Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 359 / 2

For smooth periodic fields, the Holder inequality and $H^1\hookrightarrow L^6$ giveSimilarly,Adding the bounds proves that the skew-symmetrized transport form extends continuously from smooth fields to ${\tilde{V}}^3$ andThe periodic Poincare inequality givesHence

Periodic integration by parts givesAdding one half of the divergence term to both transport forms yieldsDensity extends the identity from smooth fields to all $u,v,w\in \tilde{V}$. In particular,without requiring $\nabla \cdot u=0$.

Take the $\tilde{H}$ inner product of the finite-dimensional equation with $u_m$. Part a gives exact cancellation of the nonlinear term:By the Cauchy-Schwarz inequality and the Poincare inequality,Thus every solution satisfiesThe bound is independent of $m$.

On ${\tilde{H}}_m$, use the inner productThe map in the question is continuous, and part a givesOn the sphere $\Vert v\Vert =\rho$ with any $\rho >R$,The Brouwer inward-pointing zero lemma therefore supplies $v^{\ast }$ in the ball with ${\Phi }_m(v^{\ast })=0$. Multiplication by $\nu \tilde{A}$ shows that this zero satisfiesThus every Galerkin system has at least one solution.

The uniform estimate gives a subsequenceThe compact periodic embedding $\tilde{V}⋐\tilde{H}$ improves this toOne may also take strong convergence in $L^3$ by the Rellich-Kondrachov compactness theorem. Combining this with weak convergence of the gradients and using the skew form when derivatives must be moved to a test function givesFor any test function in a fixed finite-dimensional subspace, the Galerkin equation therefore passes to the limit. Density then givesThis is a weak solution of the steady skew-symmetrized Navier-Stokes equation.

Initially $u\in H^1(\Omega )$, so in three dimensionsBoth terms intherefore belong to $L^{3/2}$. The equation becomesPeriodic elliptic regularity givesThe Sobolev embedding then yieldsConsequently each product in $\tilde{B}(u,u)$ belongs to $L^2$, becauseA second application of elliptic regularity now givesEvery term in the equation belongs to $\tilde{H}$, and hence