Tangential contact between free and fixed boundaries for variational solutions to variable coefficient Bernoulli type Free boundary problems

In this paper, we show that given appropriate boundary data, the free boundary and the fixed boundary of minimizers of functionals of type \eqref{functional} contact each other in a tangential fashion. We prove this result via classification of the global profiles, adapting the ideas from \cite{KKS06}.


Introduction
Objective of the paper of to study the behavior of free boundary near the fixed boundary of domain, for minimizers of Bernoulli type functionals with Hölder continuous coefficients.
A is an elliptic matrix with Hölder continuous entries, and Λ(v) = λ + χ {v>0} + λ − χ {v≤0} .We prove that if the value of boundary data and its derivative at a point are equal to zero (i.e. it satisfies the (DPT) condition mentioned below), then the contact of free boundary and the fixed boundary is tangential.Boundary interactions of free boundaries have gained significant attention in recent years.Whenever there are two medias involved, interactions of their respective diffusions can be modeled by free boundary problems.Often, free boundary of solution and fixed boundary of set come in contact.In applications, the Dam problem [3] and Jets, Wakes and Cavities [7] model phenomenas which involve understanding of free boundary and fixed boundary.
Very recently, works of Indrei [12], [13] study interactions of free boundaries and fixed boundaries for fully non-linear obstacle problems.We refer to [10] where authors shed more light into angle of contact between fixed boundary and free boundary for one phase Bernoulli problem.
As it is common by now, our strategy in this article is to classify blow up of minimizers by using the ideas from [14].We prove that the blow up and also their positivity sets converge to a global solution in P ∞ as defined in [14].In Section 2, we list the assumptions and set some notations and then in Section 3, we prove that blow ups of minimizers converge to that of global solutions (c.f.Definition 2.6).In the last section we prove our main result.

Setting up the problem
We consider the following class of function which we denote as P r (α, M, λ, D, µ).Before definition, we set the following notation For x ∈ R N we denote x ∈ R N −1 as the projection of x on the plane {x N = 0}, we denote the tangential gradient ∇ as follows We define the affine space set H 1 φ (B + R ) as follows, For a given function v ∈ H 1 (B + 2 ), we denote F (v) as F (v) := ∂{v > 0} and Id is the notation for N × N identity matrix.Definition 2.1.A function u ∈ H 1 (B + 2/r ) is said to belong to the class P r (α, M, λ ± , D, µ) if there exists A ∈ C α (B + 2/r ) N ×N , φ ∈ C 1,α (B + 2/r ), λ ± > 0, 0 < µ < 1 and D > 0 such that (P1) ). (P6) There exists 0 < r 0 such that for all 0 < ρ ≤ r 0 we have Remark 2.2.In fact, the functions u ∈ P r (α, M, λ ± , D, µ) carry more regularity than being only a Sobolev function.They are Hölder continuous in (B + 2/r ) (c.f.Lemma 3.2).In the absence of ambiguity on values of α, M, λ ± , D, µ we use the notation P r in place of For the coefficient matrix A, A r (x) is defined as follows One can check that if u ∈ P 1 (α, M, λ ± , D, µ), then u r ∈ P r (α, M, λ ± , D, µ).Indeed if u ∈ P 1 and u minimizer the functional J (c.f.(P4)) Then by simple change of variables we can check that u r ∈ H 1 φr (B + 2/r ) (this verifies (P5)) and u r minimizes Moreover, if A and φ satisfy the conditions (P1), (P2) for r = 1, then A r and φ r satisfy (P1), (P2) for r. (P3) and (P6) remains invariant under the change variables.Therefore u r ∈ P r .
In order to study the blow-up limits (lim r→0 u r ) of functions u ∈ P 1 (α, M, λ ± , D, µ), we define a class of global solutions P ∞ (C, λ ± ).Let us set the following notation before giving the definition Here (Λ(s) = λ + χ {s>0} + λ − χ {s≤0} ) and for every v Our main result intends to show that for a minimizer u of J(•; A, λ + , λ − , B + 2 ) with A, λ ± and u satisfying the properties (P1)-(P6), the free boundary of every such minimizer touches the flat part of fixed boundary tangentially at the origin.For this, we prove that as we approach closer and closer to the origin, the free boundary points cannot lie outside any cone which is perpendicular to the flat boundary and has its tip at the origin.The main result in this paper is stated below.Theorem 2.4.There exists a constant ρ 0 and a modulus of continuity σ such that if |x|} Here σ depends only on α, M, λ ± , D, µ.

