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 131

Proof As it has been sketched in Remark 2.3.6, it is easy to see that for all a,b > 0

there results

N−2

so for a = 1 we obtain the claim. The proof of the second part is obvious, and the

third part easily derives from the fact that u

and v

are radial functions. 

We now deﬁne the number

S =inf



u



u ∈D



β(|x|)|u|

∗

dx =1



and we show that

S is attained.

We start the argument by setting, in case β(0)>0,

=β(0)

and we ﬁrst compare

S to S

−1

Lemma 3.4.15 We have

S<S

−1

Proof Let w be a nonnegative radial minimizer for S

−1

, that is, w ∈D

\{0}, w ≥0,



∗

dx =1 and w

−1

From (b

) we see that



β(|x|)w

∗

dx >



∗

dx =1.

We deﬁne now

μ =





β(|x|)w

∗



1/2

∗

so that μ>1, and v =w/μ. Then



β(|x|)v

∗

dx =1

and therefore

S ≤v

w

−1



We can now repeat the arguments of the previous section, and we give the details

only in case something different has to be done. As a ﬁrst thing, arguing as we did

in Lemmas 3.4.5 and 3.4.6, we obtain the following result.

Lemma 3.4.16 There exist a minimizing sequence {u

}

⊂D

for S and a function

u ∈D

such that

132 3 Minimization Techniques: Lack of Compactness

• u

≥0, u ≥0.

• u

uin D

1,2

) and in L

∗

), and u

(x) →u(x) a.e. in R

• for all j ∈N the sequences





|x|<1/j

|∇u



and





|x|>j

|∇u



admit ﬁnite limits as k →∞.

We continue with the analogue of Lemma 3.4.7.

Lemma 3.4.17 The following limits exist and are ﬁnite:

1. ν

=lim

R→0

(limsup



|x|<R

β(|x|)|u

∗

dx),

∞

=lim

R→+∞

(limsup



|x|≥R

β(|x|)|u

∗

dx),

2. μ

=lim

R→0

(limsup



|x|<R

|∇u

dx),

∞

=lim

R→+∞

(limsup



|x|≥R

|∇u

dx).

Moreover, ν

,ν

∞

∈[0, 1] and

3. if β(0) =0 then ν

=ν

∞

=0.

Proof The argument is exactly the same as that of Lemma 3.4.7. We have only to

prove the last claim. Assume β(0) =0 and take ε>0 and r

> 0 such that

β(r) ≤ε ∀r ∈[0,r

]∪[R

, +∞)

which is possible by (b

). Deﬁne

M =sup





∗



Then, for all 0 <R

and all R

we have



|x|<R

β(|x|)|u

∗

dx ≤Mε and



|x|>R

β(|x|)|u

∗

dx ≤Mε,

so that ν

≤Mε and ν

∞

≤Mε. This holds for all ε>0 and the claim follows. 

This allows us to prove the same limit relations on the above quantities as we did

in the previous section. We collect all the results in a single statement.

Lemma 3.4.18 There results

1 ≤



β(|x|)|u|

∗

dx +ν

+ν

∞

;

S ≥



|∇u|

dx +μ

+μ

∞

;

Sν

2/2

∗

≤μ

, Sν

2/2

∗

∞

≤μ

∞

Lemma 3.4.19 Only the following three cases can occur: either

3.4 Problems with Critical Exponent 133

•



β(|x|)u

∗

dx =1 and ν

=ν

∞

=0,

• or ν

=1 and



β(|x|)u

∗

dx =ν

∞

=0,

• or ν

∞

=1 and



β(|x|)u

∗

dx =ν

=0.

Proof Same as that of Lemma 3.4.11. 

The only result that requires a new proof is the next one.

Lemma 3.4.20 There results

=ν

∞

=0 and



β(|x|)|u|

∗

dx =1.

Proof We argue by contradiction. If the claim is not true, then by the previous

lemma we have



β(|x|)|u|

∗

dx =0,

that is β(|x|)|u|

∗

= 0 a.e., and ν

= 1orν

∞

= 1. Hence, by Lemma 3.4.17,we

have β(0) =β

> 0, so, by Lemma 3.4.15, S<S

−1

. We claim that

lim



∗

dx =1. (3.23)

Indeed we have

1 =



β(|x|)|u

∗

dx =



∗

dx +



(β(|x|) −β

)|u

∗

dx,

so we have to prove that

lim



(β(|x|) −β

)|u

∗

dx =0.

