UCSC
AMS/ECON 11B
Autonomous Differential Equations
c
2007
by Yonatan Katznelson, UCSC
1. Introduction
A differential equation is called autonomous if it has the form
dy
(1.1)
= F (y).
dt
In an autonomous differential equation, the rate of change of the unknown function y
depends directly (and only) on the function itself.
Examples of autonomous differential equations include
a. the equation for Newton’s law of cooling
dτ
= k(TA − τ ),
(1.2)
dt
where τ (t) is the temperature at time t of an object submerged in a medium of constant
ambient temperature TA ;
b. the exponential model for population growth
dy
(1.3)
= ky,
dt
where y(t) is the size of a population at time t;
c. the logistic model for population growth
dy
(1.4)
= ky(M − y),
dt
where y(t) is again the size of a population at time t, and M is the carrying capacity of
the population, i.e., the (theoretical) maximum size of the population.
2. Dependence on initial conditions: a case study
A differential equation, like (1.1), together with an initial condition y(0) = y0 give rise
to a (unique) particular solution of the differential equation.1 When we change the initial
conditions, the corresponding solutions can exhibit significantly different behavior, even
though the differential equation itself doesn’t change.
Case study. Consider the logistic initial value problem
dy
(2.1)
= 0.1y(10 − y),
y(0) = y0 .
dt
The nature of the solutions of this equation depend both on the equation and on the initial
value, as illustrated below. We begin by deriving the general form of the solution.
First, separate the variables and integrate both sides of the resulting equation to obtain
y = t + C.
(2.2)
ln 10 − y 1The questions of whether an initial value problem of this kind has a solution, and whether that solution
is unique are important, but are beyond the scope of this course. If the function F (y) is continuous, then
the existence and uniqueness are guaranteed, with certain restrictions.
1
2
Exponentiating both sides of this leads to the equation
y t+C
= Aet ,
10 − y = e
where A = eC . At this point, it is convenient to solve for the constant A in terms of the
initial value y(0) = y0 . Plugging t = 0 into both sides of the last equation gives
y0 ,
A=
10 − y0 and the solution takes the form
y y0 t
=
10 − y 10 − y0 e .
Dropping2 the absolute value signs (| |) from both sides of the equation gives
y
y0
=
et .
10 − y
10 − y0
Solving the last equation for y gives
(2.3)
y=
10y0 et
.
10 − y0 + y0 et
If y0 6= 0, then we can divide both the numerator and denominator by y0 et to obtain the
solution
10
(2.4)
y=
,
1 + be−t
10 − y0
where b =
.
y0
Case 1. If y0 = 0, then a quick glance at (2.3) shows that the solution is the constant
function y = 0.
Case 2. If y0 = 10, then it is just as easy to see that the solution is the constant function
y = 10.
* For the remaining cases, we use the second form of the solution, (2.4). In all of
these cases (y0 6= 0, 10), we have
lim y(t) = 10
t→∞
and
lim y(t) = 0,
t→−∞
since limt→∞ (1 + be−t ) = 1 and limt→−∞ (1 + be−t ) = ∞.
Case 3. If y0 < 0, then b < −1 (why?) and the solution takes the form
10
.
(2.5)
y(t) =
1 − |b|e−t
In this case ln |b| > 0 and as t approaches ln |b|, the denominator approaches 0.
More precisely, as t approaches ln |b| from above, the denominator approaches 0
2This requires some justification, since for example |u − 1| = |1 − u| for all u, but u − 1 6= 1 − u, unless
u = 1. The justification is that the function y is continuous, so the function y/(10 − y) is also continuous
(as long as y 6= 10). This means that for y very close to y0 (i.e., t very close to 0), the two ratios must have
the same sign.
3
from above, and as t approaches ln |b| from below the denominator approaches 0
from below. This means that
lim
t→ln |b|+
y(t) = ∞
and
lim
t→ln |b|−
y(t) = −∞.
