Existence of viscosity solutions to two-phase problems for fully nonlinear equations with distributed sources

: In this paper we construct a viscosity solution of a two-phase free boundary problem for a class of fully nonlinear equation with distributed sources, via an adaptation of the Perron method. Our results extend those in [Ca ﬀ arelli, 1988], [Wang, 2003] for the homogeneous case, and of [De Silva, Ferrari, Salsa, 2015] for divergence form operators with right hand side.


Introduction
In the last years the regularity theory for two phase problems governed by uniformly elliptic equations with distributed sources has reached a considerable level of completeness (see for instance the survey paper [10]) extending the results in the seminal papers [2,4] (for the Laplace operator) and in [17,18] (for concave fully non linear operators) to the inhomogeneous case, through a different approach first introduced in [7].
In particular the papers [15] and [8] provides optimal Lipschitz regularity for viscosity solutions and their free boundary for a large class of fully nonlinear equations.
Existence of a continuous viscosity solution through a Perron method has been established for linear operators in divergence form in [3] (homogeneous case) and in [9] (inhomogeneous case), and for a class of concave operators in [19]. The main aim of this paper is to adapt the Perron method to extend the results of [19] to the inhomogeneous case. Although we are largely inspired by the papers [3] and [9], the presence of a right hand side and the nonlinearity of the governing equation presents several delicate points, significantly in Section 6, which require new arguments.
We now introduce our class of free boundary problems and their weak (or viscosity) solutions.
Let Sym n denote the space of n × n symmetric matrices and let F : Sym n → R denote a positively homogeneous map of degree one, smooth except at the origin, concave and uniformly elliptic, i.e. such that there exist constants 0 < λ ≤ Λ with where M = max |x|=1 |Mx| denotes the (L 2 , L 2 )-norm of the matrix M.
As we said, the main aim of this paper is to adapt Perron's method in order to prove the existence of a weak (viscosity) solution of the above f.b.p., with assigned Dirichlet boundary conditions For any u continuous in Ω we say that a point x 0 ∈ F (u) is regular from the right (resp. left) if there exists a ball B ⊂ Ω + (u) (resp. B ⊂ Ω − (u)) such that B ∩ F (u) = x 0 . In both cases, we denote with ν = ν(x 0 ) the unit normal to ∂B at x 0 , pointing toward Ω + (u). Definition 1.1. A weak (or viscosity) solution of the free boundary problem (1.1) is a continuous function u which satisfies the first two equality of (1.1) in viscosity sense (see Appendix A), and such that the free boundary condition is satisfied in the following viscosity sense: (i) (supersolution condition) if x 0 ∈ F is regular from the right with touching ball B, then, near x 0 , with equality along every non-tangential direction, and α ≤ G(β, x 0 , ν(x 0 )); (ii) (subsolution condition) if x 0 ∈ F is regular from the left with touching ball B, then, near x 0 , with equality along every non-tangential direction, and α ≥ G(β, x 0 , ν(x 0 )); We will construct our solution via Perron's method, by taking the infimum over the following class of admissible supersolutions S.
is regular from the left, with touching ball B, then The last ingredient we need is that of minorant subsolution.
(b) every x 0 ∈ F (u) is regular from the right, with touching ball B, and near x 0 where ω(r) → 0 as r → 0 + , and Our main result is the following. (b) the set {w ∈ S : w ≥ u, w = g on ∂Ω} is not empty, then u = inf{w : w ∈ S, w ≥ u} is a weak solution of (1.1) such that u = g on ∂Ω.
Once existence of a solution is established, we turn to the analysis of the regularity of the free boundary.
Theorem 1.5. The free boundary F (u) has finite (n − 1)-dimensional Hausdorff measure. More precisely, there exists a universal constant r 0 > 0 such that for every r < r 0 , for every x 0 ∈ F (u), Moreover, the reduced boundary F * (u) of Ω + (u) has positive density in H n−1 measure at any point of F(u), i.e. for r < r 0 , r 0 universal for every x ∈ F (u). In particular Using the results in [8] we deduce the following regularity result.
Notation. Constants c, C and so on will be termed "universal" if they only depend on λ, Λ, n, Ω, f i ∞ and g.

