MATH 219Fall 2014

Lecture 4

Content: Differences between linear and nonlinear equations (section 2.4)).

Suggested Problems:

2.4: 2, 4, 6, 7, 8, 10, 13, 14, 15, 22, 23, 24, 25, 26Recall that a first order ODE is called linear if it can be written in the form

dy

dt+ p(t)y = q(t)

for some functions p(t) and q(t). Although this is a correct definition, it doesntgive much insight about the word linear and there is not much hint about thedifferences between linear and nonlinear equations. In this lecture we will studythese differences.

1 Principle of Superposition

First, consider a first order linear ODE where q(t) = 0, namely

y + p(t)y = 0.

Such an equation is called homogenous. There is an unfortunate clash of nameswith the homogenous equations defined in lecture 2. The two types of equationsare completely different.

Definition 1.1 Say y1(t) and y2(t) are two functions of t. A function of the formc1y1(t) + c2y2(t) is called a linear combination of y1 and y2, where c1 and c2 areconstants.

Proposition 1.1 Suppose that y1(t) and y2(t) are both solutions of the linear ho-mogenous equation y+ p(t)y = 0. Then any linear combination of y1 and y2 is alsoa solution of the same equation.

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Proof: Since y1 and y2 are solutions, we have y1 + p(t)y1 = 0 and y

2 + p(t)y2 = 0.

But then

(c1y1 + c2y2) + p(t)(c1y1 + c2y2) = c1(y1 + p(t)y1) + c2(y

2 + p(t)y2)

= c10 + c20

= 0.

Therefore c1y1 + c2y2 is also a solution of the ODE. A linear combination is also called a superposition of the relevant functions. Thefact above is commonly referred to as the principle of superposition.

What if q(t) 6= 0? Then the principle holds in a modified form. We say thaty + p(t)y = 0 is the homogenous equation associated to y + p(t)y = q(t).

Proposition 1.2 Suppose that y1 and y2 are solutions of the first order linear ODEy + p(t)y = q(t). Then y1 y2 is a solution of the associated homogenous equationy + p(t)y = 0.

Proof: We assume that y1 + p(t)y1 = q(t) and y2 + p(t)y2 = q(t). Then

(y1 y2) + p(t)(y1 y2) = (y1 + p(t)y1) (y2 + p(t)y2)= q(t) q(t)= 0.

The following variant is also useful.

Proposition 1.3 Suppose that y1 and y2 are solutions of the first order linear ODEy + p(t)y = q(t). Then c1y1 + c2y2 is also a solution if c1 + c2 = 1.

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Proof: We assume that y1 + p(t)y1 = q(t) and y2 + p(t)y2 = q(t). Then

(c1y1 + c2y2) + p(t)(c1y1 + c2y2) = c1(y1 + p(t)y1) + c2(y

2 + p(t)y2)

= c1q(t) + c2q(t)

= (c1 + c2)q(t)

= q(t).

Example 1.1 The principle of superposition fails in general for non-linear equa-tions. For instance, consider the nonlinear ODE y = y2. It is separable and we canwrite:

dy

y2=

dt

1y

= t+ c

y(t) = 1t+ c

But a linear combination of two solutions of this form is almost never a solution.For instance, 1

tand 1

t1 are solutions. But

c11t c2 1

t 1 = (c1 + c2)t c1

t(t 1)is not a solution unless c1 = 0, c2 = 1 or c1 = 1, c2 = 0.

2 Existence and Uniqueness Theorems

A first order ODE typically has infinitely many solutions because of the constantc appearing in the solution. Our experience so far is that if an additional initialcondition y(t0) = y0 is given then the constant and therefore the solution is uniquelydetermined. One can ask whether this is always the case. The answer is given bythe existence-uniqueness theorem. Let us start from the case of linear equationssince the statement looks more intuitive.

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Theorem 2.1 Suppose that we have an initial value problem y + p(t)y = q(t) andy(t0) = y0 where the functions p(t), q(t) are continuous on an open interval (a, b)and t0 (a, b). Then this initial value problem has a unique solution y(t) which isvalid over (at least) the whole open interval (a, b).

