First Order Nonlinear Equations
The most general nonlinear first order ordinary differential equation we could imagine would
be of the form
FÝt, yÝtÞ, y v ÝtÞÞ = 0. Ý1Þ
In general we would have no hope of solving such an equation. A less general nonlinear
equation would be one of the form
y v ÝtÞ = FÝt, yÝtÞÞ, Ý2Þ
but even this more general equation is often too difficult to solve. We will consider then,
equations of the form
y v ÝtÞ = FÝyÝtÞÞ. Ý3Þ
Equation (3) is said to be an autonomous differential equation, meaning that the nonlinear
function F does not depend explicitly on t. The equation (2) is nonautonomous because F
does contain explicit t dependence. The equations,
y v ÝtÞ = yÝtÞ 2 and y v ÝtÞ = yÝtÞ 2 + t 2 ,
are examples of autonomous and nonautonomous equations, respectively.
We will consider some examples of nonlinear first order equations first and then state
some general principles that will make it clear why autonomous equations are easier to deal
with than nonautonomous ones. We will first recall a few of the properties that we have
observed about linear problems. We saw in several examples that solutions to linear
problems tend to be smooth functions, even when the coefficients and forcing term are
discontinuous. The only thing that seemed to lead to a ”blow up” or singularity in the
solution (i.e., a point where the solution becomes undefined) was a singularity in a
coefficient or forcing term. Thus, when there are no singularities in the inputs of the
problem, there will be no singularities in the solution and the solution will satisfy the
equation for all t. Another way to say this is that there are no spontaneous singularities in
the solution to a linear ODE. Solution singularities can only result from input singularities.
In addition, the general solution of a linear equation is a 1-parameter family of functions
which satisfies the equation for every choice of the parameter and which contains all
possible solutions to the equation. That is, there are no solutions to the equation that can’t
be written in the form of the general solution for some choice of the parameter. We have
not yet proved this last statement but will prove it later. When an initial condition is
combined with the equation, it follows that there is a unique value of the parameter in the
general solution that causes the initial condition to be satisfied. Thus the initial value
problem will always have a unique solution. Summarizing these features of linear problems,
t no spontaneous singularities
t 1-parameter family of general solutions
t initial value problem has a unique solution
We will now give some examples that will show that for nonlinear problems, none of these
things need be true.
Example 1 Consider the initial value problem,
y v ÝtÞ = yÝtÞ 2 , yÝ0Þ = A > 0.
Writing X dy = X dt
?yÝtÞ ?1 = t ? C 0 ,
yÝtÞ = 1 .
C0 ? t
Then yÝ0Þ = A implies C 0 = 1/A and the solution of the initial value problem is
yÝtÞ = A .
1 ? At
Note that this function has a singularity at t = 1/A so that, even though there is nothing in
the equation to suggest it, the solution develops a spontaneous singularity.
Example 2 Consider the initial value problem,
y v ÝtÞ = 2 yÝtÞ , yÝt 0 Þ = 0, t 0 > 0.
Writing X = X dt,
yÝtÞ = t ? C 0
yÝtÞ = Ýt ? C 0 Þ 2 .
The functions yÝtÞ solve the differential equation for each value of the constant C 0 .
However, this cannot be called the general solution of the equation since the zero function
yÝtÞ = 0 also solves the equation but the zero function does not equal Ýt ? C 0 Þ 2 for any value
of C 0 .
Notice also that
yÝtÞ = Ýt ? t 0 Þ 2
solves the intial value problem, while at the same time, for every t 1 > t 0 , the functions
0 if t 0 < t < t 1
Ýt ? t 1 Þ 2 if t > t1
also solve the initial value problem. These two examples illustrate that solutions to nonlinear
differential equations may behave quite differently from solutions to linear problems. This
does not mean that all nonlinear problems are necessarily badly behaved, it only means
that we should not assume that we cannot suppose that they are necessarily as well
behaved as linear problems.
Autonomous First Order Equations
The simplest possible model for population growth is the equation
P v ÝtÞ = kPÝtÞ, PÝ0Þ = P 0
where the constant k denotes the growth rate of the population. If k is positive, the
population described by this equation grows rapidly to infinity while, if k is negative, it
decays steadily to zero. In an effort to make a more realistic model for growth of population
size, we may suppose that k is not a constant but depends on P. In particular, if we
suppose kÝPÞ = rÝP K ? PÝtÞÞ for some positive constants r, P K , then we obtain the
P v ÝtÞ = r ÝP K ? PÝtÞÞPÝtÞ, PÝ0Þ = P 0 . Ý4Þ
Here FÝPÞ = r ÝP K ? PÝtÞÞPÝtÞ does not depend explicitly on t. It is not difficult to solve this
differential equation but we are going to see that it is possible to completely understand the
behavior this equation predicts without actually solving the equation. In fact, it is generally
harder to see the predicted behavior from the solution than from the qualitative analysis we
are going to describe.
