ANNALES
POLONICI MATHEMATICI
85.3 (2005)
Dirihlet problems without onvexity assumption
by
Aleksandra Orpel
(Šód¹)
Abstrat. We deal with the existene of solutions of the Dirihlet problem for sublinear and superlinear partial dierential inlusions onsidered as generalizations of the
EulerLagrange equation for a ertain integral funtional without onvexity assumption.
We develop a duality theory and variational priniples for this problem. As a onsequene
of the duality theory we give a numerial version of the variational priniples whih enables
approximation of the solution for our problem.
1. Introdution. Let us state our notations and hypotheses.
Rn having a loally
Lipshitz boundary. Assume that the funtions G : Ω × R → R and H :
Ω × Rn → R satisfy the Carathéodory ondition and H(y, ·) is Gateaux
dierentiable and onvex for a.e. y ∈ Ω . Suppose additionally that there
exist onstants b1 , b2 , b3 , b4 > 0, r > 1, q ≥ 3, q > n + 1 and funtions
k1 , k2 ∈ L1 (Ω, R), k3 , k4 ∈ Lq−1 (Ω, R) suh that
Hypothesis
(H).
Let
Ω
be a bounded domain in
b2 q
b1 r
|x| + k1 (y) ≤ G(y, x) ≤
|x| + k2 (y),
r
q
b3 2
b4 2
|z| + k3 (y) ≤ H(y, z) ≤
|z| + k4 (y)
2
2
n
for a.e. y ∈ Ω and all x ∈ R, z ∈ R .
d
d
n
Let Hz (y, z) =
dz1 H(y, z), . . . , dzn H(y, z) for z = [z1 , . . . , zn ] ∈ R and
let ∂x G(y, x) denote the subdierential of the funtion G(y, ·) for y ∈ Ω . We
shall onsider the Dirihlet problem for the partial dierential inlusion
(1.1)
− div Hz (y, ∇x(y)) ∈ ∂x G(y, x(y))
for a.e.
y ∈ Ω,
whih is a generalization of the membrane equation. We are looking for a
x ∈ W01,2 (Ω, R) of this problem suh that Hz (·, ∇x(·))
2
divergene that is an element of L (Ω, R).
nonzero weak solution
has a distributional
Mathematis Subjet Classiation : 35J20, 35J25.
Key words and phrases : Dirihlet problem, duality, variational priniple, Euler
2000
Lagrange equation.
[193℄
194
A. Orpel
There have been numerous papers onerning similar problems. If we
assume the dierentiability of
G
with respet to the seond variable, in-
lusion (1.1) beomes an ellipti partial dierential equation in divergene
form disussed e.g. in [5℄, where
G ∈ C(Ω × R),
x and Ω is a
the right-hand side is independent of
in [2℄, or in [4℄, where
bounded
n-dimensional
polyhedral domain. In [6℄ N. Grenon has proved the existene of a solution
x ∈ W01,p (Ω, R) ∩ L∞ (Ω, R), p > 1,
− div A(y, x, Dx) = H(y, x, Dx)
(1.2)
where
for the PDE
Ω
is an open set in
Rn , n ≥ 1.
in
Ω,
This follows from the existene of
a solution of an assoiated symmetrized semilinear problem. The basi assumptions in [6℄ are the following:
A : Ω × R × Rn → Rn and H : Ω × R × Rn → R
n
funtions suh that for a.e. y ∈ Ω , all x ∈ R and ξ ∈ R ,
(a)
(1.3)
|A(y, x, ξ)| ≤ β(|x|)|ξ|p−1 + b(y),
(1.4)
|H(y, x, ξ)| ≤ γ(|x|){|ξ|p + d(y)},
β, γ are positive and loally bounded, b
′
Lp (Ω, R), p′ = p/(p − 1), and d ∈ L1 (Ω, R);
(b) for a.e. y ∈ Ω and all x ∈ R,
where
hA(y, x, ξ) − A(y, x, ξ ′ ), ξ − ξ ′ i > 0
and there exists
α>0
for all
suh that for a.e.
α|ξ|p ≤ hA(y, x, ξ), ξi
is a positive element of
ξ, ξ ′ ∈ Rn
y∈Ω
and all
for all
are Carathéodory
suh that
ξ 6= ξ ′ ,
x ∈ R,
ξ ∈ Rn ,
ki , θi ∈ C(R+ , R+ ), nonnegative fi ∈
max{n/p, 1} < q ≤ ∞, i = 1, 2, with θi (0) > 0 suh that
α{k1 (x)|ξ|p + θ1 (x)f1 (y)}
for all x ≥ 0,
H(y, x, ξ) ≤
α{−k2 (−x)|ξ|p − θ2 (−x)f2 (y)} for all x ≤ 0,
() there are nondereasing
Lq (Ω, R+ ),
for
y∈Ω
and
ξ ∈ Rn .
Let us note that for
A(y, x, ξ) = Hz (y, ξ)
and
H(y, x, ξ) = Gx (y, x),
(1.2) gives (1.1). In spite of this fat, we annot use the results of [6℄. In
G satises the Carathéodory
Gx (y, ·) is not neessarily ontinuous. We also do not
assume any additional estimate on Gx (see (1.3), (1.4) and ()).
There are a lot of results onerning the ase when H has the speial form
H(y, z) = 12 |z|2 for y ∈ Ω and z ∈ Rn (see e.g., [7℄, [8℄, [14℄). In [17℄ and [13℄
the general ase desribed by hypothesis (H),
ondition only, so that
the existene of a lassial solution of (1.1) is disussed under the following
assumptions:
Gx (·, ·) ∈ C(Ω × R, R), Gx
satises an additional estimate on
Dirihlet problems without onvexity assumption
Ω × R and the following relation
and r ≥ 0 suh that for |x| ≥ r ,
between
G
and
Gx
195
holds: there exist
µ>0
0 < µG(y, x) ≤ xGx (y, x).
