Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers 2

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}

share|improve this answer
    
+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
add comment

Here is a solution to the problem.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.