Stability of advection equation using fourier transform

We want to study $$v_t + a v_x=0$$ using the one-sided, implicit BTFS scheme on $$(- \infty, \infty)$$.

The scheme is $$u_k^n = u_k^{n-1} - R (u_{k+1}^n - u_k^n )$$ where $$R = \dfrac{a \Delta t}{\Delta x}$$. Applying the discrete fourier transform, we obtain $$\hat{u}^{n} = \hat{u}^{n-1} - R e^{i \xi} \hat{u}^n + R \hat{u}^n$$ Thus, we see that $$\hat{u}^n ( 1 - R e^{i \xi } + R ) = \hat{u}^{n-1}$$ So, the symbol of the scheme is $$\rho( \xi ) = \frac{1}{1 + R - R e^{i \xi} } = \frac{1}{1 + R - R \cos \xi - R i \sin \xi}$$ thus, $$|\rho(\xi)|^2 = \frac{1}{(1+R-R\cos \xi )^2 + R^2 \sin^2 \xi }$$ Isn't the above always less than $$1$$, thus implying scheme is unconditionally stable?

Observe that \begin{align*} (1 + R - R\cos\xi)^2 + R^2\sin^2\xi & = (1+R)^2 - 2R(1+R)\cos\xi + R^2 \\ & = 1 + 2R + 2R^2 - 2R(1+R)\cos\xi \\ & = 1 + 2R(1+R) - 2R(1+R)\cos\xi \\ & = 1 + 2R(1+R)(1- \cos\xi). \end{align*} Since $$(1-\cos\xi)\ge 0$$ for any $$\xi$$, we have $$|\rho|^2\le 1$$ for all $$\xi$$ as long as $$R(1+R)\le 0$$, i.e. $$-1\le R\le 0$$.
• but if $R(R+1) \leq 0$, then $|p|^2$ is more than 1 – James Mar 3 '19 at 22:59
• It's not. Let $A = 2R(1+R)(1-\cos\xi)$. Since the term $(1-\cos\xi)$ is nonnegative for any $\xi$, $A$ has the same sign as $R(1+R)$. Since $|\rho|^2 = 1 + A$, $|\rho|^2\le 1$ if $A\le 0$, or $R(1+R)\le 0$. – Chee Han Mar 3 '19 at 23:44