Badiale M., Serra E. Semilinear Elliptic Equations for Beginners: Existence Results via the Variational Approach
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3.3 A Serious Loss of Compactness 111
The case u
2
< |u|
p
p
+ λ|u|
r
r
cannot occur, since arguing as usual, and taking
t ∈(0, 1) such that tu∈N, we would have
m ≤ I(tu)
=t
2
1
2
−
1
r
u
2
+t
p
1
r
−
1
p
|u|
p
p
<
1
2
−
1
r
u
2
+
1
r
−
1
p
|u|
p
p
≤lim inf
k
I(u
k
) =m,
a contradiction.
We now turn to the case l =0. We are going to use some of the results of Sect. 3.1.
First we introduce in H
1
(R
N
) the norm
u
2
α
=
R
N
|∇u|
2
dx +α
R
N
|u|
2
dx,
where α is the positive number defined in (q
2
). Of course u
α
in an Hilbertian
norm equivalent to u. Then we define a functional
I
α
:H
1
(R
N
) →R as I
α
(u) =
1
2
u
2
α
−
1
p
|u|
p
p
−λ
1
r
|u|
r
r
,
and the associated Nehari manifold
N
α
=
u ∈H
1
(R
N
) |u =0,I
α
(u)u =0
=
u ∈H
1
(R
N
)
u =0, u
2
α
=|u|
p
p
+λ|u|
r
r
.
Finally we define
m
α
= inf
u∈N
α
I
α
(u).
The intuitive idea behind the introduction of this auxiliary problem is that if l =0,
then all the mass of u
k
escapes to infinity. But close to infinity q(x) “looks like α”,
which is sup
R
N
q: it should not be convenient, if u
k
has to minimize I ,tohaveits
mass located where q is large. Let us see that this is indeed true. First we compare
m and m
α
.
Lemma 3.3.7 There results m<m
α
.
Proof By the results of Sect. 3.1 we know that N
α
=∅, m
α
> 0 and that there exists
u
0
∈N
α
such that I
α
(u
0
) =m
α
. We also know that u
0
≥0.
By (q
2
) and (q
3
) we infer that there exist δ
1
> 0 and a ball B
R
(x
1
) such that
q(x) ≤α −δ
1
∀x ∈B
R
(x
1
).
Since u
0
does not vanish identically, there exist δ
2
> 0, a ball B
R
(x
2
) and a set
A ⊆B
R
(x
2
) of positive measure such that
u
0
(x) ≥δ
2
a.e. in A.
112 3 Minimization Techniques: Lack of Compactness
We now define a function u
1
∈H
1
(R
N
) as
u
1
(x) =u
0
(x −x
1
+x
2
).
By the invariance of integrals with respect to translations we have
I
α
(u
1
) =I
α
(u
0
) =m
α
and I
α
(u
1
)u
1
=I
α
(u
0
)u
0
=0.
Notice that if x ∈B
R
(x
1
), then x −x
1
+x
2
∈B
R
(x
2
); therefore u
1
(x) ≥δ
2
a.e. in a
set A
⊆B
R
(x
1
) of positive measure. Then
B
R
(x
1
)
(α −q(x))u
2
1
dx ≥
A
δ
1
δ
2
2
=Cδ
1
δ
2
2
,
where C is the measure of A
.
Since q(x) ≤α for every x, we obtain the estimate
R
N
(α −q(x))u
2
1
dx ≥
B
R
(x
1
)
(α −q(x))u
2
1
dx ≥Cδ
1
δ
2
2
> 0,
so that
R
N
q(x)u
2
1
dx < α
R
N
u
2
1
dx,
which implies
R
N
|∇u
1
|
2
dx +
R
N
q(x)u
2
1
dx <
R
N
|∇u
1
|
2
dx +α
R
N
u
2
1
dx,
that is,
u
1
2
< u
1
2
α
.