Blow-up analysis
The following is a classical result (c.f.[1, Remark 4.2]) , we present the proof for the case of variable coefficients.Lemma 3.1.Given a strictly elliptic matrix and bounded A(x) and a non-negative continuous function w such that div(A(x)∇w) = 0 in {w > 0} ∩ B + 2 , then w ∈ H 1 loc (B + 2 ) and div(A(x)∇w) ≥ 0 in weak sense in B + 2 . Proof.
By the choice of η and ellipticity of the matrix A, we obtain using Young's inequality after choosing of δ > 0 very small and rearranging the terms in the equation above, since η = 1 in D we finally get As ε → 0, we obtain Since w ∈ C(B + 2 ) therefore, w is uniformly bounded in supp(η) and therefore Therefore, we have |∇w||∇ϕ| dx The last term goes to zero as ε → 0. Therefore, we can say that This concludes the proof.
Remark 3.3.Since every function u ∈ P 1 is continuous.Therefore, the positivity set {u > 0} is an open set.
Proof.The claim follows directly from Lemma 3.2 and Lemma 3.1. .
Proof.Let w be such that in the last inequality, we have used (P1).Now, we prove that the term ∇w L ∞ (B + 1 ) is uniformly bounded.
Remark 3.6.We can check that for every u ∈ P 1 , u r ∈ P r and u r is A r -subharmonic and satisfies (3.1) Proof.Since u r ∈ P r , from (P4) we can say that u r is a minimizer of J(•; A r , λ ± , B + 2R ) with boundary data φ r .Here A r and φ r satisfy the conditions (P1) and (P2).Precisely speaking, u r is minimizer of the following functional ).From minimality of u r and the choice of h, we have We use ellipticity of A, we get expanding the left hand side, we get ˆB+ by choosing ε = 1 8 we are left with the following, ˆB+ Thus we obtain a uniform bound on ´B+ R |∇h| 2 dx.
Lemma 3.8 (Compactness).Let r j → 0 + , and a sequence {v j } ∈ P 1 .Then the blow-ups u j := (v j ) r j (as defined in (2.3)) converges (up to subsequece) uniformly in B + R and weakly in H 1 (B + R ) to some limit for any R > 0.Moreover, if u 0 is such a limit of u j in the above mentioned topologies, then u 0 belongs to P ∞ .
Proof.We fix R > 0, since v j ∈ P 1 , therefore u j ∈ P r j and as argued in the proof of previous Lemma, the functions u j are minimizers of the functional J(•; A j , λ ± , B + R ) for j sufficiently large that R < 1 r j .We set the notation for the functional J j as We also denote the boundary values for u j ∈ P r j as φ j .Here the sequences A j ∈ C α (B + 2/r j ) N ×N and φ j ∈ C 1,α (B 2/r j ) satisfy the condition (P1), (P2) with r = r j , Λ(v) := λ + χ {v>0} + λ − χ {v≤0} .We set the following notation for the functional J 0 From Lemma 3.2, we know that u j ∈ C α 0 (B + 2/r j ) which implies C α 0 (B + R ).In particular Hence, u j is a uniformly bounded and equicontinuous sequence in B + R , we can apply Arzela Ascoli theorem to show that u j uniformly converges to a function Thus u 0 satisfies (G2) and (G3) inside the domain B + R .Also, from Lemma 3.7 we have ˆB+ then, by the linear growth condition (c.f.Remark 3.6), u j also satisfies Hence, passing to the limit, we have Thus (3.8) and (3.10) imply that u j is a bounded sequence in H 1 (B + R ).Hence, up to a subsequence, u j u 0 weakly in H 1 (B + R ).We rename the subsequence again as u j .We have found a blow-up limit up to a subsequence u 0 and have shown that u 0 satisfies (G1), (G2) and (G3) in B + R .In order to show that u 0 ∈ P ∞ , it only remains to verify that u 0 satisfies (G4), i.e. u 0 is a local minimizer of J 0 (•; B + R ) for all R > 0 (c.f.(3.7)).For that, we first claim that ˆB+ Indeed, let us look separately at the term J j (u j ) on the right hand side of the above equation We rewrite the first term as follows From (P1) and (P2) we have for all Hence, the first term on the right hand side of (3.12) tends to zero as j → ∞.Thus, from (3.12) and by weak lower semi-continuity of H 1 norm, we have ˆB+ For the second term, we claim that To see this, we first show that for almost every x ∈ B + R , we have Then by the uniform convergence of u j to u 0 , we can easily see that u j (x 0 ) attains the sign of u 0 (x 0 ) for sufficiently large value of j.Hence, (3.15) holds in {u 0 > 0} ∪ {u 0 < 0}.Now, assume x 0 ∈ B + R ∩ {u 0 = 0}.Then left hand side of (3.15) is equal to Regarding RHS of (3.15), we see that Since λ − < λ + (c.f.(P3)), the right hand side in the equation above is always greater than or equal to λ − .Then Thus, (3.15) is proven for all x ∈ B + R and hence (3.14) holds by Fotou's lemma.
