## Exercises 3.4.e

Exercises 3.4.e

For:=rln(x)+x-1

x

i) Sketch the different vector field types that appear when you vary .

r

We’re going to do this a little differently from how we’ve done the others. To draw the vector field, you need to be able to draw whether is greater than or less than zero, which is the same as asking, in this case, whether . Let’s first ask this question for positive This is actually the same as comparing So let’s plot this for different values of positive

x

rln(x)isgreaterthanorlessthan1-x

r.

ln(x)and.

1-x

r

randdrawarrowsonthediagrams:

In[]:=

pl1=GraphicsGridPartitionPlotLog[x],,{x,0,4},PlotRange->{{0,4},{-5,5}},AxesLabelEvaluateStyle[#,14]&/@"x","ln(x) and ",AspectRatio->1,PlotLabel->Style["r = "<>ToString[#],14]&/@Range[0.25,2.75,0.5],3,ImageSize->800

1-x

#

1-x

r

Out[]=

We see that for we always have that >ln(x) and so <0, and conversely for , so we have an unstable fixed point:

x<0

1-x

r

x

x>0

In[]:=

GraphicsGridPartitionShowPlotLog[x],,{x,0,6},PlotRange->{{0,6},{-5,5}},AxesLabelEvaluateStyle[#,14]&/@"x","ln(x) and ",AspectRatio->1,PlotLabel->Style["r = "<>ToString[#],14],Graphics[{Thick,Arrow[{{0.9,0},{0.1,0}}],Arrow[{{1.1,0},{3.1,0}}]}],Graphics[Circle[{1,0},0.1]]&/@Range[0.25,2.75,0.5],3,ImageSize->1000

1-x

#

1-x

r

Out[]=

For r=0, we get something different, because we can’t do the dividing by r trick, so we just have to plug in r=0 into the original equation and we get =x-1 which has a very simple phase portrait:

x

In[]:=

Show[Plot[x-1,{x,0,4}],Graphics[{Thick,Arrow[{{0.9,0},{0.1,0}}],Arrow[{{1.1,0},{3.1,0}}]}],Graphics[Circle[{1,0},0.1]],AspectRatio->1,AxesLabelEvaluate[Style[#,18]&/@{"x",""}]]

x

Out[]=

How about for ? We have to be a little bit more careful here. We can again compare then that tells us that <0, which tells us that >0 (because r is negative). Let’s plot the graphs and see what we find:

r<0

ln(x)and,butnowifthisvalueis<0

1-x

r

x

r

x

In[]:=

GraphicsGridPartitionPlotLog[x],,{x,0,4},PlotRange->{{0,4},{-2,2}},AxesLabelEvaluateStyle[#,14]&/@"x","ln(x) and ",AspectRatio->1,PlotLabel->Style["r = "<>ToString[#],14]&/@Range[-2,-0.25,0.2],4,ImageSize->1000

1-x

#

1-x

r

Out[]=

Now we have something interesting going on. Most of the time there are two points of intersection, apart from around where it seems that there’s one. We’ll come to that in a moment. Let’s figure out about the arrows first. When , this means that , which means that , so >0, so for instance in the top left plot, the region between 0 and 1 satisfies this inequality so we will have an arrow going to the right. Let’s fill this in on all diagrams:

r=-1

ln(x)<

1-x

r

ln(x)+<0

x-1

r

rln(x)+x-1>0

x

Now taking the arrows and fixed points alone and plotting them as a vector field:

Now for the bifurcation diagram we need to flip the axes, and we would get something that looks like:

And so the equation close to the critical value of x can be written as

Now the left hand side is really

so let's define

and finally we have

which is precisely the normal form of a transcritical bifurcation.