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On optimal solutions of boundary value problems for differential equations
NonlinearAnalysis, Theory. Methods & Applications. Vol. 26, No. 3, pp. 509-524, 1996 Copyright © 1995 ElsevierScienceLtd Printed in Great Britain. All rights reserved 0362-546X/96 $9.50+ .00
Pergamon
0362-546X(94)00294-0 ON OPTIMAL
SOLUTIONS OF BOUNDARY VALUE FOR DIFFERENTIAL EQUATIONS
PROBLEMS
DARIUSZ BIELAWSKI Institute of Mathematics, University of Gdafisk, ul. Wita Stwosza 57, 80-952 Gdafisk, Poland
(Received 18 February 1994; received in revised f o r m 26 August 1994; received f o r publication 19 October 1994) Key words and phrases: Picard problem, Nicoleni problem, Floquet problem, hyperbolic boundary value problems, existence of solutions, existence and uniqueness of solutions, equivalent norms. O. I N T R O D U C T I O N
We present a general construction of equivalent norms in Banach spaces. It appears that some maps which arise in boundary value problems for differential equations are contractive under these norms. Using completely continuous multivalued regulators Lasota and Opial showed in [1] that a vector field is a homeomorphism which means that the corresponding map has exactly one fixed point. In [2] they proved existence of fixed point for a multivalued map with convex and closed values and a suitable multivalued regulator. To construct equivalent norms we also use multivalued regulators. We prove two general theorems (theorems 1 and 2) without assuming that regulators are completely continuous. Using our method we extend classical results for the Nicoletti boundary value problem due to Lasota and Olech (see [3]) and for the Floquet boundary value problem due to Kasprzyk and Myjak (see [4]). We are also concerned with the Picard boundary value problem for the system of second order ordinary differential equations and some boundary value problems for hyperbolic partial differential equations which were posed in [5]. Obtained results seem to be the best possible in certain classes of the right-hand sides of considered systems. 1. A C O N S T R U C T I O N
OF EQUIVALENT
NORMS
In this paper (E, I1" [[) denotes a real or complex Banach space. By n ( E ) we understand the family of all nonempty subsets of E and by/~(0, r) the closed ball about 0 of radius r in E. For a multivalued map H : E ~ n ( E ) and c~ > 0 we will say that D C E is (H, u)-invariable if H(D) C ~D.
THEOREM 1. Suppose that h: E ~ E, a multivalued map H : E --- n(E), a e (0, 1) and D C E satisfy the following conditions H(l~x) = l~H(x)
for x 6 E, /~ ~ R (or C),
H ( x + y) C H ( x ) + H ( y ) H ( D ) C ~D,
/~(0, 1) C D,
h(x) - h ( y ) ~ H ( x - y)
Then, there exists a norm/1" space (E, I1" IIH).
IIH equivalent
(1.1)
for x, y ~ E,
(1.2)
H ( D ) is bounded,
(1.3)
for x, y c E.
(1.4)
to I1" II and such that h is contractive in the Banach 509
510
D. BIELAWSKI
P r o o f . For every x ~ E we define the minimal (H, o0-invariable set containing x by D(x) = A {C : C C E, x ~ C , H ( C ) C otC].
F r o m this definition it follows that x ~ D(x) and H ( D ( x ) ) C oLD(x). F r o m (1.1) and (1.3) we obtain
n(~-~llxlln(o) U B(O, [Ixll)) c ~ - ' l l x
IH(H(D)) U
n(l]xllO)
c IIxIIH(D) c ~(~-111xlIH(D)U B(O, Ilxll)). Since x ~
c~-~tlxItH(D) U B(O, Itxll),
f r o m the above inclusions we get
D(x) C
~-'IixlIH(D) U B(0, Ilxll).
(1.5)
Therefore, D(x) is b o u n d e d for every x e E. Define
llxll. = suplllyll : y ~ D(x)]. Using (1.1)-(1.3) one can check that II" IIH is a n o r m in the space E. We show for instance that IIx + ylIH -< Ilxll. ÷ Ilyll~, for x, y e E. Observe that D ( x + y) C D ( x ) + D ( y ) . Indeed, x + y ~ D(x) + D ( y ) and by (1.2) H ( D ( x ) + D ( y ) ) C H ( D ( x ) ) + H ( D ( y ) ) C ~¢(D(x) + D ( y ) ) .
Therefore,
tlx ÷ yll. = suplllzll
:z ~
D(x + y)l ~ sup[llzll
:z e
D(x) + D ( y ) I
-< suplllzll :z ~O(x)l + suplllzll :z ~ O(y)l = Ilxll. + IlYlI..
I1" II and I1" II.
N o w we show that the n o r m s
are equivalent. Since x ~ D(x), we have
Ilxtl. = suplllyll : y ~ O(x)} > Ilxll. O n the other hand, f r o m (1.5) we get
Ilxll. -< m a x ( l , - 1
suplllytl : y ~ n(o)l)llxll.
It remains to prove that for x, y e E
IIh(x) - h(y)[l~ <- ~llx - YlIH.
(1.6)
F r o m (1.4) we deduce that h(x) - h ( y ) ~ H ( x - y) C H ( D ( x - y)) C ¢xD(x - y).
Hence D(h(x) - h(y)) C a D ( x - y). This and the definition o f 1[. [in yield (1.6), which finishes the p r o o f . E x a m p l e 1. Consider the C a u c h y problem x ' U ) = f ( t , x(t)),
I
t ~ [0, p],
(1
x(0) = a,
where a e R" and f : [0,p] x R" ~ R" is a continuous m a p such that for a fixed K > 0 I f ( t , u) - f ( t , v)[ < K l u - v[
for t ~ [0,p], u, v ~ R".
Hyperbolic boundary value problems
511
Let E be the Banach space C([0,p], R n) of all continuous m a p s f r o m [0,p] to R n with the n o r m Ilxll = s u p l l x ( t ) l : t • [0,p]l. set o~ = 1 - e x p ( - K p ) . Define a h o m e o m o r p h i s m P : E --+ E by P(x)(t) = exp(Kt)x(t) and m a p s h: E -+ E, H : E ~ n(E) by h(x)(t) = a +
f(6, x(s)) ds, ~0
H(x) = c~P(/~(0,
IIP-l(x)ll)).
Now, the p r o b l e m (1.7) can be written as x = h(x), where x • E. Observe that for every x, y • E we have lIP
~ o hop(x)
-
P-~
o h op(y)ll
- o~llx- yll.
This implies that the condition (1.4) holds. After setting D = P(/~(0, 1)) it is possible to check that all remaining hypotheses of t h e o r e m 1 are satisfied and, therefore, h is contractive under the n o r m I[" ]In. Hence, the p r o b l e m (1.7) has exactly one continuously differentiable solution on
[0,p].
One can give an explicit expression for II'[In- Note that for every x • E we have H(oc-IH(x)) C H(x), x • u - i l l ( x ) and H(x) C H(D(x)) C teD(x). Thus, D(x) = c~-lH(x) and
Ilxll,, This means that
= sup[lly[[ : y • o~-'H(x)l = e x p ( K p ) s u p l e x p ( - K t ) l x ( t ) l : t
I1" IIH is the
• [0, PII.
classical Bielecki n o r m (see [6]) multiplied by the constant exp(Kp).
Example 2. Let h: E --, E be a linear b o u n d e d operator. Suppose that p < 1, where p is the spectral radius o f h (p = lim,,~=]]hm]]l/'~). Choose c~ • ( p , 1). Define H : E ~ n(E) by H(x) = [h(x)] and D C E by oo
D = ~_) a - m h " ( B ( 0 , 1)). m=0
F r o m t h e o r e m 1 it follows that h is contractive under the n o r m H" [[/4. This n o r m is equivalent to []. I[ and can be written as Ilxll~ = supl~-mllhm(x)ll
: m = 0, 1 ....
}.
THEOREM 2. Suppose that c~ • (0, 1), D C E, H : E - - * n(E) satisfy (1.1)-(1.3) and that a multivalued m a p G: E ~ n(E) has b o u n d e d image. Then, every completely continuous h: E - . E such that h(x) • H(x) + G(x)
for x • E,
(1.8)
has at least one fixed point. Proof. Sets D(x) and the n o r m I]']]n we define as in the p r o o f of t h e o r e m 1. Write M = supIIIYI[H :Y • G(x)] and R = M / ( I - ~). We claim that h(Bt4(0, R)) C /~H(0, R),
(1.9)
where/~H(0, R) = IY ~ E : [lY[[H -- R]. T o show (1.9) notice that H(/~/4(0, R)) C od)H(0, R).
(1.10)
512 In fact,
D. B1ELAWSKI for
[[xl[/4 - R and y ~ H ( x ) , we have y ~ H ( D ( x ) ) C o~D(x). This m e a n s that IlylIH <-- ~IIxIIH. F r o m (1.8) and (1.10) we conclude that
D ( y ) C o~D(x) and so
IlxllH --- e = IIh(x)llH ~ ~ g
+ M = R,
which proves (1.9). Since h is c o m p a c t o n / ~ u ( 0 , R), it has a fixed point. 2. THE PICARD PROBLEM Consider the Picard p r o b l e m x"(t) = f(t, x(t), x'(t)),
I
t ~ [0, p],
(2. 1)
x ( p ) = b,
x(O) = a,
where a, b ~ R" and f : [0, p] z R" x R" ~ R" is continuous. By a solution o f (2.1) we m e a n any x ~ Cz([O,p], R"), which satisfies (2.1).
THEOREM 3. Suppose that M > 0 and L 0, L I > 0 satisfy the inequality p2 ~Lo
+ PLtTr < 1.
Then: (i) if for all x, y, £, ~ e R", t e [0, p] ]f(t,x,y)
- f(t,£,~)[
<_ L o l x - £t + L , [ y - .~[,
(2.2)
then the p r o b l e m (2.1) has exactly one solution; (ii) if for all t 6 [0, p], x, y ~ R n If(t,x,y)l
<- L o l x
(2.3)
+ L~Iy] + M ,
then the p r o b l e m (2.1) has at least one solution; (iii) statements (i) and (ii) are not true for L0 = Or/P) 2. Statement (i) was proved by Picard under stronger a s s u m p t i o n ( L o p Z / 8 ) + ( L i p ~ 2 ) < 1 (see [7, p. 423]). T h e o r e m 3 is p r o b a b l y not optimal. H o w e v e r , this result is o p t i m a l if we restrict ourselves to the p r o b l e m x " = f ( t , x ) with the Picard b o u n d a r y value conditions x(O) = a, x ( p ) = b. P r o o f . Write c~ = ( L o p 2 / g 2) + (Lip~re). By E we denote the space C ( [ 0 , p ] , R n) o f all continuous m a p s f r o m [0,p] to R" with the n o r m ][x[[ = sup{]x(t)[ : t ~ [0,p]J. Define S: E ~ E , T: E --. E , h: E -~ E , H : E --, n ( E ) , G: E --* n ( E ) and D C E by S(x)(t) = P-
T(x)(t) =
1 P
(
(P
o sx(s) ds - .,, ( p - s)x(s) ds t)
,Jo
sx(s) ds + t
i" t
(p
(2.4)
,
s)x(s) ds
)
,
(2.5)
Hyperbolic boundary value problems
513
h(y)(t) = f ( t , (1 - (t/p))a + (t/p)b + T(y)(t), (1/p)(b - a) + S(y)(t)), H ( y ) = {z • E : lz(t)[ <- Lot T(y)(t)[ + L~]S(y)(t)II, G(y) = [z • E :
Iz(t)[ < M + Lo(lal + Ibl) + Z l l b - a l / P L
Is 1
D =
• E :
ly(t)l 2 d t _< p
0
1
.
