# Existence of Periodic Solutions for Nonlinear Fully Third-Order Differential Equations

Authors: Yunfei Zhang; Minghe Pei
Journal: Journal of Function Spaces
 (2020)

---

## Abstract

In this paper, we study the existence of periodic solutions to nonlinear fully third-order differential equationx‴+ft,x,x′,x″=0,t∈ℝ≔−∞,∞, where f:ℝ4⟶ℝ is continuous and T-periodic in t. By using the topological transversality method together with the barrier strip technique, we obtain new existence results of periodic solutions to the above equation without growth restrictions on the nonlinearity. Meanwhile, as applications, an example is given to demonstrate our results.


---

## Body

## 1. Introduction

In this paper, we consider the existence of periodic solutions for nonlinear fully third-order differential equation(1)x‴t+ft,xt,x′t,x″t=0,t∈ℝ,where f:ℝ4⟶ℝ is continuous and T-periodic with respect to t.The third-order periodic problem arises in many areas of applied mathematics and physics, and so it has been extensively studied by many authors via various methods, for instance, see [1–22] and the references therein. Among a substantial number of works, we mention that the upper and lower solutions method is used in [5, 6, 15, 16], Leray–Schauder continuation theorem is used in [1, 10, 20, 21], Leray–Schauder degree theory or the Schauder-fixed-point theorem is used in [10, 12, 13, 18], Mawhin coincidence degree theory is used in [2, 3, 8, 17], and fixed-point theorem in cone or fixed-point index theory is used in [7, 9, 11, 19, 22]. However, to the best of our knowledge, there is no work that refers to periodic solutions of equation (1) using the topological transversality method.Recently, Kelevedjiev and Todorov [23] have used the topological transversality method and barrier strip technique to study various third-order two-point boundary value problems. But, they did not consider the third-order periodic boundary value problem.Motivated and inspired by the aforementioned works, the aim of this paper is to establish new existence results of periodic solutions to equation (1) by using the topological transversality method together with barrier strip technique. It is worth mentioning that our results do not need any growth restrictions on the nonlinearity. In addition, compared with the corresponding ones in the known literature, the barrier strip technique we use to estimate a prior bounds of periodic solutions is essentially new.This work is organized as follows. In Section2, we first introduce some notations and lemmas and then estimate a prior bounds of periodic solutions of equation (1) by using barrier strip technique. Finally, by using the topological transversality method, we establish the existence results of periodic solutions to equation (1). As applications of our main results, an example is given in the last section.


