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First, I have to give you our definitions:

Consider $\Omega:=\mathbb{R}^n\setminus\overline{B}_R(0)$ with $R>0$ and $n>1$. The function $\phi\colon\Omega\to G:=B_R(0)\setminus\left\{0\right\}$ with $$ y=\phi(x):=\frac{R^2}{\| x\|^2}x\text{ for }\| x\|>R $$ is called Kelvin-transformation (relating to the sphere $S_R(0)$). Let $u\colon\Omega\to\mathbb{R}$. The function defined by $$ v(y):=\frac{\| x\|^{n-2}}{R^{n-2}}u(x)\text{ for }\lVert x\rVert >R~~~(*) $$ is called Kelvin-transform of u.

Alternatively, the Kelvin-transform of $u$ can be definied by $$ v(z):=\frac{R^{n-2}}{\lVert z\rVert^{n-2}}u\left(\frac{R^2}{\lVert z\rVert^2}z\right)\text{ for }\lVert z\rVert <R. $$

Let $u$ be harmonic in $\Omega$. Then its Kelvin-transform is harmonic in $G$. Show that for the case $n=2$.

To be honest, I do not know, what is to show here, because if I take the alternative definition of the Kelvin-transform, for $n=2$ it is $$ v(z)=u\left(\frac{R^2}{\lVert z\rVert^2}z\right), z\in G $$ and then $$ \Delta v(z)=\Delta u\left(\frac{R^2}{\lVert z\rVert^2}z\right)=0, $$ because $x:=\frac{R^2}{\lVert z\rVert^2}z$ is in $\Omega$ and $u$ is harmonic in $\Omega$.

So I only had to use that the inverse of the Kelvin-transformation is given by $$ \phi^{-1}\colon G\to\Omega, y\longmapsto\frac{R^2}{\lVert y\rVert^2}y. $$

Is that the proof? It seems TOO easy...

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  • 1
    $\begingroup$ You have $(\Delta v)(z) = \bigl(\Delta (u\circ \phi)\bigr)(z)$, it is not obvious that that becomes $\bigl((\Delta u)\circ \phi\bigr)(z) \cdot \text{ something with }\phi$, you have to calculate to see that. $\endgroup$ – Daniel Fischer Dec 10 '13 at 14:50
  • $\begingroup$ How can I for example calculate $\frac{\partial}{\partial z_1}(u(\frac{R^2}{\lVert z\rVert^2}z))$? $\endgroup$ – math12 Dec 10 '13 at 15:47
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Generally, to compute the partial derivatives of a composite function, you use the chain rule,

$$\frac{\partial (u\circ \phi)}{\partial z_i}(z) = \sum_{k=1}^n \bigl(\partial_k u\bigr)(\phi(z))\cdot \frac{\partial \phi_k}{\partial z_i}(z).$$

To compute the Laplacian, you need second order derivatives,

$$\frac{\partial^2 (u\circ\phi)}{\partial z_i^2}(z) = \sum_{k,m = 1}^n \bigl(\partial_m\partial_k u\bigr)(\phi(z))\cdot \frac{\partial \phi_k}{\partial z_i}(z)\cdot\frac{\partial \phi_m}{\partial z_i}(z) + \sum_{k=1}^n \bigl(\partial_k u\bigr)(\phi(z))\cdot \frac{\partial^2 \phi_k}{\partial z_i^2}(z),$$

which can result in a fair amount of computation. If we collect terms, we can write somewhat more succinctly

$$\Delta (u\circ\phi)(z) = \sum_{k,m=1}^n \bigl(\partial_m\partial_k u\bigr)(\phi(z))\langle \nabla \phi_k(z)\mspace{-3mu}\mid\mspace{-3mu} \nabla \phi_m(z)\rangle + \sum_{k=1}^n \bigl(\partial_k u\bigr)(\phi(z))\bigl(\Delta \phi_k\bigr)(z).\tag{1}$$

For the general case $n \neq 2$, you additionally have a factor $k_n(z) = \dfrac{R^{n-2}}{\lVert z\rVert^{n-2}}$, which by the product rule leads to

$$\begin{align} \Delta v(z) &= \Delta \bigl( k_n(z)\cdot (u\circ\phi)(z)\bigr)\\ &= \bigl(\Delta k_n\bigr)(z)\cdot (u\circ\phi)(z) + 2 \langle \nabla k_n(z)\mspace{-3mu}\mid\mspace{-3mu} \nabla (u\circ\phi)(z)\rangle + k_n(z)\cdot \bigl(\Delta (u\circ\phi)\bigr)(z). \end{align}$$

The properties of $\phi$ and $k_n$ (the latter is harmonic in $\mathbb{R}^n\setminus\{0\}$, by the way) make the computation less ugly than it looks in abstract form, when organised in the right way, but it's still offering ample opportunities to make a mistake.

