NONLINEAR MULTIVALUED BOUNDARY VALUE PROBLEMS

Differential Inclusions, Control and Optimization 21 (2001 ) 127–148

NONLINEAR MULTIVALUED BOUNDARY VALUE PROBLEMS

Ralf Bader Center of Mathematics Technical University Muenchen Arcisstr. 21, D-80333 Muenchen, Germany

e-mail: bader@appl-math.tu-muenchen.de and

Nikolaos S. Papageorgiou

National Technical University, Department of Mathematics Zografou Campus, Athens 15780, Greece

e-mail: npapg@math.ntua.gr

Abstract

In this paper, we study nonlinear second order differential inclusions with a multivalued maximal monotone term and nonlinear boundary conditions. We prove existence theorems for both the convex and nonconvex problems, when domA 6= R

and domA = R

, with A being the maximal monotone term. Our formulation incorporates as special cases the Dirichlet, Neumann and periodic problems. Our tools come from multivalued analysis and the theory of nonlinear monotone operators.

Keywords and phrases: usc and lsc multifunction, measurable selection, Leray-Schauder alternative theorem, Sobolev space, compact embedding, maximal monotone map, coercive map, surjective map, convex and nonconvex problem, nonlinear boundary conditions.

2000 Mathematics Subject Classification: 34B15.

∗

This work was partially done while the first author visited the NTUA. Support was

provided by a scholarship offered by the Deutsche Forschungsgemeinschaft (DFG), Bonn

(Germany).

1. Introduction

In this paper, we deal with the following multivalued boundary value problem:

( ¡

kx

(t)k

x

(t) ¢

∈ A(x(t)) + F (t, x(t), x

(t)) a.e. on T = [0, b]

(ϕ(x

(0)), −ϕ(x

(b))) ∈ ξ(x(0), x(b)), 2 ≤ p < ∞.

) (1)

Here A : R

→ 2

is a maximal monotone map, F : T ×R

×R

→ 2

is a multifunction and ϕ : R

→ R

is defined by ϕ(r) = krk

r. In a recent paper (see Bader-Papageorgiou [2]), the authors studied (1) with F single- valued and proved two existence theorems. This work extends the results investigated in Bader-Papageorgiou [2] to the multivalued boundary prob- lem. At the same time it extends the set-valued results of Erbe-Krawcewicz [7], Frigon [9], Halidias-Papageorgiou [11] and Kandilakis-Papageorgiou [14], where the inclusion is semilinear (i.e. p = 2) and A = 0. Moreover, our boundary conditions are general nonlinear boundary conditions, which get a unified treatment of the classical boundary value problems, such as the Dirichlet problem, the Neumann problem and the periodic problem. Fi- nally, we should also mention the recent works on problems involving the one dimensional p-Laplacian, by Boccardo-Drabek-Giachetti-Kucera [3], Dang- Oppenheimer [4], Del Pino-Elgueta-Manasevich [5], Del Pino-Manasevich- Murua [6], Fabry-Fayyad [8], Guo [10], and Manasevich-Mawhin [15]. We point out that in all these works F is single-valued, A = 0, the boundary conditions are among the classical ones (Dirichlet, Neumann and periodic) and with the exception of Manasevich-Mawhin, they all deal with the scalar problem (i.e. N = 1).

Our approach is based on notions and results from multivalued analysis and from the theory of nonlinear operators of monotone type. They lead to an eventual application of a generalized version of the Leray-Schauder alternative principle, proved recently by Bader [1]. In Section 2, we recall the basic definitions and facts from multivalued analysis and the theory of monotone operators, which we will need in the sequel. Our main sources are the books of Hu-Papageorgiou [13] and Zeidler [17].

2. Mathematical preliminaries

Let (Ω, Σ) be a measurable space and X a seperable Banach space. We

introduce the following notations: P

(X) = {A ⊆ X : A is nonempty,

closed (and convex)} and P

(X) = {A ⊆ X : A is nonempty, (weakly-) compact (and convex)}. A multifunction F : Ω → P

(X) is said to be measurable, if for all x ∈ X, ω 7→ d(x, F (ω)) is measurable. Also we say that F : Ω → 2

\ {∅} is graph measurable, if GrF = {(ω, x) ∈ Ω × X : x ∈ F (ω)} ∈ Σ × B(X), with B(X) being the Borel σ-field of X. For P

(X)- valued multifunctions, measurability implies graph measurability, while the converse is true if Σ is complete (i.e. Σ = ˆ Σ = the universal σ-field). Recall that if µ is a measure on Σ and Σ is µ-complete, then Σ = ˆ Σ. Now let (Ω, Σ, µ) be a finite measure space. Given a multifunction F : Ω → 2

\ {∅}

and 1 ≤ p ≤ ∞, we introduce the set S

= {f ∈ L

(Ω, X) : f (ω) ∈ F (ω) µ-a.e.}. In general, this set may be empty. It is easy to check that if ω 7→ inf{kxk : x ∈ F (ω)} ∈ L

(Ω), then S

6= ∅.

