Von Neumann Analysis
Von Neumann analysis is used to verify the stability of a finite-difference scheme. We will only consider one time dimension, but any number of spatial dimensions.
The procedure, in principle, is to perform a spatial Fourier transform along all spatial dimensions, thereby reducing the finite-difference scheme to a time recursion in terms of the spatial Fourier transform of the system. The system is then stable if this time recursion is at least marginally stable as a digital filter.
Let's apply von Neumann analysis to the finite-difference scheme for
the ideal vibrating string Eq.(D.3):
![$\displaystyle y_{n+1,m}= y_{n,m+1}+ y_{n,m-1}- y_{n-1,m} \protect$](http://www.dsprelated.com/josimages_new/pasp/img4453.png)
![$ m$](http://www.dsprelated.com/josimages_new/pasp/img6.png)
where
![$ k=2\pi/\lambda$](http://www.dsprelated.com/josimages_new/pasp/img3203.png)
![$ m$](http://www.dsprelated.com/josimages_new/pasp/img6.png)
![$ Y_n(k)e^{jkX}$](http://www.dsprelated.com/josimages_new/pasp/img4493.png)
![$ y(n,m)$](http://www.dsprelated.com/josimages_new/pasp/img4494.png)
![$ z$](http://www.dsprelated.com/josimages_new/pasp/img76.png)
A method equivalent to checking the pole radii, and typically used
when the time recursion is first order, is to compute the
amplification factor as the complex gain in
the relation
![$\displaystyle Y_{n+1}(k) = G(k)Y_n(k).
$](http://www.dsprelated.com/josimages_new/pasp/img4496.png)
![$ \vert G(k)\vert\leq 1$](http://www.dsprelated.com/josimages_new/pasp/img4497.png)
![$ k$](http://www.dsprelated.com/josimages_new/pasp/img89.png)
Since the finite-difference scheme of the ideal vibrating string is so
simple, let's find the two poles. Taking the z transform of Eq.(D.8)
yields
![$\displaystyle zY(z,k) = 2c_k Y(z,k) - z^{-1}Y(z,k)
$](http://www.dsprelated.com/josimages_new/pasp/img4498.png)
![$\displaystyle z^2 - 2c_k z - 1 = 0
$](http://www.dsprelated.com/josimages_new/pasp/img4499.png)
![$\displaystyle z = c_k \pm \sqrt{c_k^2 - 1}.
$](http://www.dsprelated.com/josimages_new/pasp/img4500.png)
![$\displaystyle \left\vert z\right\vert^2 = c_k^2 \pm (c_k^2 - 1) =
\left\{\begi...
...eq 1 \\ [5pt]
[1,1], & \left\vert c_k\right\vert\leq 1 \\
\end{array}\right..
$](http://www.dsprelated.com/josimages_new/pasp/img4501.png)
![$ \left\vert c_k\right\vert\leq 1$](http://www.dsprelated.com/josimages_new/pasp/img4502.png)
![$\displaystyle z = c_k \pm j\sqrt{1-c_k^2} = \cos(kX) \pm j\sin(kX) = e^{\pm jkX}.
$](http://www.dsprelated.com/josimages_new/pasp/img4503.png)
![$ k\in[-\pi/X,\pi/X]$](http://www.dsprelated.com/josimages_new/pasp/img4504.png)
![$ \vert c_k\vert\leq1$](http://www.dsprelated.com/josimages_new/pasp/img4505.png)
![$ k$](http://www.dsprelated.com/josimages_new/pasp/img89.png)
![$ \,$](http://www.dsprelated.com/josimages_new/pasp/img196.png)
In summary, von Neumann analysis verifies that no spatial Fourier components in the system are growing exponentially with respect to time.
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