Initially $u\in H^1(\Omega)$, so in three dimensions
$$
u\in L^6,\qquad \nabla u\in L^2.
$$
Both terms in
$$
\widetilde B(u,u)
=(u\mathbin\cdot\nabla)u
+\frac12(\nabla\mathbin\cdot u)u
$$
therefore belong to $L^{3/2}$. The equation becomes
$$
-\nu\Delta u=f-\widetilde B(u,u)
\quad\hbox{with right-hand side in }L^{3/2}.
$$
Periodic <elliptic regularity> gives
$$
u\in W^{2,3/2}.
$$
The Sobolev embedding then yields
$$
u\in W^{1,3}\cap L^6.
$$
Consequently each product in $\widetilde B(u,u)$ belongs to $L^2$, because
$$
L^6\cdot L^3\subset L^2.
$$
A second application of <elliptic regularity> now gives
$$
u\in H^2_{\rm per}(\Omega)=D(\widetilde A).
$$
Every term in the equation belongs to $\widetilde H$, and hence
$$
\boxed{
u\in D(\widetilde A),
\qquad
\nu\widetilde Au+\widetilde B(u,u)=f
\quad\hbox{in }\widetilde H}.
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-359/2/c/ii/solution