Bounded data means bounded solution to parabolic PDE

Question

Let $u_0 \in L^\infty(\Omega)$ and $f \in L^\infty((0,T)\times\Omega).$ Consider $$u_t - \Delta u = f$$ $$u|_{\partial\Omega} = 0$$ $$u(0) = u_0$$ or more generally replace $\Delta$ with a suitable elliptic operator $A$. How does one show that $u \in L^\infty((0,T)\times\Omega)$?

(My question stems from this paper: http://www.mat.uniroma2.it/~porretta/papers/Blanchard-Porretta-JDE.pdf. See Theorem A.1 in the appendix (page 425). It is a different nonlinear equation but this should still be true).

Thanks

Answer 1

If you have a classical solution, evaluate at the point of maximum of $u$. The laplacian has the appropriate sign and then you get $$ \frac{d}{dt}\|u(t)\|_{L^\infty}\leq \|f\|_{L^\infty}. $$ Then integrate in time.

Answer 2

Let $\Omega\subset\mathbb{R}^n,\,n\geqslant 2,$ be a bounded domain with Lipschitz boundary. For a cylinder $\,Q_T=\Omega\times(0,T),\,$ denote by $S_T$ its side surface $\partial\Omega\times (0,T)$, with notation $\Gamma_T$ standing for its parabolic boundary $\{(x,t)\colon\; x\in \overline{\Omega},t=0\}\cup S_T$. Obviously, solution $u$ to the inhomogeneous heat equation $\,u_t-\Delta u=f\,$ in $\,Q_T\,$ with innhomogeneous boundary condition $u|_{S_T}=\psi$ can be represented in the form $u=v+w\,$ where $$ \begin{cases} v_t-\Delta v=0\;\;\text{in}\;\;Q_T\,,\\ v|_{\partial\Omega}=\psi,\quad t\in [0,T],\\ v|_{t=0}=u_0,\quad x\in\overline{\Omega}, \end{cases} \qquad \begin{cases} w_t-\Delta w=f\;\;\text{in}\;\;Q_T\,,\\ w|_{\partial\Omega}=0,\quad t\in [0,T],\\ w|_{t=0}=0,\quad x\in\overline{\Omega}. \end{cases} $$ Let $v\in H^1(Q_T)$ be a weak solution satisfying the integral identity $$ \int\limits_{Q_T}v_t\varphi_t\,dxdt+\int\limits_{Q_T}\nabla v\cdot\nabla\varphi\,\,dxdt =0\quad \forall\varphi\in H^1(Q_T)\colon\; \varphi|_{S_T}=0\tag{1} $$ with initial and boundary conditions $$ v|_{t=0}=u_0\in H^{\frac{1}{2}}(\Omega),\quad v|_{S_T}=\psi\in H^{\frac{1}{2}}(S_T). $$ For solution $v$, denote $m\overset{\rm def}{=} \underset{\Gamma_T}{\rm ess\,sup\,}{v}\,$  with the finite essential supremum taken over parabolic boundary $\Gamma_T\,$ w.r.t.  the $n$-dimensional Lebesgue mesure, and let $\eta_m$ be a Lipschitz function of the form $$ \eta_m(\xi)= \begin{cases} 0,\quad \xi\leqslant m,\\ \xi-m, \quad \xi>m. \end{cases} $$
