Let $P$ be a harmonic function
show that : $$\Delta P = 4\frac{\partial^2 P}{\partial z\partial \overline{z}}$$
I have no idea how to even start, please if someone can push me in the right way that would be great.
Harmonic function 2
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harmonic-functions
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1What is the domain and codomain of $P$? – 2017-02-28
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0the domain is $D$ (it is an open subset of $\mathbb{R}^2$) and the codomain is $\mathbb{R}$ – 2017-02-28
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0If P is harmonic, then it also satisfies the Cauchy-Riemann equations. By taking the appropriate derivates and doing some algebra you should get the conclusion. – 2017-02-28
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0Do you know the definition of $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial\bar{z}}$ in terms of $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$? Also, do you know the definition of $\Delta$? – 2017-02-28
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0@ Michael Albanese yes and yes the thing that is bothering me is $\partial z\partial \overline{z}$ – 2017-02-28
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0$$\frac{\partial^2 P}{\partial z\partial\bar{z}} = \frac{\partial}{\partial z}\left(\frac{\partial P}{\partial\bar{z}}\right)$$ – 2017-02-28
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0Compute $\frac{\partial}{\partial \bar{z}}\left(\frac{\partial P}{\partial z}\right).$ – 2017-02-28
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0@ Michael Albanese it makes much more sense to me now so I should be able to solve it thank you for shedding light – 2017-02-28
1 Answers
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$$ x = \frac{z+\bar{z}}{2}, \quad y = \frac{z-\bar{z}}{2i}. $$ $$ \begin{align} \frac{\partial P}{\partial \bar{z}} &= \frac{\partial x}{\partial \bar{z}} \frac{\partial P}{\partial x} + \frac{\partial y}{\partial \bar{z}} \frac{\partial P}{\partial y} = \frac{1}{2} \frac{\partial P}{\partial x} + \frac{i}{2} \frac{\partial P}{\partial y} \\ \frac{\partial^2 P}{\partial z \partial \bar{z}} &= \frac{\partial x}{\partial z} \left(\frac{1}{2} \frac{\partial^2 P}{\partial x^2} - \frac{i}{2} \frac{\partial^2 P}{\partial x \partial y} \right) + \frac{\partial y}{\partial z} \left(\frac{1}{2} \frac{\partial^2 P}{\partial y\partial x} - \frac{i}{2} \frac{\partial^2 P}{\partial y^2} \right) \\ &= \frac{1}{4} \left( \frac{\partial^2 P}{\partial x^2} - i \frac{\partial^2 P}{\partial x \partial y} \right) + \frac{i}{4} \left( \frac{\partial^2 P}{\partial y\partial x} - i \frac{\partial^2 P}{\partial y^2} \right) \\ &= \frac{1}{4} \Delta P + \frac{i}{4} \left( \frac{\partial^2 P}{\partial y\partial x} - \frac{\partial^2 P}{\partial x\partial y} \right), \end{align} $$ and the second term vanishes since $P$ is twice-differentiable.