(1.5)
A ondition similar to (1.5) is also used in [3℄. Numerous papers onern
G
similar problems for
being a polynomial with respet to
x
(see [15℄, [19℄).
In many papers the right-hand side of the equation is ontinuous ([3℄, [10℄,
x (see [16℄). Here we point out that weaker
G (R ∋ x →
7 G(y, x) is not neessarily onvex and
[18℄) or onvex with respet to
assumptions made on
ontinuous) are still suient to onlude the existene of a solution for (1.1).
To this end (1.1) is onsidered as the generalized EulerLagrange equation
J
for the funtional
given by
\
J(x) = {H(y, ∇x(y)) − G(y, x(y))} dy.
(1.6)
Ω
We see that under hypothesis (H),
W01,q (Ω, R),
type or nd subsets
dual
JD
J
is not, in general, bounded on
so that we must look for ritial points of (1.6) of minmax
X
and
X d,
on whih the ation funtional
J
or its
is bounded. We shall apply another approah and hoose speial
sets over whih we will alulate the minimum of
ulty in this approah is to show that
X 6= ∅.
J
and
JD .
The main dif-
Of ourse, we have at our
disposal the Morse theory and its generalizations, saddle points theorems,
mountain pass theorems (see e.g. [10℄, [11℄, [13℄, [17℄, [15℄), but none of these
methods exhausts all ritial points of
J.
Moreover our assumptions are
not strong enough to use, for example, the Mountain Pass Theorem:
G
is
a Carathéodory funtion so it is not suiently smooth, we assume neither
G nor additional relations onerning Gx
1
onsequene, J is not C on a suiently large subset
onvexity of
G (see (1.5)), in
W01,q (Ω, R) and it
and
of
does not satisfy, in general, the (PS)-ondition. We shall develop a duality
theory that permits us to omit, in our proof of the existene of ritial points,
the deformation lemmas, the Ekeland variational priniple or PS type onditions. Our approah also enables a numerial haraterization of solutions
of our problem.
For
A>0
and
p1 ∈ Lq−1 (Ω, Rn )
let
X := {x ∈ W01,q−1 (Ω, R); div Hz (·, ∇x(·)) ∈ Lq (Ω, R)
′
and
with
kHz (·, ∇x(·)) − p1 kLq−1 (Ω,Rn ) ≤ A}
q ′ = q/(q − 1).
Remark
1. By the assumption
1,q−1
theorem, we have W0
(Ω, R)
⊂
n < q−1
Lq (Ω, R).
and the Sobolev embedding
196
A. Orpel
X0 ⊂ X with
x
e ∈ X0 suh that
Consider sets
there exists
\
the following property: for eah
x ∈ X0 ,
{hx(y), − div(Hz (y, ∇e
x(y)))i − G∗ (y, − div(Hz (y, ∇e
x(y))))} dy
(1.7)
Ω
=
\
G(y, x(y)) dy.
Ω
X0 exist. In Setion 5
y ∈ Ω , z ∈ Rn and
assumptions onerning G whih
What is laking is the fat that nonempty sets
we shall onsider (1.1) for
k ∈ C 1 (Ω, R),
H(y, z) =
1
2
2 k(y)|z| for
and formulate a sequene of
X0 . We will show that, in this ase, it is suient to
q−1 norm on
assume the onvexity of G and the boundedness of Gx in the L
1
a ball only. Sine J is C on the ball only and we have the loal estimate
on Gx , the existene result for (1.1) annot be derived from the Mountain
Pass Theorem. We also give an example of G satisfying all these assumptions.
yield a nonempty set
Throughout the paper we shall assume hypothesis (H) and
Hypothesis
(H1).
For any suh
X0 ,
There exists a nonempty set X0 satisfying (1.7).
we dene
X
to be the union of
solutions of problem (1.1) whih belong to
Remark
2. For all
X0
and the set of all
X.
x ∈ X,
∂x G(y, x(y)) 6= ∅
and
G(y, x(y)) = G∗∗ (y, x(y))
Ω.
a.e. on
Proof.
X,
This follows from the denition of
the properties of subdier-
ential and the Fenhel formula.
Let
X d := {p ∈ Lq−1 (Ω, Rn );
(1.8)
there exists
x∈X
suh that
p(y) = Hz (y, ∇x(y))
Sine
q−1>2
Remark
and
Ω
is bounded,
3. For every
x ∈ X,
for a.e.
y ∈ Ω}.
X d ⊂ L2 (Ω, Rn ).
there exists
− div p(y) ∈ ∂x G(y, x(y))
p ∈ Xd
for a.e.
satisfying
y ∈ Ω.
Proof. Fix x ∈ X . Then there exists xe ∈ X suh that (1.7) holds. Taking
p(y) = Hz (y, ∇e
x(y)) for a.e. y ∈ Ω we see that p ∈ X d ,
\
\
Ω
Ω
{hx(y), − div(p(y))i − G∗ (y, − div p(y))} dy =
and, in onsequene, the required relation is satised.
G(y, x(y)) dy,
Dirihlet problems without onvexity assumption
Remark
4. The denitions of
suh that for all
X
and
Xd
197
imply that there exists
M >0
X d,
p∈
kpkLq−1 (Ω,Rn ) ≤ M.