We then have
I
(u
1
)u
1
=u
1
2
−|u
1
|
p
p
−λ|u
1
|
r
r
< u
1
2
α
−|u
1
|
p
p
−λ|u
1
|
r
r
=I
α
(u
1
)u
1
=I
α
(u
0
)u
0
=0.
Hence, by usual arguments, there exists t ∈(0, 1) such that tu
1
∈N, so that
m ≤ I(tu
1
) =t
2
1
2
−
1
r
u
1
2
+t
p
1
r
−
1
p
|u
1
|
p
p
<
1
2
−
1
r
u
1
2
+
1
r
−
1
p
|u
1
|
p
p
<
1
2
−
1
r
u
1
2
α
+
1
r
−
1
p
|u
1
|
p
p
=
1
2
−
1
r
u
0
2
α
+
1
r
−
1
p
|u
0
|
p
p
=I
α
(u
0
) =m
α
.
Now the last step: if all the mass of u
k
escapes to infinity, then I(u
k
) is too large
(at least m
α
), and u
k
cannot be a minimizing sequence.
Lemma 3.3.8 The case l =0 cannot occur.
3.3 A Serious Loss of Compactness 113
Proof If l =0, then u =0, which implies in particular that u
k
→0inL
2
loc
(R
N
).We
first prove the following claim:
lim
k→+∞
R
N
|q(x) −α|u
2
k
dx =0. (3.7)
To prove (3.7)wefixε>0 and we take R
ε
> 0 such that
|q(x) −α|≤ε ∀|x|≥R
ε
;
this is possible by (q
3
). We can then estimate
R
N
|q(x) −α|u
2
k
dx =
|x|≤R
ε
|q(x) −α|u
2
k
dx +
|x|>R
ε
|q(x) −α|u
2
k
dx
≤C
|x|≤R
ε
u
2
k
dx +Mε,
where
C = max
x∈R
N
|q(x) −α| and M =sup
k
R
N
u
2
k
dx.
When k →∞we obtain
lim sup
k
R
N
|q(x) −α|u
2
k
dx ≤εM
for every ε>0, because u
k
→0inL
2
loc
(R
N
). As this holds for every ε>0, (3.7)is
proved.
From (3.7) we deduce that
lim
k→+∞
u
k
α
= lim
k→+∞
u
k
. (3.8)
Now we know that
u
k
2
α
≥u
k
2
=|u
k
|
p
p
+λ|u
k
|
r
r
,
so that, by usual arguments, we can prove that for each k there is t
k
≥ 1 such that
t
k
u
k
∈N
α
, namely
t
2
k
u
k
2
α
=t
p
k
|u
k
|
p
p
+λt
r
k
|u
k
|
r
r
. (3.9)
We know that the sequences {|u
k
|
p
}, {|u
k
|
r
} and {u
k
α
} are bounded and that
lim
k
|u
k
|
p
p
=β>0.
Therefore from (3.9) we deduce that {t
k
}
k
is bounded and, up to a subsequence, we
can assume t
k
→t
0
with, of course, t
0
≥1. We also know that
u
k
2
α
=u
k
2
+ε
k
,
with ε
k
→0, and that
λ|u
k
|
r
r
=u
k
2
−|u
k
|
p
p
.
114 3 Minimization Techniques: Lack of Compactness
Substituting in (3.9) we get
t
2
k
−t
r
k
u
k
2
+t
k
ε
k
=
t
p
k
−t
r
k
|u
k
|
p
p
.
Passing to the limit, and setting γ =lim
k
u
k
2
> 0, we obtain
t
2
0
−t
r
0
γ =
t
p
0
−t
r
0
β;
so t
0
> 1 gives a contradiction because 2 <r<p, and then necessarily t
0
=1. Now
we can write
m
α
≤I
α
(t
k
u
k
) =t
2
k
1
2
−
1
r
u
k
2
α
+t
p
k
1
r
−
1
p
|u
k
|
p
p
=t
2
k
1
2
−
1
r
u
k
2
+t
2
k
1
2
−
1
r
ε
k
+t
2
k
1
r
−
1
p
|u
k
|
p
p
+
t
p
k
−t
2
k
1
r
−
1
p
|u
k
|
p
p
=t
2
k
I(u
k
) +o(1).
