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I'm a beginner in PDE's and numerical methods, so please go slow :-D

I'm trying to show that the Lax-Wendroff scheme is stable for $|a\lambda| < 1$. The scheme is this:

$v_m^{n+1} = v_m^n - \frac{a \lambda }{2}\left(v_{m+1}^n - v_{m - 1}^n\right) + \frac{a^2 \lambda ^2 }{2}\left(v_{m+1}^n - 2v_m^n + v_{m - 1}^n\right)$

And I want to show that this scheme is stable without relying on Fourier methods... Instead I want to analyze the linear system that we can construct using the scheme.

So we can write out:

$v_1^{n+1} = v_1^n - \frac{a \lambda }{2}\left(v_{2}^n - v_{0}^n\right) + \frac{a^2 \lambda ^2 }{2}\left(v_{2}^n - 2v_1^n + v_{0}^n\right)$

$v_2^{n+1} = v_2^n - \frac{a \lambda }{2}\left(v_{3}^n - v_{1}^n\right) + \frac{a^2 \lambda ^2 }{2}\left(v_{3}^n - 2v_2^n + v_{1}^n\right)$

$v_3^{n+1} = v_3^n - \frac{a \lambda }{2}\left(v_{4}^n - v_{2}^n\right) + \frac{a^2 \lambda ^2 }{2}\left(v_{4}^n - 2v_3^n + v_{2}^n\right)$

...

So this gives a system of equations which we can write as a matrix equation:

$v^{n+1} = Av^{n} + f$ where $v^{n+1} = \begin{pmatrix}v_1^{n+1}\\v_2^{n+1}\\\vdots\\ v_{m-1}^{n+1}\end{pmatrix}$, $v^{n} = \begin{pmatrix}v_1^{n}\\v_2^{n}\\\vdots\\ v_{m-1}^{n}\end{pmatrix}$, f contains the boundary conditions and is $f = \begin{pmatrix}\frac{a \lambda + a^2 \lambda^2}{2} v_0^n\\0\\0\\\vdots\\0\\ \frac{-a \lambda + a^2 \lambda^2}{2} v_m^n\end{pmatrix}$ and A is the tridiagonal matrix with main diagonal having all entries being $1 - a^2 \lambda^2$, the super-diagonal having entries $\frac{a^2\lambda^2 - a \lambda}{2}$, and the sub-diagonal having entries $\frac{a^2\lambda^2 + a \lambda}{2}$.

Now, to make sure the scheme is stable, my understanding is that we want $||A|| \le 1$ where the norm can be any vector-induced norm.

So we will look at when $||A||_{\infty} \le 1$.

We see that $||A||_{\infty} = |\frac{a^2 \lambda^2 + a \lambda}{2}| + |1 - a^2\lambda^2| + |\frac{a^2\lambda^2 - a\lambda}{2}|$

Now we have several cases, depending on where $a\lambda$ lies on the number line.

If $a \lambda < -1$, then $||A||_{\infty} = \frac{a^2 \lambda^2 + a \lambda}{2} - \left(1 - a^2\lambda^2\right) + \frac{a^2\lambda^2 - a\lambda}{2}$.

If $-1 \le a \lambda < 0$, then $||A||_{\infty} = -\left(\frac{a^2 \lambda^2 + a \lambda}{2}\right) + \left(1 - a^2\lambda^2\right) + \frac{a^2\lambda^2 - a\lambda}{2}$.

If $0 \le a \lambda \le 1$, then $||A||_{\infty} = \left(\frac{a^2 \lambda^2 + a \lambda}{2}\right) + \left(1 - a^2\lambda^2\right) - \left(\frac{a^2\lambda^2 - a\lambda}{2}\right)$.

If $1 < a \lambda$, then $||A||_{\infty} = \left(\frac{a^2 \lambda^2 + a \lambda}{2}\right) - \left(1 - a^2\lambda^2\right) + \left(\frac{a^2\lambda^2 - a\lambda}{2}\right)$.

The first and last case, we boil down to $||A||_\infty = 2a^2\lambda^2 - 1$, and given $|a\lambda| > 1$, we see that $||A||_\infty > 1$ in these cases.

If $-1 \le a \lambda < 0$, then $||A||_{\infty} = 1 - a\lambda\left(a\lambda + 1\right)$ and we'd also have $||A||_\infty > 1$.

If $0 \le a \lambda \le 1$, then $||A||_{\infty} = 1 + a\lambda\left(-a\lambda + 1\right)$ and again we'd also have $||A||_\infty > 1$.

So it seems like no matter what, we'd have the scheme being unstable.

What is the mistake here?

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