Remarks on exponential stability for a coupled system of elasticity and thermoelasticity with second sound

Abstract

We study the large time behavior of solutions to a linear transmission problem in one space dimension. The problem at hand models a thermoelastic material with second sound confined by a purely elastic one. We shall characterize all equilibrium states of the considered system and prove that every solution approaches one designated equilibrium state with an exponential rate as time goes to infinity. Hereto, we apply methods from the theory of strongly continuous semigroups. In particular, we obtain uniform resolvent bounds for the underlying generator. This removes the largeness assumption of elastic wave speeds imposed in Meng and Wang (Anal Appl (Singap) 13, 2015) for having an exponential energy decay rate when the problem only has the trivial equilibrium. In an appendix, we provide a similar exponential stability result for the case where heat conduction is modeled using Fourier’s law.

Appendix A: Uniform exponential stability in the case of Fourier’s law

We demonstrate in this appendix how exponential stability can be obtained, by the same method as above, when the thermoelastic material is modeled in the classical way using Fourier’s law for heat conduction. We present all steps but shall avoid repeated proofs and explanations. From now on, the heat flux q is given by $q = -k \theta _x$, see e.g., [19], and with constants $a, b , m, k > 0$, the system (1) becomes

$$\begin{aligned} u_{tt} - a u_{xx} + m \theta _x= & {} 0 \quad \text{ in } \quad [L_1, _2]\times {\mathbb {R}}_+, \nonumber \\ \theta _t - k^2 \theta _{xx} + mu_{xt}= & {} 0 \quad \text{ in } \quad [L_1, L_2]\times {\mathbb {R}}_+, \nonumber \\ v_{tt} - b v_{xx}= & {} 0 \quad \text{ in } \quad [0, L_1] \cup [L_2, L_3] \times {\mathbb {R}}_+. \end{aligned}$$

(A.1)

The only difference to (2)–(4) is that, instead of the boundary condition for q, now $\theta _x(L_1, t) = \theta _x(L_2, t) = 0$ is imposed for $t \ge 0$ and that the initial condition for q is dropped. The state space is given by

$$\begin{aligned} {\mathscr {H}}_F:= & {} \Big \{ [u^1, v^1, \theta , u^2, v^2]' \in H^1(L_1, L_2) \times H^1((0, L_1) \cup (L_2, L_3)) \\&\times L^2(L_1, L_2) \times L^2(L_1, L_2) \times L^2((0,L_1) \cup (L_2,L_3)) \\ \text{ such } \text{ that } v^1(0)= & {} v^1(L_3) = 0 \text{ and } u^1(L_i) = v^1(L_i) \text{ for } i= 1,2 \Big \}, \end{aligned}$$

equipped, using (7), with the norm $\Vert \cdot \Vert _{{\mathscr {H}}_F}$ induced by the inner product

$$\begin{aligned} \langle U, {\tilde{U}} \rangle _{{\mathscr {H}}_F} := \int _{L_1}^{L_2} \left( a u^1_x \overline{\tilde{u^1_x}} + u^2 \overline{\tilde{u^2}} + \theta \overline{{\tilde{\theta }}} \right) \, \mathrm{d}x + \left[ \int _0^{L_1} + \int _{L_2}^{L_3}\right] \left( b v^1_x \overline{\tilde{v^1}_x} + v^2 \overline{\tilde{v^2}} \right) \, \mathrm{d}x. \end{aligned}$$

Let the linear operator ${\mathscr {A}}_F :{\mathscr {H}}_F \supseteq D({\mathscr {A}}_F) \rightarrow {\mathscr {H}}_F$ be defined for $U = [u^1, v^1, \theta , u^2, v^2]'$ by

$$\begin{aligned} {\mathscr {A}}_FU := [u^2, v^2, k^2\theta _{xx} -mu^2_x, au^1_{xx} -m\theta _x, bv^1_{xx}]', \end{aligned}$$

on the domain

$$\begin{aligned} D({\mathscr {A}}_F)= & {} \Big \{ U \in {\mathscr {H}}_F \cap \big [ H^2(L_1, L_2) \times H^2((0, L_1) \cup (L_2, L_3)) \times H^2(L_1, L_2) \\&\times H^1(L_1, L_2) \times H^1((0, L_1) \cup (L_2, L_3)) \big ] \,\, \Big | \,\, u^2(L_i) = v^2(L_i), \\&v^2(0) = v^2(L_3) = 0, \, au^1_x(L_i) - m\theta (L_i) = bv^1_x(L_i), \, \theta _x(L_i) = 0, \, i = 1,2 \Big \}, \end{aligned}$$

such that (A.1) as well as the initial and boundary conditions transform into the following Cauchy problem in ${\mathscr {H}}_F$,

