# Periodic Solutions of Duffing Equation with an Asymmetric Nonlinearity and a Deviating Argument

Authors: Zaihong Wang; Jin Li; Tiantian Ma
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study the existence of periodic solutions of the second-order differential equationx′′+ax+-bx-+g(x(t-τ))=p(t), where a,b are two constants satisfying 1/a+1/b=2/n, n∈N, τ is a constant satisfying 0≤τ<2π, g,p:R→R are continuous, and p is 2π-periodic. When the limits limx→±∞g(x)=g(±∞) exist and are finite, we give some sufficient conditions for the existence of 2π-periodic solutions of the given equation.


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## Body

## 1. Introduction

In this paper, we are concerned with the existence of periodic solutions of the second-order differential equation with an asymmetric nonlinearity and a deviating argument:(1)x′′+ax+-bx-+g(x(t-τ))=p(t),
where a,b are two constants satisfying 1/a+1/b=2/n, n∈N, τ is a constant satisfying 0≤τ<2π, g,p:R→R are continuous, and p is 2π-periodic.In recent years, the periodic problem of the second-order differential equation with a deviating argument has been widely studied because of its background in applied sciences (see [1–6] and the references cited therein).In case whenτ=0 and a=b=n2, (1) becomes
(2)x′′+n2x+g(x)=p(t).
Assume that the limits
(g)limx→±∞g(x)=g(±∞)
exist and are finite. Lazer and Leach [7] proved that (2) has one 2π-periodic solution provided that the function
(3)Ψ(θ)=2[g(+∞)-g(-∞)]-∫02π‍p(t)sinn(t+θ)dt
is of constant sign.In case whenτ=0 and a,b satisfy the equation 1/a+1/b=2/n, n∈N, (1) becomes
(4)x′′+ax+-bx-+g(x)=p(t).
Equation (4) was first introduced by Fučík [8]. Lately, the periodic problem of (4) was widely studied in the literature (see [9–13] and the references cited therein). To deal with the existence of periodic solutions of (4), Dancer [9] introduced a 2π/n-periodic function
(5)Φ(θ)=2n[g(+∞)a-g(-∞)b]-∫02π‍p(t)c(t+θ)dt,
where c(t) is a 2π/n-periodic function defined by
(6)c(t)={1asin(at),0≤t≤πa,-1bsin[b(t-πa)],πa≤t≤2πn.
Obviously, c(t) is a periodic solution of the equation x′′+ax+-bx-=0 satisfying the initial value x(0)=0, x′(0)=1. It was proved in [9] that (4) has at least one 2π-periodic solution provided that Φ has a constant sign in [0,2π/n).In the present paper, we will deal with the periodic solutions of (1) under condition (g). Owing to the appearance of the asymmetric nonlinearity ax+-bx-, the methods in [4, 5] are no longer valid. To overcome this difficulty, we embed (1) into an operator equation with the form Lx=N(x,λ) instead of Lx=λNx as in [4, 5]. We first prove a continuation lemma and then apply this continuation lemma to prove the existence of periodic solution of (1).Let us denote(7)ν=τ(mod2πn).
Obviously, we have
(8)0≤ν<2πn.
We obtain the following result.Theorem 1.
Assume that condition(g) holds and 0≤ν≤min{π/a,π/b}. Then (1) has at least one 2π-periodic solution provided that either
(9)ng(-∞)(1-cosaνa-1+cosbνb)+ng(+∞)(1+cosaνa-1-cosbνb)≠∫02π‍p(t)c(t+θ)dt,∀θ∈[0,2π]
or
(10)n(g(-∞)-g(+∞))(sinaνa+sinbνb)≠∫02π‍p(t)s(t+θ)dt,∀θ∈[0,2π]
holds, where the function s is defined by s(t)=c′(t), t∈R.Remark 2.
In the case whenmin{π/a,π/b}≤ν<2π/n, we can obtain the similar sufficient conditions. For brevity, we omit the detailed description.Remark 3.
