# Well-Posedness and Stability Result of the Nonlinear Thermodiffusion Full von Kármán Beam with Thermal Effect and Time-Varying Delay

Authors: Abdelbaki Choucha; Djamel Ouchenane; Salah Mahmoud Boulaaras; Bahri Belkacem Cherif; Mohamed Abdalla
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this work, we consider a new full von Kármán beam model with thermal and mass diffusion effects according to the Gurtin-Pinkin model combined with time-varying delay. Heat and mass exchange with the environment during thermodiffusion in the von Kármán beam. We establish the well-posedness and the exponential stability of the system by the energy method under suitable conditions.


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## Body

## 1. Introduction and Preliminaries

In this paper, we are concerned with the following problem:(1)wtt−d1ux+12wx2wxx+d2wxxxx+μ1wt+μ2wtx,t−τt=0,utt−d1ux+12wx2x−δ1θx−δ2Px=0,cθt+dPt−∫0∞β1σθxxt−σdσ−δ1utx=0,dθt+rPt−∫0∞β2σPxxt−σdσ−δ2utx=0,where
(2)x,σ,t∈0,L×ℝ+×0,∞.Here,τt>0 represents the time-varying delay, and d1, d2, δ1, δ2, c, d, r, and μ1 are positive constants; μ2 is a real number, and β1 and β2 are the relaxation functions, with the initial data
(3)wx,0=w0x,wtx,0=w1x,ux,0=u0x,utx,0=u1x,θx,0=θ0x,Px,0=P0x,wtx,t−τ0=f0x,t−τ0,where
(4)x,t∈0,L×0,τ0,and Neumann-Dirichlet boundary conditions
(5)wx,t=ux,t=Px,t=0,x=0,L,∀t≥0,wxx,t=θx,t=0,x=0,L,∀t≥0.The case of time-varying delay in the wave equation has been studied recently by Nicaise et al. [1]; they proved the exponential stability under the condition
(6)μ2<1−dμ1,where d is a constant that satisfies
(7)τ′t≤d<1,∀t>0.For the wave equation with a time-varying delay, in [1], the authors consider the system
(8)utt−Δu=0,ux,t=0,dudvx,t=μ1utx,t+μ2utx,t−τt,where the time-varying delay τt>0 satisfies
(9)0≤τt≤τ¯,∀t>0,(10)τ′t≤1,∀t>0,(11)τt∈W2,∞0,T,∀T>0.They proved the exponential stability under suitable conditions.The purpose of this work is to study problem (1)–(5), with a delay term appearing in the control term at the first equation, introducing the time-varying delay term β2wtx,t−τt; thermal and mass diffusion effects make the problem different from those considered in the literature (see [2–30]).This paper is organized as follows: in the rest of this section, we put the preliminaries necessary for problem (1); in Section 2, we establish the well-posedness. As for Section 3, we prove the exponential stability result by the energy method and Lyapunov function.In order to prove the existence of a unique solution of problem (1)–(5), we introduce the new variable
(12)zx,ρ,t=wtx,t−τtρ.Then, we obtain(13)τtztx,ρ,t+1−τ′tρzρx,ρ,t=0,zx,0,t=wtx,t.And it is more convenient to work in the history space setting by introducing the so-called summed past history ofθ and P defined by (see [31–36])
(14)ηtσ=∫0σθt−ζdζ,νtσ=∫0σPt−ζdζ,t,σ∈0,∞×ℝ+.Differentiating (14)1 and (14)2, we get
(15)ηttσ+ησtσ=θt,νttσ+νσtσ=Pt,with the boundary and initial conditions
(16)ηt0=νt0=0,t≥0,η0σ=η0σ,ν0σ=ν0σ,σ≥0.We set(17)η0σ=∫0σθ¯0τdτ,ν0σ=∫0σP¯0τdτ,σ∈ℝ+.Concerning the memory kernelsβ1 and β2, we set
(18)βσ=−β1′σ,λσ=−β2′σ.Assumingβ1∞=β2∞=0, then from (14), we infer
(19)∫0∞β1σθt−σdσ=−∫0∞β1′σηtσdσ,∫0∞β2σPt−σdσ=−∫0∞β2′σνtσdσ,and therefore,
(20)∫0∞β1σθxxt−σdσ=∫0∞βσηxxtσdσ,∫0∞β2σPxxt−σdσ=∫0∞λσνtσdσ.Consequently, the problem is equivalent to(21)wtt−d1ux+12wx2wxx+d2wxxxx+μ1wt+μ2zx,1,t=0,utt−d1ux+12wx2x−δ1θx−δ2Px=0,cθt+dPt−∫0∞βσηxxtσdσ−δ1utx=0,dθt+rPt−∫0∞λσνxxtσdσ−δ2utx=0,ηtt+ησt=θ,νtt+νσt=P,τtztx,ρ,t+1−τ′tρzρx,ρ,t=0,where
(22)x,σ,ρ,t∈0,L×ℝ+×0,1×0,∞,with the initial and boundary conditions
(23)wx,t=wxx,t=ux,t=Px,t=θx,t=0,x=0,L,wx,0=w0x,wtx,0=w1x,ux,0=u0x,utx,0=u1x,θx,0=θ0x,Px,0=P0x,zx,ρ,0=f0x,−ρτ0,η0x,σ=η0x,σ,ν0x,σ=ν0x,σ,x,σ∈0,1×ℝ+,(24)∀x,ρ,σ,t∈0,L×0,1×0,∞×0,∞,where the function τt satisfies (7), (11), and the condition
(25)0<τ0<τt<τ¯,∀t>0.In this paper, we establish the well-posedness and prove the exponential stability by using the variable of Kato under some restrictions and assumptions:(H1).
