Frobenius Solution To The Hypergeometric Equation - Solution Around Infinity

Solution Around Infinity

Finally, we study the singularity as x → ∞. Since we can't study this directly, we let x = s−1. Then the solution of the equation as x → ∞ is identical to the solution of the modified equation when s = 0. We had

\begin{align} & x(1-x)y''+\left\{ \gamma -(1+\alpha +\beta )x \right\}y'-\alpha \beta y=0 \\ & \frac{dy}{dx}=\frac{dy}{ds}\times \frac{ds}{dx}=-s^2\times \frac{dy}{ds}=-s^2y' \\ & \frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left( \frac{dy}{dx} \right)=\frac{d}{dx}\left( -s^2 \times \frac{dy}{ds} \right)=\frac{d}{ds}\left( -s^2 \times \frac{dy}{ds} \right)\times \frac{ds}{dx} = \left( (-2s)\times \frac{dy}{ds}+(-s^{2})\frac{d^{2}y}{ds^{2}} \right) \times (-s^{2})=2s^{3}y'+s^{4}y'' \end{align}

Hence, the equation takes the new form

which reduces to

Let

\begin{align} P_{0}(s) &=-\alpha \beta, \\ P_{1}(s) &=((2-\gamma )s^{2}+(\alpha +\beta -1)s), \\ P_{2}(s) &=(s^{3}-s^{2}). \end{align}

As we said, we shall only study the solution when s = 0. As we can see, this is a singular point since P2(0) = 0. To see if it's regular,

\begin{align} \lim_{s \to a} \frac{\left( s-a \right)P_{1}(s)}{P_{2}(s)} & =\lim_{s \to 0} \frac{\left( s-0 \right)((2-\gamma )s^{2}+(\alpha +\beta -1)s)}{(s^{3}-s^{2})} \\ &= \lim_{s \to 0} \frac{((2-\gamma )s^{2}+(\alpha +\beta -1)s)}{s^{2}-s} \\ & =\lim_{s \to 0} \frac{((2-\gamma )s+(\alpha +\beta -1))}{s-1}=1-\alpha -\beta. \\ \lim_{s \to a} \frac{\left( s-a \right)^{2}P_{0}(s)}{P_{2}(s)} &=\lim_{s \to 0} \frac{\left( s-0 \right)^{2}\left( -\alpha \beta \right)}{(s^{3}-s^{2})}=\lim_{s \to 0} \frac{\left( -\alpha \beta \right)}{s-1}=\alpha \beta. \end{align}

Hence, both limits exist and s = 0 is a regular singular point. Therefore, we assume the solution takes the form

with a0 ≠ 0.

Hence,

\begin{align} y'&=\sum\limits_{r=0}^{\infty }{a_{r}(r+c)s^{r+c-1}}\\ y''&=\sum\limits_{r=0}^{\infty }{a_{r}(r+c)(r+c-1)s^{r+c-2}} \end{align}

Substituting in the modified hypergeometric equation we get

And therefore:

i.e.,

In order to simplify this equation, we need all powers to be the same, equal to r + c, the smallest power. Hence, we switch the indices as follows

\begin{align} & \sum_{r=1}^{\infty }{a_{r-1}(r+c-1)(r+c-2)s^{r+c}}-\sum_{r=0}^{\infty }{a_{r}(r+c)(r+c-1)s^{r+c}} +(2-\gamma )\sum_{r=1}^{\infty }{a_{r-1}(r+c-1)s^{r+c}}+ \\ & \qquad \qquad + (\alpha +\beta -1)\sum_{r=0}^{\infty }{a_{r}(r+c)s^{r+c}}-\alpha \beta \sum_{r=0}^{\infty }{a_{r}s^{r+c}}=0 \end{align}

Thus, isolating the first term of the sums starting from 0 we get

\begin{align} & a_{0}\left\{ -(c)(c-1)+(\alpha +\beta -1)(c)-\alpha \beta \right\}s^{c}+\sum_{r=1}^{\infty }{a_{r-1}(r+c-1)(r+c-2)s^{r+c}} -\sum_{r=1}^{\infty }{a_{r}(r+c)(r+c-1)x^{r+c}}+\\ & \qquad \qquad + (2-\gamma )\sum_{r=1}^{\infty }{a_{r-1}(r+c-1)s^{r+c}} +(\alpha +\beta -1)\sum_{r=1}^{\infty }{a_{r}(r+c)s^{r+c}}-\alpha \beta \sum_{r=1}^{\infty }{a_{r}s^{r+c}}=0 \end{align}

Now, from the linear independence of all powers of s (i.e., of the functions 1, s, s2, ...), the coefficients of sk vanish for all k. Hence, from the first term we have

which is the indicial equation. Since a0 ≠ 0, we have

Hence, c1 = α and c2 = β.

Also, from the rest of the terms we have

Hence,

But

\begin{align} (r+c)(r+c-\alpha -\beta )+\alpha \beta &=(r+c-\alpha )(r+c)-\beta (r+c)+\alpha \beta \\ & =(r+c-\alpha )(r+c)-\beta (r+c-\alpha ). \end{align}

Hence, we get the recurrence relation

Let's now simplify this relation by giving ar in terms of a0 instead of ar−1. From the recurrence relation,

\begin{align} a_1 &=\frac{(c)(c+1-\gamma )}{(c+1-\alpha )(c+1-\beta )}a_{0} \\ a_2 &=\frac{(c+1)(c+2-\gamma )}{(c+2-\alpha )(c+2-\beta )}a_{1}=\frac{(c+1)(c)(c+2-\gamma )(c+1-\gamma )}{(c+2-\alpha )(c+1-\alpha )(c+2-\beta )(c+1-\beta )}a_{0} = \frac{(c)_{2}(c+1-\gamma )_{2}}{(c+1-\alpha )_{2}(c+1-\beta )_{2}}a_{0} \end{align}

As we can see,

Hence, our assumed solution takes the form

We are now ready to study the solutions corresponding to the different cases for c1 − c2 = α − β.

Read more about this topic:  Frobenius Solution To The Hypergeometric Equation

Famous quotes containing the words solution and/or infinity:

    “Give a scientist a problem and he will probably provide a solution; historians and sociologists, by contrast, can offer only opinions. Ask a dozen chemists the composition of an organic compound such as methane, and within a short time all twelve will have come up with the same solution of CH4. Ask, however, a dozen economists or sociologists to provide policies to reduce unemployment or the level of crime and twelve widely differing opinions are likely to be offered.”
    —Derek Gjertsen, British scientist, author. Science and Philosophy: Past and Present, ch. 3, Penguin (1989)

    “The poetic notion of infinity is far greater than that which is sponsored by any creed.”
    —Joseph Brodsky (b. 1940)