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Let $P : D\subset \mathbb{R}^2 \to \mathbb{R}$ be a harmonic function

show in both cases that $Q$ is harmonic

first case $Q = x\frac{\partial P}{\partial x}+y\frac{\partial P}{\partial y}$ $$ \Delta Q = \frac{\partial^2 Q}{\partial x^2}+\frac{\partial^2 Q}{\partial y^2}$$

$$=\frac{\partial}{\partial x} (\frac{\partial P}{\partial x})+\frac{\partial}{\partial y} (\frac{\partial P}{\partial y}) = \frac{\partial^2 P}{\partial x^2}+\frac{\partial^2 P}{\partial y^2} = \Delta P = 0 \implies Q\; \text{is harmonic}$$ please check my notations I'm not familiar with partial derivatives and I wanna make sure I wrote everything correctly.

second case : $Q = y\frac{\partial P}{\partial x}-x\frac{\partial P}{\partial y}$

$$ \Delta Q = \frac{\partial^2 Q}{\partial x^2}+\frac{\partial^2 Q}{\partial y^2}$$ $$= \frac{\partial}{\partial x} (-\frac{\partial P}{\partial y})+\frac{\partial}{\partial y} (\frac{\partial P}{\partial x}) = -\frac{\partial^2 P}{\partial x \partial y} + \frac{\partial^2 P}{\partial y \partial x} $$ since P is a harmonic function it's twice continuously differentiable which means I can use Schwarz's theorem so $\Delta Q = 0 \implies Q$ is harmonic.

this is my first time dealing with harmonic functions Am I doing it right ? especially the notations.

the_firehawk
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1 Answers1

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I don't think so, no. Here's an example with $Q$ as in the first case.

$$\frac{\partial}{\partial x}Q= \frac{\partial}{\partial x}\left(x\frac{\partial P}{\partial x}+y\frac{\partial P}{\partial y}\right)= \frac{\partial}{\partial x}\left(x\frac{\partial P}{\partial x}\right)+\frac{\partial}{\partial x}\left(y\frac{\partial P}{\partial y}\right)\\ =\frac{\partial P}{\partial x}+x\frac{\partial^2 P}{\partial x^2}+y\frac{\partial^2 P}{\partial x\partial y}$$

Here, we have used that $(f+g)'=f'+g'$ and that $(fg)'=f'g+fg'$. Do you think you can finish it on your own now, bearing in mind this example and the fact that

$$\frac{\partial^2}{\partial x^2}Q=\frac{\partial}{\partial x}\left(\frac{\partial}{\partial x}Q\right)\text{?}$$

Fimpellizzeri
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  • @Fimpelliizieri thank you for the enlightenment. now I should have no problem finishing the computations. – the_firehawk Mar 07 '17 at 21:33
  • Glad to have helped! – Fimpellizzeri Mar 07 '17 at 21:43
  • I've found that $\Delta Q= x\frac{\partial^3 P}{\partial x^3}+y\frac{\partial^3 P}{\partial y^3} + x\frac{\partial^3 P}{\partial y^2 \partial x}+y\frac{\partial^3 P}{\partial x^2 \partial y}$

    is that correct ? if yes how can I continue ?

     – the_firehawk Mar 07 '17 at 21:59
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    Notice that $x\frac{\partial^3}{\partial x^3}P+x\frac{\partial^3}{\partial y^3}P=x\cdot\frac{\partial}{\partial x}\left(\frac{\partial^2}{\partial x^2}P+\frac{\partial^2}{\partial y^2}P\right)$. Since $P$ is harmonic, what can you conclude? Can you do something similar for the other factor? – Fimpellizzeri Mar 07 '17 at 22:55