c 2014 International Press u+b u+au=0 (1.1)

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1 COMMUN. MATH. SCI. Vol. 2, No. 4, pp c 204 International Press THE HARNACK INEQUALITY FOR SECOND-ORDER ELLIPTIC EQUATIONS WITH DIVERGENCE-FREE DRIFTS MIHAELA IGNATOVA, IGOR KUKAVICA, AND LENYA RYZHIK Abstract. We consider an elliptic equation with a divergence-free drift b. We prove that an inequality of Harnack type holds under the assumption b L n/2+δ where δ>0. As an application we provide a one-sided Liouville s theorem provided that b L n/2+δ R n ). Key words. Harnack inequality, Liouville theorem, regularity, drift-diffusion equations. AMS subject classifications. 35B53, 35B65, 35J5, 35Q35.. Introduction In this paper, we consider elliptic equations of the form u+b u+au=0.) in a domain Ω R n. Here ax) is a given function and bx) is a prescribed divergence free vector field, i.e., divb = 0. The qualitative properties of solutions to elliptic and parabolic equations in divergence form with low regularity of the coefficients have been studied extensively, starting with the classical papers of De Giorgi [5], Nash [2], and Moser []. We are mostly interested in the improved regularity for divergence free drifts b, which arise in fluid dynamics models cf. [, 2, 6, 9, 5, 0, 6]). As can be easily seen from a simple scaling argument, the natural Lebesgue spaces for the coefficients in the equation for the local regularity theory to hold are a L n/2 and b L n, and, indeed, regularity properties of solutions for the case a L n/2+δ, where δ>0, and b L n have been known since the work of Stampacchia [4]. It is well known that a strong divergence free flow may induce better regularity and decay of solutions of elliptic and parabolic problems by means of improved mixing; see, for instance, [4] and references therein. It is also known that a divergence free-drift of critical regularity can still lead to regular solutions [2, 3]. The question we study in this paper is whether the divergence free condition on b allows one to relax the regularity assumptions on b given by Stampacchia. Let us recall some recent results in this direction. In a recent paper [3], Nazarov and Ural tseva significantly relaxed the classical regularity assumptions for divergencefree b by establishing the Harnack inequality and the Liouville theorem for weak solutions to.) if b belongs to a Morrey space Mq n/q with n/2<q n, which lies between L n and BMO. In [6], Friedlander and Vicol proved the Hölder continuity of weak solutions to drift-diffusion equations with a drift in BMO. In [5], Seregin et al. established the Liouville theorem and the Harnack inequality for elliptic and parabolic equations with divergence free drifts b lying in the scale invariant space Received: July 9, 202; accepted in revised form): May 20, 203. Communicated by Alexander Kiselev. Department of Mathematics, Stanford University, Stanford, CA 94305, USA mihaelai@stanford. edu). Department of Mathematics, University of Southern California, Los Angeles, CA 90089, USA kukavica@usc.edu). Department of Mathematics, Stanford University, Stanford, CA 94305, USA ryzhik@stanford. edu). 68

