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Ann. I. H. Poincaré – AN 30 (2013) 959–960
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Erratum
Erratum to “Fading absorption in non-linear elliptic equations”
[Ann. I. H. Poincaré – AN 30 (2) (2013) 315–336]
✩Moshe Marcus
a,∗, Andrey Shishkov
baDepartment of Mathematics, Technion Haifa, Israel bInstitute of Appl. Math. and Mech., NAS of Ukraine, Donetsk, Ukraine
Received 21 December 2012; accepted 22 December 2012 Available online 23 January 2013
The purpose of this note is to correct an error that occurred in the proof of Theorem 1.2 of the paper ‘Fading absorption in non-linear elliptic equations’ which appeared in Ann. I. H. Poincaré – AN, 2013.
The theorem itself is correct as stated. However Proposition 3.1 (used in its proof) and relation (3.18) are wrong.
We restate Proposition 3.1 and provide a modified argument to replace the part of the proof from (3.18) to the end.
LetUj,j=1,2, . . .be the unique solution of the boundary value problem
−Uj+ ¯hUjq=0 inRN+, Uj
x,0
=γj x
forx∈RN−1 (0.1)
dominated by the harmonic function
RN−1P (x, y)γj(y) dy. Hereh¯andγjare given by (1.3) and (3.2) respectively.
Proposition 3.1 is replaced by:
Proposition 3.1.Under the assumptions of Theorem1.2,
jlim→∞Uj(0, xN)= ∞ ∀xN>0. (0.2)
Proof. The proof is based on (3.17) and the inequalityujUj inΩj. This inequality follows from the comparison principle and the fact thath¯aj inΩj whileujUj on∂Ωj. This inequality and (3.17) yield
uj−1 x,0
Uj x, τj
∀jj0, x< rj−1. Therefore by the comparison principle applied inΩj−1,
uj−1
x, xN
Uj
x, xN+τj
∀jj0, x∈Ωj−1. (0.3)
Letj > j0and 0kj−j0. Using (0.3), (3.17) and induction onkwe obtain, uj−k−1
x, xN Uj
x, xN+ k i=0
τj−i
∀x∈Ωj−k−1. (0.4)
DOI of original article:http://dx.doi.org/10.1016/j.anihpc.2012.08.002.
✩ This research was supported by The Israel Science Foundation grant No. 91/10.
* Corresponding author.
E-mail addresses:[email protected](M. Marcus),[email protected](A. Shishkov).
0294-1449/$ – see front matter ©2013 Elsevier Masson SAS. All rights reserved.
http://dx.doi.org/10.1016/j.anihpc.2012.12.006
960 M. Marcus, A. Shishkov / Ann. I. H. Poincaré – AN 30 (2013) 959–960
By (1.10) and (3.16) ∞j=0τj = ∞and supm0τm= ¯τ <∞. Therefore, if b >τ¯ then, for every j jb, there exists an integerλj=λj(b)such that 0b− jλj+1τm=:δjτλj andλj→ ∞asj → ∞. Hence by (0.4)
uλj x, δj
Uj
x, δj+ j λj+1
τm
=Uj x, b
∀x: x< rλj. (0.5)
Applying the comparison principle touλj inΩλj and using (3.12) we find that the inequality 0δj τλj implies uλj(0, δj)4α1uλj(0, τλj). Therefore (0.5) and (3.17) imply,
1 4αA−λ1
j−1 1
4αγλj−1(0) 1
4αuλj(0, τλj)Uj(0, b).
Finally, as limk→∞Ak=0, this inequality implies (0.2) forxN>τ¯. It is easy to see that if (0.2) fails for somexN>0 then it fails for all larger values ofxN. Therefore (0.2) holds for everyxN>0. 2
Completion of proof of Theorem 1.2. Letvj be the solution of (3.3) whereγj is replaced byΓj =A−j1rjN−1δ0on
∂Ωj∩ [xN=0]. As in Section 3.1, the functionv˜j defined byv˜j(y)=Ajvj(rjy),y ∈D0 satisfies the boundary value problem (3.9) withγ˜ replaced byΓ˜=δ0. Denote the solution of this problem byv. Next we apply Lemma 3.1˜ tov˜ inD0∩ [xN> b](b a fixed positive number). We conclude that choosingβ >0 sufficiently large 0< c(β)
˜ v(y,β)
φ1(y) 1 for|y|<1. Hence, ifγj(x):=vj(x, rjβ)thenc(β) γj(x)
A−j1φ1(x/rj)1 in the ball|x|< rj. Obviously uj(x):=vj(x, xN+rjβ)satisfies (3.3) withγj replaced byγj. Proceeding as in Section 3.2 we obtain a sequence {τj}satisfying (3.16) and
γj−1 x
uj x, τj
, xrj, jj0.
LetUj (resp.Vj) be defined in the same way asUj except thatγj is replaced byγj (respectivelyΓj) extended by zero for|x|rj. Then Proposition 3.1applies to{Uj}so that
jlim→∞Uj(0, xN)= ∞. (0.6)
FurthermoreVjvj inΩj so thatVj(x, rjβ) > γj(x),|x|< rj. By the comparison principle,Vj(x, xN+rjβ) Uj(x)inRN+. Hence
jlim→∞Vj(0, xN)= ∞ ∀xN>0. 2
