0387986995Basic TheoryC

8	I. FUNDAMENTAL THEOREMS OF ODES

is considered on the set j = {t E R : t A 0), the general solution is y(t) = cl +In Jtj for t > 0 and y(t) = c2 + In JtJ for t < 0, where ct and c2 are independent arbitrary

constants. If the Heaviside function

H(t) = {1	if
-I	if
is used, this general solution is given by

t > 0, t < 0

y(t) =	c1 - c2 H(t) + c'+ c2 + In JtI.
	2	2

Equation (E) may be looked at in terms of the generalized functions (or the distributions of L. Schwartz [Sc]). The transformation y = u + In Jti, changes equation

(E) to	t- = 0.
(E')

As a differential equation on the generalized functions, (E') is reduced to

(E")	j = cb(t),

where c is an arbitrary constant and b(t) is the Dirac delta function. Integrating

(E"), we obtain u(t) = cH(t) + c, where c is another arbitrary constant. Therefore, the general solution of (E) is y(t) = cH(t) + c + in Jtt. For more information on differential equations on the generalized functions, see [Ko] for examples.

The following example might be a little strange. Let us construct solutions of the differential equation

(E) L + 2y(cost + cos(2t)) = 0

as a formal Fourier series

+'

(S)	y(t) _	am eimc
		m=-oo
Inserting (S) into (E) we obtain
(R)	imam + am-1 + am+1 + am-2 + am+2 = 0		(m E Z),

where Z is the set of all integers. Equation (R) is a fourth-order difference equation on am. Therefore, solution (S) contains four independent arbitrary constants.

Of course, if (S) is restricted to be a convergent series, three of these four constants should be eliminated. A similar phenomenon may be observed if solutions of a certain differential equation are expanded in positive and negative powers of

+oo

independent variables such as F, amxm. For such an example, see [Dw]. m=-oo

1-2. Existence without the Lipschitz condition

To treat the initial-value problem (P) without the Lipschitz condition, we need some preparation.

2. EXISTENCE WITHOUT THE LIPSCHITZ CONDITION

9

Definition 1-2-1. A set.F of functions f (t) defined on an interval I is said to be equicontinuous on I if for every e > 0, there exists a b(e) > 0 such that ] f(tl) -

f (t2)I < e for every f E .F whenever it, - t2I < b(E), tl E :r and t2 E Z.

For example, if the set F consists of all functions f such that f is continuous on a closed interval It - tol < a, f is continuous in it - to, < a, and If (t)l < L in it - toy < a, then.F is erquioontinuous on the interval I = {t : It - tol < a). In fact,

since At 0 - f(t2) = J t l f'(s)ds for t1 E T and t2 E Z, we have ] f (tl) - f (t2)l < E

t2

whenever It1 - t2l < b(c) = L.

Definition 1-2-2. A set F of functions is said to be bounded on the interval Z, if for every t E I there exists a non-negative number M(t) such that lf (t)I < M(t) for every t E Z and f E.F.

Lemma 1-2-3 (C. Arzela [Ar]-G. Ascoli [As]). On a bounded interval Z, let.F be an infinite, bounded, and equicontinuous set of functions. Then, F contains an infinite sequence which is uniformly convergent on Z.

Proof.

Let Q be the set of all rational numbers contained in the interval Z. Then,

(1)Q is dense in Z; i.e., for every 6 > 0 and every t E Z, there exists a rational number r(t, b) E Q such that it - r(t, b)1 < b,

(2)Q is an enumerable set; i.e., Q = {r, : j = 1, 2, 3, ... }.

Note that the choice of the rational number r(t, 6) is not unique. We prove this lemma in two steps.

Step 1. The set .F contains an infinite sequence { fh	: h = 1, 2.... } such that
o
lim fh(r) exists for every rational number r E Q.
h
Proof.
Choose from F a sequence of subsequences F.	1, 2.... }, j =
1, 2, ... , such that
(i) .Fj +1 C .Fj for every j,
(ii) t lim o f',,e(rl) = c, exists for every j.

To do this, first look at functions in F at t = r1. Using the boundedness of the set {f(r1) : f E .F}, we choose an infinite subsequence F1 = {f 1.t : t = 1, 2,... }

from .F so that lim f1,t(r1) = cl exists. Next, look at the sequence F1 at t = r2.

1 +00

Using the boundedness of the set { f l,t(r2) : t = 1, 2.... }, we choose an infinite subsequence F2 = {j 2 t : t = 1,2.... }from F1 so that lim h,t(r2) = c2 exists.

Continuing this way, we can choose subsequences .F,+1 from Y. successively.

