How to transform the Thomas Fermi differential equation?

Question

We had an exercise without a solution that defined the thomas fermi differential equation as $$\frac{d^2y(t)}{dt^2} = \frac{y^{3/2}(t)}{t^{1/2}}$$ with initla values $$y(0)=y_0\neq0, \frac{dy(0)}{dt}=z_0.$$

The goal was to transform this differential equation to a system of differential equations of first order, which in addition satisfies the Lipschitz-condition. The following hint was given:

Use the substitution: $$s=t^{1/2},y(t)=w(s),u(s)=\frac{w'(s)}{s}.$$

But now I am really lost and do not know what to do. What are the essential steps in the substitution and how would I apply a similar method to another problem of this kind?

These kind of problems are new to me. Thanks in advance!

Answer 1

We begin by computing the derivatives of $w$ with respect to $s$:\begin{align} \frac{dw}{ds}&=\frac{dy}{dt}\,\frac{dt}{ds}=2\,s\,\frac{dy}{dt},\\ \frac{d^2w}{ds^2}&=2\,\frac{dy}{dt}+4\,s^2\,\frac{d^2y}{dt^2}=u+4\,s\,w^{3/2}. \end{align} The first equation of the new system will be $w'=s\,u$. To get the second one we compute $u'$: \begin{align} u'&=-\frac{1}{s^2}\,w'+\frac{1}{s}\,\frac{d^2w}{ds^2}\\ &=-\frac{1}{s}\,u+\frac{1}{s}\bigl(u+4\,s\,w^{3/2}\bigr)\\ &=4\,s\,w^{3/2}. \end{align}

Answer 2

You have $w(s)=y(s^2)$ so that $$w'(s)=2sy'(s^2)=su(s),$$ thus $u(s)=2y'(s^2)$ and $$u'(s)=4sy''(s^2)=4y(s^2)^{3/2}=4w(s)^{3/2}.$$

Answer 3

This may not be exactly what you're looking for, but you may find it interesting given the types of problems you are solving.

Use a group transformation $t'=\lambda t$ and $y'=\lambda ^\beta y$ with the unit transformation occuring at $\lambda =1$. (The primes denote transformed variables.) Then $\frac{d^2 y'}{dt'^2}=\lambda^{\beta -2} \frac{d^2 y}{dt^2}$. Apply this to your DEQ and equate the coefficients and you'll find that $\beta -2 = \frac{3}{2}\beta - \frac{1}{2}$ which results in $\beta = -3$. Thus the group invariant to your Thomas Fermi equation is $t'=\lambda t$ and $y'=\lambda^{-3} y$, $\lambda_o = 1$. Stabilizers of this group will partition the DEQ and reduce your order.

To find the stabilizers take partial derivatives with respect to $\lambda$ of t', y', $\dot{y}'$ and $\ddot{y}'$ evaluated at $\lambda =1$ and then use the method of characteristics. $$\frac{dt}{t}=\frac{dy}{\beta y}=\frac{d\dot{y}}{(\beta -1)\dot{y}}=\frac{d\ddot{y}}{(\beta -2)\ddot{y}}=etc...$$ Independent integrations of these fractions give rise to constants of integration, and because constants are stabilizers for the group of polynomials these constants will partition your DEQ. For your invariant group these stabilizers are $S_0 = t^3 y$, $S_1 = t^4 \dot{y}$, and $S_2 = t^5 \ddot{y}$. The Thomas Fermi equation then becomes $S_2 = S_0^{\frac{3}{2}}$.

Since $t\frac{dS_0}{dt}=S_1+3S_0$ and $t\frac{dS_1}{dt}=S_2+4S_1=S_0^{\frac{3}{2}}-4S_1$, the order is now reduced: $$\frac{dS_1}{dS_0}=\frac{S_0^{\frac{3}{2}}-4S_1}{S_1+3S_0}$$ This can be solved numerically but if you want an analytic solution look at the separatrix between the poles where $\frac{dS_0}{dt}=\frac{dS_1}{dt}=0$, resulting in $S_2=(-3)(-4)S_0=12S_0$. Thus $12S_0=S_0^{\frac{3}{2}}$ resulting in $y=\frac{144}{t^3}$ which solves your equation.

Lawrence Dresner has a more complete treatment of this problem in his Applications of Lie's Theory of Ordinary and Partial Differential Equations.