Regularity for the fully nonlinear parabolic thin obstacle problem
aa r X i v : . [ m a t h . A P ] A p r REGULARITY FOR THE FULLY NONLINEAR PARABOLIC THINOBSTACLE PROBLEM
GEORGIANA CHATZIGEORGIOU
Abstract.
We prove C ,α regularity (in the parabolic sense) for the viscosity solution of a bound-ary obstacle problem with a fully nonlinear parabolic equation in the interior. Following the methodwhich was first introduced for the harmonic case by L. Caffarelli in 1979, we extend the results ofI. Athanasopoulos (1982) who studied the linear parabolic case and the results of E. Milakis andL. Silvestre (2008) who treated the fully nonlinear elliptic case. Introduction
In the present work we intent to study the regularity of the viscosity solution of the followingthin obstacle problem in a half-cylinder, F ( D u ) − u t = 0 , in Q +1 u y ≤ , on Q ∗ u ≥ ϕ, on Q ∗ u y = 0 , on Q ∗ ∩ { u > ϕ } u = u , on ∂ p Q +1 \ Q ∗ (1.1)where, F is a uniformly elliptic operator on S n with ellipticity constants λ and Λ and ϕ : Q ∗ → R , u : ∂ p Q +1 \ Q ∗ → R are given functions. Function ϕ is the so-called obstacle and u ≥ ϕ on ∂ p Q ∗ for compatibility reasons. Our aim is to prove that u is in H α up to the flat boundary Q ∗ . Themain theorem of this paper follows (notations’ details can be found in subsection 2.1). Theorem 1.
Let P = ( x , t ) ∈ Q ∗ / be a free boundary point, there exist universal constants < α < , C > , < r << and an affine function R ( X ) = A + B · ( X − ( x , , where A = u ( P ) , B = Du ( P ) so that | u ( X, t ) − R ( X ) | ≤ C (cid:16) | X − ( x , | + | t − t | / (cid:17) α , for any ( X, t ) ∈ Q + r ( P ) . Mathematics Subject Classification (2010).
Keywords.
Parabolic thin obstacle problem; Fully non-linear parabolic equations; Free boundary problems;Regularity of the solution.
The classical obstacle problem as well as the thin obstacle problem are originated in the contextof elasticity since model the shape of an elastic membrane which is pushed by an obstacle (whichmay be very thin) from one side affecting its shape and formation. The same model appears incontrol theory when trying to evaluate the optimal stopping time for a stochastic process withpayoff function. Important cases of obstacle type problems occur when the operators involvedare fractional powers of the Laplacian as well as nonlinear operators since they appear, amongothers, in the analysis of anomalous diffusion, in quasi-geostrophic flows, in biology modeling flowsthrough semi-permeable membranes for certain osmotic phenomena and when pricing Americanoptions regulated by assets evolving in relation to jump processes.Thin (or boundary) obstacle problem (or Signorini’s problem) was extensively studied in theelliptic case. For Laplace equation and more general elliptic PDEs in divergence form the problemcan be also understood in the variational form, that is as a problem of minimizing a suitablefunctional over a suitable convex class of functions which should stay above the obstacle on a partof the boundary (or on a sub-manifold of co-dimension at least 1) of the domain of definition. The C ,α -regularity of the weak solution for the harmonic case was proved first in 1979 by L. Caffarelliin [7] who treats also the divergence case for regular enough coefficients. Results for more generaldivergence-type elliptic operators can be found in [24]. For optimal regularity and regularity of thefree boundary in the case of linear elliptic equations we refer to [2] and [5] where the harmonic caseis studied and to [15], [14], [16] for the case of variable coefficients. Similar results exist also for thecase of fractional Laplacians. Regularity of the solution for the classical (thick) obstacle problemwas studied in [22], then via the extension problem introduced in [10] the thin obstacle problemwas treated in [9]. Finally, for fully nonlinear elliptic operators, regularity of the viscosity solutionwas proved in [18] (see also [13]) while for optimal and free boundary regularity the only existingwork is [20].The corresponding regularity theory for thin obstacle problems of parabolic type is much lessdeveloped. The C ,α -regularity of the weak solution was obtained in 1982 by I. Athanasopoulos in[6] who studied the case of heat equation and the case of smooth enough linear parabolic equation.The case of more general linear parabolic operators was examined in [23] and [1]. Optimal andfree boundary regularity for the caloric case have been obtained very recently in [4] (see also [12]).Finally for the case of parabolic operators of fractional type we refer the reader to [3] and [8].In this paper our purpose is to combine the techniques of [7], [6] and [18] adapting them in ourfully nonlinear parabolic framework. To achieve this we need up to the boundary H¨older estimatesfor viscosity solutions of nonlinear parabolic equations with Neumann boundary conditions (as [17]is used in [18]). This type of estimates have developed recently by the author and E. Milakis in[11]. EGULARITY FOR THE FULLY NONLINEAR PARABOLIC THIN OBSTACLE PROBLEM 3
The paper is organized as follows. In Section 2 we give a list of notations used throughout thistext. We discuss also the assumptions we make on the data of our problem and finally we provea reflection property which is useful in our approach. In Section 3 we examine the semi-concavityproperties of our solution. We prove Lipschitz continuity in space variables, a lower bound for u t and for the second tangential derivatives of u (semi-convexity) and an upper bound for thesecond normal derivative of u (semi-concavity). All these bounds are universal and hold up tothe flat boundary Q ∗ . The boundedness of the first and second normal derivatives ensures theexistence of u y + on Q ∗ . Our first intention is to prove that u y + ≤ Q ∗ (which apriori holdsonly in the viscosity sense). To achieve this we use the penalized problem defined and studied inSection 4. Finally in Section 5 we prove the main theorem. To do so we obtain first an estimate inmeasure (Lemma 12) for u y + on Q ∗ and subsequently we see how such a property can be carriedinside Q +1 (Lemma 13). An iterative application of the above two properties gives the regularityof u y + on Q ∗ around free boundary points (Lemma 14) and then our problem can be treated as anon-homogeneous Neumann problem. 2. Preliminaries
Notations.
We denote X = ( x, y ) ∈ R n , where x ∈ R n − and y ∈ R and P = ( X, t ) ∈ R n +1 ,where X are the space variables and t is the time variable. The Euclidean ball in R n and theelementary cylinder in R n +1 will be denoted by B r ( X ) := { X ∈ R n : | X − X | < r } , Q r ( X , t ) := B r ( X ) × ( t − r , t ]respectively. We define the following half and thin-balls in R n , for r > , x ∈ R n − B + r ( x ) := B r ( x , ∩ { y > } , B ∗ r ( x ) := B r ( x , ∩ { y = 0 } and the following half and thin-cylinders in R n +1 , for r > , x ∈ R n − , t ∈ R Q + r ( x , t ) := Q r ( x , , t ) ∩ { y > } , Q ∗ r ( x , t ) := Q r ( x , , t ) ∩ { y = 0 } . Note that, Ω ◦ , Ω , ∂ Ω will be the interior, the closure and the boundary of the domain Ω ⊂ R n +1 ,respectively, in the sense of the Euclidean topology of R n +1 . We define also the parabolic interiorto be, int p (Ω) := { ( X, t ) ∈ R n +1 : there exists r > Q ◦ r ( X, t ) ⊂ Ω } and the parabolic boundary, ∂ p (Ω) := Ω \ int p (Ω) . Let us also define the parabolic distance for P = ( X, t ) , P = ( Y, s ) ∈ R n +1 , p ( P , P ) := max {| X − Y | , | t − s | / } . Note that in this case Q r ( P ) will be the set { P ∈ R n +1 : p ( P, P ) < r, t < t } .Next we define the corresponding parabolic H¨older spaces. For a function f defined in a domainΩ ⊂ R n +1 we set, GEORGIANA CHATZIGEORGIOU [ f ] α ;Ω := sup P ,P ∈ Ω ,P = P | f ( P ) − f ( P ) | p ( P , P ) α , h f i α +1;Ω := sup ( X,t , ( X,t ∈ Ω t = t | f ( X, t ) − f ( X, t ) || t − t | α +12 . Then we say that, • f ∈ H α (Ω) if || f || H α (Ω) := sup Ω | f | + [ f ] α ;Ω < + ∞ . • f ∈ H α +1 (Ω) if || f || H α +1 (Ω) := sup Ω | f | + n X i =1 sup Ω | D i f | + n X i =1 [ D i f ] α ;Ω + h f i α +1;Ω < + ∞ . • f ∈ H α +2 (Ω) if || f || H α +2 (Ω) := sup Ω | f | + n X i =1 sup Ω | D i f | + sup Ω | f t | + n X i,j =1 sup Ω | D ij f | + [ f t ] α ;Ω + n X i,j =1 [ D ij f ] α ;Ω + n X i =1 h D i f i α +1;Ω < + ∞ . Due to the nonlinear character of our problem, we will mainly prove H α +1 -regularity results in thepunctual sense at a point. We say that u is punctually H α +1 at a point P ∈ Ω if there exists R P ( X ) = A P + B P · ( X − X ), where A P ∈ R and B P ∈ R n and some cylinder Q r ( P ) ⊂ Ω,so that for any 0 < r < r , | u ( X, t ) − R P ( X ) | ≤ K r α , for every ( X, t ) ∈ Q r ( P )for some constant K > S n denotes the class of symmetric n × n real matrices.2.2. Problem Set-up.
