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):

$$y_{n+1,m}= y_{n,m+1}+ y_{n,m-1}- y_{n-1,m} \protect$$

There is only one spatial dimension, so we only need a single 1D Discrete Time Fourier Transform (DTFT) along $m$ [454]. Using the shift theorem for the DTFT, we obtain

$$Y_{n+1}(k) = (e^{jkX} + e^{-jkX})Y_n(k) - Y_{n-1}(k)$$

$$= 2\cos(kX)Y_n(k) - Y_{n-1}(k)$$

$$\isdef 2c_kY_n(k) - Y_{n-1}(k) \protect$$ (D.8)

where $k=2\pi/\lambda$ denotes radian spatial frequency (wavenumber). (On a more elementary level, the DTFT along $m$ can be carried out by substituting $Y_n(k)e^{jkX}$ for $y(n,m)$ in the finite-difference scheme.) This is now a second-order difference equation (digital filter) that needs its stability checked. This can be accomplished most easily using the Durbin recursion [452], or we can check that the poles of the recursion do not lie outside the unit circle in the $z$ plane.

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 $G(k)$ in the relation

$$Y_{n+1}(k) = G(k)Y_n(k).$$

The finite-difference scheme is then declared stable if $\vert G(k)\vert\leq 1$ for all spatial frequencies $k$ .

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

$$zY(z,k) = 2c_k Y(z,k) - z^{-1}Y(z,k)$$

yielding the following characteristic polynomial:

$$z^2 - 2c_k z - 1 = 0$$

Applying the quadratic formula to find the roots yields

$$z = c_k \pm \sqrt{c_k^2 - 1}.$$

The squared pole moduli are then given by

$$\left\vert z\right\vert^2 = c_k^2 \pm (c_k^2 - 1) = \left\{\begin{array}{ll} [2c_k^2-1, 1], & \left\vert c_k\right\vert\geq 1 \\ [5pt] [1,1], & \left\vert c_k\right\vert\leq 1 \\ \end{array} \right..$$

Thus, for marginal stability, we require $\left\vert c_k\right\vert\leq 1$ , and the poles become

$$z = c_k \pm j\sqrt{1-c_k^2} = \cos(kX) \pm j\sin(kX) = e^{\pm jkX}.$$

Since the range of spatial frequencies is $k\in[-\pi/X,\pi/X]$ , we spontaneously have $\vert c_k\vert\leq1$ for all $k$ . Therefore, we have shown by von Neumann analysis that the finite-difference scheme Eq.(D.3) for the ideal vibrating string is stable.

In summary, von Neumann analysis verifies that no spatial Fourier components in the system are growing exponentially with respect to time.