Download Assignment 2

Computer Simulation and Applications
Assignment No 2
André Deutz
September 9, 2008
Due Date: September 16, 2008
Number of credit points is 8. Refer any questions you might have to [email protected].
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Assignments
1. Study the dynamics of the following maps. In each case analyze the map
numerically (by writing a program in your favorite language or using a
calculator or using a spreadsheet) and with cobwebbing – in that case
plot the graph by, for instance, gnuplot. Find out whether the maps have
sinks or sources. Determine also the basin of attraction for the sinks. For
each of the initial value (that is for each element of the domain), describe
the behavior of the orbit (that is to what limit does the orbit tend to).
Possible limit values are, of course, the real numbers. Next to the real
numbers we consider infinity also as a possible limit value (that is, in that
case the orbit tends to infinity).
a f1 (x) = x2 , range=domain=R
√
b f2 (x) = x, range=domain=non-negative reals.
c f3 (x) = sin(x), where x is in radians and range=domain=R.
d f4 (x) = cos(x), where x is in radians and range=domain=R.
2. Consider the map f5 (x) = 2x(1 − x) with domain and range equal to R.
Verify the results about this map stated in the lecture yourself. That is,
compute its fixed points. Using the theorem of the lecture classify them.
Compute the basin for each of the sinks. Again state for each initial value
what the behavior of its orbit is. Again use of computing tools in case the
algebraic analysis is too tough.
3. Consider the map f6 (x) = (3x − x3 )/2, domain=range=all reals. Try to
analyze this map as you have done in the previous cases. Also try to find
a large set of initial conditions that do not converge to any sink of f6 by
solving the inequality |f6 (x) − 0| > |x − 0|: this identifies points whose
distance to 0 increases on each iteration.
4. Consider the map f7 (x) = 4x(1−x), domain=range=[0, 1]. Try to analyze
this map as you have done in the previous cases.
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5. The map f8 (x) = 3x2 − 5x on R has fixed points at x = 0 and x = 3. Find
a period-two orbit of f8 by solving f82 (x) = x for x.
6. Let p be a fixed point of a map f. Given some ² > 0, find a geometric
condition under which all points x in N² (p) are in the basin of p. Use
cobwebbing (cobweb plot analysis) to explain your reasoning.
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Assignments suggested by the lecture of September 9
1. In order to motivate the study of maps, we discussed in the lecture at least
three situations in which they emerge. One of them is the discretization
of differential equations. The other two were the Poincaré map and the
logistic growth model.
Consider the differential equation Michael alluded to in the lecture of
September 2:
x0 (t) = k · x(t)
, where k is some constant and x0 (t) is the derivative of the function x(t)
with respect to t.
Recall the definition of the derivative: x0 (t) = limτ →0
this definition we can loosely say that
x(t+τ )−x(t)
.
τ
Using
x(t + τ ) − x(t)
≈ x0 (t)
τ
, where τ is a very small positive number. Using this approximation derive
the rule f in the following:
x(tn + τ ) = f ((x(tn ))
. Remark: by setting tn+1 = tn + τ we will have our map (or our discrete
dynamical system).
Aside: Of course, this could also be done for more general situations such
as differential equations (say) of the form:
x0 (t) = F (t, x(t))
where F is some continuous function. F : D → R and D ⊆ R2 .
2. In the lecture we discussed the following theorem.
THM 1 Let f be a smooth map on R, and assume that p is a fixed point
of f . Then:
a If |f 0 (p)| < 1, then p is a sink.
b If |f 0 (p)| > 1, then p is a source.
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We remarked in the lecture that we can readily proof this. We also remarked that proof is instructive because we learn from the proof what the
rate of convergence is of f k (x) to p. So we will proceed to give the proof
of 2a:
Let b be any number between |f 0 (p)| and 1. For example, b could be chosen
to be (|f 0 (p)| + 1)/2. Since,
lim
x→p
|f (x) − f (p)|
= |f 0 (p)|
|x − p|
, there is a neighborhood N² (p) for some ² > 0 such that
|f (x) − f (p)|
<b
|x − p|
for any x in N² (p), x 6= p.
In other words, f (x) is closer to p than x is by at least a factor of b (which
is less than 1). This implies two things:
a if x ∈ N² (p), then f (x) ∈ N² (p); this means that if x is within ² from
p, then so is f (x), and by repeating the argument, so are f 2 (x), f 3 (x),
and so forth. (Remember f (p) = p)
b Second, it follows that
|f k (x) − p| ≤ bk |x − p|
, for all k ≥ 1 .
Thus p is a sink. ¤ We now state two questions you are invited to answer:
a Show that the inequality |f k (x) − p| ≤ bk |x − p| holds for k = 2.
Then carry out the mathematical induction argument to show that
it holds for all k = 1, 2, 3, . . ..
b Use the ideas of the proof in 2a of THM 1 to prove Part 2b.
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