Toronto Math Forum
APM3462016F => APM346Lectures => Chapter 8 => Topic started by: Luyu CEN on November 23, 2016, 04:26:33 PM

http://www.math.toronto.edu/courses/apm346h1/20169/PDEtextbook/Chapter8/S8.1.html#mjxeqneq8.1.9
If we follow from equation(7) http://www.math.toronto.edu/courses/apm346h1/20169/PDEtextbook/Chapter8/S8.1.html#mjxeqneq8.1.7
(9) should be
\begin{equation}
\sin^2(\phi)
\Phi'' +\sin(\phi)\cos(\phi)\Phi' = \bigl(l(l+1)\sin^2(\phi) m^2\bigr)\Phi,
\end{equation}
instead of
\begin{equation}
\sin^2(\phi)
\Phi'' +2\sin(\phi)\cos(\phi)\Phi' = \bigl(l(l+1)\sin^2(\phi) m^2\bigr)\Phi,
\end{equation}
There will be a coefficient 2 if we substitute $\Phi(\phi)=L(\cos(\phi))$ into the equation and use the chain rule to get its derivatives but I think not now.

Thanks. There are also more serious omissions. Corrected.