## Fast Growth or Decay Estimates of Thermoelastic Equations in an External Domain

Li Yuanfei,, Shi Jincheng, Zhu Huishan, Huang Shiqi

 基金资助: 广东省普通高校重点项目（自然科学）.  2019KZDXM042广州华商学院科研团队项目.  2021HSKT01

 Fund supported: Key Projects of Colleges and Universities in Guangdong Province(Natural Science).  2019KZDXM042the Science Research Team Project in Guangzhou Huashang College.  2021HSKT01

Abstract

The Thermoelastic equation of type Ⅲ defined on the outer region of a three-dimensional sphere is considered. It is assumed that the solution of the equation satisfies certain boundary conditions at the boundary of the sphere. By setting an arbitrary parameter greater than zero in the energy expression after making certain constraints on the boundary conditions, the fast growth rate or decay rate of the solution with the distance from the origin is obtained by using energy analysis method and differential inequality technology. In the case of decay, the continuous dependence of solutions on coefficients is proved.

Keywords： Type Ⅲ thermoelastic equation ; Decay rate ; Growth rate ; Continuous dependence

Li Yuanfei, Shi Jincheng, Zhu Huishan, Huang Shiqi. Fast Growth or Decay Estimates of Thermoelastic Equations in an External Domain. Acta Mathematica Scientia[J], 2021, 41(4): 1042-1052 doi:

## 2 二择性定理

$$${{{\mathit{\boldsymbol{ u }}}}}_{tt}-\mu\Delta {{{\mathit{\boldsymbol{ u }}}}}-(\mu+\lambda)\nabla({\rm div} {{{\mathit{\boldsymbol{ u }}}}})+\beta\nabla\theta = 0, \ ({{{\mathit{\boldsymbol{ x }}}}}, t)\in \Omega(R_0) \times (0, T),$$$

$$$\theta_{tt}-\kappa\Delta\theta-\delta\Delta\theta_t+\beta {\rm div}{{{\mathit{\boldsymbol{ u }}}}}_{tt} = 0, \ ({{{\mathit{\boldsymbol{ x }}}}}, t)\in \Omega(R_0) \times (0, T),$$$

$$${{{\mathit{\boldsymbol{ u }}}}}(x_1, x_2, x_3, 0) = {{{\mathit{\boldsymbol{ u }}}}}_t(x_1, x_2, x_3, 0) = 0, \ {{{\mathit{\boldsymbol{ x }}}}}\in\Omega(R_0),$$$

$$$\theta(x_1, x_2, x_3, 0) = \theta_t(x_1, x_2, x_3, 0) = 0, \ {{{\mathit{\boldsymbol{ x }}}}}\in \Omega(R_0),$$$

$$${{{\mathit{\boldsymbol{ u }}}}}(x_1, x_2, x_3, t) = {{{\mathit{\boldsymbol{ g }}}}}(x_1, x_2, x_3, t), \ ({{{\mathit{\boldsymbol{ x }}}}}, t)\in B(R_0)\times (0, T),$$$

$$$\theta(x_1, x_2, x_3, t) = h(x_1, x_2, x_3, t), \ ({{{\mathit{\boldsymbol{ x }}}}}, t)\in B(R_0)\times (0, T),$$$

$\begin{eqnarray} && \frac{1}{2}{\rm e}^{-\tau t}\int_{\Omega(r)/\Omega(r_0)}[\textbf{w}_{t}^2+\mu|\nabla\textbf{w}|^2+(\mu+\lambda)|{\rm div}\textbf{w}|^2+\theta_{t}^2 +\kappa|\nabla\theta|^2]{\rm d}A{\rm d}\xi\\ &&+\frac{1}{2}\tau\int_0^t\int_{\Omega(r)/\Omega(r_0)}{\rm e}^{-\tau\eta}[\textbf{w}_{\eta}^2+\mu|\nabla\textbf{w}|^2 +(\mu+\lambda)|{\rm div}\textbf{w}|^2+\theta_{\eta}^2+\kappa|\nabla\theta|^2]{\rm d}A{\rm d}\xi {\rm d}\eta\\ && +\delta\int_0^t\int_{\Omega(r)/\Omega(r_0)}{\rm e}^{-\tau\eta}|\nabla\theta_\eta|^2{\rm d}A{\rm d}\xi {\rm d}\eta\geq {\cal F}(r_0, t)[{\rm e}^{M\tau(r-r_0)}-1], \end{eqnarray}$

