Time continuity in $H^1_0$ for weak solution to $u_t - \Delta u = u \log |u|$

Question

I don't know how to prove this example. let $\Omega \subset \mathbb{R}^n$ is a bounded with smooth boundary. Assume that $u=u(x, t)$ on $\Omega \times [0, T)$ satisfies: \begin{equation} u\in L^{\infty}(0, T; H_0^1(\Omega)), u_t\in L^2(0, T; L^2(\Omega)) \end{equation} \begin{equation} (u_t, v)_{L^2}+(\nabla u, \nabla v)_{L^2} = (u\log|u|, v)_{L^2}, \mbox{for any}\ v\in H_0^1(\Omega), t\in(0, T). \end{equation} then $\|u(t_1)-u(t_2)\|_{H_0^1}\to 0$ as $t_1\to t_2$.

Answer 1

By the Sobolev inequality, $$ \|u\|_p \le C_1 \|u\|_{H_0^1} ,$$ for any $p>2$ satisfying $\frac1p \ge \frac12 - \frac1n$. Also $$ \| u \log |u| \|_2 \le C_2 \| u\|_p .$$

Let $\phi_n$ be the eigenvalues of the Laplacian on $\Omega$ with Dirichlet boundary conditions. Then every function in $L^2$ can be written as $$ u = \sum_{k=1}^\infty (u, \phi_k) \phi_k .$$ Let $$ S_n u = u_n = \sum_{k=1}^n (u, \phi_k) \phi_k .$$ Thus $\Delta S_n = S_n \Delta$, and $S_n^2 = S_n$.

Now put $-\Delta u_n$ into the weak PDE to get $$ \frac12 \frac{\partial}{\partial t} {\|\nabla u_n\|}_2^2 + {\|\Delta u_n\|}_2^2 \le {\|\Delta u_n \|}_2 {\| u \|}_{H^1_0} \le \frac12 {\|\Delta u_n \|}_2^2 + 2{\| u \|}_{H^1_0}^2 ,$$ that is $$ \frac12 {\|\nabla u_n(t)\|}_2^2 - \frac12 {\|\nabla u_n(0)\|}_2^2 + \frac12 \int_0^t {\|\Delta u_n(s) \|}_2^2 \, ds \le 2 \int_0^t {\| u(s) \|}_{H^1_0}^2 \, ds .$$ Let $n \to \infty$. Then $$ {\|u\|}_{L^\infty H^1_0}^2 + {\| u\|}_{L^2 H^2_0} ^2 \le {\| u(0)\|}_{H^1_0}^2 + 4 {\| u \|}_{L^2 H^1_0}^2 .$$ So $u \in L^2 H^2_0$.

Now put $v = \Delta u_n(t_1) - \Delta u_n(t_0)$ into the weak PDE, and integrate with respect to $t$ from $t_0$ to $t_1$. \begin{align} \|u_n(t_1) - u_n(t_0)\|_{H^1_0}^2 &= \int_{t_0}^{t_1} ( v,\partial_t u(t))_{L^2} \, dt \\&= - \int_{t_0}^{t_1} ( \nabla v, \nabla u(t) )_{L^2} dt + \int_{t_0}^{t_1} ( v, u(t)\log|u(t)| )_{L^2} dt \\&\le \| v \|_{H_0^2} \int_{t_0}^{t_1} \|u(t)\|_{H_0^2} \, dt + C_1 C_2 \|v\|_{H_0^2} \int_{t_0}^{t_1} \|u(t)\|_{H_0^1} \, dt . \end{align} (I got a bit lazy at the end, so you might want to double check it.) And $$ \int_{t_0}^{t_1} \|u(t)\|_{H_0^1} \, dt \to 0 $$ as $t_1 \to t_2$.