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Let $\|\cdot\|$ denote the $\mathbb R^2$euclidean norm.

Let a such that $\|\mathbf a\|=1$

Let $B$ denote the open unit disk in $\mathbb R^2$

Define $\displaystyle \begin{array}{ccccc} f & : & B & \to & R \\ & & \mathbf x& \mapsto & \frac{1-\|\mathbf x\|^2}{\|\mathbf a-\mathbf x\|^2} \\ \end{array}$

Prove that $\Delta f=0$ (where $\Delta$ denotes the Laplace operator)

In $\mathbb R^2$, $\displaystyle\Delta=\frac{\partial^2}{\partial x^2 } +\frac{\partial^2}{\partial y^2 } $, so the question may boil down to a very tedious computation.

I don't know how to get a neat proof for this.

Thanks for providing insight about this.

Community
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Gabriel Romon
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1 Answers1

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If you know what a holomorphic function is and that the real part of any holomorphic function is harmonic (i.e. satisfies $\Delta f=0$), you can prov it as follows.

Identify $\mathbb R^2$ with $\mathbb C$, and accordingly, call your variable $z$ rather than $\mathbf x$. Then a not tedious at all computation gives you that $$f(z)={\rm Re}\left(\frac{1+\bar a z}{1-\bar a z}\right). $$ Since the function inside the brackets is holomorphic in the unit disk, the result follows.

Another proof runs as follows. Still identifying $\mathbb R^2$ with $\mathbb C$, write $z=re^{i\theta}$ and $a=e^{i\alpha}$. Then you can check (not too tedious) that $$f(re^{i\theta})=\sum_{n=-\infty}^{+\infty} r^{\vert n\vert} e^{i(\theta-\alpha)}\, .$$ Using the expression for the Laplace operator in polar coordinates (and differentiating under the $\Sigma$, which is allowed), you get the result.

Note. The function $f$ is quite famous; try an internet search for Poisson kernel.

Etienne
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  • Would you mind expanding on why $f(z) = \text{Re} \frac{1 + \bar a z}{1 - \bar a z} = \frac{1 - \Vert \mathbf x \Vert^2}{\Vert \mathbf a - \mathbf x \Vert^2}$? – Robert Lewis May 21 '14 at 18:32
  • I think I should also add I like your approach to this problem! Cheers! – Robert Lewis May 21 '14 at 18:38
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    @RobertLewis Write $\frac{1+\bar az}{1-\bar az}=\frac{(1+\bar az)(1-a\bar z)}{\vert 1-\bar az\vert^2}$, expand the product and observe that ${\rm Re}(\bar az-a\bar z)=0$. Note also that $\vert 1-\bar a z\vert=\vert a-z\vert$. – Etienne May 21 '14 at 18:48
  • @RobertLewis Yes, exactly. Cheers. – Etienne May 21 '14 at 19:16
  • Yes, OK, so the numerator becomes $1 - z \bar z$ since $a \bar a = 1$, and we can deal with the denominator via the observation that $\vert 1 - \bar a z \vert^2 = \vert a \vert^2 \vert 1 - \bar a z \vert^2 = \vert a - z \vert^2$, again using $\vert a \vert^2 = \bar a a = 1$. Gotcha! And thanks. And +1! – Robert Lewis May 21 '14 at 19:18