An Elliptic Free Boundary Arising From the Jump of Conductivity
aa r X i v : . [ m a t h . A P ] J un AN ELLIPTIC FREE BOUNDARY ARISING FROM THE JUMPOF CONDUCTIVITY
SUNGHAN KIM, KI-AHM LEE, AND HENRIK SHAHGHOLIANA bstract . In this paper we consider a quasilinear elliptic PDE,div( A ( x , u ) ∇ u ) =
0, where the underlying physical problem givesrise to a jump for the conductivity A ( x , u ), across a level surface for u . Our analysis concerns Lipschitz regularity for the solution u ,and the regularity of the level surfaces, where A ( x , u ) has a jumpand the solution u does not degenerate.In proving Lipschitz regularity of solutions, we introduce anew and unexpected type of ACF-monotonicity formula with twodi ff erent operators, that might be of independent interest, andsurely can be applied in other related situations. The proof ofthe monotonicity formula is done through careful computations,and (as a byproduct) a slight generalization to a specific type ofvariable matrix-valued conductivity is presented. C ontents
1. Introduction 21.1. The model equation 21.2. Applications 21.3. Methodology and Approach 31.4. Relation to existing literatures 51.5. Organization of the paper 52. Existence and Free Boundary Condition 63. Regularity of Solutions 9
S. Kim has been supported by NRF (National Research Foundation of Korea)Grant funded by the Korean Government (NRF-2014-Fostering Core Leaders of theFuture Basic Science Program). K. Lee was supported by the National ResearchFoundation of Korea (NRF) grant funded by the Korea government (MSIP) (No.NRF-2015R1A4A1041675). K. Lee also holds a joint appointment with the ResearchInstitute of Mathematics of Seoul National University. H. Shahgholian has beensupported in part by Swedish Research Council. This project is part of an STINT(Sweden)-NRF (Korea) research cooperation program.The authors are grateful to referees’ comments that led to a substantial improve-ment of the first submitted draft. John Andersson is acknowledged for invaluableinput, specially for clarifications of several crucial aspects in his work with coau-thor [AM] .
4. Preliminaries properties of the Free Boundary 135. Further Properties of the Free Boundary 156. C ,α Regularity of Nondegenerate Free Boundaries 217. Analysis on Matrix Coe ffi cient Cases 31Appendix A. ACF Monotonicity Formula 33References 371. I ntroduction The model equation.
In this paper, we consider weak solutions u of( L ) div( A ( x , u ) ∇ u ) = Ω , where Ω is a bounded domain in R n ( n ≥
2) and A : R n × R → R isdefined by(1.1) A ( x , s ) : = a − ( x ) + ( a + ( x ) − a − ( x )) H ( s ) = a − ( x )(1 − H ( s )) + a + ( x ) H ( s ) , where H is heaviside function and a ± ∈ C ( R n ) satisfying the followingstructure conditions: • there is a λ > λ ≤ min { a − ( x ) , a + ( x ) } , max { a − ( x ) , a + ( x ) } ≤ λ , ∀ x ∈ R n ; • there is a modulus of continuity ω such that(1.3) max {| a − ( x ) − a − ( y ) | , | a + ( x ) − a + ( y ) |} ≤ ω ( | x − y | ) , ∀ x , y ∈ R n . We call u a weak solution of ( L ), if u ∈ W , loc ( Ω ) ∩ L ( Ω ) satisfies Z A ( x , u ) ∇ u ∇ φ = , ∀ φ ∈ W , ( Ω ) . Applications.
Heat or electric conduction through certain ma-terials are usually, due to complex properties of the materials, veryhard to model. For instance, mathematical modeling of composites,consisting of materials with di ff erent conductivity properties, is onesuch problem which has been subject for intense (mathematical) stud-ies. Applications of such models can be found in problems related totransmissions [B], and inverse and discontinuous conductivity [AI],as well as other related problems [BC], where there is a jump in theconductivity.The conductivity problem becomes substantially complicated whenthe materials also undergo a phase transition, which can cause abruptchanges in the conductivity, due to a threshold of the heat or electric FREE BOUNDARY ... CONDUCTIVITY 3 current. The discontinuity in the conduction that arise from a struc-tural phase change in crystalline materials has been considered inapplied literature specially in relation to transport in solids, such asnanowires. Relevant discussions in applied literature can be foundin [BDHS], [WLSXMZ]. The mathematical problem of jump in con-ductivity across level surfaces have also been considered recently in[ACS], [AM] and [AT].The simplest model of a material-dependent conductivity can bewritten as A ( x , u ) ∇ u , where now A ( x , u ) has a discontinuity in the u -variable, which represents the heat, or electric charge. In this paper,we have chosen a simple elliptic model, where the model equation iswritten as A ( x , u ) = a + ( x ) χ { u > } + a − ( x ) χ { u ≤ } .It should be remarked that heat conduction in certain materials,which also undergo a phase transition, may cause a change in con-ductivity only at the phase, in terms of latent heat on the boundarybetween two-phases. Typical examples are melting ice or flame prop-agation. The mathematical study of such problems, usually entitledBernoulli free boundaries, have been carried out in a large scale in thelast few decades, e.g., [AC], [ACF], [CS] and the references therein.Our problem, although qualitatively di ff erent, carries features remi-niscent of that of Bernoulli type problem for the latent heat.1.3. Methodology and Approach.
As appearing in (1.1), the jumpof A ( x , u ) in the second argument across { u = } naturally induces afree boundary condition, which can be formally represented as a + ( x ) u + ν = a − ( x ) u − ν , where ν points towards level sets where u increases.Assuming the continuity (1.3) of our coe ffi cients, it is not hard tosee that our limiting equation after blowing up the solution at a point z is of the formdiv(( a − ( z ) + ( a + ( z ) − a − ( z )) H ( v )) ∇ v ) = , so that the function w , defined by w ( x ) : = a + ( z ) v + ( x ) − a − ( z ) v − ( x ) , becomes harmonic. Hence, the heart of our analysis lies in seekinga way to bring back the strong properties of our limit profiles to theoriginal solutions before blowup.Our first result concerns with the Lipschitz regularity of weak so-lutions to ( L ). In this direction, we develop a new monotonicity For this reason, it seems natural to call our limit profile a “broken” harmonicfunction.
SUNGHAN KIM, KI-AHM LEE, AND HENRIK SHAHGHOLIAN formula of the ACF type, involving two di ff erent operators. A keyissue in deriving an ACF type monotonicity formula is to make surethat the concentrated energy on each component ( I ( r , u + ) and I ( r , u − )in Appendix A) can be compared on the same sphere, so that theeigenvalue inequality and the Friedland-Hayman inequality can beapplied to both components simultaneously. In this direction, onepossible choice of two di ff erent operators is that the associated coordi-nate systems can be simultaneously rotated into concentric balls; seeTheorem A.4. One may notice that the two distinct operators associ-ated with ( L ) certainly enjoy this property. As the ACF monotonicitybeing adapted to linear scaling, we may get rid of any possibilitythat a sublinearly scaled blowup limit at a free boundary point endsup with a nontrivial function, proving the Lipschitz regularity of oursolutions.Our analysis on the regularity of our free boundary is conductedby making an exclusive use of measure µ = div( a + ( x ) ∇ u + ), whichturns out to carry essential information on our free boundary points.Our main interest is in twofold; first, smoothness of our free bound-ary around nondegenerate points, see Definition 6.2, and second,characterization of µ in terms of the ( n − ff measure. We first make a crude observation that rapidly vanishingpoints (Lemma 5.5) carries null µ -measure, which allows us to focuson vanishing points with finite orders. Essentially, finitely vanish-ing points admit doubling conditions, which stablize the blowupsequence and after all give us a nontrivial limit. With this stability athand, we may follow the arguments in [AM] to deduce that µ -almostevery free boundary point has a two plane solution as its blowuplimit, consequently yielding a flatness condition.Assuming H ¨older regularity on each part of our coe ffi cients, weinvoke a linearization technique to improve the flatness condition inan inductive way, where the technique itself is a well established the-ory that by now can be considered as classical. Through an iterationargument, we deduce that our free boundary around a flat point canbe trapped locally uniformly by two C ,α graphs, and thus provingits C ,α regularity. Equipped with the C ,α regularity theory, we arealso able to rule out degenerate points (Definition 6.2) to have anyflatness condition, which well matches our intuition, and finally con-clude that the support of µ consists of nondegenerate points, fromwhich we deduce its σ -finiteness of ( n − ff measure. FREE BOUNDARY ... CONDUCTIVITY 5
Relation to existing literatures.
The model equation with vary-ing coe ffi cients have not been treated earlier, even when the coe ffi -cient on each component is assumed to be smooth. To the authors’best knowledge, the only existing literature so far is that of [AM],which treats the conductivity jump with constant matrices. The au-thors of [AM] do not prove Lipschitz regularity for solutions, as thisis not true in general. For the parabolic case, Ca ff arelli and Stefanelliproved in [Ca-St] that solutions to similar problems, with conduc-tivity jump, cannot be Lipschitz regular in space. In a recent result[CDS], the Lipschitz regularity is proved only in two dimensionalcase for a similar problem, using geometric methods. According tothe authors of [CDS], one can construct in higher dimensions non-Lipschitz solutions to a Bernoulli type free boundary problem withtwo di ff erent operators and a nonlinear Bernoulli boundary gradientcondition. Thus, it should be remarked that the problem under ourconsideration introduces substantial di ffi culty, not undertaken ear-lier in the literatures, and we believe that our results in this paper areoriginal.To the best of our knowledge, our result on the Lipschitz regu-larity of solutions is completely new. As mentioned above, evenin the scalar case and for constant-matrix valued conductivity thereare no such results in the literature, since this in general is not true.Besides, the way we derive the Lipschitz regularity has its indepen-dent value in the regularity theory for elliptic PDEs, since we invokea free boundary technique to detect the optimal regularity of weaksolutions to PDEs whose coe ffi cients have specified discontinuity. Itshould also be stressed that as mentioned earlier, our newly devel-oped monotonicity formula can be extended even for certain matrixvalued conductivities of specific types, and thus will probably havean impact in solving other related problems where there are varyingoperators on each phase of a free boundary problem with a phasetransition.It is noteworthy that finding counterexamples to Lipschitz regular-ity for the matrix coe ffi cient case seems a challenging problem. Alsofinding optimal conditions for Lipschitz regularity of the matrix caseremains unanswered. It seems however not obvious whether Lips-chitz regularity is a necessity for smoothness of the free boundary ornot.1.5. Organization of the paper.
