'THS C Date 1,0L lllllllllllWIll!UlHllJlllllllllHllHllllllllHUlllllllllll 01716 3563 This is to certify that the dissertation entitled Regularity of Interfaces for Solutions of the Degenerate Parabolic p-Laplacian Equation presented by Youngsang Ko has been accepted towards fulfillment of the requirements for Ph. D. degree in Mathematics géfim VMajor professor 7/4944? l / MS U is an Affirmative Action/Equal Opportunity Institution 0-12771 LIBRARY Michigan State University PLACE IN RETURN BOX to remove this checkout from your record. TO AVOID FINES return on or before date due. [RTE DUE MTE DUE DATE DUE 18! W14 Cl,“ Regularity of Interfaces for Solutions of the Degenerate Parabolic p—Laplacian Equation By Youngsang Ko A DISSERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1997 ABSTRACT 01’“ Regularity of Interfaces for Solutions of the Degenerate Parabolic p—Laplacian Equation By Youngsang Ko In this dissertation, we study interfaces of the parabolic p-Laplacian equation at = Apu for p > 2. We establish their CL“ regularity after a large time under some suitable conditions on the initial data. It is well known that the solution u and Va 6 C“ for all t > 0, a: 6 RN. Hence Vu = 0 on the interface, which unfortunately does not give information on the interface itself. The key idea is to study a new function u = fin??? which has the same interface as u. The lower and upper bounds for v, and |Vv| can be estimated near the interface by constructing some lower and upper solutions. Consequently some standard theory on strongly parabolic equations can be used for 22. Finally, the Holder continuity of the normal direction of the interface can be obtained by using dilation arguments and some special monotonicity of 2). In addition, some useful and interesting properties of v are also obtained along the way to prove the main theorem. To my family, especially to my mother. iii ACKNOWLEDGMENTS I would like to express my sincere gratitude to Professor Zhengfang Zhou, my dissertation advisor. All of my work has been done under his patient guidance and many valuable suggestions. I would also like to express my appreciation to my committee members Profes- sor D. R. Dunninger, Professor M. Frazier, Professor T. Parker and Professor C. Y. Wang for their time and valuable suggestions. Finally I would like to thank my wife Jin-Sook Choi, daughter Sarah and son Sachan for sharing the difficulties while we were in East Lansing. iv TABLE OF CONTENTS 1 Introduction 2 Preliminaries 3 Nondegeneracy of V1) near the interface 4 Proof of Main Theorem BIBLIOGRAPHY 39 56 78 CHAPTER 1 Introduction Let x = (2:1,...,:1:N) denote any point in R” and let Vu(z, t) = (ux,,...,u$N) be the usual gradient of u with respect to space. For notational simplicity we will use Putz, t) to denote (um, . . . , um”, 21,). We consider the Cauchy problem for the degenerate parabolic p-Laplacian equation (1.1) ut = div(|Vu|p‘2Vu), x E RN,t > 0, with the initial condition (1.2) u(x, 0) = u0(:1:). We will assume throughout this thesis that p > 2 and 110(2):) is nonnegative, integrable, and of compact support. Equation (1.1) appears in a number of ap- plications to describe the evolution of diffusion processes, in particular the flow of non-Newtonian fluids in a porous medium in which it stands for the density [20]. There are, in general, no classical solutions of (1.1) and (1.2) because of the degeneracy of the equation at points where Vu = 0. Nevertheless, it is well known [10] that the Cauchy problem has a unique weak solution u 2 0 satisfying (i) u E C([O,oo), L1(RN)) flL°°(RN x [7, 00)) for any 7' > 0, (ii) u, Vu 6 03,, for some a > 0, (iii) u E C°°(Sl), where Q = {(27, t) 6 R" x (0,oo)|Vu 5f 0}. Solutions of the Cauchy problem (1.1) and (1.2) have an important special property: finite propagation speed. If no has compact support, so does u(., t) for each positive time t. Therefore there is an interface that separates the region where u > 0 from the region where u = 0. Let Q = {($,t) 6 RN x [0, 00) I u(:r:,t) > 0}, and {2(7) 2 Qfl{t = T} C R”. Then the interface is I‘ = aflfflt > 0}. The purpose of this thesis is to study the regularity of this interface I‘. The regularity of I‘ for this equation was studied by [9] and [20]. They proved that the interface is Lipschitz continuous for large times, and is globally Lipschitz if the initial data no satisfies certain nondegeneracy conditions. On the other hand, L.A. Caffarelli and NJ, Wolanski [6] have established the C' 1’“ regularity for the interface of the degenerate porous medium equation uzAum, m>1,$ERN,tE 0,00, (13) t ( ) ( ) u(a:, O) = u0(:r:). This equation is, at least formally, quite similar to (1.1). The idea of this thesis is to adapt the methods of L.A. Caffarelli and NJ, Wolanski to obtain 01’“ regularity of the interface for the p—Laplacian equation (1.1). Establishing C 1’“ regularity is equivalent to showing that the normal direction u(:r, t) of the interface is a 0" function locally. The following important observation from [6] is a key. Proposition 1.1 ([6]) Let u be a function in 82 x (—2, 2) with a Lipschitz interface through (0,0). If there exist positive constants J, S and a sequence of monotone cones K (11],, 6k) in R” x R with axis in the direction of 12k and aperture 8;, such that 9,, _>_ (1 — S)Gk_1 + 5% and DTv(:c,t) 2 0 for (x,t) E BJ:c x (—J",J"),T E K(Vk,9k), then the interface of v is 01*“ near (0,0). The idea of [6] is to start from a solution u of (1.3) and construct a new function u that has the same interface as u such that 1) satisfies the conditions of Proposition 1.1. For the porous medium equation, the flux is given by V(u"‘). Consequently the velocity of the particles is fjwum-l). Naturally the velocity of the particles should tell us how the interface moves as time progresses. It is no surprise that —l a new function u = —u"‘ was used in that case [6]. The interface for u is m m—l obviously the same as that for 22. But 2) has better properties for the study of the regularity of the interface. For example, '0 is Lipschitz continuous while u is not. Furthermore, one can estimate the lower and upper bounds for lel and vt near the interface by using a dilation argument: If v(a:, t) is a solution, so is 711-11013, ht). Also after a suitable dilation, the equation for 1) becomes a nondegenerate equation near the boundary, so some estimates can be derived by the theory of nondegenerate parabolic equations. The most difficult and important step in [7] is obtaining that a positive lower bound on IVvl near the interface. For the parabolic p—Laplacian, the regularity of its solutions is better than solutions of porous medium equation since u and Vu are in C“. One might expect that this alone might give better regularity of the interface for p—Laplacian. But a function f 6 C“ with V f E C“ does not automatically have a 01'“ interface. For example, let g1 and 92 be two continuously differentiable functions on R+ such that 91 (t) > 92(t) for all t > 0, g; (t) > 1 and g§(t) < —1 and g1 (t) is NOT Hiilder continuous. Then min{|$ - 91(t)|1+aa l2: — g2(t)|l+a} if 92(t) < IL‘ < .91“): 0 otherwise f($,t) = has an interface :1: = gl(t) that is