Hypergeometric Differential Equation

 
  If you ever have taken an undergraduate level quantum physics class based on Griffiths' textbook, and if you have solved many problem sets of his textbook, you might've confronted a potential in a form of $ \mathrm{sech}^2 $ of which name is "Pöschl–Teller" potential. This potential manifests an interesting property, "reflectionless", which is tautological that its transmission coefficient is exactly 1. However, Griffiths just presents us the exact wave function and let us to check whether this eigen-function is really reflectionless. ISN'T IT ANNOYING? Of course it is not that easy to solve the Schrodinger's equation under this potential. But not so much difficult to solve it either!

  If we are equipped with some knowledges about the "hypergeometric" differential equation, then we can derive the eigenstates of this potential pretty easily. Hence, in this post, i'm gonna talk about the general solution of the hypergeometric differential equation and some properties in which the general solutions of this ODE inhere.

0. Hypergeometric Differential Equation

Consider the hypergeometric differenctial equation.

\[x\,(1-x) \, y'' \;+\; \left[ \,c-(a+b+1)\,x\, \right] \,y' \;-\; ab\,y \;=\; 0 \qquad \qquad \forall a, b, c \in \mathbb{R} ,\quad c>0\]
Note that the ODE above has singular points at $ x=0, 1, \infty $.

Since $ x=0 $ is a regular singularity, it ensures at least one solution is in the form of Frobenius series. Hence, we may write down the first solution as \[y(x) = \sum_{n=0}^{\infty} a_n x^{n+s}\] and by plugging it into the differential equaiton above, we obtain \[\sum_{n=0}^{\infty} \Biggl[ \, x(1-x) a_n (n+s)(n+s-1) x^{n+s-2} + [ c-(a+b+1) x ] a_n (n+s) x^{n+s-1} -ab \, a_n x^{n+s} \,\Biggr] = 0 \; .\]
By sorting each term to have same order of $ x $, the equation above is reduced into \[\sum_{n=0}^{\infty} x^{n+s-1} \Biggl[\, a_n (n+s)(n+s-1+c) - a_{n-1} \Bigl[ (n+s-1)(n+s-2) + (n+s-1)(a+b+1) + ab \Bigr] \Biggr] = 0 \; .\]

For $ n=0 $, only the first term contributes and we obtain the secular equation. \[s(s-1+c) = 0 \qquad \Rightarrow \qquad s=0 \; , \;\; s=1-c\]
For $ n \ge 1 $, we should take into account both terms and therefore we obtain a recursion relation. \[a_n (n+s)(n+s-1+c) = a_{n-1} \Bigl[ (n+s-1)(n+s-2) + (n+s-1)(a+b+1) + ab \Bigr]\]

1. First Solution ( $ s=0 $ )

For $ s=0 $, the recursion relation is written as

\[a_{n+1} = \frac{(n+a)(n+b)}{(n+1)(n+c)}\, a_n \, . \]

Let us assume that $ a_0 = 1 $ for convenience. Then,

\[y_{1}(x) = \sum_{n=0}^{\infty} a_{n} x^{n} = 1 + \frac{ab}{c} \frac{x}{1!} + \frac{a(a+1)\,b(b+1)}{c(c+1)} \frac{x^2}{2!} + \frac{a(a+1)(a+2)\,b(b+1)(b+2)}{c(c+1)(c+2)} \frac{x^3}{3!} + \cdots\]

Therefore, applying mathematical induction, we can generalize our consequence into

\[y_{1}(x) \equiv F\,(a,b,c:x) = \frac{\Gamma(x)}{\Gamma(a)\,\Gamma(b)} \sum_{n=0}^{\infty} \frac{\Gamma(a+n)\,\Gamma(b+n)}{\Gamma(c+n)}\, \frac{x^n}{n!}\]

and we refer $ F\,(a,b,c:x) $ as Hypergeometric Function.

Observe that

• $ F\,(a,b,c:x) $ converges in the region $ |x|<1 $.

• $ F\,(a,b,c:x) $ converges at $ x=1 $ if $ c> a+b $.

• $ F\,(a,b,c:x) $ converges at $ x=-1 $ if $ c>a+b-1 $.

• $ F\,(a,b,c:x) $ has a symmetry in the exchanging parameters $ a $ and $ b $. $ \quad \rightarrow \quad F\,(a,b,c:x) = F\,(b,a,c:x) $

2. Second Solution ($ \,s=1-c\, $)

For $ s=1-c $, \[a_{n} (n+1-c)\,n = a_{n-1} \Bigl[ \, (n-c)(n+a+b-c) + ab \Bigr]\]

\[\overset{n\,\rightarrow\,n+1}{\Longrightarrow} \qquad a_{n+1}(n+1)(n+2-c) = a_n \Bigl[\, (n+1-c)(n+a+b-c+1)+ab \,\Bigr] = a_{n} (n+1-c+a)(n+1-c+b)\]

Thus, a recursion relation for the second solution is given by \[a_{n+1} = \frac{(n+a-c+1)(n+b-c+1)}{(n+1)(n+2-c)}\, a_n\] and it becomes to the recursion relation of the first solution above if we re-scale constants as below. \[a \rightarrow a-c+1 \; , \quad b \rightarrow b-c+1 \; , \quad c \rightarrow 2-c\]

Hence, we can write the second solution as

\[y_{2}(x) = \sum_{n=0}^{\infty} b_n x^{1-c+n} = (\text{const.})\cdot x^{1-c}\, F\,(a-c+1,\, b-c+1,\, 2-c : x) \, . \]

Therefore, the general solution to the hypergeometric equation which is valid for $ |x|<1 $ is given by

\[\therefore \quad {y(x) \;=\; A\cdot F\,(a,b,c:x) \;+\; B\cdot x^{1-c}\,F\,(a-c+1,\, b-c+1,\, 2-c : x)}\] with $ c $, $ a-b $, $ c-a-b $ are all non-integers!


If the solution is to be regular at $ x=0 $, then one requires $ B=0 $.

If $ c $ is an integer, two solutions will coincide or one will blow up. In such the case, we should apply other methods such as the Wronskian method in order to derive the second solution.

Some Properties of the Hypergeometric Function

There are some interesting properties on the hypergeometric function which can be derived easily.

\[\lim_{n\rightarrow \infty} \, F\,\left(1,m,1 : \frac{x}{m} \right) = e^x \;, \quad F\,\left(m+1, -m, 1 : \frac{1-x}{2}\right) = P_{m}(x) \qquad \forall m \in \mathbb{Z}\]
or
\[F\,\left( \frac{1}{2}, -\frac{1}{2}, \frac{1}{2} : \sin^2{x} \right) = \cos{x}\]

You may check other properties of the hypergeometric fuinction on Arfken’s or Riley’s textbooks.

References

• George B. Arfken, Mathematical Methods for Physicists, Academic Press

• K. F. Riley, Mathematical Methods for Physics and Engineering, Cambridge University Press