Take the $L^2$ inner product of the first Galerkin equation with $\omega_m$. The <orthogonal projection> may be removed against $\omega_m\in\widetilde H_m$, and <skew-symmetry of incompressible transport> cancels the nonlinear term. Hence
$$
\nu\|\nabla\omega_m\|_2^2+\gamma\|\omega_m\|_2^2
=(g,\omega_m).
$$
The <Cauchy-Schwarz inequality> and <Young inequality> give
$$
\nu\|\nabla\omega_m\|_2^2+\frac{\gamma}{2}\|\omega_m\|_2^2
\leq\frac1{2\gamma}\|g\|_2^2.
$$
Thus both quantities requested in the first estimate are bounded by, for example,
$$
\boxed{R_1^2=\frac{\|g\|_2^2}{\gamma}}.
$$

Next take the inner product with $-\Delta\omega_m$. Periodicity and incompressibility give
$$
\begin{aligned}
\left|((u_m\mathbin\cdot\nabla)\omega_m,-\Delta\omega_m)\right|
&=\left|\int_\Omega\partial_j(u_m)_i\,\partial_i\omega_m\,\partial_j\omega_m\,dx\right|\\
&\leq \|\nabla u_m\|_2\|\nabla\omega_m\|_4^2.
\end{aligned}
$$
The given curl identity and the two-dimensional <Gagliardo-Nirenberg inequality> imply
$$
\|\nabla u_m\|_2\leq C\|\omega_m\|_2,
\qquad
\|\nabla\omega_m\|_4^2
\leq C\|\nabla\omega_m\|_2\|\Delta\omega_m\|_2.
$$
Applying Young's inequality to this term and to $(g,-\Delta\omega_m)$ yields
$$
\frac{\nu}{2}\|\Delta\omega_m\|_2^2
+\gamma\|\nabla\omega_m\|_2^2
\leq\frac{C}{\nu}
\left(\|g\|_2^2+\|\omega_m\|_2^2\|\nabla\omega_m\|_2^2\right).
$$
The first estimate bounds the right-hand side independently of $m$. Therefore one may choose a constant $R_2=R_2(\|g\|_2,\gamma,\nu,\mu_1)$ such that
$$
\boxed{\nu\|\Delta\omega_m\|_2^2\leq R_2^2,
\qquad
\gamma\|\nabla\omega_m\|_2^2\leq R_2^2}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 359 / 2 / a / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-359/2/a/i/solution