The goal of this paper was to improve some known results of fixed points by using w-distances and properties of locally symmetric H-transitivity of binary relations. Also, we gave the application of the obtained results for finding the solution of nonlinear matrix equations. Finally, we gave a numerical example to demonstrate the applicability of our results.
Citation: Koti N. V. V. V. Prasad, Vinay Mishra, Zoran D. Mitrović, Ahmad Aloqaily, Nabil Mlaiki. Fixed point results for generalized almost contractions and application to a nonlinear matrix equation[J]. AIMS Mathematics, 2024, 9(5): 12287-12304. doi: 10.3934/math.2024600
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The goal of this paper was to improve some known results of fixed points by using w-distances and properties of locally symmetric H-transitivity of binary relations. Also, we gave the application of the obtained results for finding the solution of nonlinear matrix equations. Finally, we gave a numerical example to demonstrate the applicability of our results.
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)+δ⟨px,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)+(2−μ)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+b∫Ωp2xdxdy. |
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, b∫Ωp2xdxdy 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 δ⟨px,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+k∫l0[∂p(ξ,t)∂ξ]2dξ)pxx+γpxxxxt−μ⟨px,ptx⟩pxx+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+κpt−λΔpt+μΔ2pt−[β+γ∫Ω|∇p|2dx+δ|∫Ω∇p⋅∇ptdx|q−2∫Ω∇p⋅∇ptdx]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,ξ∈R and q≥2.
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 q≥2 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,κ∈R and q≥2; 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|m−2pt=|p|r−2p,m≥2,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={p∈N+:J(p)<d}, |
with
N+={p∈V:I(p)>0}∪{0},I(p)=‖p‖2V+(ap,p)−‖p‖rr, |
J(p)=12‖p‖2V+12(ap,p)−1r‖p‖rrandd=infp∈V∖{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={w∈H2(Ω):w=0on{0,π}×(−d,d)}, |
with the scalar product
(p,q)=∫Ω[ΔpΔq+(1−μ)(2pxyqxy−pxxqyy−pyyqxx)]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 f∈L2(Ω), then there is a unique p∈V such that, for all q∈V, we have
(p,q)=∫Ωfq. | (2.1) |
The function p∈V 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)+(2−μ)pxxy(x,±d)=0. | (2.2) |
Lemma 2.2. [20] Let p∈V and 1≤r<+∞. Then, we have
‖p‖rr≤Cr‖p‖rV, | (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}j≥1 of strictly positive numbers that diverge to +∞, and that the set of eigenfunctions {wj}j≥1 of (2.2) is a complete system in V.
The energy related to (1.1) is given as follows
E(t)=12‖pt(t)‖2+12‖p(t)‖2V−a2‖px(t)‖2+b4‖px(t)‖4, | (2.4) |
which satisfies the following identity
E′(t)=−α‖pt‖2V−δ(12ddt‖px‖2)2≤0, | (2.5) |
This indicates that the energy decreases with time t and E(t)≤E(0),∀t≥0.
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)ds≤CE(t),∀t≥0. |
Then
E(t)≤E(0)e1−tC,∀t≥C. |
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:={w∈H1(Ω):w=0 on {0,π}×(−d,d)},C∞∗(Ω):={w∈C∞(¯Ω):∃ε>0,w(x,y)=0 if x∈[0,ε]∪[π−ε,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 V↪W is compact and the optimal embedding constant satisfies
Λ1:=minw∈V||w||2V||w||2W. |
Lemma 2.7. [17] Assume that 0≤a≤Λ1; then, E(t)≥0.
