Local Hölder continuity for doubly nonlinear parabolic equations
aa r X i v : . [ m a t h . A P ] S e p LOCAL H ¨OLDER CONTINUITY FOR DOUBLYNONLINEAR PARABOLIC EQUATIONS
TUOMO KUUSI, JUHANA SILJANDER AND JOS´E MIGUEL URBANO
Abstract.
We give a proof of the H¨older continuity of weak solutionsof certain degenerate doubly nonlinear parabolic equations in measurespaces. We only assume the measure to be a doubling non-trivial Borelmeasure which supports a Poincar´e inequality. The proof discriminatesbetween large scales, for which a Harnack inequality is used, and smallscales, that require intrinsic scaling methods. Introduction
We consider the regularity issue for nonnegative weak solutions of thedoubly nonlinear parabolic equation ∂ ( u p − ) ∂t − ∇ · ( |∇ u | p − ∇ u ) = 0 , ≤ p < ∞ . (1.1)This equation is a prototype of a parabolic equation of p -Laplacian type. Itssolutions can be scaled by nonnegative factors, but in general we cannot adda constant to a solution so that the resulting function would be a solutionto the same equation.The purpose of this paper is to obtain a clear and transparent proof forthe local H¨older continuity of nonnegative weak solutions of (1.1). Our workis a continuation to [17], where Harnack’s inequality for the same equationis proved. See also [21], [11], [10] and [24]. However, since we cannot addconstants to solutions, the Harnack estimates do not directly imply the localH¨older continuity. To show that our proof is based on a general principle, weconsider the case where the Lebesgue measure is replaced by a more generalBorel measure, which is merely assumed to be doubling and to support aPoincar´e inequality. In the weighted case, parabolic equations have earlierbeen studied in [1], [2] and [20]. See also [8].This kind of doubly nonlinear equations have been considered by Vespri[23], Porzio and Vespri [19], and Ivanov [14], [15]. The known regularityproofs are based on the method of intrinsic scaling, originally introducedby DiBenedetto, and they seem to depend highly on the particular formof the equation. However, the passage from one equation to another isnot completely clear. For other parabolic equations, the problem has beenstudied at length, see [4], [3], [7] and [22], and the references therein.The difficulty with equation (1.1) is that there is a certain kind of di-chotomy in its behavior. Correspondingly, the proof has been divided in Mathematics Subject Classification.
Primary 35B65. Secondary 35K65, 35D10.
Key words and phrases.
H¨older continuity, Caccioppoli estimates, intrinsic scaling, Har-nack’s inequality.Research of JMU supported by CMUC/FCT and project UTAustin/MAT/0035/2008. two complementary cases:Case I : 0 ≤ ess inf u << ess osc u and Case II : u p − u t ≈ Cu t . In large scales, i.e., in Case I, the scaling property of the equation domi-nates and, consequently, the reduction of the oscillation follows immediatelyfrom Harnack’s inequality. In small scales, on the other hand, the oscillationis already very small and thus the solution itself is between two constants,the infimum and the supremum, whose difference is negligible. Correspond-ingly, the nonlinear time derivative term, which formally looks like u p − u t ,behaves like u t and we end up with a p -parabolic type behavior. However,also in this case, we still need to modify the known arguments. In particular,the energy estimates are not available in the usual form and we need to usemodified versions as in [5], [15] and [25].Our argument also applies to doubly nonlinear equations of p -Laplaciantype that are of the form ∂ ( u p − ) ∂t − ∇ · A ( x, t, u, ∇ u ) = 0 , with A ( x, t, · , · ) satisfying the usual structure assumptions. For expositorypurposes, we only consider (1.1).Very recently, a direct geometric method to obtain local H¨older continuityfor parabolic equations has been developed in [6] and [9]. Despite the effort,the general picture remains unclear.2. Preliminaries
Let µ be a Borel measure and Ω be an open set in R d . The Sobolev space H ,p (Ω; µ ) is defined to be the completion of C ∞ (Ω) with respect to theSobolev norm k u k ,p, Ω = (cid:18)Z Ω ( | u | p + |∇ u | p ) dµ (cid:19) /p . A function u belongs to the local Sobolev space H ,ploc (Ω; µ ) if it belongsto H ,p (Ω ′ ; µ ) for every Ω ′ ⋐ Ω. Moreover, the Sobolev space with zeroboundary values H ,p (Ω; µ ) is defined as the completion of C ∞ (Ω) withrespect to the Sobolev norm. For more properties of Sobolev spaces, see e.g.[13].Let t < t . The parabolic Sobolev space L p ( t , t ; H ,p (Ω; µ )) is thespace of functions u ( x, t ) such that, for almost every t , with t < t < t , thefunction u ( · , t ) belongs to H ,p (Ω; µ ) and Z t t Z Ω ( | u | p + |∇ u | p ) dν < ∞ , where we denote dν = dµ dt .The definition of the space L ploc ( t , t ; H ,ploc (Ω; µ )) is analogous. OCAL H ¨OLDER CONTINUITY 3
Definition 2.1.
