Moving from Two-dimensional to Three-dimensional Cylindrical
Now, we will go one step further and consider the Laplace equation in a three-dimensional space, since we clearly live in a three-dimensional world, and many applications will involve that. A natural coordinate system may be the three-dimensional Cartesian one, so that the Laplace equation reads
\begin{align}
\nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} = 0 \tag{1}
\end{align}
However, this is a straightforward extension of the two-dimensional version and not much is new, so we will not delve into this. Another frequently used three-dimensional coordinate system is the cylindrical coordinates \((r,\theta,z)\), e.g. in a setup of a hollow pipe. The appropriate forms of the Laplacian operator and hence the Laplace equation in this case are
\begin{align}
\nabla^2 u = \frac{1}{r} \frac{\partial}{\partial r} (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{2}
\end{align}
This is roughly the same as the two-dimensional polar form with the addition of the \(z\) coordinate.
Solving Laplace Equation in Cylindrical Coordinates
Even though we are now in a three-dimensional space, we may like to try generalizing the previous method of Separation of Variables for this first. If it is the case, then we should let
\begin{align}
u(r,\theta,z) = R(r)\Theta(\theta)Z(z) \tag{3}
\end{align}
just like before, with \( R(r), \Theta(\theta), Z(z) \) being a function of \(r, \theta, z\) only, respectively. Substituting this into (2) then gives
\begin{align}
\frac{1}{r} \frac{\partial}{\partial r} (rR’\Theta Z) + \frac{1}{r^2}R\Theta^{\prime\prime}Z + R\Theta Z^{\prime\prime} &= 0 \\
\frac{1}{rR} \frac{\partial}{\partial r} (rR’) + \frac{1}{r^2}\frac{\Theta^{\prime\prime}}{\Theta} + \frac{Z^{\prime\prime}}{Z} &= 0 \tag{4}
\end{align}
where we have divided by \(R\Theta Z\). The first two terms depend on \(r\) and \(\theta\) but the last term depends on \(z\) only, so we can let
\begin{align}
\frac{1}{rR} \frac{\partial}{\partial r} (rR’) + \frac{1}{r^2}\frac{\Theta^{\prime\prime}}{\Theta} &= -k^2 \tag{5} \\
\frac{Z^{\prime\prime}}{Z} &= k^2 \tag{6}
\end{align}
Writing the separation constant in the form \( k^2 \) is for convenience that we will see soon, and we will do the same from now on. (6) is a simple second-order constant-coefficient homogeneous ODE that has the general solution of
\begin{align}
Z^{\prime\prime} -k^2Z &= 0 \\
Z &= Ee^{-kz} + Fe^{kz} \tag{7}
\end{align}
Meanwhile, multiplying (5) by \( r^2 \) gives
\begin{align}
\frac{r}{R} \frac{\partial}{\partial r} (rR’) + \frac{\Theta^{\prime\prime}}{\Theta} +k^2r^2 &= 0 \tag{8}
\end{align}
We can utilize Separation of Variables once again, as the second term depends on \( \theta \) only while the remaining terms depend on \( r \) only. This yields
\begin{align}
\frac{r}{R} \frac{\partial}{\partial r} (rR’) + k^2r^2 &= m^2 \tag{9} \\
\frac{\Theta^{\prime\prime}}{\Theta} &= -m^2 \tag{10}
\end{align}
(10) can be easily solved as a simple harmonic equation
\begin{align}
\Theta^{\prime\prime} + m^2 \Theta &= 0 \\
\Theta &= C\cos(m\theta) + D\sin(m\theta) \tag{11}
\end{align}
where \( m = 0, 1, 2, \ldots \) has to be a non-negative integer so as to ensure the single-valuedness of \(\Theta\) along \( -\pi \leq \theta \leq \pi \). In the special case of \( m = 0 \), it becomes \( \Theta = C + D\theta \). Finally, (9) can be rewritten as
\begin{align}
r\frac{\partial}{\partial r} (rR’) + (k^2r^2 -m^2)R &= 0 \tag{12}
\end{align}
Compared to (1) of this tutorial, we identify it as the Bessel equation with the replaced argument \( r \rightarrow kr \), and \( \nu = m \), so its general solution will be
