The multiphase Muskat problem with equal viscosities in two dimensions

We study the two-dimensional multiphase Muskat problem describing the motion of three immiscible fluids with equal viscosities in a vertical homogeneous porous medium identified with $\mathbb{R}^2$ under the effect of gravity. We first formulate the governing equations as a strongly coupled evolution problem for the functions that parameterize the sharp interfaces between the fluids. Afterwards we prove that the problem is of parabolic type and establish its well-posedness together with two parabolic smoothing properties. For solutions that are not global we exclude, in a certain regime, that the interfaces come into contact along a curve segment.


Introduction and the main results
The mathematical model.In this paper we study the two-dimensional multiphase Muskat problem describing the motion of three incompressible fluids with positive constant densities in a vertical porous medium identified with R 2 .Such three-phase flows are of great interest not only from a mathematical point of view but also in many other areas of science and technology, such as petroleum extraction and environmental engineering, cf.e.g.[4,8].In this paper we restrict our attention to the particular case when the three fluids have equal viscosities, which we denote by µ > 0. We further assume that the fluid phases are separated at each time instant t≥ 0 by sharp interfaces which we described as being the graphs Γ c∞ f (t) := {(x, c ∞ + f (t, x)) : x ∈ R} and Γ h (t) := {(x, h(t, x)) : x ∈ R}.Here c ∞ is a fixed positive constant and the functions f (t), h(t) : R → R are unknown and are assumed to satisfy f (t) + c ∞ > h(t) during the motion.The fluid with density ρ i occupies the domain and Ω 1 (t) := {(x, y) : y > f (t, x) + c ∞ }, Ω 3 (t) := {(x, y) : y < h(t, x)}.
In the fluid layers the dynamic is governed by the equations where p i (t) is the pressure and v i (t) := (v 1 i (t), v 2 i (t)) denotes the velocity field of the fluid located at Ω i (t).The positive constant k is the permeability of the homogeneous porous medium and g is the Earth's gravity.The equation (1.1a) 1 is Darcy's law which is oftenly used to model flows in porous media, cf.e.g.[7], and (1.1a) 2 describes the conservation of mass in all phases.
We supplement (1.1a) with the following boundary conditions at the free interfaces on ∂Ω i (t) ∩ ∂Ω i+1 (t), i = 1, 2. (1.1b) Here ν i (t) denotes the unit normal at ∂Ω i (t) ∩ ∂Ω i+1 (t) pointing into Ω i (t) and • | • is the inner scalar product in R 2 .Additionally, we impose the far-field boundary conditions which state that far away the flow is nearly stationary.Finally, in order to describe the motion of the free interfaces, we set their normal velocity equal to the normal component of the velocity field at the free boundary, that is and we impose the initial condition We call the closed system (1.1) the multiphase Muskat problem.
For the multiphase Muskat problem (1.1) considered herein much lees is known.This setting has been studied before only in a three-dimensional setting in [20] where the authors established a local existence and uniqueness result in H k (R 2 ) with k ≥ 4.Moreover, it is shown in [20] for solutions which are not global but bounded in C 1+γ (R 2 ), γ ∈ (0, 1), that the fluid interfaces cannot touch along a curve segment when the time approaches the maximal existence time, excluding thus the occurrence of so-called squirt singularities.A related scenario has been considered in two-dimensions, but in a periodic setting and with one of the fluids being air at uniform pressure, in [24,25] where the well-posedness and the stability of equilibria are investigated.
Similarly as in the three dimensional case [20], we show herein that problem (1.1) can be expressed as a nonlinear, nonlocal, and strongly coupled evolution problem with nonlinearities described by contour integrals, cf.(1.10) below.The equivalence of the formulations (1.1) and (1.10) is rigorously established in Theorem 1.1 below, by making use of the results from Appendix A. The analysis in Appendix A, where in particular we extend Privalov's theorem to contour integrals over unbounded graphs (see Theorem A.3), also motivates the choice of homogeneous Sobolev spaces in the study of the Muskat problem [2].Our second main result stated in Theorem 1.2 establishes the well-posedness of the problem in the subcritical Sobolev spaces H s (R) 2 with s ∈ (3/2, 2).It also provides two parabolic smoothing properties.Finally, in Proposition 1.3 we show for bounded solutions with finite existence time that the fluid interfaces intersect when the time approaches the maximal existence time at least in one point, but we exclude also in this two-dimensional scenario the formation of squirt singularities.