We say in such cases that the solution ‘blows up in finite time’.
Case 4. If 0 < y0 < 10, then b > 0 (why?). It follows that, in this case, 1 + be−t > 1 for
all t, so 0 < y(t) < 10 for all t.
Case 5. If 10 < y, then −1 < b < 0 (why?). The solution has the same form as (2.5), and
we see the same phenomenon that we observed in Case 3, except that ln |b| < 0
in this case, so the ‘blow up’ occurs on the negative side of the t-axis.
These five cases are illustrated in the figure below. The green line is the graph of the
solution with y(0) = 2; the blue line is the graph of the constant solution y = 10; the red line
is the graph of the solution with y(0) = −1; and the purple line is the graph of the solution
with y(0) = 12. The dotted purple and red lines are the respective vertical asymptotes of
the solutions that ‘blow up’.
24
16
8
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
-8
-16
-24
Figure 1. Typical solutions of the differential equation y 0 = 0.1y(10 − y).
3. Equilibrium solutions
Definition. A solution y(t) of an autonomous differential equation is called an equilibrium
solution if it is constant, i.e., if y(t) = C. For example, the two solutions y = 0 and y = 10
of (2.1) are eqilibrium solutions.
Such solutions are called equilibrium solutions because the dynamical system being described by the differential equation is in equilibrium, i.e., all the forces acting in the system
are in balance so there is no change.
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If y(t) = C is a solution of (1.1), then on the one hand, dy/dt = 0 (since the derivative
of a constant function is 0), and on the other hand dy/dt = F (y) = F (C). This gives us
the following fact.
Fact 1. y(t) = C is an equilibrium solution of the differential equation
only if F (C) = 0.
dy
= F (y) if and
dt
In the case study of the previous section, we found the equilibrium solutions by plugging
special initial values into the general solution of the differential equation. This is not
necessary, at least in principle. The fact above says that finding equilibrium solutions
involves solving an algebraic equation, not a differential equation.
Examples.
a. The function τ (t) = TA is an equilibrium solution of the cooling equation (1.2). This is
to be expected: if the temperature of the object is at the ambient temperature, it will
remain at that temperature for all time.
b. The function y(t) = 0 is an equilibrium solution of the exponential equation (1.3). Once
again this makes sense: if the population size is 0, there is no way for it to grow (at least
in the context of biological or chemical systems).
c. The logistic population model (1.4) has two equilibrium solutions, y(t) = 0 and y(t) = M ,
where M is the carrying capacity of the population in question.
d. The equation dy/dt = ey has no equilibrium solutions since the function ey is never 0.
e. Find the equilibrium solutions of the differential equation
(3.1)
dy
= 0.1y(10 − y) − 1.6.
dt
This amounts to finding the solutions of the equation
0.1y(10 − y) − 1.6 = 0.
Rearranging the terms of this quadratic equation gives
−0.1y 2 + y − 1.6 = 0,
and using the quadratic formula, yields the two equilibrium solutions y = 8 and y = 2.
4. Stability of equilibria
Consider the (generic) autonomous differential equation
dy
= F (y).
dt
If F (y0 ) = 0, then y = y0 is an equilibrium solution, and if y(0) = y0 , then the solution
never moves away from this value. In other words, the internal dynamics of the system
being modeled by the differential equation y 0 = F (y) keep everything in balance, and y
doesn’t change.
Suppose now that a system in equilibrium is subjected to an external perturbation that
moves y away from the equilibrium value y0 , but doesn’t change the internal dynamics of
the system. What will happen to y over time after the perturbation?
This type of question arises in applications quite frequently. For example, suppose that
y(t) is the number of trout in a small lake, and assume that the trout population is growing
(4.1)
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logistically. I.e., y satisfies the differential equation
dy
= ky(M − y),
dt
where M is the carrying capacity of the trout population, and k > 0. In particular, y = M
is an equilibrium of this system. If the trout population is in equilibrium, so that y = M ,
and someone dumps α more trout into the lake (α > 0), what will happen to the trout
population in the lake over time, assuming that the dynamics of population growth don’t
change?