N o t e t h a t T(x) = y iff y • Cz([0, p ] , R ' ) , y " = x o n [0, Pl a n d y(0) = y ( p ) = 0. S i m i l a r l y , S(x) = y i f f y • C l ( [ 0 , p ] , R ' ) , y ' = x o n [ 0 , p l a n d I~ y(s) d s = 0. A f t e r setting y(t) = x " ( O it is p o s s i b l e t o w r i t e the p r o b l e m (2.1) in an e q u i v a l e n t f o r m y = h(y), w h e r e y • E . If y = h(y), t h e n x(t) = (1 - (t/p))a + (t/p)b + T(y)(t) is a s o l u t i o n o f (2.1). C o n s i d e r the real H i l b e r t s p a c e ~ = Lz([0, p ] , R) a n d t w o o p e r a t o r s S, T: N ~ 3£ given b y (2.4) a n d (2.5), r e s p e c t i v e l y . F o r m = 1, 2 . . . . d e f i n e
era(t) = sin(rant~p),
t • [0,p].
O b s e r v e t h a t [em : m = 1, 2 . . . . } is a c o m p l e t e o r t h o g o n a l set in N . F u t h e r m o r e , pZ
T(e,~)(t) =
~
era(t),
m --- 1,2 . . . . ,
S(e,.)(t) = - p c o s ( m n t / p ) , mn
m = 1, 2 . . . . .
Hence,
IIf bl = p2/~z
Ilgll
and
=
P/n.
(2.6)
We claim that H ( D ) C (xD. In fact, i f y = (y~ . . . . . Yn) • D and z • H ( y ) , from the definition o f H and (2.6) we obtain
t.t 'P Iz(t)b z dt '/1/2 <_
o (Lo[ T(y)(t)l + L~[S(y)(t)l) 2 d t ) 1/2
0
<- Lo
[T(y)(t)l z dt
+ L1 IS(y)(t)l z dt \,10
',,,) 0 = Lo
(¢(yj)(t)) 2 dt j
l
p2 (
,'p
ly(012 d t
0
(~q(yj)(t)) 2 d t
j
-< v ,i:, to -- ~
+ L1
0 dt\l/2
) +
_< c~x/p.
1 0
p ( 7~ j = l
i
~p
,~i/z
514
D. B I E L A W S K !
We next show that H ( x + y) C H(x) + H ( y ) for x, y ~ E. I f z ~ H ( x + y), we define ~ and ~Eby
z(t)(to] T(x)(t)[ + t,lS(x)(t)[) ~(t) = Lol T(x)(t)l + L~lS(x)(t)[ + Zo] T(y)(t)l + Z~lS(y)(t)[' z(t)(Lol T(y)(t)l + LjlS(y)(t)[) ~(t) = Lo[ T(x)(t)l + Llla(x)(t)l + tol T(y)(t)[ + L~[S(y)(t)l (we set z~(t) --- 2.(t) -- 0 when the denominator equals 0). It is possible to check that ~ e H(x), ~ H ( y ) and z = ~ + ~. Now we prove that H(D) is bounded. In fact, for z ~ H(y), where y ~ D we have
Iz(t)l --- LolT(y)(t)l + LllS(y)(t)[ <- LoP
ly(t)[ d t + E l 0
<- (LoP
+
LI)
ly(t)t z dt
ly(t)l dt 0
~ LoP 2 + L i p .
~
0
Observe that (1.1) is satisfied and B(0, 1) C D. To prove (i) notice that the inequality (2.2) implies (1.4). Therefore, from theorem 1 we deduce that in this case the problem (2.1) has exactly one solution. To prove (ii) define hH, hG : E ~ E by
h(y)tt)(LolT(y)(t)] + L~IS(y)(t)I) ht4(y)(t) = t o l T ( y ) ( t ) I + Z~lS(y)(t)[ + Z0([a[ + Ibl) + (Z~lb - al/p) + M ' h(y)(t)(Lo(]a] + Ibl) + (Zllb - al/p) + M ) ha(y)(t) = LolT(y)(t)l + L1[S(y)(t)I + Lo([al + [b[) + (L~lb - al/p) + M " Note that h n + ho = h and that the inequality (2.3) implies h n ( y ) ~ H(y), h o ( y ) ~ G(y) for y ~ E. From theorem 2 we conclude that in this case the problem (2.1) has at least one solution. To show (iii) consider the problem
x"(t) = - - ~ x ( t )
+ rl(t),
t e [0,p],
k. x(O) = x(p) = o, where x e C2([0,p], R) and r / e C([0,p], R). Note that the right-hand side of the above problem satisfies (2.2) and (2.3) with constants L o = nZ/p z and Ll = 0. After setting x" = y we can write the above problem as 2l.2
y = --p~ T(y) + I1,
y e C([0,p], R),
(2.7)
where T is given by (2.5). Observe that for r/ = 0 (2.7) has a nontrivial solution y(t) = sin(nt/p). From the Fredholm alternative it follows that there exists t / ~ C([0,p], R) such that the problem (2.7) has no solution. This completes the proof.
Hyperbolic boundary value problems
515
3. T H E F L O Q U E T P R O B L E M
Consider the Floquet problem x~k)(t) = f ( t , x ( t ) , x ' ( t ) . . . . . x ~ k - n ( t ) ) ,
I
t e [0,p],
(3.1)
x(0) + 2x(p) = ro . . . . . X(k-l)(0) + 2X(k-1)(P) = r k - 1 ,
where ~ > 0, ro . . . . . rk_ 1 • R n and f : [0,p] x (R") k --. R n is continuous. By a solution o f (3.1) we m e a n any x • Ck([0,p], R") which satisfies (3.1). Set 0 = p / ~ / r t z + lnZ2. THEOREM 4. Suppose that M > 0 and Lo, L I . . . . . L k_ ~ > 0 satisfy the inequality + . . . . ~- OLk_t < 1.
OkLo + O k - ' L l
(3.2)
Then: (i) if for all t • [0, p], x o . . . . . xk- l, Yo . . . . . Yk- t • R" I f ( t , xo . . . . . x k _ , ) - f ( t , Yo . . . . . Y k - , ) [ <- Lo[X0 - Yol + "'" + L k _ l l x t - ,
- Yk-~[,
(3.3)
then the problem (3.1) has exactly one solution; (ii) if for all t • [0, p], Xo, ..., x k_ L e R" I f ( t , Xo . . . . . x k _ , ) [ -
tolXol + "'" + Lk-llXk-al + M,
(3.4)
then the problem (3.1) has at least one solution; (iii) the sign " < " in (3.2) c a n n o t be replaced by " _ " . For k = 1 this t h e o r e m was proved by Kasprzyk and M y j a k in [4]. First, we prove the following lemma. I.EMMA 1. Let ~ be the complex Hilbert space o f all x e L2([0, p], C) with the scalar p r o d u c t defined by (x, Y)~c =
22t/Px(t)y(t) dt. ~0
Let T: ~ ~ £ be given by 7"(x)(t)
Then I1711 = p/V
-
1
t" x ( s )
1 + 2 ,o
ds
tP x ( s ) d s . - -2 1 + )~ j t
(3.5)
2 + 1n22 = 0.
P r o o f . For m e Z define era(t) = exp((ln(1/2) + (2m - 1 ) i n ) t / p ) ,
t e [0,p].
Note that tem : m e Z] is a complete o r t h o g o n a l set in £ . O u r assertion is due to the fact that 7~(em) = p(ln(1/2) + (2m - 1)iTr)-lem
for m e Z.
516
D. BIELAWSKI
P r o o f o f theorem 4. Write a = O~Lo + o k - i L l + "" + O t k _ l . By E we denote the space C([0, p], R") with the supremum norm. Define T: E ---, E b y (3.5) and h: E --' E, H : E --' n(E), G: E ---, n(E), D C E by f ( t , r° + "'" + T k - l(rk-x)(t) + T g ( y ) ( t ) , . .
h(y)(t)
rk-------~l + T ( y ) ( t ) )
1 +~.
"'1 + 2
'
H ( y ) = {z • E : ] z ( t ) [ - ZolTk(y)(t)[ + "" + Lk-IIT(y)(t)[I,
Iz I, l"
• E : lz(t)[ < M + ~
G(y)=
D =
• E :
j=0
~ LjlTt-J(rt)(t)
l=j
il
,
1
2 2t/p [y(t)[ 2 dt __. p(1 + 2 2) .
o
Note that T(x) = y i f f y • CI([O,p], R"), y' = x on [0,pl and y(0) + 2y(p) = 0. After setting y = x (~) it is possible to write the problem (3.1) as y = h(y), where y • E. I f y = h(y), then
x(t) = ro + ... + rk-X(r~-l)(t) + T~(y)(t) 1+2 is a solution of (3.1). We claim that H ( D ) C e~D. In fact, if y • D and z • H ( y ) , from the definition of H and lemma 1, we get
(I~ '~2t/plz(t)[2 dt) w2 -<
(ip(l p
~.2'*(Lol Tk(y)(t)l + ... + Lk_l[ T(y)(t)I) 2 dt
0
<- Lo
2zt/"[ Tk(y)(t)[ 2 dt
)1/2
+ ... + Lk_~
\dO
(Sp
221/P IT(y)(t)[ 2 dt
0
,~2t/p('Fk(yj)(t))2 dt
= Lo
+ "'" + Lg-1
j 1 0
< Lo Ok =
)1/2
(____~lp22t/p(yj(t))2dt )1/2+ ... + Zk_lO j 1 0
0 22t/Ply(t)12dt/1/2 <- ax/p(1 +
221ml"(yj)(t)) z d
j 1 0 (~=
l~]L2t/p(yj(t))2dt) 1/2
j 1
22).
To prove (i) notice that the inequality (3.3) implies (1.4). It is also possible to check that all remaining hypotheses of theorem 1 hold and, therefore, in this case the problem (3.1) has exactly one solution. To prove (ii) observe that the inequality (3.4) implies (1.8). F r o m theorem 2 we conclude that the problem (3.1) has at least one solution.
