# A Novel Implementation of Krasnoselskii’s Fixed-Point Theorem to a Class of Nonlinear Neutral Differential Equations

Authors: Ali Rezaiguia; Abdelkader Moumen; Abdelaziz Mennouni; Mohammad Alshammari; Taher S. Hassan
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this work, we examine a class of nonlinear neutral differential equations. Krasnoselskii’s fixed-point theorem is used to provide sufficient conditions for the existence of positive periodic solutions to this type of problem.


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## Body

## 1. Introduction

In recent years, differential equations have garnered considerable interest (cf. [1, 2] and references therein). Important types of these problems include differential equations with delay. For instance, in [1, 3–10], the authors employed a variety of techniques to determine the existence of positive periodic solutions. The uniqueness and positivity of a first-order nonlinear periodic differential equation are investigated in [11]. The authors of [12] discussed nearly periodic solutions to nonlinear Duffing equations. Among them, the fixed-point principle has established itself as a critical tool for studying the existence and periodicity of positive solutions. Numerous studies, including [4, 6, 11], examined this method.In this work, we investigate the following fourth-order nonlinear neutral differential equation:(1)d4dt4xt−gt,xt−τt=−atxt+ft,xt−τt.Under the assumptions:(i)
a,τ∈Cℝ,0,∞(ii)
g∈Cℝ×0,∞,ℝ and f∈Cℝ×0,∞,0,∞(iii)
a,τ,gt,x,ft,x are ω-periodic in t, ω is a positive constantKrasnoselskii’s fixed-point theorem offers sufficient conditions for the existence of positive periodic solutions to the aforesaid problem.Neutral differential equations are employed in various technological and natural science applications. For example, they are widely employed to investigate distributed networks with lossless transmission lines (see [7]). Therefore, their qualitative qualities are significant.It is worth noting that Krasnoselskii’s fixed-point theorem was proposed in 2012 in [4] to show the existence of positive periodic solutions to the nonlinear neutral differential equation with variable delay of the form
(2)ddtxt−gt,xt−τt=rtxt−ft,xt−τt.The same researchers evaluated the existence of positive periodic solutions for two types of second-order nonlinear neutral differential equations with variable delay the following year in [5].
(3)d2dt2xt−gt,xt−τt=±atxt±ft,xt−τt,where Krasnoselskii’s fixed-point theorem is also used as a tool. The authors of [10] investigated the following third-order nonlinear neutral differential equations with variable delay.
(4)d3dt3xt−cxt−τt=−atxt+ft,xt−τt.The existence of positive periodic solutions is demonstrated using Krasnoselskii’s fixed-point theorem. In [3], the authors investigated the fourth-order nonlinear neutral differential equations with variable delay of the form
(5)d4dt4xt−gt,xt−τt=atxt−ft,xt−τt.Krasnoselskii’s fixed-point theorem is used to derive some sufficient conditions for the existence of positive periodic solutions to the aforementioned problem.The remainder of this paper is organized as follows: in the next Section, we deliver the definitions and lemmas required to prove our main results. In particular, we state some Green’s function properties related to the problem (1). Section 3 establishes some necessary conditions for the existence of positive solutions to our problem (1).


