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Regularity results in free boundaryproblems

GOHAR ALEKSANYAN

Doctoral ThesisStockholm, Sweden 2016

TRITA-MAT-A 2016:10ISRN KTH/MAT/A-16/10-SEISBN 978-91-7729-161-9

KTHInstitutionen för Matematik

100 44 StockholmSWEDEN

Akademisk avhandling som med tillstånd av Kungl Tekniska högskolan fram-lägges till offentlig granskning för avläggande av Filosofie doktorsexamen imatematik fredagen den 2 december 2016 kl 13:00 i sal D3, Kungl Tekniskahögskolan, Lindstedtsvägen 5, Stockholm.

c© Gohar Aleksanyan, 2016

Tryck: Universitetsservice US AB

iii

In memory of my father and teacher,Hakob Aleksanyan

v

Abstract

This thesis consists of three scientific papers, devoted to the regu-larity theory of free boundary problems. We use iteration arguments toderive the optimal regularity in the optimal switching problem, and toanalyse the regularity of the free boundary in the biharmonic obstacleproblem and in the double obstacle problem.

In Paper A, we study the interior regularity of the solution to theoptimal switching problem. We derive the optimal C1,1-regularity ofthe minimal solution under the assumption that the zero loop set is theclosure of its interior.

In Paper B, assuming that the solution to the biharmonic obstacleproblem with a zero obstacle is sufficiently close-to the one-dimensionalsolution 1

6(xn)3+, we derive the C1,α-regularity of the free boundary,

under an additional assumption that the noncoincidence set is an NTA-domain.

In Paper C we study the two-dimensional double obstacle problemwith polynomial obstacles p1 ≤ p2, and observe that there is a newtype of blow-ups that we call double-cone solutions. We investigatethe existence of double-cone solutions depending on the coefficients ofp1, p2, and show that if the solution to the double obstacle problem withobstacles p1 = −|x|2 and p2 = |x|2 is close to a double-cone solution,then the free boundary is a union of four C1,α-graphs, pairwise crossingat the origin.

vi

Sammanfattning

Denna avhandling består av tre vetenskapliga artiklar som ägnasåt regularitetsteori för frirandsproblem. Vi använder iterationsargumentför att härleda den optimala regulariteten hos optimalväxlingsproblemetoch för att analysera regulariteten av den fria randen i det biharmoniskahinderproblemet och i dubbelhinderproblemet.

I Artikel A studerar vi den lokala regulariteten hos lösningen till op-timalväxlingsproblemet. Under antagandet att nolloopmängden är till-slutningen av sitt inre visar vi C1,1-regularitet för den minimala lös-ningen.

I Artikel B, antaget att lösningen till det biharmoniska hinderpro-blemet med ett nollhinder ligger tillräckligt nära den endimensionelllösning, härleder vi C1,α-regularitet för den fria randen, under ett extraantagande att positivitetsmängden är ett NTA-område.

I Artikel A och Artikel B utför vi analys i Rn för n ≥ 2, medan i Arti-kel C studerar vi ett tvådimensionellt hinderproblem. Det huvudsakligaobjektet för undersökningen i Artikel C är så kallade dubbelkonlösning-ar. Vi visar att om lösningen till dubbelhinderproblemet med hinderp1 = −|x|2 och p2 = |x|2 är nära en dubbelkonlösning, då består denfria randen en union av fyra C1,α-grafer, parvis korsande i origo.

Contents

Contents vii

I Part: Introduction and summary

1 The obstacle problem 7

2 The double obstacle problem 112.1 Summary of Paper C . . . . . . . . . . . . . . . . . . . . . . . 12

3 The optimal switching problem 173.1 Summary of Paper A . . . . . . . . . . . . . . . . . . . . . . . 18

4 The biharmonic obstacle problem 234.1 Summary of Paper B . . . . . . . . . . . . . . . . . . . . . . . 24

References 27

II Part: Scientific papers

Paper AOptimal regularity in the optimal switching problemAnn. I. H. Poincaré-AN (2015), http://dx.doi.org/10.1016/j.anihpc.2015.06.001

Paper BRegularity of the free boundary in the biharmonic obstacle problemPreprint: http://arxiv.org/abs/1603.06819

Paper CAnalysis of blow-ups for the double obstacle problem in dimension two

vii

Acknowledgements

First and foremost I would like to thank my advisor, John Andersson, forthe mathematics I have learned from him. This thesis exists due to him. Johnhas been a great source of knowledge and inspiration to me. I am grateful foreverything.

Many thanks go to my co-advisor, Erik Lindgren, for his careful proofread-ing of the articles included in the thesis and for all the discussions. I wantto thank Diogo Aguiar Gomes and Henrik Shahgholian for introducing me tothe optimal switching problem, and for fruitful discussions on the topic.

During the years of my doctoral studies I found many friends among thefellow graduate students and young scientists. They have been very kind andsupportive; it is my pleasure and my duty to thank Mariusz Hynek, JevgenijaPavlova, my cheerful officemate Ludvig af Klinteberg, Gustav Sædén Ståhland many others.

I am grateful to my family for their love. Thank you for letting me travelthe world and impatiently awaiting each of my arrivals at Zvartnots interna-tional airport. My special thanks go to my brother, Sargis Aleksanyan, for allour conversations about life and mathematics.

ix

CONTENTS 1

Common notations

N the set of natural numbersR the set of real numbersRn the n-dimensional Euclidiean spacex · y ∑n

i=1 xiyi, the scalar product of vectors x, y ∈ Rn

Br(x) y ∈ Rn : |y − x| < r, the open ball in Rn

∂A the boundary of the set A ⊂ Rn

A the closure of the set A ⊂ Rn

A0 the interior of the set A ⊂ Rn

Ac the complement of the set A ⊂ Rn

a+ (a+) max(a, 0), for a ∈ Ra− (a−) max(−a, 0), for a ∈ R∇ the gradient vector, ∇u :=

(∂u∂x1, ..., ∂u

∂xn

)for a scalar function u

∆ the Laplace operator, ∆u := ∑ni∂2u∂x2ifor a scalar function u

∆2 the biharmonic operator, ∆2u := ∆(∆u)D2u the Hessian matrix for uD3u the tensor of partial derivatives of third order

Part I

Introduction and summary

3

CONTENTS 5

IntroductionThis thesis consists of three articles, and all of them are devoted to the regu-larity theory of free boundary problems.

The aim of Part I is to provide a brief overview of the obstacle problemand of the free boundary problems that are investigated in Part II. In Part Iwe also summarize the results obtained in the thesis, outlining the backgroundand techniques used. The core of the thesis is Part II, which is a combinationof three papers, named Papers A, B and C.

In Paper A we investigate the optimal regularity of solutions to the optimalswitching problem (see also Chapter 3). The problem is related both to theclassical obstacle problem (Chapter 1) and to the double obstacle problem(Chapter 2), while the regularity of solutions to the optimal switching problemis not covered by the known regularity theory for any of the above mentionedproblems. In fact, we are dealing with a system of obstacle-type equations,where the obstacles depend on the solution.

Paper A also includes an example of a homogeneous global solution to thetwo dimensional double obstacle problem, where the free boundary is a cross.This example shows a new type of behavior of the free boundary for the doubleobstacle problem, which does not happen in the case of the obstacle problem.Paper C analyses the blow-ups of the solution to the double obstacle problemin dimension two, with an emphasis on the so-called double-cone solutions.

In Paper B we use a linearization argument and show that if the solutionto the biharmonic obstacle problem is close to a halfspace solution, then thefree boundary is locally a C1,α-graph. The method does not require the useof comparison principles and regularity of the solution. For the biharmonicobstacle problem there are no optimal growth or nondegeneracy propertiesknown. The comparison principle for harmonic functions plays an importantrole in the analysis of solutions to the obstacle problem. While there areno comparison principles for biharmonic functions, as one can easily see bylooking at the following one-dimensional example; f(x) = x2 in R, f is anonnegative biharmonic function, obtaining a minimum at an interior pointx = 0. Hence we make an additional assumption that the noncoincidenceset is a non-tangentially accessible (NTA) domain. Chapter 4 contains a briefintroduction to the biharmonic obstacle problem and an overview of Paper B.

Chapter 1

The obstacle problem

The obstacle problem is a motivating example in the theory of variational in-equalities and free boundary problems. In this chapter we give the variationalformulation of the problem and a short survey on the known regularity the-ory of the obstacle problem. This presentation is far from being exhaustive.We have selected the properties of the obstacle problem which have a strongrelation to the problems studied in the main text.

The chapter is based on the article [5]. The text is also supported by thebook [11].

Let Ω ⊂ Rn be a given domain with a smooth boundary, and let ϕ ∈ C2(Ω)be a given function, called an obstacle. We are looking for the minimizeru ∈ W 1,2(Ω) to the Dirichlet functional

J [u] =∫

Ω|∇u(x)|2dx, (1.1)

over admissible functions u ≥ ϕ in Ω, with boundary condition u = g on∂Ω, where g is a continuous function, satisfying g ≥ ϕ on ∂Ω. Since J is astrictly convex functional over a convex admissible set, there exists a uniqueminimizer. The minimizer u is called the solution to the obstacle problem.A variational argument easily verifies that u is a superharmonic function inΩ, and therefore it has a lower semi-continuous equivalent, which we call u.Denote by

Ωu := x ∈ Ω;u(x) > ϕ(x), (1.2)

then Ωu is an open set, called the noncoincidence set. Denote the free bound-ary for the obstacle problem by

Γu := ∂Ωu ∩ Ω. (1.3)

We see that Γu depends on the solution u which is not known a priori, ex-plaining the usage of the word ‘free’.

7

8 CHAPTER 1. THE OBSTACLE PROBLEM

In the set Ωu, the function u does not touch the obstacle ϕ, and it locallyminimizes the Dirichlet energy, (1.1). Therefore u is a harmonic function inΩu. Hence we obtain the following Euler-Lagrange equation for the obstacleproblem,

min(−∆u, u− ϕ) = 0. (1.4)

It has been shown (see for instance [5]) that the solution to the obstacleproblem u is as regular as the obstacle ϕ up to C1,1, which is the best regularitythat the solution may achieve, since ∆u jumps from zero to ∆ϕ across thefree boundary.

Next we turn to to the regularity of the free boundary in the obstacleproblem. First let us rewrite equation (1.4) in a more convenient form. Denoteby v := u− ϕ, then according to (1.4), v is a nonnegative function solving

∆v = −∆ϕχv>0, (1.5)

where χA is the characteristic function of a set A ⊂ Rn. It follows from (1.4)that ∆ϕ ≤ 0 in the coincidence set u = ϕ. Taking ∆ϕ ≡ −1, we study thenormalized obstacle problem,

∆u = χu>0. (1.6)

It is easy to verify that u minimizes the following energy functional

J [u] =∫

Ω

12 |∇u(x)|2 + u(x)dx, (1.7)

over nonnegative functions with given boundary condition.Let u be the solution to the normalized obstacle problem in Ω ⊂ Rn,

B1(0) ⊂⊂ Ω. For a fixed x0 ∈ Γu∩B1 and r < 1 small, satisfying Br(x0) ⊂ B1,consider the following rescalings,

ux0,r(x) := u(rx+ x0)r2 . (1.8)

The abbreviation ur := u0,r is used. Let us note that the rescalings ux0,r solvethe obstacle problem (1.6) in the ball B 1

r(0).

By using comparison principle, we can show the following quadratic growthestimate

supBr(x0)

u ≤ Cnr2, (1.9)

where x0 ∈ Γu, B2r(x0) ⊂ Ω, and Cn is a dimensional constant. Hence therescalings ux0,r are bounded,

supB1

ux0,r ≤ Cn, (1.10)

9

for small r > 0.Furthermore, at free boundary points the solution cannot decay faster than

quadratically. Let, x0 ∈ Ωu ∩ Ω, then

supBr(x0)

u ≥ 12nr

2, (1.11)

which can be derived by applying the maximum principle for the harmonicfunction v = u − 1

2n |x − x0|2 in Ωu ∩ Br(x0). Inequality (1.11) implies thatrescalings ux0,r possess the following nondegenaracy property,

supB1

ux0,r ≥1

2n. (1.12)

From (1.10) we can obtain that ‖ux0,r‖C1,1(B1/2) is uniformly bounded.Hence through a subsequence ux0,r converges to a function u0 in C1,α(B1/2),where 0 < α < 1. Let us emphasize that, at this point we do not have anytools which would guarantee the existence of the limr→0+ ux0,r in any functionspace. Any limit of the rescaling ux0,r over a sequence rj → 0+ as j → ∞ iscalled a blow-up of the solution u. The blow-ups are defined in Rn. Further-more, the blow-ups are global solutions, i.e. solutions in Rn to the normalizedobstacle problem (1.6). It follows from the nondegenaracy property (1.12) andthe C1,α-convergence that the blow-ups are not identically zero.

The analysis of the possible blow-ups and the proof of the uniqueness ofthe blow-up are far from being easy, and we cannot include it in this shortoverview of the obstacle problem. Let us only state the following importanttheorem.

Theorem 1.1. Let u be the solution to the normalized obstacle problem, (1.6),and let x0 ∈ Γu be a free boundary point. Then the blow-up of u at x0 is unique,that is the limit u0(x) = limj→∞

u(rjx+x0)r2j

does not depend on the sequencerj → 0+. Furthermore, there are only two types of possible blow-ups; eitheru0 is a half-space solution,

u0(x) = 12(x · e)2

+, where e ∈ Rn is a unit vector, (1.13)

or u0 is a polynomial solution,

u0(x) = 12(x · Ax), where A is an n× n nonnegative

symmetric matrix with trA = 1.(1.14)

10 CHAPTER 1. THE OBSTACLE PROBLEM

If ux0,r → 12(x ·e)2

+, then the vector e is the approximate unit normal to Γuat x0, and x0 is called a regular point. It has been shown that almost everyfree boundary point is regular, and that the set of regular points is a relativelyopen subset of Γu. Furthermore, in a neighborhood of a regular point, the freeboundary is a C1,α-graph, moreover, it is real analytic.

If the blow-up of u at x0 is a polynomial, then x0 is called a singular point.It has been shown that the set of singular points is a closed set contained ina lower dimensional C1 manifold.

Chapter 2

The double obstacle problem

Let Ω be a bounded open set in Rn with smooth boundary. The solution tothe double obstacle problem in Ω is the minimizer of the functional

J(u) =∫

Ω|∇u(x)|2dx

over functions u ∈ W 1,2(Ω), ψ1 ≤ u ≤ ψ2, satisfying the boundary conditionu = g on ∂Ω. For the problem to be well defined we assume that ψ1 ≤ ψ2 inΩ, and ψ1 ≤ g ≤ ψ2 on ∂Ω. The functions ψ1 and ψ2 are called respectivelythe lower and the upper obstacles.

If u < ψ2, then u solves the obstacle problem with ψ1, and therefore−∆u ≥ 0 by (1.4). Also observe that if u > ψ1, then −u solves the obstacleproblem with −ψ2, and ∆u ≥ 0 by (1.4). Therefore we may conclude that usatisfies the following inequalities,

ψ1 ≤ u ≤ ψ2, ∆u ≥ 0 if u > ψ1 and ∆u ≤ 0 if u < ψ2. (2.1)

It has been shown that the solution to the double obstacle problem is locallyC1,1 under the assumption ψi ∈ C2(Ω), see for instance [3, 8]. Hence we canrewrite (2.1) as follows,

ψ1 ≤ u ≤ ψ2 and ∆u = ∆ψ1χu=ψ1 + ∆ψ2χu=ψ2 −∆ψ1χψ1=ψ2 a.e.,

where χA is the characteristic function of a set A ⊂ Rn. Let us observe that∆u = ∆ψ1 in the set u = ψ1∩ u < ψ2, hence (2.1) implies that ∆ψ1 ≤ 0,and ψ1 is a superharmonic function. By a similar argument, we obtain thatψ2 is a subharmonic function in the set u = ψ2 ∩ u > ψ1.

Since the optimal regularity of the solution is known, our main interestbecomes the analysis of the free boundary. Denote by

Ω1 := u > ψ1, Ω2 := u < ψ2, and Ω12 := Ω1 ∩ Ω2, (2.2)

11

12 CHAPTER 2. THE DOUBLE OBSTACLE PROBLEM

then Ω = Ω1 ∪ Ω2 ∪ Λ, where Λ := x ∈ Ω : ψ1(x) = ψ2(x). We call Ω12the noncoincidence set. It follows from (2.1) that ∆u ≥ 0 and ∆u ≤ 0 in Ω12.Hence u is a harmonic function in the noncoincidence set, as in the ”single”obstacle problem.

Define the free boundary for the double obstacle problem

Γu := ∂Ω12 ∩ Ω, (2.3)

and letΓi := ∂Ωi ∩ Ω, i = 1, 2, then Γu ⊂ Γ1 ∪ Γ2. (2.4)

Assume thatψ1 ≤ ψ2, and ψi ∈ C2(Ω). (2.5)

If ψ1 < ψ2, then Γ1∩Γ2 = ∅, and x0 ∈ Γu implies that x0 ∈ Γ1 or x0 ∈ Γ2. Letx0 ∈ Γ1, then there exists a small ball Br(x0) such that Br(x0)∩u = ψ2 = ∅.Hence u solves the obstacle problem with obstacle ψ1 in the ball Br(x0), andthe known regularity theory for the obstacle problem can be applied to analysethe free boundary in a neighborhood of x0. Therefore we assume that

Λ = x ∈ Ω : ψ1(x) = ψ2(x) 6= ∅. (2.6)

Let x0 ∈ Γu be a free boundary point, if x0 /∈ Λ, then in a neighborhoodof x0, we have ψ1 < ψ2, this case has already been discussed above.

Assume that x0 ∈ Γu∩Λ. First let us observe that x0 cannot be an interiorpoint of Λ. Otherwise, if x0 ∈ Λ0, then there exists Br(x0) ⊂ Λ, which is thesame as ψ1 = ψ2 in Br(x0). Since ψ1 ≤ u ≤ ψ2, we obtain u = ψ1 = ψ2

in Br(x0), hence x0 /∈ Γu. So we are interested in the behavior of the freeboundary at the points x0 ∈ ∂Λ ⊂ Γu. First we want to understand thebehavior of the free boundary when x0 ∈ Λ is an isolated point. Paper Cpartially answers this question in dimension two with polynomial obstacles.

2.1 Summary of Paper CIn Paper C we study the following normalized double obstacle problem indimension two,

∆u = λ1χu=p1 + λ2χu=p2 (2.7)

with polynomial obstacles p1 ≤ p2, where λ1 = ∆p1 ≤ 0 and λ2 = ∆p2 ≥ 0are constants. We assume that the obstacles p1 and p2 meet at a single pointx0, i.e. p1(x) = p2(x) iff x = x0.

The work is inspired by the following example in R2,

u0(x) = x21 sgn(x1) + x2

2 sgn(x2). (2.8)

2.1. SUMMARY OF PAPER C 13

It is easy to see that u0 solves the double obstacle problem (2.1), with obstaclesp1(x) = −x2

1 − x22 and p2(x) = x2

1 + x22. Observe that Λ = 0, and the free

boundary consists of two lines crossing at the origin. The function u0 is amotivational example for double-cone solutions, which are the novelty of thispaper.

Let u be the solution to the normalized double obstacle problem (2.7).Assume that x0 = 0, we can always come to this situation with a changeof variables. By subtracting a first order polynomial from p1, p2 and u, andtaking into account that u ∈ C1,1, we get

u(0) = p1(0) = p2(0) = 0, and ∇u(0) = ∇p1(0) = ∇p2(0) = 0.

Hence

p1(x) = a1x21 + 2b1x1x2 + c1x

22 and p2(x) = a2x

21 + 2b2x1x2 + c2x

22.

Furthermore, we may assume that b1 = b2 = 0, by rotating the coordinatesystem and subtracting a harmonic polynomial from p1, p2 and from u, thusobtaining

p1(x) = a1x21 + c1x

22 and p2(x) = a2x

21 + c2x

22. (2.9)

Observe that p1, p2 are second order homogeneous polynomials; which means

p1(rx) = r2p1(x), and p2(rx) = r2p2(x). (2.10)

As in the classical obstacle problem, any limit of u(rx)r2 as r → 0+, is called

a blow-up of the solution u to the double obstacle problem at the origin.We show that the blow-ups of a solution to the normalized double obstacleproblem are homogeneous of degree two functions via Weiss’ monotonicityformula. This result is not surprising at all, since we already saw in Theorem1.1 that the blow-ups of the solution to the obstacle problem are homogeneousdegree two functions.

Knowing that the blow-ups are homogeneous global solutions, we makea complete characterization of possible blow-ups in dimension n = 2. Inparticular we see that there exist blow-ups of a new type. We call thesesolutions double-cone solutions, since the noncoincidence set is a union of twocones with a common vertex at the origin. We show that there exist double-cone solutions if and only if the following polynomial

P = P (x1, x2) ≡ p1(x1, x2) + p2(x2, x1) = (a1 + c2)x21 + (a2 + c1)x2

2 (2.11)

has zeroes other than x = 0. If P ≡ 0, there are infinitely many double-conesolutions, and if P 6≡ 0, but P = 0 on a line, there are finitely many double-cone solutions. This result is quite surprising and unexpected. In the nextparagraph we describe how (2.11) affects the stability of the free boundary.

14 CHAPTER 2. THE DOUBLE OBSTACLE PROBLEM

Let ε be an arbitrary number, |ε| << 1. Then for polynomials p1(x) =−x2

1 − x22, p2(x) = x2

1 + x22 there exist infinitely many double-cone solutions.

While when we look at the double obstacle problem with p1 = −x21 − x2

2 andp2 = (1 − ε)x2

1 + (1 + ε)x22 there are only four double-cone solutions, and

for p1 = −x21 − x2

2 and p2 = (1 + ε)x21 + (1 + ε)x2

2 there are none. Thisproperty reveals the instability of the double obstacle problem in the sensethat changing the obstacles slightly, may change the solution and the freeboundary significantly.

In this paper we also study the regularity of the free boundary in the casewhen P ≡ 0. Since in this case we have infinitely many rotational invariantdouble-cone solutions, we can use a flatness improvement argument to showthat the free boundary is locally a union of four C1,α-graphs. We have alreadyused a similar argument in Paper B, see also Chapter 4.

The case P ≡ 0, can be reduced to the case p1 = −|x|2, and p2 = |x|2. Ageneral double-cone solution in this case can be written in polar coordinatesas follows,

µ = µφ1,φ2(r, θ) :=

r2, if − φ2 ≤ 2θ ≤ φ1

r2 cos(2θ − φ1), if φ1 ≤ 2θ ≤ π + φ1

r2 cos(2θ + φ2), if − π − φ2 ≤ 2θ ≤ −φ2

−r2, otherwise,

(2.12)

where 0 ≤ φ1, φ2 ≤ π are parameters describing a double-cone solution µ.The main result in Paper C is the following theorem.

Theorem 2.1 (Theorem 4.7 in Paper C). Let u be the solution to the two-dimensional double obstacle problem with obstacles p1 = −x2

1 − x22 and p2 =

x21 + x2

2. Assume that ‖u − µ‖L2(B2) = δ is sufficiently small, where µ is adouble-cone solution, and is not a halfspace solution. Then in a small ball Br0

the free boundary consists of four C1,γ- graphs meeting at the origin, denotedby Γ+

1 , Γ−1 , Γ

+2 , Γ

−2 . Neither Γ1 = Γ+

1 ∪ Γ−1 nor Γ2 = Γ+2 ∪ Γ−2 has a normal

at the origin. The curves Γ+1 and Γ+

2 cross at a right angle, the same is truefor Γ−1 and Γ−2 .

We show that if the solution is close to a double-cone solution in B1,then the blow-up at the origin is unique. Furthermore, employing the knownregularity theory for the free boundary in the obstacle problem, we derivethat the free boundary Γ for the double obstacle problem is a union of fourC1,γ-graphs meeting at the origin. Neither Γ1 nor Γ2 is flat at the origin, andthey meet at right angles, see Figure 2.1.

2.1. SUMMARY OF PAPER C 15

x1

x2

Γ+2

Γ+1

Γ−2

Γ−1

∆u = 0

u = p2

u = p1

∆u = 0

S1

S2

x0

ν(0) ν(x0)ν(0)

Figure 2.1: The behavior of the free boundary, with obstacles touching at a single point

We mentioned that if P 6≡ 0, but P = 0 on a line, there are finitely manydouble-cone solutions. We aim to describe the behavior of the free boundaryin this case in a future publication.

Chapter 3

The optimal switching problem

Let Ω ⊂ Rn be a bounded domain with smooth boundary. Consider a powerplant with several modes (states) of energy production in Ω. When runningthe production mode 1 ≤ i ≤ m, we pay a running cost f i(x) depending alsoon the present position x ∈ Ω. We can switch from mode i to another mode1 ≤ j ≤ m in order to minimize the cost of energy production, but we haveto pay a switching cost ψij(x) depending also on the present state x. Assumethat ψii = 0, which is reasonable; if we do not switch, there is no need to paya switching cost.

The solution to the optimal switching problem is a vector valued functionu = (u1, u2, ..., um) and it represents the cost of an optimal strategy for energyproduction. As it has been discussed in the literature [10], [4], the negativeof the solution to the optimal switching problem, −u, solves the followingsystem:

min(−Liui + f i,minj

(ui − uj + ψij)) = 0, in Ω, (3.1)

where Li is an elliptic operator, corresponding to the mode i. Equation (3.1)says that if minj(ui − uj + ψij) > 0, it is not optimal to switch to any othermode j 6= i, and we continue running mode i.

For the optimal switching problem to be well defined, we need to imposethe nonnegative loop condition: Let i0, i1, . . . , il = i0 be any loop of lengthl, i.e. including l number of states. Assume that (u1, u2, ..., um) is a solutionto system (3.1), then ui − uj + ψij ≥ 0 for any i, j ∈ 1, 2, ...,m, and aftersumming the equations over the loop, we get

l∑j=1

ψij−1,ij ≥ 0. (3.2)

Condition (3.2) is a necessary assumption for the existence of a solution to(3.1). Without (3.2) we can make the energy production cost arbitrary small

17

18 CHAPTER 3. THE OPTIMAL SWITCHING PROBLEM

by looping. For example, if l = 2 and ψ12(x) + ψ21(x) < 0 for some x, thenat the point x we can switch from mode 1 to mode 2, and immediately switchback to mode 1, thus paying a negative cost. This way we can decrease thecost as much as we want by switching many times. Therefore the assumptionψ12 + ψ21 ≥ 0 is needed in order to obtain a finite cost function.

The uniqueness and C1,1-regularity of the solution to system (3.1) havebeen studied in the literature under the assumption that the switching costsψi are nonnegative constants, [7], [10], [4].

In Paper A we consider a system, arising in a model optimal switchingproblem with only two states, i. e. m = 2, and L1 = L2 = ∆,min(−∆u1 + f 1, u1 − u2 + ψ1) = 0

min(−∆u2 + f 2, u2 − u1 + ψ2) = 0,(3.3)

with given Dirichlet boundary conditions ui = gi on ∂Ω. The switching costsψ1, ψ2 are given functions, satisfying the nonnegative loop assumption (3.2),which is the same as

ψ1 + ψ2 ≥ 0, (3.4)

since m = 2. Denote by

L := x ∈ Ω;ψ1(x) + ψ2(x) = 0,

and call it a zero loop set.

3.1 Summary of Paper AIn Paper A our aim is to investigate if solutions to (3.3) are C1,1. Assume that

f 1, f 2 ∈ Cα, and ψ1, ψ2 ∈ C2,α, for some 0 < α < 1. (3.5)

Then there exist v1, v2 ∈ C2,αloc solving the Poisson equation ∆vi = f i in Ω.

Denote ui0 := ui− vi, then ui0 is as regular as ui up to C2,α, and (u10, u

20) solves

the following system min(−∆u1, u1 − u2 + ϕ1) = 0,min(−∆u2, u2 − u1 + ϕ2) = 0,

(3.6)

where ϕ1 = v1 − v2 + ψ1 and ϕ2 = v2 − v1 + ψ2 are the new switching costfunctions preserving the loop condition, since ϕ1 + ϕ2 ≡ ψ1 + ψ2.

We see that u1 solves the obstacle problem with the obstacle u2 −ϕ1, andu2 is not known apriori. On the other hand u2 solves the obstacle problem with

3.1. SUMMARY OF PAPER A 19

the obstacle u1−ϕ2. Thus (3.7) is a system of obstacle-type equations, wherethe obstacles depend on the solution (u1, u2). Hence the regularity of thesolution to system (3.6) cannot be obtained directly by the known regularitytheory for the obstacle problem. We see that (u1, u2) cannot have a betterregularity than C1,1 which is the best regularity for the obstacle problem.

If L = ∅, that is ϕ1+ϕ2 > 0, then for any x ∈ Ω, u1(x)−u2(x)+ϕ1(x) > 0or u2(x) − u1(x) + ϕ2(x) > 0. Otherwise, if u1(x) − u2(x) + ϕ1(x) = 0 andu2(x)−u1(x) +ϕ2(x) = 0, we obtain ϕ1(x) +ϕ2(x) = 0, contradicting L = ∅.Observe that if u1−u2 +ϕ1 > 0, then u1 is a harmonic function, and u2 solvesthe obstacle problem with the obstacle u1 − ϕ2. Hence we may conclude thatu1 is an infinitely differentiable function, and u2 ∈ C1,1

loc .Thus, if L = ∅, we obtain (u1, u2) ∈ C1,1

loc by the regularity theory for theobstacle problem. This result is interesting, but not enough, since we want toknow if the C1,1-regularity holds in case L 6= ∅.

In L we have that ϕ1 = −ϕ2, hence we obtain from (3.6) that u1−u2+ϕ1 ≥0 and u2 − u1 − ϕ1 ≥ 0, which implies u1 − u2 + ϕ1 ≡ 0 in L . Take anysubharmonic function u1 ∈ W 2,p, for n < p < ∞ and let u2 = u1 + ϕ1, then(u1, u2) solves (3.6) in L . Hence system (3.6) may not have a unique solutionif L 6= ∅, and (3.6) admits solutions that are not C1,1. Instead we considerthe following system,

min(−∆u1, u1 − u2 + ϕ1) = 0,min(−∆u2, u2 − u1 + ϕ2) = 0,min(−∆u1,−∆u2) = 0,

(3.7)

which already has a unique solution. By using comparison principles, we showthat the solution to system (3.7) is the smallest among the solutions to (3.6).

