Let us consider
$f(t,x)$ and
$g(t,x)$, two functions assumed to be twice continuously differentiable with respect to the time variable
t. By direct calculations, it is easy to deduce that
$\frac{d}{dt}(f\dot{g}\dot{f}g)=\dot{f}\dot{g}+f\ddot{g}\ddot{f}g\dot{f}\dot{g}=f\ddot{g}\ddot{f}g.$
For the sake of simplicity, the spatial argument and the time argument of the functions $f(t,x)$ and $g(t,x)$ are omitted because there is no likelihood of confusion.
In the above equality, we substitute the functions
$f(t,x)$ and
$g(t,x)$ by the functions
${U}_{i}(x,t)$ and
${V}_{i}(x,t)$, respectively, which also are assumed to be twice continuously differentiable with respect to the time variable, and then we obtain the following well known Lagrange identity:
$\begin{array}{r}{\int}_{B}\varrho (x)[{U}_{i}(x,t){\dot{V}}_{i}(x,t){\dot{U}}_{i}(x,t){V}_{i}(x,t)]\phantom{\rule{0.2em}{0ex}}dV\\ \phantom{\rule{1em}{0ex}}={\int}_{0}^{t}{\int}_{B}\varrho (x)[{U}_{i}(x,s){\ddot{V}}_{i}(x,s){\ddot{U}}_{i}(x,s){V}_{i}(x,s)]\phantom{\rule{0.2em}{0ex}}dV\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{2em}{0ex}}+{\int}_{B}\varrho (x)[{U}_{i}(x,0){\dot{V}}_{i}(x,0){\dot{U}}_{i}(x,0){V}_{i}(x,0)]\phantom{\rule{0.2em}{0ex}}dV.\end{array}$
(9)
Let us denote by
$({u}_{i}^{(\alpha )},{\theta}^{(\alpha )})$ (
$\alpha =1,2$) two solutions of the mixed initial boundary value problem defined by (8), (6) and (7) which correspond to the same boundary data and same initial data, but to different body forces and heat supplies,
$({f}_{i}^{(\alpha )},{r}^{(\alpha )})\phantom{\rule{1em}{0ex}}(\alpha =1,2),$
respectively.
We introduce the following notations:
$\begin{array}{r}{v}_{i}={u}_{i}^{(2)}{u}_{i}^{(1)},\phantom{\rule{2em}{0ex}}\chi ={\theta}^{(2)}{\theta}^{(1)},\\ {t}_{ij}={\sigma}_{ij}^{(2)}{\sigma}_{ij}^{(1)},\phantom{\rule{2em}{0ex}}{m}_{ijk}={\mu}_{ijk}^{(2)}{\mu}_{ijk}^{(1)},\phantom{\rule{2em}{0ex}}\eta ={S}^{(2)}{S}^{(1)},\\ {p}_{i}={q}_{i}^{(2)}{q}_{i}^{(1)},\phantom{\rule{2em}{0ex}}{\mathcal{F}}_{i}={f}_{i}^{(2)}{f}_{i}^{(1)},\phantom{\rule{2em}{0ex}}\mathcal{P}={r}^{(2)}{r}^{(1)}.\end{array}$
(10)
The constitutive equations become
$\begin{array}{r}{t}_{ij}={A}_{ijrs}{v}_{r,s}+{B}_{ijpqr}{v}_{p,qr}+{E}_{ij}\chi ,\\ {\mu}_{ijk}={B}_{rsijk}{v}_{r,s}+{C}_{ijkmnr}{v}_{r,mn}+{D}_{ijk}\chi ,\\ \eta ={E}_{ij}{v}_{i,j}{D}_{ijk}{v}_{k,ij}+\frac{c}{{T}_{0}}\chi {b}_{i}{\chi}_{,i},\\ {p}_{i}={\theta}_{0}({b}_{i}\dot{\chi}{k}_{ij}{\chi}_{,j}).\end{array}$
(11)
So, we deduce that the differences
$({v}_{i},\chi )$ satisfy the following equations and conditions:

the equations of motion
$\begin{array}{rl}\varrho {\ddot{v}}_{i}=& {A}_{ijrs}{v}_{r,sj}+{B}_{ijpqr}{v}_{r,pqj}+{E}_{ij}{\chi}_{,j}\\ +{B}_{mnsji}{v}_{m,nsj}+{C}_{sjimnr}{v}_{r,mnsj}+{D}_{sji}{\chi}_{,sj}+\varrho {\mathcal{F}}_{i};\end{array}$
(12)

the equations of energy
$c\dot{\chi}+{\theta}_{0}({E}_{ij}{\dot{v}}_{i,j}+{D}_{ijk}{\dot{v}}_{k,ij})={({k}_{ij}{\chi}_{,j})}_{,i}+\varrho \mathcal{P};$
