I'm working on a heat diffusion problem, $$ \frac{\partial T}{\partial t}=\vec{\nabla}\cdot\left(\kappa T^{5/2}\,\vec{\nabla}T\right) $$

I know that, after discretization, the time step for the 1D case becomes, $$ dt= \frac{\rho\,dx^2}{\kappa T_{max}^{5/2}} $$ Does this imply/mean that the maximum wave speed is $$ \frac{dx}{dt} = \frac{\kappa T_{max}^{5/2}}{\rho\,dx} $$ or would I have to analyze this differently to get the wavespeed?

  • $\begingroup$ What is $\rho$ here? $\endgroup$ – Geoff Oxberry Oct 23 '13 at 23:00
  • $\begingroup$ Density. This is being attached to a hydro code $\endgroup$ – Kyle Kanos Oct 24 '13 at 2:14
  • $\begingroup$ What do you mean by 'the time step becomes'? I suspect you interprete the CFL condition in the wrong way... $\endgroup$ – Jan Oct 24 '13 at 8:08
  • $\begingroup$ @Jan: This time step was computed in Reale 1995. I admit I removed a factor of order unity, but that's the essence of $dt$ for this problem. All I am asking is if it's valid to use $dx/dt$ to get the wavespeed, or do I have to do something different. $\endgroup$ – Kyle Kanos Oct 24 '13 at 13:09
  • $\begingroup$ Since $dt$ and $dx$ is your personal choice, I doubt that you can use them to get a characteristic of the actual continuous equation. $\endgroup$ – Jan Oct 24 '13 at 13:21

For the heat equation, the wave speed is infinite and this is reflected by the fact that you are solving a globally coupled differential equation if you are using an implicit time stepping scheme.


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