On the Solvability of a Nonlocal Boundary Value Problem for Systems of Differential Equations with Involution

Abstract

We study a nonlocal boundary value problem for a system of functional-differential equations with involution and an additional parameter. The proposed approach is based on a parameterization method developed by D. Dzhumabaev. The original boundary value problem is transformed into an equivalent Cauchy problem posed at the midpoint of the interval, together with a system of algebraic equations for the unknown parameters. By exploiting the symmetry and antisymmetry properties of the solution, we reduce the resulting Cauchy problem to a system of coupled Volterra integral equations of the second kind. This reduction makes it possible to derive explicit solution representations and to establish necessary and sufficient conditions for unique solvability in terms of the invertibility of the matrix generated by the boundary conditions. Finally, a first-order functional-differential equation is analyzed to demonstrate the applicability of the proposed method and to illustrate the obtained solvability conditions.

1. Introduction

Differential equations with deviating arguments constitute an important class of mathematical models arising in biology, economics, physics, and many other applied sciences. They are widely used to describe evolutionary and dynamic processes, including interactions between biological populations and models of agricultural production. Consequently, numerous problems in these areas are formulated in terms of differential and integro-differential equations with deviating arguments.

The theory of differential equations with involution has been extensively developed in the literature. Fundamental contributions to the solvability theory were presented in the monographs of D. Przeworska-Rolewicz [1] and J. Wiener [2]. J. Wiener also investigated the existence of solutions to partial differential equations with involution by applying the method of separation of variables. The properties of involutive transformations and the corresponding operators were further studied by N. Karapetiants and S. Samko [3]. Green’s functions for one-dimensional differential equations with involution were constructed by A. Cabada and F. Tojo [4].

In recent years, the theory has been further extended in several directions. In particular, Ashyralyev and co-authors [5] investigated numerical methods and boundary value problems for differential equations with involution. Sarsenbi [6] studied nonlinear second-order boundary value problems with involution and established existence and uniqueness results. Mixed problems for nonlinear parabolic equations with involution were considered by Yuldashev [7].

The spectral properties of differential operators with involution have also been extensively studied. Sadybekov and co-authors [8] investigated spectral problems under periodic-type boundary conditions, whereas Kozhanov and Bzheumikhova [9] analyzed eigenvalues and eigenfunctions of differential operators with involution. Related asymptotic and well-posedness issues were considered in [10].

Direct and inverse initial-boundary value problems for a time-fractional heat equation with involution perturbation were investigated by Al-Salti et al. [11]. The involution concept was further generalized and applied to fractional conformable Poisson equations in [12].

Boundary value problems with nonlocal and integral conditions have been extensively investigated over the past decades for ordinary differential equations and related operator equations [13,14,15,16]. Such problems arise naturally in heat transfer, control theory, mechanics, and other applications, and numerous existence, uniqueness, and solvability results have been obtained. Against this background, the study of nonlocal boundary value problems for differential equations with involution represents a natural extension of the classical theory, since the presence of an involution substantially changes the structure of the problem and requires new analytical techniques [17,18].

The parameterization method developed by D. Dzhumabaev [19] is a constructive approach for investigating boundary value problems for ordinary differential and integro-differential equations and systems. The central idea of the method is to transform the original boundary value problem into an equivalent Cauchy problem with parameters together with a finite-dimensional system of linear algebraic equations for the introduced parameters [20,21,22]. As a consequence, necessary and sufficient conditions for the unique solvability of boundary value problems can be established.

More recently, the parameterization method has been extended to other classes of differential equations and boundary value problems, including systems of partial differential equations and pseudo-parabolic equations [23,24].

Although the parameterization method was recently applied to integrodifferential equations with involution in [25,26], the problems considered there differ substantially from the present one. In [25], an integrodifferential equation with multipoint boundary conditions was studied, whereas in [26], the equation additionally contains a derivative under the integral sign. In contrast, the present paper investigates a system of functional-differential equations with variable matrix coefficients, an additional parameter, and nonlocal integral boundary conditions. These essential differences require a different reduction procedure.

Despite the substantial development of the theory of differential equations with involution, many questions concerning nonlocal boundary value problems for systems with variable coefficients remain open. In particular, the applicability of D. Dzhumabaev’s parameterization method to this class of problems has not been sufficiently investigated.

The main contributions of the present paper can be summarized as follows:

• A constructive extension of the parameterization method is proposed for a class of nonlocal boundary value problems for systems of functional-differential equations with involution, variable matrix coefficients, and an additional parameter;
• The proposed approach transforms the original boundary value problem into an equivalent Cauchy problem with parameters, which is then reduced to a coupled system of Volterra integral equations of the second kind, providing a constructive framework for the solvability analysis;
• The invertibility of the matrix associated with the boundary conditions serves as the basis for deriving necessary and sufficient conditions ensuring the unique solvability of the boundary value problem.

An important feature of the proposed approach is that the equivalent Cauchy problem is posed at the midpoint of the interval. This formulation enables the symmetry and antisymmetry properties of the solution to be exploited, thereby simplifying the analysis. As a result, the solvability analysis is reduced to verifying the invertibility of a finite-dimensional matrix associated with the boundary conditions.

We consider the following nonlocal boundary value problem for a system of differential equations with involution posed on the interval $\left[0,T\right]$:

$${\displaystyle \frac{dx\left(t\right)}{dt}}=A_0\left(t\right)x\left(t\right)+A_1\left(t\right)x\left(T-t\right)+B\left(t\right)\lambda +f\left(t\right),$$

(1)

$$C_{0}x\left(0\right)+C_{1}x\left(T\right)=d_1,{\displaystyle \underset{0}{\overset{T}{\int }}}K\left(s\right)x\left(s\right)ds=d_2,d_1,d_2\in {\mathbb{R}}^n.$$

(2)

Here, $A_0\left(t\right)$, $A_1\left(t\right)$, $K\left(t\right),$ and $B\left(t\right)$ are continuous matrix functions on $\left[0,T\right]$ matrix functions of the order $n\times n$, while $C_0$ and $C_1$ are constant matrices of the same order, and the n- dimensional vector function $f\left(t\right)$ is continuous on $\left[0,T\right].$

We say that a pair $\left(x\left(t\right),\lambda \right)\in C^1\left(\left[0,T\right],{\mathbb{R}}^n\right)\times {\mathbb{R}}^n$ is a solution to Problems (1) and (2) if it satisfies Equation (1) together with the boundary conditions (2).