(13)

the initial conditions
${v}_{i}(x,0)=0,\phantom{\rule{2em}{0ex}}{\dot{v}}_{i}(x,0)=0,\phantom{\rule{2em}{0ex}}\chi (x,0)=0,\phantom{\rule{1em}{0ex}}x\in \overline{B};$
(14)

the boundary conditions
$\begin{array}{r}{v}_{i}(x,t)=0\phantom{\rule{1em}{0ex}}\text{on}\partial {B}_{1}\times [0,{t}_{0}),\phantom{\rule{2em}{0ex}}{t}_{ji}(x,t){n}_{j}=0\phantom{\rule{1em}{0ex}}\text{on}\partial {B}_{1}^{c}\times [0,{t}_{0}),\\ \chi (x,t)=0\phantom{\rule{1em}{0ex}}\text{on}\partial {B}_{2}\times [0,{t}_{0}),\phantom{\rule{2em}{0ex}}{p}_{i}(x,t){n}_{i}=0\phantom{\rule{1em}{0ex}}\text{on}\partial {B}_{2}^{c}\times [0,{t}_{0}).\end{array}$
(15)
We are now in a position to prove the first basic result.
Theorem 1 For the differences $({v}_{i},\chi )$ of two solutions of the mixed initial boundary value problem (8), (6)
and (7),
the Lagrange identity becomes $\begin{array}{r}2{\int}_{B}\varrho {v}_{i}(t){\dot{v}}_{i}(t)\phantom{\rule{0.2em}{0ex}}dV+{\int}_{B}\frac{1}{{\theta}_{0}}{k}_{ij}({\int}_{0}^{t}{\chi}_{,i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )({\int}_{0}^{t}{\chi}_{,j}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )\phantom{\rule{0.2em}{0ex}}dV\\ \phantom{\rule{1em}{0ex}}={\int}_{0}^{t}\phantom{\rule{0.2em}{0ex}}ds{\int}_{B}\varrho [{v}_{i}(2ts){\mathcal{F}}_{i}(s){v}_{i}(s){\mathcal{F}}_{i}(2ts)]\phantom{\rule{0.2em}{0ex}}dV\\ \phantom{\rule{2em}{0ex}}+{\int}_{0}^{t}{\int}_{B}\frac{\varrho}{{\theta}_{0}}[\chi (s){\int}_{0}^{2ts}\mathcal{P}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi \\ \phantom{\rule{2em}{0ex}}\chi (2ts){\int}_{0}^{s}\mathcal{P}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi ]\phantom{\rule{0.2em}{0ex}}dV\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}t\in [0,\frac{{t}_{0}}{2}).\end{array}$
(16)
Proof Because of the linearity of the problem defined by (8), (6) and (7), we deduce that the differences
$({v}_{i},\chi )$ represent the solution of a mixed initial boundary value problem analogous to (8), (6) and (7), namely the problem consisting of equations (
12) and (
13) with loads
${\mathcal{F}}_{i}$, respectively
$\mathcal{P}$, the initial conditions (14) and boundary conditions (15). By setting
${U}_{i}(x,s)={v}_{i}(x,s),\phantom{\rule{2em}{0ex}}{V}_{i}(x,s)={v}_{i}(x,2ts),\phantom{\rule{1em}{0ex}}s\in [0,2t],t\in [0,\frac{{t}_{0}}{2}),$
then the identity (9), after some straightforward calculation, becomes
$2{\int}_{B}\varrho {v}_{i}(t){v}_{i}(t)\phantom{\rule{0.2em}{0ex}}dV={\int}_{0}^{t}\phantom{\rule{0.2em}{0ex}}ds{\int}_{B}\varrho [{v}_{i}(2ts){\ddot{v}}_{i}(s){\ddot{v}}_{i}(2ts){v}_{i}(s)]\phantom{\rule{0.2em}{0ex}}dV,$
(17)
where we have used the fact that the initial and boundary data are null.
We shall eliminate the inertial terms on the righthand side of the relation (17) by means of the equations of motion for the differences $({v}_{i},\chi )$.
