Research article

On the exponential decay of a Balakrishnan-Taylor plate with strong damping

  • Received: 03 January 2024 Revised: 30 March 2024 Accepted: 08 April 2024 Published: 18 April 2024
  • MSC : 35B40, 35L75, 74K10

  • In this manuscript, we study a thin and narrow plate equation that models the deck of a suspension bridge that is subject to a Balakrishnan-Taylor damping and a strong damping. First, by using the Faedo Galerkin method, we prove the existence of both global weak and regular solutions. Second, we prove the exponential stability of the energy for regular solutions by combining the multiplier method and a well-known result of Komornik.

    Citation: Zayd Hajjej. On the exponential decay of a Balakrishnan-Taylor plate with strong damping[J]. AIMS Mathematics, 2024, 9(6): 14026-14042. doi: 10.3934/math.2024682

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  • In this manuscript, we study a thin and narrow plate equation that models the deck of a suspension bridge that is subject to a Balakrishnan-Taylor damping and a strong damping. First, by using the Faedo Galerkin method, we prove the existence of both global weak and regular solutions. Second, we prove the exponential stability of the energy for regular solutions by combining the multiplier method and a well-known result of Komornik.



    A rectangular thin and narrow plate that models the deck of a suspension bridge is considered in the domain Ω=(0,π)×(d,d), where d<<π. The nonlocal evolution equations that describe how the plate is deformed are as follows:

    {ptt(x,y,t)+Δ2p(x,y,t)(ϕ(p)+δpx,pxt)pxx+αΔ2pt=0 in Ω×(0,+),p(0,y,t)=pxx(0,y,t)=p(π,y,t)=pxx(π,y,t)=0(y,t)(d,d)×(0,+),pyy(x,±d,t)+μpxx(x,±d,t)=0(x,t)(0,π)×(0,+),pyyy(x,±d,t)+(2μ)pxxy(x,±d,t)=0(x,t)(0,π)×(0,+),p(x,y,0)=p0(x,y),pt(x,y,0)=p1(x,y) in Ω, (1.1)

    where δ,α>0 and ϕ which introduces a nonlocal effect is given by

    ϕ(p)=a+bΩp2xdxdy.

    The constant μ is the Poisson ratio which is generally in the range of (1,12) due to physical reasons (see [1] for more details). It has a value of about 0.3 for metals and between 0.1 and 0.2 for concrete. Due to this, we suppose that 0<μ<12. The constant b>0 is determined by the elasticity of the deck's material, bΩp2xdxdy determines the plate's geometric nonlinearity as a result of its stretching, and a>0 is the constant for prestressing. Specifically, if the plate is compressed, we have that a>0 and if the plate is stretched, one has that a<0.

    The model (1.1) describes the vibrations of the deck of a suspension bridge in the presence of a Balakrishnan-Taylor damping (the term δpx,pxt) and a strong damping (the term αΔ2pt).

    Note that , denotes the usual scalar product in L2(Ω).

    Let us recall some works in the literature that are related to our problem. For one dimensional problems, in [2], the author considered the following equation

    ptt+αpxxxx(β+kl0[p(ξ,t)ξ]2dξ)pxx+γpxxxxtμpx,ptxpxx+δpt=0,in(0,l)×(0,+), (1.2)

    where the constants α,k,γ,μ are positive, and the constants β and δ have no restrictions on their sign. Here, l denotes the beam's length. The author established existence, uniqueness, and regularity theorems for the situations in which the beam's ends are clamped or hinged. Regarding higher dimensions, consider the work of Emmrich and Thalhammer [3], who provided a general model for describing nonlinear extensible beams with weak, viscous, strong, and Balakrishnan-Taylor damping, as follows:

    ptt+αΔ2p+ξp+κptλΔpt+μΔ2pt[β+γΩ|p|2dx+δ|Ωpptdx|q2Ωpptdx]Δp=h (1.3)

    in Ω×(0,T), where Ω is a bounded domain and T>0, the constants α and γ are positive, and λ,μ, and δ are nonnegative, whereas β,κ,ξR and q2.