2. Application of the Parameterization Method

Let us set

$$\mu =x\left({\displaystyle \frac{T}{2}}\right),x\left(t\right)=u\left(t\right)+\mu .$$

Under this transformation, Boundary Value Problems (1) and (2) become the following equivalent parameter-dependent problem:

$${\displaystyle \frac{du\left(t\right)}{dt}}=A_0\left(t\right)u\left(t\right)+A_1\left(t\right)u\left(T-t\right)+B\left(t\right)\lambda +f\left(t\right)+\left(A_0\left(t\right)+A_1\left(t\right)\right)\mu ,$$

(3)

$$u\left({\displaystyle \frac{T}{2}}\right)=0,$$

(4)

$$C_{0}u\left(0\right)+C_{1}u\left(T\right)+\left(C_0+C_1\right)\mu =d_1,$$

(5)

$${\displaystyle \underset{0}{\overset{T}{\int }}}K\left(s\right)u\left(s\right)ds+{\displaystyle \underset{0}{\overset{T}{\int }}}K\left(s\right)ds\mu =d_2.$$

(6)

Lemma 1. 

Let $A_0\left(·\right),A_1\left(·\right),B\left(·\right),K\left(·\right)\in C\left(\left[0,T\right],{\mathbb{R}}^{n\times n}\right),$ $f\left(·\right)\in C\left(\left[0,T\right],{\mathbb{R}}^n\right).$ Then, Problems (1) and (2) and (3)–(6) are equivalent. More precisely, their solutions are connected by the transformation

$$x\left(t\right)=u\left(t\right)+\mu ,\mu =x\left({\displaystyle \frac{T}{2}}\right).$$

Proof. 

Suppose that $\left(x\left(t\right),\lambda \right)$ is a solution to Problems (1) and (2). Set

$$u\left(t\right)=x\left(t\right)-x\left({\displaystyle \frac{T}{2}}\right),\mu =x\left({\displaystyle \frac{T}{2}}\right).$$

Then, $\left(u\left(t\right),\mu ,\lambda \right)$ is a solution to Problems (3)–(6).

$${\displaystyle \frac{du\left(t\right)}{dt}}-A_0\left(t\right)u\left(t\right)-A_1\left(t\right)u\left(T-t\right)-B\left(t\right)\lambda -f\left(t\right)-\left(A_0\left(t\right)+A_1\left(t\right)\right)\mu$$

$$={\displaystyle \frac{dx\left(t\right)}{dt}}-A_0\left(t\right)x\left(t\right)-A_1\left(t\right)x\left(T-t\right)-B\left(t\right)\lambda -f\left(t\right)=0.$$

Thus, the triple $u\left(t\right),$ $\mu$, $\lambda$ satisfies (3).

Since $\mu =x\left({\displaystyle \frac{T}{2}}\right),$ we obtain

$$u\left({\displaystyle \frac{T}{2}}\right)=x\left({\displaystyle \frac{T}{2}}\right)-\mu =\mu -\mu =0.$$

Boundary Conditions (5) and (6) follow from

$$C_{0}u\left(0\right)+C_{1}u\left(T\right)+\left(C_0+C_1\right)\mu -d_1=C_{0}x\left(0\right)+C_{1}x\left(T\right)-d_1=0,$$

$${\displaystyle \underset{0}{\overset{T}{\int }}}K\left(s\right)u\left(s\right)ds+{\displaystyle \underset{0}{\overset{T}{\int }}}K\left(s\right)ds\mu -d_2={\displaystyle \underset{0}{\overset{T}{\int }}}K\left(s\right)\left(x\left(s\right)-\mu \right)ds+{\displaystyle \underset{0}{\overset{T}{\int }}}K\left(s\right)ds\mu -d_2$$

$$={\displaystyle \underset{0}{\overset{T}{\int }}}K\left(s\right)x\left(s\right)ds-d_2=0.$$

Conversely, suppose that $\left(u\left(t\right),\mu ,\lambda \right)$ is a solution to Problems (3)–(6). Set

$$x\left(t\right)=u\left(t\right)+\mu .$$

Then, the pair $\left(x\left(t\right),\lambda \right)$ is a solution to Problems (1) and (2). Indeed, since $x^{\prime }\left(t\right)=u^{\prime }\left(t\right),$ we have

$${\displaystyle \frac{dx\left(t\right)}{dt}}-A_0\left(t\right)x\left(t\right)-A_1\left(t\right)x\left(T-t\right)-B\left(t\right)\lambda -f\left(t\right)$$

$$={\displaystyle \frac{du\left(t\right)}{dt}}-A_0\left(t\right)u\left(t\right)-A_1\left(t\right)u\left(T-t\right)-B\left(t\right)\lambda -f\left(t\right)-\left(A_0\left(t\right)+A_1\left(t\right)\right)\mu =0.$$

Thus, $\left(x\left(t\right),\lambda \right)$ satisfies Equation (1).

We still must verify that the pair $\left(x\left(t\right),\lambda \right)$ satisfies the boundary conditions (2):

$$C_{0}x\left(0\right)+C_{1}x\left(T\right)-d_1=C_{0}u\left(0\right)+C_{1}u\left(T\right)+\left(C_0+C_1\right)\mu -d_1=0,$$

$${\displaystyle \underset{0}{\overset{T}{\int }}}K\left(s\right)x\left(s\right)ds-d_2={\displaystyle \underset{0}{\overset{T}{\int }}}K\left(s\right)\left(u\left(s\right)+\mu \right)ds-d_2$$

$$={\displaystyle \underset{0}{\overset{T}{\int }}}K\left(s\right)u\left(s\right)ds+{\displaystyle \underset{0}{\overset{T}{\int }}}K\left(s\right)ds\mu -d_2=0.$$

The Lemma is proved. □

Thus, solving the original boundary value problem is reduced to solving Cauchy Problems (3) and (4) together with the boundary conditions (5) and (6) for the parameters $\mu$ and $\lambda$.

