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Existence and uniqueness theorems for some functional differential equations of neutral type
JOURNAL
OF DIFFERENTIAL
EQUATIONS
Existence Some
Functional
11,
607-623 (1972)
and Uniqueness Differential
Equations
ROGER D
Department of Mathematm,
Theorems of
for Neutral
Type*
N~SSBAUM
Rutgers Unzverszty, New Brunswzck, New Jersey 08903 Recewed September
13, 1971
INTRODUCTION
This article is motivated
by the simple functional
x’(t) = kx’(mt)
differential
0 < t <; H, s(O) = 0.
+ g(t),
equation
(1)
In this equation k and m are constants, 0 < M < 1, and g is a C” function. The work of Cruz and Hale [2] or of Badoev and Sadovskn [I] can be applied to this equation, and both approaches yield the same answer: If 1k/m / < 1, there is a unique absolutely contmuous function with L1 derivative satisfying (1), and if 1k/m 1 > 1, no information on existence or umqueness IS provided. Nevertheless, If / k/m 1 > 1 one can easily show directly that (1) has an absolutely contmuous solution on [0, H] and in fact has an infinite-dimensional subspace of solutions (see [ 131). Th us it is clear that the Cruz-Hale or Badoev-Sadovskii approaches miss a lot of information in this case, and the question is if some of this information can be retrieved. The answer is yes. We shall show below (as a very special case) that if n > 2, / kmn-l ( < 1 and km’ f 1 for 0 < j < n - 2, there exrsts a unique solutron of (1) in C”([O, HI). It has been known for some time that the solutrons of even linear functional differential equations may have nasty discontmuitres m their derivatives, and this has encouraged existence and uniqueness theorems in settings like the Sobolev spaces IP”, 1 <. p < 00. The moral of our work here is that for certain “nice” neutral functional different equations (NFDE’s) one may lose a great deal of information (both existence and uniqueness) by looking only for absolutely continuous solutions. For some NFDE’s obtaining sharp existence and uniqueness results demands a closer examination of drfferentrabrlity properties of the equation than has been customary. The organization of this paper is as follows: In the first section we consider a class of nonlinear NFDE’s for which the standard results m the literature * Partially
supported
by NSFGP20228.
607 0
1972 by Academtc Press, Inc.
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grve weak and inadequate exrstence and uniqueness theorems. We prove a local existence theorem, some continuation theorems, and a few other results. In the second sectron we apply the nonlinear theorems to obtain theorems for linear NFDE’s. Surprisingly, we obtain sharp results for the linear theory in this way. 1. Let H be a posrtive real number and let f : [0, H] x R x R ---f R be a function which has continuous partials of all orders up to n - 1, 71 > 2. Let fL , i = 1, 2, 3, denote the partial off with respect to its z-th variable. If m is a positrve integer, let Cm([O, H]) d enote the Banach space of m times continuously differentiable functrons x from [0, H] to R, and define
C([O, H]) is the Banach space of continuous the usual sup norm. Let 8, + E C-l([O, H]) for t E [0, H] and 0 < 4(t) < t for t E (0, H]. there exists a > 0 and x E C”([O, a]) = X equations: x’(t) = f(t, x(qt)),
x’(e(t)>>,
functrons from [0, H] to R in and assume that 0 < 0(t) < t We want to investigate whether which satisfies the following
O
(2)
x(0) = xg . We also want to investigate how far solutions of (2) can be extended and when closed, bounded sets of solutrons of (2) in X are compact. The basic tool we shall use is the “measure of noncompactness,” an idea due to Kuratowski [7]: DEFINITION [7]. Let (X, p) be a metric space and A a bounded subset of X. The measure of noncompactness of A 5 y(A) = inf{d > 0: there exists a finite number of sets S, , Sa ,..., S, such that A = ur=r S, and diameter
(SJ G 4. DEFINITION [7]. Let (XI , pr) and (X, , pa) be metric spaces, let G be a subset of X and let f : G --f X, be a continuous map. We say that f is a “k-setcontraction” if for every bounded set A C G, f(A) is bounded and y,(f(A)) < ky,(A), where yz denotes the measure of noncompactness in X, . The basic results we shall need are contained in the following propoation, which was proved by G. Darbo [3]: PROPOSITION 1 [3]. Let X be a Banach space and y the measure of noncompactness induced by the norm.
EXISTENCE
AND
UNIQUENESS
THEOREMS
609
(1) If A and B are bounded subsets of X and A + B = (a + b : a E A, b E W, y(A + B) < ~(4 (2)
+ Y(B)-
If A is a bounded subset of X and Z(A)
= convex closure of A,
rW4
= r(A). (3) If G is a closed, bounded convex subset of X and f: G + k-set-contraction with k < 1, f has a$xed point.
G zs a
Of course the reason for the nomenclature “measure of noncompactness” is that if M is a subset of a complete metric space X, y(M) = 0 iff A/I has compact closure in X. Before consrdering (2) we need a few simple lemmas. For a > 0 and m > 1 let X = P’([O, a] and let Y = C([O, a]) (with the norms already mentioned). Let y denote the measure of noncompactness m X and y1 that in IT. LEMMA
1. Let S be a bounded set an X. Then if J : X -+ Y is dejned by
Jx = x(") 9 Y(S)
Proof.
= n(J(W
Take E > 0. Since S is bounded, A = {(x(O), x’(O),..., x’m-l)(o)) : x E sj
is a bounded set in R”. Take as the norm in R”, /I v 11= maxsG,G;,-, 1vu,I, where v = (vs , vI ,..., v&, and let vz, 1 < i < N, be an c/2 net in A with respect to this norm. For 1 < i < N let s, = (x E s : 11(x(O),x’(O),..., x(+1)(0)) - 0%jj < 42). Suppose that n(J(S))
= d. Then ,(J(S,))
< d, and we can write
J(S,) = ilj T,,, 3 pl
where diameter (Tz,3) < d + E. Let S,,, = {x E S, : Jx E T,.,}. We claim that diameter (S,,,) < d + 2~, for if
x, y E s,,, , (su$, I .yt)
- Y’m’(t)l G d + E
and
so that /j x - y /I < d + 2~. Since S = US,,, , this shows that y(S) < d + 2r, and since E > 0 is arbitrary, y(S) < K( J(S)). The opposite inequality is obvious. Q.E.D. 505/11/3-11
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LEMMA 2. Let R and S be bounded subsets of Y (Y as in Lemma 1) and let T = (xy :x E R, y E S}. Let M, = sup{// x )/ :x E R} and let Mz = SUP{IIYII :Y E sl- Then ~~07 d Jfd‘9
+ Ma(R).
Proof. Suppose y,(R) = dl and yr(S) = d, . Then grven E > 0, R = (Jz, R, with diam(R,) < dl + E and S = (J,“=, S, , diam(S,) < d, + E. It follows that T = (JI,> Ri,S, , where R,S, ={xy:xx~R,,y~SJ. If zr = xryr
and xa = s,y,
are elements of R,S, ,
This imphes that diam(R,S,) < Mid, arbitrary, rl(T) < Mid, + n/r,d, .
+ Mzd, + (Ml + M&,
and since E 1s Q.E.D.
LEMMA 3. Let S, and S, be bounded subsets of Y and assume that y,(S,) = yl(S,) = 0. Let g : [0, a] x Iw x Iw ---, [w be a continuous map and dejne s = {z : x(s) = g(s, x(s), y(s)), x E s, ) y E As,). Then yl(S)
= 0.
Proof. This is an easy consequence of the Ascolr-Arzela uniform continuity of g on bounded sets.
theorem and the Q.E.D.
The way in which we shall use these Lemmas 1s as follows: Suppose S 1s a bounded subset of X = C*([O, HI), n 3 2. Then by the Ascoli-Arzela theorem {X(J) : s E S} is precompact in Y = C([O, H]) for 0 < j < n - 1. If 0 and 4 are continuous maps of [0, H] to [0, H] it 1s also clearly true that {XC” 0 4 : x E S> and {X(J) o B : x E S> are precompact in C( [0, HI). By Lemma 3, if g : [0, H] X [w X Iw + [w is a continuous function, {,z : z(t) = g(f, x(&), x’(+t)), x E S} is precompact in Y. By Lemma 2, {z : z(t) = g(t, s(h),
x’(~t))(x(‘)(Ot))(x(k)(~t)), x E S} is precompact in Y ifj s< n - 1 and R :< n - 1 and yl({z : z(t) = g(t, x@), x’(~t))x’“‘(~t), < sup{/ g(t, x&Q), x’(&))l = x’“‘(&),
x E As> : 0 < t & H, x E S}y,({z
x E S}).
We can now prove our first nontrivial
: z(t) (3)
result.
EXISTENCE AND UNIQUENESS THEOREMS
611
PROPOSITION 2. Let f : [0, H] x R x R -+ R be a function with n - 1 continuouspartials, n 3 2. Assume that 0,4 E C+l([O, HI), that 0 < 6(t) < t for t E [0, H] and that 0 G+(t) < t for t E (0, H]. Let X = C”([O, H]) and Y = C([O, HI). Suppose B zs a closed, bounded subset of X and define F: B ---f X by (Fx)(t) = x(0) + $,f (s, x(&), x’(&)) ds. Assume that for all x E B, / f3(0, s(O),X'(O))(+'(O))~-l 1 < c < 1. Then S = {x E B :Fx = x} is compact (possibly empty). Also, there exists Y > 0, r & H, such that if x = {x 1 [0, r] : x E X}, 4 = {x 1 [0, r] : x E Bj and P = F 1B, P is a k-set-contraction for some k < 1.
