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CHAPTER 4. THEORY 79<br />

given by<br />

Kmax<br />

−Kmax<br />

|u(x, k ′ )|dk ′ =<br />

≤<br />

Kmax<br />

−Kmax<br />

Kmax<br />

−Kmax<br />

(1)|u(x, k ′ )|dk ′<br />

≤ 2KmaxuX.<br />

1 2 dk ′<br />

1<br />

2 Kmax<br />

−Kmax<br />

|u(x, k ′ )| 2 dk ′<br />

1<br />

2<br />

We will now get an estimate for T (x, k − k ′ ) for u ∈ L 2 ([0,L] × [−Kmax,Kmax]). By<br />

Equation (1.16), we know that U(x) is the sum of the electrostatic potential up(x)and<br />

the potential barriers ∆c(x). The dependence of T (x, k −k ′ )onu is through Poisson’s<br />

equation (Equation (1.13)). We can solve Poisson’s equation explicitly through the<br />

use of a Green’s function [13]. The Green’s function for Poisson’s equation in one<br />

dimension for the domain [0,L]isgivenby<br />

Therefore, the exact solution up(x) isgivenby<br />

L<br />

up(x) =<br />

0<br />

⎧<br />

⎪⎨ z(<br />

J(x, z) =<br />

⎪⎩<br />

x − 1), if 0 ≤ z ≤ x<br />

L<br />

x( z<br />

. (4.4)<br />

− 1), if x ≤ z ≤ L<br />

L<br />

2 q<br />

J(x, z) Nd(z) −<br />

ɛ<br />

1<br />

Kmax<br />

u(z,k<br />

2π −Kmax<br />

′ )dk ′<br />

<br />

dz.

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