four inequalities
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Short worked note, dated 6.6.15 in its heading, from Phil's Lagrange multipliers files. Assuming energies ε3 > ε2 > ε1, it rewrites the linear relations for N1 and N2 in terms of N3, then examines the four inequalities 0 ≤ N1, N2 ≤ M one at a time. These combine into a single range for N3, which is evaluated for an example with u = 1.5. Many fractions are lost in extraction, so some bounds are unclear.
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The four inequalities problem PhL 6.6.15
Assume,
ε3 > ε2 > ε1 .
From (6.3.1),
N1 = + (ε2- ε1)-1[ ε2M - U + (ε3- ε2)N3] = N1(N3)
N2 = - (ε2- ε1)-1[ ε1M - U + (ε3 - ε1)N3] = N2(N3) . (6.3.1)
Rewrite these as
N1 = + (ε2- ε1)-1[(ε2-u)M + (ε3- ε2)N3] = N1(N3)
N2 = - (ε2- ε1)-1[ (ε1-u)M + (ε3 - ε1)N3] = N2(N3) . (6.3.1)
Rewrite again as
N1 = + (ε2- ε1)-1[(ε2-u)M + (ε3- ε2)N3] = N1(N3)
N2 = +(ε2- ε1)-1[ (u-ε1)M - (ε3 - ε1)N3] = N2(N3) . (6.3.1)
There are four inequalities
0 ≤ N1 ≤ M
0 ≤ N2 ≤ M
Examine the four inequalities one at a time.
Inequality #1:
N1 ≥ 0 (ε2-u)M + (ε3- ε2)N3 ≥ 0 N3 ≥ M
This is only a constraint if u > ε2,
Inequality #2:
N2 ≥ 0 (u-ε1)M - (ε3 - ε1)N3 ≥ 0 N3 ≤ M
Since ε1 ≤ u ≤ ε3, we always have u ≥ ε1 so in general this is always a constraint.
Inequality #3:
N1 ≤ M (ε2-u)M + (ε3- ε2)N3 ≤ (ε2-ε1)M
(ε3- ε2)N3 ≤ (ε2-ε1)M + (u-ε2)M = (u-ε1)M
N3 ≤ M
Since ε1 ≤ u ≤ ε3, we always have u ≥ ε1 so in general this is always a constraint.
Inequality #4:
N2 ≤ M (u-ε1)M - (ε3 - ε1)N3 ≤ (ε2-ε1)M
(ε3 - ε1)N3 ≥ (u-ε1)M + (ε1-ε2)M = (u-ε2)M
N3 ≥ M
This is only a constraint if u > ε2.
Summary
N3 ≥ M N3 ≥ M
N3 ≤ M N3 ≤ M
Combine these to get
N3 ≥ max ( M, M) only a constraint if u > ε2
N3 ≤ min( M, M)
In the first line, the second denominator is larger, so this second term is smaller, and we get
N3 ≥ M as the summary of both constraints
In the second line, the second denominator is larger, so this second term is smaller, so
N3 ≤ M
So this entire mess can be summarized in this simple way
M ≤ N3 ≤ M
In my application this says
(u-2) M ≤ N3 < (1/2) (u-1) M
If I choose u = 1.5 for the example, we get
-(1/2) M ≤ N3 < (1/2) (1/2) M
and this boils down to
0 ≤ N3 ≤ M/4