## Analysis

Our first goal was to find the steady states of this model. We tried setting both equations equal to zero and solving the system,
x' = c*(y+x-1/3*x^3)
y' = -(x-a+b*y)/c
However, this was unfruitful for x. However, for y, we came up with the steady state expression y=-(x-a)/b.
Plugging this expression into the original x' equation, we get: x'=c*(-(x-a)/b+x-1/3*x^3). Then we take the derivative of this expression and set it equal to zero, then solve for the roots. The roots we get are (1/b*(b*(-1+b))^(1/2), -1/b*(b*(-1+b))^(1/2)). These roots are not dependent on either parameter a or c. They are only dependant on b. We substitute the roots back into our steady state equation and solve that for a. A plot of that equation :

This is a graph of the limit point bifurcation.

Because the system depends on c as well and that isn't represented by this limit point bifurcation, we took the Jacobian of the system. Then we took the trace of the Jacobian and set it equal to zero. We then solved for x. Then we plugged our new definition of x into our original steady state equation and solved for a. There are two ranges for which c is significant. When 0<=c<=1 is one range, and when c>1 is another. We first analyzed the Hopf bifurcation for c=1.

#### c=1

We plotted this hopf bifurcation function on the same plot as the limit point bifurcation. From this graph, we named the three regions A, B, and C and examined the phase portraits of the original equations with the parameters equal to points in each respective region.

##### Region A

a=0.4, b=0.2, c=1

From this phase portrait, we can see that the solution is periodic. No matter what the initial conditions are, the curve will spiral towards the periodic solution.

The above is from xrk and it shows the system's response to different initial conditions.

###### Region B

a=7, b=1, c=1

Region B has a stable fixed point at approximately (2.8, 4.2).

###### Region C

a=0.5, b=3, c=1

For region C, there are two stable fixed points at approximately (-1.2, 0.5) and (1.4, -0.36). Because there are two stable fixed points, there must be a saddle point somewhere in between them.

#### c=2

We plugged our value of x from setting the trace of the Jacobian of the original system equal to zero into our steady state equation and also substituted in c=2. Since there were two roots for the trace of the Jacobian, there are two curves. By plotting these new a,b curves with our previous bifurcation diagram, we found that there are now six more regions to be analyzed. This is the graph of the c=2 bifurcation plot.

Then we analyzed each region by plugging values within the region into the original system and doing a phaseportrait for each.

###### Region D

a=0.1, b=1, c=2

For Region D there is a periodic solution. No matter what the initial conditions are, they will go to this solution.

###### Region E

a=0.2, b=1.2, c=2

The fixed point is at approximately (0.9,-0.6).

###### Region F

a=0.06, b=1.3, c=2

For Region F, the fixed point is at (0.9, -0.7).

###### Region G

a=0.1, b=1.6, c=2

For Region G there are two fixed points. One is at (1.1, -0.6) and the other is at (-0.9, 0.6).

###### Region H

a=0.02, b=1.2, c=2

Region H also has a periodic solution.

###### Region I

a=1, b=2.9, c=2

Region I has two fixed points. One is at (-0.9, 0.6) and the other is (1.7, -0.2).

To see how we analyzed this model, our Maple worksheet is available.