Burgers’ Riemann Problem

Burgers’ Riemann Problem#

As discussed with advection, the physics of the equation comes into play when we determine what the correct state is on the interface given a left and right state. This is the Riemann problem. For Burgers’ equation, this Riemann problem is more complex.

Shock speed#

If the characteristics intersect in the $x$ - $t$ plane, then it is not possible to trace backwards in time to see where the information came from—this is the condition of a shock. The shock speed is computed through the Rankine-Hugoniot jump conditions.

Consider the following space-time diagram:

../_images/rh.png — Fig. 4 A rightward moving shock in the $x$ - $t$ plane separating two state#

This shows a left and right state, $u_{l}$ and $u_{r}$ , separated by a rightward moving shock in the $x$ - $t$ plane (the dark line). Notice that:

At time $t^{n}$ , the state in our integral $x \in [x_{l}, x_{r}]$ is entirely $u_{r}$ .
As time evolves (think about moving upward in this figure), the state becomes a mix of states $u_{l}$ and $u_{r}$ .
Finally at time $t^{n + 1}$ the state is entirely $u_{l}$ in $x \in [x_{l}, x_{r}]$ .
The shock speed is clearly $S = Δ x / Δ t$ in this figure.

To determine the speed, we start with Burgers’ equation in conservative form:

u_{t} = - [f (u)]_{x}

and integrate our conservation law over space and time (and normalize by $Δ x$ ):

\frac{1}{Δ x} \int_{x_{l}}^{x_{r}} d x \int_{t^{n}}^{t^{n + 1}} d t u_{t} = - \frac{1}{Δ x} \int_{x_{l}}^{x_{r}} d x \int_{t^{n}}^{t^{n + 1}} d t [f (u)]_{x}

Recognizing that at $t = t^{n}$ , $u = u_{r}$ and at $t = t^{n + 1}$ , $u = u_{l}$ , the left side becomes:

\frac{1}{Δ x} \int_{x_{l}}^{x_{r}} {u (t^{n + 1}) - u (t^{n})} d x = u_{l} - u_{r}

Now for the right side. We see that all along $x = x_{l}$ , the flux is $f = f (u_{l})$ for $t \in [t^{n}, t^{n + 1}]$ . Likewise, all along $x = x_{r}$ , the flux is $f = f (u_{r})$ in the same time interval. Therefore, our expression becomes:

(u_{l} - u_{r}) = - \frac{Δ t}{Δ x} [f (u_{r}) - f (u_{l})]

Using $S = Δ x / Δ t$ , we see:

S = \frac{f (u_{r}) - f (u_{l})}{u_{r} - u_{l}}

and taking $f (u) = u^{2} / 2$ , we get:

S = \frac{1}{2} (u_{l} + u_{r})

Sampling the solution#

Now that we understand the shock speed, we need to determine what the state is on the interface.

For an interface in our domain, $u_{i + 1 / 2}$ , we need to solve the Riemann problem $u_{i + 1 / 2} = R (u_{i + 1 / 2, L}, u_{i + 1 / 2, R})$ . We do this first by looking at whether the flow is converging or diverging. For converging flow, $u_{i + 1 / 2, L} > u_{i + 1 / 2, R}$ , we need to consider a shock; otherwise we consider a rarefaction. We write this as:

\begin{array}{r} u_{i + 1 / 2} = R (u_{i + 1 / 2, L}, u_{i + 1 / 2, R}) = {\begin{array}{cc} u_{shock} & if u_{i + 1 / 2, L} > u_{i + 1 / 2, R} \\ u_{rare} & otherwise \end{array} \end{array}

where $u_{s}$ is the shock case and $u_{r}$ is the rarefaction case.

For the shock, we look at the direction the shock is moving and choose the appropriate state:

\begin{array}{r} u_{shock} = {\begin{array}{cc} u_{i + 1 / 2, L} & if S > 0 \\ u_{i + 1 / 2, R} & if S < 0 \end{array} \end{array}

For the rarefaction, we do:

\begin{array}{r} u_{rare} = {\begin{array}{cc} u_{i + 1 / 2, L} & if u_{i + 1 / 2, L} > 0 \\ u_{i + 1 / 2, R} & if u_{i + 1 / 2, R} < 0 \\ 0 & otherwise \end{array} \end{array}

Burgers’ Riemann Problem

Contents

Burgers’ Riemann Problem#

Shock speed#

Sampling the solution#