Abstract

CHAPTER 4. THEORY 76
If we write out and rearrange the terms, set the constant C = 2πm∗
hτ , assume k = 0,
then we have
∂f
∂x +C
k
f(x, k) =C
k
If we note that
f0(x, k)
Kmax
−Kmax f0(x, k ′ )dk ′
∂
∂x
f(x, k)e Cx
k
Kmax
−Kmax
= ∂f Cx
e
∂x
then by multiplying each side by e Cx
k ,wehave
∂
∂x
f(x, k)e Cx
k
= C
k
We will now consider two cases:
Case 1: k>0
f(x, k ′ )dk ′ − 4Cτ
hk
k + C
k
f(x, k)e Cx
k ,
f0(x, k)e Cx
k
Kmax
−Kmax f0(x, k ′ )dk ′
Kmax
f(x, k
−Kmax
′ )dk ′ − 4Cτ
hk
Kmax
−Kmax
Kmax
−Kmax
f(x, k ′ )T (x, k−k ′ )dk ′ .
When k>0, we have boundary conditions for f(x, k)atx =0,thatisf(0,k)=f1(k).
So, we have
f(x, k)e Cx
k − f(0,k)=
x
Therefore, we can integrate from 0 to x to get:
0
dz ∂
∂z
f(z,k)e Cz
k
Cx
−
f(x, k) =e k f1(k)+G1(f)+G2(f)
.
e Cx
k f(x, k ′ )T (x, k−k ′ )dk ′ .