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Show that a harmonic function satisfies the formal differential equation: $$\frac{\partial^2u}{\partial z \partial \overline{z}}=0$$

Question

How is $\frac{\partial u}{\partial z}$ even defined when $u$ isn't constant. It is clear that for any non-constant complex variabled, real valued, analytic function $u$ we have that $\frac{\partial u}{\partial z}$ is not defined. This comes from Cauchy-Riemann equations. So how does this question make sense?

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  • $\begingroup$ This might help. $\endgroup$ – mattos Sep 26 '17 at 5:19
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    $\begingroup$ They are defined, but maybe not the way you think. By definition, $$\eqalign{\dfrac{\partial u}{\partial z} &= \frac{1}{2} \left(\dfrac{\partial u}{\partial x} - i \dfrac{\partial u}{\partial y}\right)\cr \dfrac{\partial u}{\partial \overline{z}} &= \frac{1}{2} \left(\dfrac{\partial u}{\partial x} + i \dfrac{\partial u}{\partial y}\right)}$$ $\endgroup$ – Robert Israel Sep 26 '17 at 5:46
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Using the notation that Robert Israel mentioned you can restate the equation as

$$\dfrac{\partial^2u}{\partial z \partial \bar{z}}=\frac{1}{4}\left[\frac{\partial}{\partial x}-i\dfrac{\partial}{\partial y} \right]\left[\frac{\partial u}{\partial x}+i\dfrac{\partial u}{\partial y} \right]=\frac{1}{4}\left[\dfrac{\partial^2u}{\partial x^2}-i\dfrac{\partial^2u}{\partial y\partial x}+i\dfrac{\partial^2u}{\partial x\partial y}+\dfrac{\partial^2u}{\partial y^2}\right].$$

Can you complete it from here?

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