Ωcδ
2
|∇((1 − φδ)v)|2
|x|2an dx
C
δ, ρ, ρ, n
Ωcδ
2
|(1 − φδ)v|n2n−2
|x|2nann−2 dx
n−2
n
C
δ, ρ, ρ, n
Ωδc
dnn−2|(1 − φδ)v|n2n−2
|x|(1+2an)n−2n dx
n−2
n
,
where in the last inequality we have used again (2.65) and the fact that Ωδc⊂ Ωcδ 2
. Now by Theorem 2.4 in[5]and (2.65) for sufficiently small δ > 0 we have
Ωδ
d|∇(φδv)|2
|x|1+2an dx+
Ωδ
d|φδv|2
|x|1+2an(1+ d2+σ)dx
C
δ, ρ, ρ, n
Ωδ
dn−2n |φδv|n2n−2
|x|(1+2an)n−2n dx
n−2
n
.
The rest of the proof is similar as in Theorem2.2, and we omit it. 2
Theorem 2.12. Let n= 3 and Ω be an exterior domain not containing the origin. Then the following inequality is valid
Ω
d
|x|2
|∇u|2+ u2 1+ d2+σ
dx C
Ω
d3u6X4(|x|ρ)
|x|6 dx
1
3
, ∀u ∈ C0∞(Ω)
where X(t)= (1 + ln t)−1. Moreover, the power 4 on X cannot be replaced by a smaller power.
Proof. The proof of the theorem is the same as in Theorem2.11. The only difference is that, we use here Lemma2.4instead of Lemma2.10. 2
3. Existence of minimizers in suitable spaces and their behavior
In this section, we assume that the set Ω is an exterior domain not containing the origin. By Theorems2.9(for n= 3) and2.8(for n 4) we note that there exists a constant λ ∈ R such that
−∞ < λ = inf
u∈C0∞(Ω)
Ω|∇u|2dx−14
Ω u2 d2dx
Ω u2 1+d2+σ dx
, (3.1)
where σ > 0.
The main goal of this section is to prove the existence of a ground state function φ∈ Hloc1 (Ω) which solves the corresponding Euler–Lagrange of (3.1) in the weak sense i.e.
−φ − φ
4d2= λ φ
1+ d2+σ in Ω. (3.2)
Also, we would like to know how this function φ behaves. The space which we use to prove the existence of φ is D1,20 (Ω; |x|, d) which is the closure of C0∞(Ω)functions under the norm
u 2D1,2
0 (Ω;|x|,d)=
Ω
d
|x|2an+1
|∇u|2+ u2 1+ d2+σ
dx,
where an=n−22 +
(n−2)2
4 −14 and σ > 0. By Theorems2.12and2.11, we have for n= 3 and n 4 respectively the following inequalities
u 2D1,2
0 (Ω;|x|,d) C
Ω
d3u6X4(|x|ρ )
|x|6 dx
1
3
, ∀u ∈ C0∞(Ω), (3.3)
where ρ= inf{|x|: x ∈ ∂Ω} and X(t) = (1 + ln t)−1.
u 2D1,2
0 (Ω;|x|,d) C
Ω
dnn−2un2n−2
|x|(2an+1)n−2n dx
n−2
n
, ∀u ∈ C0∞(Ω), (3.4)
where an=n−22 +
(n−2)2 4 −14.
Theorem 3.1. Let n= 3 and let Ω be an exterior domain not containing the origin. Then there exists a ground state function φ ∈ Hloc1 (Ω) such that φ solves the problem (3.2) in the weak sense.
Proof. Let η∈ C2(Ω)be a function such that η(x)= d12(x)near the boundary, say, d(x) ε0, and η(x)= |x|−12 away from the boundary, say|x| R > 2R0= 2 supx∈∂Ω|x| and c1 η c2
otherwise, where c1, c2are positive constants. Then v W1
0(Ω;|x|,d)is equivalent with the norm v 2=
Ω
η2
|∇v|2+ v2 1+ d2+σ
dx.
Also, by (3.3), we have the following inequality
Ω
η2
|∇v|2+ v2 1+ d2+σ
dx C
Ω
η6v6X4
|x|
ρ
dx
1
3
, (3.5)
where ρ= inf{|x|: x ∈ ∂Ω}. Changing the variables by u = ηv in (3.1) we have the equivalent problem
−∞ < λ = inf
v∈C∞0 (Ω)
Ωη2|∇v|2dx−
Ω(ηη+14dη22)v2dx
Ω η2v2 1+d2+σ dx
. (3.6)
For R and δ sufficiently large and small respectively we get
where in the last two inequalities we have used Hölder inequality and inequality (3.5). Also since ηη+14ηd22 ∈ L∞(Ω), we have
which implies
Finally we combine the estimates (3.7), (3.8) and (3.9) to deduce that for any ε > 0 there exist Mεsuch that re-alizes the infimum in (3.6). To this end let wk be a minimizing sequence normalized by
Ω v2
1+d2+σdx= 1. Then using (3.10) we can easily obtain (by (3.6)) that the sequence wk is bounded i.e. supk wk < N. Therefore there exists a subsequence still denoted by wk such that it converges to W01(Ω; |x|, d)-weakly to ψ1. Clearly by embedding theorems for R > 0 and δ > 0 large and small enough respectively we have
and the result follows by lower semicontinuity of the gradient term of numerator in (3.6). 2 Theorem 3.2. Let n 4 and Ω be an exterior domain not containing the origin. Then there exists a function φ∈ Hloc1 (Ω) such that φ solves the problem(3.2) in the weak sense.
