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Can someone help with 1.6: Exploration, page 15 in above document? I am curious for a solution. I found this problem very intereesting.

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Which problem? 1,2 ,3 or what? – Babak S. Feb 1 '13 at 10:38
this exploration problem is 9 parts. I needed help with all if you can. THanks – mary Feb 1 '13 at 19:11
@mary: Have you reviewed the Chapter and tried any parts of it? Are there parts which you do not understand? Have you done any of (1) through (9)? Regards – Amzoti Feb 2 '13 at 17:05
I reviewed the chapter, tried out parts but was making some mistakes. Can you put a solution here so I can check over my work? – mary Feb 2 '13 at 21:03
@mary: if you don't put the at-user, I would not receive notice, so it is always good to add. For me, it is at-amzoti, followed by your comments. Regards – Amzoti Feb 4 '13 at 0:03
up vote 4 down vote accepted

The source of this problem is "Differential Equations, Dynamical Systems, And an Introduction to Chaos", 2nd edition, by Hirsch, Smale and Devaney.

This is problem number 1.6: Exploration: A Two-Parameter Family.

Consider the family of differential equations

$$\tag 1 x' = f_{a, b}(x) = ax -x^{3} -b$$

which depends on two parameters, $a$ and $b$.

The goal is to explore and put together a complete picture of the two-dimensional parameter plane (the $ab$-plane) for this differential equation.

(Part-1) First, fix $a = 1$. Use the graph of $f_{1, b}$ to construct the bifurcation diagram for this family of DEQs depending on $b$.

$$\tag 2 x' = f_{1, b}(x) = x -x^{3} -b$$

Varying $b$ corresponds to a vertical shift of the $x$-axis in the plot of the graph of $f_{1, b}(x) = x -x^3 -b$.

For $b = 0$, $(2)$ becomes

$$\tag 3 x' = f_{1, 0}(x) = x -x^{3}$$

To find the equilibrium points, we set $x' = g(x) = x - x^{3} = x(1 - x^{2}) = x(1 + x)(1-x)$, which gives us three equilibrium points for this equation that are $-1, 0$ and $1$.

To classify these, we find:

$$g'(x)= 1 - 3x^{2} = \begin{cases} 1,&\text{if }x = 0\\\\ -2,&\text{if } x = \pm 1\\\\ \end{cases}$$

Hence, $0$ is a source (because $g'(x) \gt 0$) and $\pm 1$ are sinks (because $g'(x) \lt 0$).

$x' = f_{a, 0}(x) = x -x^{3}$ is positive in the interval $(- \infty, -1)$, negative on $(-1, 0)$, positive on $(0, 1)$ and negative on $(1, + \infty)$.

Therefore, for the $b = 0$ case, you can draw a phase-portrait where the orbits are the open intervals $(- \infty, -1)$, $(-1, 0)$, $(0, 1)$, $(1, + \infty)$, and the points $\{-1\}$, $\{0\}$ and $\{1\}$, which have a stable orbit structure.

Next, we need to consider:

$$\tag 4 x' = f_{a, b}(x) = x - x^{3} - b = 0$$

From a geometric point of view, changing $b$ in $x' = f_{1, b}(x) = x - x^{3} -b$ simply shifts the graph of $x - x^{3}$ up and down. So the more negative $b$ (the more to the left), the graph of $f_{1, b}$ crosses the $x$-axis at a higher point. $f_{1, b}$ is clearly increasing through that point, so we have a source for large, negative values of $b$.

A bifurcation occurs when we increase $b$ enough so that the graph of $f_{1, b}$ now touches zero on its local maximum. From this we see also that since $0$ is a maximum, the phase line of $x$ must decrease through that equilibrium point.

Increasing $b$ even more, we have that the zero-crossing occurs at three different places: increasing, then decreasing, then increasing again. So the order (from the bottom to top) is source, sink, and source again.

