Subsection 7.3.4 Two Dimensional Laplace’s Equation
The two-dimensional Laplace equation is a partial differential equation that describes the behavior of a scalar function in two dimensions. The general form of the two-dimensional Laplace equation is:
\begin{equation*}
\nabla^{2}u =0
\end{equation*}
where u is the scalar function and \(\nabla^{2}\) is the Laplacian operator. In two dimensions, the Laplacian operator can be expressed as the sum of the second partial derivatives with respect to the spatial coordinates x and y:
\begin{equation*}
\nabla^{2}u = \frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^2}
\end{equation*}
This equation states that the sum of the second partial derivatives of the function with respect to both x and y is equal to zero. The Laplace equation is a special case of the more general Poisson equation, where a non-zero source term is present on the right-hand side:
\begin{equation*}
\nabla^{2}u =f(x, y)
\end{equation*}
Subsubsection 7.3.4.1 In Cartesian Coordinates
Consider a thin plate with insulated surfaces bounded by the lines \(x=0\text{,}\) \(x=\lambda {,}y=0 \text{,}\) and \(y=\infty\text{.}\) Let at steady state, the temp. \(u(x,y)\) at any point \(p(x,y)\) satisfies the equation
\begin{equation}
\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}} =0\tag{7.3.59}
\end{equation}
assuming the edges \(x=0\) and \(x=\lambda \) is kept at temperature zero and also the lower edge \(y=0\) is kept at tempemperature \(f(x)\) and the edge \(y=\infty\) at zero temperature.
Then solution of equation (7.3.59) is given by
\begin{equation}
u(x,y) = X(x)Y(y)\tag{7.3.60}
\end{equation}
Now three possible solutions of equation (7.3.59) are
\begin{equation}
u=\left(C_{1}e^{px}+C_{2}e^{-px}\right)\left(C_{3}\cos py+C_{4}\sin py\right)\tag{7.3.61}
\end{equation}
\begin{equation}
u=\left(C_{5}\cos px+C_{6}\sin px\right)\left(C_{7}e^{py}+C_{8}e^{-py}\right)\tag{7.3.62}
\end{equation}
\begin{equation}
u=\left(C_{9}x+C_{10}\right)\left(C_{11}y+C_{12}\right)\tag{7.3.63}
\end{equation}
Of these, we have to choose that solution which is consistant with the physical nature of the problem. The solution (7.3.61) and (7.3.63) can not satisfy the boundary conditions. Therefore, possible solution is only (7.3.62). From the boundary condition, \(u(0,y) =0\text{,}\) we have from solution (7.3.62),
\begin{equation*}
0= C_{5}\left(C_{7}e^{py}+C_{8}e^{-py}\right)\Rightarrow C_{5} =0.
\end{equation*}
therefore, solution (7.3.62) reduces to
\begin{equation}
u=C_{6}\sin px\left(C_{7}e^{py}+C_{8}e^{-py}\right) \tag{7.3.64}
\end{equation}
using \(u(\lambda,y) =0\text{,}\) we have -
\begin{equation*}
0= C_{6}\sin p\lambda\left(C_{7}e^{py}+C_{8}e^{-py}\right)\quad \because C_{6} \neq 0,
\end{equation*}
and
\begin{equation*}
\sin p\lambda = 0=\sin n\pi \quad \therefore p=\frac{n\pi}{\lambda}, \quad n=1,2,3,\cdots
\end{equation*}
also to satisfy the condition \(u(x,\infty)=0, \quad C_{7} =0\text{.}\) Hence equation (7.3.64) takes the form \(u(x,y) = C_{6}C_{8}\sin px \cdot e^{-py}\text{.}\) Or,
\begin{equation}
u(x,y) = \sum\limits_{n=1}^{\infty}b_{n}\sin\frac{n\pi x}{\lambda}\cdot e^{-\frac{n\pi y}{\lambda}}\tag{7.3.65}
\end{equation}
But the initial condition \(u(x,0) =f(x)\text{,}\) we have
\begin{equation}
f(x)=\sum\limits_{n=1}^{\infty}b_{n}\sin\frac{n\pi x}{\lambda}\tag{7.3.66}
