Momentum, angular, kinetic/internal and total balance algebra

Fix $V$ and $t_0\in I$. Its compact boundary is covered by finitely many relatively compact $C^1$ graph charts. On smaller compact chart pieces the graph functions have bounded derivative. Cover each two-dimensional base by squares of side $\delta$; the graph over each square has Euclidean diameter at most $C\delta$, hence lies in an axis-parallel cube of side $C^{\mathrm{\prime }}\delta$. There are $O({\delta }^{-2})$ such squares, so the total cube volume is $O(\delta )$ and tends to zero. Thus for every positive bound the boundary has a finite axis-parallel box cover of smaller total volume, so $K=\bar{V}$ is compact Jordan in the sense of [L1]. The map ${\chi }_{t_0}$ is a diffeomorphism on a neighborhood of $K$, so ${\chi }_{t_0}(K)=\overline{P_{t_0}}$ and $\mathrm{\partial }{P}_{t_0}={\chi }_{t_0}(\mathrm{\partial }V)$. Its derivative is bounded on finitely many compact chart neighborhoods of $\mathrm{\partial }V$, so it is Lipschitz there. The image of each graph-covering square has diameter at most a constant times $\delta$, so it lies in an axis-parallel cube with volume $O({\delta }^3)$; summing over $O({\delta }^{-2})$ squares gives arbitrarily small finite cube covers of $\mathrm{\partial }{P}_{t_0}$. Thus ${\chi }_{t_0}(K)$ is also compact Jordan. In particular, the volume integral over $P_{t_0}$ equals the corresponding integral over ${\chi }_{t_0}(K)$ for every continuous integrand because their difference is on the volume-null boundary. The fields are $C^1$ on neighborhoods of the swept closures. Pulling a common such neighborhood back by the continuous map $(t,a)\mapsto (t,\chi (t,a))$ gives an open set containing $\{t_0\}\times K$; compactness supplies an interval around $t_0$ and an open reference neighborhood $U_0$ of $K$ on whose product the fields are defined. Restrict the motion to that neighborhood and interval, with reference body $V$ and reference domain $U_0$, translate time so that $t_0$ is zero, and apply [L1] to $K$ and each component of $f=\rho u$. The transport formula gives $\frac{d}{dt}{\int }_{P_t}\rho udx={\int }_{P_t}({\mathrm{\partial }}_t(\rho u)+\mathrm{div}(\rho u\otimes u))dx$ at $t_0$. Componentwise product differentiation and [A1] give ${\mathrm{\partial }}_t(\rho u)+\mathrm{div}(\rho u\otimes u)=\rho D_{t}u$. The linear-momentum balance [A2], traction representation [F1], and divergence theorem [F2] now give ${\int }_{P_{t_0}}(\rho D_{t}u-\rho b-\mathrm{div}\sigma )dx=0$. Every sufficiently small open spatial ball $Q$ with compact closure in ${\mathrm{\Omega }}_{t_0}$ is a material image $P_{t_0}$: its inverse image under ${\chi }_{t_0}$ is a bounded $C^1$ domain compactly contained in $B$. If a component of this continuous residual were positive or negative at an interior point, it would retain that strict sign on a small ball and have a nonzero integral, a contradiction. Hence $\rho D_{t}u=\rho b+\mathrm{div}\sigma$ pointwise.

For each component $i$, expand the left side using the product rule: ${\mathrm{\partial }}_t(\rho u_i)+{\sum }_j{\mathrm{\partial }}_{x_j}(\rho u_i{u}_j)=\rho ({\mathrm{\partial }}_t{u}_i+u\cdot {\mathrm{\nabla }}_x{u}_i)+u_i({\mathrm{\partial }}_t\rho +\mathrm{div}(\rho u))\mathrm{.}$ The last term vanishes by [A1], and the remaining term is $\rho D_t{u}_i$. Substitution of step 1.1 gives the displayed conservative momentum equation. Conversely, if that equation holds, the same expansion and [A1] give $\rho D_{t}u=\rho b+\mathrm{div}\sigma$. Integrate this identity over $P_t$, use [F2] rowwise on $\sigma$, and use [L1] componentwise on $\rho u$; the transport derivative turns its left side into $\frac{d}{dt}{\int }_{P_t}\rho udx$, recovering the linear-momentum integral balance in [A2].

