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All the Steps in Transforming the Laplace Equation ($\displaystyle\Delta f=0$)
from Rectangular to Cylindrical and to Spherical Coordinates

(Copyright©2023 by Daniel B. Sedory.)

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Obtaining the Second Order Partial Differentials:  $\partial^2/\partial y^2$

  The second order partial derivative of the function $f$ (with respect to $y$) is also found by substituting some of the right-sides of the equations already shown there into the first order differentials highlighted below, and then making use of the chain rule, $\colorbox{lightblue}{$\frac{d}{dx}(uw) = u\frac{dw}{dx} + w\frac{du}{dx}$}$, as before:

$\newcommand{\pdif}[2]{\frac{\partial #1}{\partial #2}}$ \begin{align*} \large{\pdif{^2 f}{y^2}} &\colorbox{yellow}{$=$}\ \large{\sin\theta\ \sin\phi\ \pdif{^2\ f}{r\ \partial y}+\ \frac{\cos\theta\ \sin\phi}{r}\ \pdif{^2 f}{\theta\ \partial y}\ +\ \frac{\cos\phi}{r \sin\theta}\ \pdif{^2 f}{\phi\ \partial y}}\\ \\ &\colorbox{yellow}{$=$}\sin\theta\ \sin\phi\ \left(\pdif{}{\ r}\ \colorbox{yellow}{$\pdif{f}{y}$}\right)+\frac{\cos\theta\ \sin\phi}{r}\ \left(\pdif{f}{\theta}\ \colorbox{yellow}{$\pdif{f}{y}$}\right)+\frac{\cos\phi}{r \sin\theta}\ \left(\pdif{f}{\phi}\ \colorbox{yellow}{$\pdif{f}{y}$}\right)\\ \\ &\colorbox{yellow}{$=$}\sin\theta\ \sin\phi\ \left[\pdif{}{\ r}\ \left(\sin\theta\ \sin\phi\ \pdif{f}{\ r}+\frac{\cos\theta\ \sin\phi}{r}\ \pdif{f}{\theta}+\frac{\cos\phi}{r \sin\theta}\ \pdif{f}{\phi}\right)\right]\textbf{...}\\ &\textbf{...}\ \ +\frac{\cos\theta\ \sin\phi}{r}\ \left[\pdif{}{\theta}\ \left(\sin\theta\ \sin\phi\ \pdif{f}{\ r}+\frac{\cos\theta\ \sin\phi}{r}\ \pdif{f}{\theta}+\frac{\cos\phi}{r \sin\theta}\ \pdif{f}{\phi}\right)\right]\textbf{...}\\ &\textbf{...}\ \ +\frac{\cos\phi}{r \sin\theta}\ \left[\pdif{}{\phi}\ \left(\sin\theta\ \sin\phi\ \pdif{f}{\ r}+\frac{\cos\theta\ \sin\phi}{r}\ \pdif{f}{\theta}+\frac{\cos\phi}{r\ \sin\theta}\ \pdif{f}{\phi}\right)\right]\ . \end{align*}

 Upon comparison of what follows to the equations for  $\partial^2 f / \partial x^2,$  they will be found to vary only in the sign (+/-) in front of some terms and some changes from $\cos$ to $\sin$ or vice versa. All of the differentiations being quite similar to those on the previous page, producing:

