Composition of a harmonic function with a holomorphic function is still harmonic

Question

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!

Answer 1

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.

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

Answer 2

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.

Answer 3

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.

Answer 4

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 $Re (f(g(z))) = u(f(z))$.