# Multiple Periodic Solutions to a Suspension Bridge Wave Equation with Damping

Authors: Shanshan Wang; Yukun An
Journal: ISRN Mathematical Analysis
 (2011)

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## 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.


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## Body

## 1. Introduction

In [1], also see [2–6], 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 (1.1)θtt-ε1θxx=3Kmlcos⁡⁡[(y-lsin⁡⁡θ)+]-δθt+h1(x,t),ytt+ε2yxxxx=-Km[(y-lsin⁡⁡θ)++(y+lsin⁡⁡θ)+]-δyt+g+h2(xt),θ(0,t)=θ(L,t)=y(0,t)=y(L,t)=yxx(0,t)=yxx(L,t)=0,
where u+=max{u,0}, ε1, ε2 are physical constants related to the flexibility of the beam, δ is the damping constant, h1 and h2 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, (1.2)θtt-ε1θxx=-6Kmcos⁡⁡θsin⁡⁡θ-δθt+h1(x,t),θ(0,t)=θ(L,t)=0,(1.3)ytt+ε2yxxxx=-2Kmy-δyt+g+h2(x,t),y(0,t)=y(L,t)=yxx(0,t)=yxx(L,t)=0.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
(1.4)utt-uxx+bsin⁡⁡u=εh(x,t),u(0,t)=u(π,t)=0,u(x,0)=u(x,π),ut(x,0)=ut(x,π),u(x,t)=u(π-x,t),u(x,t)=u(x,π-t).
And it proves that (1.4) has at least two solutions in the subspace H of L2. Where H is defined as
(1.5)H={u∈L2(Ω)u(x,t)=u(π-x,t),u(x,t)=u(x,π-t),uisπ-periodicint}.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,δ≠0, and study the following problem:
(1.6)utt-uxx+δut+bsin⁡⁡u=εh(x,t),u(0,t)=u(π,t)=0,u(x,0)=u(x,π),ut(x,t)=ut(x,π),u(x,t)=u(π-x,t).


## 2. Preliminaries

LetN={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 ∥u∥1.Define the operatorLδu=utt-uxx+δut:H→H by (2.1)D(Lδ)={u∈H∣u(x,t)=∑ΛumnΦmn,∑Λ(((2n+1)2-4m2)2+4m2δ2)|umn|2<∞},Lδu=∑Λ((2n+1)2-4m2+2mδ)umnΦmn,forallu∈H.We know that the eigenvalues and corresponding eigenfunctions ofLδ are (2.2)λmn=(2n+1)2-4m2+2mδi,(m,n)∈Λ,Φmn(x,t)=e2mtisin⁡⁡(2n+1)x,(m,n)∈Λ.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.
Lδ-1 exists, Lδ-1:H→H is compact, and ∥Lδ-1∥=1.Proof.
Because we are restricted to the subspaceH of L2, and λmn≠ 0, we easily know Lδ-1 exists.
We proveLδ-1:H→H is compact below. We find that
(2.3)Lδ-1u=∑Λ1(2n+1)2-4m2+2mδiumnΦmn,
for all u=∑ΛumnΦmn∈H. For any (m,n)∈Λ, we have
(2.4)|(2n+1)2-4m2+2mδi|2≥1,
then
(2.5)‖Lδ-1u‖2=∑Λ|1(2n+1)2-4m2+2mδiumn|2≤∑Λ|umn|2=‖u‖2.
Hence,
(2.6)‖Lδ-1u‖≤‖u‖.
On the other hand,
(2.7)‖Lδ-1u‖12=∑Λ|1+(2n+1)2+4m2((2n+1)2-4m2)2+4m2δ2umn|2,
while
(2.8)1+(2n+1)2+4m2((2n+1)2-4m2)2+4m2δ2=1+(2n+1)2+4m2(2n+2m+1)2(2n-2m+1)2+4m2δ2≤1+(2n+1)2+4|m|2(2n+2|m|+1)2+4|m|2δ2≤(2n+2|m|+1)2+1(2n+2|m|+1)2≤2.
Hence,
(2.9)‖Lδ-1u‖12≤2∑Λ|u|2=2‖u‖.
By (2.6) and (2.9), we can find that the operator Lδ-1:H→H is compact since the embedding H1→L2 is compact.
Finally, we prove∥Lδ-1∥= 1. By (2.2) and
(2.10)|λmn|2=|(2n+1)2-4m2+2mδi|2≥1.
Set u=Φ00, such that, ∥Lδ-1u∥/∥u∥=1. Therefore,
(2.11)‖Lδ-1‖=1.
Hence, we complete the proof of this lemma.Definition 2.2.
One says thatu∈H is a solution to (1.6) if
(2.12)u=Lδ-1(εh-bsin⁡⁡u).
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 operatorT1 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.
LetI1, I2 be open, bounded intervals in R, and define Q:=I1×I2. Let B be a subspace of Lp(Q), p≥1, and define ∥u∥≔∥u∥Lp. Consider the problem
(2.13)Lu+f(u)=εh(x,t),
where L, f, and h satisfy the following:(H1)
L-1 is compact;(H2)
∥L-1∥≤ 1;(H3)
f(0)=0;(H4)
f is Lipschitz with Lipschitz constant M;(H5)
h∈B and h≤1;(H6)
the Gâteaux  derivativedf(0,u) exists and satisfies df(0,u)=ρu, where ρ>0 and -ρ is not an eigenvalue of L.Define T0: B→B by
(2.14)T0(u)=u+ρL-1(u),
and T1:B→B by
(2.15)T1(u)=u-L-1(εh-f(u)).
Then for ε sufficiently small, there exists γ>0 such that
(2.16)deg⁡(T1,Bγ(0),0)=deg⁡(T,Bγ(0),0).


