Weak–smooth uniqueness on a smooth interval

Fix $t\in [0,T]$ and choose smooth time cutoffs converging to the indicator of $[0,t]$ in the weak identity for $u$, with test field equal to the cutoff times $v$. The terms are integrable because $u\in L^{\mathrm{\infty }}H\cap L^{2}V$ and $v,v_t,\mathrm{\nabla }v,q$ are bounded; $t\mapsto (u(t),v(t))$ is continuous because $u\in C_{w}H$ and $v$ is norm-continuous. Passing the cutoff limit gives $(u(t),v(t))-(u_0,v(0))={\int }_0^t[(u,v_t)+(u\otimes u,\mathrm{\nabla }v)-\nu (\mathrm{\nabla }u,\mathrm{\nabla }v)]ds\mathrm{.}$ Using $v_t=-\mathrm{div}(v\otimes v)-\mathrm{\nabla }q+\nu \mathrm{\Delta }v$, the pressure pairing vanishes because $u$ is distributionally divergence free. Periodic integration by parts then rewrites the integrand as $(v\otimes v,\mathrm{\nabla }u)+(u\otimes u,\mathrm{\nabla }v)-2\nu (\mathrm{\nabla }u,\mathrm{\nabla }v)$.

Testing the smooth equation by $v$, periodic integration by parts gives $\frac{1}{2}\frac{d}{dt}\mathrm{\parallel }v{\mathrm{\parallel }}_2^2+\nu \mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_2^2=0$: the transport integral is $\int \mathrm{div}(v\mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}2)=0$, the pressure term is $-\int q\mathrm{div}v=0$, and the Laplacian term is $-\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_2^2$. Integrating gives the energy equality for $v$. Add it to the weak energy inequality for $u$ and subtract twice the cross identity of step 1.1. The viscous terms combine to $2\nu {\int }_0^t\mathrm{\parallel }\mathrm{\nabla }(u-v){\mathrm{\parallel }}_2^{2}ds$. With $w=u-v$, incompressibility and periodic integration by parts give $(v\otimes v,\mathrm{\nabla }u)+(u\otimes u,\mathrm{\nabla }v)=(w\otimes w,\mathrm{\nabla }v)$: expand $u=v+w$; the $v\otimes v$ term has zero integral, the $v_i{w}_j{\mathrm{\partial }}_j{v}_i$ term is a divergence, and $(v\otimes v,\mathrm{\nabla }w)=-(w_i{v}_j,{\mathrm{\partial }}_j{v}_i)$ cancels the remaining mixed term. Hence $D(t)+2\nu {\int }_0^t\mathrm{\parallel }\mathrm{\nabla }w{\mathrm{\parallel }}_2^{2}ds\le D(0)-2{\int }_0^t(w\otimes w,\mathrm{\nabla }v)ds\mathrm{.}$ Since $\mathrm{\mid }(w\otimes w,\mathrm{\nabla }v)\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_{\mathrm{\infty }}D(s)$, dropping the nonnegative gradient integral proves the stated integral inequality.

Set $a(t)=2\mathrm{\parallel }\mathrm{\nabla }v(t){\mathrm{\parallel }}_{\mathrm{\infty }}\in L^1(0,T)$ and $G(t)=D(0)+{\int }_0^{t}a(s)D(s)ds$. Step 2.1 gives $0\le D\le G$. By [F2], $G$ is absolutely continuous and $G^{\mathrm{\prime }}=aD\le aG$ almost everywhere. The function $A(t)={\int }_0^{t}a(s)ds$ is absolutely continuous; by [F3], $e^{-A}G$ is absolutely continuous with derivative $e^{-A}(G^{\mathrm{\prime }}-aG)\le 0$ almost everywhere. The absolutely-continuous fundamental theorem in [F3] then implies $e^{-A(t)}G(t)\le G(0)=D(0)$, and therefore $D(t)\le D(0)e^{A(t)}$. When $D(0)=0$, this forces $D(t)=0$ for every $t\in [0,T]$, proving uniqueness against the smooth solution.