Citation: Kasia A. Pawelek, Sarah Tobin, Christopher Griffin, Dominik Ochocinski, Elissa J. Schwartz, Sara Y. Del Valle. Impact of A Waning Vaccine and Altered Behavior on the Spread of Influenza[J]. AIMS Medical Science, 2017, 4(2): 217-232. doi: 10.3934/medsci.2017.2.217
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The importance of bridges is undeniable and their presence in human daily life goes back a long time. With the presence of the bridges, road and railway traffic runs without any interruption over rivers and hazardous areas, time and fuel are saved, congestion on roads is minimized, distances between places are reduced, and many accidents have been avoided, as the bridges have reduced the number of bends and zig-zags in roads. As a result, many economies have grown and many societies have become connected. However, bridges have brought some challenges, such as collapse and instability due to natural hazards such as wind, earthquakes, etc. To overcome these difficulties, engineers and scientists have made efforts to find the best designs and possible models. Our aim in this work is to investigate the following plate problem
{utt+Δ2u+α(t)g(ut)=u|u|β,inΩ×(0,T),u(0,y,t)=uxx(0,y,t)=u(π,y,t)=uxx(π,y,t)=0,(y,t)∈(−d,d)×(0,T),uyy(x,±d,t)+σuxx(x,±d,t)=0,(x,t)∈(0,π)×(0,T),uyyy(x,±d,t)+(2−σ)uxxy(x,±d,t)=0,(x,t)∈(0,π)×(0,T),u(x,y,0)=u0(x,y),ut(x,y,0)=u1(x,y), in Ω×(0,T), | (1.1) |
where Ω=(0,π)×(−d,d), d,β>0, g:R→R and α:[0,+∞)→(0,+∞) is a nonincreasing differentiable function, u is the vertical displacement of the bridge and σ is the Poisson ratio. This is a weakly damped nonlinear suspension-bridge problem, in which the damping is modulated by a time dependent-coefficient α(t). Firstly, we prove the local existence using the Faedo-Gherkin method and Banach fixed point theorem. Secondly, we prove the global existence by using the well-depth method. Finally, we establish an explicit and general decay result, depending on g and α, for which the exponential and polynomial decay rate estimates are only special cases. The proof is based on the multiplier method and makes use of some properties of convex functions, including the use of the general Young inequality and Jensen's inequality.
The famous report by Claude-Louis Navier [1] was the only mathematical treatise of suspension bridges for several decades. Another milestone theoretical contribution was the monograph by Melan [2]. After the Tacoma collapse, engineers felt the necessary to introduce the time variable in mathematical models and equations in order to attempt explanations of what had occurred. As a matter of fact, in Appendix VI of the Federal Report [3], a model of inextensible cables is derived and the linearized Melan equation was obtained. Other important contributions were the works by Smith-Vincent and the analysis of vibrations in suspension bridges presented by Bleich-McCullough-Rosecrans-Vincent [4]. In all these historical references, the bridge was modelled linearly as a beam suspended to a cable. Hence, all the equations were linear. Mathematicians have not shown any interest in suspension bridges until recently. McKenna, in 1987, introduced the first nonlinear models to study them from a theoretical point of view, and he was followed by several other mathematicians (see [5,6]). McKenna's main idea was to consider the slackening of the hangers as a nonlinear phenomenon, a statement which is by now well-known also among engineers [7,8]. The slackening phenomenon was analyzed in various complex beam models by several authors (see [9,10,11]). Motivated by the wonderful book of Rocard [12], where it was pointed out that the correct way to model a suspension bridge is through a thin plate, Ferrero-Gazzola [13] introduced the following hyperbolic problem:
{utt(x,y,t)+ηut+Δ2u(x,y,t)+h(x,y,u)=f,inΩ×R+,u(0,y,t)=uxx(0,y,t)=u(π,y,t)=uxx(π,y,t)=0,(y,t)∈(−ℓ,ℓ)×R+,uyy(x,±ℓ,t)+σuxx(x,±ℓ,t)=0,(x,t)∈(0,π)×R+,uyyy(x,±ℓ,t)+(2−σ)uxxy(x,±ℓ,t)=0,(x,t)∈(0,π)×R+,u(x,y,0)=u0(x,y),ut(x,y,0)=u1(x,y), in Ω×R+, | (2.1) |
where Ω=(0,π)×(−ℓ,ℓ) is a planar rectangular plate, σ is the well-known Poisson ratio, η is the damping coefficient, h is the nonlinear restoring force of the hangers and f is an external force. After the appearance of the above model, many mathematicians showed interest in investigating variants of it, using different kinds of damping with the aim to obtain stability of the bridge modeled through the above problem. Messaoudi [14] considered the following nonlinear Petrovsky equation
utt+Δ2u+aut|ut|m−2=bu|u|p−2, | (2.2) |
and proved the existence of a local weak solution, showed that this solution is global if m≥p and blows up in finite time if p>m and the energy is negative. Wang [15] considered the equation
utt+δut+Δ2u+au=u|u|p−2, | (2.3) |
where a=a(x,y,t) together with the above initial and boundary conditions. After showing the uniqueness and existence of local solutions, he gave sufficient conditions for global existence and finite-time blow-up of solutions. Mukiawa [16] considered a plate equation modeling a suspension bridge with weak damping and hanger restoring force. He proved the well-posedness and established an explicit and general decay result without putting restrictive growth conditions on the frictional damping term. Messaoudi and Mukiawa [17] studied problem (2.3), where the linear frictional damping was replaced by nonlinear frictional damping and established the existence of a global weak solution and proved exponential and polynomial stability results. Audu et al. [18] considered a plate equation as a model for a suspension bridge with a general nonlinear internal feedback and time-varying weight. Under some conditions on the feedback and the coefficient functions, the authors established a general decay estimate. For more results related to the existence of work on similar problems, we mention the work of Xu et al. [19], in which they proved the local existence of a weak solution by the Galerkin method and the global existence by the potential well method. He et al. [20] considered the following Kirchhoff type equation
−(a+b∫Ω|∇u|2dx)Δu=f(u)+h,in Ω, | (2.4) |
where Ω⊆R3 is a bounded domain or Ω=R3, 0≤h∈L2(Ω) and f∈C(R,R). The authors proved the existence of at least one or two positive solutions by using the monotonicity trick, and nonexistence criterion is also established by virtue of the corresponding Pohoaev identity. Recently, Wang et al. [21] considered the fractional Rayleigh-Stokes problem where the nonlinearity term satisfied certain critical conditions and proved the local existence, uniqueness and continuous dependence upon the initial data of ε-regular mild solutions. More results in this direction can be found in [22,23,24,25,26,27]. The paper is organized as follows. In Section 3, we present some preliminaries and essential lemmas. We prove the local existence in Section 4 and the global existence in Section 5. The statement and the proof of our stability result will be given in Section 6.
In this section, we present some material needed in the proofs of our results. First, we introduce the following space
H2∗(Ω)={w∈H2(Ω):w=0on{0,π}×(−d,d)}, | (3.1) |
together with the inner product
(u,v)H2∗=∫Ω(ΔuΔv+(1−σ)(2uxyvxy−uxxvyy−uyyvxx))dx. | (3.2) |
It is well known that (H2∗(Ω),(⋅,⋅)H2∗) is a Hilbert space, and the norm ‖.‖2H2∗ is equivalent to the usual H2, see [13]. Throughout this paper, c is used to denote a generic positive constant.
