Low dimension anomalies and solvability in higher dimensions for some perturbed Pohozaev equation G. Mancini∗ Dipartimen...
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Low dimension anomalies and solvability in higher dimensions for some perturbed Pohozaev equation G. Mancini∗ Dipartimento di Matematica, Universit´a di Roma Tre Abstract In this talk, we will present a new solvability condition in higher dimension for a Brezis Nirenberg type equation. This result, which we believe to be sharp in some sense, has been obtained in collaboration with Adimurthi
1
Introduction
Very surprising ”low dimension phenomena” have been observed by Brezis and Nirenberg in their pioneering work [BN] concerning perturbations of the ”Pohozaev equation”
(P )
N +2
−∆u = u N −2 + p(u) u>0 in Ω u=0 in ∂Ω
in Ω
Here, Ω ⊂ 3, (P ) has a solution ∀Ω (ii)if N = 3, (P ) is not solvable in general (in fact (P ) has a solution in a ball Br iff r is large enough) As for (i) it follows from the very general sufficient condition they found (assuming p ≥ 0, but this is non really relevant)
(∗) lim →0
4−N 2
Z
|x|≤ √1
P ((
N −2 1 ) 2 ) = +∞ + |x|2
Rs N where P (s) = 0 p(t)dt. Of course, p(s) = s N −2 does not satisfy (∗) if N = 3. This example shows two facts : 0∗
Supported by M.U.R.S.T.
1
- anomalies may occurr in dimension 3 - condition (∗) is sharp in dimension 3. In view of this example, two related questions naturally arise: - is condition (∗) sharp also in higher dimensions ? - do anomalies occur also in higher dimensions ? We also refer to [PS], [BG] for related problems and interpretation of ”low dimension phenomena”. In a recent paper [AY], Adimurthi and Yadava gave an adfirmative answer to the question above in dimensions 4, 5, 6: they exibited a class of compactely supported perturbations PN such that (k) if N = 3, 4 and p ∈ PN , there are anomalies (i.e. there are solutions on Br iff r is large) R (kk) if N = 5, 6, and p ∈ PN and, in addition, P ( |x|N1 −2 ) = 0, then there are anomalies. (kkk) if N ≥ 7, and p ∈ PN , then (P ) is solvable on any ball. Of course, in cases (k) − (kk) Rcondition (∗) is violated: in case (k), because obviously P ( |x|N1 −2 ) < ∞, while, in case (kk), because R R N −2 1 2 ) = 0() (see Lemma 4.1 below). P ( |x|N1 −2 ) = 0 implies |x|≤ √1 P (( +|x| 2)
In the attempt to extend the existence result in (kkk) to more general perturbations and domains, Adimurthi and myself [AM] realized that condition (∗) is not sharp in dimension N ≥ 7. To be more precise, let us first give a closer look to condition (∗) , limiting ourselves, for sake of simplicity, to perturbations p in the class P := C0∞ ((0, ∞)). If we denote, for a given p ∈ P
Ik = Ik (p) :=
Z
0 for some k = 0, 1... then (∗) holds iff N > 2k + 4.In fact, 4−N R N −2 1 2 )= if I0 = I1 = ... = Ik−1 = 0 then lim→0 2 |x|≤ √1 P (( +|x| 2) k N −2 4+2k−N R d 1 2 ) lim→0 2 P (( +|x| 2) I1 = − N 2−2 p( |x|N1 −2 ) |x|1N ⇒ (∗) fails ∀N . On the other hand, we will prove that, if
(∗∗) I0 = 0 and
Z
1 N
|∇H|2 −
Z
p(
1 1 ) >0 |x|N −2 |x|N
then(P ) has a solution on any Ω if N ≥ 7. Here, H is the finite energy solution of − ∆u = p(
1 ) in N1 p( |x|N1 −2 ) |x|1N , then - if N ≥ 7, (P ) has a solution ∀Ω - if N = 3, 4, 5, 6, (P ) has a solution if Ω contains a large ball. 3
By standard arguments, a non zero critical point of 1 ˜ E(u) := 2
Z
N −2 |∇u| − 2N 2
Z
+
(u )
2N N −2
−
Z
P (u+ ) u ∈ H01 (Ω)
˜ enjoies the mountain pass is a solution of (P ). The energy functional E geometry, and, again by standard arguments, a mountain pass level is critical N
S 2 (S := best Sobolev constant). provided it is strictly smaller than N In fact it is enough to prove that, for some V ∈ H01 (Br ) , Br ⊂ Ω, it results N
S 2 max E(tV ) < (2.1) t≥0 N R R R 2N −2 where E(u) = 21 R|∇u|2 − N2N |u| N −2 − P (u) , u ∈ H01 (Br ) and P is s the even extension of 0 p(t)dt, s ≥ 0. This is because, after extending V equal to zero outside Br , it results ˜ E(t|V |) ≤ E(tV ).
3
Choise of the test function and basic estimates
To get (2.1), we will choose V = V , for suitably small, where V ∈ H01 (Br ) satisfies N +2
− ∆V = UN −2 + p(U ) on Br ⊂ Ω
(3.1)
N −2
Here U (x) = − 2 U ( x ) p N −2 cN 2 , cN = N (N − 2) U (x) = ( 1+|x| 2) N +2
Recall that −∆U = UN −2 in