ON A BOUNDARY VALUE PROBLEM FOR NONLINEAR FUNCTIONAL DIFFERENTIAL EQUATIONS ROBERT HAKL Received 21 doc

Efficient conditions sufficient for the ability and unique solvability of the problem considered are established.. Another type of 2.1 is an equation whereH is a linear operator.. Other cond

FUNCTIONAL DIFFERENTIAL EQUATIONS

ROBERT HAKL

Received 21 August 2004 and in revised form 1 March 2005

We consider the problemu
(t) = H(u)(t) + Q(u)(t), u(a) = h(u), where H, Q : C([a, b]; R)

→ L([a, b]; R) are, in general, nonlinear continuous operators, H ∈Ᏼαβ

ab( 0,g1,p0,p1), and

h : C([a, b]; R) → R is a continuous functional Efficient conditions sufficient for the ability and unique solvability of the problem considered are established

solv-1 Notation

The following notation is used throughout the paper:

N is the set of all natural numbers.

R is the set of all real numbers, R+=[0, +∞[, [x]+=(1/2)( | x |+x), [x] − =(1/2)( | x | − x).

C([a, b]; R) is the Banach space of continuous functions u : [a, b] → R with the norm

 u  C =max{|u(t) |:t ∈[a, b] }.

C([a, b]; R) is the set of absolutely continuous functions u : [a, b] → R.

L([a, b]; R) is the Banach space of Lebesgue integrable functions p : [a, b] → R with the

norm p  L =a b | p(s) | ds.

L([a, b]; R+)= { p ∈ L([a, b]; R) : p(t) ≥0 fort ∈[a, b] }.

ᏹabis the set of measurable functionsτ : [a, b] →[a, b].

᏷ab is the set of continuous operators F : C([a, b]; R) → L([a, b]; R) satisfying the

Carath`eodory condition, that is, for eachr > 0 there exists q r ∈ L([a, b]; R+) such that

F(v)(t)  ≤ q r(t) fort ∈[a, b], v ∈ C

[a, b]; R

, v  C ≤ r. (1.1)

K([a, b] × A; B), where A ⊆ R2,B ⊆ R, is the set of functions f : [a, b] × A → B

satisfy-ing the Carath`eodory conditions, that is, f ( ·, x) : [a, b] → B is a measurable function for

allx ∈ A, f (t, ·) : A → B is a continuous function for almost all t ∈[a, b], and for each

r > 0 there exists q r ∈ L([a, b]; R+) such that

f (t, x)  ≤ q r(t) fort ∈[a, b], x ∈ A,  x  ≤ r. (1.2)

Copyright©2006 Hindawi Publishing Corporation

Boundary Value Problems 2005:3 (2005) 263–288

DOI: 10.1155/BVP.2005.263

2 Statement of the problem

We consider the equation

whereH ∈Ᏼαβ

ab( 0,g1,p0,p1) (seeDefinition 2.1) andQ ∈᏷ab By a solution of (2.1) weunderstand a functionu ∈
C([a, b]; R) satisfying the equality (2.1) almost everywhere in[a, b].

Definition 2.1 We will say that an operator H belongs to the setᏴαβ

M =max

v(t) +:t ∈[a, b]

ab( 0,g1,p0,p1) contains all the positively homogeneous operatorsH and,

in particular, those defined by the formula

wherep, q0∈ L([a, b]; R), τ i,ν i ∈ᏹab (i =1, 2),τ1(t) ≤ τ2(t), ν1(t) ≤ ν2(t) for t ∈[a, b],

and f ∈ K([a, b] × R2;R), can be rewritten in form (2.1) withH ∈Ᏼ00

ab([p]+, [p] −, [p]+,[p] −)

Another type of (2.1) is an equation whereH is a linear operator In that case the

results presented coincide with those obtained in [5,6] Other conditions ing the solvability of (2.1), (2.8) with a linear operator H can be found, for example,

guarantee-in [10,11,13,15] Conditions for the solvability and unique solvability of other types ofboundary value problems for (2.1) with a linear operatorH are established, for example,

bound-to fill this gap in a certain way More precisely, inSection 3, we establish unimprovable

efficient conditions sufficient for the solvability and unique solvability of problem (2.1),(2.8) InSection 4, some auxiliary propositions are proved Sections5and6are devoted

to the proof of the main results and the examples demonstrating their optimality, tively

and, for t ∈[a, b], the inequalities

(3.4)–(3.6) can be replaced by the nonstrict one (seeRemark 6.1)