So, in view of equation (
12), we have
$\begin{array}{c}\varrho [{v}_{i}(2ts){\ddot{v}}_{i}(s){\ddot{v}}_{i}(2ts){v}_{i}(s)]\hfill \\ \phantom{\rule{1em}{0ex}}=\{{v}_{i}(2ts)[{A}_{ijrs}{v}_{r,s}(s)+{B}_{ijpqr}{v}_{r,pq}(s)+{E}_{ij}\chi (s)\hfill \\ \phantom{\rule{2em}{0ex}}+{B}_{rskji}{v}_{r,sk}(s)+{C}_{kjipqr}{v}_{r,pqk}(s)+{D}_{kji}{\chi}_{,k}(s)]{\}}_{,j}\hfill \\ \phantom{\rule{2em}{0ex}}\{{v}_{i}(s)[{A}_{ijrs}{v}_{r,s}(2ts)+{B}_{ijpqr}{v}_{r,pq}(2ts)+{E}_{ij}\chi (2ts)\hfill \\ \phantom{\rule{2em}{0ex}}+{B}_{rskji}{v}_{r,sk}(2ts)+{C}_{kjipqr}{v}_{r,pqk}(2ts)+{D}_{kji}{\chi}_{,k}(2ts)]{\}}_{,j}\hfill \\ \phantom{\rule{2em}{0ex}}{A}_{ijrs}{v}_{r,s}(s){v}_{i,j}(2ts){B}_{ijpqr}{v}_{r,pq}(s){v}_{i,j}(2ts){E}_{ij}\chi (s){v}_{i,j}(2ts)\hfill \\ \phantom{\rule{2em}{0ex}}{B}_{rskji}{v}_{r,sk}(s){v}_{i,j}(2ts){C}_{kjipqr}{v}_{r,pqk}(s){v}_{i,j}(2ts){D}_{kji}{\chi}_{,k}(s){v}_{i,j}(2ts)\hfill \\ \phantom{\rule{2em}{0ex}}+{A}_{ijrs}{v}_{r,s}(2ts){v}_{i,j}(s)+{B}_{ijpqr}{v}_{r,pq}(2ts){v}_{i,j}(s)+{E}_{ij}\chi (2ts){v}_{i,j}(s)\hfill \\ \phantom{\rule{2em}{0ex}}+{B}_{rskji}{v}_{r,sk}(2ts){v}_{i,j}(s)+{C}_{kjipqr}{v}_{r,pqk}(2ts){v}_{i,j}(s)+{D}_{kji}{\chi}_{,k}(2ts){v}_{i,j}(s)\hfill \\ \phantom{\rule{2em}{0ex}}+\varrho [{\mathcal{F}}_{i}(s){v}_{i}(2ts){\mathcal{F}}_{i}(2ts){v}_{i}(s)].\hfill \end{array}$
After we use the symmetry relations (5), this equality takes on the form
We integrate by parts equality (18), and after using boundary conditions (15), we get the equality
Now we integrate equation (
13) on the interval
$[0,s]$ and take into account the zero initial data in (14) so that we obtain the relation
$\begin{array}{r}{E}_{ij}{v}_{i,j}(s)+{D}_{kji}{v}_{i,kj}(s)\\ \phantom{\rule{1em}{0ex}}=\frac{c}{{\theta}_{0}}\chi (s)\frac{1}{{\theta}_{0}}{({\int}_{0}^{s}{\chi}_{,j}(z)\phantom{\rule{0.2em}{0ex}}dz)}_{,i}\frac{\varrho}{{\theta}_{0}}{\int}_{0}^{s}\mathcal{P}(z)\phantom{\rule{0.2em}{0ex}}dz,\phantom{\rule{1em}{0ex}}s\in [0,{t}_{0}).\end{array}$
(20)
After we multiply in equality (20) by
$\chi (2ts)$ and use a similar result obtained for
${E}_{ij}{v}_{i,j}(2ts)+{D}_{kji}{v}_{i,kj}(2ts)$, multiplied by
$\chi (s)$, we find
If we introduce (21) into (19), we obtain
Based on the symmetry of the tensor
${k}_{ij}$, we get
On the other hand, integrating by parts, we obtain
$\begin{array}{r}{\int}_{0}^{t}\phantom{\rule{0.2em}{0ex}}ds{\int}_{B}\frac{1}{{\theta}_{0}}{k}_{ij}[{\chi}_{,i}(s){\int}_{0}^{2ts}{\chi}_{,j}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi {\chi}_{,j}(2ts){\int}_{0}^{s}{\chi}_{,i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi ]\phantom{\rule{0.2em}{0ex}}dV\\ \phantom{\rule{1em}{0ex}}={\int}_{0}^{t}\phantom{\rule{0.2em}{0ex}}ds{\int}_{B}\frac{1}{{\theta}_{0}}{k}_{ij}\frac{d}{ds}\left[({\int}_{0}^{s}{\chi}_{,i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )({\int}_{0}^{2ts}{\chi}_{,j}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )\right]\phantom{\rule{0.2em}{0ex}}dV.\end{array}$
(24)
From (23) and (24) we deduce
$\begin{array}{r}{\int}_{0}^{t}\phantom{\rule{0.2em}{0ex}}ds{\int}_{B}\frac{1}{{\theta}_{0}}{k}_{ij}[{\chi}_{,i}(2ts){\int}_{0}^{s}{\chi}_{,j}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi {\chi}_{,i}(2ts){\int}_{0}^{2ts}{\chi}_{,j}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi ]\phantom{\rule{0.2em}{0ex}}dV\\ \phantom{\rule{1em}{0ex}}={\int}_{B}\frac{1}{{\theta}_{0}}{k}_{ij}\left[({\int}_{0}^{s}{\chi}_{,i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )({\int}_{0}^{2ts}{\chi}_{,j}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )\right]\phantom{\rule{0.2em}{0ex}}dV.\end{array}$