C'1 but not CL“. It also should be pointed out that even though the solution it of Equation (1.3) is only C“, its interface is 01’“ for suitable no. It is conjectured in [6] that the interface could even be smooth if 51(0) is a smooth convex set and no satisfies some extra technical conditions. One such example is given by Barenblatt solutions [10]. It was reported in the AMS Detroit meeting recently that C°° regularity of the interface for the porous medium equation has been established by Panagiota Daskalopoulous and Richard Hamilton. For p—Laplacian, we will follow a similar strategy, taking 2) to be the function _ 1 _ (1.4) v(:z:,t) = ;_ 2u5’3%(:r:,t). A straightforward computation shows that — 2 (1.5) v. = p Impv + IVvl”. The motivation for this choice is that equation (1.5) has a nice dilation property and we can establish the lower and upper bounds for v; and |Vv| (this is done in chapter 2). The exponent E comes from the fundamental solution(again, see chapter 2). This function u was also used in [9, 20]. Before we state our main result, let us make a definition to facilitate later statements. DEFINITION 1.1 We say that a function u) : RN -—) R is in the class C" if u) satisfies the following conditions: (i) D = {2: I w(:z:) > O} is a C1 domain and D C BR, where 8;; denotes the ball of radius R centered at 0; (ii) w 6 01(5),- (iii) There exist a strip S C D along the boundary 8D and positive constants 1:1 and [:2 such that (61$ leIS [:2 in S; (iv) There exists a constant a > 0 such that w 2 a in D\S; (1)) There exists a constant 1:0 > 0 such that (3,,w) Z —k01 in the sense of distributions. The main result of this thesis is the following theorem. Bi Theorem 1.2 Let v0 = fluff—l E C”, then there exists To 2 0 depending only on p, N and the constants in C“ with the following properties: (i) u(x, t) > 0 for all (x,t) E BI+R x [T0,oo); and (ii) For any To < t < t0 < t, there is an a > 0 depending only on p, N, Li and constants in C" such that the interface near the point (x0, to) E I‘ is 01’“. The thesis will be organized as follows. In Chapter 2, we will state and prove some preliminary results on the interface I‘ by using a comparison principle. In particular, we will prove the Lipschitz continuity of I‘. In Chapter 3, we establish lower bounds for at and |Vv| near the boundary by carefully studying the equation for v and by using the general theory on degenerate parabolic equations. Finally, Theorem 1.2 is proved in Chapter 4 by using the homogeneous structure of v and by constructing some lower and upper solutions. CHAPTER 2 Preliminaries In this chapter we state some preliminary results which will be useful later. Most of these were established in [9], [20]. Here we will summarize the relevant results and sketch some proofs; complete details can be found in [9, 20]. First, it is well known [20] that equation (1.1) has the following comparison principle. Proposition 2.1 Let ul and u2 be two nonnegative solutions of (1.1) on ST, where ST = R” x (0,T) with initial values cm, W 6 L1(RN) respectively. Then (pl 3 902 on R” implies ul S U2 on ST. PROOF. Let (pf = (p,- * p5, i = 1, 2, where p5 is a convolution kernel whose support is the ball of center 0 and radius 6, and let uf, i = 1,2 be the solutions of the following approximate problems of (1.1),(1.2) ut = div((|Vu|2 + 6)L;_2Vu), u(x, 0) = (pf. Then by the comparison principle for strongly parabolic equations, we have (2.1) u? S ug in ST. By the uniqueness of solutions of (1.1),(1.2) u1 2 Pin) u‘ls, U2 = (1531(1) ug. Hence Proposition 2.1 follows by letting 6 —> O in (2.1). Remark 2.1 The conclusion of Proposition 2.1 can be extended to a domain of the form Q X (0,T) with 052 being smooth. There are two types of solutions which are frequently used for comparison functions for (1.1). One is a separable solution ATP 1 1w (2.2) u(x,t) = 9(r)f(t) = #fim , for any T > 0, where r = |x| is the usual norn in R” , and 1 2.3 A = ( ) N (p - 2) + p- The other is the Barenblatt solution x — x _ _ (2.4) Bk,p(x,t,§:,fl = kpN[s(t)]—AN[1 _ (L__l)p/(t — i), t2 t. One can check easily that Bk,p(x, t, x, D = M6(x — x) where 6 (x — x) as the Delta measure concentrated at 2‘: and M = [RN Bk,p(x,t,x,fldx. The next proposition gives the monotonicity of the support {2(t). Proposition 2.2 The solution of (1.1) and (1.2) satisfies (2-5) “t Z — (p _ 2)t in the sense of distributions. PROOF. Let u,(x, t) = ru(x,r”'2t). Then u, is a solution of (1.1) with initial value ur(x,0) = ru(x,0) = ru0(x). Since u,(x,0) S u0(x) for r 6 (0,1), by Proposition 2.1, we have u,(x, t) g u(x, t). It follows that u(x, rp"2t) — u(x, t) g (1 — r)u(x, rp_2t) and therefore u(x, rp‘zt) — u(x, t) r — 1 > —— p-zt . (av-2 — 1)t - (1 — rP'2)tu($’T ) Letting r —> 1', we get lim u(x,t+ s) — u(x,t) _ —u(x,t). 3—)0- s (p — 2)t Similarly, we can choose r > 1 and r -> 1+ to obtain lim u(x, t + s) — u(x,t) Z -u(x,t) 340+ s (p — 2)t and the proposition is proved. As a consequence of (2.5), the function utiv+27 is non-decreasing for every x 6 RN. Hence u(x,to) > 0 and t1 > t0 imply u(x, t1) > 0. Thus the support Q(t) is monotonically increasing: Q(t°) C 52(t1) whenever t1 > to. We also have the following monotonicity prOperty for solutions of (1.1) and (1.2) for any fixed time t. The version for the porous medium equation was proved in [7]. Proposition 2.3 Let R0 be a positive number such that 9(0) C BRO. Then (1) For x1,x2 6 R” with |x1|,|x2| > R0 and (2.6) cos(x2 — x1,xl) 2 u(x2, t) S u(xl, t) for any positive t, where (x2 —x1, x1) is the angle from the vector x2 — x1 to x1; 11 (2) For every r,t > 0 we have (2.7) iIIBIfu(x, t) Z sup u(x,t). aBr+2Ro The proof of Proposition 2.3 requires the following Lemma. Lemma 2.4 Let D = suppuo(x) be compact and D C {x E RNIxN > 0}. Then u(y,-TN) t) Z ’U.(g, —$Na t) for all y 6 Riv—13 (EN > 07 LE (0,T) PROOF- Set ”(130 = u(y, “$N,t). Then v(y, 0, t) = u(y,0, t) for each y 6 RN-1 and t E (0, T), and v(y,xN,0) = u(y, —xN,0) = 0 S u(y,x~,0) for each y 6 RN—1 and xN > 0. Hence by Remark 2.1, we obtain u(y, —xN,t) = v(y,xN,t) g u(y,xN,t) in RN—1 x R+ x (0, T). Proof of Proposition 2.3. Let H denote the hyperplane in R” which bisects the line segment between x1 and x2 orthogonally. That is, H = {x e RN|($,$1- x2): ($(w1+x2,x‘>— 252)}, 12 where (-, ) denotes the usual scalar product in R”. It is easy to verify that 1 |(X1,X1 - X2) + (X2,Xl - X2)I 2 IX2 —x1| dist(H,0) = Note that by (2.6), (x1,x2 — x1) = |xl||x2 — xllcos(xl,x2 - x1) 2 Rolx2 — x1 and (x2,x2 — x1) 2 (x1,x2 — x1) > Rolx2 — x1, we have dist(H,0) > R0. Therefore x1 and suppuo (x) are in the same half-space with respect to H. Moreover, x2 is the reflection of x1 with respect to H. Thus by Lemma 2.4 and a suitable translation and a rotation, u(xl, t) Z u(x2, t). For the second part, if x1 6 B, and x2 E 63...sz then by the same argument as above we have 1 (r + 2R0)2 — r2 2 2(r + R0) dist(H, 0) Z 2 R0. Hence the same argument gives u(xl, t) Z u(x2, t). 