Apply [L1] componentwise to the angular-momentum density $\rho (x\times u)$. Since $D_{t}x=u$ and $u\times u=0$, $D_t(x\times u)=x\times D_{t}u$, so step 1.1 makes its time derivative ${\int }_{P_t}\rho x\times D_{t}u$. By [A2] the same derivative is ${\int }_{P_t}x\times (\rho b)+{\int }_{\mathrm{\partial }{P}_t}x\times (\sigma \nu )$. The flux matrix for $x\times (\sigma \nu )$ has entries $M_{ij}={\sum }_{k,l}{ϵ}_{ikl}{x}_k{\sigma }_{lj}$; applying [F2] rowwise changes the boundary integral to ${\int }_{P_t}\mathrm{div}(x\times \sigma )$. Substituting $\rho D_{t}u=\rho b+\mathrm{div}\sigma$ from step 1.1 and cancelling the body-force terms gives zero integral of $\mathrm{div}(x\times \sigma )-x\times \mathrm{div}\sigma$ over every material subdomain. The same fixed-sign-ball argument as in step 1.1 makes this continuous residual zero pointwise. Componentwise differentiation gives ${\sum }_j{\mathrm{\partial }}_{x_j}{M}_{ij}={\sum }_{j,l}{ϵ}_{ijl}{\sigma }_{lj}+(x\times \mathrm{div}\sigma {)}_i$. Thus ${\sum }_{j,l}{ϵ}_{ijl}{\sigma }_{lj}=0$ for every $i$. For $i=1,2,3$ these equations are respectively ${\sigma }_{32}={\sigma }_{23}$, ${\sigma }_{13}={\sigma }_{31}$, and ${\sigma }_{21}={\sigma }_{12}$, so $\sigma ={\sigma }^T$.

Dot $\rho D_{t}u=\rho b+\mathrm{div}\sigma$ from step 1.1 with $u$. The component product rule gives $u\cdot \mathrm{div}\sigma =\mathrm{div}({\sigma }^{T}u)-\sigma :L$, and differentiation of $k=\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}2$ gives $D_{t}k=u\cdot D_{t}u$. By the product rule and continuity equation [A1], ${\mathrm{\partial }}_t(\rho k)+\mathrm{div}(\rho ku)=\rho D_{t}k$. Substitution yields ${\mathrm{\partial }}_t(\rho k)+\mathrm{div}(\rho ku-{\sigma }^{T}u)=\rho b\cdot u-\sigma :L$.

Apply [L1] to $f=\rho E$ in [A3]. Rewrite the boundary power as $(\sigma \nu )\cdot u=({\sigma }^{T}u)\cdot \nu$, then use [F2] on ${\sigma }^{T}u$ and $q$. The integral balance becomes the integral of the continuous residual ${\mathrm{\partial }}_t(\rho E)+\mathrm{div}(\rho Eu-{\sigma }^{T}u+q)-\rho b\cdot u-\rho r$ over every $P_t$; the fixed-sign-ball argument from step 1.1 makes this residual zero pointwise. This is the conservative total-energy equation. Conversely, integrating that equation over $P_t$ and using [F2] and [L1] recovers [A3].

Conversely, suppose the conservative momentum equation holds and $\sigma ={\sigma }^T$. Then ${\sum }_{j,l}{ϵ}_{ijl}{\sigma }_{lj}=0$, so the component calculation in step 2.2 gives $\mathrm{div}(x\times \sigma )=x\times \mathrm{div}\sigma$. The product rule, continuity equation [A1], and $u\times u=0$ give ${\mathrm{\partial }}_t(\rho x\times u)+\mathrm{div}(\rho (x\times u)\otimes u)=x\times ({\mathrm{\partial }}_t(\rho u)+\mathrm{div}(\rho u\otimes u))$. Substituting the momentum equation and stress-divergence identity yields ${\mathrm{\partial }}_t(\rho x\times u)+\mathrm{div}(\rho (x\times u)\otimes u)=x\times (\rho b)+\mathrm{div}(x\times \sigma )$. Integrate this local angular identity over $P_t$, use [F2] on the flux matrix $x\times \sigma$, whose boundary flux is $x\times (\sigma \nu )$, and use [L1] on the material density $\rho (x\times u)$. This recovers the angular-momentum integral balance in [A2]; the linear-momentum integral balance has already been recovered in step 2.1.

Subtract the kinetic equation in step 2.3 from the total-energy equation in step 2.4 and use $E=e+k$. The body-work terms cancel, leaving ${\mathrm{\partial }}_t(\rho e)+\mathrm{div}(\rho eu+q)=\sigma :L+\rho r$. The product rule and continuity equation [A1] give ${\mathrm{\partial }}_t(\rho e)+\mathrm{div}(\rho eu)=\rho D_{t}e$, so rearrangement yields $\rho D_{t}e=\sigma :L-\mathrm{div}q+\rho r$. Conversely, adding this internal-energy equation to the kinetic equation in step 2.3 recovers the total-energy equation in step 2.4.