(25)
We now substitute (25) in (22), and so we are led to equality (16). With this, the proof of Theorem 1 is completed. □
Remark It is important to note that the identity (16) is just like in the classical thermoelasticity (see [17]).
The identity (16) constitutes the basis on which we shall prove the uniqueness and the continuous dependence results.
We proceed first to obtain the uniqueness of the solution of the mixed initial boundary value problem defined by (8), (6) and (7).
Theorem 2
Assume that the conductivity tensor
${k}_{ij}$
is positive definite in the sense there exists a positive constant
${k}_{0}$
such that
${k}_{ij}{\xi}_{i}{\xi}_{j}\ge {k}_{0}{\xi}_{i}{\xi}_{i},\phantom{\rule{1em}{0ex}}\mathrm{\forall}{\xi}_{i}.$
Also, we suppose that the symmetry relations (5) are satisfied. If $\partial {B}_{2}$ is not empty or $c(x)\ne 0$ on B, then the mixed initial boundary value problem in thermoelastodynamics of nonsimple materials has at most one solution.
Proof Suppose, by contrary, that our mixed problem defined by (8), (6) and (7) has two solutions $({u}_{i}^{(\alpha )},{\theta}^{(\alpha )})$ ($\alpha =1,2$) that correspond to the same initial and boundary data, to the same body force and the same heat supply.
If we denote by
${v}_{i}={u}_{i}^{(2)}{u}_{i}^{(1)},\phantom{\rule{2em}{0ex}}\chi ={\theta}_{i}^{(2)}{\theta}_{i}^{(1)},$
(26)
then we shall prove that
$\begin{array}{r}{v}_{i}(x,t)=0,\phantom{\rule{2em}{0ex}}{\psi}_{i}(x,t)=0,\\ \delta (x,t)=0,\phantom{\rule{2em}{0ex}}\chi (x,t)=0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}(x,t)\in B\times [0,{t}_{0}).\end{array}$
(27)
It is clear that the differences
$({v}_{i},{\psi}_{i},\delta ,\chi )$ from (26) also represent a solution of our problem but with null body force and null heat supply. If we write the identity (16) for this particular case, we have
$2{\int}_{B}\varrho {v}_{i}(t){\dot{v}}_{i}(t)\phantom{\rule{0.2em}{0ex}}dV+{\int}_{B}\frac{1}{{\theta}_{0}}{k}_{ij}({\int}_{0}^{t}{\chi}_{,i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )({\int}_{0}^{t}{\chi}_{,j}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )\phantom{\rule{0.2em}{0ex}}dV=0.$
Now, we integrate this equality on the interval
$[0,s]$,
$s\in [0,{t}_{0}/2)$ and obtain
${\int}_{B}\varrho {v}_{i}(s){v}_{i}(s)\phantom{\rule{0.2em}{0ex}}dV+{\int}_{0}^{s}{\int}_{B}\frac{1}{{\theta}_{0}}{k}_{ij}({\int}_{0}^{\tau}{\chi}_{,i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )({\int}_{0}^{\tau}{\chi}_{,j}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )\phantom{\rule{0.2em}{0ex}}dV\phantom{\rule{0.2em}{0ex}}d\tau =0.$
Taking into account the properties of
ϱ and
${k}_{ij}$, the above identity proves that
${v}_{i}(x,t)=0,\phantom{\rule{2em}{0ex}}{\chi}_{,i}(x,t)=0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}(x,t)\in B\times [0,{t}_{0}/2).$
(28)
If $\partial {B}_{2}$ is not empty, considering the boundary conditions (7), then from (28) we deduce that (27) holds. If $a(x)\ne 0$, from the equation of energy (written for the differences), we get $\dot{\chi}=0$. However, χ vanishes initially, such that (27) again holds true.