13 Since x1 and x2 are arbitrary, (2.7) is established. It follows immediately from (2.6) that u(-,t) is non-increasing along the ray {x = /\x1, /\ > 1} if |x1| > R0, and the free boundary F can be represented by a spherical coordinates in the form r = f(0,t) for 0 6 SN‘1 r > R0. It was also shown in [9] and [20] that the initial behavior of the interface is determined by the local properties of no as follows. Pr0position 2.5 Let (2.8) B(x) = glam—”’55 [W nanny) Then we have u(x, t) > 0 for everyt > 0 if and only if B(x) = 00. Moreover, there exists a constant C = C(p, N) > 0 such that u(x, t) = 0 if 0 < t < C[B(x)]2"’. PROOF. Suppose B(x) = 00. From the Harnack principle (see Corollary 1 in [11]) we have that N(E—2)+g R—N—Fg‘i 812(3) u0(x)dx S C(t-fi5 + tFR—N_F§7[u(x,t)] v ). 14 Hence, if u(x, t) = 0 for some t > 0 then B (a?) S Cit—9+3; this contradicts B (x) = 00. Now assume B(x) < 00. From Theorem 1 in [12] we know that sup u(x, t) 3 ct" Nir§25+r p357 [13(3)] W p for all p > R > 0 and 0 < t < T(R), where In($) = salvo—”‘3'33 / Uo(y)dy P>R 39(1) and TR 2 c[B(x)]—”+2. Taking R —) 0 we are done. In order to estimate the growth of the interface, we define the function (2.9) d(x) = sup{r > 0| [3 ( )uo(y)dy = 0}, i.e., d(x) measures the distance from x to the initial support of u. Since B(x) S Inolrx(n~)(d($))(—N+35), Proposition 2.5 yields 15 Corollary 2.6 Let x E R” with d(x) > 0. Then “(23, t) = 0 if 0 s t g Cl(d(x))N(P-2)+pluOli:p, where C1 is a positive constant depending only on N and p. Using Corollary 2.6, we see that if suppuo C BRO, then for every t > 0, Q(t) is contained in the ball BR“) with (2.10) 12(1) 3 02011011721) 7—” ,3. +» + R), which gives the upper bound of (2(t). It turns out that this gives a very accurate information on the size of 9(t). We set (2.11) RM(t) = sup{|x| | x E Q(t)}, R4,,(t) = inf{|x| | x 6 89(0}. Proposition 2.7 For the solution of (1.1)-(1.2), we have (2-12) R140) S Rm(t)+2Ro, (2.13) RM(t) ~ C(p,N,uo)t-7—>—~p-‘2+p, as t—>oo. where C(p, N, uo) is a constant depending only on p, N and |u0|L1. PROOF. Clearly the first inequality follows from part (2) of PrOposition 2.3. Let V(z. t, C) = Waco, N){C — (Int—tom? 16 where C > 0 is a constant, a = W, 00(1), N) = (p; 21%[No - 2) win—1‘2. Then V(x, t, C) is a solution of (1.1) from (2.4). Without loss of generality, we may assume that uo(0) > 0. Hence we can choose to, 01 > 0 such that ”l _NL —L 2 to "00091 N){C'1 — (lxlto p)?" l? 3 110(37)- It follows from Proposition 2.1 that 2;; u(x,t) 2 (t+to>-fi'co(p,1v){cl- (|$|(t+t0)_"+“);fi}i-2- This implies the second part of Proposition 2.7. Proposition 2.7 shows that if t is large, 9(t) contains a ball of center 0 and radius of the order t", A = Thus there exists a time t(uo) < 00 such that 1 N (p-2)+p° for t > t(u0) we have R c 9(t). Let T0 = inf{t > 0 | "BE; c Q(t)}. Then we have the following facts which were proven in [9] and [20]. Proposition 2.8 The interface F can be represented as r=f(0,t) for t>T0, OESN"1, 17 where f is a Lipschitz continuous function of 0 and t. PROOF. First we can show that f is Lipschitz continuous with respect to 6. We know that there exists a large To such that B Re (0) C Q“) for all t > To. Then it follows from Pr0position 2.3 that for every (x, t) E F, [El > R > R0, t > To, we have u(x, t) > 0 for every x in a small cone K. of the form (1 + €)R l'fl } K6 = {x: Ix—x'l < e and cos(x—x,x) g — and u(x, t) = 0 in a cone K; of the form 5 In} K;={x:[x—x| <6 and cos(x—x,x) 2 Therefore, if (x,t) E I‘ and [x — ‘xl < e, we have x 6 RN — (Ke U K;). This implies that f is Lipschitz continuous in 6. To prove that r = f (6, t) is Lipschitz continuous with respect to t, we need the following estimates which were proven in [9], [20]. Lemma 2.9 There exists a constant C(R0,u0,To) > 0 such that _!&+_1 t m. [VU(CC, t)l S C(ROa ”01T0) if I22! 3 Rmt 2 To, where u = with 18 PROOF. We define a family of functions uk(x, t) = kNuch, kN"t) k > 0. Then uk is a solution of (1.1) with initial value uk(x,0) = kNuOch). From the proof of Proposition 2.7, there exist constants C1, C2, T 0 > 0 such that [7. it}? u(x,t) s (t + Toricupavncz — (1qu + Torn) “($1102 (15+ To)—%Co(p,N){C1-(|x|(t+ Tolv’l’l-‘Tlglé‘l1i It follows that 1 1 L‘.‘ u(x, t) S k”(k”"t + To)‘ICo(p, N){02 — (klxl(k”"t + Tormfiw l 1 rd uk(:L‘,t) Z kN(kNut + T0)—;CO(P,N){C'1-(k[x[(k~”t+ T0)—N—”)PET}i—2, Thus there exist t2 > t1 > O,a,)6 > 0 independent of k such that if t1 3 t 3 t2, |x| 3 R), then a S U): S 3- Hence by the interior estimates, we have IVukISC z'f tIStStm lxlSRo 19 for some constant C independent of k. This means that IVu(x.k""t1)l s Ole—(NH) if le 3 mo. Putting t = kN"t1, we have men s Gigi—ii 2'1 m s [grit—m... This completes the proof. By using the same family of solutions as in Lemma 2.9, we have the following estimate of supru| in terms of u after a large time. Corollary 2.10 For every 6 > 0 there exists T1 2 T1 (e, R0, uo) such that |Vu(x,t)| S eu(x,t) for all |x| S R0 andt 2 T1. PROOF. We consider again the family of solutions uk(x, t) = kNuUtx, kN“t) as in Lemma 2.9. We know that u), converge to a fundamental solution "a uniformly with respect to t Z r for every T(see Theorem 3 in [21]). Therefore there exists k0 such that 1 c > 2a(:1:,t) Z uk(x,t) Z u(x,t) > E > 0 [\DIi—t 20 for |x| S 2R0, % S t S 2, k 2 kg, and for some 0 depending only on no, N and p. Since u), are uniformly bounded in R” x (%, 2), we obtain [Vukl S CI for t = 1, |x| S R0, and for some cl, and this implies [Vu(y, 19”") S cit-(NH) for all |y| < R0. Note that u(x, kN") 2 ck'N for all |x| S R0. Setting t = k”“ we conclude that IVu(x, t)| S ct—TVIFu(x, t) if [x[ S R0,t 2 k”", and k 2 k0. This completes the proof. Next Lemma gives an estimate on ut. Lemma 2.11 There exists a time to > 0 such that ut E L°°(RN x (r, 00)) for r > to and on RN X (to, 00) (2.14) 2(t — t0)ut(x, t) + x - Vu(x, t) — u(x, t) S 0. PROOF. We consider the approximate problem of (1.1),(1.2) (2.15) 11. = div(([Vu[2 + a)’€—’Vu) (2.16) u(x, 0) = (p. 21 It is well known that (2.15),(2.16) has a solution u5 with the property that for any 3 > 0, u5 E C°°(RN x (s, 00)) {]L°°(RN x (s, 00)) and |u5(x,t) -— u(x,t)[C1(K) —) 0 as 6 —) 0, for every compact subset K of ST. For every 6 > 0 we define a family of solutions of (2.15) u6(x, t) = u6((1+e)x,(1 +e)2t+to), 6 1+6 where to is a large time to be chosen below. We want show first that for every 6 > 0 and x E R” uf($10) S ”3:0(110): u6($,to), if to is large enough. For this purpose we write the difference uf(x, 0) — u5 (x, to) as uf(x, 0) — u5(x,t0) 2 ((1+ e)”1—1)u6((1 + e)x, to) + u‘5((1 + e)x,to), (2.17) —u6(x,to) = — u6((1 + e)x, to) + u6((1+ €)x, to) — u5(x, to). 1+6 Notice that Proposition 2.3 still holds for the solution u". We have u6((1 + 6)x,to) S u‘s(x,to) if |x| > R0. 22 Since u6 2 0, we conclude that ui(x,0) Su6($,to) if le 2120. In case |x| < R0, by Lemma 2.9 for to large enough we have C > u(x,to) > %, [Vul S 25120 in 330(0)- It follows that there exists a constant C1 such that C1 > u6(x,to) > i, [Vu5| S 1 in 830(0) C1 ZClRo provided that 6 is small enough. Therefore, since 6 Iu"((1+ onto) — “6(331toll s elxIIVu“(€,to)I s 55:, we obtain from (2.17) uf(x, 0) S 115:0(2, 0). Hence by the maximum principle, we conclude that u6(x, t) S uf=0(x,t) for x 6 RN, t > to. 6 Differentiation of uf (x, t) with respect to e at e = 0 gives (2.18) u6(x,t + to) —- 2tuf(x,t + to) — x . Vu5(x,t + to) 2 0. 