Proof. Let η∈ C2(Ω)be a function such that η(x)= d12(x)near the boundary, say, d(x) ε0,
0(Ω;|x|,d)is equivalent with the norm v 2=
Also, by (3.4), we have the following inequality
Changing the variables by u= ηv in (3.1) we have the equivalent problem
−∞ < λ = inf
The rest of the proof is the same as in Theorem3.1and we omit it. 2
Theorem 3.3. The asymptotic behavior of φ in Theorems3.1and 3.2is like d12 near to the boundary and like|x|−anaway from the boundary, where an=n−22 +
(n−2)2 4 −14.
Proof. Assume first n 4. It is well known that the eigenfunction φ ∼ d12 (see Lemma 7 in [4]for a lower bound and see Appendix in[8]for an upper bound). Thus we will focus away from the boundary such that ψ1is the minimizer of (3.15). For|x| > R where R is large enough, ψ1solves the problem
Lψ1= − div for C1>0 and|x| large enough. On the other hand the first eigenfunction ψ1of L satisfies
Now since both function ψ1and 1+ C1|x|−σ are smooth away from the boundary, we can
By the above inequality, equality (3.15) and the fact that M >−λ, we have g+= 0 and the lower bound follows.
For the upper bound we first note by (2.64) that
where in the last inequality we have chosen R big enough. Thus by (3.17) and (3.18) we have
C(Ω,n) 2
BRc |u|n−22n
|x|2nann−2
dxn−2n
BRc 1
(1+d2+σ)n2 dxn
2
Ω |u|n2n−2
|x|2nann−2
dxn−2
n
− 1 → ∞, as R → ∞, ∀u ∈ C0∞ BRc
.
(3.19) Since 14( 1
d2|x|−|x|13)4|x|2R20d2, for|x| > R0= supx∈∂Ω|x| we have two cases:
Case 1: If 0 < σ < 1 then as before we see that (L−|x|2an(1λ+d2+σ))(1− C1|x|−σ) 0 for C1>0 big enough and (L−|x|2an(1λ+d2+σ))ψ1= 0. We next choose ε > 0 big enough so that g(x)= εψ1− (1 − C1|x|−σ) 0 on ∂BRc.
Case 2: If σ 1 we note that (L − |x|2an(1λ+d2+σ))(1− C1|x|−1) 0 for C1 >0 big enough and (L− |x|2an(1λ+d2+σ))ψ1= 0. We next choose ε > 0 big enough so that g(x) = εψ1− (1 − C1|x|−1) 0 on ∂BRc.
Thus in both cases, since g+is a test function we have
BRc
1
|x|2an∇g∇g+dx+1 4
1
|x|2an+2− 1 d2|x|2an
gg+dx−
BRc
λ
|x|2an(1+ d2+σ)gg+dx 0,
from which it follows
BRc
|x|12an|∇g+|2+14(|x|2an+21 − 1
d2|x|2an)g+2dx
BRc
g+2
|x|2an(1+d2+σ)
λ.
This contradicts with (3.19) unless g+= 0 from which follows the upper bound for φ.
For n= 3 the only difference is that in (3.19) we use (3.5) instead of (3.14). 2 4. Hardy–Sobolev inequalities in domains above the graphs ofC1,1functions
In this section we will prove Hardy–Sobolev type inequalities in domains above the graphs of C1,1functions. More precisely, let μ > 0 and let Γ: Rn−1→ R satisfy the conditions |∇Γ | < μ and Γ ∈ C1,1(Rn−1). We call the set
Ω=
x, xn
∈ Rn: xn> Γ x
,
domain above the graph of a C1,1function.
The half spaceRn+= {(x, xn)∈ Rn: xn>0} is an example of a domain above the graph of C1,1 function. Especially, we have the Hardy–Maz’ya–Sobolev inequality in half space (for n 3)
Rn+
|∇u|2dx−1 4
Rn+
u2
xn2dx C(n)
Rn+
|u|n2n−2dx
n−2
n
, ∀u ∈ C0∞
Rn+
. (4.1)
We note here that the inequality (4.1) is valid for n= 3 without using some logarithmic function as in exterior domains. Thus, the proof of Hardy–Maz’ya–Sobolev inequality in domain above the graphs of C1,1functions is different from the proof in exterior domains.
Set d(x)= infy∈∂Ω|x − y| and δ(x) = xn− Γ (x). Then we can easily prove that kδ(x) d(x) δ(x), where k =1+μ1 . Then we have the following theorem.
