Exponential stabilization of swelling porous systems with thermoelastic damping

Lemma 2.1

The energy E of system (1.13) defined by (2.6), satisfies

$$E^{\prime }(t)=-{\mu }_2{\int }_0^L{\theta }_x^2\mathit{dx}\le 0,\forall t\ge 0\text{.}$$

Proof

By respectively multiplying (1.13)₁, (1.13)₂, and (1.13)₃ by $u_t,v_t$ , and $\theta$ , then using integration by parts with respect to x on the product, we get

$$\frac{{\rho }_1}{2}\frac{d}{\mathit{dt}}{\int }_0^L{u}_t^2\mathit{dx}+\frac{a_1}{2}\frac{d}{\mathit{dt}}{\int }_0^L{u}_x^2\mathit{dx}+a_2\frac{d}{\mathit{dt}}{\int }_0^L{u}_x{v}_x\mathit{dx}-a_2{\int }_0^L{u}_x{v}_{\mathit{xt}}\mathit{dx}=0,\forall t\ge 0,$$

(2.10)

$$\frac{{\rho }_2}{2}\frac{d}{\mathit{dt}}{\int }_0^L{v}_t^2\mathit{dx}+\frac{{\mu }_1}{2}\frac{d}{\mathit{dt}}{\int }_0^L{v}_x^2\mathit{dx}+a_2{\int }_0^L{u}_x{v}_{\mathit{xt}}\mathit{dx}+{\beta }_2{\int }_0^L{v}_{\mathit{xt}}\theta \mathit{dx}=0,\forall t\ge 0,$$

(2.11)

$$\frac{\rho }{2}\frac{d}{\mathit{dt}}{\int }_0^L{\theta }^2\mathit{dx}+{\mu }_2{\int }_0^L{\theta }_x^2\mathit{dx}-{\beta }_2{\int }_0^L{v}_{\mathit{xt}}\theta \mathit{dx}=0,\forall t\ge 0\text{.}$$ Summing (2.9)–(2.11) and considering (2.6), yields (2.8). square

Lemma 2.2

The functional ${\mathcal{F}}_1$ , defined by (2.2), satisfies

$$\begin{array}{cc}{\mathcal{F}}_1^{\prime }(t)⩽-\hfill & \frac{a_2^2}{2}{\int }_0^L{u}_x^2\mathit{dx}+\left(\frac{a_2^2{\rho }_2}{{\rho }_1}+{\left(\frac{a_1{\rho }_2}{{\rho }_1}-{\mu }_1\right)}^2\right){\int }_0^L{v}_x^2\mathit{dx}+{\beta }_2^2{c}_p{\int }_0^L{\theta }_x^2\mathit{dx},\forall t\ge 0\text{.}\hfill \end{array}$$

Proof

Taking the derivative of (2.2) and using (1.13), we see that, for all $t\ge 0$ ,

$${\mathcal{F}}_1^{\prime }(t)=-a_2^2{\int }_0^L{u}_x^2\mathit{dx}+\frac{a_2^2{\rho }_2}{{\rho }_1}{\int }_0^L{v}_x^2\mathit{dx}+a_2\left(\frac{a_1{\rho }_2}{{\rho }_1}-{\mu }_1\right){\int }_0^L{u}_x{v}_x\mathit{dx}-a_2{\beta }_2{\int }_0^L{u}_x\theta \mathit{dx}\text{.}$$ By first utilizing Young’s inequality followed by Poincaré’s inequality, we get,

(2.14)

$$a_2\left(\frac{a_1{\rho }_2}{{\rho }_1}-{\mu }_1\right){\int }_0^L{u}_x{v}_x\mathit{dx}⩽\frac{a_2^2}{4}{\int }_0^L{u}_x^2\mathit{dx}+{\left(\frac{a_1{\rho }_2}{{\rho }_1}-\mu \right)}^2{\int }_0^L{v}_x^2\mathit{dx},\forall t\ge 0,$$

