Regular Singular Point at \( x = \infty \)
There is, not so obviously at first sight, a third regular singular point at \(x = \infty\) for the hypergeometric equation. For this, we can study the behavior of its solution at the point of infinity by making the substitution \( t = 1/x \), so that we may look at \( t = 0\), which will then correspond to \( x = \infty\) exactly. Subsequently, we have
\begin{align}
\frac{dy}{dx} &= \frac{dy}{dt}\frac{dt}{dx} \\
&= \frac{dy}{dt} (-\frac{1}{x^2}) = -t^2 \frac{dy}{dt} \tag{1} \\
\frac{d^2y}{dx^2} &= \frac{d}{dx}(\frac{dy}{dx}) \\
&= \frac{dt}{dx} \frac{d}{dt}(-t^2 \frac{dy}{dt}) \\
&= -t^2 (-2t\frac{dy}{dt} -t^2\frac{d^2y}{dt^2}) \tag{2}
\end{align}
By plugging these into the hypergeometric equation ((1) of this tutorial), we have
\begin{align}
\begin{aligned}\frac{1}{t}(1-\frac{1}{t}) (-t^2 (-2t\frac{dy}{dt} -t^2\frac{d^2y}{dt^2})) \\
+ [c-(a+b+1)\frac{1}{t}]( -t^2 \frac{dy}{dt}) -aby \end{aligned} &= 0 \\
-2t(1-t)\frac{dy}{dt} -(1-t)t^2\frac{d^2y}{dt^2} -t[ct-(a+b+1)]\frac{dy}{dt} -aby &= 0 \\
\frac{d^2y}{dt^2} + (\frac{2}{t} + \frac{ct-(a+b+1)}{t(1-t)}) \frac{dy}{dt} + \frac{ab}{t^2(1-t)}y &= 0 \\
\frac{d^2y}{dt^2} + \frac{(c-2)t-(a+b-1)}{t(1-t)} \frac{dy}{dt} + \frac{ab}{t^2(1-t)}y &= 0 \tag{3}
\end{align}
The regular singular point at \( t = 0 \), i.e. \( x = \infty \), is now clear with \( 1/t \) and \( 1/t^2 \) in front of \( dy/dt \) and \( y \). The indicial equation reads
\begin{align}
m^2 -(1+(a+b-1)) m + ab &= 0 \\
m^2 -(a+b)m + ab &= 0 \\
(m-a)(m-b) &= 0 \tag{4}
\end{align}
which has the roots \( m = a,b \) as the exponents. The general solution of the hypergeometric equation at \(t = 0\) or \(x = \infty\) then can be found by making a change of variable \( y = t^a w \) (or similarly \( y = t^b w \)), where \( dy/dx = t^{a}dw/dx + at^{a-1}w \) and \( d^2y/dx^2 = t^{a}d^2w/dx^2 +2at^{a-1}dw/dx + a(a-1)t^{a-2} w\): Substituting these into (3) gives
\begin{align}
\begin{aligned}(t^{a}\frac{d^2w}{dx^2} +2at^{a-1}\frac{dw}{dx} + a(a-1)t^{a-2}w) \\
+ \frac{(c-2)t-(a+b-1)}{t(1-t)} (t^{a}\frac{dw}{dx} + at^{a-1}w) + \frac{ab}{t^2(1-t)}t^a w \end{aligned} &= 0 \\
\begin{aligned} t^2\frac{d^2w}{dx^2} +2at\frac{dw}{dx} + \frac{a(a-1)(1-t)}{1-t}w \\
+ \frac{[(c-2)t-(a+b-1)]t}{1-t} \frac{dw}{dx} \\
+ \frac{a[(c-2)t-(a+b-1)]}{1-t}w + \frac{ab}{1-t} w \end{aligned} &= 0 \\
\begin{aligned} t^2\frac{d^2w}{dx^2} +\frac{2a(1-t)t}{1-t}\frac{dw}{dx} -\frac{a(a-1)t}{1-t}w \\
+ \frac{[(c-2)t-(a+b-1)]t}{1-t} \frac{dw}{dx} + \frac{a(c-2)t}{1-t}w \end{aligned} &= 0 \\
\begin{aligned}t(1-t)\frac{d^2w}{dx^2} + ((a-b+1)-(2a-c+2)t) \frac{dw}{dx}\\
-a(a-c+1)w\end{aligned} &= 0 \tag{5}
\end{align}
This is just another hypergeometric equation with \(b \to a-c+1, c \to a-b+1\), so \( w = F(a,a-c+1,a-b+1;t) \) and \( y_1 = t^aF(a,a-c+1,a-b+1;t) = x^{-a} F(a,a-c+1,a-b+1;1/x) \). In a similar vein, \( y_2 = x^{-b} F(b,b-c+1,b-a+1;1/x) \). This requires that \(a-b\) is not an integer.
