1. Introduction. In this paper we study the following nonlinear bound- ary value problem in R

POLONICI MATHEMATICI LXXIII.1 (2000)

Quasilinear vector differential equations with maximal monotone terms and nonlinear boundary conditions

by Ralf Bader (M¨ unich) and Nikolaos S. Papageorgiou (Athens)

Abstract. We consider a quasilinear vector differential equation which involves the p-Laplacian and a maximal monotone map. The boundary conditions are nonlinear and are determined by a generally multivalued, maximal monotone map. We prove two exis- tence theorems. The first assumes that the maximal monotone map involved is everywhere defined and in the second we drop this requirement at the expense of strengthening the growth hypothesis on the vector field. The proofs are based on the theory of operators of monotone type and on the Leray–Schauder fixed point theorem. At the end we present some special cases (including the classical Dirichlet, Neumann and periodic problems), which illustrate the general and unifying features of our work.

1. Introduction. In this paper we study the following nonlinear bound- ary value problem in R

:

(1)

 (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 < ∞.

Here A : R

→ 2

is a maximal monotone operator and ϕ : R

→ R

is defined by ϕ(r) = krk

r.

Recently nonlinear boundary value problems involving the p-Laplacian have been studied by several authors. We mention the works of Boccardo–

Dr´ abek–Giachetti–Kuˇcera [2], Dang–Oppenheimer [3], del Pino–Elgueta–

Man´asevich [4], del Pino–Man´ asevich–Mur´ ua [5], Fabry–Fayyad [6], Guo [7], Kandilakis–Papageorgiou [11] and Man´asevich–Mawhin [12]. In all these papers A ≡ 0 and with the exception of Man´asevich–Mawhin, all the other

2000 Mathematics Subject Classification: Primary 34B15.

Key words and phrases: monotone operator, maximal monotone operator, demicon- tinuous operator, coercive operator, resolvent operator, Yosida approximation, projection theorem, measurable selection, Dirichlet, Neumann and periodic problems.

This work was done while the first author visited NTUA. Support was provided by a scholarship offered by Deutsche Forschungsgemeinschaft (DFG), Bonn (Germany).

[69]

works deal with the scalar problem (i.e. N = 1). It should be mentioned that Dang–Oppenheimer (scalar problem) and Man´asevich–Mawhin (vec- tor problem) work with a more general p-Laplacian-like differential opera- tor. Boccardo–Dr´ abek–Giachetti–Kuˇcera and del Pino–Elgueta–Man´ asevich deal with the Dirichlet problem, del Pino–Man´ asevich–Mur´ ua, Fabry–Fay- yad and Man´asevich–Mawhin examine the periodic problem and the first of these three works proves a multiplicity result, Guo considers both the periodic and the Neumann problem, Kandilakis–Papageorgiou consider the Neumann problem and finally Dang–Oppenheimer investigate all three problems (Dirichlet, Neumann and periodic problem). Also in the works of Boccardo–Dr´ abek–Giachetti–Kuˇcera, del Pino–Elgueta–Man´ asevich, del Pino–Man´ asevich–Mur´ ua, Fabry–Fayyad, Guo (the Neumann problem) and Kandilakis–Papageorgiou, the right hand side function f is independent of x

. Most of these works use degree-theoretic techniques, while Kandila- kis–Papageorgiou base their proof on variational arguments (critical point theory for nonsmooth functionals, since they do not assume continuity of f (t, ·)).

Our formulation here unifies the basic vector boundary value problems (Dirichlet, Neumann and periodic problem) and goes beyond them. Also the presence in (1) of the maximal monotone multivalued operator A incorpo- rates second order systems with convex potential. Such systems were studied by Mawhin–Willem [14] (see Section 1.7). In the book of Mawhin–Willem the potential function is nonautonomous, but smooth. The fact that in our case A can be multivalued allows us to include also problems with non- smooth (convex) potential. Moreover, in Theorem 6 we do not require that dom A = {x ∈ R

: A(x) 6= ∅} = R

and so our formulation incorporates second order variational inequalities (see Example (a), Section 4).

Our method of proof is based on the theory of operators of monotone type, which leads to an eventual application of the Leray–Schauder fixed point theorem.

2. Preliminaries. Since our approach uses the theory of nonlinear operators of monotone type, in this section we recall some basic definitions and facts from this theory that we will need. The basic references are the books of Hu–Papageorgiou [10] and Zeidler [16].

Let X be a reflexive Banach space and X

its topological dual. A pos-

sibly multivalued map A : D ⊆ X → 2

is said to be monotone if for all

x, y ∈ D and all x

∈ A(x), y

∈ A(y), we have (x

− y

, x − y) ≥ 0, where

(·, ·) denotes the duality brackets for the pair (X, X

). If (x

− y

, x − y) = 0

implies x = y, then we say that A is strictly monotone. A monotone map

for which the inequalities (x

− y

, x − y) ≥ 0 for all y ∈ D and all y

∈ A(y)

imply x ∈ D and x

∈ A(x) is said to be maximal monotone. It is clear

from this definition that A is maximal monotone if and only if its graph Gr A = {[x, x

] ∈ X × X

: x

∈ A(x)} is maximal with respect to in- clusion among the graphs of monotone maps. If A is maximal monotone, then for any x ∈ D, the set A(x) is nonempty, closed and convex. More- over, Gr A is demiclosed, i.e. if [x

, x

] ∈ Gr A, n ≥ 1, and either x

→ x in X and x

→ x

in X

, or x

→ x in X and x

→ x

in X

, then [x, x

] ∈ Gr A.

