Not every appearance of Laplacian in a PDE yields smoothing effect. For example, in the backwards heat equation $u_t=-\Delta u$ it has the opposite effect of reducing the smoothness of solution as time increases. And in the Scrödinger equation $iu_t =\Delta u$ the smoothness of $u$ neither improves nor gets worse.
One way of seeing this is to imagine $u$ decomposed into harmonics, $$u(x,t)=\sum_n A_n(t)e^{i \omega_n x}$$
The smoothness of $u$ with respect to $x$ is measured by how quickly $|A_n(t)|$ decays as $n\to\infty$. (E.g., $|A_n(t)|=O(n^{-k-1-\epsilon})$ gives you $k$ continuous derivatives, because the series remains uniformly convergent after being differentiated $k$ times.)
Example. For the heat equation $u_t=\Delta u$ we have
$$ \sum_n A_n'(t)e^{i \omega_n x} = - \sum_n \omega_n^2 A_n (t)e^{i \omega_n x}$$
hence $A_n(t) = A_n(0)\exp(-\omega_n^2 t)$. Since $|\omega_n|$ is typically $\Omega(n^\alpha)$ for some $\alpha>0$, it follows that for every $t>0$ the decay of $A_n(t)$ is very fast (superexponential), which implies $u$ is infinitely differentiable in $x$.
Repeating the above for $iu_t =\Delta u$ you will see that no decay of $A_n(t)$ takes place, and therefore no gain of smoothness is to be expected.
Summary:
- Laplacian is a diagonal operator in the basis of complex exponentials, which gives large negative coefficients to high-frequency harmonics.
- Thus, evolving a function in the direction of Laplacian reduces said high-frequency harmonics, smoothing the functions.