We consider that the solution u of (1.1) can be recovered as the minimumviscosity supersolution of F ( D v ) − v t ≤ , in Q +1 v y ≤ , on Q ∗ v ≥ ϕ, on Q ∗ v ≥ u , on ∂ p Q +1 \ Q ∗ (2.1)with u t locally bounded by above in Q +1 (note that under suitable assumptions on F we have that u t does exist in Q +1 once F ( D u ) − u t = 0 in Q +1 in the viscosity sense).To get the desired regularity we make the following assumptions on F and u . • Assumptions on F . First we assume that F is convex on S n so we have interior H α -estimates EGULARITY FOR THE FULLY NONLINEAR PARABOLIC THIN OBSTACLE PROBLEM 5 for the viscosity solutions (see [26]). Moreover considering the following extension of F in R n × n F ( M ) = F (cid:18) M + M τ (cid:19) , for M ∈ R n × n we assume that F is continuously differentiable in R n and we denote by F ij := ∂F∂m ij . We can easilysee that F ij ( M ) = F ji ( M ) for any M . Indeed, let H ij denote the matrix with elements (cid:16) H ijh (cid:17) kl = , if k = i or j = l,h, if k = i and j = l where h ∈ R and observe that (cid:16) H ijh (cid:17) τ = H jih then F ij ( M ) = lim h → F (cid:18) M + H ijh + ( M + H ijh ) τ (cid:19) − F ( M ) h = lim h → F (cid:18) M + H jih + ( M + H jih ) τ (cid:19) − F ( M ) h = F ji ( M ) . We suppose also that F in = 0, for any i = 1 , . . . , n − F ni = 0 as well). Finally, we assumefor convenience that F ( O ) = 0 which can be easily removed (subtracting a suitable paraboloid). • Assumptions on u . Note that we intend to examine the regularity up to flat boundary Q ∗ (and not up to ∂ p Q +1 \ Q ∗ ) thus we may assume that u > ϕ on ∂ p Q ∗ . Therefore if v ∈ C (cid:16) Q +1 (cid:17) isthe viscosity solution of F ( D v ) − v t = 0 , in Q +1 v y = 0 , on Q ∗ v = u , on ∂ p Q +1 \ Q ∗ (2.2)then due to the continuity of v and ϕ and the compactness of ∂ p Q ∗ we see that there exists some0 < ρ < v > ϕ on Q ∗ \ Q ∗ − ρ . Then using an ABP-type estimate (see Theorem 5 in [11])we get that u > ϕ on Q ∗ \ Q ∗ − ρ thus u y = 0 on Q ∗ \ Q ∗ − ρ , in the viscosity sense. • Assumptions on ϕ . We assume that ϕ ∈ H α ( Q ∗ ).We denote by ∆ ∗ := { ( x, t ) ∈ Q ∗ : u ( x, , t ) = ϕ ( x, t ) } the contact set , by Ω ∗ := { ( x, t ) ∈ Q ∗ : u ( x, , t ) > ϕ ( x, t ) } the non-contact set and by Γ = ∂ ∆ ∗ ∩ Q ∗ the free boundary . We assumethat ∆ ∗ = ∅ since otherwise we would have a Neumann boundary value problem for which theregularity is known (see [11]). Note that around the points of int(∆ ∗ ) and around the points ofΩ ∗ we can treat our problem as Dirichlet or Neumann problem respectively. Finally, we denoteby K := || u || L ∞ ( Q +1 ) + || ϕ || H α ( Q ∗ ) and in the following a constant C >
K, n, λ,
Λ and ρ will be called universal . GEORGIANA CHATZIGEORGIOU
Reflection Principle.
Here we show a reflection property which will be useful in severaltimes in our approach. We remark that since F in = F ni = 0 for any i = 1 , . . . , n − M = ( m ij ) ∈ R n × n if we denote by ¯ M the matrix with elements¯ m ij := m ij , if i, j < n or i = j = n − m ij , if i < n and j = n or i = n and j < n we have that F ( M ) = F ( ¯ M ). Observe that Pucci’s extremal operators have this property as well.Indeed, M and ¯ M have the same eigenvalues since,det( ¯ M − lI n ) = det M n − − lI n − d τ d m nn − l ! = det M n − − lI n − − d τ − d m nn − l ! = det( M − lI n )where d := ( − m n , . . . , − m nn − ), hence M ± ( ¯ M , λ,
Λ) = M ± ( M, λ,
Λ).
Proposition 2. (Reflection Principle). Let u ∈ C ( Q +1 ∪ Q ∗ ) and satisfies F ( D u ) − u t ≤ in Q +1 and u y ≤ on Q ∗ in the viscosity sense. Consider the reflected function, u ∗ ( x, y, t ) = u ( x, y, t ) , if y ≥ u ( x, − y, t ) , if y < , for ( X, t ) ∈ Q . Then F ( D u ∗ ) − u ∗ t ≤ in Q in the viscosity sense.Proof. We observe that u ∗ ∈ C ( Q ) and that F ( D u ∗ ) − u ∗ t ≤ Q +1 . Also it can be easily verified F ( D u ∗ ) − u ∗ t ≤ Q − (regarding the observation we made above). To get that this is true in Q as well it remains to study what happens across Q ∗ . To do so we approximate by suitablesupersolutions, by considering v γ ( X, t ) := u ∗ ( X, t ) − γ | y | for γ >
0. Then we have that F ( D v γ ) − ( v γ ) t ≤ Q +1 ∪ Q − and we will show that v γ cannot betouched by below by any test function at any point of Q ∗ . Indeed, let φ be a test function in Q that touches v γ by below at some point P = ( x , , t ) ∈ Q ∗ . Our purpose is to use the viscosityNeumann condition to get a contradiction. We have that φ ( X, t ) + γy touches u by below at P in some Q + ρ ( P ) ⊂ Q +1 . Then φ y ( P ) + γ ≤
0, that is φ y ( P ) ≤ − γ < . But on the other hand, φ ( X, t ) − γy touches u ∗ by below at P in some Q − ρ ( P ) ⊂ Q − . A change of variables implies that u ( X, t ) ≥ φ ( x ′ , − y, t ) + γy , for ( X, t ) ∈ Q + ρ ( P ). Then − φ y ( P ) + γ ≤
0, that is φ y ( P ) ≥ γ >
0, acontradiction. Therefore such a test function cannot exist.Consequently, since F ( D v γ ) − ( v γ ) t ≤ Q +1 ∪ Q − and no test function can touch v γ by belowat any point of Q ∗ in a neighborhood in Q , that is F ( D v γ ) − ( v γ ) t ≤ Q in the viscositysense. Finally, we observe that, | v γ − u ∗ | = | γ || y | ≤ | γ | → γ →
0, which means that v γ → u ∗ , EGULARITY FOR THE FULLY NONLINEAR PARABOLIC THIN OBSTACLE PROBLEM 7 as γ → Q . So, we can consider for k ∈ N the sequence { v k } and use the closednessof viscosity supersolutions to complete the proof. (cid:3) Note that an analogous result holds for subsolutions. That is, if v ∈ C ( Q +1 ∪ Q ∗ ) which satisfies F ( D v ) − v t ≥ Q +1 and v y ≥ Q ∗ in the viscosity sense, then F ( D v ∗ ) − v ∗ t ≥ Q .3. Semi-concavity properties
In this section we obtain bounds for the first and second derivatives of the solution. A firstapplication of these bounds will ensure the existence of u y + on Q ∗ . Proposition 3.
