SEMILINEAR BIHARMONIC PROBLEMS WITH A SINGULAR TERM.
MAYTE PÉREZ-LLANOS AND ANA PRIMO
Abstract. The aim of this work is to study the optimal exponent p to have solvability of
problem

u
2
p

 ∆ u = λ |x|4 + u + cf in Ω,
u > 0 in Ω,

 u = −∆u = 0 on ∂Ω,
where p > 1, λ > 0, c > 0, and Ω ⊂ RN , N > 4, is a smooth and bounded domain such that
0 ∈ Ω.
We consider f > 0 under some hypothesis that we will precise later.
1. Introduction
The main concern of this work is to determine a critical threshold exponent p to have solvability
of the following problem,
 2
u
p
in Ω,
 ∆ u = λ |x|4 + u + cf
(1)
u>0
in Ω,

u = −∆u = 0
on ∂Ω,
where p > 1, λ > 0, c > 0, and Ω ⊂ RN , N > 4, is a smooth and bounded domain such that 0 ∈ Ω.
There exists a large literature dealing with the case λ = 0. The differential equation ∆2 u = f
is called the Kirchhoff-Love model for the vertical deflection of a thin elastic plate. For a more
elaborate history of the biharmonic problem and the relation with elasticity from an engineering
point of view one may consult a survey of Meleshko [19]. In particular, the Kirchhoff-Love model
with Navier conditions, u = −∆u = 0 on ∂Ω, corresponds to the hinged plate model and allows
rewriting these fourth order problems as a second order system.
Among nonlinear problems for fourth order elliptic equations with λ = 0, there exist several
results concerning to semilinear equations with power type nonlinear sources (see for example [17]).
N +4
A crucial role is played by the critical (Sobolev) exponent, N
−4 , which appears whenever N > 4
(see [5]).
u
The case λ > 0 is quite different. The singular term
is related to the Hardy inequality:
|x|4
Date: January 24, 2014.
2010 Mathematics Subject Classification. 35J91 , 35B25.
Keywords. Semilinear biharmonic problems, Hardy potential, Navier conditions, existence and nonexistence
results.
Authors are supported by project MTM2010-18128, MEC, Spain.
1
2
MAYTE PÉREZ-LLANOS AND ANA PRIMO
Let u ∈ C 2 (Ω) and N > 4, then it holds that
Z
ΛN
Ω
u2
dx 6
|x|4
Z
|∆u|2 dx,
Ω
N 2 (N − 4)2 is optimal (see Theorem 2.2 below).
16
First of all, it is not difficult to show that any positive supersolution of (1) is unbounded near
the origin and then additional hypotheses on p are needed to ensure existence of solutions. We
will say that problem (1) blows-up completely if the solutions to the truncated problems (with the
λ
λ
weight
instead of the Hardy type term
) tend to infinity for every x ∈ Ω as n → ∞.
|x|4
|x|4 + n1
The main objective of this work is to explain the influence of the Hardy type term on the
existence or nonexistence of solutions and to determine the threshold exponent p+ (λ) to have a
complete blow-up phenomenon if p > p+ (λ).
The corresponding elliptic semilinear case with the Laplacian operator was studied in [8, 14],
where the authors show the existence of a critical exponent p∗ (λ) > 1 such that the problem has
no local distributional solution if p > p∗ (λ). Furthermore, they prove the existence of solutions
with p < p∗ (λ) under some suitable hypothesis on the datum.
In fourth order problems, Navier boundary conditions play an important role to prove existence
results. The problem can be rewritten as a second order system with Dirichlet boundary conditions. By classical elliptic theory, we easily prove a Maximum Principle. As a consequence, we
deduce a Comparison Principle that allows us to prove the existence of solutions for p < p+ (λ) as
a limit of approximated problems.
where ΛN =
The paper is organized as follows.
In Section 2 we briefly describe the natural functional framework for our problem and the
embeddings we will use throughout the paper.
Section 3 is devoted to some definitions and preliminary results. First, we describe the radial
solutions to the homogeneous problem that allow us to know the singularity of our supersolutions
near the origin and prove nonexistence results with local arguments. The notion of solution we
are going to consider in the nonexistence results is local in nature, we just ask the regularity
needed to give distributional sense to the equation. In this section, we prove a Moreau type
decomposition and a Picone inequality that will be used in the nonexistence proofs and which
are interesting themselves. However, to prove existence of solutions to (1) with L1 data, we will
consider the solution obtained as limit of approximations. For this, we use a comparison result
that immediately follows from Navier boundary conditions.
In Section 4 we prove the existence of a threshold exponent for existence of solutions, namely,
when we consider an exponent over the critical, there do not exist positive solutions even in the
sense of distributions. Furthermore, in Section 5, we prove that a complete blow-up phenomenon
occurs over the critical exponent.
Section 6 deals with the complementary interval of powers. In this range it is shown that, under
some suitable hypotheses on f , problem (1) has a positive solution.
SEMILINEAR BIHARMONIC PROBLEMS WITH A SINGULAR TERM.
3
2. Functional Framework
We briefly describe the natural framework to treat the solutions to the problem in consideration.
Let Ω ⊂ RN denote a bounded and smooth domain. We define the Sobolev space
W k,p (Ω) = {u ∈ Lp (Ω),
Dα u ∈ Lp (Ω)
for all 1 6 |α| 6 k},
endowed with the norm
kukW k,p (Ω)
(2)
 p1

Z
Z
p

=
|u| dx +
Ω
X
|Dα u|p dx ,
Ω 16|α|6k
or equivalently
X
kukW k,p (Ω) = kukp +
kDα ukp ,
16|α|6k
where k · kp denotes the usual norm in Lp (Ω) for 1 6 p < ∞.
Taking the closure of C0k (Ω) in W k,p (Ω) gives rise to the following Sobolev space W0k,p (Ω), with
the norm
 p1

