Our first result is a reciprocity one. To this end, we need the convolution product for two continuous functions. So, if φ and ψ are two scalar functions on \([0,\infty )\times D\) continuous in time, then their convolution product, denoted by ∗, is defined by
$$ (\varphi *\psi ) (t,x)= \int _{0}^{t}\varphi (t - \tau , x)\psi ( \tau , x) \,d\tau . $$
Now we introduce the functions \(p(t)\) and \(r(t)\), useful in the following, defined by
$$\begin{aligned} p(t)=1, \qquad r(t)=(p*p) (t)=t,\quad t\in [0,\infty ), \end{aligned}$$
(14)
and we consider the following writing convention:
$$\begin{aligned} \hat {\varphi}(t,x)= \int _{0}^{t}\varphi (\tau , x)\,d\tau =(p* \varphi ) (t,x). \end{aligned}$$
(15)
To obtain a more accessible form of the energy equation (3), we consider the function ω defined on \([0,\infty )\times D\) by the relation
$$\begin{aligned} \omega = \hat {S} +\vartheta _{0}(\eta _{0}-a), \end{aligned}$$
(16)
where Ŝ is defined as in (15).
Also, for a function u of class \(C^{0,1}\) on \([0,\infty )\times D\), we define the functions β and γ by
$$\begin{aligned} \beta u=u+\alpha u,\qquad \gamma u= p*u+\alpha u, \end{aligned}$$
(17)
where α is from Eqs. (4).
In the following proposition, we formulate the energy equation from (3) in a different manner.
Proposition 1
If the functions \(q_{m}\in C^{1,0}\) and \(\eta \in C^{0,1}\) satisfy the equation of energy (3) and the initial condition \(\eta (0, x)=\eta ^{0}(x)\), \(x\in D\), then they satisfy the equation
$$\begin{aligned} \hat {q}_{m,m} +\omega =\vartheta _{0}(\eta -a)\quad \forall (t,x) \in [0,\infty )\times D. \end{aligned}$$
(18)
The reciprocal statement is also true.
Proof
Both statements are easy to prove; just have to take into account the writing convention (15). □
A reciprocity relation refers to the connection between two external data systems
$$ {\mathcal{D}}^{(\nu )}= \bigl\{ f_{m}^{(\nu )},g_{m}^{(\nu )},S^{(\nu )}, \bar {v}_{m}^{(\nu )},\bar {\phi}_{m}^{(\nu )}, \bar {t}_{m}^{(\nu )}, \bar {m}_{k}^{(\nu )}, \bar {\vartheta}^{(\nu )},\bar {q}^{(\nu )},v_{m}^{0,( \nu )},v_{m}^{1,(\nu )}, \phi _{m}^{0,(\nu )},\phi _{m}^{1,(\nu )}, \eta ^{0,(\nu )}, \vartheta ^{0,(\nu )} \bigr\} , $$
and the solutions corresponding to these data systems
$$ {\mathcal{S}}^{(\nu )}= \bigl\{ v_{m}^{(\nu )},\phi _{m}^{(\nu )}, \vartheta ^{(\nu )},\tau _{mn}^{(\nu )},\sigma _{mn}^{(\nu )},\eta{( \nu )}, q_{m}^{(\nu )} \bigr\} . $$
In both systems, \(\nu =1,2\).
To simplify the writing of reciprocity relations, we need the following notations:
$$\begin{aligned} \Gamma _{\nu \mu}(s,r) =& \int _{\partial D} \biggl[t_{m}^{( \nu )}(s,x)v_{m}^{(\mu )}(r,x) + m_{k}^{(\nu )}(s,x)\phi _{m}^{( \mu )}(r,x) - \frac{1}{\vartheta _{0}}\beta \bar {q}_{m}^{(\nu )}(s,x) \vartheta ^{(\mu )}(r,x) \biggr]\,dA \\ &{}+ \int _{D} \biggl[f_{m}^{(\nu )}(s,x)v_{m}^{(\mu )}(r,x) + g_{m}^{( \nu )}(s,x)\phi _{m}^{(\mu )}(r,x) - \frac{1}{\vartheta _{0}}\beta \omega ^{(\nu )}(s,x)\vartheta ^{(\mu )}(r,x) \biggr]\,dV \\ &{}+ \int _{D} \bigl[\varrho \ddot{v}_{m}^{(\nu )}(s,x)v_{m}^{(\mu )}(r,x)+I_{mn} \ddot{\phi}_{m}^{(\nu )}(s,x)\phi _{n}^{(\mu )}(r,x) \\ & {}- h\dot{\vartheta}^{(\nu )}(s,x)\vartheta ^{(\mu )}(r,x)- \alpha d\vartheta ^{(\nu )}(s,x)\dot{\vartheta}^{(\mu )}(r,x) \bigr] \,dV \\ &{}+\frac{1}{\vartheta _{0}} \int _{D}\beta \bar {q}_{m}^{(\nu )}(s,x) \vartheta _{,m}^{(\mu )}(r,x)\,dV,\quad \nu ,\mu =1,2, \end{aligned}$$
(19)
where we used the following convention:
$$\begin{aligned}& t_{m}^{(\nu )}=\tau _{mk}^{(\nu )}n_{k}, \qquad m_{k}^{(\nu )}=\sigma _{kl}^{( \nu )}n_{l}, \\& \omega ^{(\nu )}=\bar {S}^{(\nu )}+\vartheta _{0} \bigl( \eta ^{0,( \nu )}-a \bigr), \qquad q^{(\nu )}=q_{k}^{(\nu )}n_{k}. \end{aligned}$$
(20)
Also, we introduce the following notations:
$$\begin{aligned} J_{\nu \mu}(s,r) =&\tau _{mn}^{(\nu )}(s)v_{m,n}^{(\mu )}(r)+ \sigma _{mn}^{( \nu )}(s)\phi _{m,n}^{(\mu )}(r) \\ &{}-\beta \bigl[\eta ^{(\nu )}(s)-a \bigr]\vartheta ^{(\mu )}(r), \quad \nu ,\mu =1,2, \end{aligned}$$
(21)
and
$$\begin{aligned} I_{\nu \mu}(s,r)= J_{\nu \mu}(s,r)+ h\dot{\vartheta}^{(\nu )}(s) \vartheta ^{(\mu )}(r)+ \alpha \,d\vartheta ^{(\nu )}(s)\dot{\vartheta}^{( \mu )}(r),\quad \nu ,\mu =1,2, \end{aligned}$$
(22)
where we suppressed the variable x.
Now we formulate and prove the first reciprocity result.
Theorem 1
If the symmetry relations (5) are satisfied, then for any \(s,r\in [0,\infty )\), we have the equality
$$\begin{aligned} \Gamma _{\nu \mu}(s,r)=\Gamma _{\mu \nu}(r,s),\quad \nu ,\mu =1,2. \end{aligned}$$
(23)
Proof
Taking into account the constitutive relations (4) and notations (21), we deduce
$$\begin{aligned} J_{\nu \mu}(s,r) =&C_{klmn}v_{m,n}^{(\nu )}(s)v_{k,l}^{(\mu )}(r) \\ &{}+B_{klmn} \bigl[v_{m,n}^{(\nu )}(s)\phi _{k,l}^{(\mu )}(r)+v_{m,n}^{( \mu )}(s)\phi _{k,l}^{(\nu )}(r) \bigr]+ A_{klmn}\phi _{m,n}^{(\nu )}(s) \phi _{k,l}^{(\nu )}(r) \\ &{}-a_{mn} \bigl[\beta v_{m,n}^{(\nu )}(s)\vartheta ^{(\mu )}(r)+\beta v_{m,n}^{( \mu )}(s)\vartheta ^{(\nu )}(r) \bigr] \\ &{}-b_{mn} \bigl[\beta \phi _{m,n}^{(\nu )}(s) \vartheta ^{(\mu )}(r)+ \beta \phi _{m,n}^{(\mu )}(s) \vartheta ^{(\nu )}(r) \bigr] \\ &{}-\alpha h \dot{\vartheta}^{(\nu )}(s)\dot{\vartheta}^{(\mu )}(r)- d \vartheta ^{(\nu )}(s) \vartheta ^{(\mu )}(r) \\ &{}-h \dot{\vartheta}^{(\nu )}(s)\vartheta ^{(\mu )}(r)- \alpha \,d \vartheta ^{(\nu )}(s) \dot{\vartheta}^{(\mu )}(r). \end{aligned}$$
(24)
Based on relations (5), (24), and (22), we easily deduce that
$$\begin{aligned} I_{\nu \mu}(s,r)=I_{\mu \nu}(r,s). \end{aligned}$$
(25)
Now we consider equations (2) and (18) and notation (21), and so we are led to the relation
$$\begin{aligned} J_{\nu \mu}(s,r) =& \biggl[\tau _{mn}^{(\nu )}(s)v_{m}^{(\mu )}(r)+ \sigma _{mn}^{(\nu )}(s)\phi _{m}^{(\mu )}(r) - \frac{1}{\vartheta _{0}}\beta \bar {q}_{n}^{(\nu )}(s)\vartheta ^{( \mu )}(r) \biggr]_{,n} \\ &{}+f_{m}^{(\nu )}(s)v_{m}^{(\mu )}(r) +g_{m}^{(\nu )}(s)\phi _{m}^{( \mu )}(r)- \frac{1}{\vartheta _{0}}\beta \omega ^{(\nu )}(s) \vartheta ^{(\mu )}(r) \\ &{}-\varrho \ddot{v}_{m}^{(\nu )}(s)v_{m}^{(\mu )}(r) -I_{mn} \ddot{\phi}_{m}^{( \nu )}(s)\phi _{m}^{(\mu )}(r) +\frac{1}{\vartheta _{0}}\beta \bar {q}_{m}^{(\nu )}(s)\vartheta _{,m}^{(\mu )}(r). \end{aligned}$$
(26)
Now we integrate over D equality (22) and then use relation (25) and the theorem of divergence, so that we are led to the equality
$$\begin{aligned} \int _{D} I_{\nu \mu}(s,r)\,dV=\Gamma _{\nu \mu}(r,s), \end{aligned}$$
which, together with (25), ensures equality (23). The proof of the theorem is complete. □
To obtain another reciprocity result, we introduce the notations
$$\begin{aligned}& F_{m}^{(\nu )} =r* \bigl[f_{m}^{(\nu )}(s)+g_{m}^{(\nu )}(s) \bigr]+ \varrho \bigl[t\dot{v}_{m}^{1,(\nu )}+v_{m}^{0,(\nu )} \bigr], \\& G_{m}^{(\nu )} =r* \bigl[g_{m}^{(\nu )}(s)+g_{m}^{(\nu )}(s) \bigr]+ \varrho \bigl[t\dot{v}_{m}^{1,(\nu )}+v_{m}^{0,(\nu )} \bigr], \\& R^{(\nu )}=-t\vartheta ^{0,(\nu )}, \quad \nu =1,2. \end{aligned}$$
(27)
With the help of Theorem 1 and notations (27), we obtain a new reciprocity result.
