Butler CC Math Friesen Butler CC Math Friesen Linearizing systems of Differential Equations Starring The Jacobian Matrix Keanu Jacobi
Sep 28, 2015
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Linearizing systems of Differential Equations
StarringThe Jacobian Matrix
Keanu Jacobi
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Given a Non-Linear System of Differential Equations
dx/dt = f(x,y)dy/dt = g(x,y)
Equilibrium points f(x0,y0) = g(x0,y0) = 0
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Linearizing Systems of Autonomous Differential Equations
Non-linear systems can be linearized by approximating them with a linear system
Near the equilibrium points the linear approximation is good.
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Transform the Equilibrium Points to the Origin
Let u = x - x0 so x = u + x0Let v = y - y0 so y = v + y0
y
x
(x0,y0) in the x-y plane becomes (0,0) in u-v plane.
v
u
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Linearizing Systems
Also, knowing that u,v,x,y are functions of t:u = x - x0 so du/dt = dx/dtv = y - y0 so dv/dt = dy/dt
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Review of the total differential dz
The total differential giving the estimated change z of a function of 2 variables is:
z dz = f/x(x0,y0) dx + f/y(x0,y0) dy
dzz
f(x0,y0)
f(x0+dx,y0+dy)
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f(x,y) = f(x0+u,y0+v) f(x0y0) + f/x(x0,y0) u + f/y(x0,y0) v g(x,y) = g(x0+u,y0+v) g(x0y0) + g/x(x0,y0) u + g/y(x0,y0) v
Total Differential: dz = f/x(x0,y0) dx + f/y(x0,y0) dy
Approximating Systems
x = u + x0y = v + y0
Moving from f(x0,y0) to f(x0+u,y0+v) is approximated using the total differential and by letting x = u and y = v:
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We replace the non-linear differential equations with their linear approximations.
f(x,y) = f(x0+u,y0+v) f(x0y0) + f/x(x0,y0) u + f/y(x0,y0) v g(x,y) = g(x0+u,y0+v) g(x0y0) + g/x(x0,y0) u + g/y(x0,y0) v
Since f(x0,y0) = g(x0,y0) = 0 we get:du/dt = f(x,y) f/x(x0,y0) u + f/y(x0,y0) v dv/dt = g(x,y) g/x(x0,y0) u + g/y(x0,y0) v
Approximating Systems
Near the equilibrium points the the u-v system is close to the original x-y system.
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Using matrix notation
=
vu
yyxg
xyxg
yyxf
xyxf
dtdvdtdu
),(),(
),(),(
0000
0000
Putting all of this together in matrix form gives the following where the partial derivative matrix is called the Jacobian Matrix:
du/dt = f/x(x0,y0) u + f/y(x0,y0) v dv/dt = g/x(x0,y0) u + g/y(x0,y0) v
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To linearize a system of diff. eq.
1. Find the equilibrium point(s)2. Use the Jacobian matrix to change each
equilibrium point in the x-y plane to the origin in the u-v plane
3. The eigenvalues at each equilibrium point determine whether you have a sink, saddle, or source there.
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Example of linearizing a system
Find the equilibrium point(s)
Linearize the non-linear system:dx/dt = xy - 2dy/dt = x - 2y
Using substitution and solving:Equilibrium points are (2,1),(-2,-1)
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Example of linearizing a system
a. at (2,1)
b. at (-2,-1)
=
vuxy
dtdvdtdu
21
=+=
=
vudtdv
vudtdu
vu
dtdvdtdu
2
2
2121
==
=
vudtdv
vudtdu
vu
dtdvdtdu
2
2
2121
Use the Jacobianmatrix to change each equilibrium point (x0,y0) in the x-y plane to the origin in the u-v plane:
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Example of linearizing a system
1. The eigenvalues at each equilibrium point determine whether you have a sink, saddle, or source there.
a. At (2,1) = 1.56, -2.56 saddle
b. At (-2,-1) = -1.5 1.32ispiral sink
02121 =
02121 =
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Comparison of non-linear phase plane and linear approx. at equilibrium point (-2,-1)
x-y phase plane u-v phase plane
u
v
x
y
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An example of linearizing gone horribly wrong
dx/dt = y - (x2 + y2)xdy/dt = -x - (x2 + y2)y
(0,0) is an equilibrium point
=
vu
yxxyxyyx
dtdvdtdu
22
22
321213
==
=
udtdv
vdtdu
vu
dtdvdtdu
0110
= +/- i (center)
At (x0,y0) = (0,0):
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Linearizing doesnt predict long-term behavior
u
x-y non-linear is a sink u-v linearized is a center
v
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System with a parameter
Linearize the non-linear system, where a is an adjustable parameter:dx/dt = y - x2dy/dt = y - a
=
vux
dtdvdtdu
1012
Find the Jacobian matrix:
Find the equilibrium point(s):(a,a), (-a,a)
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System with a parameter
01012 =
aFor the equilibrium point (a,a) we have eigenvalues = 1, -2aFor the equilibrium point(-a,a) we have eigenvalues = 1, 2a
01012 =
a
When a < 0 we have eigenvalues that are imaginary. When a > 0 we have real eigenvalues. a is a bifurcation value - the nature of the solutions changes.
=
vux
dtdvdtdu
1012
Looking at the discriminant of (-2a-)(1-) = 2+(2a-1)-2a = 0 there are repeated roots at a = so this could also be considered a bifurcation value.
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System with a parameter
a = 1/4 (bifurc.) equil. points (1/2,1/4)
a = 1 equilibrium points (1,1)
a = 1/16 equilibrium points (1/4,1/16)
a = 0 (bifurc.) equilibrium point (0,0)
a = -1 no equilibrium points
dx/dt = y - x2dy/dt = y - a
Phase Plane Plotter
HPG System Solver
Linearizing systems of Differential EquationsGiven a Non-Linear System of Differential EquationsLinearizing Systems of Autonomous Differential EquationsTransform the Equilibrium Points to the OriginLinearizing SystemsThe total differential giving the estimated change z of a function of 2 variables is: Approximating SystemsWe replace the non-linear differential equations with their linear approximations.Using matrix notationTo linearize a system of diff. eq.Example of linearizing a systemExample of linearizing a systemExample of linearizing a systemComparison of non-linear phase plane and linear approx. at equilibrium point (-2,-1)An example of linearizing gone horribly wrongLinearizing doesnt predict long-term behaviorSystem with a parameterSystem with a parameterSystem with a parameter