Conditional wall energy and smooth uniqueness

Define the flux $F_c=\frac{1}{2}\rho \mathrm{\mid }u{\mathrm{\mid }}^{2}u$. The wall-domain definition [F1] gives C2 spatial regularity for $u$, and the C2 boundary is a bounded C1 domain by [F2]; hence this flux is C1 up to the boundary and [F4] applies. The product rule gives $\mathrm{div}{F}_c=\rho (u\cdot \mathrm{\nabla }u)\cdot u+\frac{1}{2}\rho \mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{div}u=\rho (u\cdot \mathrm{\nabla }u)\cdot u$. The resulting boundary flux uses the surface measure [F3] and equals $\frac{1}{2}\rho {\int }_{\mathrm{\partial }\mathrm{\Omega }}\mathrm{\mid }u{\mathrm{\mid }}^{2}u\cdot ndS$, which vanishes under either wall law: no-slip has $u=0$, and impermeable slip has $u\cdot n=0$.

The tensor $\sigma$ is C1 up to the boundary by [F1] and the stated regularity of $u,p$. Applying [F4] to the vector field with components $(F_{\sigma }{)}_j={\sum }_i{u}_i{\sigma }_{ij}$ yields ${\int }_{\mathrm{\Omega }}u\cdot \mathrm{div}\sigma dx={\int }_{\mathrm{\partial }\mathrm{\Omega }}u\cdot \sigma ndS-{\int }_{\mathrm{\Omega }}\sigma :\mathrm{\nabla }udx$. The C1-domain and surface-measure conventions [F2], [F3] match this use. Since $\sigma =-pI+2\eta D$, $\sigma :\mathrm{\nabla }u=-p\mathrm{div}u+2\eta D:\mathrm{\nabla }u=2\eta \mathrm{\mid }D{\mathrm{\mid }}^2$: the pressure term vanishes, and the symmetric matrix $D$ is orthogonal to the skew-symmetric part of $\mathrm{\nabla }u$.

Take the inner product of the momentum equation with $u$ and integrate. The time term is $K^{\mathrm{\prime }}(t)$, the convective term vanishes by step 1.1, and step 1.2 gives the stress work. In the no-slip case the boundary work is zero. In the slip case $u\cdot n=0$ and $(\sigma n{)}_{\tan }=-\beta u_{\tan }$ imply $u\cdot \sigma n=-\beta \mathrm{\mid }{u}_{\tan }{\mathrm{\mid }}^2$. Substitution gives exactly the two stated energy balances.

Let $u,\tilde{u}$ be two solutions with the same data and set $w=\tilde{u}-u$, $q=\tilde{p}-p$, and ${\sigma }_w=-qI+2\eta D(w)$. Subtracting their equations gives $\rho (w_t+\tilde{u}\cdot \mathrm{\nabla }w+w\cdot \mathrm{\nabla }u)=\mathrm{div}{\sigma }_w$. Pair with $w$ and integrate. The transport term is zero by [F4], $\mathrm{div}\tilde{u}=0$, and the common impermeable/no-slip trace. As in step 1.2, ${\sigma }_w:\mathrm{\nabla }w=2\eta \mathrm{\mid }D(w){\mathrm{\mid }}^2$. The C1 boundary and surface convention [F2], [F3] and the C1 flux regularity from [F1] justify this integration; the boundary term is zero for no-slip and is $-\beta \mathrm{\mid }{w}_{\tan }{\mathrm{\mid }}^2$ for the same linear slip traction law. Therefore $\frac{1}{2}\frac{d}{dt}\mathrm{\parallel }w{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }w{\mathrm{\parallel }}_2^2$.

Since $w(0)=0$, multiplying the inequality in step 2.2 by $\exp (-2{\int }_0^t\mathrm{\parallel }\mathrm{\nabla }u(s){\mathrm{\parallel }}_{\mathrm{\infty }}ds)$ gives a nonincreasing nonnegative function initially zero. Thus $\mathrm{\parallel }w(t){\mathrm{\parallel }}_2=0$ and the two continuous velocity fields agree everywhere on $[0,T]\times \bar{\mathrm{\Omega }}$.