# Spectral Analysis for a Wave/Plate Transmission System

Authors: Chengqiang Wang
Journal: Advances in Mathematical Physics
 (2019)

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## Abstract

We are concerned with the transmission system of a 1D damped wave equation and a 1D undamped plate equation. Our result reads as follows: the spectrum of the infinitesimal generator of the semigroup associated with the system in question consists merely of an infinite sequence of eigenvalues which are all located in the open left half of the complex plane; the sequence of eigenvalues has the imaginary axis and another vertical line to the left of the imaginary axis as its asymptote lines; the energy of the system under consideration decreases to zero as time goes to infinity.


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## Body

## 1. Introduction

Due to their wide applicability, spectral properties of coupled systems of partial differential equations have been extensively investigated in the literature. Guo and Yung [1] conducted spectral analysis for a 1D thermoelastic system and obtained an energy decay estimate from the spectral property; Lu and Wang [2] studied the spectral property of a wave/Schröodinger transmission system and proved that the associated energy decreases to zero as time approaches infinity, to name just a few of them. In this paper, we are concerned with the spectral property of the transmission system(1)∂t2u-∂x2u+a∂tu=0,∀x,t∈0,1×0,+∞,∂t2w+∂x4w=0,∀x,t∈-1,0×0,+∞,u1,t=∂x2w0,t=w-1,t=∂x2w-1,t=0,∀t∈0,+∞,u0,t=w0,t,∂xu0,t=-∂x3w0,t,∀t∈0,+∞,u·,0=u0,∂tu·,0=u1,∀x∈0,1,w·,0=w0,∂tw·,0=w1,∀x∈-1,0, in which a>0 and the energy in the wave and in the plate can be transferred throughout the interference point x=0 by the relation u(0,t)=w(0,t), ∂xu(0,t)=-∂x3w(0,t).One of our motivations to conduct the spectral analysis is to understand better the asymptotic behavior of solutions to (1). Recently, the wave/plate system (or wave/plate transmission system in the case of multidimensions) has been investigated in several references for their stabilization and/or asymptotic property. Ammari and Nicaise [3] proved under a certain geometric condition that the energy of the transmission system of a damped wave equation and a damped plate equation decays exponentially by a multiplier argument. Zhang and Zhang [4] studied the stabilization of transmission coupled wave and Euler-Bernoulli equations on Riemannian manifolds by introducing nonlinear feedback. Hassine [5] proved by the Carleman estimate an energy decay estimate for the transmission system of an undamped Euler-Bernoulli plate and a wave equation with a localized Kelvin-Voigt damping. Gong, Yang and Zhao [6] studied the stabilization of a wave/plate transmission system via a Riemannian geometric approach. Inspired by the afore-mentioned results, and in view of the presence of the damping in the wave equation, we are interested in the large-time behavior of solutions to system (1). And therefore, we are tempted to study the spectral property of the infinitesimal generator of the semigroup associated with system (1). The controllability of wave/plate transmission systems was also investigated in the literature; see [7].By introducingv(x,t)=w(-x,t) into (1), we are led to the system(2)∂t2u-∂x2u+a∂tu=0,∀x,t∈0,1×0,+∞,∂t2v+∂x4v=0,∀x,t∈0,1×0,+∞,u1,t=∂x2v0,t=v1,t=∂x2v1,t=0,∀∈0,+∞,u0,t=v0,t,∂xu0,t=∂x3v0,t,∀t∈0,+∞,u·,0=u0,∂tu·,0=u1,∀x∈0,1,v·,0=v0,∂tv·,0=v1,∀x∈0,1.Thus, our main goal boils down to analyzing the spectral property of the infinitesimal generator A of the semigroup associated with system (2). Here A is defined by (3)Aφ,φ~,ψ,ψ~=φ~,φ′′-aφ~,ψ~,-ψ4,∀φ,φ~,ψ,ψ~∈DA,DA=φ,φ~,ψ,ψ~∈H;φ~1=ψ~1=φ~0-ψ~0=ψ′′0=ψ′′1=φ′0-ψ′′′0=0,where H is the closure of D(A) in H1(0;1)×L2(0;1)×H2(0;1)×L2(0;1). Endowed with the sesquilinear form (4)H2∋φ,φ~,ψ,ψ~,f,f~,g,g~↦∫01φ′f′¯+φ~f~¯+ψ′′g′′¯+ψ~g~¯dx∈Cwhere z¯ denotes the complex conjugate of z∈C; H is a Hilbert space. H is the natural finite energy space for system (2).We shall prove that0∈ρ(A) by using the idea of Green’s functions, thereby proving that A has compact resolvents, shall prove that σ(A) is contained in the open left half of the complex plane by proving that A is dissipative in H and that iR∩σ(A)=∅;, and shall obtain the spectral asymptotics for A.The rest of the paper is planned as follows. In Section2, we analyze the spectral property of A. In Section 3, we conclude the paper by some remarks.


