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

By Luis Caffarelli, Sandro Salsa

Unfastened or relocating boundary difficulties seem in lots of components of research, geometry, and utilized arithmetic. a regular instance is the evolving interphase among a high-quality and liquid part: if we all know the preliminary configuration good sufficient, we must always be capable of reconstruct its evolution, particularly, the evolution of the interphase. during this ebook, the authors current a sequence of principles, tools, and strategies for treating the main easy problems with this type of challenge. specifically, they describe the very basic instruments of geometry and genuine research that make this attainable: homes of harmonic and caloric measures in Lipschitz domain names, a relation among parallel surfaces and elliptic equations, monotonicity formulation and pressure, and so on. The instruments and ideas awarded the following will function a foundation for the research of extra complicated phenomena and difficulties. This publication comes in handy for supplementary interpreting or might be an excellent autonomous examine textual content. it really is appropriate for graduate scholars and researchers drawn to partial differential equations.

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3. STRONG RESULTS 47 since, for proper choices of c we can make Ncε (F (u)) ∩ BR ⊂ {0 < u < ε} ∩ BR or vice versa. It follows that the quantities a), b) and c) are all comparable to Rn−1 . Finally, let {Brj (xj )}, xj ∈ F (u), a finite covering of F (u)∩BR by balls of radius rj < ε, that approximates H n−1 (F (u) ∩ BR ). Let r < min rj and {Br (xkj )} a finite overlapping covering for F (u) ∩ Brj (xj ). Then, on one hand |∂Br (xkj )| ≤ cRn−1 k,j by the argument above with ε = r. On the other hand |∂Br (xkj )| ≥ crjn−1 k again by the above discussion with R = rj .

C(λ) 1), then F (vϕ ) is a Lipschitz graph with Lipschitz constant λ ≤ λ + c1 sup |∇ϕ| . Proof. a) Notice that Ω+ (vϕ ) contains the set K = {|x − y1 |2 < ϕ(x)2 } since for |x − y1 | < ϕ(x), we have vϕ(x) (x) > u(y1 ) > 0. The boundary of K is a C 2 -surface, since along ∂K ∇(|x − y1 |2 − ϕ(x)2 ) = 2(x − y1 − ϕ(x)∇ϕ(x)) = 0 because |∇ϕ| < 1. Now, x1 ∈ ∂K so that a) is proven. 4. 13) holds, we have, since y1 = x1 + ϕ(x1 )ν, vϕ (x) ≥ u(y) = α x + ϕ(x)ν − y1 , ν + = α x − x1 + [ϕ(x) − ϕ(x1 )]ν, ν = α x − x1 , ν + ∇ϕ(x1 ) + + o(|y − y1 |) + + o(|x − x1 |) + o(|x − x1 |) .

T. 11, to conclude that, in a neighborhood of the free boundary, Dτ u ≥ 0 along every τ ∈ Γ(θ, en ) with θ ≤ 1 2 arcotan L or θ ≥ π 4 − 1 2 artan L ≡ θ0 . We call Γ(θ, en ) the monotonicity cone. t. the same direction en . We may suppose that this happens in the whole cylinder C1 , by using, if necessary, the invariance by elliptic dilations of the problem. We will call δ0 = π 2 − θ the defect angle because it measures how far are the level sets of u from being flat. 2. , Den u and |∇u| are equivalent.

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