Project Details: Applications of Linear, Constant Coefficient Systems of
ODEs and Their Phase Portraits
MATH527 Fall 2018
For the project this semester, you will choose a physical, biological, economic, or social system that can
be modeled by a linear, constant coefficient 2-by-2 system of differential equations. Your project will have
two main pieces:
• Analysis of a single differential equation - either a (1) single first-order linear differential equation
with constant coefficients OR (2) a single second-order linear differential equation with constant
coefficients.
• Analysis of a system of differential equations - linear, constant coefficient 2-by-2 system of differential
equations.
Most applications are of the first type: when one unknown quantity is considered, the model is a single
first-order linear differential equation with constant coefficients; when that quantity interacts with another
quantity, the model is a 2-by-2 linear, constant coefficient 2-by-2 system of differential equations. Examples
include:
• Pond or Lake Pollution
• Home Heating
• Drug delivery/diffusion (Compartmental Model)
• Pesticide in Trees and Soil
• Price-Inventory
• Chemical Reactions
• Any other that fits the above description exactly.
You can use this chapter of a differential equations text by G.B. Gustafson from the University of Utah to
get descriptions and details of some of these systems: www.math.utah.edu/~gustafso/2250systems-de.
pdf
Some applications are of the second type: the model is a single second-order linear differential equation
with constant coefficients. It can be analyzed as such with the methods of Lebl Chapter 2. However, the
second-order equation can be converted to a 2-by-2 system and analyzed with the methods of Lebl Chapter
3. Examples include:
• Spring-mass systems (used in sample project; cannot be chosen for project)
• LRC Circuits
• Any other that fits the above description exactly.
1. Outline of Project Write-up
(1) Introduction
(2) Single Equation
(a) Derivation of equation and meaning of parameters (include units)
(b) Meaning and Relevance of Homogeneous vs Non-Homogenous
(c) General Solution
(d) An Initial Value Problem and a Particular Solution (Typically Non-Homogeneous)
(e) Behavior: Discuss solution in the language of the application
(3) System of ODEs
(a) Derivation of equation and meaning of parameters (including units)
(b) Meaning and Relevance of Homogeneous vs Non-Homogenous
(c) Homogenous System
(i) Pick two sets of values for parameters that will give two of the three possibilities for
eigenvalues: (1) distinct, real; (2) complex conjugate pair; (3) repeated (or multiple)
eigenvalues (I suggest using computational aids for this.)
(ii) For each of the two situations:
(A) Discuss how realistic the parameter values are. (At least one set of values should be
realistic.)
(B) Give general solution
1
2
(C) Draw a phase portrait
(D) Give two (meaningfully different) initial conditions and their particular solutions
• Describe the particular solutions of each IVP in language of the application
(d) Non-homogeneous
(i) Meaning of the inhomogeneity
(ii) Find the general solution using Wolframalpha or another symbolic computational aid.
(iii) Plot the general solution on Desmos.com
(iv) Give two situations (by changing initial conditions or the forcing) that lead to meaningfully different behaviors
• Draw the trajectory of each of the two particular solutions (using Desmos plots)
• Describe the particular solution of each IVP in language of the application
(4) Appendix
(a) Hand-written work for finding general solution of single equation
(b) Hand-written work for finding general solutions of 2-by-2 systems
2. Dates
• Tues. Nov 6th: Your chosen application must be reported to your TA. (Also, you should vote.)
• Tues. Nov. 13th: TAs will address some questions about the project.
• Tues. Nov. 27th: Project is due in recitation.
3. Comments
• A sample project, using the spring-mass system, will be posted. It will include examples of the
inputs for 3 d (ii) and (iii).
• All work for finding solutions should be put in a neat, hand-written appendix.
• Phase plots may be neatly hand-drawn or copied and pasted from a computer plotting tool.
• Discussions should be typed, and manageable mathematical symbols should be typed. However, long
or complicated mathematical expressions can be hand-written.
• The responses for many parts listed in the outline should be brief. A sentence or two will suffice
for many pieces. Derivations in 2 (a) and 3 (a) should be a short paragraph. Descriptions of the
behavior of solutions in the language of your application (2(e), 3 c (ii) (D), 3 d (iv)) should be your
longest written sections, but still do not need to be longer than 4-5 sentence paragraphs.