Let D = X and assume that A : X → X

is single-valued. We say that A is demicontinuous if x

→ x in X implies A(x

) → A(x), i.e. A is sequen-

tially continuous from X into X

(here X

denotes the Banach space X

equipped with the weak topology). A monotone, demicontinuous map A : X → X

is maximal monotone. A map A : D ⊆ X → 2

is said to be coer- cive if D is bounded or D is unbounded and inf{kx

k

: x

∈ A(x)} → ∞ as kxk → ∞, x ∈ D (here k·k and k·k

are the norms of X and X

respectively).

A basic theorem says that a maximal monotone, coercive operator is surjec- tive. In particular, a monotone, demicontinuous coercive map A : X → X

is surjective.

Suppose X

is strictly convex and A : D ⊆ X → 2

is maximal mono- tone. We know that for every x ∈ D, A(x) is nonempty, closed and con- vex. Thus we can define the single-valued map A

: D ⊆ X → X

by A

(x) = proj(0; A(x)), i.e. A

(x) is the unique element of A(x) with min- imal norm. We call A

the minimal section; it is important in the the- ory of maximal monotone operators. Recall that a reflexive Banach space can be equivalently renormed so that both X and X

are strictly convex (Asplund’s theorem, see Hu–Papageorgiou [10]). If X = H = Hilbert space, A : D ⊆ H → 2

is maximal monotone and λ > 0, we intro- duce J

= (I + λA)

(the resolvent of A) and A

= λ

(I − J

) (the Yosida approximation of A). We have dom J

= dom A

= H for all λ > 0.

Moreover, for every λ > 0, J

is nonexpansive (i.e. Lipschitz with con- stant 1), A

(x) ∈ A(J

(x)) for all x ∈ H, A

(·) is monotone and Lipschitz with constant λ

(hence maximal monotone), and kA

(x)k ≤ kA

(x)k for all x ∈ D. In addition, lim

A

(x) = A

(x) for all x ∈ D and lim

J

(x) = proj(x; D) for all x ∈ H. A well known maximal mono- tone operator is the subdifferential ∂ϕ of a convex, lower semicontinuous, proper (i.e. not identically +∞) function ϕ. Recall that ∂ϕ : X → 2

is de- fined by ∂ϕ(x) = {x

∈ X

: (x

, y−x) ≤ ϕ(y)−ϕ(x) for all y ∈ Y }. Finally if C ⊆ X is nonempty, then we denote by σ(·, C) : X

→ R = R ∪ {+∞}

the support function of C, i.e. σ(x

, C) = sup{(x

, c) : c ∈ C}, which is sublinear and weakly lower semicontinuous. So if C is closed and convex, then C = {x ∈ X : (x

, x) ≤ σ(x

, C) for all x

∈ X

}.

Let Y and Z be Banach spaces. An operator K : Y → Z (not necessarily

linear) is said to be:

(a) completely continuous if y

→ y in Y implies K(y

) → K(y) in Z;

(b) compact if K is continuous and maps bounded sets in Y into relatively compact sets in Z.

In general these two notions are not comparable. However, if Y is reflex- ive, then complete continuity of K implies compactness. Moreover, if K is linear and Y is reflexive, then the two notions are equivalent.

As already mentioned, our proof ultimately relies on a fixed point ar- gument which makes use of the “Leray–Schauder fixed point theorem”. For the convenience of the reader we recall this result here (see Zeidler [15], Theorem 6A, p. 245).

Theorem 1. If Y is a Banach space, K : Y → Y is compact and there exists r > 0 such that y = λK(y) with 0 < λ < 1 implies kyk ≤ r (a priori bound), then K has a fixed point (i.e. there exists y ∈ Y such that y = K(y)).

Finally recall that if 2 ≤ p < ∞ and a, c ∈ R, then 2

|a − c|

≤ (|a|

a − |c|

c)(a − c).

Our hypotheses on the data of (1) are the following:

H(A)

. A : R

→ 2

is a maximal monotone map such that dom A = {x ∈ R

: A(x) 6= ∅} = R

and 0 ∈ A(0).

Remark. The hypothesis 0 ∈ A(0) is not an essential restriction, since we can always have it by translating things if necessary. Also since dom A = R

, A

(·) is bounded on compact subsets of R

.

We can weaken the conditions on A (at the expense of strengthening the growth hypothesis on f (t, x, ·)) as follows:

H(A)

. A : R

→ 2

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

Remark. Again it is enough to assume that 0 ∈ dom A and by transla- tion we will have 0 ∈ A(0).

H(f )

. f : T × R

× R

→ R

is a function such that (i) for all (x, y) ∈ R

× R

, t 7→ f (t, x, y) is measurable;

(ii) for almost all t ∈ T, (x, y) 7→ f (t, x, y) is continuous;

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

,

(f (t, x, y), 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[(f (t, x, y), x)

+ kyk

: kx − x

k + ky − y

k < δ] ≥ ξ > 0;

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

,

kf (t, x, y)k ≤ γ

(t, kxk) + γ

(t, kxk)kyk

with sup

γ

(t, r) ≤ η

(t) a.e. on T , η

∈ L

(T ) (1/p + 1/q = 1) and sup

γ

(t, r) ≤ η

(t) a.e. on T , η

∈ L

(T ).

Remark. Hypothesis H(f )(iv) is a slight extension of the classical Nagu- mo–Hartman condition for continuous vector fields (see Hartman [8], p.

