For spatial accuracy in 2-D Laplace equation, a 9-point stencil is better than a 5-point one.

$$\partial_tq= r\left(\partial^2_x q + \partial^2_y q\right)$$

for FTCS (forward-time, central-space) scheme to converge for 5 point stencil, $r/2\ge \delta t/ \delta x^2$

What would be the condition for 9 point stencil?


1 Answer 1


There are actually a few different 9 point stencils in use, but they can all be written as a linear combination of the standard and skewed 5 point stencils. Performing the usual von Neumann analysis for a general Fourier mode, $e^{ikx+ily}$, produces an equation like $$a_{n+1} = a_n \left(1-\lambda+\frac{\alpha\lambda}{2}\left[\cos\theta_1+\cos\theta_2\right]+\lambda\frac{1-\alpha}{2}\left[\cos(\theta_1+\theta_2)+\cos(\theta_1-\theta_2)\right]\right)$$ where $\lambda=\frac{r\Delta t}{\Delta x^2}$. Hence the stability criterion is still $\lambda\leq \frac{1}{2}$, with the derivation and proof left as an exercise to the reader. This shouldn't be too much of a surprise, since for the "worst case" of a plane wave aligned with the coordinate system, both the 5 point and 9 point stencils collapse back down to the 1D case.


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