If
${t}_{0}$ is infinite, then the proof of Theorem 2 is complete. If
${t}_{0}$ is finite, then we set
${v}_{i}(x,\frac{{t}_{0}}{2})={\dot{v}}_{i}(x,\frac{{t}_{0}}{2})=0,\phantom{\rule{2em}{0ex}}\chi (x,\frac{{t}_{0}}{2})=0$
and repeat the above procedure on the interval $[{t}_{0}/2,{t}_{0}/2+{t}_{0}/4]$ such that we extend the conclusion (27) on $B\times [0,3{t}_{0}/4)$, and so on.
Finally, we obtain (27) on $B\times [0,{t}_{0})$ and this concludes the proof of Theorem 2. □
We are ready to state and prove the continuous dependence theorem with regard to body force and heat supply on the compact subintervals of the interval $[0,{t}_{0})$ for the solution of the mixed initial boundary value problem defined by the system of equations (8), the initial conditions (6) and the boundary conditions (7).
Theorem 3 Suppose the same conditions as in Theorem 2.
Let $({u}_{i}^{(\alpha )},{\theta}^{(\alpha )})$ (
$\alpha =1,2$)
be two solutions of our mixed problem which correspond to the same initial and boundary data but to different body force and different heat supply,
$({\mathcal{F}}_{i}^{(\alpha )},{\mathcal{P}}^{(\alpha )})$ (
$\alpha =1,2$),
where ${\mathcal{F}}_{i}={f}_{i}^{2}{f}_{i}^{1},\phantom{\rule{2em}{0ex}}\mathcal{P}={r}^{2}{r}^{1}.$
Moreover,
we suppose that there exists ${t}_{\ast}\in (0,{t}_{0})$ such that $\begin{array}{r}{\int}_{0}^{{t}_{\ast}}{\int}_{B}\varrho {\mathcal{F}}_{i}(t){\mathcal{F}}_{i}(t)\phantom{\rule{0.2em}{0ex}}dV\phantom{\rule{0.2em}{0ex}}dt\le {M}_{1}^{2},\phantom{\rule{2em}{0ex}}{\int}_{0}^{{t}_{\ast}}{\int}_{B}\frac{\varrho}{{\theta}_{0}}{({\int}_{0}^{t}\mathcal{P}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )}^{2}\phantom{\rule{0.2em}{0ex}}dV\phantom{\rule{0.2em}{0ex}}dt\le {M}_{2}^{2},\\ {\int}_{0}^{{t}_{\ast}}{\int}_{B}\varrho {v}_{i}(t){v}_{i}(t)\phantom{\rule{0.2em}{0ex}}dV\phantom{\rule{0.2em}{0ex}}dt\le {K}^{2},\phantom{\rule{2em}{0ex}}{\int}_{0}^{{t}_{\ast}}{\int}_{B}\frac{\varrho}{{\theta}_{0}}{\chi}^{2}(t)\phantom{\rule{0.2em}{0ex}}dV\phantom{\rule{0.2em}{0ex}}dt\le {Q}^{2}.\end{array}$
(29)
Then we have the following estimate:
$\begin{array}{r}{\int}_{B}\varrho {v}_{i}(s){v}_{i}(s)\phantom{\rule{0.2em}{0ex}}dV+{\int}_{0}^{s}{\int}_{B}\frac{1}{{\theta}_{0}}{k}_{ij}({\int}_{0}^{t}{\chi}_{,i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )({\int}_{0}^{t}{\chi}_{,j}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )\phantom{\rule{0.2em}{0ex}}dV\phantom{\rule{0.2em}{0ex}}dt\\ \phantom{\rule{1em}{0ex}}\le {t}_{\ast}K{[{\int}_{0}^{{t}_{\ast}}{\int}_{B}\varrho {\mathcal{F}}_{i}(t){\mathcal{F}}_{i}(t)\phantom{\rule{0.2em}{0ex}}dV\phantom{\rule{0.2em}{0ex}}dt]}^{1/2}\\ \phantom{\rule{2em}{0ex}}+{t}_{\ast}Q{\left[{\int}_{0}^{{t}_{\ast}}{\int}_{B}\frac{\varrho}{{\theta}_{0}}{({\int}_{0}^{t}\mathcal{P}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )}^{2}\phantom{\rule{0.2em}{0ex}}dV\phantom{\rule{0.2em}{0ex}}dt\right]}^{1/2},\end{array}$
(30)
where ${v}_{i}(s)$ and $\chi (s)$ are the differences defined in (26) and $s\in [0,{t}_{\ast}/2)$.