To this aim, ﬁx ε>0 and take r

> 0 such that β(r) −β

≤ε for all r ∈[0,r

]

and all r ∈[R

, +∞), and deﬁne

M =sup





∗



Then

0 ≤



(β(|x|) −β

)|u

∗



|x|<r

(β(|x|) −β

)|u

∗

dx +



<|x|<R

(β(|x|) −β

)|u

∗



|x|>R

(β(|x|) −β

)|u

∗

≤2Mε +



<|x|<R

(β(|x|) −β

)|u

∗

dx.

134 3 Minimization Techniques: Lack of Compactness

In the bounded set A

={x ∈R

< |x|<R

}the functions β(|x|)|u

∗

are uni-

formly bounded and they converge pointwise to β(|x|)|u|

∗

= 0, so that, by domi-

nated convergence,

lim



<|x|<R

(β(|x|) −β

)|u

∗

dx =0,

which yields

0 ≤lim sup



(β(|x|) −β

)|u

∗

dx ≤2Mε.

Since this holds for all ε>0,

lim



(β(|x|) −β

)|u

∗

dx =0,

which concludes the proof of (3.23). Now we can ﬁnish the proof of the lemma.

Deﬁne μ



∗

dx)

1/2

∗

and v

/μ

.Wehave



∗

dx =1

and therefore

−1

≤v



u



→S.

This shows that S

−1

≤S, a contradiction. 

Since



β(|x|)|u|

∗

dx =1,

arguing as in Chap. 2 we can conclude that there exists u ∈D

\{0}, u ≥0, such that



∇u ·∇vdx=



β(|x|)u

∗

−1

vdx (3.24)

for all v ∈ D

. We now want to prove that this equation is satisﬁed for all v in

1,2

). This is carried out in the following lemma, which is a particular case of

the “principle of symmetric criticality” by Palais [38],seealso[48].

Lemma 3.4.21 Equation (3.24) holds for every v ∈D

1,2

Proof Let us consider the functional I :D

1,2

) →R deﬁned by

I(u)=

u

−

∗



β(|x|)|u|

∗

and let I

be its restriction to D

.By(3.24)wehaveI



(u) = 0. We want to prove

that also I



(u) = 0, that is, ∇I(u)=0inD

1,2

). For this, we claim that ∇I(u)

is a radial function.

3.4 Problems with Critical Exponent 135

To prove the claim, we take a rotation of R

with matrix R and we denote its

transpose by S. Of course, S =R

−1

We set f

(x) = f(Rx) for all f ∈ D

1,2

) and we prove that for all such R

and for all f, g ∈D

1,2

) there results



∇f

·∇gdx=



∇f ·∇g

dx. (3.25)

We check (3.25) for smooth f,g, the general case being easily obtained by density.

An easy computation gives

∇f

(x) =S∇f(Rx),

and hence

∇f

(x) ·∇g(x) =S∇f(Rx)·∇g(x) =∇f(Rx)·R∇g(x).

Integrating and changing variables yields



∇f

(x) ·∇g(x)dx =



∇f(Rx)·R∇g(x)dx =



∇f(x)·R∇g(Sx) dx



∇f(x)·∇g

(x) dx.

This means

|g) =(f |g

where (·|·) is the scalar product in D

1,2

To prove that ∇I(u) is radial we have to prove that, for all rotations R,

∇I(u)=(∇I(u))

that is

(∇I(u)|v) =



(∇I(u))





for all v ∈D

1,2

). Recalling that u is a radial function, we compute



(∇I(u))





=(∇I(u)|v

) =I



(u)v

=(u|v

) −



β(|x|)u

∗

−1

dx =(u

|v) −



β(|x|)u

∗

−1

vdx

=(u|v) −



β(|x|)u

∗

−1

vdx=I



(u)v =(∇I(u)|v).

This shows that ∇I(u) is radial.

Now it is easy to conclude the proof. We have

0 =I



(u)(∇I(u))=I



(u)(∇I(u))=(∇I(u)|∇I(u))=∇I(u)

so ∇I(u)=0, namely I



(u)v =0 for all v ∈D

1,2

). 

136 3 Minimization Techniques: Lack of Compactness

Remark 3.4.22 Hypothesis (b

) can be weakened in the sense that it is not neces-

sary to assume β(r) > β(0) for all r>0 but it is enough that β(r) ≥ β(0) for all

r>0 and β(r

)>β(0) for some r

> 0. Indeed, if we assume this weaker con-

dition, as we know that the minimizer for S is everywhere strictly positive (see

Remark 3.4.12), in Lemma 3.4.15 we obtain



β(|x|)w

∗

dx >



∗

dx,

and this is all we need to conclude.