2. Duality. The aim of this setion is to develop a duality whih de-
sribes the onnetions between the ritial values of
the dual funtional
JD :
Xd
→R
\
J
and the inmum of
dened as follows:
JD (p) = {−H (y, p(y)) + G∗ (y, − div p(y))} dy,
(2.1)
∗
Ω
H ∗ (y, ·) and G∗ (y, ·) (y ∈ Ω )
G(y, ·), respetively.
where
and
denote the Fenhel onjugate of
To this end we need the perturbation
\
Jx : Lq (Ω, R) → R
of
J
H(y, ·)
given by
Jx (g) = {−H(y, ∇x(y)) + G(y, g(y) + x(y))} dy.
Ω
Jx (0) = −J(x) for all x ∈ X .
x ∈ X dene a onjugate Jx# : X d → R
It is lear that
For every
(2.2)
Jx# (p)
=
of
Jx
by
\
{hg(y), div p(y)i − G(y, g(y) + x(y)) + H(y, ∇x(y))} dy
sup
g∈Lq (Ω,R) Ω
\
= {G∗ (y, div p(y)) + H(y, ∇x(y)) − hx(y), div p(y)i} dy.
Ω
Now we show that for all
p ∈ X d,
sup (−Jx# (−p)) = −JD (p).
(2.3)
x∈X
X d . From (1.8) we obtain the existene of
p∈
p(·) = Hz (·, ∇x(·))
Indeed, x
\
a.e. on
Ω
{h∇x(y), p(y)i − H(y, ∇x(y))} dy =
(2.4)
Ω
so that
(2.5)
x∈X
satisfying
and, in onsequene,
\
H ∗ (y, p(y)) dy,
Ω
\
{h∇x(y), p(y)i − H(y, ∇x(y))} dy
\
Ω
≤ sup {h∇x(y), p(y)i − H(y, ∇x(y))} dy
x∈X Ω
≤
sup
\
v∈L2 (Ω,Rn ) Ω
=
{hv(y), p(y)i − H(y, v(y))} dy
\
\
Ω
Ω
H ∗ (y, p(y)) dy = {h∇x(y), p(y)i − H(y, ∇x(y))} dy.
198
A. Orpel
This implies
\
sup (−Jx# (−p)) = sup {h∇x(y), p(y)i−H(y, ∇x(y))−G∗ (y, − div p(y))} dy
x∈X
x∈X
\
Ω
= {H ∗ (y, p(y)) − G∗ (y, − div p(y))} dy = −JD (p),
Ω
as laimed.
We also need another relation:
sup (−Jx# (−p)) = −J(x)
(2.6)
p∈X d
for eah
x ∈ X.
To prove this, x
suh that for a.e.
x∈X
and use Remark 3 to nd
p ∈ Xd
y ∈ Ω,
− div p(y) ∈ ∂x G(y, x(y)),
and further
\
\
Ω
Ω
{hx(y), − div p(y)i − G∗ (y, − div p(y))} dy =
(2.7)
G(y, x(y)) dy.
By arguments similar to those in the proof of (2.5), we obtain
(2.8)
sup
p∈X d
\
{hx(y), − div p(y)i − G∗ (y, − div p(y))} dy
Ω
=
\
\
Ω
Ω
G∗∗ (y, x(y)) dy =
G(y, x(y)) dy,
where the last equality is due to Remark 2. By (2.8) and (2.2),
sup (−Jx# (−p))
p∈X d
= sup
p∈X d
\
\
{hx(y), − div p(y)i − G∗ (y, − div p(y)) − H(y, ∇x(y))}
Ω
= {−H(y, ∇x(y)) + G(y, x(y))} dy = −J(x).
Ω
Now we have the following duality priniple:
Theorem 2.1.
inf J(x) = inf JD (p).
x∈X
Proof.
p∈X d
From (2.6) and (2.3),
sup (−J(x)) = sup sup (−Jx# (−p)) = sup sup (−Jx# (−p)) = sup (−JD (p)),
x∈X
x∈X p∈X d
whih yields the assertion.
p∈X d x∈X
p∈X d
Dirihlet problems without onvexity assumption
199
3. Variational priniples. Now we use the duality from the previ-
ous setion to establish relations between the existene of minimizers of
JD
and
J.
We also give a variational priniple for minimizing sequenes
of both funtionals. This result enables numerial haraterization of minimizing sequenes of
JD
and approximation of its inmum. To this end we
need a kind of perturbation of
JD p :
L2 (Ω, Rn )
→R
JD .
For eah
p ∈ Xd
dene the funtional
as
\
JDp (h) = {H ∗ (y, p(y) + h(y)) − G∗ (y, − div p(y))} dy.
Ω
Assume that p ∈ X d is a minimizer of JD , i.e., JD (p) =
inf p∈X d JD (p). Then there exists x ∈ X whih is a minimizer of J ,
Theorem 3.1.
J(x) = inf J(x),
(3.1)
x∈X
and suh that ∇x ∈ ∂JDp (0). Moreover
(3.2)
Jx# (−p) + JDp (0) = 0,
(3.3)
Jx# (−p) − J(x) = 0.
Proof.
By (1.8) for some
\
x ∈ X,
{h∇x(y), p(y)i − H(y, ∇x(y))} dy =
Adding
T
Ω
\
H ∗ (y, p(y)) dy.
Ω
∗
Ω {−G (y, − div p(y))} dy to both sides we obtain (3.2).
After some alulation, we see that
∗
JD
(∇x) = Jx# (−p)
p
(where
∗
JD
p
denotes the Fenhel onjugate of
JDp ).
Thus, by (3.2) and the
∇x ∈ ∂JDp (0).