## 2. Main Results

Throughout this section, the following assumptions are used:H0: there exists M>0 such that, for any T-periodic function x∈C2ℝ,(2)signxt⋅∫0Tft,xt,x′t,x″tdt>0,ifmint∈0,Txt>M.H0′: there exists M>0 such that, for any T-periodic function x∈C2ℝ,(3)signxt⋅∫0Tft,xt,x′t,x″tdt<0,ifmint∈0,Txt>M.H1: there exists M1>0 such that(4)yft,x,y,0<0,∀t,x∈0,T×ℝ,y>M1.H2: there are constants Li,i=1,2,3,4 with L2>L1>0 and L3<L4<0 such that ft,x,y,z does not change its sign for t,x,y,z∈D1 and for t,x,y,z∈D2, respectively, here(5)D1=0,T×−M0,M0×−M1,M1×L1,L2,D2=0,T×−M0,M0×−M1,M1×L3,L4,and M0=M+TM1.Let us introduce some notations of the topological transversality method for the convenience of the reader. LetU be a convex subset of a Banach space Y and Ω⊂U be open in U. Denote by H∂ΩΩ¯,U the set of compact operators F:Ω¯⟶U which are fixed-point-free on ∂Ω. We say that F∈H∂ΩΩ¯,U is essential if every operator in H∂ΩΩ¯,U which agrees with F on ∂Ω has a fixed point in Ω.The next two lemmas can be found in [24].Lemma 1.
Ifp∈Ω and F∈H∂ΩΩ¯,U is a constant operator, Fy=p for y∈Ω¯ then is essential.Lemma 2.
Let(i)
F∈H∂ΩΩ¯,U be essential(ii)
H:Ω¯×0,1⟶U be a compact homotopy, H⋅,0=F, and Hy,λ≠y for y∈∂Ω and λ∈0,1
Then,H⋅,1 is essential, and therefore, it has a fixed point in Ω.We note that the existence ofT-periodic solutions for equation (1) is equivalent to the solvability of the following third-order periodic boundary value problem (for short PBVP):(6)x‴t+ft,xt,x′t,x″t=0,0≤t≤T,x0=xT,x′0=x′T,x″0=x″T,in C20,T.We first consider the family of following PBVPs:(7)x‴t+λft,xt,x′t,x″t=0,0≤t≤T,(8)x0=xT,x′0=x′T,x″0=x″T,where λ∈0,1.A priori bounds for the solutions of PBVP (7), (8) are presented in the following lemmas.Lemma 3.
Suppose thatH0(or H0′) and H1 hold. Let xt be a solution of PBVPs (7) and (8) for some λ∈0,1. Then,(9)xt≤M0,∀t∈0,T,(10)x′t≤M0,∀t∈0,T.Proof.
At first, we show that (10) holds. Indeed, suppose on the contrary that there exist t1∈0,T such that x′t1>M1. We may assume that x′t1>M1. Let t2∈0,T be such that(11)x′t2=maxt∈0,Tx′t>M1.
Without loss of generality, we assume thatt2∈0,T; then, x″t2=0 and x‴t2≤0. It follows from condition H1 that(12)0≥x′t2x‴t2=−x′t2λft2,xt2,x′t2,0>0,which is a contradiction. This means that (10) holds.
Next, we prove that (9) holds. Indeed, integrating the equation in (7) from 0 to T, we obtain that(13)∫0Tft,xt,x′t,x″tdt=0.
This together with the conditionH0(or H0′) implies that there exists ξ∈0,T such that xξ≤M. It follows from (10) that, for t∈0,T,(14)xt=xξ+∫ξtx′sds≤M+∫0Tx′sds≤M0,which means that (9) holds. This completes the proof of the lemma.Lemma 4.
Suppose thatH0(or H0′), H1, and H2 hold. Let xt be a solution of PBVPs (7) and (8) for some λ∈0,1. Then,(15)x″t≤M2≔maxL1,−L4,∀t∈0,T.Proof.
We estimatex″t by using the barrier strip technique. From condition H2, ft,x,y,z does not change its sign for t,x,y,z∈D1 and for t,x,y,z∈D2, respectively. For the sake of certainty, without loss of generality, we assume that(16)ft,x,y,z≤0,onD1,ft,x,y,z≤0,onD2.
Notice thatx′0=x′T, and from Rolle’s mean value theorem, there exists η∈0,T such that x″η=0. Let(17)S0=t∈0,η:L1<x″t≤L2,S1=t∈0,η:L3≤x″t<L4.
We now assert that the setsS0 and S1 are empty. We shall complete the proof in two steps.Step 1.
Show thatS0=∅. Suppose on the contrary that there exist some t0∈S0. Then, L1<x″t0≤L2, and t0∈0,η. Since x″t is continuous on 0,η, there exist t0≤t1<t2<η such that(18)L1<x″t2<x″t1≤L2,x″t2≤x″t≤x″t1,∀t∈t1,t2,and so t1,t2⊂S0. Consequently, from assumption (16), we have(19)x‴t=−λft,xt,x′t,x″t≥0,∀t∈t1,t2⊂S0,and thus,(20)x″t2≥x″t1,which contradicts (18). This implies that S0=∅.Step 2.
Prove thatS1=∅. By contradiction, assume that there exist some t0′∈S1. Then, L3≤x″t0′<L4, and t0′∈0,η. We now assert that(21)x″0≥L4.
Indeed, ifx″0≥L4, then from the fact x″0=x″T, it follows that x″T≥L4. Notice that x″η=0, and it follows from the continuity of x″t on η,T that there exist η<η1<η2≤T such that(22)L3≤x″η2<x″η1<L4,x″η2≤x″t≤x″η1,∀t∈η1,η2.
Hence, from Lemma3, we have(23)t,xt,x′t,x″t∈D2,∀t∈η1,η2,and so from (16), it follows that(24)x‴t=−λft,xt,x′t,x″t≥0,∀t∈η1,η2.
Therefore,(25)x″η2≥x″η1,which contradicts (22). This means that (21) holds.
Thus, from (21) and the continuity of x″t on 0,η, there exist 0<t1′<t2′<t0′ such that(26)L3≤x″t2′<x″t1′<L4,x″t2′≤x″t≤x″t1′,∀t∈t1′,t2′,and thus, t1′,t2′⊂S1. It follows from assumption (16) that(27)x‴t=−λft,xt,x′t,x″t≥0,∀t∈t1′,t2′⊂S1,and hence,(28)x″t1′≤x″t2′,which contradicts (26). This implies that S1=∅.
