Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

2 Answers 2

up vote 3 down vote accepted

Let us consider the Cauchy Riemann conditions

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

and

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

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

share|improve this answer
    
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
add comment

Hint: See Cauchy–Riemann equations.

share|improve this answer
    
This is true since v satisfies laplace equation. Right? –  ddabir Jul 12 '13 at 18:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.