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}U(t)= & {} {\mathscr {A}}_F U(t), \quad t \ge 0, \nonumber \\ U(0)= & {} U_0 \in {\mathscr {H}}_F. \end{aligned}$$

(A.2)

As in Proposition 2.1, ${\mathscr {A}}_F$ is densely defined and using in particular the boundary conditions prescribed via ${\mathscr {H}}_F$ and $D({\mathscr {A}}_F)$ one can verify that ${\mathscr {A}}_F^*$ is given by $D({\mathscr {A}}_F^*) = D({\mathscr {A}}_F)$ and

$$\begin{aligned} {\mathscr {A}}_F^* [u^1, v^1, \theta , u^2, v^2]' = [-u^2, -v^2, k^2\theta _{xx} + mu^2_x, -au^1_{xx} + m\theta _x, -bv^1_{xx}]'. \end{aligned}$$

Repeating these calculations, one finds $({\mathscr {A}}_F^*)^* = {\mathscr {A}}_F$, which implies that ${\mathscr {A}}_F$ is closed. Moreover, for $U \in D({\mathscr {A}}_F) = D({\mathscr {A}}_F^*)$ it holds that

$$\begin{aligned} \begin{aligned} \mathfrak {R}\langle {\mathscr {A}}_FU, U \rangle _{{\mathscr {H}}_F} = \mathfrak {R}\langle {\mathscr {A}}_F^*U, U \rangle _{{\mathscr {H}}_F} = - k^2 \int _{L_1}^{L_2} \theta _x \overline{\theta _x} \, \mathrm{d}x \le 0. \end{aligned} \end{aligned}$$

(A.3)

Hence, ${\mathscr {A}}_F$ and ${\mathscr {A}}_F^*$ are dissipative and, by the Lumer and Phillips Theorem, ${\mathscr {A}}_F$ generates a strongly continuous contraction semigroup $(T_F(t))_{t \ge 0}$ on ${\mathscr {H}}_F$. Moreover, $\sigma ({\mathscr {A}}_F) \subseteq \{\mathfrak {R}\lambda \le 0\}$ consists only of eigenvalues due to the compact embedding $D({\mathscr {A}}_F) \hookrightarrow {\mathscr {H}}_F$.

Remark A.1

For $U \in D({\mathscr {A}}_F)$, $l \in {\mathbb {R}}$ and $F = (il - {\mathscr {A}}_F)U$, we have like in Remark 2.2 that $k^2\Vert \theta _x\Vert _{L^2(L_1,L_2)}^2\le \Vert U\Vert _{{\mathscr {H}}_F} \Vert F\Vert _{{\mathscr {H}}_F}$.

The statement of Proposition 2.2 and its proof remain true if one replaces q by $-k\theta _x$ at some points or deletes some expressions which do not appear anymore.

Proposition A.1

It is $0 \in \sigma ({\mathscr {A}}_F)$, $i {\mathbb {R}} \cap \sigma ({\mathscr {A}}_F) = \{0\}$ and the stationary solutions of (A.2) are characterized by $\hbox {ker}({\mathscr {A}}_F) = \hbox {span}_{{\mathscr {H}}_F}\left\{ [\zeta ^1, \zeta ^2, \zeta ^3, 0, 0]' \right\}$, where $\zeta ^1, \zeta ^2$ and $\zeta ^3$ are the same as defined in Proposition 2.2.

Because $\hbox {ker}({\mathscr {A}}_F) = \hbox {ker}({\mathscr {A}}_F^*)$, the following result can be proved like Lemma 2.1.

Lemma A.1

${\mathscr {H}}_F = \hbox {ker}({\mathscr {A}}_F) \oplus \mathrm{range}({\mathscr {A}}_F)$.

