4
$\begingroup$

Let $\Omega_T=(0,T) \times \Omega$, where $\Omega$ a bounded smooth domain of $\mathbb{R}^n$ and $T>0$. Let $a\in L^\infty(\Omega)$ and consider the heat equation $$u_t=\Delta u + a(x)u, \;\; (t,x)\in \Omega_T ,$$ $$u|_{\partial \Omega}=0,$$ $$u(0,\cdot)=u_0.$$ Assume that the initial condition $u_0 \in H^2(\Omega)\cap H^1_0(\Omega) \cap L^\infty(\Omega)$, can we prove that the solution $u$ is such that $u_t \in L^\infty(\Omega_T)$ ?

I found some old results which use much of regularity on $a$ and $u_0$ based on maximum principles. Is there any other ways to prove such results under weaker assumptions. Thank you.

$\endgroup$
2
$\begingroup$

For the heat equation $Lu=u_t-\Delta u=0$ to guarantee boundedness of $u_t$ as $t$ tends to zero one has to demand more regularity from the initial function, e.g. $u_0\in C^{1,1}(\bar\Omega)$ (the first order derivatives are uniformly Lipschitz in $\Omega$).

As for the low order term, differentiating wrt $t$ we have that $u_t$ satisfies the same equation. So if $u_0,\Delta u_0\in L_\infty(\Omega)$ then $u_t|_{t=0}\in L_\infty(\Omega)$ and $u_t$ is bounded.

Denote $a_0=\|a\|_{L_\infty(\Omega)}$ and $v(t)=\|u\|_{L_\infty(\Omega_t)}$. For the first BVP $u_t-\Delta u=f$, $u|_{t=0}=u_0$ with zero boundary condition it follows that $$ v(t)\le t a_0 v(t)+\|u_0\|_{L_\infty(\Omega)}. $$ From here for $T_0=1/(2a_0)$ it follows that $v(T_0)\le 2\|u_0\|_{L_\infty(\Omega)}$. For arbitrary $T$ the estimate $v(T)\le C\|u_0\|_{L_\infty(\Omega)}$ follows from step by step argument, where $C$ depends upon $T$ as well as on $a_0$. For the derivative it gives the estimate $$ \|u_t\|_{L_\infty(\Omega_T)}\le C(T,a_0)(\|u_0\|_{L_\infty(\Omega)}+\|\Delta u_0\|_{L_\infty(\Omega)}). $$

$\endgroup$
  • $\begingroup$ Thank you @Andrew, I'm looking for further assumptions which imply the boundedness result. I think it is sufficient to see the boundedness of the solution and deduce the result for the time derivative. $\endgroup$ – S. Maths Dec 26 '18 at 14:32
  • $\begingroup$ Okey, in my case $Lu=u_t- \Delta u=f=au$. What is the assumption on $f$ to obtain the boundedness for $u$. I need some references on such results. Thanks. $\endgroup$ – S. Maths Dec 26 '18 at 18:24
  • $\begingroup$ I found that the assumption is $f\in L^\infty(\Omega_T)$ and this is the result we look for. See this link math.stackexchange.com/questions/776017/… $\endgroup$ – S. Maths Dec 26 '18 at 18:30
  • 1
    $\begingroup$ For the heat equation in a smooth bounded domain there exists the Green function $G$ of the first BVP. The solution with zero initial condition can be written as $$ u(x,t)=\int_0^t \int_\Omega G(x,y,t-\tau)f(y,\tau)\,dyd\tau. $$ Also one can get the estimate $0<G(x,y,t)<Z(x-y,t)$, where $Z$ is the fundamental solution of the heat equation. From there the estimate $$ |u(x,t)|\le t \|f\|_{L_\infty(\Omega_t)} $$ follows immediately. $\endgroup$ – Andrew Dec 26 '18 at 19:13
  • 1
    $\begingroup$ I've edited the answer. $\endgroup$ – Andrew Dec 27 '18 at 18:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.