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

In three dimensions, the Sobolev inequality gives $\Vert v{\Vert }_{L^6(U)}\le C_U\Vert v{\Vert }_{H_0^1(U)}$. Thereforeand hence

Using $w^3-{\tilde{w}}^3=(w-\tilde{w})(w^2+w\tilde{w}+{\tilde{w}}^2)$, the Holder inequality and the same Sobolev embedding give at each timeTaking the $L^2$ norm in time supplies the required estimate with the factor $T^{1/2}$.

The assumed linear estimate and the first bound giveChoose $b>2\alpha \Vert \psi {\Vert }_{H^1}$ and then choose $\tau >0$ so small that $\alpha \beta {\tau }^{1/2}{b}^3\le b/2$. This proves that $A$ maps $X_{b,\tau }$ into itself.

For $w,\tilde{w}\in X_{b,\tau }$, linearity of the heat equation and the second cubic estimate giveShrinking $\tau$ again makes $2\alpha \gamma {\tau }^{1/2}{b}^2<1$. Thus $A$ is a contraction of the closed ball $X_{b,\tau }$ in the Banach space $L^{\infty }((0,\tau );H_0^1(U))$.

The Banach fixed-point theorem gives a unique $u\in X_{b,\tau }$ with $A(u)=u$. By the definition of $A$, this fixed point is a weak solution of the linear heat problem with forcing $f=-u^3$. It therefore satisfieswith the prescribed initial and homogeneous Dirichlet data. Hence the nonlinear heat equation has a local weak solution in $L^{\infty }((0,\tau );H_0^1(U))$.