PlanetPhysics/Laplace Equation in Cylindrical Coordinates

Laplace Equation in Cylindrical Coordinates

Solutions to the Laplace equation in cylindrical coordinates have wide applicability from fluid [[../Mechanics/|mechanics]] to electrostatics. Applying the method of [[../SeparationOfVariables/|separation of variables]] to Laplace's [[../DifferentialEquations/|partial differential equation]] and then enumerating the various forms of solutions will lay down a foundation for solving problems in this coordinate [[../SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence/|system]]. Finally, the use of [[../BesselEquation2/|Bessel functions]] in the solution reminds us why they are synonymous with the cylindrical [[../Bijective/|domain]].

Separation of Variables

Beginning with the [[../LaplacianInCylindricalCoordinates/|Laplacian in Cylindrical Coordinates]], apply the [[../QuantumSpinNetworkFunctor2/|operator]] to a potential [[../Bijective/|function]] and set it equal to zero to get the [[../FluorescenceCrossCorrelationSpectroscopy/|Laplace equation]]

${\mathrm{\nabla }}^2\mathrm{\Phi }=\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial \mathrm{\Phi }}{\partial r}\right)+\frac{1}{r^2}\frac{{\partial }^2\mathrm{\Phi }}{\partial {\theta }^2}+\frac{{\partial }^2\mathrm{\Phi }}{\partial z^2}=0.$

First expand out the terms

${\mathrm{\nabla }}^2\mathrm{\Phi }=\frac{1}{r}\frac{\partial \mathrm{\Phi }}{\partial r}+\frac{{\partial }^2\mathrm{\Phi }}{\partial r^2}+\frac{1}{r^2}\frac{{\partial }^2\mathrm{\Phi }}{\partial {\theta }^2}+\frac{{\partial }^2\mathrm{\Phi }}{\partial z^2}=0.$

Then apply the method of separation of variables by assuming the solution is in the form

$\mathrm{\Phi }(r,\theta ,z)=R(r)P(\theta )Z(z).$

Plug this into (2) and note how we can bring out the functions that are not affected by the derivatives

$\frac{P(\theta )Z(z)}{r}\frac{\partial R(r)}{\partial r}+P(\theta )Z(z)\frac{{\partial }^{2}R(r)}{\partial r^2}+\frac{R(r)Z(z)}{r^2}\frac{{\partial }^{2}P(\theta )}{\partial {\theta }^2}+R(r)P(\theta )\frac{{\partial }^{2}Z(z)}{\partial z^2}=0.$

Divide by $R(r)P(\theta )Z(z)$ and use short hand notation to get

$\frac{1}{Rr}\frac{\partial R}{\partial r}+\frac{1}{R}\frac{{\partial }^{2}R}{\partial r^2}+\frac{1}{P{r}^2}\frac{{\partial }^{2}P}{\partial {\theta }^2}+\frac{1}{Z}\frac{{\partial }^{2}Z}{d{z}^2}=0.$

"Separating" the z term to the other side gives

$\frac{1}{Rr}\frac{dR}{dr}+\frac{1}{R}\frac{d^{2}R}{d{r}^2}+\frac{1}{P{r}^2}\frac{d^{2}P}{d{\theta }^2}=-\frac{1}{Z}\frac{d^{2}Z}{d{z}^2}.$

This equation can only be satisfied for all values if both sides are equal to a constant, $\lambda$, such that

$-\frac{1}{Z}\frac{d^{2}Z}{d{z}^2}=\lambda$

$\frac{1}{Rr}\frac{dR}{dr}+\frac{1}{R}\frac{d^{2}R}{d{r}^2}+\frac{1}{P{r}^2}\frac{d^{2}P}{d{\theta }^2}=\lambda .$

Before we can focus on solutions, we need to further separate (4), so multiply (4) by $r^2$

$\frac{r}{R}\frac{dR}{dr}+\frac{r^2}{R}\frac{d^{2}R}{d{r}^2}+\frac{1}{P}\frac{d^{2}P}{d{\theta }^2}=\lambda r^2.$

Separate the terms

$\frac{r}{R}\frac{dR}{dr}+\frac{r^2}{R}\frac{d^{2}R}{d{r}^2}-\lambda r^2=-\frac{1}{P}\frac{d^{2}P}{d{\theta }^2}.$

As before, set both sides to a constant, $\kappa$

$-\frac{1}{P}\frac{d^{2}P}{d{\theta }^2}=\kappa$

$\frac{r}{R}\frac{dR}{dr}+\frac{r^2}{R}\frac{d^{2}R}{d{r}^2}-\lambda r^2=\kappa .$

Now there are three [[../DifferentialEquations/|differential equations]] and we know the form of these solutions. The differential equations of (3) and (5) are [[../DifferentialEquations/|ordinary differential equations]], while (6) is a little more complicated and we must turn to Bessel functions.

