0387986995Basic TheoryC

Assumption 1. Let t and u be two real variables and let g(t,u) be a real-valued and continuous function of (t, u) on a domain D in the (t, u)-plane. Also let .y(t) be the maximal solution of the initial-value problem

du

on an interval lo = It : a < t < b}, where (a, uo) E D and (t, i,b(t)) E D on To.

We first prove the basic theorem given below.

Theorem 111-5-1. Assume that

(1)g(t, u) > 0 on D and uo > 0,

(2)an R"-valued function ¢'(t) is continuously differentiable on a subinterval I =

it: a< t <,r) of 4,

Proof.

Let a be a positive number and let 0(t, e) be any solution of the initial-value problem

If e > 0 is sufficiently small, i/i(t, e) exists on Zo and lim 0(t, e) = P(t) uniformly on

To (cf. Theorem III-4-5). Let us make the following observations:

where t > s. Suppose that l (s)I = r'(s, e) for some s E Z. Then, there exists a positive number b(e) such that

(cf. Figure 19).

FIGURE 19.

Note that

we can apply Theorem 111-5-1 to fi(t) with g(u) = C + Mu. Note that the initial-

value problem = C + Mu, u(a) = uo has the unique solution

We still assume that Assumption 1 is satisfied by the function g(t, u) and O(t) and that uo > 0 and g(t, u) > 0 on D. Consider a differential equation

dy

(111.5.1) dt = f (t, yI

under the following assumption:

(a) the function f (t, yj is continuous on a domain f2 in the (t, y7)-space,

(0) (t, Iyj) E D if (t, y-) E S2,

Proof

Suppose that problem (III.6.2) has a nontrivial solution j(t) on Z. This means that (a) ,-a< 6. for some a E Z. Choose a positive number 6 so that f <

Let us make the following two observations.

Observation 1. Note that the set A(i3) = {(t, u) : a < t < b, 0 < u < B} is a subset ofd and that u = 0 is a solution of the differential equation

on the interval a < t < b. Using Theorem 111-2-7, we can construct a nontrivial solution uo(t) of (111.6.3) on the interval a < t < b so that (t, uo(t)) E A(f3) on a<t<b(cf. Figure 20).

Observation 2. The nontrivial solution uo(t) of (111.6.3) which was constructed in Observation 1 can be continued on the interval a < t < a so that

To show this, we first remark that for a given positive number e, any solution t(i(t, c) of the initial-value problem

satisfies the condition tfi(t, e) < on any subinterval a < t < a of I if it exists on this subinterval (cf. Figure 21).

Then, any solution ty(t, e) of problem (III.6.5) can be continued on the interval

a < t < a as a solution of the differential equation = G(t, u) + e. Since the set

{t(i(-, e) : 0 < e < 1) is bounded and equicontinuous on every closed subinterval of a < t < or, we can select a sequence {Ek : k = 1,2.... } of positive numbers such

of a < t < a. It is easy to show that j)(t) is a solution of the initial-value problem

du = g(t, u), u(a) = uo(a) which satisfies condition (111.6.4).

Thus, we constructed a nontrivial solution u0(t) of differential equation (111.6.3)

such that (t,uo(t)) E 0 on I and 0 < uo(t) < ]5(t)l on a < t < a. Since

Urn ±L = 0, this contradicts condition (5) of Assumption 1. 0

t-a r(t)

From Theorem III-6-1, we derive the following result concerning uniqueness of solutions.

Theorem 111-6-2. Suppose that Assumption 1 is satisfied and that

(i)an 1R"-valued function f (t, y-) is continuous on a domain D in the (t, yl -space,

(ii)f satisfies the condition J f (t, 91) - f (f, 92)1 < g(t, Iyi - 92{) whenever t E 1,

Iy, - 921:5 w(t), (t, yi) E V, and (t, y2) E D.

Let Q(t) be a solution of the differential equation

dt = f (t, y) on I = {t : a < t < b}

such that the set Do = {(t, y-) 4(t)1 < w(t), t E 1} is contained in D. Then, the problem

has only the solution (t) itself.

Proof.

Set l(t) = z(t) + .(t). Then, problem (111-6.6) becomes

Let us apply Theorem III-6-2 to some initial-value problems.

Example 1 (The Osgood condition (cf. [Os])). Consider a real-valued function g(t, u) = h(t)p(u) in the case when

(1) h(t) is continuous and h(t) > 0 on an interval 1 = {t : a < t < b},

(2)p(u) is continuous on an interval 0 ,.5 u < K, where K is a positive number,

(3)p(0) = 0 and p(u) > O for 0 < u < K,

(4)r h(t)dt < +oc,

a+

(}(5) jo+ p(u) = +oo.

Assume that

(1) an R"-valued function f (t, y) is continuous on a domain V in the (t, yam)-space,

whenever t E Z, I171 - g2 l < K, (t, y"1) E D, and (t, y2) E D.

on the interval Zo = {t : a < t < b}. Then, p1(t) = ¢'2(t) on Zo.

Proof.

It is sufficient to show that Assumption 1 is satisfied by three functions r(t) = 1,

condition is evidently satisfied, since ¢2(a) = 41(a) and r(t) = 1. Conditions (1),

(2), (3), and (4) of Assumption 1 are also evidently satisfied. Therefore, it suffices to prove that condition (5) is also satisfied.

Let u(t) be a real-valued function such that

(A)0<u(t) <Kon asubinterval r<t<aof I={t:a<t<b},

(B)lira+ u(t) = 0,

t

(C) dutt) = h(t)p(u(t)) on r < t < o.

Then,

This contradicts condition (5).

Example 2 (The Lipschitz condition). If we choose h(t) = L and p(u) = u, where

L is a positive constant, then the Osgood condition becomes the Lipschitz condition

(cf. Assumption 2 of §I-1).

Example 3 (The Nagumo condition). Define r(t), w(t). and g(t, u) by

where A is a non-negative constant and K is a positive constant. Assume also that

is given by u(t) = c(t - a)', where c is an arbitrary constant. This implies that

Assumption 1 is satisfied by r(t), w(t), and g(t,u). In particular, choosing A = 1, we derive the following result due to M. Nagumo (Nal and Na2J. Assume that

dY

Suppose that 1(t) and y+,2(t) are two solutions of the initial-value problem dt = f (t, y), g(a) = # such that

on the interval I= {t: a<t<b}. Then, $1(t) = &t) on 1.

Remark 111-6-3. The sufficient conditions of Osgood's [Os] and M. Nagumo's

[Nal and Na21 for uniqueness are also explained in [CL. pp. 48-601 and [Hart, pp.

31-351.

Problem 4. Show that the initial-value problem

between y1 and y2. Therefore, we can use the Lipschitz condition for r - 0, whereas we can use the Nagumo condition at r = 0.

Remark 111-6-4. In order to apply Theorem 111-6-2 to the initial-value problems in Examples 1 and 3, it was assumed that f (t, y-) is continuous at the initial point

(a,'qj. However, the reader must notice that problem (111.6.6) is more general than an initial-value problem. Actually, we can apply Theorem 111-6-2 even when f (t, yr) is not continuous at t = a. For example, the Nagumo condition (ii) of

Example 3 is satisfied by the function sin 1 t I at a = 0. Therefore, the problem