Mark ALLENLuis CAFFARELLIAlexis VASSEUR

(Dedicated to Haim Brezis on the occasion of his 70th birthday)

1 Introduction

In this paper we study both existence and regularity for solutions to a porous medium equation.The pressure is related to the density via a nonlocal operator.This diffusion takes into account long-range effects.The time derivative is nonlocal and fractional and therefore takes into account the past.In the typical derivation of the porous medium equation(see[16])the equation one considers is

with u(t,x)≥ 0.By Darcy’s law in a porous medium,v= ??p arises as a potential where p is the pressure.According to a state law p=f(u).In our case,we consider a potential which takes into account long-range interactions,namely p=(?Δ)?σu.A porous medium equation with a pressure of this type

has been recently studied.For 0< σ <1 with σ,existence of solutions was shown in[6]while regularity and further existence properties were studied in[5].Uniqueness for the range≤σ<1 was shown in[19].Another model of the porous medium equation

was introduced by Caputo in[7].In the above equation,is the Caputo derivative and the diffusion is local.Solvability for a more general equation was recently studied in[17].The fractional derivative takes into account models in which there is “memory”.The Caputo derivative has also been recently shown(see[8–9])to be effective in modeling problems in plasma transport.See also[13,18]for further models that utilize fractional equations in both space and time to account for long-range interactions as well as the past.

The specific equation we study is

The operatoris of Caputo-type and is defined by

When K(t,s,x)=(t?s)?1?α,this is exactly the Caputo(or Marchaud)derivative(see Section 2),which we denote by.We assume the following bounds on the kernel K:

Our kernel in time then can be thought of as having “bounded,measureable coefficients”.We also require the following relation on the kernel:

The relation(1.4)allows us to give a weak(in space and in time)formulation of(1.2).This weak formulation is given in Section 2.

In this paper,we also restrict ourselves to the range 0<σ<.In[5],the use of a transport term was made to work in the range<σ<1.We have not yet found the correct manner in which to prove our results for<σ<1 when dealing with the nonlocal fractional time derivative

1.1 Accounting for the past

Nonlocal equations are effective in taking into account long-range interactions and taking into account the past.However,the nonlocal aspect of the equation provides both advantages and disadvantages in studying local aspects of the equation.One advantage is that there is a relation between two points built into the equation.Indeed,we utilize two nonlocal terms that are not present in the classical porous medium equation to prove Lemma 5.3.One disadvantage of nonlocal equations is that when rescaling of the form v(t,x)=Au(Bt,Cx),the far away portions of u cannot be discarded,and hence v begins to build up a “tail”.Consequently,the usual test function(u?k)+or F((u?k)+)for some function F and a constant k is often insufficient.One must instead consider F((u?φ)+),where φ is a constant close by but has some“tail” growth at infinity.This difficulty of course presents itself with the Caputo derivative.One issue becomes immediately apparent.If we choose F((u?φ)+)as a test function,then

The second term will no longer be identically zero if φ is not constant.When using energy methods,this second term can be treated as part of the right-hand side,and hence it becomes natural to consider an equation of the form(1.2)with the right-hand side.The main challenge with accomodating a nonzero right-hand side is that the natural test function lnu used in[5–6]is no longer available since the function u can evaluate zero.Indeed,if the initial data for a solution is compactly supported,then the solution is compactly supported for every time t>0(see Remark 4.2).We choose as our basic test function uγfor γ >0.For small σ,we will have to choose small γ.We then can accomodate the right-hand side as well as avoiding delicate integrability issues involved when using lnu as a test function.Using careful analysis,it is still most likely possible to utilize lnu as a test function for our equation(1.2)with zero right-hand side,but we find it more convenient to use uγand prove the stronger result that includes the right-hand side.Our method using uγshould also work for the equation(1.1)to be able to prove existence and regularity with the right-hand side.One benefit of accomodating the right-hand side in L∞is that we obtain immediately regularity up to the initial time for smooth initial data(see Theorem 1.2).

1.2 Overview of the main results

We will prove our results for a class of weak solutions(2.4)later formulated in Section 2.Our first main result is existence.We use an approximating scheme as in[6]as well as discretizing in time as in[1].We prove the following theorem.

Theorem 1.1Let 0≤ u0(x),f(t,x)≤ Ae?|x|for some A ≥ 0.Assume further that u0∈C2.Then there exists a solution u to(2.4)in(0,∞)×Rnthat has initial data u(0,x)=u0(x).

Remark 1.1Our constructions are made via recursion over a finite time interval(0,T1).Since our constructions are made via recursion,if T2=mT1for m∈N,it is immediate that if uiis the solution constructed on(0,Ti),then u2=u1on(0,T1).

Remark 1.2For technical reasons seen in the proof of Lemma 4.4,when n=1,we make the further restriction 0<σ<

The main result of the paper is an interior H¨older regularity result.As expected,the H¨older norm will depend on the distance from the interior domain to the initial time t0.However,by assuming the intial data u0is regular enough(for intance C2)then we obtain regularity up to the initial time.This is a benefit of allowing the right-hand side.By extending the values of our solution u(t,x)=u(0,x)for t<0,we have(2.4)on(?∞,∞)× Rnwith the right-hand side in L∞.The right-hand side f for t≤ 0 will not necessarily satisfy f≥ 0.However,this nonnegativity assumption on f was only necessary to guarantee the existence of a solution u≥0.It is not a necessary assumption to prove regularity.From Remark 1.1,the solution constructed on(?∞,T)will agree with the original solution over the interval(0,T).