Theorem 4.1. Let n 3 and Ω be the domain above the graph of C1,1function which satisfies
−d 0. Then there exists a positive constant C which depends only on n and μ, such that
Ω
|∇u|2dx−1 4
Ω
u2
d2dx C(n, μ)
Ω
|u|n2n−2dx
n−2
n
, ∀u ∈ C0∞(Ω). (4.2)
Proof. Set u= d12vthen (4.2) becomes equivalent to
Ω
d|∇v|2dx−1 2
Ω
du2dx C
Ω
dn−2n |u|n−22n dx
n−2
n
.
Since−d 0 and kδ(x) d(x) δ(x); k =1+μ1 , it is enough to prove
Ω
δ|∇v|2dx C(μ, n)
Ω
δnn−2|v|n2n−2dx
n−2
n
.
But by inequality (4.1) if we set u= xn12vwe have that
Rn+
yn|∇yv|2dy C(n)
Rn+
y
n−2n
n |v|n2n−2dy
n−2
n
, ∀v ∈ C0∞
Rn+
. (4.3)
Now set in (4.3) xi= yi for i= 1, . . . , n − 1 and xn= yn+ Γ (y)then∇yv= ∇xv+ vxn∇xΓ and vyn= vxn, thus,
C(μ)|∇xv| |∇yv| c(μ)|∇xv| and by (4.3) we have
Ω
δ|∇v|2dx C(μ)
Ω
δnn−2|v|n2n−2dx
n−2n ,
which is the desired result. 2
Lemma 4.2. Let a, b, p and q be such that 1 p < n, p < q npn−p and b= a − 1 +qqp−pn.
Then for any η > 0, there holds:
λη−1−λλ xanv
L
np
n−p(Rn+)+ (1 − λ)ηxan−1v
Lp(Rn+)xbnv
Lq(Rn+), ∀u ∈ C0∞
Rn+
where
0 < λ=n(q− p)
qp 1. (4.4)
Proof. For p∗=nnp−p and λ=n(qqp−p) we use Hölder inequality to obtain:
Rn+
xnqbvqdx=
Rn+
xnaλqvλqdqb−aλ|v|q(1−λ)dx
xapn ∗|v|p∗λq
p∗
Rn+
xnp(a−1)|v|pdx
(1−λ)qp
⇔ xnbv
Lq(Rn+)xnavλ
L
np n−p(Rn+)
xna−1v1−λ
Lp(Rn+).
Now use the fact that xλy1−λ λη−1−λλ x+ (1 − λ)ηy, for any η > 0, to reach to the desired result. 2
Theorem 4.3. Let n 3. Then the following inequality is valid
Rn+
xn|∇u|2dx
Rn+
xnu2(nn−1+1)dx
nn+1−1
, ∀u ∈ C0∞
Rn+
where C is a positive constant which depends only on dimension n.
Proof. By Lemma4.2, if we choose p= 1 and η = 1 we have the following inequality
xbnv
Lq(Rn+) λxnav
L
n
n−1(Rn+)+ (1 − λ)xna−1v
L1(Rn+), (4.5) where λ=n(qq−1), 1 < qn−1n and b= a − 1 +q−1q n.
Now, for any a = 0 we have
Rn+
xna−1|v| dx = 1 a
Rn+
∇xna∇xn|v| dx = −1 a
Rn+
xan∇xn∇|v| dx 1
|a|
Rn+
xna|∇v| dx. (4.6)
By Sobolev inequality and (4.6) we have
Sndxnav
L
n n−1(Rn+)
Rn+
∇xnavdx a
Rn+
xna−1|v| dx +
Rn+
xna|∇v| dx
2
Rn+
xna|∇v| dx, (4.7)
where Sn= nπ12(Γ (1+n2))−1n see[19], p. 189. Thus by (4.5), (4.6) and (4.7) we have
Rn+
xbqn |v|q
1
q
2λ
Sn +1− λ
|a|
Rn+
xna|∇v| dx. (4.8)
Now, replace v by usin the above inequality to obtain
Rn+
xbqn |u|sq
1
q
2λ
Sn +1− λ
|a|
s
Rn+
xna|u|s−1|∇u| dx
2λ
Sn +1− λ
|a|
s
Rn+
xna|∇u|2dx
1
2
Rn+
xna|u|2s−2dx
1
2
and the result follows, if we choose a= 1, q = n+1n λ=n+1n and s=n2n−1. 2 Finally, we prove a Hardy–Sobolev type inequality which is of independent interest.
Theorem 4.4. Let n 3 and Ω be the domain above the graph of C1,1function which satisfies
−d 0. Then there exists a positive constant C which depends only on n and μ, such that
Ω
|∇u|2dx−1 4
Ω
u2
d2dx C(μ, n)
Ω
du2(nn−1+1)dx
n−1
n+1
, ∀u ∈ C0∞(Ω). (4.9)
Proof. The proof is the same as in Theorem 4.1. The only difference is that we use Theo-rem4.3. 2