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Let $f=u+iv:\mathbb C\to\mathbb C$ be analytic. Then is it true that $\dfrac{\delta^2 v}{\delta x^2}+\dfrac{\delta^2 v}{\delta y^2}=0?$

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up vote 3 down vote accepted

Let us consider the Cauchy Riemann conditions

$\frac {\partial u} {\partial x}$ = $\frac {\partial v} {\partial y}$


$\frac {\partial u} {\partial y}$=-$\frac {\partial v} {\partial x}$

So you are asking if $\dfrac{\delta^2 v}{\delta x^2}+\dfrac{\delta^2 v}{\delta y^2}=0$

Lets find $\dfrac{\delta^2 v}{\delta x^2}$

$\frac \partial {\partial x}$($\frac {\partial v} {\partial x}) $= $\frac \partial {\partial x}$(-$\frac {\partial u} {\partial y}$) = -$\frac {\partial u} {\partial x \partial y}$

Nowlets find $\dfrac{\delta^2 v}{\delta y^2}$

$\frac \partial {\partial y}$($\frac {\partial v} {\partial y}) $= $\frac \partial {\partial y}$($\frac {\partial u} {\partial x}$) = $\frac {\partial u} {\partial y \partial x}$

So we end up with

$\frac {\partial u} {\partial y \partial x}$-$\frac {\partial u} {\partial x \partial y}$=0 (mixed derivatives are equal)

So your statement is correct

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You think I can not work it out for him. I left it to him to do some work. – Mhenni Benghorbal Jul 13 '13 at 3:21

Hint: See Cauchy–Riemann equations.

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This is true since v satisfies laplace equation. Right? – ddabir Jul 12 '13 at 18:52

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