Proof We will use the identity (16). On the righthand side of this identity, we employ the Schwarz inequality for each integral.
For instance, we have
$\begin{array}{c}{\int}_{0}^{t}{\int}_{B}\varrho {v}_{i}(2ts){\mathcal{F}}_{i}(s)\phantom{\rule{0.2em}{0ex}}dV\phantom{\rule{0.2em}{0ex}}ds\hfill \\ \phantom{\rule{1em}{0ex}}\le {[{\int}_{0}^{t}{\int}_{B}\varrho {\mathcal{F}}_{i}(s){\mathcal{F}}_{i}(s)\phantom{\rule{0.2em}{0ex}}dV\phantom{\rule{0.2em}{0ex}}ds]}^{1/2}{[{\int}_{0}^{{t}_{\ast}}{\int}_{B}\varrho {v}_{i}(2ts){v}_{i}(2ts)\phantom{\rule{0.2em}{0ex}}dV\phantom{\rule{0.2em}{0ex}}ds]}^{1/2}\hfill \\ \phantom{\rule{1em}{0ex}}={[{\int}_{0}^{t}{\int}_{B}\varrho {\mathcal{F}}_{i}(s){\mathcal{F}}_{i}(s)\phantom{\rule{0.2em}{0ex}}dV\phantom{\rule{0.2em}{0ex}}ds]}^{1/2}{[{\int}_{t}^{2t}{\int}_{B}\varrho {v}_{i}(s){v}_{i}(s)\phantom{\rule{0.2em}{0ex}}dV\phantom{\rule{0.2em}{0ex}}ds]}^{1/2}\hfill \\ \phantom{\rule{1em}{0ex}}\le K{[{\int}_{0}^{{t}_{\ast}}{\int}_{B}\varrho {\mathcal{F}}_{i}(s){\mathcal{F}}_{i}(s)\phantom{\rule{0.2em}{0ex}}dV\phantom{\rule{0.2em}{0ex}}ds]}^{1/2},\hfill \end{array}$
where, at last, we use the substitution $2ts\to s$.
We proceed analogously with other integrals in the identity (16). Finally, we integrate the resulting inequality over $[0,s]$, $s\in [0,{t}_{\ast}/2]$ and we obtain the inequality (30) and the proof of Theorem 3 is complete. □
In the following theorem, we use the estimate (30) in order to deduce a continuous dependence result upon initial data.
Theorem 4 Assume that the symmetry relations (5)
are satisfied.