Let Y, Z be Hausdorff topological spaces. A multifunction G : Y → 2

\ {∅} is said to be lower semicontinuous (lsc for short) (resp. upper semicontinuous (usc for short)), if for all C ⊆ Z closed, then the set G

(C) = {y ∈ Y : G(y) ⊆ C}(resp. G

(C) = {y ∈ Y : G(y) ∩ C 6= ∅}) is closed in Y . An usc multifunction G has a closed graph (i.e. GrG = {(y, z) ∈ Y × Z : z ∈ G(y)} is closed), while the converse is true if G is locally compact. Also if Z is a metric space, then G is lsc if and only if for every y

→ y in Y , we have G(y) ⊆ limG(y

) = {z ∈ Z : lim d(z, G(y

)) = 0} = {z ∈ Z : z = lim z

, z

∈ G(y

), n ≥ 1}.

Let X be a reflexive Banach space and X

its dual. A map A : D ⊆ X → 2

is said to be monotone, if for all (x, x

), (y, y∗) ∈ GrA, we have (x

− y

, x − y) ≥ 0 (by (·, ·) we denote the duality brackets for the pair (X, X

)). When (x

− y

, x − y) = 0 implies that x = y, then we say that A is strictly monotone. The map A is said to be maximal monotone, if (x

− y

, x − y) ≥ 0 for all (x, x

) ∈ GrA, imply that (y, y∗) ∈ GrA. So according to this definition the graph of A is maximal with respect to the inclusion among the graphs of all monotone maps from X into 2

. It is easy to see that a maximal monotone map A has a demiclosed graph, i.e. GrA is sequentially closed in X × X

or in X

× X

(here by X

and X

we denote the spaces X and X

furnished with their respective weak topologies). If A : X → X

is every where defined, single-valued map, we say that A is demicontinuous, if x

→ x in X implies that A(x

) → A(x) in X

. A map A : X → X

which is monotone and demicontinuous, is maximal monotone.

Also a map A : D ⊆ X → 2

is said to be coercive, if D ⊆ X is bounded

or if D is unbounded and

→ ∞ as kxk → ∞, x ∈ D. A

maximal monotone, coercive map is surjective.

Let Y, Z be Banach spaces and K : Y → Z a generally nonlinear map.

We say

(a) K is completely continuous, if y

→ y in Y implies K(y

) → K(y) in Z;

(b) K is compact, if K is continuous and maps bounded sets into relatively compact sets.

In general, these are two distinct notions. However, if Y is reflexive, then complete continuity implies compactness. Moreover, if Y is reflexive and K is linear, then the two notions are equivalent.

Finally, we will need the following generalization of the Leray- Schauder principle. Let X, Y be Banach spaces, G : X → P

(Y ) an usc multifunc- tion from X into Y

and K : Y → X a completely continuous map. We set Φ = K ◦ G. We have the following alternative principle (see Bader [1]):

Proposition 1. If X, Y and Φ are as above and Φ is compact, then the set S = {x ∈ X : x ∈ λΦ(x) for some 0 < λ < 1}

is unbounded or otherwise Φ has a fixed point.

3. An auxiliary problem

In this section, we consider the following“regular” approximation to problem (1):

( ¡

kx

(t)k

x

(t) ¢

∈ A

(x(t)) + F (t, x(t), x

(t)) a.e. on T (ϕ(x

(0)), −ϕ(x

(b))) ∈ ξ(x(0), x(b)), λ > 0.

) (2)

Here for every λ > 0, A

: R

→ R

is the Yosida approximation of the maximal monotone map A. First we will establish the existence of solutions for problem (2), when F takes convex values (“convex problem”). For this purpose we introduce the following hypothesis on the data of (2).

H(A)

: A : R

→ 2

is a maximal monotone map such that 0 ∈ A(0).

Remark. In fact, it is enough to assume that 0 ∈ domA = {x ∈ R

: A(x) 6= ∅} and then by translation we can have 0 ∈ A(0).

