Form of Hypergeometric Equation
Hypergeometric Equations are second-order linear ODEs in the form of
\begin{align}
x(1-x) y^{\prime\prime} + [c-(a+b+1)x]y’ -aby = 0 \tag{1}
\end{align}
where \( a,b,c, \) are given parameters. It can be immediately seen that \( x = 0,1 \) are the two regular singular points of (1) through dividing by \(x(1-x)\). The coefficients of (1) are set up to produce a specific form of solutions, which we will derive below. Consider the regular singular point of \(x = 0\) first, it is not hard to see that, according to this tutorial, \( p_{-1} = c \) and \( q_{-2} = 0 \), and hence the indicial equation is
\begin{align}
m^2 + (c-1)m &= 0 \\
m(m+c-1) &= 0 \tag{2}
\end{align}
with the roots \( m_1 = 0, m_2 = 1-c \). If \(m_2 = 1-c\) is not a positive integer, i.e. \( c \) is not \(0\) or a negative integer, then one Frobenius solution with \( m_1 = 0 \) (which reduces to a series solution) will be present:
\begin{align}
y_1 &= x^0 \sum_{n=0}^{\infty} a_nx^n \\
&= \sum_{n=0}^{\infty} a_nx^n = a_0 + a_1x + a_2x^2 + \cdots \tag{3}
\end{align}
Substituting this (equivalently (1)-(3) in this tutorial) into (1) will then lead to
\begin{align}
\begin{aligned} x(1-x) \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2} \\
+ [c-(a+b+1)x] \sum_{n=1}^{\infty} na_nx^{n-1} -ab\sum_{n=0}^{\infty} a_nx^n \end{aligned} &= 0 \\
\begin{aligned} x \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2} -x^2\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}\\
+ c\sum_{n=1}^{\infty} na_nx^{n-1} -(a+b+1)x \sum_{n=1}^{\infty} na_nx^{n-1} -ab\sum_{n=0}^{\infty} a_nx^n \end{aligned} &= 0 \\
\begin{aligned} \sum_{n=2}^{\infty} n(n-1)a_nx^{n-1} -\sum_{n=2}^{\infty} n(n-1)a_nx^{n}\\
+ c\sum_{n=1}^{\infty} na_nx^{n-1} -(a+b+1) \sum_{n=1}^{\infty} na_nx^{n} -ab\sum_{n=0}^{\infty} a_nx^n \end{aligned} &= 0 \\
\begin{aligned} \sum_{n=1}^{\infty} (n+1)na_{n+1}x^{n} -\sum_{n=2}^{\infty} n(n-1)a_nx^{n}\\
+ c\sum_{n=0}^{\infty} (n+1)a_{n+1}x^{n} -(a+b+1) \sum_{n=1}^{\infty} na_nx^{n} -ab\sum_{n=0}^{\infty} a_nx^n \end{aligned} &= 0 \tag{4}
\end{align}
with appropriate indices adjusted by \(1\). Further filling up empty leading terms for some summations and equating the coefficients, we have the recurrence relation of
\begin{align}
\begin{aligned}(n+1)na_{n+1} -n(n-1)a_n \\
+ c(n+1)a_{n+1} -(a+b+1)na_n -ab a_n\end{aligned} &= 0 \\
(n+1)(c+n)a_{n+1} – (n^2+(a+b)n+ab) a_n &= 0 \\
a_{n+1} &= \frac{(a+n)(b+n)}{(c+n)(n+1)}a_n \tag{5}
\end{align}
We can write down the first few terms:
\begin{align}
\begin{aligned}
a_1 &= \frac{ab}{c} a_0 \\
a_2 &= \frac{(a+1)(b+1)}{(c+1)(2)} a_1 = \frac{a(a+1)b(b+1)}{(1)(2)c(c+1)} a_0 \\
a_3 &= \frac{(a+2)(b+2)}{(c+2)(3)} a_2 = \frac{a(a+1)(a+2)b(b+1)(b+2)}{(1)(2)(3)c(c+1)(c+2)} a_0 \\
