Nonlinear oblique derivative problems for singular degenerate parabolic equations on a general domain

Transcription

1 Nonlinear Analysis 57 (2004) Nonlinear oblique derivative problems for singular degenerate parabolic equations on a general domain Hitoshi Ishii a; ;1, Moto-Hiko Sato b;2 a Department of Mathematics, School of Education, Waseda University, Nishi-Waseda 1-6-1, Shinuku-ku, Tokyo , Japan b Common Subject Division, Muroran Institute of Technology, Muroran , Japan Abstract We establish comparison and existence theorems of viscosity solutions of the initial-boundary value problem for some singular degenerate parabolic partial dierential equations with nonlinear oblique derivative boundary conditions. The theorems cover the capillary problem for the mean curvature ow equation and apply to more general Neumann-type boundary problems for parabolic equations in the level set approach to motion of hypersurfaces with velocity depending on the normal direction and curvature.? 2004 Elsevier Ltd. All rights reserved. MSC: 35K60; 35K65 Keywords: Capillary problem; Mean curvature ow; Viscosity solutions; Singular degenerate parabolic equations; Neumann-type boundary conditions 1. Introduction In this paper we are concerned with the following boundary value problem: u t + F(t; x; u; Du; D 2 u)=0 in Q =(0;T) ; (1.1) B(x; Du)=0 in S (1.2) Corresponding author. addresses: ishii@edu.waseda.ac.jp (H. Ishii), motohiko@mmm.muroran-it.ac.jp (M.-H. Sato). 1 Supported in part by Grant-in-Aid for Scientic Research (No , No ) of JSPS. 2 Supported in part by Grant-in-Aid for Scientic Research (No ) of JSPS X/$ - see front matter? 2004 Elsevier Ltd. All rights reserved. doi: /j.na

2 1078 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) where is a bounded domain in R n and T 0. Here u t and Du and D 2 u denote, respectively, the gradient and Hessian of u. Weassume throughout this paper that is a bounded domain in R n with C 1 boundary. We deal with Eqs. (1.1) in a class of singular degenerate parabolic equations which includes the mean curvature ow equation. See [3,6,9] for level set approaches for curvature ows. The boundary condition (1.2) is a type of fully nonlinear oblique derivative boundary conditions, which will be made precise later. A typical example we have in mind is the capillary = a(x) Du ; where a(x) is a smooth function on with values in (0; 1). This boundary condition appears, in fact, as the interpretation of the standard capillary condition in the level set approach to motion of hypersurfaces by curvatures. See, for instance [5,12], see also [7]. In the case when F is continuous in its variables, there are already a lot of comparison and existence results for viscosity solutions of second order degenerate parabolic PDE with boundary condition (1.2). We refer for this to [1,8,10] and references therein. In the case of singular PDE like the mean curvature ow equation, Giga and Sato [7,11] have established comparison and existence results for viscosity solutions under the Neumann condition. When is a half space Sato has established comparison and existence theorems under the capillary condition in [12]. Our aim in this paper is to establish comparison and existence theorems concerning viscosity solutions of (1.1) and (1.2). Our results in this paper extends the results obtained in [7,12] to general C 1 bounded domains and fully nonlinear oblique boundary conditions. A variant of these results has been already announced and utilized in studying the stability of motion of hypersurface driven by curvature in the work [5] byei et al., where this paper is referred to as Capillary problem for singular degenerate parabolic equations. (We hope this change of title does not cause any confusion.) After we had almost completed this work we noticed the work by Barles [2] which treated (1.1) and (1.2). (We would like to thank Mariko Arisawa who brought [2] to our notice.) Compared with our results here, his results [2] cover more general F and B, but less general domains. This dierence is similar to that between [1] and [8]. This paper is organized as follows. In Section 2 we state and prove our comparison result. In Section 3 we establish our existence result. In Section 4 we explain how to build test functions which are needed in the proof of the comparison and existence theorems. In Section 5 we discuss a few examples including the capillary problem for the mean curvature ow equation. 2. A comparison theorem We start by listing our assumptions on F and B. Henceforth, for p; q R n \{0} we write p = p p and (p; q) =[( p q ) 1 p q ] 1. Here and henceforth we use

