# Periodic Solutions and Nontrivial Periodic Solutions for a Class of Rayleigh-Type Equation with Two Deviating Arguments

Authors: Meiqiang Feng
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

The Rayleigh equation with two deviating argumentsx′′(t)+f(x'(t))+g1(t,x(t-τ1(t)))+g2(t,x(t-τ2(t)))=e(t) is studied. By using Leray-Schauder index theorem and Leray-Schauder fixed point theorem, we obtain some new results on the existence of periodic solutions, especially for the existence of nontrivial periodic solutions to this equation. The results are illustrated with two examples, which cannot be handled using the existing results.


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## Body

## 1. Introduction

Consider the Rayleigh equation with two deviating arguments in the form of(1)x′′(t)+f(x′(t))+g1(t,x(t-τ1(t)))+g2(t,x(t-τ2(t)))=e(t),
where
(2)f∈C(R,R),gi∈C(R×R,R),i=1,2,e,τi∈C(R,R),i=1,2,gi(t+T,x)=gi(t,x),τi(t+T)=τi(t),i=1,2,e(t+T)=e(t).The dynamic behaviors of Rayleigh equation have been widely investigated due to their applications in many fields such as physics, mechanics, and the engineering technique fields. For example, an excess voltage of ferroresonance, known as some kind of nonlinear resonance having long duration that arises from the magnetic saturation of inductance in an oscillating circuit of a power system, and a boosted excess voltage can give rise to some problems in relay protection. To probe this mechanism, a mathematical model was proposed in [1–3], which is a special case of the Rayleigh equation with two delays. This implies that (1) can represent analog voltage transmission. In a mechanical problem, f usually represents a damping or friction term, gi represents the restoring force, e is an externally applied force, and τi is the time lag of the restoring force (see [4]). Some other examples in practical problems concerning physics and engineering technique fields can be found in [5–7].At the same time, the periodic solutions for Rayleigh equations with two deviating arguments have been studied by authors [8–10] under the assumption of
(3)f(0)=0orf(t,0)=0.It is not difficult to see that ifg1(t,0)+g2(t,0)≢e(t), then the periodic solution obtained in [8–10] must be nontrivial. But if g1(t,0)+g2(t,0)≡e(t), then the periodic solution obtained in [8–10] may be trivial under the assumption of f(0)=0 or f(t,0)=0. And if the periodic solution is unique, then it must be trivial. Thus, it is worth discussing the existence of the nontrivial periodic solutions of Rayleigh equations with two deviating arguments in this case.The main purpose of this paper is to establish sufficient conditions for the existence of periodic solution, especially for the existence of nontrivial periodic solutions of (1) by using the Leray-Schauder index theory. We remark that our methods are different from those used in [8–10] to some degree. In particular, two examples are also given to illustrate the effectiveness of our results.For ease of exposition, throughout this paper, we assume thatT>0.On the other hand, the following assumptions are used in this paper.(H1)
f∈C(R,R), gi∈C(R2,R), τi∈C(R,R), i=1,2, e∈C(R,R), gi(t+T,x)=gi(t,x), τi(t+T)=τi(t), i=1,2, and e(t+T)=e(t).(H2)
f(0)=0, and there exists γ>0 such that xf(x)≥γ|x|2, for all x∈R (or xf(x)≤-γ|x|2, for all x∈R).(H3)
gi is differentiable with respect to t, and there exist ai≥0, bi>0, i=1,2, such that
(4)|git′(t,x)|≤ai+bi|x|,∀(t,x)∈R2,i=1,2.(H4)
There existli>0 such that |gi(t,x1)-gi(t,x2)|≤li|x1-x2|, for all t∈R, x1,x2∈R, i=1,2.(H5)
There existsd>0, such that for |x|>d, we have
(5)(g1(t,x)+g2(t,x)-e(t))signx>0,∀t∈R.(H6)
There exist integersni such that δi≜maxt∈[0,T]|τi(t)-niT|≤T, i=1,2.The main results of the present paper are as follows.Theorem 1.
If (H1)–(H6) hold, and (b1+b2)γ-1T2+22(l1δ1+l2δ2)π2γ-1<2π2, then (1) has at least one periodic solution.The proof of Theorem1 will be given in Section 3.Remark 2.
Ifg1(t,0)+g2(t,0)≢e(t), then the periodic solution obtained by Theorem 1 must be nontrivial. If g1(t,0)+g2(t,0)≡e(t), we could not conclude whether or not the periodic solution is nontrivial.
So we give the following conditions:(H7)
gix′(t,x), i=1,2 exist and are continuous, g1x′(t,0)+g2x′(t,0)≡const.<0.(H8)
limx→0(f(x)/x)=r≥γ.Theorem 3.
If (H1)–(H8) hold, and g1(t,0)+g2(t,0)≡e(t), (b1+b2)γ-1T2+22(l1δ1+l2δ2)π2γ-1<2π2, then (1) has at least one nontrivial periodic solution.The proof of Theorem3 will be given in Section 4.


