Solving Burger’s equation

We are now in a position to modify our advection solver to do Burgers’ equation.

Boundary conditions

For the problems we are going to consider, we want outflow boundary conditions. These have the property that the ghost cells simply copy the value just inside the domain, so for the left boundary, we would do:

\[u_{\mathrm{lo}-1} = u_{\mathrm{lo}}\]

\[u_{\mathrm{lo}-2} = u_{\mathrm{lo}}\]

and similar for the right boundary.

This is the only change needed to the FVGrid class (along with renaming the data array u instead of a).

Interface states and fluxes

We get the interface state by reconstructing \(u\) to each interface, getting left and right states and then solve the Riemann problem. Once we have the interface state, \(u_{i+1/2}\), we can compute the flux:

\[F_{i+1/2} = \frac{1}{2} u_{i+1/2}^2\]

and then the divergence of the fluxes:

\[-\frac{1}{\Delta x} (F_{i+1/2} - F_{i-1/2}) = -\frac{1}{\Delta x} \left ( \frac{1}{2} u_{i+1/2}^2 - \frac{1}{2} u_{i-1/2}^2 \right )\]

Main driver

The biggest difference for our main driver between advection and Burgers’ equation is that we need to recompute the timestep each step, since \(u\) changes in space and time.

Test problems

We will do 2 different test problems: a shock and a rarefaction

Steepening into a shock

Consider the profile:

\[\begin{split}u(x, t=0) = \left \{ \begin{array}{cc} 1 & \mbox{if } x < \frac{1}{3} \\ 1 + \frac{1}{2} \sin\left (6\pi (x - \frac{1}{3}) \right ) & \mbox{if } \frac{1}{3} \le x < \frac{2}{3} \\ 1 & \mbox{if } x \ge \frac{2}{3} \end{array} \right . \end{split}\]

This velocity is positive everywhere, but not uniform. This places a single wavelength of a sine wave in the middle of third of our domain. As time evolves, the higher velocity will steepen into a shock.

Rarefaction

Now consider the profile:

\[\begin{split}u(x, t=0) = \left \{ \begin{array}{cc} 1 & \mbox{if } x < \frac{1}{2} \\ 2 & \mbox{if } x \ge \frac{1}{2} \end{array} \right . \end{split}\]

This again is positive everywhere, but the velocity in the righthalf of the domain is larger, so it will spread out, creating a rarefaction to connect the left and right states.

Measuring the shock speed

Finally, let’s consider a simple shock:

\[\begin{split}u(x, t=0) = \left \{ \begin{array}{cc} 2 & \mbox{if } x < \frac{1}{2} \\ 1 & \mbox{if } x \ge \frac{1}{2} \end{array} \right . \end{split}\]

This will immediate create a shock moving to the right. We know from the jump conditions that the shock speed should be:

\[S = \frac{1}{2} (u_l + u_r) = \frac{3}{2}\]

for this case.

Run this problem for a time \(T = 0.2\) and measure the shock speed by estimating the position of the discontinuity at two points in time (like \(t = 0.1\) and \(t = 0.2\)) and differencing.