\end{equation}
where
\begin{equation*}
b_{n} = \frac{2}{\lambda}\int\limits_{0}^{\lambda}f(x)\sin\frac{n\pi x}{\lambda}\,dx
\end{equation*}
If the given plate is of finite length \(a\) and the temp. at this end is \(u=f(x) \text{,}\) then the boundary conditions are \(u(0,y) =0=u(\lambda,y)=u(x,0)\) and \(u(x,a)=f(x) \text{.}\)
Let the solution of this problem is
\begin{equation}
u=\left(C_{1}\cos px+C_{2}\sin px\right)\left(C_{3}e^{py}+C_{4}e^{-py}\right)\tag{7.3.67}
\end{equation}
\begin{equation*}
0=C_{1}\left(C_{3}e^{py}+C_{4}e^{-py}\right) \quad \Rightarrow C_{1} =0
\end{equation*}
hence eqn. (7.3.67) reduces to
\begin{equation}
u=C_{2}\sin px\left(C_{3}e^{py}+C_{4}e^{-py}\right) \tag{7.3.68}
\end{equation}
on putting \(x=\lambda\) and \(u=0\text{,}\) we have -
\begin{equation*}
0=C_{2}\sin p\lambda\left(C_{3}e^{py}+C_{4}e^{-py}\right)
\end{equation*}
or,
\begin{equation*}
\sin p\lambda =0=\sin n\pi \quad \text{as}\quad C_{2} \neq 0
\end{equation*}
or,
\begin{equation*}
p=\frac{n\pi}{\lambda}, \quad n=1,2,3,\cdots
\end{equation*}
Now equation (7.3.68) becomes -
\begin{equation}
u=C_{2}\sin\frac{n\pi x}{\lambda}\left(C_{3}e^{\frac{n\pi y}{\lambda}}+C_{4}e^{-\frac{n\pi y}{\lambda}}\right)\tag{7.3.69}
\end{equation}
\begin{equation*}
0= C_{2}\sin\frac{n\pi x}{\lambda}(C_{3}+C_{4}) \quad \Rightarrow C_{3}+C_{4} =0
\end{equation*}
or, \(C_{4} =-C_{3}\) therefore, equation (7.3.69) becomes
\begin{equation*}
u=C_{2}C_{3}\sin\frac{n\pi x}{\lambda}\left(e^{\frac{n\pi y}{\lambda}}-e^{-\frac{n\pi y}{\lambda}}\right)
\end{equation*}
or,
\begin{equation*}
u=\sum\limits_{n=1}^{\infty}b_{n}\sin\frac{n\pi x}{\lambda}\left(e^{\frac{n\pi y}{\lambda}}-e^{-\frac{n\pi y}{\lambda}}\right)\times\frac{2}{2}
\end{equation*}
or,
\begin{equation}
u=\sum\limits_{n=1}^{\infty}c_{n}\sin\frac{n\pi x}{\lambda}\sin\frac{n\pi y}{\lambda} \tag{7.3.70}
\end{equation}
\([\because \quad 2b_{n} =c_{n}]\text{.}\) on putting \(y=a\) and \(u=f(x)\) in equation (7.3.70), we get -
\begin{equation}
f(x) =\sum\limits_{n=1}^{\infty}c_{n}\sin\frac{n\pi x}{\lambda}\sin\frac{n\pi y}{\lambda}\tag{7.3.71}
\end{equation}
where
\begin{equation*}
c_{n}\sinh \frac{n\pi a}{\lambda} = \frac{2}{\lambda}\int\limits_{0}^{\lambda}f(x)\sin\frac{n\pi x}{\lambda}\,dx.
\end{equation*}
Subsubsection 7.3.4.2 In Cylindrical Coordinates
The Laplace equation in two dimensional case is
\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}}=0 \tag{7.3.72}
\end{equation}
Let
\begin{equation}
u(r,\theta) = R(r)\Theta(\theta) \tag{7.3.73}
\end{equation}
be the solution of equation (7.3.72), so that
\begin{equation}
\Theta\frac{\partial^{2}R}{\partial r^{2}}+\frac{\Theta}{r}\frac{\partial R}{\partial r}+\frac{R}{r^{2}}\frac{\partial^{2}\Theta}{\partial \theta^{2}}\tag{7.3.74}
\end{equation}
or,
\begin{equation}
\frac{1}{R}\left(r^{2}\frac{\partial^{2}R}{\partial r^{2}}+r\frac{\partial R}{\partial r}\right) =-\frac{1}{\Theta}\frac{\partial^{2}\Theta}{\partial \theta^{2}}=n^{2} \quad \text{(say)} \tag{7.3.75}
\end{equation}
\begin{equation}
r^{2}\frac{\partial^{2}R}{\partial r^{2}}+r\frac{\partial R}{\partial r}=-n^{2}R=0 \tag{7.3.76}
\end{equation}
This is a homogenous linear differential equation. Its solution can be found by putting \(r=e^{z}\) or, \(\log r=z\) or,
\begin{equation*}
\frac{1}{r}\frac{\partial r}{\partial z} =1.
