Saddle-Node Bifurcation

Module 1: Low-Dimensional Dynamics

Reading

Compulsory: Strogatz (2024), Section 3.7.
Recommended: (murray2002mathematical?), Chapter 1 (continuous population models for single species).

A bifurcation is a qualitative change in the dynamics as a parameter crosses a critical value: fixed points appear, disappear or change stability. In one dimension, the basic way to create or destroy fixed points is the saddle-node bifurcation (Strogatz 2024, sec. 3.1).

Normal form

The prototype is

\[ \dot x = \mu + x^2 . \tag{1}\]

$\mu < 0$: two fixed points $x^* = \pm\sqrt{-\mu}$. The negative one is stable, the positive one unstable.
$\mu = 0$: one half-stable fixed point at $x^* = 0$.
$\mu > 0$: no fixed points.

As $\mu$ increases through 0, the two fixed points approach each other, collide and annihilate. Near any saddle-node bifurcation of a smooth 1D system, a Taylor expansion reduces the dynamics to Equation 1, which is why it is called a normal form (Strogatz 2024, sec. 3.1).

Conditions

At a saddle-node bifurcation of $\dot x = f(x; \mu)$, a fixed point is also a point of tangency:

\[ f(x^*; \mu_c) = 0, \qquad \frac{\partial f}{\partial x}(x^*; \mu_c) = 0 . \tag{2}\]

The linearization gives no information at $\mu_c$: the decay rate $|f'(x^*)|$ goes to zero, and perturbations relax more and more slowly as $\mu \to \mu_c$. This is called critical slowing down.

Bifurcation diagram of the budworm

Apply Equation 2 to the budworm model with $k$ as the control parameter and $r = 0.56$ fixed. Figure 1 plots the positive fixed points against $k$.

The diagram has an S shape. The middle branch is unstable. At each fold, a stable branch meets the unstable branch in a saddle-node bifurcation.

You can find the folds numerically by checking where the number of equilibria changes:

counts = [len(budworm_equilibria(0.56, k)) for k in np.linspace(2, 16, 1400)]
k_grid = np.linspace(2, 16, 1400)
changes = k_grid[1:][np.diff(counts) != 0]
print("Number of equilibria changes near k =", np.round(changes, 2))

Number of equilibria changes near k = [6.45 9.94]

Exercise

Show that the saddle-node conditions Equation 2 for $f(x) = rx(1 - x/k) - x^2/(1+x^2)$ can be solved for $r$ and $k$ in terms of the double root $x$.

Solution

For $x \neq 0$ divide $f = 0$ by $x$: $r(1 - x/k) = x/(1+x^2)$. Tangency of the two curves means equal slopes: $-r/k = (1 - x^2)/(1 + x^2)^2$. Substituting $r/k$ into the first equation gives

\[ r = \frac{2x^3}{(1 + x^2)^2}, \qquad k = \frac{2x^3}{x^2 - 1}, \qquad x > 1, \]

as in Strogatz (2024), Section 3.7. These curves are used on the next page.

Strogatz, Steven H. 2024. Nonlinear Dynamics and Chaos: With Applications to Physics, Biology, Chemistry, and Engineering. Chapman; Hall/CRC.

--- title: "Saddle-Node Bifurcation" subtitle: "Module 1: Low-Dimensional Dynamics" format: html --- ::: {.callout-note} ## Reading - **Compulsory:** @strogatz2024nonlinear, Section 3.7. - **Recommended:** @murray2002mathematical, Chapter 1 (continuous population models for single species). ::: A bifurcation is a qualitative change in the dynamics as a parameter crosses a critical value: fixed points appear, disappear or change stability. In one dimension, the basic way to create or destroy fixed points is the saddle-node bifurcation [@strogatz2024nonlinear, Section 3.1]. ## Normal form The prototype is $$ \dot x = \mu + x^2 . $$ {#eq-saddle-node} - $\mu < 0$: two fixed points $x^* = \pm\sqrt{-\mu}$. The negative one is stable, the positive one unstable. - $\mu = 0$: one half-stable fixed point at $x^* = 0$. - $\mu > 0$: no fixed points. As $\mu$ increases through 0, the two fixed points approach each other, collide and annihilate. Near any saddle-node bifurcation of a smooth 1D system, a Taylor expansion reduces the dynamics to @eq-saddle-node, which is why it is called a normal form [@strogatz2024nonlinear, Section 3.1]. ## Conditions At a saddle-node bifurcation of $\dot x = f(x; \mu)$, a fixed point is also a point of tangency: $$ f(x^*; \mu_c) = 0, \qquad \frac{\partial f}{\partial x}(x^*; \mu_c) = 0 . $$ {#eq-sn-conditions} The linearization gives no information at $\mu_c$: the decay rate $|f'(x^*)|$ goes to zero, and perturbations relax more and more slowly as $\mu \to \mu_c$. This is called critical slowing down. ## Bifurcation diagram of the budworm Apply @eq-sn-conditions to the budworm model with $k$ as the control parameter and $r = 0.56$ fixed. @fig-bif-k plots the positive fixed points against $k$. ```{python} #| echo: false #| label: fig-bif-k #| fig-cap: "Bifurcation diagram of the budworm model for r = 0.56. Solid: stable fixed points. Dashed: unstable. Saddle-node bifurcations occur near k = 6.45 and k = 9.94." #| fig-width: 7 #| fig-height: 4 import matplotlib.pyplot as plt import numpy as np from amlab.odes_1d.budworm_analysis import budworm_equilibria, budworm_slope r = 0.56 ks, xs, stable = [], [], [] for k in np.linspace(2, 16, 1400): for x in budworm_equilibria(r, k): ks.append(k) xs.append(x) stable.append(budworm_slope(x, r, k) < 0) ks, xs, stable = map(np.array, (ks, xs, stable)) fig, ax = plt.subplots(figsize=(7, 4)) ax.plot(ks[stable], xs[stable], "k.", ms=2, label="stable") ax.plot(ks[~stable], xs[~stable], ".", color="tab:red", ms=1, label="unstable") ax.set_xlabel("k") ax.set_ylabel("x*") ax.legend() ax.grid(alpha=0.3) plt.tight_layout() plt.show() ``` The diagram has an S shape. The middle branch is unstable. At each fold, a stable branch meets the unstable branch in a saddle-node bifurcation. You can find the folds numerically by checking where the number of equilibria changes: ```{python} counts = [len(budworm_equilibria(0.56, k)) for k in np.linspace(2, 16, 1400)] k_grid = np.linspace(2, 16, 1400) changes = k_grid[1:][np.diff(counts) != 0] print("Number of equilibria changes near k =", np.round(changes, 2)) ``` ::: {.callout-note} ## Exercise Show that the saddle-node conditions @eq-sn-conditions for $f(x) = rx(1 - x/k) - x^2/(1+x^2)$ can be solved for $r$ and $k$ in terms of the double root $x$. ::: ::: {.callout-tip collapse="true"} ## Solution For $x \neq 0$ divide $f = 0$ by $x$: $r(1 - x/k) = x/(1+x^2)$. Tangency of the two curves means equal slopes: $-r/k = (1 - x^2)/(1 + x^2)^2$. Substituting $r/k$ into the first equation gives $$ r = \frac{2x^3}{(1 + x^2)^2}, \qquad k = \frac{2x^3}{x^2 - 1}, \qquad x > 1, $$ as in @strogatz2024nonlinear, Section 3.7. These curves are used on the next page. ::: ## Next