Theorem 1.2Let u be a solution to(2.4)obtained via approximation from Theorem 1.1 on[0,T]×Rnwith 0≤ u0(x),f(t,x)≤ Ae?|x|.Assume also u0∈ C2.Then u is Cβcontinuous on[0,T]× Rn(for some exponent βdepending on α,Λ,n,σ)with a constant that depends on the L∞norm of u and f,the C2norm of u0,and on T.

1.3 Future Directions

We prove existence and regularity for solutions obtained via limiting approximations.In this paper,we do not address the issue of uniqueness.As mentioned earlier,uniqueness for(1.1)for the range≤σ<1 was shown in[19].The issue of uniqueness for(1.2)is not trivial because of the nonlinear aspect of the equation as well as the lack of a comparison principle.The equation(2.4)which we consider also should present new difficulties because of the weak/veryweak formulation in time as well as the minimal“bounded,measurable” assumption(1.3)on the kernel K(t,s,x).An interesting problem would be to then address the issue of uniqueness for solutions of(2.4).

Theorems 1.1–1.2 can most likely be further refined by making less assumptions on u(0,x),assuming the right-hand side f∈Lpas was done for a similar problem in[17],and proving the estimates uniform as σ → 0 and recovering H¨older continuity for the local diffusion problem.Also,as mentioned earlier the theorems can be improved to include the range≤σ<1.

Finally,just like in the local porous medium equation(see[16])as well as for(1.1),the equation(2.4)has the property of finite propagation(see Remark 4.2).Therefore,it is of interest to study the free boundary?{u(t,x)>0}.

1.4 Outline

The outline of this paper will be as follows.In Section 2,we state basic results for the Caputo derivative.We also give the weak formulation of the equation we study.In Section 3,we state some results for the discretized version ofthat we will use to prove the existence of solutions.In Section 4,we use the estimates and follow the approximation method from[6]combined with the method of discretization and the estimates presented in[1]to prove existence.In Section 5,we state the main lemmas that we will need to be able to prove H¨older regularity.In Section 6,we prove the most technically difficult Lemma 5.1 of the paper.This Lemma 5.1 most directly handles the degenerate nature of the problem.In Section 7,we prove an analogue of Lemma 5.1.In Section 8,we prove the final lemmas we need which give a one-sided decrease in oscillation from above.The one-sided decrease in oscillation combined with Lemma 5.1 is enough to prove the H¨older regularity and this is explained in Section 9.

1.5 Notations

We list here the notations that will be used consistently throughout the paper.The following letters are fixed throughout the paper and always refer to the following:

(1)α—the order of the Caputo derivative.

(2) σ—the order of inverse fractional Laplacian(?Δ)?σ.We use σ for the order because s will always be a variable for time.

(3)a—the initial time for which our equation is defined.

(4)—the Caputo derivative as defined in Section 2.

(5)—the Caputo-type fractional derivative with“bounded,measurable”coefficients with bounds(1.3)and relation(1.4).

(6)Λ—the constant appearing in(1.3).

(7)—the discretized version ofas defined in(3.1).

(8)?refers to the time length of the discrete approximations as defined in Section 3.

(9)n will always refer to the space dimension.

(10)Γm—the parabolic cylinder(?m,0)×Bm.

(11)Wβ,p—the fractional Sobolev space as defined in[10].

(12)Lp(0,T;Wβ,p)—the set of functions u(t,x)for whichdt<∞.

(13)u±—the positive and negative parts respectively so that u=u+?u?.

(14)—the extension(t)=u(?j)for?j?1

2 Caputo Derivative

In this section,we state various properties of the Caputo derivative that will be useful.The proofs and methods of proof were shown in[1].The Caputo derivative for 0<α<1 is defined by

By using integration by parts,we have

For the remainder of the paper,we will drop the subscript a when the initial point is understood.We now recall some properties of the Caputo derivative that were proven in[1].

For a function g(t)defined on[a,t],it is advantageous to define g(t)for t

utilized in[1].(See also[14]under Marchaud derivative and[2–3]for properties of this one-sided nonlocal derivative.)This looks very similar to(?Δ)αexcept the integration only occurs for s

Proposition 2.1Let g,h∈C1(a,T).Then

Formula(2.1)is based on the following formal computation:

In the above computation to utilize the cancellation we only need a kernel K(t,s)satsifying

To make the above computation rigorous,we will use the discretization in Section 3.An alternative,equivalent integration by parts formula is to extend g(t)=g(a)for t

with cα= αΓ(1 ? α)?1.Now(2.2)and(2.3)will both imply each other,so(2.3)is an alternative way of handling the initial condition g(a).Furthermore,both(2.2)and(2.3)are weak formulations for the Caputo derivative that only require that g ∈((a,T)).(2.3)will work for any kernel K(t,s)satisfying the relation(1.4).In view of(2.3),we now give the exact formulation of our weak solutions.We assume the bounds(1.3)and the relation(1.4)on the kernel K(t,s,x).For smooth initial data u0∈C2,we assign u(t,x)=u(a,x)for t

We say that u is a weak solution,if for any φ ∈(?∞,T)×Rn,we have

We will also utilize a fractional Sobolev norm that arises from the fractional derivative.