J : X → R. By Theorem 2.1, to
JD (p) ≥ J(x), and this follows from
properties of the subdierential, we have the inlusion
We now show that
x
is a minimizer of
prove (3.1), it is suient to show that
the equalities
(3.4)
JDp (0) = −JD (p),
(3.2) and (2.8):
JD (p) = Jx# (−p) ≥ inf (Jx# (−p))
= − sup
\
p∈X d
{hx(y), − div p(y)i − G∗ (y, − div p(y)) − H(y, ∇x(y))} dy
p∈X d Ω
\
= − {G(y, x(y)) − H(y, ∇x(y))} dy = J(x).
Ω
Finally, (3.3) is a simple onsequene of (3.2) and the fat that
−JD (p) = −J(x).
JDp (0) =
200
A. Orpel
If p ∈ X d satises JD (p) = inf p∈X d JD (p) then there
exists x ∈ X whih is a solution of the Dirihlet problem for (1.1):
Corollary 3.2.
− div[Hz (y, ∇x(y))] ∈ ∂x G(y, x(y))
(3.5)
for a.e. y ∈ Ω .
Proof. By Theorem 3.1 there exists x ∈ X for whih (3.2) and (3.3) hold.
\
Hene
{H ∗ (y, p(y)) + H(y, ∇x(y)) − h∇x(y), p(y)i} dy = 0
Ω
\
and
{G∗ (y, − div p(y)) + G(y, x(y)) − hx(y), − div p(y)i} dy = 0.
Ω
Using the properties of the Fenhel onjugate, we obtain, for a.e.
y ∈ Ω,
∗
H (y, p(y)) + H(y, ∇x(y)) − h∇x(y), p(y)i = 0,
∗
G (y, − div p(y)) + G(y, x(y)) − hx(y), − div p(y)i = 0
so that
p(y) = Hz (y, ∇x(y))
for a.e.
y ∈ Ω.
and
− div p(y) ∈ ∂x G(y, x(y))
This implies (3.5).
Now we prove a numerial version of the above variational priniple.
We give a result on minimizing sequenes that is analogous to the previous
theorem.
Theorem 3.3.
Xd
→R
suh that
Let {pn }n∈N ⊂ X d be a minimizing sequene for JD :
c := inf JD (pn ) > −∞.
(3.6)
n∈N
Then for any n ∈ N there exists xn ∈ X satisfying
(3.7)
∇xn ∈ ∂JDpn (0),
inf J(xn ) = inf J(x).
n∈N
x∈X
Moreover for all n ∈ N,
(3.8)
JDpn (0) + Jx#n (−pn ) = 0,
and for eah ε > 0, there exists n0 ∈ N suh that for all n > n0 ,
(3.9)
Jx#n (−pn ) − J(xn ) ≤ ε,
(3.10)
|JD (pn ) − J(xn )| ≤ ε.
Proof. As in the proof of Theorem 3.1, for any n ∈ N there exists
xn ∈ X satisfying (3.8) and ∇xn ∈ ∂JDpn (0). By Theorem 2.1, to prove
that {xn }n∈N ⊂ X is a minimizing sequene of J on X , it sues to show
Dirihlet problems without onvexity assumption
201
that
inf J(xn ) = c,
(3.11)
n∈N
beause from (3.11) and Theorem 2.1 it follows that
inf J(xn ) = c = inf JD (pn ) = inf JD (p) = inf J(x).
n∈N
n∈N
x∈X
p∈X d
It is lear that by Theorem 2.1 and (3.6), for all
n ∈ N,
J(xn ) ≥ c.
(3.12)
ε > 0. From (3.6) there exists n0 ∈ N suh that ε > JD (pn ) − c for all
n > n0 . Therefore, the equalities JDpn (0) = −JD (pn ), (3.8) and (2.6) imply
that for all n > n0 ,
Fix
c + ε > JD (pn ) = Jx#n (−pn ) ≥ inf (Jx#n (−p)) = J(xn ).
p∈X d
Thus, by (3.12),
c = inf n∈N J(xn ).
Conditions (3.9) and (3.10) are satised beause of the last assertion and
the fat that
Jx#n (−pn ) ≤ c + ε
for all
n > n0 .
As a onsequene of the previous theorem we obtain
Suppose that {pn }n∈N ⊂ X d is a minimizing sequene
for JD on
and inf n∈N JD (pn ) = c > −∞. Then there exists a minimizing
sequene {xn }n∈N ⊂ X for J with
Corollary 3.4.
Xd
Hz (y, ∇xn (y)) = pn (y)
(3.13)
for a.e. y ∈ Ω and every n ∈ N. Moreover
(3.14)
lim
n→∞
\
{G∗ (y, − div pn (y))+G(y, xn (y))+hdiv pn (y), xn (y)i} dy = 0.
Ω
4. The existene of solutions for the Dirihlet problem. This
setion is devoted to the existene of a solution of (1.1) whih is a minimizer
of
J.
First we all a relevant lemma from [12℄:
Let Ω ⊂ Rn and let F : Ω → R be a onvex and lower
semiontinuous funtion suh that for eah u ∈ Ω the following inequalities
hold :
Lemma 4.1.
−b ≤ F (u) ≤ a
1 q
|u| + c,
q
for some onstants a > 0, b, c ≥ 0, q > 1. Then for all v ∈ ∂F (u),
′
|v| ≤ (q ′ aq /q (|u| + b + c) + 1)q−1 ,
where q ′ = q/(q − 1).
202
A. Orpel
There exists x0 ∈ X suh that
Theorem 4.2.
