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: Δu:=(∂2∂x2+∂2∂y2)u=0.
∂:=∂z:=12(∂x−i∂y) and ¯∂:=∂¯z=12(∂x+i∂y). Moreover, the 1-form corresponding to F can be written as dF=∂F+¯∂F=∂F∂zdz+∂F∂¯zd¯z.
Written slightly more explicitly: ∂F∂z=12(∂F∂x+1i∂F∂y)∂F∂¯z=12(∂F∂x−1i∂F∂y).
If u is harmonic on Ω then u(z0)=u(x0+iy0)=12πr∮∂Dr(z0)uds=1πr2∬Dr(z0)u(x,y)dxdy.
proof (?):
Define F(r):=12πr∮Dr(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=12πr∮∂Dr(z0)(x−x0r)ux(x,y)+(y−y0r)uy(x,y)ds=12πr∮∂Dr(z0)∂u∂wdsw=[x−x0r,y−y0r]=12πr∭Dr(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*}
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.
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*}
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.
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By CR, \begin{align*} u_x = v_y && u_y = -v_x .\end{align*}
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Differentiate with respect to x: \begin{align*} u_{xx} = v_{yx} && u_{yx} = -v_{xx} .\end{align*}
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Differentiate with respect to y: \begin{align*} u_{xy} = v_{yy} && u_{yy} = -v_{xy} .\end{align*}
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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.
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):
Show that if u,v are harmonic conjugates, then
- u^2-v^2 is harmonic
- uv is harmonic.
- u_x is harmonic.
solution:
Write f=u+iv, which is analytic.
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f^2 is analytic, and f^2 = (u+iv)^2 = u^2 - v^2 + i (2uv), which necessarily has harmonic components.
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Covered by the first case.
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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:
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*}