MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've encountered the following problem which is causing me some trouble :

Let $(M^2,g)$ be a smooth compact Riemannian surface (with constant curvature for instance) and consider a smooth function $u:M\times [0,1) \to \mathbb{R}$ which is a solution of the following PDE : $\partial_t u(x,t)=A(x,t)\Delta u(x,t)$ where $A$ is a function which is smooth on $M\times [0,1)$ and satisfies : $C_1|1-t|^\alpha \leq A(x,t)\leq C_2$.

It is not hard to see that $u$ is uniformly bounded with the maximum principle, a simple computation also shows that $\int_M |\nabla u(x,t)|^2dv_g$ decreases with $t$, which gives $L^p$ convergence of $u(.,t)$ as $t$ goes to $1$.

My question is : does $u$ have a continuous extension to $M\times[0,1]$ ?

Any insight or reference on this kind of problem is welcome.



Some further observations :

  • For the application I have in mind, I would be happy with just a sequence $t_i$ going to $1$ such that $u(.,t_i)$ uniformly converges.

  • The statement in the previous item is true if we replace $M$ by an interval, thanks the Rellich-Kondrachov compactness theorem, in fact, with $M$ of dimension 2,we are exactly at the critical exponent given by Rellich-Kondrachov Theorem.

  • The statement is true if $A$ depends only on $t$ (not on $x$), just put $t'=\int_0^tA(\tau)d\tau$ and you get that $u(x,t')$ satisfies the usual heat equation.

Thanks again.

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.