Special Functions – Hypergeometric functions – Differential equations

The hypergeometric differential equation is

\[z(1-z)y''+\left\{c-(a+b+1)z\right\}y'-aby=0,\quad a,b,c\in\mathbb{C}.\]

It is clear that \(z=0\) and \(z=1\) are the only singular points of this differential equation. Let \(p(z)=\displaystyle\frac{c-(a+b+1)z}{z(1-z)}\) and \(q(z)=-\displaystyle\frac{ab}{z(1-z)}\), then

\[zp(z)=\frac{c-(a+b+1)z}{1-z}\quad\text{and}\quad z^2q(z)=-\frac{abz}{1-z}\]

are both analytic at \(z=0\). So we conclude that \(z=0\) is a regular singular point of the differential equation. Note that

\[\lim\limits_{z\to 0}zp(z)=c\quad\text{and}\quad\lim\limits_{z\to 0}z^2q(z)=0,\]

which leads to the indicial equation \(r(r-1)+cr=0\;\Longleftrightarrow\;r(r-1+c)=0\).

We also have that

\[(z-1)p(z)=-\frac{c-(a+b+1)z}{z}\quad\text{and}\quad(z-1)^2q(z)=\frac{ab(z-1)}{z}\]

are both analytic at \(z=1\), which implies that \(z=1\) is a regular singular point of the differential equation as well. In that case we have

\[\lim\limits_{z\to 0}zp(z)=a+b-c+1\quad\text{and}\quad\lim\limits_{z\to 0}z^2q(z)=0,\]

which leads to the indicial equation \(r(r-1)+(a+b-c+1)r=0\;\Longleftrightarrow\;r(r+a+b-c)=0\).

This implies that there exist solutions of the hypergeometric differential equation of the form

\[y_1(z)=\sum_{n=0}^{\infty}a_nz^n\quad\text{and}\quad y_2(z)=z^{1-c}\sum_{n=0}^{\infty}b_nz^n.\]

These two solutions are linearly independent at least for \(c\notin\mathbb{Z}\). We also have solutions of the hypergeometric differential equation of the form

\[y_3(z)=\sum_{n=0}^{\infty}c_n(1-z)^n\quad\text{and}\quad y_4(z)=(1-z)^{c-a-b}\sum_{n=0}^{\infty}d_n(1-z)^n.\]

These two solutions are linearly independent at least for \(c-a-b\notin\mathbb{Z}\).

In the course Differential Equations we have already seen that

\[a_n=\frac{a(a+1)\cdots(a+n-1)b(b+1)\cdots(a+b-1)}{1\cdot2\cdots n\,c(c+1)\cdots(c+n-1)}c_0=\frac{(a)_n(b)_n}{(c)_nn!}c_0,\] \[b_n=\frac{(a-c+1)(a-c+2)\cdots(a-c+n-1)(b-c+1)(b-c+2)\cdots(b-c+n-1)}{1\cdot2\cdots n\,(2-c)(3-c)\cdots(n+1-c)}b_0 =\frac{(a-c+1)_n(b-c+1)_n}{(2-c)_nn!}b_0,\] \[c_n=\frac{a(a+1)\cdots(a+n-1)b(b+1)\cdots(b+n-1)}{1\cdot2\cdots n\,(a+b-c+1)(a+b-c+2)\cdots(a+b-c+n)}c_0 =\frac{(a)_n(b)_n}{(a+b-c+1)_nn!}c_0\]

and

\[d_n=\frac{(c-a)(c-a+1)\cdots(c-a+n-1)(c-b)(c-b+1)\cdots(c-b+n-1)}{1\cdot2\cdots n\,(c-a-b+1)(c-a-b+2)\cdots(c-a-b+n)}d_0 =\frac{(c-a)_n(c-b)_n}{(c-a-b+1)_nn!}d_0.\]

Hence we have:

\[\sum_{n=0}^{\infty}\frac{(a)_n(b)_n}{(c)_nn!}z^n={}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{c}\,;\,z\right)\quad\text{and}\quad z^{1-c}\sum_{n=0}^{\infty}\frac{(a-c+1)_n(b-c+1)_n}{(2-c)_nn!}z^n=z^{1-c}{}_2F_1\left(\genfrac{}{}{0pt}{}{a-c+1,b-c+1}{2-c}\,;\,z\right)\]

are solutions around \(z=0\). Further we have:

\[\sum_{n=0}^{\infty}\frac{(a)_n(b)_n}{(a+b-c+1)_nn!}(z-1)^n={}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{a+b-c+1}\,;\,1-z\right)\]

and

\[(1-z)^{c-a-b}\sum_{n=0}^{\infty}\frac{(c-a)_n(c-b)_n}{(c-a-b+1)_nn!}(1-z)^n =(1-z)^{c-a-b}{}_2F_1\left(\genfrac{}{}{0pt}{}{c-a,c-b}{c-a-b+1}\,;\,1-z\right)\]

are solutions around \(z=1\).