• Derivations and realistic values for parameters should be cited.
• For 3 c (i), you may want to use a computational phase portrait creator, like those found at http://
mathlets.org/mathlets/linear-phase-portraits-matrix-entry/ and http://parasolarchives.
com/tools/phaseportrait. You can try different parameters and see the phase portraits associated
with them to choose your two sets of parameters.
• I use the phrase ‘meaningfully different’ in a couple of places. What I mean, for example, is not
to choose initial conditions (0, 1) and (0, 1.1). The behaviors will be nearly identical. Choose (0, 1)
and (−1, 0) or (1, 10) and (10, 1). It will depend on your system, but there should be something
interesting to say about the behavior for the two ‘meaningully different’ choices.
Sample Project: Spring-Mass via Systems of ODEs and Phase Portraits
MATH527 Fall 2018: Sample Produced by Instructor, John McClain
1. Introduction
I study a damped spring-mass system as a single second-order equation and as a system of first-order
equations. I discuss phase portraits and trajectories in the case of underdamped and critically damped
unforced systems and in the case of a forced, underdamped system.
2. Single Equation
2.1. Derivation and Meaning of Parameters. Let us assume that there are at most three forces acting
on the bob: the force that is due to gravity, the force produced by the spring, and possibly a drag force.
The force that is due to gravity is given by mass × gravity (mg), and the force produced by the spring
is proportional to the amount (length) that the spring has been stretched or compressed (k∆L) from its
natural length (this is called Hooke’s law ). The constant k in the spring force is called the spring constant,
and it has units dyne/cm or Newton/m. The drag or damping force is proportional to the velocity of the
mass, but opposes the motion. The proportionality constant, c, has units of N · s/m. According to Newton’s
second law of motion, the equation of motion of the bob is
mass × acceleration of the bob = sum of all forces acting on the bob
(1)
md00 = k∆L − mg = k(∆l − d) − mg − cd0 + fext (t) = −kd − cd0 + fext (t)
This gives the differential equation:
md00 + cd0 + kd = fext (t).
2.2. Meaning and Relevance of Homogeneous vs Non-Homogeneous. If there is an external force
fext (t), the equation is non-homogeneous. If there is no external force, so that fext (t) = 0, the equation is
homogenous. The external force models a time-dependent force on the bob, like a motor attached to it.
2.3. General Solution. I picked the following values of the parameters to study: m = 1 kg, c = 6 N · s/m,
and k = 9 N/m. I picked an external of fext = 12 cos 3t N. The general solution is:
d(t) = c1 e−3t + c2 te−3t + 2 cos 3t.
2.4. IVP and a Paricular Solution. I picked the initial condition that the spring is pushed at 10 m/s
upward from 2, m above equilibrium. So, d(0) = 2 m. and d0 (0) = 10. The particular solution is:
y(t) = 10te−3t + 2 cos 3t.
2.5. Behavior of This Solution. The bob in this case moves upward at first to its highest point of about
2.8 m above equilibrium, then back downward to below equilibrium. It moves back up to just slightly above
2 m above equilibrium. Note that 2 m is the amplitude of the forcing. After only one cycle of the bob
moving up, down, and then back up, its motion appears to be completely determined by the forcing. The
effect of the initial condition has decayed to a low level, leaving a motion that is dominated by oscillations
in sync with the sinusoidal external force.
1
2
3. System of ODEs
3.1. Derivation of the System of Equations Model and Meaning of Parameters. To get a system
of equations from the single, second-order differential equation, we define y1 (t) to be the displacement from
equilibrium. Then, we define y2 (t) to be the velocity, the derivative of displacement. Then, the single,
second-order equation, md00 + cd0 + kd = fext (t), becomes the system of equations:
y10 = y2
y20 = −
k
c
fext (t)
y1 − y2 +
m
m
m
or as a matrix vector system:
0
y =
0
1
k
−m
c
−m
!
y+
where y =
!
0
,
fext (t)
m
y1
y2
!
=
d
!
.
v
The parameters here have the same meanings as for the single equation. They now show up as ratios
only.