23 If we let 6 —> 0, then we obtain (2.14) for u. Combining (2.14) and Proposition 2.7, we obtain ut E L°°(RN x (7', 00)) for any 7' > to. Now we are ready to prove the remaining part of Proposition 2.8. PROOF. We have already proved in that I‘ can be represented as T=f(91t) if t>T01 and f is a Lipschitz continuous function of the variable 0. Thus it remains to prove that f is Lipschitz continuous in t uniformly in 0 E S N ‘1. Notice that by Proposition 2.3, x . Vu‘s = rug < 0 if le > R0 and that w; 2 0. By (2.18), we have d u6(r,0,t) 1 5 6 5 _ —_ <.—.— 2 —T — < dS( t_T0 )—2(t_T0)(rur+ (t 0)ut u)-—O’ where r = mes-‘1, r1 > R0, t = To + (t1 — To)e""1, s 2 t1 > To and u6 is the solution of (2.15), (2.16). It follows that u6(r, 0, t) < u5(r1, 0, t1) i t>t t—To '- tl—To f 1 and u(r, 0, t) < u(r1,0,t1) t—To ‘ t1 —To 2f t>t1. 24 Hence if u(r1,0,t1) = 0, we get t-To t—To f(91t)S7'1€$—tl =7'1t1_T0 — i.e., f(0,t) - f(6.t1) _<. no, 1,): t1 — if t>t1>T>T0. 1 _ Finally we define a family of solutions to (1.1) in RN x (To, 00): u.(x, 1) = ———L—,___Tu((1 + at, (1 + e)t + T) (1+ cw for e > 0 and T > To. We want to show that for every 6 6 (0,1) and x 6 B”, 11603.0) E u(SET)- To do this we write the difference u£(x, 0) — u(x, T) as _—1E —1]u((1+ e)x,T) + u((l + (5)5137")- u(x,T). uex, —ux,T= < 0) ( > 0+6)!” If |x| > R0, then from Proposition 2.3, u((l + e)x,T) — u(x,T) S 0 and — I] u((l + e)x,T) S 0. 25 Now we consider the case [xl S R0. From Corollary 2.10, we have u((1+ aw) — u(x,T) s elwlqu(€x,T)| _<. 6302—23743th 3 finger) with some constant c, where E E (1, 1 + 6). Hence we have u€(x,0) — u(x,T) S 0 6 for all e 6 (0,1), and differentiating u,(x, t) with respect to e we have p__2u(x,t+T)-i-x-Vu(x,t+T)+tu¢(x,t+T) S 0. Replacing t by t + T we obtain (2.19) (t — T)ut(x, t) S p _ ;u(x, t) — x - Vu(x, t). Hence ut is bounded. Now if h is a fixed positive constant such that t — T 2 h, (2.19) can be written in the form d _ [g—lgt t—t d_t[e P-2>hu(roe_’-"L,6,t)] S 0 for T + 1 2 t > t1 > T and 9 fixed. Therefore, if u(ro, 0, t1) = 0, then u(roe%,0,t)=0 for t>t1. 26 This gives f(62t) S f(01 t1)e-t:h_tl, andwegetforT 0. Furthermore the following was established in [9], [20]. Proposition 2.12 There exist constants A, B > 0 such that A—p p—l (2.20) v(x, t) + x - Vv(x, t) + (At + B)vt 2 0 in the sense of distributions. PROOF. We consider a family of solutions (2.21) v‘(x, t) = (1 + ’46)}: v((1 + e)x, (1 + Ae)t + Be). (1+ c)P-1 The idea is to show that v‘(x, t) Z v(x, t) 27 for small 6, and then by differentiating v‘ with respect to e we have 24-19 p—l v(x,t) +x . Vv + (At + B)vt Z 0. We approximate v0 by v3 = v0 :1: p5(x) + 6““, where p5(x) is a convolution kernel and a will be chosen later. v5(x, t) = fiflu" (x, t))5_:% is the solution to — 2 vf : :_ 1v‘sdiv(|Vv‘5|p_2Vv‘s) + [Vv‘5|p with v5(x,0) = v3(x). We note that v‘5 2 6" > 0 and v5 E C°°. Suppose that If there is no confusion, we omit 6 in various expressions. If 6 is sufficiently small, then v(x,0)2 in 9129\5 NIQ and |Vv(x,0)|2% in S. In fact, this inequality is also true in a neighborhood U65 of 09 of the form U... = {x e RNIdist(x, on) < cd} for a constant c E (0, é). Now we consider several different regions. 28 (i) First we consider the region where IVvol > 541(in particular, S U U65). We have (v‘(x, 0) — v(x, 0)) _ [(1 + EA —p) v((1 +e)x, Be) — v(x,0) i—tmlr—I IV e 2p—1 if e is small. Hence from the mean value theorem, v((1+ e)x, Be) + th((1 + e)x, 0e) + x - Vv({, 0), where 0 6 (0, B) and (5 lies in the line segment from x to (1 + e)x. If e is small enough, then v((1 + e)x, Be) E v(x,0), vt((1 + e)x, 0e) E vt(x, 0), Vv(§, 0) ’—_‘-’ Vv(x,0). Therefore, there exists c > 0 depending only on N, p and v3 such that I.>-1—A_p “2p—1 v(x, 0) + th(x, 0) + x ' Vv(x, 0) — ce for some c. Using the equation — 2 p 1vApv + IVvl”, 29 we get 1 A 2 > (2p—:_l— + §__1BAPU($’ 0)) v(x, 0) + BIVv(x, 0)|” + x - Vv(x,0) - ce Since |x| S R + 6 and Avg 2 —Nko, we have 1A p— 2 > ________ P‘l_ _. _ . I (219:1) 1 p- 1BNko) v(x, 0) + IVv(x,0)| (Blel R 6) ce If we choose A and B such that B>iR- and lU-E—Z-BNk0>O, [cf—1 2p—1 p-l then k1 Bid,” >_ _ _ _ I._ 4( 4 R 6) ce>0 for small e and 6. (ii) Next, we consider the region {21 = Q \ S. We only need to consider those points where IVvo($)l s % In this case we see that a I.> — — BNk ———— , —(2p—1 p—l “)2 4 66 30 . A— _ and If W21; — fiBN/Co 2 Rkl/a, then 1520 for small e. (iii) Next we consider the region 523 = {x E R” | dist(x, Q) 2 6}. In Q3, v(x, 0) = 6". Since v 2 6“ from the maximum principle, we have IE _>_ 0. (iv) Finally, we consider the region 04 = {x 6 RN |c6 S dist(x,fl) S 6} with 0 < c < 1. In this case we select a particular family of cutoff functions {p5} satisfying pa($)=0 z'f |x| 22 p100) = 21(0) if livl S <5 - 51“ for some 7 6 (0,1), 0 S P5017) S {96(0), .06 E C°°~ Now suppose dist(x, Q) E (6 — 61”, 6); then IVv3($)l |/\ [am a IVvo(y)|pa(x - y)dy S Ckz/B no p5(x — y)dy. 6 31 Now we observe that / pm - My 5 «26+ 3509 and hence |va(x)| g ckgle—ii. Thus 16 > lg — ?'——2.BNko 6° - (R + 26)]C267L5L1 '- CC, 2 p — 1 p — 1 In particular, if 0 < < 7 < 1 and e is small, then N2+1 IC 2 0. Finally, we consider those points x such that c6 S dist(x,Q) S 6 — 6”“. Recall that V2110 Z —k01 in the sense of distributions. We know that 1A— p p— 2 Ie Z 21)- 1 +p—1 BIVvI” 2a,-J-(Vv)vx,zj v5 +B|Vv(x,0)|p — |x||Vv5(x,0)| — ce, 32 where vain,- Cit-AV”) = 511' + (P - mm- Since v0 6 W1’1(S), we have that IVU($10)| = If moms — may Bang 2 1., f 105(13 — may Ban!) 2 (3,9167%. Also since V2v(x, 0) Z —kOI, we have A =/ A — d —/ v . — d ’0 Hm vo(y)pa($ yly 6(an)( vo 1010.106 may and A > k/ — d —Nk/ — I ”l .. 18606010de 3!) 0y o B 105(23 yldy 500 _ 1(N—1) 7(N+l 1+ 2 — Nk06 2 IV 1:16 Observe that at x E (90, Vvo = cu. Hence we obtain vmg ij (p — 2) W, [m vow/w — 11m... dy 33 vxgvxj = (P — WW [8609(Uolx.-x,-PJ($ — y)dy _ _ Elia] . _ (P 2)[V’U[2 Banan(v0)xiVJp6(x 3/)de 2 —Ck06’7-N_&Ll for some c. Therefore, combining all these together, we obtain g:iBIVvlp_2aij(Vv)vx.-xj”($1 O) _ (R + 25)[V’Ul IV :1;ch [Ratliff—2 (kla-HJ—H ”2" — snarl?) 6" —(R + 26)? — ce. Hence for sufficiently small 6, we have Mb 2 +2 I6 2 Bckf-16_l+a+7( )+ (R + 26¢;l — ce. Therefore if we choose 0 < a < 1 and 7 so that 2a 1—a 1+N<7 To, then A B 3 f(6’, t) 2 f (0. To) (Eli—0%) 35 PROOF. The curves (x(s), t(s)) above can be written as Hence if 0 = 39L and r0 = |x(0)|, then v(ro, 0, To) > 0 implies Ix(0)| At+B 7‘1 W((m) 1“)” Therefore we obtain At + B )% NM) 2 7’0 (m for any r0 such that v(ro, 0, To) > 0. Consequently At+B )% f(9,t) 2 f(9,To) (m Proposition 2.15 Let v0 6 C“. Then for every point (x,t) 6 ST there exist posi- tive constants A, B, C > 0 depending only on v0, N, p, t and R1 = sup{dist(x, y)|y E D} such that (2.23) v(x,t) 2 v(x,t)fcu'J) for every (x, t) satisfying t < t < t+ e for some e = e(A, B) and s | 1' £0 I: l 51' 36 PROOF. Let us take A, B and C = ’13—} as in Proposition 2.12 with R = 3R1. Let We first prove that for every (x, t) E U, there exists a point y 6 RN such that z-y=<:;:z>* and Q is contained in the ball BR(y). Clearly, for every (x, t) there exists a point y 6 R” such that (2.24) holds. We now prove that D C BR(y). By (2.24), if (x, t) E U then we have If- |(At+B)% < Ix—mI-x— l y AZ+B — y _ R1At+B+lx—yl Hence we have _ At+B % t—t |$_y|[(Ai+B) _1 —R1AZ+B 1 Writing (%)x as a power of t — t and using the fact that e is small, we get 1|? I t—t < t—t 2 yAi+B" 1At+B' Thus |x— y| S 2R1, and this gives [y — z| S 3R1 = R for every 7. E D. Notice that 37 the curve (2.24) can be written in the form xey=@—xma+r,tey=fimu+BnM—B] and it passes through (x, t). Thus (2.23) follows from Proposition 2.13. Theorem 2.16 Let v0 6 C“. Then the interface I‘ of (1.1), (1.2) can be written as t = S (x) for x 6 R” \D and S is a Lipschitz continuous function. PROOF. We first show that I‘ can be written as t = S (x) If the assertion is not true, then there exists two points (x°,t1), (x0, t2) 6 I‘ with t1 < t2 < t1 + 6. Then there exists a point x1 6 R" such that v(x1,t2) > 0 and [x—xll R1 1 1 <———, t0 2f [517—23 1. Then there exists a constant c depending on p, N, Ixol, to, A and B of Proposition 2.12 such that (3.1) sup{v(x,t) | [x — x0] < h, 0 S t — t0 S t(x°)h} > ch where t(xo) = 537:?