2 682 HARNACK INEQUALITY FOR ELLIPTIC EQUATIONS BMO. All these spaces share the same scaling properties as L n and are thus the natural candidates for good regularity theory. In the present paper, we establish the Harnack inequality and the one-sided Liouville theorem for Lipschitz generalized solutions to.) when ax) and bx) lie in the space L q Ω) with n/2<q n, and b is divergence free. Our results also hold for weak solutions provided that the drift b satisfies certain additional assumptions cf. equation 27) in [3]). More precisely, we establish a Harnack-type inequality sup uy) C inf uy),.2) y B Rx) y B Rx) for all R>0 see Theorem 2.), and use this estimate to establish the one-sided Liouville theorem when a=0 in Theorem 2.3. The constant C in.2) depends on the L q -norms of a and b, where q>n/2, but not on the solution u. Note that the L n/2 -normisnotscaleinvariant: Ifwesetb l x)=/l)bx/l), then b l L n/2=l b L n/2. Because of this, one can not expect the constant C to be independent of R, and, indeed, the constant given explicitly in Remark 2.2 blows up as R 0. The paper is organized as follows. In Section 2, we state our main results, Theorems 2. and 2.3. The proof is based on two auxiliary results, Lemmas 2.4 and 2.5. We first show see Lemma 2.4) that weak solutions of.) are locally bounded by employing the classical Moser iteration technique. Then, in Lemma 2.5, we derive a weak Harnack inequality, the proof of which is inspired by the proof of Han and Lin [8, Theorem 4.5] for elliptic equations without lower-order coefficients. Our main results, Theorems 2. and 2.3, are direct consequences of Lemmas 2.4 and The main results Our first result is the Harnack inequality. Theorem 2. Harnack inequality). Let u be a nonnegative Lipschitz solution to the elliptic equation.) in Ω, that is, j u) j ϕ)+ b j j u)ϕ+ auϕ=0 Ω Ω for any Lipschitz function ϕ 0 in Ω such that ϕ=0 in Ω c. Assume that a L q Ω), b L q Ω) for n/2<q, q n, and that divb=0 in the sense of distributions. Then for any B R Ω we have Ω supu Cinfu. 2.) B R B R Here C is a constant depending on n, q, q, R, a L q B R), and b L q B R). Remark 2.2. From the proof we can deduce that C=Cn,q, q) +R 2 n/q a Lq B R)) /2 n/q) +R n/ q b L q B R)) /2 n/ q) ) Cn)MR, 2.2) where M R =+R 2 n/q a L q B R)+R n/ q b L q B R). Theorem 2. has the following consequence when Ω=R n.

3 M. IGNATOVA, I. KUKAVICA, AND L. RYZHIK 683 Theorem 2.3 One-sided Liouville s theorem). Let ax) 0 and bx) be as in Theorem 2.. Then any nonnegative Lipschitz solution u to the elliptic equation.) in R n is equal to a constant. We note that [3] provides a two-sided Liouville s theorem under the same assumptions, that is, the only solutions of.) that are bounded both from above and from below are constants. However, the one-sided Liouville s theorem in [3] requires b to belong to a Morrey space which is in the same scaling class as L n. Proof. Without loss of generality, we may assume that u is a nonnegative Lipschitz solution to.) with inf R nu=0. Then for every ɛ>0, we have inf BR u ɛ for any sufficiently large ball B R. By Theorem 2., sup BR u Cinf BR u Cɛ for all sufficiently large R > 0. Observe that the constant C given explicitly by 2.2) depends on R but remains bounded as R. Therefore, the assertion is established. Theorem 2. is an immediate consequence of the following two lemmas that compare sup BθR u and inf BθR u to u L p R) with some small p>0 and 0<θ<τ <. Here, for 0<p< we still use the notation though it is not a norm. u L p B)= ) /p u p, B Lemma 2.4. Assume that u is a nonnegative Lipschitz subsolution to the equation u+b u+au=0 2.3) with a L q Ω), b L q Ω) for n/2<q, q n, and divb 0 in the sense of distributions. Then for any B R Ω, p>0, and 0<θ<τ <, ) /2 γ0) ) ) /2 γ0) n/p supu C + R 2 n/q a L q B R) + R n/ q b L q B R) B θr R n/p u Lp R), 2.4) where C=Cn,p, q,θ,τ) is a positive constant and γ 0 0,2) is such that γ 0 =n/ q for n 3 and γ 0 = q/2 /2 )) for n=2 with 2 >2 q/ q ). Lemma 2.5. Assume that u is a nonnegative Lipschitz supersolution to.) satisfying the assumptions of Theorem 2.. Then for any B R Ω and 0<θ<τ < there exists a small positive number p 0 =p 0 n,q, q,θ,τ,r,m R ) such that R u p0 where C=Cn,q, q,θ,τ,r,m R ) is a positive constant and ) /p0 C inf B θr u, 2.5) M R =+R 2 n/q a L q B R)+R n/ q b L q B R). Proof. [Proof of Theorem 2..] We apply Lemmas 2.4 and 2.5, which are valid for any Lipschitz solution u 0 to.) and a small number p=p 0 as in Lemma 2.5.