We prove the existence of weak so-lutions of ( L ) in Section 2 and give a weak formulation of our freeboundary condition. Section 3 is devoted to the optimal regularity SUNGHAN KIM, KI-AHM LEE, AND HENRIK SHAHGHOLIAN of our solutions (Theorem 3.3). In Section 4 we introduce the mea-sure µ = L + u + and derive some of its basic properties. In section5 we observe further properties of our free boundary by means of µ and prepare for Section 6. Section 6 is devoted to prove that ourfree boundary is a C ,α graph in a neighborhood of µ almost everypoint (Theorem 6.1), and that the support of µ has σ -finite ( n −
1) di-mensional Hausdor ff measure (Theorem 6.3). In Section 7 we extendall the prescribed results for certain class of varying matrix valuedconductivity. Finally in Appendix A we give a proof for the newlydeveloped ACF monotonicity formula with two di ff erent operators(Theorem A.4).Throughout this paper, we say a constant to be universal if itdepends only on n , λ and ω , those appearing in (1.2) and (1.3).Moreover, any constant will be considered positive universal, un-less otherwise stated. Also given a function u , u + : = max { u , } and u − : = − min { u , } , while by the subscript notation (e.g., u ± ) we justdenote two di ff erent quantities.2. E xistence and F ree B oundary C ondition Let us begin with the existence theory of our problem ( L ). Proposition 2.1.
Given g ∈ W , ( Ω ) , there exists a weak solution u of ( L ) in Ω such that u − g ∈ W , ( Ω ) .Proof. Let ψ : R → R be defined by ψ ( t ) : = t > , t if 0 ≤ t ≤ , t < , and define ψ ε and Ψ ε for each ε > ψ ε ( t ) : = ψ (cid:18) t ε (cid:19) and Ψ ε ( t ) : = Z t −∞ ψ ε ( s ) ds . Set A ε ( x , s ) : = a − ( x ) + ( a + ( x ) − a − ( x )) ψ ε ( s ). Since x A ε ( x , s ) ismeasurable on Ω and s A ( x , s ) is Lipschitz on R , there exists aweak solution u ε ∈ W , ( B ) fordiv( A ε ( x , u ε ) ∇ u ε ) = Ω , with u ε − g ∈ W , ( Ω ). It also follows from the uniform ellipticity of A ε ( x , s ), with ellipticity constants being λ and λ (see (1.2)) indepen-dent of ε >
0, we will have a uniform estimate(2.1) k u ε k W , ( Ω ) ≤ C g W , ( Ω ) , FREE BOUNDARY ... CONDUCTIVITY 7 for some universal C > Ψ ε ( u ε )in the norm k·k W , ( Ω ) . Due to the fact that Ψ ε ( s ) ≤ s for s > Ψ ε ( s ) ≡ s ≤
0, we obtain Z Ψ ε ( u ε ) ≤ Z u ε . Then we use the chain rule to derive Z |∇ Ψ ε ( u ε ) | = Z | ψ ε ( u ε ) ∇ u ε | ≤ Z |∇ u ε | . Therefore,(2.2) k Ψ ε ( u ε ) k W , ( Ω ) ≤ k u ε k W , ( Ω ) ≤ C g W , ( Ω ) . Now we are in position to take limits. From (2.1) there is a subse-quence ε k ց u ∈ W , ( Ω ) such that u ε k → u strongly in L ( Ω ) , ∇ u ε k → ∇ u weakly in L ( Ω ) . (2.3)Note that the strong L convergence implies u − g ∈ W , ( Ω ).Next we apply the estimate (2.2) along this subsequence. It allowsus to extract a further subsequence, which we still denote by ε k ,and some v ∈ W , ( Ω ) such that Ψ ε k ( u ε k ) → v strongly in L ( Ω ) and ∇ Ψ ε k ( u ε k ) → ∇ v weakly in L ( Ω ). However, since Ψ ε ( s ) → max { s , } uniformly in R , we must have v = u + . Also recall again that ∇ Ψ ε ( u ε ) = ψ ε ( u ε ) ∇ u ε . Thus, rephrasing the convergence of { Ψ ε k ( u ε k ) } , we arriveat Ψ ε k ( u ε k ) → u + strongly in L ( Ω ) ,ψ ε ( u ε ) ∇ u ε → ∇ u + weakly in L ( Ω ) . (2.4)Now let us verify that the limit function u is a weak solution fordiv( A ( x , u ) ∇ u ) = Ω . Let φ ∈ C ∞ c ( Ω ) be a given. Then the weak convergence in (2.3) and(2.4) implies respectively thatlim k →∞ Z a − ∇ u ε k ∇ φ = Z a − ∇ u ∇ φ, SUNGHAN KIM, KI-AHM LEE, AND HENRIK SHAHGHOLIAN and thatlim k →∞ Z ( a + − a − ) ψ ε k ( u ε k ) ∇ u ε k ∇ φ = Z ( a + − a − ) ∇ u + ∇ φ = Z ( a + − a − ) H ( u ) ∇ u ∇ φ, where we have used that ∇ u + = ∇ u a.e. on { u > } and ∇ u + = { u ≤ } . Combining these two limits, we get Z A ( x , u ) ∇ u ∇ φ = lim k →∞ Z A ε k ( x , u ε k ) ∇ u ε k ∇ φ = , as desired. (cid:3) Remark . Note that we have not made any assumption on theregularity of a ± in Proposition 2.1. We also make another remark inthe proof that the weak star convergence of ψ ε k ( u ε k ) → H ( u ) in L ∞ loc ( Ω )is in general not true. Notation . From now on u is a nontrivial weak solution of ( L ) in Ω , unless otherwise stated. Also Ω + ( u ) : = { u > } ∩ Ω , Ω − ( u ) : = { u < }∩ Ω and Γ ( u ) : = ∂ { u > }∩ Ω . Moreover, L ± are the operators definedby L ± u = div( a ± ( x ) ∇ u ). We will also use the notation A z ( s ) : = A ( z , s )to emphasize that the z -argument is being fixed.Although the free boundary condition is already dictated by theequation, it is by no means elementary to show that it holds at allfree boundary points. Here we shall show a weak version of the freeboundary condition, which holds for all free boundaries and not onlynondegenerate ones. Lemma 2.4.
Suppose that a ± are Dini continuous, and let u be a weaksolution for ( L ) in Ω . Assume further that sets { u > ε }∩ Ω and { u < − ε }∩ Ω have locally finite perimeter for each ε > . Then (2.5) lim ε ց Z ∂ { u >ε }∩ Ω a + |∇ u | η = lim ε ց Z ∂ { u < − ε }∩ Ω a − |∇ u | η, for any η ∈ C ∞ c ( Ω ) .Proof. Since u ∈ W , ( Ω ), ∇ u = { u = } . Then0 = Z Ω A ( x , u ) ∇ u ∇ η = lim ε ց Z { u >ǫ } a + ∇ u ∇ η + lim ε ց Z { u < − ǫ } a − ∇ u ∇ η. FREE BOUNDARY ... CONDUCTIVITY 9
From the Dini continuity of a + , we know that u ∈ C ( Ω + ( u )). More-over, from the assumption that { u > ε }∩ Ω has locally finite perimeter,we may adopt integration by part and deduce that Z { u >ε }∩ Ω a + ∇ u ∇ η = Z ∂ { u >ε }∩ Ω a + ∇ u · ν { u >ε }∩ Ω η, where ν { u >ε }∩ Ω is the measure theoretic outward unit normal of ∂ { u >ε }∩ Ω ; for the definition of measure theoretic unit normals, see Section5 of [EG]. In particular, if x is a point on ∂ { u > ε } ∩ Ω such that ∇ u ( x ) ,
0, then ν { u >ε }∩ Ω ( x ) = −∇ u ( x ) / |∇ u ( x ) | . Therefore, one mayrephrase the surface integral on the right hand side by Z ∂ { u >ε }∩ Ω a + ∇ u · ν { u >ε }∩ Ω η = − Z ∂ { u >ε }∩ Ω a + |∇ u | η. By a similar reason, one may also derive that Z { u < − ε }∩ Ω a − ∇ u ∇ η = Z ∂ { u < − ε }∩ Ω a − |∇ u | η. Summing all up, we arrive at0 = − lim ε ց Z ∂ { u >ε }∩ Ω a + |∇ u | η + lim ε ց Z ∂ { u < − ε }∩ Ω a − |∇ u | η, as desired. (cid:3) Remark . The assumption that { u > ε } ∩ Ω and { u < − ε } ∩ Ω haslocally finite perimeters is met, under the circumstance that a + and a − are Lipschitz continuous; see Theorem 5.2.1 in [HL]. It is not easy,however, to derive the same conclusion even with H ¨older regular a + and a − . 3. R egularity of S olutions In this section, we prove interior regularity of our solutions to ( L ).We know from the De Giorgi theory that solutions of ( L ) is locallyH ¨older continuous for some universal exponent 0 < α <
1, evenwhen our coe ffi cients a ± are assumed to be bounded measurable. Ourfirst observation is that as we assume the continuity assumption (1.3),our solutions become locally H ¨older continuous for any exponentarbitrarily close to 1. Lemma 3.1.
For each α ∈ (0 , there exists some C α > such that forD ⋐ Ω and z ∈ { u = } ∩ D, | u ( x ) | ≤ C α k u k L ∞ ( D ) | x − z | α d α in D , where d = dist( D , ∂ Ω ) .Proof. By a scaling argument, it su ffi ces to show it only for Ω = B , D = B / and u such that k u k L ∞ ( D ) = α ∈ (0 ,
1) forwhich the statement of this proposition is true. Thus, we only needto consider α ∈ ( ¯ α, u j of ( L )in B , x j ∈ Γ ∩ B / and r j < / r j ց B rj ( x j ) | u j | = jr α j and sup B r ( x j ) | u j | ≤ jr α for all r j ≤ r ≤ / . Consider v j ( x ) : = u j ( r j x + x j ) jr α j , x ∈ B / r j . Then v j solves div( A j ( x , v j ) ∇ v j ) = B / r j , where A j ( x , s ) : = A ( r j x + x j , s ), and thatsup B | v j | = B R | v j | ≤ R α for all 1 ≤ R ≤ / (4 r j ) . Notice that λ ≤ A j ( x , s ) ≤ λ for all ( x , s ) ∈ R n × R . Thus theuniform supremum estimate above yields that there is some v ∈ W , loc ( R n ) ∩ C ¯ α loc ( R n ) such that v j → v locally uniformly in R n , ∇ v j → ∇ v weakly in L loc ( R n ) , (3.1)along a subsequence, which we still denote by j , and that(3.2) sup B | v | = B R | v | ≤ R α , ∀ R ≥ . Let us derive the equation which v satisfies. Recall that x j ∈ B / ,so there is x ∈ ¯ B / to which x j is convergent along a subsequence.Let us again denote this convergent subsequence by j . Then thecontinuity assumption (1.3) on a ± and the weak convergence in (3.1) We make a remark here that only continuity is used.
