# Boundary Value Problem of Nonlinear Hybrid Differential Equations with Linear and Nonlinear Perturbations

Authors: Said Melliani; Abdelati El Allaoui; Lalla Saadia Chadli
Journal: International Journal of Differential Equations
 (2020)

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## Abstract

The aim of this paper is to study a boundary value problem of the hybrid differential equation with linear and nonlinear perturbations. It generalizes the existing problem of second type. The existence result is constructed using the Leray–Schauder alternative, and the uniqueness is guaranteed by Banach’s fixed-point theorem. Towards the end of this paper, an example is provided to illustrate the obtained results.


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## Body

## 1. Introduction

The application of differential equations in different real domains has increased the importance of this theory which is still under development. Hybrid differential equations are a subfield of differential equations which also has enough importance. Recently, it has attracted the attention of several mathematicians [1–5]. In [6], the authors studied the following hybrid differential equation with linear perturbation:(1)ddtxt−ft,xt=gt,xt,x0=x0,where the existence of solutions to these problems has been ensured using Dhage’s theorem.Motivated by the abovementioned problem, we consider the following boundary value problem for hybrid differential equation:(2)ddtxtft,xt−gt,xt=ht,xt,t∈I=0,a,a>0,x0f0,x0+αxafa,xa=g0,x0+αga,xa+β,where f∈CI×ℝ;ℝ/0andg,h∈CI×ℝ;ℝ are given functions and α,β∈ℝ such that α≠−1.The proposed problem can be considered as generalization of problem (1) which becomes a special case if we take f=1, and also, the novelty is at the level of the relationship between the boundary values. Using Banach’s fixed-point theorem, we show the existence and the uniqueness of the solution of the proposed problem.The fixed-point theorems used for hybrid differential equations with perturbation of first or second type are those based on the composition of the solution as sum or product of two operators such as the Dhage case. For our case, we have a mixed problem which brings together the two types, where we thought of using Leray–Schauder’s Fixed-Point Theorem as a second existence result for which we will have a single operator.


## 2. Preliminaries

First, we recall some basic results used in this paper. We start by recalling Leray–Schauder alternative.Lemma 1 (see [7]).
LetΠ:Y⟶Y be a completely continuous operator and(3)PΠ=y∈Y:y=δΠyfor some0<δ<1.Then, either the setPΠ is unbounded or Π has at least one fixed point.Now, we recall the following lemmas on which we will base ourselves to build the solution of our problem.Lemma 2 (see [6]).
Suppose thatx↦x−gt,x is increasing in ℝ for each t∈I. Then, for any h:I⟶ℝ+, the function x∈CJ,ℝ+ is a solution of the hybrid differential equation(4)ddtxt−gt,xt=ht,t∈I,x0=x0∈ℝ,if and only if x satisfies the following hybrid integral equation:(5)xt=x0−g0,x0+gt,xt+∫0th,sds,t∈J.


