11.4: Digression to harmonic functions

Last updated
Save as PDF

Page ID: 6540

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$

Theorem $\PageIndex{1}$

If $u$ and $v$ are harmonic conjugates and $g = u + iv$ has $g'(z_0) \ne 0$, then the level curves of $u$ and $v$ through $z_0$ are orthogonal.

Note

We proved this in an earlier topic using the Cauchy-Riemann equations. Here will make an argument involving conformal maps.

Proof

First we’ll examine how $g$ maps the level curve $u(x, y) = a$. Since $g = u + iv$, the image of the level curve is $w = a + iv$, i.e it’s (contained in) a vertical line in the $w$-plane. Likewise, the level curve $v(x, y) = b$ is mapped to the horizontal line $w = u + ib$.

Thus, the images of the two level curves are orthogonal. Since $g$ is conformal it preserves the angle between the level curves, so they must be orthogonal.

$屏幕快照 2020-09-13 下午2.01.44.png$
$g = u + iv$ maps level curves of $u$ and $v$ to grid lines.