Proof. Using Remark 2.6, we obtain the following inequality
||w||2W≤Λ−11||w||2V, for all w∈V. | (2.7) |
Since
||px||2≤∫Ω|∇p|2dxdy≤Λ−11||p||2V, |
then we have
−a2||px||2≥−a2Λ−11||p||2V,∀p∈V, |
and consequently
12||p||2V−a2||px||2≥12||p||2V(1−aΛ−11). |
So, if 0≤a≤Λ1 we conclude that 12||p||2V−a2||px||2≥0 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
p∈L∞(0,T;V),pt∈L∞(0,T;L2(Ω))∩L2(0,T;V),andptt∈L2(0,T;H(Ω)), |
constitue a weak solution of (1.1) when
⟨ptt,w⟩2,−2+(p,w)+∫Ω(−a+b‖px‖2+δ⟨px,pxt⟩)pxwxdxdy+α(pt,w)=0,∀w∈V,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 0≤a≤Λ1 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)k≥1 (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)=k∑j=1ajwj(x,y),pk1=k∑j=1bjwj(x,y), |
such that pk0,pk1∈Vk and
pk0→p0inV,andpk1→p1inL2(Ω). | (3.2) |
We are looking for a solution of the form
pk(x,y,t)=k∑j=1cj(t)wj(x,y), | (3.3) |
that solves the following in Vk:
⟨pktt,wj⟩2,−2+(pk,wj)+∫Ω(−a+b‖pkx‖2+δ⟨pkx,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 k≥1, the above problem (3.4) yields a solution pk on [0,tk), where 0<tk≤T. Now, we multiply (3.4) by c′j(t) and sum over j=1,...,k to obtain
ddtEk(t)=−α‖pkt‖2V−δ(12ddt‖pkx‖2)2≤0, | (3.5) |
where
Ek(t)=12‖pkt(t)‖2+12‖pk(t)‖2V−a2‖pkx(t)‖2+b4‖pkx(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)+α∫t0‖pkt‖2Vds+δ∫t0(12ddt‖pkx‖2)2ds≤Ek(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:
‖pk‖2V,‖pkt(t)‖2,∫t0‖pkt‖2V≤C. | (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:
{pk⇀pweaklystarinL∞(0,T;V),pkt⇀ptweaklystarinL∞(0,T;L2(Ω))∩L2(0,T;V),pk⟶pinL2(Q)stronglyanda.e,‖pkx‖2pkxx⇀X1weaklystarinL∞(0,T;L2(Ω)),⟨pkx,pkxt⟩pkxx⇀X2weaklystarinL∞(0,T;L2(Ω)), | (3.10) |
where Q=Ω×(0,T).
Step 2. Here, we will prove that X1=‖px‖2pxx and X2=⟨px,pxt⟩pxx by following the same arguments as in [2,28].
For the first one, the following lemma is required.
Lemma 3.3. Suppose that p,q∈V. We have
⟨‖px‖2pxx−‖qx‖2qxx,p−q⟩≤0. |
Proof. One has
⟨‖px‖2pxx−‖qx‖2qxx,p−q⟩=‖px‖2(⟨px,qx⟩−‖px‖2)+‖qx‖2(⟨px,qx⟩−‖qx‖2)≤‖px‖2(‖px‖‖qx‖−‖px‖2)+‖qx‖2(‖px‖‖qx‖−‖qx‖2)=−(‖px‖−‖qx‖)(‖px‖3−‖qx‖3)≤0. |
Now, let q∈L2(0,T;V). From Lemma 3.3, one obtains that
∫T0⟨‖pkx‖2pkxx−‖qx‖2qxx,pk−q⟩dt≤0. |
By following the same steps as in [28], we derive that
X1=‖px‖2pxx. |
Next, to prove that X2=⟨px,pxt⟩pxx, we note first note that
⟨pkx,pkxt⟩=−⟨pxx,pkt⟩−⟨pk−p,pkxxt⟩. |
It is clear that ⟨pxx,pkt⟩⟶⟨pxx,pt⟩ in L∞(0,T). Besides, from (3.8), we have
∫T0|⟨pk−p,pkxxt⟩|dt≤(∫T0‖pk−p‖2dt)12(∫T0‖pkxxt‖2dt)12≤C(∫T0‖pk−p‖2dt)12. |
Hence ⟨pk−p,pkxxt⟩⟶0 in L1(0,T); thus, we deduce that
⟨pkx,pkxt⟩⟶⟨px,pxt⟩inL1(0,T). | (3.11) |
Now, let φ∈L1(0,T;L2(Ω)). Then
∫T0⟨px,pxt⟩⟨pxx,φ⟩dt=∫T0⟨pkx,pkxt⟩⟨pkxx,φ⟩dt+∫T0[⟨px,pxt⟩−⟨pkx,pkxt⟩]⟨pkxx,φ⟩dt+∫T0⟨pxx−pkxx,⟨px,pxt⟩φ⟩dt. | (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,pxt⟩pxx.