A function u ∈ L ploc ( t , t ; H ,ploc (Ω; µ )) is a weak solution ofequation (1.1) in Ω × ( t , t ) if it satisfies the integral equality Z t t Z Ω (cid:18) |∇ u | p − ∇ u · ∇ φ − u p − ∂φ∂t (cid:19) dν = 0 (2.2)for every φ ∈ C ∞ (Ω × ( t , t )).Next, we recall a few definitions and results from analysis on metric mea-sure spaces. The measure µ is doubling if there is a universal constant D ≥ µ ( B ( x, r )) ≤ D µ ( B ( x, r )) , for every B ( x, r ) ⊂ Ω. Here B ( x, r ) = { y ∈ R d : | y − x | < r } denotes the standard open ball in R d . Let 0 < r < R < ∞ . A simpleiteration of the doubling condition implies that µ ( B ( x, R )) µ ( B ( x, r )) ≤ C (cid:18) Rr (cid:19) d µ , where d µ = log D . A doubling measure in Ω also satisfies the followingannular decay property. There exist constants 0 < α < c ≥ µ ( B ( x, r ) \ B ( x, (1 − δ ) r )) ≤ cδ α µ ( B ( x, r )) , (2.3)for all B ( x, r ) ⊂ Ω and 0 < δ < , p )-Poincar´e inequality if thereexist constants P > τ ≥ − Z B ( x,r ) | u − u B ( x,r ) | dµ ≤ P r − Z B ( x,τr ) |∇ u | p dµ ! /p , for every u ∈ H ,ploc (Ω; µ ) and B ( x, τ r ) ⊂ Ω. Here, we denote u B ( x,r ) = − Z B ( x,r ) u dµ = 1 µ ( B ( x, r )) Z B ( x,r ) u dµ. The word weak refers to the constant τ , that may be strictly greater thanone. In R d with a doubling measure, the weak (1 , p )-Poincar´e inequality withsome τ ≥ , p )-Poincar´e inequality with τ = 1, see Theorem3.4 in [12]. Hence, we may assume that τ = 1.On the other hand, the weak (1 , p )-Poincar´e inequality and the doublingcondition imply a weak ( κ, p )-Sobolev-Poincar´e inequality with κ = d µ pd µ − p , < p < d µ , p, p ≥ d µ , (2.4)where d µ is as above. In other words, Poincar´e and doubling imply theSobolev inequalities. More precisely, there are constants C > τ ′ ≥ (cid:16) − Z B ( x,r ) | u − u B ( x,r ) | κ dµ (cid:17) /κ ≤ Cr − Z B ( x,τ ′ r ) |∇ u | p dµ ! /p , (2.5) KUUSI, SILJANDER AND URBANO for every B ( x, τ ′ r ) ∈ Ω. The constant C depends only on p , D and P .For the proof, we refer to [12]. Again, by Theorem 3.4 in [12] we may take τ ′ = 1 in (2.5).For Sobolev functions with the zero boundary values, we have the fol-lowing version of Sobolev’s inequality. Suppose that u ∈ H ,p ( B ( x, r ); µ ).Then − Z B ( x,r ) | u | κ dµ ! /κ ≤ Cr − Z B ( x,r ) |∇ u | p dµ ! /p . (2.6)For the proof we refer, for example, to [18].Moreover, by a recent result in [16], the weak (1 , p )-Poincar´e inequalityand the doubling condition also imply the (1 , q )-Poincar´e inequality for some q < p , that is − Z B ( x,r ) | u − u B ( x,r ) | dµ ≤ Cr − Z B ( x,r ) |∇ u | q dµ ! /q . (2.7)Consequently, also (2.5) holds with p replaced by q . We also obtain the( q, q )-Poincar´e inequality for some q < p .In the sequel, we shall refer to data as the set of a priori constants p , d , D , and P .Our main result is the following theorem. Theorem 2.8.
Let ≤ p < ∞ and assume that the measure is doubling,supports a weak (1 , p ) -Poincar´e inequality and is non-trivial in the sense thatthe measure of every non-empty open set is strictly positive and the measureof every bounded set is finite. Moreover, let u ≥ be a weak solution ofequation (1.1) in R d . Then u is locally H¨older continuous. We will use the following notation for balls and cylinders, respectively: B ( r ) = B (0 , r )and Q t ( s, r ) = B ( r ) × ( t − s, t ) . For simplicity, we will also denote Q ( s, r ) = Q ( s, r ) = B ( r ) × ( − s, . Recall Harnack’s inequality from [17].
Theorem 2.9.
Let < p < ∞ and suppose that the measure µ is dou-bling and supports a weak (1 , p ) -Poincar´e inequality. Moreover, let u ≥ be a weak solution to (1.1) in R d . Then there exists a constant H = H ( p, d, D , P , ( t − ( s − r p )) /r p ) ≥ such that ess sup Q t ( r p ,r ) u ≤ H ess inf Q s ( r p ,r ) u, where s > t + r p .Proof. See [17]. (cid:3)
In addition, in [17] it is also proved that all solutions of equation (1.1) arelocally bounded. In the sequel, we will assume this knowledge without anyfurther comments.