Theorem 3.2 Let there exist c ∈ R+such that, on the set C([a, b]; R), inequality (3.2) is fulfilled and, on the set { v ∈ C([a, b]; R) : | v(a) | ≤ c } , the inequality

then the problem (2.1), (2.8) has at least one solution.

Theorem 3.2is unimprovable in the sense that neither of the strict inequalities in (3.4),(3.8), and (3.9) can be replaced by the nonstrict one (seeRemark 6.4)

Theorem 3.3 Assume that the operators H z , z ∈ { v ∈ C([a, b]; R) : v(a) = h(v) } , defined

by the formula

H z(v)(t)def= H(v + z)(t) − H(z)(t) for t ∈[a, b] (3.10)

belong to the setᏴαβ

ab( 0,g1,p0,p1) Let, moreover, for all v and w from the set C([a, b]; R), the inequality

Theorem 3.4 Let the operators H z , z ∈ { v ∈ C([a, b]; R) : v(a) = h(v) } , defined by (3.10) belong to the setᏴαβ

ab( 0,g1,p0,p1) Assume also that, on the set C([a, b]; R) the inequality (3.11) is fulfilled, and, on the set { v ∈ C([a, b]; R) : | v(a) | ≤ | h(0) |} , the inequality

Remark 3.5 The inclusions H z ∈Ᏼαβ

ab( 0,g1,p0,p1), whereH z are defined by (3.10), arefulfilled, for example, ifH is a strongly bounded linear operator In this case, the opti-

mality of obtained results was proved in [21] (see Remark 4.2 on page 97 and Remark12.2 on page 243 therein) More precisely, Theorems3.3and3.4are unimprovable in thesense that neither of the strict inequalities (3.4)–(3.6), (3.8), and (3.9) can be replaced bythe nonstrict one

The following corollary gives conditions sufficient for the solvability of problem (2.7),(2.8)

Corollary 3.6 Let there exist c ∈ R+such that on the set C([a, b]; R) the inequality (3.2)



p(s)

b a

then problem (2.7), (2.8) has at least one solution.

Corollary 3.7 Let inequality ( 3.11) be fulfilled on the set C([a, b]; R) If, moreover, (3.15) and (3.16) hold, then problem (2.6), (2.8) has a unique solution.

Remark 3.8 Corollaries3.6and3.7are unimprovable in the sense that neither of the strictinequalities (3.15) and (3.16) can be replaced by the nonstrict one Indeed, ifτ1≡ τ2and

ν1≡ ν2, then (2.6) and (2.7) are differential equations with deviating arguments In thatcase, the optimality of obtained results was established in [21] (see Remark 4.2 on page

97 and Proposition 10.1 on page 190 therein)

Corollary 3.9 Let there exist c ∈ R+such that on the set C([a, b]; R) the inequality (3.2)

is fulfilled and

f (t, x, y) sgn x ≤ q

t, | x | for t ∈[a, b], x, y ∈ R. (3.17)

If, moreover, (3.15) and

b a

hold, then the problem (2.7), (2.8) has at least one solution.

The following corollary gives conditions sufficient for the unique solvability of lem (2.7), (2.8)

prob-Corollary 3.10 Let inequality ( 3.11) be fulfilled on the set C([a, b]; R) and, in addition,

τ1≡ τ2andν1≡ ν2, (2.7) is a differential equation with deviating arguments and, in thiscase, the optimality of obtained results is proved in [21] (see Remark 12.2 on page 243therein)

4 Auxiliary propositions

First we formulate a result from [25] in a suitable for us form

Lemma 4.1 Let there exist a number ρ > 0 such that, for every δ ∈]0, 1[, an arbitrary tion u ∈
C([a, b]; R) satisfying

func-u
(t) = δ

H(u)(t) + Q(u)(t)

for t ∈[a, b], u(a) = δh(u), (4.1)

admits the estimate

Then problem (2.1), (2.8) has at least one solution.

