Holder continuity for two-phase flows in porous media

全文

(1)MATHEMATICAL METHODS IN THE APPLIED SCIENCES Math. Meth. Appl. Sci. 2006; 29:1261–1289 Published online 13 March 2006 in Wiley InterScience (www.interscience.wiley.com). DOI: 10.1002/mma.724 MOS subject classication: 35 B 27; 35 B 45; 35 K 55; 35 K 65; 76 S 05. Holder continuity for two-phase ows in porous media Li-Ming Yeh∗; † Department of Applied Mathematics; National Chiao Tung University; Hsinchu 30050; Taiwan; R.O.C.. Communicated by W. Sproig SUMMARY This work is to prove the Holder continuity of the solutions of the degenerate dierential equations describing two-phase, incompressible, immiscible ows in porous media. The dierential equations allow degeneracy at two end points and the assumption on mild degeneracy is not required in this study. The regularity result is proved by an alternative argument. Uniqueness of the weak solutions of the dierential equations is a direct consequence from this Holder continuity. Copyright ? 2006 John Wiley & Sons, Ltd. KEY WORDS:. two-phase ow; global pressure; alternative argument. 1. INTRODUCTION Holder continuity of the solutions of the degenerate dierential equations describing two-phase, incompressible, immiscible ows in porous media is concerned. The existence of weak solutions of the dierential equations is well-known (see References [1–7] and references therein). However, the regularity of the weak solutions has not been wellestablished. In this work, we show Holder continuity of the weak solutions for the dierential equations. Uniqueness of the weak solutions is then a direct consequence from this result. If ⊂ N (N = 3 in reality) is a porous medium, equations for the two-phase ows in porous media in global pressure formulation are (see References [2,4]),. ∗ Correspondence. to: Li-Ming Yeh, Department of Applied Mathematics, National Chiao Tung University, Hsinchu 30050, Taiwan, R.O.C. [email protected]. † E-mail:. Contract=grant sponsor: National Science Council; contract=grant number: NSC 91-2115-M-009-003. Copyright ? 2006 John Wiley & Sons, Ltd.. Received 15 November 2004.

(2) 1262 for t¿0,. L.-M. YEH. w o ∇ (S) = 0 @t S − ∇ · Kw (S)∇(P − Ew ) − K . (1). −∇ · (K(S)∇P − Kw (S)∇Ew − Ko (S)∇Eo ) = 0. (2). Here is porosity, K is absolute permeability eld, S ∈ [0; 1] is water saturation, i (i = w; o) is phase mobility of i-phase and is a nonnegative monotone function of S, ( ≡ w + o ) is the total mobility, P denotes global pressure, Ei (i = w; o) is a function depending on density, gravity, and position, and is capillary pressure and is a nonnegative decreasing function of S. In practice, (w o =) (0) = (w o =) (1) = 0 [1,2,4]. So Equation (1) is a degenerate parabolic equation with degeneracy at two end points (that is, degeneracy appears at S = 0; 1). Boundary @ of the porous medium includes 1 and 2 satisfying 1 ∩ 2 = ∅ and. 1 ∪ 2 = @. The initial and boundary conditions are given by ⎧ w o ⎪ ⎪ K(w (S)∇(P − Ew ) − ∇ (S)) · n = 0 for x ∈ 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ K((S)∇P − w (S)∇Ew − o (S)∇Eo ) · n = 0 for x ∈ 1 ⎨ (3) for x ∈ 2 S = Sb ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ P = Pb for x ∈ 2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ S(0; x) = S (x) for x ∈ init. where n is the unit vector outward normal to 1 . Regularity results of the weak solutions for porous media problems in nondegenerate case are well-known (see References [2,4,6] and reference therein). Continuity of saturation S for (1)–(3) in interior region of T ( ≡ (0; T ) × ) had been shown in Reference [1] if mild degeneracy was assumed at one end point. Holder continuity of S with degeneracy only at one end point was considered in Reference [5]. In this work, we prove Holder continuity of S for the case where equations are degenerate at two end points and no mild degeneracy is assumed. Though initial and boundary values of saturation are assumed to be away from the two end points 0 and 1 (see A5 below), the saturation inside the domain T still can reach 0 and 1. The process of proof is rst to derive a uniform Holder estimate for the solutions of regularized problems of (1)–(3). Then by compactness principle we get Holder continuity of the solution of (1)–(3). Rest of the paper is organized as follows: Notation and main result are stated in Section 2. In Sections 3–5, we shall derive a uniform Holder estimate for the solutions of the regularized problems of (1)–(3) by an alternative argument [8]. More precisely, in Section 3 we state some auxiliary results needed in Section 4. Holder estimate of the solutions for the regularized equations in the interior region is given in Section 4. Holder estimate of the solutions for the regularized equations on the parabolic boundary can be proved by a similar argument as that for interior region and is sketched in Section 5. Proof of main result is in Section 6. Copyright ? 2006 John Wiley & Sons, Ltd.. Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(3) HOLDER CONTINUITY FOR TWO-PHASE FLOWS. 1263. 2. NOTATION AND MAIN RESULT We shall use the following notation: ⎧ T. i ≡ (0; T ) × i ; i = 1; 2 ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ U ≡ { ∈ H () : | 2 = 0} ⎪ ⎪ ⎪ ⎪ V(T ) ≡ L∞ (0; T ; L2 ()) ∩ L2 (0; T ; U) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ dual X ≡ dual space of X ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ Lq;r (T ) ≡ Lr (0; T ; Lq ()) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2 2 ⎨ (k1 is given in A4 below) ≡ − N k1 ⎪ ⎪ ⎪ ⎪ 2 ⎪ ⎪ ; ≡ min ⎪ ⎪ N +2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −w o ⎪ ⎪ J(z) ≡ (z); z ∈ (0; 1) ⎪ ⎪ ⎪ ⎪. ⎪ z ⎪ ⎪ ⎪ ⎪ R (z) ≡ J() d; z ∈ (0; 1) ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎩ XD is a characteristic function of D. (4). Denition 2.1 Boundary @ of the bounded domain belongs to H∗m , m¿1, if (1) in the neighbourhood U (x) of each boundary point x ∈= 1 ∩ 2 there exists a homeomorphic transformation ˜ |¿c¿0 (d x=dx ˜ is the Jacobian of the transformax(x) ˜ = (x˜1 (x); x˜2 (x); : : : ; xÑ (x)) ∈ C m , |d x=dx tion) such that xÑ (@ ∩ U (x)) = 0; xÑ ( ∩ U (x))¿0, i.e. i (i = 1; 2) can be locally straightened, ˜ with (2) in the neighbourhood of each point x ∈ 1 ∩ 2 there exists a transformation x˜ = x(x) the same properties mapping it at the neighbourhood of the edge of a cube in variable x. ˜ Next we make the following assumptions: A1. @ ∈ H∗1 , A2. w (resp. o ) : [0; 1] → [0; 1] is continuous and increasing (resp. decreasing), w (0) = o (1) = 0, w o (z)|z ∈ (0;1) = 0, inf z∈[0;1] (z)¿0, A3. : (0; 1] → + 0 is onto, decreasing, and a locally Lipschitz continuous function, and inf z∈(0;1] | (z)|¿0, (w o =) (z) ∈ L∞ ((0; 1)), A4. 0¡K ∈ W 1;∞ (), Ew ; Eo ∈ L∞ (0; T ; W 1;∞ ()), Pb ∈ L∞ (0; T ; W 1;k1 ()), k1 ¿N , A5. Sb ; Sinit ∈ L2 (0; T ; H 1 ()) ∩ C 0;k2 (T ), @t (Sb ) ∈ L1 (T ), Sb ; Sinit ∈ (k2 ; 1 − k2 ), Sinit | 2 = Sb | T2 (t = 0), ∈ (k3 ; k4 ), A6. maxz∈[0;1] |(z) − 1. | + |(k4 =k3 ) − 1|¡k5 (k5 is small and depends only on , K), A7. w o (z) ∝ z |1 − z | J(z) and J(z) ∝ z m |1 − z |m1 for z ∈ (0; #) ∪ (1 − #; 1), where ki (i = 1; : : : ; 5); m; m1 are positive constants, and # ∈ (0; 1=8) is a number such that J is increasing (resp. decreasing) in (0; #) (resp. (1 − #; 1)). Some remarks about the assumptions are given below: A1 will be used to derive the regularity of water saturation S and global pressure P around the edge 1 ∩ 2 . From A3, Copyright ? 2006 John Wiley & Sons, Ltd.. Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(4) 1264. L.-M. YEH. | (z)| may tend to innity as z → 0 or 1, a property for a capillary pressure function [2,4]. (w o =) (z) ∈ L∞ ((0; 1)) allows equation (1) to be a degenerate parabolic equation, a characteristic of the porous medium equation (PME) [2,4]. A5 means initial and boundary values of the saturation are away from the two end points 0 and 1. However, the saturation inside the domain T still can reach 0 and 1. The assumption max |(z) − 1| ¡ k5 in A6 z∈[0;1]. is used in Lemma 3.1 and | kk43 − 1| ¡ k5 is used in Lemma 4.4. The explicit restriction of k5 can be found in Reference [2, p. 224, Theorem 4.2] and in the proof of Lemma 4.4. A7 gives restrictions on water (resp. oil) mobility function and capillary pressure around the neighbourhood of 0 (resp. 1). The assumptions on the two end points 0 and 1 are similar, so one can expect water saturation in the neighbourhood of the two end points 0 and 1 has similar properties. These properties are crucial in the regularity proof of saturation in this work. Since we assume m; m1 are positive constants in A7, no mild degeneracy is required. Our main result is Theorem 2.1 Under A1–A7, saturation S of (1)–(3) is Holder continuous in T . Under A1–A7 as well as ∈ W 1;∞ () and Pb ; Ew ; Eo ∈ L∞ (0; T ; C 1;k ()), we have P ∈ L∞ (0; T ; W 1;∞ ()) by Theorem 2.1 and Corollary 8.35 of Reference [9]. Therefore, by Theorem 2.3 [7], we get uniqueness of weak solution of (1)–(3).. 3. SOME AUXILIARY LEMMAS We rst derive regularized equations of (1)–(3). Let be a small number satisfying 0¡¡k2 =4. Extend i (i = w; o) constantly and continuously to and dene ˜ i ; ˜ as z− ˜ i (z) ≡ i 0:5 ; 0:5 − . ˜ (z) ≡ ˜ w (z) + ˜ o (z). (5). ; Sb such that By A5, there exist smooth functions Sinit. ; Sb ∈ Sinit. k2 k2 ;1 − 2 2. . ;. Sinit | 2 = Sb | T2 (t = 0). ⎧ ⎨ Sinit ; Sb → Sinit ; Sb. in L2 (0; T ; H 1 ()) ∩ C 0;k2 (T ). ⎩ @ (S ) → @ (S ) t t b b. in L1 (T ). (6). as → 0. (7). The regularized problem is: Find {S ; P } satisfying @t S ∈ dual L2 (0; T ; U). (8). 6S 61 − . (9). Copyright ? 2006 John Wiley & Sons, Ltd.. Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(5) HOLDER CONTINUITY FOR TWO-PHASE FLOWS. 1265. R(S ) − R(Sb ); P − Pb ∈ L2 (0; T ; U). . T. T. (10). @t S +. T. K(˜ w (S )∇(P − Ew ) + ∇R(S ))∇ = 0. K ˜ (S )∇P ∇ − i∈{w;o}. (11). T. K ˜ i (S )∇Ei ∇ = 0. (12). S (0; x) = Sinit. (13). for any ; ∈ L2 (0; T ; U): It is easy to see that for each xed , (11) is a nondegenerate parabolic equation and (11)–(12) imply, if So ≡ 1 − S ,. T. @t So +. T. K(˜ o (1 − So )∇(P − Eo ) − ∇R(1 − So ))∇ = 0. (14). By References [5,7,10,11], it is known Lemma 3.1 Under A1–A6 and (6)–(7), (8)–(13) has a weak solution {S ; P } for each . Moreover,