OCAL H ¨OLDER CONTINUITY 5 Constructing the setting
Our proof is based on the known classical argument of reducing the os-cillation, see [4], [7] and [22]. However, the equation under study has someintrinsic properties which are not present, for instance, in the case of the p -parabolic equation. In large scales, the scaling property dominates andthe oscillation reduction follows easily from Harnack’s inequality. In thiscase, the equation resembles the usual heat equation.In small scales, in turn, the equation changes its behavior to look morelike the evolution p -Laplace equation. Indeed, when we zoom in by reducingthe oscillation, the infimum and the supremum get closer and closer to eachother. Consequently, the weight u p − in the time derivative term starts tobehave like a constant coefficient and we end up with a p -parabolic typebehavior. Resembling this divide between large and small scales, the proofhas to be divided in two cases.We study the (local) H¨older continuity in a compact set K and we choosethe following numbers accordingly. Let µ − ≤ ess inf K u and µ +0 ≥ ess sup K u, and define ω = µ +0 − µ − . Furthermore, choose µ − small enough so that(2 H + 1) µ − ≤ ω (3.1)holds. We will construct an increasing sequence { µ − i } and a decreasingsequence { µ + i } such that µ + i − µ − i = ω i = σ i ω for some 0 < σ <
1. Moreover, these sequences can be chosen so thatess sup Q i u ≤ µ + i and ess inf Q i u ≥ µ − i , for some suitable sequence { Q i } of cubes. Consequently,ess osc Q i u ≤ ω i . The cubes here will be chosen so that their size decreases in a controllableway, from which we can deduce the H¨older continuity. Observe also that if(2 H + 1) µ − j ≤ ω j fails for some j , the above sequences have been chosen so that µ + j µ − j < H + 2 (3.2)for all j ≥ j .We are studying the local H¨older continuity in a compact set K . Our aimis to show that the oscillation around any point in K reduces whenever wesuitably decrease the size of the set where the oscillation is studied. KUUSI, SILJANDER AND URBANO
The next step is to iterate this reduction process. We end up with asequence of cylinders Q i . For all purposes, in the sequel, it is enough tostudy the cylinder Q := Q ( ηr p , r ) instead of the set K . Indeed, for anypoint in K we can build the sequence of suitable cylinders, but since we canalways translate the equation, we can, without loss of generality, restrict thestudy to the origin.The equation (1.1) has its own time geometry too, that we need to respectin the arguments. This is important, especially in small scales, when theequation resembles the evolution p -Laplace equation. We will use a scalingfactor η = 2 λ ( p − in the time direction, where λ ≥ Fundamental estimates
We start the proof of Theorem 2.8 by proving the usual energy estimatein a slightly modified setting, which overcomes the problem that we cannotadd constants to solutions, see [5], [15] and [25]. We introduce the auxiliaryfunction J (( u − k ) ± ) = ± Z u p − k p − (cid:16) ξ / ( p − − k (cid:17) ± dξ = ± ( p − Z uk ( ξ − k ) ± ξ p − dξ =( p − Z ( u − k ) ± ( k ± ξ ) p − ξ dξ. Hence, we have ∂∂t J (( u − k ) ± ) = ± ∂ ( u p − ) ∂t ( u − k ) ± . (4.1)In the sequel, we will also need the following estimates. Clearly, J (( u − k ) + ) = ( p − Z ( u − k ) + ( k + ξ ) p − ξ dξ ≤ p −
12 ( k + ( u − k ) + ) p − ( u − k ) ≤ p − u p − ( u − k ) (4.2)and J (( u − k ) + ) ≥ ( p − k p − Z ( u − k ) + ξ dξ ≥ ( p − k p − ( u − k ) . (4.3)Observe that the assumption p ≥ J (( u − k ) − ) = ( p − Z ( u − k ) − ( k − ξ ) p − ξ dξ ≥ ( p − u p − ( u − k ) − . (4.4) OCAL H ¨OLDER CONTINUITY 7
Moreover, J (( u − k ) − ) = ( p − Z ( u − k ) − ( k − ξ ) p − ξ dξ ≤ ( p − k p − Z ( u − k ) − ξ dξ = ( p − k p − ( u − k ) − . (4.5)Now we are ready for the fundamental energy estimate. Lemma 4.6.