Canonicity of Hypergeometric Equation
Now we are going to derive the canonicity of the hypergeometric equation: Its form is completely determined by the facts that it has three regular singular points at \(x = 0,1,\infty\), and at least one of the exponents is zero at both \(x=0,1\). Or in other words, if a second-order linear ODE exhibits these properties, it must be the hypergeometric equation. Moreover, we will also show that all second-order linear ODEs with three regular singular points within the real \(x\)-axis and ordinary points elsewhere (including the point at infinity) can be transformed into the hypergeometric form.
For the first part, it is mainly based on the reference Differential Equations with Applications and Historical Notes by Simmons. Suppose the second-order linear ODE is in the general form of
\begin{align}
\frac{d^2y}{dx^2} + P(x)\frac{dy}{dx} + Q(x)y = 0 \tag{6}
\end{align}
Then again by a change of variable \( x = 1/t\) with (1) and (2), (6) becomes
\begin{align}
-t^2 (-2t\frac{dy}{dt} -t^2\frac{d^2y}{dt^2}) + P(\frac{1}{t})(-t^2 \frac{dy}{dt}) + Q(\frac{1}{t})y &= 0 \\
\frac{d^2y}{dt^2} + (\frac{2}{t}-\frac{P(1/t)}{t^2})\frac{dy}{dt} + \frac{Q(1/t)}{t^4}y &= 0 \tag{7}
\end{align}
Now given that \(x = \infty\) is a regular singular point, it means that the functions \(\frac{1}{t}P(1/t)\) and \(\frac{1}{t^2}Q(1/t)\) have to be analytic at \(t = 0\). Further invoke the assumption that \( x = 0,1 \) are also two regular singular points and any other points are ordinary, so \( xP(x) \) will be analytic at \(x = 0\), \((x-1)P(x)\) is analytic at \(x=1\). And combining these, \(x(x-1)P(x)\) must be analytic at all values of \(x\), such that we can write
\begin{align}
x(x-1)P(x) = \sum_{n=0}^{\infty} a_n x^n \tag{8}
\end{align}
Substituting \( x = 1/t\) into (8) yields
\begin{align}
\frac{1}{t}(\frac{1}{t}-1)P(\frac{1}{t}) &= \sum_{n=0}^{\infty} a_n (\frac{1}{t})^n \\
\frac{1}{t}P(\frac{1}{t}) &= \frac{t}{1-t}\sum_{n=0}^{\infty} a_n (\frac{1}{t})^n \\
&= \frac{t}{1-t} (a_0 + a_1\frac{1}{t} + a_2\frac{1}{t^2} + \cdots) \\
&= \frac{1}{1-t} (a_0t + a_1 + a_2\frac{1}{t} + \cdots) \tag{9}
\end{align}
We previously have argued that \(\frac{1}{t}P(1/t)\) is analytic at \( t = 0 \), so we have \(a_2 = a_3 = \cdots = 0\), and (8) becomes
\begin{align}
x(x-1)P(x) &= a_0 + a_1x \\
P(x) &= \frac{a_0 + a_1x}{x(x-1)} \\
&= \frac{A(x-1) + Bx}{x(x-1)} = \frac{A}{x} + \frac{B}{x-1} \tag{10}
\end{align}
as partial fractions for certain computable constants \(A\) and \(B\). In a similar vein, \(x^2(x-1)^2Q(x)\) is analytic for all \(x\). Hence
\begin{align}
x^2(x-1)^2Q(x) = \sum_{n=0}^{\infty} b_n x^n \tag{11}
\end{align}
and let \( x = 1/t \) again to get
