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    EXISTENCE RESULTS FOR OPERATOR EQUATIONS

    INVOLVING DUALITY MAPPINGS

    VIA THE MOUNTAIN PASS THEOREM

    JENICA CRINGANU

    We derive existence results for operator equations having the form

    Ju= Nfu,

    by using the Mountain Pass Theorem. Here J is a duality mapping on a realreflexive and smooth Banach space, compactly imbedded in a Lq-space, and Nfis the Nemytskii operator generated by a Caratheodory functionf which satisfiessome appropriate growth conditions.

    AMS 2000 Subject Classification: 35J20, 35J60.

    Key words: Duality mapping, Mountain Pass Theorem, Nemytskii operator, p-Laplacian.

    1. INTRODUCTION

    In [9] the existence of the weak solution in W1,p

    0 (), 1 < p < +, for

    the Dirichlet problem

    pu= f(x, u) in RN,(1.1)

    u= 0 on (1.2)

    was obtained using (among other methods) the Mountain Pass Theorem.

    It is well known that the operator p : W1,p0 () W

    1,p() is in

    fact the duality mapping on W1,p0 () corresponding to the gauge function(t) = tp1. In this paper we generalize the results from [9] by consideringoperator equations of the form

    (1.3) J

    u= Nf

    u,

    where, instead of the special duality mapping (p), we consider the case ofan arbitrary duality mapping. More precisely, we consider equation (1.1) inthe functional framework below.

    MATH. REPORTS 11(61), 1 (2009), 1120

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    12 Jenica Crnganu 2

    (H1) Xis a real reflexive and smooth Banach space, compactly imbedded in

    Lq(), where 1< q 0 is constant and b Lq

    ().

    By solution of equation (1.3) we mean an element u Xwhich satisfies

    (1.5) Ju= (iNfi)u,

    where i is the compact imbedding of X into Lq() and i : Lq

    () X isits adjoint.

    Notice that by (1.4) the operator Nf is well-defined, continuous and

    bounded from Lq() into Lq

    () (see, e.g., [8]), so that iNfi is also well-defined and compact from X into X.

    Moreover, the functional : Lq() R, defined by(u) =

    F(x, u)dx,

    whereF(x, s) =s0 f(x, t) dt is C

    1 on Lq(), hence on Xand (u) =Nfuforall u X.

    Notice also thatF is a Caratheodory function and there exists a constantc1> 0 and a function c L

    q(), c 0, such that

    (1.6) |F(x, s)| c1|s|q +c(x), x , s R.

    Let us remark that (1.5) is equivalent to

    (1.7) Ju, v= Nf(iu), ivLq(),Lq ()=

    f(x, u)vdx, v X.

    By Asplunds theorem, Ju = (u) for each u X, where (u) =u0 (t)dt and : X P(X) is the subdifferential of in the sense of

    convex analysis, i.e.,

    (u) ={x X : (v) (u) x, v u for all v X}

    (see, e.g., [4], [5] or [12]). Since is convex and X is smooth, is Gateauxdifferentiable on X and (u) = {(u)} for all u X. So, Ju = (u)

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    3 Existence results for operator equations involving duality mappings 13

    for all u X and, by continuity of J (see (H2)), we have C1(X,R).

    Consequently, the functional F :X R defined by

    F(u) = (u) (u) =

    u0

    (t)dt

    F(x, u)dx, u X,

    is C1 on X and F(u) = (u) (u) =Ju Nfu for all u X. Therefore,u Xis a solution of equation (1.3) if and only ifu is a critical point for F,i.e.,F(u) = 0.

    To prove that Fhas at least a critical point in Xwe use the MountainPass Theorem, which we recall below.

    Theorem 1.1. LetXbe a real Banach space and assumeI C1(X,R)satisfies the Palais-Smale(PS) condition. SupposeI(0) = 0 and that

    (i) there are constants, >0 such thatI|u= ;(ii) there is an elemente X, e> such thatI(e) 0.