Definition 4.2 We will say that an operator H ∈᏷ab belongs to the setᐁ, if there exists

a numberr > 0 such that for any q ∗ ∈ L([a, b]; R+),c ∈ R+, andδ ∈]0, 1], every function

u ∈
C([a, b]; R) satisfying the inequalities | u(a) | ≤ c and

u
(t) − δH(u)(t)  ≤ q ∗(t) fort ∈[a, b] (4.3)admits the estimate

Proof Let r be the number appearing inDefinition 4.2 According to (3.1), there exists

Now assume that a functionu ∈
C([a, b]; R) satisfies (4.1) for someδ ∈]0, 1[ Then,

according to (3.2),u satisfies the inequality | u(a) | ≤ c By (3.3) we obtain that the equality (4.3) is fulfilled withq ∗(t) = q(t,  u  C) fort ∈[a, b] Hence, by the condition

in-H ∈ ᐁ and the definition of the number ρ, we get estimate (4.2)

Sinceρ depends neither on u nor on δ, it follows fromLemma 4.1that problem (2.1),

Leth i ∈ L([a, b]; R+) (i =1, 2, 3, 4),α, β ∈[0, 1[ For an arbitrary fixedt ∈[a, b],

con-sider the systems of inequalities

Proof Assume on the contrary that, for every n ∈ N, there exist q ∗ n ∈ L([a, b]; R+),c n ∈

R+,δ n ∈]0, 1], and u n ∈
C([a, b]; R) such that

According to (4.18) and (4.19), for everyn ∈ N, the points s n,t n ∈[a, b] can be chosen in

the following way:

(i) ifM n =0, then lett n = a and let s n ∈[a, b] be such that

n =1 and { t n }+∞

n =1 are bounded, and, over, for everyn ∈ N we have either

Consequently, the pair (M0,m0) is a solution of system ((4.6)t)s0 withh1≡ g0,h2≡ g1,

h3≡ p0,h4≡ p1,α = α, and β = β However, inequality (4.27) contradicts inclusion (4.8)

If (4.26) is fulfilled, then it can be shown analogously that (M0,m0) is a solution ofinequalities ((4.7)t)t0withh1≡ g0,h2≡ g1,h3≡ p0,h4≡ p1,α = α, and β = β Also in this

Lemma 4.6 Let condition ( 3.4) be satisfied and let for t ∈[a, b] the inequalities (3.5) and (3.6) be fulfilled Then the inclusion (4.8) holds.

Proof Assume on the contrary that there exists t0∈[a, b] such that either the system

((4.6)t)t0or the system ((4.7)t)t0withh1≡ g0,h2≡ g1,h3≡ p0,h4≡ p1,α = α, and β = β

has a nontrivial solution

First suppose that (M0,m0) is a nontrivial solution of ((4.6)t)t0 Put

IfM0=0, thenm0> 0, and from (4.32), in view of (3.4) and (4.31), we get a contradiction

m0< m0 Ifm0=0, thenM0> 0, and from (4.33), in view of (3.4) and (4.31), we get acontradictionM0< M0 Therefore assume that

which, in view of (4.31), contradicts (3.5) witht = t0

Suppose thatβ =0 Since the functionx → x1− β A − x, defined on [0, + ∞[, A ∈ R+,achieves the maximal value at the pointx =(1− β)1/βA1/β, from (4.37) we obtain

The last inequality results in

Furthermore, since the functionsG0andG1are nondecreasing in [a, b] and the function

P1is nonincreasing in [a, b], the equality (4.42), on account of (4.46), results in

However, since the functionx → x1− β A − x, defined on [0, + ∞[ with A > 0, is

nonde-creasing in [0, ((1− β)A)1/β], from (4.36) and (4.37), by virtue of (4.47), we obtain

In analogous way it can be shown that assuming (M0,m0) to be a nontrivial solution