(2.15)

$$-a_2{\beta }_2{\int }_0^L{u}_x\theta \mathit{dx}⩽\frac{a_2^2}{4}{\int }_0^L{u}_x^2\mathit{dx}+{\beta }_2^2{c}_p{\int }_0^L{\theta }_x^2\mathit{dx},\forall t\ge 0\text{.}$$ The substitution of estimates (2.14) and (2.15) into (2.13) gives (2.12). square

Lemma 2.3

The functional ${\mathcal{F}}_2$ , defined by (2.3), satisfies, for any ${\epsilon }_1>0$ and some constant $m_0>0$ ,

$${\mathcal{F}}_2^{\prime }(t)⩽-\frac{m_0}{2}{\int }_0^L{v}_x^2\mathit{dx}+{\epsilon }_1{\int }_0^L{u}_t^2\mathit{dx}+\left({\rho }_2+\frac{a_2^2{\rho }_1^2}{4a_1^2{\epsilon }_1}\right){\int }_0^L{v}_t^2\mathit{dx}+\frac{{\beta }_2^2{c}_p}{2m_0}{\int }_0^L{\theta }_x^2\mathit{dx},\forall t\ge 0\text{.}$$

Proof

Differentiating ${\mathcal{F}}_2$ and adapting (1.13)₁ and (1.13)₂ for terms $u_{\mathit{tt}}$ and $v_{\mathit{tt}}$ , respectively, we get

$${\mathcal{F}}_2^{\prime }(t)=-\left({\mu }_1-\frac{a_2^2}{a_1}\right){\int }_0^L{v}_x^2\mathit{dx}+{\rho }_2{\int }_0^L{v}_t^2\mathit{dx}-\frac{a_2{\rho }_1}{a_1}{\int }_0^L{u}_t{v}_t\mathit{dx}-{\beta }_2{\int }_0^L{v}_x\theta \mathit{dx},\forall t\ge 0\text{.}$$ Implementing Young’s inequality for any ${\sigma }_1>0,{\epsilon }_1>0$ , followed by Poincaré’s inequality, we achieve

(2.18)

$$-\frac{a_2{\rho }_1}{a_1}{\int }_0^L{u}_t{v}_t\mathit{dx}⩽{\epsilon }_1{\int }_0^L{u}_t^2\mathit{dx}+\frac{a_2^2{\rho }_1^2}{4a_1^2{\epsilon }_1}{\int }_0^L{v}_t^2\mathit{dx},$$

(2.19)

$$-{\beta }_2{\int }_0^L{v}_x\theta \mathit{dx}⩽{\sigma }_1{\int }_0^L{v}_x^2\mathit{dx}+\frac{{\beta }_2^2{c}_p}{4{\sigma }_1}{\int }_0^L{\theta }_x^2\mathit{dx}\text{.}$$ Direct substitution of (2.18) and (2.19) into (2.17) gives $${\mathcal{F}}_2^{\prime }(t)⩽-\left({\mu }_1-\frac{a_2^2}{a_1}-{\sigma }_1\right){\int }_0^L{v}_x^2\mathit{dx}+{\epsilon }_1{\int }_0^L{u}_t^2\mathit{dx}+\left({\rho }_2+\frac{a_2^2{\rho }_1^2}{4a_1^2{\epsilon }_1}\right){\int }_0^L{v}_t^2\mathit{dx}+\frac{{\beta }_2^2{c}_p}{4{\sigma }_1}{\int }_0^L{\theta }_x^2\mathit{dx},\forall t\ge 0\text{.}$$ Using the fact that $a_1{\mu }_1>a_2^2$ , we have $m_0={\mu }_1-\frac{a_2^2}{a_1}>0$ . So, by taking ${\sigma }_1=\frac{m_0}{2}$ , we end up with estimate (2.16). square

