A.3. The iterative sequence

Date: 2015-10-07; view: 474.

The integral equation (A.68), or its equivalent form (A.128), will now be solved iteratively by introducing

As in Figs. (4.87), (4.88) and (4.89), f_n (x) and its associated energy are determined by

(A.137)

and

(A.138)

When n = 0, we set

Introduce

(A.140)

and

(A.141)

From (A.59) and

(A.142)

we derive

(A.143)

and

(A.144)

For n = 1, we have from Figs. (A.139), (A.140) and (A.141),

(A.145)

(A.146)

and

(A.147)

For small μ², since E_a−E_b and M₀−N₀ are both O (μ²), we find

(A.148)

in agreement with (A.117), given by the simple two-level formula.

Next, we examine the integration for f_n (x). Consider first the region

(A.137) can be written as

(A.150)

Introduce

(A.152)

and

(A.153)

When n = 0, we set

From (A.150), or more conveniently by using (x|G|z) given by (A.130), one can readily verify that, for α < x < γ,

(A.155)

etc. These solutions can also be readily derived by directly using the differential equation satisfied by ψ_n (x) = χ (x)f_n (x):

(A.156)

where in accordance with (A.14), . For α < x < γ, we have

and therefore

(A.157)

Introduce S_n (ξ) and C_n (ξ) to be polynomials in ξ, with

(A.158)

From Figs. (A.153), (A.157) and (A.158), we find

(A.159)

where the dot denotes , so that , etc. At x = γ, we have ξ = 0, and therefore

(A.160)

For n = 0, S₀ (ξ) = 1 and C₀ (ξ) = 0. Therefore, for n = 1, (A.159) becomes

(A.161)

Assuming S₁ and C₁ to be both polynomials of ξ, we can readily verify that S₁ is a constant and C₁ is proportional to ξ. Using (A.161) and the boundary condition (A.160), we can establish the first equation in (A.155), and likewise the other equations for n > 1.

To understand the structure of v₁ (ξ), v₂ (ξ), v₃ (ξ), … , we may turn to the exact solution ψ (x) given by (A.123). In analogy to (A.153), we define v (ξ) through

(A.162)

Thus, for α < x < γ,

(A.163)

From Figs. (A.13), (A.14) and (A.118), we have

(A.164)

In terms of

(A.165)

we write

(A.166)

with ξ given by (A.151), as before. It is straightforward to expand v (ξ) as a power series in :

(A.167)

To compare the above series with v_n (ξ) of (A.155), we can neglect O ( ^{n + 1}) in (A.167). The replacements of all linear -terms by _n, ²-terms by _{n − 1 n}, ³-terms by _{n−2 n − 1 n}, etc. lead from (A.167) to v_n (ξ). It is of interest to note that the expansion (A.167) of v (ξ) in power of has a radius of convergence

| |<1.

(A.168)

On the other hand, the iterative sequence {v_n (ξ)} is always convergent, on account of the Hierarchy Theorem. The main difference between Figs. (A.155) and (A.167) is that in (A.155) each iterative _n is determined by the fraction (A.143).

In a similar way, we can derive ψ_n (x) in other regions, −α < x < α and −γ < x < −α. The results for n = 1 are given in Table 2. The functions ₁ (x) and ξ (x) are discontinuous from region to region. The constants κ_II and ρ_II are determined by requiring ψ₁ (x) and and to be continuous at x = α. In region I, when x = α+, we have

(A.169)

and

(A.170)

where the constant

(A.171)

with

(A.172)

In region II, when x = α−

(A.173)

and

(A.174)

where the constant

(A.175)

The constants κ_II and ρ_II are determined by

(A.176)

Likewise, the constants κ_III and ρ_III are given by

(A.177)

and the constants κ_IV and E₁ are determined by

(A.178)

Table 2.

The n = 1 iterative solution ψ₁ (x) in the four regions: I (α < x < γ), II (0 < x < α), III (−α < x < 0), and IV (−γ < x < −α)

Region	1 (x)	ξ (x)	ψ1 (x)
I		ka (−x + γ)
II		qax
III		−qbx
IV		pb (x + γ)

The constants , κ_II, κ_III, κ_IV, ρ_II, and ρ_III are given by Figs. (A.176), (A.177) and (A.178).