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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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(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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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