\begin{align*} \large{\pdif{^2 f}{y^2}} &\colorbox{yellow}{$=$}\sin\theta\ \sin\phi\ \bigg(\sin\theta\ \sin\phi\ \pdif{^2 f}{\ r^2}+\frac{\cos\theta\ \sin\phi}{r}\ \pdif{^2 f}{r\ \partial\theta}-\frac{\cos\theta\ \sin\phi}{r^2}\ \pdif{f}{\theta}\textbf{...}\\ &\textbf{...}\ \ +\frac{\cos\phi}{r \sin\theta}\ \pdif{^2 f}{r\ \partial\phi}-\frac{\cos\phi}{r^2\sin\theta}\ \pdif{f}{\phi}\bigg)\textbf{...}\\ \\ &\textbf{...}\ \ +\frac{\cos\theta\ \sin\phi}{r}\ \bigg(\sin\theta\ \sin\phi\ \pdif{^2 f}{r\ \partial\theta}+\cos\theta\ \sin\phi\ \pdif{f}{r}+\frac{\cos\theta\ \sin\phi}{r}\ \pdif{^2 f}{\theta^2}\textbf{...}\\ &\textbf{...}\ \ -\frac{\sin\theta\ \sin\phi}{r}\ \pdif{f}{\theta}+\frac{\cos\phi}{r\ \sin\theta}\ \pdif{^2 f}{\theta\ \partial\phi}-\frac{\cos\theta\ \cos\phi}{r\ \sin^2\theta}\ \pdif{f}{\phi}\bigg)\textbf{...}\\ \\ &\textbf{...}\ \ +\frac{\cos\phi}{r \sin\theta}\ \bigg(\sin\theta\ \sin\phi\ \pdif{^2 f}{r\ \partial\phi}+\sin\theta\ \cos\phi\ \pdif{f}{r}+\frac{\cos\theta\ \sin\phi}{r}\ \pdif{^2 f}{\theta\ \partial\phi}\textbf{...}\\ &\textbf{...}\ \ +\frac{\cos\theta\ \cos\phi}{r}\ \pdif{f}{\theta}+\frac{\cos\phi}{r\ \sin\theta}\ \pdif{^2 f}{\phi^2}-\frac{\sin\phi}{r\ \sin\theta}\ \pdif{f}{\phi}\bigg)\ . \end{align*}

And after multiplying through to eliminate the parentheses, we get:

\begin{align*} \large{\pdif{^2 f}{y^2}} &\colorbox{yellow}{$=$}\sin^2\theta\ \sin^2\phi\ \pdif{^2 f}{\ r^2}+\colorbox{lightgreen}{$\frac{\sin\theta\ \cos\theta\ \sin^2\phi}{r}\ \pdif{^2 f}{r\ \partial\theta}$}-\colorbox{lightgreen}{$\frac{\sin\theta\ \cos\theta\ \sin^2\phi}{r^2}\ \pdif{f}{\theta}$}\ \textbf{...}\\ &\textbf{...}\ \ +\frac{\colorbox{pink}{$\sin\theta$}\ \colorbox{lightgreen}{$\sin\phi\ \cos\phi$}}{\colorbox{lightgreen}{$r$}\ \colorbox{pink}{$\sin\theta$}}\ \colorbox{lightgreen}{$\pdif{^2 f}{r\ \partial\phi}$}-\frac{\sin\phi\ \cos\phi}{r^2}\ \pdif{f}{\phi}+\colorbox{lightgreen}{$\frac{\sin\theta\ \cos\theta\ \sin^2\phi}{r}\ \pdif{^2 f}{r\ \partial\theta}$}\ \textbf{...}\\ &\textbf{...}\ \ +\frac{\cos^2\theta\ \sin^2\phi}{r}\ \pdif{f}{r}+\frac{\cos^2\theta\ \sin^2\phi}{r^2}\ \pdif{^2 f}{\theta^2}-\colorbox{lightgreen}{$\frac{\cos\theta\ \sin\theta\ \sin^2\phi}{r^2}\ \pdif{f}{\theta}$}\ \textbf{...}\\ &\textbf{...}\ \ +\colorbox{lightgreen}{$\frac{\cos\theta\ \sin\phi\ \cos\phi}{r^2\ \sin\theta}\ \pdif{^2 f}{\theta\ \partial\phi}$}-\frac{\cos^2\theta\ \sin\phi\ \cos\phi}{r^2\ \sin^2\theta}\ \pdif{f}{\phi}+\colorbox{lightgreen}{$\frac{\sin\phi\ \cos\phi}{r}\ \pdif{^2 f}{r\ \partial\phi}$}\ \textbf{...}\\ &\textbf{...}\ \ +\frac{\cos^2\phi}{r}\ \pdif{f}{r}+\colorbox{lightgreen}{$\frac{\cos\theta\ \sin\phi\ \cos\phi}{r^2\ \sin\theta}\ \pdif{^2 f}{\theta\ \partial\phi}$}+\frac{\cos\theta\ \cos^2\phi}{r^2\ \sin\theta}\ \pdif{f}{\theta}\ \textbf{...}\\ &\textbf{...}\ \ + \frac{\cos^2\phi}{r^2\ \sin^2\theta}\ \pdif{^2 f}{\phi^2}-\frac{\sin\phi\ \cos\phi}{r^2\ \sin^2\theta}\ \pdif{f}{\phi}\ .\end{align*}

