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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