# Euler Equations, Sod shock tube & conservation

Conservation of momentum?

I am considering the Euler equations in conservative form and solving the Sod shock tube problem I have written a Godunov finite volume type solver. It solves for density ρ, momentum ρu, and total energy E; therefore, I would expect all of these quantities to be conserved wrt time. Density and total energy are indeed conserved, however, momentum is not. The Euler equations are given by $$\frac{\partial}{\partial t}\begin{bmatrix} \rho \\ \rho u \\ E \end{bmatrix} + \frac{\partial}{\partial x}\begin{bmatrix} \rho u \\ \rho u^2 + p \\ u(E +p) \end{bmatrix} = 0$$

where pressure is related to the conserved quantities by $p = (\gamma - 1)(E - 0.5 \rho u^2)$

The Sod shock problem splits the domain into two regions separated by a density and pressure discontinuity with initial velocity zero. That is, $\rho_L = 1, \rho_R = .125; p_L=1, p_R =.1; u_L=u_R =0.$ These initial conditions imply that ρu=0, therefore, momentum should be zero throughout the simulation.

The solution profiles are well known and can be found here. We see that ρ >0 and u > 0, therefore, there is no way for momentum ρu = 0 (which it should be from the intial conditions). As a result, I do not even see why it is reasonable to expect that momentum would be conserved. A paper by Sod which surveys some methods for solution, on page 20, list a table which shows momentum to be increasing linearly. I generally do not work in this area, so maybe I am missing something basic. Can anyone shed some light on this? Thanks!

• The imbalance of pressure in the initial condition is being converted into momentum. – Joce Jul 1 '14 at 7:15
• Note that the code used to display the animation is not very stable and there a lot of numerical instabilities in these simulations which are non-physical. Just something to be aware of... – MoonKnight Jul 9 '14 at 9:26

Formally, this is mathematically described using the integral formulation, $$\frac{\partial}{\partial t}\iiint_V\rho \mathbf u\,dV=-\oint_S\left(\rho \mathbf u\,d\mathbf S\right)\mathbf u-\oint_S p\,d\mathbf S+\iiint_V\rho\mathbf f_{body}\,dV+\mathbf F_{surf}$$ where $\mathbf f_{body}$ are body forces and $\mathbf F_{surf}$ are surface forces, $d\mathbf S$ is the cell surface and $dV$ its volume; all other variables take their normal meaning.
In the case of Eulerian hydrodynamics, $\mathbf f_{body}=\mathbf F_{surf}=0$. Then we can use the divergence theorem to obtain (for a 1D flow), $$\frac{\partial\rho u}{\partial t}+\frac{\partial}{\partial x}\left(\rho u^2+p\right)=0$$ So this equation does contain the conservation law, just not quite how you were expecting it. You may also want to read over the Wikipedia article on the Rankine-Hugoniot jump conditions, as this might explain a bit more clearly the quote I give at the top.