## 2. Preliminaries

For a fixedω>0, we consider a set Pω of continuous scalar functions x which are periodic in t, with period ω. We recall that xt−τ and xt are in Pω and Pω,. is a Banach space with the supremum norm [13, 14].
(6)x≔supt∈ℝxt=supt∈0,ωxt.Define(7)Pω+≔x∈Pω,x>0,m≔inft∈0,ωat,M≔supt∈0,ωat,β≔M4.Lemma 1.
The equation(8)d4dt4xt+Mxt=ht,ht∈Pω+
has a uniqueω-periodic solution
(9)xt=∫tt+ωGt,shsds,where
(10)Gt,s=14γ3Asinhγs−t−ω2sinγs−t−ω2s∈t,t+ω,+Bcoshγs−t−ω2cosγs−t−ω2,(11)A≔sinγω/2coshγω/2−cosγω/2sinhγω/2coshγω−cosγω,(12)B≔cosγω/2sinhγω/2+sinγω/2coshγω/2coshγω−cosγω.Proof.
First, it is evident that the homogeneous equation associated with (8) has a solution
(13)xt=c1eγ1+it+c2eγ−1+it+c3eγ−1−it+c4eγ1−it,whereγ=22β.
Using the parameter variation method, we obtain(14)c1′t=hte−tγ1+i−81−iγ3,c2′t=htetγ1−i81+iγ3,c3′t=htetγ1+i81−iγ3,c4′t=hte−tγ1−i−81+iγ3.
Keeping in mind thatxt, x′t, x′′t, and x′′′t are periodic functions, we obtain
(15)c1t=tt+ω−eγ1+iω−s81−iγ31−eγ1+iωhsds,c2t=tt+ω−eγ1−is81+iγ31−eγ1−iωhsds,c3t=tt+ω−eγ1+is81−iγ31−eγ1+iωhsds,c4t=tt+ω−eγ1−iω−s81+iγ31−eγ1−iωhsds.
Hence,(16)xt=c1teγ1+it+c2teγ−1+it+c3teγ−1−it+c4teγ1−it=∫tt+ωGt,shsds,where Gt,s is identified by (10).Lemma 2.
Assume that(17)∫tt+ωGt,shsds=1M,maxat:t∈0,ω<4πω4.
Then,(18)0<α1<Gt,s<α2,forallt∈0,ω,s∈t,t+ω.Proof.
The definition ofGt,s gives
(19)∫tt+ωGt,sds=14γ3∫tt+ωAsinhγs−t−ω2sinγs−t−ω2ds+∫tt+ωBcoshγs−t−ω2cosγs−t−ω2ds=A16γ4eγs−t−ω/2eγ2s−2t+ω−1cosγs−t−ω2++eγ2s−2t+ω+1sinγs−t−ω2tt+ω++B16γ4eγs−t−ω/2eγ2s−2t+ω+1sinγs−t−ω2−−eγ2s−2t+ω−1cosγs−t−ω2tt+ω=14γ4Acoshγω2sinγω2−sinhγω2cosγω2+14γ4Bcoshγω2sinγω2+sinhγω2cosγω2=14γ42coshγω/2sinγω/22+sinhγω/2cosγω/22coshγω−cosγω=14γ421/2coshγω−1/2cosγωcoshγω−cosγω=14γ4=1β4=1M.
On the other hand, it is simple to demonstrate thatd/dsGt,s=0 only if s=t+ω/2.
Hence,(20)Gt,t=Gt,t+ω=14γ3sinhγω/2coshγω/2+sinγω/2cosγω/2coshγω/2−cosγω/2=α2,Gt,t+ω2=14γ3cosγω/2sinhγω/2+sinγω/2coshγω/2coshγω−cosγω=α1.
Since(21)at:t∈0,ω<4πω4,we get
(22)0<γω2<π2.
So,(23)sinγω2>0,1>cosγω2>0,sinhγω2>0.
Consequently,(24)0<α1<Gt,s<α2,forallt∈0,ω,s∈t,t+ω.Lemma 3.
If(25)maxat:t∈0,ω<4πω4,Ft,x>0.
Then,x∈Pω solves equation (1) if and only if
(26)xt=gt,xt−τt+∫tt+ωGt,sM−asxs+fs,xs−τs−Mgt,xs−τsds.Proof.
Letx∈Pω be a solution of (1). Equation (1) reads as
(27)d4dt4xt−gt,xt−τt+Mxt−gt,xt−τt=−atxt+ft,xt−τt+Mxt−gt,xt−τt=M−atxt+ft,xt−τt−Mgt,xt−τt.
According to Lemma1, we obtain
(28)xt−gt,xt−τt=∫tt+ωGt,sM−atxt+ft,xt−τtds−∫tt+ωGt,sMgt,xt−τtds,which implies that
(29)xt=gt,xt−τt+∫tt+ωGt,sM−asxs+fs,xs−τs−Mgt,xs−τsds.
This completes the proof. Let us define the two operatorsT1,T2:Pω⟶Pω as follows:
(30)T1φt≔gt,φt−τt,T2φt≔∫titi+ωGti,sM−asφs+fs,φs−τs−Mgt,φs−τsds.
