※ 引述《womack79 (糖做的老虎)》之铭言： : 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? : 谢谢 (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. (可参考 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(和可微分必连续), 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. 这不一定对,例如: f(z) = y + ix 因为 Cauchy-Riemann equations 不成立(这是可解析的必要条件) u(x,y) = y, v(x,y) = x =>u = 0 = v , but u = 1 ≠ -v = -1 x y y x 所以你会发现到,要加上"符合Cauchy-Riemann equations"才对, 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 (在这种情况下,叫做 v 是 u 的 harmonic conjugate) (f(z) = v(x,y) + i u(x,y)可不见得解析!! 你可用以上的例子去验证) 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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