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.
Best Answer
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:
$$ \frac{1}{h} \leq \frac{1}{2} $$