Proof. Since F(S) = S, F”(S) = S for all integers m > 1 and thus y(F”‘(S)) = y(S). Therefore to show S is compact (it is closed because F is clearly a continuous function), it suffices to show that for some constant k < 1 and some integer vz, y(F”(S)) < KY(S). Before going further we make some observations which will be useful later in the proof. If s, = sup(fl(t) : 0 < t < H}(@ = the j-th iterate of 4) note that s, -+ 0, because $(t) < t for t > 0. Let C, = sup{\ f (t, x(&t), x’(#+lt))(#‘(~ct))n-l j : 0 < t f H, x E B}, i > 0. Since lim,, x(s) == x(O) and lim,,, x’(s) =: x’(0) uniformly for x E B and because F(t) < s, , it follows that hm,,, C, = /f (0, x(O), x’(O))($‘(O))+l / < c. It follows that if we define k, = Hi=, C, , there exists an integer m such that k, < 1 forj > m. For an integer m let S, = F”(S). We wish to estimate r(Sm). By Lemma 1, rG%J = rdJ&d = r~Ur;‘(Sm-I>)- BY a direct calculation one can see that there exist continuous functions gl,,,k : [0, H] x lR x R --f 52 and h 1,3,1e: [0, H] -+ [0, H] indexed by integers j, k, 1 such that 0 < j < n - 1, 0 < k < n - 1 and 0 < 1 < l(j, k) such that for any x E X,
+ f3(t, x(q, 4+(t~kw)~-~
x(yw
It follows then by the remarks before this proposition rl(JF(Sm-1))
= df
(4) that
3(t , x,-l(et), x~-,(+t))(+f(t))“-l&~l(4t> : k1 E&A)
9 corl({~~Q~t): G-1 E&n-S). If we now write that
S,-,
= F(S,-,)
Yb%J
G
Gcl~l({4z2(+2t)
(5)
and apply the same calculation,
: %-2E
&-21).
we find
612
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Continuing in this way we eventually find that
By our previous remarks k,-1 < I for m large enough, so we have proved the first part of the proposition. Since lim,,, x(s) = x(O) uniformly for x E B and similarly for x’, there exists r > 0 such that sup{1MY 4% +W#‘(W1 (7) -f3(0, x(O), x’(O))(~‘(O))‘+~ / : 0 < t < r, x E B) < (1 - c)/2. For this r, if B and 2 are as above and A is any bounded subset of B, r(-P(A)) = r1U%4> = n((f3(t, x(&), d(&))(+‘(t))“-’ < sup{/f,(t, @‘t), x’(+t))(4’(t))“-1 x
T4W'
*c
LX(~)(#) : x E A, 0 < t G r>) I : x E A, 0 9 t < r>
:*~-49 d [Cl + WlA4
(8)
It follows that P is a k-set-contraction with k = (1 + c)/2.
Q.E.D.
Our next result is a slight sharpening of Proposition 2 which will be useful later for the linear theory. PROPOSITION 3. Assume that f, 8, and 46are as in Proposition 2. Suppose in addition that there exists a function k: [0, H] -+ R such that for all (x,, , y,,) E R, IfJt, x,, , yo)j < k(t) and 1k(0)($‘(O))“-l ) < 1. Then if M is any compact subsetof X and B is any closed,bounded subsetof X, S = {x E B : x - FX E M) is compact.
Proof. In analogy with Proposition 2, let Sl = F(S) + M and generally S, = F(S,-,) + M for j > 1. Since S, > S, r(S,) > y(S), so to prove the proposition it suffices to show that there exists c* < 1 and an integer m such that y(S,) < c*r(S). The proof mimics that of Proposition 2, but since we no longer know that S, C B, the stronger assumptions are needed. Let C, = supoGtGH [ k(+t)[$‘(+t)]+l 1 and note that lim,,, C, = 1k(0)[4’(O)]+1 1 ( 1. It follows that there exists an integer m such that ~~~oC3~c*<1forr~m-1. Using the results of Proposition 1 we find r(&J
< rPVm-1)) = ?4if3(t, G
GYlGG#J
+ Y(M) = r(W,-1))
x,-l(et),x~-l(~t)>(q(t))"-'X~l(~t) : Xm-1E
sn-3).
: %-1E
s!-1:
(9)
EXISTENCE
AND
UNIQUENESS
613
THEOREMS
Since xmel = F(x,-.J + y for some xm-s E S,-s and some y E M and since y1((y(“) 0$ : y E M}) = 0, we find that
C,r,({x%
a c : G-1 E &4H
=
Co~#%n-z)(a)~~
=
COyl({f#,
<
coc,yl({+'"'
: xm-2
x,-2(e$t),
0 $"
'rn2
E &-21)
xL-2(Qt))[~(~t)l"-1X~~2(~2t)
:
: %-2
x,-2 E s,-2)).
E %n-21)
(10)
Continuing in this we find that
(11) Q.E.D. We can now estabhsh a local existence theorem for our original FDE (2). In the following theorem, fs indicates partial differentiation with respect to the third variable. THEOREM 1. Let f : [0, H] x Iw x [w --+ R be a Cn-l function, n 9 2. Let 8, q5 : [0, H] 3 R be Cn-l functions such that 0 < O(t) < t for t E [0, H] and 0 < C(t) < t for t E (0, H]. Let x,, E R be given and assume there exists x1 E R such that Xl
=
f(O,
x0
9 XI),
f3(0, x0 I xJ($‘(W
f
1
for1 0 and a function x E C*([O, r]) such that x’(t) = f (t, x(&), x’(+)) for 0 < t < r and x(0) = x0, Also, for any b > 0, b < H, and any closed, bounded set B C C”([O, b]), S = {x E B : x’(t) = f (t, x(h), x’(+t)) for t E [0, b], x(0) = x0, x’(0) = x1} is compact (possibly empty). Proof. Let B, = {x E B : x(0) = x,, , x’(0) = x1} and consider (Fx)(t) = x(0) + $f (s, x(@s),~‘(4s)) ds for 0 < t < b. The elements of S are precisely the fixed points of F in B, , so S is compact by Proposition 2. We next claim that if for some r > 0 and some x E C”([O, r]), x(t) = (Fx)(t) for 0 < t < r and x(0) = x0, x’(0) = xl , then x(?)(O)is uniquely determined for 2 < j < n. For example we have xv3
= fl(o, x0 , x1) + f,(o, x0 , xl)waxl
+ [f3(op x0 , x,)d~0)]xv),
(12)
and since f3(0, x0, x,)$‘(O) # 1, X(~)(O)is uniquely determined, X(~)(O)= xL . Generally, if x(?)(O) = x3 is determined for 2 < j < m, m < n, @+l)(O) = (function of x0 , xi , .. ., x,) + [ f3(0, x0 , xl)(+‘(0))~]x(“+l)(O),
(13)
614
NUSSBAUM
and since 1 # fs(0, x0, x~)($‘(O))~ for 1 ,< m < 72,@‘+1)(O) is uniquely determined, ~@+l)(0) = x,+i , If we define Kb = (x E C”([O, b] : x(j)(O) = x, for 0 < j < n}, it is not hard to see that F : K, -+ Kb . Select a fixed number R > 0 and let V, = {x E Kb : 11x - p 11< R}, where p(t) = Cy=,, (x,/j!)tj. By Proposition 2 there exists b* > 0 such that F ( V,, is a K-set-contraction, K < 1, and in fact the same proof shows F j V, is K-set-contraction for 0 < b < b *. If we can find b > 0, b < b* such that F( V,) C V, , then by Proposition 1, F will have a fixed point, and we will be done. Thus consider [jFx - p 11= SUP~<~<~j(Fx)fa’(t) - x, / for x E V, . As was remarked in the proof of Proposition 2, one can write (i++“‘(t)
=
C c g,& ,,h
x(W x’(+)) h,&> ~(~‘(et) x’“‘(+)
t f3(& x(@f), x’W))(#(W1
x’“‘(@),
(14)
where the g,,, and h,,, are certain continuous functions. As usual, since lim,, X(J)(S):= X, uniformly for all x E VH and j < n , one can see that for b small enough,
and
- f3(0, x0 , x1)(+‘(O))-
x@‘(O)1 f
w
R.
It follows that for this b, sup 1I(Fx)‘“‘(t)
- x, /j =
0
Thus we have F(VJ C V, .
sup l/(F~)(~‘(t) - (Fx)(“‘(O)ll
< R.
(17)
OCt
Q.E.D.
Remark 1. Suppose all the hypotheses of Theorem 1 hold except the assumption that fa(O, x,, , x1)(4’(O))? # 1 for 1 < j < n - 2. Assume instead that 0 = f(0, x0 , 0), that 8f/%(O, x0 , 0) = 0 for 1 < j < n - 1 and that If&4 x0 3O)(+‘(O))+i ( < 1. Then if we define x, = 0 for 1 < j < n, it is not hard to see that if x E Cn([O, b]) and G(O) = x, for 0 < j < n, then (Fx)cj’(O) = x, for 0 < j < n. If we now use the above proof, it follows that the equation (2) has a solution x E Cn([O, b]) for some b > 0.
EXISTENCE
AND
UNIQUENESS
THEOREMS
615
Remark 2. If in addition to the hypotheses of Theorem 1 one assumes that all the partials off of order n - 1 are Lipschitz in their second and third variables on bounded subsets of [0, H] x R x R, then one can show that if x and y are solutions of (*) with x’(O) = y’(0) = x1 , x and y must agree where they are both defined. In this case one can use the contraction mapping principle rather than Darbo’s theorem. We want to make some additional assumptions on f which will guarantee that the local solution given by Theorem 1 can be extended to a solution x E P([O, HI). We begin with a simple lemma. LEMMA 4. Let f, 8, and 4 be as in thejirst paragraph of this section. Suppose that x E C”([O, a]), y E C”([O, b])(O < a < b) and x and y satisfy (2). Assume that there exists 6 > 0 such that x(t) = y(t) for t E [0,6]. Then x(t) = y(t) for 0 < t < a. Furthermore, ;fz E Cn([O, d]), 0 < d < H, and x satisfies (2), there exists E > 0 and z1 E C”([O, d + ~1) such that z, extends x and satisJies (2).
