3
$\begingroup$

I want to learn how to apply certain ODE theory to PDE. If we have a Banach space ODE $$x'(t) = f(t, x(t), p),$$ $$x(0) = x_0$$ where the equation is over same domain $t \in (a,b)$, then via the implicit function theorem, there is a unique solution around neighborhoods of the initial value and the parameter, and the solution depends continuously on the initial value and the parameter $p$. This is according to Zeidler's book in Section 4.11

Is there a way this ODE theory can be directly applied to PDE theory for some parabolic equation to get continuous dependence? Or does one just adapt the proof?

$\endgroup$

1 Answer 1

2
$\begingroup$

Yes, you can get some results on some PDE's directly with abstract ODE methods. However, in my experience with nonlinear PDE's it tends to be preferable to deal directly with the iteration/fixed point method, because the abstraction you get by using ODE method is very thin, and so it does not add much insight. Removing this layer makes things more transparent, and it is instructive when you study more advanced techniques for more complicated PDE's.

$\endgroup$
3
  • $\begingroup$ Out of curiosity, which fixed-point theorem do you/people tend to use most often in nonlinear PDEs? $\endgroup$
    – Bloop
    Commented Aug 14, 2012 at 18:23
  • $\begingroup$ @Bloop: For local existence theory, Banach's fixed point theorem is mostly used. It can be tricky to see if the equation is amenable to such methods though. Sometimes you need a clever variable transformation. Sometimes you have to modify the equation so you can use this theorem, and then approximate the original equation by such easier equations. In some rare cases people also use the Nash-Moser iteration. $\endgroup$
    – timur
    Commented Aug 14, 2012 at 20:41
  • $\begingroup$ @timur: Could you give some references for the continuous and particularly smoothness and even analyticity (possessing power series expansion) of the parabolic PDE solution on the PDE parameters? $\endgroup$
    – Hans
    Commented Oct 24, 2015 at 7:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .