Kurganov-Tadmor scheme

Nereus · January 20, 2012, 11:24

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

. 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?

duri · January 21, 2012, 08:09

The derivative represents eigen value of jacobian A. If you are solving 1D buger's equation in max() term is correct.

January 20, 2012, 11:24	Kurganov-Tadmor scheme	#1
Nereus New Member Nereus Join Date: Sep 2011 Posts: 18 Rep Power: 14	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 . 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?

January 21, 2012, 08:09		#2
duri Senior Member duri Join Date: May 2010 Posts: 245 Rep Power: 16	The derivative represents eigen value of jacobian A. If you are solving 1D buger's equation in max() term is correct.