Nonlinear mechanical systems
Old mass-spring systems
Math 216
Differential Equations
Old model. We have been assuming the the response of the spring
was constant regardless of its elongation or compression:
mx 00 + cx 0 + kx = F (t)
Kenneth Harris
kaharri@umich.edu
x(t) is the displacement from the equillibrium position,
m is the mass on the spring,
Department of Mathematics
University of Michigan
c is damping,
k is the spring elasticity, and obeys Hooke’s law Fspring = −kx
November 17, 2008
Kenneth Harris (Math 216)
Math 216 Differential Equations
where m, c, k > 0.
F is the external force applied to the system.
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Nonlinear mechanical systems
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Nonlinear mechanical systems
Nonlinear elasticity
Nonlinear elasticity
New model. In actual systems, there is a progressive stiffening or
weakening of the spring as it is elongated or compressed. This is a
nonlinear response.
We assume the spring reacts symmetrically, the simplest modification:
Nonlinear model. We will consider nonlinear models without external
forces:
mx 00 + cx 0 + kx − βx 3 = 0
Fspring = −kx + βx 3
Normal Form. We convert this to a first-order system by introducing
velocity, y = y (t):
so, the nonlinear mass-spring system is
mx 00 + cx 0 + kx − βx 3 = F (t)
where m, c, k > 0.
dx
dt
dy
dt
β = 0: original linear equation.
β < 0: spring becomes increasingly stiffer as it is elongated or
depressed (hard spring),
β > 0: spring becomes increasingly less resilient as it is elongated
or depressed (soft spring).
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Kenneth Harris (Math 216)
= y
= −
c
k
β
y − x + x3
m
m
m
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Nonlinear mechanical systems
Nonlinear mechanical systems
Linearization
Critical point at origin
Nonlinear model. We start without damping:
dx
dt
dy
dt
dx
= y,
dt
= y
= −
Linearization at (0, 0).
k
β
x
x + x3 =
βx 2 − k
m
m
m
J(x, y ) =
Critical points. (0, 0) is a critical point.
− mk
Kenneth Harris (Math 216)
1
0
0
J(0, 0) =
− mk
k
m
1
0
(natural
Analysis. The linearization has a stable center at the origin. The
critical point of the nonlinear system at (0, 0) could be either a stable
center or spiral point (stable or unstable). We need to analyze
solutions to obtain further information.
Linearization. The Jacobian is
0
2
+ 3β
mx
− mk
0
2
+ 3β
mx
The characteristic equation is λ2 + ω 2 , where ω 2 =
frequency). The eigenvalues are λ = ±ωi.
β < 0 (hard spring): (0, 0) is the only critical point.
q
β > 0 (soft spring): (± βk , 0) are also critical points.
dy
k
β
= − x + x3
dt
m
m
1
0
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Nonlinear mechanical systems
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Nonlinear mechanical systems
Solutions at origin
Conservation of energy
dx
= y,
dt
dy
k
β
= − x + x3
dt
m
m
Position-velocity solutions to
mx 00 + kx − βx 3 = 0
Solutions. We use the chain rule to rewrite as a first order system
(with independent variable x and dependent variable y )
dy
=
dx
dy
dt
dx
dt
=
− mk x +
y
β 3
mx
satisfy the conservation of energy
1
1
1
my 2 + kx 2 − βx 4 = E
2
2
4
.
where E is the total energy,
Separating variables
Kinetic energy (due to inertia) is 12 my 2 ,
3
my dy = −kx + βx dx.
Potential energy (due to spring potential) is 12 kx 2 − 41 βx 4
Integrating
1
1
1
my 2 + kx 2 − βx 4 = E
2
2
4
Analysis. Position-velocity trajectories are along paths of constant
energy.
where E is a an arbitrary constant of integration.
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Hard springs without damping
Hard springs without damping
Hard spring
Trajectories
Hard springs get stiffer as they are stretched or compressed:
Solutions obey the conservation of energy (where x is position and y
is velocity)
1
1
1
my 2 + kx 2 − βx 4 = E
2
2
4
Like the linear system, these solutions are periodic. (Recall, the linear
system is when β = 0.)
Fspring = −kx − |β|x 3
where β < 0.
The nonlinear undamped mass-spring system is
mx 00 + kx + |β|x 3 = 0
where m, c, k > 0.
Trajectories in the Position-Velocity plane are similar to the linear
system with two exceptions
As a system of equations:
dx
= y,
dt
dy
k
|β| 3
=− x+
x
dt
m
m
Critical point. The only critical point is at (0, 0), which is a center in
the linearization.
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1
Trajectories are flatter quartic ovals (not quadratic ellipses),
2
The period and amplitude of the nonlinear system depends on the
initial position x0 q
and initial velocity y0 . (In the linear system, the
period is T = 2π
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k
m,
Kenneth Harris (Math 216)
Hard springs without damping
so independent of the initial conditions.)
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Hard springs without damping
Example
Position-Velocity plots
Position-Velocity plots at different energy levels
Trajectories at constant energy
6
E=32
Example. Hard spring: β = −8, m, k = 2:
4
00
E=16
3
2x + 2x + 8x = 0.
