Discussion
The function \(u(r, \theta)\) satisfying the Laplace equation
\[ \begin{aligned} &\frac{\partial^{2} u}{\partial r^{2}} + \frac{1}{r}\frac{\partial u}{\partial r} \\ &+ \frac{1}{r^{2}}\frac{\partial^{2} u}{\partial \theta^{2}} = 0, \end{aligned} \]
\(e \lt r \lt e^{2}\), subject to the conditions \(u(e, \theta) = 1,\ u(e^{2}, \theta) = 0\) is
Answer: C
The solution is \(\ln(e^{2}/r)\).
The boundary values do not depend on \(\theta\), so look for a radial solution. With \(u_{\theta\theta} = 0\) the equation becomes \((ru')' = 0\), whose solutions are \(u = A + B\ln r\). Applying the conditions:
\[ \begin{aligned} A + B &= 1 \quad (r = e) \\ A + 2B &= 0 \quad (r = e^{2}) \\ B &= -1,\ A = 2 \end{aligned} \]
So \(u = 2 - \ln r = \ln(e^{2}/r)\); check: \(u(e) = 2 - 1 = 1\) and \(u(e^{2}) = 0\). \(\ln(e/r)\) gives 0 and -1 at the two radii, and \(\ln(e/r^{2})\) gives -1 and -3. The series vanishes at \(r = e^{2}\), but at \(r = e\) it becomes \(\sum \sin n\theta\), not the constant 1, and its terms \((r - e^{2})\sin n\theta\) do not satisfy Laplace's equation.
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