    The authors proved the existence of a weak solution for (1.3) under hinged or clamped boundary conditions by using time discretization in both cases, i.e., when λ,μ>0 and q2 or when λ=μ=0 and q=2. Under the conditions of applying κ=λ=0, μ>0 and q=2 in (1.2), Clark [4] established the existence, uniqueness, and asymptotic behavior of the solutions in N-dimensional bounded and unbounded domains. In [5], You proved that there are global solutions in the cases in which κ=μ=0, λ>0, q>2 and Ω=(0,1). Also, he gave results on the existence of inertial manifolds and the finite-dimensional stabilization. Subsequently Tavares et al. [6] studied the problem (1.3) for λ=μ=0,κR and q2; they established the existence of a unique mild (and strong) solution and analyzed the long-time dynamics of solutions (in the mentioned case) when κ>0 and β is bounded from below by a negative expression, as well as with the existence of nonlinear source.

    Now, let us mention some works on suspension bridges. In [7,8], the existence of nonlinear oscillations was proved. The deck of a suspension bridge has been modeled in a simple form in [9]. See also Gazzola's book [10] and recent results [11,12,13] for additional details. The bending and stretching energies of the model presented in [9] were examined by Al-Gwaiz et al. in [14]. We mention also the recent work of [15] in which the authors provide a new model for a suspension bridge.

    Recently, many researchers have been interested in studying the stability of a plate model for the deck of a suspension bridge. Messaoudi and Mukiawa [16] showed an exponential decay in the presence of both a global frictional damping and a nonlinear term. In [17] (resp. in [18]), the authors studied the same problem as in [16] but with linear (resp. nonlinear) local damping distributed around a neighborhood of the boundary, and they proved an exponential decay estimate of the associated energy.

    Liu and Zhuang [19] expanded the work of [20] and proved, without considering the relation between m and r, that the solutions of the equation, i.e.,

    ptt+Δ2p+ap+|pt|m2pt=|p|r2p,m2,r>2,

    exist globally if and only if there exists a real number t0[0,Tmax) such that p(t0)W and the energy at the time t0 is less than such a constant that depends on r and Cr (Cr is defined in (2.3)), where

    Tmax=sup{T>0:p=p(t)exists on[0,T]}

    and

    W={pN+:J(p)<d},

    with

    N+={pV:I(p)>0}{0},I(p)=p2V+(ap,p)prr,
    J(p)=12p2V+12(ap,p)1rprrandd=infpV{0}maxλ>0J(λp).

    Moreover, the energy decay results were obtained, and when r>m a blow-up result was established. Later, in [21], the authors established the existence of a global weak solution and proved a stability result under the conditions of an external force f and a nonlinear frictional damping. Finally, we cite the work [22], in which the author studied the same problem as described here, but it was subject to different types of damping, i.e., one of memory type (of the form t0g(s)Δ2p(s)ds) and a nonlinear localized frictional damping (of the form a(x,y)|pt|mpt). The author proved the existence of global solutions as well as a general stability result. For other results concerning partially hinged plate equations, we refer the reader to the recent papers [23,24,25].

    Motivated by all mentioned works, our current paper investigates the exponential stability of solutions to system (1.1) with a strong damping and a Balakrishnan-Taylor damping. As mentioned at the end of the paper, the Balakrishnan-Taylor damping (alone) is insufficient to deduce exponential stability. For this reason, we chose to add another damping to obtain the uniform stability.

    The structure of the paper is as follows. In the next section, we present some fundamental preliminaries that will be used to prove our main results. In the third section, the well-posedness of the problem (1.1) is proved. We show the exponential stability of system (1.1) in the last section.

    Here and in the sequel, we use to denote the usual norm in L2(Ω).

    We define the space

    V={wH2(Ω):w=0on{0,π}×(d,d)},

    with the scalar product

    (p,q)=Ω[ΔpΔq+(1μ)(2pxyqxypxxqyypyyqxx)]dxdy.

