Two-phase free boundary problems and the Friedland-Hayman inequality Thomas Beck Fordham University tbeck7@fordham.edu November 3, 2020 Thomas Beck (Fordham University) Two-phase free boundary problems November 3, 2020 1 / 20
A two-phase free boundary problem Let Ω ⇢ R n be an open, bounded, convex domain, with K ⇢ ∂ Ω closed. Consider the functional ˆ | r v | 2 + 1 { v > 0 } d x . J [ v ] = Ω Here v 2 H 1 ( Ω ), with v = u 0 2 C 1 ( K ) on K , and 1 { v > 0 } is the indicator function of the set { v > 0 } . We assume that u 0 takes positive and negative values on K (two-phase). It is straightforward to establish the existence of the minimizer u 2 H 1 ( Ω ). Aim Determine what further regularity the minimizer u has. Application to the irrotational flow of two ideal fluids, and other applications in fluid mechanics, electromagnetism, and optimal shape design. Thomas Beck (Fordham University) Two-phase free boundary problems November 3, 2020 2 / 20
Properties of the minimizer Aim Ω | r v | 2 + 1 { v > 0 } has. ´ Determine what regularity the minimizer u of J [ v ] = Formally, the Euler-Lagrange equations J 0 [ u ] = 0 are 1) u is harmonic in the positive phase Ω + = { u > 0 } and non-negative phase Ω � = { u 0 } ; 2) ∂ ν u = 0 on the Neumann part of the boundary ∂ Ω \ K ; 3) u satisfies the gradient jump condition so | r u + ( x ) | 2 � | r u � ( x ) | 2 = 1 on Γ = ∂ Ω + \ ∂ Ω � (the free boundary). Cartoon picture of the two-phase minimizer: But a priori, u is only in H 1 ( Ω ) and so a major goal of the regularity theory is to show that 3) holds in a suitable sense. Thomas Beck (Fordham University) Two-phase free boundary problems November 3, 2020 3 / 20
Properties of the minimizer Using the fact that u is a minimizer it is (fairly) straightforward to show that it satisfies the following properties: (Alt-Ca ff arelli-Friedman ’84, Gurevich ’99, Raynor ’08) 1) u is subharmonic in Ω and harmonic in the two phases Ω + = { u > 0 } Ω � = { u 0 } (that is, ∆ u is a positive measure supported on the free boundary) 2) u is H¨ older continuous (up to the boundary) for some exponent α > 0 3) ∂ ν u = 0 weakly on the Neumann boundary ∂ Ω \ K The key idea behind proving these properties is to combine u minimizing the functional with harmonic replacement. The first major step in the regularity theory is to determine if u is Lipschitz continuous . Thomas Beck (Fordham University) Two-phase free boundary problems November 3, 2020 4 / 20
Lipschitz continuity of minimizers Theorem (Alt-Ca ff arelli-Friedman (ACF), ’84) The minimizer is Lipschitz continuous in the interior of Ω . Why is Lipschitz continuity a key step in the regularity theory? It allows a rescaling of u by dilation and to study the blow-up limit i u ( x 0 + rx ) � u ( x 0 ) u 0 ( x ) = lim . r r ! 0 This is used by ACF to show that minimizer and free boundary are smooth. Question Is the minimizer u Lipschitz continuous up to the Neumann boundary? u may only be H¨ older continuous at the intersection of K with the free boundary (Gurevich ’99). Convexity is a natural (and close to necessary) restriction on Ω for a positive answer. Thomas Beck (Fordham University) Two-phase free boundary problems November 3, 2020 5 / 20
The Alt-Ca ff arelli-Friedman functional Theorem (Alt-Ca ff arelli-Friedman (ACF), ’84) The minimizer is Lipschitz continuous in the interior of Ω . To prove this interior Lipschitz regularity they introduced the following functional: ! ! | r u + | 2 | r u � | 2 1 1 ˆ ˆ Φ ( t ) = | x � x 0 | n � 2 d x | x � x 0 | n � 2 d x t 2 t 2 B t ( x 0 ) B t ( x 0 ) Here x 0 is an interior point on the free boundary and t > 0. Proposition (Monotonicity of the ACF functional, ’84) The functional Φ ( t ) is a monotone increasing function of t , and so in particular Φ ( t ) is uniformly bounded by Φ (1) for all 0 < t 1 . This proposition is the key step in their proof of Lipschitz continuity. Remark In the one phase case, Lipschitz continuity can be obtained without using the functional (Alt-Ca ff arelli ’81, Raynor ’08). Thomas Beck (Fordham University) Two-phase free boundary problems November 3, 2020 6 / 20
The Alt-Ca ff arelli-Friedman functional Proposition (Alt-Ca ff arelli-Friedman, ’84) The functional Φ ( t ) " ✓ 1 ◆ ✓ 1 | r u + | 2 | r u � | 2 ◆ ˆ ˆ Φ ( t ) = | x | n � 2 d x | x | n � 2 d x t 2 t 2 B t B t Bt is a monotone increasing function of t . i Idea of the proof: By direct calculation, ∂ B 1 | r u + | 2 d σ ∂ B 1 | r u � | 2 d σ ´ ´ Φ 0 (1) Φ (1) = + � 4 | r u + | 2 | r u − | 2 ´ ´ | x | n − 2 d x | x | n − 2 d x B 1 B 1 and also ˆ ˆ ˆ | r u ± | 2 � | ∂ r u ± | 2 + λ ± (1) | u ± | 2 , ∂ B 1 ∂ B 1 ∂ B 1 ◆ 1 / 2 ✓ ˆ ◆ 1 / 2 | r u ± | 2 ✓ ˆ + n � 2 ˆ ˆ ( u ± ) 2 ( ∂ r u ± ) 2 ( u ± ) 2 . | x | n � 2 2 B 1 ∂ B 1 ∂ B 1 ∂ B 1 Here λ + (1) is the first Dirichlet eigenvalue of { u > 0 } \ ∂ B 1 . Thomas Beck (Fordham University) Two-phase free boundary problems November 3, 2020 7 / 20