Lemma 2.1Let u be defined on[a,T].We have for two constants c1,c2depending on α,|T?a|,

The following estimate will be needed for our choice of cut-o fffunctions.

Lemma 2.2Let

with ν < α.Then

for t ∈ R.Here,cν,α,Λis a constant depending only on α,ν,Λ.

Finally,we point out that if g=g+?g?the positive and negative parts respectively,then

3 Discretization in Time

To prove existence of solutions to(2.4),we will discretize in time.The discretization also allows us to make the computations involving the fractional derivative rigorous.This section contains properties of a discrete fractional derivative which we will utilize.

For future reference,we denote the discrete fractional derivative with kernel K as

We writewhen a=0.The following is the discretized argument for the cancellation that appears in the formal computation of the version of(2.1)that we will need.

Lemma 3.1Assume g(a)=0 and define g(t):=0 for t

Let(t):=g(?j)for?j?1

ProofFor notational simplicity,we assume a=0.

Inequality(3.2)follows from the estimates in[1].

Lemma 3.1 combined with the estimates in[1]can be used to show.

Lemma 3.2Let u(0)=0 and assume u≥0.For fixed 0

This next lemma is analogous to Lemma 2.2 and was shown in[1].

Lemma 3.3Let h be as in Lemma 2.2.Then for 0

for t∈?Z and a

This last estimate we will use often the following lemma.

Lemma 3.4Let F be a convex function with F??≥γ,F?≥0,F(0)=0.Assume g ≥ 0,g(a)=0.Then there exists c depending on α,Λ,such that

ProofAssume for notational simplicity a=0.Since F is convex,

The result then follows from applying Lemma 3.1.

Finally,we point out that if g is a limit of?g?which are discretized problems with the assumptions in Lemma 3.4,it follows that

4 Existence

In this section,we prove the existence of weak solutions following the construction given in[6].We will also discretize in time.We first consider a smooth approximation of the kernel(?Δ)?σas Kζ.We start with the smooth classical solution to the elliptic problem

with u≡0 on?BR.g,f≥0 are smooth.The signs of f,g guarantee the solution is nonnegative.δ,d>0 are constants.The nonlocal part is computed in the expected way by extending u=0 on.To find such a solution,we first consider the linear problem

forwith v ≥ 0.With fixed d,δ,R,ζ>0,one can apply Schauder estimate theory to conclude

with u≥0.The map T:v→u is then a compact map.The set{v}with v≥0 andis a closed convex set,and hence we can apply the fixed point theorem(see[12,Corollary 11.2]),to conclude there is a solution to(4.1).By bootstrapping,we conclude u is smooth.

Now we use the existence of solutions to(4.1)to obtain(via recursion)solutions to the discretized problem

with u(0,x)=u0(x)an initially defined smooth function with compact support.?=for some k∈N.We will eventually let k→∞,so that?→0.

For the next two lemmas,we will utilize the solution to

which as in[11]is given by

where Eαis the Mittag-Leffler function of order α.We will utilize in the next lemmas two specific instances of(4.3).We define Y1(t)to be the solution to(4.3)with Y(0)=supu(0,x),c=0,h=2Λf.We define Y2(t)to be the solution to(4.3)with c=CΛ?1,Y2(0)=2 and h=0.The constant C will be chosen later.

Lemma 4.1Let u be a solution to(4.2).Let Y1(t)be defined as above.Then there exists?0depending only on T,α,?f?L∞ such that if?≤ ?0,then

ProofSince Y1(t)is an increasing function

Depending on T and?f?L∞,there exists?0,such that if?≤ ?0,then

We use(u(t,x)?Y1(t))+as a test function.Since(u?Y)+(0)=0,it follows from(2.5)and Lemma 3.4 that

We define

where Hζ(x,y)=ΔKζ.We have the identity(see[5])

Then for?small enough and j>0,

Thus(u?Y1)+≡0.

Lemma 4.2Let u be a solution to(4.2)in[0,T]×BR.Assume that

If A is large,there exists constants

such that if μ < μ0, δ< δ0, ζ< ζ0,?

for any t=?j.

ProofAs before,there exists?0depending only on T,α,such that for?≤ ?0,we have

Since u is smooth and hence continuous,u≤ LY2(?j)e?|x|for some L>A.We lower L ≥ A until it touches u for the first time.Since u=0 on?BR,this cannot happen on the boundary.Since u is smooth,this cannot happen at a point(?j,0).Also,LY2≥2A≥2u(0,x),so this cannot occur at the initial time.We label a point of touching as(tc,rc).We compute the operator in nondivergence form and write Kζ(u)=p and use the estimates in[6]to conclude for?small enough that

where in the equation the bar above means evaluation at rc.Then using again the estimates from[6],for small enough ζ,we have a universal constant M depending only on n,σ,such that

Now recalling also that LY2(tc)e?rc≤ Y1(T),

Choosing δ,d small enough,the above inequality implies

If we choose now C>2MY1(T)+4,we obtain a contradiction.We note that C will only depend on n,σ,?u(0,x)?L∞,?f?L∞.