Since $\hbox {ker}({\mathscr {A}}_F)$ is non-trivial, the operator ${\mathscr {A}}_{F,0}$ is introduced as the restriction of ${\mathscr {A}}_F$ to $\hbox {range}({\mathscr {A}}_F)$ with domain $D({\mathscr {A}}_{F, 0}) = D({\mathscr {A}}_F) \cap \hbox {range}({\mathscr {A}}_F)$, and by Lemma A.1 it holds that $\sigma ({\mathscr {A}}_{F, 0}) = \sigma ({\mathscr {A}}_F)\setminus \{0\}$. In order to apply Theorem 3.1, which will lead to exponential stability for $(T_F(t))_{t \ge 0}$ in the same way as demonstrated for $(T(t))_{t \ge 0}$, we argue by contradiction and assume that (16) is not true for ${\mathscr {A}}_{F, 0}$. Then there exists a sequence $(l_n)_{n \in {\mathbb {N}}} \subseteq {\mathbb {R}}\setminus \{0\}$ with $|l_n| \rightarrow +\infty$, as $n \rightarrow +\infty$ and, dropping the index n, a sequence $U_l \in D({\mathscr {A}}_{F, 0})$ with $\Vert U_l\Vert _{{\mathscr {H}}_F} = 1$ for all $l \in \{l_n \, | \, n \in {\mathbb {N}} \}$ such that in ${\mathscr {H}}_F$ we have

$$\begin{aligned} F_l = [f^1_l, g^1_l, h_l, p_l, f^2_l, g^2_l]' := (il - {\mathscr {A}}_{F, 0})U_l \rightarrow 0, \quad |l| \rightarrow +\infty . \end{aligned}$$

Then, for $|l| \rightarrow +\infty$, it holds that

$$\begin{aligned}&il u^1_l - u^2_l = f^1_l \longrightarrow 0 \quad \text{ in } \quad H^1(L_1,L_2), \end{aligned}$$

(A.4a)

$$\begin{aligned}&il v^1_l - v^2_l = g^1_l \longrightarrow 0 \quad \text{ in } H^1((0, L_1) \cup (L_2, L_3)),\end{aligned}$$

(A.4b)

$$\begin{aligned}&il \theta _l - k^2\theta _{l,\, xx} + mu^2_{l,\, x} = h_l \longrightarrow 0 \quad \text{ in } \quad L^2(L_1,L_2),\end{aligned}$$

(A.4c)

$$\begin{aligned}&il u^2_l - au^1_{l,\, xx} + m \theta _{l,\, x} = f^2_l \longrightarrow 0 \quad \text{ in } L^2(L_1,L_2), \end{aligned}$$

(A.4d)

$$\begin{aligned}&il v^2_l - bv^1_{l,\, xx} = g^2_l \longrightarrow 0 \quad \text{ in } L^2((0,L_1) \cup (L_2,L_3)). \end{aligned}$$

(A.4e)

For the model with Cattaneo’s law, in the context of Sect. 3, the equation (17d) prevented one to conclude $\theta _{l,\, x} \rightarrow 0$ in $L^2$ as $|l| \rightarrow + \infty$. However, here this convergence holds due to Remark A.1, which leads to simplifications at some points. We continue as in Sect. 3 and observe from (A.4a) and (A.4b) that $\Vert u^1_l\Vert _{L^2(L_1, L_2)} \rightarrow 0$, $\Vert v^1_l\Vert _{L^2((0,L_1) \cup (L_2,L_3))} \rightarrow 0$ as $|l| \rightarrow +\infty$, as well as from (A.4c) and (A.4d) that

$$\begin{aligned} |l| \Vert \theta _l\Vert _{H^{-1}(L_1, L_2)}\le & {} C (\Vert \theta _{l,\, x}\Vert _{L^2(L_1, L_2)} + \Vert u^2_l\Vert _{L^2(L_1, L_2)} + \Vert F_l\Vert _{{\mathscr {H}}_F} ),\\ |l| \Vert u^2_l\Vert _{H^{-1}(L_1, L_2)}\le & {} C (\Vert u^1_{l,\, x}\Vert _{L^2(L_1, L_2)} + \Vert \theta _l\Vert _{L^2(L_1, L_2)} + \Vert F_l\Vert _{{\mathscr {H}}_F} ). \end{aligned}$$