Passing to the limit we then obtain
m
α
≤m,
a contradiction which concludes the proof.
As we anticipated, the hard case occurs when l ∈(0,β), because this means that
the weak limit u is not identically zero, namely that only part of the mass escapes to
infinity. A careful analysis will show that this is enough to conclude. We define
l
1
=|u|
r
r
.
We begin by choosing an increasing sequence {R
j
}
j
of positive numbers such
that R
j+1
>R
j
+ 1, so that in particular R
j
→+∞, and such that the following
properties hold:
• l −
1
j
≤
B
R
j
|u|
p
dx ≤l, hence also
B
c
R
j
|u|
p
dx ≤
1
j
,
• l
1
−
1
j
≤
B
R
j
|u|
r
dx ≤l
1
, hence also
B
c
R
j
|u|
r
dx ≤
1
j
,
•
B
c
R
j
u
2
dx ≤
1
j
.
Then we define
C
j
=B
R
j
+1
\B
R
j
={x ∈R
N
|R
j
≤|x| <R
j
+1}.
Since we have compactness on bounded subsets of R
N
, for every j there results
lim
k
B
R
j
|u
k
|
p
dx =
B
R
j
|u|
p
dx, lim
k
C
j
|u
k
|
p
dx =
C
j
|u|
p
dx,
lim
k
B
R
j
|u
k
|
r
dx =
B
R
j
|u|
r
dx, lim
k
C
j
|u
k
|
r
dx =
C
j
|u|
r
dx
3.3 A Serious Loss of Compactness 115
and
lim
k
C
j
|u
k
|
2
dx =
C
j
|u|
2
dx.
Moreover, since
C
j
|u|
p
dx ≤
B
c
R
j
|u|
p
dx ≤
1
j
,
C
j
|u|
r
dx ≤
B
c
R
j
|u|
r
dx ≤
1
j
and
C
j
|u|
2
dx ≤
B
c
R
j
|u|
2
dx ≤
1
j
,
we can extract a subsequence u
k
j
in order to have, for every j ∈N,
l −
2
j
≤
B
R
j
|u
k
j
|
p
dx ≤l +
1
j
,
C
j
|u
k
j
|
p
dx ≤
2
j
,
l
1
−
2
j
≤
B
R
j
|u
k
j
|
r
dx ≤l
1
+
1
j
,
C
j
|u
k
j
|
r
dx ≤
2
j
,
C
j
|u
k
j
|
2
dx ≤
2
j
.
We take {u
k
j
} as a new minimizing sequence renaming it {u
j
}
j
.
We also consider, for every j , a function ψ
j
∈C
∞
(R
N
) such that
• 0 ≤ψ
j
(x) ≤1 for every x,
• ψ
j
(x) =1if|x|≤R
j
,
• ψ
j
(x) =0if|x|≥R
j
+1,
•|∇ψ
j
(x)|≤C for every x
and we define auxiliary functions
u
j
=ψ
j
u
j
, and u
j
=(1 −ψ
j
)u
j
.
Of course, u
j
,u
j
≥0 and u
j
=u
j
+u
j
for every j .
Lemma 3.3.9 The following properties hold as j →∞:
1. u
j
tends to u weakly in H
1
(R
N
) and strongly in L
p
(R
N
) and in L
r
(R
N
), while
u
j
tends to 0 weakly in H
1
(R
N
).
2.
R
N
|u
j
|
p
dx =
R
N
|u
j
|
p
dx +
R
N
|u
j
|
p
dx +o(1).
3.
R
N
|u
j
|
r
dx =
R
N
|u
j
|
r
dx +
R
N
|u
j
|
r
dx +o(1).
4. u
j
2
≥u
j
2
+u
j
2
+o(1).