Obviously, ifν = 0 or τ = 2kπ/n,  k=0,1,2,…,n-1, then the first inequality of Theorem 1 reduces to the condition as in [8]; namely,
(11)2n[g(+∞)a-g(-∞)b]≠∫02π‍p(t)c(t+θ)dt,∀θ∈[0,2π].Throughout this paper, we always useR to denote the real number set. For a multivariate function ζ depending on r, the notation ζ=o(1) always means that, for r→∞, ζ→0 holds uniformly with respect to other variables, whereas ζ=O(1) (or ζ=O(r-1)) always means that ζ (or r·ζ) is bounded for r large enough. For any continuous 2π-periodic function ϕ(t), we always set ∥ϕ∥∞=max0≤t≤2π|ϕ(t)|.


## 2. Preliminary Lemmas

We now embed (1) into a family of equations with one parameter λ∈[0,1],
(12)x′′+ax+-bx-+(1-λ)ψ(x′)+λg(x(t-τ))=λp(t),
where ψ:R→R is continuous and satisfies the sign condition as follows:
(13)ψ(x)x>0,∀x∈R,x≠0.Lemma 4.
Suppose that there exist two positive constantsM1 and M2 such that, for any 2π-periodic solution x(t) of (12), the following conditions hold:
(14)∥x∥∞<M1,∥x′∥∞<M2.
Then (1) has at least one 2π-periodic solution.Proof.
We follow an argument in [14] to prove Lemma 4. At first, we introduce some notations. Let X and Y be two Banach spaces defined by
(15)X={x∈C1(R,R):x(t+2π)=x(t),∀t∈R},Y={y∈C(R,R):y(t+2π)=y(t),∀t∈R},
with the norms
(16)∥x∥X=max{∥x∥∞,∥x′∥∞},∥y∥Y=∥y∥∞.
Define a linear operator by
(17)L:D(L)⊂X⟶Y,Lx=x′′,
where D(L)={x∈X:x′′∈C(R,R)}, and a nonlinear operator N:X×[0,1]→Y,
(18)N(x,λ)(t)=-(ax+-bx-)-(1-λ)ψ(x′)-λg(x(t-τ))+λp(t).
It is easy to see that
(19)KerL=R,ImL={y∈Y:∫0T‍y(t)dt=0}.
It follows that L is a Fredholm mapping of index zero.
Let us define two continuous projectorsP:X→KerL and Q:Y→Y by setting
(20)Px=x(0),Qy=1T∫0T‍y(t)dt.
Set LP=L|D(L)∩KerP→ImL. Then LP is an algebraic isomorphism, and we define KP:ImL→D(L) by
(21)KP=LP-1.
Clearly, we have that, for any y∈ImL,
(22)(KPy)(t)=-tT∫0T‍(t-s)y(s)ds+∫0t‍(t-s)y(s)ds.
For any open bounded set Ω⊂X, we can prove by standard arguments that KP(I-Q)N and QN are relatively compact on the closure Ω-. Therefore, N is L-compact on Ω-.
It is noted that (12), together with the 2π-periodic boundary condition, is equivalent to the operator equation
(23)Lx=N(x,λ).
Let Ω⊂X be the open bounded set defined by
(24)Ω={x∈X:∥x∥∞<M1,∥x′∥∞<M2}.
From (14), we have
(25)Lx≠N(x,λ),forx∈∂Ω∩D(L),λ∈[0,1].
Since L is a Fredholm operator with index zero and N is L-compact on Ω-×[0,1], we get from the homotopic invariance of the coincidence degree that
(26)DL(L-N(·,1),Ω)=DL(L-N(·,0),Ω).
Next, we will compute DL(L-N(·,0),Ω). To this end, we introduce an auxiliary operator S:Ω-×[0,1]→Y defined by
(27)S(x,μ)=-(ax+-bx-)-ψ(x′)-μx′.
Clearly, S is L-compact on Ω-×[0,1] and
(28)S(x,0)=N(x,0),forx∈Ω-.
Now, we will prove that
(29)Lx≠S(x,μ),forx∈∂Ω∩domL,μ∈[0,1].
Obviously, it follows from (25) and (28) that
(30)Lx≠S(x,0),forx∈∂Ω.
On the other hand, if x∈domL is a solution of Lx=S(x,μ), then x satisfies the equation as follows:
(31)x′′+μx′+(ax+-bx-)+ψ(x′)=0.