4 684 HARNACK INEQUALITY FOR ELLIPTIC EQUATIONS The rest of the paper contains the proofs of Lemma 2.4 and Lemma 2.5. Both lemmas are proved using the Moser iteration, with the general strategy based on the proof of the Harnack inequality in [8]. 3. The proof of Lemma 2.4 Let u be a nonnegative Lipschitz subsolution of 2.3) in Ω, i.e., j u) j ϕ)+ b j j u)ϕ+ auϕ 0 3.) Ω Ω for any Lipschitz function ϕ 0 in Ω such that ϕ=0 in Ω c. For simplicity of presentation, we assume a=0. Also, we assume without loss of generality that R=. The proof consists of a priori estimates which can be made rigorous as in [7, 8]. First, we obtain an a priori bound on the L p -norm of u on a smaller ball B r, in terms of an L p2 -norm of u on a larger ball B r2 with r <r 2 but p >p 2. Then an iterative procedure is used to bring the gap between r and r 2 to zero and simultaneously send p to infinity. Let β 0 and ηx) be a Lipschitz cut-off in the ball such that 0 ηx). We use β/2+)u β+ η 2γ as a test function in 3.) to obtain ) ) β β 2 + j u) j u β+ )η 2γ u β+ j u) j η 2γ ) ) β b j u β+ j u)η 2γ ) Let w=u β/2+ so that j w=β/2+)u β/2 j u. By 3.2), we get β+ w 2 η 2γ 2γ w j w)η 2γ j η) b j w j w)η 2γ. 3.3) β/2+ For the first term in the right side we have 2γ w j w)η 2γ j η=γ w 2 η 2γ η+2γ )η 2γ 2 η 2), 3.4) Ω while for the second b j w j w)η 2γ = 2 j b j )w 2 η 2γ +γ b j w 2 η 2γ j η γ b j w 2 η 2γ j η, 3.5) as divb 0. Therefore, we get w 2 η 2γ β/2+ β+ γ + β/2+ β+ γ w 2 η 2γ η+2γ )η 2γ 2 η 2) b j w 2 η 2γ j η. 3.6) Next, set γ 0 =n/ q. Then, as q>n/2, we have γ 0 0,2) and, in addition q +γ γ 0 = 3.7) 2

5 M. IGNATOVA, I. KUKAVICA, AND L. RYZHIK 685 for n 3 where 2 =2n/n 2). If n=2 then we may choose 2 >2 q/ q ) arbitrarily and γ 0 = q/2 /2 )) so that γ 0 0,2) and 3.7) are also satisfied. Let γ=/2 γ 0 ) so that γγ 0 =2γ. Then, by Hölder s inequality we have using 3.7) b j w 2 η 2γ j η b j wη γ γ0 w 2 γ0 j η 3.8) b L q wη γ γ0 L 2 w η /2 γ0) 2 γ0 L 2, as 0 η. By the Gagliardo-Nirenberg inequality, this leads to b j w 2 η 2γ j η 2 wηγ ) 2 L 2+C b 2/2 γ0) w η /2 γ0) 2 L q L2. 3.9) By 3.6), 3.4), and 3.9), we obtain u β/2+ η γ ) 2 C u β+2 η 2γ η +C u β+2 η 2γ 2 η 2 +C b 2/2 γ0) u β/2+ η) /2 γ0) 2 L q L2. 3.0) By Sobolev embedding used in the left side of 3.0), we get u β/2+ η γ L 2χ C /2 u β+2 η η ) 2γ +C u β+2 η 2γ 2 η 2 ) /2 +C b /2 γ0) L q u β/2+ η) /2 γ0) L 2, 3.) where χ=n/n 2) if n 3 and χ>2 is arbitrary if n=2. Now, let η C 0 Ω) be such that η in B ri+, η 0 in B c r i, η C/r i r i+ ), and η C/r i r i+ ) 2. Then we have u β/2+ C L 2χ B ri+ ) u β/2+ L2 B r i r ri ) i+ C + r i r i+ ) /2 γ0) b /2 γ0) L q B ri ) uβ/2+ L2 B ri ). 3.2) The main point of 3.2) is that, since χ>, we have a bound on a higher norm of u on a smaller ball in terms of the lower norm of u on a larger ball. We now apply the estimate 3.2) iteratively on pairs of balls B ri+ B ri, and also let β i +. More precisely, we choose β i =2χ i ) and r i =θ+τ θ)2 i for i=0,,2,..., so that r i r i+ =τ θ)2 i+). We obtain u L 2χ i+ B ri+ ) C2 i+ ) ) C2 i+ /2 γ0) /χ i τ θ + τ θ b L q B ri ) u L2χi B ri ) C /χi 2 i+)/γχi ) τ θ) + ) ) /χ i /2 γ0) τ θ) b L q B ri ) u L 2χ i B ri ), 3.3)

