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[SOLVED] Sturm-Liouiville

dwsmith

Well-known member
Feb 1, 2012
1,673
Given
$$
\lambda^2\int_{-1}^1y_ny_mdx = \lambda^2\int_{-1}^1y_my_ndx
$$
Why does this imply that either $m = n$ or the integral is 0?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
You've marked this thread as solved. I'm a bit puzzled by the question myself, though. The two integrals are equal, period, because multiplication is commutative. The equality of those two integrals implies nothing whatsoever. Did you mean, instead, that
$$\lambda^{2}\int_{-1}^{1}y_{n}y_{n}\,dx=\lambda^{2}\int_{-1}^{1}y_{m}y_{m}\,dx?$$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
You've marked this thread as solved. I'm a bit puzzled by the question myself, though. The two integrals are equal, period, because multiplication is commutative. The equality of those two integrals implies nothing whatsoever. Did you mean, instead, that
$$\lambda^{2}\int_{-1}^{1}y_{n}y_{n}\,dx=\lambda^{2}\int_{-1}^{1}y_{m}y_{m}\,dx?$$
I was looking at a paper about the Legendre polynomials, and I didn't understand what they were doing. I then realized they should have indexed there lambdas.


We can rewrite $(1 - x^2)y_{\ell}'' - 2xy_{\ell}' + \left(\lambda_{\ell}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell}$ as
\begin{alignat}{3}
\frac{d}{dx}\left[(1 - x^2)y_{\ell}'\right] + \left(\lambda_{\ell}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell} & = & 0\notag\\
\frac{d}{dx}\left[(1 - x^2)y_{\ell}'\right] & = & - \left(\lambda_{\ell}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell}\notag\\
y_n\frac{d}{dx}\left[(1 - x^2)y_{\ell}'\right] & = & - \left(\lambda_{\ell}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell}y_n\notag\\
\int_{-1}^1y_n\left[(1 - x^2)y_{\ell}'\right]'dx & = & - \int_{-1}^1\left(\lambda_{\ell}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell}y_ndx
\end{alignat}
Integrating (2) by parts where $u = y_n$ and $dv = [(1 - x^2)y_{\ell}']'$, we have
\begin{alignat*}{3}
\underbrace{\left.(1 - x^2)y_n'y_{\ell}\right|_{-1}^1}_{ = 0} - \int_{-1}^1(1 - x^2)y_n'y_{\ell}'dx & = & - \int_{-1}^1\left(\lambda_{\ell}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell}y_ndx\\
\int_{-1}^1(1 - x^2)y_n'y_{\ell}'dx & = & \int_{-1}^1\left(\lambda_{\ell}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell}y_ndx
\end{alignat*}
Since we could have started with $y_n$ and the choice of $y_{\ell}$ was arbitrary, we would also have
$$
\int_{-1}^1(1 - x^2)y_n'y_{\ell}'dx = \int_{-1}^1\left(\lambda_{n}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell}y_ndx.
$$
Therefore, we have that
\begin{alignat*}{3}
\int_{-1}^1\left(\lambda_{n}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell}y_ndx & = & \int_{-1}^1\left(\lambda_{\ell}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell}y_ndx\\
\int_{-1}^1\left(\lambda_{n}^2y_{\ell}y_n - \frac{m^2}{1 - x^2}y_{\ell}y_n - \lambda_{\ell}^2y_{\ell}y_n + \frac{m^2}{1 - x^2}y_{\ell}y_n\right)dx & = & 0\\
(\lambda_n^2 - \lambda_{\ell}^2)\int_{-1}^1y_{\ell}yndx & = & 0
\end{alignat*}
When $n\neq\ell$, $\lambda_n^2 - \lambda_{\ell}^2$ is non-zero; therefore, $\int_{-1}^1y_{\ell}yndx = 0$.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
Ah, yes; the standard argument that for an Hermitian operator (corresponding to the Sturm-Liouville DE), eigenfunctions corresponding to different eigenvalues are orthogonal.