Critical Points and Determining What Happens
In this blog entry we are working with a system of two equations:
x' = f(x,y)
y' = g(x,y)
where x and y are functions of a independent variable, say t for example. Well treat t as a time variable.
Today's blog will cover a three step process:
1. Finding Critical Points
2. Determining the Jacobian Matrix
3. Finding the Eiegenvalues of that Jacobian Matrix
1. Finding Critical Points
It is relatively easy to find the critical points of a system. Add that needs to be done is to set x' = 0 and y' = 0. The next step is to solve for x and y.
2. Determining the Jacobian Matrix
The Jacobian matrix of the system is
Hint: If the system is
x' = Ax + By
y' = Cx + Dy
Then the system's critical point is (0, 0) and the Jacobian Matrix is
J = [ [A, B] [C, D] ]
3. Determine the Eigenvalues
As we recall the eigenvalues of a 2 × 2 matrix can be determined by solving the following characteristic equation:
λ^2 - trace * λ + determinant = 0
Where:
* trace is the sum of the matrix's diagonal elements
* determinant is the determinant of the matrix
The roots are crucial and determine the behavior of the critical point.
Types of critical points:
* source: individual curve solutions x(t) and y(t) are trajectories going away from the critical point
* sink: individual curve solutions x(t) and y(t) are trajectories go towards the critical point
* saddle point: the critical point acts as a sink for some trajectories and a source for other trajectories
* center: trajectories orbit around the critical point, most likely circular or elliptical orbits
Determining the type of critical points:
* λ1 and λ2 are real and positive: critical point is a source.
* λ1 and λ2 are real and negative: critical point is a sink
* λ1 and λ2 are real and have opposite signs: critical point is a saddle
* λ is a double root and it's positive: critical point is a source
* λ is a double root and it's negative: critical point is a sink
* λ = S ± Ti and S is positive: point is a source, trajectories are spiral
* λ = S ± Ti and S is negative: point is a sink, trajectories are spiral
* λ = ± Ti: critical point is a center
Source: http://kaharris.org/teaching/216/Lectures/lec28/lec28.pdf
Let's work with some examples.
1.
x' = x + 4y
y' = 2x + 3y
Critical Point:
0 = x + 4y
0 = 2x + 3y
x = -4y
0 = 2(-4y) + 3y which implies y = 0 and x = 0
Our critical point is (0,0). The Jacobian matrix is J = [ [1, 4] [2, 3] ].
trace(J) = 1 + 3 = 4
det(J) = 1 * 3 - 2 * 4 = -5
The eigenvalues are:
λ^2 - 4*λ - 5 = 0
(λ - 5)(λ + 1) = 0
λ1 = 5, λ2 = -1
Real and opposite signs, the critical point (0,0) is a saddle point.
2.
x' = x + 2y - 6
y' = 6x - 3y + 24
Critical Point:
0 = x + 2y - 6
0 = 6x - 3y + 24
Which leaves is with the system of
x + 2y = 6
6x - 3y = -24
Solving the system yields the critical point (-2, 4).
With
f = x + 2y - 6
df/dx = 1
df/dy = 2
g = 6x - 3y + 24
dg/dx = 6
dg/dy = -3,
The Jacobian matrix is
J = [ [1, 2] [6, -3] ]
The eigenvalues of J are 3 and -5. Since the eigenvalues are real and are of opposite signs, the critical point (-2, 4) is a saddle point.
3.
x' = 2x + 3y
y' = x - 2y^2
Finding the critical points:
0 = 2x + 3y
0 = x - 2y^2
x = 2y^2
0 = 2(2y^2) + 3y
0 = 4y^2 + 3y
0 = y * (4y + 3)
which implies that we have two critical points, one with y = 0 and the other with y = -3/4.
If y = 0, then x = 2*0^2 = 0. One critical point is (0,0).
If y = -3/4, then x = 2*(-3/4)^2 = 9/8. The other critical point is (9/8, -3/4).
With
f = 2x + 3y, df/dx = 2, df/dy = 3
g = x - 2y^2, dg/dx = 1, dg/dy = -4y
Critical Point (0,0):
J = [ [2, 3] [1, 0] ]
And the eigenvalues of J are -1 and 3.
The critical point (0,0) is a saddle point.
Critical Point (9/8, -3/4):
J = [ [2, 3] [1, 3] ]
And the eigenvalues are (5 ± √13)/2. Approximate values are 4.302776 and 0.697224. The critical point (9/8, -3/4) is a source.
Hope this helps and we'll talk next time. Please feel free to post questions and comments.
Eddie
This blog is property of Edward Shore. 2013
