Differential equations – Series solutions of linear differential equations – The hypergeometric differential equation

Consider the hypergeometric differential equation \(x(1-x)y''(x)+(\gamma-(\alpha+\beta+1)x)y'(x)-\alpha\beta y(x)=0\) with \(\alpha,\beta,\gamma\in\mathbb{R}\). Note that both \(x=0\) and \(x=1\) are singular points. Further we have:

\[\lim\limits_{x\to0}x\cdot\frac{\gamma-(\alpha+\beta+1)x}{x(1-x)}=\lim\limits_{x\to0}\frac{\gamma-(\alpha+\beta+1)x}{1-x}=\gamma \quad\text{and}\quad\lim\limits_{x\to0}x^2\cdot\frac{-\alpha\beta}{x(1-x)}=\lim\limits_{x\to0}\frac{-\alpha\beta x}{1-x}=0.\]

This implies that \(x=0\) is a regular singular point with indicial equation \(r(r-1)+\gamma r=0\) or equivalently \(r(r-1+\gamma)=0\). So the indices are \(r=0\) and \(r=1-\gamma\). Similarly we have:

\[\lim\limits_{x\to1}(x-1)\cdot\frac{\gamma-(\alpha+\beta+1)x}{x(1-x)}=\lim\limits_{x\to1}\frac{\gamma-(\alpha+\beta+1)x}{-x}=\alpha+\beta-\gamma+1 \quad\text{and}\quad\lim\limits_{x\to1}(x-1)^2\cdot\frac{-\alpha\beta}{x(1-x)}=\lim\limits_{x\to1}\frac{-\alpha\beta(x-1)}{-x}=0.\]

This implies that \(x=1\) is a regular singular point with indicial equation \(r(r-1)+(\alpha+\beta-\gamma+1)r=0\) or equivalently \(r(r+\alpha+\beta-\gamma)=0\). So the indices are \(r=0\) and \(r=\gamma-\alpha-\beta\).

So there exist solutions in the form of an ordinary power series about \(x=0\):

\[y(x)=\sum_{n=0}^{\infty}c_nx^n\quad\Longrightarrow\quad y'(x)=\sum_{n=1}^{\infty}nc_nx^{n-1} \quad\Longrightarrow\quad y''(x)=\sum_{n=2}^{\infty}n(n-1)c_nx^{n-2}.\]

Substitution then gives:

\[x(1-x)\sum_{n=2}^{\infty}n(n-1)c_nx^{n-2}+(\gamma-(\alpha+\beta+1)x)\sum_{n=1}^{\infty}nc_nx^{n-1}-\alpha\beta\sum_{n=0}^{\infty}c_nx^n=0\]

or equivalently

\[\sum_{n=2}^{\infty}n(n-1)c_nx^{n-1}-\sum_{n=2}^{\infty}n(n-1)c_nx^n+\gamma\sum_{n=1}^{\infty}nc_nx^{n-1} -(\alpha+\beta+1)\sum_{n=1}^{\infty}nc_nx^n-\alpha\beta\sum_{n=0}^{\infty}c_nx^n=0.\]

Hence:

\[\sum_{n=0}^{\infty}(n+1)nc_{n+1}x^n-\sum_{n=0}^{\infty}n(n-1)c_nx^n+\gamma\sum_{n=0}^{\infty}(n+1)c_{n+1}x^n -(\alpha+\beta+1)\sum_{n=0}^{\infty}nc_nx^n-\alpha\beta\sum_{n=0}^{\infty}c_nx^n=0.\]

This implies that \((n+1)(n+\gamma)c_{n+1}-\{n(n-1)+(\alpha+\beta+1)n+\alpha\beta\}c_n=0\) for \(n=0,1,2,\ldots\). Hence:

\[c_n=\frac{(n-1+\alpha)(n-1+\beta)}{n(n-1+\gamma)}c_{n-1}=\frac{\alpha(\alpha+1)\cdots(\alpha+n-1)\beta(\beta+1)\cdots(\beta+n-1)}{n!\,\gamma(\gamma+1)\cdots(\gamma+n-1)}c_0, \quad n=1,2,3,\ldots.\]

Hence: \(y(x)=1+\displaystyle\sum_{n=1}^{\infty}\frac{\alpha(\alpha+1)\cdots(\alpha+n-1)\beta(\beta+1)\cdots(\beta+n-1)}{n!\,\gamma(\gamma+1)\cdots(\gamma+n-1)}x^n\) is a solution.