Proof 1. If B is any bounded set in R
N
, we know that u
j
= u
j
in B, for every
j big enough, from which we can immediately conclude that u
j
uin H
1
(R
N
)
(hence in L
p
(R
N
) and L
r
(R
N
)). This also implies u
j
0inH
1
(R
N
), L
p
(R
N
)
and L
r
(R
N
).
116 3 Minimization Techniques: Lack of Compactness
Moreover,
R
N
|u
j
|
p
dx ≥
B
R
j
|u
j
|
p
dx =
B
R
j
|u
j
|
p
dx ≥l −
2
j
,
and on the other hand, recalling that 0 ≤u
j
≤u
j
,
R
N
|u
j
|
p
dx =
B
R
j
+1
|u
j
|
p
dx ≤
B
R
j
+1
|u
j
|
p
dx
=
B
R
j
|u
j
|
p
dx +
C
j
|u
j
|
p
dx ≤l +
1
j
+
2
j
=l +
3
j
.
Thus we obtain that
lim
j
R
N
|u
j
|
p
dx =l =
R
N
|u|
p
dx,
and since we have weak convergence we can conclude (see e.g. [11]) that u
j
→ u
in L
p
(R
N
). In the same way we get u
j
→u in L
r
(R
N
).
2. It is immediate to check that |u
j
|
p
≥|u
j
|
p
+|u
j
|
p
; hence,
R
N
|u
j
|
p
−|u
j
|
p
−|u
j
|
p
dx ≥0.
On the other hand,
R
N
|u
j
|
p
dx =
B
R
j
|u
j
|
p
dx +
C
j
|u
j
|
p
dx +
B
c
R
j
+1
|u
j
|
p
dx
=
B
R
j
|u
j
|
p
dx +
C
j
|u
j
|
p
dx +
B
c
R
j
+1
|u
j
|
p
dx
≤
R
N
|u
j
|
p
dx +
C
j
|u
j
|
p
dx +
R
N
|u
j
|
p
dx
≤
R
N
|u
j
|
p
dx +
2
j
+
R
N
|u
j
|
p
dx,
so that
0 ≤
R
N
|u
j
|
p
−|u
j
|
p
−|u
j
|
p
dx ≤
2
j
,
which proves 2.
3. The identity for the L
r
norms can be proved as in 2.
4. Exactly as above, |u
j
|
2
≥|u
j
|
2
+|u
j
|
2
, and since q(x) ≥0, we obtain
R
N
q(x)u
2
j
dx ≥
R
N
q(x)|u
j
|
2
dx +
R
N
q(x)|u
j
|
2
dx.
Also,
R
N
|∇u
j
|
2
dx =
R
N
|∇u
j
|
2
dx +
R
N
|∇u
j
|
2
dx +2
R
N
∇u
j
·∇u
j
dx.
3.3 A Serious Loss of Compactness 117
Then we have
R
N
∇u
j
·∇u
j
dx =
R
N
u
j
∇ψ
j
+ψ
j
∇u
j
·
(1 −ψ
j
)∇u
j
−u
j
∇ψ
j
dx
=
R
N
u
j
(1 −ψ
j
)∇ψ
j
·∇u
j
dx −
R
N
u
2
j
|∇ψ
j
|
2
dx
+
R
N
ψ
j
(1 −ψ
j
)|∇u
j
|
2
dx −
R
N
ψ
j
u
j
∇u
j
·∇ψ
j
dx
≥
R
N
u
j
(1 −ψ
j
)∇ψ
j
·∇u
j
dx −
R
N
u
2
j
|∇ψ
j
|
2
dx
−
R
N
ψ
j
u
j
∇u
j
·∇ψ
j
dx.