## 2. Preliminaries

In this section, to establish the periodic solutions of (1), we provide some background definitions and some well-known results, which are crucial in our arguments.LetX be a real Banach space, and let A:X→X be a completely continuous operator.Definition 4 (see [11, 12]).
Ifx0 is an isolated fixed point of A, then the fixed point index at x0 of A is defined by
(6)i(A,x0)=deg(I-A,Ω,θ),
where Ω is a neighborhood of x0, which satisfies that x0 is the unique fixed point in Ω of A.Definition 5 (see [11, 12]).
If there exitsx0∈X with x0≠θ such that Ax0=λx0, λ∈R, then λ is called an eigenvalue of operator A and x0 is called the eigenfunction of operator A corresponding to λ.Definition 6.
Letu:R→R be continuous. u(t) is said to be periodic on R if
(7)u(t+T)=u(t),∀t∈R.Lemma 7 (see [11, 12], index theorem of Leray-Schauder).
Suppose thatx0 is a fixed point of A, A is Fréchet differentiable at x0, and 1 is not the eigenvalue of A′(x0). Then x0 is an isolated fixed point of A, and
(8)i(A,x0)=i(A′(x0),0)=(-1)β,
where β is equal to the sum of the algebraic multiplicities of all of the eigenvalues λi>1 of A′(x0).Lemma 8 (see [11, 12], fixed point theorem of Leray-Schauder).
LetX be a real Banach space and A:X→X be a completely continuous operator. If
(9){x:x∈X,x=λAx,0<λ<1}
is bounded, then A has a fixed point x*∈Ω, where
(10)Ω={x:x∈X,∥x∥≤l},l=sup{x:x∈X,x=λAx,0<λ<1}.Lemma 9.
Supposex(t)∈C1[0,T] and x(0)=x(T)=0. Then ∫0T‍|x(t)|2dt≤(T2/π2)∫0T‍|x′(t)|2dt.Lemma 10 (see [13]).
Let0≤α≤T be constant, s∈C(R,R) be periodic with period T, and maxt∈[0,T]|s(t)|≤α. Then for any u∈C1(R,R) which is periodic with period T, we have
(11)∫0T‍|u(t)-u(t-s(t))|2dt≤2α2∫0T‍|u′(t)|2dt.


