Form of Confluent Hypergeometric Equation
The Confluent Hypergeometric Equation is a degenerate case of the hypergeometric equation, which has the form of
\begin{align}
xy^{\prime\prime} + (c-x)y’ -ay =0 \tag{1}
\end{align}
It clearly has a regular singular point at \(x = 0\). But what is not obvious is that the point at infinity \( x = \infty\) is an essential singularity. This can be seen by applying the substitution \( t=1/x\) and (1),(2) in the last tutorial:
\begin{align}
\frac{1}{t}(-t^2(-2t\frac{dy}{dt} -t^2\frac{d^2y}{dt^2})) + (c-\frac{1}{t})(-t^2 \frac{dy}{dt}) -ay &=0 \\
2t^2\frac{dy}{dt} + t^3\frac{d^2y}{dt^2} + (t-ct^2)\frac{dy}{dt} -ay &=0 \\
\frac{d^2y}{dt^2} + (\frac{1}{t^2}-\frac{c-2}{t})\frac{dy}{dt} -\frac{a}{t^3}y &=0 \tag{2}
\end{align}
\( x = \infty \) now becomes an essential singularity (\(1/t^2\) and \(1/t^3\) for \(y’\) and \(y\)) because the confluent hypergeometric equation merges (confluence) the original two regular singular points \( x = 1, \infty \). This is done by letting \( x = z/b \) in the old hypergeometric equation and taking the limit \( b \to \infty \). Then both \( x = 1, \infty \) correspond to \( z = \infty \), \( dy/dx = (dy/dz) (dz/dx) = b (dy/dz) \) and similarly \( d^2y/dx^2 = b^2 d^2y/dz^2 \). The hypergeometric equation ((1) of this post) is then transformed into
\begin{align}
\lim_{b\to\infty}[\frac{z}{b}(1-\frac{z}{b}) (b^2\frac{d^2y}{dz^2}) + [c-(a+b+1)\frac{z}{b}](b\frac{dy}{dz}) -aby] &= 0 \\
\lim_{b\to\infty}[z(1-\frac{z}{b}) \frac{d^2y}{dz^2} + [c-(\frac{a}{b}+1+\frac{1}{b})z]\frac{dy}{dz} -ay] &= 0 \\
z(1)\frac{d^2y}{dz^2} + (c-(0+1+0)z)\frac{dy}{dz} -ay &= 0 \\
z\frac{d^2y}{dz^2} + (c-z)\frac{dy}{dz} -ay &= 0 \tag{3}
\end{align}
which is the same as (1) just with \(x\) now being \(z\). Similarly, the solutions of the confluent hypergeometric equation will be the ones in the old hypergeometric equation where we let \(x \to x/b\) and \(b \to \infty \): from (7) of this post:
\begin{align}
y_1 &= \lim_{b\to \infty} [1+\sum_{n=1}^{\infty} \frac{a(a+1)\cdots(a+n-1)b(b+1)\cdots(b+n-1)}{n!c(c+1)\cdots(c+n-1)} (\frac{x}{b})^n] \\
&= 1+\sum_{n=1}^{\infty} \lim_{b\to \infty} \frac{b(b+1)\cdots(b+n-1)}{b^n} \frac{a(a+1)\cdots(a+n-1)}{n!c(c+1)\cdots(c+n-1)} x^n \\
&= 1+\sum_{n=1}^{\infty} (1)^n \frac{a(a+1)\cdots(a+n-1)}{n!c(c+1)\cdots(c+n-1)} x^n \\
&= 1 + \frac{a}{c}\frac{x}{1!} + \frac{a(a+1)}{c(c+1)}\frac{x^2}{2!} + \cdots \tag{4}
\end{align}
(assumed possible to interchange the limit and summation) The quantity in (4) is then known as the confluent hypergeometric function \( M(a,c;x) \) (or Kummer function). In the same way for the hypergeometric function, the second solution is given by (below (15) of this post) \( y_2 = x^{1-c}M(a-c+1,2-c;x) \), provided that \( c \) is not an integer.
Properties of Confluent Hypergeometric Function
First, let’s take a look at some special values that the confluent hypergeometric function may take: (again they follow from the textbook Mathematical Methods for Physics and Engineering by Riley, Hobson, and Bence.)
\begin{align}
\begin{aligned}
M(a,a;x) &= e^x \\
M(1,2;2x) &= e^x\frac{\sinh(x)}{x} = \frac{e^{2x}-1}{2x} \\
M(\frac{1}{2}, \frac{3}{2}; -x^2) &= \frac{\sqrt{\pi}}{2x}\text{erf}(x)
\end{aligned} \tag{5}
\end{align}
Here we briefly verify the second of them by working on (4):
\begin{align}
M(1,2;2x) &= 1+\sum_{n=1}^{\infty} \frac{(1)(2)\cdots(n)}{n!(2)(3)\cdots(n+1)} (2x)^n \\
&= 1+\sum_{n=1}^{\infty} \frac{(2x)^n}{(n+1)n!} = 1+\sum_{n=1}^{\infty} \frac{(2x)^n}{(n+1)!} \\
&= \frac{1}{2x}(2x+\sum_{n=1}^{\infty} \frac{(2x)^{n+1}}{(n+1)!}) = \frac{1}{2x} \sum_{n=0}^{\infty} \frac{(2x)^{n+1}}{(n+1)!} \\
&= \frac{1}{2x}(e^{2x}-1) \tag{6}
\end{align}
as we identify the summation with the Taylor series of \( e^{2x} \) minus \(1\).