For any < δ < , (A) | u x i | , | u y | ≤ C , in Q +1 − δ , for any i = 1 , . . . , n − u x i x i , u t ≥ − C , in Q +1 − δ , for any i = 1 , . . . , n − u yy ≤ C , in Q +1 − δ where the constant C > depends only on K, n, λ, Λ , ρ and δ . Note that since F is convex, we have that u x i x j and u t exist in Q +1 in the classical sense byinterior estimates (see [26]) . Proof.
For (A) , we thicken the obstacle ϕ . First, we extend ϕ as a solution inside Q +1 and Q − (followingthe idea of Theorem 1(a) in [2], see also Proposition 2.1 in [13]), that is we consider the viscositysolutions of the Dirichlet problems F ( D ˜ ϕ ) − ˜ ϕ t = 0 , in Q +1 ˜ ϕ = ϕ, on Q ∗ ˜ ϕ = −|| u || L ∞ ( Q +1 ) , on ∂ p Q +1 \ Q ∗ and F ( D ˜ ϕ ) − ˜ ϕ t = 0 , in Q − ˜ ϕ = ϕ, on Q ∗ ˜ ϕ = −|| u || L ∞ ( Q +1 ) , on ∂ p Q − \ Q ∗ . For any 0 < δ < ϕ is smooth enough we obtain, using Theorem 12 of [11], that ˜ ϕ isLipschitz in Q − δ with a constant that depends only on K, n, λ,
Λ and δ . Moreover using maximumprinciple we can obtain that u ∗ ≥ ˜ ϕ in Q , where u ∗ denotes the even reflection of u in y inside Q . Finally, Proposition 2 ensures that F ( D u ∗ ) − u ∗ t ≤ Q and that F ( D u ∗ ) − u ∗ t = 0 in Q ∩ { u ∗ > ˜ ϕ } in the viscosity sense. Therefore u ∗ satisfies a thick obstacle problem in Q withobstacle ˜ ϕ which is Lipschitz in Q − δ . In particular, we get that u ∗ ∈ H ( Q − δ ) with a constantthat depends only on K, n, λ,
Λ and δ (see [21], [19]) which gives (A) .For (B) , we denote by d := min { ρ, δ } and we consider the set ˜ Q + := Q +1 − d \ Q +1 − d . We observethat u y = 0 on ˜ Q ∗ in the viscosity sense, since ˜ Q ∗ ⊂ Q ∗ \ Q ∗ − ρ . Thus up to the boundary H α -estimates (see Theorem 23 in [11]) can be applied in ˜ Q + and we get H α -estimates for u x i x i and u t GEORGIANA CHATZIGEORGIOU on ∂ p Q +1 − d \ Q ∗ − d . In particular we have uniform bounds for the corresponding difference quotients,that is,(3.1) u ( x + he i , y, t ) + u ( x − he i , y, t ) − u ( x, y, t ) h ≥ − C where { e i } ≤ i ≤ n is the normal basis of R n and(3.2) u ( x, y, t − h ) − u ( x, y, t ) h ≥ − C for ( X, t ) ∈ ∂ p Q +1 − d \ Q ∗ − d , h > d ) and C >
K, n, λ, Λ , ρ and δ .We study (3.1) first in order to bound u x i x i , for i = 1 , . . . , n −
1. We observe that v ( x, y, t ) := u ( x + he i , y, t ) + u ( x − he i , y, t )2 + Ch ≥ u ( x, y, t ) , on ∂ p Q +1 − d \ Q ∗ − d . Moreover, for ( x, t ) ∈ Q ∗ − d , v ( x, , t ) = u ( x + he i , , t ) + u ( x − he i , , t )2 + Ch ≥ ϕ ( x + he i , t ) + ϕ ( x − he i , t )2 + Ch ≥ ϕ ( x, t )changing C if necessary depending on K . We observe also that the convexity of F ensures that F ( D v ) − v t ≤ Q +1 − d in the viscosity sense. Finally note that v y ≤ Q ∗ − d in the viscositysense (which can be obtained as Proposition 11 in [11]). That is v is a viscosity supersolution of(2.1) in Q +1 − d , thus v ≥ u in Q +1 − d . Therefore u ( x + he i , y, t ) + u ( x − he i , y, t ) − u ( x, y, t ) h ≥ − C in Q +1 − d and C >
K, n, λ, Λ , ρ and δ . Next we study (3.2) in a similar way inorder to bound u t . Observe that w ( x, y, t ) := u ( x, y, t − h ) + Ch ≥ u ( x, y, t ) , on ∂ p Q +1 − d \ Q ∗ − d . Moreover, for ( x, t ) ∈ Q ∗ − d , w ( x, , t ) = u ( x, , t − h ) + Ch ≥ ϕ ( x, t − h ) + Ch ≥ ϕ ( x, t )changing C if necessary depending on K . Finally, note that F ( D w ) − w t = 0 in Q +1 − d and w y ≤ Q ∗ − d . That is w is a viscosity supersolution of (2.1) in Q +1 − d , thus w ≥ u in Q +1 − d . Therefore u ( x, y, t − h ) − u ( x, y, t ) h ≥ − C in Q +1 − d and C >
K, n, λ, Λ , ρ and δ . EGULARITY FOR THE FULLY NONLINEAR PARABOLIC THIN OBSTACLE PROBLEM 9
For (C) , we will use (B) and the equation. Define a ij ( X, t ) := Z F ij (cid:0) hD u ( X, t ) (cid:1) dh and we observe that ddh (cid:2) F (cid:0) hD u ( X, t ) (cid:1)(cid:3) = P ni,j =1 F ij (cid:0) hD u ( X, t ) (cid:1) u x i x j ( X, t ). That is, n X i,j =1 a ij ( X, t ) u x i x j ( X, t ) = Z n X i,j =1 F ij (cid:0) hD u ( X, t ) (cid:1) u x i x j ( X, t ) dh = F (cid:0) D u ( X, t ) (cid:1) . Thus, P ni,j =1 a ij ( X, t ) u x i x j ( X, t ) − u t ( X, t ) = 0 in Q +1 . Also, we have that a ij = a ji and that a in = a ni = 0, for any 1 ≤ i ≤ n −
1, from our assumptions on F . Additionally we may observethat using the ellipticity of F we have that for any M ∈ S n and h > λh ≤ F (cid:0) M + H iih (cid:1) − F ( M ) ≤ Λ h so taking h → + we have that λ ≤ F ii ( M ) ≤ Λ. In particular, λ ≤ a ii ( X, t ) ≤ Λ, for any(
X, t ) ∈ Q +1 , i = 1 , . . . , n . So if A n − ( X, t ) := ( a ij ( X, t )) i,j =1 ,...,n − ∈ S n − we have a nn ( X, t ) u yy ( X, t ) = − n − X i,j =1 a ij ( X, t ) u x i x j ( X, t ) + u t ( X, t )= − tr (cid:0) A n − ( X, t ) D n − u ( X, t ) (cid:1) + u t ( X, t )= − tr (cid:2) A n − ( X, t ) (cid:0) D n − u ( X, t ) + C I n − (cid:1)(cid:3) + tr ( CA n − ( X, t )) + u t ( X, t )where
C > (B) , thustr (cid:2) A n − ( X, t ) (cid:0) D n − u ( X, t ) + C I n − (cid:1)(cid:3) = tr [ A n − ( X, t )] tr (cid:2) D n − u ( X, t ) + C I n − (cid:3) ≥ CA n − ( X, t )) = C P n − i =1 a ii ( X, t ) ≤ C Λ( n − a nn ( X, t ) ≥ λ . Hence we have that u yy ≤ C Λ( n −
