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I am unsure of some notation and also how a particular identity was derived. I read in a paper that because $\Delta = 4 \partial \bar \partial$ we have

$$ I = \int_\Omega \frac{1}{z-\zeta}\Delta u(\zeta) d\zeta = -2i \int_{\partial \Omega} \frac{1}{z-\zeta}\partial u(\zeta) d\zeta. $$

First of all, what is $\partial \bar \partial$ and how is the Laplacian $\Delta = 4 \partial \bar \partial$?

Secondly, how has the identity been derived..it seems to be due to some integration by parts type formula, but what exactly?

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  • $\begingroup$ It's a simple application of Gauß Theorem. $\endgroup$ – Von Neumann Nov 15 '18 at 10:06
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Identifying the complex plane with $\mathbb R^2$ with $z=x+iy$ the definition of $\partial$ and $\bar{\partial}$ is $$\partial=\frac{1}{2} \left(\partial_x+i\partial_y \right)$$ $$\bar{\partial}=\frac{1}{2} \left(\partial_x-i\partial_y \right).$$ Then you indeed have $$4\partial \bar{\partial}= \partial_x^2+\partial_y^2 = \Delta$$


Note that you can avoid using complex number as in $\mathbb R^2$ a formulation can be: $$\partial=\frac{1}{2} \begin{pmatrix} \partial_x \\ \partial_y\end{pmatrix}=\frac {1}{2}\nabla$$ $$\bar{\partial}=\frac{1}{2} \begin{pmatrix} \partial_x \\ -\partial_y\end{pmatrix}=\frac {1}{2} J \nabla^\perp=\frac{1}{2} J \operatorname{curl}$$ with $ J =\begin{pmatrix} 0&-1\\1&0\end{pmatrix}$ correspond to $i$.


In these terms Green's theorem can be rewritten as $$\int_{\partial \Omega} g(\zeta) d \zeta = \int_\Omega 2i \bar{\partial} g(\zeta) d \zeta$$

We use this formula with $g(\zeta)=\frac{1}{z-\zeta} \partial u(\zeta)$.

As $\zeta \mapsto \frac{1}{z-\zeta}$ is holomorphic you have $\bar{\partial} \frac{1}{z-\zeta}=0$ so $$\bar{\partial} \left( \frac{1}{z-\zeta} \partial u(\zeta) \right)=\frac{1}{z-\zeta} \bar{\partial} \partial u(\zeta)=\frac{1}{z-\zeta}\frac{1}{4} \Delta u(\zeta)$$ from where you obtain: $$\int_{\partial \Omega} \frac{1}{z-\zeta} \partial u(\zeta) d \zeta= \frac{2i}{4} \int_\Omega \frac{1}{z-\zeta} \Delta u(\zeta) d\zeta$$ whic is the claimed result as $(i/2)^{-1}=-2i$.

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  • $\begingroup$ Thanks..it seems there might be a sign error in the paper as it says that $4 \hat \partial \partial = \Delta$ which is the opposite to what you have. $\endgroup$ – sonicboom Nov 29 '18 at 9:18
  • $\begingroup$ There is several small mistakes in my answer but it seems that the sign is the same. I will edit it to correct these mistakes to finish the computation. $\endgroup$ – Delta-u Nov 29 '18 at 12:23

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