3. The Solution to the Cauchy Problem

Consider Equation (3) at the points $t^{*}=T-t$; then,

$${\left.{\displaystyle \frac{du\left(t\right)}{dt}}\right|}_{t=T-t}=A_0(T-t)u(T-t)+A_1(T-t)u\left(t\right)+B(T-t)\lambda$$

$$+ f(T-t)+\left(A_0(T-t)+A_1(T-t)\right)\mu$$

Suppose that the coefficient matrices $A_0\left(t\right)$ and $A_1\left(t\right)$ satisfy the symmetry condition with respect to $T/2$:

$$A_i\left(t\right)=A_i\left(T-t\right),i=0,1,t\in \left[0,T\right].$$

Set

$$v\left(t\right)=u\left(t\right)+u\left(T-t\right),w\left(t\right)=u\left(t\right)-u\left(T-t\right).$$

This yields the following Cauchy system:

$${\displaystyle \frac{dw\left(t\right)}{dt}}=\left(A_0\left(t\right)+A_1\left(t\right)\right)v\left(t\right)+B_{+}\left(t\right)\lambda +f_{+}^{*}\left(t\right),$$

(7)

$${\displaystyle \frac{dv\left(t\right)}{dt}}=\left(A_0\left(t\right)-A_1\left(t\right)\right)w\left(t\right)+B_{-}\left(t\right)\lambda +f_{-}^{*}\left(t\right),$$

(8)

$$w\left({\displaystyle \frac{T}{2}}\right)=0,$$

(9)

$$v\left({\displaystyle \frac{T}{2}}\right)=0,$$

(10)

where $B_{\pm }\left(t\right)=B\left(t\right)\pm B(T-t),$ $f_{+}^{*}\left(t\right)=f\left(t\right)+f\left(T-t\right)+2\left(A_0\left(t\right)+A_1\left(t\right)\right)\mu ,$ and $f_{-}^{*}\left(t\right)=f\left(t\right)-f\left(T-t\right)$.

For fixed parameters $\mu$ and $\lambda ,$ the Cauchy problem (7)–(10) can be equivalently written in integral form:

$$w\left(t\right)={\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{+}\left(\xi \right)v\left(\xi \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{B}_{+}\left(\xi \right)\lambda d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{f}_{+}^{*}\left(\xi \right)d\xi ,$$

$$v\left(t\right)={\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{-}\left(\xi \right)w\left(\xi \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{B}_{-}\left(\xi \right)\lambda d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{f}_{-}^{*}\left(\xi \right)d\xi ,$$

where $A_{+}\left(t\right)=A_0\left(t\right)+A_1\left(t\right),A_{-}\left(t\right)=A_0\left(t\right)-A_1\left(t\right)$.

Therefore,

$$w\left(t\right)={\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{+}\left(\xi \right)\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{\xi }{\int }}}{A}_{-}\left(\tau \right)w\left(\tau \right)d\tau \right)d\xi$$

$$+\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{+}\left(\xi \right)\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{\xi }{\int }}}{B}_{-}\left(\tau \right)d\tau \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{B}_{+}\left(\xi \right)d\xi \right)\lambda$$

$$+{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{+}\left(\xi \right)\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{\xi }{\int }}}{f}_{-}^{*}\left(\tau \right)d\tau \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{f}_{+}^{*}\left(\xi \right)d\xi ,$$

(11)

$$v\left(t\right)={\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{-}\left(\xi \right)\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{\xi }{\int }}}{A}_{+}\left(\tau \right)v\left(\tau \right)d\tau \right)d\xi$$

$$+\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{-}\left(\xi \right)\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{\xi }{\int }}}{B}_{+}\left(\tau \right)d\tau \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{B}_{-}\left(\xi \right)d\xi \right)\lambda$$

$$+{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{-}\left(\xi \right)\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{\xi }{\int }}}{f}_{+}^{*}\left(\tau \right)d\tau \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{f}_{-}^{*}\left(\xi \right)d\xi .$$

(12)

To simplify the representation of the Volterra integral Equations (11) and (12), we introduce the kernels

$$K_{\pm }\left(t,\tau \right)=\left({\displaystyle \underset{\tau }{\overset{t}{\int }}}{A}_{+}\left(\xi \right)d\xi \right)A_{-}\left(\tau \right),K_{\mp }\left(t,\tau \right)=\left({\displaystyle \underset{\tau }{\overset{t}{\int }}}{A}_{-}\left(\xi \right)d\xi \right)A_{+}\left(\tau \right),$$

together with the functions

$$g_{+}\left(t\right)=\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{+}\left(\xi \right)\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{\xi }{\int }}}{B}_{-}\left(\tau \right)d\tau \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{B}_{+}\left(\xi \right)d\xi \right)\lambda$$

$$+{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{+}\left(\xi \right)\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{\xi }{\int }}}{f}_{-}^{*}\left(\tau \right)d\tau \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{f}_{+}^{*}\left(\xi \right)d\xi ,$$

$$g_{-}\left(t\right)=\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{-}\left(\xi \right)\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{\xi }{\int }}}{B}_{+}\left(\tau \right)d\tau \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{B}_{-}\left(\xi \right)d\xi \right)\lambda$$

$$+{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{-}\left(\xi \right)\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{\xi }{\int }}}{f}_{+}^{*}\left(\tau \right)d\tau \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{f}_{-}^{*}\left(\xi \right)d\xi .$$

Then, the Volterra integral Equations (11) and (12) take the compact form:

$$w\left(t\right)={\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{K}_{\pm }\left(t,\tau \right)w\left(\tau \right)d\tau +g_{+}\left(t\right),$$

(13)

$$v\left(t\right)={\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{K}_{\mp }\left(t,\tau \right)v\left(\tau \right)d\tau +g_{-}\left(t\right).$$

(14)

Equations (13) and (14) are Volterra integral equations of the second kind with continuous kernels $K_{\pm }(t,\tau ),K_{\mp }(t,\tau )$ and continuous right-hand sides $g_{\pm }\left(t\right)$. Consequently, they admit unique solutions in $C\left[{\displaystyle \frac{T}{2}},T\right]$.

Lemma 2. 

Assume that

$$A_i\left(t\right)=A_i\left(T-t\right),i=0,1,t\in \left[0,T\right].$$

The solution to Cauchy problems (7)–(10) satisfies the conditions $w\left(t\right)=-w\left(T-t\right)$, $v\left(t\right)=v\left(T-t\right)$ on the interval $[0,T]$.

Proof. 

Suppose that $\left(w\left(T-t\right),v\left(T-t\right)\right)$ is a solution to Cauchy problems (7)–(10) on $\left[0,T\right]$. Since the coefficients are symmetric and $A_{\pm }\left(t\right)=A_0\left(t\right)\pm A_1\left(t\right),$ Cauchy problems (7)–(10) are invariant with respect to the substitution $T-t$. Hence, the functions $w\left(T-t\right),v\left(T-t\right)$ also solve Cauchy problems (7)–(10).