$\begin{eqnarray} && \frac{1}{2}{\rm e}^{-\tau t}\int_r^\infty\int_{B(\xi)}[\textbf{w}_{t}^2+\mu|\nabla\textbf{w}|^2+(\mu+\lambda)|{\rm div}\textbf{w}|^2+\theta_{t}^2 +\kappa|\nabla\theta|^2]{\rm d}A{\rm d}\xi\\ &&+\frac{1}{2}\tau\int_0^t\int_r^\infty\int_{B(\xi)}{\rm e}^{-\tau\eta}[\textbf{w}_{\eta}^2+\mu|\nabla\textbf{w}|^2 +(\mu+\lambda)|{\rm div}\textbf{w}|^2+\theta_{\eta}^2+\kappa|\nabla\theta|^2]{\rm d}A{\rm d}\xi {\rm d}\eta\\ && +\delta\int_0^t\int_r^\infty\int_{B(\xi)}{\rm e}^{-\tau\eta}|\nabla\theta_\eta|^2{\rm d}A{\rm d}\xi {\rm d}\eta\leq[-{\cal F}(R_0, t)]{\rm e}^{-M\tau(r-R_0)}. \end{eqnarray}$

利用Hölder不等式和Young不等式, 可得

$\begin{eqnarray} |I_1|&\leq&\mu\Big(\int_0^t\int_{B(r)}{\rm e}^{-\tau\eta}|\nabla{{{\mathit{\boldsymbol{ w }}}}}|^2{\rm d}A{\rm d}\eta \int_0^t\int_{B(r)}{\rm e}^{-\tau\eta}{{{\mathit{\boldsymbol{ w }}}}}_{\eta}^2{\rm d}A{\rm d}\eta\Big)^\frac{1}{2}\\ &\leq&\frac{\sqrt{\mu}}{\tau}\Big(\int_0^t\int_{B(r)}\mu\tau {\rm e}^{-\tau\eta}|\nabla{{{\mathit{\boldsymbol{ w }}}}}|^2{\rm d}A{\rm d}\eta \int_0^t\int_{B(r)}\tau {\rm e}^{-\tau\eta}{{{\mathit{\boldsymbol{ w }}}}}_{\eta}^2{\rm d}A{\rm d}\eta\Big)^\frac{1}{2}\\ &\leq&\frac{\sqrt{\mu}}{2\tau}\int_0^t\int_{B(r)}\tau {\rm e}^{-\tau\eta}(\mu|\nabla{{{\mathit{\boldsymbol{ w }}}}}|^2 +{{{\mathit{\boldsymbol{ w }}}}}_{\eta}^2){\rm d}A{\rm d}\eta\\ &\leq&\frac{\sqrt{\mu}}{\tau}[\frac{\partial}{\partial r}{\cal F}(r, t)]. \end{eqnarray}$

$\begin{eqnarray} |I_2|&\leq&(\mu+\lambda)\Big(\int_0^t\int_{B(r)}{\rm e}^{-\tau\eta}({\rm div}{{{\mathit{\boldsymbol{ w }}}}})^2{\rm d}A{\rm d}\eta \int_0^t\int_{B(r)}{\rm e}^{-\tau\eta}|{{{\mathit{\boldsymbol{ w }}}}}_\eta|^2{\rm d}A{\rm d}\eta\Big)^\frac{1}{2} \\ &\leq&\frac{\sqrt{\mu+\lambda}}{\tau}[\frac{\partial}{\partial r}{\cal F}(r, t)], \end{eqnarray}$

