Closed smooth incompressible kinetic-energy dissipation

The pointwise model equation in [A1] holds on the same fixed spatial region used in the energy integral (or on its periodic lift). Dot ${\rho }_0{D}_{t}u=\mathrm{div}\sigma$ with $u$. The component product rule gives $u_i{\mathrm{\partial }}_{x_j}{\sigma }_{ij}={\mathrm{\partial }}_{x_j}({\sigma }_{ij}{u}_i)-{\sigma }_{ij}{\mathrm{\partial }}_{x_j}{u}_i$, while $D_t(\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}2)=u\cdot D_{t}u$. Since ${\rho }_0$ is constant and $\mathrm{div}u=0$, this yields ${\mathrm{\partial }}_t\left(\frac{{\rho }_0\mathrm{\mid }u{\mathrm{\mid }}^2}{2}\right)+\mathrm{div}(\frac{{\rho }_0\mathrm{\mid }u{\mathrm{\mid }}^2}{2}u-{\sigma }^{T}u)=-\sigma :\mathrm{\nabla }u\mathrm{.}$

Fix $t$ and put $e={\rho }_0\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}2$ and $F=eu-{\sigma }^{T}u$. In the periodic case, [F2] integrates each ${\mathrm{\partial }}_j{F}_j$ first in $x_j$; opposite endpoint values agree, so ${\int }_{\mathrm{\Omega }}\mathrm{div}F=0$. In the bounded-domain case, $F$ is $C^1$ on a neighborhood of $\bar{\mathrm{\Omega }}$ and vanishes on $\mathrm{\partial }\mathrm{\Omega }$. Here is a direct zero-flux argument requiring no additional choice axiom. Use a finite cubic mesh of side $h$ in a fixed containing box and let $P_h$ be the union of the closed mesh cubes contained in $\mathrm{\Omega }$. Let $B_h$ be the union of mesh cubes at distance at most $4\sqrt{3}h$ from $\mathrm{\partial }\mathrm{\Omega }$. Since this compact boundary has content zero by [F1], $\mathrm{vol}(B_h)\to 0$: for any finite open box cover of the boundary, all sufficiently small $B_h$ lie in that cover, and such covers have arbitrarily small total volume. Also $\mathrm{\Omega }\setminus P_h\subseteq B_h$. Applying [F2] on each cube of $P_h$ and cancelling common faces expresses ${\int }_{P_h}\mathrm{div}F$ as the sum of the exposed face integrals. Each exposed face is within $2\sqrt{3}h$ of the boundary. The derivative of $F$ is bounded by a constant $M$ on a compact neighborhood of that boundary, so the segment mean-value estimate [F3] and $F{\mathrm{\mid }}_{\mathrm{\partial }\mathrm{\Omega }}=0$ give $\mathrm{\mid }F\mathrm{\mid }\le 2\sqrt{3}Mh$ on these faces for sufficiently small $h$. Their number is at most six times the number of cubes in $B_h$. Thus [F4] bounds the absolute sum of face integrals by $12\sqrt{3}M\mathrm{vol}(B_h)\to 0$. The omitted volume integral is bounded by ${\sup }_{\bar{\mathrm{\Omega }}}\mathrm{\mid }\mathrm{div}F\mathrm{\mid }\mathrm{vol}(B_h)\to 0$. Hence ${\int }_{\mathrm{\Omega }}\mathrm{div}F=0$ also in this case.

On every compact time interval, $e$ and $e_t$ are continuous on its product with $\bar{\mathrm{\Omega }}$. For fixed $t$ and small nonzero $k$, the mean-value estimate gives ${\sup }_{x\in \bar{\mathrm{\Omega }}}\mid \frac{e(t+k,x)-e(t,x)}{k}-e_t(t,x)\mid \le {\sup }_{\begin{array}{c}\mathrm{\mid }s-t\mathrm{\mid }\le \mathrm{\mid }k\mathrm{\mid }\\ x\in \bar{\mathrm{\Omega }}\end{array}}\mathrm{\mid }{e}_t(s,x)-e_t(t,x)\mathrm{\mid }⟶0,$ by [F3]. Every spatial integral exists by [F1] or rectangular continuity. Multiplying the displayed uniform error by the finite volume and using [F4] proves $E_k^{\mathrm{\prime }}(t)={\int }_{\mathrm{\Omega }}{e}_t(t,x)dx$; the same estimate without the quotient proves continuity of this derivative. Integrating step 1.1 spatially and applying step 2.1 therefore gives $E_k^{\mathrm{\prime }}(t)=-{\int }_{\mathrm{\Omega }}\sigma :\mathrm{\nabla }udx$.

Using the Frobenius contraction and $\mathrm{\mid }D{\mathrm{\mid }}^2=D:D$ defined in the Statement, write $\mathrm{\nabla }u=D+W$, where $W=(\mathrm{\nabla }u-\mathrm{\nabla }{u}^T)\mathrm{/}2$ is skew-symmetric. A symmetric matrix is Frobenius-orthogonal to a skew-symmetric matrix, so $D:W=0$. The stress law gives $\sigma :\mathrm{\nabla }u=(-p{I}_3+2\mu D):(D+W)=-p\mathrm{div}u+2\mu \mathrm{\mid }D{\mathrm{\mid }}^2=2\mu \mathrm{\mid }D{\mathrm{\mid }}^2$. Thus $\frac{d}{dt}{E}_k=-2\mu {\int }_{\mathrm{\Omega }}\mathrm{\mid }D{\mathrm{\mid }}^{2}dx$. The spatial dissipation integral is continuous in time by the same uniform-continuity estimate as step 3.1. Applying [F2] in time from $t_1$ to $t_2$ proves the stated energy equality and nonincrease because $\mu \ge 0$.