# On some boundary value problems with non-local and periodic conditions

## Abstract

In this work, we concern non-local and periodic boundary value problems. We will prove the existence of at least one solution of these problems such that the functions satisfy the growth condition. Hence, we will study the existence of at least one solution for a boundary value problem with periodic and integrable conditions.

## Introduction

Differential equations with non-local conditions were considered in many works (see , , , and ). Also, anti-periodic problems can be found in  and .

Here, we study the existence of at least one solution for the boundary value problem with non-local and periodic conditions:

$$\left\{ \begin{array}{c}x^{\prime\prime}(t)~=~f(t,~x(t),~x^{\prime}(t))~~~\text{a.e.}~t~\in~(0,2\pi),\\ x(0)~=~x(2\pi)~~\text{and}~~\sum_{k=1}^{m}~a_{k}~x(\tau_{k})~=~x_{0} \end{array}\right.$$
(1)

where x0R, 0<τ1<τ2<<τm<2π and ak≠0 for all k=1,2,,m.

Also, the boundary value problem with integral and periodic conditions:

$$\left\{ \begin{array}{c}x^{\prime\prime}(t)~=~f(t,~x(t),~x^{\prime}(t))~~~\text{a.e.}~~t~\in~(0,2\pi),\\ x(0)~=~x(2\pi)~~\text{and}~~\int_{0}^{2\pi}~x(t)~dt~=~x_{0} \end{array}\right.$$
(2)

will be considered.

Problem (2) was studied in , but the author has not shown the equivalence between the differential problem (2) and the integral equation equivalent with it.

Here, we prove, by using nonlinear alternative of Leray-Schauder type, the existence of at least one solution for problem (1) such that the function f:I×R×RR, I=[0,2π] satisfies the growth conditions.

## Preliminaries

### Theorem 1

(Nonlinear alternative of Leray-Schauder type)  Let E be a Banach space and Ω be a bounded open subset of E, 0Ω and $$T:\bar {\Omega }\rightarrow E$$ be a completely continuous operator. Then, either there exists xΩ,λ>1 such that T(x)=λx, or there exists a fixed point $$x^{\ast } \in \bar {\Omega }$$.

Denote by C(I) the space of all continuous functions defined on the interval I with norm

$$||u||_{C}~=~\sup_{t \in I}~|u(t)|$$

and by L1(I) the space of all Lebesgue integrable functions on the interval I with norm

$$||u||_{L_{1}}~=~\int_{I}~|u(t)|~dt.$$

The growth condition on the function f means that

$$|f(t,u)|~\leq ~a(t)~+~b~|u|,$$

where a(t)L1, b is a nonnegative constant.

## Main results

Let the function f:I×R×RR satisfy the following assumptions:

1. (1)

f:I×R×RR is measurable in tI for any (u1,u2)R×R

2. (2)

f:I×R×RR is continuous in (u1,u2)R×R for any tI

3. (3)

There exist two positive constants b1,b2 and a function c(t)L1(I) such that

$$|f(t,~u_{1},~u_{2})|~\leq~c(t)~+~b_{1}~|u_{1}|~+~b_{2}~|u_{2}|.$$

### Lemma 1

Let the assumptions (1)–(3) be satisfied. If the solution of the boundary value problem (1) exists, then it can be represented by

$$\begin{array}{@{}rcl@{}} x(t)&=&A~\left(x_{0}~-~\sum_{k=1}^{m}~a_{k}~\int_{0}^{\tau_{k}}~(\tau_{k}~-~s)~y(s)~ds\right)\\ &+&\left(t~-~A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right)~\left(\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds\right)~+~\int_{0}^{t}~(t~-~s)~y(s)~ds,\\ \end{array}$$

where

$$\begin{array}{@{}rcl@{}} y(t)&=&f\left(t,y_{1}(t),y_{2}(t)\right),\\ y_{1}(t)&=&A~\left(x_{0}~-~\sum_{k=1}^{m}~a_{k}~\int_{0}^{\tau_{k}}~(\tau_{k}~-~s)~y(s)~ds\right)\\ &+&\left(t~-~A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right)~\left(\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds\right)\\ &+&\int_{0}^{t}~(t~-~s)~y(s)~ds\\ \text{and}~~~y_{2}(t)&=&\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds~+~\int_{0}^{t} y(s)~ds,~~t\in I. \end{array}$$
(3)

