I'm studying the Taylor-Sedov self-similar solution to the problem of a strong explosion in a homogenoeus atmosphere. The problem is discussed in Landau & Lifschitz VI (in the 2nd edition it's §106). In these notes there's a reproduction of Landau's derivation, that solves analitically the problem (I believe that the original solution is due to Sedov).

In a nutshell ($R$ is the radius of the shock, $r$ is the radial coordinate, the quantities with $_0$ are the unperturbed quantities in front of the shock). enter image description here

The last integral is derived by Landau and in the notes above by energetical considerations (that is, by requiring that the energy within a radius $\eta R$ with $\eta \in (0,1)$ stays constant in time).

Now, the EDOs above for $f,\phi,\psi$ were solved numerically by G.I. Taylor in 1941. See here. I'm comparing the analytical solution with Taylor's numerical solution, and looking at the table of values at page 164, I realized that the integral above, is far from conserved. The identity is satisfied at $\eta = 1$ but for $\eta = 0.5$ the difference beetween LHS and RHS is about $62$!

I am confused: is the above integral an additional constraint, following from physical considerations, which allows to solve analitically the equations of motion? I believe that it isn't so, the solution to this problem should be unique and so has to satisfy the relation.

Thanks in advance for any help.

For clarity, in Landau's book the adimensionalization is done in a slitghly different manner: $$u=\dfrac{2r}{5t}V=U\cdot V,\quad \rho = \rho _0 G,\quad c^2=\gamma\dfrac {p}{\rho}=\dfrac{4r^2}{25t^2}Z=U^2Z, $$ and the integral writes as: $$Z=\dfrac{\gamma(\gamma-1)(1-V)V^2}{2(\gamma V-1)}.$$ Going back to Taylor's variables, one finds: $$V=\phi,\quad G=\psi,\quad Z=f/\psi$$ and so the above relation.


I had made a silly mistake in the conversion from $Z,V,G$ to $\phi, f,\psi$ coordinates. The correct relations are: $$\eta ^2 Z = f/\psi,\qquad\eta V = \phi.$$ With these substitutions, the integral reads: $$f=\dfrac{\gamma(\gamma-1)}{2}\dfrac{\eta-\phi}{\gamma \phi -\eta}\phi ^2 \psi$$ and is perfectly conserved along my numerical solution.

  • $\begingroup$ If I can do anything to improve my question, let me know. $\endgroup$ – pppqqq Feb 10 '15 at 11:52

On page 38 of the lecture notes, the author says,

Since the energy is independent of time, the surface integral must be zero. The terms in the integrand are independent of angle and therefore the integrand itself must be zero

where the integral in question is on page 37, $$ \int_\Omega\left[\rho V\left(\frac12V^2+h\right)-\left(\varepsilon+\frac12\rho V^2\right)\frac{dr}{dt}\right]r^2d\Omega $$ Then, on page 41

Since we have first integral of the equations, the number of independent equations is now reduced from three to two.

So the energy integral method itself introduces a constraint based on the fact that energy is conserved. Hence the reduction from 3 to 2 equations.

Zel'dovich states

Using a brilliant method, which employed the energy integral, Sedov succeeded in finding an exact analytic solution to the equations of self-similar motion.

which enhances the point that it was done to find the analytic solution.

Obviously the two solutions should match at $\eta=1$ because it is a boundary condition. At less than that, I am not convinced the relationship, $$ Z=\frac{\gamma(\gamma-1)}{2}\frac{V^2(1-V)}{\gamma V-1} $$ is satisfied. Note that as $\eta\to0$, $V\to0$. When $V\to0$, the RHS above becomes negative (occurs at $\eta=0.92$ using Taylor's table). Since $Z$ represents the ratio of pressure and density, both of which are non-negative, I do not see how this condition can be valid for $\eta\neq1$.

Both sets of solutions are valid to some degree of accuracy. Frank Timmes, along with James Kamm, solved the Sedov problem again (for generic coordinate systems (dimension $j$) and with generic external ambient profile, $\rho(r)=\rho_0r^{-\omega}$) and gets similar numbers as Sedov for the $j=2,\,\omega=0$ case (off in the 3rd or 4th decimal place in some places, see the Kamm, Bolstead (RIP), & Timmes paper in the above link).
Taylor, on the other hand, used an approximate solver for the ODE. This is where any difference between the two lies: numerical methods vs analytic solution. I've plotted the three solutions (Taylor, Timmes & Sedov) for the density (scaled to the peak) for the $j=2,\,\gamma=1.4$ case, so you can see the minor divergence enter image description here

The fact that $f(\eta)/\psi(\eta)$ does not match the RHS for $\eta<1$ does not mean that the solution is wrong, it only means that the ratio is only valid for $\eta=1$.

  • $\begingroup$ Hi @Kyle Kanos, thank you. I totally agree with your last point, but I am now confused. I'm assuming that the Cauchy problem for "$\psi$, $\phi$, $f$", or $V,Z,G$, with the Rankine-Hugoniot relations as initial conditions has a unique solution. Better, that the fluidodynamical problem, with the datum of $E$, the energy released at $t=0$, has a unique solution. So it seems to me mandatory that the numerical solution converges to the analytical one obtained by the energy trick. Why it isn't so? $\endgroup$ – pppqqq Feb 11 '15 at 7:37
  • $\begingroup$ @pppqqq: I've updated my answer in relation to your comment. $\endgroup$ – Kyle Kanos Feb 11 '15 at 13:51
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    $\begingroup$ I've just discovered that I had made a silly mistake in the conversions beetween Sedov's variables $V,Z,G$ and Taylor's $\psi, \phi, f$; the correct relationships are $V=\phi/\eta$, $\eta^2 Z=f/\psi$. With this correction, my solution coincides exactly with Timmes' and the integral is perfectly conserved. Thank you very much for pointing me those papers! $\endgroup$ – pppqqq Feb 11 '15 at 15:42

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