Regularity results for a penalized boundary obstacle problem

In this paper we are concerned with a two-penalty boundary obstacle problem of interest in thermics, fluid dynamics and electricity. Specifically, we prove existence, uniqueness and optimal regularity of the solutions, and we establish structural properties of the free boundary.


Introduction
In this paper we study a penalized boundary obstacle problem of interest in thermics, fluid mechanics, and electricity.Given a domain Ω in R n , n ≥ 2, with sufficiently regular boundary Γ and unit outer normal ν, we consider the following stationary problem: (1.1) ∆u = f in Ω, Here u + = max{u, 0}, u − = − min{u, 0} ≥ 0, k + and k − are non-negative constants, and p > 1.Our goal is to establish optimal regularity of the solutions, and to study properties of the free boundary (∂{u > h} ∪ ∂{u < h})∩Γ.We begin by observing that in the limiting case k + = k − = 0, u is clearly the solution of a classical Neumann problem.The other limiting case, when k + = 0 and k − = +∞ (or equivalently k + = +∞ and k − = 0), is more interesting.The boundary condition, in fact, becomes and u is a solution of the Signorini problem, also known as the thin obstacle problem.The Signorini problem has received a resurgence of attention in the last decade, due to the discovery of several families of powerful monotonicity formulas, which in turn have allowed to establish the optimal regularity of the solution, a full classification of free boundary points, smoothness of the free boundary at regular points, and the structure of the free boundary at singular points.We refer the interested reader to [AC1], [ACS], [CSS], [GP], [DSS], [KPS], see also the survey [DS] and the references therein.
The general scheme of a solution to the Signorini problem provides a road map for the solution of problem (1.1), but there are two new substantial difficulties.The first one is due to the non-homogeneous nature of the boundary condition in (1.1), which in particular implies that this problem does not admit global homogeneous solutions of any degree.This is in stark contrast with the Signorini problem, where the existence and classification of such solutions play a pivotal role.Moreover, in the thin obstacle problem it is readily seen that continuity arguments force u to be always above h (hence the nomenclature), whereas the case h(x) > u(x) is no longer ruled out in (1.1).Allowing for both constants k + , k − to be finite (even when one of the two vanishes) de facto destroys the one-phase character of the problem.In order to focus the attention on these new aspects, it is useful to understand first a simplified local version of (1.1), posed in the upper half ball Here An alternate perspective is given by the associated energy.We seek to minimize over all v ∈ W 1,q (B 1 ) with q = max{2, p} and v − g ∈ W 1,q 0 (B 1 ) for given boundary data g.Here k± = 2k ± /p, and x = (x ′ , x n ).In this context we think of the data in (1.2) as extended to all of B 1 by even reflection.A minimizer to this energy will be symmetric about Γ and u will correspond to the restriction to B + 1 .
Our first main result is the following: Theorem 1.1.Let g ∈ W 1,q (Ω), 0 ≤ k ± < ∞, k + = k − , and p > 1.Then there exists a unique minimizer u ∈ W 1,q (B 1 ) of the energy ) for every α < p − 1, and there exists a constant In the case p = 2, we can in fact establish that the regularity is optimal at points where the gradient does not vanish.
Theorem 1.2.Let u be the (unique As an immediate consequence of the regularity of the solution and of the implicit function theorem, we obtain the following result on the regularity of the free boundary.
Definition 1.3.The regular set of the free boundary is defined as We next turn our attention to the study of the singular set.
Definition 1.5.For and we define the set of singular points with frequency µ ∈ N as where p x 0 µ is a homogeneous polynomial of degree µ as in Theorem 7.1.Finally, we introduce The structure of the singular set is described in the following result.
Theorem 1.6.Let u be a solution to (1.2).Then for every µ ∈ N and d = 0, 1, . . ., n − 2, the set The proof of Theorem 1.6 follows the ideas of the corresponding result in [GP] for the Signorini problem.It hinges on the monotonicity (or almost-monotonicity) of a perturbed Almgren functional and a Monneau-type functional (see Theorem 4.1 and Corollary 6.2).From these results we infer the growth rate and nondegeneracy of the solution near the free boundary.In turn, these properties allow to prove uniqueness and continuous dependance on the singular point of the blow-up limits.The rest of the proof is based on the Whitney's extension and the implicit function theorems.