    We note that (V,(,)) is a Hilbert space, and that the norm .V is equivalent to the H2 norm (see [9, Lemma 4.1]).

    Moreover, we denote by H(Ω) the dual space of V, and we indicate by .,.2,2 the associated duality. We have the following:

    Lemma 2.1. [9] If 0<μ<12 and fL2(Ω), then there is a unique pV such that, for all qV, we have

    (p,q)=Ωfq. (2.1)

    The function pV satisfying (2.1) is known as the weak solution to the following stationary problem

    {Δ2p=f,p(0,y)=p(π,y)=pxx(0,y)=pxx(π,y)=0,pyy(x,±d)+μpxx(x,±d)=pyyy(x,±d)+(2μ)pxxy(x,±d)=0. (2.2)

    Lemma 2.2. [20] Let pV and 1r<+. Then, we have

    prrCrprV, (2.3)

    for some positive constant Cr=Cr(Ω,r).

    Remark 2.3. Let f=λp in (2.2). Then, Theorem 3.4 in [9] asserts that the set of eigenvalues of (2.2) may be ordered in an increasing sequence {λj}j1 of strictly positive numbers that diverge to +, and that the set of eigenfunctions {wj}j1 of (2.2) is a complete system in V.

    The energy related to (1.1) is given as follows

    E(t)=12pt(t)2+12p(t)2Va2px(t)2+b4px(t)4, (2.4)

    which satisfies the following identity

    E(t)=αpt2Vδ(12ddtpx2)20, (2.5)

    This indicates that the energy decreases with time t and E(t)E(0),t0.

    We recall the following theorem (see [26, Theorem 8.1]) that will be useful in the proof of the main result.

    Theorem 2.4. Let E:[0,)[0,) be a non-increasing function and assume that there exists a constant C>0 such that

    tE(s)dsCE(t),t0.

    Then

    E(t)E(0)e1tC,tC.

    Remark 2.5. We remark that the energy is nonnegative if a<0, and this case is equivalent to a stretched plate. However, this scenario is not applicable to real-world bridges [27]. When a>0, which is the utmost likely situation for bridges, the energy E(t) may be negative. This issue can be solved by following some ideas from [14, Section 3]. To do this, we define

    W:={wH1(Ω):w=0 on {0,π}×(d,d)},C(Ω):={wC(¯Ω):ε>0,w(x,y)=0 if x[0,ε][πε,π]}.

    endowed with the following norm

    ||p||W:=(Ω|p|2dxdy)1/2, (2.6)

    where W is a normed space.

    Remark 2.6. [14] W is defined as the completion of C(Ω) according to the norm ||||W. It is clear that the embedding VW is compact and the optimal embedding constant satisfies

    Λ1:=minwV||w||2V||w||2W.

    Lemma 2.7. [17] Assume that 0aΛ1; then, E(t)0.

    Proof. Using Remark 2.6, we obtain the following inequality

    ||w||2WΛ11||w||2V, for all wV. (2.7)

    Since

    ||px||2Ω|p|2dxdyΛ11||p||2V,

    then we have

    a2||px||2a2Λ11||p||2V,pV,

    and consequently

    12||p||2Va2||px||212||p||2V(1aΛ11).

    So, if 0aΛ1 we conclude that 12||p||2Va2||px||20 and therefore E(t)0. This is in agreement with the hypothesis of Theorem 4 in [27].

    Definition 3.1. Let T be a positive number. The functions

    pL(0,T;V),ptL(0,T;L2(Ω))L2(0,T;V),andpttL2(0,T;H(Ω)),

    constitue a weak solution of (1.1) when

    ptt,w2,2+(p,w)+Ω(a+bpx2+δpx,pxt)pxwxdxdy+α(pt,w)=0,wV,p(x,y,0)=p0(x,y),pt(x,y,0)=p1(x,y), (3.1)

    for almost everywhere t[0,T].

    Theorem 3.2. Suppose that 0aΛ1 and let (p0,p1)V×L2(Ω). Then, the problem (1.1) has a unique global weak solution on [0,T] for any T>0.

    Proof. We divide our proof into 4 steps.