Proof. Let c = sup{t : X(S) = y(s) for 0 < s < t}, c > 6. We have to show that c = a. Suppose not, so c < a. Since t - 4(t) > 0 for t E [c, H], there exists 7 > 0, 7 < a - c, such that t - 4(t) >, 27 for t E [c, W]. For t E [c, c + 71 and u E iw define a Cl function g by g(t, u) = f (f, u, x’(@)) = f (t, u, y’($t)). It follows by the standard theory for retarded functional differential equations that there exists q’ > 0, q’ < 7 and a unique function u E Cl([c, c + 7’1) such that u j [O, c] = x j [O, c] and u’(t) = g(t, u(t)) for t E [c, c + 7’1. Since x’(t) = g(t, x(t)) and y’(t) = g(t, y(t)) for t E [c, c + 7’1, x(t) = y(t) = u(t) for t E [c, c + 7’1, a contradiction. To prove the second half of the theorem let z E C”([O, d]) satisfy (2) and let uj = ,&j(d) = the j-th left-hand derivative of x at d, 0 < j < n. For E < min(q, H - d), let K, = (X E C”([O, d + c]) : x / [O, d] = x) and define F: K, ---f K, by (Fx) / [0, d] = z and (Fx)(t) = x(d) + slf (s, LX(&), x’($s) ds for t E [d, d + ~1. One has to check that Fx E C”([O, d + e]) for x E K,, but this follows without trouble because +(t) < d for t E [d, d + ~1. Take R > 0, let p(t) = x(t) for 0 .< t < d and p(t) = Cy=, (u,/j)(t - d), for t E [d, d + E] and define V, = {x E K, : /I x - p j( < R). We claim that for E small enough F: I’, + V, . We have
IIFx -P II = tE;;;+f,l(F4(n)(t) - (Fx)(“)(d)l, and just as in the proof of Theorem 1, to show /IFx - p jj < R for all x E VG, it suffices to show that there exists c < 1 such that
SUP{1 f&S 4% -
x’(4t))(~‘(t))“-1x’n’(+t)
f3(d,x(0d),
x’(+d)(qS(d))“-‘x’“‘(+d)[
: t E [d, d + E], x E V’,: < CR.
(I 8)
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However, in our casethis is trivial because ~c(“)($t) = P)($t) for t E [d, d + ~1, so the above supremum actually approaches 0 as E > 0. Thus for c small enough, F : V, -+ V, . To complete the proof it suffices to observe that F(V,) is compact, for by the usual argument, y(F(V,)) = yd{w E C(P, d + 4) : w = (W”),
x E VJ)
= n(iw ECC4 d + 4 : w(t) = f3(t,x(h), x’(~t))($6’(t))“-?z’“y+t)}) = 0.
(19)
By the Schauder fixed-point theorem, F has a fixed point z, in V,, and a1 Q.E.D. gives the desired extensron of x. THEOREM 2. Letf:[O,H] x 172x aB-+[W,8:[O,H]-,aBand~:[O,HJ-,[W be Cn-l functions, n > 2. Assume that 0 < e(t) < t for t E [0, H] and 0 < 4(t) < t for t E (0, H]. Suppose that for any constant C > 0, there exist constants Ml and M, (depending on C) such that If(t, x, y)/ < Ml + Mzl x 1 for (t, x,y) E [0, H] x !R x [-C, C]. Then $ u E C”([O, b]), 0 < b < H, sutis$es (2), there exists a unique function x E C”([O, H]) such that x j [0, b] = u and x satisfies (2).
Proof. Using the first part of Lemma 4 and a standard argument there exists a unique maximal solution of (2) such that x ( [0, b] = u and x is n times continuously differentiable on its domain. The function x is defined on an interval [0, r) or [0, Y]. If x is defined on [0, r], r = H, and we are done. Otherwise, by Lemma 4, there exists E > 0 and an extension f E C”([O, Y + 61) which satisfies (2), contradicting the maximality of x. Thus to complete the proof it suffices to show that if x is defined on [0, r), x has a Cn extension to [0, r]. Since +([O, R]) C [0, s] for some s < Y, we have I x’(+t)l < C for t E [O, r) and If(t, 4% x’(+t))l < K + J&l de(t)1 for t E [0, r). Therefore we find for t E [0, r)
1@>I G I xo I + j;'"' I x'(s)1ds < I x,, I + j: 11% x(84, x'(+s))lds < I xc,I + Md + M, jt I x@)I ds.
(20)
0
By Gronwall’s
inequality
this implies
1x(&)1 < const for t E [0, r),
1x’(t)1 = lf(t, x(et), x’(+t))l G const It follows that lim,,, fimt,,Jlt, xw,4#4) of (2), we find lim,,, O
for
t
SO
E [0, r).
x(t) = x0 exists; so we find that lim,,,- x’(t) = = f(y, 4w, x’W> exists. Differentiating both sides x(z)(t) exists and generally lim,,,- xW,t) exists for Q.E.D.
617
EXISTENCE AND UNIQUENESS THEOREMS
We have assumed above that $ has a unique fixed point on [0, H], and this has played an important part in the proofs. We shall now assume that C$has a second fixed point t* > 0 and examine the behavior of the solution x as t approaches t* from the left. In the following theorem, +-‘(t*) denotes the left-hand derivative of & at t*. THEOREM 3. Let f : [0, t*] x [w x R + R and 0 : [0, t*] + R be continuous functions and let $ : [0, t*] -+ R be continuously differentiable. Assume that 0 < O(t) < t for t E [0, t*], that 0 < 4(t) < t for t E (0, t*), and that +(t*) = t*. Assume that there exists s < t* and constants Ml , M, , MS such that for t E [s, tJ and (x, y) E R x IF!4’(t) > 0 and
If (t, x, Y)I < Ml + Mzl x I + M,I Y I> where j M&5-‘(t*)i < 1. Then zf x : [0, t*) -+ R is any absolutely continuous function such that x’(t) = f (t, x(&), x’($t)) for &most all t E [s, t*), lim,,,, x(t) exzsts. Proof. By the continuity of (b’ there exists s*, s < s* < t*, such that 1M.&‘(t)1 < c < 1 for t E [s*, P]. By taking s* closer to t* we can also assume that M2(t* - s*) < 1 and c/[l - M2(t* - s*)] = h < 1. Since 4 is one-to-one on [s*, t*] and since+(t) < t for t E [s*, t*], let s1E [s*, t*] be the unique number such that +(sr) = s* = s,,. Generally, let s, be the unique number in [s*, t*] such that $(sn) = s,-r . The sequence {sz} is strictly monotone increasing and limn+m s, = t*. Define a, = ,,,+I 1x’(t)1 dt. To show that lim,,, x(t) exists, it suffices to show that Ji: j x’(t)\ dt exists or, equrvalently, that ZN = xf=, a, is bounded as N --f co. Define M = supastGs* j x(t)(. We then find for n > 1
a, = ,I:”
j x’(t)] dt = j::”
< M&,+,
If (t, x(et), xl(+))\
dt
- s,) + M2 j”” / x(&)1 dt + Ma ,I:” sn
I x’(+t)l dt.
(21)
By a change of variables we find that M3 j::”
j x’(#t)l dt < c ,I:.,
Fort E[sn, s,+,lwe see that I x(Wl < I +*)I +
1x’(s)/ ds = ~a,-, .
I x(&)1 < M if O(t) < s* and
j;r)I x’(r)1dr < M + 2=0 f a,
if
e(t) > s*.
(22)
618
NUSSBAUM
In either case we find that / x(&)1 < M + z,“=,, uL for t E [So , sn+r]. Plugging in these estimates we find that
an < Jqs,+1 - sn)+ w&2+1- 4w + 421+ %-1 . Adding these inequalities right we find that
for 1 < n ,( N and replacing
Z;, by 2,
(23) on the
.zNG ao+ adze+, - s*) + hh(~~+~ - s*)w + 44 + 4-1 . Transposing (1 - M&N+1
the ZN term on the right to the left side and dividing - s*)) > 1 - M,(t* - s*) gives
(24) by
zl, < [a0+ hfl(~N+l- S*) + M&,~+I - s*)][l - nf&* - s*)l-’ + ‘&v-l < [a, + (Ml + M,)(t* - s*)][l - &(t* - S*>l-’ + &v--1 .
(25)
If we denote the first term on the rrght by R, we have that .Z’, < R + kZNMI . Iterating this inequality N times gives N-l
ZIN <
c
kIR + kNuo < R/(1 - k) + a,, .
Q.E.D.
,=O
The reader may have noted that we have avoided considering the case when f, 6’ and + are only continuous (n = 1). Actually, we can also obtain results in this case, though the methods requrred are somewhat different. We mention the following result, though for reasons of brevity we shall not prove it here. THEOREM. Let f : [0, H] >c DBx [w -+ Iw, t3 : [0, H] -+ Iw and+ : [0, H] + Iw be continuousfunctions. Assume thutfor t E [0, H], 0 < l?(t) < t and 0 <4(t) < t. (Notice that q5 may now have $.xed points other than 0.) Assume that for all x, u, v E Iw, ]f(t, x, u) -f (t, x, v)j < k(t)1 u - v I, where k is a continuous function such that k(t) < 1 for all t such that d(t) = t. Finally, ussnme that there exist constants MI and n/l, such that /f (t, x y)l < M, + M2j x 1 + k(t)/ y / for all (t, x, y) E [0, H] x [w x 02. Thengiven x0 E Iw, there exists x E Cl([O, H]) such that x(O) = x0 and x’(t) = f (t, x(&), x’($t)) for 0 < t < H. Furthermore, if S = {x E C1[O, H] : x(O) = x0 , x’(t) = f (t, x(&), x’(@)) for 0 ,< t < H), S is a compact subset of Cl([O, HI). II. In this section we shall apply our nonlinear theorems following linear problem: Find x E C”([O, H]) (n > 2) such that x’(t) = k(t) 4C(t)> x(0) = 0.