Z
kukW k,p (Ω) = 
0
X
|Dα u|p dx ,
Ω 16|α|6k
equivalent to (2).
In fact, using interpolation theory one can get rid of the intermediate derivatives and find that
Z
p1
Z
p
k p
kukW k,p (Ω) =
|u| dx +
|D u| dx
,
Ω
Ω
defines a norm which is equivalent to (2), and similarly
Z
p1
(3)
kukW k,p (Ω) =
|Dk u|p dx
,
0
Ω
see for instance [2].
The following embeddings will be useful in the forthcoming arguments.
Theorem 2.1 (Rellich-Kondrachov’s Theorem). Let k ∈ N+ and 1 6 p < +∞. Suppose that
Ω ∈ RN is a Lipschitz domain. Then,
Np
W k+j,p (Ω) ,→ W j,p (Ω) and W k,p (Ω) ,→ Lq (Ω), for all 1 6 q 6
= p∗ ,
N − kp
with the convention that
Np
N −kp
= +∞ if kp > N . Moreover, this embedding is compact if q <
Np
N −kp .
The statement of the previous theorem also holds if we replace W m,p (Ω) with its proper subspace
W0m,p (Ω).
In particular, the natural space for our problem is
V = {u ∈ W 2,2 (Ω), such that u = 0, −∆u = 0 on ∂Ω} ⊂ W 2,2 (Ω) ∩ W01,2 (Ω).
4
MAYTE PÉREZ-LLANOS AND ANA PRIMO
Let us briefly discuss that for certain domains, H := W 2,2 (Ω) ∩ W01,2 (Ω) is a Hilbert space (so it
is V as a closed subset of W 2,2 (Ω) ∩ W01,2 (Ω)), endowed with the following scalar product
Z
(4)
hu, viH =
∆u∆v dx,
Ω
which induces the norm
(5)
kukH = k∆ukL2 (Ω) ,
equivalent to (3) with k = p = 2. Note that in (5) we ignore not only the intermediate derivatives,
but also some of the highest order derivatives in a larger space than W02,2 (Ω).
Trivially, for any u ∈ W 2,2 (Ω) ∩ W01,2 (Ω) it holds that
!2
2 X
2
N N N
X
1 X ∂2u
1
∂2u
∂2u
2 2
|D u| =
>
>
= (∆u)2 .
∂xi xj
∂xi xi
N i=1 ∂xi xi
N
i,j=1
i=1
To show the reverse inequality we refer to Theorem 2.2 in [3], where it is shown that under certain
conditions on the domain, in particular for smooth domains, there exists a positive constant,
independent of u, such that
kukW 2,2 (Ω) 6 Ck∆ukL2 (Ω) .
From Rellich-Kondrachov’s Theorem it follows that,
2N
= 2∗
W 2,2 (Ω) ,→ W 1,r (Ω) with r 6
N −2
(6)
2N
= 2∗ ,
W 2,2 (Ω) ,→ Lq (Ω) with q 6
N −4
where the embeddings are compact with strict inequalities on r and q. There exists an extensive
literature about this field. For more details and properties of higher order spaces and some applications we refer for example to the books [14, 17] and references therein. See also the classic books
k,p
[2, 18]. We denote by Wloc
(Ω) as the set of functions belonging to W k,p (Ω0 ), for any Ω0 ⊂⊂ Ω.
Furthermore, we need to understand deeply the behavior of the linear operator:
λ
∆2 (·) − 4 (·) : W 2,2 (Ω) ∩ W01,2 (Ω) −→ W0−2,2 (Ω)
|x|
N 2 (N − 4)2 It is well known that this operator is coercive if 0 6 λ < ΛN , noting by ΛN =
.
16
This coercivity is a direct consequence of the the optimality of the constant ΛN in the Rellich
inequality, stated in the next theorem. The proof of this result is due to Rellich in [25], but see
also [13] for alternative proofs. For convenience of the reader, we include here an elementary proof
of this inequality (based in a proof of Marı́a J. Esteban for the Laplacian operator, see also [20]).
Theorem 2.2. For any u ∈ W 2,2 (Ω) ∩ W01,2 (Ω) and N > 4, it holds that
Z
Z
u2
(7)
ΛN
dx 6
|∆u|2 dx,
4
Ω |x|
Ω
N 2 (N − 4)2 where ΛN =
is optimal.
16
SEMILINEAR BIHARMONIC PROBLEMS WITH A SINGULAR TERM.
5
Proof. We consider the following identities that follow from integrating by parts
2
Z
Z
Z
x
x u2
u∆u
2
2
=
∆u
(∆u)
dx
+
2λ
dx
0 6
+
λu
dx
+
λ
4
2
|x|
|x|3 L2 (Ω)
Ω
Ω |x|
Ω |x|
Z
Z
Z
Z
u2
u
(8)
= (∆u)2 dx + λ2
dx
−
2λ
h∇u,
∇
idx
=
(∆u)2 dx
4
2
|x|
|x|
Ω
Ω
Ω
Ω
Z
Z
Z
2
2
u
|∇u|
h∇u, xiu
+λ2
dx − 2λ
dx + 4λ
dx.
4
2
|x|
|x|
|x|4
Ω
Ω
Ω
Noticing that
N
N
X
X
∂u
∂(uxi )
xi =
− N u,
∂xi
∂xi
i=1
i=1
we can write
Z
Ω
h∇u, xiu
dx
|x|4
Z
h∇
=−
Ω
Z
=−
Ω
u
|x|4
Ω
u2
dx
|x|4
Ω
u2
dx.
|x|4
Z
, xiudx − N
h∇u, xiu
dx − (N − 4)
|x|4
Z
Hence,
Z
Ω
h∇u, xiu
(N − 4)
dx = −
|x|4
2
Z
Ω
u2
dx.
|x|4
On the other hand, thanks to Caffarelli-Kohn-Nirenberg inequalities (see [10]) with p = 2, γ = 1,
2 Z
Z
|∇u|2
N −4
u2
dx
>
dx.
2
4
2
Ω |x|
Ω |x|
Substituting these last two expressions in (8) we obtain
Z
Z
N (N − 4)
u2
0 6 (∆u)2 dx + λ2 + 2λ
dx.
4
4
Ω
Ω |x|
The polinomial p(λ) = −λ2 − 2λ N (N4−4) reaches its maximum at λ =
N (N −4)
4
and (7) follows. This constant ΛN is optimal but not achieved. For literature on this topic see for instance
[13, 16] and the references therein.
3. Some definitions and Preliminary results
For 0 6 λ < ΛN , let us consider the homogenous equation,
u
(9)
∆2 u − λ 4 = 0 in RN .
|x|
6
MAYTE PÉREZ-LLANOS AND ANA PRIMO
We denote by
α± (λ) =
N −4
2
±
1
2
β± (λ) =
N −4
2
±
1
2
q
p
(N 2 − 4N + 8) − 4 (N − 2)2 + λ
q
p
(N 2 − 4N + 8) + 4 (N − 2)2 + λ
(10)
the four real roots which give the radial solutions |x|−α± , |x|−β± to equation (9). It follows that
that β− (λ) 6 α− (λ) 6 α+ (λ) 6 β+ (λ).
N 2 (N − 4)2
,
For 0 6 λ < ΛN , then α− ∈ [0, N 2−4 ) and α+ ∈ ( N 2−4 , N − 4] and if λ = ΛN =
16
N −4
then α− = α+ =
. Moreover, notice that if 0 6 λ < ΛN , then β− 6 −2, and β+ > (N − 2).
2
<
β− (λ)
α− (λ)
β− (ΛN )
−2
0
α+ (λ)
><
N −4
2
β+ (λ)
>
(N − 4)
(N − 2)
β+ (ΛN )
Exponents of the radial solutions to the homogeneous problem.
We also remark that |x|−β− is the unique bounded solution and |x|−α− is the unique singular
solution that belongs to the Sobolev Space W 2,2 (Ω) for λ < ΛN .
We give the notion of solution we are going to consider in the nonexistence results.
Definition 3.1. We say that u ∈ L1loc (Ω) is a supersolution (subsolution) to
∆2 u = g(u) in Ω,
u = ∆u = 0 on ∂Ω,
in the sense of distributions if g ∈ L1loc (Ω) and for all positive φ ∈ C0∞ (Ω), we have that
Z
Z
∆u ∆φ dx > (6)
g(u) φ dx.
Ω
Ω
If u is a supersolution and subsolution in the sense of distributions, then we say that u is a
distributional solution to (1).
u
p
1
In particular, in problem (1), we consider g(u) = (λ |x|
4 + u + cf ) ∈ Lloc (Ω).
In the previous Definition 3.1 we just ask the regularity needed to give distributional sense to
the equation. As we will prove below, Remark 3.11, if u is a supersolution to (1) in the sense of
distributions, then g must satisfy a regularity condition. We will use this general framework to
prove nonexistence of positive solutions u satisfying −∆u > 0 in Ω.
We note here that from Navier boundary conditions, we can straightforward deduce a Strong
Maximum Principle and as a consequence, a comparison result.
Lemma 3.2. Strong Maximum Principle Let us consider u to be a nontrivial supersolution to
2
in Ω,
∆ u=0
(11)
u = −∆u = 0
on ∂Ω.
SEMILINEAR BIHARMONIC PROBLEMS WITH A SINGULAR TERM.
7
Then −∆u > and u > 0 in Ω.
Proof. Considering the change of variables −∆u = v, if u is a supersolution to (11), then v is a
supersolution to
−∆v = 0
in Ω,
(12)
v=0
on ∂Ω.
Applying the known Strong Maximum Principle to the Laplacian operator, it immediately follows
that v > 0 in Ω and then u > 0 in Ω.
Lemma 3.3. Comparison Principle Let

