Introduction

Differential equations with non-local conditions were considered in many works (see [1], [2], [3], and [4]). Also, anti-periodic problems can be found in [5] and [6].

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 [7], 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) [8] Let E be a Banach space and Ω be a bounded open subset of E, 0∈Ω 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

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

Integral representation

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 [9]) 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 [10])

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

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.