+ b(t) +W>
+ g(t),
to the
O
(26)
EXISTENCE
AND
UNIQUENESS
THEOREMS
619
The more general problem with mitral value x,, = s(O) can be reduced to the above form by taking u(t) = x(t) - x0 and gl(t) = g(t) + by,, . For the moment we shall assume that K, b, g, 0 and 4 are C-l functions on [0, H] (n > 2), that 0 < e(t) < t for t E [0, H] and that 0 < C(t) < t for t E (0, H]. To facilitate discussion of (26) we need some further notation. Let k’ = {X E C”([O, H]) : X(O) = 0} and let W = {x E V : x(‘)(O) = 0 for 0 < j < n}. Define L : V - V by (Lx)(t) = J”: (k(s)x’(+s) + b(s)x(&)) ds. It = is clear that (26) is solvable iff h(t) = Jig(s) ds is an element of R(I -L) the range of I -L, and this solution is unique iff N(I -L) = the null space of I - L is 0 dimensional. Thus it suffices to study N(I -L) and R(I - L). First, we need a trivial lemma (see [5, p. 3441). LEMMA 5. Let E denote the set of polynomials equal to n. Then V = E 8 W.
in k’ of degree less than or
Proof. Ifp(t) = C,:, a$? E E n W, thenpo)(O) =j! a, = 0 for 1
Proof. For 0 < r < 1, define f,. : [0, H] x R x IF8-+ R by fr(t, x, y) = rk(t)y + rb(t)x and note that / af,./ay j = 1rk(t)j. Since j rk(O)l ](+‘(O))+l1 < 1, it follows by Proposition 3 that for any compact set M C I -, {x E V : x - rLx E M and 1)x /j < l} is compact. By Lemma 3.1 of [14], this implies that R(I - rL) IS closed and N(I - rL) is finite dimensional, i.e., 1 - rL is a semi-Fredholm map for 0 < r < 1 (see [6] for definitions). By the continuity of the index for semi-Fredholm maps, index (I -L) = index 1 = 0, so I -L : V - V is a Fredholm map of index zero. Since W is a closed subspace of I; it is obviously also true that for any compact subset MC W, {x E W : x - rLx E M, j/ x/I < 1) is compact. Thus since L : W--t W, the same reasoning shows that (I - L) ] W is Fredholm of index zero. Let h be any element of IV, To show there exists x E W such that (I- L)x = h, it suffices to show that there exists x E W such that x’(t) = k(t)x’($t) + b(t)x(Bt) + g(t) for 0 < t < II, g(t) = (dh/‘dt)(t). Define and notice that 0 = p(O, 0, 0) and fx + g(t)
620
NUSSBAUM
(8j/St~)(O, 0,O) = 0 for 1 < j < n - 1. It follows from Remark 1 after Theorem 1 that there exists 6 > 0 and x E C*([O, S]) such that x(j)(O) = 0 for 0 < j < n and x’(t) = f(t, x(&), x’($t)) for 0 < t < 6. However, jmeets the conditions of Theorem 2 so x can be extended to an element of C”([O, H]) such that x’(t) =f(t, x(&), x’(+t)) for 0 < t < H. This implies that (1 -L) 1 W is onto W, and since (I -L) 1 W is Fredholm of index zero, (I- L) 1 W is one to one and onto. It remains to investigate dim(N(.Z - L)). Let E, = {p E E : (I - L)p E W}, where E is as in Lemma 5. We claim that dim(&) = dim N(I -L). To see this, forp E E, define S(p) E N(I- L) by S(p) = [(I -L) 1w]-l(1L)p - p; S is one to one, since S(p) = S(g) for p, q E E. implies that p - q E W or p = q. It follows that dim(E,,) < dim N(I - L). Conversely, if x E N(I - L) and x=u+p for UGW and PGE, then PEE,,, smce (I-L)p= we can define T(x) =p E E,; -(I - L)u E W. Thus, for XEN(I--L) Tisonetoone,because~fu+p,v+pEN(I-~)foru,v~Wandp~E, (I - L)(u - U) = 0, so u = v. Therefore we have dim N(I - L) < dim(E,,), so the two are equal. If p E E and p(t) = C,“=, (u,fi!)t’, p E E,, iff qfj-l)(O) = 0 for 1 < j < n, where q(t) = p’(t) - k(t)p’($t) - b(t)p(&). It 1s easy to see that q(j-l)(0) = zi=, c,,P(~)(O), where the c,, are constants and c,, = (1 - C~J-l). The assumptions we have made imply that 0 < m < 1 and 1cmn-l 1 < 1, so that cm3 = 1 for at most onej, 0 < j ,( 1z- 1. The matrix (c,,) is triangular, so if cj3 # 0 for 1 ,( j < n, (cZ3)is nonsingular and dim(E,) = 0. This shows thatifcm~#1forO~j,(n-2and~cm”-r~<1,I--Lisonetoone and onto I/. If cjj = 0 for some j, 1 < j < n - 1, one can see directly that dim(E,,) = 1; in fact in this case a, , aa ,..., a,..., are zero, a, is arbitrary, and Q.E.D. a,,, ,a*., a, are linear functions of a, . Remurlz 3. If b = 0 and the hypotheses of Theorem 4 hold (m particular cm3 # 1 for 0 < j < 71- 2), one can in some sense explicitly solve (26). First assume that g(?)(O) = 0 for 0
n-1 &> = g(t)- c (dw)/~!) t3 and let 12-l
p(t) = 1 (gCJ)(O)/j!)ti. 3=0
EXISTENCE
AND
UNIQUENESS
THEOREMS
621
Then solving (26) is equivalent to solving the followmg two equations on
PI HI: %‘(4= W~l’W) + g1(4, Xl(O)= 0, x2’(t) = k(t)x,‘(+t) + p(t), Equation 0 < j < will be wM44 following
“z(O) = 0.
(27) cw
(27) can be solved as described above, since go)(O) = 0 for n - 1. To solve (28) let q(t) = J& b,t*, where the numbers 6, selected later. By replacing x2 by xs - 4 and defining h(t) = - 4’(t) + f-e>> one can easily show that (28) is equivalent to the Eq. (29):
By our previous remarks (29) can be solved if b, , b, ,..., b, can be selected such that ho)(O) = 0 for 0 < j < n - 1. If one evaluates ho)(O) one finds that because cm? # 1 for 0 < j ,< n - 1, there exist unique numbers h , b, ,***, b, such that h(j)(O) = 0 for 0 < j < n - 1. Remark 4. If 1cm”-r 1 > 1, then N(I - L) is, in general, infinite dimensional. For example, consider the following equations: x’(t) = cx’(mt) + g(t),
o
x(0) = 0.
(30)
Assume that c and m are constants, 0 < m < 1, 1cmn-l / > 1 and g E C+r([O, 11). Let y,, be any C” function with compact support in (m, 1). For t EI~ = [m2, m], define yr(t) = (l/c)ya(t/m) and generally if y,(t) has been defined for t E I, = [rnj+l, ml], define y,+,(t) = (l/c)y,(t/m) for t E 1,+r . If we define y(t) = y,(t) for t E I, , then clearly y is a C” map from (0, l] to R. Also, using the assumption 1crn+l 1 > 1, one can show that lim,,,yo)(t) = 0 for 0 < j < n - 1, so y E C+l([O, l]), if we set y(0) = 0. If we define x(t) = $ y(s) ds, x E P([O, l]), x satisfies the homogeneous equation (30) with g = 0, and the collection of such x is infinite dimensional. If cm3 # 1 for 0 < j < n - 2 and 1cm+l / > 1, one can always find x E Cn([O, 11) satisfyin (30). We outline the proof and leave the details to the f reader. If g(t) = CyzO up is a polynomial, (30) can be solved explicitly: X(t) = 2ITr,’ [a,/(j + I)(1 - c@)lt ?+I. For general g, we can assume that g(‘)(l) = 0 for 0 < j < n - 1 by subtracting the polynomial n-1
2 LPYl)/M - 1Y
622
NUSSBAUM
(for which we can solve the equation). Now we use an Idea of D. Orth [13]: for t E [mP+l, mfl), define
x(t) = f (g
J:-, g(s)ds*
2=1
Using the fact that g(‘)(l) = 0 for 0 < j < n - 1, one can prove that x so defined is n times continuously drfferentiable on (0, 11; and usmg the assumption that 1cmn-l 1 > 1, one can show that lim,,, x(‘)(t) exists for 0 < j < n. Finally, one can check that x(t) - x(0) satisfies (30). We conclude our discussion with a trivial apphcation to the linear case of Theorem 3. THEOREM 5. Let k, b, g and 19: [0, t*] --) R be continuous functions and let 4 : [0, t*] -+ [w be a C’ function. Assume that 0 < 6(t) < t fey t E [0, t*], 0 < 4(t) < t joy t E (0, t*) and $(t*) = t*. Suppose that 1k(t*)/$‘(t*)l < 1. Then if x : [0, t*) + IF!is an absolutely continuous function such that x’(t) = w)x’(4t) + wxw + g(t) f OY almost all t E [s, t*), where 0 ,< s < t* and 4’(t) > 0 joy t E [s, t*], llmt_tt* x’(t) exists.
Proof. Take 6 > 0 so small that if n/r, = sup{/ h(t)/ : t E [t* - 6, t*]}, M&‘(t*) < 1. Then if we define f(t, x, y) = b(t)x + K(t)y + g(t), Theorem 5 follows immediately from Theorem 3. Theorem 5 is the best possible result, as we can see by the following simple example: Example. Consider the equation x’(t) = 2x’(F) + tZ, x(0) = 0, for 0 < t < 1. One can verify directly that x(t) = f
01% = 2*+1 + 1
+ f 2” $ , kl
satisfies the equation and is infinitely lim,,, x(t) = co.
differentiable
on [0, 1). However,
REFERENCES 1. A. L. BADOEV AND B. N. SADOVSKII, An example of a denslfymg operator m the theory of differential equations with a deviating argument of neutral type, So&et Math. Dokl. 10 (1969), 724-728 [Dokl. Akad. Nauk. SSSR 186 (1969), 1239-12421. 2. M. A. CRUZ AND J. K. HALE, Existence, umqueness and continuous dependence for hereditary systems, Ann. IMat. Pura. Appl. 85 (1970), 63-82. 3. G. DARBO, Punt1 umti m trasformaziom a condlmimo non compatto, Rend. Sem. Mat. Univ. Padova 24 (1955), 84-92.
EXISTENCE
AND
UNIQUENESS
THEOREMS
623
4. R. D. DRIVER, Existence and contmuous dependence of solutions of a neutral functlonal dlfferentlal equation, Arch. Rat. Mech. Anal. 19 (1965), 149-166. 5. N. DUNFORD AND J. T. SCHWARTZ, “Linear Operators,” Part I, Intersclence, New York, 1958. 6. T. KATO, “Perturbation Theory for Linear Operators,” Springer-Verlag, New York, 1966. 7. C. KURATOWSKI, Sur les espaces complets, Fund. Math. 15 (1930), 301-309. 8. W. R. MELVIN, “A class of neutral functional dlfferentlal equations,” Ph.D. Dlssertatlon, Brown Umverslty, Providence, RI, 1971. 9. R. D. NUSSBAUM, The fixed pomt index and asymptotic fixed pomt theorems for k-set-contractions, Bull. Amer. Math. Sot. 75 (1969), 490-495. 10. R. D. NUSSBAUM, The fixed point mdex for local condensing maps, Ann. Mat. Pma Appl. 89 (1971), 217-258. 11. R. D. NUSSBAUM, Degree theory for local condensing maps, J. Math. Anal. Appl., to appear. 12. R. D. NUSSBAUM, A generahzation of the Ascoh theorem and an apphcatlon to functional differential equations, J. Muth. Anal. Appl. 35 (1971), 600-610. 13. D. ORTH, A numerical method for some neutral functlonal dlfferentlal equations, to appear. preserved under addltlon of a 14. B. YOOD, Propertles of lmear transformations completely contmuous transformation, Duke Math. J. 18 (1951), 599-612.