Theorem 2
If the symmetry relations (5) are satisfied and \({\mathcal{S}}^{(\nu )}\) is the solution corresponding to the external data system \({\mathcal{S}}^{(\nu )}\), \(\nu =1,2\), then we have the following equality:
$$\begin{aligned}& \int _{\partial D}r* \biggl[t_{m}^{(1)}(s)*v_{m}^{(2)}+m_{k}^{(1)}* \phi _{k}^{(2)} -\frac{1}{\vartheta _{0}} q^{(1)}*\gamma \vartheta ^{(2)} \biggr]\,dA \\& \qquad {}+ \int _{D} \biggl[F_{m}^{(1)}*v_{m}^{(2)}+G_{m}^{(1)}* \phi _{m}^{(2)} - \frac{1}{\vartheta _{0}}p*\vartheta ^{(2)}* \gamma \omega ^{(1)} \biggr]\,dV \\& \qquad {}-\frac{\alpha}{\vartheta _{0}} \int _{\partial D}p* q^{(1)}*R^{(2)}\,dA - \frac{\alpha}{\vartheta _{0}} \int _{D}\omega ^{(1)}*R^{(2)}\,dV \\& \qquad {}+ \int _{D} \bigl[(h-\alpha d)R^{(1)}*\vartheta ^{(2)}+\alpha p*K_{mn} \vartheta _{,n}^{(1)}*R_{,m}^{(2)} \bigr]\,dV \\& \quad = \int _{\partial D}r* \biggl[t_{m}^{(2)}(s)*v_{m}^{(1)}+m_{k}^{(2)}* \phi _{k}^{(1)} -\frac{1}{\vartheta _{0}} q^{(2)}*\gamma \vartheta ^{(1)} \biggr]\,dA \\& \qquad {}+ \int _{D} \biggl[F_{m}^{(2)}*v_{m}^{(1)}+G_{m}^{(2)}* \phi _{m}^{(1)} - \frac{1}{\vartheta _{0}}p*\vartheta ^{(1)}* \gamma \omega ^{(2)} \biggr]\,dV \\& \qquad {}-\frac{\alpha}{\vartheta _{0}} \int _{\partial D}p* q^{(2)}*R^{(1)}\,dA - \frac{\alpha}{\vartheta _{0}} \int _{D}\omega ^{(2)}*R^{(1)}\,dV \\& \qquad {}+ \int _{D} \bigl[(h-\alpha d)R^{(2)}*\vartheta ^{(1)}+\alpha p*K_{mn} \vartheta _{,n}^{(2)}*R_{,m}^{(1)} \bigr]\,dV. \end{aligned}$$
(28)
Proof
We use the reciprocity relation (23) in which we replace r with τ and s with \(t - \tau \), and then integrate the relation obtained over the interval \([0,t]\), so that we obtain the relation
$$\begin{aligned} F_{\nu \mu}(t) =& \int _{\partial D} \biggl[t_{m}^{(\nu )}(s)*v_{m}^{( \mu )}+m_{k}^{(\nu )}* \phi _{k}^{(\mu )} -\frac{1}{\vartheta _{0}} p*q^{( \nu )}*\beta \vartheta ^{(\mu )} \biggr]\,dA \\ &{}+ \int _{D} \biggl[f_{m}^{(\nu )}*v_{m}^{(\mu )}+g_{m}^{(\nu )}* \phi _{m}^{( \mu )} -\frac{1}{\vartheta _{0}}\omega ^{(\nu )}*\beta \vartheta ^{( \mu )} \biggr]\,dV \\ &{}- \int _{D} \bigl[\varrho \ddot{v}_{m}^{(\nu )}*v_{m}^{(\mu )} + I_{mn}\ddot{\phi}_{m}^{(\nu )}*\phi _{n}^{(\mu )} - h \dot{\vartheta}^{(\nu )}*\vartheta ^{(\mu )} - \alpha d \vartheta ^{( \nu )}*\dot{\vartheta}^{(\mu )} \bigr]\,dV \\ &{}+\frac{1}{\vartheta _{0}} \int _{D} p*q_{m}^{(\nu )}*\beta \vartheta _{,m}^{( \mu )}\,dV, \end{aligned}$$
(29)
where
$$ F_{\nu \mu}(t)= \int _{0}^{t} \Gamma _{\nu \mu}(s,t-s)\,ds. $$
