The structure of the singular set in the thin obstacle problem for degenerate parabolic equations

We study the singular set in the thin obstacle problem (1.1) for degenerate parabolic equations with weight $|y|^a$ for $a \in (-1,1)$. Such problem arises as the local extension of the obstacle problem for the fractional heat operator $(\partial_t - \Delta)^s$ for $s \in (0,1)$, see the recent work [2]. Our main result establishes the complete structure and regularity of the singular free boundary. To achieve it, we prove new Weiss and Monneau type monotonicity formulas which generalise those in [15], where the case $a=0$ was treated.

For a detailed explanation of (1.1) and the relevant notation we refer the reader to Section 2. We say that (1.1) is a thin obstacle problem since the function ψ is supported on the codimension one manifold {y = 0} × (−1, 0) in the space-time variables (X, t), with X = (x, y) ∈ R n × (0, ∞). An important motivation for (1.1) is provided by its connection to the obstacle problem for the nonlocal heat operator with the fractional parameter s ∈ (0, 1) related to a ∈ (−1, 1) by the equation a = 1 − 2s. The passage from (1.2) to (1.1) rests on the extension procedure for the operator (∂ t −∆) s , developed independently by Nyström and Sande in [25] and by Stinga and Torrea in [28], which constitutes the parabolic counterpart of the celebrated extension procedure of Caffarelli and Silvestre in [11]. For instance, when s = 1/2 the problem (1.2) arises in the modelling of semipermeable membranes in the process of osmosis (for this and related problems see the classical monograph [16]). In such case, by taking a = 0 in (1.1), we see that (1.2) is equivalent to a lower-dimensional obstacle problem of Signorini type for the standard heat equation. We recall that in the paper [15] three of us and T. To developed an extensive analysis for this parabolic Signorini problem and established the optimal regularity of the solution, together with the H 1+α -regularity of the so-called regular free boundary, and a structure theorem for the singular part of the free boundary. We also refer to [3] for quasiconvexity results for certain generalised versions of the Signorini problem studied in [15].
In the present paper, and in the forthcoming work [8], we develop an analysis similar to the one in [15], but for the general case −1 < a < 1 in (1.1). In the first part of our program, which is the content of this paper, we provide a systematic classification of free boundary points. The main tool is a new monotonicity formula of Almgren-Poon type, which we utilise in the analysis of the blowup limits of appropriate rescalings. We also establish new monotonicity formulas of Weiss-and Monneau-type, which we employ to establish a structure theorem for the singular set.
Although the work in [15] has served as a road map for our analysis, the more general setting of the present paper requires delicate technical adaptations in the derivation of the above mentioned monotonicity formula of Almgren-Poon type, of Gaussian estimates, as well as in the subsequent blowup analysis. The novel complications are caused by the presence of the degenerate weight y a in (1.1). Such difficulty had already been tackled in the elliptic case in [22]. However, in the parabolic case one is faced with additional complications stemming from lower regularity properties of the solution in the time variable, and also from the fact that the relevant W 2,2 estimates need to be in the Gaussian space, because of the nature of the Almgren-Poon frequency. As the interested reader will see, all of these steps are substantially more involved than in the elliptic case.
In connection with our results we recall that in their recent work [2], Athanasopoulos, Caffarelli and Milakis show that, at a local level, the fractional obstacle problem (1.2) is equivalent to one of type (1.1) (under appropriate initial and boundary conditions). Based on such correspondence, the authors focus their attention on (1.1), establishing the optimal interior regularity of the solution, as well as the C 1,α regularity of the free boundary near certain non-singular points (which we call hyperbolic regular points, see Remark 7.4 for more details).
While the present work assumes the results in [2], it is developed in total independence from it. First of all, our focus is the analysis of the singular free boundary in (1.1). Secondly, in the forthcoming article [8], which complements and provides a foundation to the content of the present work, we establish localized versions of the regularity estimates in [2], both for the solution and for the free boundary.
To provide the reader with some further perspectives for the objectives of the present paper, we mention that our results are inspired by those obtained in [1], [4], [10] and [20] in the time-independent case. In such context, when a = 0 the structure of the singular set was first studied by two of us in [20]. In that paper the analysis hinges on new monotonicity formulas of Weiss and Monneau type. More recently, the results in [20] have been extended to the whole range a ∈ (−1, 1) in the cited work [22]. We also mention the recent interesting paper [13], where for the time-independent Signorini problem (a = 0) a finer stratification result for the singular set is obtained using a variant of Weiss' epiperimetric inequality, and the work [17] for a further refined analysis of the structure of the singular set under certain geometric assumption on the obstacle. A parabolic version of such epiperimetric inequality (again, when a = 0) has been very recently established in [27], where it has also been shown that such an inequality, combined with the results in [15], provides a finer structure theorem of the singular set in the parabolic thin obstacle problem. Finally, we also mention the recent interesting work [5] for related results on the nodal sets of solutions to degenerate parabolic equations such as that in (1.1).
In closing, we say something about the organization of the present paper. In Section 2 we introduce some basic notations and gather some known results which are relevant to our work. In Section 3 we introduce the class of global solutions S F (S + 1 ) of our thin obstacle problem with a thin obstacle ψ. In particular, we show how to effectively "subtract" the obstacle by maximally using its regularity, thus converting the original problem into one with zero thin obstacle, but with a non-homogeneous right hand side. In Section 4 we establish a generalized Almgren-Poon type monotonicity formula for solutions to the degenerate parabolic Signorini problem (1.1). Section 5 contains W 2,2 -type estimates in the Gaussian space. Such estimates are instrumental to the study of blowups carried out in Section 6, which is the most technical part of the paper. There, we prove the existence and homogeneity of blowups at free boundary points where the separation rate of the solution from the thin obstacle dominates the "truncation" terms in the generalized monotonicity formula. In Section 7 we establish a basic Liouville type theorem, which is used in Section 8 to classify the free boundary points according to the homogeneity of the blowup. In Section 9 we give a characterization of the so-called singular points (i.e., points where the free boundary is asymptotically negligible). Section 10 contains new Weiss-and Monneau-type monotonicity formulas which generalize those in [20], [22] and [15]. Finally, we briefly outline in Section 11 how to combine the Weiss-and Monneau-type monotonicity formulas with the results established in the previous sections, following the circle of ideas introduced in [15] for the case a = 0, to conclude uniqueness of blowups and obtain a structure theorem for the singular set (see Theorem 11.2). The paper ends with an appendix where we prove some of the auxiliary results stated in Section 4, that are crucial in the proof of our Almgren-Poon type monotonicity formula.

Notations and Preliminaries
In this section we introduce the basic notation and collect some background material which will be used throughout our work. We indicate with x = (x 1 , . . . , x n ) a generic point in R n , by (x, t) a point in the space-time R n × R, whereas the letter y will denote the "extension variable" on the half-line (0, ∞). The generic point in R n+1 + = R n × (0, ∞) will be denoted by X = (x, y). At times, we will tacitly use the same notation to indicate the generic point in R n+1 , i.e., without the restriction that y be > 0. For instance, given r > 0 we respectively denote by B r and B r the Euclidean balls centered at the origin with radius r in the variables x ∈ R n and X = (x, y) ∈ R n+1 . We also let B + r = {X = (x, y) ∈ B r | y > 0}. We denote by respectively the parabolic cylinders in the thin space (x, t) ∈ R n × R and thick space (X, t) ∈ R n+1 × R. We will indicate by the parabolic half-cylinder in the thick space. The notation ∂ p Q r will be used to denote the usual parabolic boundary of the thin cylinder Q r . We observe that the second order degenerate parabolic equation in (1.1) (the extension equation) can also be written in nondivergence form in the following way where we have denoted by B (a) y = ∂ 2 y + (a/y)∂ y the generator of the Bessel semigroup on (R + , y a dy). Using the representation (2.1) of the extension problem it was proved in [25] and [28], see also Section 3 in [6] for details, that one has in L 2 (R n+1 Moreover, it is shown in [2] that, at a local level, the fractional obstacle problem (1.2) is equivalent to the following lower-dimensional (or thin) obstacle problem for the local degenerate parabolic operator It is important to note that the obstacle ψ in (2.3) is thin in the sense that it lives on the thin (lower-dimensional) manifold Q 1 in space-time R n ×(−1, 0]. Throughout the paper, we will routinely refer to the thin obstacle problem above as the (parabolic) Signorini problem.
