Question

7. Consider a family of maps f :R-R, where f(r)= 2+c, cE R. a) Let c 0. Find all the fixed points of f and analyze the map by drawing a cobweb. Check stability of the fixed points b) Find and classify all the fixed points of f as a function of c. c) Find the values of c at which the fixed points bifurcate, and classify those bifurcations. d) For which values of c is there an attracting cycle of period 2?

Answer 1

First, we need to understand what is a fixed point of a function. The simple definition of a fixed point is f(x)=x, so if we apply the function f to the value x, the result is the same value we entered.

So, in order to know the fixed points of this cuadratic function we apply the definition:

f(x) c

a) in this case c=0, so  $x^2=x$

-T=

(1)0

so there are 2 fixed points, x=0 and x=1

By a theorem, we know that a fixed point is attracting or repelling if the derivative of the function evaluated in the fixed point (in module) is greater or lesser than 1 (in case it is one, the theorem does not conclude anything).

ATTRACTING

< 1

REPELLING

So we know that

2.r

in x=0 --> So x=0 is an attracting fixed point.

2 00<1 f(O=

then in x=1--> So x=1 is a repelling fixed point.

= 2> 1 f(1)=2*1

b) Now we need to keep the constant, so

f(x) c

$x^2+c=x$

c- = 0

now we need to apply the cuadratic formula$x=\frac{-b\pm {\sqrt{b^2-4ac}}}{2a}$

$x=\frac{1\pm {\sqrt{1^2-4*1*c}}}{2*1}$

1 t 1-4* c 2.

So there are two fixed points, one with the plus and other with the minus. Now we evaluate the derivative in these points in order to check the stability

$|f'(x)|=|2*x|=|2*\frac{1\pm {\sqrt{1-4*c}}}{2}|=|{1\pm {\sqrt{1-4*c}}}|$

so we need to know for what values of c this module is greater than 1:

$|{1\pm {\sqrt{1-4*c}}}|=1$

first we see the form with the - and we yield these values

$\sqrt{1-4*c}=0$ or $\sqrt{1-4*c}=2$

$c=\frac{1}{4}$ or  $c=-\frac{3}{4}$

So if c is smaller than -3/4 or greater than 1/4 then the result becomes greater than 1, so fixed points are repelling. In case c is in between this two values, then the fixed points become attractive.

For the other case of the solver (with the + sign) the analysis is analogous, and you will find a certain interval at which the fixed points are attractive other at which they are repulsive.

c) There are many types of bifurcation, and without a greater knowledge of the subject and the contents that are supposed tobe given in those courses, i may give an answer that is not completely right.

But I think that the bifurcation are given by the values found earlier for the cuadratic formula. So there will be 4 bifurcations (two for the part with the - sign and another pair for the part with the + sign) and there will be a certain behaviour in the different intervals. Then you will need to clasify the "edges" of those intervals following the guidelines given to you in class.