The third equation in system (3.7) has a natural implication in the optimalswitching setting; it says that we always run one of the modes i = 1 or i = 2,even when we can switch for free. Let us also observe that the extra (third)equation in (3.7) holds automatically if L = ∅.

Now let us study the relation between the system (3.7) and the doubleobstacle problem. It is easy to see that U = u1 − u2 solves the followingdouble obstacle problem

−ϕ1 ≤ U ≤ ϕ2, −∆U ≤ 0 a.e. if U > −ϕ1, −∆U ≥ 0 a.e. if U < ϕ2 (3.8)

with obstacles −ϕ1 ≤ ϕ2. Assuming that ϕi ∈ C2,α, by the known regularitytheory for the double obstacle problem, we obtain that U ∈ C1,1

loc . Thus thedifference u1 − u2 ∈ C1,1, while the question if u1 ∈ C1,1

loc and u2 ∈ C1,1loc is not

covered by the regularity theory of the double obstacle problem.


Let us sketch the main results in Paper A. Consider system (3.7), and letΩ1 := −∆u1 > 0, Ω2 := −∆u2 > 0, and Ω12 := Ω \ Ω1 ∪ Ω2 be disjointopen sets. Then

−∆u1 = ∆ϕ1 > 0,−∆u2 = 0 in Ω1

−∆u2 = ∆ϕ2 > 0,−∆u2 = 0 in Ω2

−∆u1 = 0,−∆u2 = 0 in Ω12.

(3.9)

In the set Ω \ Ω1 we get −∆u1 = 0, and therefore u2 solves the obstacleproblem

min(−∆u2, u2 − u1 + ϕ2) = 0,with a C2,α-obstacle u1−ϕ2. Hence u2 ∈ C1,1

loc (Ω \Ω1). By a similar argumentwe obtain u1 ∈ C1,1

loc (Ω \ Ω2).It remains to study the regularity of the solution in a neighborhood of the

set ∂Ω1 ∩ ∂Ω2 ∩ Ω. Note that ∂Ω1 ∩ ∂Ω2 ⊂ L , since u1 − u2 + ϕ1 = 0 in Ω1and u2 − u1 + ϕ2 = 0 in Ω2.

Let us study the regularity of (u1, u2) in the interior of L , denoted byL 0. We have already observed before that u2 = u1 + ϕ1 in L . Hence weobtain from the third equation in system (3.7) that min(−∆u1,−∆u2) =min(−∆u1,−∆u1 −∆ϕ1) = 0, therefore u1 solves the following equation,

−∆u1 = (∆ϕ1)+ in L 0, (3.10)

and u2 = u1 + ϕ1 in L . By the classical regularity theory for elliptic equa-tions, solutions to equation (3.10) are locally C2,α, if ∆ϕ1 ∈ Cα. Thus in aneighborhood of the points x ∈ ∂Ω1 ∩ ∂Ω2 ∩L 0, the solution (u1, u2) ∈ C2,α.

The main difficulty is to study the regularity of (u1, u2) at the pointsx0 ∈ ∂Ω1 ∩ ∂Ω2 ∩ ∂L , called ‘meeting’ points.

Let u be a twice continously differentiable function in B1(x0), havingHölder continuous second order derivatives, i. e. u ∈ C2,α(B1(x0)), for some0 < α < 1. Let px0 be the second degree Taylor polynomial for u at x0, then

supx∈Br(x0)

|u(x)− px0(x)| ≤ Cn‖D2u‖Cαr2+α, (3.11)

where Cn is a dimensional constant. Estimate (3.11) can be obtained easilyby using the mean-value property for continously differentiable functions andthat D2u ∈ Cα.

On the other hand if u is a continuous function in Ω, and for every x0 ∈ Ωthere exists a polynomial px0 such that

supx∈Br(x0)

|u(x)− px0(x)| ≤Mr2+α, (3.12)

3.1. SUMMARY OF PAPER A 21

for a uniform constant M > 0, then u is twice continously differentiable, and‖D2u‖Cα ≤ cnM .

Our main result in Paper A is the following theorem.

Theorem 3.1 (Theorem 4 in Paper A). Assume ϕ1, ϕ2 ∈ C2,α, and L = L 0.Then the solution to the system (3.7), (u1, u2) is C2,α-regular on ∂Ω1 ∩ ∂Ω2 ∩∂L ∩ Ω, in the sense that for every x0 ∈ ∂Ω1 ∩ ∂Ω2 ∩ ∂L ∩ Ω, there existsecond order polynomials p1

x0, p2x0, such that

supx∈Br(x0)

|ui(x)− pix0(x)| ≤ Cr2+α (3.13)

where the constant C > 0 depends only on the given data.

The idea of the proof comes from [2], where the authors derive the optimalregularity for the no-sign obstacle problem. Let u ∈ W 2,p, for p <∞, if ∆u isa bounded function, it does not imply that D2u is bounded, but it does implythat D2u ∈ BMO. First, we show that ∆u1 and ∆u2 are bounded functions,then use the BMO-argument to derive D2ui ∈ L∞.

Let 0 ∈ ∂Ω1 ∩ ∂Ω2 ∩ ∂L 0, be a meeting point. Since ϕi ∈ C2,α, we get∆ϕ1(0) + ∆ϕ2(0) = 0. According to (3.9), ∆ϕ1 ≥ 0 in Ω1 and ∆ϕ2 ≥ 0 in Ω2,hence ∆ϕ1(0) = ∆ϕ2(0) = 0.

Employing the BMO-estimates for D2u1 and D2u2, we show that thepolynomials

p1r(x) := (u1)r + (∇u1)r · x+ 1

2x · (D2u1)r · x and

p2r(x) := (u2)r + (∇u2)r · x+ 1

2x · (D2u2)r · x

converge to harmonic polynomials, denoted respectively by p10 and p2

0. We alsodescribe the rate of convergence,

supx∈Br(0)

|pir(rx)− pi0(rx)| ≤ Cr2+α,

where C is just a constant. Considering the following rescalings;

vir(x) = ui(rx)− pi0(rx)r2+α ,

we study the corresponding system for (v1r , v

2r). Taking into account that

∆ϕi(rx)rα

is uniformly bounded for i = 1, 2, we show that rescalings (v1r , v

2r) are

uniformly bounded in the ball B1, which is equivalent to (3.13).


It follows from Theorem A that the solution to system (3.7) is locally C1,1

if L = L 0.In the end of the article we justify our assumption x0 ∈ ∂L 0 with a

counterexample. By considering a particular system in R2, with L = 0,we find an explicit solution, which is not C1,1 in any neighborhood of theorigin. This example reveals that the regularity of the solution to the optimalswitching problem depends on the topological properties of the zero loop set.We saw that at the so-called meeting points |D2ui(rx)| is of order rα. While if0 is an isolated point of L , we may have |D2u(rx)| ≈ − ln r as r → 0+. Thisis a new result, and it shows how different the optimal switching problem isfrom the obstacle or double obstacle problems.

Let us also mention that Paper C was born from the example in Paper A,although we gave the summary of Paper C in the previous chapter.

Chapter 4

The biharmonic obstacle problem

Let Ω ⊂ Rn be a given domain, and ϕ ∈ C2(Ω), ϕ ≤ 0 on ∂Ω be a givenfunction, called an obstacle. Consider the problem of minimizing the followingfunctional

J [u] =∫

Ω(∆u(x))2 dx, (4.1)

over all functions u ∈ W 2,20 (Ω), such that u ≥ ϕ. The functional J admits

a unique minimizer, called the solution to the biharmonic obstacle problemwith obstacle ϕ.

A variational argument easily verifies that ∆u is a weakly subharmonicfunction. Furthermore, ∆u is a harmonic function in the noncoincidence setu > ϕ. Therefore the solution satisfies the following variational inequality

∆2u ≥ 0, u ≥ ϕ, ∆2u · (u− ϕ) = 0.

It has been shown in [6] and [9] that the solution u ∈ W 3,2loc (Ω), ∆u ∈

L∞loc(Ω), and moreover u ∈ W 2,∞loc (Ω). Furthermore, in the paper [6], the

authors show that in dimension n = 2 the solution u ∈ C2(Ω) and that thefree boundary Γu := ∂u = ϕ lies on a C1-curve in a neighbourhood of thepoints x0 ∈ Γu, such that ∆u(x0) > ∆ϕ(x0).

The setting of our problem is slightly different from the one in [6] and[9]. We consider a zero-obstacle problem with general nonzero boundary con-ditions. We look for a minimizer to the functional (4.1) over the admissibleset

A :=u ∈ W 2,2(Ω), u ≥ 0, u = g > 0, ∂u

∂ν= f on ∂Ω

.

The minimizer u exists, and it is unique. The minimizer is called the solutionto the biharmonic obstacle problem. As in the obstacle problem, we denotethe free boundary by

Γu := ∂Ωu ∩ Ω, where Ωu := u > 0.

23

24 CHAPTER 4. THE BIHARMONIC OBSTACLE PROBLEM

There are several important questions regarding the biharmonic obstacleproblem that remain open. For example, the optimal regularity of the solution,the characterization of blow-ups at free boundary points, etc..

In Paper B we focus on the regularity of the free boundary for an n-dimensional biharmonic obstacle problem, assuming that the solution is closeto the one-dimensional solution 1

6(xn)3+. We saw in Chapter 1 that the solution

to the obstacle problem at a regular point has a unique blow-up of the form12(x+

n )2 in some coordinate sytem. A flatness improvement argument showsthat if the solution to the obstacle problem is close to the halfspace solution12(x+

n )2, then the free boundary is locally a graph of a C1,α-function. We expectto obtain a similar result for the biharmonic obstacle problem assuming thatthe solution is close to the one-dimensional solution c(x+

n )3. We choose c = 1/6so that∇∆

(16x

3n

)= en, but we could have as well as chosen a different positive

constant.

4.1 Summary of Paper BIn [1], using flatness improvement argument, the authors show that the freeboundary in the p-harmonic obstacle problem is a C1,α graph in a neighbor-hood of the points where the solution is almost one-dimensional. We applythe same technique in order to study the regularity of the free boundary inthe biharmonic obstacle problem.

In the first step we study one-dimensional solutions. Assume that u0solves the biharmonic obstacle problem with zero obstacle in an open inter-val (−a, a) ⊂ R, and let the origin be a free boundary point. By a directcomputation we see that u0 is one of the following functions; cx3

+, cx3− or cx3,

where c is a positive constant. If u(x) = u0(xn) = c(xn)3+, for x ∈ Rn, then

the noncoincidence set for u is the halfspace x ∈ Rn, xn > 0, and the freeboundary is the plane xn = 0.

Let us choose c = 16 , and introduce the class B%

κ(ε) of solutions to thebiharmonic obstacle problem, that are close to the one-dimensional solution16(x+

n )3.

Definition 4.1. Let u ≥ 0 be the solution to the biharmonic obstacle problemin a domain Ω ⊂ Rn, B2 ⊂⊂ Ω and assume that 0 ∈ Γu is a free boundarypoint. We say that u ∈ B%

κ(ε), if the following assumptions are satisfied:

1. u is almost one dimensional, that is

‖∇′u‖W 2,2(B2) ≤ ε,

where ∇′ := ∇− en ∂∂xn

.

4.1. SUMMARY OF PAPER B 25

2. The set Ωu := u > 0 is a non-tangentially accessible (NTA) domainwith constants r0 = M−1 = %.

3. There exists 2 > t > 0, such that u = 0 in B2 ∩ xn < −t.

4. We have the following normalization

‖D3u‖L2(B1) = 16∥∥∥D3(xn)3

+

∥∥∥L2(B1)

= |B1|12

2 12

:= ωn, (4.2)

and we also assume that

‖D3u‖L2(B2) < κ, (4.3)

where κ > 16

∥∥∥D3(xn)3+

∥∥∥L2(B2)

= 2n2ωn.

Assumption 1 states that the solution is flat in the xn-direction. This canbe compared to the flatness assumption in [1].

We show that the precise value of the parameter t in assumption 3 is notvery important. The normalization in point 4 corresponds to our choice ofc = 1

6 , and it can always be achieved by a renormalization.For the biharmonic obstacle problem there are no optimal growth or non-

degeneracy properties known, that could help us avoid additional assumptionson the set Ωu. Hence we make assumption 2 in the definition of B%

κ(ε). TheNTA-domain assumption is not very strong, and it holds for a wide class ofdomains. However, it is not easy to verify directly that the noncoincidence setis an NTA-domain.

Evidently 16(xn)3

+ ∈ B%κ(ε), for any ε > 0 and % > 0. Our first step is to

show that if u ∈ B%κ(ε), with ε > 0 small, then u ≈ 1

6(xn)3+ in W 3,2(B1). The

proof of the last statement is a few pages long, and we will not discuss it inthe introduction.

Let us sketch briefly our flatness implies regularity argument for the bi-harmonic obstacle problem. The proofs are long and technical, and we referto Paper B for the details.

Denote by

ur(x) := u(rx)r3 , for 0 < r < 1. (4.4)

By choosing a new coordinate system, we prove that if ε is small enough, andu ∈ B%

κ(ε), then

‖∇′us‖W 2,2(B1) ≤ γ‖∇′u‖W 2,2(B2) ≤ γε (4.5)

in a normalized coordinate system with respect to u, where 1/4 < s < γ < 1/2are fixed constants. Inequality (4.5) says that in a smaller ball the rescaled

26 CHAPTER 4. THE BIHARMONIC OBSTACLE PROBLEM

function us is γ-times more flat than u, this property is called flatness im-provement. From inequality (4.5) we obtain

Us := ωnus(x)‖D3us‖L2(B1)

∈ B%κ(βε), (4.6)

in our new (normalized) coordinate system, where γ < β < 1 is a fixed number.Then we argue inductively to show that Usk ∈ B%

κ(Cβkε).We obtain good decay estimates for almost flat solutions, which imply the

convergence of the sequence of normalized coordinate systems. There exists aunit vector η0 ∈ Rn, such that

‖∇′η0usk‖W 2,2(B1)

‖D3usk‖L2(B1)≤ Cβkε (4.7)

for 0 < s < β < 1, where ∇′η := ∇ − η(η · ∇), for η ∈ Rn, |η| = 1. Via astandard iteration argument it follows that η0 is the approximate unit nor-mal to the free boundary at the origin. Applying the same argument at freeboundary points x ∈ Γu ∩B1, we see that Γu has an approximate unit normalat the point x, denoted by ηx. Finally employing our estimates for |ηx − η0|,we obtain the C1,α-regularity of the free boundary in B1, where α = lnβ

ln s .Our main result in Paper B is the following theorem.

Theorem 4.2 (Theorem 3.12 in Paper B). Assume that u ∈ B%κ(ε), with

an ε > 0 small. Then Γu ∩ B1 is a C1,α-graph for any 0 < α < 1 and theC1,α-norm of the graph is bounded by Cε.

From the C1,α-regularity of the free boundary it follows that ∆u ∈ C1,α

up to the free boundary. We also show that u is C3,α up to the free boundary.Therefore we obtain that a solution u ∈ B%

κ(ε) is locally C2,1, which is thebest regularity that a solution may achieve, as we can see by looking at theone-dimensional example (xn)3

+.We also provide a two-dimensional counterexample to the C2,1-regularity,

showing that without our flatness assumptions there exists a solution that isC2, 1

2 but is not C2,α for any α > 12 . Hence C2, 1

2 is the best regularity thata solution may achieve in dimension n ≥ 2. Unlike the classical obstacleproblem, the solution to the biharmonic obstacle problem in dimension n ≥ 2is not as regular as in dimension one.

References

[1] John Andersson, Erik Lindgren, and Henrik Shahgholian. Almost ev-erywhere regularity for the free boundary of the normalized p-harmonicobstacle problem.

[2] John Andersson, Erik Lindgren, and Henrik Shahgholian. Optimal reg-ularity for the no-sign obstacle problem. Communications on Pure andApplied Mathematics,, 66:245–262, 2013.

[3] John Andersson, Henrik Shahgholian, and Georg S. Weiss. Double obsta-cle problems with obstacles given by non-C2 Hamilton-Jacobi equations.Arch. Ration. Mech. Anal. 206, no. 3, 779–819., 2012.

[4] Stavros A. Belbas and Suzanne M. Lenhart. Nonlinear PDEs for stochas-tic optimal control with switchings and impulses. Appl. Math. Optim.,14:215–227, 1986.

[5] Luis A. Caffarelli. The obstacle problem revisited. J. Fourier Anal. Appl.4 (1998), no. 4-5, 383–402.

[6] Luis A. Caffarelli and Avner Friedman. The obstacle problem for thebiharmonic operator. Ann. Scuola Norm. Sup. Pisa, Cl. Sci. (4) 6, no.1:151–184., 1979.

[7] Lawrence C. Evans and Avner Friedman. Optimal stochastic switchingand the Dirichlet problem for the Bellman equation. Trans. Amer. Math.Soc., 253:365–389, 1979.

[8] Alessio Figalli and Henrik Shahgholian. An overview of unconstrained freeboundary problems. Philos. Trans. A 373 (2015), no. 2050, 20140281,11 pp.

[9] Jens Frehse. On the regularity of the solution of the biharmonic varia-tional inequality. Manuscripta Math. 9 (1973), 91–103.

27

28 REFERENCES

[10] Suzanne M. Lenhart and Stavros A. Belbas. A system of nonlinear partialdifferential equations arising in the optimal control of stochastic systemswith switching costs. SIAM J. Appl. Math., 43:465–475, 1983.

[11] Arshak Petrosyan, Henrik Shahgholian, and Nina Uraltseva. Regularity offree boundaries in obstacle-type problems. Graduate Studies in Mathemat-ics, 136. American Mathematical Society, Providence, RI, 2012. x+221pp. ISBN: 978-0-8218-8794-3.

Part II

Scientific papers

29

Paper A

Ann. I. H. Poincaré-AN (2015),http://dx.doi.org/10.1016/j.anihpc.2015.06.001

31

Optimal regularity in the optimal switching

problem

Gohar Aleksanyan

Abstract

In this article we study the optimal regularity for solutions to thefollowing weakly coupled system with interconnected obstacles

min(−∆u1 + f1, u1 − u2 + ψ1) = 0

min(−∆u2 + f2, u2 − u1 + ψ2) = 0,

arising in the optimal switching problem with two modes.We derive the optimal C1,1-regularity for the minimal solution under

the assumption that the zero loop set L := ψ1 + ψ2 = 0 is the closureof its interior. This result is optimal and we provide a counterexampleshowing that the C1,1-regularity does not hold without the assumptionL = L 0.

1 Introduction

We consider the following system of weakly coupled equations of obstacle type

min(−∆u1 + f1, u1 − u2 + ψ1) = 0

min(−∆u2 + f2, u2 − u1 + ψ2) = 0,(1)

with given Dirichlet boundary conditions ui = gi on ∂Ω. These type of systemsarise in optimal switching problems with two switching modes. Here f1 andf2 are the running cost functions corresponding to the switching modes. Thefunctions ψ1 and ψ2 are the costs of switching from one mode to the other.More details on the optimal switching problem are provided in Section 2.1.

The uniqueness and C1,1-regularity of the solutions to such systems havebeen studied in the literature under the assumption that the switching costs arenonnegative constants, [4], [7], [2]. Obstacle type weakly coupled systems withfirst order Hamiltonians and nonconstant switching costs have been studiedin [3], [6]. In the paper [3], Section 5 the authors investigate the speed ofconvergence of the solutions to a penalized system, they also show that thesolution of the first order Hamilton-Jacobi obstacle type system is Lipschitzcontinuous, under the assumption that each of the switching costs is boundedfrom below by a positive constant.

In our paper we make only the nonnegative loop assumption. This is anecessary condition for the system to be well-defined. Indeed, let (u1, u2) be asolution to (1), then u1 − u2 + ψ1 ≥ 0 and u2 − u1 + ψ2 ≥ 0, which implies

ψ1(x) + ψ2(x) ≥ 0. (2)

1

In the optimal switching setting, the condition (2) prevents the agent frommaking arbitrary gains by looping, in the sense that ψ1(x) + ψ2(x) is the costof switching from one mode to the other and immediately switching back. Wedenote the set where it is possible to switch for free by

L = x ∈ Ω | ψ1(x) + ψ2(x) = 0,

and call it free switching or zero loop set.By using the penalization/regularization method we derive the existence of

solutions, showing that through a subsequence the solutions of the penalizedsystem converge to the minimal solution (u1

0, u20) to (1). Then we see that the

solution ui0 ∈ C1,γ , for every 0 < γ < 1 and

‖∆ui0‖L∞(Ω) ≤ maxi‖∆ψi‖L∞(Ω) + 3 max

i‖f i‖L∞(Ω). (3)

The aim of the paper is to investigate if the solutions are C1,1, which is thebest regularity that we can hope that the solutions achieve. The structure of oursystem shows that at some subdomains of Ω, the regularity of the solutions canbe derived by already known C1,1-regularity results for the obstacle problem.In our discussion we see that the main point is to describe the regularity at socalled meeting points lying on ∂L , the boundary of the zero loop set.

In the main theorem, Theorem 4, we show that at the meeting points x0 ∈∂L 0 ∩ Ω the solutions are C2,α, under the assumption that f i ∈ Cα and ψi ∈C2,α. By L 0 we denote the interior of the set L , and by pointwise C2,α

regularity we mean uniform approximation with a second order polynomial withthe speed r2+α.

The idea of the proof is the same as in deriving the optimal regularity for theno-sign obstacle problem in [1]. The proof is based on the BMO-estimates forD2u1

0 and D2u20 following from the estimate (3). At the point x0, we consider

r2+α-th order rescalings of ui0 denoted by vir, and show that these are uniformlybounded in W 2,2(B1). Then, looking at the corresponding system for (v1

r , v2r),

we conclude that the rescalings are uniformly bounded in the ball B1.In the end we justify our assumption 0 ∈ ∂L 0 with a counterexample: We

consider a particular system in R2, where the zero loop set L = 0, then wefind an explicit solution, that is not C1,1.

The paper is structured as follows: In Section 2 we provide some backgroundmaterial. In Section 3 we use the penalization method to derive the existenceof strong solutions, and observe that these are actually minimal solutions. Themain results are presented in the last section, where we prove that the minimalsolution is locally C1,1 if the zero loop set is the closure of its interior, andprovide a counterexample to C1,1-regularity when ψ1 +ψ2 has an isolated zero.

1.1 Acknowledgements

I would like to thank Prof John Andersson, Prof Diogo Gomes and Prof HenrikShahgholian for several fruitful discussions. In particular, I would like to thankJohn Andersson for the idea of the proof of C2,α-regularity at the meetingpoints.

2

2 Background material

In this section we state some known results, which we use in our discussion,without giving any proofs.

2.1 Optimal switching problem

Let Ω ⊂ Rn be a bounded domain with a smooth boundary. We consider anagent that can be anywhere in Ω and in one of a finite number m of states. Forevery 1 ≤ i ≤ m, the agent moves in Ω according to a diffusion

dx = bi(x)dt+ σi(x)dWt,

where Wt is a Brownian motion in a suitable probability space, bi : Ω → Rnand σi : Ω → Rn×m are smooth functions. The generator of the diffusions isdenoted by Liv = 1

2σiσTi : D2v + bi ·Dv.

The agent can switch from any diffusion mode to another. At every instantt the agent pays a running cost f i(t)(x), depending on the present state i(t)and position x. Additionally, when changing state i to state j he incurs in aswitching cost −ψij(x). Finally, when the diffusion reaches the boundary andthe agent is in state i, the process is stopped and a cost −gi(x) is incurred.As it is traditional in optimal switching setting, we consider the problem ofmaximizing a certain profit (the negative of the cost) functional

ui(x) = maxi(t),i(0)=i

E[ ˆ T∂Ω

0

f i(t)(x(t))dt

−∑

t≤T∂Ω

ψi(t−),i(t+)(x(t)) + gi(T∂Ω)(x(T∂Ω)

],

where T∂Ω denotes the exit time of Ω. Additionally, the convention ψii = 0 isassumed.

As it has been discussed in the literature [7], [2], the corresponding valuefunction ui solves the following system:

min(−Liui + f i,minj

(ui − uj + ψij)) = 0. (4)

with boundary conditions ui = gi on ∂Ω.For the optimal switching problem to be well defined, we need to impose

the nonnegative loop condition: Let i0, i1, . . . , il = i0 be any loop of length l,i.e. including l number of states. Assume that (u1, u2, ..., um) is a solution tosystem (4), then ui−uj+ψij ≥ 0 for any i, j ∈ 1, 2, ...,m, then after summingthe equations over the loop, we get

l∑

j=1

ψij−1,ij ≥ 0.

This condition is a necessary assumption for the existence of a solution to (3),and it prevents the agent from making arbitrary gains by looping.

In this paper we consider a system, arising in a model optimal switchingproblem with only two states.

3

2.2 The Poisson equation, Calderon-Zygmund estimates

We start by recalling the definition of the Holder space Ck,γ . Let us denote thecontinuity norm

‖u‖C(Ω) = supx∈Ω|u(x)|,

and the Holder seminorm

[u]C0,γ(Ω) = supx,y∈Ω,x 6=y

|u(x)− u(y)||x− y|γ .

Definition 1. The Holder space Ck,γ(Ω) consists of all functions u ∈ Ck(Ω)such that

‖u‖Ck,γ(Ω) :=∑

|α|≤k‖Dαu‖C(Ω) +

∑

|α|=k[Dαu]C0,γ(Ω) <∞.

The next theorem states the known regularity of the solution to the Poissonequation ∆u = f , under the assumption that f is Holder continuous, and canbe found in the book [5].

Theorem 1. Assume that f ∈ Cγ , then there exists a classical solution to thePoisson equation

∆u = f in Ω.

Moreover, the solution is locally C2,γ(Ω), and for every Ω′ b Ω

‖u‖C2,γ(Ω′) ≤ Cn,γ(Ω′)(‖u‖C(Ω) + ‖f‖C0,γ(Ω)

),

where the constant Cn,γ(Ω′) depends on diamΩ′ and dist(Ω′, ∂Ω).

Next let us recall the definition of BMO spaces, and then state the Calderon-Zygmund estimates for the Poisson equation ∆u = f , when f ∈ Lp, 1 < p ≤ ∞.

Definition 2. We say that a function f ∈ L2(Ω) is in BMO(Ω) if

‖f‖2BMO(Ω) := supx∈Ω,r>0

1

rn

ˆ

Br(x)∩Ω

|f(y)− (f)r,x|2dy + ‖f‖2L2(Ω) <∞,

where (f)r,x is the average of f in Br(x) ∩ Ω.

The proofs of the following results can be found in [5] when p < ∞ and in[9] when p =∞.

Theorem 2. Consider the equation

∆u = f in B2R.

If f ∈ Lp(B2R) for 1 < p <∞, then the solution u ∈W 2,p(BR), and

‖D2u‖Lp(BR) ≤ Cp,n(‖f‖Lp(B2R) + ‖u‖L1(B2R)

)

If f ∈ L∞(B2R), then in general u /∈W 2,∞(BR), but

‖D2u‖BMO(BR) ≤ C∞,n(‖f‖L∞(B2R) + ‖u‖L1(B2R)

),

here Cp,n, C∞,n are dimensional constants.

4

2.3 The obstacle problem

In this section we state the regularity of the solution to the following obstacleproblem,

min(−∆u+ f, u− ψ) = 0 in Ω

with boundary conditions u− g ∈W 1,20 (Ω).

Here we will omit the variational formulation of the problem, the first regu-larity results and will state the C1,1-regularity of the solutions referring to thebook [8].

In order to be consistent with the assumptions in our paper, we will assumethat f ∈ Cα and the obstacle ψ ∈ C2,α, although these assumptions can beweakened.

Theorem 3. Assume that f ∈ Cα and ψ ∈ C2,α, and u solves the obstacleproblem

min(−∆u+ f, u− ψ) = 0 a.e. in Ω.

Then u ∈ C1,1(Ω′) for every Ω′ b Ω, and

‖u‖C1,1(Ω′) ≤ C(‖u‖L∞(Ω) + ‖f‖C0,α(Ω) + ‖ψ‖C2,α(Ω)

),

where the constant C depends on the dimension and on the subset Ω′ b Ω.

3 Existence of C1,α solutions

We consider the system (1) with boundary conditions ui = gi on ∂Ω, gi ∈C2. Then we also need to impose the following compatibility condition on theboundary data:

g1 − g2 + ψ1 ≥ 0, and g2 − g1 + ψ2 ≥ 0 on ∂Ω. (5)

Clearly, without the compatibility conditions, there are no solutions to (1)achieving the boundary data.

We are interested in deriving C1,1-regularity for the solutions to our system,which is the best regularity one can expect. Throughout our discussion we willassume that

f1, f2 ∈ Cα(Ω), and ψ1, ψ2 ∈ C2,α(Ω), (6)

for some 0 < α < 1. These are natural assumptions, since f being boundedor continuous, is not enough for its Newtonian potential to be C1,1. We alsoprovide a one-dimensional counterexample to the existence of solutions in casethe switching costs are not smooth.

Example 1 (Diogo Gomes). Consider the following system in the interval(−1, 1) with zero Dirichlet boundary conditions,

min(−(u1)xx, u1 − u2 + (1− |x|) cos

(π

1−|x|

))= 0,

min(−(u2)xx, u2 − u1 + (1− |x|)(1− cos

(π

1−|x|

))= 0.

Then the value function of the corresponding optimal control problem is notfinite.

5

Proof. In our example the running costs are identically zero, the switching costssatisfy the nonnegative loop assumption ψ1(x) + ψ2(x) > 0 in (−1, 1), and thecompatibility condition on the boundary ψ1(±1) = ψ2(±1) = 0.

The example illustrates that when the switching costs are not smooth, thenthe negative values give infinity growth to the value function of the correspond-ing optimal control problem. In order to show this, we choose optimal controlsi(t) as follows: the switching occurs at times tk where π

1−|x(tk)| = πk: Whenπ

1−|x(tk)| = πk = π(2n+1), n ∈ N0 , we switch from regime 1 to regime 2 gaining1

2n+1 and for the values π1−|x(tk)| = πk = 2πn we switch back from regime 2 to

1 paying zero cost, and so

ui(x) ≥ −∑

0≤t≤T∂Ω

ψi(tk),i(tk+1)(x(t)) =∑ 1

2n+ 1.

