Griffiths introduction to electrodynamics, third edition
problem 5.54
Prove the uniqueness theorem
sol:
用題目給的hint
∫▽．[U X (▽ X V)] dτ
=∫{(▽ X V)．(▽ X U) - U．[▽ X (▽ X V)]} dτ
=∮[ U X (▽ X V)]．da
X : cross product
V U a 上都有向量符號 難畫 略
As always, suppose we have two solutions, B1 (and A1) & B2 (and A2).
define B3≡B2 - B1 (and A3≡A2 - A1)
so we have ▽ X A3 = B3
and ▽ X B3 = ▽ X B1 - ▽ X B2 = μo J -μo J = 0
set U = V = A3 in above identity:
∫{(▽ X A3)．(▽ X A3) - A3．[▽ X (▽ X A3)]} dτ
=∫{(B3)．(B3) - A3．[▽ X B3]} dτ
=∫(B3)^2 dτ
=∮[A3 X (▽ X A3)]．da = ∮(A3 X B3)．da
but either A is specified(A3=0) or else B is specified(B3=0), at the surface
In either case ∮(A3 X B3)．da = 0
so ∫(B3)^2 dτ = 0
and hence B1 = B2 qed.
證明到這裡告一段落
http://farside.ph.utexas.edu/teaching/em/lectures/node62.html
上面是我在網路上找到的uniqueness theorem
在第(699)式
▽ X B3 = 0
...either B3 or A3 is zero
這裡開始，寫出來跟老師討論時，說:這個假設是有問題的
(不能確定其中之一的值是0 ?!?!)
所以底下的就不用看了...
Griffiths的也這樣假設，請問這問題出在哪呢???
感謝看完這篇 <(_ _)>
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