Need help determining bifurcation points

Question

So I have to find the bifurcation points of the system: $\dot{x}=(ax-x^3+x^5)(x-a+2)$, where $a\in\mathbb{R}$ is a parameter.

Attempt:
I know that a bifurcation point is the point, where there is a change in stability or number of fixed points. I have tried visualising the graph, and have come to the conclusion, that there are:
4 fixed points for $a\leq 0$.
6 fixed points for $0<a\leq 0.2$.
2 fixed points for $0.2<a<2$.
1 fixed point for $a=2$
2 fixed points for $2<a$.
The change in stability happens at the same time as the number of fixed points changes.

From what I have learned, I'm pretty sure that one bifurcation point is $(a,x)=(2,0)$, and I think that a transcritical bifurcation happens at this point.

I think there is another bifurcation point, when we go from 4 to 6 to 2 points. I just don't know exactly what that point is? $a=0$? $a=0.2$? It confueses me, that the change seems to happen before and after the interval $0<a\leq 0.2$. Normally the change should happen at a single point?

All help is appreciated!

Answer 1

Another way in which the problem can be addressed without referring to graphs is to consider that the function in the differential equation $ \ \dot{x} \ = \ f(x) \ $ is a sixth-degree polynomial for which the partial factorization is $ \ f(x) \ = \ x·( \ x - [a-2] \ )·(a - x^2 + x^4) \ \ . \ $ Since the coefficients are real, there can be up to six real zeroes, two of which are $ \ x \ = \ 0 \ $ and $ \ x \ = \ a - 2 \ \ . \ $ The number of remaining real zeroes depends upon the value of $ \ a \ $ in the biquadratic equation $ \ x^4 - x^2 + a \ = \ 0 \ \ ; \ $ we find that $$ x^2 \ \ = \ \ \frac{1 \ \pm \ \sqrt{1 \ - \ 4a}}{2} \ \ , $$ as also given by Gregory.

Since the equilibria ("fixed points") of this system are represented by real zeroes of $ \ f(x) \ \ , \ $ we need to examine the discriminant of the biquadratic equation. For $ \ 1 - 4a \ < \ 0 \ \Rightarrow \ a \ > \ \frac14 \ \ , \ \ x^2 \ $ has no real values and for $ \ a \ = \ \frac14 \ \ , \ $ the zeroes are $ \ x \ = \ \pm \frac{1}{\sqrt2} \ \ . \ $

For $ \ a \ < \ \frac14 \ \ , \ $ we must be a bit more thorough. For $ \ 0 \ < \ \sqrt{1 - 4a} \ < \ 1 \ \Rightarrow \ 0 \ < \ a \ < \ \frac14 \ \ , \ \ x^2 \ > \ 0 \ $ has two possible values, permitting four real zeroes of the biquadratic equation. At $ \ a \ = \ 0 \ \ , \ $ we have $ \ x \ = \ \pm 1 \ $ and $ \ x \ = \ 0 \ \ $ (a double zero). With $ \ \sqrt{1 - 4a} \ > \ 1 \ \Rightarrow \ a \ < \ 0 \ \ , \ \ x^2 \ $ is only positive in the case of $ \ x^2 \ = \ \frac{1 \ + \ \sqrt{1 \ - \ 4a}}{2} \ \ , $ giving us two real zeroes.

We have established that $ \ a \ = \ 0 \ $ and $ \ a \ = \ \frac14 \ \ $ are "special" values of this parameter, but we must also consider that the location of one of the zeroes of $ \ f(x) \ $ also depends upon $ \ a \ \ . \ $ So for $ \ a \ > \ \frac14 \ \ , \ $ $ \ f(x) \ $ has the two (single) zeroes $ \ x \ = \ 0 \ $ and $ \ x \ = \ a - 2 \ \ , \ $ except at $ \ a \ = \ 2 \ \ , \ $ when $ \ x \ = \ 0 \ $ becomes a double zero (the polynomial is $ \ x^2·(x^4 - x^2 + 2) \ \ ) \ . \ $ As we will see, these zeroes with multiplicity larger than one are important. For the other cases we've discussed, the $ \ a - 2 \ $ zero is negative and smaller than any other zero.

