Abstract

This paper is concerned with the existence of multiple periodic solutions for a suspension bridge wave equation with damping. By using Leray-Schauder degree theory, the authors prove that the damped wave equation has multiple periodic solutions.

1. Introduction

In [1], also see [26], the author considered a horizontal cross-section of the center span of a suspension bridge and proposed a partial differential equation model for the torsional motion of the cross-section and treated the center span of the bridge as a beam of length 𝐿 and width 2𝑙 suspended by cables. Consider the horizontal cross-section of mass 𝑚 located at position 𝑥 along the length of the span. She treated this cross-section as a rod of length 2𝑙 and mass 𝑚 suspended by cables. Let 𝑦(𝑥,𝑡) denote the downward distance of the center of gravity of the rod from the unloaded state and let 𝜃(𝑥,𝑡) denote the angle of the rod from horizontal at time 𝑡. Assume that the cables do not resist compression, but resist elongation according to Hooke’s law with spring constant 𝐾. Then the torsional and vertical motion of the span satisfy 𝜃𝑡𝑡𝜀1𝜃𝑥𝑥=3𝐾𝑚𝑙cos(𝑦𝑙sin𝜃)+𝛿𝜃𝑡+1𝑦(𝑥,𝑡),𝑡𝑡+𝜀2𝑦𝑥𝑥𝑥𝑥𝐾=𝑚(𝑦𝑙sin𝜃)++(𝑦+𝑙sin𝜃)+𝛿𝑦𝑡+𝑔+2(𝑥𝑡),𝜃(0,𝑡)=𝜃(𝐿,𝑡)=𝑦(0,𝑡)=𝑦(𝐿,𝑡)=𝑦𝑥𝑥(0,𝑡)=𝑦𝑥𝑥(𝐿,𝑡)=0,(1.1) where 𝑢+=max{𝑢,0}, 𝜀1, 𝜀2 are physical constants related to the flexibility of the beam, 𝛿 is the damping constant, 1 and 2 are external forcing terms, and 𝑔 is the acceleration due to gravity. The spatial derivatives describe the restoring force that the beam exerts, and the time derivatives 𝜃𝑡 and 𝑦𝑡 represent the force due to friction. The boundary conditions reflect the fact that the ends of the span are hinged.

Throughout the paper [1] the author assumes that the cables never lose tension; that is, it is assumed that (𝑦±sin𝜃)0. In this case, we see that (1.1) becomes uncoupled, and the torsional and vertical motions satisfy, respectively, 𝜃𝑡𝑡𝜀1𝜃𝑥𝑥=6𝐾𝑚cos𝜃sin𝜃𝛿𝜃𝑡+1𝑦(𝑥,𝑡),𝜃(0,𝑡)=𝜃(𝐿,𝑡)=0,(1.2)𝑡𝑡+𝜀2𝑦𝑥𝑥𝑥𝑥=2𝐾𝑚𝑦𝛿𝑦𝑡+𝑔+2(𝑥,𝑡),𝑦(0,𝑡)=𝑦(𝐿,𝑡)=𝑦𝑥𝑥(0,𝑡)=𝑦𝑥𝑥(𝐿,𝑡)=0.(1.3)

In paper [1], removing the damping term; that is, let 𝛿=0, changing variables, and imposing boundary and periodicity conditions, the author rewrites (1.2) as 𝑢𝑡𝑡𝑢𝑥𝑥𝑢+𝑏sin𝑢=𝜀(𝑥,𝑡),(0,𝑡)=𝑢(𝜋,𝑡)=0,𝑢(𝑥,0)=𝑢(𝑥,𝜋),𝑢𝑡(𝑥,0)=𝑢𝑡(𝑥,𝜋),𝑢(𝑥,𝑡)=𝑢(𝜋𝑥,𝑡),𝑢(𝑥,𝑡)=𝑢(𝑥,𝜋𝑡).(1.4) And it proves that (1.4) has at least two solutions in the subspace 𝐻 of 𝐿2. Where 𝐻 is defined as 𝐻=𝑢𝐿2(Ω)𝑢(𝑥,𝑡)=𝑢(𝜋𝑥,𝑡),𝑢(𝑥,𝑡)=𝑢(𝑥,𝜋𝑡),𝑢𝑖𝑠𝜋periodicin𝑡.(1.5)

Notice that (1.4) is particular in no damping and the selection of 𝐻. Hence, in [1] the author left a problem which is relevant to this case.

