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Unformatted text preview: y Differential Equations&Mathematica
Authors: Bruce Carpenter, Bill Davis and Jerry Uhl ©20012007
Publisher: Math Everywhere, Inc.
Version 6.0 2 DE.08 Linearizations
LITERACY What you need to know when you're away from the machine. · L.1)
Here is a system:
x£ @tD =  [email protected] [email protected] + 1
y£ @tD = [email protected] [email protected]  [email protected] a) Explain how you know that this system is nonlinear. b) Explain how you know that 8x, y< = 81, 1< is an equilibrium point of this system. 1 2 1 c) What information do you seek when you linearize at this equilibrium point? d) Linearize this system at 81, 1< and write down the resulting coefficient matrix.
e) The eigenvalues of a cleared matrix
ab
A=
cd
are given by
1
2 and
1
2 where [email protected] +
[email protected]  [email protected] = a + d 1 2 x 1
Is the equilibrium point a repeller or an attractor?
b) Here is the flow of the linearization of the system in part a): [email protected]  4 [email protected]
[email protected]  4 [email protected] [email protected] = a d  b c.
Use these formulas and a cheap calculator and apply them to the matrix you came up with
in part d) above to determine whether the equilibrium point 81, 1< is an attractor, a repeller
(or neither) of the given system.
and · L.2)
a) What do you mean when you say that 8a, b< is an equilibrium point of a given system
x£ @tD = [email protected]@tD, [email protected]
y£ @tD = [email protected]@tD, [email protected]? Linearized at equilibrium
y b) Given that 8a, b< is an equilibrium point of a given system
x£ @tD = [email protected]@tD, [email protected]
y£ @tD = [email protected]@tD, [email protected],
come up with exact formulas for the solutions [email protected] and [email protected] with starter data [email protected] = a and
[email protected] = b. · L.3)
What good is the fact that near the equilibrium point of a given system, the linearization
approximates the given system very, very well?
· L.4)
a) Here is a scaled plot of the flow of a certain system
x£ @tD = [email protected]@tD, [email protected]
y£ @tD = [email protected]@tD, [email protected]
centered on an equilibrium point:
x£ @tD ã  4.3 [email protected]@tDD + 0.17 [email protected]
y£ @tD ã 1.2  0.5 [email protected]  1.29 [email protected] 2 1 1 1
Here are both flows together: 1 2 x Linearized and Original Flows y y 3.5 3.0
2
2.5
1
2.0 1 1 2 x 1.5 1.5
1
How does this plot signal that the original system was not linear?
c) Take another look at the plot of both flows. What principle about a system and its
linearization at an equilibrium point does this plot display?
d) You are given starter data and you go to the linearization to come up with approximate
formulas for the solutions of the original nonlinear system.
If you are given starter data [email protected], [email protected]< = 82, 2.5<, would you trust the formula you get
from the linearization to be a
trustworthy approximations of the real thing? Why or why not? 2.0 2.5 3.0 3.5 x This makes it clear that the equilibrium point is a repeller. But the eigenvalues of the
linearization at the equilibrium point are
0.4 I and  0.4 I.
So the linearization predicts neither a repeller nor an attractor. Does this make you want to
change your answers to any of the previous parts? Why or why not?
· L.8)
What are Lyapunov's rules? If you are given starter data [email protected], [email protected]< = 80.5, 2.5<, would you trust the formula you get
from the linearization to be a
trustworthy approximations of the real thing? Why or why not? · L.5)
The damped pendulum oscillator is modeled by the second order differential equation y! @tD + L M +
= 0.
L
where
g is gravity, L is the length of the pendulum, M is the mass of the bob, and b is friction.
Convert this system to a system of two first order differential equations, linearize it, and
then convert the linearization back into a single second order differential equation.
· L.6)
Use your experience to help write a few words about when the pendulum oscillator is well
approximated by its linearization and when it is not.
· L.7)
a) The coefficient matrix of the linearization of a certain system at an equilibrium point has
eigenvalues
0.3 + 1.3 I and 0.3  1.3 I.