It is clear that $\eta_m(v)\in H^1(Q_T)$ while $\eta_m(v)_{S_T}=0$ for any solution $v$. Taking in $(1)$ the test function $\varphi=\eta_m(v)$ yields $$ \int\limits_{Q_T}\frac{\partial\,}{\partial t}\Phi_m(v)\,dxdt+ \int\limits_{Q_T}\eta'_m(v)|\nabla v|^2dxdt=0\tag{2} $$ due to the chain rule $\nabla\varphi=\eta'_m(v)\nabla v$, where a nonnegative function $\Phi_m$ is defined as $$ \Phi_m(\xi)= \begin{cases} 0,\quad \xi\leqslant m,\\ \frac{1}{2}(\xi-m)^2, \quad \xi>m. \end{cases} $$ But $\,\Phi_m(v)|_{t=0}=0$, hence by $(2)$ follows $$ \int\limits_{\Omega}\Phi_m\bigl(v(\cdot,T)\bigr)\,dx+\int\limits_{Q_T}|\nabla \eta_m(v)|^2dxdt=0$$ since $\eta'_m=(\eta'_m)^2$. Therefore $\nabla \eta_m(v)=0\,$ a.e.  in $Q_T$ which implies that $\eta_m(v)=C(t)\,$ a.e.  in $Q_T\,$. But $\eta_m(v)|_{S_T}=0$, i.e., $\,C(t)=0\,$ a.e.  in $(0,T)\,$, and hence $\eta_m(v)=0\,$ a.e.  in $Q_T\,$. The latter implies the weak essential maximum principle $$ \underset{Q_T}{\rm ess\,sup\,}{v}\leqslant m=\underset{\Gamma_T}{\rm ess\,sup\,}{v} \tag{3} $$ with the essential supremum taken over the cylinder $Q_T\,$ w.r.t.  the $(n+1)$-dimensional Lebesgue mesure. To establish $$ \underset{\Gamma_T}{\rm ess\,inf\,}{v}\leqslant\underset{Q_T}{\rm ess\,inf\,}{v} \tag{4} $$ it suffices to substitute $v$ in $(3)$ by $-v$. Inequalities $(3)$ and $(4)$ yield the weak essential maximum principle for the modulus $$ \underset{Q_T}{\rm ess\,sup\,}{|v|}\leqslant\underset{\Gamma_T}{\rm ess\,sup\,}{|v|} \tag{5} $$ To estimate solution $w$, notice that it can be constructed using Duhamel's principle, i.e., in the form $$ w(x,t)=\int\limits_0^t h(x,t,\tau)\,d\tau $$ with function $h=h(x,s,\tau)$ defined for every $\tau\in (0,T)$ with finite norm $\|f(\cdot,\tau)\|_{L^{\infty}(\Omega)}$ as solution of the intial boundary value problem $$ \begin{cases} h_{s}-\Delta h=0\;\;\text{in}\;\;\Omega\times (\tau,T)\,,\\ h|_{\partial\Omega}=0,\quad s\in [\tau,T],\\ h|_{s=0}=f(x,\tau),\quad x\in\overline{\Omega}. \end{cases} $$ Otherwise, i.e., for $\tau\in (0,T)$ with the infinite norm $\|f(\cdot,\tau)\|_{L^{\infty}(\Omega)}\,$, a zero option $h(\cdot,\cdot,\tau)=0$ is chosen without loss of generality. Hence for almost every $s\in (\tau,T)$ by $(5)$ follows $$ \underset{\Omega}{\rm ess\,sup\,}{|h(\cdot,s,\tau)|} \leqslant\underset{\Omega}{\rm ess\,sup\,}{|f(\cdot,\tau)|} $$ for almost all $\tau\in (0,s)$, whence follows $$ \underset{\Omega}{\rm ess\,sup\,}|w(\cdot,t)|\leqslant \int\limits_0^t \underset{\Omega}{\rm ess\,sup\,}|h(\cdot,t,\tau)|\,d\tau \leqslant\int\limits_0^t \underset{\Omega}{\rm ess\,sup\,}|f(\cdot,\tau)|\,d\tau $$ for almost all $t\in (0,T)$. For a weak solution $u\in H^1(Q_T)$ to the inhomogeneous heat equation $u_t-\Delta u=f$, thus holds the following weak essential maximum principle for the modulus $$ \underset{Q_T}{\rm ess\,sup\,}|u|\leqslant\underset{\Gamma_T}{\rm ess\,sup\,}{|u|} +\int\limits_0^T \underset{\Omega}{\rm ess\,sup\,}|f(\cdot,t)|\,dt. $$