E=8
2
E=4
E=1
0
Solution. Position x and velocity y solutions satisfy constant energy E:
-2
y 2 + x 2 + 2x 4 = E
-4
-6
-6
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-4
-2
0
2
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4
6
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Hard springs without damping
Soft springs without damping
Trajectories
Soft spring
Plot of position at different energies: 0.3125 and 20.
Period decreases, amplitude increases at higher energy.
Soft springs weaken as they are stretched or compressed:
Position plots at different energies
Fspring = −kx + βx 3
2
E = 0.3125
E = 20
where β > 0,
1.5
so, the nonlinear undamped mass-spring system is
1
mx 00 + kx − βx 3 = 0
position
0.5
where m, c, k , β > 0.
0
The behavior of the spring depends on displacement x.
−0.5
When kx > βx 3 spring’s force is directed toward equillibrium
position.
−1
−1.5
When kx = βx 3 spring exerts no force.
−2
−2.5
When kx < βx 3 spring’s force repulses from equillibrium position.
0
2
4
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8
10
time
12
14
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16
18
20
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Soft springs without damping
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Soft springs without damping
Soft spring
Analysis of other critical points
Soft spring model
mx 00 + kx − βx 3 = 0
dx
= y,
dt
where m, c, k , β > 0.
q
Linearization at (± βk , 0).
As a system of equations:
dx
= y,
dt
dy
k
β
= − x + x3
dt
m
m
dy
k
|β| 3
x
=− x+
x =−
k − βx 2
dt
m
m
m
J(x, y ) =
− mk
0
2
+ 3β
mx
1
0
s
J(±
k
, 0) =
β
0
2k
m
1
0
The characteristic equation is λ2 − 2k
m:
there are two real eigenvalues of opposite signs.
Critical point. Thereqare critical points at (0, 0) (a center in the
linearization) and (± βk , 0).
Trajectories obey the conservation of energy
q
Analysis. The nonlinear system has saddlepoints at (± βk , 0).
1
1
1
my 2 − kx 2 + βx 4 = E
2
2
4
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Soft springs without damping
Soft springs without damping
Saddlepoints
Separatices
Linearization. Eigenvectors correspond to the lines which separate different
behavior of the system. These lines separate different behavior in
trajectories.
0
Separatrices are the nonlinear deformations of the eigenvectors for
the saddlepoint in the linearization.
q
They occur at the same energy level as the critical points (± βk , 0).
0
E=
0
1 k2 1 k2
1 k2
−
=
2 β
4 β
4 β
The separatrices separate different kinds of behavior of trajectories in
the nonlinear system.
0
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Soft springs without damping
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Soft springs without damping
Separatrices
Trajectories
Separatrices separate the kinds of trajectories in the phase plane .
Separatrices at E=
Several trajectories from each of the three regions, with critical points.
Trajectories at constant energy
k2
6
4Β
4
II
2
III
III
0
H-
k
Β
,0L
I
H
k
Β
,0L
-2
-4
II
-6
-6
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-4
-2
0
2
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4
6
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Soft springs without damping
Soft springs without damping
Separatrices: Three regions
Example
There are three regions of behavior determined by the separatrices.
Example. Soft spring: β = 2, m = 2, k = 8:
I Periodic. Trajectories where 0 ≤ E < 8. Mass oscillates around
central equillibrium.
2x 00 + 8x − 2x 3 = 0.
II Unbounded. Trajectories where E > 8. Velocity is too great and
carries mass beyond the spring threshold. (In physical
applications: spring breaks!!)
Critical points. (0, 0), (2, 0), (−2, 0)
III Unbounded. Trajectories where E < 0. Physically impossible in
mass-spring. Mass too far away from central equillibrium point
and spring acts repulsively.
q
Note. The critical points (± βk , 0) occur where the mass is sufficiently
far that the spring exerts no force on the mass. Beyond this distance
spring repulses the mass (no physical analogy with mass-spring).
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Solution. Position x and velocity y at constant energy E:
1
y 2 + 4x 2 − x 4 = E
2
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Soft springs without damping
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Soft springs without damping
Differential fields with separatices
Amplitude and period
Separatrices plotted on with critical points
x’=y
y ’ = − 4 x + x3
Periodic behavior. At sufficiently low energy, trajectories are periodic.
Like the hard spring model, both amplitude and period depend upon
the initial position and velocity.
4
3
2
1
y
However,
0
Amplitude increases at higher energy.
−1
Period decreases at higher energy.
−2
This makes sense: when the mass is farther from the equillibrium the
spring’s force is weaker.
−3
−4
−5
−4
−3
−2
−1
0
1
2
3
4
x
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Soft springs without damping
Degenerate Example
Trajectories
Example
Plot of position at different energies: 0.1592 and 3.5.
Period increases, amplitude increases at higher energy.
Example.
Position plots at different energies
1
E = 0.1592
E = 3.5
0.8
x0
=
4xy
0
=
x2 − y2
y
0.6
0.4
Critical points. The only critical point is (0, 0).
position
0.2
0
Jacobian.