OF DIFFERENTIAL
EQUATIONS
Existence Some
Functional
11,
607-623 (1972)
and Uniqueness Differential
Equations
ROGER D
Department of Mathematm,
Theorems of
for Neutral
Type*
N~SSBAUM
Rutgers Unzverszty, New Brunswzck, New Jersey 08903 Recewed September
13, 1971
INTRODUCTION
This article is motivated
by the simple functional
x’(t) = kx’(mt)
differential
0 < t <; H, s(O) = 0.
+ g(t),
equation
(1)
In this equation k and m are constants, 0 < M < 1, and g is a C” function. The work of Cruz and Hale [2] or of Badoev and Sadovskn [I] can be applied to this equation, and both approaches yield the same answer: If 1k/m / < 1, there is a unique absolutely contmuous function with L1 derivative satisfying (1), and if 1k/m 1 > 1, no information on existence or umqueness IS provided. Nevertheless, If / k/m 1 > 1 one can easily show directly that (1) has an absolutely contmuous solution on [0, H] and in fact has an infinite-dimensional subspace of solutions (see [ 131). Th us it is clear that the Cruz-Hale or Badoev-Sadovskii approaches miss a lot of information in this case, and the question is if some of this information can be retrieved. The answer is yes. We shall show below (as a very special case) that if n > 2, / kmn-l ( < 1 and km’ f 1 for 0 < j < n - 2, there exrsts a unique solutron of (1) in C”([O, HI). It has been known for some time that the solutrons of even linear functional differential equations may have nasty discontmuitres m their derivatives, and this has encouraged existence and uniqueness theorems in settings like the Sobolev spaces IP”, 1 <. p < 00. The moral of our work here is that for certain “nice” neutral functional different equations (NFDE’s) one may lose a great deal of information (both existence and uniqueness) by looking only for absolutely continuous solutions. For some NFDE’s obtaining sharp existence and uniqueness results demands a closer examination of drfferentrabrlity properties of the equation than has been customary. The organization of this paper is as follows: In the first section we consider a class of nonlinear NFDE’s for which the standard results m the literature * Partially
supported
by NSFGP20228.
607 0
1972 by Academtc Press, Inc.
608
NUSSBAUM
grve weak and inadequate exrstence and uniqueness theorems. We prove a local existence theorem, some continuation theorems, and a few other results. In the second sectron we apply the nonlinear theorems to obtain theorems for linear NFDE’s. Surprisingly, we obtain sharp results for the linear theory in this way. 1. Let H be a posrtive real number and let f : [0, H] x R x R ---f R be a function which has continuous partials of all orders up to n - 1, 71 > 2. Let fL , i = 1, 2, 3, denote the partial off with respect to its z-th variable. If m is a positrve integer, let Cm([O, H]) d enote the Banach space of m times continuously differentiable functrons x from [0, H] to R, and define
C([O, H]) is the Banach space of continuous the usual sup norm. Let 8, + E C-l([O, H]) for t E [0, H] and 0 < 4(t) < t for t E (0, H]. there exists a > 0 and x E C”([O, a]) = X equations: x’(t) = f(t, x(qt)),
x’(e(t)>>,
functrons from [0, H] to R in and assume that 0 < 0(t) < t We want to investigate whether which satisfies the following
O
(2)
x(0) = xg . We also want to investigate how far solutions of (2) can be extended and when closed, bounded sets of solutrons of (2) in X are compact. The basic tool we shall use is the “measure of noncompactness,” an idea due to Kuratowski [7]: DEFINITION [7]. Let (X, p) be a metric space and A a bounded subset of X. The measure of noncompactness of A 5 y(A) = inf{d > 0: there exists a finite number of sets S, , Sa ,..., S, such that A = ur=r S, and diameter
(SJ G 4. DEFINITION [7]. Let (XI , pr) and (X, , pa) be metric spaces, let G be a subset of X and let f : G --f X, be a continuous map. We say that f is a “k-setcontraction” if for every bounded set A C G, f(A) is bounded and y,(f(A)) < ky,(A), where yz denotes the measure of noncompactness in X, . The basic results we shall need are contained in the following propoation, which was proved by G. Darbo [3]: PROPOSITION 1 [3]. Let X be a Banach space and y the measure of noncompactness induced by the norm.
EXISTENCE
AND
UNIQUENESS
THEOREMS
609
(1) If A and B are bounded subsets of X and A + B = (a + b : a E A, b E W, y(A + B) < ~(4 (2)
+ Y(B)-
If A is a bounded subset of X and Z(A)
= convex closure of A,
rW4
= r(A). (3) If G is a closed, bounded convex subset of X and f: G + k-set-contraction with k < 1, f has a$xed point.
G zs a
Of course the reason for the nomenclature “measure of noncompactness” is that if M is a subset of a complete metric space X, y(M) = 0 iff A/I has compact closure in X. Before consrdering (2) we need a few simple lemmas. For a > 0 and m > 1 let X = P’([O, a] and let Y = C([O, a]) (with the norms already mentioned). Let y denote the measure of noncompactness m X and y1 that in IT. LEMMA
1. Let S be a bounded set an X. Then if J : X -+ Y is dejned by
Jx = x(") 9 Y(S)
Proof.
= n(J(W
Take E > 0. Since S is bounded, A = {(x(O), x’(O),..., x’m-l)(o)) : x E sj
is a bounded set in R”. Take as the norm in R”, /I v 11= maxsG,G;,-, 1vu,I, where v = (vs , vI ,..., v&, and let vz, 1 < i < N, be an c/2 net in A with respect to this norm. For 1 < i < N let s, = (x E s : 11(x(O),x’(O),..., x(+1)(0)) - 0%jj < 42). Suppose that n(J(S))
= d. Then ,(J(S,))
< d, and we can write
J(S,) = ilj T,,, 3 pl
where diameter (Tz,3) < d + E. Let S,,, = {x E S, : Jx E T,.,}. We claim that diameter (S,,,) < d + 2~, for if
x, y E s,,, , (su$, I .yt)
- Y’m’(t)l G d + E
and
so that /j x - y /I < d + 2~. Since S = US,,, , this shows that y(S) < d + 2r, and since E > 0 is arbitrary, y(S) < K( J(S)). The opposite inequality is obvious. Q.E.D. 505/11/3-11
610
NUSSBAUM
LEMMA 2. Let R and S be bounded subsets of Y (Y as in Lemma 1) and let T = (xy :x E R, y E S}. Let M, = sup{// x )/ :x E R} and let Mz = SUP{IIYII :Y E sl- Then ~~07 d Jfd‘9
+ Ma(R).
Proof. Suppose y,(R) = dl and yr(S) = d, . Then grven E > 0, R = (Jz, R, with diam(R,) < dl + E and S = (J,“=, S, , diam(S,) < d, + E. It follows that T = (JI,> Ri,S, , where R,S, ={xy:xx~R,,y~SJ. If zr = xryr
and xa = s,y,
are elements of R,S, ,
This imphes that diam(R,S,) < Mid, arbitrary, rl(T) < Mid, + n/r,d, .
+ Mzd, + (Ml + M&,
and since E 1s Q.E.D.
LEMMA 3. Let S, and S, be bounded subsets of Y and assume that y,(S,) = yl(S,) = 0. Let g : [0, a] x Iw x Iw ---, [w be a continuous map and dejne s = {z : x(s) = g(s, x(s), y(s)), x E s, ) y E As,). Then yl(S)
= 0.
Proof. This is an easy consequence of the Ascolr-Arzela uniform continuity of g on bounded sets.
theorem and the Q.E.D.
The way in which we shall use these Lemmas 1s as follows: Suppose S 1s a bounded subset of X = C*([O, HI), n 3 2. Then by the Ascoli-Arzela theorem {X(J) : s E S} is precompact in Y = C([O, H]) for 0 < j < n - 1. If 0 and 4 are continuous maps of [0, H] to [0, H] it 1s also clearly true that {XC” 0 4 : x E S> and {X(J) o B : x E S> are precompact in C( [0, HI). By Lemma 3, if g : [0, H] X [w X Iw + [w is a continuous function, {,z : z(t) = g(f, x(&), x’(+t)), x E S} is precompact in Y. By Lemma 2, {z : z(t) = g(t, s(h),
x’(~t))(x(‘)(Ot))(x(k)(~t)), x E S} is precompact in Y ifj s< n - 1 and R :< n - 1 and yl({z : z(t) = g(t, x@), x’(~t))x’“‘(~t), < sup{/ g(t, x&Q), x’(&))l = x’“‘(&),
x E As> : 0 < t & H, x E S}y,({z
x E S}).
We can now prove our first nontrivial
: z(t) (3)
result.
EXISTENCE AND UNIQUENESS THEOREMS
611
PROPOSITION 2. Let f : [0, H] x R x R -+ R be a function with n - 1 continuouspartials, n 3 2. Assume that 0,4 E C+l([O, HI), that 0 < 6(t) < t for t E [0, H] and that 0 G+(t) < t for t E (0, H]. Let X = C”([O, H]) and Y = C([O, HI). Suppose B zs a closed, bounded subset of X and define F: B ---f X by (Fx)(t) = x(0) + $,f (s, x(&), x’(&)) ds. Assume that for all x E B, / f3(0, s(O),X'(O))(+'(O))~-l 1 < c < 1. Then S = {x E B :Fx = x} is compact (possibly empty). Also, there exists Y > 0, r & H, such that if x = {x 1 [0, r] : x E X}, 4 = {x 1 [0, r] : x E Bj and P = F 1B, P is a k-set-contraction for some k < 1.