\begin{align}
\frac{1}{t^2}(\frac{1}{t}-1)^2Q(\frac{1}{t}) &= \sum_{n=0}^{\infty} b_n (\frac{1}{t})^n \\
\frac{1}{t^2}Q(\frac{1}{t}) &= \frac{t^2}{(1-t)^2}\sum_{n=0}^{\infty} b_n (\frac{1}{t})^n \\
&=\frac{t^2}{(1-t)^2} (b_0 + b_1\frac{1}{t} + b_2\frac{1}{t^2} + b_3\frac{1}{t^3} + \cdots) \\
&=\frac{1}{(1-t)^2} (b_0t^2 + b_1t + b_2 + b_3\frac{1}{t} + \cdots) \tag{12}
\end{align}
By the same analyticity argument as above, \( b_3 = b_4 = \cdots = 0\), thus we will obtain the partial fractions of
\begin{align}
x^2(x-1)^2Q(x) &= b_0+b_1x+b_2x^2 \\
Q(x) &= \frac{b_0+b_1x+b_2x^2}{x^2(x-1)^2} \\
&= \frac{C}{x} + \frac{D}{x^2} + \frac{E}{x-1} + \frac{F}{(x-1)^2} \tag{13}
\end{align}
Now observe that \(\frac{1}{t^2}Q(1/t)\) is analytic at \( t = 0 \) as stated before, hence \(x^2Q(x)\) has to be bounded for large \(x\), so by (13)
\begin{align}
x^2Q(x) &= x^2(\frac{C}{x} + \frac{D}{x^2} + \frac{E}{x-1} + \frac{F}{(x-1)^2}) \\
&= Cx + D + E\frac{x^2}{x-1} + F\frac{x^2}{(x-1)^2} \tag{14}
\end{align}
The \( D \) and \( Fx^2/(x-1)^2 \) terms will be finite asymptotically as \(x \to \infty\) so they are not of concern. But for the remaining asymptotically linear terms \( Cx + Ex^2/(x-1) \) to stay bounded, note that
\begin{align}
x^2(\frac{C}{x} + \frac{E}{x-1}) &= x^2 (\frac{C(x-1)+Ex}{x(x-1)}) \\
&= x^2 (\frac{(C+E)x-C}{x(x-1)}) \tag{15}
\end{align}
and we need \(C+E = 0\), so (13) can be rewritten as
\begin{align}
Q(x) &= \frac{C}{x} + \frac{D}{x^2} -\frac{C}{x-1} + \frac{F}{(x-1)^2} \\
&= -\frac{C}{x(x-1)} + \frac{D}{x^2} + \frac{F}{(x-1)^2} \tag{16}
\end{align}
Therefore, (6) should look like
\begin{align}
\frac{d^2y}{dx^2} + (\frac{A}{x} + \frac{B}{x-1})\frac{dy}{dx} + (\frac{D}{x^2} + \frac{F}{(x-1)^2} -\frac{C}{x(x-1)})y = 0 \tag{17}
\end{align}
The indicial equations for the three regular singular points \( x = 0,1,\infty \) are then
\begin{align}
\begin{aligned}
m^2 + (A-1)m + D &= 0 \\
m^2 + (B-1)m + F &= 0
\end{aligned} \tag{18}
\end{align}
and by substituting \( x = 1/t \), referring to (7) and inspecting \( t = 0 \) for \(x = \infty\):
\begin{align}
\frac{d^2y}{dt^2} + (\frac{2}{t}-\frac{At + B\frac{t}{1-t}}{t^2})\frac{dy}{dt} + \frac{Dt^2 + F\frac{t^2}{(1-t)^2} -C\frac{t^2}{1-t}}{t^4} y &= 0 \tag{19}
\end{align}
so
\begin{align}
m^2 + (1-A-B)m + (D+F-C) = 0 \tag{20}
\end{align}
Denote the exponents for the regular singular points at \(x = 0,1,\infty\) by \(\alpha_1, \alpha_2\), \(\beta_1, \beta_2\), and \(\gamma_1, \gamma_2\) respectively. Then, we can infer the sums and products of these roots from the quadratic equations in (18) and (20):
\begin{align}
\begin{aligned}
\alpha_1 + \alpha_2 &= 1 -A & & \alpha_1\alpha_2 = D \\