Then the conclusion follows from the divergence of harmonic series.

3.1 Penalization method

In this section we approximate the system (1) with a smooth penalized system.Let us take any smooth nonpositive function β : R→ (−∞, 0], such that

β(s) = 0 for s ≥ 0,

β(s) < 0 for s < 0 and

0 < β′(s) ≤ 1 for s < 0,

lims→−∞

β(s) = −∞

Next we consider the following penalization function βε(s) = β(s/ε), for s ∈R, ε > 0, and the corresponding penalized system

−∆u1

ε + f1 + βε(u1ε − u2

ε + ψ1) = 0

−∆u2ε + f2 + βε(u

2ε − u1

ε + ψ2) = 0,(7)

with boundary conditions uiε = gi on ∂Ω.For ε > 0 fixed, the penalized system (7) can be solved by several methods.

In the paper [4] the authors use nonlinear functional analysis methods in orderto derive the existence of classical solutions, that is uiε ∈ C2(Ω), assuming thatthe switching costs are positive constants. The proof is rather technical, howeverit works line for line in our case with variable switching costs, therefore we omitit.

Lemma 1. Under the assumptions (5) and (6) the solutions to the penalizedsystem (7), uiε satisfy the following estimates for every ε > 0

i.)−max

i‖f i‖L∞ ≤ −∆uiε ≤ max

i‖∆ψi‖L∞ + 3 max

i‖f i‖L∞ .

ii.)u1ε − u2

ε + ψ1 ≥ −Cε and u2ε − u1

ε + ψ2 ≥ −Cε

6

In ii.) the constant C > 0 depends only on the given data and can be computedexplicitly in terms of β.

Proof. For our convenience, let us denote θ1ε = u1

ε−u2ε+ψ1 and θ2

ε = u2ε−u1

ε+ψ2,and observe that θ1

ε and θ2ε cannot be negative at the same time according to

the nonnegative loop assumption.Now let us fix ε > 0, and consider the function βε(θ

iε(x)), x ∈ Ω. It is

bounded from above by 0, our aim is to prove that βε(θiε(x)) is bounded from

below. Let x0 = x0(ε) be a point of minimum for the function βε(θ1ε(x)),

moreover without loss of generality, we may assume that

mini=1,2;x∈Ω

βε(θiε(x)) = βε(θ

1ε(x0)) < 0.

If x0 ∈ ∂Ω, then βε(θ1ε(x0)) = 0 according to (5). Therefore x0 ∈ Ω is an

interior point, and βε(θ1ε(x0)) < 0. Then θ1

ε(x0) < 0, and since θ1ε + θ2

ε ≥ 0,we get θ2

ε(x0) ≥ 0 consequently βε(θ2ε(x0)) = 0. Since βε is nondecreasing and

βε(t) < 0 if and only if t < 0, we get that

mini=1,2;x∈Ω

θiε(x) = θ1ε(x0).

This implies that θ1ε = u1

ε − u2ε + ψ1 achieves its minimum at an interior point

x0, hence ∆u1ε − ∆u2

ε + ∆ψ1 ≥ 0 at x0. The last inequality together with−∆u2

ε(x0) + f2(x0) = 0 shows that

βε(θ1ε(x0)) = ∆u1

ε(x0)− f1(x0) =

∆u1ε(x0)−∆u2

ε(x0) + f2(x0)− f1(x0) ≥ −∆ψ1(x0) + f2(x0)− f1(x0).

The estimate above is true for any ε > 0, and therefore it proves the rightinequality in i.). The left inequality in i.) is a direct consequence of −βε ≥ 0.

In order to prove ii.), we recall that lims→−∞ β(s) = −∞, and βε(s) =

β(s/ε), hence βε(θiε) is bounded imples that

θiεε is uniformly bounded from below

by a negative constant −C ≤ 0. This finishes the proof of point ii.) in ourlemma.

Using the Sobolev embedding theorem and Calderon-Zygmund estimates,we can conclude that the functions uiε are uniformly bounded in W 2,p for every1 < p < ∞. Therefore through a subsequence uiε converges to a function ui0locally weakly in W 2,p and strongly in C1,γ for every 0 < γ < 1.

Now we proceed to prove the existence of solutions to system (1).

Proposition 1. Let (u10, u

20) = limε→0(u1

ε, u2ε) through a subsequence weakly in

W 2,p and strongly in C1,γ . Then (u10, u

20) solves the following system

min(−∆u10 + f1, u1

0 − u20 + ψ1) = 0,

min(−∆u20 + f2, u2

0 − u10 + ψ2) = 0,

min(−∆u10 + f1,−∆u2

0 + f2) = 0

(8)

in a strong sense, i.e. ui0 − uj0 + ψi ≥ 0 and if we have a strict inequality atsome point then ui0 satisfies ∆ui0 = f i in a neighborhood of that point, and−∆ui0 + f i ≥ 0 a.e. .

7

Proof. The property ii.) in Lemma 1, together with the strong convergencein C1,γ shows that u1

0 − u20 + ψ1 ≥ 0 and u2

0 − u10 + ψ2 ≥ 0. If u1

0(x0) −u2

0(x0) + ψ1(x0) > 0, then the strict inequality u1ε − u2

ε + ψ1 > 0 holds in asmall ball Br(x0), centered at x0 for ε > 0 small enough. Then it follows that−∆u1

ε + f1 = 0 in Br(x0), and we know that ‖∆u1ε‖L∞ is uniformly bounded,

therefore through a subsequence, ∆u1ε → ∆u1

0 a.e. as ε → 0, consequently−∆u1

0 + f1 = 0 a.e. in Br(x0). Moreover, since f1 ∈ Cα, we know that u10 is a

classical solution to −∆u10 + f1 = 0 in the ball Br(x0).

The solutions of the penalized system satisfy the equation

min(−∆u1ε + f1,−∆u2

ε + f2) = 0.

After passing to a limit through a subsequence, we get the following

min(−∆u10 + f1,−∆u2

0 + f2) = 0 a.e..

Proposition 1 shows that there exists (u10, u

20), ui0 ∈ W 2,p,∀p < ∞ solving

(1) in a strong sense. According to Lemma 1, ui0 has the following property

‖∆ui0‖L∞ ≤ maxi‖∆ψi‖L∞ + 3 max

i‖f i‖L∞ , (9)

which will be relevant for deriving further regularity of solutions.Furthermore, Proposition 1 tells us that the solution we get via the penal-

ization method, solves an extra equation, which turns out to be very importantin the discussion of the uniqueness.

3.2 Uniqueness

It has been shown in the paper [7] that if there are no zero loops, then thesolution to the system (1) is unique. Here we give a counterexample showingthat the uniqueness does not hold in case there are zero loops

Example 2 (Diogo Gomes). The following system

min(−∆u1 −M,u1 − u2 + ψ) = 0

min(−∆u2 +M,u2 − u1 − ψ) = 0,(10)

with given boundary conditions ui = gi, g1−g2 +ψ = 0 on ∂Ω, admits infinitelymany solutions, provided 2M > ‖∆ψ‖L∞ .

Moreover, (10) admits solutions u1, u2 /∈ C1,1.

Proof. Let (u1, u2) be a solution to the system (10). Since both u1−u2 +ψ ≥ 0and u2−u1−ψ ≥ 0, it follows that u1−u2+ψ ≡ 0, therefore−∆u1 = −∆u2+∆ψ.

Now let us take any u1 ∈W 2,p, p > n, u1 = g1 on ∂Ω, such that −∆u1−M ≥0 a.e.. Then the function u2 = u1 +ψ satisfies the boundary conditions u2 = g2

on ∂Ω, and −∆u2 + M ≥ 0 a.e. since 2M > ‖∆ψ‖L∞ . Thus we get infinitelymany solutions of the form (u1, u1 + ψ), which may not be C1,1.

8

We observe that if the zero loop set is empty, then the equation min(−∆u1 +f1,−∆u2 + f2) = 0 is satisfied automatically. Under the nonnegative loopassumption, we saw that there exists a solution to system (1) also solving system(8). Next we show that the system (8) has a unique solution, which is actuallythe minimal solution to (1).

Proposition 2. The system (8) has a unique solution (u10, u

20) in W 2,p for every

p <∞.

Proof. Let us assume that (u1, u2) is a solution to system (8), then the differenceU = u1 − u2 solves the following double-obstacle problem in Ω:

−∆U + f1 − f2 ≤ 0 a.e. if U > −ψ1

−∆U + f1 − f2 ≥ 0 a.e. if U < ψ2

−ψ1 ≤ U ≤ ψ2,

with boundary conditions U = g1 − g2 on ∂Ω.It is well-known that the solution to the double-obstacle problem with given

boundary data is unique in W 2,p. Indeed, let V be another solution, thenwithout loss of generality, we may assume that maxx∈Ω(U − V ) = U(x0) −V (x0) > 0. Then in a small ball Br(x0), one has U − V > 0, and U − V hasa maximum at x0. The inequality U > V ≥ −ψ1 implies that U > −ψ1 inBr(x0), hence −∆U − f1 + f2 ≤ 0. Similarly, V < ψ2 in Br(x0) and therefore−∆V − f1 + f2 ≥ 0. After combining the inequalities −∆U − f1 + f2 ≤ 0 and−∆V − f1 + f2 ≥ 0, we see that U − V is a subharmonic function in the ballBr(x0). Recalling that U − V has a maximum at an interior point x0, we get acontradiction to the maximum principle for subharmonic functions.

Now let us assume that (v1, v2) is another solution to system (8), thenu1 − u2 ≡ v1 − v2 in Ω. Denote h = u1 − v1 ≡ u2 − v2 in Ω, then h = 0on ∂Ω.

Now let us plugg-in v1 = u1 − h and v2 = u2 − h to the equation

0 = min(−∆v1 + f1,−∆v2 + f2) =

min(−∆u1 + f1,−∆u2 + f2) + ∆h = ∆h a.e. .

Then it follows that ∆h = 0 a.e. in Ω, h ∈W 2,p(Ω), for every 1 < p <∞, henceh is a harmonic function. Then the difference ui − vi is a harmonic function inΩ, vanishing on the boundary, therefore ui− vi ≡ 0, according to the maximumprinciple for harmonic functions.

Corollary 1. The solution to the system (8) is the minimal solution to system(1), that is if (v1, v2) solves (1), then u1

0 ≤ v1 and u20 ≤ v2.

Proof. Assume (v1, v2) solves (1) with given boundary conditions, and let ω =min(−∆v1 +f1,−∆v2 +f2) , then ω ≥ 0 a.e., ω ∈ L∞. Let h be the solution to∆h = ω in Ω with zero Dirichlet boundary conditions on ∂Ω. Then accordingto the weak maximum principle for subharmonic functions, we get that h ≤ 0in Ω.

Now we note that the pair (v1 + h, v2 + h) solves the system (8) with thesame boundary conditions as (u1

0, u20), hence v1 +h = u1

0 and v2 +h = u20. Then

the minimality of (u10, u

20) follows from nonpositivity of h.

9

From now on we will be interested in studying the regularity for the minimalsolutions. As the Example 2 shows, there is no hope to get C1,1-regularity fornon-minimal solutions.

4 Optimal regularity of the solutions

In this section we prove that the solution to the system (8) is locally C1,1, ifL = L 0. In particular we study the regularity of the solutions on ∂L , theboundary of the zero-loop set.

Before proceeding to the discussion of C1,1-regularity, let us rewrite oursystem in a more convenient way. We have assumed that f1, f2 ∈ Cα, thereforethere exist v1, v2 ∈ C2,α

loc solving the Poisson equation ∆vi = f i in Ω. Recallthat (u1

0, u20) is the solution to system (8), and define ui = ui0 − vi, then ui0 is

as regular as ui up to C2,α, and (u1, u2) solves the following system

min(−∆u1, u1 − u2 + ϕ1) = 0,

min(−∆u2, u2 − u1 + ϕ2) = 0,

min(−∆u1,−∆u2) = 0.

(11)

Here ϕ1 = v1 − v2 + ψ1 and ϕ2 = v2 − v1 + ψ2 are the new switching costfunctions preserving the loop condition ϕ1 + ϕ2 ≡ ψ1 + ψ2, and ϕ1, ϕ2 ∈ C2,α

loc .From now on we will be focused on studying the regularity of (u1, u2) solving

the system (11).We define the open set Ω1 := ∪Br(x0, u

1), where the union is taken over theballs Br(x0, u

1), such that −∆u1 > 0 a.e. in Br(x0, u1). Similarly we define the

set Ω2 corresponding to the function u2, and let Ω12 = Ω \ Ω1 ∪ Ω2. Then Ω1,Ω2 and Ω12 are disjoint open sets, and since ϕ1, ϕ2 ∈ C2,α,

−∆u1 = ∆ϕ1 > 0,−∆u2 = 0 in Ω1,

−∆u2 = ∆ϕ2 > 0,−∆u1 = 0 in Ω2,

−∆u1 = 0,−∆u2 = 0 in Ω12.

In the set Ω \ Ω1 we get −∆u1 = 0, and the function u2 solves the obstacleproblem min(−∆u2, u2−u1 +ϕ2) = 0, with a C2,α obstacle u1−ϕ2. Thereforeu2 is locally C1,1 in Ω \ Ω1. Similarly we get that u2 is locally C1,1 in Ω \ Ω2.

Next we need to study the regularity of the solution in a neighborhood ofthe set ∂Ω1 ∩ ∂Ω2 ∩ Ω. Let us note that it is contained in the zero loop set,∂Ω1 ∩ ∂Ω2 ⊂ L , since u1 − u2 + ϕ1 ≡ 0 in Ω1 and u2 − u1 + ϕ2 ≡ 0 in Ω2. Inthe interior of the zero loop set the system (11) reduces to the equation

−∆u1 = (∆ϕ1)+, u2 = u1 + ϕ1 in L 0. (12)

From the classical theory, solutions to the equation (12) are locally C2,α if∆ϕ1 ∈ Cα. So in a neighborhood of the points x ∈ ∂Ω1 ∩ ∂Ω2 and x ∈ L 0, thesolution is C2,α.

It remains to study the regularity of (u1, u2) at the points x0 ∈ ∂Ω1 ∩ ∂Ω2 ∩∂L , called a ”meeting” point. In this section we show that u1 and u2 areactually C2,α-regular at such points.

For simplicity, let us study the system locally in the unit ball B1, assumingthat 0 ∈ ∂L ∩ ∂Ω1 ∩ ∂Ω2. We can always come to such a situation with achange of variables.

10

4.1 Blow-up procedure

Assume that (u1, u2) solves system (11) in the unit ball B1, and 0 ∈ ∂L = ∂L 0

is a meeting point, and let us study the regularity of the functions u1 and u2 at0.

Definition 3. For a function u ∈ W 2,2, define Π(u(x), r) = pr(x), wherepr(x) = x ·Ar · x+ br · x+ cr is a second order polynomial with the matrix Ar,vector br and scalar cr minimizing the following expression

minA,b,c

ˆ

Br

(|D2u− 2Ar|2 + |∇u− br|2 + |u− cr|2

)dx.

Then Π(u(rx), 1) = pr(rx), and it is easy to see that

pr(x) =1

2x · (D2u)r · x+ (∇u)r · x+ (u)r,

where (u)r := (u)r,0, and (u)r,x0is the average of u over the ball Br(x0) = x ∈

Rn | |x− x0| < r,(u)r,x0

=1

|Br|

ˆ

Br(x0)

u.

Let us recall that (u1, u2) ∈ W 2,2 is the solution to (11) in the unit ballB1, and denote Air = 1

2 (D2ui)r, bir = (∇ui)r, cir = (ui)r, for i ∈ 1, 2, then

Π(ui(rx), 1) = r2x ·Air · x+ rbir · x+ cir.Next for any 0 < r < 1, we define

vir(x) =ui(rx)−Π(ui(rx), 1)

S(r),

where S(r) is chosen such that maxi ‖D2vir‖L2(B1) = 1. Our aim is to describethe rate of convergence of S(r) as r goes to zero.

It follows immediately from our definition of S(r), and BMO-estimates thatS(r)r2 is uniformly bounded from above. In order to show this, let us recall that‖∆ui‖L∞ ≤ maxi ‖∆ϕi‖L∞ , hence D2ui ∈ BMO locally, with the followingestimate

‖D2ui‖BMO(B 12

) ≤ C(maxi‖∆ϕi‖L∞ + ‖ui‖L2(B1)).

Without loss of generality, we may assume that ‖D2v1r‖L2(B1) = 1, for a

fixed r > 0, thenS(r)

r2= ‖D2u1(rx)− 2A1

r‖L2(B1).

A change of variable will give us

(S(r)

r2

)2

=1

rn

ˆ

Br

|D2u1(y)− 2A1r|2dy =

1

rn

ˆ

Br

|D2u1(y)− (D2u1)r|2dy ≤ C(

maxi‖∆ϕi‖L∞ + ‖u1‖L2(B1)

)2

,

therefore ∀r < 12

S(r)

r2≤ C

(maxi‖∆ϕi‖L∞ + max

i‖ui‖L2(B1)

):= C0. (13)

11

So S(r) has at least quadratic decay as r → 0. Next we improve the estimate,showing that actually S(r) ≤ C0r

2+α for r > 0 small enough.

Proposition 3. Let ϕ1, ϕ2 ∈ C2,α for some 0 < α < 1, and L = L 0, then the

function S(r)r2+α is uniformly bounded as r goes to zero

S(r)

r2+α≤ C

(maxi‖∆ϕi‖L∞ + max

i‖ui‖L2(B1)

), (14)

where C is a dimensional constant.

Proof. Let us start with an important observation: The assumptions 0 ∈ ∂L ∩∂Ω1 ∩ ∂Ω2, L = ϕ1 + ϕ2 = 0, L = L 0 and ϕ1, ϕ2 ∈ C2,α imply that∆ϕ1(0) + ∆ϕ2(0) = 0. On the other hand ∆ϕ1 ≥ 0 in Ω1 and ∆ϕ2 ≥ 0 in Ω2,therefore ∆ϕ1(0) = ∆ϕ2(0) = 0.

Next we show that ϕi ∈ C2,α together with ∆ϕi(0) = 0, provide the growthestimate (14). The proof is based on an argument of contradiction, assume thatS(r)r2+α is not bounded, then there exists a sequence rk → 0 as k → ∞, such

that S(rk) = kr2+αk and S(r) ≤ kr2+α for all r ≥ rk. Our aim is to study the

convergence of the sequence vik := virk as k → ∞. For that we will need somebasic properties of the functions vir, where 0 < r < 1.

According to the definition of S(r), ‖D2vir‖L2(B1) ≤ 1 for r < 1. Thenapplying Poincare’s inequality for the function ∇vir in the unit ball B1, we get‖∇vir‖L2(B1) ≤ Cn for every 0 < r < 1, since (∇vir)1 = 0. Next we study theaverage of vir in the unit ball

(vir)1 =(ui)r − n

2 r2 · tr(D2ui)r

ffl

B1x2

1dx− rffl

B1x · (∇ui)rdx− (ui)r

S(r)

= −αnr2(∆ui(rx)

)1

S(r),

where αn = n2

ffl

B1x2

1dx is a dimensional constant. Now let us recall that (u1, u2)

solves (11), and therefore

0 ≤ −∆ui(rx) ≤ max(0,∆ϕi(rx)) ≤ ‖ϕi‖C2,αrα|x|α,

since ∆ϕi(0) = 0, ϕi ∈ C2,α. Hence we get 0 ≤ (vir)1 ≤ Cn r2+α

S(r) ‖ϕi‖C2,α , where

Cn > 0 is a dimensional constant. Next we apply Poincare’s inequality one moretime, ‖vir − (vir)1‖L2(B1) ≤ Cn‖∇vir‖L2(B1). Therefore we may conclude that

‖∇vir‖L2(B1) ≤ C, ‖vir‖L2(B1) ≤ C(

1 + ‖ϕ‖C2,α

r2+α

S(r)

), (15)

for every 0 < r < 1, where the constant C > 0 depends only on the dimension.Next by using (15) we want to estimate the ‖vik‖W 2,2(BR) for R < 1/rk as

k →∞. Let us start by looking at the expressions |Ai2lrk−Ai2l−1rk

|, where l ∈ Nand rk < 1 are chosen such that s = rk2l−1 ≤ 1

4 . It follows from Minkowski’s

12

inequality, that

|Ai2s −Ais| ≤(

B1

|D2ui(xs)− 2Ais|2) 1

2

+

(

B1

|D2ui(xs)− 2Ai2s|2) 1

2

≤ S(s)

s2+ 2

n2

B 12

|D2ui(2xs)− 2Ai2s|2

12

≤ S(s)

s2+ 2

n2S(2s)

4s2.

Hence|Ai2lrk −A

i2l−1rk

| ≤ k(1 + 2n2 +α)(rk2l−1)α,

provided rk2l−1 ≤ 14 .

Now let us take any m ∈ N such that 2m+1rk ≤ 1, then

(ˆ

B2m

|D2vik(x)|2dx) 1

2

=r2k

S(rk)

(ˆ

B2m

|D2ui(rkx)− 2Airk |2dx) 1

2

≤ 2mn2

krαk

(ˆ

B1

|D2ui(2mrkx)− 2Airk |2dx) 1

2

≤ 2mn2

krαk

((ˆ

B1

|D2ui(2mrkx)− 2Ai2mrk |2dx) 1

2

+ |Ai2mrk −Airk |)

≤ 2mn2

krαk

S(2mrk)

(2mrk)2+

m∑

j=1

|Ai2jrk −Ai2j−1rk

|

≤ 2mn2

krαk

k2mαrαk + k2nrαk

m∑

j=1

2α(j−1)

≤ 2n+12m(n2 +α).

For every R < 12rk

we can find an m ∈ N such that 2m−1 ≤ R < 2m, andthen applying the estimates above, we get

ˆ

BR

|D2vik(x)|2dx ≤ CnRn+2α,

for every R < 12rk

, where Cn is a dimensional constant. Then we can also

show that ‖∇virk‖L2(BR) and ‖virk‖L2(BR) are bounded by a constant dependingonly on R. Indeed, applying the corresponding estimates for Airk , and the firstinequality in (15), we get

|bi2lrk − bi2l−1rk

| ≤ Cn,αk(rk2l)1+α,

and therefore|biRrk − birk | ≤ Cn,αk(rkR)1+α.

Then the Poincare’s inequality in a ball BR implies that

‖∇vik − (∇vik)R‖L2(BR) ≤ CnR‖D2vik‖L2(BR),

and‖vik − (vik)R‖L2(BR) ≤ CnR‖∇vik‖L2(BR),

13

where

(∇vik)R =rk

S(rk)(∇uirk − rkAirk · x− birk)R

=rk

S(rk)

((∇ui)Rrk − birk

)=

rk

kr2+αk

(biRrk − birk

),

and

(vik)R =1

S(rk)

(ciRrk − cirk +

n

2r2k(∆ui)rk(x2

1)R

).

Next let us observe that the second inequality in (15), with the correspondingestimates for Air and bir imply that

|ciRrk − cirk | ≤ Ck(Rrk)2+α.

Then it follows from the triangle’s inequality that

‖∇vik‖L2(BR) ≤ C(Rn2 +1+α +R

n2

rk

kr2+αk

|biRrk − birk |)≤ CnR

n2 +1+α,

and also

‖vik‖L2(BR) ≤ C(R‖∇vik‖L2(BR) +R

n2 (vik)R

)≤

C

(Rn2 +2+α +R

n2|ciRrk − cirk |S(rk)

+ ‖ϕ‖C2,αRn2 +2 r

2+αk

S(rk)

)≤ C ′Rn

2 +2+α.

Therefore we have shown that the sequence vik is locally uniformly boundedin W 2,2, hence through a subsequence, vik converges weakly in W 2,2(BR), andstrongly in W 1,2(BR), denote vi0 = limk→∞ vik for i = 1, 2. Then the weakconvergence of the second order derivatives implies that

ˆ

BR

|D2vi0(x)|2dx ≤ lim supk→∞

ˆ

BR

|D2vik(x)|2dx,

and thereforeˆ

BR

|D2vi0(x)|2dx ≤ CnRn+2α. (16)

Next we describe further properties of the limit functions, v10 and v2

0 . Recallthat

vik(x) =ui(rkx)−Π(ui(rkx), 1)

S(rk),

then we have

−∆vik(x) =r2k

S(rk)

(−∆ui(rkx) + trAirk

).

Let us denote

q1k(x) =

p1rk

(rkx)− p2rk

(rkx) + ϕ1(rkx)

S(rk), and

q2k(x) =

p2rk

(rkx)− p1rk

(rkx) + ϕ2(rkx)

S(rk).

14

Then (v1k, v

2k) is a strong solution to the following system

min(−∆v1k −

trA1rk

krαk, v1k − v2

k + q1k) = 0

min(−∆v2k −

trA2rk

krαk, v2k − v1

k + q2k) = 0

min(−∆v1k −

trA1rk

krαk,−∆v2

k −trA2

rk

krαk) = 0,

therefore

−∆v1k(x) =

trA1rk

krαk+ ∆ϕ1(rkx)

krαk, if rkx ∈ Ω1

trA1rk

krαk, otherwise ,

and

−∆v2k(x) =

trA2rk

krαk+ ∆ϕ2(rkx)

krαk, if rkx ∈ Ω2

trA2rk

krαk, otherwise .

Then ∆ϕi(0) = 0, for i = 1, 2 together with ϕi ∈ C2,α, implies that

‖∆ϕi(rk·)‖L2(BR)

krαk≤ Cn

kRn2 +α‖ϕi‖C2,α → 0, as k →∞,

for i = 1, 2, and for any fixed 1 ≤ R <∞.We have that vik vi0 weakly in W 2,2(BR) and vik → vi0 in W 1,2(BR),

therefore ∆vik ∆vi0 weakly in L2(BR), but ∆vik =trAirkkrαk

+ ∆ϕi(rkx)krαk

χΩi(rkx),

and ‖∆ϕi(rkx)krαk

χΩi(rkx)‖L2(BR) → 0. Thus we may conclude that the sequence

of numberstrAirkkrαk

converges, and denote ai := limk→∞trAirkkrαk

. Then ‖∆vik −ai‖L2(BR) → 0 as k →∞ for every 1 ≤ R <∞. Therefore both −∆v1

0 − a1 ≡ 0and −∆v2

0 − a2 ≡ 0 in Rn.

We have shown that vi0(x) − ai |x|2

2n is a harmonic functions in Rn. Hencethe matrix D2vi0 has harmonic entries Dkvi0, where k is a multiindex, |k| = 2.Next we can apply the estimates of the derivatives for harmonic functions andinequality (16), to get

|∇Dkvi0(x0)| ≤ R−n2−1‖Dkvi0‖L2(BR(x0))

≤ R−n2−1‖Dkvi0‖L2(B2R) ≤ C ′R−1+α,

provided R > |x0|. Letting R → ∞, we see that the derivatives of Dkvi0 arevanishing, hence D2vi0 is a constant matrix, and therefore vi0, i ∈ 1, 2 is asecond order polynomial.

According to our construction, vi0 are orthogonal to the second order polyno-mials in L2(B1)-sense, hence both v1

0 and v20 must be identically zero. Then the

constants a1 = a2 = 0, and ‖∆vik‖L2(B1) → 0 as k → ∞, the latter contradictsto the condition maxi ‖D2vik‖L2(B1) = 1.

4.2 C2,α- regularity at the meeting points

We start by showing that the approximating polynomials pir converge to a poly-nomial pi0, and describe the rate of convergence.

15

Lemma 2. Let (u1, u2) be a solution to (11), and assume that ϕi ∈ C2,α. Letthe polynomials pir be as in the Definition 3, then there exists a polynomial pi0such that

supx∈Br|pir(x)− pi0(x)| ≤ Cr2+α. (17)

Proof. The condition ‖D2vir‖L2(B1) ≤ 1 with the inequality (14) implies that

(ˆ

B1

|D2ui(rx)−D2pir|2dx) 1

2

≤ S(r)

r2≤ C0r

α (18)

Recall that Air = D2pir, then using the triangle inequality, and that Air is

minimizing ‖D2ui(rx)− A‖L2(B1) over matricies A ∈ Rn2

, we get |Air − Air2| ≤

C0rα for all 0 < r < 1. By taking r = 2−n, we see that Ai2−n is a Cauchy

sequence;|Ai2−n −Ai2−n−m | ≤ Σm−1

k=0 |Ai2−n−k −Ai2−n−k−1 | ≤Σm−1k=0 (2−α)n+k = 2−αnΣm−1

k=0 2−αk,

from the convergence of the series Σ(2−α)n, it follows that Ai2−n converges tosome matrix Ai0 as n goes to infinity. The inequality also provides the rate ofconvergence; for a fixed n, by letting m go to infinity, we see that |Ai2−n−Ai0| ≤C02−nα.

Moreover, we get the estimate

|Air −Ai0| ≤ Crα,

for 0 < r < 1, by choosing n so that 2−n−1 < r ≤ 2−n.Next we proceed to describe the rate of convergence of bir and cir. We know

that bir = (∇ui)r, and cir = (ui)r, taking into account that ui ∈ C1,γ , we seethat bir → ∇ui(0) and cir → u(0) as r → 0. Our aim is to show that actually

|bir −∇ui(0)| ≤ Cr1+α and |cir − ui(0)| ≤ Cr2+α.

Let us recall that (∇vir)1 = 0, and therefore Poincare’s inequality impliesthat

‖∇vir‖L2(B1) ≤ C‖D2vir‖L2(B1) ≤ C ′.

Hence (ˆ

B1

|∇ui(rx)−∇pir(rx)|2dx) 1

2

≤ C ′S(r)

r≤ Cr1+α. (19)

Taking into account that bir is minimizing ‖∇ui(rx)− rx ·Air− b‖L2(B1) overb ∈ Rn, and applying triangle’s inequality we get |bir −∇ui(0)| ≤ Cr1+α.

