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Two questions, one on harmonic functions

nacho

Active member
Sep 10, 2013
156
Could I get some hints on how to evaluate these question.
The question asking to find where $f(re^{i\theta})$ is differentiable doesn't seem to involved,
however would I use C-R equations, or would it just be for wherever $r \neq 0$. Although that is given in the domain, so I'm assuming they want us to use C-R equations.
If i were to use C-R equations, then I would have to convert to cartesian coordinates, correct?
How would I do that.


As for the second question,
could I get any hints on what theorems may be relevant. I am not sure how to approach this.

Thanks
 

Attachments

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
In the first question you are asked to find the domain of differentiability of the function

\(\displaystyle f(z)=\log(z^2) \)

Hint : choose a suitable branch cut .
 

chisigma

Well-known member
Feb 13, 2012
1,704
Could I get some hints on how to evaluate these question.
The question asking to find where $f(re^{i\theta})$ is differentiable doesn't seem to involved,
however would I use C-R equations, or would it just be for wherever $r \neq 0$. Although that is given in the domain, so I'm assuming they want us to use C-R equations.
If i were to use C-R equations, then I would have to convert to cartesian coordinates, correct?
How would I do that.


As for the second question,
could I get any hints on what theorems may be relevant. I am not sure how to approach this.

Thanks
Question 1: with a little of patience You can derive the C-R equations in polar coordinates...

$\displaystyle \frac{\partial{u}}{\partial{r}} = \frac{1}{r}\ \frac {\partial {v}}{\partial{\theta}} $

$\displaystyle \frac{\partial{v}}{\partial{r}} = - \frac{1}{r}\ \frac {\partial {u}}{\partial{\theta}} \ (1)$

Here $\displaystyle u(r, \theta) = \ln r^{2}$ and $v(r, \theta) = 2\ \theta$, so that is...

$\displaystyle \frac{\partial{u}}{\partial{r}} = \frac{2}{r},\ \frac{\partial{u}}{\partial{\theta}} = 0,\ \frac{\partial{v}}{\partial{r}} = 0,\ \frac{\partial{v}}{\partial{\theta}} = 2$ so that the (1) are satisfied everywhere with the only exception of the point r = 0...


Kind regards


$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Could I get some hints on how to evaluate these question.
The question asking to find where $f(re^{i\theta})$ is differentiable doesn't seem to involved,
however would I use C-R equations, or would it just be for wherever $r \neq 0$. Although that is given in the domain, so I'm assuming they want us to use C-R equations.
If i were to use C-R equations, then I would have to convert to cartesian coordinates, correct?
How would I do that.


As for the second question,
could I get any hints on what theorems may be relevant. I am not sure how to approach this.

Thanks
Question 2: a function u(x,y) is said to be harmonic if its second derivatives are continous and is...

$\displaystyle u_{xx} = - u_{yy}\ (1)$

An important theorem extablishes that if u is harmonic in a simply connected domain G, then it exist a function v(x,y) so that f(x + i y) = u(x,y) + i v(x,y) is analytic in G [v is said to be the harmonic coniugate of u...].

If we suppose that G is a disk centered in w with radious r and call $\gamma$ its contour, the the Cauchy Integral Formula extablishes that...

$\displaystyle f(w) = \frac{1}{2\ \pi\ i}\ \int_{\gamma} \frac{f (z)}{z - w}\ d z\ (2)$

With the substitution $z - w = r\ e^{i\ \theta}$ the (2) becomes...


$\displaystyle f(w) = \frac{1}{2\ \pi}\ \int_{0}^{2\ \pi} f(w + r\ e^{i\ \theta})\ d \theta\ (3)$

... and, taking the real part of f we have...

$\displaystyle u(w) = \frac{1}{2\ \pi}\ \int_{0}^{2\ \pi} u (w + r\ e^{i\ \theta})\ d \theta\ (4)$

Kind regards

$\chi$ $\sigma$