    Step 1. In this step, we will prove some convergence results for the sequence (pk)k1 (defined below) and its derivative.

    We start by applying the Faedo-Galerkin approach. By Remark 2.3, we may consider {wj}j=1 as a basis of V and let Vk=span{w1,w2,...,wk} be subspace of V with finite dimensions, which is spanned by the first k vectors. Let

    pk0(x,y)=kj=1ajwj(x,y),pk1=kj=1bjwj(x,y),

    such that pk0,pk1Vk and

    pk0p0inV,andpk1p1inL2(Ω). (3.2)

    We are looking for a solution of the form

    pk(x,y,t)=kj=1cj(t)wj(x,y), (3.3)

    that solves the following in Vk:

    pktt,wj2,2+(pk,wj)+Ω(a+bpkx2+δpkx,pkxt)pkx(wj)xdxdy+α(pkt,wj)=0,j=1,...,k,pk(x,y,0)=pk0(x,y),pkt(x,y,0)=pk1(x,y). (3.4)

    It is easy to check that, for any k1, the above problem (3.4) yields a solution pk on [0,tk), where 0<tkT. Now, we multiply (3.4) by cj(t) and sum over j=1,...,k to obtain

    ddtEk(t)=αpkt2Vδ(12ddtpkx2)20, (3.5)

    where

    Ek(t)=12pkt(t)2+12pk(t)2Va2pkx(t)2+b4pkx(t)4. (3.6)

    Now, we integrate (3.5) over (0,t), where 0<t<tk; we also note, from (3.2), that (pk0) and (pk1) are respectively bounded in V and L2(Ω); we then obtain

    Ek(t)+αt0pkt2Vds+δt0(12ddtpkx2)2dsEk(0)C, (3.7)

    where C is a positive constant that does not depend on t and k, and that may vary from line to line.

    Hence, we get the following bounds:

    pk2V,pkt(t)2,t0pkt2VC. (3.8)

    As a result, one obtains that tk=T and we have the following:

    {(pk)isboundedinL(0,T;V),(pkt)isboundedinL(0,T;L2(Ω))L2(0,T;V). (3.9)

    Hence, there exists a subsequence of (pk), still denoted by (pk), that verifies the following:

    {pkpweaklystarinL(0,T;V),pktptweaklystarinL(0,T;L2(Ω))L2(0,T;V),pkpinL2(Q)stronglyanda.e,pkx2pkxxX1weaklystarinL(0,T;L2(Ω)),pkx,pkxtpkxxX2weaklystarinL(0,T;L2(Ω)), (3.10)

    where Q=Ω×(0,T).

    Step 2. Here, we will prove that X1=px2pxx and X2=px,pxtpxx by following the same arguments as in [2,28].

    For the first one, the following lemma is required.

    Lemma 3.3. Suppose that p,qV. We have

    px2pxxqx2qxx,pq0.

    Proof. One has

    px2pxxqx2qxx,pq=px2(px,qxpx2)+qx2(px,qxqx2)px2(pxqxpx2)+qx2(pxqxqx2)=(pxqx)(px3qx3)0.

    Now, let qL2(0,T;V). From Lemma 3.3, one obtains that

    T0pkx2pkxxqx2qxx,pkqdt0.

    By following the same steps as in [28], we derive that

    X1=px2pxx.

    Next, to prove that X2=px,pxtpxx, we note first note that

    pkx,pkxt=pxx,pktpkp,pkxxt.

    It is clear that pxx,pktpxx,pt in L(0,T). Besides, from (3.8), we have

    T0|pkp,pkxxt|dt(T0pkp2dt)12(T0pkxxt2dt)12C(T0pkp2dt)12.

    Hence pkp,pkxxt0 in L1(0,T); thus, we deduce that

    pkx,pkxtpx,pxtinL1(0,T). (3.11)

    Now, let φL1(0,T;L2(Ω)). Then

    T0px,pxtpxx,φdt=T0pkx,pkxtpkxx,φdt+T0[px,pxtpkx,pkxt]pkxx,φdt+T0pxxpkxx,px,pxtφdt. (3.12)

    The last two terms on the right side of (3.12) go to zero as k+ when applying (3.8) and (3.11). Since φ is arbitrary, we conclude that X2=px,pxtpxx.

