Transcript Section 2.1 MODELING VIA SYSTEMS
Section 2.1
MODELING VIA SYSTEMS
A tale of rabbits and foxes
Suppose you have two populations: rabbits and foxes.
R(t)
represents the population of rabbits at time
t.
F(t)
represents the population of foxes at time
t
.
• What happens to the rabbits if there are no foxes?
Try to write a DE.
• What happens to the foxes if there are no rabbits?
Try to write a DE.
• What happens when a rabbit meets a fox?
• If
R
is the number of rabbits and
F
is the number of foxes, the number of “rabbit-fox interactions” should be proportional to what quantity?
The predator-prey system
A system of DEs that might describe the behavior of the populations of predators and prey is
dR
2
R
1.2
RF dt dF
F
0.9
RF dt
1. What happens if there are no predators? No prey?
2. Explain the coefficients of the
RF
terms in both equations.
R =
0 and
F
= 0?
4. Are there other situations in which both populations are constant?
5. Modify the system so that the prey grows logistically if there are no predators.
Exercises
Page 164, 1-6. I will assign either system (i) or (ii).
Graphing solutions
Here are some solutions to
dR
2
R
1.2
RF dt dF
F
0.9
RF dt P
(0) = 0 prey predators
R
(0) = 0 predators prey
A startling picture!
Here’s what happens if we start with
R
(0) = 4 and
F
(0) = 1.
predators prey
The phase plane
Look at PredatorPrey demo.
This is the graph of the
parametric equation
(x,y) = (R(t), F(t))
for the IVP.
R
(0) = 4
F
(0) = 1
Exercises
• p. 165 #7a, 8ab • Look at GraphingSolutionsQuiz in the Differential Equations software (hard!)
Spring break!
Now for something completely different… Suppose a mass is suspended on a spring.
• Assume the only force acting on the mass is the force of the spring.
• Suppose you stretch the spring and release it. How does the mass move?
Quantities:
y(t)
– – – = the position of the mass at time
y
(0) = resting
y(t
) > 0 when the spring is stretched
y(t
) < 0 when the spring is compressed
t.
Newton’s Second Law: force = mass acceleration
F
m d
2
y dt
2 Hooke’s law of springs: the force exerted by a spring is proportional to the spring’s displacement from rest.
F s
ky k
is called the
spring constant
powerful the spring is.
and depends on how
DE for a simple harmonic oscillator
Combine Newton and Hooke:
F s
ky
m d
2
y dt
2 Sooo….
d
2
y
dt
2
k m y
0
harmonic oscillator
. It is a
second-order
DE because it contains a second derivative (duh).
How to solve it!
Now we do something really clever. We don’t have any methods to solve second-order DEs. Let
v(t)
= velocity of the mass at time
t.
Then
v(t) = dy/dt
and
dv/dt = d
2
y/dt
2 . Now our DE becomes a system: Comes from our assumption
dy dt dv dt
v
k m y
Comes from the original DE
Exercises
p. 167 #19 • Rewrite the DE as a system of first-order DEs.
• Do (a) and (b).
• Check (b) using the MassSpring tool.
• Do (c) and (d).
Homework
(due 5pm Thursday)
• Read 2.1
• Practice: p. 164-7, #7, 9, 11, 15, 17, 19 • Core: p. 164-7, #10, 16, 20, 21
Some of the problems in this section are really wordy. You don’t have to copy them into your HW.