Local smooth balances with continuous body sources

Fix an admissible $V$ and let $K=\bar{V}$. By [F4], $K$ is compact Jordan and its open-region integrals equal the corresponding closure integrals. Apply [F3] componentwise to $f=\rho u$; the derivative of its material integral is ${\int }_{P_t}({\mathrm{\partial }}_t(\rho u)+{\mathrm{div}}_x(\rho u\otimes u))dx\mathrm{.}$ Pointwise mass continuity and the product rule give ${\mathrm{\partial }}_t(\rho u)+{\mathrm{div}}_x(\rho u\otimes u)=\rho D_{t}u\mathrm{.}$

Substitute the first balance in [A1] into step 1.1 and apply [F5] to $\sigma$, using countable choice from [F6]. This yields ${\int }_{P_t}(\rho D_{t}u-\rho b-{\mathrm{div}}_x\sigma )dx=0.$ The integrand is continuous because $b$ is continuous and the other fields are $C^1$. Every sufficiently small open ball $Q⋐{\mathrm{\Omega }}_t$ is such a material region: its inverse image under ${\chi }_t$ is a bounded $C^1$ region compactly contained in $B$. If a component of the continuous integrand were nonzero at a point, it would have a strict fixed sign on a smaller ball and its integral there could not vanish. Thus $\rho D_{t}u=\rho b+{\mathrm{div}}_x\sigma$ pointwise on ${\mathcal{O}}_B$.

For each component $i$, ${\mathrm{\partial }}_t(\rho u_i)+{\sum }_j{\mathrm{\partial }}_{x_j}(\rho u_i{u}_j)=\rho ({\mathrm{\partial }}_t{u}_i+u\cdot {\mathrm{\nabla }}_x{u}_i)+u_i({\mathrm{\partial }}_t\rho +{\mathrm{div}}_x(\rho u))\mathrm{.}$ The last term is zero by [F2]. Substituting step 2.1 and moving $\mathrm{div}\sigma$ to the flux gives the conservative momentum equation.

Apply [F3] to $\rho (x\times u)$ and use $D_{t}x=u$ and $u\times u=0$. The transport derivative is ${\int }_{P_t}x\times (\rho D_{t}u)dx$. Substitute step 2.1 and compare with the angular balance in [A1]; applying [F5] to the moment-stress flux gives ${\int }_{P_t}({\mathrm{div}}_x(x\times \sigma )-x\times {\mathrm{div}}_x\sigma )dx=0.$ This residual is continuous, so the fixed-sign-ball argument of step 2.1 makes it zero pointwise. With ${ϵ}_{ikl}$ the alternating symbol and $M_{ij}={\sum }_{k,l}{ϵ}_{ikl}{x}_k{\sigma }_{lj}$, direct differentiation gives $({\mathrm{div}}_x(x\times \sigma )-x\times {\mathrm{div}}_x\sigma {)}_i={\sum }_{j,l}{ϵ}_{ijl}{\sigma }_{lj}\mathrm{.}$ For $i=1,2,3$ these equations are respectively ${\sigma }_{32}={\sigma }_{23}$, ${\sigma }_{13}={\sigma }_{31}$, and ${\sigma }_{21}={\sigma }_{12}$, hence $\sigma ={\sigma }^{\mathsf{T}}$.

Apply [F3] to $\rho E$ in the total-energy balance [A1]. Rewrite $(\sigma n)\cdot u=({\sigma }^{\mathsf{T}}u)\cdot n$ and apply [F5] to ${\sigma }^{\mathsf{T}}u$ and $q$. The integral of the continuous residual ${\mathrm{\partial }}_t(\rho E)+{\mathrm{div}}_x(\rho Eu-{\sigma }^{\mathsf{T}}u+q)-\rho b\cdot u-\rho r$ is zero on every $P_t$. The fixed-sign-ball argument from step 2.1 proves that this residual vanishes pointwise. Conversely, integrating this local equation and using [F3] and [F5] recovers the material total-energy balance.

If $\sigma$ is symmetric, the component identity in step 3.2 gives ${\mathrm{div}}_x(x\times \sigma )=x\times {\mathrm{div}}_x\sigma$. The product rule and $u\times u=0$ give ${\mathrm{\partial }}_t(\rho x\times u)+{\mathrm{div}}_x(\rho (x\times u)\otimes u)=x\times ({\mathrm{\partial }}_t(\rho u)+{\mathrm{div}}_x(\rho u\otimes u))\mathrm{.}$ Use step 3.1 and the stress-divergence identity, then integrate over $P_t$. The transport formula [F3] and divergence theorem [F5] recover the angular-momentum integral balance in [A1].

Dot $\rho D_{t}u=\rho b+\mathrm{div}\sigma$ from step 2.1 with $u$. The component product rule gives $u\cdot \mathrm{div}\sigma =\mathrm{div}({\sigma }^{\mathsf{T}}u)-\sigma :D_{x}u$. Since $\sigma$ is symmetric, its contraction with the skew part of $D_{x}u$ is zero, so $\sigma :D_{x}u=\sigma :D$. Also $D_{t}k=u\cdot D_{t}u$. Combining these identities with [F2] yields ${\mathrm{\partial }}_t(\rho k)+{\mathrm{div}}_x(\rho ku-{\sigma }^{\mathsf{T}}u)=\rho b\cdot u-\sigma :D\mathrm{.}$

Subtract the kinetic equation in step 4.2 from the total-energy equation in step 3.3 and use $E=e+k$. The body-work terms cancel and leave ${\mathrm{\partial }}_t(\rho e)+{\mathrm{div}}_x(\rho eu+q)=\sigma :D+\rho r$. Expanding the mass-transport term using [F2] gives $\rho D_{t}e=\sigma :D-{\mathrm{div}}_{x}q+\rho r$. Conversely, adding this internal-energy equation to the kinetic equation in step 4.2 recovers the total-energy equation in step 3.3.