Subsection 7.3.6 Three Dimensional Laplace’s Equation
The Laplace equation arises in various physical problems, including electrostatics, heat conduction, fluid flow, and potential theory. Solving the 3D Laplace equation involves finding a function \(u(x)\) that satisfies the equation \(\nabla^{2}u =0\text{.}\) This typically involves specifying appropriate boundary conditions or constraints on the function at the boundaries of the domain. The solutions to the Laplace equation are harmonic functions, which have many interesting mathematical properties.
Subsubsection 7.3.6.1 In Rectangular Coordinates
The equation is
\begin{equation}
\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}+\frac{\partial^{2}u}{\partial z^{2}} =0\tag{7.3.90}
\end{equation}
Let
\begin{equation}
u=X(x)Y(y)Z(z) \tag{7.3.91}
\end{equation}
be the solution of equation (7.3.90) Then
\begin{equation}
\frac{1}{X}\frac{\partial^{2}X}{\partial x^{2}}+\frac{1}{Y}\frac{\partial^{2}Y}{\partial y^{2}}+\frac{1}{Z}\frac{\partial^{2}Z}{\partial z^{2}} =0\tag{7.3.92}
\end{equation}
In equation (7.3.92) all are independent to each other hence each may be seperately equal to some constant, i.e.,
\begin{equation*}
\frac{1}{X}\frac{\partial^{2}X}{\partial x^{2}} =n^{2} \quad \Rightarrow X=A\cos nx+B\sin nx,
\end{equation*}
or,
\begin{equation*}
\frac{1}{Y}\frac{\partial^{2}Y}{\partial y^{2}} =-m^{2} \quad \Rightarrow Y=C\cos my+D\sin my,
\end{equation*}
and
\begin{equation*}
\frac{1}{Z}\frac{\partial^{2}Z}{\partial z^{2}}=p^{2} \quad \Rightarrow Z=Ee^{pz}+Fe^{-pz}
\end{equation*}
Thus, the general solution of equation (7.3.90) is
\begin{equation*}
u=(A\cos nx+B\sin nx)
\end{equation*}
\begin{equation}
(C\cos my+D\sin my)(Ee^{pz}+Fe^{-pz})\tag{7.3.93}
\end{equation}
where \(p^{2}=m^{2}+n^{2}\text{.}\)
Alternative solution of equation (7.3.90) is
\begin{equation*}
u=(Ae^{nx}+Be^{-nx})
\end{equation*}
\begin{equation}
(Ce^{my}+De^{-my})(E\cos{pz}+F\sin{pz})\tag{7.3.94}
\end{equation}
Subsubsection 7.3.6.2 In Cylindrical Coordinates
We have,
\begin{equation}
\frac{\partial^{2}u}{\partial r^{2}}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}u}{\partial \theta^{2}}+\frac{\partial^{2}u}{\partial z^{2}} =0\tag{7.3.95}
\end{equation}
Let
\begin{equation}
u(r,\theta,z) =R(r)\Theta(\theta)Z(z) \tag{7.3.96}
\end{equation}
be the solution of eqn.(7.3.95). Hence,
\begin{equation}
\frac{1}{R}\frac{\partial^{2}R}{\partial r^{2}}+\frac{1}{rR}\frac{\partial R}{\partial r}+\frac{1}{r^{2}\Theta}\frac{\partial^{2}\Theta}{\partial \theta^{2}}+\frac{1}{Z}\frac{\partial^{2}Z}{\partial z^{2}} =0\tag{7.3.97}
\end{equation}
As variables are separated, we may take -
\begin{equation*}
\frac{1}{Z}\frac{\partial^{2}Z}{\partial z^{2}}=\lambda^{2}
\end{equation*}
or,
\begin{equation}
\frac{\partial^{2}Z}{\partial z^{2}}-\lambda^{2}Z =0 \quad \Rightarrow Z=Ae^{\lambda z}+Be^{-\lambda z}\tag{7.3.98}
\end{equation}
and
\begin{equation}
\frac{r^{2}}{R}\frac{\partial^{2}R}{\partial r^{2}}+\frac{r}{R}\frac{\partial R}{\partial r}+\lambda^{2}r^{2}=-\frac{1}{\Theta}\frac{\partial^{2}\Theta}{\partial \theta^{2}}=\mu^{2}\tag{7.3.99}
\end{equation}
also,
\begin{equation}
\frac{1}{\Theta}\frac{\partial^{2}\Theta}{\partial \theta^{2}}+\mu^{2}\Theta =0 \quad \Rightarrow \Theta = C\cos \mu \theta +D\sin \mu\theta\tag{7.3.100}
\end{equation}
and
\begin{equation}
r^{2}\frac{\partial^{2}R}{\partial r^{2}}+r\frac{\partial R}{\partial r}+(\lambda^{2}r^{2}-\mu^{2})R=0\tag{7.3.101}
