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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 $D_0$?

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

Thank you so much!

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5 Answers 5

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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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    $\begingroup$ 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. $\endgroup$
    – Dimitris
    Apr 15, 2014 at 1:18
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    $\begingroup$ @user31373 The second derivative step should produce one more term $$f_z(g)g_{z\overline{z}}$.It seems that this term is not always zero.Moreover $\Delta u=4_{z\head{z}}$ $\endgroup$
    – Daniel S.
    May 8, 2014 at 14:36
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Let $\phi(u,v)$ be harmonic in $D$. Let $w=f(z)=u(x,y)+iv(x,y)$ be analytic in $D_0$ defining a mapping $D_0\to D$. We have $$\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 $\phi_{uu}+\phi_{vv}=0$ implies $\phi_{xx}+\phi_{yy}=0$. We conclude that $\phi(x,y)$ is a harmonic funtion.

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  • $\begingroup$ 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. $\endgroup$ May 31, 2012 at 7:11
  • $\begingroup$ @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. $\endgroup$
    – user31373
    May 31, 2012 at 15:09
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The term "harmonic" will mean "real and harmonic" below. I will use the following:

1) Compositions of holomorphic functions are holomorphic.

2) The real and imaginary parts of a holomorphic function are harmonic.

3) If $D\subset \mathbb C$ is an open disc, and $u$ is harmonic in $D,$ then there exists $v$ harmonic in $D$ such that $u+iv$ is holomorphic in $D.$

1) is just the chain rule. 2) follows from the Cauchy-Riemann equations. If you are unfamiliar with 3), I can add a proof later. I'll assume it for now.

Thm: Suppose $\Omega_1, \Omega_2 \subset \mathbb C$ are open. Assume $u$ is harmonic on $\Omega_2$ and $f: \Omega_1 \to \Omega_2$ is holomorphic. Then $u\circ f$ is harmonic on $\Omega_1.$

Proof: Let $a \in \Omega_1.$ Because harmonicity is a local property, it suffices to show $u\circ f$ is harmonic in a neighborhood of $a.$

Choose an open disc $D$ centered at $f(a)$ contained in $\Omega_2.$ Then by 3) above, there exists a $v$ harmonic in $D$ such that $u+iv$ is holomorphic in $D.$ Let $\omega = f^{-1}(D).$ Then $\omega $ is an open neighborhood of $a$ contained in $\Omega_1,$ and $f:\omega \to D.$ By 1), $(u+iv)\circ f$ is holomorphic in $\omega.$ By 2), the real part of $(u+iv)\circ f$ is harmonic in $\omega.$ But this real part is precisely $u\circ f,$ and we're done.

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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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  • $\begingroup$ A typo in the Theorem creates ambiguity: should "with is" be read as "if it is" or "if and only if it is"? $\endgroup$
    – user31373
    May 31, 2012 at 15:11
  • $\begingroup$ @Leonid: fixed. $\endgroup$ May 31, 2012 at 15:42
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I was wondering the same question and reached here.

Here is how I thought about it:

Let $u$ be harmonic. And let $v$ be its harmonic conjugate. Then $f(z) = u(z) + iv(z)$ is analytic. If $g$ is another analytic function such that range of $g$ is contained in the domain of $f$ then $$f(g(z)) = u(f(z)) + iv(g(z))$$ is also analytic with $Re (f(g(z))) = u(f(z))$.

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