Does anybody know how the derivatives of the coefficient functions in equation (4) of the paper "High-Order Difference Schemes for Two-Dimensional Elliptic Equations" by M. M. Gupta, R. P. Manohar and J. W. Stephenson are discretised? Assuming that the nine-point stencil should be kept, second-order central differences are a "natural" choice, I assume. But how can this lead to a fourth-order accurate scheme?
Thanks for your answers.
Article High-Order Difference Schemes for Two-Dimensional Elliptic Equations
Thank you for your answer. The reason why I was particularly interested in the initially cited paper is that the stencil they use contains only 3x3 points (in 2D) for fourth order accuracy. I am not so much interested in high order, but more in the stability of the scheme when the convection-type terms dominate. Looking at the schemes you mention, I understood that their stencil is much wider?
The key idea follows from looking at the standard three-point finite difference approximation for the f'' and noting that the error term is proportional to h^2 f'''' , where h is the step size. One then applies a second-order finite difference approximation to evaluate f'''' , thus yielding a fourth-order error.
The wikipedia article on Numerov's method shows the derivation in one dimension, http://en.wikipedia.org/wiki/Numerov's_method , and from that you can get an ide of how the stencil is constructed in two or more dimensions.
Thank you both very much for your answers. I think that I understood now how this method works due to your remarks and the link.
It's true that for the method from Gupta et al., also the coefficients corresponding to (i \pm 1, j \pm 1) are non-zero. But in my case this is not a problem, since these points are included anyway for the cross derivatives. Extending the stencil beyond one neighbouring point, on the other side, would imply heavy changes in the implementation.
- Badges
- Science topic
- Similar topics
- Linguistics
- Composition Studies
- Publication