\end{equation}
Now put \(\lambda r=s \) so that
\begin{equation*}
\frac{\partial }{\partial r} = \frac{\partial }{\partial s}\cdot \frac{\partial s}{\partial r}=+\lambda \frac{\partial }{\partial s}
\end{equation*}
and
\begin{equation*}
\frac{\partial^{2} }{\partial r^{2}} =\lambda^{2} \frac{\partial^{2} }{\partial s^{2}}
\end{equation*}
therefore equation (7.3.101) reduces to
\begin{equation*}
\frac{s^{2}}{\lambda^{2}}\lambda^{2}\frac{\partial^{2} R}{\partial s^{2}}+\frac{s}{\lambda}\lambda \frac{\partial R}{\partial s}+(s^{2}-\mu^{2})R =0
\end{equation*}
or,
\begin{equation*}
s^{2}\frac{\partial^{2} R}{\partial s^{2}}+s\frac{\partial R}{\partial s}+(s^{2}-\mu^{2})R =0
\end{equation*}
which is Bessel’s equation and it’s general solution is given as
\begin{equation*}
R=EJ_{\mu}(s)+FJ_{-\mu}(s)
\end{equation*}
or,
\begin{equation}
R=EJ_{\mu}(\lambda r)+FJ_{-\mu}(\lambda r)\tag{7.3.102}
\end{equation}
since \(u(r,\theta,z)\) be finite at \(r \to 0 \text{,}\) we have
\begin{equation*}
s=\lambda r =0 \Rightarrow J_{-\mu} \to \infty
\end{equation*}
or,
\begin{equation}
\therefore R= EJ_{\mu}(\lambda r)\tag{7.3.103}
\end{equation}
Hence,
\begin{equation*}
u(r,\theta,z) = (Ae^{\lambda z}+Be^{-\lambda z})(C\cos \mu \theta +D\sin \mu\theta)[EJ_{\mu}(\lambda r)]
\end{equation*}
Subsubsection 7.3.6.3 In Spherical Coordinates
\begin{equation*}
\frac{\partial^{2}u}{\partial r^{2}}
\end{equation*}
\begin{equation}
+\frac{2}{r}\frac{\partial u}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}u}{\partial \theta^{2}}+\frac{\cot\theta}{r^{2}}\frac{\partial u}{\partial \theta}+\frac{1}{r^{2}\sin ^{2}\theta}\frac{\partial^{2}u}{\partial \phi^{2}} =0\tag{7.3.104}
\end{equation}
or,
\begin{equation*}
r^{2}\frac{\partial^{2}u}{\partial r^{2}}
\end{equation*}
\begin{equation}
+2r\frac{\partial u}{\partial r}+\frac{1}{\sin\theta}\frac{\partial }{\partial \theta}\left(\sin\theta\frac{\partial u}{\partial \theta}\right)+\frac{1}{\sin ^{2}\theta}\frac{\partial^{2}u}{\partial \phi^{2}} =0\tag{7.3.105}
\end{equation}
Suppose,
\begin{equation}
u(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi)\tag{7.3.106}
\end{equation}
be its solution. Then,
\begin{equation*}
\left[\frac{r^{2}}{R}\frac{\partial^{2}R}{\partial r^{2}}+\frac{2r}{R}\frac{\partial R}{\partial r}+\frac{1}{\Theta\sin\theta}\frac{\partial }{\partial \theta}\left(\sin\theta\frac{\partial \Theta}{\partial \theta}\right)\right]\sin ^{2}\theta
\end{equation*}
\begin{equation}
= -\frac{1}{\Phi}\frac{\partial^{2}\Phi}{\partial \phi^{2}} = \lambda^{2} \quad \text{(say)}\tag{7.3.107}
\end{equation}
considering eqn.(7.3.107),
\begin{equation}
\frac{1}{\Phi}\frac{\partial^{2}\Phi}{\partial \phi^{2}} = -\lambda^{2} \quad \Rightarrow \Phi=Ce^{\pm i\lambda \phi}\tag{7.3.108}
\end{equation}
and
\begin{equation*}
\frac{1}{R}\frac{\partial }{\partial r}\left(r^{2}\frac{\partial R}{\partial r}\right)
\end{equation*}
\begin{equation}
-\frac{1}{\Theta\sin\theta}\frac{\partial }{\partial \theta}\left(\sin\theta\frac{\partial \Theta}{\partial \theta}\right)+\frac{\lambda^{2}}{\sin^{2}\theta} =n(n+1) \quad \text{(say)}\tag{7.3.109}
\end{equation}
or,
\begin{equation}
\frac{\partial }{\partial r}\left(r^{2}\frac{\partial R}{\partial r}\right)-n(n+1)R =0\tag{7.3.110}
\end{equation}
and
\begin{equation}
\frac{1}{\sin\theta}\frac{\partial }{\partial \theta}\left(\sin\theta\frac{\partial \Theta}{\partial \theta}\right)+\left[n(n+1)-\frac{\lambda^{2}}{\sin^{2}\theta}\right]\Theta =0\tag{7.3.111}
