Exponential stabilty for a Timoshenko system with nonlocal delay

ABSTRACT

The purpose of this paper is to study the Timoshenko system with the nonlocal time-delayed condition. The well-posedness is proved by Hille-Yosida theorem. Exploring the dissipative properties of the linear operator associated with the full damped model, we obtain the exponential stability by using Gearhart-Huang-Prüss theorem.

Keywords:
Timoshenko system; nonlocal time-delayed condition; exponential stability

RESUMO

O objetivo deste artigo é estudar o sistema de Timoshenko com uma condição de retardo de tempo não local. A boa colocação é provada através do teorema de Hille- Yosida. Explorando as propriedades dissipativas do operador linear associado ao modelo totalmente amortecido, obtemos a estabilidade exponencial usando o Teorema de Gearhart- Huang-Prüss.

Palavras-chave:
sistema de Timoshenko; condição de retardo não local; estabilidade exponencial

1 INTRODUCTION

The history of nonlocal problems with integral conditions for partial differential equations is re- cent and goes back to ⁴. In particular, a review of the progress related to the nonlocal models with integral type was given in ³ with many discussions about physical justifications, advan- tages, and numerical applications. For a nonlocal hyperbolic equation with integral conditions of the 1st kind, we cite ²⁰. Dissipative properties associated with the Timoshenko system have been studied by several authors by considering the dissipative mechanism of frictional or vis- coelastic type. An interesting problem was brought out when the dissipation acts in different ways on the domain. For the case of terms of time-varying delay in the internal feedbacks, the stability result of the Timoshenko system can be found in ¹⁰. On the other hand, for the case of delay and boundary feedback, we can see in ²⁴. The Timoshenko beam system with delay in the boundary control was studied in ²⁵ where the exponential stabilization result is proved via a test of exact observability of the system. Distributed delay in the boundary control was considered in ¹³. Distributive delay in a Timoshenko-type system of thermoelasticity of type III was considered in [8] and then, in ⁹, the same problem was dealt with constant delay. The Timoshenko system with second sound and the internal distributed delay was investigated in ².

The transmission problem with delay in porous-elasticity was considered in ²¹. For a nonlinear Timoshenko system with delay, we cite ⁵ and reference therein. As far as we know, there is no result for the Timoshenko system with nonlocal delay.

Let $\Omega =(0,L)$ be an interval on ℝ. In this paper, $\phi =\phi (x,t)$ describe the small transverse displacement of the beam and $\psi =\psi (x,t)$ the rotation angle of a filament of the beam in Ω, respectively at the time t. For a constant $c>0$ and

$F,G:(0,c)\to \mathrm{ℝ}$

bounded functions, we define the nonlocal time delayed integral of the 1st kind condition by

${\int }_0^{c}F\left(s\right){\phi }_t(x,t-s)ds,{\int }_0^{c}G\left(s\right){\psi }_t(x,t-s)ds.$ (1.1)

These conditions (1.1) are called nonlocal because the integral is not a pointwise relation, so it provides a problem with distributed delay. Well-posedness and exponential stability for this kind of nonlocal time-delayed for a wave equation were studied in ^{17), (22} by different techniques.

Let b, k, α, β be positive constants. The Timoshenko system with frictional damping and nonlocal time-delayed condition is given by

${\rho }_1{\phi }_{tt}-k{({\phi }_x+\psi )}_x+\alpha {\phi }_t+{\int }_0^{c}F\left(s\right){\phi }_t(x,t-s)ds=0,\text{in}\Omega \times (0,\infty ),$ (1.2)

${\rho }_2{\psi }_{tt}-b{\psi }_{xx}+k({\phi }_x+\psi )+\beta {\psi }_t+{\int }_0^{c}G\left(s\right){\psi }_t(x,t-s)ds=0,\text{in}\Omega \times (0,\infty ),$ (1.3)

$\phi (x,0)={\phi }_0\left(x\right)\text{and}\psi (x,0)={\psi }_0\left(x\right),x\in \Omega ,$ (1.4)

${\phi }_t(x,0)={\phi }_1\left(x\right)\text{and}{\psi }_t(x,0)={\psi }_1\left(x\right),x\in \Omega ,$ (1.5)

${\phi }_t(x,-s)=f_0(x,-s),x\in \Omega ,s\in (0,c),$ (1.6)

${\psi }_t(x,-s)=g_0(x,-s),x\in \Omega ,s\in (0,c).$ (1.7)

We consider the Dirichlet boundary conditions as follows

$\phi (0,t)=\phi (L,t)=0\text{and}\psi (0,t)=\psi (L,t)=0,t>0.$ (1.8)