    ThenIpossesses a critical valuec . Moreover,

    c= infg

    maxug([0,1])

    I(u)

    where ={g C([0, 1], X) :g(0) = 0, g(1) =e}. (It is obvious that each criti-cal pointu at levelc (I(u) = 0, I(u) =c) is a nontrivial one.)

    For the proofsee, e.g., [2], [11] or [13].Let us recall that the functional I C1(X,R) is said to satisfy the

    (PS) condition if any sequence (un) X for which (I(un)) is bounded and

    I(un) 0 as n , possesses a convergent subsequence.

    2. THE MAIN EXISTENCE RESULT

    First, we establish some preliminary results.

    Proposition 2.1. If (un) X is bounded andF(un) 0 as n ,

    then(un) has a convergent subsequence.

    Proof. This result is proved in [9], Lemma 2.

    Theorem 2.1. Assume that there are >1 ands0 > 0 such that

    (2.1) F(x, s) sf(x, s), x , |s| s0,

    (2.2) lim inf t

    t0(s)ds

    t(t) >

    1

    .

    ThenF satisfies the(PS) condition.

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    14 Jenica Crnganu 4

    Proof. According to Proposition 2.1 it is sufficient to show that any

    sequence (un) Xfor which (F(un)) is bounded andF(un) 0, is bounded.As in the proof of Theorem 15 in [9], there exists a constant k > 0

    such that

    (2.3) (un) 1

    f(x, un) undx k.

    By F(un) 0, as in the proof of the same theorem, there exists n0 Nsuch that Jun, un

    f(x, un)un

    un, () n n0,which implies

    (2.4) 1

    (un)un +1

    f(x, un)undx 1

    un, () n 1.

    From (2.3) and (2.4) we obtain

    (un) 1

    (un)un

    1

    un k, () n n0,

    so that un0

    (s)ds 1

    (un)un

    1

    un k, () n n0

    or, equivalently,

    (2.5) (un)un un

    0 (s)ds(un)un

    1

    1(un)

    k.

    By condition (2.2) there are constants > 1 and t0 > 0 such thatt0(s)ds

    t(t) , t t0.

    Ifun there exists n1 N such thatun t0 for n n1 and thenun0 (s)ds

    un(un) , n n1.

    From (2.5) we obtain

    (un)un

    1

    1

    (un)

    k, n n1.

    a contradiction (because > 1 and (un) ).Consequently, (un) is bounded in X.

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    5 Existence results for operator equations involving duality mappings 15

    Theorem 2.2. Assume that there exists >1 such that

    (2.6) limt

    t0(s)ds

    t = 0

    and either

    (i) there existss1 > 0 such that

    (2.7) 0< F(x, s) sf(x, s), x , s s1,

    or

    (ii) there existss1 < 0 such that

    (2.8) 0< F(x, s) sf(x, s), x , s s1.

    ThenF is unbounded from below.

    Proof. We are going to prove the sufficiency of condition (i) (similararguments may be used if (ii) holds).

    Let u X, u > 0 (in fact i(u) > 0) be such that meas (M1(u)) > 0,where

    M1(u) ={x :u(x) s1} (in facti(u)(x) s1).

    We shall prove that F(u) as .For 1, from the proof of Theorem 16 in [9] we have

    F(x,u)dx k1(u) k2,

    wherek1(u)> 0, k2 > 0 are constants. Therefore,

    F(u) = (u)

    F(x,u)dx (u) k1(u) +k2

    =

    u0 (t)dt

    k1(u)

    +k2.

    By condition (2.6) there are constants (0, k1(u)) and t0 > 0 such thatt0(s)ds

    t u , t t0,

    and then

    F(u) u0 (t)dt

    (u) u k1(u) +k2

    ( k1(u)) +k2

    for u t0. Since k1(u)< 0, we have F(u) as .