Which after combining some of the terms, leaves us with:

\begin{align*} \large{\pdif{^2 f}{y^2}} &\colorbox{yellow}{$=$}\sin^2\theta\ \sin^2\phi\ \pdif{^2 f}{\ r^2}+\frac{\colorbox{lightgreen}{$2$}\ \sin\theta\ \cos\theta\ \sin^2\phi}{r}\ \pdif{^2 f}{r\ \partial\theta}-\frac{\colorbox{lightgreen}{$2$}\ \sin\theta\ \cos\theta\ \sin^2\phi}{r^2}\ \pdif{f}{\theta}\ \textbf{...}\\ &\textbf{...}\ \ +\frac{\colorbox{lightgreen}{$2$}\ \sin\phi\ \cos\phi}{r}\ \pdif{^2 f}{r\ \partial\phi}-\frac{\sin\phi\ \cos\phi}{r^2}\ \pdif{f}{\phi}+\frac{\cos^2\theta\ \sin^2\phi}{r}\ \pdif{f}{r}\ \textbf{...}\\ &\textbf{...}\ \ +\frac{\cos^2\theta\ \sin^2\phi}{r^2}\ \pdif{^2 f}{\theta^2}+\frac{\colorbox{lightgreen}{$2$}\ \cos\theta\ \sin\phi\ \cos\phi}{r^2\ \sin\theta}\ \pdif{^2 f}{\theta\ \partial\phi}-\frac{\cos^2\theta\ \sin\phi\ \cos\phi}{r^2\ \sin^2\theta}\ \pdif{f}{\phi}\ \textbf{...}\\ &\textbf{...}\ \ +\frac{\cos^2\phi}{r}\ \pdif{f}{r}+\frac{\cos\theta\ \cos^2\phi}{r^2\ \sin\theta}\ \pdif{f}{\theta}+\frac{\cos^2\phi}{r^2\ \sin^2\theta}\ \pdif{^2 f}{\phi^2}-\frac{\sin\phi\ \cos\phi}{r^2\ \sin^2\theta}\ \pdif{f}{\phi}\ .\end{align*}

 

Obtaining the Second Order Partial Differentials:  $\partial^2/\partial z^2$

  Again, as noted above, the second order partial derivative of the function $f$ (this time, with respect to $z$) is found by substituting some of the right-sides of the equations already shown on the previous page into the first order differentials highlighted below, and then making use of the chain rule as before:

\begin{align*} \large{\pdif{^2 f}{z^2}} &\colorbox{yellow}{$=$}\ \large{\cos\theta\ \pdif{^2\ f}{r\ \partial z}-\ \frac{\sin\theta}{r}\ \pdif{^2 f}{\theta\ \partial z}}\\ \\ &\colorbox{yellow}{$=$}\cos\theta\ \left(\pdif{}{\ r}\ \colorbox{yellow}{$\pdif{f}{z}$}\right)-\frac{\sin\theta}{r}\ \left(\pdif{}{\theta}\ \colorbox{yellow}{$\pdif{f}{z}$}\right)\\ \\ &\colorbox{yellow}{$=$}\cos\theta\ \left[\pdif{}{\ r}\ \left(\cos\theta\ \pdif{f}{\ r}\ -\ \frac{\sin\theta}{r}\ \pdif{f}{\theta}\right)\right]\textbf{...}\\ &\textbf{...}\ \ -\frac{\sin\theta}{r}\ \left[\pdif{}{\theta}\ \left(\cos\theta\ \pdif{f}{\ r}\ -\ \frac{\sin\theta}{r}\ \pdif{f}{\theta}\right)\right]\ .\end{align*}

After carrying out the differentiations, we obtain:

\begin{align*} \large{\pdif{^2 f}{z^2}} &\colorbox{yellow}{$=$}\ \cos\theta\ \bigg(\cos\theta\ \pdif{^2 f}{\ r^2}-\frac{\sin\theta}{r}\ \pdif{^2 f}{r\ \partial\theta}+\frac{\sin\theta}{r^2}\ \pdif{f}{\theta}\bigg)\textbf{...}\\ &\textbf{...}\ \ -\frac{\sin\theta}{r}\ \bigg(\cos\theta\ \pdif{^2 f}{r\ \partial\theta}-\sin\theta\ \pdif{f}{r}-\frac{\sin\theta}{r}\ \pdif{^2 f}{\theta^2}-\frac{\cos\theta}{r}\ \pdif{f}{\theta}\bigg)\ .\end{align*}
\begin{align*} \large{\pdif{^2 f}{z^2}} &\colorbox{yellow}{$=$}\cos^2\theta\ \pdif{^2 f}{\ r^2}-\colorbox{lightgreen}{$\frac{\sin\theta\ \cos\theta}{r}\ \pdif{^2 f}{r\ \partial\theta}$}+\colorbox{lightgreen}{$\frac{\sin\theta\ \cos\theta}{r^2}\ \pdif{f}{\theta}$}\textbf{...}\\ &\textbf{...}\ \ -\colorbox{lightgreen}{$\frac{\sin\theta\ \cos\theta}{r}\ \pdif{^2 f}{r\ \partial\theta}$}+\frac{\sin^2\theta}{r}\ \pdif{f}{r}+\frac{\sin^2\theta}{r^2}\ \pdif{^2 f}{\theta^2}+\colorbox{lightgreen}{$\frac{\sin\theta\ \cos\theta}{r^2}\ \pdif{f}{\theta}$}\ .\end{align*}
\begin{align*} \large{\pdif{^2 f}{z^2}} &\colorbox{yellow}{$=$}\cos^2\theta\ \pdif{^2 f}{\ r^2}-\frac{\colorbox{lightgreen}{$2$}\ \sin\theta\ \cos\theta}{r}\ \pdif{^2 f}{r\ \partial\theta}\textbf{...}\\ &\textbf{...}\ \ +\frac{\colorbox{lightgreen}{$2$}\ \sin\theta\ \cos\theta}{r^2}\ \pdif{f}{\theta}+\frac{\sin^2\theta}{r}\ \pdif{f}{r}+\frac{\sin^2\theta}{r^2}\ \pdif{^2 f}{\theta^2}\ .\end{align*}

 

The Sum of All Three Second Order Partial Differentials

  Here we sum up all three of the previously derived terms of the second order differentials of $f$ with respect to $x$, $y$ and $z$:

\begin{align*}&\large{\pdif{^2 f}{x^2}+\pdif{^2 f}{y^2}+\pdif{^2 f}{z^2}}\colorbox{yellow}{$=$}\\ &\colorbox{yellow}{$=$}\sin^2\theta\ \cos^2\phi\ \pdif{^2 f}{\ r^2}+\frac{2\ \sin\theta\ \cos\theta\ \cos^2\phi}{r}\ \pdif{^2 f}{r\ \partial\theta}-\frac{2\ \sin\theta\ \cos\theta\ \cos^2\phi}{r^2}\ \pdif{f}{\theta}\ \textbf{...}\\ &\textbf{...}\ \ \colorbox{pink}{$-\frac{2\ \sin\phi\ \cos\phi}{r}\ \pdif{^2 f}{r\ \partial\phi}+\frac{\sin\phi\ \cos\phi}{r^2}\ \pdif{f}{\phi}$}+\frac{\cos^2\theta\ \cos^2\phi}{r}\ \pdif{f}{r}\ \textbf{...}\\ &\textbf{...}\ \ +\frac{\cos^2\theta\ \cos^2\phi}{r^2}\ \pdif{^2 f}{\theta^2}\colorbox{pink}{$-\frac{2\ \cos\theta\ \sin\phi\ \cos\phi}{r^2\ \sin\theta}\ \pdif{^2 f}{\theta\ \partial\phi}+\frac{\cos^2\theta\ \sin\phi\ \cos\phi}{r^2\ \sin^2\theta}\ \pdif{f}{\phi}$}\ \textbf{...}\\ &\textbf{...}\ \ +\frac{\sin^2\phi}{r}\ \pdif{f}{r}+\frac{\cos\theta\ \sin^2\phi}{r^2\ \sin\theta}\ \pdif{f}{\theta}+\frac{\sin^2\phi}{r^2\ \sin^2\theta}\ \pdif{^2 f}{\phi^2}\colorbox{pink}{$+\frac{\sin\phi\ \cos\phi}{r^2\ \sin^2\theta}\ \pdif{f}{\phi}$}\ \textbf{...}\\ &\textbf{...}\ \ +\sin^2\theta\ \sin^2\phi\ \pdif{^2 f}{\ r^2}+\frac{2\ \sin\theta\ \cos\theta\ \sin^2\phi}{r}\ \pdif{^2 f}{r\ \partial\theta}-\frac{2\ \sin\theta\ \cos\theta\ \sin^2\phi}{r^2}\ \pdif{f}{\theta}\ \textbf{...}\\ &\textbf{...}\ \ \colorbox{pink}{$+\frac{2\ \sin\phi\ \cos\phi}{r}\ \pdif{^2 f}{r\ \partial\phi}-\frac{\sin\phi\ \cos\phi}{r^2}\ \pdif{f}{\phi}$}+\frac{\cos^2\theta\ \sin^2\phi}{r}\ \pdif{f}{r}\ \textbf{...}\\ &\textbf{...}\ \ +\frac{\cos^2\theta\ \sin^2\phi}{r^2}\ \pdif{^2 f}{\theta^2}\colorbox{pink}{$+\frac{2\ \cos\theta\ \sin\phi\ \cos\phi}{r^2\ \sin\theta}\ \pdif{^2 f}{\theta\ \partial\phi}-\frac{\cos^2\theta\ \sin\phi\ \cos\phi}{r^2\ \sin^2\theta}\ \pdif{f}{\phi}$}\ \textbf{...}\\ &\textbf{...}\ \ +\frac{\cos^2\phi}{r}\ \pdif{f}{r}+\frac{\cos\theta\ \cos^2\phi}{r^2\ \sin\theta}\ \pdif{f}{\theta}+\frac{\cos^2\phi}{r^2\ \sin^2\theta}\ \pdif{^2 f}{\phi^2}\colorbox{pink}{$-\frac{\sin\phi\ \cos\phi}{r^2\ \sin^2\theta}\ \pdif{f}{\phi}$}\textbf{...}\\ &\textbf{...}\ \ +\cos^2\theta\ \pdif{^2 f}{\ r^2}-\frac{2\ \sin\theta\ \cos\theta}{r}\ \pdif{^2 f}{r\ \partial\theta}\textbf{...}\\ &\textbf{...}\ \ +\frac{2\ \sin\theta\ \cos\theta}{r^2}\ \pdif{f}{\theta}+\frac{\sin^2\theta}{r}\ \pdif{f}{r}+\frac{\sin^2\theta}{r^2}\ \pdif{^2 f}{\theta^2}\ .\end{align*}