(%1)p’1(1 — |x°|‘2)P/2. 39 40 PROOF. We will use the comparison principle to prove this proposition. Let x1 = (1+ Ego—In”. Then for x E Q(t°) we have (3-2) Ia: - xll _>_ (1 - MOI—91”- In fact , if |x| S 1, then we obviously have MID: wlz‘ lavl - 31| Z lxll - |x| 2 _>_ (1 - MOI—2)“- For |x| > 1, x E 9(t0) implies that v(x,to) > 0 = v(x°,t°). Hence by Proposition 2.3 — > cos(x — x0, x0), i.e., (x0, x — x0) = |x0||x — xolcos(x — x0, x0) S [x — x 0'. |w°l Therefore hx0 _ 12 = _ o___2 Ix ml Irv 2: wall h2 (x—x0 x0) 02 a = ——h [x x| + 4 [130' h2 |x—x°[ > 02 —h _ [x xl + 4 Imol h 2 h2 1 _ ([(L' $0l 2|$0|) +—4— _leIQ] h: 1 , — 4 I$OI2 which implies (3.2). 41 Now set wx :p—l AIx-xll” 71—1 (’3 ( 1) 19 fi- (t -t" with _ 2pcp—lh—I(E;Ll)p—l (l-lzvol“"')‘”/2 ’ where c is a constant we will choose later. Then we know from (2.2) that w is a non-negative solution of _2 A wt: p——wpr+|Vw|p, 0 — 4— _— -Ahw+B [ uwmA 1” and fl 1 = _____.__ > , A{A|t°| +3 (2%} - 0 if h < 2|x°|(2711——l — 1) and fl 2 flfglfir—B. Therefore when t = t0 + g, l l h (3-5) f(00,t) > f(001t°)[9(h)lX _>_ f(901 t°)[8(h)17 = IJIOI + 5- 0 Since x1 = (1 + )x0 = (IxOI + %)00, 60 = {go—I, we should have h 2|x0| 1 o A (3.6) v(x ,t + 3) > 0, which contradicts to our previous estimate (3.4). Therefore sup v(x, t) > ch lx-xolsh 09—10 St(x°)h if h < 2|$0[(22‘1——1 —— 1) and c is small enough. Note that the only condition on c is that p—l _Al—x0’2P/21 fl = ( )p1( l l ) _1 p 21’ cP Ato+B |x°| 44 i.e., < 1.1—.0 - Ix0I-21m'a1p—1 [(Ato + B)2p|xonpi—x :2 Therefore c can be chosen independent of h and (x, t) in regions of the form |x| 2 R, t 5? and Proposition 3.1 will be true for h S 2R(2fi - 1). Actually we have the following stronger version of Proposition 3.1. Corollary 3.2 Let v0 6 C“ and (x1,t1) be such that [xl[_>_R>1, T0 Eh for some point in a small neighborhood of (x0, to). Finally we will use Proposition 2.3 and Proposition 2.12 to see that v(x1,t1) 2 Ev(x2,t2). 45 Let a = %min(1, t — T o) and let c be the constant in Proposition 3.1 for the region [2:] 2 R and t S i. Let us choose a constant k large enough so that 1 h /\ p—l toztl—A1——T——>tl—_ .: 17—1 1( RM(t)2)l€— 0" Al 2PcP-1( p ) ’ where RM (f) is given (2.11) and is estimated in Proposition 2.6 in terms of t and [[uOHL1. Note that k can be chosen independent of h S 1. Let now x0 be such that (20,110) e r and l—i’rl = 1%) = 01. That is, |x0| = f(61, to). Then we have |x°|— |1:1| 2 g if k is large enough independently of h. In fact, since |x1| = f (01,t1) — h, we have [550' " [371' = “911150) — f(911t1)+ h- Note that f is non-decreasing in t from Proposition 2.13,we see that [xol S 2|x1| for h S 1S lel. Recall Proposition 2.8 concludes that Mal — to) if t0 > T0 to —To f(01,t‘) — f(01,t°) _<. c3f(01,t°)(t1—t°) if To + 1 21° 2; where g = t — 521°,c3 = c3(§,To). (i) when To + 26 < t1, we have 0l_|$1l = f(011t0) — f(611t1) + h Ix f(011t0)1 0 > _— _ _ h tO—To(t t) l > h— 2|z|(t1—t°) _ tO—To 46 2'371' 1 0 - h tl—(To+a)(t t) 22(5) —-2h > __ _ _ _ h 6 A10 RU) )k h h = _ _>_ h ”Is-k 1f1—%2%,where 2Rt - 1:). a00—12(24). (ii) whentl _<_To+2agTo+1, we haveto ST0+1andt° 2111—222 12— (E—To)= NIH HH- 1 Ol-lxll Z h_[f(011tl)_f(011t0)l [x 2 h — c3f(01,t°)(t1 — t0) 2 h — 2222110:1 — t”) = h — 203A1(1— R(D_2)%|x1| 2 h — 20.91136)“ — R(?)’2)% 2 h — 6% 2 g 6 = 263A1R(L)(1 — Raf—2)- Both (i) and (ii) are satisfied for A: large enough independent of h. As |x°| _>_ |x1| Z R and t0 S t1 S t, the constant c works for (x0, to). Now we see that the ball of center x0 and radius %(1 — |x1[‘2)1/ 2 is contained in the cone with vertex 47 x1 of directions in which v is decreasing. That is h (3.8) [x — x0]2 S (h)2(1 — [x1|_2) implies (x1,x — x1) 2 [x — xll. In fact, if (x1,x — x1) < Ix — xll, then Isa—mo)? = 12—2 —2|—:.,)2 lx—m‘l Imll > |x—x1|2+,u2—2)u = FUSE — SBII) where F()\) = A2 — I—ZlilA + [12 with ,u = |x0| — |x1| 2 [5. Hence lx-wolzzminFW: F-—( fl)= u2(1-I$1|'2)>(kh -)(1- -|-’v1| )- lxll Therefore (3.8) holds. Let now and let us apply proposition 3.1 in the cylinder |x — xol S h, 0 S t — t0 S A1(1— [x°|_2)§h Then we have a. point (x2,t2) in that cylinder for which v(xz, t2) > ch. Since 2 t1 _ t0, t2—t°sx\1(1—lw°""2|)5 sltxu—Rmr )g 48 and Ixol = f(01,t°) S f(61,f) S RMG), we have t2 S t1, and |le Z |xol—lx2—xol Z MOI—52 |x1| >R> 1. Recall that Proposition 2.3 implies that 2 r6110: ,t) < 23-sz (9r _0. - - a 2 A—Zfit Hence Proposrtlon 2.12 concludes that a(v(x ,t)ev 1 ) Z 0. Hence ’U($2, t1) 2 ”($2, t2)e-’-:Ef(tl_t2) 2 v(x2,t2)e:_:¥(z—9 2 ch. Since 1:2 belongs to the cone with vertex 2:1 of directions where 'u is decreasing, we have v(x1,t‘)2 v(x2,t‘) 2 ah, where E depends only on A, p, N, R} and L. The Holder continuity of gradient of solutions of (1.1) and the following general class of quasilinear equations (3.9) at — divA(x,t, Vu) = B(x, t, u, Vu) in QT, was discussed in [8],[10] and [24], where the functions AzflTxRN—HR, 49 BzflTxRxRN—H‘Z, satisfy the structure conditions: (51) CoID’UIP — 900 S A ° Du S CIIDUP + (P1, (32) a3? um,- Due,- 2 CoIDUIP‘Z’Zj-V: IDux,|2 - «p0 (53) Zr- llaiA l 113:2 It was shown that for any compact subset K of QT, the Holder coefficient and exponent of Vu depend only on K, p and N. In particular from [10] we have Proposition 3.3 ( The degenerate case p > 2 ) Let u be a weak solution in RT of (3.9). Then there exist constants ”y > 1 and a 6 (0,1) that can be determined a priori only in terms of N and p such that, for every compact subset K of QT, IVu(x° to) _ VU($1 t1)|.