0387986995Basic TheoryC

38	II. DEPENDENCE ON DATA
(iii) the partial derivative	t8tn	exists and is continuous in (t, r, Ft,

n) on the domain D = {(t, r, 1, ... , {n) : t E To, t1 < T < t2,

-')(T)I < p (1 = 1,2,...,n)};

(iv) u = Lo (t, r,			is the unique solution of the initial-value problem
dnU	rn	f	`	-1
				tt
dtn	- `J= axf		,u(n-2)(7)	(t,7,6,... ,Sn)) dtj_1
U(T) = -6, u'(T)				u(n_1) =	,Ln).

11-7. Assume that f (t, X1, x2) is a real-valued, continuous, and continuously differentiable function of (t, xl, x2) on an open set A in the (t, x1, x2}space. Assume also that O0(t) is a real-valued solution of the second-order differential equation x" = f (t, x, x') and (t, Oo(t), 00'(t)) E A on the interval Zo = (t : 0 < t < 1).

Set ¢0(0) = a and 0o(0) = b. Denote by t(t,/3) the unique solution of the initial value-problem x" = f (t, x, x'), x(0) = a, x'(0) = ,B, where lb - 81 is sufficiently

small. Show that (1,b) > 0 if of	0 for t C-
013

11-8. Let g(t) be a real-valued and continuous function oft on the interval 0 < t <

1.Also, let A be a real parameter.

(1)Show that, if Q(t, A) is a real-valued solution of the boundary-value problem

Ez + (g(t) + A)u = 0, u(0) = 0, u(1) = 0,

and if 8 &2 (t, A) is continuous on the region A = {(t, A) : 0 < t < 1, a <

A < b}, then d(t, A) is identically equal to zero on A, where a and b are real numbers.

(2) Does the same conclusion hold if 0(t,,\) is merely continuous on A?

11-9. Let a(x, y) and b(x, y) be two continuously differentiable functions of two variables (x, y) in a domain Do = {(x, y) : IxI < a:, IyI < Q} and let F(x, y, z) be a continuously differentiable function of three variables (x, y, z) in a domain Do = {(x, y, z) : Ixl < a:, IyI < 3, Izl < 7}. Also, let x = f (t, i7), y = 9(t, ij), and z = h(t,17) be the unique solution of the initial-value problem

= a(x,y),	L = b(x,y),		= F(x,y,z),
dt
x(0) = 0,	y(0) = n,	z(0) = c(+1).

where c(q) is a differentiable function of 11 in the domain 14 = (q: In1 < p). Assume that (t, rl) = (d(x, y), O(x, y)) is the inverse of the relation (x, y) = (f (t,17), g(t, rt)), where we assume that O(z, y) and i,1(x, y) are continuously differentiable with re- spect to (x, y) in a domain Al = {(x, y) : IxI < r, IyI < p}. Set H(x,y) = h(p(x, y), tp(x, y)). Show that the function H(x, y) satisfies the partial differential equation

a(x, y) + b(x, y) M = F(x, y, H)

and the initial-condition H(0,8 y) = c(y).

EXERCISES II

39

Hint. Differentiate H(x,y) with respect to (x,y).

11-10. Let F(x, y, z, p, q) be a twice continuously differentiable function of

(x,y,u,p,q) for (x,y,u,p,q) E 1R5. Also, let

x = x(t, s),

y = y(t, s),

z = z(t, s),

p = P(t, s),

q = q(t, s)

be the solution of the following system:

dx -

dy _ dt

dz

dt =

dp dt dq dt

OF

(x+ y, z, p, q),

OF

flq(T, y, z, p, q),

OF	OF
pij (x,y,z,P,q) + gaq(x,y,z,P,q),
OF	OF
8x (T, y, z, P, q) - P 8z (x, y, z, p, q),
OF	OF
8y	(x, y, z, P, q) - q az (x, y, z. P, q)

satisfying the initial condition

x(O,s) = xo(s),	y(O,s) = yo(s),	z(O,s) = zo(s),
p(O,s) = Po(s),	q(O,s) = qo(s),

where x0 (s), yo(s), zo(s), po(s), and qo(s) are differentiable functions of s on R such that

	F(xo(s),yo(s),zo(s),Po(s),go(s))=0,	dzo(s)=Po(s)-		(s)+go(s)dyo(s)
		ds	ds	ds

on R. Show that

F(x(t, s), y(t, s), z(t, s), p(t, s), q(t, s)) = 0,

at (t, s) = At, s) 5 (t, s) + q(t, s) (t, s),

flz(t,s)	= P(t,s)L(t,s)	+ q(t,s)ay(t,8)
fls	as	as

as long as the solution (x, y, z, p, q) exists.

Comment. This is a traditional way of solving the partial differential equation

F(x, y, z, p, q) = 0, where p = 8 and q =

For more details, see (Har2, pp.

131-143].