433).

When dom A 6= R

, we have to strengthen the growth condition on f (t, x, ·). More precisely, we will need the following hypothesis:

H(f )

. f : T × R

× R

→ R

is a function that satisfies (i)–(iv) of H(f )

and

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

, we have kf (t, x, y)k ≤ 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 let r/0 = ∞ for r > 0).

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) dom ξ = {(a, d) ∈ R

× R

: a = d}.

H

. For all (a, d) ∈ dom ξ and all (a

, d

) ∈ ξ(a, d),

(A

(a), a

)

+ (A

(d), d

)

≥ 0 for all λ > 0.

Remark. If ξ = ∂ψ with ψ : R

× R

→ R convex (hence locally Lipschitz), then if we denote by ∂

ψ, i = 1, 2, the partial subdifferential of ψ(a, d) with respect to a (resp. d), then ∂ψ(a, d) ⊆ ∂

ψ(a, d) × ∂

ψ(a, d).

In this setting the condition that (A

(a), a

)

≥ 0 and (A

(d), d

)

≥ 0 for all (a

, d

) ∈ ξ(a, d), (a, d) ∈ dom ξ, is equivalent to saying that ψ(J

(a), d) ≤ ψ(a, d) and ψ(a, J

(d)) ≤ ψ(a, d) respectively (see Hu–Papa- georgiou [10]).

3. Auxiliary results. We start by solving the following auxiliary prob- lem:

(2)

 −(kx

(t)k

x

(t))

+ kx(t)k

x(t) = h(t) a.e. on T, (ϕ(x

(0)), −ϕ(x

(b))) ∈ ξ(x(0), x(b)), 2 ≤ p < ∞.

Here h ∈ L

(T, R

) and 1/p + 1/q = 1. By a solution of problem (2) we

mean a function x ∈ C

(T, R

) such that kx

(·)k

x

(·) ∈ W

(T, R

)

and x satisfies (2).

Proposition 2. If ξ : R

×R

→ 2

is a maximal monotone map with (0, 0) ∈ ξ(0, 0), then problem (2) has a unique solution x ∈ C

(T, R

).

P r o o f. Given v, w ∈ R

, we consider the following two-point boundary value problem:

(3)

 −(kx

(t)k

x

(t))

+ kx(t)k

x(t) = h(t) a.e. on T, x(0) = v, x(b) = w.

Let η(t) = (1 − t/b)v + (t/b)w, so that η(0) = v, η(b) = w. We introduce the function y(t) = x(t) − η(t) and rewrite problem (3) as a homogeneous Dirichlet problem for y:

(4)

 



−(k(y + η)

(t)k

(y + η)

(t))

+ k(y + η)(t)k

(y + η)(t)

= h(t) a.e. on T, y(0) = y(b) = 0.

We solve (4) for y and then x = y + η will be the solution of (3). To solve (4), let V

: W

(T, R

) → W

(T, R

) be defined by

hV

(u), zi =

b

\

0

ku

+ η

k

(u

+ η

, z

)

dt +

b

\

0

ku + ηk

(u + η, z)

dt

for all u, z ∈ W

(T, R

). Here h·, ·i are the duality brackets for the pair (W

(T, R

), W

(T, R

)). Then for u, z ∈ W

(T, R

), we have hV

(u) − V

(z), u − zi

=

b

\

0

ku

+ η

k

(u

+ η

, u

− z

)

dt +

b

\

0

ku + ηk

(u + η, u − z)

dt

−

b

\

0

kz

+ η

k

(z

+ η

, u

− z

)

dt −

b

\

0

kz + ηk

(z + η, u − z)

dt.

Note that (5)

b

\

0

[ku

+ η

k

(u

+ η

, u

− z

)

− kz

+ η

k

(z

+ η

, u

− z

)

] dt

≥

b

\

0

(ku

+ η

k − kz

+ η

k)(ku

+ η

k

− kz

+ η

k

) dt

≥ 2

b

\

0

 ku

+ η

k − kz

+ η

k  

dt.

Similarly we obtain (6)

b

\

0

[ku + ηk

(u + η, u − z)

− kz + ηk

(z + η, u − z)

] dt

≥ 2

b

\

0

 ku + ηk − kz + ηk  

dt.

From (5) and (6), we infer that hV

(u) − V

(z), u − zi ≥ 0, i.e. V

(·) is monotone. In fact, V

is strictly monotone. Indeed, if hV

(u) − V

(z), u − zi

= 0, then

b

\

0

 ku

+ η

k − kz

+ η

k  

dt +

b

\

0

 ku + ηk − kz + ηk  

dt = 0,

hence

k(u

+ η

)(t)k = k(z

+ η

)(t)k = k

(t), k(u + η)(t)k = k(z + η)(t)k = k

(t) for almost all t ∈ T . So we have

0 =

b

\

0

k

(t)

ku

(t) − z

(t)k

dt +

b

\

0

k

(t)

ku(t) − z(t)k

dt, thus u

= z

and u = z, and so V

is strictly monotone.

Also using the extended dominated convergence theorem (see for example Hu–Papageorgiou [10], Theorem A.2.54, p. 907), we can easily check that V

is demicontinuous. Moreover,

hV

(u), ui =

b

\

0

ku

+ η

k

(u

+ η

, u

)

dt +

b

\

0

ku + ηk

(u + η, u)

dt

≥ ku

+ η

k

− ku

+ η

k

kη

k

+ ku + ηk

− ku + ηk

kηk

≥ ku + ηk

− µ

ku + ηk

for some µ

> 0.

