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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

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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 $$ \underset{Q_T}{\rm ess\,sup\,}|u|\leqslant\underset{\Gamma_T}{\rm ess\,sup\,}{|u|} +\int\limits_0^T \underset{\Omega}{\rm ess\,sup\,}|f(\cdot,t)|\,dt. $$

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  • $\begingroup$ Hello, Is there other methods without using maximum principle ? Thank you. $\endgroup$
    – S. Maths
    Dec 24 '18 at 22:30
  • $\begingroup$ Is there any similar result for $u_t -\Delta u + a(x) u=0$ where $a\in L^\infty(\Omega)$ ? $\endgroup$
    – S. Maths
    Dec 27 '18 at 13:48
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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.

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  • $\begingroup$ Thanks. But you won't get classical solution (I guess we need at least $u \in C^2$ to do what you suggest) without more regularity on the data I believe. $\endgroup$
    – user142819
    Apr 30 '14 at 18:58
  • $\begingroup$ Actually you need $C^1([0,T]\times\Omega)$ for the approximated solutions. $\endgroup$
    – guacho
    Apr 30 '14 at 21:29
  • $\begingroup$ Such "approximated solutions" obviously require approximated data $u_0\in C^1(\Omega)$ inconsistent with $L^{\infty}$-norm unless $\,u_0\in C(\Omega)$. Besides, $u\in C^1([0,T]\times\Omega)$ is obviously not enough to decide whether the Laplacian does possess the appropriate sign. $\endgroup$
    – mkl314
    May 4 '14 at 21:15
  • $\begingroup$ Since $\,u|_{\partial\Omega}=0$, given any $t>0$, a positive maximum $M(t)\overset{\rm def}{=}\underset{x\in\overline{\Omega}}{\rm max}\,u(x,t)\,$ is attained generally at some subset of interior points $\,a(t)\subset\Omega$, while $\,\|u(\cdot,t)\|_{L^{\infty}(\Omega)}=M(t)=u\bigl(a(t),t)\bigr)$. Despite the fact that $\nabla_x u\bigl(a(t),t\bigr)=0$, it seems unlikely that one can easily establish differentiability of a composite function $u\bigl(a(t),t\bigr)$ with a set-valued $a$. The case of a multi-valued $M'(t)$ cannot be a priori excluded, which makes the approach overcomplicated. $\endgroup$
    – mkl314
    May 5 '14 at 0:18
  • $\begingroup$ Well, I don't know it it's seems unlikely or not. You can NOT establish differentiability. What you do is to prove that M(t) is Lipschitz, ergo, it is differentiable a.e.. Once you have this, you prove that $M'(t)=\partial_t u(x_t,t)$. Of course that $M(t)$ may be attained in more than one point, however, this is not relevant since at every point will decay. This "overcomplicated" approach is widely spread (check this, for instance, uam.es/personal_pdi/ciencias/acordoba/documentos/articulos/…) $\endgroup$
    – guacho
    May 6 '14 at 17:07

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