Meanwhile, we can also derive the integral representation of the confluent hypergeometric function. We recall (7) of this tutorial with \(a\) and \(b\) swapped as enabled by the symmetry of the original hypergeometric function, and then we do the same of letting \(x \to x/b\) and \(b \to \infty \) as above:
\begin{align}
M(a,c;x) &= \lim_{b\to\infty} F(a,b,c;\frac{x}{b}) \\
&= \lim_{b\to\infty} \frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)} \int_0^1 t^{a-1}(1-t)^{c-a-1} (1-t\frac{x}{b})^{-b} dt \\
&= \frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)} \int_0^1 t^{a-1}(1-t)^{c-a-1} \lim_{b\to\infty} (1-\frac{tx}{b})^{-b} dt \\
&= \frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)} \int_0^1 t^{a-1}(1-t)^{c-a-1} e^{tx} dt \tag{7}
\end{align}
(again assumed that the limit can be put inside the integral) where we recall the elementary limit \( \lim_{n\to\infty} (1 + x/n)^{n} = e^x \).
Finally, we can obtain relationships similar to (8) and (9) of this tutorial for the confluent hypergeometric equation instead. We will make the change of variable \( t = 1-u \) for (7) to obtain
\begin{align}
M(a,c;x) &= \frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)} \int_0^1 (1-u)^{a-1}u^{c-a-1} e^{(1-u)x} d(1-u) \\
&= \frac{\Gamma(c)}{\Gamma(c-a)\Gamma(a)} \int_0^1 u^{c-a-1}(1-u)^{a-1} e^{x}e^{-ux} du \\
&= e^{x} \frac{\Gamma(c)}{\Gamma(c-a)\Gamma(a)} \int_0^1 u^{(c-a)-1}(1-u)^{c-(c-a)-1} e^{u(-x)} du \\
&= e^{x} M(c-a,c;-x) \tag{8}
\end{align}
On the other hand, differentiating (4) straightforwardly produces
\begin{align}
M'(a,c;x) &= \frac{d}{dx} [1+\sum_{n=1}^{\infty} \frac{a(a+1)\cdots(a+n-1)}{n!c(c+1)\cdots(c+n-1)} x^n] \\
&= \sum_{n=1}^{\infty} \frac{na(a+1)\cdots(a+n-1)}{n!c(c+1)\cdots(c+n-1)}x^{n-1} \\
&= \sum_{n=1}^{\infty} \frac{a(a+1)\cdots(a+n-1)}{(n-1)!c(c+1)\cdots(c+n-1)}x^{n-1} \\
&= \sum_{n=0}^{\infty} \frac{a(a+1)\cdots(a+n+1-1)}{n!c(c+1)\cdots(c+n+1-1)}x^{n} \\
&= \frac{a}{c}\sum_{n=0}^{\infty} \frac{(a+1)\cdots(a+1+n-1)}{n!(c+1)\cdots(c+1+n-1)}x^{n} \\
&= \frac{a}{c}M(a+1,c+1;x) \tag{9}
\end{align}
where in the middle we have shifted the index by \(1\) and recognized the new confluent hypergeometric function with parameters raised by \(1\).
Exercise
Directly derive the confluent hypergeometric function (4) by applying the Frobenius method to the confluent hypergeometric equation (1) without resorting to the limiting process.
Answer
Following the idea outlined in this tutorial, we first obtain the indicial equation, where particularly \( p_{-1} = c \):
\begin{align}
m^2 + (c-1)m &= 0 \\
m(m-(1-c)) &= 0
\end{align}
Hence, the roots/exponents will be \( m_1 = 0, m_2 = 1-c \). The first one is reduced to the usual series solution, and we can plug (1)-(3) of this tutorial into (1) to get
\begin{align}
x \sum_{n=2}^{\infty} n(n-1) a_nx^{n-2} + (c-x) \sum_{n=1}^{\infty} n a_nx^{n-1} -a\sum_{n=0}^{\infty} a_n x^n &= 0 \\
\sum_{n=2}^{\infty} n(n-1) a_nx^{n-1} + c\sum_{n=1}^{\infty} n a_nx^{n-1} -\sum_{n=1}^{\infty} n a_nx^n -a\sum_{n=0}^{\infty} a_n x^n &= 0 \\
\sum_{n=1}^{\infty} (n+1)n a_{n+1}x^{n} + c\sum_{n=0}^{\infty} (n+1) a_{n+1}x^{n} -\sum_{n=1}^{\infty} n a_nx^n -a\sum_{n=0}^{\infty} a_n x^n &= 0
\end{align}
By comparing the coefficients, we obtain the recurrence relation when \( n \geq 1 \):
\begin{align}
(n+1)na_{n+1} + c(n+1)a_{n+1} -na_n -aa_n &= 0 \\
a_{n+1} &= \frac{(a+n)a_n}{(c+n)(n+1)}
\end{align}
So the general form of the terms is
\begin{align}
a_n &= \frac{(a+n-1)}{(c+n-1)n}a_{n-1} \\
&= \frac{(a+n-1)(a+n-2)}{(c+n-1)(c+n-2)n(n-1)}a_{n-2} \\
&= \frac{(a+n-1)(a+n-2)\cdots a}{(c+n-1)(c+n-2)\cdots c n!}a_0
\end{align}
and the desired series is (absorbed the \(a_0\) factor)
\begin{align}
y_1 = 1 + \sum_{n=1}^{\infty}\frac{a(a+1)\cdots(a+n-1)}{c(c+1)\cdots (c+n-1)n!}x^n
\end{align}
which is consistent with the confluent hypergeometric function indicated by (4). The process for deriving \( y_2 \) is similar.
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