Compared to the two-phase Muskat problem, cf.[19], new difficulties arise from the fact that the coupling terms in (1.10) are of highest order.However, based on the mapping properties established in Section 2, we prove that the linearized operator, which is represented as a 2 × 2 matrix, see Section 3, has lower order off-diagonal entries.A similar feature has been evinced for the Muskat problem investigated in [25].The benefit of this weak coupling at the level of the linearization is that only the diagonal terms need to be considered when establishing parabolicity for the problem.Once this is done, we can make use of the abstract parabolic theory from [31] in the study of this multiphase Muskat problem.
Notation.Given k, n ∈ N and an open set Ω ⊂ R n , we denote by C k (Ω) the space consisting of real-valued k-time continuously differentiable functions on Ω, and UC k (Ω) is the subspace of C k (Ω) having functions with uniformly continuous derivatives up to order k as elements.Moreover, BUC k (Ω) is the Banach space of functions with bounded and uniformly continuous derivatives up to order k.Finally, given α ∈ (0, 1), we set Given Banach spaces X and Y , the space C1− (X, Y ) consists of all locally Lipschitz maps from X to Y .Moreover, we write bounded, and symmetric.Solving the fixed time problem.A remarkable property of (1.1) is the fact that the equations (1.1a)-(1.1c)are linear and have constant coefficients.This property enables us to identify the velocity field in terms of the a priori unknown functions f and h by means of contour integrals.Such an approach has been followed in the context of the Muskat problem at least at formal level already in the 80's, cf.[22].For the clarity of the exposition we omit in this part the time dependence and write (•) ′ for the the x-derivative of functions that depend only on x.In Theorem 1.1 below we provide, under suitable regularity constraints, an explicit formula for the velocity field in terms of X := (f, h).Our approach generalizes the one followed in [34] in the context of the two-phase Muskat problem and strongly relies on results from Appendix A.
with constants Proof.We devise the proof into two steps.
In the notation from Appendix A, see (A.1), it holds that and, according to Theorem A.3, we also have v i ∈ BUC r−3/2 (Ω i ) for 1 ≤ i ≤ 3.Moreover, Lemma A.4 yields that (1.2) 2 and (1.2) 5 hold true.In view of Lemma A.1 we further get and where PV is the principal value and, setting X := (f, h), we defined (1.7) The formulas (1.5) and (1.6) now show the validity of (1.2) 4 .We next define pressures p i : Ω i → R, 1 ≤ i ≤ 3, by the formula where v i =: (v 1 i , v 2 i ), c i ∈ R are constants, and with

Taking advantage of ∂
4, we deduce that p i ∈ C 1 (Ω i ) and that (1.2) 1 is satisfied.The regularity properties established for v i now imply that indeed Uniqueness.It remains to show that the system (1.2) has, when setting the gravity constant g equal to zero, only the trivial solutions defined by v = (v 1 , v 2 ) = 0 and p = c ∈ R. To begin, we note that (1.2) 2) 3 , and (1.2) 4 , we get v ∈ BUC(R 2 ).Stokes' theorem then yields (1.9) We next set , where ψ i : Ω i → R are given by It follows immediately that Ψ ∈ C(R 2 ).Additionally, using Stokes's theorem and (1.2) 2 we may show that and (1.9) then yields ∆Ψ = 0 in D ′ (R 2 ).Consequently, Ψ is the real part of a holomorphic function u : 2) 5 , Liouville's theorem yields u ′ = 0, hence v = 0.Moreover, in view of (1.2) 1 , we now obtain that ∇p = 0 in R 2 , meaning that p is constant in R 2 .This completes our arguments.
The contour integral formulation and the main results.Concerning the multiphase Muskat problem (1.1), Theorem 1.1 implies that if, at any given time t ≥ 0, f (t) and h(t) belong to H r (R), with r ∈ (3/2, 2), and are given by (1.5) and (1.6).Recalling also (1.1d), we can thus formulate the moving boundary problem (1.1) as an autonomous evolution problem for the pair X := (f, h) which reads as where the nonlinear operator Φ = (Φ 1 , Φ 2 ) is defined by and (1.12) The constants Θ i , i = 1, 2, are introduced in (1.4) and, given u ∈ H r (R), with r ∈ (3/2, 2) which is fixed in the remaining part, we denote by B(u) the linear operator where the operators B 0 m,1 , m = 0, 1, as well as C 1 , C ′ 1 , D 1 , and D ′ 1 are defined in (2.1)-(2.2) and (2.3) below.We shall treat (1.10) as a fully nonlinear evolution problem in H r−1 (R) 2 .To this end we prove in Corollary 2.7 below that Φ is smooth, that is where Moreover, our analysis (see Proposition 3.2 below) will disclose that (1.10) is of parabolic type in the phase space O r , that is the Fréchet derivative ∂Φ(X) at any X ∈ O r , viewed as an unbounded operator in H r−1 (R) 2 with domain H r (R) 2 , is the generator of an analytic semigroup in L(H r−1 (R) 2 ).In the notation introduced in [5] the latter property writes as The properties (1.14) and (1.16) enable us to use the parabolic theory presented in [31] to establish the following results for (1.10).
Theorem 1.2.Let r ∈ (3/2, 2).Given X 0 ∈ O r , the multiphase Muskat problem (1.10) has a unique maximal solution X := X( • ; X 0 ) such that with T + = T + (X 0 ) ∈ (0, ∞] denoting the maximal time of existence.Moreover, we have: The solution depends continuously on the initial data; The next result shows, for bounded solutions with T + < ∞, that the fluid interfaces intersect in at least one point along a sequence t n → T + .Moreover, using the same strategy as in [17,20], we exclude for such solutions that the two fluid interfaces collapse along a curve segment. ) be a maximal solution to (1.10) with T + < ∞ and such that, for some M > 0, X(t) H r ≤ M for all t ∈ [0, T + ).Then, there exists x 0 ∈ R with the property that (1.17) Moreover, for each x 0 satisfying (1.17) and for each δ > 0, we have The proofs of Theorem 1.2 and Proposition 1.3 are postponed to the very end of Section 3.

Mapping properties
In this section we introduce the operators B 0 m,1 , m = 0, 1, and 12) in a more general context and study the properties of these operators.The main goal is to establish the smoothness property (1.14), see Corollary 2.7 below.
Motivated by the formulas (1.5) and (1.6), we introduce the family B n,m , n, m ∈ N, of singular integral operators, where, given Lipschitz continuous maps u 1 , . . ., u m , v 1 , . . ., v n : R → R and ω ∈ L 2 (R), the operator B n,m is defined by Here we use the notation introduced in (1.7).Furthermore, we define The operators B n,m were introduced in [34], but they are also important in the study of the Stokes problem, cf.[32].It is important to point out that B 0,0 = πH, where H denotes the Hilbert transform.We now recall some important properties of these operators.
, there exists a constant C that depends only on n, m, r, and max 1≤i≤m u i H r such that The evolution equation (1.10) consists actually of an equation for f and one for h which are coupled.The coupling terms contain highest (first) order derivatives of both variables and they are expressed by using the aforementioned operators We next introduce these operators as elements of a larger family of operators enjoying similar properties. Given for ω ∈ L 2 (R) and x ∈ R, where we used again the notation introduced in (1.7).Since X i ∈ O r for 1 ≤ i ≤ m, these operators are no longer singular.However, the kernels of D 1 and D ′ 1 behave for large s similarly as that of the truncated Hilbert transform We recall that, given δ > 0, H δ is a Fourier multiplier with symbol [ξ → m δ (ξ)] given by Since m δ ∞ ≤ 2 for all δ > 0, it follows that We next study the mapping properties of the operators Then, there exists a constant C that depends only on m, c 0 , and . (2.8) Therefore, when considering C m (the case C ′ m is similar), we get, by making use of Minkowski's integral inequality and (2.7) follows.
It is not difficult to extend the proof of Lemma 2.2 in the context of the operators D m and D ′ m with m ≥ 2. The case m = 1 is however more subtle and requires a different strategy which uses the estimate (2.5).
Then, there exists a constant C that depends only on r, m, c 0 , and max 1≤i≤m X i H r such that Proof.The proof in the case m ≥ 2 follows along the lines of the proof of Lemma 2.2.We now consider the operator D 1 (the estimate for D ′ 1 follows similarly).Let δ > 0 be as defined in (2.8) (with m = 1) and set With H δ denoting the truncated Hilbert transform, see (2.4), we have for x ∈ R that Recalling (2.5), we conclude that (2.10) is satisfied.
