Let $M$ be a compact $2d$-manifold in $\mathbb R^3$ and take $T>0\;$. We set $M_T:=M \times (0,T)$ and we consider the following parabolic PDE

\begin{align} \partial t u-\Delta_M u&=f \quad \text{ in } M_T \tag 1 \\ u(\cdot,0)&=u_0 \quad \text{in } M \tag 2 \end{align}


  • The right-hand side of $(1)$ is a product function of the form $f=g\times h \in L^\infty(M_T)\times C^2(M)$
  • The initial data can be as regular and smooth we wish, for instance $C^4(M)$

EDIT: $h$ cannot be constant. It is a function depending only on space variable $x$.

Goal: To find how regular and smooth $u$ is.


  1. What is the exact regularity of $f$? In order to estimate the regularity of $u$, I should first clarify this part. However although I know I should expect something "good" for $f$, I don't know how "good" exactly $f$ can be. Is this product function $C^2$? Is there any helpful "rule" in order to recognize "easily" where a product function lies?
  2. I am aware of theorems that provide $W^{2,1}_p$-estimates for solutions of problem $(1),(2)$ as long as $f \in L^p$ and $u_0\in W^{2,p}$ but I don't know any theorems or references where the right-hand side has better regularity than $L^p$. Is there any reference for that or it follows by a bootstrap argument? For example, if $f$ was $C^1(M_T)$, then is there any reference/theorem proving that $u$ is $C^3$ in space and $C^1$ in time? Or would this follow by a bootstrap argument? How would this argument work then?

I hope I made my questions clear enough because I always have a hard time getting my head around regularity issues. I would appreciate if someone could answer these questions and I apologize in advance if they sound too silly\elementary.

Many many thanks for the time!

  • $\begingroup$ There is no way $f$ can have better regularity than $L^\infty$: there is no conditions on $h$. $\endgroup$ Nov 18, 2021 at 12:35
  • $\begingroup$ @ArcticChar What do you mean there is no condition on $h$? You mean for time? $\endgroup$ Nov 18, 2021 at 12:38
  • $\begingroup$ Let say $h=1$. Then $f=g$ is in $L^\infty$ and not in $C^1$. $\endgroup$ Nov 18, 2021 at 12:44
  • $\begingroup$ @ArcticChar well $h$ can not be constant. I will add it in the post $\endgroup$ Nov 18, 2021 at 12:47
  • $\begingroup$ Does it matter? Whenever $h$ is non-zero in a neigborhood, $g = f/h$. So if $g$ is not continuous, $f$ cannot be continuous there too. $\endgroup$ Nov 18, 2021 at 12:55

1 Answer 1


Bootstrapping. The bootstrapping method applies only if $f$ depends on $u$. More precisely, if improved regularity of $u$ implies improved regularity of $f$. The way you described your problem, there's no hope for this, so I wouldn't expect a bootstrapping argument here. Your boots don't have straps for us to pull.

Regularity of $f$. The way it is now, $f$ is clearly $L^\infty$ (and nothing more).

Regularity of $u$. You already noted that $u \in W^{2,p}$ for any $p < \infty$. By Morrey's embedding, also $u \in C^{1,\alpha}$ for any $\alpha < 1$. And I wouldn't expect more.

It would be tempting to think that $$ f \in L^\infty \Longrightarrow \nabla^2 u \in L^\infty \quad \text{and/or} \quad f \in C^0 \Longrightarrow u \in C^2. $$ But these implications are false. In fact (see Gilbarg and Trudinger's book, problem 4.9) there's $f \in C^0$ such that any solution $v$ of $-\Delta v = f$ has unbounded second derivatives. If you take $u(t,x) = v(t,x)$, you obtain a solution $u \notin C^2$ of $(\partial_t-\Delta) u = f$.

  • $\begingroup$ First of all, thanks a lot for the answer. Since I am not familiar with this "bootstrapping" technique, I might erroneous ommited that indeed in my case $f$ depends on $u$. So then, since $u$ is uniformly continuous (due to the Hölder regularity), could we probably say more about the regularity of $f$ or at least have these straps for pulling as you suggested? $\endgroup$ Nov 18, 2021 at 14:30
  • $\begingroup$ Well, if $f$ in fact is a function of $u$, that's a whole new question. I'm not an expert, but I think the answer depends on the details. $\endgroup$ Nov 18, 2021 at 14:34
  • $\begingroup$ Ok that's fair. Thanks a lot $\endgroup$ Nov 18, 2021 at 14:34

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