User:Warrickball/Lane–Emden equation



In astrophysics, the Lane–Emden equation is a dimensionless form of Poisson's equation for the gravitational potential of a self-gravitating, spherically symmetric, polytropic fluid. It is named after astrophysicists Jonathan Homer Lane and Robert Emden. The equation reads


 * $$ \frac{1}{\xi^2} \frac{d}{d\xi} \left({\xi^2 \frac{d\theta}{d\xi}}\right) + \theta^n = 0 $$

where $$\xi$$ is a dimensionless radius and $$\theta$$ is related to the density (and thus the pressure) by $$\rho=\rho_c\theta^n$$ for central density $$\rho_c$$. The index $$n$$ is the polytropic index that appears in the polytropic equation of state,


 * $$ P = K \rho^{1 + \frac{1}{n}}\, $$

where $$P$$ and $$\rho$$ are the pressure and density, respectively, and $$K$$ is a constant of proportionality. The standard boundary conditions are $$\theta(0)=1$$ and $$\theta'(0)=0$$. Solutions thus describe the run of pressure and density with radius and are known as polytropes of index $$n$$.

Applications
Physically, hydrostatic equilibrium connects the gradient of the potential, the density, and the gradient of the pressure, whereas Poisson's equation connects the potential with the density. Thus, if we have a further equation that dictates how the pressure and density vary with respect to one another, we can reach a solution. The particular choice of a polytropic gas as given above makes the mathematical statement of the problem particularly succinct and leads to the Lane–Emden equation. The equation is a useful approximation for self-gravitating gaseous spheres such as stars, but typically it is a rather limiting assumption.

From hydrostatic equilibrium
Consider a self-gravitating, spherically symmetric fluid in hydrostatic equilibrium. Mass is conserved and thus described by the continuity equation


 * $$ \frac{dm}{dr} = 4\pi r^2 \rho $$

where $$\rho$$ is a function of $$r$$. The equation of hydrostatic equilibrium is


 * $$ \frac{1}{\rho}\frac{dP}{dr} = -\frac{Gm}{r^2} $$

where $$m$$ is also a function of $$r$$. Differentiating again gives


 * $$\begin{align}

\frac{d}{dr}\left(\frac{1}{\rho}\frac{dP}{dr}\right) &= \frac{2Gm}{r^3}-\frac{G}{r^2}\frac{dm}{dr} \\ &=-\frac{2}{\rho r}\frac{dP}{dr}-4\pi G\rho \end{align}$$

where we have used the continuity equation to replace the mass gradient. Multiplying both sides by $$r^2$$ and collecting the derivatives of $$P$$ on the left, we can write


 * $$ r^2\frac{d}{dr}\left(\frac{1}{\rho}\frac{dP}{dr}\right)+\frac{2r}{\rho}\frac{dP}{dr}

=\frac{d}{dr}\left(\frac{r^2}{\rho}\frac{dP}{dr}\right)=-4\pi Gr^2\rho $$

Dividing both sides by $$r^2$$ yields, in some sense, a dimensional form of the desired equation. If, in addition, we substitute for the polytropic equation of state with $$P=K\rho_c^{1+\frac{1}{n}}\theta^{n+1}$$ and $$\rho=\rho_c\theta^n$$, we have


 * $$ \frac{1}{r^2}\frac{d}{dr}\left(r^2K\rho_c^\frac{1}{n}(n+1)\frac{d\theta}{dr}\right)=-4\pi G\rho_c\theta^n$$

Gathering the constants and substituting $$r=\alpha\xi$$, where


 * $$\alpha^2=(n+1)K\rho_c^{\frac{1}{n}-1}/4\pi G$$,

we have the Lane–Emden equation,


 * $$ \frac{1}{\xi^2} \frac{d}{d\xi} \left({\xi^2 \frac{d\theta}{d\xi}}\right) + \theta^n = 0 $$

From Poisson's equation
Equivalently, one can start with Poisson's equation,


 * $$ \nabla^2\Phi=\frac{1}{r^2}\frac{d}{dr}\left({r^2\frac{d\Phi}{dr}}\right) = -4\pi G\rho $$

We can replace the gradient of potential using hydrostatic equilibrium, via


 * $$ \frac{d\Phi}{dr}=\frac{1}{\rho}\frac{dP}{dr} $$

which again yields the dimensional form of the Lane–Emden equation.

