# If $u$ is harmonic, then $u\left(\frac{y_1}{|y|^2}, -\frac{y_2}{|y|^2}\right)$ is also harmonic

Let $$u$$ harmonic in $$\Omega=\{x\in \mathbb{R}^2:|x|>1\}$$. Show that $$v(y) = u\left(\frac{y_1}{|y|^2}, -\frac{y_2}{|y|^2}\right)$$ is harmonic in $$U=\{y\in\mathbb{R}^2: 0<|y|<1\}$$. Suggestion: to simplify the calculations, define $$g_1 = y_1|y|^{-2}, g_2 = -y_2|y|^{-2}$$ and show that, first, $$(g_1)_{y_1} = (g_2)_{y_2}$$ and also that $$(g_1)_{y_2} = -(g_2)_{y_1}$$.

I guess $$(g_1)_{y_1}$$ is the partial of $$g_1$$ with respect to $$y_1$$ and so on.

Recall that

$$p = \frac{1}{|y|^2} = \frac{1}{y_1^2 + y_2^2} \implies \frac{\partial p}{\partial y_1}= -1\frac{1}{(y_1^2 + y_2^2)^2}2y_1 = -1\frac{1}{|y|^4}2y_1,$$

$$\frac{\partial g_1}{\partial y_1} = |y|^{-2} -1y_1\frac{1}{|y|^4}2y_1,$$

$$\frac{\partial g_1}{\partial y_1} = |y|^{-2} -1y_2\frac{1}{|y|^4}2y_1,$$

which are not the same. What is wrong?

Suppose I did carry the calculations right. Then I must show that $$\Delta v(y) = 0$$ for $$y$$ with $$0<|y|<1$$.

$$\frac{\partial v}{\partial y_i} = \partial_1 u\frac{\partial g_1}{\partial y_i} + \partial_2 u\frac{\partial g_2}{\partial y_i},$$

$$\frac{\partial^2 v}{\partial y_i^2} = \partial_1^2 u\frac{\partial g_1}{\partial y_i} + \partial_1 u\frac{\partial^2 g_1}{\partial y_i^2} + \partial_2^2 u\frac{\partial g_2}{\partial y_i} + \partial_2 u\frac{\partial^2 g_2}{\partial y_i^2},$$

so

$$\Delta v(y) = \partial_1^2 u\frac{\partial g_1}{\partial y_1} + \partial_1 u\frac{\partial^2 g_1}{\partial y_1^2} + \partial_2^2 u\frac{\partial g_2}{\partial y_1} + \partial_2 u\frac{\partial^2 g_2}{\partial y_1^2} + \\ \partial_1^2 u\frac{\partial g_1}{\partial y_2} + \partial_1 u\frac{\partial^2 g_1}{\partial y_2^2} + \partial_2^2 u\frac{\partial g_2}{\partial y_2} + \partial_2 u\frac{\partial^2 g_2}{\partial y_2^2}=\\(\partial_1^2 u + \partial_2^2 u)(g_1)_{y_1} + (-\partial_1^2 u + \partial_2^2 u)(g_1)_{y_1} + \\ \partial_1 u\frac{\partial^2 g_1}{\partial y_1^2} + \partial_2 u\frac{\partial^2 g_2}{\partial y_1^2} + \partial_1 u\frac{\partial^2 g_1}{\partial y_2^2} + \partial_2 u\frac{\partial^2 g_2}{\partial y_2^2},$$

which must be $$0$$. I do not think I should take the second derivatives of $$g_1$$ and $$g_2$$, because it would be huge. Maybe the conditions $$(g_1)_{y_1} = (g_2)_{y_2}$$, $$(g_1)_{y_2} = -(g_2)_{y_1}$$ imply something about the second derivatives that I can use. I do not know. The only thing I can see being $$0$$ is $$(\partial_1^2 u + \partial_2^2 u)(g_1)_{y_1}$$, because $$u$$ is harmonic.

