Badiale M., Serra E. Semilinear Elliptic Equations for Beginners: Existence Results via the Variational Approach
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3.4 Problems with Critical Exponent 121
The hard part in the proof of the main result of this section will be devoted to a
careful analysis of minimizing sequences to understand the consequences of spread-
ing or concentration of mass.
Anyway, the first thing to do if one wants to solve (3.13), is to get rid of the
invariance under translations. This can be done by symmetrization, as in Sect. 3.1.
We recall that we have defined
D
r
=
u ∈D
1,2
(R
N
) |u is radial
.
We are going to prove the following theorem.
Theorem 3.4.1 There exists a nonnegative and nontrivial u ∈D
r
that solves weakly
Problem (3.13), namely such that
R
N
∇u ·∇vdx=
R
N
u
2
∗
−1
vdx (3.15)
for every v in D
1,2
(R
N
).
Remark 3.4.2 It can be proved, with essentially the same arguments as in Exam-
ple 1.3.21, that the functional
u →
R
N
|u|
2
∗
dx
(and hence the functional I ) is differentiable on D
1,2
(R
N
). This does not follow
directly from Example 1.3.21, because the H
1
norm is larger that the D
1,2
norm. In
other words, quantities that are o(v
H
1
) are not, generally, o(v
D
1,2
). A slightly
more general case is reported in Exercise 4.
In the proof of Theorem 3.4.1 we will need the pointwise estimate that we have
proved in Lemma 3.1.2.
Now we can start with the main argument. We want to minimize the functional I
on the unit sphere of L
2
∗
(R
N
). This amounts to setting
S =inf
u
2
u ∈D
1,2
(R
N
),
R
N
|u|
2
∗
dx =1
. (3.16)
Remark 3.4.3 The value S is known as the best Sobolev constant. It is the largest
positive constant S such that
S|u|
2
2
∗
≤u
2
for every u ∈D
1,2
(R
N
).
Thanks to the Sobolev inequalities, we know that S>0. We will show that the
infimum S is attained by some u ∈D
1,2
(R
N
).
We begin to carry out the analysis of minimizing sequences by selecting one that
enjoys some special properties, as we show in the next three lemmas. To begin with,
we need a minimizing sequence made by radial functions.
122 3 Minimization Techniques: Lack of Compactness
Lemma 3.4.4 There exists a sequence {v
k
}
k
⊂D
r
such that v
k
≥0,
|v
k
|
2
∗
=1 and v
k
2
→S.
Proof Let {w
k
}
k
⊂ D
1,2
(R
N
) be a minimizing sequence for S; of course it is not
restrictive to assume that w
k
≥0. Define v
k
=w
∗
k
≥0 (see Theorem 3.1.5). Then
v
k
∈D
r
, |v
k
|
2
∗
=|w
k
|
2
∗
=1 and v
k
2
≤w
k
2
,
so that {v
k
}
k
⊂D
r
is a minimizing sequence for S.
Lemma 3.4.5 There exist a minimizing sequence {u
k
}
k
⊂D
r
for S and a function
u ∈D
r
with the following properties:
• u
k
≥0, u ≥0.
•
|x|<1
|u
k
|
2
∗
dx =
|x|≥1
|u
k
|
2
∗
dx =
1
2
.
• u
k
uin D
1,2
(R
N
) and in L
2
∗
(R
N
), and u
k
(x) →u(x) a.e. in R
N
.
Proof Let {v
k
}
k
⊂D
r
be a minimizing sequence for S, with v
k
≥0. Since
R
N
|v
k
|
2
∗
dx =1,
there exists R
k
> 0 such that
|x|<R
k
|v
k
|
2
∗
dx =
|x|≥R
k
|v
k
|
2
∗
dx =
1
2
.
If we define
u
k
(x) =R
N−2
2
k
v
k
(R
k
x),
it is immediate to check that u
k
∈ D
r
and that u
k
=v
k
and |u
k
|
2
∗
=|v
k
|
2
∗
.