Multiplying both sides of (31) by x′ and integrating over [0,2π], we get
(32)μ∫02π‍x′2(t)dt+∫02π‍ψ(x′)x′dt=0.
If μ>0, then we infer from (13) and (32) that x′(t)≡0 for every t∈[0,2π]. Furthermore, x(t)≡c for every t∈[0,2π], where c is a constant. Consequently, we have x(t)≡0, and then x∈Ω.
From the homotopic invariance of the coincidence degree, we have(33)DL(L-S(·,0),Ω)=DL(L-S(·,1),Ω).
In the following, we will compute DL(L-S(·,1),Ω). To this end, we use the equality [15] as follows:
(34)|DL(L-S(·,1),Ω)|=|dB(-QS(·,1)|KerL,Ω∩KerL,0)|,
which holds provided that the following conditions are satisfied,
(35)Lx≠λS(x,1),∀x∈∂Ω∩domL,λ∈(0,1],(36)QS(x,1)≠0,∀x∈∂Ω∩KerL.
In what follows, we will prove that conditions (35) and (36) are satisfied. In fact, if x∈∂Ω∩domL is a solution of Lx=λS(x,1), then x(t) satisfies the equation as follows:
(37)x′′(t)+λx′+λ(ax+-bx-)+λψ(x′)=0.
Using the same method as before, we can get x∈Ω. This is a contradiction. To check condition (36), we notice that if x∈∂Ω∩KerL, then x(t)=c′ with |c′|=M1. Hence, we have that, for x∈∂Ω∩KerL,
(38)QS(x,1)=1T∫0T‍(-ax++bx-)dt=-ac′or-bc′≠0.
Finally, we can easily calculate the Brouwer degree dB(-QS(·,1)|KerL,Ω∩KerL,0) and obtain
(39)dB(-QS(·,1)|KerL,Ω∩KerL,0)=1.
Therefore, we have
(40)DL(L-N(·,1),Ω)≠0.
Consequently, the equation
(41)Lx=N(x,1)
has at least one 2π-periodic solution. Equivalently, (1) has at least one 2π-periodic solution.Remark 5.
In (12), if ψ satisfies the following condition,
(42)xψ(x)<0,∀x∈R,x≠0,
then the conclusion of Lemma 4 still holds. This claim can be proved by using the same method as the one used for proving Lemma 4. In fact, we only need to modify the term -μx′ in the auxiliary operator S(x,μ) to the term μx′.


## 3. Periodic Solutions of Duffing Equation with a Deviating Argument

At first, we choose a continuous functionψ:R→R satisfying
(43)limx→±∞ψ(x)=ψ(±∞),
where ψ(±∞)∈R are constants. Moreover, ψ satisfies condition (13).Considering the equivalent system of (12),
(44)x′=y,y′=-(ax+-bx-)-λg(x(t-τ))-(1-λ)ψ(x′)+λp(t).
Let x(t) be any (possible) 2π-periodic solution of (12). Write y(t)=x′(t). Then, (x(t),y(t)) is a 2π-periodic solution of system (44).In what follows, we will introduce a transformation. To this end, let us denote byc(t) a solution of equation x′′+ax+-bx-=0 satisfying the initial condition c(0)=0, c′(0)=1. Obviously, c(t) is 2π/n-periodic. The derivative of c(t) will be denoted by s(t)=c′(t). It is easy to check that the following properties are satisfied:(1)
c(t+2π/n)=c(t), s(t+2π/n)=s(t).(2)
c′(t)=s(t), s′(t)=-(ac+(t)-bc-(t)).(3)
s(t)2+ac+(t)2+bc-(t)2=1,  ∀t∈R.Let us define a mapping Φ:(θ,ρ)∈S1×(0,+∞)→(x,y)∈R2∖{0} as follows:
(45)x=ρ1/2c(θn),y=ρ1/2s(θn),
where S1=R/2πZ.Under the transformationΦ, if |x(t)|+|y(t)|≠0, ∀t∈[0,2π], then the 2π-periodic solution (x(t),y(t)) of system (44) can be expressed in the form (ρ(t),θ(t)) satisfying the equations as follows:
(46)dρdt=-2λρ1/2×(g(ρ1/2(t-τ)c(θ(t-τ)n))s(θn)-p(t)s(θn))-2(1-λ)ρ1/2ψ(ρ1/2s(θn))s(θn),dθdt=n+nλρ-1/2×(g(ρ1/2(t-τ)c(θ(t-τ)n))c(θn)-p(t)c(θn))+n(1-λ)ρ-1/2ψ(ρ1/2s(θn))c(θn).