### Proof

Let y=x′′(t)=f(t,x,x). □

Integrating both sides, we obtain

$$x^{\prime}(t)~-~x^{\prime}(0)~=~\int_{0}^{t}~y(s)~ds.$$

Integrating again, we get

$$\begin{array}{@{}rcl@{}} x(t)~=~x(0)&+&t~x^{\prime}(0)~+~\int_{0}^{t}~\int_{0}^{s}~y(\theta)~d\theta~ds\\ ~=~x(0)&+&t~x^{\prime}(0)~+~\int_{0}^{t}~(t~-~s)~y(s)~ds. \end{array}$$

From the boundary condition, we obtain

$$x^{\prime}(0)~=~\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds,$$

then

$$\begin{array}{@{}rcl@{}} x^{\prime}(t)&=&x^{\prime}(0)~+~\int_{0}^{t}~y(s)~ds\\ &=&\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds~+~\int_{0}^{t}~y(s)~ds. \end{array}$$
(4)

Now,

$$\begin{array}{@{}rcl@{}} x(t)&=&x(0)~+~t~x^{\prime}(0)~+~\int_{0}^{t}~(t~-~s)~y(s)~ds,\\ \text{then}~~~~x(\tau_{k})&=&x(0)~+~\tau_{k}~x^{\prime}(0)~+~\int_{0}^{\tau_{k}}~(\tau_{k}~-~s)~y(s)~ds\\ \text{and}~~~\sum_{k=1}^{m}~a_{k}~x(\tau_{k})&=&x(0)\sum_{k=1}^{m}~a_{k}~~+~\sum_{k=1}^{m}~a_{k}~\tau_{k}~x^{\prime}(0)~+~\sum_{k=1}^{m}~a_{k}~ \int_{0}^{\tau_{k}}~(\tau_{k}~-~s)~y(s)~ds. \end{array}$$

Take $$A=(\sum _{k=1}^{m}~a_{k})^{-1}$$, then

$$x(0)~=~A~\left(x_{0}~-~\sum_{k=1}^{m}~a_{k}~\tau_{k}~x^{\prime}(0)~-~\sum_{k=1}^{m}~a_{k}~\int_{0}^{\tau_{k}}~(\tau_{k}~-~s)~y(s)~ds\right).$$

Substituting the values of x(0) and x(0) in x(t), we get

$$\begin{array}{@{}rcl@{}} x(t)&=&A~\left(x_{0}~-~\sum_{k=1}^{m}~a_{k}~\int_{0}^{\tau_{k}}~(\tau_{k}~-~s)~y(s)~ds\right)\\ &+&\left(t~-~A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right)~\left(\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds\right)~+~\int_{0}^{t}~(t~-~s)~y(s)~ds.\\ \end{array}$$
(5)

Inserting (4) and (5) in x′′(t) = f(t, x(t), x(t)), we get

$$\begin{array}{@{}rcl@{}} y(t)&=&f\left(t,y_{1}(t),y_{2}(t)\right)\\ &=&f\left(t,~A~\left(x_{0}~-~\sum_{k=1}^{m}~a_{k}~\int_{0}^{\tau_{k}}~(\tau_{k}~-~s)~y(s)~ds\right)\right.\\ &+&\left(t~-~A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right)~\left(\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds\right)\\ &+&\left.\int_{0}^{t}~(t~-~s)~y(s)~ds,~\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds~+~\int_{0}^{t} y(s)~ds\right),~~t\in[0,2\pi]. \end{array}$$

### Existence of solution

Define the operator T by

$$\begin{array}{@{}rcl@{}} T~y(t)&=&f\left(t,~y_{1}(t),~y_{2}(t)\right),~~t\in I \end{array}$$

where

$$\begin{array}{@{}rcl@{}} y_{1}(t)&=&A~\left(x_{0}~-~\sum_{k=1}^{m}~a_{k}~\int_{0}^{\tau_{k}}~(\tau_{k}~-~s)~y(s)~ds\right)\\ &+&\left(t~-~A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right)~\left(\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds\right)\\ &+&\int_{0}^{t}~(t~-~s)~y(s)~ds \end{array}$$

and

$$\begin{array}{@{}rcl@{}} y_{2}(t)&=&\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds~+~\int_{0}^{t} y(s)~ds. \end{array}$$

Firstly, we prove that the functional Eq. (3) has at least one solution yL1(I); in order to do that, we will show that the operator T has a fixed point yL1(I).