1.1.Structure of the paper.The paper is organized as follows.In Section 2 we describe some applications to problems of semi-permeable membranes and of temperature control, which motivate the study of (1.1).In Section 3 we establish existence and uniqueness of solutions, and prove Theorems 1.1 and 1.2.In Section 4 we prove the monotonicity of the perturbed functional of Almgren type, and infer some properties of the solution as a consequence.In Section 5 we introduce the Almgren rescalings, and discuss their blow-up limits.In Section 6 we prove the almost-monotonicity of a Monneau-type functional, and establish nondegeneracy of solutions.Finally, Section 7 is devoted to the proof of Theorem 1.6.

Motivation
2.1.Semi-permeable membranes.Following [DL,Section 2.2.2], we briefly describe the process of osmosis through semi-permeable walls.By Ω we denote a domain in R n , n ≥ 2, with sufficiently regular boundary ∂Ω.The region Ω consists of a porous medium occupied by a viscous fluid which is only slightly compressible, and we denote its field of pressure by u(x).We assume that a portion Γ of ∂Ω consists of a semi-permeable membrane of finite thickness, i.e. the fluid can freely enter in Ω, but the outflow of fluid is prevented.Combining the law of conservation of mass with Darcy's law, one finds that u satisfies the equation where f = f (x, t) is a given function.When a fluid pressure h(x), for x ∈ Γ, is applied to Γ on the outside of Ω, one of two cases holds: In the former, the semi-permeable wall prevents the fluid from leaving Ω, so that the flux is null.If we let ν denote the outer unit normal to Γ, we then have (2.1) ∂u ∂ν = 0.
In the latter case, the fluid enters Ω.It is reasonable to assume the outflow to be proportional to the difference in pressure, so that (2.2) where k > 0 measures the conductivity of the wall.Combining (2.1) and (2.2), we obtain the boundary condition In our model (1.1), we allow for fluid flow to occur both into and out of Ω with different permeability constants, under the assumption that the flux in each direction is proportional to a power of the pressure.

Temperature control.
An alternate interpretation of the model is as a boundary temperature control problem, which we only briefly outline here.We assume that a continuous medium occupies a region Ω in R n , with boundary Γ and outer unit normal ν.Given a reference temperature h(x), for x ∈ Γ, it is required that the temperature at the boundary u(x, t) deviates as little as possible from h(x).To this end, thermostatic controls are placed on the boundary to inject an appropriate heat flux when necessary.The controls are regulated as follows: (i) If u(x, t) = h(x), no correction is needed and therefore the heat flux is null.
(ii) If u(x, t) = h(x), a quantity of heat proportional to the difference between u(x, t) and h(x) is injected.We can thus write the boundary condition as where if u > h More in general, one can assume that Φ(u) is a continuous and increasing function of u.For further details, we refer to [DL,Section 2.3.1],see also [AC1] for the limiting case k − = 0 and k + = +∞ and [ALP] for the case p = 1 in (1.3).

Optimal regularity of solutions
We begin this section by proving existence and uniqueness of minimizers to (1.3).We let 0 (B 1 )}.Lemma 3.1.There exists a unique minimizer u ∈ K for the energy J(v) given by (1.3).
Proof.Throughout this proof we will pass to subsequences whenever necessary without comment.Let u l be a minimizing sequence.Then ∇u l 2 is clearly bounded owing to the form of the energy itself.By using the Poincaré inequality on u l − g we deduce that the sequence u l is bounded in the W 1,2 (B 1 ) norm.Thus there exists a weak limit u which is necessarily in K.We may assume that u l → u in L 2 and a.e.The weak convergence of u l to u in W 1,2 and the strong convergence in This clearly follows from the property of weak convergence and, because of the strong L 2 convergence, the inequality must fall on the gradient part of the norm.
To prove that u is a minimizer we must show then that It will suffice to demonstrate this for u − ; the result for u + is proved in an analogous fashion.