1) + 1 λ , in Q +1 − δ . (cid:3) For any ( x, t ) ∈ Q ∗ we define σ ( x, t ) := lim y → + u y ( x, y, t ) . Note that Proposition 3 ensures the existence of the above limit for any ( x, t ) ∈ Q ∗ . Indeed,we consider the function v ( X, t ) = u y ( X, t ) − Cy , for ( X, t ) ∈ Q +1 . Then using (A) and (C) ofProposition 3 we obtain that v > − C and v y = u yy − C ≤ Q +1 − δ , that is, v is monotonedecreasing in y and bounded by below, thus lim y → + v ( x, y, t ) exists for ( x, t ) ∈ Q ∗ − δ , for any0 < δ < Furthermore we remark that the existence of the above limit ensures the existence oflim y → + u ( x,y,t ) − u ( x, ,t ) y , that is u y + exists on Q ∗ and equals to σ (note also that u y is continuous in y up to Q ∗ ). Thereafter the viscosity condition u y ≤ Q ∗ suggests that one should have(3.3) σ ≤ , on Q ∗ . Although we know that u y + = σ on Q ∗ in the classical sense, we cannot use the viscosity condition toget (3.3) since we do not know if u y + is continuous in ( x, , t ). To obtain (3.3) we use a penalizationtechnique introduced in the next section.4. A penalized problem
We focus now on showing (3.3) by approximating u by suitable classical solutions. So for any k ∈ N we consider the penalized problem F (cid:0) D u ( k ) (cid:1) − (cid:0) u ( k ) (cid:1) t = 0 , in Q +1 (cid:0) u ( k ) (cid:1) y = − k (cid:0) ϕ − u ( k ) (cid:1) + := g ( k ) , on Q ∗ u ( k ) = u , on ∂ p Q +1 \ Q ∗ . (4.1)Note that (4.1) is not a free boundary problem. Using ABP-estimate and a barrier argument weobtain estimates for u ( k ) and g ( k ) (Lemmata 4 and 5) which are independent of k . Then we will beable to treat (4.1) as a non-homogeneous Neumann problem and, using suitable H¨older estimates,we obtain the uniform convergence of u ( k ) to u (Proposition 6) and the existence of (cid:0) u ( k ) (cid:1) y in theclassical sense (Lemma 8). This last property means that the viscosity condition for u ( k ) holds inthe classical sense. This makes the penalized problem very useful in proving (3.3) (see Lemma 9below). Note also that for any k ∈ N , we have that u ( k ) > ϕ on Q ∗ \ Q ∗ − ρ by comparing u ( k ) withthe solution v of (2.2) (see Theorem 5 in [11]). Lemma 4 (Independent of k estimate for u ( k ) ) . For any k ∈ N , (4.2) || u ( k ) || L ∞ ( Q +1 ) ≤ max {|| u || L ∞ ( ∂ p Q +1 \ Q ∗ ) , || ϕ || L ∞ ( Q ∗ ) } . Proof.
First, by Theorem 5 of [11] we haveinf Q +1 u ( k ) ≥ inf ∂ p Q +1 \ Q ∗ u ( k ) = inf ∂ p Q +1 \ Q ∗ u since (cid:0) u ( k ) (cid:1) y ≤ Q ∗ . Hence it remains to bound sup Q +1 u ( k ) .Assume that sup Q +1 u ( k ) > sup ∂ p Q +1 \ Q ∗ u and let ( X , t ) ∈ Q +1 be such that u ( k ) ( X , t ) = sup Q +1 u ( k ) =: M . From maximum principle(see [25], Corollary 3.20) we know that || u ( k ) || L ∞ (cid:16) Q +1 (cid:17) ≤ || u ( k ) || L ∞ ( ∂ p Q +1 ), thus we can choose EGULARITY FOR THE FULLY NONLINEAR PARABOLIC THIN OBSTACLE PROBLEM 11 ( X , t ) = ( x , , t ) ∈ Q ∗ . Then by Hopf’s lemma we obtain that u ( k ) y ( x , t ) < − k (cid:0) ϕ ( x , t ) − u ( k ) ( x , , t ) (cid:1) <
0, that is M = u ( k ) ( x , , t ) < ϕ ( x , t ) ≤|| ϕ || L ∞ ( Q ∗ ). (cid:3) Lemma 5 (Independent of k estimate for g ( k ) ) . For any k ∈ N , (4.3) || g ( k ) || L ∞ ( Q ∗ ) ≤ C ( K, n, λ, Λ , ρ ) . Proof.
Note that g ( k ) ≤ Q ∗ , so we need to obtain only a lower bound. Let ( x , t ) ∈ Q ∗ be suchthat g ( k ) ( x , t ) = min Q ∗ g ( k ) and we may assume that g ( k ) ( x , t ) <
0. Recall also that u ( k ) > ϕ on Q ∗ \ Q ∗ − ρ which implies that g ( k ) = 0 on Q ∗ \ Q ∗ − ρ , that is, ( x , t ) ∈ Q ∗ − ρ .We intend to turn the obstacle ϕ into a suitable test function that touches u ( k ) by below at( x , t ) and then to use the viscosity condition (cid:0) u ( k ) (cid:1) y = g ( k ) to bound g ( k ) ( x , t ). We denote by M := inf Q +1 u − sup Q ∗ ϕ and observe that M ≤
0, indeedinf Q +1 u ≤ inf Q ∗ u ≤ u ( x ∗ , , t ∗ ) = ϕ ( x ∗ , t ∗ ) ≤ sup Q ∗ ϕ where ( x ∗ , t ∗ ) is any point of ∆ ∗ . Keep also in mind that by Lemma 4, M ≤ inf Q +1 u ( k ) − sup Q ∗ ϕ .We consider b to be the solution of the following Dirichlet boundary value problem M − (cid:0) D b, λn , Λ (cid:1) − b t = (Λ n + 1) || ϕ || H α ( Q ∗ ) , in Q + ρ b = M, on ∂ p Q + ρ \ Q ∗ ρ b = 0 , on Q ∗ ρ/ b ( x, , t ) = Mρ (cid:16) max n | x | , | t | o − ρ (cid:17) , on Q ∗ ρ \ Q ∗ ρ/ . Note that Mρ (cid:16) max n | x | , | t | o − ρ (cid:17) = 0 on ∂ p Q ∗ ρ/ and Mρ (cid:16) max n | x | , | t | o − ρ (cid:17) = M on ∂ p Q ∗ ρ .Hence the Dirichlet data on ∂ p Q + ρ is a continuous function. Moreover applying regularity resultsfor Dirichlet problems in Q + ρ/ , we obtain that b ∈ H α (cid:16) Q + ρ/ (cid:17) with the corresponding estimatedepending only on ρ, n, λ, Λ , K , in particular, | Db (0 , | ≤ C ( K, n, λ, Λ , ρ ).Next, we consider the functionΦ( X, t ) = u ( k ) ( x , , t ) − ϕ ( x , t ) + ϕ ( x, t ) + b (( X, t ) − ( x , , t ))for ( X, t ) ∈ Q + ρ ( x , t ) ⊂ Q +1 . We have that Φ( x , , t ) = u ( k ) ( x , , t ). On ∂ p Q + ρ ( x , t ) \ Q ∗ ρ ( x , t ), Φ( X, t ) ≤ inf Q +1 u ( k ) ≤ u ( k ) ( X, t ), since g ( k ) ( x , t ) < b = M . Also on Q ∗ ρ ( x , t ),Φ( x, , t ) ≤ u ( k ) ( x, , t ) − ϕ ( x, t ) + ϕ ( x, t ) = u ( k ) ( x, , t ), using that b ≤ ∂ p Q + ρ , g ( k ) ( x , t ) ≤ g ( k ) ( x, t ) for any ( x, t ) ∈ Q ∗ and g ( k ) ( x , t ) <
0. That is we have that Φ ≤ u ( k ) on ∂ p Q + ρ ( x , t ).Note also that if we extend ϕ in Q +1 by ϕ ( X, t ) = ϕ ( x, t ) and l i , i = 1 , . . . , n denote the eigenvalues of D ϕ ∈ S n then M − (cid:18) D ϕ, λn , Λ (cid:19) − ϕ t ≥ − Λ n || D ϕ || ∞ − | ϕ t | ≥ − (Λ n + 1) || ϕ || H α ( Q ∗ ) . That is, M − (cid:0) D b + D ϕ, λn , Λ (cid:1) − b t − ϕ t ≥ . Thus, u ( k ) − Φ ∈ S p (cid:0) λn , Λ (cid:1) in Q + ρ ( x , t ). Applyingmaximum principle we have that Φ ≤ u ( k ) in Q + ρ ( x , t ). In other words, Φ touches u ( k ) by belowat ( x , t ). Hence Φ y ( x , , t ) ≤ g ( k ) ( x , t ). On the other hand, Φ y ( x , , t ) = b y (0 ,
0) whichcompletes the proof. (cid:3)
Proposition 6. u ( k ) → u uniformly in Q +1 .Proof. We split our proof into two steps:
Step 1.
