0387986995Basic TheoryC

As mentioned in §111-2, under this assumption, every solution of differential equation (111.3.1) exists on the interval To = {t : a < t < b} if (to,y-(to)) E ft for some to E I. Define the sets 1(A) and SS(A) for a subset A of ) by Definition III-2-1.

The main concern of this section is to prove the existence of solution curves on the boundary of R(A). We start with the following basic lemma.

Lemma 111-3-1. Suppose that (1) the set A consists of one point (c1, t) (i. e.,

A = {(c,, f)}), (ii) a < cl < co < c2 < b, and (iii) ij is on the boundary of Sc2(A). Let B = Then, SI(B) contains at least a boundary point of S0(A).

Proof

A contradiction will be derived from the assumption that SI(B) does not contain any boundary points of SS (A). Note that SI(B) n 4 (A) 54 0. Set S, = 4 (B) n S,,. (A), S2 = {y : y' E Se. (B) and Ii 0 SS(A)}. Then, S., (B) = 81 U S2 and

S, n S2 = 0. It is known that S1 is a nonempty compact set. Also, $2 is closed and bounded, since S,,, (5) is compact and does not contain any boundary points of

SS (A). If S2 is not empty, the set Sc.(B) contains a point in Sc,, (A) and another point which does not belong to S,, (A). Then, S,.(8) contains a boundary point of

Sc. (A), since S,, (B) is connected. This is a contradiction. Therefore, if it is proved that S2 is not empty, the proof of Lemma 111-3-1 will be completed (cf. Figure 11).

To prove that S2 is not empty, observe first that Sam({(a2,()}) nS.,,(A) = 0 if

0 SS,(A) (cf. Figure 12).

FIGURE 11. FIGURE 12.

Let rj be on the boundary of $, (A). Then, there exists a sequence of points

{Sk V Sc, (A) : k = 1, 2, ... ) and a sequence (¢k : k = 1, 2.... ) of solutions of

lim bk, (t) = Q(t) exists uniformly on Zo. The limit function is a solution of

++oo

(111.3.1) such that ¢(c2) = tj. Hence, d(co) E Sc,(B). Since S, (B) does not contain any boundary points of S,(A), we must have ¢'(co) if $,,(A). This implies that

S2 is not empty. 0

The following theorem is the main result in this section which is due to M.

Hukuhara (Hukl( (see also (Huk2J and [HN31).

Theorem 111-3-2. Suppose that

(1)a<cl <c2 <b,

(2)is on the boundary of

Then, there exists a solution of differential equation (M.3.1) such that

(i)(Cl) = E, (C2) = 6,

(ii)is on the boundary of R({(ci,l;))) for cl < t < c2.

Proof.

(Cf. Lemma 111-3-1 and Figure 13.)

FIGURE 13.

Choose a sequence {ok : k = 1, 2, ... } of subdivisions of the interval cl < t < c2 such that

f Ok : C1 = Tk,O < Tk.1 < ... < Tk,2k-I < Tk,2" = C2,

The family {4k k = 1, 2,... } is bounded and equicontinuous on 10. Hence,

(Qo) (Tk,t,. k(Tk,t)) is on the boundary of R({(c1i,)}), where f 2k - 1 and k = 1,2,....

Since the set {Tk,t : e = 1,2,... 2k - 1; k = 1,2.... } is dense on the interval Cl < t < c2, the curve (t,$(t)) is on the boundary of for cl < t <

C2- 0

Remark 111-3-3. In general, the conclusion of Theorem 111-3-2 does not hold on an interval larger than cl < t < c2.

111-4. Maximal and minimal solutions

Consider a scalar differential equation

where t and y are real variables, and assume that f is real-valued, bounded, and continuous on a region H = {(t,y) : a < t < b, -oo < y < +oo}. Choose c in the interval I = (t : a < t < b}. Then, Sr ({ (c, r,) }) is compact and connected for every r E I and every real number r) (cf. Theorem III-2-4). This implies that there exist two numbers 41(r) and ¢2(r) such that ST({(c,17)}) = { y E R : bl(r) < y < &2(r)}. Hence, R({(c,n)}) = {(t,y) E R2 : ¢1(t) < y <

02(t), a < t < b} (cf. Figure 14). The two boundary curves (t,4i(t)) and (t,02(t)) of R({(c,rl)}) are solution curves of differential equation (111.4.1) (cf. Theorem

111-3-2). Every solution ¢(t) of (111.4.1) such that 0(c) = n satisfies the inequalities

01(t) < ¢(t) < ¢2(t) on Z. The solution 02(t) (respectively 01(t)) is called the maximal (respectively minimal) solution of the initial-value problem

dy

on the interval Z. In this section, we explain the basic properties of the maximal and minimal solutions. Before we define the maximal and minimal solutions more precisely, let us make some observations.

FIGURE 14.

