Question 3.5: An infinitely long rectangular metal pipe (sides a and b) is...

This is a genuinely three-dimensional problem,

\frac{\delta ^{2}V}{\delta x^{2}}+ \frac{\delta ^{2}V}{\delta y^{2}}+\frac{\delta ^{2}V}{\delta z^{2}}=0,       (3.43)

subject to the boundary conditions

\begin{cases}(i) V = 0\quad when\quad y = 0,\\(ii) V = 0\quad when\quad y = a,\\(iii) V = 0\quad   when\quad     z = 0,\\(iv) V = 0 \quad  when\quad   z = b,\\(v) V → 0 as\quad x →\infty ,\\(vi) V = V_{0}(y, z)   when\quad    x = 0.\end{cases}           (3.44)

As always, we look for solutions that are products:

V(x, y, z) = X(x)Y (y)Z(z).          (3.45)

Putting this into Eq. 3.43, and dividing by V, we find

It follows that

Our previous experience (Ex. 3.3) suggests that C_{1} must be positive,latex]C_{2}[/latex] and C_{3} negative. Setting C_{2}=-k^{2} and C_{3}=-l^{2}, we have C_{1}=k^{2}+l^{2}, and hence

\frac{1}{X}\frac{d^{2}X}{dx^{2}}=(k^{2}+l^{2})X,\frac{1}{Y}\frac{d^{2}Y}{dy^{2}}=-k^{2}Y, \frac{1}{Z}\frac{d^{2}Z}{dy^{2}} =-l^{2}Z.      (3.46)

Once again, separation of variables has turned a partial differential equation into ordinary differential equations. The solutions are

Boundary condition (v) implies A = 0, (i) gives D = 0, and (iii) yields F = 0, whereas (ii) and (iv) require that k = nπ/a and l = mπ/b, where n and m are positive integers. Combining the remaining constants, we are left with

V(x, y, z)=Ce^{−π\sqrt{(n/a)^{2}+(m/b)^{2}} x} \sin(nπy/a) \sin(mπz/b).       (3.47)

This solution meets all the boundary conditions except (vi). It contains two unspecified integers (n and m), and the most general linear combination is a double sum:

V(x, y, z)=\sum\limits_{n=1}^{\infty }{}\sum\limits_{m=1}^{\infty }{} C_{n.m}e^{−π\sqrt{(n/a)^{2}+(m/b)^{2}} x} \sin(nπy/a) \sin(mπz/b).      (3.48)

We hope to fit the remaining boundary condition,

V(0, y, z)=\sum\limits_{n=1}^{\infty }{}\sum\limits_{m=1}^{\infty }{} C_{n.m} \sin(nπy/a) \sin(mπz/b)=V_{0}(y,z).     (3.49)

by appropriate choice of the coefficients C_{n,m}. To determine these constants, we multiply by \sin(\acute{n}πy/a) \sin(\acute {m}πz/b), where \acute {n} and \acute {m} are arbitrary positive integers, and integrate:

Quoting Eq. 3.33, the left side is (ab/4)C_{n^′,m^′ }, so

\int_{0}^{a}{sin(n\pi y/a)sin(n^\prime \pi y/a)} \begin{cases} 0, & if \quad n^\prime \neq n \\ \frac{a}{2} & n^\prime = n \end{cases}         (3.33)

C_{n,m}=\frac{4}{ab}\int_{0}^{a}{\int_{0}^{b}{V_0(y,z)sin(n\pi y/a)sin(m\pi y/a)dydz} }              (3.50)

Equation 3.48, with the coefficients given by Eq. 3.50, is the solution to our problem.
For instance, if the end of the tube is a conductor at constant potential V_0 ,

=\begin{cases} 0, & if \quad n \quad or\quad m \quad is \quad even \\ \frac{16V_0}{\pi ^2nm} & n \quad or\quad m \quad is \quad odd \end{cases}           (3.51)

In this case

V(x,y,z)=\frac{16V_0}{\pi ^2nm} \sum\limits_{n,m=1,3,5,…}^{\infty }{\frac{1}{nm}e^{-\pi \sqrt{(n/a)^2+(m/b)^2} x }sin(n\pi y/a)sin(m\pi z/b) }         (3.52)

Notice that the successive terms decrease rapidly; a reasonable approximation would be obtained by keeping only the first few.