(6) P

(7) L∞ (0;T ;W 1;k1 ()) is bounded by a constant which is independent of , S is Holder continuous in T , and ⎧ ⎨S → S ⎩. pointwise and in Lr (T ); r¡∞. R(S ); P → R(S); P. in L2 (0; T ; H 1 ()). as → 0. where {S; P } is a weak solution of (1)–(3). We shall prove that the Holder norm of S is actually bounded by a constant which is independent of . If that is so, then by Lemma 3.1, we obtain S is Holder continuous and complete the proof of Theorem 2.1. From now on, A1–A7 will be assumed throughout this paper, and is xed and dropped for convenience of presentation. Given any constant ¿0, dene the cube K ≡ x ∈ N : max |xi |¡ (15) 16i6N. ˆ Also, for ¿0 a given number, dene For xˆ ∈ N , xˆ + K denotes the cube of centre x. Q(; ) ≡ (−; 0) × K . For (tˆ; x) ˆ ∈ N +1 , let (tˆ; x)+ ˆ Q(; ) be the cylinder congruent to Q(; ); ˆ + Q(; ), we introduce piecewise smooth i.e. (tˆ; x) ˆ + Q(; ) = (tˆ − ; tˆ) × {xˆ + K }. In (tˆ; x) cut-o functions (t; x) and (x) such that both satisfy ˜ ∈ [0; 1];. |∇˜|¡∞;. Copyright ? 2006 John Wiley & Sons, Ltd.. ˜ x) = 0 (t;. for x ∈= xˆ + K. (16). Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(8) 1266. L.-M. YEH. For ∈ L1 (T ) and 0¡h¡T , introduce the Steklov average h by ⎧ t+h ⎪ ⎨1 (; x) d if 0¡t 6T − h h (t; x) ≡ h t ⎪ ⎩ 0 if t¿T − h (11) implies the following equation:. @t Sh ’ + K(∇R(S) + ˜ w (S)∇(P − Ew ))h ∇’ = 0 K. (17). K. T where K is any compact subset of , 0¡t¡T − h, ’ ∈ H01 (K ) ∩ L∞ loc (). Domain includes three subregions:. ⎧ T 1 ≡ {(t; x) ∈ T : S(t; x)¡#} ⎪ ⎪ ⎪ ⎪ ⎨ # # T T 2 ≡ (t; x) ∈ : ¡S(t; x)¡1 − ⎪ 4 4 ⎪ ⎪ ⎪ ⎩ T 3 ≡ {(t; x) ∈ T : 1 − #¡S(t; x)}. (18). Next we derive energy and logarithmic estimates for the interior region T . For (tˆ; x) ˆ ∈ T ± T xed, let and be small so that (tˆ; x)+ ˆ Q(; ) ⊂ . Dene a set Dj; () as, for ∈ (tˆ− ; tˆ) and for every level j, ± Dj; () ≡ {x ∈ xˆ + K : (S − j)± (; x)¿0}. where.

(9). (S − j)+ ≡. S −j. if S − j¿0. 0. otherwise.

(10). (S − j)− ≡. and. (19). j−S. if S − j¡0. 0. otherwise. Lemma 3.2 ˆ Q(; ) ⊂ T1 There is a constant d1 (independent of ; ; ; j) such that for every cylinder (tˆ; x)+ and every level j, we have. 2 2 (S − j)+ + J(S)|∇(S − j)+ |2 sup (t;ˆ x)+Q(;) ˆ. ˆ ˆ t−¡t¡ tˆ x+K. . 6d1. (S − j)2+ 2 (tˆ − ; x) + x+K ˆ . + (t;ˆ x)+Q(;) ˆ. (S − j)2+ @t +. tˆ. ˆ t−. (t;ˆ x)+Q(;) ˆ. J(S)(S − j)2+ |∇|2 . + |Dj; ()|(1−2=k1 ) d. (20). where is a piecewise smooth cut-o function satisfying (16). Equation (20) also holds if (tˆ; x) ˆ + Q(; ) is a subset of T2 or T3 . But for T3 case, S in (20) should be replaced by So ≡ 1 − S. Copyright ? 2006 John Wiley & Sons, Ltd.. Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(11) HOLDER CONTINUITY FOR TWO-PHASE FLOWS. 1267. Proof Without loss of generality, let (tˆ; x) ˆ = (0; 0). Take ’ = (Sh − j)+ 2 in (17) and integrate the resulting equation over (−; t) for t ∈ (−; 0) to obtain. t. −. K. @t Sh ’ +. t. −. K. (K J(S)∇S)h ∇’. + −. t. K. (K ˜ w (S)∇(P − Ew ))h ∇’ = 0. (21). We now estimate each of the terms in (21). First, we integrate by parts in t, let h → 0+ , and apply Lemma 3.2 of Chapter 1 [8] to see t . t 2 2 2 t (S − j)+ |− − (S − j)2+ @t 2 @t Sh (Sh − j)+ = lim h→0+ − K 2 − K 2 K Next, by Holder inequality, t (K J(S)∇S)h ∇((Sh − j)+ 2 ) lim+ h→0. ¿. −. 1 2. K. t.

(12). −. K J(S)∇(S − j)+

(13) 2L2 (K ) − c1. t. −.

(14). K J(S)(S − j)+ ∇

(15) 2L2 (K ). Because Q(; ) ⊂ T1 , by (5) and A7, t (K ˜ w (S)∇(P − Ew ))h ∇((Sh − j)+ 2 ) lim+ h→0. 6. −. 1 4. K. −. + c3. t.

(16). t. −. K J(S)∇(S − j)+

(17) 2L2 (K ) + c3. K. t. −.