## 3.  Proof of Theorem1

In this section, we will use Lemma8 to prove Theorem 1.Proof.
Let(12)X={x:x∈C(R,R),x(t+T)=x(t)},Y={x:x∈C1(R,R),x(t+T)=x(t)}.
ThenX and Y are real Banach spaces endowed with the norm
(13)∥x∥∞=maxt∈[0,T]|x(t)|,∥x∥=∥x∥∞+∥x′∥∞,
respectively.
Choosingm>0 with m≠(2kπ/T)2(k=1,2,…), then
(14)x′′(t)+mx(t)=0
has only trivial solution in Y.
In fact, it is easy to see that the general solution ofx′′(t)+mx(t)=0 is
(15)x(t)=c1sinmt+c2cosmt.
By the periodic boundary conditions we obtain thatx=0 is its unique solution in Y. Then for h∈X,
(16)-x′′(t)-mx(t)=h(t)
has unique solution x∈Y. Writing x=Kh, then K:X→Y is a completely continuous operator.
Define an operatorG:Y→X by
(17)(Gx)(t)=f(x′(t))+g1(t,x(t-τ1(t)))+g2(t,x(t-τ2(t)))-mx(t)-e(t),x∈Y.
ThenG:Y→X is continuous and bounded. Let A=KG:Y→Y. Then A is also a completely continuous operator. By Lemma 8, if
(18){x:x∈Y,x=λAx,0<λ<1}
is bounded in Y, then A has a fixed point in Y. Thus, (1) has periodic solution.
Now suppose thatx∈Y, 0<λ<1 satisfying x=λAx. Then x(t) is a solution of
(19)x′′(t)+λf(x′(t))+λg1(t,x(t-τ1(t)))+λg2(t,x(t-τ2(t)))+(1-λ)mx(t)=λe(t).
Lett_, t¯ be the maximum point and minimum point of x(t) on [0,T], respectively. Then
(20)x′(t_)=0,x′′(t_)≤0,x′(t¯)=0,x′′(t¯)≥0.
Noticingf(0)=0, we have
(21)λg1(t_,x(t_-τ1(t_)))+λg2(t_,x(t_-τ2(t_)))+(1-λ)mx(t_)-λe(t_)≥0,λg1(t¯,x(t¯-τ1(t¯)))+λg2(t¯,x(t¯-τ2(t¯)))+(1-λ)mx(t¯)-λe(t¯)≤0,
and hence, there exists ξ∈[t_,t¯] (or [t¯,t_]) such that
(22)λg1(ξ,x(ξ-τ1(ξ)))+λg2(ξ,x(ξ-τ2(ξ)))-λe(ξ)+(1-λ)mx(ξ)=0,
which implies that
(23)λ[g1(ξ,x(ξ-τ1(ξ)))+g2(ξ,x(ξ-τ2(ξ)))-e(ξ)]×signx(ξ)+(1-λ)m|x(ξ)|=0.
From(H5) we know that |x(ξ)|≤d (see [10]). So we have
(24)|x(t)|=|x(ξ)+∫ξt‍x′(s)ds|≤d+T∥x′∥L2,
and then,
(25)∥x∥∞≤d+T∥x′∥L2,
where ∥·∥L2 is the norm of L2[0,T].
Multiplying (19) with x′(t) and integrating from 0 to T, then we have
(26)λ∫0T‍f(x′(t))x′(t)dt=-λ∫0T‍g1(t,x(t-τ1(t)))x′(t)dt+λ∫0T‍g2(t,x(t-τ2(t)))x′(t)dt+λ∫0T‍e(t)x′(t)dt.
By(H2) we know that
(27)∫0T‍f(x′(t))x′(t)dt≥γ∫0T‍|x′(t)|2dt.
By Hölder's inequality, from (26) and (27) we have
(28)γ∫0T‍x′2(t)dt≤|∫0T‍g1(t,x(t-τ1(t)))x′(t)dt|+|∫0T‍g2(t,x(t-τ2(t)))x′(t)dt|+∥e∥L2∥x′∥L2≤∫0T‍|g1(t,x(t-τ1(t)))-g1(t,x(t))||x′(t)|dt+|∫0T‍g1(t,x(t))x′(t)dt|+∫0T‍|g2(t,x(t-τ2(t)))-g2(t,x(t))||x′(t)|dt+|∫0T‍g2(t,x(t))x′(t)dt|+∥e∥L2∥x′∥L2.