Consider $({u}_{i}^{1},{\theta}^{1}),\phantom{\rule{2em}{0ex}}({u}_{i}^{1}+{v}_{i},{\theta}^{1}+\chi )$
two solutions of the mixed initial boundary value problem defined by (8), (6)
and (7)
which correspond to the same body force and heat supply and to the same boundary data,
but to different initial data $({u}_{0i}^{1},{u}_{1i}^{1},{\theta}^{1}),\phantom{\rule{2em}{0ex}}({u}_{0i}^{2},{u}_{1i}^{2},{\theta}^{2}),$
where
${u}_{0i}^{2}={u}_{0i}^{1}+{a}_{i}^{0},\phantom{\rule{2em}{0ex}}{u}_{1i}^{2}={u}_{1i}^{1}+{a}_{i}^{1},\phantom{\rule{2em}{0ex}}{\theta}^{2}={\theta}^{1}+{d}^{0}.$
Here the perturbations $({a}_{i}^{0},{a}_{i}^{1},{d}^{0})$ obey the following restrictions:
${\int}_{B}\varrho ({a}_{i}^{0}{a}_{i}^{0}+{a}_{i}^{1}{a}_{i}^{1})\phantom{\rule{0.2em}{0ex}}dV\le {M}_{3}^{2},\phantom{\rule{2em}{0ex}}{\int}_{B}\frac{{T}_{0}}{\varrho}{\eta}_{0}^{2}\phantom{\rule{0.2em}{0ex}}dV\le {M}_{4}^{2},$
where we used the notation
${\eta}_{0}(x)=\frac{c(x)}{{\theta}_{0}}{d}^{0}(x){E}_{ij}(x){a}_{i,j}^{0}(x){D}_{ijk}(x){a}_{k,ij}^{0}(x).$
Using perturbation ${v}_{i}$ and χ,
we define the functions ${U}_{i}(x,t)$ and $\mathrm{\Theta}(x,t)$ by ${U}_{i}(x,t)={\int}_{0}^{t}{\int}_{0}^{s}{v}_{i}(x,\tau )\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{2em}{0ex}}\mathrm{\Theta}(x,t)={\int}_{0}^{t}{\int}_{0}^{s}\chi (x,\tau )\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds.$
(31)
If the functions $({U}_{i},\mathrm{\Theta})$ satisfy the conditions (29),
then we have the following estimate:
$\begin{array}{r}{\int}_{B}\varrho {U}_{i}(t){U}_{i}(t)\phantom{\rule{0.2em}{0ex}}dV+{\int}_{0}^{t}{\int}_{B}\frac{1}{{\theta}_{0}}{k}_{ij}({\int}_{0}^{s}{\mathrm{\Theta}}_{,i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )({\int}_{0}^{s}{\mathrm{\Theta}}_{,j}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )\phantom{\rule{0.2em}{0ex}}dV\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}\le {t}_{\ast}K{[({t}_{\ast}+{t}_{\ast}^{2}/2){\int}_{B}\varrho {a}_{i}^{0}{a}_{i}^{0}\phantom{\rule{0.2em}{0ex}}dV+({t}_{\ast}^{2}/2+{t}_{\ast}^{3}/3){\int}_{B}\varrho {a}_{i}^{1}{a}_{i}^{1}\phantom{\rule{0.2em}{0ex}}dV]}^{1/2}\\ \phantom{\rule{2em}{0ex}}+{t}_{\ast}^{7/2}Q\frac{1}{\sqrt{20}}{\left({\int}_{B}\frac{{T}_{0}}{\varrho}{\eta}_{0}^{2}\phantom{\rule{0.2em}{0ex}}dV\right)}^{1/2},\phantom{\rule{1em}{0ex}}t\in [0,\frac{{t}_{\ast}}{2}].\end{array}$
(32)
Proof Integrating by parts in (31), we deduce
${U}_{i}(x,t)={\int}_{0}^{t}(ts){v}_{i}(x,s)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{2em}{0ex}}\mathrm{\Theta}(x,t)={\int}_{0}^{t}(ts)\chi (x,s)\phantom{\rule{0.2em}{0ex}}ds.$
It is easy to prove that the difference functions
$({v}_{i},\chi )$ satisfy the equations of motion and the equation of energy as in (8), but with null loads
Also, by direct calculations, we deduce that the difference functions satisfy the initial conditions in the form
${v}_{i}(x,0)={a}_{i}^{0}(x),\phantom{\rule{2em}{0ex}}{\dot{v}}_{i}(x,0)={a}_{i}^{1}(x),\phantom{\rule{2em}{0ex}}\chi (x,0)={d}^{0}(x),\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in B.$
Then, a straightforward calculation proves that the functions
$({U}_{i},\mathrm{\Theta})$ defined in (31) satisfy the equations of motion and the equation of energy as in (8), but with the following body force and heat supply:
$\begin{array}{c}{f}_{i}(x,t)={a}_{i}^{0}(x)+t{a}_{i}^{1}(x),\hfill \\ r(x,t)=\frac{{\theta}_{0}}{\varrho}t[\frac{c(x)}{{\theta}_{0}}{d}^{0}(x){E}_{ij}(x){a}_{i,j}^{0}(x){D}_{ijk}(x){a}_{k,ij}^{0}(x)].\hfill \end{array}$
By using these specifications, the estimate (32) follows from (30) and Theorem 4 is concluded. □
Finally, we obtain a continuous dependence result of the solution to problems (8), (6) and (7) upon the thermoelastic coefficients, again as a consequence of Theorem 3.
