Harmonic Functions

Delbar and the Laplacian

definition (Laplacian and Harmonic Functions):

A real function of two variables $u(x, y)$ is harmonic iff it is in the kernel of the Laplacian operator: $\begin{align*} \Delta u \coloneqq\qty{{\frac{\partial ^2}{\partial x^2}\,} + {\frac{\partial ^2}{\partial y^2}\,}}u = 0 .\end{align*}$

definition (del and delbar operators):

$\begin{align*} {\partial}\coloneqq{\partial}_z \coloneqq{1\over 2}\qty{{\partial}_x - i {\partial}_y} \quad \text{ and } \quad { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu} \coloneqq{\partial}_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} ={1\over 2}\qty{ {\partial}_x + i{\partial}_y} .\end{align*}$ Moreover, the 1-form corresponding to $F$ can be written as $\begin{align*} dF = {\partial}F + { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}F = {\frac{\partial F}{\partial z}\,} \,dz+ {\frac{\partial F}{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu }\,}\,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu .\end{align*}$

Written slightly more explicitly: $\begin{align*} {\frac{\partial F}{\partial z}\,} = {1\over 2}\qty{{\frac{\partial F}{\partial x}\,} + {1\over i}{\frac{\partial F}{\partial y}\,} } && {\frac{\partial F}{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu }\,} = {1\over 2}\qty{{\frac{\partial F}{\partial x}\,} - {1\over i}{\frac{\partial F}{\partial y}\,} } .\end{align*}$

proposition (Mean Value Property):

If $u$ is harmonic on $\Omega$ then $\begin{align*} u(z_0) = u(x_0 + i y_0) = {1\over 2\pi r} \oint_{{{\partial}}{\mathbb{D}}_r(z_0)} u \,ds= {1\over \pi r^2} \iint_{{\mathbb{D}}_r(z_0)} u(x, y) \,dx\,dy .\end{align*}$

proof (?):

Define $\begin{align*} F(r) \coloneqq{1\over 2\pi r} \oint_{{\mathbb{D}}_r(z_0)} u\,ds= {1\over 2\pi} \int_{[-\pi, \pi]} u(z_0 + re^{it} ) \,dt .\end{align*}$

Then differentiate: $\begin{align*} F'(r) &= {1\over 2 \pi} \int_{[-\pi, \pi]} \cos(t) u_x(z_0 +re^{it} ) + \sin(t) u_y(z_0 + re^{it}) \,dt\\ &= {1\over 2\pi r} \oint_{{{\partial}}{\mathbb{D}}_r(z_0)}\qty{x-x_0\over r} u_x(x, y) + \qty{y-y_0\over r}u_y(x, y) \,ds\\ &= {1\over 2\pi r} \oint_{{{\partial}}{\mathbb{D}}_r(z_0)} {\frac{\partial u}{\partial w}\,} \,ds\qquad w = {\left[ {{x-x_0\over r}, {y-y_0\over r}} \right]} \\ &= {1\over 2\pi r} \iiint_{{\mathbb{D}}_r(z_0)} \Delta u \,dx\,dy\\ &= 0 ,\end{align*}$ where we’ve used Green’s theorem and that $\Delta u = 0$ . so $F$ is constant. Take $r\to 1$ to obtain $\begin{align*} F(r) = {1\over 2\pi} \int_{[-\pi, \pi]} u(z_0 + re^{it}) \,ds\longrightarrow\int_{[-\pi, \pi]} u(z_0) \,ds= u(z_0) = u(x_0, y_0) .\end{align*}$

remark:

An important use: if $u$ satisfies the mean value property on every disc and is continuous, then $u$ is automatically harmonic.

Exercises: Harmonic Functions

exercise (Holomorphic iff delbar vanishes):

Show that $f$ is holomorphic iff ${ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}f = 0$ .

solution:

$\begin{align*} 2{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}f &\coloneqq({\partial}_x + i {\partial}_y) (u+iv) \\ &= u_x + iv_x + iu_y - v_y \\ &= (u_x - v_y) + i(u_y + v_x) \\ &= 0 && \text{by Cauchy-Riemann} .\end{align*}$

exercise (Holomorphic functions have harmonic components):

Show that if $f = u+iv$ is holomorphic then $u, v$ are harmonic.

solution (?):

Idea: use Cauchy-Riemann, take further derivatives, and use equality of partials.

