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?
Best Answer
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. $-1\le R\le 0$.