### Theorem 2

Let the function f:I×R×RR satisfy the assumptions (1)–(3) and the following assumption:

• Every solution y(.)L1(I) to the equation

$$\begin{array}{@{}rcl@{}} y(t)&=&\gamma~f\left(t,~y_{1}(t),~y_{2}(t)\right)~~ \text{a.e. on}~I,~ \gamma ~\in ~(0,1) \end{array}$$

satisfies $$||y||_{L_{1}}\not =r$$ (r is arbitrary but fixed).

Then the operator T has a fixed point yL1(I), which is a solution to Eq. (3).

### Proof

Let y be an arbitrary element in the open set $$B_{r}= \{y:||y||_{L_{1}}< r, r=\frac {||c||_{L_{1}}+2\pi b_{1} |A| |x_{0}|}{1-\left (8 \pi ^{2} b_{1}+2\pi ~b_{1}~|A|~|\sum _{k=1}^{m}~a_{k}~\tau _{k}|+4~\pi ~b_{2}\right)}>0\}$$. Then from the assumptions (1)–(3), we have

{\begin{aligned} ||Ty||_{L_{1}}&=\int_{0}^{2\pi}~|Ty(t)|~dt\\[-1pt] &=\int_{0}^{2\pi}\left|f\left(t,y_{1}(t),y_{2}(t)\right)\right|~dt\\[-1pt] &\leq\int_{0}^{2\pi}\left[|c(t)|~+~b_{1}~|y_{1}(t)|~+~b_{2}~|y_{2}(t)|\right]~dt\\[-1pt] &\leq||c||_{L_{1}}~+~b_{1}~\int_{0}^{2\pi}|y_{1}(t)|~dt~+~b_{2}~\int_{0}^{2\pi}|y_{2}(t)|~dt\\[-1pt] &\leq||c||_{L_{1}}~+~b_{1}~\int_{0}^{2\pi}~\left|A~\left(x_{0}~-~\sum_{k=1}^{m}~a_{k}~\int_{0}^{\tau_{k}}~(\tau_{k}~-~s)~y(s)~ds\right)\right.\\[-1pt] &+\left(t~-~A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right)~\left(\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds\right)\\[-1pt] &+\int_{0}^{t}~(t~-~s)~y(s)~ds\left|~dt~+~b_{2}~\int_{0}^{2\pi}~\left|\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds~+~\int_{0}^{t} y(s)~ds\right|~dt\right.\\[-1pt] &\leq||c||_{L_{1}}~+~b_{1}~\int_{0}^{2\pi}~\left|A~\left(x_{0}~-~\sum_{k=1}^{m}~a_{k}~\int_{0}^{\tau_{k}}~(\tau_{k}~-~s)~y(s)~ds\right)\right|~dt\\[-1pt] &+b_{1}\int_{0}^{2\pi}\left|\left(t~-~A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right) \left(\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds\right)\right|~dt\\[-1pt] &+b_{1}\int_{0}^{2\pi}\left|\int_{0}^{t}~(t~-~s)~y(s)~ds\left|~dt~+~b_{2}~\int_{0}^{2\pi} \right|\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds\right|~dt\\[-1pt] &+b_{2}~\int_{0}^{2\pi}~\int_{0}^{t}~|y(s)|~ds~dt\\[-1pt] &\leq||c||_{L_{1}}~+~b_{1}~\int_{0}^{2\pi}~|A~x_{0}|~dt~+~b_{1}~\int_{0}^{2\pi}~\left|A~\sum_{k=1}^{m}~a_{k}~\int_{0}^{\tau_{k}}~(\tau_{k}~-~s)~y(s)~ds\right|~dt\\[-1pt] &+b_{1}~\int_{0}^{2\pi}~t~\left|\frac{1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds\right|~dt\\[-1pt] &+b_{1}~\int_{0}^{2\pi}~\left|A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