By interpolation and Remark A.1, noting that now $\Vert \theta _{l,\, x}\Vert _{L^2(L_1, L_2)} \rightarrow 0$ for $|l| \rightarrow +\infty$, it holds

$$\begin{aligned} \Vert \theta _l\Vert _{L^2(L_1, L_2)}^2\le & {} \Vert \theta _l\Vert _{H^{-1}(L_1, L_2)}\Vert \theta _l\Vert _{H^{1}(L_1, L_2)} \nonumber \\\le & {} \frac{1}{|l|} C (\Vert \theta _{l,\, x}\Vert _{L^2(L_1, L_2)} + \Vert u^2_l\Vert _{L^2(L_1, L_2)} + \Vert F_l\Vert _{{\mathscr {H}}_F}) \nonumber \\&\times (\Vert \theta _l\Vert _{L^2(L_1, L_2)} + \Vert \theta _{l,\, x} \Vert _{L^2(L_1, L_2)}) \nonumber \\\rightarrow & {} 0, \quad |l| \rightarrow +\infty . \end{aligned}$$

(A.5)

Moreover, one has

$$\begin{aligned} \Vert u^2_l\Vert _{L^2(L_1, L_2)}^2\le & {} \Vert u^2_l\Vert _{H^{-1}(L_1, L_2)}\Vert u^2_l\Vert _{H^{1}(L_1, L_2)} \nonumber \\\le & {} \frac{C}{|l|} (\Vert u^1_{l,\, x}\Vert _{L^2(L_1, L_2)} + \Vert \theta _l\Vert _{L^2(L_1, L_2)} + \Vert F_l\Vert _{{\mathscr {H}}_F}) \Vert u^2_l\Vert _{H^1(L_1, L_2)}\nonumber \\= & {} \frac{C}{|l|} (\Vert u^1_{l,\, x}\Vert _{L^2(L_1, L_2)} + \Vert \theta _l\Vert _{L^2(L_1, L_2)} + \Vert F_l\Vert _{{\mathscr {H}}_F}) \Vert il u^1_l - f_l \Vert _{H^1(L_1, L_2)}.\nonumber \\ \end{aligned}$$

(A.6)

Now, we can conclude similar to Lemma 3.1 that $u^1_{l,\, x}$ and $u^2_l$ converge to zero. Hereto, by using (A.4a), (A.4c), (A.4d), as well as the boundary condition for $\theta _{l,\, x}$, we obtain

$$\begin{aligned} 0= & {} \int _{L_1}^{L_2} \left[ \theta _l \overline{u^1_{l,\, x}} + \frac{1}{il} \left( k^2 \theta _{l,\, x} \overline{u^1_{l,\, xx}} + m(ilu^1_{l,\, x} - f^1_{l,\, x})\overline{u^1_{l,\, x}} -h_l \overline{u^1_{l,\, x}} \right) \right] \, \mathrm{d}x\nonumber \\= & {} \int _{L_1}^{L_2} \left[ \theta _l \overline{u^1_{l,\, x}} \right. \nonumber \\&\left. + m|u^1_{l,\, x}|^2 + \frac{1}{il} \left( \frac{k^2}{a} \theta _{l,\, x} (-il\overline{u^2_l} + m \overline{\theta _{l,\, x}} - \overline{f^2_l}) - mf^1_{l,\, x}\overline{u^1_{l,\, x}} - h_l \overline{u^1_{l,\, x}} \right) \right] \, \mathrm{d}x.\nonumber \\ \end{aligned}$$

(A.7)