&\vdots
\end{aligned} \tag{6}
\end{align}
So the first series/Frobenius solution for (1) will be
\begin{align}
y_1 &= a_0 (1 + \frac{ab}{1!c}x + \frac{a(a+1)b(b+1)}{2!c(c+1)}x^2 \\
&\quad + \frac{a(a+1)(a+2)b(b+1)(b+2)}{3!c(c+1)(c+2)} x^3 + \cdots ) \\
&= 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)} x^n \tag{7}
\end{align}
(absorbed the \(a_0\) factor). With the help of the Gamma Function, which has the property of
\begin{align}
\Gamma(n+1) = n\Gamma(n) \tag{8}
\end{align}
When \( n \) is a positive integer, \( \Gamma(n) = (n-1)! \) coincide. (7) can be compactly rewritten as
\begin{align}
y_1 = \frac{\Gamma(c)}{\Gamma(a)\Gamma(b)} \sum_{n=0}^{\infty} \frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(c+n)} \frac{x^n}{n!} \tag{9}
\end{align}
because
\begin{align}
&\quad (\Gamma(a))a(a+1)\cdots(a+n-1) \\
&= (\Gamma(a+1))(a+1)\cdots(a+n-1) \\
&= \cdots = (\Gamma(a+n-1))(a+n-1) = \Gamma(a+n) \tag{10}
\end{align}
and similarly for \(b, c\). (9) is then defined to be the Hypergeometric Function, denoted by \(F(a,b,c;x)\) with \(a,b,c\) as the parameters and \(x\) the variable.
Basic Properties of Hypergeometric Function
The convergence behavior of the Hypergeometric Function can be simply inferred using the Ratio test and (5):
\begin{align}
\left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right| &= \left|\frac{(a+n)(b+n)}{(c+n)(n+1)}\right||x| \to |x| \tag{11}
\end{align}
when \(n \to \infty\) (the leading fraction goes to \(1\) asymptotically). This means that \(F(a,b,c;x)\) converges and is analytic over \(|x| < 1\), given the same restriction that \( c \) is not \(0\) or a negative integer. For reference, it also converges at \(x=1\) if \(c > a+b\) and \(x = -1\) if \(c > a+b-1\). This can be expected from the fact that there is another singularity at \(x = 1\) “blocking” this supposed solution near \( x = 0\). Also, it is easy to notice that the Hypergeometric Function is symmetric in the parameters \( a \) and \( b \), i.e. \( F(a,b,c;x) = F(b,a,c;x) \) remains unchanged when \( a \) and \( b \) are swapped, as the same for the Hypergeometric Equation (1). When \( a \) or \( b \) is zero or a negative integer, then (5) will terminate and (7)/(9) is truncated to a finite polynomial.
Finally, we can understand why it bears the name of Hypergeometric Function. It generalizes the usual geometric series as a special case of it when \( a = 1, c = b \):
\begin{align}
F(1,b,b;x) &= \frac{\Gamma(b)}{\Gamma(1)\Gamma(b)} \sum_{n=0}^{\infty} \frac{\Gamma(1+n)\Gamma(b+n)}{\Gamma(b+n)} \frac{x^n}{n!} \\
&= \sum_{n=0}^{\infty} \frac{n!}{n!} x^n = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + \cdots \tag{12}
\end{align}
as \(\Gamma(1+n) = n!\).