<_. fyLo(lx _ 331' + max{1 L2 4M t0 —t1|) , , dist(K; I‘) for every pair of points (512‘, t‘) E K,i = 0,1, where L = supnT |Vu|. But it can be seen from the proof that L can be replaced by sup K ldul. Now consider the following equation which by a straightforward computation is equivalent to (1.4) (3.10) ”t = div(: : iv|Vv|p’2Vv) + p—iIIVvF’ H 50 Let __ 2 1 p 1v|Vv|”‘2Vv, B(x,t,v,Vv) = 5—}:IVvl". A(x, t, Vv) = Then since |Vv| S L < 00, A, B clearly satisfy the structure conditions and Vv is Holder continuous in a compact subset K of Q, where 0 < c1 S v S c2 < 00 for some constant c1, c2. Theorem 3.4 Let v0 6 C”. Then there exists a neighborhood of F in each strip of the form To < t S t S t and a positive constant fl depending on the strip and on the initial value v0 such that |Vv(x,t)|,vt _>_ 3 if v(x, t) > 0 and (x, t) belongs to that neighborhood. PROOF. Let (x1,t1) 6 F and let W = B(x1,6h) x (t1 — 6h, t1 + 6h) for some small h be a neighborhood of (x1, t1) such that there exists a cone Q of directions in which v is increasing for each point in W (from 3.8). We may assume that the axis of the cone is the xN-axis. Then there is a function f (x, t) such that I‘ n W can be represented by IBN = f (575, t) where if = (x1,- . -,xN_1) with f Lipschitz. Let (x0,t°) E W be such that x‘,’,, — f (.750, to) = h. Since the cone Q can be taken to contain the radial direction of each point in W, v(x°,t°) 2 ch by Corollary 3.2. Note that |Vv| S L < oo in R” x (T, 00) for some T > O see([13]). 51 123,,(F, f(F, to) + y, to) 2 g for some y 6 (53-h, h). In fact, suppose on the contrary that va(-x—°, f (F, to) + y,t°) < E for every y 6 (55h, h). Let 1 o o 1 — — o o vh(x,t) = Ev(x + hx,t + ht) = Ev(x0 + hx,xN + th,t + ht). Then vh(0,0) = %v(x°,t°), vh(0,—1,0) = %v(-x5,f(-xfi,t°),t°) = 0, vh(0, —1 + %,O) = #6213, + f(xfi,t°),t°). Note %(x,t) = v,,,,,(x0 + hx,t0 + ht). Then, 31%.“), 7), 0) < E, for all n 6 (—1 + 5%, 0) by assumption. Therefore BxN nl |/\ 1)};(0, 0, 0) — 1)},“), —1,0) 0 = f 622,, (0,71, 0)dn —16xN _1+2—EL' 6’0}; 0 Ovh = —0, ,Od / —0, ,0d [_1 6x~( 77 )77+ -1+&8m~( 7? )n E E +50“? _i<5 4L ' lot I/\ b. ' 2 + [\DIOI wlcl b! (Eh—Uh) such that vx~(-x_°,f(F,t°) + y, to) 2 3. Let (F,f(x0 to) + y, to) = (z°,t°). Then Hence there exists 1) E (—1 + &,0) such that 82:3:- 2 g, i.e., there exists y E Eh — E S 29v " f($0at0) = y. Now choose M > 0 large enough so that 7(fi)“ S % where "y and a are constants in Proposition 3.3. Let K1 = {(x, t) : |x—z°| S Ah, |t—t°| S sh} where A and s are small enough so that K1 c QT and dist(K1;I‘) 2 (2M + max(LL§Z,1)\/§E))M. 52 Let maxK, |Vv| 2 L1 and let |Vv(x1,t1)| 2 L1, where (x1,t1) 6 K1. Then L1 2 NIOI and -2 2Ah + max(1, Lf‘z‘wgfi ”View + w"): — IVvatJWII S W dist(K1;P) )0 Therefore |Vv(_xa,z?v + Ah, t°)| 2 %L1 2 5155, and 2%, + Ah -— f(F,t°) = y + Ah. Next, let (zl,t°) = (F,z?v+Ah,t°) and K2 = {(x,t) : lx—zll S Ah, |t—t°| S sh}. Then K1 0 K2 76 ¢, measK1 = measK2 and dist(K2;I‘) 2 dist(K1;I‘). Let maxK2 |Vv| 2 L2. Then L2 _>_ 5%.Let |Vv(x2,t2)| : L2 for some (x2,t2) 6 K2. Then again by Proposition 3.3 we have — o 0 L2 5 |Vv(x°,z~ +2Ah,t )I 2 2 2 23. And ZR, + 2Ah — f (F, to) = y + 2Ah. Hence inductively, for each positive integer n we can define a sequence of compact sets Kn with dist(Kn;I‘) Z (2Ah + max(L%2,1)\/s_h))M and a point (z",t°) = (Ry + nAh + f(x_°,t°),t°) which belongs to that set. Now let no be the largest integer such that noA S 1 — &. And let Km, and (2"0‘1,t°) 6 Km, be the corresponding compact set and a point respectively. Let Ln0 = maxKno IVvI. Then Lno _>_ 7.3%. And let (x"°,t"°) be a point in Km, such that |Vv(x"°,t"°)| = Lno. Then |Vv(x°,t°)| Z 275—.” = fll. And this inequality is valid for every (x0, to) in a neighborhood of (x1, t1) contained in W. 53 Thus it remains to show that v, is bounded away from zero. By Proposition 2.12 we have A —p p— 1 v(x, t) + x - Vv(x, t) + (At + B)vt Z 0 for some constants A and B. Also by (2.6) and (3.8) we have 51 R+1 D_,v = |Vv|cos(—r, Vv) Z \/(R +1)2 -— R2 = c > 0. Therefore 5:12 > P“ _ At + B Rc— $0 >____._ — At+B’ —Tvr — v ”t where C is a bound of v in W. Then since for (x,t) E W = B(x1,6h) x (t1 — 6h, t1 + 6h), v(x,t) = v(x, t) -— v(‘if,f(x,t),t) S |Vv|6h S 6Lh. Hence C S 6Lh. Therefore, vt 2 ,82 > O in W if h is small enough. Hence IVvI, v, 2 fl = min(fl1,fl2) in W. Then by Theorem 3.4 and [13], we have Corollary 3.5 Let u is a solution of (1.1), (1.2) in 86 x (—6,6). Let (0,0) 6 1". Then for v = EuK—f the following statements hold: (i) v is Lipschitz continuous. That is, there exists L > 0 such that |Vv|, v, S L. (ii) D71}, 0,2) 2 ,8 > 0 ifv > 0, where V is a unit direction in R”. (iii) Apv 2 —CI in 86 x (—6,6). 54 PROOF. Let (x1,t1) e I‘ with |x1| 2 R+1,To < t g t1 _<_ i < 00. Let 6ho = min(tl — t, t — t1). Then there exist constants h S ho, ,8, L > 0 depending only on ho and a direction 71‘ in R” such that for vh(x, t) = %v(x1 + hx, t1 + ht), Dfi’Uh(.’L',t), Bat—Uh Z :6) Ivvhla gull S L in 86 x (—6, 6) by Theorem 3.4 and [13]. Also it was shown that there is a constant 01 depending only on To such that Apv 2 —Cl if t _>_ To. In particular, Apvh(x, t) = hA,,,v(x1 + hx, t1 + ht) Z —Clh Z —01 in B6 X (—6,6)(h S 1). Next, we notice that equation (1.4) can be written as the following equation: p—2 v, = p _ 1v(|vv|P-2Av + V(|vv|P-2vv) + |vv|P p_ 2 p—2 vxivx' = 61" - J x-x- p. p _ leVvl ( J + (p 2) |Vv|2)v , J + |Vv| Hence if v is a solution of (1.4) then v is also a solution of 1115va Will2 — 2 (3.11) wt = p UIVUIP—2(6ij + (p — 2) 2 p _ 1 )wx,,,-j + IVvI . 55 Therefore in any compact subdomain (21 of Q where v and |Vv| are bounded above and bounded away from zero (since the coefficients and free terms are bounded and continuous) (3.11) becomes a uniformly parabolic equation and llDz‘jvlleml) < 00 if v is a solution of (1.4) in (21 by Theorem 12.2 on page 224 of [23]. CHAPTER 4 Proof of Main Theorem To prove our main theorem we need the following sequence of propositions. Proposition 4.1 Let v be as in Corollary 3.5 and V = fi7+ fiat/+1. Then there exists 60 > 0 such that DTv Z 0 in 86 x (—-6, 6) for any r in ROI, 90), where I“((1/’90) = {T | (N!) .<_ 90} PROOF. Since by Corollary 3.5, Duv = Wvlcos(u, 6V) 2 fl for some fl > O and Wvl S \/2L, we have cos(u, vv) 2 ——fl— = fll > 0. fiL 56 57 Let 90 = arcsinfll. If (V,r) S 90 then (r, 6v) S %. Therefore D,v(x,t) Z 0 (x, t) E Be X (—6, 6), r E K0490). Proposition 4.2 Let v be as before and r be such that D,v(x, t) 2 0 for (x,t) 6 B4 x (—4,4). Then v((x, t) + 7r)> (1 + 6)v(x, t) for (x, t) E 32 x (—4(%:1)e,4(l’+1)e)fl(v = 6), where e, 'y > 0 are sufficiently small and 6 = C'ye”1cos(r,@v(i7, ——$I%Ile)). PROOF. It suffices to show that there exists C > 0 such that 13.2; 2 Ccos(r, we, —2r)) in B2 x (—2r, 2r) fl(v = c) with r = gig—:25. Now consider a non-negative function g = DTv in B4 x (—4, 4). Then g is a solution of the equation 9. = 5:1vIVv|”2[621+(p 2)“ M I?” +[—(’;__ 1) vIVvlp‘4Av + (1’ pm” _ 4)v|Vv|p‘6(V2vVv) - Vv]Vv - W —1 2(p-2) p—1 - 2 +p|Vv|”’2Vv - Vg + |Vv|p’4(V2vVv) - Vg + %—_—1(Apv)g. This is a uniformly parabolic equation with bounded coefficients in any bounded region where v and |Vv| is bounded away from zero. We choose a region in such a way that it contains the set (v = e) and use Harnack’s inequality there [23]. 58 Consider a new system of coordinates in R” such that e N = v. Since D3,, 2 ,6 > 0, there exists a Lipschitz continuous representation of F, (4.1) xN = f(x, t), x = (x1, - --,xN_1). We use c to denote a Lipschitz constant for f. Let ist—fcfao) Q: |x|S4 ”mag—11. and 91 II E |/\ to ltl S 11%1-15 We see that 32 x (—4(%:1)e,4(%:1)e)fl(v Z c) is contained in $21. In fact, for |t| S #6, |x| S 2 and xN — f(x, t) < f we have, by the mean value theorem, v(x,xN,t) = v(x,xN,t)—v(x,f(x,t),t) : va(-fa¢at)(xlv_f(fat)) S L i: :6 To apply Harnack’s inequality, we need a lower and an upper bound for v in 9. For (x,t) E Q, v(x,t) = v(x,xN,t) — v(x,f('f,t),t) 59 A T.