We now give some Sobolev estimates.Because we have the right-hand side,we choose not to use ln(u)as the test function.For 0< γ <1,we use(u+d)γ?dγas a test function.The function

will satisfy the conditions in Lemma 3.4.We now assume that u is a solution to(4.2)with assumptions as in Lemma 4.2,so that|u|≤ Me?|x|for some large M.As discussed in the introduction,we can extend u(?j,x)=u(0,x)for j<0,and u will be a solution to(4.2)on(?∞,T)×Rnwith the right-hand side

for j≤0.This right-hand side is not necessarily nonnegative.However,we only required the nonnegativity of the right-hand side to guarantee that our solution is nonnegative.In this case,we already know our solution is nonnegative.We fix a smooth cut-offφ(t)with φ(t)≥ M for t≤ ?2 and φ(t)=0 for t≥ ?1.We now take our test function as?F?([u(t,x)? φ(t)]+).We define

We define=u(t)for?j?1

We now consider the nonlocal spatial term.We will also use the following property:For an increasing function V and a constant l,

We have for the nonlocal spatial terms,

From Proposition 10.1,ifthen

Then

For the local spatial term,we have

Now combining the previous estimates with the right-hand side term f,we have for a certain constant C depending on n,σ,α,Λ,γ,M,T which can change line by line,

The second to last inequality comes from Proposition 10.2.The value C is independent of ζ,d,R,?,δ if

Then as ζ,d → 0,we have uniform control and obtain the estimate

Notice that the constant C only depends on the exponential decay of f,u0and on σ,α,n,T,but not on R,δ.Letting d,ζ→ 0,we obtain

We now give a compactness result.

Lemma 4.3Assume for any v∈F,

Then F is totally bounded in Lp([0,T]×BR)for 1≤p≤2.

ProofWe utilize the proof provided in[10]for compactness in fractional Sobolev spaces.We will show the result for p=2,and it will follow for p<2,since BRis a bounded set.We divide T into k increments.(This k is unrelated to the number k for the?approximations.)Let l=.We define

From[10],

The above estimate is uniform for any vj.We now utilize that[0,T]is a finite measure space as well as Minkowski’s inequality:The norm of the sum is less than or equal to the sum of the norm,

It then follows from the result in[10]that for every j and λ>0,there exists finitely many{β1,···,βMj},such that for any fixed j and v ∈ F,there exists βi∈ {β1,···,βMj},such that

Then combining the above estimate with(4.8),we obtain

Since l,λ can be chosen arbitrarily small,F is totally bounded.

The following result will guarantee that?(?Δ)?σu ∈ Lpas δ→ 0,R → ∞.

Lemma 4.4Let u be a solution to(4.6)with the right-hand side f and u0both satisfying the exponential bound(4.4).Then

with the constant C depending only on the exponential bounds in(4.4),n,γ,T.

Proofu is extended to be zero outside of BR.We utilize inequality(4.5)and the lifting property for Besov spaces.We use the following results found in[15].To begin,we have the following characterization of homogenous Besov spaces:

We also have the lifting property of the Riesz potential for the homogeneous Besov spaces

To bound u in the nonhomogeneous Besov space,we recall

for any 1≤q<.From the exponential bounds(4.4)and growth we have that u is uniformly in Lqfor all 1≤q≤∞.Letting

(or let σ

Using again the characterization of homogeneous Besov spaces in(4.9),we obtain the result.

Corollary 4.1Let ukbe a sequence of solutions to(4.6)with R → ∞ and δ→ 0.For fixed ρ>0,there exists a subsequence and limit with

Furthermore,for any compactly supported φ,

ProofThe strong and weak convergence is an immediate result of the bound(4.5)and Lemma 4.3.For γsmall enough depending on σ,then

Then from Lemma 4.4,we have that

and in particular

Then it is immediate from the weak and strong convergence that u0is a solution.

We now show the proof of Theorem 1.1.

Proof of Theorem 1.1We first assume f,u0smooth and satisfying the exponential bounds(4.4).Consider solutions u?to(4.10)over a finite interval(0,T).As before,as?→0,there exists a subsequence and a limit u?→u0with the weak convergence as in(4.11)and strong convergence over compact sets for 1≤p≤2 just as in Lemma 4.3.Then for fixed φ∈,that u0is a solution follows from this convergence.The spatial piece and right-hand side is straightforward to show,and the nonlocal time piece is taken care of as in[1].We now consider a sequence of solution{uj}with{fj}∈C∞with fj→ f in weak?L∞.Then again there exists a limit solution u with the right-hand side f.From Remark 1.1 and Lemma 4.1,we can let T→∞.

Remark 4.1Theorem 1.1 provides a solution u with initial data u0in the sense of(2.4).However,since the initial data is C2,we have later from Theorem 1.2 that u is actually continuous up to the initial time u(0,x).