(18). K J(S)(S − j)+ ∇

(19) 2L2 (K ). (|∇P |2 + |∇Ew |2 )X{(S−j)+ ¿0}. where X{(S−j)+ ¿0} is the characteristic function (see (4)). Now we combine the above results with Holder inequality and Lemma 3.1 to obtain (20). If (tˆ; x) ˆ + Q(; ) ⊂ T2 , Equation (21) corresponds to a uniform (independent of ) parabolic equation. So we may repeat above argument to obtain (20). If (tˆ; x)+ ˆ Q(; ) ⊂ T3 , we use (14) and repeat above argument. Note that Equation (11) around the end point 0 has similar properties as (14) around the other end point 1. Therefore we can easily get the same inequality (20) as (tˆ; x) ˆ + Q(; ) ⊂ T1 case except replacing S in (20) by So . Let us dene a function in (tˆ; x) ˆ + Q(; ) as Hj (Hj ; (S − j)+ ; ) ≡ ln+ Hj − (S − j)+ + . (22). where Hj ≡ sup(t;ˆ x)+Q(;) (S − j)+ , ln+ ’ ≡ max{0; ln ’}, 0¡ ¡ min{1; Hj }. ˆ Copyright ? 2006 John Wiley & Sons, Ltd.. Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(20) 1268. L.-M. YEH. Lemma 3.3 ˆ Q(; ) ⊂ T1 There is a constant d1 (independent of ; ; ; j) such that for every cylinder (tˆ; x)+ and every level j. 2 (Hj ; (S − j)+ ; )(t; x)2. sup ˆ ˆ t−¡t¡ tˆ x+K. . 6. (Hj ; (S − j)+ ; )(tˆ − ; x) + d1 2. x+K ˆ . +. J(S) (Hj ; (S − j)+ ; )|∇|2. 2. |1 + ln+ (Hj = )|. 2. tˆ. ˆ t−. (t;ˆ x)+Q(;) ˆ. + |Dj; ()|(1−2=k1 ). d. (23). where is a piecewise smooth cut-o function satisfying (16). Equation (23) also holds if (tˆ; x) ˆ + Q(; ) is a subset of T2 or T3 . But for T3 case, S in (23) should be replaced by So ≡ 1 − S. Proof Let (tˆ; x) ˆ = (0; 0). For convenience, set Hj (x) ≡ (Hj ; (x − j)+ ; ) = ln Hj − (x − j)+ + +. (24). By (22) and (24), we have 06 (Sh )6 ln+ (Hj = ); 06 (Sh )61= ; 0¡ ¡1. Since T 2 2 ) (Sh ) in (17) and integrate ( 2 ) (Sh ) = 2(1 + )( )2 ∈ L∞ loc ( ) by (22), we take ’ = ( the resulting equation over (−; t) for t ∈ (−; 0) to see. t. −. K. @t Sh (. t. ) +. t. −. K. (K J(S)∇S)h ∇((. + −. 2 2. K. 2 2. ) ). 2 2. (K ˜ w (S)∇(P − Ew ))h ∇((. ) )=0. (25). Each term of (25) is estimated similarly as that for (21). First. lim+. h→0. t. −. Next. @t Sh (. K. t. −. ¿. t. −. ) = lim+ h→0. lim. h→0+. 2 2. K. (K J(S)∇S)h ∇((. K. t. −. 2 2. K. ) )=. @t. t. −. 2 2. =. 2. (S)2 ()|t−. K J(S)∇S ∇((. K. K J(S)|∇S |2 (1 + )( )2 2 − c1. Copyright ? 2006 John Wiley & Sons, Ltd.. K. t. −. 2 2. ) ). K. K J(S) |∇|2. Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(21) HOLDER CONTINUITY FOR TWO-PHASE FLOWS. 1269. Since Q(; ) ⊂ T1 , by (5) and A7,. t. (K ˜ w (S)∇(P − Ew ))h ∇((. lim. h→0+. 6. −. 1 4. K. t. −. K. + c1. −. ) ). K J(S)|∇S |2 (1 + )( )2 2 +. t. 2 2. K. t. −. K. K J(S) |∇|2. (|∇Ew |2 + |∇P |2 )(1 + )( )2 2. Then, combine above estimates with Holder inequality and Lemma 3.1 to get (23). If (tˆ; x) ˆ + Q(; ) ⊂ T2 , (25) is in uniform (independent of ) parabolic equation case and we may repeat above argument to obtain (23). If (tˆ; x) ˆ + Q(; ) ⊂ T3 , we use (14) and repeat above argument to get (23) except that S in (23) should be replaced by So . 4. HOLDER ESTIMATE OF REGULARIZED EQUATIONS IN INTERIOR REGION In this section, we give a Holder estimate for S in the interior region of T . Let O( ⊂ T ) be a closure of O and let ∗ ; $ be two small positive numbers satisfying | ∗ |$ 6 min{1; J(#=2); J(1 − #=2)}. (26). ˆ ∈ O and ¡ ∗ . Dene + ≡ sup(t;ˆ x)+Q(. as well as (tˆ; x)+ ˆ Q( 2−$ ; 2 ) ⊂ T for any (tˆ; x) 2−$ ;2 ) S, ˆ − + − S, and ! ≡. −. . By (9), ≡ inf (t;ˆ x)+Q(. 2−$ ˆ ;2 ) !=0. or. 0¡!6#=2. or. #=26!61. (27). ˆ Q( 2−$ ; 2 ). If ! = 0, S is constant in (tˆ; x)+ ˆ Q( 2−$ ; 2 ). So S is Holder continuous in (tˆ; x)+ The other two cases of (27) are discussed in Sections 4.1, 4.2 separately. 4.1. For 0¡!6#=2 case Since !6#=2, we have + ¡# or #=4¡ − 6 + ¡1 − (#=4) or 1 − #¡ − . By (18), ⎧ T 1 if + ¡# ⎪ ⎪ ⎪ ⎪ ⎨ # # (tˆ; x) ˆ + Q( 2−$ ; 2 ) ⊂ T2 if ¡ − 6 + ¡1 − ⎪ 4 4 ⎪ ⎪ ⎪ ⎩ T − 3 if 1 − #¡. (28). 4.1.1. For + ¡# case. Set M to be a constant satisfying M=26 + 6M 6# and dene 1=

(22) ≡ J(M ). Then

(23) ¿ −$ or

(24) ¡ −$ . The former (i.e.

(25) ¿ −$ ) implies, by A7, !6c $=m. (29). for a constant c. The latter (i.e.

(26) ¡ −$ ) means Q(

(27) 2 ; ) ⊂ Q( 2−$ ; 2 ) and is discussed below. Note (tˆ; x) ˆ + Q( 2−$ ; 2 ) ⊂ T1 here. Copyright ? 2006 John Wiley & Sons, Ltd.. Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(28) 1270. L.-M. YEH. 4.1.1.1. First alternative Lemma 4.1 ˆ ; + ; −; ) For any ‘1 ¿2, there is a 1 ∈ (0;1) (depending on ‘1 and data but independent of tˆ; x; 2 2−$ so that if Q(

(29) ; ) ⊂ Q( ; 2 ) and ⎧ ! ⎪ ˆ + Q(

(30) 2 ; ) : S(t; x)¡ − + ‘1 6 1 |Q(

(31) 2 ; )| ⎨ (t; x) ∈ (tˆ; x) 2 (30) ! −2 ⎪ ⎩

(32) N 61 2‘1 ˆ + Q(

(33) | =2|2 ; =2). See (4) for . then S(t; x)¿ − + (!=2‘1 +4 ) for (t; x) ∈ (tˆ; x) Proof After translation we assume (tˆ; x) ˆ = (0; 0). Dene n ≡ ( =2)+( =2n+1 ); n = 0; 1; 2; : : : : Construct a family of nested cylinders Q(

(34) 2n ; n ) and let n be a piecewise smooth cut-o functions in Q(

(35) 2n ; n ) such that ⎧ 0¡n (t; x)61 for (t; x) ∈ Q(

(36) 2n ; n ) ⎪ ⎪ ⎪ ⎪ ⎪ in Q(

(37) 2n+1 ; n+1 ) n = 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ on the parabolic boundary of Q(

(38) 2n ; n ) =0 ⎪ ⎪ n ⎪ ⎪ ⎪ 2n+2 ⎨ |∇n |6c (31). ⎪ ⎪ 2 ⎪ ⎪ 2(n+2) ⎪ ⎪ ⎪ |

(39) n |6c ⎪ ⎪. ⎪ ⎪ ⎪ 2 ⎪ ⎪ ⎪ 1 2(n+1) ⎪ ⎩ 06@t n 6

(40). Dene in Q( 2−$ ; 2 ). ⎧ ! − ⎪ ≡ max S;. + ⎪ ⎪ ⎪ 2‘1 +4 ⎪ ⎪ ⎨ ! ! jn ≡ − + ‘1 +1 + ‘1 +1+n ; n = 0; 1; 2; : : : 2 2 ⎪ S ⎪ ⎪ ⎪ ! − ⎪ ⎪ − F (S) ≡ − max ;. + j n ⎩ ‘1 +4 jn. Then. 2. −. d. ⎧ 1 ! 7! ! 2 − ⎪ ⎪ F (S) = | ( − j ) | + + S −. − n − ⎪ ⎪ 2 2‘1 +4 2‘1 +1+n 2‘1 +4 − ⎪ ⎪ ⎪ ⎪ ⎨ @t F(S) = − ( − jn )− @t S ! 2 ⎪ ⎪ F (S) 6 c ‘1 ⎪ ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎩ ( − j ) 6(S − j ) 6 ! n − n − ‘1. (32). 2. Copyright ? 2006 John Wiley & Sons, Ltd.. Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(41) HOLDER CONTINUITY FOR TWO-PHASE FLOWS. 1271. We now take = − ( − jn )− n2 XQ(

(42) 2n ; n ) in (11) and estimate each term as follows: By (31)–(32),. − ( − jn )− n2 @t S = F(S)n2 (0) − F(S)@t n2 Q(

(43) 2n ; n ). ¿. K n. −. Q(

(44) 2n ; n ). 2 1 c 2n ! |( − jn )− |2 n2 (0) − ‘1 X{(−jn )− ¿0} 2

(45) 2 Q(

(46) 2n ; n ). Q(

(47) 2n ; n ). ¿. K n. K ∇R(S)∇(( − jn )− n2 ). Q(

(48) 2n ; n ). K J()|∇(n ( − jn )− )|2. −. Q(

(49) 2n ; n ). K( −. jn )− ∇R(S)∇n2. 2 c 2n ! − ‘1 X{(−jn )− ¿0}

(50) 2 Q(

(51) 2n ; n ). If − 6 12 + , then + 62!. So J(M=2‘1 +6 )6J( + =2‘1 +5 )6J(). If 12 + ¡ − , then J(M=4)6J( + =2)6J(). Therefore, by A7, we have J()¿(1=

(52) )|1=2‘1 +6 |m in Q(

(53) 2n ; n ). Note 06R(S) − R( − )6(S − − )J(S) for S ∈ ( − ; jn ). By (31)–(32), (28)1 , and A7,. K( − jn )− ∇R(S)∇n2 = K( − jn )− ∇(R(S) − R( − ))∇n2 Q(

(54) 2n ; n ). =− 1 6 4. Q(

(55) 2n ; n ). Q(

(56) 2n ; n ). 1 4. Q(

(57) 2n ; n ). (R(S) − R( − ))∇(K( − jn )− ∇n2 ). 6. Q(

(58) 2n ; n ). K J(S)|∇( −. jn )− |2 n2. 2 c 2n ! + ‘1 X{(−jn )− ¿0}

(59) 2 Q(

(60) 2n ; n ). K ˜ w (S)∇(P − Ew )∇(( − jn )− n2 ). Q(

(61) 2n ; n ). K J(S)|∇( − jn )− |2 n2. 2 0 c 2n ! + ‘1 X{(−jn )− ¿0} d + d1 |Dj−n ; n ()|(1−2=k1 ) d

(62) 2 −

(63) 2n Q(

(64) 2n ; n ). where d1 = d1 (K;

(65) ∇P; ∇Ew

(66) L∞ (0;T ;Lk1 ()) ). Combining above estimates, it is not di cult to see m . 1 1 sup n2 ( − jn )2− dx + ‘1 +6 |∇(n ( − jn )− )|2

(67) 2 Q(

(68) 2n ; n ) −

(69) 2n ¡t¡0 K n 2 0 c 2n ! 6 ‘1 X{(−jn )− ¿0} d + d1 |Dj−n ; n ()|(1−2=k1 ) d

(70) 2 −

(71) 2n Q(

(72) 2n ; n ) Copyright ? 2006 John Wiley & Sons, Ltd.. (33). Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(73) 1272. L.-M. YEH. We claim. lim. n→∞. Q(

(74) 2n ; n ). X{(S−jn )− ¿0} d = 0. If so, since jn j∞ ¿ − + (!=2‘1 +4 ), this would imply 2 (t; x) ∈ Q

(75) ; : S(t; x)¡ − + ! =0 ‘ +4 2 2 21. and proves the lemma. Change variable z = t=

(76) which transforms Q(

(77) 2n ; n ) into Qn ≡ Q( 2n ; n ) ˆ ·) ≡ (z

(78) ; ·), ˆn (z; ·) ≡ n (z

(79) ; ·), Dn (z) ≡ {x ∈ K n : (z; ˆ x)¡jn } = [− 2n ; 0] × K n . Dene (z; 0 and |Dn | ≡ − 2 |Dn (z)| dz. Equation (33) can be written as n. . n 2 2 m(‘1 +6) 2 ! ˆ ˆ