Furthermore, using Poincare’s inequality once again, we see that

(ˆ

B1

|ui(rx)− pir(rx) + r2(∆ui)r|2dx) 1

2

≤ C ′S(r) ≤ Cr2+α, (20)

then|cir − ui(0)| ≤ Cr2+α,

16

by using that cir is minimizing ‖ui(rx)+r2(∆ui)r− 12r

2x·Air ·x−rbir ·x−c‖L2(B1)

over c ∈ R.Finally, after combining our estimates for Air, b

ir and cir, we get (17), where

pi0(x) =1

2x ·Ai0 · x+∇ui(0) · x+ ui(0),

for i = 1, 2.

Corollary 2. Under the assumptions of Lemma 2 it follows that

‖ui(rx)− pir(rx)‖W 2,2(B1) ≤ Cn,α(

maxi‖ϕi‖C2,α + max

i‖ui‖L2

)r2+α, (21)

where Cn,α is a dimensional constant.

Proof. Let us recall that ‖∆ui(rx)‖L2(B1) ≤ C‖∆ϕi‖Cαrα, then the statementfollows from the inequalities (18), (19) and (20).

Now we are ready to prove the main theorem.

Theorem 4. Assume ϕ1, ϕ2 ∈ C2,α, and L = L 0 then the solution to thesystem (11), (u1, u2) is C2,α-regular on ∂Ω1 ∩ ∂Ω2 ∩ ∂L ∩Ω, in the sense thatfor every x0 ∈ ∂Ω1 ∩ ∂Ω2 ∩ ∂L ∩ Ω, there exist second order polynomials p1

x0,

p2x0

, such that

supx∈Br(x0)

|ui(x)− pix0(x)| ≤ Cr2+α (22)

where the constant C > 0 depends only on the given data.

Proof. Without loss of generality, we may assume x0 = 0, and consider thefollowing rescalings

vir(x) =ui(rx)− pi0(rx)

r2+α,

then according to Lemma 2 and Corollary 2, ‖vir‖W 2,2(B1) ≤ C.The pair (v1

r , v2r) solves the following system

min(−∆v1r − trA1

0

rα , v1r − v2

r + q1r) = 0


0

rα , v2r − v1

r + q2r) = 0


0

rα ,−∆v2r − trA2

0

rα ) = 0,

where

q1r(x) =

p10(rx)− p2

0(rx) + ϕ1(rx)

r2+α, q2r(x) =

p20(rx)− p1

0(rx) + ϕ2(rx)

r2+α.

Then

−∆vir =

trAi0rα + ∆ϕi(rx)

rα if rx ∈ ΩitrAi0rα otherwise.

We assumed that 0 ∈ ∂Ω1 ∩ ∂Ω2 ∩ ∂L , then ∆ϕi(0) = 0, i = 1, 2. Hence

|∆ϕ1(rx)rα | ≤ ‖ϕ1

r‖C2,α(B1)|x|α. We know that ‖∆vir‖L2(B1) is bounded, thereforetrAi0 = 0, and ∆vir(x) is uniformly bounded.

17

We have that ‖vir‖L2(B1) ≤ C and we saw that ‖∆vir‖L∞(B1) ≤ ‖ϕi‖C2,α(B1).Using the Calderon-Zygmund estimates, we conclude that ‖vir‖C1,γ is uniformlybounded. In particular, |vir(x)| ≤ C ′(C0 + ‖ϕi‖C2,α(B1)) for every x ∈ B1 and

r ≤ 12 .

Recall that we set C0 = C(maxi ‖∆ϕi‖L∞ + maxi ‖ui‖L2(B1)

), and vir(x) =

ui(rx)−pi0(rx)r2+α . Then we get the desired inequality

supx∈Br|ui(x)− pi0(x)| ≤ C

(maxi∈1,2

‖ui‖L2(B1) + maxi∈1,2

‖ϕi‖C2,α(B1)

)r2+α.

4.3 A counterexample in case the zero-loop set has anisolated point

Here we give a counterexample, showing that if the zero loop set has an isolatedpoint, then the solution may not be C1,1.

We consider the following system in R2

min(−∆u1, u1 − u2 + ϕ) = 0

min(−∆u2, u2 − u1 + ϕ) = 0,(23)

with ϕ = 14 |x|2.

Then the difference U = u1−u2 solves the following double-obstacle problemin R2

−∆U =

1, if U = −ϕ−1, if U = ϕ

0, if − ϕ < U < ϕ,

(24)

and −∆u1 = (−∆U)+ and −∆u2 = (∆U)+.Now let us consider a function w defined as follows

w =

− 14 |x|2, if x1 > 0, x2 > 0

14 (x2

1 − x22), if x1 < 0, x2 > 0

14 (x2

2 − x21), if x1 > 0, x2 < 0

14 |x|2, if x1 < 0, x2 < 0.

Then w ∈ C1,1 also solves the double-obstacle problem (24), therefore we chooseU ≡ w.

Next we write u1 explicitly in polar coordinates

u1(r, θ) =

− 1

4r2 − 1

2π r2θ cos 2θ − 1

2π r2 ln r sin 2θ, if 0 < θ ≤ π

2

− 14r

2 cos 2θ + 12π r

2θ cos 2θ + 12π r

2 ln r sin 2θ, otherwise ,

here the function r2θ cos 2θ+ r2 ln r sin 2θ ∈ C1,γ for every 0 < γ < 1, solves theLaplace equation in R2\0, but is not C1,1 near the origin.

Therefore u1 is a C1,γ function in the unit ball in R2 but it is not C1,1 inthe neighborhood of the origin, since |∂2u1

∂r2 | ≈ |ln r| → ∞ as r → 0. Moreover,−∆u1(r, θ) = χ0<θ≤π2 = χu1>0, provided r > 0 is small enough.

18

Next we take u2(r, θ) = u1(r, θ)− w, then

u2(r, θ) =

− 12π r

2θ cos 2θ − 12π r

2 ln r sin 2θ, if 0 < θ ≤ π2

− 12r

2 cos 2θ + 12π r


2 ln r sin 2θ, if π2 < θ ≤ π

− 14r

2 − 14r

2 cos 2θ + 12π r


2 ln r sin 2θ, if π < θ ≤ 3π2

12π r


2 ln r sin 2θ, if 3π2 < θ ≤ 2π

Neither u1 nor u2 is a C1,1 function. However, it is easy to see that (u1, u2)solves (23), and it is minimal, since min(−∆u1,−∆u2) = 0 a.e..

References

[1] John Andersson, Erik Lindgren, and Henrik Shahgholian. Optimal regularityfor the no-sign obstacle problem. Communications on Pure and AppliedMathematics, 66:245–262, 2013.

[2] Stavros A. Belbas and Suzanne M. Lenhart. Nonlinear PDEs for stochasticoptimal control with switchings and impulses. Appl. Math. Optim., 14:215–227, 1986.

[3] Filippo Cagnetti, Diogo Gomes, and Hung Vinh Tran. Adjoint methods forobstacle problems and weakly coupled systems of PDE. ESAIM ControlOptim. Calc. Var., 19:754–779, 2013.

[4] Lawrence C. Evans and Avner Friedman. Optimal stochastic switching andthe Dirichlet problem for the Bellman equation. Trans. Amer. Math. Soc.,253:365–389, 1979.

[5] David Gilbarg and Neil S. Trudinger. Elliptic partial differential equationsof second order. Springer-Verlag, Berlin, 2001, Reprint of the 1998 edition.

[6] Diogo Gomes and Antonio Serra. Systems of weakly coupled Hamilton-Jacobi equations with implicit obstacles. Canad. J. Math., 64:1289–1309,2012.

[7] Suzanne M. Lenhart and Stavros A. Belbas. A system of nonlinear partialdifferential equations arising in the optimal control of stochastic systemswith switching costs. SIAM J. Appl. Math., 43:465–475, 1983.

[8] Arshak Petrosyan, Henrik Shahgholian, and Nina Uraltseva. Regularity offree boundaries in obstacle-type problems. American Mathematical Society,Providence, RI, 2012.

[9] Elias M. Stein. Harmonic analysis: real-variable methods, orthogonality, andoscillatory integrals. Princeton University Press, Princeton, NJ, 1993.

19

Paper B

Preprint: http://arxiv.org/abs/1603.06819

53

Regularity of the free boundary in the biharmonic

obstacle problem

Gohar Aleksanyan ∗

Abstract

In this article we use flatness improvement argument to study the reg-ularity of the free boundary for the biharmonic obstacle problem with zeroobstacle. Assuming that the solution is almost one-dimensional, and thatthe non-coincidence set is an non-tangentially accessible (NTA) domain,we derive the C1,α-regularity of the free boundary in a small ball centeredat the origin.

From the C1,α-regularity of the free boundary we conclude that thesolution to the biharmonic obstacle problem is locally C3,α up to the freeboundary, and therefore C2,1. In the end we study an example, showing

that in general C2, 12 is the best regularity that a solution may achieve in

dimension n ≥ 2.

Keywords: free boundary problem; biharmonic operator; obstacle problemAMS classification: 35R35

Contents

1 Introduction 2

2 The obstacle problem for the biharmonic operator 32.1 Existence, uniqueness and W 3,2-regularity of the solution . . . . 32.2 C1,α-regularity of the solution . . . . . . . . . . . . . . . . . . . . 8

3 Regularity of the free boundary 113.1 One-dimensional solutions . . . . . . . . . . . . . . . . . . . . . . 113.2 The class B%

κ(ε) of solutions to the biharmonic obstacle problem 123.3 Linearization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.4 Properties of solutions in a normalized coordinate system . . . . 193.5 C1,α-regularity of the free boundary . . . . . . . . . . . . . . . . 23

4 On the regularity of the solution 284.1 C2,1-regularity of the solutions in B%

κ(ε) . . . . . . . . . . . . . . 28

4.2 In general the solutions are not better that C2, 12 . . . . . . . . . 29

References 30

∗KTH, Department of Mathematics, 100 44 Stockholm, SwedenE-mail: [email protected]

1

1 Introduction

Let Ω ⊂ Rn be a given domain, and ϕ ∈ C2(Ω), ϕ ≤ 0 on ∂Ω be a givenfunction, called an obstacle. Then the minimizer to the following functional

J [u] =

ˆ

Ω

(∆u(x))2dx, (1.1)

over all functions u ∈ W 2,20 (Ω), such that u ≥ ϕ, is called the solution to

the biharmonic obstacle problem with obstacle ϕ. The solution satisfies thefollowing variational inequality

∆2u ≥ 0, u ≥ ϕ, ∆2u · (u− ϕ) = 0.

It has been shown in [2] and [4] that the solution u ∈W 3,2loc (Ω), ∆u ∈ L∞loc(Ω),

and moreover u ∈ W 2,∞loc (Ω). Furthermore, in the paper [2], the authors show

that in dimension n = 2 the solution u ∈ C2(Ω) and that the free boundaryΓu := ∂u = ϕ lies on a C1-curve in a neighbourhood of the points x0 ∈ Γu,such that ∆u(x0) > ∆ϕ(x0).

The setting of our problem is slightly different from the one in [2] and [4].We consider a zero-obstacle problem with general nonzero boundary conditions.Let Ω be a bounded domain in Rn with smooth boundary. We consider theproblem of minimizing the functional (1.1) over the admissible set

A :=

u ∈W 2,2(Ω), u ≥ 0, u = g > 0,

∂u

∂ν= f on ∂Ω

.

The minimizer u exists, it is unique. The minimizer is called the solution tothe biharmonic obstacle problem. We will denote the free boundary by Γu :=∂Ωu ∩ Ω, where Ωu := u > 0.

There are several important questions regarding the biharmonic obstacleproblem that remain open. For example, the optimal regularity of the solution,the characterization of blow-ups at free boundary points, etc. In this article wefocus on the regularity of the free boundary for an n-dimensional biharmonicobstacle problem, assuming that the solution is close to the one-dimensionalsolution 1

6 (xn)3+. In [1], using flatness improvement argument, the authors show

that the free boundary in the p-harmonic obstacle problem is a C1,α graph ina neighborhood of the points where the solution is almost one-dimenional. Weapply the same technique in order to study the regularity of the free boundaryin the biharmonic obstacle problem.

In Section 2 we study the basic properties of the solution in the new setting,and show that it is locally in W 3,2 ∩ C1,α.

In Section 3 we introduce the class B%κ(ε) of solutions to the biharmonic ob-

stacle problem, that are close to the one-dimensional solution 16 (xn)3

+. Following

[1], we show that if ε is small enough, then there exists a rescaling us(x) = u(sx)s3 ,

such that‖∇′us‖W 2,2(B1) ≤ γ‖∇′u‖W 2,2(B2) ≤ γε

in a normalized coordinate system, where ∇′η := ∇− η(η · ∇),∇′ := ∇′en , andγ < 1 is a constant. Repeating the argument for the rescaled solutions, usk , weshow that there exists a unit vector η0 ∈ Rn, such that

‖∇′η0usk‖W 2,2(B1)

‖D3usk‖L2(B1)≤ Cβkε (1.2)

2

for 0 < s < β < 1. Then the C1,α-regularity of the free boundary in a neigh-borhood of the origin follows via a standard iteration argument.

From the C1,α-regularity of the free boundary it follows that ∆u ∈ C1,α upto the free boundary. We move further and show that u is C3,α up to the freeboundary. Thus a solution u ∈ B%

κ(ε) is locally C2,1, which is the best regularitythat a solution may achieve. We provide a two-dimensional counterexample tothe C2,1-regularity, showing that without our flatness assumptions there existsa solution that is C2, 12 but is not C2,α for α > 1

2 . Hence C2, 12 is the bestregularity that a solution may achieve in dimension n ≥ 2.

Acknowledgements

I wish to thank my advisor John Andersson for suggesting this research projectand for all his help and encouragement.I am grateful to Erik Lindgren for reading preliminary versions of this manuscriptand thereby improving it significantly.

2 The obstacle problem for the biharmonic op-erator

In this section we show that there exists a unique solution to the biharmonicobstacle problem. Furthermore we show that the solution is locally W 3,2∩C1,α.

2.1 Existence, uniqueness and W 3,2-regularity of the solu-tion

Let us start with the proof of existence and uniqueness of the minimizer offunctional (1.1). Throughout the discussion we denote by BR(x0) the open ballin Rn, centered at x0 ∈ Rn, with radius R > 0, and BR := BR(0), B+

R := xn >0 ∩BR.

Lemma 2.1. Let Ω be an open bounded subset of Rn with a smooth boundary.Then the functional (1.1) admits a unique minimizer in the set A .

Proof. Here we use the standard terminology from [3]. First we show thatthe functional J is weakly lower semicontinuous, i.e. given a sequence ukconverging weakly to a function u ∈W 2,2(Ω), then

lim infk→∞

J [uk] ≥ J [u]. (2.1)

Upon passing to a subsequence, we may assume that

lim infk→∞

J [uk] = limk→∞

J [uk].

According to the definition of weak convergence in W 2,2(Ω), ∆uk converges to∆u weakly in L2(Ω), hence

limk→∞

ˆ

∆uk∆u =

ˆ

(∆u)2,

3

and the inequalityˆ

(∆u)2 +

ˆ

(∆uk)2 − 2

ˆ

∆uk∆u =

ˆ

(∆uk −∆u)2 ≥ 0

impliesˆ

(∆uk)2 ≥ 2

ˆ

∆uk∆u−ˆ

(∆u)2,

after passing to the limit as k →∞, we get the desired inequality, (2.1).Next we take a minimizing sequence uk ⊂ A , and show that it converges

weakly to some function u in W 2,2(Ω) through a subsequence, and that u is anadmissible function. Define

m := infv∈A

ˆ

(∆v)2,

thenlimk→∞

J [uk] = m.

Let us note that J [uk] = ‖∆uk‖2L2 , so ∆uk is bounded in L2, and since uk−ω = 0

and ∂(uk−ω)∂n = 0 on ∂Ω in the trace sense for any fixed ω ∈ A , the sequence

is bounded in W 2,2(Ω). Hence it has a subsequence which converges weaklyin W 2,2, we will keep the notation, call it uk. We want to show that thelimit function u ∈ A . According to the Sobolev embedding theorem ukconverges to u strongly in L2 up to a subsequence, hence upon passing to a newsubsequence uk → u a.e. in Ω. The latter proves that u ≥ 0 a.e..

It remains to show that u satisfies the boundary conditions. For any ω ∈ A ,uk − ω ∈ W 2,2

0 (Ω), since W 2,20 (Ω) is a closed, linear subspace of W 2,2(Ω), it

is weakly closed, according to Mazur’s theorem ([3], pp. 471 and 723). Thisproves that u− ω ∈W 2,2

0 (Ω) and therefore u ∈ A .According to (2.1), m ≥ J [u], but the reversed inequality is also true since u

is admissible and according to our choice of the sequence uk. Thus m = J [u],and u is a minimizer.

The uniqueness of the minimizer follows from the convexity of the functional:assuming that both u and v are minimizers, it follows that u+v

2 is also admissible,so

J

[u+ v

2

]≥ J [u] + J [v]

2,

but the reversed inequality is also true with equality if and only if ∆u = ∆v.Thus if u and v are both minimizers in A then ∆(u−v) = 0 and u−v ∈W 2,2

0 (Ω),which implies that u = v in Ω.

Now we turn our attention to the regularity of the solution to the biharmonicobstacle problem.

Proposition 2.2. Let u be the solution to the biharmonic obstacle problem inthe unit ball B1, then

‖∆u‖W 1,2(B 12

) ≤ C‖u‖W 2,2(B1),

where the constant C depends only on the space dimension.

4

Proof. The proof is based on a difference quotient method. Let e1, e2, ..., enbe the standard basis in Rn. For a fixed i ∈ 1, 2, ..., n denote

ui,h(x) := u(x+ hei), for x ∈ B1−h. (2.2)

Take a nonnegative function ζ ∈ C∞0 (B 34), such that ζ ≡ 1 in B 1

2. Then for

small values of the parameter t > 0, the function u+ tζ2(ui,h − u) is admissiblefor the biharmonic obstacle problem in B1. Indeed, u + tζ2(ui,h − u) = u(1 −tζ2) + tζ2ui,h ≥ 0 if t > 0 is small, and obviously it satisfies the same boundaryconditions as the minimizer u. Hence

ˆ

B1

(∆(u+ tζ2(ui,h − u))

)2 ≥ˆ

B1

(∆u)2. (2.3)

Assuming that h < 14 , the inequality will still hold if we replace the integration

over the ball B1 by B1−h, since ζ is zero outside the ball B 34.

It is clear that ui,h is the solution to the biharmonic obstacle problem inB1−h, and ui,h + tζ2(u− ui,h) is an admissible function. Hence

ˆ

B1−h

(∆(ui,h + tζ2(u− ui,h))

)2 ≥ˆ

B1−h

(∆ui,h)2. (2.4)

After dividing both sides of the inequalities (2.3) and (2.4) by t, and taking thelimit as t→ 0, we get

ˆ

B1−h

∆u∆(ζ2(ui,h − u)

)≥ 0, (2.5)

andˆ

B1−h

∆ui,h∆(ζ2(u− ui,h)

)≥ 0. (2.6)

We rewrite inequalities (2.5) and (2.6) explicitly, that is

ˆ

B1−h

∆u((ui,h − u)∆ζ2 + ζ2∆(ui,h − u) + 2∇ζ2∇(ui,h − u)

)≥ 0, and

ˆ

B1−h

∆ui,h((u− ui,h)∆ζ2 + ζ2∆(u− ui,h) + 2∇ζ2∇(u− ui,h)

)≥ 0.

After summing the inequalities above, we obtainˆ

B1−h

ζ2(∆(ui,h − u))2 ≤

ˆ

B1−h

(ui,h − u)∆ζ2∆(u− ui,h) + 4

ˆ

B1−h

∇ζ∇(ui,h − u)ζ∆(u− ui,h).

Dividing both sides of the last inequality by h2, we get

ˆ

B1−h

ζ2(∆ui,h −∆u)2

h2≤ˆ

B1−h

(ui,h − u)

h2∆ζ2∆(u− ui,h)

+4

ˆ

B1−h

∇ζ (∇ui,h −∇u)

hζ

(∆u−∆ui,h)

h.

(2.7)

5

First let us study the first integral on the right side of (2.7)

∣∣∣∣∣

ˆ

B1−h

(ui,h − u)

h2∆ζ2∆(u− ui,h)

∣∣∣∣∣

=

∣∣∣∣∣

ˆ

B1−h

∆u

((ui,h − u)

h2∆ζ2 − (u− ui,−h)

h2∆ζ2

i,−h

)∣∣∣∣∣

=

∣∣∣∣∣

ˆ

B1−h

∆u∆ζ2

(ui,h − 2u+ ui,−h

h2

)+ ∆u

(∆ζ2 −∆ζ2

i,h

h

)(u− ui,−h

h

)∣∣∣∣∣≤ C‖∆u‖L2(B1)‖u‖W 2,2(B1),

(2.8)where we applied Holder’s inequality, and used the fact that the L2-norm of thefirst and second order difference quotients of a function u ∈W 2,2 are uniformlybounded by its W 2,2-norm.

Next we estimate the absolute value of the second integral in (2.7)

∣∣∣∣∣

ˆ

B1−h

∇ζ(∇ui,h −∇u

h

)ζ

(∆u−∆ui,h

h

)∣∣∣∣∣

≤ 8

ˆ

B1−h

|∇ζ|2 |∇ui,h −∇u|2

h2+

1

8

ˆ

B1−h

ζ2 (∆(ui,h − u))2

h2,

(2.9)

where we applied Cauchy’s inequality.Combining inequalities (2.7), (2.8) and (2.9), we obtain

ˆ

B1−h

ζ2 (∆(ui,h − u))2

h2≤ C‖u‖2W 2,2(B1).

According to our choice of function ζ,

ˆ

B 12

(∆(ui,h − u))2

h2≤ˆ

B1−h

ζ2 (∆(ui,h − u))2

h2,

so the L2-norm of the difference quotients of ∆u is uniformly bounded in B 12

hence ∆u ∈W 1,2(B 12), and

‖∆u‖W 1,2(B 12

) ≤ C‖u‖W 2,2(B1),

where the constant C depends only on the function ζ, and can be computedexplicitly, depending only on the space dimension.

Corollary 2.3. Assume that Ω is a bounded open set in Rn. Then the solutionto the obstacle problem is in W 3,2(K) for any K ⊂⊂ Ω, and

‖u‖W 3,2(K) ≤ C‖u‖W 2,2(Ω), (2.10)

where the constant C depends on the space dimension n and on dist(K, ∂Ω).

6

Proof. It follows from Proposition 2.2 by a standard covering argument that

‖∆u‖W 1,2(Ω′) ≤ CΩ′‖u‖W 2,2(Ω),

for any Ω′ ⊂⊂ Ω.Let K ⊂⊂ Ω′ ⊂⊂ Ω, according to the Calderon-Zygmund inequality ([6],

Theorem 9.11),

‖D3u‖L2(K) ≤ CK(‖∆u‖W 1,2(Ω′) + ‖u‖W 2,2(Ω′)

).

Then it follows that u is in W 3,2 locally, with the estimate (2.10).

Lemma 2.4. Let u be the solution to the biharmonic obstacle problem in Ω.Take K ⊂⊂ Ω, and a function ζ ∈ C∞0 (K), ζ ≥ 0, then

ˆ

Ω

∆uxi∆(ζuxi) ≤ 0, (2.11)

for all i = 1, 2, ..., n.

Proof. Fix 1 ≤ i ≤ n, denote ui,h(x) := u(x+hei), where 0 < |h| < dist(K, ∂Ω),hence ui,h is defined in K. Let us observe that the function u+ tζ(ui,h − u) iswell defined and nonnegative in Ω for any 0 < t < 1

‖ζ‖L∞ , and it satisfies the

same boundary conditions as u. Thereforeˆ

Ω

(∆(u+ tζ(ui,h − u)))2 ≥

ˆ

Ω

(∆u)2,

after dividing the last inequality by t, and taking the limit as t→ 0, we obtainˆ

K

∆u∆(ζ(ui,h − u)) ≥ 0. (2.12)

Note that ui,h is the solution to the biharmonic obstacle problem in K, andui,h + tζ(u− ui,h) is an admissible function, hence

ˆ

K

(∆(ui,h + tζ(u− ui,h)))2 ≥

ˆ

K

(∆ui,h)2,

after dividing the last inequality by t, and taking the limit as t→ 0, we obtainˆ

K

∆ui,h∆(ζ(u− ui,h)) ≥ 0. (2.13)

Inequalities (2.12) and (2.13) imply that

ˆ

K

(∆ui,h −∆u)∆(ζ(ui,h − u)) ≤ 0, (2.14)

dividing the last inequality by h2, and taking into account that u ∈ W 3,2loc , we

may pass to the limit as |h| → 0 in (2.14), and conclude that

ˆ

K

∆uxi∆(ζuxi) ≤ 0.

7

2.2 C1,α-regularity of the solution

It has been shown in [2], Theorem 3.1 that ∆u ∈ L∞loc for the solution to thebiharmonic obstacle problem with nonzero obstacle and zero boundary condi-tions. In this section we show that the statement remains true in our setting,with a quantitative estimate of ‖∆u‖L∞ .

Lemma 2.5. The solution to the biharmonic obstacle problem satisfies the fol-lowing equation in the distribution sense

∆2u = µu, (2.15)

where µu is a positive measure on Ω.

Proof. For any nonnegative test function η ∈ C∞0 (Ω), the function u + εη isobviously admissible for any ε > 0. Hence J [u+ εη] ≥ J [u], consequently

ˆ

ε2(∆η)2 + 2ε∆u∆η ≥ 0,

and after dividing by ε and letting ε go to zero, we obtain

ˆ

∆u∆η ≥ 0,

for all η ∈ C∞0 (Ω), η ≥ 0, so ∆2u ≥ 0 in the sense of distributions.Let us consider the following linear functional defined on the space C∞0 (Ω),

Λ(η) =

ˆ

Ω

∆u∆η.

Then Λ is a continuous linear functional on C∞0 (Ω), hence it is a distribution.According to the Riesz theorem, a positive distribution is a positive measure,let us denote this measure by µ := µu. Then ∆2u = µu in the sense that

ˆ

Ω

∆u∆η =

ˆ

Ω

ηdµu.

for every η ∈ C∞0 (Ω).

Corollary 2.6. There exists an upper semicontinuous function ω in Ω, suchthat ω = ∆u a.e. in Ω.

Proof. For any fixed x0 ∈ Ω, the function

ωr(x0) :=

Br(x0)

∆u(x)dx

is decreasing in r > 0, since ∆u is subharmonic by Lemma 2.5. Define ω(x) :=limr→0 ωr(x), then ω is an upper semicontinuous function. On the other handωr(x)→ ∆u(x) as r → 0 a.e., hence ω = ∆u a.e. in Ω.

The next lemma is a restatement of the corresponding result in [2], Theorem2.2.

8

Lemma 2.7. Let Ω ⊂ Rn be a bounded open set with a smooth boundary, andlet u be a solution to the biharmonic obstacle problem with zero obstacle. Denoteby S the support of the measure µu = ∆2u in Ω, then

ω(x0) ≥ 0, for every x0 ∈ S. (2.16)

Proof. The detailed proof of Lemma 2.7 can be found in the original paper [2]and in the book [5](pp. 92-94), so we will provide only a sketch, showing themain ideas.

Extend u to a function in W 2,2loc (Rn), and denote by uε the ε-mollifier of u.

Let x0 ∈ Ω, assume that there exists a ball Br(x0), such that uε ≥ α > 0 inBr(x0). Let η ∈ C∞0 (Br(x0)), η ≥ 0 and η = 1 in Br/2(x0). Then for anyζ ∈ C∞0 (Br/2(x0)) and 0 < t < α

2‖ζ‖∞ the function

v = ηuε + (1− η)u± tζ

is nonnegative and it satisfies the same boundary conditions as u. Hence

ˆ

(∆u)2 ≤ˆ

(∆(ηuε + (1− η)u± tζ))2,

after passing to the limit in the last inequality as ε→ 0, we obtain

ˆ

(∆u)2 ≤ˆ

(∆u± t∆ζ)2,

Thereforeˆ

∆u∆ζ = 0,

for all ζ ∈ C∞0 (Br/2(x0)), hence ∆2u = 0 in Br/2(x0) and x0 /∈ S. It followsthat if x0 ∈ S, then there exists xm ∈ Ω, xm → x0, and εm → 0, such that

uεm(xm)→ 0, as m→∞.

Then by Green’s formula,

uεm(xm) =

∂Bρ(xm)

uεmdHn−1 −ˆ

Bρ(xm)

∆uεm(y)V (xm − y)dy,

where ρ < dist(x0, ∂Ω) and −V (z) is Green’s function for Laplacian in the ballBρ(0). Hence

lim infm→∞

ˆ

Bρ(xm)

∆uεm(y)V (xm − y)dy ≥ 0,

Then it follows from the convergence of the mollifiers and the upper semiconti-nuity of ω, that ω(x0) ≥ 0, for any x0 ∈ S.

Knowing that ∆u is a subharmonic function, and ω ≥ 0 on the support of∆2u, we can show that ∆u is locally bounded (Theorem 3.1 in [2]).

9

Theorem 2.8. Let u be the solution to the biharmonic obstacle problem withzero obstacle in Ω, B1 ⊂⊂ Ω. Then

‖∆u‖L∞(B1/3) ≤ C‖u‖W 2,2(Ω), (2.17)

where the constant C > 0 depends on the space dimension n and on dist(B1, ∂Ω).

Proof. The detailed proof of the theorem can be found in the original paper [2],Theorem 3.1, and in the book [5], pp. 94-97. Here we will only provide a sketchof the proof.

Let ω be the upper semicontinuous equivalent of ∆u and x0 ∈ B1/2, then

ω(x0) ≤

B1/2(x0)

∆u(x)dx,

since ω is a subharmonic function. Applying Holder’s inequality, we obtain

ω(x0) ≤ |B1/2|−12 ‖∆u‖L2(B1). (2.18)

It remains to show that ∆u is bounded from below in B1/2. Let ζ ∈ C∞0 (B1),ζ = 1 in B2/3 and 0 ≤ ζ ≤ 1 elsewhere. Referring to [5], p.96, the followingformula holds for any x ∈ B1/2

ω(x) = −ˆ

B1/2

V (x− y)dµ−ˆ

B1\B1/2

ζ(y)V (x− y)∆2udy + δ(x), (2.19)

where V is Green’s function for the unit ball B1, and δ is a bounded function,

‖δ‖L∞(B1/2) ≤ Cn‖∆u‖L2(B1). (2.20)

Denote

V (x) :=

ˆ

B1/2

V (x− y)dµ(y),

then V is a superharmonic function in Rn, and the measure υ := ∆V is sup-ported on S0 := B1/2 ∩ S, moreover according to Lemma 2.7, (2.16)

V (x) ≤ −ω(x) + δ(x) ≤ δ(x) on S0.