Remark 4.2In this section,we have shown how the estimates in[6]work for equations of the form(2.4).In the same way,one can show that the method of“true(exaggerated)supersolutions” as shown in[6]for σ

5 Continuity:Method and Lemmas

In this section,we outline the method used to prove H¨older regularity of solutions to(2.4).We follow the method used in[5],which is an adaption of the ideas originally used by De Giorgi.We prove a decrease in oscillation on smaller cylinders,and then utilize the scaling property that if u is a solution to(2.4),then v(t,x)=A(Bt,Cx)is also a solution to(2.4)if A=BαC2?2σ.Because of the degenerate nature of the problem,the decrease in oscillation will only occur from above.Since we do not have a decrease in oscillation from below,we will need a lemma that says in essence that if the solution u is aboveon most of the space time,then u is a distance from zero on a smaller cylinder.To prove the lemmas in this section,we will use energy methods,and thus we will want to use as a test function F(u)for some F.If u is a solution to(2.4),then

and it is not clear that?F(u)will be a valid test function.We therefore prove the lemmas for the approximate problems

for some large R>0 and small δ>0 with u ≡ 0 onIt is actually only necessary to prove the energy inequalities that we will utilize with constants uniform as δ→0 and R→∞.We could also prove the lemmas for the approximate problems(4.6).However,for notational convenience and to make the proofs more transparent,we have chosen to let?→0.Because our solution is a limit of discretized solutions,we then are allowed to make the formal computations involved witheven though u may not be regular enough forto be defined.One simply proves the energy inequalities(and hence the lemmas)for the discretized solutions as was done in[1].

Because of the one-sided nature of our problem,we prove the lemmas for solutions to the equation with the modified term div(D(u)?(?Δ)?σu),where D(u)=d1u+d2.We assume 0≤d1,d2≤2 and either d1=1 or d2≥.As will be seen later,when d2≥,the proofs are simpler because the problem is no longer degenerate.We now define the exact class of solutions for which we prove the lemmas of this section.u is a solution if u≡0 on×(0,T)and for every φ ∈((?∞,T)×BR),we have

By Lemmas 4.3–4.4,the lemmas stated in this section will be true when R → ∞ and δ→ 0.

Before stating the lemmas,we define the following function for small 0<τ<

We now state the lemmas we will need.

Lemma 5.1Let u be a solution to(5.2)with the right-hand side|f|≤1 and R>4 and assume

Given μ0∈ ?0,?and τ0<,there exists κ >0 depending on μ0,τ0,σ,α,n,such that if

then u ≥ μ0on the smaller cylinder Γ1.

We have a similar lemma from the above.

Lemma 5.2Under the same assumptions as Lemma 5.1,given μ1∈ ?0,?and τ0<,there exists κ >0 depending on μ1,τ0,σ,α,n,such that if

then u ≤ 1? μ1on the smaller cylinder Γ1.

Lemma 5.2 in essence states that if the set of points for which u≥is very small,then u pulls down from 1 in a smaller neighborhood.To prove the regularity,we will need a stronger result that in essence states that if the set of points for which u ≥is not all of Γ4,then u still pulls down from 1 in a smaller neighborhood.This is given by the following.

Lemma 5.3Under the same assumptions as Lemma 5.1,assume further for fixed k0,

There exists λ0>0 depending on κ0,such that if|f|≤,then u ≤ 1 ? μ2on Γ1for some μ2depending on κ0.

We will choose κ0to equal the κin Lemma 5.1.

6 Pulling the Solution up from Zero

In this section,we provide the proof of Lemma 5.1.This lemma is the most technical to prove.We first prove the lemma in the most difficult case when D(u)=u+d with 0≤d≤2.Afterwards,we show how the proof is much simpler when D(u)=d1u+d2with d2≥and 0≤d1≤2.

We will need the following technical lemma.The proof is found in the appendix.

Lemma 6.1Let u,φ be two functions,such that 0≤u≤φ≤1.Let 0<γ<1 be a constant.If|u(x)? u(y)|≥ 4|φ(x)? φ(y)|,then

Also,if

then

If instead we assume|u(x)? u(y)|≤ 4|φ(x)? φ(y)|,then

Remark 6.1When 0≤u≤φ≤3,Lemma 6.1 will hold with new constants by applying the lemma to

We will use a sequence of cut-o fffunctions{φk}which will be chosen to be smooth cut-offfunctions in space,and smooth increasing cut-o fffunctions in time.We recall that for small 0<τ<

We now recall the construction of a sequence of smooth radial cut-offs θkfrom[5]that satisfy

(1)θk(x)≤ θk?1(x)≤ ···≤ θ0(x),

(2)with m≥2,

(3)θk?1? θk≥ (1? μ0)2?kin the support of θk,

(4)θk→ μ0χB2as k→ ∞,

(5)the support of θkis contained in the set where θk?1achieves its maximum.

We also have θ0≡ 1 on B3and the support of θ0is contained in B4.As a cut-o ffin time,we consider a sequence{ξk}satisfying

(1)ξk(t)≤ ξk?1(t),

(2)(t)≤ Ck,

(3)ξk→ χ{t>?2}as k→ ∞,

(4) ξk=maxξk=1 on the interval[?2?2?k,0],

(5)the support of ξkis contained in the set where ξk?1achieves its maximum.