If \(\gamma\notin\mathbb{Z}\) then for \(0 < x < 1\) there also exist a solution in the form of a generalized power series about \(x=0\):

\[y(x)=x^{1-\gamma}\sum_{n=0}^{\infty}c_nx^n=\sum_{n=0}^{\infty}c_nx^{n+1-\gamma}\quad\Longrightarrow\quad y'(x)=\sum_{n=0}^{\infty}(n+1-\gamma)c_nx^{n-\gamma}\quad\Longrightarrow\quad y''(x)=\sum_{n=0}^{\infty}(n-\gamma)(n+1-\gamma)c_nx^{n-1-\gamma}.\]

Substitution then gives:

\[x(1-x)\sum_{n=0}^{\infty}(n-\gamma)(n+1-\gamma)c_nx^{n-1-\gamma}+(\gamma-(\alpha+\beta+1)x)\sum_{n=0}^{\infty}(n+1-\gamma)c_nx^{n-\gamma} -\alpha\beta\sum_{n=0}^{\infty}c_nx^{n+1-\gamma}=0\]

or equivalently

\begin{align*} &\sum_{n=0}^{\infty}(n-\gamma)(n+1-\gamma)c_nx^{n-\gamma}-\sum_{n=0}^{\infty}(n-\gamma)(n+1-\gamma)c_nx^{n+1-\gamma} +\gamma\sum_{n=0}^{\infty}(n+1-\gamma)c_nx^{n-\gamma}\\[2.5mm] &{}\hspace{25mm}{}-(\alpha+\beta+1)\sum_{n=0}^{\infty}(n+1-\gamma)c_nx^{n+1-\gamma}-\alpha\beta\sum_{n=0}^{\infty}c_nx^{n+1-\gamma}=0. \end{align*}

The coefficient of \(x^{-\gamma}\) is: \(-\gamma(1-\gamma)c_0+\gamma(1-\gamma)c_0=0\). Further we have:

\begin{align*} &\sum_{n=0}^{\infty}(n+1-\gamma)(n+2-\gamma)c_{n+1}x^{n+1-\gamma}-\sum_{n=0}^{\infty}(n-\gamma)(n+1-\gamma)c_nx^{n+1-\gamma} +\gamma\sum_{n=0}^{\infty}(n+2-\gamma)c_{n+1}x^{n+1-\gamma}\\[2.5mm] &{}\hspace{25mm}{}-(\alpha+\beta+1)\sum_{n=0}^{\infty}(n+1-\gamma)c_nx^{n+1-\gamma}-\alpha\beta\sum_{n=0}^{\infty}c_nx^{n+1-\gamma}=0. \end{align*}

This implies that \((n+1)(n+2-\gamma)c_{n+1}-\{(n-\gamma)(n+1-\gamma)+(\alpha+\beta+1)(n+1-\gamma)+\alpha\beta\}c_n=\) for \(n=0,1,2,\ldots\). Hence:

\begin{align*} c_n&=\frac{(n+\alpha-\gamma)(n+\beta-\gamma)}{n(n+1-\gamma)}c_{n-1}\\[2.5mm] &=\frac{(\alpha-\gamma+1)(\alpha-\gamma+2)\cdots(\alpha-\gamma+n)(\beta-\gamma+1)(\beta-\gamma+2)\cdots(\beta-\gamma+n)} {n!\,(2-\gamma)(3-\gamma)\cdots(n+1-\gamma)}c_0,\quad n=1,2,3,\ldots. \end{align*}

So \(y(x)=x^{1-\gamma}+x^{1-\gamma}\displaystyle\sum_{n=1}^{\infty}\frac{(\alpha-\gamma+1)(\alpha-\gamma+2)\cdots(\alpha-\gamma+n)(\beta-\gamma+1)(\beta-\gamma+2)\cdots(\beta-\gamma+n)} {n!\,(2-\gamma)(3-\gamma)\cdots(n+1-\gamma)}x^n\) is also a solution.

