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.
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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:
where x0∈R, 0<τ1<τ2<⋯<τm<2π and ak≠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 Leray-Schauder 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 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 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 L1(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)∈L1, 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 (u1,u2)∈R×R
-
(2)
f:I×R×R→R is continuous in (u1,u2)∈R×R for any t∈I
-
(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
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∈L1(I); in order to do that, we will show that the operator T has a fixed point y∈L1(I).
Theorem 2
Let the function f:I×R×R→R 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 y∈L1(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 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 ℵ be a bounded subset of Br. Then T(ℵ) is bounded in L1(I). Now we show that (Ty)h→Ty in L1(I) as h→0, uniformly with respect to Ty∈T ℵ.
Indeed:
Since
we have that f in L1(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∈L1(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 x∈C1(I).
Proof
Let x(t) be a solution of (5)
by differentiation, we obtain
Since Theorem 2 proved that y∈L1(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 non-local boundary value problem (1) is equivalent to the integral Eq. (5). Hence problem (1) has at least one solution x∈C1(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∈C1(I), and its solution is given by
Also,
Proof
If we take ak=tk−tk−1,τk∈(tk−1,tk) and 0<t1<t2<...<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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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 El-Salam, S. On some boundary value problems with non-local and periodic conditions. J Egypt Math Soc 27, 38 (2019). https://doi.org/10.1186/s42787-019-0040-y
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DOI: https://doi.org/10.1186/s42787-019-0040-y