Ten of the terms cancel out immediately. However, in order to reduce all of the remaining terms to (Equation S), trigonometric identities, such as, $\sin^2 x+\cos^2 x= 1$, and algebra must be used after combining like terms:

\begin{align*}&\large{\pdif{^2 f}{x^2}+\pdif{^2 f}{y^2}+\pdif{^2 f}{z^2}}\colorbox{yellow}{$=$}\\ &\colorbox{yellow}{$=$}\bigg[\sin^2\theta\ \cos^2\phi+\sin^2\theta\ \sin^2\phi+\cos^2\theta\bigg]\colorbox{lightblue}{$\pdif{^2 f}{\ r^2}$}\textbf{...}\\ &\textbf{...}\ \ +\colorbox{pink}{$\bigg[\frac{2\ \sin\theta\ \cos\theta\ \cos^2\phi}{r}+\frac{2\ \sin\theta\ \cos\theta\ \sin^2\phi}{r}-\frac{2\ \sin\theta\ \cos\theta}{r}\bigg]\ \pdif{^2 f}{r\ \partial\theta}$}\textbf{...}\\ &\textbf{...}\ \ +\bigg[-\frac{2\ \sin\theta\ \cos\theta\ \cos^2\phi}{r^2}+\frac{\cos\theta\ \sin^2\phi}{r^2\ \sin\theta}-\frac{2\ \sin\theta\ \cos\theta\ \sin^2\phi}{r^2}\textbf{...}\\ &\textbf{...}\ \ +\frac{\cos\theta\ \cos^2\phi}{r^2\ \sin\theta}+\frac{2\ \sin\theta\ \cos\theta}{r^2}\bigg]\ \colorbox{lightblue}{$\pdif{f}{\theta}$}\textbf{...}\\ &\textbf{...}\ \ +\bigg[\frac{\cos^2\theta\ \cos^2\phi}{r}+\frac{\sin^2\phi}{r}+\frac{\cos^2\theta\ \sin^2\phi}{r}+\frac{\cos^2\phi}{r}+\frac{\sin^2\theta}{r}\bigg]\colorbox{lightblue}{$\pdif{f}{r}$}\textbf{...}\\ &\textbf{...}\ \ +\bigg[\frac{\cos^2\theta\ \cos^2\phi}{r^2}+\frac{\cos^2\theta\ \sin^2\phi}{r^2}+\frac{\sin^2\theta}{r^2}\bigg]\colorbox{lightblue}{$\pdif{^2 f}{\theta^2}$}\textbf{...}\\ &\textbf{...}\ \ +\bigg[\frac{\colorbox{lightgreen}{$\sin^2\phi$}}{r^2\ \sin^2\theta}+\frac{\colorbox{lightgreen}{$\cos^2\phi$}}{r^2\ \sin^2\theta}\bigg]\colorbox{lightblue}{$\pdif{^2 f}{\phi^2}$}\ .\end{align*}