It is no difficult to verify that for arbitrary continuous functions r and p on \([0,\infty )\times D\), we have the following equalities:
$$\begin{aligned}& r*\dot{\vartheta}^{(\nu )}*\vartheta ^{(\mu )}= p* \bigl(p* \dot{\vartheta}^{(\nu )} \bigr)*\vartheta ^{(\mu )}= p* \bigl( \vartheta ^{(\nu )}-\vartheta _{0}^{(\nu )} \bigr)* \vartheta ^{(\mu )} \\ & \hphantom{r*\dot{\vartheta}^{(\nu )}*\vartheta ^{(\mu )}}=p*\vartheta ^{(\nu )}*\vartheta ^{(\mu )}+R^{(\nu )}* \vartheta ^{( \mu )}, \\ & r*\beta \vartheta ^{(\mu )}= p* \bigl(\gamma \vartheta ^{(\mu )}- \alpha \vartheta _{0}^{(\mu )} \bigr)= p*\gamma \vartheta ^{(\mu )}+ \alpha R^{(\mu )}, \\& r* \ddot{v}_{m}^{(\nu )}=v_{m}^{(\nu )}-tv_{m}^{1,(\nu )}-v_{m}^{0,( \nu )}, \\& r* \ddot{\phi}_{m}^{(\nu )}=\phi _{m}^{(\nu )}-t \phi _{m}^{1,(\nu )}- \phi _{m}^{0,(\nu )}, \\& l*\gamma h=l*(p*h+\alpha h)=h*\gamma l. \end{aligned}$$
(30)
Based on (30) and (29), we deduce
$$\begin{aligned} r*F_{\nu \mu}(t) =& \int _{\partial D}r* \biggl[t_{m}^{(\nu )}(s)*v_{m}^{( \mu )}+m_{k}^{(\nu )}* \phi _{k}^{(\mu )} -\frac{1}{\vartheta _{0}} q^{( \nu )}*\gamma \vartheta ^{(\mu )} \biggr]\,dA \\ &{}+ \int _{D} \biggl[F_{m}^{(\nu )}*v_{m}^{(\mu )}+G_{m}^{(\nu )}* \phi _{m}^{( \mu )} -\frac{1}{\vartheta _{0}}p*\vartheta ^{(\mu )}* \gamma \omega ^{( \nu )} \biggr]\,dV \\ &{}- \int _{D} \bigl[\varrho v_{m}^{(\nu )}*v_{m}^{(\mu )} + I_{mn} \phi _{m}^{(\nu )}*\phi _{n}^{(\mu )} - (h+\alpha d)p* \vartheta ^{( \nu )}*\vartheta ^{(\mu )} \bigr]\,dV \\ &{}+ \int _{D} r*K_{mn}\vartheta _{,n}^{(\nu )}* \bigl(p*\vartheta _{,m}^{( \mu )}+\alpha \vartheta _{,m}^{(\mu )} \bigr)\,dV \\ &{}-\frac{\alpha}{\vartheta _{0}} \int _{\partial D} p*q^{(\nu )}*R^{( \mu )}\,dA- \frac{\alpha}{\vartheta _{0}} \int _{D} \omega ^{(\nu )}*R^{( \mu )}\,dA \\ &{}+ \int _{D} \bigl[hR^{(\nu )}*\vartheta ^{(\mu )}+ \alpha \,dR^{(\mu )}* \vartheta ^{(\nu )}+\alpha p* K_{mn} \vartheta _{,n}^{(\nu )}*R_{,m}^{( \mu )} \bigr] \,dV. \end{aligned}$$
(31)
Finally, from relations (5), (23), (29), and (31) we obtain the desired relation (28). □
To simplify the relations that follow, we will use the notation
$$\begin{aligned} G(s,r) =& \int _{D} \biggl[f_{m}(s)v_{m}(r)+g_{m}(s) \phi _{m}(r)- \frac{1}{\vartheta _{0}}\omega (s)\beta \vartheta (r) \biggr]\,dV \\ &{}+ \int _{\partial D} \biggl[t_{m}(s)v_{m}(r)+m_{k}(s) \phi _{m}(r)- \frac{1}{\vartheta _{0}}\bar {q}(s)\beta \vartheta (r) \biggr] \,dA,\quad s,r \in [0,\infty ). \end{aligned}$$
(32)
The reciprocity relation (23) is also the basis of the following theorem.