Problem 1. “Under appropriate hypotheses on the forcing term, does a similar result hold for the damped equation?”
Motivated by this problem, in this paper, we suppose that the damping is present, that is, 𝛿0, and study the following problem: 𝑢𝑡𝑡𝑢𝑥𝑥+𝛿𝑢𝑡𝑢+𝑏sin𝑢=𝜀(𝑥,𝑡),(0,𝑡)=𝑢(𝜋,𝑡)=0,𝑢(𝑥,0)=𝑢(𝑥,𝜋),𝑢𝑡(𝑥,𝑡)=𝑢𝑡(𝑥,𝜋),𝑢(𝑥,𝑡)=𝑢(𝜋𝑥,𝑡).(1.6)

2. Preliminaries

Let 𝑁={0,1,} and 𝑍 be the set of integers, Λ=𝑁×𝑁. Let Ω=(0,𝜋)×(0,𝜋) and 𝐿2(Ω) be usual space of square integrable functions with usual inner product (,) and corresponding norm . For the Sobolev space 𝐻1(Ω), we denote the standard inner product by (𝑢,𝑣)1=(𝑢,𝑣)+(𝑢𝑥,𝑣𝑥)+(𝑢𝑡,𝑣𝑡) and norm by 𝑢1.

Define the operator 𝐿𝛿𝑢=𝑢𝑡𝑡𝑢𝑥𝑥+𝛿𝑢𝑡𝐻𝐻 by 𝐷𝐿𝛿=𝑢𝐻𝑢(𝑥,𝑡)=Λ𝑢𝑚𝑛Φ𝑚𝑛,Λ(2𝑛+1)24𝑚22+4𝑚2𝛿2||𝑢𝑚𝑛||2,𝐿<𝛿𝑢=Λ(2𝑛+1)24𝑚2𝑢+2𝑚𝛿𝑚𝑛Φ𝑚𝑛,forall𝑢𝐻.(2.1)

We know that the eigenvalues and corresponding eigenfunctions of 𝐿𝛿 are 𝜆𝑚𝑛=(2𝑛+1)24𝑚2Φ+2𝑚𝛿𝑖,(𝑚,𝑛)Λ,𝑚𝑛(𝑥,𝑡)=𝑒2𝑚𝑡𝑖sin(2𝑛+1)𝑥,(𝑚,𝑛)Λ.(2.2)

In order to seek the solutions of (1.6), we first investigate the properties of operator 𝐿𝛿. We have the following Lemma.

Lemma 2.1. 𝐿𝛿1 exists, 𝐿𝛿1𝐻𝐻 is compact, and 𝐿𝛿1=1.

