2
$\begingroup$

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.

Thanks

Improve this question
$\endgroup$
5
  • 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... $\endgroup$
    – Anton Petrunin
    Commented Jul 18, 2012 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. $\endgroup$
    – Sajjad Lakzian
    Commented Jul 18, 2012 at 13:51
  • $\begingroup$ If $t_0$ is arbitrarily small then you can do this by the implicit function theorem. $\endgroup$
    – George Lowther
    Commented Jul 19, 2012 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. $\endgroup$
    – Sajjad Lakzian
    Commented Jul 19, 2012 at 16:18
  • $\begingroup$ There is an issue though. Using the implicit function theorem, I get a solution on an open subset of M ...... $\endgroup$
    – Sajjad Lakzian
    Commented Jul 20, 2012 at 0:23

1 Answer 1

Reset to default
2
$\begingroup$

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 $|g|\leq\psi$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Dear Timur, I fix a point $p$ and then require $\Delta = \Delta \psi$ only at ONE point, namely the chosen point $p$. $\endgroup$
    – Sajjad Lakzian
    Commented Jul 19, 2012 at 0:55

You must log in to answer this question.