H(F )

: F : T × R

× R

→ P

(R

) is a multifunction such that

(i) for all (x, y) ∈ R

× R

, t → F (t, x, y) is graph measurable;

(ii) for almost all t ∈ T, (x, y) → F (t, x, y) has a closed graph;

(iii) for almost all t ∈ T , all x, y ∈ R

and all v ∈ F (t, x, y), we have (v, x)

≥ −akxk

− γkxk

kyk

− c(t)kxk

with a, γ ≥ 0, 1 ≤ r, s < p and c ∈ L

(T );

(iv) there exists M > 0 such that if kx

k > M and (x

, y

)

= 0, then we can find δ > 0 and ξ > 0 such that for almost all t ∈ T

inf [(v, x)

+ kyk

: kx − x

k + ky − y

k < δ, v ∈ F (t, x, y)] ≥ ξ > 0;

(v) for almost all t ∈ T , all x, y ∈ R

and all v ∈ F (t, x, y) kvk ≤ γ

(t, kxk) + γ

(t, kxk)kyk

with sup

γ

(t, r) ≤ η

(t) a.e. on T , η

∈ L

(T ) (

+

= 1) and sup

γ

(t, r) ≤ η

(t) a.e. on T , η

∈ L

(T ).

Remark. Hypothesis H(F )

(iv) is an appropriate extension of the Nagumo-Hartman condition (see Hartman [12], p. 432–433).

H(ξ): ξ : R

× R

→ 2

is a maximal monotone map such that (0, 0) ∈ ξ(0, 0) and one of the following holds:

(i) for every (a

, d

) ∈ ξ(a, d) we have (a

, a)

≥ 0 and (d

, d)

≥ 0;

or

(ii) domξ = {(a, d) ∈ R

× R

: a = d}.

Proposition 2. If hypotheses H(A)

, H(F )

and H(ξ) hold, then problem (2) has a solution x ∈ C

(T, R

).

P roof. Let

D = {x ∈ C

(T, R

) : kx

(·)k

x

(·) ∈ W

(T, R

), (ϕ(x

(0)), −ϕ(x

(b))) ∈ ξ(x(0), x(b))}

and let V : D ⊆ L

(T, R

) → L

(T, R

) be defined by V (x)(·) = − ¡

kx

(·)k

x

(·) ¢

, x ∈ D.

From Proposition 3 of Bader-Papageorgiou [2], we know that V is maximal monotone. Also let ˆ A

: L

(T, R

) → L

(T, R

) be the Nemitsky operator corresponding to the Yosida approximation A

, i.e. A ˆ

(x)(·)

= A

(x(·)) and let J : L

(T, R

) → L

(T, R

) be defined by J(x)(·)

= kx(·)k

x(·). Consider the map K

= V + ˆ A

+ J. Note that ˆ A

and J are both monotone, continuous, thus maximal monotone. So from Theorem III. 3.3, p. 334, of Hu-Papageorgiou [13], we have that K

is maximal monotone. Because 0 = A

(0), we have

(K

(x), x)

≥ (V (x), x)

+ (J(x), x)

,

where by (·, ·)

we denote the duality brackets for the pair (L

(T, R

), L

(T, R

)). Using Green’s identity, the fact that if x ∈ D, then (ϕ(x

(0)),

−ϕ(x

(b))) ∈ ξ(x(0), x(b)) and hypothesis H(ξ), we obtain (V (x), x)

= −

Z

0

¡ (kx

(t)k

x

(t))

, x(t) ¢

R^N

dt

= ¡

−kx

(b)k

x

(b), x(b) ¢

R^N

+ ¡

kx

(0)k

x

(0), x(0) ¢

R^N

+ kx

k

≥ kx

k

. Also (J(x), x)

= kxk

. So we obtain

(K

(x), x)

≥ kx

k

+ kxk

= kxk

,

where k · k

denotes the norm in the Sobolev space W

(T, R

). From this last inequality we infer that K

is coercive. But recall (see Section 2) that a maximal monotone and coercive operator is surjective. So R(K

) = L

(T, R

). Also it is clear that J is strictly monotone and so it follows that K

is injective. Hence we can define the operator K

: L

(T, R

) → D ⊆ W

(T, R

).

Claim 1. K

: L

(T, R

) → D ⊆ W

(T, R

) is completely continuous.

Note that by virtue of the reflexivity of L

(T, R

), complete continuity implies compactness. Assume that y

→ y in L

(T, R

) and set x

= K

(y

), n ≥ 1, and x = K

(y). We have

y

= V (x

) + ˆ A

(x

) + J(x

).

⇒ (y

, x

)

= (V (x

), x

)

+ ( ˆ A

(x

), x

)

+ (J(x

), x

)

⇒ kx

k

≤ ky

k

kx

k

.

From the last inequality it follows that {x

}

⊆ W

(T, R

) is bounded.

Hence by passing to a subsequence if necessary, we may assume that x

→ z

in W

(T, R

) and from the compact embedding of W

(T, R

) into L

(T, R

), we also have x

→ z in L

(T, R

). Then

(y

, x

− z)

, ( ˆ A

(x

), x

− z)

and (J(x

), x

− z)

→ 0 as n → ∞,

⇒ lim(V (x

), x

− z)

= 0.

