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Find the equilibrium temperature on a half-disk of radius 1 when the temperature is held to 1 degree on the curved edge, while the straight edge is insulated.

For this problem I think the solution uses Laplace equation in polar coordinates using method of separation of variables. However, I'm not sure how to derive the boundary conditions.

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Let $u(r,\theta)$ be the solution in polar coordinates, $0<r\le1$, $0\le\theta\le\pi/2$.

  • The temperature is held to 1º on the curved edge: $u(1,\theta)=1$, $0\le\theta\le\pi/2$.
  • The straight edge is insulated. This is a little more complicated. In cartesian coordinates it would be $u_y(x,0)=0$. Since $u_y=u_r\sin\theta+u_\theta\cos\theta$, this translates into $u_\theta(r,0)=u_\theta(r,\pi)=0$, $0<r\le1$.
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  • $\begingroup$ Could you explain why the insulated condition means $u_y(x,0) = 0$? $\endgroup$
    – user90593
    Feb 5, 2015 at 11:22
  • $\begingroup$ $u_y(x,0)$ represents the heat flux through the line $y=0$ (Newton's law: heat flux is proportional to the gradient of the temperature). Insulation means that there is no heat flux through that line, so $u_y=0$. $\endgroup$
    – Julián Aguirre
    Feb 5, 2015 at 11:47
  • $\begingroup$ Thank you. Would $0 \leq \theta \leq \bf{\pi}$ since the domain is a half disk? $\endgroup$
    – user90593
    Feb 5, 2015 at 11:49
  • $\begingroup$ Yes. ${}{}{}{}{}{}$ $\endgroup$
    – Julián Aguirre
    Feb 5, 2015 at 11:51
  • $\begingroup$ I'm a little confused about your $u_y$. I get $u_y = u_r / \sin \theta + u_\theta / (r\cos\theta)$. $\endgroup$
    – user90593
    Feb 5, 2015 at 16:43

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