Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 12 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.

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

share|cite|improve this answer
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. – Dimitrios Nt Apr 15 '14 at 1:18
@user31373 The second derivative step should produce one more term $$f_z(g)g_{z\overline{z}}$.It seems that this term is not always zero.Moreover $\Delta u=4_{z\head{z}}$ – Daniel S. May 8 '14 at 14:36

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 $D$

share|cite|improve this answer
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

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.

share|cite|improve this answer
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

Your Answer


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.