Math5535: Identifying 1D bifurcations

Raj Saha

September 18, 2013

In this text, we will discuss how to identify 1D bifurcations when given x = f(x)and how to solve for the bifurcation point (rc, xc).

Ex. 1: x = r − 3x2.

Solution:

We recognize that there is no value of x for which x = 0 always. So we can rule outTranscritical or Pitchfork bifurcations1.

When r = 0, x = −3x2, which is the normal form of a saddle node bifurcation. Fixedpoints occur when

x = 0⇒ 3x2 = r

x? = ±√

r

3

Once again we see that the system can have two fixed points for r > 0, one for r = 0,and none for r < 0. Thus we have a saddle node bifurcation at rc = 0, xc = 0.

1This is a simple enough case that we can readily see the normal form of a saddle node bifurcation,but nonetheless it is a good first step to see if a fixed point always exists for a given value of x.

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Ex. 2: x = r + 12x− x/(1 + x)

Solution:

We recognize that there are is no value of x for which x = 0 always, and we can ruleout Transcritical or Pitchfork bifurcations.

At fixed points,

x = 0 = r +1

2x− x

1 + x

0 = (1 + x)(r +1

2x)− x

0 =1

2x2 + (r − 1

2)x + r

x? =−(r − 1

2)±

√(r − 1

2)2 − 4 · 1

2· r

2 · 12

= −(r − 1

2)±

√(r − 1

2

)2− 2r

The number of fixed points can be either one (when (r − 1/2)2 − 2r = 0), or two.Thus we can infer that the system will undergo a saddle node bifurcation.

Solving for the bifurcation point (rc, xc) can be done either geometrically (equatingderivatives of two parts of x) or algebraically (by letting the discriminant equal tozero).

Using the algebraic method,

(r − 1/2)2 − 2r = 0

rc =3

2±√

2

and

xc = −(r − 1

2)

= −1±√

2

2

Ex. 3: x = rx− ln(1 + x)

Solution:We recognize that x = 0 always leads to x = 0, regardless of the value of r. Thusthe system could undergo either Transcritical or Pitchfork bifurcation at x? = 0.

As this function is difficult to solve algebraically, we expand it around x = 0 usingTaylor Series:

rx− ln(1 + x) = rx− (x− x2

2+ O(x3))

≈ rx− x +x2

2

The fixed points are given by,

rx− x +x2

2= 0

x(r − 1) = −x2

2x? = 0

x? = 2(1− r)

Since there could be two fixed points at the most, and one of which always exists,the system would undergo a transcritical bifurcation.

At the bifurcation point, xc = 0, thus

2(1− rc) = 0

rc = 1

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Ex. 4: x = x + rx/(1 + x2)

Solution:

We recognize that x = 0 always leads to x = 0, regardless of the value of r. Thusthe system could undergo either Transcritical or Pitchfork bifurcation at x? = 0.

The fixed points are given by,

x +rx

1 + x2= 0

x(1 + x2) = −rxx? = ±

√−(r + 1)

Thus the system can have three fixed points (two for r < −1 and another at x? = 0),so it would undergo a pitchfork bifurcation at x? = 0.

The bifurcation points can be solved for using either of the methods discussed earlier.The algebraic method is considerably easier here.√

−(r + 1) = 0

rc = −1

and xc = 0.

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