We will also assume that the solution of (2.3) has the following minimal regularity: . (Hereafter, we indicate with H α,α/2 the classical parabolic Hölder spaces, see e.g. [15].) This regularity follows for instance from global semiconvexity assumptions in the parabolic fractional obstacle problem in [2], with norms depending on the initial data, but can also be obtained directly for solutions of (2.3) in the form of interior estimates independent of initial data; see the forthcoming paper [8].
We next consider, for any a > −1, the Cauchy problem with Neumann boundary condition This corresponds to Brownian motion reflected at y = 0. Consider the following classes of functions . As it was observed in (22.8) of [18] membership in C 1 (a) (0, ∞) imposes, in particular, the weak Neumann condition For an analytic proof of the next result we refer the reader to Proposition 22.3 in [18].
, the Cauchy problem (2.4) admits the following solution where for y, η, t > 0 we have denoted by For t ≤ 0 we set p (a) (y, η, t) ≡ 0.
In (2.7) we have denoted by I ν (z) the modified Bessel function of the first kind and order ν ∈ C defined, in the complex plane cut along the negative real axis, by the series , |z| < ∞, | arg z| < π.
By (2.12) and the Remark 2.9 above, we obtain From (2.15) and (ii) ′ we have for every t > 0 We denote by G a (X, t) = G a (X, |t|), t < 0, the Neumann fundamental solution of the backward operator L ⋆ a = y a ∂ ∂t + div X (y a ∇ X ). This means that G a satisfies the equation in R n+1 From (2.12), for X ∈ R n+1 + and t < 0 we have the reproducing property We now consider the parabolic dilations in R n+1 × R defined by .
A function f : R n+1 × R → R is said to be homogeneous of degree κ ∈ R with respect to (2.21) A C 1 function is κ-homogeneous with respect to (2.21) if and only if one has Zf = κf . For instance, since from (iii) ′ above we see that we thus have For later use we notice that for every (X, t) such that t = 0, (2.22) can be rewritten Further, we indicate with |(X, t)| = |X| 2 + |t|, the standard parabolic pseudo-distance from the origin in the variables (X, t) ∈ R n+1 × R. Notice that such function is positively homogenous of degree one with respect to the dilations (2.21).
In closing, for every r > 0 we introduce the sets We emphasize that the + sign in the notation S + r refers to the variable y > 0 and not to the time variable t, which is instead negative for points in such set. The following simple lemma will be used in the subsequent sections. Proof. By (2.23) we have where in the second equality we have made a change of variables Y = X/r, τ = t/r 2 , for which y a dXdt = r n+a+3 η a dY dτ , and in the last equality we have used (2.17).

Classes of solutions
In this section we make some critical reductions on the problem (2.3). Hereafter, when for a function f (X, t) we write ∂ a y f on {y = 0} we mean ∂ a y f (x, 0, t) = lim y→0 + y a f y (x, y, t).
As a first step, we reduce the problem (2.3) to one with zero obstacle at the expense of introducing a nonzero right-hand side in the governing equation. The most straightforward way to do so is by considering the difference (Later on, we will make a more refined construction to take advantage of a possible higher regularity of ψ.) Since U solves (2.3), we have in Q + For later purposes it is important that we note here that the functionF , being independent of the variable y, is automatically even in such variable. If we now assume that ψ ∈ C 1,1 x and ψ ∈ C 0,1 t , then we clearly haveF ∈ L ∞ (R n+1 × R). We thus see that the function W satisfies We next want to extend (3.2) to a problem in a strip S + 1 .
Pick a cut-off function ζ = ζ(X) ∈ C ∞ 0 (B 1 ), 0 ≤ ζ ≤ 1, and such that ζ ≡ 1 in B 3/4 and ζ(X) = ζ ⋆ (|X|), and let Clearly, V is supported in Q + 1 . Since ζ is smooth and symmetric in y, the function V will satisfy on the thin set S 1 the same Neumann condition as W . Furthermore, we have ζ y = O(|y|) near the thin set {y = 0}, which implies that y −a div(y a ∇ζ) = O(1) and ζ y V y = O(1) up to the thin set. Therefore, if we let then F ∈ L ∞ (S + 1 ) and V solves the problem Recalling now the minimal regularity assumptions imposed on the solutions of (2.3), we are ready to introduce a central class of solutions in this paper.
Definition 3.1 (Solutions in strips). Given a function for F ∈ L ∞ (S + 1 ), we say that Suppose now the obstacle ψ in (2.3) is of class H ℓ,ℓ/2 (Q 1 ) with ℓ = k + γ ≥ 2, k ∈ N, 0 < γ ≤ 1. We then make the following more refined construction that takes advantage of the higher regularity of ψ. Let q k (x, t) be the parabolic Taylor polynomial of ψ at the origin of parabolic degree k. Then we have , for any multi-index α and j ≥ 0 with |α| + 2j ≤ k. We then extend the polynomial q k into R n+1 × R as an a-caloric polynomial, even in y, with the help of the following lemma. . For a given polynomial q(x, t) in R n ×R, there exists a unique polynomialq(x, y, t) in R n+1 × R, which satisfies Moreover, if q(x, t) is parabolically homogeneous of degree κ, thenq has the same homogeneity.
Proof. The proof is similar to that of Lemma 4.3 in [15] for the case a = 0 and Lemma 5.2 in [22] for the stationary case. For a given polynomial q(x, t), let , c 0 = 1.
Note that the sum above runs over a finite range of k, with 2k not exceeding the parabolic degree of q(x, t). It is clear thatq and thatq is even in y. Further, using that (with a convention that c −1 = 0) it is straightforward to check that Hence,q is the required a-caloric extension of q, even in y. We next show the uniqueness of such extension. By linearity of L a , it is sufficient to show that the only extension of q = 0 isq = 0. Note that for any such extension, both q and ∂ y aq vanish on {y = 0}. Now from the strong unique continuation property (which follows by applying the arguments as in Lemma 7.7 in [6]) we can conclude thatq ≡ 0.
Let now q k be the parabolic Taylor polynomial of ψ of parabolic degree k andq k be the corresponding a-caloric extension as in Lemma 3.2. Then consider where U is as in (2.3). It is easy to see that U k solves the thin obstacle problem with the thin obstacle ψ k . Consider then where ζ is a cut-off as in (3.3). Then V k is a global solution to the Signorini problem (3.4) corresponding to F k which is given by Now since ζ ≡ 1, in a neighborhood of 0, from (3.6) we obtain that F k satisfies when ℓ ≥ 2. If ℓ ≥ 3 we will also have and for ℓ ≥ 4 we will gain The additional growth condition as in (3.9)-(3.11) will allow a finer classification of free boundary points with the help of Almgren-Poon type monotonicity formulas that we prove in the next section.

