有誰告訴我觀念~極坐標的微分

u=u(x,y), x= rcosθ, y=rsinθ
∂u/∂r=∂u/∂x* ∂x/∂r + ∂u/∂y * ∂y/∂r 沒錯啊!

∂x/∂r= cosθ
∂r/∂x= cosθ =>∂x/∂r=∂r/∂x 也沒錯!
∂y/∂r=∂r/∂y也對!

n-dim空間, 哪有極坐標呢?極坐標本來就討論(x, y), (r,θ)的關係 !
除非版大所指為球坐標類似的坐標系

2009-07-16 00:35:07 補充：
應該是∂u/∂r=∂u/∂x1 *∂x1/∂r+∂u/∂x2 *∂x2/∂r+ ...+ ∂u/∂xn * ∂xn/∂r
"形式"而已

2009-07-16 12:16:13 補充：
1. ∂u/∂xi與∂u/∂r *∂xi/∂r不同, xi視為r, θ1, θ2, ...函數, 除了對 r 偏導函數外,
仍須對θ1,θ2, ...偏導
2. ∂r/∂xi = xi/r 對!
r^2 = x1^2+...+xn^2 => 對xi偏導函數得 2r*∂r/∂xi = 2xi ,so, ∂r/∂xi= xi/r

2009-07-17 12:36:17 補充：
如果U(x1,..,xn)=u(r)
=>∂U/∂xi=∂u/∂r *∂r/∂xi=∂u/∂r * xi/r沒錯!

1. x = r cos P , y= r sin P---change of variables (x,y) to (r,P), which should result U(x,y)=u(r,P), not U(x,y)=u(r) directly.
2. When using U(x,y)=u(r), we have tacitly agree that u (orU) is radially symmetric, i. e. u is independent of angle P. This assumption is frequently used in finding a simple solution in PDE.
3. 不就變成 dr/dx = dx/dr 了嗎?????---Nothing is wrong. Just an coincident here in polar coordinates for n=2. P_x=x_P=cosP.
4. n維的 u----Like n=2, we need to start from the change of variables(x1,x2,x3,...xn) to (r,p1,p2,p3,...p(n-1)) [one radial variable, (n-1)angle variables]: by the relations
x1=rcosp1
x2=rsinp1cosp2
x3=rsinp1sinp2cosp3
....
x(n-1)=rsinp1sinp2...sinp(n-2)cosp(n-1)
xn =rsinp1sinp2...sinp(n-2)sinp(n-1)
[notice that the last two is a pair, different from other members; angles p1 through p(n-2) each ranges from 0 to pi, while angle pn ranges from 0 to 2pi]
the above polar coordinates for general n will automatically include n=2, n=3 (球座標)as special cases. Please try it to convince yourself.　　Thus we arrive at U(x_1,x_2,..................,x_n)=u(r, p1,p2,..p(n-1))and U(x_1,x_2,..................,x_n)=u(r) by the radial symmetry. I am sure you then know how to calculate u_r.
5.我再來要修PDE 自己先自修----good for you祝你成功.

2009-07-16 08:26:18 補充：
In 3. P_x=x_P=cosP should be r_x=x_r=cosP.