A rectangular pipe, running parallel to the z-axis (from -∞ to +∞), has three grounded metal sides, at y = 0, y = a, and x = 0. The fourth side, at x = b, is maintained at a specified potential V_{0}(y) .
(a) Develop a general formula for the potential inside the pipe
(b) Find the potential explicitly, for the case V_{0}(y)=V_{0} (aconstant).
(a) \frac{\partial^{2} V}{\partial x^{2}}+\frac{\partial^{2} V}{\partial y^{2}}=0 , with boundary conditions
\left\{\begin{array}{l}\text { (i) } V(x, 0)=0, \\\text { (ii) } V(x, a)=0, \\\text { (iii) } V(0, y)=0, \\\text { (iv) } V(b, y)=V_{0}(y) .\end{array}\right\}As in Ex. 3.4, separation of variables yields
V(x, y)=\left(A e^{k x}+B e^{-k x}\right)(C \sin k y+D \cos k y) .
Here \text { (i) } \Rightarrow D=0, \text { (iii) } \Rightarrow B=-A,( ii ) \Rightarrow k a \text { is an integer multiple of } \pi :
V(x, y)=A C\left(e^{n \pi x / a}-e^{-n \pi x / a}\right) \sin (n \pi y / a)=(2 A C) \sinh (n \pi x / a) \sin (n \pi y / a).
But (2AC) is a constant, and the most general linear combination of separable solutions consistent with (i), (ii), (iii) is
V(x, y)=\sum_{n=1}^{\infty} C_{n} \sinh (n \pi x / a) \sin (n \pi y / a) .
It remains to determine the coe!cients C_{n} so as to fit boundary condition (iv):
\sum C_{n} \sinh (n \pi b / a) \sin (n \pi y / a)=V_{0}(y) . \text { Fourier's trick } \Rightarrow C_{n} \sinh (n \pi b / a)=\frac{2}{a}\int\limits_{0}^{a}{} V_{0}(y) \sin (n \pi y / a) d y .
Therefore
C_{n}=\frac{2}{a \sinh (n \pi b / a)} \int\limits_{0}^{a}{} V_{0}(y) \sin (n \pi y / a) d y .
(b) C_{n}=\frac{2}{a \sinh (n \pi b / a)} V_{0} \int\limits_{0}^{a}{} \sin (n \pi y / a) d y=\frac{2 V_{0}}{a \sinh (n \pi b / a)} \times\left\{\begin{array}{c}0, \text { if } n \text { is even, } \\\frac{2 a}{n \pi}, \text { if } n \text { is odd. }\end{array}\right\}
V(x, y)=\frac{4 V_{0}}{\pi} \sum_{n=1,3,5, \ldots} \frac{\sinh (n \pi x / a) \sin (n \pi y / a)}{n \sinh (n \pi b / a)} .