We prove equicontinuity of u ( k ) . For, it is enough to obtain an independent of k modulusof continuity of u ( k ) in Q +1 . Note that Lemma 4 gives a uniform L ∞ -bound for u ( k ) in Q +1 . AlsoLemma 5 gives a uniform L ∞ -bound for g ( k ) , thus using Theorem 6 in [11] we get a uniform H α -estimate for u ( k ) in Q +1 − ρ . So it remains to get a uniform modulus of continuity in Q +1 \ Q +1 − ρ .Note that (cid:0) u ( k ) (cid:1) y =0 on Q ∗ \ Q ∗ − ρ . Thus if we extend u ( k ) in Q \ Q − ρ considering its evenreflection ˜ u ( k ) with respect to y we have that ˜ u ( k ) ∈ S p ( λ, Λ) (see Proposition 2). We observealso that ˜ u ( k ) | ∂ p Q = u is independent of k and smooth enough and ˜ u ( k ) | ∂ p Q − ρ satisfy uniform H α -estimate. Thus using global H α -estimates for Dirichlet problems we get the desired uniformmodulus in Q +1 \ Q +1 − ρ . Step 2.
Arzel´a-Ascoli lemma implies that every subsequence of { u ( k ) } has a subsequence thatconverges uniformly in Q +1 . We claim that every uniformly convergent subsequence of { u ( k ) } mustconverge to u , then we should have that u ( k ) → u uniformly in Q +1 . To prove this claim let v be theuniform limit of { u ( k m ) } in Q +1 . If we show that v satisfies problem (1.1) then v = u by uniqueness.The closedness result of Proposition 31 in [11] gives immediately that F ( D v ) − v t = 0 in Q +1 and v y ≤ Q ∗ in the viscosity sense. Additionally, v = u on ∂ p Q +1 \ Q ∗ . It remains to check that(1) v y = 0 on Q ∗ ∩ { v > ϕ } , in the viscosity sense.(2) v ≥ ϕ on Q ∗ .For (1) let ( x , t ) ∈ Q ∗ be so that v ( x , , t ) > ϕ ( x , t ). From the continuity of v and ϕ , thereexists some small δ > v ( x, , t ) > ϕ ( x, t ) for any ( x, t ) ∈ Q ∗ δ ( x , t ). Next we use theuniform convergence of u ( k m ) to v . Take ε := min Q ∗ δ ( v − ϕ ) > n ∈ N so that | u ( k m ) − v | < ε in Q ∗ δ ( x , t ) for any m ≥ n . Hence u ( k m ) − v > − ε ≥ − v + ϕ , that is u ( k m ) > ϕ , so (cid:0) u ( k m ) (cid:1) y = 0 in Q ∗ δ ( x , t ) for any m ≥ n . Since F (cid:0) D u ( k m ) (cid:1) − (cid:0) u ( k m ) (cid:1) t = 0 in Q + δ ( x , t ) againfrom the closedness result of Proposition 31 in [11] we get that v y = 0 on Q ∗ δ ( x , t ).For (2) we assume that there exists some ( x , t ) ∈ Q ∗ such that v ( x , , t ) < ϕ ( x , t ) toget a contradiction. Again using the convergence we have that there exists n ∈ N so that EGULARITY FOR THE FULLY NONLINEAR PARABOLIC THIN OBSTACLE PROBLEM 13 u ( k m ) ( x , , t ) − v ( x , , t ) < ϕ ( x , t ) − v ( x , , t ) for any m ≥ n . Hence g ( k m ) ( x , , t ) = − k m (cid:0) ϕ ( x , t ) − u ( k m ) ( x , , t ) (cid:1) , that is, ϕ ( x , t ) − u ( k m ) ( x , , t ) = − k m g ( k m ) ( x , , t ) for any m ≥ n and g ( k ) is bounded independently of k by Lemma 5. By taking m → ∞ we get that ϕ ( x , t ) = v ( x , , t ) which is a contradiction. (cid:3) Proposition 6 gives the following.
Lemma 7.
For any < δ < , Du ( k ) → Du uniformly in K δ := Q − δ ∩ { y > δ } .Proof. Note first that from interior H α -estimates for viscosity solutions of F ( D v ) − v t = 0we know the existence of Du ( k ) , Du in K δ and a uniform H α -estimate for Du ( k ) (recall that || u ( k ) || L ∞ ( Q +1 ) are uniformly bounded). Therefore using Arzel´a-Ascoli lemma we get that everysubsequence of { Du ( k ) } has a subsequence that converges uniformly in K δ . Then by standard cal-culus we know that any uniformly convergent subsequence of { Du ( k ) } should converge to Du . (cid:3) Lemma 8.
For any < δ < , u ( k ) ∈ H α (cid:16) Q +1 − δ (cid:17) . Although the H α -estimates of the above may depend on k , Lemma 8 ensures the existence andregularity of (cid:0) u ( k ) (cid:1) y on Q ∗ in the classical sense. Proof.
Using Lemma 5 and Theorem 6 in [11] we get a uniform H α -estimate for u ( k ) in Q +1 − δ whichmeans that g ( k ) = − k (cid:0) ϕ − u ( k ) (cid:1) + is H α on Q ∗ − δ . Then applying Theorem 17 in [11] we get thedesired. (cid:3) Now we proceed in proving (3.3).