We now define

We use the convention for negative part that u=u+?u?.We also write:=(u ? φk)?.We now consider the convex function

Because of the degenerate nature of our equation,we will want to utilize the test function

The basic idea of the proof of Lemma 5.1 is as follows.We first seek to obtain an energy inequality of the form

The first step of the proof consists of obtaining the first term on the left-hand side(or the energy)in time.The second step of the proof consists of obtaining the second term on the left-hand side(or the energy)in space.The third step of the proof consists of bounding all remaining terms by the right-hand side and thus obtaining the inequality.In the fourth step,we utilize the fractional Sobolev embedding theorem to obtain the inequality

where Ukis the set on which φkobtains its maximum.We then utilize Tchebychev’s inequality to obtain

with p>b.This nonlinear recursion relation allows us to conclude the proof.

Proof of Lemma 5.1First Step:Obtaining the energy piece in timeWe note that for 0≤x≤1,F(x)is convex,F?(x)≥0,and F??(x)≥γ.From the convexity and second derivative estimate,we also conclude for 0≤x,y≤1

We now considerand rewriteTo obtain an energy in time,we first consider

In the first inequality,we used that φkis increasing in t and positive for t ≥ ?4 as well as(s)=0 for s≤ ?4,and in the second inequality we used(6.5).Term(2)is half of what we will need for the Sobolev embedding(see Lemma 2.1).To gain the other half,we consider term(1).For c,Ckdepending on Λ,αand the Lipschitz constant of φk,we have

The last inequality coming from(6.6).

Now

and in this proof,we ignore this term which will be on the left hand side.Now for the term involving φk,we have

Then utilizing the embedding theorem for fractional Sobolev spaces(see[10])combined with the above inequalities,we obtain

After integrating in the spatial variable,we have

Second Step:Obtaining the energy piece in spaceWe now turn our attention to the elliptic portion of the problem.We recall from[5]the identity

We multiply by our test function(6.4)and integrate by parts.On the left-hand side of the equation,we have

We now focus on(1)which will give us the energy term we need.For the term(?Δ)?σu,we rewriteThen we rewrite(1)=(1a)+(1b)+(1c).We focus on the term(1b).We rewrite

The term(1bi)will give us the energy term in space that we will need,

We define the set

It is clear that Akcontains the set Vk×Vk,where we define Vkas theset on which θkachieves its maximum.From Lemma 6.1 and Remark 6.1,we have

We now label Ukas the set where φkachieves its maximum.Notice that Uk=[?2?2?K,0]×Vk.To utilize the fractional Sobolev embedding on Vk×Vk,we also will need an Lpnorm ofon Vk.We utilize half of the integral ofthat we gained from the following fractional time term:

The inequality comes from the fact that 0≤≤ 1.Now from the fractional Sobolev embedding(see[10]),

This is the helpful spatial term on the left-hand side that we will return to later.

Third Step:Bounding the remaining termsWe will now show that everything left in our equation can be bounded by

We will denote

For the remainder of term(1bi),we have

The last inequality is due to the Lipschitz constant of φkwhen x,y are close,and the tail growth of φkwhen x,y are far apart.

We now control the term(1bii).Again,we split the region of integration over AkandUsing H¨older’s inequality(provided 2σ >and therefore we must choose γsmall when σis small)as well as the Lipschitz and sup bounds on φk,we have

The first term is absorbed into the left-hand side and the second term is controlled exactly as before.

We now consider the integration over

We now turn our attention to the term(1c).By Lemma 6.1,we have

Both of the above terms are handled exactly as before by using Lemma 6.1 and splitting the region of integration over Akand

The term(1a)is

The factor of 2 comes form the symmetry of the kernel.We will utilize this nonnegative term shortly.

We now consider the term(2)which we recall as

In the above,L= ?(?Δ)?σ,and we have

We again writeTo control the term involving φk,we integrate over the two sets{|x?y|≤ 8}and{|x?y|>8}.We use that|φk(x)?φk(y)|≤ Ck|x?y|when|x?y|≤ 8,and|φk(x)? φk(y)|≤ |x?y|τwhen|x?y|>8,as well as the bound|?φk|≤ Ck,to obtain

We now use the same set decomposition with,the inequality||≤ 1 as well as H¨older’s inequality

The third term is bounded by

provided τ<1?2σ as well as the first term provided again that 2σ >.The second term can be bounded as before by splitting the region of integration over Akandand absorbing the region over Akinto the left-hand side.

We now turn our attention to the last term involving.We first remark that the integral becomes

We first consider the set|x?y|>8.Since

When|x?y|<8,we make the further decomposition

to absorb the integral by the nonnegative quantity(6.10).In the complement,when

we useand integrate in y

The remainder of the terms are bounded byBy multiplying by the term χ{u<φk}and integrating,we end up in the worst case with

since m≥2.The last term to consider is the local spatial term.We use the Cauchy-Schwarz inequality

Retaining the energy from(6.8)on the left-hand side and moving everything else to the right-hand side which is bounded by(6.9),our energy inequality becomes

Fourth Step:The nonlinear recursion relationWe now(as in[1])use H¨older’s inequality twice with the relations

For a function v,we obtain

We now choose

so that if r=2?2σ,then

We now use H¨older’s inequality one more time to obtain

We choose

so that

where we used Minkowski’s inequality in the last inequality.Substituting u?φkfor v in(6.12)and utilizing(6.8),we obtain

We first recall that φk?1≥ φk+(1 ? μ0)2?k.We now utilize Tchebychev’s inequality

Combining the above inequality with(6.13),we conclude

If we define

then

Since p>b,if M0is sufficiently small(depending on C and),we obtain that Mk→0,and hence u≥μ0.