(26)μ2≤1−dμ1.(H2). The symmetric matrix Λ is positive definite, where
(27)Λ=cddr.That is,∣Λ∣=cr−d2>0 implies that
(28)c∫0Lθ2dx+2d∫0LθPdx+r∫0LP2dx>0.Condition (28) is needed to stabilize the system when diffusion effects are added to thermal effects (see, e.g., [31–38] for more information on this). By virtue of cr>d2, we deduce that d/c<r/d. Let, then, ζ be a number chosen in such a way that
(29)dc<ζ<rd.Thus, Young’s inequality leads to(30)2d∫0LθPdx≤dζ∫0Lθ2dx+dζ∫0LP2dx.(H3). We assume the following set of hypotheses on μ and λ:
(31)β,λ∈C1ℝ+∩L1ℝ+,βσ,λσ≥0,β′σ,λ′σ≤0,∀σ∈ℝ+,β′σ+α1βσ≤0,λ′σ+α2λσ≤0,forsomeα1,α2>0,∀σ∈ℝ+,(32)β0=∫0∞βσdσ≔β0>0,λ0=∫0∞λσdσ≔λ0>0.Letf be a memory kernel satisfying the assumptions (31) and (32).Now, we consider the weighted Hilbert spaces(33)Mf=L2ℝ+,H010,L=Φ:ℝ+→H010,L/∫0L∫0∞fσΦx2σdσdx<∞,equipped with the inner product
(34)<Φ,Ψ>Mf=∫0L∫0∞fσΦxσΨxσdσdx,and the norm
(35)∥Φ∥Mf2=<Φ,Φ>Mf=∫0L∫0∞fσΦx2σdσdx.We also introduce the linear operatorT on Mf defined by
(36)TΦ=−Φσ,Φ∈DT,with
(37)DT=Φ∈Mf/Φσ∈Mf,Φ0=0,where Φσ is the distributional derivative of Φ with respect to the internal variable σ, and then, the operator T is the infinitesimal generator of a C0-semigroup of contractions. Following Ref. [39], there holds
(38)<TΦ,Φ>Mf=<−Φσ,Φ>Mf=−12∫0∞fσddσ∫0LΦx2σdxdσ,∀Φ∈DT.Integration by parts yields(39)∫0∞fσddσ∫0LΦx2σdxdσ=fσ∫0LΦx2σdx0∞−∫0∞f′σ∫0LΦx2σdxdσ.Hence, from (31), we obtain
(40)<TΦ,Φ>Mf=12∫0∞f′σ∫0LΦx2σdxdσ≤0.As a direct consequence, we deduce from (32) and (40) that
(41)<Tη,η>Mβ=12∫0∞β′σ∫0Lηx2σdxdσ≤−α12∫0∞βσ∫0Lηx2σdxdσ=−α12∥ηx∥Mβ2,<Tν,ν>Mλ=12∫0∞λ′σ∫0Lνx2σdxdσ≤−α22∫0∞λσ∫0Lνx2σdxdσ=−α22∥νx∥Mλ2,for all η,ν∈DT. Finally, we define the operator Lf:DLf→L20,L by
(42)LfΦ=∫0∞fσΦxxσdσ,with the domain
(43)DLf=Φ∈Mf/∫0∞fσΦxxσdσ∈L20,L,Φ0=0.


## 2. Well-Posedness

In this section, we give sufficient conditions that guarantee the well-posedness of this problem. Let(44)U=w,wt,u,ut,θ,ηt,P,νt,zT.For the sake of simplicity, we writeη=ηtσ and ν=νtσ and the new dependent variables φ=ωt and ψ=ut; then, (21)–(23) can be written as
(45)U′=AtU+FU,U0=w0,w1,u0,u1,θ0,η0,P0,ν0,f0.,−ρτ0T,with the linear problem
(46)U′=AtU,U0=w0,w1,u0,u1,θ0,η0,P0,ν0,f0.,−ρτ0T,where the time-varying operator A is defined by
(47)AtwφuψθηPνz=φ−d2wxxxx−μ1φ−μ2zx,1,tψd1uxx+δ1θx+δ2Px−1α1dδ2−rδ1ψx−rLβη+dLλνθ+Tη−1α2dδ1−cδ2ψx+dLβη−cLλνP+Tντ′tρ−1τtzρ,(48)FU=0d1ux+12wx2x0d12wx200000.The energy spaceH is defined as
(49)H=H40,L∩H020,L×H010,L×H20,L∩H020,L×H010,L×L20,L×Mβ×L20,L×Mλ×L20,L,0,1,and the domain of A is
(50)DAt=U∈H/φ=z.,0,θ,P∈H010,L,Lβη,Lλν∈L20,L,η,ν∈DT.We equipH with the inner product
(51)<U,U¯>H=∫0Lφφ¯+d1uxu¯x+ψψ¯+d2wxxw¯xxdx+∫0L∫01zx,ρ,tz¯x,ρ,tdρdx+<Λθ,PT,θ¯,P¯T>+<η,η¯>Mβ+<ν,ν¯>Mλ,with the existence and the uniqueness in the following result.Theorem 1.