    Remark 2.1. By Theorem 2.2, sinceFis unbounded from below, for each >0 there exists e X withe> such that F(e) 0.

    Now we are in a position to prove the main result of this section.

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    Theorem2.3. Assume that hypotheses(H1), (H2)and(H3)hold. More-

    over, assume that(i) there are >1 ands0>0 such that

    (2.9) 0< F(x, s) sf(x, s), x , |s| s0;

    (ii) lim inft

    t0(s)ds

    t(t) >

    1

    ;

    (iii) limt

    t0(s)ds

    t = 0;

    (iv) there arer1, r2 (1, ), r1 < r2, andc0> 0 such that

    F(u) c2ur1X c3u

    r2X () u X withuX< c0,

    wherec2, c3 > 0 are constants.Then equation (1.3) has at least a nontrivial solution inX.

    Proof. Let us notice that condition (iv) is suggested by [10].It is sufficient to show that F has at least a nontrivial critical point

    u X. To do it, we shall use Theorem 1.1. Clearly, F(0) = 0. By (i), (ii) andTheorem 2.1,Fsatisfies the (PS) condition. By (iv) we have

    F(u) ur1X

    c2 c3ur2r1X

    for all u X, withuX< c0,hence there are constants , >0, < c0 such

    that F(u) >0 provided that uX= is small enough.Finally, from (i), (iii) and Theorem 2.2 (see also Remark 2.1) there is an

    element e X,e> , such that F(e) 0, and the proof is complete.

    Remark 2.2. A sufficient condition in order to have (iv) is the existenceof real numbers r1, r2, 1< r1 < r2, such that

    (iv1) lims0

    s0 (t)dt

    |s|r1>0,

    (iv2) lims0

    f(x, s)

    |s|r21 0 is small enough while,

    by (iv2),

    F(x, u)dx c3ur2 ,

    provided that u > 0 is small enough (here, c2, c3 > 0 are constants). Con-sequently,F(u) c2u

    r1 c3ur2 for u> 0 sufficiently small.

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    7 Existence results for operator equations involving duality mappings 17

    3. EXAMPLES

    Example3.1 Consider the Dirichlet problem (1.1), (1.2), where 1 < p p,

    + ifNp.

    Let us remark that thep-Laplacian operator p: W1,p0 () W

    1,p()defined by

    pu= Ni=1

    xi

    |u|p2

    u

    xi

    , u W1,p0 (),

    or, equivalently,

    pu, v=

    |u|p2u v, u, v W1,p0 (),

    is the duality mapping

    J: W1,p0 () W

    1,p()

    corresponding to the gauge function (t) =tp1 (see, e.g., [9] or [12]).

    On the other hand, the Nemytskii operatorNfis continuous and boundedfrom Lq() into Lq

    ().By solution of the Dirichlet problem (1.1), (1.2) we mean an element

    u W1,p0 () which satisfies

    (3.2) pu=

    iNfi

    u

    in W1,p

    (), or, equivalently,

    |u|p2uv=

    f(x, u)v, v W1,p0 (),

    where i is the compact imbedding of W1,p0 () into Lq() and i : Lq

    ()

    W1,p

    () is its adjoint. Consequently, (3.2) may be equivalently written as

    (3.3) Ju= Nfu

    (here, by Nfwe mean iNfi).

    We shall formulate sufficient conditions for equation (3.3) to admit a non-

    trivial solution, via Theorem 2.3. TakeX=W1,p0 () with RN, N2, a

    bounded domain with smooth boundary, 1 < p

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    18 Jenica Crnganu 8

    t 0, f : R R a Caratheodory function which satisfies the conditions

    below. The growth condition (3.1). There are > p and s0 > 0 such that

    (3.4) 0< F(x, s) sf(x, s), x , |s| s0.