Set fh = fh,h (h = 1, 2, 3.... ). Then, ft, E fl for every h >) since fh = fh,h E

.Fh C .Fj if h > j. Therefore,	lim fh(rj) = c, exists for every j, i.e.,	lira fh(r)
-	h-+oo	h-+oo

exists for every rational number r E Q.

10 I. FUNDAMENTAL THEOREMS OF ODES

Step 2. The sequence { fh : h = 1, 2.... } of Step 1 converges uniformly on the interval I.

Proof

For a given positive number f and a rational number r E Q, there exist a positive number b(e) and a positive integer N(r, e) such that

lfh(t) - fh(r)I <c	whenever	it - rl < b(e),
Ifh (r) - ft(r)I < E	whenever	h > N(r, E) and I > N(r, E),
1J(r) - f (t)I < E	whenever	it - rl < b(e).
Now, observe that

Ifh(t) - ft(t)1 <- LAW - fh(r)I + Ifh(r) - ft(r)I + I h(r) - f!(t)I,

where t E I and r r= Q. Therefore, by choosing r = r (t, b ( 3)) , we obtain

(1.2.1) Ifh(t) - ft(t)I < E

whenever h > N (r I t, 6(3 bigg)) 3 I and t > N (r (t, b (3 ) . 3 )

Since the interval I is bounded, we can cover I by a ite number of open intervals 11, I2, ... ,1,(,) in such a way that the length of each interval I1 is not

greater than b (11) , where p(E) is a positive integer depending on E. For every t,

/E \

choose a rational number rt(e) E 1t f1 Q. For all t E It, set r t,b1 3) I = rt(E)

and set

1

N(e) = 1 maxE N (r,(.E), 3 )

Then, (1.2.1) is satisfied whenever h > N(e) and I > N(E). 0

To construct a solution of the initial-value problem (P) without the Lipschitz con- dition, approximate the given differential equation by another equation which satisfies the Lipschitz condition. The unique solution of such an approximate problem is called an E-approximate solution. Using the following lemma, an E-approximate solution is found.

Lemma 1-2-4. Suppose that f (t, y) is continuous on a rectangular region

R={(t,y): It - tol a, Iy-col<b}.

Then, for every positive number f, there exists a function FF(t, y) such that

(i) F( is continuous for It - toI < a and all g,

(ii) 1 has continuous partial derivatives of all orders with respect to yl, y2,... , y f o r It - tot < a and all y",

(iii) I FE(t,y)I <c** max If(r,r1)I for It -to l<aand all y",

X

(iv) IFe(t,y)-f(t,y)I r=<e on

.

Proof.

We prove this lemma in three steps.

2. EXISTENCE WITHOUT THE LIPSCHITZ CONDITION

11

Step 1. There exists a function P(t, yj such that

(1) F is continuous for It - tot < a and all y,

(2) IF(t, y)I < c >SERI If (-r,	for It - tot < a and all f,
(3)
{f(t	on R,
F(t, yJ = 0	for It-tot <a and If-cal>b+1.

The construction of such an f is left to the reader as an exercise. Since f is uniformly continuous on R, properties (1), (2), and (3) of f imply that

(4) for every positive number e, there exists a positive number b(e) such that

IF(t,f1)-F(t,f2)i <e

whenever It - tot < a, Ig, - f2l < b(e).

Step 2. For every positive number b, there exists a real-valued function p(t, 6) of a real variable t such that

(a) p has continuous derivatives of all orders with respect to for -cc < t < +oo,

(b)p(4, b) > 0 for -co < t < +oc,

(c)p(.,b)=0forI{I>b,

(d) f(.5)de = 1.

,o

The construction of such a function p is left to the reader as an exercise.

Set 0(g,6) = p(yi, b)p(y2, b) ... p(y,,, b). Then

(a')	has continuous partial derivatives of all orders with respect to y1, y2,.,. , y
	for ally",
(b') t(#, 6) 0 for all
(c') 0(f, b) = 0 for Iyj > 6,
(d') r+00 ... f+00 '(y, 6)dy1dy2 ... d1M = 1.
	f
Step 3. Set
	4(t, YJ = 1-00+- ...	J+oo (y- 1, b(E))F(t, ff)d ... dnn,
		00

where b(e) is defined in (4) of Step 1. Then, F,(t, y-) satisfies all of the requirements (i), (ii), (iii), and (iv) of Lemma 1-2-4.

Proof

Properties (i) and (ii) are evident. Property (iii) follows from the estimate

I e(t,M1<max{IF(t,10i: all ;t} j

00

< iTMaExRIf(T,nl1.

