Regularity of Solutions to the Muskat Equation

In this paper, we show that if a solution to the Muskat problem in the case of different densities and the same viscosity is sufficiently smooth, then it must be analytic except at the points where a turnover of the fluids happens.


Introduction
The Muskat problem is a free boundary problem studying the interface between fluids in the porous media [30]. It can also describe the Hele-Shaw cell [34]. The density function ρ follows the active scalar equation Here D 1 (t) and D 2 (t) are open domains with D 1 (t) ∪ D 2 (t) ∪ ∂ D 1 (t) = R 2 . The velocity field v in (1.1) satisfies Darcy's law, (1.2) and the incompressibility condition where p is the pressure and μ is the viscosity. κ, g are the permeability constant and the gravity force.
We focus on the problem where two fluids have different densities ρ 1 , ρ 2 and the same viscosity μ.
Given an initial interface at time 0, (1.3) is divided into three regimes. When the interface is a graph and the heavier fluid is on the bottom as in Fig. 1a, it is in a stable regime. When heavier fluid is above the boundary as in Fig. 1b, it is in a stable regime when time flows backward. Thus, given any initial data, (1.3) can be solved for small negative time t. In both regimes, shown in Fig. 1a, b, (1.3) can not be solved in the wrong direction unless the initial interface is real analytic. The third regime, shown in Fig. 1c, it highly unstable because the heavier fluid lies on top near point S 1 while the lighter fluid lies on top near point S 2 . Note two turnover points T 1 and T 2 where the interface has a vertical tangent. For generic initial data in the turnover regime, (1.3) has no solutions either as time flows forward or backward.
In the third regime, there are several examples from the literature (eg. [9,10,19,20]), but they are all real analytic solutions. Without the real analytic assumption, due to the spatially non-consistent parabolic behavior, the existence is usually false and the uniqueness is unknown. To address this gap, this paper studies to what extent the solution of (1.3) is analytic.
Moreover, for the analytic solutions, one can prove an energy estimate on an analyticity region that shrinks when time increases. That energy estimate implies uniqueness in the class of analytic solutions. [10]. Therefore, the investigation towards analyticity can serve as a first step to deal with the uniqueness.
Our method concerning the analyticity is not limited to the Muskat problem. A simplified version of our method can be used to show the analyticity of the solution to a kind of non-local differential equations (see Section 10). This approach is new to our best knowledge.
In our forthcoming work [35], we focus on the degenerate analyticity near the turnover points. The existence and uniqueness are crucially related to the way the real-analyticity degenerates at those points. Given an extra assumption, we have the following theorem in [35]: values of α of these two turnover points. If we assume that the solution satisfies the following three conditions: (1.4) ∂ α f 1 (α, t) = 0 except at Z 1 (t), Z 2 (t), (1.5) and then when −t 0 < t < t 0 , f (·, t) can be analytically extended to region

Background
In order to make the equation well-defined, the arc-chord condition is introduced, saying that The Rayleigh-Taylor coefficient σ is used to characterize the three regimes in Fig. 1 and is defined as (1.7) σ ≥ 0 is corresponding to the stable regime and σ ≤ 0 the backward stable regime. When σ changes sign, it is in the unstable regime.
When the heavier liquid is above the lighter liquid, the equation is ill-posed when time flows forward [22].
A solution that starts from a stable regime and develops turnover points was first discovered in [10]. That solution still exists for a short time after turnover due to the analyticity when the turnover happens. Moreover, breakdown of smoothness can happen [9]. There are also examples where the solutions transform from stable to unstable and go back to stable [19] and vice versa [20].
Weak solutions and a special kind of weak solutions: mixing solutions of (1.1) have also been studied. They do not satisfy (1.3) and can develop a mixing zone. Weak solutions do not have uniqueness [18]. In all three regimes, there are infinitely many mixing solutions ( [8,11,12,26,33,37]).