Therefore, by the facts thatx″η=0 and the continuity of x″t on 0,η, we obtain(29)L4≤x″t≤L1,∀t∈0,η.
In particular,L4≤x″0≤L1. Notice x″0=x″T, and we have(30)L4≤x″T≤L1.
We now let(31)S2=t∈η,T:L1<x″t≤L2,S3=t∈η,T:L3<x″t≤L4.
Notice that, from (30), using the similar arguments on S0=∅ and S1=∅, we can show that(32)S2=∅,S3=∅.
Hence,(33)L4≤x″t≤L1,∀t∈η,T.
From this, together with (29), it follows that(34)L4≤x″t≤L1,∀t∈0,T,which implies that(35)x″t≤maxL1,−L4=M2,∀t∈0,T.
This completes the proof of the lemma.
Now, denoteY=C20,T×ℝ the Banach space equipped with the norm x,r=x∞+x′∞+x″∞+r. Set(36)U=x,r∈Y:x0=xT=0,r∈ℝ,Ω=x,r∈U:x∞<2M0+1,x′∞<M1+1,x″∞<M2+1,r<M0+1.
Then,U is a closed and convex subset of Y, and Ω is an open subset of U.
We now give two lemmas which will be used in the proof of our main theorem.Lemma 5.
Suppose thatH0 holds. Let the operator F1:Ω¯⟶U be defined by(37)F1x,r=0,r−∫0Tfτ,xτ+r,x′τ,x″τdτ.Then,F1 is essential.Proof.
DefineH:Ω¯×0,1⟶U by(38)Hx,r,λ=0,λr−λ∫0Tfτ,xτ+r,x′τ,x″τdτ.
Then,H⋅,⋅,1=F1⋅,⋅ and Hx,r,0=0,0∈Ω for x,r∈Ω¯. Thus, it follows from Lemma 1 that Hx,r,0 is essential. Meanwhile, by a standard argument, it is easy to show that Hx,r,λ is compact.
We now show that(39)Hx,r,λ≠x,r,∀x,r∈∂Ω,λ∈0,1.
Obviously,Hx,r,0≠x,r for all x,r∈∂Ω. Suppose that Hx0,r0,λ0=x0,r0 for some x0,r0∈∂Ω and λ0∈0,1. Then, x0t≡0 on 0,T, and so(40)∫0Tfτ,r0,0,0dτ=1−1λr0.
Hence, fromH0, we can deduce that −M≤r0≤M, which contradicts 0,r0∈∂Ω. This implies that (39) holds. Therefore, from Lemma 2, F1·,·=H·,·,1 is essential. This completes the proof of the lemma.Lemma 6.
Suppose thatH0′ holds. Let the operator F2:Ω¯⟶U be defined by(41)F2x,r=0,r+∫0Tfτ,xτ+r,x′τ,x″τdτ.Then,F2 is essential.Proof.
The proof is similar to the proof of Lemma5 and hence is omitted.Theorem 1.
Suppose thatH0, H1, and H2 hold. Then, equation (1) has at least one T-periodic solution x=xt satisfying (9), (10), and (15).Proof.
At first, we define operatorA:Ω¯×0,1⟶C20,T by(42)Ax,r,λ=λ∫0TKt,s∫0sfτ,xτ+r,x′τ,x″τdτ−1T∫0TT−τfτ,xτ+r,x′τ,x″τdτds,where(43)Kt,s=1TtT−s,0≤t≤s≤T;sT−t,0≤s≤t≤T.
It is easy to check that, for eachx,r,λ∈Ω¯×0,1, Ax,r,λ is the unique solution of the following boundary value problem:(44)u‴+λft,xt+r,x′t,x″t=0,t∈0,T,u0=uT=0,u′0=u′T.
Furthermore, by a standard argument, it is easy to show that the operatorAx,r,λ is completely continuous.
We now define operatorG1:Ω¯×0,1⟶U by(45)G1x,r,λ=Ax,r,λ,r−∫0Tfτ,xτ+r,x′τ,x″τdτ.
Suppose thatx1,r1 is a fixed point of G1·,·,1. Then,(46)x1t=∫0TKt,s∫0sfτ,x1τ+r1,x1′τ,x1′′τdτ−1T∫0TT−τfτ,x1τ+r1,x1′τ,x1′′τds,∫0Tfτ,x1τ+r1,x1′τ,x1′′τdτ=0.
It follows thatx10=x1T=0, x1′0=x1′T, and(47)−x1′′t=∫0tfτ,x1τ+r1,x1′τ,x1′′τdτ−1T∫0TT−τfτ,x1τ+r1,x1′τ,x1′′τdτ,t∈0,T,and thus, by (46),(48)x1′′0=1T∫0TT−τfτ,x1τ+r1,x1′τ,x1′′τdτ=x1′′T.
Setx2t=x1t+r1 for t∈0,T. It is easy to see that x2t is a solution of PBVP (6), and validity of (9), (10), and (15) now follows from Lemmas 3 and 4. Therefore, to prove the existence of solutions of PBVP (6) satisfying (9), (10), and (15), it is sufficient to show that the operator G1·,·,1 has at least one fixed point. Since G1·,·,0=F1·,· and F1 is essential by Lemma 5, for the existence of a fixed point of G1·,·,1, it is sufficient to verify (i) and (ii) of Lemma 2. Notice that operator A is completely continuous, then operator G1 is continuous, and also G1Ω¯×0,1 is compact in U. Thus, (i) of Lemma 2 is satisfied. Let G1x0,r0,λ0=x0,r0 for some x0,r0∈∂Ω and λ0∈0,1. If λ0=0, then x0,r0∉∂Ω, which has been proved in the proof of Lemma 5. Let λ0∈0,1. Then,(49)x0t=λ0∫0TKt,s∫0sfτ,x0τ+r0,x0′τ,x0″τdτ−1T∫0TT−τfτ,x0τ+r0,x0′τ,x0″τdτds,∫0Tfτ,x0τ+r0,x0′τ,x0″τdτ=0.
Hence,(50)−x0″′t=λ0ft,x0t+r0,x0′t,x0″t,t∈0,T,x00=x0T=0,x0′0=x0′T,and so(51)x0″0=1T∫0TT−τfτ,x0τ+r0,x0′τ,x0″τdτ=x0″T.
Setxt=x0t+r0 for t∈0,T. We can see that xt is a solution of PBVPs (7) and (8) with λ=λ0. It follows from Lemmas 4 and 5 that(52)x0+r0∞=x∞≤M0,x0′∞=x′∞≤M1<M1+1,x0′′∞=x′′∞≤M2<M2+1.
Sincex00=0, the first inequality of (52) yields(53)r0≤M0<M0+1,and thus,(54)x0∞≤M0+r0≤2M0<2M0+1.
Hence,x0,r0∉∂Ω, and so (ii) of Lemma 2 is verified. This completes the proof of the theorem.Theorem 2.
Suppose thatH0′, H1, and H2 hold. Then, equation (1) has at least one T-periodic solution x=xt satisfying (9), (10), and (15).Proof.
The proof is similar as that for Theorem1 except that(55)G2x,r,λ≔Ax,r,λ,r+∫0Tfτ,xτ+r,x′τ,x″τdτand F2 are used in place of G1x,r,λ and F1, respectively, and hence is omitted. This completes the proof of the theorem.


