A geometric approach to free boundary problems by Luis Caffarelli, Sandro Salsa

By Luis Caffarelli, Sandro Salsa

Loose or relocating boundary difficulties seem in lots of components of study, geometry, and utilized arithmetic. a customary instance is the evolving interphase among a great and liquid part: if we all know the preliminary configuration good adequate, we must always be ready to reconstruct its evolution, particularly, the evolution of the interphase. during this ebook, the authors current a sequence of rules, tools, and strategies for treating the main easy problems with this sort of challenge. particularly, they describe the very basic instruments of geometry and actual research that make this attainable: houses of harmonic and caloric measures in Lipschitz domain names, a relation among parallel surfaces and elliptic equations, monotonicity formulation and pressure, and so forth. The instruments and concepts awarded right here will function a foundation for the research of extra complicated phenomena and difficulties. This booklet turns out to be useful for supplementary analyzing or could be an exceptional self sufficient research textual content. it truly is appropriate for graduate scholars and researchers attracted to partial differential equations.

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It follows that 4 u2 (x) − u1 (x + (1 + μb)εσ) = w(x) + u1 (x + εσ) − u1 (x + (1 + μb)εσ) cμb εu2 (0) ≥ cbεu2 (0) − R ≥ c¯bεu2 (0) if μ = μ(R, n) is chosen small. 4. 5 in B1/6 (x0 ) to u1 (x) = u(x − τ ) and u2 (x) = u(x) . 7). Let y ∈ Bε (x) and notice that if τ ∈ Γ( θ2 , en ) and τ¯ = τ − (y − x) τ − τ | = |x − y| ≤ |τ | sin θ2 . Also then α(τ, τ¯) ≤ 2θ , since |¯ |¯ τ | ≥ |τ | − |τ | sin since θ 2 1 θ ≥ |τ | 2 2 < π4 . 8, we deduce that inf B1/8 (x0 ) Dτ¯ u ≥ c0 ν, τ¯ |∇u(x0 )| ≥ c ν, τ¯ u(x0 ) τ | cos α(ν, τ¯) ≥ c1 |¯ sup u B1/8 (x0 ) ≥ bε sup u B1/8 (x0 ) where b = b(τ ) = C cos( θ2 + α(ν, τ )).

Let 0 ≤ u1 ≤ u2 be harmonic functions in BR = BR (0). 8) Proof. 6)) in BR−ε . 6), in B 3 R , 4 w(x) ≥ cw(0) ≥ cbεu2 (0) . Shauder estimates and Harnack inequality again give |∇u1 (x)| ≤ c c u1 (0) ≤ u2 (0) R R in B 3 R . It follows that 4 u2 (x) − u1 (x + (1 + μb)εσ) = w(x) + u1 (x + εσ) − u1 (x + (1 + μb)εσ) cμb εu2 (0) ≥ cbεu2 (0) − R ≥ c¯bεu2 (0) if μ = μ(R, n) is chosen small. 4. 5 in B1/6 (x0 ) to u1 (x) = u(x − τ ) and u2 (x) = u(x) . 7). Let y ∈ Bε (x) and notice that if τ ∈ Γ( θ2 , en ) and τ¯ = τ − (y − x) τ − τ | = |x − y| ≤ |τ | sin θ2 .

LIPSCHITZ FREE BOUNDARIES ARE C 1,γ 68 with β¯ 1 + |∇ϕ(x1 )| α ¯ ≥G 1 − |∇ϕ(x1 )| . 18) Proof. Let vϕ (x1 ) = u(y1 ). 14) hold with α ≥ G(β). 16) ν¯ = α ¯ = α|ν + ∇ϕ(x1 )| , β¯ = β|ν + ∇ϕ(x1 )| . 18). 18) says that vϕ is “almost” a subsolution, due to the fact that ∇ϕ is not identically zero. We shall later perturb vϕ to make it a subsolution. 7, such that vεϕt carries the monotonicity gain from B1/8 (x0 ) to the free boundary as t goes from 0 to 1. This means that we want ϕt = 1 along, say ∂B1 , ϕt ≈ 1 + ctb on ∂B1/8 (x0 ) and ¯tb in B1/2 .

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