I simply want to solve the elliptic equation: $$ -\kappa \nabla^2 u + u = f $$ where $f\in [0,1]$. When using continuous Galerkin with Lagrange elements, I have noticed that $\kappa$ has to be greater than a certain value related to the mesh resolution to avoid values of $u$ under 0 or over 1. I would like to avoid these over/undershoots without having to increase the mesh resolution. I was advised to solve this equation using finite volume, which solved the problem. However, its implementation with non-cartesian second-order meshes is onerous. Therefore I decided to switch to stabilized methods: $$ (k\nabla u, \nabla v) + (u,v) + (\alpha h^2 \nabla u, \nabla v) = (f,v) $$ This seems to work given that I am artificially increasing the diffusion coefficient which will vanish with refinement. However, I haven't been able to find literature on appropriate values for $\alpha$. Could anybody give me some pointers? Thanks.

  • $\begingroup$ What kind of BC's and load $f$ you are using? In the limit $\kappa \rightarrow 0$ the solution is in $L^2$ so when $\kappa$ is very small then the solution can have large changes very quickly and this causes the overshoots when approximated with $H^1$ conforming elements. Whether these changes exists or not depends on your loading and BC's. $\endgroup$
    – knl
    Sep 9, 2018 at 18:11
  • $\begingroup$ I have zero Neuman boundary conditions everywhere. $f$ is usually a function between 0 and 1, it can be discontinuous as well. $\endgroup$
    – balborian
    Sep 9, 2018 at 19:21
  • $\begingroup$ Can't you solve the equation assuming $\kappa=1$ and then just the solution? $\endgroup$
    – nicoguaro
    Sep 10, 2018 at 2:59
  • $\begingroup$ Could you be more specific? $\endgroup$
    – balborian
    Sep 10, 2018 at 16:25
  • $\begingroup$ The equation that you are solving is linear. So you can write it in nondimensional form, withouth $\kappa$ appearing on it. $\endgroup$
    – nicoguaro
    Sep 11, 2018 at 1:28

1 Answer 1


The problem you want to solve, with a small $\kappa$ is "singularly perturbed", i.e., it generally has boundary and internal layers. The way you can think of it is that if $\kappa$ were zero, then you'd get $u=f$, and for $\kappa>0$, you get a smoothed version of $f$ with a smoothing length scale $L$ that is related to $\kappa$ in some way -- I believe that it is $L\propto\sqrt{\kappa}$.

This works well if you have that the mesh size $h$ is on the order of or smaller than $L$, but if $h>L$, then you are effectively trying to resolve a discontinuous function. For these kinds of functions, you will get Gibbs' phenomenon unless you use a scheme that either (i) is nonlinear or (ii) has sufficient artificial diffusion so that the length scale that results from the physical diffusion $\kappa$ plus the artificial diffusion is at least on the same order as the mesh size.

Both finite volume methods (which are basically piecewise constant "discontinuous Galerkin" schemes) and the addition of $\alpha h^2$ to $\kappa$ fall into the second class. An alternative is to use something like a nonlinear shock capturing method, though this makes the problem substantially more complicated to solve.

  • $\begingroup$ Thanks, but I am trying to find a relation between $\alpha$ and $\kappa$ to ensure the solution is between 0 and 1. I have done more research and $\alpha$ has to be such that the Discrete Maximum Principle is satisfied. $\endgroup$
    – balborian
    Sep 11, 2018 at 16:42
  • 1
    $\begingroup$ Based on what I wrote, you have to satisfy that $\sqrt{\kappa + \alpha h^2}$ is approximately equal to (a multiple of) $h$. So an $\alpha$ on the order of 1 or 10 will likely work, though $\alpha = \max\left\{\frac{ch^2-\kappa}{h^2},0\right\}$ is probably the better choice with $c$ in the range of 1 to 10. $\endgroup$ Sep 11, 2018 at 19:35
  • $\begingroup$ Do you have resources to back your claim of the length scale $L$ being proportional to $\sqrt{\kappa}$? $\endgroup$
    – balborian
    Sep 14, 2018 at 23:05
  • $\begingroup$ Search for "singularly perturbed elliptic problem". I'm pretty sure I got this right, but you may want to check. $\endgroup$ Sep 17, 2018 at 2:56

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