Harmonic Functions

Delbar and the Laplacian

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*}

\begin{align*} {\partial}\coloneqq{\partial}_z \coloneqq{1\over 2}\qty{{\partial}_x - i {\partial}_y} \quad \text{ and } \quad { \overline{{\partial}}} \coloneqq{\partial}_{\overline{z}} ={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 + { \overline{{\partial}}}F = {\frac{\partial F}{\partial z}\,} \,dz+ {\frac{\partial F}{\partial \overline{z}}\,}\,d\overline{z} .\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 \overline{z}}\,} = {1\over 2}\qty{{\frac{\partial F}{\partial x}\,} - {1\over i}{\frac{\partial F}{\partial y}\,} } .\end{align*}

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*}

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*}

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

Exercises: Harmonic Functions

Show that $$f$$ is holomorphic iff $${ \overline{{\partial}}}f = 0$$.

\begin{align*} 2{ \overline{{\partial}}}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*}

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

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*}

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

Write $$f=u+iv$$ for $$v$$ a harmonic conjugate of $$u$$, then $$f$$ is holomorphic on $${\mathbf{C}}$$. 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.

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

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

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*}

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

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:

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*}