Exercises 3.4.e
Exercises 3.4.e
For:=rxsinh(x)
x
i) Sketch the different vector field types that appear when you vary .
r
So, for this example we can’t solve the equation analytically. This is what’s known as a transcendental equation, so we have to be a little smarter. You can look up the graph of and you will get:
0=rxsinh(x)
y=sinh(x)
In[]:=
Plot[Sinh[x],{x,4,4},AxesLabel>{Style["x",18],Style["y",18]},AspectRatio>1,PlotStyle>Blue]
Out[]=
From this immediately it should give you hints of why we might expect a pitchfork bifurcation. This looks quite a bit like the graph of It’s not quite the same, but it has some qualitative features. OK, so to work out the vector fields, we have to plot for different values of see whether in which case>0, or vice versa. Clearly for all negative values of be a single point of intersection (at and for and for . For instance:
y=.
3
x
y=rx
rand
rx>sinh(x)
x
r,therewill
x=0)
rx>sinh(x)
x<0
rx<sinh(x)
x>0
In[]:=
Show[Plot[{Sinh[x],x},{x,4,4},AxesLabel>{Style["x",18],Style["y",18]},PlotStyle>{Blue,Red}],Graphics[{Thick,Arrowheads[0.02],Arrow[{{4,0},{0.1,0}}],Arrow[{{4,0},{0.1,0}}],Disk[{0,0},{0.1,0.3}]}],ImageSize>700,AspectRatio>1]
Out[]=
Things don’t start getting interesting until the gradient of is as large as the gradient of , which happens when:
y=rx
y=sinh(x)(atx=0)
d(rx)
dx
d(sinh(x))
dx

x=0
At this point we will have:
In[]:=
Show[Plot[{Sinh[x],x},{x,4,4},AxesLabel>{Style["x",18],Style["y",18]},PlotStyle>{Blue,Red}],Graphics[{Thick,Arrowheads[0.02],Arrow[{{4,0},{0.1,0}}],Arrow[{{4,0},{0.1,0}}],Disk[{0,0},{0.1,0.3}]}],ImageSize>700,AspectRatio>1]
Out[]=
So it’s still stable at When however, we get something else happening. Let’s look at
r=1.
r>1
r=2:
In[]:=
sols=x/.FindRoot[2x==Sinh[x],{x,#}]&/@{4,0,4}Show[Plot[{Sinh[x],2x},{x,4,4},AxesLabel>{Style["x",18],Style["y",18]},PlotStyle>{Blue,Red}],Graphics[{Thick,Arrowheads[0.02],Arrow[{{4,0},{sols[[1]]0.1,0}}],Arrow[{{0.1,0},{sols[[1]]+0.1,0}}],Arrow[{{0.1,0},{sols[[3]]0.1,0}}],Arrow[{{4,0},{sols[[3]]+0.1,0}}],Circle[{0,0},{0.1,0.7}],Disk[{sols[[1]],0},{0.1,0.7}],Disk[{sols[[3]],0},{0.1,0.7}]}],ImageSize>700,AspectRatio>1]
Out[]=
{2.17732,0.,2.17732}
Out[]=
Let’s make sure that we can put this into the appropriate Normal Form. One thing which will help us is to know that:
and therefore its Maclaurin expansion is given by:
Which is the Normal Form of a supercritical pitchfork bifurcation.