$$ u_{j+1,m} = \frac{1}{2}\sigma(\sigma-1)u_{j,m+1}-(\sigma^1-1)u_{j,m}+\frac{1}{2}\sigma(\sigma+1)u_{j,m-1}\\ \sigma = \frac{c\Delta t}{\Delta x}\\ u_t+cu_x = 0 $$ I need find the Neumann Stability of the above scheme. My work is below:
$$ u_{j+1,m} = \lambda e^{ikx_m} \hspace{5mm} u_{j,m+1} = e^{ik(x_m+\Delta x)} \hspace{5mm} u_{j,m} = e^{ikx_m} \hspace{5mm} u_{j,m-1} = e^{ik(x_m-\Delta x)}\\ \lambda e^{ikx_m} = \frac{1}{2}\sigma(\sigma-1)e^{ik(x_m+\Delta x)}-(\sigma^1-1)e^{ikx_m}+\frac{1}{2}\sigma(\sigma+1)e^{ik(x_m-\Delta x)}\\ \lambda = \sigma^2\cos(k\Delta x)-i\sigma\sin(k\Delta x)+1-\sigma^2\\ \left|\lambda\right| \leq 1 \implies \left|\lambda\right|^2 \leq 1\\ \left|\lambda\right|^2 = (1-\sigma^2(1-\cos(k\Delta x)))^2+\sigma^2\sin^2(k\Delta x) $$
I know the above work is correct but I have been struggling to get past this point to find the correct solution.
$$ \left|\lambda\right|^2 = 2\sigma^2\cos(k\Delta x) -2\sigma^4\cos(k\Delta x)+\sigma^4\cos^2(k\Delta x) -2\sigma^2+\sigma^4+1+\sigma^2\sin(k\Delta x)\leq 1 $$ I know $$ -1 \leq \cos(k\Delta x) \leq 1 \hspace{5mm} \text{and} \hspace{5mm} 0 \leq \cos^2(k\Delta x) \leq 1 $$
Applying these conditions to the equation for $\left|\lambda\right|^2$ that I found gives me the final solution of a constant. This would show unconditional stability, but the correct answer should be conditional stability. The correct answer should also be the same as the CFL condition. Can someone please help show me where I keep going wrong?
So let's start with $$ \rho \equiv |\lambda|^2 = (1 - \sigma^2 + \sigma^2 \cos \alpha)^2 + \sigma^2 \sin^2 \alpha. $$ I've denoted $\alpha = k \Delta x$. First, let's introduce $\mu = \cos \alpha$ and get rid of $\sin \alpha$: $$ \rho = (1 - \sigma^2 + \sigma^2 \mu)^2 + \sigma^2 (1 - \mu^2). $$ When $\alpha$ runs for all possible values, $\mu$ runs only from $-1$ to $1$. So basically we need to find for which $\sigma$ the following inequality holds $$ \max_{\mu \in [-1, 1]} \rho \leq 1. \tag{*} $$ Observe that $\rho$ is a quadratic polynomial in $\mu$: $$ \rho(\mu) = (\sigma^2 - 1) \sigma^2 \mu^2 + 2(1-\sigma^2) \sigma^2 \mu + (1 - \sigma^2)^2 + \sigma^2. $$ so we could find the maximum pretty easy. The maximum could be attained only at $\mu = \pm 1$ or at the root of $\frac{d\rho}{d\mu} = 0$ if it lies in $\mu \in [-1, 1]$. $$ \rho(\mu = 1) = 1, \qquad \rho(\mu = -1) = (1 - 2\sigma^2)^2. $$ It is apparent now that $|\sigma| > 1$ violates (*), since $1 < \rho(\mu = -1) \leq \max_{\mu \in [-1, 1]} \rho(\mu)$. So the scheme is unstable when $|\sigma| > 1$.
The case $\sigma = \pm 1$ is also trivial: $\rho(\mu) = 1$. The scheme is stable.
Now let's consider $|\sigma| < 1$. Since $(\sigma^2 - 1) \sigma^2 < 0$ the $\rho(\mu)$ is a concave function. It's extremum is a maximum. It is located at the root of $\rho'(\mu) = 0$ $$ 2(\sigma^2 - 1)\sigma^2 \mu + 2 (1 - \sigma^2) \sigma^2 = 0. $$ The maximum always is at $\mu = 1$, so $$ \max_{\mu \in [-1, 1]} \rho(\mu) = \rho(\mu = 1) = 1. $$ Summarizing, the scheme is stable for $|\sigma| \leq 1$.
Here's some plots of $\rho(\mu)$ for different $\sigma$: