In three dimensions, the <Sobolev inequality> gives $\lVert v\rVert_{L^6(U)}\leq C_U\lVert v\rVert_{H_0^1(U)}$. Therefore
$$
\lVert w^3\rVert_{L^2(U_T)}^2
=\int_0^T\lVert w(t)\rVert_6^6dt
\leq C_U^6T\lVert w\rVert_{L_t^\infty H_x^1}^6,
$$
and hence
$$
\boxed{\lVert w^3\rVert_{L^2(U_T)}
\leq C_U^3T^{1/2}\lVert w\rVert_{L_t^\infty H_x^1}^3.}
$$

Using $w^3-\widetilde w^3=(w-\widetilde w)(w^2+w\widetilde w+\widetilde w^2)$, the <Holder inequality> and the same Sobolev embedding give at each time
$$
\lVert w^3-\widetilde w^3\rVert_2
\leq C_U^3\lVert w-\widetilde w\rVert_{H^1}
\big(\lVert w\rVert_{H^1}^2+\lVert\widetilde w\rVert_{H^1}^2\big).
$$
Taking the $L^2$ norm in time supplies the required estimate with the factor $T^{1/2}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 105 / 4 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-105/4/a/solution