Theorem 3
If the symmetry relations (5) are satisfied and \({\mathcal{S}}^{(\nu )}\) is the solution corresponding to the external data system \({\mathcal{S}}^{(\nu )}\), \(\nu =1,2\), then we have the following equality:
$$\begin{aligned}& \frac{d}{dt} \biggl\{ \int _{D} [\varrho v_{m}v_{m}+I_{mn} \phi _{m} \phi _{n}+\alpha K_{mn}\bar { \vartheta}_{,m}\bar {\vartheta}_{,n} ]\,dV \biggr\} \\& \qquad {}+\frac{d}{dt} \biggl\{ \int _{0}^{t} \int _{D} \bigl[(\alpha d-h) \vartheta ^{2}+K_{mn} \bar {\vartheta}_{,n}\bar {\vartheta}_{,m} \bigr]\,dV\,ds \biggr\} \\& \quad = \int _{0}^{t}\bigl[G(t - \tau , t + \tau )-G(t + \tau , t - \tau )\bigr]\,d\tau \\& \qquad {}+ \int _{D} \bigl[ \varrho \bigl(\dot{v}_{m}(2t)v_{m}(0) + \dot{v}_{m}(0)v_{m}(2t) \bigr] + I_{mn} \bigl[\dot{\phi}_{m}(2t) \phi _{m}(0) + \dot{\phi}_{m}(0)\phi _{m}(2t) \bigr) \bigr]\,dV \\& \qquad {}+ \int _{D} \bigl[(\alpha d-h)\vartheta (0)\vartheta (2t)+\alpha K_{mn} \bar {\vartheta}_{,n}(2t)\vartheta _{,m}(0) \bigr]\,dV. \end{aligned}$$
(33)
Proof
By equality (23) we obtain
$$\begin{aligned} \int _{0}^{t}\Gamma _{11}(t + \tau , t - \tau )\,d\tau = \int _{0}^{t} \Gamma _{11}(t - \tau , t + \tau )\,d\tau , \end{aligned}$$
(34)
which by relations (23) and (32) results in
$$\begin{aligned}& \int _{0}^{t}\Gamma _{11}(t + \tau , t - \tau )\,d\tau \\& \quad = \int _{0}^{t} G(t + \tau , t - \tau )\,d\tau \\& \qquad {}- \int _{0}^{t} \int _{D} \bigl[\varrho \ddot{v}_{m}(t + \tau )v_{m}(t - \tau )+I_{mn}\ddot{\phi}_{m}(t + \tau ) \phi _{m}(t - \tau ) \bigr]\,dV \,d\tau \\& \qquad {}+ \int _{0}^{t} \int _{D} \bigl[h\dot{\vartheta}(t + \tau )\vartheta (t - \tau )+ \alpha d \vartheta (t + \tau )\dot{\vartheta}(t - \tau ) \bigr]\,dV \,d\tau \\& \qquad {}+ \int _{0}^{t} \int _{D}K_{mn}\bar {\vartheta}_{,n}(t+s) \bigl[ \dot{\bar{\vartheta}}_{,m}(t - \tau )+\alpha \ddot{\bar{ \vartheta}}_{,m}(t - \tau ) \bigr]\,dV\,d\tau . \end{aligned}$$
(35)
Analogously, we deduce
$$\begin{aligned}& \int _{0}^{t}\Gamma _{11}(t - \tau , t + \tau )\,d\tau \\& \quad = \int _{0}^{t} G(t - \tau , t + \tau )\,d\tau \\& \qquad {}- \int _{0}^{t} \int _{D} \bigl[\varrho \ddot{v}_{m}(t - \tau )v_{m}(t + \tau )+I_{mn}\ddot{\phi}_{m}(t - \tau ) \phi _{m}(t + \tau ) \bigr]\,dV \,d\tau \\& \qquad {}+ \int _{0}^{t} \int _{D} \bigl[h\dot{\vartheta}(t - \tau )\vartheta (t + \tau )+ \alpha \,d \vartheta (t - \tau )\dot{\vartheta}(t + \tau ) \bigr]\,dV \,d\tau \\& \qquad {}+ \int _{0}^{t} \int _{D}K_{mn}\bar {\vartheta}_{,n}(t-s) \bigl[ \dot{\bar{\vartheta}}_{,m}(t + \tau )+\alpha \ddot{\bar{ \vartheta}}_{,m}(t + \tau ) \bigr]\,dV\,d\tau . \end{aligned}$$
(36)
After usual integration by parts, we deduce the following formulas:
$$\begin{aligned}& \int _{0}^{t}\ddot{u}(t + \tau )v(t - \tau )\,d\tau = \dot{u}(2t)v(0)- \dot{u}(t)v(t)+ \int _{0}^{t}\dot{v}((t - \tau )\dot{u}(t + \tau )\,d \tau , \\& \int _{0}^{t}\ddot{v}(t - \tau )u(t + \tau )\,d\tau = \dot{v}(t)u(t)- \dot{v}(0)v(2t)+ \int _{0}^{t}\dot{v}((t - \tau )\dot{u}(t + \tau )\,d \tau , \\& \int _{0}^{t} u(t + \tau )v(t - \tau )\,d\tau =- v(0)u(2t)+u(t)v(t)+ \int _{0}^{t}\dot{u}((t + \tau )u(t - \tau )\,d\tau . \end{aligned}$$
