Review of Lecture 5
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Models could involve just one or two equations (e.g. orbit
calculation), or hundreds of equations (as in climate
modeling).
To model a vertical cannon shot:
F =
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Expressed in terms of altitude x = r − R, we have
mv
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dv
GMm
= −m
2
r
dt
dv
GMm
=−
dx
(R + x)2
Max altitude
ξ=
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v02 R 2
.
2GM − v02 R
Escape velocity
r
vescape =
2GM
R
Autonomous equations
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The general first-order equation is
dy
= F (y , t)
dt
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If F (t, y ) actually depends only on y we have
dy
= f (y ).
dt
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Such an equation is called autonomous
Autonomous equations
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The general first-order equation is
dy
= F (y , t)
dt
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If F (t, y ) actually depends only on y we have
dy
= f (y ).
dt
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Such an equation is called autonomous
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For example the simplest equations for growth or decay:
y 0 = ky
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Growth with k > 0, decay with k < 0, since y = Ae kt
Modeling population growth
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With a constant birth rate and constant death rate, we get for
the population y ,
y 0 = ry .
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Here r is the difference of the birth rate and the death rate.
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With r > 0 the model predicts exponential growth.
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In a finite environment, that can’t be right for long.
The logistic equation
We need a revised model that allows y to be bounded when y
grows large.
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One such model is the logistic equation:
y
0
y
y =r 1−
K
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For y small it’s almost y 0 = ry .
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For y large y 0 will be negative, so y will decrease.
The logistic equation
We need a revised model that allows y to be bounded when y
grows large.
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One such model is the logistic equation:
y
0
y
y =r 1−
K
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For y small it’s almost y 0 = ry .
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For y large y 0 will be negative, so y will decrease.
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Is there an equilibrium solution?
Is there an equilibrium solution?
y
y
y0 = r 1 −
K
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To find an equilibrium solution, set y 0 = 0:
y
0=r 1−
y
K
There are two equilibrium solutions: y = 0 and y = K .
Is there an equilibrium solution?
y
y
y0 = r 1 −
K
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To find an equilibrium solution, set y 0 = 0:
y
0=r 1−
y
K
There are two equilibrium solutions: y = 0 and y = K .
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K is called the “carrying capacity” of the environment.
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Solutions can’t “cross” or “touch” an equilibrium solution,
because through each point (t, y ) there is a unique solution.
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So if another solution touched or crossed an equilibrium
solution, it would have to coincide with that solution.
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As engineers you’ll probably take that on authority or faith.
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The curious or skeptical can read section 2.8 of the text
Graphical investigation
In class we will graph the solutions and see that all solutions
approach the equilibrium solution y = K for large t.
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Solutions with y0 < K increase
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Solutions with y > 0 decrease
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If you are viewing these slides at another time, refer to pp.
80-81, and use MathXpert Grapher or other software to graph
the solutions.
Concavity
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Recall from calculus that the second derivative y 00 determines
the concavity (upwards or downwards), and the inflection
points (where y 00 = 0).
We can do some of this analysis for any autonomous equation
y 0 = f (y ).
y 00 =
d 0
y
dt
Concavity
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Recall from calculus that the second derivative y 00 determines
the concavity (upwards or downwards), and the inflection
points (where y 00 = 0).
We can do some of this analysis for any autonomous equation
y 0 = f (y ).
y 00 =
=
d 0
y
dt
d
f (y )
dt
Concavity
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Recall from calculus that the second derivative y 00 determines
the concavity (upwards or downwards), and the inflection
points (where y 00 = 0).
We can do some of this analysis for any autonomous equation
y 0 = f (y ).
d 0
y
dt
d
=
f (y )
dt
dy
= f 0 (y )
dt
y 00 =
Concavity
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Recall from calculus that the second derivative y 00 determines
the concavity (upwards or downwards), and the inflection
points (where y 00 = 0).
We can do some of this analysis for any autonomous equation
y 0 = f (y ).
d 0
y
dt
d
=
f (y )
dt
dy
= f 0 (y )
dt
= f 0 (y )f (y )
y 00 =
Thus the signs of f and f 0 tell us about concavity
After-the-lecture viewers, see pages 80-81 to replace graphs done
in class.
The phase line
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If we think of t as time and only use one space dimension,
then our graph becomes a moving point on a line.
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It is customary to draw that line vertically.
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The direction field from the 2-d graph now becomes just an
arrow up or down at every point on the line. This line, with
arrows, is the “phase line”.
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In our case, the arrows point up at points below y = K , and
down at points above y = K .
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at y = K the arrow has zero length. That is a “critical point.”
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There is another critical point at y = 0, so technically the
arrow doesn’t point up there, but it does point up at points
just below or above there
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At the critical point y = K the arrows do change direction.
After-lecture viewers, see Fig. 2.5.3.
Determining qualitative behavior
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We can determine the qualitative behavior of the solutions
from the graph of f alone.
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We did not need a computer or an explicit solution, although
in class a computer was used first.
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We could have started with f , found the critical points and
the phase line, sketched the direction field and solutions using
the concavity information.
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This procedure was illustrated at the whiteboard.
