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Probably a trivial question. Let's say I have the following system of equations:

\begin{cases} f\left(x,y,p\right)=0\\ \\ y=g\left(x\right) \end{cases}

where $p$ is a parameter, and I want to study its saddle-node bifurcations (when an eigenvalue goes to zero), and its Hopf bifurcations (when two eigenvalues have the form $\lambda = \pm i\omega$). I can think about two methods to calculate the eigenvalues $\lambda$:

  1. Use the Jacobian matrix of the whole $2 \times 2$ system:

\begin{equation} \det\left(J-\lambda I\right)=\left|\begin{array}{cc} \frac{\partial f\left(x,y,p\right)}{\partial x}-\lambda & \frac{\partial f\left(x,y,p\right)}{\partial y}\\ -\frac{dg\left(x\right)}{dx} & 1-\lambda \end{array}\right| \end{equation}

  1. Replace the second equation into the first, obtaining $f\left(x,g\left(x\right),p\right)=0$, and calculate the Jacobian matrix (i.e. the derivative) of this reduced $1 \times 1$ system:

\begin{equation} \lambda=\frac{df\left(x,g\left(x\right),p\right)}{dx} \end{equation}

Are the two methods equivalent? Thanks in advance for your help!

P.S.: in the second approach I get only one eigenvalue, which means the system cannot undergo Hopf bifurcations. Notwithstanding, also the first method in principle could generate only real eigenvalues, so it's not clear to me if the two approaches are equivalent or not.

EXAMPLE

In the dynamical system studied here by means of the first approach, we see there are no Hopf bifurcations, since the real part of $\lambda$ is always different from zero. Moreover, there is a saddle-node bifurcation for $\mu = 1$. Now, if I use the second approach, and I replace for example the first equation into the second, I get $\mu+x^{2}-2x=0$, and therefore $\lambda=2x-2$, so $\lambda =0$ for $x=1$, which according to the first equation gives $\mu =1$. So the two methods are equivalent in this example.

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  • $\begingroup$ Where all time derivatives have gone?? $\endgroup$
    – Evgeny
    Commented Nov 15, 2015 at 19:47
  • $\begingroup$ Let's say the first equation is $\frac{dx}{dt}=f\left(x,y,p\right)$ while the second is $\frac{dy}{dt}=y-g\left(x\right)$, but they are both set to zero because saddle-node and Hopf bifurcations are evaluated at the equilibrium points. $\endgroup$ Commented Nov 15, 2015 at 19:52
  • $\begingroup$ Ah, ok, I got it then. What were you considerations when you've decided to make this modification to the method? $\endgroup$
    – Evgeny
    Commented Nov 16, 2015 at 10:15
  • $\begingroup$ My idea was basically to try to reduce the size of the Jacobian matrix through this trick, because this should simplify the calculation of the eigenvalues. Intuitively,since the local bifurcations depend only on the equations at equilibrium, I'd expect that two methods should be equivalent. Not sure though. It would be nice to find a counter example to my reasoning. $\endgroup$ Commented Nov 16, 2015 at 10:55

2 Answers 2

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For a counter example to your approach take $f(x,y) = x + y$ and $g(x) = x^2$. So $$\frac{df(x, g(x))}{dx} = \frac{d(x+x^2)}{dx} = 2x+1 \neq 0$$ for the both equilibrium points $(0,0)$ and $(-1,1)$.

$f(x_0, g(x_0))=0$ doesn't mean that $f'(x_0, g(x_0)) = 0$ also, as seen in the example.

Also, total derivative of $f$ w.r.t $x$ doesn't make sense for linearization. You need to calculate both $\partial f / \partial x$ and $\partial f / \partial g$ for multivariable functions.

So, you cannot possibly determine the bifurcation points by using the "reduced" system. You need to calculate the eigenvalues of the Jacobian and find critical $p$ such that $\operatorname{Re} \lambda = 0$ to find the stability changes.

Edit: The thing is equilibrium points change with $p$. In my example if we take $f(x,y,p) = x+y+p$, then the equilibrium points would be determined by the roots of the polynomial $x^2+x+p$, whenever they are real. So a bifurcation would occur at $p=1/4$, which is a transition from 2 equilibrium points to none.

In general, a bifurcation occurs whenever the number of equilibrium points change or the behaviour of one of the equilibrium points change. To find for which $p$ this occurs, you need to find $p$ such that $\operatorname{Re} \lambda = 0$. Even so, this $p$ is a candidate, so you need to check if this is indeed a bifurcation.

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  • $\begingroup$ Thanks obareey. Indeed a counter example is what I am looking for, but there is something I don't understand in your explanation. Your first equation should be something like $f\left(x,y,p\right)=x+y+p$. So I think you forgot to add the bifurcation parameter, which can be adjusted to some value $\overline{p}$ such that $x+y+\overline{p}=0$ at the equilibrium points. It should be possible to have derivative equal to zero at the equilibrium points. $\endgroup$ Commented Nov 16, 2015 at 9:06
  • $\begingroup$ @user2983638 $p$ is a free parameter, you cannot "adjust" it however you want, but you can analyze what happens if $p$ has certain values. This is what bifurcation is all about. See my updated answer. $\endgroup$
    – obareey
    Commented Nov 16, 2015 at 11:11
  • $\begingroup$ Yes sure :) What I meant is that if you don't consider $p$, you cannot reconcile the two equations $x+x^2=0$ and $\frac{df\left(x,g\left(x\right)\right)}{dx}=2x+1=0$. However, what you say is correct: at $p=1/4$ the number of solutions changes from $0$ to $2$, but again I get this result with both the approaches I have proposed, because in both the cases I get $\lambda=0$ at the equilibrium point. As you say I should check also the remaining conditions, notwithstanding it's not clear to me why this should be a counter example. $\endgroup$ Commented Nov 16, 2015 at 14:41
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Ok I have found a possible answer, at least for the saddle-node bifurcations. For the total derivative formula, we get:

\begin{equation} \lambda=\frac{df\left(x,g\left(x\right)\right)}{dx}=\frac{\partial f\left(x,y\right)}{\partial x}+\frac{\partial f\left(x,y\right)}{\partial y}\frac{dy}{dx}=\frac{\partial f\left(x,y\right)}{\partial x}+\frac{\partial f\left(x,y\right)}{\partial y}\frac{dg\left(x\right)}{dx} \end{equation}

so if $\lambda =0$ we get:

\begin{equation} \frac{\partial f\left(x,y\right)}{\partial x}=-\frac{\partial f\left(x,y\right)}{\partial y}\frac{dg\left(x\right)}{dx} \end{equation}

Now, if we replace it inside the $2 \times 2$ Jacobian matrix, we get:

\begin{equation} \left|\begin{array}{cc} -\frac{\partial f\left(x,y\right)}{\partial y}\frac{dg\left(x\right)}{dx}-\lambda & \frac{\partial f\left(x,y\right)}{\partial y}\\ -\frac{dg\left(x\right)}{dx} & 1-\lambda \end{array}\right|=\lambda^{2}+\left(\frac{\partial f\left(x,y\right)}{\partial y}\frac{dg\left(x\right)}{dx}-1\right)\lambda=\lambda^{2}-\left(\frac{\partial f\left(x,y\right)}{\partial x}+1\right)\lambda = 0 \end{equation}

so we obtain again that $\lambda = 0$ is a solution. Probably the remaining solution of the characteristic polynomial is always non-zero at the saddle-node point, which should prove that at least the saddle-node bifurcations are the same in the two methods.

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