Let us consider the function

$$z_1\left(t\right)=w\left(t\right)+w\left(T-t\right).$$

Using the integral representation of Solutions (11) and (12), performing the substitution $T-\tau$, and taking into account the symmetry $A_{\pm }\left(t\right)=A_{\pm }\left(T-t\right)$, we conclude that $z\left(t\right)$ is governed by the following homogeneous Volterra integral equation of the second kind:

$$z_1\left(t\right)={\int }_{{\displaystyle \frac{T}{2}}}^t{K}_{\pm }\left(t,\tau \right)z_1\left(\tau \right)d\tau ,t\in \left[{\displaystyle \frac{T}{2}},T\right]$$

Via the uniqueness theorem for homogeneous Volterra integral equations of the second kind with continuous kernels, we conclude that $z_1\left(t\right)=0.$ Hence,

$$w\left(t\right)=-w\left(T-t\right).$$

Because the transformation $t\mapsto T-t$ exchanges the intervals $\left[0,{\displaystyle \frac{T}{2}}\right]$ and $\left[{\displaystyle \frac{T}{2}},T\right]$, this identity holds throughout $\left[0,T\right]$.

The proof is analogous for $v\left(t\right)=v\left(T-t\right)$. The proof is complete. □

Lemma 3. 

Let $\left(w\left(t\right),v\left(t\right)\right)$ be the solution to Cauchy problems (7)–(10). Then, the function

$$u\left(t\right)={\displaystyle \frac{1}{2}}\left(w\left(t\right)+v\left(t\right)\right)$$

is the unique solution to the Cauchy problem (3) and (4).

Proof. 

Let $w\left(t\right)$ and $v\left(t\right)$ denote the solution to the Cauchy problem (7)–(10). Then

$u^{\prime }\left(t\right)={\displaystyle \frac{1}{2}}\left(w^{\prime }\left(t\right)+v^{\prime }\left(t\right)\right)$ and

$${\displaystyle \frac{du\left(t\right)}{dt}}-A_0\left(t\right)u\left(t\right)-A_1\left(t\right)u\left(T-t\right)-B\left(t\right)\lambda -f\left(t\right)-\left(A_0\left(t\right)+A_1\left(t\right)\right)\mu$$

$$={\displaystyle \frac{1}{2}}\left({\displaystyle \frac{dw\left(t\right)}{dt}}+{\displaystyle \frac{dv\left(t\right)}{dt}}\right)-{\displaystyle \frac{A_0\left(t\right)}{2}}\left(w\left(t\right)+v\left(t\right)\right)-{\displaystyle \frac{A_1\left(t\right)}{2}}\left(w\left(T-t\right)+v\left(T-t\right)\right)-B\left(t\right)\lambda$$

$$-f\left(t\right)-\left(A_0\left(t\right)+A_1\left(t\right)\right)\mu ={\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{dw\left(t\right)}{dt}}-\left(A_0\left(t\right)+A_1\left(t\right)\right)v\left(t\right)-B_{+}\left(t\right)\lambda -f_{+}^{*}\left(t\right)\right]$$

$$+{\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{dv\left(t\right)}{dt}}-\left(A_0\left(t\right)-A_1\left(t\right)\right)w\left(t\right)-B_{-}\left(t\right)\lambda -f_{-}^{*}\left(t\right)\right]=0.$$

The Lemma is proved. □

The solution to integral Equations (13) and (14) can be written in the form [27]

$$w\left(t\right)=g_{+}\left(t\right)+{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{R}_{\pm }\left(t,\tau \right)g_{+}\left(\tau \right)d\tau ,$$

$$v\left(t\right)=g_{-}\left(t\right)+{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{R}_{\mp }\left(t,\tau \right)g_{-}\left(\tau \right)d\tau ,$$

where $R_{\pm }\left(t,\tau \right)$ and $R_{\mp }\left(t,\tau \right)$ denote the resolvent kernels associated with the Volterra integral Equations (13) and (14), respectively.

Lemma 3 yields the following representation of the solution to Cauchy problems (3) and (4):

$$u\left(t\right)={\displaystyle \frac{1}{2}}\left(g_{+}\left(t\right)+g_{-}\left(t\right)\right)+{\displaystyle \frac{1}{2}}{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}\left(R_{\pm }\left(t,\tau \right)g_{+}\left(\tau \right)+R_{\mp }\left(t,\tau \right)g_{-}\left(\tau \right)\right)d\tau .$$

(15)

Using the definitions of $g_{\pm }\left(t\right)$, we obtain

$$g_{+}\left(t\right)=\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{+}\left(\xi \right)\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{\xi }{\int }}}{B}_{-}\left(\tau \right)d\tau \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{B}_{+}\left(\xi \right)d\xi \right)\lambda$$

$$+{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{+}\left(\xi \right)\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{\xi }{\int }}}{f}_{-}^{*}\left(\tau \right)d\tau \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{f}_{+}^{*}\left(\xi \right)d\xi ,$$

$$g_{-}\left(t\right)=\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{-}\left(\xi \right)\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{\xi }{\int }}}{B}_{+}\left(\tau \right)d\tau \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{B}_{-}\left(\xi \right)d\xi \right)\lambda$$

$$+{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{-}\left(\xi \right)\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{\xi }{\int }}}{f}_{+}^{*}\left(\tau \right)d\tau \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{f}_{-}^{*}\left(\xi \right)d\xi ,$$

$$f_{+}^{*}\left(t\right)=f\left(t\right)+f\left(T-t\right)+2A_{+}\left(t\right)\mu ,f_{-}^{*}\left(t\right)=f\left(t\right)-f\left(T-t\right),$$

which can be represented as

$$g_{+}\left(t\right)=g_{+}^0\left(t\right)+2g_{+}^1\left(t\right)\mu +g_{+}^2\left(t\right)\lambda ,$$

$$g_{-}\left(t\right)=g_{-}^0\left(t\right)+2g_{-}^1\left(t\right)\mu +g_{-}^2\left(t\right)\lambda ,$$

where $g_{+}^0\left(t\right)={\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{+}\left(\xi \right)\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{\xi }{\int }}}{f}_{-}\left(\tau \right)d\tau \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{f}_{+}\left(\xi \right)d\xi$, $g_{+}^1\left(t\right)={\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{+}\left(\xi \right)d\xi ,$

$$g_{+}^2\left(t\right)=\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{+}\left(\xi \right)\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{\xi }{\int }}}{B}_{-}\left(\tau \right)d\tau \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{B}_{+}\left(\xi \right)d\xi \right),$$

$$g_{-}^0\left(t\right)={\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{-}\left(\xi \right)\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{\xi }{\int }}}{f}_{+}\left(\tau \right)d\tau \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{f}_{-}\left(\xi \right)d\xi ,$$

$$g_{-}^1\left(t\right)={\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{-}\left(\xi \right)\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{\xi }{\int }}}{A}_{+}\left(\tau \right)d\tau \right)d\xi ,$$

$$g_{-}^2\left(t\right)=\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{A}_{-}\left(\xi \right)\left({\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{\xi }{\int }}}{B}_{+}\left(\tau \right)d\tau \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}{B}_{-}\left(\xi \right)d\xi \right).$$

