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\begin{document}
\section{Laplace Equation in Cylindrical Coordinates}
Solutions to the Laplace equation in cylindrical coordinates have wide applicability from fluid \htmladdnormallink{mechanics}{http://planetphysics.us/encyclopedia/Mechanics.html} to electrostatics.
Applying the method of \htmladdnormallink{separation of variables}{http://planetphysics.us/encyclopedia/SeparationOfVariables.html} to Laplace's \htmladdnormallink{partial differential equation}{http://planetphysics.us/encyclopedia/DifferentialEquations.html} and then enumerating the various forms
of solutions will lay down a foundation for solving problems in this coordinate \htmladdnormallink{system}{http://planetphysics.us/encyclopedia/SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence.html}. Finally, the use of \htmladdnormallink{Bessel functions}{http://planetphysics.us/encyclopedia/BesselEquation2.html} in the solution reminds us why they are synonymous with the cylindrical \htmladdnormallink{domain}{http://planetphysics.us/encyclopedia/Bijective.html}.
\subsection{Separation of Variables}
Beginning with the \htmladdnormallink{Laplacian in Cylindrical Coordinates}{http://planetphysics.us/encyclopedia/LaplacianInCylindricalCoordinates.html}, apply the \htmladdnormallink{operator}{http://planetphysics.us/encyclopedia/QuantumSpinNetworkFunctor2.html} to a potential \htmladdnormallink{function}{http://planetphysics.us/encyclopedia/Bijective.html} and set it equal to zero to get the \htmladdnormallink{Laplace equation}{http://planetphysics.us/encyclopedia/FluorescenceCrossCorrelationSpectroscopy.html}
\begin{equation}
\nabla^{2} \Phi = \frac{1}{r} \frac{\partial \Phi}{\partial r}\left(r \frac{\partial\Phi}{\partial r}\right) + \frac{1}{r^2} \frac{\partial^2\Phi}{\partial \theta^2} + \frac{\partial^2 \Phi}{\partial z^2} = 0.
\end{equation}
First expand out the terms
\begin{equation}
\nabla^{2} \Phi = \frac{1}{r} \frac{\partial \Phi}{\partial r} + \frac{\partial^2\Phi}{\partial r^2} + \frac{1}{r^2} \frac{\partial^2\Phi}{\partial \theta^2} + \frac{\partial^2 \Phi}{\partial z^2} = 0.
\end{equation}
Then apply the method of separation of variables by assuming the solution is in the form
$$ \Phi \left ( r,\theta,z \right) = R(r)P(\phi)Z(z).$$
Plug this into (2) and note how we can bring out the functions that are not affected by the derivatives
$$ \frac{P(\phi) Z(z)}{r} \frac{\partial R(r)}{\partial r} + P(\phi) Z(z) \frac{\partial^2 R(r)}{\partial r^2} + \frac{R(r) Z(z)}{r^2} \frac{\partial^2 P(\phi)}{\partial \theta^2} + R(r) P(\phi) \frac{\partial^2 Z(z)}{\partial z^2} = 0.$$
Divide by $R(r) P(\phi) Z(z)$ and use short hand notation to get
$$\frac{1}{Rr} \frac{\partial R}{\partial r} + \frac{1}{R} \frac{\partial^2R}{\partial r^2} + \frac{1}{Pr^2} \frac{\partial^2P}{\partial \theta^2} + \frac{1}{Z} \frac{\partial^2 Z}{dz^2} = 0.$$
``Separating'' the z term to the other side gives
$$\frac{1}{Rr} \frac{d R}{d r} + \frac{1}{R} \frac{d^2R}{d r^2} + \frac{1}{Pr^2} \frac{d^2P}{d \theta^2} = - \frac{1}{Z} \frac{d^2 Z}{dz^2}.$$
This equation can only be satisfied for all values if both sides are equal to a constant, $\lambda$, such that
\begin{equation}
-\frac{1}{Z} \frac{d^2 Z}{dz^2} = \lambda
\end{equation}
\begin{equation}
\frac{1}{Rr} \frac{d R}{d r} + \frac{1}{R} \frac{d^2R}{d r^2} + \frac{1}{Pr^2} \frac{d^2P}{d\theta^2} = \lambda.