\begin{align}
R = AJ_m(kr) + B Y_m(kr) \tag{13}
\end{align}
If the cylindrical domain includes the axial \( r = 0 \), then we require \(B = 0\) as the Bessel functions of the second kind \( Y_m \) are singular there. Meanwhile, the allowed values of \( k \) will usually be decided by the given B.C along the circumference. The complete solution basis is therefore
\begin{align}
u(r,\theta,z) &= (AJ_m(kr) + B Y_m(kr))(C\cos(m\theta) + D\sin(m\theta))(Ee^{-kz} + Fe^{kz}) \tag{14}
\end{align}
and the full solution will be the superposition of them. If the B.C.s are in the general form of
\begin{align}
\left\{\begin{aligned}
u(r,\theta,0) &= 0 \\
u(a,\theta,z) &= 0 \\
u(r,\theta,L) &= f(r,\theta) \\
\end{aligned}\right. \tag{15}
\end{align}
so that only the top surface is imposed a varying B.C. while other surfaces are held at zero, then it is clear that the exponential part of the solution becomes
\begin{align}
Z = \sinh(kz) \tag{16}
\end{align}
And the radial part of the solution will abide
\begin{align}
R = J_m(k_{mn}\frac{r}{a}) \tag{17}
\end{align}
where \( k = k_{mn}/a \), i.e. \( k_{mn} = ka \) denotes the \(n\)-th zero of the Bessel function of the first kind of order \(m\), \( J_m \). So the solution now looks like
\begin{align}
u(r,\theta,z) &= \sum_{m=0}^{\infty}\sum_{n=1}^{\infty}(C_{mn}\cos(m\theta) + D_{mn}\sin(m\theta))J_m(k_{mn}\frac{r}{a})\sinh(k_{mn}\frac{z}{a}) \tag{18}
\end{align}
To determine the coefficients \( C_{mn}, D_{mn} \), we first need to decompose the top B.C. into Fourier modes azimuthally with varying radial parts for matching:
\begin{align}
u(r,\theta,L) = f(r,\theta) = \sum_{m=0}^{\infty} (C_m(r) \cos(m\theta) + D_m(r)\sin(m\theta)) \tag{19}
\end{align}
Then, comparing the Fourier modes in (18) and (19) at \( z = L \) leads to
\begin{align}
C_m(r) &= \sum_{n=1}^{\infty} C_{mn}J_m(k_{mn}\frac{r}{a})\sinh(k_{mn}\frac{L}{a}) \tag{20} \\
D_m(r) &= \sum_{n=1}^{\infty} D_{mn}J_m(k_{mn}\frac{r}{a})\sinh(k_{mn}\frac{L}{a}) \tag{21} \\
\end{align}
Subsequently, we make use of the orthonormality of the Bessel function as noted by this tutorial (refer to (31)) to retrieve \( C_{mn} \) as a Bessel series:
\begin{align}
C_{mn} &= \frac{\int_{0}^{a} \frac{r}{a} C_m(r) J_m(k_{mn}\frac{r}{a}) d\frac{r}{a}}{(\int_{0}^{a} \frac{r}{a} [J_m(k_{mn}\frac{r}{a})]^2 d\frac{r}{a})\sinh(k_{mn}\frac{L}{a})} \\
&= \frac{2\int_{0}^{a} r C_m(r) J_m(k_{mn}\frac{r}{a}) dr}{a^2[J_{m+1}(k_{mn})]^2\sinh(k_{mn}\frac{L}{a})} \tag{22}
\end{align}
and similarly for \( D_{mn} \). Putting back into (18), the solution is done.
Exercise
Consider a conducting cylindrical shell with a height of \( L \) and radius of \( a \). The top cap is held at a constant potential \( V_0 \) while all other surfaces are grounded and have zero potential. Find the distribution of potential within the interior of the cylindrical shell. Reference: this PDF.
Answer
Since the entire top cap has a constant potential \( V_0 \), there will be no \(\theta\) dependence and \( m = 0 \) in (19), with \( C_m(r) = V_0 \) and no \( D_m \) part. Subsequently, (22) becomes
\begin{align}
C_{0n} &= \frac{2\int_{0}^{a} rV_0 J_0(k_{0n}\frac{r}{a}) dr}{a^2[J_{1}(k_{0n})]^2\sinh(k_{0n}\frac{L}{a})} \\
&= \frac{2V_0[\frac{r}{k_{0n} a}J_1(k_{0n}\frac{r}{a})]_0^a}{[J_{1}(k_{0n})]^2\sinh(k_{0n}\frac{L}{a})} \\
&= \frac{2V_0}{k_{0n} J_{1}(k_{0n})\sinh(k_{0n}\frac{L}{a})}
\end{align}
where we used (12) in this tutorial, and hence the full solution is
\begin{align}
V &= \sum_{n=1}^{\infty} \frac{2V_0}{k_{0n} J_{1}(k_{0n})\sinh(k_{0n}\frac{L}{a})} J_0(k_{0n}\frac{r}{a})\sinh(k_{0n}\frac{z}{a})
\end{align}
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