We formulate equation (26) in Lemma 3 as follows:
(31)φt=T1φt+T2φt.Remark 4.
Any solution to equation (31) is a solution to problem (1).
Let us introduce the following hypotheses, which are assumed hereafter:
The functiongt,x is Lipschitz continuous in x. That is to say, there exists a positive constant k such that
(32)gt,x−gt,y≤kx−y,forallt∈0,ω,x,y∈Pω.Lemma 5.
Assume that (32) holds and
(33)k<1.
Then,T1 is a contraction.Proof.
It is evident thatT1φ is continuous for all φ∈D. Moreover,
(34)T1φt+ω=T1φt.
So, for allφ,ψ∈D, we have
(35)T1φt−T1ψt=gt,φt−τt−gt,ψt−τt≤supt∈0,ωgt,φt−τt−gt,ψt−τt≤kφ−ψ.
Thus,(36)T1φ−T1ψ≤kφ−ψ.
Consequently, it follows from (33) that T1:Pω⟶Pω is a contraction.Lemma 6.
Assume thatM<4π/ω4 and 0<Ft,x≤C. Then, T2 is completely continuous.Proof.
Firstly, we show thatT2 is continuous. To this end, let yn be a sequence such that yn⟶y in Pω. We have
(37)T2ynt−T2yt≤∫tt+ωGt,sM−asyns−ys+fs,yns−τs−fs,ys−τs+Mgt,yns−τs−Mgt,ys−τsds.
It follows from the continuity off and g that
(38)T2yn−T2y⟶0asn⟶∞.
Thus,T2 is continuous.
Secondly, we prove thatT2 maps bounded sets into bounded sets in Pω,.. To this end, let Br=x∈Pω,x<r be a bounded ball in Pω,., we have
(39)T2xt=∫tt+ωGt,sM−asxs+fs,xs−τs−Mgt,xs−τsds≤∫tt+ωGt,sM−asxs+fs,xs−τs−Mgt,xs−τsds.
From Lemma2 and since Ft,x≤C, we get
(40)T2xt≤α2∫tt+ωM−mr+fs,xs−τs−Mgt,xs−τsds≤α2ωM−mr+C.
The estimation ofT2x implies
(41)T2x≤α2ωM−mr+C.
This shows thatT2 is uniformly bounded.Finally, we prove thatT2 sends bounded sets into equicontinuous sets. Let t1, t2∈0,ω, t1<t2, and Br be a bounded set of Pω. For all i∈1,2, we have
(42)T2xti=∫titi+ωGti,sM−asxs+fs,xs−τs−Mgt,xs−τsds,i=1,2.DenoteT3=T2xt2−T2xt1. So, we obtain
(43)T3=∫t2t2+ωGt2,sM−asxs+fs,xs−τs−Mgt,xs−τsds−∫t1t1+ωGt1,sM−asxs+fs,xs−τs−Mgt,xs−τsds=∫t2t1+ωGt2,sM−asxs+fs,xs−τs−Mgt,xs−τsds+∫t1+ωt2+ωGt2,sM−asxs+fs,xs−τs−Mgt,xs−τsds,−∫t1t2Gt1,sM−asxs+fs,xs−τs−Mgt,xs−τss−∫t2t1+ωGt1,sM−asxs+fs,xs−τs−Mgt,xs−τsds=∫t2t1+ωGt2,s−Gt1,sM−asxs+fs,xs−τs−Mgt,xs−τsds+∫t1+ωt2+ωGt2,sM−asxs+fs,xs−τs−Mgt,xs−τsds−∫t1t2Gt1,sM−asxs+fs,xs−τs−Mgt,xs−τsds≤M−mr∫t2t1+ωGt2,s−Gt1,sds+∫t1+ωt2+ωGt2,sds+∫t1t2Gt1,sds.Ast2⟶t1, the right-hand side of the above inequality tends to zero. By the Arzela-Ascoli theorem, we conclude that T2 is a completely continuous operator. This completes the proof. This section will be concluded by referring to Krasnoselskii’s fixed-point theorem (see [9]).Theorem 7. (Krasnoselskii).
LetD be a closed convex nonempty subset of a Banach space B,.. Suppose that T1 and T2 map D into B such that
(i)
x,y∈D, implies T1x+T2y∈D(ii)
T1 is a contraction mapping(iii)
T2 is completely continuous
Then, there existsz∈D with z=T1z+T2z.