The trace operator T : ) is a bounded linear operator, since the half ball is a Lipschitz domain.Furthermore, in this setting it is a compact operator, and thus takes weakly convergent sequences to strongly convergent ones.Suppressing the T u l notation and simply writing u l we then have that From this we may assume that u l → u a.e. on Γ.But then clearly (u − l ) p → (u − ) p a.e. and applying Fatou's Lemma we have which completes the proof of existence.Uniqueness follows by observing that (f + g) ± ≤ f ± + g ± , and then applying standard arguments.
Next, we recall the definition of a weak solution (see [L]): Definition 3.2.We say that u is a weak solution to It is easy to show that the minimizer u is a weak solution to our problem.
Lemma 3.3.The minimizer u is a weak solution to (1.2).That is, ) vanishing on (∂B 1 ) + .Proof.This is a standard variational fact.See for example the proof of Lemma 4.1 in [ALP].
We now turn to the regularity of the solution.Our strategy will be to first prove an initial Hölder regularity which will when then improve on.The first step is an energy estimate for u.
Lemma 3.5.Let u be the minimizer of (1.3).Then we have for any Proof.We first prove the corresponding estimate for u In turn this implies that Expanding yields At this point standard energy arguments imply A similar argument implies the same inequality with u + ; together they yield the energy estimate for u.
Next, we use the energy estimate to prove an initial Hölder modulus of continuity for u.This regularity is much lower than optimal, but it will allows us to bootstrap to obtain higher regularity.
Proof.Let B r := B r (x) for r < 1/4 and x ∈ B 1 , and let v be the harmonic replacement of u in B r .Set Γ r = B r ∩ Γ.By minimality we have However, since v is harmonic we have and thus Next, since v is the harmonic lifting of u, |v| ≤ |u| in B r .In turn, the computation used in Lemma 3.5 demonstrated that u ± are subharmonic, and therefore |u| = u + + u − is as well.Thus, by the maximum principle, sup B 1 |u| ≤ sup ∂B 1 |u| = sup g, the given boundary data in (1.2).In particular, with C independent of x and v. From this fact, combined with (3.2) and (3.3), we infer At this point, we can mimic the derivation in [AP,Theorem 3.1] to deduce that In turn Morrey's Dirichlet Growth Theorem (see for instance [J,Corollary 9.1.6])implies the desired Hölder-1/2 regularity inside B 1/2 .
We have reached the proof of our main result: Proof of Theorem 1.1.Existence and uniqueness follow from Lemma 3.1.Thus, we need only to show the desired regularity.From Lemma 3.3 we know that u is a weak solution to our problem on B + 1 .Moreover, by Lemma 3.6 u is C 0,1/2 (B 1/2 ).The '± ′ operation preserves Hölder regularity (with the same Hölder norm) so u ± ∈ C 0,1/2 (B 1/2 ) and in particular on the thin region Γ.This implies that is Hölder continuous of order γ, although γ will in general not be 1/2.Nevertheless, this implies that u is a weak solution to an oblique derivative problem with Hölder continuous boundary data, namely −k − (u − ) p−1 + k + (u + ) p−1 .Regularity theory for such a problem (see e.g.[L,Proposition 5.53]) then yields that u must be C 1,γ up to the boundary, with But in turn this implies that u is Lipschitz up to the boundary, in which case (3.4) is Hölder continuous of order p − 1 when p ≤ 2; if p > 2 this is to be interpreted as differentiablity with a Hölder modulus of continuity.Applying the regularity theory once again we have the result of the theorem.Now suppose that g does not change sign.We aim to show that u does not change sign either, in which case u ± = u (and thus u ± is a smooth as u is) and the regularity result above can be bootstrapped to prove that u is smooth.To this end, suppose that g ≥ 0, but u attains a minimum value which is negative, say u(z) = m < 0. Then z must lie on Γ.In particular, z ∈ Γ R = Γ ∩ B R for some 0 < R < 1.Now, trivial modifications to the above arguments allow to show u ∈ C 1,α (B R ), and therefore we can assume that the restriction of u to Γ R is C 1,α .Next, we apply the Hopf Lemma.Since u is harmonic in the interior we must have ∂u ∂ν (z) < 0.