This means that {u
k
}
k
is a minimizing sequence too. Of course u
k
≥0, and a simple
change of variables shows that
|x|<1
|u
k
|
2
∗
dx =
|x|≥1
|u
k
|
2
∗
dx =
1
2
.
Since {u
k
}
k
is bounded in D
1,2
(R
N
), up to subsequences, u
k
uin D
1,2
(R
N
) and
in L
2
∗
(R
N
). Moreover, for every ball B, the sequence u
k
is bounded in H
1
(B),
so that, again up to subsequences, it converges strongly in L
2
(B) and also satisfies
u
k
(x) →u(x) almost everywhere in R
N
. The proof is complete.
Next, we force the sequence u
k
to have the technical property stated in the fol-
lowing lemma.
Lemma 3.4.6 There exists a minimizing sequence {u
k
}
k
for S that, in addition to
those of the previous lemma, satisfies also the following properties: for every j ∈N,
j ≥1, the sequences
|x|<1/j
|∇u
k
|
2
dx
k
and
|x|>j
|∇u
k
|
2
dx
k
admit finite limits as k →∞.
3.4 Problems with Critical Exponent 123
Proof Let {u
k
}
k
be the sequence defined in the previous lemma. Since the numerical
sequences
|x|<1
|∇u
k
|
2
dx and
|x|>1
|∇u
k
|
2
dx
are bounded, there exists a subsequence {u
k,1
}
k
such that the sequences
|x|<1
|∇u
k,1
|
2
dx and
|x|>1
|∇u
k,1
|
2
dx
admit finite limits. From {u
k,1
}
k
we can extract a new subsequence {u
k,2
}
k
such that
the sequences
|x|<1/2
|∇u
k,2
|
2
dx and
|x|>2
|∇u
k,2
|
2
dx
admit finite limits. Proceeding in this way, for every j ≥ 2 we have a sequence
{u
k,j
}
k
such that the sequences
|x|<1/j
|∇u
k,j
|
2
dx and
|x|>j
|∇u
k,j
|
2
dx
admit finite limits as k →∞and {u
k,j
}
k
is a subsequence of {u
k,j−1
}
k
. Then we
relabel as u
k
the “diagonal”, namely the sequence u
k
=u
k,k
. This is a minimizing
sequence that satisfies the conditions of the previous lemma, because it has been
extracted from the original minimizing sequence. Moreover, for every given j,ex-
cept the first j terms, the sequence is a subsequence of {u
k,j
}
k
, and therefore the
numerical sequences
|x|<1/j
|∇u
k
|
2
dx and
|x|>j
|∇u
k
|
2
dx
are subsequences of a convergent sequence, and therefore they are convergent too.
In the next lemma we introduce some quantities that we will use in the main
argument. The reader can recognize the effort in trying to gain information on the
spreading or on the concentration of u
k
,ask →∞.
Lemma 3.4.7 The following limits exist and are finite:
1. ν
0
=lim
R→0
+
(limsup
k
|x|<R
|u
k
|
2
∗
dx),
ν
∞
=lim
R→+∞
(limsup
k
|x|≥R
|u
k
|
2
∗
dx).
2. μ
0
=lim
R→0
+
(limsup
k
|x|<R
|∇u
k
|
2
dx),
μ
∞
=lim
R→+∞
(limsup
k
|x|≥R
|∇u
k
|
2
dx).
Moreover
3. ν
0
,ν
∞
∈[0, 1/2].
124 3 Minimization Techniques: Lack of Compactness
Proof Defining the functions
f(R)=lim sup
k
|x|<R
|u
k
|
2
∗
dx and g(R) =lim sup
k
|x|≥R
|u
k
|
2
∗
dx,
it is immediate to see that f is bounded and nondecreasing, while g is bounded and
nonincreasing. Therefore
lim
R→0
+
f(R)=ν
0
and lim
R→+∞
g(R) =ν
∞
exist and are finite (and nonnegative).