From $\Vert U_l\Vert _{{\mathscr {H}}_F} = 1$, $\Vert F_l\Vert _{{\mathscr {H}}_F} \rightarrow 0$ as $|l| \rightarrow +\infty$, (A.5) and Remark A.1, one can deduce

$$\begin{aligned} \begin{aligned}&\int _{L_1}^{L_2} \left[ \theta _l \overline{u^1_{l,\, x}} + \frac{1}{il} \left( \frac{k^2}{a} \theta _{l,\, x} (-il\overline{u^2_l} + m \overline{\theta _{l,\, x}} - \overline{f^2_l}) - mf^1_{l,\, x}\overline{u^1_{l,\, x}} - h_l \overline{u^1_{l,\, x}} \right) \right] \\&\quad \mathrm{d}x \rightarrow 0, \, |l| \rightarrow +\infty , \end{aligned} \end{aligned}$$

which implies together with (A.7) that $\Vert u^1_{l,\, x}\Vert _{L^2(L_1,L_2)} \rightarrow 0$ as $|l| \rightarrow +\infty$ and, by employing Eq. (A.6), one also has $\Vert u^2_l\Vert _{L^2(L_1,L_2)} \rightarrow 0$ as $|l| \rightarrow +\infty$.

In order to obtain controls for $\Vert v^1_{l,\, x}\Vert _{L^2((0, L_1) \cup (L_2, L_3))}$ and $\Vert v^2_l\Vert _{L^2((0, L_1) \cup (L_2, L_3))}$, we first multiply (A.4d) with $\overline{u^1_{l,\, x}}(L_2 - x)$, integrate over $(L_1, L_2)$, take real parts, incorporate (A.4a) and deduce

$$\begin{aligned} 0= & {} - \int _{L_1}^{L_2} \left[ \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}x} |u^2_l|^2 + \frac{a}{2} \frac{\mathrm{d}}{\mathrm{d}x} |u^1_{l,\, x}|^2 \right] (L_2 - x) \, \mathrm{d}x \nonumber \\&- \underset{=: {\tilde{R}}_4}{\underbrace{ \mathfrak {R}\int _{L_1}^{L_2} \left[ u^2_l \overline{f^1_{l\, x}} - m \theta _{l,\, x} \overline{u^1_{l,\, x}} + f^2_l\overline{u^1_{l,\, x}} \right] (L_2 - x) \, \mathrm{d}x }}. \end{aligned}$$

(A.8)

Repeating this calculation, but multiplying (A.4d) now with $\overline{u^1_{l,\, x}}(L_1 - x)$, results in

$$\begin{aligned} 0= & {} -\int _{L_1}^{L_2} \left[ \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}x} |u^2_l|^2 + \frac{a}{2} \frac{\mathrm{d}}{\mathrm{d}x} |u^1_{l,\, x}|^2 \right] (L_1 - x) \, \mathrm{d}x \nonumber \\&- \underset{=: {\tilde{R}}_5}{\underbrace{ \mathfrak {R}\int _{L_1}^{L_2} \left[ u^2_l \overline{f^1_{l\, x}} - m \theta _{l,\, x} \overline{u^1_{l,\, x}} + f^2_l\overline{u^1_{l,\, x}} \right] (L_1 - x) \, \mathrm{d}x }}, \end{aligned}$$

(A.9)

and summing (A.8) and (A.9), as well as using integration by parts, leads to

$$\begin{aligned} \int _{L_1}^{L_2} \left[ |u^2_l|^2 + a |u^1_{l,\, x}|^2 \right] \, \mathrm{d}x + {\tilde{R}}_4 + {\tilde{R}}_5 = \frac{L_2-L_1}{2} \sum \limits _{i=1,2} \left[ |u^2_l(L_i)|^2 + a|u^1_{l,\, x}(L_i)|^2 \right] . \end{aligned}$$

(A.10)

Both ${\tilde{R}}_4$ and ${\tilde{R}}_5$ converge to zero for $|l| \rightarrow +\infty$ by the usual argument. In a similar manner as before, by multiplying (A.4e) with $\overline{v^1_{l,\, x}}x$ and integrating over $(0, L_1)$, followed by multiplying (A.4e) with $\overline{v^1_{l,\, x}}(L_3-x)$ and integrating over $(L_2, L_3)$, one obtains

