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Consider the elliptic operator $Lu = - \Delta u + \langle \nabla u , X \rangle + c \, u $ acting on functions on a closed Riemannian manifold $M$. Here $\Delta$ denotes the Laplace-Beltrami operator, $X$ is an arbitrary smooth vector field, and $c \geq 0$ is a smooth function on $M$ which does not vanish identically. Does $L$ have a so-called `principal eigenvalue' $\lambda_1 > 0$, whose corresponding (unique up to scaling) eigenfunction does not change sign?

A similar statement holds for smooth domains in $\mathbb{R}^n$, as shown for instance in Evans' PDE book, chapter 6. Moreover, in this paper it is sated that this fact is equivalent to the operator satisfying a maximum principle (which is indeed the case for the above $L$).

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  • $\begingroup$ Yes, this is a consequence of the Krein-Milman theorem. $\endgroup$
    – leo monsaingeon
    Commented Jun 2, 2020 at 8:35
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    $\begingroup$ Do you mean the Krein-Rutman theorem perhaps? do you know a reference for the above result in the closed manifolds case? $\endgroup$
    – Ramiro Lafuente
    Commented Jun 2, 2020 at 9:08
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    $\begingroup$ Oooops, indeed I meant Krein-Rutman, sorry. I don't have any reference for manifolds, but I'm pretty sure the standard proof should carry through $\endgroup$
    – leo monsaingeon
    Commented Jun 2, 2020 at 9:10
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    $\begingroup$ Here's an outline of the general Krein-Rutman based strategy that was suggested by @leomonsaingeon: (i) Choose a space to work on - for instance, $L^2$ over the manifold. (ii) Show that all spectral value of $L$ have real part $\ge \varepsilon$ for some $\varepsilon > 0$. (iii) Show compactness of the resolvent (for instance by showing that the domain of $L$ embeds compactly into $L^2$) (iv) Use the maximum principle to show that $L^{-1}$ a positive operator. [to be continued] $\endgroup$
    – Jochen Glueck
    Commented Jun 2, 2020 at 10:16
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    $\begingroup$ [continuation] (v) Apply the Krein-Rutman theorem to $L^{-1}$ in order to see that its spectral radius is in the spectrum and that it has a positive eigenvector. (vi) Use the spectral mapping theorem for resolvents to get back to the operator $L$. $\endgroup$
    – Jochen Glueck
    Commented Jun 2, 2020 at 10:16

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