LNCS 2950 - Aspects of Molecular Computing (Frontmatter Pages)

−2α �
kNNk = −2αMN,
k
α � �
kNiNj = α
k i+j=k
�
(i + j)NiNj =2αMN,
i,j
−β �
k(k − 1)Nk = −β �
m(m − 1)Nm,
k
m
Splicing to the Limit 197
2β � �
kNm = β �
�
2 �
�
k Nm = β �
m(m − 1)Nm.
k
m>k
m
kk
m
k 0, then N(t) is defined for all t ≥ 0andN → ¯ N as t →∞.
We want to find the limiting values of the quantities Nk. Firstwehavea
simple result in differential equations.
Lemma 1. Suppose a and b are continuous functions on [t0, ∞) and a(t) →
ā>0 and b(t) → ¯ b as t →∞. Then any solution of Y ′ = −aY + b satisfies
limt→∞ Y (t) = ¯ b/ā.