    Step 3. In this step, we will show that p is the unique solution of the system (3.1).

    By integrating (3.4) over (0,t), we obtain

    Ωpktwdxdy+t0(pk,w)ds+t0Ω(a+bpkx2+δpkx,pkxt)pkxwxdxdyds+αt0(pkt,w)ds=Ωpk1w,wV. (3.13)

    Recall (3.10) and let k; we get

    ΩptwdxdyΩp1w=t0(p,w)dst0Ω(a+bpx2+δpx,pxt)pxwxdxdydst0α(pt,w)ds. (3.14)

    This means that (3.14) holds true for any wV. Since the terms on the right hand side of (3.14) are absolutely continuous, then (3.14) is differentiable for almost everywhere t0. It holds that

    ptt,w2,2+(p,w)+Ω(a+bpx2+δpx,pxt)pxwxdxdy+α(pt,w)=0,wV. (3.15)

    Regarding the initial conditions, from (3.10), and by using Lions' lemma [29], we can simply get

    pkpinC([0,T),L2(Ω)). (3.16)

    pk(x,y,0) then makes sense and pk(x,y,0)p(x,y,0) in L2(Ω). Noting that

    pk(x,y,0)=pk0(x,y)p0(x,y)inV,

    we get

    p(x,y,0)=p0(x,y). (3.17)

    Besides, as in [30], we multiply (3.4) by ϕC0(0,T) and integrate over (0,T); we get the following for any jk:

    T0Ωpkt(t)wϕ(t)dxdydt=T0(pk,w)ϕ(t)dtT0Ω(a+bpkx2+δpkx,pkxt)pkxwxϕ(t)dxdydtαT0(pkt,w)ϕ(t)dt. (3.18)

    As k+, we have the following for any wV and any ϕC0(0,T)

    T0Ωpt(t)wϕ(t)dxdydt=T0(p,w)ϕ(t)dtT0Ω(a+bpx2+δpx,pxt)pxwxϕ(t)dxdydtαT0(pt,w)ϕ(t)dt, (3.19)

    which implies that (see [30])

    pttL2(0,T;H(Ω)).

    Given that ptL2(0,T;L2(Ω)), we conclude that ptC(0,T;H(Ω)).

    pkt(x,y,0) therefore makes sense and

    pkt(x,y,0)pt(x,y,0)inH(Ω).

    However

    pkt(x,y,0)=pk1(x,y)p1(x,y)inL2(Ω).

    So,

    pt(x,y,0)=p1(x,y). (3.20)

    For the uniqueness, assume that p and ¯p verify (3.15), (3.17), and (3.20). So, by integrating by parts, q=p¯p satisfies

    Ωqtt(x,t)wdxdy+(q,w)+aΩqxxwdxdybΩ(px2pxx¯px2¯pxx)wdxdyΩδ(px,pxtpxx¯px,¯pxt¯pxx)wdxdy+α(qt,w)=0,wV,q(x,y,0)=qt(x,y,0)=0. (3.21)

    Then, (3.21) holds true for any wC0(Ω×(0,T)) by the density it is also valid for any wL2(Ω×(0,T)).

    If we test (3.21) with qt, we get

    ddt{12qt2+12q2V}+aΩqxxqtdxdybΩ(px2pxx¯px2¯pxx)qtdxdyΩδ(px,pxtpxx¯px,¯pxt¯pxx)qtdxdy+αqt2V=0. (3.22)

    By using Young's inequality, we get

    aΩqxxqtdxdya2qt2+a2qxx2C(qt2+q2V). (3.23)

    Next, it is easy to see that

    px2pxx¯px2¯pxx=px2qxxp+¯p,qxx¯pxx, (3.24)

    and

    Ωδ(px,pxtpxx¯px,¯pxt¯pxx)qtdxdy=δ(qt,¯pxx)2δqt,¯pxxpt,qxxδpt,pxxqt,qxx. (3.25)

