Suppose $f:\mathbb{C}\to\mathbb{C}$ is an analytic function and $f:=u+iv$. Then is it always true that $u_{xx}+u_{yy}=v_{xx}+v_{yy}=0$ ?
A doubt about analytic functions
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complex-analysis
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1You know that analytic functions satisfy the Cauchy-Riemann equations, right? Take those and try to proceed from there. – 2012-08-20
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0But for that we need $v_{xy}$ and $v_{yx}$ to be continous, is that always true ? – 2012-08-20
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3An analytic function is a power series, and thus its derivatives are all smooth. – 2012-08-20
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0I got the answer. In wikipedia I saw that analytic functions are infinitely differentiable. Thanks for all your comments. – 2012-08-20
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0@pritam, we need to assume $u_{xy}=u_{yx}$ and $v_{xy}=v_{yx}$, right? I'd like to know in what condition $f_{xy}≠f_{yx}$ and when they coincide. – 2012-08-20
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0@lab $f_{xy}=f_{yx}$ when they are both continous. See here :http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives#Clairaut.27s_theorem – 2012-08-20
1 Answers
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Analytic function satisfies Cauchy Riemann equation.
i.e.
$u_{x}=v_{y}$ & $u_{y}=-v_{x}$
so, $u_{xx}=v_{xy}$ & $u_{yy}=-v_{yx}$
so, $u_{xx}+u_{yy}=0\ when\ v_{xy}=v_{yx}$