Dot the momentum equation with $\mathbf u$, multiply the temperature equation by $\sigma\operatorname{Ra}\theta$, and integrate over $V$. The pressure term vanishes by incompressibility, while <skew-symmetry of incompressible transport> removes both nonlinear advection terms. Integration by parts and the boundary conditions give
$$
\boxed{
\frac{dE}{dt}=\sigma\int_V
[2\operatorname{Ra}w\theta-|\nabla\mathbf u|^2
-\operatorname{Ra}|\nabla\theta|^2],dV},
$$
where
$$
E=\frac12\int_V
(|\mathbf u|^2+\sigma\operatorname{Ra}\theta^2)\,dV.
$$
For $\operatorname{Ra}=0$, this reduces to
$$
\frac{dE}{dt}=-\sigma\int_V|\nabla\mathbf u|^2\,dV\leq0,
$$
so viscosity monotonically dissipates kinetic energy.

Solution

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

 Solution

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