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

(1)
f:I×R×R→R is measurable in t∈I for any (u_{1},u_{2})∈R×R

(2)
f:I×R×R→R is continuous in (u_{1},u_{2})∈R×R for any t∈I

(3)
There exist two positive constants b_{1},b_{2} and a function c(t)∈L_{1}(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 y∈L_{1}(I); in order to do that, we will show that the operator T has a fixed point y∈L_{1}(I).
Theorem 2
Let the function f:I×R×R→R satisfy the assumptions (1)–(3) and the following assumption:
Then the operator T has a fixed point y∈L_{1}(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}\leftf\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] &\leqc_{L_{1}}~+~b_{1}~\int_{0}^{2\pi}y_{1}(t)~dt~+~b_{2}~\int_{0}^{2\pi}y_{2}(t)~dt\\[1pt] &\leqc_{L_{1}}~+~b_{1}~\int_{0}^{2\pi}~\leftA~\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] &\leqc_{L_{1}}~+~b_{1}~\int_{0}^{2\pi}~\leftA~\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] &\leqc_{L_{1}}~+~b_{1}~\int_{0}^{2\pi}~A~x_{0}~dt~+~b_{1}~\int_{0}^{2\pi}~\leftA~\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}~\leftA~\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] &\leqc_{L_{1}}~+~2\pi~b_{1}~A~x_{0}~+~b_{1}~\int_{0}^{2\pi}~\leftA~\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}~\leftA~\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\pis)~y(s)~ds\\[1pt] &\leqc_{L_{1}}~+~2\pi~b_{1}~A~x_{0}~+~b_{1}~\int_{0}^{2\pi}~\leftA~\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}~\leftA~\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] &\leqc_{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] &\leqc_{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 B_{r} into L_{1}. □
Also, from assumption (2), we deduce that T maps B_{r} continuously into L_{1}(I).
Now, we will use Kolmogorov compactness criterion (see [9]) to show that T is compact. So, let ℵ be a bounded subset of B_{r}. Then T(ℵ) is bounded in L_{1}(I). Now we show that (Ty)_{h}→Ty in L_{1}(I) as h→0, uniformly with respect to Ty∈T ℵ.
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 L_{1}(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. t∈I. So, T(ℵ) 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 y∈L_{1}(I).
Theorem 3
If the assumptions of Theorem 2 are satisfied, then the periodic and nonlocal boundary value problem (1) has at least one solution x∈C^{1}(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 y∈L_{1}(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
$$\begin{array}{@{}rcl@{}} x(0)&\,=\,&A\left(x_{0}\,\,\sum_{k=1}^{m}a_{k}\int_{0}^{\tau_{k}}~(\tau_{k}s)~y(s)~ds\right)\,+\,\left(\,\,~A\sum_{k=1}^{m}a_{k}\tau_{k}\right)\left(\frac{1}{2\pi} \int_{0}^{2\pi}(2\pis)~y(s)~ds\right)\\ \end{array} $$
(6)
and
$$ \begin{aligned} x(2\pi)&=A~\left(x_{0}~~\sum_{k=1}^{m}~a_{k}~\int_{0}^{\tau_{k}}~(\tau_{k}~~s)~y(s)~ds\right)\\ &+\left(2\pi~~A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right)~\left(\frac{1}{2\pi}~\int_{0}^{2\pi}~(2\pi~~s)~y(s)~ds\right)~+~\int_{0}^{2\pi}~(2\pi~~s)~y(s)~ds\\ &=A\left(x_{0}\sum_{k=1}^{m}a_{k}\int_{0}^{\tau_{k}}~(\tau_{k}s)~y(s)~ds\right)+\left(~A\sum_{k=1}^{m}a_{k}\tau_{k}\right)\left(\frac{1}{2\pi} \int_{0}^{2\pi}(2\pis)~y(s)~ds\right).\\ \end{aligned} $$
(7)
From (6) and (7), we get x(0)=x(2π). □
Also,
$$ {\begin{aligned} x(\tau_{k})&=A~\left(x_{0}~~\sum_{k=1}^{m}~a_{k}~\int_{0}^{\tau_{k}}~(\tau_{k}~~s)~y(s)~ds\right)\\ &+\left(\tau_{k}~~A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right)~\left(\frac{1}{2\pi}~\int_{0}^{2\pi}~(2\pi~~s)~y(s)~ds\right) +\int_{0}^{\tau_{k}}~(\tau_{k}~~s)~y(s)~ds,\\ a_{k}~x(\tau_{k})&=a_{k}~A~\left(x_{0}~~\sum_{k=1}^{m}~a_{k}~\int_{0}^{\tau_{k}}~(\tau_{k}~~s)~y(s)~ds\right)\\ &+a_{k}~\left(\tau_{k}A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right)\left(\frac{1}{2\pi}\int_{0}^{2\pi}(2\pis)y(s)~ds\right) +a_{k}~\int_{0}^{\tau_{k}}(\tau_{k}s)y(s)~ds,\\ \sum_{k=1}^{m}~a_{k}~x(\tau_{k})&=\sum_{k=1}^{m}~a_{k}~A~\left(x_{0}~~\sum_{k=1}^{m}~a_{k}~\int_{0}^{\tau_{k}}~(\tau_{k}~~s)y(s)~ds\right)\\ &+\sum_{k=1}^{m}~a_{k}\left(\tau_{k}A~\sum_{k=1}^{m}~a_{k}~\tau_{k}\right)\left(\frac{1}{2\pi}\int_{0}^{2\pi}(2\pis)y(s)~ds\right) +\sum_{k=1}^{m}~a_{k}\int_{0}^{\tau_{k}}(\tau_{k}s)y(s)~ds\\ &=x_{0}~. \end{aligned}} $$
Then the periodic and nonlocal boundary value problem (1) is equivalent to the integral Eq. (5). Hence problem (1) has at least one solution x∈C^{1}(I).
Theorem 4
If f:I×R×R→R satisfies the assumptions of Theorem 2, then the boundary value problem (2) has at least one solution x∈C^{1}(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 a_{k}=t_{k}−t_{k−1},τ_{k}∈(t_{k−1},t_{k}) and 0<t_{1}<t_{2}<...<2π, we get
$$\sum_{k=1}^{m}(t_{k} t_{k1})x(\tau_{k})~=~x_{0}. $$
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.