## 3. Existence of Positive Periodic Solutions

We will examine the existence of positive periodic solutions to problem (1) using Krasnoselskii’s fixed-point theorem. For this purpose, we consider B,.=Pω,. and for some positive constant K and L. Moreover, define the set D=φ∈Pω:K≤φ≤L, which is a closed convex and bounded subset of the Banach space Pω.By looking at the three casesgt,x<0,gt,x=0, and gt,x>0 for all t∈ℝ,x∈D, we can prove the existence of a positive periodic solution of (1).


### 3.1. The Casegt,x<0

We assume that there exist nonpositive constantsk1 and k2 such that
(44)k1≤gt,x≤k2,torallt∈0.ω,x∈D.Theorem 8.
Assume thatM<4π/ω4 and the function f satisfies
(45)K−k1α1ω≤ft,xt−τt≤Lα2ω−M−mL+Mk1.
Then, problem (1) has a positive ω-periodic solution x in the subset D.Proof.
Let us start by proving that(46)T1φ+T2ϕ∈D,forallφ,ϕ∈D.
In fact,(47)T1φ+T2ϕ=gt,φt−τt+∫tt+ωGt,sM−asφs+fs,φs−τs−Mgt,φs−τsds≤α2ωM−mL−Mk1+α2∫tt+ωfs,φs−τsds≤α2ωM−mL−Mk1+α2ωLα2ω−M−mL+Mk1=L.
On the other hand,(48)T1φ+T2ϕ=gt,φt−τt+∫tt+ωGt,sM−asφs+fs,φs−τs−Mgt,φs−τsds,≥k1+α1ω∫tt+ωfs,φs−τsds,≥k1+α1ωK−k1α1ω=K,which leads to
(49)T1φ+T2ϕ∈D,forallφ,ϕ∈D.
We conclude from Lemma5 that T1 is a contraction. Also, Lemma 6 implies that the operator T2 is completely continuous.
We deduce from Krasnoselskii’s fixed-point theorem (see [15], p.~31) that T=T1+T2 has a fixed point φ∈D which is a solution to (31). As a result of Remark 4, φ is a solution to problem (1). This completes the proof.


### 3.2. The Casegt,x=0

Theorem 9.
Assume thatM<4π/ω4, and
(50)Kα1ω≤ft,xt−τt≤Lα2ω−M−mL,forallt∈0.ω,x∈D.
Then, equation (1) has a positive ω-periodic solution x in the subset D.Proof.
According to [16], we have T1=0. Similarly to the proof of Theorem 8, we show that (1) has a nonnegative ω-periodic solution x∈D. Since Ft,x>0, it is easy to see that xt>0; i.e., (1) has a positive ω-periodic solution x∈D.


### 3.3. The Casegt,x>0

We assume that there exist nonnegative constantsk3 and k4 such that
(51)k3≤gt,x≤k4,torallt∈0.ω,x∈D.Theorem 10.
Assume thatM<4π/ω4 and the function f satisfies
(52)K−k1α1ω≤ft,xt−τt≤L−k4α2ω−M−mL+Mk3.
Then, problem (1) has a positive ω-periodic solution x in the subset D.Proof.
According to Lemma5, it follows that the operator T1 is a contraction, and from Lemma 6, the operator T2 is completely continuous.
Now, we prove that(53)T1φ+T2ϕ∈D,forallφ,ϕ∈D.
We have(54)T1φ+T2ϕ=gt,φt−τt+∫tt+ωGt,sM−asφs+fs,φs−τs−Mgt,φs−τsds≤k4+α2ωM−mL−Mk3+α2∫tt+ωfs,φs−τsds≤k4+α2ωM−mL−Mk3+α2ωL−k4α2ω−M−mL+Mk3=L.
Also,(55)T1φ+T2ϕ=gt,φt−τt+∫tt+ωGt,sM−asφs+fs,φs−τs−Mgt,φs−τsds≥k3+α1ω∫tt+ωfs,φs−τsds≥k3+α1ωK−k3α1ω=K.
Thus,(56)T1φ+T2ϕ∈D,forallφ,ϕ∈D.
By Krasnoselskii’s theorem (see [15], p. 31), we deduce that T=T1+T2 has a fixed point which is a solution to (31), so problem (1) has a positive ω-periodic solution x in the subset D.