We begin by determining if the equation has any critical points. These are values P D of
P such that FÝP D Þ = 0, (which means that if there is any time t = T D at which PÝT D Þ = P D ,
then PÝtÞ = P D for all t ³ T D ). Equation (4) has two critical points, namely P = 0 and P = P K .
In addition, for any solution curve starting at P 0 with 0 < P 0 < P K , we have
P v Ý0Þ = r ÝP K ? P 0 ÞP 0 > 0, which means the curve starts out with P increasing. In fact, we
will show that P is increasing along this solution curve for all t > 0. The argument goes like
this. If there is a point on this solution curve where P is not increasing, then there has to be
a time t 0 > 0 where P v Ýt 0 Þ = 0 (i.e., if P is decreasing, it has to first stop increasing). Since
this is a point on a solution curve, if P v Ýt 0 Þ = 0 then either PÝt 0 Þ = 0 or else PÝt 0 Þ = P K . But
PÝt 0 Þ = 0 is not possible since P started at P 0 with 0 < P 0 < P K , and P has been increasing
since then. On the other hand, PÝt 0 Þ = P K is not possible either since if t 0 is the first time
after t = 0 where PÝt 0 Þ = P K , then PÝtÞ < P K for t < t 0 which would imply P v Ýt 0 Þ > 0. But
this contradicts equation (4) which asserts that P v Ýt 0 Þ = 0 if PÝt 0 Þ = P K . The only
remaining possibility is that a solution curve which originates at P 0 with 0 < P 0 < P K , must
approach the horizontal asymptote P = P K , increasing steadily as t tends to K. In the same
way, we can argue that any solution curve that originates at P 0 with P 0 > P K , must
approach the horizontal asymptote P = P K , from above, decreasing steadily as t tends to
K. A plot showing three different solution curves is shown below.
If we define a asymptotically stable critical point P = P D to mean a value P D such that
FÝP D Þ = 0, and such that any for trajectory, PÝtÞ, originating at PÝ0Þ near P D , it follows that
PÝtÞ ¸ P D as t ¸ K. A critical point for which the distance between PÝtÞ and P D increases
as t ¸ K, even for PÝ0Þ arbitrarily near P D , is said to be unstable. Equation (4) has critical
points at P = 0 and P = P K . T
By examining the figure above, we see that the critical point at P = 0 is unstable, while the
critical point at P = P K is asymptotically stable.
Now we state some general results about autonomous nonlinear equations. If we
consider the equation
y v ÝtÞ = FÝyÝtÞÞ, Ý5Þ
1. the critical points of (5) are the values y D for which FÝy D Þ = 0
2. the critical point, y D is stable if F v Ýy D Þ < 0
3. the critical point, y D is unstable if F v Ýy D Þ > 0
4. distinct trajectories of (5) can never cross
Here 1 is just the definition of critical point. Points 2 and 3 can be proved in a similar
manner so let us prove 3. The figure below shows an F(y) having a zero at y D (hence y D is a
critical point for (5)) with F v Ýy D Þ > 0. Since the derivative of F at y D is positive and FÝy D Þ = 0,
it follows that at any y > y D , we have FÝyÞ > 0 which implies, in turn through (5) that y v ÝtÞ is
positive (i.e., y is increasing when y > y D Þ . Similarly, at any y < y D , we have FÝyÞ < 0 which
implies, in turn through (5) that y v ÝtÞ is negative (i.e., y is decreasing when y < y D Þ . Then
solution curves of (5) which originate near y D move away from y D rather than towards it. This
is what it means for the critical point to be unstable. Point 2 is proved by a similar argument.
To see that point 4 must be true suppose that y = y 1 ÝtÞ and y = y 2 ÝtÞ are two different
solutions of (5) whose graphs cross at some time t 0 . To say the graphs ”cross” at t = t 0
means that y 1 Ýt 0 Þ = y 2 Ýt 0 Þ, and y v1 Ýt 0 Þ ® y v2 Ýt 0 Þ.; i.e., the graphs go through the same point
but have different slopes there. But
y 1 Ýt 0 Þ = y 2 Ýt 0 Þ
y v1 Ýt 0 Þ = FÝy 1 Ýt 0 ÞÞ = FÝy 2 Ýt 0 ÞÞ = y v2 Ýt 0 Þ,
hence the slopes cannot be different. This contradiction shows that solution curves are
either identical or else they never cross. We say that the family of solution curves is