6 686 HARNACK INEQUALITY FOR ELLIPTIC EQUATIONS whereγ =min{2 γ 0,}. Byiteration, lettingi +, weconcludethattheestimate 2.4) holds for p 2. Now, let p 0,2). We have just shown that supu C τ θ) + τ θ) b L q )) ) /2 γ0) n/2 u L 2 ) B θ C τ θ) + τ θ) ) ) /2 γ0) n/2 p/2 b L q ) u L ) u p/2 L p B, 3.4) τ) which leads to supu B θ 2 u L )+C τ θ) + τ θ) /2 γ0) b L q )) ) n/p u L p ). A standard iteration argument cf. [8, Lemma 4.3]) then implies supu C τ θ) + τ θ) b L q )) ) /2 γ0) n/p u Lp ), 3.5) B θ and the proof of Lemma 2.4 is complete. 4. Proof of Lemma 2.5 We assume without loss of generality that R=. The proof is similar in spirit to that of Lemma 2.4: We obtain an a priori bound and use it iteratively. Let u be a nonnegative Lipschitz supersolution to.). We assume without loss of generality that u>0, and consider v=/u. The function v satisfies v+b v av 0 in Ω, 4.) or equivalently j v) j ϕ)+ b j j v)ϕ avϕ 0 4.2) for any function ϕ C 0 Ω) such that ϕ 0 in Ω. By Lemma 2.4, it follows that for any 0<θ<τ < and p>0, we have with C=Cn,p,q, q,τ,θ,m ). Therefore, we have sup B θ v C v Lp ) 4.3) infu ) /p ) /p u p u p u p. 4.4) B θ C We claim that there exists p 0 >0 such that u p0 u p0 C 4.5) with a constant C=Cn,q, q,τ,m ), which would finish the proof of Lemma 2.5.

7 M. IGNATOVA, I. KUKAVICA, AND L. RYZHIK 687 Reduction to an exponential bound. In order to prove 4.5) for some sufficiently small p 0 >0, denote logu) Bτ = logu, and set We shall show that there exists p 0 >0 such that w=logu logu) Bτ. 4.6) e p0 w C, 4.7) where C=Cτ), which in turn implies 4.5). Indeed, if we assume that 4.7) holds, then e p0logu logu)bτ ) C 4.8) and e p0logu logu)bτ ) C. 4.9) Therefore, we have e p0logu)bτ e p0logu C and e B p0logu)bτ τ e p0logu C. Multiplying these two inequalities then leads to 4.5). An L 2 -bound for w. We now prove 4.7). First, we establish bounds on the L 2 -norm of w. The function w satisfies w 2 w+b w+a in. 4.0) Fix τ 0,), and let +τ)/2 r <r 2. Also, let η C0Ω) with 0 η be a cutoff such that η on B r, η 0 on Br c 2, and η C/r 2 r ). Multiplying 4.0) by η 2 and integrating over, we obtain w 2 η 2 2 j w)η j η)+ b j j w)η 2 + aη 2 B 2 η w L 2 η L 2+ b j j w)η 2 + a L q η 2 L q, 4.) where /q+/q =. Integrating by parts in the drift term and using divb=0 gives b j j wη 2 = 2 b j wη j η 2 b L q wη η L q L, 4.2) where / q+/ q =. By the Gagliardo-Nirenberg inequality, we have wη L q C wη α L wη) α 2 L2, 4.3) with α=n/2 n/ q if q 2 and α=0 otherwise. Using Young s inequality and wη) α L 2 C η w α L 2+C w η α L2, 4.4)