−0.2
−0.4
−0.6
4x
−2y
0 0
J(0, 0) =
0 0
We cannot use the linearization to get any information about the critical
point (0, 0), since det(J(0, 0)) = 0 and so J(0, 0) is not invertible.
−0.8
−1
4y
J(x, y ) =
2x
0
2
4
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8
10
time
12
14
16
18
20
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Degenerate Example
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Degenerate Example
Phase plot
Ray solutions
Phase plot shows six solutions lying on rays. (The mauve-colored dashed
lines.) These are separatrices. Linear systems have at most four.
Lets look at the ray solutions of
x 0 = 4xy ,
x’=4xy
y ’ = x2 − y2
y0 = x2 − y2
4
Case 1. If x = 0 then x 0 = 0 (so, x never changes!) and y 0 = −y 2 .
We can solve y 0 = −y 2 using separation of variables:
3
2
y (t) =
y
1
1
t −c
0
where c is an arbitrary constant.
−1
Analysis. One kind of trajectory lives on the y -axis:
−2
−3
x(t) = 0,
−4
−4
−3
Kenneth Harris (Math 216)
−2
−1
0
x
1
Math 216 Differential Equations
2
3
y (t) =
1
.
t −c
4
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Degenerate Example
Lorenz Equation
Ray solutions
Lorenz system
Lets look at the ray solutions of
x 0 = 4xy ,
Case 2. Suppose y = mx and lets find m. So,
dx
= 4mx 2 ,
dt
Example. In the late fifties the meteorologist Edward Lorenz came
upon the following nonlinear autonomous three-dimensional system
when modeling atmospheric turbulence beneath a thunderhead
(where air is cooled from above and warmed from below).
y0 = x2 − y2
dy
dx
= m and
dy
= (1 − m)x 2
dt
x 0 = −ax + ay
y 0 = rx − y − xz
By the chain rule
z 0 = −bz + xy
m=
dy
=
dx
dy
dt
dx
dt
=
(1 − m)x 2
1−m
=
4mx 2
4m
where a, b, r are positive constants.
Motivation. What drove Lorenz was the search for an equation which
would model some of the unpredictable behavior which we normally
associate with the weather.
Thus, 5m2 − 1 = 0, and m = ± √15 .
Analysis. A second kind of trajectory lives on the lines with slope
m = ± √15 .
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Lorenz Equation
Math 216 Differential Equations
Basic properties of the Lorenz system
Example. Fix a, b > 0 in the system
Example. a = 10, b = 83 , r = 28:
x 0 = −ax + ay
x 0 = −10x + 10y
y 0 = rx − y − xz
y 0 = 28x − y − xz
8
z 0 = − z + xy
3
z 0 = −bz + xy
Critical points. When r > 1, there are three critical points:
p
p
p
p
(0, 0, 0), ( b(r − 1), b(r − 1), r −1), (− b(r − 1)), − b(r − 1), r −1)
The critical points are unstable for most values of r .
Trajectories stay close to the origin, regardless of where the initial
point is chosen. They wrap crazily around the second two unstable
critical points.
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Lorenz Equation
Basic properties of the Lorenz system
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Trajectories. All trajectories eventually crazily orbit the unstable
critical points:
√
√
√
√
(6 2, 6 2, 27), (−6 2, −6 2, 27)
producing a butterfly-like pattern.
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Lorenz Equation
Lorenz Equation
Phase plot
Basic properties of the Lorenz system
3-D Phase plot shows the butterfly-like trajectory.
Example. a = 10, b = 83 , r = 28:
The Lorenz attractor with a = 10, b = 2.6667, r = 28
50
x 0 = −10x + 10y
45
y 0 = 28x − y − xz
8
z 0 = − z + xy
3
40
35
z
30
25
Chaos. Even very small perturbations of the starting position lead to
dramatically different orbits. In fact, the disturbance is compounded
exponentially, leading to the butterfly effect:
20
15
10
A butterfly flapping its wings in Brazil can produce a tornado in
Texas. (due to Lorenz)
5
This strong dependence of outcomes on very slightly differing initial
conditions is a hallmark of the mathematical behavior known as chaos.
0
50
0
−20
−50
−15
0
−5
−10
y
5
10
15
20
x
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Math 216 Differential Equations
Lorenz Equation
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Lorenz Equation
Time plot of x-coordinate
Time plot of z-coordinate
The initial value of the Dashed green trajectory differed by 0.0001
√ in the
x-coordinate only. Note that the x-value is oscillating around ±6 2.
The initial value of the Dashed green trajectory differed by 0.0001 in the
x-coordinate only. Note that the z-value is oscillating around 27.
The Lorenz attractor with a = 10, b = 2.6667, r = 28
45
15
40
10
35
5
30
0
25
z
x
The Lorenz attractor with a = 10, b = 2.6667, r = 28
20
−5
20
−10
15
−15
10
−20
275
280
285
290
295
5
275
300
t
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Math 216 Differential Equations
280
285
290
295
300
t
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