The appearance of parameter h in the equation implies that \Psi depends
on the parameter. Sometimes we use the notation \Psi(x, t; h), where we use
';' to separate variable(s) and parameter(s). We expect that when h varies
a little, the solution won't change much. So that we can use the solution
of h=0 to deduce the solution for small h. It resembles to Tayler expansion
and is called asymptotic expansion.
Let \Psi(x, t; h)
=\Psi(x, t; 0)+ h\Psi_h(x, t; 0)+O(h^2)
=A+h B+ O(h^2)
and substitute it to the original equaiton & BC, we obtain
W [A+h B+ O(h^2)] =h [A+h B+O(h^2)]^2, where W denotes the wave operator.
(#) WA+ h (WB-A^2)+ O(h^2) = 0
(ICBC1)[A+hB](0,t)=e^{i4t}
(ICBC2)[A+hB]_x(0, t)=-4ie^{i4t}
Equation (#) along with BCs should hold for arbitray h, so we arrive at
0) Equation of order h^0
WA=0
A(0,t)=e^{i4t}
A_x(0, t)=-4i e^{i4t}
1) Equation of order h^1
WB=A^2
B(0,t)=0
B_x(0,t)=0
What you have obtained is the soluton for 0). You just need to carry on 1).
※ 引述《konayuki (粉雪)》之銘言：
: 我把題目做成圖檔
: http://ppt.cc/ycUv
: 從第二個和第三個條件的式子，可知道ψ=exp(4it-4ix)
: 我原本以為這就快到終點了
: 但帶回第一個式，式子的左邊一算出來是0......
: 到底是怎麼回事呢
: 另外題目下面說
: Calculate ψ(x,t) for x>=0 to first order accuracy in h.
: 這又是?
: 代表ψ(x,t)是一個數列?? h的意義又是?
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