3 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) the notation: a b = min{a; b} and a b = max{a; b}. We denote by S n the space of real n n symmetric matrices which is equipped with the usual ordering 6 and with the usual norm. That is, for X; Y S n, we write X 6 Y,if (X Y ); 6 0 for all R n, and X = max{ X : R n ; =1}. More generally, for any real n m matrix M, M will denote the norm M = max{ Mx : x R m ; x =1}. (F1) F C([0;T] R (R n \{0}) S n ). (F2) There exists a constant R such that for each (t; x; p; X ) [0;T] (R n \{0}) S n the function u F(t; x; u; p; X ) u is non-decreasing on R. (F3) For each R 0 there exists a continuous function! R :[0; ) [0; ) satisfying! R (0) = 0 such that if X; Y S n and 1 ; 2 [0; ) satisfy X 0 I I Y I I then F(t; x; u; p; X ) F(t; y; u; q; Y ) + 2 ( I 0 0 I! R ( 1 ( x y 2 +(p; q) 2 )+ 2 + p q + x y ( p q + 1)): for all t [0;T]; x;y ; u R, with u 6 R, and p; q R n \{0}. (F4) F is continuous at (t; x; u; 0; 0) for any (t; x; u) [0;T] R in the sense that F (t; x; u; 0; 0) = F (t; x; u; 0; 0) (2.1) holds. Here F and F denote, respectively, the upper and lower semi-continuous envelopes of F, which are dened on [0;T] R R n S n. (B1) B C(R n R n ) C 1;1 (R n (R n \{0})). (B2) For each x R n the function p B(x; p) is positively homogeneous of degree one in p, i.e., B(x; p)=b(x; p) for all 0 and p R n \{0}. (B3) There exists a positive constant such that (z);d p B(z; p) for all and p R n \{0}. Here (z) denotes the unit outer normal vector of at We may assume in what follows that the function! R in (F3) is non-decreasing on [0; ). Theorem 2.1. Suppose that (F1) (F4) and (B1) (B3) hold. Let u USC([0;T) ) and v LSC([0;T) ) be, respectively, viscosity sub and supersolutions of (1.1) and (1.2). If u(0;x) 6 v(0;x) for x, then u 6 v on (0;T). Remark 2.2. Assumptions (F1) and (F3) guarantee the degenerate ellipticity of F on [0;T]. Indeed, for any X; Y S n satisfying X 6 Y, 0, and for ; R n we have X; Y; = X ( ); +2 X ( ); + (X Y ); ) ; 6 X 2 +2 X 6 ( X + 1 ) X 2 + X 2 ;