This continues until $b$ makes the graph touch zero at its local minimum, which means the point is again neither a source nor a sink, and $x$ increases through that point. Finally shifting by a large positive $b$ makes the graph of $f_{1, b}$ again cross at only one place, the lower branch, where it is increasing. So those points are sources.

So, lets find the values where these occur. The critical case occurs when the horizontal line $b$ is just tangent to either the local minimum or maximum of the cubic; then we have a saddle-node bifurcation. To find the values of $b$ at which the bifurcation occurs, note that the cubic has a local maximum when $\frac{d}{dx}(x - x^{3}) = 1 - 3x^{2} = 0$. Hence,

$$x_{max} = \frac{1}{\sqrt{3}}$$

and the value of the cubic at the local maximum is

$$x_{max} - (x_{max})^{3} = \frac{1}{\sqrt{3}} - (\frac{1}{\sqrt{3}})^{3} = \frac{2}{3 \sqrt{3}}$$

Similarly, the value at the minimum is the negative of this quantity. Hence saddle-node-bifurcations occur when $ b = \pm \frac{1}{\sqrt{3}}$.

The flow continues to have a stable orbit structure for small values of $b$, that is, for $- b_1 \lt b \lt b_1$, where $b_1 = \frac{2}{3 \sqrt{3}}$ is the local maximum value and $- b_1$ is the local minimum value of $f_{1, b}(x)$.

For $b = -b_1$ or $b = b_1$, the equation is at a bifurcation point. The parameter values $b \lt - b_1$ and $b \gt b_1$, the equation has a stable orbit structure.

Now, you should be able to draw the entire bifurcation diagram for $f_{1, b}(x) = x - x^{3} -b$.


If we start the system $f_{1, b}(x) = x - x^{3} -b$ with a very large negative value of $b$, after a long time, regardless of the initial condition of $x_0$, the system will be near a stable equilibrium state on the left of the cubic. If we continuously increase the value of $b$, since the system was near a stable state when we began to vary $b$, it will stay near this stable state for small variations in $b$. In fact, as we increase the parameter $b$, the system will follow that state equilibria on the left until $b = b_1$.

At this point, the system will jump to a different stable equilibrium state on the right leg of the cubic. As we continue to increase the parameter $b$, the system will follow the stable equilibria on the right. The system will experience a jump at two different values of $b$, depending on which direction the parameter $b$ is varied.

This phenomenon is called a Hysteresis loop and can occur in many physical systems.

Parts 2 through 4 are a variation of the above - so you should work them.

Parts 5 through 9 are the DEQ in it's full glory for $a, b$. This leads to what is called a Fold - Cusp system.

Hints for remaining parts:


$$f_{a, b}(x) = 0~~ \text{and} ~~ \frac{\partial}{\partial x} f_{a, b}(x) = 0 $$

This yields:

$$ ax -x^{3} - b = 0 ~~ \text{and} ~~ a - 3x^{2} = 0 $$

Eliminating $x$ from the resulting simultaneous equations, results in a cusp in the $(a, b)$-plane given by: $4 a^{3} = 27 b^{2}$.

This system has a lot of interesting behaviors:

  • We already showed hysteresis when fixing $a = 1$.

  • At $b = 0$, we have a pitchfork bifurcation for $x' = ax - x^{3}$.

  • At $b = -1$, we have a nicely disguised super-critical saddle-node bifurcation!

Drawing the $(a, b, x)$-space cusp bifurcation diagram has all of those hidden gems! Explore!!!

Lastly, you might find Section 1.2 of THE CUSP–HOPF BIFURCATION paper helpful to read.


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Very good solution m8 love the detail. – Faust7 Apr 11 '13 at 2:50
@Faust7: Thank you! That is a very nice problem! – Amzoti Apr 11 '13 at 2:52
Great answer, great feedback! What more... ;-) +1 – amWhy May 4 '13 at 2:16
@amWhy: That is actually a fun, beautiful problem from that author - all sorts of things to explore and many concepts used. Regards – Amzoti May 4 '13 at 2:17

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