1. The linear operator A 0 is a closed operator.
Proof. Let u r , r = 1, 2, . . . , be a sequence in D(A 0 ), u r → u 0 and A 0 u r → f . Then, Au r → f and since A is a closed operator, we get that u 0 ∈ D(A) and Au 0 = f . Moreover, since Υ 1 , . . . , Υ n are bounded functionals on X, we get This is that u 0 ∈ D(A 0 ) and so Au 0 = A 0 u 0 = f . Hence, A 0 is a closed operator. Theorem 1. Let A 0 be the linear operator defined by (9), z = (z 1 , . . . , z n ) a basis of ker A, and A −1 the inverse of the correct operator A in (8). Then: (i) The operator A 0 is injective if and only if det Υ(z) = 0.
(ii) In addition, under (i), the operator A 0 is correct and the unique solution to the boundary value problem (9), for every f ∈ X, is given by That is u 0 = zc ∈ ker A 0 and as a consequence A 0 is not injective.
(ii) Let det Υ(z) = 0. Then, from statement (i) follows that the opeartor A 0 is injective and hence there exists the inverse operator A −1 0 : R(A 0 ) ⊆ X → X. Since by Lemma 1 the operator A 0 is closed, it is implied that A −1 0 is a closed operator. Furthermore, A is a correct restriction of the linear operator A and therefore the general solution of the problem Au = f , for every f ∈ X, may be written as follows where c = col(c 1 , . . . , c n ) is a vector of arbitrary constants. By requiring u to satisfy the boundary conditions in (9), we have Substitution of c into (11) yields (10), which is the unique solution of the boundary value problem A 0 u = f . In addition, it follows that R(A 0 ) = X and since A −1 0 is a closed operator with D(A −1 0 ) = R(A 0 ) = X, it is implied that A −1 0 is bounded on X. This proves that the operator A 0 is correct.

Non-Homogeneous Boundary Conditions
Next, we consider the complete non-homogeneous boundary value problem where A 1 : X → X is a restriction of the n-th order linear operator A in (1), the components of the vector Υ = col(Υ 1 , . . . , Υ n ) ∈ [X n−1 n ] * are as in (3), (4), b = col(β 1 , . . . , β n ) ∈ R n , and f ∈ X. It is noted that the operator A 1 is not linear, since its domain is a nonlinear set.
We state and prove the next theorem for the existence and construction of the unique solution of the boundary value problem (12). Theorem 2. Let A 1 be the operator defined by (12), z = (z 1 , . . . , z n ) a basis of ker A, A −1 the inverse of the correct operator A in (8), and A 0 the operator defined in (9). Then, the operator A 1 is injective if and only if A 0 is injective. In this case, for every f ∈ X and b ∈ R n , the unique solution of (12) is given by Proof. Suppose A 0 is injective. Then, ker A 0 = {0} and det Υ(z) = 0 by Theorem 1. Let u 1 , u 2 ∈ D(A 1 ) and A 1 u 1 = A 1 u 2 = f . That is, from which we get by taking into account that Az = 0 and (5). From (14), it is implied that From Theorem 1, we have Since A 0 is injective, it is concluded that u 1 = u 2 and therefore A 1 is an injective operator. Conversely, suppose A 1 is injective. Let u ∈ ker A 1 , which means It follows that u = zc, where c is a vector of constants, and Since A 1 is injective, the system (16) has only one solution, that is det Υ(z) = 0, and hence by Theorem 1 A 0 is injective. Finally, under the hypothesis that A 1 is injective, for any u ∈ D(A 1 ) that solves the completely nonhomogeneous problem A 1 u = f , we have This means that u − zΥ −1 (z)b ∈ D(A 0 ) and The solution to this problem follows from Theorem 1, namely

Composition of Operators
In this Section, we investigate boundary value problems for special differential operators, specifically the k-th power of an operator and the product of two operators, with general homogeneous boundary conditions.

