Jim Lambers
MAT 285
Spring Semester 2012-13
Lecture 11 Notes
These notes correspond to Section 2.5 in the text.
Autonomous Equations
A first-order ODE is said to be autonomous if it is of the form
dy
= f (y).
dt
That is, the derivative of the dependent variable y can be expressed as a function of only y, and
not the independent variable t. An autonomous equation can easily be seen to be separable, and
its solution satsifies the equation
Z y(t)
1
du = t − t0 ,
f (u)
y0
where y(t0 ) = y0 is the initial condition. However, even if the integral of 1/f is not easily computed,
it is still possible to obtain a detailed understanding of the qualitative behavior of the solution for
different initial values.
We recall that an equilibrium solution is a value of y such that dy/dt = 0 for all t in the
domain of the solution. Therefore, any value of y such that f (y) = 0 is an equilibrium solution.
Once all equilibrium solutions are found, we can then determine how the solution behaves for nonequilibrium values of y. For example, suppose that y1 is the only equilibrium solution. If f (y) < 0
for y > y1 , and f (y) > 0 for y < y1 , then it follows that for any initial value y0 , the solution y(t)
will converge to y1 as t → ∞, because the sign of dy/dt = f (y) will cause y(t) to approach the
line y = y1 . We say that y1 is a stable equilibrium solution, as the solution y(t) does not deviate
significantly from the equilibrium state when the initial value y0 deviates in either direction from
y1 .
Now, suppose that the opposite situation occurs: f (y) > 0 for y > y1 , and f (y) < 0 for y < y1 .
Then, we say that y1 is an unstable equilibrium solution, because for any initial value y0 other than
y1 , the solution y(t) will not converge to y1 as t → ∞; rather, it will exhibit exponential deviation
from y1 . If there is another equilbrium solution, then y(t) may approach that value instead. The
reason why y1 is called an unstable solution is that if y0 deviates from y1 at all, then no matter
how small the deviation, the solution y(t) will exhibit further deviation from y1 as t increases.
Finally, it is possible that an equilibrium solution y1 may be neither stable nor unstable, but
rather semi-stable. This is the case if solutions approach y1 when the initial value y0 deviates from
y1 in one direction, but not in the other. That is, y1 appears to be stable for y > y1 and unstable
for y < y1 , or vice versa. This occurs when f (y1 ) = 0 and f 0 (y1 ) = 0 but f 00 (y1 ) 6= 0, which causes
f to touch, but not cross, the y-axis at y1 . Therefore, dy/dt has the same sign for y on either side
of y1 , leading to stability on one side of the line y = y1 but instability on the other.
To help determine whether equilbrium solutions are stable, unstable, or semi-stable, it is helpful
to draw a phase line. A phase line is a representation of the real number line, drawn vertically
and corresponding to the y-axis with the positive y-axis pointing up. First, all equilibrium solutions are marked on the number line. Then, in the intervals that have equilibrium solutions as
endpoints (including semi-infinite intervals), arrows are used to indicate the sign of f (y): arrows
point downward where f (y) < 0 and upward where f (y) > 0.
1
Then, the arrows can be used to determine stability of the equilibrium solutions. If the arrows
above and below an equilibrium solution y1 both point toward y1 , then y1 is stable. If both arrows
point away from y1 , then it is unstable. If one arrow points toward y1 and the other points away
from y1 , then y1 is semi-stable.
Example We consider an initial value problem involving exponential growth,
dy
= ry,
dt
y(0) = y0 ,
where r is a positive constant. The ODE in this problem can be used as a simple model of population
growth. The initial value problem has the solution
y(t) = y0 ert .
With f (y) = ry, the only equilibrium solution is y = 0. When y > 0, f (y) > 0, and when y < 0,
f (y) < 0, so in either case, the solution y(t) is diverging from the equilibrium solution. Therefore,
y = 0 is an unstable equilibrium point. The phase line for this problem features an upward arrow
on the positive y-axis and a downward arrow on the negative y-axis, both diverging from the point
y = 0 that is marked on the phase line as an equilibrium solution. 2
Example We consider the logistic equation
dy
y
=r 1−
y,
dt
K
y(0) = y0 ,
where r and K are positive constants. This is a modification of the previous equation that attempts
to model population growth more realistically. The constant K is referred to as a saturation level,
or environmental carrying capacity, of a species. The growth rate, rather than being a constant r,
is now r(1 − y/K), so that when y reaches K, the growth rate becomes 0, and if y > K, the growth
rate becomes negative.
We will solve this equation, but first we will determine the behavior of its equilibrium solutions,
which are y = 0 and y = K. When y < 0, dy/dt < 0, so solutions with y0 < 0 diverge from y = 0.
When 0 < y < K, dy/dt > 0, so solutions with 0 < y0 < K diverge from y = 0 but converge toward
y = K. Finally, when y > K, dy/dt < 0, so solutions with y0 > K converge toward y = K. We
conclude that y = 0 is an unstable equilibrium point, and that y = K is a stable equilibrium point.
Now, we solve the equation. First, we rewrite it as
r+
K
dy
= 0,
y(y − K) dt
and then integrate with respect to t from 0 to T to obtain
Z y(T )
Z T
K
r dt +
du = 0,
u(u − K)
0
y0
where we have used the substitution u = y(t) in the second integral. Using partial fraction decomposition, we obtain
Z y(T )
1
1
rT +
− du = 0,
u
−
K
u
y0
which yields
rT + ln |y(T ) − K| − ln |y(T )| − ln |y0 − K| + ln |y0 | = 0.
2
Using the properties of logarithmic functions, and then exponentiating, we obtain
y0 (y(T ) − K) −rT
.
y(T )(y0 − K) = e
Rearranging to solve for y(T ) yields
|y0 (y(T ) − K)| = e−rT |y(T )(y0 − K)|,
Assume that 0 < y0 < K. Then 0 < y(T ) < K as well, which allows us to drop the absolute value
signs. We then have
y0 y(T ) − y0 K = e−rT y(T )(y0 − K),
which can be rearranged to obtain
y(T ) =
y0 K
.
y0 + e−rT (K − y0 )
It can be seen from this explicit solution that as T → ∞, y(T ) → K, which is consistent with our
previous analysis.
If we assume that y0 < 0, or y0 > K, then we actually obtain the same formula for y(T ). In the
case of y0 > K, the solution will approach K as T increases, as expected since y = K is a stable
equilibrium point. However, if y0 < 0, then there is a finite time T at which the denominator will
equal zero, resulting in a vertical asymptote at which y → −∞. In Figure 1, the solution is plotted
for several values of y0 , in the case of r = 1 and K = 1, and the phase line is shown as well.
Figure 1: Solutions (left plot) and phase line (right plot) for y 0 = (y − 1)y
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