Then, the solution to Cauchy problem (15) can be written as

$$u\left(t\right)={\displaystyle \frac{1}{2}}\left(g_{+}^0\left(t\right)+g_{-}^0\left(t\right)\right)+{\displaystyle \frac{1}{2}}{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}\left(R_{\pm }\left(t,\tau \right)g_{+}^0\left(\tau \right)+R_{\mp }\left(t,\tau \right)g_{-}^0\left(\tau \right)\right)d\tau$$

$$+\left(g_{+}^1\left(t\right)+g_{-}^1\left(t\right)+{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}\left(R_{\pm }\left(t,\tau \right)g_{+}^1\left(\tau \right)d\tau +R_{\mp }\left(t,\tau \right)g_{-}^1\left(\tau \right)\right)d\tau \right)\mu$$

$$+ {\displaystyle \frac{1}{2}}\left(g_{+}^2\left(t\right)+g_{-}^2\left(t\right)+{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}\left(R_{\pm }\left(t,\tau \right)g_{+}^2\left(\tau \right)d\tau +R_{\mp }\left(t,\tau \right)g_{-}^2\left(\tau \right)\right)d\tau \right)\lambda ,$$

or as

$$u\left(t\right)=u_0\left(t\right)+u_1\left(t\right)\mu +u_2\left(t\right)\lambda ,$$

(16)

where $u_0\left(t\right)={\displaystyle \frac{1}{2}}\left(g_{+}^0\left(t\right)+g_{-}^0\left(t\right)+{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}\left(R_{\pm }\left(t,\tau \right)g_{+}^0\left(\tau \right)+R_{\mp }\left(t,\tau \right)g_{-}^0\left(\tau \right)\right)d\tau \right)$,

$$u_1\left(t\right)=g_{+}^1\left(t\right)+g_{-}^1\left(t\right)+{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}\left(R_{\pm }\left(t,\tau \right)g_{+}^1\left(\tau \right)d\tau +R_{\mp }\left(t,\tau \right)g_{-}^1\left(\tau \right)\right)d\tau ,$$

$$u_2\left(t\right)={\displaystyle \frac{1}{2}}\left(g_{+}^2\left(t\right)+g_{-}^2\left(t\right)+{\displaystyle \underset{{\displaystyle \frac{T}{2}}}{\overset{t}{\int }}}\left(R_{\pm }\left(t,\tau \right)g_{+}^2\left(\tau \right)d\tau +R_{\mp }\left(t,\tau \right)g_{-}^2\left(\tau \right)\right)d\tau \right).$$

Substituting the obtained solution to the Cauchy problem (16) into Boundary Conditions (5) and (6), we obtain

$$\left\{\begin{array}{c}\left(C_0{u}_1\left(0\right)+C_1{u}_1\left(T\right)+C_0+C_1\right)\mu +\left(C_0{u}_2\left(0\right)+C_1{u}_2\left(T\right)\right)\lambda =d-C_0{u}_0\left(0\right)-C_1{u}_0\left(T\right),\hfill \\ \left({\displaystyle \underset{0}{\overset{T}{\int }}}K\left(s\right)u_1\left(s\right)ds+{\int }_0^{T}K\left(s\right)ds\right)\mu +{\displaystyle \underset{0}{\overset{T}{\int }}}K\left(s\right)u_2\left(s\right)ds\lambda =d_2-{\displaystyle \underset{0}{\overset{T}{\int }}}K\left(s\right)u_0\left(s\right)ds.\hfill \end{array}\right.$$

(17)

Using Q to denote the coefficient matrix of system (17), $\gamma ={(\mu ,\lambda )}^{\prime }$ to denote the vector of unknown parameters, and b to denote the right-hand side vector, we can write System (17) in the compact form:

$$Q\gamma =b.$$

(18)

Remark 1. 

The matrix Q is the coefficient matrix of the algebraic system (17), which is obtained by substituting solution (16) to the equivalent Cauchy problem into the boundary condition (5) and the nonlocal integral condition (6). Consequently, system (17) determines the unknown parameters μ and λ. Therefore, the unique solvability of the original boundary value problem is reduced to verifying the invertibility of the matrix Q.

4. Results

Theorem 1. 

Let

$$A_0\left(·\right),A_1\left(·\right),B\left(·\right),K\left(·\right),f\left(·\right)\in C\left(\left[0,T\right]\right)$$

and

$$A_i\left(t\right)=A_i\left(T-t\right),i=0,1,t\in \left[0,T\right].$$

Then, the boundary value problems (1) and (2) are uniquely solvable on $\left[0,T\right]$ if and only if the associated matrix Q is invertible.

Proof of Theorem 1. 

Sufficiency: Let the matrix Q be invertible. Then, the system (18)

$$Q\gamma =b$$

has a unique solution $\gamma ={\left(\mu ,\lambda \right)}^{\prime }$. Since Cauchy Problems (3) and (4) have a unique solution, the solution to the boundary value problems (1) and (2) is determined by the pair

$$x\left(t\right)=u_0\left(t\right)+\left(u_1\left(t\right)+I,u_2\left(t\right)\right)\gamma .$$

Therefore, the boundary value problem has a unique solution.

Necessity: Let the boundary value problems (1) and (2) have a unique solution. If Q is non-invertible and system $Q\gamma =b$ is solvable, then system (18) has at least two solutions ${\gamma }_1$ and ${\gamma }_2$. By Lemma 1, the boundary value problems (1) and (2) and (3)–(6) are equivalent. Therefore, problems (3)–(6) admit two distinct solution pairs

$$\left(u_0\left(t\right)+\left(u_1\left(t\right)+I,u_2\left(t\right)\right){\gamma }_1,{\lambda }_1\right),\left(u_0\left(t\right)+\left(u_1\left(t\right)+I,u_2\left(t\right)\right){\gamma }_2,{\lambda }_2\right),$$

which contradicts the uniqueness of the solution.

If, however, it is unsolvable, then the boundary value problem has no solution at all, which also contradicts unique solvability.

Therefore, the matrix Q is invertible. □

Consequence: If $det\left(Q\right)\ne 0$, then the solution to the boundary value problem (1) and (2) has the form

$$x\left(t\right)=u_0\left(t\right)+\left(u_1\left(t\right)+I,u_2\left(t\right)\right)\left(Q^{-1}b\right).$$

Theorem 2. 

Suppose that the matrix Q is non-invertible. Then, the boundary value problems (1) and (2) are solvable if and only if

$$b\perp ker\left(Q^{*}\right).$$

Proof of Theorem 2. 