¢,t)($~ — f(f.t)) _ va [B . IV film and 8(L + 1) 3p v(x,t) = v(x, t) — v(x,0) + v(x,0) — v(0,0) S L|t| + lel S L(4 + 6). Now applying Harnack’s inequality to the function g = DTv in 9 yields, noting that (eN, —flL—+lle) E {21 when 6 is small, pp D,v(x,t) Z AD,v(eN, —4(L + 1)€). fin Therefore D,v(x,t) Z AD,v(eN, —4(Lfl: 1)e) = A(r, <7v((eN, —4(L,B: 1)e) = A|vv((eN, —4(Lfl: 1)e)|cos(r,@v((e1v, —4(Lfl: 1)€)> 4(L +1) 2 Aflcos(r, vv((eN, - fl” 6)). To show that v((x, t) + 7r) 2 (1 + d)v(x,t) for v(x, t) < e, we need a technical lemma. Lemma 4.3 Let v be a solution of (1.5) satisfying Corollary 3.5 in B5 x (—5, 5). Then there exists 02 > 0 such that ”Diij_C2i ’l,j=1,°",N, 60 in 32 x (—2r, 2r) ifr < min(c, 1), 0 being the Lipschitz constant off in (4.1). PROOF. Let f be the representation of I‘ in (4.1). And let (x°,t°) E 32 x (—2r, 2r) n(v > O) and let h = mail —f(F,t0) = 33% — f('fo,t0) +f(:—E-6)O) _ f(FaO) — f(020)' Then h S 2(1 + c + or) S (1 + c)(1 + r). Let us consider the region IIxN — f(r.t°)) — (.9, - foo—W)» s m Rh: . Iii—56' SW, 0 h |t—t| gm. It is straightforward to check that Rh is contained in B5 x (-5, 5). By the change of the variables S) E Rh, 61 where g(f) = %{f(o:—°+ hit") — f(?°,t°)}. Let w(x,t) = %v(m0 + hx,t0 + ht), (x,t) E (2. Then vh(:1:, t) 2 %fl in Q. In fact, let (:12, t) E Q and (y, s) 6 Rh be such that y=$0+hx, s=t0+ht. Then yN _ f(y)8) = yN _ f(37)t0) +f(gito) — f(i18) 2 an — fat”) — gh + f(@’,t°) — Io, s) 1 1 2 g}: — Zh :2 1h. Thus v(y, s)_ i—flh and consequently 1 . w(x,t) 2 Zfl 2n (2. Since 12,, is also a solution of a uniformly parabolic equation vt = 1;:ivdiv(|Vv|p_2Vv)+IVvIpv 2 = 2:1 levl”2{6z-J+(p— 2)” IV: '2 inIJ‘} + IVUIP' 62 Since leulP‘2{6ij + (p — 2)%’T?vxixj}, lvvlv e 00%(12) (see [23]), H Dijvh IILw(R)S C for some constant C depending on fl, L. Hence lDij’U(-T°,t°)| S ail 0 Since v(zo, to) S Lh we have |v(x°,t0)D,-jv(a:°,t°)| S CL for every (23”, to) 6 82 x (—2r, 27‘) 0(2) > 0), which implies that uDiJ-v 2 —CL = Cg in the sense of distributions in B2 x (—2r, 2r). Lemma 4.4 Let u be a solution of (1.5) in B4 x (-—4,4). Let k be a C1 function defined in 32 such that k E 0 in B1; IVkI S 26, —2eI S sz S 26] (I: unit NxN matrix), It 2 0 and k E c if |x| = 2. Let 6 < 1 and y(:c, t) = v(x,t) + 6[v(:1:, t) + 2751—1305 + a) — k(a:)]+ if |x| S 2,t 6 (—§(%:—lle,—a). Then y is a subsolution of (1.4) in 82 x (—§%fle, —a) fl(v S 6) when 6 is small enough. PROOF. Let 9(3) 2 63+. Then g(s) S s+g'(s) for s S e, g’(s) S 6 < 1 and g”(3) 2 0 in the sense of distributions. Set 63 13 Ly = Fifi/Apt! + IVyI” " yt- We want to show that Ly Z 0. Note that fl? = 1 I I lit ( +9)vt+—2(L+1)9 yx. = (1+ g')vx.- - 97:».- W = (1+ 9')Vv - g’Vf, IVyI2 = (1+ g’)2|V'Ul2 - 29’(1 + 9’)Vv ° Vf + (9')2lVf|2, yxixj : gnvl‘gvxj _ gnvngxj + (1 + g’)v$g$j + gflfx.‘ ij — gflvl‘j fx.‘ — glfxng Ay = (1+ 9’)Av - g'Af + 9"IV(f - v)l2. Hence IVyley = {(1 + 9')2|V'vl2 - 2g'(1+ 9')Vv - Vf + (9’)2|Vf|2} ><{(1 + 9’)Av - g’Af + 9"(|V(v - f)|2} = (1 + 9')3|VUI2AU + 9"(1 + 9')2|V’U|2|V(’U - f)|2 +(1+ 9’){-29’(1 + 9’)Vv - Vf + (g')2|Vf|2}Av + 9’0 where G = 0(6). And yzgyxjyxng = {(1 + g’)vxg _ QIfIL'g}{(1 + ‘9’)ij _ g’ij} X{g”(v$iv$j _ vxiij — ”13 f2.- + fngxj) + (1 + 931%in — glf-Tixj} = {(1 + 9’)2vxivxj _ 9’(1 + 9’)('Ux.-f:rj + ”$1 fxi) + (9’)2fxiij} 64 X{9”(vxsvxj - ”rift; "’ v-ij-Ti + fziij) + (1 + g’)v$i$j + 9,0} IV (1 + g’)3v$iv$j "m,- + 9”“ + 9’)2(fi4 — 6M) “9,0 + g’)v-Ti3j {(1 + 9’) ('02:;ij + ”1; fan) " g’fxsfzj} + g'G IV (1 + g’)3v$iv$j ”m,- “ 9,0 + gl)v$i1‘j {(1 + 9,) (”@ij + ”my fan) +g’fx.‘ fmj} + g’G if e is small enough , where M = M (N, L) is a finite constant. Therefore Apy = IVyIP—4{|Vy|2Ay + (P _ 2)y$iy$jy1i$j} IVylp-4(1+ g')3{IVv|2Av + (p — 2)vxivxjvxixj} IV -g'(1+ g’)|Vy|”'4{[2(1 + 9')Vv - Vf - g'lVfVlAv +(p - 2)[(l + 9’) (1’3:ij + ”I; fzi) _ g’f-Tiijlvxivxj ”xi-1a} +g’G. NOW Will = |(1+ 9')Vv - g'Vfl Z (1 + 9')|Vv| — Q'IVfl = IVvl + 9'(|V’U| - lVf|)- Hence |Vy| Z |Vv| if e is small enough and also we have IVyl S (1+ 9’)|Vv| + g’IVfl S (1+ 9')|Vv| + 29’6 < (1+ g')|Vu| + 2c. 65 Then by Lemma 4.3 and g S 622+, we have _2 I L3! = :_ 1(12+9)Al»3/+|Vy|"’-(1+g')vt g'+gG p - 2 IVs/l”—4 p — 1|VvIP”4 fl? —2(L +1) fl? _%L+D IV - 2 (1 + g')3vApv — (1 + 9')“. + Iva? + iii—1913.2; 9’+g’G- Now ifv—f+2—(f:—1)(t+a) <6, we havegSg’e and gApy >g(1 + g')3[lvy ]” 4Apv +g'G _ W?) | 3|__Vy -4 , > — 1 " 9G 2 9'0. Hence Ivylp-4 I 3 p— 2 > __ _ :0 Ly ._ IV'U lp—4(1+g) (p_1'UAp'U 'Ut)+|Vy| IVyIP—4 I _ 16p I I +{IV'U |p_14( +g’) 1}(1+g)vt 2(L+1)9+QG- Now IVyl" = |V7yl’”"’|Vz/|4 = IVylp"4{(1 + 9')4|V'v|4} + 9’0 = IVy|”"4{g’(1 + 9')3|V'v|4 + (1 + g’)3|Vv|4} + g’G IVyI‘”'4 lvvlp_4 (1 + 9’)3|V’Ulp + g’(]. + g’)3|Vy|p"4|V,U|4 + gIG 66 Hence Ly 2 {:——Vy:}P-4(1++g§—'—')3(—_§-vzxpv+IVvIP—va +{< :——Vy:)P-4(1 + 9')? — 1}(1 + gov. + g'<1+ 9’)3|Vy|”“‘|Vvl“ IBP _m = {(%)p’4(1 + 9')2 - 1}(1 + 9’)vt + g’(1 + .(1’)3|Vy|”‘4|Vv|4 fl? 2(L +1) 9! + gIG g' + 9’0- Now let us consider the following two cases: (i) p 2 4. Since {g1} 21,th 0, '6? 2(L + 1) Ly 2 g'{(1 + 9')?” - 9’ + G} 2 0 if e is small enough. (ii) 2 < p < 4. Since lVyl < (1+ 9'an + 29% 1S|——Vvl_ lvvl and 0<4—p<2, we have (IVy|)p— ->4 >{ |V’UI }2. IVUI (1+ 9 ’)IVv| + 29’ 6 Hence Ly 2 {(1%) (1+9 )—1}(1+g)vt+g(1+g)|Vy|” 41w“ 67 ___(”’__ 2(L + 1) { ((1+g’)IV'U|)2 ((1 EPQ’HV'UI + 251’s)2 ‘uL+n 2{(1 + 9’)|Vv| + 9’6} -29’6 = 1 I (1 + 9')|Vv| + 2g’e (1+ 9')|Vv| + 29,€( + 9 )1): +g'{(1 +g’)3|Vy|”"4|Vv|"- —L—":2( ——1)9 +0} 1 _ _L [(1+ g’)|Vv| + 29%]? 2(L +1) W H” G >0 1 +g')L+zg'e]2 _ 2(L+ 1) + }— 9’ + 9’0 IV — 1}(1+ g')vt + g’(1 + g')3|Vy|p—4|Vvl4 9’ + 9’0 IV g'{(1+9')3|Vvlp + 0} IV g'{(1 + 9')3 [( if e is small enough. Hence in the set where v + Edi—”(t + 01) — k is less than c, Ly Z 0 if e is small enough. Since in the set 82 x (—%e, —a) 0(2) < e),v + 27—51—15“ + a)— k < e, and we obviously have the relation Ly Z 0 is valid there. Proposition 4.5 Let u and 112 be Lipschitz continuous solutions of — 2 ct = p lvApv + IVvl” in B4 x (—4,4). p— with v satisfying Corollary 3.5 in B4 x (—4, 4). Suppose that w 2 v, 0 < 6 < 1 and w 2 (1 + 6)?) in 32 X (—8(%:1)e,8(L+1)e)fl(v—— 6). Then ife is small enough, 4L+n ML+D fin e, fip w2(1+6)v le(— e)fl(vSe). PROOF. Let y = v + 6(2) + 4(L+1)(E 4(L+1j [3? 27457;)“ + or) — k)... with k as in Lemma 4. 4 and a E (— 6). Then since for t— — —flc,v(x,t)+ 2—(%(t+a) —k S O fiP 68 in 82, by the assumption we have w 2 y on the parabolic boundary of the set 82 x (—-§-§%,+—ll-e, —a) 0(1) S e), 8(L + 1) ,6? wa in ng(— c,—a)fl(1)Sc). Since 11: E 0 in Bl, w(x, —a) 2 (1+ 6)1)(:r, —a) in Bl fl(1) S e). Since a E (—4(Lfi:1) e, 4(15:96) is arbitrary, 4(L + 1) 4(L + 1) fl? 6, 3p 6) 0(1) S c). w2(1+6)1) in le(— Proposition 4.6 Let 1) and u) be as in Proposition 4.5 and 1) S e in 82,- x (-—2r, 21') with O < r < min{%, agig—1213}. Then, for any vector fi 6 RN, w(x,t) Z (1 + 6)1)(:L‘ + (t + 2r)¢(x)fi, t) in B2, x (—2r, 2r) if¢ is a smooth function such that suppq) C B2,; ()5, Iqul, |V2¢| S 116; (b 2 0 with u small enough. PROOF. Let y(:):, t) = (1+6)v($+(t+2r)¢(x)17, t). Then by Proposition 4.5, w 2 y on the parabolic boundary of Bzr x (-2r, 27'). (We recall that 32,- x (—2r, 2r) C Bl x ((—4(%:1)e, 4(%:1)6) 0(1) S 6)). We only need to prove that y is a subsolution of (1.4) in B2,. x (—2r, 2r) if u is small enough. This proves our proposition. Let eN = fi. Then yt = yxi : Vy= IVs/I2 = yxng _ lVyley = 69 (1+ 5){vt + amt}, (1+ 6)(1)x, + 1),,” (t + 21045“) (1+ 6){V1) + v,” (t + 2r)V¢} (1+ 6){A1) + 2(t + 2r)Vv,~ ~V¢ +vx~z~ (t + 2r)2|V¢|2 + 1),,” (t + 2r)A¢} (1+ 6)"’{|V1)|2 + 1):” (t + 2r)2|V¢|2 + 2(t + 21)va V1) - V45} (1 + 6){vx.x, + 1);,” (t + 2r)¢.,. + 1),.” (t + 2r)¢mj +113”,- (t + 2r)¢.