Let (7), (11), and (25) be satisfied and assume that (26)–(31) hold. Then, for all U0∈DA0, there exists a unique solution U of problem (21)–(23) satisfying
(52)U∈C0,+∞,DA0∩C10,+∞,H.In order to prove Theorem1, we will use the variable norm technique developed by Kato in [40]. The following theorem is proved in [40].Theorem 2.
Assume that(1)
DA0 is a dense subset of H(2)
DAt=DA0,∀t>0(3)
For allt∈0,T, At generates a strongly continuous semigroup on H and the family A=At:t∈0,T is stable with stability constants C and m independent of t; i.e., the semigroup Stss≥0 generated by At satisfies
(53)∥Stsu∥H≤Cems∥u∥H,∀u∈H,s≥0.(4)
dtAt∈L∗∞0,T,BDA0,H, where L∗∞0,T,BDA0,H is the space of equivalent classes of essentially bounded, strongly measurable functions from 0,T into the set BDA0,H of bounded operators from DA0 into H
Then, problem (46) has a unique solution
(54)U∈C0,T,DA0∩C10,T,H,for any initial datum in DA0.Proof.
To prove Theorem1, we use the method in [1] with the necessary modification.
(1)
First, we show thatDA0 is dense in H
LetF=f1,f2,f3,f4,f5,f6,f7,f8,f9∈H be orthogonal to all elements of DA0 with respect to the inner product .,.H:
(55)0=<U,F>H=∫0Lφf2+ψf4+d2wxxf1xx+d1uxf3xdx+∫0L∫01zx,ρ,tf9dρdx+<Λθ,PT,f5,f7T>+<η,f6>Mβ+<ν,f8>Mλ.
For allU=w,φ,u,ψ,θ,η,P,ν,zT∈DA0, our goal is to prove that fi=0,∀i=1,⋯,9. Let us first take z∈D0,L×0,1 and w=φ=ψ=u=θ=q=ϕ=0, so the vector U=0,0,0,0,0,0,0,0,zT∈DA0, and therefore, from (55), we deduce that
(56)∫0L∫01zx,ρf7dρdx=0.
SinceD0,L×0,1 is dense in L20,L×0,1, it follows then that f9=0.
Similarly, letφ∈H010,L; then, U=0,φ,0,0,0,0,0,0,0T∈DA0, which implies from (55) that
(57)∫0Lφf2dx=0.
So, as above,f2=0.
And letU=w,0,0,0,0,0,0,0,0T; then, we obtain from (55) that
(58)∫0Lwxxf1xxdx=0.
It is obvious thatU=w,0,0,0,0,0,0,0,0T∈DA0 only if w∈H40,L∩H020,L is dense in H020,L, with respect to the inner product
(59)<g,h>H020,L∫0Lgxxhxxdx.
We getf1=0. By the same ideas as above, we can also show that f3=0.
Foru∈DAt, we get from (55) that
(60)∫0Luxf3xdx=0,and by the density of DAt in H010,L, we obtain f3=0.
Forψ∈DAt, we get from (55) that
(61)∫0Lψf4dx=0,and by the density of DAt in H10,L, we obtain f4=0.
Next, letU=0,0,0,0,θ,0,0,0,0T; then, we obtain from (55) that
(62)∫0Lθf5dx=0.
It is obvious thatU=0,0,0,0,θ,0,0,0,0T∈DA0 only if θ∈L20,L is dense in L20,L; we get f5=0; for η∈Mβ, we get from (55) that
(63)∫0L∫0∞βσηxf6xdσdx=0,which gives f6=0. Similarly, for P and ν. This completes the proof of (1).
(2)
With our choice,DAt is independent of t; consequently,
(64)DAt=DA0,∀t>0.(3)
Now, we show that the operatorAt generates a C0-semigroup in H for a fixed t. We define the time-dependent inner product on H:(65)<U,U¯>t=∫0Lφφ¯+d1uxu¯x+ψψ¯+d2wxxw¯xxdx+ξτt∫0L∫01zx,ρ,tz¯x,ρ,tdρdx+<Λθ,PT,θ¯,P¯T>+<η,η¯>Mβ+<ν,ν¯>Mλ,where ξ satisfies
(66)∣μ2∣1−d≤ξ≤2μ1−∣μ2∣1−d,thanks to hypothesis (26).
Let us set(67)κt=τ′t2+11/22τt.
In this step, we prove the dissipativity of the operatorA¯t=At−τtI.
For a fixedt and U=w,φ,u,ψ,θ,η,P,ν,zT∈DAt, we have
(68)<AtU,U>t=−μ1∫0Lφ2dx−μ2∫0Lφzx,1,tdx+<Tη,η>Mβ+<Tν,η>Mλ−ξ∫0L∫011−τ′tρzx,ρ,tzρx,ρ,tdρdx.
