Two-dimensional uniqueness and conditional three-dimensional weak–strong uniqueness

For $d=2$, [F1] identifies each velocity component with a solenoidal energy weak velocity, and [F2] gives equality of those velocities for the common data.

Now let $d=3$, let $u$ be any energy weak-solution velocity, and assume $v$ satisfies the additional hypotheses in the Statement. Testing the weak equation for $u$ by time cutoffs of $v$ is allowed because $v\in C^1([0,T];V_{\sigma })$; the pressure term vanishes. The scalar pairing $(u,v{)}_2$ has an integrable distributional derivative by [F3], so [F7] makes it absolutely continuous; its derivative is $(u,v_t{)}_2+(f,v{)}_2-a_{\nu }(u,v)-c(u;u,v)$. Both weak solutions have initial trace $u_0$. Testing the strong equation for $v$ by $v$ gives its energy equality since $c(v;v,v)=0$. Combining that equality, the cross-pairing identity, and the weak energy inequality for $u$ yields, for almost every $t$, $\frac{1}{2}\mathrm{\parallel }u(t)-v(t){\mathrm{\parallel }}_2^2+\nu {\int }_0^t\mathrm{\parallel }\mathrm{\nabla }(u-v){\mathrm{\parallel }}_2^{2}ds\le {\int }_0^t\mathrm{\mid }c(u-v;v,u-v)\mathrm{\mid }ds$. Here $a_{\nu }(u,u)+a_{\nu }(v,v)-2a_{\nu }(u,v)=a_{\nu }(u-v,u-v)$, which is bounded below by the viscous term using [F4]. The convection terms reduce as $c(v;v,u)+c(u;u,v)=c(w;u,v)=c(w;v+w,v)=-c(w;v,w)$ for $w=u-v$, by skewness and [F3]; the strong equation for $v$ may be paired with $w\in L^2(0,T;V_{\sigma })$ and integrated.

By [F3], $\mathrm{\mid }c(w;v,w)\mathrm{\mid }=\mathrm{\mid }{\int }_D(w\cdot \mathrm{\nabla })v\cdot w\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}\mathrm{\parallel }w{\mathrm{\parallel }}_2^2$. Dropping the nonnegative viscous term in step 1.2 gives $y(t)\le 2{\int }_0^{t}b(s)y(s)ds$ for almost every $t$, where $y(t)=\mathrm{\parallel }w(t){\mathrm{\parallel }}_2^2$ and $b(t)=\mathrm{\parallel }\mathrm{\nabla }v(t){\mathrm{\parallel }}_{L^{\mathrm{\infty }}}\in L^1(0,T)$.

Since $u,v\in L^{\mathrm{\infty }}(0,T;H_{\sigma })$, $y\in L^{\mathrm{\infty }}(0,T)$; thus $Y(t)={\int }_0^{t}b(s)y(s)ds$ and $B(t)=2{\int }_0^{t}b(s)ds$ are absolutely continuous, with $Y(0)=0$. Step 2.1 gives $Y^{\mathrm{\prime }}(t)=b(t)y(t)\le 2b(t)Y(t)=B^{\mathrm{\prime }}(t)Y(t)$ almost everywhere. By [F5], $e^{-B}Y$ is absolutely continuous and has derivative $e^{-B}(Y^{\mathrm{\prime }}-B^{\mathrm{\prime }}Y)\le 0$ almost everywhere. It starts at zero and is nonnegative by [F6], so it is identically zero; hence $Y=0$ and step 2.1 implies $y=0$ almost everywhere. The difference $w=u-v$ is weakly continuous into $H_{\sigma }$; for any $t$, choose $t_n\to t$ from the full-measure set where $w(t_n)=0$. Weak continuity gives $(w(t),h{)}_2=0$ for every $h\in H_{\sigma }$, so $w(t)=0$.

In either case let $p_1,p_2$ be the associated pressures. Subtracting the full weak equations after velocity equality gives ${\int }_0^{T}b(v,p_1-p_2)\eta dt=0$ for every $v\in V$ and $\eta \in C_c^{\mathrm{\infty }}(0,T)$. For any $q\in Q$, choose $v=B_{D}q$ using [F8]. Then $(p_1-p_2,q{)}_2$ is zero as a time distribution and hence almost everywhere. The space $Q$ is separable: take a countable dense family in $L^2(D)$ from [F8] and subtract each mean. Outside the union of the null sets for that family all these pairings vanish; density implies $p_1(t)=p_2(t)$ in $Q$. Thus the zero-mean pressures agree almost everywhere, completing both uniqueness claims.