Almgren-Poon type monotonicity formula
In this section we establish a monotonicity formula which play an essential role in our classification of free boundary points. We consider a function U ∈ S F (S + 1 ). We recall, see (3.5), that this means in particular that U solves the equation in S + We will also assume that the function F satisfies for some ℓ ≥ 2 and a constant C ℓ . Recall that such a hypothesis can be ensured by a reduction argument at the end of Section 3 when the obstacle is of class H ℓ,ℓ/2 . For t < 0 we introduce the quantity We also define the quantities for t < 0 where Z is the vector field in (2.22) above. Henceforth, we will routinely drop the indication of the variables (X, t) and of the (n + 1)-dimensional Lebesgue measure dX in all integrals involved. We will need the following result connecting d(U, t) and i(U, t).
Proof. To prove (4.6) we observe that, by the differential equation (3.5) satisfied by U in This gives The following computation can be justified rigorously by first considering the region and then let ε → 0 + . One should keep in mind that the outer normal on ∂R ε is ν = −e n+1 . Integrating by parts we find where in the last equality we have used (2.20). We thus have From this equation, from (2.25) and (4.5), we obtain Combining this equation with (4.7), we conclude that (4.6) holds.
Next, we introduce the following Steklov-type averaged versions of the quantities h(U, t), d(U, t), and i(U, t): and We now define two initial frequencies of U that will each prove useful in the computations. Remark 4.2. We remark that if U ∈ S 0 (S + 1 ) is homogeneous of degree κ with respect to the dilations (2.21), then we have In fact, since F ≡ 0 we have I(U, r) = D(U, r) from Lemma 4.3. But then, keeping in mind that ZU = κU , we find from (4.10) This proves the claim.
Using Lemma 4.1 we immediately obtain the following alternative expression for I(U, r).
One has for every r ∈ (0, 1) We now list the two main first variation formulas for H(U, r) and I(U, r). We observe that combining Lemma 4.4 with the former identity in (4.11), for every r ∈ (0, 1) such that H(U, r) > 0 we can write (4.12) N (U, r) = rH ′ (U, r) 2H(U, r) .
We will need the following result.
Lemma 4.5. For every r ∈ (0, 1) such that H(U, r) > 0 one has If H(U, r) > 0 this implies For later use in the proof of Theorem 4.8 we also record the following notable consequence of the above computation Lemma 4.6 (First variation of the energy). For a.e. r ∈ (0, 1) we have Combining Lemma 4.3 with Lemma 4.6 we immediately obtain.
Lemma 4.7 (First variation of the total energy). For a.e. r ∈ (0, 1) we have With the statement of Lemmas 4.4 and 4.7 in place we now establish a basic monotonicity formulas which plays a central role in our classification of free boundary points.
Then, for every σ ∈ (0, 1) there exist a constant C > 0, depending on n, a, C ℓ and σ, such that the function is monotone nondecreasing on (0, 1). In particular, the following limit exists Proof. We begin by introducing the set As it is well known by now in order to prove the theorem it suffices to verify the monotonicity of the function r → Φ σ (U, r) in the set E ℓ,σ . Now in E ℓ,σ we have where in the last equality we have used Lemma 4.4 and the former identity in (4.11). We now make the crucial observation that, thanks to Lemma 4.5, we can say that r → N (U, r) + 2 > 0 in E ℓ,σ . Therefore, to complete the proof it will thus suffice to show that we have in E ℓ,σ Finally, this is equivalent to proving that for every r ∈ E ℓ,σ we have for some constantC > 0 depending on n, a, C ℓ , σ. Then, the thesis of the theorem will follow with C =C/(1 − σ) > 0. We thus turn to proving (4.15). Using the first equation in (4.11) we find where in the last equality we have used (4.10). We thus obtain Cauchy-Schwarz inequality now gives Substituting in the above we find Expanding the first integral in the right-hand side of the latter inequality, and returning to the definitions of H(U, r) and I(U, r), we find This gives Using the Cauchy-Schwarz inequality we find .
To proceed, we note that for r ∈ E ℓ,σ we have in particular H(U, r) > 0, and thus we are in the conditions of Lemma 4.5. In particular, we trivially infer from (4.13) Using this bound in (4.16) gives This estimate obviously implies At this point we observe that (4.2), (2.23) and a simple rescaling argument imply Similarly, we obtain Now, if r ∈ E ℓ,σ we have H(U, r) > r 2ℓ−2+2σ and thus from (4.17), (4.18) and the above estimate for (2 + N (U, r)) ′ we find Since by Lemma 4.5 we know that 2 + N (U, r) > 0, this proves (4.15), thus completing the proof.

Gaussian estimates
In this section we establish some uniform second derivative estimates in Gaussian spaces that play a crucial role in the blowup analysis in Section 6.
. Then, for any 0 < ρ < 1 there exists a constant C(n, ρ) > 0 such that the following estimates hold: Proof. We closely follow the ideas in Appendix A in [15] and in [6]. In the rest of the proof, whenever we refer to the weak formulation of (4.1) we mean that, given η ∈ W 1,2 (R n+1 Having clarified this, we divide the proof into four steps. Step 1: Let 0 < ρ < 1 be fixed. We first establish (5.1), which represents a Caccioppoli type energy estimate in Gaussian space. We begin by noting that, since U is in S F (S + 1 ), there exists R > 0 such that U (·, t) is supported in B R for every t ∈ (−1, 0). Letρ be such that ρ <ρ < 1, and fix r ∈ [ρ,ρ]. We fix a cut-off functionτ 0 ∈ C ∞ 0 (R n+1 ) such that τ 0 ≡ 0 outside B R . Corresponding to suchτ 0 , for every k ∈ N we define a homogeneous function of degree k in S 1 by letting (5.4) τ k = |t| k/2τ 0 (X/ |t|). Using the test function η = U τ 2 1 G a in (5.3), for 0 < δ < r sufficiently small we obtain In (5.5) we have used the hypothesis that U ∂ a y U = 0 on the thin set {y = 0}, see (3.5), and the reproducing property (2.20). Since from (5.5) we have Handling the term Z(U 2 ) in (5.6) requires some care. For this, we argue as on p.92 in the Appendix of [15], making the change of variables t = −λ 2 , X = λY , and exploiting the homogeneity of G a and of τ 1 . After some work, we find From (5.6), (5.7) and Young's inequality, we obtain Integrating (5.8) with respect to r ∈ [ρ,ρ], then letting the support ofτ 0 sweep R n+1 , and δ → 0, we conclude that the estimate (5.1) holds.
Step 2: We next establish (5.2). We begin with the following second derivative estimate for tangential derivatives With ρ,ρ, r, δ as in Step 1, for a given i ∈ {1, . . . , n} and ε > 0, we also let is compactly contained in the interior of the set {∂ a y U = 0}, a standard difference quotient argument as in Section 5 in [6], allows us to assert that ∇U x i , ∂ t U x i ∈ L 2 loc (·, y a dXdt) up to {y = 0} in A ε (we stress that here we crucially use the fact that F t ∈ L ∞ (S + 1 )). Once we know this, with η as in (5.10), we use η x i as a test function in the weak formulation (5.3). Integrating by parts with respect to x i and by a limiting type argument (i.e., by first integrating in the region {y > β}, and then letting β → 0), we obtain and in the second term in the first integral in the left-hand side we have used (2.20). We stress that η x i is not a legitimate test function. Nevertheless, the computation in (5.11) can be justified by using as a test function difference quotients of the form instead of η x i , and then finally let h → 0. We also note that for the difference quotients the integration by parts with respect to x i is equivalently replaced by an identity of the following type which holds for arbitrary compactly supported functions f, g and is a consequence of a standard change of variable formula. In (5.11), we have also used that, since ∂ a y U = 0 on the set A ε , we have (∂ a y U ) x i = 0 on A ε . Letting ε → 0 in (5.11), we find (5.12) To handle the term with Z(U 2 x i ) we argue again as in the opening of page 92 in [15], obtaining in the right-hand side of (5.12) can be estimated by Young's inequality as follows (5.14) In the last step we have used the fact that the following inequality holds at every time level: for any v ∈ W 1,2 (R n+1 , G a y a dX) one has We mention that (5.15) corresponds to the inequality (8.17) in [6]. The remaining integrals in the right-hand side of (5.12) can be estimated in a similar way. Using Young's inequality we find Combining the estimates (5.14)-(5.16), and subtracting from the left-hand side the integral we obtain (5.17) As before, we now integrate over r ∈ [ρ,ρ], let the support ofτ 0 exhaust the whole of R n+1 , then let δ → 0, and also using the previously established estimate (5.1), we finally deduce that (5.9) holds.