Definition 4.7 We will say that an operator H ∈᏷abbelongs to the setᐂ, if there exists

a numberr > 0 such that for any q ∗ ∈ L([a, b]; R+),c ∈ R+, andδ ∈]0, 1], every function

u ∈
C([a, b]; R) satisfying the inequalities | u(a) | ≤ c and



u
(t) − δH(u)(t)

sgnu(t) ≤ q ∗(t) fort ∈[a, b] (4.49)admits the estimate (4.4)

Lemma 4.8 Let there exist c ∈ R+such that on the set C([a, b]; R) the inequality (3.2) is isfied and on the set { v ∈ C([a, b]; R) : | v(a) | ≤ c } the inequality (3.7) is fulfilled If, more- over, H ∈ ᐂ, then the problem ( 2.1), (2.8) has at least one solution.

sat-Proof Let r be the number appearing inDefinition 4.7 According to (3.1), there exists

Now assume that a functionu ∈
C([a, b]; R) satisfies (4.1) for someδ ∈]0, 1[ Then,

according to (3.2),u satisfies the inequality | u(a) | ≤ c By (3.7) we obtain that the equality (4.49) is fulfilled withq ∗(t) = q(t,  u  C) fort ∈[a, b] Hence, by the condition

in-H ∈ ᐂ and the definition of the number ρ we get the estimate (4.2)

Sinceρ depends neither on u nor on δ, it follows fromLemma 4.1that the problem

Leth i ∈ L([a, b]; R+) (i =1, 2, 3, 4),α, β ∈[0, 1[ For arbitrarily fixedt ∈[a, b] consider

the systems of inequalities

Definition 4.9 Let h i ∈ L([a, b]; R+) (i =1, 2, 3, 4) andα, β ∈[0, 1[ We will say that a tuple (h1,h2,h3,h4) belongs to the setᏮab(α, β), if for every t ∈[a, b] the systems ((4.51)t)t

4-and ((4.52)t)thave only the trivial solution

Proof Assume on the contrary that for every n ∈ N there exist q n ∗ ∈ L([a, b]; R+),c n ∈ R+,

δ n ∈]0, 1], and u n ∈
C([a, b]; R) such that the inequalities (4.9),

q n(t) sgn v n(t) ≤ q ∗ n(t)

u n

C

fort ∈[a, b], n ∈ N. (4.55)

Forn ∈ N define numbers M nandm nby (4.18) Evidently,M n ≥0,m n ≥0 forn ∈ N,

and on account of (4.13), the inequality (4.19) holds

According to (4.18) and (4.19), for everyn ∈ N the points σ n,s n,ξ n,t n ∈[a, b] can be

chosen in the following way:

(i) ifM n =0, then letξ n = a, t n = a, s n ∈[a, b] be such that (4.20) is fulfilled, and let

By virtue of (4.13) and (4.18) we have that the sequences{ M n }+∞

n =1 and { m n }+∞

n =1 arebounded Obviously, also the sequences{ s n }+∞

Therefore, without loss of generality we can assume that there existM0,m0∈ R+ands0,

t0∈[a, b] such that (4.24) is fulfilled, and either

a ≤ σ n ≤ s n ≤ ξ n ≤ t n ≤ b forn ∈ N, (4.62)or

a ≤ ξ n ≤ t n ≤ σ n ≤ s n ≤ b forn ∈ N. (4.63)Furthermore, on account of (4.19) we have (4.27)

The integration of (4.14) fromσ ntos nand fromξ n tot n, respectively, by virtue of(4.58), for everyn ∈ N yields

the inequality (4.27) contradicts the inclusion (4.53)

Lemma 4.11 Let the inequalities ( 3.4), (3.8), and (3.9) be fulfilled Then the inclusion (4.53) holds.

Proof Assume on the contrary that there exists t0∈[a, b] such that either the system

((4.51)t)t0 or the system ((4.52)t)t0 withh1≡ g0,h2≡ g1,h3≡ p0,h4≡ p1,α = α, and

β = β has a nontrivial solution.