Cauchy Problems (3) and (4) have a unique solution. Therefore, the solvability of the boundary value problems (1) and (2) follows from the solvability of the system of linear equations $Q\gamma =b$. From the theory of systems of linear equations, it is known that the system $Q\gamma =b$ is solvable if and only if $b\perp ker\left(Q^{*}\right)$. □

To illustrate the proposed approach, we consider the following model example.

Example 1. 

Let us consider, on the interval $\left[0,1\right]$, the boundary value problem

$$y^{\prime }\left(x\right)=a_{0}y\left(x\right)+a_{1}y\left(1-x\right)+\lambda f\left(x\right),$$

(19)

$$b_{0}y\left(0\right)+b_{1}y\left(1\right)=d_0,$$

(20)

$${\displaystyle \underset{0}{\overset{1}{\int }}}y\left(x\right)dx=d_1,$$

(21)

where $a_0,a_1,b_0,b_1,d_0,d_1\in R$, and $f\in C\left[0,1\right].$

We must find the pair $\left(y\left(x\right),\lambda \right)$, where $y\in C^1\left[0,1\right],\lambda \in \mathbb{R}.$

Let us introduce the notation $\mu =y\left(1/2\right)$ and perform the substitution $y\left(x\right)=u\left(x\right)+\mu$; from the boundary value problems (19)–(21), we proceed to the following equivalent boundary value problem with the parameter

$$u^{\prime }\left(x\right)=a_{0}u\left(x\right)+a_{1}u\left(1-x\right)+f^{*}\left(x\right),$$

(22)

$$u\left({\displaystyle \frac{1}{2}}\right)=0,$$

(23)

$$b_{0}u\left(0\right)+b_{1}u\left(1\right)+\left(b_0+b_1\right)\mu =d_0,$$

(24)

$${\displaystyle \underset{0}{\overset{1}{\int }}}u\left(x\right)dx+\mu =d_1,$$

(25)

where $f^{*}\left(x\right)=\lambda f\left(x\right)+\left(a_0+a_1\right)\mu .$

Let us consider the Equation (22) for $x^{*}=1-x$. Then,

$$u^{\prime }\left(1-x\right)=a_{0}u\left(1-x\right)+a_{1}u\left(x\right)+f^{*}(1-x).$$

(26)

Add and subtract Equations (22) and (26). Let us set $v\left(x\right)=u\left(x\right)+u\left(1-x\right),$ $w\left(x\right)=u\left(x\right)-u\left(1-x\right)$; then, we obtain the Cauchy problem for the system of differential equations:

$$w^{\prime }\left(x\right)=\left(a_0+a_1\right)v\left(x\right)+f_{+}^{*}\left(x\right),$$

(27)

$$v^{\prime }\left(x\right)=\left(a_0-a_1\right)w\left(x\right)+f_{-}^{*}\left(x\right),$$

(28)

$$w\left({\displaystyle \frac{1}{2}}\right)=0,$$

(29)

$$v\left({\displaystyle \frac{1}{2}}\right)=0,$$

(30)

where $f_{+}^{*}\left(x\right)=f^{*}\left(x\right)+f^{*}\left(1-x\right),$ $f_{-}^{*}\left(x\right)=f^{*}\left(x\right)-f^{*}\left(1-x\right)$. The solution to Cauchy problems (27)–(30) is equivalent to the following system of integral equations:

$$w\left(x\right)=\left(a_0^2-a_1^2\right){\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\left(x-\xi \right)w\left(\xi \right)d\xi +\left(a_0+a_1\right){\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\left(x-\xi \right)f_{-}^{*}\left(\xi \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}{f}_{+}^{*}\left(\xi \right)d\xi ,$$

(31)

$$v\left(x\right)=\left(a_0^2-a_1^2\right){\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\left(x-\xi \right)v\left(\xi \right)d\xi +\left(a_0-a_1\right){\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\left(x-\xi \right)f_{+}^{*}\left(\xi \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}{f}_{-}^{*}\left(\xi \right)d\xi .$$

Let us introduce the notation $v=a_0^2-a_1^2,$ $v_1=a_0+a_1.$ Let us consider Equation (31) for $x=t+{\displaystyle \frac{1}{2}}$; then,

$$w\left(t+{\displaystyle \frac{1}{2}}\right)=v{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{t+{\displaystyle \frac{1}{2}}}{\int }}}\left(t+{\displaystyle \frac{1}{2}}-\xi \right)w\left(\xi \right)d\xi$$

$$+ v_1{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{t+{\displaystyle \frac{1}{2}}}{\int }}}\left(t+{\displaystyle \frac{1}{2}}-\xi \right)f_{-}^{*}\left(\xi \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{t+{\displaystyle \frac{1}{2}}}{\int }}}{f}_{+}^{*}\left(\xi \right)d\xi .$$

After the substitution $s=\xi -{\displaystyle \frac{1}{2}}$, we obtain

$$W\left(t\right)=v{\displaystyle \underset{0}{\overset{t}{\int }}}\left(t-s\right)W\left(s\right)ds+v_1{\displaystyle \underset{0}{\overset{t}{\int }}}\left(t-s\right)F_{-}^{*}\left(s\right)ds+{\displaystyle \underset{0}{\overset{t}{\int }}}{F}_{+}^{*}\left(s\right)ds,$$

(32)

where $W\left(t\right)=w\left(t+{\displaystyle \frac{1}{2}}\right)$, $F_{-}^{*}\left(t\right)=f_{-}^{*}\left(t+{\displaystyle \frac{1}{2}}\right)$, $F_{+}^{*}\left(t\right)=f_{+}^{*}\left(t+{\displaystyle \frac{1}{2}}\right)$.

Applying the Laplace transform to Equation (32), we use the following properties:

$$\mathcal{L}\left\{{\displaystyle \underset{0}{\overset{t}{\int }}}\left(t-s\right)g\left(s\right)ds\right\}={\displaystyle \frac{1}{p^2}}\tilde{g}\left(p\right),\mathcal{L}\left\{{\displaystyle \underset{0}{\overset{t}{\int }}}g\left(s\right)ds\right\}={\displaystyle \frac{1}{p}}\tilde{g}\left(p\right),$$

and then

$$\tilde{W}\left(p\right)=v{\displaystyle \frac{1}{p^2}}\tilde{W}\left(p\right)+v_1{\displaystyle \frac{1}{p^2}}{\tilde{F}}_{-}\left(p\right)+{\displaystyle \frac{1}{p}}{\tilde{F}}_{+}\left(p\right),$$

(33)

where $\tilde{W}\left(p\right)=\mathcal{L}\left\{W\left(t\right)\right\},{\tilde{F}}_{\pm }\left(p\right)=\mathcal{L}\left\{F_{\pm }^{*}\left(t\right)\right\}$.