FREE BOUNDARY ... CONDUCTIVITY 11 imply that0 = lim j →∞ Z A j ( x , v j ) ∇ v j ∇ φ = lim j →∞ "Z a + ( r j x + x j ) ∇ v + j ∇ φ − Z a − ( r j x + x j ) ∇ v − j ∇ φ = Z a + ( x ) ∇ v + ∇ φ − Z a − ( x ) ∇ v − ∇ φ = Z A x ( v ) ∇ v ∇ φ, for any φ ∈ C ∞ c ( R n ). Thus, v is a weak solution to(3.3) div( A x ( v ) ∇ v ) = R n . Now define w : = a + ( x ) v + − a − ( x ) v − in R n . It immediately follows from (3.3) and the definition of A x ( v ) that w is harmonic in the entire space R n . From (3.2), we have | w ( x ) | ≤ λ | v ( x ) | ≤ λ | x | α for | x | ≥ . Hence, the Liouville theorem implies that w is a constant functionin R n . Now from the uniform convergence in (3.1), we have v (0) = lim j →∞ v j (0) =
0. Thus, w ≡ w (0) = R n . However, from the equality in (3.2),sup B | w | = sup B ( a + ( x ) v + + a − ( x ) v − ) ≥ λ sup B | v | = λ > , a contradiction. (cid:3) By assigning higher regularity on ω , we are able to improve Lemma3.1 to Lipschitz regularity. Lemma 3.2.
Suppose that a + and a − satisfy (1.3) with a Dini continuous ω . There exists some C > such that for D ⋐ Ω and z ∈ { u = } ∩ D, | u ( x ) | ≤ C k u k L ∞ ( D ) | x − z | d in D , where d = dist( D , ∂ Ω ) . A modulus of continuity ω is said to be Dini continuous, if R ω ( r ) r dr < ∞ . Proof.
As in the proof of Lemma 3.1, we simplify the setting andargue by contradiction. Suppose that there exist solutions u j of ( L ) in B , x j ∈ Γ ∩ B / and r j < / r j ց B rj ( x j ) | u j | = jr j and sup B r ( x j ) | u j | ≤ jr for all r j ≤ r ≤ / . We know that x j → x for some x ∈ ¯ B / up to a subsequence, and letus overwrite the indices of this subsequence by j .Consider v j ( x ) : = u j ( r j x + x j ) jr j , x ∈ B / r j . Then following the argument in the proof of Lemma 3.1, there issome v ∈ W , loc ( R n ) ∩ C α loc ( R n ) with a universal α ∈ (0 ,
1) such that (3.1)is true along a subsequence, which we still denote by j , and that(3.4) sup B | v | = . Moreover, v is a weak solution to (3.3). Since ∇ v j → ∇ v weakly in L loc ( R n ), | x | − n ∇ v ± j → | x | − n ∇ v ± weakly in L loc ( R n ) . As a result, for each R ≥ Z B R |∇ v ± | | x | n − ≤ lim j →∞ Z B R |∇ v ± j | | x | n − . Now apply to u j the ACF estimate, that is, Proposition A.3, to arriveat Φ ( R , v + j , v − j ) = j Φ ( Rr j , x j , u + j , u − j ) ≤ Cj , for all j such that r j ≤ r R , where r and C are chosen as in PropositionA.3; see (A.1) for the definition of Φ . Thus, Φ ( R , v + , v − ) ≤ lim j →∞ Φ ( R , v + j , v − j ) ≤ lim j →∞ Cj = , for each R ≥
1. This implies that either ∇ v + or ∇ v − vanishes a.e. in R n .Without losing any generality we may assume ∇ v + = R n . Then the continuity of v + implies that v + ≡ R n . In view of(3.3), this implies that v − is harmonic in the entire space R n . Thenthe maximum principle implies that v − ≡ R n . Finally, we obtain v ≡ R n , which violates (3.4). (cid:3) Now we are ready to state our first main result.
FREE BOUNDARY ... CONDUCTIVITY 13
Theorem 3.3.
Let a + and a − satisfy (1.3) with a Dini continuous ω and letu be a weak solution to ( L ) in Ω . Then u ∈ W , ∞ loc ( Ω ) and for any D ⋐ Ω , (3.5) k∇ u k L ∞ ( D ) ≤ C k u k L ( Ω ) d n + , where C depends only on n, λ and ω , and d = dist( D , ∂ Ω ) .Proof. Fix D ⋐ Ω with d = dist( D , ∂ Ω ) > D such that D ⋐ ˜ D ⋐ Ω and dist( D , ∂ ˜ D ) = dist( ˜ D , ∂ Ω ) = d /
2. Write M = k u k L ( Ω ) .By the local boundedness of weak solutions (Theorem 8.17 in [GT]),we know that K : = k u k L ∞ ( ˜ D ) ≤ cM / d n / for a constant c depending onlyon n and λ .Choose any z ∈ D such that u ( z ) , r = dist( z , { u = } ∩ D ).If r ≥ d /
2, then the standard regularity theory yields that |∇ u ( z ) | ≤ C K / d .On the other hand, if r < d /
2, then we consider the scaled function v ( x ) : = u ( rx + z ) / r in B . Suppose that u ( z ) >
0. Then v is a nonneg-ative weak solution for div( a + ( rx + z ) ∇ v ) = B . Since a + is Dinicontinuous, the standard regularity theory yields that v ∈ C ( B ).Then the Harnack inequality implies that |∇ v (0) | ≤ C v (0). By meansof Lemma 3.2, we have v (0) = u ( z ) / r ≤ C K / d .A similar argument applies to the case u ( z ) <
0. Using K ≤ cM / d n + ,we finish the proof. (cid:3)
4. P reliminaries properties of the F ree B oundary It is noteworthy that the zero set of u may have vanishing points ofinfinite order, or even contain an open subset while keeping the solu-tion from being trivial. This is true for general uniformly elliptic par-tial di ff erential equations with only H ¨older continuous coe ffi cients;e.g., see [M]. For this reason, we may expect the same situation to oc-cur in our case; note that if v solves div( a ( x ) ∇ v ) =
0, then v + /β + − v − /β − for some distinct β ± > L ) with a ± : = a ( x ) /β ± , and has thesame zero level set with v . Nevertheless, we still the following lemmadue to the Harnack inequality. Lemma 4.1.
Let u be a weak solution of ( L ) in Ω . Then Γ ( u ) = ∂ { u < }∩ Ω .Proof. Let z ∈ Γ ( u ) and let us show that z ∈ ∂ { u < } ∩ Ω as well.Take any open ball B centered at z and the concentric open ball B ′ with half the radius. By the choice of z , we have sup U u > U u ≤ U = B and B ′ . Consider v = u − inf B u which isnonnegative on B . Since v satisfies div( A ( x , u ) ∇ v ) = B , it follows from the Harnack inequality for bounded measurable coe ffi cientsthat sup B ′ v ≤ C inf B ′ v with some universal C >
1, whence(4.1) 0 < sup B ′ u ≤ C inf B ′ u − ( C −
1) inf B u ≤ − ( C −
1) inf B u . Therefore, we can choose y ∈ B such that u ( y ) <
0. Since B was chosento be an arbitrary open ball centered at z , we get that z ∈ ∂ { u < } ∩ Ω .It implies that Γ ( u ) ⊂ ∂ { u < } ∩ Ω . By a similar argument we obtainthe reverse inclusion, so we conclude that Γ ( u ) = ∂ { u < } ∩ Ω . (cid:3) We are interested in the measure µ defined by d µ = L + u + ; i.e., Z φ d µ : = Z a + ∇ u + ∇ φ, ∀ φ ∈ C ∞ c ( Ω ) . Let us make a basic observation on the measure µ . Lemma 4.2.
The measure µ is a positive Radon measure with spt( µ ) ⊂ Γ ( u ) . Moreover, µ is locally finite and satisfies for any B r ( z ) ⊂ Ω , (4.2) µ ( B r ( z )) r n − ≤ C k u k L ( B r ( z )) , where C > is a constant depending only on n and λ .Proof. If φ ∈ C ∞ c ( Ω + ( u )), then since ∇ u + = ∇ u a.e. in Ω + ( u ), we have Z a + ∇ u + ∇ φ = Z a + ∇ u ∇ φ = , from which we deduce that spt( µ ) ⊂ Γ ( u ).Next we prove that µ is a positive measure. Suppose that φ ∈ C ∞ c ( Ω ) is nonnegative on Ω . Let us choose ψ ε for each ε > ≤ ψ ε ≤ , ψ ′ ε ≥ , ψ ε ( t ) = t ≤ ε/ ψ ε ( t ) = t ≥ ε. Then since ψ ε ( u ) φ ∈ W , ( Ω + ( u )), we deduce from the weak formula-tion of ( L ) that0 = Z a + ∇ u ∇ ( ψ ε ( u ) φ ) = Z a + ψ ′ ε ( u ) |∇ u | φ + Z a + ψ ε ( u ) ∇ u ∇ φ. Owing to the fact that ψ ′ ε ≥
0, we have that Z a + ψ ε ( u ) ∇ u ∇ φ ≤ . Following the argument in the proof of Proposition 2.1, we derivethat ψ ε ( u ) ∇ u → ∇ u + weakly in L ( Ω ) as ε →
0, which in turn yields
FREE BOUNDARY ... CONDUCTIVITY 15 that Z a + ∇ u + ∇ φ ≤ . as desired.Along with the observation that L + u + ≥ Ω in the sense of dis-tribution, the inequality (4.2) follows immediately from the interiorenergy estimate. We omit the details. (cid:3) Next lemma shows that there is no di ff erence to define µ by L − u − . Lemma 4.3.
There holds Z φ d µ = Z a − ∇ u − ∇ φ, ∀ φ ∈ C ∞ c ( Ω ) . Proof.
For any φ ∈ C ∞ c ( Ω ), Z φ d µ + − Z a − ∇ u − ∇ φ = − Z ( a + ∇ u + − a − ∇ u − ) ∇ φ = − Z ( a + H ( u ) + a − (1 − H ( u ))) ∇ u ∇ φ = − Z A ( x , u ) ∇ u ∇ φ = . (cid:3)
5. F urther P roperties of the F ree B oundary In this section, we investigate some measure theoretic properties inregard of our free boundary, in preparation of the analysis in Section6. A key observation is that for µ almost every free boundary point,we can find a blowup limit which is a two plane solution, i.e., afunction in form of α x + n − β x − n , after rotation. This idea was used andplayed a key role in [AM]. Although the main arguments in thissection are similar to those in Section 4 of [AM], we will providesome details, for the readers convenience. Definition 5.1.
