# von Neumann analysis: computation of maximum value of amplification factor

In this question I adress the stability property of a numerical scheme (a FDM described below) to the problem

\begin{align} u_{t} &= \alpha u_{xx}+\beta u_{xxxx}, & &x\in\mathbb{R},\;t>0, \\[3mm] u(x,0) &= \sin x, & &x\in\mathbb{R}, \\[3mm] u(x+2\pi,t)&=u(x,t), & &x\in\mathbb{R},\;t>0. \end{align}

where $$\beta \leq0$$ and $$\alpha >0$$. It can be shown for this setting that the problem is well-posed. I discretize the problem by a central finite difference in space and forward Euler in time:

$$\begin{eqnarray} u_{j}^{n+1} &=& \underbrace{\frac{\Delta t\beta}{\Delta x^{4}}}_{\equiv b_{2}}u_{j+2}(t) + \underbrace{\left( \frac{\Delta t\alpha}{\Delta x^{2}} - \frac{4\Delta t\beta}{\Delta x^{4}} \right)}_{\equiv b_{1}} u_{j+1}(t) \\ &+& \underbrace{\left(1 + \frac{\Delta t\beta}{\Delta x^{4}} -\frac{\Delta t\alpha}{\Delta x^{2}}\right)}_{\equiv b_{0}}u_{j}(t) + \underbrace{\left( \frac{\Delta t\alpha}{\Delta x^{2}} - \frac{4\Delta t\beta}{\Delta x^{4}} \right)}_{\equiv b_{-1}}u_{j-1}(t) \nonumber \\ &+& \underbrace{\frac{\Delta t\beta}{\Delta x^{4}}}_{\equiv b_{-2}} u_{j-2}(t) \nonumber. \end{eqnarray}$$

To determine stability of the scheme we want

$$\max_{k}\vert g_{k}\vert = \max_{k}\vert \sum_{l=-2}^{2}b_{l}\exp(\text{i}kl\Delta x)\vert \leq 1+\theta\Delta t,\quad \text{for some \theta>0}.$$

My problem is that I need a little help with the algebra, which gets messy. I obtain

\begin{align} &\vert g_{k}\vert = \left\vert b_{2}e^{2\text{i}k\Delta x} + b_{1}e^{\text{i}k\Delta x} + b_{0}e^{0\text{i}k\Delta x} + b_{-1}e^{-\text{i}k\Delta x} + b_{-2}e^{-2\text{i}k\Delta x} \right\vert = \\[3mm] &= \left\vert 2b_{2}\cos(2k\Delta x) + 2b_{1}\cos(k\Delta x) + b_{0} \right\vert. \end{align}

From here I am wondering, is there anymore simplifications I can make before taking an upper bound to this result? Can I simply continue by making

\begin{align} &\left\vert 2b_{2}\cos(2k\Delta x) + 2b_{1}\cos(k\Delta x) + b_{0} \right\vert \leq 2\vert b_{2}\vert\vert\cos(2k\Delta x)\vert + 2b_{1}\vert\cos(k\Delta x)\vert +\vert b_{0}\vert \leq \\[3mm] &\leq -2\frac{\Delta t\beta}{\Delta x^{4}}\vert\cos(2k\Delta x)\vert +2\left( \frac{\Delta t\alpha}{\Delta x^{2}} - \frac{4\Delta t\beta}{\Delta x^{4}} \right)\vert\cos(k\Delta x)\vert + 1 -\frac{\Delta t\beta}{\Delta x^{4}}+\frac{\Delta t\alpha}{\Delta x^{2}} \end{align}

and obtain a reasonably tight upper bound?