Here ν is the outer normal vector, which at the point z is −e n .Thus However, the boundary condition along Γ is given by which holds in a classical sense since u is C 1,α in a neighborhood of z.But u(z) < 0, and therefore the boundary condition at z is u xn = −k − u − (z) < 0, a contradiction.We have thus shown that, if g ≥ 0, u cannot be negative along Γ.As a consequence, u is non-negative everywhere, so that u ± = u and higher regularity follows by bootstrapping.
A similar argument shows that if g ≤ 0 then u ≤ 0 everywhere, which again implies higher regularity.
We now show that, at least in the case p = 2, the regularity obtained in Theorem 1.1 is optimal at points where the gradient of u is non-vanishing.
Proof of Theorem 1.2.We argue by contradiction, and assume that u ∈ C 1,1 (0), with ∇u(0) = 0.After rearrangements, it follows from the extension theorem in [CS], and the semigroup property of −(−∆u) s , that Thanks to Theorem 1.1, we know that u has a unique differential P 1 = ∇u(0).Without loss of generality, we may assume that P 1 is also a superdifferential for u − (if not, consider u + ).Trivially, P 2 = 0 is a subdifferential for u − .Furthermore, it follows from our C 1,1 assumption that (u − ) τ τ (0) ≤ C for some constant C, for any direction τ .We now consider A straightforward computation yields In addition, u − (x) ≤ ψ(x), with equality at x = 0. Applying the comparison principle, we infer But we showed above that −(−∆) 1/2 u − (0) ≥ −C.We have thus reached a contradiction.

Monotonicity of a perturbed Almgren frequency functional
In this section we establish some properties of the solution around free boundary points.For u solution to (1.2), we define the coincidence set Λ(u) = {(x ′ , 0) | u(x ′ , 0) = 0}, and the free boundary F (u) = ∂Λ(u).in the Signorini problem, the monotonicity of the Almgren's Frequency Functional (4.1) plays a fundamental role in the study of both the solution and the free boundary.In our setting, N(r) may fail to be monotone, but a suitable perturbation is.From this point on, we will assume p ≥ 2.

Proof. Let
We begin by observing We also have which we can rewrite as (4.4) Using integration by parts we note that Applying this fact in (4.4) we obtain For the sake of brevity we will define A direct computation, together with (4.4) and (4.5), yields ´(∂Br ´(∂Br) + u 2 dσ(x) .
Collecting terms we have ´(∂Br) + uu ν dσ(x) ´(∂Br) + u 2 dσ(x) Clearly the first term in (4.6) is non-negative, whereas the last one vanishes.On the other hand, from (1.2) we know (4.7) − ´(∂Br) + uu n u ´(∂Br) + u 2 ≥ ´(∂Br) + u 2 ´(∂Br) + uu ν − ´(∂Br) + uu ν ´(∂Br) + u 2 ≥ 0 by the Cauchy-Schwartz inequality.Hence N (r) ≥ 0, and the proof is complete.We now state some consequences of Theorem 4.1.The first result shows that, even if the Almgren's Frequency Functional N(r) in (4.1) fails to be monotone, it still has a limit as r → 0 + , and in fact its limit coincides with the one of Ñ(r).In order to prove this result, we will need the following trace-type inequality.Since its proof is an easy modification of classical arguments, we will report it, for the sake of completeness, in the Appendix.
On the other hand, if we let k = max{k + , k − } and 0 Applying (4.8) we get (4.10) Using the notations introduced in the proof of Theorem 4.1, we thus obtain Hence, (4.11) and the desired conclusion follows from (4.9) and (4.11).
Next, we introduce the quantity The following hold: (a) The function r → r −2µ ϕ(r) is nondecreasing for 0 < r < 1/2.In particular, Here C is the constant appearing in (4.10).
Proof.We begin the proof of (a) by computing Hence, taking (4.7) into account, we have In the last inequality we have used Theorem 4.1 and the fact that p ≥ 2.
Thanks to Corollary 4.3, there exists To conclude we integrate the inequality over (r, R).
