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- Jan 26, 2012

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

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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.

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

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, $\int_{-1}^1y_{\ell}yndx = 0$.

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- Jan 26, 2012

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