# Convergence of heat flow on non-compact manifolds?

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 $$M$$?

• eigenfunction/eigenvalue in what sense? $L^2$? Commented Apr 17, 2023 at 20:46
• yeah, that's right Commented Apr 17, 2023 at 20:46
• Convergence in which sense? $L^2$-convergence follows from the spectral theorem. Do you want something stronger? Commented Apr 19, 2023 at 9:34
• @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? Commented Apr 19, 2023 at 20:33
• @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. Commented Apr 20, 2023 at 6:44

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 $$m\in\mathbb N$$.

• 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. Commented Apr 20, 2023 at 12:05
• 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. Commented Apr 20, 2023 at 12:24
• So $-\Delta-\lambda_1$ is positive semi-definite, does it still have a unique extension? Commented Apr 20, 2023 at 12:28
• Yes, essential self-adjointness is stable under additive bounded (self-adjoint) perturbations. Commented Apr 20, 2023 at 12:55
• 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. Commented Apr 20, 2023 at 13:54