Second Solution to Hypergeometric Equation
(2) indicates that there exists a second Frobenius solution to (1) about the regular singular point \(x = 0\) if the exponent \( m_2 = 1-c \) is not zero or a negative integer, or in other words \( c \) is not a positive integer, with the form of
\begin{align}
y_2 &= x^{1-c} \sum_{n=0}^{\infty} a_nx^n = \sum_{n=0}^{\infty} a_nx^{n+1-c} \tag{13}
\end{align}
The first, direct way to obtain \( y_2 \) is by substitution into (1) again ((4)-(6) in this tutorial). This leads to
\begin{align}
\begin{aligned} x(1-x) \sum_{n=0}^{\infty} (n+1-c)(n-c)a_nx^{n-1-c} \\
+ [c-(a+b+1)x] \sum_{n=0}^{\infty} (n+1-c)a_nx^{n-c} -ab\sum_{n=0}^{\infty} a_nx^{n+1-c} \end{aligned} &= 0 \\
\begin{aligned} \sum_{n=0}^{\infty} (n+1-c)(n-c)a_nx^{n-c} -\sum_{n=0}^{\infty} (n+1-c)(n-c)a_nx^{n+1-c} \\
+ c\sum_{n=0}^{\infty} (n+1-c)a_nx^{n-c} -(a+b+1)\sum_{n=0}^{\infty} (n+1-c)a_nx^{n+1-c} \\
-ab\sum_{n=0}^{\infty} a_nx^{n+1-c} \end{aligned} &= 0 \\
\begin{aligned} \sum_{n=0}^{\infty} (n+2-c)(n+1-c)a_{n+1}x^{n+1-c} -\sum_{n=0}^{\infty} (n+1-c)(n-c)a_nx^{n+1-c} \\
+ c\sum_{n=0}^{\infty} (n+2-c)a_{n+1}x^{n+1-c} -(a+b+1)\sum_{n=0}^{\infty} (n+1-c)a_nx^{n+1-c} \\
-ab\sum_{n=0}^{\infty} a_nx^{n+1-c} \end{aligned} &= 0 \tag{14}
\end{align}
with \(n\) replaced by \(n+1\) in the appropriate summations (the new \(n = -1\) terms cancel out each other). Then, comparing the coefficients yields a new recurrence relation of
\begin{align}
\begin{aligned}(n+2-c)(n+1-c)a_{n+1} -(n+1-c)(n-c)a_n \\
+ c(n+2-c)a_{n+1} -(a+b+1)(n+1-c)a_n -aba_n\end{aligned} &= 0 \\
\begin{aligned}(n+2-c)(n+1)a_{n+1} \\
-((a+b+(n+1-c))(n+1-c) + ab)a_n\end{aligned} &= 0 \\
\begin{aligned}(n+2-c)(n+1)a_{n+1} \\
-(a+n+1-c)(b+n+1-c)a_n\end{aligned} &= 0 \\
\Rightarrow a_{n+1} = \frac{(a-c+n+1)(b-c+n+1)}{(n+2-c)(n+1)}a_n \tag{15}
\end{align}
Notice (15) is the same as (5) but with the replacements \( a \rightarrow a-c+1, b \rightarrow b-c+1, c \to 2-c\), and thus the second solution to the Hypergeometric Equation at \(x = 0\) also follows a Hypergeometric Function: \( y_2 = x^{1-c} F(a-c+1,b-c+1,2-c;x) \).
