If we impose a more restrictive growth we collect some results to improve the statement of [4, Theorem 1.1] concerning the multiplicity of positive solutions for (1.6).
Remark 6.1. Assume N 4 and 1 < p < (N + 2)/(N − 2). Then, there exists α0(p, N )such that for all α > α0(p, N ), problem (1.6) has at least[N/2] + 1 non-rotational equivalent positive classical solutions. Precisely:
(1) one radial positive solution;
(2) the[N/2] − 1 solutions given by [4, Theorem 1.1];
(3) the (positive) ground state solution, which is non-radial provided α > α0(p, N )but, which is foliated Schwarz symmetric; see the proof in [28, Theorem 3.1 and p. 281].
Similarly, assuming more restrictive growth, we can improve the statement of Theorem 1.7 concerning the multiplicity of positive solutions for (1.5) under Dirichlet boundary condition.
Remark 6.2. If N= 4 assume p > 1. If N 5 assume 1 < p < (N + 4)/(N − 4). Then, there exists α0(p, N ) >0 such that for all large α > α0(p, N ), problem (1.5) under Dirichlet bound-ary condition (Bu = u,∂u∂ν) has at least[N/2] + 1 non-rotational equivalent positive classical solutions. Precisely:
(1) one radial positive solution;
(2) the[N/2] − 1 solutions given by Theorem 1.7;
(3) the (positive) ground state solution, which is non-radial provided α > α0(p, N )is sufficiently large but, which is foliated Schwarz symmetric (see the proof in [5, Theorem 2]).
In addition all these solutions satisfy u > 0 in B and u changes sign in B.
Now we comment on results about the existence of multiple positive solutions of (1.5) un-der Navier boundary condition. Firstly, analyzing the pairs (p, l) that satisfy (H1)–(H2) from Theorem 1.8 we present a multiplicity result there contained.
Remark 6.3. Consider (1.5) under Navier boundary condition (Bu = u, u). Then Theorem 1.8 guarantees that for all large α:
(1) If N= 4, 5 and p > 3, then (1.5) has [N/2] − 1 non-radial positive classical solutions.
(2) If N = 6, 7, 8 and 4 < p + 1 < 2(N − 1)/(N − 5), then (1.5) has [N/2] − 1 non-radial positive classical solutions.
(3) If N 9 assume 2(N+ 6)
N+ 1 < p+ 1 <2(4N− 1)
4N− 21 in case 4N− 1
5 ∈ N and
2[(N + 1)/5]
[(N − 4)/5] < p+ 1 <2[(4N + 4)/5]
[(4N − 16)/5] in case4N− 1 5 ∈ N./
(a) If N is odd and 4N5−1∈ N, then (1.5) has [(4N − 1)/5] −N+12 non-radial positive clas-sical solutions.
(b) If N is odd and 4N5−1∈ N, then (1.5) has [(4N − 1)/5] −/ N−12 non-radial classical positive solutions.
(c) If N is even and4N5−1∈ N, then (1.5) has [(4N −1)/5]−N2 non-radial classical positive solutions.
(d) If N is even and 4N5−1 ∈ N, then (1.5) has [(4N − 1)/5] −/ N−22 non-radial classical positive solutions.
In case (N − 4)(p + 1) < 2N, it is proved in [7] that the (positive) ground state solution of (1.5) under Navier boundary condition is foliated Schwarz symmetric. Therefore, each pair (p, l) satisfying (H1)–(H2) from Theorem 1.8, (N− 4)(p + 1) < 2N and with α sufficiently large, produces a non-radial positive classical solution which is non-rotational equivalent to the (positive) ground state solution. This analysis helps to restate Remark 6.3 in the same sense that Theorem 1.7 was restated at Remark 6.2.
Acknowledgments
We thank the referees for their careful reading of the first version of this manuscript and their valuable suggestions.
Appendix A. Proofs of Propositions 4.13 and 4.14
Let us start by proving the following proposition.
Proposition A.1. Let N 4 and let l be an integer such that 2 N − l l. Let 2 < p + 1 < 2l, where 2l := 2(l + 1)/(l − 3) if l 4 and 2l is any number in the range (2,∞) if l < 4. Set ql:= N − (N − l + 1)22
l. Given α0>2l2ql (see Theorem 1.4), there exists C(p, N, l) > 0 such that
m,α,l C(p, N, l)α(N−l−1)1p+1−p+2pp+1+3, ∀α α0, (A.1) with m,α,l(under both Dirichlet and Navier boundary conditions) as defined by (4.5).
Proof. Let u: RN → R be a regular function such that u(y, z) = u(|y|, |z|) where y = (y1, . . . , yl), z= (z1, . . . , zN−l)and x= (y, z). Given such a function u let v(s, t), for s > 0 and t > 0 be the function defined by v(s, t):= u(y, z) with s = |y| and t = |z|. Under these variables we have
u(x)= vss(s, t )+l− 1
s vs(s, t )+ vt t(s, t )+N− l − 1
t vt(s, t )=: Luv(s, t ).
Next using polar coordinates in the variables (s, t), s = ρ cos θ and t = ρ sin θ, setting
Then w∈ Cc∞( A)where
First observe that from (A.4)
Now observe that there exist positive constants C1, C2such that
C1N−l−1 H (θ) C2N−l−1, ∀(r, θ) ∈ A. (A.6)
and, in order to simplify notation, we rewrite the last inequality in the variables (ρ, θ ) (this is not a change of variables, just a change of notation)
B
|x|α|u|p+1dx CN−l+1
A
ψ(ρ, θ )p+1ρN−1dρ dθ, with = N
α+ N. (A.7)
Now, using (A.2), we estimate
B
|u|2dx= C
A
Buw(ρ, θ )2ρN−1H (θ ) dρ dθ.
