# Download Superlinear Parabolic Problems: Blow-up, Global Existence by Pavol Quittner, Philippe Souplet PDF By Pavol Quittner, Philippe Souplet

This publication is dedicated to the qualitative learn of strategies of superlinear elliptic and parabolic partial differential equations and structures. This classification of difficulties comprises, particularly, a couple of reaction-diffusion structures which come up in numerous mathematical versions, specially in chemistry, physics and biology. The publication is self-contained and up to date, taking unique care at the didactical coaching of the fabric. it truly is dedicated to difficulties which are intensively studied yet haven't been handled up to now intensive within the publication literature.

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Additional resources for Superlinear Parabolic Problems: Blow-up, Global Existence and Steady States (Birkhäuser Advanced Texts Basler Lehrbücher)

Example text

1. Let p ≥ pS . 2) possesses a unique positive solution Uα ∈ C 2 ([0, ∞)) satisfying Uα (0) = α. This solution is decreasing and we have Uα (r) = αU1 (α(p−1)/2 r). 4) If p > pS , then r2/(p−1) Uα (r) → cp as r → ∞. If p = pS , then U1 (r) = n(n − 2) n(n − 2) + r2 (n−2)/2 . 5) Let α1 > α2 > 0. If p ≥ pJL , then U∗ (r) > Uα1 (r) > Uα2 (r) for all r > 0. If pS < p < pJL , then Uα1 and Uα2 intersect inﬁnitely many times and Uα1 , U∗ intersect inﬁnitely many times as well. If p = pS , then Uα1 , Uα2 intersect once and Uα1 , U∗ intersect twice.

Without loss of generality we may assume e = e1 . ✻ zn D en 2 λ Σ(λ) ✲ z1 Figure 4: Moving planes. 26). Given λ ∈ [0, 1/2), set Σ(λ) := {z ∈ D : z1 > λ}, z λ := (2λ − z1 , z2 , . . , zn ). The point z λ is the reﬂection of z with respect to the hyperplane {z1 = λ} and Σ(λ) is called a cap. We next deﬁne w(z; λ) := v(z λ ) − v(z) for z ∈ Σ(λ) 46 I. Model Elliptic Problems (the parameter λ will be omitted in w when no risk of confusion arises). Then ∆w = ∆v(z λ ) − ∆v(z) = −|z λ |γ v p (z λ ) + |z|γ v p (z) = |z|γ − |z λ |γ v p (z λ ) − |z|γ v p (z λ ) − v p (z) .

5. We have S0 (Ω1 ) = S0 (Ω2 ) for any open sets Ω1 , Ω2 ⊂ Rn . If Ω = Rn , then S0 (Ω) is not attained. Proof. Let Ω1 , Ω2 ⊂ Rn be open. Since S0 (Ω) = S0 (x + Ω) for any x ∈ Rn , we may assume 0 ∈ Ω1 ∩ Ω2 . Denote wR (x) := w(Rx). Let ε > 0 and 0 = u1 ∈ W01,2 (Ω1 ), S0 (u1 , Ω1 ) < S0 (Ω1 ) + ε. Setting u˜1 (x) := ˜1 (x) = 0 if x ∈ / Ω1 , we have u ˜1 ∈ W01,2 (Rn ) = W 1,2 (Rn ) and u(x) if x ∈ Ω1 , u supp (˜ uR ˜R 1 ) ⊂ Ω2 if R is suﬃciently large. Let u2 be the restriction of u 1 to Ω2 . 1,2 Then u2 ∈ W0 (Ω2 ), u2 = 0, and n uR u 1 , Rn ) S0 (Ω2 ) ≤ S0 (u2 , Ω2 ) = S0 (˜ 1 , R ) = S0 (˜ = S0 (u1 , Ω1 ) < S0 (Ω1 ) + ε.