Miroslawa Zima
Positive operators in Banach spaces and their applications
WYDAWNICTWO UNIWERSYTETU RZESZOWSKIEGO RZESZ6W 2005
Opiniowali prof. dr hab. J6ZEF BANAS prof. dr hab. JAROSLAV ZEMANEK
© Copyright by Wydawnictwo Uniwersytetu Rzeszowskiego Rzesz6w 2005
ISBN 83-7338-290-9 200 WYDAWNICTWO UNIWERSYTETU RZESZOWSKIEGO 35-310 Rzesz6w, ul. Cwiklinskiej 2, skr. poczt. 155, tel. 872 16 95 Ark. wyd. 4 Format B5 Wydanie I Zam. 90/2005 Ark. druk. 6,5 WYKONANO W DRUKARNI UR W RZESZOWIE
Contents Introduction 1 Positive operators in partially ordered Banach spaces 1.1 Cones in Banach spaces . . . . . . . . . . . 1.2 Positive linear operators . . . . . . . . . . . 1.3 Cone compression and expansion operators 2 Applications of positive operators to differential and integral equations 2.1 Initial and boundary value problems for functional - differential equations . . . . 2.2 Boundary value problems for second order differential equations . . . . . . . . . . . . . 2.3 Hammerstein integral equation on the half-line 2.4 Boundary value problems for differential equations in Banach spaces . . . . . . . . . . . . . . . . . . . . Bibliography
7 9 9 16 32
49
49 64 73
85
97
Introduction The origins of theory of partially ordered normed spaces and closely related to it cone theory in Banach spaces can be found in the papers of L. V. Kantorovic ([31]) and M. G. Krein ([38]) published in 1936 and 1940. Later on several significant results on positive operators were obtained for example by M.A. Krasnoselskii, V. Ya. Stetsenko, P. P. Zabreiko, H. Schaefer, V. Lakshmikantham, D. Guo and W. V. Petryshyn.· Let us mention that by a positive operator we mean a mapping which leaves invariant a cone in a Banach space.
The theory of positive operators is very important in many topics of functional analysis. The methods based on this theory are effective, precise and have a broad spectrum of applications. In the present monograph we focus our attention on a few selected problems from this area. In the first section of Chapter 1 we recall some basic definitions and results from the cone theory in Banach spaces. Subsequently we study the properties of the spectral radius and the local spectral radius of positive bounded linear operators and their applications to the fixed point theory. We are mainly concerned with the subadditivity and submultiplicativity of the spectral radius. Moreover, the relations between the spectral radius and its local counterpart are discussed. In Chapter 1 we also study positive operators compressing or expanding a cone in a Banach space. In 1960, Krasnoselskii proved a theorem on the existence of a positive fixed point for a cone expanding or compressing completely continuous operator (see for example [35]). Krasnoselskii's theorem was then generalized in many directions. Particularly, in 1988, D. Guo proved its norm type version using the concept of the fixed point index of positive completely
8
Introduction
continuous operators. The principal significance of cone expansion and compression theorems derives from the fact that they provide not only the existence of a fixed point but some additional information on its properties as well. Our purpose is to present a few further improvements of Krasnoselskii-Guo theorem for strict set contractions and condensing operators. We study the existence of at least one fixed point for cone compressions and expansions. The multiplicity theorems and their applications can be found in [23], [24], [25], [41] and [65]. For example, in [41], Leggett and Williams presented criteria which guaranteed the existence of at least three fixed points for positive completely continuous operator. It should be emphasized that we deal with the singlevalued operators only. Some extensions of Krasnoselskii's type theorems for multivalued mappings can be found for instance in [53]. It is worth pointing out that Theorems 1.3.11 and 1.3.12 from the last section of Chapter 1 have been recently extended to the case of multivalued mappings ([80]). Chapter 2 is devoted to the examples of applications of the abstract results from Chapter 1 to the theory of differential and integral equations. We study the existence and uniqueness of solutions of nonlinear integral equations and initial value problems for differential equations with deviated argument. Using the properties of the spectral radius and a generalization of Banach's contraction principle (Theorem 1.2.12) we establish criteria which guarantee that the equations under consideration have a unique solution. We are also concerned with boundary value problems for second order differential equations and Hammerstein integral equations. In order to obtain the existence of positive solutions we apply a variety of fixed point theorems on cone expansion and compression. It is noteworthy that there is a vast amount of literature concerning positive operators and their applications. Let us mention here only a few monographs of general interest. For instance, the detailed discussion on cone theory in Banach spaces can be found in [17], [24] and [37]. A comprehensive survey of results on
positive bounded linear operators is presented in [36] and [65]. Finally, numerous applications of Krasnoselskii's type theorems to boundary value problems and integral equations are discussed in [1].
Chapter 1
Positive operators in partially ordered Banach spaces 1.1
Cones in Banach spaces
The purpose of this section is to provide some background material from the theory of cones in Banach spaces. We restrict our attention to these cones which are essential for our further considerations. Namely, we will discuss the basic properties and a few examples of normal, solid and generating cones. A comprehensive treatment is given for instance in [17], [24] and [37].
Definition 1.1.1. A nonempty subset P, P is called a cone if P is closed, convex and
=f. {0}, of a
real Banach space E
(i) Ax E P for all x E P and A 2: 0, (ii) x, -x E P implies x = 0.
Obviously, P +PC P. It is well-known that every cone induces a partial ordering in E as follows: for x, y E E we say that x :::5 y if and only if y - x E P. Moreover, we will write
x-
Chapter 1. Positive operators
10 and X
-t y
X ~
if y -
P.
Now we pass on to the cones which possess some additional properties. First, we discuss the concept of a normal cone.
Definition 1.1.2. A cone P is said to be normal if there exists "( > 0 such that if () j x :::S y, then
llxll
:S:
'YIIYII·
The smallest number "( satisfying the above condition is called the normal constant of P. Obviously, "( 2:: 1.
In other words, a cone is normal if the norm in E is semimonotonic. The notion of a normal cone is very useful and will be often used later on.
Lemma 1.1.1. A cone P is normal if and only if there exists 6 > 0 such that for all x,y E P
llx + Yll 2:: 6llxll· Proof. Assume that P is a normal cone. Let x, y E P. Since () j x j x+y,
the normality of P implies 1
llx + Yll 2:: -llxll. "( Suppose now that there exists 6 > 0 such that
llx + Yll 2:: 6llxll for all x, y E P. If () j x j y then y - x E P. In consequence
IIYII = llx + Y- xll 2:: 6llxll, which means that P is normal.
0
1.1. Cones in Banach spaces
11
Some further necessary and sufficient conditions for a cone to be normal can be found for example in [24, Theorem 1.1.1] and [37]. Now we will give an example of a normal cone. Example 1.1.1. In the Banach space C[a, b] of all continuous functions on [a, b] with the norm
llxll = tE[a,b] max lx(t)l consider the set P
= {x
E
C[a, b] : x(t)
2: 0 on [a, b]}.
It is easy to check that P is a cone in C[a, b]. The relation given by P is the following: x ~ y if and only if x(t) ::;: y(t) on [a, b]. Thus, if () ~ x ~ y then llxll ::;: IIYII· This means that P is a normal cone. Clearly, its normal constant 1 = 1.
The following example shows that not every cone is normal. Example 1.1.2. Consider the Banach space C 1 [0, 1] of continuously differentiable functions on [0, 1] with the norm
llxll = tE[O,lj max lx(t)l + tE[O,Ij max lx'(t)l. Let P
= {x
E
C 1 [0, 1] : x(t) 2: 0 on [0, 1]}.
Obviously, Pis a cone in C 1 [0, 1]. Suppose that there exists 1
() ~ x ~ y implies Put Xn(t) = sinmrt + 1 and Yn(t)
() ~ Xn ~ Yn,
= 2 on
> 0 such that
llxll :S: IIIYII[0, 1]. Then
llxnll = 2 + mr
and
llYn I = 2
for every n E N. In consequence, 2 + mr ::;: 21 for each n E N, a contradiction. Thus P is not a normal cone.
12
Chapter 1. Positive operators
The following property, mentioned in the papers [12), [52), [53), is valid for any cone in E.
Lemma 1.1.2. Let P be a cone in E. Then for every u E P\ {0} there exists
a positive number a(u) such that
llx + ull ~ a(u)llxll for all x
E
P.
Proof. Fix u E P \ {0}. Suppose, for the contrary, that for every n EN there exists Xn E P such that
(1) If the sequence {llxnll} is bounded then (1) implies
lim
n->oo
Xn
=
-u.
t.
Obviously, -u P. This contradicts the fact that P is closed. If {llxnll} is unbounded from above then we can choose a subsequence {xnk} c P\ {0} such that
Then from ( 1) we get for every k E N
Hence
1-~<...!:.. llxnkll
for every k E N, which is impossible.
nk
0
It is not difficult to show that a(u) ::; 1, see [53). So, to be precise, we can define a(u) as the largest positive number satisfying Lemma 1.1.2.
13
1.1. Cones in Banach spaces Corollary 1.1.1. [52] Let P be a cone in E and u E P \ {0}. If a(u) such that
> 0 is
llx + ul! ?:: a(u)llxll for all x E P, then for every A > 0
llx + Aull ?:: a(u)llxll· Proof. Let A> 0. Then by Lemma 1.1.2 1
1
llx +Au II= All-;xx + ull ?:: Aa(u)ll-;xxll = a(u)l!xl!. 0
Example 1.1.3. Let P be the cone considered in Example 1.1.2, that is,
P = {x Fix u(t)
E
C 1[0, 1]: x(t)?;: 0, t
=1 on [0, 1]. Then a(u)
E
[0, 1]}.
= 1.
Next, we will discuss another important class of cones.
Definition 1.1.3. A cone P is called solid if its interior is nonempty, that is intP f 0.
Observe that the cone considered in Example 1.1.1 is solid. Its interior consists of all functions with positive minimum on [a, b]. The same conclusion can be drawn for the cone from Example 1.1.2.
Lemma 1.1.3. Let P be a solid cone in E and xo E int P. Then for each x E E there exists f3(x) > 0 such that -f3(x)xo :5 x :5 f3(x)xo.
Chapter 1. Positive operators
14
Proof. Since xo E int P, there exists r > 0 such that B(xo, r) C P, where B(x0 , r) denotes the open ball centered at xo with radius r. Let x E E \ {0}. Then r r
xo- 2llxllx, xo + 2llxllx
E
B(xo,r).
Hence
0
Corollary 1.1.2. If P is a solid cone in E and xo E int P, then for each x E E there exists 6(x) > 0 such that
xo- 6(x)x
E
P.
Proof. By Lemma 1.1.3, for every x E E there exists ,B(x) ,B(x)xo- x E P. Hence
> 0 such that
1
xo- ,B(x)xEP
0
which proves our assertion.
Definition 1.1.4. A cone P is said to be generating (or reproducing) if E = P- P, where P- P = {x E E: x = u- v, u,v E P}.
Now we will study the relations between solid and generating cones.
Lemma 1.1.4. [24] If P is a solid cone, then P is generating.
Proof. Let xo E int P. Then there exists r the proof of Lemma 1.1.3 we have
> 0 such that B(xo, r)
r r xo- 2llxllx, xo + 2llxllx
E
B(xo,r)
C
P. As in
15
1.1. Cones in Banach spaces for every x E E \ {0}. Put u
llxll ( r ) =-;:xo + 2llxllx
llxll (
and
v = -;:-
r ) xo - 2 llxll x .
Then u, v E p and X= u- v, so X E p- P. Obviously, E = P- P, which completes the proof.
eE
p- P. Thus
0
It follows from the above lemma that the cones considered in Examples 1.1.1 and 1.1.2 are generating. In particular, for any x E C[a, b] we have x = u- v, where
u(t) = max{x(t), 0}, v(t) =max{ -x(t), 0}, t
E
[a, b].
We complete this section with an example of a generating cone with empty interior. Example 1.1.4. Let co denote the Banach space of all real sequences x = {xi} convergent to zero with the supremum norm llxll =sup lxil· iEN
The set
P=
{X E
Co :
Xi ~
is a cone in CQ. Observe that every x E x = u- v, where
0, i E N'}
eo
can be expressed in the form
Therefore P is generating. We will show that P is not solid. Let x ={xi} E P. Since li!ni--+oo Xi = 0, we have
1\
v 1\-. <0. Xi-~
r>O i 0 EN i>io
For arbitrary r
z
> 0 define i = 1, 2, ... , io,
Xi,
Yi=
{
x·-r. t 2
.
- - .- , Z
z
. > ZQ.
16
Chapter 1. Positive operators
Then y = {yi} E
eo
and
1\ lxi- Yil ~ ~iEN
Hence y E B(x, r). But y t. P because Yi arbitrary, we conclude that int P = 0.
1.2
< 0 for i > io. Since x and r are
Positive linear operators
In this section we will study the properties of the spectral radius of bounded linear operators. Particularly, we will be concerned with the estimations of the spectral radius and local spectral radius for the sum and composition of operators acting in partially ordered Banach spaces.
Definition 1.2.1. Let P be a cone in E and A: E---+ E be a linear operator. We say that A is positive if A(P) C P. It is an easy consequence of the above definition that A(P) c P is equivalent to the fact that A is increasing, that is, if x :::5 y, then Ax :::5 Ay.
Definition 1.2.2. Let P be a cone in E and uo E P \ {0}. A positive linear operator A is said to be uo-upper bounded if for every x E P there is jj(x) > 0 such that Ax :::5 jj(x)uo. The proof of the following result is similar to that of Lemma 1.1.3.
Lemma 1.2.1. Suppose that P is a solid cone in E, uo E int P and A is a positive bounded linear operator. Then A is uo-upper bounded. Let us denote by C(E) the Banach space of all bounded linear operators from E into E.
Definition 1.2.3. Let A E C(E). The number (2)
r(A) = lim
is called the spectral radius of A.
n-+oo
IIAnlll/n
1.2. Positive linear operators
17
It is interesting to note that the limit lim IIAnlll/n
n-><Xl
always exists and
=
lim IIAnlll/n
n--->oo
inf IIAnlll/n.
nEN
This is the consequence of the obvious inequality
which means that the sequence an= IIAnll is submultiplicative, that is,
It can be also proved that for given
E
> 0 there exists a norm 11·11" equivalent
to II · II such that
r(A) :S IIAIIc :S r(A)
+E
(see [17], [36]).
Remark . The spectral radius can be also defined in a different way. Let E be a complex Banach space and A E .C(E). Denote by a(A) the spectrum of A. Then the number
(3)
r(A) = max{IAI: A E a(A)}
is called the spectral radius of A. For a real Banach space r(A) = r(A) by definition, where A stands for the complexification of A. Recall that the complexification E of a real Banach space E is defined as the complex Banach space consisting of all pairs (x, y), denoted by x + iy, x, y E E. The addition and the scalar multiplication are defined in the natural way. The norm II · II E in E is given by llx + iyllt
=
max II (cos t)x
tE[0,27r]
+ (sin t)yll-
By the complexification of A E .C(E) we mean the operator by
A(x + iy) =Ax+ iAy, x, y E E.
A E .C(E)
defined
Chapter 1. Positive operators
18
It is easy to show that IIAII_g = IIAII. The spectrum of A is defined as follows: u(A) = u(A). It can be shown that
The proof of the above equality (well-known as Gelfand's formula) can be found for instance in [50] and [56]. In our considerations we will deal with the real Banach spaces only and use the formula (2) mainly.
From (2) we get immediately the following properties of r(A):
a) r(A)::::; IIAII, b) r(.XA) = I.XIr(A) for all .X E R, c) r(Ak) = [r(A)]k for all kEN.
In general, it is not easy to find r(A) by (2). Thus, in order to calculate r(A) we often use the notion of the local spectral radius of a bounded linear operator.