Proof. Because we are restricted to the subspace 𝐻 of 𝐿2, and 𝜆𝑚𝑛 0, we easily know 𝐿𝛿1 exists.
We prove 𝐿𝛿1𝐻𝐻 is compact below. We find that 𝐿𝛿1𝑢=Λ1(2𝑛+1)24𝑚2𝑢+2𝑚𝛿𝑖𝑚𝑛Φ𝑚𝑛,(2.3) for all 𝑢=Λ𝑢𝑚𝑛Φ𝑚𝑛𝐻. For any (𝑚,𝑛)Λ, we have ||(2𝑛+1)24𝑚2||+2𝑚𝛿𝑖21,(2.4) then 𝐿𝛿1𝑢2=Λ||||1(2𝑛+1)24𝑚2𝑢+2𝑚𝛿𝑖𝑚𝑛||||2Λ||𝑢𝑚𝑛||2=𝑢2.(2.5) Hence, 𝐿𝛿1𝑢𝑢.(2.6) On the other hand, 𝐿𝛿1𝑢21=Λ|||||1+(2𝑛+1)2+4𝑚2(2𝑛+1)24𝑚22+4𝑚2𝛿2𝑢𝑚𝑛|||||2,(2.7) while 1+(2𝑛+1)2+4𝑚2(2𝑛+1)24𝑚22+4𝑚2𝛿2=1+(2𝑛+1)2+4𝑚2(2𝑛+2𝑚+1)2(2𝑛2𝑚+1)2+4𝑚2𝛿21+(2𝑛+1)2+4|𝑚|2(2𝑛+2|𝑚|+1)2+4|𝑚|2𝛿2(2𝑛+2|𝑚|+1)2+1(2𝑛+2|𝑚|+1)22.(2.8) Hence, 𝐿𝛿1𝑢212Λ|𝑢|2=2𝑢.(2.9) By (2.6) and (2.9), we can find that the operator 𝐿𝛿1𝐻𝐻 is compact since the embedding 𝐻1𝐿2 is compact.
Finally, we prove 𝐿𝛿1= 1. By (2.2) and ||𝜆𝑚𝑛||2=||(2𝑛+1)24𝑚2||+2𝑚𝛿𝑖21.(2.10) Set 𝑢=Φ00, such that, 𝐿𝛿1𝑢/𝑢=1. Therefore, 𝐿𝛿1=1.(2.11) Hence, we complete the proof of this lemma.

Definition 2.2. One says that 𝑢𝐻 is a solution to (1.6) if 𝑢=𝐿𝛿1(𝜀𝑏sin𝑢).(2.12)
To establish the existence of multiple periodic solutions to (1.6), we use Leray-Schauder degree theory to prove the existence of multiple zeros of a related operator 𝑇1. To compute the degree of 𝑇1, we continuously deform it to a linear operator 𝑇0, the Gâteaux derivative of 𝑇1, and compute its degree via a direct calculation.
It is not difficult to show that the homotopy property of Leray-Schauder degree ensures that the degree of an operator 𝑇1 is preserved as 𝑇1 is continuously deformed to its Fréchet derivative under appropriate hypotheses. However, the nonlinear term in (1.6), 𝑓(𝑢)=sin𝑢, is not Fréchet differentiable in 𝐿2 at 𝑢=0.
There is a theorem in paper [1], in which, the author shows that, under certain conditions on the nonlinear term 𝑓 and the differential operator 𝐿, Leray-Schauder degree is indeed preserved under homotopy from the operator 𝑇1 to its Gâteaux derivative 𝑇0. This result can be used to establish multiplicity of solutions to equations of the form (1.6). The result follows.

Lemma 2.3. Let 𝐼1, 𝐼2 be open, bounded intervals in 𝑅, and define 𝑄=𝐼1×𝐼2. Let 𝐵 be a subspace of 𝐿𝑝(𝑄), 𝑝1, and define u=u𝐿𝑝. Consider the problem 𝐿𝑢+𝑓(𝑢)=𝜀(𝑥,𝑡),(2.13) where 𝐿, 𝑓, and satisfy the following:(H1)𝐿1 is compact;(H2)𝐿1≤ 1;(H3)𝑓(0)=0;(H4)𝑓 is Lipschitz with Lipschitz constant 𝑀;(H5)𝐵 and 1;(H6) the Gâteaux derivative 𝑑𝑓(0,𝑢) exists and satisfies 𝑑𝑓(0,𝑢)=𝜌𝑢, where 𝜌>0 and 𝜌 is not an eigenvalue of 𝐿.Define 𝑇0: 𝐵𝐵 by 𝑇0(𝑢)=𝑢+𝜌𝐿1(𝑢),(2.14) and 𝑇1𝐵𝐵 by 𝑇1(𝑢)=𝑢𝐿1(𝜀𝑓(𝑢)).(2.15) Then for 𝜀 sufficiently small, there exists 𝛾>0 such that 𝑇deg1,𝐵𝛾(0),0=deg𝑇,𝐵𝛾(0),0.(2.16)

3. Result and Proof

The main result of this paper is as follows.