Here the initial data

$\left({\phi }_0\right(x),{\psi }_0(x\left)\right)\in H_0^1(0,1)\times H_0^1(0,1),\left({\phi }_1\right(x),{\psi }_1(x\left)\right)\in L^2(0,1)\times L^2(0,1),$

$f_0(x,-s),g_0(x,-s)$ belong to suitable spaces and

${\int }_0^{c}F\left(s\right)ds<\alpha \text{and}{\int }_0^{c}G\left(s\right)ds<\beta .$

We use the Sobolev spaces with its properties as in Adams ¹ and the semigroup theory ( see Pazy ¹⁸). In this paper, we apply the semigroup technique for dissipative systems (see Liu and Zheng ¹⁴), that is different from some others in the literature, for example, like as the energy method (see Rivera ²³), the direct method (see Kormonik ^{11), (12}) and the Nakao’s method (see ¹⁵). This manuscript is organized as follows. In Section 2, we deal with the semigroup setting where we prove the well-posedness of the system. In section 3, we show the exponential stability by using the Gearhart-Huang-Prüss theorem, ^{6), (7), (19}.

2 SEMIGROUP SETUP

As in Nicaise and Pignotti ¹⁷ we introduce the new variables

$$\begin{gathered} z(x,\rho ,t,s)={\phi }_t(x,t-\rho s),(x,\rho ,t,s)\in Q, \\ y(x,\rho ,t,s)={\psi }_t(x,t-\rho s),(x,\rho ,t,s)\in Q, \end{gathered}$$

where $Q=\Omega \times (0,1)\times (0,\infty )\times (0,c)$

The new variables z, y satisfy

$$\begin{gathered} s{z}_t(x,\rho ,t,s)+z_{\rho }(x,\rho ,t,s)=0, \\ s{y}_t(x,\rho ,t,s)+y_{\rho }(x,\rho ,t,s)=0. \end{gathered}$$

Moreover, using the approach as in ¹⁶, the equations

$\lambda sz(x,\rho ,t,s)+z_{\rho }(x,\rho ,t,s)=f,\text{with}\lambda >0,f\in L^2\left(Q\right),$ (2.1)

$\lambda sy(x,\rho ,t,s)+y_{\rho }(x,\rho ,t,s)=g,\text{with}\lambda >0,g\in L^2\left(Q\right)$ (2.2)

has unique solution

$z(x,\rho ,t,s)=z(x,0,t,s)e^{-\lambda \rho s}+e^{-\lambda \rho s}{\int }_0^{\rho }{e}^{\lambda \sigma s}f(x,\sigma ,t,s)d\sigma ,$ (2.3)

$y(x,\rho ,t,s)=y(x,0,t,s)e^{-\lambda \rho s}+e^{-\lambda \rho s}{\int }_0^{\rho }{e}^{\lambda \sigma s}g(x,\sigma ,t,s)d\sigma ,$ (2.4)

respectively. The problem (1.2)-(1.7) is equivalent to

${\rho }_1{\phi }_{tt}-k({\phi }_x+\psi {)}_x+\alpha {\phi }_t+{\int }_0^{c}F(s\left)z\right(x,1,t,s)ds=0,(x,t)\in \Omega \times (0,\infty ),$ (2.5)

${\rho }_2{\psi }_{tt}-b{\psi }_{xx}+k({\phi }_x+\psi )+\beta {\psi }_t+{\int }_0^{c}G\left(s\right)y(x,1,t,s)ds=0,(x,t)\in \Omega \times (0,\infty ),$ (2.6)

$s{z}_t(x,\rho ,t,s)+z_{\rho }(x,\rho ,t,s)=0,(x,\rho ,t,s)\in Q,$ (2.7)

$s{y}_t(x,\rho ,t,s)+y_{\rho }(x,\rho ,t,s)=0,(x,\rho ,t,s)\in Q,$ (2.8)

$\phi (x,0)={\phi }_0\left(x\right)\text{and}\psi (x,0)={\psi }_0\left(x\right),x\in \Omega ,$ (2.9)

${\phi }_t(x,0)={\phi }_1\left(x\right)\text{and}{\psi }_t(x,0)={\psi }_1\left(x\right),x\in \Omega ,$ (2.10)

$z(x,\rho ,0,s)=f_0(x,-\rho s),(x,\rho ,s)\in \Omega \times (0,1)\times (0,c),$ (2.11)

$\phi (x,\rho ,0,s)=g_0(x,-\rho s),(x,\rho ,s)\in \Omega \times (0,1)\times (0,c),$ (2.12)

with the Dirichlet boundary condition (1.8) and $z(x,\rho ,t,s)=y(x,\rho ,t,s)=0$ on the boundary $x=0,L$.