Lemma 3.1. [15]Let u∈H2∗(Ω) and assume that 1≤p<∞, then, there exists a positive constant Ce=Ce(Ω,p)>0 such that
‖u‖p≤Ce‖u‖H2∗(Ω). |
Lemma 3.2. (Jensen's inequality)Let ψ:[a,b]⟶R be a convex function. Assume that the functions f:(0,L)⟶[a,b] and r:(0,L)⟶R are integrable such that r(x)≥0, for any x∈(0,L) and ∫L0r(x)dx=k>0. Then,
ψ(1k∫L0f(x)r(x)dx)≤1k∫L0ψ(f(x))r(x)dx. | (3.3) |
We consider the following hypotheses:
(H1). The function g:R→R is nondecreasing C0 function satisfying for ε,c1,c2>0,
c1|s|≤|g(s)|≤c2|s|, if |s|≥ε,|s|2+g2(s)≤G−1(sg(s)), if |s|≤ε, | (3.4) |
where G:R+→R+ is a C1 function which is linear or strictly increasing and strictly convex C2 function on [0,ε] with G(0)=0 and G′(0)=0. In addition, the function g satisfies, for ϑ>0,
(g(s1)−g(s2))(s1−s2)≥ϑ|s1−s2|2. | (3.5) |
(H2). The function α:R+→R+ is a nonincreasing differentiable function such that ∫∞0α(t)dt=∞.
Remark 3.3. Hypothesis (H1) implies that sg(s)>0, for all s≠0 and it was introduced and employed by Lasiecka and Tataru [28]. It was shown there that the monotonicity and continuity of g guarantee the existence of the function G with the properties stated in (H1).
Remark 3.4. As in [28], we use Condition (3.5) to prove the uniqueness of the solution.
The following lemmas will be of essential use in establishing our main results.
Lemma 3.5. [29] Let E:R+→R+ be a nonincreasing function and γ:R+→R+ be a strictlyincreasing C1-function, with γ(t)→+∞ as t→+∞. Assume that there exists c>0 such that
∞∫Sγ′(t)E(t)dt≤cE(S)1≤S<+∞. |
Then there exist positive constants k and ω such that
E(t)≤ke−ωγ(t). |
Lemma 3.6. [30] Let E:R+→R+ be a differentiable and nonincreasing function and χ:R+→R+ be a convex and increasing function such that χ(0)=0. Assume that
∫+∞sχ(E(t))dt≤E(s),∀s≥0. | (3.6) |
Then, E satisfies the following estimate
E(t)≤ψ−1(h(t)+ψ(E(0))),∀t≥0, | (3.7) |
where ψ(t)=∫1t1χ(s)ds, and
{h(t)=0,0≤t≤E(0)χ(E(0)),h−1(t)=t+ψ−1(t+ψ(E(0)))χ(ψ−1(t+ψ(E(0)))),t>0. |
In this section, we state and prove the local existence of weak solutions of problem (1.1). Similar results can be found in [31,32]. To this end, we consider the following problem
{utt(x,y,t)+Δ2u(x,y,t)+α(t)g(ut)=f(x,t),inΩ×(0,T),u(0,y,t)=uxx(0,y,t)=u(π,y,t)=uxx(π,y,t)=0,(y,t)∈(−d,d)×(0,T),uyy(x,±d,t)+σuxx(x,±d,t)=0,(x,t)∈(0,π)×(0,T),uyyy(x,±d,t)+(2−σ)uxxy(x,±d,t)=0,(x,t)∈(0,π)×(0,T),u(x,y,0)=u0(x,y),ut(x,y,0)=u1(x,y), in Ω×(0,T), | (4.1) |
where f∈L2(Ω×(0,T)) and (u0,u1)∈H2∗(Ω)×L2(Ω). Then, we prove the following theorem:
Theorem 4.1. Let (u0,u1)∈H2∗(Ω)×L2(Ω). Assume that(H1) and (H2) hold. Then, problem (4.1) has a unique local weaksolution
u∈L∞([0,T),H2∗(Ω)),ut∈L∞([0,T),L2(Ω)),utt∈L∞([0,T),H(Ω)), |
where H(Ω) is the dual space of H2∗(Ω).
Proof. Uniqueness: Suppose that (4.1) has two weak solutions (u,v). Then, w=u−v satisfies
{wtt(x,y,t)+Δ2w(x,y,t)+α(t)g(ut)−α(t)g(vt)=0,inΩ×(0,T),w(0,y,t)=wxx(0,y,t)=w(π,y,t)=wxx(π,y,t)=0,(y,t)∈(−d,d)×(0,T),wyy(x,±d,t)+σwxx(x,±d,t)=0,(x,t)∈(0,π)×(0,T),wyyy(x,±d,t)+(2−σ)wxxy(x,±d,t)=0,(x,t)∈(0,π)×(0,T),w(x,y,0)=wt(x,y,0)=0, in Ω×(0,T). | (4.2) |
Multiplying (4.2) by wt and integrating over (0,t), we get
12ddt[∫Ω(w2t+|Δw|2)dx]+α(t)∫Ω(g(ut)−g(vt))(ut−vt)dx=0. | (4.3) |
Integrating (4.3) over (0,t), we obtain
∫Ω(w2t+|Δw|2)dx+2α(t)∫t0∫Ω(g(ut)−g(vt))(ut−vt)dxds=0. | (4.4) |
Using Condition (3.5) and (H2), for a.e.x∈Ω, we have
∫Ω(w2t+|Δw|2dx)=0, | (4.5) |
We conclude u=v=0 on Ω×(0,T), which proves the uniqueness of the solution of problem (4.1). Existence: To prove the existence of the solution for problem (4.1), we use the Faedo-Galerkin method as follows: First, we consider {vj}∞j=1 an orthonormal basis of H2∗(Ω) and define, for all k≥1, a sequence vk in Vk=span{v1,v2,...,vk}⊂H2∗(Ω), given by