Observe that with π6< θ <π3
Buw(ρ, θ )2 C
w2ρρ+ 1
ρ2wρ2+ 1
ρ4wθ θ2 + 1 ρ4w2θ
.
Hence, using that w(ρ, θ )= ψ(ρ1,θ)and proceeding with steps similar to the ones above we obtain
B
|u|2dx CN−l−3
A
ψρρ2 (ρ, θ )ρ4+ ψρ2(ρ, θ )ρ2+ ψθ θ2 (ρ, θ )
+ ψθ2(ρ, θ )
ρ−1dρ dθ, = N
α+ N. (A.8)
From (A.7) and (A.8), it follows that there exists C > 0 such that
m,α,l C
α+ N N
(N−l−1)1−pp+1+2p+3p+1
Cα(N−l−1)1p−p+1+2pp+3+1, ∀α α0, (A.9)
since
(N− l − 1)1− p
p+ 1+ 2p+ 3
p+ 1>0. (A.10)
Observe that (A.10) is a straightforward consequence of
(l− 3)(p + 1) < 2(l + 1) and N − l l.
Observe also that (l− 3)(p + 1) < 2(l + 1) comes directly from p + 1 < 2l. 2 On the other hand the following result also holds.
Proposition A.2. Let N 4 and let l be an integer such that 2 N − l l. Let 2 < p + 1 < 2l, where 2l:= 2(l + 1)/(l − 3) if l 4 and 2l is any number in the range (2,∞) if l < 4. There exists α0(p, N ) and C(p, N ) >0 such that:
m,α,rad,DBC C(p, N)αp+12 +3, ∀α α0(p, N ), (A.11) m,α,rad,NBC C(p, N)αp+12 +2, ∀α α0(p, N ). (A.12) Here m,α,rad,DBCand m,α,rad,NBCare as defined by (4.3).
Proof. Let u∈ Hrad, with Hradas defined by (4.1). Associated to u, let v be as in Theorem 2.2 and Remark 2.1 such that u(x)= v(|x|). Set =NN+α and
w(ρ)= v
exp(−ρ)
. (A.13)
So w∈ C1([0, ∞)), w (in the classical sense) exists a.e. in (0,∞) and w(0) = 0. In addition,
v
exp(−ρ)
= −−1exp(ρ)w (ρ) and v
exp(−ρ)
= −2exp(2ρ)
w (ρ)+ w (ρ)
. (A.14)
Now observe that
B
u(x)p+1|x|αdx= ωN
1
0
v(t )p+1tα+N−1dt
= ωN
∞
0
w(ρ)p+1exp(−Nρ) dρ. (A.15)
On the other hand, from (A.14) we have that
B
|u|2dx= ωN
1
0
v (t )+N− 1 t v (t )
2tN−1dt
= ωN−3
∞
0
w (ρ)− (N − 2)w (ρ)2exp
−(N − 4)ρ dρ
= ωN−3
∞
0
exp
−(N − 2)ρ
w (ρ) 2exp(N ρ) dρ. (A.16)
Therefore, from (A.15) and (A.16), we have that, for every u∈ Hradand u= 0:
Case A: Dirichlet boundary condition. Observe that in this case z(0)= 0 and therefore
w(t ) =
Using the Cauchy–Schwarz inequality in the inner integral we can estimate
w(t ) ∞
Observing that
Therefore, from (A.17) and (A.21) it follows that there exists a constant C(p, N ) > 0 such that
This finishes the proof of (A.11).
Case B: Navier boundary condition. Observe that in this case z does not necessarily vanish at zero and therefore the identity z(s)=s
0z (ρ) dρdoes not necessarily holds. For this reason we apply z(s)= −∞
As in the previous case we use the Cauchy–Schwarz inequality and we obtain
w(t ) ∞
Therefore, there exists α0(p, N ) >0 and C(p, N ) > 0 such that
∞
0
w(t )p+1exp(−Nt) dt
C(p, N)
∞
0
exp
−(N − 2)ρ
w (ρ) 2exp(N ρ) dρ
(p+1)/2
−p+12 (A.22)
for all α > α0(p, N ).
Finally, from (A.17) and (A.22) it follows that there exists a constant C(p, N ) > 0 such that
m,α,rad C(p, N)−(p+12 +2)= C(p, N)
N
N+ α
−(p+12 +2)
C(p, N)α(p+12 +2), ∀α > α0(p, N ).
This finishes the proof of (A.12). 2
End of proofs of Propositions 4.13 and 4.14. Now we just compare some exponents. Proposi-tion 4.13 is a straightforward consequence of ProposiProposi-tion A.1 and (A.11), since p > 1 and N > l.
Also, the proof of Proposition 4.14 follows from Proposition A.1 and (A.12).
Remark A.3. The change of variables w(ρ)= v(ρ), which has been used in the proof of [29, Theorem 4.1] to treat (1.6), in place of (A.13) induces
m,α,rad,DBC C(p, N)αp+12 +2, ∀α α0(p, N ) and m,α,rad,NBC C(p, N)αp+12 +2, ∀α α0(p, N ).
This turns out to be our estimate in the case of Navier boundary condition (cf. (A.12)), but weaker in the case of Dirichlet boundary condition and not sufficient for our purposes.
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