# Elliptic Partial Differential Equations by Lucio Boccardo, Gisella Croce

By Lucio Boccardo, Gisella Croce

Elliptic partial differential equations is among the major and such a lot energetic components in arithmetic. In our publication we learn linear and nonlinear elliptic difficulties in divergence shape, with the purpose of delivering classical effects, in addition to more moderen advancements approximately distributional strategies. hence the ebook is addressed to master's scholars, PhD scholars and an individual who desires to start examine during this mathematical box.

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Elliptic Partial Differential Equations

Elliptic partial differential equations is likely one of the major and such a lot energetic parts in arithmetic. In our ebook we learn linear and nonlinear elliptic difficulties in divergence shape, with the purpose of supplying classical effects, in addition to newer advancements approximately distributional options. therefore the publication is addressed to master's scholars, PhD scholars and a person who desires to commence study during this mathematical box.

Extra info for Elliptic Partial Differential Equations

Example text

In Ω. 12 (Maximum Principle). Let u ∈ H01 (Ω). (1) Assume that Ω M(x)∇u · ∇v ≤ 0 for every positive v ∈ H01 (Ω). Then u ≤ 0. (2) Assume that Ω M(x)∇u · ∇v ≥ 0 for every positive v ∈ H01 (Ω). Then u ≥ 0. Proof. (1) Choose v = u+ , that is, the positive part of u as a test function. Then M(x)∇u+ · ∇u+ = Ω M(x)∇(u+ − u− ) · ∇u+ ≤ 0 . Ω Using the ellipticity of M , u+ = 0. (2) Choose v = u− , that is, the negative part of u as a test function. Then M(x)∇(u+ − u− ) · ∇u− = − 0≤ Ω M(x)∇u− · ∇u− . Ω The ellipticity of M gives u− = 0.

8. Using the previous theorem one sees immediately that there exists a solution u to u ∈ H01 (Ω) ∩ Lp−1 (Ω) : eu − 1 ∈ L1 (Ω) : −div(M(x)∇u) + |u|p−2 u = f u ∈ H01 (Ω), −div(M(x)∇u) + eu − 1 = f . 9. In Chapters 10 and 11, we shall again study approximating problems to get a priori estimates and then pass to the limit. 1) under the following hypotheses: 2N (1) f ∈ L N+2 (Ω); (2) g : R → R is increasing and continuous; there exists γ > 0 such that |g(s)| ≤ γ|s|a , a≤ N +2 . N −2 We recall that (1) M is elliptic, that is, there exists α > 0 such that M(x)ξ · ξ ≥ α|ξ|2 , ∀ ξ ∈ RN ; (2) M is bounded, meaning there exists β > 0 such that |M(x)| ≤ β , ∀ x ∈ Ω.

E. in E . From Beppo Levi’s theorem and from Step II , one has +∞ |f | = lim sn = lim n→∞ E meas{x ∈ E : |sn (x)| > t} dt n→∞ E 0 +∞ = lim dt n→∞ 0 χ{x∈E:|sn (x)|>t} . 8) 22 Preliminaries of real analysis Lebesgue’s theorem implies that χ{x∈E:|sn(x)|>t} = lim n→∞ E χAt ∩E . 9) E We set gn (t) = χ{x∈E:|sn(x)|>t} . 9) means that gn (t) → meas(At ∩ E) if n → ∞. To prove the result it suffices to prove that +∞ +∞ gn (t) dt = lim n→+∞ 0 meas(At ∩ E) dt . e. in (0, +∞); moreover |gn (t)| ≤ meas(E); it is sufficient to prove that meas(At ∩ E) belongs to L1 ((0, +∞)), and then apply Lebesgue’s theorem.