Let us look at solutions to the linear heat equation on $\mathbb{R}$: $$ u_t = u_{xx}.$$ Is it true that solutions to the equation with nice enough initial datum are analytic after a certain time $T >0$?
I know that, if one starts with the datum $$ u_0(x) = \frac 1 {x^2 +1}, $$ the resulting solution is not analytic in a neighbourhood of $(0,0)$ (this example is due to Kowalevskaya).
Let us suppose for simplicity that the initial datum $u_0$ is smooth and of compact support. By the representation formula for solutions, we know that $$ u(x,t) = \frac C{\sqrt t} \int_{\mathbb{R}} \exp\left(- \frac{(x-y)^2}{4t} \right)u_0(y) dy, $$ for some positive constant $C$. It looks like maybe one could do a series expansion and get analyticity from this representation formula.
In case the assertion is false, could you provide an example in which analyticity fails after every positive time?
Any help will be appreciated.
Yes, for initial data with compact support the solution is real analytic in the half-plane $\{(x,t):t>0\}$.
One approach is to consider a function of two complex variables $z=x+iy$, $\zeta = t+is$: $$u(z,\zeta ) = \frac C{\sqrt{t+is}} \int_{\mathbb{R}} \exp\left(- \frac{(x+iy-w)^2}{4(t+is)} \right)u_0(w) dw,$$ The integrand is holomorphic with respect to $z$ and $\zeta$ as long as $t>0$. Integration preserves holomorphicity in this setup (integrating against a finite measure, everything clearly converges). Hence, the function is locally represented by a convergent power series.
Working with power series expansion would be mostly reproving that theorem about holomorphic integrals.
A different approach is to consider the Fourier transform with respect to $x$: it is $\hat u_0$ multiplied by a Gaussian. One can then figure out what $u_0$ needs to be in order for the Paley–Wiener theorem to apply. This should give a result for some $u_0$ that are not compactly supported; on the other hand, this is only analyticity with respect to $x$.