4 1080 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) that is, ( X ) I I I 0 X + X : 0 Y I I 0 I Hence, from (F3) we get F(t; x; u; p; Y ) 6 F(t; x; u; p; X )+! R ( X ): for any (t; x; u; p) [0; T] R (R n \{0}). Sending 0, we get F(t; x; u; p; Y ) 6 F(t; x; u; p; X ): We need the next lemma for the proof of Theorem 2.1. We refer the reader to Lemma 3.4 in [8] for a proof of Lemma 2.3. Lemma 2.3. Assume that (B1) and (B3) hold. For any (0; 1) there exists a function C satisfying the properties: (x) 0 for all x and D p B(x; p); D (x) 1 for all (x; (R n \{0}). Proof of Theorem 2.1. We may assume by replacing T 0 by a smaller number if necessary that u and v is bounded above on [0;T). For any constant A max x u(0;x) ( v(0;x)), if we choose a constant B 0 large enough, then the functions f(t; x)= A Bt and g(t; x)=a + Bt are, respectively (viscosity) sub and supersolutions of (1.1) and (1.2). For such functions f and g, weset ũ(t; x)=u(t; x) f(t; x) and ṽ(t; x)=v(t; x) g(t; x); and observe that ũ and ṽ are, respectively, sub and supersolutions of (1.1) and (1.2) and that ũ(0;x) 6 ṽ(0;x) for x. If we can show that ũ 6 ṽ on [0;T) for any such f and g, then we see that u 6 v on [0;T). This observation allows us to assume that u and v are bounded. Also, the standard technique reduces the proof to the case when =0 in (F2). Indeed, if 0, then the functions û(t; x) =e t u(t; x) and ˆv(t; x) =e t v(t; x) are, respectively, sub and supersolutions of (1.1) and (1.2) with F(t; x; r; p; X ) replaced by the function r +e t F(t; x; e t r; e t p; e t X ): Thus we may assume that the function r F(t; x; r; p; X ) is non-decreasing in R for each (t; x; p; X ) [0;T] (R n \{0}) S n. In view of Lemma 2.3, we may choose a function C 2 so that (x) 0 for all x and D p B(x; p);d (x) 1 for all (x; (R n \{0}). By virtue of Theorem 4.4 in Section 4, there are a function w C 1;1 ( ) and positive constants d and C such that for all (x; y), { x y 4 6 w(x; y) 6 C x y 4 ; (2.2) D x w(x; y) D y w(x; y) 6 C x y 3 ;

5 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) { B(x; Dx w(x; y)) 0 if B(y; D y w(x; y)) 0 if { Dx w(x; y)+d y w(x; y) 6 C x y 4 ; (D x w(x; y); D y w(x; y)) 6 C x y if 0 x y 6 d; (2.3) (2.4) and for a.e. (x; y) ; { } I I I 0 D 2 w(x; y) 6 C x y 2 + x y 4 : (2.5) I I 0 I Note here that it is implicit in (2.4) that if 0 x y 6 d, then D x w(x; y) 0 and D y w(x; y) 0. We argue by contradiction and thus suppose that m 0 := sup{u(t; x) v(t; x): (t; x) [0;T) } 0: (2.6) For 0; 0; 0 we dene (t; x; y)= + w(x; y)+( (x)+ (y)); T t (t; x; y)=u(t; x) v(t; y) (t; x; y) for (t; x; y) [0;T). From (2.6) we infer that for suciently small 0 and 0, the function attains a maximum greater than m 0 =2. Fix such and, and choose a maximum point (ˆt; ˆx; ŷ) of. Note that and (ˆt; ˆx; ŷ) depend on ; ;. It is now well-known (see, e.g. [4]) that lim 0 lim lim 0 (ˆt; ˆx; ŷ)=m 0 ; (2.7) lim sup{w(ˆx; ŷ): 0 1; 0 1} =0: (2.8) We will pass to the limit as 0, in this order. Thus, in view of (2.7) and (2.8), we may assume that ˆt 0, u(ˆt; ˆx) v(ˆt;ŷ), and ˆx ŷ 6 d. Note that B(ˆx; ˆp) 0 if (2.9) B(ŷ; ˆq) 0 if (2.10) where ˆp = D x w(ˆx; ŷ)+d (ˆx) and ˆq = D y w(ˆx; ŷ) D (ŷ). Indeed, for ŷ, ifd x w(ˆx; ŷ) = 0, then we have for some ˆ R n \{0}, B(ˆx; ˆp)=B(ˆx; 0) + D p B(ˆx; ˆ);D (ˆx) : Next, consider the case where D x w(ˆx; ŷ) 0. Suppose for the moment that D x w(ˆx; ŷ)+ sd (ˆx) = 0 for some s 0. Let r 0 be the smallest among such s. Then we have

