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5.1 EQUILIBRIUM POINT ANALYSIS
dx/dt = y
dy/dt = -x + (1 - x2)y
Determine all equilibrium points
dx/dt = y
dt/dt = -x + y - x2y
What happens near (0,0)?
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dx/dt = y
dy/dt = -x + y
What are eigenvalues? What type of
system do we have?
The linear system and the Van der Pol
system have vector fields that are similar
near the origin.
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Since systems of the linear system spiral away
from the origin, solutions of the Van der Pol
equations that start near the origin also spiral
away.
This technique is called linearization. Near
the equilibrium point we approximate the
nonlinear system by an appropriate linear
system.
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Competing Species Model
What does an increase in either population
mean for the population of the species?
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What happens if y = 0?
What happens if x = 0?
Find all equilibrium points.
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Equilibrium points are (0,0), (2,0), (0,3),
and (1,1).
What does the point (1,1) say about the
system?
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Linear systems always have an equilibrium
point at (0,0). We will use a change of
variables to move (1,1) to the origin.
u = x-1 and v = y-1
Rewrite the system in terms of the new
variables.
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Note: (0,0) is an equilibrium point.
Which terms are smaller near the
origin?
Approximate with a linear system near
(0,0).
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Find the eigenvalues and determine what
type of equilibrium point we have at (0,0).
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Linearization:
dx/dt = f(x,y)
dy/dt = g(x,y)
u = x - x0
v = y - y0
du/dt = f(x0
+ u, y0 + v)
dv/dt = g(x0 + u, y0 + v)
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We want the best "linear approximation" of the
function. For functions of two variables, the
best linear approximation at a particular point
is given by the tangent plane.
Thus:
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Write as:
This is called the linearized system at the
equilibrium point
The 2x2 matrix of partial derivatives is called
the Jacobian matrix.
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Example:
dx/dt = x(1- x - y)
dy/dt = y(-2x - y + 100)
a. find and classify all equilibria in the first
quadrant.
b. sketch the phase portrait of the system
near each equilibrium point.
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Classifications:
Sink
Spiral sink
Source
Spiral source
Saddle
This classification of equilibrium points for
nonlinear systems only deals with initial
positions near the equilibrium point.
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Nonpolynomial Case
dx/dt = 3 sinx +y
dy/dt = 4x + cosy - 1
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Bifucations
dx/dt = x2 - a
dy/dt = -y(x2 +1)
a. Show that the system has no equilibrium
points if a < 0.
b. Show that the system has two equilibrium
points if a > 0.
c. Show that the system has exactly one
equilibrium point if a = 0.
d. Find the linearization of the equilibrium
point for a = 0 and compute the eigenvalues
of this linear system.
(a = 0 is a bifurcation value)
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Example:
dx/dt = y - x2
dy/dt = y - x - a
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Linearization Failure
Two cases in which the long-term behavior of
solutions near an equilibrium point of the
nonlinear system and its linearization can
differ.
1. The linearized system is a center.
2. The linearized system has zero as an
eigenvalue.