The outline of the Proof of Theorem 1.1
Inspired by the instant analyticity results in the stable case [10,27,29], our first idea is localization. If locally the lighter liquid is over the heavier one, we let the time go forward, and if locally the heavier one is over the lighter one, we let the time go backward.
Since it leads to lots of difficulties by the standard method due to the localization, we use a new idea to prove analyticity except at turnover points. The idea is to make a C 1 continuation of the parametrized interface α → ( f 1 (α, t), f 2 (α, t)) to complex α and then prove the C 1 continuation satisfies the Cauchy-Riemann equation. To do so, we break the complex region into curves α + ic(α)γ t with γ ∈ [−1, 1]. On each such curve, we solve an equation for ( f 1 , f 2 ). We then show that when γ varies, our solutions on the curve fit together into an C 1 function of α + iβ. Finally, we prove that C 1 function satisfies the Cauchy-Riemann equation, thus producing the desired analytic continuation.
In Section 3, we define a cut off function λ(α) and focus on f c (α, t) = λ(α) f (α, t). We then localize the equation such that the modified R-T condition has a fixed sign. In order to make use of the sign, if the sign is positive, we let the time go forward. If the sign is negative, we let the time go backward.
In Section 5, for each fixed γ , we use the energy estimate and the Galerkin method to show the existence of the solution z(α, γ , t). The main term is controlled by Gårding's inequality, where we use a lemma from [9]. This part is similar as to [9,10].
In Sections 6, 7, and 8, we verify that the z(α, 0, t) coincides with the f c (α, t) and that z(α, γ , t) is also smooth enough with respect to γ .
In Section 9, we derive some lemmas about the Cauchy-Riemann operator and use those lemmas to show analyticity of z(α, β c(α)t γ, t) by checking that it satisfies the Cauchy-Riemann equations.

Remark 1.3.
In [10], the analyticity domain can be chosen as a strip, and the analyticity follows directly from existence. Since our c(α) is supported in a small region, we do not have such good behavior.
For any vector function z = (z 1 , z 2 ) ∈ H k : z 1 ∈ H k and z 2 ∈ H k .

The Localization
This step is to localize the equation such that the R-T coefficient has a fixed sign. Without loss of generality, we study the behavior at origin and let ρ 2 −ρ 1 2 = 1. Let λ(α) ∈ C 100 (−∞, ∞) satisfying λ(α) ≥ 0 and Here δ is a sufficiently small number such that when α ∈ [−2δ, 2δ], ∂ α f 1 (α, 0) has a fixed sign. Without loss of generality, we assume Then we have Here T is the torus of 2π .

Change the contour
Let c(α) satisfy Here c(α) is defined such thatf , λ can be analytically extended to the complex and for any γ ∈ [−1, 1].

The equation on the curve
Let z(α, γ , t) be the solution of the equation (4.4) with initial data z(α, we have We drop the analyticity assumption of f c from now. Notice thatf and λ can still be analytically extended to D A as in (4.2) and (4.3).

Energy estimate
We first assume z is of finite Fourier modes here and do the energy estimate. The idea of the energy estimate is similar as in [9,10].
Here we omit the dependency of z on γ and t, and the dependency off on t for the sake of simplicity. Let where T is the torus of length 2π and For the L 2 norm, we have Here C is a bounded function depending on δ, δ c and . We will keep using the same notation C in the following proof. Now we take 5th derivative and have Here O i terms contain at most 5th derivative on both z andf . Before we show the explicit form of O i , we introduce some notations. Let When we write X i (α, t), we mean We claim that we can write O i as following three types, by separating the highest order term in the derivative. Here we omit the dependency on γ and t.
Then we have the following lemmas: we could use Lemma 11.4 to get the result for O 1,i . Moreover, we have we then use Lemma 11.2 to get the estimate for O 2,i . O 3,i can be bounded easily.

Lemma 5.2. We have
It is also of K −1 type. We have Then the result follows from Lemma 11.2.
Then we are left to deal with T 1 + T 2 . By using the same notation as in Lemma 5.2, we have Moreover, we could further split the T 2,2 and have We can do the integration by parts in T 2,2,2 (z) to get that Therefore we have In conclusion, we have where is (− ) 1 2 on the Torus T of length 2π and C( z Ar c + z U ).
where B.T. ≤ C( z Ar c + z U ). Next we show a lemma for controlling the main terms.
We could do the integration by parts to I 2,1 and have Moreover, We can still do the integration by parts to the I 1,2 and have Moreover, for any Hence . Now we are left to control I 1,M + I 2,s . We have Now we use a lemma from [9, Section 2.4].
Lemma 5.4. Let a, b be real valued functions on T, a(α) ≥ |b(α)| and satisfying a, b ∈ C 2 (T). Then we have Then, from Lemma 5.4, we have and Moreover, when t = 0, α ∈ [−2δ, 2δ], from (5.10), (5.11) and (5.5) we have From ( Then let Therefore, we could let z Ũ = z U + z Ar c + z RT . From (5.12), and the following Lemma 5.5, we have Then z(α, γ , t) Ũ is bounded for sufficiently small time t 1 . We claim that the bound and the time can be chosen such that it holds for all γ ∈ [−1, 1].
From (5.1), it is easy to get Then T erm 2,1 C( z Ũ ). Hence Then we have the estimate We also introduce a corollary here to be used in a later section.