\end{equation}
Before we can focus on solutions, we need to further separate (4), so multiply (4) by $r^2$
$$\frac{r}{R} \frac{d R}{d r} + \frac{r^2}{R} \frac{d^2R}{d r^2} + \frac{1}{P} \frac{d^2P}{d\theta^2} = \lambda r^2.$$
Separate the terms
$$\frac{r}{R} \frac{d R}{d r} + \frac{r^2}{R} \frac{d^2R}{d r^2} - \lambda r^2 = - \frac{1}{P} \frac{d^2P}{d \theta^2}.$$
As before, set both sides to a constant, $\kappa$
\begin{equation}
- \frac{1}{P} \frac{d^2P}{d \theta^2}= \kappa
\end{equation}
\begin{equation}
\frac{r}{R} \frac{dR}{d r} + \frac{r^2}{R} \frac{d^2R}{d r^2} - \lambda r^2 = \kappa.
\end{equation}
Now there are three \htmladdnormallink{differential equations}{http://planetphysics.us/encyclopedia/DifferentialEquations.html} and we know the form of these solutions. The differential equations of
(3) and (5) are \htmladdnormallink{ordinary differential equations}{http://planetphysics.us/encyclopedia/DifferentialEquations.html}, while (6) is a little more complicated and we must turn to Bessel functions.
\subsection{Axial Solutions ($z$)}
Following the guidelines setup in [Etgen] for linear homogeneous differential equations, the first step in solving
$$\frac{d^2 Z}{dz^2} + Z \lambda = 0$$
is to find the roots of the characteristic polynomial
$$C(r) = r^2 + \lambda = 0$$
$$r = \pm \sqrt{ -\lambda}.$$
Although, one can go forward using the \htmladdnormallink{square}{http://planetphysics.us/encyclopedia/PiecewiseLinear.html} root, here we will introduce another constant, $\gamma$ to imply
the following cases. So if we want real roots, then we want to ensure a negative constant
$$ \lambda = -\gamma^2 $$
and if we want complex roots, then we want to ensure a positive constant
$$ \lambda = \gamma^2 ,$$
{\bf Case 1}: $\lambda \le 0$ and real roots $(\lambda = -\gamma^2)$.
For every real root, there will be an exponential in the general solution. The real roots are
$$ r_1 = \gamma $$
$$ r_2 = -\gamma $$
$$ r_3 = 0 .$$
Therefore, the solutions for these roots are
$$ h_1(z) = C_1e^{\gamma z} $$
$$ h_2(z) = C_2e^{-\gamma z} $$
$$ h_3(z) = C_3z e^0 = C_3z.$$
Combining these using the principle of superposition, gives the general solution,
\begin{equation}
Z_{\gamma}(z) = C_1e^{\gamma z} + C_2e^{-\gamma z} + C_3z + C_4.
\end{equation}
{\bf Case 2}: $\lambda > 0$ and complex roots $(\lambda = \gamma^2)$.
The roots are
$$ r_4 = i \gamma $$
$$ r_5 = -i \gamma $$
and the corresponding solutions
$$ h_4(z) = C_5e^0 \cos (\gamma z) = C_5\cos (\gamma z) $$
$$ h_5(z) = C_6e^0 \sin (\gamma z)= C_6\sin (\gamma z) .$$
Combining these into a general solution yields
$$Z_{\gamma}(z) = C_5\cos (\gamma z) + C_6\sin (\gamma z) + C_7.$$
\subsection{Azimuthal Solutions ($\theta$)}
Azimuthal solutions for
$$ \frac{d^2P}{d \theta^2} + \kappa P = 0$$
are in the most general sense obtained similarly to the axial solutions with the characteristic polynomial
$$C(r) = r^2 + \kappa = 0$$
$$r = \pm \sqrt{- \kappa}.$$
Using another constant, $\nu$ to ensure positive or negative constants, we get two cases.