k-th Power of an Operator
The k-th power of an operator A k is defined as the composition of the operator with itself, repeated k times, i.e., If A : X → X is an n-th order linear differential operator with D(A) = X n then A k : X → X is a kn-th order linear operator with D(A k ) = X kn .
Let the boundary value problem where the operator A k 0 : X → X, the operators A and A 0 are defined as in (1) and (9), respectively, the components of the vector Υ = col(Υ 1 , . . . , Υ n ) are as in (3), (4) where now Υ i ∈ [X kn−1 ] * , i = 1, . . . , n, and f ∈ X. We state the following theorem.  (18), for any f ∈ X, is given by Proof. (i) Let det Υ(z) = 0. Then, by Theorem 1, the operator A 0 is injective. Further, the operator A k 0 is injective as a composition of injective operators. Conversely, let A k 0 be injective. Then, ker A k 0 = {0}, and from the well known relation, which holds for any linear operator A 0 , follows that ker A 0 = {0}, i.e., A 0 is injective. Then, by Theorem 1, we have det Υ(z) = 0.
(ii) Let det Υ(z) = 0. Then, by Theorem 1, the operator A 0 is correct. Observe that the problem (18), for any f ∈ X, by setting Au = v 1 , Av 1 = v 2 , . . . , Av k−2 = v k−1 , Av k−1 = f can be decomposed into the k boundary value problems: By applying Theorem 1 successively, we get Finally, since R(A −k 0 ) = X and A −k 0 is bounded as a composition of bounded operators, it is concluded that the operator A k 0 is correct.
For the important category of boundary value problems for k = 2, we state the following corollary, which follows immediately from Theorem 3.

Corollary 1. The boundary value problem
is correct if and only if det Υ(z) = 0 and the unique solution, for every f ∈ X, is given by

Product of Two Operators
Here, we are looking at yet another special boundary value problem, which is the generalization of Corollary 1. In particular, we consider the boundary value problem where A, A 0 are defined as in (1) and (9), respectively, andÃ 0 : X → X is a restriction of A defined byÃ (ii) Furthermore, under (24), the operator A 0Ã0 is correct and the unique solution to the boundary value problem (22), for any f ∈ X, is given by Proof. (i)-(ii) By settingÃ 0 u = v, the problem (22) may be decomposed into the following two boundary value problems: By Theorem 1, the boundary value problem (26) is correct if and only if det Υ(z) = 0 and its unique solution is given by Similarly, the boundary value problem (27) is correct if and only if detΥ(z) = 0 and its solution is Substitution of (28) into (29) yields which is (25). Thus, the operator A 0Ã0 is correct if and only if det Υ(z) = 0 and detΥ(z) = 0, and the unique solution of A 0Ã0 u = f is given explicitly by (25).

Examples
To explain the implementation of the results presented in the previous section and to show the efficiency of the proposed solution routine, we solve two example problems. Example 1. Consider the differential Equation [29] u (x) + u(x) = 0, 0 < x < π 2 , with the constraints Comparing with (12), it is natural to take respectively. Observe that the only solution of Au = 0 is u ≡ 0, z 1 = cosx and z 2 = sinx are two linearly independent solutions of Au = 0, and the matrix is non-singular. Thus, by applying Theorem 2, we get the unique solution Example 2. Let the second order ordinary differential equation subjected to non-local boundary conditions Observe that this problem can be written as follows: Comparing now with (22), we take Finally, note that z(x) = x 3 is a fundamental solution of the homogeneous equation Au = 0 and that Υ(z) = 0,Υ(z) = 0.

Discussion
A method for constructing closed-form solutions to boundary value problems for ordinary differential equations with general multipoint and integral boundary conditions has been presented. Ready to use solution formulae in a symbolic form have been derived for some classes of boundary value problems. Specifically, we considered the following boundary value problems: where the operators A 0 , A 1 ,Ã 0 are restrictions of the n-th order linear differential operator A in (1) and Υ,Υ are vectors of linear bounded functionals as in (4) and (23), respectively, describing general non-local boundary conditions.
The proposed methodology can be specialized to other categories of boundary value problems and extended to some classes of partial differential equations.
Author Contributions: Conceptualization, E.P., S.Z. and I.F.; validation, S.Z. and I.F.; formal analysis, E.P. All authors have read and agreed to the published version of the manuscript.
Funding: This research received no external funding.