uk(x,t)=Σkj=1aj(t)vj(x), |
for all x∈Ω and t∈(0,T) and satisfies the following approximate problem
{∫Ωuktt(x,t)vjdx+∫ΩΔuk(x,t)Δvjdx+α(t)∫Ωg(ukt)vj=∫Ωf(x,t)vjdx,inΩ×(0,T),uk(x,y,0)=uk0(x,y),ukt(x,y,0)=uk1(x,y), in Ω×(0,T), | (4.6) |
for all j=1,2,...,k,
uk(0)=uk0=Σki=1⟨u0,vi⟩vi, ukt(0)=uk1=Σki=1⟨u1,vi⟩vi, | (4.7) |
such that
uk0⟶u0∈H2∗(Ω),uk1⟶u1∈L2(Ω). | (4.8) |
For any k≥1, problem (4.6) generates a system of k nonlinear ordinary differential equations. The ODE's standard existence theory assures the existence of a unique local solution uk for problem (4.6) on [0,Tk), with 0<Tk≤T. Next, we have to show, by a priori estimates, that Tk=T,∀k≥1. Now, multiplying (4.6) by a′j(t), using Green's formula and the boundary conditions, and then summing each result over j we obtain, for all 0<t≤Tk,
12ddt[∫Ω(|ukt|2+(Δuk)2)dx]+α(t)∫Ωuktg(ukt)dx=∫Ωf(x,t)ukt(x,t)dx. | (4.9) |
Then, integrating (4.9) over (0,t) leads to
12∫Ω(|ukt|2+|Δuk|2)dx+∫t0∫Ωα(s)uktg(ukt)dxds=12∫Ω(|uk1|2+|Δuk0|2)dx+∫t0∫Ωf(x,t)ukt(x,t)dxds. | (4.10) |
From the convergence (4.8), using the fact that f∈L2(Ω×(0,T)), and exploiting Young's inequality, then (4.10) becomes, for some C>0, and for any t∈[0,tk)
12∫Ω[|ukt|2+|Δuk|2dx]+∫t0∫Ωα(s)uktg(ukt)dxds≤12∫Ω[|uk1|2+|Δuk0|2]dx+ε∫t0∫Ω|ukt|2dxds+Cε∫t0∫Ω|f(x,s)|2dxds≤Cε+εsup(0,Tk)∫Ω|ukt|2dx. | (4.11) |
Therefore, we obtain
12sup(0,Tk)∫Ω|ukt|2dx+12sup(0,Tk)∫Ω|Δuk|2dx+12sup(0,Tk)∫tk0∫Ωα(s)ukt(x,s)g(ukt(x,s))dxds≤Cε+εsup(0,Tk)∫Ω|ukt|2dx. | (4.12) |
Choosing ε=14, estimate (4.12) yields, for all Tk≤T and C>0,
sup(0,Tk)∫Ω|ukt|2dx+sup(0,Tk)∫Ω|Δuk|2dx+sup(0,Tk)∫tk0∫Ωα(s)ukt(x,s)g(ukt(x,s))dxds≤C. | (4.13) |
Consequently, the solution uk can be extended to (0,T), for any k≥1. In addition, we have
(uk) is bounded in L∞((0,T),H2∗(Ω))and(ukt) is bounded in L∞((0,T),L2(Ω)). |
Therefore, we can extract a subsequence, denoted by (uℓ) such that, when ℓ→∞, we have
uℓ→u weakly * in L∞((0,T),H2∗(Ω))anduℓt→ut weakly * in L∞((0,T),L2(Ω)). |
Next, we prove that g(uℓt) is bounded in L2((0,T);L2(Ω)). For this purpose, we consider two cases:
Case 1. G is linear on [0,ε]. Then using (H1) and Young's inequality, we get
∫Ωg2(uℓt)dx≤c∫Ωuℓtg(uℓt)dx−∫Ω|uℓt|2dx≤c4δ0∫Ω|uℓt|2dx+δ0∫Ωg2(uℓt)dx, | (4.14) |
for a suitable choice of δ0 and using the fact that uℓt is bounded in L2((0,T),L2(Ω)), we obtain
∫T0∫Ωg2(uℓt)dxdt≤c. | (4.15) |
Case 2. G is nonlinear. Let 0<ε1≤ε such that
sg(s)≤min{ε,G(ε)} for all |s|≤ε1. | (4.16) |
Then, one can show that
{s2+g2(s)≤G−1(sg(s))for all|s|≤ε1c′1|s|≤|g(s)|≤c′2|s|for all|s|≥ε1. | (4.17) |
Define the following sets
Ω1={x∈Ω:|uℓt|≤ε1},andΩ2={x∈Ω:|uℓt|>ε1}. | (4.18) |
Then, using (4.17) and (4.18) leads for some c′2>0,
∫Ωg2(uℓt)dx=∫Ω2g2(uℓt)dx+∫Ω1g2(uℓt)dx≤c′2∫Ω2|uℓt|2dx+∫Ω1(|uℓ|2t+g2(uℓt))dx−∫Ω1|uℓt|2dx≤c′2∫Ω2|uℓt|2dx+∫Ω1G−1(uℓtg(uℓt))dx. | (4.19) |
Let
Jℓ(t):=∫Ω1uℓtg(uℓt)dx, |
Eℓ(t)=12(‖uℓt‖22+‖uℓ‖2H2∗(Ω))−1β+2‖uℓ‖β+2β+2, | (4.20) |
and
(Eℓ)′(t)=−α(t)∫Ωuℓtg(uℓt)dx≤0. | (4.21) |
Using (4.19) and Jensen's inequality, we obtain
∫Ωg2(uℓt)dx≤c∫Ω|uℓt|2dx+G−1(Jℓ(t))=c∫Ω|uℓt|2dx+G′(ε0Eℓ(t)Eℓ(0))G′(ε0Eℓ(t)Eℓ(0))G−1(Jℓ(t)). | (4.22) |
Using the convexity of G (G′ is increasing), we obtain for t∈(0,T),
G′(ε0Eℓ(t)Eℓ(0))≥G′(ε0Eℓ(T)Eℓ(0))=c. |
Let G∗ be the convex conjugate of G in the sense of Young (see [33], pp. 61–64), then, for s∈(0,G′(ε)],
G∗(s)=s(G′)−1(s)−G[(G′)−1(s)]≤s(G′)−1(s). | (4.23) |
Using the general Young inequality
AB≤G∗(A)+G(B),ifA∈(0,G′(ε)],B∈(0,ε], |
for
A=G′(ε0Eℓ(t)Eℓ(0))andB=G−1(Jℓ(t)), |
and using the fact that Eℓ(t)≤Eℓ(0), we get
∫Ωg2(uℓt)dx≤c∫Ω|uℓt|2dx+cε0Eℓ(t)Eℓ(0)G′(ε0Eℓ(t)El(0))−C(Eℓ)′(t)≤c∫Ω|uℓt|2dx+cε0Eℓ(t)Eℓ(0)G′(ε0Eℓ(t)Eℓ(0))−C(Eℓ)′(t)≤c∫Ω|uℓt|2dx+c−C(Eℓ)′(t). | (4.24) |
