Blow up solutions for a Liouville equation with singular by Esposito P.

By Esposito P.

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Extra resources for Blow up solutions for a Liouville equation with singular data

Sample text

We can choose ν1 ∈ (0, 1) which in turn implies ||L−1 ρ η||E = O r1 and ν2 ∈ (1, 2) in such a way that (ν1 , 1 − ν1 ) ∩ (ν2 − 1, 2 − ν2 ) = ∅ and we can suppose that δ is fixed to belong in this set. 13 we get 1−δ + r22−δ . ||L−1 (0,0,0) η||E = O r1 Let us define σ = ρ ) 2(1− 4α+5 2σν1 4α+5 2ν1 → 0 as ρ → 0, (37) 1 + 1 and let us choose ri = ρ σνi . In this way, we have ρ2 r25 =ρ 5 2(1− 2σν ) 2 = → 0 as ρ → 0 and 2 ri−νi ρ2 r14α+5 r11−δ + r22−δ = 2 ρ i=1 as ρ → 0. 36 1−δ−ν1 σν1 +ρ 2−δ−ν2 σν2 →0 (38) Now, taking the derivative with respect to z¯ in the expression for η = p|2α f (z)ev(ρ,0,0) and using (13), (35) and (36), we can conclude ||∂z¯ v(ρ, 0, 0) + ρ 2 |z − v(ρ, 0, 0) + ρ2 |z − p|2α f (z)ev(ρ,0,0) ||0,β,B(q,2)\B(q,1) = O(ρ2 ) (39) v(ρ, 0, 0) + ρ2 |z − p|2α f (z)ev(ρ,0,0) ||0,β,ν2 −2,B(q,1) = O r22−ν2 , (40) and ||∂z¯ where we use in a crucial way the fact that |z − p| 2α f (z)e−P0 (z) − 1 = O(|z − q|3 ).

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36 1−δ−ν1 σν1 +ρ 2−δ−ν2 σν2 →0 (38) Now, taking the derivative with respect to z¯ in the expression for η = p|2α f (z)ev(ρ,0,0) and using (13), (35) and (36), we can conclude ||∂z¯ v(ρ, 0, 0) + ρ 2 |z − v(ρ, 0, 0) + ρ2 |z − p|2α f (z)ev(ρ,0,0) ||0,β,B(q,2)\B(q,1) = O(ρ2 ) (39) v(ρ, 0, 0) + ρ2 |z − p|2α f (z)ev(ρ,0,0) ||0,β,ν2 −2,B(q,1) = O r22−ν2 , (40) and ||∂z¯ where we use in a crucial way the fact that |z − p| 2α f (z)e−P0 (z) − 1 = O(|z − q|3 ). Acnowledgements: I express my deep gratitude to Gabriella Tarantello, I am indebted to her for the stimulating discussions and the helpful comments about the manuscript.