But as we have already mentioned V is maximal monotone and so according to Remark III. 6.3, p. 365, of Hu-Papageorgiou [13], we have that V is also generalized pseudomonotone, which implies that V (x

) → V (z) in

L

(T, R

). So in the limit as n → ∞ we obtain

y = V (z) + ˆ A

(z) + J(z)

⇒ z = K

(y), i.e. z = x.

Moreover, note that {u

= kx

k

x

}

⊆ W

(T, R

) is bounded and so we may assume that u

→ u in W

(T, R

). From the compact embed- ding of W

(T, R

) into C(T, R

), we have that u

→ u in C(T, R

). Since ϕ : R

→ R

is a homeomorphism, we have that ϕ

(u

) = x

→ ϕ

(u) in L

(T, R

). Therefore ϕ

(u) = x

and so x

→ x in W

(T, R

) which proves the claim. Next let N

: W

(T, R

) → 2

be the multivalued Nemitsky operator corresponding to F , i.e. N

(x) = S

.

Claim 2. N

has values in P

(L

(T, R

)) and is usc from W

(T, R

) into L

(T, R

)

. Note that hypotheses H(F )

(i) and (ii), do not imply joint measurability of F and so it is not immediately clear that N

has nonempty values. Let x ∈ W

(T, R

). We can find {s

}

, {r

}

step functions such that ks

(t)k ≤ kx(t)k, kr

(t)k ≤ kx

(t)k, s

(t) → x(t) and r

(t) → x

(t) a.e. on T . By virtue of hypothesis H(F )

(i), for every n ≥ 1, t 7→ F (t, s

(t), r

(t)) is measurable. So we can apply the Yankov-von Neumann-Aumann selection theorem (see Hu-Papageorgiou [13], Theorem II. 2.14, p. 158) and obtain f

: T → R

a measurable map such that f

(t) ∈ F (t, s

(t), r

(t)) a.e. on T . Evidently {f

}

⊂ L

(T, R

) is bounded (see hypothesis H(F )

(v)) and so we may assume that f

→ f in

L

(T, R

). Invoking Proposition VII. 3.9, p. 694, of Hu-Papageorgiou [13], we obtain

f (t) ∈ conv lim F (t, s

(t), r

(t)) ⊂ F (t, x(t), x

(t)) a.e. on T,

the last inclusion following from the fact that for almost all t ∈ T, GrF (t, ·, ·) is closed (see hypothesis H(F )

(i)). So f ∈ N

(x) and we have established that N

has nonempty values. It is clear that the values of N

are bounded, closed, convex, hence they belong to P

(L

(T, R

)). Next we will show the upper semicontinuity of N

from W

(T, R

) into L

(T, R

)

.

From hypothesis H(F )

(v), we see that N

is locally compact into L

(T, R

)

and so by virtue of Proposition I. 2.23, p. 43, of Hu- Papageorgiou [13] and the fact that on bounded sets in L

(T, R

)

, the weak topology is metrizable, it suffices to show that GrN

is sequentially closed in W

(T, R

) × L

(T, R

)

. So let (x

, f

) ∈ GrN

, n ≥ 1, and assume that x

→ x in W

(T, R

) and f

→ f in L

(T, R

). We may assume that x

(t) → x(t) and x

(t) → x

(t) a.e. on T and so as above via Proposition VII. 3.9, p. 694, of Hu-Papageorgiou [13], we show that (x, f ) ∈ GrN

. This completes the proof of the claim. Let N

= −N

+ J.

Evidently N

: W

(T, R

) → P

(L

(T, R

)) is usc from W

(T, R

) into L

(T, R

)

. We consider the multivalued operator

K

◦ N

: W

(T, R

) → P

(W

(T, R

)).

This map is usc and maps bounded sets into relatively compact ones (i.e.

K

◦ N

is compact). So in order to be able to apply Proposition 1 and obtain a fixed point of K

◦ N

, we need to prove the following claim:

Claim 3. The set

S = {x ∈ W

(T, R

) : x ∈ βK

◦ N

(x), 0 < β < 1}

is bounded.

Let x ∈ S. We have K

µ 1 β x

¶

∈ N

(x),

⇒ V µ 1

β x

¶ + ˆ A

µ 1 β x

¶ + J

µ 1 β x

¶

∈ −N

(x) + J(x),

⇒ µ

V ( 1 β x), x

¶

pq

+ µ

A ˆ

( 1 β x), x

¶

pq

+ µ

J( 1 β x), x

¶

pq

= (−f, x)

+ (J(x), x)

, f ∈ N

(x).