Another way to derive this is to make a change of variable \( y = x^{1-c}z\). Then
\begin{align}
\begin{aligned}
y’ &= x^{1-c}z’ + (1-c)x^{-c}z \\
y^{\prime\prime} &= x^{1-c}z^{\prime\prime} + 2(1-c)x^{-c}z’ -c(1-c)x^{-1-c}z
\end{aligned} \tag{16}
\end{align}
Plugging these into (1) then gives
\begin{align}
\begin{aligned}x(1-x) (x^{1-c}z^{\prime\prime} + 2(1-c)x^{-c}z’ -c(1-c)x^{-1-c}z) \\
+ [c-(a+b+1)x](x^{1-c}z’ + (1-c)x^{-c}z) -abx^{1-c}z\end{aligned} &= 0 \\
\begin{aligned}x(1-x) (z^{\prime\prime} + \frac{2(1-c)}{x}z’ -\frac{c(1-c)}{x^2}z) \\
+ [c-(a+b+1)x](z’ + \frac{1-c}{x}z) -abz\end{aligned} &= 0 \\
\begin{aligned} x(1-x) z^{\prime\prime} + 2(1-c)(1-x)z’ -\frac{c(1-c)(1-x)}{x}z \\
+ [c-(a+b+1)x]z’ + \frac{(1-c)[c-(a+b+1)x]}{x}z) -abz\end{aligned} &= 0 \\
\begin{aligned} x(1-x) z^{\prime\prime} + [(2-2c)-x(2-2c)+c-x(a+b+1)]z’ \\
– [(1-c)(a+b+1)-c(1-c)+ab]z \end{aligned} &= 0 \\
\begin{aligned} x(1-x) z^{\prime\prime} + [(2-c)-x(a+b-2c+3)]z’ \\
– [(1-c)^2+(1-c)(a+b)+ab]z \end{aligned} &= 0 \\
\begin{aligned} x(1-x) z^{\prime\prime} + [(2-c)-((a-c+1)+(b-c+1)+1)x]z’ \\
– (a-c+1)(b-c+1)z \end{aligned} &= 0 \tag{17}
\end{align}
which is just the same as (1) but again with the replacements \( a \rightarrow a-c+1, b \rightarrow b-c+1, c \to 2-c\). So (17) has the series solution of \( F(a-c+1,b-c+1,2-c;x) \) and we arrive at the same conclusion that the second Frobenius solution of (1) is \( y_2 = x^{1-c} F(a-c+1,b-c+1,2-c;x) \).
Therefore, provided that \( c \) is not an integer, the general solution to the Hypergeometric Equation near the regular singular point \(x = 0\) will be
\begin{align}
y = AF(a,b,c;x) + Bx^{1-c} F(a-c+1,b-c+1,2-c;x) \tag{18}
\end{align}
When \( c \) is an Integer
If \( c \) is an integer then (18) requires some modifications. There are two cases to consider. First, if \( c \) is a positive integer, then the problem comes from the second exponent \( m_2 = 1-c \) becoming either zero or a negative integer, and the second Frobenius solution is blocked. Fortunately, if \( a \text{ or } b = 1, 2, \ldots, c-1 \), then (18) can still be valid (the numerator in (15) will become \(0\) before or when \(n\) reaches \(c-2\) and the denominator becomes \(0\) too), provided that we interpret \( y_2 \) by taking the limit of
\begin{align}
F(-m,b,-m-l;x) = \lim_{c \to -m-l} (\lim_{a \to -m} F(a,b,c;x)) \tag{19}
\end{align}
and putting back \( m = c-a-1 \), \( l = a-1 \). During computation (e.g. by (9)), we can make use of the fact about the Gamma Function that
\begin{align}
\lim_{m \to a}\frac{\Gamma(m+n)}{\Gamma(m)} = a(a+1)\cdots(a+n-1) \tag{20}
\end{align}
and similarly for the part involving \(c\). Meanwhile, if now \( c \) is still a positive integer but \(a,b\) are not taking the specified values above, then \(y_2\) will have the form of (8) as stated in this tutorial. The full expression for this is too tedious to state here. Finally, if \( c = 0, -1, -2, \ldots \) is zero or a negative integer, so that the second exponent \( m_2 = 1-c = 1, 2, 3, \ldots \) is a positive integer, then the solutions will be the same as the above discussion, but with \( a, b, c \) replaced by \(a -c + 1, b -c+1, 2-c\) respectively, and consequently multiplied by \( x^{1-c} \). (This can be verified by making a change of the dependent variable \( y = x^{1-c}w \) in (1).) In the case of (18), it simply means that the roles of the two complementary solutions \( y_1, y_2 \) are swapped so that if the interpretation (19) is needed it is applied to the other solution instead.