Asymptotic developments
In this section we show that positive solutions of F(D 2 u) = f (with f continuous up to the boundary) have asymptotically linear behavior at any boundary point which admits a touching ball, either from inside or from outside the domain. We need the following preliminary result.
Let u be the solution to where g E is a cut-off function, g E = 1 on E. If r is sufficiently small then there exists a positive constant We have By [11] we have that v(e 1 /2) ≥ Cδ, for some constant C = C(n, λ, Λ, σ), and by the Boundary Harnack principle applied to v and u 1 (x) = x 1 we get that, in B + 1/2 , for some positive constants c 0 and c 1 , Put z(x) = 1 2 min a 11 (x 1 − x 2 1 )r. The function z is positive in B + 1 and L u z = − a 11 min a 11 r ≤ −r.

Lemma 2.2.
Let Ω 1 be a bounded domain with 0 ∈ ∂Ω 1 and Let u be non-negative and Lipschitz in Ω 1 ∩ B 2 , such that F(D 2 u) = f in Ω 1 ∩ B 2 and that u = 0 in ∂Ω 1 ∩ B 2 . Then there exists α ≥ 0 such that Since u is Lipschitz, a simple computation implies that The functions u k are defined in B + 1 and, by assumption of homogeneity on F, we have F(D 2 u k (x)) = F(r k D 2 u(r k x)) = r k F(D 2 u(r k x)) = r k f (r k x) ≤ r k f ∞ .
We deduce that u k is a supersolution of (2.1). By comparison and Lemma 2.1, there exists C > 0, not depending on k, such that Choosing k, k in such a way that α k + C > α and k > 2/r k we obtain α k > α, a contradiction.
Proof. By assumption, we have that Then we can extend u as the zero function on B + 1 \ Ω 1 so that it is a Lipschitz, non-negative solution to Reasoning in a similar way as in Lemma 2.2, we define α k = inf{β : u(x) ≤ βx 1 in B 1/k }, k ≥ 1. Then 0 ≤ α k < +∞ (u is Lipschitz), and α k α ≥ 0, with u(x) ≤ αx 1 + o(|x|) in B + 1 . Again, let us suppose by contradiction that As before, such inequality propagates by Lipschitz continuity: Defining the elliptic, homogeneous operator F * (M) = −F(−M), we have that the functions As a consequence, a contradiction can be obtained by reasoning as in Lemma 2.2. Let u be non-negative and Lipschitz in Proof. In both cases, we use the smooth change of variable whereF is still a uniformly elliptic operator. As a consequence the lemma follows by arguing as in the proofs of Lemmas 2.2, 2.3, with minor changes.
We conclude this section by providing a uniform estimate from below of the development coefficient α, in case the touching ball is inside the domain.
Lemma 2.5. Let u ∈ C(B r (re 1 )), r ≤ 1, be such that Moreover, assume that u(re 1 ) ≥ Cr, for some C > 0. Then where c only depends on λ, Λ, n. We are in a position to apply Lemma A.2, which provides as x → −e 1 , and the lemma follows.
Remark 2.6. Notice that the above results can be applied both to F(D 2 u + ) = f 1 in Ω + (u) and to

