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**Web Bonus Problems / Re: Exam Week**

« **on:**April 14, 2018, 02:12:26 PM »

Writing the change of coordinates Sheng applied to the boundary conditions I found, $ w|_{\pi=0}=0$ and $ w|_{r=1}=2\sin^2\theta $ I can apply separation of variables to the function $ w = P(\theta)R(r) $ which I can solve for this half disk as

$$ w = \sum A r^n \sin(n\theta) $$ and therefore

$$ A = \frac{4}{\pi}\int_{0}^{\pi} \sin^2(\theta) \sin(n\theta) d \theta =\frac{ (4 (2 \cos(π n) - 2))}{(π (n^3 - 4 n)) }$$

and this is $ \frac{-8}{(π (n^3 - 4 n)) } $ when n is odd. So I have

$$ w = \sum_{n \text{ odd}} \frac{-8}{(π (n^3 - 4 n)) } r^n \sin(n\theta) $$

and therefore I can write $$ u = w + v = \sum_{n \text{ odd}} \frac{-8}{(π (n^3 - 4 n)) } r^n \sin(n\theta) + r^2 \cos^2(\theta) - r^2 sin^2(\theta) $$ which can be written most concisely as

$$ u = r^2 \cos(2 \theta) + \sum_{n \text{ odd}} \frac{-8}{(π (n^3 - 4 n)) } r^n \sin(n\theta) $$

$$ w = \sum A r^n \sin(n\theta) $$ and therefore

$$ A = \frac{4}{\pi}\int_{0}^{\pi} \sin^2(\theta) \sin(n\theta) d \theta =\frac{ (4 (2 \cos(π n) - 2))}{(π (n^3 - 4 n)) }$$

and this is $ \frac{-8}{(π (n^3 - 4 n)) } $ when n is odd. So I have

$$ w = \sum_{n \text{ odd}} \frac{-8}{(π (n^3 - 4 n)) } r^n \sin(n\theta) $$

and therefore I can write $$ u = w + v = \sum_{n \text{ odd}} \frac{-8}{(π (n^3 - 4 n)) } r^n \sin(n\theta) + r^2 \cos^2(\theta) - r^2 sin^2(\theta) $$ which can be written most concisely as

$$ u = r^2 \cos(2 \theta) + \sum_{n \text{ odd}} \frac{-8}{(π (n^3 - 4 n)) } r^n \sin(n\theta) $$