Remark 2.1 It is possible that a = , b = + or both.

Proof: We already derived a formula for the general solution in section 2:

y =

(t)q(t)dt

(t)

where (t) = ep(t)dt. If p(t) and q(t) are both continuous on (a, b) then they are

integrable. Same holds for (t). Also, (t) 6= 0 since it is an exponential. Thereforey(t) is defined for all t (a, b). What about the constant? Suppose that F (t) is oneof the antiderivatives of (t)q(t). Then we can rewrite the solution as

y =F (t) + c

(t)

If we put y(t0) = y0 then we see that c = (t0)y0 F (t0), therefore it is uniquelydetermined by the initial condition. Hence the solution is unique. What about non-linear equations? The existence-uniqueness theorem still holds butthere are some changes. First of all, we have to replace the condition about p(t)and q(t) by something else. More seriously, the solution obtained might exist on asmaller interval then expected:

Theorem 2.2 Suppose that we have an initial value problem y = f(t, y) andy(t0) = y0 where f(t, y) and

fy

are both continuous on a rectangle (a, b) (c, d)containing (t0, y0). Then there exists a unique solution of this initial value problem,defined possibly for t belonging to a smaller subinterval of (a, b).

The proof of this theorem is beyond the scope of these lectures.

Example 2.1 We want to see that the condition fy

really plays an important rolein the theorem. For this purpose consider the ODE

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y = y1/3

together with the initial condition y(0) = 0. Then f(t, y) = y1/3 is continuouseverywhere, however f

y= 1

3y2/3 is not continous on the line y = 0. This line

contains the initial point (0, 0), so the conditions of the existence-uniqueness theoremare not satisfied. Let us indeed see that there are multiple solutions to this initialvalue problem. The equation is separable:

dy

y1/3=

dt

3

2y2/3 = t+ c

y = (

2

3(t+ c)

)3/2

The condition y(0) = 0 gives c = 0. But there are still at least two solutions ofthe initial value problem, namely y1(t) = (

23t)3/2 and y2(t) = (23t)3/2. Also, in the

derivation we assumed that y 6= 0. The function y3 = 0 is yet a third solution. (Infact, one can construct infinitely many solutions to this initial value problem. Canyou see how?)

It is also instructive to notice that if the ODE is linear, then f(t, y) = q(t) p(t)y.In this case f

y= p(t). Therefore the continuity of this partial derivative implies

the continuity of p(t). If f is also continuous, then we deduce that q(t) must alsobe continuous. Hence the hypothesis reduces to the assumptions of the existence-uniqueness theorem in the linear case.

3 Finite Time Blow-up

In the previous section it was mentioned that the solution of a non-linear initial valueproblem may not extend to the largest interval on which the relevant functions arecontinuous. This phenomenon is called finite time blow-up, and is especiallydisturbing in any theoretical or practical consideration of differential equations. Letus see that this can indeed happen, through an example.

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Example 3.1 Consider the non-linear ODE y = y2 together with the initial con-dition y(0) = y0 > 0. We obtained the general solution to this ODE to be

y(t) = 1t+ c

.

Since y(0) = y0, we have y0 = 1c , so c = 1y0 . Then

y(t) = 1t 1

y0

.

Now, the conditions of the existence-uniqueness theorem are satisfied for this ODE.Indeed, f(t, y) = y2 and f

y= 2y are continuous on all of R2. However, the solution

is defined only on the interval (, 1y0

) and not on all of (,) (We say thesolution blows up at 1

y0. Notice that the place of the blow-up depends on the initial

condition, hence it can not be quickly predicted from the ODE only. One actuallyhas to solve the ODE in order to understand what happens.

4 Other Differences

There are some other differences between linear and nonlinear equations

The general solution of a linear first order ODE can be written explicitly, sincewe have a formula for the solution that involves an integrating factor. This isnot the case for nonlinear ODEs. Even if one can find all solutions, not allsolutions need to correspond to special values of a constant in some generalformula.

In solving nonlinear ODEs we sometimes need to leave the solution in an im-plicit form. This is never necessary for a first order linear ODE.

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