Let us denote   (ρ0,θ0)=(ρ(0),θ(0)). From now on, we always assume that g is bounded. From the first equation of (46) we get that
(47)dρ1/2dt=-λg(ρ1/2(t-τ)c(θ(t-τ)n))s(θn)+λp(t)s(θn)-(1-λ)ψ(ρ1/2s(θn))s(θn).
Therefore, we have
(48)ρ(t)1/2=ρ01/2+O(1).
Furthermore, we get
(49)ρ(t)-1/2=ρ0-1/2+O(ρ0-1).
From the second equation of (46), we have
(50)dθdt=n+O(ρ0-1/2).
As a result,
(51)θ(t)=θ0+nt+O(ρ0-1/2).
Substituting (51) in (47), we obtain that, for t∈[0,2π],
(52)dρ1/2dt=-λg(ρ01/2c(t-τ+θ0n)+O(1))s(t+θ0n)+λp(t)s(t+θ0n)-(1-λ)ψ(ρ1/2s(θn))s(θn)+O(ρ0-1/2).
Consequently,
(53)ρ1/2(2π)=ρ01/2-λ∫02π‍g(ρ01/2c(t-τ+θ0n)+O(1))s(t+θ0n)dt-(1-λ)∫02π‍ψ(ρ01/2s(t+θ0n)+O(1))s(t+θ0n)dt+λ∫02π‍p(t)s(t+θ0n)dt+O(ρ0-1/2).
Similarly, substituting (51) in the second equality of (44), we get that, for t∈[0,2π],
(54)dθdt=n+nλρ0-1/2g(ρ01/2c(t-τ+θ0n)+O(1))×c(t+θ0n)-nλρ0-1/2p(t)c(t+θ0n)+n(1-λ)ρ0-1/2ψ(ρ01/2s(t+θ0n)+O(1))×c(t+θ0n)+O(ρ0-1).
Therefore, we have
(55)θ(2π)=θ0+2nπ+nλρ0-1/2×∫02π‍g(ρ01/2c(t-τ+θ0n)+O(1))c(t+θ0n)dt+n(1-λ)ρ0-1/2×∫02π‍ψ(ρ01/2s(t+θ0n)+O(1))c(t+θ0n)dt-nλρ0-1/2∫02π‍p(t)c(t+θ0n)dt+O(ρ0-1).
Write
(56)ψ1(θ0)=∫02π‍g(ρ01/2c(t-τ+θ0n)+O(1))s(t+θ0n)dt,ψ2(θ0)=∫02π‍g(ρ01/2c(t-τ+θ0n)+O(1))c(t+θ0n)dt,ψ3(θ0)=∫02π‍ψ(ρ01/2s(t+θ0n)+O(1))s(t+θ0n)dt,ψ4(θ0)=∫02π‍ψ(ρ01/2s(t+θ0n)+O(1))c(t+θ0n)dt.
Recalling that ν=τ(mod(2π/n)) and 0≤ν<2π/n, we have the following estimates.Lemma 6.
Assume that condition(g) holds. Then, for ρ0→+∞,
(57)ψ1(θ0)={-n(g(+∞)-g(-∞))(sinaνa+sinbνb)+o(1),forν≤πa≤πb,orν≤πb≤πa,n(g(+∞)-g(-∞))×[sinb(ν-π/a)b-sinbνb]+o(1),forπa<ν≤πb,n(g(+∞)-g(-∞))×[sina(ν-π/b)a-sinaνa]+o(1),forπb<ν≤πa,n(g(+∞)-g(-∞))×[sina(ν-π/b)a+sinb(ν-π/a)b]+o(1),forπa≤πb<νorπb≤πa<ν.Proof.