Since ˆ A

is monotone and 0 = ˆ A

(0), it follows that

³ A ˆ

(

x), x

´

pq

≥ 0.

Also µ

V ( 1 β x), x

¶

pq

= − Z

0

1 β

¡ (kx

(t)k

x

(t))

, x(t) ¢

R^N

dt

= − µ 1

β

kx

(b)k

x

(b), x(b)

¶

R^N

+ µ 1

β

kx

(0)k

x

(0), x(0)

¶

R^N

+ 1

β

kx

k

≥ 1

β

kx

k

(from the boundary conditions and hypothesis H(ξ)).

So we obtain

1

β

kx

k

+ 1

β

kxk

≤ −(f, x)

+ kxk

⇒ kx

k

≤ −β

(f, x)

+ (β

− 1)kxk

≤ −β

(f, x)

(since 0 < β < 1).

(3)

From hypothesis H(F )

(iii), we have

−β

(f, x)

= β

Z

0

− (f (t), x(t))

dt

≤ β

akxk

+ β

γ Z

0

kx(t)k

kx

(t)k

dt + β

kck

kxk

. Let τ = p − r, µ =

abd µ

=

(

+

= 1). Applying H¨older’s inequality with this pair of conjugate exponents, we obtain

Z

0

kx(t)k

kx

(t)k

dt ≤ µZ

0

kx(t)k

dt

¶

µZ

0

(kx

(t)k

dt

¶

≤ kxk

kx

k

. It follows that

−β

(f, x)

≤ β

akxk

+ β

γkxk

kx

k

+ β

kck

kxk

. (4)

We will show that for every x ∈ S, kxk

≤ M with M as in hypothesis

H(F )

(iv). For this purpose we introduce the function r(t) = kx(t)k

and let t

∈ T be the point where r attains its maximum on T . Suppose that M

< r(t

) and first suppose that t

∈ (0, b). We have 0 = r

(t

) = pkx(t

)k

(x(t

), x

(t

))

and so (x(t

), x

(t

))

= 0 (unless x ≡ 0 in which case we trivially have kxk

≤ M ). Then from hypothesis H(F )

(iv), we know that there exists δ > 0 and ξ > 0 such that

inf £

(v, x)

+ kyk

: kx − x(t

)k + ky − x

(t

)k < δ, v ∈ F (t, x, y) ¤

≥ ξ > 0.

Because x ∈ S, we have x ∈ D and so kx

(·)k

x

(·) ∈ W

(T, R

) ⊆ C(T, R

). Because ϕ is a homeomorphism, it follows that ϕ

(kx

(·)k

x

(·)) = x

(·) ∈ C(T, R

). Also x ∈ W

(T, R

) ⊆ C(T, R

). Thus for δ > 0 as above, we can find δ

> 0 such that if t ∈ (t

, t

+ δ

], we have

kx(t) − x(t

)k + kx

(t) − x

(t

)k < δ.

Therefore for almost all t ∈ (t

, t

+ δ

] and all v ∈ F (t, x(t), x

(t)), we have β

(v, x(t))

+ β

kx

(t)k

≥ β

ξ

⇒ β

(f (t), x(t))

+ β

kx

(t)k

≥ β

ξ.

(5)

Recall that V

µ 1 β x

¶ + ˆ A

µ 1 β x

¶ + J

µ 1 β x

¶

= −f + J(x)

⇒ − ¡

kx

(t)k

x

(t) ¢

+ β

A

µ 1

β x(t)

¶

= −β

f (t) + (β

− 1)kx(t)k

x(t) a.e. on T ,

⇒ β

f (t) = ¡

kx

(t)k

x

(t) ¢

− β

A

µ 1

β x(t)

¶

+ (β

− 1)kx(t)k

x(t) a.e. on T . Using this in (5), we obtain

¡ (kx

(t)k

x

(t))

, x(t) ¢

R^N

− β

µ

A

( 1

β x(t)), x(t)

¶

R^N

+ (β

− 1)kx(t)k

+ β

kx

(t)k

≥ β

ξ a.e. on (t

, t

+ δ

].

Note that (β

− 1)kx(t)k

≤ 0 (since 0 < β < 1) and (A

(

x(t)), x(t))

≥ 0 (since A

is monotone and A

(0) = 0). Thus after integration over [t

, t], t ∈ (t

, t

+ δ

], we obtain

Z

t0

¡ (kx

(s)k

x

(s))

, x(s) ¢

R^N

ds + β

Z

t0

kx

(s)k

ds ≥ β

ξ(t − t

).

