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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.

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I will just point out that the problem you specify is exactly the same as the transient 1D heat equation only with different letters. In the equation you are analyzing just switch c with T, x with t and y with x and you will get the 1D heat equation.

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  • $\begingroup$ But can we actually do it knowing that $x$ is rather discretized in space not in time, i.e. in the problem itself $x$ is space not time. Furthermore, look at the discretization scheme, you can clearly see that it is completely different from the one you'd obtain by discretizing 1D heat equation because you'd have time and space there, while here we have space and space again, hence the discretized scheme I've posted. As a result, I suspect that you are wrong. $\endgroup$ – Alexander Shukaev Apr 14 '13 at 23:49
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    $\begingroup$ The mathematical analysis doesn't care about the words "space" and "time". They are all just variables of differentiation. $\endgroup$ – Bill Barth Apr 15 '13 at 0:03

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