Corollary 4.5.Let u be a solution to (1.2).Then, for all Proof.We note that (u + ) 2 is a positive subharmonic function in the domain.Hence, by Corollary 4.4(a).A similar estimate holds for (u − ) 2 .

Almgren rescalings and blow-ups
The next step in our analysis is to study blow-up sequences around a free boundary point x 0 ∈ F (u). Without loss of generality, we may assume x 0 = 0. We define, for 0 < r < 1, the Almgren rescalings We note that v r L 2 ((∂B 1 ) + ) = 1.Moreover, for R 0 = R 0 (δ) as in Corollary 4.4(b), a fixed R > 1, and every r > 0 such that rR ≤ R 0 , we have, thanks to Corollaries 4.3 and 4.4(b), Hence, any sequence {v r j }, with r j → 0 + as j → ∞, is equibounded in H 1 loc (R n ), and by Theorem 1.1, it is also bounded in C 1,α loc (R n ).Thus, there exists a subsequence, denoted by v j , and a function v * (which we will refer to as the Almgren blow-up), such that uniformly on every compact subset of R n .We note that the fact v j L 2 ((∂B 1 ) + ) = 1 in particular implies that the blow-up is nontrivial.In addition, by rescaling, and therefore we have (keeping in mind that u(0) = 0) µ > 0. A similar rescaling argument, in fact, shows that for any ρ > 0 (5.2) We now assume p ≥ 3.An application of (4.8) yields Since [ϕ(r; u)] 1/2 ≤ Cr µ by Corollary 4.4(a) and sup B + r |u| ≤ Cr µ by Corollary 4.5, we conclude that the last term in (5.4) goes to zero as r → 0 + .Hence, if we extend v * by even reflection across {x n = 0}, then v * is harmonic in B 1 .The same conclusion can be reached in the case 2 ≤ p < 3 by applying Hölder's inequality in the last integral in (5.3).Now, it is well known (see, for instance [PSU,Section 9.3.1])that a function harmonic in B 1 and satisfying (5.2) is necessarily an homogeneous harmonic polynomial of degree µ ∈ N (since we have already ruled out the possibility µ = 0).If we also assume ∇ x ′ u(0) = 0, from the uniform convergence of ∇v j to ∇v * we deduce µ ≥ 2. We have thus proved the following.
Theorem 5.1.Let u be a solution to (1.2), with u(0) = 0 and ∇ x ′ u(0) = 0.If v r is as in (5.1), then for any sequence r j → 0 + there exists a subsequence {v j } of {v r j } and a function v * such that ) and in C 1 (B + 1 ).Furthermore, the even reflection of v * across {x n = 0} is an homogeneous harmonic polynomial of degree µ = N(0+; u) ∈ N, µ ≥ 2.

A Monneau-type monotonicity formula
Our next step consists in establishing almost-monotonicity of a functional of Monneau type.Using the notations introduced in the proof of Theorem 4.1, we define the Weiss functional Theorem 6.1.Let u be as in Theorem 5.1, and let p µ be an harmonic polynomial, homogeneous of degree µ and even in x n .If we define the Monneau functional as then there exists C > 0 such that Proof.Let p µ be an harmonic polynomial, homogeneous of degree µ and even in x n .Since N(r; p µ ) = µ, we have W µ (r; p µ ) = 0. We now rewrite Integrating by parts in the first integral, keeping in mind that p µ is harmonic and ∂pµ ∂xn = 0 on Γ r , we obtain Noting that ∇p µ • x = µp µ , we infer We now observe the following facts: ) In (6.5) we have used the boundary condition in (1.2), Corollary 4.5, and the fact that p µ is homogeneous of degree µ.As a consequence, the constant in (6.5) will depend on sup |u| and p µ L 1 (Γ) .Using (6.3)-(6.5) in (6.2), we obtain An application of (6.8) yields d dr thus concluding the proof.
Corollary 6.4.Let u and µ be as in Theorem 5.1.The set Σ µ (u) (see Definition 1.5) is of type F σ , i.e. it is the union of countably many closed sets.
Proof.The proof follows the lines of the one of Lemma 1.5.3 in [GP], and it is omitted.