right|~\left|\frac{1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds\right|~dt\\[-1pt] &+b_{1}\int_{0}^{2\pi}\int_{s}^{2\pi}~(t~-~s)~|y(s)|~dt~ds~+~b_{2}~\int_{0}^{2\pi}~\int_{0}^{2\pi}~(1-\frac{s}{2\pi})~|y(s)|~ds~dt\\[-1pt] &+b_{2}~\int_{0}^{2\pi}~\int_{s}^{2\pi}~|y(s)|~dt~ds\\[-1pt] &\leq||c||_{L_{1}}~+~2\pi~b_{1}~|A|~|x_{0}|~+~b_{1}~\int_{0}^{2\pi}~\left|A~\sum_{k=1}^{m}~a_{k}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds\right|~dt\\[-1pt] &+b_{1}~\int_{0}^{2\pi}~t~\int_{0}^{2\pi}~(1-\frac{s}{2\pi})~|y(s)|~ds~dt+b_{1}~\int_{0}^{2\pi}~\left|A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right|~ \int_{0}^{2\pi}~(1-\frac{s}{2\pi})~|y(s)|~ds~dt\\[-1pt] &+b_{1}\int_{0}^{2\pi}\frac{(t~-~s)^{2}}{2}|_{s}^{2\pi}~|y(s)|~ds~+~b_{2}~\int_{0}^{2\pi}~\int_{0}^{2\pi}~|y(s)|~ds~dt\\[-1pt] &+b_{2}~\int_{0}^{2\pi}~(2\pi-s)~|y(s)|~ds\\[-1pt] &\leq||c||_{L_{1}}~+~2\pi~b_{1}~|A|~|x_{0}|~+~b_{1}~\int_{0}^{2\pi}~\left|A~\sum_{k=1}^{m}~a_{k}\left|~\int_{0}^{2\pi}~(2\pi)~|y(s)|~ds~dt\right.\right.\\[-1pt] &+b_{1}~\int_{0}^{2\pi}~t~\int_{0}^{2\pi}~|y(s)|~ds~dt+b_{1}~\int_{0}^{2\pi}~\left|A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right|~ \int_{0}^{2\pi}~|y(s)|~ds~dt\\[-1pt] &+b_{1}\int_{0}^{2\pi}\frac{(2\pi~-~s)^{2}}{2}~|y(s)|~ds~+~b_{2}~||y||_{L_{1}}~\int_{0}^{2\pi}~dt+2\pi~b_{2}~\int_{0}^{2\pi}~|y(s)|~ds\\[-1pt] &\leq||c||_{L_{1}}~+~2\pi~b_{1}~|A|~|x_{0}|~+~(2\pi)^{2}~b_{1}~||y||_{L_{1}}~+~b_{1}~||y||_{L_{1}}~\frac{(2\pi)^{2}}{2}\\[-1pt] &+2\pi~b_{1}~|A|~|\sum_{k=1}^{m}~a_{k}~\tau_{k}|~||y||_{L_{1}}~+~\frac{(2\pi)^{2}}{2}~b_{1}~||y||_{L_{1}}~+~2\pi~b_{2}~||y||_{L_{1}}~+~2\pi~b_{2}~||y||_{L_{1}}\\[-1pt] &\leq||c||_{L_{1}}~+~2\pi~b_{1}~|A|~|x_{0}|~+~4~\pi^{2}~b_{1}~||y||_{L_{1}}~+~2\pi^{2}~b_{1}~||y||_{L_{1}}\\[-1pt] &+2\pi~b_{1}~|A|~|\sum_{k=1}^{m}~a_{k}~\tau_{k}|~||y||_{L_{1}}~+~2\pi^{2}~b_{1}~||y||_{L_{1}}~+~4~\pi~b_{2}~~||y||_{L_{1}}\\[-1pt] &=||c||_{L_{1}}~+~2\pi~b_{1}~|A|~|x_{0}|~+~8~\pi^{2}~b_{1}~||y||_{L_{1}}+2\pi~b_{1}~|A|~|\sum_{k=1}^{m}~a_{k}~\tau_{k}|~||y||_{L_{1}}~+~4~\pi~b_{2}~~||y||_{L_{1}}. \end{aligned}}

The above inequality means that the operator T maps Br into L1. □

Also, from assumption (2), we deduce that T maps Br continuously into L1(I).

Now, we will use Kolmogorov compactness criterion (see ) to show that T is compact. So, let be a bounded subset of Br. Then T() is bounded in L1(I). Now we show that (Ty)hTy in L1(I) as h→0, uniformly with respect to TyT .