Lemma 2.4

The functional ${\mathcal{F}}_3$ , defined by (2.4), satisfies, for any ${\epsilon }_2,{\epsilon }_3>0$ , the estimate

$$\begin{array}{cc}{\mathcal{F}}_3^{\prime }(t)⩽-\hfill & \frac{{\beta }_2}{2}{\int }_0^L{v}_t^2\mathit{dx}+{\epsilon }_2{\int }_0^L{u}_x^2\mathit{dx}+{\epsilon }_3{\int }_0^L{v}_x^2\mathit{dx}+\left(\frac{\rho {\beta }_2{c}_p}{{\rho }_2}+\frac{{\rho }^2{a}_2^2{c}_p}{4{\epsilon }_2{\rho }_2^2}+\frac{{\rho }^2{\mu }_1^2{c}_p}{4{\epsilon }_3{\rho }_2^2}+\frac{{\mu }_2^2}{2{\beta }_2}\right){\int }_0^L{\theta }_x^2\mathit{dx}\text{.}\hfill \end{array}$$

Proof

Employing (1.13), we obtain, for all $t\ge 0$ ,

$$\begin{array}{cc}{\mathcal{F}}_3^{\prime }(t)=-\hfill & {\beta }_2{\int }_0^L{v}_t^2\mathit{dx}+\frac{\rho {\beta }_2}{{\rho }_2}{\int }_0^L{\theta }^2\mathit{dx}+\frac{\rho a_2}{{\rho }_2}{\int }_0^L{u}_x\theta \mathit{dx}+\frac{\rho {\mu }_1}{{\rho }_2}{\int }_0^L{v}_x\theta \mathit{dx}-{\mu }_2{\int }_0^L{v}_t{\theta }_x\mathit{dx}\text{.}\hfill \end{array}$$ Similar to the proof of the previous two lemmas, by enforcing Young’s inequality for any ${\epsilon }_2,{\epsilon }_3>0$ , and Poincaré’s inequality, the last three integrals of (2.21), give

(2.22)

$$\frac{\rho a_2}{{\rho }_2}{\int }_0^L{u}_x\theta \mathit{dx}⩽{\epsilon }_2{\int }_0^L{u}_x^2\mathit{dx}+\frac{{\rho }^2{a}_2^2{c}_p}{4{\epsilon }_2{\rho }_2^2}{\int }_0^L{\theta }_x^2\mathit{dx},$$

(2.23)

$$\frac{\rho {\mu }_1}{{\rho }_2}{\int }_0^L{v}_x\theta \mathit{dx}⩽{\epsilon }_3{\int }_0^L{v}_x^2\mathit{dx}+\frac{{\rho }^2{\mu }_1^2{c}_p}{4{\epsilon }_3{\rho }_2^2}{\int }_0^L{\theta }_x^2\mathit{dx},$$

(2.24)

$$-{\mu }_2{\int }_0^L{v}_t{\theta }_x\mathit{dx}⩽\frac{{\beta }_2}{2}{\int }_0^L{v}_t^2\mathit{dx}+\frac{{\mu }_2^2}{2{\beta }_2}{\int }_0^L{\theta }_x^2\mathit{dx}\text{.}$$ The combination of (2.21)–(2.24) gives (2.20). square

Lemma 2.5

The functional ${\mathcal{F}}_4$ , defined by (2.5), satisfies

$${\mathcal{F}}_4^{\prime }(t)⩽-{\int }_0^L{u}_t^2\mathit{dx}+\frac{2a_1}{{\rho }_1}{\int }_0^L{u}_x^2\mathit{dx}+\frac{{\mu }_1}{4{\rho }_1}{\int }_0^L{v}_x^2\mathit{dx},\forall t\ge 0\text{.}$$

