→ kafai:你那麼有空爬文… 推140.112.239.182 11/08
→ JGU:我是昨天才知道這個方法... 推 61.230.40.184 11/08
→ kafai:不用複習微分方程吧？XD 推140.112.239.182 11/08
(這是一個數學系的朋友告訴我的方法,有點嚇人)
※ 引述《kafai (雲^^)》之銘言：
the followings Ψ noted as f
since f" + k^2 * f = 0 (*) k ≠ 0
assume f(0) = A , f'(0)/k = B
let g(x) = f(x) - ( A*cos(kx) + B*sin(kx) ) (想要證明 g(x) = 0 )
then g'(x) = f'(x) - ( -kA*sin(kx) + kB*cos(kx) )
g"(x) = f"(x) - ( -k^2*A*cos(kx) - k^2*B*sin(kx) )
= -k^2*f(x) + k^2*( A*cos(kx) + B*sin(kx) )
= -k^2 * g(x)
g(0) = f(0) - A = 0 , g'(0) = f'(0) - kB = 0
let h(x) = (g'(x))^2 + (k*g(x))^2 , then h(0) = 0
h'(x) = g"(x) * 2g'(x) + k*g'(x) * 2k*g(x)
= 2g'(x) ( g"(x) + k^2*g(x) ) = 0
So h(x) is a constant => h(x) = 0 => g'(x) = g(x) = 0
=> f(x) = A*cos(kx) + B*sin(kx) = f(0)*cos(kx) + (f'(0)/k)*sin(kx)
: since f(0)=0,
: f(0) = A' = 0
: f = B' * sin(kx)
: i.e. Ψ = A * sin(kx)
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