6 1082 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) for some ˆ R n \{0}, 0=B(ˆx; D x w(ˆx; ŷ)+rd (ˆx)) = B(ˆx; D x w(ˆx; ŷ)) + r D p B(ˆx; ˆ);D (ˆx) r: This contradiction shows that D x w(ˆx; ŷ) +sd (ˆx) 0 for all s 0. Moreover, the above computation, with in place of r, shows that B(ˆx; ˆp). Thus (2.9) is valid. Similarly, we can show that (2.10) holds, the details of which we leave to the reader. We apply the maximum principle for semi-continuous functions (see [4]), to nd matrices X; ˆ Yˆ S n such that X ˆ 0 I I I 0 6 3C ˆx ŷ 2 + C 1 ( ˆx ŷ 4 + ) ; 0 Yˆ I I 0 I where C is the constant from (2.5) and C 1 = C sup x D 2 (x), and such that (T ˆt) + F (ˆt; ˆx; û; ˆp; X ˆ) F (ˆt;ŷ; ˆv; ˆq; Y ˆ) 6 0; 2 where û = u( ˆt; ˆx) and ˆv = v( ˆt;ŷ). From the latter, noting that û ˆv and using the monotonicity of F (t; x; u; p; X ) in the variable u, weget T 2 + F (ˆt; ˆx; û; ˆp; X ˆ) F (ˆt;ŷ; û; ˆq; Y ˆ) 6 0: We now intend to send 0 while keeping and xed. There is a sequence { j } (0; 1), t (0;T), x; y, u R, and X; Y S n such that j 0asj and such that ˆt t, û u, ˆx x, ŷ y, Xˆ X, and Yˆ Y along the sequence = j as j. We then have X 0 I I I 0 6 3C x y 2 + C 1 x y 4 (2.11) 0 Y I I 0 I and T 2 + F (t; x; u; p; X ) F (t; y; u; q; Y ) 6 0; (2.12) where p = D x w(x; y) and q = D y w(x; y). Noting that x y 6 d, we now divide our considerations into two cases. The rst case is when x =y. In this case, we nd from (2.2) that p =q = 0. Also, by (2.11) we have X ; 0 Y 0 0 which is equivalent to that X 6 0 and Y 6 0. Combining these, we get T 2 + F (t; x; u; 0; 0) F (t; y; u; 0; 0) 6 0: Here, we have used as well the degenerate ellipticity of F and F. Then, using (F4), we obtain =T 2 6 0, which is an obvious contradiction.

7 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) The other case is, of course, when 0 x y 6 d. In view of (2.4), we have p 0 and q 0.WesetR =sup [0;T) ( u + v ) and! =! R, where! R is the function from (F3). In view of (F1) and (F3), from (2.11) weget 0 T 2!(r 1 + r 2 ); where r 1 =3C x y 2 ( x y 2 + ( p; q) 2 )+C 1 x y 4 ; r 2 = p q + x y ( p q +1): We now use (2.4) and (2.2) to obtain r 1 6 (3C(1 + C 2 )+C 1 ) x y 4 ; r 2 6 2C x y 4 + x y : In view of (2.2) and (2.8), letting,weget=t 2 6 0, a contradiction. Thus we conclude that u 6 v on Q. 3. An existence theorem We next show the existence of a viscosity solution of the initial-boundary value problem u t + F(t; x; u; Du; D 2 u)=0 in Q; (3.1) B(x; Du)=0 on S; (3.2) u(0;x)= g(x) for x ; (3.3) where g C is a given function. The main result in this section is the following. Theorem 3.1. Assume that (F1) (F4) and (B1) (B3) hold. Then for each g C there is a (unique) viscosity solution u C([0;T) ) of (3.1) and (3.2) satisfying (3.3). Proof. The uniqueness assertion is an immediate consequence of Theorem 2.1. We use the Perron method (see [4]) to show the existence of a continuous viscosity solution of (3.1) (3.3). For this, the rst step is to build sub and supersolutions of (3.1) and (3.2) satisfying (3.3). If we introduce the new unknown û(t; x) =e t u(t; x), where R is the constant from (F2), then the problem (3.1) (3.3) is reduced to the case when = 0. Hence, we may assume in the following proof that =0.