 ∆2 u
u

−∆u
u and v satisfy the following
>
>
>
∆2 v in Ω,
v on ∂Ω,
−∆v on ∂Ω.
Then, −∆v 6 −∆u and v 6 u in Ω.
Proof. It is sufficient to apply to w = u − v, a supersolution to (11), the previous Strong Maximum
Principle.
Notice that from the associated system (12), we can also easily obtain a Weak Harnack’s type
inequality.
Lemma 3.4. Weak Harnack inequality Let u be a positive distributional supersolution to (11),
then for any BR (x0 ) ⊂⊂ Ω, there exists a positive constant C = C(θ, ρ, q, R), 0 < q < NN−2 ,
0 < θ < ρ < 1, such that
kukLq (BρR (x0 )) 6 C ess inf u.
BθR (x0 )
Related to comparison of solutions we have the following decomposition due to Moreau, see [21].
For details of the proof considering more general operators, we refer for instance to [15, 17].
However, for completeness, we perform the proof in the case of the bilaplacian operator in the
Hilbert Space H = W 2,2 (Ω) ∩ W01,2 (Ω), with the scalar product defined in (4), see [14, 17].
Proposition 3.5. Let u ∈ H.RThen, there exist unique u1 , u2 ∈ H such that u = u1 +u2 , satisfying
u1 > 0, and u2 6 0 in Ω and Ω ∆u1 ∆u2 dx = 0.
Proof. The arguments of this proof are based on the fact that W 2,2 (Ω) ∩ W01,2 (Ω) is a Hilbert
Space endowed with the following scalar product
Z
hu, viH =
∆u∆v dx.
Ω
Then, if we consider the cone of the a.e positive functions defined in Ω,
C = {v ∈ H,
such that v(x) > 0 a.e in Ω}
the corresponding dual cone with respect to the scalar product above is defined as
Z
∗
C = {w ∈ H, such that
∆u∆wdx 6 0, for every u ∈ C}.
Ω
Let us take as u1 the orthogonal projection of u ∈ H on C, namely, let u1 be such that
Z
Z
|∆(u − u1 )|2 dx = min
|∆(u − ν)|2 dx.
Ω
ν∈C
Ω
8
MAYTE PÉREZ-LLANOS AND ANA PRIMO
Letting u2 = u − u1 , for all t > 0 and ν ∈ C, it holds that
Z
Z
|∆(u − u1 )|2 dx 6
|∆(u − (u1 − tν))|2 dx
Ω
Ω
Z
Z
Z
=
|∆(u − u1 )|2 dx − 2t
∆(u − u1 )∆νdx + t2
|∆ν|2 dx.
Ω
Ω
Ω
Therefore,
Z
(13)
∆(u − u1 )∆νdx 6 t
2t
2
Z
BR (0)
|∆ν|2 dx.
BR (0)
Choosing t > 0, simplifying and making then t → 0, we obtain that
Z
∆u2 ∆νdx 6 0, for any ν ∈ C, hence u2 ∈ C ∗ .
Ω
In particular, we can put ν = u1 , thus
Z
∆u1 ∆u2 dx 6 0.
Ω
Arguing analogously, for some t ∈ [−1, 0) in (13), and letting t tend to 0, we get the reverse
inequality, and hence
Z
∆u1 ∆u2 dx = 0.
Ω
Let us show next the uniqueness. Assume that u = u1 + u2 = v1 + v2 with u1 , v1 ∈ C and
u2 , v2 ∈ C ∗ . Then
0
= hu1 − v1 + u2 − v2 , u1 − v1 + u2 − v2 iH
= ku1 − v1 kH + ku2 − v2 kH + 2hu1 − v1 , u2 − v2 iH
= ku1 − v1 kH + ku2 − v2 kH − 2hu1 , v2 iH − 2hv1 , u2 iH
> ku1 − v1 kH + ku2 − v2 kH .
This implies that u1 = v1 and u2 = v2 as desired.
To conclude the proof we show that every function w ∈ C ∗ is non positive and, in particular,
u2 6 0. For every arbitrary nonnegative h ∈ C0∞ (Ω), consider the solution to the following problem
2
∆ v=h
in Ω,
v = 0, −∆v = 0 on ∂Ω.
By the Maximum Principle v ∈ C. But then,
Z
Z
0>
∆v∆wdx =
hw dx,
Ω
for every nonnegative function h ∈
we wanted to prove.
C0∞ (Ω).
Ω
By density, we conclude that w(x) 6 0 a. e. x ∈ Ω as
To show existence of solutions to (1) with L1 data, we will consider the solution obtained as
limit of approximations (see [6, 7, 11, 23] for the Laplacian operator case).
SEMILINEAR BIHARMONIC PROBLEMS WITH A SINGULAR TERM.
9
We denote
(
s,
Tk (s) =
k sign(s),
|s| 6 k
|s| > k,
the usual truncation operator. Consider f ∈ L1 (Ω), f > 0, and let {fn } be a sequence of functions
such that, fn (x) 6 f (x), x ∈ Ω, and fn → f in L1 (Ω).
Let un ∈ W 2,2 (Ω) ∩ W01,2 (Ω) ∩ L∞ (Ω) be the weak solution to problem,
 2
in Ω,
 ∆ u n = fn
un > 0
in Ω,
(14)