We prove next that the operators there exists a positive constant C that depends only on r, m, c 0 , and max 1≤i≤m X i H r such that (2.14) Proof.Let {τ ξ } ξ∈R denote the group of right translations and assume first that ω ∈ C ∞ 0 (R).Given 0 = ξ ∈ R, the formula (2.12) leads us to Recalling (2.11), we may pass to the limit ξ → 0 on the right of the latter equation and conclude All the terms on the right are well-defined (and belong to However, using integration by parts we can rewrite this term as respectively Combining the last three identities, Lemma 2.2, Lemma 2.3, and using a standard density argument we get that (2.14) holds for E m .The claim for the operator E ′ m follows similarly.Finally, the Lipschitz continuity property is obtained from (2.14) and (2.12)-(2.13).
Since our goal is to establish the smoothness of Φ, cf.(1.14), we next prove that operators E m and E ′ m with E m ∈ {C m , D m } depend smoothly on the variable X ∈ O r .This requires some additional notation.Given Y := (u, v) ∈ H r (R) 2 , we set , and E ∈ {C, D}, we set ds for x ∈ R, where j = 0 if E = C and j = 1 for E = D. We point out that, given E ∈ {C, D} and m ≥ 1, we have E 0,m,0 (X) = E m (X) and E ′ 0,m,0 (X) = E ′ m (X).Hence the latter formulas extend our previous notation introduced in ( where for each S ⊂ {1, . . ., n} we set S c := {1, . . ., n} \ S.Moreover, letting X j := (f j , h j ), m + 1 ≤ j ≤ m + p, it holds that Recalling Lemma 2.5, we deduce for (2.17) The estimate (2.16) shows that In the next lemma we establish the Fréchet differentiability of E n m,p .
Lemma 2.6.Given n, m, p ∈ N, m ≥ 1, and X ∈ O r , the operator E n m,p is Fréchet differentiable in X and its Fréchet derivative is given by Consequently, for each n, m, p ∈ N, m ≥ 1, we have elementary algebraic manipulations lead us to the following identity Hence, for all Y sufficiently close to X in H r (R) 2 it follows from Lemma 2.5, by arguing as in the derivation of (2.16), that and the claim follows.
We complete this section by establishing (1.14).

The generator property and the proof of the main results
In the first part of this section we show that the evolution problem (1.10) is parabolic in O r by establishing the generator property (1.16).In the second part we prove our main results.With respect to the first task, let X = (f, h) ∈ O r be fixed.We can represent the Fréchet derivative ∂Φ(X) as the matrix operator Though the coupling terms in (1.11)-(1.12)involve highest order derivatives of both unknowns, Lemma 2.6 and (2.16) show that the off-diagonal entry ∂ h Φ 1 (X) can be treated as being a perturbation.Indeed, recalling Lemma 2.6, we obtain in particular for E ∈ {C, D} that for all Y = (u, v) ∈ H r (R) 2 .Using this formula, it follows from (1.11) that and (2.16) yields In view of [5, Theorem I.1.6.1] and of the property v which holds for any given arbitrary small ν > 0, we conclude that (1.16) is satisfied provided the diagonal entries are analytic generators, that is To establish the generator property for ∂ f Φ 1 (X) we follow the strategy from [1, Section 4] (see also [23,27] where similar arguments are used in other contexts).To this end we first deduce from Lemma 2.1 (ii) that the mapping is smooth.In view of this property, of (3.1), and of (2.16) we infer from (1.11) that where, according to Lemma 2.5, Moreover, T lot is a lower order operator, more precisely We next consider the continuous path Observe that Ψ(1) = ∂ f Φ 1 (X) and Ψ(0) is the Fourier multiplier with symbol [ξ → Θ 1 |ξ|].The next step is to approximate Ψ(τ ) locally by certain Fourier multipliers, see Lemma 3.1 below.Therefore, we choose for each ε ∈ (0, 1), a finite ε-localization family, that is a family To each finite ε-localization family we associate a second family • supp χ ε j is an interval of length 3ε and with the same midpoint as supp π ε j , |j| ≤ N − 1. Lemma 3.1.Let X ∈ O r be fixed and chose r ′ ∈ (3/2, r).Given ν > 0, there exist ε ∈ (0, 1), a finite ε-localization family {π ε j : −N + 1 ≤ j ≤ N }, a positive constant K = K(ε, X), and bounded operators A j,τ ∈ L(H r (R), H r−1 (R)), j ∈ {−N + 1, . . ., N } and τ ∈ [0, 1], such that , and u ∈ H r (R).The operators A j,τ are defined by , where x ε j ∈ supp π ε j , |j| ≤ N − 1, and with functions α τ , β τ given by Proof.As shown in the proof of [1,Theorem 4.3] (in a more general context), if ε is chosen sufficiently small, then for all τ ∈ [0, 1] and u ∈ H r (R) we have We next recall, see e.g.[1, Eq. 2.1], there exists a constant C > 0 such that Using this estimate together with the identity if ε is sufficiently small, respectively, in view of the fact that a(X) vanishes at infinity, for all u ∈ H r (R).These estimates together with (3.3) lead us to (3.4).
We are now in a position to prove Theorem 1.2.

The mean value theorem together with the inequality a
, and [31, Theorem 8.1.1]applied in the context of (1.10) with r replaced by r ′ ensures that X 1 = X 2 in [0, T ], which contradicts our assumption.This unique local solution can be extended up to a maximal existence time T + = T + (X 0 ), see [31,Section 8.2].
The continuous dependence of the solution on the initial data stated at (i) follows from [31,Proposition 8.2.3].
The proof of (ii) uses a parameter trick which was successfully applied also to other problems, cf., e.g., [1,6,26,34,39].Since the details are very similar to those in [1, Theorem 1.2 (ii)] we omit them.
To prove (iii) we assume there exists a maximal solution X = X(•; X 0 ) to (1.10) with T + < ∞ and such that Arguing as above, we deduce for some fixed r ′ ∈ (3/2, r), that X : [0, T + ) → O r ′ is Hölder continuous.Applying [31,Theorem 8.1.1]to (1.10) (with r replaced by r ′ ) we may extend the solution X to an interval [0, T ′ ) with T + < T ′ and such that 2 Given α ∈ (0, 1), T > 0, and a Banach space X, let B((0, T ], X) denote the Banach space of all bounded functions from (0, T ] into X.The Banach space C α α ((0, T ], X) is then defined as Moreover, the parabolic smoothing property established at (ii) (with r replaced by r ′ ) implies that X ∈ C 1 ((0, T ′ ), H r (R) 2 ), and this contradicts the maximality property of X.This completes the proof.
We conclude this section with the proof of Proposition 1.3.
Proof of Proposition 1.3.Since X(t) H r ≤ M for all t ∈ [0, T + ), (1.1d) and Lemma A.2 imply there exists C > 0 such that dX(t) dt ) has a bounded derivative.Moreover, we also have that We next show that (x n ) is bounded.To this end we infer from the convergence The latter inequality together with (3.9) implies that (x n ) is indeed bounded.We may thus assume (after eventually subtracting a subsequence), that Finally, since X 2 (t n , x 0 ) → X * (x 0 ), together with the latter identity we conclude that c ∞ + f (t n , x 0 ) − h(t n , x 0 ) → 0, and therefore In order to prove the second claim we argue by contradiction and assume there exists x 0 ∈ R and δ > 0 such that lim inf Since (1.10) is invariant under horizontal translations we may assume without loss of generality that x 0 = 0. Hence, there exists a sequence (t n ) with t n ր T + and Recalling Lemma A.2, we find a constant Then R is a positive increasing function with R(t) → δ for t → T + .We further define the surface area The restrictions v ± := v| Ω ± : Ω ± → R 2 of the function v defined in (A.1) extend continuously up to Γ f and, given x ∈ R, we have (A.4) Proof.Let z 0 := (x 0 , f (x 0 )) ∈ Γ f .In order to prove that v + can be extended continuously in z 0 we consider the polygonal path Γ 1 ⊂ Ω + defined by the segments and ] and oriented counterclockwise.
Here we set D := 1 + 2 f ′ ∞ + max{f (x 0 − 1), f (x 0 + 1)}.Moreover, we let and we define the closed curve Γ := Γ 0 + Γ 1 which is again oriented counterclockwise.Additionally, we define the function ϕ ∈ BUC α (Γ) by setting where ϕ ± := ϕ(x 0 ± 1, f (x 0 ± 1)).It is not difficult to check that Given z ∈ Ω + which is sufficiently close to z 0 , it then holds Since z 0 ∈ Γ 0 , Lebegue's dominated convergence shows that Additionally, according to the Plemelj formula, cf.e.g.[30], it holds that These two convergences imply that v + can indeed be extended continuously in z 0 , the value of the extension in z 0 being as given in formula (A.4).Finally, the corresponding claim for v − follows by arguing similarly.