If the α additional trout were dumped into the lake at time t0 , then y(t0 ) = M + α. Since
the internal dynamics haven’t changed, it follows that
dy = ky(t0 )(M − y(t0 )) = k(M + α)(−α) < 0
dt t=t0
so the trout population will start to move down from M + α. In fact, as long as y > M , we
have y 0 < 0 and the population continues to decrease in size for as long as y(t) > M .
Suppose, on the other hand, that β of the trout are removed from the lake at time t0 , so
that y(t0 ) = M − β (where β < M ). Then
dy = ky(t0 )(M − y(t0 )) = k(M − β)(β) > 0
dt t=t0
so the trout population will increase up from M − β, and in fact will continue to increase
in size for as long as y(t) < M .
In both cases, it appears that over time the trout population will tend to return to the
equilibrium size M .
Definition. An equilibrium solution y = y0 is called stable if the solution y(t) tends to
return to the equilibrium value over time after (small) perturbations. We can express this
more formally as follows.
dy
= F (y) is stable if the
The equilibrium solution y = y0 of the differential equation
dt
solution y(t) of the initial value problem
dy
= F (y);
y(0) = y1
dt
satisfies
lim y(t) = y0
t→∞
when |y1 − y0 | is sufficiently small.
Example. Take another look at the case study in §2. The analysis there shows that if
y(0) > 10 or if 0 < y(0) < 10, then the corresponding solution of (2.1) satisfies
lim y(t) = 10.
t→∞
This means that the equilibrium solution y(t) = 10 is stable.
On the other hand, if you perturb the other equilibrium solution y(t) = 0 away from 0,
then the new solution stays away from 0 for all t > 0. This follows because if y(0) > 0, the
solution moves towards y = 10 (and away from 0), and if y(0) < 0 then the solution moves
towards −∞. So the solution y(t) = 0 is unstable.
As it turns out, however, we don’t always need to have the explicit solutions of an
autonomous differential equation in order to determine which of its equilibrium solutions
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are stable. It is often enough to study the behavior of the function F (y) around the
equilibrium value to determine whether the equilibrium is stable.
Suppose that y = y0 is an equilibrium solution of the differential equation (4.1). Then, in
particular, F (y0 ) = 0. There are now several possibilities for the behavior of F (y) around
the point y0 . To keep the discussion simple, I’ll assume that F is continuous.
I. F (y) = 0 for all y in some interval (y0 − a, y0 + a) around y0 . In this case, y = y1
is an equilibrium solution for every y1 in (y0 − a, y0 + a). So perturbing the solution
y = y0 a small amount just moves you from one equilibrium solution to another, and
none of these equilibria are stable, since you never return to the original equilibrium
solution.
II. F (y) > 0 for all y in some interval (y0 −a, y0 +a) around y0 . In this case, if we perturb
the equilibrium y = y0 a little below y0 , e.g., by setting y(0) = y1 for y0 − a < y1 < y0 ,
then y 0 (0) = F (y1 ) > 0, and in fact y 0 (t) > 0 for as long as y(t) < y0 . This forces y(t)
to approach y0 as t → ∞.3
On the other hand, if we perturb the equilibrium y = y0 a little above y0 , e.g., by
setting y(0) = y2 for y0 < y2 < y0 + a, then y 0 (0) = F (y2 ) > 0, and in fact y 0 (t) > 0
for as long as y(t) < y0 + a. This forces y(t) to move away from y0 as t → ∞. In this
scenario, either y(t) approaches some other equilibrium solution, or tends to infinity.
In either case, y(t) does not return to y0 .
In conclusion, if F (y) > 0 for all y in some interval (y0 − a, y0 + a) around y0 , then
y = y0 is not stable (though some people say that it is semistable, since it is stable
on one side).