(6)

From Green’s identity, we have Z

t0

¡ (kx

(s)k

x

(s))

, x(s) ¢

R^N

ds = kx

(t)k

(x

(t), x(t))

− kx

(t

)k

(x

(t

), x(t

))

− Z

t0

kx

(s)k

ds

= kx

(t)k

(x

(t), x(t))

− Z

t0

kx

(s)k

ds (since (x(t

), x

(t

))

= 0).

Using this equality in (6), we have

kx

(t)k

(x

(t), x(t))

+ (β

− 1) Z

t0

kx

(s)k

ds ≥ β

ξ(t − t

)

⇒ kx

(t)k

(x

(t), x(t))

≥ β

ξ(t − t

) (since 0 < β < 1), t ∈ (t

, t

+ δ

]

⇒ (x(t), x

(t))

> 0

⇒ r

(t) > 0 for all t ∈ (t

, t

+ δ

], i.e. r is strictly increasing on (t

, t

+ δ

].

This contradicts the choice of t

∈ T . So kx(t

)k ≤ M and this proves the desired bound when t ∈ (0, b). Suppose t

= 0. Then r

(0) ≤ 0 and so (x(0), x

(0))

≤ 0. If condition H(ξ) (i) is satisfied we have (x(0), x

(0))

≥ 0 and so (x(0), x

(0))

= 0, i.e. r

(0) = 0.

Thus we can proceed as before. If hypothesis H(ξ) (ii) holds, we have

x(0) = x(b) and r

(0) ≤ 0 ≤ r

(b). So (x

(0), x(0))

≤ 0 ≤ (x

(b), x(b))

,

while from the fact that (ϕ(x

(0)), −ϕ(x

(b))) ∈ ξ(x(0), x(b)), we have that

(x

(b), x(b))

≤ (x

(0), x(0))

(recall that (0, 0) ∈ ξ(0, 0)). Thus finally,

we have 0 = (x

(0), x(0))

= (x

(b), x(b))

and so r

(0) = r

(b) = 0 and

we can repeat the previous argument. Similarly, we can analyze the case

t

= b. Hence we have proved that for all x ∈ S, kxk

≤ M .

Using (4) in (3) we obtain

kx

k

≤ β

akxk

+ β

γkxk

kx

k

+ β

kck

kxk

≤ c

+ c

kx

k

(τ < p) for some c

, c

> 0.

So {x

}

is bounded in L

(T, R

). Since kxk

≤ M for all x ∈ S, we infer that S ⊆ W

(T, R

) is bounded. Applying Proposition 1, we obtain x ∈ D such that x ∈ K

◦N

(x) ⇒ K

(x) ∈ N

(x) ⇒ x solves the auxiliary problem (2).

In the above result F was convex-valued. We can still have an existence theorem for (2), even if F is not necessarily convex-valued. More precisely, we assume on F as follows:

H(F )

: F : T × R

× R

→ P

(R

) is a multifunction such that (i) (t, x, y) ∈ R

× R

, t → F (t, x, y) is graph measurable;

(ii) for almost all t ∈ T, (x, y) → F (t, x, y) is lsc;

and conditions (iii), (iv) and (v) of hypothesis H(F )

hold.

Proposition 3. If hypothesis H(A)

, H(F )

and H(ξ) hold then there exists a solution x ∈ C

(T, R

) to problem (2).

P roof. As before let N

: W

(T, R

) → 2

be the multivalued Nemitsky operator corresponding to F , i.e. N

(x) = S

. By virtue of hypothesis H(F )

and the Yankov-von Neumann-Aumann selection theorem (see Hu-Papageorgiou [13], Theorem II. 2.14, p. 158), we have that N

has values in P

(L

(T, R

)).

Claim. N

is lsc from W

(T, R

) to L

(T, R

). Let C ⊆ L

(T, R

) closed. We will show that N

(C) = {x ∈ W

(T, R

) : N

(x) ⊆ C}

is closed. To this end, let x

∈ N

(C) and assume that x

→ x in W

(T, R

). By passing to a subsequence if necessary, we may assume that x

(t) → x(t) for all t ∈ T and x

(t) → x

(t) a.e. on T . Let f ∈ N

(x) and let f

∈ N

(x

) such that kf − f

k

≤ d(f, N

(x

)) +

.

From Hu-Papageorgiou [13], p. 237, we have

d(f, N

(x

)) = µZ

0

d(f (t), F (t, x

(t), x

(t)))

dt

¶

Since F (t, ·, ·) is lsc, from Proposition I. 2.26, p. 45, of Hu-Papageorgiou [13], we have that for all v ∈ R

, (x, y) 7→ d(v, F (t, x, y)) is upper semicontinuous.