un = ∆un = 0 on ∂Ω.
Then un verifies
Z
Z
(−∆un )(−∆φ) dx =
Ω
fn φ dx,
∀φ ∈ W 2,2 (Ω) ∩ W01,2 (Ω).
Ω
Noting by −∆un = vn such that −∆vn = fn , we obtain that {vn } is bounded in W01,q (Ω), for
N
. Furthermore, for all k > 0, {Tk (vn )} is bounded in W01,2 (Ω) (see [6]).
any 1 6 q <
N −1
N
, such that
Consequently, up a subsequence, there exists v ∈ W01,q (Ω), for all 1 6 q <
N −1
N
vn * v, weakly in W01,q (Ω), ∀q ∈ [1,
).
N −1
With these previous properties one can prove that ∇vn → ∇v a.e. in Ω. Therefore by Fatou’s
N
and Vitali’s Theorems it follows that vn → v strongly in W01,q (Ω), ∀q ∈ [1,
) and Tk (vn ) →
N −1
1,2
Tk (v) strongly in W0 (Ω), for all k > 0. Hence v is a positive solution of the problem −∆v = f .
Taking into account that −∆un = vn and applying Embedding Theorem 2.1, we conclude that
there exists u ∈ W 2,q (Ω), for all 1 6 q < NN−1 , with v = −∆u, such that
N
).
N −1
By the Strong Maximum Principle for the truncated problems (14), we find that for f > 0, then
un > 0 and u > 0. This shows that u is a positive solution of the problem
 2
in Ω,
 ∆ u=f
u>0
in Ω,

u = −∆u = 0 on ∂Ω.
un → u strongly in W 2,q (Ω),
∀q ∈ [1,
The positive supersolution obtained above allows us to construct a positive solution to (1). This
is the core of the following lemma.
Lemma 3.6. Let ū ∈ L1loc (Ω̃) be a positive distributional supersolution to (1) with λ 6 ΛN ,
f ∈ L1 (Ω), f > 0, and Ω̃ ⊃⊃ Ω, such that −∆ū > 0 in the distributional sense. Then, there exists
a positive minimal solution v ∈ L1 (Ω) ∩ Lp (Ω) to problem (1) obtained as limit of approximations.
Proof. Let ū be a positive supersolution to (1) with λ 6 ΛN . For fn = Tn (f ), we construct
recursively a sequence
{vn } ∈ L1 (Ω) ∩ Lp (Ω),
10
MAYTE PÉREZ-LLANOS AND ANA PRIMO
starting with
 2
in Ω,
 ∆ v1 = f 1
v1 > 0
in Ω,

v1 = −∆v1 = 0 on ∂Ω.
By the comparison principle in Lemma 3.3, it follows that v1 6 ū in Ω. By iteration, we define for
n > 1,

vn−1
p

+ fn−1
in Ω,
∆2 vn = λ 4 1 + vn−1


|x| + n
(15)
vn > 0
in Ω,



vn = −∆vn = 0
on ∂Ω.
As above, it follows that 0 6 v1 6 . . . 6 vn−1 6 vn 6 ū
v(x) = limn→∞ vn (x), which verifies that 0 6 v 6 ū and

v
p
2

 ∆ v =λ 4 +v +f

|x|
v>0



v = −∆v = 0
in Ω, so we obtain the pointwise limit,
in Ω,
in Ω,
on ∂Ω.
Moreover, v has the regularity of a solution obtained as limit of approximations to (1) in Ω.
We state some inequalities we will use along the paper. We start formulating an extension of a
well-known Picone identity, that in the case of regular functions and the Laplacian operator was
obtained by Picone in [24] (see also [4] and [1] for an extension to positive Radon measures and
the p-Laplacian with p > 1).
Lemma 3.7. Let u, v ∈ C 2 (Ω) satisfy v > 0 and −∆v > 0 in Ω. The following functionals,
2
u
L(u, v) =
∆u − ∆v ,
v
2
u
2
R(u, v) = |∆u| − ∆v∆
,
v
verify that R(u, v) > L(u, v) > 0 in Ω.
Proof. Notice that,
u2
R(u, v) = |∆u| − ∆v∆
v
u
u2
u
u
u
2
2
= |∆u| − 2 ∆u∆v + 2 |∆v| − 2∆v∇
∇u + 2 ∆v∇
∇v
v
v
v
v
v
2
2
∆v
u
u
∇u − ∇v > L(u, v) > 0.
=
∆u − ∆v − 2
v
v
v
2
Lemma 3.8. Let v ∈ W 2,2 (Ω) be such that v > δ > 0 in Ω. Then for all u ∈ C0∞ (Ω),
Z
Z
∆2 v 2
|∆u|2 dx >
u dx.
Ω
Ω v
SEMILINEAR BIHARMONIC PROBLEMS WITH A SINGULAR TERM.
11
Proof. Since v ∈ W 2,2 (Ω) and v > δ > 0 in Ω, there exists a family of regular functions vn such
that
δ
vn → v in W 2,2 (Ω), vn ∈ C 2 (Ω), vn → v, a.e, and vn > in Ω.
2
Furthermore, ∆2 is a continuous operator from W 2,2 (Ω) to W −2,2 (Ω), see [17]. Hence ∆2 vn →
∆2 v in W −2,2 (Ω). By the previous lemma applied to vn , we deduce
2
u
(16)
|∆u|2 > ∆vn ∆
.
vn
Thanks to (16), integrating by parts,
2
Z
Z
Z
∆2 vn 2
u
u dx =
dx 6
|∆u|2 dx.
∆vn ∆
vn
Ω vn
Ω
Ω
Using the hypothesis on vn and Lebesgue dominated convergence theorem, we obtain
Z
Z
∆2 v 2
|∆u|2 dx, u ∈ C0∞ (Ω),
u dx 6
Ω v
Ω
and the result follows.
2,2
(Ω) ∩ W01,2 (Ω)
2
Theorem 3.9. Picone’s inequality. Consider u, v ∈ W
such that ∆ v = µ, where
µ is a positive bounded Radon measure, v = ∆v = 0 on ∂Ω and v 0. Then,
Z
Z
∆2 v 2
|∆u|2 dx >
u dx.
Ω
Ω v
1
Proof. The Strong Maximum principle yields that v > 0 in Ω. Set v̂m (x) = v(x) + m
, m ∈ N.
Note that ∆2 v̂m = ∆2 v and v̂m → v in W 2,2 (Ω) and almost everywhere. The general result
follows from the previous lemma and a density argument. Namely, let un ∈ C0∞ (Ω) be such that
un → u in W 2,2 (Ω) ∩ W01,2 (Ω). By Lemma 3.8, it holds that
Z
Z
Z
∆2 v 2
∆2 v̂m 2
un dx =
un dx 6
|∆un |2 dx.
Ω v̂m
Ω v̂m
Ω
Taking limits m → ∞, n → ∞ and applying Fatou’s Lemma we obtain the desired result.
In the following result we find some estimates on the supersolutions to our problem (1). They
will be used in the next section to show nonexistence of solutions when p is above the threshold
p+ (λ).
Lemma 3.10. Assume that u is a nonnegative function defined in Ω such that u 6≡ 0, u ∈ L1loc (Ω)
u
u
∈ L1loc (Ω). If u satisfies −∆u > 0, ∆2 u − λ 4 > 0 in the sense of distributions, then
and
|x|4
|x|
there exists a small ball Bη (0) ⊂ Ω such that
u > ĉ|x|−α−
∗
and
− ∆u > c∗ |x|−(α− +2)
in Bη (0),
∗
where ĉ = ĉ(η), c = c (η) and α− is defined in (10).
Proof. Since u is a nonnegative function satisfying −∆u > 0 in the sense of distributions and
λ > 0, using the Strong Maximum Principle 3.2, it is not difficult to obtain that u > δ > 0 and
−∆u > ν > 0 in a small ball Bη (Ω).
12
MAYTE PÉREZ-LLANOS AND ANA PRIMO
Fixed η > 0, in order to get the above estimate, let v ∈ W 2,2 (Bη (0)) be the unique solution to:

v
2


 ∆ v − λ |x|4 = 0
v>0



v = δ, −∆v = ν,
(17)
in Bη (0),
in Bη (0),
on ∂Bη (0).
By a direct computation we obtain that v(r) = C1 r−β− + C2 r−α− with C1 and C2 satisfying
(
C1 η −β− + C2 η −α− = δ,
C1 β− (N − 2 − β− )η −β− −2 + C2 α− (N − 2 − α− )η −α− −2 = ν.
Since u is a positive supersolution to problem (17) and using a Comparison Principle, we conclude
that u > v and −∆u > −∆v in Bη (0), so u > ĉ|x|−α− in Bη (0), and −∆u > c∗ |x|−(α− +2) in Bη (0).
We search a solution u = A|x|−α of the associated radial elliptic problem in BR (0),
∆2 u − λ
u
= up in BR (0).
|x|4
Hence by a direct computation, we obtain that
|x|−α−4 α4 − 2α3 (N − 4) + (N 2 − 10N + 20)α2 + 2(N − 2)(N − 4)α − λ = Ap−1 |x|−pα in BR (0).
Therefore,
α=
4
and Ap−1 = α4 − 2α3 (N − 4) + (N 2 − 10N + 20)α2 + 2(N − 2)(N − 4)α − λ.
p−1
It is clear that α4 − 2α3 (N − 4) + (N 2 − 10N + 20)α2 + 2(N − 2)(N − 4)α − λ > 0 if and only if:
• α− < α < α+ , which means p− (λ) < p < p+ (λ) with p+ (λ) = 1+
4
4
and p− (λ) = 1+
.
α−
α+
• α > β+ which implies p < p− (λ).
We will see that if we perturb the bilaplacian operator with the Hardy potential, the critical power
is p+ (λ).
Some properties of p− (λ) and p+ (λ) are,
p+ (λ) →
N +4
as λ → ΛN ,
N −4
p− (λ) →
N +4
as λ → ΛN ,
N −4
p+ (λ) → ∞ as λ → 0,
p− (λ) →
N
N −4
as λ → 0.
SEMILINEAR BIHARMONIC PROBLEMS WITH A SINGULAR TERM.
13
6
p+ (λ)
p− (λ)
λ
ΛN
-
It is clear that p+ (λ) and p− (λ) are respectively decreasing and increasing functions on λ and
then,
1 < p− (λ) 6 2∗ − 1 6 p+ (λ).
2N
= 2∗ .
Recall that W 2,2 (Ω) ,→ Lq (Ω) with q 6
N −4
Remark 3.11. Notice that if w satisfies
w
∆2 w − λ 4 > g, −∆w > 0, in the sense of distributions in Ω̃,
|x|
with g ∈ L1loc (Ω̃), g(x) > 0, Ω̃ ⊃⊃ Ω, and λ 6 ΛN , then g must satisfy g|x|−α− ∈ L1loc (Ω).
Indeed, we can construct a sequence {ϕn } as follows. Take ϕ1 such that ∆2 ϕ1 = 1 in Ω,
ϕ1 > 0 in Ω and ϕ1 = ∆ϕ1 = 0 on ∂Ω. The rest of the sequence is determined by

ϕn

∆2 ϕn+1 = λ 4 1 + 1
in Ω,


|x| + n

ϕn+1 > 0
in Ω,




ϕn+1 = −∆ϕn+1 = 0
on ∂Ω,
ϕ
such that ϕn 6 ϕn+1 6 ϕ, where ϕ is the positive solution to ∆2 ϕ − λ 4 = 1 in Ω, with
|x|
ϕ = ∆ϕ = 0 on ∂Ω. Thanks to Lemma 3.10 there exists a positive constant c and a small ball
BR (0) ⊂ Ω such that ϕ(x) > c|x|−α− in BR (0), where α− is defined in (10).
As in Lemma 3.6, we can construct a minimal solution to problem

w
2

 ∆ w − λ |x|4 = g in Ω,
w>0
in Ω,


w = −∆w = 0
on ∂Ω,
obtained as limit of approximations. Then we observe that
Z
Z
ϕn
2
∞>
wdx =
w ∆ ϕn+1 − λ 4 1 dx
|x| + n
Ω
Ω
Z
Z
Z
w
2
>
ϕn ∆ w − λ 4 dx =
gϕn dx >
gϕn dx.
|x|
Ω
Ω
BR (0)
14
MAYTE PÉREZ-LLANOS AND ANA PRIMO
Therefore, {gϕn } is an increasing sequence uniformly bounded in L1loc (Ω). The result follows by
the Monotone Convergence Theorem and
Z
Z
c
|x|−α− g dx 6
gϕ dx < ∞.
BR (0)
BR (0)
Since we are considering positive solutions to problem (1), then by setting g = up + cf , we
obtain that
Z
(up + cf )|x|−α− dx < ∞ for all BR (0) ⊂⊂ Ω.
BR (0)
This necessary condition will be useful in the forthcoming arguments.
4. Nonexistence results
We devote this section to show that p+ (λ) is the threshold exponent for existence of solutions,
namely, when we consider an exponent over the critical, there do not exist positive solutions even
in the sense of distributions. To this end we argue ad contrarium, we assume that there exists a
positive supersolution to (1) and this will yield a contradiction with Hardy inequality.
Theorem 4.1. If λ > ΛN and p > 1, or 0 < λ 6 ΛN and p > p+ (λ), then the problem (1) has
no positive distributional supersolution u satisfying −∆u > 0 in the sense of distributions. In case
that f ≡ 0, the unique nonnegative distributional supersolution is u ≡ 0.
Proof. Without loss of generality, we can assume f ∈ L∞ (Ω). Arguing by contradiction, we suppose
that ũ is a positive supersolution satisfying −∆ũ > 0 in the sense of distributions. By Lemma 3.6,
the problem (1) has a minimal solution u obtained as a limit of solutions un of truncated problems
(15). We consider some different cases.
Case λ > ΛN .
It suffices to consider ũ as a positive distributional supersolution to the problem