\beta_1 + \beta_2 &= 1 -B & & \beta_1\beta_2 = F \\
\gamma_1 + \gamma_2 &= A+B-1 & & \gamma_1\gamma_2 = D+F-C \\
\end{aligned} \tag{21}
\end{align}
Adding up the first column of (21) easily leads to
\begin{align}
\alpha_1+\alpha_2+\beta_1 + \beta_2+\gamma_1 + \gamma_2 = 1 \tag{22}
\end{align}
and applying (21) again to (17), we have
\begin{align}
\begin{aligned}\frac{d^2y}{dx^2} + (\frac{1-\alpha_1-\alpha_2}{x} + \frac{1-\beta_1-\beta_2}{x-1})\frac{dy}{dx} \\ + (\frac{\alpha_1\alpha_2}{x^2} + \frac{\beta_1\beta_2}{(x-1)^2} +\frac{\gamma_1\gamma_2-\alpha_1\alpha_2-\beta_1\beta_2 }{x(x-1)})y \end{aligned}= 0 \tag{23}
\end{align}
Finally, we impose the last requirement that one of the exponents is zero at each of \(x=0,1\), let’s say \( \alpha_1 = \beta_1 = 0 \), then (23) is reduced to
\begin{align}
\frac{d^2y}{dx^2} + (\frac{1-\alpha_2}{x} + \frac{1-\beta_2}{x-1})\frac{dy}{dx} +\frac{\gamma_1\gamma_2}{x(x-1)}y &= 0 \\
x(1-x)\frac{d^2y}{dx^2} + [(1-\alpha_2)(1-x) -(1-\beta_2)x]\frac{dy}{dx} -\gamma_1\gamma_2y &= 0 \\
x(1-x)\frac{d^2y}{dx^2} + [(1-\alpha_2) + (\alpha_2+\beta_2-2)x]\frac{dy}{dx} -\gamma_1\gamma_2y &= 0 \\
x(1-x)\frac{d^2y}{dx^2} + [(1-\alpha_2) -(\gamma_1 + \gamma_2 + 1)x]\frac{dy}{dx} -\gamma_1\gamma_2y &= 0 \tag{24}
\end{align}
where in the last line we recall (22). This can be promptly identified with the Hypergeometric Equation with \( a = \gamma_1\), \( b = \gamma_2 \), \(c = 1-\alpha_2\), which shows that indeed, given the conditions in the beginning, the equation must be of the hypergeometric type.
The second result is the possibility of transforming a second-order linear ODE with three singular points to a hypergeometric equation. To establish this, we will show the equivalent that a hypergeometric equation can be transformed into such ODEs in the opposite way. (Then we just need to apply a reverse transformation.) This part is largely based on the reference Mathematical Methods of Physics by Mathews and Walker. Denote the three regular singular points by \(s_1, s_2, s_3\), then the change of variable
\begin{align}
x = \frac{(w-s_1)(s_2-s_3)}{(w-s_3)(s_2-s_1)}\tag{25}
\end{align}
will send the points \( x = 0,1,\infty \) in the hypergeometric equation to \( w = s_1, s_2, s_3 \) respectively. Note that
\begin{align}
\frac{dy}{dx} &= \frac{dy}{dw}\frac{1}{\frac{dx}{dw}} \\
&= \frac{dy}{dw} \frac{1}{\frac{(s_2-s_3)[(w-s_3) -(w-s_1)]}{(s_2-s_1)(w-s_3)^2}} \\
&= \frac{dy}{dw} \frac{(s_2-s_1)(w-s_3)^2}{(s_2-s_3)(s_1-s_3)} \tag{26} \\
\frac{d^2y}{dx^2} &= \frac{d}{dx} (\frac{dy}{dw} \frac{(s_2-s_1)(w-s_3)^2}{(s_2-s_3)(s_1-s_3)}) \\
&= \frac{dw}{dx} [\frac{d^2y}{dw^2} \frac{(s_2-s_1)(w-s_3)^2}{(s_2-s_3)(s_1-s_3)} + \frac{dy}{dw} \frac{d}{dw} \frac{(s_2-s_1)(w-s_3)^2}{(s_2-s_3)(s_1-s_3)}] \\
