Do certain PDEs guarantee analytic solutions at $t>0$ even if the initial condition at $t=0$ is not analytic?

Question

Are there theorems that guarantee that applying a PDE on a function that is not necessarily analytic at time $t=0$, creates an analytic function at $t>0$? For example, consider the PDE $$ \frac{\partial f}{\partial t} = \frac{\triangle f}{2}, $$ where $f=f(x,t)$, $f\colon \mathbb{R}^n\times \mathbb{R}\to \mathbb{R}$, $f$ is integrable as a function of $x$ for any fixed $t$ and $\triangle f$ is its Laplacian. The solution at $t=1$ is $$ f(x,1) = \int f(a,0)\frac{1}{\sqrt{2\pi}}\exp(-(x-a)^2/2)da $$ To see this, $f(x,t)$ can be viewed as the density of the diffusion process $dx_t = dB_t$ with the initial condition that $x_0$ has density $f(x,0)$. Notice that $f$ is analytic. If we take the PDE that corresponds to the gradient of the logarithm of $f$, call it $s$, we get the following PDE: $$ \frac{\partial s}{\partial t} = J_s s + \frac{\triangle s}{2} $$ where $J_s$ is the Jacobian of $s$ and $\triangle s$ is a vector whose $i$'th entry is the Laplacian of that $i$th entry of $s$. Notice that if $s$ is the gradient of the logarithm of some integrable function $f$ then the solution of $s$ at $t=1$ is also analytic: indeed, one could obtain its formula by differentiating the logarithm of the the solution for $f$ and see that it is analytic. It seems plausible that $s(x,1)$ is analytic also if $s$ is not the gradient of some function, however, I am not sure how to prove this.