Laplace Equation on a Quarter Disk
Consider the temperature distribution \( T \) on a thin flat piece of metal shaped like a quarter disk of radius \( R \), such that \( r < R, \) and \( 0 < \theta < \frac{\pi}{2}. \)
The flat edges at \( \theta = 0 \) and \( \theta = \frac{\pi}{2} \) are insulated, which implies
$$ \partial_{\theta} T(r,0) = 0, \quad \partial_{\theta} T\left(r,\tfrac{\pi}{2}\right) = 0 $$while the curved edge is held at a constant temperature given by
$$ T(R,\theta) = 4\cos(6\theta) $$We solve for the steady-state temperature inside the disk, which satisfies Laplace's equation:
$$ \nabla^2 T = 0 $$ $$ \frac{1}{r}\frac{\partial}{\partial r} \left( r\frac{\partial}{\partial r} T(r,\theta) \right) + \frac{1}{r^2}\frac{\partial^2}{\partial \theta^2} T(r,\theta) = 0 $$Assume a separated solution of the form \( T(r,\theta) = f(r)g(\theta). \)
$$ \frac{1}{f(r)} \frac{1}{r} \frac{\partial}{\partial r} \left(rf'(r)\right) + \frac{1}{r^2}\frac{g''(\theta)}{g(\theta)} = 0 $$Since the radial and angular terms depend on different variables, both sides must equal the same constant.
The angular equation becomes
$$ g''(\theta) = \lambda g(\theta) $$with boundary conditions
$$ g'(0) = 0, \quad g'(\tfrac{\pi}{2}) = 0 $$Assume a solution of the form \( g(\theta) = e^{k\theta} \)
$$ k^2 e^{k \theta} = \lambda e^{k \theta} \quad \Rightarrow \quad k^2 = \lambda $$i) \( \lambda > 0 \)
$$ k = \pm \sqrt{\lambda} $$ $$ g(\theta) = c_1 e^{\sqrt{\lambda}\theta} + c_2 e^{-\sqrt{\lambda}\theta} $$ $$ g'(\theta) = c_1\sqrt{\lambda} e^{\sqrt{\lambda}\theta} - c_2\sqrt{\lambda} e^{-\sqrt{\lambda}\theta} $$Applying the boundary conditions,
$$ g'(0) = \sqrt{\lambda}(c_1 - c_2) = 0 \quad \Rightarrow \quad c_1 = c_2 $$ $$ g'(\tfrac{\pi}{2}) = c_1\sqrt{\lambda}\left( e^{\sqrt{\lambda}\pi/2} - e^{-\sqrt{\lambda}\pi/2} \right) = 0 $$This holds \( \, \Leftrightarrow \, c_1 = 0 \), since \( \, e^{\sqrt{\lambda}\pi} \neq 1 \, \). Therefore positive eigenvalues are incompatible with the insulated boundary conditions.
ii) \( \lambda = 0 \)
$$ k = 0 \quad \Rightarrow \quad g''(\theta) = 0 $$ $$ g(\theta) = c_1\theta + c_2 $$ $$ g'(\theta) = c_1 $$Applying the boundary conditions,
$$ g'(0) = c_1 = 0, \quad g'(\tfrac{\pi}{2}) = c_1 = 0 $$ $$ g(\theta) = c_2 $$So \( \lambda = 0 \) corresponds to a constant angular solution.
iii) \( \lambda < 0 \)
$$ k = \pm i\sqrt{-\lambda} \quad |\lambda| = - \lambda $$Let \( \lambda = -b^2 \) where \( b \ge 0 \)
$$ g(\theta) = c_1 e^{i b \theta} + c_2 e^{-i b \theta} $$ $$ g'(\theta) = c_1 ib e^{ib\theta} - c_2 ib e^{-ib\theta} $$ $$ g'(0) = ib(c_1 - c_2) = 0 \quad \Rightarrow \quad c_1 = c_2 $$ $$ g'(\tfrac{\pi}{2}) = c_1 ib \left( e^{ib\pi/2} - e^{-ib\pi/2} \right) = 0 $$This holds \(\Leftrightarrow\) either \( \, c_1 = 0 \, \) or \( \, e^{i b \pi} = 1 \)
$$ \Rightarrow \quad b \pi = 2\pi n, \quad n = 0, 1, 2, ... $$ $$ b = 2n, \quad \lambda_n = - (2n)^2 $$Therefore,
$$ g(\theta) = c_1 \left( e^{ib\theta} + e^{-ib\theta} \right) = 2c_1\cos(b\theta) $$so the angular eigenfunctions are
$$ g_n(\theta) = c_n \cos(2n\theta), \quad n = 0,1,2,\dots $$Negative values of \( n \) do not produce new angular modes since \( \cos(-x) = \cos x \).
The insulated boundary conditions therefore select only even cosine modes.
The radial equation is
$$ \frac{1}{r}\frac{f'(r) + rf''(r)}{f(r)} + \frac{\lambda}{r^2} = 0 $$This is a Cauchy-Euler equation, so we look for power-law solutions of the form \( f(r) \propto r^p. \) This gives
$$ p(p-1) + p + \lambda = 0 $$ $$ p^2 - 4n^2 = 0 $$ $$ p = \pm 2n $$For \( n = 0 \), we need a second solution. These are
$$ r^0 = 1, \quad \ln r $$Since the terms \( r^{-2n} \) and \( \ln r \) are singular at \( r = 0 \), we discard them.
For \( n = 0 \), \( \, f(r) \) is a constant. Therefore, the general solution is
$$ T(r,\theta) = \sum_{n=0}^{\infty} a_n r^{2n}\cos(2n\theta) $$Apply the boundary condition at \( r = R \)
$$ T(R,\theta) = \sum_{n=0}^{\infty} a_n R^{2n}\cos(2n\theta) = 4\cos(6\theta) $$So only the \( n = 3 \) term survives, \( \displaystyle a_3 = \frac{4}{R^6} \)
$$ \therefore \, T(r,\theta) = 4\left(\frac{r}{R}\right)^6 \cos(6\theta) $$The temperature profile therefore inherits the same angular structure as the boundary condition, while the radial factor determines how the solution varies toward the origin.