Proof

Using (1.13), it is clear that $${\mathcal{F}}_4^{\prime }(t)=-{\int }_0^L{u}_t^2\mathit{dx}+\frac{a_1}{{\rho }_1}{\int }_0^L{u}_x^2\mathit{dx}+\frac{a_2}{{\rho }_1}{\int }_0^L{u}_x{v}_x\mathit{dx},\forall t\ge 0\text{.}$$ Using Young’s inequality, we get $$\begin{array}{c}\hfill \frac{a_2}{{\rho }_1}{\int }_0^L{u}_x{v}_x\mathit{dx}⩽\frac{a_1}{{\rho }_1}{\int }_0^L{u}_x^2\mathit{dx}+\frac{a_2^2}{4a_1{\rho }_1}{\int }_0^L{v}_x^2\mathit{dx}\\ \hfill ⩽\frac{a_1}{{\rho }_1}{\int }_0^L{u}_x^2\mathit{dx}+\frac{{\mu }_1}{4{\rho }_1}{\int }_0^L{v}_x^2\mathit{dx}\mathrm{using}{a}_1{\mu }_1>a_2^2\text{.}\end{array}$$ Consequently, we end up with (2.25). square

Theorem 2.6

The problem (1.13) is exponentially stable, that is, for some constants $k_0,k_1>0$ , the energy of system (1.13) defined by (2.6) satisfies

$$E(t)⩽k_0{e}^{-k_{1}t}\forall t\ge 0\text{.}$$

Proof

Recall the Lyapunov functional defined by (2.1), that is $$\mathcal{P}(t)≔\mathcal{C}E(t)+{\displaystyle \sum _{i=1}^4}{\mathcal{C}}_i{\mathcal{F}}_i(t),\forall t\ge 0,$$ and using the estimates (2.8), (2.12), (2.16)–(2.25), together with setting $${\mathcal{C}}_4=2,{\epsilon }_1=\frac{1}{{\mathcal{C}}_2},{\epsilon }_2=\frac{a_2^2{\mathcal{C}}_1}{4{\mathcal{C}}_3}{\epsilon }_3=\frac{m_0{\mathcal{C}}_2}{4{\mathcal{C}}_3},$$ we conclude that $$\begin{array}{c}\hfill {\mathcal{P}}^{\prime }(t)⩽-{\int }_0^L{u}_t^2\mathit{dx}-\left[\frac{m_0}{4}{\mathcal{C}}_2-\left(\frac{a_2^2{\rho }_2}{{\rho }_1}+{\left(\frac{a_1{\rho }_2}{{\rho }_1}-{\mu }_1\right)}^2\right){\mathcal{C}}_1-\frac{{\mu }_1}{2{\rho }_1}\right]{\int }_0^L{v}_x^2\mathit{dx}\\ \hfill -\left[{\mu }_2\mathcal{C}-{\beta }_2^2{c}_p{\mathcal{C}}_1-\frac{{\beta }_2^2{c}_p}{2m_0}{\mathcal{C}}_2-\left(\frac{{\mu }_2^2}{2{\beta }_2}+\frac{\rho {\beta }_2{c}_p}{{\rho }_2}+\frac{{\rho }^2{c}_p{\mathcal{C}}_3}{{\rho }_2^2{\mathcal{C}}_1}+\frac{{\rho }^2{\mu }_1^2{c}_p{\mathcal{C}}_3}{m_0{\rho }_2^2{\mathcal{C}}_2}\right){\mathcal{C}}_3\right]{\int }_0^L{\theta }_x^2\mathit{dx}\\ \hfill -\left[\frac{a_2^2}{4}{\mathcal{C}}_1-\frac{4a_1}{{\rho }_1}\right]{\int }_0^L{u}_x^2\mathit{dx}-\left[\frac{{\beta }_2}{2}{\mathcal{C}}_3-\left({\rho }_2+\frac{a_2^2{\rho }_1^2{\mathcal{C}}_2}{4a_1^2}\right){\mathcal{C}}_2\right]{\int }_0^L{v}_t^2\mathit{dx},\forall t\ge 0\text{.}\end{array}$$ By consecutively letting $$\left\{\begin{array}{c}{\mathcal{C}}_1=4a_2^{-2}\left(\frac{4a_1}{{\rho }_1}+1\right),\hfill \\ {\mathcal{C}}_2=4m_0^{-1}\left(\left(\frac{a_2^2{\rho }_2}{{\rho }_1}+{\left(\frac{a_1{\rho }_2}{{\rho }_1}-{\mu }_1\right)}^2\right){\mathcal{C}}_1+\frac{{\mu }_1}{2{\rho }_1}+1\right),\hfill \\ {\mathcal{C}}_3=2{\beta }_2^{-1}\left(\left({\rho }_2+\frac{a_2^2{\rho }_1^2{\mathcal{C}}_2}{4a_1^2}\right){\mathcal{C}}_2+1\right),\hfill \\ \mathcal{C}={\mu }_2^{-1}\left({\beta }_2^2{c}_p{\mathcal{C}}_1+\frac{{\beta }_2^2{c}_p}{2m_0}{\mathcal{C}}_2+\left(\frac{{\mu }_2^2}{2{\beta }_2}+\frac{\rho {\beta }_2{c}_p}{{\rho }_2}+\frac{{\rho }^2{c}_p{\mathcal{C}}_3}{{\rho }_2^2{\mathcal{C}}_1}+\frac{{\rho }^2{\mu }_1^2{c}_p{\mathcal{C}}_3}{m_0{\rho }_2^2{\mathcal{C}}_2}\right){\mathcal{C}}_3+c_p\right)\hfill \end{array}\right.$$ and using Poincaré’s inequality $$-c_p{\int }_0^L{\theta }_x^2\mathit{dx}⩽-{\int }_0^L{\theta }^2\mathit{dx},\forall t\ge 0,$$ it turns out that

