Exercises 3.4.e

For:

x
=rx-sinh(x)
i) Sketch the different vector field types that appear when you vary
r
.
​
So, for this example we can’t solve the equation
0=rx-sinh(x)
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
y=sinh(x)
and you will get:
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
y=
3
x
.
It’s not quite the same, but it has some qualitative features. OK, so to work out the vector fields, we have to plot
y=rx
for different values of
rand
see whether
rx>sinh(x)
in which case

x
>0
, or vice versa. Clearly for all negative values of
r,therewill
be a single point of intersection (at
x=0)
and
rx>sinh(x)
for
x<0
and
rx<sinh(x)
for
x>0
. For instance:
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
y=rx
is as large as the gradient of
y=sinh(x)(atx=0)
, which happens when:
d(rx)
dx
=
d(sinh(x))
dx
⟹r=cosh(x)
|
x=0
=1
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
r=1.
When
r>1
however, we get something else happening. Let’s look at
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.
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