Hyperbolicboundary value problems
517
TO prove (iii) suppose that tgkLo + O~-IL~ + ... + OLk_~ = 1, write ln(l/2) + #t and consider the problem x(~) = L o 6 l ( T k ) x - L o 3 ( y k ) y + ... + Lk_~(R(7)x ~k-~) - Lk_13(y)y (k-l) + r/l,
I
y(k) = Lo3(yk)x + L o ( R ( y k ) y + ... + Lk_~3(7)x(k-~) + L k _ I 6 ~ ( 7 ) y ( k - l ) + tlZ, X('D(O) + )tx(J)(p) = O,
y(J)(O) + )ty(J)(p) -= O,
j = 0 ..... k -
1,
where rtl, qz E 6"([0, p], R) and x, y e CtC([0, p], R). Notice that the right-hand side o f this system satisfies conditions (3.3) and (3.4). After setting z ~- x (e~ + iy (k) and 1/= r/~ + in 2 we can write the above problem as z = ? ~ L o T ~ ( z ) + ...
+ 7Lk-~T(z)
+ t]
(3.6)
(here we consider T given by (3.5) in the complex Banach space C([0,p], C) with the usual supremum norm). For ~/ = 0 (3.6) has a nontriviat solution exp((ln(1/2) + i n ) t / p ) . From the Fredholm alternative it follows that there exists r / ~ C([0, p], C) such that (3.6) has no solution. This completes the proof. 4. THE N1COLETTt PROBLEM Consider the Nicoletti problem I x(k~(t) = f ( t , x ( t ) , x ' ( t ) . . . . . x (k- 1)(t)),
Xj(t}(ttj)
rtj,
t E [0,p],
(4. 1)
I = 0 . . . . , k - 1, j = 1 . . . . . n,
where i t = ( t t l , . . . , b , ) ¢ [ O , p ] " , rt= (rtl,...,rt,)ER n for t = 0 . . . . . k - 1 and f : [0, p] × (R") k --. It n is continuous. By a solution o f (4.1) we mean any x e Ck([0, p], R") which satisfies (4.1). Set co = 2 p / n . THEOREM 5. Suppose that M > 0 and L o, L1 . . . . . L~.I > 0 satisfy the inequality ~okLo + ook-lL~ + ... + w L ~ _ Z < 1.
(4.2)
Then: (i) if for all t ~ [0, p], Xo . . . . . x k - ~ , Yo . . . . . y~_~ e R " If(t, Xo,---,xk-1)
- f ( t , Yo . . . . . Ye-~)I <- Lolxo - Yol + "'" + L k - l i X k - 1 -- Y k - l t ,
(4.3)
then the problem (4.1) has exactly one solution; (ii) if for all t e [0, p], Xo, . . . , x k _ t e R" I f ( t , Xo . . . . . xk- ,)I <- Lolxol + "'" + L k - ijxk_,l + M ,
then the problem (4.1) has at least one solution; Oil) there exist {to] such that the sign " < " in (4.2) cannot be replaced by " - " .
(4.4)
518
D. B I E L A W S K I
F o r k = 1 this t h e o r e m was p r o v e d by L a s o t a a n d Olech in [3]. F i r s t we p r o v e the f o l l o w i n g lemma. LEMMA 2. F o r every c o n t i n u o u s x : [0, p] ~ R a n d c e [0, p] we have
l (Six,s,4
dt -< oa2
,0
1p
xE(s) ds.
P r o o f . I f c > 0, c o n s i d e r the real H i l b e r t space ~ c o n t i n u o u s a d j o i n t o p e r a t o r s A , B: 3E --, 9]Z given b y A ( x ) ( t ) = ~. x(s) ds, J0
O b s e r v e t h a t em e ~
(4.5)
0
= L2([0, c], R) a n d two c o m p l e t e l y
B(x)(t) =
x(s) ds. t
(m = 1 , 2 . . . . ) d e f i n e d by em(t) = sin
(2m-
1)~t
2c
are eigenvectors o f the c o m p l e t e l y c o n t i n u o u s s e l f - a d j o i n t A B : ~g --' 91"(.. Thus, {era : m = 1 , 2 . . . . } is a c o m p l e t e o r t h o g o n a l set in ~ . N o t e t h a t 2c
B(em)(t) -
Hence
IIBII =
(2m-
(2m - l)Trt - - , 2c
cos
1)~
m = 1,2 .....
2c/r~ a n d for e v e r y x e C ( [ 0 , p l , R)
t (I 0
x(s) ds
xZ(s) ds.
dt <_
t
0
Similarly, one can show t h a t for x e C([O, p], R) c x(s) ds
x2(s) ds.
dt <_ . . . .
F r o m these two inequalities we get (4.5). This c o m p l e t e s the p r o o f o f l e m m a 2. P r o o f o f t h e o r e m 5. W r i t e a = cokL0 + ~ok-lL t + -.. + ¢ o L k _ 1 . The B a n a c h space (E, I1"11) we define as in Section 3. Define T~: E--* E (l = 0 . . . . . k - 1), h : E ~ E , H , G : E ~ n(E) a n d D C E by
(T~(y)(t))j =
yj(s) ds,
j = 1, . . . , n,
tO
h ( y ) ( t ) = f ( t , r o + To(rO(t) + ... + To . . . . . + TO . . . . . H ( y ) = {z ¢ E :
Tk_2(rk_l)(t)
Tk_l(Y)(t) . . . . . r k - l + Tk_l(Y)(t)),
Iz(t)l - LolTo . . . . .
T~_,(y)(t)l + ... + L k - l l T k - ~ ( Y ) ( t ) l } ,
G ( y ) = tz e E : Iz(t)l -< M + t01rol + tolTo(rD(t)l + "" + t o ITo . . . . .
D =
Tk_2(rk-O(t)l + .." +
e E:
ly(t) ,0
dt _< p
.
Irk-~l},
Hyperbolic boundary value problems
519
Note that T~(x) = y i f f y e C~([0,p], R"), y ' = x on 10,p] and yj(t o) = 0 f o r j = 1 . . . . . n. After setting y = x (*) it is possible to write the p r o b l e m (4.1) as y = h(y), where y e E. I f y = h(y), then x(t) = ro + To(rl)(t) + "'" + To . . . . .
Tk_2(rk_l)(t) + To . . . . .
Tk_l(y)(t)
is a solution o f (4.1). F r o m l e m m a 2 it follows that Ti . . . . .
Tk_I(D) C ~k-~D
for 1 = 0 . . . . . k - 1.
Using this inclusion one can show similarly as in Section 3 that H ( D ) C o~D. T o prove (i) notice that the inequality (4.3) implies (1.4). It is also possible to check that all remaining hypotheses o f t h e o r e m 1 hold and, therefore, in this case the p r o b l e m (4.1) has exactly one solution. T o prove (ii) observe that the inequality (4.4) implies (1.8). F r o m t h e o r e m 2 we conclude that the p r o b l e m (4.1) has at least one solution. To show (iii) suppose that ~ k L 0 + ~ k - 1Lj + ... + ~ L k - i = I and define linear J : R z ~ R z by J(u, v) = (v, - u ) . Consider the p r o b l e m x k) = Lodk(x) + ... + L ~ _ i J ( x O'-t)) + q,
I
xf)(to) = O,
l= 0 ..... k-
l , j = 1,2,
where x e C k ( [ 0 , p ] , 112), q E C([0, p], Rz), t 0 = 0 if l + j is odd and t o = p if / + j is even. Notice that the right-hand side of this system satisfies (4.3) and (4.4). After setting y = x (k) we can write the above p r o b l e m as y = LoJk(To . . . . .
Tk-1(Y)) + "'" + L ~ _ I J ( T k - I ( Y ) ) + ft.
(4.6)
For q = 0 (4.6) has a nontrivial solution y ( t ) = (sin(k)(rrt/2p), cos(k)(nt/2p)). F r o m the F r e d h o l m alternative it follows that there exists q e C([0, p ] , R z) such that (4.6) has no solution. This completes the p r o o f . 5. H Y P E R B O L I C
BOUNDARY VALUE PROBLEMS
Consider the hyperbolic p r o b l e m l u~ = f ( x , y, u),
(x, y) e [0, p] X [0, q],
u(O,y) + 2 ~ u ( p , y ) = ~u(y),
(5.1)
u(x,O) + ,~2u(x,q) = q~(x),
where 2~, 22 > 0 , f : [0,p] x [0, q] x R" ~ R" is continuous, ~0: [0,p] --' !1", ~: [0, q] ---, R" are continuously differentiable and ~0(0) + 2~0(p) = qJ(0) + 2zq/(q). By a solution o f (5.1) we m e a n a continuous u: [0, p] x [0, q] ~ R n with continuous derivatives ux, Uy, U~y = Uyx which satisfies (5.1) on [0, p] × [0, q]. Set 0F = p/x/~& + In~ ~ and 02 = q/x/r& + In 2 2z. THEOREM 6. Suppose that M > 0 and L > 0 satisfies the condition ~/zr 2 + l n 2 2 t ~ 2 L<
Pq
+ ]nZAz
(5.2)
520
D. B I E L A W S K I
Then: (i) if for all x • [0, p], y • [0, q], u, v • R"
l f i x , y, u ) - f i x , ) ,
v)t < - L [ u -
v[,
(5.3)
then the problem (5.1) has exactly one solution; (ii) if for all x • [0, p], y • [0, ql, u • R"
I f ( x , y , u)] -< Llut + M ,
(5.4)
then the problem (5.1) has at least one solution; (iii) the sign " < " in (5.2) c a n n o t be replaced by " _ " .
Proof. Write cz = 0102L. By E we denote the space C([0, p] x [0, q], R") with the s u p r e m u m n o r m . Define S, T: E ---, E by
S(v)(x,y)
-
- -
1 + 21 .o
T(v)(x,y)-1+
v(s, y) ds
1 2~ [-'v J0 v(x,
t)dt
1 + 21
f, x
v(s, y)
1+2222 o~I'"v(x,
ds,
t) dt.
Note that S ( v ) = u iff U, U x • E , ux = v on [0,p] x [0, q] and u ( O , y ) + 2 1 u ( p , y ) = 0 for y • [0, q]. Similarly, T(v) = u iff u, Uy • E, Uy = v on [0,p] x [0, q] and u(x, O) + 22u(x, q) = 0 for x • [0, p]. Using the Fubini t h e o r e m one can check that S o T = T o S. Define also h: E --, E, H, G: E --* n(E) and D C E by
h(v)(x,y) = f
(co(x) + q/(y) x , y , 1 + 2-----~ 1 + 2~
co(O) + 21(p(p) (1 + ,ll)(1 + ,~2) + r o S ( v ) ( x , y )
) ,
,
H(v) = [w • E : Iw(x, y)l <_ L[ T o S(v)(x, y)l 1, G(v) = I w e E :
D=
/
I w ( x , y ) l <-L(I¢(x)I
veE: ,0
+ [~u(y)l +
Ico(o)l
q )~21x/PR~Y/qlv(x,y)12dxdy<_pq(1
+ Ico(p)l)+ MI, + 221)(1 + 2 ~z) .