## 3.1. The Casegt,x<0

We assume that there exist nonpositive constantsk1 and k2 such that
(44)k1≤gt,x≤k2,torallt∈0.ω,x∈D.Theorem 8.
Assume thatM<4π/ω4 and the function f satisfies
(45)K−k1α1ω≤ft,xt−τt≤Lα2ω−M−mL+Mk1.
Then, problem (1) has a positive ω-periodic solution x in the subset D.Proof.
Let us start by proving that(46)T1φ+T2ϕ∈D,forallφ,ϕ∈D.
In fact,(47)T1φ+T2ϕ=gt,φt−τt+∫tt+ωGt,sM−asφs+fs,φs−τs−Mgt,φs−τsds≤α2ωM−mL−Mk1+α2∫tt+ωfs,φs−τsds≤α2ωM−mL−Mk1+α2ωLα2ω−M−mL+Mk1=L.
On the other hand,(48)T1φ+T2ϕ=gt,φt−τt+∫tt+ωGt,sM−asφs+fs,φs−τs−Mgt,φs−τsds,≥k1+α1ω∫tt+ωfs,φs−τsds,≥k1+α1ωK−k1α1ω=K,which leads to
(49)T1φ+T2ϕ∈D,forallφ,ϕ∈D.
We conclude from Lemma5 that T1 is a contraction. Also, Lemma 6 implies that the operator T2 is completely continuous.
We deduce from Krasnoselskii’s fixed-point theorem (see [15], p.~31) that T=T1+T2 has a fixed point φ∈D which is a solution to (31). As a result of Remark 4, φ is a solution to problem (1). This completes the proof.


## 3.2. The Casegt,x=0

Theorem 9.
Assume thatM<4π/ω4, and
(50)Kα1ω≤ft,xt−τt≤Lα2ω−M−mL,forallt∈0.ω,x∈D.
Then, equation (1) has a positive ω-periodic solution x in the subset D.Proof.
According to [16], we have T1=0. Similarly to the proof of Theorem 8, we show that (1) has a nonnegative ω-periodic solution x∈D. Since Ft,x>0, it is easy to see that xt>0; i.e., (1) has a positive ω-periodic solution x∈D.


## 3.3. The Casegt,x>0

We assume that there exist nonnegative constantsk3 and k4 such that
(51)k3≤gt,x≤k4,torallt∈0.ω,x∈D.Theorem 10.
Assume thatM<4π/ω4 and the function f satisfies
(52)K−k1α1ω≤ft,xt−τt≤L−k4α2ω−M−mL+Mk3.
Then, problem (1) has a positive ω-periodic solution x in the subset D.Proof.
According to Lemma5, it follows that the operator T1 is a contraction, and from Lemma 6, the operator T2 is completely continuous.
Now, we prove that(53)T1φ+T2ϕ∈D,forallφ,ϕ∈D.
We have(54)T1φ+T2ϕ=gt,φt−τt+∫tt+ωGt,sM−asφs+fs,φs−τs−Mgt,φs−τsds≤k4+α2ωM−mL−Mk3+α2∫tt+ωfs,φs−τsds≤k4+α2ωM−mL−Mk3+α2ωL−k4α2ω−M−mL+Mk3=L.
Also,(55)T1φ+T2ϕ=gt,φt−τt+∫tt+ωGt,sM−asφs+fs,φs−τs−Mgt,φs−τsds≥k3+α1ω∫tt+ωfs,φs−τsds≥k3+α1ωK−k3α1ω=K.
Thus,(56)T1φ+T2ϕ∈D,forallφ,ϕ∈D.
By Krasnoselskii’s theorem (see [15], p. 31), we deduce that T=T1+T2 has a fixed point which is a solution to (31), so problem (1) has a positive ω-periodic solution x in the subset D.


## 4. Conclusion

In this work, we established sufficient conditions for the existence of positive periodic solutions to the fourth-order nonlinear neutral differential equations with variable delay. Our proof relies on Krasnoselskii’s fixed-point theorem, which is an excellent tool when the conditions of the Banach or Schauder fixed-point theorems are not fulfilled.



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