3.4.3 A Nonexistence Result

If one tries to attack the analogue of problem (3.13)onabounded domain , namely

the problem



−u =|u|

∗

−2

u in ,

u =0on∂,

(3.26)

one realizes quickly that the methods used in Sect. 3.4.1 break down.

The question that one has to face is then: are there more powerful techniques that

allow one to solve (3.26), or does something really happen when the growth of the

nonlinearity reaches the critical value 2

∗

In this section we show, via a classical argument, that no method can produce a

nontrivial solution to (3.26), for example when  is a ball. In other words, when

power nonlinearities are considered, the value 2

∗

represents a threshold beyond

which existence of nontrivial solutions may fail. For the sake of simplicity we limit

ourselves to the nonexistence of solutions of constant sign; the general case requires

some technical arguments that are beyond the scope of this book.

We begin by describing the class of sets  to which the argument applies.

Deﬁnition 3.4.23 We say that an open subset  of R

is starshaped with respect to

a point x

∈ if for every x ∈, the segment joining x

to x, namely the set

{λx +(1 −λ)x

|λ ∈[0, 1]}

is entirely contained in

.

One says that  is strictly starshaped with respect to x

∈ if for every x ∈ ,

the segment

{λx +(1 −λ)x

|λ ∈[0, 1)}

is entirely contained in .

Normally one assumes that 0 ∈, and requires  to be strictly starshaped with

respect to 0. In this case the deﬁnition reads

x ∈

 implies λx ∈ ∀λ ∈[0, 1).

3.4 Problems with Critical Exponent 137

Convex sets are starshaped with respect to every internal point, and strict convexity

implies strict starshapedness. However it is easy to see that there are starshaped sets

that are not convex (those that look like a star, exactly).

The main property that we are going to use is contained in the following lemma.

The reader can prove it as an exercise or consult [18].

Lemma 3.4.24 Assume that  ⊂R

is smooth and strictly starshaped with respect

to 0. Let ν(x) denote the outward normal to ∂ at x. Then

ν(x) ·x>0 ∀x ∈∂.

The main nonexistence result is the following.

Theorem 3.4.25 Assume that  ⊂R

, with N ≥3, is open, bounded, smooth and

strictly starshaped with respect to 0. Then the problem

⎧

⎨

⎩

−u =|u|

∗

−2

u in ,

u>0 in ,

u =0 on ∂

(3.27)

has no solutions in H

().

The proof of this theorem is based on a regularity result, which asserts that any

weak H

() solution is in fact of class C

on  (see [45]), and on an integral iden-

tity established in [39] that we now describe. If one wants to bypass the regularity

argument, one can read the above theorem simply as follows: problem (3.27) has no

smooth solutions. Of course the same result holds for negative solutions, with the

same proof.

Theorem 3.4.26 (Pohožaev identity [39]) Let f : R → R be continuous and let

F(t) =



f(s)ds. Assume that  ⊂ R

is open and bounded. If u ∈ C

() is a

solution of the problem



−u =f(u) in ,

u =0 on ∂,

(3.28)

then

N −2



|∇u|

dx −N



F(u)dx =−



∂



∂u

∂ν



ν(x) ·xdσ. (3.29)

Proof The proof consists in multiplying the equation in (3.28)by∇u ·x and inte-

grating by parts by means of the Green’s formula (Theorem 1.2.5). We begin with

the left-hand side. We have

−



u ∇u ·xdx=



∇u ·∇(∇u ·x)dx −



∂

∂u

∂ν

∇u(x) ·xdσ. (3.30)

Now

138 3 Minimization Techniques: Lack of Compactness

∂

∂x

(∇u ·x) =

∂

∂x





i=1

∂u

∂x





i=1



∂

∂x

∂u

∂x





i=1

∂

∂x

∂u

∂x

so that

∇u ·∇(∇u ·x) =



j=1

∂u

∂x





i=1

∂

∂x

∂u

∂x





i=1

∂

∂x





j=1



∂u

∂x





+|∇u|

∇



|∇u|



·x +|∇u|

∇



|∇u|



·∇



|x|



+|∇u|

Using again the Green’s formula we obtain



∇u ·∇(∇u ·x)dx



|∇u|

dx +



∇



|∇u|



·∇



|x|





|∇u|

dx +



∂

|∇u|

∂

∂ν



|x|



dσ −



|∇u|

2 −N



|∇u|

dx +



∂

|∇u|

ν(x) ·xdσ, (3.31)

because (

|x|

) =N .