− div(Hz (y, ∇x0 (y)) ∈ ∂x G(y, x0 (y))
for a.e. y ∈ Ω . Moreover x0 is a minimizer of J on X :
J(x0 ) = inf J(x).
x∈X
Proof.
By hypothesis (H) for every
\
p ∈ Xd
we obtain
(4.1) JD (p) = [−H ∗ (y, p(y)) + G∗ (y, − div p(y))] dy
≥
≥
where
β=
Ω
′
b1−q
2
q′
′
b1−q
2
q′
′
kdiv pkqLq′ (Ω,R) −
′
kdiv pkqLq′ (Ω,R) −
\
1
kpk2L2 (Ω,Rn ) + (k3 (y) − k2 (y)) dy
2b3
\
Ω
1
βM 2 + (k3 (y) − k2 (y)) dy,
2b3
Ω
[vol(Ω)]1−2/(q−1). This implies that
JD
is bounded below on
X d.
G and H we see
Pea = {p ∈ X d ; e
a ≥ JD (p)}
is not empty. Now we an hoose a minimizing sequene {pm }m∈N ⊂ Pe
a
d
for JD . By (4.1) the sequene {div pm }m∈N ⊂ X is bounded in the norm
k · kLq′ (Ω,R) . Moreover, by the denition of X d , {pm }m∈N is bounded in
Lq−1 (Ω, Rn ). Thus, passing to a subsequene if neessary, we dedue that
pm ⇁ p0 as m → ∞, where p0 ∈ Lq−1 (Ω, Rn ), and div pm ⇁ z as m → ∞,
q′
where z ∈ L (Ω, R) (⇁ denotes weak onvergene). So
Taking into aount the growth onditions imposed on
at one that for
e
a ∈ R
large enough the set
\
hp0 (y), ∇h(y)i dy = lim
m→∞
Ω
\
hpm (y), ∇h(y)i dy
Ω
= − lim
m→∞
for any
h∈
\
\
Ω
Ω
hdiv pm (y), h(y)i dy = − hz(y), h(y)i dy
C0∞ (Ω, R), hene
\
(hp0 (y), ∇h(y)i + hz(y), h(y)i) dy = 0
Ω
h ∈ C0∞ (Ω, R), and nally, by the EulerLagrange lemma, div p0 (y) =
z(y) for a.e. y ∈ Ω .
q−1 (Ω, Rn ); kp − zk
Let B(p1 ; A) = {z ∈ L
1
Lq−1 (Ω,Rn ) ≤ A}. Sine
q−1
{pm }m∈N ⊂ B(p1 ; A) ⊂ L (Ω, Rn ) and pm ⇁ p0 , and B(p1 , A) is weakly
q−1 (Ω, Rn ), we have
sequentially losed as a onvex, losed subset of L
for all
p0 ∈ B(p1 ; A).
(4.2)
Moreover, by (4.1),
{pm }m∈N
satises the assumptions of Corollary 3.4.
Therefore there exists some sequene
{xm }m∈N ⊂ X
minimizing
J
on
X
Dirihlet problems without onvexity assumption
and suh that for all
m ∈ N,
Hz (y, ∇xm (y)) = pm (y)
(4.3)
203
for a.e.
y ∈ Ω.
Taking into aount hypothesis (H), we an use the Fenhel formula and
rewrite (4.3) as follows:
∇xm (y) ∈ ∂p H ∗ (y, pm (y))
(4.4a)
for a.e.
y ∈ Ω,
∗
where ∂p H (y, p) is the subdierential of the funtion
Rn ∋ p 7→ H ∗ (y, p),
y ∈ Ω . By Lemma 4.1 and the boundedness of {pm }m∈N in Lq−1 (Ω, Rn ) we
q−1 (Ω, Rn ) and {x }
see that {∇xm }m∈N is bounded in L
m m∈N is bounded in
1,q−1
W0
(Ω, R). Thus, we may hoose a subsequene still denoted by {xm }m∈N
1,q−1
weakly onvergent to x0 ∈ W0
(Ω, R). Hene, by the RellihKondrashov
theorem,
lim kxm − x0 kLq (Ω,R) = 0.
m→∞
As
to
{xm }m∈N tends strongly to x0 in Lq (Ω, R) and {div pm }m∈N tends weakly
′
div p0 in Lq (Ω, R) we get
\
\
Ω
Ω
hdiv pm (y), xm (y)i dy = hdiv p0 (y), x0 (y)i dy
lim
(4.5)
m→∞
\
and
(4.6)
[G(y, xm (y)) + G∗ (y, − div pm (y))] dy
lim inf
m→∞
Ω
= lim
m→∞
\
\
G(y, xm (y)) dy + lim inf
m→∞
Ω
\
G∗ (y, − div pm (y)) dy
Ω
≥ [G(y, x0 (y)) + G∗ (y, − div p0 (y))] dy,
T
Ω
Lq (Ω, R) ∋ x 7→ Ω G(y, x(y)) dy and
T
′
Lq (Ω, R) ∋ z 7→ Ω G∗ (y, z(y)) dy . Combining
whih follows from the ontinuity of
weak lower semiontinuity of
(4.5), (4.6) and (3.14) we see that
\
{G∗ (y, − div p0 (y)) + G(y, x0 (y)) + hdiv p0 (y), x0 (y)i} dy ≤ 0.
(4.7)
Ω
Thus, by the properties of the Fenhel transform, we have equality in (4.7),
and, in onsequene, for a.e.
y ∈ Ω,
∗
G (y, − div p0 (y)) + G(y, x0 (y)) + hdiv p0 (y), x0 (y)i = 0.
Finally, we obtain
− div p0 (y) ∈ ∂x G(y, x0 (y))
(4.8)
for a.e.
y ∈ Ω.