v
v
2

in Ω,

 ∆ v − ΛN |x|4 = (λ − ΛN ) |x|4 + g



v>0
in Ω,
v = −∆v = 0
on ∂Ω,
where g(x) = v p + cf . By Remark 3.11, necessarily
N −4
ũ
(λ − ΛN,s ) 4 + g |x|− 2 ∈ L1 (Br (0))
|x|
for all Br (0) ⊂⊂ Ω. In particular, this implies
(λ − ΛN,s )
N −4
ũ
|x|− 2 ∈ L1 (Br (0)),
4
|x|
and hence, applying Lemma 3.10,
(λ − ΛN,s )|x|−N ∈ L1 (Br (0)),
which is a contradiction. Therefore, there does not exist a positive supersolution if λ > ΛN,s .
Case 0 < λ 6 ΛN and p > p+ (λ).
SEMILINEAR BIHARMONIC PROBLEMS WITH A SINGULAR TERM.
15
u
> 0 in D0 (Ω).
|x|4
Therefore, by Lemma 3.10, there exists a positive constant C and a small ball BR (0) ⊂ Ω such
that u(x) > C|x|−α− in BR (0), where α− is defined in (10).
Let us consider the positive solutions un to the truncated problems (15). Since un > 0 in Ω,
|φ|2
−∆un > 0 in Ω, using
with φ ∈ C0∞ (BR (0)) as a test function in the approximated problems
un
(15), and applying Picone’s inequality,
Z
Z
Z
upn−1 2
φ2
2
φ dx 6
|∆φ|2 dx.
∆ un dx 6
un
BR (0) un
BR (0)
BR (0)
Notice that the minimal positive solution u satisfies −∆u > 0 and ∆2 u − λ
Passing to the limit as n → ∞, by Fatou’s Lemma and thanks to Lemma 3.10, we deduce that
Z
Z
φ2
dx
6
|∆φ|2 dx.
C
(p−1)α−
|x|
BR (0)
BR (0)
Since p > p+ (λ), then (p − 1)α− > 4 and we obtain a contradiction with the Rellich inequality.
Case λ < ΛN and p = p+ (λ). This case is more delicate because we consider the threshold
exponent. In the sequel, the constant C may vary from line to line.
u
Since the mimimal positive solution u satisfies −∆u > 0 and ∆2 u − λ 4 > 0 in D0 (Ω), then
|x|
thanks to Lemma 3.10, there exists η small enough and ĉ = ĉ(η) such that u(x) > ĉ|x|−α− in
Bη (0), where α− is defined in (10). Without loss of generality, from now on, we can fix η < e−1 .
Next we define
b
−b
1
1
w = A|x|−a log
in Bη (0), with A = log
|x|
η
where a = α− , and b ∈ (0, 1) will be chosen below conveniently small. Since λ < ΛN , then
w ∈ W 2,2 (Bη (0)). By a direct but tedious computation, we get
b−4
b−3
w
1
1
∆2 w − λ 4 = Ab log |x|
|x|−a−4 C1 (a, b, N ) + Ab log |x|
|x|−a−4 C2 (a, b, N )
|x|
b−2
b−1
1
1
+Ab log |x|
|x|−a−4 C3 (a, b, N ) + Ab log |x|
|x|−a−4 C4 (a, b, N )
b
1
|x|−a−4 a4 − 2a3 (N − 4) + a2 (N 2 − 10N + 20) + 2a(N − 2)(N − 4) − λ .
+A log |x|
Notice that a4 − 2a3 (N − 4) + a2 (N 2 − 10N + 20) + 2a(N − 2)(N − 4) − λ = 0, thus the term of
b
order (log(1/|x|)) |x|−a−4 cancels. Then, since we are taking η < e−1 , the leader term has order
b−1
(log(1/|x|))
|x|−a−4 , so one can conclude that
b−1
w
1
(18)
∆2 w − λ 4 6 AbC log
|x|−a−4 in Bη (0),
|x|
|x|
for some C = C(a, b, N ) small if b 1. On the other hand, we observe that
bp+ (λ)
1
wp+ (λ) = Ap+ (λ) |x|−p+ (λ)a log
in Bη (0).
|x|
16
MAYTE PÉREZ-LLANOS AND ANA PRIMO
Consider the function
−p+ (λ)b
1
.
h(x) = A−(p+ (λ)−1) log
|x|
Since |x| < e−1 in Bη (0), we have that h > 0 and h ∈ L∞ (Bη (0)) being
−b
1
khk∞ = η α− (p+ (α)−1) log
.
η
Furthermore, h satisfies
hwp+ (λ) = A|x|−a−4 ,
and then thanks to (18),
(19)
∆2 w − λ
w
6 Cbhwp+ (λ) in Bη (0).
|x|4
Let us construct one supersolution of (19). Define u1 = c1 u.
Observe that w = η −α− on ∂Bη (0). It is easy to check that there exists C 0 > 0 such that
−∆w 6 C 0 η −(α+2) on ∂Bη (0). It immediately follows that u1 > c1 ĉw on ∂Bη (0) and −∆u1 >
c1 c∗
c∗
C 0 (−∆w) on ∂Bη (0). Noticing C2 = min{ĉ, C 0 }, it is sufficient to take c1 C2 > 1 ensuring that
u1 > w on ∂Bη (0) and −∆u1 > −∆w on ∂Bη (0).
1−p (λ)
We claim that u1 > w in Bη (0). Indeed, we can choose b sufficiently small such that c1 + >
bkhkL∞ (Bη (0)) , so that
u1
1−p (λ) p (λ)
p (λ)
in Bη (0).
∆2 u1 − λ 4 > c1 + u1+ > bhu1+
|x|
Define by $ = w − u1 . This function verifies
$
p (λ)
.
∆2 $ − λ 4 6 Cbh wp+ (λ) − u1+
|x|
Now we need to use Proposition 3.5, a type Moreau decomposition. To this end we define
θ = $ + ϕ, with ϕ satisfying
 2
∆ ϕ=0
in Bη (0),