{\bf Case 1}: $\kappa \le 0$ and real roots $(\kappa = -\nu^2)$.
The solutions for these roots are then
$$ h_1(z) = C_1e^{\nu \theta} $$
$$ h_2(z) = C_2e^{-\nu \theta} $$
$$ h_3(z) = C_3\theta e^0 = C_3\theta.$$
Combining these for the general solution,
\begin{equation}
P_{\nu}(\theta) = C_1e^{\nu \theta} + C_2e^{-\nu \theta} + C_3\theta + C_4.
\end{equation}
{\bf Case 2}: $\kappa > 0$ and complex roots $(\kappa = \nu^2)$.
The roots are
$$ r_4 = i \nu $$
$$ r_5 = -i \nu $$
and the corresponding solutions
$$ h_4(\theta) = C_5e^0 \cos (\nu \theta) = C_5\cos (\nu \theta) $$
$$ h_5(\theta) = C_6e^0 \sin (\nu \theta)= C_6\sin (\nu \theta) .$$
Combining these into a general solution
$$P_{\nu}(\theta) = C_5\cos (\nu \theta) + C_6\sin (\nu \theta) + C_7.$$
For the first glimpse at simplification, we will note a restriction on $\kappa$
that is used when it is required that the solution be periodic to ensure $P$ is single valued
$$P(\theta) = P(\theta + 2 n \pi).$$
Then we are left with either the periodic solutions that occur with complex roots or the zero case. So not only
$$(\kappa = \nu^2)$$
but also $\nu$ must be an integer, i.e.
\begin{equation}
P_{\nu}(\theta) = C_5 \cos (\nu \theta) + C_6 \sin (\nu \theta) + C_7 \,\,\,\,\,\,\,\,\,\,\, \nu = 0, 1, 2, ...
\end{equation}
Note, that $\nu = 0$, is still a solution, but to be periodic we can only have a constant
$$P(\theta) = C_4.$$
\subsection{Radial Solutions ($r$)}
The radial solutions are the more difficult ones to understand for this problem and are solved using a \htmladdnormallink{power}{http://planetphysics.us/encyclopedia/Power.html} series.
The two \htmladdnormallink{types}{http://planetphysics.us/encyclopedia/Bijective.html} of solutions generated based on the choices of constants from the $\theta$ and $z$ solutions (excluding non-periodic solutions for $P$)
leads to the Bessel functions and the modified Bessel functions. The first step for both these cases is to transform (6)
into the Bessel differential equation.
{\bf Case 1}: $\lambda < 0$ $(\lambda = -\gamma^2)$, $\kappa > 0$ $(\kappa = \nu^2)$.
Substitute $\gamma$ and $\nu$ into the \htmladdnormallink{radial equation}{http://planetphysics.us/encyclopedia/RadialEquation.html} (6) to get
\begin{equation}
\frac{r}{R} \frac{dR}{d r} + \frac{r^2}{R} \frac{d^2R}{d r^2} + \gamma^2 r^2 - \nu^2 = 0.
\end{equation}
Next, use the substitution
$$x = \gamma r$$
$$r = \frac{x}{\gamma}.$$
Therefore, the derivatives are
$$dx = \gamma dr$$
$$dr = \frac{dx}{\gamma}$$
and make a special note that
$$\frac{d^2}{dx^2} = \frac{d}{dx} \frac{d}{dx}$$
so
$$dx^2 = dx*dx = \gamma^2 dr^2$$
$$dr^2 = \frac{dx^2}{\gamma^2}.$$
Substituting these relationships into (10) gives us
$$\frac{x \gamma}{\gamma R} \frac{dR}{dx} + \frac{x^2 \gamma^2}{\gamma^2 R} \frac{d^2R}{dx^2} + x^2 - \nu^2 = 0.$$
Finally, multiply by $R/x^2$ to get the \emph{Bessel differential equation}
\begin{equation}
\frac{d^2R}{dx^2} + \frac{1}{x} \frac{dR}{dx} + \left(1 - \frac{\nu^2}{x^2} \right )R = 0.