Definition 1.2.4. Let A
E
C(E) and x E E. The number
(4)
r(A, x) =lim sup IIAnxll 1/n n-+oo
is called the local spectral radius of A at x.
It is easy to verify that r(A, x) ::::; r(A) for every x E E. Let us point out that the limit lim IIAnxlll/n n-+oo
may not exist. Unlike the sequence {IIAnll} the sequence {IIAnxll} for fixed x E E is, in general, not submultiplicative.
1.2. Positive linear operators
19
Example 1.2.1. [40] Fork, n EN consider the sequence
where 2n-l ~ k < 2n. Observe that {ak} is a decreasing sequence. Moreover, if kn = 2n-l then lim (a )1/kn = e-2 n->oo
while for
k~ =
kn
'
2n - 1 we get
Therefore the sequence {a~/k} is divergent. For x = {xk}, defined by
where k E N, we have 00
00
L lxkl = L(ak- ak+I) = a1. k=l Hence
for
X E
X=
k=l
l 1. Let A be the left shift on l 1' that is,
{xk} E l 1. Then
and
00
IIAnxll =
L lxn+kl = an+l k=l
for all n EN. As a result, the limit limn_, 00 11Anxll 1/n does not exist.
We now turn to a few important theorems involving r(A) and its local counterpart r(A, x). We first recall the following result mentioned in [64]. Theorem 1.2.1. Let A E .C(E). If there exists a;::: 0 such that r(A, x) ~ a for all x E E, then r(A) ~ a.
20
Chapter 1. Positive operators
Proof. By assumption, n--+oo for all x E E. Hence for an arbitrary e > 0 and for each x E E there is a number 6(x) ~ 1 such that for all n EN
By the Banach-Steinhaus theorem there exists 6 ~ 1 such that
for all n E N. This implies that lim sup IIAnlll/n:::;; a+ e.
n--+oo
As the sequence {IIAnlll/n} is convergent, we get r(A) :::;; a+e, which completes the proof, since e > 0 is arbitrary. D
In (13], Danes obtained an improvement of Theorem 1.2.1. Namely, he proved that the set of elements x E E for which r(A, x) = r(A) is of the second category in E. In particular,
(5)
r(A)
=
max{r{A, x) : x
E
E}.
The proof of (5) is also given in (21] and (40]. In (21], Forster and Nagy obtained the following strengthening of (5) for operators which are positive with respect to a generating cone. Theorem 1.2.2. If P is a generating cone in E and A E C(E) is positive, then
r(A) = max{r(A, x) : x
E
P}.
Proof By (5), there is xo E E such that r(A, xo) = r(A). Since P generating, there exist uo, vo E P such that xo = uo - vo. Hence
IS
21
1.2. Positive linear operators for all n E N. Therefore
r(A,xo) :S max{T(A,uo),r(A,vo)} :S r(A), which implies that r(A)
= max{r(A, x) : x
E P}.
0
We now give two sufficient conditions for the equality r(A, x) = r(A) to be fulfilled. The discussed results refer to the positive bounded linear operators.
Theorem 1.2.3. [22] Let P be a generating and normal cone in E, uo E P\ {B} and A E C(E) be a positive uo-upper bounded operator. Then r(A, uo) = r(A).
Proof Clearly, r(A, uo) :S r(A). So we only need to show that r(A) :S r(A, uo). Let u E P. Then there exists a positive number f3(u) such that Au :j {3(u)uo. Since A is positive, and thus increasing, we obtain
for all n EN. By assumption, Pis a normal cone, hence
for all n EN. In consequence, we get r(A, u) ::; r(A, uo) for every u E P. By Theorem 1.2.2,
r(A) = max{r(A,u): u which completes the proof.
E
P} :S r(A,uo) 0
It is worth to mention that Theorem 1.2.3 can be also easily derived from
Theorem 1.2.1. As we have already shown, from the assumptions of Theorem 1.2.3 it follows that r(A, u) :S r(A, uo) for every u E P. Let x E E. Since
Chapter 1. Positive operators
22
Pis generating, there exist u, v E P such that x = u- v. As in the proof of Theorem 1.2.2 this gives r(A,x):::; max{r(A,u),r(A,v)}.
Thus r(A, x) :::; r(A, uo) for every x E E. By Theorem 1.2.1, r(A) :::; r(A, uo) and the assertion of Theorem 1.2.3 follows. As a corollary of Theorem 1.2.3 we get the following result (see [22], [36], [57]). Theorem 1.2.4. Let P be a solid and normal cone in E and A E C(E) be a positive operator. Then for every x E int P r(A, x) = r(A). Proof. Let x E int P. By Lemma 1.2.1, A is x-upper bounded. From Lemma 1.1.4 it follows that Pis a generating cone. In view of Theorem 1.2.3, r(A, x) = r(A). D
We now pass on to the estimations of the spectral radius of the sum and composition of bounded linear operators. In the monograph [55] the following classical theorem is given. Theorem 1.2.5. If A, B E C(E) and AB = BA then
(6)
r(AB) :::; r(A)r(B)
and r(A +B) :::; r(A)
(7)
+ r(B).
Proof. Since A and B are commutative we get (AB)n
=
An nn for every n E N.
Thus r(AB)
=
lim
n-+oo
:::; lim
n->oo
II(ABtll 1/n
=
lim IIAnBnlll/n
n-+oo
(IIAnJilfniiBnJilfn)
= T(A)r(B).
1.2. Positive linear operators
23
In order to prove (7), we first observe that for all n EN
II(A + B)nll =II
t, (;)Ainn-ill·
Let us take arbitrary numbers p and q such that p > r(A) and q > r(B). Then there exists an integer number K > 0 such that for n ;?: K
IIAnlll/n < P
and
IIBnlll/n < q.
Hence, there exist constants s and t such that for all n E N
Thus, for n
> 2K we get
Ln (;)piqn-i (t)n-i _
+.
q
z=n-K+l
where M=
max
O:$i$K-1
(
8
p
)i +1+
max
O:$i$K-1
(t)i q
Then Hence Since the numbers p and q were chosen arbitrary, the inequality (7) follows.
D
Chapter 1. Positive operators
24
In [13], Dane8 obtained the following analogue of Theorem 1.2.5 for the local spectral radius.
Theorem 1.2.6. [13, Lemma 1] If A, B E C(E) and AB = BA then for all xEE
(8)
r(AB,x):::; r(A,x)r(B)
and
(9)
r(A + B, x):::; r(A, x)
+ r(B).
Example 1.2.2. In the Euclidean space JR2 consider the operators Ax= (x2, 0), Bx = (0, x1), where x = (x1, x2). Obviously, AB and x E JR2 we have
"I BA and r(A)
=
r(B) = 0. For all n EN
Consequently, r(AB) = 1, r(A +B) = 1 and the inequalities (6) and (7) are not fulfilled in this case. Moreover, it is clear that (8) does not hold for any x E JR2 with x1 "I 0 and (9) is satisfied only for x = (0, 0).
The above example demonstrates that the condition AB = BA in Theorems 1.2.5 and 1.2.6 is essential. However, it occurs that the assumption of commutativity can be weakened if A and B are positive operators. We now discuss some results of this kind. It is easy to check that r(AB) = r(BA) for any A and B belonging to C(E). Thus, in what follows we will estimate r(AB) only.
Theorem 1.2. 7. [73] Let P be a normal and generating cone in E and A, B E C(E) be positive operators. If BAx ~ ABx for every x E P then the inequalities (6} and (7} hold.
25
1.2. Positive linear operators Proof. By assumption it follows that
for all x E P and n EN. Since P is a normal cone, we get
As a result
r(AB, x) ~ r(A)r(B) for every x E P. By Theorem 1.2.2, (6) is fulfilled. To prove (7), observe that
for every n EN and x E P. Since P is a normal cone, we get
This gives
JJ(A + BtxJJ
~ -yJJxll ~ (7)piqn-iM,
where p, q and Mare defined as in the proof of Theorem 1.2.5. Therefore, for every x E P r(A + B, x) ~ r(A) + r(B) which implies (7).
0
Next, we recall a result concerning the spectral radius of the sum of positive operators obtained in [71]. It can be stated as follows. Theorem 1.2.8. Let P be a normal cone in E and A, B E .C(E) be positive operators. Suppose that there exists xo E P such that r(A+B,xo) = r(A+B) and
for j = 1, 2, ... , k = 0, 1, ... . Then (7) is fulfilled.
Chapter 1. Positive operators
26
Proof. In view of our assumptions we get for all n EN (A+ Btxo ::S
t
(7)Ainn-ix 0 .
t=O
Hence
II(A + Btxoll
:S 'Y
t, (7)
IIAiiiiiBn-illllxoll·
Proceeding analogously to the proof of Theorem 1.2.5 we obtain
r(A
+ B, xo)
::; r(A)
+ r(B), 0
which gives (7).
Remark . For the operators A and B considered in Example 1.2.2 the assumptions of Theorem 1.2.8 are not satisfied. Indeed, suppose that Pis a cone in the Euclidean space JR2 . Let A(P) C P, B(P) C P and xo = (x~, xg) E P. Then BABxo = (0, x~) E P and AB 2 xo = (0, 0).
Thus BABxo ::S AB 2 xo if and oply if (0, -x?) E P, which gives x? = 0. Moreover,
BAxo = (0, xg) E P and ABxo = (x~, 0) = (0, 0). Hence BAxo ::S ABxo if and only if (0, -xg) E P. This implies xg = 0. Thus,
xo
= (0, 0), but then r(A + B, xo)
-=/=
r(A +B).
We now discuss some generalizations of Theorem 1.2.6 for positive operators. Since r (AB, x) may be different from r ( B A, x), we will estimate both of them. Theorem 1.2.9. [73] Assume that Pis a normal cone in E and A, BE C(E) are positive operators. If ABx ::S BAx for every x E P, then
r(A + B, x) ::; r(A, x) r(AB, x) ::; r(A, x)r(B) and for every x E P.
+ r(B),
r(BA, x) ::; r(A, x)r(B)
27
1.2. Positive linear operators
The proof is similar to that of Theorem 1.2. 7 so we can omit it. It is of interest to know whether r(B) can be replaced by r(B, x) in the case r(B, x) < r(B), that is, if it is possible to obtain sharper estimations for r(A + B, x), r(AB, x) and r(BA, x). We present here two results of this type.
Theorem 1.2.10. [76] Let P be a normal cone in E and A, B positive operators. Suppose that A is uo-upper bounded and
E
.C(E) be
(10) for i = 1, 2, ... , and j = 0, 1, 2,. . . . Then r(A + B, uo) :::; r(A, tto)
(11)
(12)
+ r(B, uo)
r(AB, uo) :::; r(A, tto)r(B, uo) and r(BA, tto) :::; r(A, uo)r(B, uo).
Proof. The method of the proof is partly based on that used in [20] for the spectral radius. Since A is uo-upper bounded and positive, for each x E P there exists f3(x) > 0 such that
for all n E N. Hence
which gives
(13) for every x E P. Let
r(A, x) :::; r(A, uo) E
> 0 and with 8B
=
r(B, uo)
+ c/2
put
00
_ '""' ..-(i+l)Bi UQ. U - L__, UB
i=O
Evidently, u is well defined. Observe that u
(14)
E
P and 8Bu = uo +Bu. Hence
Chapter 1. Positive operators
28 Thus, r(A, uo) :::; r(A, u) which with (13) gives r(A, u) = r(A, uo).
Next, put 8A
= r(A, uo) + e/2 and define 00
_ ""' _.-(i+l)Ai
v-
L.., UA i=O
u.
It is easy to verify that v E P, u ~ 8Av, Av ~ 8Av and uo ~ 8A8Bv. Moreover, by (10) and (14) we have 00
Bv =
L
00
&:;;-(i+l) BAiu
i=O
~
00
L
&:;;-(i+l) Ai Bu
i=O
~
L
&:;;-(i+l) 8BAiu =
8Bv.
i=O
Then
(A+ B) 2 uo ~ 8A8B(8A
+ 8B)(A + B)v ~ 8A8B(8A + 8B) 2 v,
and generally for all n E N
Therefore and consequently r(A
+ B,uo):::; 8A + 8B,
which gives {11), since e > 0 is arbitrary. To prove (12), observe that
and Thus, for all n EN. Then
29
1.2. Positive linear operators which gives The last inequality clearly implies
r(AB, uo)
~
r(A, uo)r(B, uo).
In the same manner we can see that
for all n E N. Therefore
r(BA, uo)
~
r(A, uo)r(B, uo). D
This completes the proof.
Theorem 1.2.11. [76] Let P be a normal cone in E and A, B E C(E) be positive operators. Suppose that for uo E P, {10} is satisfied and
Buo :j r(B, uo)uo. Then the inequalities {11} and {12} hold.
Proof. By assumption,
for all i EN. Therefore, for 00
_ ~ _.-(i+l)Bi
U- ,L-oB
UQ,
i=O
where 8B is defined as in the proof of Theorem 1.2.10, we obtain
which gives r(A, u) ~ r(A, uo). The same type of analysis as that in the proof D of Theorem 1.2.10 establishes the result.
Chapter 1. Positive operators
30 Example 1.2.3. Consider the Banach space
P
=
{x E co:
Obviously, Pis a normal cone in
Bx
CQ.
= (x2
.
co and the set
Xi~ O,i EN}.
Let
+ x3, 2x3, 2x4, ... ) 1
2
and uo = {1/3'}. Then r(A, uo) = 9 , r(B, uo) = 3· Clearly, Buo ::5 r(B, uo)uo. Moreover, fori= 1, 2, ... , j = 0, 1, ... , we get
. . BA' B 3 uo
.
1
1
2
2
= 23 ( 3-f+2i+2 + 3-f+2i+3' 3-f+2i+3' 3-f+2i+4' · · ·)
and i
A B
j+l
. -
uo - 2
1
j+l
1
1
( 3-f+2i+2' 3-f+2i+3' 3i+2i+4' · · · ) ·
Thus, (10) is fulfilled. By Theorem 1.2.11, r(A + B, uo) ~ ~ and r(AB, uo) ~ 2 2 7 • Observe that in this case we have r(A) = 1 and r(B) = 2.
Some other results concerning the subadditivity and submultiplicativity of spectral radius and local spectral radius can be found for example in [20], [58], [66], [73] and [76]. Several interesting properties of r(A) and r(A, x) are also discussed in [10], [17], [21], [36], [37], [40], [50], [58]. Finally, we give a fixed point theorem for a nonlinear operator which satisfies generalized Lipschitz condition with respect to a positive bounded linear operator with the spectral radius less than 1. Similar results are discussed for example in [36] and [63].
Theorem 1.2.12. [70] Let E be a Banach space with a binary relation ::5 and a mapping m: E---+ E. Assume that:
(i) the relation ::5 is transitive,
1.2. Positive linear operators (ii) 0 ~ m(x) and
31
llm(x)ll = llxll
for all
X E
E,
(iii) the norm 11·11 in E is monotonic, that is, for x, y E E, if 0 ~ x
llxll
~
~
y then
IIYII-
Moreover, let the operators F : E properties:
~
E, A : E
~
E be given with the following
(iv) A E £(E), r(A) < 1 and A is increasing with respect to e ~ X ~ y then Ax ~ Ay'
(v) m(Fx- Fy)
~
~'
that is, if
Am(x- y) for all x, y E E.