Examples of how the terms above can be reduced:

\begin{align*}&\quad\bigg[\sin^2\theta\ \cos^2\phi+\sin^2\theta\ \sin^2\phi+\cos^2\theta\bigg]\ \pdif{^2 f}{\ r^2}\\ &\colorbox{yellow}{$=$}\bigg[\sin^2\theta\ \bigg(\cos^2\phi+\sin^2\phi+\frac{\cos^2\theta}{\sin^2\theta}\bigg)\bigg]\ \pdif{^2 f}{\ r^2}\\ &\colorbox{yellow}{$=$}\bigg[\sin^2\theta\ \bigg(1+\frac{\cos^2\theta}{\sin^2\theta}\bigg)\bigg]\ \pdif{^2 f}{\ r^2}\\ &\colorbox{yellow}{$=$}(\sin^2\theta+\cos^2\theta)\ \pdif{^2 f}{\ r^2}\colorbox{yellow}{$=$}(1)\ \pdif{^2 f}{\ r^2}\ .\\ \\ &\quad\bigg[-\frac{2\ \sin\theta\ \cos\theta\ \cos^2\phi}{r^2}+\frac{\cos\theta\ \sin^2\phi}{r^2\ \sin\theta}-\frac{2\ \sin\theta\ \cos\theta\ \sin^2\phi}{r^2}\textbf{...}\\ &\textbf{...}\ \ +\frac{\cos\theta\ \cos^2\phi}{r^2\ \sin\theta}+\frac{2\ \sin\theta\ \cos\theta}{r^2}\bigg]\ \pdif{f}{\theta}\\ &\colorbox{yellow}{$=$}\bigg[-2\ \sin\theta\ \cos\theta \bigg(\frac{\colorbox{lightgreen}{$\cos^2\phi$}}{r^2}+\frac{\colorbox{lightgreen}{$\sin^2\phi$}}{r^2}-\frac{1}{r^2}\bigg)\textbf{...}\\ &\textbf{...}\ \ +\frac{\cos\theta\ \sin^2\phi}{r^2\ \sin\theta}+\frac{\cos\theta\ \cos^2\phi}{r^2\ \sin\theta}\bigg]\ \pdif{f}{\theta}\\ &\colorbox{yellow}{$=$}\bigg[\colorbox{pink}{$0$}+\cos\theta\ \bigg(\frac{\colorbox{lightgreen}{$\sin^2\phi$}}{r^2\ \sin\theta}+\frac{\colorbox{lightgreen}{$\cos^2\phi$}}{r^2\ \sin\theta}\bigg)\bigg]\ \pdif{f}{\theta}\\ &\colorbox{yellow}{$=$}\frac{cos\ \theta}{r^2\ \sin\theta}\ \pdif{f}{\theta}\ .\end{align*}

After applying similar algebraic operations to all the terms, the equation does in fact reduce to:

$$\large{\nabla^2 f} =\colorbox{lightblue}{$\pdif{^2 f}{\ r^2}\ +\ \frac{2}{r}\ \pdif{f}{\ r}\ +\frac{1}{r^2}\ \pdif{^2 f}{\ \theta^2}\ +\frac{cos\ \theta}{r^2\ \sin\theta}\ \pdif{f}{\ \theta}\ +\ \frac{1}{r^2\ \sin^2\ \theta}\ \pdif{^2 f}{\phi^2}$} = 0\quad\textbf{(Equation S)}$$

 

 

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First Posted on: 23 November 2023 (2023.11.23).
Updated on: 24 NOV 2023 (2023.11.24); made a sign correction and finished the page.

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