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: Δu:=(2x2+2y2)u=0.

definition (del and delbar operators):

:=z:=12(xiy) and ¯:=¯z=12(x+iy). Moreover, the 1-form corresponding to F can be written as dF=F+¯F=Fzdz+F¯zd¯z.

Written slightly more explicitly: Fz=12(Fx+1iFy)F¯z=12(Fx1iFy).

proposition (Mean Value Property):

If u is harmonic on Ω then u(z0)=u(x0+iy0)=12πrDr(z0)uds=1πr2Dr(z0)u(x,y)dxdy.

proof (?):

Define F(r):=12πrDr(z0)uds=12π[π,π]u(z0+reit)dt.

Then differentiate: F(r)=12π[π,π]cos(t)ux(z0+reit)+sin(t)uy(z0+reit)dt=12πrDr(z0)(xx0r)ux(x,y)+(yy0r)uy(x,y)ds=12πrDr(z0)uwdsw=[xx0r,yy0r]=12πrDr(z0)Δudxdy=0, 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 { \overline{{\partial}}}f = 0.

solution:

\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*}

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 {\mathbf{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 {\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.

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*}

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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*}

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