• I get $$(g_{1})_{y_{1}} = (g_{2})_{y_{2}} = \frac{y_{2}^{2} - y_{1}^{2}}{\lvert y \rvert^{4}}$$ To see \begin{align} g_{1}&=\frac{y_{1}}{y_{1}^{2}+y_{2}^{2}} \\ \implies (g_{1})_{y_{1}}&=\frac{1}{y_{1}^{2}+y_{2}^{2}}+y_{1} \cdot (-1)(2y_{1})\frac{1}{(y_{1}^{2}+y_{2}^{2})^{2}} \\ &=\frac{1}{| y |^{2}}-\frac{2y_{1}^{2}}{| y |^{4}} \\ &=\frac{| y |^{2}-2y_{1}^{2}}{| y |^{4}} \\ &=\frac{y_{2}^{2}-y_{1}^{2}}{| y |^{4}} \end{align} Similarly for $(g_{2})_{y_{2}}$. Also, the equality of derivatives is just the Cauchy-Riemann equations. – mattos Sep 19 '18 at 12:13

﻿$$\def\peq{\mathrel{\phantom{=}}{}}$$First,\begin{align*} \frac{∂g_1}{∂y_1}(y) &= \frac{1}{y_1^2 + y_2^2} + y_1 \cdot \left( -\frac{1}{(y_1^2 + y_2^2)^2} \right) \cdot 2y_1 = \frac{y_2^2 - y_1^2}{(y_1^2 + y_2^2)^2},\\ \frac{∂g_2}{∂y_2}(y) &= -\frac{1}{y_1^2 + y_2^2} - y_1 \cdot \left( -\frac{1}{(y_1^2 + y_2^2)^2} \right) \cdot 2y_1 = \frac{y_2^2 - y_1^2}{(y_1^2 + y_2^2)^2}, \end{align*}\begin{align*} \frac{∂g_1}{∂y_2}(y) &= y_1 \cdot \left( -\frac{1}{(y_1^2 + y_2^2)^2} \right) \cdot 2y_2 = -\frac{2y_1 y_2}{(y_1^2 + y_2^2)^2},\\ \frac{∂g_2}{∂y_1}(y) &= -y_2 \cdot \left( -\frac{1}{(y_1^2 + y_2^2)^2} \right) \cdot 2y_1 = \frac{2y_1 y_2}{(y_1^2 + y_2^2)^2}, \end{align*} thus $$f_1 := \dfrac{∂g_1}{∂y_1} = \dfrac{∂g_2}{∂y_2}$$, $$f_2 := \dfrac{∂g_1}{∂y_2} = -\dfrac{∂g_2}{∂y_1}$$.
Next, note that$$h_1 := \frac{∂f_1}{∂y_1} = \frac{∂^2 g_2}{∂y_1 ∂y_2} = -\frac{∂f_2}{∂y_2},\quad h_2 := \frac{∂f_1}{∂y_2} = \frac{∂^2 g_1}{∂y_1 ∂y_2} = \frac{∂f_2}{∂y_1},$$ then\begin{align*} \frac{∂v}{∂y_1}(y) &= \frac{∂u}{∂x_1}(g_1(y), g_2(y)) f_1(y) - \frac{∂u}{∂x_2}(g_1(y), g_2(y)) f_2(y),\\ \frac{∂v}{∂y_2}(y) &= \frac{∂u}{∂x_1}(g_1(y), g_2(y)) f_2(y) + \frac{∂u}{∂x_2}(g_1(y), g_2(y)) f_1(y), \end{align*}\begin{align*} \frac{∂^2 v}{∂y_1^2}(y) &= \left( \frac{∂^2 u}{∂x_1^2}(g_1(y), g_2(y)) f_1(y) - \frac{∂^2 u}{∂x_1 ∂x_2}(g_1(y), g_2(y)) f_2(y) \right) f_1(y) + \frac{∂u}{∂x_1}(g_1(y), g_2(y)) h_1(y)\\ &\peq - \left( \frac{∂^2 u}{∂x_1 ∂x_2}(g_1(y), g_2(y)) f_1(y) - \frac{∂^2 u}{∂x_2^2}(g_1(y), g_2(y)) f_2(y) \right) f_2(y) - \frac{∂u}{∂x_1}(g_1(y), g_2(y)) h_2(y)\\ &= \frac{∂^2 u}{∂x_1^2}(g_1(y), g_2(y)) f_1^2(y) + \frac{∂^2 u}{∂x_2^2}(g_1(y), g_2(y)) f_2^2(y) - 2\frac{∂^2 u}{∂x_1 ∂x_2}(g_1(y), g_2(y)) f_1(y) f_2(y)\\ &\peq + \frac{∂u}{∂x_1}(g_1(y), g_2(y)) h_1(y) - \frac{∂u}{∂x_1}(g_1(y), g_2(y)) h_2(y), \end{align*}\begin{align*} \frac{∂^2 v}{∂y_2^2}(y) &= \left( \frac{∂^2 u}{∂x_1^2}(g_1(y), g_2(y)) f_2(y) + \frac{∂^2 u}{∂x_1 ∂x_2}(g_1(y), g_2(y)) f_1(y) \right) f_2(y) - \frac{∂u}{∂x_1}(g_1(y), g_2(y)) h_1(y)\\ &\peq + \left( \frac{∂^2 u}{∂x_1 ∂x_2}(g_1(y), g_2(y)) f_2(y) + \frac{∂^2 u}{∂x_2^2}(g_1(y), g_2(y)) f_1(y) \right) f_1(y) + \frac{∂u}{∂x_1}(g_1(y), g_2(y)) h_2(y)\\ &= \frac{∂^2 u}{∂x_1^2}(g_1(y), g_2(y)) f_2^2(y) + \frac{∂^2 u}{∂x_2^2}(g_1(y), g_2(y)) f_1^2(y) - 2\frac{∂^2 u}{∂x_1 ∂x_2}(g_1(y), g_2(y)) f_1(y) f_2(y)\\ &\peq - \frac{∂u}{∂x_1}(g_1(y), g_2(y)) h_1(y) + \frac{∂u}{∂x_1}(g_1(y), g_2(y)) h_2(y), \end{align*} which implies\begin{align*} Δv(y) &= \frac{∂^2 u}{∂x_1^2}(g_1(y), g_2(y)) f_1^2(y) + \frac{∂^2 u}{∂x_2^2}(g_1(y), g_2(y)) f_2^2(y)\\ &\peq + \frac{∂^2 u}{∂x_1^2}(g_1(y), g_2(y)) f_2^2(y) + \frac{∂^2 u}{∂x_2^2}(g_1(y), g_2(y)) f_1^2(y)\\ &= \left( \frac{∂^2 u}{∂x_1^2}(g_1(y), g_2(y)) + \frac{∂^2 u}{∂x_2^2}(g_1(y), g_2(y)) \right)(f_1^2(y) + f_2^2(y))\\ &= Δu(g_1(y), g_2(y))(f_1^2(y) + f_2^2(y)) = 0. \end{align*}
Let $\tilde{u}$ be the harmonic conjugate of $u$, so that $u(z)+i\tilde{u}(z)$ is holomorphic on $\{|z| > 1\}$. Then $v(z)+i\tilde{u}(\frac{\overline{z}}{|z|^2}) = u(\frac{\overline{z}}{|z|^2})+i\tilde{u}(\frac{\overline{z}}{|z|^2})$ is antiholomorphic on $\{|z| > 1\}$ and thus $v$ is harmonic on $\{|z| > 1\}$.
Is your new $$u$$ even well-defined?! Because the division by $$|y|^2$$ takes a point to the perimeter of the unit circle, where the original $$u$$ is not assigned a value, at least immediately.