Exact Solutions
There are only three values of the polytropic index $$n$$ that lead to exact solutions.

$$n=0$$
If $$n=0$$, the equation becomes


 * $$ \frac{1}{\xi^2} \frac{d}{d\xi} \left({\xi^2 \frac{d\theta}{d\xi}}\right) + 1 = 0 $$

Re-arranging and integrating once gives


 * $$ \xi^2\frac{d\theta}{d\xi} = C_1-\frac{1}{3}\xi^3 $$

Dividing both sides by $$\xi^2$$ and integrating again gives


 * $$ \theta(\xi)=C_0-\frac{C_1}{\xi}-\frac{1}{6}\xi^2 $$

The boundary conditions $$\theta(0)=1$$ and $$\theta'(0)=0$$ imply that the constants of integration are $$C_0=1$$ and $$C_1=0$$.

$$n=1$$
When $$n=1$$, the equation can be expanded in the form


 * $$ \frac{1}{\xi^2} \frac{d^2\theta}{d\xi^2}+\frac{2}{\xi}\frac{d\theta}{d\xi} + \theta = 0 $$

Multiplying both sides by $$\xi^2$$ gives a spherical Bessel differential equation with $$k=1$$ and $$n=0$$. The solution, after application of the boundary conditions, is


 * $$ \theta(\xi)=\frac{\sin\xi}{\xi} $$

$$n=5$$
After a sequence of substitutions, it can be shown that the Lane–Emden equation has a further solution


 * $$ \theta(\xi)=\frac{1}{\sqrt{1+\xi^2/3}} $$

when $$n=5$$. This solution is infinite in radial extent.

Numerical Solutions
In general, solutions are found by numerical integration. Many standard methods require that the problem is formulated as a system of first-order ordinary differential equations. For example,


 * $$ \frac{d\theta}{d\xi}=-\frac{\phi}{\xi^2} $$
 * $$ \frac{d\phi}{d\xi}=\theta^n\xi^2 $$

Here, $$\phi(\xi)$$ is interpreted as the dimensionless mass, defined by $$m(r)=4\pi\alpha^3\rho_c\phi(\xi)$$. The relevant boundary conditions are $$\phi(0)=0$$ and $$\theta(0)=1$$. The first equation represents hydrostatic equilibrium and the second mass conservation.

Homology-invariant equation
It is known that if $$\theta(\xi)$$ is a solution of the Lane–Emden equation, then so is $$C^{2/n+1}\theta(C\xi)$$. Solutions that are related in this way are called homologous; the process that transforms them is homology. If we choose variables that are invariant to homology, then we can reduce the order of the Lane–Emden equation by one.

A variety of such variables exist. A suitable choice is


 * $$U=\frac{d\log m}{d\log r}=\frac{\xi^3\theta^n}{\phi}$$

and


 * $$U=\frac{d\log P}{d\log r}=(n+1)\frac{\phi}{\xi\theta}$$

We can differentiate the logarithms of these variables with respect to $$\xi$$, which gives


 * $$\frac{1}{U}\frac{dU}{d\xi}=\frac{1}{\xi}(3-n(n+1)^{-1}V-U)$$

and
 * $$\frac{1}{V}\frac{dV}{d\xi}=\frac{1}{\xi}(-1+U+(n+1)^{-1}V)$$.

Finally, we can divide these two equations to eliminate the dependence on $$\xi$$, which leaves


 * $$\frac{dV}{dU}=-\frac{V}{U}\left(\frac{U+(n+1)^{-1}V-1}{U+n(n+1)^{-1}V-3}\right)$$

This is now a single first-order equation.

Topology of the homology-invariant equation
The homology-invariant equation can be regarded as the autonomous pair of equations


 * $$\frac{dU}{d\log\xi}=-U(U+n(n+1)^{-1}V-3)$$

and


 * $$\frac{dV}{d\log\xi}=V(U+(n+1)^{-1}V-1)$$.

The behaviour of solutions to these equations can be determined by linear stability analysis. The critical points of the equation (where $$dV/d\log\xi=dU/d\log\xi=0$$) and the eigenvalues and eigenvectors of the Jacobian matrix are tabulated below.