Hypergeometric equation in a particular case

Question

I have a question to make in relation to the solution of the hypergeometric differential equation. Let us consider the aforesaid equation,

\begin{equation} y(1-y)h'' + [c-(1+a+b)y]h' -abh=0, \end{equation} where $h(y)$ and the prime stands for derivatives with respect to $y$. I would like to know what is the general solution when the following relation amongst the parameters holds: $a+b-c=0$. This case is always excluded in the textbooks and I did not find a book discussing such a delicate point. Perhaps, there is only one solution or the equation becomes another well-known ODE, I do not really know. Anyway, under the condition $c=a+b$ the new differential equation becomes \begin{equation} y(1-y)h'' + [(a+b)-(1+a+b)y]h' -abh=0. \end{equation} It looks quite similar to the hypergeometric differential equation but now it only has two parameters $a$ and $b$.

A second point that I would like to discover is the behavior of that general solution near $y=1$ and $y \rightarrow \infty$.

Any help in this matter it will be much appreciated.

Answer 1

Hypergeometric equation and solution: $$y(1-y)h'' + [a+b-(1+a+b)y]h' -abh=0$$ $$\Rightarrow h(y)=c_2 (-1)^{1-a-b} y^{1-a-b} \, _2F_1(1-a,1-b;2-a-b;y)+c_1 \, _2F_1(a,b;a+b;y),$$ with $c_1$ and $c_2$ arbitrary constants. So there are two independent solutions.

A few special cases ($B_y$ is the incomplete Beta function) $$a=1:\;\;h(y)=c_1 \, _2F_1(1,b;b+1;y)+(-1)^{-b} c_2 y^{-b}$$ $$a=0:\;\;h(y)=(-1)^{-b} (b-1) c_2 B_y(1-b,0)+c_1$$ $$a=-1:\;\;h(y)=(-1)^{-b} c_2 y^{2-b} \, _2F_1(2,1-b;3-b;y)+\frac{c_1 (1-b y+b)}{b-1}$$

Limit $y\rightarrow 1$: $$h(y)\rightarrow\log |y-1| \left(\frac{c_2 (-1)^{-a-b} \Gamma (2-a-b)}{\Gamma (1-a) \Gamma (1-b)}-\frac{c_1 \Gamma (a+b)}{\Gamma (a) \Gamma (b)}\right)$$

Limit $y\rightarrow\infty$:

$$h(y)\rightarrow (-1)^{-a} y^{-a} \Gamma (b-a)\frac{ c_1 \Gamma (1-a)^2 \Gamma (a+b)+c_2 \Gamma (b)^2 \Gamma (2-a-b)}{\Gamma (1-a)^2 \Gamma (b)^2}+(-1)^{-b} y^{-b} \Gamma (a-b)\frac{ c_2 \Gamma (a)^2 \Gamma (2-a-b)+c_1 \Gamma (1-b)^2 \Gamma (a+b)}{\Gamma (a)^2 \Gamma (1-b)^2}$$

Answer 2

Mathematica 12.0 answers

DSolve[y*(1 - y)*h''[y] + ((a + b) - (1 + a + b)*y) h'[y] - a*b*h[y] == 0, h[y], y]

$ \left\{\left\{h(y)\to c_2 (-1)^{-a-b+1} y^{-a-b+1} \, _2F_1(1-a,1-b;-a-b+2;y)+c_1 \, _2F_1(a,b;a+b;y)\right\}\right\}$

Next,

AsymptoticDSolveValue[y*(1 - y)*h''[y] + ((a + b) - (1 + a + b)*y) h'[y] - a*b*h[y] == 0, 

h[y], {y, Infinity, 1}]

$$ c_1 \left(-\frac{a b y^{-b-1}}{a b+(b+1) (-a-b+1)+(b+1) b}+\frac{b y^{-b-1}}{a b+(b+1) (-a-b+1)+(b+1) b}+y^{-b}\right)+c_2 \left(\frac{a y^{-a-1}}{(a+1) (-a-b+1)+a b+a (a+1)}-\frac{a b y^{-a-1}}{(a+1) (-a-b+1)+a b+a (a+1)}+y^{-a}\right)$$

In order to obtain the asymptotics as $y\to 1$, we need put the value of $h(y)$ at $y=1$, e.g.

AsymptoticDSolveValue[{y*(1 - y)*h''[y] + ((a + b) - (1 + a + b)*y) h'[y] - a*b*h[y] == 0,  

 h[1] == 1}, h[y], {y, 1, 1}]

$ 1-a b (y-1)$

Maple 2018.2 answers

dsolve(y*(1-y)*(diff(h(y), y, y))+(a+b-(1+a+b)*y)*(diff(h(y), y))-a*b*h(y) = 0, h(y));

$ h \left( y \right) ={\it \_C1}\,{\mbox{$_2$F$_1$}(a,b;\,a+b;\,y)}+{ \it \_C2}\,{y}^{-a-b+1}{\mbox{$_2$F$_1$}(1-b,1-a;\,2-a-b;\,y)} $

and

series(_C1*hypergeom([a, b], [a+b], y)+_C2*y^(-a-b+1)*hypergeom([1-b, 1-a], [2-a-b], y), y = 1, 1);

$$ -{\frac {{\it \_C1}\,\Gamma \left( a+b \right) \left( \ln \left( -1+ y \right) +i{\it csgn} \left( i \left( -1+y \right) \right) \pi+2\, \gamma+\Psi \left( a \right) +\Psi \left( b \right) \right) }{\Gamma \left( a \right) \Gamma \left( b \right) }}-{\frac {{\it \_C2}\, \Gamma \left( 2-a-b \right) \left( \ln \left( -1+y \right) +i{\it csgn} \left( i \left( -1+y \right) \right) \pi+2\,\gamma+\Psi \left( 1-b \right) +\Psi \left( 1-a \right) \right) }{\Gamma \left( 1-b \right) \Gamma \left( 1-a \right) }}+O \left( -1+y \right) $$

Unfortunately,

series(_C1*hypergeom([a, b], [a+b], y)+_C2*y^(-a-b+1)*hypergeom([1-b, 1-a], [2-a-b], y), y = infinity, 1);

Error, (in asympt) unable to compute series