Here k · k

denotes the norm of the Sobolev space W

(T, R

). So V

is coercive. Therefore V

, being monotone, demicontinuous (hence maximal monotone) and coercive, is surjective. So there exists y ∈ W

(T, R

) such that V

(y) = h. Evidently by the strict monotonicity of V

, this y is unique.

Let ψ ∈ C

(T, R

) and denote by (·, ·)

the duality brackets for the pair (L

(T, R

), L

(T, R

)). We have hV

(y), ψi = (h, ψ)

, hence

b

\

0

ky

+ η

k

(y

+ η

, ψ

)

dt +

b

\

0

ky + ηk

(y + η, ψ)

dt =

b

\

0

(h, ψ)

dt

and so

−h(ky

+ η

k

(y

+ η

))

, ψi + hky + ηk

(y + η), ψi

= hh, ψi (by Green’s identity).

Since (ky

+ η

k

(y

+ η

))

∈ W

(T, R

) (see the representation the- orem for the space W

(T, R

) in Adams [1], Theorem 3.10, p. 50) and since C

(T, R

) is dense in the predual space W

(T, R

) (recall that W

(T, R

)

= W

(T, R

)), we conclude that

−(ky

+ η

k

(y

+ η

))

+ ky + ηk

(y + η) = h and hence

ky

+ η

k

(y

+ η

) ∈ W

(T, R

).

Let x = y + η ∈ C

(T, R

) with kx

k

x

∈ W

(T, R

). This is the unique solution of (3).

Now let s : R

× R

→ C

(T, R

) be the map which to each (v, w) ∈ R

× R

assigns the unique solution s(v, w) = x ∈ C

(T, R

) of (3). Then let ̺ : R

× R

→ R

× R

be defined by

̺(v, w) = (−ks(v, w)

(0)k

s(v, w)

(0), ks(v, w)

(b)k

s(v, w)

(b)).

Claim 1. ̺ is monotone.

Let x = s(α, β) and x

= s(α

, β

). Using Green’s identity, we have



̺(α, β) − ̺(α

, β

),

 α − α

β − β



R^2N

=(kx

(b)k

x

(b) − kx

(b)k

x

(b), β − β

)

− (kx

(0)k

x

(0) − kx

(0)k

x

(0), α − α

)

=

b

\

0

kx

(t)k

(x

(t), x

(t) − x

(t))

dt

−

b

\

0

kx

(t)k

(x

(t), x

(t) − x

(t))

dt

+

b

\

0

((kx

(t)k

x

(t))

− (kx

(t)k

x

(t))

, x(t) − x

(t))

dt.

Note that

b

\

0

kx

(t)k

(x

(t), x

(t) − x

(t))

dt

−

b

\

0

kx

(t)k

(x

(t), x

(t) − x

(t))

dt ≥ 0

(see Section 2). Also because x = s(α, β) and x

= s(α

, β

), we have

b

\

0

((kx

(t)k

x

(t))

− (kx

(t)k

x

(t))

, x(t) − x

(t))

dt

=

b

\

0

(kx(t)k

x(t) − kx

(t)k

x

(t), x(t) − x

(t))

dt ≥ 0.

Thus finally we obtain



̺(α, β) − ̺(α

, β

),

 α − α

β − β



R^2N

≥ 0, which proves the claim.

Claim 2. ̺ : R

× R

→ R

× R

is continuous.

Assume that α

→ α and β

→ β in R

and set x

= s(α

, β

), n ≥ 1, x = s(α, β). As before we introduce η

(t) = (1 − t/b)α

+ (t/b)β

, n ≥ 1, and η(t) = (1 − t/b)α + (t/b)β and set y

= x

− η

. We have

−

b

\

0

((ky

+ η

k

(y

+ η

))

, y

)

dt

+

b

\

0

ky

+ η

k

(y

+ η

, y

)

dt =

b

\

0

(h(t), y

(t))

dt, that is,

b

\

0

ky

+ η

k

(y

+ η

, y

)

dt +

b\

0

ky

+ η

k

(y

+ η

, y

)

dt

=

b

\

0

(h(t), y

(t))

dt (by Green’s identity), therefore

ky

+ η

k

− k

ky

+ η

k

+ ky

+ η

k

− k

ky

+ η

k

≤ khk

ky

+ η

k

+ k

for some k

, k

, k

> 0, and so {x

= y

+ η

}

⊆ W

(T, R

) is bounded. Thus {kx

k

x

}

, {kx

k

x

}

⊆ L

(T, R

) are both bounded. Moreover, from (3) it follows that {kx

k

x

}

⊆ W

(T, R

) is bounded. So we may assume that x

→ u in W

(T, R

) and kx

k

x

→ v in W

(T, R

) as n → ∞. In particular x

→ u in C(T, R

) (recall

that W

(T, R

) is compactly embedded in C(T, R

)) and (kx

k

x

)

→ v

in L

(T, R

) as n → ∞. Thus in the limit as n → ∞, we obtain

 −v

(t) + ku(t)k

u(t) = h(t) a.e. on T, u(0) = α, u(b) = β.

Since kx

k

x

→ v in W

(T, R

), we have kx

k

x

→ v in C(T, R

) (from the compact embedding of W

(T, R

) in C(T, R

)). Also ϕ

exists and is continuous because ϕ(r) = krk

r is a homeomorphism.