Integrating (4.24) over (0,T), we obtain
∫T0∫Ωg2(uℓt)dxdt≤c∫T0∫Ω|uℓt|2dxdt+cT−C(Eℓ(T)−Eℓ(0)). | (4.25) |
Using (4.21) and the fact that uℓt is bounded in L2((0,T);L2(Ω)), we conclude that g(uℓt) is bounded in L2((0,T);L2(Ω)). So, we find, up to a subsequence, that
g(uℓt)⇀χ in L2((0,T);L2(Ω)). | (4.26) |
Now, we have to show that χ=g(ut). In (4.6), we use uℓ instead of uk and then integrate over (0,t) to get
∫Ωuℓtvjdx−∫Ωuℓ1vjdx+∫t0∫ΩΔuℓΔvjdxds+∫t0∫Ωα(s)g(uℓt)vjdxds=∫t0∫Ωfvjdxds,j<ℓ. | (4.27) |
As ℓ→+∞, we easily check that
∫Ωutvjdx−∫Ωu1vjdx+∫t0∫ΩΔuΔvjdxds+∫t0∫Ωα(s)χvjdxds=∫t0∫Ωfvjdxds,j≥1. | (4.28) |
Hence, for v∈H2∗(Ω), we have
∫Ωutvdx−∫Ωu1vdx+∫t0∫ΩΔuΔvdxds+∫t0∫Ωα(s)χvdxds=∫t0∫Ωfvdxds. | (4.29) |
Since all terms define absolute continuous functions, we get, for a.e.t∈[0,T] and for v∈H2∗(Ω), the following
ddt∫Ωutvdx+∫ΩΔuΔvdx+∫Ωα(t)χvdx=∫Ωfvdxds. | (4.30) |
This implies that
utt+Δ2u+α(t)χ=f,inD′(Ω×(0,T)). | (4.31) |
Using (H1), we see that
Xℓ:=∫T0∫Ωα(s)(uℓt−v)(g(uℓt)−g(v))dxdt≥0,v∈L2((0,T);L2(Ω)). | (4.32) |
So, by using (4.6) and replacing uk by uℓ, we get
Xℓ=∫T0∫Ωfuℓtdxdt+12∫Ω(|uℓt|2+|Δuℓ|2)dx−12∫Ω|uℓt(x,T)|2dx−12∫Ω|Δuℓt(x,T)|2dx−∫T0∫Ωα(t)g(uℓt)vdxdt−∫T0∫Ωα(t)g(v)(uℓt−v)dxdt. | (4.33) |
Taking ℓ→+∞, we obtain
0≤limsupℓXℓ≤∫T0∫Ωfutdxdt+12∫Ω(|u1|2+|Δu0|2)dx−12∫Ω|ut(x,T)|2dx−12∫Ω|Δut(x,T)|2dx−∫T0∫Ωα(t)χvdxdt−∫T0∫Ωα(t)g(v)(ut−v)dxdt. | (4.34) |
Replacing v by ut in (4.30) and integrating over (0,T), we obtain
∫T0∫Ωfutdxdt=12∫Ω(|ut(x,T)|2dx+|Δu(x,T)|2)dx−12∫Ω|u1|2dx−12∫Ω|Δu0|2dx+∫T0∫Ωα(t)χutdxdt. | (4.35) |
Adding of (4.34) and (4.35), we get
0≤limsupℓXℓ≤∫T0∫Ωα(t)χutdxdt−∫T0∫Ωα(t)χvdxdt−∫T0∫Ωα(t)g(v)(ut−v)dxdt. | (4.36) |
This gives
∫T0∫Ωα(t)(χ−g(v))(ut−v)dxdt≥0,v∈L2((0,T),L2(Ω)). | (4.37) |
Hence,
∫T0∫Ωα(t)(χ−g(v))(ut−v)dxdt≥0,v∈L2(Ω×(0,T)). | (4.38) |
Let v=λw+ut, where λ>0 and w∈L2(Ω×(0,T)). Then, we get
−λ∫T0∫Ωα(t)(χ−g(λw+ut))wdxdt≥0,w∈L2(Ω×(0,T)). | (4.39) |
For λ>0, we have
λ∫T0∫Ωα(t)(χ−g(λw+ut))wdxdt≤0,w∈L2(Ω×(0,T)). | (4.40) |
As λ→0 and using the continuity of g with respect of λ, we get
λ∫T0∫Ωα(t)(χ−g(ut))wdxdt≤0,w∈L2(Ω×(0,T)). | (4.41) |
Similarly, for λ<0, we get
λ∫T0∫Ωα(t)(χ−g(ut))wdt≥0,w∈L2(Ω×(0,T)). | (4.42) |
This implies that χ=g(ut). Hence, (4.30) becomes
∫Ω(uttv+ΔuΔv+α(t)g(ut)v)dx=∫Ωfvdx,v∈L2((0,T);H2∗(Ω)). | (4.43) |
which gives
utt+Δ2u+α(t)g(ut)=f,inD′(Ω×(0,T)). | (4.44) |
To handle the initial conditions of problem (4.1), we first note that
uℓ⇀uweakly * inL∞(0,T;H2∗(Ω))uℓt⇀utweakly * inL∞(0,T;L2(Ω)). | (4.45) |
Thus, using Lion's Lemma and (4.6), we easily obtain uℓ→u∈C([0,T];L2(Ω)). Therefore, uℓ(x,0) makes sense and uℓ(x,0)→u(x,0)∈L2(Ω). Also, we see that
uℓ(x,0)=uℓ0→u0(x)∈H2∗(Ω). |
Hence, u(x,0)=u0(x). As in [34], let ϕ∈C∞0(0,T), and replacing uk by uℓ, we obtain from (4.6) and for any j≤ℓ
{−∫T0∫Ωuℓt(x,t)vj(x)ϕ′(t)dxdt=−∫T0∫ΩΔuℓ(x,t)Δvj(x)ϕ(t)dxdt−∫T0∫Ωα(t)g(uℓt)vj(x)ϕ(t)dxdt+∫T0∫Ωf(x,t)vj(x)ϕ(t)dxdt. | (4.46) |
As ℓ→+∞, we have for any ϕ∈C∞0((0,T)),
{−∫T0∫Ωut(x,t)vj(x)ϕ′(t)dxdt=−∫T0∫ΩΔu(x,t)Δvj(x)ϕ(t)dxdt−∫T0∫Ωα(t)g(ut)vj(x)ϕ(t)dxdt+∫T0∫Ωf(x,t)vj(x)ϕ(t)dxdt, | (4.47) |
for all j≥1. This implies that
−∫T0∫Ωut(x,t)v(x)ϕ′(t)dxdt=∫T0∫Ω[−Δ2u(x,t)−α(t)g(ut)+f(x,t)]v(x)ϕ(t)dxdt, | (4.48) |
for all v∈H2∗(Ω). This means that utt∈L∞((0,T);H(Ω)) and u solves the equation
utt+Δ2u+α(t)g(ut)=f. | (4.49) |
Thus
ut∈L∞((0,T);L2(Ω)),utt∈L∞((0,T);H(Ω)). |
Consequently, ut∈C((0,T);H(Ω)). So, uℓt(x,0) makes sense and follows that
uℓt(x,0)→ut(x,0) in H(Ω) |
and since
uℓt(x,0)=uℓ1(x)→u1(x) in L2(Ω), |
then
ut(x,0)=u1(x). |
This ends the proof of Theorem 4.1.
Now, we proceed to establish the local existence result for problem (1.1).