Is this enough to tell you that the equilibrium point is definitely a repeller?
b y @tD
£ g [email protected]@tDD b) The coefficient matrix of the linearization of a certain system at an equilibrium point
has eigenvalues
 0.3 + 1.3 I and  0.3  1.3 I.
Is this enough to tell you that the equilibrium point is definitely a attractor?
c) The coefficient matrix of the linearization of a certain system at an equilibrium point has
eigenvalues
0.4 I and  0.4 I.
Is this enough to tell you that trajectories of this system oscillate around the equilibrium
point on closed curves?
d) Here is the flow of a nonlinear system in the vicinity of an equilibrium point: · L.9)
You are dealing with a gradient system
x£ @tD = [email protected]@tD, [email protected]
y£ @tD = [email protected]@tD, [email protected]
This means that there is a function f @x, yD with
[email protected], yD = !x f @x, yD
and
[email protected], yD = !y f @x, yD;
i.e.
[email protected], yD, [email protected], yD< = ıf @x, yD (gradient).
Remember that the ıf @x, yDpoints in the direction of greatest initial increase of f @x, yD as
8x, y< leaves 8a, b<. a) When you linearize a gradient system at an equilibrium point 8xx, yy< and find that the
eigenvalues of the resulting matrix are both negative, which do you expect:
i) f @xx, yyD > f @x, yD for 8x, y< nearby 8xx, yy<; i.e. the equilibrium point is a local
maximizer of f @x, yD:
ii) f @xx, yyD < f @x, yD for 8x, y< nearby 8xx, yy<; i.e. the equilibrium point is a local
minimizer f @x, yD;
iii) f @xx, yyD < f @x, yD for some 8x, y< nearby 8xx, yy< and f @xx, yyD > f @x, yD for some
other 8x, y< nearby 8xx, yy<; i.e. the equilibrium point is a saddle point of f @x, yD. b) When you linearize a gradient system at an equilibrium point 8xx, yy< and find that the
eigenvalues of the resulting matrix are both positive, which do you expect:
i) f @xx, yyD > f @x, yD for 8x, y< nearby 8xx, yy<; i.e. the equilibrium point is a local
maximizer of f @x, yD:
ii) f @xx, yyD < f @x, yD for 8x, y< nearby 8xx, yy<; i.e. the equilibrium point is a local
minimizer f @x, yD;
iii) f @xx, yyD < f @x, yD for some 8x, y< nearby 8xx, yy< and f @xx, yyD > f @x, yD for some
other 8x, y< nearby 8xx, yy<; i.e. the equilibrium point is a saddle point of f @x, yD. c) When you linearize a gradient system at an equilibrium point 8xx, yy< and find that one
of the eigenvalues of the resulting matrix is positive and the other eigenvalue is negative,
which do you expect:
i) f @xx, yyD > f @x, yD for 8x, y< nearby 8xx, yy<; i.e. the equilibrium point is a local
maximizer of f @x, yD:
ii) f @xx, yyD < f @x, yD for 8x, y< nearby 8xx, yy<; i.e. the equilibrium point is a local
minimizer f @x, yD;
iii) f @xx, yyD < f @x, yD for some 8x, y< nearby 8xx, yy< and f @xx, yyD > f @x, yD for some
other 8x, y< nearby 8xx, yy<; i.e. the equilibrium point is a saddle point of f @x, yD. · L.10) L.10)
Most math books give you formulas to plug and chug with,
but math books that math profs call "cookbooks" are those
that don't do much of a job explaining where the formulas come from.
Many differential equations books fall into the "cook book" category. You have a friend who was looking at a math cook book and thought that the cook book
said that to check the nature of an equilibrium point 8p, q< of a given system
x£ @tD = [email protected]@tD, [email protected]
y£ @tD = [email protected]@tD, [email protected]
you look at the eigenvalues of the matrix
!x [email protected], yD !y [email protected], yD
!x [email protected], yD !y [email protected], yD Was your friend right? · ê. 8x Ø p, y Ø q< ...
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This note was uploaded on 09/21/2011 for the course MATH 285 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Staff
 Differential Equations, Equations

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