We next prove in Lemma A.2 that v is bounded in R 2 \ Γ f .In fact we bound in Lemma A.2 the L ∞ -norm of v by a constant that depends explicitly on the norms of the functions f and ω.
Lemma A.2.There exists a constant C, which is independent of f and ω, such that Proof.We devise the proof in several steps.
Step 1.In this step we provide bounds for the restrictions of v ± to Γ f .Given x ∈ R, it follows from (A.4) and Hölder's inequality that where Concerning I 2 , we have Gathering these estimates we conclude that Step 2. Given z = (x, y) ∈ R 2 , we set d(z) := dist (z, Γ f ).We next prove that , |s|} for all s ∈ R and together with Hölder's inequality we conclude from (A.1) that Step 3. In this final step we prove that We first consider the case when z ∈ Ω + .We associate to z a point z Γ = (x 0 , f Let Γ = Γ 0 + Γ 1 and ϕ be as defined in the proof of Lemma A.1 (with z Γ instead of z 0 ).Recalling (A.7), we have where Hölder's inequality leads us to where p ′ ∈ (1, ∞) is the adjoint exponent to p, that is p −1 + p ′ −1 = 1.In order to estimate T 3 we first note that The first relation follows from Cauchy's integral formula.The second identity is a direct consequence of Plemelj's formula, cf.e.g.[30].Using these two identities we get where The arguments used to estimate T 2 lead us to ), the latter arguments show that (A.9) holds for z ∈ Ω + .Arguing along the same lines it is easy to see that (A.9) is satisfied also for z ∈ Ω − .The claim (A.6) follows now from (A.7)-(A.9).We now extend in Theorem A.3 Privalov's theorem to the setting considered herein where the contour integral in (A.1) is defined over an unbounded graph.
Proof.We only establish the Hölder continuity of v + =: (v 1 + , v 2 + ) (that of v − follows by using similar arguments).We devise the proof in several steps.
Step 1.Let z, z ′ ∈ Ω + satisfy |z − z ′ | > 1/8.Then according to Lemma A.1 and Lemma A.2 we have |v Step 2. Given z ∈ R 2 , we set again d(z) := dist (z, Γ f ).Assume now that z, z ′ ∈ Ω + are chosen such that |z − z ′ | ≤ 1/8.Then, letting S zz ′ := {(1 − t)z + tz ′ : t ∈ [0, 1]} denote the segment that connects z and z ′ there exists at least a point ζ ∈ S zz ′ such that We distinguish two cases. Step we obtain in view of these inequalities Step 2b.We now consider the second case when |z − z Assuming there exits a constant C > 0 such that and the claim then follows. Step It remains to estimate the term T 3 which, in view of (A.10), can be written as where, letting z Γ be defined by the relation d(z) = |z − z Γ |, we set Moreover, given ξ ∈ Γ 1 , we have min{|ξ − z|, |ξ − z 0 |} ≥ 3/4 and together with (A.5) we get In order to estimate S 3 we let η := |z − z 0 | ∈ (0, 1/4], we set z 0 =: (x 0 , f (x 0 )), and we introduce the curve Γ η := {(s, f (s)) : |s − x 0 | ≤ 2η}.It then holds where and, after identifying the real and imaginary parts of the integral, we get Hence, we have shown that S 3c ≤ C|z 0 − z| α and the proof is completed.
We conclude this section with the following result.
Lemma A.4.It holds that Proof.The relations (A.12) follow by direct computation.We next prove that v + vanishes at infinity (the claim for v − follows by arguing along the same lines).We divide the proof in two steps.
Step 1.We first show that v + (x, f (x)) → 0 for |x| → ∞.Recalling Lemma A.1 and (2.2), we write Because ω ∈ BUC α (R) ∩ L p (R), the last term on the right vanishes at infinity.We next prove that, given n, m ∈ N, we also have This establishes (A.14).

2 ) 1
and (1.5)-(1.6), it then immediately follows that (p 1 − p 2 )| Γ c∞ f and (p 2 − p 3 )| Γ h are constants.Hence, for a suitable choice of c i , we may achieve that (1.2) 3 are satisfied.Therewith we established the existence of at least a solution to (1.2).