Thomas–Fermi equation

In mathematics, the Thomas–Fermi equation for the neutral atom is a second order non-linear ordinary differential equation, named after Llewellyn Thomas and Enrico Fermi,  which can be derived by applying the Thomas–Fermi model to atoms. The equation reads


 * $$\frac{d^2y}{dx^2} = \frac{1}{\sqrt x} y^{3/2}$$

subject to the boundary conditions


 * $$y(0)=1, \quad \quad \begin{cases}y(\infty)=0 \quad \text{for neutral atoms}\\

y(x_0)=0 \quad \text{for positive ions}\\ y(x_1)-x_1y'(x_1)=0\quad \text{for compressed neutral atoms}\end{cases}$$

If $$y$$ approaches zero as $$x$$ becomes large, this equation models the charge distribution of a neutral atom as a function of radius $$x$$. Solutions where $$y$$ becomes zero at finite $$x$$ model positive ions. For solutions where $$y$$ becomes large and positive as $$x$$ becomes large, it can be interpreted as a model of a compressed atom, where the charge is squeezed into a smaller space. In this case the atom ends at the value of $$x$$ for which $$dy/dx=y/x$$.

Transformations
Introducing the transformation $$z=y/x$$ converts the equation to


 * $$\frac{1}{x^2}\frac{d}{dx}\left(x^2\frac{dz}{dx}\right) - z^{3/2}=0$$

This equation is similar to Lane–Emden equation with polytropic index $$3/2$$ except the sign difference. The original equation is invariant under the transformation $$x\rightarrow c x, \ y\rightarrow c^{-3} y$$. Hence, the equation can be made equidimensional by introducing $$y=x^{-3} u$$ into the equation, leading to


 * $$x^2 \frac{d^2u}{dx^2} - 6x\frac{du}{dx} + 12 u = u^{3/2}$$

so that the substitution $$x=e^t$$ reduces the equation to


 * $$\frac{d^2u}{dt^2} - 7\frac{du}{dt} +12 u = u^{3/2}.$$

Treating $$w =du/dt$$ as the dependent variable and $$u$$ as the independent variable, we can reduce the above equation to


 * $$w \frac{dw}{du} - 7w = u^{3/2}-12 u.$$

But this first order equation has no known explicit solution, hence, the approach turns to either numerical or approximate methods.

Sommerfeld's approximation
The equation has a particular solution $$y_p(x)$$, which satisfies the boundary condition that $$y\rightarrow 0$$ as $$x\rightarrow\infty$$, but not the boundary condition y(0)=1. This particular solution is


 * $$y_p(x) = \frac{144}{x^3}.$$

Arnold Sommerfeld used this particular solution and provided an approximate solution which can satisfy the other boundary condition in 1932. If the transformation $$x=1/t, \ w = yt$$ is introduced, the equation becomes


 * $$t^4\frac{d^2w}{dt^2} = w^{3/2}, \quad w(0)=0, \ w(\infty)\sim t.$$

The particular solution in the transformed variable is then $$w_p(t)= 144 t^4$$. So one assumes a solution of the form $$w=w_p(1+\alpha t^\lambda)$$ and if this is substituted in the above equation and the coefficients of $$\alpha$$ are equated, one obtains the value for $$\lambda$$, which is given by the roots of the equation $$\lambda^2 + 7\lambda -6=0$$. The two roots are $$\lambda_1 = 0.772, \ \lambda_2 = -7.772$$, where we need to take the positive root to avoid the singularity at the origin. This solution already satisfies the first boundary condition ($$w(0)=0$$), so, to satisfy the second boundary condition, one writes to the same level of accuracy for an arbitrary $$n$$


 * $$W=w_p(1+\beta t^\lambda)^n = [144 t^3(1+\beta t^\lambda)^n]t.$$

The second boundary condition will be satisfied if $$ 144t^3(1+\beta t^\lambda)^n = 144 t^3 \beta^n t^{\lambda n}(1+\beta^{-1}t^{-\lambda})^n\sim 1$$ as $$t\rightarrow\infty$$. This condition is satisfied if $$\lambda n + 3 =0, \ 144 \beta^n =1$$ and since $$\lambda_1\lambda_2=-6$$, Sommerfeld found the approximation as $$\lambda = \lambda_1, \ n =-3/\lambda_1 = \lambda_2/2$$. Therefore, the approximate solution is


 * $$y(x) = y_p(x) \{1+[y_p(x)]^{\lambda_1/3}\}^{\lambda_2/2}.$$

This solution predicts the correct solution accurately for large $$x$$, but still fails near the origin.

Solution near origin
Enrico Fermi provided the solution for $$x\ll 1$$ and later extended by Edward B. Baker. Hence for $$x\ll 1$$,



\begin{align} y(x) = {} & 1 - Bx + \frac{1}{3} x^3 - \frac {2B} {15} x^4 + \cdots {} \\[6pt] & \cdots + x^{3/2} \left[\frac 4 3 - \frac{2B} 5 x + \frac{3B^2}{70} x^2 + \left(\frac{2}{27} + \frac{B^3}{252}\right) x^3 + \cdots\right] \end{align} $$

where $$B\approx 1.588071$$.

It has been reported by Salvatore Esposito that the Italian physicist Ettore Majorana found in 1928 a semi-analytical series solution to the Thomas–Fermi equation for the neutral atom, which however remained unpublished until 2001. Using this approach it is possible to compute the constant B mentioned above to practically arbitrarily high accuracy; for example, its value to 100 digits is $$B = 1.588 071 022 611 375 312 718 684 509 423 950 109 452 746 621 674825616765677418166551961154309262332033970138428665$$.