## 3. Existence Result

Before presenting the existence results, we pose the following hypotheses:(i)
The mapx⟼xft,x−gt,x is increasing in ℝ for each t∈I.(ii)
There exist positive constantsνf and μg, such that(6)ft,x≥νf,gt,x≤μg.(iii)
There exists positive constantsλf,λg,  andλh such that(7)ft,x−ft,y≤λfx−y,gt,x−gt,y≤λgx−y,ht,x−ht,y≤λhx−y,for each t∈I and x,y∈ℝ.DenoteC:=CI,ℝ, the space of all continuous mapping defined on I into ℝ endowed with the norm x=supt∈Ixt.Lemma 3.
Leth∈CI,ℝ, then x is an integral solution of (2) if and only if it satisfies the following integral equation:(8)xt=β1+αft,xt−α1+αft,xt∫0ah,s,xsds+gt,xtft,xt+1ft,xt∫0th,s,xsds,for each t∈I.Proof.
Suppose thatx is a solution for (2), then we obtain(9)xtft,xt=x0f0,x0−g0,x0+gt,xt+∫0ths,xsds,fort∈I.
Then,(10)αxafa,xa=αx0f0,x0−αg0,x0+αga,xa+α∫0ahs,xsds.
Hence,(11)x0f0,x0+αxafa,xa=1+αx0f0,x0−αg0,x0+αga,xa+α∫0ahs,xsds.
By using the second equation in (2), we obtain(12)x0f0,x0−g0,x0=β1+α−α1+α∫0ahs,xsds.
By replacing in (9), we obtain(13)xt=β1+αft,xt−α1+αft,xt∫0ahs,xsds+gt,xtft,xt+1ft,xt∫0ths,xsds,t∈I.
The other implication is trivial.
Now, we can give the definition of an integral solution of problem (2).Definition 1.
An integral solution of problem (2) is a function x∈C which satisfies the following:(1)
The mapt⟼xft,x−gt,x is continuous for each x∈ℝ, and(2)
x satisfies the following integral equation:(14)xt=β1+αft,xt−α1+αft,xt∫0ahs,xsds+gt,xtft,xt+1ft,xt∫0ths,xsds,for each t∈I.
To reduce the form of mathematical expressions, consider the following notations:(15)π1=β+akhα+μg+akh1+α,π2=νf1+α,π3=aλhα+1+α,ω=λfβ+νfλg+λfμg1+α+aνfλh+λfλhr+khα+1+ανf21+α,where r is a real number which will be defined later.
Now, we can provide our first existence result.Theorem 1.
Suppose thatA0−A3 are satisfied. In addition, assume that the following condition is verified:(16)maxω,π3π2<1.Then, the problem (2) has a unique solution.Proof.
First, we define the following closed ball:(17)Br=x∈PC:x≤r,where(18)r≥π1π2−π3.
Also, we define the following operatorΠ on C by(19)Πxt=β1+αft,xt−α1+αft,xt∫0ahs,xsds+gt,xtft,xt+1ft,xt∫0ths,xsds,for each t∈I.
The proof will be made in two steps:(i)
ΠBr⊆Br. Indeed, for x∈Br and t∈I, we have(20)Πxt≤β1+αft,xt+α1+αft,xt∫0ahs,xs−hs,0+hs,0ds+gt,xtft,xt+1ft,xt∫0ths,xs−hs,0+hs,0ds≤βvf1+α+αvf1+α∫0aλhxs+khds+μgvf+1vf∫0tλhxs+khds≤β+akhανf1+α+μg+akhνf+aλhα+1+ανf1+αr≤π1π2+π3π2r.(ii)
Hence, according to (18), we obtain(21)Πxt≤1−π3π2r+π3π2r≤r.(iii)
Then,(22)Πx≤r.(iv)
Π is a contraction:
Forx,y∈Br and t∈I, we have(23)Πxt−Πyt≤β1+αft,xt−β1+αft,yt+α1+αft,xt∫0ahs,xsds−α1+αft,xt∫0ahs,ysds+α1+αft,xt∫0ahs,ysds−α1+αft,yt∫0ahs,ysds+gt,xtft,xt−gt,ytft,xt+gt,ytft,xt−gt,ytft,yt+1ft,xt∫0ths,xsds−1ft,xt∫0ths,ysds+1ft,xt∫0ths,ysds−1ft,yt∫0ths,ysds≤βνf21+αft,xt−ft,yt+ανf1+α∫0ahs,xs−hs,ysds+ανf21+α∫0ahs,ys−hs,0+hs,0dsft,xt−ft,yt+1νfgt,xt−gt,yt+μgνf2ft,xt−ft,yt+1νf∫0ths,xs−hs,ysds+1νf2∫0ths,ys−hs,0+hs,0dsft,xt−ft,yt≤λfβνf21+α+aλhανf1+α+aλfαλhr+khνf21+α+λgνf+λfμgνf2+aλhνf+aλfλhr+khνf2x−y≤ωx−y.which shows Π is a contraction.