Let D be a domain in R n and v ∈ L ( D ). Define afunction v z , r in B by v z , r ( x ) : = v ( rx + z ) r n k v k L ( B r ( z )) if k v k L ( B r ( z )) > , , provided that B r ( z ) ⊂ D . Let u be a nontrivial weak solution to ( L ) and z ∈ Γ ( u ). By definition, u z , r L ( B ) ≤ < r < dist( z , ∂ Ω ). Therefore, there is a v ∈ L ( B ) and a sequence r j ց u z , r j → v weakly in L ( B )as j → ∞ . In particular, k v k L ( B ) ≤ u z , r is a weak solution todiv( A ( rx + z , u z , r ) ∇ u z , r ) = B . Utilizing the local energy estimate, we deduce that u z , r j → v weaklyin W , loc ( B ) as j → ∞ up to a subsequence, and thus strongly in L loc ( B ). Likewise, by means of the interior H ¨older estimate, we have0 < γ < u z , r j → v in C γ ′ loc ( B ) as j → ∞ up to a subsequence,for any 0 < γ ′ < γ . As a result, v ∈ W , loc ( B ) ∩ C γ loc ( B ) ∩ L ( B ).As we follow the argument in the proof of Lemma 3.1, we observethat v is a weak solution todiv( A z ( v ) ∇ v ) = B . By defining a function w in B by w ( x ) : = a + ( z ) v + ( x ) − a − ( z ) v − ( x ) , w becomes harmonic in B . It is noteworthy that the zero level surfaceof v coincides with that of w , whence, inherits nice properties of nodalsets of harmonic functions. For instance, v possesses the uniquecontinuation property. Definition 5.2.
Given z ∈ Γ ( u ), define Blo ( u , z ) by the class of allpossible strong L loc ( B ) limits of { u z , r } r > .The following lemma tells us that any free boundary point ofa blowup limit admits a convergent sequence of the original freeboundary points, unless the blowup limit is trivial. Lemma 5.3.
Let S ⊂ Γ ( u ) with µ ( S ) > . Then the following is true for µ -a.e. z ∈ S: For any v ∈ Blo ( u , z ) \ { } and x ∈ ∂ { v > } ∩ B , (5.1) lim r → dist( z + rx , S ) r = . Proof.
Since µ is a Radon measure (Lemma 4.2) and µ ( S ) > µ -a.epoint is a density point of S ; i.e., for µ -a.e. z ∈ S , we have(5.2) lim r → µ ( S ∩ B r ( z )) µ ( B r ( z )) = . A function is said to have the unique continuation property, if it does not vanishon any open subset, unless it vanishes in the entire domain.
FREE BOUNDARY ... CONDUCTIVITY 17
We prove that all density points of S satisfies (5.1).Suppose towards a contradiction that there is some v ∈ Blo ( u , z )and x ∈ ∂ { v > } ∩ B such that (5.1) fails. Then we may take a smallnumber ε > r ε > ε + | x | < z + rx , S ) r ≥ ε for any 0 < r ≤ r ε . In particular, we have that B ε r ( z + rx ) ∩ S = ∅ .Select 0 < ρ < ε + | x | < ρ . Then B ε r ( z + rx ) ⊂ B ρ r ( z ). Onthe other hand, from (5.2) one may deduce thatlim r → µ ( B ρ r \ S ) µ ( B ρ r ( z )) = . Combining these two observations, we arrive at(5.4) lim r → µ ( B ε r ( z + rx )) µ ( B ρ r ( z )) = . Now let us take a sequence r j ց u z , r j → v stronglyin L loc ( B ). As observed earlier, we may without loss of generality saythat u z , r j → v weakly in W , loc ( B ). As we denote by µ j the measure d µ j = div( a + ( r j x + z ) ∇ u z , r j ) and by µ the measure d µ = a + ( z ) ∆ v + , onemay easily deduce from the weak W , loc ( B ) convergence that µ j → µ locally in B in the sense of Radon measure. As both B ε ( x ) and B ρ being compactly contained in B , it followsfrom the above convergence that µ v ( B ε ( x )) = lim j →∞ µ u z , rj ( B ε ( x )) < ∞ and µ v ( B ρ ) = lim j →∞ µ u z , rj ( B ρ ) < ∞ . Moreover, we have µ v ( B ε ( x )) >
0. Otherwise, we have µ v ( B ε ( x )) =
0, which implies that v + is anonnegative harmonic function in B ε ( x ), and thus, by the maximumprinciple, v + ≡ B ε ( x ). Due to Lemma 4.3, however, one may alsodeduce from µ v ( B ε ( x )) = v − is a nonnegative harmonic functionin B ε ( x ), whence showing that v − ≡ B ε ( x ). It follows that v ≡ B ε ( x ), and thus by the unique continuation property of v , we arriveat v ≡ B , a contradiction to our initial choice of v . Similarly, wemay prove µ v ( B ρ ) >
0. However, (5.4) implies that µ v ( B ε ( x )) µ v ( B ρ ) = lim j →∞ µ u z , rj ( B ε ( x )) µ u z , rj ( B ρ ) = lim j →∞ µ ( B ε r j ( z + r j x )) µ ( B ρ r j ( z )) = , That is, Z φ d µ j → Z φ d µ for any φ ∈ C ∞ c ( B ) . a contradiction. Thus, the lemma is proved. (cid:3) With the previous lemma at hand, we observe that blowup at afree boundary point of a blowup limit indeed belongs to a limit ofthe original blowup sequence. This is certainly true for harmonicfunctions. Moreover, it becomes intuitively clear that only nonde-generate points enjoy this property, unless degenerate points carrysome positive measure.
Lemma 5.4.
Let < ρ < . The following is true for µ -a.e. z ∈ Γ ( u ) : Forany v ∈ Blo ( u , z ) and x ∈ ∂ { v > } ∩ B − ρ , there holds v x ,τρ ∈ Blo ( u , z ) forany < τ ≤ .Proof. Fix 0 < ρ <
1. Note that the statement is clearly true if x = u z , r i → v in L loc ( B ) as i → ∞ , u z ,ρ r i = ( u z , r i ) ,ρ → v ,ρ in L ( B ) as i → ∞ . Therefore, we only need to consider x ∈ ∂ { v > } ∩ ( B ρ \ { } ).In addition, one may argue in a similar manner as above anddeduce that if v x ,ρ ∈ Blo ( u , z ), then v x ,τρ ∈ Blo ( u , z ) for any 0 < τ ≤ ffi ces to prove the statement of Lemma 5.4 for τ = j , k , l ) of positive integer, define S j , k , l by the subset of Γ ( u ) such that for each z ∈ S j , k , l , dist( z , ∂ Ω ) ≥ l and there correspond v z ∈ Blo ( u , z ) and x z ∈ ∂ { v > } ∩ B ρ such that(5.5) u z , r − v zx z ,ρ L ( B / (2 j ) ) > k , ∀ r ≤ l . Note that for any z ∈ S j , k , l , the corresponding v z is not identically zeroin B because any scaled version of a trivial function was defined tobe trivial.Assume to the contrary that µ ( S j , k , l ) > j , k and l , and let us write S = S j , k , l .Set A : = { v zx z ,ρ : z ∈ S } . Note that A is uniformly bounded in W , ( B / (2 j ) ). As W , ( B / (2 j ) ) being compactly embedded in L ( B / (2 j ) ), A can be covered by a countable collection { G m } ∞ m = of balls in L ( B / (2 j ) )with diam( G m ) ≤ k ; i.e., w − φ L ( B / (2 j ) ≤ k whenever w , φ ∈ G m .As A being covered by { G m } ∞ m = , we may decompose S into acountable union of S m , m = , , · · · , where S m : = { z ∈ S : v zx z ,ρ ∈ G m } .From the assumption that µ ( S ) >
0, there exists some m such that µ ( S m ) > S the set S m .Owing to Lemma 5.3, we may choose z ∈ S such that (5.1) is true.Since | x z | < ρ , we are able to take 0 < σ < | x z | + (1 − ρ ) < FREE BOUNDARY ... CONDUCTIVITY 19 σ . Now we denote by { r i } ∞ i = the sequence along which u z , r i → v zx z ,ρ strongly in L loc ( B ), and thus, in L ( B σ ), as i → ∞ . By means of (5.1),one may assign a point z i ∈ S for each i = , , · · · such that0 = lim i →∞ | z i − ( z + r i x z ) | r i = lim i →∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z i − zr i − x z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Take a su ffi ciently large i such that | z i − zr i | ≤ σ − (1 − ρ ) for all i ≥ i .Then z i − zr i + ρ y ∈ B σ for any y ∈ B , which implies that u z i ,ρ r i = ( u z , r i ) zi − zri ,ρ → v zx z ,ρ strongly in L ( B ) . Hence, by making i even larger if necessary, we have(5.6) u z i ,ρ r i − v zx z ,ρ L ( B ) ≤ k , ∀ i ≥ i , where k is the integer chosen in the beginning of this proof.However, we observe that(5.7) v z i x zi ,ρ − v zx z ,ρ L ( B / (2 j ) ≤ k , ∀ i = , , · · · , since all z i ’s and z belong to S = S m ; recall that diam( G m ) ≤ k .Combining (5.7) with (5.6), we arrive at u z i ,ρ r i − v z i x zi ,ρ L ( B / (2 j ) ≤ k , ∀ i ≥ i . This contradicts our assumption in (5.5) as we further enlarge i sothat r i ≤ l for all i ≥ i . This completes the proof. (cid:3) Next we prove that degenerate points with vanishing order greaterthan or equal to 2 have µ -measure zero. Lemma 5.5.