ϕ>0
in Bη (0),
 −∆ϕ = ∆$ on ∂Bη (0),


ϕ = −$
on ∂Bη (0).
Since $ 6 0 on Bη (0) and −∆$ 6 0 on Bη (0), it follows that θ > $ and θ ∈ W 2,2 (Bη (0)) ∩
W01,2 (Bη (0)) satisfies
(
p (λ)
∆2 θ − λ |x|θ 4 6 Cbh wp+ (λ) − u1+
,
(20)
θ = −∆θ = 0
on ∂Bη (0).
It suffices to show that θ 6 0 in Bη (0).
Thanks to Proposition 3.5, we can decompose it as θ = v1 + v2 being v1 , v2 ∈ W 2,2 (Bη (0)) ∩
1,2
W0 (Bη (0)) such that v1 > 0 and v2 6 0. Taking v1 as a test function in (20) yields
Z
Z
Z
θv1
p (λ)
∆θ∆v1 dx − λ
dx 6
bh(wp+ (λ) − u1+ )v1 dx.
4
Bη (0)
Bη (0) |x|
Bη (0)
SEMILINEAR BIHARMONIC PROBLEMS WITH A SINGULAR TERM.
17
R
Taking into account that Bη (0) ∆v1 ∆v2 dx = 0, we deduce that
Z
Z
Z
θv1
p (λ)
|∆v1 |2 dx − λ
dx
6
bh(wp+ (λ) − u1+ )v1 dx.
4
|x|
Bη (0)
Bη (0)
Bη (0)
Using now that v2 6 0 and θ 6 v1 , together with Rellich inequality in the left hand side, it follows
that
Z
Z
p (λ)
2
(21)
|∆v1 | dx 6 cλ
bh(wp+ (λ) − u1+ )v1 dx,
Bη (0)
Bη (0)
ΛN
where cλ = ΛN
−λ > 0 (recall that λ < ΛN ). Then by convexity, the fact that $ 6 θ 6 v1 and
(21), we obtain that
Z
Z
|∆v1 |2 dx 6 cλ
p+ (λ)bhwp+ (λ)−1 v12 dx.
Bη (0)
Bη (0)
Notice that taking b enough small, there exists 0 < < 1 such that
hwp+ (λ)−1 6
ΛN
|x|−4 in Bη (0),
cλ p+ (λ)b
so applying again Rellich inequality, we get that
Z
Z
Z
|∆v1 |2 dx 6 cλ
p+ (λ)bhwp+ (λ)−1 v12 dx 6 Bη (0)
Bη (0)
|∆v1 |2 dx.
Bη (0)
It implies that kv1 kW 2,2 (Bη (0))∩W 1,2 (Bη (0)) = 0, hence v1 = 0, and then θ = v2 6 0 as we wanted to
0
show.
Once we have that u1 > w in Bη (0) we reach the desired contradiction just observing that, as
before,
Z
Z
p+ (λ)−1 2
u1
φ dx 6
|∆φ|2 dx.
Bη (0)
Bη (0)
But then
(p+ (λ)−1)b 2
Z
φ
1
dx
6
|∆φ|2 dx,
log
4
|x|
|x|
Bη (0)
Bη (0)
Z
which contradicts the optimality of the constant in Rellich inequality.
N +4
N −4
and p+ (ΛN ) =
. If ũ
2
N −4
is a supersolution to (1), then thanks to Lemma 3.10 and Remark 3.11, we have that
Z
Z
p
−α− (p+ (ΛN )+1)
C
|x|
dx 6
|x|−α− ũp+ (ΛN ) dx < ∞.
Case p = p+ (λ) and λ = ΛN . In this case we have that α− =
Br (0)
Br (0)
Since α− (p+ (ΛN ) + 1) = N , we reach the desired contradiction.
18
MAYTE PÉREZ-LLANOS AND ANA PRIMO
5. Complete blow up results
The nonexistence result obtained above for p > p+ (λ) is very strong in the sense that a complete
blow-up phenomenon occurs in two different senses.
a) If un is the solution to the approximated problem with p > p+ (λ), where the Hardy
potential is substituted by the bounded weight (|x|4 + n1 )−1 , then un (x) → ∞ as n → ∞.
b) If un is the solution to the approximated problem with p = pn < p+ (λ) and pn → p+ (λ)
as n → ∞, then un (x) → ∞ as n → ∞.
5.1. Blow up for the approximated problems when λ > ΛN and p > 1, or 0 < λ 6 ΛN
and p > p+ (λ). We get a blow-up behavior for the following approximated problems.
Theorem 5.1. Let un ∈ L1 (Ω) ∩ Lp (Ω) be a positive solution to the problem

upn

 ∆2 u n =
+ λan (x)un + c f
in Ω,
1 + n1 upn
(22)

