Question 1: Consider the wave equation $$ u_t + c(x) u_x = 0 , $$ where $x\in \Omega \subset \Bbb R$ and $c(x)$ is a function of $x$.

(a) Show that the Lax-Wendroff scheme for this PDE is given by $$ u_j^{n+1} = u_j^n - c_j \Delta t \frac{D_x u_j^n}{2 \Delta x} + \frac{c_j^2 \Delta t^2}{2} \frac{\delta_x^2 u_j^n}{\Delta x^2} + \frac{c_j \Delta t^2}{8 \Delta x ^2} (D_x c_j)(D_x u_j^n) , $$ where $D_x$ is the first central difference operator, $\delta_x^2$ is the second central difference operator, and $\Delta t$ and $\Delta x$ are the mesh-spacing in $t$ and $x$, respectively. The $j$ and $n$ are space and time indices, respectively, and $u_j^n$ is the grid function such that $u_j^n\approx u(x_j,t_n)$ and $c_j \approx c(x_j)$.

I would also like to answer the following question

Assuming $c(x)$ is a constant, and given that the initial value problem is well posed, carry out a von Neumann Stability analysis and hence show that the scheme is convergent provided $$\bigg \lvert \frac{c\Delta t}{ \Delta x} \bigg \rvert \leq1$$

I have tried to do the stability analysis but I am not getting anywhere near what the question requires. Could someone please check my working and provide some advice.

The scheme can be written as

$$u_{j}^{n+1} = u_j^n - \frac{p}{2}(u_{j+1}-u_{j-1}) + \frac{p^2}{2}(u_{j+1}^n-2u_j^n+u_{j-1}^n) + \frac{p^2}{8}(u_{j+1}-u_{j-1}), \quad \text{where $p = \frac{c\Delta t}{5 \Delta x}$}$$

I think this is how the scheme should be written. However, I am not sure about the $D_{x}c_{j}$ term, since it is a constant, should I just ignore the $D_x$ operator?

For the stability analysis, using a trial solution of the form $u_{j}^{n} = A \xi^{n}e^{i \omega j}$ gives

$$A \xi^{n+1} e^{i \omega j} = A \xi^{n} e^{i \omega j} - \frac{p}{2}\left[ A \xi^{n} e^{i \omega (j+1)} - A \xi^{n} e^{i \omega (j-1)} \right] + \frac{p^{2}}{2} \left[ A \xi^{n} e^{i \omega (j+1)} - 2 A \xi^{n} e^{i \omega j} + A \xi^{n} e^{i \omega (j-1)} \right] + \frac{p^2}{8}\left[ A \xi^{n} e^{i \omega (j+1)} + A \xi^{n} e^{i \omega (j-1)} \right]$$

Dividing through and by $u_j^n = A \xi^{n} e^{i \omega j}$ I get

$$\xi = 1 + \frac{p}{2}\left[ e^{i\omega} - e^{-i\omega} \right] - \frac{p^{2}}{2} \left[ e^{i\omega} - 2 + e^{-i\omega} \right] + \frac{p^{2}}{8} \left[ e^{i \omega} + e^{-i \omega} \right] = 0$$

Than, using Eulers identity

$$\implies \xi = 1 + \frac{p^{2}}{2}\cdot 2 \cos \omega - \frac{p^{2}}{2} (2 \cos \omega - 2) - \frac{p^{2}}{8}\cdot 2 \cos \omega$$

Can anyone help me from here?

  • 1
    $\begingroup$ I tried to fix your post, but there were so many typos and errors that I'm not even sure I have fixed it correctly. Can you check to make sure the post is as you wanted it. $\endgroup$ – mattos Apr 19 '18 at 0:42
  • 1
    $\begingroup$ Also, $e^{i \omega} - e^{-i \omega} = 2i \sin \omega$ not $2 \cos \omega$. $\endgroup$ – mattos Apr 19 '18 at 0:43
  • $\begingroup$ Yes u edited it correctly, thank you so much! $\endgroup$ – italy Apr 19 '18 at 1:02

Recall how the Lax-Wendroff method is obtained in the constant-speed case [1]:

  • a Taylor series in time is written: $$ u(x,t_{n+1}) = u(x,t_{n}) + \Delta t\, u_t(x,t_{n}) + \frac{1}{2}\Delta t^2\, u_{tt}(x,t_{n}) + \dots $$
  • the time derivatives are eliminated using the PDE: $u_t = -c u_x$ and $u_{tt} = c^2 u_{xx}$.
  • the spatial derivatives are replaced by central finite-difference approximations.