Let D ( u ) be a subset of Γ ( u ) such that for each z ∈ D ( u ) , (5.8) lim r → k u k L ( B r ( z )) r n + < ∞ . Then µ ( D ( u )) = .Proof. Owing to (4.2), we have, for any z ∈ D ( u ),(5.9) lim r → µ ( B r ( z )) r n < ∞ . Hence, the measure µ restricted on D ( u ) is absolutely continuouswith respect to L n , the n -dimensional Lebesgue measure. Let us consider a subset D ′ ( u ) of D ( u ) consisting of all z at whichlim r → k u k L ( B r ( z )) r n + < ∞ . Following the argument right above, one may also observe thatthe measure µ restricted on D ′ ( u ) is absolutely continuous with re-spect to H n + , the ( n + ff measure. Evidently, H n + ( D ′ ( u )) =
0, from which it follows that µ ( D ′ ( u )) = ffi ce to prove L n ( D ( u ) \ D ′ ( u )) = µ ( D ( u )) =
0. Suppose towards a contradiction that L n ( D ( u ) \ D ′ ( u )) >
0. Then there exists a density point z ∈ D ( u ) ∩ D ′ ( u ) with respect to L n . As D ( u ) ∩ D ′ ( u ) ⊂ Γ ( u ), it follows that(5.10) lim r → L n ( Γ ( u ) ∩ B r ( z )) L n ( B r ( z )) = . However, as z being chosen from D ( u ) \ D ′ ( u ), we havelim r → k u k L ( B r ( z )) r n + = ∞ , by which we may take a sequence r j ց k u k L ( B rj ( z )) ≤ k u k L ( B rj / ( z )) . Taking a further subsequence of { r j } ∞ j = if necessary, we obtain astrong L loc ( B ) limit v of { u z , r j } ∞ j = ; see the discussion before Lemma5.3. With (5.11) at hand, v satisfies k v k L ( B / ) ≥ − . Without loss of generality, we can take x ∈ ¯ B / such that v ( x ) ≥ c n .Then the local H ¨older regularity of v allows us to choose a smallconstant 0 < δ < such that sup B δ ( x ) v ≥ c n . Utilizing the locallyuniform convergence, we may take a su ffi ciently large j δ such thatsup B δ ( x ) u z , r j ≥ c n for all j ≥ j δ , which su ffi ces to prove that B δ ( x ) ⊂ Ω + ( u z , r j ) ∩ B for all j ≥ j δ . Alternatively, we have B δ r j ( r j x + z ) ⊂ Ω + ( u ) ∩ B r j ( z ) for any j ≥ j δ , whencelim j →∞ L n ( B r j ( z ) \ Γ ( u )) L n ( B r j ( z )) ≥ lim j →∞ L n ( B δ r j ( r j x + z )) L n ( B r j ( z )) = δ n > , which violates (5.10). Hence, the proof is finished. (cid:3) We finish this section by proving that for µ almost every pointof the free boundary has a two plane solution as its blowup limit.Due to the preceding lemma, we only need to consider those points FREE BOUNDARY ... CONDUCTIVITY 21 which have vanishing order less than 2. This will give us a doublingcondition, as observed in the proof of Lemma 5.5, so that we mayfind a nontrivial blowup limit.
Lemma 5.6.
For µ -a.e. z ∈ Γ ( u ) , there exists a function v ∈ Blo ( u , z ) suchthat v ( x ) = β + ( x · ν ) + − β − ( x · ν ) − in B , where β ± = β/ a ± ( z ) for some β > and ν is a unit vector in R n .Proof. With Lemma 5.5 at hand, it su ffi ces to prove this lemma for z ∈ Γ ( u ) \ D ( u ).Since z < D ( u ), one may follow the argument in the proof of Lemma5.5 and obtain a nontrivial v ∈ Blo ( u , z ). As discussed in the begin-ning of this section, the function w , defined by w ( x ) : = a + ( z ) v + ( x ) − a − ( z ) v − ( x ), is harmonic in B . Hence, we may find x ∈ ∂ { w > } ∩ B where |∇ w ( x ) | >
0. Taking ν : = ∇ w ( x ) |∇ w ( x ) | and β = |∇ w ( x ) | , one mayeasily deduce that as ρ ց w x ,ρ → β ( x · ν ) strongly in L loc ( B ) , or alternatively, v x ,ρ → β + ( x · ν ) + − β − ( x · ν ) − strongly in L loc ( B ) . The proof now follows by Lemma 5.4. (cid:3) C ,α R egularity of N ondegenerate F ree B oundaries This section is devoted to the investigation of higher regularityof our free boundaries. According to Lemma 5.6, it is natural tostart with those points which have two plane solutions as blowuplimits. We may prove later that our free boundary satisfies a flatnesscondition (e.g., (6.1)) around such points, and it turns out that such aflatness condition can be improved in an inductive way (Lemma 6.6).After all, we observe that those points admit a small neighborhoodin which the free boundary is a C ,α graph. Theorem 6.1.
Let a + and a − satisfy (1.3) with an α -H¨older continuous ω ,and u be a weak solution to ( L ) in Ω . Then for µ -a.e. z ∈ Γ ( u ) , there isr > such that Γ ( u ) ∩ B r ( z ) is a C ,α graph; here the radius r and the C ,α character of the graph may depend on u, z and the H¨older norm of a ± . For further analysis on the measure theoretic regularity of our freeboundary, let us define the classes of nondegenerate and degeneratepoints.
Definition 6.2.
Let u be a weak solution to ( L ) in Ω . Define N ( u )and S ( u ) by the classes of nondegenerate points and respectivelydegenerate points, i.e., N ( u ) : = ( z ∈ Γ ( u ) : lim r → k u k L ( B r ) r n + > ) , and S ( u ) : = Γ ( u ) \ N ( u ) = ( z ∈ Γ ( u ) : lim r → k u k L ( B r ) r n + = ) . Theorem 6.1 implies that our free boundary points are essentiallynondegenerate, which opens up a way to characterize the measure µ with respect to ( n − ff measure. Theorem 6.3.
Under the assumption of Theorem 6.1, spt( µ ) has σ -finite ( n − -dimensional Hausdor ff measure. More specifically, N ( u ) has σ -finite ( n − -dimensional Hausdor ff measure, while S ( u ) has µ -measure zero. Inaddition, the conclusion of Theorem 6.1 holds for every z ∈ N ( u ) , and N ( u ) is relatively open in Γ ( u ) . Let us observe an elementary fact for the future reference.
Lemma 6.4.
Let D be a domain in R n , ≤ p ≤ ∞ and f , g ∈ L p ( D ) . Thenfor any given positive numbers β + and β − , there holds ( β + f + − β − f − ) − ( β + g + − β − g − ) L p ( D ) ≤ ( β + + β − ) f − g L p ( D ) . Proof.
Set δ : = f − g L p ( D ) . We only need to consider the inequalityon D ∩ { f ≥ } ∩ { g ≤ } and D ∩ { f ≤ } ∩ { g ≥ } . Due to the symmetryof the proof, let us only focus on the former set, which we denote by E , for notational convenience. Then 0 ≤ − g ≤ f − g and 0 ≤ f ≤ f − g on E , which implies that g L p ( E ) ≤ δ and f L p ( E ) ≤ δ . Therefore, ( β + f + − β − f − ) − ( β + g + − β − g − ) L p ( E ) = β + f − β − g L p ( E ) ≤ ( β + + β − ) δ, and thus the proof is finished. (cid:3) In what follows we are going to use Lemma 6.4 with p = ∞ .Let us introduce a notation for two plane solutions. Definition 6.5.
Given β > z ∈ R n and a vector ν in R n , define afunction P z β,ν on R n by P z β,ν ( x ) : = β a + ( z ) ( x · ν ) + − β a − ( z ) ( x · ν ) − , In particular, we denote P z β,ν by P β,ν . FREE BOUNDARY ... CONDUCTIVITY 23
Lemma 6.6.
Let u be a weak solution to ( L ) in B satisfying ∈ Γ ( u ) , k∇ u k L ∞ ( B ) ≤ and (6.1) u − P β,ν L ( B ) ≤ ε, for some β, ε > and a nonzero vector ν ∈ R n . There are (small) positiveuniversal constants η and ¯ r such that if a + and a − satisfy (6.2) | a ± ( x ) − a ± (0) | ≤ εη | x | α in B , then there exists a sequence { ν k } ∞ k = of vectors in R n such that (6.3) 1¯ r n / u − P β,ν k L ( B ¯ rk ) ≤ ε ¯ r k (1 + α ) , ∀ k = , , , · · · , and that with a universal constant c > , (6.4) | ν k − ν k − | ≤ c β ε ¯ r k α , ∀ k = , , · · · . Proof.
From (6.1) observe that the initial case for (6.3) is satisfied bysimply setting ν = ν .Now we suppose that (6.3) and (6.4) are met for some k ≥
0. Thendefine a function u k in B by u k ( x ) : = u (¯ r k x ) / ¯ r k . Since k∇ u k L ∞ ( B ) ≤ k∇ u k k L ∞ ( B ) = k∇ u k L ∞ ( B ¯ rk ) ≤ v k and w k defined in B by v k ( x ) : = a + (0) u + k ( x ) − a − (0) u − k ( x ) , and respectively by w k ( x ) : = v k ( x ) − β ( x · ν k ) ε ¯ r k α . By the induction hypothesis on (6.3) and Lemma 6.4 we have(6.5) k w k k L ( B ) ≤ λ . On the other hand, one may observe from ( L ) that w k solves(6.6) ∆ w k = ε ¯ r k α div( σ k ( x ) ∇ u k ) in B in the sense of distribution, where σ k ( x ) : = ( a + (0) − a + (¯ r k x )) H ( u k ) + ( a − (0) − a − (¯ r k x ))(1 − H ( u k )) . Since k∇ u k k L ∞ ( B ) ≤
1, we deduce from (6.2) that(6.7) 1 ε ¯ r k α k σ k ∇ u k k L ∞ ( B ) ≤ η. Now consider the harmonic replacement h of w k ; i.e., ∆ h = B with h = w on ∂ B . With (6.7) at hand, we may apply the global L ∞ ,and thus, L estimate (e.g., Theorem 8.15 in [GT]) to (6.6) and obtain(6.8) k w k − h k L ( B ) ≤ c η. Due to (6.5) and (6.8), it follows that k h k L ( B ) ≤ c ; here we assumethat η <
1, which will be fulfilled later. By the interior estimatesfor derivatives of harmonic functions, we know that |∇ h (0) | ≤ c and D h L ∞ ( B / ) ≤ c . Let us denote by l ( x ) the linear function ∇ h (0) · x .Then, the Taylor expansion yields that if 0 < r ≤ ,1¯ r n / k h − h (0) − l k L ( B ¯ r ) ≤ c r . On the other hand, since w k (0) =
0, it follows from (6.8) that | h (0) | = | h (0) − w k (0) | ≤ c η . Therefore, as we denote the linear function ∇ h (0) · x by l ( x ), we derive that 1¯ r n / k w k − l k L ( B ¯ r ) ≤ λ r + α , as we choose ¯ r small enough such that 2 c ¯ r ≤ λ ¯ r + α , and accordinglyselect η so as to satisfy 8 c η ≤ λ ¯ r + α .Let us define ν k + : = ν k + ε ¯ r k α β − ∇ h (0). Then as we rephrase theabove inequality in terms of v k and then apply Lemma 6.4, we arriveat 1¯ r ( k + n / u − P β,ν k + L ( B ¯ rk + ) ≤ ε ¯ r ( k + + α ) . This inequality is exactly (6.3) with k replaced by k +
1. Owing to thefact that |∇ h (0) | ≤ c , the inequality in (6.4) is also true again for k + k . The proof is finished by the induction principle. (cid:3) An immediate consequence of the preceding lemma is that if u isclose to a two plane solution around a free boundary point, then thefree boundary can be locally trapped in between two C ,α graph. Weskip the proof; one can easily deduce it by making use of Lemma 6.4. Corollary 6.7.
