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 $0<|y|<1$.

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

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


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

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.

  • $\begingroup$ 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. $\endgroup$ – 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*}

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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 $\{|z| > 1\}$.

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  • 1
    $\begingroup$ Thank you so much for your answer, but I'm not allowed to use any results about complex analysis. $\endgroup$ – Paprika Sep 18 '18 at 20:28

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.

Secondly, consider the mean/average property of harmonic maps.

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