The confluent hypergeometric differential equation is given by

\[zy''+(c-z)y'-ay=0,\quad a,c\in\mathbb{C}.\]

Note that \(z=0\) is the only singular point of this differential equation. Let \(p(z)=\displaystyle\frac{c-z}{z}\) and \(q(z)=-\displaystyle\frac{a}{z}\), then

\[zp(z)=c-z\quad\text{and}\quad z^2q(z)=-az\]

are both analytic at \(z=0\). So we conclude that \(z=0\) is a regular singular point of the differential equation. Note that

\[\lim\limits_{z\to0}zp(z)=c\quad\text{and}\quad\lim\limits_{z\to0}z^2q(z)=0,\]

which leads to the indicial equation \(r(r-1)+cr=0\;\Longleftrightarrow\;r(r-1+c)=0\). Hence, the indices are \(r_1=0\) and \(r_2=1-c\). This implies that for \(c\notin\mathbb{Z}\) we have two linearly independent solutions of the form

\[y_1(z)=\sum_{n=0}^{\infty}a_nz^n\quad\text{and}\quad y_2(z)=z^{1-c}\sum_{n=0}^{\infty}b_nz^n.\]

If we substitute

\[y(z)=\sum_{n=0}^{\infty}a_nz^n,\quad y'(z)=\sum_{n=0}^{\infty}na_nz^{n-1}\quad\text{and}\quad y''(z)=\sum_{n=0}^{\infty}n(n-1)a_nz^{n-2}\]

into the differential equation, we obtain

\[\sum_{n=0}^{\infty}n(n-1)a_nz^{n-1}+c\sum_{n=0}^{\infty}na_nz^{n-1}-\sum_{n=0}^{\infty}na_nz^n-a\sum_{n=0}^{\infty}a_nz^n=0\]

or equivalently

\[\sum_{n=0}^{\infty}\left[\left\{(n+1)n+c(n+1)\right\}a_{n+1}-(n+a)a_n\right]z^n=0.\]

This leads to the recurrence relation

\[(n+1)(n+c)a_{n+1}=(n+a)a_n,\quad n=0,1,2,\ldots\]

with solution

\[a_n=\frac{(a)_n}{(c)_n\,n!}a_0,\quad n=0,1,2,\ldots.\]

Hence we have a solution of the form

\[y(z)=\sum_{n=0}^{\infty}a_nz^n=a_0\sum_{n=0}^{\infty}\frac{(a)_n}{(c)_n}\frac{z^n}{n!} =a_0\,{}_1F_1\left(\genfrac{}{}{0pt}{}{a}{c}\,;\,z\right).\]

If we substitute

\[y(z)=z^{1-c}\sum_{n=0}^{\infty}b_nz^n=\sum_{n=0}^{\infty}b_nz^{n+1-c},\] \[y'(z)=\sum_{n=0}^{\infty}(n+1-c)b_nz^{n-c}\quad\text{and}\quad y''(z)=\sum_{n=0}^{\infty}(n+1-c)(n-c)b_nz^{n-1-c}\]

into the differential equation, we obtain

\[\sum_{n=0}^{\infty}(n+1-c)(n-c)b_nz^{n-c}+c\sum_{n=0}^{\infty}(n+1-c)b_nz^{n-c} -\sum_{n=0}^{\infty}(n+1-c)b_nz^{n+1-c}-a\sum_{n=0}^{\infty}b_nz^{n+1-c}=0\]

or equivalently

\[\sum_{n=0}^{\infty}\left[\left\{(n+2-c)(n+1-c)+c(n+2-c)\right\}b_{n+1}-(n+a+1-c)b_n\right]z^{n+1-c}=0.\]

This leads to the recurrence relation

\[(n+2-c)(n+1)b_{n+1}=(n+a+1-c)b_n,\quad n=0,1,2,\ldots\]

with solution

\[b_n=\frac{(a+1-c)_n}{(2-c)_n\,n!}b_0,\quad n=0,1,2,\ldots.\]

Hence we have a solution of the form

\[y(z)=z^{1-c}\sum_{n=0}^{\infty}b_nz^n=b_0\,z^{1-c}\sum_{n=0}^{\infty}\frac{(a+1-c)_n}{(2-c)_n}\frac{z^n}{n!} =b_0\,z^{1-c}\,{}_1F_1\left(\genfrac{}{}{0pt}{}{a+1-c}{2-c}\,;\,z\right).\]

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Last modified on May 15, 2021