Sturm-Liouiville: Why Does Integral Equal 0 or m=n?

[Dustinsfl]

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]

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
$$\lambda^{2}\int_{-1}^{1}y_{n}y_{n}\,dx=\lambda^{2}\int_{-1}^{1}y_{m}y_{m}\,dx?$$

 

[Dustinsfl]

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]

Ah, yes; the standard argument that for an Hermitian operator (corresponding to the Sturm-Liouville DE), eigenfunctions corresponding to different eigenvalues are orthogonal.

 

[blue_raver22]

The Sturm-Liouville equation is a fundamental concept in mathematics and physics, and it is used to solve various differential equations. In this context, the given equation represents an eigenvalue problem, where $\lambda$ is the eigenvalue and $y_n$ and $y_m$ are the eigenfunctions corresponding to the eigenvalues $n$ and $m$, respectively.

The fact that the integral on the left-hand side of the equation is equal to the integral on the right-hand side implies that the eigenfunctions $y_n$ and $y_m$ are orthogonal over the interval $[-1, 1]$. This means that the integral of the product of $y_n$ and $y_m$ over the interval is equal to 0, unless $n = m$.

This is a result of the orthogonality property of eigenfunctions, which states that the inner product of two distinct eigenfunctions is equal to 0. Therefore, the given equation implies that either $m = n$, in which case the integral is non-zero, or $m \neq n$, in which case the integral is equal to 0.

In conclusion, the given equation is a manifestation of the fundamental orthogonality property of eigenfunctions, which explains why the integral is either 0 or equal to the eigenvalue squared. This result is crucial in solving differential equations and has many important applications in various fields of science and engineering.