## 2. Spectral Property ofA

In this section, we analyze in detail the spectral property ofA. We first note that(5)ReAφ,φ~,ψ,ψ~,φ,φ~,ψ,ψ~H=-aφ~L20,12,∀φ,φ~,ψ,ψ~∈DA,which implies that A is a dissipative operator in H, and that λ∉σ(A) whenever Reλ>0.Proposition 1.
0 belongs to the resolvent set of A, or in symbols, 0∈ρ(A).Proof.
The statement is equivalent to the well-posedness of the boundary value problem (BVP)(6)-φ′′=f,ψ4=g,x∈0,1,φ1=ψ′′0=ψ1=ψ′′1=0,φ0=ψ0,φ′0=ψ′′′0.By utilizing the idea of Green’s functions, we know that φ and ψ can be written as(7)φx=-∫01c11x-ξ+c12+Hx-ξc~11x-ξ+c~12fξdξ+∫01c13x-ξ+c14gξdξ,∀x∈0,1,(8)ψx=-∫01c21x-ξ3+c22x-ξ2+c23x-ξ+c24fξdξ+∫01c25x-ξ3+c26x-ξ2+c27x-ξ+c28+Hx-ξc~25x-ξ3+c~26x-ξ2+c~27x-ξ+c~28gξdξ,∀x∈0,1,where the constants c11, c12, c~11, c~12, c13, c14, c21, c22, c23, c24, c15, c26, c27, c28, c~25, c~26, c~27, and c~28, are such that (9)-φ′′=f⟹c~12=0,2c~11=1,ψ4=g⟹c~28=0,4c~27=0,12c~26=0,24c~25=1,φ1=0⟹c111-ξ+c12+c~111-ξ+c~12=0,c131-ξ+c14=0,ψ′′0=0⟹-6c21ξ+2c22=0,-6c25ξ+2c26=0,ψ1=0⟹c211-ξ3+c221-ξ2+c231-ξ+c24=0,c251-ξ3+c261-ξ2+c271-ξ+c28+c~251-ξ3+c~261-ξ2+c~271-ξ+c~28=0,ψ′′1=0⟹6c211-ξ+2c22=0,6c251-ξ+2c26+6c~251-ξ+2c~26=0,φ0=ψ0⟹-c11ξ+c12--c21ξ3+c22ξ2-c23ξ+c24=0,-c13ξ+c14--c25ξ3+c26ξ2-c27ξ+c28=0,φ′0=ψ′′′0⟹c11-6c21=0,c13-6c25=0,∀ξ∈0,1.By applying Cramer’s rule, we can deduce(10)c~12=0,c~11=12,c~28=0,c~27=0,c~26=0,c~25=124,c11=0,c12=0,c13=ξ-14,c14=ξ-124,c21=0,c22=0,c23=ξ-12,c24=-ξ-122,c25=ξ-124,c26=ξξ-18,c27=16ξ3-14ξ2+13ξ-14,c28=112ξ3+112ξ2-512ξ2+14,∀ξ∈0,1.Plug this into (7) and (8), to yield(11)φx=1-x2∫0xfξdξ+12∫x11-ξfξdξ+14∫01ξ-1x-ξ+ξ-12gξdξ,x∈0,1,(12)ψx=12∫01ξ-1x-ξ+ξ-12fξdξ+124∫0xξ-1x-ξ3+3ξξ-1x-ξ2+4ξ3-6ξ2+8ξ-6x-ξ+2ξ3+2ξ2-10ξ+6+x-ξ3gξdξ+124∫x1ξ-1x-ξ3+3ξξ-1x-ξ2+4ξ3-6ξ2+8ξ-6x-ξ+2ξ3+2ξ2-10ξ+6gξdξ,x∈0,1,We deduce from this expression that φH2(0,1)2⩽C[fL2(0,1)2+gL2(0,1)2] and ψH4(0,1)2⩽C[fL2(0,1)2+gL2(0,1)2] for some C>0. This means that BVP (6) is well-posed, i.e., 0∈ρ(A).Remark 2.
Since the embeddingD(A)⊂H is compact, A has compact resolvents. And therefore, σ(A) consists merely of a sequence of eigenvalues.
Next, we shall prove thatσ(A) is contained in the open left half complex plane.Proposition 3.
σ ( A ) is contained in the open left half of complex plane, or in symbols, σ(A)⊂C-, where C-=λ∈C;Reλ<0.Proof.