Taking into account that V (+∞) <∞, the authors in [2] apply Evans maximumprinciple, [8] to the superharmonic function V − V (+∞), and conclude that

V (x) ≤ ‖δ‖L∞(B1/2) in Rn. (2.21)

It follows from equation (2.19) that

ω(x) ≥ −‖δ‖L∞(B1/2) − cnµu(B1) + δ(x), (2.22)

for any x ∈ B1/3.Let η ∈ C∞0 (Ω) be a nonnegative function, such that η = 1 in B1 and

0 ≤ η ≤ 1 in Ω. Then

µu(B1) ≤ˆ

Ω

ηdµu =

ˆ

Ω

∆u∆η ≤ ‖∆u‖L2(Ω)‖∆η‖L2(Ω),

10

and η can be chosen such that ‖∆η‖L2(Ω) ≤ C(dist(B1, ∂Ω)). Hence

µu(B1) ≤ C‖∆u‖L2(Ω), (2.23)

where the constant C > 0 depends on the space dimension and on dist(B1, ∂Ω).Combining the inequalities (2.18) and (2.22) together with (2.23), (2.20), we

obtain (2.17).

Corollary 2.9. Let u be the solution to the biharmonic obstacle problem in Ω.Then u ∈ C1,α

loc , for any 0 < α < 1, and

‖u‖C1,α(K) ≤ C‖u‖W 2,2(Ω), (2.24)

where the constant C depends on the space dimension and dist(K, ∂Ω).

Proof. It follows from Theorem 2.8 via a standard covering argument, that

‖∆u‖L∞(K) ≤ C‖u‖W 2,2(Ω).

Then inequality (2.24) follows from the Calderon-Zygmund inequality and theSobolev embedding theorem.

According to Corollary 2.9, u is a continuous function in Ω, and thereforeΩu := u > 0 is an open subset of Ω. We define the free boundary

Γu = ∂Ωu ∩ Ω. (2.25)

It follows from our discussion that the measure µu = ∆2u is supported on Γu.

3 Regularity of the free boundary

In this section we investigate the regularity of the free boundary Γu, under theassumption that the solution to the biharmonic obstacle problem is close to theone-dimensional solution 1

6 (xn)3+.

3.1 One-dimensional solutions

Here we study the solution to the biharmonic obstacle problem in the interval(0, 1) ⊂ R.

Example 3.1. The minimizer u0 of the functional

J [u] =

ˆ 1

0

(u′′(x))2dx, (3.1)

over nonnegative functions u ∈ W 2,2(0, 1), with boundary conditions u(0) =1, u′(0) = λ < −3 and u(1) = 0, u′(1) = 0, is a piecewise 3-rd order polynomial,

u0(x) =λ3

33

(x+

3

λ

)3

−, x ∈ (0, 1), (3.2)

hence u0 ∈ C2,1(0, 1).

11

Proof. Let u0 be the minimizer to the given biharmonic obstacle problem. If0 < x0 < 1, and u0(x0) > 0, then

´

u′′0η′′ = 0, for all infinitely differentiable

functions η compactly supported in a small interval centered at x0. Hence

the minimizer u0 has a fourth order derivative, u(4)0 (x) = 0 if x ∈ u0 > 0.

Therefore u0 is a piecewise polynomial of degree less than or equal to three.Denote by γ ∈ (0, 1] the first point where the graph of u0 hits the x-axes. Ouraim is find the explicit value of γ. Then we can also compute the minimizer u0.

Observe that u0(γ) = 0, and u′0(γ) = 0, since u′0 is an absolutely continuousfunction in (0, 1). Taking into account the boundary conditions at the points 0and γ, we can write u0(x) = ax3 + bx2 + λx+ 1 in (0, γ), where

a =λγ + 2

γ3, b = −2λγ + 3

γ2.

We see that the point γ is a zero of second order for the third order polynomialu0, and u0 ≥ 0 in (0, γ]. That means the third zero is not on the open interval(0, γ), hence γ ≤ − 3

λ .Consider the function

F (γ) :=

ˆ γ

0

(u′′(x))2dx,

then F (γ) = 4γ3 (λ2γ2 + 3λγ + 3). Hence F ′(γ) = − 4

γ4 (λγ + 3)2, showing that

the function F is decreasing, so it achieves minimum at the point γ = − 3λ .

Therefore we may conclude that

u0(x) =λ3

33

(x+

3

λ

)3

−, x ∈ (0, 1), (3.3)

and γ = − 3λ is a free boundary point. Observe that u′′(γ) = 0, and u′′ is

a continuous function, but u′′′ has a jump discontinuity at the free boundarypoint γ = − 3

λ .

The example above characterizes one-dimensional solutions. It also tells usthat one-dimensional solutions are C2,1, and in general are not C3.

3.2 The class B%κ(ε) of solutions to the biharmonic obstacle

problem

Without loss of generality, we assume that 0 ∈ Γu, and study the regularity ofthe free boundary, when u ≈ 1

6 (xn)3+.

Let us start by recalling the definition of non-tangentially accessible domains,[7].

Definition 3.2. A bounded domain D ⊂ Rn is called non-tangentially accessible(abbreviated NTA) when there exist constants M , r0 and a function l : R+ 7→ Nsuch that

1. D satisfies the corkscrew condition; that is for any x0 ∈ ∂D and anyr < r0, there exists P = P (r, x0) ∈ D such that

M−1r < |P − x0| < r and dist(P, ∂D) > M−1r. (3.4)

12

2. Dc := Rn \D satisfies the corkscrew condition.

3. Harnack chain condition; if ε > 0 and P1, P2 ∈ D, dist(Pi, ∂D) > ε, and|P1 − P2| < Cε, then there exists a Harnack chain from P1 to P2 whoselength l depends on C, but not on ε, l = l(C). A Harnack chain fromP1 to P2 is a chain of balls Brk(xk), k = 1, ..., l such that P1 ∈ Br1(x1),P2 ∈ Brl(xl), Brk(xk) ∩Brk+1

(xk+1) 6= ∅, and

Mrk > dist(Brk(xk), ∂D) > M−1rk. (3.5)

Let us define rigorously, what we mean by u ≈ 16 (xn)3

+.

Definition 3.3. Let u ≥ 0 be the solution to the biharmonic obstacle problemin a domain Ω, B2 ⊂⊂ Ω and assume that 0 ∈ Γu is a free boundary point. Wesay that u ∈ B%

κ(ε), if the following assumptions are satisfied:

1. u is almost one dimensional, that is

‖∇′u‖W 2,2(B2) ≤ ε,

where ∇′ := ∇− en ∂∂xn

.

2. The set Ωu := u > 0 is an NTA domain with constants r0 = M−1 = %,and with a function l, indicating the length of a Harnack chain.

3. There exists 2 > t > 0, such that u = 0 in B2 ∩ xn < −t.

4. We have the following normalization

‖D3u‖L2(B1) =1

6

∥∥D3(xn)3+

∥∥L2(B1)

=|B1|

12

212

:= ωn, (3.6)

and we also assume that

‖D3u‖L2(B2) < κ, (3.7)

where κ > 16

∥∥D3(xn)3+

∥∥L2(B2)

= 2n2 ωn.

In the notation of the class B%κ(ε) we did not include the length function l,

since later it does not appear in our estimates. For the rest of this paper wewill assume that we have a fixed length function l. Later on in Corollary 3.5we will see that the precise value of the parameter t in assumption 3 is not veryimportant, and therefore we also omit the parameter t in our notation.

Evidently 16 (xn)3

+ ∈ B%κ(ε), for any ε > 0 and % > 0. Next we show that if

u ∈ B%κ(ε), with ε > 0 small, then u ≈ 1

6 (xn)3+ in W 3,2(B1).

From now on κ > 2n2 ωn and 1 > % > 0 are fixed parameters.

Lemma 3.4. There exists a modulus of continuity σ = σ(ε) ≥ 0, such that

∥∥∥∥u(x)− 1

6(xn)3

+

∥∥∥∥W 3,2(B1)

≤ σ(ε), (3.8)

for any u ∈ B%κ(ε).

13

Proof. We argue by contradiction. Assume that there exist σ0 > 0 and a se-quence of solutions, uj ∈ B%

κ(εj), such that

‖∇′uj‖W 2,2(B2) = εj → 0,

but ∥∥∥∥uj(x)− 1

6(xn)3

+

∥∥∥∥W 3,2(B1)

> σ0 > 0. (3.9)

According to assumption 4 in Definition 3.3, ‖D3uj‖L2(B2) < κ and accord-ing to assumption 2 the functions uj are vanishing on an open subset of B2.Therefore it follows from the Poincare inequality that ‖uj‖W 3,2(B2) ≤ C(%, n)κ.Hence up to a subsequence uj u0 weakly in W 3,2(B2), uj → u0 strongly inW 2,2(B2) and according to Corollary 2.9, uj → u0 in C1,α(B3/2). Hence

‖∇′u0‖W 1,2(B2) = limj→∞

‖∇′uj‖W 1,2(B2) ≤ limj→∞

εj = 0.

This implies that u0 is a one-dimensional solution (depending only on the vari-able xn). Example 3.1 tells us that one-dimensional solutions in the interval(−2, 2) have the form

u0(xn) = c1(xn − a1)3− + c2(xn − a2)3

+,

where c1, c2 ≥ 0 and −2 ≤ a1 ≤ a2 ≤ 2 are constants. According to assumption3 in Definition 3.3, u0 = c(xn − a)3

+. In order to obtain a contradiction toassumption (3.9), we need to show that uj → u0 = 1

6 (xn)3+ in W 3,2(B1). The

proof of the last statement can be done in two steps.Step 1: We show that

uj → c(xn − a)3+ in W 3,2(B1). (3.10)

Denote ujn := ∂uj

∂xn∈ W 2,2(B2), j ∈ N0, and let ζ ∈ C∞0 (B 3

2) be a nonnegative

function, such that ζ ≡ 1 in B1. According to Lemma 2.4,

0 ≥ˆ

B2

∆(ζujn)∆ujn =

ˆ

B2

ujn∆ζ∆ujn +

ˆ

B2

ζ(∆ujn)2 + 2

ˆ

B2

∇ζ∇ujn∆ujn,

and therefore

lim supj→∞

ˆ

B2

ζ(∆ujn)2 ≤ − limj→∞

ˆ

B2

ujn∆ζ∆ujn − 2 limj→∞

ˆ

B2

∇ζ∇ujn∆ujn

= −ˆ

B2

u0n∆ζ∆u0

n − 2

ˆ

B2

∇ζ∇u0n∆u0

n =

ˆ

B2

ζ(∆u0n)2,

(3.11)

where in the last step we used integration by parts.On the other hand, since ∆ujn ∆u0

n weakly in L2(B2), it follows that

lim infj→∞

ˆ

B2

ζ(∆ujn)2 ≥ˆ

B2

ζ(∆u0n)2. (3.12)

Therefore, we may conclude from (3.11) and (3.12) that

limj→∞

ˆ

B2

ζ(∆ujn)2 =

ˆ

B2

ζ(∆u0n)2.

14

Hence we obtain

∂∆uj

∂xn→ ∂∆u0

∂xnin L2(B1).

Similarly ∂∆uj

∂xi→ 0 in L2(B1), for i = 1, ..., n− 1. Knowing that

‖∇∆uj −∇∆u0‖L2(B1) → 0, and ‖uj − u0‖W 2,2(B2) → 0,

we may apply the Calderon-Zygmund inequality, and conclude (3.10). Recallingthat ‖D3uj‖L2(B1) = ωn, we see that

‖D3u0‖L2(B1) = ωn. (3.13)

Since u0 = c(xn − a)3+ ≥ 0, it follows that

‖D3u0‖2L2(B1) = c2Ln(B1 ∩ xn > a) > 0,

hencec > 0 and a < 1. (3.14)

Step 2: We show that a = 0 and c = 16 . Taking into account that uj → u0

in C1,α and uj(0) = 0, we conclude that u0(0) = 0, thus a ≥ 0. Assume thata > 0. Since 0 ∈ Γj , and Ωj is an NTA domain, there exists Pj = P (r, 0) ∈ Ωj ,for 0 < r < min(%, a/2) as in the corkscrew conditon,

%r < |Pj | < r and dist(Pj , ∂Ωj) > %r.

Therefore up to a subsequence Pj → P0, hence r% ≤ |P0| ≤ r, Br′(P0) ⊂ Ωj ,for all j large enough, where 0 < r′ < r% is a fixed number. Since we havechosen r < a/2, we may conclude that

Br′(P0) ⊂ xn < a ∩ Ωj .

Thus ∆uj is a sequence of harmonic functions in the ball Br′(P0), and therefore

∆uj → 0 locally uniformly in Br′(P0), (3.15)

according to (3.10).Let Q := en, then u0(Q) = c(1 − a)3 > 0, since uj → u0 uniformly in

B3/2, we see that uj(Q) > 0 for large j, and Q ∈ Ωj . Therefore there exists

a Harnack chain connecting P0 with Q; Br1(x1), Br2(x2), ..., Brl(xl) ⊂ Ωj ,

whose length l does not depend on j. Denote by Kj := ∪iBri(xi) ⊂⊂ Ωj , andlet V j ⊂⊂ Kj ⊂⊂ Ωj where V j is a regular domain, such that dist(Kj , ∂V j)and dist(V j , ∂Ωj) depend only on r and %.

Let wj+ be a harmonic function in V j , with boundary conditions wj+ =

(∆uj)+ ≥ 0 on ∂V j , then wj+ −∆uj is a harmonic function in V j , and wj+ −∆uj = (∆uj)− ≥ 0 on ∂V j , hence

0 ≤ wj+ −∆uj ≤ ‖(∆uj)−‖L∞(V j) in V j . (3.16)

Let us observe that ∆uj → ∆u0 = 6c(xn − a)+ implies that ‖(∆uj)−‖L2(B2) →0. Since (∆uj)− is a subharmonic function in Ωj , and V j ⊂⊂ Ωj it follows that

‖(∆uj)−‖L∞(V j) ≤ C(n, r, %)‖(∆uj)−‖L2(B2) → 0.

15

So wj+ is a nonnegative harmonic function in V j , and by the Harnack in-equality

CH infBrl (x

l)wj+ ≥ sup

Brl (xl)

wj+ ≥ wj+(en) ≥ ∆uj(en) ≥ 1

2∆u0(en) = 3c(1− a),

if j is large, where CH is the constant in Harnack’s inequality, it depends on %and r but not on j. Denote C(a, c) := 3c(1 − a) > 0 by (3.14). Applying theHarnack inequality again, we see that

CH infBrl−1

(xl−1)wj+ ≥ sup

Brl−1(xl−1)

wj+ ≥ infBrl (x

l)wj+ >

C(a, c)

CH.

Inductively, we obtain that

CH infBr1 (x1)

wj+ ≥ supBr1 (x1)

wj+ >C(a, c)

Cl−1H

, (3.17)

where l does not depend on j. Hence wj+(P0) ≥ C(a,c)

ClHfor all j large, and

according to (3.16),

limj→∞

∆uj(P0) ≥ C(a, c)

ClH> 0,

the latter contradicts (3.15). Therefore we may conclude that a = 0.Recalling that ‖D3u0‖L2(B1) = ωn, we see that c = 1

6 , but then we obtain

uj → 16 (xn)3

+ in W 3,2(B1) which is a contradiction, since we assumed (3.9).

Lemma 3.4 has an important corollary, which will be very useful in our laterdiscussion.

Corollary 3.5. Let u be the solution to the biharmonic obstacle problem, u ∈B%κ(ε). Then for any fixed t > 0 we have that u(x) = 0 in B2 ∩ xn < −t,

provided ε = ε(t) > 0 is small.

Proof. Once again we argue by contradiction. Assume that there exist t0 > 0and a sequence of solutions uj ∈ B%

κ(εj), εj → 0, such that xj ∈ B2 ∩ Γj ,and xjn < −t0. For 0 < r < min(%, t0/2) choose P j = P (r, xj) ∈ Ωj as in thecorkscrew condition,

r% < |xj − P j | < r, Br%(Pj) ⊂ Ωj .

Upon passing to a subsequence, we may assume that P j → P 0. Fix 0 < r′ < r%,then for large j

Br′(P0) ⊂⊂ Ωj ∩ xn < 0.

Hence ∆uj is a sequence of harmonic functions in Br′(P0). According to Lemma

3.4, uj → 16 (xn)3

+, and therefore ∆uj → 0 in Br′(P0), and ∆uj(en) → 1.

Since Ωj is an NTA domain, there exists a Harnack chain connecting P 0 withQ := en ∈ Ωj ; Br1(x1), Br2(x2), ..., Brk(xk) ⊂ Ωj , whose length does notdepend on j. Arguing as in the proof of Lemma 3.4, we obtain a contradictionto ∆uj → 0 in Br′(P

0).

16

3.3 Linearization

Let uj be a sequence of solutions in Ω ⊃⊃ B2, uj ∈ B%κ(εj), and assume that

εj → 0 as j →∞. It follows from Lemma 3.4, that up to a subsequence

uj → 1

6(xn)3

+ in W 2,2(B2) ∩ C1,αloc (B2). (3.18)

Let us denote

δji :=

∥∥∥∥∂uj

∂xi

∥∥∥∥W 2,2(B2)

.

Without loss of generality we may assume that δji > 0, for all j ∈ N. Indeed,

if δji = 0 for all j ≥ J0 large, then uj does not depend on the variable xi, andthe problem reduces to a lower dimensional case. Otherwise we may pass to asubsequence satisfying δji > 0 for all j.

Denote

vji :=1

δji

∂uj

∂xi, for i = 1, ..., n− 1, (3.19)

then ‖vji ‖W 2,2(B2) = 1. Therefore up to a subsequence vji converges to a functionv0i weakly in W 2,2(B2) and strongly in W 1,2(B2). For the further discussion we

need strong convergence vji → v0i in W 2,2, at least locally.

Lemma 3.6. Assume that uj is a sequence of solutions in Ω ⊃⊃ B2, uj ∈B%κ(εj), εj → 0. Let vji be the sequence given by (3.19), and assume that vji v0

i

weakly in W 2,2(B2), strongly in W 1,2(B2), for i = 1, ..., n− 1, then

∆2v0i = 0 in B+

2 , v0i ≡ 0 in B2 \B+

2 . (3.20)

Furthermore,‖vji − v0

i ‖W 2,2(B1) → 0. (3.21)

Proof. Denote by Ωj := Ωuj , Γj := Γuj . It follows from Corollary 3.5 thatv0i ≡ 0 in B2 \ B+

2 , hence v0i = |∇v0

i | = 0 on xn = 0 ∩ B2 in the trace sense.Moreover, if K ⊂⊂ B+

2 is an open subset, then K ⊂ Ωj for large j by (3.18).

Hence ∆2vji = 0 in K, and therefore ∆2v0i = 0 in B+

2 , and (3.20) is proved.Now let us proceed to the proof of the strong convergence. Let ζ ∈ C∞0 (B 3

2)

be a nonnegative function, such that ζ ≡ 1 in B1. It follows from (3.20) that

0 =

ˆ

B 32

∆v0i∆(ζv0

i ) =

ˆ

B 32

v0i∆ζ∆v0

i +

ˆ

B 32

ζ(∆v0i )2 + 2

ˆ

B 32

∇ζ∇v0i∆v0

i .

(3.22)According to Lemma 2.4

0 ≥ˆ

B 32

∆(ζvji )∆vji =

ˆ

B 32

vji∆ζ∆vji +

ˆ

B 32

ζ(∆vji )2 + 2

ˆ

B 32

∇ζ∇vji∆vji ,

(3.23)and therefore

lim supj→∞

ˆ

B 32

ζ(∆vji )2 ≤ − lim

j→∞

ˆ

B 32

vji∆ζ∆vji − 2 limj→∞

ˆ

B 32

∇ζ∇vji∆vji

= −ˆ

B 32

v0i∆ζ∆v0

i − 2

ˆ

B 32

∇ζ∇v0i∆v0

i ,

17

where we used that vji → v0i in W 1,2(B2) and ∆vji ∆v0

i in L2(B2).From the last inequality and (3.22) we may conclude that

lim supj→∞

ˆ

B 32

ζ(∆vji )2 ≤

ˆ

B 32

ζ(∆v0i )2. (3.24)

On the other hand

lim infj→∞

ˆ

B 32

ζ(∆vji )2 ≥

ˆ

B 32

ζ(∆v0i )2 (3.25)

follows from the weak convergence ∆vji ∆v0i in L2(B2), and we may conclude

from (3.24) and (3.25) that

limj→∞

ˆ

B 32

ζ(∆vji )2 =

ˆ

B 32

ζ(∆v0i )2.

Hence we obtain ‖∆vji −∆v0i ‖L2(B1) → 0, and therefore vji → v0

i in W 2,2(B1)according to the Calderon-Zygmund inequality.

In the next proposition we use Lemma 3.6 in order to estimate ‖∇′u‖W 2,2(B1)

by ‖∇′∆u‖L2(B1), for a solution u ∈ B%κ(ε).

Proposition 3.7. For any δ > 0, there exists a constant c(δ) ≥ 1, and ε(δ) > 0,such that the inequality

‖∇′u‖W 2,2(B1) ≤ c(δ) ‖∇′∆u‖L2(B1) + δ ‖∇′u‖W 2,2(B2) , (3.26)

holds for any u ∈ B%κ(ε), if ε ≤ ε(δ).

Proof. According to the Cauchy-Schwarz inequality, it is enough to show thatfor any i ∈ 1, ..., n− 1, the inequality

∥∥∥∥∂u

∂xi

∥∥∥∥W 2,2(B1)

≤ c(δ)∥∥∥∥∂∆u

∂xi

∥∥∥∥L2(B1)

+ δ

∥∥∥∥∂u

∂xi

∥∥∥∥W 2,2(B2)

holds for u ∈ B%κ(ε), provided ε is small enough. We argue by contradiction.

Assume that there exists a sequence of solutions uj ∈ B%κ(εj), such that εj → 0,

but ∥∥∥∥∂uj

∂xi

∥∥∥∥W 2,2(B1)

> j

∥∥∥∥∂∆uj

∂xi

∥∥∥∥L2(B1)

+ δ0

∥∥∥∥∂uj

∂xi

∥∥∥∥W 2,2(B2)

(3.27)

for some δ0 > 0.Let vji be the corresponding sequence, given by (3.19). The inequality (3.27)

implies that

‖∆vji ‖L2(B1) ≤1

j, and ‖vji ‖W 2,2(B1) ≥ δ0, for any j. (3.28)

According to Lemma 3.6, vji → v0i in W 2,2(B1), and v0

i = 0 in B1 \ B+1 . After

passing to the limit in (3.28), we obtain

∆v0i = 0 in B1 and ‖v0

i ‖W 2,2(B1) ≥ δ0.So v0

i is a harmonic function in the unit ball B1, vanishing in B1 \ B+1 , hence

v0i ≡ 0 in B1. This contradicts ‖v0

i ‖W 2,2(B1) ≥ δ0.

18

3.4 Properties of solutions in a normalized coordinate sys-tem

Let us define

ur,x0(x) :=

u(rx+ x0)

r3, for x0 ∈ Γu, x ∈ B2, r ∈ (0, 1), (3.29)

and ur := ur,0.First we would like to know how fast ‖∇′∆ur‖L2(B1) decays with respect to

‖∇′∆u‖L2(B1), for r < 1. In particular, it is well known that an inequality

‖∇′∆u(sx)‖L2(B1) ≤ τ‖∇′∆u‖L2(B1), (3.30)

for some 0 < s, τ < 1 would provide good decay estimates for ‖∇′∆u(skx)‖L2(B1),k ∈ N. By choosing a suitable coordinate system, we succeed to show a weakerversion of the desired inequality. This weaker version of (3.30) is good enoughto perform an iteration argument to prove the regularity of the free boundary.

First let us observe that 16 (η · x)3

+ ∈ B%κ(ε) if |η − en| ≤ Cnε, for some

dimensional constant Cn.

Definition 3.8. Let u be the solution to the biharmonic obstacle problem. Wesay that the coordinate system is normalized with respect to u, if

infη∈Rn,|η|=1

∥∥∥∥∇′η∆

(u(x)− 1

6(η · x)3

+

)∥∥∥∥L2(B1)

=

∥∥∥∥∇′en∆

(u(x)− 1

6(xn)3

+

)∥∥∥∥L2(B1)

,

where ∇′η := ∇− (η · ∇)η, and ∇′ := ∇′en .

A minimizer η always exists for a function u ∈ B%κ(ε), and since ∇′−η =

∇′η, −η is also a minimizer, thus we always choose a minimizer satisfying thecondition en · η ≥ 0. A normalized coordinate system always exists by choosingη = en in the new coordinate system.

Let us also observe that for every η ∈ Rn,

∇′η∆

(u(x)− 1

6(η · x)3

+

)= ∇′η∆u(x)

and ∥∥∇′η∆u∥∥2

L2(B1)= ‖∇∆u‖2L2(B1) − ‖η · ∇∆u‖2L2(B1) .

Lemma 3.9. Assume that u ∈ B%κ(ε) solves the biharmonic obstacle problem

in a fixed coordinate system with basis vectors e1, ..., en. Let e11, ..., e

1n be a

normalized coordinate system with respect to u, and assume that e1n · en ≥ 0.

Then|en − e1

n| ≤ C(n)‖∇′∆u‖L2(B1) ≤ C(n)ε,

if ε is small, where C(n) > 0 is a dimensional constant.

19

Proof. According to Definition 3.8,

‖∇′e1n∆u‖L2(B1) = ‖∇∆u− e1n(e1

n · ∇∆u)‖L2(B1) ≤ ‖∇′∆u‖L2(B1). (3.31)

It follows from the triangle inequality that

∥∥∥∥∂∆u

∂xn− (en · e1

n)2 ∂∆u

∂xn

∥∥∥∥L2(B1)

≤∥∥∥∥∂∆u

∂xn− (en · e1

n)(e1n · ∇∆u)

∥∥∥∥L2(B1)

+

∥∥∥∥(en · e1n)(e1

n · ∇∆u)− (en · e1n)2 ∂∆u

∂xn

∥∥∥∥L2(B1)

≤ ‖∇′e1n∆u‖L2(B1) + (en · e1n)‖e1

n · ∇′∆u‖L2(B1) ≤ 2‖∇′∆u‖L2(B1),

(3.32)

according to (3.31), and taking into account that 0 ≤ en · e1n ≤ 1.

Note that Lemma 3.4 implies that∥∥∥∂∆u∂xn

∥∥∥L2(B1)

is uniformly bounded from

below, and therefore by choosing ε > 0 small, we may conclude from (3.32) that

1− (en · e1n)2 ≤ 2‖∇′∆u‖L2(B1)∥∥∥∂∆u

∂xn

∥∥∥L2(B1)

≤ C(n)‖∇′∆u‖L2(B1).

Since 0 ≤ en · e1n ≤ 1, we get

0 ≤ 1− en · e1n ≤ 1− (en · e1

n)2 ≤ C(n)‖∇′∆u‖L2(B1). (3.33)

Denote by (e1n)′ := e1

n − en(en · e1n). It follows from the triangle inequality

and (3.31) that

‖(e1n)′(e1

n · ∇∆u)‖L2(B1) ≤ ‖∇′∆u− (e1n)′(e1

n · ∇∆u)‖L2(B1)

+‖∇′∆u‖L2(B1) ≤ ‖∇′e1n∆u‖L2(B1) + ‖∇′∆u‖L2(B1) ≤ 2‖∇′∆u‖L2(B1).

Hence

|(e1n)′| ≤ 2‖∇′∆u‖L2(B1)

‖e1n · ∇∆u‖L2(B1)

.

Let us choose ε > 0 small, then∥∥e1n · ∇∆u

∥∥L2(B1)

is bounded from below by a

dimensional constant according to Lemma 3.4 and inequality (3.33). Thereforewe obtain

|(e1n)′| ≤ C(n)‖∇′∆u‖L2(B1). (3.34)

Note that|en − e1

n| ≤ |1− en · e1n|+ |(e1

n)′|.Applying inequalities (3.33) and (3.34) we obtain the desired inequality,

|en − e1n| ≤ C(n)‖∇′∆u‖L2(B1) ≤ C(n)ε,

and the proof of the lemma is now complete.

Lemma 3.9 provides an essential estimate, which will be useful in our laterdiscussion.

20

Proposition 3.10. For any fixed 0 < s < τ < 1 and δ > 0, there existsε = ε(δ, τ, s), such that if 0 < ε ≤ ε(δ, τ, s), then for any u ∈ B%

κ(ε)

‖∇′∆us‖L2(B1) ≤ τ ‖∇′∆u‖L2(B1) + δ ‖∇′u‖W 2,2(B2) , (3.35)

in a normalized coordinate system with respect to u.

Proof. According to the Cauchy-Schwarz inequality, it is enough to show thatthe inequality

∥∥∥∥∂∆us∂xi

∥∥∥∥L2(B1)

≤ τ∥∥∥∥∂∆u

∂xi

∥∥∥∥L2(B1)

+ δ

∥∥∥∥∂u

∂xi

∥∥∥∥W 2,2(B2)

holds for any i ∈ 1, ..., n − 1, provided ε is small enough. We argue bycontradiction. Assume that there exist 0 < s < τ < 1 and δ0 > 0 for whichthere exists a sequence of solutions uj ⊂ B%

κ(εj) in a normalized coordinatesystem, such that εj → 0, as j →∞, but for some i ∈ 1, 2, ..., n− 1

∥∥∥∥∂∆ujs∂xi

∥∥∥∥L2(B1)

> τ

∥∥∥∥∂∆uj

∂xi

∥∥∥∥L2(B1)

+ δ0

∥∥∥∥∂uj

∂xi

∥∥∥∥W 2,2(B2)

. (3.36)

Let vji be given by (3.19), then according to Lemma 3.6, vji → v0i in

W 2,2(B1). Dividing equation (3.36) by δji , we see that ‖∆vji (sx)‖L2(B1) ≥ δ0.Hence

‖∆vji ‖L2(B1) ≥ sn2 ‖∆vji (sx)‖L2(B1) ≥ s

n2 δ0,

and after passing to the limit as j →∞, we obtain ‖∆v0i ‖L2(B1) ≥ s

n2 δ0.