    Then, using (3.24), we infer that

    bΩ(px2pxx¯px2¯pxx)qtdxdyCqxxqtC(qt2+q2V). (3.26)

    Besides, from (3.25), one derives the following:

    Ωδ(px,pxtpxx¯px,¯pxt¯pxx)qtdxdyδ(qt,¯pxx)2+Cqxxqtδ(qt,¯pxx)2+C(qt2+q2V). (3.27)

    By using (3.23), (3.26), and (3.27) we can deduce that

    ddt{12qt2L2(Ω)+12q2V}+δ(qt,¯pxx)2+αqt2VC(qt2+q2V). (3.28)

    By Gronwall's inequality, we obtain

    qt2+q2VCeCt(qt(0)2+q(0)2V),

    which gives that q=0 and thus p=¯p.

    The following theorem gives an additional regularity result.

    Theorem 3.4. Suppose that 0aΛ1 holds true and let (p0,p1)X×V, with X=H4(Ω)V. Then there is a unique function p=p(x,y,t) that satisfies the initial conditions (3.17) and (3.20), and that it satisfies

    pL(0,T;X),ptL(0,T;V)L2(0,T;X),andpttL2(0,T;L2(Ω)),

    and

    ptt+Δ2p(ϕ(p)+δpx,pxt)pxx+αΔ2pt=0,inL2(0,T;L2(Ω)). (3.29)

    Proof. Let {φj}j=1 be a basis of X (this basis exists since X is a separable Hilbert space). The solutions pk can be written in the form (3.3) and verify (3.4) as well as the following initial conditions

    pk0p0inX,andpk1p1inV.

    It is easy to see that the bounds defined in (3.8) are satisfied. If we test (3.4) with Δ2pkt and integrate by parts, we get

    12ddt(pkt2V+Δ2pk2)+αΔ2pkt2=apkxx,Δ2pkt+bpkx2pkxx,Δ2pkt+δpkx,pkxtpkxx,Δ2pkt.

    Therefore, by integrating by parts over (0,t), it follows that

    pkt2V+Δ2pk2+2αt0Δ2pkt2dsC1+C2t0|pkxx,Δ2pkt|dsC1+C2t0pkxx2ds+αt0Δ2pkt2ds.

    By using (3.9), we derive that

    pkt2V,Δ2pk2,andt0Δ2pkt2dsC.

    We proceed as in Theorem (3.2) to prove the existence of a unique pL(0,T;X) that satisfies (3.17), (3.20), (3.29), and ptL(0,T;V)L2(0,T;X). It follows from (3.29) that pttL2(0,T;L2(Ω)).

    This section's major result is as follows:

    Theorem 4.1. Let 0a<Λ1. Then, there are two constants K>0 and λ>0 such that the energy defined in (2.4) verifies

    E(t)Keλt,t0. (4.1)

    Proof. We will work with regular solutions and by using standard density arguments; the decay holds true even for weak solutions. Multiplying (1.1) by p and integrating over Ω×(s,T), for 0<s<T, we get

    TsΩ(pttp+pΔ2pϕ(p)pxxpδpx,pxtpxxp+αpΔ2pt)dxdydt=0. (4.2)

    Using Lemma 2.1 and integration by parts, we obtain

    TsΩ(ptp)tdxdydtTsΩp2tdxdydt+Tsp2Vdt+TsΩϕ(p)p2xdxdydt+TsΩδpx,pxtp2xdxdydt+αTs(p,pt)dt=0. (4.3)

    This yields

    TsE(t)dt+TsΩ(ptp)t32TsΩp2t+12Tsp2Va2Tspx2+3b4Tspx4+δTsΩpx,pxtp2xdxdydt+αTs(p,pt)dt=0. (4.4)

    Then, we obtain

    TsE(t)dtTsΩ(ptp)t+32TsΩp2t+a2Tspx2δTsΩpx,pxtp2xdxdtαTs(p,pt)dt. (4.5)

