4
$\begingroup$

I was wondering is there a sufficient condition (or sufficient and necessary condition) for the existence of positive solutions of the following linear PDE on a closed manifold $(M, g)$,

\begin{equation*} -\Delta u +\nabla u\nabla f +hu=0. \end{equation*} where $f, h\in C^{\infty}(M)$.

I got some necessary conditions using the Stokes formula, but I couldn't find a statement for sufficient condition, or sufficient and necessary condition. Thank you very much for any suggestions.

$\endgroup$
8
$\begingroup$

Your operator is self adjoint on $ L^2 (M , e^{-f}) $: If your operator is $L$. Then we have for all fonctions u,v $\int uLv e^{-f}=\int (<du, dv>+huv) e^{-f}$

Hence , the equation $ Lu=0$ has a positive solution if and only if 0 is the lowest eigenvalue of $L $.

Let's me elaborate on this

This is a consequence of the so called Barta's lemma : If $u>0$ then a integration by parts formula implies that for every v: $$\int_M [|d(uv)|^2+h(uv)^2]e^{-f}= \int_M (Lu)uv^2+|dv|^2 u^2e^{-f}$$ Hence if there is a positive function $u$ such that $Lu\ge 0$, we get that the spectrum of $L$ is non negative. (This is the content of Barta's lemma: J. Barta, Sur la vibration fondamentale d'une membrane, {\em C. R. Acad. Sci. Paris} \textbf{204} (1937), 472--473.)

This Barta's lemma implies that if the equation $Lu=0$ has a positive solution, then $0$ is the lowest eigenvalue of $L$.

The other implication follows from the quadratic form : $$Q(v)=\int_M [|dv|^2+hv^2]e^{-f}= \int_M (Lv)v e^{-f}$$ The lowest eigenvalue , called it $\lambda_0$, of $L$ has a variationnal formulation : $$\lambda_0 =\inf\{ Q(v), v\in W^{1,2}, \int_M v^2 e^{-f}=1\}$$

And the eigenspace associated to $\lambda_0$ is precisely $$\{v\in C^\infty(M), Q(v)=\lambda_0\int_M v^2 e^{-f}\}$$ (the fact that these eigenfunctions are smooth is a consequence of Elliptic regularity).

But if $v\in W^{1,2}$ then $|v|\in W^{1,2}$ and $Q(|v|)\le Q(v)$ , $\int_M |v|^2 e^{-f}=\int_M v^2 e^{-f}$, hence if $u$ is an eigenvalue of $L$ associated to $\lambda_0$ then $|u|$ is also an eigenvalue and by linearity $u+|u|$ and $|u|-u$ also. But the unique continuation properties implies that if $|u|\not =u$ , then the function $|u|+u$ vanishes on the open subset where $u$ is non positive hence we must have $u+|u|=0$, hence $u$ is non negative. So that there is always a non negative eigenfunction associated to $\lambda_0$. The fact that such a non negative eigenfunction is positive is a consequence of the Harnack inequality (see Gilbarg-Trudinger subsection 8.8).

There is another approach that is more abstract (in the context of Dirichlet form see : The Allegretto-Piepenbrink Theorem for Strongly Local Dirichlet Forms, by D. Lenz, P. Stollmann, I.Veselic publish in {\em Documenta Mathematica} \textbf{14} (2009) 167--189).

For complete non compact manifold, the existence of a positive solution of $Lu=0$ is equivalent to the fact that the spectrum of $L$ is non negative. This is called the Allegretto-Piepenbrink principle and has been adapted to manifold by Fischer-Colbrie and Schoen (see W.F. Moss and J. Piepenbrink. Positive Solutions of Elliptic Equations. {\em Pacific J. Math.}, \textbf{75}:219--226, 1978.

D. Fischer-Colbrie and R. Schoen. The Structure of Complete Stable Minimal Surfaces in 3-manifolds of Non-negative Scalar Curvature.{\em Comm. Pure Appl. Math.}, \textbf{XXXIII}:199--211, 1980.

and the lemma 3.10 in the beautiful book : S. Pigola, M. Rigoli, and A. Setti. {\em Vanishing and Finiteness Results in Geometric Analysis.} Birkhäuser, 2008.

$\endgroup$
3
  • $\begingroup$ Why? Can you elaborate on your points? I am really sorry for not getting it. $\endgroup$ – András Bátkai Mar 20 '14 at 20:24
  • $\begingroup$ Thank you very much! If the average of $h$ is positive, i.e., $\int_M h>0$, could we prove that $0$ is the lowest eigenvalue of $L$? $\endgroup$ – user38600 May 5 '14 at 18:35
  • $\begingroup$ is there any uniqueness for the positive solution on complete(non-compact) manifold in Fischer-Colbrie and Schoen's theorem? Thanks! I checked their paper, they didn't address this kind of problem. (sorry that I can't comment ubik's answer) $\endgroup$ – BewSMA May 6 '14 at 12:52

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.