For \(0 < x < 1\) there also exist solutions in the form of a power series about \(x=1\):

\[y(x)=\sum_{n=0}^{\infty}c_n(1-x)^n\quad\Longrightarrow\quad y'(x)=-\sum_{n=1}^{\infty}nc_n(1-x)^{n-1} \quad\Longrightarrow\quad y''(x)=\sum_{n=2}^{\infty}n(n-1)c_n(1-x)^{n-2}.\]

Substitution then gives:

\[x(1-x)\sum_{n=2}^{\infty}n(n-1)c_n(1-x)^{n-2}-(\gamma-(\alpha+\beta+1)x)\sum_{n=1}^{\infty}nc_n(1-x)^{n-1}-\alpha\beta\sum_{n=0}^{\infty}c_n(1-x)^n=0.\]

Now we use that \(x=1-(1-x)\):

\begin{align*} &\sum_{n=2}^{\infty}n(n-1)c_n(1-x)^{n-1}-\sum_{n=2}^{\infty}n(n-1)c_n(1-x)^n-(\gamma-\alpha-\beta-1)\sum_{n=1}^{\infty}nc_n(1-x)^{n-1}\\[2.5mm] &{}\hspace{25mm}{}+(\alpha+\beta+1)\sum_{n=1}^{\infty}nc_n(1-x)^n-\alpha\beta\sum_{n=0}^{\infty}c_n(1-x)^n=0 \end{align*}

or equivalently

\begin{align*} &\sum_{n=0}^{\infty}(n+1)nc_{n+1}(1-x)^n-\sum_{n=0}^{\infty}n(n-1)c_n(1-x)^n-(\gamma-\alpha-\beta-1)\sum_{n=0}^{\infty}(n+1)c_{n+1}(1-x)^n\\[2.5mm] &{}\hspace{25mm}{}-(\alpha+\beta+1)\sum_{n=1}^{\infty}nc_n(1-x)^n-\alpha\beta\sum_{n=0}^{\infty}c_n(1-x)^n=0. \end{align*}

This implies that \((n+1)(n+\alpha+\beta-\gamma+1)c_{n+1}-\{n(n-1)+(\alpha+\beta+1)n+\alpha\beta\}c_n=0\) for \(n=0,1,2,\ldots\). Hence:

\[c_n=\frac{(n-1+\alpha)(n-1+\beta)}{n(n+\alpha+\beta-\gamma)}c_{n-1}=\frac{\alpha(\alpha+1)\cdots(\alpha+n-1)\beta(\beta+1)\cdots(\beta+n-1)} {n!\,(\alpha+\beta-\gamma+1)(\alpha+\beta-\gamma+2)\cdots(\alpha+\beta-\gamma+n)}c_0,\quad n=1,2,3,\ldots.\]

Hence: \(y(x)=1+\displaystyle\sum_{n=1}^{\infty}\frac{\alpha(\alpha+1)\cdots(\alpha+n-1)\beta(\beta+1)\cdots(\beta+n-1)} {n!\,(\alpha+\beta-\gamma+1)(\alpha+\beta-\gamma+2)\cdots(\alpha+\beta-\gamma+n)}(1-x)^n\) is a solution.

If \(\gamma-\alpha-\beta\notin\mathbb{Z}\) then for \(0 < x < 1\) there also exists a solution in the form of a generalized power series about \(x=1\):

\begin{align*} &y(x)=(1-x)^{\gamma-\alpha-\beta}\sum_{n=0}^{\infty}c_n(1-x)^n=\sum_{n=0}^{\infty}c_n(1-x)^{n+\gamma-\alpha-\beta} \quad\Longrightarrow\quad y'(x)=-\sum_{n=0}^{\infty}(n+\gamma-\alpha-\beta)c_n(1-x)^{n+\gamma-\alpha-\beta-1}\\[2.5mm] &{}\hspace{25mm}\Longrightarrow\quad y''(x)=\sum_{n=0}^{\infty}(n+\gamma-\alpha-\beta)(n+\gamma-\alpha-\beta-1)(1-x)^{n+\gamma-\alpha-\beta-2}. \end{align*}

Substitution then gives:

\begin{align*} &x(1-x)\sum_{n=0}^{\infty}(n+\gamma-\alpha-\beta)(n+\gamma-\alpha-\beta-1)(1-x)^{n+\gamma-\alpha-\beta-2}\\[2.5mm] &{}\hspace{25mm}{}-(\gamma-(\alpha+\beta+1)x)\sum_{n=0}^{\infty}(n+\gamma-\alpha-\beta)c_n(1-x)^{n+\gamma-\alpha-\beta-1} -\alpha\beta\sum_{n=0}^{\infty}c_n(1-x)^{n+\gamma-\alpha-\beta}=0. \end{align*}