Observe that(69)∫0L∫011−τ′tρzx,ρ,tzρx,ρ,tdρdx=12∫0L∫011−τ′tρddρz2dρdx=τ′t2∫0L∫01z2x,ρ,tdρdx+12∫0Lz2x,1,t1−τ′t−z2x,0,tdρdx,<Tη,η>Mβ+<Tν,η>Mλ=+12∫0∞β′σ∫0Lηx2σdxdσ+12∫0∞λ′σ∫0Lνx2σdxdσ≤−α12∥ησ∥Mβ2−α22∥νσ∥Mλ2,whereupon
(70)<AtU,U>t=−μ1∫0Lφ2dx−μ2∫0Lφzx,1,tdx−α12∥ησ∥Mβ2−α22∥νσ∥Mλ2−ξτ′t2∫0L∫01z2x,ρ,tdρdx−ξ2∫0Lz2x,1,t1−τ′tdx+ξ2∫0Lφ2dx.
By using Young’s inequality and (7), we get
(71)<AtU,U>t≤−μ1+∣μ2∣21−d+ξ2∫0Lφ2dx+∣μ2∣1−d2−ξ1−d2∫0Lz2x,1,tdx−α12∥ησ∥Mβ2−α22∥νσ∥Mλ2+κt<U,U>t,under condition (66) which allows to write
(72)−μ1+μ221−d+ξ2≤0,μ21−d2−ξ1−d2≤0.
Consequently, the operatorA¯t=At−κtI is dissipative.
Now, we prove the subjectivity of the operatorI−At for fixed t>0.
Letf1,f2,f3,f4,f5,f6,f7,f8,f9T∈H; we seek U=w,φ,u,ψ,θ,η,P,ν,zT∈DAt solution of the following system:
(73)w−φ=f1,φ+d2wxxxx+μ1φ+μ2z.,1,t=f2,u−ψ=f3,ψ−d1uxx−δ1θx−δ2Px=f4,α1θ+dδ2−rδ1ψx−rLβη+dLλν=α1f5,η−θ−Tη=f6,α2P+dδ1−cδ2ψx+dLβη+−cLλν=α2f7,ν−P−Tν=f8,z−τ′tρ−1τtzρ=f9.
Suppose that we have foundw and u. Then,
(74)w−φ=f1,u−ψ=f3.
Furthermore, by (73), we can find z as
(75)zx,0=φx,x∈0,L.
Following the same approach as in [1], we obtain, by using the last equation in (73),
(76)zx,ρ=φxe−ρτt+τte−ρτt∫01f9x,yeyτtdy,ifτ′t=0,zx,ρ=φxeηρt+eηρt∫01τt1−τ′tyf9x,ye−ηytdy,ifτ′t≠0,where ηyt=τt/τ′tln1−τ′tρ. Whereupon, from (74), we obtain
(77)zx,ρ=φxe−ρτt−f1e−ρτt+τte−ρτt∫01f9x,yeyτtdy,ifτ′t=0,zx,ρ=φxeηρt−f1eηρt+eηρt∫01τt1−τ′tyf9x,ye−ηytdy,ifτ′t≠0.
Integrating (73)6 and (73)8 with η0=ν0=0, we have
(78)ησ=1−e−σθ+∫0σes−σf6sds,νσ=1−e−σP+∫0σes−σf8sds.
Substituting (73)1,3,6,8,9 into the others, we obtain the following system. Now, we have to find w, u, θ, and P as solutions of the equations:
(79)w+d2wxxxx+μ1φ+μ2z.,1,t=f2+f1+β1f1,u−d1uxx−δ1θx−δ2Px=f4+f3,α1θ−rCβθxx+dCλPxx+dδ2−rδ1ux=h3,α2P+dCβθxx−cCλPxx+dδ1−cδ1ux=h4.
Solving (79), we get
(80)μ3w+d2wxxxx=h1,u−d1uxx−δ1θx−δ2Px=h2,α1θ−rCβθxx+dCλPxx+dδ2−rδ1ux=h3,α2P+dCβθxx−cCλPxx+dδ1−cδ1ux=h4,where
(81)μ3=1+μ1+e−τt,h1=f2+1+μ1f2−μ2z0,h2=f4+f3,h3=α1f5+dδ2−rδ1f3x+r∫0∞βσ∫0σes−σf6xxsdsdσ−d∫0∞λσ∫0σes−σf8xxsdsdσ,h4=α2f7+dδ1−cδ2f5x−d∫0∞βσ∫0σes−σf6xxsdsdσ+c∫0∞λσ∫0σes−σf8xxsdsdσ.
From (77), we have
(82)zx,1=wxe−τt+z0x,ifτ′t=0,wxeηρt+z0x,ifτ′t≠0,where x∈0,L and
(83)z0x=−f1e−ρτt+τte−ρτt∫01f9x,yeyτtdy,ifτ′t=0,−f1eηρt+eηρt∫01τt1−τ′tyf9x,ye−ηytdy,ifτ′t≠0.