-, + (t + 262131.151, 1).-m} (1+ 6)3{|V1)|2 + 1):”(t + 21~)2|V¢|2 + 2(t + 2r)u,,~vv - W1} ><{A1) + (t + 2r)[2V1)zN ~V¢ + um” (t + 2r)IV¢|2 + 1)” A¢]} (1+ 6)3|V1)|2A1) + 6G +(1 + 6)3(t + 2r){v,N A1)[(t + 2r)1)_.,,,,|Vd)|2 + 2% - 17¢] + (t + 2r) XUxNIN IV¢I2HVUI2 + U311) (t + 27l2lv¢|2 + 2(t + 2r)1)xNV’U ' V¢]} where C = 0(a). And similarly ya.- yxj 311:ng (1 + 6)3(1)Ii + 1);.” (t + 2r)¢x,) - (1);]. + 1)“, (t + 270(1),”) '[vzixj + (t + 2T)vxi$N¢-’Bj + (t + 2T)v1Nij¢$i] +(1 + 6)3{(t + 2T)2¢$i¢$j vaxN + (t + 27‘)va (bin-w} Now 70 _ 3 — (1+ 6) oxivmjvxfij + 60 +(1 + (5)3v2ng (t + 2T)[vxiv$1v 0511' + vxj ”IN $.17; + "Ugh, ¢$g¢$j ¢$i$j]} +(1 + 6)3(t + 210(1):“ + 1),,” (t + 279%,) - (1);]. + 1)“, (t + 2r)quj) X[vx,‘.’t~ ¢$j + vxjx~¢zg + (t + 2r)¢$g¢1‘j vaINl° |Vv| S IVvl — L(t + 2r)u6 + 6)[|V1)| — L(t + 2r)u6] = (1 + 6)[|V1)| — L(t + 2r)u6] |/\ Will 3 (1 + 6)|V1)|[1 + (t + 2r)u6]. if ,u is small enough. Hence by Lemma 4.3 11%?) = lVylp’4{y|Vy|2Ay + .101 - 2)yx.yx,-yx.x,-} Z IVyIP"4(1+ 6)41){|V1)|2A1) + (p — 2)(V21)V1))V1)} + 66'. But, IVyI‘D = IVyl'D’4IVyI4 = IVylp‘4(1 + 6)4|V1)|4 + 60. Therefore Ly P P ”6+":3 ”d 2 17—11%?) + IVyl” - 111 1 —2(1 + 5)4v|vy|P-4{|vv|2m + (p — 2)(V21)V1))V1)} H 1 + 6)4|Vy|"’—4|V1)|4 — (1 + 6)vt + 66' MA ———(1 + 6)41)IVyIP’4IVvl4’pIVvlp-4{le|2Av + (p — 2)(V21)V1))V1)} 1 71 + (1+ 6)4|Vy|”"4|Vvl4‘p|Vv|p — (1 + 6)vt + 6G lVyl le | + (1 +6){(1 +6)3[-—— (1+ 6)“1— —]”'4(1;—:—21)Ap1) + IVvI” — vt) MlVyl le | {(1+6) 3[:—V—VZ :]P’4+6(1+6)3[—— slVyl le | ____p-4] — 1}'Ut +6G lVyl le | 14 4x) + <1 + 6))(1 + argylr- 1111+ 60 ]” 4 —(1 +6)}vt +6G' > 6(1 +6)3 [— Hence for p 2 4, Will [l—Vv I]'""‘4(1 + 6)2 2 (1 +6)2 21 For 2 < p < 4, since Zlvyl 11—4) 2 WM 2 1 (1+6) [I—V Il >(1+6) [|——vy I1 2 [1+(H2TWF, 24|Vy|p_ 1 _ _2+(t+2r)u6 —(t+2r)u6 _ (1+5) [Iv )1 12[1+(t+2r)u6]2 _1+(t+2r)u61+(t+2r)u6_6G' Hence Ly 2 6{(1+ 6) [:—-;y:l”‘4fl + G} > 0 if u is small enough. Corollary 4.7 Let u and w be as in Proposition 4.6. Then the distance between their free boundaries is at least 13—603 > O in B, x (—r, r) where 0 is such that a) Z 0 in B, and Co is a bound of IVwI. Moreover, the distance between their level surface is at least @643 and 0 can be estimated from below by gr2u6 with u as in Proposition 4.6. 72 PROOF. Suppose v(x, t) = A; then w(x, t) > v(x + t + 2r)(b(:1:)fi, t) Z v(x, t) + fl(t + 2r)¢(:1:) 2 A+fir6 if t2 -r,:1:€ 3,. Suppose dist((z, t); (w = A)) S 5%); then w(x, t) S x\ + BN9, which implies that dist((v 2: A); (w = A) > —. Since we can assume 4) 2 gr2u6 in B,, the second assertion follows. Now we are ready to prove our main proposition. Proposition 4.8 Let u be a solution of (1.5) satisfying Corollary 3.5 in B5 x (—5,5). Then there exist J > 0,5 > 0 depending on the constants in Corollary 3.5 and a monotone family of cones I‘m/1,6,.) such that 9;, Z (1 — S)9k_1 + 5%71 and for any 7' E fro/1,6,), (a), t) 6 BJ): x (—J", J"), D,v(x, t) Z 0. PROOF. Let w(z, t) = v((x, t)+'yr) with r e L((V, 90),where L((V, 90) is the cone defined in Proposition 4.1. Then v and w satisfy the hypothesis of Proposition 4.5 with 6 = Cye‘lcos(r, Vv(fi, —2r), 73 with e and r as determined by Pr0position 4.6 and Lemma 4.3. Therefore, by Corollary 4.7 for any A > 0, dist((v(x, t) + 77') = A); (v(x, t) = A)) 2 grfl in B, x (-r, r). We know that 9 Z gr2u6, where u is the constant in Proposition 4.6. Thus A dist((v(x, t) + 477') = A); (v(x, t) = A)) Z Cycosb’, Vv(fi, —2r)), where C depends only on 6, L and C1. So, D.v(:r, t) = 113% ”((35) t) + 777) - v(x, t) '7 2 flCcos(r, Vv (71', —2r)) in B, x (—r, r) 0(2) > 0). Let H = {(11), t)|((a:, t),vv(v, —2r)) 2 0}. Then 6H = {(513, t)|((a:, t), {u(y, —2r)) = 0}. So dist(r, 6H) = cos(r, vvw, —2r)), and D,v(a:, t) Z Cdist(r, 6H), where C is a constant depends only on 6, L and C1. Let p(r) = C |r|cos(r, VVW, —2r)). If q E B(r, p(r)), the ball with center 1' and radius p(r), then qu(:r, t) 2 0 for any (x, t) E B, x (—r, r). In fact let C be such that cos(r, va, t)) 2 Ccos(r, Vv(‘17, —2r)) 74 in B, x (—r, r) (C = C/L). Then if q E B(r,p(r)) then sin(r,q) S Elm < TI— Ccos(r, VVW, -2r)). Hence sin(r, q) S cos(r, Vv(x, t)). Hence qu(a:,t) = |Vv|cos(q,Vv(x,t)) Z |Vv|sin(r, q) 2 0. So let S ~ = Ur€K(u,Go)B(T) 9(7)); then q 6 36* => qu(:c,t) Z O in B, X (—r, r). Then by Lemma 16 of [5],.5'5. contains an intermediate cone K (14, 91). That is there exist V1 6 RN“, 81 2 80 + S(g — 60) such that K0490) c I‘m/1,9,) c SC. where S depends only on C and 90 and we can choose it in such a way that it will be the same as we obtain by replacing 80 by any 9 2 90 in the definition of San. Let v,/5(x, t) = f:v(%rz,%rt) for (x, t) 6 B5 x (—5,5); Then v = vg satisfies equation (1.4) and the conditions of Corollary 3.5 with the same constants )8, L and Cl. Also, for any r 6 fat/1,61), D,v%(x,t) 2 O in B5 x (—5,5). So we can repeat the argument above and deduce that, in B, x (—r, r), qug (113, t) Z 0 75 for any q E L((Vg, 82) D K(V1,91) with 82 Z 61+S(-§—91); That is D,v(a:, t) Z 0 in B(gp x (—(§)2, (92) for any r E K012, 62). Rescaling and repeating the argument above we end up with a monotone family of cones K(V),,Gk) such that 8), 2 91,4 + S(g — 91,4) and D,v(:c, t) Z 0 in B(gy. x (—(g)", (9") for any r E [((1/(”60. Proof of Theorem 1.2. Let (2:0, to) be the point as in Theorem 1.2. Then by dilation we may assume (x°,t°) E I‘nBl x (—1,1). Let w(x,t) = 1)),(230 + 51:, t0 + t). Then w(0,0) = vh(.1:1, t1) and w satisfies the conditions of Corollary 3.5 in B5 x (—5, 5) and by Proposition 4.8, there exists a monotone family of cones K (V1,, 6),) such that DJ 2 0 in By. x (—J’°, J") ifr E K(uk,6k) with G), 2 Gk_1+ S(%7r — 91—1) J and S' > 0 depend only on constants in Corollary 3.5. Since G), —> é-ir as k —-) 00, the free boundary of w is differentiable at (0, 0) i.e., I‘ is differentiable at (1:0, to). In fact, 9*? Z "23'” +0" Slek—l 2%71 + (1 — S)(-‘2§71 + (1 — S)@k_1) : §W+§jl—2:—S—)7r+(l—Slzek—224~ Z §7T(1+(l-S)+---+(1—S)"’1)+(1—S)"(—)0 _ 7r 1—(1—S)" k — 251—(1—s)+(1 S)Oo 7r = 5(1 — (1 -—- S)’°) + (1 —— S)"Oo 5 - (1— SH; — ea) 76 Also, Vk+1 - Vk|2 = 2 _ 2(Vk+1, Vic) = 2(1 _ C056) _<_ 26)2 _<_ 2(32E — 69,.)2 s a; — eo)4<1 — s>44. Let us write v(x”, to) = limkaoo 11],. Then we have (4.2) |V(:c°,t°) — ukl S CA,“ with 0 < A < 1. Assume now (2:, t) E I‘ and suppose 7. slk' T Iz-xOI < (5)4. 11—141 <( Then D,u)(y, s) 2 0 ifw > 0 for any r E L((z/k, 9k) and (y, 3) close to (x, t). Thus (Vv(y, s),1/),) S %1r — 8),. Since Vv(y, 8) WVWBJ) as (mu—40914)) we deduce that (u(x,t), 1),.) _<_ -;-7r — e,c g (1 — S)"(:,12-7r — ea). 77 Consequently |u(:z:,t) — ukl S CA" with the same constants C and A as in (4.2). Therefore |1/(a',t) — u($1,t1)| S 2CA". 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