Axial Solutions (z)

Following the guidelines setup in [Etgen] for linear homogeneous differential equations, the first step in solving

$\frac{d^{2}Z}{d{z}^2}+Z\lambda =0$

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 [[../PiecewiseLinear/|square]] 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

$\lambda ={\gamma }^2,$

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

$r_1=\gamma$, $r_2=-\gamma$.

Therefore, the solutions for these roots are

$h_1(z)=C_1{e}^{\gamma z}$, $h_2(z)=C_2{e}^{-\gamma z}$.

Combining these using the principle of superposition, gives the general solution,

$Z_{\gamma }(z)=C_1{e}^{\gamma z}+C_2{e}^{-\gamma z}.$

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 are

$h_4(z)=C_5{e}^0\cos (\gamma z)=C_5\cos (\gamma z)$, and $h_5(z)=C_6{e}^0\sin (\gamma z)=C_6\sin (\gamma z).$

Combining these into a general solution yields

$Z_{\gamma }(z)=C_5\cos (\gamma z)+C_6\sin (\gamma z).$

Azimuthal Solutions (θ)

Azimuthal solutions for

$\frac{d^{2}P}{d{\theta }^2}+\kappa P=0$

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.

Case 1: $\kappa \le 0$ and real roots $(\kappa =-{\nu }^2)$.

The solutions for these roots are then

$h_1(z)=C_1{e}^{\nu \theta }$ $h_2(z)=C_2{e}^{-\nu \theta }$ $h_3(z)=C_3\theta e^0=C_3\theta .$

Combining these for the general solution,

$P_{\nu }(\theta )=C_1{e}^{\nu \theta }+C_2{e}^{-\nu \theta }+C_3\theta +C_4.$

Case 2: $\kappa >0$ and complex roots $(\kappa ={\nu }^2)$.

The roots are

$r_4=i\nu$, and $r_5=-i\nu$

and the corresponding solutions

$h_4(\theta )=C_5{e}^0\cos (\nu \theta )=C_5\cos (\nu \theta )$, and $h_5(\theta )=C_6{e}^0\sin (\nu \theta )=C_6\sin (\nu \theta ).$

Combining these into a general solution

$P_{\nu }(\theta )=C_5\cos (\nu \theta )+C_6\sin (\nu \theta ).$

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

$P(\theta )=P(\theta +2n\pi ).$

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.

$P_{\nu }(\theta )=C_5\cos (\nu \theta )+C_6\sin (\nu \theta )\nu =0,1,2,...$

Note, that $\nu =0$, is still a solution, but to be periodic we can only have a constant

$P(\theta )=C_4.$

Radial Solutions (r)

The radial solutions are the more difficult ones to understand for this problem and are solved using a [[../Power/|power]] series. The two [[../Bijective/|types]] 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.

Case 1: $\lambda <0$ $(\lambda =-{\gamma }^2)$, $\kappa >0$ $(\kappa ={\nu }^2)$.

Substitute $\gamma$ and $\nu$ into the [[../RadialEquation/|radial equation]] (6) to get

$\frac{r}{R}\frac{dR}{dr}+\frac{r^2}{R}\frac{d^{2}R}{d{r}^2}+{\gamma }^2{r}^2-{\nu }^2=0.$

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

$\frac{d^2}{d{x}^2}=\frac{d}{dx}\frac{d}{dx}$

so

$d{x}^2=dx*dx={\gamma }^{2}d{r}^2$ $d{r}^2=\frac{d{x}^2}{{\gamma }^2}.$

Substituting these relationships into (10) gives us

$\frac{x\gamma }{\gamma R}\frac{dR}{dx}+\frac{x^2{\gamma }^2}{{\gamma }^{2}R}\frac{d^{2}R}{d{x}^2}+x^2-{\nu }^2=0.$

Finally, multiply by $R/x^2$ to get the Bessel differential equation

$\frac{d^{2}R}{d{x}^2}+\frac{1}{x}\frac{dR}{dx}+(1-\frac{{\nu }^2}{x^2})R=0.$

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.