Now we notice that since u = 0on∂,wehave∇u(x) =

∂u

∂ν

(x)ν(x) for every

x ∈∂, so that |∇u|=|

∂u

∂ν

|and ∇u ·x =

∂u

∂ν

ν(x)·x on ∂. Taking this into account

and inserting (3.31)into(3.30) we see that

−



u ∇u ·xdx=

2 −N



|∇u|

dx +



∂

|∇u|

ν(x) ·xdσ

−



∂

∂u

∂ν

∇u(x) ·xdσ

2 −N



|∇u|

dx −



∂



∂u

∂ν



ν(x) ·xdσ. (3.32)

The treatment of the right-hand side is much easier. Indeed we have

f(u)∇u ·x =∇(F (u)) ·x =∇(F (u)) ·∇



|x|



so that by the Green’s formula,

3.4 Problems with Critical Exponent 139



f(u)∇u ·xdx=



∇(F (u)) ·∇



|x|





∂

F(u)

∂

∂ν



|x|



dσ −N



F(u)dx

=−N



F(u)dx, (3.33)

since F(u(x))= F(0) = 0on∂. Putting together (3.32) and (3.33), we conclude

that multiplying the equation by ∇u ·x yields

N −2



|∇u|

dx −N



F(u)dx =−



∂



∂u

∂ν



ν(x) ·xdσ,

as we wanted to prove. 

Proof of Theorem 3.4.25 Let u ∈ H

() solve (3.27). By regularity results (see

[45]), we can assert that u ∈C

(). Applying Theorem 3.4.26, with

f(u)=|u|

∗

−2

u and F(u)=

∗

|u|

∗

yields

N −2



|∇u|

dx −

∗



|u|

∗

dx =−



∂



∂u

∂ν



ν(x) ·xdσ.

Multiplying the equation in (3.27)byu and integrating shows that



|∇u|

dx =



|u|

∗

dx,

which we can insert in the preceding formula to obtain



N −2

−

∗





|u|

∗

dx =−



∂



∂u

∂ν



ν(x) ·xdσ.

But

N −2

−

∗

by deﬁnition of 2

∗

, so that



∂



∂u

∂ν



ν(x) ·xdσ =0.

Since  is strictly starshaped with respect to 0, we have ν(x) ·x>0 everywhere on

∂, and then, necessarily,

∂u

∂ν

(x) =0 ∀x ∈∂.

We conclude by integrating the equation in (3.27) and using again the Green’s for-

mula. We obtain



|u|

∗

−2

udx =−



u dx =−



∂

∂u

∂ν

dσ =0.

Since u is positive in , this is impossible. 

140 3 Minimization Techniques: Lack of Compactness

Remark 3.4.27 What about non starshaped domains? This question has been the ob-

ject of intense research starting in the eighties. Celebrated results (see the references

at the end of the chapter) ﬁrst showed that if the set  is an annulus, or has a small

hole, then problem (3.27) has a solution. The most general result states that if 

has nontrivial topology in some appropriate sense, then the problem is still solvable.

So for a while it seemed that the relevant point was the topology of . But then

other results showed that the problem still has a solution when  has trivial topol-

ogy (e.g., is contractible), but looks like a noncontractible domain, for example an

annulus with a thin radial slice removed. Summing up, the solvability of problems

with critical growth is a very delicate matter and depends subtly on the data of the

problem.

3.5 Exercises

1. Assume that N ≥3. Prove that there exists a constant C>0 such that



|x|

dx ≤C



|∇u|

dx ∀u ∈D

1,2

). (3.34)

This is the celebrated Hardy inequality. Hint: assume ﬁrst that u ∈ C

∞

)

and notice that left-hand side is ﬁnite. Write, for λ>0 and x ∈R

(x) =−



+∞

dλ

(λx) dλ =−2



+∞

u(λx)∇u(λx) ·xdλ.

Then divide by |x|

and integrate over R

setting y =λx. Conclude by density.

2. Use the Hardy inequality (3.34) to prove that the expression

(u|v) =



∇u ·∇vdx+



|x|

uv dx

deﬁnes a scalar product in D

1,2

that induces a norm equivalent to the standard

one. Then

(a) Let u=(u|u)

1/2

. For every u ∈ D

1,2

) and every λ>0 deﬁne

(x) =λ

N−2

u(λx).

Prove that u

=u for every λ>0.

(b) Deﬁne

C =inf



u



u ∈D

1,2



|u|

∗



Prove that C>0.

3. Using the notation and the results of Exercise 2, ﬁnd a radial non negative and

non trivial solution of the problem

⎧

⎪

⎨

⎪

⎩

−u +

|x|

u =|u|

∗

−2

u ∈D

1,2