Under the circumstance of Lemma 6.6, there exists a unitvector e in R n such that r n / u − P β, e L ( B r ) ≤ C ε r + α , and in particular, Γ ( u ) ∩ B ⊂ { x ∈ B : | x · e | ≤ C εβ − | x | + α } . FREE BOUNDARY ... CONDUCTIVITY 25
Lemma 6.8.
Let u be a weak solution to ( L ) in B with k u k L ( B ) ≤ .Suppose that u satisfies (6.9) u − P β,ν L ( B / ) ≤ δ and (6.10) | a ± ( x ) − a ± (0) | ≤ δ | x | α in B . for some positive numbers β and δ < , and a unit vector ν ∈ R n . Thenfor < ρ ≤ satisfying ρ + α ≤ c β , there holds for any z ∈ B ρ ∩ Γ ( u ) ,B ρ ( z ) ⊂ B / and (6.11) u ( ρ · + z ) ρ − P z β,ν L ( B ) ≤ c δρ . Proof.
From (6.9) and Lemma 6.4, we obtain(6.12) k v − l k L ( B / ) ≤ δλ , where l ( x ) : = β ( x · ν ). As observed in the proof of Lemma 6.6, ∆ ( v − l ) = div( σ ( x ) ∇ u ) in B / , where σ ( x ) : = ( a + (0) − a + ( x )) H ( u ) + ( a − (0) − a − ( x ))(1 − H ( u )) in B . Dueto the assumption that k u k L ( B ) ≤
1, we observe from Theorem 3.3that k∇ u k L ∞ ( B / ) ≤ c . In combination with (6.10), we derive that(6.13) k σ ∇ u k L ∞ ( B / ) ≤ c δ. Utilizing (6.12) and (6.13), we deduce from the local boundness ofweak solutions (Theorem 8.18 in [GT]) that k v − l k L ∞ ( B / ) ≤ c (cid:16) k v − l k L ( B / ) + k σ ∇ u k L ∞ ( B / ) (cid:17) ≤ c δ. Making use of Lemma 6.4 once again, we arrive at(6.14) u − P β,ν L ∞ ( B / ) ≤ c δ. From (6.14), we deduce that for any z ∈ Γ ( u ) ∩ B / ,(6.15) β | z · ν | max { a + (0) − , a − (0) − } ≤ c δ. Therefore, one may deduce as in Lemma 6.4 that if x ∈ B , then(6.16) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) P β,ν x + z ρ ! − P β,ν ( x ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ c δρ . On the other hand, as we restrict z ∈ Γ ( u ) ∩ B ρ , we derive from(6.10) that for any x ∈ B , | P z β,ν ( x ) − P β,ν ( x ) | ≤ βδρ α λ ≤ c δρ , (6.17)by choosing ρ su ffi ciently small such that βρ + α /λ ≤ c ; here wehave used the assumption that ν is a unit vector to estimate | x · ν | ≤ (cid:3) We are now in position to prove Theorem 6.1.
Proof of Theorem 6.1.
For definiteness, let ω in (1.3) to be ω ( r ) = ω r α with ω > < α < µ -a.e. z ∈ Γ ( u ), there is a sequence r j → u z , r j → P β,ν in L loc ( B ) for some β > ν ∈ R n .Now fix δ > ffi ciently large j such that u z , r j − P β,ν L ( B / ) ≤ δ, ∀ j ≥ j δ . By definition, for any j = , , · · · , u z , r j L ( B ) = u z , r j solvesdiv( A ( r j x + z , u z , r j ) ∇ u z , r j ) = B . Since ω is H ¨older, and thus, Dini continuous, we may invoke Theo-rem 3.3 and assert that(6.18) ∇ u z , r j L ∞ ( B / ) ≤ L , where L > n , λ , ω and α . It is also clear that(6.19) | a ± ( r j x + z ) − a ± ( z ) | ≤ ω r α j | x | α in B . Select an integer k δ larger than j δ such that(6.20) ω r α k δ ≤ δ. Then applying Lemma 6.8 to u z , r j and a ± ( r j · + z ), we obtain 0 < ρ < with ρ ≤ c β , such that for any ξ ∈ B ρ ( z ) ∩ Γ ( u ), B ρ ( ξ ) ⊂ B / and(6.21) u z , r k δ ( ρ · + ξ ) ρ − P ξβ,ν L ( B ) ≤ c δρ . Now let us fix ξ ∈ B ρ ( z ) ∩ Γ ( u ), and for notational convenience,define a function ˜ u on B by˜ u ( x ) : = u z , r j ( ρ x + ξ ) ρ L , FREE BOUNDARY ... CONDUCTIVITY 27 where L is chosen from (6.18), so that we have(6.22) k∇ ˜ u k L ∞ ( B ) ≤ L u z , r k δ L ( B ρ ( ξ )) ≤ . Observe that ˜ u is a weak solution todiv( ˜ A ( x , ˜ u ) ∇ ˜ u ) = B , where ˜ A ( x , ˜ u ) = ˜ a + ( x ) H ( ˜ u ) + ˜ a − ( x )(1 − H ( ˜ u )) with˜ a ± ( x ) : = a ± ( r k δ ρ x + r k δ ξ + z ) in B . Therefore, it follows from (6.20) that(6.23) | ˜ a ± ( x ) − ˜ a ± (0) | ≤ ω r α k δ ρ α | x | α ≤ δρ α | x | α in B . Moreover, in view of (6.21), we have that(6.24) ˜ u − P β L ,ν L ( B ) ≤ c δρ L . Owing to the inequalities in (6.22), (6.23) and (6.24), ˜ u and ˜ a ± fallunder the situation of Lemma 6.6; more specifically, we replace β and ε in (6.1) with β/ L and respectively c δρ L , and make a further restrictionon ρ such that ρ + α ≤ c η L as well, where η is the constant in (6.2).Hence, we deduce from Corollary 6.7 that there exists a nonzerovector e ξ ∈ R n such that Γ ( ˜ u ) ∩ B ⊂ ( x ∈ B : | x · e ξ | ≤ c ε L β | x | + α ) . In terms of u z , r k δ , we see that for any ξ ∈ Γ ( u z , r k δ ) ∩ B ρ , Γ ( u z , r k δ ) ∩ B ρ ( ξ ) ⊂ ( x ∈ B ρ ( ξ ) : | ( x − ξ ) · e ξ | ≤ c δρ + α β | x − ξ | + α ) . Since the constant c δ/ ( ρ + α β ) is independent on ξ , we conclude that Γ ( u z , r j ) ∩ B ρ , and thus, Γ ( u ) ∩ B r k δ ρ ( z ) is a C ,α graph, proving thetheorem. (cid:3) Now we are left with proving Theorem 6.3. Before we begin, let usmake an important observation on the nondegenerate part N ( u ) ofthe free boundary, thanks to the monotonicity of the ACF formula. Lemma 6.9. z ∈ N ( u ) if and only if there exist (small) positive numbers η and r , both depending on z, such that r n / k u k L ( B r ( z )) ≥ η r for any < r ≤ r . Proof.
The ‘if’ part is obvious by the definition of N ( u ), whence weprove the ‘only if’ part.Without loss of any generality, let us assume that k u k L ( Ω ) = z ∈ N ( u )which admits two sequences r j , t j ց r − n / j k u k L ( B rj / ( z )) ≥ η r j for some η > t − n / j k u k L ( B tj ( z )) ≤ j t j for j = , , · · · .Define u j ( x ) : = u ( r j x + z ) / r j and v j ( x ) : = u ( t j x + z ) / t j , in which casewe have u j L ( B ) ≥ η . Without loss of any generality let us assumethat sup B u j ≥ c η . As shown in the proof of Lemma 4.1, it followsthat inf B u j ≤ − c η . The uniform Lipschitz regularity (Theorem 3.3)implies that there are some balls B ± j ⊂ B such that u j ≥ c η in B + j and u j ≤ − c η in B − j with L n ( B ± j ) ≥ δ >
0. Thus, the Poincar´e inequalityyields that c ,δ η ≤ u ± j L ( B ) ≤ c ,δ ∇ u ± j L ( B ) . By the monotonicity of the ACF formula (Proposition A.2), Φ (0 + , z , u + , u − ) = lim j →∞ Φ ( r j , z , u + , u − ) = lim j →∞ Φ (1 , u + j , u − j ) ≥ c δ η ;see (A.1) for the definition of Φ .On the other hand, v j L ( B ) ≤ j implies that ∇ v j W , ( B / ) ≤ c j ,and thus, v j → W , ( B / ). Then Φ (0 + , z , u + , u − ) = lim j →∞ Φ ( t j / , z , u + , u − ) = lim j →∞ Φ (1 / , v + j , v − j ) = , a contradiction. (cid:3) Remark . We may only assume Dini continuity on a + and a − tohave Lemma 6.9, since its proof only involves the monotonicity ofthe ACF formula.By Lemma 6.9, we may decompose the class N ( u ) into a countableunion of N j , k ( u ) for j , k = , , · · · , where z ∈ N j , k ( u ) if1 r n / k u k L ( B r ( z )) ≥ j r for any 0 < r ≤ k . Note that for any z ∈ N j , k ( u ), we have dist( z , ∂ Ω ) ≥ k . Lemma 6.11.
For any pair ( j , k ) of positive integers, H n − ( N j , k ( u )) ≤ C j , k k u k L ( Ω ) . Proof.