The function u + is Lipschitz continuous
In this section we adapt the strategy developed in [3], in order to show that u + is locally Lipschitz. To this aim we need to use the following almost-monotonicity formula, provided in [6,14].
Then there exist universal constants C 0 and r 0 , such that the functional Lemma 3.2. Let w ∈ S. There existsw ∈ S such that Proof. Let w ∈ S and Ω + = Ω + (w). We define Since u + ∈ V we obtain that V is not empty and that u ≤w ≤ w. Moreoverw is a solution of the obstacle problem (see [13]) In particular, regularity results for the obstacle problem for fully nonlinear equations imply thatw is C 1,1 in Ω + (see [13]). To conclude thatw ∈ S, we need to show that the free boundary conditions in Definition 1.2 hold true. Let x 0 ∈ F (w): if x 0 ∈ F (w) too, then the free boundary condition follows from the fact thatw ≤ w; otherwise, x 0 ∈ Ω + is an interior zero ofw, and the free boundary condition follows by the C 1,1 regularity ofw.
, and let x 0 ∈ F (w) be regular from the right. Then u admits developments Proof. If x 0 is not regular from the left, then by definition of S the asymptotic developments hold with α = β = 0 and there is nothing to prove. On the other hand, if x 0 is also regular from the left, then the asymptotic developments and the free boundary condition hold true by definition of S and by Lemma 2.4. Also in this case, if α = 0 then there is nothing else to prove, thus we are left to deal with the case α > 0.
Proposition 3.4. For every D ⊂⊂ Ω there exists a constant L D , depending only on D, G, u and S, such that for every x, y ∈ D, x y, and for every w ∈ S with F(D 2 w) = f 1 in Ω + (w).
Proof. Let x 0 ∈ Ω + (w) ∩ D such that We will show that there exists M > 0, not depending on w, such that w(x 0 ) r ≤ M, and the lemma will follow by Schauder estimates and Harnack inequality. By contradiction, let M large to be fixed and let as assume that Then Lemma 2.5 applies and we obtain Then x 0 is regular from the right and Lemma 3.3 applies, with α M ≤ α ≤ G(β, x 0 , ν(x 0 )), providing . This provides a contradiction for M sufficiently large.

The function u is Lipschitz continuous
Now we turn to the Lipschitz continuity of u.
To prove that u is Lipschitz continuous we use the double replacement technique introduced in [9]. Let w ∈ S with w(x 0 ) < 0 and , Working on Ω 1 , we define (which is non empty, for R sufficiently small, as u + ∈ V 1 ). Then solves the obstacle problem (see [13]) On the other hand, working on B, let (which is non empty, as w − ∈ V 2 ). Again, Under the above notation, the double replacementw of w, relative to B, is defined as There exists > ε 0 (depending on dist(x 0 , ∂Ω) and u) such that: 1. the double replacementw of w, relative to B h ε (x 0 ), satisfies u ≤w ≤ w in Ω; Proof. The inequality w 1 ≤ w in Ω 1 follows by the maximum principle, while On the other hand, provided ε is sufficiently small (depending on the Lipschitz constant of u), we have that u < 0 in B := B h ε (x 0 ) and u * ∈ V 1 , so that w 1 ≥ u; finally, by the maximum principle in {w 2 > 0}, also −w 2 ≥ u, and part 1. follows. Turning to part 2., assume by contradiction that ∂{w 2 > 0} ∩ B h ε (x 0 ) ∅. Then, by the regularity properties of the obstacle problem (4.2) (see [13]), we obtain that Since −w 2 (x 0 ) ≤ w(x 0 ) = −h, we obtain a contradiction for ε sufficiently small. Then w 2 > 0 in B h ε (x 0 ), w 2 solves the equation by (4.2), and the remaining part of 2. follows by standard Schauder estimates and Harnack inequality.
Coming to part 3., the fact thatw satisfies (a) in Definition 1.2 follows by equations (4.1), (4.2) and by part 2. above, and we are left to check the free boundary conditions. Forx ∈ F (w), three possibilities may occur. Ifx ∈ F (w) then, sincew ≤ w, thenw has the correct asymptotic behavior both whenx is regular and when it is not (recall that G(0, ·, ·) > 0. Ifx ∈ ∂{w 1 > 0} ∩ Ω 1 , then we can use again the regularity of the obstacle problem (4.1) to obtain the correct asymptotic behavior. We are left to the final case, whenx ∈ ∂B ∩ Ω + (w). By Proposition 3.4, let us denote with L the Lipschitz constant of w in B dist(x 0 ,∂Ω)/2 (x 0 ). Theñ Then Lemma A.1 applies, yielding for universal c 1 , c 2 . Going back tow we obtaiñ On the other hand, we can apply Lemma 2.5 to (−w 2 ) ε , obtaining for universal c 1 , c 2 , and thusw Comparing (4.3) and (4.4) we have that, choosing ε small so that the free boundary condition holds true.
There exist an non-increasing sequence {w k } ⊂ S,w k ≥ u, and ε > 0, depending on dist(x 0 , ∂Ω) and u, such that the following hold: Proof. Let u(x 0 ) = −h < 0, {w k } ⊂ S be such that w k u in some neighborhood of x 0 and {w k } ⊂ S be the corresponding double replacements, as in Lemma 4.2. Then first three points are direct consequence of the lemma above, and we are left to prove thatw k u uniformly on B h /4 . By equicontinuity,w k →w in B h/2 (x 0 ), and suppose by contradiction thatw(x 1 ) > u(x 1 ) for some x 1 ∈ B h/4 (x 0 ). Then consider a new sequence {v k } k converging to u at x 1 and define {ũ k } k as the double replacement of {min{ṽ k ,w k }} k in B h/2 (x 0 ). Thenũ k →ũ,ũ ≤w in B h/2 (x 0 ),ũ(x 0 ) =w(x 0 ) and u(x 1 ) <w(x 1 ). Since F(D 2w ) = F(D 2ũ ) = f 2 in B h/2 (x 0 ), this contradicts the strong maximum principle. Proof. The first part follows from the previous corollary. By compactness, it is enough to prove the second part for balls B ε (x 0 ) ⊂ Ω − (u), with ε small. Let w k u uniformly in B 2ε (x 0 ) ⊂ Ω − (u), and let Let φ be such that with a and ε positive and sufficiently small so that ∇φ(e 1 ) · e 1 < inf x,ν G(0, x, ν) (this is possible by explicit calculations, see for instance Lemma A.1); notice that this condition insure that φ, extended to zero in B 1 , is a supersolution in B 2 (when c universal is suitably chosen). Since u ε ≤ 0 in B 2 , for k sufficiently large w k ≤ a/2 in B 2 . Let us definē Then, by Lemma 4.1, the functionw