The number of equilibria (real zeroes) are thus

$ \mathbf{ a \ > \ 2 \ \ : } \quad \quad 0 \ \ \ , \ \ \ a - 2 \ > \ 0 \ \ ; $

$ \mathbf{ a \ = \ 2 \ \ : } \quad \quad 0 \ \ $ (double zero) ;

$ \mathbf{ \frac14 \ < \ a \ < \ 2 \ \ : } \quad \quad a - 2 \ < \ 0 \ \ \ , \ \ \ 0 \ \ ; $

$ \mathbf{ a \ = \ \frac14 \ \ : } \quad \quad -\frac74 \ \ \ , \ \ \ -\frac{1}{\sqrt2} \ \ \ , \ \ \ 0 \ \ \ , \ \ \ \frac{1}{\sqrt2} \ \ $ (with the second and fourth on this list being double zeroes ; the polynomial is $ \ \frac14·x·\left(x + \frac74 \right)·(2x^2 - 1)^2 \ \ ) \ \ $ ;

$ \mathbf{ 0 \ < \ a \ < \ \frac14 \ \ : } \quad \quad a - 2 \ \ \ , \ \ \ -\sqrt{\frac{1 \ + \ \sqrt{1 \ - \ 4a}}{2}} \ \ \ , \ \ \ -\sqrt{\frac{1 \ - \ \sqrt{1 \ - \ 4a}}{2}} \ \ \ , \ \ \ 0 \ \ \ , \ \ \ \sqrt{\frac{1 \ - \ \sqrt{1 \ - \ 4a}}{2}} \ \ \ , \ \ \ \sqrt{\frac{1 \ + \ \sqrt{1 \ - \ 4a}}{2}} \ \ ; $

$ \mathbf{ a \ = \ 0 \ \ : } \quad \quad -2 \ \ \ , \ \ \ -1 \ \ \ , \ \ \ 0 \ \ $ (triple zero) $ \ \ \ , \ \ \ 1 \ \ $ (the polynomial is $ \ x^3·(x+2)·(x+1)·(x-1) \ \ ) \ \ $ ;

$ \mathbf{ a \ < \ 0 \ \ : } \quad \quad a - 2 \ \ \ , \ \ \ -\sqrt{\frac{1 \ + \ \sqrt{1 \ - \ 4a}}{2}} \ \ \ , \ \ \ 0 \ \ \ , \ \ \ \sqrt{\frac{1 \ + \ \sqrt{1 \ - \ 4a}}{2}} \ \ . $

What we may observe from this analysis is that the zeroes of multiplicity greater than one at critical values of $ \ a \ $ are the "bifurcation points" for the system. "Near" $ \ a \ = \ 2 \ \ , \ $ the double zero at $ \ x \ = \ 0 \ \ $ "splits" into $ \ 0 \ $ and $ \ a - 2 \ \ . \ $ The double zeroes that emerge at $ \ a \ = \ \frac14 \ \ $ each "split" into two zeroes as $ \ a \ $ is decreased. Finally, as $ \ a \ $ decreases from small positive values to zero, the zeroes $ \ \pm \sqrt{\frac{1 \ - \ \sqrt{1 \ - \ 4a}}{2}} \ $ "merge" with $ \ x \ = \ 0 \ \ . $

So, as you believed, there is a transcritical bifurcation at $ \ x \ = \ 0 \ $ for $ \ \mathbf{a \ = \ 2} \ \ . \ $ There is a (supercritical) pitchfork bifurcation at $ \ \mathbf{a \ = \ 0} \ \ $ in which the equilbrium at $ \ x \ = \ 0 \ $ changes character as it "splits from" or "merges with" the two nearest equilibria. Finally, there is a "saddle-node" bifurcation at $ \ \mathbf{a \ = \ \frac14} \ \ $ with four equilibria "splitting from" or "merging into" two and vanishing for $ \ a \ > \ \frac14 \ \ . $ (So this system has "something for everybody"...)

$$ \ \ $$

Although it isn't specifically asked for in the problem, we can say a bit about the types of the equilibria. One way we can obtain this information is to differentiate the original differential equation to produce an expression for $ \ \ddot{x} \ \ , \ $ and determine its sign at an equilibrium point for varying values of $ \ a \ \ . \ $ As this is a bit cumbersome for the sixth-degree polynomial, we could instead look at the signs of $ \ \dot{x} \ $ on "either side" of equilibria. As this is still daunting without the use of graphs, we can look at the properties of the polynomial $ \ f(x) \ $ in the vicinity of bifurcation points.

At $ \ a \ = \ 2 \ \ , \ $ we have $ \ x^2·(x^4 - x^2 + 2) \ \ , \ $ which "opens upward" and has a global minimum at $ \ x \ = \ 0 \ \ . \ $ For $ \ a \ = \ 2^{-} \ \ , \ $ the curve "deforms" slightly with one zero at $ \ x \ = \ a - 2 \ < \ 0 \ $ and the other zero at $ \ x \ = \ 0 \ \ . \ $ So $ \ \dot{x} \ $ goes from positive to negative at $ \ a - 2 \ $ and from negative back to positive at $ \ x \ = \ 0 \ \ , \ $ making $ \ x \ = \ a - 2 \ $ a stable equilibrium and $ \ 0 \ $ an unstable one. For $ \ a \ = \ 2^{+} \ \ , \ $ the direction of the changes in $ \ \dot{x} \ $ is reversed, so $ \ 0 \ $ becomes stable and $ \ a - 2 \ > \ 0 \ $ is now unstable. This confirms the transcritical character of the bifurcation at $ \ a \ = \ 2 \ \ . $ [At $ \ a \ = \ 2 \ $ then, $ \ x \ = \ 0 \ $ is a "saddle point" or "semi-stable" equilibrium.]