It is clear from the above formula thatz0 depends only on f1,f9. Consequently, problem (80) is equivalent to
(84)ζw,u,θ,P,w^,u^,θ^,P^=Γw^,u^,θ^,P^,where the bilinear form ζ:H020,L×H010,L×L20,L×L20,L2→ℝ and the linear form Γ:H020,L×H010,L×L20,L×L20,L→ℝ are defined by
(85)ζw,u,θ,P,w^,u^,θ^,P^=∫0Lμ3ww^+d2wxxw^xx+uu^+d1uxu^xdx+α1∫0Lθθ^dx+α2∫0LPP^dx+rCβ∫0Lθxθ^xdx+cCλ∫0LPxP^xdx−dCβ∫0LθxP^xdx−dCλ∫0LPxθ^xdx+dδ2−rδ1∫0Luxθ^dx+dδ1−cδ2∫0LuxP^dx+∫0Lδ1θ+δ2Pu^xdx,Γw^,u^,θ^,P^=∫0Lh1w^dx+∫0Lh2u^dx+∫0Lh3θ^dx+∫0Lh4P^dx.
Now, forH1=H020,L×H010,L×L20,L×L20,L, equipped with the norm
(86)∥w,u,θ,P∥H12=∥w∥22+∥wxx∥22+∥u∥22+∥ux∥22+∥θ∥22+∥θx∥22+∥Px∥22+∥P∥22,then, we have
(87)Bw,u,θ,P,w,u,θ,P=μ3∫0Lw2dx+d2∫0Lwxx2dx+∫0Lu2dx+d1∫0Lux2dx+α1∫0Lθ2dx+α2∫0LP2dx+rCβ∫0Lθx2dx+cCλ∫0LPx2dx−dCβ+dCλ∫0LPxθxdx+dδ2−rδ1∫0Luxθdx+dδ1−cδ2∫0LuxPdx+∫0Lδ1θ+δ2Puxdx.
Then, for someM0>0,
(88)Bw,u,θ,P,w,u,θ,P≥M0∥w,u,θ,P∥H12.
Thus,B is coercive.
By Cauchy-Schwarz’s and Poincaré’s inequalities, we obtain(89)Bw,u,θ,P,w^,u^,θ^,P^≤M1∥w,u,θ,P∥H12∥w^,u^,θ^,P^∥H12.
Similarly, we get(90)Γw^,u^,θ^,P^≤M2∥w^,u^,θ^,P^∥H12.
Consequently, applying the Lax-Milgram theorem, problem (84) admits a unique solution w,u,θ,P∈H1, for all w^,u^,θ^,P^∈H1. Applying the classical elliptic regularity, it follows from (80) that w,u,θ,P∈H1.
Therefore, the operatorI−At is surjective for any fixed t>0. Since κt>0 and
(91)I−A¯t=1+κtI−At,we deduce that the operator I−A¯t is also surjective for any t>0.
To complete the proof of (3), it suffices to show that(92)∥U∥t∥U∥s≤ec/2τ0t−s,∀t,s∈0,T,where U=w,φ,u,ψ,θ,η,P,ν,zT and .t is the norm associated with the inner product (56).
Fort,s∈0,T, we have from (56) that
(93)∥U∥t2−∥U∥s2ec/τ0∣t−s∣=1−ec/τ0∣t−s∣∫0Lφ2+d2wxx2+d1ux2+ψ2dx+1−ec/τ0∣t−s∣<Λθ,PT,θ,PT>+1−ec/τ0∣t−s∣∥η∥Mβ2+∥ν∥Mλ2+ξτt−τsec/τ0∣t−s∣∫0L∫01z2x,ρ,tdρdx.
It is clear that1−ec/τ0∣t−s∣≤0. Now, we will prove that τt−τsec/τ0∣t−s∣≤0 for c>0. To do this, we have
(94)τt=τs+τ′at−s,where a∈s,t, which implies
(95)τtτs≤1+τaτst−s.
By using (11), we deduce that
(96)τtτs≤1+cτ0t−s≤ec/τ0t−s,which proves (92); therefore, this completes the proof of (3).
(4)
It is clear that(97)ddtAtU=00000000τ′′tτ′tρ−τ′tτ′tρ−1τ2tzρ.
Then, by (11) and (25), (4) holds exactly as in [1]. Consequently, from the above analysis, we deduce that the problem
(98)U¯t=A¯tU¯t,U¯t0=U0,has a solution U¯∈C0,∞,H, and if U0∈DA0, then
(99)U¯∈C0,∞,DA0∩C10,∞,H.
Now, let(100)Ut=eϑtU¯t,with ϑt=∫0tκsds; then, by using (98), we have
(101)Utt=κteϑtU¯t+eϑtU¯tt=κteϑtU¯t+eϑtA¯tU¯t=eϑtκtU¯t+A¯tU¯t=eϑtAtU¯t=AtUt.
Consequently,Ut is the unique solution of (46).
It remains to prove that the operatorF defined in (48) is locally Lipschitz in H.
LetU1=w1,φ1,u1,ψ1,θ1,η1,P1,ν1,z1T∈H and U2=w2,φ2,u2,ψ2,θ2,η2,P2,ν2,z2T∈H. Then, we have
(102)∥FU1−FU2∥=d1R2+K2,where
(103)R=u1x+12w1x2w1x−u2x+12w2x2w2x,K=12w1x2−w2x2.