Therefore ϕ

(kx

k

x

) = x

→ ϕ

(v) in L

(T, R

) (in fact in C(T, R

)), hence ϕ

(v) = u

and so v = ϕ(u

) = ku

k

u

. Thus finally we have

 −(ku

(t)k

u

(t))

+ ku(t)k

u(t) = h(t) a.e. on T, u(0) = α, u(b) = β.

Hence u = s(α, β) = x, i.e. s is continuous from R

× R

into C

(T, R

).

From the continuity of s we deduce at once the continuity of ̺.

Claim 3. ̺ is coercive.

We have, with x = s(α, β),

̺(α, β),

R^2N

= kx

(b)k

(x

(b), β)

− kx

(0)k

(x

(0), α)

=

Tb

0

((kx

(t)k

x

(t))

, x(t))

dt + kx

k

≥ kxk

+ kx

k

− khk

kxk

where we have used Green’s identity and (3).

From the mean value theorem for integrals (see for example Hewitt–

Stromberg [9], Theorem 21.69, p. 420), we can find t

∈ T such that kx(t

)kb =

Tb

0

kx(t)k dt. So for every t ∈ T , kx(t)k ≤ kx(t

)k +

t

\

t0

kx

(s)k ds ≤ 1

b kxk

+ b

kx

k

≤ k

kxk

for some k

> 0 and hence

≤ k

kxk

for some k

> 0. Therefore we have

̺(α, β),

R^2N

≥ kxk

− khk

kxk

k

kxk

, i.e. ̺ is coercive.

From Claims 1, 2 and 3, it follows that ̺ is maximal monotone.

Now let σ = ̺ + ξ : R

× R

→ 2

. Then σ is maximal monotone

(see Hu–Papageorgiou [10], Theorem III.3.3, p. 334) and coercive (recall

that ̺ is coercive by Claim 3 and (0, 0) ∈ ξ(0, 0) by hypothesis H(ξ)). Thus σ is surjective. So we can find (α, β) ∈ R

× R

such that 0 ∈ σ(α, β). Let x = s(α, β). Evidently this is the desired solution of (2).

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.

Proposition 3. If ξ : R

× R

→ 2

is a maximal monotone map with (0, 0) ∈ ξ(0, 0), then V : D ⊆ L

(T, R

) → L

(T, R

) is maximal monotone.

P r o o f. First note that for every x, y ∈ D, we have (V (x)− V (y), x− y)

≥

b

\

0

(kx

(t)k

− ky

(t)k

)(kx

(t)k− ky

(t)k) dt ≥ 0, hence V is monotone.

To prove the maximality of V it suffices to show R(V + J) = L

(T, R

) with J : L

(T, R

) → L

(T, R

) defined by J(x)(·) = kx(·)k

x(·). In- deed, assume for the moment that V + J is surjective and suppose that y ∈ L

(T, R

) and v ∈ L

(T, R

) satisfy

(V (x) − v, x − y)

≥ 0 for all x ∈ D.

Let x

∈ D be such that v + J(y) = V (x

) + J(x

). Thus we can write 0 ≤ (V (x

) − V (x

) − J(x

) + J(y), x

− y)

= (J(y) − J(x

), x

− y)

. But it is easy to see that J is strictly monotone. So from this last inequality it follows that y = x

∈ D and v = V (x

), which proves the maximality of V . Therefore we need to show that R(V + J) = L

(T, R

). But this is an immediate consequence of Proposition 2.

Next let λ > 0 and let A

: R

→ R

be the Yosida approximation of A. Note that as indicated in the remark following hypothesis H(A), we may assume without loss of generality that 0 ∈ A(0), which in turn implies that 0 = A

(0) for all λ > 0. Let b A

: L

(T, R

) → L

(T, R

) be the Nemytski˘ı operator corresponding to A

, i.e. b A

(x)(·) = A

(x(·)) for all x ∈ L

(T, R

). Clearly b A

is monotone, continuous (Krasnosel’ski˘ı’s theorem), thus maximal monotone. We consider the following auxiliary boundary value problem:

(7)

 (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)).

Proposition 4. If hypotheses H(A)

, H(f )

(or H(f )

) and H(ξ) hold, then problem (7) has a solution x ∈ C

(T, R

).

P r o o f. We do the proof when H(f )

holds, the other case being similar.

Let K

= V + b A

+ J : D ⊆ L

(T, R

) → L

(T, R

), λ > 0. Note that K

is maximal monotone (see Proposition 3 and Theorem III.3.3, p. 334, of Hu–Papageorgiou [10]). Also since 0 = A

(0), we have

(K

(x), x)

≥ (V (x), x)

+ (J(x), x)

. But

(V (x), x)

= −

b

\

0

((kx

(t)k

x

(t))

, x(t))

dt

= (−kx

(b)k

x

(b), x(b))

+ (kx

(0)k

x

(0), x(0))

+ kx

k

≥ kx

k

.

Here we have used Green’s identity and the fact that (ϕ(x

(0)), −ϕ(x

(b))) ∈ ξ(x(0), x(b)) and (0, 0) ∈ ξ(0, 0). Also (J(x), x)

= kxk

. Thus finally

(K

(x), x)

≥ kx

k

+ kxk

= kxk

,

thus K

is coercive, hence surjective. Moreover, from the strict monotonicity of J, it follows that K

is injective. So K

: L

(T, R

) → D ⊆ W

(T, R

) is well defined.

Claim 1. K

is compact from L

(T, R

) into W

(T, R

).