By CR, $\begin{align*} u_x = v_y && u_y = -v_x .\end{align*}$
Differentiate with respect to $x$ : $\begin{align*} u_{xx} = v_{yx} && u_{yx} = -v_{xx} .\end{align*}$
Differentiate with respect to $y$ : $\begin{align*} u_{xy} = v_{yy} && u_{yy} = -v_{xy} .\end{align*}$
Clairaut’s theorem: partials are equal, so $\begin{align*} u_{xx} - v_{yx} = 0 \implies u_{xx} + u_{yy} = 0 \\ \\ v_{xx} + u_{yx} = 0 \implies v_{xx} + v_{yy} = 0 \\ \\ .\end{align*}$

exercise (Bounded harmonic function is constant):

Show that if $u$ is harmonic on ${\mathbb{R}}^2$ and bounded, then $u$ is constant.

solution (Using Liouville):

Write $f=u+iv$ for $v$ a harmonic conjugate of $u$ , then $f$ is holomorphic on ${\mathbb{C}}$ . Now $e^f = e^{u+iv} = e^u e^{iv}$ and thus $\begin{align*} {\left\lvert {e^f} \right\rvert}\leq {\left\lvert {e^u} \right\rvert} {\left\lvert {e^{iv}} \right\rvert} = {\left\lvert {e^u} \right\rvert} \end{align*}$ is bounded, so $f$ is a bounded entire function and thus constant by Liouville.

solution (Using the mean value property):

figures/2021-12-19_20-20-29.png

exercise (Proving functions are harmonic using components of holomorphic functions):

Show that if $u,v$ are harmonic conjugates, then

$u^2-v^2$ is harmonic
$uv$ is harmonic.
$u_x$ is harmonic.

#complex/exercise/completed

solution:

Write $f=u+iv$ , which is analytic.

$f^2$ is analytic, and $f^2 = (u+iv)^2 = u^2 - v^2 + i (2uv)$ , which necessarily has harmonic components.
Covered by the first case.
$f'$ is analytic and one can write $f' = u_x + iv_x$ , which has harmonic components.

As an alternative to show that $uv$ is harmonic directly by showing it’s in the kernel of the Laplacian. A computation: $\begin{align*} \Delta(uv) &= (uv)_{xx} + (uv)_{yy} \\ &= (u_{xx}v + uv_{xx} + 2u_x v_x) + (u_{yy}v + uv_{yy} + 2u_y v_y) \\ &= (u_{xx} + u_{yy}) v + (v_{xx} + v_{yy}) u + 2(u_xv_x + u_yv_y) \\ &= (u_{xx} + u_{yy}) v + (v_{xx} + v_{yy}) u + 2(-u_x u_y + u_y u_x) && v_x = -u_y,\, v_y = u_x \\ &= (u_{xx} + u_{yy}) v + (v_{xx} + v_{yy})u \\ &= \Delta(u)v + \Delta(v)u \\ &= 0 .\end{align*}$

exercise (Finding harmonic conjugates):

Find a harmonic conjugate for $\begin{align*} u(x, y) = x^3 - 3xy^2 -x -y .\end{align*}$

#complex/exercise/completed

concept:

The standard procedure for harmonic conjugates:

Start with $u$
Take ${\frac{\partial }{\partial x}\,}$ to get $u_x$
Apply CR to get $u_x = v_y$
Take $\int \,dy$ to get $v$ , which is essentially the solution up to an unknown $f(x)$ .
Take ${\frac{\partial }{\partial x}\,}$ to get $v_x$ which involves $f_x$
Apply CR to set $v_x = -u_y$ and solve for $f_x$
Compute $\int f_x \,dx$ to obtain $f(x)$ .

My quick mnemonic:

Link to Diagram

solution:

First, check that $u$ is actually harmonic: $\begin{align*} \Delta u = {\frac{\partial }{\partial x}\,}(3x^2-3y^2-1) + {\frac{\partial }{\partial y}\,}(-6xy - 1) = 6x + (-6x) = 0 .\end{align*}$

Standard procedure: integrate $v_y=u_x$ with respect to $x$ , $\begin{align*} v_y = u_x = 3x^2 - 3y^2 - 1 \implies v = \int u_x \,dy= 3x^2y - y^3 - y + f_1(x) .\end{align*}$ Now differentiate $v$ with respect to $x$ and set $v_x = -u_y$ : $\begin{align*} v_x = 6xy + (f_1)_x = -u_y = 6xy + 1\implies f_1 = x + c_1 .\end{align*}$ Thus $\begin{align*} v(x, y) = 3x^2y - y^3 - y + x + c_1 .\end{align*}$