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Von Neumann analysis

  1. Oct 16, 2014 #1
    I'm trying to refresh some numerical science stuff. Von Neumann analysis, if I take a slimmed down equation, convection. [tex] \frac{∂u}{∂t}+a ∇ u =0[/tex] If I'm using Euler forward, [tex] \frac{u^{n+1}-u^n}{\Delta t}+\frac{a}{2h} \left( u_{j+1}^n -u_{j-1}^n \right) =0[/tex] For [itex]\hat{u}^n = G^n\hat{u}^0[/itex] a growth factor [itex]|G|\leq1[/itex] is sufficient. This gives, [tex] \frac{u^{n+1}-u^n}{\Delta t} + \beta h \left( \frac{u_{j+1}^n-2u_j^n + u_{j-1}^n}{h^2} \right)[/tex] With the smallest possible diffusion but still stable, [itex]\beta \geq \frac{a}{2}[/itex] becomes [itex]\beta = \frac{a}{2}[/itex]. If this is plugged into the equation, [tex] \frac{u^{n+1}-u^n}{\Delta t} + \frac{a}{2h} \left( {u_{j+1}^n-2u_j^n + u_{j-1}^n} \right)[/tex] Now, according to some old notes I have the result is, [tex] \frac{u^{n+1}-u^n}{\Delta t} + \frac{1}{h} \left( {au_j^n -a u_{j-1}^n} \right)[/tex] Well this is somewhat vague for me. Specially the last step where the defraction term is finalised.

    Any pointers of help would be appreciated, specially that last step.
     
  2. jcsd
  3. Oct 21, 2014 #2

    K41

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    I faintly remember that sometimes you could simplify things using a Taylor series going forward in j and a taylor series going backwards in j, then subtracting (or adding) one from the other. At least something along those lines. Considering the j-1 term remained, I'd consider doing the taylor series over the j+1 term (eq1) and the j term (eq2) and then multiplying eq2 by a factor of 2 and then adding (or subtracting) eq1 from eq2 and see what happens. You can probably ignore higher order terms which are negligible (or your only looking at a first order scheme).
     
    Last edited: Oct 21, 2014
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