Who will win?
A stretch of desert is populated by two species of animals, roadrunners and coyotes. The populations r ( t ) and c ( t ) of roadrunners and coyotes t years from now can be modelled by
r ( t + 1 ) c ( t + 1 ) = = 5 r ( t ) − 4 c ( t ) 2 r ( t ) − c ( t )
If there are 30 road runners and 20 coyotes initially (at time t = 0 ), what will happen in the long run? Find t → ∞ lim c ( t ) r ( t ) .
Enter 666 if you come to the conclusion that no such limit exists.
Hint We are told that the roadrunners and coyotes live in two separate oases that are far apart. Initially there are 10 of each species in the "Oasis of Stability", while there are 20 coyotes and 10 roadrunners in the "Oasis of Fertility". Since the given equations are linear, you can study the evolution of each population separately and then add up the result. Just find out what happens in one year!
The answer is 2.
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4 solutions
Yes, that's the approach I had in mind: eigenvectors and eigenvalues (+1)
Let x ( t ) = c ( t ) r ( t ) . Then from the given dynamics, we have the following recursion x ( t + 1 ) = 2 x ( t ) − 1 5 x ( t ) − 4 = f ( x ( t ) ) , where f ( x ) = 2 5 − 4 3 x − 1 / 2 1 Thus x ( t ) = f ( t ) ( x ( 0 ) ) where the superscript implies composition of the function. Hence we are in the framework of Banach's fixed point theorem , if we can argue that (1) f : [ 3 / 2 , 2 ] → [ 3 / 2 , 2 ] and (2) f is a contraction in the interval [ 3 / 2 , 2 ] . (3) x ( 0 ) ∈ [ 3 / 2 , 2 ] . These three conditions can be verified easily. Thus we solve for the fixed point equation in that interval f ( x ) = x to get the answer lim t → ∞ x ( t ) = 2 .
Yes, that's a sensible approach! (+1)
Small typo: In the formula for f ( x ) it should be x − 1 / 2 in the denominator.
The use of Banach's fixed point theorem is certainly optional here since you can actually find the fixed point algebraically.
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Typo corrected. Banach's fixed point theorem is used to show that the limit exists. The fixed point is computed by solving the fixed point equation.
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Right. Since we can find the fixed point, we don't need Banach's theorem to show that it exists ;)
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@Otto Bretscher – I don't understand your comment. Fixed points might exist (here are two fixed points actually) without the limit being any one of the fixed points. For example say x ( t + 1 ) = x ( t ) 2 and x ( 0 ) = 1 0 . We have x ( t ) → ∞ as t → ∞ while the fixed points are 0 , 1 .
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@Abhishek Sinha – As I said, using Banach is optional, but there are several less fancy ways to argue that might be more accessible to younger members.
One way is to point out that (a) 2 is a fixed point , (b) f ( x ) is increasing on [ 1 . 5 , 2 ] and (c) f ( x ) ≥ x on [ 1 . 5 , 2 ] . Graphically, this method can be illustrated by means of a cobweb.
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@Otto Bretscher – Yes. I agree that here Banach is a sledge-hammer to kill a mosquito, but it is good to test it out sometimes :P
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@Abhishek Sinha – I know that you love Banach's fixed point theorem just as much as I love cobwebs ;)
@Abhishek Sinha – Isn't the real point that it is not so much the result of the BFPT that we need (that there is a unique fixed point; in this case this could be established by trivial algebra), but rather the details of the standard proof of the BFPT (namely, that f ( n ) ( x ) converges to the unique fixed point for any initial choice of x )?
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@Mark Hennings – For a contraction, the convergence is trivial once we have the fixed point. I don't see how the proof of BFPT is of any use to us whatsoever.
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@Otto Bretscher – Yes, the convergence is trivial for a contraction. For that matter, the BFPT is pretty trivial, particularly if we restrict attention to a bounded interval of the reals, and not more generally to a complete metric space. (the sequence f ( n ) ( x ) is then a Cauchy sequence, so convergent).
My point was that talking about the BPFT is not really relevant. We do not need the result, since the existence of the unique fixed point is an elementary matter in this case. What we need to answer the question are the (trivial) observations about contraction mappings which ensure convergence, which are like the details of the proof of the BFPT, and not the BFPT itself.
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@Mark Hennings – It depends on your standards for triviality, I suppose. My students in the big calculus "service" courses would frown upon a proof like this (they would most likely not have seen Cauchy sequences). The convergence argument is easily accessible, on the other hand, particularly when supported by a pretty picture of a cobweb. In a situation like this, I would not even bring up contractions.
@Mark Hennings – Also, it depends on what you are including in the statement of the theorem. This wikipedia article includes the convergence result as part of the statement of BFPT.
@Mark Hennings – Yes. That is precisely my point.
By fetching few terms of both the sequences one can easily make out that r ( t ) = 2 0 × 3 t − 1 + 1 0 , and c ( t ) = 1 0 × 3 t − 1 + 1 0 . Hence l i m t → ∞ r ( t ) / c ( t ) = l i m t → ∞ ( 2 0 + 1 0 / 3 t − 1 ) / ( 1 0 + 1 0 / 3 t − 1 ) = 2. Easy and simple!!
True! (+1)
We can also use trial and error method :
r(1)/c(1) = 70/40 = 1.75.
r(2)/c(2) = 190/100 = 1.9.
r(3)/c(3) = 550/280 = 1.96.
r(4)/c(4) = 1630/820 = 1.98.
From above values, we can see the value approaches to 2 .
A table alone is not a fully convincing way to find a limit. It is conceivable that the "trend" could change later one.
If x ( t ) is the vector ( c ( t ) r ( t ) ) , then we have x ( t + 1 ) = ( 5 2 − 4 − 1 ) x ( t ) . The matrix ( 5 2 − 4 − 1 ) has eigenvalues 1 and 3 , with associated eigenvectors ( 1 1 ) and ( 1 2 ) . Since x ( 0 ) = ( 2 0 3 0 ) = 1 0 ( 1 1 ) + 1 0 ( 1 2 ) we deduce that x ( t ) = 1 0 ( 1 1 ) + 1 0 × 3 t ( 1 2 ) so that c ( t ) r ( t ) = 1 0 ( 1 + 3 t ) 1 0 ( 1 + 2 × 3 t ) = 1 + 3 − t 2 + 3 − t which converges to 2 as t → ∞ .
If this model persists, this stretch of desert won't remain deserted long; it will have an acute case of cartoon character overpopulation!