3.2. Meaning and Relevance of Homogeneous vs Non-Homogeneous. Again, the inhomogeneity
models an external force on the bob, like that supplied by a AC motor producing sinusoidally-varying forces
on the bob. The inhomogeneity for the system must have the form
!
0
f (t) =
fext (t)
m
because the force only appears in the equation for y20 which comes from Newton’s 2nd Law. If there is no
external force, the system is homogeneous.
3.3. Homogeneous Systems. I pick two sets of parameters. One {m = 1 kg, c = 1 N ·s/m, k = 17/4 N/m}
leads to complex eigenvalues, while the other {m = 4 kg, c = 12 N · s/m, k = 9 N/m} leads to a repeated
eigenvalue. (The second set of parameters is the same as I used in my singl equation.) fext (t) = 0 for all t,
so the system is homogeneous.
3.3.1. Complex Eigenvalues. With the parameters {m = 1 kg, c = 1 N · s/m, k =
17
4
N/m}, the system can
be written:
0
1
−17/4
−1
0
y =
!
y.
These parameters are realistic, representing a very large strong spring, with a sizeable bob on it. The
damping is fairly strong, but the system is still underdamped.
The general solution to this system can be written:
y = c1 e−t/2
or
cos 2t
− 21 cos 2t − 2sin2t
!
+ c2 e−t/2
sin 2t
− 21 sin 2t + 2 cos 2t
!
3
Figure 1. Phase Portrait for Damped System, {m = 1 kg, c = 1 N · s/m, k =
y = e−t/2
N/m}
!
c1 cos 2t + c2 sin 2t
(− c21 + 2c2 ) cos 2t − (2c1 +
17
4
c2
2 ) sin 2t
3.3.2. Phase Portrait. Fig. 1 shows a phase portrait for this system. All solutions spiral toward the origin
because the two eigenvalues are complex with negative imaginary parts.
3.3.3. Initial Conditions and Particular Solutions with Description of Behavior: I looked at the two initial
conditions:
0
y(0) =
!
2
and
1
2
− 14
y(0) =
.
!
In the first case, the bob is initially at equilibrium and is pushed upward with a velocity of 2 m/s. The
particular solution is
y=e
!
sin 2t
−t/2
2 cos 2t −
1
2
.
sin 2t
The trajectory is shown in Fig. 2 on the left. The bob initially moves upward above equilibrium. It hits
its highest point (of about 0.65 m above equilibrium) when its velocity hits 0. It then moves back toward
equilibrium, reaching its largest downward velocity (of about -1.05 m/s) before it hits equilibrium. It then
moves past equilibrium, hitting its lowest point (of about 0.35 m) when its velocity hits 0. It moves back
toward equilibrium, again hitting a maximum of velocity of about 0.45 m/s) before it reaches equilibrium.
It crosses equilibrium and moves up to a maximum of displacment that is much smaller than its previous
maximum. It continues oscillating up and down through equilibrium, with its maximum and minimum
displacements and velocities in each cycle approacing 0.
In the second, case, the bob is initially
− 14
1
2
m above equilibrium and is pushed with a downward velocity of
m/s. The particular solution is
1
2
−t/2
y=e
1
4
cos 2t
cos 2t − sin 2t
!
.
The trajectory is shown in Fig. 2 on the right. This trajectory is very similar to the one described previously.
The bob starts above equilibrium moving in a downward direction. Just like for the other initial condition,
4
Figure 2. Trajectories for {m = 1 kg, c = 1 N ·s/m, k =
17
4
N/m} Left: Initial displacement
is 0 and initial velocity is 2 m/s. Right: Initial displacement is 0.5 m and initial velocity is
-0.25 m/s.
Figure 3. Phase Portrait for Damped System, {m = 4 kg, c = 12 N · s/m, k = 9 N/m}
the bob oscillates up and down through equilibrium and its amplitude of oscillation decreases over time,
until it is nearly at rest at the equilibrium point.
3.4. Repeated Eigenvalues. With the parameters {m = 4 kg, c = 12 N · s/m, k = 9 N/m}, the system can
be written:
0
y =
0
1
−9/4
−3
!
y.
These parameters are realistic, representing a very large strong spring, with a sizeable bob on it. The
damping is very strong. The system is critically damped.