12 I. FUNDAMENTAL THEOREMS OF ODES

Property (iv) follows from the estimate

I A, (t, yi - F'(t, Y) I			j+ (y-;),b(E)){F(t,n7)-F(t,)}dg1...dg
=l	r+oo
			00	l
_	...	r		(9 - n,b(E)) {F(t,nj - F(t,y-) } dm ... dn.
E tI ...			r-	n, b(E))drh ... dnn l = E.
	Il ff	JWl1 <_a(E)

This completes the proof of Lemma I-2-4.

Now, we prove the fundamental existence theorem without the Lipschitz condi- tion.

Theorem 1-2-5. If f (t, y-) is continuous on a region

P- = {(t,y): It-tol < a, I#-col < b}

and M = ma c I f (t, )l, there exists a function fi(t) such that

(i)(to)

(ii)4'(t) = f (t, 0(t)) on It - toy < a,

when	if M = 0,
a
/ l	if M > 0.
minla,i )

Proof.

We prove this theorem in four steps.

Step 1. For every positive number e, construct a function FE (t, y1 which satisfies all of the requirements given in Lemma 1-2-4. Using Theorem I-1-4, we construct a function 0 (t) such that

(1)S(,E(to) =_4o,

(2)P. (t, E(t)) on It - toI < o.

Furthermore,

(3)(t,4(t)) E 1 on It - toj < a (cf. Lemma 1-1-1 and Remark 1-1-3).

Step 2. The set F = {¢E : E > 0} is bounded and equicontinuous on It - toI < a.

Proof

Property (3) of functions ¢E implies that F is bounded on It - tol < a and that lFE(t, jE(t))I < M on It - tot < a. Hence, property (2) of ¢ implies that

I4(t1) - 4(t2)1:5 ljtlt2 iE(s)dsl < Mlt1- t21

if Iti - tol < a and It2 - t0I < a (cf. Lemma 1-1-2). Therefore, for a given positive number p, we have I0E(t0 - 0E(t2)l < µ whenever Itl - toI < a, It2 - toI < a, and

It1 - t21 < -E. This shows that Jr is equicontinuous on It - tol < a.

2. EXISTENCE WITHOUT THE LIPSCHITZ CONDITION

13

Step 3. Using Lemma 1-2-3 (Arzeli -Ascoli) choose a sequence {ee : I = 1, 2,. .. } of

positive numbers such that	lim e, = 0 and that the sequence	l = 1, 2.... }
	r+oo
converges uniformly on It - tot < or as t -- +oo. Then, set

	,(t) = lim ,,(t)	on It - to, < a.
	+oo
Step 4. Observe that
(t) = c"o +	fP, (s, ,(s))ds
=ca+	f(s,0,(s))ds+J {f (s,t (s))-f(s,4(s))}ds
	eo	to
and that
(s)) - f(s,Z (s))Ids		< e It - tol	on It - to; < a.
Ifca

This is true for all e > 0. Letting a - 0, we obtain

(t} = c"o + f L f (s, 0(s))ds. u

This completes the proof of Theorem 1-2-5. 0

Example 1-2-6. Theorem 1-2-5 applies to the initial-value problem

dy

(P)	xyl/s,	y(3) = 0.
	dx

However, Theorem I-1-4 does not apply to (P). In fact y(x) = 0 is a solution of

Problem (P) and also

0

for

y(x) _ (2(x2_ 9)154

for

5

x < 3,

x > 3

is a solution of Problem (P). Note that the right-hand side of the differential

equation (P) does not satisfy the Lipschitz condition on any y-interval containing

y=0.

Two other methods of proving Theorem 1-2-5 are summarized in the following two remarks.

Remark 1-2-7. Let every entry of an n-dimensional vector f (t, y-) be a real-valued continuous function of n + 1 independent variables t and y = (yl, ... , y,,) on a

rectangular region R = {(t, y-) : It - tot < a, Iy"-cod < b}. Assume that I f(t,yj 1 < M

14	I. FUNDAMENTAL THEOREMS OF ODES

on R, where M is a positive number. Set a = min (a, M ) . Since f (t, yI is

uniformly continuous on R, there exists a positive n umber p(e) for every given

positive number a such that I f (tl, y1) - f (t2,1%2)I < E Whenever Itl - t2[ < p(c) and

Iyl - 1%2I 5 p(e) for (xl, WI) E R and (x2, y2) E R. Let 0(e) : to = ro < r1 <

< rn(c) = to + a be a subdivision of the interval to < t < to + a such that

n(c)-1		p(E)
m ax (I rj - rj+ll) < nun P(E),		M	. Set
	co			for	t = to,
91(t) =	1/e(rj-1) + J (r)-1,1/c(rj_1))(t - 7-j-1)			for	rj_1 < t < TI,

(j = 1,2,... ,n(6)).

