# Real analyticity of solution of heat equation

Consider the heat equation $\partial_t u - \Delta u = 0, u(0, x) = u_0$ on a complete (non-compact) Riemannian manifold $M$, may be even $\mathbb{R}^n$. I was wondering, what are some known sufficient conditions under which the solution to the heat equation becomes instantaneously real analytic in time? This is mainly a reference request: the only reference I know is K. Yosida's paper An Abstract Analyticity in Time for Solutions of a Diffusion Equation'' (available here). Also, with regard to the main result of Yosida's paper, I have the further question: does one need to have a spatial bound on the solution (like $L^\infty$-bound on $u$, or $u \in L^2$, as in Yosida)? The reason I am asking about $L^\infty$ bound on $u$ is, in Tychonoff's example, (see this MO post), $u$ grows rapidly at infinity, and the solution constructed turns out to be not real analytic.

• I guess the manifold $M$ had better be real analytic (with real analytic transition functions); otherwise your question is not even well defined. – Fan Zheng Oct 4 '15 at 0:01
• @FanZheng Aren't all smooth manifolds real analytic with the two structures being compatible? I thought this was a result of Whitney. – HSM Oct 4 '15 at 12:00
• @FanZheng Besides, even if they weren't, I don't see clearly how that affects this issue. I only want real analyticity with respect to the variable $t$. May be I am not thinking clearly. – HSM Oct 4 '15 at 12:03

In the case of $\mathbb R^n$, the analyticity-in-time of the solution, with $u_0\in\mathcal S'$ (tempered distributions), stems from that of $e^{-(t+i\tau)|\xi|^2}$ which implies that of $u(t+i\tau,x)=\mathcal F^{-1}_{\xi\to x}[\hat u_0(\xi)e^{-(t+i\tau)|\xi|^2}]$ as a function of the complex variable $t+i\tau$, $t>0$.
Consider the closed quadratic form $a(u):=\int_M |\nabla u|^2 \ d\sigma$ for $u\in H^1(M)$. The associated operator is $\Delta$ and it generates an analytic semigroup by the general theory of linear semigroups. Voilà.
• The point is that on e.g. $\mathbb{R}^n$ the Tychonoff solution exists, and that is not analytic in time. So it is not quite as straightforward as "voila". – Glen Wheeler Sep 13 at 0:22