(80) n ( − jn )−

(81) V(Qn ) 6c2 |D | +

(82) n 2‘1 . . 0. − 2n. |Dn ()|. (1−2=k1 ). d. (34). Since ˆn (ˆ − jn )− vanishes on the lateral boundary of Qn , by Corollary 3.1 of Chapter 1 [8] we have, by (34), |Dn+1 | ! 2 6 |jn − jn+1 |2 |Dn+1 |6

(83) (ˆ − jn )−

(84) 2L2 (Qn+1 ) 2(n+2) ‘1 . 2. 2. 6

(85) ˆn (ˆ − jn )−

(86) 2L2 (Qn ) 6c

(87) ˆn (ˆ − jn )−

(88) 2V(Qn ) |Dn |2=(N +2) 6 c2. m(‘1 +6). . 2n ! 2 |Dn |(N +4)=(N +2) +

(89) |Dn |2=(N +2) 2‘1 . . 0. − 2n. |Dn ()|. (1−2=k1 ). d. (35). 0 Dene Yn ≡ |Dn |= |Qn |, Zn ≡ 1= |K n |( − 2 |Dn ()|(1−2=k1 ) d)1=(1+) , where is dened in (4). n Divide (35) by |Qn+1 | to obtain, by (30)2 ,. Yn+1 6c2m(‘1 +6) 42n (Yn1+(2=(N +2)) + Yn2=(N +2) Zn1+ );. n = 0; 1; 2; : : :. (36). Next by the embedding of Proposition 3.3 of Chapter 1 [8] Zn+1 (jn − jn+1 )2 6.

(90) ˆn (ˆ − jn )−

(91) 2Lq; r (Qn ) |K n+1 |. 6c −N

(92) ˆn (ˆ − jn )−

(93) 2V(Qn ). where r = 2(1 + ) ∈ [2; ∞), q = r=(1 − 2=k1 ) ∈ [2; 2N =(N − 2)]. Therefore Zn+1 6c2m(‘1 +6) 42n (Yn + Zn1+ ); Copyright ? 2006 John Wiley & Sons, Ltd.. n = 0; 1; 2; : : :. (37). Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(94) HOLDER CONTINUITY FOR TWO-PHASE FLOWS. 1273. By (36)–(37) and Lemma 4.2 of Chapter 1 [8], it follows that limn→∞ Yn = limn→∞ Zn = 0, provided Y0 + Z01+ 6 ≡ |c2m(‘1 +6) |−(1+)= 4−2(1+)=. 2. (38). See (4) for . By (30)1 and Z0 6Y02=q , if we select 1 satisfying 1 + 12(1+)=q = , then (38) holds. So we complete the proof of the lemma. Lemma 4.2 For any ‘1 ¿2, there exist numbers 1 ∈ (0; 1); 1 ∈ (1=2; 1) and I1 1 (depending upon ‘1 and given data but independent of tˆ, x, ˆ , + , − , ) so that if Q(

(95) 2 ; ) ⊂ Q( 2−$ ; 2 ) and ! ˆ + Q(

(96) 2 ; ) : S(t; x)¡ − + ‘1 6 1 |Q(

(97) 2 ; )| (39) (t; x) ∈ (tˆ; x) 2 then either !6I1 N=(2+m). (40). or ess osc(t;ˆ x)+Q(

(98) | =2| 2 ; =2) S 61 !. ˆ Proof Assume (40) is violated. By A7 and Lemma 4.1, inf. 2 ; =2) (t;ˆ x)+Q(

(99) | =2| ˆ. S ¿ − +. ! 2‘1 +4. (41). (‘1 +4) ))!. So the From (41) and denition of + , we get ess osc(t;ˆ x)+Q(

(100) | =2| 2 ; =2) S 6(1 − (1=2 ˆ (‘1 +4) lemma follows with 1 = 1 − (1=2 ).. 4.1.1.2. Second alternative. In this subsection, we shall x ‘1 (so 1 is xed as well) and assume that (39) of Lemma 4.2 is violated, i.e. for the subcylinder Q(

(101) 2 ; ) ⊂ Q( 2−$ ; 2 ) ! ˆ + Q(

(102) 2 ; ) : S(t; x)¡ − + ‘1 ¿ 1 |Q(

(103) 2 ; )| (42) (t; x) ∈ (tˆ; x) 2 Since + − (!=2‘1 )¿ − + (!=2‘1 ) for all ‘1 ¿2, (42) implies ! ˆ + Q(

(104) 2 ; ) : S(t; x)¿ + − ‘1 6(1 − 1 )|Q(

(105) 2 ; )| (t; x) ∈ (tˆ; x) 2. (43). In view of (43), we will study the behaviour of S near its supremum + and work with the truncated function (S − j)+ for the levels j = + − (!=2‘1 +n ); n¿0. Lemma 4.3 Under (43), there is a time level t ∗ ∈ [tˆ −

(106) 2 ; tˆ −

(107) 2 1 =2] so that ! 1 − 1 ∗ + |K | x ∈ K : S(t ; xˆ + x)¿ − ‘1 6 2 1 − 1 =2 See Lemma 4.2 for ‘1 ; 1 and (15) for K . Copyright ? 2006 John Wiley & Sons, Ltd.. Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(108) 1274. L.-M. YEH. Proof If not, for all t ∈ [tˆ −

(109) 2 ; tˆ −

(110) 2 1 =2], ! 1 − 1 |K | x ∈ K : S(t; xˆ + x)¿ + − ‘1 ¿ 2 1 − 1 =2. (44). Equation (44) implies |{(t; x) ∈ (tˆ; x) ˆ + Q(

(111) 2 ; ) : S ¿ + − (!=2‘1 )}|¿(1 − 1 )|Q(

(112) 2 ; )|, contradicting (43). The above lemma asserts that at some time level t ∗ , the set where S is close to its supremum occupies only a portion of the cube K . The next lemma claim that this indeed occurs for all time levels near the top of the cylinder (tˆ; x) ˆ + Q(

(113) 2 ; ) ⊂ T1 . Lemma 4.4 Under assumptions of Lemma 4.3, there is a positive number n (depending only on the given data) such that either ! −2 (45) (‘1 +n)

(114) N ¿1 2 or, for all t ∈ [tˆ −

(115) 2 1 =2; tˆ ], 2 ! 1 x ∈ K : S(t; xˆ + x)¿ + − ‘2 6 1 − |K | 2 2 where ‘2 ≡ ‘1 + n. Proof For convenience, we set xˆ = 0. Because of (28)1 , we use (23) of Lemma 3.3 written over the box (t ∗ ; tˆ) × K for the function (S − j)+ with level j = + − (!=2‘1 ). The number in (22) is taken to be = !=2‘1 +n . Thus we have Hj + (46) (t; x) = ln Hj − (S − ( + − (!=2‘1 )))+ + (!=2‘1 +n ) + where Hj = sup(t;0)+Q(

(116). − (!=2‘1 )))+ . Note = 0 on {S¡ + − (!=2‘1 )} and 2 ; ) (S − ( ˆ. 06 6 ln+. Hj 2‘1 +n 6n ln 2 !. (47). The cut-o function x → (x) is taken so that (16) holds, = 1 in the cube K(1−) ; ∈ (0; 1) and |∇|6( )−1 . With these choices, Lemma 3.3 gives. sup t ∗ 6t6tˆ K(1−). 2 (t; x) dx6. K. 2 (t ∗ ; x) dx + d. tˆ t∗. ! −2 ‘1 +n tˆ + Hj 2 + (1−2=k1 ) + ‘1 +n 1 + ln |Dj; ()| d . 2. Copyright ? 2006 John Wiley & Sons, Ltd.. !. K. J(S) (t; x). ( )2. (48). t∗. Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(117) HOLDER CONTINUITY FOR TWO-PHASE FLOWS. 1275. Using Lemma 4.3, (47), tˆ − t ∗ 6

(118) 2 , and (45), we estimate the right-hand side of (48) and get . 1 − 1 k4 |K | + c n|K | 2 (t; x) dx6 |n ln 2|2 (49) sup k 1 −. =2 2 ∗ 3 1 t 6t6tˆ K(1−). where c is a constant depending on given data. Left-hand side of (49) is estimated below by integrating over the smaller set {x ∈ K(1−) : S(t; x)¿ + − (!=2‘1 +n )}. On this set, !=2‘1 = |(n − 1) ln 2|2 2 ¿ ln2 (50) !=2‘1 +n−1 Equations (49) and (50) imply that, for all t ∈ (t ∗ ; tˆ), 2 c ! k4 n 1 − 1 + |K | + x ∈ K(1−) : S(t; x)¿ − ‘1 +n 6 2 k3 n − 1 1 − 1 =2 n2 Also note, for each xed t ∈ (t ∗ ; tˆ), ! ! x ∈ K : S ¿ + − ‘1 +n 6 x ∈ K(1−) : S ¿ + − ‘1 +n + N|K | 2 2 By A6 and letting small and n large (depending on the given data), above estimates imply the lemma. Next we focus on the cylinder (tˆ; x) ˆ + Q(

(119) 2 1 =2; ). Introduce following notation: ⎧

(120) 1 ≡ 1

(121) =2 ⎪ ⎪ ⎪ ⎨ Bi (t) ≡ {x ∈ xˆ + K : S(t; x)¿ + − !=2i } ⎪ ⎪ ⎪ ⎩ Gi ≡ {(t; x) ∈ (tˆ; x) ˆ + Q(

(122) 1 2 =2; ) : S(t; x)¿ + − !=2i }. (51). The information of Lemma 4.4 will be employed to deduce that the set where S is close to ˆ + Q(

(123) 1 2 =2; ) can be made arbitrary small. its supremum + within the cylinder (tˆ; x) Lemma 4.5 Under assumptions of Lemma 4.4, for any 2 ∈ (0; 1) there is a number ‘3 (depending on ˆ ; + ; − ; ) such that if 1 ; 2 ; ‘1 ; ‘2 and given data but independent of tˆ; x; ! −2 ‘3

(124) N 61. 2. (52). then |G‘3 |6 2 |Q(

(125) 1 2 =2; )|. Proof Set (tˆ; x) ˆ = (0; 0). Use (20) of Lemma 3.2 written over box Q(

(126) 1 2 ; 2 ) for the functions (S − j)+ . The levels j are given by j = + − (!=2i ), where ‘2 6i6‘3 and ‘3 is to be chosen. We take a cut-o function such that (16) holds, equals 1 in Q((

(127) 1 =2) 2 ; ), vanishes on Copyright ? 2006 John Wiley & Sons, Ltd.. Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(128) 1276. L.-M. YEH. the parabolic boundary of Q(

(129) 1 2 ; 2 ), and satises |∇|61= ; 06@t 62(

(130) 1 2 )−1 . Neglect the rst term on the left-hand side of the energy estimate (20) to obtain. Q((

(131) 1. =2) 2 ; ). 6d. J(S)|∇(S − j)+ |2. 1

(132) 1 2. Q(

(133) 1 2 ;2 ). (S −. j)2+. . 0. + −

(134) 1 2. + |Dj;2. ()|(1−2=k1 ). d. (53). By (51)1 and (52), we estimate left- and right-hand sides of (53) to obtain 1

(135). c ! 2

(136) 1 2. ; |∇S | 6c J(S)|∇(S − j)+ | 6 2 i Q.

