Getting a bound on solution of PDE in $L^\infty(0,T;L^2(\Omega))$?

Question

Let $$\varphi(s) = \begin{cases} s &: s < 0\\ 0 &: s \in [0,1]\\ s-1 &: s > 1 \end{cases}$$ Note that $\varphi$ is Lipschitz. Consider where $f \in L^2(0,T;L^2(\Omega)$, $$u_t - \Delta (\varphi(u)) = f \text{ in $\Omega \times (0,T)$}$$ $$u|_{\partial\Omega} =0$$ $$u(0) = u_0$$ Apparently if I take $\varphi(u)$ as a test function, I can obtain an estimate on $u$ in $L^\infty(0,T;L^2(\Omega))$, but I can't see this. Can someone help me? Thanks.

Answer 1

Since $\varphi$ is just Lipschitz, second order weak derivatives of $\varphi(u)$ may not exist no matter how smooth be a solution $u$. That is why $u$ is to be treated only as a weak solution. So, let $u\in H^1\bigl(\Omega \times (0,T))\bigr)$ be a weak solution of the problem, i.e., let $u$ possessing traces $u|_{t=0}=u_0$ and $u|_{\partial\Omega}=0$ satisfy the integral identity $$ \int\limits_{\Omega}u_t v\,dx+\int\limits_{\Omega}\nabla\varphi(u)\cdot \nabla v\,dx=\int\limits_{\Omega}fv\,dx \quad \forall\, v\in H^1\bigl(\Omega\times (0,T))\bigr)\colon\,v|_{\partial\Omega}=0 $$ a.e. on $(0,T)$. The required bound follows by a routine trick of introducing a function $\Phi\in C^1(\mathbb{R})$ such that $\Phi'=\varphi$ where $$ \Phi(s)= \begin{cases} \frac{s^2}{2}, &s<0,\\ 0, &s\in [0,1],\\ \frac{(s-1)^2}{2}, &s>1. \end{cases}\tag{$\ast$} $$ Notice that $\varphi^2(s)=2\Phi(s)\leqslant s^2\;\forall\,s\in\mathbb{R}$. Choosing a test function $v=\varphi(u)$ yields $$ \frac{d\,}{dt}\int\limits_{\Omega}\Phi(u)\,dx+\int\limits_{\Omega}|\nabla\varphi(u)|^2dx =\int\limits_{\Omega}f\varphi(u)\,dx $$ a.e. on $(0,T)$, whence by the Cauchy–Bunyakovski inequality follows $$ \frac{d\,}{dt}\int\limits_{\Omega}\Phi(u)\,dx\leqslant \|f\|_{L^2(\Omega)} \|\varphi(u)\|_{L^2(\Omega)}=\|f\|_{L^2(\Omega)}\biggl(2\int\limits_{\Omega} \Phi(u)\,dx\biggr)^{1/2} $$ a.e. on $(0,T)$, implying the bound $$ \biggl(2\int\limits_{\Omega}\Phi\bigl(u(x,t)\bigr)\,dx\biggr)^{1/2}\leqslant \int\limits_0^T\|f\|_{L^2(\Omega)}dt+\|u_0\|_{L^2(\Omega)} $$ a.e. on $(0,T)$. The rest is easy due to the definition $(\ast)$ of function $\Phi$.

Remark. At points $x,t$ where $0\leqslant u(x,t)\leqslant 1$ solution's smoothness w.r.t. $x$ is not generally to be expected higher than that of a solution $u(x,t)\overset{\rm def}{=}\psi(x)$ with some $L^2(\Omega)$ function $\psi\,\colon \Omega\to [0,1]$ vanishing near the boundary $\partial\Omega$ and possessing no weak derivatives, which solves the problem corresponding to $f=0$ and $u_0=\psi$.

Answer 2

Well, let me forget $f$. Then, notice that $$ \int u_t\phi(u)=-\int (\nabla \phi(u))^2, $$ where we used the boundary conditions and the definition of $\phi$ to remove the boundary term $$ \int_{\partial\Omega}\phi(u)\nabla\phi(u)nds=0. $$ Now $$ \int u_t\phi(u)=\int u_t u\textbf{1}_{u<0}+\int u_t (u-1)\textbf{1}_{u>1}=\int u_t u\textbf{1}_{u<0}+\int u_t u\textbf{1}_{u>1}-\int u_t \textbf{1}_{u>1}. $$ Now $$ 0.5\frac{d}{dt}|u|_{L^2}=\int uu_tdx=\left(\int uu_t\textbf{1}_{u>1}+\int uu_t\textbf{1}_{0\leq u\leq 1}+\int uu_t\textbf{1}_{u>1}\right). $$ So $$ 0.5\frac{d}{dt}|u|_{L^2}=\int u_t\phi(u)+\int u_t\textbf{1}_{u>1}+\int uu_t\textbf{1}_{0\leq u\leq 1}, $$ $$ 0.5\frac{d}{dt}|u|_{L^2}=-\int (\nabla \phi(u))^2+\int u_t\textbf{1}_{u>1}+\int uu_t\textbf{1}_{0\leq u\leq 1}. $$ Now, if $u>1$, $u_t=\Delta (u-1)=\Delta u$, so $$ \int u_t\textbf{1}_{u>1}=0. $$ On the other hand, if $|u|<1,$ $\partial_t u=0$. Putting all together we conclude, right? Is that correct?