When Does a Laser Model Show a Transcritical Bifurcation?

[Jamin2112]

A simple model of a laser (URGENT!)

Homework Statement

Need to turn this in 2 hrs from now. Yes, I'm a slacker.

(a) Suppose that N evolves much more quickly than n so that we can make the quasi-static approximation N ≈ 0. Given this approximation, re-write our equations as a 1-D dynamical system for n.

(b) Show that for p > p_c, the "trivial" fixed point n* = 0 is unstable, and that for p < p_c, n* = 0 is stable. No need to explicitly determine p_c.

(c) What type of bifurcation occurs as the pump strength increases past the laser threshold p_c?

(d) For what range in the parameters (G, k, f, p) do you expect the quasi-static approximation to be good?

Homework Equations

"This will give some good intuition and background information on the physics involved. The problem we will consider is as follows. Let N be the number of excited atoms in our laser, and let n be the number of photons (particles of light) that ... [etc.] ...

n'(t) = GnN - kn
N'(t) = -GnN - fN + p,

where G is the gain coefficient ... [etc.] ...

The Attempt at a Solution

Part (a) is easy: I have n'(t) = Gnp/(Gn+f) - k. Note that this means n'(t) = 0 if n = 0 or n = p/k - f/g.

So then (n'(t))'(n) = Gpf/(Gn+f)² - k. Plugging in n*=0 to that gives Gp/f - k. I'm assuming that the p_c in the problem means p_c = kf/G ? Because then we have a situation where p > p^c makes Gp/f - k > 0, and p < p_c makes Gp/f - k < 0.

I've noticed also that p = fk/g makes (n'(t))'(n) = 0 when I plug in the other fixed point, n* = p/k - f/g. I'm still trying to figure out the significance of this.

 

[mfb]

I'm assuming that the p_c in the problem means p_c = kf/G ?

Right, as this point is the border between a stable an an unstable point n*=0.

I've noticed also that p = fk/g makes (n'(t))'(n) = 0 when I plug in the other fixed point, n* = p/k - f/g. I'm still trying to figure out the significance of this.

Saddle point? Could be stable.

 

[Jamin2112]

Right, as this point is the border between a stable an an unstable point n*=0.

That's why I hate these deceptive homework problems. When it says "No need to explicitly determine p_b", I get the impression that p_c is a value that's extremely difficult to solve for.

 

[Jamin2112]

Saddle point? Could be stable.

I think it would be considered a saddle point since there's only one fixed point when p = p_c, but then another appears out of nowhere when p ≠ p_c.

EDIT: It would be a "transcritical bifurcation" since the stability at n=0 changes as p is varied from p=p_c to p>p_c.