Step 3: Our next objective is to establish the following second derivative estimate in the normal direction: For this we make use of the following conjugate equation which is satisfied in R n+1 For a given ε > 0, we consider the test function in the weak formulation of (5.19). We note that since ∂ a y U ≤ 0 on the thin set {y = 0}, thanks to the Hölder continuity of y a U y up to {y = 0}, the function η is compactly supported in the region {y > 0}, and therefore it is a legitimate test function. With w = y a U y , we thus have (5.20) Similarly to (5.13), we now obtain Using Young's inequality, we estimate Finally, the last term in the left-hand side of (5.20) is estimated in the following way. First, Young's inequality (and the trivial observation that |y| ≤ |X|) gives where c 0 can be chosen arbitrarily small. To control the second integral in the right-hand side of (5.22) we argue similarly to (5.15), but with a replaced by −a. Inserting the ensuing estimate in (5.22), the resulting inequality becomes (5.23) for some constant C 1 > 0. At this point we choose c 0 small enough such that C 1 c 0 < 1 4 . Combining the estimates (5.21) and (5.23), and then subtracting the integral from the left hand side of (5.20), we finally obtain Integrating again in r ∈ [ρ,ρ], by a limiting argument and finally letting ε → 0, we deduce the following estimate where ρ <ρ < 1.
As a next step, we obtain an estimate similar to (5.24) for ∇w − in L 2 (·, G a y −a dXdt). Given ε > 0, we consider the function v ε = (w + ε) − , where w = y a U y as before. Let C ε = supp v ε , and denote byC ε the reflected portion of C ε across {y = 0}. Since U ∂ a y U = 0 on {y = 0}, by the continuity of y a U y up to {y = 0}, we see that C ε ∩ {y = 0} is contained in the interior of {(x, 0) | U (x, 0) = 0}. Therefore near C ε ∩ {y = 0}, if U is oddly reflected, then the extended U solves the following equation whereF is the odd extension of F across {y = 0}. Therefore, in the set D ε the function v = |y| a U y is an even extension of w across {y = 0}, and it solves the following conjugate Using the test function η = (w + ε) − τ 2 2 G a in the weak formulation of (5.25), arguing as in (5.20)-(5.24), and finally letting ε → 0, we obtain the following estimate By combining (5.24) and (5.26), and using the previously established estimate (5.1), we finally have which completes the proof of Step 3.
Step 4: At this point, using the equation satisfied by U , the corresponding estimate for U t claimed in (5.2) follows from (5.9) and (5.18). This finishes the proof of the lemma.
Remark 5.2. We note that assumption that F t L ∞ (S + 1 ) be finite in Lemma 5.1 is not restrictive since it can always be achieved if the obstacle is sufficiently regular.
We also need the following estimate in our blowup analysis in Section 6.
. Then, for any 0 < ρ < 1, one has Proof. First, we even reflect U 1 , F 1 , U 2 , F 2 across {y = 0}. Consider now U = U 1 − U 2 . We claim that the following holds in S 1 , We note that it is clear that the differential inequalities (5.28) are respectively satisfied in S + 1 and S − 1 . Therefore, we only need to show the inequality near a point (x 0 , 0, t 0 ) ∈ Q 3/4 . Suppose U (x 0 , 0, t 0 ) > 0. Since it must necessarily be U 1 (x 0 , 0, t 0 ) > 0, we infer the existence of a sufficiently small δ > 0, such that ∂ a y U 1 ≡ 0 in Q δ (x 0 , 0, t 0 ). This implies that L a U 1 = |y| a F 1 in Q δ (x 0 , 0, t 0 ). On the other hand, . Now by using a standard argument as in the proof of Lemma 2.1 in [26], we can deduce that L a U + ≤ |y| a (F 1 − F 2 ) + in Q 3/4 and hence in S 1 . The argument for U − is similar and thus we can assert that (5.28) holds. Now given the validity of (5.28), by arguing as in Step 1 as in the proof of Lemma 5.1 by using η = U ± τ 2 1 G a as a test function in the weak formulation for U ± to conclude that the weighted Caccioppoli type estimate (5.27) holds.

Existence and homogeneity of blowups
Throughout this section, we assume that U ∈ S F (S + 1 ), where F satisfies (4.2) for some ℓ ≥ 2. Towards the end of the section, we will need ℓ ≥ 4 and require the following additional bounds to hold for and some positive constant C ℓ We note that this hypothesis can be ensured by taking an obstacle of class H ℓ,ℓ/2 and following the reduction argument at the end of Section 3.
We now state our first relevant result.
Furthermore, if κ < ℓ − 1 + σ, then there exists r 0 = r 0 (U ) > 0 such that for every r ∈ (0, r 0 ) In particular, we have in such case Proof. The proof of (6.3) and (6.4) follows the lines of that of Lemma 7.1 in [15] in the case a = 0, and we thus refer the reader to that source for details. In order to establish (6.5) we note explicitly that from (4.9), (4.11) and Lemma 4.3, we have Keeping in mind that the hypothesis (4.2) implies (4.18), Cauchy-Schwarz's inequality gives This shows that lim Later in the paper we will need to work with two different families of rescalings, which we now introduce. For κ > 0 we define the κ-homogeneous rescalings of U as (6.7)Ũ r = U • δ r r κ .
We note that the rescaled functions U r solve (6.8) where (6.9) F r = r 2 F (rX, r 2 t) H(U, r) 1/2 We have the following crucial result whose elementary verification we leave to the reader. In particular, we have for the parabolic Almgren rescalings Lemma 6.4. Let U ∈ S F (S + 1 ) where F satisfies (4.2) for some ℓ ≥ 2. Then we have that U L ∞ (B + Ar ×(−r 2 /4,0)) ≤ CH(U, r) 1/2 + Cr ℓ for some C universal depending also on ℓ, A where r < 1.
Proof. It suffices to prove the claim for U + and U − since both of them satisfy L a V ≤ |y| a F L ∞ (Qr) in Q r after even reflection for any r > 0.
A simple calculation shows that such can be ensured. For simplicity of notations, in the ensuing computations below, we let C 0 = C n,γ,a,ℓ . We then note that the following holds where in the last inequality, we used change of variables and the homogeneity property of G a . Now given (X 0 , t 0 ) = (x 0 , y 0 , t 0 ) ∈ B + Ar × (−r 2 /4, 0) and using thatṼ has a polynomial growth at infinity, we can use a Tychonoff type comparison to deduce the following submean value type estimate where p (a) is as in (2.7) and s 0 < t 0 < 0.