Theorem 4.2. Let (u0,u1)∈H2∗(Ω)×L2(Ω) begiven. Then problem (1.1) has a unique local weak solution
u∈L∞([0,T),H2∗(Ω)),ut∈L∞([0,T),L2(Ω)),utt∈L∞([0,T),H(Ω)). |
Remark 4.3. In this remark, we point out four cases regarding the solution of problem (1.1):
1) If β=0, g is linear and (u0,u1)∈(H4(Ω)∩H2∗(Ω))×H2∗(Ω), then problem (1.1) has a unique classical solution
u∈C2([0,T),H2∗(Ω)),ut∈C1([0,T),L2(Ω)),utt∈C([0,T),H(Ω)). |
2) If β=0, g is linear and (u0,u1)∈H2∗(Ω)×L2(Ω), then problem (1.1) has a unique weak solution
u∈C1([0,T),H2∗(Ω)),ut∈C([0,T),L2(Ω)),utt∈L∞([0,T),H(Ω)). |
3) If β>0 or g is nonlinear and (u0,u1)∈H2∗(Ω)×L2(Ω), then problem (1.1) has a unique weak solution
u∈L∞([0,T),H2∗(Ω)),ut∈L∞([0,T),L2(Ω)),utt∈L∞([0,T),H(Ω)). |
4) If β>0 or g is nonlinear and (u0,u1)∈(H4(Ω)∩H2∗(Ω))×H2∗(Ω), then problem (1.1) has a unique strong solution
u∈L∞([0,T),H4(Ω)∩H2∗(Ω)),ut∈L∞(([0,T),H2∗(Ω)),utt∈L∞([0,T),L2(Ω)). |
Proof. To prove Theorem 4.2, we first let v∈L∞([0,T),H2∗(Ω)) and ˜f(v)=|v|βv. Then, by the embedding Lemma 3.1, we have
||˜f(v)||22=∫Ω|v|2(β+1)dx<+∞. | (4.50) |
Hence,
˜f(v)∈L∞([0,T),L2(Ω))⊂L2(Ω×(0,T)). |
Therefore, for each v∈L∞([0,T),H2∗(Ω)), there exists a unique solution
u∈L∞([0,T),H2∗(Ω)),ut∈L∞([0,T),L2(Ω)) |
satisfying the following nonlinear problem
{utt+Δ2u+α(t)g(ut)=˜f(v),inΩ×(0,T),u(0,y,t)=uxx(0,y,t)=u(π,y,t)=uxx(π,y,t)=0,(y,t)∈(−d,d)×(0,T),uyy(x,±d,t)+σuxx(x,±d,t)=0,(x,t)∈(0,π)×(0,T),uyyy(x,±d,t)+(2−σ)uxxy(x,±d,t)=0,(x,t)∈(0,π)×(0,T),u(x,y,0)=u0(x,y),ut(x,y,0)=u1(x,y), in Ω×(0,T), | (4.51) |
Now, let
WT={w∈L∞((0,T),H2∗(Ω))/wt∈L∞((0,T),L2(Ω))}, |
and define the map K:WT⟶WT by K(v)=u. We note that WT is a Banach space with respect to the following norm
||w||WT=||w||L∞((0,T),H2∗(Ω))+||wt||L∞((0,T),L2(Ω)). |
Multiply (4.51) by ut and integrate over Ω×(0,t), we get for all t≤T,
12∫Ωu2tdx+12∫Ω|Δu|2dx+∫t0∫Ωα(s)utg(ut)dxds=12∫Ωu21dx+12∫Ω|Δu0|2dx+∫t0∫Ω|v|βvutdxds. | (4.52) |
Using Young's inequality and the embedding Lemma 3.1, we have
∫Ω|v|βvutdx≤ε4∫Ωu2tdx+4ε∫Ω|v|2(β+1)dx≤ε4∫Ωu2tdx+4Ceε||v||2(β+1)H2∗. | (4.53) |
Thus, (4.52) becomes
12∫Ωu2tdx+12∫Ω|Δu|2dx≤λ0+εT4sup(0,T)∫Ωu2tdx+Ceε∫T0||v||2(β+1)H2∗dt, | (4.54) |
where λ0=12||u1||22+12||Δu0||22 and Ce is the embedding constant. Choosing ε such that εT2=14, we get
||u||2WT≤λ+Tb||v||2(β+1)WT. |
Suppose that ||v||WT≤M and for M2>λ and T≤T0<M2−λbM2(β+1), we conclude that
||u||2WT≤λ+TbM2(β+1)≤M2. |
Therefore, we deduce that K:B⟶B, where
B={w∈L∞((0,T),H2∗(Ω))/wt∈L∞((0,T),L2(Ω));||w||WT≤M}. |
Next, we prove, for T0(even smaller), K is a contraction. For this purpose, let u1=K(v1) and u2=K(v2) and set u=u1−u2, then u satisfies the following
utt+Δ2u+α(t)g(u1t)−α(t)g(u2t)=|v1|βv1−|v2|βv2. | (4.55) |
Multiplying (4.55) by ut and integrating over Ω×(0,t) we get, for all t≤T,
12∫Ωu2tdx+12∫Ω|Δu|2dx+∫t0∫Ω(α(t)g(u1t)−α(t)g(u2t))(u1t−u2t)dxds=∫t0∫Ω(˜f(v1)−˜f(v2))utdxds. | (4.56) |
Using (3.5) and (H2), we have
12∫Ωu2tdx+12∫Ω|Δu|2dx≤∫t0∫Ω(˜f(v1)−˜f(v2))utdxds. | (4.57) |
Now, we evaluate
Λ:=∫Ω|˜f(v1)−˜f(v2)||ut|dx=∫Ω|˜f′(ξ)||v||ut|dx, | (4.58) |
where v=v1−v2, ξ=τv1+(1−τ)v2, 0≤τ≤1, and ˜f′(ξ)=(β+1)|ξ|β.
Young's inequality implies
Λ≤δ2∫Ωu2tdx+2δ∫Ω|˜f′(ξ)|2|v|2dx≤δ2∫Ωu2tdx+2(β+1)2δ∫Ω|αv1+(1−α)v2|2β|v|2dx≤δ2∫Ωu2tdx+Cδ(|v|2nn−2)n−2n(|αv1+(1−α)v2|nβ)2n. | (4.59) |
Using the embedding Lemma 3.1, we arrive at
Λ≤δ2∫Ωu2tdx+CδCe||v||2H2∗(||v1||2βH2∗+||v2||2βH2∗)≤δ2∫Ωu2tdx+4CδCeM2β||v||2βH2∗. | (4.60) |
Therefore, (4.57) takes the form
12||u||2WT≤δT02||u||2WT+CδM2βT0||v||2βWT. | (4.61) |
Choosing δ sufficiently small, we see that
||u||2WT≤4CδM2βT0||v||2βWT=γ0T0||v||2βWT. | (4.62) |
Taking T0 small enough so that,
||u||2WT≤ν||v||2βWT,for some0<ν<1. | (4.63) |
Thus, K is a contraction. The Banach fixed point theorem implies the existence of a unique u∈B satisfying K(u)=u. Thus, u is a local solution of (1.1).
Uniqueness: Suppose that problem (1.1) has two weak solutions (u,v). Taking, w=u−v, that satisfies the following equation, for all t∈(0,T),
{wtt−Δ2w+α(t)g(ut)−α(t)g(vt)=u|u|β−v|v|βw(0,y,t)=wxx(0,y,t)=w(π,y,t)=wxx(π,y,t)=0,(y,t)∈(−d,d)×(0,T),w(x,0)=wt(x,0)=0, in Ω. | (4.64) |
Multiplying (4.64) by wt and integrating over Ω×(0,t), we obtain
12∫Ωw2tdx+12∫Ω|Δw|2dx+∫t0∫Ω(α(t)g(ut)−α(t)g(vt))(ut−vt)dxds=∫t0∫Ω(u|u|β−v|v|β)wtdxds. | (4.65) |
Using (3.5) and (H2) implies that
12∫Ωw2tdx+12∫Ω|Δw|2dx≤∫t0∫Ω(u|u|β−v|v|β)wtdxds. | (4.66) |
By repeating the same above estimates, we obtain
∫Ω(w2tdx+|Δw|2)dx=0. | (4.67) |
This gives w≡0. The proof of the uniqueness is completed.
In this section, we prove that problem (1.1) has a global solution. For this purpose, we introduce the following functionals. The energy functional associated with problem (1.1) is
E(t)=12(‖ut‖22+‖u‖2H2∗(Ω))−1β+2‖u‖β+2β+2. | (5.1) |
Direct differentiation of (5.1), using (1.1), leads to
E′(t)=−α(t)∫Ωutg(ut)dx≤0. | (5.2) |
J(t)=12‖u‖2H2∗(Ω)−1β+2‖u‖β+2β+2 | (5.3) |
and
I(t)=‖u‖2H2∗(Ω)−‖u‖β+2β+2. | (5.4) |
Clearly, we have
E(t)=J(t)+12‖ut‖22. | (5.5) |
Lemma 5.1. Suppose that (H1) and (H2) hold and (u0,u1)∈H2∗(Ω)×L2(Ω), such that
0<γ=Cβ+2e(2(β+2)βE(0))β2<1,I(u0)>0, | (5.6) |
then I(u(t))>0,∀t>0.
