The non-linear Poisson equation in one-dimension,

$$ 0 = \frac{\partial^2u}{\partial x^2} - f(u) $$

can be discretised as to give,

$$ u_{j-1} -2u_{j} + u_{j+1} = h^2 f(u_j) $$

where $h$ is the step size of the mesh.

Is there any advantage (in general or not) to write the non-linear source term as the average value of the solution variable $\bar{u_j}$ over the neighbouring mesh points?

For example,

$$f(u_j) \rightarrow f(\bar{u_j})$$


$$ \bar{u_j} = \frac{1}{2}\left( u_{j-1} + u_{j+1} \right) $$

I have noticed in practice that this sometimes improves the solution stability when using relaxation methods.

  • $\begingroup$ The gain in stability is probably due to the smoothing out of high frequencies. $\endgroup$ – Jan Feb 5 '14 at 8:50
  • $\begingroup$ Yes, high frequency components was (still is!) the main problem I was seeing. $\endgroup$ – boyfarrell Feb 5 '14 at 9:05

For your example equation, taking the average approach, the local consistency error $$ \frac{1}{h^2}[u(x-h) - 2 u(x) + u(x+h)]-f(\frac{1}{2}[u(x-h)+u(x+h)]) = \frac{1}{2}f_uu_{xx}h^2 + hot. $$ will be of order $2$ (instead of order $3$). ($hot.$ means higher order terms)

Therefore, if your overall approximation is of order $1$, e.g. if you use upwind somewhere, then it doesn't matter.

If you use higher order schemes, then this approach will limit of your convergence rate.

| cite | improve this answer | |
  • $\begingroup$ So to keep the local consistency error as low as possible I should stick with the original approach (not to average)? $\endgroup$ – boyfarrell Feb 5 '14 at 9:07
  • $\begingroup$ Yes, I would do so. The main thing is to be consistent. If you run into stability issues, then you may think of such stabilizations. $\endgroup$ – Jan Feb 5 '14 at 9:17

If you look at it from a Galerkin perspective, the right side could also be approximated by


the Kepler formula, I believe. This would be valid for first sampling $f(u)$ and then approximating by a piecewise linear function. If the sampled $u$ is approximated piecewise linearly, then the non-linearity of $f$ will destroy this property, giving no general rule.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.