For simplicity let us assume that k u k L ( Ω ) =
1. Fix a pair ( j , k )of positive integers. Let us take a compact subset D of Ω such that k < dist( D , ∂ Ω ) < k . FREE BOUNDARY ... CONDUCTIVITY 29
Select a smooth cuto ff function φ on Ω such that φ ≡ D ,dist(spt( φ ) , ∂ Ω ) ≥ k and |∇ φ | ≤ Ck . Given η >
0, consider a function ψ η on R defined by ψ η ( t ) = t ≥ η , ψ η ( t ) = η − t if 0 ≤ t < ε and ψ η ( t ) = t <
0. By Lemma 4.2, we derive that(6.25) Z a + ∇ u + ∇ ( ψ η ( u + ) φ ) ≤ , by using ψ η ( u + ) φ as a test function.Noting that Z a + ∇ u + ∇ ( φ η ( u + ) φ ) = η − Z { < u + <η } a + |∇ u + | φ + Z ψ η ( u + ) a + ∇ u + ∇ φ, and using (1.2), we may derive from (6.25) that λη − Z { < u + <η } |∇ u + | φ ≤ λ − Z |∇ u + ||∇ φ | . Thus, from Theorem 3.3 and the construction of φ , it follows that Z { < u + <η }∩ D |∇ u + | ≤ ck η. One may also notice that the above argument can also be applied to u − , whence we conclude that(6.26) Z {| u | <η }∩ D |∇ u | ≤ ck η. Now let us take a countable open cover { B i } ∞ i = of N j , k ( u ), wherefor each i , B i is a ball of radius ε with its center at N j , k ( u ). By theVitali covering lemma, we may also choose this open cover to havefinite overlapping times, whose finiteness, say N , depends only onthe dimension n . Moreover, using the nondegeneracy of u on N j , k ( u )and the local Lipschitz regularity in Theorem 3.3, it is not hard toobserve that each B i contains subballs B ± i ⊂ Ω ± ( u ) with radius c j , k ε where u ± ≥ ε j . Owing to this fact, we are able to use the Poincar´einequality to deduce that(6.27) c j , k ε L n ( B i ) ≤ Z B i ( u ± ) ≤ ˜ c j , k ε Z B i |∇ u ± | Now let us take ε ≤ k . Then dist( B i , ∂ Ω ) ≥ k , since we knowthat dist( N j , k ( u ) , ∂ Ω ) ≥ k . Because of Theorem 3.3, we deduce that B i ⊂ {| u | ≤ ck ε }∩ D , where D is the compact set chosen in the beginning of this proof. Hence, inserting η = ck ε in (6.26), we derive from (6.27)that ∞ X i = L n ( B i ) ≤ ˆ c j , k ∞ X i = Z B i |∇ u | ≤ ˆ c j , k N Z {| u |≤ ck ε }∩ D |∇ u | ≤ c ˆ c j , k Nk ε. Therefore, ∞ X i = diam( B i ) n − ≤ C j , k , and the conclusion of this lemma follows from the arbitrary choiceof ε ≤ k . (cid:3) Remark . As in the proof of Lemma 6.9, that of Lemma 6.11 onlyrequires Dini continuity of a + and a − , since the local Lipschitz regu-larity of u in Theorem 3.3 played an essential role.We are now ready to prove Theorem 6.3. Proof of Theorem 6.3.
Suppose towards a contradiction that µ ( S ( u )) >
0. Then by Theorem 6.1, we can find z ∈ S ( u ) such that B r ( z ) ∩ Γ ( u ) isa C ,α graph for some r >
0. In particular, from the proof of Theorem6.1 and Corollary 6.7, we observe that for a su ffi ciently small r , wehave 1 ρ u z , r ( ρ · ) → P β, e strongly in L ( B ) as ρ → , for some β > e ∈ R n . Therefore,lim ρ → k u k L ( B ρ r ( z )) ( ρ r ) n + = k u k L ( B r ( z )) r n + P β, e L ( B ) > , a contradiction against the assumption that z ∈ S ( u ).Thus, µ ( S ( u )) =
0. Then it follows from N ( u ) = Γ ( u ) \ S ( u ) that N ( u ) has full µ -measure. Due to Lemma 6.11, we know that N ( u ) has σ -finite ( n − ff measure, and so does spt( µ ).Finally, let us prove that N ( u ) is relatively open in Γ ( u ). Fix z ∈ N ( u ).Then Blo ( u , z ) contains a two plane solution, say P β,ν , for some β > ν ∈ R n , due to the uniform Lipschitz regular-ity (Theorem 3.3) and the nondegeneracy of z . Hence, we may gothrough the proof of Theorem 6.1 and conclude that B r ( z ) ∩ Γ ( u ) is a C ,α graph. Then we apply the Hopf lemma to ( L ) in B r ( z ) ∩ Ω ± ( u ),and deduce that for all ξ ∈ B r ( z ) ∩ Γ ( u ), ∂ u ± ∂ν ± ( ξ ) >
0, where ν ± is theinward unit normal to Ω ± ( u ) at ξ . Clearly, ξ ∈ N ( u ), which provesthat B r ( z ) ∩ Γ ( u ) ⊂ N ( u ), as desired. (cid:3) FREE BOUNDARY ... CONDUCTIVITY 31
7. A nalysis on M atrix C oefficient C ases Here we shall extend our main results to (a special type of) matrixcoe ffi cient cases. Let a + and a − be functions on R n satisfying (1.2)and (1.3). Under this situation, consider a symmetric ( n × n )-matrixvalued mapping P on R n satisfying(7.1) λ I ≤ P ( x ) ≤ λ I , ∀ x ∈ R n , and(7.2) P ( x ) − P ( y ) ≤ ω ( | x − y | ) , ∀ x , y ∈ R n , where λ and ω are the same quantities appearing in (1.2) and respec-tively (1.3). Define symmetric ( n × n )-matrix valued mappings A + and A − on R n by(7.3) A + ( x ) : = a + ( x ) P ( x ) and A − ( x ) : = a − ( x ) P ( x ) , and then a function A : R n × R → R by(7.4) A ( x , s ) : = A + ( x ) H ( s ) + A − ( x )(1 − H ( s )) , where H is the Heaviside function. Set Ω to be a bounded domain in R n and let u be a weak solution to( P ) div( A ( x , u ) ∇ u ) = Ω . Remark . Alternatively, we may consider uniformly elliptic Dinicontinuous matrices A + and A − such that A + ( x ) = f ( x ) A − ( x ) for a Dinicontinuous real valued f .The existence of weak solutions to ( P ) can be proved by the sameargument in the proof of Proposition 2.1, since there we only usethe interior energy estimate for weak solutions to elliptic PDE withbounded measurable coe ffi cients.It is noteworthy that the limiting equation of ( P ) at a point z ∈ Ω isdiv(( A − ( z ) + ( A + ( z ) − A − ( z )) H ( v )) ∇ v ) = , whence the function w , defined by w ( x ) : = a + ( z ) v + ( x ) − a − ( z ) v − ( x ),solves div( P ( z ) ∇ w ) = . That is, w is a harmonic function up to a bilinear transformation.Hence, the arguments throughout Section 3 – 6 are expected to gothrough with weak solutions to ( P ) as well.The first result in concern with weak solutions to ( P ) is the interiorLipschitz regularity of the associated weak solutions. Theorem 7.2.
Assume that a ± and P satisfy (1.3) and respectively (7.2) with a Dini continuous ω and let u be a bounded weak solution to ( P ) in Ω .Then u ∈ W , ∞ loc ( Ω ) and for any D ⋐ Ω , k∇ u k L ∞ ( D ) ≤ C k u k L ( Ω ) d n + , where C is a constant depending only on n, λ and ω and d = dist( D , ∂ Ω ) .Proof. Note that Lemma 3.1 can be extended to weak solutions of ( P )in an obvious way. Moreover, by applying Theorem A.4 instead ofProposition A.3, Lemma 3.2 continues to hold. Hence one may followexactly the same argument as in Theorem 3.3 to derive Theorem 7.2.We omit the details. (cid:3) Next we define the measure µ by d µ = div( A + ( x ) ∇ u + ), and inves-tigate the regularity of our free boundary. Let us hereafter followNotation 2.3 and Definition 6.2 for the definition of Ω + ( u ), Ω − ( u ), Γ ( u ), N ( u ) and S ( u ). Theorem 7.3.
Let a ± and P satisfy (1.3) and respectively (7.2) with an α -H¨older continuous ω , and u be a weak solution to ( P ) in Ω . Then for µ -a.e. z ∈ Γ ( u ) , there is r > such that Γ ( u ) ∩ B r ( z ) is a C ,α graph; here theradius r and the C ,α norm of the graph may depend on u, z and the H¨oldernorm of a ± and P.Proof. The arguments in Section 4 and 5 can be easily be generalizedto weak solutions to ( P ). Observe that the function v , defined by v ( x ) : = a + (0) u + ( x ) − a − (0) u − ( x ), is a weak solution todiv( P (0) ∇ v ) = div( σ ( x ) P ( x ) ∇ u ) , where σ is defined exactly the same as in the proof of Lemma 6.6. Un-der the assumptions on Theorem 7.3, we have k σ ∇ u k L ∞ ( B r ) ≤ λ − ω r α for some ω > < α <
1. Thus, Lemma 6.6 can also be extendedto u . The rest of the proof follows similarly with that of Theorem 6.1,and we skip the details. (cid:3) Finally we state our result on the measure theoretic regularity of Γ ( u ) in the matrix coe ffi cient case. Theorem 7.4.
Under the assumption of Theorem 7.3, spt( µ ) has σ -finite ( n − -dimensional Hausdor ff measure. More specifically, N ( u ) has σ -finite ( n − -dimensional Hausdor ff measure, while S ( u ) has µ -measure zero. Inaddition, the conclusion of Theorem 7.3 holds for every z ∈ N ( u ) , and N ( u ) is relatively open in Γ ( u ) . FREE BOUNDARY ... CONDUCTIVITY 33
Proof.
The proof of Lemma 6.9 works with weak solutions to ( P )by invoking Theorem 7.2 and Theorem A.4 instead of Theorem 3.3and respectively Proposition A.2. The proof of Lemma 6.11 can begeneralized to the matrix coe ffi cient case in an obvious manner. Thus,one may follow the same arguments in the proof of Theorem 6.3 toderive Theorem 7.4. We leave out the details to the reader. (cid:3) A ppendix A. ACF M onotonicity F ormula For r > z ∈ R n and u ∈ W , ( B r ( z )), set I ( r , z , u ) : = Z B r ( z ) |∇ u | | x | n − dx , and define the ACF functional Φ ( r , u , v ) by(A.1) Φ ( r , z , u , v ) : = r − I ( r , z , u ) I ( r , z , v ) = r Z B r ( z ) |∇ u | | x | n − dx Z B r ( z ) |∇ v | | x | n − dx . For the notational convenience, let us abbreviate I ( r , , u ) and Φ ( r , , u , v )by I ( r , u ) and respectively Φ ( r , u , v ).Let ω be a modulus of continuity and λ a positive constant. Define L ( λ, ω ) by the class consisting of all elliptic operators L = div( a ( x ) ∇ )such that λ ≤ a ( x ) ≤ λ − and | a ( x ) − a ( y ) | ≤ ω ( | x − y | ) for any x , y ∈ B .In what follows, any moduli of continuity ω is assumed to be Dinicontinuous; i.e., R ω ( r ) r dr < ∞ . Given such an ω , define ψ : (0 , → [0 , ∞ ) by ψ ( r ) = ω ( r ) + Z r ω ( ρ ) ρ d ρ + Z r ω ( ρ ) ρ ! d ρ. Here we follow the approach in [CJK] and [MP].