Taking these formulas into account, from (35) we obtain
$$\begin{aligned}& \int _{0}^{t}\Gamma _{11}(t + \tau , t - \tau )\,d\tau \\& \quad = \int _{0}^{t} G(t + \tau , t - \tau )\,d\tau \\& \qquad {}- \int _{D} \bigl[\varrho \bigl(\dot{v}_{m}(2t)v_{m}(0)- \dot{v}_{m}(t)v_{m}(t) \bigr)+I_{mn} \bigl(\dot{\phi}_{m}(2t)\phi _{n}(0)-\dot{\phi}_{m}(t) \phi _{n}(t) \bigr) \bigr]\,dV \,d\tau \\& \qquad {}- \int _{0}^{t} \int _{D} \bigl[h \bigl(\vartheta (2t)\vartheta (0)- \vartheta ^{2}(t) \bigr)+ \alpha d \bigl(\vartheta ^{2}(2t)- \vartheta (0)\vartheta (2t) \bigr) \bigr]\,dV \,d\tau \\& \qquad {}- \int _{0}^{t} \int _{D} \bigl[\varrho \dot{v}_{m}(t - \tau )\dot{v}_{m}(t + \tau )+I_{mn}\dot{\phi}_{m}(t - \tau ) \dot{\phi}_{n}(t + \tau ) \bigr]\,dV \,d\tau \\& \qquad {}+ \int _{0}^{t} \int _{D} \bigl[h\vartheta (t + \tau )\dot{\vartheta}(t - \tau )+\alpha d \dot{\vartheta}(t + \tau )\dot{\vartheta}(t - \tau ) \bigr]\,dV \,d \tau \\& \qquad {}+ \int _{0}^{t} \int _{D}K_{mn} \bigl[\dot {\bar { \vartheta}}_{,n}(t + \tau )\bar {\vartheta}_{,m}(t - \tau )+ \alpha \vartheta _{,n}(t + \tau )\vartheta _{,m}(t - \tau ) \bigr]\,dV\,d\tau \\& \qquad {}+ \int _{D}K_{mn} \bigl[\bar {\vartheta}_{,n}(2t) ( \bar {\vartheta}_{,m}+\alpha \dot {\bar {\vartheta}}_{,m} ) (0)- \alpha \bar {\vartheta}_{,n}(2t)\vartheta _{,m}(0) \bigr]\,dV \end{aligned}$$
(37)
Similarly, from (36) we have
$$\begin{aligned}& \int _{0}^{t}\Gamma _{11}(t - \tau , t + \tau )\,d\tau \\& \quad = \int _{0}^{t} G(t - \tau , t + \tau )\,d\tau \\& \qquad {}- \int _{D} \bigl[\varrho \bigl(\dot{v}_{m}(t)v_{m}(t) - \dot{v}_{m}(0)v_{m}(2t) \bigr) + I_{mn} \bigl( \dot{\phi}_{m}(t) \phi _{n}(t) - \dot{\phi}_{m}(0) \phi _{n}(2t) \bigr) \bigr]\,dV \,d\tau \\& \qquad {}- \int _{0}^{t} \int _{D} \bigl[\varrho \dot{v}_{m}(t - \tau )\dot{v}_{m}(t + \tau )+I_{mn}\dot{\phi}_{m}(t - \tau ) \phi _{n}(t + \tau ) \bigr]\,dV \,d\tau \\& \qquad {}- \int _{0}^{t} \int _{D} \bigl[h \dot{\vartheta}(t - \tau )\vartheta (t + \tau )+ \alpha d \vartheta (t - \tau )\dot{\vartheta}(t + \tau ) \bigr]\,dV \,d\tau \\& \qquad {}+ \int _{0}^{t} \int _{D}K_{mn}\bar {\vartheta}_{,n}(t - \tau ) \dot {\bar {\vartheta}}_{,m}(t + \tau )\,dV\,d\tau - \int _{D}\alpha K_{mn} \dot {\bar { \vartheta}}_{,m}\bar {\vartheta}_{,m}\,dV \\& \qquad {}+ \int _{0}^{t} \int _{D}\alpha K_{mn}\vartheta _{,m}(t + \tau ) \vartheta _{,n}(t - \tau )\,dV\,ds. \end{aligned}$$
(38)
Clearly, taking into account relations (5), (34), (37), and (38), we arrive at the desired relation (33). □
Now we can approach the problem of the uniqueness of the solution of problem \({ \mathcal{P}}\).
Theorem 4
We suppose that:
-
the symmetry relations (5) are satisfied;
-
ϱ and \(\alpha d-h\) are strictly positive;
-
the tensor \(K_{mn}\) is positive semidefinite;
-
\(\alpha \ge 0\).
Then the mixed problem \({\mathcal{P}}\) admits at most one solution.
