Consider the heat equation $\partial_t u= \Delta u+\lambda_1 u$ on a noncompact 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 $M$?

$\begingroup$ eigenfunction/eigenvalue in what sense? $L^2$? $\endgroup$– Willie WongApr 17 at 20:46

$\begingroup$ yeah, that's right $\endgroup$– StudentApr 17 at 20:46

$\begingroup$ Convergence in which sense? $L^2$convergence follows from the spectral theorem. Do you want something stronger? $\endgroup$– MaoWaoApr 19 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? $\endgroup$– OverflowianApr 19 at 20:33

$\begingroup$ @Overflowian See my answer. Regarding exponential decay, an estimate of the form $\lVert u_tu_\infty\rVert_2\leq e^{\alpha t}\lVert u_0u_\infty\rVert_2$ is equivalent to a gap of size at least $\alpha$ between $\lambda_1$ and the rest of the spectrum. $\endgroup$– MaoWaoApr 20 at 6:44
1 Answer
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 LaplaceBeltrami operator on a complete manifold is essentially selfadjoint on $C_c^\infty$. Since $\lambda_1$ is the bottom of the spectrum of $\Delta$, the operator $(\Delta+\lambda_1)$ has a unique positive selfadjoint 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 $$ \lVert u_tu_\infty\rVert_{C^m(\Omega)}\lesssim_{m,\Omega}\lVert u_tu_\infty\rVert_{H^{2k}(\Omega)}\lesssim_{k,\Omega}\lVert A^k(u_tu_\infty)\rVert_{L^2(\Omega)}+\lVert u_tu_\infty\rVert_{L^2(\Omega)} $$ for every sufficiently nice bounded domain $\Omega$ in $M$.
Therefore, $u_t\to u_\infty$ in $C^m$ on compact subsets for every $m\in\mathbb N$.

$\begingroup$ 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. $\endgroup$ Apr 20 at 12:05

1$\begingroup$ 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. $\endgroup$– MaoWaoApr 20 at 12:24

$\begingroup$ So $\Delta\lambda_1$ is positive semidefinite, does it still have a unique extension? $\endgroup$ Apr 20 at 12:28

2$\begingroup$ Yes, essential selfadjointness is stable under additive bounded (selfadjoint) perturbations. $\endgroup$– MaoWaoApr 20 at 12:55

1$\begingroup$ 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. $\endgroup$ Apr 20 at 13:54