On the regular singularities of a second order differential equation

[Rulonegger]

Homework Statement

The only singularities of the differential equation
$$y''+p(x)y'+q(x)y=0$$
are regular singularities at x=1 of exponents $\alpha$ and $\alpha'$, and at x=-1 of exponents $\beta$ and $\beta'$, the point at infinity being an ordinary point.
Prove that $\beta=-\alpha$ and $\beta'=-\alpha'$ and that the differential equation is
$$(x^2+1)^2 y''+2(x-1)(x-\alpha-\alpha') y'+4\alpha\alpha' y=0$$
where $c_1,c_2$ are constants.

Homework Equations

The solution to find is given
$$y(x)=c_1\left(\frac{x-1}{x+1}\right)^{\alpha}+c_2\left(\frac{x-1}{x+1}\right)^{\alpha'}$$

The Attempt at a Solution

I understand that if at x=1 is a regular singular point, it means that the factor $(x-1)$ appears in the denominator of p(x) at the first power (at most), and at the second power in the denominator of q(x) (at most), then
$$p(x)=\frac{f(x)}{(x-1)^{\alpha}(x+1)^{\beta}}$$
$$q(x)=\frac{g(x)}{(x-1)^{\alpha'}(x+1)^{\beta'}}$$
with $\alpha,\beta \leq 1$ and $\alpha',\beta' \leq 2$, and $f(x),g(x)$ without $x=\pm 1$ as a root. Assuming that the point at infinity is an ordinary point, it means that f(x) and g(x) are finite as $x\rightarrow\infty$ (for example f and g being constant). Then, for example, f and g could be a sine or cosine function of x, and $\alpha=\beta=1$ and $\alpha'=\beta'=2$, and it makes that the differential equation has only 2 singularities, both being regular, and with the point at infinity being an ordinary point. Therefore, i don't understand how is to be "proven" that $\alpha$ must be $-\beta$ and so on, and then proving the closed form of the differential equation, without any traces of $f(x)$ neither as $g(x)$.

 

[mighty2000]

Thank you for your post. I can see that you have a good understanding of regular singularities and ordinary points. The proof for \beta=-\alpha and \beta'=-\alpha' can be done using the Frobenius method, which is commonly used for solving differential equations with regular singularities. The Frobenius method involves assuming a solution of the form y(x)=x^r\sum_{n=0}^{\infty}a_nx^n and substituting it into the differential equation. By comparing coefficients of different powers of x, you can obtain a recurrence relation for the coefficients a_n, which will eventually lead to the values of \alpha and \beta. Similarly, you can assume a solution of the form y(x)=x^r\ln(x)\sum_{n=0}^{\infty}a_nx^n for the case of x=\pm 1 being a regular singular point of exponent 1.

As for the closed form of the differential equation, you can also obtain it using the Frobenius method. By substituting the solution y(x)=c_1\left(\frac{x-1}{x+1}\right)^{\alpha}+c_2\left(\frac{x-1}{x+1}\right)^{\alpha'} into the differential equation, you can obtain a recurrence relation for the coefficients c_1 and c_2. By solving this recurrence relation, you can obtain the closed form of the differential equation.

I hope this helps. Let me know if you need any further clarification. Keep up the good work in your studies!