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:

    “All the followers of science are fully persuaded that the processes of investigation, if only pushed far enough, will give one certain solution to each question to which they can be applied.... This great law is embodied in the conception of truth and reality. The opinion which is fated to be ultimately agreed to by all who investigate is what we mean by the truth, and the object represented in this opinion is the real.”
    —Charles Sanders Peirce (1839–1914)

    “We must not suppose that, because a man is a rational animal, he will, therefore, always act rationally; or, because he has such or such a predominant passion, that he will act invariably and consequentially in pursuit of it. No, we are complicated machines; and though we have one main spring that gives motion to the whole, we have an infinity of little wheels, which, in their turns, retard, precipitate, and sometime stop that motion.”
    —Philip Dormer Stanhope, 4th Earl Chesterfield (1694–1773)