The Alt-Ca ff arelli-Friedman functional Setting ˆ ˆ z ± = w ± = | ∂ r u ± | 2 , | u ± | 2 , ∂ B 1 ∂ B 1 therefore gives z + + λ + (1) w + z � + λ � (1) w � Φ 0 (1) 2 w + + 2 w � � 4 . Φ (1) � ( z + w + ) 1 / 2 + n � 2 ( z � w � ) 1 / 2 + n � 2 It then becomes a calculus exercise to minimize the right hand side over z ± , w ± � 0, " # r r Φ 0 (1) ( n � 2) 2 ( n � 2) 2 � n � 2 + λ + (1) � n � 2 Φ (1) � 2 + + + λ � (1) � 2 . 2 4 2 4 Question Is this right hand side positive? Thomas Beck (Fordham University) Two-phase free boundary problems November 3, 2020 8 / 20
The Friedland-Hayman inequality To answer this, consider the following eigenvalue problem on S n � 1 . Definition Given disjoint subsets E ± of S n � 1 , define λ ( E ± ) to be the first Dirichlet eigenvalue of E ± . iii. ¥ En : Call r ( n � 2) 2 α ( E ± ) = � n � 2 + λ ( E ± ) + 2 4 the characteristic exponent of E ± . - I E Theorem (Friedland-Hayman ’76, Beckner-Kenig-Pipher ’88) The characteristic exponents α ( E ± ) satisfy α ( E + ) + α ( E � ) � 2 . Equality if and only if E ± are hemispheres. Thomas Beck (Fordham University) Two-phase free boundary problems November 3, 2020 9 / 20
The Alt-Ca ff arelli-Friedman functional Theorem (Friedland-Hayman ’76, Beckner-Kenig-Pipher ’88) The characteristic exponents α ( E ± ) satisfy α ( E + ) + α ( E � ) � 2 . Equality if and only if E ± are hemispheres. The lower bound on Φ 0 (1) / Φ (1) can be written as Φ 0 (1) Φ (1) � 2( α + (1) + α � (1) � 2) . So the monotonicity of Φ follows from the Friedland-Hayman inequality! Strict monotonicity unless { u > 0 } \ B t , { u 0 } \ B t are hemispheres. Remark The characteristic exponent α ( E ± ) is the positive homogeneities of the harmonic extensions of the eigenfunctions to the cone generated by E ± . Thomas Beck (Fordham University) Two-phase free boundary problems November 3, 2020 10 / 20
Regularity near the convex boundary So, the Friedland-Hayman inequality directly gives the monotonicity of Φ ( t ) and leads to the interior Lipschitz regularity of the minimizer. Question Can we extend the Lipschitz continuity to the convex Neumann boundary? A natural change of functional for x 0 2 ∂ Ω is ! ! | r u + | 2 | r u � | 2 1 ˆ 1 ˆ Ψ ( t ) = | x � x 0 | n � 2 d x | x � x 0 | n � 2 d x . t 2 t 2 B t ( x 0 ) \ Ω B t ( x 0 ) \ Ω Just as in the interior case, Lipschitz regularity reduces to the boundedness of Ψ ( t ). Bek . Following the calculation in the interior case gives Ψ 0 (1) / Ψ (1) � 2( α + (1) + α � (1) � 2) � Error , with the Error term on ∂ Ω measuring the non-conic nature of the boundary. But the characteristic exponents are now di ff erent! Thomas Beck (Fordham University) Two-phase free boundary problems November 3, 2020 11 / 20
A variant of the Friedland-Hayman inequality Definition Let W ⇢ S n � 1 be a geodesically convex subset of S n � 1 . Given disjoint subsets W ± of W , define µ ( W ± ) to be the first eigenvalue of W ± with Neumann boundary conditions on ∂ W ± \ ∂ W and Dirichlet boundary conditions otherwise. Again, call r ( n � 2) 2 α ( W ± ) = � n � 2 + µ ( W ± ) + 2 4 the characteristic exponent of W ± . Theorem (B-Jerison-Raynor ’20) The characteristic exponents α ( W ± ) satisfy α ( W + ) + α ( W � ) � 2 . Remark (Work in preparation with David Jerison) Equality precisely when W ⇢ S n � 1 has antipodal points. Thomas Beck (Fordham University) Two-phase free boundary problems November 3, 2020 12 / 20
A variant of the Friedland-Hayman inequality Theorem (B-Jerison-Raynor ’20) The characteristic exponents α ( W ± ) satisfy α ( W + ) + α ( W � ) � 2 . On S 1 the eigenvalues can be computed explicitly to prove the theorem (Gemmer-Moon-Raynor ’18). w - The key steps in the proof of the original Friedland-Hayman inequality: . 1) A symmetrization argument to reduce to studying Dirichlet eigenvalues of spherical caps; 2) Obtain a lower bound for spherical caps either by a direct numerical calculation or comparing to Gaussian eigenvalues. Step 1) breaks down in our Dirichlet-Neumann case. Thomas Beck (Fordham University) Two-phase free boundary problems November 3, 2020 13 / 20
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