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I'm given two solutions to Legendre's equation:


$$Q_0=\frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)$$

I'm trying to explain why their overlap integral (i.e. $\int_{-1}^{1} P_1 Q_0 dx$) is non-zero. I computed it and it is indeed non-zero, but I'm having a difficult time justifying why that is. I'm thinking it has something to do with that fact that the $P_n$ and $Q_n$ solutions are constructed w.r.t different weight functions. Or perhaps it has something to do with the completeness of solutions. Any thoughts?

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my thoughts are with the different weight functions. –  Dylan Sabulsky Nov 27 '12 at 2:25
What do you mean different weight functions? –  Pragabhava Nov 27 '12 at 2:29
I think a simple way of looking at this is that both $P_1$ and $Q_0$ are odd functions so the product $P_1Q_0$ is even and thus the integral is non-zero (since the product is not zero everywhere). –  aawaldrop Nov 27 '12 at 4:19
@Pragabhava: I'm not exactly sure. I just know the $P_n$ solutions can be formed from the Sturm-Liouville form of the Legendre operator and have the S-L form uses a weight function. –  Alex Nov 27 '12 at 4:46
@aawaldrop: I don't think that's correct. It isn't that simple. The solutions corresponding to a particular eigenvalue are, in general, orthogonal (for a self-adjoint operator). But, for some reason, that doesn't apply here. I'm trying to figure out why... –  Alex Nov 27 '12 at 4:48
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2 Answers

up vote 1 down vote accepted

As has been shown, the integral is not zero. This is okay. Since $P$ and $Q$ obey different boundary conditions they are eigenfunctions of different Sturm-Liouville systems, so we should not expect them to be orthogonal.

Consider a more familiar example, $$\begin{array}{l} y'' + n^2 y = 0 \\ y(0) = y(\pi) = 0. \end{array}$$ The unnormalized eigenfunctions are $f_{n} = \sin n x$, where $n \in \mathbb{N}$. Sturm-Liouville theory tells us the eigenfunctions must be orthogonal, and of course they are. The related system $$\begin{array}{l} y'' + n^2 y = 0 \\ y'(0) = y'(\pi) = 0 \end{array}$$ has eigenfunctions $g_{n} = \cos n x$. Again, the eigenfunctions are orthogonal. However, Sturm-Liouville theory has nothing to say about whether $f_m$ and $g_n$ are orthogonal, and in fact they are not in general. For example, $$\int_0^\pi dx\, \sin x \cos 2x = -\frac{2}{3}.$$

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Awesome. That's what I suspected. What are the corresponding Sturm-Liouville systems for the P and Q solutions? –  Alex Nov 27 '12 at 22:44
@Alex: The differential equation is the Legendre differential equation. The restrictions on $P$ are that $P$ be finite and nonzero at $x=\pm 1$. The restrictions on $Q$ are that $1/Q$ vanish at $x = \pm 1$. Have a look here for a short discussion of the boundary conditions on $P$. Cheers! –  user26872 Nov 27 '12 at 23:05
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Since $Q_0(x)$ is pointymmetric $$ Q_0(x)=\frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)=-\frac{1}{2} \ln\left(\frac{1+(-x)}{1-(-x)}\right)=-Q_0(-x) $$ as is $P_1(x)=-P_1(-x)$, there product is symmetric $$ P_1(x)Q_0(x)=(-1)^2P_1(-x)Q_0(-x). $$

Since your limits are also symmetric we'll get $$ \int_{-1}^{1} P_1(x) Q_0(x) dx=\int_{-1}^{0} P_1(x) Q_0(x) dx+\int_{0}^{1} P_1(x) Q_0(x) dx=2\int_{0}^{1} P_1(x) Q_0(x) dx $$ and since $P_1(x)$ and $ Q_0(x)$ are both positive on $[0,1]$, your integral is non-zero.

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Right. I agree. But the question is why... Typically eigenfunctions corresponding to distinct eigenvalues for a self-adjoint operator are orthogonal, but they obviously aren't in this case. I'm trying to find out essentially why the theorem doesn't apply. And I think it has something to do with the weight functions. –  Alex Nov 27 '12 at 16:42
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