\end{equation}
Delving into all the nuances of solving Bessel's differential equation is beyond the scope of this article, however, the curious
are directed to Watson's in depth treatise [Watson]. Here, we will just present the results as we did for the previous
differential equations. The general solution is a linear combination of the Bessel function of the first kind $J_{\nu}(r)$
and the Bessel function of the second kind $Y_{\nu}(r)$. Remebering that $\nu$ is a positive integer or zero.
\begin{equation}
R_{\nu}(r) = C_1 J_{\nu}(\gamma r) + C_2 Y_{\nu}(\gamma r) + C_3
\end{equation}
Bessel function of the first kind:
$$J_{\nu}(x) = \sum_{m=0}^{\infty} \frac{ (-1)^m ( \frac{-1}{2}x)^{\nu + 2m} }{ m! (m + \nu)! }.$$
Bessel function of the second kind (using Hankel's \htmladdnormallink{formula}{http://planetphysics.us/encyclopedia/Formula.html}):
$$Y_{\nu}(x) = 2J_{\nu}(x) \left (\eta + ln \left( \frac{x}{2}\right ) \right ) - \left( \frac{x}{2} \right)^{-\nu} \sum_{m=0}^{\nu-1} \frac{(\nu - m - 1)!}{m!} \left ( \frac{x}{2} \right)^{2m}$$
$$ \,\,\,\,\, -\sum_{m=0}^{\infty} \frac{ (-1)^m (\frac{x}{2})^{\nu + 2m} }{m! (\nu + m)!} \left \{ \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{m}+ \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{\nu + m} \right \}.$$
For the unfortunate person who has to evaluate this function, note that when $m = 0$, the singularity is taken care of by replacing the series in brackets by
$$ \left \{ \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{\nu} \right \}.$$
Some solace can be found since most physical problems need to be analytic at $x = 0$ and therefore $Y_{\nu}(x)$ breaks down at $ln(0)$.
This leads to the choice of constant $C_2$ to be zero.
{\bf Case 2}: $\lambda > 0$ $(\lambda = \gamma^2)$, $\kappa > 0$ $(\kappa = \nu^2)$.
Using the previous method of substitution, we just get the change of sign
\begin{equation}
\frac{d^2R}{dx^2} + \frac{1}{x} \frac{dR}{dx} - \left(1 + \frac{\nu^2}{x^2} \right)R = 0.
\end{equation}
This leads to the modified Bessel functions as a solution, which are also known as the pure imaginary Bessel functions. The
general solution is denoted
\begin{equation}
R_{\nu}(r) = C_1 I_{\nu}(\gamma r) + C_2 K_{\nu}(\gamma r) + C_3
\end{equation}
where $I_{\nu}$ is the modified Bessel function of the first kind and $K_{\nu}$ is the modified Bessel function of the
second kind
$$I_{\nu}(\gamma r) = i^{-\nu} J_{\nu}(i \gamma r)$$
$$K_{\nu}(\gamma r) = \frac{\pi}{2} i^{\nu+1} \left ( J_{\nu}(i \gamma r) + iY_{\nu}(i \gamma r) \right ) .$$
\subsection{Combined Solution}
Keeping track of all the different cases and choosing the right terms for \htmladdnormallink{boundary}{http://planetphysics.us/encyclopedia/PiecewiseLinear.html} conditions is a daunting task when one attempts
to solve Laplace's equation. The short hand notation used in [Kusse] and [Arfken] will be presented here
to help organize the choices as a reference. It is important to remember that these solutions are only for
the single valued azimuth cases $(\kappa = \nu^2)$.
Once the separate solutions are obtained, the rest is simple since our solution is separable
$$ \Phi \left ( r,\theta,z \right) = R(r)P(\theta)Z(z).$$
so we just combine the individual solutions to get the general solutions to the Laplace equation in cylindrical coordinates.