By the reflexivity of L

(T, R

), to establish the claim, it suffices to show that if u

→ u in L

(T, R

), then K

(u

) → K

(u) in W

(T, R

). Let x

= K

(u

), n ≥ 1. We have

(V (x

), x

)

+ ( b A

(x

), x

)

+ (J(x

), x

)

= (u

, x

)

,

hence kx

k

+ kx

k

≤ ku

k

kx

k

, and thus {x

}

⊆ W

(T, R

) is bounded. Thus we may assume that x

→ x in W

(T, R

) and x

→ x in L

(T, R

) (because W

(T, R

) is compactly embedded in L

(T, R

)).

Since (x

, u

) ∈ Gr K

, n ≥ 1, and K

is maximal monotone (it has a demiclosed graph in L

(T, R

)× L

(T, R

), see Section 2), we have (x, u) ∈ Gr K

. So u = V (x) + b A

(x) + J(x). Moreover

lim(V (x

) + b A

(x

) + J(x

), x

− x)

= lim(u

, x

− x)

= 0 and hence lim(V (x

), x

− x)

= 0 (since ( b A

(x

), x

− x)

→ 0 and (J(x

), x

− x)

→ 0).

But, by Proposition 3, we know that V is maximal monotone, hence it

is also generalized pseudomonotone (see Hu–Papageorgiou [10], Definition

III.6.2 and Remark III.6.3, p. 365). Therefore V (x

) → V (x) in L

(T, R

).

Note that {kx

k

x

}

⊆ W

(T, R

) and

kx

k

x

→ kx

k

x

in W

(T, R

), hence

kx

k

x

→ kx

k

x

in C(T, R

).

So acting with ϕ

we obtain x

→ x

in L

(T, R

) as n → ∞. Since x

→ x in L

(T, R

), we conclude that x

→ x in W

(T, R

) and this proves the claim.

Let N : W

(T, R

) → L

(T, R

) be the Nemytski˘ı operator corre- sponding to f , i.e. N (x)(·) = f (·, x(·), x

(·)). Clearly N is continuous (see hypothesis H(f )). Let N

= −N + J and consider the following abstract fixed point problem:

(8) x = K

N

(x).

Observe that N

: W

(T, R

) → L

(T, R

) is continuous, bounded and K

: L

(T, R

) → W

(T, R

) is compact. Hence K

N

: W

(T, R

)

→ W

(T, R

) is compact. So according to Theorem 1, in order to solve (8), it suffices to prove the following claim:

Claim 2. S = {x ∈ W

(T, R

) : x = βK

N

(x), 0 < β < 1} is bounded.

Let x ∈ S. We have K

((1/β)x) = N

(x) and so V

 1 β x

 + b A

 1 β x

 + J

 1 β x



= −N (x) + J(x).

Hence

 V

 1 β x

 , x



pq

+

 A b

 1 β x

 , x



pq

+

 J

 1 β x

 , x



pq

= −(N (x), x)

+ (J(x), x)

. Note that ( b A

((1/β)x), x)

≥ 0 (recall that 0 = b A

(0) and b A

is mono- tone). Moreover,

 V

 1 β x

 , x



pq

= −

b\

0

1

β

((kx

(t)k

x

(t))

, x(t))

dt

= −

 1

β

kx

(b)k

x

(b), x(b)



R^N

+

 1

β

kx

(0)k

x

(0), x(0)



R^N

+ 1

β

kx

k

≥ 1

β

kx

k

(from the boundary conditions).

So we have 1

β

kx

k

+ 1

β

kxk

≤ −(N (x), x)

+ kxk

and hence

(9) kx

k

≤ −β

(N (x), x)

+ (β

− 1)kxk

≤ −β

(N (x), x)

(since 0 < β < 1). Using hypothesis H(f )(iii), we have

−β

(N (x), x)

= β

b

\

0

−(f (t, x(t), x

(t)), x(t))

dt

≤ β

akxk

+ β

γ

b

\

0

kx(t)k

kx

(t)k

dt + β

kck

kxk

.

Let τ = p−r and set µ = p/r, µ

= p/τ (1/µ +1/µ

= 1). Apply H¨older’s inequality to obtain

b

\

0

kx(t)k

kx

(t)k

dt ≤ 

0

kx(t)k

dt 



0

kx

(t)k

dt 

≤ kxk

kx

k

. It follows that

(10) −β

(N (x), x)

≤ β

akxk

+ β

γkxk

kx

k

+ β

kck

kxk

. Next we will show that for all x ∈ S we have kxk

≤ M (with M > 0 as in hypothesis H(f )(iv)). To this end let r(t) = kx(t)k

and let t

∈ T be the point where r(·) attains its maximum. Suppose that r(t

) > M

and assume first that 0 < t

< b. Then 0 = r

(t

) = pkx(t

)k

(x

(t

), x(t

))

, hence (x

(t

), x(t

))

= 0. By hypothesis H(f )(iv), we can find δ, ξ > 0 such that

inf[(f (t, x, y), x)

+ kyk

: kx − x(t

)k + ky − x

(t

)k < δ] ≥ ξ > 0.

Note that since x ∈ S we see that kx

(·)k

x

(·) ∈ W

(T, R

) ⊆ C(T, R

), hence ϕ

(kx

(·)k

x

(·)) = x

(·) ∈ C(T, R

). Thus we can find δ

> 0 such that if t

< t ≤ t

+ δ

, then

kx(t) − x(t

)k + kx

(t) − x

(t

)k < δ.