Then by Fatou’s lemma, we have

lim d(f, N

(x

)) ≤ µZ

0

d(f (t), F (t, x(t), x

(t)))

dt

¶

= 0

and therefore kf − f

k

→ 0 as n → ∞. Because f

∈ C and C is closed, we have f ∈ C. Since f ∈ N

(x) was arbitrary, we infer that x ∈ N

(C) and so N

is lsc as claimed. Apply Theorem II. 8.7, p. 245, of Hu-Papageorgiou [13], to obtain u : W

(T, R

) → L

(T, R

) continuous such that u(x) ∈ N

(x) for every x ∈ W

(T, R

). Consider the following single-valued boundary value problem

( ¡

kx

(t)k

x

(t) ¢

∈ A

(x(t)) + u(x)(t) a.e. on T (ϕ(x

(0)), −ϕ(x

(b))) ∈ ξ(x(0), x(b)), λ > 0.

) (7)

As in the proof of Proposition 2, using the Leray-Schauder alternative theorem (the single-valued version of Proposition 1), we obtain a solution x ∈ C

(T, R

) to problem (7). Evidently this solves problem (2).

4. Existence Theorems

In this case using Propositions 2 and 3, we obtain existence theorems for the convex and nonconvex problems, for two different situations. In the first one we assume that domA = R

, while in the other we are able to assume that domA 6= R

at the expense of slightly strengthening the growth hypothesis on F (t, x, y). This case is of special interest since it incorporates variational inequalities. We start with the analysis of the problem in which A is defined everywhere, i.e. domA = {x ∈ R

: A(x) 6= ∅} = R

. More precisely, we assume:

H(A)

: A : R

→ 2

is a maximal monotone map with domA = R

and 0 ∈ A(0).

Theorem 4. If hypotheses H(A)

, H(F )

and H(ξ) hold, then problem (1)

has a solution x ∈ C

(T, R

).

P roof. Let λ

→ 0, λ

> 0 and let x

∈ C

(T, R

) be solutions to the approximate problem (2) (Proposition 2). From the proof of Proposition 2, we know that kx

k

≤ M for all n ≥ 1. We have

V (x

) + ˆ A

(x

) = −f

, f

∈ N

(x

)

⇒ (V (x

), x

)

+ ( ˆ A

(x

), x

)

= −(f

, x

)

.

Using Green’s identity and the boundary conditions on the first term of the left hand side and exploiting the monotonicity of ˆ A

and the fact that 0 = ˆ A

(0), we obtain

kx

k

≤ kf

k

kx

k

≤ c

kf

k

for some c

> 0 and all n ≥ 1

⇒ kx

k

≤ c

¡

kη

k

+ kη

k

kx

k

¢

, (hypothesis H(F )

(v))

⇒ {x

}

⊆ L

(T, R

) is bounded,

⇒ {x

}

⊆ W

(T, R

) is bounded.

So we may assume that x

→ x in W

(T, R

) and x

→ x in L

(T, R

).

From Proposition III. 2.29, p. 325, of Hu-Papageorgiou [13], we know that kA

(x

(t))k ≤ kA

(x

(t))k. Since domA = R

, from Theorem III. 1.21, p. 306 of Hu-Papageorgiou [13], we know that A

is bounded on compact sets. Note that since {x

}

⊆ W

(T, R

) is bounded, from the com- pact embedding of W

(T, R

), we have that {x

}

is relatively com- pact in C(T, R

) and so sup

kA

(x

(t))k ≤ c

for some c

> 0. Hence kA

(x

(t))k ≤ c

for all t ∈ T and all n ≥ 1. Thus we may assume that ˆ A

(x

) → u in L

(T, R

) as n → ∞. Repeating the argument in Claim 1 in the proof of Proposition 2, we obtain x

→ x in W

(T, R

) and kx

k

x

→ kx

k

x

in W

(T, R

). Since f

∈ N

(x

), n ≥ 1, from Hypothesis H(F )

(v), we see that {f

}

⊆ L

(T, R

) is bounded and so we may assume that f

→ f in L

(T, R

). As in the proof of Propo- sition 2, using Proposition VII. 3.9, p. 694, of Hu-Papageorgiou [13], we obtain f ∈ N

(x). Taking n → ∞, we obtain

(kx

(t)k

x

(t))

= u(t) + f (t) a.e. on T .

Also since kx

k

x

→ kx

k

x

in W

(T, R

), we have kx

k

x

→

kx

k

x

in C(T, R

) and so ϕ

(kx

(t)k

x

(t)) = x

(t) → x

(t) =

ϕ

(kx

(t)k

x

(t)) for all t ∈ T . Also, at least for a subsequence, we have x

→ x in C(T, R

). Since (ϕ(x

(0)), −ϕ(x

(b)) ∈ ξ(x

(0), x

(b)) and because Grξ is closed we conclude that (ϕ(x

(0)), −ϕ(x

(b)) ∈ ξ(x(0), x(b)).

