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Consider the heat equation $\partial_t u= \Delta u+\lambda_1 u$ on a non-compact complete manifold $M$ (with nonpositive curvature) where $\lambda_1$ is the first eigenvalue and we start with some smooth initial data $u_0$ at $t=0.$ Then does the heat flow converge to the first eigenfunction on $M$?

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  • $\begingroup$ eigenfunction/eigenvalue in what sense? $L^2$? $\endgroup$ Commented Apr 17, 2023 at 20:46
  • $\begingroup$ yeah, that's right $\endgroup$
    – Student
    Commented Apr 17, 2023 at 20:46
  • $\begingroup$ Convergence in which sense? $L^2$-convergence follows from the spectral theorem. Do you want something stronger? $\endgroup$
    – MaoWao
    Commented Apr 19, 2023 at 9:34
  • $\begingroup$ @MaoWao I'd love to see a proof of that in an answer, is the convergence rate exponential? Does one also have convergence $C^0$ on compact subsets? $\endgroup$ Commented Apr 19, 2023 at 20:33
  • $\begingroup$ @Overflowian See my answer. Regarding exponential decay, an estimate of the form $\lVert u_t-u_\infty\rVert_2\leq e^{-\alpha t}\lVert u_0-u_\infty\rVert_2$ is equivalent to a gap of size at least $\alpha$ between $\lambda_1$ and the rest of the spectrum. $\endgroup$
    – MaoWao
    Commented Apr 20, 2023 at 6:44

1 Answer 1

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The following answer only applies if $u_0\in L^2$, and it mostly relies on abstract operator theoretic arguments. I am sure more could be said by exploiting the geometric structure.

The Laplace-Beltrami operator on a complete manifold is essentially self-adjoint on $C_c^\infty$. Since $\lambda_1$ is the bottom of the spectrum of $-\Delta$, the operator $-(\Delta+\lambda_1)$ has a unique positive self-adjoint extension, which I denote by $A$. The heat flow is given by $e^{-tA}u_0$.

Let $E$ denote the spectral measure of $A$ and $d\mu(\lambda)=d\langle u_0,E(\lambda) u_0\rangle$. Note that $\mu$ is a finite measure on $\mathbb R_+$ with $\mu(\mathbb R_+)=\lVert u_0\rVert_2^2$. By the spectral theorem we have $$ \lVert A^k(e^{-tA}-1_{\{0\}}(A))u_0\rVert_2^2=\int_{[0,\infty)}\lambda^{2k}(e^{-t\lambda}-1_{\{0\}}(\lambda))^2\,d\mu(\lambda). $$ For $t\geq t_0$ (and $\lambda\geq 0$) we have $$ \lambda^{2k}(e^{-t\lambda}-1_{\{0\}}(\lambda))^2\leq \lambda^{2k}e^{-2t\lambda}\lesssim_k t_0^{-2k}. $$ Thus $\lVert A^k(e^{-tA}-1_{\{0\}}(A))u_0\rVert_2^2\to 0$ as $t\to\infty$ by the dominated convergence theorem.

The operator $1_{\{0\}}(A)$ is the projection onto $\ker(\Delta+\lambda_1)$. In particular, if $\lambda_1$ is an eigenvalue of $-\Delta$, then $1_{\{0\}}(A)u_0$ is an eigenfunction of $-\Delta$. In the following I will write $u_t$ for $e^{-tA}u_0$ and $u_\infty$ for $1_{\{0\}}(A)u_0$. By what we have seen before, $u_t\to u_\infty$ in $L^2$.

By elliptic regularity and Sobolev embedding, if $m\in\mathbb N$ and $2k\geq m+\dim(M)/2+1$, then $$ \lVert u_t-u_\infty\rVert_{C^m(\Omega)}\lesssim_{m,\Omega}\lVert u_t-u_\infty\rVert_{H^{2k}(\Omega)}\lesssim_{k,\Omega}\lVert A^k(u_t-u_\infty)\rVert_{L^2(\Omega)}+\lVert u_t-u_\infty\rVert_{L^2(\Omega)} $$ for every sufficiently nice bounded domain $\Omega$ in $M$.

Therefore, $u_t\to u_\infty$ in $C^m$ on compact subsets for every $m\in\mathbb N$.

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  • $\begingroup$ I'm a bit confused by signs here, you seems to say that the spectrum of $-\Delta $ satisfies $\sigma(-\Delta)\subset (\lambda_1,+\infty)$. So is the definition of $\lambda_1 $ given by $\inf \sigma(-\Delta)$?. I thought that it was the smallest positive eigenvalue and in particular it should be contained in the spectrum. $\endgroup$ Commented Apr 20, 2023 at 12:05
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    $\begingroup$ I took $\lambda_1$ to mean the bottom of the spectrum of $-\Delta$ (which is actually contained in the spectrum because it is closed). Of course, if $\lambda_1$ is embedded in the spectrum, this is no longer true -- then you can have approximate eigenfunctions of the bottom of the spectrum for which $e^{t(\Delta+\lambda_1)}u_0$ blows up in $L^2$ norm. $\endgroup$
    – MaoWao
    Commented Apr 20, 2023 at 12:24
  • $\begingroup$ So $-\Delta-\lambda_1$ is positive semi-definite, does it still have a unique extension? $\endgroup$ Commented Apr 20, 2023 at 12:28
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    $\begingroup$ Yes, essential self-adjointness is stable under additive bounded (self-adjoint) perturbations. $\endgroup$
    – MaoWao
    Commented Apr 20, 2023 at 12:55
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    $\begingroup$ For your question that is completely unimportant. The proof that @MaoWao provided works equally well on compact manifolds, possibly with positive curvature. But on compacts since the spectrum is discrete you will have spectral gap and the convergence is exponential. $\endgroup$ Commented Apr 20, 2023 at 13:54

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