SOLUTION OF THE HYPERGEOMETRIC DIFFERENTIAL EQUATION IN THE DEGENERATE CASE

V. E. Kornilov

(Presented by the Department of Higher Mathematics)

The hypergeometric differential equation

\[ z(1-z)\frac{d^{2}u}{dz^{2}}+\bigl[c-(a+b+1)z\bigr]\frac{du}{dz}-abu=0 \tag{1} \]

belongs to the degenerate case when one of the numbers \(a, b, c-a, c-b\) is an integer [1], p. 99.

A summary of complete solutions of equation (1) in the degenerate case is given in the reference mathematical literature [1], pp. 82–85, where the summary table lists 29 cases in which the degenerate solution is a rational function or a function of the form

\[ u_{1}(z)=b^{\lambda}(1-z)^{\mu}p_{n}(z), \tag{2} \]

where \(p_{n}(z)\) is a polynomial of degree \(n\) such that \(p_{n}(0)\ne 0\) and \(p_{n}(1)\ne 0\) ([1], p. 81).

In cases 1–4 of the table mentioned, the second linearly independent solution is represented by a function of the form

\[ u_{1}=F(1+m,b;c;z),\quad c\ne 0,-1,\ldots \tag{3} \]

In the present article, the question is studied of representing functions of the form (3) by means of combinations of a rational function and a function of the form (2) with the incomplete beta function [2], p. 308:

\[ F(1+A,b;c;z)= \]

\[ F(1,b+m;c;z)= \frac{(c-1)z^{1-c}}{(1-z)^{b+m+1-c}} \int_{0}^{z} t^{c-2}(1-t)^{b+m-c}\,dt,\quad c>1. \tag{4} \]

In view of the fact that in the remaining cases 5–29 of the table given in the reference book [1] the second solutions are special cases of functions of type (3), in general the second solution of differential equation (1) is a combination of function (4) with a rational function and a function of the form (2), or the solution will be one of these functions. Under analytic continuation, some solutions contain logarithms; such solutions can also be obtained by evaluating integrals of the form (4).

1. In the paper [3] the following formula was obtained: \(F(1+m,b;c;z)=\)

\[ =\frac{c-1}{(1)_m}\sum_{\kappa=1}^{m}(-1)^{\kappa-1}C_m^\kappa \left[\sum_{p=0}^{\kappa-1}\frac{(b-\kappa)_{m+p}}{(b-c+m-\kappa+1)_{p+1}}(1-z)^p\right]+ \]

\[ +\frac{(b)_m}{(1)_m}F(-m,1-b;c-b-m;1-z)\cdot F(1,b+m;c;z); \]

\[ c-b\ne 1,\ldots,m. \tag{5} \]

Applying formula (4) to the last function in equality (5), we obtain:

\[ F(1+m,b;c;z)= \]

\[ =\frac{c-1}{(1)_m}\sum_{\kappa=1}^{m}(-1)^{\kappa-1}C_m^\kappa \left[\sum_{p=0}^{\kappa-1}\frac{(b-\kappa)_{m+p}}{(b-c+m-\kappa+1)_{p+1}}(1-z)^p\right]+ \]

\[ +\frac{(b)_m(c-1)z^{1-c}}{(1)_m(1-z)^{b+m+1-c}} F(-m,1-b;c-b-m;1-z)\times \]

\[ \times\int_{0}^{z}t^{c-2}(1-t)^{b+m-c}\,dt; \qquad c-b\ne 1,\ldots,m;\quad c>1. \tag{6} \]

The equality (6) obtained is a representation of the function (3) by means of the incomplete beta function.

If \(c-b=p=1,\ldots,m\), then, in view of [1], p. 113, (2), the original function will be as follows:

\[ F(1+m,b;b+p;z)=(1-z)^{p-m-1}F(p,b+p-m-1;b+p;z), \tag{7} \]

and to the function located on the right-hand side of equality (7) the formulas (5) and (6) are already applicable.