    The terms on the right hand side of (4.5) can be estimated as follows. Using Lemma 2.2 and Young's inequality, we infer that

    |TsΩ(ptp)t||Ωpt(s)p(s)|+|Ωpt(T)p(T)|12Ωp2t(s)+12Ωp2t(T)+12Ωp2(s)+12Ωp2(T)E(s)+E(T)+Cp(s)2V+Cp(T)2VCE(s), (4.6)

    where C is a generic positive constant. For the second term, thanks to Lemma 2.2 we have

    32TsΩp2tCTspt2VCαTs(E(t))dtCαE(s). (4.7)

    The third term on the right hand side of (4.5) may be estimated as follows:

    a2Tspx2aΛ112Tsp2VaΛ11TsE(t)dt. (4.8)

    Thanks to Young's inequality and (2.5), we deduce, for any ε>0, that

    |δTsΩpx,pxtp2xdxdt|CεTs(ddtpx2)2dt+δε4Tspx4dtCεTs(E(t))dt+δεbTsE(t)dtCεE(s)+δεbTsE(t)dt, (4.9)

    and

    |αTs(p,pt)dt|αTspVptVCεTspt2Vdt+αε2Tsp2VdtCεE(s)+αεTsE(t)dt. (4.10)

    By inserting (4.6)–(4.10) into (4.5), and choosing ε such that 1aΛ11(δb+α)ε>0, we conclude that there exists a positive constant C1 satisfying

    TsE(t)dtC1E(s),s>0.

    Let T+, and thanks to Theorem 2.4, we get the desired inequality (4.1).

    Remark 4.2. As remarked in [31] (for extensible beams), the Balakrishnan-Taylor damping does not seem sufficient to provide a "good" stability result to our problem (1.1). In fact, if a=b=α=0 in (1.1), we have the following equation:

    ptt+Δ2pδpx,pxtpxx=0,inΩ×(0,+). (4.11)

    The corresponding energy for system (4.11) is given by

    E(t)=12pt2+12p2V,

    which satisfies

    E(t)=δ[px,pxt]20,t>0. (4.12)

    Hence, the system is dissipative. One can ask about its stability. If the system is strongly stable, that is, E(t)0 as t+, then, using the fact that px,pxt=pxx,pt, we get

    |px,pxt|CE(t)0,ast+. (4.13)

    This indicates that the Balakrishnan-Taylor damping gets less and less effective as t+. In addition, it is clear, from (4.12) and (4.13), that

    E(t)|px,pxt|CE(t)Cδ.

    This subsequently gives

    E(t)E2(t)C2δ. (4.14)

    Integrating the inequality (4.14) over (0,t), it follows that

    E(t)1δC2t+E(0)1,t>0, (4.15)

    which means that the energy is bounded from below polynomially, and consequently the Balakrishnan-Taylor damping term δpx,pxtpxx (alone) is no longer enough to ensure exponential stability. In conclusion, we need to add another damping term, like a strong damping of the form αΔ2pt, to recover exponential decay for system (4.11).

    This paper describes the study of a plate equation that is subject to a Balakrishnan-Taylor damping and a strong damping. This equation models the deformation of the deck of a suspension bridge. First, we have proved the existence of weak solutions and regular ones by using the Faedo-Galerkin approach. Second, by using the multiplier techniques, we proved the exponential decay of energy in our model. We also showed that if the plate equation is subject only to Balakrishnan-Taylor damping, then the exponential stability of this model cannot be reached. In conclusion, the Balakrishnan-Taylor damping is not enough to stabilize (exponentially) the deck. Hence, the need to add another damping.

    Regarding future works, we can change the type of damping by considering, for example, structural damping (of the form Δpt). Also, we can study a coupled Balakrishnan-Taylor plate with only one strong damping.

    The author declares he has not used Artificial Intelligence (AI) tools in the creation of this article.

    This work was supported by Researchers Supporting Project number (RSPD2024R736), King Saud University, Riyadh, Saudi Arabia.

    The author declares that there is no conflict of interest.



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