Let u ≥ be a weak solution of (1.1) and let k ≥ . Thenthere exists a constant C = C ( p ) > such that ess sup t Using (4.1) and integrating by parts, ± Z τ τ Z Ω ∂ ( u p − ) ∂t ( u − k ) ± ϕ p dν = Z τ τ Z Ω ∂∂t J (( u − k ) ± ) ϕ p dν = (cid:20)Z Ω J (( u ( x, t ) − k ) ± ) ϕ p ( x, t ) dµ (cid:21) τ t = τ − p Z τ τ Z Ω J (( u − k ) ± ) ϕ p − ∂ϕ∂t dν. So we obtain Z Ω J (( u ( x, τ ) − k ) ± ) ϕ p ( x, τ ) dµ + Z τ τ Z Ω |∇ ( u − k ) ± ϕ | p dν ≤ C Z Ω J (( u ( x, τ ) − k ) ± ) ϕ p ( x, τ ) dµ + C Z τ τ Z Ω ( u − k ) p ± |∇ ϕ | p dν + C Z τ τ Z Ω J (( u − k ) ± ) ϕ p − (cid:18) ∂ϕ∂t (cid:19) + dν. (4.9)Now we can drop the second term from the left hand side, let τ → t , choose τ such that Z Ω J (( u ( x, τ ) − k ) ± ) ϕ p ( x, τ ) dµ ≥ 12 ess sup t Lemma 4.11. Let u ≥ be a weak solution of equation (1.1) . Then thereexists a constant C = C ( p ) > such that ess sup t Proof. Choose φ ± ( u ) = ∂∂u ( ψ ± ( u )) ϕ p in the definition of weak solution and observe that( ψ ± ) ′′ = (1 + ψ ± )( ψ ′± ) . (4.12)The parabolic term will take the form Z t t Z Ω ∂∂t u p − φ ± ( u ) dν = Z t t Z Ω ∂∂t Z u p − k p − φ ± ( s / ( p − ) ds dν = "Z Ω Z u p − k p − φ ± ( s / ( p − ) ds dµ t t = (cid:20) ( p − Z Ω Z uk φ ± ( r ) r p − dr dµ (cid:21) t t . Now an integration by parts gives Z uk φ ± ( r ) r p − dr = Z uk ( ψ ± ( r )) ′ r p − drϕ p = ϕ p (cid:2) ψ ± ( r ) r p − (cid:3) uk − ( p − Z uk ψ ± ( r ) r p − drϕ p = ψ ± ( u ) u p − ϕ p − ( p − Z uk ψ ± ( r ) r p − drϕ p . In the plus case, we have Z uk φ + ( r ) r p − dr ≥ ψ ( u ) u p − ϕ p − ψ ( u )( u p − − k p − ) ϕ p = ( p − ψ ( u ) k p − ϕ p and trivially Z uk φ + ( r ) r p − dr ≤ ψ ( u ) u p − ϕ p , since p ≥ 2. Similar estimates are true also for the minus case.On the other hand, by using (4.12) together with Young’s inequality, weobtain |∇ u | p − ∇ u · ∇ φ ± = |∇ u | p − ∇ u · ∇ (( ψ ± ( u )) ′ ϕ p ) ≥ |∇ u | p (1 + ψ ± )( ψ ′± ) ϕ p − p |∇ u | p − ψ ± ψ ′± ϕ p − |∇ ϕ |≥ |∇ u | p (1 + ψ ± )( ψ ′± ) ϕ p − Cψ ± ( ψ ′± ) − p |∇ ϕ | p , almost everywhere, from which the claim follows. (cid:3) We will need the following notations in the next lemma, which is the mostcrucial part of the argument. Let r n = r r n +1 , Q ± n = B n × T ± n = B ( r n ) × ( t ∗ − γ ± r pn , t ∗ )and A ± n = (cid:8) ( x, t ) ∈ Q ± n : ± u ( x, t ) > ± k ± n (cid:9) , for n = 0 , , , . . . .Recall the definitions µ + i − µ − i = ω i = σ i ω , where µ + i ≥ ess sup Q i u and µ − i ≤ ess inf Q i u , for i ≥ 1. Observe, however, that we have to choose µ +0 ≤ ess sup K u and µ − ≥ ess inf K u, where the infimum and the supremum are taken over K instead of Q . Thisis because we need the argument to be independent of the initial cylinder Q . Now we are ready to prove the fundamental lemma everything dependsupon. Lemma 4.13. Let < ε ± ≤ , ( k + n ) n be an increasing sequence and ( k − n ) n a decreasing sequence, both of nonnegative real numbers. Suppose u ≥ isa weak solution of equation (1.1) , ( u − k ± n ) ± ≤ ε ± ω i and | k ± n +1 − k ± n | ≥ ε ± ω i n +2 . In addition, assume further u ≥ C k − n (4.14) and µ + i ≤ k + n , n = 1 , , . . . (4.15) for the minus and plus cases, respectively. Then there exist constants C − = C ( D , P , C , p ) > and C + = C ( D .P , p ) > such that ν ( A ± n +1 ) ν ( Q ± n +1 ) ≤ C n +1 ± Γ ± (cid:18) ν ( A ± n ) ν ( Q ± n ) (cid:19) − p/κ (4.16) for every n = 0 , , , . . . . Here κ is the Sobolev exponent as in (2.4) and Γ ± = 1 γ ± (cid:18) k ± n ε ± ω i (cid:19) p − γ ± (cid:18) ε ± ω i k ± n (cid:19) p − + 1 ! − p/κ . Proof. Choose the cutoff functions ϕ ± n ∈ C ∞ ( Q ± n ) so that 0 ≤ ϕ ± n ≤ ϕ ± n = 1 in Q ± n +1 and |∇ ϕ ± n | ≤ C n +1 r and (cid:12)(cid:12)(cid:12)(cid:12) ∂ϕ ± n ∂t (cid:12)(cid:12)(cid:12)(cid:12) ≤ C p ( n +1) γ ± r p . (4.17)Denote in short v n = ( u − k n ) ± , k n = k ± n , ε = ε ± and Q n = B n × T n = Q ± n , A n = A ± n , γ = γ ± , ϕ n = ϕ ± n . OCAL H ¨OLDER CONTINUITY 11 By H¨older’s inequality, together with the Sobolev inequality (2.6), weobtain − Z Q n +1 v − p/κ )+ pn dν ≤ ν ( Q n ) ν ( Q n +1 ) − Z Q n v − p/κ )+ pn ϕ p (1 − p/κ )+ pn dν ≤ C − Z T n (cid:18) − Z B n v n ϕ pn dµ (cid:19) − p/κ (cid:18) − Z B n ( v n ϕ n ) κ dµ (cid:19) p/κ dt ≤ Cr p (cid:18) ess sup T n − Z B n v n ϕ pn dµ (cid:19) − p/κ − Z Q n |∇ ( v n ϕ n ) | p dν. (4.18)Here, we applied the doubling property of the measure ν giving ν ( Q n ) ν ( Q n +1 ) ≤ C. We continue by studying the term involving the essential supremum. Bythe assumption u ≥ C k − n and (4.4), we obtain( u − k − n ) − ≤ p − u − p J (( u − k − n ) − ) ≤ C ( k − n ) − p J (( u − k − n ) − ) . On the other hand, the lower bound (4.3) gives immediately( u − k + n ) ≤ C ( k + n ) − p J (( u − k + n ) + ) . Using these estimates together with the energy estimate, Lemma 4.6, yieldsess sup T n − Z B n v n ϕ pn dµ ≤ C ( k n ) − p ess sup T n − Z B n J ( v n ) ϕ pn dµ ≤ C ( k n ) − p γr pn − Z Q n (cid:18) v pn |∇ ϕ n | p + J ( v n ) ϕ p − n (cid:18) ∂ϕ n ∂t (cid:19) + (cid:19) dν. Furthermore, the estimates (4.2) and (4.5) imply J (( u − k n ) + ) ≤ C ( k + n ) p − ( u − k n ) , J (( u − k n ) − ) ≤ C ( k − n ) p − ( u − k n ) − . For the plus case, we used (4.15). Next, using (4.17), we arrive at r pn − Z Q n (cid:18) v pn |∇ ϕ n | p + J ( v n ) ϕ p − n (cid:18) ∂ϕ n ∂t (cid:19) + (cid:19) dν ≤ C np − Z Q n (cid:18) v pn + ( k n ) p − γ v n (cid:19) dν ≤ C np ( εω i ) p γ (cid:18) εω i k n (cid:19) − p ! ν ( A n ) ν ( Q n ) , (4.19)where the last inequality follows from the fact that ( u − k n ) ± ≤ ε ± ω i . Thus,we concludeess sup T n − Z B n v n ϕ pn dµ ≤ C np ( εω i ) γ (cid:18) εω i k n (cid:19) p − + 1 ! ν ( A n ) ν ( Q n ) . (4.20) Furthermore, since − Z Q n |∇ ( v n ϕ n ) | p dν ≤ C − Z Q n |∇ v n | p ϕ pn dν + C − Z Q n v pn |∇ ϕ n | p dν, applying again the energy estimate and (4.19) leads to − Z Q n |∇ ( v n ϕ n ) | p dν ≤ C np ( εω i ) p γ (cid:18) εω i k n (cid:19) − p ! ν ( A n ) ν ( Q n ) . (4.21)To finish the proof, note first that( u − k ± n ) ± χ { ( u − k ± n ) ± > } ≥ ( u − k ± n ) ± χ { ( u − k ± n +1 ) ± > } ≥| k ± n +1 − k ± n |≥ − ( n +2) ε ± ω i . It then follows that − Z Q n +1 v − p/κ )+ pn dν ≥ (cid:16) − ( n +2) εω i (cid:17) − p/κ )+ p ν ( A n +1 ) ν ( Q n +1 ) . (4.22)Inserting estimates (4.20), (4.21), and (4.22) into (4.18) concludes the proof. (cid:3) Remark . If we have the extra knowledge that ( u − k − n ) − = 0, almosteverywhere in B ( r ) at a given time level, we can choose the test functions tobe independent of time and the cylinder Q − n so that the length in the timedirection stays constant, and the bottom of the cylinder stays at the giventime level. In this case, by choosing a time independent test function, theright hand side of the energy estimate simplifies so that we can get rid ofthe term +1 in the formulation andΓ ± = γ ± (cid:18) ε ± ω i k ± n (cid:19) p − ! − p/κ in (4.16). This will get us the required extra room in the end of the firstalternative of Case II.Furthermore, in the previous lemma, we chose the radii of the cylinder as r n = r r n +1 . However, the factor 2 in the denominator can naturally be replaced by anygreater number.We start the proof by considering Case I. Here, we use the previous lemmaonly in the plus case. Consequently, the first case does not depend on theconstant C .We recall a lemma on the fast geometric convergence of sequences from[4]. Lemma 4.24. Let ( Y n ) n be a sequence of positive numbers, satisfying Y n +1 ≤ Cb n Y αn (4.25) where C, b > and α > . Then ( Y n ) n converges to zero as n → ∞ provided Y ≤ C − /α b − α . (4.26) OCAL H ¨OLDER CONTINUITY 13 On several occasions in the sequel, we use this lemma, together with thefundamental estimate Lemma 4.13, to conclude that a ratio of the form Y n = ν ( A ± n ) ν ( Q ± n )converges to zero and consequently that ν ( A ± n ) → n → ∞ . This willultimately lead to a reduction of the oscillation which is our final goal.Once a recursive inequality of type (4.16) has been established, the con-vergence to zero of ν ( A ± n ) follows from the condition ν ( A ± ) ν ( Q ± ) ≤ α ± , with α ± = Γ − / (1 − p/κ ) ± C − / (1 − p/κ )+1 − (1 − p/κ ) ± , (4.27)where the constants C ± and Γ ± are the constants from the previous lemma.Note that an explicit value of α ± only follows after fixing C ± and Γ ± .5. The Case I Now we assume that (3.1) holds. Our aim is to show that the measuresof certain distribution sets tend to zero and that the local H¨older continuityfollows from this.We start by studying the subcylinder Q ( r p , r ) ⊂ Q ( ηr p , r ). Let γ ± = 1, ε ± = 2 − and k + n = µ +0 − ω − ω n +2 . Observe that, after fixing these quantities, the constant α +0 can be fixed aswell.We will study two different alternatives which are considered in the fol-lowing two lemmata, respectively. Lemma 5.1. Let λ > be sufficiently large and let u ≥ be a weak solutionof equation (1.1) . Furthermore, assume ν (cid:18) { ( x, t ) ∈ B ( r ) × ( − r p , − r p λ ) : u ( x, t ) ≥ µ − + ω } (cid:19) = 0 . (5.2) Then there exists a constant σ ∈ (0 , such that ess osc Q (( r ) p , r ) u ≤ σω . Proof. By the choices preceding the statement of this lemma, we have( u − k + n ) + ≤ ε + ω . The assumption (3.1) implies µ +0 = µ − + ω ≤ ω . Thus 1 ≤ k + n ε + ω ≤ . Plug these in Lemma 4.13 to deduce ν ( A + n +1 ) ν ( Q + n +1 ) ≤ C n +1 (cid:18) ν ( A + n ) ν ( Q + n ) (cid:19) − p/κ . On the other hand, by (5.2) we have the trivial estimate ν ( A +0 ) ν ( Q +0 ) ≤ λ ≤ α +0 , choosing λ > ν ( A + n ) ν ( Q + n ) → n → ∞ . This implies ess sup Q (( r ) p , r ) u ≤ µ +0 − ω . So, if this alternative occurs, we choose µ +1 = µ +0 − ω µ − = µ − . These choices yield ess osc Q (( r ) p , r ) u ≤ (cid:18) − (cid:19) ω as required, with σ = 34 . (cid:3) For the second possibility, we have the following lemma. Lemma 5.3. Let u ≥ be a weak solution of equation (1.1) and suppose ν (cid:18) { ( x, t ) ∈ B ( r ) × ( − r p , − r p λ ) : u ( x, t ) ≥ µ − + ω } (cid:19) > . (5.4) Then there exists a constant σ = σ ( H ) ∈ (0 , such that ess osc Q (cid:16)(cid:16) r λ (cid:17) p , r λ (cid:17) u ≤ σω . Proof. By assumption (5.4), we haveess sup B ( r ) × ( − r p , − rpλ ) u ≥ µ − + ω . OCAL H ¨OLDER CONTINUITY 15 Now we can use Harnack’s inequality (Theorem 2.9), together with the CaseI assumption (3.1), to deduceess inf Q (cid:16)(cid:16) r λ (cid:17) p , r λ (cid:17) u ≥ H ess sup B ( r ) × ( − r p , − rpλ ) u ≥ µ − H + ω H ≥ µ − + µ − H − ω H + 1 + ω H ≥ µ − + ω H (2 H + 1) . Observe that the constant H depends on λ , but this does not matter since λ depends only on the data.Now, if we end up in this alternative, we choose µ − = µ − + ω H (2 H + 1)and µ +1 = µ +0 . We also obtain ess osc Q (cid:16)(cid:16) r λ (cid:17) p , r λ (cid:17) u ≤ ω − ω H (2 H + 1) = σω , with σ = 1 − H (2 H + 1) , as required. (cid:3) The case II In Case II the equation looks like the evolution p -Laplace equation. Inthis case, we need to use the scaling factor η in the time geometry of ourcylinders. The difficulty is now that we cannot use the Harnack principleanymore, as the lower bound it gives might be trivial. Indeed, the infimumcan be larger than the lower bound Harnack’s inequality gives. On the otherhand, we have the following kind of elliptic Harnack’s inequality.Suppose that j is the first index for which assumption (3.1) does nothold. Then we have ω j ≤ µ + j ≤ (2 H + 2) µ − j . (6.1)Clearly, this Harnack’s inequality is valid also for every subset of the initialcylinder Q j = Q ( ηr p , r ) and, consequently, for every j ≥ j .Recall, that ω j = σω j − and ω j (2 H + 2) ≤ µ − j ≤ µ − j − + (1 − σ ) ω j − ≤ (2 − σ ) ω j − ≤ − σσ ω j . (6.2) Thus, we obtain σ (2 H + 2)(2 − σ ) ≤ σ − σ µ − j ω j ≤ Q (cid:18) C (2 H + 2) p − ηr p , r (cid:19) ⊂ Q C µ − j ω j ! p − ηr p , r ⊂ Q ( ηr p , r ) , where C = σ p − / (2 − σ ) p − . We will consider the cylinder Q := Q C µ − j ω j ! p − ηr p , r . By the above calculation, we have shrunk the cylinder by a factor which iscontrollable by the data.In the sequel, we will denote θ = C µ − j ω j ! p − . (6.3)Recall the definitions Q ± n = Q t ∗ ( γ ± r pn , r n ) = B n × T n = B ( r n ) × ( t ∗ − γ ± r pn , t ∗ )and A ± n = { ( x, t ) ∈ Q ± n : ± u > ± k ± n } . Now the proof will follow the classical argument of DiBenedetto, see [4]and [22], and is again divided into two alternatives. In the first one, weassume that there is a suitable cylinder for which the set where u is closeto its infimum is very small. In the second alternative, we assume that thisdoes not hold true.6.1. The First Alternative. We first suppose that there exists a constant α ∈ (0 , 1) (to be determined in the course of the next lemma, dependingonly on the data) such that ν (cid:16) { ( x, t ) ∈ Q − : u < µ − j + ω j } (cid:17) ≤ α ν ( Q − ) , for a cylinder Q − = Q t ∗ ( θr p , r ) ⊂ Q (2 λ ( p − θr p , r ) . Our aim is to use Lemma 4.13 to conclude for the reduction of the oscil-lation. Lemma 6.4. For every s > , ν (cid:16)n ( x, t ) ∈ Q (cid:16) θ (cid:16) r (cid:17) p , r (cid:17) : u ( x, t ) < µ − j + ω j s o(cid:17) ≤ C λ ( p − s − s − ν (cid:16) Q (cid:16) θ (cid:16) r (cid:17) p , r (cid:17)(cid:17) , where θ is as in (6.3) . OCAL H ¨OLDER CONTINUITY 17 Proof. We start by using Lemma 4.13, with the choices r n = r r n +1 , k − n = µ − j + ω j ω j n +2 ,ε − = 1 / γ − = θ . We also need the assumption (6.1) to deduce that u ≥ C ( µ − j + 2(2 H + 2) µ − j ) ≥ C ( µ − j + 2 ω j ) ≥ C k − n , (6.5)with C = 3(2 H + 2). This knowledge is needed in Lemma 4.13. Now, afterfixing ε − , γ − , k − n and C we can fix α − , see (4.27).We also obtain, using (6.1) and (6.3), the bounds2 p − C ≤ γ − (cid:18) k − n ε − ω j (cid:19) p − ≤ p − C (2 H + 2) p − and thus we can conclude ν ( A − n +1 ) ν ( Q − n +1 ) ≤ C n +1 (cid:18) ν ( A − n ) ν ( Q − n ) (cid:19) − p/κ . By the assumption of this alternative, together with the lemma of fastgeometric convergence (Lemma 4.24), we have u > k almost everywhere