Thus,Π is a contraction. Then, the existence and uniqueness of the solution is guaranteed by Banach’s fixed-point theorem.
Now, we present the second existence result using Leray–Schauder alternative.Theorem 2.
Suppose thatA0 and A1 are satisfied. In addition, assume that there exist γ1,γ2>0, such that(24)ht,x≤γ1+γ2x,for eacht,x∈I×ℝ.Also,νf>aγ2. Then, problem (2) has at least one solution.Proof.
LetA⊆C be a bounded subset.
Then, there existsηh>0 such that(25)ht,xt≤ηh.for each x∈A and t∈I.
The proof will be given in several steps:(i)
Π is uniformly bounded:
Forx∈A and t∈I, we have(26)Πxt≤β1+αft,xt+α1+αft,xt∫0ahs,xsds+gt,xtft,xt+1ft,xt∫0ths,xsds≤β+aηhανf1+α+μg+aηhνf.
Then,Π is uniformly bounded.(ii)
Π is equicontinuous:
For0<τ1<τ2<a and x∈A, we have(27)Πxτ2−Πxτ1≤β1+αfτ1,xτ1−β1+αfτ2,xτ2+α1+αfτ1,xτ1−α1+αfτ2,xτ2∫0ahs,xsds+gτ1,xτ1fτ1,xτ1−gτ2,xτ2fτ1,xτ1+gτ2,xτ2fτ1,xτ1−gτ2,xτ2fτ2,xτ2+1fτ1,xτ1∫0τ1hs,xsds−1fτ1,xτ1∫0τ2hs,xsds+1fτ1,xτ1∫0τ2hs,xsds−1fτ2,xτ2∫0τ2hs,xsds≤β+aηhανf21+αfτ1,xτ1−fτ2,xτ2+1νfgτ1,xτ1−gτ2,xτ2+μgνf2fτ1,xτ1−fτ2,xτ2+ηhνfτ2−τ1+τ2ηhνf2fτ1,xτ1−fτ2,xτ2⟶0,ast2⟶t1.
Hence,Π is equicontinuous.(iii)
PΠ is bounded:
We denote by(28)PΠ=x∈ℝ:x=δΠx,0≤δ≤1.
Letx∈PΠ and t∈I, we have(29)xt≤β+aηhανf1+α+μgνf+aνfγ1+γ2x,which implies that(30)x≤β+aηhανf1+α+μgνf+aνfγ1+γ2x.
Then,(31)x≤β+aηhα+μg+aγ11+ανf−aγ21+α.
Thus, all assumptions of Lemma1 are satisfied. So, Π has at least one fixed point which is a solution for our problem.
Now, we give an example to illustrate the obtained results.Example 1.
Consider the following problem:(32)ddt8xt+1−cost4xtxt−16et11+xt=15+17xt,t∈0,π,x0+16xπ+xπxπ=14811+x0+13eπ11+xπ+1.
This problem can be written as (2), where(33)ft,xt=8+1−cost4xt,gt,xt=16et11+xt,ht,xt=15+17xt,α=16andβ=1.
We can easily verify that(34)νf=8,λf=14,μg=16,λg=16,λh=γ2=17,kh=γ1=15,α=16,β=1,a=π.
We haveνf>aγ2; then, from Theorem 2, this problem has a at least one solution.
Now, we know that a solution exists and we seek if it is unique. We can easily verify that by using the following:(35)π1=1136+33π5,π2=136,π3=33π7.
We taker=1 which satisfies condition (10). Then,(36)ω=2038704+1419π38080<1,π3<π2.
According to Theorem1, we can deduce that our problem has a unique solution.


## 4. Conclusion

In this paper, we treated a hybrid differential equation with linear and nonlinear perturbations where we showed the existence and uniqueness of the solution using Banach’s fixed-point theorem, and also, we used the Leray–Schauder theorem as another way to guarantee the existence of the solution, which it was used for the first time, to our knowledge, for this type of problem. In the future research, we hope to generalize this problem either at the level of the order of the derivative or the initial condition by taking inspiration from [8, 9].



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