The function u + is non-degenerate
In this section we will show that u + is non-degenerate, in the sense of the following result. Moreover where all the constants C depend on d(x, ∂Ω) and on u.
Corollary 5.2. F (w k ) → F (u) locally in Hausdorff distance and χ {w k >0} → χ {u>0} in L 1 loc . The proof of the above result will follow by the two following lemmas.

Lemma 5.3. Let u be a Lipschitz function in
If there exists c > 0 such that for every x ∈ Ω ∩ B 1/2 (5.1) then there exists a constant C > 0 such that for all r ≤ r 0 universal.
We will show that, for δ > 0 to be fixed, there exists x 1 ∈ B ε (x 0 ) such that Then, iterating the procedure, one can conclude as in [9, Lemma 5.1]. Assume by contradiction that (5.2) does not hold. Then, defining the elliptic, homogeneous operator Let r(L) = 1 − c/(4L); using the Harnack inequality we have that there exists C(L) such that provided both δ and ε are sufficiently small (depending on c, L and f ∞ ). In terms of u, the previous inequality writes as u ≥ cε 2 in B r(L)ε (x 0 ).
Notice that ρ(x) > 0 for every x, since any point in F (u) is regular from the right by assumption. Thus, recalling that u + has linear growth bounded below by inf x,ν G(0, x, ν), and noticing that B 3r/4 (x 0 )∩Ω + (u) contains a ball of radius comparable with r (at least for a suitable choice ofr): whereC only depends on u.
On the other hand, in case dist(x 0 , Ω + (u)) ≥ r 2 , we have u ≤ 0 in B r/2 (x 0 ). By Corollary 4.4 we can find {w k } k ⊂ S converging uniformly to u on some D ⊃ B r (x 0 ). By scaling we need to findC universal such that u r (0) ≥C. Let us assume by contradiction that u r (0) <C.

Then by Harnack inequality
and, for k sufficiently large, Now, reasoning as in the proof of Corollary 4.4, let φ be such that with a and r positive and sufficiently small so that ∇φ(e 1 /4) · e 1 < inf x,ν G(0, x, ν), in such a way that φ, extended to zero in B 1/4 , is a supersolution in B 1/2 . Then, choosingC < (a − r f 1 ∞ )/C we obtain that w r k < φ on ∂B 1/2 and then the functions . This is in contradiction with the fact that u(x 0 ) > 0, and the lemma follows.