Example
Solve the following hypergeometric equation
\begin{align}
x(1-x)y^{\prime\prime} + (\frac{3}{2} -\frac{x}{2})y’ + \frac{1}{2}y = 0 \tag{21}
\end{align}
It is not hard to spot that \( c = 3/2 \), and \(a+b+1 = 1/2\) and \( ab = -1/2\). This means that \(a, b\) are roots to the equations \( x^2 + (1/2)x -1/2 = 0 \), that is, we can take \( a = 1/2, b = -1\). By (18), the general solution is
\begin{align}
y &= AF(\frac{1}{2},-1,\frac{3}{2};x) \\
&\quad + Bx^{1-\frac{3}{2}} F(\frac{1}{2}-\frac{3}{2}+1,-1-\frac{3}{2}+1,2-\frac{3}{2};x) \\
&= AF(\frac{1}{2},-1,\frac{3}{2};x) + Bx^{-\frac{1}{2}} F(0,-\frac{3}{2},\frac{1}{2};x) \tag{22}
\end{align}
with, according to (7)
\begin{align}
&\quad F(\frac{1}{2},-1,\frac{3}{2};x) \\
&= 1+\frac{(1/2)(-1)}{(3/2)}x + \frac{(1/2)(3/2)(-1)(0)}{2!(3/2)(5/2)}x^2 + \cdots \\
&= 1 -\frac{1}{3}x + 0 = 1 -\frac{1}{3}x \tag{23}
\end{align}
and it is obvious that \( F(0,-\frac{3}{2},\frac{1}{2};x) = 1 \). So the final solution will be
\begin{align}
y = A(1 -\frac{1}{3}x) + \frac{B}{\sqrt{x}} \tag{24}
\end{align}
Exercise
Find the solution to the following hypergeometric equation about the regular singular point \( x = 0\):
\begin{align}
x(1-x)y^{\prime\prime} -(x+2)y’ + y = 0 \tag{25}
\end{align}
Answer
This time, we can see that \(a = 1, b = -1\), however \( c = -2 \) is an integer so we have to refer to the dedicated discussion above. The solution will follow that of (18):
\begin{align}
y &= AF(1,-1,-2;x) \\
&\quad+ Bx^{1-(-2)} F(1-(-2)+1,-1-(-2)+1,2-(-2);x) \\
&= AF(1,-1,-2;x) + Bx^{3} F(4,2,4;x)
\end{align}
but with (19) applied to the first hypergeometric function \( F(1,-1,-2;x) \) (\(b\) taking the position of \(a\) here):
\begin{align}
&\quad F(1,-1,-2;x) \\
&= \lim_{c \to -2} (\lim_{b \to -1} \frac{\Gamma(c)}{\Gamma(1)\Gamma(b)} \sum_{n=0}^{\infty} \frac{\Gamma(1+n)\Gamma(b+n)}{\Gamma(c+n)} \frac{x^n}{n!}) \\
&= \lim_{c \to -2} (\Gamma(c) \sum_{n=0}^{\infty} \frac{\Gamma(1+n)((-1)(0)\cdots(-1+n-1))}{\Gamma(c+n)} \frac{x^n}{n!}) \\
&= \lim_{c \to -2} \left(\Gamma(c) (\frac{\Gamma(1)}{\Gamma(c)} + \frac{(\Gamma(2))(-1)}{\Gamma(c+1)}x + (0))\right) \\
&= 1 + \frac{(1)(-1)}{(-2)}x = 1 + \frac{x}{2}
\end{align}
where (20) is also used. On the other hand, by (7)
\begin{align}
&\quad F(4,2,4;x) \\
&= 1+\sum_{n=1}^{\infty} \frac{4(4+1)\cdots(4+n-1)2(2+1)\cdots(2+n-1)}{n!4(4+1)\cdots(4+n-1)} x^n \\
&= 1 + \sum_{n=1}^{\infty} \frac{2(2+1)\cdots(n+1)}{n!} x^n \\
&= 1 + \sum_{n=1}^{\infty} \frac{(n+1)!}{n!} x^n = 1 + \sum_{n=1}^{\infty} (n+1) x^n \\
&= 1 + 2x + 3x^2 + \cdots = \frac{1}{(1-x)^2}
\end{align}
So the desired solution will be
\begin{align}
y &= A(1 + \frac{x}{2}) + B\frac{x^3}{(1-x)^2}
\end{align}
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