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+∫t0∫Ω(−a+b‖pkx‖2+δ⟨pkx,pkxt⟩)pkxwxdxdyds+α∫t0(pkt,w)ds=∫Ωpk1w,∀w∈V. | (3.13) |
Recall (3.10) and let k→∞; we get
∫Ωptwdxdy−∫Ωp1w=−∫t0(p,w)ds−∫t0∫Ω(−a+b‖px‖2+δ⟨px,pxt⟩)pxwxdxdyds−∫t0α(pt,w)ds. | (3.14) |
This means that (3.14) holds true for any w∈V. Since the terms on the right hand side of (3.14) are absolutely continuous, then (3.14) is differentiable for almost everywhere t≥0. It holds that
⟨ptt,w⟩2,−2+(p,w)+∫Ω(−a+b‖px‖2+δ⟨px,pxt⟩)pxwxdxdy+α(pt,w)=0,∀w∈V. | (3.15) |
Regarding the initial conditions, from (3.10), and by using Lions' lemma [29], we can simply get
pk→pinC([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 ϕ∈C∞0(0,T) and integrate over (0,T); we get the following for any j≤k:
−∫T0∫Ωpkt(t)wϕ′(t)dxdydt=−∫T0(pk,w)ϕ(t)dt−∫T0∫Ω(−a+b‖pkx‖2+δ⟨pkx,pkxt⟩)pkxwxϕ(t)dxdydt−α∫T0(pkt,w)ϕ(t)dt. | (3.18) |
As k→+∞, we have the following for any w∈V and any ϕ∈C∞0(0,T)
−∫T0∫Ωpt(t)wϕ′(t)dxdydt=−∫T0(p,w)ϕ(t)dt−∫T0∫Ω(−a+b‖px‖2+δ⟨px,pxt⟩)pxwxϕ(t)dxdydt−α∫T0(pt,w)ϕ(t)dt, | (3.19) |
which implies that (see [30])
ptt∈L2(0,T;H(Ω)). |
Given that pt∈L2(0,T;L2(Ω)), we conclude that pt∈C(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)+a∫Ωqxxwdxdy−b∫Ω(‖px‖2pxx−‖¯px‖2¯pxx)wdxdy−∫Ωδ(⟨px,pxt⟩pxx−⟨¯px,¯pxt⟩¯pxx)wdxdy+α(qt,w)=0,∀w∈V,q(x,y,0)=qt(x,y,0)=0. | (3.21) |
Then, (3.21) holds true for any w∈C∞0(Ω×(0,T)) by the density it is also valid for any w∈L2(Ω×(0,T)).
If we test (3.21) with qt, we get
ddt{12‖qt‖2+12‖q‖2V}+a∫Ωqxxqtdxdy−b∫Ω(‖px‖2pxx−‖¯px‖2¯pxx)qtdxdy−∫Ωδ(⟨px,pxt⟩pxx−⟨¯px,¯pxt⟩¯pxx)qtdxdy+α‖qt‖2V=0. | (3.22) |
By using Young's inequality, we get
−a∫Ωqxxqtdxdy≤a2‖qt‖2+a2‖qxx‖2≤C(‖qt‖2+‖q‖2V). | (3.23) |
Next, it is easy to see that
‖px‖2pxx−‖¯px‖2¯pxx=‖px‖2qxx−⟨p+¯p,qxx⟩¯pxx, | (3.24) |
and
∫Ωδ(⟨px,pxt⟩pxx−⟨¯px,¯pxt⟩¯pxx)qtdxdy=−δ(⟨qt,¯pxx⟩)2−δ⟨qt,¯pxx⟩⟨pt,qxx⟩−δ⟨pt,pxx⟩⟨qt,qxx⟩. | (3.25) |
Then, using (3.24), we infer that
b∫Ω(‖px‖2pxx−‖¯px‖2¯pxx)qtdxdy≤C‖qxx‖‖qt‖≤C(‖qt‖2+‖q‖2V). | (3.26) |
Besides, from (3.25), one derives the following:
∫Ωδ(⟨px,pxt⟩pxx−⟨¯px,¯pxt⟩¯pxx)qtdxdy≤−δ(⟨qt,¯pxx⟩)2+C‖qxx‖‖qt‖≤−δ(⟨qt,¯pxx⟩)2+C(‖qt‖2+‖q‖2V). | (3.27) |
By using (3.23), (3.26), and (3.27) we can deduce that
ddt{12‖qt‖2L2(Ω)+12‖q‖2V}+δ(⟨qt,¯pxx⟩)2+α‖qt‖2V≤C(‖qt‖2+‖q‖2V). | (3.28) |
By Gronwall's inequality, we obtain
‖qt‖2+‖q‖2V≤CeCt(‖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 0≤a≤Λ1 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
p∈L∞(0,T;X),pt∈L∞(0,T;V)∩L2(0,T;X),andptt∈L2(0,T;L2(Ω)), |
and
ptt+Δ2p−(ϕ(p)+δ⟨px,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
pk0→p0inX,andpk1→p1inV. |
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(‖pkt‖2V+‖Δ2pk‖2)+α‖Δ2pkt‖2=−a⟨pkxx,Δ2pkt⟩+b‖pkx‖2⟨pkxx,Δ2pkt⟩+δ⟨pkx,pkxt⟩⟨pkxx,Δ2pkt⟩. |
Therefore, by integrating by parts over (0,t), it follows that
‖pkt‖2V+‖Δ2pk‖2+2α∫t0‖Δ2pkt‖2ds≤C1+C2∫t0|⟨pkxx,Δ2pkt⟩|ds≤C1+C2∫t0‖pkxx‖2ds+α∫t0‖Δ2pkt‖2ds. |
By using (3.9), we derive that
‖pkt‖2V,‖Δ2pk‖2,and∫t0‖Δ2pkt‖2ds≤C. |
We proceed as in Theorem (3.2) to prove the existence of a unique p∈L∞(0,T;X) that satisfies (3.17), (3.20), (3.29), and pt∈L∞(0,T;V)∩L2(0,T;X). It follows from (3.29) that ptt∈L2(0,T;L2(Ω)).