Therefore for almost all t ∈ (t

, t

+ δ

] we have

(11) β

(f (t, x(t), x

(t)), x(t))

+ β

kx

(t)k

≥ β

ξ.

Now from the equation V (β

x) + b A

(β

x) + J(β

x) = −N (x) + J(x) it

follows that

−(kx

(t)k

x

(t))

+ β

A

β

x(t)

= −β

f (t, x(t), x

(t))(β

− 1)kx(t)k

x(t) a.e. on T . Using this last equality in (11), we have

((kx

(t)k

x

(t))

, x(t))

− β

(A

(β

x(t)), x(t))

+ (β

− 1)kx(t)k

+ β

kx

(t)k

≥ β

ξ and hence

t

\

t0

((kx

(s)k

x

(s))

, x(s))

ds + β

t

\

t0

kx

(s)k

ds ≥ β

ξ(t − t

).

Using Green’s identity for the first integral and the fact that (x

(t

), x(t

))

= 0, we have

kx

(t)k

(x

(t), x(t))

−

t

\

t0

kx

(s)k

ds + β

t

\

t0

kx

(s)k

ds

≥ β

ξ(t − t

).

Therefore kx

(t)k

(x

(t), x(t))

≥ β

ξ(t − t

) (since 0 < β < 1) and so r

(t) > 0 for all t ∈ (t

, t

+ δ

]. Thus r(t) > r(t

) for t

< t ≤ t

+ δ

, a contradiction to the choice of t

. So kx(t)k ≤ M for all t ∈ T and all x ∈ S, provided t

∈ (0, b).

Next let t

= 0. Then r

(0) ≤ 0 and so kx(0)k

(x

(0), x(0))

≤ 0, hence (x

(0), x(0))

≤ 0. First suppose that hypothesis H(ξ)(i) is in effect.

We have kx

(0)k

(x

(0), x(0))

≥ 0, hence (x

(0), x(0))

≥ 0 and finally (x

(0), x(0))

= 0 and so we can proceed as before. Now suppose that hypothesis H(ξ)(ii) is in effect. We may assume that x 6= 0. Then x(0) = x(b) and r

(0) ≤ 0 ≤ r

(b), which implies that

kx(0)k

(x

(0), x(0))

≤ 0 ≤ kx(b)k

(x

(b), x(b))

and so

(x

(0), x(0))

≤ 0 ≤ (x

(b), x(b))

.

Also from the boundary conditions, we see that kx

(b)k

(x

(b), x(b))

≤ kx

(0)k

(x

(0), x(0))

. Thus we obtain either x

(0) = x

(b) = 0, from which it follows that r

(0) = r

(b) = 0, or (x

(0), x(0))

= (x

(b), x(b))

, from which again it follows that r

(0) = r

(b) = 0. So we always have r

(0) = 0 and proceed as before. Similarly we treat the case t

= b. Hence we have established that kxk

≤ M for all x ∈ S. Then from (10) we have

β

(−N (x), x)

≤ k

+ k

kx

k

for some k

, k

> 0.

Using this estimate in (9), we obtain

kx

k

≤ k

+ k

kx

k

(τ < p),

which yields that {x

}

⊆ L

(T, R

) is bounded, and therefore {x}

⊆ W

(T, R

) is bounded. This proves Claim 2.

Now we can apply Theorem 1 to obtain x ∈ D such that x = K

N

(x), that is, K

(x) = N

(x) and thus x is a solution of (7).

4. Main results, special cases. In this section we state and prove the main existence theorems concerning problem (1) and present some special cases of interest, which illustrate the generality of our results.

Theorem 5. If hypotheses H(A)

, H(f )

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

(T, R

).

P r o o f. Let λ

→ 0, λ

> 0 and let x

∈ C

(T, R

) be solutions of the auxiliary problem (7). As in the proof of Proposition 4, we can show that kx

(t)k ≤ M for all t ∈ T and all n ≥ 1. Also we have V (x

) + b A

(x

) =

−N (x

), which yields

(V (x

), x

)

+ ( b A

(x

), x

)

= −(N (x

), x

)

and therefore

kx

k

≤ kN (x

)k

kx

k

≤ k

kN (x

)k

for some k

> 0.

Here we have used Green’s identity and the boundary conditions in the first term (V (x

), x

)

and the fact that b A

(·) is monotone with 0 = b A

(0) in the second term ( b A

(x

), x

)

. Then invoking hypothesis H(f )

(v), we obtain

kx

k

≤ k

(kη

k

+ kη

k

kx

k

),

showing that {x

} ⊆ L

(T, R

) is bounded, and so {x

} ⊆ W

(T, R

) is bounded; hence we may assume x

→ x in W

(T, R

) and x

→ x in L

(T, R

). Also note that kA

(x

(t))k ≤ kA

(x

(t))k (since dom A = R

, see Hu–Papageorgiou [10], Proposition III.2.29, p. 325). Moreover, because dom A = R

, A

is bounded on compact sets. Note that since {x

}

is bounded in W

(T, R

), it is relatively compact in C(T, R

). Hence, sup[kA

(x

(t))k : t ∈ T, n ≥ 1] ≤ k

for some k

> 0. Thus we may assume that A

(x

(·)) → u in L

(T, R

) as n → ∞. Now arguing as in the proof of Proposition 4 (Claim 1), we see that x

→ x

in L

(T, R

) and so x

→ x in W

(T, R

). Moreover, {kx

(·)k

x

(·)}

⊆ W

(T, R

) is bounded (since (kx

(t)k

x

(t))

= A

(x

(t)) + f (t, x

(t), x

(t)) a.e.

on T ). Thus we may assume that kx

(·)k

x

(·) → v in W

(T, R

) and kx

(·)k

x

(·) → v in C(T, R

). Then ϕ

(kx

(·)k

x

(·)) = x

→ ϕ

(v) in L

(T, R

) and so ϕ

(v) = x

and v = kx

(·)k

x

(·). Therefore

(kx

(t)k

x

(t))

= u(t) + f (t, x(t), x

(t)) a.e. on T

and (ϕ(x

(0)), −ϕ(x

(b)) ∈ ξ(x(0), x(b)) (since Gr ξ is closed because of the maximal monotonicity of ξ).