Since∫0x(t)‍gi(t,v)dv, i=1,2 are T-periodic, differentiable, and
(29)ddt∫0x(t)‍gi(t,v)dv=gi(t,x(t))x′(t)+∫0x(t)‍git′(t,v)dv,i=1,2,
we have
(30)∫0T‍gi(t,x(t))x′(t)dt=-∫0T‍dt∫0x(t)‍git′(t,v)dv,i=1,2.
Combining (28) and (30) with (H3) and (H4) we obtain
(31)γ∫0T‍x′2(t)dt≤l1∫0T‍|x(t)-x(t-τ1(t))||x′(t)|dt+l2∫0T‍|x(t)-x(t-τ2(t))||x′(t)|dt+∫0T‍dt∫0|x(t)|‍(a1+b1|v|)dv+∫0T‍dt∫0|x(t)|‍(a2+b2|v|)dv+∥e∥L2∥x′∥L2≤l1∥x′∥L2(∫0T‍|x(t)-x(t-τ1(t)-n1T)|2dt)1/2+l2∥x′∥L2(∫0T‍|x(t)-x(t-τ2(t)-n2T)|2dt)1/2+(a1+a2)∫0T‍|x(t)|dt+b1+b22∫0T‍|x(t)|2dt+∥e∥L2∥x′∥L2.
By Lemma9 we have
(32)∫0T‍|x(t)|2dt=∫ξξ+T‍|x(t)|2dt=∫ξξ+T‍|x(t)-x(ξ)|2dt+2x(ξ)∫ξξ+T‍x(t)dt-|x(ξ)|2T≤T2π2∫ξξ+T‍|x′(t)|2dt+2d∫0T‍|x(t)|dt-|x(ξ)|2T≤T2π2∫0T‍|x′(t)|2dt+2d∫0T‍|x(t)|dt.
By (32) and Hölder's inequality we have
(33)∫0T‍|x(t)|dt≤T(∫0T‍|x(t)|2dt)1/2≤T(T2π2∫0T‍|x′(t)|2dt+2d∫0T‍|x(t)|dt)1/2≤T3/2π(∫0T‍|x′(t)|2dt)1/2+2dT(∫0T‍|x(t)|dt)1/2≤T3/2π(∫0T‍|x′(t)|2dt)1/2+14∫0T‍|x(t)|dt+2dT.
From (33) we have
(34)∫0T‍|x(t)|dt≤4T3/23π(∫0T‍|x′(t)|2dt)1/2+8dT3.
By(H6) and Lemma 10 we have
(35)(∫0T‍|x(t)-x(t-τi(t)-niT)|2dt)1/2≤2δi(∫0T‍|x′(t)|2dt)1/2,i=1,2.
Thus, it follows from (31), (32), (34), and (35) that
(36)γ∥x′∥L22≤(2l1δ1+2l2δ2+(b1+b2)T22π2)∥x′∥L22+[4T3/23π[(a1+a2)+(b1+b2)d]+∥e∥L2]∥x′∥L2+8dT3[(a1+a2)+(b1+b2)d].
Combining this with(b1+b2)γ-1T2+22(l1δ1+l2δ2)π2γ-1<2π2, we know that there exists c1 such that ∥x′∥L2≤c1. Then
(37)∥x∥∞≤d+Tc1≜M1.
Multiplying (19) with x′′(t) and integrating from 0 to T, then we have
(38)∥x′′∥L22≤|-λ∫0T‍g1(t,x(t-τ1(t)))x′′(t)dt-λ∫0T‍g2(t,x(t-τ2(t)))x′′(t)dt-(1-λ)m∫0T‍x(t)x′′(t)dt+λ∫0T‍e(t)x′′(t)dt|≤(g1M1+g2M1)T∥x′′∥L2+mM1T∥x′′∥L2+∥e∥L2∥x′′∥L2,
where
(39)g1M1=maxt∈[0,T],∥x∥∞≤M1|g1(t,x(t))|,g2M1=maxt∈[0,T],∥x∥∞≤M1|g2(t,x(t))|.
Thus,(40)∥x′′∥L2≤(g1M1+g2M1)T+mM1T+∥e∥L2≜M2.
Selectingη∈[0,T] such that x′(η)=0, then we have
(41)|x′(t)|≤∫0T‍|x′′(t)|dt≤TM2.
Thus, from (35) and (41) we know that ∥x∥≤M1+TM2≜M. It follows that
(42){x:x∈Y,x=λAx,0<λ<1}
is bounded. Therefore, by Lemma 8, we obtain that A has a fixed point x*∈Ω, where Ω={x:x∈Y,∥x∥≤M}, and hence, it follows that (1) has a periodic solution.