8 688 HARNACK INEQUALITY FOR ELLIPTIC EQUATIONS we obtain b j j wη 2 4 η w 2 L 2+C b 2/2 α) L q From 4.) and 4.5), it follows that wη 2 2α)/2 α) L η 2/2 α) 2 L +C b L q wη α L w η α 2 L 2 η L. 4.5) η w 2 L 2 2 η w 2 L 2+ C r 2 r ) 2 C + wη 2 2α)/2 α) r 2 r ) 2/2 α) b 2/2 α) L q L 2 C + r 2 r ) +α b L q w L 2+C a Lq. 4.6) Absorbing the factor η w 2 L 2 and using Young s inequality, we get w 2 L 2 B r ) 2 w 2 L 2 B r2 ) + C r 2 r ) 2+2α b 2L q+c a L q+ C r 2 r ) 2 2 w 2 L 2 B r2 ) + CM r 2 r ) 2+2α, 4.7) where M =+ a Lq )+ b 2 L q ). Now, since w=0, by the Poincaré inequality and 4.7), we obtain w 2 L 2 B r ) 2 w 2 L 2 B r2 ) + CM r 2 r ) 2+2α 4.8) for all +τ)/2 r <r 2, which implies cf. [8, Lemma 4.3]) B +τ)/2 w 2 C τ M, 4.9) where the constant C τ depends on τ 0,). Also, since +τ)/2 τ, we conclude by the Poincaré inequality w 2 C w 2 C τ M. 4.20) B +τ)/2 B +τ)/2 Bounds on the higher norms of w. Next, we need to estimate w β for all β. As in the proof of Lemma 2.4 the idea is to bound first the higher norms of w on smaller balls in terms of the lower norms of w on larger balls and then use the iteration process. We multiply 4.0) by w 2β η 2γ and integrate over in order to obtain w 2β w 2 η 2γ 2β w 2β 2 w w 2 η 2γ +2γ w 2β j w)η 2γ j η)

9 M. IGNATOVA, I. KUKAVICA, AND L. RYZHIK 689 2γ b j w 2β wη 2γ j η)+ a w 2β η 2γ. 2β+ 4.2) Here we utilized divb=0 and j w =w j w/ w. For the first term on the right side of 4.2) we use 2β w 2β 4 w 2β +8β) 2β, 4.22) while for the second 2γ w 2β j w)η 2γ j η) w 2β w 2 η 2γ +Cγ 2 w 2β η 2γ 2 η ) 4 This leads to w 2β w 2 η 2γ C8β) 2β w 2 η 2γ +Cγ 2 w 2β η 2γ 2 η 2 + Cγ b w 2β+ η 2γ η +C a w 2β η 2γ. 4.24) β+ Let τ r<r +τ)/2. We now choose a cutoff η C0Ω) with 0 η such that η on B r, η 0 on BR c, and η C/R r). By 4.7), we have for the first term on the right side of 4.24) 8β) 2β w 2 η 2γ 8β) 2β w 2 C τ 8β) 2β M. 4.25) B +τ)/2 On the other hand, for the left side of 4.24), we use w β+ η γ ) 2 2γ 2 w 2β+2 η 2γ 2 η 2 +2β+) 2 w 2β w 2 η 2γ. 4.26) Hence, we obtain w β+ η γ ) 2 Cγ 2 w 2β+2 η 2γ 2 η 2 +Cβ+) 2 8β) 2β M +Cγ 2 β+) 2 w 2β η 2γ 2 η 2 B +Cγβ+) b w 2β+ η 2γ η +Cβ+) 2 a w 2β η 2γ. 4.27) For the third term on the right side we utilize ) β+) 2 w 2β β+)2β+2 w 2β β+)/β + 8β) 2β + w 2β+2, 4.28) β+ β+)/β which gives Cγ 2 β+) 2 w 2β η 2γ 2 η 2

