Re: phi(x)=a sin(kx) 的原因

[Deatheye (寢ます)]

※ 引述《JGU (ROYGBIV)》之銘言：
: (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)
我怎麼記的好像不用這麼麻煩.......^^"

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