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Is it true or not : if $u(z)$ is harmonic, $u(\overline{z})$ is also harmonic.

My try :

$u(z)=u(x,y)$ is harmonic Define $s=-y$

Let $U := u(\overline{z})=u(x,-y)=u(x,s)$ : $$\frac{\partial U}{\partial x}=\frac{\partial u}{\partial x} \Rightarrow \frac{\partial^2 U}{\partial x^2}=\frac{\partial^2 u}{\partial x^2}$$ And $$\frac{\partial U}{\partial y}=\frac{\partial u}{\partial x} \frac{\partial x}{\partial y} + \frac{\partial u}{\partial s} \frac{\partial s}{\partial y} = - \frac{\partial u}{\partial s} $$ Similarly $$\frac{\partial^2 U}{\partial y^2}= \left[ \frac{\partial }{\partial y} ( \frac{\partial U}{\partial y}) \right] =- \left[ \frac{\partial }{\partial y} ( \frac{\partial u}{\partial s}) \right] = ... = - \left[ - \frac{\partial }{\partial s} ( \frac{\partial u}{\partial s}) \right] = \frac{\partial^2 u}{\partial s^2} $$ Hence $$\frac{\partial^2 U}{\partial x^2} + \frac{\partial^2 U}{\partial y^2} = \frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2 u}{\partial s^2}= 0 \ \ \ (*)$$

Therefore $u(\overline{z})$ is also harmonic.

*My Questions is : Is my try problematic?Does $(*)$ needs justification? *

the8thone
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  • @John Your Example is a good one if we were talking about analyticity of $u(\overline{z})$, am I right ?! – the8thone Dec 19 '13 at 02:22
  • I don't see anything wrong upon a first reading. You also may want to check out the answer to this question for verification: http://math.stackexchange.com/questions/101224/why-are-uz-and-u-barz-simultaneously-harmonic?rq=1 – tylerc0816 Dec 19 '13 at 02:52
  • @tylerc0816 Thanks for your comment , My problem with my proof is kind of strange, I am worrying because in $(*)$, I have $\frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2 u}{\partial s^2}$ NOT $ \frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2 u}{\partial y^2}$ ..... – the8thone Dec 19 '13 at 03:09
  • It has been already discussed at http://math.stackexchange.com/questions/101224/why-are-uz-and-u-barz-simultaneously-harmonic?rq=1 and http://math.stackexchange.com/questions/474972/uz-harmonic-if-and-only-if-u-overlinez-harmonic. But the recent answers are better. – Supriyo Dec 28 '13 at 16:38

2 Answers2

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Let $f(z)$ be a function which is analytic on the given domain and such that $\mbox{Re}(f)=u$. Then $\overline{f(\bar{z})}$ is Analytic and $\mbox{Re}(\overline{f(\bar{z})})=u(\bar{z})$. Hence, $u(\bar{z})$ is harmonic.

The proof that $\overline{f(\bar{z})}$ is Analytic is simple: it follows immediately from the definition of the derivative that if $f$ is differentiable at $z_0$ then $\overline{f(\bar{z})}$ is differentiable at $\bar{z_0}$.

N. S.
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  • to extend a harmonic function $u$ to an analytic function $f$ whose real part is $u$ , we need the domain be simply connected , right ? If that's the case, your proof does not work in general. – the8thone Dec 19 '13 at 03:51
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    @Roozbeh-unity From what I remember, you don't need simply connected. And even if it is needed, analyticity is a local property. So for every $z_0 \in D$ you can pick a simply connected open neighborhood $U$ of $z_0$ and do the above proof in that neighborhood. This way you get that $u(\bar{z})$ is locally harmonic, but this is the same as being harmonic. – N. S. Dec 19 '13 at 03:56
  • Simply connected domains are the only ones where every (real-valued) harmonic function is the real part of a holomorphic function [$\log \lvert z-a\rvert$ for $a$ in a "hole"]. But the local argument is all we need, of course. – Daniel Fischer Dec 19 '13 at 09:46
2

A funciton $\phi$is harmonic if

$$ \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} = 0$$

It can be shown that $$ \frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2} = \frac{\partial}{\partial z}\frac{\partial}{\partial \bar{z}}$$

This means for a function of a complex variable to be harmonic it needs to be an analytic function of just $z$ or an analytic function of just $\bar{z}$. Analytic so that the derivative exists and a function of just one or the other so that the result is zero.