We only give the proof for the case0≤ν≤π/a≤π/b<2π/n. The other cases can be treated similarly. Since s(t) is 2π/n-periodic, it follows from the expression of ψ1(θ0) that
(58)ψ1(θ0)=∫02π‍g(ρ01/2c(t-τ+θ0n)+O(1))s(t+θ0n)dt=∫02π‍g(ρ01/2c(u)+O(1))s(u+τ)du=∫02π‍g(ρ01/2c(u)+O(1))s(u+ν)du.
From the dominated convergent theorem, we have that, for ρ0→∞,
(59)ψ1(θ0)=ng(+∞)∫0π/a‍s(u+ν)du+ng(-∞)×∫π/a2π/n‍s(u+ν)du+o(1)=ng(+∞)[∫0π/a-ν‍s(u+ν)du+∫π/a-νπ/a‍s(u+ν)du]+ng(-∞)[∫π/a2π/n-ν‍s(u+ν)du+∫2π/n-ν2π/n‍s(u+ν)du]+o(1)=-ng(+∞)(sinaνa+sinbνb)+ng(-∞)(sinaνa+sinbνb)+o(1)=-n(g(+∞)-g(-∞))(sinaνa+sinbνb)+o(1).Lemma 7.
Assume that condition(g) holds. Then, for ρ0→+∞,
(60)ψ2(θ0)={ng(+∞)(1+cosaνa-1-cosbνb)+ng(-∞)(1-cosaνa-1+cosbνb)+o(1),forν≤πa≤πb,orν≤πb≤πa,ng(+∞)[cosbν-cosb(ν-π/a)b]+ng(-∞)×[(2a-2b)-cosbν-cosb(ν-π/a)b]+o(1),forπa<ν≤πb,ng(+∞)[(2a-2b)-cosa(ν-π/b)-cosaνa]+ng(-∞)[cosa(ν-π/b)-cosaνa]+o(1),forπb<ν≤πa,ng(+∞)[1-cosa(ν-π/b)a-1+cosb(ν-π/a)b]+ng(-∞)[1+cosa(ν-π/b)a-1-cosb(ν-π/a)b]+o(1),forπa≤πb<νorπb≤πa<ν.Proof.
We also only give the proof for the case0≤ν≤π/a≤π/b<2π/n. The other cases can be treated similarly. Since c(t) is 2π/n-periodic, it follows from the expression of ψ2(θ0) and the dominated convergent theorem that, for ρ0→∞,
(61)ψ2(θ0)=∫02π‍g(ρ01/2c(t-τ+θ0n)+O(1))c(t+θ0n)dt=∫02π‍g(ρ01/2c(u)+O(1))c(u+τ)du=∫02π‍g(ρ01/2c(u)+O(1))c(u+ν)du=ng(+∞)∫0π/a‍c(u+ν)du+ng(-∞)∫π/a2π/n‍c(u+ν)du+o(1)=ng(+∞)×[∫0π/a-ν‍c(u+ν)du+∫π/a-νπ/a‍c(u+ν)du]+ng(-∞)×[∫π/a2π/n-ν‍c(u+ν)du+∫2π/n-ν2π/n‍c(u+ν)du]+o(1)=ng(+∞)(1+cosaνa-1-cosbνb)+ng(-∞)(1-cosaνa-1+cosbνb)+o(1).Lemma 8.
Assume that condition (43) holds. Then, for ρ0→+∞,
(62)ψ3(θ0)=2[ψ(+∞)-ψ(-∞)]+o(1),ψ4(θ0)=n[ψ(+∞)+ψ(-∞)](1a-1b)+o(1).Proof.
From the expression ofψ3(θ0) and the dominated convergent theorem we have that, for ρ0→∞,
(63)ψ3(θ0)=∫02π‍ψ(ρ01/2s(t+θ0n)+O(1))s(t+θ0n)dt=∫02π‍ψ(ρ01/2s(u)+O(1))s(u)du=nψ(+∞)(∫0π/2a‍s(u)du+∫π/a+π/2b2π/n‍s(u)du)+nψ(-∞)(∫π/2aπ/a‍s(u)du+∫π/aπ/a+π/2b‍s(u)du)+o(1)=2[ψ(+∞)-ψ(-∞)]+o(1).