Now using (4.3) we infer that
\
{H ∗ (y, pm (y)) + H(y, ∇xm (y)) − hpm (y), ∇xm (y)i} dy = 0.
Ω
204
A. Orpel
Analysis similar to that in the proof of (4.8) now shows that
p0 (y) = Hz (y, ∇x0 (y))
(4.9)
for a.e.
y ∈ Ω.
(4.8) and (4.9) imply
− div Hz (y, ∇x0 (y)) ∈ ∂x G(y, x0 (y))
From (4.2) and (4.9) we have
x0 ∈ X
for a.e.
y ∈ Ω.
and further, by the last equality,
x0 ∈ X .
\
To prove the last assertion, it is suient to note that
inf J(x) = lim inf J(xm ) = lim inf {H(y, ∇xm (y)) − G(y, xm (y))} dy
m→∞
x∈X
= lim inf
m→∞
≥
\
\
m→∞
Ω
H(y, ∇xm (y)) dy − lim
Ω
H(y, ∇x0 (y)) dy −
Ω
\
m→∞
\
G(y, xm (y)) dy
Ω
G(y, x0 (y)) dy = J(x0 ),
T
Ω
Lq (Ω, T
R) ∋ x 7→ Ω G(y, x(y)) dy , weak
2
n
lower semiontinuity of L (Ω, R ) ∋ z 7→
Ω H(y, z(y)) dy and the fats that
xm → x0 in Lq (Ω, R) and ∇xm ⇁ ∇x0 in L2 (Ω, Rn ) as m → ∞.
whih is due to the ontinuity of
Remark
5. If
0 ∈
/ Xd
(that is,
kp1 kLq−1 (Ω,Rn ) > A)
then the above
theorem gives the existene of a nonzero solution of (1.1).
5. Appliations. We shall apply our theory to derive the existene of
solutions of the Dirihlet problem for a ertain lass of partial dierential
equations.
Now we reall the relevant theorems from [5℄ and [1℄:
Let Ω be a C 1,1 domain in Rn . If f ∈ Lp (Ω, R) with
1 < p < ∞, and k ∈ C 1 (Ω, R), k 0 ≥ k(y) ≥ k0 > 0 for all y ∈ Ω , then the
Dirihlet problem
div(k(y)∇u(y)) = f (y) for a.e. y ∈ Ω ,
Theorem 5.1.
u ∈ W01,p (Ω, R),
has a unique solution u ∈ W 2,p (Ω, R).
1,1 domain in Rn , 1 < p < ∞ and k ∈
Theorem 5.2. Let Ω be a C
1
C (Ω, R) and k 0 ≥ k(y) ≥ k0 > 0 for any y ∈ Ω . Then there exists a
onstant ec (independent of u) suh that
(5.1)
for all u ∈
kukW 2,p (Ω,R) ≤ e
ckdiv(k∇u)kLp (Ω,R)
W01,p (Ω, R)
∩ W 2,p (Ω, R).
Let Ω be a bounded doand let p, m ≥ 1, k ≤ m, with pk > n. The following inequality
Theorem 5.3 (Sobolev Imbedding Theorem).
main in
Rn
Dirihlet problems without onvexity assumption
205
holds for u ∈ W0m,p (Ω, R) with a onstant c depending on m, k and p, but
not otherwise on Ω or on u:
max
sup |D α u(y)| ≤ ckukW0m,p (Ω,R) .
0≤|α|≤2−k y∈Ω
Assume that
q ∈ [3, ∞), Ω ∈ C 1,1 is a bounded domain in Rn , with n + 1 < q ;
k ∈ C 1 (Ω, R), k 0 ≥ k(y) ≥ k0 for all y ∈ Ω ;
G : Ω × R → R is dierentiable with respet to the seond variable on
R for a.e. y ∈ Ω ;
there exists z ∈ Lq−1 (Ω, R) with the following properties. Let z ∈
W 2,q−1 (Ω, R) denote the solution of the equation
− div(k(y)∇z(y)) = Gx (y, z(y)) for a.e. y ∈ Ω
(the existene of z follows from Theorem 5.1) and let e
c be the onstant
introdued in Theorem 5.2 for W01,q−1 (Ω, R) ∩ W 2,q−1 (Ω, R). Then
there exist onstants S1 > 0, b1 , b2 > 0, 1 < r < q and funtions
k1 , k2 ∈ L1 (Ω, R) suh that
(a) for all x ∈ R and a.e. y ∈ Ω ,
Theorem 5.4.
1.
2.
3.
4.
b2 q
b1 r
|x| + k1 (y) ≤ G(y, x) ≤
|x| + k2 (y),
r
q
(5.2)
(b)
()
(5.3)
B(z; e
cS1 ) ∋ x 7→
T
Ω
G(y, x(y)) dy
is onvex , where
B(z; e
cS1 ) := {u ∈ W02,q−1 (Ω, R); ku − zkW 2,q−1 (Ω,R) < e
cS1 },
0
for any x ∈ B(z; ecS1 ),
kGx (·, x(·)) − Gx (·, z(·))kLq−1(Ω,R) < S1 .
Then there exists a solution
x0 ∈ {u ∈ W01,q−1 (Ω, R); kk∇u − k∇zkLq−1 (Ω,Rn ) ≤ k 0 e
cS1 }
of the Dirihlet problem for the PDE
(5.4)
− div(k(y)∇x(y)) = Gx (y, x(y))
for a.e. y ∈ Ω
suh that div(k(·)∇x0(·)) ∈ L (Ω, R).