## 3. Result and Proof

The main result of this paper is as follows.Theorem 3.1.
Leth∈H with ∥h∥≤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.
LetL=Lδ and f(u)=bsinu, it is easy to know that L and f satisfy the conditions (H1–H5) in Lemma 2.3.
Reply Lemma2.3, we define T0: H→H by
(3.1)T0(u)=u+bLδ-1(u),
and T1: H→H by
(3.2)T1(u)=u-Lδ-1(εh-bsin⁡⁡(u)).
And note that zeros of T1 correspond to solutions of (1.6). To prove the theorem, we will show the following:(C1)
there existsR0>0 such that for R>R0, deg(T1,BR(0),0)=1;(C2)
there existsγ∈(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(3.3)Tβu=u-βLδ-1(εh-bsin⁡⁡(u)),
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 0∈¯Tβ(∂BR(0)) for all β∈[0,1].
Fixβ∈[0,1] and suppose u∈H solves Tβu=0. We will show that u is bounded above by some R0>0 and that this bound is independent of β.
SinceTβu=0, we have
(3.4)‖u‖=β‖Lδ-1(εh-bsin⁡⁡u)‖≤β[ε0+b‖sin⁡⁡u‖]≤[ε0+bm(Ω)1/2]<[ε0+b2π]<R0,
if we choose R0>ε0+b2π.
Thus, forR>R0, we have
(3.5)deg⁡(T1,BR(0),0)=deg⁡(T0,BR(0),0)=1,
and (C1) above holds.
To establish (C2), letε<ε0; we will determine the value of ε0 later. For μ∈[0,1] define
(3.6)Tμu=u+(1-μ)Lδ-1(bu)-μLδ-1(εh-bsin⁡⁡u),
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>0(3.7)deg⁡(T1,Bγ(0),0)=deg⁡(T0,Bγ(0),0)=-1.
Observe that forL=Lδ and f(u):=bsinu, hypotheses (H1)–(H5) of Lemma 2.3 are satisfied. To verify hypothesis (H6), we need to show that
(3.8)df(0,u)=bu.
By definition of the Gâteaux derivative,
(3.9)df(0,u)=ddtf(0+tu)|t=0=lim⁡h⟶0⁡f((t+h)u)-f(tu)h|t=0=lim⁡h⟶0⁡bsin⁡⁡(hu)h.
We will show that the limit above (in H) is bu.
Note first that inR we have
(3.10)lim⁡h→0⁡=sin⁡⁡(hu)h=lim⁡h→0⁡sin⁡⁡(hu)huu=u,
and hence
(3.11)|sin⁡⁡(hu)h-u|2⟶0,
as h→0. Invoking the convexity of ω2, we have
(3.12)|sin⁡⁡(hu)h-u|2≤4[12|sin⁡⁡(hu)h|2+12|u|2]≤4u2.
Sinceu∈L2, |(sin(hu)/h)-u|2 is dominated in L1; thus by the dominated convergence theorem,
(3.13)‖bsin⁡⁡(hu)h-bu‖⟶0,
as h→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
(3.14)deg⁡(T0,Bγ(0),0)=deg⁡(T1,Bγ(0),0).
Finally, we will show that(3.15)deg⁡(T0,Bγ(0),0)=deg⁡(I+bLδ-1,Bγ(0),0)=-1.
Consider the finite dimensional subspaceMN=span{Φmn}0N of H and recall that, by compactness, bLδ-1 can be approximated in operator norm by the operators BN:MN→MN given by
(3.16)BN(u)=b∑-NN∑0NcmnλmnΦmn.
By definition of Leray-Schauder degree, for N sufficientlylarge,
(3.17)deg⁡(T0,Bγ(0),0)=deg⁡(I+BN,Bγ(0)⋂MN,0)=∑u∈(I+BN)-1(0)sgn⁡JI+BN(u),
where JΦ(u) is the Jacobian determinant of Φ at u.
SinceI+BN can be identified with an (2N+1)2×(2N+1)2 diagonal matrix whose entries are 1+b/λmn, we have
(3.18)deg⁡(I+BN,Bγ(0)⋂MN,0)=sgn⁡∏m=-NN∏n=0N(1+bλmn).
Now we consider the following two cases.(D1)
Ifλmn contains imaginary part, suppose a pair of conjugate complex numbers are a±ci(c≠0), then,(3.19)(1+ba+ci)(1+ba-ci)=(a+b)2+c2a2+c2>0.(D2)
Ifλmn is real, then m=0, here λmn=(2n+1)2.
Sinceb∈(-25+4δ2,-9+4δ2), and 0<δ2<14, the only negative value of 1+b/(2n+1)2 occurs at λ00. Therefore,
(3.20)deg⁡(I+BN,Bγ(0)⋂MN,0)=-1.
The proof of the theorem is complete.



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*Source: 154806-2011-12-27.xml*