(137) 2 2 Gi Q((

(138) 1 =2) 2 ; ). 2. 2. (54). where c above depends on given data only. Lemma 4.4 implies, for t ∈ (−

(139) 1 2 ; 0), 2 ! 1 x ∈ K : S(t; x)¡ + − i = |K | − |Bi (t)|¿ |K |. 2. 2. (55). See (51) for Bi (t). Next we use (55) and Lemma 2.2 of Chapter 1 [8] applied to the function S(t; ·) for all time −(

(140) 1 =2) 2 6t 60 and for the levels k = + − (!=2i ), l = + − (!=2i+1 ), l − k = !=2i+1 to obtain c N +1 ! |Bi+1 (t)|6 2 i+1 2 1 |K |. Bi (t)\Bi+1 (t). |∇S |. (56). Integrate (56) over (−(

(141) 1 =2) 2 ; 0) to get c. ! |Gi+1 |6 2 i+1 2 1. c. |∇S |6 2 |Gi \ Gi+1 | |∇S |2 1 Gi \Gi+1 Gi. (57). By (54), (57) gives c |Gi+1 | 6 4 1 2. Q

(142) 1 2 ; |Gi \ Gi+1 | 2. (58). Above inequalities are valid for all ‘2 6i6‘3 . We add them for i = ‘2 ; ‘2 + 1; : : : ; ‘3 − 1. The right-hand side of (58) can be majorized by a convergent series bounded above by 4 2 2 |Q((

(143) 1 =2) 2 ; )|. Therefore (‘3 − ‘2 )|G‘3 |2 6(c= √ 1 )|Q((

(144) 1 =2) ; )| . To prove the lemma we 2 divide by ‘3 − ‘2 and take ‘3 so large that c= 1 ‘3 − ‘2 6 2 . Next we show that indeed S is strictly below its supremum + in a smaller box coaxial with (tˆ; x) ˆ + Q(

(145) 1 2 =2; ) and with the same vertex. Copyright ? 2006 John Wiley & Sons, Ltd.. Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(146) HOLDER CONTINUITY FOR TWO-PHASE FLOWS. 1277. Lemma 4.6 ˆ ; + ; − ; ) so Under assumptions of Lemma 4.4, there is a number ‘4 (independent of tˆ; x; that if ! −2 (59) ‘4

(147) 1 N 61 2 ˆ + Q((

(148) 1 =2)| =2|2 ; =2). then S(t; x)6 + − (!=2‘4 +4 ) for (t; x) ∈ (tˆ; x) Proof Set (tˆ; x) ˆ = (0; 0). We use (20) of Lemma 3.2 written over the boxes Q(

(149) 1 2n =2; n ) to the function (S − jn )+ , where. n =. + ; 2 2n+1. jn = + −. ! ! − 2‘4 +1 2‘4 +1+n. for n = 0; 1; 2; : : :. The cut-o functions n are taken to satisfy ⎧ ⎪ 0¡n (t; x)61 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ n = 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ n = 0 n+2 2 ⎪ ⎪ ⎪ |∇n |6c ⎪ ⎪. ⎪ ⎪ ⎪ ⎪ ⎪ 2 ⎪ ⎪ ⎪ 1 2(n+2) ⎪ ⎪ ⎩ 06@t n 6

(150) 1. . for (t; x) ∈ Q(

(151) 1 2n =2; n ) in Q(

(152) 1 2n+1 =2; n+1 ) on the parabolic boundary of Q(

(153) 1 2n =2; n ) (60). With these choices, (20) of Lemma 3.2 gives. sup −

(154) 1 2n =2¡t¡0. K n. n2 (S − jn )2+ +. . 2n ! 2 1 6c ‘4 . 2

(155) 1. 1

(156) 1. Q(

(157) 1 2n =2; n ). |∇(n (S − jn )+ )|2. Q(

(158) 1 2n =2; n ). X{(S−jn )+ ¿0} +. . 0. −

(159) 1 2n =2. |Dj+n ; n ()|(1−(2=k1 )) d. (61). Next, in the cylinders Q(

(160) 1 2n =2; n ) we change variable z = 2t=

(161) 1 which maps Q(

(162) 1 2n =2; n ) ˆ ·) ≡ S(z

(163) 1 =2; ·), ˆn (z; ·) ≡ n (z

(164) 1 =2; ·), Dn (z) ≡ {x ∈ K n : into Qn ≡ [− 2n ; 0] × K n . Setting (z; 0 ˆ (z; x)¿jn }, and |Dn | ≡ − 2 |Dn (z)| dz. Inequality (61) can be rewritten as, by (59), n.

(165) ˆn (ˆ −. . ! jn )+

(166) 2V(Qn ) 6c ‘4 2. 2 . Copyright ? 2006 John Wiley & Sons, Ltd.. . 2n 2 |Dn | + −N . . 0. − 2n. |Dn ()|. (1−2=k1 ). d. (62). Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(167) 1278. L.-M. YEH. Equation (62) and Corollary 3.1 of Chapter 1 [8] give ! 2 ‘4 |Dn+1 |6|jn − jn+1 |2 |Dn+1 |6

(168) (ˆ − jn )+

(169) 2L2 (Qn+1 ). 1. 22(n+2) 2. 6

(170) ˆn (ˆ − jn )+

(171) 2L2 (Qn ) 6c

(172) ˆn (ˆ − jn )+

(173) 2V(Qn ) |Dn |2=(N +2) ! 2 6c ‘4 . 2. 22n |Dn |1+(2=(N +2)) + −N |Dn |2=(N +2). 2. . 0. |Dn ()|. (1−2=k1 ). − 2n. d. 0 Dene Yn ≡ |Dn |= |Qn |; Zn ≡ (1= |K n |)( − 2 |Dn ()|(1−2=k1 ) d)1=(1+) , we have the recursive n inequalities as Lemma 4.1, ⎧ ⎨ Yn+1 6c42n (Yn1+(2=(N +2)) + Yn2=(N +2) Zn1+ ) ⎩. Zn+1 6c4 (Yn + 2n. Zn1+ ). for n = 0; 1; 2; : : :. It follows from these with the aid of Lemma 4.2 of Chapter 1 [8] that Yn and Zn tend to zero 2 as n → ∞, provided Y0 + Z01+ 6 ≡ c−(1+)= 4−2(1+)= . See (4) for . Then continuing as the argument of the proof of Lemma 4.1 and using Lemma 4.5, one can choose ‘4 and complete the proof of this lemma. The results of the second alternative imply Lemma 4.7 Under assumptions of Lemma 4.3, there are numbers 2 ∈ (1=2; 1) and I2 1 (independent S 62 !. of tˆ; x; ˆ ; + ; − ; ) such that either !6I2 N=(2+m) or ess osc(t;ˆ x)+Q((

(174) 2 ˆ 1 =2)| =2| ; =2) We combine Lemmas 4.2, 4.7 with (29) into Lemma 4.8 If 6 ∗ , 0¡!6#=2, and + ¡#, there exist constants ≡ max{1 ; 2 } ∈ (1=2; 1). and. I ≡ max{I1 ; I2 }. that are determined by given data and independent of tˆ; x, ˆ , + , − , , such that either min{N=(2+m);$=m} ; #=2} or ess osc(t;ˆ x)+Q((

(175) S 6!. !6 min{I. 2 ˆ 1 =2)| =2| ; =2) 4.1.1.3. Holder estimate. Using notations of Lemma 4.8, we dene ⎧ M0 ≡ + ¿! ⎪ ⎪ ⎪ ⎪ ⎨ # min{N=(2+m);$=m} }; !1 ≡ min max{!; I. 2 ⎪ ⎪ ⎪ ⎪ ⎩ M ≡ max{! ; sup S} 2 1 1 (t;ˆ x)+Q((

(176) ˆ 1 =2)| =2| ; =2) Copyright ? 2006 John Wiley & Sons, Ltd.. (63). Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(177) HOLDER CONTINUITY FOR TWO-PHASE FLOWS. 1279. Lemma 4.9 There is a constant d¿1 (determined by given data) such that J(M0 )6dJ(M1 ). Proof If !1 = #=2, then J(M0 )6dJ(M1 ) by A7. If !1 ¡#=2, then !1 ¿!. Therefore if !1 ¡#=2 and − 6 12 + , then M0 62!62!1 =62M1 =. So J(M0 )6dJ(M1 ). If !1 ¡#=2 and − ¿ 12 + , then M0 =2 = + =2¡ − 6M1 . Therefore, J(M0 )6J(2M1 )6dJ(M1 ). Let us estimate from below the length of the cylinder Q((

(178) 1 =2)| =2|2 ; =2) for which the conclusion of Lemma 4.8 holds. We have, by (51),

(179) 1 2 where 1 ≡. 2 =. 2. 2 2 1. 21 1 ¿ = 4J(M0 ) 2 4dJ(M1 ) 2 J(M1 ). ( 1 =16d) ≡ A . So Q( 21 = J(M1 ); 1 ) ⊂ Q((