First suppose that (M0,m0) is a nontrivial solution of ((4.51)t)t0 Define functionsG0,

G1,P0, andP1by (4.31) Then, according to the assumptions, (M0,m0) satisfies

IfM0=0, thenm0> 0, and from (4.68), in view of (3.4) and (4.31), we get a contradiction

m0< m0 If m0=0, thenM0> 0, and from (4.69), in view of (3.4) and (4.31), we get

a contradictionM0< M0 Therefore assume that (4.34) holds In this case, according to(2.4), we have (4.35) Thus, on account of (3.4), (4.31), and (4.34), from (4.68) and (4.69)

However, on account of (4.31) and (4.34), the last inequality contradicts (3.8)

In analogous way it can be shown that assuming (M0,m0) to be a nontrivial solution

5 Proofs

4.8,4.10, and4.11

Proof of Theorem 3.3 From the conditions (3.11) and (3.12) it follows that the tions (3.2) and (3.3) are satisfied withc = | h(0) |andq ≡ | q ∗ |, and so the assumptions of

condi-Theorem 3.1are fulfilled Therefore, the problem (2.1), (2.8) has at least one solution Itremains to show that the problem (2.1), (2.8) has no more than one solution

Letu, v ∈
C([a, b]; R) be solutions of (2.1), (2.8) Then, in view of (2.8) and (3.11),

we have

u(a)  = h(u) sgn u(a) ≤h(0), v(a)  = h(v) sgn v(a) ≤h(0). (5.1)

Put

Then, in view of (2.8), (3.11), and (5.2), we obtain

w(a)  =  u(a) − v(a)  =  h(u) − h(v)

Proof of Theorem 3.4 From the conditions (3.11) and (3.13) it follows that the conditions(3.2) and (3.7) are satisfied withc = | h(0) |andq ≡ | Q(0) | Consequently, the assump-

tions of Theorem 3.2are fulfilled Therefore, the problem (2.1), (2.8) has at least onesolution It remains to show that the problem (2.1), (2.8) has no more than one solution.Letu, v ∈
C([a, b]; R) be solutions of (2.1), (2.8) Then, in view of (2.8) and (3.11), theinequalities (5.1) are fulfilled Definew by (5.2) Then, on account of (2.8), (3.11), and(5.2), (5.3) holds, and, according to (3.10) and (5.1)–(5.3),w is a solution of the problem

w
(t) = H v(w)(t) + Q v(w)(t), w(a) =0, (5.5)where

Q v(w)(t) = Q(w + v)(t) − Q(v)(t) fort ∈[a, b]. (5.6)Furthermore, by virtue of (3.13), (5.1), (5.2), and (5.6),

Q v(w)(t) sgn w(t) =Q(u)(t) − Q(v)(t)

sgn

u(t) − v(t)

≤0 fort ∈[a, b], (5.7)and, with respect to the inequalities (3.4), (3.8), (3.9), andLemma 4.11, we have the inclu-sion (4.53) Therefore, according to the assumptionH v ∈Ᏼαβ

ab( 0,g1,p0,p1),Lemma 4.10,

Proof of Corollary 3.6 To prove the corollary it is sufficient to show that the assumptions

ofTheorem 3.1are fulfilled

Define operatorsH and Q by the equalities

g0≡[p]+, p0≡[p]+, g1≡[p] −, p1≡[p] − (5.10)ThenH ∈Ᏼαβ

ab( 0,g1,p0,p1), the condition (3.14) yields (3.3), and (3.15) implies (3.4) Itremains to verify that fort ∈[a, b] the inequalities (3.5) and (3.6) hold

According to (5.10) and since α =0 andβ =0, the inequalities (3.5) and (3.6) areequivalent Assume on the contrary that there existst0∈[a, b] such that



. (5.11)Now, since

Thus, in view of (5.13) and (5.14), (5.11) yieldsb

4

b a

Proof of Corollary 3.7 To prove the corollary it is sufficient to show that the assumptions

ofTheorem 3.3are fulfilled

Define operatorH by (5.8) and functionsg0,g1,p0,p1by (5.10) Putα =0,β =0, and

Q(v)(t)def= q0(t) fort ∈[a, b], v ∈ C

[a, b]; R

In analogous way it can be shown that assuming (M0,m0)... g1,h3≡ p0,h4≡ p1,α = α, and

β = β has a nontrivial solution.