Lemma 9. σ ≤ on Q ∗ .Proof. For k ∈ N (fixed), we consider the solution u ( k ) of (4.1). We denote by v := (cid:0) u ( k ) (cid:1) y whichexists in the classical sense and it is continuous in Q +1 ∪ Q ∗ (due to Lemma 8). Then v ≤ Q ∗ and if 0 < δ < ρ , then v = 0 on Q ∗ \ Q ∗ − δ . Moreover we can use Theorem 15 of [11] in Q +1 − δ \ Q +1 − δ to obtain that v ≤ M, on ∂ p Q +1 − δ \ Q ∗ − δ where M > k .Next we apply a barrier argument to v . We define the function b to be the viscosity solution of M + (cid:0) D b, λn , Λ (cid:1) − b t = 0 , in Q +1 − δ b = M, on ∂ p Q +1 − δ \ Q ∗ − δ b = 0 , on Q ∗ − δ b ( x, , t ) = Mδ (cid:16) max n | x | , | t | o − δ (cid:17) , on Q ∗ − δ \ Q ∗ − δ . We remark that v ≤ b on ∂ p Q +1 − δ \ Q ∗ − δ and v ≤ ≤ b on Q ∗ − δ . Finally we know that v ∈ S p (cid:0) λn , Λ (cid:1) in Q +1 − δ , then v − b ∈ S p (cid:0) λn , Λ (cid:1) in Q +1 − δ . Using maximum principle we get that v ≤ b in Q +1 − δ and note that function b does not depend on k . On the other hand (cid:0) u ( k ) (cid:1) y → u y as k → ∞ pointwise in Q +1 − δ by Lemma 7. Hence u y ≤ b in Q +1 − δ . Finally, we observe that b = 0 on Q ∗ − δ , for any 0 < δ < ρ and we take y → + . (cid:3) Regularity of the solution
As we have mentioned at the points of Ω ∗ the regularity is known, therefore at these points theviscosity Neumann condition holds in the classical sense, thus σ = 0 in Ω ∗ .In this section we concentrate in studying the regularity of σ around free boundary points in orderto treat our problem as a non-homogeneous Neumann boundary value problem around these points.To achieve this we show first Lemma 14, an H α -estimate for σ in universal neighborhoods of pointsof Ω ∗ . Lemma 14 is based on Lemmata 12 and 13 and on semi-concavity of u in y . Lemma 12 saysthat considering a non-contact point P ∈ Q ∗ / , we can find a universal neighborhood of P whichcontains a small universal thin-cylinder where σ decays proportionally to its radius. Finally Lemma13 says that the information we have inside this small thin cylinder can be carried to a suitableset inside Q +1 and then is carried back in a parabolic neighborhood of P using semi-concavity in y . An iterative application of the above gives Lemma 14.We start with Lemma 11 which is important in proving Lemma 12. The following simple remarkis useful. Remark 10.
For P := ( x , t ) ∈ Ω ∗ , K := 2 K and ˜ ϕ P ( x, t ) := ϕ ( x , t ) + Dϕ ( x , t ) · ( x − x ) − K ( t − t ) + K | x − x | . we have that ˜ ϕ P > ϕ in Q ∗ ∩ { t ≤ t } \ { ( x , t ) } . Indeed, let Φ = ˜ ϕ P − ϕ . Then we observe that Φ( x , t ) = 0 and (a) D Φ( x, t ) = Dϕ ( x , t ) + 2 K ( x − x ) − Dϕ ( x, t ), thus D Φ( x , t ) = 0. (b) D Φ( x, t ) = 2 K I n − − D ϕ ( x, t ) >
0, that is Φ is convex with respect to x . (c) Φ t ( x, t ) = − K − ϕ t ( x, t ) <
0, that is Φ is monotone decreasing with respect to t .Then (b) (through integration) gives that Φ( x, t ) − Φ( x , t ) > ( x − x ) · D Φ( x , t ) = 0 for x = x . Thus by (a) we have that Φ( x, t ) > Φ( x , t ) = 0, for x = x . On the other hand (c) gives that Φ( x, t ) > Φ( x, t ) for any t < t and any x . Combining the above we get thatΦ( x, t ) > Φ( x , t ) = 0, for any x = x and any t < t . Lemma 11.
For P = ( x , t ) ∈ Ω ∗ , K := 2 K and C > nλ [Λ( n −
1) + 1] we define h P ( x, y, t ) := ϕ ( x , t ) + Dϕ ( x , t ) · ( x − x ) − K ( t − t ) + K | x − x | − C K y . EGULARITY FOR THE FULLY NONLINEAR PARABOLIC THIN OBSTACLE PROBLEM 15
We consider any set of the form
Θ := ˜Θ × ( t , t ] ⊂ Q , with P ∈ Θ , ˜Θ ⊂ R n a bounded domaincontaining x and < t < t . Then sup ∂ p Θ ∩{ y ≥ } ( u − h P ) ≥ . Proof.
Let w := u − h p then we have that w ( x , , t ) = u ( x , , t ) − ϕ ( x , t ) >
0, since ( x , t ) ∈ Ω ∗ . Moreover, w ∈ S p (cid:0) λn , Λ (cid:1) in Q +1 . Indeed, we note that ( h P ) ij = 0 for i = j , ( h P ) ii = 2 K for i < n , ( h P ) nn = − C K and ( h P ) t = − K . Then M + (cid:0) D h P , λn , Λ (cid:1) − ( h P ) t < − K < w y = 0 on Ω ∗ in the classical sense. Indeed, it isenough to note that ( h P ) y = − K n Λ λ y , that is ( h P ) y = 0 on Q ∗ .Now we denote by w ∗ the extension of w in Q considering its even reflection with respect to y and we have that w ∗ ∈ S p (cid:0) λn , Λ (cid:1) in Q \ ∆ ∗ (see Proposition 2). Then maximum principle givesthat sup ∂ p (Θ \ ∆ ∗ ) ∩{ y ≥ } w = sup ∂ p (Θ \ ∆ ∗ ) w ∗ ≥ sup Θ \ ∆ ∗ w ∗ ≥ w ( x , , t ) > x , t ) ∈ Θ \ ∆ ∗ . Finally we observe that ∂ p (Θ \ ∆ ∗ ) ∩ { y ≥ } ⊂ ( ∂ p Θ ∩ { y ≥ } ) ∪ (∆ ∗ ∩ { t ≤ t } ). On the other hand, h P = ˜ ϕ P > ϕ on Q ∗ ∩ { t ≤ t } \ { ( x , t ) } from Remark 10and ϕ = u on ∆ ∗ , that is w < ∗ ∩ { t ≤ t } and the proof is complete. (cid:3) Lemma 12.
For γ > we define Ω ∗ γ := { ( x, t ) ∈ Q ∗ : σ ( x, t ) > − γ } . Let ( x , t ) ∈ Ω ∗ ∩ Q ∗ / , thenthere exist constants < ¯ C < ¯¯ C < which depend only on K , n, λ, Λ , ρ so that for any < γ < there exists a thin-cylinder Q ∗ ¯ Cγ (¯ x, ¯ t ) so that Q ∗ ¯ Cγ (¯ x, ¯ t ) ⊂ Q ∗ ¯¯ Cγ ( x , t ) ∩ Ω ∗ γ . Proof.
Let ( x , t ) ∈ Ω ∗ ∩ Q ∗ / , we apply Lemma 11 withΘ := B ∗ C γ ( x ) × ( − C γ, C γ ) × (cid:0) t − ( C γ ) , t (cid:3) where 0 < C << C << P = ( x , y , t ) ∈ ∂ p Θ ∩ { y ≥ } sothat(5.1) u ( P ) − h P ( P ) ≥ . We split into two cases.
