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I noticed this question while reading several pdfs of lecture notes, and I'm having trouble figuring it out. Can anyone help?

If $f$ is a harmonic function in a domain $D \subset \mathbb{C}$, and $g$ is a conformal mapping of a domain $D_0$ onto $D$, is $f \circ g$ harmonic in $D_0$?

Thank you so much!

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3 Answers

up vote 10 down vote accepted

The answer is yes, and we only need $g$ to be holomorphic. One can prove this by directly computing the Laplacian of $f\circ g$ using the Chain Rule. I'd rather use $z$ and $\bar z$ than $x$ and $y$ for this purpose.

$$\Delta(f\circ g)=\frac{1}{4}(f\circ g)_{z\bar z} = \frac{1}{4}[(f_z\circ g) g']_{\bar z} = \frac{1}{4}(f_{z\bar z}\circ g) \overline{g'} g'= [(\Delta f)\circ g]|g'|^2=0$$

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One has to be more careful with the computation of the derivative since $f$ generally is a function of $z,\overline{z}$, so the chain rule should give you more. The result is correct though. –  Dimitris Dallas Apr 15 at 1:18
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Let $\phi(x,y)$ be harmonic in $D$. Let $w=f(z)=u(x,y)+iv(x,y)$ be analytic in $D$ defining a mapping $D\to D_0$. Let $\Phi(u,v)=\phi(x,y)$ $$\phi_x=\Phi_u u_x+\Phi_v v_x$$ $$\phi_y=\Phi_u u_y+\Phi_v v_y$$ $$\phi_{xx}=\Phi_{uu}(u_x)^2+\Phi_{uv} u_x v_x +\Phi_u u_{xx} +\Phi_{vv} (v_x)^2+\Phi_{vu} v_x u_x +\Phi_v v_{xx}$$ $$\phi_{yy}=\Phi_{uu}(u_y)^2+\Phi_{uv} u_y v_y +\Phi_u u_{yy} +\Phi_{vv} (v_y)^2+\Phi_{vu} v_y u_y +\Phi_v v_{yy}$$ $$\phi_{xx}+\phi_{yy}=[(u_x)^2+(v_x)^2][\Phi_{uu}+\Phi_{vv}]$$ because $u_{xx}+v_{yy}=0$, $v_{xx}+v_{yy}=0$, $u_xv_x=-u_yv_y$ Hence $$\Delta \Phi = \frac{1}{|f'(z)|^2}\Delta \phi$$ So if $f'(z)\ne 0$ in $D$, then $\Phi$ is harmonic n $D$

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This is not quite what was asked: you have $\phi = \Phi \circ f$ and conclude that if $\phi$ is harmonic and $f$ is analytic then $\Phi$ is harmonic. The question went the other way: if $\phi$ is harmonic and $g$ is analytic then $\phi \circ g$ is harmonic. –  Robert Israel May 31 '12 at 7:11
    
@RobertIsrael In the original question, $g$ is assumed to be conformal (which I read as biholomorphic), so Valentin's answer is sufficient to answer that. Of course, the assumption of invertibility is extraneous. –  user31373 May 31 '12 at 15:09
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Your question can be interpreted in the greater context of "maps preserving harmonic functions".

Definition Let $(M,g)$ and $(N,h)$ be Riemannian manifolds. A mapping $\Phi:M\to N$ is said to be a harmonic morphism if whenever $u:N\to\mathbb{R}$ is a harmonic function (solving $\triangle_h u = 0$ where $\triangle_h$ is the Laplace-Beltrami operator for the Riemannian metric $h$) the composition $u\circ \Phi$ is a harmonic function on $M$.

Theorem A mapping is a harmonic morphism if and only if it is a harmonic map which is weakly horizontally conformal.

(Don't worry too much about the undefined terms in the above theorem.)

Corollary If $M$ and $N$ have the same number of dimensions, then

  • If dimension is 2, $\Phi$ is a harmonic morphism if and only if $\Phi$ is conformal.
  • If the dimension is bigger than 2, $\Phi$ is a harmonic morphism if and only if $\Phi$ is a conformal map with a constant coefficient of conformality.

For reference, see this paper of Bent Fuglede's.

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A typo in the Theorem creates ambiguity: should "with is" be read as "if it is" or "if and only if it is"? –  user31373 May 31 '12 at 15:11
    
@Leonid: fixed. –  Willie Wong May 31 '12 at 15:42
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