## 4. Proof of Theorem3

In this section, we will use Lemma7 to prove Theorem 3.Proof.
LetX,Y,A be defined and same as in Section 3. We prove that A is Fréchet differentiable at 0 and(43)(A′(0)x)(t)=K[rx′(t)+(g1x′(t,0)+g2x′(t,0))x(t)-mx(t)].
In fact, byg1(t,0)+g2(t,0)≡e(t), we have A0=0. Let
(44)(Bx)(t)=K[rx′(t)+(g1x′(t,0)+g2x′(t,0))x(t)-mx(t)].
We have(45)∥Ax-Bx∥=∥[rx′(t)+(g′1x(t,0)+g′2x(t,0))x(t)-mx(t)]K[f(x′(t))+g1(t,x(t-τ1(t)))+g2(t,x(t-τ2(t)))-mx(t)-e(t)x′(t)]-K[rx′(t)+(g1x′(t,0)+g2x′(t,0))x(t)-mx(t)]x′(t)∥≤∥K∥∥[f(x′(t))x′(t)-r]x′(t)+g1(t,x(t-τ1(t)))+g2(t,x(t-τ2(t)))-e(t)-(g1x′(t,0)+g2x′(t,0))x(t)[f(x′(t))x′(t)-r]∥≤∥K∥∥[f(x′(t))x′(t)-r]+g1x′(t,θ1x(t-τ1(t)))-g1x′(t,0)+g2x′(t,θ2x(t-τ2(t)))-g2x′(t,0)[f(x′(t))x′(t)-r]∥∥x∥,
where θi=θi(t)∈(0,1), i=1,2. Thus,
(46)lim∥x∥→0∥Ax-A0-Bx∥∥x∥=0,
which implies that A′(0) exists and A′(0)=B.
Now, we prove that1 is not the eigenvalue of A′(0).
In fact, if there existsx∈Y satisfying A′(0)x=x, then x is a solution of
(47)-x′′(t)-mx(t)=rx′(t)+(g1x′(t,0)+g2x′(t,0))x(t)-mx(t).
Integrating (47) from 0 to T, we obtain
(48)∫0T‍(g1x′(t,0)+g2x′(t,0))x(t)dt=0.
By(H7), there exists t0∈[0,T] such that x(t0)=0.
Multiplying (47) with x′(t) and integrating it from 0 to T, we have
(49)∫0T‍r|x′|2dt=-∫0T‍(g1x′(t,0)+g2x′(t,0))x(t)x′(t)dt.
By(H7) we have
(50)∫0T‍r|x′|2dt=0,
which implies that x′(t)≡0, and hence x(t)≡0. It follows that 1 is not the eigenvalue of A′(0). By Lemma 7 we have
(51)i(A,0)=i(A′(0),0)=(-1)β.
Next we prove thatβ=1. We can prove that(52)-x′′(t)-mx(t)=λ(rx′(t)+(g1x′(t,0)+g2x′(t,0))x(t)-mx(t))
has only one eigenvalue in (0, 1) and the algebraic multiplicity of it is 1.
In fact, it is easy to see thatλ∈(0,1) is the eigenvalue of (52) if and only if m+λ(g1x′(t,0)+g2x′(t,0)-m)=0, that is,
(53)λ=λ0=mm-g1x′(t,0)-g2x′(t,0)∈(0,1).
In this case, (52) degenerates to
(54)x′′(t)+mm-g1x′(t,0)-g2x′(t,0)rx′(t)=0.
It is not difficult to see that all the solutions of (54) are x(t)≡const. Therefore the geometric multiplicity of λ0=m/(m-g1x′(t,0)-g2x′(t,0)) is 1.
If(I-λ0A′(0))2x(t)=0, then (I-λ0A′(0))x(t)=c=const, and hence, x(t) satisfies
(55)-x′′(t)-mx(t)=λ0(rx′(t)+(g1x′(t,0)+g2x′(t,0))x(t)-mx(t))+mc,
which implies that
(56)-x′′(t)=λ0rx′(t)+mc.
Integrating (56) from 0 to T, we obtain c=0, which implies that x(t)≡const. It follows that the algebraic multiplicity of λ0=m/(m-g1x′(t,0)-g2x′(t,0)) is also 1, and therefore, β=1. So i(A,0)=-1. Letting δ>0 small enough, then we have
(57)deg(I-A,Bδ,0)=-1.
On the other hand, by the proof of Theorem1, we know that there exists M>0 such that if x∈Y, 0<λ<1 satisfying x=λAx, then ∥x∥≤M. Suppose Ax≠x for x∈Y,∥x∥=M. By the homotopy invariance of topological degree, we have
(58)deg(I-A,BM,0)=1,
where BM={x∈Y:∥x∥<M}. It follows from (57) and (58) that
(59)deg(I-A,BM∖B-δ,0)=2,
which implies that A has a fixed point x in BM∖B-δ and x is a nontrivial periodic solution of (1).