Proof. Since I(u0)>0, then there exists (by continuity) Tm<T such that I(u(t)≥0, ∀t∈[0,Tm]; which gives
J(t)=12‖u‖2H2∗(Ω)−1β+2‖u‖β+2β+2=β2(β+2)‖u‖2H2∗(Ω)+1β+2I(t)≥β2(β+2)‖u‖2H2∗(Ω). | (5.7) |
By using (5.2), (5.5) and (5.7), we have
‖u‖2H2∗(Ω)≤2(β+2)βJ(t)≤2(β+2)βE(t)≤2(β+2)βE(0),∀t∈[0,Tm]. | (5.8) |
The embedding theorem, (5.6) and (5.8) give, ∀t∈[0,Tm],
‖u‖β+2β+2≤Cβ+2e‖u‖β+2H2∗(Ω)≤Cβ+2e‖u‖βH2∗(Ω)‖u‖2H2∗(Ω)≤γ‖u‖2H2∗(Ω)<‖u‖2H2∗(Ω). | (5.9) |
Therefore,
I(t)=‖u‖2H2∗(Ω)−‖u‖β+2β+2>0,∀t∈[0,Tm]. |
By repeating this procedure, and using the fact that
limt→TmCβ+2e(2(β+2)βE(t))β2≤γ<1, |
Tm is extended to T.
Remark 5.2. The restriction (5.6) on the initial data will guarantee the nonnegativeness of E(t).
Proposition 5.3. Suppose that (H1) and (H2) hold. Let (u0,u1)∈H2∗(Ω)×L2(Ω) be given, satisfying (5.6). Thenthe solution of (1.1) is global and bounded.
Proof. It suffices to show that ‖u‖2H2∗(Ω)+‖ut‖22 is bounded independently of t. To achieve this, we use (5.2), (5.4) and (5.5) to get
E(0)≥E(t)=J(t)+12‖ut‖22≥β−22β‖u‖2H2∗(Ω)+12‖ut‖22+1βI(t)≥β−22β‖u‖2H2∗(Ω)+12‖ut‖22, | (5.10) |
since I(t) is positive. Therefore
‖u‖2H2∗(Ω)+‖ut‖22≤CE(0), |
where C is a positive constant, which depends only on β.
In this section, we state and prove our stability result. For this purpose, we establish some lemmas.
Lemma 6.1. (Case: G is linear) Let u be the solution of (1.1). Then, for T>S≥0, the energy functionalsatisfies
∫TSα(t)E(t)dt≤cE(S). | (6.1) |
Proof. We multiply (1.1) by αu and integrate over Ω×(S,T) to get
0=∫TSα(t)∫Ω(uutt+uΔ2u+α(t)ug(ut)−|u|β+2)dxdt=∫TSα(t)∫Ω((uut)t−u2t+α(t)ug(ut)−|u|β+2)dxdt+∫TSα(t)‖u‖2H2∗(Ω)dt=∫TSα(t)ddt(∫Ωuutdx)dt+∫TSα(t)∫Ωu2tdxdt+∫TSα(t)‖u‖2H2∗(Ω)dt−2∫TSα(t)∫Ωu2tdxdt+∫TSα2(t)∫Ωug(ut)dxdt−∫TSα(t)‖u‖β+2β+2dt. | (6.2) |
Adding and subtracting the following terms
γ∫TSα(t)‖u‖2H2∗(Ω)dt+(1+γ)∫TSα(t)‖ut‖22dt, where γ is defined in (5.6), |
to (6.2), and recalling (5.9), we arrive at
∫TSα(t)ddt(∫Ωuutdx)dt+(1−γ)∫TSα(t)(‖u‖2H2∗(Ω)+‖ut‖22)dt−(2−γ)∫TSα(t)∫Ωu2tdxdt+∫TSα2(t)∫Ωug(ut)dxdt=−∫TSα(t)(γ‖u‖2H2∗(Ω)−‖u‖β+2β+2)dt≤0. | (6.3) |
Integrating the first term of (6.3) by parts and using (5.1), then (6.3) becomes
(1−γ)∫TSαEdt≤(1−γ)∫TSα(‖u‖2H2∗(Ω)+‖ut‖22)dt≤−[α∫Ωuutdx]TS+∫TSα′∫Ωuutdxdt+(2−γ)∫TSα∫Ωu2tdxdt−∫TSα2∫Ωug(ut)dxdt. | (6.4) |
Now, we estimate the terms in the right-hand side of (6.4) as follows:
1) Estimate for −[α∫Ωuutdx]TS.
Using Lemma 3.1 and Young's inequality, we obtain
∫Ωuutdx≤12∫Ω(u2+u2t)dx≤c‖u‖H2∗(Ω)+‖ut‖22≤cE(t), | (6.5) |
which implies that
−[α∫Ωuutdx]TS≤c[−α(T)E(T)+α(S)E(S)]≤cα(S)E(S)≤cE(S). | (6.6) |
2) Estimate for ∫TSα′∫Ωuutdxdt.
The use of (6.5) and (H2) leads to
∫TSα′∫Ωuutdxdt≤c|∫TSα′Edt|≤cE(S)|∫TSα′dt|≤cE(S). | (6.7) |
3) Estimate for ∫TSα(∫Ωu2tdx)dt.
Using (H1), (5.2) and recalling that G is linear, we have
∫TSα(∫Ωu2tdx)dt≤1c1∫TSα(t)∫Ωutg(ut)dxdt≤−∫TScE′(t)dt≤cE(S). | (6.8) |
4) Estimate for −∫TSα2(t)∫Ωug(ut)dxdt.
Using (H1), Lemma 3.1, Holder's inequality and recalling G is linear, we obtain
α2(t)∫Ωug(ut)dx≤α2(t)(∫Ω|u|2dx)12(∫Ω|g(ut)|2dx)12≤α32(t)‖u‖H2∗(Ω)(α(t)∫Ωutg(ut)dx)12≤cα(t)E12(t)(−E′(t))12. | (6.9) |
Applying Young's inequality to E12(t)(−E′(t))12 with p=2 and p∗=2, to get
α2(t)∫Ωug(ut)dx≤cα(t)(εE(t)−CεE′(t))≤cεαE(t)−CεE′(t), | (6.10) |
which implies that
∫TSα2(t)(∫Ω(−ug(ut))dx)dt≤cε∫TSα(t)E(t)dt+CεE(S). | (6.11) |
Combining the above estimates and taking ε small enough, we get (6.1).
Lemma 6.2. (Case: G is nonlinear) Let u be the solution of (1.1). Then, for T>S≥0, the energy functionalsatisfies
∫TSα(t)˜ϕ(E(t))dt≤c˜ϕ(E(S))+c∫TSα(t)˜ϕ(E)E∫Ω(|ut|2+|ug(ut)|)dxdt, | (6.12) |
where ˜ϕ:R+→R+ is any convex, increasing and of class C1[0,∞) function such that ˜ϕ(0)=0.