(180) 1 =2)| =2|2 ; =2).. Lemma 4.10 There are constants a; A ∈ (0; 1); ∈ (1=2; 1) and I (depending on given data but independent of tˆ; x; ˆ ; + ; − ; ) and it is possible to construct a sequence, for n = 0, . !0 ≡ !;. M0 ≡ + ; 0 ≡ ;. E0 ≡ Q. 20 ; 0 J(M0 ). . (64). and, for n ∈ N,

(181). !n ; Eˆ n−1 ; Mn ≡ max !n ;. . sup (t;ˆ x)+ ˆ Eˆn−1. . 2n ; n S ; n ≡ A n−1 ; En ≡ Q J(Mn ). . such that, for all n ∈ N, ⎧ a ⎪ ⎨ ess osc(t;ˆ x)+ ˆ Eˆn−1 S 6!n ≡ min {max{!n−1 ; I n−1 }; #=2} ⎪ ⎩ E ⊂ Eˆ n n−1 ⊂ En−1. (65). Proof This is proved by induction. Let (tˆ; x) ˆ = (0; 0). Take a = min{N=(2 + m); $=m}, A is the one dened in the remark after Lemma 4.9, and ; I are the ones in Lemma 4.8. For i = 0 case, !0 ; M0 ; 0 ; E0 are given in (64). By (63), Lemma 4.9, and remark after Lemma 4.9, we get !1 ; Eˆ 0 ; M1 ; 1 ; E1 and it is easy to see that they satisfy (65). Assume that the sequence is obtained up to i = n. We reset + = supEˆn−1 S; − = inf Eˆn−1 S; ! = + − − . By (65), !6!n . Copyright ? 2006 John Wiley & Sons, Ltd.. Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(182) 1280. L.-M. YEH. (1) If Mn = !n and supEˆn−1 S 6Mn =2, we pick ⎧ !n+1 ≡ min{max{!n ; I an }; #=2} ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ Eˆ n = En ⎪ ⎪ ⎪ ⎪ ⎨ Mn+1 ≡ max{!n+1 ; supEˆn S } ⎪ ⎪ ⎪. n+1 ≡ A n ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪. 2n+1 ⎪ ⎩ En+1 ≡ Q ; n+1 J(Mn+1 ). (66). Since ess oscEˆn S 6!n =26!n 6!n+1 , (65)1 holds. Also note if !n+1 = #=2, then J(Mn )6J(Mn+1 ), and if !n+1 ¡#=2, then J(Mn ) = J(!n )6J(!n+1 =)6dJ(Mn+1 ). Therefore, we have En+1 ⊂ Eˆ n . So we prove (65)2 . (2) If Mn = !n and Mn =26 supEˆn−1 S 6Mn , we repeat the proofs of Lemmas 4.1–4.8 to see that if !n+1 ≡ min{max{!n ; I an }; #=2}, Eˆ n ≡ Q(( 1 =4J(Mn ))| n =2|2 ; n =2) then, by (63), ⎧ ess oscEˆn S 6 min{max{!; I an }; #=2}6!n+1 ⎪ ⎨ . 2n ⎪ ⎩ Eˆ n ⊂ Q ; n = En J(Mn ) Dene Mn+1 ; n+1 ; En+1 as (66) and argue as Lemma 4.9 to get En+1 ⊂ Eˆ n . So we prove (65). (3) If Mn = supEˆn−1 S, one repeats the proofs of Lemmas 4.1–4.9 as case (2), and it is easy to get (65). Lemma 4.11 There exist constants a; A; ∈ (0; 1), ∈ (1=2; 1), and I (depending on given data but independent of tˆ; x, ˆ , + , − , ) such that if 6Aa , then a S 6c(; a; A; I )|!0 + 0 | ess osc 2 ˆ. 0 (t;x)+Q( ˆ =J(M0 );) for all cylinders Q(2 = J(M0 ); ) and 0¡6 ∗0 6 0 (see (64) for !0 ; M0 ; 0 ). Here ∗0 depends on 0 ; !0 , and given data only. Proof Let (tˆ; x) ˆ = (0; 0). By (65), !n+1 6!n + I an . By iteration, !n 6n !0 + I a0. n−1. i Aa(n−i). i=0. Since 6Aa , !n 6n !0 + nI | 0 An |a Copyright ? 2006 John Wiley & Sons, Ltd.. (67) Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(183) HOLDER CONTINUITY FOR TWO-PHASE FLOWS. 1281. For any ∈ (0; 0 A), there exists an integer n (depending on ) such that. 0 A(n +1) 66 0 An. (68). Equation (68) implies ⎧ 1 ⎪ ⎪ n 6 ln 6n + 1 ⎪ ⎨ ln A. 0. 1 ⎪ ⎪ ⎪ ⎩ n 6−1 . 0. ln where 1 ≡ ln A. We take ∗0 ¡ 0 so small that, for 6 ∗0 , ⎧ 0 a=2 ⎪ ⎪ ⎨ ln 6 0 ⎪ ⎪ ⎩ ! 6! n. (69). (70). 0. Here ∗0 depends on 0 ; !0 , and given data. Equations (68), (69)1 , and (70)1 imply that if 6 ∗0 , a 6c(a; A; I ) 0a=2 a=2 n I | 0 An |a 6I A−a ln (71) | ln A| 0 Therefore, (67), (69)2 , and (71) imply !n 6c(; a; A; I )|!0 + a0 ||= 0 | where = min{1 ; a=2}. Note Q(2 = J(M0 ); ) ⊂ En by (68), (70)2 , and the construction of En in Lemma 4.10. So we prove the lemma. 4.1.2. For #=4¡ − 6 + ¡1 − #=4 and 1 − #¡ − cases. For #=4¡ − 6 + ¡1 − #=4 case, (tˆ; x) ˆ + Q( 2−$ ; 2 ) ⊂ T2 by (28)2 . Equation (11) corresponds to a uniform (independent of ) parabolic equation in (tˆ; x) ˆ + Q( 2−$ ; 2 ). So we may repeat above argument to obtain Lemma − ˆ + Q( 2−$ ; 2 ) ⊂ T3 by (28)3 . Since Equation (11) around 4.11. For 1 − #¡ case, (tˆ; x) the end point 0 has similar properties as (14) around the other end point 1 (see remark in Section 2), we use (14) and repeat above argument to get the same conclusion as Lemma 4.11. In summary, Lemma 4.11 holds for 0¡!6#=2 case. 4.2. For #=26!61 case In this section, we shall prove that if #=26!61 and if is small enough, lower (resp. upper) bound of saturation S is greater than 0 (resp. less than 1), and the lower and upper bounds are independent of . If this is the case, we are in case (28)2 again and Lemma 4.11 can be shown by following the argument in Section 4.1. Dene

(184) −1 ≡ sup − 66 + J(). Since !¿#=2, by (26), | ∗ |$ 6 min{J(#=2); J(1 − #=2)}6

(185) −1 6 sup J() 0661. So Q(

(186) 2 ; ) ⊂ Q( 2−$ ; 2 ) for all 6 ∗ . Copyright ? 2006 John Wiley & Sons, Ltd.. Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(187) 1282. L.-M. YEH. Lemma 4.12 Under 6 ∗ and !¿#=2, there is a 3 ∈ (0; 1) (depending on given data but independent of tˆ; x; ˆ ; + ; − ; ) such that (1) if ⎧ ! −2 ⎪ −‘5 ⎪ ‘ ¿ 2; 2 6 #; ‘5 N 61 ⎨ 5 2 ⎪ ! ⎪ ⎩ (t; x) ∈ (tˆ; x) ˆ + Q(

(188) 2 ; ) : S(t; x)¡ ‘5 6 3 |Q(

(189) 2 ; )| 2 then S(t; x)¿!=2‘5 +4 a.e. in (tˆ; x) ˆ + Q(

(190) | =2|2 ; =2), and (2) if ⎧ ! −2 ⎪ −‘ N ⎪ ⎨ ‘5 ¿2; 2 5 6#; ‘5 61 2 ⎪ ! ⎪ ⎩ (t; x) ∈ (tˆ; x) ˆ + Q(

(191) 2 ; ) : S(t; x)¿1 − ‘5 6 3 |Q(

(192) 2 ; )| 2 then S(t; x)61 − (!=2‘5 +4 ) a.e. in (tˆ; x) ˆ + Q(

(193) | =2|2 ; =2). Proof Proof is almost same as that of Lemma 4.1. Lemma 4.13 If (6T ), ‘(¿2 + ‘0 ¿4) ∈ N, and k2 =2‘0 6#, solutions of Lemma 3.1 satisfy sup |{x ∈ : S(t; x)6w or 1 − w6S(t; x)}|6 t6. c0 |c0 |‘−‘0 (‘ − ‘0 )(‘−‘0 )f‘. (72). where lim‘→∞ f‘ = 1, w ≡ k2 =2‘ and c0 is a constant independent of ; ‘; . Proof Dene Lw , L&;w , X w as. Lw (z) ≡. L&;w (z) ≡. ⎧ ⎪ ⎪0 ⎪ ⎨ ⎪ ⎪ ⎪ ⎩. z − 2w. for w6z 62w. −w. for z 6w. ⎧ 0 ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