Theorem 5 Assume that the symmetry relations (5)
are satisfied and consider $({u}_{i}^{1},{\theta}^{1}),\phantom{\rule{2em}{0ex}}({u}_{i}^{1}+{v}_{i},{\theta}^{1}+\chi )$
two solutions of the mixed initial boundary value problem defined by (8), (6)
and (7)
which correspond to the same body force and heat supply and to the same boundary and initial data,
but to different thermoelastic coefficients $\begin{array}{c}({A}_{ijrs}^{(1)},{B}_{ijpqr}^{(1)},{C}_{ijkmnr}^{(1)},{E}_{ij}^{(1)},{D}_{ijk}^{(1)},{b}_{i}^{(1)},{k}_{ij}^{(1)},{c}^{(1)}),\hfill \\ ({A}_{ijrs}^{(1)}+{\mathcal{A}}_{ijrs},{B}_{ijpqr}^{(1)}+{\mathcal{B}}_{ijpqr},{C}_{ijkmnr}^{(1)}+{\mathcal{C}}_{ijkmnr},\hfill \\ \phantom{\rule{1em}{0ex}}{E}_{ij}^{(1)}+{\mathcal{E}}_{ij},{D}_{ijk}^{(1)}+{\mathcal{D}}_{ijk},{b}_{i}^{(1)}+{\mathcal{B}}_{i},{k}_{ij}^{(1)}+{\mathcal{K}}_{ij},{c}^{(1)}+\mathcal{C}).\hfill \end{array}$
Suppose that the perturbations $({v}_{i},\chi )$ satisfy the conditions (29).
Then any solution $({u}_{i},\theta )$ of the initial boundary value problem defined by (8), (6)
and (7)
that satisfies the condition ${\int}_{0}^{{t}_{\ast}}{\int}_{B}({u}_{i,j}{u}_{i,j}+{u}_{i,jk}{u}_{i,jk}+{\dot{u}}_{i,j}{\dot{u}}_{i,j}+{\theta}_{,j}{\theta}_{,j}+{\theta}_{,jk}{\theta}_{,jk}+{\dot{\theta}}^{2})\phantom{\rule{0.2em}{0ex}}dV\phantom{\rule{0.2em}{0ex}}ds\le {M}_{5}^{2}$
depends continuously on the thermoelastic coefficients on the interval
$[0,{t}_{\ast}/2]$
in
${\int}_{B}\varrho {v}_{i}(t){v}_{i}(t)+{\int}_{0}^{t}{\int}_{B}\frac{1}{{\theta}_{0}}{k}_{ij}({\int}_{0}^{s}{\chi}_{,i}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )({\int}_{0}^{s}{\chi}_{,j}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi )\phantom{\rule{0.2em}{0ex}}dV\phantom{\rule{0.2em}{0ex}}ds.$
Proof A straightforward calculation proves that the perturbations
$({v}_{i},\chi )$ of two solutions verify the equations of motion and the equation of energy with the following body force load and heat supply:
$\begin{array}{c}\begin{array}{rl}\varrho {\mathcal{F}}_{i}=& {\mathcal{A}}_{ijrs}{u}_{r,sj}^{(2)}+{\mathcal{B}}_{ijpqr}{u}_{r,pqj}^{(2)}+{\mathcal{E}}_{ij}{\theta}_{,j}^{(2)}\\ +{\mathcal{B}}_{rskji}{u}_{r,skj}^{(2)}+{\mathcal{C}}_{kjipqr}{u}_{r,pqkj}^{(2)}+{\mathcal{D}}_{kji}{\theta}_{,kj}^{(2)},\end{array}\hfill \\ \begin{array}{r}\varrho \mathcal{P}={\left({\mathcal{K}}_{ij}{\theta}_{,j}^{(2)}\right)}_{,i}{\theta}_{0}(\frac{\mathcal{C}}{{\theta}_{0}}{\dot{\theta}}^{2}{\mathcal{E}}_{ij}{u}_{i,j}^{(2)}{\mathcal{D}}_{ijk}{u}_{k,ij}^{(2)}).\end{array}\hfill \end{array}$
Thus the problem is analogous to the problem from Theorem 4. Therefore, according to the estimates (32) and (30), we obtain the desired result. □