I am given the advection equation $$ u_t=u_x $$ and then the transformation of coordinates $$ x=x(\xi,\theta), \qquad t=\theta $$ which leads us to the transformed equation $$ x_{\xi} u_{\theta} - u_{\xi} x_{\theta} = u_{\xi} $$ Further, i have to derive the central difference approximation for $u_x$ (at point $x_i$). I derived it via Taylor expansions as $$ u_x|_i \approx \frac{u_{i+1} - u_{i-1}}{x_{i+1}- x_{i-1}} = -\frac{1}{x_{i+1}-x_{i-1}} u_{i-1} - \frac{1}{x_{i+1}-x_{i-1}} u_{i+1} $$ Next, i have to work out local truncation error $\tau$ in terms of the transformation derivatives $x_\xi$, $x_{\xi \xi}$,$\dots$ and solution derivatives $u_{xx}$, $u_{xxx}$,$\dots$ and write it in the form $$ \tau = E\cdot H^2 + O(H^3) $$ where $H=\Delta \xi$, the constant stepsize in transformed variable $\xi$.

I get $$ \tau = \frac{1}{x_{i+1}- x_{i-1}} \big( -\frac{(\Delta x_{i-1})^2}{2} + \frac{(\Delta x_{i})^2}{2} \big) u_{xx} + \frac{1}{x_{i+1}- x_{i-1}} \big( \frac{(\Delta x_{i-1})^3}{6} + \frac{(\Delta x_{i})^3}{6} \big) u_{xxx} + \dots $$ where $\Delta x_i = x_{i+1} - x_i$.

But here my problem arises. I don't know where the $x_\xi,x_{\xi \xi},\dots$ are hidden and where do i find $\Delta \xi$. Any help appreciated.

  • $\begingroup$ Is this question a homework problem? If so, please state that explicitly. $\endgroup$ – Geoff Oxberry May 8 '13 at 23:06
  • $\begingroup$ Yes. I added tag $\endgroup$ – Uroš May 9 '13 at 7:04

I have it. I just expand grid function around $x_i$: $$ x_{i+1}(\xi,\theta) = x_i + \Delta \xi x_\xi + \frac{(\Delta \xi)^2}{2}x_{\xi \xi} + \dots $$ and $$ x_{i-1}(\xi,\theta) = x_i - \Delta \xi x_\xi + \frac{(\Delta \xi)^2}{2}x_{\xi \xi} + \dots $$ Then i use $$ (\Delta x_i)^2 - (\Delta x_{i-1})^2 = (\Delta x_{i} - \Delta x_{i-1})(\Delta x_{i} + \Delta x_{i-1}) $$ and $$ (\Delta x_i)^3 + (\Delta x_{i-1})^3 = (\Delta x_{i} + \Delta x_{i-1})((\Delta x_{i})^2 - \Delta x_i \Delta x_{i-1} + (\Delta x_{i-1})^2) $$ and $$ \Delta x_{i} + \Delta x_{i-1} = x_{i+1}-x_{i-1} $$ After calculation you get to the solution.

| cite | improve this answer | |

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.