,0
1
After setting v = u~, the problem (5.1) is equivalent to v = h(v), where v • E. If v = h(v), then
u(x, y) -
C0(x)
1 + 2z
q/(y)
+ - -
1 + 21
¢(0) + 21CO(p) + T o S(v)(x, y) 22)
(1 + 21)(1 +
is a solution o f (5.1). The inclusion H(D) C ~ D follows f r o m the following two inequalities
,0
'~ 'q 22,x/p2}"~qls(v)(x,y)lZ d x d y _< Jr 2 _~ ~nz 2~ ,o ~o ,0
,~, ,~2 , 'q~2x/P~2Y/qlT(v)(x,y)12 d x d y <-- ~z + 1n222 . o jo 2a(/P2zzY/qlv(x,Y)12 d x d y , ,0 ,0
Hyperbolic boundary value problems
521
Using l e m m a 1 we show for instance the first is
]L2x/p)L2Y/qIS(u)(x,y)l 2 d x dy = ~i"q~2y/qti~ IP~I o oq o <_
'q(
2~y/qo 2
= O~
o o
~2x/PIS(u)(x, y)l 2 dxI
Sop
dy
2zx/ulv(x,y)12 d x d y
A~x/P).22Y/qlv(x,y)12dxdy.
N o w we prove that H(u + v) C H(u) + H(v) for u, v e E. If w e H(u + v), we define ~,, ~ b y
w(x, Y)I T(S(u))(x, y)[ ¢v(x, y) = IT(S(u))(x, Y)I + IT(S(v))(x, Y)I ' w(x, y)[ T(S(v))(x, y)[ ~'(x, y) = I T(S(u))(x, Y)I + I T(S(v))(x, Y)I (we set ~ ( x , y ) = ~ ( x , y ) = 0 if the d e n o m i n a t o r equals 0). It is possible to check that w = ¢v + f¢, ¢v ~ H ( u ) a n d ~ e H ( v ) .
T o show (i) we use t h e o r e m 1. T o prove (ii) define hn, hc : E -, E by h(v)(x, y)LI T(S(v))(x, y)[ h . ( v ) ( x , y ) = L(l~p(x)l + I~(Y)I + 1¢(0)1 + I~P(R)I + Ir(S(v))(x,y)l) + M '
h(v)(x,y)(L(l~(x)l + ]~(Y)I + [¢p(O)l + [ ¢ ( P ) I ) + M ) hG(v)(x'Y) = L(l~(x)l + I~'(Y)I + 10,(0)l + [¢(p)l + IT(S(v))(x,y)[) + M" Notice that hn + hc = h and that the inequality (5.4) implies ht-t(v) ~ H(v), hG(v) ~ G(v) for v ~ E . F r o m t h e o r e m 2 we conclude that the p r o b l e m (5.1) has a solution. T o show (iii), write y = 0 n ( 1 / 2 0 + in)(ln(1/22) + in)/(pq) and consider on [0,p] × [0, q] the p r o b l e m Uxy = ~ ( y ) u - 3(y)v + ~h,
I
v~ = 3(y)u + 6~(y)v + ~h,
u(O,y) + 2 ~ u ( p , y ) = v(O,y) + 2 1 v ( p , y ) = O, u(x, O) + 22u(x, q) = v(x, O) + 2zv(x, q) = O,
where u, v: [0,pl × [0, q] ~ R and e l , r/z e C([0,p] × [0, q], R). Notice that the right-hand side of this system satisfies (5.3) and (5.4) with L = (01 02) -1 After setting w = U~y + iVxy and r/ = r/1 + it/2 we can write the a b o v e p r o b l e m as
w = vT(S(w)) + rl
(5.5)
(here we consider S and T in the complex Banach space C([0, p] × [0, q], C) with the s u p r e m u m norm). For r/ = 0 (5.5) has a nontrivial solution
w(x, y) = e x p ( ( l n ( l / 2 0 + in)x/p)exp((ln(1/Az) + in)y/q).
522
D. BIELAWSKI
F r o m the F r e d h o l m alternative it follows that there exists r / e C([0,p] × [0, q], C) such that (5.5) has no solution. This completes the p r o o f Consider the hyperbolic p r o b l e m
l Uxy = f ( x , y, u),
(x, y) • [0, p] × [0, q],
uj(Sg, y) = q/j(y),
uj(x, tj) = ~pj(x),
(5 o6~ j = 1 . . . . . n,
where f : [0,p] × [0, q] × R n ~ R" is continuous, ¢0: [0,p] --, R n, ~,: [0, q] --, R" are continuously differentiable, sj e [0,p], ti e [0, q] and ~0j(s~) = ~j(tj) f o r j = 1 . . . . . n. By a solution o f (5.1) we m e a n a continuous u: [0, p] × [0, q] --, R" with continuous derivatives ux, Uy, Uxy = Uyx which satisfies (5.6) on [0, p] × [0, q]. The above p r o b l e m is a generalization of the well-known D a r b o u x p r o b l e m (see [8] and the references given there). THEOREM 7. Suppose that M > 0 and L > 0 satisfy the condition ~2 t
~ _m
4t~q"
(5.7)
Then: (i) if for all x ~ [0, p], y • [0, q], u, v ~ R n
[f(x,y, u)-f(x,y,
v) I --< L l u -
v]
(5.8)
then the p r o b l e m (5.6) has exactly one solution; (ii) if for all x ~ [0,p], y e [0, q], u • R ~
] f ( x , y , u)[ _< Llu[ + M , then the p r o b l e m (5.6) has at least one solution; (iii) there exist [sj] and [tj] such that the sign " < "
(5.9)
in (5.7) cannot be replaced by "_<".
Proof. Write o~ = 4pqL/Tz z and r = (r I . . . . . r,,), where rj -- ej(sj) = ~;(tj). By E we denote the space C([0, p] x [0, q ] , R n) with the s u p r e m u m n o r m . Define S, T:E---, E, h : E - - * E, H, G : E -, n(E) and D C E by
S(v)(x, Yb =
vAs, y) ds, ~,3j
T(v)(x, y)j =
Vg(X,t) dt, ,9
h(v)(x, y) = f ( x , y, ~p(x) + qJ(y) - r + T o S(v)(x, y)), H(v) = [w e E : Jw(x,y) G(v) = [ w ~ E : D=
<_ L [ T o S(v)(x,y)[I,
jw(x,y)[ < L(I~o(x)[ + v•E:
I~u(y)l
+ [r[) + M ] ,
Iv(x,y)lZdxdy<_pq ,0
0
.
Hyperbolic boundary value problems
523
N o t e that S(v) = u iff u, u x • E, ux = v on [ 0 , p ] × [0, q] a n d u~(sj,y) = 0 for y • [0, q], j = 1 . . . . . n. Similarly, T(v) = u iff u, Uy • E, Uy = v o n [ 0 , p ] x [0, q] a n d uj(x, t~) = 0 for x • [0, p], j = 1 . . . . . n. Using the F u b i n i t h e o r e m one can check t h a t S o T = T o S. A f t e r setting v = u.. the p r o b l e m (5.6) is equivalent to v = h(v), where v • E . I f v = h(v), t h e n u(x, y) = to(x) + qKY) - r + T ° S(v)(x, y) is a s o l u t i o n o f (5.6). F r o m l e m m a 1 it follows that
I'P !'q ,0
IS(v)tx, y)l z d x d y <_ ~p- 2 .
0
o .o "Lip i"q ]v(x,
y ) l2 d x dy,
4q2"t'Pti'qo Iv(x, y)l z dx dy. t'P t'q [T(v)(x'Y)IZ d x d y <- ~2-~t, •0 , O Using these two inequalities one can check that H(D) C o~D. A s b e f o r e , to show (i) we use t h e o r e m 1 a n d to show (ii) t h e o r e m 2. T o p r o v e (iii) c o n s i d e r on [0, p] × [0, q] the p r o b l e m U_rv = 4 ~ q U + /71,
v~ = 4p--~ u + rl2 , \ u(O, y) = u(x, O) = v(p, y) = v(x, q) = O, where u, v: [0,p] x [0, q] ~ R a n d r/ = (r/1 , rh) • C([0, p] x [0, q], R2). N o t i c e t h a t the righth a n d side o f this system satisfies (5.8) a n d (5.9) with L = rEZ/(4pq). A f t e r setting w = (Uxy, V~y) the a b o v e p r o b l e m is equivalent to 7T2
w = --T(S(wO 4pq
+ q.
(5.10)
F o r r/ = 0 (5.10) has a n o n t r i v i a l s o l u t i o n
w(x, y) = (cos(zrx/2p) cos(zry/2q), sin(zrx/2p) sinUzy/2q)). F r o m the F r e d h o l m a l t e r n a t i v e it follows that there exist r / e C ( [ 0 , p ] x [0, q], C) such t h a t (5.10) has no solution. This c o m p l e t e s the p r o o f .
Acknowledgement--The
author wishes to express his thanks to Professor Tadeusz Pruszko for suggesting example 2 and his help during preparation of this paper. REFERENCES
1. LASOTA A. & OPIAL Z., On the existence and uniqueness of solutions of nonlinear functional equations, Bull. Acad. Pol. Sci 15, 797-800 (1967). 2. LASOTA A. & OPIAL Z., Fixed point theorems for multi-valued mappings and optimal control problems, Bull. Acad. Pol. Sci. 16, 645-649 0968). 3. LASOTA A. & OLECH C., An optimal solution of Nicoletti's boundary value problem, Ann. Pol. Math. 18, 131-139 (1966). 4. KASPRZYK S. & MYJAK J., On the existence of solutions of the Floquet problem for ordinary differential equations, Zeszyty Nauk. Uniw. Jagiellon. Prace Mat. 13, 35-39 (1969).
524
D. BIELAWSKI
5. PRUSZKO T., Some applications of the topological degree theory to multi-valued boundary value problems, Dissnes Math. 229, 1-52 (1984). 6. BIELECKI A., Une remarque sur la m6thode de Banach-Caccioppoli-Tikhonov dans la th~orie des ~quations differentieUes ordinaires, Bull. Acad. Pol. Sci. 4, 261-264 (1956). 7. HARTMAN P., Ordinary Differential Equations. Birkh~user, New York (1982). 8. KISYIqSKI J. & PELCZAR A., Comparison of solutions and successive approximations in the theory of the equation O2z/Ox Oy = f ( x , y, z, OZ/Ox, Oz/Oy), Dissnes Math. 76, 1-77 (1970).