Now let b A : b D ⊆ L

(T, R

) → L

(T, R

) be defined by A(x) = {g ∈ L b

(T, R

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

D = {x ∈ L b

(T, R

) : there is g ∈ L

(T, R

) satisfying

g(t) ∈ A(x(t)) a.e. on T }.

We claim that b A is maximal monotone. Since monotonicity is clear, we start to show that R( b A + J) = L

(T, R

), where as before J : L

(T, R

) → L

(T, R

) is defined by J(x)(·) = kx

(·)k

x

(·). To this end let h ∈ L

(T, R

) be given and let

S(t) = {(x, a) ∈ R

× R

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

(we consider a finite-valued representative of h). The map A + ϕ is maximal monotone (see Hu–Papageorgiou [10], Theorem III.3.3, p. 334) and coercive.

Thus if we choose R(t) = kh(t)k

+ 1 large enough, then by Theorem III.6.28, p. 371, of Hu–Papageorgiou [10], we see that S(t) 6= ∅ for all t ∈ T . Moreover, R ∈ L

(T )

,

Gr S = {(t, x, a) ∈ T × R

× R

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

(because A(x) is closed (in fact compact), due to the maximality of A). So Gr S ∈ L × B(R

) × B(R

) (L = Lebesgue σ-field of T ). Invoking the Yankov–von Neumann–Aumann selection theorem (see Hu–Papageorgiou [10], Theorem II.2.14, p. 158) we obtain measurable functions x, a : T → R

such that (x(t), a(t)) ∈ S(t) a.e. on T , hence a(t)+ ϕ(x(t)) = h(t) a.e. on T . Evidently a ∈ L

(T, R

). This proves the surjectivity of b A + J.

Then we argue as in the proof of Proposition 3. Namely suppose that for y ∈ L

(T, R

) and v ∈ L

(T, R

) we have

(a − v, x − y)

≥ 0 for all x ∈ b D and all a ∈ b A(x).

Let x = x

where x

∈ b D is such that v + J(y) = a

+ J(x

), a

∈ b A(x

) (here we use the surjectivity of b A + J). Then we can write

(a

− a

− J(x

) + J(y), x

− y)

≥ 0,

hence (J(y) − J(x

), x

− y)

≥ 0 and so y = x

∈ b D and v = a

∈ b A(x

), which proves the maximality of b A.

Recall that A

(x

(t)) ∈ A(J

(x

(t))) a.e. with J

: R

→ R

being the resolvent operator corresponding to A. Note that

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 → ∞. Hence J

(x

(·)) → x(·) in L

(T, R

) (dominated convergence theorem). Since (J

(x

(·)), A

(x

(·))) ∈ Gr b A and the latter is demiclosed (since b A is maximal monotone), we have (x, u) ∈ Gr b A, hence u(t) ∈ A(x(t)) a.e. on T, u ∈ L

(T, R

). This proves that x ∈ C

(T, R

) is a solution of (1).

Remark. An interesting byproduct of the above proof is that b A : D ⊆ L

(T, R

) → 2

is maximal monotone when dom A = R

, a fact well known for p = q = 2, but for which we have been unable to find a proof in the literature when p 6= 2.

Now we state and prove an existence theorem for problem (1) for the case when dom A is not all of R

.

Theorem 6. If hypotheses H(A)

, H(f )

, H(ξ) and H

hold , then prob- lem (1) has a solution x ∈ C

(T, R

).

P r o o f. As in the proof of Theorem 5, let λ

→ 0, λ

> 0 and let x

∈ C

(T, R

) be the solutions of the auxiliary problem (7). Again {x

} ⊆ W

(T, R

) is bounded and so we may assume that x

→ x in W

(T, R

).

Also for all n ≥ 1 we have V (x

) + b A

(x

) = −N (x

) and hence (V (x

), A b

(x

))

+ k b A

(x

)k

= −(N (x

), b A

(x

))

.

Note that b A

(x

) ∈ C(T, R

) for all n ≥ 1. We have (V (x

), b A

(x

))

= −

b

\

0

((kx

(t)k

x

(t))

, A

(x

(t)))

dt

= − kx

(b)k

(x

(b), A

(x

(b)))

+ kx

(0)k

(x

(0), A

(x

(0)))

+

b

\

0

kx

(t)k



x

(t), d

dt A

(x

(t))



R^N

dt (Green’s identity). Since A

(·) is Lipschitz continuous, it is differentiable almost everywhere (Rademacher’s theorem). Also from the monotonicity of A

(·) we have



y, A

(x + ty) − A

(x) t



R^N

≥ 0,

i.e. (y, A

(x)y)

≥ 0 for every x which is a point of differentiability of A

(·) and every y ∈ R

. Moreover, from Marcus–Mizel [13] we know that

d

dt A

(x

(t)) = A

(x

(t))x

(t).