Proof. We multiply (1.1) by α(t)˜ϕ(E)Eu and integrate over Ω×(S,T) to get
0=∫TSα(t)˜ϕ(E)E∫Ω((uut)t−u2t+α(t)ug(ut)−|u|β+2)dxdt+∫TSα(t)˜ϕ(E)E‖u‖2H2∗(Ω)dt=∫TSα(t)˜ϕ(E)Eddt(∫Ωuutdx)dt+∫TSα(t)˜ϕ(E)E‖u‖2H2∗(Ω)dt+∫TSα(t)˜ϕ(E)E∫Ωu2tdxdt−2∫TSα(t)˜ϕ(E)E∫Ωu2tdxdt+∫TSα2(t)˜ϕ(E)E∫Ωug(ut)dxdt−∫TSα(t)˜ϕ(E)E‖u‖β+2β+2. | (6.13) |
Adding and subtracting to (6.13) the following terms
γ∫TSα(t)˜ϕ(E)E‖u‖2H2∗(Ω)dt+(1+γ)∫TSα(t)˜ϕ(E)E‖ut‖22dt, where γ is defined in (5.6), |
we arrive at
(1−γ)∫TSα(t)˜ϕ(E)dt≤−∫TSα(t)˜ϕ(E)Eddt(∫Ωuutdx)dt+(2−γ)∫TSα(t)˜ϕ(E)E∫Ωu2tdxdt−∫TSα2(t)˜ϕ(E)E∫Ωug(ut)dxdt−∫TSα˜ϕ(E)E(γ‖u‖2H2∗(Ω)−‖u‖β+2β+2). | (6.14) |
Using (5.9), it is easy to deduce that −∫TSα˜ϕ(E)E(γ‖u‖2H2∗(Ω)−‖u‖β+2β+2)dt≤0.
Integrating by parts in the first term, in the right-hand side of (6.14), we get
(1−γ)∫TSα(t)˜ϕ(E)dt≤−[α(t)˜ϕ(E)E∫Ωuutdx]TS+∫TS∫Ωut(α′(t)˜ϕ(E)Eu+α(t)(˜ϕ(E)E)′u)dxdt+(2−γ)∫TSα(t)˜ϕ(E)E∫Ωu2tdxdt−∫TSα2(t)˜ϕ(E)E∫Ωug(ut)dxdt. | (6.15) |
Using Cauchy Schwarz' inequality, Lemmas 3.1 and 5.1, we obtain
∫Ωuutdx≤(∫Ω|u|2dx)12(∫Ω|ut|2dx)12≤c‖u‖H2∗(Ω)‖ut‖2≤cE(t). | (6.16) |
Using (6.16), the properties of α(t) and the fact that the function s→˜ϕ(s)s is non-decreasing and E is non-increasing, we have
∫TSα′(t)˜ϕ(E)E∫Ωuutdxdt≤c∫TSα′(t)˜ϕ(E)EEdt≤c˜ϕ(E(S))∫TSα′(t)dt≤c˜ϕ(E(S)). | (6.17) |
Similarly, we get
∫TSα(t)(˜ϕ(E)E)′∫Ωuutdxdt≤E(S)∫TSα(t)(˜ϕ(E)E)′dt≤E(S)[α(t)˜ϕ(E)E]TS−E(S)∫TSα′(t)˜ϕ(E)Edt≤E(S)(α(T)˜ϕ(E(T))E(T)−α(S)˜ϕ(E(S))E(S))−E(S)˜ϕ(E(S))E(S)∫TSα′(t)dt≤E(S)α(T)˜ϕ(E(T))E(T)−˜ϕ(E(S))(α(T)−α(S))≤E(S)α(S)˜ϕ(E(S))E(S)+˜ϕ(E(S))α(S)≤c˜ϕ(E(S)). | (6.18) |
A combination of (6.15)–(6.18) leads to (6.12).
In order to finalize the proof of our result, we let
˜ϕ(s)=2ε0sG′(ε20s), andG1(s)=G(s2), |
where ε0>0 is small enough and G∗ and G∗1 denote the dual functions of the convex functions G and G1 respectively in the sense of Young (see, Arnold [33], pp. 64).
Lemma 6.3. Suppose G is nonlinear, then the following estimates
G∗(˜ϕ(s)s)≤˜ϕ(s)s(G′)−1(˜ϕ(s)s) | (6.19) |
and
G∗1(˜ϕ(s)√s)≤ε0˜ϕ(√s). | (6.20) |
hold, where ˜ϕ is defined earlier in Lemma 6.2.
Proof. Since G∗ and G∗1 are the dual functions of the convex functions G and G1 respectively, then
G∗(s)=s(G′)−1(s)−G[(G′)−1(s)]≤s(G′)−1(s) | (6.21) |
and
G∗1(s)=s(G′1)−1(s)−G1[(G′1)−1(s)]≤s(G′1)−1(s). | (6.22) |
Using (6.21) and the definition of ˜ϕ, we obtain (6.19). For the proof of (6.20), we use (6.22) and the definitions of G1 and ˜ϕ to obtain
˜ϕ(s)√s(G′1)−1(˜ϕ(s)√s)≤2ε0√sG′(ε20s)(G′1)−1(2ε0√sG′(ε20s))=2ε0√sG′(ε20s)(G′1)−1(G′1(ε0√s))=2ε20sG′(ε20s)=ε0˜ϕ(√s). | (6.23) |
Now, we state and prove our main decay results.