The function u is a supersolution
This section is devoted to the proof that u satisfies the supersolution condition (i) in Definition 1.1. Thanks to Lemma 2.4 we only need to prove that, whenever u admits asymptotic developments at x 0 ∈ F (u), with coefficients α and β, then α ≤ G(β, x 0 , ν x 0 ). To do that, we need to distinguish the two cases β > 0 and β = 0.
Proof. Since β > 0, then F (u) is tangent at x 0 to the hyperplane π : x − x 0 , ν = 0 in the following sense: for any point Otherwise we get a contradiction to the asymptotic development of u.
Let {w k } k ⊂ S be uniformly decreasing to u, as in Corollary 4.4. By the non-degeneracy of u + we have that, for k large, w k can not remain strictly positive near x 0 . Let d k = d H (F (w k ) , F (u)) be the Hausdorff distance between the two free boundaries. In the ball B 2 √ d k (x 0 ), F (u) is contained in a strip parallel to π of width o √ d k and, since d k → 0, F (w k ) is contained in a strip S k of width Consider now the points x k = x 0 − √ d k ν and let B k = B r k (x k ) be the largest ball contained in Ω − (w k ) with touching point z k ∈ F (w k ). Then z k ∈ S k and, since w k ≥ u, from the asymptotic developments of w k and u we have Passing to the limit we infer lim sup β k ≤ β.
Reasoning in the same way on the other side th the points y k = x 0 + √ d k (and the same z k , which are regular from the left), we get α ≤ lim inf α k .
To treat the case β = 0 we need the following preliminary lemma.
Proof. As before, let {w k } k ⊂ S be uniformly decreasing to u, with w k that is not strictly positive near x 0 , for k large. The first part of the proof is exactly as in Lemma 6.3 of [9], until equation (6.2) below. For the reader's convenience, we recall such argument here. For each k we denote with the largest ball centered at x 0 + ν/m contained in Ω + (w k ), touching F (w k ) at x m,k where ν m,k is the unit inward normal of F (w k ) at x m,k . Then up to proper subsequences we deduce that λ m,k → λ m , x m,k → x m , ν m,k → ν m and B λ m (x 0 + ν/m) touches F (u) at x m , with unit inward normal ν m . From the behavior of u + , we get that Now since w k ∈ F , near x m,k in B m,k : (by Lemma 2.5 the touching occurs at a regular point, for m, k large.) We know that and m → 0, as m → ∞. We have to show that We assume by contradiction thatβ > 0. Acting as in [9, Lemma 6.3] we obtain, for r small, where Φ r x m,k , w k = Ψ r x m,k , w + k Ψ r x m,k , w − k . By concavity we have that ∆w ± k ≥ −M where M = c min ( f 1 ∞ , f 2 ∞ ). Lemma 6.2 implies where L is the uniform Lipschitz constant of {w + k } k (recall Lemma 3.4). Taking the lim inf as m, k → ∞ and using the uniform convergence of w k to u we infer 0 < c n α 2β2 ≤ C 1 (n, M, L) Recalling that, by assumption, as r → 0, and we get a contradiction.

The function u is a subsolution
In this section we want to show that u is a subsolution according to Definition 1.1. Note that, if x 0 ∈ F (u) is a regular point from the left with touching ball B ⊂ Ω − (u), then near to x 0 in Ω\B. Indeed, even if β = 0, then Ω + (u) and Ω − (u) are tangent to { x − x 0 , ν = 0} at x 0 since u + is non-degenerate. Thus u has a full asymptotic development as in the next lemma. We want to show that α ≥ G(β, x 0 , ν). We follow closely [3] and [9].
Proof. Assume by contradiction that α < G(β, x 0 , ν). We construct a supersolution w ∈ S which is strictly smaller than u at some point, contradicting the minimality of u. Let u 0 be the two-plane solution, i.e.

