Example Execution

Build ASGarD with -DASGARD_USE_MATLAB=ON or no output will be displayed!

• 1D Diffusion equation https://github.com/project-asgard/asgard/blob/develop/src/pde/pde_diffusion1.hpp

Example PDE using the 1D Diffusion Equation with a manufactured solution.

PDE: $\frac{df}{dt} = \frac{d^2 f}{dx^2} -\frac{\pi^2}{4}\exp(-\frac{\pi^2}{2})\cos(\frac{\pi x}{2})$

Analytic solution: $\cos(\frac{\pi \cdot x}{2}) \exp(-\frac{\pi^2 \cdot t}{2})$

First, run ASGarD selecting the diffusion_1 PDE, implicit time advance, 1 time step, and real-space output every 1 time step.

./asgard -p diffusion_1 -i -n 1 --plot_freq 1

This matches the analytic solution at t=0. $f(0,0) = 1$ and $f(1,0) = 0$.

Next, run for 1000 time-steps. After each calculation the figure will be updated.

./asgard -p diffusion_1 -i -n 1000 --plot_freq 10

The final figure shows $f(0,t)$ decreasing with time, matching the analytic solution.

Lastly, let's increase the precision of this calculation. Increasing the number of legendre basis polynomials (-d) and the starting (-l) and maximum (-m) hierarchical levels (resolution) results in a denser grid.

./asgard -p diffusion_1 -n 1 -d 3 -l 4 -m 4 --plot_freq 1

./asgard -p diffusion_1 -n 1000 -d 3 -l 4 -m 4 --plot_freq 10

Lastly, let's enable adaptivity with the --adapt option and set --thresh 1e-4.

 ./asgard -p diffusion_1 -n 1 -d 3 -l 5 -m 10 --adapt --thresh 1e-4 --plot_freq 1

The refined grid contains 12 elems or 36 DOF. Compared with the full grid solution

 ./asgard -p diffusion_1 -n 1 -d 3 -l 5 -m 10 -f --plot_freq 1

DOF / FG DOF = 36/96 = 0.375

• 2D Continuity Equation

Consider the continuity equation in two dimensions

$\frac{\partial f}{\partial t} + \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} = 0$ //Add source1 + source 2

with analytic solution

$f(x,y,t) = \cos(\pi x)\cos(2 \pi y)\sin(2t)$

Note that this solution is a function in $(x,y)$ that is scaled in time.

Use the -p continuity_2 option to select the 2D continuity equation. Let's start with a higher precision calculation. Notice that in 2 dimensions and the starting (-l) hierarchical levels (resolution) takes a string containing two values.

First, start with -n 1 to get the initial values at t=0.

./asgard -p continuity_2 -n 1 -d 3 -l "4 4" -m 4 --plot_freq 1

Note we are in agreement with of analytic solution of 0.

Next, we run for 627 time steps, which should display the greatest amplitude.

./asgard -p continuity_2 -n 627 -d 3 -l "4 4" -m "4 4"

The slices are slightly offset from 0. At $y=0.125$, we see the amplitude matches $\cos(2 \cdot \pi \cdot 0.125) = 0.707$ At $x=0.0625$, we see the amplitude matches $\cos(\pi \cdot 0.0625) = 0.981$.

432 DOF Relative difference (numeric-analytic) [wavelet]: 1.05497 %

2304 DOF Relative difference (numeric-analytic) [wavelet]: 0.0653147 %

• 6D Continuity Equation $f = f(x,y,z,v_x,v_y,v_z)$ where

$\frac{\partial f}{\partial t} + \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} + 3\frac{\partial f}{\partial z} + 4\frac{\partial f}{\partial v_x} + 3\frac{\partial f}{\partial v_y} + 2\frac{\partial f}{\partial v_z} = S$ //Add source

./asgard -p continuity_6 -n 60 -d 3 

degrees of freedom: 24786 Relative difference (numeric-analytic) [wavelet]: 0.158717 %

./asgard -p continuity_6 -n 60 -d 3 -f

degrees of freedom: 2985984 Relative difference (numeric-analytic) [wavelet]: 0.00313217 %

• Vlasov Problem

We now solve the Sod Shock Tube example of the Vlasov-Lenard-Bernstein Equation

$\frac{\partial f}{\partial t} + v \frac{\partial f}{\partial x} = \nu \frac{\partial}{\partial v}( (v-u_f) f + \theta_f \frac{\partial f}{\partial v}) $

where $u_f$ and $\theta_f$ are the bulk-velocity and temperature of the distribution $f(x,v,t)$ and $\nu>0$ is the collision frequency.