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First Order Nonlinear Equations

The most general nonlinear first order ordinary differential equation we could imagine wouldbe of the form

Ft,yt,y vt = 0. 1

In general we would have no hope of solving such an equation. A less general nonlinearequation would be one of the form

y vt = Ft,yt, 2

but even this more general equation is often too difficult to solve. We will consider then,equations of the form

y vt = Fyt. 3

Equation (3) is said to be an autonomous differential equation, meaning that the nonlinearfunction F does not depend explicitly on t. The equation (2) is nonautonomous because Fdoes contain explicit t dependence. The equations,

yvt = yt2 and y vt = yt2 + t2,

are examples of autonomous and nonautonomous equations, respectively.

We will consider some examples of nonlinear first order equations first and then statesome general principles that will make it clear why autonomous equations are easier to dealwith than nonautonomous ones. We will first recall a few of the properties that we haveobserved about linear problems. We saw in several examples that solutions to linearproblems tend to be smooth functions, even when the coefficients and forcing term arediscontinuous. The only thing that seemed to lead to a blow up or singularity in thesolution (i.e., a point where the solution becomes undefined) was a singularity in acoefficient or forcing term. Thus, when there are no singularities in the inputs of theproblem, there will be no singularities in the solution and the solution will satisfy theequation for all t. Another way to say this is that there are no spontaneous singularities inthe 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 functionswhich 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 cantbe written in the form of the general solution for some choice of the parameter. We havenot yet proved this last statement but will prove it later. When an initial condition iscombined with the equation, it follows that there is a unique value of the parameter in thegeneral solution that causes the initial condition to be satisfied. Thus the initial valueproblem will always have a unique solution. Summarizing these features of linear problems,we have

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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,

yvt = yt2, y0 = A > 0.

Writing X dyy2

= X dt

leads to

?yt?1 = t? C0,

or

yt = 1C0 ? t

.

Then y0 = A implies C0 = 1/A and the solution of the initial value problem isgiven by,

yt = A1 ? At

.

Note that this function has a singularity at t = 1/A so that, even though there is nothing inthe equation to suggest it, the solution develops a spontaneous singularity.

Example 2 Consider the initial value problem,

yvt = 2 yt , yt0 = 0, t0 > 0.

Writing X dy2 yt

= X dt,

we find

yt = t? C0

or

yt = t? C02.

The functions yt solve the differential equation for each value of the constant C0.However, this cannot be called the general solution of the equation since the zero function

yt = 0 also solves the equation but the zero function does not equal t? C02 for any value

of C0.

Notice also that

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yt = t? t02

solves the intial value problem, while at the same time, for every t1 > t0, the functions

yt =0 if t0 < t < t1

t? t12 if t > t1

also solve the initial value problem. These two examples illustrate that solutions to nonlineardifferential equations may behave quite differently from solutions to linear problems. Thisdoes not mean that all nonlinear problems are necessarily badly behaved, it only meansthat we should not assume that we cannot suppose that they are necessarily as wellbehaved as linear problems.

Autonomous First Order Equations

The simplest possible model for population growth is the equation

Pvt = kPt, P0 = P0

where the constant k denotes the growth rate of the population. If k is positive, thepopulation described by this equation grows rapidly to infinity while, if k is negative, itdecays steadily to zero. In an effort to make a more realistic model for growth of populationsize, we may suppose that k is not a constant but depends on P. In particular, if wesuppose kP = rPK ? Pt for some positive constants r, PK, then we obtain theautonomous equation

Pvt = rPK ? PtPt, P0 = P0. 4

Here FP = rPK ? PtPt does not depend explicitly on t. It is not difficult to solve thisdifferential equation but we are going to see that it is possible to completely understand thebehavior this equation predicts without actually solving the equation. In fact, it is generallyharder to see the predicted behavior from the solution than from the qualitative analysis weare going to describe.

We begin by determining if the equation has any critical points. These are values PD ofP such that FPD = 0, (which means that if there is any time t = TD at which PTD = PD,then Pt = PD for all t TD). Equation (4) has two critical points, namely P = 0 and P= PK.In addition, for any solution curve starting at P0 with 0 < P0 < PK, we have

Pv0 = rPK ? P0P0 > 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 likethis. If there is a point on this solution curve where P is not increasing, then there has to bea time t0 > 0 where Pvt0 = 0 (i.e., if P is decreasing, it has to first stop increasing). Sincethis is a point on a solution curve, if Pvt0 = 0 then either Pt0 = 0 or else Pt0 = PK. But

Pt0 = 0 is not possible since P started at P0 with 0 < P0 < PK, and P has been increasingsince then. On the other hand, Pt0 = PK is not possible either since if t0 is the first timeafter t = 0 where Pt0 = PK, then Pt < PK for t < t0 which would imply Pvt0 > 0. Butthis contradicts equation (4) which asserts that Pvt0 = 0 if Pt0 = PK. The only

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remaining possibility is that a solution curve which originates at P0 with 0 < P0 < PK, mustapproach the horizontal asymptote P = PK, increasing steadily as t tends to K. In the sameway, we can argue that any solution curve that originates at P0 with P0 > PK, mustapproach the horizontal asymptote P = PK, from above, decreasing steadily as t tends toK. A plot showing three different solution curves is shown below.

If we define a asymptotically stable critical point P = PD to mean a value PD such that

FPD = 0, and such that any for trajectory, Pt, originating at P0 near PD, it follows thatPt PD as t K. A critical point for which the distance between Pt and PD increasesas t K, even for P0 arbitrarily near PD, is said to be unstable. Equation (4) has criticalpoints at P = 0 and P= PK. T

By examining the figure above, we see that the critical point at P = 0 is unstable, while thecritical point at P = PK is asymptotically stable.

Now we state some general results about autonomous nonlinear equations. If weconsider the equation

yvt = Fyt, 5

then

1. the critical points of (5) are the values yD for which FyD = 0

2. the critical point, yD is stable if FvyD < 0

3. the critical point, yD is unstable if FvyD > 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 similarmanner so let us prove 3. The figure below shows an F(y) having a zero at yD (hence yD is acritical point for (5)) with FvyD > 0. Since the derivative of F at yD is positive and FyD = 0,

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it follows that at any y > yD, we have Fy > 0 which implies, in turn through (5) that y vt ispositive (i.e., y is increasing when y > yD . Similarly, at any y < yD, we have Fy < 0 whichimplies, in turn through (5) that y vt is negative (i.e., y is decreasing when y < yD . Thensolution curves of (5) which originate near yD move away from yD rather than towards it. Thisis 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 = y1t and y = y2t are two different

solutions of (5) whose graphs cross at some time t0. To say the graphs cross at t = t0means that y1t0 = y2t0, and y1

v t0 y2v t0.; i.e., the graphs go through the same point

but have different slopes there. But

y1t0 = y2t0

y1v t0 = Fy1t0 = Fy2t0 = y2

v t0,

hence the slopes cannot be different. This contradiction shows that solution curves areeither identical or else they never cross. We say that the family of solution curves iscoherent.

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