Indeed:

$$\begin{array}{@{}rcl@{}} ||(Ty)_{h}~-~Ty||_{L_{1}}&=&\int_{0}^{2\pi}~|~(Ty)_{h}(t)~-~(Ty)(t)~|~dt\\ &=&\int_{0}^{2\pi}~\left|\frac{1}{h}~\int_{t}^{t+h}~(Ty)(s)~ds~-~(Ty)(t)\right|~dt\\ &\leq&\int_{0}^{2\pi}~\left(~\frac{1}{h}~\int_{t}^{t+h}~|~(Ty)(s)~-(Ty)(t)~|~ds~\right)~dt\\ &\leq&\int_{0}^{2\pi}~\frac{1}{h}~\int_{t}^{t+h}~|~f(s,y_{1}(s),~y_{2}(s))~-~f(t,y_{1}(t),~y_{2}(t))~|~ds~dt. \end{array}$$

Since

{\begin{aligned} ||f||_{L_{1}}&\leq&||c||_{L_{1}}+2\pi~b_{1}~|A|~|x_{0}|+8~\pi^{2}~b_{1}~||y||_{L_{1}}+2\pi~b_{1}~|A|~|\sum_{k=1}^{m}~a_{k}~\tau_{k}|~||y||_{L_{1}}+4~\pi~b_{2}~~||y||_{L_{1}}, \end{aligned}}

we have that f in L1(I). So, we have (see )

$$\frac{1}{h}~\int_{t}^{t+h}~|~f(s,y_{1}(s),~y_{2}(s))~-~f(t,y_{1}(t),~y_{2}(t))~|~ds~\rightarrow~0,$$

for a.e. tI. So, T() is relatively compact, that is, T is a compact operator.

Now from assumption (4) and Theorem 1, we get that T has a fixed point yL1(I).

### Theorem 3

If the assumptions of Theorem 2 are satisfied, then the periodic and non-local boundary value problem (1) has at least one solution xC1(I).

### Proof

Let x(t) be a solution of (5)

$$\begin{array}{@{}rcl@{}} x(t)&=&A~\left(x_{0}~-~\sum_{k=1}^{m}~a_{k}~\int_{0}^{\tau_{k}}~(\tau_{k}~-~s)~y(s)~ds\right)\\ &+&\left(t~-~A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right)~\left(\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds\right)~+~\int_{0}^{t}~(t~-~s)~y(s)~ds, \end{array}$$

by differentiation, we obtain

$$\begin{array}{@{}rcl@{}} x^{\prime}(t)&=&\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds~+~\int_{0}^{t}~y(s)~ds. \end{array}$$

Since Theorem 2 proved that yL1(I), then by differentiating again, we get

$$x^{\prime\prime}(t)~=~y(t)~=~f(t,~x(t),~x^{\prime}(t)).$$

Substituting respectively by x=0 and x=2π in (5), we get

$$\begin{array}{@{}rcl@{}} x(0)&\,=\,&A\left(x_{0}\,-\,\sum_{k=1}^{m}a_{k}\int_{0}^{\tau_{k}}~(\tau_{k}-s)~y(s)~ds\right)\,+\,\left(\,-\,~A\sum_{k=1}^{m}a_{k}\tau_{k}\right)\left(\frac{-1}{2\pi} \int_{0}^{2\pi}(2\pi-s)~y(s)~ds\right)\\ \end{array}$$
(6)

and

\begin{aligned} x(2\pi)&=A~\left(x_{0}~-~\sum_{k=1}^{m}~a_{k}~\int_{0}^{\tau_{k}}~(\tau_{k}~-~s)~y(s)~ds\right)\\ &+\left(2\pi~-~A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right)~\left(\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds\right)~+~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds\\ &=A\left(x_{0}-\sum_{k=1}^{m}a_{k}\int_{0}^{\tau_{k}}~(\tau_{k}-s)~y(s)~ds\right)+\left(-~A\sum_{k=1}^{m}a_{k}\tau_{k}\right)\left(\frac{-1}{2\pi} \int_{0}^{2\pi}(2\pi-s)~y(s)~ds\right).\\ \end{aligned}
(7)

