# Problem of the Week #29 - October 15th, 2012

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#### Chris L T521

##### Well-known member
Staff member
I realized that I had posted solutions last night to the POTWs, but forgot to create the new ones last night...I guess that not sleeping well the night before travelling all day can make you do these kinds of things. Anyways, thanks to those who participated in last week's POTW!! Here's this week's problem!

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Problem: Prove that if $r$ and $\theta$ are polar coordinates, then the functions $r^n\cos(n\theta)$ and $r^n\sin(n\theta)$, where $n$ is an integer, are harmonic as functions of $x$ and $y$.

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#### Chris L T521

##### Well-known member
Staff member
I'm going to be lenient about solutions to this problem; Sudharaka and dwsmith answered it correctly if one could use the polar form of the Laplace equation to show that the given two functions were harmonic. However, this question was looking for you to show that the polar functions were harmonic as functions of $x$ and $y$.

You can find Sudharaka's solution below.

The Laplace's equation in polar coordinates could be written as,

$\nabla^2 f = {1 \over r} {\partial \over \partial r} \left( r {\partial f \over \partial r} \right) + {1 \over r^2} {\partial^2 f \over \partial \theta^2}$

We shall show that both $$f_{1}=r^n\cos(n\theta)$$ and $$f_{2}=r^n\sin(n\theta)$$ satisfies the Laplace's equation.

\begin{eqnarray}

\nabla^2 f_{1}&=&{\cos(n\theta) \over r} {\partial \over \partial r} \left(nr^{n}\right) - n r^{n-2}{\partial \over \partial \theta}\sin(n\theta)\\

&=&n^2 r^{n-2}\cos(n\theta) - n^2 r^{n-2}\cos(n\theta)\\

&=&0\\

\end{eqnarray}

Similarly,

\begin{eqnarray}

\nabla^2 f_{2}&=&{\sin(n\theta) \over r} {\partial \over \partial r} \left(nr^{n}\right) + n r^{n-2}{\partial \over \partial \theta}\cos(n\theta)\\

&=&n^2 r^{n-2}\sin(n\theta) - n^2 r^{n-2}\sin(n\theta)\\

&=&0\\

\end{eqnarray}

Therefore both $$f_{1}$$ and $$f_{2}$$ are harmonic functions.

Q.E.D.

Here's the solution I was looking for.

Let $f=r^n\cos(n\theta)$ and $g=r^n\sin(n\theta)$. Observe that in rectangular coordinates, \begin{aligned} r^n\cos(n\theta) &= \tfrac{1}{2}r^n\left[(\cos(n\theta)+i\sin(n\theta))+(\cos(n\theta)-i\sin(n\theta))\right]\\ &= \tfrac{1}{2}r^n\left[(\cos\theta+i\sin\theta)^n+(\cos\theta-i\sin\theta)^n\right]\qquad(\text{by DeMoivre's theorem})\\ &= \tfrac{1}{2}(r\cos\theta+ir\sin\theta)^n+\tfrac{1}{2}(r\cos\theta-ir\sin\theta)^n\\ &= \tfrac{1}{2}(x+iy)^n+\tfrac{1}{2}(x-iy)^n.\end{aligned}
It now follows that
$\frac{\partial^2 f}{\partial x^2} = \tfrac{1}{2}(n^2-n)\left[(x+iy)^{n-2}+(x-iy)^{n-2}\right]$
and
$\frac{\partial^2f}{\partial y^2}=\tfrac{1}{2}(n^2-n)i^2\left[(x+iy)^{n-2}+(x-iy)^{n-2}\right] = -\tfrac{1}{2}(n^2-n)\left[(x+iy)^{n-2}+(x-iy)^{n-2}\right].$
Therefore,
$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \tfrac{1}{2}(n^2-n)\left[(x+iy)^{n-2}+(x-iy)^{n-2}\right] - \tfrac{1}{2}(n^2-n)\left[(x+iy)^{n-2}+(x-iy)^{n-2}\right] = 0.$
Thus, as a function of $x$ and $y$, $r^n\cos(n\theta)$ is harmonic. In a similar fashion (work omitted), $r^n\sin(n\theta)$ is also harmonic as a function of $x$ and $y$. Q.E.D.

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