The next-easiest bifurcation to discuss is $ \ a \ = \ 0 \ \ . \ $ The associated polynomial $ \ x^3·(x+2)·(x+1)·(x-1) \ \ $ changes from positive to negative at $ \ x \ = \ -2 \ $ and the triple zero at $ \ x \ = \ 0 \ $ and from negative to positive at $ \ x \ = \ -1 \ $ and $ \ x \ = \ 1 \ \ . \ $ For $ \ a \ = \ 0^{+} \ \ , \ $ two additional zeroes appear symmetrically around $ \ x \ = \ 0 \ \ ; \ $ as all of the zeroes are now single, we must have the "deformed" polynomial change from negative to positive at $ \ x \ = \ 0 \ $ and from positive to negative at the "new" zeroes. The equilibrium at $ \ x \ = \ 0 \ $ switches from stable to unstable and the new equilibria are stable, while the equilbria at $ \ x \ = \ \pm 1 \ $ remain unstable and the one at $ \ x \ = \ -2 \ $ remains stable.

Finally, there is the bifurcation at $ \ a \ = \ \frac14 \ $ with associated polynomial $ \ \frac14·x·\left(x + \frac74 \right)·(2x^2 - 1)^2 \ \ . \ $ The curve again "opens upward" and, for $ \ a \ = \ \frac14^{-} \ \ , \ $ the six zeroes are all single, so we must have $ \ \dot{x} \ $ change from positive to negative for the first $ \ ( \ a - 2 \ ) \ \ , $ third, and fifth zeroes, and from negative to positive for the second, fourth $ \ ( \ 0 \ ) \ \ , $ and sixth. So $ \ x \ = \ a - 2 \ $ is stable and $ \ x \ = \ 0 \ $ is unstable; we also have the pairs $ \ -\sqrt{\frac{1 \ + \ \sqrt{1 \ - \ 4a}}{2}} \ \ , \ \ -\sqrt{\frac{1 \ - \ \sqrt{1 \ - \ 4a}}{2}} \ $ (unstable-stable) and $ \ \sqrt{\frac{1 \ - \ \sqrt{1 \ - \ 4a}}{2}} \ \ \ , \ \ \ \sqrt{\frac{1 \ + \ \sqrt{1 \ - \ 4a}}{2}} \ $ (stable-unstable). These pairs "merge" into saddle-points at $ \ x \ = \ \pm \frac{1}{\sqrt2} \ $ for $ \ a \ = \ \frac14 \ \ , \ $ which then diaappear ("annihilate") for $ \ a \ > \ \frac14 \ \ . $

Answer 2

Lets take this step-by-step:

• Find the fixed points as a function of our parameters (e.g. $a$).
• Investigate the changes of the locations as we vary the parameters
• Determine the stability of the fixed points.
• Investigate the nature of these fixed points as we vary the parameters.

The fixed points: $$ x_c = 0, a-2, \pm \sqrt{\frac{1 \pm \sqrt{1 - 4a}}{2}} $$

When $a < 0$, we have four solutions. Our first (pitchfork) bifurcation occurs at $a = 0$ where two new solutions emerge. We now have six solutions for $0 < a < 1/4$, where we reach our next double (saddle-node) bifurcation and lose two solutions. Down to only two fixed points, one more (transcritical) bifurcation occurs at $a = 2$.

Stability can be found by linearizing about the critical points in the usual way.

Answer 3

One extra tip that augments the prior answer. The easy way to locate the fixed points is to first use the equation $\dot x=0$ by setting each factor equal to zero, and then (the sneaky part) plot the solution set for each factor by expressing $a$ as a function of $x$. (This is much easier than solving for $x$ as a function of $a$.)

In your example the first factor give the equation (i) $a=x+2$.

The second factor gives either (ii) $a= x^2- x^4$ or (iii) $x=0$ with no restriction on $a$. Plot all these solutions on a single picture and this will reveal how the number of solutions varies as you adjust the variable $a$.

In the graph below (i) is blue, (iii) is the faint green vertical axis, and (ii) is the quartic.

Horizontal lines correspond to setting $a$ constant. As you vary the height ( the value of $a$), the number of solutions changes.