&= \frac{d^2y}{dw^2} \frac{(s_2-s_1)^2(w-s_3)^4}{(s_2-s_3)^2(s_1-s_3)^2} + 2\frac{dy}{dw} \frac{(s_2-s_1)^2(w-s_3)^3}{(s_2-s_3)^2(s_1-s_3)^2} \tag{27}
\end{align}
Plugging all these into the Hypergeometric equation leads to
\begin{align}
\begin{aligned} \frac{(w-s_1)(s_2-s_3)}{(w-s_3)(s_2-s_1)}(1-\frac{(w-s_1)(s_2-s_3)}{(w-s_3)(s_2-s_1)}) \\
[\frac{d^2y}{dw^2} \frac{(s_2-s_1)^2(w-s_3)^4}{(s_2-s_3)^2(s_1-s_3)^2} + 2\frac{dy}{dw} \frac{(s_2-s_1)^2(w-s_3)^3}{(s_2-s_3)^2(s_1-s_3)^2}]\\
+ [c-(a+b+1)\frac{(w-s_1)(s_2-s_3)}{(w-s_3)(s_2-s_1)}]\frac{dy}{dw} \frac{(s_2-s_1)(w-s_3)^2}{(s_2-s_3)(s_1-s_3)} -aby \end{aligned} &= 0 \\
\begin{aligned} \frac{(s_1-s_3)(w-s_2)}{(s_1-s_2)(w-s_3)} \\
[\frac{d^2y}{dw^2} \frac{(s_2-s_1)(w-s_3)^3(w-s_1)}{(s_2-s_3)(s_1-s_3)^2} + 2\frac{dy}{dw} \frac{(s_2-s_1)(w-s_3)^2(w-s_1)}{(s_2-s_3)(s_1-s_3)^2}]\\
+ [c\frac{(s_2-s_1)(w-s_3)^2}{(s_2-s_3)(s_1-s_3)}-(a+b+1)\frac{(w-s_3)(w-s_1)}{(s_1-s_3)}]\frac{dy}{dw} -aby \end{aligned} &= 0 \\
\begin{aligned} -\frac{d^2y}{dw^2} \frac{(w-s_3)^2(w-s_2)(w-s_1)}{(s_2-s_3)(s_1-s_3)} -2\frac{dy}{dw} \frac{(w-s_3)(w-s_2)(w-s_1)}{(s_2-s_3)(s_1-s_3)} \\
+ [c\frac{(s_2-s_1)(w-s_3)^2}{(s_2-s_3)(s_1-s_3)}-(a+b+1)\frac{(w-s_3)(w-s_1)}{(s_1-s_3)}]\frac{dy}{dw} -aby \end{aligned} &= 0 \tag{28}
\end{align}
It just needs some care to see that the three points \( s_1, s_2, s_3 \) retain the same role of regular singular points in \( w \) via analyzing and dividing by the factors.
Exercise
Show that the following form of second-order ODE
\begin{align}
(x-A)(x-B) y^{\prime\prime} + (C+Dx)y’ + Ey = 0 \tag{29}
\end{align}
particularly with \(x = A,B\) as its regular singular points, can be converted to a standard hypergeometric equation by using the linear transformation
\begin{align}
t = \frac{x-A}{B-A} \tag{30}
\end{align}
Answer
Now \( t=0,1 \) correspond to \( x = A,B\) which are expected to remain as the regular singular points. Applying the given transformation should produce
\begin{align}
\begin{aligned} ([A+t(B-A)]-A)([A+t(B-A)]-B) \frac{1}{(B-A)^2}y^{\prime\prime} \\
+ (C+D[A+t(B-A)])\frac{1}{B-A}y’ + Ey \end{aligned} = 0 \\
\frac{t(B-A)(t-1)(B-A)}{(B-A)^2}y^{\prime\prime}+ (\frac{C}{B-A}+D[\frac{A}{B-A}+t])y’ + Ey = 0 \\
t(1-t)y^{\prime\prime} -(\frac{C}{B-A}+D[\frac{A}{B-A}+t]) y’ -Ey = 0
\end{align}
This is clearly a hypergeometric equation defined by the relation \( E = ab, -(C+DA)/(B-A) = c, D = a+b+1 \) in which \( a,b,c \) can be inverted. This means that the original ODE may be solved in terms of the hypergeometric function at \( x = A,B \) with the appropriate transformation (30).
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