This section's major result is as follows:
Theorem 4.1. Let 0≤a<Λ1. Then, there are two constants K>0 and λ>0 such that the energy defined in (2.4) verifies
E(t)≤Ke−λt,∀t≥0. | (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
∫Ts∫Ω(pttp+pΔ2p−ϕ(p)pxxp−δ⟨px,pxt⟩pxxp+αpΔ2pt)dxdydt=0. | (4.2) |
Using Lemma 2.1 and integration by parts, we obtain
∫Ts∫Ω(ptp)tdxdydt−∫Ts∫Ωp2tdxdydt+∫Ts‖p‖2Vdt+∫Ts∫Ωϕ(p)p2xdxdydt+∫Ts∫Ωδ⟨px,pxt⟩p2xdxdydt+α∫Ts(p,pt)dt=0. | (4.3) |
This yields
∫TsE(t)dt+∫Ts∫Ω(ptp)t−32∫Ts∫Ωp2t+12∫Ts‖p‖2V−a2∫Ts‖px‖2+3b4∫Ts‖px‖4+δ∫Ts∫Ω⟨px,pxt⟩p2xdxdydt+α∫Ts(p,pt)dt=0. | (4.4) |
Then, we obtain
∫TsE(t)dt≤−∫Ts∫Ω(ptp)t+32∫Ts∫Ωp2t+a2∫Ts‖px‖2−δ∫Ts∫Ω⟨px,pxt⟩p2xdxdt−α∫Ts(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
|−∫Ts∫Ω(ptp)t|≤|∫Ωpt(s)p(s)|+|∫Ωpt(T)p(T)|≤12∫Ωp2t(s)+12∫Ωp2t(T)+12∫Ωp2(s)+12∫Ωp2(T)≤E(s)+E(T)+C‖p(s)‖2V+C‖p(T)‖2V≤CE(s), | (4.6) |
where C is a generic positive constant. For the second term, thanks to Lemma 2.2 we have
32∫Ts∫Ωp2t≤C∫Ts‖pt‖2V≤Cα∫Ts(−E′(t))dt≤CαE(s). | (4.7) |
The third term on the right hand side of (4.5) may be estimated as follows:
a2∫Ts‖px‖2≤aΛ−112∫Ts‖p‖2V≤aΛ−11∫TsE(t)dt. | (4.8) |
Thanks to Young's inequality and (2.5), we deduce, for any ε>0, that
|−δ∫Ts∫Ω⟨px,pxt⟩p2xdxdt|≤Cε∫Ts(ddt‖px‖2)2dt+δε4∫Ts‖px‖4dt≤Cε∫Ts(−E′(t))dt+δεb∫TsE(t)dt≤CεE(s)+δεb∫TsE(t)dt, | (4.9) |
and
|−α∫Ts(p,pt)dt|≤α∫Ts‖p‖V‖pt‖V≤Cε∫Ts‖pt‖2Vdt+αε2∫Ts‖p‖2Vdt≤CεE(s)+αε∫TsE(t)dt. | (4.10) |
By inserting (4.6)–(4.10) into (4.5), and choosing ε such that 1−aΛ−11−(δb+α)ε>0, we conclude that there exists a positive constant C1 satisfying
∫TsE(t)dt≤C1E(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+Δ2p−δ⟨px,pxt⟩pxx=0,inΩ×(0,+∞). | (4.11) |
The corresponding energy for system (4.11) is given by
E(t)=12‖pt‖2+12‖p‖2V, |
which satisfies
E′(t)=−δ[⟨px,pxt⟩]2≤0,∀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⟩|C≥√−E′(t)C√δ. |
This subsequently gives
−E′(t)E−2(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 −δ⟨px,pxt⟩pxx (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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