CONCENTRATING SOLUTIONS FOR A PLANAR ... - CAPDE

10 PIERPAOLO ESPOSITO, MONICA MUSSO, AND ANGELA PISTOIA
Hence, by the choice of v ∞ , w 0 and w 1 and expansion (2.10), we obtain that
∆v + v p = O( 1 p 4 log 6 (|y| + 2)
(1 + |y| 2 ) 2 ) in |y| ≤ Ce p 8 .
As before, we will see now that a proper choice of the parameter µ will automatically
imply that this approximation for v is also good for the boundary
condition to be satisfied. Indeed, observe that by (2.7), (2.9) and Lemma
2.1 we get: for i = 1, 2
(
P w i ( x − ξ )
) = w i ( x − ξ ) − 2πC i H(x, ξ) + C i log δ + O(δ) in C 1 (¯Ω)(2.11)
δ
δ
(
P w i ( x − ξ )
) = −2πC i G(x, ξ) + O(δ) in C 1
δ
loc (¯Ω \ {ξ}),
provided ξ is bounded away from ∂Ω. If we take µ as a solution of
(
log(8µ 4 ) = 8πH(ξ, ξ) 1 − C 0
4p − C )
1
4p 2 + log δ (
C 0 + C )
1
,
p p
we get that
u(x) = e 2(p−1)
p
p p
p−1 µ 2
p−1
[P U δ,ξ (x) + 1 p P (
w 0 ( x − ξ
δ
)
) + 1 (
p 2 P
w 1 ( x − ξ
δ
)]
)
is a good first approximation in order to construct a solution for (1.1) with
just one concentration point.
Let us remark that µ bifurcates, as p gets large, by ¯µ = e − 3 4 e 2πH(ξ,ξ) , solution
of equation
More precisely,
log(8µ 4 ) = 8πH(ξ, ξ) − C 0
4
= 8πH(ξ, ξ) − 3 + log 8.
µ = e − 3 4 e 2πH(ξ,ξ) (1 + O( 1 p )).
Let us see now how things generalize if we want to construct a solution to
problem (1.1) which exhibits m points of concentration. Let ε > 0 fixed and
take an m−uple ξ = (ξ 1 , . . . , ξ m ) ∈ O ε , where
O ε = {ξ = (ξ 1 , . . . , ξ m ) ∈ Ω m : dist (ξ i , ∂Ω) ≥ 2ε, |ξ i − ξ j | ≥ 2ε, i ≠ j} .
Define
U ξ (x) =
where
m∑
j=1
γµ
1
γ = p p
2
p−1
j
[P U δj ,ξ j
(x) + 1 (
p P w 0 ( x − ξ )
j
) + 1 (
δ j p 2 P w 1 ( x − ξ ) ]
j
) ,
δ j
p−1 e − p
2(p−1)
and δ j = µ j e − p 1
4 ,
C ≤ µ j ≤ C. (2.12)