Defining $U=(\phi ,\psi ,u,v,z,y{)}^T,u={\phi }_t$ and $v={\psi }_t$, we formally get that U satisfies the Cauchy problem

$$\begin{gathered} U_t=\mathcal{A}U,t>0, \\ U\left(0\right)=U_0=({\phi }_0,{\psi }_0,{\phi }_1,{\psi }_1,f_0,g_0{)}^T, \end{gathered}$$ (2.13)

where the operator 𝒜 is defined by

$\mathcal{A}U=\left[\begin{array}{c}u\\ v\\ \frac{1}{{\rho }_1}\left[k({\phi }_x+\psi {)}_x-\alpha u-{\int }_0^{c}F(s\left)z\right(x,1,t,s)ds\right]\\ \frac{1}{{\rho }_2}\left[b{\psi }_{xx}-k({\phi }_x+\psi )-\beta v-{\int }_0^{c}G\left(s\right)y(x,1,t,s)ds\right]\\ -s^{-1}{z}_{\rho }(x,\rho ,t,s)\\ -s^{-1}{y}_{\rho }(x,\rho ,t,s)\end{array}\right].$

We introduce the energy space

$\mathcal{H}=H_0^1(\Omega {)}^2\times L^2(\Omega {)}^2\times L^2{\left(\Omega \times (0,1)\times (0,c)\right)}^2$

equipped with the inner product

$$\begin{gathered} ⟨U,\overline{U}{⟩}_{\mathcal{H}}={\int }_{\Omega }\left[k({\phi }_x+\psi )({\overline{\phi }}_x+\overline{\psi })+{\rho }_{1}u\overline{u}+{\rho }_{2}v\overline{v}+b{\psi }_x{\overline{\psi }}_x\right]dx \\ +{\int }_{\Omega }{\int }_0^1{\int }_0^{c}sF\left(s\right)z\overline{z}dsd\rho dx+{\int }_{\Omega }{\int }_0^1{\int }_0^{c}sG\left(s\right)y\overline{y}dsd\rho dx, \end{gathered}$$

for $U=(\phi ,\psi ,u,v,z,y{)}^T$ and $\overline{U}=(\overline{\phi },\overline{\psi },\overline{u},\overline{v},\overline{z},\overline{y}{)}^T$.

The domain of 𝒜 is defined by

$D\left(\mathcal{A}\right)={\left[H^2\left(\Omega \right)\cap H_0^1\left(\Omega \right)\right]}^2\times H_0^1(\Omega {)}^2\times L^2{\left(\Omega \times (0,1);H^1\left(\Omega \right)\right)}^2.$

Clearly, D(𝒜) is dense in ℋ and independent of time $t>0$. Next, we will prove that the operator 𝒜 is dissipative.

Lemma 2.1.For$U=(\phi ,\psi ,u,v,z,y)\in D\left(\mathcal{A}\right)$. Assume that

${\int }_0^{c}F\left(s\right)ds<\alpha \text{and}{\int }_0^{c}G\left(s\right)ds<\beta$ (2.14)

then we have

$$\begin{gathered} a⟨\mathcal{A}U,U{⟩}_{\mathcal{H}}\le -\left(\alpha -{\int }_0^{c}F\left(s\right)ds\right){\int }_{\Omega }{\left|u\right|}^{2}dx \\ -\left(\beta -{\int }_0^{c}G\left(s\right)ds\right){\int }_{\Omega }{\left|v\right|}^{2}dx \\ \le 0. \end{gathered}$$ (2.15)

Proof.

$$\begin{gathered} ⟨\mathcal{A}U,U{⟩}_{\mathcal{H}}=k{\int }_{\Omega }(u_x+v)({\phi }_x+\psi )dx \\ +{\int }_{\Omega }\left[k({\phi }_x+\psi {)}_x-\alpha u-{\int }_0^{c}F(s\left)z\right(x,1,t,s)ds\right]udx \\ +{\int }_{\Omega }\left[b{\psi }_{xx}-k({\phi }_x+\psi )-\beta v-{\int }_0^{c}G\left(s\right)y(x,1,t,s)ds\right]vdx \\ +b{\int }_{\Omega }{v}_x{\psi }_{x}dx-{\int }_{\Omega }{\int }_0^1{\int }_0^{c}F\left(s\right)z_{\rho }(x,\rho ,t,s)z(x,\rho ,t,s)dsd\rho dx \\ -{\int }_{\Omega }{\int }_0^1{\int }_0^{c}G\left(s\right)y_{\rho }(x,\rho ,t,s)y(x,\rho ,t,s)dsd\rho dx. \end{gathered}$$

Integrating by parts on Ω,

$$\begin{gathered} ⟨\mathcal{A}U,U{⟩}_{\mathcal{H}}=-\alpha {\int }_{\Omega }{\left|u\right|}^{2}dx-{\int }_{\Omega }{\int }_0^{c}F\left(s\right)z(x,1,t,s)udsdx \\ -\beta {\int }_{\Omega }{\left|v\right|}^{2}dx-{\int }_{\Omega }{\int }_0^{c}G\left(s\right)y(x,1,t,s)vdsdx \\ -{\int }_{\Omega }{\int }_0^1{\int }_0^{c}F\left(s\right)z_{\rho }(x,\rho ,t,s)z(x,\rho ,t,s)dsd\rho dx \\ -{\int }_{\Omega }{\int }_0^1{\int }_0^{c}G\left(s\right)y_{\rho }(x,\rho ,t,s)y(x,\rho ,t,s)dsd\rho dx. \end{gathered}$$ (2.16)

Taking into account $z(x,0,t,s)={\phi }_t(x,t)=u,y(x,0,t,s)={\psi }_t(x,t)=v$ we have

$$\begin{gathered} {\int }_{\Omega }{\int }_0^1{\int }_0^{c}F\left(s\right)z_{\rho }(x,\rho ,t,s)z(x,\rho ,t,s)dsd\rho dx \\ ={\int }_{\Omega }{\int }_0^c{\int }_0^{1}F\left(s\right)\frac{1}{2}\frac{\partial }{\partial \rho }\left|z\right(x,\rho ,t,s){|}^{2}d\rho dsdx \\ =\frac{1}{2}{\int }_{\Omega }{\int }_0^{c}F(s\left)\right|z(x,1,t,s){|}^{2}dsdx-\frac{1}{2}{\int }_0^{c}F\left(s\right)ds{\int }_{\Omega }{\left|u\right|}^{2}dx \end{gathered}$$ (2.17)

and

$$\begin{gathered} {\int }_{\Omega }{\int }_0^1{\int }_0^{c}G\left(s\right)y_{\rho }(x,\rho ,t,s)y(x,\rho ,t,s)dsd\rho dx \\ =\frac{1}{2}{\int }_{\Omega }{\int }_0^{c}G\left(s\right)\left|y\right(x,1,t,s){|}^{2}dsdx-\frac{1}{2}{\int }_0^{c}G(s)ds{\int }_{\Omega }{\left|v\right|}^{2}dx. \end{gathered}$$ (2.18)

Inserting (2.17) and (2.18) into (2.16), apply Young’s inequality and simplifying the terms, we obtain

$⟨\mathcal{A}U,U{⟩}_{\mathcal{H}}\le -\left(\alpha -{\int }_0^{c}F\left(s\right)ds\right){\int }_{\Omega }|u{|}^{2}dx-\left(\beta -{\int }_0^{c}G\left(s\right)ds\right){\int }_{\Omega }|v{|}^{2}dx.$

Finally, by the assumption (2.14) we conclude the proof. □

The well-posedness of (2.5)-(2.12) is ensured by the following theorem.

Teorema 2.1.For$U_0\in \mathcal{H}$, there exists a unique weak solution U of (2.13) satisfying

$U\in C\left(\right(0,\infty );\mathcal{H}).$ (2.19)

Moreover, if$U_0\in D\left(\mathcal{A}\right)$, then

$U\in C\left(\right(0,\infty );D(\mathcal{A}\left)\right)\cap C^1\left(\right(0,\infty );\mathcal{H}).$ (2.20)

Proof. We will use the Hille-Yosida theorem. Since 𝒜 is dissipative and D(𝒜 ) is dense in ℋ , it is sufficient to show that 𝒜 is maximal; that is, $I-\mathcal{A}$ is surjective. Given $H=(h_1,h_2,\dots ,h_6)\in \mathcal{H}$, we must show that there exists $U=(\phi ,\psi ,u,v,z,y)\in D\left(\mathcal{A}\right)$ satisfying $(I-\mathcal{A})U=H$ which is equivalent to

$\phi -u=h_1,$ (2.21)

$\psi -v=h_2,$ (2.22)