Then F has a unique fixed point x* E E. Moreover, for each x E E, lim Fnx = x*.
n-+oo
Proof. From (i), (iv) and (v) we get for all x, y E E
and generally
In view of (ii) and (iii)
for all x, y E E and n E N. On the other hand, since
r(A) = inf
nEIII
IIAnlll/n < 1,
there exists no E N such that llAno II < 1. Thus, pno is a strict contraction in E. By Banach's contraction principle, pno has a unique fixed point x* E E. This implies that x* is a unique fixed point of F in E. Indeed, from
Chapter 1. Positive operators
32
pnox* = x* we get pno+lx* = Fx*. Hence pno(Fx*) = Fx* , which implies Fx* = x*. Suppose that there exists y* E E, y* "I x*, such that Fy* = y*. Then pnoy* = y*, which is impossible. Therefore x* is a unique fixed point of Fin E. Moreover, r(A) < 1 implies that for each x E E the series 00
L
iJFnx - pn+lxii
n=l
is convergent. Thus {Fnx} is a Cauchy sequence for each x E E. Since E is complete, { Fnx} is convergent. The standard argument shows that for each xEE lim Fnx = x*. n--+oo
0
Evidently, Theorem 1.2.12 is a generalization of Banach's contraction principle. A survey of numerous extensions of Banach's principle for the operators acting in· complete metric spaces can be found for example in [17, Chapter 5] and [33].
Remark . In the applications of Theorem 1.2.12 presented in Chapter 2 the relation ~ will coincide with the one given by an appropriate cone in E.
1.3
Cone compression and expansion operators
This section is devoted to the existence of fixed points in cones for positive operators in partially ordered Banach spaces. As in the case of linear operators, we say that a nonlinear operator F : E ~ E is positive if it leaves invariant a cone P C E (see Definition 1.2.1). The most widely known result on the existence of fixed points in cones is Krasnoselskii's theorem on cone expansion and compression. In this section we will present some of its refinements for completely continuous operators, strict set contractions and condensing mappings. We begin with recalling some definitions and basic facts from the fixed point index theory on cones.
1.3. Cone compression and expansion operators
33
Definition 1.3.1. We say that an operator F: D --4 E, DC E, is completely continuous if it is continuous and maps bounded subsets of E into precompact sets.
Definition 1.3.2. Let B denote the family of all bounded subsets of E. The Kuratowski measure of noncompactness a(O) of a set nEB is defined as the infimum of the numbers c > 0 such that n admits a finite covering by the sets of diameter smaller than c, that is
c·u ni n
a(O)
= inf{c
> 0:
n
with diam(Oi) < c, 1 :S i :S n}.
i=l
It is well known that if
n,
01, 02 E B and ..\ E R then:
1) a(conv(O)) = a(O), where conv(O) denotes the closure of the convex hull of n in E,
3) a(.Xn) = I.XIa(n),
5) a(O) = 0 if and only if
7) a(O) = a(n)
(see [5], [17], [26]).
n is precompact,
Chapter 1. Positive operators
34
Definition 1.3.3. A map F : D . . . _. E, D C E, is said to beak-set contraction if it is continuous, maps bounded subsets of D into bounded subsets of E and there exists k 2: 0 such that a(F(S)) -:::; ka(S) for every bounded subsetS of D. A k-set contraction is called a strict set contraction if k < 1.
Obviously, every completely continuous operator is a 0-set contraction, that is strict set contraction with k = 0, and conversely. Definition 1.3.4. A map F : D . . . _. E, D c E, is said to be condensing if it is continuous, maps bounded subsets of D into bounded subsets of E and a(F(S)) < a(S) whenever S c Dis bounded and a(S) -=1- 0.
It is evident that every strict set contraction is a condensing mapping. The converse conclusion is not true (see [17, Example 9.3]).
In the sequel, we will use some properties of the fixed point index for positive condensing mappings. We restrict our attention to the basic facts from the fixed point index theory. More extended treatment of the concept of the fixed point index for positive operators is given for example in [24] and
[26]. Theorem 1.3.1. Let X be a nonempty closed convex subset of E and 0 be a bounded and relatively open subset of X. Then for every condensing operator F : 0 . . . _. X which has no fixed points on 80 there exists a uniquely defined integer ix(F, 0) satisfying the following conditions:
(a) if Fx := Xo for all (normalization),
X
E
0
and some fixed
Xo
E 0, then ix(F, 0)
= 1
(b) ix(F, 0) = ix(F, 01) +ix(F, Oz) whenever 01 and Oz are disjoint open subsets of 0 such that F has no fixed points on 0\ (01 UOz) (additivity),
35
1.3. Cone compression and expansion operators
(c) ix(H(t,·),n) is independent oft E [0,1] whenever H: [0,1] x 0---+ X is condensing and H(t, x) =1- X for (t, x) E [0, 1] X an (homotopy invariance), (d) if Xo is a closed and convex subset of X and F(O) c Xo then ix(F, n) ix0 (F, n n Xo) {permanence).
The number ix(F, n) is called the fixed point index ofF on to X.
=
n with respect
Theorem 1.3.2. Under the assumptions of Theorem 1.3.1 we have:
(e) if no is an open subset of n such that F has no fixed points in then ix(F, n) = ix(F, no) {excision property),
(f) if ix (F, n) =f. 0 then there exists x* E n such that Fx*
=
0\ no
x* {solution
property).
Remark . From Definitions 1.3.1, 1.3.3 and 1.3.4 it follows that the properties (a) - (f) apply, in particular, to strict set contractions and completely continuous operators.
In what follows, for our purpose, we will take X = P, where P is a cone in E. Let us also mention two important consequences of Theorems 1.3.1 and 1.3.2.
Lemma 1.3.1. [24] Let n be a bounded open subset of E and P be a cone in E. Let () E n and F : P n 0 ---+ P be a condensing operator. Suppose that Fx =f. /-LX for all x E P nan and 1-L 2: 1. Then ip(F, P n n) = 1.
36
Chapter 1. Positive operators
Lemma 1.3.2. [24] Let n be a bounded open subset of E and P be a cone in E. Suppose that F1 : P n n --t P and F2 : P n an --t P are completely continuous operators such that
and
for all X E p
nan
and t
~
0. Then ip(H, p
n n) = 0.
In the next three corollaries of Lemma 1.3.2, subset of E and Pis a cone in E.
Corollary 1.3.1. [24] If F : P exists uo E P \ {0} such that
n n --t x- Fx
for all X E P n an and t
~
Proof. Put Flx = Fx for X E p Lemma 1.3.2, ip(F, P n n) = 0.
Corollary 1.3.2. [78] If F : P exists uo E P \ {0} such that
and t
~
P is completely continuous and there
f tuo
n n) = o.
nn
and F2X
= uo for X
E p
nan. By 0
n0
x - Fx
n an
a bounded open
0, then
ip(F, P
for all X E P
n denotes
--t
P is completely continuous and there
f t( uo + Fx)
0, then
ip(F,Pnn) =
o.
1.3. Cone compression and expansion operators
37
Proof Put FIX = Fx for X E p n fi and F2x = Uo + Fx for X E p Then .f]_ and F2 are completely continuous. Moreover, we have
nan.
Indeed, if this were not the case, then for each n E N we could choose Xn E p n an such that 1 lluo + Fxnll < -. n Hence lim (uo
n~oo
+ Fxn)
0,
=
and in consequence, lim Fxn = -uo,
n-+oo
which is impossible, because {Fxn} C P, Pis closed and -uo 1.3.2, ip(F, P n n) = o.
Corollary 1.3.3. [24] Let F : P
~
P. By Lemma
o
n fi - P be completely continuous. If Fx =I J.LX
for all X E P
n an
and J.L E (0, 1] and
in£ IIFxll xEPnan
> 0,
then ip(F,Pnn)=O.
Proof Set Flx
= Fx for X E p
n fi
and F2x
= Fx for X E p
nan.
Then
inf IIF2xll > 0. xEPnan Suppose that there exist xo
E
P n an and to
xo - Fxo
=
~
0 such that
toFxo.
Hence
Fxo = (1
+ to)- 1xo.
Put J.Lo = (1 + to)- 1. Then J.Lo E (0, 1] and Fxo = J.Loxo which contradicts our assumption. Application of Lemma 1.3.2 completes the proof. 0
Chapter 1. Positive operators
38
We now turn to some fixed point theorems for positive operators compressing or expanding a cone in a Banach space. First we recall Krasnoselskii's theorem.
Theorem 1.3.3. [35] Let P be a cone in E and let F : P -+ P be a completely continuous operator with F(B) =B. Suppose there exist r, R, 0 < r < R, such that one of the following conditions is satisfied:
(i) Fx f-_ x for x E P with 0 < llxll : : ; r and (1 and x E P with llxll 2 R (cone compression),
+ c)x f-_
Fx for all
(ii) Fx f-_ x for x E P with llxll 2 R and (1 + c)x f-_ Fx for all x E P with 0 < llxll : : ; r (cone expansion).
Then F has a fixed point x* E P with r ::::;
llx* I : : ;
E
E
>0
> 0 and
R.
Some refinements of Krasnoselskii's theorem are given in [17], [24], [65]. Let us recall a few of them. The following result, usually called Krasnoselskii's theorem on cone expansion and compression of the norm type, is due to Guo.
Theorem 1.3.4. [24] Let n1 and n2 be bounded open sets in E such that () E D1 and D1 c D2. Suppose that F : P n (D2 \ DI) -+ P is a completely continuous operator such that either:
p n onl and
(i)
IIFxll : : ; llxll
for
X
E
(ii)
llxll : : ; IIFxll
for
X
Ep n onl
llxll : : ; IIFxll
for
X
E
IIFxll : : ; llxll
for
X
Ep n 8n2
p n 8n2
or and
is satisfied. Then F has a fixed point in P n (D2 \ n1).
39
1.3. Cone compression and expansion operators
Proof. By the Dugundji extension theorem (see [65, Proposition 16.3]), F can be extended to a completely continuous operator (also denoted by F) on P. We may assume that F has no fixed points on (P n ani) U (P n an2). Let (i) be satisfied. Since IIFxll :::; llxll for all X E p n ani, we have Fx =I JLX for all X E p n anl and JL 2: 1. Indeed, suppose for the contrary, that there exist Xo E p n anl and JLo > 1 such that Fxo = JLoXo. Then
IIFxoll = JLollxoll > llxoll which contradicts (i). Thus, by Lemma 1.3.1, ip(F, P also follows that inf
xEPnan2
IIFxll
Moreover, if there existed Xo E p we would have
2:
inf
xEPn8Sh
n an2
n n1)
= 1. From (i) it
llxll > 0.
and JLo E (0, 1) such that Fxo = JLoXo,
IIFxoll = JLollxoll < llxoll, in contradiction with (i). Therefore Fx =I JLX for all x E Pnan 2 and JL E (0, 1] and by Corollary 1.3.3, ip(F, P n n2) = 0. From the additivity of the fixed point index we get
ip(F, P
n (n2 \ fh))
= ip(F, P
n n2)- ip(F, P n n1) =
-1.
Hence, in view of the solution property of the fixed point index, F has a fixed point in p n (n2 \ 'fh). If (ii) holds the proof is similar. 0
It is well-known that each of the norm type assumptions of Theorem 1.3.4 can be replaced by a condition formulated in terms of the relation given by P (see [24]). Namely, we can suppose for instance that
X
t Fx
for
X EP
n an1
and
Fx
tX
for
X EP
n an2
holds instead of (i) of Theorem 1.3.4. Now we discuss some further improvements of Theorem 1.3.3 for completely continuous operators. We begin with the result due to Leggett and Williams. In 1980, they established the following generalization of Krasnoselskii's theorem.
Chapter 1. Positive operators
40
Theorem 1.3.5. [42] Let P be a cone in E. For R > 0 put Pn = {x E P: !!x!! :S R}. Let F: Pn ~ P be a completely continuous operator with F(O) = 0. Suppose there exist r, 0 < r < R, and uo E P \ {0} such that Fx ~ x if x E P(uo) and l!xl!
= r,
where P(uo)
=
{x E P: >.uo j x
Suppose further that for each (1
E:
for some
>. > 0}.
>0
+ c)x ~ Fx
if x E P and l!xl! = R.
Then F has a fixed point x in P with r :S l!xl! :SR.
Some extensions of Theorem 1.3.5 can be found in [2] and [24]. Here we recall the result from [2] only. Theorem 1.3.6. Let P be a cone in a Banach space E and r1, r2 > 0, r1 i= r2, with R = max{r1,r2}, r = min{rt,r2}. Let F: Pn ~ P be a completely continuous operator such that:
(i) I!Fxl! :S l!xl! for x E P and l!xl!
=
r1,
(ii) there exists uo E P\ {0} such that Fx ~ x for x E P(uo) and l!xl! = r2. Then F has at least one fixed point x* E P with r :S l!x* II :S R.
Evidently, the above theorem also holds with (i) replaced by x ~ Fx
for all
x E P
and
l!xl!
=
Indeed, if x ~ Fx for all x E P and l!xl! = r1, then Fx with l!xl! = r1 and J.L 2: 1. By Lemma 1.3.1,
r1.
i=
J.LX for all x E P
1.3. Cone compression and expansion operators
41
where Pr 1 = {x E P: llxll < ri}. As we have shown in the proof of Theorem 1.3.4_, the condition IIFxll ~ llxll for x E P and llxll = r1 also implies ip(F, PrJ = 1. This observation leads to a natural question .. Is it possible to replace the assumption (ii) of Theorem 1.3.6 by the condition of the norm type? The following result gives us a positive answer to this question.
Theorem 1.3. 7. [78] Let P be a cone in E and uo E P \ {0}. Suppose that !11 and !12 are bounded open sets in E such that 0 E !11 and ?'h c !12. Let F : P n (02 \ !11) -> P be a completely continuous operator s7tch that one of the following conditions holds:
(i)
(ii)
llxll
~ a(uo)IIFxll for x E
I!Fxll
~
llxll
for x E
P(uo)nDnl
Pn8n1
and
llxll
and
IIFxll
~
llxll
for x E
Pn8n2,
~ a(uo)IIFxll for x E P(uo)nan2,
where P(uo)
=
{x E P: >.uo ::::S x for some>.> 0}
and a(uo) is such that
(15)
llx + uoll
~ a(uo)lixll
Then F has a fixed point in the set P
for all
x E P.
n (02 \ !11).
Proof. First, observe that in view of Lemma 1.1.2 there exists a constant a( u 0 ), a(uo) > 0, satisfying (15). We may assume that F has no fixed points on the set (P n ant) u (P n 8!12)· Suppose that (i) is fulfilled. Since IIFxll ~ llxll for all x E P n 8!12, it follows from Lemma 1.3.1 that
(16) We will show that for all t
~
0 and x E P
x- Fx
n 8!11
f:. t(uo + Fx).
Chapter 1. Positive operators
42
Suppose, contrary to our claim, that there exist Xo Ep that xo- Fxo = to(uo + Fxo).
n anl
and to > 0 such
From the above equality it follows that xo E P(uo). Furthermore, by Corollary 1.1.1
llxoll = 11(1 + to)Fxo
+ touoll
2: a(uo)ll(1 + to)Fxoll > a(uo)IIFxoll·
This contradicts (i). Thus, by Corollary 1.3.2 ip(F, P
(17)
n nl) = o.
From (16), (17) and the additivity of the fixed point index we get ip(F, P
n (n2 \fit)) = 1.
This implies that F has a fixed point in the set P proof is similar.
n (f22 \fit).
If (ii) holds the D
In the remainder of this section we will be concerned with some refinements of Krasnoselskii's and Leggett-Williams theorems for strict set contractions and condensing operators. In what follows, we will use a counterpart of Corollary 1.3.1 for condensing operators. Lemma 1.3.3. Let P be a cone in E, n be a bounded open subset of P and F : 0---+ P be a condensing mapping. Assume that there exists uo E P \ {0} such that x- Fx =f .Xuo
for all X E
an
and A ;::: 0. Then ip(F, n) =
o.
Proof. Suppose, contrary to our claim, that
(18)
ip(F, n)
=f o.