    (3.5) lim sups0

    f(x, s)

    |s|p2s< 1 uniformly in x ,

    where 1= inf

    vp1,pvp0,p

    :v W1,p0 (), v= 0

    is the first eigenvalue ofp

    in W1,p0 ().

    SinceW1,p0 () is a reflexive and smooth Banach space, compactly imbed-ded in Lq(), hypothesis (H1) of Theorem 2.3 is satisfied. Since J =p :

    W1,p0 () W

    1,p() is continuous and satisfies condition (S+) (see, e.g., [9])hypothesis (H2) of Theorem 2.3 is satisfied, too. By the growth condition(3.1), hypothesis (H3) of Theorem 2.3 is also satisfied. Since > p, conditions(ii) and (iii) are satisfied. Finally, condition (3.5) implies condition (iv) inTheorem 2.3 (see, e.g., [9]).

    Consequently, Theorem 2.3 applies and gives the already known result(see, e.g., [9]) on the existence of a nontrivial solution for problem (1.1), (1.2).

    Example3.2. Consider the Neumann problempu+ |u|

    p2u= f(x, u) in ,(3.6)

    |u|p2u

    n= 0 on ,(3.7)

    where 1 < p < , RN, N 2, is a bounded domain with smoothboundary and f : R R is a Caratheodory function which satisfies thegrowth condition (3.1).

    By solution of the Neumann problem (3.6), (3.7) we mean an elementu W1,p() which satisfies

    (3.8) |u|

    p2

    uv+ |u|

    p2

    uv= f(x, u)v, v W

    1,p

    ().

    Assume X=W1,p() is endowed with the norm

    |||u|||p1,p= up0,p+ |u|

    p0,p, u W

    1,p(),

    which is equivalent to the standard norm on the space W1,p() (see [6]).

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    9 Existence results for operator equations involving duality mappings 19

    In this case, the duality mappingJon W1,p(), ||| |||1,pcorrespondingto the gauge function (t) =tp1 is defined (see [6]) by

    J:

    W1,p(), ||| |||1,p

    W1,p(), ||| |||1,p

    ,

    Ju= pu+ |u|p2u, u W1,p().

    It is easy to see that u W1,p() is a solution of problem (3.6), (3.7) in thesense of (3.8) if and only if

    (3.9) Ju= Nfu

    (here, byNfwe also meaniNfi, whereiis the compact imbedding of (W

    1,p(),

    ||| |||1,p) into Lq() and i :Lq

    ()

    W1,p(), ||| |||1,p

    is its adjoint).We shall formulate sufficient conditions for equation (3.9) to admit a

    nontrivial solution, via Theorem 2.3.Take X = W1,p() with RN, N 2, a bounded domain with

    smooth boundary, 1< p p and s0 > 0 such that

    (3.10) 0< F(x, s) sf(x, s), x , |s| s0.

    (3.11) lim sups0

    f(x, s)

    |s|p2s< 1 uniformly in x ,

    where1= inf

    |||v|||p1,p

    vp0,p:v W1,p(), v= 0

    .

    Since

    W1,p(), ||| |||1,p

    is a reflexive and smooth Banach space, com-pactly imbedded inLq() (see [6]), hypothesis (H1) of Theorem 2.3 is satisfied.Since J :

    W1,p(), ||| |||1,p

    W1,p(), ||| |||1,p

    is continuous and satisfies

    condition (S+) (see, e.g., [6]), hypothesis (H2) of Theorem 2.3 is satisfied, too.By the growth condition (3.1), hypothesis (H3) of Theorem 2.3 is also satisfied.Since > p, conditions (ii) and (iii) are satisfied. Finally, condition (3.11) im-plies condition (iv) in Theorem 2.3 (see, e.g., [6]). Consequently, Theorem 2.3applies and gives the already known result (see, e.g., [6]) on the existence of anontrivial solution for problem (3.6), (3.7).

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    Received 14 November 2008 University of Galat i

    Department of Mathematics

    800008 Galati, Romania

    [email protected]