Using \(x=1-(1-x)\) we obtain:

\begin{align*} &\sum_{n=0}^{\infty}(n+\gamma-\alpha-\beta)(n+\gamma-\alpha-\beta-1)c_n(1-x)^{n+\gamma-\alpha-\beta-1} -\sum_{n=0}^{\infty}(n+\gamma-\alpha-\beta)(n+\gamma-\alpha-\beta-1)c_n(1-x)^{n+\gamma-\alpha-\beta}\\[2.5mm] &{}\hspace{25mm}{}-(\gamma-\alpha-\beta-1)\sum_{n=0}^{\infty}(n+\gamma-\alpha-\beta)c_n(1-x)^{n+\gamma-\alpha-\beta-1} -(\alpha+\beta+1)\sum_{n=0}^{\infty}(n+\gamma-\alpha-\beta)c_n(1-x)^{n+\gamma-\alpha-\beta}\\[2.5mm] &{}\hspace{50mm}{}-\alpha\beta\sum_{n=0}^{\infty}c_n(1-x)^{n+\gamma-\alpha-\beta}=0. \end{align*}

Then the coefficient of \(x^{\gamma-\alpha-\beta-1}\) equals: \((\gamma-\alpha-\beta)(\gamma-\alpha-\beta-1)c_0-(\gamma-\alpha-\beta-1)(\gamma-\alpha-\beta)c_0=0\). Further we have:

\begin{align*} &\sum_{n=0}^{\infty}(n+\gamma-\alpha-\beta+1)(n+\gamma-\alpha-\beta)c_{n+1}(1-x)^{n+\gamma-\alpha-\beta} -\sum_{n=0}^{\infty}(n+\gamma-\alpha-\beta)(n+\gamma-\alpha-\beta-1)c_n(1-x)^{n+\gamma-\alpha-\beta}\\[2.5mm] &{}\hspace{25mm}{}-(\gamma-\alpha-\beta-1)\sum_{n=0}^{\infty}(n+\gamma-\alpha-\beta+1)c_{n+1}(1-x)^{n+\gamma-\alpha-\beta} -(\alpha+\beta+1)\sum_{n=0}^{\infty}(n+\gamma-\alpha-\beta)c_n(1-x)^{n+\gamma-\alpha-\beta}\\[2.5mm] &{}\hspace{50mm}{}-\alpha\beta\sum_{n=0}^{\infty}c_n(1-x)^{n+\gamma-\alpha-\beta}=0. \end{align*}

This implies that \((n+1)(n+\gamma-\alpha-\beta+1)c_{n+1}-\{(n+\gamma-\alpha-\beta)(n+\gamma-\alpha-\beta-1)+(\alpha+\beta+1)(n+\gamma-\alpha-\beta)+\alpha\beta\}c_n=0\) for \(n=0,1,2,\ldots\). Hence:

\begin{align*} c_n&=\frac{(n-1+\gamma-\alpha)(n-1+\gamma-\beta)}{n(n+\gamma-\alpha-\beta)}c_{n-1}\\[2.5mm] &=\frac{(\gamma-\alpha)(\gamma-\alpha+1)\cdots(\gamma-\alpha+n-1)(\gamma-\beta)(\gamma-\beta+1)\cdots(\gamma-\beta+n-1)} {n!\,(\gamma-\alpha-\beta+1)(\gamma-\alpha-\beta+2)\cdots(\gamma-\alpha-\beta+n)}c_0,\quad n=1,2,3,\ldots. \end{align*}

Hences:

\[y(x)=(1-x)^{\gamma-\alpha-\beta}+\displaystyle(1-x)^{\gamma-\alpha-\beta}\sum_{n=1}^{\infty}\frac{(\gamma-\alpha)(\gamma-\alpha+1)\cdots(\gamma-\alpha+n-1)(\gamma-\beta)(\gamma-\beta+1)\cdots(\gamma-\beta+n-1)} {n!\,(\gamma-\alpha-\beta+1)(\gamma-\alpha-\beta+2)\cdots(\gamma-\alpha-\beta+n)}(1-x)^n\]

is a solution.

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Last modified on May 13, 2021