40	H. DEPENDENCE ON DATA

II-11. Let H(t, x, y, p, q) be a twice continuously differentiable function of

(t, x, y, p, q) in RI. Show that we can solve the partial differential equation

Oz	Z 8z\	=0
	+ H t, x, y,-,

by using the system of ordinary differential equations

dx		8H	dy	8H
dt =		8p'	t =	q ,
dp		8H	(LL dq	_ _ 8H
dt		8x '	dt		W,
dz			8H	8H	du	OH
	= u + p-p + q8q,				dt	= --6T.

CHAPTER III

NONUNIQUENESS

We consider, in this chapter, an initial-value problem

(P)	dy	= f(t,y1,
	dt

without assuming the uniqueness of solutions. Some examples of nonuniqueness are given in §III-1. Topological properties of a set covered by solution curves of problem

(P) are explained in §§III-2 and 111-3. The main result is the Kneser theorem

(Theorem III-2-4, cf. [Kn] ). In §1II-4, we explain maximal and minimal solutions and their continuity with respect to data. In §§III-5 and 111-6, using differential inequalities, we derive a comparison theorem to estimate solutions of (P) and also some sufficient conditions for the uniqueness of solutions of (P). An application of the Kneser theorem to a second-order nonlinear boundary-value problem will be given in Chapter X (cf. §X-1).

111-1. Examples

In this section, four examples are given to illustrate the nonuniqueness of solutions of initial-value problems. As already known, problem (P) has the unique solution if f'(t, y) satisfies a Lipschitz condition (cf. Theorem I-1-4). Therefore, in

order to create nonuniqueness, f (t, y-) must be chosen so that the Lipschitz condition is not satisfied.

Example III-1-1. The initial-value problem

	(III.1.1)	dy	= yt/3			y(to) = 0
		dt
	has at least three solutions
	(S.1.1)	y(t) = 0			(-oo < t < +oc),
					3/2
	(S.1.2)	y(t) _	[3	t - to)		,	t > to,
					I]
		0,					t < to,
	and
			[3(t2		- to),	3/2
	(S.1.3)	y(t)					t > to,
		0,					t < to

41

42	111. NONUNIQUENESS

(cf. Figure 1). Actually, the region bounded by two solution curves (S.1.2) and

(S.1.3) is covered by solution curves of problem (I11.1.1). Note that, in this case, solution (S.1.1) is the unique solution of problem (P) for t < to. Solutions are not unique only fort ? to.

Example 111-1-2. Consider a curve defined by

(111.1.2)

y = sin t

(-oo < t < +oo)

and translate (111.1.2) along a straight line of slope 1. In other words, consider a family of curves

(111.1.3)

y = sin(t - c) + c,

where c is a real parameter. By eliminating c from the relations

dt = c os(t - c), y - t = sin(t - c) - (t - c),

we can derive the differential equation for family (111.1.3). In fact, since sin u - u is strictly decreasing, the relation v = sinu - u can be solved with respect to u to obtain u = G(v) - v, where G(v) is continuous and periodic of period 27r in v, G(2n7r) = 0 for every integer n, and G(v) is differentiable except at v = 2n7r for every integer n. The differential equation for family (111.1.3) is given by

dy

(II1.1.4) dt = 1OS[G(y-t)-(y-t)1.

Since G(2n7r) = 0 and cos(-2n7r) = 1 for every integer n, differential equation

(111. 1.4) has singular solutions y = t + 2mr, where n is an arbitrary integer. These lines are envelopes of family (111.1.3) (cf. Figure 2).

	FIGURE 1.	FIGURE 2.
Example 111-1-3. The initial-value problem
(111.1.5)	dt =	y(to) = 0
has at least two solutions
(S.3.1)	y(t) = 0	(-oo < t < +oo),

	1. EXAMPLES	43
and
	t,0)2,	t > to ,
	4 (t -
(S.3.2)	y(t)	t < to
	- 4(t - to)2,

(cf. Figure 3). The region bounded by two solution curves (S.3.1) and (S.3.2) is covered by solution curves of problem (111.1.5).

FIGURE 3.

Consider the following two perturbations of problem

dy

=

+ E,

y(to) = 0

dy _	y2		y(to) = 0,
	y2 + C2	lyl

where a is a real positive parameter. Each of these two differential equations satisfies the Lipschitz condition. In particular, the unique solution of problem (111. 1.6) is given by

	J4(t-to+2VI-E)2 - E,	t > to
(S.3.3)	y(t) =
	-4(t-to-2f)2 + E,	t < to

(cf. Figure 4). On the other hand, (S.3.1) is the unique solution of problem (111.1.7). Figure 5 shows shapes of solution curves of differential equation (111.1.7). Note that nontrivial solution of (111.1.7) is an increasing function of t, but it does not reach y = 0 due to the uniqueness.

44	III. NONUNIQUENESS

y

FIGURE 4.

FIGURE 5.

Generally speaking, starting from a differential equation which does not satisfy any uniqueness condition, we can create two drastically different families of curves by utilizing two different smooth perturbations. In other words, a differential equation without uniqueness condition can be regarded as a branch point in the space of differential equations (cf. [KS]).