Adding and subtracting the termu1x+1/2w1x2w2x inside the norm ∣R∣, we find
(104)∣R∣≤∥w1x−w2x∥L∞0,L∣u1x+12w1x2∣+∥w2x∥L∞∣u1x−u2x∣+12∥w2x∥L∞∣w1x+w2x∣w1x−w2xL∞0,L.
Using the embedding ofH10,L into L∞0,L, from (104), one has
(105)∣R∣≤k1∥U1∥H,∥U2∥H∥U1−U2∥.
Using once again the embedding ofH10,L into L∞0,L, one also sees that
(106)∣K∣≤k2∥U1∥H,∥U2∥H∥U1−U2∥.
Combining (102), (105), and (106), consequently, FU is locally Lipschitz continuous in H. This ends the proof of Theorem 1.


## 3. General Decay

In this section, we state and prove the stability of system (21)–(23) using the multiplier technique under the assumptions (26)–(31).We define the energy functionalE by
(107)Et=12∫0Lwt2+ut2+d2wxx2+d1ux+12wx22+cθ2+rP2dx+d<θ,P>+12∥η∥Mβ2+12∥ν∥Mλ2+ξ2∫0L∫01τtz2x,ρ,tdρdx,where
(108)∣μ2∣1−d≤ξ≤2μ1−μ21−d.The following lemma shows that the energy is decreasing.Lemma 3.
Assume that (26)–(31) hold and the hypotheses (7), (11), and (25) are satisfied. Then, for ∀C≥0,
(109)E′t≤−C∫0Lwt2dx+∫0Lz2x,1,tdx−α14∥η∥Mβ2+14∫0∞β′σ∥ηxσ∥2dσ−α24∥ν∥Mλ2+14∫0∞λ′σ∥νxσ∥2dσ≤0.Proof.
Multiplying the equations of (21) by wt, ut, θ, η, P, ν, and ξz, respectively, then by integration by parts, we get
(110)12ddt∫0Lwt2+ut2+d2wxx2+d1ux+12wx22+cθ2+rP2dx+ddtd<θ,P>+12∥η∥Mβ2+12∥ν∥Mλ2+ξ2ddt∫0L∫01τtz2x,ρ,tdρdx=−μ1∫01ωt2dx−μ2∫0Lwtzx,1,tdx+12∫0∞β′σ∥ηxσ∥2dσ+12∫0∞λ′σ∥νxσ∥2dσ+ξ2∫0L∫01τ′tz2x,ρ,tdρdx−ξ∫0L∫011−τ′tρzx,ρ,tzρx,ρ,tdρdx≤−μ1∫01ωt2dx−μ2∫0Lwtzx,1,tdx−α14∥η∥Mβ2−α24∥ν∥Mλ2+14∫0∞β′σ∥ηxσ∥2dσ+14∫0∞λ′σ∥νxσ∥2dσ−ξ2∫0L∫01ddρ1−τ′tρz2x,ρ,tdρdx=−μ1∫01ωt2dx−μ2∫0Lwtzx,1,tdx−α14∥η∥Mβ2−α24∥ν∥Mλ2+14∫0∞β′σ∥ηxσ∥2dσ+14∫0∞λ′σ∥νxσ∥2dσ+ξ2∫0Lz2x,0,t−z2x,1,tdx+ξτ′t2∫0Lz2x,1,tdx.
From (110), we find
(111)E′t≤−μ1−ξ2∫0Lwt2dx+ξτ′t2−ξ2∫0Lz2x,1,tdx−μ2∫0Lwtzx,1,tdx−α14∥η∥Mβ2−α24∥ν∥Mλ2+14∫0∞β′σ∥ηxσ∥2dσ+14∫0∞λ′σ∥νxσ∥2dσ.
Using Young’s inequality, we have(112)−μ2∫0Lwtzx,1,tdx≤∣μ2∣21−d∫0Lwt2dx+∣μ2∣1−d2∫0Lz2x,1,tdx.
Inserting (112) into (111), we get
(113)E′t≤−μ1−ξ2−μ221−d∫0Lwt2dx+ξ2τ′t−1+μ21−d2∫0Lz2x,1,tdx−α14∥η∥Mβ2+14∫0∞β′σ∥ηxσ∥2dσ−α24∥ν∥Mλ2+14∫0∞λ′σ∥νxσ∥2dσ.
Then, by using (7), (28)–(31), and (108), we obtain (109).In the following, we state and prove our stability result; we introduce and prove several lemmas.Lemma 4.
The functional(114)F1t≔∫0Lutu+12wtw+β14w2dx,satisfies, for any ε1>0,
(115)F1′t≤−d1∫0Lux+12wx22dx−d24∫0Lwxx2dx+∫0Lut2dx+12∫0Lwt2dx+2ε1∫0Lux2dx+δ124ε1∫0Lθ2+δ224ε1∫0LP2+c∫0Lz2x,1,tdx.Proof.
By differentiatingF1, then by integration by parts, we obtain
(116)F1′t=∫0Lut2dx+12∫0Lwt2dx−12d1∫0Lux+12wx2wx2dx−d1∫0Luxux+12wx2dx−μ22∫0Lwzx,1,tdx−d22∫0Lwxx2dx+δ1∫0Lθuxdx+δ2∫0LPuxdx.