The general solution to this system can be written:
−3t/2
y(t) = e
c1 + c2 (1 + t)
− 23 c1 + c2 (− 12 − 23 t)
3.4.1. Phase Portrait. Fig. 3 shows a phase portrait for this system.
!
.
5
Figure 4. Trajectories for {m = 4 kg, c = 12 N ·s/m, k = 9 N/m} Left: Initial displacement
is 2 m below equilibrium and initial velocity is 7 m/s. Right: Initial displacement is 10 m
below equilibrium and initial velocity is 10 m/s.
3.4.2. Initial Conditions and Particular Solutions with Description of Behavior: I looked at the two initial
conditions:
−2
y(0) =
!
and
7
−10
y(0) =
10
!
.
In the first case, the bob starts 2 m below equilibrium with an upward velocity of 7 m/s. The particular
solution is:
−3t/2
y=e
4t − 2
!
7 − 6t
.
The trajectory is shown in Fig. 4 on the left. It shows that the bob starts below the equilibrium with
upward velocity. It moves upward through the equilibrium to its maximum displacement of about 0.46 m
when its velocity is 0. Then it moves back down toward equilibrium, with negative velocity, but its velocity
approaches 0 and it nevers reaches equilibrium.
In the second case, the bob starts 10 m below equilibrium with an upward velocity of 10 m/s. The
particular solution is:
−5t − 10
y = e−3t/2
10 +
!
.
15
2 t
The trajectory is shown in Fig. 4 on the right. The bob starts far (10 m) below equilibrium with an
upward velocity of 10 m/s. It approaches equilibrium from below as its velocity drops to 0, never reaching
equilibrium.
3.5. Non-homogeneous System.
3.5.1. Meaning of Inhomogeneity: I used the same parameters as in my first homogeneous system: {m =
1 kg, c = 1 N · s/m, k = 17/4 N/m}. I chose a sinusoidal external forcing for the inhomogeneity, fext (t) =
16 cos 4t, modeling a motor attached to the bob pushing it up and down smoothly over time.
!
!
0
0
f (t) =
=
fext (t)
16 cos 4t
m
The system in matrix-vector form is then:
0
y =
0
1
− 17
4
−1
!
y+
0
16 cos 2t
!
.
6
Figure 5. Trajectories for {m = 1 kg, c = 1 N · s/m, k =
17
4
N/m, fext (t) = 16 cos 4t N }
Left: Initial displacement is 2 m below equilibrium and initial velocity is -7 m/s. Right:
Initial displacement is 0 and initial velocity is 0.
From wolframalpha.com, a particular solution is:
yp =
64(16 sin 4t−47 cos 4t)
2465
4t+16 cos 4t
− 256(47 sin2465
!
.
Adding this to my general solution to the homogeneous system, the general solution to the non-homogeneous
system is:
−t/2
y(t) = e
64(16 sin 4t−47 cos 4t)
2465
256(47 sin 4t+16 cos 4t
c2
)
sin
2t
+
2
2465
c1 cos 2t + c2 sin 2t +
(− c21 + 2c2 ) cos 2t − (2c1 +
!
Fig. 5 shows, on the left, a trajectory satisfying the initial condition
!
−2
.
y(0) =
−7
In this case, the bob starts below equilibrium with a downward velocity. It moves down almost 1 m to its
lowest point (when the velocity is 0 as was always the case without forcing). It starts moving back toward
equilibrium with positive velocity but before it reaches it, its velocity becomes negative and it starts moving
down away from equilibrium again. But then the velocity increases and it begins moving toward equilibrium
again. From then on, it moves up and down past equilibrium approaching a perfectly sinusoidal motion (like
an undamped system) with angular frequency matching that of the motor, 4 rad/s. The effect of the initial
conditions decay, and the system follows the motor.
Fig. 5 shows, on the right, a trajectory satisfying the initial condition
!
0
y(0) =
.
0
In this case, the bob is released from rest at the equilibrium position. So, the motor has no initial
conditions to ‘overcome.’ Thus, the motion just follows the motor from the start; the bob moves is perfect
sinusoidal motion with the same frequency as the motor.
4. Appendix
Handwritten calculations. Full detail is not required, but it needs to be enough for us to see that you
have solved the systems. The calculations can be in done with parameters, rather than specific values.
Purchase answer to see full
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