The same argument works for μ
0
and μ
∞
. The fact that ν
0
≤1/2 and ν
∞
≤1/2
comes from
|x|<1
|u
k
|
2
∗
dx =
|x|≥1
|u
k
|
2
∗
dx =
1
2
.
The following lemmas establish some inequalities among the numbers intro-
duced in the above result that will be essential in the proof of Theorem 3.4.1.
Loosely speaking, the next result establishes that as k →∞, the original mass of
u
k
in not less than the sum of the contributions of the weak limit, of the part that
spreads (ν
∞
) and of the part that concentrates (ν
0
).
Lemma 3.4.8 Let u be the weak limit of the minimizing sequence u
k
constructed
above. Then
1 ≤
R
N
|u|
2
∗
dx +ν
0
+ν
∞
.
Proof For every k ∈N and for every R>1wehave
1 =
R
N
|u
k
|
2
∗
dx =
|x|<1/R
|u
k
|
2
∗
dx +
1/R<|x|<R
|u
k
|
2
∗
dx +
|x|≥R
|u
k
|
2
∗
dx.
In the bounded open set
{x ∈R
N
1/R < |x|<R}
we have |u
k
(x)|
2
∗
→|u(x)|
2
∗
almost everywhere and moreover
|u
k
(x)|
2
∗
≤CR
N
,
by Lemma 3.1.2. Then, by dominated convergence,
lim
k
1/R<|x|<R
|u
k
|
2
∗
dx =
1/R<|x|<R
|u|
2
∗
dx.
Then we have
1 =lim sup
k
|x|<1/R
|u
k
|
2
∗
dx +
1/R<|x|<R
|u
k
|
2
∗
dx +
|x|≥R
|u
k
|
2
∗
dx
≤
1/R<|x|<R
|u|
2
∗
dx +limsup
k
|x|<1/R
|u
k
|
2
∗
dx +limsup
k
|x|≥R
|u
k
|
2
∗
dx.
3.4 Problems with Critical Exponent 125
As R →+∞we obtain
1 ≤
R
N
|u|
2
∗
dx +ν
0
+ν
∞
.
A similar behavior (but with the opposite inequality) holds also for the D
1,2
norm.
Lemma 3.4.9 Let u be the weak limit of the minimizing sequence u
k
constructed
above. Then
S ≥
R
N
|∇u|
2
dx +μ
0
+μ
∞
.
Proof Setting
ε
k
=u
k
2
−S,
for every k and every j there results
S +ε
k
=
R
N
|∇u
k
|
2
dx
=
|x|<1/j
|∇u
k
|
2
dx +
1/j <|x|<j
|∇u
k
|
2
dx +
|x|>j
|∇u
k
|
2
dx. (3.17)
The functional
J(u)=
1/j <|x|<j
|∇u|
2
dx
is continuous and convex in D
1,2
(R
N
), so that it is weakly lower semicontinuous.
Recalling that ε
k
→0 and that the two numerical sequences
|x|<1/j
|∇u
k
|
2
dx
k
and
|x|>j
|∇u
k
|
2
dx
k
admit limits, if in (3.17) we take the liminf with respect to k, we obtain
S =lim inf
k
|x|<1/j
|∇u
k
|
2
dx +
1/j <|x|<j
|∇u
k
|
2
dx +
|x|>j
|∇u
k
|
2
dx
=lim inf
k
1/j <|x|<j
|∇u
k
|
2
dx +lim
k
|x|<1/j
|∇u
k
|
2
dx +lim
k
|x|>j
|∇u
k
|
2
dx
≥
1/j <|x|<j
|∇u|
2
dx +lim
k
|x|<1/j
|∇u
k
|
2
dx +lim
k
|x|>j
|∇u
k
|
2
dx.
Letting now j →∞we see that
S ≥
R
N
|∇u|
2
dx +μ
0
+μ
∞
.
From the definition of S one has, for every v ∈D
1,2
(R
N
),
S
R
N
|v|
2
∗
dx
2/2
∗
≤
R
N
|∇v|
2
dx. (3.18)
126 3 Minimization Techniques: Lack of Compactness
The next lemma is of fundamental importance. In some sense it says that the
inequality (3.18) holds also for the “components” of u
k
that spread or concentrate.
Lemma 3.4.10 There results
Sν
2/2
∗
0
≤μ
0
and Sν
2/2
∗
∞
≤μ
∞
.
Proof Let ϕ :R →R be a C
∞
function such that ϕ(t) =1ift ∈[0, 1] and ϕ(t) =0
if t ≥2. For every R>0 and every x ∈R
N
,set
ϕ
R
(x) =ϕ
|x|
R
.
Then ϕ
R
∈C
∞
0
(R
N
),
⎧
⎨
⎩
ϕ
R
(x) =1if|x|≤R,
ϕ
R
(x) =0if|x|≥2R,
|∇ϕ
R
(x)|≤C/R for every x ∈R
N
for some C independent of x and R.
For every v ∈ D
1,2
(R
N
), the function ϕ
R
v is also in D
1,2
(R
N
). Indeed ϕ
R
v
belongs to L
2
∗
(R
N
), and we only have to check that its gradient is in L
2
.But
∇
(
ϕ
R
v
)
=v∇ϕ
R
+ϕ
R
∇v,
and obviously |ϕ
R
∇v|∈L
2
(R
N
). On the other hand,
R
N
|v|
2
|∇ϕ
R
|
2
dx =
R<|x|<2R
|v|
2
|∇ϕ
R
|
2
dx
≤
R<|x|<2R
|v|
2
∗
N−2
N
R<|x|<2R
|∇ϕ
R
|
N
dx
2/N
≤
R
N
|v|
2
∗
N−2
N
C
R
N
R<|x|<2R
dx
2/N
≤C|v|
2
2
∗
< +∞.
Thus, the gradient of ϕ
R
v is the sum of two L
2
functions.
We now consider, for every k and for every R, the function ϕ
R
u
k
∈ D
1,2
(R
N
).
Certainly,
S
R
N
|ϕ
R
u
k
|
2
∗
dx
2/2
∗
≤
R
N
|∇(ϕ
R
u
k
)|
2
dx. (3.19)
We analyze what happens to this inequality when we compute some limits. As
far as the left-hand side is concerned, we have
|x|<R
|u
k
|
2
∗
dx ≤
R
N
|ϕ
R
u
k
|
2
∗
dx ≤
|x|<2R
|u
k
|
2
∗
dx,
so that
lim sup
k
|x|<R
|u
k
|
2
∗
dx ≤lim sup
k
R
N
|ϕ
R
u
k
|
2
∗
dx ≤lim sup
k
|x|<2R
|u
k
|
2
∗
dx.
3.4 Problems with Critical Exponent 127
Now the first and the last term in this inequality tend to ν
0
, when R tends to zero
(Lemma 3.4.7), and therefore so does the central term, namely
lim
R→0
lim sup
k
R
N
|ϕ
R
u
k
|
2
∗
dx
=ν
0
.
We turn to the study of the right-hand side of (3.19). First of all,
|∇(ϕ
R
u
k
)|
2
=u
2
k
|∇ϕ
R
|
2
+ϕ
R
|∇u
k
|
2
+2u
k
ϕ
R
∇ϕ
R
·∇u
k
.
Repeating the above argument, we deduce that
R
N
|u
k
|
2
|∇ϕ
R
|
2
dx ≤C
R<|x|<2R
|u
k
|
2
∗
dx
N−2
N
.
