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When one considers the 1-D diffusion equation in cartesian coordinates

$$\frac{\partial u}{\partial t}=\chi\frac{\partial^2u}{\partial x^2},$$

one finds that the amplification factor for the Crank Nicolson scheme (with central differences in the spatial derivatives) is

$$A=\frac{1-2F\sin^2(k\Delta x/2)}{1+2F\sin^2(k\Delta x/2)},$$

where $F=\chi\Delta t/(\Delta x)^2$. Meaning that this method is unconditionally stable, i.e., the method is stable whatever $F$ is.

Now, let us consider the equation in cylindrical coordinates with a source term:

$$\frac{\partial u}{\partial t}=\chi\frac{1}{r}\frac{\partial}{\partial r}\left[r\frac{\partial u}{\partial r}\right]+S_{ext}$$

with boundary conditions $\partial_rT(r=0)=0$ and $T(r=1)=T_a$.

When $S_{ext}=0$, I get that the amplification factor is given by

$$A=\frac{1-\frac{F}{2}\left[4\sin^2\left(\frac{k\Delta r}{2}\right)-\frac{\sin(k\Delta r)}{q}\right]}{1+\frac{F}{2}\left[4\sin^2\left(\frac{k\Delta r}{2}\right)-\frac{\sin(k\Delta r)}{q}\right]}$$

where $q$ is the cell number. I am not sure if this is the correct result, namely it is strange to me the fact that it depends on $q$. Indeed, for small $q$ it is easy to see that $A$ may become larger than 1 for some of the wavenumbers $k$ and independently of $F$ .

Questions:

1. Does this mean that the method may become unstable?

2. When onde considers $S_{ext}\neq0$ the same result for $A$ applies?

3. And what about the accuracy in the $S_{ext}\neq0$ case? Because I have found that, with increasing $S_{ext}$, the deviation from the theoretical result increases a lot before the steady-state is reached.

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  • $\begingroup$ I am not sure what is your numerical scheme here. How do you deal with $r$ and $1/r$ in your scheme ? And what is $K_{ext}$ ? $\endgroup$
    – Héhéhé
    Commented Jul 19, 2019 at 23:11
  • $\begingroup$ I used the Crank Nicolson scheme. Regarding the $1/r$ at $r=0$, I make use of the boundary condition $\partial_ rT(r=0)=0$. Should I insert the discretized equation in my question? About the $K_{ext}$, that was a typo: I should have written $S_{ext}$. $\endgroup$
    – AJHC
    Commented Jul 20, 2019 at 8:47
  • $\begingroup$ Yeah I think you should insert it, because it is still not clear for me how you discretize $r$ and $1/r$. $\endgroup$
    – Héhéhé
    Commented Jul 20, 2019 at 9:30

1 Answer 1

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The Von Neumann stability analysis only works for linear PDEs with constant coefficients. It is not the case for your problem because of the $r$ and $1/r$ terms.

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  • $\begingroup$ And do you know what type of stability analysis could be done in this case? $\endgroup$
    – AJHC
    Commented Jul 20, 2019 at 10:02
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    $\begingroup$ I guess energy method could work, but I am not an expert. $\endgroup$
    – Héhéhé
    Commented Jul 20, 2019 at 11:01

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