So it remains to show that u(t) ∈ A(x(t)) a.e. on T . To this end, let A : ˆ ˆ D ⊆ L

(T, R

) → 2

be defined by

A(x) = {g ∈ L ˆ

(T, R

) : g(t) ∈ A(x(t)) a.e. on T } for all

x ∈ ˆ D = {x ∈ L

(T, R

) : there is g ∈ L

(T, R

) satisfying g(t) ∈ A(x(t)) a.e. on T }.

We show that ˆ A is maximal monotone. Let J : L

(T, R

) → L

(T, R

) be defined by J(x)(·) = kx(·)k

x(·). We will show that R( ˆ A + J) = L

(T, R

). To this end, let h ∈ L

(T, R

) be given and let

Γ(t) = {(x, a) ∈ R

× R

: a + ϕ(x) = h(t), a ∈ A(x), kxk ≤ r(t)}

with r(t) = kh(t)k

+ 1. Note that A + ϕ is maximal monotone and so by Theorem III. 6.28, p. 371, of Hu-Papapgeorgiou [13], we have that Γ(t) 6= ∅ a.e. on T . Moreover,

GrΓ

= {(t, x, a) ∈ T × R

× R

: a + ϕ(x) = h(t), d(a, A(x)) = 0, kxk ≤ r(t)}

∈ L × B(R

) × B(R

), with L being the Lebesgue σ-field of T .

Invoking the Yankov-von Neumann-Aumann selection theorem, we obtain measurable maps x, a : T → R

such that (x(t), a(t)) ∈ Γ(t) a.e. on T . Evidently x ∈ L

(T, R

) and a ∈ L

(T, R

). So R( ˆ A + J) = L

(T, R

).

Now we will show that this surjectivity implies the maximality of the mono- tone operator ˆ A. Indeed suppose y ∈ L

(T, R

), v ∈ L

(T, R

) and assume that

(a − v, x − y)

≥ 0 for all x ∈ ˆ D, a ∈ ˆ A(x).

Since R( ˆ A + J) = L

(T, R

), we can find x

∈ ˆ D such that v + J(y) =

a

+ J(x

), a

∈ ˆ A(x

). Then (a

− a

+ J(y) − J(x

), x

− y)

≥ 0 ⇒

(J(y) − J(x

), x

− y)

≥ 0 ⇒ x

= y ∈ ˆ D (since J is strictly monotone)

and v = a

∈ ˆ A(x

) i.e. (y, v) ∈ Gr ˆ A and this proves the maximality of ˆ A.

Let J

: R

→ R

be the resolvent map. Then for all t ∈ T

kJ

(x

(t)) − x(t)k ≤ kJ

(x

(t)) − J

(x(t))k + kJ

(x(t)) − x(t)k

≤ kx

(t) − x(t)k + kJ

(x(t)) − x(t)k → 0 as n → ∞.

Recall that A

(x

(t)) ∈ A(J

(x

(t))) for all t ∈ T . We have (J

(x

(·)), A

(x

(·))) ∈ Gr ˆ A

and J

(x

) → x in L

(T, R

), ˆ A

(x

) → u in L

(T, R

). Since ˆ A is maximal monotone, in the limit we have (x, u) ∈ Gr ˆ A, i.e. u(t) ∈ A(x(t)) a.e. on T . This proves that x ∈ C

(T, R

) is a solution of (1).

In the same way, using this time Proposition 3, we can have a“nonconvex”

existence theorem.

Theorem 5. If the hypotheses H(A)

, H(F )

and H(ξ), then problem (1) has a solution x ∈ C

(T, R

).

Now we will prove existence theorems for the case when domA 6= R

. As we have already mentioned, this requires a slightly stronger growth condition.

More precisely, the new hypotheses on F are the following:

H(F )

: F : T ×R

×R

→ P

(R

) is a multifunction such that hypothesis H(F )

(i) – (iv) hold and

(v) for almost all t ∈ T , all x, y ∈ R

and all v ∈ F (t, x, y) kvk ≤ γ

(t, kxk) + γ

(t, kxk)kyk

with sup

γ

(t, r) ≤ η

(t) a.e. on T , η

∈ L

(T ) and sup

γ

(t, r) ≤ η

(t) a.e. on T , η

∈ L

(T ) (as usual

= ∞ for r > 0).

Similarily, we can have the hypothesis of the nonconvex problem.

H(F )

: F : T × R

× R

→ P

(R

) is a multifunction such that hypothesis H(F )

(i) – (iv) and H(F )

(v) hold.

Also we will need a compatibility condition between the maps A and ξ.