Thus, the following scheme is obtained: $$ u_j^{n+1} = u_j^n - c\Delta t \frac{D_x u_j^n}{2 \Delta x} + \frac{c^2\Delta t^2}{2} \frac{\delta_x^2 u_j^n}{\Delta x^2} \, . $$ This method can be adapted to the variable velocity case.

Let us analyze the stability of the Lax-Wendroff scheme above (constant-speed case). Assuming a perturbation of the form $u_j^{n} = \xi^n \text{e}^{\text i k x_j}$, one has \begin{aligned} \xi &= 1 - c \Delta t \frac{\text{e}^{\text i k \Delta x} - \text{e}^{-\text i k \Delta x}}{2 \Delta x} + \frac{c^2 \Delta t^2}{2} \frac{\text{e}^{\text i k \Delta x} - 2 + \text{e}^{-\text i k \Delta x}}{\Delta x^2} \\ &= 1 - \text{i} \kappa \sin (k\Delta x) + \kappa^2 \left(\cos (k\Delta x) - 1\right) . \end{aligned} where $\kappa = c\frac{\Delta t}{\Delta x}$ is the Courant number. Thus, the squared modulus of the amplification factor is \begin{aligned} |\xi|^2 &= (1 - \kappa^2 (1-\cos(k\Delta x)))^2 + \kappa^2 (1-\cos^2(k\Delta x))\\ &= 1 - 2\kappa^2 (1-\cos(k\Delta x)) + \kappa^2 (1-\cos^2(k\Delta x)) + \kappa^4 (1-\cos(k\Delta x))^2 \\ &= 1 - \kappa^2 (1-\kappa^2) (1-\cos(k\Delta x))^2 \\ &= 1 - 4\kappa^2 (1-\kappa^2) \sin^4\left(\tfrac{1}{2}k\Delta x\right) . \end{aligned} Finally, the Lax-Wendroff scheme is Neumann-stable provided that $|\xi|^2 \leq 1$, which implies that the CFL condition $|\kappa| \leq 1$ is satisfied.

[1] R.J. LeVeque, Finite Volume Methods for Hyperbolic Problems, Cambridge University Press, 2002.

  • $\begingroup$ Do you ignore the last term while doing the stability analysis ? $\endgroup$ – italy Apr 24 '18 at 15:48
  • $\begingroup$ Oh okay, thank you $\endgroup$ – italy Apr 24 '18 at 16:17
  • $\begingroup$ I am trying the get this amplification factor but I don’t seem to get the same answer... $ 1 -pi sin \omega - p^{2} (1-cos \omega )$ from here I know I would have to take the complex conjugate was of this, $ \xi ^{2} = 1 +p^{2}(1-cos^{2} \omega)-2p^{2}(1-cos \omega) p^{4}(1-cos \omega)^{2} $ I have tried to simplify this but I get anything meaningful can you suggest anything from here? Thank you $\endgroup$ – italy Apr 24 '18 at 19:07
  • $\begingroup$ The amplification factor I am looking for should be $\xi = 1-2p^{2} sin^{2} \frac{\omega}{2} -2ip sin \frac{\omega}{2} cos\frac{\omega}{2} $ $\endgroup$ – italy Apr 24 '18 at 19:16
  • 1
    $\begingroup$ @italy Thanks for the catch: the square was not at the right place in my answer. Now, a few intermediate steps have been added. Note that $\cos u = \cos 2\frac{u}{2} = 1 - 2\sin^2 \frac{u}{2}$ and $\sin u = \sin 2\frac{u}{2} = 2\sin \frac{u}{2}\cos \frac{u}{2}$. $\endgroup$ – Harry49 Apr 24 '18 at 22:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.