We also have
R
N
u
2
j
|∇ψ
j
|
2
dx ≤C
C
j
u
2
j
dx ≤
2C
j
=o(1),
and
R
N
ψ
j
u
j
∇u
j
·∇ψ
j
dx
≤
R
N
ψ
j
u
j
|∇u
j
||∇ψ
j
|dx
≤
R
N
|∇u
j
|
2
dx
1/2
R
N
u
2
j
|∇ψ
j
|
2
dx
1/2
≤M
2C
j
1/2
=o(1),
where M =sup
j
(
R
N
|∇u
j
|
2
dx)
1/2
.
In the same way we obtain
R
N
(1 −ψ
j
)u
j
∇u
j
·∇ψ
j
dx
≤o(1),
and putting all the estimates together we deduce that
R
N
|∇u
j
|
2
dx ≥
R
N
|∇u
j
|
2
dx +
R
N
|∇u
j
|
2
dx +o(1),
and finally,
u
j
2
≥u
j
2
+u
j
2
+o(1).
Lemma 3.3.10 There results
1
2
−
1
r
u
2
+
1
r
−
1
p
|u|
p
p
≤m.
Proof Since u
k
uin H
1
(R
N
) and in L
p
(R
N
),
118 3 Minimization Techniques: Lack of Compactness
1
2
−
1
r
u
2
+
1
r
−
1
p
|u|
p
p
≤
1
2
−
1
r
lim inf
k
u
k
2
+
1
r
−
1
p
lim inf
k
|u
k
|
p
p
≤lim inf
k
I(u
k
) =m.
We now prove that u ∈ N. We know that u = 0, so we just have to check that
u
2
=|u|
p
p
+λ|u|
r
r
. First we rule out the easy inequality.
Lemma 3.3.11 It cannot be
u
2
< |u|
p
p
+λ|u|
r
r
.
Proof If u
2
< |u|
p
p
+ λ|u|
r
r
, then, by usual arguments, there is some t ∈ (0, 1)
such that tu ∈N, so that
m ≤ I(tu)=t
2
1
2
−
1
r
u
2
+t
p
1
r
−
1
p
|u|
p
p
<
1
2
−
1
r
u
2
+
1
r
−
1
p
|u|
p
p
≤m,
a contradiction.
Then, with some more effort, the opposite case.
Lemma 3.3.12 It cannot be
u
2
> |u|
p
p
+λ|u|
r
r
.
Proof Using Lemma 3.3.9 we can write
u
j
2
+u
j
2
≤u
j
2
+o(1) =|u
j
|
p
p
+λ|u
j
|
r
r
+o(1)
=|u
j
|
p
p
+λ|u
j
|
r
r
+|u
j
|
p
p
+λ|u
j
|
r
r
+o(1). (3.10)
Assume for contradiction that u
2
> |u|
p
p
+λ|u|
r
r
and pick δ>0 such that
u
2
> |u|
p
p
+λ|u|
q
q
+δ.
Since u
j
uin H
1
(R
N
),
lim inf
j
u
j
2
≥u
2
,
while u
j
→u in L
p
(R
N
) and in L
r
(R
N
). From this we deduce that, for large j’s,
u
j
2
> |u
j
|
p
p
+λ|u
j
|
r
r
+δ
1
,
for some δ
1
∈(0,δ). Together with (3.10), this implies
u
j
2
≤|u
j
|
p
p
+λ|u
j
|
r
r
−δ
2
,
3.4 Problems with Critical Exponent 119
for large j’s and for some δ
2
∈(0,δ
1
). Recalling that u
j
0inH
1
(R
N
) and argu-
ingasinLemma3.3.8, we obtain
lim
j
u
j
2
=lim
j
u
j
2
α
,
so that
u
j
2
α
≤|u
j
|
p
p
+λ|u
j
|
r
r
−δ
3
,
for large j’s and for some δ
3
∈ (0,δ
2
). Hence, by usual arguments, we can find
s
j
∈(0, 1) such that s
j
u
j
∈N
α
. So we obtain
m
α
≤I
α
(s
j
u
j
) =s
2
j
1
2
−
1
r
u
j
2
α
+s
p
j
1
r
−
1
p
|u
j
|
p
p
≤
1
2
−
1
r
u
j
2
+
1
r
−
1
p
|u
j
|
p
p
+o(1)
≤
1
2
−
1
r
u
j
2
+u
j
2
+
1
r
−
1
p
|u
j
|
p
p
+|u
j
|
p
p
+o(1)
≤
1
2
−
1
r
u
j
2
+
1
r
−
1
p
|u
j
|
p
p
+o(1) =I(u
j
) +o(1).