In what follows, using Young’s and Poincaré’s inequalities, we obtain (115).Then, we have the following lemma.Lemma 5.
The functional(117)F2t≔∫0LutΦdx,where −δ1Φx=cθ+dP, with Φ0=ΦL=0, satisfies
(118)F2′t≤−∫0Lut2dx+ε2∫0Lux+12wx22dx+c∥η∥Mμ2+c1+1ε2∫0Lθ2dx+c1+1ε2∫0LP2dx.Proof.
For direct computations, we have(119)F2′t=∫0LuttΦdx⏟f1t+∫0LutΦtdx⏟f2t.
Using Young’s inequality and integrating by parts, we obtain(120)f1t≤ε2∫0Lux+12wx22dx+c1+1ε2∫0Lθ2dx+c1+1ε2∫0LP2dx,(121)f2t=−1δ1∫0Lut∂x−1∫0∞βσηxxσdσ+δ1utxdx=−1δ1∫0Lut∫0∞βσηxσdσ+δ1utdx≤−∫0Lut2dx+c∥η∥Mβ2.
From (120) and (121), we obtain (118).Lemma 6.
Assuming that assumptions (31) and (32) hold, the functional
(122)F3t≔−∫0∞βσ∫0Lcθ+dPηdxdσ⏟G1−∫0∞λσ∫0Ldθ+rPνdxdσ⏟G2,satisfies
(123)F3′t≤−c^∫0Lθ2dx−r^∫0LP2dx+β0∥η∥Mβ2+λ0∥ν∥Mλ2+cε3∫0Lut2dx−Cβ0∫0∞β′σ∥ηxσ∥2σdσ−Cλ0∫0∞λ′σ∥νxσ∥2σdσ,where
(124)c^=12β0c−β0+λ0dζ,r^=12λ0r−μ0+λ0dζ,and ζ>0 satisfies (29).Proof.
We take the derivative ofF3=G1+G2, which gives
(125)G1′t=−∫0∞βσ∫0Lcθ+dPtηdxdσ−∫0∞βσ∫0Lcθ+dPηtdxdσ=−∫0∞βσ∫0Lcθt+dPtηdxdσ+c∫0∞βσ∫0Lθησdxdσ+d∫0∞βσ∫0LPησdxdσ−cβ0∫0Lθ2dx−d∫0∞βσ∫0LPθdxdσ.
The first term on the right-hand side of (125) is
(126)−∫0∞βσ∫0Lcθ+dPtηdxdσ=−δ1∫0∞βσ∫0Lutxηdxdσ−∫0L∫0∞βσηxxdσ∫0∞βσηdσdx,and can be controlled in the following way:
(127)−δ1∫0∞βσ∫0Lutxηdxdσ≤Cε3∥η∥Mβ2+cε3∫0Lut2dx,(128)−∫0L∫0∞βσηxxdσ∫0∞βσηdσdx≤β0∥η∥Mβ2.
Moreover, by integration by parts, we get(129)c∫0∞βσ∫0Lθησdxdσ=c−∫0∞β′σ∫0Lθηdxdσ≤cμ04∫0Lθ2dx−Cβ0∫0∞β′σ∥ηxσ∥2dσ,where Cβ0>0. Similarly, we obtain
(130)d∫0∞βσ∫0LPησdxdσ=c−∫0∞β′σ∫0LPηdxdσ≤rλ04∫0LP2dx−Cβ0′∫0∞β′σ∥ηxσ∥2dσ,where C′β0>0. Using (29), we get
(131)−d∫0∞βσ∫0LθPdxdσ≤β0d2ζ∫0Lθ2dx+β0dζ2∫0LP2dx.
Then, we obtain(132)G1′t≤β02dζ−3c2∫0Lθ2dx+12β0dζ+rλ02∫0LP2dx+cε3∫0Lut2dx−Cβ0∫0∞β′σ∥ηxσ∥2dσ+β0+Cε3∥η∥Mβ2,where Cβ0=Cβ0+C′β0. Then, using the same arguments, we find
(133)G2′t≤12λ0dζ+β0c2∫0Lθ2dx+λ02dζ−3r2∫0LP2dx+cε3∫0Lut2dx−Cλ0∫0∞λ′σ∥νxσ∥2dσ+λ0+Cε3∥ν∥Mλ2.
Adding (127) and (133), we obtain (123).
We chooseζ in such a way that
(134)c^=12β0c−β0+λ0dζ>0,r^=12λ0r−β0+λ0dζ>0,which implies
(135)dc<β0+λ0β0dc<ζ<λ0β0+λ0rd<rd.
Then,ζ satisfies (29).Now, let us introduce the following functional.Lemma 7.
The functional(136)F4t≔ξτt∫0L∫01e−2τtρz2x,ρ,tdρdx,satisfies
(137)F4′t≤−2F4t−η1∫0Lz2x,1,tdx+ξ∫0Lwt2dx,where η1 is a positive constant.Proof.
By differentiatingF4, with respect to t, we have
(138)F4′t=ξτ′t∫0L∫01e−2τtρz2x,ρ,tdρdx+ξτt∫0L∫01−2τ′tρe−2τtρz2+e−2τtρztzdρdx.