8 1084 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) According to Theorem 4.4 in Section 4, there is a function w C 1;1 ( ) having the following properties: B(x; D x w(x; y)) 0 for (x; ; (3.4) B(y; D y w(x; y)) 6 0 for (x; (3.5) x y 4 6 w(x; y) 6 C x y 4 for x; y (3.6) and for some constant C 0. By replacing C with a larger number if necessary, we may assume that Dw(x; y) 6 C for all (x; y), D 2 w(x; y) 6 C for a.e. (x; y), and g(x) 6 C for all x. This second inequality implies that for (x; y; X ) S n, if (D x w(x; y);x) J 2;+ w( ;y)(x); then X CI; (3.7) if (D x w(x; y);x) J 2; w( ;y)(x); then X 6 CI; (3.8) if (D y w(x; y);x) J 2;+ w(x; )(y); then X CI; (3.9) if (D y w(x; y);x) J 2; w(x; )(y); then X 6 CI: (3.10) Here, we have used the notation: for xed y and for the function u on dened by u(x)=w(x; y), we write J 2;± w( ;y)(x) for J 2;± u(x). Similarly, for xed x and for the function v on dened by v(y)=w(x; y), we write J 2;± w(x; )(y) for J 2;± v(y). Inequalities (3.7) (3.10) are more or less well known. We just show briey that (3.7) holds. The rst thing to note is that Dxw(x; 2 y) 6 C for a.e. x and for all y. Let (D x w(x; y);x) J 2;+ w( ;y)(x) and C 2 () be a function such that the function w(z; y) (z) ofz attains a strict maximum at x and such that D (x) =D x w(x; y) and D 2 (x)=x. Using Jensen s Lemma (see Lemma A.3 in [4]), we nd sequences {a j }; {z j } R n, which both converge to zero, such that for each j N, the function w(z; y) (z) a j ;z of z attains a local maximum at z j and it has Taylor s expansion of second order at z j. Furthermore, Jensen s Lemma says that the points z j can be chosen so that Dxw(z 2 j ;y) 6 C for all j N. Then, we have CI 6 Dxw(z 2 j ;y) 6 CI for all j N. Since z j is a local maximizer of z w(z; y) (z) a j ;z, we have Dxw(z 2 j ;y) 6 D 2 (z j ) for all j N. Thus we see that D 2 (z j ) CI for all j N, and, by sending j, we conclude that X CI. Next, we x a unit vector e R n and observe that for X S n, X 0 I I 0 6 X ; 6 X I 0 X 0 I and from (F3) that for (t; x; u; p; X ) [0;T] [ R; R] (R n \{0}) S n and R 0, F(t; x; u; p; X ) F(t; x; u; e; 0)! R ( X + p e ) F(t; x; u; e; 0)! R ( X + p +1); F(t; x; u; p; X ) 6 F(t; x; u; e; 0) +! R ( X + p +1):

9 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) We may assume here that! R is continuous on [0; ). Then we get from the above F (t; x; u; p; X ) F(t; x; u; e; 0)! R ( X + p +1); (3.11) F (t; x; u; p; X ) 6 F(t; x; u; e; 0) +! R ( X + p + 1) (3.12) for all (t; x; u; p; X ) [0;T] [ R; R] R n S n and R 0. Fix any 0 1 and choose a constant () 0 so that g(x) g(y) 6 + () x y 4 for all x; y : (3.13) Let be a positive function of (0; 1) to be xed later, and we dene functions V ± on [0;T) parametrized by (0; 1); y, respectively, by V + (t; x; ; y)=g(y)+ + ()w(x; y)+()t; V (t; x; ; y)=g(y) ()w(y; x) ()t: It is clear from (3.13) and (3.6) that for any y, V (t; x; ; y) 6 g(x) 6 V + (t; x; ; y) for all x : We intend to select the function so that V + and V are viscosity super and subsolutions of (3.1) and (3.2), respectively. To do this, x (t; x) (0; T) and let (a; p; X ) P 2; V + (t; x), where a R; p R n ; X S n. (See [4] for the denition of parabolic semi-jets P 2;±.) We rst consider the case when We then observe (see (2.15) of [4] for a closely related observation) that p = ()D x w(x; y)+r(x) for some r 0. Hence, using (B3) and (3.4), we see that B(x; p) 0. This shows that B(x; p) (a + F (t; x; V + (t; x; ; y);p;x)) 0: We next consider the case when x. By our choice (in particular recall (3.8)) of C, we have p 6 C; X 6 ()CI; V + (t; x; ; y) g(y) C: Using (F2), with = 0, the degenerate ellipticity of F, and (3.11) we nd that F (t; x; V + (t; x; ; y);p;x) F (t; x; C; p; ()CI) F(t; x; C; e; 0)! C ((()+1)C +1): We x () = max F(t; x; C; e; 0) F(t; x; C; e; 0 +! C ((()+1)C +1); (t;x) Q and observe that a = () and hence B(x; p) (a + F (t; x; V + (t; x; ; y);p;x)) a + F (t; x; V + (t; x; ; y);p;x) 0: Consequently, with the above choice of function, for each (; y) (0; 1) the function V + ( ; ; y) is a viscosity supersolution of (3.1) and (3.2). Similar considerations to the above show that, with the above choice of, for each (; y) (0; 1) the function V ( ; ; y) is a viscosity subsolution of (3.1) and (3.2).