Indeed, (6.12) can be justified using a variation argument as in [14]. Let τ R be a cut-off such that τ R ≡ 1 on B R and vanishes outside B 2R . Then w =Ṽ τ R solves (6.13) L a w ≤ 2y a (|∇Ṽ ||∇τ R | + y a |Ṽ ||∇ 2 τ R |) for t < 0. Fix t 0 < 0. Then, for a given s 0 < 0 define N (y, y 0 , −s 0 − t 2 ) by K(X, X 0 , −t 2 ). (Note that here, (x 0 , y 0 ) refers to the point X 0 .) Then φ(t) is absolutely continuous since w is Lipschitz in t, and therefore by differentiating under the integral sign, we obtain that for a.e. t Now using that (6.15) y a ∂ t K = div(y a ∇K) y a K y → 0 as y → 0 (where the latter limit follows from the explicit representation of K), integrating by parts the second integral in (6.14) and by using (6.13) (which holds in the weak sense at almost every time level t), we deduce that the following bound holds, By applying the Fundamental theorem of Calculus to the absolutely continuous function φ(t), this implies that Now, by letting t → √ −s 0 and using the Dirac property, we obtain, via shift of time variables (more precisely by letting t 0 + s 0 + s as the new time variable s) in the second integral on the right hand side in (6.16), that the following holds, To proceed further, we fix b > 0 small such that −1/2 < t 0 < −b and s 0 < −b. Let also |X 0 | ≤ A and R is sufficiently large (say R ≥ 100A + 1). Note that G R is supported in the region where |X| ≥ R. From the representation of p (a) as in (2.7) and the asymptotic behavior of the modified Bessel function I 1−a 2 , we note that the following claim holds (similar to Claim 7.8 in [15] when a = 0).
In order to prove the claim, we need the following asymptotics for the Bessel function (see for instance [23]) which asserts that there exists c(ν) > 0 and a universal constant C = C(ν) such that Moreover, we also have for |X| ≥ 100A + 1 and |X 0 | ≤ A, We also note that from (2.7), p (a) can be rewritten as (6.20) p (a) (y, y 0 , t 0 − s) = e Now using (6.18), (6.19) and (6.20) with ν = 1−a 2 , we subdivide the proof of case (a) in Claim 6.5 into two cases. When  Next, when yy 0 we obtain and proceeding similarly to the proof of 2 • of Claim 7.8 in [15], by also uning the fact that t 0 − s ≥ −s/8 > c 0 b. Now from the fact thatṼ , ∇Ṽ have at most polynomial growth at infinity, by letting R → ∞ and by using the bounds in case (a) of Claim 6.5, we deduce that the second integral in (6.17) goes to 0. Also, by using the bounds in case (b) of Claim 6.5 and Cauchy-Schwarz inequality, we deduce that the first integral in (6.17) converges to the integral in the right hand side of (6.12) as R → ∞ and consequently we can assert that the sub-mean value type estimate (6.12) holds. Now in (6.12), for t 0 ∈ (−r 2 /4, 0), in (6.12), we take s 0 ∈ (−r 2 /2, −r 2 /4]. Then by arguing as in the proof of Claim 6.5 and by additionally using the fact that |y 0 |, |x 0 | ≤ Ar, |s 0 |, |t 0 + s 0 | ∼ r 2 , we can assert that the following holds x,x 0 2|t 0 +s 0 | e C 4 yy 0 2|t 0 +s 0 | where C 4 is universal and depends also on a. Note that this follows by letting t 0 + s 0 as our new s in which case −s 0 becomes t 0 − s. Then we can argue as in the proof of Claim 6.5 to prove the validity of (6.23). Therefore we havẽ Then again by using Cauchy-Schwartz, we obtaiñ x,x 0 |s| e C 4 yy 0 |s| 1/2 (6.24) Now, using the fact that |X 0 | ≤ Ar and that |s| ∼ r, we obtain for some C 10 , Subsequently, by integrating from s ∈ (−r 2 , −r 2 /2) and applying the Cauchy-Schwartz inequality as in the proof of Lemma 9.3 in [15], we obtaiñ ≤ CH(Ṽ , r) 1/2 ≤ CH(U, r) 1/2 + Cr ℓ .
We also need the following two lemmas. Lemma 6.6. With κ as in Lemma 6.1, let κ ′ be such that κ < κ ′ < ℓ − 1 + σ and H(U, r) ≥ r 2ℓ−2+2σ . Then there exists r U > 0 depending on κ ′ , σ such that there exists r 0 > 0 depending on U, κ ′ such that for 0 < r < r U < r 0 we have Then, by integrating from ρr to r and exponentiating, we obtain The second estimate in (6.26) follows similarly by integrating from r to rR and by noting that rR < r U .
Lemma 6.7. Under the hypothesis of Lemma 6.6 above, with F r as in (6.9), we have that for any R ≥ 1 such that 0 < Rr ≤ r 0 , for some universal constant C, with κ ′ as in Lemma 6.6.
Proof. We have that where we used the fact since Rr ≤ r U < r 0 , implying that H(U, Rr) ≥ (Rr) 2ℓ−2+2σ .
From this point on, we need to assume ℓ ≥ 4 as well as that F satisfies the bounds (4.2), (6.1), (6.2), unless stated otherwise. We then have the following theorem concerning the existence and homogeneity of blowups. Theorem 6.8 (Existence and homogeneity of blowups). Let U ∈ S F (S + 1 ) with F satisfying (4.2), (6.1), (6.2) for some ℓ ≥ 4. Fix σ ∈ (0, 1). Suppose κ = Φ ℓ,σ (U, 0 + ) < ℓ − 1 + σ For r > 0, let U r denote the Almgren rescalings as in Definition 6.2. Then (i) For every R > 0, there exists r R,U > 0 such that (ii) There exists a sequence r j → 0 and a function U 0 ∈ S + ∞ such that (iii) U 0 is parabolically homogeneous of degree κ and is a global solution of the thin obstacle problem, i.e., Proof. We first note that U r solves (6.8). Therefore by taking r R,U = r U R , the first estimate in (i) in Theorem 6.8 follows from the Gaussian estimates in Lemma 5.1, the fact that H Ur (1) = 1 and Lemma 6.7.
We now establish the second estimate in (i). In order to do so, we first note that Lemma 6.4 coupled with the fact that H(U, r) ≥ r 2ℓ−2+2σ for small enough r implies that U r is bounded up to {t = 0} and consequently we can assert that U r is uniformly bounded in L 2 (Q + R , y a dXdt) for any R > 0 and r sufficiently small. Now since F satisfies (4.2) and for small enough r, H(U, r) ≥ r 2l−2+2σ , we can deduce that {F r } is uniformly bounded for small enough r. Therefore it follows from the W 2,2 estimates in [8] applied to U r , which solves the Signorini problem corresponding to F r , that such W 2,2 estimates are stable up to {t = 0} and hence we can assert that the second estimate in (i) holds. Now we note that in view of Lemma 5.3 and Lemma 6.7, in order to establish (ii), it suffices to show the existence of U 0 and the convergence (6.29) for a subsequence r j → 0. Since κ ≥ 0, for all small enough r, say r ≤ r 1 , we have Integrating the above inequality from rδ to r we obtain and consequently S + δ y a U 2 r G a ≤ δ, 0 < r < r 1 .
We now choose A > 2 large enough such that for all −1 < t < 0, Now by using the uniform gradient estimates from (i), i.e., S + R y a |t||∇U r | 2 G a < C R and the inequalities as in (6.31) and (6.32), we can argue as in (7.17)-(7.26) in [6], which crucially uses the estimate (6.33), to conclude that the following holds, (6.34) for some universal C, which also depends on R. Now in the set we have that G a is bounded from above and below. Therefore from the uniform Gaussian estimates as in (i), we have that {U r } is uniformly bounded in W 2,1 2 (E, y a dXdt) and thus we can extract out a subsequence which converges strongly to some U 0 in L 2 (E, y a dXdt) and consequently in L 2 (E, y a G a dXdt). Hence the claim in (ii) now follows in a standard way by a Cantor diagonalization argument by letting δ → 0 and A → ∞.
