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Let $d$ denote a positive integer. Let $f$ be a positive function on $\mathbb{R}^d$. We also assume that $f$ is bounded above and below. That is, there exists $C>0$ such that $C^{-1}\le f(x)\le C$, $x \in \mathbb{R}^d.$ We now consider the following heat equation with initial data $u_0$ (here, $u_0\colon \mathbb{R}^d \to \mathbb{R}$ is bounded measurable): \begin{equation*} \left\{ \begin{alignedat}{2} \frac{\partial u}{\partial t}(t,x)&=f(x)^{-1}\Delta u (t,x),\quad (t,x)\in (0,\infty) \times \mathbb{R}^d, \\ u(0,x)&=u_0(x),\quad x \in \mathbb{R}^d. \end{alignedat} \right. \end{equation*}

I know that this equation possesses a unique solution $u(t,x)$. It is also known that $x \mapsto u(t,x)$ is Hölder continuous. When the function $f$ belongs to $C^1(\mathbb{R}^d)$, can we show that the map $x \mapsto u(t,x)$ also belongs to $C^1(\mathbb{R}^d)$?

I think it is known whether this is true or not. Please let me know if you have any references.

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    $\begingroup$ Yes, this is true for $t>0$ under the assumption that $f$ is uniformly continuous. It is in section 3 of the book by A. Lunardi "Analytic semigroups and optimal regularity in parabolic problems". $\endgroup$ Commented Jan 4, 2022 at 14:50
  • $\begingroup$ @GiorgioMetafune Thank you for your comment. I'll take a look at the book right away. $\endgroup$
    – sharpe
    Commented Jan 4, 2022 at 15:53

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