I can not prove the following problem.

In area $\Omega$, we have $v\in C^2(\Omega\times[T,\infty))\bigcup C^1(\partial\Omega\times[T,\infty))$ satisfy the following equation:

$$\frac{\partial v}{\partial t}-\Delta v \leq \alpha -v$$ and $$\frac{\partial v}{\partial n}=0,\ v(x,0)=v_0(x)$$

Prove that ${\rm lim}_{t_0\rightarrow \infty}{\rm sup}_{t\geq t_0}v(x,t)\leq \alpha$

I have try to use Proofs by Contradiction but I can only proof that if the consequence is wrong, the maximal of $v$ will not be attain in the inner of area. Thanks a lot.


If $v(x,t) \leq \alpha$ for all $t< t_0$ and if $v(x_0,t_0)=\alpha$ then $$ \Delta v (x_0,t_0)\leq 0 $$ so that $$\frac{\partial v}{\partial t} (x_0,t_0)\leq \Delta v(x_0,t_0)+ \alpha - v(x_0,t_0) =\Delta v(x_0,t_0)\leq 0 $$

  • $\begingroup$ we can only prove that $\frac{\partial v}{\partial t}\geq 0$, we can not get $\frac{\partial v}{\partial t}> 0$, And if the maximal is in boundary, we cannot get anything from the $\Delta v$ $\endgroup$ – Siqi He Dec 2 '13 at 8:24
  • $\begingroup$ I corrected. If at $(x_0,_0)$ $v$ has value $\alpha$ then it cannot increase more wrt time $t$. $\endgroup$ – HK Lee Dec 3 '13 at 14:28

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