$$\begin{aligned} 0= & {} \frac{b}{2} \Vert v^1_{l,\, x}\Vert _{L^2((0, L_1) \cup (L_2, L_3))}^2 + \frac{1}{2}\Vert v^2\Vert _{L^2((0, L_1) \cup (L_2, L_3))}^2 \nonumber \\&- \frac{1}{2} L_1 [|v^2_l(L_1)|^2 + b|v^1_{l,\, x}(L_1)|^2] - \frac{1}{2} (L_3 - L_2) [|v^2_l(L_2)|^2 + b|v^1_{l,\, x}(L_2)|^2]\nonumber \\&+ \underset{=: {\tilde{R}}_6}{\underbrace{\mathfrak {R}\int _{L_2}^{L_3} \left[ v^2_l\overline{g^1_{l,\, x}} + v^2_l\overline{g^1_{l,\, x}} \right] (L_3 - x) \, \mathrm{d}x - \mathfrak {R}\int _0^{L_1} \left[ v^2_l\overline{g^1_{l,\, x}} + g^2_l\overline{v^1_{l,\, x}} \right] x \, \mathrm{d}x}}.\nonumber \\ \end{aligned}$$

(A.11)

Note that, due to the usual Sobolev embedding, $|\theta _l(x)| \le C \Vert \theta _l\Vert _{H^1(L_1, L_2)}$ for all $x \in [L_1, L_2]$, where $C > 0$ is independent of $l \in \{l_n \, | \, n \in {\mathbb {N}} \}$. Therefore, there exist $C_1, C_2 > 0$, independent of l, such that for

$$\begin{aligned} I := \Vert v^1_{l,\, x}\Vert _{L^2((0, L_1) \cup (L_2, L_3))}^2 + \Vert v^2\Vert _{L^2((0, L_1) \cup (L_2, L_3))}^2 , \end{aligned}$$

one can infer from (A.11) and the transmission conditions that

$$\begin{aligned} I\le & {} C_1 \sum \limits _{i=1,2} \left[ |u^2_l(L_i)|^2 + |u^1_{l,\, x}(L_i)|^2 + |\theta _l(L_i)|^2 + |u^1_{l,\, x}(L_i)||\theta _l(L_i)| \right] + C_1|{\tilde{R}}_6| \nonumber \\\le & {} C_2 \sum \limits _{i=1,2} \left[ |u^2_l(L_i)|^2 + |u^1_{l,\, x}(L_i)|^2 + \Vert \theta _l\Vert _{H^1(L_1, L_2)}^2 + |u^1_{l,\, x}(L_i)|\Vert \theta _l\Vert _{H^1(L_1, L_2)} \right] \nonumber \\&+ C_1|{\tilde{R}}_6|, \end{aligned}$$

(A.12)

where ${\tilde{R}}_6$ vanishes as $|l| \rightarrow +\infty$. Furthermore, (A.10) allows to infer $|u^2_l(L_i)| + |u^1_{l,\, x}(L_i)| \rightarrow 0$, as $|l| \rightarrow +\infty$, for $i=1,2$ and Remark A.1 together with (A.5) implies $\Vert \theta _l \Vert _{H^1(L_1, L_2)} \rightarrow 0$, as $|l| \rightarrow +\infty$. As a result, from (A.12) it follows that $I \rightarrow 0$, as $|l| \rightarrow +\infty$. In total, all terms contained in $\Vert U_l\Vert _{{\mathscr {H}}_F}$ are seen to converge to zero for large |l| and we arrive at

$$\begin{aligned} \Vert U_l\Vert _{{\mathscr {H}}_F} \rightarrow 0, \quad |l| \rightarrow +\infty , \end{aligned}$$

which is in contradiction with $\Vert U_l\Vert _{{\mathscr {H}}_F} = 1$.

Above we verified the requirements of Theorem 3.1 for ${\mathscr {A}}_{F,0}$ and therefore conclude the following exponential stability result, which can be proved in the same way as Theorem 3.2.

Theorem A.1

Let $U_0 \in {\mathscr {H}}_F$, according to Lemma A.1 be uniquely decomposed as $U_0 = W_0 + V_0$, with $W_0 \in \hbox {ker}({\mathscr {A}}_F)$ and $V_0 \in \hbox {range}({\mathscr {A}}_F)$. There exist constants $M > 0$ and $\omega > 0$, which are independent of $U_0$, such that

$$\begin{aligned} \Vert T_F(t) U_0 - W_0\Vert _{{\mathscr {H}}_F} \le M \hbox {e}^{-\omega t} \Vert U_0 \Vert _{{\mathscr {H}}_F} \quad (t \ge 0). \end{aligned}$$