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I'm trying to solve the Laplace equation: $u_{xx} + u_{yy} = 0$ on the disk ${r < a}$ with the boundary condition $u = sin^3(\theta)$

All I note is that I can use the identity $\sin(3 \theta) = 3 \sin(\theta) - 4 {\sin}^3(\theta)$

I was thinking if I can use an integrating factor as one of the steps?

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up vote 2 down vote accepted

The simplest way to do this is via a complex detour. (I'll do it for $a = 1$, it's straight-forward to adapt the solution to other radii.) Note that if $z = re^{i\theta}$, then $\newcommand{\imag}{\operatorname{Im}}\imag{z} = r\sin \theta$. The real and imaginary parts of a holomorphic function is harmonic, and since

$$\sin^3 \theta = \frac34\sin\theta - \frac14 \sin 3\theta$$

it makes sense to look at $$f(z) = \frac34 z - \frac14 z^3$$

If $|z| = 1$, then $\imag(f(z)) = \frac34\sin\theta - \frac14\sin 3\theta$, so the solution you want is

\begin{align} u = \imag(f(x+iy)) &= \imag\Big( \frac34(x+iy) - \frac14(x+iy)^3 \Big) \\ &= \frac34 y - \frac34 x^2 y + \frac14 y^3 \end{align}

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+1 but I'll add a version without a complex detour: $\Delta (r^n \sin n\theta)=0$ for $n\in\mathbb N$, which can be calculated via polar coordinates. – user53153 Jan 27 '13 at 0:07

Here is a solution to the problem.

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Thanks, the website is nice, but it has too much details and I'm not sure what to filter for this problem – mary Nov 19 '12 at 7:40
The solution Mhenni provided is your problem with $h\left(\theta\right) = \sin^3 \theta$. – Eric Angle Nov 19 '12 at 19:21
Can you please write it out here? I still cannot see what part I need – mary Nov 19 '12 at 19:24

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