$\begin{eqnarray} |I_3|&\leq&\beta\Big(\int_0^t\int_{B(r)}{\rm e}^{-\tau\eta}\theta_\eta^2{\rm d}A{\rm d}\eta \int_0^t\int_{B(r)}{\rm e}^{-\tau\eta}|{{{\mathit{\boldsymbol{ w }}}}}_{\eta}|^2{\rm d}A{\rm d}\eta\Big)^\frac{1}{2} \\ &\leq&\frac{\beta}{\tau}[\frac{\partial}{\partial r}{\cal F}(r, t)], \end{eqnarray}$

$\begin{eqnarray} |I_4|&\leq&\kappa\Big(\int_0^t\int_{B(r)}{\rm e}^{-\tau\eta}\theta_\eta^2{\rm d}A{\rm d}\eta \int_0^t\int_{B(r)}{\rm e}^{-\tau\eta}|\nabla\theta|^2{\rm d}A{\rm d}\eta\Big)^\frac{1}{2} \\ &\leq&\frac{\sqrt{\kappa}}{\tau}[\frac{\partial}{\partial r}{\cal F}(r, t)], \end{eqnarray}$

$\begin{eqnarray} |I_5|&\leq&\delta\Big(\int_0^t\int_{B(r)}{\rm e}^{-\tau\eta}\theta_\eta^2{\rm d}A{\rm d}\eta \int_0^t\int_{B(r)}{\rm e}^{-\tau\eta}|\nabla\theta_\eta|^2{\rm d}A{\rm d}\eta\Big)^\frac{1}{2}\\ &\leq&\frac{\sqrt{\delta}}{\tau}[\frac{\partial}{\partial r}{\cal F}(r, t)]. \end{eqnarray}$

$\begin{eqnarray} |{\cal F}(r, t)|\leq\frac{1}{M\tau}[\frac{\partial}{\partial r}{\cal F}(r, t)], \ r\geq R_0, \end{eqnarray}$

$$$\frac{\partial}{\partial r}\Big\{{\cal F}(r, t){\rm e}^{-M\tau r}\Big\}\geq0, \ r\geq r_0.$$$

$$$\frac{\partial}{\partial r}\Big\{{\cal F}(r, t){\rm e}^{M\tau r}\Big\}\geq0, \ r\geq R_0.$$$

$$$\nabla\cdot{{{\mathit{\boldsymbol{ w }}}}}_{tt}-\mu\Delta \nabla\cdot{{{\mathit{\boldsymbol{ w }}}}}-(\mu+\lambda)\Delta({\rm div} {{{\mathit{\boldsymbol{ w }}}}})+\beta\Delta\theta_t = 0,$$$

$$$\nabla\theta_{tt}-\kappa\Delta\nabla\theta-\delta\Delta\nabla\theta_t+\beta \nabla{\rm div}{{{\mathit{\boldsymbol{ w }}}}}_{t} = 0.$$$

$$$\int_0^t\int_r^\infty\int_{B(\xi)}{\rm e}^{-\tau\eta}[\nabla\theta_{\eta\eta}-\kappa\Delta\nabla\theta-\delta\Delta\nabla\theta_\eta+\beta \nabla({\rm div}{{{\mathit{\boldsymbol{ w }}}}}_{\eta})]\nabla\theta_{\eta}{\rm d}A{\rm d}\xi {\rm d}\eta = 0,$$$

$$$\int_0^t\int_r^\infty\int_{B(\xi)}{\rm e}^{-\tau\eta}[\nabla\cdot{{{\mathit{\boldsymbol{ w }}}}}_{tt}-\mu\Delta \nabla\cdot{{{\mathit{\boldsymbol{ w }}}}}-(\mu+\lambda)\Delta({\rm div} {{{\mathit{\boldsymbol{ w }}}}})+\beta\Delta\theta_t]\nabla\cdot{{{\mathit{\boldsymbol{ w }}}}}_\eta {\rm d}A{\rm d}\xi {\rm d}\eta = 0.$$$