Proof
We do the proof by reduction to the absurd. Suppose that problem \({\mathcal{P}}\) admits two solutions \((v_{m}^{(1)}, \phi _{m}^{(1)}, \vartheta ^{(1)} )\) and \((v_{m}^{(2)}, \phi _{m}^{(2)}, \vartheta ^{(2)} )\). Due to the linearity of problem \({\mathcal{P}}\), the difference between the two solutions
$$ (v_{m}, \phi _{m}, \vartheta )= \bigl(v_{m}^{(1)}-v_{m}^{(2)}, \phi _{m}^{(1)}-\phi _{m}^{(2)}, \vartheta ^{(1)}-\vartheta ^{(2)} \bigr) $$
is also the solution of problem \({\mathcal{P}}\), but corresponding to zero initial data and homogeneous boundary conditions. In this situation, equality (33) reduces to
$$\begin{aligned}& \int _{D} [\varrho v_{m}v_{m}+I_{mn} \phi _{m} \phi _{n}+\alpha K_{mn} \bar { \vartheta}_{,m}\bar {\vartheta}_{,n} ]\,dV \\& \quad {}+ \int _{0}^{t} \int _{D} \bigl[(\alpha d-h)\vartheta ^{2}+K_{mn} \bar {\vartheta}_{,n}\bar {\vartheta}_{,m} \bigr]\,dV\,ds=0, \quad t\in [0, \infty ). \end{aligned}$$
(39)
Based on the assumptions of the theorem, we have \(\varrho >0\), \(\alpha d-h>0\), and \(\alpha \ge 0\), and the tensor \(K_{mn}\) is positive semidefinite, so that inequality (39) implies
$$ v_{m}=0,\qquad \phi _{m}=0,\qquad \vartheta =0, $$
which completes the proof of the theorem. □
As the last our main result, we approach a variational principle for a Cosserat thermoelastic body. In fact, we extend Lebon’s principle, formulated in the theory of classical thermoelasticity.
To this aim, we first consider a so-called thermoelastic state of the media with the content
$$ {\mathcal{A}} = (v_{m}, \phi _{m}, \vartheta , e_{mn}, \varepsilon _{mn}, \tau _{mn}, \sigma _{mn}, q_{m}, S ), $$
and denote by H the set of states of the media of this form. If we consider the usual addition and multiplication of states with scalars, then we deduce that H is endowed with a structure of linear space.
Now we introduce the functional \({\mathcal{F}}\) on H by
$$\begin{aligned} {\mathcal{F}}(t,{\mathcal{A}}) =& \int _{D} p* \biggl[ C_{klmn} e_{kl}*e_{mn}+B_{mnkl}e_{kl}* \varepsilon _{mn}+ A_{klmn} \varepsilon _{kl}* \varepsilon _{mn} \\ &{}-\varrho v_{m}*v_{m}+I_{mn}\phi _{m}*\phi _{n}- \frac{1}{\vartheta _{0}}r*K_{mn}\vartheta _{,m}*\vartheta _{,n}-r*q_{m}* \vartheta _{,m} \\ &{}+ [\varrho v_{m}-r*\tau _{mn,n} - F_{m} ]*v_{m}+ (I_{mn} \phi _{m} - p * \sigma _{mn,n} - \epsilon _{mjk}\tau _{jk} - G_{m} )*\phi _{m} \\ &{}+ \frac{\vartheta _{0}}{a}r* (S-a_{mn}e_{mn}-b_{mn} \varepsilon _{mn} )* (S-a_{kl}e_{kl}-b_{kl} \varepsilon _{kl} ) \\ &{} -r* (\tau _{mn}*e_{mn}+\sigma _{mn}* \varepsilon _{mn} ) - (p* S+r*q_{m,m}-R )*\vartheta \biggr]\,dV \\ &{}+ \int _{\Sigma _{1}}r*t_{m}*{\tilde{v}}_{m}\,dA+ \int _{\Sigma _{1}^{c}}r* (t_{m}-{\tilde{t}}_{m} )*v_{m}\,dA \\ &{}+ \int _{\Sigma _{2}}r*m_{k}*{\tilde{\phi}}_{k}\,dA+ \int _{\Sigma _{2}^{c}}r* (m_{k}-{\tilde{m}}_{k} )*\phi _{k}\,dA \\ &{}+ \int _{\Sigma _{3}}r*q*{\tilde{\vartheta}}\,dA+ \int _{\Sigma _{3}^{c}}r* (q-{\tilde{q}} )*\vartheta \,dA, \quad t\in [0, \infty ), \end{aligned}$$
(40)
for any \({\mathcal{A}} = (v_{m}, \phi _{m}, \vartheta , e_{mn}, \varepsilon _{mn}, \tau _{mn}, \sigma _{mn}, q_{m}, S ) \in H \).