Example 111-1-4. The general solution of the differential equation

	2
(1II.1.8)	d) +y2=1
is given by
(111.1.9)	y = sin(t + c),

where c is a real arbitrary constant. Also, y = 1 and y = -1 are two singular solutions. Two solution curves (111.1.9) with two different values of c intersect each other. Hence, the uniqueness of solutions is violated (cf. Figure 6).

This phenomenon may be explained by observing that (111. 1.8) actually consists of two differential equations:

(III.1.10)	dy = 1 - yz	and	dy=-	l-y2.
	dt		dt

Each of these two differential equations satisfies the Lipschitz condition for lyI < 1. Figures 6-A and 6-B show solution curves of these two differential equations, respectively.

y=I

y=-I

FIGURE 6.

FIGURE 6-A.

FIGURE 6-B.

Observe that each of these two pictures gives only a partial information of the

complete picture (Figure 6).

2. THE KNESER THEOREM

45

We can regard differential equation (111.1.8) as a differential equation

(111.1.11)	dy =w
	dt
on the circle
(111.1.12)	w2 + y2 = 1.

If circle (111.1.12) is parameterized as y = sinu, w = cosu, differential equation

(III.1.11) becomes

(III.1.13) d = 1 or cos u = 0 .

Solution curves u = t of (111.1.13) can be regarded as a curve on the cylinder

{(t, y, w) : y =sin u, w = cos u, -oc < u < +oo (mod 2w), -oo < t < +oo}

(cf. Figure 7). Figure 6 is the projection of this curve onto (t,y)-plane.

FIGURE 7.

In a case such as this example, a differential equation on a manifold would give a better explanation. To study a differential equation on a manifold, we generally use a covering of the manifold by open sets. We first study the differential equation on each open set (locally). Putting those local informations together, we obtain a global result. Each of Figures 6-A and 6-B is a local picture. If these two pictures are put together, the complete picture (Figure 6) is obtained.

111-2. The Kneser theorem

We consider a differential equation

(III.2.1)

dt = f(t,y-)

under the assumption that the R"-valued function f is continuous and bounded on a region

(111.2.2)

S2 = {(t, y-) : a:5 t < b , Iy1 < +oo }.

Under this assumption, every solution of differential equation (III.2.1) exists on the interval Zfl = {t : a < t < b} if (to, y"(to)) E f2 for some to E Zo (cf. Theorem 1-3-2 and Corollary I-3-4). The main concern in this section is to investigate topological properties of a set which is covered by solution curves of differential equation

(111.2.1).

Let d be the distance between two compact sets F1 and F2.

2. THE KNESER THEOREM

47

Theorem 111-2-4 ((Kn]). If A is compact and connected, then SS(A) is also compact and connected for every c E To.

Proof.

The compactness of SS(A) was already explained. So, we prove the connectedness only.

Case 1. Suppose that A consists of a point (r, t), where we assume without any loss of generality that r < c. A contradiction will be derived from the assumption that there exist two nonempty compact sets F1 and F2 such that

(111.2.3)

SS(A)=F1uF2,

F1nF2=0.

If t'1 E F1 and 2 E F2, there exist two solutions 1 and 4'2 of (111.2. 1) such that

01(r) _ , 01(c) _ 1, and 42(r) _ ., 02(c) _ 6, (cf. Figure 9).

	Set	1(r+11)	for 05µ5c-r,
	h(µ) _
		{ &T + JJUJ)	f o r	- , r )

Let { fk(t, y-) : k = 1, 2,... } be a sequence of R"-valued functions such that

(a)the functions fk (k = 1, 2, ...) are continuously differentiable on 0,

(b)I fk(t, y1I < M for (t, y) E f2, where M is a positive number independent of (t, yl and k,

(c)ra+A = f'uniformly on each compact set in Q

k El oo
(cf. Lemma I-2-4). For each k, let &(t,µ) be the unique solution of the initial-
value problem	= fk(t, y-), y(-r+ Iµ4) = h(µ). The solution 15k(t, µ) is continuous
for t E Zo and iµl < c - r. It is easy to show that the family		k =
1, 2,... ; jµj 5 c - r} is bounded and equicontinuous on the interval Io.

Note that the functions Z(rk(c, µ) (k = 1, 2, ...) are continuous in µ for 1µI <

c - r and that r'k(c,c - r) = ¢'1(c) E F1 and t k(c,-(c - r)) = &c) E F2.

Since &(c,µ) is continuous in p, there exists, for each k, a real number µk such that IµkI < c-r and

distance(!& (c, µk), F1) = d.	Since the family	ilk) : k = 1, 2, ... } is bounded
and equicontinuous on the interval Io, there exists a subsequence
j = 1,2.... } such that (i)	lim k, _ +oo, (ii)	lim µk, = po exists, and (iii)
	i--+00	)-.+oo

urn 1Pk, (t, µk1) = 0(t) exists uniformly on To. It is easy to show that s-+oo

fi(t) = K(AO) + ft f (s, ¢(s))ds,

+ {µol

lµo1 < c - r, distance(¢(c), F1) = 2

Hence, (c) E $.(A) but * (c) 4 F1 U F2. This is a contradiction (cf. Figure 10).