3
$\begingroup$

I remember seeing in the book by Kreiss "Time-dependent partial differential equations and their numerical solution" that if some elliptic differential operator satisfies $$(Lu,u)\leq K(u.u)$$ for the equation $u_t=Lu+f$ with some boundary condition then it can be shown that, the equation is stable, that is continuously depends on the initial data. However, when I think of $L:=u''$ and $u=-\sin(nx)$ on $(0,\pi)$ as a solution of $u_t=u''+n^2\sin(nx)$, I have boundary conditions as zero at both ends independently of $n$. Then we can calculate \begin{equation} (Lu,u)=(u^{\prime\prime},u) = \int u^{\prime\prime} udx = n^2 \int \sin^2(nx)dx \end{equation} and \begin{equation} (u,u)= \int u^2dx = \int \sin^2(nx)dx \end{equation} Clearly, the inequality $(Lu,u)\leq K(u,u)$ doesn't hold as $n$ increases, even though second derivative is a proper elliptic operator. What am I missing here? I did not put the proof here but it is just a few lines, however the example above contradicts to the statement.

When does the estimate $(Lu,u)\leq K(u.u)$ hold then?

$\endgroup$
  • $\begingroup$ In how far is this a contradiction? The theorem of Kreiss appears to give a sufficient but not a necessary condition. You should clarify your question (I don't the theorem, however.) $\endgroup$ – shuhalo Sep 24 '12 at 9:10
5
$\begingroup$

I get $u''=-n^2 u$, and the condition holds with $K=0$.

By the way, Kreiss' conclusion holds for any linear operator; no ellipticity must be assumed.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ wait a second, if $u=-\sin(nx)$ then $u'=-n\cos(nx)$ and $u'=n^2\sin(nx)$. I chose ``-'' sign on purpose, the problem is in the magnitude, that the second derivative get this $n^2$ factor...It is good that the argument is for the general operator but I don't have a book handy to verify the conditions of the argument and want to have a precise statement for which functions it is satisfied. $\endgroup$ – Kamil Sep 25 '12 at 0:48
  • $\begingroup$ @Medan: You made an elementary mistake. You get an additional factor $-1$. when you express $u''$ in terms of $u$, or when you multiply $u''u$. You get the same result when you begin with $u=\pm \sin (nx)$ with either sign. $\endgroup$ – Arnold Neumaier Sep 25 '12 at 7:12
  • $\begingroup$ Arnold, but is this condition sufficient or also necessary? $\endgroup$ – Kamil Sep 26 '12 at 0:35
  • $\begingroup$ The condition is sufficient only, but it is the one one usually tries to meet when checking for continuous dependence. No other easily verifiable condition is known. $\endgroup$ – Arnold Neumaier Sep 26 '12 at 8:02

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.