whose equilibrium points occur at the roots of f(x)=0, or
Noting that the minimum of the parabola that represents the vector field occurs at x=-r/2, we can graph a few representative flows.
SEE HARD COPY FOR FIGURES
We see that 
are bifurcaton points and a saddle node (blue
sky) bifurcation occurs at both values. For |r|>2, there are two
equilibrium points, the smaller is stable and the larger is unstable.
For |r|<2, there are no equilibrium points. The bifurcation diagram
looks like this:
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shows that the vector field is given by a parabola with zeros (fixed points) at x=0 and x=(r- 1)/r. A few sample phase portraits are essential here.
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Notice that the bifurcation takes place at r=1, but that the fixed points are undefined at r=0. The bifurcation diagram, shown below, does not conform strictly with any of the prototype forms, but resembles the transcritical bifurcation most closely.
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We see that the phase portraits consist of cubics with zeros at x=0
and 
.
The bifurcation occurs at r=0 since there is one fixed point for r<0
and three fixed points
for r>0. Several phase portraits are shown below.
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The bifurcation diagram shows a singularity (fixed points are undefined) at r=0. This bifurcation doesn't fit any of the normal forms exactly. The fact that two stable fixed points coexist with an unstable fixed point for r>0 gives it the appearance of a pitchfork bifurcation.
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We see that the phase portraits consist of cubics with zeros at x=0
for all r and at 
when r>0.
The bifurcation occurs at r=0 since there is one fixed point for r<0
and three fixed points
for r>0. Several phase portraits are shown below.
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The bifurcation diagram (above right) shows a standard pitchfork bifurcation.
, which has
equilibrium points at 
provided 
. Some
typical phase portraits are given in the three left figures below.
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The bifurcation diagram (right figure above) shows two equilibrium points, one stable and one unstable, for r>0 and no equilibrium points for r<0. This is a standard saddle-node bifurcation.
- Adding the three ODEs we see that x'(t) + y'(t) + z'(t) = (x(t) + y(t) + z(t))' = 0 which implies that x(t) + y(t) + z(t) is constant for all times. Call this constant N, the total number of people ``in the system'' at all times.
- Since
, it follows that 
. Therefore, separating variables,
Integrating once, we find that
. If we
assume that initially there are no dead people (z(0) = 0), then
. This relationship between x(t) and z(t) holds for all
times 
. In particular, it follows that
that is, in the steady state the healthy people do not vanish.
- We can now make some substitutions in the z equation and reduce
the three ODEs to a since ODE in z. We have that
- It helps to scale (nondimensionalize) the ODE. Letting
and 
, where T is to be determined, we find
that
Letting
, we can write
- The graphs show that if b>1, then
has a maximum at
. If 0<b<1, then has a maximum at some
. Since 
and , z'(t)
will have a maximum at the corresponding time 
. Since
(by the third ODE), y(t) has a maximum at the same
time as z'(t). This says that the death rate is a maximum at the
same time that the number of infected people is a maximum. - The above graphs and arguments show that if 0<b<1, then
increases initially until 
and then
decreases to zero. This means that the death rate increases for a while
before decreasing; this is the case of an epidemic. - The above graphs and arguments show that if b>1, then
and decreases monotonically to zero. The death rate
decreases steadily and this is the non-epidemic case.
- The second ODE reads
. If
, then
and the infected population starts out
with a zero growth rate. If 0<b<1, then 
which means
y'(0)>0 and the infected population increases initially. If b>1,
then 
which means y'(0)<0 and the infected population
decreases. The condition b=1 gives the critical size 
that determines if there are enough healthy people to start an epidemic. - The model assumes a self-contained population with no influx of healthy people. Any epidemic model would need to have a supply of healthy people. An AIDS model would also need a category for non-symptomatic carriers of the HIV.
The equations can be integrated directly to give a conserved quantity. Dividing y'(t) by x'(t) gives us the single equation
where C is an arbitrary constant. In other words, the quantity 
, which seems to have no physical meaning, is conserved
along any trajectory. If initial conditions 
are specified,
then the solution is
We see that if y(0)=0 (no infected people initially), then the system
remains in that state. However, if y(0)>0 and 
, then the
infected population increases while the healthy population decreases.
The epidemic peaks when 
and then the infected population
decreases (due to deaths). If y(0)>0 and 
, then the
infected population decreases from the start and an epidemic does not
occur. In either case, 
and
, which says that
steady state population of healthy people, 
, is always
positive (and not zero).
, by 
to give
We can integrate this equation once to obtain 
.
Using the initial conditions 
(initial angular displacement of 
and initial velocity of zero),
the constant C can be evaluated and we have
The fact we will need is that
If the pendulum swings from 
to 
, it passes
through a quarter of a period, taking time T/4 to do so. We can
integrate the previous expression with respect to t to determine the
period T:
We choose the minus branch of the square root, since for this part of
the period ( 
), 
.
The remaining task is to evaluate this integral. We first use the
identity 
to write
Now change variables from 
to 
by defining 
. Note that when , then 
and when , then 
. Doing the change of
variable give us
where 
and K(k) is the complete elliptic
integral of the first kind.
Notice that letting 
(hence 
),
we recover the linear period 
. For k<<1, the integrand can be
expanded in a binomial series
( 
) and integrated term-by-term:
We see that the effect of the nonlinear terms is to increase the period
of the pendulum over the period of the linear pendulum (which is
).