1.3. Cone compression and expansion operators Let >.o
43
> 0 be such that
,
(19)
AO
>
sup{llxll + IIFxll : x lluoll
E
0}
·
FortE [0, 1] and x E 0 define H(t, x) = t>.ouo+Fx. Clearly, H is a condensing mapping and for all X E an and t E [0, 1] we have X f= H(t, x). Thus, by the homotopy invariance property
ip(H(1, ·), 0) = ip(H(O, ·), 0).
(20)
Since H(O, x) = Fx, it follows from (18), (20) and the solution property of the fixed point index that there exists x* E 0 such that x* = >.ouo + Fx*. Hence
>. 0
~
llx*ll + IIFx*ll lluoll
which contradicts (19).
0
We can now recall some extensions of Krasnoselskii's theorem for strict set contractions. Theorem 1.3.8. [26] Let P be a cone in E and Pr,s = {x E P: r ~ llxll ~ s} with 0 < r < s. Suppose that F : P r,s -+ P is a strict set contraction such that one of the following conditions is satisfied:
(i) Fx (ii) x
i
i
x for x E P with
Fx for x E P with
llxll =
r and x
llxll = r
i
and Fx
Then F has a fixed point x* E P such that r <
Fx for x E P with
llxll =
i
llxll = s.
x for x E P with
s,
llx*ll < s.
The next result, which can also be regarded as a refinement of Theorem 1.3.4, is due to Li.
Chapter 1. Positive operators
44
Theorem 1.3.9. [43] Let P be a cone in E, Br = {x E E : llxll < r}, Bn = {x E E: llxll < R}, 0 < r < R. Let the norm in E be monotonic, that is, if () :j x :j y then llxll:::; IIYII· Suppose that F: Pn Bn---+ Pis a strict set contraction which satisfies one of the following conditions:
(i)
IIFxll :::; llxll
for
(ii)
llxll :::; IIFxll
for x E P n 8Br and
X E
p
n 8Br and llxll :::; IIFxll for
IIFxll :::; llxll
X E
p
n 8Bn,
for x E P n 8Bn.
Then F has a fixed point in P n (Bn \ Br)·
Similar results are also presented in [59]. Now we pass on to the generalizations of Leggett-Williams theorem.
Theorem 1.3.10. Let P be a normal and solid cone in E and uo E int P. Let 01 and 02 be bounded open subsets of E such that () E fh and 01 c 02. Suppose that F : P n 02 ---+ P is a condensing mapping such that either
(i) 'Y llxll
:::; IIFxll
for
X
EP(uo) n anl
and
IIFxll :::; llxll
for
:::; IIFxll
X
X
Ep n an2
or (ii)
IIFxll :::; llxll
for
X
Ep n anl
and 'Y llxll
for
EP(uo) n an2
is satisfied, where 1 denotes the normal constant of P. Then F has a fixed point in the set P n (02 \ OI).
Proof. Suppose that (i) holds. We may assume that F has no fixed points on (P n 8S11) U (P n 8S12). As in the proof of Theorem 1.3.4, by Lemma 1.3.1 and the assumption IIFxll :::; llxll for x E P n 8S12, we have
1.3. Cone compression and expansion operators
45
Next, we will prove that for all t 2:: 0 and x E P n 801 (21)
x - Fx -/=- tuo.
To obtain a contradiction, suppose that there exist xo E P n 801 and to > 0 such that (22)
xo - Fxo
= touo.
Then xo E P(uo) and touo E int P. Thus, by (22)
xo - Fxo E int P. By Corollary 1.1.2, there exists r5 E (0, 1) such that
xo - Fxo - 6xo E P. Hence () :::S Fxo :::S (1 - 6)xo, and since P is a normal cone we obtain
IIFxoll
:=::;
1'(1- r5)llxoll
< l'llxoll,
contrary to (i). By (21) and Lemma 1.3.3 ip(F, P
n 01) = o.
From the additivity of the fixed point index we get
This implies that F has a fixed point in the set P the proof is similar.
n (02 \ "fh).
If (ii) is fulfilled
D
In a similar way we can prove the following result. Theorem 1.3.11. Let P be a cone in E and uo E P \ {e}. Let 01 and 02 be bounded open subsets of E such that () E 01 and 01 C 02. Suppose that F : P n 02 ____, P is a condensing mapping such that either
(i) llxll < a(uo)IIFxll for x E P(uo)n801 and IIFxll
:=::;
llxll for x E Pn802
Chapter 1. Positive operators
46
or
(ii) IIFxll ::; llxll for x
E
Pn8n1 and llxll < a(uo)IIFxll for x
is satisfied. Then F has a fixed point in the set P
E
P(uo)nan2
n (0"2 \ OI).
Proof. Suppose that (i) holds. It is enough to observe that for all x E Pn8n 1 and .X 2: 0 we have x =1- .Xuo + Fx. Indeed, if xo = AoUo Corollary 1.1.1
+ Fxo
llxoll
=
for some Xo E p n anl and .Xo > 0 then by
II.Xouo + Fxoll 2: a(uo)IIFxoll,
contrary to (i). The rest of the proof follows the usual reasoning.
0
We finish this section with a Leggett-Williams norm type theorem for strict set contractions.
Theorem 1.3.12. Let P be a cone in E and uo E P \ {0}. Let S11 and S12 be bounded open subsets of E such that 0 E S11 and 01 c S12. Suppose that F : P n 02 ---4 P is a k-set contraction, k E [0, 1), such that either
(i) llxll ::; a(uo)IIFxll for x E P(uo)nanl and IIFxll ::; llxll for x E Pn8n2 or
· (ii) IIFxll ::; llxll for x E Pn8n1 and llxll ::; a(uo)IIFxll for x E P(uo)n8n2 is satisfied. Then F has a fixed point in the set P n (02 \ S11).
47
1.3. Cone compression and expansion operators
Proof. For k = 0 the assertion follows from Theorem 1.3.7. Let k E (0, 1). Suppose that (i) holds and Fx f. x for x E (Pnani) u (Pn8fh). By Lemma
1.3.1,
ip(F, P n n2) = 1. It suffices to prove that ip(F, P
n S11) =
0. To do this, we first show that
x f. .>.uo + (.>.M + 1)Fx
(23) for all ). ~ 0 and x E P inequality
n 8S1r,
where M is a positive constant satisfying the
M < 1- k inf{lluo + MFxll: X E p n ITl}. k sup{llxll + IIFxll: X E p n nl} Observe that such an M always exists. To see this, put 1-k
L=-----------=ksup{llxll
and suppose that for every M
M
~
+ IIFxll: X
E p
n nl}
> 0 we have
Linf{iluo
+ MFxii: x
Then for each n E N we can choose
Xn
E P
1
1
n
n
E
Pn 01}.
n 'fh such that
- ~ Llluo + -Fxnll· This gives
~ (~ + IIFxnll) ~ lluoll for all n E N. Since F is a strict set contraction, the sequence {IIFxnll} is bounded, and we get uo = (), a contradiction. Next, suppose contrary to (23), that there exist Xo E P such that
xo
=
n 8S11 and .>.o > 0
.>.ouo +(>.oM+ 1)Fxo.
This implies xo E P(uo). Moreover, by Corollary 1.1.1, llxoll = II.Xouo +(>.oM+ 1)Fxoll ~ a(uo)ll(.>.oM + 1)Fxoll which contradicts (i). Thus (23) holds.
> a(uo)IIFxoll,
Chapter 1. Positive operators
48 Next, choose .X* (24)
> 0 such that n nl} n 01}
1- k '* sup{llxll + IIFxll: _ X E_ p_-=~ -->A>-::.._:.;c:......:.:....._....;.;_____;c:___
kM
inf{lluo + M Fxll
:X E
p
and put k = (.X* M + 1)k. Obviously k E (0, 1). Consider H: [0, 1] x (Pn01)---+ P given by H(t,x) = t.X*uo + (t.X*M + 1)Fx. Clearly, H is a k-set contractive map. In view of (23), xi- H(t, x) for (t, x) E [0, 1] x (P n 801). From the homotopy invariance of the fixed point index we get (25)
ip(H(1, ·), P n ol)
=
ip(H(o, ·), P n 01).
Suppose that ip(H(1, ·), P n 01) =f. 0. Then from the solution property of the fixed point index there exists x* E Pn0 1 such that x* = .X*uo+(.X* M +1)Fx*. Therefore x* - Fx* = .X *uo + .X* M Fx*. Hence llx*- Fx*ll
=
.X*IIuo + MFx*ll·
Consequently, .X*< sup{llxll + IIFxll: X E p n nl} - inf{lluo + MFxll: x E P n 01}' contrary to (24). Thus, ip(H(1, ·), P n 01) = 0 which together with (25) gives
ip(F, P
n OI) =
0.
The assertion now follows from the properties of the fixed point index.
0
It is noteworthy here that Theorems 1.3.10 - 1.3.12 improve the result established in [79] for a normal cone and a completely continuous operator. Remark . As we have already mentioned, Krasnoselskii's theorem was generalized in many ways. In this monograph we restrict our attention to the single valued mappings. We emphasize that there are several multivalued versions of cone expansion or compression theorems. For instance, in [53] Petryshyn obtained a multivalued analogue of Krasnoselskii's theorem for an upper semicontinuous, strict set contractive map with nonempty, closed and convex values. We point out that Theorems 1.3.11 and 1.3.12 have been recently extended to the case of multivalued mappings ([80]).
Chapter 2
Applications of positive operators to differential and integral equations 2.1
Initial and boundary value problems for functional - differential equations
In this section we study initial and boundary value problems for nonlinear functional-differential equations. We deal with the equations of neutral type mainly. Using the properties of the spectral radius of positive bounded linear operator and Theorem 1.2.12 we obtain some existence and uniqueness results. Let us mention that functional-differential and functional-integral equations were studied by many authors. The examples of techniques applied to the equations of such type are the following: Banach's contraction principle ([3]), measure of noncompactness ([4]) and Borsuk's antipodal theorem ([18]). First consider the following initial value problem of neutral type (26)
x'(t)
f(t, x(h(t)), x'(H(t))), t
(27)
x(O)
xo, xo E R
E
[0, T], T > 0
By a .solution of (26)-(27) we will mean a function which is absolutely continuous on [0, T] (AC for short), satisfies the equation (26) a.e. on [0, T] and the initial condition (27). In what follows, L 1 [0, T] denotes the Banach
Chapter 2. Applications
50
space of Lebesgue integrable functions on [0, T] with the norm
1T
llxll =
lx(t)idt.
Theorem 2.1.1. [73] Suppose that:
1° h : [0, T]
----+
[0, T] is a continuous function,
2° H : [0, T]
----+ [0, T] is a monotonic and AC-function such that 0 IH'(t)l S 1 a. e. on [0, T] and H- 1 ([0, h(H(t))]) C [0, h(t)],
3° (t,x,y)
f(t,x,y) is a real function defined on the set [O,T] x JR 2 , Lebesgue measurable with respect to t for all (x, y) E JR 2 and satisfying the Lipschitz condition ----+
4° r(A1) + L2M < 1, where M = (min[o,T] IH'(t)l)- 1 and r(A1) denotes the spectral radius of the operator fh(t)
(A1x)(t)
= L1 lo
x(s)ds,
t E
[O,T],
in the space L 1 [0, T],
5° there exists a function g : [0, T] ----+ JR+ Lebesgue integrable on [0, T] such that lf(t, 0, O)l S g(t) a. e. on [0, T]. Then the problem (26}-(27} has a unique solution defined on [O,T].
Proof. To prove the assertion we will apply Theorem 1.2.12. Let E = L 1 [0, T] and m(x)(t) = lx(t)l fortE [0, T] and x E L 1 [0, T]. We say that x ~ y if and only if x(t) y(t) a.e. on [0, T]. It is easy to verify that the problem (26)-(27) is equivalent to the integral-functional equation
s
fh(t)
z(t) = f(t, xo
+ Jo
z(s)ds, z(H(t))),
t E [0, T],
2.1. Initial and boundary value problems where
x(t)
=
xo
51
+lot z(s)ds.
For z E £ 1 [0, T] consider the operator rh(t)
(Fz)(t) = f(t, xo
+ Jo
z(s)ds, z(H(t))),
t E
[O,T].
Obviously, in view of 1° - 3° and 5°, the operator F maps the space £ 1 [0, T] into itself. Moreover, by 3°, we have for all w, z E L 1 [0,T] rh(t)
i(Fw)(t)- (Fz)(t)i ::::; L1 Jo
iw(s)- z(s)ids + L2iw(H(t))- z(H(t))i.
Hence
(28)
i(Fw)(t)- (Fz)(t)i ::::; (A1 + A2)(lw- zi)(t).
where
1
h(t)
(A1z)(t) = L1
z(s)ds
()
and In view of 1° and 2°, A 1 + A 2 is a bounded positive linear operator mapping £ 1 [0, T] into itself. The inequality (28) means that the assumption (v) of Theorem 1.2.12 is satisfied for A = A1 + A2. What is left to show is that r(A 1 + A2) < 1. To prove this we apply Theorem 1.2.7. Let
P = {x E £ 1 [0, T] : x(t) ~ 0 a.e. on
[0, T]}.
Clearly, P is a normal and generating cone in £ 1 [0, T]. Notice that the relation given by P is the same as the one defined at the beginning of our proof. By 2° we get for z E P
r'*l z(H(s))ds = (A1A2z)(t)
rh(H(t))
(A2A1z)(t) = L1L2 Jo
z(s)ds::::; L1L2 Jo
a.e. on [0, T]. This means that A2A1z :::S A1A2z for every z E P. Therefore, in view of Theorem 1.2. 7
r(A1
+ A2)
::::; r(A!)
+ r(A2).
Since r(A2) ::::; L 2M, it follows from 4° that r(A1 + A2) < 1. By Theorem 1.2.12, F has a unique fixed point in £ 1 [0, T]. This completes the proof. D
Chapter 2. Applications
52
Remark . In order to calculate the spectral radius of the operator rh(t)
(A1z)(t) = L1 Jo
z(s)ds,
t E
[0, TJ,
in the space £ 1[0, T], we can use Theorem 1.2.3. Indeed, since the cone
P
=
{x
E £ 1 [0, T]:
x(t) ~ 0 a.e. on
[0, T]}
is normal and generating, and zo(t) = 1 a.e. on [0, T] belongs toP and A1 is positive and zo-upper bounded, we get in view of Theorem 1.2.3
r(A) =lim sup IIAfzoll 1/n. n->oo In particular, if T
~ 1
and h(t) = t 01 , where a
(Aizo)(t) and generally, for n EN, n
=
LI
E
(0, 1], then
t0/(01+1) a+1
,
> 1, we have tan+Otn-1 + ... +01
(An Z 1
) ( t) =
°
L n -:-----:-:---::------:----:--::------,,-------,-
1 (a+1)(a 2 +a+1)· ... ·(an- 1 +an- 2 + ... +a+1)"
Hence ran+O/n-l+ ... +a+1 II An Z II = L n -:-----:-:---::------:----,---:-----:------:1 (a+ 1)(a2 +a+ 1) · ... ·(an+ an- 1 + ... +a+ 1)" 1
°
Obviously, IIAfzoll
> 0 for all n EN. It is easy to show that . IIA~+ 1 zoll hm n->oo IIAn zo II = £1(1- a). 1
Since in this case
we obtain
r(Al) = £1(1- a).
53
2.1. Initial and boundary value problems
In a similar way we can show that the spectral radius of the generalized Volterra operator
(Vx)(t)
=lata x(s)ds,
t
E
[0, T], T 2:: 1,
in the space LP[O, T], p 2:: 1, is equal to 1- a. Recently, Ignat Domanov has found the spectrum a(V) of the operator
(Vx)(t)
[v'i
=
Jo
x(s)ds,
t E
[0, 1],
in the space £ 2 [0, 1]. Namely, he has proved that
a(V)
=
{0} U {
1 1 1 } 2' 4' ... ' 2n'...