Case 1. If | x − x | = C γ or t = t − ( C γ ) . Then using (5.1) and Remark 10 we have in thefirst occasion that u ( P ) ≥ ϕ ( x , t ) + Dϕ ( x , t ) · ( x − x ) − K ( t − t ) + K | x − x | + K | x − x | − Kn Λ λ y ≥ ϕ ( x , t ) + K C γ ) − Kn Λ λ ( C γ )
26 GEORGIANA CHATZIGEORGIOU and similarly in the second occasion that u ( P ) ≥ ϕ ( x , t ) + K ( C γ ) − K n Λ λ ( C γ ) .Thus in any case(5.2) u ( x , y , t ) ≥ ϕ ( x , t ) + C γ where C > C , C (choosing 0 < C < q λ n Λ C ).Now take any ( x , t ) ∈ Q ∗ C γ ( x , t ), for C to be chosen. We intend to transfer the information(5.2) from ( x , y , t ) to ( x , t ) through integration and using the bounds of Proposition 3 forsuitable derivatives. We denote by τ = x − x | x − x | ∈ R n − and we assume that ( x − x ) · D n − ( u − ϕ )( P ) ≥ ϕ in Q +1 where ϕ ∗ ( x, y, t ) = ϕ ( x, y )). We notice that Z | x − x | Z e ( u − ϕ ) ττ ( x + τ h, y , t ) dhde = ( u − ϕ )( x , y , t ) − ( u − ϕ )( x , y , t ) − | x − x | ( u − ϕ ) τ ( x , y , t )and Z t t ( u − ϕ ) t ( x , y , h ) dh = ( u − ϕ )( x , y , t ) − ( u − ϕ )( x , y , t ) . Combining the above we get Z | x − x | Z e ( u − ϕ ) ττ ( x + τ h, y , t ) dhde − Z t t ( u − ϕ ) t ( x , y , h ) dh = ( u − ϕ )( x , y , t ) − ( u − ϕ )( x , y , t ) − | x − x | ( u − ϕ ) τ ( x , y , t ) . (5.3)On the other hand using (B) of Proposition 3 we have Z | x − x | Z e ( u − ϕ ) ττ ( x + τ h, y , t ) dhde ≥ − C | x − x | ≥ − C ( C γ ) and − Z t t ( u − ϕ ) t ( x , y , h ) dh ≥ − C ( t − t ) ≥ − C ( C γ ) . Therefore returning to (5.3) we have that( u − ϕ )( x , y , t ) ≥ ( u − ϕ )( x , y , t ) + ( x − x ) · D n − ( u − ϕ )( x , y , t ) − C ( C γ ) ≥ C γ − C ( C γ ) > < C < C C .Now (to get a contradiction) we assume that ( x , t ) / ∈ Ω ∗ γ , that is σ ( x , t ) ≤ − γ <
0. Then( x , t ) ∈ ∆ ∗ , that is u ( x , , t ) = ϕ ( x , t ). Similarly as before we want to transfer this information EGULARITY FOR THE FULLY NONLINEAR PARABOLIC THIN OBSTACLE PROBLEM 17 from ( x , , t ) to ( x , y , t ) via integration of u yy and using (C) of Proposition 3. We have Cy ≥ Z y Z e u yy ( x , h, t ) dhde = u ( x , y , t ) − u ( x , , t ) − y σ ( x , t )then, u ( x , y , t ) − ϕ ( x , t ) ≤ Cy + y ( − γ ) ≤ y γ ( CC − <
0, choosing 0 < C ≤ C . This is acontradiction regarding (5.4). Case 2. If y = C γ . Then using (5.1) and Remark 10 we have(5.5) u ( x , y , t ) ≥ ϕ ( x , t ) − K n Λ λ C γ . We take any ( x , t ) ∈ Q ∗ C γ ( x , t ). Assuming that ( x − x ) · D n − ( u − ϕ )( P ) ≥ u − ϕ )( x , C γ, t ) ≥ − CC γ > − C C γ (5.6)where 0 < C < CC .Now (to get a contradiction) we assume that σ ( x , t ) ≤ − γ <
0. Then u ( x , , t ) = ϕ ( x , t ).Similarly as in Case 1 we get that u ( x , C γ, t ) − ϕ ( x , t ) ≤ C γ ( CC − < − C C γ , choosing0 < C < − CC and C small enough. This is a contradiction regarding (5.6).In any case we have that there exists 0 < C << ρ, n, λ, Λ, K so that if( x , t ) ∈ Q ∗ C γ ( x , t ) with ( x − x ) · D n − ( u − ϕ )( x , y , t ) ≥ Q ∗ C γ ( x , t )) then ( x , t ) ∈ Ω ∗ γ . Moreover choosing 1 > ¯¯ C > C + C it is easy to check that Q ∗ C γ ( x , t ) ⊂ Q ∗ ¯¯ Cγ ( x , t ). By choosing a thin cylinder Q ∗ ¯ Cγ (¯ x, ¯ t ) inside Q ∗ C γ ( x , t ) ∩ { ( x − x ) · D n − ( u − ϕ )( x , y , t ) ≥ } the proof is complete. (cid:3) Now maximum principle and a barrier argument give the following important property.
Lemma 13.
Consider the set K := B ∗ × (0 , × ( − , and assume that w ∈ C ( K ) satisfies inthe viscosity sense . M − (cid:0) D w, λ, Λ (cid:1) − w t ≤ , in K w ≥ , in K . Suppose that there exists some neighborhood Q ∗ δ (¯ x, ¯ t ) ⊂ Q ∗ so that lim inf y → + w ( x, y, t ) ≥ , for any ( x, t ) ∈ Q ∗ δ (¯ x, ¯ t ) . Then, there exists ε = ε ( δ, n, λ, Λ) > so that w ( x, y, t ) ≥ ε, for any ( x, y, t ) ∈ B ∗ / × (cid:20) , (cid:21) × (cid:20) − δ , (cid:21) . Proof.
For any P ′ = ( x ′ , t ′ ) ∈ Q ∗ − δ we define the auxiliary function M − (cid:0) D b P ′ , λn , Λ (cid:1) − ( b P ′ ) t = 0 , in K b P ′ = 0 , on ∂ p K \ Q ∗ δ ( P ′ ) b P ′ ( x, , t ) = 1 − δ max {| x − x ′ | , √ | t − t ′′ | } , on Q ∗ δ ( P ′ )where t ′′ := t ′ − δ . Applying regularity results for Dirichlet-type boundary value problems (see[26]) we have that b P ′ is Lipschitz in K with the corresponding constant depending only on δ anduniversal quantities (but not on P ′ ).We claim that(5.7) b P ′ > , in B ∗ / × (cid:20) , (cid:21) × (cid:20) − δ , (cid:21) := K . Indeed, note first that b P ′ ≥ ∂ p K , thus by maximum principle b P ′ ≥ K . We suppose thatthere exists some ( x , y , t ) ∈ K with b P ′ ( x , y , t ) = 0 which means that b P ′ attains its minimumover K at ( x , y , t ). Then strong maximum principle gives that b P ′ = 0 on K ∩ { t ≤ t } . Notethat t ≥ − δ > t ′ − δ then there exists ( x, t ) ∈ Q ∗ δ ( P ′ ) such that t < t , that is b P ′ ( x, , t ) > t < t which is a contradiction.Now let ε ( P ′ , δ, n, λ, Λ) := min K b P ′ > ε ( δ, n, λ, Λ) := inf P ′ ∈ Q ∗ − δ ε ( P ′ , δ, n, λ, Λ) ≥ . We want to show that ˜ ε >
0. We assume that ˜ ε = 0, then there exists { P ′ j := ( x ′ j , t ′ j ) } j ∈ N ⊂ Q ∗ − δ so that ε ( P ′ j , δ, n, λ, Λ) → j → ∞ . Also for any j ∈ N there exists ( X j , t j ) ∈ K so that ε ( P ′ j , δ, n, λ, Λ) = b P ′ j ( X j , t j ). We notice also that { P ′ j } , { ( X j , t j ) } are both bounded sequences andtherefore there exist convergent subsequences (for which we use the same indices for simplicity).That is P ′ j → P ′∞ ∈ Q ∗ − δ , ( X j , t j ) → ( X ∞ , t ∞ ) , as j → ∞ . On the other hand b P ′ j are equicontinuous and uniformly bounded in K , thus there exist auniformly convergent subsequence in K , that is b P ′ j → b ∞ uniformly in K as j → ∞ . To get thecontradiction it is enough to show that(5.8) b ∞ = b P ′∞ , in K . Indeed, if (5.8) holds then by uniform convergence we have that b P ′ j ( X j , t j ) → b P ′∞ ( X ∞ , t ∞ ) as j → ∞ , thus b P ′∞ ( X ∞ , t ∞ ) = 0 which contradicts (5.7) since ( X ∞ , t ∞ ) ∈ K . Now to obtain (5.8),using uniqueness, it is enough to show that b ∞ solves the same Dirichlet problem as b P ′∞ in K .From closedness of viscosity solutions we know that M − (cid:0) D b ∞ , λn , Λ (cid:1) − ( b ∞ ) t = 0 in K . Also b ∞ = 0 on ∂ p K \ Q ∗ . Thus it remains to check the following two EGULARITY FOR THE FULLY NONLINEAR PARABOLIC THIN OBSTACLE PROBLEM 19 (1) b ∞ ( x, , t ) = 1 − δ max {| x − x ′∞ | , √ | t − t ′′∞ | } on Q ∗ δ ( P ′∞ )(2) b ∞ = 0 on Q ∗ \ Q ∗ δ ( P ′∞ ).For ( x, t ) such that | x − x ′∞ | < δ and | t − t ′′∞ | < δ we can choose an integer m = m ( x, t, δ ) > δ so that ( x, t ) ∈ Q ∗ δ − m ( P ′∞ ). Also there exists integer N = N ( x, t, δ ) ∈ N so that for any j ≥ N , | x ′ j − x ′∞ | < m and | t ′′ j − t ′′∞ | < m . Then for any j ≥ N , using that m > δ , we have that( x, t ) ∈ Q ∗ δ ( P ′ j ), that is b P ′ j ( x, , t ) = 1 − δ max {| x − x ′ j | , √ | t − t ′′ j | } and taking j → ∞ we obtain(1) at ( x, t ). Note that for ( x, t ) such that | x − x ′∞ | = δ or t ′∞ − δ = t or t = t ′∞ we use thecontinuity of b ∞ . Finally for ( x, t ) ∈ Q ∗ \ Q ∗ δ ( P ′∞ ), we follow a similar argument as before bychoosing m = m ( x, t, δ ) > δ so that ( x, t ) ∈ Q ∗ \ Q ∗ δ + m ( P ′∞ ).Now if ¯ P = (¯ x, ¯ t ) the given point we have that b ¯ P ≥ ˜ ε in K . We use maximum principle to getthis bound for w as well. So let v = w − b ¯ P then v ∈ S p (cid:0) λn , Λ (cid:1) in K . Moreover if ( x, t ) ∈ Q ∗ δ ( ¯ P ) thenlim inf y → + v ( x, y, t ) ≥ − b ¯ P ( x, , t ) ≥ x, t ) ∈ ∂ p K \ Q ∗ δ ( ¯ P ) then lim inf y → + v ( x, y, t ) ≥ w ≥