By using the observation (5), it suffices to show that iR∩σ(A)=∅. Indeed, 0∉σ(A). Otherwise, there exists, by Remark 2, a nonzero quadruple (φ,φ~,ψ,ψ~) such that A(φ,φ~,ψ,ψ~)=(φ~,φ′′-aφ~,ψ~,-ψ(4))=0(φ,φ~,ψ,ψ~). We have necessarily φ~=0, ψ~=0, φ′′=0 and ψ(4)=0. And therefore, φ and ψ assume the forms φ(x)=c11x+c12 and ψ(x)=c21x3+c22x2+c23x+c24, respectively. Recalling the boundary conditions to which φ and ψ are subject, we have further (13)φ1=0⟹c11+c12=0,,ψ1=0⟹c21+c22+c23+c24=0,φ0-ψ0=0⟹c12-c24=0,ψ′′0=0⟹2c22=0,ψ′′1=0⟹6c21+2c22=0,φ′0-ψ′′′0=0⟹c11-6c21=0.It is ready to check that(14)det11000000111101000-100020000620010-6000=-12≠0.Then c11=c12=c21=c22=c23=c24=0, that is, φ=0 and ψ=0. This, together with the observation that φ~=0 and ψ~=0, contradicts the fact that the quadruple (φ,φ~,ψ,ψ~) is nonzero. That is, 0∈ρ(A).
It remains to show thatiμ∉σ(A) for every μ∈R∖{0}. Let (φ,φ~,ψ,ψ~)∈D(A) be such that A(φ,φ~,ψ,ψ~)=iμ(φ,φ~,ψ,ψ~) for some μ∈R∖{0}. We have 〈A(φ,φ~,ψ,ψ~),(φ,φ~,ψ,ψ~)〉H=iμ(φ,φ~,ψ,ψ~)H2 which implies Re〈A(φ,φ~,ψ,ψ~),(φ,φ~,ψ,ψ~)〉H=0. This, together with (5), implies φ~(x)=0, x∈[0,1]. Thus, the equation A(φ,φ~,ψ,ψ~)=iμ(φ,φ~,ψ,ψ~) is reduced to the boundary value problem(15)ψ4=μ2ψin0,1,ψ0=ψ′′0=ψ′′1=ψ′′′0=0.The solution to BVP (15) can be written as (16)ψx=c^1cosh⁡xμ+c^2sinh⁡xμ+c^3cos⁡xμ+c^4sin⁡xμwhere c^1,c^2,c^3, and c^4 are chosen so that (17)c^1+c^3=0,c^1-c^3=0,c^1cosh⁡μ+c^2sinh⁡μ-c^3cos⁡μ-c^4sin⁡μ=0,c^2-c^4=0.We deduce from the first two equations that c^1=c^3=0. Substitute this into the third equation, use the fact that sinh⁡(|μ|)>|μ|>sin⁡(|μ|) for μ∈R∖{0}, and conduct some routine calculations, to obtain c^2=c^4=0. Therefore, ψ(x)=0, x∈[0,1]. To sum, (φ,φ~,ψ,ψ~)=0. This means that iμ∉σ(A) for every μ∈R∖{0}. The proof is complete.Proposition 4.
Letλ∈R∖{0}. Then λ∈σ(A) if and only if λ satisfies the equation Σ(λ)=0, where(18)Σλ=λiλsinh⁡λλ+acosh⁡iλsinh⁡iiλ-icosh⁡iiλsinh⁡iλ+2iλλ+acosh⁡λλ+asinh⁡iiλsinh⁡iλ.Proof.
We should note thatλ∈σ(A) if and only if the following boundary value problem admits a nonzero solution: (19)λ2φ-φ′′+aλφ=0in0,1,λ2ψ+ψ4=0in0,1,φ1=ψ′′0=ψ1=ψ′′1=0,φ0=ψ0,φ′0=ψ′′′0.By the classical theory of ODEs, (φ,ψ) can be written as (20)φx=c1cosh⁡1-xλλ+a+c2sinh⁡1-xλλ+a,ψx=c3cosh⁡1-xiλ+c4sinh⁡1-xiλ+c5cosh⁡1-xiiλ+c6sinh⁡1-xiiλ.The coefficients c1,c2,c3,c4,c5,c6 are determined by the boundary conditions on φ and ψ. More precisely, we have(21)φ1=0⟹c1=0,ψ1=0⟹c3+c5=0,ψ′′1=0⟹c3iλ-c5iλ=0,ψ′′0=0⟹c4iλsinh⁡iλ-c6iλsinh⁡iiλ=0,φ0-ψ0=0⟹c2sinh⁡λλ+a-c4sinh⁡iλ-c6sinh⁡iiλ=0,φ′0-ψ′′′0=0⟹-c2λλ+acosh⁡λλ+a+c4iλiλcosh⁡iλ+c6λiλcosh⁡iiλ=0.By the related classical theory from Linear Algebra, (21) has a nontrivial solution if and only if (22)det⁡10000000101000iλ0-iλ0000iλsinh⁡iλ0-iλsinh⁡iiλ0sinh⁡λλ+a0-sinh⁡iλ0-sinh⁡iiλ0-λλ+acosh⁡λλ+a0iλiλcosh⁡iλ0λiλcosh⁡iiλ=2iλ2Σλ=0.By recalling that λ≠0, we infer that the proof is complete.Theorem 5.
