4
$\begingroup$

Are there any existence results for parabolic PDE of the type $$u_t - Au = f$$ in some Gelfand triple setting ($V \subset H \subset V^*$) with $A$ an operator that it is not quite monotone but close: $$(Au-Av, u-v) \geq -C|u-v|_{H}^p$$ and satisfies some coercivity and hemicontinuity conditions, etc?

$\endgroup$
2
  • $\begingroup$ Are you thinking of a linear $A$? Should $V$ be a Hilbert space as in the usual notion of Gel'fand triple? $\endgroup$ Feb 18, 2014 at 8:34
  • $\begingroup$ @DelioM. $A$ is nonlinear for what I am thinking. $V$ can be taken to be a Hilbert space if necessary.. I thought Banach was enough but if not let it be Hilbert. $\endgroup$
    – weasd
    Feb 18, 2014 at 9:13

1 Answer 1

1
$\begingroup$

What you need is the theory of subdifferentials of convex lower semicontinuous functionals. You can find a lot in these nice lecture notes, chapter 8.

$\endgroup$
3
  • $\begingroup$ Thank you for these. In the existence proof in chapter 8, the convergence $u_m' \rightharpoonup u'$ in $L^2(0,T;H)$ is used to identify $\chi = \epsilon'(u)$. Unfortunately I don't have this convergence of the time deriviatives at all. Wonder if you have any idea how to get round this? $\endgroup$
    – weasd
    Feb 20, 2014 at 12:04
  • $\begingroup$ I am not sure to understand. That convergence is automatic if your functional satisfies the assumption of that theorem (coercivity etc.). And those assumptions are of course formulated thinking of a relatively general parabolic setting. How comes that that convergence fails? Do you have an oscillating behaviour as $m\to\infty$? $\endgroup$ Feb 21, 2014 at 9:59
  • $\begingroup$ In my case I do not have all the assumptions of that theorem. I was hoping to modify the proof of the theorem for my setting. In my case I have only the monotonicity property I mentioned above and my operator $A$ is not gradient system like in that notes, and so I am unable to find a good bound on the $u_m'$.. $\endgroup$
    – weasd
    Feb 21, 2014 at 19:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.