${\rho }_{1}u-k({\phi }_x+\psi {)}_x+\alpha u+{\int }_0^{c}F(s\left)z\right(x,1,t,s)ds={\rho }_1{h}_3,$ (2.23)

${\rho }_{2}v-b{\psi }_{xx}+k({\phi }_x+\psi )+\beta v+{\int }_0^{c}G\left(s\right)y(x,1,t,s)ds={\rho }_2{h}_4,$ (2.24)

$sz(x,\rho ,t,s)+z_{\rho }(x,\rho ,t,s)=s{h}_5,$ (2.25)

$sy(x,\rho ,t,s)+y_{\rho }(x,\rho ,t,s)=s{h}_6.$ (2.26)

Suppose that we have found φ and ψ with the appropriated regularity. Therefore, (2.21) and (2.22) give

$u=\phi -h_1,$ (2.27)

$v=\psi -h_2.$ (2.28)

It is clear that u,$u,v\in H_0^1\left(\Omega \right)$.

From (2.1),(2.3) it follows that equation (2.25) has a unique solution given by

$z(x,\rho ,s)=u\left(x\right)e^{-\rho s}+s{e}^{-\rho s}{\int }_0^{\rho }{e}^{\sigma s}{h}_5(x,\sigma ,s)d\sigma$

and from (2.2),(2.4) it follows that equation (2.26) has a unique solution given by

$y(x,\rho ,s)=v\left(x\right)e^{-\rho s}+s{e}^{-\rho s}{\int }_0^{\rho }{e}^{\sigma s}{h}_6(x,\sigma ,s)d\sigma .$

So, from (2.27) and (2.28),

$$\begin{gathered} z(x,\rho ,s)=\phi \left(x\right)e^{-\rho s}-h_1\left(x\right)e^{-\rho s}+s{e}^{-\rho s}{\int }_0^{\rho }{e}^{\sigma s}{h}_5(x,\sigma ,s)d\sigma , \\ y(x,\rho ,s)=\psi \left(x\right)e^{-\rho s}-h_2\left(x\right)e^{-\rho s}+s{e}^{-\rho s}{\int }_0^{\rho }{e}^{\sigma s}{h}_6(x,\sigma ,s)d\sigma , \end{gathered}$$

and, in particular,

$$\begin{gathered} z(x,1,s)=\phi \left(x\right)e^{-s}+z_0(x,s), \\ y(x,1,s)=\psi \left(x\right)e^{-s}+y_0(x,s), \end{gathered}$$

where $z_0(x,s),y_0(x,s)\in L^2\left(\Omega \times (0,c)\right)$ defined by

$$\begin{gathered} z_0(x,s)=-h_1\left(x\right)e^{-s}+s{e}^{-s}{\int }_0^1{e}^{\sigma s}{h}_5(x,\sigma ,s)d\sigma , \\ {y}_0(x,s)=-h_2\left(x\right)e^{-s}+s{e}^{-s}{\int }_0^1{e}^{\sigma s}{h}_6(x,\sigma ,s)d\sigma . \end{gathered}$$

By (2.23), (2.24), (2.27) and (2.28), we see that the functions φ and ψ satisfy the following system

$\left\{\begin{array}{l}\lambda \phi -k{\left({\phi }_x+\psi \right)}_x=g_1,\\ \eta \psi -b{\psi }_{xx}+k\left({\phi }_x+\psi \right)=g_2,\end{array}\right.$ (2.29)

where

$$\begin{gathered} \lambda ={\rho }_1+\alpha +{\int }_0^c{e}^{-s}F\left(s\right)ds,\eta ={\rho }_2+\beta +{\int }_0^c{e}^{-s}G\left(s\right)ds, \\ {g}_1={\rho }_1{h}_1+\alpha h_1+{\rho }_1{h}_3-{\int }_0^{c}F\left(s\right)z_0(x,s)ds, \\ {g}_2={\rho }_2{h}_2+\beta h_2+{\rho }_2{h}_4-{\int }_0^{c}G\left(s\right)y_0(x,s)ds. \end{gathered}$$

Solving the system (2.29) is equivalent to finding $\left(\phi ,\psi \right)\in {\left[H^2\left(\Omega \right)\cap H_0^1\left(\Omega \right)\right]}^2$ such that

$\left\{\begin{array}{l}\lambda {\int }_{\Omega }\phi \tilde{\phi }dx+k{\int }_{\Omega }\left({\phi }_x+\psi \right){\tilde{\phi }}_{x}dx={\int }_{\Omega }{g}_1\tilde{\phi }dx,\\ \\ \eta {\int }_{\Omega }\psi \tilde{\psi }dx+b{\int }_{\Omega }{\psi }_x{\tilde{\psi }}_{x}dx+k{\int }_{\Omega }\left({\phi }_x+\psi \right)\tilde{\psi }dx={\int }_{\Omega }{g}_2\tilde{\psi }dx,\end{array}\right.$ (2.30)

for all $\left(\tilde{\phi },\tilde{\psi }\right)\in H_0^1\left(\Omega \right)\times H_0^1\left(\Omega \right)$.