Theorem 3.1. Let 𝐻 with 1, and let 𝑏(25+4𝛿2,9+4𝛿2), 0<𝛿2<14. Then there exists 𝜀0>0 such that if |𝜀|<𝜀0, (1.6) has at least two solutions in 𝐻.

Proof. Let 𝐿=𝐿𝛿 and 𝑓(𝑢)=𝑏sin𝑢, it is easy to know that 𝐿 and 𝑓 satisfy the conditions (H1–H5) in Lemma 2.3.
Reply Lemma 2.3, we define 𝑇0: 𝐻𝐻 by 𝑇0(𝑢)=𝑢+𝑏𝐿𝛿1(𝑢),(3.1) and 𝑇1: 𝐻𝐻 by 𝑇1(𝑢)=𝑢𝐿𝛿1(𝜀𝑏sin(𝑢)).(3.2) And note that zeros of 𝑇1 correspond to solutions of (1.6). To prove the theorem, we will show the following:(C1) there exists 𝑅0>0 such that for 𝑅>𝑅0, deg(𝑇1,𝐵𝑅(0),0)=1;(C2) there exists 𝛾(0,𝑅0) such that deg(𝑇1,𝐵𝛾(0),0)=1.Then, since deg(𝑇1,𝐵𝛾(0),0)0, there exists a zero of 𝑇1 (i.e., a solution of (1.6)) in 𝐵𝛾(0). Moreover, by the additivity property of degree, deg(𝑇1,𝐵𝑅(0)𝐵𝛾(0),0)0 and hence (1.6) has a second solution in the annulus 𝐵𝑅(0)𝐵𝛾(0).
To establish (C1), define 𝑇𝛽𝑢=𝑢𝛽𝐿𝛿1(𝜀𝑏sin(𝑢)),(3.3) or 𝛽[0,1], and note that this definition of 𝑇1 is consistent with our previous definition. Note also that 𝑇0 is simply the identity map; hence, for any 𝑅>0 we have deg(𝑇0,𝐵𝑅(0),0)=1. The homotopy property of degree ensures that deg(𝑇𝛽,𝐵𝑅(0),0) is constant provided that 0𝑇𝛽(𝜕𝐵𝑅(0)) for all 𝛽[0,1].
Fix 𝛽[0,1] and suppose 𝑢𝐻 solves 𝑇𝛽𝑢=0. We will show that u is bounded above by some 𝑅0>0 and that this bound is independent of 𝛽.
Since 𝑇𝛽𝑢=0, we have 𝐿𝑢=𝛽𝛿1𝜀(𝜀𝑏sin𝑢)𝛽0𝜀+𝑏sin𝑢0+𝑏𝑚(Ω)1/2<𝜀0+𝑏2𝜋<𝑅0,(3.4) if we choose 𝑅0>𝜀0+𝑏2𝜋.
Thus, for 𝑅>𝑅0, we have 𝑇deg1,𝐵𝑅𝑇(0),0=deg0,𝐵𝑅(0),0=1,(3.5) and (C1) above holds.
To establish (C2), let 𝜀<𝜀0; we will determine the value of 𝜀0 later. For 𝜇[0,1] define 𝑇𝜇𝑢=𝑢+(1𝜇)𝐿𝛿1(𝑏𝑢)𝜇𝐿𝛿1(𝜀𝑏sin𝑢),(3.6) and note again that this definition of 𝑇1 is consistent with our previous definitions. We will again apply the homotopy property of degree (via Lemma 2.3) and a standard degree calculation to show that for some 𝛾>0𝑇deg1,𝐵𝛾𝑇(0),0=deg0,𝐵𝛾(0),0=1.(3.7)