σ ( A ) is contained in the open left half complex plane C- and consists merely of eigenvalues {λ1k,λ2l} of A. And moreover, the eigenvalues are distributed as follows: (23)λ1k=-a±i4k2π2-a22+Ok-1/2ask→+∞withk∈N,λ2l=±π2l+142i+Ol-1/2asl→+∞withl∈Z.Proof.
Recall thatσ(A)⊂C-. We shall first consider the situation λ∈D1∪D2 where D1={λ∈C-;argλ∈[3π/4,π]} and D2={λ∈C-;argλ∈[π/2,3π/4]}. For the sake of simplification, we would use λ=|λ|eiθ for complex numbers. We next study the eigenvalues of A in D1∪D2.
Case  1 (λ∈D1). By applying Taylor’s expansion, we have(24)eλλ+a=eλλ+a=e-λ1+a/λ=e-λ1+a/2λ+Oλ-2=e-λ-a/2+Oλ-1for λ with sufficiently large modules. Besides, we have(25)eiλ=exp⁡λeiθ/2-3π/4=exp⁡λcosθ2-3π4expiλsin⁡θ2-3π4,(26)eiiλ=exp⁡λcos⁡θ2-π4exp⁡iλsin⁡θ2-π4.Therefore,(27)limλ→+∞eλλ+a=+∞,limλ→+∞eiλ=+∞,limλ→+∞eiiλ=+∞.By analyzing the definition (18) of Σ(λ), we have (28)limλ∈D1,λ→+∞Σλλiλeλλ+a+iλ+iiλ=1-i8≠0.This, together with (27), implies that λ∈D1 cannot be a solution to the equation Σ(λ)=0, or equivalently, λ cannot be an eigenvalue of A, whenever its module |λ| is sufficiently large.
Case  2 (λ∈D2). We deduce from (24), (25), and (26) that (29)e-λλ+a=eλ+a/2+Oλ-1⩽C1,for some C1>0, (30)e-iλ=exp⁡-λcos⁡θ2-3π4⩽1,e-iiλ=exp⁡-λcos⁡θ2-π4⩽e-2λ/2.Divide the both hand sides of the equation Σ(λ)=0 by λiλeλ(λ+a)+iλ+iiλ, and conduct some routine calculations, to obtain (31)1-e-2λλ+ai-e-2iλ=Oλ-1/2as |λ|→+∞ with λ∈D2. Note that the equation 1-e-2λ(λ+a)i-e-2iλ=0 with the restriction λ∈D2 has the solutions λ^1k=(-a+i4k2π2-a2)/2 with k∈N, and λ^2l=π2l+1/42i with l∈Z. By applying Rouché’s theorem (see [1]) and Proposition 4, we obtain the spectral asymptotics for A: λ1k=(-a+i4k2π2-a2)/2+O(k-1/2) as k→∞ with k∈N, and λ2l=π2l+1/42i+O(|l|-1/2) as l→∞ with l∈Z.
Lastly, noting thatλ∈σ(A) iff λ¯∈σ(A), we know that there are eigenvalues of A which are distributed as follows: λ1k=(-a-i4k2π2-a2)/2+O(k-1/2) as k→∞ with k∈N, and λ2l=-π2l+1/42i+O(|l|-1/2) as l→∞ with l∈Z. This, together with the afore-conducted analysis, implies that the proof is complete.Theorem 6.
The natural energy functional associated with system (1)(32)Et=12∫01∂tux,t2+∂xux,t2dx+12∫-10∂twx,t2+∂x2wx,t2dxdecreases to 0 as t→+∞, (u,v) is a solution to system (1).Proof.
Let us define the natural energy functional associated with(33)E~t=12∫01∂tux,t2+∂xux,t2dx+12∫01∂tvx,t2+∂x2vx,t2dx,t∈0,+∞.Recalling the relation v(x,t)=w(-x,t), and by applying integration by change-of-variable, we have E(t)=E~(t) for all t∈[0,+∞). But we have also (34)u·,t,∂tu·,t,v·,t,∂tv·,t=etAu0,u1,v0,v1,∀t∈0,+∞.By, we have (35)limt→+∞etAu0,u1,v0,v1H2=0.And therefore,(36)limt→+∞Et=limt→+∞E~t=12limt→+∞u·,t,∂tu·,t,v·,t,∂tv·,tH2=12limt→+∞etAu0,u1,v0,v1H2=0.The proof is complete.


