Here is the statement of the problem:

Consider the Legendre Equation

$$ (*)\qquad (1-x^2)y''-2xy'+2y=0 $$

(a) Find two linearly independent solutions about $x=0$, solving completely any relevant recurrence relations.

(b) Compute the radius of convergence for each fundamental solution in part (a).

(c) Show that $x=1$ is a regular singular point of $(*)$.

(d) Assume that $y=(x-1)^s\sum_{n=0}^\infty a_n(x-1)^n$ (with $s \in \mathbb R$) is a solution near $x=1$, solve the indicial equation satisfied by $s$. Find one fundamental solution of $(*)$ near $x=1$.

Where I am:

Ok, this problem has been bugging me for a few days now. First off...

For (a): Since $x=0$ is an ordinary point of $(*)$, we may assume that a general solution of our ODE about this point is of the form:

$$ y=\sum_{n=0}^\infty a_nx^n. $$

So, taking some derivatives, plugging into our ODE, I found the following recurrence relation (assuming all of my arithmetic was correct):

$$ a_n = \frac{a_{n-2}(n^2-7n+12)}{n(n-1)} $$

Now, note that $a_0$ and $a_1$ are undefined here due to a zero in the denominator. Ok, fine. So, I started with $a_2$:

$$ a_2 = \frac{a_{0}(4-14+12)}{2} = -a_0 $$

Ok, sure. It's easy to see that, beyond that, all other terms vanish. Therefore, our solution becomes the following:

$$ y=\sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 = a_0 + a_1x -a_0x^2 = a_0(1-x^2)+a_1x $$

Sure, that seems fine; it's a nice little polynomial. But, how do I find the other solution? I'm completely clueless here.

Now, for part (b). Well, the radius of convergence for the solution that I found is infinity, I suppose, because it's just a polynomial. I suppose that once I find the other solution, this part will make more sense.

For part (c): This is easy. Just take some limits, and whatnot. I only included it here for context.

I haven't started part (d) yet because I'm too frustrated with the first part. I'm going to work on it now, though.

If anyone can I help me out here, I'd greatly appreciate it. Thanks!

  • 2
    $\begingroup$ For part (a), you have no condition on either $a_0$ or $a_1$. They are therefore independent of each other. So your expression is actually two solutions in one already. Congratulations! $\endgroup$ – bob.sacamento Jul 1 '15 at 18:14
  • $\begingroup$ Oh... you're right! Look at that. Hmm. The problem is that my instructor said that one function would be a polynomial and that the other would not be analytic, so I must've done something wrong... $\endgroup$ – thisisourconcerndude Jul 1 '15 at 18:19
  • $\begingroup$ Yep. I think $1-x^2$ is not a solution. $\endgroup$ – bob.sacamento Jul 1 '15 at 21:32

$$(1-x^2)y''-2xy'+2y=0$$ Since $y=x$ is an obvious solution, we look for other solutions on the form $y=xf(x)$

$y'=f+xf'$ and $y''=2f'+xf''$

$(1-x^2)(2f'+xf'')-2x(f+xf')+2xf=0$ $$2(1-2x^2)f'+x(1-x^2)f''=0$$


The integration leads to : $$f'=c_1\frac{1}{x^2(1-x^2)}$$ Integrating again leads to : $$f=c_1\left(-\frac{1}{x}+\frac{1}{2}\ln\left| \frac{1-x}{1+x} \right| \right)+c_2$$ $$y=c_1 \left(-1+\frac{x}{2}\ln\left| \frac{1-x}{1+x} \right| \right)+c_2 x$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.