Lemma A.1.
Let L ∈ L ( λ, ω ) and suppose that u is a nonnegative weaksolution for Lu ≥ in B . Then there are c = c ( n , λ ) > and r = r ( n , λ, ω ) such thatI ( r , u ) ≤ (1 + c ψ ( r )) r n − Z ∂ B r u | ∂ ν u | + n − r n − Z ∂ B r u ! for any < r ≤ r .Proof. Without loss of generality, we may assume that a (0) = c will denote a dimensional constant whichmay vary from one line to another. Write L = div( a ( x ) ∇ ) and b ( x ) = − a ( x ). Then since ∆ u ≥ div( b ( x ) ∇ u )in the sense of distribution, we have ∆ ( u ) ≥ |∇ u | + u div( b ( x ) ∇ u ) inthe same sense, which implies(A.2) I ( r , u ) ≤ Z B r | x | n − ∆ u − Z B r u | x | n − div( b ∇ u ) . Since ∆ | x | − n is a nonpositive measure in R n ,12 Z B r | x | n − ∆ u ≤ r n − Z ∂ B r u ∂ ν u + n − r n − Z ∂ B r u , (A.3)Hence, we only need to estimate the second term of the inequality(A.2). By means of integration by part, one can easily derive that − Z B r u | x | n − div( b ∇ u ) ≤ ψ ( r ) r n − Z ∂ B r u | ∂ ν u | + I ( r , u ) ! + (2 − n ) Z B r bu ∇ u · x | x | n dx . (A.4)Let us split the last term of the right hand side of (A.4) into two parts,namely, Z B r bu ∇ u · x | x | n dx = Z r Z ∂ B ρ bv ρ ∂ ν u ρ n − d H n − d ρ + Z r Z ∂ B ρ m ( ρ ) b ∂ ν u ρ n − d H n − d ρ, (A.5)where m ( ρ ) : = > ∂ B ρ u and v ρ : = u − m ( ρ ). By the Poincar´e inequalityand the H ¨older inequality, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z r Z ∂ B ρ bv ρ ∂ ν u ρ n − d H n − d ρ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ψ ( r ) Z r κ n ρ n − Z ∂ B ρ |∇ θ u | Z ∂ B ρ | ∂ ν u | d ρ ≤ κ n ψ ( r )2 I ( r , u ) . (A.6)On the other hand, the estimate for R r R ∂ B ρ m ( ρ ) b ∂ ν u ρ n − d H n − d ρ can bedone as follows. First using the equation Lu ≥
0, it is easy to see that m ( ρ ) ≤ m ( τ ) − n ω n Z B τ \ B ρ b ∇ u · x | x | n dx whenever 0 < ρ < τ <
1. However, by the H ¨older inequality, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z B τ \ B ρ b ∇ u · x | x | n dx (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ Z τ ω ( σ ) σ n − Z ∂ B σ |∇ u | ! d σ ≤ ( n ω n ) / ψ ( r ) I ( τ, u ) / . FREE BOUNDARY ... CONDUCTIVITY 35
Using these two inequalities, we obtain (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z r Z ∂ B ρ m ( ρ ) b ( x ) ∂ ν u ρ n − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ m ( r ) + ψ ( r )2( n ω n ) / I ( r , u ) / ! ( n ω n ) / ψ ( r )2 I ( r , u ) / ≤ ψ ( r ) r n − Z ∂ B r u + I ( r , u ) ! , (A.7)where in the last inequality we have used Young’s inequality and thefact that m ( r ) = n ω n r n − Z ∂ B r u ! ≤ n ω n r n − Z ∂ B r u . Collecting the estimates (A.6) and (A.7), we proceed in (A.5) as (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z B r bu ∇ u · x | x | n dx (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ c ψ ( r ) r n − Z ∂ B r u + I ( r , u ) ! , and hence, in (A.4), we get (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z B r u | x | n div( b ∇ u ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ c ψ ( r ) r n − Z ∂ B r u | ∂ ν u | + ( n − r n − Z ∂ B r u + I ( r , u ) ! . Inserting this inequality and (A.3) into (A.2), we arrive at(1 − c ψ ( r )) I ( r , u ) ≤ (1 + c ψ ( r )) r n − Z ∂ B r u | ∂ ν u | + n − r n − Z ∂ B r u ! . As a final step, we choose r su ffi ciently small such that c ψ ( r ) ≤ forany r ∈ (0 , r ], and then set c = c , by which the proof is complete. (cid:3) Let us define a function g on [0 ,
1) by g ( r ) : = Z r ψ ( ρ ) d ρ. Proposition A.2.
Let L ± ∈ L ( λ, ω ) and u ± ∈ W , ( B ) ∩ C ( ¯ B ) be suchthat u ± ≥ , u + u − = and L ± u ± ≥ in B , where the last equality is valid in the sense of distribution. Then thefunctional r e ¯ cg ( r ) Φ ( r , u + , u − ) is monotone increasing in (0 , ¯ r ] , for some positive ¯ c and ¯ r, both dependingonly on n, λ and ω . Proof.
Since u ± and ω are being fixed throughout the proof, we sim-plify our notations Φ ( r , u + , u − ) and I ( r , u ± ) by Φ ( r ) and respectively I ± ( r ). Denote by λ ± ( r ) the principal eigenvalue of the spherical Lapla-cian in Σ ± ( r ) : = { u ± > } ∩ ∂ B r . The eigenvalue inequality reads(A.8) λ ± ( r ) Z Σ ± ( r ) u ± ≤ Z Σ ± ( r ) |∇ θ u | . Next write by β ± ( r ) the characteristic constant for Σ ± ( r ). Note that(A.9) β ± ( r )( β ± ( r ) + n − = λ ± ( r ) . We will use the Friedland-Hayman inequality,(A.10) β + ( r ) + β − ( r ) ≥ . Again for brevity, let u , I ( r ), Σ ( r ), α ( r ) and λ ( r ) respectively standfor either u + , I + ( r ), Σ + ( r ), β + ( r ) and λ + ( r ) or u − , I − ( r ), Σ − ( r ), β − ( r ) and λ − ( r ). Applying Young’s inequality and (A.8) together with (A.9), wederive that2 r Z Σ ( r ) u | ∂ ν u | + ( n − Z Σ ( r ) u ≤ ( α ( r ) + n − Z Σ ( r ) u + r α ( r ) Z Σ ( r ) | ∂ ν u | ≤ r α ( r ) Z Σ ( r ) |∇ u | . On the other hand, since ∇ u ∈ L ( B ), I ( r ) is absolutely continuous for r ∈ (0 , r ∈ (0 , I ′ ( r ) = r n − R Σ ( r ) |∇ u | .In view of Lemma A.1 and the inequality above,(A.11) I ( r ) ≤ + c ψ ( r )2 α ( r ) rI ′ ( r ) , for a.e. r ∈ (0 , r ), where c and r are chosen as in Lemma A.1.As I ± ( r ) being absolutely continuous in r , we have, for a.e. r ∈ (0 , ddr ( e ¯ cg ( r ) Φ ( r )) = e ¯ cg ( r ) Φ ( r ) " rI ′ + ( r ) I + ( r ) + rI ′− ( r ) I − ( r ) − + ¯ c ψ ( r ) . By means of (A.10) and (A.11), we arrive at ddr ( e ¯ cg ( r ) Φ ( r )) ≥ e ¯ cg ( r ) Φ ( r ) " β + ( r ) + β − ( r ))1 + c ψ ( r ) − + ¯ c ψ ( r ) ≥ , provided that we have choose r such that c ψ ( r ) < < r ≤ r and then ¯ c ≥ c . The proof is completed. (cid:3) As a result, we also get a useful estimate for the ACF functional.
FREE BOUNDARY ... CONDUCTIVITY 37
Proposition A.3.
Let L ± , u ± and r as in Proposition A.2. Then Φ ( r , u + , u − ) ≤ C k u + k L ( B ) k u − k L ( B ) , for any < r ≤ r , where C depends only on n, λ and ω .Proof. It is not hard to see that I ( r , u ± ) ≤ N k u ± k L ( B ) for some N = N ( n , λ, ω ). With Proposition A.2 at hand, we obtain Φ ( r , u + , u − ) ≤ e ¯ c ( g ( r ) − g ( r )) Φ ( r , u + , u − ) , for 0 < r ≤ r . The proof is now finished by taking C = e ¯ cg ( r ) N / r . (cid:3) Let us make a further generalization to Proposition A.2 and A.3.
Theorem A.4.
The conclusions of Proposition A.2 and A.3 remain to holdwith L ± = div ( A ± ( x ) ∇ ) , where A + and A − are symmetric matrix valuedfunctions on B satisfying the following structure conditions:(i) There is < λ < such that λ I ≤ A ± ( x ) ≤ λ − I in B . (ii) There is a Dini continuous modulus of continuity ω such that | A ± ( x ) − A ± ( y ) | ≤ ω ( | x − y | ) for any x , y ∈ B . (iii) There exists a positive number κ such thatA + (0) = κ A − (0) . Proof.
One may notice that Lemma A.1 can be straightforwardly ex-tended to the matrix coe ffi cient case, provided that the associatedmatrix is symmetric uniformly elliptic and Dini continuous. Next inthe proof of Proposition A.2, the condition (iii) allows us to applyLemma A.1 for u + and u − in the same coordinate, thus not a ff ectingthe Friedland-Hayman inequality. The rest of the proof follows inexactly the same way of those in Proposition A.2 and A.3, and hence,we omit the details. (cid:3) R eferences [AC] Alt, H. W.; Ca ff arelli, L. A. Existence and regularity for a minimum problem withfree boundary . Univ., Sonderforschungsbereich, (1980), 105-144.[ACF] Alt H. W.; Ca ff arelli, L. A.; Friedman A. Variational problems with two phasesand their free boundaries . Trans. Amer. Math. Soc. (2) (1984), 431-461.[ACS] Athanasopolous, I.; Ca ff arelli, L. A.; Salsa, S. The free boundary in an inverseconductivity problem . J. reine angew. Math. (2001), 1-31.[AI] Alessandrini, G.; Isakov, V.
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