From (33), we obtain

$$\tilde{W}\left(p\right)={\displaystyle \frac{v_1{\tilde{F}}_{-}\left(p\right)+p{\tilde{F}}_{+}\left(p\right)}{p^2-v}}.$$

From the Laplace transform table, we have the following, for $v>0:$

$${\mathcal{L}}^{-1}\left\{{\displaystyle \frac{1}{p^2-v}}\right\}={\displaystyle \frac{1}{\sqrt{v}}}\sinh \left(\sqrt{v}t\right),{\mathcal{L}}^{-1}\left\{{\displaystyle \frac{p}{p^2-v}}\right\}=\cosh \left(\sqrt{v}t\right),$$

for $v<0:$

$${\mathcal{L}}^{-1}\left\{{\displaystyle \frac{1}{p^2-v}}\right\}={\displaystyle \frac{1}{\sqrt{-v}}}\sin \left(\sqrt{-v}t\right),{\mathcal{L}}^{-1}\left\{{\displaystyle \frac{p}{p^2-v}}\right\}=\cos \left(\sqrt{-v}t\right).$$

Then, for $v>0:$

$$W\left(t\right)={\displaystyle \frac{v_1}{\sqrt{v}}}{\displaystyle \underset{0}{\overset{t}{\int }}}\sinh \left(\sqrt{v}\left(t-s\right)\right)F_{-}^{*}\left(s\right)ds+{\displaystyle \underset{0}{\overset{t}{\int }}}\cosh \left(\sqrt{v}\left(t-s\right)\right)F_{+}^{*}\left(s\right)ds,$$

for $v<0:$

$$W\left(t\right)={\displaystyle \frac{v_1}{\sqrt{-v}}}{\displaystyle \underset{0}{\overset{t}{\int }}}\sin \left(\sqrt{-v}\left(t-s\right)\right)F_{-}^{*}\left(s\right)ds+{\displaystyle \underset{0}{\overset{t}{\int }}}\cos \left(\sqrt{-v}\left(t-s\right)\right)F_{+}^{*}\left(s\right)ds.$$

Returning to the original variables $x=t+{\displaystyle \frac{1}{2}}$, we obtain, for $v>0:$

$$w\left(x\right)={\displaystyle \frac{v_1}{\sqrt{v}}}{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\sinh \left(\sqrt{v}\left(x-s\right)\right)f_{-}^{*}\left(s\right)ds+{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\cosh \left(\sqrt{v}\left(x-s\right)\right)f_{+}^{*}\left(s\right)ds,$$

for $v<0:$

$$w\left(x\right)={\displaystyle \frac{v_1}{\sqrt{-v}}}{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\sin \left(\sqrt{-v}\left(x-s\right)\right)f_{-}^{*}\left(s\right)ds+{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\cos \left(\sqrt{-v}\left(x-s\right)\right)f_{+}^{*}\left(s\right)ds.$$

Similarly, we find for $v>0:$

$$v\left(x\right)={\displaystyle \frac{v_2}{\sqrt{v}}}{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\sinh \left(\sqrt{v}\left(x-s\right)\right)f_{+}^{*}\left(s\right)ds+{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\cosh \left(\sqrt{v}\left(x-s\right)\right)f_{-}^{*}\left(s\right)ds,$$

for $v<0:$

$$v\left(x\right)={\displaystyle \frac{v_2}{\sqrt{-v}}}{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\sin \left(\sqrt{-v}\left(x-s\right)\right)f_{+}^{*}\left(s\right)ds+{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\cos \left(\sqrt{-v}\left(x-s\right)\right)f_{-}^{*}\left(s\right)ds,$$

where $v_2=a_0-a_1.$

For $v=0:$

$$w\left(x\right)=v_1{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\left(x-\xi \right)f_{-}^{*}\left(\xi \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}{f}_{+}^{*}\left(\xi \right)d\xi ,$$

$$v\left(x\right)=v_2{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\left(x-\xi \right)f_{+}^{*}\left(\xi \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}{f}_{-}^{*}\left(\xi \right)d\xi .$$

The solution to Cauchy problems (22) and (23) can be written in the following form, for $v>0:$

$$u\left(x\right)={\displaystyle \frac{a_0}{\sqrt{v}}}{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\sinh \left(\sqrt{v}\left(x-s\right)\right)f^{*}\left(s\right)ds-{\displaystyle \frac{a_1}{\sqrt{v}}}{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\sinh \left(\sqrt{v}\left(x-s\right)\right)f^{*}\left(1-s\right)ds$$

$$+{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\cosh \left(\sqrt{v}\left(x-s\right)\right)f^{*}\left(s\right)ds,$$

for $v<0:$

$$u\left(x\right)={\displaystyle \frac{a_0}{\sqrt{-v}}}{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\sin \left(\sqrt{-v}\left(x-s\right)\right)f^{*}\left(s\right)ds-{\displaystyle \frac{a_1}{\sqrt{-v}}}{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\sin \left(\sqrt{-v}\left(x-s\right)\right)f^{*}\left(1-s\right)ds$$

$$+{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\cos \left(\sqrt{-v}\left(x-s\right)\right)f^{*}\left(s\right)ds,$$

for $v=0:$

$$u\left(x\right)=a_0{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\left(x-\xi \right)f^{*}\left(\xi \right)d\xi -a_1{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\left(x-\xi \right)f^{*}(1-\xi )d\xi +{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}{f}^{*}\left(\xi \right)d\xi .$$

Let $v=a_0^2-a_1^2=0.$ Since, $f^{*}\left(x\right)=\lambda f\left(x\right)+\left(a_0+a_1\right)\mu$,

$$a_0{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\left(x-\xi \right)\left(a_0+a_1\right)\mu d\xi -a_1{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\left(x-\xi \right)\left(a_0+a_1\right)\mu d\xi$$

$$=\left(a_0^2-a_1^2\right)\mu {\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\left(x-\xi \right)d\xi =0,$$

we get

$$u\left(x\right)=\lambda a_0{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\left(x-\xi \right)f\left(\xi \right)d\xi -\lambda a_1{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\left(x-\xi \right)f(1-\xi )d\xi$$

(34)

$$+ \lambda {\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}f\left(\xi \right)d\xi +\mu \left(a_0+a_1\right)\left(x-{\displaystyle \frac{1}{2}}\right)\lambda {\displaystyle \underset{0}{\overset{1}{\int }}}{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{\tau }{\int }}}f\left(\xi \right)d\xi d\tau$$