We now prove Lemma 5.1 in the case when D(x)=(d1x+d2)with d1≤2 and d2≥.This is actually much simpler,because we can utilize the test function?(u?φk)?as when dealing with a linear equation.

ProofWe choose as the test function ?(u? φk)?.Notice that

The fact that d2≥gives a nondegenerate linear term which we utilize.From the computations in[1],we then have

and even more importantly

Notice that in the above inequality,we have the power|·|2rather than|·|2+γ.

We now show how to bound the terms invovling d1u,

Then multiplying(1)by ?(?Δ)?σu and integrating over Rn,we have

Now

We write u=?uφk.We break up our set into the two regions

We notice that on the set Fk,we have

Then integrating over Fk,we have for the term(1a)with right term

This is the nonnegative energy piece which we actually do not need having obtained a better piece in(6.14).All of the remaining terms in(1)can be bounded by breaking up the region of integration over Fk,.Over Fk,we use H¨older’s inequality with p=2 rather than with p=2+γ and absorb the small pieces by the term in(6.14).We use the same methods as before to bound the integration over.Bounding the term(2)is done as before with slightly easier computations.

The local spatial term is bounded in the usual manner.

7 Pulling the Solution Down

In this section,we prove Lemma 5.2.We will need the following estimate that is analogous to Lemma 6.1.

Lemma 7.1Let u,φ be two functions such that≤φ≤u≤1.Let 0<γ<1 be a constant.If|u(x)? u(y)|≥ 4|φ(x)? φ(y)|,then

Also,if

then 0≤ (u?φ)?(x)?(u?φ)?(y).

If instead we assume|u(x)? u(y)|≤ 4|φ(x)? φ(y)|,then

The proof is similar to the proof of Lemma 6.1.In this case,since u>φ one uses the bound above on u and the fact that φ is bounded by below.

Proof of Lemma 5.2The proof is nearly identical.We mention the differences.We only consider D(u)=u,since the modifications for handling D(u)=d1u+d2have already been shown in the proof of Lemma 5.1.We consider a similar test function

We then utilize

This time we consider the same test functions θk(x)in the space variable,but this time we multiply only by a single cut-o ffin time ξ0(t).We define our φkas

To obtain the same estimate in time,we only need to recognize that φkis now decreasing in time and bounded by below by

The negative constant comes from the fact thatis Lipschitz.Then everything proceeds as before.Since our cut-o ffis bounded by below,our Lpnorm in time occurs over all of(?∞,0).We obtain as before

The spatial portion of the problem is handled exactly as before.

8 Decrease in Oscillation

We define

We point out that F is Lipschitz,compactly supported in[?3,0]× B3and equal toin[?2,0]×B2.We also define for 0< λ <,

and zero otherwise.The value of ν will be determined later.Finally,we define for i∈ {0,1,2,3,4}

Then≤ φ0≤ ···≤ φ4≤ 1 in Γ4.

Lemma 8.1Let κ be the constant defined in Lemma 5.2.Let u be a solution to(5.2).There exists a small constant ρ >0 depending only on n,σ,α and λ0depending only on n,σ,α,ρ,such that for any solution u defined in(a,0)×Rnwith a

with λ ≤ λ0,and f ≤ λ3,then if

then

ProofWe will show the computations for D(u)=u.The general situation is handled as before as in Lemma 5.1.

First Step:Revisiting the energy inequalityWe return again to the energy inequality.This time,however,we will make use of the nonnegative terms.We seek to obtain a bound on the right-hand side of the form

We now consider the test function as in(7.3),but with cut-offφ1.If u> φi,thenu≤1,and so

To take care of the piece in time,we first note that φ1is Lipschitz in time for t ∈ [0,4]with Lipschitz constant 2λ.Then as before

For(T1),we return to the inequality(7.4)and utilize the Lipschitz nature of,to obtain

The nonnegative piece(T2)will be utilized in the second step of this proof.For(T3),we note that since φiis decreasing,we have

Clearly,for t≤0 from the Lipschitz nature of F.For?4≤ t≤0,we have

We therefore pick ν small enough thatα?νν>2,

Our energy inequality becomes

Since f≤ λ3,everything is bounded on the right-hand side by Cλ2.

We now turn our attention to the elliptic portion.We consider the terms(1a),(1bi),(1bii),(1c),(2)as the analogous terms for those defined in the proof of Lemma 5.1.As before we obtain a nonnegative energy from the term(1bi).Everything else we will absorb into this energy or bound byThe term from(1bi)overis bounded as follows:

We have the following inequality from the computations given in[4]:

withIn particular,(8.1)will hold for γ=0.For the term(1bii),we break up the region of integration into A1andOn A1,we use H¨older’s inequality as before

The last term is absorbed into the left-hand side.The other term is bounded again from(8.1).The term(1c)is bounded in exactly the same way.(1a)is nonnegative and will be utilized later.We now turn our attention to the term(2).We rewriteThe term involvingwith|x?y|≤ η is absorbed by the nonnegative term(1a)on the left-hand side.We now utilize the inequalities:

Then

These terms are all bounded as before.Notice that we have λ2on the outside.Then all nonlocal terms on the right-hand side are bounded by Cλ2+γ1+γ.The local term div(D(u)?u)is handled in the usual manner by use of the Cauchy-Schwarz inequality.

