# 285lect12 - 1 Lecture 12 Example 1 Consider a population P...

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1 Lecture 12 Example 1 Consider a population P ( t ) of unsophisticated animals that rely solely on chance of encounter to meet mates for reproductive purposes. We can assume that such encounter occurs at a rate that is proportional to the product P= 2 of males and 2 of females. So, the births occurs at a rate proportional to ( 2) ( 2) , i.e., &P = kP 2 (per unit of time) ) & = kP . Let the death rate be constant. Then P 0 = ( & & ± ) P = ( kP & ± ) P = kP ( P & a ) ; where a = ± k : First solve the foregoing equation: dP P ( P & a ) = kdt ) Z dP P ( P & a ) = kt + C: 1 P ( P & a ) = & 1 a & 1 P & 1 P & a ± ) Z dP P ( P & a ) = & 1 a ²Z dP P & Z dP P & a ³ = & 1 a [ln P & ln j P & a j ] = & 1 a ln ´ ´ ´ ´ P P & a ´ ´ ´ ´ : Finally, ln ´ ´ ´ ´ P P & a ´ ´ ´ ´ = & kat & aC ) P P & a = ± e aC e & kat = Ae & kat : Let t = 0 . Then A = P 0 P 0 & a : 1

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So, P P & a = P 0 P 0 & a e & akt ) P = ( P & a ) P 0 P 0 & a e & akt ) P & 1 & P 0 P 0 & a e & akt ± = & a P 0 P 0 & a e & akt ) P ( t ) = a P 0 P 0 & a e & akt P 0 P 0 & a e & akt & 1 : Consider 3 cases: A. P 0 = a . Note that P ( t ) = a , is a solution of the initial value problem P 0 = kP ( P & a ) ; P (0) = a: By the existence and uniqueness theorem, P ( t ) = a , t ± 0 , is the only solution of the above initial value problem.
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## This note was uploaded on 01/26/2012 for the course MATH 285 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.

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285lect12 - 1 Lecture 12 Example 1 Consider a population P...

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