Substituting (34) into Conditions (24) and (25), we obtain

$$\left({\displaystyle \frac{1}{2}}\left(b_1-b_0\right)\left(a_0+a_1\right)+\left(b_1+b_0\right)\right)\mu$$

$$+(\left(b_0{a}_0-b_1{a}_1\right){\displaystyle \underset{0}{\overset{{\displaystyle \frac{1}{2}}}{\int }}}\xi f\left(\xi \right)d\xi +\left(b_0{a}_1-b_1{a}_0\right){\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{1}{\int }}}\left(1-\xi \right)f\left(\xi \right)d\xi$$

$$- b_0{\displaystyle \underset{0}{\overset{{\displaystyle \frac{1}{2}}}{\int }}}f\left(\xi \right)d\xi +b_1{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{1}{\int }}}f\left(\xi \right)d\xi )\lambda =d_0,$$

(35)

$$\mu +[{\displaystyle \frac{a_0-a_1}{2}}\left({\displaystyle \underset{0}{\overset{{\displaystyle \frac{1}{2}}}{\int }}}{\xi }^{2}f\left(\xi \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{1}{\int }}}{\left(1-\xi \right)}^{2}f\left(\xi \right)d\xi \right)$$

$$-{\displaystyle \underset{0}{\overset{{\displaystyle \frac{1}{2}}}{\int }}}\xi f\left(\xi \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{1}{\int }}}\left(1-\xi \right)f\left(\xi \right)d\xi ]\lambda =d_1.$$

(36)

Let us introduce the notation

$$A=\left({\displaystyle \frac{1}{2}}\left(b_1-b_0\right)\left(a_0+a_1\right)+\left(b_1+b_0\right)\right),$$

$$B=\left(b_0{a}_0-b_1{a}_1\right){\displaystyle \underset{0}{\overset{{\displaystyle \frac{1}{2}}}{\int }}}\xi f\left(\xi \right)d\xi +\left(b_0{a}_1-b_1{a}_0\right){\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{1}{\int }}}\left(1-\xi \right)f\left(\xi \right)d\xi$$

$$- b_0{\displaystyle \underset{0}{\overset{{\displaystyle \frac{1}{2}}}{\int }}}f\left(\xi \right)d\xi +b_1{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{1}{\int }}}f\left(\xi \right)d\xi ,$$

$$C=\left[{\displaystyle \frac{a_0-a_1}{2}}\left({\displaystyle \underset{0}{\overset{{\displaystyle \frac{1}{2}}}{\int }}}{\xi }^{2}f\left(\xi \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{1}{\int }}}{\left(1-\xi \right)}^{2}f\left(\xi \right)d\xi \right)-{\displaystyle \underset{0}{\overset{{\displaystyle \frac{1}{2}}}{\int }}}\xi f\left(\xi \right)d\xi +{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{1}{\int }}}\left(1-\xi \right)f\left(\xi \right)d\xi \right],$$

Then, the systems (35) and (36) can be written in the form

$$A\mu +B\lambda =d_0,$$

$$\mu +C\lambda =d_1.$$

Therefore, for the unique solvability of the boundary value problems (19)–(21), for $v=a_0^2-a_1^2=0,$ it is necessary and sufficient that the following condition holds: $AC-B\ne 0.$

In particular, if $f\left(x\right)={\displaystyle \frac{1}{2}}-x$ and $b_0=-b_1\ne 0,a_0\ne -a_1$, then from the system, we obtain

$$\lambda =24d_1-{\displaystyle \frac{24d_0}{b_1\left(a_0+a_1\right)}},\mu =d_1$$

and

$$u\left(x\right)=\lambda a_0{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\left(x-\xi \right)f\left(\xi \right)d\xi -\lambda a_1{\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}\left(x-\xi \right)f(1-\xi )d\xi$$

$$+ \lambda {\displaystyle \underset{{\displaystyle \frac{1}{2}}}{\overset{x}{\int }}}f\left(\xi \right)d\xi +\mu \left(a_0+a_1\right)\left(x-{\displaystyle \frac{1}{2}}\right).$$

Therefore,

$$y\left(x\right)=-{\displaystyle \frac{\lambda \left(a_0+a_1\right)}{6}}{\left(x-{\displaystyle \frac{1}{2}}\right)}^3-{\displaystyle \frac{\lambda }{2}}{\left(x-{\displaystyle \frac{1}{2}}\right)}^2$$

$$+\left(a_0+a_1\right)d_1\left(x-{\displaystyle \frac{1}{2}}\right)+d_1,\lambda =24d_1-{\displaystyle \frac{24d_0}{b_1\left(a_0+a_1\right)}}.$$

Substituting the obtained solutions for $v>0$ and $v<0$ into the boundary conditions (24) and (25) also leads to systems of linear equations with respect to $\mu$ and $\lambda$. The invertibility of the matrix of the corresponding system makes it possible to establish necessary and sufficient conditions for the solvability of the original problem.

5. Conclusions

In this paper, D. Dzhumabayev’s parameterization method is extended to a class of nonlocal boundary value problems for systems of functional-differential equations with involution and an additional parameter. After introducing a parameter defined by the value of the solution at the midpoint of the interval, we transform the original boundary value problem into an equivalent parameter-dependent Cauchy problem.

By exploiting the symmetric and antisymmetric components of the solution, we decompose the Cauchy problem into two independent Cauchy problems, each admitting a unique solution. Substituting the obtained solution into the nonlocal boundary conditions leads to a finite-dimensional system of linear algebraic equations with respect to the introduced parameters.

A representation of the solution is obtained in terms of the resolvent kernels of the corresponding integral equations. Necessary and sufficient conditions for the unique solvability of the boundary value problem are established in terms of the invertibility of the matrix arising from the boundary conditions. For a model first-order equation with involution, explicit solution formulas are derived, and the cases corresponding to various values of the parameter are examined.

The scope of the proposed method is determined by the possibility of reducing the original boundary value problem to an equivalent Cauchy problem with parameters. Since the associated Cauchy problem has a unique solution via the classical theory of ordinary differential equations, the solvability of the original boundary value problem is completely characterized by the invertibility of the matrix arising from the boundary conditions. Therefore, the proposed method is applicable to various classes of functional-differential and integro-differential equations with involution and nonlocal conditions that admit such a reduction.

The constructive nature of the proposed method also makes it promising for the investigation of more general functional-differential models, including higher-dimensional problems and more complex nonlocal boundary value problems that admit a similar reduction.