Proof. First we reall the denition of X :
q′
X = {x ∈ W01,q−1 (Ω, R); div(k∇x) ∈ Lq (Ω, R)
′
and
Let
cS1 }.
kk∇x − k∇zkLq−1 (Ω,Rn ) ≤ k 0 e
X0 := {x ∈ W02,q−1 (Ω, R); kdiv(k∇x) + Gx (·, z(·))kLq−1(Ω,R) < S1 }.
206
A. Orpel
We show the inlusion
have
X0 ⊂ B(z; e
cS1 ).
Fix
x ∈ X0 .
Using Theorem 5.2, we
kx − zkW 2,q−1 (Ω,R) ≤ e
ck div(k∇x − k∇z)kLq−1 (Ω,R)
(5.5)
0
so that
x ∈ B(z; e
cS1 ).
Now we prove that
exists
x ∈ X0
X0
≤e
ckdiv k∇x + Gx (·, z(·))kLq−1(Ω,R) < e
cS1 ,
has the following property: for every
x ∈ X0 , there
suh that
\
{hx(y), − div(k(y)∇x(y))i − G∗ (y, − div(k(y)∇x(y))} dy
(5.6)
Ω
=
\
G(y, x(y)) dy.
Ω
x ∈ X0 . Sine Gx (·, x(·)) ∈ Lq−1 (Ω, R), by Theorem 5.1 there
1,q−1
(Ω, R) ∩ W 2,q−1 (Ω, R) of the Dirihlet
unique solution x0 ∈ W0
To this end x
exists a
problem for the equation
− div(k(y)∇x0 (y)) = Gx (y, x(y))
(5.7)
Thus, by (5.3), (5.7) and the inlusion
a.e. on
X0 ⊂ B(z; e
cS1 ),
Ω.
we obtain
kdiv(k(y)∇x0 (y)) + Gx (·, z(·))kLq−1(Ω,R)
= kGx (·, x(·))) − Gx (·, z(·))kLq−1(Ω,R) < S1 .
This implies
x0 ∈ X 0
and, in onsequene,
and the onvexity of the funtional
φ(u) =
T
G(y, u(y)) dy
+∞
for
Ω
for
we obtain
x0 ∈ B(z; e
cS1 ).
Thus, by (5.7)
u ∈ B(z; e
cS1 ),
2,q−1
u ∈ W0
(Ω, R) \ B(z; e
cS1 ),
− div(k∇x0 ) ∈ ∂φ(x)
and the properties of the subdierential yield (5.6).
Moreover, from (5.5), for all
x ∈ X0 ,
we have
kk∇x − k∇zkLq−1 (Ω,Rn ) ≤ k 0 kx − zkW 2,q−1 (Ω,R) ≤ k 0 e
cS1 .
(5.8)
0
X0 6= ∅ (z ∈ X0 ), X0 ⊂ X and X0 has the required property.
we dene X to be the union of X0 and the set of all solutions of
d
(5.4) whih belongs to X . Let X be given by
Summarizing,
Now
problem
X d := {p ∈ Lq−1 (Ω, Rn );
there exists
x∈X
suh that
p(y) = k(y)∇x(y)
Now Theorem 4.2 yields the existene of a solution
with
′
div(k(·)∇x0 (·)) ∈ Lq (Ω, R),
of the PDE
div(k(y)∇x(y)) + Gx (y, x(y)) = 0.
for a.e.
x0 ∈
y ∈ Ω}.
W01,q−1 (Ω, R),
Dirihlet problems without onvexity assumption
Remark
6. It worth noting that if
kk∇zkLq−1 (Ω,Rn ) > k 0 e
cS1
207
then the
above theorem states the existene of a nonzero solution of (5.4).
Proof.
0 ∈ X,
If the solution desribed by Theorem 5.4 is the zero funtion, then
whih implies the estimate
kk∇zkLq (Ω,R) ≤ k 0 e
cS1 .
But it ontradits our assumption.
As a onsequene of Theorem 5.4 we obtain
Suppose that
q ∈ [3, ∞) and Ω ∈ C 1,1 is a bounded domain in Rn , with n + 1 < q ;
k ∈ C 1 (Ω, R), k 0 ≥ k(y) ≥ k0 for all y ∈ Ω ;
G : Ω × R → R is dierentiable with respet to the seond variable on
R for a.e. y ∈ Ω ;
there exist onstants b1 , b2 > 0, 1 < r < q and funtions k1 , k2 ∈
L1 (Ω, R) suh that for all x ∈ R and a.a. y ∈ Ω ,
Theorem 5.5.
1.
2.
3.
4.
b2 q
b1 r
|x| + k1 (y) ≤ G(y, x) ≤
|x| + k2 (y);
r
q
5.
is a subset of N suh that for eah i ∈ I there exists zi ∈ Lq−1 (Ω, R)
with the following properties :
i
(a) there exists S1 > 0 suh that
I
kGx (·, x(·)) − Gx (·, zi (·))kLq−1 (Ω,R) < S1i
for any x ∈ B(z i ; T
c)
S1i e
is onvex ,
where z i ∈ W02,q−1 (Ω, R) satises the equality
− div(k(y)∇z i(y)) = Gx (y, zi (y)) for a.e. y ∈ Ω
(the existene of z i follows from Theorem 5.1).
Then for all i ∈ I there exists a solution xi of the Dirihlet problem for the
PDE
− div(k(y)∇x(y)) = Gx (y, x(y)) for a.e. y ∈ Ω
(5.9)
′
suh that div(k(·)∇xi(·)) ∈ Lq (Ω, R) and xi ∈ Bi , where
(b)
B(z i ; S1i e
c ) ∋ x 7→
Ω
G(y, x(y)) dy
cS1i }.
