→ ntust661:沒有正規化 12/03 19:33
→ ntust661:記得極座標的Laplacian 12/03 19:33
→ endlesschaos:可以再解釋得清楚一點嗎？謝謝 12/03 20:47
題目是這樣的
→
If a vector F is given by
→ → → →
F = (x^2 + y^2 + z^2)^n * (x i + y j + z k )
→
find ▽^2 F
method Ⅰ.
→ d^2 d^2 d^2 →
▽^2 F = (------ + ------ + ------) F
dx^2 dy^2 dz^2
d → → → →
其中 ---- F = 2nx * (x^2 + y^2 + z^2)^(n-1) * (x i + y j + z k )
dx
→
+ (x^2 + y^2 + z^2)^n i
d^2 → → → →
=> ------ F = 2n * (x^2 + y^2 + z^2)^(n-1) * (x i + y j + z k )
dx^2
→ → →
+ 2nx * (n-1) * (x^2 + y^2 + z^2)^(n-2) * 2x * (x i + y j + z k)
→
+ 2nx * (x^2 + y^2 + z^2)^(n-1) i
→
+ n * (x^2 + y^2 + z^2)^(n-1) * 2x i
→ → → → →
令 r = x i + y j + z k , r = ｜ r ｜
d^2 → → →
=> ------ F = 2n * r^(2n-2) r + 4n(n-1) * x^2 * r^(2n-4) r
dx^2
→
+ 4n * r^(2n-2) * x i
d^2 d^2 d^2 →
故所求 = (------ + ------ + ------) F
dx^2 dy^2 dz^2
→ →
= 6n * r^(2n-2) r + 4n(n-1) * (x^2 + y^2 + z^2) * r^(2n-4) r
→ → →
+ 4n * r^(2n-2) * (x i + y j + z k )
→
= 2n * (2n + 3) * r^(2n-2) r
--
method Ⅱ.
→ →
F = r^(2n) r
→ →
▽^2 F = ▽．▽ F
→
= ▽．▽[r^(2n) r ]
→ →
= ▽．{ ▽[r^(2n)] r + r^(2n) ▽ r }
→ →
= ▽．[ 2n * r^(2n - 1) ▽r r + r^(2n) ▽ r ]
→
r → →→ →→ →→
其中 ▽r = --- , ▽ r = I = i i + j j + k k
r
→ → →
=> 原式 = ▽．[2n * r^(2n - 2) r r + r^(2n) ▽ r ]
→ → → →
= 2n * ▽[r^(2n - 2)]．r r + 2n * r^(2n - 2) ▽．r r
→ → → →
+ 2n * r^(2n - 2) r ▽．r + ▽[r^(2n)]．▽ r + r^(2n) ▽．▽ r
→
r → → →
= 2n(2n - 2) * r^(2n - 3) * ---．r r + 2n * r^(2n - 2) * 3 r
r
→
→ r →
+ 2n * r^(2n - 2) r * 3 + 2n * r^(2n - 1) * ---．▽ r + 0
r
→ → →
= 4n(n-1) * r^(2n - 2) r + 12n * r^(2n - 2) r + 2n * r^(2n - 2) r
→
= 2n(2n + 5) * r^(2n - 2) r
解答證明 method Ⅰ 的答案才是對的
想請問一下 method Ⅱ 當中步驟有哪裡有錯呢？
感謝各位
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◆ From: 114.34.133.34
※ 編輯: endlesschaos 來自: 114.34.133.34 (12/02 23:26)