10 690 HARNACK INEQUALITY FOR ELLIPTIC EQUATIONS C8β) 2β γ 2 η 2γ 2 η 2 +Cγ 2 C8β)2β γ 2 M R r) 2 +Cγ 2 w 2β+2 η 2γ 2 η 2 w 2β+2 η 2γ 2 η 2, 4.29) as M. The last two terms in 4.27) are estimated as follows. First, we have a w 2β η 2γ = a w β+ η γ ) 2β/β+) η 2γ/β+) a L q w β+ η γ 2β/β+) L 2βq /β+), 4.30) where /q+/q =. Now, we use the Gagliardo-Nirenberg inequality w β+ η γ L 2βq /β+) C w β+ η γ α L 2 w β+ η γ ) α L 2 4.3) with α=n/2 n/2βq /β+)) if 2βq /β+) 2, and α=0 otherwise. Note that by the assumption q>n/2 we have α<. If α>0 we obtain by Young s inequality a w 2β η 2γ C a L q w β+ η γ 2 α)β/β+) L w β+ η γ ) 2αβ/β+) 2 L 2 ) β+)/αβ η γ ) 2αβ/β+) 2β+)) 2αβ/β+) w β+ L 2 ) β+)/β α)+). +C 2β+)) 2αβ/β+) a L q w β+ η γ 2 α)β/β+) L 4.32) 2 If α=0 the same inequality holds with the first term on the far right side omitted. As α [0,) and 2αβ/β+) 2, this implies a w 2β η 2γ 2β+)) 2 w β+ η γ ) 2 L 2 +Cβ+) 2α a α L q w β+ η γ α2 L ) Here we denoted α =β+)/β α)+) and α 2 =2β α)/β α)+). Observe that α and α is smaller than a constant independent of β, while 0<α 2 <2 with α 2 2 as β. For the last remaining term in 4.27), we have Cγβ+) b w 2β+ η 2γ η B =Cγβ+) b w β+ η γ) 2β+)/β+) η γ/β+) η. 4.34) Let us choose γ=β+. Then, the above expression becomes Cγβ+) b w β+ η γ) 2β+)/β+) η Cβ+) 2 b L q w β+ η γ 2β+)/β+) L 2β+) q /β+) η L, 4.35)

11 M. IGNATOVA, I. KUKAVICA, AND L. RYZHIK 69 where / q+/ q =. Once again we apply the Gagliardo-Nirenberg inequality w β+ η γ L 2β+) q /β+) C w β+ η γ ᾱ L 2 w β+ η γ ) ᾱl ) with ᾱ=n/2 n/2β+) q /β+)) if 2β+) q /β+) 2 and ᾱ=0 otherwise. Thus, by Young s inequality, we have Cγβ+) b w β+ η γ) 2β+)/β+) η Cβ+) 2 b L q w β+ η γ ᾱ)2β+)/β+) L 2 w β+ η γ ) ᾱ2β+)/β+) L η 2 L 4 w β+ η γ ) 2 L 2+Cβ+)2ᾱ b ᾱ w β+ η γ L q ᾱ2 L. 4.37) 2 R r)ᾱ Here we denoted ᾱ =2β+2)/2β ᾱ)+2 ᾱ) and ᾱ 2 =22β+) ᾱ)/2β ᾱ)+2 ᾱ). Note that, as in 4.3), we have ᾱ and ᾱ is less than a constant independent of β, while 0<ᾱ 2 <2, and ᾱ 2 2 when β. Putting together 4.27), 4.29), 4.3), and 4.37), we obtain w β+ η γ ) 2 L 2 B R) Cβ+)2 R r) 2 w β+ 2 8β) 2β M L 2 B R) +Cβ+)2 R r) 2 +Cβ+) 2α+2 a α L q B R) w β+ α2 L 2 B R) + Cβ+)2ᾱ b ᾱ R r)ᾱ L q B R) w β+ ᾱ2 L 2 B. 4.38) R) Using Sobolev embedding, we may rewrite 4.38) in the form w β+ 2 L 2χ B Cβ+)2κ M α r) w β+ 2 R r)ᾱ+2 L 2 B R) +8β)2β + w β+ α2 L 2 B + w β+ ᾱ2 R) L 2 B R) ) 4.39) where κ=max{α +,ᾱ }, α=max{α,ᾱ /2}, and χ=n/n 2) if n 3 and χ> if n=2. The estimate 4.39) is analogous to 3.2): A higher norm of w on a smaller ball is bounded in terms of a lower norm of w on a larger ball. The case 0 β<. We now show that 4.39) remains valid for 0 β< by considering two separate cases: 0 β</2 and /2 β<. The estimates obtained here are similar to the ones already derived for β, and we point out only the main steps. First, for 0 β</2, we use as a test function w 2 +) β w 2 η 2γ for 4.0) in order to obtain w B 2 w 2 w w w 2 +) βη2γ C B 2 w 2 +) βη2γ +2γ +Cγ w 2 j w w 2 +) βη2γ j η b w 2 +) β+/2 η 2γ j η B + a w 2 +) β w 2 η 2γ 4.40)

