滿足 steady ( $\partial\vec{V}/\partial t=0$ ) 的 potential flow (incompressible, and irrotational) $\vec{V}$ 看似限制很多沒什麼用處, 但事實上應用還是很廣, 後面的課程會繼續介紹.

怎麼找到這樣的 potential flow 呢? 其實只要找到一個滿足 Laplace’s equation 的 scalar field $\varphi$ (i.e. $\nabla^2\varphi=0$) 其定義的 gradient field $\nabla\varphi$ 就是這樣的 potential flow $\vec{V}$ 了!

首先 gradient field $\nabla\varphi$ 一定是 irrotational 再來如果滿足 Laplace’s equation 則會有 incompressible (上圖左下部份證明) 而 steady 只要設定跟 $t$ 無關即可

考慮複數 $z=x+iy$, 並令複數函數 $\Phi(z)$ 由兩個實數函數 $\varphi(x,y),\psi(x,y)$ 分別代表實部虛部所組成.

則只要 $\Phi(z)$ 是 analytic function ([註1]), 那麼 $\varphi(x,y),\psi(x,y)$ 就自動滿足 Laplace’s equation 了. (老師說這是因為 analytic function 要滿足 Cauchy-Riemann equation [註2, 4], 而 Cauchy-Riemann 會使得 Laplace’s equation 被滿足, 見[註3])

$$ \underbrace{\Phi(z)}{\text{complex potential}}=\underbrace{\varphi(x,y)}{\text{potential func.}}+i\underbrace{\psi(x,y)}_{\text{stream func.}} \\ ,\text{where}\quad z=x+yi $$

所以 vector fields $\nabla\varphi,\nabla\psi$ 就滿足 steady incompressible, and irrotational 了!

[註1]: 高微課本 elementary classical analysis p360 說明什麼是 real analytic function!

[註2]: for complex function 是否是可微的有一個充要條件稱 Cauchy–Riemann equations From wiki (一個解說的中文 [YouTube])

[註3]: 檢查 $\nabla^2 u=0?$ 式 (3) 是因為 Cauchy-Riemann equation 的關係

$$ \begin{align} \nabla^2 u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2} \\ = \frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x}\right) + \frac{\partial}{\partial y}\left(\frac{\partial u}{\partial y}\right) \\ = \frac{\partial}{\partial x}\left({\color{orange}{\frac{\partial v}{\partial y}}}\right)+\frac{\partial}{\partial y}\left({\color{orange}{-\frac{\partial v}{\partial x}}}\right) \\ =\frac{\partial^2 v}{\partial x \partial y} - \frac{\partial^2 v}{\partial x \partial y} = 0 \end{align} $$

Let $z=x+iy$, $f(z)=u(x,y)+iv(x,y)$

且令 $\Delta z=(x+\Delta x)+i(y+\Delta y)$, 則

$$ f(z+\Delta z)=u(x+\Delta x,y+\Delta y)+iv(x+\Delta x,y+\Delta y) $$

根據定義

$$ \frac{df}{dz}(z)=\lim_{\Delta z\rightarrow0}\frac{f(z+\Delta z)-f(z)}{\Delta z} \\ =\lim_{\Delta x\rightarrow0}\lim_{\Delta y\rightarrow0}\frac{u(x+\Delta x,y+\Delta y)-u(x,y)+i[v(x+\Delta x,y+\Delta y)-v(x,y)]}{\Delta x+i\Delta y} $$

可以先讓 $\Delta y\rightarrow 0$ 得到

$$ \lim_{\Delta x\rightarrow0}\frac{u(x+\Delta x,y)-u(x,y)+i[v(x+\Delta x,y)-v(x,y)]}{\Delta x} \\ =\lim_{\Delta x\rightarrow0}\frac{u(x+\Delta x,y)-u(x,y)}{\Delta x}+i\lim_{\Delta x\rightarrow0}\frac{v(x+\Delta x,y)-vx,y)}{\Delta x} \\ =\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x} $$

或先讓 $\Delta x\rightarrow 0$ 得到

$$ \lim_{\Delta y\rightarrow0}\frac{u(x,y+\Delta y)-u(x,y)+i[v(x,y+\Delta y)-v(x,y)]}{i\Delta y} \\ =\lim_{\Delta y\rightarrow0}\frac{u(x,y+\Delta y)-u(x,y)}{i\Delta y}+\lim_{\Delta y\rightarrow0}\frac{v(x,y+\Delta y)-v(x,y)}{\Delta y} \\ =\frac{1}{i}\frac{\partial u}{\partial y}+\frac{\partial v}{\partial y} = \frac{\partial v}{\partial y} - i\frac{\partial u}{\partial y} $$

所以

$$ {\color{orange}{\frac{df}{dz}(z) = \frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x} = \frac{\partial v}{\partial y} - i\frac{\partial u}{\partial y}}} $$

因此得到 Cauchy-Riemann equation