Approximation for the picard theorem
Now we approximate the problem and have the following equations, and initial value z n (α, γ , 0) = ϕ n * f (α, 0). Here the convolution of ϕ n is the projection to the finite Fourier modes of α. By the Picard theorem, for any fixed γ ∈ [−1, 1], there exists solutions in C 1 ([0, t n ], H 5 α (T)). Moreover, by the structure of our approximation, we have z n = ϕ n * z n , and for 1 ≤ j ≤ 5, | cosh( ((ϕ n * z n 2 )(α) +f 2 (α))) − cos( ((ϕ n * z n with ((ϕ n * z n 2 )(α) +f 2 (α)) and ((ϕ n * z n 1 )(α) +f 1 (α)) from (5.13), (5.14), and letting Then we can use the similar energy estimate and the compactness argument to show there exist a solution Since the energy estimate has a bound for all γ ∈ [−1, 1], we have a existence time t 1 that holds for all γ . Now we abuse the notation and write T (z) as T (z(α, γ , t), γ , t). We have the following lemma:  Proof. We only show (5.19) and the left can be shown in the same way. From (5.1), we have It is trivial that T 1 satisfying the (5.19) since c(α) is sufficiently smooth. Moreover, the ∂ αf ∈ H 5 (T) and is more regular than ∂ α g(α) 1+ic (α)γ t . Hence we only consider T 2 . For T 2 , we have Moreover, we can use the notation from (5.4), (5.2), and get ] is the biggest integer less than j+1 2 . Then from (5.22), we could use Lemma 11.2 to bound T erm 2,1 . Moreover, since j Then we could use Lemma 11.4 to bound T erm 2,2 .

The Uniqueness
In this section we show there exists sufficiently 0 < t 2 ≤ t 1 such that for 0 ≤ t ≤ t 2 , we have z(α, 0, t) = f c (α, t).
Let z 0 (α, t) = z(α, 0, t). From (5.1) and (5.5), we have Then we have the equation for the difference: We first control T erm 2 , we have Since the component of ∇ K is of −2 type, we have When t = 0, we have z 0 = f c , then Moreover, we have the following lemma: Then we have the result.

The Continuity of z with Respect to γ
We first show z(α, For the sake of simplicity, we further shrink the time t 1 tot 1 such that for all This is not necessary but helps to simplify our estimate in this section. Now we estimate the difference, we have α, γ , t), η, t)dη = T erm 1 + T erm 2 .

The Differentiability of z with Respect to γ
Now we show the differentiability. We define a new function w(α, γ , t). It satisfies the equation that dz dγ would satisfy if it is differentiable. Let w be the solution of the equation with initial value w(α, γ , 0) = 0. Here D z T (z(α, γ , t), γ , t)[w] is the Gateaux derivative.
As in the existence of z(α, γ , t), we first show the energy estimate. First, from (7.4), we have Moreover, It has the similar structure as (7.6) and (7.8). The only difference between the first two terms in (7.6) and (8.2) is that ∂ α w(α, γ ) takes the place of z(α,γ )−z(α,γ ) In (7.8), and the third term of (7.6), w(α, γ ) takes the place of z(α,γ )−z(α,γ ) ). Therefore we could use the similar estimate and have As in the existence of z(α, γ , t), we could do the similar energy estimate to the approximation of the equation with initial value w n (α, γ , 0) = 0. Then from the Picard theorem and compactness argument, there exists 0 ≤ t 3 ≤ t 1 , such that We claim there is an uniform t 3 holds for all γ ∈ [−1, 1]. Moreover, we have the following lemma: Proof. It is easy to get these bounds since z(α, .
Since the component of ∇ 1 K is of −2 type, we could use Lemma 11.2 to bound T erm 1,5 and have Then from Lemma 11.3, and (7.15), we have T erm 1,6 L 2 (T) |γ − γ |.
Moreover from the interpolation theorem, we have Then from (8.8), we have .
In conclusion we have (8.9)

The Analyticity
In this section, we want to show f (α, t) is a real analytic function near 0 for each fixed t, 0 < t < t 3 . We first show that it is enough to prove that Proof. From the uniqueness (6.5), we have z(α, 0, t) = f c (α, t). Then Then we have Then Then ∂ α f c , ∂ γ f c are continuous. Therefore we have the analyticity of f c near 0. We also have f (α, t) = f c (α, t) when |α| ≤ δ. Then we have the result.
Before we prove A 0 (z) = 0, we introduce some general lemmas.
Proof. First, for the right hand side, we have From the two equalities above, we have the result.
Here D h is the Gateaux derivative.

Using the Energy Estimate to Show the Analyticity
Following an idea similar to that in the previous sections, we introduce a way to study the analyticity of the solution to some differential equations, which is, to our knowledge, a new method.