Problem 6.5
Let us start with the scheme given in Eqns 6.46-6.47:
\[\begin{align*}
v_j^{n+1} &= \frac{1}{2} (v_{j+1}^n + v^n_{j-1}) + \frac{\Delta t}{\rho_j} \frac{\sigma^n_{j+1} - \sigma^n_{j-1}}{2\Delta x} \\
\sigma_j^{n+1} &= \frac{1}{2} (\sigma_{j+1}^n + \sigma^n_{j-1}) + \Delta t \, \mu_j \frac{v^n_{j+1} - v^n_{j-1}}{2\Delta x}
\end{align*}\]
Now we assume a periodic solution as in 6.34:
\[\begin{align*}
v^n_j &= v_0 A^n e^{ikj \Delta x} \\
\sigma^n_j &= \sigma_0 A^n e^{ikj \Delta x}
\end{align*}\]
We can write these conditions as
\[\begin{align*}
v_0 A^{n+1} e^{ikj \Delta x} &= \frac{1}{2} ( v_0 A^n e^{ik(j+1) \Delta x} + v_0 A^n e^{ik(j-1)\Delta x} ) \\
&+ \frac{\Delta t}{\rho_j} \frac{ \sigma_0 A^n e^{ik(j+1) \Delta x} - \sigma_0 A^n e^{ik(j-1) \Delta x} }{2\Delta x} \\
\sigma_0 A^{n+1} e^{ikj \Delta x} &= \frac{1}{2} (\sigma_0 A^n e^{ik(j+1) \Delta x} + \sigma_0 A^n e^{ik(j-1) \Delta x}) \\
&+ \Delta t \, \mu_j \frac{v_0 A^n e^{ik(j+1) \Delta x} - v_0 A^n e^{ik(j-1) \Delta x} }{2\Delta x}
\end{align*}\]
Let us divide by \(A^n \, e^{ikj\Delta x} \)
\[\begin{align*}
v_0 A = \frac{1}{2} ( v_0 e^{ik \Delta x} + v_0 e^{-ik\Delta x} ) + \frac{\Delta t}{\rho_j} \frac{ \sigma_0 e^{ik \Delta x} - \sigma_0 e^{- ik \Delta x} }{2\Delta x} \\
\sigma_0 A = \frac{1}{2} (\sigma_0 e^{ik \Delta x} + \sigma_0 e^{- ik \Delta x}) + \Delta t \, \mu_j \frac{v_0 e^{ik \Delta x} - v_0 e^{- ik \Delta x} }{2\Delta x}
\end{align*}\]
We can then use the identity that
\[\begin{align*}
\frac{1}{2} \left( e^{i \phi} + e^{- i \phi} \right) &= \cos(\phi) \\
\frac{1}{2} \left( e^{i \phi} - e^{- i \phi} \right) &= i\, \sin(\phi)
\end{align*}\]
to simplify this to
\[\begin{align*}
v_0 A &= v_0 \cos( k\Delta x) + \frac{i \Delta t}{\rho_j \, \Delta x} \sigma_0 \sin (k \Delta x) \\
\sigma_0 A &= \sigma_0 \cos (k \Delta x) + \frac{i \Delta t \mu_j}{\Delta x} v_0 \sin (k \Delta x)
\end{align*}\]
This set of linear equations may be written as, for a given \(\mu_j = \mu\) and \(\rho_j = \rho\):
\[\begin{equation*}
\begin{bmatrix}
\cos( k\Delta x) - A && \frac{i \Delta t}{\rho \, \Delta x} \sin (k \Delta x) \\
\frac{i \Delta t \mu}{\Delta x} \sin (k \Delta x) && \cos (k \Delta x) - A
\end{bmatrix} \begin{bmatrix}v_0 \\ \sigma_0 \end{bmatrix} = 0
\end{equation*}\]
For this system of equations to have solutions, the determinant of the \(2 \times 2\) matrix must be zero:
\[\begin{equation*}
\cos^2(k \Delta x) - 2 A \cos(k\Delta x) + A^2 - \frac{i^2 (\Delta t)^2 \, \mu}{\rho (\Delta x)^2} \sin^2(k\Delta x) = 0
\end{equation*}\]
Substituting the Courant number squared
\[\begin{equation*}
C^2 = \frac{\mu (\Delta t)^2}{\rho (\Delta x)^2}
\end{equation*}\]
and the fact that \(i^2 = -1\) we get
\[\begin{equation*}
\cos^2(k \Delta x) - 2 A \cos(k\Delta x) + A^2 + C^2 \sin^2(k\Delta x) = 0
\end{equation*}\]
This is a quadratic equation in \(A\) such that solutions can be found with a quadratic formula:
\[\begin{equation*}
A^2 - 2 A \cos(k\Delta x) + \cos^2(k \Delta x) + C^2 \sin^2(k\Delta x) = 0
\end{equation*}\]
such that the solution for \(A\) is
\[\begin{equation*}
A = \frac{2\,\cos(k\Delta x) \pm \sqrt{ 4 \cos^2(k\Delta x) - 4(\cos^2(k \Delta x) + C^2 \sin^2(k\Delta x)) }}{2}
\end{equation*}\]
which reduces to
\[\begin{equation*}
A = \frac{2\,\cos(k\Delta x) \pm \sqrt{ -4\, C^2 \sin^2(k\Delta x) }}{2}
\end{equation*}\]
and therefore to
\[\begin{equation*}
A = \cos(k\Delta x) \pm i\, C \sin(k\Delta x)
\end{equation*}\]
We require A to be less than or equal to 1, so that the \(A^n\) component of the periodic velocity or stress does not grow exponentially with the timestep. Evidently the real component of \(A\), \(\cos(k\Delta x)\), is bounded between [-1, 1] due to the range of the cosine function. Since the sine function has a similar range, it is clear that \( C \le 1\) such that the imaginary component of \(A\) remains bounded for any positive \(A^n\).
