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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?

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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.) – Martin Sep 24 '12 at 9:10

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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.

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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. – Kamil Sep 25 '12 at 0:48
@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. – Arnold Neumaier Sep 25 '12 at 7:12
Thanks! Silly mistake that made me doubt the statement! – Kamil Sep 26 '12 at 0:17
Arnold, but is this condition sufficient or also necessary? – Kamil Sep 26 '12 at 0:35
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. – Arnold Neumaier Sep 26 '12 at 8:02

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