We now prove the claim in (iii). Now given any compact subset K of S ∞ = R n+1 + × (−∞, 0], we have that U r 's are uniformly bounded in W 2,2 (K, y a dXdt) which follows from the second estimate in i). Then we can apply the local regularity estimates as in [8] to assert that ∇ x U r , y a ∂ y U r ∈ H s,s/2 (K), ∂ t U r ∈ L ∞ (K) uniformly in r. This follows from the fact that the conditions (4.2), (6.1), (6.2) imply that F r , ∇ X F r , and ∂ t F r are locally uniformly bounded in S ∞ . Thus, for ∂ t F r we have for small r > 0, which gives the uniform bound on compact subsets of S ∞ . The above uniform regularity of U r is enough to pass to the limit in the Signorini problem and infer that U 0 solves the Signorini problem in (6.28). Moreover by lower semicontinuity, we can assert that the following estimate holds for U 0 , for any R > 0. Now given the estimate (6.35), we can now justify the Poon type computations for U 0 by using truncations as in the appendix of [6]. Here, we note that the intermediate calculations for the corresponding truncated functionals can be justified by using the fact that ∇ x U 0 , y a ∂ y U 0 ∈ H s,s/2 (K), ∂ t U 0 ∈ L ∞ (K) for any compact subset K of R n+1 + × (−∞, 0]. Therefore, we can finally assert that the following holds for any r > 0. Now because of (ii), we have that Now, by Lemma 6.6, since H(U r , ρ) ≥ ρ 2κ ′ for any κ ′ > κ, therefore H(U 0 , ρ) = 0 for any ρ > 0. Now we can conclude from (6.37) that i.e., (6.38) N (U 0 , ·) ≡ κ and hence the right hand side in (6.36) vanishes since N ′ (U 0 , ·) ≡ 0. Then from the equality in the Cauchy-Schwartz inequality we get (6.39) ZU 0 = κ 0 U 0 Now again since I(U 0 , r) has the following alternate representation we get by using (6.38) and (6.39) that κ 0 = κ and consequently we have that U 0 is parabolically homogeneous of degree κ. This finishes the proof of (iii).
Finally we note that (iv) follows from the second estimate in (i) (which is uniform in r) and hence the local regularity estimates developed in [8] implies that for any compact subset K of R n+1 + × (−∞, 0], we have that ∇ x U r , y a ∂ y U r ∈ H s (K), ∂ t U r ∈ L ∞ (K) uniformly in r. Consequently by Arzela-Ascoli, we have that for a subsequence of {r j } as in (ii), U r j , ∇ x U r j and y a ∂ y U r j converges uniformly in K to U 0 , ∇ x U 0 and y a ∂ y U 0 respectively and thus (iv) holds. Note that ∂ t U 0 ∈ L ∞ (K) follows from the uniform convergence of U r j to U 0 and the uniform time Lipschitz bounds for U r j 's.

Homogeneous global solutions and regular points
In this section, we show that the frequency limit at a free boundary point is either κ = 1 + s or κ ≥ 2. Furthermore, we show that the free boundary is regular near points where κ = 1 + s. Theorem 7.1. Let ℓ ≥ 4 and σ > 0. Then with κ as in Theorem 6.8, we have that Proof. Since (0, 0) ∈ Γ * (U ), we have ∇U (0, 0, 0) = 0. By using the boundedness of U t , we can reduce the regularity considerations to the elliptic case as in [8] (see also [2]) and from the regularity results in the elliptic case as in [10], we obtain that for some universal C, the following holds, Hence, where the second inequality in (7.1) follows from change of variable and homogeneity property of G a . Now with κ as in Theorem 6.8, we obtain from the non-degeneracy Lemma 6.6 that for any κ ′ > κ. Then (7.1) and (7.2) together imply The conclusion now follows by letting κ ′ → κ.
Next, we state our gap lemma.
Proof. Assume on the contrary that κ < 2. This also implies κ < ℓ − 1 + σ. Then it follows that for any κ ′ such that κ < κ ′ < 2, we have that for small enough 0 < r < r 0 , i.e., As before, note that U r solves the Signorini problem corresponding to F r as in (6.9). Using (4.2) and (7.3), it follows that F r → 0 as r → 0 as well as that since U t is bounded and r 2 H(U, r) → 0 as r → 0, because of (7.3) with κ ′ < 2. This implies that U 0 is a time independent global solution of the Signorini problem and is homogeneous of degree κ which is less than 2. Then it follows from the classification result in Theorem 5.7 in [10] that κ = 1 + s.
We now show that at near points with frequency κ = 1 + s, the free boundary is H 1+α regular.
Since U (·, 0) solves the elliptic Signorini problem with bounded right hand side (because of the boundedness of U t ), when the parabolic frequency limit κ at (0, 0) equals 1 + s, we show that the elliptic Almgren frequency at 0 ∈ Γ(U (·, 0)) (say κ 0 ) also equals 1 + s and consequently it follows that the free boundary is H 1+α,(1+α)/2 regular near (0, 0) ∈ R n × (−∞, 0] in x, t for some α > 0. We refer to [8] for a rigorous justification of H 1+α,(1+α)/2 regularity of the free boundary in space and time near such an elliptic regular point for U (·, 0). This later result crucially uses the elliptic epiperimetric inequality developed in [21] coupled with the boundedness of U t . The corresponding result can be stated as follows.
Proof. Let κ < κ ′ < 2. For 0 < δ < 1, split the integral in the definition of H(U, r) as follows (7.5) H(U, r) = 1 with the idea of estimating the second integral in the right hand side of (7.5). By writing we note that in the region {(X, t) | t < 0, |X| ≥ |t| δ/2 , |t| ≤ r 2 }, we have the bound ≤ e − 1 8r 2−2δ . Therefore, using (7.6) and the boundedness of U , the second integral on the right hand side of (7.5) can be estimated as follows Consequently, we obtain where C also depends on the global bounds of U . Next note that 0 < r < r 1 , where r 1 = r 1 (δ) is small enough, we have c 0 r 2κ ′ and therefore we can deduce from (7.3), combined with (7.8) and (7.9), that r 2 ,0)) . Since κ ′ < 2 and U t is bounded, we obtain by letting r δ as our new r that the following inequality holds for all r ≤r, with small enoughr =r(δ), Now since δ can be chosen arbitrarily close to 1 and κ ′ can be chosen arbitrarily close to κ, by letting κ ′ δ as our new κ ′ , we deduce that there exists r 2 small enough depending also on κ ′ such that the following inequality holds for r ≤ r 2 , We now claim that (7.10) implies that 0 ∈ Γ * (U (·, 0)) is a regular free boundary point for the corresponding elliptic problem. If not, then it follows from [10] that the the elliptic Almgren frequency limit κ 0 for U (·, 0) as in [10] or [9] is bigger than or equal to 2. This follows from the classification result for global time independent solutions as in Theorem 5.7 in [10]. Then from the estimate in Lemma 6.5 in [10], we obtain that U (·, 0) separates from the free boundary at a rate which is atleast quadratic and this is a contradiction to the estimate (7.10) above since κ ′ < 2. Therefore the elliptic frequency limit κ 0 has to be equal 1 + s and then the regularity result for the free boundary in [8] implies that the free boundary is H 1+α,(1+α)/2 regular near (0, 0) in space and time for some α > 0.
Remark 7.4. The proof of Theorem 7.3 can be viewed as the consolidation of "parabolic" and "elliptic" approaches to the definition of regular points. Namely, we say that (x 0 , t 0 ) ∈ Γ * (U ) is a parabolic regular point if the parabolic frequency κ U (x 0 , t 0 ) = Φ ℓ,σ (U, 0 + ) = 1 + s. We say that (x 0 , t 0 ) is an elliptic regular point if x 0 ∈ Γ(U (·, t 0 )) with elliptic frequency κ U (·,t 0 ) = 1 + s. The proof of Theorem 7.3 shows that these two notions are in fact equivalent. These points can also be defined as free boundary points where the quantities are bounded away from zero and infinity, i.e., 0 < L ell < ∞, 0 < L par < ∞, for elliptic and parabolic regular points, respectively. There is also the third approach, taken by the authors of [2], which we call "hyperbolic." Namely, we say that (x 0 , t 0 ) ∈ Γ * (U ) is a hyperbolic regular point, if the quantity , is bounded away from zero and infinity, i.e., 0 < L hyp < ∞. It is proved in [2] that near such points the free boundary is C 1,α regular in space and time. Because of this regularity, it is easy to see that hyperbolic regular points are also elliptic (and equivalently) parabolic regular. The converse statement that elliptic (or parabolic) regular points are hyperbolic is not immediately obvious. However, we should point out that in the case when s = 1/2 (or equivalently a = 0), the converse statement does hold because of the higher regularity of the free boundary near (parabolic) regular points, see [7].
We close this section with a Liouville type result for the operator L a which will be used subsequently in the classification of singular points.
Proof. The proof is similar to that of the elliptic case as in [10] and is based on induction in the degree k. The following elementary fact will be used: Fact: If ∇ x v, ∇ t v are polynomials and v(0, y) is a polynomial, then v is a polynomial.
Suppose k ≤ 1. We first note that the following estimate holds We note that the estimate (7.11) follows from the Hölder regularity result of [12] coupled with the translation invariance of the equation in (x, t). Now since L a v = 0, (7.11) implies (7.12) sup So by letting r → ∞ (when k = 1), we obtain v yy + a y v y ≡ 0 We also obtain from (7.11) by letting r → ∞ that v is time independent. Then by repeating the arguments as in the proof of Lemma 2.7 in [10] which only uses the symmetry of v in y, we can assert that v = bx + c in this case. Now for general k (assuming the assertion of the Lemma holds up to k − 1), it follows from the following rescaled estimate the induction hypothesis and the the fact that ∇ x v and v t solves the same equation, that ∇ x v is a polynomial of order k − 1 and v t is a polynomial of order k − 2. Also from (7.12) we obtain v yy + a y v y ≤ C(|X| 2 + |t|) Now similar to the elliptic case, it follows from the observation (7.14) v yy + a y v y = |y| −a ∂ y (|y| a v y ) that v yy + a y v y solves the same equation as v (Note that w = |y| a v y solves the conjugate equation L −a w = 0, see for instance [6], and therefore v yy + a y v y being the twice conjugate of v solves the same PDE as v). Then, from the estimate in (7.12), the fact that w = v yy + a y v y solves L a w = 0 and the induction hypothesis, we deduce that v yy + a y v y is a polynomial of order at most k − 2. In particular, for (x, t) = (0, 0), v yy (0, y, 0) + a y v y (0, y, 0) is an even polynomial p(y) = a 0 + a 2 y 2 + · · · + a 2d y 2d . Then, by using the expression for v yy + a y v y as in (7.14) and integrating twice, we obtain v(0, y) = c + by|y| −a + a 0 2(1 + a) y 2 + a 2 2(3 + a) y 4 + · · · + a 2d (2d + 2)(2d + 1 + a) y 2d+2 .
Next, using the evenness of v, we conclude that b = 0 and hence v(0, y) is a polynomial. Finally, by using the fact that ∇ x v, v t are polynomials, we deduce that v is a polynomial.

Classification of free boundary points
Let the obstacle ψ be of class H ℓ,ℓ/2 and let V k be as in (3.8). Now given σ < 1, by repeating the arguments in the proof of Lemma 10.1 in [15] we can show that the limit is independent of the cut-off τ in the definition of V k . Therefore if we denote κ in (8.1) by κ ℓ,σ U (since this quantity is independent of the cut-off τ and consequently independent of U k ), then we have the following consistency result for the truncated frequencies whose proof is exactly the same as in the case for a = 0 as in [15] (See proposition 10.3 in [15]). (8.2) κ ℓ,σ U (0, 0) = min{κl ,σ U (0, 0), ℓ − 1 + σ} whenever ℓ ≤l and ℓ − 1 + σ ≤l − 1 +σ. Consequently in view of (8.2), if ψ ∈ Hl ,l 2 , then is well defined and we can define this quantity in (8.3) above to be κ U (x, t) can be defined for any (x, t) ∈ Γ * (U ). Then similar to Definition 10.6 in [15], if ψ ∈ H ℓ,ℓ/2 , then for κ ∈ [1 + s, ℓ), we define U (x, t) = κ} In view of (8.2), we have the following result on the consistency of classification of free boundary points analogous to proposition 10.7 in [15].
Similarly, we also have the following characterization of points on the extended free boundary Γ * (U ) but which are not on the free boundary Γ(U ).
Proof. The first part is nothing but Lemma 7.2.
Suppose now (x 0 , t 0 ) ∈ Γ * (U ) \ Γ(U ) and that the frequency limit κ (ℓ) U < ℓ. By translation we may assume that (x 0 , t 0 ) = (0, 0). Then there exists a small δ > 0 such that U = ψ on Q δ (0, 0). Let V = V k be as in (3.8). Since κ < ℓ, by Theorem 6.8, there exists a blow up U 0 of V over a sequence r j → 0. Since V = 0 on Q δ , we obtain that U 0 vanishes on {y = 0}. Then we have that the odd extensionŨ 0 is L a caloric and homogeneous of degree κ. Then we have that |y| a (Ũ 0 ) y is L −a caloric, symmetric and homogeneous of degree κ − 1 + a. Then from the Liouville theorem Lemma 7.5, it follows that κ − 1 + a is an integer and moreover since U 0 satisfies the Signorini condition, we have that −∂ y a U 0 is a non-negative polynomial on {y = 0}. Therefore there are two possibilities, either ∂ y a U 0 is identically zero on {y = 0} or κ − 1 + a is an even integer. The former is not possible because U 0 and ∂ y a U 0 vanishing identically on {y = 0} implies U 0 ≡ 0 because of strong unique continuation property. This follows from the proof of Lemma 7.7 in [6]. Therefore we have κ − 1 + a is even and consequently κ is of the form 2m + 1 − a. This finishes the proof of the proposition.

Singular points
In this section we define the singular free boundary points as the points of zero Lebesgues density of the coincidence set Λ(U ).
Proof. (i) ⇒ (ii) Note that the rescalings U r satisfy L a U r = |y| a F r − 2(∂ a y U )H n+1 Λ(Ur) in S 1/r , in the sense of distributions, after an even reflection in y variable. Since U r are uniformly bounded in W 2,1 2 (Q + 2R , |y| a dXdt) for small r by Theorem 6.8, ∂ a y U r are uniformly bounded in L 2 (Q R ). On the other hand, if 0 ∈ Σ(U ), then where ℓ 0 = ℓ−(1−σ)/2 ∈ (κ, ℓ) and we have used the fact that H U (r) ≥ r 2ℓ 0 for 0 < r < r U , by Lemma 6.6. Hence, any blowup U 0 is caloric in Q R for any R > 0, meaning that it is caloric in the entire strip S ∞ = R n+1 ×(−∞, 0]. On the other hand, by the charecterization of blowups in Theorem 6.8(iii), U 0 is homogeneous in S ∞ and therefore has a polynomial growth at infinity. Then by the Liouville-type Lemma 7.5 we can conclude that U 0 must be a parabolically homogenous polynomial p κ of a certain integer degree κ. Note that p κ = U 0 ≡ 0 by construction. The properties of U also imply that that p κ (x, 0, t) ≥ 0 for all (x, t) ∈ S ∞ and and p κ (x, −y, t) = p κ (x, y, t) for all (x, y, t) ∈ S ∞ . In other words, U 0 = p κ ∈ P + κ . (ii) ⇒ (iii) Let p κ be a blowup of U at the origin. Since p κ is a polynomial, clearly κ ∈ N. Assume now, towards the contradiction, that κ is odd. Then, the nonnegativity of p κ on R n × {0} × {−1} implies that p κ vanishes there identically, implying that p κ ≡ 0 on S ∞ . Now, using the even symmetry in y and the fact that L a p κ = 0, we are going to infer that p κ ≡ 0, contrary to the assumption that p κ is nonzero. From even symmetry in y, we represent p κ (x, y, t) = (α,k,j)∈Z n + ×Z + ×Z + |α|+2k+2j=κ c α,k,j x α y 2k t j , Now, for (α, k, j) such that |α| + 2k + 2j = κ, consider the partial derivative ∂ α x ∂ j t p κ . Since ∂ x i and ∂ t are derivatives in directions tangential to the thin space, we conclude that We now prove by indiction in k, that c α,k,j = 0 for k = 0, 1, . . . , ⌊κ/2⌋. When k = 0, we have |α| + 2j = κ and therefore ∂ α x ∂ j t p κ ≡ α!j!c α,0,j and from vanishing of ∂ α x ∂ j t p κ on S ∞ , we conclude that c α,0,j = 0. Suppose now we know that c α,k ′ ,j = 0 for 0 ≤ k ′ < k ≤ ⌊κ/2⌋ and show that it holds also for k. Indeed, one consequence from the inductive assumption is that ∂ α x ∂ j t p κ (x, y, t) = α!j!c α,k,j y 2k , which is a-caloric if and only if c α,k,j = 0. Hence, we can conclude that p κ ≡ 0, contrary to our assumption. Thus, we must have κ ∈ {2m | m ∈ N}.
(iii) ⇒ (ii) The proof of this implication is stated as a separate Liouville-type result in Lemma 9.3 below.
(ii) ⇒ (i) Suppose that 0 is not a singular point and that over some sequence r = r j → 0 + we have H n+1 (Λ(U r ) ∩ Q 1 ) ≥ δ > 0. From the second estimate in (i) in Theorem 6.8, the local regularity estimates developed in [8] and Arzela-Ascoli, by taking a subsequence if necessary, we may assume that U r j converges locally uniformly to a blowup U 0 . We claim that . Since U 0 = p κ is a polynomial, vanishing on a set of positive H n+1 -measure on S ∞ , it follows that U 0 vanishes identically on S ∞ . But then, repeating the argument at the end of the step (ii) ⇒ (iii), we conclude that U 0 ≡ 0, a contradiction. Thus, 0 is a singular point.
The implication (iii) ⇒ (ii) in Proposition 9.2 is a consequence of the Liouville-type result Lemma 7.5 which is the parabolic counterpart of Lemma 1.3.3 in [20].
This, in turn, is a particular case of the following lemma, analogous to Lemma 1.3.4 in [20] in the elliptic case, which stems from Lemma 7.6 in [24].
If v is parabolically 2m-homogeneous, m ∈ N, and has a polynomial growth at infinity, then L a v = 0 in S ∞ .
Note that we have the following identity (similar to that of G a ) We have div(|y| a ∇v), [v X, ∇P − P X, ∇v ] Ψ dX dt.

Weiss and Monneau type monotonicity formulas
In this section we establish two families of monotonicity formulas that play a crucial role in our analysis of singular points. The elliptic ancestors of these formulas were first obtained in [20] in the study of the Signorini problem corresponding to a = 0 (or s = 1/2), and were subsequently generalized to all a ∈ (−1, 1) (all s ∈ (0, 1)) in [22]. In the parabolic setting and still for the case a = 0 such formulas were first proved in [15]. Theorems 10.1 and 10.5 below respectively extend to all values a ∈ (−1, 1) Theorems 13.1 and 13.4 in [15].
In the following statement the quantities H(U, r) and D(U, r) are those defined in (4.8) and (4.9) respectively. Theorem 10.1 (Weiss type monotonicity formula in Gaussian space). Let U ∈ S F (S + 1 ) with F satisfying (4.2) for some ℓ ≥ 2 and a constant C ℓ .
In the sequel we will need the following results. with C = C(κ, σ, n) > 0.
This gives The following computation can be justified rigorously considering the region and then let ε → 0 + . One should keep in mind that the outer normal on ∂R ε is ν = −e n+1 . We obtain where in the last equality we have used (2.20) and the fact that U ∂ a y U G a = 0.
The vanishing of this integral is proved as follows. We write R n ×{0} U ∂ a y U G a = (R n ×{0})∩{U >0} U ∂ a y U G a + (R n ×{0})∩{U =0} U ∂ a y U G a .
We conclude that From this formula, (2.25) and (4.5) we conclude that Combining this equation with (4.7), we conclude that (4.6) holds.
Proof of Lemma 4.4. Using our assumptions on U we can proceed as in the proof of Lemma 6.5 in [6]. We thus skip most details and only refer to the relevant changes. The first step is to recognize that for t ∈ (−1, −δ) one has (12.5) h ′ (U, t) = 1 t R n+1 + U ZU G a y a = 2 t i(U, t).
Again the proof of (12.5) can be rigorously justified by integrating on the region R ε , where we know that (12.3) holds, and then let ε → 0 + using (2.20), (2.18), (2.19) and the assumptions on U on the thin set {y = 0}.
Using ( At this point we can argue as in the proof of Lemma 6.5 in [6] to pass to the limit as δ → 0 + in (12.7) and reach the desired conclusion for H ′ (U, r).

Recalling that
if we differentiate with respect to λ and set λ = 1 in the previous identity we find Consider the vector fields X i = ∂ ∂x i , i = 1, . . . , n, X n+1 = ∂ ∂y . One easily verifies that the commutator [X i , Z] = X i , i = 1, . . . , n + 1. This gives (using summation convention) Substituting in the latter equation and integrating by parts and recalling that the outer unit normal on ∂R n+1 + is −e n+1 , we find 2td ′ (U, t) = −2t We have thus proved the following formula for t ∈ (−1, −δ) Substituting now (12.9) in (12.8) we obtain We claim that on the thin set {y = 0} we have (12.11) ∂ a y U ZU = 0 a.e. with respect y a dXdt. We first note that U restricted to {y = 0} is locally Lipschitz continuous in x, t. We also have that for a.e t, since ∇U t (·, t) ∈ L 2 loc (R n+1 + , y a dX), therefore U t has a L 2 loc trace at {y = 0}. Moreover by a standard weak type argument using test functions, we can show that such a trace is infact bounded because of the lipschitz continuity of U in t and coincides with the weak time derivative of U at {y = 0}. Now on the set {U > 0}, we have that lim y→0 y a U y = 0, hence a.e. we have lim y→0 y a U y ZU = 0 on {U > 0}.