.
Thus r(V) = ~' which coincides with our result. Now, we will study the following two point boundary value problem for first order implicit differential equation:
(29)
x'(t)
f(t, x(h(t)), x'(t)), t
(30)
x(O)
ax(T)
E
[0, T], T > 0,
+ b,
where a, bE JR. Clearly, the problem (29)-(30) becomes an initial value problem if a = 0 and a periodic value problem if a = 1 and b = 0. In what follows we will assume that a -1- 0 and a -1- 1. Let us mention that similar problems were studied for example in the papers [3], [39] and [46]. Particularly, in [39], the authors obtained some existence results for the periodic boundary value problem
{
x'(t)
=
f(t,x(t)),
x(O)
=
x(2n)
by means of a monotone iterative technique. We say that x : [0, T] ~ lR is a solution of (29)-(30) if x is an AC-function on [0, T], satisfies the equation (29) a.e. on [0, T] and the boundary condition (30).
Chapter 2. Applications
54
Theorem 2.1.2. Assume that:
1° h: [0, T]---+ [0, T] is a continuous function such that h(t)
~
t on [0, T],
2° (t,x,y) ---+ f(t,x,y) is a real function defined on the set [O,T] x JR2 , Lebesgue measurable with respect to t for all (x, y) E JR 2 and satisfying the Lipschitz condition
for all (t,xl,Yl), (t,x2,y2) E [O,T] x JR 2, where L1, L2
> 0,
4° there exists a function g : [0, T] ---+ JR+ essentially bounded on [0, T] such that lf(t,O,O)I ~ g(t) a.e. on [O,T]. Then the problem {29}-(30} has a unique solution defined on [0, T].
Proof. We can proceed analogously to the proof of Theorem 2.1.1 with E = V)()[O, T]. Particularly, changing the unknown function x in (29)-(30) by putting x(t) = -b1-a
we get
(31) z(t)
+ -a-
1T
1-a o
1T
z(s)ds
b a +z(s)ds 1-a 1-a o
= f(t, -
+
+
1h(t) o
loto z(s)ds,
z(s)ds, z(t)),
t E
[O,T].
It is easy to see that the integral-functional equation (31) is equivalent to (29)-(30). Moreover, we need to show that the spectral radius of the bounded linear operator (Az)(t) = L1
,_a_,a lo{T 1-
z(s)ds + L1
r z(s)ds + L2z(t),
lo
where z E L 00 [0, T] and t E [0, T], is less than 1. To this end observe that A is positive with respect to the cone P
= {x
E
L00 [0, T]: x(t) :2: 0 a.e. on [0, T]}
2.1. Initial and boundary value problems and A = A1
55
+ A2, where (A1z)(t) = L1
1-a-1a lo
{T z(s)ds + L1
1-
t
lo
z(s)ds
and
Obviously, r(A2) = L2. In order to calculate r(A1) observe that the cone P is normal and generating and A1 is a positive, linear and completely continuous operator (see [62]). Therefore r(A1) = A*, where A* is the maximal positive eigenvalue of A1 (see [17, Theorem 19.2], [37, Theorem 8.1]). It is easy to check that the eigenvalues of A1 coincide with these A for which the problem
{
Az'(t)
=
L1z(t)
(1 + ll~a I) z(O)
=
z(T)
has a nontrivial solution. Consequently,
r(A) = 1
By 3°, r(A)
L1T ln(1+ll~a~)·
< 1. The rest of the proof runs as before. 0
Next, consider the Darboux problem of neutral type (32)
{
Zxy
=
f(;;,y,z(h(x,y)),zxy(H(x,y))),
z(x,O) = 0, x E I, where I= [0, T], T
h(x,y)
=
(x,y)
E / 2,
z(O,y) = 0, y E I,
> 0 and
(h1(x,y),h2(x,y)), H(x,y)
=
(H1(x,y),H2(x,y)).
By the solution of (32) we mean a function z: I 2 ----+ lR such that z = z(x, y) is an AC-function in view of each variable x and y, Zx is an AC-function with
Chapter 2. Applications
56
respect to y for a.e. x E I, zy is an AC-function with respect to x for a.e. y E I, Zxy(x, y) = f(x, y, z(h(x, y)), Zxy(H(x, y))) a.e. on I 2 , z(x, 0) = 0 for x E I and z(O,y) = 0 for y E I. We will show that under suitable assumptions on the functions f, h and H, the problem (32) has a unique solution in the space L 1 (I 2 ) of Lebesgue measurable functions z : I 2 ----? lR such that [ iz(x, y)idxdy < oo, 1[2
with the norm
liz II
=
[
}p
iz(x, y)idxdy.
Theorem 2.1.3. [9] Suppose that:
1° h : I 2
----?
I 2 is a continuous function,
2° U, V c IR 2 are given open sets containing I 2 , H: U----? Vis a diffeomorphism with the property H(I 2)
c
I 2,
3° IH'(x, y)l ::; 1 a. e. on I 2 , where H'(x, y) denotes the Jacobian of H, and H- 1(D(h(H(x,y)))) c D(h(x,y)) for (x,y) E I 2 , where D(x,y) = {(t,s) E I 2 : 0::;
t::;
x, 0::; s::; y},
4° (x,y,w,z)----? f(x,y,w,z) is a real function defined on I 2 x IR 2 , Lebesgue measurable with respect to (x, y) E I 2 for every (w, z) E IR 2 and satisfying the Lipschitz condition
5° r(AI) + L2M < 1, where M = (min1ziH'(x, y)l)- 1 and r(AI) denotes the spectral radius of the operator (A1u)(x, y) = L1 [ JD(h(x,y))
in the space L 1 (I 2 ),
u(t, s)dtds
2.1. Initial and boundary value problems
57
6° there exists a function g : ! 2 ____, JR+ which is Lebesgue integrable and lf(x,y,O,O)I:::; g(x,y) a.e. on ! 2 . Then the problem (32) has a unique solution defined on 1 2 .
Proof. First, we verify that the problem (32) is equivalent to the following functional-integral equation
w(x,y)
(:{3)
=
f(x,y, { JD(h(x,y))
w(~,ry)d~dry,w(H(x,y))),
Let z: ! 2 ____, IR he a solution of (32). Put Zxy(x,y) By the definition of a solution of (32) we have
r
} D(h(x,y))
w(~, 17)d~d'f7 =
r
j D(h(x,y))
=
(x,y) E ! 2 .
w(x,y) for (x,y) E ! 2 .
Z~r1 (~, 17)d~d'f7
l
) - rhz(x,y) [ rhl(x,y) a ( a a~ ary z(~, "7) d~ dry Jo - Jo
rh2(x,y) a ary z(h1(x, y), ry)dr1
= Jo
= z(h1(.r,y),h2(x,y))- z(h1(x,y),O) = z(h1 (x, y), h2(x, y)) = z(h(x, y)). Thus w : 1 2 ____, IR is a solution of (33). On the other hand, let w : ! 2 be a solution of (33) in the space £ 1 (! 2 ). Then, putting
we have by the Tolstov theorem (see [60])
Zxy(x, y)
=
] a2 rx [ [Y axay lo lo w(~, ry)dry d~,
____,
lR
Chapter 2. Applications
58
for almost all (x,y) E 1 2. Hence Z:cy(H(x,y)) = w(H(x,y)) for almost all (x,y) E 1 2 . Consequently, z: 12 -JR is a solution of (32). Consider the operator
(Fw)(x, y) = f(x, y, { 1n(h(x,y))
w(~, ry)~dry, w(H(x, y))),
where wE £ 1 (1 2 ) and (x,y) E 1 2 . Since the function
(x,y)-
r
1n(h(x,y))
w(~,1J)~d1]
is continuous on 1 2 and the function (x, y) - w(H(x, y)) is Lebesgue integrable on 1 2 , the function
w(~, ry)~dry, w(H(x, y)))
(x, y)- f(x, y, { 1 D(h(x,y))
is Lebesgue measurable on 1 2 . Moreover, by 4° and 6° we get for wE £ 1 (1 2 )
j(Fw)(x,y)j
~ Ltl
{ 1 D(h(x,y))
~ Lt I r
1n(h(x,y))
w(~,ry)d~dryj +L2jw(H(x,y))j + lf(x,y,O,O)j
w(~, ry)d~drfl + L2lw(H(x, y))l + g(x, y)
a.e. on 1 2 • This implies F(L 1 (J 2 )) c £ 1 (1 2 ). To apply Theorem 1.2.12 put m(w)(x,y) = Jw(x,y)J for (x,y) E 1 2 • For v, wE £ 1(1 2 ) we say that v j w if and only if v(x,y) ~ w(x,y) a.e. on 12 . In view of 4°, for v, wE L 1 (J2 ) and (x, y) E 12 we obtain
j(Fv)(x,y)- (Fw)(x,y)J
~ £1
{ 1n(h(x,y))
jv(~,7J)- w(~,ry)j~d7J
+ L2jv(H(x, y))- w(H(x, y))J. Therefore
l(Fv)(x, y)- (Fw)(x, y)J
~
((A1
+ A2)lv- wj)(x, y),
where
(A1v)(x,y)
= £1 { 1n(h(x,y))
v(~,ry)~d7J
2.1. Initial and boundary value problems
59
and
This means that the assumption (v) of Theorem 1.2.12 is fulfilled. To prove that r(A1 + A2) < 1 we again apply Theorem 1.2. 7. Consider the cone
Obviously, the cone P is normal and generating and A1, A2 are positive operators. As in the proof of Theorem 2.1.1 it follows from 2° and 3° that A2A1w :::5 A1A2w for any wE P. Thus, by Theorem 1.2.7, r(A1 + A2) :::; r(AI) + r(A2). From 5° we get r(A1 + A2) < 1. Application of Theorem 1.2.12 completes the proof.
0
Remark . The problem (32) was considered for instance in the papers [8] and [9]. Particularly, in [9] we applied the Banach fixed point theorem to prove the existence and uniqueness of a solution of (32) in the space L 1 (!2 ). To obtain the uniqueness we assumed that L1T2 + L2M < 1, which is in general more restrictive than 5°. For example, if h(x, y) :::; (x, y) a.e. on ! 2 , that is h1(x,y) :::; x and h2(x,y) :::; y a.e. on ! 2 , we have r(A1) = 0 and the assumption 5° of Theorem 2.1.3 becomes L2M < 1. This example demonstrates that the application of Theorem 1.2.12 and the properties of spectral radius of positive operators instead of classical contraction principle allows us to relax the assumptions on the Lipschitz constants.
Next, we will deal with the following integral-functional equation (34)
z(x, y) = ¢(x, y)
+
r
JD(x,y)
!(~, TJ, z(h(~, TJ)))d~d'fJ,
where (x, y) E ! 2 , I = [0, T] and T > 0. We will show that the equation (34) has exactly one solution under assumptions of Caratheodory type and Lipschitz condition in the space L 1 (!2 ). Theorem 2.1.4. [72] Suppose that:
Chapter 2. Applications
60
1° (x, y, z) - f(x, y, z) is a real function defined on 12 xlR, Lebesgue measurable with respect to (x, y) E 12 for all z E lR and satisfying the Lipschitz condition if(x, y, zl)- f(x, y, z2)l :S k(x, y)iz1 - z2l for all (x, y, z1), (x, y, z2) E 12 x IR, where the function k: 12 is Lebesgue measurable and essentially bounded on 12 ,
[0, oo)
2° f (x, y, 0) is Lebesgue integrable on 1 2 ,
3° U, V c JR2 are given open sets containing ! 2 and h a diffeomorphism with the property h(I2 ) c ! 2 ,
4° the function ¢ : ! 2
U -
V is
lR is Lebesgue integrable on ! 2 ,
-
5° r(A) < 1, where r(A) denotes the spectral radius of the operator (Az)(x, y)
=!,
D(x,y)
k(~, 17)z(h(~, ry))d~dr7
Then the equation {34} has a unique solution in L 1 (J2 ).
Proof. The proof is similar to that of Theorem 2.1. 3. For z E L 1 ( / the operator (Fz)(x, y)
=
¢(x, y)
+ { Jo(x,y)
where (x, y) E / 2 • Let M
=
2)
consider
f(~, ry, z(h(~, ry))~dry,
esssupJ2 k(x, y). Then in view of our assumptions
2.1. Initial and boundary value problems
61
we have
I(Fz)(x,y)l::; l¢(x,y)l
+ +
::; 1¢(x, y)l
r
lf(~,1J,z(h(~,1J)))- !(~,1J,O)Id~d1J
r
If(~. 1], o)ld~d1J
1D(x,y)
1D(x,y)
+ f
1D(x,y)
+ ::; 1¢(x,y)l +
+ ::; 1¢(x, y)l
k(~, 7J)Iz(h(~, 1J))I~d1J
r
If(~. 1], o)ld~d1J
1D(x,y)
.
mlll[2
~( x, y )I 1[2 f lz(h(~,1J))IIh'(~,1J)I~d1J
If(~. 1], ~)l~d1J
r
1D(x,y)
+ mlllJ2 . ~( x, y )I 1[2 f lz(~, 11)ld~d11 +
r
1D(x,y)
If(~. 1], o)ld~d1J,
where h' denotes the Jacobian of h. This means that F maps the space L 1 (J 2 ) into itself. Moreover, in view of 1°, for all (x,y,w), (x,y,z) E J 2 x JR, we get
I(Fw)(x, y)- (Fz)(x, y)l ::; { 1D(x,y)
k(~, 1J)Iw(h(~, r1))- z(h(~, 17))l~dr7.
Let us define
(35)
(Az)(x, y)
=
{ 1D(x,y)
k(~, 17)z(h(~, 1J))~d1J.
Obviously, A : L 1 (J 2 ) - L 1 (J 2 ) is a bounded linear operator. By 5°, r(A) < 1. The rest of the proof follows directly from Theorem 1.2.12. 0
With connection to the equation (34), let us study the following Picard
Chapter 2. Applications
62 problem Zxy =
f(x,y,z(h(x,y))),
{
(36)
z(x, 0) = e(x), z(w(y), y) = r(y), where (x, y) E I 2 . By the solution of (36) we mean a function z: I 2 ----? lR such that z = z(x, y) is an AC-function in view of each variable x andy, zx is an AC-function with respect to y for a.e. x E I, zy is an AC-function with respect to x for a.e. y E I, Zxy = f(x, y, z(h(x, y))) a.e. on I 2 , z(x, 0) = e(x) for x E I and z(w(y), y) = r(y) for y E I. Theorem 2.1.5. Assume that the conditions 1° - 3° and 5° of Theorem 2.1.4 are fulfilled and moreover:
6° the functions e,
T:
7° the function w : I w(O) = 0.
I----? lR are absolutely continuous and e(O) = r(O),
----?
I is absolutely continuous and monotonic on I and
Then the problem {36} has exactly one solution defined on I 2 .
Proof. It is easy to show that under the above assumptions the problem (36) is equivalent to the equation z(x, y) = e(x) + r(y) - e(w(y)) +
f
}Dw(x,y)
where Dw(x,y) = {(~,TJ) E JR2 define the operator (37)
(Fz)(x, y) = e(x)
:
f(~, TJ, z(h(~, 1J)) )d~d1J,
w(y) ~ ~ ~ x, 0 ~ 1J ~ y}. For z E L 1 (I 2 )
+ r(y)- e(w(y)) + f }Dw(x,y)
where (x,y) E I 2 . Obviously, F(L 1 (I2 )) (x, y, w), (x, y, z) E I 2 x lR we get
I(Fw)(x, y)- (Fz)(x, y)l
~
{ }Dw(x,y)
f(~, TJ, z(h(~, TJ)))d~d1J,
c L 1 (I2 ).
