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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?

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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. $-1\le R\le 0$.

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  • $\begingroup$ but if $R(R+1) \leq 0$, then $|p|^2$ is more than 1 $\endgroup$ – James Mar 3 '19 at 22:59
  • $\begingroup$ 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 $R(1+R)\le 0$. $\endgroup$ – Chee Han Mar 3 '19 at 23:44

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