The name is absent



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)
a
i2


(16)


Combined with (15) gives:

ai(t)      ʌ / . /                     

—2- = 2x(t)(nιCι + n2C2)
ai


1
a
i (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)
(x
2(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




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