\end{equation*}
or,
\begin{equation}
\frac{\partial r}{\partial z} =r\tag{7.3.77}
\end{equation}
Also,
\begin{equation*}
\frac{\partial R}{\partial r} =\frac{\partial R}{\partial z}\cdot\frac{\partial z}{\partial r} = \frac{1}{r}\frac{\partial R}{\partial z}
\end{equation*}
or,
\begin{equation}
r\frac{\partial R}{\partial r}=\frac{\partial R}{\partial z}= DR \tag{7.3.78}
\end{equation}
assume
\begin{equation*}
D\equiv \frac{\partial }{\partial z}.
\end{equation*}
again,
\begin{equation*}
\frac{\partial^{2}R}{\partial r^{2}} = \frac{\partial }{\partial r}\left(\frac{\partial R}{\partial r}\right) = \frac{\partial }{\partial z}\left(\frac{1}{r}\frac{\partial R}{\partial z}\right)\frac{\partial z}{\partial z}
\end{equation*}
\begin{equation*}
= \frac{1}{r}\left[\frac{1}{r}\frac{\partial^{2}R}{\partial z^{2}}- \frac{1}{r^{2}}\frac{\partial r}{\partial z}\frac{\partial R}{\partial z}\right] = \frac{1}{r}\left[\frac{1}{r}D^{2}R- \frac{1}{r^{2}}\cdot r\cdot DR\right]
\end{equation*}
or,
\begin{equation*}
r^{2}\frac{\partial^{2}R}{\partial r^{2}}=D^{2}R-DR=D(D-1)R
\end{equation*}
from eqn.(7.3.76),
\begin{equation*}
D(D-1)R+DR-n^{2}R =0
\end{equation*}
or,
\begin{equation*}
D^{2}R -DR+DR-n^{2}R =0
\end{equation*}
or,
\begin{equation}
(D^{2}-n^{2})R=0\tag{7.3.79}
\end{equation}
or,
\begin{equation*}
D=\pm n
\end{equation*}
or,
\begin{equation*}
R=C_{1}e^{nz} +C_{2}e^{-nz}
\end{equation*}
\begin{equation}
\therefore R=C_{1}r^{n}+C_{2}r^{-n} \tag{7.3.80}
\end{equation}
\begin{equation*}
[e^{nz}=e^{n\log r}=e^{\log r^{n}} = r^{n}].
\end{equation*}
and
\begin{equation}
\frac{\partial^{2}\Theta}{\partial \theta^{2}}+n^{2}\Theta =0 \quad \Rightarrow \Theta =C_{3}\cos n\theta + C_{4}\sin n\theta \tag{7.3.81}
\end{equation}
Hence,
\begin{equation}
u_{n}=\left(C_{1}r^{n}+C_{2}r^{-n}\right)\left(C_{3}\cos n\theta+C_{4}\sin n\theta\right)\tag{7.3.82}
\end{equation}
If \(n=0\text{,}\) we have from equation (7.3.81),
\begin{equation}
\frac{\partial^{2}\Theta}{\partial \theta^{2}} =0 \quad \Rightarrow \Theta = A_{o}\theta+B_{o}\tag{7.3.83}
\end{equation}
from equation (7.3.79), we get -
\begin{equation}
D^{2}R =0 \quad \Rightarrow R = C_{o}z+D_{o} = C_{o}\log r + D_{o}\tag{7.3.84}
\end{equation}
Hence,
\begin{equation}
u_{o}=(A_{o}\theta+B_{o})(C_{o}\log r + D_{o})\tag{7.3.85}
\end{equation}
But the general single - valued solution of equation (7.3.72) for all possible value of \(n\) may be written as
\begin{equation*}
u=a_{o}\log r+\sum\limits_{n=1}^{\infty}r^{n}(a_{n}\cos n\theta +b_{n}\cos n\theta)
\end{equation*}
\begin{equation}
+\sum\limits_{n=1}^{\infty}r^{-n}(p_{n}\cos n\theta +q_{n}\cos n\theta)+c_{o}\tag{7.3.86}
\end{equation}
