推 womack79 :謝謝! 04/18 01:02
推 ntust661 :推啦^^ 04/18 02:15
推 ntust661 :另外大大小筆誤，是Clairaut's 不是 Chairaut's XD 04/18 03:02
※ 引述《womack79 (糖做的老虎)》之銘言：
: f(z) = u(x,y) + iv(x,y)
: 如果 u(x,y) and v(x,y) are harmonic function
: imply f(z) analytic?
: 相反地, 如果 f(z) is analytic
: imply u(x,y) and v(x,y) are harmonic function?
: 謝謝
(1) If f(z) is analytic, then u(x,y) and v(x,y) are harmonic.
Proof:
If f(z) is analytic, then its derivative are analytic for all orders.
(可參考 James Ward Brown & Ruel V. Churchill的 Complex variable
and applications 8th 的 Sec 52)
Then, all patial derivative of u(x,y) and v(x,y) exist for all orders.
Besides, u (x,y) = v (x,y)
x y
(Cauchy-Riemann Equations)
u (x,y) = -v (x,y)
y x
Then,
u (x,y) = v (x,y)
xx yx
u (x,y) = - v (x,y)
yy xy
By Chairaut's Theorem(和可微分必連續), v (x,y) = v (x,y)
xy yx
So, u (x,y) + u (x,y) = 0
xx yy
Similarly, v (x,y) + v (x,y) = 0
xx yy
Therefore, u(x,y) and v(x,y) are harmonic.
(2) If u(x,y) and v(x,y) are harmonic, then f(z) = u(x,y) + iv(x,y) is
analytic.
這不一定對,例如:
f(z) = y + ix
因為 Cauchy-Riemann equations 不成立(這是可解析的必要條件)
u(x,y) = y, v(x,y) = x =>u = 0 = v , but u = 1 ≠ -v = -1
x y y x
所以你會發現到,要加上"符合Cauchy-Riemann equations"才對, 且要注意
Cauchy-Riemann equations的形式:
u = v , u = - v and f(z) = u(x,y) + iv(x,y)
x y y x
(在這種情況下,叫做 v 是 u 的 harmonic conjugate)
(f(z) = v(x,y) + i u(x,y)可不見得解析!! 你可用以上的例子去驗證)
f(z) = z = x + iy => u(x,y) = x, v(x,y) = y
和 f(z) = y + ix, 但不可解析
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