 u = −∆u = 0
on ∂Ω,
n
n
1
|x|4 +
Then un (x0 ) → ∞, ∀x0 ∈ Ω.
with f 6= 0, and an (x) =
1
n
. We consider λ > ΛN and p > 1, or 0 < λ 6 ΛN and p > p+ (λ).
Proof. Without loss of generality, we can assume that f ∈ L∞ (Ω). The existence of a positive
solution to problem (22) follows using classical sub-supersolution arguments. Thanks to the monotonicity of the nonlinear term and the coefficient an we can assume the existence of minimal
solution un such that un 6 un+1 for all n > 1. Therefore to get the blow-up result we have just to
show the complete blow-up for the family of minimal solutions denoted by un .
Applying weak Harnack inequality in Lemma 3.4, we conclude that for any BR (x0 ) ⊂⊂ Ω, there
exists a positive constant C = C(θ, ρ, R), which may vary from line to line, with 0 < θ < ρ < 1,
such that
(23)
kun kL1 (BρR (x0 )) 6 C ess
inf
BθR (x0 )
un 6 Cun (x0 ) 6 C.
We claim that, in particular, there exists r > 0 and a positive constant C = C(r), such that
Z
un dx 6 C, uniformly in n ∈ N.
Br (0)
If 0 ∈ BρR (x0 ), the claim follows directly by (23) for any Br (0) ⊂⊂ BρR (x0 ). If 0 6∈ BρR (x0 ), let
us consider a smooth curve Γ ∈ Ω joining x0 with the origin. We define
1
r = min dist(x, ∂Ω),
2 x∈Γ
to ensure that Br (x) ⊂⊂ Ω, for every x ∈ Γ. Applying Weak Harnack inequality, we get
Z
(24)
un dx 6 Cun (x0 ) 6 C.
Br (x0 )
Therefore, thanks to the uniform integral estimate (24), we conclude that ∀D ⊂ Br (x0 ),
Z
Z
1
1
C
infD un 6
un (x) dx 6
un (x) dx 6
.
|D| D
|D| Br (x0 )
|D|
SEMILINEAR BIHARMONIC PROBLEMS WITH A SINGULAR TERM.
19
Now, we take x1 ∈ Br (x0 ) ∩ Γ and D1 = Br (x0 ) ∩ Bθr (x1 ), with 0 < θ < 1. Once more, Weak
Harnack inequality yields,
Z
C
un dx 6 C ess inf un 6 ess inf un 6
.
D1
|D1 |
Brθ (x1 )
Br (x1 )
Recursively, in a finite number of steps, we conclude the claim.
Furthermore, by the monotone convergence theorem, there exists u > 0 such that un ↑ u strongly
in L1 (Br (0)).
Let us take ϕ, the solution to the problem
2
∆ ϕ = χBr (0)
in Ω,
ϕ = −∆ϕ = 0
on ∂Ω,
as a test function in (22). Then we have
Z
Z
Z
Z
0
upn
an (x)un ϕ dx + c
C >
f ϕ dx.
un (x) dx =
1 p ϕ dx + λ
Ω
Ω
Br (0)
Ω 1 + n un
By the monotone convergence theorem and Fatou’s Lemma,
upn
1 + n1 upn
an (x)un
→ up in L1loc (Br (0)),
%
u
in L1loc (Br (0)).
|x|4
Thus it follows that u is a very weak supersolution to (1) in Br1 (0) ⊂⊂ Br (0), a contradiction with
Theorem 4.1.
5.2. Blow up when pn → p+ (λ). We prove now another strong blow-up result when the power
pn ↑ p+ (λ).
Theorem 5.2. Assume that pn satisfies pn < p+ (λ) and pn → p+ (λ) as n → ∞ and f 0. Let
un ∈ L1loc (Ω) be a very weak supersolution to the problem
(
un
∆2 un > λ 4 + upnn + f in Ω,
|x|
(25)
un = −∆un = 0 on ∂Ω.
Then un (x0 ) → ∞, ∀ x0 ∈ Ω.
Proof. Without loss of generality we can assume that f ∈ L∞ (Ω). Suppose by contradiction that
there exists a subsequence denoted by pn and a supersolution un such that for some point x0 ∈ Ω
4
we have un (x0 ) → C0 < ∞, ∀n ∈ N. It is not restrictive taking pn (λ) = 1 +
.
α− + n1
As before, thanks to the weak Harnack inequality, we can deduce the existence of r > 0 and a
positive constant C = C(r) such that
Z
un (x) dx 6 C, uniformly in n ∈ N.
Br (0)
If un ∈ L1loc (Ω) is a supersolution to problem (25), then it can be shown as in Lemma 3.6 that
there exists a minimal solution to (25) in Ω1 ⊂⊂ Ω with 0 ∈ Ω1 ⊂⊂ Ω obtained by approximation.
20
MAYTE PÉREZ-LLANOS AND ANA PRIMO
Let us denote by vn 6 un this minimal solution. Then vn solves
(
vn
∆2 vn = λ 4 + vnpn + f in Ω1 ,
|x|
(26)
vn = −∆vn = 0 on ∂Ω1 .
Take φ, the solution to the problem
∆2 φ = 1 in Ω1 ,
φ = −∆φ = 0 on ∂Ω1 ,
as a test function in (26). Since vn 6 un , there results that
Z
Z
C>
vn (x) dx =
gn (x)φdx,
Ω1
Ω1
vn
where gn (x) = λ 4 + vnpn + f . Thus,
|x|
Z
gn φ dx 6 C for all n.
Ω1
This implies that gn is uniformly bounded in L1loc (Ω1 ) and then gn * µ in the sense of measures.
Using the usual change of variables ∆2 vn = (−∆)(−∆vn ) = −∆wn in (26) we get,
(
vn
(−∆)wn = λ 4 + vnpn + f in Ω1 ,
|x|
(27)
vn = un = 0 on ∂Ω1 .
Now take Tk (wn ) · ϕ with ϕ ∈ C0∞ (Ω1 ) as a test function in (27). Invoking the previous boundedness, we get
Z
Z
Z
Z
|∇Tk (wn )|2 ϕdx +
Θk (wn )(−∆ϕ)dx =
gn (x)ϕTk (wn )dx 6 k
gn (x)ϕdx 6 C,
Ω1
Ω1
Z
where Θk (s) =
Ω1
Ω1
s
Tk (σ)dσ.
0
1,2
Hence, Tk (wn ) * Tk (w) weakly in Wloc
(Br (0)). Applying the renormalized theory, the Embedding Theorem 2.1, and the fact that −∆vn = wn , we conclude that there exists a non-negative
N
function v such that vn → v strongly in Lqloc (Br (0)), q <
.
N −3
∞
Let ψ ∈ C0 (Br (0)) be a nonnegative function. By Fatou Lemma it follows that
Z
Z
Z
Z
vψ
lim
gn (x)ψdx >
v p+ (λ) ψdx + λ
dx
+
f ψdx.
4
n→∞ B (0)
Br (0)
Br (0) |x|
Br (0)
r
Therefore, using ψ as a test function in (26) and passing to the limit as n → ∞, we conclude
that
Z
Z
Z
Z
vψ
v ∆2 ψ dx >
v p+ (λ) ψdx + λ
dx
+
f ψdx.
4
Br (0)
Br (0)
Br (0) |x|
Br (0)
Hence v is a positive distributional supersolution to (1) and then we reach a contradiction.
SEMILINEAR BIHARMONIC PROBLEMS WITH A SINGULAR TERM.
21
6. Existence of solutions: p < p+ (λ)
The goal of this section is to prove that, under some suitable hypotheses on f , problem (1) has
a positive solution, if we consider the complementary interval of powers, namely, 1 < p < p+ (λ).
For the existence result, we first analyze the case f ≡ 0.
We distinguish some cases:
(A) 0 < λ < ΛN .
N +4
and Ω is a bounded domain, there exists a positive solution to
N −4
problem (1) with f ≡ 0 using the classical Mountain Pass Theorem in the Sobolev
space W01,2 (Ω) ∩ W 2,2 (Ω) (see for example [5, 17, 22]).
N +4
< p < p+ (λ), there exists a positive solution in RN . By setting w(x) =
(II) If
N −4
4
A|x|−α with α =
and Ap−1 = α4 − 2α3 (N − 4) + (N 2 − 10N + 20)α2 + 2(N −
p−1
2)(N − 4)α − λ, then w is a supersolution to (1) in any domain Ω. It is clear that if
w
∈ L1loc (RN ) and
p− (λ) < p < p+ (λ), then Ap−1 > 0. Notice that w ∈ Lploc (RN ),
|x|4
the solution could be verified in distributional sense. This is the way in which the
critical powers p− (λ) and p+ (λ) appear .
N +4
(B) If λ = ΛN , then p+ (ΛN ) =
= p− (ΛN ). Define the Hilbert space H(Ω) as the
N −4
completion of C0∞ (Ω) with respect to the norm
Z
φ2
2
||φ|| = (|∆φ|2 − ΛN 4 )dx.
|x|
Ω
(I) If 1 < p <
Invoking the improved Hardy-Sobolev inequality (see for example [9, 16]), then classical
variational methods in the space H(Ω) allow us to prove the existence of a positive solution
w to problem (1) with f ≡ 0.
c0
Remark 6.1. In the presence of a source term f 0, if f (x) 6 |x|
4 with c0 > 0 and sufficiently
small, by similar arguments as above we can show the existence of a supersolution. Then, the
existence of a minimal solution to problem (1) follows for all p < p+ (λ).
Aknowledgements. The authors would like to thank I. Peral for the motivation to study this
problem and B. Abdellaoui for helpful comments and discussions.
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Dpto de Matemáticas, Universidad Autónoma de Madrid,
Campus Cantoblanco, 28049 Madrid, Spain.
E-mail addresses: [email protected], [email protected].