By adding (3.13) and (3.14) and [6, Theorem 3.127] we obtain (3.11).Now we will use (3.11) prove the minimality of u 0 for the functional J 0 (•; B + R ) (c.f.3.7).Pick any w ∈ H 1 (B + R ) such that, u 0 − w ∈ H 1 0 (B + R ).We construct an admissible competitor w δ j to compare the minimality of u j for the functional J j (•; B + R ).Then we intend to use (3.11).In this direction, we define two cutoff functions η δ : R N → R and θ : R → R as follows, we can take |∇η δ | ≤ C(N ) δ .We define θ j (x) = θ( x N d j ), for a sequence d j → 0, which we be suitably chosen in later steps of the proof.Let w δ j be a test function defined as Since, the function w δ j − w = (1 − η δ )(u j − u 0 ) + η δ θ j φ j is continuous in B + R and is pointwise equal to zero on ∂B + R , which is a Lipschitz surface in R N .Therefore, u j − w δ j ∈ H 1 0 (B + R ).For further steps, the reader can refer to the Figure 1.Let Ω δ,j = B + R ∩ {θ j = 0} ∩ {η δ = 1}, and R δ,j = B + R \ Ω δ,j by observing w δ j = w on Ω δ,j we see that From the above discussions, we have and similarly Given u j ∈ P r j and w δ j − u j ∈ H 1 0 (B + R ), from the minimility of u j for the functional J j we have and from (3.11) we obtain ˆB+ from the same reasoning as for the justification of (3.13), we have To obtain the claim above, we prove that From the definition of w δ j , we know that Let us consider the first term on the right hand side.We know that Regarding the second term, since |u j − u 0 | tends to zero in L 2 (B + R ) as j → ∞, therefore the second term also tends to zero as j → ∞.We write Lastly, we claim that We know that |{θ j = 0} ∩ B + R | → 0 as j → ∞, hence from (3.28), the first and second term in (3.27) tend to zero as j → ∞.The last term in (3.23) also tends to zero as j → ∞, indeed from (P1) we have if we choose a sequence d j → 0 + such that we also have From the equations (3.17), (3.18), and (3.22) we obtain that the right hand side of (3.21) is equal to J 0 (w; . Since the inequality above (which corresponds to (G4)) and other verified properties of u 0 (i.e.(G1), (G2) and (G3) in B + R ) hold for every R > 0, therefore u 0 ∈ H 1 loc (R N + ) satisfies all the properties in the Definition 2.3.Hence u 0 ∈ P ∞ .