By using the last equation of (21), we have
(139)τt∫0L∫01e−2τtρztzdρdx=∫0L∫01τ′tρ−1e−2τtρzρzdρdx=12∫0L∫01ddρτ′tρ−1e−2τtρz2dρdx+τt∫0L∫01τ′tρ−1e−2τtρz2dρdx−τ′t2∫0L∫01e−2τtρz2dx.
Using (137)–(139), we get
(140)F4′t=−2ξτt∫0L∫01e−2τtρz2x,ρ,tdρdx+ξ∫0Lz2x,0,tdx−ξ1−τ′te−2τt∫0Lz2x,1,tdx.
Then, by using (7), (25), and the fact that zx,0,t=wtx,t and setting η1=ξ1−de−2τ¯, we obtain (137).We are now ready to prove the following result.Theorem 8.
Assume (26)–(31) hold; there exist positive constants C1 and C2 such that the energy functional given by (107) satisfies
(141)Et≤C2e−C1t,∀t≥0.Proof.
We define a Lyapunov functional(142)Lt≔NEt+∑i=1i=3NiFit+F4t,where N and Ni, i=1,2,3, are positive constants to be selected later.
By differentiating (142) and using (109), (115), (118), (123), and (137), including the relation
(143)∫0Lux2dx=∫0Lux2+12wx2−12wx2dx≤2∫0Lux+12wx22dx−12∫0Lwx4dx≤2∫0Lux+12wx22dx−L4∫0Lwxx2dx,we get
(144)L′t≤−d1−2ε1N1−ε2N2∫0Lux+12wx22dx−N2−N1−cε3N3∫0Lut2dx−d24−L2ε1N1∫0Lwxx2dx−CN−12N1−ξ∫0Lωt2dx−c^N3−δ124ε1N1−c1+1ε2N2∫0Lθ2dx−r^N3−δ224ε1N1−c1+1ε2N2∫0LP2dx−CN−cN1+η1∫0Lz2x,1,tdx−2F4t−α14N−cN2−μ0N3∥η∥Mβ2−α24N−λ0N3∥ν∥Mλ2+14N−Cμ0N3∫0∞β′σ∥ηxσ∥2dσ+14N−Cλ0N3∫0∞λ′σ∥νxσ∥2dσ.
First, we chooseε1 small enough such that
(145)d1−2ε1>0,d24−L2ε1>0.
By setting(146)ε2=d1−2ε1N12N2,ε3=2cN3N2,we obtain
(147)L′t≤−12d1−2ε1N1∫0Lux+12wx22dx−12N2−N1∫0Lut2dx−d24−L2ε1N1∫0Lwxx2dx−CN−12N1−ξ∫0Lωt2dx−c^N3−δ124ε1N1−c1+N2N1N2∫0Lθ2d−r^N3−δ224ε1N1−c1+N2N1N2∫0LP2dx−CN−cN1+η1∫0Lz2x,1,tdx−2F4t−α14N−cN2−μ0N3∥η∥Mβ2−α24N−λ0N3∥ν∥Mλ2+14N−Cμ0N3∫0∞β′σ∥ηxσ∥2dσ+14N−Cλ0N3∫0∞λ′σ∥νxσ∥2dσ.
Next, we carefully choose our constants so that the terms inside the brackets are positive.
We chooseN2 large enough such that
(148)k1=12N2−N1>0.
Then, we chooseN3 large enough such that
(149)k2=r^N3−δ224ε1N1−c1+N2N1N2>0,k3=r^N3−δ224ε1N1−c1+N2N1N2>0.
Thus, we arrive at(150)L′t≤−k0∫0Lux+12wx22dx−k1∫0Lut2dx−k4∫0Lwxx2dx−CN−c∫0Lωt2dx−k2∫0Lθ2dx−k3∫0LP2dx−CN−c∫0Lz2x,1,tdx−2F4t−α14N−c∥η∥Mβ2−α24N−c∥ν∥Mλ2+14N−c∫0∞β′σ∥ηxσ∥2dσ+14N−c∫0∞λ′σ∥νxσ∥2dσ,where k0=1/2d1−2ε1N1 and k4=d2/4−L/2ε1N1.
On the other hand, we let(151)Tt=∑i=1i=3NiFit+F4t.
Exploiting Young’s, Cauchy-Schwarz’s, and Poincaré’s inequalities, we get(152)Tt≤c∫0Lωt2+ut2+ux+12wx22+ωxx2+θ2+P2dx+c∥η∥Mβ2+c∥ν∥Mλ2+c∫0L∫01z2x,ρ,tdρdx.
Then,(153)Tt≤cEt.
Consequently, we obtain(154)Tt=Lt−NEt≤cEt,that is,
(155)N−cEt≤Lt≤N+cEt.
Now, we chooseN large enough such that
(156)N−c>0,α14N−c>0,α24N−c>0,14N−c>0>0,CN−c>0.
Exploiting (107), estimates (150) and (155), respectively, give
(157)L′t≤−a1Et,for some a1>0, and
(158)c1Et≤Lt≤c2Et,∀t≥0,for some c1,c2>0; we have
(159)Lt~Et.
A combination with (157) and (158) gives
(160)L′t≤−C1Lt,∀t≥0,where C1=a1/c2.
Finally, by simple integration of (159) and (160), we obtain the result (141).



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