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On some boundary value problems with nonlocal and periodic conditions
Journal of the Egyptian Mathematical Society volume 27, Article number: 38 (2019)
Abstract
In this work, we concern nonlocal 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 nonlocal conditions were considered in many works (see [1], [2], [3], and [4]). Also, antiperiodic problems can be found in [5] and [6].
Here, we study the existence of at least one solution for the boundary value problem with nonlocal and periodic conditions:
where x_{0}∈R, 0<τ_{1}<τ_{2}<⋯<τ_{m}<2π and a_{k}≠0 for all k=1,2,⋯,m.
Also, the boundary value problem with integral and periodic conditions:
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 LeraySchauder type, the existence of at least one solution for problem (1) such that the function f:I×R×R→R, I=[0,2π] satisfies the growth conditions.
Preliminaries
Theorem 1
(Nonlinear alternative of LeraySchauder 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 x∈∂Ω,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
and by L_{1}(I) the space of all Lebesgue integrable functions on the interval I with norm
The growth condition on the function f means that
where a(t)∈L_{1}, b is a nonnegative constant.
Main results
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
where
Proof
Let y=x^{′′}(t)=f(t,x,x^{′}). □
Integrating both sides, we obtain
Integrating again, we get
From the boundary condition, we obtain
then
Now,
Take \(A=(\sum _{k=1}^{m}~a_{k})^{1}\), then
Substituting the values of x^{′}(0) and x(0) in x(t), we get
Inserting (4) and (5) in x^{′′}(t) = f(t, x(t), x^{′}(t)), we get
Existence of solution
Define the operator T by
where
and
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:

Every solution y(.)∈L_{1}(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 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
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:
Since
we have that f in L_{1}(I). So, we have (see [10])
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)
by differentiation, we obtain
Since Theorem 2 proved that y∈L_{1}(I), then by differentiating again, we get
Substituting respectively by x=0 and x=2π in (5), we get
and
From (6) and (7), we get x(0)=x(2π). □
Also,
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
Also,
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
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):
we obtain (8):
This completes the proof.
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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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Abd ElSalam, S. On some boundary value problems with nonlocal and periodic conditions. J Egypt Math Soc 27, 38 (2019). https://doi.org/10.1186/s427870190040y
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DOI: https://doi.org/10.1186/s427870190040y
Keywords
 Non linear boundary value problems
 Nonlocal boundary value problems
 Periodic conditions
AMS Subject classification
 34B15
 34B10
 34C25