12 692 HARNACK INEQUALITY FOR ELLIPTIC EQUATIONS where for the drift term we also utilized divb=0 and b j j w w 2 +) β w 2 η 2γ = b j j φη 2γ = 2γ b j φη 2γ j η 4.4) with φx)= x 0 y2 +) β y 2 dy Cx 2 +) β+/2. Now, since w w 2 +) β for 0< β</2, we have by 4.9) w w B 2 w 2 +) βη2γ w 2 η 2γ CM. 4.42) For the second term on the right side of 4.40), we get 2 2γ w 2 j w w 2 +) βη2γ j η w B 2 w 2 w 2 +) βη2γ +Cγ B 2 w 2 η 2γ 2 η 2 w 2 +) β w 2 B 2 w 2 w 2 +) βη2γ + CM R r) 2, 4.43) where we used w 2 +) β and 4.20). Since w 2 +) β+)/2 η γ ) 2 2γ 2 w 2 +) β+ η 2γ 2 η 2 +2β+) 2 w 2 +) β w 2 w 2 η 2γ, this leads to w 2 +) β+)/2 η γ ) 2 C w 2 +) β+ η 2γ 2 η 2 + CM R r) 2 +C b w 2 +) β+/2 η 2γ j η B +C a w 2 +) β w 2 η 2γ. 4.44) Using estimates similar to 4.33) and 4.37) for the last two terms, and Sobolev s inequality for the left side of 4.44), we obtain w 2 +) β+)/2 2 L 2χ B r) CM α R r)ᾱ+2 w 2 +) β+)/2 2 L 2 B R) + + w 2 +) β+)/2 α2 L 2 B R) + w 2 β+)/2 +) ᾱ2 L 2 B R) ), 4.45) where α, α 2, and ᾱ 2 are defined as above. Hence, we conclude 4.39) for 0<β</2. Finally, if /2 β<, we use as a test function w 2 +ɛ) β w 2 η 2γ for 4.0). Then, we proceed exactly as in the case of β. We conclude 4.39) by letting ɛ 0.

13 M. IGNATOVA, I. KUKAVICA, AND L. RYZHIK 693 The iteration process. Next, we consider the iteration process. Let β i =χ i and r i =τ+ τ)/2 i+ for i=0,,2,... From 4.39), we get w χi 2 L 2χ B ri+ ) Cχ2κi 2 ᾱ+2)i+2) M w α χi 2 L 2 Bri ) +8χi ) 2χi ) + w χi α2 L 2 + w χi ᾱ2 B ri ) L 2 B ri ) Cχ 2κi 2 ᾱ+2)i+2) M α w χi 2 L 2 B ri ) +8χi ) 2χi) 4.46) for all i=0,,2,... For the second inequality in 4.46) we also used α 2,ᾱ 2 2, so that w α2 L p + w 2 Lp and w ᾱ2 L p + w 2 L p. Taking /2χi ) power on both sides of 4.46) gives w L 2χ i+ B ri+ ) C /2χi) χ κi/χi 2 ᾱ+2)i+2)/2χi) M α/2χi ) This leads to the inequality w L 2χ i B ri ) +8χi). 4.47) w L 2χ i+ B ri+ ) CM ) α/2χi) 2χ) κi+2)/χi w L 2χ i B ri ) +8χi) 4.48) for all i=0,,2,..., since α and ᾱ +2 2κ. NotethatifasequenceY i satisfiesy i+ C i Y i +χ i )withc i and i= C i K, then we have by induction i ) Y i C C 2...C i Y 0 + χ j Y K 0 + χi CY 0 +χ i ) 4.49) χ j= for all i=0,,2,... Thus, we obtain w L 2χ i+ B ri+ ) CMCn) CM +χ i+) CM Cn) χ i+ 4.50) for all i=0,,2,..., as i j= j/χj C. Finally, for any β there exists i {0,,2,...} such that Thus, we have w β+ Therefore, for all β, we obtain by taking 2χ i β+ 2χ i+. 4.5) ) /β+) C w L2χi+ B ri+ ) CMCn) β+). 4.52) Bτ p 0 w ) β+) β+)! ) β+) p β+ 0 CM Cn) e 4.53) 2 β+) p 0 = 2CM Cn) e. 4.54)

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