III. F (y) < 0 for all y in some interval (y0 − a, y0 + a) around y0 . The analysis of this
case is analogous to the previous case, but with the effects of perturbing up and down
reversed. I.e., if we perturb the equilibrium solution to a slightly higher value, then the
resulting system returns to the equilibrium solution, but if we perturb the equilibrium
to a lower value, then the resulting system moves away from the equilibrium forever.
So the equilibrium is not stable in this case.
IV. F (y) > 0 for y0 −a < y < y0 and F (y) < 0 for y0 < y < y0 +a. In this case, perturbing
the equilibrium to a slightly lower value has the same effect that it did in case II, and
perturbing the equilibrium to a slightly higher value has the same effect that it had
in case III. Either way, the perturbed system returns toward the equilibrium, so the
equilibrium is stable in this case.
V. F (y) < 0 for y0 − a < y < y0 and F (y) > 0 for y0 < y < y0 + a. In this case,
perturbing the equilibrium to a slightly lower value has the same effect that it did
in case III, and perturbing the equilibrium to a slightly higher value has the same
effect that it had in case II. Either way, the perturbed system moves away from the
equilibrium forever, so the equilibrium is not stable.
Conclusion. The equilibrium solution y = y0 of the system (4.1) is stable if and only if
F (y) > 0 to the left of y0 and F (y) < 0 to the right of y0 .
If the function F (y) is differentiable, then this conclusion yields a very simple criterion
for stability. Namely, if F (y0 ) = 0 and y = y0 is stable, then the function F (y) must be
decreasing at y0 .
3This last statement requires an argument that goes beyond the scope of the course.
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Fact 2. If F (y0 ) = 0 and F 0 (y0 ) < 0, then y = y0 is a stable equilibrium solution of the
differential equation
dy
= F (y).
dt
If F 0 (y0 ) > 0, then y = y0 is not a stable equilibrium.
Examples.
a. Returning once more to the logistic differential equation (1.4), we see that y = M is
a stable equilibrium and y = 0 is unstable, because for F (y) = ky(M − y), we have
F 0 (y) = −2ky + kM , so F 0 (M ) = −kM < 0 and F 0 (0) = kM > 0.
b. Consider the equilibrium solutions, y = 8 and y = 2, of the differential equation (3.1)
that we found in example e. of §3. For that equation, F 0 (Y ) = 1 − 0.2y, and it follows
from Fact 2 that y = 8 is a stable equilibrium of the system because F 0 (8) = −0.6 < 0
and y = 2 is an unstable equilibrium because F 0 (2) = 0.6 > 0.
c. Find the equilibrium solutions of the differential equation
dy
= 0.05y 3 − 0.2y 2 − 1.25y + 5,
dt
and determine the stability or lack thereof of each one.
We begin by solving the equation 0.05y 3 − 0.2y 2 − 1.25y + 5 = 0, which is easier than
you may think, because the cubic on the left factors nicely. Namely,
0.05y 3 −0.2y 2 −1.25y+5 = 0.05y 2 (y−4)−1.25(y−4) = (0.05y 2 −1.25)(y−4) = 0.05(y−5)(y+5)(y−4).
This means that the solutions of the equation are, in ascending order, y1 = −5, y2 = 4
and y3 = 5, and the equilibrium solutions of the differential equation above are therefore
y(t) = −5, y(t) = 4 and y(t) = 5.
d
(0.05y 3 − 0.2y 2 − 1.25y + 5) = 0.15y 2 − 0.4y − 1.25, and
Next,
dy
• 0.15 · (−5)2 − 0.4 · (−5) − 1.25 = 4.5 > 0, so y(t) = −5 is not stable;
• 0.15 · 42 − 0.4 · 4 − 1.25 = −0.45 > 0, so y(t) = 4 is stable;
• 0.15 · 52 − 0.4 · 5 − 1.25 = 0.5 > 0, so y(t) = −5 is not stable.