10 1086 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) Next, we dene functions f ± on [0;T) by f + (t; x) = inf {V + (t; x; ; y): 0 1; y }; f (t; x)=sup{v (t; x; ; y): 0 1; y }: Then we easily deduce that f ± are continuous on [0;T), f (t; x) 6 g(x) 6 f + (t; x) for all (t; x) [0;T) and f ± (0;x)=g(x) for all x and that f + and f are viscosity super and subsolutions of (3.1) and (3.2), respectively. Now we conclude by the Perron method together with Theorem 2.1 that if we dene the function u on [0;T) by u(t; x)=sup{v(t; x): v S }; where S denotes the set of functions v on [0;T) such that v is a viscosity subsolution of (3.1) and (3.2) and such that f 6 v 6 f + on [0;T), then u is a continuous function on [0;T), which is a consequence of Theorem 2.1, and a viscosity solution of (3.1) and (3.2). Noting that u satises (3.3), we conclude the proof. 4. Construction of a test function Since is a bounded C 1 domain, there are a compact neighborhood N and a vector eld C(N; R n ) such that =1 on N and = Under the assumptions (B1) (B3), we may assume in addition that (x);d p B(x; p) 0 for all (x; p) N (R n \{0}): (4.1) We x such a pair of N and, and we write for. For any vector eld C(N; R n ) and constant 0 we write W (; )={(x; ) N (R n \{0}): (x); 6 }: Lemma 4.1. Assume (B1) (B3). Let N and be as above. Then there are positive constants ; C, and a function u C 1;1 (W (; )) C(W (; )) such that for all (x; ) W (; ), 2 6 u(x; ) 6 C 2 ; (4.2) u(x; )= 2 u(x; ) for all [0; ); (4.3) B(x; D u(x; ))=0: (4.4)

