Cauchy-Riemann Equation & Harnomic function
2020-2학기, 대학에서 ‘응용복소함수론’ 수업을 듣고 공부한 바를 정리한 글입니다. 지적은 언제나 환영입니다 :)
Cauchy-Riemann Equations
“주어진 복소함수 $f(z) = u(x, y) + i v(x, y)$에 대해, 실수부 $u(x, y)$와 허수부 $v(x, y)$를 보고 analytic 함수임을 보장할 수 있을까?”
$f$ is analytic in a domain $D$
$\iff$ the first partial derivatives of $u$ and $v$ satisfy
the Cauchy-Riemann equations: $u_x = v_y$, $u_y = -v_x$
proof.
$f$ is differentiable $\implies$ $u$, $v$ satisfy the Cauchy-Riemann Equations.
For
\[f'(z) = \lim_{\Delta z \rightarrow 0} \frac{f(z+\Delta z) - f(z)}{\Delta z}\]Think of two direction $\Delta x$ and $i \Delta y$, then
\[\begin{aligned} f'(z) &= \lim_{\Delta x \rightarrow 0} \frac{f(z+\Delta x) - f(z)}{\Delta x} \\ &= \lim_{\Delta x \rightarrow 0} \frac{u(x+\Delta x, y) + v(x+\Delta x, y) i - u(x, y) - v(x, y) i}{\Delta x} \\ &= \lim_{\Delta x \rightarrow 0} \frac{u(x+\Delta x, y) - u(x, y) + v(x+\Delta x, y) i - v(x, y) i}{\Delta x} \\ &= u_x + v_x i \end{aligned}\]마찬가지로 $i \Delta y$에 대해서 $u(x, y)$, $v(x, y)$에 대해 미분하면 아래와 같다.
\[\begin{aligned} f'(z) &= \lim_{\Delta y \rightarrow 0} \frac{f(z+\Delta y) - f(z)}{\Delta y} \\ &= \lim_{\Delta y \rightarrow 0} \frac{u(x, y+\Delta y) - u(x, y) + v(x, y+\Delta y) i - v(x, y) i}{i \Delta y} \\ &= \frac{u_y}{i} + v_y \end{aligned}\]이때, differentiable하기 위해선 극한이 존재해야 하므로,
\[u_x + u_y = \frac{u_y}{i} + v_y\]따라서
\[\begin{aligned} u_x &= v_y \\ u_y &= -v_x \end{aligned}\]$u, v$ have continuous partial derivatives satisfy the Cauchy-Riemann equations $\implies$ $f$ is analytic.
Let $\Delta u = u(x_0 + \delta x, y_0 + \delta y) - u(x_0, y_0)$.
Then, because the first order partial derivative of $u$ are continuous at $(x_0, y_0)$,
\[\Delta u = u_x (x_0, y_0) \Delta x + u_y (x_0, y_0) \Delta y + \epsilon_1 \Delta x + \epsilon_2 \Delta y\]where $\epsilon_1$ and $\epsilon_2$ tend to zero as $\Delta x$, $\Delta y$ approach to zero.
Likewise, we can think of $\Delta v$, then
\[\begin{aligned} f'(z) &= \lim_{\Delta z \rightarrow 0} \frac{f(z+\Delta z) - f(z)}{\Delta z} \\ &= \lim_{(\Delta x + i \Delta y) \rightarrow 0} \frac{\Delta u + i \Delta v}{\Delta x + i \Delta y} \\ &= \lim_{(\Delta x + i \Delta y) \rightarrow 0} \frac{(u_x \Delta x + u_y \Delta y) - i(v_x \Delta x + v_y \Delta y) + E}{\Delta x + i \Delta y} \\ &= \lim_{(\Delta x + i \Delta y) \rightarrow 0} \frac{(u_x \Delta x + u_y \Delta y) - i(-u_y \Delta x + u_x \Delta y) + E}{\Delta x + i \Delta y} \quad (\textrm{C-R equation}) \\ &= \lim_{(\Delta x + i \Delta y) \rightarrow 0} \frac{u_x (\Delta x + i \Delta y) + u_y (-i\Delta x + \Delta y)}{\Delta x + i \Delta y} \\ &= u_x - i u_y \quad (\textrm{C-R equation}) \\ &= u_x + i v_x \end{aligned}\]example.
Supp. that $f(z)$ is analytic in a domain $D$, and $\left| f(z) \right| = k = \textrm{constant}$ in $D$. Show that $f(z)$ is constant.
Polar Coordinate
Cauchy-Riemann equation for polar coordinate
\[u_r = \frac{1}{r}u_\theta, \quad v_r = - \frac{1}{r} u_{\theta}\]Harmonics functions & Laplace’s Equation
A real-valued function $H(x, y)$ is harmonic in a domain $D$, if it satisfies “Laplace equation”
\[\nabla^2 H = H_{xx} + H_{yy} = 0\]Theorem.
If $f(z)=u(x, y) + i v(x, y)$ is analytic in a domain $D$,
then both $u$ and $v$ are Harmonic functions.
If two harmonic functions $u$ and $v$ satisfy the Cauchy-Riemann equations in a domain $D$,
then $f(z) = u(x, y) + i v(x, y)$ is analytic in $D$.
In this case, $v$ is called a “harmonic conjugate function of $u$” in $D$.