(194). X w (z) ≡. for 2w6z. for &(2w)6z. z − &(2w). for &(w)6z 6&(2w). &(w) − &(2w). for z 6&(w). 1. for w6z 62w. 0. otherwise. Copyright ? 2006 John Wiley & Sons, Ltd.. Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(195) HOLDER CONTINUITY FOR TWO-PHASE FLOWS. where. &(z) ≡. z. 0:5. ˜ w () d; ˜. 1283. z ∈ (0; 1). (73). ˜ By 2w6k2 =2 and (10), both Then X w (z) = Lw (z) = L&;w (&(z)), (d=dz)&(z) = (˜ w = )(z). Lw (S); L&;w (&(S)) ∈ L2 (0; T ; U). Take = Lw (S); = L&;w (&(S)) in (11)–(12) to obtain, by A4,. Lw (S)@t S − Kw (S)X w (S)∇ (S)∇S 6c1 K ˜ w X w (S)|∇S | (74) . . . . where constant c1 is independent of w; . If Lw (S)@t S ¿0, (74) implies . ˜ w X w K ˜ w X w (S)|∇S |6c2 K (S) K ˜ w X w (S)|∇S | |d =dS | where constant c2 is independent of w; . Equations (74)–(75) imply. ˜ w X w Lw (S)@t S 6c3 K (S) |d =dS | S Dene Zw ≡ 2w Lw (z) dz. Equation (76) implies. ˜ w X w @t Zw = Lw (S)@t S 6c3 K (S) |d =dS | . (76). (77). Equation (77) and A7 yield that, if 06t1 6t2 6T , t2 t2 @t Zw 6c4 Z2w t1. . t1. . where c4 is independent of t1 ; t2 ; w; . Dene F(w; ) ≡ (1=w2 ) supt6 imply that, for 06t1 6t2 6T , F(w; t2 ) − F(w; t1 )6c5 (t2 − t1 )F(2w; t2 ). (75). (78) . Zw (t; ·). A5 and (78). (79). where c5 is independent of t1 ; t2 ; w; . By induction and (5), one obtains, for j ∈ N; jh6T , k2 k2 j−1 ‘−‘0 F ; jh 6(‘ − ‘0 + 1) |c5 h| F ; jh (80) 2‘ 2‘ 0 If j = (‘ − ‘0 )=log(‘ − ‘0 ) and = jh in (80), then |c5 |‘−‘0 k2 k2 6 F ; F ; 2‘ (‘ − ‘0 )(‘−‘0 )f‘ 2‘ 0. (81). where f‘ → 1 as ‘ → ∞. Dene B(t) ≡ {x ∈ : S(t; x)6w = k2 =2‘ }. Equation (81) implies . k2 c6 |c5 |‘−‘0 k2 F ; sup XB(t) 6c6 F ‘ ; 6 2 (‘ − ‘0 )(‘−‘0 )f‘ 2‘0 t6 where c6 is independent of ; ‘; w; . So proof for one end point of (72) is completed. Proof of the other end point of (72) can be proceeded in the same manner, so we skip it. Copyright ? 2006 John Wiley & Sons, Ltd.. Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(196) 1284. L.-M. YEH. By Lemmas 4.12 and 4.13, one can conclude: Corollary 4.1 There is a ‘6 (depending on given data and O but independent of ) such that if ‘¿‘6 ,. = |#=21+‘ |2=N , and #=26!, then # 2‘+5. 6S(t; x)61 −. # 2‘+5. for any (t; x) ∈ (tˆ; x) ˆ + Q(

(197) | =2|2 ; =2) and (tˆ; x) ˆ ∈ O. Similar to the case where 0¡!6#=2 and #=4¡ − 6 + ¡1 − (#=4) hold (i.e. (28)2 case), Corollary 4.1 implies that Equation (11) corresponds to a uniform (independent of ) parabolic ˆ ∈ O. So we may repeat the argument in Section 4.1 equation in (tˆ; x)+ ˆ Q(

(198) | =2|2 ; =2) for (tˆ; x) to get the same conclusion as Lemma 4.11 for !¿#=2 case. Combining (27), Lemma 4.11, remark in Section 4.1.2, Corollary 4.1 with a standard covering argument, we get uniformly Holder continuous for S over any compact subset O of T . 5. HOLDER ESTIMATE OF REGULARIZED EQUATIONS ON PARABOLIC BOUNDARY In this section, we give a Holder estimate for S on parabolic boundary of T . Basically, the estimate can be shown by following the proof for the interior region. Therefore we only give a sketch of the proof. The idea to treat parabolic boundary can also be found in Reference [8]. The boundary consists of Dirichlet boundary ( T2 ), Neumann boundary ( T1 ), edge ( T1 ∩ T2 ), and initial boundary (T = 0). They are discussed below: 5.1. Dirichlet boundary, Neumann boundary, and edge We rst derive a result similar to Lemmas 3.2 and 3.3 for Dirichlet boundary. Fix (tˆ; x) ˆ ∈ T2 and consider the cylinder (tˆ; x) ˆ + Q(; ), where and are so small that tˆ − ¿0 and {(tˆ; x) ˆ + Q(; )} ∩ @T ⊂ T2 . Then we modify the interior quantities in (19), (22) as Dˆ ± } ∩ : (S − j)± (; x)¿0} j; () ≡ {x ∈ {xˆ + K. (Hˆ ± j ; (S. Hˆ ± j − j)± ; ) ≡ ln ± Hˆ j − (S − j)± + +. (82) (83). ˆ± where Hˆ ± ˆ Q(; ), we introduce T (S −j)± and 0¡ ¡ min {1; H j }. In (tˆ; x)+ ˆ j ≡ sup{(t;ˆ x)+Q(;)}∩ a piecewise smooth cut-o function (t; x) → (t; x) satisfying (16). We observe that for all t ∈ (tˆ − ; tˆ); x → (t; x) vanishes on the boundary of xˆ + K and not on the boundary of {xˆ + K } ∩ . Local energy estimate for S near (tˆ; x) ˆ are obtained by taking ’ = ± (Sh − j)± 2 in (17), integrating over (tˆ − ; tˆ) and letting h → 0+ . Such a choice of testing function is admissible if for a.e. t ∈ (tˆ − ; tˆ), (Sh (·; t) − j)± 2 (t; ·) ∈ H01 ({xˆ + K } ∩ ) Copyright ? 2006 John Wiley & Sons, Ltd.. (84). Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(199) HOLDER CONTINUITY FOR TWO-PHASE FLOWS. 1285. Since x → (t; x) vanishing on the boundary of {xˆ + K } and not on the boundary of {xˆ + K } ∩ , condition (84) needs to be veried whether for a.e. t ∈ (tˆ − ; tˆ), (Sh − j)± = 0 in the sense of the trace on boundary of {xˆ + K } ∩ This can be realized for the function (Sh − j)+ if j is chosen to satisfy j¿. sup T {(t;ˆ x)+Q(;)}∩. ˆ 2. Sb. (85). Analogously, the function −(Sh − j)− can be taken as testing function in (17) if j6. inf. T {(t;ˆ x)+Q(;)}∩. ˆ 2. Sb. (86). With these choices of j, we may repeat calculation in all analogous to those of Lemma 3.2 and derive energy inequality for S near T2 . Analogous considerations hold for a version of the logarithmic estimates along the lines of Lemma 3.3. We summarize Lemma 5.1 There is a constant d (independent of ; ; ; j) such that for every (tˆ; x) ˆ ∈ T2 , every cylinder ˆ + Q(; )} ∩ @T ⊂ T2 , and every {(tˆ; x) ˆ + Q(; )} ∩ T ⊂ T1 satisfying tˆ − ¿0 and {(tˆ; x) level j satisfying ⎧ for the function (Sh − j)+ T Sb ⎨ j¿ sup{(t;ˆ x)+Q(;)}∩. ˆ 2 (87) ⎩ j6 inf T Sb for the function (S − j) h − {(t;ˆ x)+Q(;)}∩. ˆ 2 the following inequalities hold:. 2 2 (S − j)± + sup ˆ ˆ }∩ t−¡t¡ tˆ {x+K. T {(t;ˆ x)+Q(;)}∩ ˆ. . 6d. {x+K ˆ }∩. (S − j)2± 2 (tˆ − ; x) +. tˆ. ˆ t−. J(S)|∇(S − j)± |2. (1−2=k1 ) |Dˆ ± d j; ()|. . + T {(t;ˆ x)+Q(;)}∩ ˆ. sup ˆ ˆ }∩ t−¡t¡ tˆ {x+K. {x+K ˆ }∩. 2 ˆ 2 (Hˆ ± j ; (S − j)± ; )(t − ; x) (x). + T {(t;ˆ x)+Q(;)}∩ ˆ. 1 + 2. (88). 2 2 (Hˆ ± j ; (S − j)± ; )(t; x) (x). . 6d. (J(S)(S − j)2± |∇|2 + (S − j)2± @t ). 2 J(S) (Hˆ ± j ; (S − j)± ; )|∇|. tˆ Hˆ ± j + ± (1−2=k ) 1 |Dˆ j; ()| d 1 + ln t− ˆ. Copyright ? 2006 John Wiley & Sons, Ltd.. (89). Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(200) 1286. L.-M. YEH. where ; are piecewise smooth cut-o functions satisfying (16). Equations (88)–(89) also ˆ + Q(; )} ∩ T is a subset of T2 and T3 . But for T3 case, S in (88)–(89) hold if {(tˆ; x) should be replaced by So and proofs of those results are required to use Equation (14). Now take a cylinder (tˆ; x) ˆ + Q( 2−$ ; 2 ), where ¿0 is so small that tˆ − 2−$ ¿0 2−$ ˆ ˆ + Q( ; 2 )} ∩ @T ⊂ T2 . Assume (tˆ; x) ˆ = (0; 0). Set + ≡ supQ( 2−$ ;2 )∩T S, and {(t ; x) − + − ≡ inf Q( 2−$ ;2 )∩T S, ! ≡ − . The Dirichlet boundary problem is studied by considering the following four cases: ⎧ k2 # k2 # ⎪ − + ⎪ ; ; 6 6 61 − min min ⎪ ⎪ ⎪ 4 4 4 4 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ + ⎪ ⎨ 6 min k2 ; #. 2 2. ⎪ ⎪ ⎪ ⎪ k2 # ⎪ ⎪ ; 6 − 1 − min ⎪ ⎪ ⎪ 2 2 ⎪ ⎪ ⎪ ⎪ ⎩. (90). Not above three cases. For (90)1 case, we dene

(201) −1 ≡ sup − 66 + J() and let Q(

(202) 2 ; ) ⊂ Q( 2−$ ; 2 ) for all 6 ∗ . Set b+ ≡ supQ(

(203) 2 ; )∩ T2 Sb ; b− ≡ inf Q(

(204) 2 ; )∩ T2 Sb ; !b ≡ b+ − b− . If the two inequalities ⎧ ! + + ⎪ ⎪ ⎨ − 2‘1 6 b ; ⎪ ! ⎪ ⎩ − + ¿ b− ; ‘1. ‘1 ¿2. (91). 2. are both true, subtracting the second from the rst gives !62!b . If (91)1 is violated, then the levels j ≡ + − (!=2i ); i¿‘1 satisfy (87)1 , and we may derive an energy estimate for (S − j)+ . Since (S − j)+ vanishes on Q(

(205) 2 ; ) ∩ @T , we may extend it to the whole Q(

(206) 2 ; ) by setting it to be zero outside T within the box Q(

(207) 2 ; ). Thus conclusion of Lemma 4.4 is satised by (S − j)+ . We then use Lemma 5.1 and argue as Lemma 4.5 to deduce that for all 2 ∈ (0; 1), there are positive numbers I; ‘3 that are dependent on given data and independent of tˆ; x; ˆ ; + ; − ; such that either !6I N=2 or ! (t; x) ∈ Q(