According to Lemma 3.6, ∆v0i is a harmonic function in xn > 0 ∩ B1.

Therefore it may be written as a sum of homogeneous orthogonal harmonicpolynomials

∆v0i (x) =

∞∑

m=0

ami (x),

where m indicates the degree of the polynomial ami .Next we show that a0

i = 0, using the definition of a normalized coordinatesystem. Consider the following direction

ηji :=(

0, ..., δji a0i , ..., 0, λ

j)

= δji a0i ei + λjen

where λj > 0, is chosen so that |ηji | = 1, and since δji → 0, we may concludethat λj → 1.

By the normalization of the coordinate system, Definition 3.8, it follows that

‖∇′ηji

∆uj‖2L2(B1) ≥ ‖∇′∆uj‖2L2(B1),

or equivalently

‖∇∆uj − ηji (ηji · ∇∆uj)‖2L2(B1) ≥ ‖∇∆uj − en(en · ∇∆uj)‖2L2(B1). (3.37)

21

Expanding (3.37) in coordinates, and taking into account that ∂∆uj

∂xi= δji∆v

ji ,

we get

∑

l 6=i,n(δjl )

2∥∥∥∆vjl

∥∥∥2

L2(B1)+ (δji )

2∥∥∥∆vji − a0

i (ηji · ∇∆uj)

∥∥∥2

L2(B1)

+

∥∥∥∥∂∆uj

∂xn− λj(ηji · ∇∆uj)

∥∥∥∥2

L2(B1)

≥n−1∑

l=1

(δjl )2∥∥∥∆vjl

∥∥∥2

L2(B1).

The substitution

ηji · ∇∆uj = a0i (δ

ji )

2∆vji + λj∂∆uj

∂xn

leads to

(δji )2

∥∥∥∥(

1− (a0i δji )

2)

∆vji − a0iλj ∂∆uj

∂xn

∥∥∥∥2

L2(B1)

+

∥∥∥∥(1− (λj)2

) ∂∆uj

∂xn− λja0

i (δji )

2∆vji

∥∥∥∥2

L2(B1)

≥ (δji )2∥∥∥∆vji

∥∥∥2

L2(B1).

Dividing the last inequality by (δji )2, and taking into account that 1− (λj)2 =

(a0i δji )

2, we obtain

(λj)2

∥∥∥∥λj∆vji − a0

i

∂∆uj

∂xn

∥∥∥∥2

L2(B1)

+ (1− (λj)2)

∥∥∥∥λj∆vji − a0

i

∂∆uj

∂xn

∥∥∥∥2

L2(B1)

≥∥∥∥∆vji

∥∥∥2

L2(B1).

Hence ∥∥∥∥λj∆vji − a0

i

∂∆uj

∂xn

∥∥∥∥2

L2(B1)

≥∥∥∥∆vji

∥∥∥2

L2(B1). (3.38)

Lemma 3.4 implies that ∂∆uj

∂xn→ χB+

1in L2(B1), and according to Lemma

3.6, ∆vji → ∆v0i in L2(B1) as j → ∞, and v0

i = 0 in B1 \ B+1 . Therefore we

may pass to the limit in the inequality (3.38), and obtain

∥∥∆v0i − a0

i

∥∥2

L2(B+1 )≥∥∥∆v0

i

∥∥2

L2(B+1 ). (3.39)

Now let us recall that ∆v0i =

∑∞m=0 a

mi , where ami are orthogonal homoge-

neous polynomials of degree m. Therefore (3.39) is equivalent to the followinginequality

∞∑

m=1

‖ami ‖2L2(B+1 )≥∞∑

m=0

‖ami ‖2L2(B+1 ),

which implies that a0i = 0.

Now let τ and s be the numbers in (3.36), and let us observe that

∆v0i (sx) =

∞∑

m=1

ami (sx) = s∞∑

m=1

sm−1ami (x),

22

and therefore for any 0 < s < 1

‖∆v0i (sx)‖2L2(B1) ≤ s2

∞∑

m=1

‖ami ‖2L2(B1) = s2‖∆v0i ‖2L2(B1).

Then according to the strong convergence ‖∆vji ‖L2(B1) → ‖∆v0i ‖L2(B1), and

taking into account that ‖∆v0i ‖L2(B1) ≥ s

n2 δ0 > 0, we obtain

‖∆vji (sx)‖L2(B1) ≤ τ‖∆vji ‖L2(B1),

for j > 1 large enough. Hence we may conclude that

∥∥∥∥∂∆uj

∂xi(sx)

∥∥∥∥L2(B1)

≤ τ∥∥∥∥∂∆uj

∂xi

∥∥∥∥L2(B1)

,

contradicting (3.36).

3.5 C1,α-regularity of the free boundary

In this section we perform an iteration argument, based on Proposition 3.7,Proposition 3.10, and Lemma 3.9, that leads to the existence of the unit normalη0 of the free boundary at the origin, and provides good decay estimates for‖∇′η0ur‖W 2,2(B1).

First we would like to verify that u ∈ B%κ(ε) imply that us ∈ B%

κ(Cε). It iseasy to check that the property of being an NTA domain is scaling invariant, inthe sense that if D is an NTA domain and 0 ∈ ∂D, then for any 0 < s < 1 theset Ds := s−1(D ∩Bs) is also an NTA domain with the same parameters as D.

Assumption 3 in Definition 3.3 holds for us according to Corollary 3.5. In-deed, let t = s in Corollary 3.5, then u(sx) = 0 if xn < −1 .

Thus us satisfies 2, 3 in Definition 3.3, but it may not satisfy 4. Instead weconsider rescaled solutions defined as follows

Us(x) :=ωnus(x)

‖D3us‖L2(B1), (3.40)

then assumption 4 also holds. Indeed, ‖D3Us‖L2(B1) = ωn by definition of Us,and

‖D3Us‖L2(B2) =ωn‖D3us‖L2(B2)

‖D3us‖L2(B1)=ωn‖D3u‖L2(B2s)

‖D3u‖L2(Bs)

≤ ωnωn(2s)

n2 + σ(ε)

ωn(s)n2 − σ(ε)

< κ,

according to Lemma 3.4 provided ε = ε(n, κ, s) is small.In the next lemma we show that Us ∈ B%

κ(τε) in a normalized coordinatesystem, then we argue inductively to show that Usk ∈ B%

κ(Cβkε), β < 1.

Lemma 3.11. Assume that u ∈ B%κ(ε) solves the biharmonic obstacle problem

in a normalized coordinate system e1, e2, ..., en. Let 0 < α < 1 be a number.

23

Then there exists a unit vector η0 ∈ Rn, such that |η0 − en| ≤ Cε, and for any0 < r < 1

‖∇′η0ur‖W 2,2(B1)

‖D3ur‖L2(B1)≤ Crαε, (3.41)

provided ε = ε(n, κ, %, α) is small enough. The constant C > 0 depends on thespace dimension n and on the given parameters α, κ, %.

Proof. Throughout e1, ..., en is a fixed coordinate system normalized withrespect to the solution u ∈ B%

κ(ε), and ∇′u = ∇′enu. We may renormalizethe coordinate system with respect to Us and denote by e1

1, ..., e1n the set of

basis vectors in the new system. Inductively, ek1 , ..., ekn, k ∈ N is a normalizedsystem with respect to Usk , and e0

i := ei. According to Lemma 3.9,

|ekn − ek−1n | ≤ C(n)

‖∇′ek−1n

∆usk‖L2(B1)

‖D3usk‖L2(B1), (3.42)

provided ‖∇′ek−1n

Usk‖W 2,2(B2) is sufficiently small.

Let us fix 14 < s < 1

2 , then for any k ∈ N0 and η ∈ Rn, |η| = 1 the followinginequality holds

‖∇′ηusk+1‖W 2,2(B2) ≤ CP (%, n)‖D2∇′ηusk+1‖L2(B2)

≤ CP (%, n)s−n2 ‖D2∇′ηusk‖L2(B1) ≤ C(%, n)‖∇′ηusk‖W 2,2(B1)

(3.43)

where we used the Poincare inequality in the first step for the function ∇′ηusk+1 .

From now on 14 < s < β < 1/2 are fixed constants, τ and γ are arbitrary

constants satisfying1

4< s < τ < γ < β <

1

2. (3.44)

Let c(δ) > 0 be the constant in Proposition 3.7, with δ > 0 to be chosenlater. Then

‖∇′u‖W 2,2(B1) ≤ c(δ)‖∇′∆u‖L2(B1) + δ‖∇′u‖W 2,2(B2). (3.45)

Denote by

λ :=c(δ)C(%, n)

γ − τ ,

where τ, γ are the constants in (3.44), and C(%, n) is the same constant as in(3.43). Take δ0 = δ

λ in Proposition 3.10, and ε ≤ ε(min(δ0, δ)), then

‖∇′∆us‖L2(B1) ≤ τ‖∇′∆u‖L2(B1) + δ0‖∇′u‖W 2,2(B2). (3.46)

According to our choice of e1n and inequality (3.43), we obtain from (3.45) and

(3.46)

λ‖∇′e1n∆us‖L2(B1) + ‖∇′us‖W 2,2(B2) ≤ λ‖∇′∆us‖L2(B1)

+C(%, s, n)‖∇′u‖W 2,2(B1) ≤ (λτ + c(δ)C(%, s, n))‖∇′∆u‖L2(B1)

+(λδ0 + C(%, s, n)δ)‖∇′u‖W 2,2(B2) = λγ‖∇′∆u‖L2(B1)

+δ(1 + C(%, s, n))‖∇′u‖W 2,2(B2).

24

Therefore, if δ ≤ 11+C(%,s,n) , then

λ‖∇′e1n∆us‖L2(B1) + ‖∇′us‖W 2,2(B2) ≤ γ(λ‖∇′∆u‖L2(B1) + ‖∇′u‖W 2,2(B2)

).

(3.47)Now let us consider the sequence of numbers Akk∈N0

, defined as follows:

A0 :=λ‖∇′∆u‖L2(B1) + ‖∇′u‖W 2,2(B2)

‖D3u‖L2(B1),

and

Ak :=λ‖∇′ekn∆usk‖L2(B1) + ‖∇′

ek−1n

usk‖W 2,2(B2)

‖D3usk‖L2(B1), for k = 1, 2, .... (3.48)

The assumption u ∈ B%κ(ε) implies that

A0 ≤λ+ 1

ωnε.

Let us also observe that

‖D3u‖L2(B1)

‖D3us‖L2(B1)=sn2 ‖D3u‖L2(B1)

‖D3u‖L2(Bs)≤ s

n2 ωn

sn2 ωn − σ(A0)

≤ β

γ.

according to Lemma 3.4, since A0 is small. Hence the inequality (3.47) impliesthat

A1 =λ‖∇′e1n∆us‖L2(B1) + ‖∇′us‖W 2,2(B2)

‖D3us‖L2(B1)≤ γA0

‖D3u‖L2(B1)

‖D3us‖L2(B1)≤ βA0.

Thus A1 ≤ βA0. We use induction to show that

Ak ≤ βkA0, for all k ∈ N0 (3.49)

for a fixed γ < β < 1/2. Assuming that (3.49) holds up to k ∈ N, we will showthat Ak+1 ≤ βk+1A0. The proof is quite long and technical.

Recalling our notation (3.40) for Usk , and notation (3.48) for Ak, we see that

‖∇′ek−1n

Usk‖W 2,2(B2) =ωn‖∇′ek−1

nusk‖W 2,2(B2)

‖D3usk‖L2(B1)≤ ωnAk ≤ βk(λ+ 1)ε. (3.50)

Hence

Usk =ωnusk(x)

‖D3usk‖L2(B1)∈ B%

κ(βkε0)

in the coordinate system ek−11 , ..., ek−1

n , and ε0 = (λ + 1)ε is small. By def-inition, ek1 , ..., ekn is a normalized coordinate system with respect to Usk ∈B%κ(βkε0).

The definition of a normalized coordinate system, and inequality (3.43) implythat

Ak+1 =λ‖∇′

ek+1n

∆usk+1‖L2(B1) + ‖∇′eknusk+1‖W 2,2(B2)

‖D3usk+1‖L2(B1)

≤λ‖∇′ekn∆usk+1‖L2(B1) + C(%, s, n)‖∇′eknusk‖W 2,2(B1)

‖D3usk+1‖L2(B1).

(3.51)

25

Applying Proposition 3.7 for the function Usk ∈ B%κ(βkε0), we obtain

‖∇′eknusk‖W 2,2(B1) ≤ c(δ)‖∇′ekn∆usk‖L2(B1) + δ‖∇′eknusk‖W 2,2(B2). (3.52)

It follows from Proposition 3.10 and our choice of the coordinate system ek1 , ..., ekn,that

‖∇′ekn∆usk+1‖L2(B1) ≤ τ‖∇′ekn∆usk‖L2(B1) + δ0‖∇′eknusk‖W 2,2(B2), (3.53)

where δ, c(δ) and δ0 are the same numbers as before.Combining inequalities (3.51), (3.52) and (3.53), we derive the following

estimate for Ak+1,

Ak+1 ≤λγ‖∇′ekn∆usk‖L2(B1) + δ(1 + C(%, n))‖∇′eknusk‖W 2,2(B2)

‖D3usk+1‖L2(B1). (3.54)

It follows from the triangle inequality and Lemma 3.9 that

‖∇′eknusk‖W 2,2(B2) ≤ ‖∇′ek−1n

usk‖W 2,2(B2) + 2|ekn − ek−1n |‖∇usk‖W 2,2(B2)

≤ ‖∇′ek−1n

usk‖W 2,2(B2) + 2C(n)‖∇′

ek−1n

usk‖W 2,2(B2)

‖D3usk‖L2(B1)‖∇usk‖W 2,2(B2)

≤ C ′(%, n)‖∇′ek−1n

usk‖W 2,2(B2).

The last inequality and (3.54) imply that

Ak+1 ≤λγ‖∇′ekn∆usk‖L2(B1) + δC ′(%, n)(1 + C(%, n))‖∇′

ek−1n

usk‖W 2,2(B2)

‖D3usk+1‖L2(B1)

≤λγ‖∇′ekn∆usk‖L2(B1) + γ‖∇′

ek−1n

usk‖W 2,2(B2)

‖D3usk+1‖L2(B1)= γAk

‖D3usk‖L2(B1)

‖D3usk+1‖L2(B1)

(3.55)if

δ ≤ γ

C ′(%, n)(1 + C(%, n)),

independent of k ∈ N.In order to complete the induction argument, we observe that

‖D3usk‖L2(B1)

‖D3usk+1‖L2(B1)=sn2 ‖D3usk‖L2(B1)

‖D3usk‖L2(Bs)

≤ sn2 ωn

sn2 ωn − σ(βkA0)

≤ β

γ,

(3.56)

according to Lemma 3.4, since Usk ∈ B%κ(βkε0) and βkε0 < ε0 is small. Finally

we obtain from (3.55) and (3.56) that

Ak+1 ≤ βAk ≤ βk+1A0, (3.57)

this completes the proof of inequality (3.49).

26

Next we can show that ekn is a Cauchy sequence by using (3.42) and (3.49).Indeed for any m, k ∈ N,

|ek+mn − ekn| ≤

m∑

l=1

|ek+ln − ek+l−1

n | ≤ C(n)m∑

l=1

‖∇′ek+l−1n

∆Usk+l‖L2(B1)

≤ C(n)m∑

l=1

Ak+l ≤C(n)A0

λ

m∑

l=1

βk+l ≤ C(n)A0

λ(1− β)βk,

hence ekn → η0, as k →∞ for some η0 ∈ Rn, |η0| = 1 and

|η0 − ekn| ≤ C(n, λ)A0βk ≤ C ′(n, λ)βkε, (3.58)

in particular |η0 − en| ≤ C ′(n, λ)ε.Combining (3.49) and (3.58), we may conclude that

‖∇′η0∆usk‖L2(B1)

‖D3usk‖L2(B1)≤ Cβkε and

‖∇′η0usk‖W 2,2(B2)

‖D3usk‖L2(B1)≤ Cβkε, (3.59)

where the constant C > 0 depends only on the space dimension and on thegiven parameters % and 0 < s < τ < γ < β < 1/2.

The inequality (3.41) follows from (3.59) via a standard iteration argument.Let 0 < α < 1 be any number, fix 1/4 < s < 1/2, and take β = sα > s. If0 < r < 1, then there exists k ∈ N0, such that sk+1 ≤ r < sk. Hence

‖∇′η0ur‖W 2,2(B1)

‖D3ur‖L2(B1)≤ C ‖∇

′η0usk‖W 2,2(B1)

‖D3usk‖L2(B1)≤ Cβkε = Csαkε ≤ C(n, %, α)rαε.

(3.60)

Now we are ready to prove the C1,α-regularity of the free boundary.

Theorem 3.12. Assume that u ∈ B%κ(ε), with an ε > 0 small. Then Γu ∩ B1

is a C1,α-graph for any 0 < α < 1 and the C1,α-norm of the graph is boundedby Cε.

Proof. Let 0 < α < 1 be any number, fix 1/4 < s < 1/2, and take β = sα > s.It follows from Lemma 3.11 that for u ∈ B%

κ(ε)

‖∇′ur‖W 2,2(B1)

‖D3ur‖L2(B1)≤ Crα → 0 as r → 0,

after a change of variable, by choosing en = η0, where η0 is the same vector asin Lemma 3.11. Then

ωnur(x)

‖D3ur‖L2(B1)→ 1

6(xn)3

+

according to Lemma 3.4.So we have shown that in the initial coordinate system,

ωnu(rx)

r3‖D3ur‖L2(B1)→ 1

6(η0 · x)3

+ in W 3,2(B1) ∩ C1,α(B1), as r → 0, (3.61)

and therefore η0 is the measure theoretic normal to Γu at the origin.

27

Now let x0 ∈ Γu ∩ B1 be a free boundary point, and consider the function

ux0,1/2(x) = u(x/2+x0)(1/2)3 , x ∈ B2, then

Ux0(x) :=

ωnux0,1/2(x)

‖D3ux0,1/2‖L2(B1)∈ B%

κ(C(n)ε).

According to Lemma 3.11, Ux0has a unique blow-up

Ur,x0(x) :=

Ux0(rx)

r3=

ωnux0,1/2(rx)

r3‖D3ux0,1/2(rx)‖L2(B1)→ 1

6(ηx0

· x)3+.

and therefore ηx0 is the normal to Γu at x0.Next we show that ηx is a Holder continuous function on Γu ∩ B1. If x0 ∈

Γu ∩B1, then sk+1 < |x0| ≤ sk, for some k ∈ N0. Hence ‖∇′η0Usk,x0‖W 2,2(B2) ≤

Cβkε, and ‖∇′ηx0Ur,x0‖W 2,2(B1) → 0 as r → 0. Applying Lemma 3.11 for the

function Usk,x0∈ B%

κ(Cβkε), we obtain

|ηx0− η0| ≤ Cβkε ≤

C

β|x0|αε. (3.62)

Furthermore, the inequality

|ηx − ηy| ≤ C|x− y|αε, for any x, y ∈ Γu ∩B1

follows from (3.62).

4 On the regularity of the solution

In this section we study the regularity of the solution to the biharmonic obstacleproblem. Assuming that u ∈ B%

κ(ε), with ε > 0 small, we derive from Theorem3.12 that u ∈ C2,1

loc (B1). In the end we provide an example showing that withoutthe NTA domain assumption, there exist solutions, that are not C2,1.

4.1 C2,1-regularity of the solutions in B%κ(ε)

After showing the C1,α-regularity of the free boundary Γu ∩ B1, we may gofurther to derive improved regularity for the solution u ∈ B%

κ(ε).

Theorem 4.1. Let u ∈ B%κ(ε) be the solution to the biharmonic obstacle problem

in Ω ⊃⊃ B2, and let 0 < α < 1 be a fixed number. Then u ∈ C2,1(B1/4),provided ε = ε(κ, %, α) is small. Furthermore, the following estimate holds

‖u‖C3,α(Ωu∩B1/4) ≤ C(n)‖u‖W 2,2(B2) ≤ C(n)κ,

where C(n) is just a dimensional constant.

Proof. According to Theorem 3.12, Γu ∩ B1 is a graph of a C1,α-function. Weknow that ∆u ∈ W 1,2(B2) is a harmonic function in Ωu := u > 0, and alsou ∈ W 3,2(B2), u ≡ 0 in Ω \ Ωu, hence ∆u = 0 on Γu = ∂Ωu ∩ B2 in the

28

trace sense. Therefore we may apply Corollary 8.36 in [6], to conclude that∆u ∈ C1,α((Ωu ∪ Γu) ∩B3/4), and

‖∆u‖C1,α(Ωu∩B3/4) ≤ C(n)‖∆u‖L∞(B1). (4.1)

It follows from the Calderon-Zygmund estimates that u ∈ W 3,p(B1/2), forany p <∞. According to the Sobolev embedding theorem, u ∈ C2,α(B1/2), forall α < 1, with the following estimate

‖u‖C2,α(Ωu∩B1/2) ≤ C(n)(‖∆u‖C1,α(Ωu∩B3/4) + ‖u‖C1,α(B3/4)

). (4.2)

Denote by uij := ∂2u∂xi∂xj

. Then uij ∈ W 1,2(B1) ∩ Cα(Ωu ∩ B3/4) is a weak

solution of ∆uij =∂fj∂xi

in Ωu∩B3/4, where fj := ∂∆u∂xj∈ Cα(Ωu∩B3/4). Taking

into account that uij = 0 on ∂Ωu ∩ B1/2, we may apply Corollary 8.36 in [6]once again and conclude that

‖uij‖C1,α(Ωu∩B1/4) ≤ C(n)(‖uij‖C0(Ωu∩B1/2) + ‖∆u‖C1,α(Ωu∩B3/4)

),

hence

‖D2u‖C1,α(Ωu∩B1/4) ≤ C ′n(‖∆u‖C1,α(Ωu∩B3/4) + ‖u‖C1,α(B3/4)

),

according to (4.2).Therefore we obtain

‖u‖C3,α(Ωu∩B1/4) ≤ ‖u‖C1,α(B3/4) + ‖D2u‖C1,α(Ωu∩B1/4)

≤ C(n)(‖∆u‖C1,α(Ωu∩B3/4) + ‖u‖C1,α(B3/4)

).

Taking into account that

‖D3u‖L∞(B1/4) ≤ ‖D3u‖C0,α(Ωu∩B1/4),

we see that u ∈ C2,1(B1/4).

4.2 In general the solutions are not better that C2, 12

Let us observe that the assumption u ∈ B%κ(ε) is essential in the proof of u ∈

C2,1(Br). The next example shows that without our flatness assumptions thereexists a solution to the biharmonic obstacle problem in R2, that do not possessC2,1- regularity.

Example 4.2. Consider the following function given in polar coordinates inR2,

u(r, ϕ) = r52

(cos

ϕ

2− 1

5cos

5ϕ

2

), r ∈ [0, 1), ϕ ∈ [−π, π) (4.3)

then u ∈ C2, 12 is the solution to the biharmonic zero-obstacle problem in theunit ball B1 ⊂ R2.

29

Proof. It is easy to check that u ≥ 0, u(x) = 0 if and only if −1 ≤ x1 ≤ 0and x2 = 0. Hence the set Ωu = u > 0 is not an NTA domain, since thecomplement of Ωu does not satisfy the corkscrew condition.

Let us show that ∆2u is a nonnegative measure supported on [−1, 0]× 0.For any nonnegative f ∈ C∞0 (B1), we compute

ˆ

B1

∆u(x)∆f(x)dx =

ˆ 1

0

ˆ π

−πr∆f(r, ϕ)∆u(r, ϕ)dϕdr

= 6

ˆ 1

0

ˆ π

−πr

32 ∆f(r, ϕ) cos

ϕ

2dϕdr = 6

ˆ 1

0

r−12 f(r, π)dr ≥ 0,

where we used integration by parts, and that f is compactly supported in B1.We obtain that u solves the following variational inequality,

u ≥ 0, ∆2u ≥ 0, u ·∆2u = 0. (4.4)

Now we show that u is the unique minimizer to the following zero-obstacleproblem: minimize the functional (1.1) over

A := v ∈W 2,2(B1), v ≥ 0, s.t. v = u,∂v

∂n=∂u

∂n, on ∂B1 6= ∅.

According to Lemma 2.1 there exists a unique minimizer, let us call it v. Itfollows from (4.4), that

ˆ

B1

∆u∆(v − u) =

ˆ

B1

(v − u)∆2u =

ˆ

B1

v∆2u ≥ 0.

Hence

ˆ

B1

(∆u)2 ≤ˆ

B1

∆u∆v ≤(ˆ

B1

(∆u)2

) 12(ˆ

B1

(∆v)2

) 12

,

where we used the Holder inequality in the last step. Therefore we obtainˆ

B1

(∆u)2 ≤ˆ

B1

(∆v)2,

thus u ≡ v, and u solves the biharmonic zero-obstacle problem in the unit ball.However ∆u = 6r

12 cos ϕ2 , which implies that u is C2, 12 , and that the expo-

nent 12 is optimal, in particular u is not C2,1.

References

[1] John Andersson, Erik Lindgren, and Henrik Shahgholian. Almost every-where regularity for the free boundary of the normalized p-harmonic obstacleproblem.

[2] Luis A. Caffarelli and Avner Friedman. The obstacle problem for the bihar-monic operator. Ann. Scuola Norm. Sup. Pisa, Cl. Sci. (4) 6, no. 1:151–184.,1979.

30

[3] Lawrence C. Evans. Partial differential equations. Graduate Studies inMathematics, 19. American Mathematical Society, Providence, RI, 2010.xxii+749 pp. ISBN: 978-0-8218-4974-3.

[4] Jens Frehse. On the regularity of the solution of the biharmonic variationalinequality. Manuscripta Math. 9 (1973), 91–103.

[5] Avner Friedman. Variational principles and free-boundary problems. AWiley-Interscience Publication. Pure and Applied Mathematics. John Wiley& Sons, Inc., New York, 1982. ix+710 pp. ISBN: 0-471-86849-3.

[6] David Gilbarg and Neil S. Trudinger. Elliptic partial differential equationsof second order. Classics in Mathematics. Springer-Verlag, Berlin, 2001,Reprint of the 1998 edition.

[7] David S. Jerison and Carlos E. Kenig. Boundary behavior of harmonicfunctions in nontangentially accessible domains. Adv. in Math. 46, no. 1,80–147., 1982.

[8] N. S. Landkof. Foundations of modern potential theory. Springer-Verlag,1972. x+424 pp.

31

Paper C

87

Analysis of blow-ups for the double obstacle problem in

dimension two

Gohar Aleksanyan

November 1, 2016

Abstract

In this article we study a normalized double obstacle problem with polynomial obstaclesp1 ≤ p2 under the assumption p1(x) = p2(x) iff x = 0. In dimension n = 2 we give a completecharacterization of blow-up solutions depending on the coefficients of the polynomials p1, p2.We see that there exists a new type of blow-ups, that we call double-cone solutions sincethe noncoincidence set is a union of two cones with a common vertex.

The main object of investigation is the double obstacle problem, having rotational invari-ant double-cone blow-up solutions, which happens if p1(x) = −x21 −x22 and p2(x) = x21 +x22.Assuming that the solution is close to a double-cone solution in the unit ball B1 ⊂ R2, weprove that in a neighborhood of the origin the free boundary is a union of four C1,γ-graphs,pairwise crossing at the origin.

Contents

1 Introduction 11.1 Summary of the results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2 Weiss’ energy functional for the double obstacle problem 3

3 Characterization of blow-ups in R2 53.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63.2 Double-cone solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

4 Uniqueness of blow-ups, Case 1 14

References 24

1 Introduction

Let Ω be a bounded open set in Rn with smooth boundary. The solution to the double obstacleproblem in Ω is the minimizer of the functional

J(v) =

ˆ

Ω

|∇v(x)|2dx

over functions v ∈ W 1,2(Ω), ψ1 ≤ v ≤ ψ2, satisfying the boundary condition v = g on ∂Ω. Forthe problem to be well defined we assume that ψ1 ≤ ψ2 in Ω and ψ1 ≤ g ≤ ψ2 on ∂Ω. Thefunctions ψ1 and ψ2 are called respectively the lower and the upper obstacles.

1

If ψ1 < ψ2 then the problem reduces locally to a single obstacle problem. Therefore we areinterested in the case when

Λ := x ∈ Ω : ψ1(x) = ψ2(x) 6= ∅. (1.1)

It is well known that the solution to the double obstacle problem satisfies the followinginequalities

ψ1 ≤ u ≤ ψ2, ∆u ≥ 0 if u > ψ1 and ∆u ≤ 0 if u < ψ2. (1.2)

It has been shown that the solution to the double obstacle problem is locally C1,1 under theassumption ψi ∈ C2(Ω), see for instance [3, 5]. Therefore we may rewrite (1.2) as

ψ1 ≤ u ≤ ψ2 and ∆u = ∆ψ1χu=ψ1 + ∆ψ2χu=ψ2 −∆ψ1χψ1=ψ2 a.e., (1.3)

where χA is the characteristic function of a set A ⊂ Rn.Let us introduce some notations that will be used throughout. Denote by

Ω1 := u > ψ1, Ω2 := u < ψ2, and Ω12 := Ω1 ∩ Ω2 (1.4)

then Ω = Ω1 ∪ Ω2 ∪ Λ, where Λ is given by (1.1). Let us observe that u is a harmonic functionin Ω12, which we call the noncoincidence set. Define the free boundary for the double obstacleproblem

Γ := ∂Ω12 ∩ Ω ⊂ Γ1 ∪ Γ2, where Γi := ∂Ωi ∩ Ω, i = 1, 2. (1.5)

Let x0 ∈ Γ be a free boundary point, if x0 ∈ Γ1 \ Γ2, or if x0 ∈ Γ2 \ Γ1, then locally weare in the setting of the classical obstacle problem. In this case the known regularity theoryfor the classical obstacle problem (see [4]) can be applied to analyse the free boundary Γ in aneighborhood of x0. Hence we are more curious about the behaviour of the free boundary atthe points x0 ∈ Γ1 ∩ Γ2 = ∂Λ. In this article we focus on the case when x0 ∈ Γ1 ∩ Γ2 is anisolated point of Λ. The work is inspired by the following example of a homogeneous of degreetwo solution in R2,

u0(x) = x21sgn(x1) + x2

2sgn(x2), (1.6)

where the obstacles p1(x) = −p2(x) = −x21 − x2

2, and Λ = 0. Example (1.6) has also beenconsidered in [1], when investigating the optimal regularity in the optimal switching problem.The optimal switching problem and the double obstacle problem are related, and we see that inboth cases the solution shows a new type of behavior at isolated points of Λ. The function u0 isa motivational example for double-cone solutions, see Definition 3.4.

