Consider a closed Riemannian manifold $(M,g)$ and a positive function $\psi: M \to R$. Fix a point $p \in M$, I have been struggling to construct a solution to the heat equation, $\partial_t u = \Delta u$, such that at some positive time $t_0>0$, I have $|u(x,t_0)| \le \psi(x)$ and $|u(p,t_0)| = \psi(p)$ and $ |\Delta u (p,t_0)| = |\Delta \psi (p)|$.

I have been considering two approaches:

1) One approach is to let $h(x,t) = \psi(x) - u(x,t)$. $h$ satisfies $\partial_t h = \Delta h - \Delta \psi$ then the question becomes: construct a solution $h$ such that at some positive time $t_0$, $h(.,t_0) \ge 0$ , $h(p,t_0) = 0$ and $\Delta h (p,t_0) = 0$. We know that the above Heat equation for $h$ has a heat kernel but I have not been able to construct such a solution ??? (We might need to use some assertions about the zero crossings of a heat type equation.)

2) Take a function $g$ that satisfies $|g(x)|\le \psi(x)$ , $|g(p)| = \psi(p)$ and $ \Delta g (p) = \Delta \psi (p)$ and solve the backward Heat equation for a short time but the problem is that the backward Heat equation is not well-posed. At least we must have $g$ analytic and furthermore satisfying some proper decay rates on its derivatives. The question is, can we always find such a function $g$ satisfying the properties we want for which the backward heat equation is solvable for a short time? (I do not really need uniqueness)

P.S.: My goal is to prove a similar thing when, $(M,g(t))$ satisfy the Ricci flow on a time interval $[0,T]$ $\psi: M \to R$ a positive obstacle and when the heat equation is the heat equation under Ricci flow namely $\partial_t u = \Delta_{g(t)} u$.

Hope somebody could help me or just give me some ideas as to how to proceed further.


  • 1
    $\begingroup$ Please correct the conditions on $u$. At the moment the condition with RHS $\psi(p)$ contradicts the one with RHS $\Delta\psi(p)$. It is not clear what did you want to say... $\endgroup$ Jul 18 '12 at 7:44
  • $\begingroup$ Thanks for comment. I fixed some of the conditions, hope my question makes sense now. by the way u is not positive necessarily. I want at some positive time t_0 the graph of |u| to lie under ψ, touching it at a chosen point p with Δu(p)=Δψ(p) or Δu(p)=−Δψ(p) depending on the sign of u near p. $\endgroup$ Jul 18 '12 at 13:51
  • $\begingroup$ If $t_0$ is arbitrarily small then you can do this by the implicit function theorem. $\endgroup$ Jul 19 '12 at 10:13
  • $\begingroup$ Dear George, Thanks for the comment, very helpful. If I let $A(x,t,u) = (u_t + \Delta u)^2 + (\Delta u (p) - \Delta \psi(p))^2 + (u(p) - \psi(p))$ and apply the implicit function theorem to $A = 0$ on an appropriate domain namely, $u(x,t) \le \psi(x)+ \lambda(d(x,p))$ where $\lambda$ is a nonnegative function and $\lambda(0)=0$. Appreciate your help. $\endgroup$ Jul 19 '12 at 16:18
  • $\begingroup$ There is an issue though. Using the implicit function theorem, I get a solution on an open subset of M ...... $\endgroup$ Jul 20 '12 at 0:23

Regarding approach 2), for given $\psi$, the condition $\Delta g=\Delta \psi$ fixes $g$ up to an additive constant. This means that after satisfying $|g(p)|=\psi(p)$ you have a very little freedom to play with the condition $|g|\leq\psi$.

  • $\begingroup$ Dear Timur, I fix a point $p$ and then require $\Delta = \Delta \psi$ only at ONE point, namely the chosen point $p$. $\endgroup$ Jul 19 '12 at 0:55

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.