We discussed in class the definition of the English word bifurcation. Here it is:

Concise Oxford Dictionary, 8th Ed., Copyright 1991 Oxford Univ. Press: /bifurcation/n. 1. a. a division into two branches; b. either or both of such branches. 2. the point of such a division.

Please read the excellent intro given here and borrowed from the UTEP ``SOS math'' project. When you have more time, you may want to also go to their general differential equations resource.

As explained in class (and perhaps too briefly in the book, in page 99), here is a quick way to look for bifurcation values of the parameter for the one-dimensional differential equation

 d y dt = fp(y)
where ``p'' is the parameter and fp(y) is continuously differentiable on y and p.
1. Solve this set of two simultaneous algebraic equations:

 fp(y)
 =
 0
 d fp dy (y)
 =
 0
for values p0 and y0. Let us call the solutions (p0,y0) the ``critical pairs'' of parameter values and dependent variables.
2. Now, for each critical pair (p0,y0), we look at the parameter p0, and determine if it is indeed a bifurcation value.

This procedure is very similar to the procedure which you followed in calc 1, when, in order to maximize a function, you first found critical points, and then studied, for each critical point, if it was a maximization point or not.

As explained in class, the justification for following this procedure works is that when the derivative in 1. is not zero, one has, for any parameter near p0, some equilibrium near y0 and of the same type (source or sink) and thus there is no change of behavior at that equilibrium point, for the given parameter values. So there is no need to look at the values where this derivative is not zero, and it is enough to look for those for which the derivative is zero. The first equation, fp(y) = 0, of course just says that y is an equilibrium.

Let us work out with this method the example of populations under constant fishing given in pages 101-103 of the book. We have the equation

 d P dt = kP æç è 1- P N ö÷ ø - C
and consider C as the parameter of interest. (Yes , one could also think of k and N as parameters. We are just studying one parameter in order to keep things simple.)

We need to solve the two equations in part 1. above, with y being ``P'' and p being ``C''. Actually, let us practice doing this with Maple. We first write the equations, including computing the derivative:

```f:=k*P*(1-(P/N))-C;
fp:=diff(f,P);
```
and we get that the derivative is
 k æç è 1- P N ö÷ ø - kP N
so we next solve:
```solve({f=0,fp=0},{P,C});
```
getting precisely one critical pair:
 ìí î P = 1 2 N,  C = 1 4 k N üý þ
So, is C0 = kN/4 a bifurcation value of the parameter? Let us see.

What are the equilibria for C < C0?

```solve(f,P);
```
gives us:
 kN+sqrt(D) 2k , kN-sqrt(D) 2k
where D = k2N2 - 4kCN. so there are no (real!) equilibrium points if D < 0, i.e. if kN < 4C, i.e., if C is bigger than C0, there is exactly one equilibrium if C = C0, and there are two equilibria if C < C0. This means that yes, C0 is a bifurcation value of the parameter (``the number or the type or equilibria changes'').

You could then go on and decide that types of equilibria we have when C < C0 and at C = C0, as done in the book.

Is this method easier than just doing as in the book? For higher dimensional problems, it does give a good systematic procedure. For the one-dim problems we study here, it depends on the problem. If the equations for the equilibrium are hard to solve, having the second equation may help a lot. To take an example I just made up, say we have

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