The name is absent



dHc  _

daj (t)

1
aj(t)

+ λ(t) = 0, j = 1, . . . , n2

(7)

dλ

л - ρλ =

dH
dx

= 2n1c1x(t) + 2n2c2x(t) + λ(t)

b-

2x(t)
(x2(t) + 1)2,

(8)

dHc _
dλ(t)

a(t) -

x2(t)
bx(t) + (x2(t) + 1)2

(9)

From (6) and (7),

λ(t)


1
ai(t)


1
aj(t)


i = 1,...,n1, j = 1,...,n2,


which implies


ai(t) = aj (t) ∀i = 1, . . . , n1 , j = 1, . . . , n2


(10)

(11)


Moreover:

λ(t)


1
ai(t)


i = 1, . . . ,n1 implies n1ai(t) =


ni
λ(t)


and therefore

n1                                                  n2

= n1ai(t), and by similar argument     = n2aj (t),

i=1                                           j=1

which means that

n1           n2

a(t) =     ai(t) +     aj(t) i = 1,...,n1, j = 1,...,n2          (12)

i=1        j=2

and

a(t) = n1ai(t) + n2aj(t) i = 1, . . . , n1, j = 1, . . . , n2              (13)

Note that λ can be expressed as a function of a via the following reasoning:

λ(t)


1
ai(t)


i = 1, . . . ,n1 and λ(t)


1

aj(t)


j = 1, . . . ,n2


⇒      ai(t)λ(t) = -1 i = 1, . . . , n1 and aj (t)λ(t) = -1 j = 1, . . . , n2

n1                                   n2

⇒ λ(t)    ai(t) = -n1 and λ(t)    aj (t) = -n2

i=1                              j=1

Adding the two together:

λ(t) ∑ ai(t) + λ(t) ∑ aj (t) = -(n + П2)
i=1              j=1



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