Theorem 5
Assume that the symmetry relations (5) are satisfied, \(a\ne 0\) in the domain D, and the thermoelastic state \({\mathcal{A}}\) is a solution of the mixed \({\mathcal{P}}\). Then the variation of the functional \({\mathcal{A}}\) is zero, i.e.,
$$\begin{aligned} \delta {\mathcal{F}}(t,{\mathcal{A}})=0,\quad t\in [0,\infty ). \end{aligned}$$
(41)
Proof
Let us consider two arbitrary thermoelastic states of the body
$$\begin{aligned}& {\mathcal{A}} = (v_{m}, \phi _{m}, \vartheta , e_{mn}, \varepsilon _{mn}, \tau _{mn}, \sigma _{mn}, q_{m}, S ), \\& {\bar {\mathcal{A}}} = ({\bar{v}}_{m}, {\bar{\phi}}_{m}, { \bar{\vartheta}}, {\bar{e}}_{mn}, {\bar{\varepsilon}}_{mn}, { \bar{\tau}}_{mn}, {\bar{\sigma}}_{mn}, {\bar{q}}_{m}, { \bar{S}} ). \end{aligned}$$
Since the space H is linear, we can conclude that \({\mathcal{A}}+\lambda \bar {{\mathcal{A}}}\in H\) for all λ, which is a real parameter.
It is no difficult to observe that
$$\begin{aligned} \delta _{\bar {{\mathcal{A}}}} {\mathcal{F}}(t,{\mathcal{A}}) =& \int _{D} \biggl\{ r* \bigl[ (C_{klmn}e_{mn}+B_{mnkl} \varepsilon _{mn} )* \bar {e}_{kl} \\ &{} + (B_{klmn} e_{mn}+A_{klmn} \varepsilon _{mn} )* \bar {\varepsilon}_{kl} \bigr] \\ &{}- r * \frac{T_{0}}{a} (a_{mn} * \bar {e}_{mn} + b_{mn} * \bar {\varepsilon}_{mn} ) (\varrho S - a_{mn}e_{mn} - b_{mn}\varepsilon _{mn} ) \\ &{}-r* (\tau _{mn}*\bar {e}_{mn}+\sigma _{mn}* \bar {\varepsilon}_{mn} ) -r* \biggl(q_{m}+ \frac{1}{T_{0}}K_{mn}\vartheta _{,n} \biggr)* \bar { \vartheta}_{,m} \\ &{} +p* \biggl[-T+\frac{T_{0}}{a}p* (\varrho S-a_{mn}e_{mn}-b_{mn} \varepsilon _{mn} ) \biggr]*\varrho \bar {S} \biggr\} \,dV \\ &{}+ \int _{D} \bigl\{ [ \varrho v_{m} - r *\tau _{mn,n} - F_{m} ] * \bar {v}_{m} \\ &{}+ ( I_{mn} \phi _{n} - p * \epsilon _{mjk}\tau _{jk} - G_{m} ) * \bar { \phi}_{m} \\ &{} - (p * \varrho S + r * q_{m,m} - R ) * \bar {\vartheta} \bigr\} \,dV \\ &{}- \int _{D} \biggl\{ h* \biggl[e_{mn}- \frac{1}{2} (u_{n,m}+u_{m,n} ) \biggr]*\bar { \tau}_{mn} \\ &{} + r* (\varepsilon _{mn} - u_{n,m} + \phi _{m} ) * \bar {\sigma}_{mn} + r * (\vartheta _{,m} - \vartheta _{,m} ) * \bar {q}_{m} \biggr\} \,dV \\ &{}+ \int _{\Sigma _{1}} r * (\bar {v}_{m}-v_{m} ) * \bar {t}_{m}\,dA + \int _{\Sigma _{1}^{c}} r * (t_{m} - {\bar{t}}_{m} ) * \bar {v}_{m}\,dA \\ &{}+ \int _{S_{2}}r* m_{k}*{\bar{\phi}}_{k}\,dA+ \int _{S_{2}^{c}}r* (m_{k}-{ \bar{m}}_{k} )*\phi _{k}\,dA \\ &{}+ \int _{S_{3}} r * ({\bar{\vartheta}} - \vartheta ) * \bar {q}\,dA + \int _{S_{3}^{c}} r * (q - {\bar{q}} ) * \bar {\vartheta}\,dA, \quad \forall t\in [0,\infty ). \end{aligned}$$
(42)
Taking into account the basic equations (13), relations (27), (33), and (34) and the boundary relations (12), from (42) we deduce that
$$\begin{aligned} \delta _{\bar {{\mathcal{A}}}} {\mathcal{F}}(t,{\mathcal{A}})=0 \quad \forall \bar {{\mathcal{A}}}\in H, \end{aligned}$$
(43)
whence it follows that
$$ \delta {\mathcal{F}}(t,{\mathcal{A}})=0, \quad t\ge 0, $$
which completes the proof of Theorem 5. □
Remark
It is not difficult to show that the statement of Theorem 5 is also valid reciprocally (see Gurtin [34]). In other words, if identity (41) is true, then the state \({\mathcal{A}}\) for which this identity holds is the unique solution of our problem. The idea of the proof, which is also suggested by Gurtin [34], is based on a particular choice of the thermoelastic state \(\bar {{\mathcal{A}}}\). In our case the thermoelastic state proposed by Lebon [21] can be successfully used.