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. Are there any other ways to prove such results under weaker assumptions?



Bounded solution for parabolic equation
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 
$$
\|u\_t\|\_{L\_\infty(\Omega\_T)}\le C(T,a\_0)(\|u\_0\|\_{L\_\infty(\Omega)}+\|\Delta u\_0\|\_{L\_\infty(\Omega)}).
$$