\end{equation}
As equation (7.3.110) being the homogenous, put \(r=e^{z} \text{,}\) hence equation (7.3.110) reduces to
\begin{equation*}
[D(D-1)+2D-n(n+1)]R=0
\end{equation*}
where, \(D\equiv \frac{\partial }{\partial z}\text{.}\) or,
\begin{equation*}
[D^{2}+D-n(n+1)]R =0
\end{equation*}
or,
\begin{equation*}
(D-n)(D+n+1)R =0
\end{equation*}
or,
\begin{equation*}
D=n,-(n+1),
\end{equation*}
\begin{equation}
\therefore R=Ae^{nz}+Be^{-(n+1)z} = Ar^{n}+Br^{-n-1}\tag{7.3.112}
\end{equation}
again, if we put \(\cos\theta =\mu \) in equation (7.3.111), then since
\begin{equation*}
\frac{\partial \Theta}{\partial\theta} = \frac{\partial \Theta}{\partial\mu} \cdot \frac{\partial \mu}{\partial\theta} =-\sin\theta \frac{\partial \Theta}{\partial\mu}
\end{equation*}
or,
\begin{equation*}
\frac{1}{\sin\theta} \frac{\partial }{\partial\theta}=-\frac{\partial }{\partial\mu}
\end{equation*}
we have,
\begin{equation*}
\frac{\partial }{\partial\mu} \left\{(1-\mu^{2})\frac{\partial \Theta}{\partial\mu} \right\}+\left\{n(n+1)-\frac{\lambda^{2}}{1-\mu^{2}}\right\}\Theta =0
\end{equation*}
Now,
\begin{equation}
(1-\mu^{2})\frac{\partial^{2} \Theta}{\partial\mu^{2}}-2\mu \frac{\partial \Theta}{\partial\mu}+\left\{n(n+1)--\frac{\lambda^{2}}{1-\mu^{2}}\right\}\Theta =0\tag{7.3.113}
\end{equation}
which is Legendre’s associated differential equational and hence has the solution,
\begin{equation}
\Theta =AP^{\lambda}_{n}(\mu)+BQ^{\lambda}_{n}(\mu)=AP^{\lambda}_{n}(\cos\theta)+BQ^{\lambda}_{n}(\cos\theta)\tag{7.3.114}
\end{equation}
Now as \(\Theta\) is the function of \(\cos\theta \text{,}\) then solution of equation (7.3.104) is in the form
\begin{equation*}
u=(Ar^{n}+Br^{-n-1})\Theta(\cos\theta)e^{\pm i\lambda\phi}
\end{equation*}
The summing over all \(n\text{,}\) the general solution of equation (7.3.104) may be written as
\begin{equation}
u(r,\theta,\phi) = \sum\limits_{n=0}^{\infty}(Ar^{n}+Br^{-n-1})\Theta(\cos\theta)e^{\pm i\lambda\phi}\tag{7.3.115}
\end{equation}
Note 1. If \(\lambda=0\text{,}\) then equation (7.3.113) reduces to Legendre’s equation,
\begin{equation*}
\Theta = P_{n}(\mu)=P_{n}(\cos\theta)
\end{equation*}
also,
\begin{equation*}
\Theta = Q_{n}(\mu)=Q_{n}(\cos\theta).
\end{equation*}
Thus,
\begin{equation*}
\Theta = C_{n}P_{n}(\mu)+D_{n}Q_{n}(\mu)
\end{equation*}
so that
\begin{equation*}
u(r,\theta,\phi) = \sum\limits_{n=0}^{\infty}\left(A_{n}r^{n}+\frac{B_{n}}{r^{n+1}}\right)\{C_{n}P_{n}(\cos\theta)
\end{equation*}
\begin{equation}
+D_{n}Q_{n}(\cos\theta)\} e^{\pm i\lambda\phi}\tag{7.3.116}
\end{equation}
In case \(D_{n}=0\) under specific boundary conditions, then \(\Theta = C_{n}P_{n}(\cos\theta)\)
\begin{equation}
\therefore u= \sum\limits_{n=0}^{\infty}\left(A_{n}r^{n} +\frac{B_{n}}{r^{n+1}}\right)C_{n}P_{n}(\cos\theta) e^{\pm i\lambda\phi}\tag{7.3.117}
\end{equation}
Note 2. If there is axial symmetry about z - axis, then \(u\) depens only upon \(r\) and \(\theta\) so that \(\phi =0 \) in equation (7.3.117),
\begin{equation}
u(r,\theta) = \sum\limits_{n=0}^{\infty}\left(A_{n}r^{n}+\frac{B_{n}}{r^{n+1}}\right)\{C_{n}P_{n}(\cos\theta)+D_{n}Q_{n}(\cos\theta)\}\tag{7.3.118}
\end{equation}
If \(D_{n}=0\text{,}\)
\begin{equation}
u(r,\theta) = \sum\limits_{n=0}^{\infty}\left(A_{n}r^{n}+\frac{B_{n}}{r^{n+1}}\right)C_{n}P_{n}(\cos\theta)\tag{7.3.119}
\end{equation}
