Von Neumann stability for convection with Euler forward

[liquidFuzz]

I'm trying to refresh some numerical science stuff. Von Neumann analysis, if I take a slimmed down equation, convection. $$\frac{∂u}{∂t}+a ∇ u =0$$ If I'm using Euler forward, $$\frac{u^{n+1}-u^n}{\Delta t}+\frac{a}{2h} \left( u_{j+1}^n -u_{j-1}^n \right) =0$$ For $\hat{u}^n = G^n\hat{u}^0$ a growth factor $|G|\leq1$ is sufficient. This gives, $$\frac{u^{n+1}-u^n}{\Delta t} + \beta h \left( \frac{u_{j+1}^n-2u_j^n + u_{j-1}^n}{h^2} \right)$$ With the smallest possible diffusion but still stable, $\beta \geq \frac{a}{2}$ becomes $\beta = \frac{a}{2}$. If this is plugged into the equation, $$\frac{u^{n+1}-u^n}{\Delta t} + \frac{a}{2h} \left( {u_{j+1}^n-2u_j^n + u_{j-1}^n} \right)$$ Now, according to some old notes I have the result is, $$\frac{u^{n+1}-u^n}{\Delta t} + \frac{1}{h} \left( {au_j^n -a u_{j-1}^n} \right)$$ Well this is somewhat vague for me. Specially the last step where the defraction term is finalised.

Any pointers of help would be appreciated, specially that last step.

 

[K41]

I faintly remember that sometimes you could simplify things using a Taylor series going forward in j and a taylor series going backwards in j, then subtracting (or adding) one from the other. At least something along those lines. Considering the j-1 term remained, I'd consider doing the taylor series over the j+1 term (eq1) and the j term (eq2) and then multiplying eq2 by a factor of 2 and then adding (or subtracting) eq1 from eq2 and see what happens. You can probably ignore higher order terms which are negligible (or your only looking at a first order scheme).