Apparently for many parabolic and elliptic PDEs the (ir-)regularity of initial data does not have any significant impact on the regularity of (weak) solutions.

Very often people when people talk about this fact they mention so-called smoothing property of Laplace operator term present in these equations. However, it seems to me that this expression is not a part of the formal PDE & analysis terminology, but rather something people tend to use informally.

Is there an actual definition of the "smoothing property of second-order derivative term"?

What is the (physical?) mechanism behind this elusive property?

I would be glad if someone could provide an intuitive and/or visual justification of this effect.


1 Answer 1


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.


  1. Laplacian is a diagonal operator in the basis of complex exponentials, which gives large negative coefficients to high-frequency harmonics.
  2. Thus, evolving a function in the direction of Laplacian reduces said high-frequency harmonics, smoothing the functions.

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