I have the maximum principle: $$\text{If } \psi\geq 0 \text{ on }\Gamma. \text{Then }L\psi\geq0\text{ implies } \psi\geq 0 \text{ in } \bar{D},$$ where $D=(0,1)\times (0,T], \Gamma$ is parabolic boundary. The operator is defined as $$Lu:=-u_{xx}+bu+du_t=f,$$ where $b,d,f$ are functions and $b\geq\beta>0,d\geq\delta>0$. I would like to show that $$||u||\leq(1+\alpha T)\max\{||Lu||,||u||_\Gamma\},$$ where $\alpha=\max\{0,(1-b)/d\}\leq1/\delta$. It is supposed to be an immediate consequence of the Maximum principle but I am stuck.

I have tried to apply the Maximum principle on $$\phi^{\pm}=\max\{||Lu||,||u||_\Gamma\}\pm u$$ but I don't seem to go anywhere.

  • $\begingroup$ What is $\Gamma$? $\endgroup$ – martini Jul 10 '14 at 16:15
  • $\begingroup$ You also haven't told us what $u$ satisfies. $\endgroup$ – parsiad Jul 10 '14 at 16:18
  • $\begingroup$ $\Gamma$ is the boundary of box $D$ without the top @par, $Lu=f$ is the differential equation satisfied by $u$. $\endgroup$ – Vaolter Jul 10 '14 at 16:54

The boundary $\Gamma$ is $$ \Gamma=\{(t,x)\in \{0\}\times (0,1)\}\cup \{(t,x)\in (0,T)\times \{0\}\}\cup \{(t,x)\in (0,T)\times \{1\}\}. $$ You consider the problem $$ -u_{xx}+bu+du_t=f \mbox{ in }(0,1)\times(0,T), \quad u=\phi \mbox{ on } \Gamma. $$ Consider $v=u+M$. It satisfies $$ -v_{xx}+bv+dv_t=f+bM \mbox{ in }(0,1)\times(0,T), \quad u=\phi+M \mbox{ on } \Gamma. $$ Choose $M$ so that $f+bM\geq0$ and $\phi+M\geq0$ and you obtain from the maximum principle you stated that $v\geq0$, in other words $$ u\geq-M=-\max(\inf(-\phi),\inf(-f/b))\geq -\max(\|\phi\|_\infty,\|\frac{f}{b}\|_\infty). $$ Symmetrically, consider now the problem satisfied by $-u$, (which means just replacing $f$ and $\phi$ by $-f$ and $-\phi$). By exactly the same argument, $$ -u\geq -N = - \max(\inf(\phi),\inf(f/b)) \geq -\max(\|\phi\|_\infty,\|\frac{f}{b}\|_\infty).. $$ so altogether $$ \|u\|_\infty\leq \max(\|\phi\|_\infty,\|\frac{f}{b}\|_\infty)\leq \max\left(1,\frac{1}{\beta}\right)\max(\|\phi\|_\infty,\|f\|_\infty). $$ This is not the constant you wanted but your constant is puzzling me. You seem to say $$ b\leq\beta,d\geq\delta\implies\max(0,\frac{b-1}{d})\leq\delta^{-1}, $$ this is not true in general, take for example $\delta<1$, and $b=\beta=1+\frac{1}{\delta^N}$, you obtain $$\max(0,\frac{b-1}{d})=\delta^{-N-1}>\delta^{-1}.$$ If you want to put some $\delta$ in the formula, you can, before you start, consider the problem satisfied by $\tilde{u}=u\exp(-\lambda t)$. Then, $f$ is changed to $\exp(-\lambda t)f$, $\phi$ is changed to $\exp(-\lambda t)\phi$, and $b$ is changed to $b+\lambda d$. The same argument then applies to $\tilde u$

Then you have instead, as long as $\beta+\lambda \delta >0$, $$ \|u\|_\infty\leq\exp(\lambda T)\|\tilde{u}\|_\infty\\ \leq \max\left(1,\frac{1}{\beta+\lambda \delta}\right)\exp(\lambda T)\max(\|\exp(-\lambda t)\phi\|_\infty,\|\exp(-\lambda t)f\|_\infty), $$ so if $f$ and $\phi$ do not depend on time, you obtain, for $\lambda>0$, $$ \|u\|_\infty\leq\max\left(1,\frac{1}{\beta+\lambda \delta}\right)\exp(\lambda T)\max(\|\phi\|_\infty,\|f\|_\infty), $$. (and a worse bound than the direct one if $-\beta/\delta<\lambda<0$).

  • $\begingroup$ why is $-\max(\inf(-\phi),\inf(-f/b))\geq -\max(||\phi||_\infty,||f/b||_\infty)$? $\endgroup$ – Vaolter Jul 16 '14 at 16:43
  • $\begingroup$ @Vaolter $|\phi|\leq \|\phi\|_{\infty}$ (a.e.), so $-\phi \geq -\|\phi\|_{\infty}$ (a.e.) idem for $f$. The max princ only asks for inequalities almost everywhere. $\endgroup$ – username Jul 16 '14 at 23:27

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