Pergamon
0362-546X(94)00294-0 ON OPTIMAL
SOLUTIONS OF BOUNDARY VALUE FOR DIFFERENTIAL EQUATIONS
PROBLEMS
DARIUSZ BIELAWSKI Institute of Mathematics, University of Gdafisk, ul. Wita Stwosza 57, 80-952 Gdafisk, Poland
(Received 18 February 1994; received in revised f o r m 26 August 1994; received f o r publication 19 October 1994) Key words and phrases: Picard problem, Nicoleni problem, Floquet problem, hyperbolic boundary value problems, existence of solutions, existence and uniqueness of solutions, equivalent norms. O. I N T R O D U C T I O N
We present a general construction of equivalent norms in Banach spaces. It appears that some maps which arise in boundary value problems for differential equations are contractive under these norms. Using completely continuous multivalued regulators Lasota and Opial showed in [1] that a vector field is a homeomorphism which means that the corresponding map has exactly one fixed point. In [2] they proved existence of fixed point for a multivalued map with convex and closed values and a suitable multivalued regulator. To construct equivalent norms we also use multivalued regulators. We prove two general theorems (theorems 1 and 2) without assuming that regulators are completely continuous. Using our method we extend classical results for the Nicoletti boundary value problem due to Lasota and Olech (see [3]) and for the Floquet boundary value problem due to Kasprzyk and Myjak (see [4]). We are also concerned with the Picard boundary value problem for the system of second order ordinary differential equations and some boundary value problems for hyperbolic partial differential equations which were posed in [5]. Obtained results seem to be the best possible in certain classes of the right-hand sides of considered systems. 1. A C O N S T R U C T I O N
OF EQUIVALENT
NORMS
In this paper (E, I1" [[) denotes a real or complex Banach space. By n ( E ) we understand the family of all nonempty subsets of E and by/~(0, r) the closed ball about 0 of radius r in E. For a multivalued map H : E ~ n ( E ) and c~ > 0 we will say that D C E is (H, u)-invariable if H(D) C ~D.
THEOREM 1. Suppose that h: E ~ E, a multivalued map H : E --- n(E), a e (0, 1) and D C E satisfy the following conditions H(l~x) = l~H(x)
for x 6 E, /~ ~ R (or C),
H ( x + y) C H ( x ) + H ( y ) H ( D ) C ~D,
/~(0, 1) C D,
h(x) - h ( y ) ~ H ( x - y)
Then, there exists a norm/1" space (E, I1" IIH).
IIH equivalent
(1.1)
for x, y ~ E,
(1.2)
H ( D ) is bounded,
(1.3)
for x, y c E.
(1.4)
to I1" II and such that h is contractive in the Banach 509
510
D. BIELAWSKI
P r o o f . For every x ~ E we define the minimal (H, o0-invariable set containing x by D(x) = A {C : C C E, x ~ C , H ( C ) C otC].
F r o m this definition it follows that x ~ D(x) and H ( D ( x ) ) C oLD(x). F r o m (1.1) and (1.3) we obtain
n(~-~llxlln(o) U B(O, [Ixll)) c ~ - ' l l x
IH(H(D)) U
n(l]xllO)
c IIxIIH(D) c ~(~-111xlIH(D)U B(O, Ilxll)). Since x ~
c~-~tlxItH(D) U B(O, Itxll),
f r o m the above inclusions we get
D(x) C
~-'IixlIH(D) U B(0, Ilxll).
(1.5)
Therefore, D(x) is b o u n d e d for every x e E. Define
llxll. = suplllyll : y ~ D(x)]. Using (1.1)-(1.3) one can check that II" IIH is a n o r m in the space E. We show for instance that IIx + ylIH -< Ilxll. ÷ Ilyll~, for x, y e E. Observe that D ( x + y) C D ( x ) + D ( y ) . Indeed, x + y ~ D(x) + D ( y ) and by (1.2) H ( D ( x ) + D ( y ) ) C H ( D ( x ) ) + H ( D ( y ) ) C ~¢(D(x) + D ( y ) ) .
Therefore,
tlx ÷ yll. = suplllzll
:z ~
D(x + y)l ~ sup[llzll
:z e
D(x) + D ( y ) I
-< suplllzll :z ~O(x)l + suplllzll :z ~ O(y)l = Ilxll. + IlYlI..
I1" II and I1" II.
N o w we show that the n o r m s
are equivalent. Since x ~ D(x), we have
Ilxtl. = suplllyll : y ~ O(x)} > Ilxll. O n the other hand, f r o m (1.5) we get
Ilxll. -< m a x ( l , - 1
suplllytl : y ~ n(o)l)llxll.
It remains to prove that for x, y e E
IIh(x) - h(y)[l~ <- ~llx - YlIH.
(1.6)
F r o m (1.4) we deduce that h(x) - h ( y ) ~ H ( x - y) C H ( D ( x - y)) C ¢xD(x - y).
Hence D(h(x) - h(y)) C a D ( x - y). This and the definition o f 1[. [in yield (1.6), which finishes the p r o o f . E x a m p l e 1. Consider the C a u c h y problem x ' U ) = f ( t , x(t)),
I
t ~ [0, p],
(1
x(0) = a,
where a e R" and f : [0,p] x R" ~ R" is a continuous m a p such that for a fixed K > 0 I f ( t , u) - f ( t , v)[ < K l u - v[
for t ~ [0,p], u, v ~ R".
Hyperbolic boundary value problems
511
Let E be the Banach space C([0,p], R n) of all continuous m a p s f r o m [0,p] to R n with the n o r m Ilxll = s u p l l x ( t ) l : t • [0,p]l. set o~ = 1 - e x p ( - K p ) . Define a h o m e o m o r p h i s m P : E --+ E by P(x)(t) = exp(Kt)x(t) and m a p s h: E -+ E, H : E ~ n(E) by h(x)(t) = a +
f(6, x(s)) ds, ~0
H(x) = c~P(/~(0,
IIP-l(x)ll)).
Now, the p r o b l e m (1.7) can be written as x = h(x), where x • E. Observe that for every x, y • E we have lIP
~ o hop(x)
-
P-~
o h op(y)ll
- o~llx- yll.
This implies that the condition (1.4) holds. After setting D = P(/~(0, 1)) it is possible to check that all remaining hypotheses of t h e o r e m 1 are satisfied and, therefore, h is contractive under the n o r m I[" ]In. Hence, the p r o b l e m (1.7) has exactly one continuously differentiable solution on
[0,p].
One can give an explicit expression for II'[In- Note that for every x • E we have H(oc-IH(x)) C H(x), x • u - i l l ( x ) and H(x) C H(D(x)) C teD(x). Thus, D(x) = c~-lH(x) and
Ilxll,, This means that
= sup[lly[[ : y • o~-'H(x)l = e x p ( K p ) s u p l e x p ( - K t ) l x ( t ) l : t
I1" IIH is the
• [0, PII.
classical Bielecki n o r m (see [6]) multiplied by the constant exp(Kp).
Example 2. Let h: E --, E be a linear b o u n d e d operator. Suppose that p < 1, where p is the spectral radius o f h (p = lim,,~=]]hm]]l/'~). Choose c~ • ( p , 1). Define H : E ~ n(E) by H(x) = [h(x)] and D C E by oo
D = ~_) a - m h " ( B ( 0 , 1)). m=0
F r o m t h e o r e m 1 it follows that h is contractive under the n o r m H" [[/4. This n o r m is equivalent to []. I[ and can be written as Ilxll~ = supl~-mllhm(x)ll
: m = 0, 1 ....
}.
THEOREM 2. Suppose that c~ • (0, 1), D C E, H : E - - * n(E) satisfy (1.1)-(1.3) and that a multivalued m a p G: E ~ n(E) has b o u n d e d image. Then, every completely continuous h: E - . E such that h(x) • H(x) + G(x)
for x • E,
(1.8)
has at least one fixed point. Proof. Sets D(x) and the n o r m I]']]n we define as in the p r o o f of t h e o r e m 1. Write M = supIIIYI[H :Y • G(x)] and R = M / ( I - ~). We claim that h(Bt4(0, R)) C /~H(0, R),
(1.9)
where/~H(0, R) = IY ~ E : [lY[[H -- R]. T o show (1.9) notice that H(/~/4(0, R)) C od)H(0, R).
(1.10)
512 In fact,
D. B1ELAWSKI for
[[xl[/4 - R and y ~ H ( x ) , we have y ~ H ( D ( x ) ) C o~D(x). This m e a n s that IlylIH <-- ~IIxIIH. F r o m (1.8) and (1.10) we conclude that
D ( y ) C o~D(x) and so
IlxllH --- e = IIh(x)llH ~ ~ g
+ M = R,
which proves (1.9). Since h is c o m p a c t o n / ~ u ( 0 , R), it has a fixed point. 2. THE PICARD PROBLEM Consider the Picard p r o b l e m x"(t) = f(t, x(t), x'(t)),
I
t ~ [0, p],
(2. 1)
x ( p ) = b,
x(O) = a,
where a, b ~ R" and f : [0, p] z R" x R" ~ R" is continuous. By a solution o f (2.1) we m e a n any x ~ Cz([O,p], R"), which satisfies (2.1).
THEOREM 3. Suppose that M > 0 and L 0, L I > 0 satisfy the inequality p2 ~Lo
+ PLtTr < 1.
Then: (i) if for all x, y, £, ~ e R", t e [0, p] ]f(t,x,y)
- f(t,£,~)[
<_ L o l x - £t + L , [ y - .~[,
(2.2)
then the p r o b l e m (2.1) has exactly one solution; (ii) if for all t 6 [0, p], x, y ~ R n If(t,x,y)l
<- L o l x
(2.3)
+ L~Iy] + M ,
then the p r o b l e m (2.1) has at least one solution; (iii) statements (i) and (ii) are not true for L0 = Or/P) 2. Statement (i) was proved by Picard under stronger a s s u m p t i o n ( L o p Z / 8 ) + ( L i p ~ 2 ) < 1 (see [7, p. 423]). T h e o r e m 3 is p r o b a b l y not optimal. H o w e v e r , this result is o p t i m a l if we restrict ourselves to the p r o b l e m x " = f ( t , x ) with the Picard b o u n d a r y value conditions x(O) = a, x ( p ) = b. P r o o f . Write c~ = ( L o p 2 / g 2) + (Lip~re). By E we denote the space C ( [ 0 , p ] , R n) o f all continuous m a p s f r o m [0,p] to R" with the n o r m ][x[[ = sup{]x(t)[ : t ~ [0,p]J. Define S: E ~ E , T: E --. E , h: E -~ E , H : E --, n ( E ) , G: E --* n ( E ) and D C E by S(x)(t) = P-
T(x)(t) =
1 P
(
(P
o sx(s) ds - .,, ( p - s)x(s) ds t)
,Jo
sx(s) ds + t
i" t
(p
(2.4)
,
s)x(s) ds
)
,
(2.5)
Hyperbolic boundary value problems
513
h(y)(t) = f ( t , (1 - (t/p))a + (t/p)b + T(y)(t), (1/p)(b - a) + S(y)(t)), H ( y ) = {z • E : lz(t)[ <- Lot T(y)(t)[ + L~]S(y)(t)II, G(y) = [z • E :
Iz(t)[ < M + Lo(lal + Ibl) + Z l l b - a l / P L
Is 1
D =
• E :
ly(t)l 2 d t _< p
0
1
.