The structure of the singular set
To continue with our analysis, we introduce the homogeneous rescalings and show existence and uniqueness of the blow-ups with respect to this family of rescalings.
Theorem 7.1.Let u and µ be as in Theorem 5.1.If v (µ) r is as in (7.1), then there exists a unique function v 0 such that ) and in C 1 (B + 1 ).Furthermore, the even reflection of v 0 across {x n = 0} is an homogeneous harmonic polynomial of degree µ.
Proof.By Corollary 4.4(a) and Lemma 6.3, there exist constants C 1 , C 2 > 0 such that From this it follows that any limit of the rescalings v (µ) r over any sequence r j → 0 + is a positive multiple of the Almgrens rescalings v r as in (5.1).By Theorem 5.1, we know that the even reflection of v 0 across {x n = 0} is an homogeneous harmonic polynomial of degree µ, and that the convergence is both in H 1 (B + 1 ) and in C 1 (B + 1 ).At this point, we only need to show uniqueness.To this end, we apply Corollary 6.2 with p µ = v 0 .We thus have where the last equality follows from the first part of the proof.In particular, we have that as r → 0 + , and not only over Hence, u ′ 0 = u 0 and the proof is complete.The next step consists in showing the continuous dependance of the blow-ups.In what follows, we denote by P µ , with µ ∈ N, the class of harmonic polynomials homogeneous of degree µ and even in x n .
Theorem 7.2.Let u be a solution to (1.2), µ ∈ N with µ ≥ 2, and x 0 ∈ Σ µ (u).Denote by p x 0 µ the blow-up of u at x 0 as in Theorem 7.1, so that u Proof.We begin by observing that P µ is a convex subset of the finite-dimensional vector space of all polynomials homogeneous of degree µ, and therefore all norms are equivalent.We choose to endow it with the norm in L 2 ((∂B 1 ) + ).We begin by fixing x 0 ∈ Σ µ (u) and ε > 0 sufficiently small.Then there is r ε = r ε (x 0 ) > 0 such that In turn, there exists Corollary 6.2 yields that M x 1 µ (r; u, p x 0 µ ) < 3ε, provided that 0 < r < r ε and r ε is small enough.Rescaling and passing to the limit as r → 0 + , we obtain and the first part of the theorem is proved.In order to establish the second part, we observe that, for |x 1 − x 0 | < δ ε and 0 < r < r ε , combining (7.2) and ( 7.3) we obtain we infer from (7.4) (7.5) v (µ) r,x 1 (x) − p x 1 µ L 2 (B + 1 ) ≤ Cε 1/2 .At this point we observe that the difference w r = v (µ) r,x 1 (x) − p x 1 µ is a weak solution to By the L ∞ − L 2 interior estimates (see, for instance [L,Theorem 5.36]), there exists a positive constant C = C(n, k + , k − ) such that, for some q > n − 1, To estimate the right-hand side in (7.6), we recall that |v Combining (7.6) with (7.5) and (7.7), we obtain (7.8) v (µ) r,x 1 (x) − p x 1 µ L ∞ ((B 1/2 ) + ) ≤ C ε for 0 < r < r ε sufficiently small, and C ε → 0 as ε → 0. To conclude, we cover the compact set K ⊂ Σ µ (u) ∩ B 1 with a finite number of balls B δε(x i 0 ) (x i 0 ) for some choice of x i 0 ∈ K, i = 1, . . ., N. Hence, for r < r K ε := min{r ε (x i 0 )|i = 1, . . ., N}, we have that (7.8) holds for all x 1 ∈ K.The desired conclusion readily follows.
Proof of Theorem 1.6.The proof of the structure of Σ d µ is centered on Corollary 6.4, Theorem 7.2, Whitney's extension theorem, and the implicit function theorem.Since the arguments are essentially identical to the ones in the proof of Theorem 1.3.8 in [GP], we omit the details and refer the interested reader to that source.We now proceed to estimate the term ´B+ r u 2 dx by adapting the proof of [L,Lemma 5.22].To this end, let We have The desired conclusion follows from combining (8.1) and (8.5).