Lax-Friedrichs Stability

**lvleph** · October 12th 2010, 10:17 AM

Define the Lax-Friedrichs scheme for $u_t + a u_x = 0,\; a>0$ as
$v^{n+1}_m = \frac{1}{2}(v_{m+1}^n + v_{m-1}^n) - \dfrac{a\lambda}{1 + a^2\lambda^2}(v_{m+1}^n - v_{m-1}^n)$ where $\lambda = \frac{k}{h}$ , $h$ is the grid spacing in the $x$ -direction, and $k$ is the grid spacing in the $t$ -direction.

To determine the stability I first transformed the Lax-Friedrichs scheme from spacial domain into the frequency domain, giving me
$g(\theta) = \frac{1}{2}\left(e^{i\theta} + e^{-i\theta}\right) - \dfrac{a\lambda}{1 + a^2\lambda^2}\left(e^{i\theta} - e^{-i\theta}\right), \text{ where } \theta = \xi h$
$g(\theta) = \cos \theta - 2\dfrac{a\lambda}{1 + a^2\lambda^2} i \sin \theta$
For this scheme to be stable we must require $|g| \le 1 \Rightarrow |g|^2 \le 1$
$\Rightarrow \cos^2 \theta + 4\dfrac{a^2\lambda^2}{(1 + a^2\lambda^2)^2} \sin^2 \theta \le 1$

And this is where I am stuck. The instructor claims that the Lax-Friedrichs scheme is unconditionally stable, but I don't see it. In the literature that I have looked at they claim that the above equation is equivalent to
$4\dfrac{a^2\lambda^2}{(1 + a^2\lambda^2)^2} \le 1$
Which doesn't seem to imply unconditional stability. I also don't necessarily see where that came from. Any help would be appreciated. Thank you in advance.

EDIT: Thanks to Ackbeet, I see how the inequalities are equivalent. (And I feel stupid for not seeing it)

EDIT2: $4a^2\lambda^2 \le 1 + 2a^2\lambda^2 +a^4\lambda^4$
$0 \le 1 - 2a^2\lambda^2 +a^4\lambda^4 \Rightarrow (a^2\lambda^2 - 1)^2 \ge 0$ Thus, Lax-Friedrichs is unconditionally stable.

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