Then, we can show that

(i)every entry of y, (t) is piecewise linear and continuous in t and (t, &(t)) E R on the interval to < t < to + a,

(ii)the set {yi : E > 0} is bounded and equicontinuous on the interval to < t <

to +a,

(iii) if we choose a decreasing sequence {c) : j = 1, 2,... } such that Jim E) = 0

-+oo

in a suitable way, the sequence {y',, (x) j = 1,2,. .. } converges to a solution of the initial-value problem

(1.2.2)	dt = f (t, J- ,	i(to) = 0o

uniformly on to < t < to + a as j -+ +oo.

For more details, see [CL, pp. 3-5].

Remark 1-2-8. We use the same assumption, M, and a as in Remark 1-2-7. Also, let i1(t) be a continuously differentiable function of t such that #(to) = co and

(t, #(t)) E R on the interval to - r < t < to, where r is a positive number. For every e such that 0 < e < r, define

		q(t)	for	to - r < t < to,
	1%c(t) =	c
		clo+ff(s,il(s_6))ds	for	to < t < to + a.
		o

Then, we can show that

(i) every entry of g, (t) is continuous in t and (t, &(t)) E R on the interval to-r <

t<to+a,

(ii) every entry of 17, (t) is continuous in t on to - r < t < to +a except a jump at

x = to,

(iii) the set {y'E : 0 < c < r} is bounded and equicontinuous on the interval

to -r<t<to +a,

(iv) if we choose a decreasing sequence {c) : j = 1, 2, ... } such that lim e) = 0

)+no

in a suitable way, the sequence {yEI (x) j = 1, 2, ... } converges to a solution of the initial-value problem (1.2.2) uniformly on xo < x < xo+a as j - +oo.

For more details, see [CL, pp. 43-44] and [Hart, pp. 10-11].

3. SOME GLOBAL PROPERTIES OF SOLUTIONS

15

Remark 1-2-9. If Ax, y-) is assumed to be measurable for each fixed y, continuous

in y for each fixed t, and I f (t, yj I is bounded by a Lebesgue-integrable function when

(t, y") E R, a result similar to Theorem 1-2-5 (Caratheodory's existence theorem)

[Ca, pp. 665-688) is obtained. Similarly, if, in addition, it satisfies an inequality similar to Lipschitz condition with the Lipschitz constant replaced by a Lebesgueintegrable function L(t) when (t, y-) E R, we obtain also a result similar to Theorem

1-1-4. (See [CL, p. 43), [Har2, p. 10], and [SC, p. 15).)

1-3. Some global properties of solutions

We can construct a solution to an initial-value problem (P) in a neighborhood of the initial point by using Theorem 1-1-4 or 1-2-5. To extend such a local solution to a larger interval of the independent variable t, the following lemma is useful.

Lemma 1-3-1. Assume that

(i)a function f(t, yi) is continuous in an open set D in the (t, y-) -space,

(ii)a function fi(t) satisfies the condition ;s (t) = f(t,a(t)), and (t, Q(t)) E 'D, in an open interval T = {t : ri < t < r2}.

Under this assumption, if lim (t,, d(t,)) = (r1, ij) E D for some sequence {t,

)too

j = 1,2.... } of points in the interval I, then lim (t, (t)) = (r1, ij). Similarly, if

lim (t,, 0(t,)) _ (T2,771 E D for some sequence {t. : j = 1,2,... } of points in the

j+oo

interval Z, then lim (t, fi(t)) = (r2i

Prof.

We shall prove the part of the lemma which concerns the left endpoint ri of the interval Z. The other part can be proven similarly.

Let U be an open neighborhood of (ri, rl. We show that (t, 0(t)) E U in an interval ri < t < r(U) for some r(U) which is determined by U. Assume that the

closure of U is contained in V and that I f (t, y-) I < M in U for some positive number

M. For every positive integer j and every positive number e, consider a rectangular region R,(e) = {(t,y-) : it - t,I < e, Iy - 4(t,)j < Me}. Then, there exist an e > 0

/ !4fe

and a j1such that (7-1,771 E Rj (e) C U. Note that e =min Ce, M I and t, - e < rt . Applying Lemma I-1-1 to the solution y" = d(t) of the initial-value problem dy = f(t,y-), y(tj) = (tj), we obtain (t,y5(t)) E Rj(E) C U on the interval r1 < t < tj. Since U is an arbitrary open neighborhood of (r1, i)), we conclude that

lim (ti,0(ti)) _ (ri,7-7) E D.