$\begin{eqnarray} \delta\int_0^t\int_r^\infty\int_{B(\xi)}{\rm e}^{-\tau\eta}|\Delta\theta_\eta|^2{\rm d}A{\rm d}\xi {\rm d}\eta\leq[-\widetilde{{\cal F}}(R_0, t)]{\rm e}^{-M\tau(r-R_0)}, \end{eqnarray}$

${{\rm{\mathit{\boldsymbol{w }}}}}, \theta $${{\rm{\mathit{\boldsymbol{ w }}}}}^*, \theta^* 是方程(2.7)–(2.12)分别对应于 \delta = \delta_1$$ \delta = \delta_2$的解. 令

$$${{{\mathit{\boldsymbol{ v }}}}}_{tt}-\mu\Delta {{{\mathit{\boldsymbol{ v }}}}}-(\mu+\lambda)\nabla({\rm div} {{{\mathit{\boldsymbol{ v }}}}})+\beta\nabla\Sigma_\eta = 0, \ ({{{\mathit{\boldsymbol{ x }}}}}, t)\in\Omega(R_0)\times (0, T),$$$

$$$\Sigma_{tt}-\kappa\Delta\Sigma-\widetilde{\delta}\Delta\theta_t-\delta_2\Delta\Sigma_t+\beta {\rm div}{{{\mathit{\boldsymbol{ v }}}}}_{t} = 0, \ ({{{\mathit{\boldsymbol{ x }}}}}, t)\in\Omega(R_0)\times (0, T),$$$

$$${{{\mathit{\boldsymbol{ v }}}}}(x_1, x_2, x_3, 0) = {{{\mathit{\boldsymbol{ v }}}}}_t(x_1, x_2, x_3, 0) = 0, \ {{{\mathit{\boldsymbol{ x }}}}}\in \Omega(R_0),$$$

$$$\Sigma(x_1, x_2, x_3, 0) = \Sigma_t(x_1, x_2, x_3, 0) = 0, \ {{{\mathit{\boldsymbol{ x }}}}}\in \Omega(R_0),$$$

$$${{{\mathit{\boldsymbol{ v }}}}}(x_1, x_2, x_3, t) = \textbf{0}, \ \Sigma(x_1, x_2, x_3, t) = 0, \ ({{{\mathit{\boldsymbol{ x }}}}}, t)\in B(R_0)\times (0, T).$$$

$\begin{eqnarray} &&\frac{1}{2}{\rm e}^{-\tau t}\int_{r}^\infty\int_{B(\xi)}[\textbf{v}_{t}^2+\mu|\nabla\textbf{v}|^2+(\mu+\lambda)|{\rm div}\textbf{v}|^2+\Sigma_{t}^2 +\kappa|\nabla\Sigma|^2]{\rm d}A{\rm d}\xi\\ &&+\frac{1}{2}\tau\int_0^t\int_{r}^\infty\int_{B(\xi)}{\rm e}^{-\tau\eta}[\textbf{v}_{\eta}^2+\mu|\nabla\textbf{v}|^2 +(\mu+\lambda)|{\rm div}\textbf{v}|^2+\Sigma_{\eta}^2+\kappa|\nabla\Sigma|^2]{\rm d}A{\rm d}\xi {\rm d}\eta\\ & & +\delta_2\int_0^t\int_{r}^\infty\int_{B(\xi)}{\rm e}^{-\tau\eta}|\nabla\Sigma_\eta|^2{\rm d}A{\rm d}\xi {\rm d}\eta\\ &\leq &\frac{2\widetilde{\delta}^2}{\tau\delta_1}[-\widetilde{{\cal F}}(R_0, t)]{\rm e}^{-\frac{M\tau(r-R_0)}{2}}[2-{\rm e}^{-\frac{M\tau(r-R_0)}{2}}]. \end{eqnarray}$