in Q t ∗ ( θ ( r/ p , r/ u − k ) − = 0and consequently ψ ( u ) := (cid:18) ln (cid:18) H − k c + H − k − ( u − k ) − (cid:19)(cid:19) + = 0almost everywhere in Q t ∗ ( θ ( r/ p , r/ t ≤ − θ (cid:16) r (cid:17) p (6.6)be a time level such that this is true for almost every x ∈ B ( r/ k = µ − j + ω j , c = ω j s and H − k = ess sup Q ( u − k ) − , where Q = Q ( ηθr p , r ). Choose ϕ ∈ C ∞ ( B ( r/ ≤ ϕ ≤ ϕ = 1 in B ( r/ 4) and |∇ ϕ | ≤ Cr . In the set { u < µ − j + ω j s } , we have ψ ≥ ( s − ln , and, on the other hand, ψ ≤ ( s − 2) ln 2 and | ψ ′ | − p ≤ (cid:16) ω j (cid:17) p − . The use of these estimates in Lemma 4.11 gives( µ − j ) p − ( s − ln · µ ( { x ∈ B ( r/ 4) : u ( x, t ) < µ − j + ω j s } ) ≤ ess sup t 2) ln 2 (cid:18) λ ω j (cid:19) p − θµ ( B ( r/ ≤ C ( s − 2) ln 2 λ µ − j ! p − µ ( B ( r/ , for almost every t ∈ ( t , η = 2 λ ( p − . The claim follows by integrating this estimate over( − θ ( r/ p , (cid:3) We conclude this alternative with the following two lemmata. Lemma 6.7. Let u ≥ be a weak solution of equation (1.1) and as-sume (6.1) holds. Then u ≥ µ − j + ω j s +1 a.e. in Q (cid:16) θ (cid:16) r (cid:17) p , r (cid:17) , where s depends only upon the data, and θ is as in (6.3) .Proof. Let r n = r r n +3 ,Q − n = B n × T = B ( r n ) × ( t , , where t is, as in the previous lemma, such that( u − k ) − ( x, t ) = 0 , t ≤ − θ (cid:16) r (cid:17) p , for a.e. x ∈ B ( r/ k − n = µ − j + ω j s +1 + ω j s + n +1 . In this case, we obtain( u − k − n ) − ≤ ε − ω j , where ε − = 12 s . Observe also that γ − = − t /r p ≤ ηθ = 2 λ ( p − θ .We will substitute these in Lemma 4.13 and, taking into account Re-mark 4.23 and estimate (6.5), we conclude as before that ν ( A − n +1 ) ν ( Q − n +1 ) ≤ C n +1 λ ( p − s ( p − ! − p/κ (cid:18) ν ( A − n ) ν ( Q − n ) (cid:19) − p/κ . OCAL H ¨OLDER CONTINUITY 19 Now choose s > λ . Then, by Lemma 4.24, we have ν ( A − n ) /ν ( Q − n ) → n → ∞ , provided ν ( A − ) /ν ( Q − ) is small enough. On the other hand, bychoosing s large enough, Lemma 6.4 guarantees that ν ( A − ) /ν ( Q − ) can bechosen to be as small as we please.This gives u ≥ µ − j + ω j s +1 a.e. in Q (cid:16) | t | , r (cid:17) and hence the lemma is proved. (cid:3) Lemma 6.8. There exists < σ < , depending only upon the data, suchthat ess osc Q ( θ ( r ) p , r ) u ≤ σω j . Proof. By the previous lemma,ess inf Q ( θ ( r ) p , r ) u ≥ µ − j + ω j s +1 , for some s > 1, which depends only upon the data and λ . Observe thathere we used the knowledge t ≤ − θ (cid:16) r (cid:17) p . If this alternative occurs, we again choose µ − j +1 := µ − j + ω j s +1 and µ + j +1 = µ + j . Finally, we get ess osc Q ( θ ( r ) p , r ) u ≤ (cid:18) − s +1 (cid:19) ω j as required, with σ = 1 − s +1 . (cid:3) Remark . Here the choice of s is possible only after λ has been deter-mined in the second alternative. Nevertheless, both of them are a prioriconstants which can be assigned explicit values depending only upon thedata.6.2. The Second Alternative. In the second alternative, the assump-tion of the first alternative is not true. In this case, for every cylinder Q t ∗ ( θr p , r ) ⊂ Q ( ηθr p , r ), we have ν (cid:16) { ( x, t ) ∈ Q t ∗ ( θr p , r ) : u ( x, t ) ≥ µ − j + ω j } (cid:17) ν ( Q t ∗ ( θr p , r )) < (1 − α ) , (6.10)where α := α − is the same constant as in the first alternative. This impliesthat, for every t ∗ ∈ ( − ( η − θr p , t with t ∗ − θr p ≤ t ≤ t ∗ − θα r p for which µ (cid:0)(cid:8) x ∈ B ( r ) : u ( x, t ) > k − (cid:9)(cid:1) ≤ − α − α µ ( B ( r )) . (6.11)Indeed, otherwise we would have ν (cid:0)(cid:8) ( x, t ) ∈ Q t ∗ ( θr p , r ) : u ( x, t ) > k − (cid:9)(cid:1) ≥ Z t ∗ − θα r p t ∗ − θr p µ (cid:0)(cid:8) x ∈ B ( r ) : u ( x, t ) > k − (cid:9)(cid:1) dt> (1 − α ) ν ( Q t ∗ ( θr p , r )) , which contradicts (6.10).This alternative is also based on Lemma 4.13. We choose λ in the defi-nition of k + n large enough so that we can force ν ( A +0 ) to be small comparedto ν ( Q +0 ).We start with forwarding the information of (6.11) in time. Lemma 6.12. There exists s ∗ > , depending only upon the data, such that µ (cid:16)n x ∈ B ( r ) : u ( x, t ) > µ + j − ω j s ∗ o(cid:17) ≤ − α − α µ ( B ( r )) . for almost all t ∈ ( t , .Proof. Let c = ω j s + n , k = µ + j − ω j s and H + k = ess sup Q ( u − k ) + , where s and n will be chosen later and Q := Q ( θηr p , r ). Our aim is againto use Lemma 4.11 to forward the information in time. We will need