Similarly, we have that, for ρ0→+∞,
(64)ψ4(θ0)=∫02π‍ψ(ρ01/2s(t+θ0n)+O(1))c(t+θ0n)dt=∫02π‍ψ(ρ01/2s(u)+O(1))c(u)du=nψ(+∞)(∫0π/2a‍c(u)du+∫(π/a)+(π/2b)2π/n‍c(u)du)+nψ(-∞)(∫π/2aπ/a‍c(u)du+∫π/a(π/a)+(π/2b)‍c(u)du)+o(1)=n[ψ(+∞)+ψ(-∞)](1a-1b)+o(1).Proof of Theorem1.
We proceed to prove Theorem1 in two different cases.
(
1
) Assume that the first inequality of Theorem 1 holds. Without loss of generality, we assume
(65)ng(-∞)(1-cosaνa-1+cosbνb)+ng(+∞)(1+cosaνa-1-cosbνb)>∫02π‍p(t)c(t+θ)dt,∀θ∈[0,2π].
Let us set
(66)η(θ)=ng(-∞)(1-cosaνa-1+cosbνb)+ng(+∞)(1+cosaνa-1-cosbνb)-∫02π‍p(t)c(t+θ)dt>0,θ∈[0,2π].
We now choose a function ψ satisfying (43) and (13). Moreover, ψ(±∞) satisfy
(67)μ=[ψ(+∞)+ψ(-∞)](1a-1b)>0.
Then we infer from Lemmas 7 and 8 that, for ρ0→∞,
(68)θ(2π)=θ0+2nπ+nρ0-1/2[λη(θ0n)+n(1-λ)μ]+o(ρ0-1/2).
Since η(θ)>0, θ∈[0,2π], and μ>0, there exists a constant γ>0 such that, for θ∈[0,2π] and λ∈[0,1],
(69)λη(θ)+n(1-λ)μ≥γ.
From (68) and (69) we have that, for ρ0→∞,
(70)θ(2π)=θ0+2nπ+o(1),θ(2π)>θ0+2nπ.
Consequently, there exists a constant M>0 such that if (ρ(t),θ(t)) is a 2π-periodic solution of system (46), then ρ(t)≤M, t∈[0,2π]. Furthermore, there exist constants M1>0 and M2>0 such that if x(t) is a 2π-periodic solution of (12), then
(71)∥x∥∞<M1,∥x′∥∞<M2.
From Lemma 4, we know that (1) has at least one 2π-periodic solution.
(
2
) We assume that the second inequality of Theorem 1 holds. Without loss of generality, we assume
(72)n(g(-∞)-g(+∞))(sinaνa+sinbνb)>∫02π‍p(t)s(t+θ)dt,∀θ∈[0,2π].
Let us set
(73)ζ(θ)=n(g(-∞)-g(+∞))(sinaνa+sinbνb)-∫02π‍p(t)s(t+θ)dt>0,θ∈[0,2π].
Similarly, we choose a continuous function ψ satisfying (43) and (13). Moreover, ψ(±∞) satisfy
(74)μ′=ψ(+∞)-ψ(-∞)>0.
Then we infer from Lemmas 7 and 8 that, for ρ0→∞,
(75)ρ1/2(2π)=ρ01/2-[λζ(θ0n)+2(1-λ)μ′]+o(1).
Since ζ(θ)>0, θ∈[0,2π], and μ′>0, there exists a constant γ′>0 such that, for θ∈[0,2π] and λ∈[0,1] and ρ0→∞,
(76)λζ(θ)+2(1-λ)μ′≥γ′.
From (75) and (76) we have that, for sufficiently large ρ0,
(77)ρ1/2(2π)≤ρ01/2-γ′2.
Consequently, there exists a constant M′>0 such that if (ρ(t),θ(t)) is a 2π-periodic solution of system (46), then ρ(t)≤M′, t∈[0,2π]. Furthermore, there exist constants M1′>0 and M2′>0 such that if x(t) is a 2π periodic solution of (12), then
(78)∥x∥∞<M1′,∥x′∥∞<M2′.
From Lemma 4 we know that (1) has at least one 2π periodic solution.



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