在方程(3.6)上乘以${{\rm{\mathit{\boldsymbol{v }}}}}_t$, 方程(3.7)上乘以$\Sigma_\eta$并在$\Omega(r)\times (0, t)$积分, 可得

$$$\int_0^t\int_r^\infty\int_{B(\xi)}{\rm e}^{-\tau\eta}[{{{\mathit{\boldsymbol{ v }}}}}_{\eta\eta}-\mu\Delta {{{\mathit{\boldsymbol{ v }}}}}-(\mu+\lambda)\nabla({\rm div} {{{\mathit{\boldsymbol{ v }}}}})+\beta\nabla\Sigma_\eta ]{{{\mathit{\boldsymbol{ v }}}}}_\eta {\rm d}A{\rm d}\xi {\rm d}\eta = 0,$$$

$$$\int_0^t\int_r^\infty\int_{B(\xi)}{\rm e}^{-\tau\eta}[\Sigma_{\eta\eta}-\kappa\Delta\Sigma-\widetilde{\delta}\Delta\theta_\eta-\delta_2\Delta\Sigma_\eta+\beta {\rm div}{{{\mathit{\boldsymbol{ v }}}}}_{\eta}]\Sigma_\eta {\rm d}A{\rm d}\xi {\rm d}\eta = 0.$$$

$\begin{eqnarray} E(r, t)& = &\frac{1}{2}{\rm e}^{-\tau t}\int_{r}^\infty\int_{B(\xi)}[{{{\mathit{\boldsymbol{ v }}}}}_{t}^2+\mu|\nabla{{{\mathit{\boldsymbol{ v }}}}}|^2+(\mu+\lambda)|{\rm div}{{{\mathit{\boldsymbol{ v }}}}}|^2+\Sigma_{t}^2 +\kappa|\nabla\Sigma|^2]{\rm d}A{\rm d}\xi\\ &&+\frac{1}{2}\tau\int_0^t\int_{r}^\infty\int_{B(\xi)}{\rm e}^{-\tau\eta}[{{{\mathit{\boldsymbol{ v }}}}}_{\eta}^2+\mu|\nabla{{{\mathit{\boldsymbol{ v }}}}}|^2 +(\mu+\lambda)|{\rm div}{{{\mathit{\boldsymbol{ v }}}}}|^2+\Sigma_{\eta}^2+\kappa|\nabla\Sigma|^2]{\rm d}A{\rm d}\xi {\rm d}\eta\\ && +\delta_2\int_0^t\int_{r}^\infty\int_{B(\xi)}{\rm e}^{-\tau\eta}|\nabla\Sigma_\eta|^2{\rm d}A{\rm d}\xi {\rm d}\eta, \end{eqnarray}$

$\begin{eqnarray} -\frac{\partial}{\partial r}E(r, t)& = &\frac{1}{2}{\rm e}^{-\tau t}\int_{B(r)}[{{{\mathit{\boldsymbol{ v }}}}}_{t}^2+\mu|\nabla{{{\mathit{\boldsymbol{ v }}}}}|^2+(\mu+\lambda)|{\rm div}{{{\mathit{\boldsymbol{ v }}}}}|^2+\Sigma_{t}^2 +\kappa|\nabla\Sigma|^2]{\rm d}A\\ &&+\frac{1}{2}\tau\int_0^t\int_{B(r)}{\rm e}^{-\tau\eta}[{{{\mathit{\boldsymbol{ v }}}}}_{\eta}^2+\mu|\nabla{{{\mathit{\boldsymbol{ v }}}}}|^2 +(\mu+\lambda)|{\rm div}{{{\mathit{\boldsymbol{ v }}}}}|^2+\Sigma_{\eta}^2+\kappa|\nabla\Sigma|^2]{\rm d}A {\rm d}\eta\\ && +\delta_2\int_0^t\int_{B(r)}{\rm e}^{-\tau\eta}|\nabla\Sigma_\eta|^2{\rm d}A{\rm d}\eta. \end{eqnarray}$