Observation 111-4-1. Assume that f (t, y) is real-valued and continuous on a domain Din the (t,y)-plane. Set lo = {t : a < t < b}. Let 01(t) and ¢2(t) be two solutions of differential equation (111.4.1) such that (t, ¢1(t)) E D and (t, 02(t)) E D for t E Zo. Note that we do not assume boundedness of f on D. Set 0(t) = max {01(t), 02(t)} for t E Zo. Then, Q(t) is also a solution of (111.4.1) on the

interval Zo.

If ¢1(to) > -02(to), then there exists a positive number 6 such that 4(t) = 01(t) for

It - tol < J. Therefore, (111.4.3) holds. Similarly, (111.4.3) holds if 02(to) > &(to).

Hence, we consider only the case when '(to) _ .01(to) = 02(to). In this case,

j = 2, depending on t. Hence, by the Mean Value Theorem on O,(t), there exists

0(t) - ¢(to) = f (r, 0,(r)). Also, 03 (r) _ t - to

¢j(to) + f (a,-Oj (a)) (7 - to) for some a E To such that a to as r - to. Since f is bounded on the two curves (t, 01(t)) and (t, 02(t)), (111.4.3) follows immediately. 0

Observation 111-4-2. Let f (t, y), D, and To be the same as in Observation

111-4-1. Consider a set F of solutions of differential equation (111.4.1) such that

(t, 0(t)) E D on To for every ¢ E F. Assuming that there exists a real number K such that 0(t) < K on lo for every 0 E F, set 00 (t) = sup{Q(t) : ¢ E F} for t E To.

Assume also that (t, 00(t)) e D on To. Then, 4'o(t) is a solution of differential equation (111.4.1) on To.

Proof.

As in Observation 111-4-1, we prove that

for each fixed to E To. Choose three positive numbers po, p, and S so that

(i) A = {(t, y) : t E To, Iy - 4o(t)I < po} C D,

(ii) we have (t, 4(t)) E A on the interval t - to I < 6, if d(t) is a solution of

I

(III.4.1) such that 0 < 0o(r) - 4i(r) < p for some r in the interval It - tol < b.

There exists a positive number Af such that I f (t, y) I < M on A.

Let us fix a point r on the interval It - tol < b. First, we prove the existence of a solution ty(t; r) of (III.4.1) such that

(111.4.5) P(r; r) = 0o (r) and yJ(t; r) < 4o(t) for it - tol < b.

To do this, select a sequence {4'k : k = 1, 2,... } from the family F so that

lira thk(r) = 40(r). We may assume that (t, hk(t)) E A on It - tol < S for

Y+ o0

k (cf. (ii) above). Then, the sequence {4'k : k = 1, 2, ... )is bounded and

equicontinuous on It - tot < J. Hence, we may assume that lim 4'k(t) = 0(t; r)

k-»+oo

exists uniformly on the interval It - tol < 6. It is easy to show that V'(t; r) is a solution of (111.4.1) and that (111.4.5) is satisfied.

Set ty(t) = max{t/,(t; r), tJ'(t; to)} for Jt - tol < b. Then, v is a solution of (111.4.1) such that (1) V)(r) = oo(r) and ty(to) = 00(to), (2) iP(t) < 4'o(t) for It - tol < b,

and (3) (t, v/ (t)) E A for It - tol < 6 (cf. Figure I5).

Y= #o(')

Y s V(1)

for some a such that Ja - tol < 6 and that or to as r -" to. Since 10(a) - tt'(td)J <

Mba - t01, (111.4.4) can be derived immediately. 0

Let us define the maximal (respectively minimal) solution of an initial-value problem

where f is real-valued and continuous on a domain V in the (t, y)-plane and the initial point (T, 1:) is fixed in D.

Definition 111-4-3. A solution tp(t) of problem (111.4.6-() is called the maximal (respectively minimal) solution of problem (111.4.6-() on an interval I = it : T < t < r' } if

(a) tp(t) is defined on I and (t, {1(t)) E D on Z,

(b) if 4(t) is a solution of problem (II14.6-{) on a subinterval r < t r" of Z,

The following two theorems are stated in terms of maximal solutions. Similar results can be stated also in terms of minimal solutions. Such details are left to the reader as an exercise. In the first of the two theorems, we consider another initial-value problem

together with problem (111-4.6-{)-

Theorem III-4-4. Suppose that the maximal solution tpt of Problem (IIL4.6-{) exists on an interval Z = {t : r < t < r'}. Then, there exists a positive number So such that the maximal solution 0,1 of problem (III 4.6-r1) exists on I for t; < rl < t + b0. Fjrthermore, tpf(t) < ip,,(t) on I for £ < 17< + 60 and limipR(t) = tb,(t)

uniformly on Z.

Proof

Set 0(p) = {(t, y) : t E Z, yt(t) < y < ot(t) + p}. For a sufficiently small positive number p, we have A(p) C 1), and, hence, if (t, y)I is bounded on A(p) for a sufficiently small positive number p.

First, we prove that for a given p > 0, there exists a positive number 6 such that, if f < q < { + b, every solution 0(t) of problem (111.4.6-1?) defined on a subinterval r < t < r" of I satisfies

(III.4.7) 0(t) < ,'(t)±p for r < t <7".