(208) 2 =2; ) : S(t; x)¿ + − ‘3 6 2 |Q(

(209) 2 =2; )| 2 An application of Lemma 4.6 now gives Lemma 5.2 There exist constants I ¿1 and ∈ (1=2; 1), dependent on given data and independent of tˆ; x; ˆ ; + ; − ; , such that either ess oscQ((

(210) =2)| =2|2 ; =2) S 6 max{!; 2!b } or !6I N=2 . If (91)2 is violated, one can use Equation (14) and repeat the above argument to get the same conclusion as Lemma 5.2, and then argue as Lemmas 4.9–4.11 to show the locally Holder continuity of S. Copyright ? 2006 John Wiley & Sons, Ltd.. Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(211) HOLDER CONTINUITY FOR TWO-PHASE FLOWS. 1287. For cases (90)2;3 , local Holder continuity can be proved by following the arguments in Section 4.1. If (90)1;2;3 are violated, then we rst show S is bounded away from the two end points by using the arguments in Section 4.2, and then follow the argument for (90)1 case to show the local Holder continuity of S. Proof of Holder continuity of S on Neumann boundary T1 and edge T1 ∩ T2 is a straightforward modication of that for Dirichlet boundary. 5.2. Initial boundary Again we rst derive a result similar to Lemmas 3.2 and 3.3. Fix (tˆ; x) ˆ ∈ T and consider ˆ ˆ ˆ ˆ + Q(t ; ) lies on the bottom of the cylindrical the cylinder (tˆ; x) ˆ + Q(t ; ). Therefore (t ; x) domain T . Consider the cuto function satisfying (16) and independent of t. Local energy estimates for S near t = 0 are derived by taking ’ = ± (Sh − j)± 2 in (17), integrating over (0; t); t ∈ (0; tˆ), and letting h → 0+ . The rst term in (17) gives. 1 1 2 2 (Sh − j)± (t; x) dx − (Sh − j)2± (0; x)2 dx 2 {x+K 2 {x+K ˆ } ˆ } If j is chosen so that j¿ sup{x+K ˆ } Sinit , then we have. 1 (Sh − j)2+ (0; x)2 dx → 0 2 {x+K ˆ }. as h → 0+. By (83), it follows that 2 (Hˆ ± j ; (Sh − j)± ; )(t; x) = 0. whenever (Sh − j)± = 0. Thus if j¿ sup{x+K ˆ } Sinit , then. + 2 (Hˆ j ; (Sh − j)+ ; )(0; x)2 (x) → 0. as h → 0+. {x+K ˆ }. Analogous considerations hold for (Sh − j)− 2 . We summarize Lemma 5.3 There is a constant d (independent of ; ; ; j) such that for every cylinder (tˆ; x)+ ˆ Q(; ) ⊂ T1 satisfying tˆ − = 0, and every level j satisfying ⎧ ⎨ j¿ sup{x+K for the function (Sh − j)+ ˆ } Sinit (92) ⎩ j6 inf for the function (Sh − j)− {x+K ˆ } Sinit the following inequalities hold:. sup ˆ ˆ }∩ t−¡t¡ tˆ {x+K. (S − j)2± 2 +. T {(t;ˆ x)+Q(;)}∩ ˆ. 6d. T {(t;ˆ x)+Q(;)}∩ ˆ. Copyright ? 2006 John Wiley & Sons, Ltd.. J(S)|∇(S − j)± |2. J(S)(S −. j)2± |∇|2. tˆ. + ˆ t−. (1−2=k1 ) |Dˆ ± j; ()|. d. (93). Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(212) 1288. L.-M. YEH. sup ˆ ˆ }∩ t−¡t¡ tˆ {x+K. 2 2 (Hˆ ± j ; (S − j)± ; )(t; x) (x). 6d. 1 + 2. T {(t;ˆ x)+Q(;)}∩ ˆ. 2 J(S) (Hˆ ± j ; (S −j)± ; )|∇|. tˆ Hˆ ± j + ± (1−2=k ) 1 |Dˆ j; ()| d 1+ ln t− ˆ. (94). where is a piecewise smooth cut-o function satisfying (16). Equations (93)–(94) hold if {(tˆ; x) ˆ + Q(; )} ∩ T is a subset of T2 and T3 . But for T3 case, S in (93)–(94) should be replaced by So and proofs need to use Equation (14). Fix (0; x) ˆ ∈ {0} × and construct the cylinder (0; x)+ ˆ Qˆ ( 2−$ ; 2 ) ≡ (0; x)+(0; ˆ. 2−$ ) × K2 ⊂ + − + − . We assume xˆ = 0. Set ≡ supQ(. ˆ 2−$ ;2 ) S, ≡ inf Q(. ˆ 2−$ ;2 ) S, ! ≡ − . We consider the following four cases: ⎧ k2 # k2 # ⎪ − + ⎪ ⎪ ⎪ min 4 ; 4 6 6 61 − min 4 ; 4 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ k2 # ⎪ ⎨ + 6 min ; 2 2 (95) ⎪ ⎪ ⎪ k2 # ⎪ ⎪ ⎪ ; 6 − 1 − min ⎪ ⎪ 2 2 ⎪ ⎪ ⎪ ⎪ ⎩ Not above three cases T. For (95)1 case, we dene

(213) −1 ≡ sup − 66 + J() and construct the box Qˆ (

(214) 2 ; ) ≡ − − + + (0;

(215) 2 ) × K ⊂ Qˆ ( 2−$ ; 2 ). Set init ≡ supK Sinit , init ≡ inf K Sinit , !init ≡ init − init , and consider the two inequalities ⎧ ! + + ⎪ ⎨ − ‘1 6 init ; 2 (96) ‘ 1 ¿2 ⎪ ⎩ − + ! ¿ − ; init 2‘1 If both of (96) hold, subtract the second from the rst to obtain !62!init . If (96)1 is violated, then the level j = + − (!=2i ), i¿‘1 satises (92)1 . By Lemma 5.3, we have energy and logarithmic estimates for the truncated functions (S − j)+ . Using the logarithmic estimate (94) and proceeding as Lemma 4.4, one can show that for any 2 ∈ (0; 1), there are positive numbers ‘2 ; I (depending on given data and independent of ) such that either !6I N=2. (97). or, for all t ∈ (0;

(216) 2 ), |{x ∈ K : S(t; x)¿ + − (!=2‘2 )}|6 2 |K |. By the energy inequality (93) and the procedure of Lemma 4.6, we conclude that if (97) does not hold, then it is easy to derive ess oscQ(

(217) | =2| 2 ; =2) S 62 !. If (96)2 is violated, we use Equation (14) and still proceed ˆ Copyright ? 2006 John Wiley & Sons, Ltd.. Math. Meth. Appl. Sci. 2006; 29:1261–1289.

(218) HOLDER CONTINUITY FOR TWO-PHASE FLOWS. 1289. as above. To summarize, going down from Qˆ ( 2−$ ; 2 ) to the smaller box Qˆ (

(219) | =2|2 ; =2), the essential oscillation of S decreases by a factor of , unless either !62!init or !6I N=2 . Lemma 5.4 ˆ ; There exist constants ∈ (1=2; 1), I (depending on given data and independent of tˆ; x; N=2 . + ; − ; ) such that either ess oscQ(

(220) | =2| 2 ; =2) S 6 max {!; 2!init } or !6I. ˆ Then we argue as Lemmas 4.9–4.11 to show locally Holder continuity of S. For cases (95)2;3 , one can follow above arguments to prove locally Holder continuity. If (95)1;2;3 are violated, then we rst show S is bounded away from the two end points by using the arguments in Section 4.2, and then follow the argument for (95)1 case to show the local Holder continuity of S. The regularity of S on the boundary @ at t = 0 can be proved by A1 and a straightforward modication of above argument, so it is skipped. 6. HOLDER CONTINUITY IN WHOLE REGION From results of Sections 4, 5 and standard covering argument, one can see that S are Holder continuous in T for all . Moreover, their Holder bounds are independent of . By Lemma 3.1, the limit function S in Lemma 3.1 is also Holder continuous in T . So we prove Theorem 2.1. ACKNOWLEDGEMENTS. This research is supported by the grant number NSC 91-2115-M-009-003 from the research program of National Science Council. REFERENCES 1. Alt HW, DiBenedetto E. Nonsteady ow of water and oil through inhomogeneous porous media. Annali dells Scouls Normals Superiors di Piss. Classe di Scienze 1985; 12(4):335–392. 2. Antontsev SN, Kazhikhov AV, Monakhov VN. Boundary Value Problems in Mechanics in Nonhomogeneous Fluids. Elsevier: Amsterdam, 1990. 3. Bourgeat A, Luckhaus S, Mikelic A. Convergence of the homogenization process for a double-porosity model of immiscible two-phase ow. SIAM Journal on Mathematical Analysis 1996; 27(6):1520–1543. 4. Chavent G, Jare J. Mathematical Models and Finite Elements for Reservoir Simulation. North-Holland: Amsterdam, 1986. 5. Chen Z. Degenerate two-phase incompressible ow 1. Existence, uniqueness and regularity of a weak solution. Journal of Dierential Equations 2001; 171:203–232. 6. Kroener D, Luckhaus S. Flow of oil and water in a porous medium. Journal of Dierential Equations 1984; 55:276–288. 7. Yeh LM. Convergence of a dual-porosity model for two-phase ow in fractured reservoirs. Mathematical Methods in the Applied Sciences 2000; 23:777–802. 8. DiBenedetto E. Degenerate Parabolic Equations. Springer: Berlin, 1993. 9. Gilbarg D, Trudinger NS. Elliptic Partial Dierential Equations of Second Order (2nd edn). Springer: Berlin, 1983. 10. Alt HW, Luckhaus S. Quasilinear elliptic-parabolic dierential equations. Mathematische Zeitschrift 1983; 183:311–341. 11. Ladyzenskaja OA, Solonnikov VA, Ural’tzeva NN. Linear and Quasilinear Equations of Parabolic Type. Translation of Mathematical Monographs, vol. 23. American Mathematical Society: Providence, RI, 1968.. Copyright ? 2006 John Wiley & Sons, Ltd.. Math. Meth. Appl. Sci. 2006; 29:1261–1289.

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