CONCENTRATING SOLUTIONS FOR A PLANAR ... - CAPDE

0, 1, 2. Hence, ˆφ ∞ j ≡ 0 for any j = 1, . . . , m contradicting lim inf n→+∞ ‖φ n ‖ i >
0.
4 th Step. We prove that there exists a positive constant C > 0 such that
any solution φ of equation Lφ = h in Ω, φ = 0 on ∂Ω, satisfies
‖φ‖ ∞ ≤ Cp‖h‖ ∗ ,
when h ∈ C 0,α (¯Ω) and we assume on φ only the orthogonality conditions
(3.5). Proceeding by contradiction as in Step 3, we can suppose further that
p n ‖h n ‖ ∗ → 0 as n → +∞ (3.9)
but we loss in the limit the condition ∫ 8
R 2 z
(1+|y| 2 ) 2 0 (y) ˆφ ∞ j
have that
19
= 0. Hence, we
ˆφ n j → C j
|y| 2 − 1
|y| 2 + 1 in C0 loc (R2 ) (3.10)
for some constants C j . To reach a contradiction, we have to show that
C j = 0 for any j = 1, . . . , m. We will obtain it from the stronger condition
(3.9) on h n .
To this end, we perform the following construction. By Lemma 2.1, we
8
find radial solutions w and t respectively of equations ∆w + w =
(1+|y| 2 ) 2
8
8
8
z
(1+|y| 2 ) 2 0 (y) and ∆t + t = in R
(1+|y| 2 ) 2 (1+|y| 2 ) 2 , such that as |y| → +∞
2
w(y) = 4 3
1
1
log |y| + O( ) , t(y) = O(
|y| |y| ),
since 8 ∫ +∞
0
t (t2 −1) 2
dt = 4 (t 2 +1) 4 3 and 8 ∫ +∞
0
t t2 −1
dt = 0.
(t 2 +1) 3
For simplicity, from now on we will omit the dependence on n. For j =
1, . . . , m, define now
u j (x) = w( x − ξ j
) + 4 δ j 3 (log δ j)Z 0j (x) + 8π 3 H(ξ j, ξ j )t( x − ξ j
)
δ j
and denote by P u j the projection of u j onto H0 1(Ω). Since u j − P u j +
4
3 log 1
|·−ξ j |
= O(δ j ) on ∂Ω (together with boundary derivatives), by harmonicity
we get
The function P u j solves
P u j = u j − 8π 3 H(·, ξ j) + O(e − p 4 ) in C 1 (¯Ω),
P u j = − 8π 3 G(·, ξ j) + O(e − p 4 ) in C 1 loc (¯Ω \ {ξ j }).
∆P u j + W (x)P u j = e U j
Z 0j +
(3.11)
(
)
W (x) − e U j
P u j + R j , (3.12)
where
R j (x) =
(
P u j − u j + 8π )
3 H(ξ j, ξ j ) e U j
.