Moreover, working as in Lemma 3.4.8, we see that
lim
k
R<|x|<2R
|u
k
|
2
∗
dx =
R<|x|<2R
|u|
2
∗
dx
so that
lim
R→0
lim
k
R<|x|<2R
|u
k
|
2
∗
dx
= lim
R→0
R<|x|<2R
|u|
2
∗
dx =0,
which yields
lim
R→0
lim
k
R
N
|u
k
|
2
|∇ϕ
R
|
2
dx
=0.
From this it also easy to see that
lim
R→0
lim
k
R
N
u
k
ϕ
R
∇ϕ
R
·∇u
k
dx
=0.
Finally, since
|x|<R
|∇u
k
|
2
dx ≤
R
N
|ϕ
R
∇u
k
|
2
dx ≤
|x|<2R
|∇u
k
|
2
dx,
arguing as above we obtain
lim
R→0
lim sup
k
R
N
|ϕ
R
∇u
k
|
2
dx
=μ
0
.
Summing up, from (3.19), taking first the lim sup over k, and then the limit as
R →0, we get to
Sν
2/2
∗
0
≤μ
0
,
the first inequality we wanted to prove.
The other inequality can be deduced in a similar way. Defining ψ : R → R to
be a C
∞
function such that ψ(t) = 0ift ∈[0, 1] and ψ(t) = 1ift ≥ 2, for every
R>0 and every x ∈R
N
we set
ψ
R
(x) =ψ
|x|
R
.
128 3 Minimization Techniques: Lack of Compactness
Again, it is immediate to check that ψ
R
∈C
∞
(R
N
),
⎧
⎨
⎩
ψ
R
(x) =0if|x|≤R,
ψ
R
(x) =1if|x|≥2R,
|∇ψ
R
(x)|≤C/R for every x ∈R
N
for some C independent of x and R. As before, ψ
R
v ∈D
1,2
(R
N
) if v ∈D
1,2
(R
N
).
Starting from the inequality
S
R
N
|ψ
R
u
k
|
2
∗
dx
2/2
∗
≤
R
N
|∇(ψ
R
u
k
)|
2
dx
and arguing as in the first part of the proof, we obtain
lim
R→+∞
lim sup
k
R
N
|ψ
R
u
k
|
2
∗
dx
=ν
∞
and
lim
R→+∞
lim sup
k
R
N
|ψ
R
∇u
k
|
2
dx
=μ
∞
,
so that also the inequality
Sν
2/2
∗
∞
≤μ
∞
is established.
The next lemma is the final step in the proof of Theorem 3.4.1.
Lemma 3.4.11 There results
ν
0
=ν
∞
=0 and
R
N
|u|
2
∗
dx =1.
Proof We know that ν
0
,ν
∞
∈[0, 1/2] and, by weak lower semicontinuity,
R
N
|u|
2
∗
dx ∈[0, 1].
Since 2
∗
> 2, for every a ∈[0, 1] we have a
2/2
∗
≥a, and equality holds if and only
if a =0ora =1.
Putting together the inequalities proved in the preceding lemmas, we have
S ≥
R
N
|∇u|
2
dx +μ
0
+μ
∞
≥S
R
N
|u|
2
∗
dx
2/2
∗
+ν
2/2
∗
0
+ν
2/2
∗
∞
≥S
R
N
|u|
2
∗
dx +ν
0
+ν
∞
≥S.
3.4 Problems with Critical Exponent 129
Thus equality must hold everywhere, and in particular
R
N
|u|
2
∗
dx
2/2
∗
+ν
2/2
∗
0
+ν
2/2
∗
∞
=
R
N
|u|
2
∗
dx +ν
0
+ν
∞
.
But then, each of the numbers
R
N
|u|
2
∗
dx, ν
0
and ν
∞
must either be 0 or 1. As
ν
0
,ν
∞
∈[0, 1/2], necessarily ν
0
=ν
∞
=0.