Solving the logistic equation
Since solutions can’t cross equilibrium solutions, if we have a
non-equilibrium solution then y is never equal to 0 or K , for any
value of t. Therefore we won’t get a zero denominator when we
separate the variables:
rdt =
dy
(1 − y /K )y
Solving the logistic equation
dy
(1 − y /K )y
Z
dy
rt =
(1 − y /K )y
rdt =
Solving the logistic equation
dy
(1 − y /K )y
Z
dy
rt =
(1 − y /K )y
Z
1
1/K
rt =
+
dy
y
1 − y /K
rdt =
Solving the logistic equation
dy
(1 − y /K )y
Z
dy
rt =
(1 − y /K )y
Z
1
1/K
rt =
+
dy
y
1 − y /K
y
rt + c = ln |y | − ln 1 − K
rdt =
Can we drop the absolute value signs?
Solving the logistic equation
dy
(1 − y /K )y
Z
dy
rt =
(1 − y /K )y
Z
1
1/K
rt =
+
dy
y
1 − y /K
y
rt + c = ln |y | − ln 1 − K
rdt =
Can we drop the absolute value signs? Yes, if 0 < y0 < K , since
then y stays in this interval.
y
rt + c = ln y − ln 1 −
K
So we need to solve this equation for y :
y
rt + c = ln y − ln 1 −
K
y
= ln
1 − y /K
So we need to solve this equation for y :
y
rt + c = ln y − ln 1 −
K
y
= ln
1 − y /K
y
e rt+c =
since e ln u = u
1 − y /K
So we need to solve this equation for y :
y
rt + c = ln y − ln 1 −
K
y
= ln
1 − y /K
y
e rt+c =
since e ln u = u
1 − y /K
y
= e c e rt = Ce r t
with C = e c
1 − y /K
So we need to solve this equation for y :
y
rt + c = ln y − ln 1 −
K
y
= ln
1 − y /K
y
e rt+c =
since e ln u = u
1 − y /K
y
= e c e rt = Ce r t
with C = e c
1 − y /K
It works if y > K too, but then we have a negative sign on the left.
Initial condition determines C
When t = 0, we have y = y0 . So we put those values into
y
1 − y /K
= e c e rt = Ce r t
with C = e c
and we get
y0
|1 − y0 /K |
= C
By putting the absolute value signs in, the equation works for all
y0 > 0.
Solving for y
So put that value for C in, and we get
y
|1 − y /K |
= Ce rt
=
y0 e rt
|1 − y0 /K |
Solving for y
So put that value for C in, and we get
y
|1 − y /K |
= Ce rt
=
y0 e rt
|1 − y0 /K |
But since 1 − y /K has the same sign as 1 − y0 /K we can take the
absolute value signs off again.
y
1 − y /K
=
y0 e rt
1 − y0 /K
Solving for y
So put that value for C in, and we get
y
|1 − y /K |
= Ce rt
=
y0 e rt
|1 − y0 /K |
But since 1 − y /K has the same sign as 1 − y0 /K we can take the
absolute value signs off again.
y
y0 e rt
=
1 − y /K
1 − y0 /K
y (1 − y0 /K ) = y0 e rt (1 − y /K )
Solving for y
So put that value for C in, and we get
y
|1 − y /K |
= Ce rt
=
y0 e rt
|1 − y0 /K |
But since 1 − y /K has the same sign as 1 − y0 /K we can take the
absolute value signs off again.
y
y0 e rt
=
1 − y /K
1 − y0 /K
y (1 − y0 /K ) = y0 e rt (1 − y /K )
y0 y0 e rt
y 1−
+
= y0 e rt
K
K
y0 y0 e rt
y 1−
+
= y0 e rt
K
K
y=
y0 e rt
1−
y0
K
+
y0 e rt
K
y0 y0 e rt
y 1−
+
= y0 e rt
K
K
y=
y
=
=
y0 e rt
1−
y0
K
+
y0 e rt
K
y0 e rt
y0
K
rt
+ y0Ke
Ky0 e rt
K − y0 + y0 e rt
1−
y
=
Ky0 e rt
K − y0 + y0 e rt
Divide numerator and denominator both by e rt :
y
=
Ky0
y0 + (K − y0 )e −rt
Qualitative conclusions check out?
Now we can confirm our qualitative conclusions:
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The limit as t goes to ∞ is K .
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Solutions starting in some interval around K converge to K .
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Solutions starting near 0 diverge from 0 (well, at least positive
ones)
Qualitative conclusions check out?
Now we can confirm our qualitative conclusions:
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The limit as t goes to ∞ is K .
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Solutions starting in some interval around K converge to K .
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Solutions starting near 0 diverge from 0 (well, at least positive
ones)
We say y = K is a stable equilibrium and y = 0 is an unstable
equilibrium.
Growth with a threshold
y
y 0 = −r 1 −
y = f (y )
T
Growth with a threshold
y
y 0 = −r 1 −
y = f (y )
T
Practice what you learned in the first part of the lecture:
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Draw the graph of f .
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Draw the phase line.
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Sketch the direction field and some solutions.