From (6) and (7), we get x(0)=x(2π). □

Also,

{\begin{aligned} x(\tau_{k})&=A~\left(x_{0}~-~\sum_{k=1}^{m}~a_{k}~\int_{0}^{\tau_{k}}~(\tau_{k}~-~s)~y(s)~ds\right)\\ &+\left(\tau_{k}~-~A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right)~\left(\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds\right) +\int_{0}^{\tau_{k}}~(\tau_{k}~-~s)~y(s)~ds,\\ a_{k}~x(\tau_{k})&=a_{k}~A~\left(x_{0}~-~\sum_{k=1}^{m}~a_{k}~\int_{0}^{\tau_{k}}~(\tau_{k}~-~s)~y(s)~ds\right)\\ &+a_{k}~\left(\tau_{k}-A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right)\left(\frac{-1}{2\pi}\int_{0}^{2\pi}(2\pi-s)y(s)~ds\right) +a_{k}~\int_{0}^{\tau_{k}}(\tau_{k}-s)y(s)~ds,\\ \sum_{k=1}^{m}~a_{k}~x(\tau_{k})&=\sum_{k=1}^{m}~a_{k}~A~\left(x_{0}~-~\sum_{k=1}^{m}~a_{k}~\int_{0}^{\tau_{k}}~(\tau_{k}~-~s)y(s)~ds\right)\\ &+\sum_{k=1}^{m}~a_{k}\left(\tau_{k}-A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right)\left(\frac{-1}{2\pi}\int_{0}^{2\pi}(2\pi-s)y(s)~ds\right) +\sum_{k=1}^{m}~a_{k}\int_{0}^{\tau_{k}}(\tau_{k}-s)y(s)~ds\\ &=x_{0}~. \end{aligned}}

Then the periodic and non-local boundary value problem (1) is equivalent to the integral Eq. (5). Hence problem (1) has at least one solution xC1(I).

### Theorem 4

If f:I×R×RR satisfies the assumptions of Theorem 2, then the boundary value problem (2) has at least one solution xC1(I), and its solution is given by

$$\begin{array}{@{}rcl@{}} x(t)&=&\frac{1}{2\pi}~\left(x_{0}~-~\int_{0}^{2\pi}~\frac{(2\pi~-~s)^{2}}{2}~y(s)~ds\right)\\ &+&\left(t~-~\ \pi\right)~\left(\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds\right)~+~\int_{0}^{t}~(t~-~s)~y(s)~ds. \end{array}$$
(8)

Also,

$$\begin{array}{@{}rcl@{}} x^{\prime}(t)&=&x^{\prime}(0)~+~\int_{0}^{t}~y(s)~ds\\ &=&\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds~+~\int_{0}^{t}~y(s)~ds. \end{array}$$

### Proof

If we take ak=tktk−1,τk(tk−1,tk) and 0<t1<t2<...<2π, we get

$$\sum_{k=1}^{m}(t_{k} -t_{k-1})x(\tau_{k})~=~x_{0}.$$

By taking the limit as m, we get $$\int _{0}^{2\pi }x(t)dt =x_{0}$$. □

Also, take the limit as m in (5):

$$\begin{array}{@{}rcl@{}} x(t)&=&A~\left(x_{0}~-~\sum_{k=1}^{m}~a_{k}~\int_{0}^{\tau_{k}}~(\tau_{k}~-~s)~y(s)~ds\right)\\ &+&\left(t~-~A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right)~\left(\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds\right)~+~\int_{0}^{t}~(t~-~s)~y(s)~ds,\\ \end{array}$$

we obtain (8):

$$\begin{array}{@{}rcl@{}} x(t)&=&\frac{1}{2\pi}~\left(x_{0}~-~\int_{0}^{2\pi}~\frac{(2\pi~-~s)^{2}}{2}~y(s)~ds\right)\\ &+&\left(t~-~\ \pi\right)~\left(\frac{-1}{2\pi}~\int_{0}^{2\pi}~(2\pi~-~s)~y(s)~ds\right)~+~\int_{0}^{t}~(t~-~s)~y(s)~ds. \end{array}$$

This completes the proof.

## Availability of data and materials

Data sharing not applicable to this article as no data sets were generated or analyzed during the current study.

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## Acknowledgements

The author is very grateful to the referee for his valuable comments and suggestions which improved the original version of the paper.

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## Author information

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The author has made each part of this paper. He read and approved the final manuscript.

### Corresponding author

Correspondence to Sheren Ahmed Abd El-Salam.

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### Competing interests

The author declares that he has no competing interests. 