Volume dilation and bulk production

For each $t$, ${\chi }_t(a)=\alpha (t)a$ maps $U$ bijectively onto the open set $\alpha (t)U$ with inverse $x\mapsto x\mathrm{/}\alpha (t)$; jointly this inverse is $C^2$ because $\alpha >0$. Differentiating $\chi$ in $a$ and $t$ gives $F=\alpha I_3$, $J={\alpha }^3>0$, and $u(t,x)=({\alpha }^{\mathrm{\prime }}\mathrm{/}\alpha )x$. Using the velocity-gradient and strain-rate definitions [F6], $L=({\alpha }^{\mathrm{\prime }}\mathrm{/}\alpha )I_3$, $D=({\alpha }^{\mathrm{\prime }}\mathrm{/}\alpha )I_3$, $\theta =3{\alpha }^{\mathrm{\prime }}\mathrm{/}\alpha$, and $D_0=0$.

Fix a bounded $C^1$ material subregion $V⋐B$ and put $K=\bar{V}$, ${\mathrm{\Omega }}_t={\chi }_t(V)$. Apply [F3] componentwise to the linear-momentum integral in [F2]; [F7] identifies the open-region integrals with their compact-closure versions. The divergence theorem [F4], under the stated choice premise [F5], converts ${\int }_{\mathrm{\partial }{\mathrm{\Omega }}_t}\sigma n$ to ${\int }_{{\mathrm{\Omega }}_t}\mathrm{div}\sigma$. The continuous residual is integrable on the compact Jordan region by [F10]: $R_m={\mathrm{\partial }}_t(\rho u)+\mathrm{div}(\rho u\otimes u)-\rho b-\mathrm{div}\sigma$ has integral zero on ${\chi }_t(K)$. Change variables by [F9], use [F7] to return to $V$, and apply [F8] to the continuous reference residual $R_m(t,\chi (t,a))J(t,a)$. Since $J>0$, $R_m=0$ pointwise. Expanding with the conventions [F6] and the continuity equation [F1] yields $\rho D_{t}u=\rho b+\mathrm{div}\sigma$.

Set $E=e+\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}2$. Apply [F3] to $\rho E$ in the total-energy balance [F2]. The surface-work integrand is $(\sigma n)\cdot u=({\sigma }^{T}u)\cdot n$; apply [F4] with [F5] to ${\sigma }^{T}u$ and $q$. The continuous total-energy residual is integrable by [F10] and has zero integral on every ${\mathrm{\Omega }}_t={\chi }_t(V)$. By [F7], [F9], and the zero-integral conclusion of [F8], it vanishes pointwise: ${\mathrm{\partial }}_t(\rho E)+\mathrm{div}(\rho Eu)=\rho b\cdot u+\mathrm{div}({\sigma }^{T}u)-\mathrm{div}q+\rho r\mathrm{.}$

Apply [F3] to $\rho s$ in the entropy inequality [F2]. By the continuity equation [F1], ${\mathrm{\partial }}_t(\rho s)+\mathrm{div}(\rho su)=\rho D_{t}s+s({\mathrm{\partial }}_t\rho +\mathrm{div}(\rho u))=\rho D_{t}s\mathrm{.}$ The surface term has the local-temperature flux $q\mathrm{/}T$; [F4] with [F5] converts it to the volume divergence. Convert each open-region integral to the compact-Jordan and then reference-domain form using [F7] and [F9]. The residual is continuous and integrable by [F10]; its nonnegative integral on every bounded $C^1$ $V⋐B$ and [F8] make it pointwise nonnegative. Since $J>0$, this gives $g=\rho D_{t}s+\mathrm{div}(q\mathrm{/}T)-\rho r\mathrm{/}T\ge 0$.

Apply [F1] to this motion. Since $J={\alpha }^3$, the density reconstruction is $\rho (t,\chi (t,a))={\rho }_0(a)\mathrm{/}{\alpha }^3$; the density remains positive because ${\rho }_0>0$ and $\alpha >0$.

Dot the local momentum equation in step 1.2 with $u$. The identity $u\cdot \mathrm{div}\sigma =\mathrm{div}({\sigma }^{T}u)-\sigma :\mathrm{\nabla }u$, the symmetry of the adopted stress [A2], and the continuity equation from [F1] give the kinetic-energy balance ${\mathrm{\partial }}_t(\rho \mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}2)+\mathrm{div}(\rho \mathrm{\mid }u{\mathrm{\mid }}^{2}u\mathrm{/}2-{\sigma }^{T}u)=\rho b\cdot u-\sigma :D$, where $D$ is the symmetric velocity gradient from [F6]. Subtract this balance from step 1.3; the $\rho b\cdot u$ and $\mathrm{div}({\sigma }^{T}u)$ terms cancel, leaving $\rho D_{t}e=\sigma :D-\mathrm{div}q+\rho r$.

From [A1], $D_{t}e=T{D}_{t}s-p{D}_t(1\mathrm{/}\rho )$. The continuity equation in [F1] gives $D_t(1\mathrm{/}\rho )=\theta \mathrm{/}\rho$, using $\theta =\mathrm{div}u$ from [F6], so $\rho T{D}_{t}s=\rho D_{t}e+p\theta$. Substitute step 2.2 and $\sigma =-p{I}_3+\tau$ from [A2]; since $I_3:D=\theta$, the pressure terms cancel and $\rho T{D}_{t}s=\tau :D-\mathrm{div}q+\rho r$.

Substitute this identity into step 1.4 and use $\mathrm{div}(q\mathrm{/}T)=\mathrm{div}q\mathrm{/}T-q\cdot \mathrm{\nabla }T\mathrm{/}{T}^2$. The definitions [F6] give $D_0:I_3=0$, $D_0:D=\mathrm{\mid }{D}_0{\mathrm{\mid }}^2$, and $\tau :D=2\mu \mathrm{\mid }{D}_0{\mathrm{\mid }}^2+\zeta {\theta }^2$; also $q=-{\kappa }_T\mathrm{\nabla }T$. Therefore $g=2\mu \mathrm{\mid }{D}_0{\mathrm{\mid }}^2\mathrm{/}T+\zeta {\theta }^2\mathrm{/}T+{\kappa }_T\mathrm{\mid }\mathrm{\nabla }T{\mathrm{\mid }}^2\mathrm{/}{T}^2$. Step 1.1 has $D_0=0$, giving the stated bulk-production formula and its uniform-temperature special case.