N o t e t h a t T(x) = y iff y • Cz([0, p ] , R ' ) , y " = x o n [0, Pl a n d y(0) = y ( p ) = 0. S i m i l a r l y , S(x) = y i f f y • C l ( [ 0 , p ] , R ' ) , y ' = x o n [ 0 , p l a n d I~ y(s) d s = 0. A f t e r setting y(t) = x " ( O it is p o s s i b l e t o w r i t e the p r o b l e m (2.1) in an e q u i v a l e n t f o r m y = h(y), w h e r e y • E . If y = h(y), t h e n x(t) = (1 - (t/p))a + (t/p)b + T(y)(t) is a s o l u t i o n o f (2.1). C o n s i d e r the real H i l b e r t s p a c e ~ = Lz([0, p ] , R) a n d t w o o p e r a t o r s S, T: N ~ 3£ given b y (2.4) a n d (2.5), r e s p e c t i v e l y . F o r m = 1, 2 . . . . d e f i n e
era(t) = sin(rant~p),
t • [0,p].
O b s e r v e t h a t [em : m = 1, 2 . . . . } is a c o m p l e t e o r t h o g o n a l set in N . F u t h e r m o r e , pZ
T(e,~)(t) =
~
era(t),
m --- 1,2 . . . . ,
S(e,.)(t) = - p c o s ( m n t / p ) , mn
m = 1, 2 . . . . .
Hence,
IIf bl = p2/~z
Ilgll
and
=
P/n.
(2.6)
We claim that H ( D ) C (xD. In fact, i f y = (y~ . . . . . Yn) • D and z • H ( y ) , from the definition o f H and (2.6) we obtain
t.t 'P Iz(t)b z dt '/1/2 <_
o (Lo[ T(y)(t)l + L~[S(y)(t)l) 2 d t ) 1/2
0
<- Lo
[T(y)(t)l z dt
+ L1 IS(y)(t)l z dt \,10
',,,) 0 = Lo
(¢(yj)(t)) 2 dt j
l
p2 (
,'p
ly(012 d t
0
(~q(yj)(t)) 2 d t
j
-< v ,i:, to -- ~
+ L1
0 dt\l/2
) +
_< c~x/p.
1 0
p ( 7~ j = l
i
~p
,~i/z
514
D. B I E L A W S K !
We next show that H ( x + y) C H(x) + H ( y ) for x, y ~ E. I f z ~ H ( x + y), we define ~ and ~Eby
z(t)(to] T(x)(t)[ + t,lS(x)(t)[) ~(t) = Lol T(x)(t)l + L~lS(x)(t)[ + Zo] T(y)(t)l + Z~lS(y)(t)[' z(t)(Lol T(y)(t)l + LjlS(y)(t)[) ~(t) = Lo[ T(x)(t)l + Llla(x)(t)l + tol T(y)(t)[ + L~[S(y)(t)l (we set z~(t) --- 2.(t) -- 0 when the denominator equals 0). It is possible to check that ~ e H(x), ~ H ( y ) and z = ~ + ~. Now we prove that H(D) is bounded. In fact, for z ~ H(y), where y ~ D we have
Iz(t)l --- LolT(y)(t)l + LllS(y)(t)[ <- LoP
ly(t)[ d t + E l 0
<- (LoP
+
LI)
ly(t)t z dt
ly(t)l dt 0
~ LoP 2 + L i p .
~
0
Observe that (1.1) is satisfied and B(0, 1) C D. To prove (i) notice that the inequality (2.2) implies (1.4). Therefore, from theorem 1 we deduce that in this case the problem (2.1) has exactly one solution. To prove (ii) define hH, hG : E ~ E by
h(y)tt)(LolT(y)(t)] + L~IS(y)(t)I) ht4(y)(t) = t o l T ( y ) ( t ) I + Z~lS(y)(t)[ + Z0([a[ + Ibl) + (Z~lb - al/p) + M ' h(y)(t)(Lo(]a] + Ibl) + (Zllb - al/p) + M ) ha(y)(t) = LolT(y)(t)l + L1[S(y)(t)I + Lo([al + [b[) + (L~lb - al/p) + M " Note that h n + ho = h and that the inequality (2.3) implies h n ( y ) ~ H(y), h o ( y ) ~ G(y) for y ~ E. From theorem 2 we conclude that in this case the problem (2.1) has at least one solution. To show (iii) consider the problem
x"(t) = - - ~ x ( t )
+ rl(t),
t e [0,p],
k. x(O) = x(p) = o, where x e C2([0,p], R) and r / e C([0,p], R). Note that the right-hand side of the above problem satisfies (2.2) and (2.3) with constants L o = nZ/p z and Ll = 0. After setting x" = y we can write the above problem as 2l.2
y = --p~ T(y) + I1,
y e C([0,p], R),
(2.7)
where T is given by (2.5). Observe that for r/ = 0 (2.7) has a nontrivial solution y(t) = sin(nt/p). From the Fredholm alternative it follows that there exists t / ~ C([0,p], R) such that the problem (2.7) has no solution. This completes the proof.
Hyperbolic boundary value problems
515
3. T H E F L O Q U E T P R O B L E M
Consider the Floquet problem x~k)(t) = f ( t , x ( t ) , x ' ( t ) . . . . . x ~ k - n ( t ) ) ,
I
t e [0,p],
(3.1)
x(0) + 2x(p) = ro . . . . . X(k-l)(0) + 2X(k-1)(P) = r k - 1 ,
where ~ > 0, ro . . . . . rk_ 1 • R n and f : [0,p] x (R") k --. R n is continuous. By a solution o f (3.1) we m e a n any x • Ck([0,p], R") which satisfies (3.1). Set 0 = p / ~ / r t z + lnZ2. THEOREM 4. Suppose that M > 0 and Lo, L I . . . . . L k_ ~ > 0 satisfy the inequality + . . . . ~- OLk_t < 1.
OkLo + O k - ' L l
(3.2)
Then: (i) if for all t • [0, p], x o . . . . . xk- l, Yo . . . . . Yk- t • R" I f ( t , xo . . . . . x k _ , ) - f ( t , Yo . . . . . Y k - , ) [ <- Lo[X0 - Yol + "'" + L k _ l l x t - ,
- Yk-~[,
(3.3)
then the problem (3.1) has exactly one solution; (ii) if for all t • [0, p], Xo, ..., x k_ L e R" I f ( t , Xo . . . . . x k _ , ) [ -
tolXol + "'" + Lk-llXk-al + M,
(3.4)
then the problem (3.1) has at least one solution; (iii) the sign " < " in (3.2) c a n n o t be replaced by " _ " . For k = 1 this t h e o r e m was proved by Kasprzyk and M y j a k in [4]. First, we prove the following lemma. I.EMMA 1. Let ~ be the complex Hilbert space o f all x e L2([0, p], C) with the scalar p r o d u c t defined by (x, Y)~c =
22t/Px(t)y(t) dt. ~0
Let T: ~ ~ £ be given by 7"(x)(t)
Then I1711 = p/V
-
1
t" x ( s )
1 + 2 ,o
ds
tP x ( s ) d s . - -2 1 + )~ j t
(3.5)
2 + 1n22 = 0.
P r o o f . For m e Z define era(t) = exp((ln(1/2) + (2m - 1 ) i n ) t / p ) ,
t e [0,p].
Note that tem : m e Z] is a complete o r t h o g o n a l set in £ . O u r assertion is due to the fact that 7~(em) = p(ln(1/2) + (2m - 1)iTr)-lem
for m e Z.
516
D. BIELAWSKI
P r o o f o f theorem 4. Write a = O~Lo + o k - i L l + "" + O t k _ l . By E we denote the space C([0, p], R") with the supremum norm. Define T: E ---, E b y (3.5) and h: E --' E, H : E --' n(E), G: E ---, n(E), D C E by f ( t , r° + "'" + T k - l(rk-x)(t) + T g ( y ) ( t ) , . .
h(y)(t)
rk-------~l + T ( y ) ( t ) )
1 +~.
"'1 + 2
'
H ( y ) = {z • E : ] z ( t ) [ - ZolTk(y)(t)[ + "" + Lk-IIT(y)(t)[I,
Iz I, l"
• E : lz(t)[ < M + ~
G(y)=
D =
• E :
j=0
~ LjlTt-J(rt)(t)
l=j
il
,
1
2 2t/p [y(t)[ 2 dt __. p(1 + 2 2) .
o
Note that T(x) = y i f f y • CI([O,p], R"), y' = x on [0,pl and y(0) + 2y(p) = 0. After setting y = x (~) it is possible to write the problem (3.1) as y = h(y), where y • E. I f y = h(y), then
x(t) = ro + ... + rk-X(r~-l)(t) + T~(y)(t) 1+2 is a solution of (3.1). We claim that H ( D ) C e~D. In fact, if y • D and z • H ( y ) , from the definition of H and lemma 1, we get
(I~ '~2t/plz(t)[2 dt) w2 -<
(ip(l p
~.2'*(Lol Tk(y)(t)l + ... + Lk_l[ T(y)(t)I) 2 dt
0
<- Lo
2zt/"[ Tk(y)(t)[ 2 dt
)1/2
+ ... + Lk_~
\dO
(Sp
221/P IT(y)(t)[ 2 dt
0
,~2t/p('Fk(yj)(t))2 dt
= Lo
+ "'" + Lg-1
j 1 0
< Lo Ok =
)1/2
(____~lp22t/p(yj(t))2dt )1/2+ ... + Zk_lO j 1 0
0 22t/Ply(t)12dt/1/2 <- ax/p(1 +
221ml"(yj)(t)) z d
j 1 0 (~=
l~]L2t/p(yj(t))2dt) 1/2
j 1
22).
To prove (i) notice that the inequality (3.3) implies (1.4). It is also possible to check that all remaining hypotheses of theorem 1 hold and, therefore, in this case the problem (3.1) has exactly one solution. To prove (ii) observe that the inequality (3.4) implies (1.8). F r o m theorem 2 we conclude that the problem (3.1) has at least one solution.