Now, we observe that solving the system (2.30) is equivalent to solve the problem

${\rm Y}\left(\left(\phi ,\psi \right),\left(\tilde{\phi },\tilde{\psi }\right)\right)=L\left(\tilde{\phi },\tilde{\psi }\right)$ (2.31)

where the bilinear form

${\rm Y}:{\left[H_0^1\left(\Omega \right)\times H_0^1\left(\Omega \right)\right]}^2\to (0,\infty )$

and the linear form

$L:H_0^1\left(\Omega \right)\times H_0^1\left(\Omega \right)\to (0,\infty )$

are defined by

$$\begin{gathered} {\rm Y}\left(\left(\phi ,\psi \right),\left(\tilde{\phi },\tilde{\psi }\right)\right)=\lambda {\int }_{\Omega }\phi \tilde{\phi }dx+k{\int }_{\Omega }\left({\phi }_x+\psi \right)\left({\tilde{\phi }}_x+\tilde{\psi }\right)dx \\ +\eta {\int }_{\Omega }\psi \tilde{\psi }dx+b{\int }_{\Omega }{\psi }_x{\tilde{\psi }}_{x}dx \end{gathered}$$

and

$L\left(\tilde{\phi },\tilde{\psi }\right)={\int }_{\Omega }{g}_1\tilde{\phi }dx+{\int }_{\Omega }{g}_2\tilde{\psi }dx.$

It is easy to verify that ϒ is continuous and coercive, and L is continuous. So applying the Lax-Milgram theorem, we deduce that for all $\left(\tilde{\phi },\tilde{\psi }\right)\in H_0^1\left(\Omega \right)\times H_0^1\left(\Omega \right)$ the problem (2.31) admits a unique solution

$\left(\phi ,\psi \right)\in H_0^1\left(\Omega \right)\times H_0^1\left(\Omega \right).$

Applying the classical elliptic regularity, it follows from (2.30) that

$\left(\phi ,\psi \right)\in H^2\left(\Omega \right)\times H^2\left(\Omega \right).$

Therefore, the operator $I-\mathcal{A}$ is surjective. As consequence of the Hille-Yosida theorem [¹⁴, Theorem 1.2.2, page 3], we have that 𝒜 generates a C ₀-semigroup of contractions $S\left(t\right)=e^{t\mathcal{A}}$ on ℋ. From semigroup theory, $U\left(t\right)=e^{t\mathcal{A}}{U}_0$ is the unique solution of (2.13) satisfying (2.19) and (2.20). The proof is complete. □

3 EXPONENTIAL STABILITY

The necessary and sufficient conditions for the exponential stability of the C ₀-semigroup of con- tractions on a Hilbert space were obtained by Gearhart ⁶ and Huang ⁷ independently, see also Pruss ¹⁹. We will use the following result due to Gearhart.

Teorema 3.2.Let ρ(𝒜) be the resolvent set of the operator 𝒜 and$S\left(t\right)=e^{t\mathcal{A}}$be the C ₀-semigroup of contractions generated by 𝒜. Then, S(t) is exponentially stable if and only if

$i\mathrm{ℝ}=\{i\zeta :\zeta \in \mathrm{ℝ}\}\subset \rho \left(\mathcal{A}\right),$ (3.1)

$\underset{\left|\zeta \right|\to \infty }{\lim \sup }||(i\zeta I-\mathcal{A}{)}^{-1}||<\infty .$ (3.2)

The main result of this manuscript is the following theorem.

Teorema 3.3. The semigroup $S\left(t\right)=e^{t\mathcal{A}}$ generated by 𝒜 is exponentially stable.

