0

Consider equilibrium solutions of $f(x; \mu) = 0$ where $f :\Bbb{R}\times\Bbb{R}\to\Bbb{R}$ is a smooth function. Suppose that $x_0$ is an equilibrium solution at $\mu_0$, meaning that $f(x_0; µ_0) = 0$. If $$f:\Bbb{R}×\Bbb{R}\to\Bbb{R}$$ is a $C^1$ function, then a necessary condition for a solution $(x_0, \mu_0)$ to be a bifurcation point of equilibria is that $\dfrac{\partial f}{\partial x}(x_0; \mu_0) =0$.

But then they explain how this condition being met doesn't guarantee bifurcations must exist. My question is, if bifurcations do not exist, why would the derivative at $x_0$, $u_0$ be not zero?

Souza
  • 889
childishsadbino
  • 417
  • Take $f(x,µ)=x^3+µ$, then at $(x_0,µ_0)=(0,0)$ the assumptions are satisfied without a bifurcation in the real roots occurring. – Lutz Lehmann Oct 19 '19 at 08:38

1 Answers1

-1

You need to match the normal form of the bifurcation. As long as $\frac{\partial f}{\partial x}$ is non-zero, the implicit function theorem gives a smooth function $x_*(\mu)$ for the stationary point. Which means that for anything interesting to be possible to happen you need $\frac{\partial f}{\partial x}(x_0,\mu_0)=0$, but it is not sufficient, as the counter-example $f(x,\mu)=x^3+\mu$ shows.

To get a fold singularity you need a local isometry of $f$ to $g(y,\lambda)=y^2+\lambda$, meaning you need additionally to the necessary condition also $\frac{\partial^2f}{\partial x^2}(x_0,\mu_0)\ne0$ and $\frac{\partial f}{\partial\mu}(x_0,\mu_0)\ne 0$.

For a pitchfork bifurcation the model is $g(y,\lambda)=y^3-\lambda y$, which implies additional conditions on second, third and mixed derivatives of $f$.

Souza
  • 889
Lutz Lehmann
  • 126,666
  • I thought bifurcations were defined as parameter values for which there was a change in number of equilibrium points. I can't seem to relate that to the statement that around a fixed point, locally I can come up with a function whose graph is the level set (f(x, u)=0). – childishsadbino Oct 19 '19 at 17:05
  • The roots/equilibrium points form that level set. If the $x$-derivative is non-zero at all equilibrium points, then in some small interval around $µ$ the roots are functions of $µ$, there is no change in the number and arrangement of the roots in that interval. – Lutz Lehmann Oct 23 '19 at 07:35