Since
R
N
|u|
2
∗
dx +ν
0
+ν
∞
≥1,
we deduce that
R
N
|u|
2
∗
dx ≥1 and hence
R
N
|u|
2
∗
dx =1.
End of the proof of Theorem 3.4.1 We have proved so far that there exists a mini-
mizing sequence for S that admits a weak limit u satisfying
R
N
|u|
2
∗
dx =1. (3.20)
By weak lower semicontinuity, we also have u
2
≤ S, and by the definition of S
and (3.20), it must be u
2
=S. Thus u is a minimizer.
Now the fact that a multiple of u weakly solves (3.13) follows from standard
arguments already used in Chap. 2. The proof of Theorem 3.4.1 is now complete.
Remark 3.4.12 The solutions of (3.13) that we have found are explicitly known.
Elementary computations show that a solution is given, for a suitable constant c>0,
by
U(x) =
c
1 +|x|
2
N−2
2
,
and it can be proved (it is not easy at all) that this function indeed solves the mini-
mization problem (3.16).
Actually all the solutions of (3.16) are known: they are the family of functions
U
λ,y
(x) =λ
N−2
2
U(λ(x −y))
for each y ∈ R
N
and each λ>0. Note that they are generated by the function u
above by translations and scalings (see e.g. [29]).
3.4.2 A Problem with a Radial Coefficient
We now use the results of the previous section to deal with a problem that depends
on a radial coefficient. It is a simple variant of (3.13).
We consider
−u =β(|x|)|u|
2
∗
−2
u,
u ∈D
1,2
(R
N
),
(3.21)
where β satisfies the following assumptions
130 3 Minimization Techniques: Lack of Compactness
(b
1
)β:[0, +∞) →[0, +∞) is continuous.
(b
2
)β(r)>β(0) for all r>0 and lim
r→+∞
β(r) =β(0).
For example, the functions β(r) =C +
r
1+r
2
or β(r) =C +re
−r
satisfy the above
assumptions for every C ≥0. Hypothesis (b
2
) can be weakened, see Remark 3.4.22
below.
We will prove the following result.
Theorem 3.4.13 Assume that (b
2
) and (b
2
) hold. Then there exists a nonnegative
and nontrivial u ∈ D
r
that solves weakly Problem (3.21), namely such that
R
N
∇u ·∇vdx=
R
N
β(|x|)u
2
∗
−1
vdx (3.22)
for every v in D
1,2
(R
N
).
To prove this result we cannot use the symmetrization technique as in the previ-
ous section, that is, we cannot apply Theorem 3.1.5. The reason is the fact that, if
v ∈ D
1,2
(R
N
) and v = u
∗
(see Theorem 3.1.5), it is true that u
2
≤v
2
but the
equality
R
N
β(|x|)u
2
∗
dx =
R
N
β(|x|)v
2
∗
dx
does not hold.
Hence we adopt a different strategy: we will study the variational problem not in
the space D
1,2
(R
N
) but directly in the space D
r
. Of course this is possible because
the problem is invariant under rotations.
We begin by introducing the following definition. For every positive number
b>0weset
S
b
=inf
u
2
u ∈D
1,2
(R
N
),
R
N
|u|
2
∗
dx =b
.
Notice that S
1
coincides with the number S defined in the previous section. We will
always write S instead of S
1
in order to be consistent with the results of Sect. 3.4.1.
The following lemma establishes the behavior of minimizers when b changes.
Lemma 3.4.14 For every b>0,
1. S
b
=b
N−2
N
S.
2. If u
1
is the minimizer for S found in Theorem 3.4.1, and v
b
(x) =b
1/2
∗
u
1
(x), the
function v
b
is a minimizer for S
b
, namely
R
N
|v
b
|
2
∗
dx =b and v
b
2
=S
b
.
3. There results
S
b
=inf
u
2
u ∈D
r
,
R
N
|u|
2
∗
dx =b
.