someestimates for doing this.Recall the definition ψ + ( u ) = Ψ( H + k , ( u − k ) + , c ) = ln + (cid:18) H + k c + H + k − ( u − k ) + (cid:19) . Trivially, we have ψ + ( u ) ≤ ln ω j s ω j s + n ! = n ln 2and, on the other hand, in the set { u > l ≡ µ + j − ω j s + n } , we get ψ + ( u ) ≥ ln ω j s ω j s + n + ω j s + n ! = ( n − 1) ln 2The last estimate we need is | ( ψ + ) ′ | − p ≤ (cid:18) c + H + k (cid:19) − p ≤ p − (cid:16) ω j s (cid:17) p − . OCAL H ¨OLDER CONTINUITY 21 Let now ϕ ∈ C ∞ ( B ( r )) be a cutoff function which is independent of timeand has the properties 0 ≤ ϕ ≤ ϕ = 1 in B ((1 − δ ) r ) and |∇ ϕ | ≤ δr , where 0 < δ < n − ln · µ ( { x ∈ B ((1 − δ ) r ) : u ( x, t ) > l } ) ≤ ess sup t Lemma 6.14. For every α ∈ (0 , , there exists λ > such that ν ( E λ ) ν ( Q ( ηθ r p , r )) ≤ α . Proof. Denote h = µ + j − ω j s +1 and k = µ + j − ω j s , where s > v = h − k, u ≥ h,u − k, k < u < h, , u ≤ k. By the previous lemma, we can choose s large enough, namely s ≥ s ∗ , sothat, for almost every t ∈ (cid:0) − ηθ r p , (cid:1) , we have µ ( x ∈ B ( r ) : v ( x, t ) = 0 } ) = µ ( { x ∈ B ( r ) : u ( x, t ) ≤ k } ) ≥ α − α µ ( B ( r ) × { t } ) ≥ α µ ( B ( r )) . OCAL H ¨OLDER CONTINUITY 23 Thus, for almost every t ∈ (cid:0) − ηθ r p , (cid:1) , we obtain v B ( r ) ( t ) = − Z B ( r ) ×{ t } v dµ ≤ (cid:16) − α (cid:17) ( h − k )and, consequently, h − k − v B ( r ) ( t ) ≥ α h − k ) . Using the ( q, q )-Poincar´e inequality for some q < p (see (2.7) and theremark after that), yields( h − k ) q µ ( E s +1 ( t )) ≤ (cid:18) α (cid:19) q Z B ( r ) ×{ t } | v − v B ( r ) ( t ) | q dµ ≤ Cr q Z B ( r ) ×{ t } |∇ v | q dµ = Cr q Z E s ( t ) \ E s +1 ( t ) |∇ u | q dµ, for almost every t ∈ (cid:0) − ηθ r p , (cid:1) . The constant (4 /α ) q above was absorbedinto the constant C . Now we integrate the above inequality over time to get( h − k ) q ν ( E s +1 ) ≤ Cr q Z E s \ E s +1 |∇ u | q dν. Next, we introduce a cutoff function ϕ ∈ C ∞ ( Q ( ηθr p , r )) such that 0 ≤ ϕ ≤ ϕ = 1 in Q (cid:16) ηθ r p , r (cid:17) and |∇ ϕ | ≤ Cr and (cid:12)(cid:12)(cid:12)(cid:12) ∂ϕ∂t (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cηθr p . Now H¨older’s inequality gives( h − k ) q ν ( E s +1 ) ≤ Cr q Z E s \ E s +1 |∇ u | p dν ! q/p ν ( E s \ E s +1 ) − q/p ≤ Cr q Z Q ( ηθr p , r ) |∇ ( u − k ) + | p ϕ p dν ! q/p ν ( E s \ E s +1 ) − q/p . By choosing λ > s ≥ s ∗ in the definition of η large enough, the firstfactor on the right hand side can be estimated by Lemma 4.6 and (4.2) as Z Q ( ηθr p , r ) |∇ ( u − k ) + | p ϕ p dν ≤ C Z Q ( ηθr p , r ) ( u − k ) p + |∇ ϕ | p dν + C Z Q ( ηθr p , r ) J (( u − k ) + ) ϕ p − (cid:12)(cid:12)(cid:12)(cid:12) ∂ϕ∂t (cid:12)(cid:12)(cid:12)(cid:12) dν ≤ Cr p (cid:16) ω j s (cid:17) p − + ( µ + j ) p − ηθ ! Z Q ( ηθr p , r ) ( u − k ) dν ≤ Cr p (cid:16) ω j s (cid:17) p ν (cid:18) Q (cid:18) ηθ r p , r (cid:19)(cid:19) . (6.15) In the last inequality we used the doubling property of the measure ν .We obtain (cid:16) ω j s +1 (cid:17) q ν ( E s +1 ) ≤ C (cid:16) ω j s (cid:17) q ν (cid:18) Q (cid:18) ηθ r p , r (cid:19)(cid:19) q/p ν ( E s \ E s +1 ) − q/p . Finally, summing s over s ∗ , . . . , λ − λ − s ∗ ) ν ( E λ ) p/ ( p − q ) ≤ Cν (cid:18) Q (cid:18) ηθ r p , r (cid:19)(cid:19) q/ ( p − q ) ν (cid:18) Q (cid:18) ηθ r p , r (cid:19)(cid:19) and hence ν ( E λ ) ≤ C ( λ − s ∗ ) ( p − q ) /p ν (cid:18) Q (cid:18) ηθ r p , r (cid:19)(cid:19) . Choosing λ large enough finishes the proof. (cid:3) Lemma 6.16. Suppose that (6.10) holds. Then there exists < σ < ,depending only upon the data, such that ess osc Q ( ηθ r p ,r ) u ≤ σω j . Proof. Let Q + n = B ( r n ) × ( − γ + r pn , ,r n = r r n +1 and A + n as before. Substituting γ + = 2 λ ( p − θ , ε + = 1 / λ and k + n = µ + j − ω j λ +1 − ω j λ + n +1 in Lemma 4.13, and using (6.1) to bound1 ≤ γ + (cid:18) k + n ε + ω j (cid:19) p − ≤ (2 H + 2) p − , yields ν ( A + n +1 ) ν ( Q + n +1 ) ≤ C n +1 (cid:18) ν ( A + n ) ν ( Q + n ) (cid:19) − p/κ . By the previous Lemma, we can choose λ large enough so that ν ( A +0 ) ν ( Q +0 )is as small as we please. 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Partial Differential Equa-tions , 7(4):289–322, 1994. Addresses:T.K.: Aalto University, Institute of Mathematics, P.O. Box 11100, FI-00076Aalto, Finland.E-mail: [email protected] J.S.: Aalto University, Institute of Mathematics, P.O. Box 11100, FI-00076Aalto, Finland.E-mail: [email protected]