Download Blow-up Theories for Semilinear Parabolic Equations by Bei Hu PDF

By Bei Hu

There is a gigantic volume of labor within the literature in regards to the blow-up habit of evolution equations. it's our purpose to introduce the idea through emphasizing the tools whereas trying to stay away from great technical computations. to arrive this target, we use the best equation to demonstrate the tools; those equipment quite often observe to extra basic equations.

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Example text

1. A function u ∈ C(Ω) ∩ C 2 (Ω) is said to be a lower (upper) solution if −Δu u ( ) f (x, u) ( )g in Ω, on ∂Ω. 1. Let Ω be a bounded domain with ∂Ω ∈ C 2+α . Assume that f ∈ C α (Ω × R), fu− ∈ L∞ (Ω × R) and g ∈ C(Ω). 2) has a 2 solution u ∈ C(Ω) ∩ C (Ω) such that u(x) u(x) u(x) on Ω. Proof. Let u0 (x) = u(x) and u0 (x) = u(x). Define c = sup fu− (x, u). Then the Ω×R function F (x, u) = cu + f (x, u) satisfies Fu (x, u) 0. B. 3) to be the solution of −Δun + cun = F (x, un−1 (x)) un = g in Ω, on ∂Ω, and un (x) to be the solution of −Δun + cun = F (x, un−1 (x)) un = g in Ω, on ∂Ω.

45, (ii) The two solutions merge into one at λ = λ∗ , and (iii) There are no solutions when λ > λ∗ . , Ω = {x ∈ Rn ; |x| < R}, are there non-radial solutions? The answer is no. In fact, the symmetry properties were studied in Gidas–Ni–Nirenberg [60] for a large class of such problems using the very powerful moving plane method. We begin with a Hopf Lemma for nonnegative solutions. 3. Let ∂Ω satisfies the interior sphere condition at x0 ∈ ∂Ω. Suppose that u ∈ C 2 (Ω) ∩ C(Ω) satisfies −aij Dij u + bi Di u + cu 0 in Ω, where aij , bi , c are continuous functions on Ω and aij satisfies the uniform ellipticity condition.

1 φ, in Ω, on ∂Ω, φ=0 Ω φ>0 φ(x)dx = 1. 6) has no solutions. Proof. We assume that a solution u exists. 10), and 0= Ω (uΔφ − φΔu)dx = Ω > Ω (−λ1 uφ + φλ exp u)dx {−λ1 uφ + φλ(u + 1)}. The right-hand side of the above inequality is positive, since λ λ1 , which is a contradiction. In this case it is natural to expect that the corresponding parabolic problem cannot have a global solution. We leave it as an exercise. 5. 6) exists if λ > 0 is small. Its proof is left as an exercise. 6. 6) may indeed have more than one positive solutions.

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