Proof. It is sufficient to verify (3.1) and (3.2). If (3.1) is not true, it means that there is a $\zeta \in \mathrm{ℝ}$ such that $\zeta \ne 0$, iζ is in the spectrum de 𝒜. From the compact immersion of D(𝒜 ) in ℋ, by spectral theory, there is a vector function

$U=(\phi ,\psi ,u,v,z,y)\in D\left(\mathcal{A}\right),\text{with}\Vert U{\Vert }_{\mathcal{H}}=1,$

such that $\mathcal{A}U=i\zeta U$, which is equivalent to

$i\zeta \phi -u=0,$ (3.3)

$i\zeta \psi -v=0,$ (3.4)

$i\zeta u-\frac{1}{{\rho }_1}\left[k({\phi }_x+\psi {)}_x-\alpha u-{\int }_0^{c}F(s\left)z\right(x,1,t,s)ds\right]=0,$ (3.5)

$i\zeta v-\frac{1}{{\rho }_2}\left[b{\psi }_{xx}-k({\phi }_x+\psi )-\beta v-{\int }_0^{c}G\left(s\right)y(x,1,t,s)ds\right]=0,$ (3.6)

$i\zeta sz(x,\rho ,t,s)+z_{\rho }(x,\rho ,t,s)=0,$ (3.7)

$i\zeta sy(x,\rho ,t,s)+y_{\rho }(x,\rho ,t,s)=0.$ (3.8)

Using (3.3) we obtain $u_x=i\zeta {\phi }_x$. Multiplying by u _x , integrating on Ω and using Young’s inequality we have

${\int }_{\Omega }|u_x{|}^{2}dx=i\zeta {\int }_{\Omega }{\phi }_x{u}_{x}dx\le -\frac{1}{2}{\zeta }^2{\int }_{\Omega }|{\phi }_x{|}^{2}dx+\frac{1}{2}{\int }_{\Omega }|u_x{|}^{2}dx,$

from where it follows that

$\frac{1}{2}{\zeta }^2{\int }_{\Omega }|{\phi }_x{|}^{2}dx+\frac{1}{2}{\int }_{\Omega }|u_x{|}^{2}dx\le 0.$ (3.9)

Applying Poincaré’s inequality in (3.9) we obtain $\phi =u=0$ a.e. in L ²(Ω). Note that (2.3) gives us $z=u{e}^{-i\zeta \rho s}$ as the unique solution of (3.7), which implies $z=0$ a.e. in $L^2\left(\Omega \times (0,1)\times (0,c)\right)$. Similarly, it is proved that $\psi =v=0$ a.e. in L ²(Ω) and $y=0$ a.e. in $L^2\left(\Omega \times (0,1)\times (0,c)\right)$. But $\phi =\psi =u=v=z=y=0$ is a contradiction with $\Vert U{\Vert }_{\mathcal{H}}=1$ and then (3.1) holds.

To prove (3.2) we use contradiction argument again. If (3.2) is not true, there exists a real sequence ζ _n , with ${\zeta }_n\to \infty$ and a sequence of vector functions $V_n\in \mathcal{H}$ that satisfies

$\frac{\Vert ({\lambda }_{n}I-\mathcal{A}{)}^{-1}{V}_n{\Vert }_{\mathcal{H}}}{\Vert V_n{\Vert }_{\mathcal{H}}}\ge n,\text{where}{\lambda }_n=i{\zeta }_n.$

Hence

$\Vert ({\lambda }_{n}I-\mathcal{A}{)}^{-1}{V}_n{\Vert }_{\mathcal{H}}\ge n\Vert V_n{\Vert }_{\mathcal{H}}.$ (3.10)

Since ${\lambda }_n\in \rho \left(\mathcal{A}\right)$ it follows that there exists a unique sequence

$U_n=({\phi }_n,{\psi }_n,u_n,v_n,z_n,y_n{)}^T\in D(\mathcal{A}),$

with unit norm in ℋ such that

$({\lambda }_{n}I-\mathcal{A}{)}^{-1}{V}_n=U_n.$

Denoting ${\xi }_n={\lambda }_n{U}_n-\mathcal{A}{U}_n$ we have from (3.10) that

$\Vert {\xi }_n{\Vert }_{\mathcal{H}}\le \frac{1}{n}$

and then ${\xi }_n\to 0$ strongly in ℋ as $n\to \infty$.

