I came across the following task recently:

Use the von-Neumann stability analysis to investigate the stability of the discrete form of $\frac{\partial c}{\partial x} = \frac{\partial^2 c}{\partial y^2}$. Use the first-order forward finite difference for the first-order derivative and the usual central difference scheme for the second-order derivative. You can use the notation $c_{i, j} = c(ih, jh)$. The corresponding mesh size $h$ is same in both $x$- and $y$-direction. Which restriction arises for mesh size $h$?

Hint: the vector $f_k(jh) = \sin(k \pi x)$, $j = 0, \dots, N$, is an eigenvector of finite difference-expression for the second-order derivative. The corresponding eigenvalue is given by $\lambda_k = \frac{2}{h^2}(\cos(\pi kh) - 1)$.

I'm completely confused. I've seen classical example when Von Neumann Stability Analysis is applied to 1D heat equation $\frac{\partial T}{\partial t} = \frac{\partial^2 T}{\partial x^2}$, and it was pretty straightforward. However, this task asks to apply this analysis to stationary problem, therefore I'm not sure how to define amplification factor. Should it be simply $1$? Secondly, now it is 2D, and I don't know how to incorporate this fact into the method. Finally, I'm confused by the hint: I understand neither what it states nor how to utilize it.

All I can currently do is discretize it:

$$ 2c_{i + 1, j} = c_{i, j - 1} + c_{i, j + 1} $$

So how do I perform the analysis properly in this case? Appreciate your help.


For your equation bellow you can make the variable change $ t = x $

$$ \frac{\partial c}{\partial x} = \frac{\partial^2 c}{\partial y^2}$$

That will turn your equation in the simple, already mention by you heat equation) $$ \frac{\partial c}{\partial t} = \alpha \frac{\partial^2 c}{\partial y^2}$$ with $ \alpha = 1 $.

So looking in the solution already posted in Wikipedia, you can go straight to the restriction in your case $ \Delta t = \Delta y = \Delta x = h $ that means the restriction found :

$$ \frac{\alpha \Delta t}{\Delta x ^2} \leq \frac{1}{2} $$

turns in your case to be:

$$ \frac{1}{h} \leq \frac{1}{2} $$


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.