1.1 Summary of the results

We consider a normalized double obstacle problem in dimension n = 2, with polynomial obstaclesp1 ≤ p2;

∆u = λ1χu=p1 + λ2χu=p2, (1.7)

where λ1 = ∆p1 < 0 and λ2 = ∆p2 > 0 are constants. Furthermore, we assume that p1 and p2

meet at a single point, i.e. p1(x) = p2(x) iff x = x0.Without loss of generality, we may assume that

p1(x) = a1x21 + c1x

22 and p2(x) = a2x

21 + c2x

22, (1.8)

wherea1 + c1 < 0, a2 + c2 > 0, and a1 < a2, c1 < c2.

2

First, we give a complete characterization of blow-up solutions depending on the coefficientsof the polynomials p1, p2, see Theorem 3.7. Our main result is Theorem 4.8: Let u solve thedouble obstacle problem with obstacles p1(x) = −x2

1 − x22 and p2(x) = x2

1 + x22. Assuming

that u has a double-cone blow-up solution at the origin, we show that the blow-up is unique.Furthermore, we prove that in a neighborhood of the origin the free boundary is a union of fourC1,γ-graphs, pairwise crossing at the origin.

The paper is structured as follows. In Section 2 we study the normalized double obstacleproblem (1.7). We show that the blow-ups of the solution to the normalized double obstacle arehomogeneous of degree two functions via Weiss’ monotonicity formula.

Knowing that the blow-up solutions are homogeneous of degree two functions, in Section 3we make a complete characterization of possible blow-ups in dimension n = 2. In particularwe see that there exist homogeneous of degree two global solutions of a new type. We callthese solutions double-cone solutions, since the noncoincidence set is a union of two cones witha common vertex at the origin. We show that there exist double-cone solutions if and only if thefollowing polynomial

P = P (x1, x2) ≡ p1(x1, x2) + p2(x2, x1) = (a1 + c2)x21 + (a2 + c1)x2

2 (1.9)

has zeroes other that x = 0. If P ≡ 0, there are infinitely many double-cone solutions, and ifP 6≡ 0, but P = 0 on a line, there are finitely many double-cone solutions.

Here we discuss an important corollary of Theorem 3.7. Let ε be an arbitrary number,|ε| << 1. Then for polynomials p1(x) = −x2

1 − x22, p2(x) = x2

1 + x21 there exist infinitely many

double-cone solutions. While when we look at the double obstacle problem with p1 = −x21 − x2

2

and p2 = (1− ε)x21 + (1 + ε)x2

2 there are only four double-cone solutions, and for p1 = −x21 − x2

2

and p2 = (1 + ε)x21 + (1 + ε)x2

2 there are none. This property of the double obstacle problemis quite surprising and unexpected. It reveals the instability of the solutions in the sense thatchanging the obstacles slightly, may change the solution and the free boundary significantly.

In Section 4 we study the free boundary Γ for the normalized double obstacle problem withobstacles p1(x) = −x2

1 − x22, p2(x) = x2

1 + x22. By using a version of a flatness improvement

argument, we show that if the solution is close to a double-cone solution in B1, then the blow-up at the origin is unique. Furthermore, employing the known regularity theory for the freeboundary in the classical problem, we derive that the free boundary Γ for the double obstacleproblem is a union of four C1,γ-graphs meeting at the origin, see Theorem 4.8. Neither Γ1 norΓ2 is flat at the origin, and they meet at right angles, see Figure 4.1.

Acknowledgements

I would like to thank my advisor, John Andersson, for his guidance and support. I am also gratefulto Erik Lindgren and Henrik Shahgholian for reading a preliminary version of the manuscriptand for their valuable feedback.

2 Weiss’ energy functional for the double obstacle problem

In this section we study the behaviour of the solutions locally at free boundary points via Weiss’monotonicity formula.

Let u be a solution to the double problem in Ω, with obstacles

ψ1 ≤ ψ2, ψ1, ψ2 ∈ C2(Ω), Λ = ψ1 = ψ2 6= ∅. (2.1)

3

Fix any x0 ∈ Γ ∩ ∂Λ and assume that B1(x0) ⊂ Ω. Denote by

ur,x0:=

u(rx+ x0)− u(x0)− r∇u(x0) · xr2

, for all 0 < r < 1, and x0 ∈ Γ.

Without loss of generality, assume that x0 = 0 and B1 ⊂ Ω. Furthermore, by subtracting a firstorder polynomial from u, we may assume that u(0) = |∇u(0)| = 0. Recalling that u ∈ C1,1

loc , weobtain ψ1(0) = ψ2(0) = u(0) = 0 and |∇ψ1(0)| = |∇ψ2(0)| = |∇u(0)| = 0. Denote by

ur(x) := ur,0 =u(rx)

r2. (2.2)

It follows from equation (1.2) and assumption (2.1), that λ1 = ∆ψ1(0) ≤ 0 and λ2 =∆ψ2(0) ≥ 0. In particular, if 0 ∈ ∂Λ, then λ1 = λ2 = 0.

Lemma 2.1. Consider the following normalized double obstacle problem

∆u = λ1χu=ψ1 + λ2χu=ψ2, in B1 (2.3)

where ψi ∈ C2(B1), and assume that

λ1 := ∆ψ1 ≤ 0 and λ2 := ∆ψ2 ≥ 0 are constants. (2.4)

Define Weiss’ energy functional for the function u and 0 < r ≤ 1 at the origin as follows

W (u, r, 0) :=1

rn+2

ˆ

Br

|∇u|2dx− 2

rn+3

ˆ

∂Br

u2dHn−1

+1

rn+2

ˆ

Br

2λ1uχu=ψ1 + 2λ2uχu=ψ2dx.(2.5)

Thend

drW (u, r, 0) = 2r

ˆ

∂B1

(durdr

)2

dHn−1 ≥ 0. (2.6)

Proof. After a change of variable in (2.5) we obtain the following scaling property for Weiss’energy functional

W (u, r, 0) = W (ur, 1, 0) =

ˆ

B1

|∇ur|2dx− 2

ˆ

∂B1

u2rdHn−1

+

ˆ

B1

2λ1urχur=ψ1r + 2λ2urχur=ψ2

rdx.(2.7)

Hence

d

drW (u, r, 0) =

d

drW (ur, 1, 0) =

ˆ

B1

d

dr|∇ur|2dx− 2

ˆ

∂B1

du2r

drdHn−1

+2

ˆ

B1

(λ1χur=ψ1

r + λ2χur=ψ2r) durdr

dx = 2

ˆ

B1

∇durdr∇urdx

−4

ˆ

∂B1

durdr

urdHn−1 + 2

ˆ

B1

(λ1χur=ψ1

r + λ2χur=ψ2r) durdr

dx.

By Green’s formulaˆ

B1

∇ur∇durdr

dx = −ˆ

B1

durdr

∆urdx+

ˆ

∂B1

durdr

∂ur∂ν

dHn−1.

4

Therefored

drW (u, r, 0) = 2

ˆ

∂B1

durdr

(∂ur∂ν− 2ur

)dHn−1

+2

ˆ

B1

durdr

(−∆ur + λ1χur=ψ1

r + λ2χur=ψ2r)dx.

(2.8)

Since u solves (2.3), equation 2.8 can be abbreviated to

d

drW (u, r, 0) = 2

ˆ

∂B1

durdr

(∂ur∂ν− 2ur

)dHn−1. (2.9)

Let us observe thatˆ

∂B1

durdr

(∂ur∂ν− 2ur

)dHn−1 =

ˆ

∂B1

durdr

(x · ∇ur − 2ur) dHn−1

=

ˆ

∂B1

r

(durdr

)2

dHn−1.

(2.10)

Equations (2.10) and (2.9) together imply the desired identity, (2.6).

3 Characterization of blow-ups in R2

Given second degree polynomials p1 ≤ p2 , satisfying (2.4), let u be the solution to the normalizeddouble obstacle problem (2.3) with p1, p2. Let 0 ∈ Γ1 ∩ Γ2 be a free boundary point. Bysubtracting a first order polynomial from p1, p2 and u, and recalling that u ∈ C1,1, we mayassume

u(0) = p1(0) = p2(0) = 0 and |∇u(0)| = |∇p1(0)| = |∇p2(0)| = 0. (3.1)

Hence p1 and p2 are homogeneous second degree polynomials.It follows from Lemma 2.1 that W (u, r, 0) is a nondecreasing absolutely continuous function

in the interval (0, 1). Hence there exists

limr→0

W (u, r, 0) := W (u, 0+, 0). (3.2)

Since u ∈ C1,1loc , we may conclude that ‖ur‖C1,1 is uniformly bounded for small r > 0.

Therefore through a subsequence ur converges in C1,α(B1). Let u0 be a blow-up of u at theorigin;

u(rjx)

r2j

→ u0 in C1,α(B1), (3.3)

for a sequence rj → 0+, as j →∞. Then (3.3) implies that for any fixed 0 < r < 1

W (u0, r, 0) = limj→∞

W (urj , r, 0)(2.7)= lim

j→∞W (u, rrj , 0)

(3.2)= W (u, 0+, 0).

Thus W (u0, r, 0) has a constant value for all 0 < r < 1, and ddrW (u0, r, 0) = 0. Note that u0 is

a global solution, i.e. solution in Rn to the double obstacle problem with the same obstacles, p1

and p2. Applying Lemma 2.1 for the solution u0, we may conclude from (2.6), that

d

dr

(u0(rx)

r2

)= 0, for any r > 0.

5

Hence u0 is a homogeneous of degree two function, which means that

u0(x) =u0(rx)

r2, for any x ∈ Rn and r > 0.

It follows that ∆u0(rx) = ∆u0(x), for any x ∈ Rn and r > 0. In other words ∆u0 is identicallyconstant on the lines passing through the origin, and therefore the free boundary is lying onstraight lines passing through the origin.

3.1 Examples

We study motivational examples of homogeneous of degree two global solutions in R2, assumingthat Λ = 0. In this section instead of our usual notation x = (x1, x2) ∈ R2, the pair (x, y)represents a point in R2. It is done to make the picture clearer, and we hope it will not beconfusing later on.

We say that u is a polynomial solution to the double obstacle problem (2.3), if u ≡ p1, u ≡ p2

or if u is a homegeneous of degree two harmonic polynomial in R2, such that p1 ≤ u ≤ p2. Letus also recall that u is called a halfspace solution, if up to a rotation of the coordinate system,∆u = ∆p1χy>0 or ∆u = ∆p2χy>0.

Example 3.1. Let us study some explicit homogeneous solutions of degree two to the doubleobstacle problem in R2, with obstacles p1(x, y) = −x2 − y2, p2(x, y) = x2 + y2.

Observe that u0 = −x2 + sgn(y)y2 and u0 = sgn(x)x2 + y2 are global solutions;

x

y

Γ2

Γ1

u0 = −x2 + y2u0 = −x2 + y2

u0 = −x2 − y2

x

y

Γ2

Γ1

u0 = x2 + y2

u0 = −x2 + y2

u0 = −x2 + y2

Figure 3.1: Examples of halfspace solutions.

Now let us look at the following two explicit solutions, which obviously are not halfspace solutions.

6

x

y

Γ2

Γ1

u0 = x2 + y2

u0 = −x2 − y2 u0 = x2 − y2

u0 = −x2 + y2

x

yy = x

Γ2

y = −x

y = −x

y = x

Γ1

u0 = x2 + y2u0 = −x2 − y2

u0 = 2xy

u0 = −2xy

Figure 3.2: New, interesting type of solutions.

We see that Γ = Γ1∪Γ2 consists of two lines meeting at right angles, and Γ1∩Γ2 = 0 = Λ.Actually there are many more solutions, for example consider the following global solutions

x

y

θ = π2

y = 2xy = −2x

Γ1

y = − x2

y = x2

Γ2

u0 = −x2 − y2

u0 = x2 + y2

u0 =−3x2+8xy+3y2

5u0 =

−3x2−8xy+3y2

5

x

y

θ = π2

y = 2xy = −2x

Γ2

y = − x2

Γ1

y = x2

u0 = x2 + y2

u0 = −x2 − y2

u0 =−3x2+8xy+3y2

5u0 =

−3x2−8xy+3y2

5

Figure 3.3: In this example we see that the cone u0 = p1 ( u0 = p2) does not have a fixed opening angle.Actually the opening angle can take any value in the closed interval [0, π].

We see that in all the examples discussed above there is one common property: in the halfplanex ≥ 0 the lines Γ1 and Γ2 intersect at a right angle, later on we will provide a rigorous argumentfor this.

Let us study two more examples, where the free boundary shows a different behavior.

Example 3.2. Let p1(x, y) = −x2 − y2 and p2(x, y) = 2x2 + 2y2. Assume that u0 is a homoge-neous of degree two solution to the double obstacle problem with obstacles p1 and p2 in R2, thenΓ2 = 0. In this case the only possible homogeneous of degree two solutions are the second orderharmonic polynomials, half-space solutions, and the polynomials p1, p2 themselves.

It is easy to verify that there is no second order harmonic polynomial in R2, satisfyingp1 ≤ q ≤ p2 and such that the polynomials p2− q and q− p1 both have roots of multiplicity two.

7

In this case we have only solutions satisfying ∆u = λ1χ(x,y)·e>0, where e is a unit vector in R2.

x

y

y = x

Γ1

u0 = −x2 − y2

u0 = −2xy

x

yy = −x

Γ1

u0 = −x2 − y2

u0 = 2xy

Figure 3.4: Examples of halfspace solutions.

Example 3.3. The following functions are global solutions to the double obstacle problem withp1(x, y) = −x2 − y2 and p2(x, y) = 2x2.

x

y Γ1y = −√

3x

Γ2

y =√

3x

u0 = −x2 − y2

u0 =x2−2

√3xy−y22

u0 =x2+2

√3xy−y22

u0 = 2x2

x

y Γ1

y = −√

3xΓ2

y =√

3x

u0 = −x2 − y2

u0 =x2−2

√3xy−y22

u0 =x2+2

√3xy−y22

u0 = 2x2

Figure 3.5: The noncoincidence set is a cone with an opening angle 2π/3 or π/3.

3.2 Double-cone solutions

Let p1 ≤ p2 be given polynomials,

pi(x) ≡ aix21 + 2bix1x2 + cix

22, for i = 1, 2, (3.4)

Consider the following normalized double obstacle problem in R2 with obstacles p1, p2;

p1 ≤ u ≤ p2, ∆u = λ1χu=p1 + λ2χu=p2, (3.5)

whereλ1 := ∆p1 = 2(a1 + c1) < 0 and λ2 := ∆p2 = 2(a2 + c2) > 0. (3.6)

8

We saw in Example 3.1 and Example 3.3 that for the double obstacle problem there existglobal solutions for which the noncoincidence set consists of two halfcones with a common vertexat the origin.

Definition 3.4. Let u solve the double obstacle problem (2.3). We say that u is a double-conesolution, if the noncoincidence set Ω12 = p1 < u < p2 consists of two halfcones S1 and S2,having a common vertex.

Remark 3.5. In Example 3.1 the first two solutions are both halfspace and double-cone solutions,since S1 and S2 have a common edge.

Our aim is to describe the possible blow-ups for a solution to the double obstacle problem inR2. In particular, we are interested to study the case when the double-cone solutions do exist.It is easy to verify that if λ1 = 0 or λ2 = 0, there are no double-cone solutions, explaining ourassumption (3.6).

A simple calculation shows that if p1 = p2 on a line, then there are no double-cone solutions.Hence we assume that p1 and p2 meet only at the origin, in other words the matrix D2(p2 − p1)is positive definite.

Without loss of generality we may assume that b1 = b2 = b in (3.4). Otherwise, if b2−b1 6= 0,we can rotate the coordinate system with an angle θ, cos 2θ

sin 2θ = a2−a1−c2+c12(b1−b2) , and obtain b1 = b2 in

the new system. Furthermore, we may subtract a harmonic polynomial h(x) = 2bx1x2 from p1,p2 and u, then consider instead the polynomials p1−h and p2−h, thus obtaining b = 0. Insteadof u, we are studying the solution u− h, but still call it u.

We saw that it is enough to study the blow-up solutions of the double obstacle problem withobstacles having the form

p1(x) = a1x21 + c1x

22 and p2(x) = a2x

21 + c2x

22. (3.7)

According to our assumption, the matrix A := D2(p2 − p1) is positive definite, hence

a2 > a1, c2 > c1. (3.8)

and by (3.6),a1 + c1 < 0, and a2 + c2 > 0. (3.9)

The following lemma is the main step to the investigation of double-cone solutions in R2.

Lemma 3.6. Let p1(x) = a1x21 + c1x

22 and p2(x) = a2x

21 + c2x

22 be given polynomials, satisfying

(3.8) and (3.9). Assume that there exists a pair (q,S), where S is an open sector in R2, withthe edges lying on the lines x2 = mx1 and x2 = kx1, and q is a harmonic homogeneous of degreetwo function in S. Moreover, assume that

p1 ≤ q ≤ p2 in S, (3.10)

and the following boundary conditions hold;

q − p1 = 0,∇(q − p1) = 0 on x2 = mx1. (3.11)

andq − p2 = 0,∇(q − p2) = 0 on x2 = kx1. (3.12)

Then, q = αx21 + 2βx1x2 − αx2

2, where α and β are real numbers solving

β2 = −(α− a1)(α+ c1) = −(α− a2)(α+ c2) ≥ 0, (3.13)

9

max(a1,−c2) ≤ α ≤ min(a2,−c1), (3.14)

andα(c1 − a1 − c2 + a2) = a1c1 − a2c2. (3.15)

The numbers m and k are given by

m =β

α+ c1and k =

β

c2 + α. (3.16)

Furthermore, the coefficients of p1 and p2 satisfy the following inequality

(a1 + c2)(c1 + a2) ≤ 0. (3.17)

Proof. Let us note that harmonic homogeneous of degree two functions in a sector are seconddegree polynomials of the form q(x) = αx2

1 + 2βx1x2 − αx22, where α and β are real numbers.

By assumption (3.10),

q − p1 = (α− a1)x21 + 2βx1x2 − (α+ c1)x2

2 ≥ 0, and

p2 − q = (a2 − α)x21 − 2βx1x2 + (c2 + α)x2

2 ≥ 0 in S.

Denote by t = x2

x1, and observe that (3.11) implies that the following quadratic polynomial

q − p1

x21

= −(α+ c1)t2 + 2βt+ α− a1

has a multiple root at the point t = m. By an elementary calculation we obtain

β2 = −(α− a1)(α+ c1), andq − p1

x21

= −(α+ c1)(t−m)2.

Hence the inequality q − p1 ≥ 0 in S implies q − p1 ≥ 0 in R2. Therefore we may conclude that

−α− c1 ≥ 0, α− a1 ≥ 0, β2 = −(α− a1)(α+ c1), and m =β

α+ c1. (3.18)

Similarly, (3.12) implies that the following quadratic polynomial

p2 − qx2

1

= (c2 + α)t2 +−2βt+ a2 − α

has a multiple root at the point t = k. Hence β2 = (a2−α)(c2 +α), and the inequality p2−q ≥ 0in S implies p2 − q ≥ 0 in R2. Therefore, by a similar argument as the one leading to (3.18), weget

c2 + α ≥ 0, a2 − α ≥ 0, β2 = (a2 − α)(c2 + α), and k =β

c2 + α. (3.19)

Let us also observe that if α = −c1, then p1(x) = q(x) implies x1 = 0, similarly, if α = −c2, thenp2(x) = q(x) implies x1 = 0. Hence (3.16) makes sense even if α = −c1 or α = −c2.

Assuming that there exists (q,S) satisfying (3.10),(3.11),(3.12), we derived (3.18) and (3.19),which in particular imply (3.13), (3.14) and (3.16). It follows from (3.13), that

α2 − a1α+ c1α− a1c1 = α2 − a2α+ c2α− a2c2,

10

hence α solves equation (3.15). As we see equation (3.15) is contained in (3.13), we stated (3.15)only for the future references.

It remains to prove the inequality (3.17), which is a necessary condition for the existence ofα, β, thus for (q,S). We discuss two cases. i) If

c1 − a1 − c2 + a2 = 0, (3.20)

it follows from equation (3.15) thata2c2 = a1c1. (3.21)

If a1 = 0, then a2 < 0 by (3.8), therefore c2 = 0, and (3.17) holds. Otherwise, if a1 6= 0, leta2 = la1, then l 6= 1 by (3.8). Hence c1 = lc2 according to (3.21). Now (3.20) implies that(l − 1)(a1 + c2) = 0, since l 6= 1, we obtain a2 + c1 = 0, and (3.17) holds.

ii) If c1 − a1 − c2 + a2 6= 0, then equation (3.15) implies that

α =a2c2 − a1c1

c2 + a1 − a2 − c1. (3.22)

By a direct computation we see that

α− a1 =(a2 − a1)(a1 + c2)

c2 + a1 − a2 − c1, and α+ c1 =

(c2 − c1)(a2 + c1)

c2 + a1 − a2 − c1,

by (3.13)

β2 = − (a2 − a1)(c2 − c1)(a1 + c2)(a2 + c1)

(c2 + a1 − a2 − c1)2≥ 0.

Taking into account (3.8), we obtain the desired inequality, (3.17).

Let us observe that if u0 is a double-cone solution (Definition 3.4), then there exist (q1,S1)and (q2,S2) as in Lemma 3.6, such that S1 ∩ S2 = ∅, and

u0 = q1 in S1 and u0 = q2 in S2. (3.23)

According to Lemma 3.6, inequality (3.17) is a necessary condition for the existence of double-cone solutions, in the next theorem we will see that (3.17) is also a sufficient condition.

Theorem 3.7. Let u0 be a homogeneous of degree two global solution to the double obstacleproblem with obstacles

p1(x) = a1x21 + c1x

22 and p2(x) = a2x

21 + c2x

22,

satisfying (3.9) and (3.8). If u0 is neither a polynomial nor a halfspace solution, then it is adouble-cone solution.

Case 1) If a2 + c1 = c2 + a1 = 0, then there are infinitely many double-cone solutions. Each of thecones S1 and S2 in (3.23) has an opening angle ϑ = π/2.

Case 2) If (a1 + c2)(c1 + a2) ≤ 0, and a1 + c2 6= a2 + c1, then there exist at most four double-conesolutions, with an opening angle ϑ, satisfying

cos2 ϑ =(a1 + c2)(a2 + c1)

(a1 + c1)(a2 + c2)∈ [0, 1). (3.24)

If (a1+c2)(c1+a2) < 0, there are four double-cone solutions, otherwise if (a1+c2)(c1+a2) =0, and a1 + c2 6= a2 + c1, there are only two.

11

Case 3) If (a1 + c2)(c1 + a2) > 0, then there are no double-cone solutions.

Proof. If u0 is neither a polynomial nor a halfspace solution, then there exists a pair (q,S), suchthat

u0 = q in S, u0 = p1 on x2 = mx1, u0 = p2 on x2 = kx1 (3.25)

where S is a sector in R2, with edges lying on the lines x2 = mx1 and x2 = kx1, and q is aharmonic homogeneous degree two function in S, satisfying (3.10). Moreover, since u0 ∈ C1,1,we obtain ∇q = ∇p1 on x2 = mx1 and ∇q = ∇p2 on x2 = kx1. Hence q takes boundaryconditions (3.11) and (3.12) on ∂S ⊂ x2 = mx1∪x2 = kx1, and therefore (q,S) satisfies theassumptions in Lemma 3.6.

According to Lemma 3.6, q = αx21 + 2βx1x2 − αx2

2, where α and β are real numbers solving(3.13) and (3.14). The numbers m and k, describing the sector S, are given by (3.16).

We are looking for all possible pairs (q,S) in terms of the parameter α. Given α, satisfying(3.14) and (3.15), we can find ±β from equation (3.13). By equation (3.16) we can identify thecorresponding sectors S.

Let us split the discussion into several cases in order to study the existence of solutions tothe equation (3.15) in variable α, satisfying inequality (3.14).

Case 1) If a2+c1 = c2+a1 = 0, as in Example 3.1. Then obviously equation (3.15) becomesan identity. Hence in this case α can be any number satisfying (3.14), that is a1 ≤ α ≤ a2. Ifα = a1, then β = 0 in view of (3.13), and according to (3.16), Γ1 = x2 = 0. In this caseΓ2 = x1 = 0, x2 ≥ 0 or Γ2 = x1 = 0, x2 ≤ 0 . Analogously if α = a2 then Γ2 = x2 = 0 andΓ1 = x1 = 0, x2 ≥ 0 or Γ2 = x1 = 0, x2 ≤ 0. Hence we obtain double-cone solutions thatare also halfspace solutions, see Figure 3.1 with α = −1 = a1 and α = 1 = a2.

Now let us fix any a1 < α < a2. It follows from (3.18) and (3.19) that

β± = ±√

(α− a1)(a2 − α),

and

m± = ∓√α− a1

a2 − α, k± = ±

√a2 − αα− a1

. (3.26)

Let us note that m±k± = −1, and therefore the lines x2 = m±x1 and x2 = k±x1 are perpendic-ular. Thus for a fixed a1 < α < a2 we obtain two polynomials

q+ := αx21 + 2β+x1x2 − αx2

2 and q− := αx21 + 2β−x1x2 − αx2

2. (3.27)

Where q+ = p1 if x2 = m+x1, q+ = p2 if x2 = k+x1, and q− = p1 if x2 = m−x1, q− = p2

if x2 = k−x1. Hence for a fixed α there are two pairs (q+,S1) and (q−,S2) forming a singledouble-cone solution u0. There are four different choices of disjoint sectors S1 and S2, satisfying(3.26). Therefore we obtain four different double-cone solutions for a fixed a1 < α < a2. Figure3.3 illustrates two of them for α = − 3

5 .In fact we obtain more double-cone solutions by ”merging” two double-cone solutions corre-

sponding to two different values of α. Consider the following example u0 = x21sgn(x1)+x2

2sgn(x2)(see the left picture in Figure 3.2). Then q1 = −x2

1 + x22 with α1 = −1 and q2 = x2

1 − x22 with

α2 = 1.Fix any a1 ≤ α1 6= α2 ≤ a2, then there are four double-cone solutions corresponding to each

of αi. From these double-cone solutions we obtain eight more double-cone solutions, such thatS1 ∩ S2 = ∅, where qi, i = 1, 2 can be either q+ or q− corresponding to αi. The solution u0 canbe described graphycally as follows:

12

x1

x2

Γ1

x2 = m1x1

x2 = k1x1

x2 = m2x1

Γ2

x2 = k2x1

u0 = q1

u0 = p1

u0 = p2

u0 = q2

S1

S2

This is a general example of a double-cone solution (3.23), where the polynomial q1, and thenumbers m1, k1 correspond to α1, similarly q2 and m2, k2 correspond to α2. We conclude that, ifu0 is a double-cone solution then the cone u0 = p1 may have any opening angle θ, 0 ≤ θ ≤ π,and the cone u0 = p2 has an angle π−θ. If θ = 0 or θ = π, then u0 is also a halfspace solution.

Finally, note that there are no homogeneous of degree two solutions u0 corresponding to threeor more different values of α, since u0 can have only an even number of (q,S), and S always hasan opening angle π/2.

Case 2) If (a1 + c2)(a2 + c1) ≤ 0, and c1 − a1 − c2 + a2 6= 0, then equation (3.15) has aunique solution,

α =a2c2 − a1c1

c2 + a1 − a2 − c1. (3.28)

From inequality (a1 + c2)(a2 + c1) ≤ 0 it easily follows that

max(a1,−c2) ≤ α ≤ min(a2,−c1). (3.29)

Referring to (3.13), we can calculate

β± = ±√

(α+ c1)(a1 − α) = ±√−(a2 − a1)(c2 − c1)(a2 + c1)(a1 + c2)

c2 + a1 − a2 − c1.

According to (3.16),

m± =β1,2

α+ c1= ∓

√− (c2 + a1)(a2 − a1)

(a2 + c1)(c2 − c1), k± =

β1,2

α+ c2= ±

√− (c1 + a2)(a2 − a1)

(a1 + c2)(c2 − c1). (3.30)

Hence we obtain two harmonic polynomials q+ and q− and four combinations of disjoint S1 andS2. Since in this case α is a fixed number, given by (3.28), there are only four double-conesolutions.

Denote by ϑi the opening angle of the cone Si, then it follows from (3.30) that

cosϑi = ± 1 + k+m+√1 + (k+)2

√1 + (m+)2

= ± 1 + k−m−√1 + (k−)2

√1 + (m−)2

=

= ±c2−c1−a2+a1

c2−c1√(a1+c1)(c2−a2+a1−c1)

(a2+c1)(c2−c1)

√(a2+c2)(c2−a2+a1−c1)

(a1+c2)(c2−c1)

= ±√

(a1 + c2)(a2 + c1)

(a1 + c1)(a2 + c2), for i = 1, 2.

13

In Example 3.3, a1 = c1 = −1, a2 = 2, c2 = 0, by a direct calculation we see that α = 12 ,

β = ±√

32 , and ϑ = π

3 or ϑ = 2π3 .

Case 3) is just an obvious corollary of inequality (3.17) in Lemma 3.6.

Let us rephrase Theorem 3.7 in a more compact algebraic form.

Corollary 3.8. Let pi = aix21 + cix

22 be given polynomials, satisfying (3.9) and (3.8). There

exist double-cone solutions for the double obstacle problem with p1, p2, if and only if the followingpolynomial

P (x) = P (x1, x2) ≡ p1(x1, x2) + p2(x2, x1) = (a1 + c2)x21 + (c1 + a2)x2

2 (3.31)

has zeroes other than x = 0. In particular, if P ≡ 0, there are infinitely many double-conesolutions. If P 6≡ 0, but P = 0 on a line, then there are finitely many double-cone solutions.

