logistic_guy
Full Member
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- Apr 17, 2024
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- 668
The steady-state temperature in a hemisphere of radius \(\displaystyle r = c\) is determined from
\(\displaystyle \frac{\partial^2 u}{\partial r^2} + \frac{2}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} + \frac{\cot \theta}{r^2}\frac{\partial u}{\partial \theta} = 0\)
\(\displaystyle 0 < r < c, \ \ \ \ 0 < \theta < \frac{\pi}{2}\)
\(\displaystyle \frac{\partial u}{\partial \theta}\bigg|_{\theta = \frac{\pi}{2}} = 0, \ \ \ \ 0 < r < c\)
\(\displaystyle u(c,\theta) = f(\theta), \ \ \ \ 0 < \theta < \frac{\pi}{2}\)
Solve for \(\displaystyle u(r,\theta)\).
\(\displaystyle \frac{\partial^2 u}{\partial r^2} + \frac{2}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} + \frac{\cot \theta}{r^2}\frac{\partial u}{\partial \theta} = 0\)
\(\displaystyle 0 < r < c, \ \ \ \ 0 < \theta < \frac{\pi}{2}\)
\(\displaystyle \frac{\partial u}{\partial \theta}\bigg|_{\theta = \frac{\pi}{2}} = 0, \ \ \ \ 0 < r < c\)
\(\displaystyle u(c,\theta) = f(\theta), \ \ \ \ 0 < \theta < \frac{\pi}{2}\)
Solve for \(\displaystyle u(r,\theta)\).