## 5. Two Examples

In this section, we give two examples to demonstrate the results obtained in previous sections.Example 1.
Consider the forced Rayleigh-type equation with period2π:
(60)x′′(t)+f(x′(t))+g1(t,x(t-τ1(t)))+g2(t,x(t-τ2(t)))=e(t),
where
(61)f(x)={ex-1,x≥0,1-e-x,x≤0,(62)g1(t,x)=19sin2tx(t-θcost)+cost,g2(t,x)=19cos2tx(t-θsint)+sint,e(t)=sint,τ1(t)=θcost,τ2(t)=θsint,θ∈(0,1).Conclusion. Equation (60) has at least one periodic solution with period 2π.Proof.
By (61) and (62), it is not difficult to see that the condition (H1) holds, T=2π, and
(63)|f(0)|=0,|g1t′(t,x)|=|19xsin2t-sint|≤19|x|+1,∀(t,x)∈R2,|g2t′(t,x)|=|-19xsin2t-cost|≤19|x|+1,∀(t,x)∈R2.
On the other hand, letd=9, γ=1, δi=θ, li=1/9, bi=1/9, and i=1,2. If θ∈(0,2/2), then
(64)xf(x)≥|x|2,∀x∈R,(b1+b2)γ-1T2+22(l1δ1+l2δ2)π2γ-1<2π2.
Hence, (H1)–(H6) are satisfied. Thus, by Theorem 1, (60) has at least one periodic solution with period 2π.Example 2.
If we replaceg1(t,x), g2(t,x), and e(t) in Example 1 by(65)g1(t,x(t-τ1(t)))=19sin2tcos(x(t-τ1(t))+12)+110x(t-τ1(t))sin12,∀(t,x)∈R2,g2(t,x(t-τ2(t)))=19cos2tcos(x(t-τ2(t))+12)+sint,∀(t,x)∈R2,e(t)=sint+(1/9)cos(1/2), then we can obtain that g1(t,0)+g2(t,0)≡e(t), for all t∈R, and
(66)g1t′(t,x)=19sin2tcos(x+12),g2t′(t,x)=-19sin2tcos(x+12)+cost,|g1t′(t,x)|≤19|x|+118,|g2t′(t,x)|≤19|x|+3718,∀(t,x)∈R2,g1x′(t,x)=-19sin2tsin(x+12)+110sin12,g2x′(t,x)=-19cos2tsin(x+12),∀t∈R,g1x′(t,0)+g2x′(t,0)=-19sin12+110sin12<0,∀t∈R.
Letd=(10/(9sin(1/2)))+((10sin(1/2))/(9cos(1/2))), γ=1, δi=θ, i=1,2, l1=19/90, l2=1/9, b1=1/9, and b2=1/9. If θ∈(0,1); then
(67)xf(x)≥|x|2,∀x∈R,(b1+b2)γ-1T2+22(l1δ1+l2δ2)π2γ-1<2π2.
Hence, (H1)–(H8) are satisfied. Thus, by Theorem 3, (60) has at least one nontrivial periodic solution with period 2π.



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