In other words, there exist double-cone solutions if and only if the matrix D2P is neitherpositive nor negative definite.

4 Uniqueness of blow-ups, Case 1

Let u be the solution to the double obstacle problem (3.5), with polynomial obstacles p1 ≤ p2,satisfying p1(x) = p2(x) iff x = 0. We study the blow-ups of u in Case 1, i.e. when thepolynomials pi are given by

p1(x) = −ax21 − cx2

2, p2(x) = cx21 + ax2

2, where a+ c > 0.

Consider the following harmonic polynomial h(x) := −a+c2 x2

1 + a−c2 x2

2, then

p1(x)− h(x) =a+ c

2(−x2

1 − x22), p2(x)− h(x) =

a+ c

2(x2

1 + x22).

Thus it is enough to study the uniqueness of blow-ups in the case

p1(x) = −x21 − x2

2, and p2(x) = x21 + x2

2. (4.1)

From now on we study the solution 2(u−h)a+c instead of u, but still call it u.

Let rj → 0+, as j →∞, and

u0(x) := limj→∞

u(rjx)

r2j

be a blow-up of u at the origin. We know that there exists

limr→0

W (u, r, 0) = limj→∞

W (urj , 1, 0) ≡W (u0, 1, 0).

Hence if u0 is another blow-up solution, then W (u0, 1, 0) = W (u0, 1, 0). Denote by

Ci := x = (x1, x2) ∈ R2;u0(x) = pi(x), (4.2)

where pi are the polynomials in (4.1).

14

Let us calculate the values of W (u0, 1, 0) for all the possible blow-up solutions u0. By defini-tion

W (u0, 1, 0) =

ˆ

B1

|∇u0|2dx− 2

ˆ

∂B1

u20dS + 2

ˆ

B1

λ1u0χu0=p1 + λ2u0χu0=p2dx

= −ˆ

B1

u0∆u0dx+ 2

ˆ

B1

λ1u0χu0=p1 + λ2u0χu0=p2dx

= λ1

ˆ

B1∩C1p1dx+ λ2

ˆ

B1∩C2p2dx.

(4.3)

After substituting λ1 = −4 and λ2 = 4 in (4.3), we obtain

W (u0, 1, 0) = 4

ˆ

B1∩C1r3drdθ + 4

ˆ

B1∩C2r3drdθ,

and we may conclude from Theorem 3.7 that

W (u0, 1, 0) =

0, if u0 is a harmonic second order polynomial

2π, if u0 ≡ p1 or u0 ≡ p2

π, if u0 is a halfspace or a double-cone solution.

(4.4)

This gives three types of possible blow-ups at a fixed free boundary point.Denote by

uj(x) :=u(rjx)

r2j

, (4.5)

and assume thatuj → u0 in C1,γ(B1). (4.6)

If u0 is a polynomial or a halfspace solution, then we are in a similar setting as in the obstacleproblem with a single obstacle. In this article we study only the case when u0 is a double-conesolution. According to Theorem 3.7, u0 can be described in terms of parameters −1 < α1, α2 < 1.Let αi = cosφi, for some 0 < φ1, φ2 < π. According to Lemma 3.10, βi = ±

√1− α2

i = ± sinφi,

mi = βicosφi−1 = ∓ tan φi

2 and ki = βicosφi+1 = ± cot φi2 . Referring to (3.27), we see that

qi(r, θ) = x21 cosφi − x2

2 cosφi ± 2x1x2 sinφi = r2(cosφi cos2 θ − cosφi sin2 θ ± sin 2θ sinφ1)

= r2(cosφi cos 2θ ± sin 2θ sinφi) = r2 cos(2θ ∓ φi).

Hence without loss of generality u0 is the following function

u0 = µ = µφ1,φ2(r, θ) :=

r2, if − φ2 ≤ 2θ ≤ φ1

r2 cos(2θ − φ1), if φ1 ≤ 2θ ≤ π + φ1

r2 cos(2θ + φ2), if − π − φ2 ≤ 2θ ≤ −φ2

−r2, otherwise.

(4.7)

For further analysis we need the following easy lemma.

Lemma 4.1. Let u and u0 be two solutions (with different boundary conditions) to the doubleobstacle problem in B1 ⊂ Rn, with given obstacles ψ1 ≤ ψ2. Then

‖u− u0‖W 1,2(B1/2) ≤ Cn‖u− u0‖L2(B1), (4.8)

where Cn is just a dimensional constant.

15

Proof. The proof is quite standard. Given a solution u to the double obstacle problem in B1,then for any ζ ∈ C2

0 (B1), the function ut(x) := u + tζ2(u0 − u) is admissible for t > 0 smallenough depending only on ζ. Hence

ˆ

B1

|∇u|2dx ≤ˆ

B1

|∇ut|2dx =

ˆ

B1

|∇u|2dx+ 2t

ˆ

B1

∇u · ∇(ζ2(u0 − u)

)dx

+t2ˆ

B1

∣∣∇(ζ2(u0 − u)

)∣∣2 dx,

after dividing the last inequality by t > 0, and taking the limit as t goes to zero, we obtain

0 ≤ˆ

B1

∇u · ∇(ζ2(u0 − u)

)dx =

ˆ

B1

(u0 − u)∇u · ∇ζ2dx+

ˆ

B1

ζ2∇u · ∇(u0 − u)dx. (4.9)

Similarly, the function u0 + tζ2(u − u0) is admissible for the double obstacle problem, havingsolution u0. Therefore

0 ≤ˆ

B1

∇u0 · ∇(ζ2(u− u0)

)dx =

ˆ

B1

(u− u0)∇u0 · ∇ζ2dx+

ˆ

B1

ζ2∇u0 · ∇(u− u0)dx. (4.10)

Choose ζ ∈ C20 (B3/4), such that 0 ≤ ζ ≤ 1 and ζ ≡ 1 in B1/2. Combining the inequalities (4.9)

and (4.10), we obtain

ˆ

B1

ζ2|∇u−∇u0|2dx ≤ −2

ˆ

B1

ζ(u− u0)(∇u−∇u0) · ∇ζdx

≤ 2

ˆ

B1

|∇ζ|2(u− u0)2dx+1

2

ˆ

B1

ζ2|∇u−∇u0|2dx,

where we used Young’s inequality in the last step. Hence

ˆ

B1/2

|∇u−∇u0|2dx ≤ˆ

B1

ζ2|∇u−∇u0|2dx ≤ 4

ˆ

B1

|∇ζ|2(u− u0)2dx ≤ Cn‖u− u0‖2L2(B1),

where Cn is just a dimensional constant, depending only on ζ. The proof of the lemma is nowcomplete.

Definition 4.2. Let u be a solution to the double obstacle problem. We say that u0 = µφ1,φ2is

a minimal double-cone solution with respect to u, if

‖u− µφ1,φ2‖L2(B1) ≤ ‖u− µφ1+τ,φ2+δ‖L2(B1), (4.11)

for any τ, δ, such that |τ | < π − φ1 and |δ| < π − φ2.

It follows from Definition 4.2, that if µφ1,φ2is a minimal double-cone solution with respect

to u, then

ˆ π/2+φi/2

φi/2

ˆ 1

0

sin(φi − 2θ) (u(r, θ)− µφ1,φ2(r, θ)) rdrdθ = 0, for i = 1, 2. (4.12)

We derive equation (4.12) by taking the partial derivatives of ‖u− µφ1+τ,φ2+δ‖L2(B1) at theorigin with respect to variables τ and δ.

16

Proposition 4.3. Let uj be a sequence of solutions to the double obstacle problem with obstaclesp1(x) = −x2

1 − x22 and p1(x) = x2

1 + x22 in Ω ⊂ R2, B2 ⊂⊂ Ω. Assume that (4.6) holds, where

u0 = µ is given by (4.7). Denote by

vj(x) :=uj(x)− µj(x)

‖uj − µj‖L2(B2), (4.13)

where µj is a minimal double-cone solution with respect to uj. Then up to a subsequence

vj v0 weakly in W 1,2(B1), and vj → v0 in L2(B1). (4.14)

Where v0 ≡ 0 in C1 ∪C2 (see (4.2)), and v0 is a harmonic function in each of the components ofthe noncoincidence set Ω12 = S1 ∪ S2. Furthermore, it follows from the minimality assumptionthat

ˆ 1

0

ˆ π/2+φ1/2

φ1/2

v0(r, θ) sin(2θ − φi)dθrdr = 0 for i = 1, 2. (4.15)

Proof. According to Lemma 4.1, ‖vj‖W 1,2(B1) ≤ Cn, where Cn is a dimensional constant. Hence(4.14) follows from the weak compactness of the space W 1,2 and from the Sobolev embeddingtheorem.

We want to show that for any K ⊂⊂ C1 ∩B1, the functions vj vanish in K for large j, sincethen we may conclude that v0 ≡ 0 in C1. Let K ⊂⊂ V ⊂⊂ C1, and d := dist(K, ∂V ). It followsfrom (4.6) that for any ε > 0 there exists j(ε), such that |uj(x) − p1(x)| ≤ ε, for any x ∈ K,

provided j ≥ j(ε) is large enough, depending only on ε. Take 0 < ε < d2

4 . Let us denote bywj := uj − p1, then 0 ≤ wj ≤ ε solves the following normalized obstacle problem with zeroobstacle

∆wj = −λ1χwj>0 = 4χwj>0 in K.

Fix x0 ∈ K, if wj(x0) > 0, then we can apply the maximum growth lemma (Lemma 5 in [4]) forthe solution to the classical obstacle problem, and obtain

d2

4> ε ≥ sup

Bd(x0)

wj ≥ d2

2, (4.16)

which is not possible, therefore wj(x0) = 0. Hence we may conclude that for j >> 1 large, vj isvanishing in K, for any K ⊂⊂ C1 ∩B1.

We obtain (4.15) by passing to the limit as j →∞ in equation (4.12) applied for the solutionsuj .

Lemma 4.4. Let v0 be the function in Proposition 4.3, then

‖v0(sx)‖L2(B1) ≤ s4‖v0‖L2(B1), (4.17)

for any 0 < s < 1.

Proof. According to Proposition 4.3, v0 is a harmonic function in the sector

S1 ∩B1 = φ1 ≤ 2θ ≤ φ1 + π ∩B1,

and v0 satisfies the following boundary conditions in the trace sense;

v0(r, φ1/2) = v0(r, φ1/2 + π/2) = 0, for all 0 < r < 1. (4.18)

17

Therefore

v0(r, θ) =∞∑

k=1

r2k (Ak cos(2kθ) +Bk sin(2kθ)) in S1 ∩B1, (4.19)

and according to (4.18), we have that

Ak cos kφ1 +Bk sin kφ1 = 0, for k = 1, 2, .... (4.20)

We claim that (4.20) implies

ˆ π/2+φ1/2

φ1/2

(Ak cos(2kθ) +Bk sin(2kθ)) sin(2θ − φ1)dθ = 0, for all k = 2, 3, .... (4.21)

The proof of (4.21) is straightforward. Fix k ≥ 2, and assume that sin kφ1 6= 0, then

Bk = −Akcos kφ1

sin kφ1,

hence we obtain

ˆ π/2+φ1/2

φ1/2

(Ak cos(2kθ) +Bk sin(2kθ)) sin(2θ − φ1)dθ

=Ak

sin kφ1

ˆ π/2+φ1/2

φ1/2

(cos(2kθ) sin kφ1 − cos kφ1 sin(2kθ)) sin(2θ − φ1)dθ

=−Ak

sin kφ1

ˆ π/2+φ1/2

φ1/2

sin(k(2θ − φ1)) sin(2θ − φ1)dθ =−Ak

2 sin kφ1

ˆ π

0

sin kt sin tdt = 0, if k 6= 1.

If sin kφ1 = 0, then cos kφ1 6= 0, and the proof of (4.21) works similarly.Our next aim is to show that A1 = B1 = 0. By using the orthogonality property (4.15) and

(4.21), and employing elementary trigonometric identities, we obtain

0 =

ˆ 1

0

ˆ π/2+φ1/2

φ1/2

v0(r, θ) sin(2θ − φ1)dθrdr

=

ˆ 1

0

ˆ π/2+φ1/2

φ1/2

∞∑

k=1

r2k+1 (Ak cos(2kθ) +Bk sin(2kθ)) sin(2θ − φ1)dθdr

=∞∑

k=1

1

2(k + 1)

ˆ π/2+φ1/2

φ1/2

(Ak cos(2kθ) +Bk sin(2kθ)) sin(2θ − φ1)dθ

=1

4

ˆ π/2+φ1/2

φ1/2

(A1 cos(2θ) +B1 sin(2θ)) sin(2θ − φ1)dθ

=1

4

ˆ π/2+φ1/2

φ1/2

−A1 sinφ1 cos2 2θ +B1 cosφ1 sin2 2θdθ =π

16(−A1 sinφ1 +B1 cosφ1).

Hence−A1 sinφ1 +B1 cosφ1 = 0. (4.22)

On the other handA1 cosφ1 +B1 sinφ1 = 0, (4.23)

18

which is (4.20) for k = 1. It is easy to see that (4.22) together with (4.23) imply A1 = B1 = 0.By (4.19),

v0(sx) = v0(sr, θ) = s4+∞∑

k=2

s2(k−2)r2k(Ak cos(2kθ) +Bk sin(2kθ)) in S1 ∩B1.

Hence we obtain‖v0(sx)‖L2(S1∩B1) ≤ s4‖v0‖L2(S1∩B1). (4.24)

Analogously,‖v0(sx)‖L2(S2∩B1) ≤ s4‖v0‖L2(S2∩B1). (4.25)

According to Proposition 4.3, v0 ≡ 0 in B1 \ (S1 ∪ S2), hence

‖v0(sx)‖L2(B1) = ‖v0(sx)‖L2(S1∩B1) + ‖v0(sx)‖L2(S2∩B1).

The desired inequality, (4.17), now follows from (4.24) and (4.25).

Corollary 4.5. For any ε > 0 and 0 < s < 1, there exists δ = δ(ε, s) such that if

‖u− µ0‖L2(B2) ≤ δ,

then‖us − µ0‖L2(B1) ≤ s‖u− µ0‖L2(B1) + ε‖u− µ0‖L2(B2), (4.26)

where µ0 is a minimal double-cone solution with respect to u.

Proof. We argue by contradiction. Let uj be a sequence of solutions to the double obstacleproblem and assume that

‖uj − µj‖L2(B1) := δj → 0,

but there exist 0 < s < 1 and ε > 0, such that

‖ujs − µj‖L2(B1) > s‖uj − µj‖L2(B1) + εδj . (4.27)

Let vj be the sequence defined by (4.13), then by (4.27)

‖vjs‖L2(B1) > s‖vj‖L2(B1) and ‖vj‖L2(B1) > ε. (4.28)

Applying Proposition 4.3, we may pass to the limit in (4.28) as j →∞, and obtain

‖v0s‖L2(B1) ≥ s‖v0‖L2(B1) and ‖v0‖L2(B1) ≥ ε,

hence‖v0s‖L2(B1) ≥ s‖v0‖L2(B1) > s2‖v0‖L2(B1),

and we derive a contradiction to Lemma 4.4.

Assume that u0 = µ given by (4.7) is a blow-up for u at the origin, that is (4.5) and (4.6)hold for a sequence rj → 0. Our aim is to show that the blow-up of u at the origin is unique;

u(rx)

r2→ u0(x), as r → 0.

19

Proposition 4.6. Let u be the solution to the double obstacle problem in Ω, B2 ⊂⊂ Ω ⊂ R2,with obstacles p1(x) = −x2

1 − x22 and p2(x) = x2

1 + x22. Assume that ‖u− µ‖L2(B2) = δ is small,

where µ is a double-cone solution, then there exists a double-cone solution u0, such that ur → u0.Furthermore, for any 0 < γ < 1,

‖ur − u0‖L2(B1) ≤ Cnrγ‖u− µ‖L2(B2), (4.29)

provided δ > 0 is small depending on γ.

Proof. Let 14 ≤ s < 1

2 be a fixed number, and τ := sγ > s. We use an induction argument toshow that for δ > 0 small enough

‖usk+1 − µk‖L2(B1) ≤ τk+1δ, and ‖µk+1 − µk‖L2(B1) ≤ 2τk+1δ, k = 0, 1, 2, 3, ... (4.30)

where by definition µk is a minimal double-cone solution with respect to usk .Let us show that (4.30) is true for k = 0. First we observe that by the triangle inequality

and the minimality assumption,

‖µ− µ0‖L2(B1) ≤ ‖u− µ‖L2(B1) + ‖u− µ0‖L2(B1) ≤ 2‖u− µ‖L2(B1). (4.31)

Note that since µk are homogeneous of degree two functions, the following relation is true

‖µk+1 − µk‖L2(B2) = 8‖µk+1 − µk‖L2(B1), for all k = 0, 1, 2, .... (4.32)

Now let us proceed to the proof of (4.30) for k = 0. According to Corollary 4.5 and (4.31),

‖us − µ0‖L2(B1) ≤ s‖u− µ0‖L2(B1) + ε‖u− µ0‖L2(B2) ≤ s‖u− µ‖L2(B1)

+ε‖u− µ‖L2(B2) + ε‖µ− µ0‖L2(B2) ≤ s‖u− µ‖L2(B1) + ε‖u− µ‖L2(B2)

+16ε‖u− µ‖L2(B1) ≤ (s+ 17ε)‖u− µ‖L2(B2) ≤ τ‖u− µ‖L2(B2),

where we take 0 < ε < τ−s17 . Thus we obtain ‖us − µ0‖L2(B1) ≤ τ‖u− µ‖L2(B2), hence

‖µ1 − µ0‖L2(B1) ≤‖us − µ1‖L2(B1)+‖us − µ0‖L2(B1) ≤ 2‖us − µ0‖L2(B1) ≤ 2τ‖u− µ‖L2(B2),

which completes the proof of (4.30) for k = 0.Let us assume (4.30) holds up to and including k, we will show that (4.30) holds for k + 1.

First note that ‖usk+1 − µk+1‖L2(B2) is small. Indeed, since 1/4 < s < 1/2, we obtain

‖usk+1 − µk‖L2(B2) =1

s2‖usk − µk‖L2(B2s) ≤ 16‖usk − µk‖L2(B1)

≤ 16‖usk − µk−1‖L2(B1) ≤ 16τkδ ≤ 16δ(4.33)

by the induction assumption. According to Corollary 4.5 for any ε > 0, we can choose 16δ > 0to be small depending on ε and s, and obtain

‖usk+2 − µk+1‖L2(B1) ≤ s‖usk+1 − µk+1‖L2(B1) + ε‖usk+1 − µk+1‖L2(B2)

≤ s‖usk+1 − µk‖L2(B1) + ε‖usk+1 − µk‖L2(B2) + ε‖µk+1 − µk‖L2(B2)

≤ s‖usk+1 − µk‖L2(B1) + 16ε‖usk − µk−1‖L2(B1) + 8ε‖µk+1 − µk‖L2(B1),

where we used (4.33) and (4.32) in the last step. Recalling our induction assumption, we obtain

‖usk+2 − µk+1‖L2(B1) ≤ (sτk+1 + 16ετk + 8ετk+1)δ ≤ τk+2δ, (4.34)

20

by choosing ε < τ(τ−s)16+8τ . It follows from the triangle inequality and the definition of minimal

double-cone solutions that

‖µk+2 − µk+1‖L2(B1) ≤ ‖usk+2 − µk+2‖L2(B1) + ‖usk+2 − µk+1‖L2(B1)

≤ 2‖usk+2 − µk+1‖L2(B1) ≤ 2τk+2δ.

The proof of the inequalities (4.30) is therefore complete.Now we are ready to show that µk is a Cauchy sequence, and therefore converges. For any

m, k ∈ N

‖µk+m − µk‖L2(B1) ≤k+m−1∑

l=k

‖µl+1 − µl‖L2(B1) ≤k+m−1∑

l=k

τ l+1δ ≤ τk+1

1− τ δ → 0,

independent of m. Hence there exists u0, such that µk → u0, furthermore

‖µk − u0‖L2(B1) ≤τk+1

1− τ δ. (4.35)

The inequalities (4.30) and (4.35) together with the triangle inequality imply that

‖usk − u0‖L2(B1) ≤ 2τkδ. (4.36)

Finally let us observe that for any 0 < r < 1 there exists a nonnegative integer k such thatsk+1 ≤ r < sk. Hence

‖ur − u0‖L2(B1) ≤ s−3‖usk − u0‖L2(B1) ≤ 2· 43τkδ ≤ 44rγδ, (4.37)

where γ = ln τln s < 1.

Corollary 4.7. Assume that µ given by (4.7) is a blow-up for u at the origin, that is (4.5) and(4.6) hold for a sequence rj → 0. Then the blow-up of u at the origin is unique;

u(rx)

r2→ µ(x), as r → 0.

Proof. Since urj → µ as j → ∞, for any δ > 0 small we can find a small ρ > 0 such that‖uρ−µ‖L2(B2) ≤ δ. Now we can apply Proposition 4.6 for the function uρ, and obtain uρr → u0

as r → 0+. Hence ur → u0 and u0 = µ.

Theorem 4.8. Let u be the solution to the two-dimensional double obstacle problem with ob-stacles p1 = −x2

1 − x22 and p2 = x2

1 + x22. Assume that ‖u − µ‖L2(B2) = δ is sufficiently small,

where µ is a double-cone solution, and is not a halfspace solution. Then in a small ball Br0 thefree boundary consists of four C1,γ- graphs meeting at the origin, denoted by Γ+

1 , Γ−1 , Γ

+2 , Γ

−2 .

Neither Γ1 = Γ+1 ∪ Γ−1 nor Γ2 = Γ+

2 ∪ Γ−2 has a normal at the origin. The curves Γ+1 and Γ+

2

cross at a right angle, the same is true for Γ−1 and Γ−2 .

Proof. The proof of the theorem is based on Proposition 4.6 and on similar estimates obtainedfor the classical obstacle problem in [2]. According to Proposition 4.6 there exists a double-conesolution u0 such that (4.29) holds. Moreover, applying Lemma 4.1 we obtain

‖ur − u0‖W 1,2(B1/2) ≤ Crγ‖u− µ‖L2(B2). (4.38)

21

x1

x2

Γ+2

Γ+1

Γ−2

Γ−1

∆u = 0

u = p2

u = p1

∆u = 0

S1

S2

x0

ν(0) ν(x0)

ν(0)

Figure 4.1: The behavior of the free boundary, with obstacles touching at a single pointHere Γ±

i are the pieces of the free boundary for u, while the dashed lines are the free boundary to thedouble-cone solution u0.

Without loss of generality we assume that u0 is given by (4.7); that is

u0 = µφ1,φ2(r, θ) :=

r2, if − φ2 ≤ 2θ ≤ φ1

r2 cos(φ1 − 2θ), if φ1 ≤ 2θ ≤ π + φ1

r2 cos(φ2 + 2θ), if − π − φ2 ≤ 2θ ≤ −φ2

−r2, otherwise

As before, we denote by Ci = u0 = pi. Let ϑi be the opening angle for Ci. By assumption,u0 is not a halfspace solution, and therefore 0 < ϑi < π.

We want to study the regularity of Γ2 = ∂u = p2 in neighborhood of the origin. We splitthe proof into two steps.

Step 1: We show that Γ2 ∩Br0 ⊂ (Q \K), for any open cones Q and K, having a commonvertex at the origin, such that K ⊂ C2 ⊂ Q and ∂K ∩ ∂C2 = ∂Q ∩ ∂C2 = 0, where r0 > 0 issmall depending on K,Q.

Let K ⊂ C2 be a cone with a vertex at the origin, such that ∂K∩∂C2 = 0. Fix 0 < % < 1/8,and denote by V := K ∩ % < x < 1/2, and σ := dist(V, ∂C2). First we will show that

ur(x) = p2(x) in V for small r > 0. Take 0 < ε < σ2

4 , then there exists rε = rσ, such that|ur(x) − u0(x)| ≤ ε if r ≤ rε. Let ω := p2 − ur, for a fixed r < rε, then 0 ≤ ω ≤ ε solves thefollowing normalized obstacle problem with zero obstacle,

∆ω = λ2χω>0 in C2. (4.39)

Fix x0 ∈ V , if ω(x0) > 0, then we can apply the maximum growth lemma (Lemma 5 in [4]) forthe solution to the obstacle problem, and obtain

σ2

4> ε ≥ sup

Bσ(x0)

ω ≥ σ2

2, (4.40)

which is not possible, hence ω(x0) = 0.

22

Thus we have shown that u(rx)r2 = p2(x) for all r < rε and any x ∈ K, such that % < |x| < 1/2.

Hence u(y) = p2(y) if %r < |y| < r2 for all r < rε, and therefore u = p2 in K ∩Br0 , r0 := rε/2.

Taking another open cone Q, with a vertex at the origin, and such that C2 ⊂ Q, ∂Q ∩ ∂C2 =0, we show that Γ2 ∩Br ⊂ Q if r is small. Let % > 0, then u0− p2 < 0 in Q \B%, and thereforeur − p2 < 0 in Q \ B% for small r > 0. Hence u < p2 in Q ∩ Br for a small fixed r > 0, andΓ2 ∩Q ∩Br = ∅.

Now we can write Γ2 = Γ+2 ∪Γ−2 , where Γ+

2 ∩Γ−2 = 0, and Γ±2 are ”squeezed” between Kand Q.

Step 2: We show that Γ+2 is a C1,α-graph up to the origin. Fix any x0 ∈ Γ2 ∩ Br0/2, and

denote by d := |x0|. Let d0 := d2 sinϑK , where ϑK is the opening angle of K. Since x0 /∈ K and

x0 ∈ Q, we see that Bd0(x0) ∩ C1 = ∅ and Bd0(x0) ∩ S2 = ∅, see Figure 4.1. Hence the functionω := p2 − u solves the following normalized obstacle problem

∆ω = λ2χω>0 in Bd0(x0). (4.41)

Now let us show that p2 − u0 is a halfspace solution for (4.41). Denote by ν(0) the unit upwardnormal to the line θ = φ1/2, as indicated in Figure 4.1;

ν(0) =

(− sin

φ1

2, cos

φ1

2

).

The following is true,

u0(x) = p2(x)− 2(ν(0) · x)2+ if u0(x) > p1(x), and x /∈ S2. (4.42)

Indeed, according to Lemma 3.6, if x ∈ S1, then

p2(x)− u0(x) = x21 + x2

2 − x21 cosφ1 + x2

2 cosφ1 − 2x1x2 sinφ1

= 2x21 sin2 φ1

2+ 2x2

2 cos2 φ1

2− 4x1x2 sin

φ1

2cos

φ1

2

= 2

(−x1 sin

φ1

2+ x2 cos

φ1

2

)2

= 2(ν(0) · x)2.

Since u0 = p2 in C2, the proof of (4.42) is complete. Hence p2−u0 is a halfspace solution for theobstacle problem in Bd0(x0), depending on the direction ν(0). Therefore we obtain

‖∇′ν(0)ω‖L2(Bd0 (x0)) = ‖∇′ν(0)

(u− u0 − 2(ν(0) · x)2

+

)‖L2(Bd0 (x0))

= ‖∇′ν(0)(u− u0)‖L2(Bd0 (x0)) ≤ 2‖∇(u− u0)‖L2(Bd0 (x0))

(4.43)

where by definition ∇′e := ∇− e(∇ · e) for a unit vector e.According to Lemma 4.1

‖u− u0‖W 1,2(Bd0 (x0)) ≤ c‖u− u0‖L2(B2),

hence by (4.43)

‖∇′ν(0)ω‖L2(Bd0 (x0)) = ‖∇(u− u0)‖L2(Bd0 (x0)) ≤ c‖u− u0‖L2(B2) ≤ cδ,

which says that ω is almost flat in the direction ν(0). According to Theorem 8.1 in [2], Γ2 ∩Bd0/2(x0) is a C1,γ- graph, and there exists a unit normal vector to Γ2 at the point x0, denote

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it by ν(x0). Furthermore, it follows from Corollary 8.1 in [2] and inequality (4.38) that

|ν(x0)− ν(0)| ≤ cd−10 ‖∇′ν(0)ω‖L2(Bd0 (x0)) ≤ cd−1

0 ‖∇(u− u0)‖L2(B2d)

= 16cd−10 d

(ˆ

B1/2

|∇u(4dy)−∇u0(4dy)|2dy) 1

2

≤ cd−10 d‖u4d − u0‖L2(B1)

≤ cd

d0dγ‖u− u0‖L2(B2),

where c stands for a general constant, and it does not depend on d. Now we may conclude that

|ν(x0)− ν(0)| ≤ cd

d0dγδ =

c|x0|γsinϑK

‖u− u0‖L2(B2),

and therefore Γ+2 is a C1,γ-graph up to the origin, for any 0 < γ < 1.

The proof of C1,γ-regularity for Γ−2 can be obtained similarly. In that case Γ−2 is almost flat

in the direction ν−2 (0) =(− cos φ2

2 ,− sin φ2

2

)6= ν(0), and we see that Γ2 is not C1 at the origin.

By a similar argument we can study Γ1, and see that Γ+1 is almost flat in the direction

ν+1 (0) =

(cos φ1

2 , sinφ1

2

). Observe that ν+

1 (0) and ν(0) are orthogonal, which means Γ+1 and Γ+

2

cross at the origin.

References

[1] Gohar Aleksanyan. Optimal regularity in the optimal switching problem. In Press;doi:10.1016/j.anihpc.2015.06.001.

[2] John Andersson. The obstacle problem. 2016.

[3] John Andersson, Henrik Shahgholian, and Georg S. Weiss. Double obstacle problems withobstacles given by non-C2 Hamilton-Jacobi equations. Arch. Ration. Mech. Anal. 206, no.3, 779–819., 2012.

[4] L. A. Caffarelli. The obstacle problem revisited. J. Fourier Anal. Appl. 4 (1998), no. 4-5,383–402.

[5] Alessio Figalli and Henrik Shahgholian. An overview of unconstrained free boundary prob-lems. Philos. Trans. A 373 (2015), no. 2050, 20140281, 11 pp.

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