Theorem 6.4. Let (u0,u1)∈H2∗(Ω)×L2(Ω). Assume that (H1) and (H2) hold. Then there exist positive constants k and c such that, for t large, the solution of (1.1) satisfies
E(t)≤ke−c∫t0α(s)ds,if G is linear, | (6.24) |
E(t)≤ψ−1(h(˜α(t))+ψ(E(0))),if G is nonlinear, | (6.25) |
where
˜α(t)=∫t0α(t)dt,ψ(t)=∫1t1χ(s)ds,andχ(s)=2ε0csG′(ε20s) |
and
{h(t)=0,0≤t≤E(0)χ(E(0)),h−1(t)=t+ψ−1(t+ψ(E(0)))χ(ψ−1(t+ψ(E(0)))),t>0. |
Proof. To establish (6.24), we use (6.1) and Lemma 3.5 for γ(t)=∫t0α(s)ds. Consequently the result follows. For the proof of (6.25), we re-estimate the terms of (6.12) as follows: we consider the following partition of Ω:
Ω1={x∈Ω:|ut|≥ε1},Ω2={x∈Ω:|ut|≤ε1}. |
So,
∫TSα(t)˜ϕ(E)E∫Ω1(|ut|2+|ug(ut)|)dxdt=∫TSα(t)˜ϕ(E)E∫Ω1|ut|2dxdt+∫TSα(t)˜ϕ(E)E∫Ω1|ug(ut)|dxdt:=I1+I2. |
Using the definition of Ω1, (3.4) and (5.2), we have
I1≤c∫TSα(t)˜ϕ(E)E∫Ω1utg(ut)dxdt≤c∫TS˜ϕ(E)E(−E′(t))dt≤c˜ϕ(E(S)). | (6.26) |
After applying Hölder's and Young's inequalities and Lemma 3.1, we obtain for some ε>0,
I2≤∫TSα(t)˜ϕ(E)E(∫Ω1|u|2dx)12(∫Ω1|g(ut)|2)12dt≤ε∫TSα(t)˜ϕ2(E)E2‖u‖H2∗(Ω)dt+c(ε)∫TSα(t)∫Ω1|g(ut)|2dt. | (6.27) |
The definition of Ω1, (3.4), (5.1), (5.2) and (6.27) lead to
I2≤ε∫TSα(t)˜ϕ2(E)Edt+c(ε)∫TSα(t)∫Ω1utg(ut)dxdt≤ε∫TSα(t)˜ϕ2(E)Edt+c(ε)E(S). | (6.28) |
Using the definition of ˜ϕ and the convexity of G, then (6.28) becomes
I2≤ε∫TSα(t)˜ϕ2(E)Edt+cE(S)=2εε0∫TSα(t)˜ϕ(E)G′(ε20E(t))dt+cE(S)≤2εε0∫TSα(t)˜ϕ(E)G′(ε20E(0))dt+cE(S)≤2cεε0∫TSα(t)˜ϕ(E)dt+cE(S). | (6.29) |
Combining (6.12), (6.26) and (6.29) and choosing ε small enough, we obtain
∫TSα(t)˜ϕ(E)dt≤c˜ϕ(E)+c∫TSα(t)˜ϕ(E)E∫Ω2(|ut|2+|ug(ut)|)dxdt. | (6.30) |
Using Young's inequality and Jensen's inequality (Eq 3.3), (Eq 3.4) and (Eq 5.1), we get
∫TSα(t)˜ϕ(E)E∫Ω2(|ut|2+|ug(ut)|)dxdt≤∫TSα(t)˜ϕ(E)E∫Ω2G−1(utg(ut))dxdt+∫TSα(t)˜ϕ(E)E‖u‖12H2∗(Ω)(∫Ω2G−1(utg(ut))dx)12dxdt≤|Ω|∫TSα(t)˜ϕ(E)EG−1(1|Ω|∫Ωutg(ut)dx)dt+∫TSα(t)˜ϕ(E)E√E√|Ω|G−1(1|Ω|∫Ωutg(ut)dx)dt. | (6.31) |
Applying the generalized Young inequality
AB≤G∗(A)+G(B) |
to the first term of (6.31), with A=˜ϕ(E)E and B=G−1(1|Ω|∫Ωutg(ut)dx), we easily see that
˜ϕ(E)EG−1(1|Ω|∫Ωutg(ut)dx)≤G∗(˜ϕ(E)E)+1|Ω|∫Ωutg(ut)dx. | (6.32) |
Then we apply it to the second term of (6.31), with A=˜ϕ(E)E√E and B=√|Ω|G−1(1|Ω|∫Ωutg(ut)dx) to obtain
˜ϕ(E)E√E√|Ω|G−1(1|Ω|∫Ωutg(ut)dx)≤G∗1(˜ϕ(E)E√E)+|Ω|G−1(1|Ω|∫Ωutg(ut)dx). | (6.33) |
Combining (6.31)–(6.33) and using (6.19) and (6.20), we arrive at
∫TSα(t)˜ϕ(E)E∫Ω2(|ut|2+|ug(ut)|)dxdt≤c∫TSα(t)(G∗1(˜ϕ(E)E√E)+G∗(˜ϕ(E)E))dt+c∫TSα(t)∫Ωutg(ut)dxdt≤c∫TSα(t)(ε0+(G′)−1(˜ϕ(E)E)E)˜ϕ(E)dt+cE(S). | (6.34) |
Using the definition of ˜ϕ and the fact that s→(G′)−1(s) is non-decreasing, we deduce that, for 0<ε0≤12,
(G′)−1(˜ϕ(E)E)E=(G′)−1(2ε0G′(ε20E))E≤ε20. | (6.35) |
Combining (6.34) and (6.35) leads to
∫TSα(t)˜ϕ(E)E∫Ω2(|ut|2+|ug(ut)|)dxdt≤cε0∫TSα(t)˜ϕ(E)dt+cE(S). | (6.36) |
Then, choosing ε0 small enough, we deduce from (6.30) and (6.36) that
∫TSα(t)˜ϕ(E(t))dt≤c(1+˜ϕ(E(S))E(S))E(S). |
Using the facts that E is non-increasing and s→˜ϕ(s)s is non-decreasing, we obtain
∫+∞Sα(t)˜ϕ(E(t))dt≤cE(S). | (6.37) |
Let ˜E=E∘˜α−1, where ˜α(t)=∫t0α(s)ds. Then we deduce from (6.37) that
∫∞S˜ϕ(˜E(t))dt=∫∞S˜ϕ(E(˜α−1(t)))dt=∫∞˜α−1(S)α(η)˜ϕ(E(η))dη≤cE(˜α−1(S))≤c˜E(S). |
Using Lemma 3.6 for ˜E and χ(s)=1c˜ϕ(s), we deduce from (3.6) the following estimate
˜E(t)≤ψ−1(h(t)+ψ(E(0))), |
which gives (6.25), by using the definition of ˜E and the change of variables.
Remark 6.5. The stability result (6.25) is a decay result. Indeed,
h−1(t)=t+ψ−1(t+ψ(E(0)))χ(ψ−1(t+ψ(E(0))))=t+c2ε0cG′(ε20ψ−1(t+r))≥t+c2ε0cG′(ε20ψ−1(r))≥t+˜c. |
Hence, limt→∞h−1(t)=∞, which implies that limt→∞h(t)=∞. Using the convexity of G, we have
ψ(t)=∫1t1χ(s)ds=∫1tc2ε0sG′(ε20s)≥∫1tcsG′(ε20)≥c[ln|s|]1t=−clnt. |
Therefore, limt→0+ψ(t)=∞ which leads to limt→∞ψ−1(t)=0.
Examples
1) Let g(s)=sm, where m≥1. Then the function G is defined in the neighborhood of zero by
G(s)=csm+12 |
which gives, near zero
χ(s)=c(m+1)2sm+12. |
So, we obtain
ψ(t)=c∫1t2(m+1)sm+12ds={ctm−12,if m>1;−clnt,if m=1, |
and then, in the neighborhood of ∞
ψ−1(t)={ct−2m−1,if m>1;ce−t,if m=1, |
Using the fact that h(t)=t as t goes to infinity, we obtain from (6.24) and (6.25)
E(t)≤{c(∫t0α(s)ds)−2m−1,if m>1;ce−∫t0α(s)ds,if m=1. |
2) Let g(s)=sm√−lns, where m≥1. Then the function G is defined in the neighborhood of zero by
G(s)=csm+12√−ln√s |
which gives, near zero
χ(s)=csm+12(−ln√s)−12(m+12(−ln√s)−14). |
Therefore, we get
ψ(t)=c∫1t1sm+12(−ln√s)−12(m+12(−ln√s)−14)ds=c∫1√t1τm−2(lnτ)−12(m+12lnτ−14)dτ={ctm−12√−lnt,if m>1;c√−lnt,if m=1, |
and then, in the neighborhood of ∞, we have
ψ−1(t)={ct−2m−1(lnt)−1m−1,if m>1;ce−t2,if m=1, |
Using the fact that h(t)=t as t goes to infinity, we obtain
E(t)≤{c(∫t0α(s)ds)−2m−1(ln(∫t0α(s)ds))−1m−1,if m>1;ce−(∫t0α(s)ds)2,if m=1, |
The authors would like to express their profound gratitude to King Fahd University of Petroleum and Minerals (KFUPM) for its continuous support. The authors also thank the referees for their very careful reading and valuable comments. This work was funded by KFUPM under Project #SB201003.
The authors declare there is no conflicts of interest.
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