Proof. Since F(S) = S, F”(S) = S for all integers m > 1 and thus y(F”‘(S)) = y(S). Therefore to show S is compact (it is closed because F is clearly a continuous function), it suffices to show that for some constant k < 1 and some integer vz, y(F”(S)) < KY(S). Before going further we make some observations which will be useful later in the proof. If s, = sup(fl(t) : 0 < t < H}(@ = the j-th iterate of 4) note that s, -+ 0, because $(t) < t for t > 0. Let C, = sup{\ f (t, x(&t), x’(#+lt))(#‘(~ct))n-l j : 0 < t f H, x E B}, i > 0. Since lim,, x(s) == x(O) and lim,,, x’(s) =: x’(0) uniformly for x E B and because F(t) < s, , it follows that hm,,, C, = /f (0, x(O), x’(O))($‘(O))+l / < c. It follows that if we define k, = Hi=, C, , there exists an integer m such that k, < 1 forj > m. For an integer m let S, = F”(S). We wish to estimate r(Sm). By Lemma 1, rG%J = rdJ&d = r~Ur;‘(Sm-I>)- BY a direct calculation one can see that there exist continuous functions gl,,,k : [0, H] x lR x R --f 52 and h 1,3,1e: [0, H] -+ [0, H] indexed by integers j, k, 1 such that 0 < j < n - 1, 0 < k < n - 1 and 0 < 1 < l(j, k) such that for any x E X,
+ f3(t, x(q, 4+(t~kw)~-~
x(yw
It follows then by the remarks before this proposition rl(JF(Sm-1))
= df
(4) that
3(t , x,-l(et), x~-,(+t))(+f(t))“-l&~l(4t> : k1 E&A)
9 corl({~~Q~t): G-1 E&n-S). If we now write that
S,-,
= F(S,-,)
Yb%J
G
Gcl~l({4z2(+2t)
(5)
and apply the same calculation,
: %-2E
&-21).
we find
612
NUSSBAUM
Continuing in this way we eventually find that
By our previous remarks k,-1 < I for m large enough, so we have proved the first part of the proposition. Since lim,,, x(s) = x(O) uniformly for x E B and similarly for x’, there exists r > 0 such that sup{1MY 4% +W#‘(W1 (7) -f3(0, x(O), x’(O))(~‘(O))‘+~ / : 0 < t < r, x E B) < (1 - c)/2. For this r, if B and 2 are as above and A is any bounded subset of B, r(-P(A)) = r1U%4> = n((f3(t, x(&), d(&))(+‘(t))“-’ < sup{/f,(t, @‘t), x’(+t))(4’(t))“-1 x
T4W'
*c
LX(~)(#) : x E A, 0 < t G r>) I : x E A, 0 9 t < r>
:*~-49 d [Cl + WlA4
(8)
It follows that P is a k-set-contraction with k = (1 + c)/2.
Q.E.D.
Our next result is a slight sharpening of Proposition 2 which will be useful later for the linear theory. PROPOSITION 3. Assume that f, 8, and 46are as in Proposition 2. Suppose in addition that there exists a function k: [0, H] -+ R such that for all (x,, , y,,) E R, IfJt, x,, , yo)j < k(t) and 1k(0)($‘(O))“-l ) < 1. Then if M is any compact subsetof X and B is any closed,bounded subsetof X, S = {x E B : x - FX E M) is compact.
Proof. In analogy with Proposition 2, let Sl = F(S) + M and generally S, = F(S,-,) + M for j > 1. Since S, > S, r(S,) > y(S), so to prove the proposition it suffices to show that there exists c* < 1 and an integer m such that y(S,) < c*r(S). The proof mimics that of Proposition 2, but since we no longer know that S, C B, the stronger assumptions are needed. Let C, = supoGtGH [ k(+t)[$‘(+t)]+l 1 and note that lim,,, C, = 1k(0)[4’(O)]+1 1 ( 1. It follows that there exists an integer m such that ~~~oC3~c*<1forr~m-1. Using the results of Proposition 1 we find r(&J
< rPVm-1)) = ?4if3(t, G
GYlGG#J
+ Y(M) = r(W,-1))
x,-l(et),x~-l(~t)>(q(t))"-'X~l(~t) : Xm-1E
sn-3).
: %-1E
s!-1:
(9)
EXISTENCE
AND
UNIQUENESS
613
THEOREMS
Since xmel = F(x,-.J + y for some xm-s E S,-s and some y E M and since y1((y(“) 0$ : y E M}) = 0, we find that
C,r,({x%
a c : G-1 E &4H
=
Co~#%n-z)(a)~~
=
COyl({f#,
<
coc,yl({+'"'
: xm-2
x,-2(e$t),
0 $"
'rn2
E &-21)
xL-2(Qt))[~(~t)l"-1X~~2(~2t)
:
: %-2
x,-2 E s,-2)).
E %n-21)
(10)
Continuing in this we find that
(11) Q.E.D. We can now estabhsh a local existence theorem for our original FDE (2). In the following theorem, fs indicates partial differentiation with respect to the third variable. THEOREM 1. Let f : [0, H] x Iw x [w --+ R be a Cn-l function, n 9 2. Let 8, q5 : [0, H] 3 R be Cn-l functions such that 0 < O(t) < t for t E [0, H] and 0 < C(t) < t for t E (0, H]. Let x,, E R be given and assume there exists x1 E R such that Xl
=
f(O,
x0
9 XI),
f3(0, x0 I xJ($‘(W
f
1
for1
= fl(o, x0 , x1) + f,(o, x0 , xl)waxl
+ [f3(op x0 , x,)d~0)]xv),
(12)
and since f3(0, x0, x,)$‘(O) # 1, X(~)(O)is uniquely determined, X(~)(O)= xL . Generally, if x(?)(O) = x3 is determined for 2 < j < m, m < n, @+l)(O) = (function of x0 , xi , .. ., x,) + [ f3(0, x0 , xl)(+‘(0))~]x(“+l)(O),
(13)
614
NUSSBAUM
and since 1 # fs(0, x0, x~)($‘(O))~ for 1 ,< m < 72,@‘+1)(O) is uniquely determined, ~@+l)(0) = x,+i , If we define Kb = (x E C”([O, b] : x(j)(O) = x, for 0 < j < n}, it is not hard to see that F : K, -+ Kb . Select a fixed number R > 0 and let V, = {x E Kb : 11x - p 11< R}, where p(t) = Cy=,, (x,/j!)tj. By Proposition 2 there exists b* > 0 such that F ( V,, is a K-set-contraction, K < 1, and in fact the same proof shows F j V, is K-set-contraction for 0 < b < b *. If we can find b > 0, b < b* such that F( V,) C V, , then by Proposition 1, F will have a fixed point, and we will be done. Thus consider [jFx - p 11= SUP~<~<~j(Fx)fa’(t) - x, / for x E V, . As was remarked in the proof of Proposition 2, one can write (i++“‘(t)
=
C c g,& ,,h
x(W x’(+)) h,&> ~(~‘(et) x’“‘(+)
t f3(& x(@f), x’W))(#(W1
x’“‘(@),
(14)
where the g,,, and h,,, are certain continuous functions. As usual, since lim,, X(J)(S):= X, uniformly for all x E VH and j < n , one can see that for b small enough,
and
- f3(0, x0 , x1)(+‘(O))-
x@‘(O)1 f
w
R.
It follows that for this b, sup 1I(Fx)‘“‘(t)
- x, /j =
0
Thus we have F(VJ C V, .
sup l/(F~)(~‘(t) - (Fx)(“‘(O)ll
< R.
(17)
OCt
Q.E.D.
Remark 1. Suppose all the hypotheses of Theorem 1 hold except the assumption that fa(O, x,, , x1)(4’(O))? # 1 for 1 < j < n - 2. Assume instead that 0 = f(0, x0 , 0), that 8f/%(O, x0 , 0) = 0 for 1 < j < n - 1 and that If&4 x0 3O)(+‘(O))+i ( < 1. Then if we define x, = 0 for 1 < j < n, it is not hard to see that if x E Cn([O, b]) and G(O) = x, for 0 < j < n, then (Fx)cj’(O) = x, for 0 < j < n. If we now use the above proof, it follows that the equation (2) has a solution x E Cn([O, b]) for some b > 0.
EXISTENCE
AND
UNIQUENESS
THEOREMS
615
Remark 2. If in addition to the hypotheses of Theorem 1 one assumes that all the partials off of order n - 1 are Lipschitz in their second and third variables on bounded subsets of [0, H] x R x R, then one can show that if x and y are solutions of (*) with x’(O) = y’(0) = x1 , x and y must agree where they are both defined. In this case one can use the contraction mapping principle rather than Darbo’s theorem. We want to make some additional assumptions on f which will guarantee that the local solution given by Theorem 1 can be extended to a solution x E P([O, HI). We begin with a simple lemma. LEMMA 4. Let f, 8, and 4 be as in thejirst paragraph of this section. Suppose that x E C”([O, a]), y E C”([O, b])(O < a < b) and x and y satisfy (2). Assume that there exists 6 > 0 such that x(t) = y(t) for t E [0,6]. Then x(t) = y(t) for 0 < t < a. Furthermore, ;fz E Cn([O, d]), 0 < d < H, and x satisfies (2), there exists E > 0 and z1 E C”([O, d + ~1) such that z, extends x and satisJies (2).
Proof. Let c = sup{t : X(S) = y(s) for 0 < s < t}, c > 6. We have to show that c = a. Suppose not, so c < a. Since t - 4(t) > 0 for t E [c, H], there exists 7 > 0, 7 < a - c, such that t - 4(t) >, 27 for t E [c, W]. For t E [c, c + 71 and u E iw define a Cl function g by g(t, u) = f (f, u, x’(@)) = f (t, u, y’($t)). It follows by the standard theory for retarded functional differential equations that there exists q’ > 0, q’ < 7 and a unique function u E Cl([c, c + 7’1) such that u j [O, c] = x j [O, c] and u’(t) = g(t, u(t)) for t E [c, c + 7’1. Since x’(t) = g(t, x(t)) and y’(t) = g(t, y(t)) for t E [c, c + 7’1, x(t) = y(t) = u(t) for t E [c, c + 7’1, a contradiction. To prove the second half of the theorem let z E C”([O, d]) satisfy (2) and let uj = ,&j(d) = the j-th left-hand derivative of x at d, 0 < j < n. For E < min(q, H - d), let K, = (X E C”([O, d + c]) : x / [O, d] = x) and define F: K, ---f K, by (Fx) / [0, d] = z and (Fx)(t) = x(d) + slf (s, LX(&), x’($s) ds for t E [d, d + ~1. One has to check that Fx E C”([O, d + e]) for x E K,, but this follows without trouble because +(t) < d for t E [d, d + ~1. Take R > 0, let p(t) = x(t) for 0 .< t < d and p(t) = Cy=, (u,/j)(t - d), for t E [d, d + E] and define V, = {x E K, : /I x - p j( < R). We claim that for E small enough F: I’, + V, . We have
IIFx -P II = tE;;;+f,l(F4(n)(t) - (Fx)(“)(d)l, and just as in the proof of Theorem 1, to show /IFx - p jj < R for all x E VG, it suffices to show that there exists c < 1 such that
SUP{1 f&S 4% -
x’(4t))(~‘(t))“-1x’n’(+t)
f3(d,x(0d),
x’(+d)(qS(d))“-‘x’“‘(+d)[
: t E [d, d + E], x E V’,: < CR.
