Download Linear Stability Analysis of Two Dimensional Systems

Linear Stability Analysis of Two Dimensional Systems
L.A. Romero
November 11, 2015
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Computing Jacobians
Example 1. Consider the vector function
q(x, y) =
x2 y
2
x + y2
(1.1)
Compute the Jacobian J(x, y) of q(x, y).
Solution 1. The Jacobian is given by
J(x, y) =
2
∂(x2 y)
∂y
∂((x2 +y 2 ))
∂y
∂(x2 y)
∂x
∂(x2 +y 2 )
∂x
!
=
2xy
2x
x2
2y
(1.2)
The Stability of Linearized Systems
When we do a linear stability analysis of a two dimensional non-linear differential equation, we end up with
a linear system of equations of the form
dz
= J0 z
dt
(2.1)
(2.2)
where
z=
x
y
and J0 is the Jacobian of our system evaluated at the equilibrium point.
The system is stable if all of the eigenvalues of J0 have a real part less than 0, and unstable if any of the
eigenvalues have a real part bigger than 0.
Example 2. Is the system
dz
= J0 z
dt
(2.3)
where
J0 =
1 2
1 0
stable or unstable ?
1
(2.4)
Solution 2. The eigenvalues of J0 are the roots to
λ−1
det
−1
−2
λ
=0
(2.5)
This gives us the equation
λ2 − λ − 2 = 0
(2.6)
λ1 = 2
(2.7)
λ2 = −1
(2.8)
This has the roots
Since one of the roots has a real part bigger than zero, the system is unstable.
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Short Cut to Stability Problems
When solving
dz
= Az
(3.1)
dt
where A is a two dimensional matrix, we derived the following shortcuts to deremining the stability.
• The system is unstable if tr(A) > 0
• The system is unstable if det(A) < 0
• The system is stable if tr(A) < 0 and det(A) > 0.
Here the trace is the sum of the diagonal elements of A. That is if
a11 a12
A=
a21 a22
(3.2)
then
tr(A) = a11 + a22
(3.3)
Example 3. Using the shortcuts, determine if the system
dz
= J0 z
dt
(3.4)
where
J0 =
1 2
1 0
(3.5)
is stable or unstable ?
Solution 3. We have
tr(J0 ) = 1
It follows that the system is unstable.
2
(3.6)
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A Linearized Stability Analysis
Example 4. Consider the system
ẋ = x2 + y
(4.1)
ẏ = xy + 1
(4.2)
Compute the stability of the equilibrium solution
x0 = 1, y0 = −1
Solution 4. The Jacobian of
(4.3)
q(x, y) =
x2 + y
xy + 1
(4.4)
J(x, y) =
2x 1
y x
(4.5)
is given by
The linearized equations are given by
dz
= J0 z
dt
(4.6)
where
J0 =
We have
2 1
−1 1
tr(J0 ) = 3
and hence the equilibrium is unstable.
Problems
exy
1. Compute the Jacobian J(x, y) of q(x, y) =
x2 + y 3
2. Compute the Jacobian J(x, y) of q(x, y) = twovecx/yxy 2
4
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3. Is the system of equations dz
=
J
z
where
J
=
stable or unstable ?
0
0
dt
−1 −2
−4 3
=
J
z
where
J
=
stable or unstable ?
4. Is the system of equations dz
0
0
dt
−2 1
5. Consider the equations ẋ = x2 y − 1, ẏ = x/y − 1 . This has an equilbrium x0 = y0 = 1.
(a) Compute J0 (x0 , y0 ) where J(x, y) is the Jacobian of our system.
(b) Is the equilibrium x0 = y0 = 1 stable or unstable ?
Answers
xy
xy
1. J(x, y) =
ye
2x
xe
3y 2
2. J(x, y) =
1/y
y2
−x/y 2
2xy
3
(4.7)
(4.8)
3. We have tr(J0 ) = 2 > 0 and hence the system is unstable.
4. We have tr(J0 ) = −3 < 0 and det(J0 ) = 2 > 0. It follows that the system is stable.
2 1
5. (a) J0 =
1 −1
(b) We have tr(J0 ) = 1 > 0, and hence the equilibrium is unstable.
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