## 3. Conclusion

We investigated in this paper the transmission system (1) of a 1D damped wave equation and a 1D undamped plate equation for its spectral property. We proved that the spectrum of the associated infinitesimal generator consists merely with eigenvalues which lie to the left of the imaginary axis and have two distinct vertical lines as its asymptote lines in which the one on the right hand side is the imaginary axis. We obtained also a byproduct: the energy E(t) of system (1) decreases to zero as time approaches infinity. The first natural question reads: Can one obtain the optimal decay rate for the energy functional E(t) defined by (32)? We would try to answer this question in our future work.Another interesting question is: Can the method be carried out for wave/plate equations in higher dimensional domains to obtain the same results as in this paper? The answer is: It seems to be possible to use the idea of this paper to conduct spectral analysis for systems posed in certain special higher dimensional domains, but would need much more complicated calculations to obtain similar results as in 1D cases. We explain here this claim in certain detail. Let us consider a very special 2D transmission coupled system of wave and plate equations, that is,(37)∂t2u-Δu+a∂tu=0,∀x,y,t∈Ω+×0,+∞,∂t2w+Δ2w=0,∀x,y,t∈Ω-×0,+∞,u1,y,t=∂x2w0,y,t=w-1,y,t=∂x2w-1,y,t=0,∀y,t∈0,1y×0,+∞,u0,y,t=w0,y,t,∂xu0,y,t=-∂x3w0,y,t,∀y,t∈0,1y×0,+∞,ux,0,t=ux,1,t=0,∀x,t∈0,1x×0,+∞,wx,0,t=∂y2wx,0,t=wx,1,t=∂y2wx,1,t=0,∀x,t∈-1,0x×0,+∞,u·,0=u0,∂tu·,0=u1,∀x,y∈Ω+,w·,0=w0,∂tw·,0=w1,∀x,y∈Ω-,where Ω-=[-1,0]x×[0,1]y, Ω+=[0,1]x×[0,1]y, Δ=∂x2+∂y2, Δ2=∂x4+∂y4, Ω=Ω+∪Ω-, u(x,y,t) and w(x,y,t). By introducing v(x,y,t)=w(-x,y,t), we obtain the auxiliary system(38)∂t2u-Δu+a∂tu=0,∀x,y,t∈Ω+×0,+∞,∂t2v+Δ2v=0,∀x,y,t∈Ω+×0,+∞,u1,y,t=∂x2v0,y,t=v1,y,t=∂x2v1,y,t=0,∀y,t∈0,1y×0,+∞,u0,y,t=v0,y,t,∂xu0,y,t=∂x3v0,y,t,∀y,t∈0,1y×0,+∞,ux,0,t=ux,1,t=0,∀x,t∈0,1x×0,+∞,vx,0,t=∂y2vx,0,t=vx,1,t=∂y2vx,1,t=0,∀x,t∈0,1x×0,+∞,u·,0=u0,∂tu·,0=u1,∀x,y∈Ω+,v·,0=v0,∂tv·,0=v1,∀x,y∈Ω+.The spectral study of (37) is equivalent to that of (38). And thus we are led to the characteristic problem (39)λ2φ-Δφ+aλφ=0,∀x,y∈Ω+,λ2ψ+Δ2ψ=0,∀x,y∈Ω+,φ1,y=∂x2ψ0,y=ψ1,y=∂x2ψ1,y=0,∀y∈0,1y,φ0,y=ψ0,y,∂xφ0,y=∂x3ψ0,y,∀y∈0,1y,φx,0=φx,1=0,∀x∈0,1x,ψx,0=∂y2ψx,0=ψx,1=∂y2ψx,1=0,∀x∈0,1x.Consider the auxiliary characteristic problems(40)-d2dy2φ+μφ=0,∀y∈0,1y,φ0=φ1=0,d4dy4ψ+ζψ=0,∀y∈0,1y,v0=d2dy2v0=v1=d2dy2v1=0. By recalling the classical results concerning PDE theory, we know that the former problem has the following pairs of nontrivial “normalized” solutions(41)μk=k2π2,φ1ky=2sin⁡kπy,and that the latter problem has the following pairs of nontrivial “normalized” solutions(42)ζk=k4π4,ψ1ky=2sin⁡kπy.Let us recall that Span({φ1k(y)})=Span({ψ1k(y)}) is dense in L2((0,1)y), and that L2(Ω+)=L2((0,1)x)⊗¯L2((0,1)y), here and hereafter, we denote ⊗¯ by the closure of the algebraic tensor product two Hilbert spaces. Let (φ,ψ,λ) be a solution triple to the problem (39). By using the results recalled just now, we know that (43)φx,y=∑k=1∞φ2kxφ1ky,ψx,y=∑k=1∞ψ2kxψ1ky.Plug this into (39), and proceed by some routine calculations, to obtain (44)∑k=1∞λ2φ2kx-φ2k′′x+μkφ2kx+aλφ2kxφ1ky=0,∀x,y∈Ω+,∑k=1∞λ2ψ2kx+φ2k4x+ζkψ2kxψ1ky=0,∀x,y∈Ω+,∑k=1∞φ2k1φ1ky=∑k=1∞ψ2k′′0ψ1ky=∑k=1∞ψ2k1ψ1ky=∑k=1∞ψ2k′′1ψ1ky=0,∀y∈0,1y,∑k=1∞φ2k0φ1ky=∑k=1∞ψ2k0ψ1ky,∑k=1∞φ2k′0φ1ky=∑k=1∞ψ2k′′′0ψ1ky,∀y∈0,1y.By utilizing the relation φ1k=ψ1k, and the fact that {φ1k(y)}={ψ1k(y)} is an orthonormal basis for L2((0,1)y), we have(45)λ2φ2kx-φ2k′′x+μkφ2kx+aλφ2kx=0,∀x∈0,1x,λ2ψ2kx+φ2k4x+ζkψ2kx=0,∀x∈0,1x,φ2k1=ψ2k′′0=ψ2k1=ψ2k′′1=0,φ2k0=ψ2k0,φ2k′0=ψ2k′′′0.By an argument similar to the one used in this paper, we can show that for each k=1,2,…, the eigenproblem (45) has a countably infinite many solutions, and eigenvalues are located in the open left half of the complex plane and can be divided into two subgroups; one subgroup has the imaginary axis as its asymptote line, while the other has a line parallel to the imaginary axis as its asymptote line.The above example shows that our method used in this paper can be carried out in certain special higher order systems. But it should be stressed again that the calculations could be much more complicated.



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