After proving that the (subsequential) limits of blow-up are global solutions, we proceed to show that the positivity sets (and hence the free boundaries) of blow-ups converge in certain sense to that of blow-up limit.For this we will need to establish that the minimizers u ∈ P r are non-degenerate near the free boundary.In the proof below, we adapt the ideas from [2].Proposition 3.9 (Non-degeneracy near the free boundary).Let u ∈ P r 0 for some r 0 > 0 and x 0 ∈ B + 2/r 0 .Then, for every 0 < κ < 1 there exists a constant c(µ, N, κ, λ ± ) > 0 such that for all B r (x 0 ) ⊂ B + 2 r 0 , we have Br(x 0 ) u + dx.We know from Lemma 3.2 that the set {u > 0} is open.Also, since u ∈ P r 0 , there exists 2/r 0 .By elliptic regularity theory, u is locally C 1,α loc ({u > 0} ∩ B + 2/r 0 ).Then, for almost every > 0, B r ∩ ∂{u > ε} is a C 1,α surface.Pick one such small ε > 0 and we consider the test function v ε given by The function v ε defined above belongs to H 1 (B r (x 0 )), thanks to [5, Theorem 3.44] [5, Theorem 3.44] is proven for C 1 domains, but the proof can also be adapted for Lipschitz domains [8,Theorem 4.6] .We intend to show that v ε is bounded in H 1 (B r (x 0 )).This ensures the existence (3.30) We can also write that w − ε = (u − ε) + on ∂B r (x 0 ) and w − ε = 0 on ∂B κr (x 0 ).Consider any sequence {x k } ⊂ B r (x 0 ) \ B κr (x 0 ) such that x k → x ∈ ∂B κr (x 0 ).By Green representation formulae for w − ε in (3.30), we have w(x k ) → w(x), indeed since where ν y is the unit outer normal vector at a point y on the boundary.We apply same arguments as above to ∇w(x) and from [11,Theorem 3.3 (vi)] on G(x, y) and therefore for x ∈ ∂B κr (x 0 ) We can easily check by respective definitions that w ≥ v ε on ∂(B r (x 0 ) \ B κr (x 0 )), moreover, by maximum principle, since div(A(x)∇w) = 0 in B r (x 0 )\B κr (x 0 ) and w ≥ ε on ∂B r (x 0 )\B κr (x 0 ), we have w > ε in B r (x 0 ) \ B κr (x 0 ).In particular w ≥ v ε on ∂D ε where By comparison principle, we know w ≥ v ε in D ε and since w = v ε = ε on ∂B κr (x 0 ) ∩ {u > ε}, hence from (3.31) Given that div(A(x)∇v ε ) = 0 in D ε , we have by divergence theorem and (3.32) justification of use of divergence theorem in D ε can be found in [2, equation (3.4)].From the calculations above, we can write putting very small ε 0 > 0 in the last inequality, we have Hence, from (3.33) and (3.34), v ε is bounded in H 1 (B r (x 0 )).Therefore, up to a subsequence, there exists a limit v = lim ε→0 v ε in weak H 1 sense, such that v satisfies the following We have second equality above because Using the ellipticity of A and shuffling the terms in the above equation, we obtain (3.36) The second to last equality in above calculation is obtained from integration by parts, its justification can be found in [2, equation (3.4)].From (3.36) and (3.32), and using the trace Let us first verify that div(A(x)∇v) = 0 in D 0 .For this let ϕ ∈ C ∞ c (D 0 ), then from continuity of u, there exists a ε 0 > 0 such that supp(ϕ) ⊂ D ε for all ε < ε 0 , also we have ˆD0 A∇v, ∇ϕ dx = ˆsupp(ϕ) A∇v, ∇ϕ dx (3.39) since supp(ϕ) ⊂ D ε , from the definition of v ε we have ˆsupp(ϕ) A∇v ε , ∇ϕ dx = 0 and we know that v is a weak limit of v ε in H 1 (B r (x 0 )), therefore from (3.39) we have ˆD0 A∇v, ∇ϕ dx = ˆsupp(ϕ) A∇v, ∇ϕ dx = lim ε→0 ˆsupp(ϕ) ∇v ε , ∇ϕ dx = 0.
Remark 3.10.In the proposition above the constant is local in nature, this means, the value of the constant depends on the choice of compact set K ⊂⊂ B + 2 where x 0 ∈ K. Lemma 3.11.Let u 0 and u k be as in Theorem 3.8.Then, for a subsequence of u k , for any R > 0 we have This in turn implies
0} and {v > 0} ⊂ {u > 0}, the integration in the set {u ≤ 0} gets cancelled from both sides and we are left with the terms mentioned below.Set D 0 .35) We verify the above properties (3.35) of v at the end of this proof.Let us use the function v as a test function with respect to minimality condition on u in B r (x 0 ), we have ˆBr(x0) A(x)∇u, ∇u + λ(u) dx ≤ ˆBr(x0) A(x)∇v, ∇v + λ(v) dx since, v = u in {u ≤