Observe that for 𝐿=𝐿𝛿 and 𝑓(𝑢)=𝑏sin𝑢, hypotheses (H1)–(H5) of Lemma 2.3 are satisfied. To verify hypothesis (H6), we need to show that 𝑑𝑓(0,𝑢)=𝑏𝑢.(3.8) By definition of the Gâteaux derivative, 𝑑𝑑𝑓(0,𝑢)=𝑑𝑡𝑓(0+𝑡𝑢)|𝑡=0=lim0𝑓((𝑡+)𝑢)𝑓(𝑡𝑢)|𝑡=0=lim0𝑏sin(𝑢).(3.9) We will show that the limit above (in 𝐻) is 𝑏𝑢.
Note first that in 𝑅 we have lim0=sin(𝑢)=lim0sin(𝑢)𝑢𝑢=𝑢,(3.10) and hence |||sin(𝑢)|||𝑢20,(3.11) as 0. Invoking the convexity of 𝜔2, we have |||sin(𝑢)|||𝑢2142|||sin(𝑢)|||2+12|𝑢|24𝑢2.(3.12)
Since 𝑢𝐿2, |(sin(𝑢)/)𝑢|2 is dominated in 𝐿1; thus by the dominated convergence theorem, 𝑏sin(𝑢)𝑏𝑢0,(3.13) as 0; therefore (3.8) holds. Moreover, by the form of eigenvalue of 𝐿𝛿 and our choice of 𝑏, 𝑏 is not an eigenvalue of 𝐿𝛿; therefore hypothesis (H6) of Lemma 2.3 holds. Thus, by Lemma 2.3, for sufficiently small 𝛾,𝜀>0, we have 𝑇deg0,𝐵𝛾𝑇(0),0=deg1,𝐵𝛾(0),0.(3.14)
Finally, we will show that 𝑇deg0,𝐵𝛾(0),0=deg𝐼+𝑏𝐿𝛿1,𝐵𝛾(0),0=1.(3.15)
Consider the finite dimensional subspace 𝑀𝑁=span{Φ𝑚𝑛}𝑁0 of 𝐻 and recall that, by compactness, 𝑏𝐿𝛿1 can be approximated in operator norm by the operators 𝐵𝑁𝑀𝑁𝑀𝑁 given by 𝐵𝑁(𝑢)=𝑏𝑁𝑁𝑁0𝑐𝑚𝑛𝜆𝑚𝑛Φ𝑚𝑛.(3.16) By definition of Leray-Schauder degree, for 𝑁 sufficientlylarge, 𝑇deg0,𝐵𝛾(0),0=deg𝐼+𝐵𝑁,𝐵𝛾𝑀(0)𝑁=,0𝑢𝐼+𝐵𝑁1(0)sgn𝐽𝐼+𝐵𝑁(𝑢),(3.17) where 𝐽Φ(𝑢) is the Jacobian determinant of Φ at 𝑢.
Since 𝐼+𝐵𝑁 can be identified with an (2𝑁+1)2×(2𝑁+1)2 diagonal matrix whose entries are 1+𝑏/𝜆𝑚𝑛, we have deg𝐼+𝐵𝑁,𝐵𝛾𝑀(0)𝑁,0=sgn𝑁𝑁𝑚=𝑁𝑛=0𝑏1+𝜆𝑚𝑛.(3.18) Now we consider the following two cases.(D1)If 𝜆𝑚𝑛 contains imaginary part, suppose a pair of conjugate complex numbers are 𝑎±𝑐𝑖(𝑐0), then, 𝑏1+𝑏𝑎+𝑐𝑖1+=𝑎𝑐𝑖(𝑎+𝑏)2+𝑐2𝑎2+𝑐2>0.(3.19)(D2)If 𝜆𝑚𝑛 is real, then 𝑚=0, here 𝜆𝑚𝑛=(2𝑛+1)2.
Since 𝑏(25+4𝛿2,9+4𝛿2), and 0<𝛿2<14, the only negative value of 1+𝑏/(2𝑛+1)2 occurs at 𝜆00. Therefore, deg𝐼+𝐵𝑁,𝐵𝛾𝑀(0)𝑁,0=1.(3.20) The proof of the theorem is complete.