## 3. An Example

In this section, we give an example to demonstrate the applications of the our results.Example 1.
Consider the nonlinear third-order differential equation(56)x‴+x′2+1x″−1−12cosxx′+x−1x2+1=sint,t∈ℝ.
Let(57)ft,x,y,z=y2+1z−1−12cosxy+x−1x2+1−sint,t,x,y,z∈ℝ4.
Then, for any2π-periodic function x∈C2ℝ, we have(58)signxt∫02πft,xt,x′t,x″tdt=signxt∫02πxt−1x2t+1dt>0,ifmint∈0,2πxt>1≔M,and so condition H0 holds. Notice that(59)limy⟶±∞yft,x,y,0=limy⟶±∞y−1−12cosxy+x−1x2+1−sint=−∞,uniformly in t,x∈0,2π×ℝ; then, there exists M1>0 such that(60)yft,x,y,0<0,∀t,x∈0,2π×ℝ,y>M1,that is, condition H1 holds. In addition, since(61)limz⟶±∞ft,x,y,z=±∞,uniformly in t,x,y∈0,2π×−M0,M0×−M1,M1, where M0=M+2πM1, it follows that condition H2 holds. Hence, by Theorem 1, third-order differential equation (56) has at least one 2π-periodic solution.



---

*Source: 6793721-2020-04-06.xml*