$\begin{eqnarray} E(r, t)& = &\widetilde{\delta}\int_0^t\int_{r}^\infty\int_{B(\xi)}{\rm e}^{-\tau\eta}\Delta\theta_\eta\Sigma_\eta {\rm d}A{\rm d}\xi {\rm d}\eta\\ &&-\int_0^t\int_{B(r)}{\rm e}^{-\omega\eta}\Big[\mu(\frac{{{{\mathit{\boldsymbol{ x }}}}}}{r}\cdot\nabla){{{\mathit{\boldsymbol{ v }}}}}{{{\mathit{\boldsymbol{ v }}}}}_{\eta} +(\mu+\lambda)({\rm div}{{{\mathit{\boldsymbol{ v }}}}}){{{\mathit{\boldsymbol{ v }}}}}_{\eta}\cdot\frac{{{{\mathit{\boldsymbol{ x }}}}}}{r} -\beta\Sigma_\eta{{{\mathit{\boldsymbol{ v }}}}}_{\eta}\cdot\frac{{{{\mathit{\boldsymbol{ x }}}}}}{r}\Big]{\rm d}A{\rm d}\eta\\ &&-\int_0^t\int_{B(r)}{\rm e}^{-\omega\eta}\Big[\kappa\nabla\Sigma\Sigma_\eta\cdot\frac{{{{\mathit{\boldsymbol{ x }}}}}}{r}+\delta_2\nabla\Sigma\Sigma_\eta\cdot\frac{{{{\mathit{\boldsymbol{ x }}}}}}{r}\Big]{\rm d}A{\rm d}\eta\\ &\doteq& J_1+J_2+J_3. \end{eqnarray}$

$\begin{eqnarray} J_1 &\leq&|\widetilde{\delta}|\Big[\int_0^t\int_{r}^\infty\int_{B(\xi)}{\rm e}^{-\tau\eta}|\Delta\theta_\eta|^2{\rm d}A{\rm d}\xi {\rm d}\eta \int_0^t\int_{r}^\infty\int_{B(\xi)}{\rm e}^{-\tau\eta}\Sigma^2_\eta {\rm d}A{\rm d}\xi {\rm d}\eta\Big]^\frac{1}{2}\\ &\leq&\frac{\widetilde{\delta}^2}{\tau\delta_1}[-\widetilde{{\cal F}}(R_0, t)]{\rm e}^{-M\tau(r-R_0)} +\frac{1}{4}\tau\int_0^t\int_{r}^\infty\int_{B(\xi)}{\rm e}^{-\tau\eta}\Sigma^2_\eta {\rm d}A{\rm d}\xi {\rm d}\eta. \end{eqnarray}$

$$$J_2\leq\frac{1}{\tau}\max\{\sqrt{\mu+\lambda}, \beta\}[-\frac{\partial}{\partial r}E(r, t)],$$$

$$$J_3\leq\frac{1}{\tau}\max\{\sqrt{\kappa}, \sqrt{\delta_2}\}[-\frac{\partial}{\partial r}E(r, t)].$$$

$\begin{eqnarray} E(r, t)\leq\frac{2\widetilde{\delta}^2}{\tau\delta_1}[-\widetilde{{\cal F}}(R_0, t)]{\rm e}^{-M\tau(r-R_0)}+\frac{2}{M\tau}[-\frac{\partial}{\partial r}E(r, t)]. \end{eqnarray}$

$\begin{eqnarray} \frac{\partial}{\partial r}\Big\{E(r, t){\rm e}^{\frac{M\tau r}{2}}\Big\}\leq\frac{\widetilde{\delta}^2}{M\delta_1}[-\widetilde{{\cal F}}(R_0, t)]{\rm e}^{-\frac{M\tau}{2}r+M\tau R_0}. \end{eqnarray}$

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