(I 8)
616
NUSSBAUM
However, in our casethis is trivial because ~c(“)($t) = P)($t) for t E [d, d + ~1, so the above supremum actually approaches 0 as E > 0. Thus for c small enough, F : V, -+ V, . To complete the proof it suffices to observe that F(V,) is compact, for by the usual argument, y(F(V,)) = yd{w E C(P, d + 4) : w = (W”),
x E VJ)
= n(iw ECC4 d + 4 : w(t) = f3(t,x(h), x’(~t))($6’(t))“-?z’“y+t)}) = 0.
(19)
By the Schauder fixed-point theorem, F has a fixed point z, in V,, and a1 Q.E.D. gives the desired extensron of x. THEOREM 2. Letf:[O,H] x 172x aB-+[W,8:[O,H]-,aBand~:[O,HJ-,[W be Cn-l functions, n > 2. Assume that 0 < e(t) < t for t E [0, H] and 0 < 4(t) < t for t E (0, H]. Suppose that for any constant C > 0, there exist constants Ml and M, (depending on C) such that If(t, x, y)/ < Ml + Mzl x 1 for (t, x,y) E [0, H] x !R x [-C, C]. Then $ u E C”([O, b]), 0 < b < H, sutis$es (2), there exists a unique function x E C”([O, H]) such that x j [0, b] = u and x satisfies (2).
Proof. Using the first part of Lemma 4 and a standard argument there exists a unique maximal solution of (2) such that x ( [0, b] = u and x is n times continuously differentiable on its domain. The function x is defined on an interval [0, r) or [0, Y]. If x is defined on [0, r], r = H, and we are done. Otherwise, by Lemma 4, there exists E > 0 and an extension f E C”([O, Y + 61) which satisfies (2), contradicting the maximality of x. Thus to complete the proof it suffices to show that if x is defined on [0, r), x has a Cn extension to [0, r]. Since +([O, R]) C [0, s] for some s < Y, we have I x’(+t)l < C for t E [O, r) and If(t, 4% x’(+t))l < K + J&l de(t)1 for t E [0, r). Therefore we find for t E [0, r)
1@>I G I xo I + j;'"' I x'(s)1ds < I x,, I + j: 11% x(84, x'(+s))lds < I xc,I + Md + M, jt I x@)I ds.
(20)
0
By Gronwall’s
inequality
this implies
1x(&)1 < const for t E [0, r),
1x’(t)1 = lf(t, x(et), x’(+t))l G const It follows that lim,,, fimt,,Jlt, xw,4#4) of (2), we find lim,,, O
for
t
SO
E [0, r).
x(t) = x0 exists; so we find that lim,,,- x’(t) = = f(y, 4w, x’W> exists. Differentiating both sides x(z)(t) exists and generally lim,,,- xW,t) exists for Q.E.D.
617
EXISTENCE AND UNIQUENESS THEOREMS
We have assumed above that $ has a unique fixed point on [0, H], and this has played an important part in the proofs. We shall now assume that C$has a second fixed point t* > 0 and examine the behavior of the solution x as t approaches t* from the left. In the following theorem, +-‘(t*) denotes the left-hand derivative of & at t*. THEOREM 3. Let f : [0, t*] x [w x R + R and 0 : [0, t*] + R be continuous functions and let $ : [0, t*] -+ R be continuously differentiable. Assume that 0 < O(t) < t for t E [0, t*], that 0 < 4(t) < t for t E (0, t*), and that +(t*) = t*. Assume that there exists s < t* and constants Ml , M, , MS such that for t E [s, tJ and (x, y) E R x IF!4’(t) > 0 and
If (t, x, Y)I < Ml + Mzl x I + M,I Y I> where j M&5-‘(t*)i < 1. Then zf x : [0, t*) -+ R is any absolutely continuous function such that x’(t) = f (t, x(&), x’($t)) for &most all t E [s, t*), lim,,,, x(t) exzsts. Proof. By the continuity of (b’ there exists s*, s < s* < t*, such that 1M.&‘(t)1 < c < 1 for t E [s*, P]. By taking s* closer to t* we can also assume that M2(t* - s*) < 1 and c/[l - M2(t* - s*)] = h < 1. Since 4 is one-to-one on [s*, t*] and since+(t) < t for t E [s*, t*], let s1E [s*, t*] be the unique number such that +(sr) = s* = s,,. Generally, let s, be the unique number in [s*, t*] such that $(sn) = s,-r . The sequence {sz} is strictly monotone increasing and limn+m s, = t*. Define a, = ,,,+I 1x’(t)1 dt. To show that lim,,, x(t) exists, it suffices to show that Ji: j x’(t)\ dt exists or, equrvalently, that ZN = xf=, a, is bounded as N --f co. Define M = supastGs* j x(t)(. We then find for n > 1
a, = ,I:”
j x’(t)] dt = j::”
< M&,+,
If (t, x(et), xl(+))\
dt
- s,) + M2 j”” / x(&)1 dt + Ma ,I:” sn
I x’(+t)l dt.
(21)
By a change of variables we find that M3 j::”
j x’(#t)l dt < c ,I:.,
Fort E[sn, s,+,lwe see that I x(Wl < I +*)I +
1x’(s)/ ds = ~a,-, .
I x(&)1 < M if O(t) < s* and
j;r)I x’(r)1dr < M + 2=0 f a,
if
e(t) > s*.
(22)
618
NUSSBAUM
In either case we find that / x(&)1 < M + z,“=,, uL for t E [So , sn+r]. Plugging in these estimates we find that
an < Jqs,+1 - sn)+ w&2+1- 4w + 421+ %-1 . Adding these inequalities right we find that
for 1 < n ,( N and replacing
Z;, by 2,
(23) on the
.zNG ao+ adze+, - s*) + hh(~~+~ - s*)w + 44 + 4-1 . Transposing (1 - M&N+1
the ZN term on the right to the left side and dividing - s*)) > 1 - M,(t* - s*) gives
(24) by
zl, < [a0+ hfl(~N+l- S*) + M&,~+I - s*)][l - nf&* - s*)l-’ + ‘&v-l < [a, + (Ml + M,)(t* - s*)][l - &(t* - S*>l-’ + &v--1 .
(25)
If we denote the first term on the rrght by R, we have that .Z’, < R + kZNMI . Iterating this inequality N times gives N-l
ZIN <
c
kIR + kNuo < R/(1 - k) + a,, .
Q.E.D.
,=O
The reader may have noted that we have avoided considering the case when f, 6’ and + are only continuous (n = 1). Actually, we can also obtain results in this case, though the methods requrred are somewhat different. We mention the following result, though for reasons of brevity we shall not prove it here. THEOREM. Let f : [0, H] >c DBx [w -+ Iw, t3 : [0, H] -+ Iw and+ : [0, H] + Iw be continuousfunctions. Assume thutfor t E [0, H], 0 < l?(t) < t and 0 <4(t) < t. (Notice that q5 may now have $.xed points other than 0.) Assume that for all x, u, v E Iw, ]f(t, x, u) -f (t, x, v)j < k(t)1 u - v I, where k is a continuous function such that k(t) < 1 for all t such that d(t) = t. Finally, ussnme that there exist constants MI and n/l, such that /f (t, x y)l < M, + M2j x 1 + k(t)/ y / for all (t, x, y) E [0, H] x [w x 02. Thengiven x0 E Iw, there exists x E Cl([O, H]) such that x(O) = x0 and x’(t) = f (t, x(&), x’($t)) for 0 < t < H. Furthermore, if S = {x E C1[O, H] : x(O) = x0 , x’(t) = f (t, x(&), x’(@)) for 0 ,< t < H), S is a compact subset of Cl([O, HI). II. In this section we shall apply our nonlinear theorems following linear problem: Find x E C”([O, H]) (n > 2) such that x’(t) = k(t) 4C(t)> x(0) = 0.
+ b(t) +W>
+ g(t),
to the
O
(26)
EXISTENCE
AND
UNIQUENESS
THEOREMS
619
The more general problem with mitral value x,, = s(O) can be reduced to the above form by taking u(t) = x(t) - x0 and gl(t) = g(t) + by,, . For the moment we shall assume that K, b, g, 0 and 4 are C-l functions on [0, H] (n > 2), that 0 < e(t) < t for t E [0, H] and that 0 < C(t) < t for t E (0, H]. To facilitate discussion of (26) we need some further notation. Let k’ = {X E C”([O, H]) : X(O) = 0} and let W = {x E V : x(‘)(O) = 0 for 0 < j < n}. Define L : V - V by (Lx)(t) = J”: (k(s)x’(+s) + b(s)x(&)) ds. It = is clear that (26) is solvable iff h(t) = Jig(s) ds is an element of R(I -L) the range of I -L, and this solution is unique iff N(I -L) = the null space of I - L is 0 dimensional. Thus it suffices to study N(I -L) and R(I - L). First, we need a trivial lemma (see [5, p. 3441). LEMMA 5. Let E denote the set of polynomials equal to n. Then V = E 8 W.