0. Therefore, w ≥ b ¯ P ≥ ˜ ε in K . (cid:3) The next lemma is a consequence of an iterative argument.
Lemma 14.
Let ( x , t ) ∈ Ω ∗ ∩ Q ∗ / , then there exists universal constants < α < , C > sothat ≥ σ ( x, t ) ≥ − C (cid:16) | x − x | + | t − t | / (cid:17) α , for any ( x, t ) ∈ Q ∗ / ( x , t ) . Proof.
Our aim is to show by induction that for any k ∈ N (5.9) u y ( X, t ) ≥ − Cθ k , for every ( X, t ) ∈ Q ∗ r k ( x , t ) × { y ∈ (0 , r k ) } where 0 < r << θ < C > k = 1 itfollows by (A) of Proposition 3 by choosing C appropriately. We assume that (5.9) holds for some k and we prove it for k + 1.We define w = u y + Cθ k − µr k + Cθ k , in Q ∗ r k ( x , t ) × { y ∈ (0 , r k ) } where 0 < µ < r < θ and µ < C we have that w ≥ Q ∗ r k ( x , t ) × { y ∈ (0 , r k ) } . Moreover, M − (cid:0) D w, λn , Λ (cid:1) − w t ≤ Q ∗ r k ( x , t ) × { y ∈ (0 , r k ) } . We observe also thatlim y → + w ( x, y, t ) = σ ( x, t ) + Cθ k − µr k + Cθ k , for ( x, t ) ∈ Q ∗ r k ( x , t ) . On the other hand applying Lemma 12 around ( x , t ) ∈ Ω ∗ ∩ Q ∗ / with γ = µr k < µr < we get that there exists Q ∗ ¯ Cµr k (¯ x, ¯ t ) ⊂ Q ∗ µr k ( x , t ) ∩ Ω ∗ µr k , where 0 < ¯ C < K , n, λ, Λ and ρ . That is, lim y → + w ( x, y, t ) ≥ x, t ) ∈ Q ∗ ¯ Cµr k (¯ x, ¯ t ) . Therefore, w satisfies the assumptions of Lemma 13 in Q ∗ r k ( x , t ) × (0 , r k ). So we apply Lemma 13 to the rescaled function W ( x, y, t ) := w ( µr k x + x , µr k y, ( µr k ) t + t ) in K and obtain that(5.10) w ≥ ε, in B ∗ µrk ( x ) × (cid:20) µr k , µr k (cid:21) × (cid:20) t − ( ¯ Cµr k ) , t (cid:21) where ε = ε ( ¯ C, n, λ, Λ) >
0, that is, u y ≥ − Cθ k + εCθ k using that r < θ and choosing µ < C .Now to fill the gap of y ∈ (cid:16) , µr k i we integrate u yy with respect to y and use (C) of Proposition3. For ( x, t ) ∈ B ∗ µrk ( x ) × h t − ( ¯ Cµr k ) , t i we have u y (cid:18) x, µr k , t (cid:19) − u y ( x, y, t ) = Z µrk y u yy ( x, h, t ) dh ≤ C µr k − C y where C > u y ( x, y, t ) ≥ − Cθ k + εCθ k − C µr k .Therefore in B ∗ µrk ( x ) × (cid:16) , µr k i × h t − ( ¯ Cµr k ) , t i we have that u y ( x, y, t ) ≥ − Cθ k + εCθ k − C µr k . We choose 0 < r < min n µ , ¯ Cµ √ o < then the above holds in B ∗ r k +1 ( x ) × (0 , r k +1 ) × (cid:2) t − ( r k +1 ) , t (cid:3) . Also using that r < θ and choosing µ < Cε C and θ > − ε we have that − Cθ k + εCθ k − C µr k ≥ − Cθ k +1 and the induction is complete.Taking y → + in (5.10) we have that for any k ∈ N σ ( x, t ) ≥ − Cθ k , for every ( x, t ) ∈ Q ∗ r k ( x , t )where 0 < r << θ < C > σ follows in a standardway. (cid:3) We are ready now to obtain the proof of the main theorem of this work.
Proof of Theorem 1.
First we use Lemma 14 to get the regularity of σ around P ∈ Γ ∗ ∩ Q ∗ / . SoLemma 14 gives that σ ( x , t ) = 0. Indeed we know that σ = 0 in Ω ∗ and since Γ ∗ = ∂ Ω ∗ ∩ Q ∗ thereexists { ( x k , t k ) } k ∈ N ⊂ Ω ∗ ∩ Q ∗ / so that ( x k , t k ) → ( x , t ) as k → ∞ . We have 0 ≥ σ ( x , t ) ≥− C (cid:0) | x − x k | + | t − t k | / (cid:1) α for any large k ∈ N . Thus taking k → ∞ we get the desired. Inaddition we have that 0 ≥ σ ( x, t ) ≥ − C (cid:0) | x − x | + | t − t | / (cid:1) α , for any ( x, t ) ∈ Q ∗ / ( x , t ).Indeed, we consider again { ( x k , t k ) } k ∈ N ⊂ Ω ∗ ∩ Q ∗ / so that ( x k , t k ) → ( x , t ) as k → ∞ . We have0 ≥ σ ( x, t ) ≥ − C (cid:0) | x − x k | + | t − t k | / (cid:1) α for any large k ∈ N and any ( x, t ) ∈ Q ∗ / ( x , t ) and welet k → ∞ .On the other hand we know that u y = σ on Q ∗ in the classical sense (thus, in the viscosity senseas well). Then once the Neumann data σ is H α we can apply Theorem 17 of [11] in Q +1 / ( x , t ) tocomplete the proof. (cid:3) EGULARITY FOR THE FULLY NONLINEAR PARABOLIC THIN OBSTACLE PROBLEM 21
Acknowledgments
This work is part of my Ph.D thesis. I would like to thank my thesis advisor, Professor EmmanouilMilakis for his guidance and fruitful discussions regarding the topics of this paper. The project wassupported by Research Promotion Foundation (Cyprus) grant no. Excellence/1216/0025.
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