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

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- Solution ii

Solution

Initially

u \in H^{1} (Ω)

, so in three dimensions

u \in L^{6}, \nabla u \in L^{2} .

Both terms in

B (u, u) = (u \cdot \nabla) u + \frac{1}{2} (\nabla \cdot u) u

therefore belong to

L^{3/2}

. The equation becomes

- ν Δ u = f - B (u, u) 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

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_{per}^{2} (Ω) = D (A) .

Every term in the equation belongs to

H

, and hence

u \in D (A), ν A u + B (u, u) = f in H .

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