After the pressure enforces incompressibility, let $L$ denote the displayed linear operator. Integration by parts gives
$$
\langle\Phi_i,L\Phi_j\rangle
=-\sigma\int_V\nabla\mathbf u_i:\nabla\mathbf u_j\,dV
-\sigma\operatorname{Ra}\int_V\nabla\theta_i\mathbin\cdot\nabla\theta_j\,dV
+\sigma\operatorname{Ra}\int_V(w_i\theta_j+\theta_iw_j)\,dV.
$$
The pressure terms vanish by incompressibility and the boundary conditions. The expression is symmetric under $i\leftrightarrow j$, so
$$
\boxed{\langle\Phi_i,L\Phi_j\rangle
=\langle L\Phi_i,\Phi_j\rangle}.
$$
Thus $L$ is a <self-adjoint operator> in the energy inner product and hence a <normal operator>. Its orthogonal eigenmodes cannot generate non-normal transient growth.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 331 / 2 / b / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-331/2/b/iv/solution