More Looney Math

Algebra Level 5

A stretch of desert is populated by two species of animals, roadrunners and coyotes, who are engaged in an endless game of rivalry and mischief. The populations r ( t ) r(t) and c ( t ) c(t) of roadrunners and coyotes t t years from now can be modelled by

r ( t + 1 ) = 0.8 r ( t ) 0.7 c ( t ) + 200 c ( t + 1 ) = 0.7 r ( t ) + 0.8 c ( t ) 170 \begin{aligned} r(t+1)&=&0.8r(t)-0.7c(t)+200 \\ c(t+1)&=&0.7r(t)+0.8c(t)-170 \end{aligned}

If there are 310 roadrunners and 200 coyotes initially (at time t = 0 t=0 ), find lim t ( r ( t ) 2 + c ( t ) 2 ) \displaystyle \lim_{t\to\infty}(r(t)^2+c(t)^2) according to this model.

If you come to the conclusion that no such (finite) limit exists, enter 666.


The answer is 666.

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1 solution

Abhishek Sinha
Mar 18, 2016

First, homogenize the dynamics by translating the state-variables appropriately, i.e. find real α , β \alpha, \beta such that with the substitution r ( t ) = r ( t ) + α r(t)=r'(t)+\alpha and c ( t ) = c ( t ) + β c(t)=c'(t)+\beta , the constant part on the right hand side vanishes. Thus, we require α = 0.8 α 0.7 β + 200 , β = 0.7 α + 0.8 β 170 \alpha=0.8 \alpha - 0.7 \beta + 200, \hspace{10pt} \beta=0.7 \alpha + 0.8 \beta -170 Solving this, we get α = 300 , β = 200 \alpha=300, \beta=200 . With the new state-variables ( r ( t ) , c ( t ) ) (r'(t),c'(t)) , by direct substitution, we have r ( t + 1 ) 2 + c ( t + 1 ) 2 = 1.13 ( r ( t ) 2 + c ( t ) 2 ) r'(t+1)^2 + c'(t+1)^2= 1.13 (r'(t)^2 + c'(t)^2) with r ( 0 ) = 10 , c ( 0 ) = 0 r'(0)=10, c'(0)= 0 . Thus it is clear that the limit diverges.

Yes, that's exactly what I had in mind! Thank you! (+1)

Otto Bretscher - 5 years, 2 months ago

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