in k’ of degree less than or
Proof. Ifp(t) = C,:, a$? E E n W, thenpo)(O) =j! a, = 0 for 1
Proof. For 0 < r < 1, define f,. : [0, H] x R x IF8-+ R by fr(t, x, y) = rk(t)y + rb(t)x and note that / af,./ay j = 1rk(t)j. Since j rk(O)l ](+‘(O))+l1 < 1, it follows by Proposition 3 that for any compact set M C I -, {x E V : x - rLx E M and 1)x /j < l} is compact. By Lemma 3.1 of [14], this implies that R(I - rL) IS closed and N(I - rL) is finite dimensional, i.e., 1 - rL is a semi-Fredholm map for 0 < r < 1 (see [6] for definitions). By the continuity of the index for semi-Fredholm maps, index (I -L) = index 1 = 0, so I -L : V - V is a Fredholm map of index zero. Since W is a closed subspace of I; it is obviously also true that for any compact subset MC W, {x E W : x - rLx E M, j/ x/I < 1) is compact. Thus since L : W--t W, the same reasoning shows that (I - L) ] W is Fredholm of index zero. Let h be any element of IV, To show there exists x E W such that (I- L)x = h, it suffices to show that there exists x E W such that x’(t) = k(t)x’($t) + b(t)x(Bt) + g(t) for 0 < t < II, g(t) = (dh/‘dt)(t). Define and notice that 0 = p(O, 0, 0) and f
620
NUSSBAUM
(8j/St~)(O, 0,O) = 0 for 1 < j < n - 1. It follows from Remark 1 after Theorem 1 that there exists 6 > 0 and x E C*([O, S]) such that x(j)(O) = 0 for 0 < j < n and x’(t) = f(t, x(&), x’($t)) for 0 < t < 6. However, jmeets the conditions of Theorem 2 so x can be extended to an element of C”([O, H]) such that x’(t) =f(t, x(&), x’(+t)) for 0 < t < H. This implies that (1 -L) 1 W is onto W, and since (I -L) 1 W is Fredholm of index zero, (I- L) 1 W is one to one and onto. It remains to investigate dim(N(.Z - L)). Let E, = {p E E : (I - L)p E W}, where E is as in Lemma 5. We claim that dim(&) = dim N(I -L). To see this, forp E E, define S(p) E N(I- L) by S(p) = [(I -L) 1w]-l(1L)p - p; S is one to one, since S(p) = S(g) for p, q E E. implies that p - q E W or p = q. It follows that dim(E,,) < dim N(I - L). Conversely, if x E N(I - L) and x=u+p for UGW and PGE, then PEE,,, smce (I-L)p= we can define T(x) =p E E,; -(I - L)u E W. Thus, for XEN(I--L) Tisonetoone,because~fu+p,v+pEN(I-~)foru,v~Wandp~E, (I - L)(u - U) = 0, so u = v. Therefore we have dim N(I - L) < dim(E,,), so the two are equal. If p E E and p(t) = C,“=, (u,fi!)t’, p E E,, iff qfj-l)(O) = 0 for 1 < j < n, where q(t) = p’(t) - k(t)p’($t) - b(t)p(&). It 1s easy to see that q(j-l)(0) = zi=, c,,P(~)(O), where the c,, are constants and c,, = (1 - C~J-l). The assumptions we have made imply that 0 < m < 1 and 1cmn-l 1 < 1, so that cm3 = 1 for at most onej, 0 < j ,( 1z- 1. The matrix (c,,) is triangular, so if cj3 # 0 for 1 ,( j < n, (cZ3)is nonsingular and dim(E,) = 0. This shows thatifcm~#1forO~j,(n-2and~cm”-r~<1,I--Lisonetoone and onto I/. If cjj = 0 for some j, 1 < j < n - 1, one can see directly that dim(E,,) = 1; in fact in this case a, , aa ,..., a,..., are zero, a, is arbitrary, and Q.E.D. a,,, ,a*., a, are linear functions of a, . Remurlz 3. If b = 0 and the hypotheses of Theorem 4 hold (m particular cm3 # 1 for 0 < j < 71- 2), one can in some sense explicitly solve (26). First assume that g(?)(O) = 0 for 0
n-1 &> = g(t)- c (dw)/~!) t3 and let 12-l
p(t) = 1 (gCJ)(O)/j!)ti. 3=0
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Then solving (26) is equivalent to solving the followmg two equations on
PI HI: %‘(4= W~l’W) + g1(4, Xl(O)= 0, x2’(t) = k(t)x,‘(+t) + p(t), Equation 0 < j < will be wM44 following
“z(O) = 0.
(27) cw
(27) can be solved as described above, since go)(O) = 0 for n - 1. To solve (28) let q(t) = J& b,t*, where the numbers 6, selected later. By replacing x2 by xs - 4 and defining h(t) = - 4’(t) + f-e>> one can easily show that (28) is equivalent to the Eq. (29):
By our previous remarks (29) can be solved if b, , b, ,..., b, can be selected such that ho)(O) = 0 for 0 < j < n - 1. If one evaluates ho)(O) one finds that because cm? # 1 for 0 < j ,< n - 1, there exist unique numbers h , b, ,***, b, such that h(j)(O) = 0 for 0 < j < n - 1. Remark 4. If 1cm”-r 1 > 1, then N(I - L) is, in general, infinite dimensional. For example, consider the following equations: x’(t) = cx’(mt) + g(t),
o
x(0) = 0.
(30)
Assume that c and m are constants, 0 < m < 1, 1cmn-l / > 1 and g E C+r([O, 11). Let y,, be any C” function with compact support in (m, 1). For t EI~ = [m2, m], define yr(t) = (l/c)ya(t/m) and generally if y,(t) has been defined for t E I, = [rnj+l, ml], define y,+,(t) = (l/c)y,(t/m) for t E 1,+r . If we define y(t) = y,(t) for t E I, , then clearly y is a C” map from (0, l] to R. Also, using the assumption 1crn+l 1 > 1, one can show that lim,,,yo)(t) = 0 for 0 < j < n - 1, so y E C+l([O, l]), if we set y(0) = 0. If we define x(t) = $ y(s) ds, x E P([O, l]), x satisfies the homogeneous equation (30) with g = 0, and the collection of such x is infinite dimensional. If cm3 # 1 for 0 < j < n - 2 and 1cm+l / > 1, one can always find x E Cn([O, 11) satisfyin (30). We outline the proof and leave the details to the f reader. If g(t) = CyzO up is a polynomial, (30) can be solved explicitly: X(t) = 2ITr,’ [a,/(j + I)(1 - c@)lt ?+I. For general g, we can assume that g(‘)(l) = 0 for 0 < j < n - 1 by subtracting the polynomial n-1
2 LPYl)/M - 1Y
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(for which we can solve the equation). Now we use an Idea of D. Orth [13]: for t E [mP+l, mfl), define
x(t) = f (g
J:-, g(s)ds*
2=1
Using the fact that g(‘)(l) = 0 for 0 < j < n - 1, one can prove that x so defined is n times continuously drfferentiable on (0, 11; and usmg the assumption that 1cmn-l 1 > 1, one can show that lim,,, x(‘)(t) exists for 0 < j < n. Finally, one can check that x(t) - x(0) satisfies (30). We conclude our discussion with a trivial apphcation to the linear case of Theorem 3. THEOREM 5. Let k, b, g and 19: [0, t*] --) R be continuous functions and let 4 : [0, t*] -+ [w be a C’ function. Assume that 0 < 6(t) < t fey t E [0, t*], 0 < 4(t) < t joy t E (0, t*) and $(t*) = t*. Suppose that 1k(t*)/$‘(t*)l < 1. Then if x : [0, t*) + IF!is an absolutely continuous function such that x’(t) = w)x’(4t) + wxw + g(t) f OY almost all t E [s, t*), where 0 ,< s < t* and 4’(t) > 0 joy t E [s, t*], llmt_tt* x’(t) exists.
Proof. Take 6 > 0 so small that if n/r, = sup{/ h(t)/ : t E [t* - 6, t*]}, M&‘(t*) < 1. Then if we define f(t, x, y) = b(t)x + K(t)y + g(t), Theorem 5 follows immediately from Theorem 3. Theorem 5 is the best possible result, as we can see by the following simple example: Example. Consider the equation x’(t) = 2x’(F) + tZ, x(0) = 0, for 0 < t < 1. One can verify directly that x(t) = f
01% = 2*+1 + 1
+ f 2” $ , kl
satisfies the equation and is infinitely lim,,, x(t) = co.
differentiable
on [0, 1). However,
REFERENCES 1. A. L. BADOEV AND B. N. SADOVSKII, An example of a denslfymg operator m the theory of differential equations with a deviating argument of neutral type, So&et Math. Dokl. 10 (1969), 724-728 [Dokl. Akad. Nauk. SSSR 186 (1969), 1239-12421. 2. M. A. CRUZ AND J. K. HALE, Existence, umqueness and continuous dependence for hereditary systems, Ann. IMat. Pura. Appl. 85 (1970), 63-82. 3. G. DARBO, Punt1 umti m trasformaziom a condlmimo non compatto, Rend. Sem. Mat. Univ. Padova 24 (1955), 84-92.
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4. R. D. DRIVER, Existence and contmuous dependence of solutions of a neutral functlonal dlfferentlal equation, Arch. Rat. Mech. Anal. 19 (1965), 149-166. 5. N. DUNFORD AND J. T. SCHWARTZ, “Linear Operators,” Part I, Intersclence, New York, 1958. 6. T. KATO, “Perturbation Theory for Linear Operators,” Springer-Verlag, New York, 1966. 7. C. KURATOWSKI, Sur les espaces complets, Fund. Math. 15 (1930), 301-309. 8. W. R. MELVIN, “A class of neutral functional dlfferentlal equations,” Ph.D. Dlssertatlon, Brown Umverslty, Providence, RI, 1971. 9. R. D. NUSSBAUM, The fixed pomt index and asymptotic fixed pomt theorems for k-set-contractions, Bull. Amer. Math. Sot. 75 (1969), 490-495. 10. R. D. NUSSBAUM, The fixed point mdex for local condensing maps, Ann. Mat. Pma Appl. 89 (1971), 217-258. 11. R. D. NUSSBAUM, Degree theory for local condensing maps, J. Math. Anal. Appl., to appear. 12. R. D. NUSSBAUM, A generahzation of the Ascoh theorem and an apphcatlon to functional differential equations, J. Muth. Anal. Appl. 35 (1971), 600-610. 13. D. ORTH, A numerical method for some neutral functlonal dlfferentlal equations, to appear. preserved under addltlon of a 14. B. YOOD, Propertles of lmear transformations completely contmuous transformation, Duke Math. J. 18 (1951), 599-612.