If \(c=c_1-p,\ 1<c_1<2,\ p=1,2,\ldots\), then the function (4), or the last function of equality (5), is first transformed according to the following formula [4], p. 17, (6):

\[ F(1,b+m;c_1-p;z)= \frac{1}{1-z}\sum_{\kappa=0}^{p-1}(-1)^\kappa \frac{(c_1-p-b-m)_\kappa}{(c_1-p)_\kappa} \left(\frac{z}{1-z}\right)^\kappa+ \]

\[ +(-1)^p\frac{(c_1-p-b-m)_p}{(c_1-p)_p} \left(\frac{z}{1-z}\right)^p F(1,b+m;c_1;z), \tag{8} \]

and then to the last function of equality (8) one may apply formulas (4) and (6). In the case \(c=p=1,\ldots,m+1\), the function (3), in view of equality (7), will be a function of the form (2),

Finally, in the case when the function (4) has parameter \(c=1+p=m+2,\ldots\), then, in view of equality (8) and the obvious equality
\(F(1,b+m;1;z)=(1-z)^{-b-m}\), we finally obtain:

\[ F(1,b+m;1+p;z)=(-1)^p\frac{(1)_p}{(1-b-m)_p} \left(\frac{1-z}{z}\right)^p\frac{1}{(1-z)^{b+m}}- \]

\[ -\frac{1}{1-z}\sum_{\kappa=0}^{p-1}(-1)^\kappa \frac{(1-b-m)_\kappa}{(1)_\kappa} \left(\frac{z}{1-z}\right)^\kappa \Bigg]; \qquad p=m+1,\ldots \tag{9} \]

1. For the analytic continuation of integrals of the type (4) and (6), it is necessary first to transform them to integrals with the smallest, in absolute value, exponents of the powers of the integrand functions. Such transformations in the general case were made in the paper [4], formulas (9)—(14).

The analytic continuation under the condition \(0<\theta<1,\ 0<\varsigma<1,\ \theta+\varsigma<1\), in view of [1], p. 115, (33), (34), is as follows:

\[ I=\int_{0}^{z} t^{\theta-1}(1-t)^{\varsigma-1}\,dt =\mathrm{B}(\theta,\varsigma)+\int_{1}^{z} t^{\theta-1}(1-t)^{\varsigma-1}\,dt = \]

\[ =\frac{1}{(-1)^{\theta}}\mathrm{B}(\theta,1-\theta-\varsigma) +\int_{\infty}^{z} t^{\theta-1}(1-t)^{\varsigma-1}\,dt . \tag{10} \]

If \(0<\theta<1,\ 0<\varsigma<1\) and \(1<\theta+\varsigma<2\), then, in view of [4], the analytic continuation of the integral \(I\) in a neighborhood of infinity is the following:

\[ I=\int_{0}^{z} t^{\theta-1}(1-t)^{\varsigma-1}\,dt =\frac{z^{\theta}(1-z)^{\varsigma-1}}{\theta+\varsigma-1} +\frac{\varsigma-1}{\theta+\varsigma-1} \int_{0}^{z}\frac{t^{\theta-1}\,dt}{(1-t)^{2-\varsigma}} = \]

\[ =\frac{z^{\theta}(1-z)^{\varsigma-1}}{\theta+\varsigma-1} +(-1)^{-\theta}\mathrm{B}(\theta,1-\theta-\varsigma) +\frac{\varsigma-1}{\theta+\varsigma-1} \int_{0}^{z}\frac{t^{\theta-1}\,dt}{(1-t)^{2-\varsigma}} . \tag{11} \]

Further, alongside the generally accepted analytic continuation directly of the hypergeometric series (3), it is simpler to carry out the analytic continuation of the last functions in equalities (5) and (8); this is equivalent to computing integrals of the type (4) and (6) in neighborhoods of the points \(0,1\), and \(\infty\).

We shall consider three cases in detail.

1) The parameters \(b,\ c\), and \(c-b\) are not integers; then, in view of (10), we obtain:

\[ I=\int_{1}^{z} t^{c-2}(1-t)^{b+m-c}\,dt =\mathrm{B}(c-1,b+m+1-c)+ \]

\[ +\int_{0}^{z} t^{c-2}(1-t)^{b+m-c}\,dt;\quad c>1,\quad b+m+2-c>1; \tag{12} \]

\[ I=\int_{0}^{z} t^{c-2}(1-t)^{b+m-c}\,dt =(-1)^{1-c}\mathrm{B}(c-1,1-b-m)+ \]

\[ +\int_{\infty}^{z} t^{c-2}(1-t)^{b+m-c}\,dt;\quad c>1,\quad 2-b-m>1. \tag{13} \]

According to equalities (4), (12), and (13), we also obtain a relation between the second integrals of equalities (12) and (13) and the hypergeometric series:

\[ F(1,b+m;\,b+m+2-c;\,1-z) =(c-b-m-1)(1-z)^{c-b-m-1}\times \]

\[ \times z^{1-c}\int_{1}^{z} t^{c-2}(1-t)^{b+m-c}\,dt;\quad b+m+2-c>1, \tag{14} \]

\[ F(1,2-c;\,2-b-m;\,1:z) =-(b+m-1)(1-z)^{c-b-m-1}\times \]

\[ \times z^{2-c}\int_{\infty}^{z} t^{c-2}(1-t)^{b+m-c}\,dt;\quad 2-b-m>1. \tag{15} \]

If the parameters \(c,\ b+m+2-c\) or \(2-b-m<1\), then formula (8) is applied to functions (4), (14), and (15) in order to increase the values of these parameters.