11 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) Remark 4.2. If we assume (B1) and (B2), then there is a constant M 0 such that for a.e. (x; p) N R n, ( p 1 B(x; p) ) D p B(x; p) ( p D 2 pb(x; p) ) 6 M: (4.5) As the proof below shows, the constants and C in Lemma 4.1 can be chosen to depend only on from (B3) and M. Proof. Our proof parallels that of Lemma 4.3 in [8]. We prove this lemma in the case when B C 2 (N (R n \{0})). The general case can be treated by a limiting argument based on smooth approximations of B, but the details will be left to the reader. Let (0; 1) be a constant to be xed later on. Choose a C 2 vector eld : N R n so that = 1 and 6 =2 onn. In view of (4.1), we may assume by replacing by a smaller positive number that (x);d p B(x; p) for all (x; p) N (R n \{0}): We dene the function u on N R n by u(x; )=sup{ p; 1 2 p p; (x) (x) 2 : p R n ; B(x; p)=0}: It is easy to check that u is positively homogeneous of degree two, i.e. u(x; ) = 2 u(x; ) for all (x; ; ) N R n [0; ). We x any z N and examine the function u(x; ) for x in a neighborhood of z. There are a neighborhood V N of z and a family {e 1 (x);:::;e n (x)} x V of orthonormal bases of R n such that e n (x)=(x) for x V and e i C 2 (V ) for i =1;:::;n.Weset B(x; p)= B(x; p 1 e 1 (x) + + p n e n (x)). Note that B C(V R n ) and B C 2 (V (R n \{0})). Setting e 1 (x) E(x)=.. ; e n (x) we have ( ) n D p B(x; p)=e(x)d p B x; p i e i (x) ; i=1 ( ) n Dp 2 B(x; p)=e(x)dpb 2 x; p i e i (x) E(x) ; i=1 where E(x) denotes the transposed matrix of E(x). According to (4.5), we have for all (x; p) V (R n \{0}), ( p 1 B(x; p) ) D p B(x; p) ( p D 2 p B(x; p) ) 6 M: Note as well that the function p B(x; p) is positively homogeneous of degree one for each x V.

12 1088 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) We now need to assume that M 6. Since ( n B(x; p) = (x);d p B x; p i e i n i=1 M 2 for all (x; p) V (R n \{0}) (4.6) 2 by (B3), there is a unique continuous function H on V R n 1 such that B(x; p) = 0 if and only if p n + H(x; p 1 ;:::;p n 1 )=0: (4.7) By the implicit function theorem, we have H C 2 (V (R n 1 \{0})). It follows from the homogeneity of B and the uniqueness of H that q H(x; q) is positively homogeneous of degree one. Also, it is easily seen that there is a constant A 0 depending only on and M such that for all (x; q) V (R n 1 \{0}), ( q 1 H(x; q) ) D q H(x; q) ( q DqH(x; 2 q) ) 6 A: (4.8) For (x; ) V R n, we have { n 1 } u(x; )=sup q i e i (x); H(x; q) e n (x); 1 2 q 2 : q R n 1 : (4.9) i=1 For (x; ) V R n, writing = n 1 i=1 y ie i (x)+te n (x) and = = 1 2, we have (x; ) W (; ) if and only if t 6 y and y 0: We set = = 1 2 ; W = {(x; y; t) V (R n 1 \{0}) R: t 6 y }; ( ) n 1 ũ(x; y; t)=u x; y i e i (x)+te n (x) for (x; y; t) V R n 1 R: From (4.9) we see that i=1 ũ(x; y; t)=sup{ q; y th(x; q) 1 2 q 2 : q R n 1 }: (4.10) It is now obvious that for each (x; y; t) V R n 1 R, the above optimization problem has a maximizer q and it satises the relation q = y td q H(x; q): (4.11) Let (x; y; t) W and q R n 1 satisfy (4.11). As in the proof of Lemma 4.3 of [8], we easily see that if 6 1=(3A), then q y 6 A t y ; 2 3 y 6 q y ; td 2 qh(x; q) ; (I + td2 qh(x; q)) 1 6 2; and that the solution q of (4.11) is unique. We write p(x; y; t) for the unique solution q of (4.11). By the implicit function theorem, we have p C 1 ( W ).