Otherwise, there exist two sequences {qk : k = 1 , 2, ... ) and {rk : k = 1, 2, ... } of

real numbers such that (1) qk > {, lim qk = t;, and r < rk < r', (2) 0&-(-r) = nk k-+oo

and Ok(Tk) = v((rk) + p, and (3) Ok(t) < vf(t) + p for r < t < rk. Furthermore, there exists a real number r(p) such that rk > r(p) > r (cf. Figure 16).

Y=W4+P

It is easy to show that the sequence {0k W : k = 1, 2, ... } is bounded and equicon- tinuous on the interval Z. Hence, we can assume without any loss of generality that

(1)tim Tk = rp exists, k-.+oo

(2)tim 10k(t) = 0(t) exists uniformly on Z.

Note that

(3)d(r) _ and O(ro) _ 'af(ro) + p,

(4)ro?r(p)>r.

It is also easy to show that 0(t) is a solution of (III.4.6-t) on the subinterval r < t < ro of Z. Since 104 is the maximal solution of (III.4.6-1:) on 7, we must have 0(t) < af(t) on the interval r < t < To. This is a contradiction, since

0(ro) = i (ro) + p. Thus, (111.4.7) holds.

Now, for a given positive number p, there exists another positive number 6(p) such that (111.4.7) holds for any solution ¢(t) of problem (III.4.6-17) if < ' < t; + 6(p). Hence, using max(4(t), 0((t)), we can extend 6(t) on the interval Z in such a way that

Let F be the set of all solutions 6(t) of (III.4.6-71) which satisfy condition (111.4.8).

Set t[i,r(t) = sup{¢(t) : 0 E F j for t E I. Then, ty,, is the maximal solution of

(III.4.6-r7) on I. Furthermore, 104(t) < 1,1(t) < ot(t) + p if t: < rt < t + 6(p).

Letting p - 0, we complete the proof of the theorem.

Assuming again that f is continuous on a domain D in the (t, y)-plane, consider initial-value problem (III.4.6-{) together with another initial-value problem

where (t, 1;) E V and f is a positive number. We prove the following theorem.

Theorem III-4-5. If the maximal solution tt'(t) of problem (111.4.6-,) on the interval I = {t : r < t < r'} exists, then, for any positive number p, there exists another positive number e(p) such that for 0 < E < E(p), every solution d(t) of (11!.4.9-E) exists on I and tp(t) < d(t) < y',(t) + p on the interval I. In particu- lar, for every sufficiently small positive number f, then exists the maximal solution tf=, (t) of problem (111.4.9-E) on I and lim tL, = tV uniformly on I.

Proof

The maximal solution tI'(t) satisfies the condition (t,>l,(t)) E D on I (cf. Definition 111-4-3). Define a function 9(t, y) by

Then, g is continuous and bounded on the domain {t E I, -oo < y < +oo}. Hence, every solution ¢(t, e) of the initial-value problem

= 9(t, y) + E, y(r) _

exists on the interval I. Note that

We prove that for a given positive number p, there exists another positive number e(p) such that if 0 < e <_ E(p), we have 0(t, e) < t1(t)+p on I. Otherwise, there exist a real number r(p) and two sequences {Ek : k = 1, 2.... ) and {rk : k = 1, 2, ... } of real numbers such that

(2) T' > rk ? T(p)> T and lim Tk = To exists, k+00

(3)0(7k, ek) = IP(rk) + p,

(4)0(t, (k) < V,(t) + p for r < t < Tk,

(5)lim 0(t, ek) = 0(t) exists uniformly on 1.

0(ro) = Ty(ro) + p, this is a contradiction. Thus, it was proved that 0(t) S 0(t, e) <

,y(t) + p on I for 0 < e < e(p). Therefore, Theorem 111-4-5 follows immediately. 0

In the proof of Theorem 111-4-5, (111.4. 11) was derived from (111.4.10) (cf. Figure

17). In a similar manner, we can prove the following result.

Lemma 111-4-6. Assume that f (t, y) is continuous on a region 12 = {(t, y) w_(t) < y < w+(t), t E TO}, where l0 = {t : a < t < b} and

(i) w+ and w_ are real-valued, continuous, and differentiable on T0, (ii) w- (t) < w+ (t) on lo.

Assume also that

dw (t) < Pt"'-(O) on TO

dt

and w_(a) w+(a). Then, every solution 0(t) of the initial-value problem

exists on To and w_ (t) < 0(t) < w+(t) on 10 (i.e. (t, y5(t)) E Q). In particular, the maximal solution 01(t) and the minimal solution 02(t) of problem (111.4.12) on lo exist and w_ (t) < 02 (t) < ¢, (t) < w+ (t) on Zo

Proof of this lemma is left to the reader as an exercise (cf. Figure 18).

Y = vKW)

v= #(z)

Y=

Using Lemma 111-4-6, we prove the following theorem due to O. Perron (Per2j.

Theorem 111-4-7. Assume that f (t, y) is continuous on a region

whereto = (t:a<t <b) and

(i) w, and w_ are teal-valued, continuous, and differentiable on 4,

(ii) j- (t) 5 w+ (t) on 4.