the followings phi noted as f
since f'' + k^2 * f = 0 (*)
let f = e ^ (ax)
so a^2 * e^(ax) + k^2 * e^(ax) =0
=> a = +/- ki
then we have
f = A * e^(ikx) + B * e^(-ikx)
= A * ( cos(kx) + i * sin(kx) ) + B * ( cos(-kx) + i * sin(-kx) )
= A'* cos(kx) + B'* sin(kx) ( A' = A+B ; B' = i*(A-B) )
since f(0)=0,
f(0) = A' = 0
f = B' * sin(kx)
i.e. phi = A * sin(kx)
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