{\bf Case 1}: $\lambda < 0$ $(\lambda = -\gamma^2)$, $\kappa > 0$ $(\kappa = \nu^2)$.
\begin{equation}
\Phi \left ( r,\theta,z \right) = \sum_{\nu} \sum_{\gamma} \left \{ \begin{array}{c}
e^{\gamma z} \\
e^{-\gamma z}
\end{array} \right .
\left \{ \begin{array}{c}
\cos (\nu \theta) \\
\sin (\nu \theta)
\end{array} \right .
\left \{ \begin{array}{c}
J_{\nu}(\gamma r) \\
Y_{\nu}(\gamma r)
\end{array} \right .
\end{equation}
{\bf Case 2}: $\lambda > 0$ $(\lambda = \gamma^2)$, $\kappa > 0$ $(\kappa = \nu^2)$.
\begin{equation}
\Phi \left ( r,\theta,z \right) = \sum_{\nu} \sum_{\gamma} \left \{ \begin{array}{c}
\cos (\gamma z) \\
\sin (\gamma z)
\end{array} \right .
\left \{ \begin{array}{c}
\cos (\nu \theta) \\
\sin (\nu \theta)
\end{array} \right .
\left \{ \begin{array}{c}
I_{\nu}(\gamma r) \\
K_{\nu}(\gamma r)
\end{array} \right .
\end{equation}
Interpreting the short hand notation is as simple as expanding terms and not forgetting the linear solutions, i.e. $(\gamma = 0)$ .
As an example, case 1, expanded out while ignoring the linear terms would give
\begin{equation}
\Phi \left ( r,\theta,z \right) = \sum_{\nu} \sum_{\gamma} \left \{ \begin{array}{c}
\, A_{\nu \gamma} e^{\gamma z} \cos (\nu \theta) J_{\nu}(\gamma r) \\
+ B_{\nu \gamma} e^{\gamma z} \cos (\nu \theta) Y_{\nu}(\gamma r) \\
+ C_{\nu \gamma} e^{\gamma z} \sin (\nu \theta) J_{\nu}(\gamma r) \\
+ D_{\nu \gamma} e^{\gamma z} \sin (\nu \theta) Y_{\nu}(\gamma r) \\
+ E_{\nu \gamma} e^{-\gamma z} \cos (\nu \theta) J_{\nu}(\gamma r) \\
+ F_{\nu \gamma} e^{-\gamma z} \cos (\nu \theta) Y_{\nu}(\gamma r) \\
+ G_{\nu \gamma} e^{-\gamma z} \sin (\nu \theta) J_{\nu}(\gamma r) \\
+ H_{\nu \gamma} e^{-\gamma z} \sin (\nu \theta) Y_{\nu}(\gamma r) \\
\end{array} \right \} .
\end{equation}
\begin{thebibliography}{9}
\bibitem{Arfken} Arfken, George, Weber, Hans, {\em Mathematical Physics}. Academic Press, San Diego, 2001.
\bibitem{Etgen} Etgen, G., {\em Calculus}. John Wiley \& Sons, New York, 1999.
\bibitem{Guterman} Guterman, M., Nitecki, Z., {\em Differential Equations, 3rd Edition}. Saunders College Publishing, Fort Worth, 1992.
\bibitem{Jackson} Jackson, J.D., {\em Classical Electrodynamics, 2nd Edition}. John Wiley \& Sons, New York, 1975.
\bibitem{Kusse} Kusse, Bruce, Westwig, Erik, {\em Mathematical Physics}. John Wiley \& Sons, New York, 1998.
\bibitem{Lebedev} Lebedev, N., {\em Special Functions \& Their Applications}. Dover Publications, New York, 1995.
\bibitem{Watson} Watson, G.N., {\em A Treatise on the Theory of Bessel Functions}. Cambridge University Press, New York, 1995.
\end{thebibliography}
\end{document}