$R_{\nu }(r)=C_1{J}_{\nu }(\gamma r)+C_2{Y}_{\nu }(\gamma r)+C_3$

Bessel function of the first kind:

$J_{\nu }(x)=\sum _{m=0}^{\mathrm{\infty }}\frac{(-1{)}^m(\frac{-1}{2}x{)}^{\nu +2m}}{m!(m+\nu )!}.$

Bessel function of the second kind (using Hankel's [[../Formula/|formula]]):

$Y_{\nu }(x)=2J_{\nu }(x)(\eta +ln\left(\frac{x}{2}\right))-{\left(\frac{x}{2}\right)}^{-\nu }\sum _{m=0}^{\nu -1}\frac{(\nu -m-1)!}{m!}{\left(\frac{x}{2}\right)}^{2m}$ $-\sum _{m=0}^{\mathrm{\infty }}\frac{(-1{)}^m(\frac{x}{2}{)}^{\nu +2m}}{m!(\nu +m)!}\{\frac{1}{1}+\frac{1}{2}+...+\frac{1}{m}+\frac{1}{1}+\frac{1}{2}+...+\frac{1}{\nu +m}\}.$

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

$\{\frac{1}{1}+\frac{1}{2}+...+\frac{1}{\nu }\}.$

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.

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

$\frac{d^{2}R}{d{x}^2}+\frac{1}{x}\frac{dR}{dx}-(1+\frac{{\nu }^2}{x^2})R=0.$

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

$R_{\nu }(r)=C_1{I}_{\nu }(\gamma r)+C_2{K}_{\nu }(\gamma r)+C_3$

where $I_{\nu }$ is the modified Bessel function of the first kind and $K_{\nu }$ is the modified Bessel function of the second kind

$I_{\nu }(\gamma r)=i^{-\nu }{J}_{\nu }(i\gamma r)$ $K_{\nu }(\gamma r)=\frac{\pi }{2}{i}^{\nu +1}(J_{\nu }(i\gamma r)+i{Y}_{\nu }(i\gamma r)).$

Combined Solution

Keeping track of all the different cases and choosing the right terms for [[../PiecewiseLinear/|boundary]] 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

$\mathrm{\Phi }(r,\theta ,z)=R(r)P(\theta )Z(z).$

so we just combine the individual solutions to get the general solutions to the Laplace equation in cylindrical coordinates.

Case 1: $\lambda <0$ $(\lambda =-{\gamma }^2)$, $\kappa >0$ $(\kappa ={\nu }^2)$.

$\mathrm{\Phi }(r,\theta ,z)=\sum _{\nu }\sum _{\gamma }\{\begin{array}{c}{e}^{\gamma z}\\ e^{-\gamma z}\end{array}\{\begin{array}{c}\cos (\nu \theta )\\ \sin (\nu \theta )\end{array}\{\begin{array}{c}{J}_{\nu }(\gamma r)\\ Y_{\nu }(\gamma r)\end{array}$

Case 2: $\lambda >0$ $(\lambda ={\gamma }^2)$, $\kappa >0$ $(\kappa ={\nu }^2)$.

$\mathrm{\Phi }(r,\theta ,z)=\sum _{\nu }\sum _{\gamma }\{\begin{array}{c}\cos (\gamma z)\\ \sin (\gamma z)\end{array}\{\begin{array}{c}\cos (\nu \theta )\\ \sin (\nu \theta )\end{array}\{\begin{array}{c}{I}_{\nu }(\gamma r)\\ K_{\nu }(\gamma r)\end{array}$

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

$\mathrm{\Phi }(r,\theta ,z)=\sum _{\nu }\sum _{\gamma }\left\{\begin{array}{c}{A}_{\nu \gamma }{e}^{\gamma z}\cos (\nu \theta )J_{\nu }(\gamma r)\\ +B_{\nu \gamma }{e}^{\gamma z}\cos (\nu \theta )Y_{\nu }(\gamma r)\\ +C_{\nu \gamma }{e}^{\gamma z}\sin (\nu \theta )J_{\nu }(\gamma r)\\ +D_{\nu \gamma }{e}^{\gamma z}\sin (\nu \theta )Y_{\nu }(\gamma r)\\ +E_{\nu \gamma }{e}^{-\gamma z}\cos (\nu \theta )J_{\nu }(\gamma r)\\ +F_{\nu \gamma }{e}^{-\gamma z}\cos (\nu \theta )Y_{\nu }(\gamma r)\\ +G_{\nu \gamma }{e}^{-\gamma z}\sin (\nu \theta )J_{\nu }(\gamma r)\\ +H_{\nu \gamma }{e}^{-\gamma z}\sin (\nu \theta )Y_{\nu }(\gamma r)\end{array}\right\}.$