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Spreading ring calculations, Part II -- Marvin

In one of my last blog posts (https://astrobear.pas.rochester.edu/trac/blog/mblank05302015) I presented some "spreading ring calculations" that allow to estimate the magnitude of numerical viscosity in disk simulations with astroBEAR. There I introduced the parameter \alpha_{\text{num}} that gives the numerical viscosity in units of the maximum \alpha-viscosity \nu_{\alpha} = c_{\text{s}} h.

In the following I show a parameter study of these calculations by varying the resolution of the simulations. However, I have performed this study in 2D to save computational time. The simulations from my previous post have a resolution (cell size) of 0.04 pc, here I additionally show simulations with 0.08, 0.02, and 0.01 pc.

The evolution of the spreading ring for all resolution levels is shown in the Figure below (upper left: 0.08 pc, upper right: 0.04 pc, lower left: 0.02 pc, lower right: 0.01 pc).

The qualiative evolution is quite similar for all the simulations, but as expected the spreading of the ring is slower with higher resolution. I also estimated the parameter \alpha_{\text{num}} following the procedure described in my previous post, and I additionally give the runtime of the simulations, all have been calculated using 64 cores:

resolution	\alpha_{\text{num}} [10^{-2}]	runtime
0.08 pc	8.11 \pm 0.21	11.3 min
0.04 pc	6.22 \pm 0.05	1 h
0.02 pc	2.53 \pm 0.03	5.3 h
0.01 pc	1.14 \pm 0.02	31.1 h

The upper right figure with a resolution of 0.04 pc can be compared with the corresponding 3D simulation of my previous post, again the qualitative evolution is similar, but the 2D simulation has a higher numerical viscosity than the 3D simulation, which has \alpha_{\text{num}} = (3.93 \pm 0.13) 10^{-2}.

The numerical viscosity decreases with increasing resolution, as expected. Furthermore, it seems to be a linear function of the resolution, which makes sense: Let us consider the following simplified transport equation: \frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} = 0 For solving this equation numerically we want to replace the spatial derivative by a difference quotient. Therefore we make a taylor expansion of the function u(x) at the point x_{i}:

u_{i+1} = u_i + \Delta x \left( \frac{\partial u}{\partial x} \right)_i + \frac{1}{2} \left( \Delta x \right)^{2} \left( \frac{\partial^{2} u} {\partial x^{2}} \right)_i + ...

Rearranging gives: \left( \frac{\partial u}{\partial x} \right)_i = \frac{u_{i+1}-u_i}{\Delta x} - \frac{1}{2} \Delta x \left( \frac{\partial^2 u}{\partial x^2} \right)_i

And inserting this into the first equation gives:

\left( \frac{\partial u}{\partial t} \right)_i + \frac{u_{i+1}-u_i}{\Delta x} = \frac{1}{2} \Delta x \left( \frac{\partial^2 u}{\partial x^2} \right)_i

The term on the right side is responsible for numerical viscosity, and is a linear function of the resolution \Delta x.

Posted: 9 years ago
Author: Marvin Blank
Categories: (none)

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150807_spreadingring_resolution.png (64.6 KB) - added by Marvin Blank 9 years ago.

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Spreading ring calculations, Part II -- Marvin

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