Bi = {u ∈ W01,q−1 (Ω, R); kk∇u − k∇z i kLq−1 (Ω,Rn ) ≤ k 0 e
If we assume additionally that Bi ∩Bj = ∅ for all i, j ∈ I , i 6= j , then xi 6= xj
for all i, j ∈ I , i 6= j and , in onsequene , #S ≥ #I , where S denotes the
set of solutions for (5.9).
Now we shall give an expliit example of
Theorem 5.4.
G satisfying
the assumptions of
208
A. Orpel
Example 5.6.
1.
2.
3.
4.
Assume that
n = 4, q = 6, Ω ⊂ R4 ;
k ∈ C 1 (Ω, R), k 0 ≥ k(y) ≥ k0 for all y ∈ Ω ;
1
z is any element of C0 (Ω, R) suh that kz 5 −6z 3 + 8z + 1kL5 (Ω,R) < 10
;
∞
b > 0, a ∈ L (Ω, R) and
3
b < kakL∞ (Ω,R) < min 1, p
.
40 5 |Ω| + 5, 5
Then there exists
a nonzero solution
x0 ∈
u∈
W01,5 (Ω, R);
kk∇u − k∇zkL5 (Ω,Rn )
3
≤ k0
5(1 + c)
of the Dirihlet problem for the PDE
(5.10)
− div(k(y)∇x(y)) =
a(y)
((x(y))5 − 6(x(y))3 + 8x(y) + 1)
(1 + c)(1 + e
c)
for a.e. y ∈ Ω , suh that div(k(·)∇x0(·)) ∈ L6/5 (Ω, R), with c being the
Sobolev onstant (Theorem 5.3 for W02,5 (Ω, R) and k = 2) and ec desribed
in Theorem 5.2 for p = 5.
Proof. It is lear that the right-hand side of (5.10) is the derivative of
G:Ω×R→R
given by
a(y)
((x2 − 4)2 (x2 − 1) + 6x)
6(1 + c)(1 + e
c)
with respet to the seond variable. For a.e. y ∈ Ω the funtion G(y, ·) is
dierentiable on R and onvex in the interval (−7/10, 7/10). Moreover for
all x ∈ R and a.e. y ∈ Ω we have
a(y)
a(y)
(3x4 − 200) ≤ G(y, x) ≤
(26x6 + 36),
6(1 + c)(1 + e
c)
6(1 + c)(1 + e
c)
G(y, x) =
G satises the required growth onditions. Sine Gx (·, z(·)) ∈ L5 (Ω, R)
1,5
2,5 (Ω, R) whih is
Theorem 5.1 leads to the existene of z ∈ W0 (Ω, R) ∩ W
so that
a solution of the PDE
− div(k(y)∇u(y)) =
From 5.2 we know that
(5.11)
a(y)
((z(y))5 − 6(z(y))3 + 8z(y) + 1).
(1 + c)(1 + e
c)
kzkC0 (Ω,R) ≤ ckzkW 2,5 (Ω,R) ≤ ce
ckdiv(k(y)∇z(y))kL5(Ω,R)
≤ (1 + c)(1 + e
c )kdiv(k(y)∇z(y))kL5(Ω,R)
a(y)
5
3
(z − 6z + 8z + 1)
= (1 + c)(1 + e
c )
(1 + c)(1 + e
c)
L5 (Ω,R)
≤ kakL∞ (Ω,R) kz 5 − 6z 3 + 8z + 1kL5 (Ω,R)
1
≤ .
10
Dirihlet problems without onvexity assumption
209
Put
3
,
5(1+c)e
c
B(z; e
cS1 ) := u ∈ W02,5 (Ω, R); ku − zkW 2,5 (Ω,R) <
S1 :=
0
T
3
.
5(1+c)
B(z; e
cS1 ) ⊂ {u ∈ C0 (Ω, R); kukC0 < 7/10},
B(z; e
cS1 ).
shall show that for all x ∈ B(z; e
cS1 ),
3
kGx (·, x(·)) − Gx (·, z(·))kL5(Ω,R) <
.
5(1 + c)e
c
Using (5.11) we have the inlusion
so that
x 7→
Now we
Ω G(y, x(y)) dy
is onvex in
Indeed,
kGx (·, x(·)) − Gx (·, z(·))kL5(Ω,R)
≤
kakL∞ (Ω,R)
{(x(y))5 − 6(x(y))3 + 8x(y)kL5 (Ω,R)
(1 + c)e
c
+ k(z(y))5 − 6(z(y))3 + 8z(y)kL5 (Ω,R) }
5
3
kakL∞ (Ω,R) p
7
7
7
11
5
≤
|Ω|
+6
+8
+
(1 + c)e
c
10
10
10
10
kakL∞ (Ω,R) p
11
3
≤
8 5 |Ω| +
≤
,
(1 + c)e
c
10
5(1 + c)e
c
as laimed.
We have just shown that all the assumptions of Theorem 5.4 are satised.
Sine zero does not satisfy the equation below, there exists a nonzero solution
x0 ∈ W01,5 (Ω, R)
of
− div(k(y)∇x(y)) =
for a.e.
y∈Ω
with
a(y)
((x(y))5 − 6(x(y))3 + 8x(y) + 1)
(1 + c)(1 + e
c)
div(k(·)∇x0 (·)) ∈ L6/5 (Ω, R).
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Faulty of Mathematis
University of Šód¹
Banaha 22
90-238 Šód¹, Poland
E-mail: orpelamath.uni.lodz.pl
Reçu par la Rédation le 17.2.2004
Révisé le 21.2.2005
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