Say you have some PDE for a function $f:[0,T]\times \mathbb{R}^n\to\mathbb C$

$$\frac{\partial f}{\partial t}=Lf,$$

for some differential operator $L$. If $L$ contains a Laplacian term $$\Delta f=\sum_{i=1}^n \frac{\partial^2 f}{\partial x_i^2},$$

why do people claim (as a rule of thumb) that this helps "smooth the solution"? From my understanding they are saying that the solution of the PDE will have better differentiability, but I am unsure why the Laplacian has this effect.

  • 3
    $\begingroup$ The concept that helps "smooth solutions" is the notion of elliptic differential operators. The Laplacian is such an operator. For these operators, there is the so-called "elliptic regularity theorem", which says how smooth the solution to the equation must be. $\endgroup$
    – C_M
    Aug 10, 2022 at 16:31
  • 2
    $\begingroup$ The expression for the Laplacian should be with second partial derivatives. It's a good question though. $\endgroup$
    – SBK
    Aug 10, 2022 at 16:47

1 Answer 1


The Laplacian itself does not smooth the solution. What smooths the solution are the dynamics $${d f\over dt} = |\kappa| \nabla f + \text{other stuff}$$ The intuitive reason is exactly the opposite: the Laplacian amplifies regions with large curvature as compared to regions that are relatively flat. The time derivative "pushes" $f$ more in regions where $f$ is rough, and the signs are such that the pushing is towards a smoother $f(x,t)$. If instead the equation was $${d f\over dt} = -|\kappa| \nabla f + \text{stuff}$$ The Laplacian would cause large curvature regions to get even rougher.


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