Second Step:Using the nonnegative pieces in space and timeWe now utilize the two nonnegative pieces.From Proposition 10.1,we have

Then we conclude that

Since we used Ψλ3,replacing φ1with φ3,we have the same inequality but with the bound

We now show how the inequality(8.2)and its analogue for φ3are enough to prove the remainder of the lemma as in[1].We note that for the proof as written in[1]to work,we need>5 which is achieved for γ small enough.We first utilize

From our hypothesis

Then the set of times Σ ∈ (?4,?2)for which|{u(t,·)has at least measure.And

so

Now,and so from Chebychev’s inequality,

The exponent on λis positive for γsmall enough.We write this as

This will be positive for λ small enough depending on n,σ,α,γ,ρ.

The proof then proceeds just as in[1],where we then utilize>5 as well as the analogue of(8.2)for φ3.

This next lemma will imply Lemma 5.3.For this next lemma,we define

Lemma 8.2Given ρ >0,there exist τ>0 and μ2,such that for any solution to(5.2)in Rn×(a,0)with a

where λ ≤ λ0with λ0as in Lemma 8.1,if

then

ProofWe consider the rescaled function.We fix τ small enough,such that

Then w satisfies equation(5.2)with D2(x)=D1(λ4x+(1 ? λ4)),where D1is the coefficient for the equation u satisfies.From our hypothesis and Lemma 8.1,

Then from Lemma 5.2,we conclude that w ≤ 1?μ1on(?1,0)×B1,and so u ≤ 1?λ4μ1=1?μ2.

9 Proof of Regularity

With Lemmas 5.1–5.3 we are ready to finish the proof of Theorem 1.2.We first mention that solutions of(2.4)satisfy the following scaling property:If u is a solution on(a,0)×Rn,then v(t,x)=Au(Bt,Cx)is a solution on(a/B,0)×Rnif A=BαC2σ?2.The method of proof is given in[5]which we now briefly outline.We take any point p=(x,t)∈Rn×(a,T)and prove that u is H¨older continuous around p.The H¨older continuity exponent will depend only on α,σ,n.The constant will depend on the L∞norm of u,f and on the C2norm of u(a,x).By translation,we assume that p=(0,0).By scaling,we assume that 0≤ u(t,x)≤ Ψ(t,x)and|f|≤for λ0as defined in Lemma 8.1.

We now take a positive constant M<,such that for 0

M will depend only on λ,μ2and τ>0.During the iteration,we have the following alternative.

Alternative 1Suppose that we can apply Lemma 5.3 repeatedly.We then consider the rescaled functions

Notice that M1

Alternative 2If at some point the assumption(5.3)fails,then we are in the situation of Lemma 5.1 and

Scaling the above situation our equations will have D(u)=d1u+d2with d2>0.We may then repeat the procedure since Lemmas 5.1 and 5.3 apply also in this situation.

10 Appendix

Proof of Lemma 6.1Since throughout this paper,we only require γsmall when σis small,we will prove the lemma for γ =for k ∈N.Now we assume without loss of generality that

We first assume that|u(x)? u(y)|≤ 4|φ(x)? φ(y)|.We need to bound

We first notice that the term above in(10.1)will be larger,if we assume that u(y)≥u(x)and φ(x)≥ φ(y)without changing|u(x)?u(y)|and|φ(x)?φ(y)|.Furthermore,the term in(10.1)will still be greater if u(y)= φ(y)and not changing|u(y)?u(x)|.We are then looking for the bound

Thus,for a constant l,we need the bound

Recalling that we are assuming 4|φ(x)? φ(y)|≥ |u(x)? u(y)|(or υ ≤ 4μ)the above term is maximized when υis largest or when υ =4μ.Now

It is clear that

To control L1we first consider when l?4μ ≤.Then l≤ 8μ and it is clearly true that

Now when l?4μ ≥,from the concavity of xγ,we have

Then

and(6.2)is proven with constant c2=14.

We now assume 4|φ(x)?φ(y)|≤ |u(x)?u(y)|,and the left-hand side of(10.2)is maximized again when 4μ = υ,and so we have

which is just(10.3)rewritten with the substitution μ =.Then

and the right-hand side of(6.1)is shown.

Now

We suppose γ=.By factoring,we have

Thus M2≤.Thus,M1is the dominant term.We then have from the convexity of xγ+1,

Proposition 10.1Let F be a function satisfying F??≥0 for x≥0.Assume also F(0)=0.If y≥x≥0,then

ProofFor fixed h>0,

Then for x,h≥0,

Let h=y?x,and multiply both sides of the equation by y?x.

Proposition 10.2Let 0<γ<1 and x,d≥0.Then

ProofFirst assume x ≤ d.From the concavity of xγ,we have

If on the other hand x>d,then

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