Hyperbolicboundary value problems
517
TO prove (iii) suppose that tgkLo + O~-IL~ + ... + OLk_~ = 1, write ln(l/2) + #t and consider the problem x(~) = L o 6 l ( T k ) x - L o 3 ( y k ) y + ... + Lk_~(R(7)x ~k-~) - Lk_13(y)y (k-l) + r/l,
I
y(k) = Lo3(yk)x + L o ( R ( y k ) y + ... + Lk_~3(7)x(k-~) + L k _ I 6 ~ ( 7 ) y ( k - l ) + tlZ, X('D(O) + )tx(J)(p) = O,
y(J)(O) + )ty(J)(p) -= O,
j = 0 ..... k -
1,
where rtl, qz E 6"([0, p], R) and x, y e CtC([0, p], R). Notice that the right-hand side o f this system satisfies conditions (3.3) and (3.4). After setting z ~- x (e~ + iy (k) and 1/= r/~ + in 2 we can write the above problem as z = ? ~ L o T ~ ( z ) + ...
+ 7Lk-~T(z)
+ t]
(3.6)
(here we consider T given by (3.5) in the complex Banach space C([0,p], C) with the usual supremum norm). For ~/ = 0 (3.6) has a nontriviat solution exp((ln(1/2) + i n ) t / p ) . From the Fredholm alternative it follows that there exists r / ~ C([0, p], C) such that (3.6) has no solution. This completes the proof. 4. THE N1COLETTt PROBLEM Consider the Nicoletti problem I x(k~(t) = f ( t , x ( t ) , x ' ( t ) . . . . . x (k- 1)(t)),
Xj(t}(ttj)
rtj,
t E [0,p],
(4. 1)
I = 0 . . . . , k - 1, j = 1 . . . . . n,
where i t = ( t t l , . . . , b , ) ¢ [ O , p ] " , rt= (rtl,...,rt,)ER n for t = 0 . . . . . k - 1 and f : [0, p] × (R") k --. It n is continuous. By a solution o f (4.1) we mean any x e Ck([0, p], R") which satisfies (4.1). Set co = 2 p / n . THEOREM 5. Suppose that M > 0 and L o, L1 . . . . . L~.I > 0 satisfy the inequality ~okLo + ook-lL~ + ... + w L ~ _ Z < 1.
(4.2)
Then: (i) if for all t ~ [0, p], Xo . . . . . x k - ~ , Yo . . . . . y~_~ e R " If(t, Xo,---,xk-1)
- f ( t , Yo . . . . . Ye-~)I <- Lolxo - Yol + "'" + L k - l i X k - 1 -- Y k - l t ,
(4.3)
then the problem (4.1) has exactly one solution; (ii) if for all t e [0, p], Xo, . . . , x k _ t e R" I f ( t , Xo . . . . . xk- ,)I <- Lolxol + "'" + L k - ijxk_,l + M ,
then the problem (4.1) has at least one solution; Oil) there exist {to] such that the sign " < " in (4.2) cannot be replaced by " - " .
(4.4)
518
D. B I E L A W S K I
F o r k = 1 this t h e o r e m was p r o v e d by L a s o t a a n d Olech in [3]. F i r s t we p r o v e the f o l l o w i n g lemma. LEMMA 2. F o r every c o n t i n u o u s x : [0, p] ~ R a n d c e [0, p] we have
l (Six,s,4
dt -< oa2
,0
1p
xE(s) ds.
P r o o f . I f c > 0, c o n s i d e r the real H i l b e r t space ~ c o n t i n u o u s a d j o i n t o p e r a t o r s A , B: 3E --, 9]Z given b y A ( x ) ( t ) = ~. x(s) ds, J0
O b s e r v e t h a t em e ~
(4.5)
0
= L2([0, c], R) a n d two c o m p l e t e l y
B(x)(t) =
x(s) ds. t
(m = 1 , 2 . . . . ) d e f i n e d by em(t) = sin
(2m-
1)~t
2c
are eigenvectors o f the c o m p l e t e l y c o n t i n u o u s s e l f - a d j o i n t A B : ~g --' 91"(.. Thus, {era : m = 1 , 2 . . . . } is a c o m p l e t e o r t h o g o n a l set in ~ . N o t e t h a t 2c
B(em)(t) -
Hence
IIBII =
(2m-
(2m - l)Trt - - , 2c
cos
1)~
m = 1,2 .....
2c/r~ a n d for e v e r y x e C ( [ 0 , p l , R)
t (I 0
x(s) ds
xZ(s) ds.
dt <_
t
0
Similarly, one can show t h a t for x e C([O, p], R) c x(s) ds
x2(s) ds.
dt <_ . . . .
F r o m these two inequalities we get (4.5). This c o m p l e t e s the p r o o f o f l e m m a 2. P r o o f o f t h e o r e m 5. W r i t e a = cokL0 + ~ok-lL t + -.. + ¢ o L k _ 1 . The B a n a c h space (E, I1"11) we define as in Section 3. Define T~: E--* E (l = 0 . . . . . k - 1), h : E ~ E , H , G : E ~ n(E) a n d D C E by
(T~(y)(t))j =
yj(s) ds,
j = 1, . . . , n,
tO
h ( y ) ( t ) = f ( t , r o + To(rO(t) + ... + To . . . . . + TO . . . . . H ( y ) = {z ¢ E :
Tk_2(rk_l)(t)
Tk_l(Y)(t) . . . . . r k - l + Tk_l(Y)(t)),
Iz(t)l - LolTo . . . . .
T~_,(y)(t)l + ... + L k - l l T k - ~ ( Y ) ( t ) l } ,
G ( y ) = tz e E : Iz(t)l -< M + t01rol + tolTo(rD(t)l + "" + t o ITo . . . . .
D =
Tk_2(rk-O(t)l + .." +
e E:
ly(t) ,0
dt _< p
.
Irk-~l},
Hyperbolic boundary value problems
519
Note that T~(x) = y i f f y e C~([0,p], R"), y ' = x on 10,p] and yj(t o) = 0 f o r j = 1 . . . . . n. After setting y = x (*) it is possible to write the p r o b l e m (4.1) as y = h(y), where y e E. I f y = h(y), then x(t) = ro + To(rl)(t) + "'" + To . . . . .
Tk_2(rk_l)(t) + To . . . . .
Tk_l(y)(t)
is a solution o f (4.1). F r o m l e m m a 2 it follows that Ti . . . . .
Tk_I(D) C ~k-~D
for 1 = 0 . . . . . k - 1.
Using this inclusion one can show similarly as in Section 3 that H ( D ) C o~D. T o prove (i) notice that the inequality (4.3) implies (1.4). It is also possible to check that all remaining hypotheses o f t h e o r e m 1 hold and, therefore, in this case the p r o b l e m (4.1) has exactly one solution. T o prove (ii) observe that the inequality (4.4) implies (1.8). F r o m t h e o r e m 2 we conclude that the p r o b l e m (4.1) has at least one solution. To show (iii) suppose that ~ k L 0 + ~ k - 1Lj + ... + ~ L k - i = I and define linear J : R z ~ R z by J(u, v) = (v, - u ) . Consider the p r o b l e m x k) = Lodk(x) + ... + L ~ _ i J ( x O'-t)) + q,
I
xf)(to) = O,
l= 0 ..... k-
l , j = 1,2,
where x e C k ( [ 0 , p ] , 112), q E C([0, p], Rz), t 0 = 0 if l + j is odd and t o = p if / + j is even. Notice that the right-hand side of this system satisfies (4.3) and (4.4). After setting y = x (k) we can write the above p r o b l e m as y = LoJk(To . . . . .
Tk-1(Y)) + "'" + L ~ _ I J ( T k - I ( Y ) ) + ft.
(4.6)
For q = 0 (4.6) has a nontrivial solution y ( t ) = (sin(k)(rrt/2p), cos(k)(nt/2p)). F r o m the F r e d h o l m alternative it follows that there exists q e C([0, p ] , R z) such that (4.6) has no solution. This completes the p r o o f . 5. H Y P E R B O L I C
BOUNDARY VALUE PROBLEMS
Consider the hyperbolic p r o b l e m l u~ = f ( x , y, u),
(x, y) e [0, p] X [0, q],
u(O,y) + 2 ~ u ( p , y ) = ~u(y),
(5.1)
u(x,O) + ,~2u(x,q) = q~(x),
where 2~, 22 > 0 , f : [0,p] x [0, q] x R" ~ R" is continuous, ~0: [0,p] --' !1", ~: [0, q] ---, R" are continuously differentiable and ~0(0) + 2~0(p) = qJ(0) + 2zq/(q). By a solution o f (5.1) we m e a n a continuous u: [0, p] x [0, q] ~ R n with continuous derivatives ux, Uy, U~y = Uyx which satisfies (5.1) on [0, p] × [0, q]. Set 0F = p/x/~& + In~ ~ and 02 = q/x/r& + In 2 2z. THEOREM 6. Suppose that M > 0 and L > 0 satisfies the condition ~/zr 2 + l n 2 2 t ~ 2 L<
Pq
+ ]nZAz
(5.2)
520
D. B I E L A W S K I
Then: (i) if for all x • [0, p], y • [0, q], u, v • R"
l f i x , y, u ) - f i x , ) ,
v)t < - L [ u -
v[,
(5.3)
then the problem (5.1) has exactly one solution; (ii) if for all x • [0, p], y • [0, ql, u • R"
I f ( x , y , u)] -< Llut + M ,
(5.4)
then the problem (5.1) has at least one solution; (iii) the sign " < " in (5.2) c a n n o t be replaced by " _ " .
Proof. Write cz = 0102L. By E we denote the space C([0, p] x [0, q], R") with the s u p r e m u m n o r m . Define S, T: E ---, E by
S(v)(x,y)
-
- -
1 + 21 .o
T(v)(x,y)-1+
v(s, y) ds
1 2~ [-'v J0 v(x,
t)dt
1 + 21
f, x
v(s, y)
1+2222 o~I'"v(x,
ds,
t) dt.
Note that S ( v ) = u iff U, U x • E , ux = v on [0,p] x [0, q] and u ( O , y ) + 2 1 u ( p , y ) = 0 for y • [0, q]. Similarly, T(v) = u iff u, Uy • E, Uy = v on [0,p] x [0, q] and u(x, O) + 22u(x, q) = 0 for x • [0, p]. Using the Fubini t h e o r e m one can check that S o T = T o S. Define also h: E --, E, H, G: E --* n(E) and D C E by
h(v)(x,y) = f
(co(x) + q/(y) x , y , 1 + 2-----~ 1 + 2~
co(O) + 21(p(p) (1 + ,ll)(1 + ,~2) + r o S ( v ) ( x , y )
) ,
,
H(v) = [w • E : Iw(x, y)l <_ L[ T o S(v)(x, y)l 1, G(v) = I w e E :
D=
/
I w ( x , y ) l <-L(I¢(x)I
veE: ,0
+ [~u(y)l +
Ico(o)l
q )~21x/PR~Y/qlv(x,y)12dxdy<_pq(1
+ Ico(p)l)+ MI, + 221)(1 + 2 ~z) .