This means that the real part and the imaginary part of said analytic function will each be harmonic functions

The following is true only if $f(z)$ is real when $z$ is real

Note that many cases nothing new is gained in making it a function of $\bar{z}$ since for certain analytic functions this is equivalent to taking the complex conjugate.

$$ f(\bar{z}) = \overline{f(z)} = \overline{u(x,y)+iv(x,y)} = u(x,y)-iv(x,y) $$

So that we don't really get fundamentally new solutions to Laplace's equation.

To obtain the first equality note that when a function is analytic within some neighborhood of a point there is a Taylor series which converges to the function in within some open set containing that point. This allows us to write,

$$f(z) = \sum_{n=0}^N a_n z^n + R_N(z)$$

Where $R_N \rightarrow 0$ as $N \rightarrow \infty$

$$ \overline{ f(z) } = \overline{\sum_{n=0}^N a_n z^n + R_N} = \overline{\sum_{n=0}^N a_n z^n} + \overline{R_N} = \sum_{n=0}^N \overline{a_n} (\overline{z})^n + \overline{R_N}$$

Since $\vert \overline{R_N} \vert = \vert R_N \vert$ we can conclude that $\overline{R_N} \rightarrow 0 $ whenever $R_N$ does. If the coefficientes $a_n$ are real then we can write $\overline{a_n} = a_n$ and conclude that,

$$ \overline{f(z)} = \sum_{n=0}^\infty a_n (\overline{z})^n = f(\bar{z})$$

Spencer
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  • Also I should point out that I agree with your proof, just thought I would share this because I found it very enlightening when I first saw it. – Spencer Dec 19 '13 at 03:04
  • Indeed it is ! Can you please justify that for analytic functions $f(\overline{z})=\overline{f(z)}$. – the8thone Dec 19 '13 at 03:05
  • My problem with my proof is kind of strange, I am worrying because in $(*)$, I have $\frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2 u}{\partial s^2}$ NOT $ \frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2 u}{\partial y^2}$ – the8thone Dec 19 '13 at 03:08
  • I'm fairly sure through an application of the chain rule that you can show that the second derivative with respect to $y$ is the same as the second derivative with respect to $-y$. – Spencer Dec 19 '13 at 03:25
  • In your proof you are using the fact that the Cauchy Riemann equations can be written as $\frac{\partial f}{\partial \overline{z}} = 0 $, Am I right? – the8thone Dec 19 '13 at 03:47
  • I suppose that you can think of it that way. My thinking was that any function of $(x,y)$ can be equivalently thought of as a function of $(z,\bar{z})$ since $x=(z+\bar{z})/2$ and $y=(z-\bar{z})/2i$ therefore $z$ and $\bar{z}$ could be thought of as independent variables. I think your interpretation is more rigorous. – Spencer Dec 19 '13 at 04:56
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    @Roozbeh-unity : it is not true that for all analytic functions $f$, $f({\overline z}) = {\overline {f(z)}}$ for all $z$. Consider $f(z)=iz$. If this claim is a crucial part of the above answer, then the answer has problems. The claim is true if $f$ is analytic and $f(x)$ is real for all real $x$. – Stefan Smith Dec 19 '13 at 05:37
  • This is true, I neglected to take the compllex conjugate of my coefficients. Its not crucial to the main portion of my answer but a comment I made after that. Thanks for pointing this out Stefan. – Spencer Dec 19 '13 at 15:33