The case $n = 2$ is a bit simpler. If we have a bit of complex analysis at our disposal, it becomes simpler still. We have

$$\Delta = 4\frac{\partial^2}{\partial w\partial \overline{w}}$$

if we identify $\mathbb{R}^2$ with $\mathbb{C}$, and using that simplifies the computation a lot when the function one composes with is holomorphic - $\dfrac{\partial \psi}{\partial \overline{w}} \equiv 0$ - or antiholomorphic - $\dfrac{\partial\psi}{\partial w}\equiv 0$:

$$\begin{align} \frac{\partial}{\partial w}\frac{\partial (u\circ\psi)}{\partial\overline{w}}(w) &= \frac{\partial}{\partial w}\left( (\partial u)(\psi(w))\cdot \frac{\partial \psi}{\partial\overline{w}}(w) + (\overline{\partial}u)(\psi(w))\cdot\frac{\partial \overline{\psi}}{\partial \overline{w}}(w)\right)\\ &= \frac{\partial}{\partial w} \left((\overline{\partial} u)(\psi(w))\cdot \overline{\frac{\partial \psi}{\partial w}(w)}\right)\\ &= (\partial \overline{\partial} u)(\psi(w)) \left\lvert \frac{\partial \psi}{\partial w}(w)\right\rvert^2 \end{align}$$

for holomorphic $\psi$, since $\overline{\psi'}$ is antiholomorphic, and similar for antiholomorphic $\psi$. The end result in either case is

$$\Delta (u\circ\psi)(w) = \bigl(\Delta u\bigr)(\psi(w))\cdot \lvert \tilde{\psi}'(w)\rvert^2,$$

where $\tilde{\psi}$ is the holomorphic one of $\psi$ and $\overline{\psi}$, which shows that the composition of a harmonic function with a holomorphic or antiholomorphic function is again harmonic.

In our case, we have - identifying $\mathbb{R}^2$ with $\mathbb{C}$ -

$$\phi(w) = \frac{R^2}{\lvert w\rvert^2}\cdot w = \frac{R^2}{\overline{w}},$$

which is antiholomorphic, and thus the result, $v = u\circ \phi$ is harmonic for harmonic $u$.

Using real methods only, we have

$$\phi(x,y) = \left(\frac{R^2 x}{x^2+y^2},\, \frac{R^2 y}{x^2+y^2}\right),$$

and thus

$$\begin{gather} \frac{\partial \phi_1}{\partial x}(x,y) = \frac{R^2}{x^2+y^2} - \frac{2R^2x^2}{(x^2+y^2)^2} = \frac{R^2(y^2-x^2)}{(x^2+y^2)^2} = - \frac{\partial\phi_2}{\partial y}(x,y),\\ \frac{\partial \phi_1}{\partial y}(x,y) = -\frac{2R^2xy}{(x^2+y^2)^2} = \frac{\partial \phi_2}{\partial x}(x,y). \end{gather}$$

From these we obtain

$$\frac{\partial^2 \phi_1}{\partial x^2} = -\frac{\partial}{\partial x}\frac{\partial\phi_2}{\partial y} = - \frac{\partial}{\partial y}\frac{\partial\phi_2}{\partial x} = - \frac{\partial}{\partial y}\frac{\partial \phi_1}{\partial y} = - \frac{\partial^2 \phi_1}{\partial y^2},$$

so $\phi_1$ is harmonic, and similar for $\phi_2$, so the second sum in $(1)$ vanishes. For the first sum, we note that $\lVert \nabla \phi_1\rVert^2 = \lVert \nabla\phi_2\rVert^2$ by symmetry, and

$$\begin{align} \langle \nabla \phi_1\mspace{-3mu}\mid\mspace{-3mu} \nabla \phi_2\rangle &= \frac{\partial \phi_1}{\partial x}\frac{\partial \phi_2}{\partial x} + \frac{\partial \phi_1}{\partial y} \frac{\partial \phi_2}{\partial y}\\ &= \frac{\partial\phi_1}{\partial x}\frac{\partial\phi_1}{\partial y} + \frac{\partial\phi_1}{\partial y}\left(-\frac{\partial\phi_1}{\partial x}\right)\\ &= 0, \end{align}$$

so the first sum simplifies to $\bigl(\Delta u\bigr)(\phi(w))\lVert\phi_1(w)\rVert^2$, and thus the desired conclusion that $v = u\circ \phi$ is harmonic if $u$ is.

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  • $\begingroup$ Wow, thanks a lot! $\endgroup$ – math12 Dec 12 '13 at 15:14

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