,0
1
After setting v = u~, the problem (5.1) is equivalent to v = h(v), where v • E. If v = h(v), then
u(x, y) -
C0(x)
1 + 2z
q/(y)
+ - -
1 + 21
¢(0) + 21CO(p) + T o S(v)(x, y) 22)
(1 + 21)(1 +
is a solution o f (5.1). The inclusion H(D) C ~ D follows f r o m the following two inequalities
,0
'~ 'q 22,x/p2}"~qls(v)(x,y)lZ d x d y _< Jr 2 _~ ~nz 2~ ,o ~o ,0
,~, ,~2 , 'q~2x/P~2Y/qlT(v)(x,y)12 d x d y <-- ~z + 1n222 . o jo 2a(/P2zzY/qlv(x,Y)12 d x d y , ,0 ,0
Hyperbolic boundary value problems
521
Using l e m m a 1 we show for instance the first is
]L2x/p)L2Y/qIS(u)(x,y)l 2 d x dy = ~i"q~2y/qti~ IP~I o oq o <_
'q(
2~y/qo 2
= O~
o o
~2x/PIS(u)(x, y)l 2 dxI
Sop
dy
2zx/ulv(x,y)12 d x d y
A~x/P).22Y/qlv(x,y)12dxdy.
N o w we prove that H(u + v) C H(u) + H(v) for u, v e E. If w e H(u + v), we define ~,, ~ b y
w(x, Y)I T(S(u))(x, y)[ ¢v(x, y) = IT(S(u))(x, Y)I + IT(S(v))(x, Y)I ' w(x, y)[ T(S(v))(x, y)[ ~'(x, y) = I T(S(u))(x, Y)I + I T(S(v))(x, Y)I (we set ~ ( x , y ) = ~ ( x , y ) = 0 if the d e n o m i n a t o r equals 0). It is possible to check that w = ¢v + f¢, ¢v ~ H ( u ) a n d ~ e H ( v ) .
T o show (i) we use t h e o r e m 1. T o prove (ii) define hn, hc : E -, E by h(v)(x, y)LI T(S(v))(x, y)[ h . ( v ) ( x , y ) = L(l~p(x)l + I~(Y)I + 1¢(0)1 + I~P(R)I + Ir(S(v))(x,y)l) + M '
h(v)(x,y)(L(l~(x)l + ]~(Y)I + [¢p(O)l + [ ¢ ( P ) I ) + M ) hG(v)(x'Y) = L(l~(x)l + I~'(Y)I + 10,(0)l + [¢(p)l + IT(S(v))(x,y)[) + M" Notice that hn + hc = h and that the inequality (5.4) implies ht-t(v) ~ H(v), hG(v) ~ G(v) for v ~ E . F r o m t h e o r e m 2 we conclude that the p r o b l e m (5.1) has a solution. T o show (iii), write y = 0 n ( 1 / 2 0 + in)(ln(1/22) + in)/(pq) and consider on [0,p] × [0, q] the p r o b l e m Uxy = ~ ( y ) u - 3(y)v + ~h,
I
v~ = 3(y)u + 6~(y)v + ~h,
u(O,y) + 2 ~ u ( p , y ) = v(O,y) + 2 1 v ( p , y ) = O, u(x, O) + 22u(x, q) = v(x, O) + 2zv(x, q) = O,
where u, v: [0,pl × [0, q] ~ R and e l , r/z e C([0,p] × [0, q], R). Notice that the right-hand side of this system satisfies (5.3) and (5.4) with L = (01 02) -1 After setting w = U~y + iVxy and r/ = r/1 + it/2 we can write the a b o v e p r o b l e m as
w = vT(S(w)) + rl
(5.5)
(here we consider S and T in the complex Banach space C([0, p] × [0, q], C) with the s u p r e m u m norm). For r/ = 0 (5.5) has a nontrivial solution
w(x, y) = e x p ( ( l n ( l / 2 0 + in)x/p)exp((ln(1/Az) + in)y/q).
522
D. BIELAWSKI
F r o m the F r e d h o l m alternative it follows that there exists r / e C([0,p] × [0, q], C) such that (5.5) has no solution. This completes the p r o o f Consider the hyperbolic p r o b l e m
l Uxy = f ( x , y, u),
(x, y) • [0, p] × [0, q],
uj(Sg, y) = q/j(y),
uj(x, tj) = ~pj(x),
(5 o6~ j = 1 . . . . . n,
where f : [0,p] × [0, q] × R n ~ R" is continuous, ¢0: [0,p] --, R n, ~,: [0, q] --, R" are continuously differentiable, sj e [0,p], ti e [0, q] and ~0j(s~) = ~j(tj) f o r j = 1 . . . . . n. By a solution o f (5.1) we m e a n a continuous u: [0, p] × [0, q] --, R" with continuous derivatives ux, Uy, Uxy = Uyx which satisfies (5.6) on [0, p] × [0, q]. The above p r o b l e m is a generalization of the well-known D a r b o u x p r o b l e m (see [8] and the references given there). THEOREM 7. Suppose that M > 0 and L > 0 satisfy the condition ~2 t
~ _m
4t~q"
(5.7)
Then: (i) if for all x ~ [0, p], y • [0, q], u, v ~ R n
[f(x,y, u)-f(x,y,
v) I --< L l u -
v]
(5.8)
then the p r o b l e m (5.6) has exactly one solution; (ii) if for all x ~ [0,p], y e [0, q], u • R ~
] f ( x , y , u)[ _< Llu[ + M , then the p r o b l e m (5.6) has at least one solution; (iii) there exist [sj] and [tj] such that the sign " < "
(5.9)
in (5.7) cannot be replaced by "_<".
Proof. Write o~ = 4pqL/Tz z and r = (r I . . . . . r,,), where rj -- ej(sj) = ~;(tj). By E we denote the space C([0, p] x [0, q ] , R n) with the s u p r e m u m n o r m . Define S, T:E---, E, h : E - - * E, H, G : E -, n(E) and D C E by
S(v)(x, Yb =
vAs, y) ds, ~,3j
T(v)(x, y)j =
Vg(X,t) dt, ,9
h(v)(x, y) = f ( x , y, ~p(x) + qJ(y) - r + T o S(v)(x, y)), H(v) = [w e E : Jw(x,y) G(v) = [ w ~ E : D=
<_ L [ T o S(v)(x,y)[I,
jw(x,y)[ < L(I~o(x)[ + v•E:
I~u(y)l
+ [r[) + M ] ,
Iv(x,y)lZdxdy<_pq ,0
0
.
Hyperbolic boundary value problems
523
N o t e that S(v) = u iff u, u x • E, ux = v on [ 0 , p ] × [0, q] a n d u~(sj,y) = 0 for y • [0, q], j = 1 . . . . . n. Similarly, T(v) = u iff u, Uy • E, Uy = v o n [ 0 , p ] x [0, q] a n d uj(x, t~) = 0 for x • [0, p], j = 1 . . . . . n. Using the F u b i n i t h e o r e m one can check t h a t S o T = T o S. A f t e r setting v = u.. the p r o b l e m (5.6) is equivalent to v = h(v), where v • E . I f v = h(v), t h e n u(x, y) = to(x) + qKY) - r + T ° S(v)(x, y) is a s o l u t i o n o f (5.6). F r o m l e m m a 1 it follows that
I'P !'q ,0
IS(v)tx, y)l z d x d y <_ ~p- 2 .
0
o .o "Lip i"q ]v(x,
y ) l2 d x dy,
4q2"t'Pti'qo Iv(x, y)l z dx dy. t'P t'q [T(v)(x'Y)IZ d x d y <- ~2-~t, •0 , O Using these two inequalities one can check that H(D) C o~D. A s b e f o r e , to show (i) we use t h e o r e m 1 a n d to show (ii) t h e o r e m 2. T o p r o v e (iii) c o n s i d e r on [0, p] × [0, q] the p r o b l e m U_rv = 4 ~ q U + /71,
v~ = 4p--~ u + rl2 , \ u(O, y) = u(x, O) = v(p, y) = v(x, q) = O, where u, v: [0,p] x [0, q] ~ R a n d r/ = (r/1 , rh) • C([0, p] x [0, q], R2). N o t i c e t h a t the righth a n d side o f this system satisfies (5.8) a n d (5.9) with L = rEZ/(4pq). A f t e r setting w = (Uxy, V~y) the a b o v e p r o b l e m is equivalent to 7T2
w = --T(S(wO 4pq
+ q.
(5.10)
F o r r/ = 0 (5.10) has a n o n t r i v i a l s o l u t i o n
w(x, y) = (cos(zrx/2p) cos(zry/2q), sin(zrx/2p) sinUzy/2q)). F r o m the F r e d h o l m a l t e r n a t i v e it follows that there exist r / e C ( [ 0 , p ] x [0, q], C) such t h a t (5.10) has no solution. This c o m p l e t e s the p r o o f .
Acknowledgement--The
author wishes to express his thanks to Professor Tadeusz Pruszko for suggesting example 2 and his help during preparation of this paper. REFERENCES
1. LASOTA A. & OPIAL Z., On the existence and uniqueness of solutions of nonlinear functional equations, Bull. Acad. Pol. Sci 15, 797-800 (1967). 2. LASOTA A. & OPIAL Z., Fixed point theorems for multi-valued mappings and optimal control problems, Bull. Acad. Pol. Sci. 16, 645-649 0968). 3. LASOTA A. & OLECH C., An optimal solution of Nicoletti's boundary value problem, Ann. Pol. Math. 18, 131-139 (1966). 4. KASPRZYK S. & MYJAK J., On the existence of solutions of the Floquet problem for ordinary differential equations, Zeszyty Nauk. Uniw. Jagiellon. Prace Mat. 13, 35-39 (1969).
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D. BIELAWSKI
5. PRUSZKO T., Some applications of the topological degree theory to multi-valued boundary value problems, Dissnes Math. 229, 1-52 (1984). 6. BIELECKI A., Une remarque sur la m6thode de Banach-Caccioppoli-Tikhonov dans la th~orie des ~quations differentieUes ordinaires, Bull. Acad. Pol. Sci. 4, 261-264 (1956). 7. HARTMAN P., Ordinary Differential Equations. Birkh~user, New York (1982). 8. KISYIqSKI J. & PELCZAR A., Comparison of solutions and successive approximations in the theory of the equation O2z/Ox Oy = f ( x , y, z, OZ/Ox, Oz/Oy), Dissnes Math. 76, 1-77 (1970).