⇒ λ(t)a(t) = -(n1 + n2)
and therefore
⇒ λ(t) = — ■■
v j a(t)
(14)
From (8)
λ(t) = 2x(t)(n1c1 + n2c2) + λ(t) b+ ρ
2x(t)
(x2(t) + 1)2.
(15)
From (10):
λ(t)
a i(t)
ai2
(16)
Combined with (15) gives:
ai(t) ʌ / .∖ / ∖
—2- = 2x(t)(nιCι + n2C2)
ai
1
ai (t)
b+ρ-
2x(t)
(x2 (t) + 1)∖
(17)
and because ai (t) = aj (t), from equation (11),
(ɪi(t) , , , , 1
= = = 2x(t)(nιcι + n2C2)--7-
ai(t)aj(t) ai(t)
2x(t)
b + ρ —---------9
( (x2 (t) + 1)2
(18)
Multiplying by (n1 + n2)ai(t)aj (t), one obtains:
(nɪai(t) + n2aj (t)) = 2ai(t)(ai(t)nι + aj (t)^)(nɪeɪ + n2C2)x(t)
(ai(t)n1 + ai(t)n2) b + ρ+
2x(t)
(x2(t) + 1)2~
Using results (11) and (13), the above is equivalent to:
t⅛(t) = 2a(t)ai(t)(n1 c1 + n2c2)x(t) — a(t) b + ρ
2x(t)
(x2(t) + 1)2
And noting that ai(t) = ^aa^n from equations (14) and (10) gives:
a(t) = 2α2(t)x(t)(cιn1 + 2n
a(t) bρ
2x(t)
(x2(t) + 1)2~
With constant loading da/dt = 0, and so:
a(t) =
2a*2 (eɪnɪ + c2n2)
nɪ + n2
x(t) — a*
b+ρ-
2x(t)
(x2 t
=0