By 1° and 7°, for all
k(~, TJ)iw(h(~, TJ))- z(h(~, TJ))Id~dTJ
~ f k(~,TJ)Iw(h(~,TJ))- z(h(~,TJ))Id~d1J. JD(x,y)
2.1. Initial and boundary V'alue problems
63
It is clear that the linear operator A which corresponds to (37) is given by (35). Therefore, by Theorem 1.2.12, F has a unique fixed point in £ 1 (1 2 ). This completes the proof. 0
Remark . Problems similar to (36) were considered in the papers [14] and [15]. Particularly, in the paper [14] the author studied the following Picard problem
{
Zxy = f(x,y,z(x,y),zx(x,y),zy(x,y)), z(x, 0) = e(x), z(w(y), y) = T(y),
under the assumptions of Caratheodory type. Using Schauder's fixed point theorem he proved the existence of a solution of the above problem.
Finally, we will deal with calculation of the spectral radius of the operator (35) in the space £ 1 (12 ). Let us put h(x,y) = (Axa, f..LYf3), k(x,y) = Kx'Yy 8 , where 'Y, t5 and K are positive numbers, 0
(Az)(x,y)
=
r
Jn(x,y)
KCr/z(A~a,wl)df.dry,
(x,y)
E
12 .
Notice that if 0 < a < 1 or 0 < f3 < 1, then the assumption 3° of Theorem 2.1.4 is not satisfied. However, as we can apply the theorem on integration by substitution in this case, the operator (38) is bounded in L 1 (I 2 ). To find r(A) we will use Theorem 1.2.3. It is easy to check that the set
is a normal and generating cone in L 1 (I 2 ). Furthermore, the function zo(x, y) = 1 a.e. on I 2 belongs toP and the operator (38) is zo-upper bounded. Thus, by Theorem 1.2.3
r(A) =lim sup 11Anzoll 1/n. n-+oo
Hence for the operator (38) we obtain
(Azo)(x, y)
x"f+l yHl
=
K - - -);-
'Y+1u+1
,
Chapter 2. Applications
64
K2 >. -y+l ,_i+l xa('Y+l)+-y+l yi3(6+l)+o+l (A zo)(x,y)= (1+1)(8+1) o{y+1)+-y+1 ,8(8+1)+8+1' 2
and generally for n EN, n
(Anzo)(x, y) =
> 1, we have Kn ).al +a2+ ... +an-1 f.Lb1 +b2+ ... +bn-1
a1a2 ... anb1b2 ... bn
where a1 = ')' + 1, an = (l:an-l + a1, b1 immediately that for n > 1
xa"yb",
= 8 + 1, bn = ,Bbn-1 + b1. It follows
Consequently, in the case of 0 < a < 1 and 0 < ,B < 1 we get
r(A)
2±.!. .ill = K).l-a f.Ll-{3
In the remaining cases r(A)
=
(1- a)(1- ,B) • ('y + 1)(8 + 1)
0.
Example 2.1.1. Consider the equation (34) with
¢(x,y) = 0, f(x,y,z) = 16xyz and h(x,y) = (y'x, .jY), where x, y E [0, 1]. Then a = ,B = 1/2, >. = f.L = -y = 8 = 1 and K = 16. Therefore we have r(A) = 1. It is easy to verify that the functions z(x, y) = C x 4 y 4 , where C E lR, are solutions of (34). The example shows that the assumption 5° of Theorem 2.1.4 is optimal in this case.
2.2
Boundary value problems for second order differential equations
In this section we will deal with the existence of positive solutions of boundary value problems (BVP for short) for second order differential equations. We begin with the application of Theorem 1.3. 7 to the following three point BVP:
x" (t)
(39)
{
+ f(t, x(t)) =
0
x(O) = 0, ax(b) = x(1), where t E [0, 1], a~ 0, bE (0, 1) and 1- ab > 0.
2.2. Boundary value problems
65
Our purpose is to obtain sufficient conditions ensuring the existence of at least one positive solution of (39). Some existence and multiplicity results for (39) can be found in [28], [30] and [47]. Particularly, in [30], Infante and Webb showed the existence of nonzero (but not necessarily positive) solutions of (39) under suitable conditions on/, when either 0::; a< 1- b or a< 0. It is easy to verify that the kernel of the integral equation which corresponds to (39) is given by
(1- 1 - a s)t, 0 < t < s < 1 l-ab - - -
{
s::;b
1-a (1 - - - t ) s 0 ::; s ::; t ::; 1 1- ab '
(40)
k(t, s) =
:b t,
11~
0 ::; t ::; s ::; 1
{
b ::; s.
s- ab s - - - t 0 ::; s ::; t ::; 1 1- ab' Observe that for bE (0, 1),
(41)
1\
a~
0 and 1- ab > 0 we have
Mk(s, s) ~ k(t, s) ~ 0,
t,sE[O,l]
where M= max
{l a(l-b)} ab ' ' 1-
and
1\
(42)
k(t, s) ~ m(t)k(s, s),
t,sE[O,l]
where m(t) = min{t, 1- t} fortE [0, 1].
Theorem 2.2.1. [79] Suppose that: 1° f : [0, 1] x [0, oo) r1 > 0 such that
-+
[0, oo) is a continuous function and there exists
66
Chapter 2. Applications for t E [0, 1] and x E [0, r1],
2° there exist toE (0, 1], c > 0, r2 > 0, r2 i- r1, and the continuous functions g : [0, 1] ---t [0, oo), h : (0, r2] ---t [0, oo) such that f(t,x) 2: g(t)h(x) fort E [0, 1] and x E (0, r2], the function h(x) = h(x)jxc is decreasing on (0, r2] and
Then the problem (39) has a positive solution on [0, 1] .
Proof. In order to apply Theorem 1.3.7, consider the Banach space C[O, 1] with the usual maximum norm and the set P = {x E C[O, 1]: x(t) 2:
m;;) llxiJ,
t E [0, 1]}.
Obviously P is a cone in C[O, 1]. Observe also that if x E P then x(t) 2: 0 on [0, 1]. Fix uo(t) = 1 on [0, 1]. Then uo E P and a(uo) = 1. In order to find P(uo) observe that if x E P(uo) then there exists..\> 0 such that x- ..\uo E P. This means that
m(t)
x(t)- ..\ 2: Mllx- ..\uoll
(43)
for every t E [0, 1]. Hence, for every t E [0, 1]
x(t) 2:
m;;) llx- ..\uoll + ..\ 2: m;;) (lixll - >.iiuoii) + ..\
= m(t) llxll + ..\(1- m(t) ). M M Since
·on [0, 1], we get (44)
x(t) >
~) llxll,
t E
[0, 1].
67
2.2. Boundary value problems
On the other hand, if (44) is fulfilled then from the continuity of x and m it follows that there exists to E [0, 1] such that
x(t)-
m;;) llxll 2: x(to)-
mzo) llxll > 0
for every t E [0, 1]. Fix
m(to)
,\ = x(to)- -ullxll. Then
llx- Auoll = llxll - .\lluoll and (43) follows. Therefore
P(uo) = {x
E
C[O, 1] : x(t) >
m;;) llxll, t
E
[0, 1]}.
For x E P and t E [0, 1] define the integral operator
(Fx)(t)
=
1 1
k(t, s)f(s, x(s))ds,
where the kernel k is given by (40). Clearly, if x E P then Fx is a continuous function on [0, 1] and every fixed point of F is a solution of the problem (39). Let t E [0, 1] and x E P. Then we have in view of (41)
r1
1
r1
Jo k(s, s)f(s, x(s))ds 2: M Jo k(t, s)f(s, x(s))ds, hence
1 1
k(s, s)f(s, x(s))ds 2:
~ IIFxll.
On the other hand, we get by (42)
(Fx)(t) =
1 1
k(t, s)f(s, x(s))ds 2: m(t)
1 1
k(s, s)f(s, x(s))ds.
Thus for every t E [0, 1] 1
(Fx)(t) 2: Mm(t)IIFxll, which means that F : P - t P. Let 01 = {x E C[O, 1] : llxll < ri} and 02 = {x E C[O, 1] : llxll < r2}. We may assume that r1 < r2. Obviously, F is
Chapter 2. Applications
68
a completely continuous operator on Pnn 2. By 1° and (41), for x E Pnan 1 and t E [0, 1] we get
(Fx)(t) ::; M
::; M
1 1
1k(s, s)f(s, x(s))ds
1
k(s, s) [M
1 1
1
k(r, r)dr] - r1ds = r1.
Hence IIFxll ::; llxll for X E p n an1. Let X E P(uo) n an2. Then and x(t) E (0, r2] for every t E [0, 1]. By 2°, we obtain
~
(Fx)(to)
=
1 1
{
}0
k(to, s)g(s)h(x(s))ds
1 k(to, s)g(s) h(xt?) xc(s)ds xc s
~ h~2 ) =
llxll = r2
1
1k(to, s)g(s) [ m;;) llxll] c ds
h(r2) [ 1
Me Jo k(to, s)g(s)mc(s)ds ~ r2.
Since a(uo) = 1, this gives IIFxll ~ llxll for X E P(uo) n an2. By Theorem 1.3.7, F has a fixed point x* E P such that r1 ::; llx*ll ::; r2. This ends the 0 proof of Theorem 2.2.1.
Example 2.2.1. Consider the problem (39) with 16(t + 1)x2 (x- 3) 2 ,
f(t, x)
(t + 1)(x- 3),
!·
(t, x)
E
[0, 1] x [0, 3]
= {
(t, x)
E [0, 1] x (3, oo)
and a = b = It is easy to verify that the assumptions of Theorem 2.2.1 are satisfied for r1 = 0 , r2 = 1, to= 1, c = 2, g(t) = 16(t + 1) and h(x) = x 2 (x- 3) 2 • Therefore the problem (39) has a positive solution x* defined on [0, 1] such that
J
Jo ::; llx*ll ::; 1.
2.2. Boundary value problems
69
Next we will be concerned with the following two point BVP:
x"(t) + f(t,x(t),x'(t))
=
0
(45)
where t E [0, 1], a1, a2 2': 0, b1, b2 > 0 and a1b2 + b1a2 + a1a2 > 0. Similar problems were studied by several authors (see for example [1], [19] and [44]). The Green's function for the homogeneous problem related to (45) is given by
(46)
1 { (a 1t
+ bi) (a2 (1 -
p
+ bi)(a2(1- t) + b2), 0 ~ s ~ t ~ 1,
G(t,s) =-
(a1s
s)
+ b2),
0~t ~s~1
where p = a1b2 + b1a2 + a1a2. It is easy to show that the function (46) has the following properties
(47)
G(t, s) 2': 0 for all t, s E [0, 1],
(48) G(t, s) + IGt(t, s)i
(49)
(50)
~
G(s, s) + IGt(s- 0, s)i fort, s E [0, 1] and t < s,
G(t, s) 2': TJ(t)G(s, s) fort, s
G(s, s) 2': d[G(t, s)
+ IGt(t, s)i]
E
[0, 1],
fort, s E [0, 1],
where Gt(s- 0, s) denotes the left-hand side derivative of (46) at (s, s),
and t E
[0, 1].
The inequality (48) also holds with Gt(s- 0, s) replaced by the right-hand side derivative Gt(s + 0, s) and t > s.
Chapter 2. Applications
70
Theorem 2.2.2. Assume that:
1° f : [0, 1] x [0, oo) x lR r1
---+
[0, oo) is a continuous function and there exists
> 0 such that f(t, x, y) :::;
[1
1
(G(s, s)
fortE [0, 1], x ~ 0 and x
+ ~max{ a1 (a2 + 1>:2), a2(a1 + b1)} )ds] - 1 r1 + IYI:::; r1.
2° there exist to E [0, 1], r2 > 0, r2 f- r1, c > 0 and the continuous functions g : [0, 1] ---+ [0, oo ), h: (0, r2] ---+ [0, oo) such that f(t, x, y) ~ g(t)h(x) for t E [0, 1], x > 0 and x decreasing on (0, r2] and
+ IYI :::;
r2, the function h(x)
=
h(x)fxc is
Then the problem {45} has a positive solution on [0, 1].
Proof. Consider the Banach space E = C 1 [0, 1] with the norm
llxll
= max tE[0,1]
and the set
{lx(t)l + lx'(t)l}
1\
P = {x E E:
x(t) ~
dry(t)llxll}.
tE[O,l]
It is easy to show that P is a cone in E. Let uo(t) = 1 on [0, 1]. Then uo E P and a(uo) = 1. Similarly as in the proof of Theorem 2.2.1 we can show that
P(uo) = {x E E:
1\
x(t) >
dry(t)llxll}.
tE[0,1]
For x
E
P define the operator (Fx)(t)
=
1 1
G(t, s)f(s, x(s), x'(s))ds, t E [0, 1].
71
2.2. Boundary value problems
By 1° and (47), (Fx)(t) 2::0 for every t E [0, 1]. Moreover, from (49) we obtain
(Fx)(t) =
1 1
2:: ry(t)
G(t, s)f(s, x(s), x'(s))ds
1 1
G(s, s)f(s, x(s), x'(s))ds.
In view of (50) we get
1 1
G(s, s)f(s, x(s), x'(s))ds 2:: d
2::
for every
T
J[l1
1
1 1
[G(T, s)
G(T, s)f(s, x(s), x'(s))dsl
+ IGt(T, s)i
+
11
1
Jf(s, x(s), x'(s))ds
Gt(T, s)f(s, x(s), x'(s))dsl]
E [0, 1]. Hence for every t E [0, 1]
(Fx)(t) 2:: dry(t)IIFxll·
Thus F(P) C P. We may assume that r1 < rz. Let 01 = {x E E: llxll < r1} and n2 = {x E E: llxll < rz}. It is clear that F is completely continuous on Pn?h. For x E Pn80 1 we get from 1° and (48)
I(Fx)(t)i
+ I(Fx)'(t)i
Hence IIFxll ~
llxll
for
X
~
1 1
[G(t, s)
E p nan].
+ IGt(t, s)i] f(s, x(s), x'(s))ds
If
X
E P(uo) n an2
then
~ q.
llxll =
rz and
Chapter 2. Applications
72
x(t) E (0, r2] for every t E [0, 1]. In view of 2° I(Fx)(to)l
+ I{Fx)'(to)l
2:
2:
1 1 1
G(to, s)f(s, x(s), x'(s))ds
1
G(to, s)g(s)h(x(s))ds
h(x(s))
[ 1
c
= Jo G(to, s)g(s) [x(s}JC [x(s)] ds 2: h(:2 ) r2
{
lo
1
G(to, s)g(s)[x(sWds
Thus, llxll :::; u(uo)IIFxll for X E P(uo) n an2. By Theorem 1.3.7, F has a fixed point in the set P n (02 \ O.I). This means that the problem (45) has D a positive solution x* such that r1 ::S llx* II :::; r2.
Example 2.2.2. Consider the problem (45) with a1 = b1 = b2 and f(t, x, y) = 3(x + lyl) 2 (x- 3) 2 + IYI(2x- 1).
= 1,
a2
=0
=
Then ry(t) = t~ 1 and d = 1/2. Fix r1 = 0.01, r2 = 0.2, to = 1, c = 2, g(t) 1 on [0, 1] and h(x) = 3x 2 (x- 3) 2 for x E (0,0.2]. It is easy to check that the assumptions of Theorem 2.2.2 are fulfilled. Therefore, the problem (45) has a positive solution x* such that 0.01 :::; llx* II :::; o.2
and
for every t E [0, 1].
73
2.3. Hammerstein integral equation on the half-line
Remark . In [2] the authors used Theorem 1.3.6 to the nonlinear integral equation
x(t)
=
1~ 7 k(t, s)f(s, x(s))ds,
t E R,
where T > 0 is a fixed constant and f is periodic in the first variable. It is worth pointing out that the method from [2] is not directly applicable to the problem considered in Example 2.2.2, because the function h(x)jx, where h(x) = 3x 2 (x- 3) 2 , is increasing on (0, 1).