-+oo

From Lemma 1-3-1, we derive the following result concerning the maximal inter- val on which a solution can be extended.

Theorem 1-3-2. If

(i) a function f (t, yj is continuous in an open set V in the (t, y-')-space,

16	I. FUNDAMENTAL THEOREMS OF ODES

(ii)a function fi(t) satisfies the condition 4 (t) = f(t, ¢(t)) and (t, fi(t)) E D, in an open interval I = {t : r1 < t < r2},

(iii)4(t) cannot be extended to the left of rl(or, respectively, to the right of r2) with property (ii),

(iv)jlimo(tj,0(ti)) = (TI,fl) (or, respectively, (r2, r))) exists for some sequence

{tj :j = 1,2....l of points in the interval Z,

then the limit point (Ti, n (or, respectively, (T2, i))) must be on the boundary of V.

Proof

We shall prove this result by deriving a contradiction from the assumption that

(r1,,7) E E) (or, respectively, (,r2, i-11 E D). Applying Lemma 1-3-1 to this situation, we obtain crlim(t,¢(t)) = (rl,r) (or, respectively, (r2ir))). Hence, by applying

Theorem 1-2-5 to the initial-value problem

	dy	= f (t, y),	y(rl) = n (or, respectively, ff(r2)
	dt

the solution 45(t) can be extended to the left of rl (or, respectively, to the right of r2). This is a contradiction. 0

The following example illustrates how to use Lemma 1-3-1.

Problem 1-3-3. Show that the solution of the initial-value problem

d2 y - 2xyL = y3 - y2	y(xo) = 17, Y (xo) = (,
2

exists at least on the interval 0 x < xo if xo > 0, q > 0, and (is any real number.

Answer.

Assume that the solution y = 4(x) of the given initial-value problem exists on an interval I = {x : £ < x < xo} for some positive number . Observe that

L2(.0')2+ 1.63 + 20-2]			= 2xo(th')2 > 0
on Z. Hence,
(O'(x))2+ 14(x)3 +	14(x)-2	< 1(0'(xo))2 + 14(x0)3 +14(x0)-2 (= Al > 0)
3	2	2	3	2
on Z. Therefore, we have
(4 (x))z < 2M,		4(x)3 < 3M,		4(x)2 >
				21Lt

Now, apply Lemma 1-3-1.

The proof of the following result is left to the reader as an exercise.

differential equation

3. SOME GLOBAL PROPERTIES OF SOLUTIONS

17

Corollary 1-3-4. Assume that f (t, yam) is continuous fort I < t < t2 and all y E R". Assume also that a function fi(t) satisfies the following conditions:

(a)and d+' are continuous in a subinterval I of the interval tl < t < t2,

(b)(t) = f (t, ¢(t)) in Z.

Then, either

(i) Qi(t) can be extended to the entire interval t1 < t < t2 as a solution of the

= f (t, y), or

(ii) limO(t) = oo for somejr in the interval ti < r < t2.

t-r

Using Corollary I-3-4, we obtain the following important result concerning a linear nonhomogeneous differential equation

dt = A(t)yy + bb(t),

where the entries of the n x n matrix A(t) and the entries of the R"-valued function b(t) are continuous in an open interval I = It : ti < t < t2}.

Theorem 1-3-5. Every solution of differential equation (1.3.1) which is defined in a subinterval of the interval I can be extended uniquely to the entire interval I as a solution of (1.3.1).

Proof

Suppose that a solution y = d(t) of (1.3.1) exists in a subinterval Z' _ It : r1 < t < r2} of the interval I such that tj < 71 < r2 < t2. Then,

At)I <- Ato)I + j {A(s)4(s) + b(s)} dsl

	o
in Z', where to is any point in the interval T. Setting
K =	'15'572
	(r2 - ri) max Ib(t)I,
L =	IA(t)I,
we obtain
	in T.
	o
Thus, the estimate
Kexp[Llt - toll < Kexp[L(r2 - ri)[ <00		in Z'

is derived by using Lemma 1-1-5. Hence, case (ii) of Corollary 1-3-4 is eliminated.

Since the right-hand side of (1.3.1) satisfies the Lipschitz condition, the extension of the solution y5(t) is unique.

For nonlinear ordinary differential equations, we cannot expect to obtain the

same result as in Theorem 1-3-5. The following example shows the local blowup of solutions of the differential equation of Problem 1-3-3.