Properties of the free boundary
In this section we prove the weak regularity properties of the free boundary. Both statements and proofs are by now rather standard and follows the papers [3] and [9] for problems governed by homogeneous and inhomogeneous divergence equations, respectively. Thus we limit ourselves to the few points in which differences from the previous cases emerge. Denote by N ε (A) an ε-neighborhood of the set A. The following lemma provides a control of the H n−1 measure of F (u) and implies that Ω + (u) is a set of finite perimeter. Lemma 8.1. Let u be our Perron solution. Let x 0 ∈ F (u) ∩ B 1 . There exists a positive universal δ 0 < 1 such that, for every 0 < ε < δ ≤ δ 0 , the following quantities are comparable: where N is the number of any family of balls of radius ε, with finite overlapping, covering Proof. From [3], it is sufficient to prove the following two equivalences: and with universal constants c 1 , c 2 , C 1 , C 2 .
Since F D 2 u = inf α L α u where L α is a uniformly elliptic operator with constant coefficients and ellipticity constant λ, Λ, we have L α u + ≥ f 1 in Ω + (u). Fix α = α 0 and set The upper bound in (8.1) follows by the Lipschitz continuity of u. The lower bound follows from sup B ε (x 0 ) u + ≥ cε, c universal, inf B ε (x 0 ) u + = 0, the Lipschitz continuity of u, and the Poincaré inequality (see [1,Lemma 1.15]).
By uniform ellipticity, since u + is Lipschitz and f 1 is bounded, we get (δ < 1) with C universal. Letting s → 0 and rescaling back, we obtain the upper bound in (8.2). For the lower bound, let V be the solution to with σ to be chosen later. By standard estimates, see for example [12], V ≤ Cσ 2−n and − A∇V, ν ∼ C * on ∂B 1 , with C * independent of σ. By Green's formula since u is Lipschitz and 0 ≤ u ε,0 ≤ ε/δ. From (8.4) and (8.5) and the fact that A δ ∇V, ν ∼ −C * on ∂B 1 we deduce that Thus using that u + is non-degenerate and choosing σ small enough (universal) we get that (δ > ε) On the other hand in {0 < u + δ < ε/δ} ∩ B 1 , Combining (8.6), (8.7) and using the ellipticity of A we get that From the estimate on V we obtain that for δ small enough for some C universal. Rescaling, we obtain the desired lower bound.
Lemma 8.1 implies that Ω + (u) ∩ B r (x), x ∈ F (u), is a set of finite perimeter. Next we show that in fact this perimeter is of order r n−1 .
Theorem 8.2. Let u be our Perron solution. Then, the reduced boundary of Ω + (u) has positive density in H n−1 -measure at any point of F (u), i.e. for r < r 0 , r 0 universal, for every x ∈ F (u).
Proof. The proof follows the lines of Corollary 4 in [3] and Theorem 8.2 in [9]. Let w k ∈ S, w k u in B 1 and L as in Lemma 8.1. Then Ω + (u) ⊂⊂ Ω + (w k ) and Lw k ≥ F D 2 w k = f 1 in Ω + (u). Let x 0 ∈ F (u). We rescale by setting Let V be the solution to (8.3). Since ∇w k,r is a continuous vector field in Ω + r (u r ) ∩ B 1 , we can use it to test for perimeter. We get w kr A∇V, ν dH n−1 .

(8.8)
Using the estimates for V and the fact that the w k are uniformly Lipschitz, we get that F * (u r )∩B 1 V A∇w k,r , ν dH n−1 ≤ C(σ)H n−1 (F * (u r ) ∩ B 1 ). (8.9) As in [3] we have, as k → ∞, F * (u r )∩B 1 w k,r A∇V, ν dH n−1 → 0, Passing to the limit in (8.8) and using all of the above we get u + r A∇V, ν dH n−1 ≤ C(σ)H n−1 (F * (u r ) ∩ B 1 ).
Also, since f r 1 is bounded, Hence choosing first σ and then r sufficiently small we get that H n−1 (F * (u r ) ∩ B 1 ) ≥C, C universal.

A. Some explicit barrier functions
For the reader's convenience we collect here some explicit barrier functions which arise frequently in our arguments. Their proof is based on comparison arguments, together with the well known chain of inequalities P − λ/n,Λ u ≤ F(D 2 u) ≤ c∆u, (A.1) where P − λ/n,Λ denotes the lower Pucci operator, and c = c(λ, Λ, n) > 0 since F is concave (see [5] for further details).
for a universal c. Then direct calculations show that, for n ≥ 3, The proof in dimension n = 2 is analogous.