2.3
Hammerstein integral equation on the half-line
This section is devoted to some existence results for the Hammerstein integral equation on the half-line. Namely, we will study the existence of positive solutions for the following equation:
x(t)
(51)
=
1
00
K(t,s)f(s,x(s))ds,
t E [O,oo)
in the Banach spaces equipped with Bielecki's norm. We begin with an application of Krasnoselskii's theorem of the norm type (Theorem 1.3.4) to the equation (51) with the kernel K given by the following formula:
K (t s) '
(52) where k
> 0. Write I
=
1 { e-ks(ekt- e-kt), 0 ~ t ~ s =-
2k
[0, oo). It is easy to show that
(53)
/\ K(t, s) 2:: 0, t,sEl
and
/\ K(t, s)e-11-t ~ K(s, s)e-ks
(54)
t,,9El
for
f..L
2:: k and t, s E I.
Moreover, for any 81, 82 such that 0 (55)
< 81 < 82 and
K(t,s) 2:: MK(s,s)e-ks,
t E [8t, 82] we have
74
Chapter 2. Applications
where s E I and
(56) In order to apply Krasnoselskii's theorem to the equation (51) we will need an appropriate function space with Bielecki's norm. Let m : I ----t (0, oo) be a continuous function. Denote by E the Banach space of all functions x continuous on I and satisfying sup{lx(t)lm(t)} < oo, I
equipped with the norm llxll = sup{lx(t)lm(t)}. I
The Arzela - Ascoli theorem fails to work in the space E, however some sufficient conditions of compactness are known (see [5]). In our considerations we will use the following modification of the compactness criterion from [67]. Theorem 2.3.1. Let 0 C E. If the functions x E 0 are almost equicontinuous on I and uniformly bounded in the sense of the norm
llxllq =sup {lx(t)l q(t)}, I
where q is a positive and continuous function on I and . m(t) lun -() = 0,
t-+oo q t
then
n
is relatively compact in E.
Recall that the functions X E n are said to be almost equicontinuous on I if they are equicontinuous in each interval [0, T], 0 < T < oo. We can now state and prove a result on positive solution of (51). We will use the space E with m(t) = e->.t, where>.> k. Theorem 2.3.2. Let K be the function given by {52}. Suppose that:
2.3. Hammerstein integral equation on the half-line 1° f : I x I--+ I is continuous and f(t, x) :::; a(t) where a, b : I --+ I are continuous functions, 2° the integrals M1 = vergent and M2 3° there exist {3 such that
J0
00
e-ksa(s)ds and M2 =
75
+ b(t)x for
J0
00
(t, x) E I x I,
e(>.-k)sb(s)ds are con-
< 2k,
> 0, {3
=1- M1 (2k - M2)- 1 , 61,62
> 0, 61 < 62 and to E I
where M is given by (56). Then the equation (51) has at least one positive solution.
Proof In the space E consider the set P = {x E E : x(t) 2: 0 on I and
min x(t) 2: M llxll} .
[
It is clear that P is a cone in E. Define the integral operator F by (Fx)(t) =
1
00
K(t, s)f(s, x(s))ds,
where x E P and t E I. Obviously, every fixed point ofF is a solution of (51). We will show that F satisfies the assumptions of Theorem 1.3.4. It is easy to prove that F maps P into E. In particular, by 1°, 2°, (53) and (54) we get for xEPandtEI I(Fx)(t)l e->.t = e->.t (57)
:::;
1
00
1
00
K(t, s)f(s, x(s))ds
e-ks K(s, s) [a(s)
+ b(s)x(s)] ds
roo e-ksa(s)ds + 2k1 llxll Joroo e(>.-k)sb(s)ds,
1 :::; 2k Jo
which implies that
Chapter 2. Applications
76
Next, observe that F(P) C P. Indeed, from 1° and (53) we see that . (Fx)(t) 2: 0 on I. Furthermore, by (54) and (55) for any t E [61, 62] and s, T E I we obtain min (Fx)(t) = min [(h,62]
[61,62]
roo K(t, s)f(s, x(s))ds 2: M loroo K(s, s)e-ks f(s, x(s))ds
lo
2: Me-kr
1
00
K(r, s)f(s, x(s))ds = Me-kr(Fx)(r).
Consequently, min (Fx)(t) 2: M IIFxll
[6~,62]
and we conclude that F maps P into itself. Fix r = M1 (2k - M2)- 1 and R = /3 and define
01
={xEE:IIxll
and
02
= {x EE: llxll < R}.
Without loss of generality we may assume that r < R. Let x E P n 02. Proceeding analogously to (57) we can prove that the functions Fx are uniformly bounded with respect to the norm
where 0 < J.L < .>.. The standard arguments show that the functions Fx are almost equicontinuous on I (see [68]). Thus, by Theorem 2.3.1, F is completely continuous on p n 02. If X E p n anl then from (57) and 2° 1
e->.t(Fx)(t) ~ 2 k (M1 hence
IIFxll
~
llxll·
If
X
E p n 802
+ rM2) =
then
and
s~p { x(t)e->.t}
=
/3.
r,
2.3. Hammerstein integral equation on the half-line Thus, fortE [61.62], we have
(Fx)(to) =
M/3:::; x(t):::; f3eM2.
1
K(to, s)f(s, x(s))ds 2:
!"2
( !"2
00
77
By 3°,
1"2 K(to, s)f(s, x(s))ds ~
0
2: }., 1 K(to,s) }.,1 K(to,T)dT
) -1 f3e>.t
0
ds
=
f3e>.to,
which implies e->.to(Fx)(to) 2: /3. This gives IIFxll 2: llxll for x E P n 802. By Theorem 1.3.4, the operator F has at least one fixed point in the set P n (02 \ 0 1 ), which means that the equation (51) has a positive solution x D such that r :::; llxll :::; R. This completes the proof.
Remark . It can be shown that under the assumptions of Theorem 2.3.2 the differential equation
x"(t)- k2 x(t) + f(t, x(t)) = 0 has a positive solution defined on I such that x(O) 0 (see [75]).
= 0 and
limt-+oo e->.tx(t)
=
We now turn to study the existence of solutions of (51) in the space of Lebesgue p-integrable functions. Let m be a positive function defined on I = [0, oo ). Consider the space
LP(I, m), p 2: 1, of Lebesgue p-integrable functions x on I with the weighting function m, i.e.
1
00
lx(t)IPm(t)dt < oo.
If we equipped it with the norm
roo
llxll = (lo lx(t)IPm(t)dt
) 1/p ,
then it becomes the Banach space. We first apply Schauder's fixed point theorem to prove the existence of a solution of (51) in LP(I,m). In what follows we will need the modification of Riesz compactness criteria in £P spaces.
Chapter 2. Applications
78
Theorem 2.3.3. [69] Let M be a nonempty bounded subset of V(I, m). Suppose that: 1° for all x EM and A E (0, oo)
lim
h-+O
(
r
A
}0
lx(t +h)- x(t)!Pm(t)dt
)1/p =
0
uniformly on M, 2° there exists a continuous function q: I---t (0, oo) such that
lim m(t) = 0, q(t)
t-+oo
and the functions x E M are uniformly bounded in the sense of the norm (58)
llxllq =
(
roo
Jo
)
lx(t)!Pq(t)dt
1/p
Then M is relatively compact in V(I, m). Consider the space V(I, q) with the norm (58), where the weighting nmction q is positive and continuous on I and
.
m(t)
hm -() q t =0.
t-+oo
Next, let Pl. P2 be positive numbers such that 1/P1 + 1/P2 = 1 and P/P1 2: 1. By £P 1 (I) we will denote the space of Lebesgue P1-integrable functions on I, with the usual norm
Finally, .£'P2 (I, m) will stand for the space of Lebesgue P2-integrable functions, tempered by the function m, with the norm
llxll2 =
(fooo lx(t)IP m(t)dt) 11P2
Theorem 2.3.4. [77] Suppose that:
2
2.3. Hammerstei11 i11tegral equatio11 1°
011
the half-li11e
79
f:
I x JR. ---t JR. is a Caratheodory function, that is, t ---t f(t, x) is measurable for all x E JR. and x ---t f(t, x) is continuous for almost all t E I,
5 a1(t) + a2ixiP/P2 for all x E JR. and a. e. t LP2 (I,m), a1(t) > 0 a.e. on I and a2 > 0,
2° if(t, x)i a1 E
E I, where
3° the function (t, s) ---t K(t, s) is measurable for (t, s) E I x I and there exists a measurable function g : I x I ---t I such that
IK(t, s)l 5 g(t, s)[m(s)] 11112 for almost all t, s E I and
lorJO [loroo [g(t,s)]Pq(t)dt]PI/P ds < oo, 4° there exists 6 all s E I
> 0 such that for ihi < 6 and almost all t
E I and almost
IK(t+h,s)-K(t,s)i5ig(t+h,s)-g(t,s)i[m(s)J 11P2 , 5° there exists r > 0 such that
Then the equation (51) has at least one solution in LP (I, m).
Proof For x E LP(I, m) and a.e. t E I define the operator
{59)
(Fx)(t) =
1
00
K(t, s)f(s, x(s))ds.
First, we will show that F maps LP(I, m) into itself. By 1° - 3°, Holder's inequality and the integral version of Minkowski's inequality (see [6], [61]) we
80
Chapter 2. Applications
get
(60)
But
Joroo [Joroo [g(t, s)]Pm(t)dt]Pi/P ds::::; Joroo [Joroo [g(t, s)]Pa q(t)dt]pJ/p ds < oo, where
a= max m(t). tEl q(t) Thus
Next, we will prove that F is continuous. For x E LP(I, m) consider the Nemytskii operator
(Nx)(t) Then
1
00
f(t, x(t)), a.e. t E I.
=
I(Nx)(t)IP2 m(t)dt
1 : :; 1
=
00
00
lf(t, x(t))IP2 m(t)dt (a1(t)
+ a2ix(t)IPIP2)P2m(t)dt
::::; (lla1ll2 + a2llxiiP/P2)P2 < oo. Thus, for every x E LP(J,m), we have Nx E LP2 (l,m). A slight modification of the well-known theorem on the Nemytskii operator acting in £P spaces
81
2.3. Hammerstein integral equation on the half-line
([34]), gives us the continuity of N. Let x E LP(I, m). Then for any {xn}, Xn E LP(I, m), convergent toxin LP(I, m), we get by 3°
IIFxn - Fxllp
~
1 [1 00
00
IK(t, s)llf(s, Xn(s))- f(s, x(s))idsr m(t)dt
~ fooo [fooo g(t,s)[m(t)] 11Pif(s,xn(s))- f(s,x(s))i[m(sWIP :5 ( fooo
t"
[fooo [g( t, s)I" [m( t)I" fp ds x
2
dsr dt
dt)
( Joroo lf(s, Xn(s))- f(s, x(s))IP m(s)ds )P/P2 2
11
oo [ oo [g(t, s)]Pm(t)dt]Pl/P ds )P/P1 IINxn- Nxll~, ~( which implies the continuity of F. Let for X E !1 we get from (60)
n=
{x
E
LP(I,m): llxll
~
r}. By 5°,
Hence F(O) C !1. Moreover, for A E (0, oo) and x E !1 we obtain in view of 2° and 4°
foAi(Fx)(t +h)- (Fx)(t)IPm(t)dt
r
~faA [laoo lg(t + h, s)- g(t, s)l[m(t)] 1/Pif(s, x(s))l [m(sWIP ds 2
~
dt
( Joroo [Joroo lg(t + h, s)- g(t, s)IPm(t)dt]pl/p ds ) P/P1
It is easy to prove that translating the functions in LP(J, m) is continuous in norm. Therefore, we get
Joroo [Joroo lg(t+h,s)-g(t,s)IPm(t)dt]Pl/P ds-tO
when h-tO.
Chapter 2. Applications
82 Hence lim
fA i(Fx)(t +h)- (Fx)(t)iPm(t)dt = 0
h-+O } 0
uniformly on fl. If X E fl, then
This means that the functions Fx are uniformly bounded with respect to the norm (58). By Theorem 2.3.3, the operator F is completely continuous. Ap0 plication of Schauder's fixed point theorem completes the proof. Next, we will study the existence of positive solutions of the equation (51) in £P(I, m).
Theorem 2.3.5. [77] Assume that:
1° f : I xI ---. I is a Caratheodory function, that is, t ---. f (t, x) is measurable for all x E I, and x ---> f (t, x) is continuous for almost all t E I., 2° for all x E I and a. e. t E I
where a1 E LP2(I,m), a1(t)
> 0 a.e. on I,
P2
> 1,
a2
> 0,
3° the function (t, s) ---. K (t, s) is measurable on I x I, and there exist measurable functions k1, k2, k3 positive a. e. on I, k1 E £P(I, q), k2 E £P 1 (I), k3 E £P(I, m) and a constant 0 < M ::::; 1 such that for almost all t, s E I
83
2.3. Hammerstein integral equation on the half-line 4° there exists
IK(t
<5
> 0 such that for 0 < Ihi <
+ h, s)- K(t, s)l
5° there exists R
and almost all t, s E I
S lk1{t +h)- k1{t)l k2(s)[m(s)] 1/P2 ,
> 0 such that llk11illk2lh (!ia11i2
6° there exist r
<5
+ a2RPIP2) S
> 0 , r =f R, and an interval J
R,
C I such that
f(t,x) 2: r ( Mllk3111 k2(s)[m(sWIP2 ds) -
1
for all x E I and a. e. t E J. Then the equation (51} has a positive solution x on I, that is, x(t) I and x E LP(I, m).
> 0 a. e. on
Proof Consider the set
P
= {x E
V(I, m): x(t) 2:
Mk3(t) llk1ll llxll a.e. on I}.
Clearly, P is a cone in LP(I, m). Analogously to the proof of Theorem 2.3.4 we can show that the operator F given by {59) maps the cone Pinto LP(I, m). Moreover, from 3° we have for x E P and a. e. t E I (Fx)(t) 2: Mk3(t)
1
00
k2(s)[m(s)] 1/P2 f(s, x(s))ds
roo
)
roo k2(s)[m(s)] 1/P f(s, x(s))ds
=
Mk (t) ( 1/p Ilk: II lo [k1(t)]Pm(t)dt lo
=
Mk (t) llk:ll
2:
~~:fit) (1
2
(roo [roo ]P )1/p Jo Jo k1(t)k2(s)[m(s)] 11P f(s,x(s))ds m(t)dt 2
00
Mk3(t)
= llk111 IIFxll·
[1
00
K(t,s)f(s,x(s))dsr m(t)dt)
1 /P
84
Chapter 2. Applications
Hence F : P - t P. It follows from 1°- 4° and Theorem 2.3.3 that F is completely continuous on P. Without loss of generality we can assume that r < R. Let fh = {x E LP(I,m): llxll < r} and fl2 = {x E LP(I,m): llxll < R}. In view of 3° and 6° we obtain for X E p n an1
IIFxll ;: : (1 (1 Mka{t)k2(s)[m(s)] 11P f(s, x(s))ds) Pm(t)dt) 1/P 00
00
2
=
(1 MP[ka(t)]Pm(t)dt) 1/P 1 k2(s)[m(s)] 1/P2 f(s, x(s))ds
=
Mllkall1 k2{s)[m(s)] 1/P f(s, x(s))ds
00
00
00
2
;: : Mllkall
i
;: : Mllkall
i k2(s)[m(s)] 1/P r ( Mllkall i
=r=
k2(s)[m(s)] 1/P2 f(s, x(s))ds 2
1
k2(r)[m(rWIP2 dr) - ds
llxll·
Thus, for X E p n an1 we have IIFxll 2: 3° and 5° we obtain similarly to (60)
llxll·
If
X
E p
n an2 then by 2°,
IIFxll ~ llk1llllk2lh (lla1ll2 + a2llxiiP/P2) ~ llk1llllk2ll1 (lla1ll2 + a2R?IP2) ~R=
llxll.
