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Kurganov-Tadmor scheme

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Old   January 20, 2012, 12:24
Default Kurganov-Tadmor scheme
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I'm a little confused regarding the implementation of the Kurganov-Tadmoor scheme, specifically with regards to the local maximum propogation speed.

a_{i\pm\frac{1}{2}}(t)=\max\left[\rho\left(\frac{\partial F(u_i(t))}{\partial u}\right),\rho\left(\frac{\partial F(u_{\pm i}(t))}{\partial u}\right)\right]

Now, say my flux is simply a 1D advective F=vu. Then, \frac{\partial F(u_i(t))}{\partial u}=v_i, right? So, does the right hand side of the equation above simply boil down to \max(v_i,v_{\pm i}), because that seems far too simple? Or, have I misunderstood the mathematics?
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Old   January 21, 2012, 09:09
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The derivative represents eigen value of jacobian A. If you are solving 1D buger's equation in max() term is correct.
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