Taking the inner product of ξ _n with U _n we have

${\lambda }_n\Vert U_n{\Vert }_{\mathcal{H}}^2-⟨\mathcal{A}{U}_n,U_n{⟩}_{\mathcal{H}}=⟨{\xi }_n,U_n{⟩}_{\mathcal{H}}.$

Using Lemma 2.1, follows that

$$\begin{gathered} {\lambda }_n\Vert U_n{\Vert }_{\mathcal{H}}^2+\left(\alpha -{\int }_0^{c}F\left(s\right)ds\right){\int }_{\Omega }{\left|u_n\right|}^{2}dx+\left(\beta -{\int }_0^{c}G\left(s\right)ds\right){\int }_{\Omega }{\left|v_n\right|}^{2}dx \\ \le ⟨{\xi }_n,U_n{⟩}_{\mathcal{H}} \end{gathered}$$

and taking the real part we have

$\left(\alpha -{\int }_0^{c}F\left(s\right)ds\right){\int }_{\Omega }{\left|u_n\right|}^{2}dx+\left(\beta -{\int }_0^{c}G\left(s\right)ds\right){\int }_{\Omega }{\left|v_n\right|}^{2}dx\le Re⟨{\xi }_n,U_n{⟩}_{\mathcal{H}}.$

As U _n is bounded and ${\xi }_n\to 0$ we obtain

$u_n\to 0\text{and}{v}_n\to 0\text{as}n\to \infty .$ (3.11)

Now, for ${\xi }_n=({\xi }_n^1,{\xi }_n^2,\dots ,{\xi }_n^6),{\xi }_n={\lambda }_n{U}_n-\mathcal{A}{U}_n$ is equivalent to

$i{\zeta }_n{\phi }_n-u_n={\xi }_n^1,$ (3.12)

$i{\zeta }_n{\psi }_n-v_n={\xi }_n^2,$ (3.13)

$i{\zeta }_n{u}_n-\frac{1}{{\rho }_1}\left[k({\phi }_{x,n}+{\psi }_n{)}_x-\alpha u_n-{\int }_0^{c}F(s\left)z_n\right(x,1,t,s)ds\right]={\xi }_n^3,$ (3.14)

$i{\zeta }_n{v}_n-\frac{1}{{\rho }_2}\left[b{\psi }_{xx,n}-k({\phi }_{x,n}+{\psi }_n)-\beta v_n-{\int }_0^{c}G\left(s\right)y_n(x,1,t,s)ds\right]={\xi }_n^4,$ (3.15)

$i{\zeta }_{n}s{z}_n(x,\rho ,t,s)+z_{\rho ,n}(x,\rho ,t,s)=s{\xi }_n^5,$ (3.16)

$i{\zeta }_{n}s{y}_n(x,\rho ,t,s)+y_{\rho ,n}(x,\rho ,t,s)=s{\xi }_n^6,$ (3.17)

where

${\xi }_n^j\to 0,$ (3.18)

for $j=1,\dots ,6$.

From (3.11), (3.12), (3.13) and (3.18), we obtain that

${\phi }_n\to 0\text{and}{\psi }_n\to 0\text{as}n\to \infty .$ (3.19)

By (2.1) and (2.3), we have that

$z_n(x,\rho ,s)=u_n\left(x\right)e^{-i{\zeta }_n\rho s}+s{e}^{-i{\zeta }_n\rho s}{\int }_0^{\rho }{e}^{i{\zeta }_n\sigma s}{\xi }_n^5(x,\sigma ,s)d\sigma$ (3.20)

is the unique solution of (3.16). Using the Euler formula for complex numbers in (3.20) we obtain

$$\begin{gathered} z_n(x,\rho ,s)=u_n\left(x\right)\left[\cos \left({\zeta }_n\rho s\right)-i\sin \left({\zeta }_n\rho s\right)\right] \\ +s\left[\cos \left({\zeta }_n\rho s\right)-i\sin \left({\zeta }_n\rho s\right)\right]{\int }_0^{\rho }\left[\cos \left({\zeta }_n\sigma s\right)+i\sin \left({\zeta }_n\sigma s\right)\right]{\xi }_n^5(x,\sigma ,s)d\sigma . \end{gathered}$$

Since $\cos \left({\zeta }_n\rho s\right)-i\sin \left({\zeta }_n\rho s\right)$ and $\cos \left({\zeta }_n\sigma s\right)+i\sin \left({\zeta }_n\sigma s\right)$ limited and by (3.11), (3.18) deducing that

$z_n(x,\rho ,t)\to 0\text{as}n\to \infty .$ (3.21)

Analogously, we conclude that

$y_n(x,\rho ,t)\to 0\text{as}n\to \infty .$ (3.22)

Finally, (3.11), (3.19), (3.21) and (3.22) give us a contradiction with $\Vert U_n{\Vert }_{\mathcal{H}}=1$. The proof is complete. □

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Publication Dates

• Publication in this collection
  30 Nov 2020
• Date of issue
  Sep-Dec 2020

History

• Received
  16 Jan 2020
• Accepted
  27 May 2020