In view of Kranoselskii's theorem of the norm type {Theorem 1.3.4), the equation {51) has a positive solution x* such that r ~ llx*ll ~ R. This com0 pletes the proof of Theorem 2.3.5. Example 2.3.1. Consider the following equation
(61)
x(t)
~ /,
00
K(t,s) ( 2 :
8
+ ( x(s)-
D')
ds, t E I,
2.4. Boundary value problems in Banach spaces
85
where
t 3 K(t,s) = -1 -e-2 8 • +s Fix m(t) = e- 3t, q(t) = e- 2t, p = 4, Pl = P2 = 2. It is easy to check that the
function
1 ( 21)
f(t, x) = 2 + t + x-
2
fulfills the assumptions of Theorem 2.3.5. Hence the equation (61) has a positive solution. It is worth pointing out that f is not an increasing function in x. Therefore the results from [49] are not applicable here.
2.4
Boundary value problems for differential equations in Banach spaces
In this section we will deal with the following two point boundary value problem for the functional-differential equation of second order
x"(t) + kx'(t) + f(t,x(h1(t)),x(h2(t))) = 0, (62)
{ ax( -1)- bx'( -1) = 0, cx(1)
+ dx'(1)
=
0,
where the function f takes values in a cone P of a real Banach space E, t E [-1, 1], k E R, a, b, c, d ~ 0 and ad+ be+ ac > 0. Observe that for h1(t) = t and h2(t) = -t we obtain the BVP involving reflection of the argument. Such problems (that is the BVPs with reflection of the argument) were considered for example in the paper [27] for k = 0 and f : [-1, 1] x R x R ---+ R and in [29] for f taking values in a real Hilbert space. The authors applied the Leray - Schauder theorem in order to establish existence results for (62) with h1(t) = t and h2(t) = -t. Our purpose is to prove the existence of positive solutions of (62). To this end, we will use Theorem 1.3.9. First, we will study some properties of the functions ~e-kt[ek(s-l)J.tl
(63)
+ c][ek(t+l)J.t2- a],
-1:::;
t:::; s:::; 1,
+ c],
-1:::;
s:::; t:::; 1,
G(t,s) = { ~e-kt[ek(s+l) J.t2 -a] [ek(t-1) J.tl
Chapter 2. Applications
86 where k
=f. 0,
f..tl = dk- c, f..t2 = bk +a and p = i[ae-kf..tl p1* ( c
(64)
G*(t, s) =
+d -
cs) (a
+ cekJ..t2]
and
+ b + at), -1st s s s 1,
{
;. (a + b + as) (c + d - ct), -1
s sst s 1,
where p* = 2ac + be + ad. Put I= [-1, 1]. It is easy to show that the function (63) fulfills the following inequalities:
1\ G(t, s) 2:: 0
(65)
t,sEl
and
1\ G(t, s) S G(s, s).
(66)
t,sEI
Moreover, for any -1 S
G(t,s) 2:: mG(s,s),
(67) where s E I and {68)
It is easily seen that m
< 1.
The function G* also satisfies the inequalities {65), (66) and {67) with m replaced by
{69)
* _ . {a + b + a81 c + d - c82 } m - mm 2a + b ' 2c + d ·
Clearly, m* < 1. From now on we will assume that P is a normal cone (with the normal constant 'Y = 1) in a real Banach space E. This means that the norm II · liE in E is monotonic with respect to P. Further, denote by C(I, E) the space of all continuous functions defined on the interval I and taking values in E, equipped with the norm
llxll
=max llx(t)IIE· tEl
2.4. Boundary value problems in Banach spaces
87
Obviously, C(I, E) is a Banach space. Let
Q = {x
E
C(I,E): fJ
~
x(t) for t E I}.
It is easy to prove that Q is a cone in C(I, E). Throughout this section we will assume that: 1°
f :I
x P x P
---+
P is a continuous function,
2° h1, h2 : I---+ I are continuous functions mapping the interval I onto itself. Next, consider the integral-functional operator
(70)
1
(Fx)(t) = /_ 1 G{t, s)f(s, x(h1(s)), x(h2(s)))ds,
where t E /, x E Q, the function G is defined by {63) and
f,
h1 and h2 satisfy
M = maxG(t, s), t,sEl
and
Br =
{x E C(I,E): ilxll :S: r}.
The following lemma is a slight modification of that given in [25). Lemma 2.4.1. Assume that for any r > 0: 3° the function f is uniformly continuous on I x ( P n T r) x (P n T r),
4° there exists a non-negative constant Lr, such that 4M Lr < 1 and
a{f(t, n, f!)) :S: Lr a(f!) for all t E I and
nc
p n T r.
Then, for any r > 0 the operator (70} is a strict set contraction on Q n Br.
Chapter 2. Applications
88
Proof. From 3° it follows that f is bounded on I the uniform continuity off (see [16])
a{f(I
X
0
X
X
(P n Tr)
X
(P n Tr)- By
0)) = maxa{f(t, 0, 0)), tEl
hence, in view of 4°
a(f(I
{71)
X
0
X
0)) :::; Lr a(O)
for every 0 c PnTr. The uniform continuity and boundedness off on the set I x (P n T r) x (P n T r) implies also continuity and boundedness of operator F on Q n Br. LetS c Q n Br. Since the functions Fx are equicontinuous and uniformly bounded for xES, we obtain (see [17])
a(F(S)) = supa(F(S)(t)), tEl
where F(S)(t) denotes the cross-section of F(S) at the point t, that is
F(S)(t)
=
{(Fx)(t) : x
E
S, t is fixed}.
Furthermore, for every t E I we get (see [48])
1
2(Fx)(t)
=
111
2 _1 G(t, s)f(s, x(h1 (s)), x(h2(s)))ds
E conv{G(t,s)J(s,x(h1(s)),x(h2(s))): s E I,x E S}
c conv {{M f(s, x(h1(s)), x(h2(s))) : s E I, xES} U {0}}. Thus, in view of the properties of the Kuratowski measure of noncompactness we obtain for t E I a(~F(S)(t))
:::; Ma({f(s,x(h1(s)),x(h2(s))): s E I,x E S}) :::; M a(f(I x S(I) x S(I)) ),
where S(I) = {x(s): s E I,x E S}. Hence, by (71) we have
a(F(S)(t)):::; 2Ma(!(I
X
S(I) x S(I))):::; 2MLra(S(I)).
Finally, proceeding as in the proof of Lemma 2 [25], we can show that
a(S(I)):::; 2a(S).
2.4. Boundary value problems in Banach spaces Hence for any S C Q
89
n Br
a(F(S)) = supa(F(S)(t))
~
4MLra(S),
tEl
which means that F is a strict set contraction on Q [l Br.
0
Remark . Obviously, Lemma 2.4.1 remains valid for the operator (70) with the function G* given by (64) and the constant
M* =max G*(t, s). t,sEl
Now we state and and prove two results on positive solutions of {62). First, consider the case k -:/:- 0. Theorem 2.4.1. Let G be given by (63) and let -1 ~ 81 < 82 ~ 1 be such that hi : [81, 82] ~ [81. 82], i = 1, 2. Suppose that the assumptions 1°- 4° are satisfied and:
5° there exists u
E P, u-:/:- (}, such that
f(t, x, y) ::5
[[~ G(s, s)ds] - 1 u
for all t E I and x, y E P such that llxiiE,
IIYIIE E [0, lluiiE],
6° there exist v E P, v-:/:- 9, llviiE-:/:- iiuiiE and toE I such that
[.1::
2
1
G(to,s)ds]- v ::5 f(t,x,y)
for all t E I and x, y E P such that llxiiE, where m is given by ( 68).
IIYIIE
Then the problem {62} has at least one positive solution.
E [m
llviiE, llviiE],
Chapter 20 Applications
90
Proof Notice that each positive solution of the problem (62) (with k a fixed point of the integral-functional operator (70), that is
=I 0)
is
1 (Fx)(t) = [ 1 G(t, s)f(s, x(h1 (s)), x(h2(s)))ds where t E I, x E C(I, E) and the function G is given by (63)0 On the other hand, if x belonging to the cone Q is a fixed point ofF, then x is a solution of (62) (see [25])0 Thus, to prove our theorem it is enough to show that F has a fixed point in Q o In the space C (I, E) consider the set
P=
{x E C(I, E) : fJ
~ x(t)
on I and
1\ 1\ mx(s) ~ x(t)}
0
tE [8u52 ] sEl
Clearly, P is a cone in C(I, E) and the norm 11°11 in C(I, E) is monotonic with respect to Po Consider the operator (70) for t E I and x E Po We will show that F satisfies the assumptions of Theorem 1.3o9o First, we will prove that F(P) C Po To this end observe that by 1° and (65) fJ ~ (Fx)(t)
(72) for every x E and s E I
P and t E Io
Moreover, it follows from (67) that for any t E [<>1. 62]
1 m(Fx)(s) = m [ 1 G(s,s)/(s,x(h1(s)),x(h2(s)))ds
~[ =
1 1 G(t, s)f(s, x(h1(s)), x(h2(s)))ds
(Fx)(t)o
Combining it with (72) we conclude that F(P) C Po Without loss of generality we may assume that lluiiE < llviiEO Fix r = lluiiE and R = llviiEO By Lemma 2.401, F is a strict set contraction on the set P n Bno Moreover, for x E P n 8Br we have fJ ~ x(h1(t)) on I and llxll = lluiiE, hence
f\ tEl
llx(h1(t))IIE:::;
lluiiE
0
2.4. Boundary value problems in Banach spaces Analogously tEl
Thus , by 5°, for any t E I we obtain 1 G(s, 1
(Fx)(t) j [
s)f(s, x(h1(s)), x(h2(s)))ds
Hence, in view of monotonicity of II·IIE we get
1\ II(Fx)(t)IIE ~ lluiiE, tEl
and as a result IIFxll ~ llxll on
p n 8Br.
Furthermore, for
X E
-
p n 8Bn we have
1\ 1\ () :::S mx(h1(s)) :::S x(h1(t)). tE(th,52] sEl
Since the norm II· liE is monotonic we obtain
f\ 1\ llmx(h1(s))IIE ~ llx(hi(t))liE, tE(5t ,82] sEl
which gives
1\
mmax llx(hl(s))IIE ~ llx(h1(t))IIE · sEl
tE(5t,52]
But h1 maps I onto itself, hence for llxll = llviiE
1\
m llviiE
~ llx(hi(t))IIE ~ llviiE ·
tE(5t,52]
In the same manner we get
f\ tE(5t,52]
m llviiE
~ llx(h2(t))IIE ~ llviiE ·
91
Chapter 2. Applications
92 Thus in view of 6°
{82
v = } 81 G(to,s)
~
( {82
} 81 G(to,r)dr
)-1
vds
1
1 G(to,s)f(s,x(h1(s)),x(h2(s)))ds 1
= (Fx)(to),
so
II(Fx)(to)IIE 2::
llviiE,
which implies
IIFxll 2:: llxll on P n 8Bn. By Theorem 1.3.9, F has a fixed point in the set P n (B n \ Br ). This means that the problem (62) has at least one positive solution x E P such that
0
This ends the proof of Theorem 2.4.1.
Next, consider the problem (62) with k = 0. Using the properties of the function G* given by (64) we can prove the following theorem in the same way as Theorem 2.4.1. Theorem 2.4.2. Let G* be given by {64} and let -1 S 81 < 82 S 1 be such that hi : [81. 82] ----t [81. 82], i = 1, 2. Suppose that 1°- 4° of Theorem 2.4.1 are satisfied and: 7° there exists u E P, u
=I (},
such that
f(t, x, y)
~ [l~ G*(s, s)ds] -
1
u
fort E I and x, y E P such that llxiiE, IIYIIE E [0, !IuilE],
93
2.4. Boundary value problems in Banach spaces 8° there exist v E P, v
=I=(), llviiE =/= iiuiiE, and to
r~o:2 G*(to,s)ds]-
1
E I such that
v::; f(t,x,y)
fort E I and x, y E P such that llxiiE, IIYIIE E [m*llviiE, llviiE], where m* is given by {69}. Then the problem {62} has at least one positive solution. Finally, we will give an example of application of Theorem 2.4.2 to the infinite system of functional-differential equations.
Example 2.4.1. Let E be the space l 00 of all bounded sequences x = {xn} with the supremum norm (73)
llxiiE =sup lxnl· nEN
Then
P =
{x
E
E :
1\ nEN
Xn
2::
0}
is a cone in E and the norm (73) is monotonic with respect to P. Consider the following BVP of an infinite system of functional-differential equations: x~(t)
+ A(t)xn(hl(t)) + B(t)xn(h2(t)) + C(t)
(74) Xn( -1)-
Xn(1)
x~( -1) = 0,
+ x~(1) =
0,
where n = 1, 2, 3, ... , t E I, x = {xn} E Q C C(I, E), the functions A, B, C: I ~ [0, oo) are continuous, w = {Wn} E P and limn-+oo Wn = 0. In our case
M* =max G*(t, s) = 1 t,sEI
94
Chapter 2. Applications
and 6 11 Assume that
max (A(t) tEl
1
+ B(t)) < -4
and minC(t) tEl
> 0.
Moreover, suppose that the functions hi satisfy 2° and hi([-~,~]) c [-~, ~], i = 1, 2. Then for <5 1 = -~, <>2 = ~ we have m* = ~ and for to= -1
1: 1
[
G*( -1, s)ds
l-1 =
2.
Consider the function
f(t, x, y)
=
A(t)x + B(t)y + C(t)
+ wJx + y,
where t E I, f = Un}, x, yEP, x = {xn}, y = {Yn}· Obviously, f is uniformly continuous on I x (P n Tr) x (P n Tr) for any r > 0. We will show that f satisfies 4°. Notice that f admits a splitting
f = 1+ 7, where
J(t, x, y)
= A(t)x + B(t)y + C(t)
and
f(t, x, y) Evidently, the function
(75)
f
= wJx + y.
is lipschitzian, hence
a(J(t, 0, 0)):::; max (A(t) tEl
+ B(t)) a(O)
for all t E I and 0 c p n Tr. To find a(f(t, 0, 0)) we will apply the following compactness criterion in the space zoo.
Lemma 2.4.2. [5] If D C zoo is bounded and limsupn_. 00 [supxED lxnl] = 0, then D is relatively compact in zoo.
2.4. Boundary value problems in Banach spaces
95
Denote
X(t) = f(t, n, n) = {f(t, x, y) : x, yEn, t is fixed } . For n E N and
X'
Since limn_. 00 Wn
y
E
nc
p
n Tr
we have
= 0, we obtain limsup [n-+oo
sup
Jfn(t,x,y)J] = 0,
f(t,x,y)EX(t)
and in consequence X(t) is relatively compact. Therefore
a(f(t, n, n)) = o.
(76)
By (75), (76) and the properties of the Kuratowski measure of noncompactness we have
a(f(t, n, n)) ::; a(J(t, n, n) + f(t, n, n)) ::; max(A(t) tEl
+ B(t)) a(n)
which means that 4° is fulfilled. Finally, we can show by simple calculation that 7° and 8° are also satisfied with u = {un}, v = {vn} E P, such that 2w~
and Vn
1
.
= -2 mm C(t), tEl
13
+ -11 maxC(t) tEl
2 ]
n = 1, 2, ....
By Theorem 2.4.2, the problem (62) has a positive solution x E
llviiE :S llxll :S lluiiE ·
P such that
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