→ Vulpix :第一步過程是錯的，連鎖律用錯了 10/11 12:00
→ swamathn :請問是在證明上面那題時用錯了嗎? 10/11 13:03
→ swamathn :感謝1F指點@@" 原來在上面那個一開始就錯了... 10/11 16:37
是 Stein的課本
p.27
題目是要我們證在 polar coordinates,the Cauchy-Riemann equations take the form
(抱歉 找不到偏微分的符號 暫時用δ代替)
δu 1 δv 1 δu -δv
--- = --- ---- --- ---- = ---
δr r δθ r δθ δr
where f=u+vi u=u(x,y) v=v(x,y)
(PS 我的問題出在下面那個小題 但這題過程會用到)
這題我的寫法是
let x = rcosθ y = rsinθ
by Cauchy-Riemann equations
we have δu δv δu δv
--- = ---- ---- = ---
δx δy δy δx
δu δu δx δu
--- = ---- --- = ---- cosθ
δr δx δr δx
and
δv δv δy δv δu δu
--- = ---- --- = ---- rcosθ= ----cosθ r = --- r
δθ δy δθ δy δx δr
=>
δu δv 1
--- = ---- ---
δr δy r
similarly
δu δu δy δu
--- = ---- --- = ---- cosθ
δθ δy δθ δy
δv δv δx δv -δu -1 δu
--- = ---- --- = ---- rcosθ= ----cosθ = --- ----
δr δx δr δx δy r δθ
#
這樣證應該沒錯吧?
但是問題就出在下個小題
Use these equations to show that the logrithm function defined by
log z = log r + iθ where z = e^iθ with -π < θ < π
is holomorphic in the region r > 0 and -π < θ < π
我的想法 也是用 Cauchy-Riemann equations推倒的東西來弄
只要能證明 log z = f = u + vi 中的 u,v 是連續可微的就可以證明他是holomorphic
(課本 p.13)
且可利用
δ 1 δ 1 δ δu δv
--- = ---(---- + --- ---- ) (課本p.12) 去做出 ---- 和 ----
δz 2 δx i δy δz δz
做法
let u = log r v = θ
δu 1 δu 1 δu 1 δu 1 δu 1
---- = ---(---- + --- ---- ) = ---(---- ---- + ---- ---- )
δz 2 δx i δy 2 δr cosθ δθ rcosθ
^^^^^ 從上題的過程來的
1 1 1 1
= ---(--- ---- + 0 ) = -------
2 r cosθ 2rcosθ
這邊就出問題了..... 當θ= π/2 or -π/2 cosθ=0
變成說複數線上不可微 跟我所想要的 負實數線上不可微 完全相反...
但是我想破頭就是想不太到哪裡有問題
還請大家多幫忙
剛剛有想到說 是不是u,v不能這樣設定? 是在這邊出錯嗎?
感謝
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