Exercise 4: The Crank-Nicolson scheme for $u_t + a u_x = 0$ is given by $$ \frac{U_{j,n+1}-U_{j,n}}{\Delta t} + \frac{a}{2}\frac{D_xU_{j,n}}{2\Delta x} + \frac{a}{2}\frac{D_xU_{j,n+1}}{2\Delta x} = 0 .$$ Show that the LTE is given by $$ \mathcal{L}_\Delta u = au_{xxx} \left(\frac{1}{6} + \frac{p^2}{12}\right) {\Delta x}^2 + O({\Delta x}^3,{\Delta t}^3) , $$ where $p = a{\Delta t}/{\Delta x}$. Find the amplification factor and find the conditions for stability.

I have expanded this about a dozen times, I do not get the correct answer. Can someone please show me.

My working

$$\frac{ u(x , t + \delta ) - u( x,t)} { \delta t} + \frac{a}{2} \frac{u( x+ \delta x,t)-u(x-\delta x,t)}{ 2 \delta x} + \frac{a}{2} \frac{u ( x + \delta , t+ \delta t ) - u ( x- \delta x, t+ \delta t)} {2 \delta x}$$

expanding using taylor series I get

$$\frac{1}{\delta t} [ u + u_{t} \delta t + \frac{u_{tt}}{2} \delta t^{2} + O ( \Delta t^{3})] $$

$$ \frac{1}{\delta x^{2}} [ u_xx \delta x^{2} + \frac{1}{12} u_{xxxx} + O(\Delta x^{6})] $$

$$ u_{xx} + u_{xxt} \delta t + O(\Delta^{2}) +\frac{1}{12}u_{xxxx} \delta x^{2} + \frac{1}{12} u_{xxxxt} \delta x^{2} \delta t $$

these are the three terms I get after expanding for each one, however after rearranging, and using $u_t = -au_x$, I still am not able to get the correct answer.

I am hoping someone can show me as I really need to know this for my exam.

Thank you,

  • $\begingroup$ Really? I have done it repeated and dont get the correct answer, clear doing it wrong, I will type my working out and put it up. Cause its so much expansions, I thouoght it would be long to type out $\endgroup$ – italy May 17 '17 at 21:36
  • $\begingroup$ I have added on my answer, can you please have look and let me know, / show me the correct way, as I really need to know how to do this. Thank you. $\endgroup$ – italy May 17 '17 at 21:50
  • $\begingroup$ anybody know how to it? $\endgroup$ – italy May 17 '17 at 23:01
  • $\begingroup$ Sorry, I'm not familiar with this, I 've just added tag Laplace to make it more visible for audience. $\endgroup$ – zwim May 18 '17 at 5:26
  • $\begingroup$ But I though you said you have jut done it and it work for you. $\endgroup$ – italy May 18 '17 at 17:12

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