Let $u$ and $v$ be two weak solutions with the same initial value, and put $w=u-v$. Bilinearity gives
$$
w_t+\nu Aw+B(P_Nu,w)+B(P_Nw,v)=0.
$$
Pair with $w$. The first transport term vanishes because $P_Nu$ is divergence free. Part a and the <Young inequality> give
$$
\begin{aligned}
|\langle B(P_Nw,v),w\rangle|
&\leq c\lambda_N^{1/4}|w|\,\|v\|\,\|w\|\\
&\leq\frac\nu2\|w\|^2
+\frac{c^2\lambda_N^{1/2}}{2\nu}
\|v\|^2|w|^2.
\end{aligned}
$$
Therefore
$$
\frac d{dt}|w|^2+\nu\|w\|^2
\leq\frac{c^2\lambda_N^{1/2}}{\nu}
\|v\|^2|w|^2.
$$
The coefficient is integrable because $v\in L^2(0,T;V)$. Since $w(0)=0$, the <Gronwall inequality> gives $w=0$ on $[0,T]$. The global weak solution is unique.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 359 / 1 / c / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-359/1/c/iv/solution