Passing to the limit this yields
m
α
≤m,
a contradiction.
End of the proof of Theorem 3.3.2 We can now conclude easily the proof of the
main result. The previous lemmas say that in any case the weak limit u satisfies
u =0, u ≥0 and
u
2
=|u|
p
p
+λ|u|
r
r
.
Therefore u ∈N and also
I(u)=
1
2
−
1
r
u
2
+
1
r
−
1
p
|u|
p
p
≤m,
so that I(u)=m and u is a minimum point for I on N. It is then easy, arguing as
in the previous sections, to prove that u is a critical point for I on H
1
(R
N
).
3.4 Problems with Critical Exponent
We discuss in this section some examples of problems where the nonlinearity has
critical growth. This is one of the most dramatic cases of loss of compactness and
has been studied intensively in the last 25 years, starting with the pioneering paper
[14].
We present here only simple cases, with the aim of showing some example of the
techniques involved.
120 3 Minimization Techniques: Lack of Compactness
3.4.1 The Prototype Problem
We consider in R
N
, with N ≥3, the equation
−u =|u|
2
∗
−2
u. (3.11)
Formally, solutions of (3.11) should arise as critical points of the functional
I(u)=
1
2
R
N
|∇u|
2
dx −
1
2
∗
R
N
|u|
2
∗
dx. (3.12)
It is not convenient to think of I as defined on H
1
(R
N
), since in this space the first
integral in I is not (the square of) a norm, and this poses serious difficulties.
We will work in the space D
1,2
(R
N
) that we introduced in Sect. 1.2. The problem
we are dealing with then becomes
−u =|u|
2
∗
−2
u,
u ∈D
1,2
(R
N
).
(3.13)
We know that D
1,2
(R
N
) ⊂ L
2
∗
(R
N
), with continuous embedding, but the em-
bedding is not compact, not even in a local sense, as we now make precise.
Clearly, since we are working in R
N
, there is a lack of compactness due to the
invariance under translations, as we discussed in earlier sections. However in this
case there is an even more serious problem, that arises from the invariance under
scalings, as follows. Let v ∈ C
∞
0
(R
N
) (hence v ∈ D
1,2
(R
N
)) be a fixed function,
and set, for λ>0,
v
λ
(x) =λ
N−2
2
v(λx).
Then it is an exercise to check that
R
N
|∇v
λ
|
2
dx =
R
N
|∇v|
2
dx for every λ>0
and
v
λ
0inD
1,2
(R
N
) for λ →0orλ →+∞.
However also
R
N
|v
λ
|
2
∗
dx =
R
N
|v|
2
∗
dx for every λ>0, (3.14)
so that {v
λ
}
λ
is not precompact in L
2
∗
(R
N
). This happens exactly because the ex-
ponent is 2
∗
.
With respect to the informal discussion on the loss of compactness carried out
in the preceding section, we have to face here new ways in which this phenomenon
takes place. Indeed, assuming to fix ideas that v(0) =0, we see that when λ →0the
functions v
λ
∈C
∞
0
(R
N
) tend to zero uniformly, while
R
N
|u|
2
∗
dx is kept constant
(see (3.14)). One can say that the “mass” spreads everywhere. On the contrary, if
λ →∞, then v
λ
(x) →0 for every x =0, while v
λ
(0) →∞, and again (3.14) holds;
this is the famous loss of compactness by concentration, that one meets in every
problem with critical growth, even on bounded domains.