I've got an inhomogeneous heat equation at hand: $$\partial_tu=\alpha\partial_x^2u+f(t,x).$$ I discretised it according to the FTCS scheme, i. e. I applied the forward Euler method for $\partial_t$ and twice the central difference for $\partial_x^2$: $$u^{n+1,m}=u^{n,m}+r\Big(u^{n,m+2}-2u^{n,m}+u^{n,m-2}\Big),$$ where $r\equiv\frac{\alpha\Delta t}{(2\Delta x)^2}$ and $u^{n,m}\equiv u(n\Delta t,m\Delta x)$.
Note that the central space difference is taken over double the lattice spacing ($4\Delta x$ instead of $2\Delta x$) in my case, compared to the standard scheme. This is because I need to calculate the first and the second derivative w. r. t. $x$ separately.
My problem: The solution shows (not-so-)slight disturbances of small wavelength, see the pictures attached: The first solution should look as on the second one.
My idea: The stability condition on the step sizes is $$\left|1-4r\sin^2\left(k\Delta x\right)\right|\leq1\;\forall\;k,$$ where $k$ is the wave number of the solution. When I plot the left hand side against $k$, one can see that regardless of my choice of $r$, there are always values of $k$ for which the condition is only marginally fulfilled (i. e. where the left hand side is equal to $1$; red dots).
May this be the source of the instability?
Annotations:
I need to compute the first derivative $\partial_xu$ separately since I'd like to manipulate it in the future before taking the second derivative. So, my routine actually looks like this:
$$u^{n,m+1}=u^{n,m}-\frac{\Delta t}{2\Delta x}\Big(v^{n,m+1}-v^{n,m-1}\Big)\\ v^{n,m}=-\frac{\alpha}{2\Delta x}\Big(u^{n,m+1}-u^{n,m-1}\Big)$$
At the boundaries $m\in\{0,M\}$ I take $u=0$ and $v$ is calculated with the forward/backward difference method rather than the central difference:
$$u^{n,0}=u^{n,M}=0\\ v^{n,0}=\frac{\alpha}{\Delta x}\Big(u^{n,1}-u^{n,0}\Big)\\ v^{n,M}=\frac{\alpha}{\Delta x}\Big(u^{n,M}-u^{n,M-1}\Big)$$
Fully expanded, this amounts to the scheme: $$u^{n+1,m}=\begin{cases}0&m\in\{0,M\},\\ u^{n,1}+2r\Big(u^{n,2}-2u^{n,1}+u^{n,0}\Big)&m=1,\\ u^{n,M-1}+2r\Big(u^{n,M}-2u^{n,M-1}+u^{n,M-2}\Big)&m=M-1,\\ u^{n,m}+r\Big(u^{n,m+2}-2u^{n,m}+u^{n,m-2}\Big)&\mathrm{else}\\\end{cases}$$