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The ODE is $−[(1−x^2)u′]′-\mu u=f(x)$ over the interval $[-1,1]$.

I want to determine whether the singular endpoint $x=1$ is of a limit circle or limit point type. To do so, we were given a Theorem from Weyl which basically says that it suffices to look at any single value of $\mu$ with Im$(\mu)=0$ and just determine the number of solutions to the homogeneous equation that are in $L^2_w[-1,1]$. But in order to use this theorem one endpoint must be regular - hence I introduced the point $x=0$ and looked at the interval $[0,1]$ rather than $[-1,1]$.

So I thought it probably easiest to look at $\mu=0$. When we do this, we can pretty easily see by inspection that two solutions of this new ODE will be: $$ u_1'(x)=0 \quad\quad\quad u_2'(x)=\frac{1}{(1-x^2)} $$ $$ \implies u_1(x)=C_1 \quad\quad\quad u_2(x)=\frac{1}{2}\ln{|\frac{x+1}{x-1}|} $$ I took out the integration constant in the second integral so my two solutions are now linearly independent. All that remains to find is whether these two solutions are in $L^2_w[0,1]$ or not.

I have a bunch of questions, however. Is introducing the intermediary point correct the way I did it? Additionally, when I went to solve the integral I was a little surprised by how complicated it was - since I am not actually interested in the value of the integral but just in showing its less than $\infty$ are there any simplification I could make? Or could I have maybe picked a different value of $\mu$ that would have made the problem easier?

Thanks for any help. I originally posted this as a separate question just asking about the integral and explaining a little of my confusion on how to proceed with solving it: Solving the integral: $\int_0^1\ln^2{(\frac{x+1}{x-1})} dx$

Steven Stadnicki
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Mjoseph
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    Thank you for the clearer question! Note that there's a big difference between $\ln\left|\frac{x+1}{x-1}\right|$ and $\ln\left(\frac{x+1}{x-1}\right)$ especially on the interval $[0,1]$ since the argument of the log in the latter is negative; this explains the issues that you were having there (and does point to $\ln\left(\frac{x+1}{1-x}\right)$ as the correct function to be using for that.) – Steven Stadnicki Apr 23 '21 at 17:39
  • ohhhhh, thank you so much - I see what you mean now. So the rest of my work and logic was correct then? – Mjoseph Apr 23 '21 at 17:41
  • Well, functional analysis has never been my strong suit; I've done a little bit of digging, though, and I suspect you've got the Weyl criterion wrong (you may be conflating different $\mu$s possibly) but I can't really speak with any confidence there. – Steven Stadnicki Apr 23 '21 at 17:49

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