$$\begin{array}{cc}{\mathcal{P}}^{\prime }(t)⩽\hfill & -{\int }_0^L\left(u_t^2+u_x^2+v_t^2+v_x^2+{\theta }^2\right)\mathit{dx},\forall t\ge 0\text{.}\hfill \end{array}$$ Meanwhile, by recalling the energy functional defined by (2.6) $$E(t)=\frac{1}{2}{\int }_0^L\left[{\rho }_1{u}_t^2+a_1{u}_x^2+{\rho }_2{v}_t^2+{\mu }_1{v}_x^2+\rho {\theta }^2+2a_2{u}_x{v}_x\right]\mathit{dx},\forall t\ge 0,$$ and using Young’s inequality, we have, for some constant $c>0$ , $$E(t)⩽c{\int }_0^L\left[u_t^2+u_x^2+v_t^2+v_x^2+{\theta }^2\right]\mathit{dx},\forall t>0$$ which implies

(2.28)

$$-{\int }_0^L\left[u_t^2+u_x^2+v_t^2+v_x^2+{\theta }^2\right]\mathit{dx}⩽-c_{1}E(t),\forall t>0,$$ where $c_1=\frac{1}{c}>0$ . Consequently, by merging (2.27) and (2.28), we have $${\mathcal{P}}^{\prime }(t)⩽-c_{1}E(t)\text{.}$$ Using (2.7), we get $${\mathcal{P}}^{\prime }(t)⩽-\frac{c_1}{{\ell }_2}\mathcal{P}(t)\text{.}$$ Solving the differential inequality, we arrive at $$\mathcal{P}(t)⩽\mathcal{P}(0)\exp \left(-\frac{c_1}{{\ell }_2}\right)\text{.}$$ Finally, using (2.7) once again, we obtain (2.26), with $k_0=\frac{{\ell }_2}{{\ell }_1}E(0)>0$ and $k_1=\frac{c_1}{{\ell }_2}>0$ . Thus, we conclude the proof of Theorem 2.6. square