Suppose that I have a parabolic PDE, (to fix notation) let's say it's of the form: $$ \frac{\partial u}{\partial t}(x,t) + \mu(x,t) \frac{\partial u}{\partial x}(x,t) + \tfrac{1}{2} \sigma^2(x,t) \frac{\partial^2 u}{\partial x^2}(x,t) -V(x,t) u(x,t) + f(x,t) = 0. $$

Let $\Omega$ be a fixed compact cube in $\mathbb{R}^{d+1}$, are there conditions guaranteeing that the PDE has a solution whose $f$ mapping $\Omega$ into $[0,1]$ (or if it's easier into $(0,1)$)?

I'm familiar with boundary conditions but this is a bit different...

  • $\begingroup$ Your question is unclear. Are you assuming that $f$ has values in $[0,1]$ or do you want conditions such that the solution $u$ has values in $[0,1]$? $\endgroup$ – Hans Engler Nov 8 '18 at 23:40
  • $\begingroup$ I want conditions such that $u$ is guaranteed to take values in $[0,1]$; assuming that $f$ takes values in $[0,1]$. $\endgroup$ – AIM_BLB Nov 8 '18 at 23:57

Added: This is a backwards heat equation. It's easy to convert this into a forward equation by replacing $t$ with $T-t$ where $T$ is the terminal time. That moves the $\frac{\partial u}{\partial t}$ term to the right hand side but otherwise does not change things.

For demonstration purposes, take the case $d = 1$ (one spatial dimension). Then $\Omega = [a,b] \times [0,T]$ and one may prescribe terminal data for $u$ at $x \in [a,b], t = T$ and additional boundary conditions at $x \in \{a,b\}, 0 \le t \le T$.

Clearly the terminal data must be such that $u(x,T) \in [0,1]$. Then a general comparison principle that works also in higher space dimensions and for a broad class of boundary conditions says: If $f \ge 0$ everywhere, then also $u \ge 0$ everywhere (and in fact $u > 0$ for $t > 0$).

To obtain conditions such that $u \le 1$, you could consider $v = 1-u$ which satisfied the same equation as $u$, but with $f$ replaced by $V - f$. Then $u \le 1$ everywhere iff $v \ge 0$ everywhere, and this is true if $V-f \ge 0$ everywhere. So if $0 \le f \le V$ and $0 \le u(x,T) \le 1$, then $0 \le u \le 1$ everywhere.

  • $\begingroup$ Is it possible to replace the initial data condition with a terminal data condition instead? Or does that ruin things? $\endgroup$ – AIM_BLB Nov 9 '18 at 15:50
  • $\begingroup$ In fact there should be terminal data since this is a backwards equation. I'll fix the answer. $\endgroup$ – Hans Engler Nov 9 '18 at 19:31
  • $\begingroup$ Hey, that's alright, thanks Hans your answer has been extremely helpful :) $\endgroup$ – AIM_BLB Nov 9 '18 at 20:45

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.