The Riemann (or Sod) ShockTube Problem
Key References
 Sod, G. A. 1978, Journal of Computational Physics, 27,
131.

Hawley, J. F., Smarr, L. L., and Wilson, J. R. 1984,
The Astrophysical Journal, 277, 296311.
"A Numerical Study of Nonspherical Black Hole Accretion. I.
Equations and Test Problems"
[See especially section IV, starting on page 300.]

Hawley, J. F., Smarr, L. L., and Wilson, J. R. 1984,
The Astrophysical Journal Supplement Series, 55, 296311.
"A Numerical Study of Nonspherical Black Hole Accretion. II.
FiniteDifferencing and Code Calibration"

Stone, J. M., and Norman, M. L. 1992,
The Astrophysical Journal Supplement Series, 80, 753790.
"ZEUS2D: A Radiation Magnetohydrodynamics Code for Astrophysical
Flows in Two Space Dimensions.
I. The Hydrodynamic Algorithms and Tests"
[See especially section 5.4, pp. 775777.]
Adopted Initial Conditions

g = 1.4

to the
left of x_{0} 
to the
right of x_{0} 
r = 1 
P = 1 
v = 0 
r = 0.125 
P = 0.1 
v = 0 
c_{sound} = [
g P /
r ]^{1/2}
= 1.183216

c_{sound} = [
g P /
r ]^{1/2}
= 1.058301

Key Positions
Key Velocities
Relationship Between Key Positions and Velocities
at any time t greater than 0.
Shock jump conditions imply :
So, if we can figure out the values of v_{post}
and P_{post}, we can determine v_{shock}.
Determining P_{post}
and v_{post}
The value of P that satisfies the following expression is the desired
value of P_{post}.
Then,
For example, with g = 1.4, P_{right} = 0.1,
and r_{right} = 0.125
(as used by Sod), I derive
P_{post} = 0.3190.
Hence, v_{post} = 0.89095.
Determining v_{shock},
r_{post}
and
r_{middle}
Hence, also,
Notice that, behind the shock (i.e., to the left of x_{4})
but to the right of the contact
discontinuity (x_{3}), the specific entropy of the fluid has been
increased over its preshock value because of heating through the shock. Hence,
( r_{post}/r_{right} )
= 2.1915 does not equal
( P_{post}/P_{right} )^{1/g}
=
2.290 .
But this adiabatic relationship between P and r must
be hold for the material that is to the left of the discontinuity
(x_{3}), that is,
Hence, r_{middle} = 0.44215.
Solution Between x_{1}
and x_{2}
Finally in the rarefaction wave,
Then, at all locations,