2) Besides the positive integer \(a=1,2,\ldots\), one of the numbers \(b,\ c\) or \(c-b=0,\ \pm 1,\ldots\). Here the second case splits into the following cases a) and b):

a) if \(c,\ b-c+1\) or \(1-b=1,2,\ldots\), then, in view of equalities (3), (7), and (9), the second solution of equation (1) will be a rational function or a function of the form (2), and analytic continuation of finite polynomials also leads to finite polynomials.

b) If \(1-c,\ c-b\) or \(b=1,2,\ldots\), then, respectively, functions (4), (14), or (15) belong to the logarithmic case. Let \(c-b=m+1\); for \(c-b=1,\ldots,m\), function (3) is transformed according to equality (7), and then we obtain the following analytic continuation of integrals of the form (4) and (6):

\[ I=\int_0^z t^{c-2}(1-t)^{-1}\,dt =\ln(1-z)+\sum_{k=1}^{\infty}\frac{(2-c)_k}{k(1)_k}\bigl[(1-z)^k-1\bigr] \]
\[ =\ln(1-z)+\int_0^1 \frac{t^{c-2}-1}{1-t}\,dt +\int_1^z \frac{t^{c-2}-1}{1-t}\,dt;\quad c>1. \tag{16} \]

and also, in view of (13) and [1], p. 18, (6),

\[ I=\int_0^z \frac{t^{c-2}\,dt}{1-t} =\frac{\pi(-1)^{1-c}}{\sin(\pi c-\pi)} +\int_{\infty}^{z}\frac{t^{c-2}\,dt}{1-t};\quad 1<c<2. \tag{17} \]

In conclusion of case 2b), it should be noted that for a rational value of the third parameter, integrals of the form (16) and (17), and other integrals belonging to this case, are expressed in elementary functions [4].

3) All parameters \(a,\ b\), and \(c=0,\ \pm1,\ldots\). In this case, if \(c=1,2,\ldots\), then function (3), in view of (7) and (9), will be a function of the form (2). If \(c=-p=0,-1,\ldots\), then we have the logarithmic case in a neighborhood of any of the three points \(0,\ 1\), and \(\infty\). Namely, function (15) is transformed, in view of [4], p. 19, (12), in the following way:

\[ \frac{z^{-p-2}}{p+2}\, F\left(1,\,2+p;\,3+p;\,\frac{1}{z}\right) =\int_{\infty}^{z} t^{-p-2}(1-t)^{-1}\,dt \]
\[ =-\sum_{k=0}^{p}\frac{z^{k-p-1}}{1+p-k} +\int_{\infty}^{z}\frac{dt}{t(1-t)} =-\sum_{k=0}^{p}\frac{z^{k-p-1}}{1+p-k} -\ln\left(1-\frac{1}{z}\right). \tag{18} \]

The analytic continuation of the second integral of equality (18) will be the following:

\[ I_2=\int_{\infty}^{z}\frac{dt}{t(1-t)} =-\ln(z-1)+\int_1^z\frac{dt}{t} =\pi\sqrt{-1}+\ln z+\int_0^z\frac{dt}{1-t}. \tag{19} \]

After replacing the integrals in equality (19) by hypergeometric functions, we also obtain

\[ I_{2}=\frac{1}{z}F\left(1,\,1;\,2;\,\frac{1}{z}\right) =-\ln(z-1)-(1-z)F(1,\,1;\,2;\,1-z)= \]

\[ =\pi\sqrt{-1}+\ln z+zF(1,\,1;\,2;\,z). \tag{20} \]

REFERENCES

1. H. Bateman and A. Erdélyi. Higher Transcendental Functions (hypergeometric function, Legendre functions). Moscow, Fizmatgiz, 1965.

2. I. M. Ryzhik. Tables of Integrals, Sums, Series, and Products. Moscow–Leningrad, Gostekhizdat, 1948.

3. V. E. Kornilov. Transformation of certain power series into continued fractions. Izv. TPI, vol. 154, 1967.

4. V. E. Kornilov. Separation of the algebraic part of integrals of binomial differentials. Izv. TPI, vol. 131, 1965.