13 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) Now, as in [8], we see that ũ C 2 ( W ), D y ũ(x; y; t)=p(x; y; t) in W, and ũ t (x; y; t)+h(x; D y ũ(x; y; t))=0 in W: Furthermore, xing =( 1 4A ) ( M ), we get M ; ũ(x; y; t) 1 2 y 2 th(x; y) 1 4 y 2 ; ũ(x; y; t) y 2 th(x; p(x; y; t)) y 2 + A t p(x; y; t) 6 2 y 2 : Hence, noting that y 2 + t 2 6 (1 + 2 ) y 2 for (x; y; t) W,weget 1 4(1 + 2 ) ( y 2 + t 2 ) 6 ũ(x; y; t) 6 2( y 2 + t 2 ) for (x; y; t) W: These readily imply that u C 2 (W (; )) C(W (; )), B(x; D u(x; ))=0 in W (; ); 1 4(1 + 2 ) 2 6 u(x; ) for all (x; ) W (; ): What remains is to observe that W (; =2) W (; ) and that the function 4(1 + 2 )u(x; ) satises (4.2) (4.4) with =2 replacing. Lemma 4.3. Assume (B1) (B3). Then there are a positive constant and a function v C 1;1 ( R n ) such that for all (x; ) R n, v(x; ) 2 ; (4.12) v(x; )= 2 v(x; ) for all [0; ); (4.13) B(x; D v(x; )) 0 if and (x); ; (4.14) B(x; D v(x; )) 6 0 if and (x); 6 : (4.15) Proof. Again, our proof is similar to that of Lemma 4.2 in [8]. We divide our arguments into two steps. Step 1: Choose 0 and u C 1;1 (W (; )) C(W (; )) as in Lemma 4.1. As before we choose a vector eld C(N ) so that = 1 and 6 =16 on N. Note that, since 6 =2 onn, we have W (; =2) W (; ): Replacing 0 by a smaller number, we may assume that 4M 6, where M is a positive constant such that (4.5) holds, so that D p B(x; p);(x) 2 for all (x; p) N (R n \{0}): We choose a C function f on R such that 0 6 f 6 1onR, f(r) =1 if r 6 1, f(r) =0 if r 2, f (r) 6 0 for r 0 and f (r) 0 for r 6 0. Also we choose

14 1090 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) a C function g on R such that 0 6 g 6 1onR, g(r)=0 if r 6 1 2, g(r)=1 if r 1, g (r) 0 for r 0 and g (r) 6 0 for r 6 0. We may assume that f (r) g (r) 6 3 for all r R. Setting a = =4, we dene functions f a, g a on N R n by ; (x) f if 0; f a (x; )= a 0 if =0; ; (x) g if 0; g a (x; )= a 0 if =0: Let b 0 be a constant to be xed later on. We dene the function v on N R n by ; (x) 2 v(x; )=f a (x; )u(x; )+g a (x; ) : b This function is well-dened since for all (x; ) N (R n \{0}), if ; (x) 6 2a ; then (x; ) W (; ): (4.16) That is, at those points where u is not dened, the function f a vanishes and the standard convention gives the meaning to v. To see (4.16), let (x; ) N (R n \{0}) and assume that ; (x) 6 2a. Then we have (x); 6 (x); a + 6 ; 2 which shows that (x; ) W (; ). It is easy to see that v(x; )= 2 v(x; ) for all (x; ; ) N R n [0; ); v C 1;1 (N R n ): Also, it is easy to observe that for any (x; ) N R n, v(x; ) 2 if ; 6 a ; u(x; ) 1 b ; 2 a2 b 2 if ; a : We assume henceforth that b 6 a 2, so that v satises (4.12) and (4.13). We intend to show that v satises (4.14) and (4.15), with a=4 in place of. Fix (x; ) N R n.if = 0, then B(x; D v(x; )) = B(x; 0) = 0. Thus we may assume in what follows that 0.By(4.12) and (4.13), we have D v(x; ) 0. We compute that ) ; 2 2 ; B(x; D v(x; )) = B (x; ud f a + f a D u + D g a + g a b b = f a B(x; D u) + D p B(x; ˆp);uD f a + ; 2 b 2 ; D g a + g a b (4.17)

15 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) for some ˆp (R n \{0}). Consider the case when ; (x) 6 a 2. We have g a(x; )=0, f a (x; )=1 and D g a (x; )=D f a (x; ) = 0. Hence, we have B(x; D v(x; )) = B(x; D u(x; ))=0: (4.18) Next consider the case when ; (x) 2a. Since f a (x; ) =0, g a (x; ) =1, and D f a (x; )=D g a (x; ) = 0, from (4.17) we have B(x; D v(x; )) = B(x; 0) + D p B(x; ˆp); 2g a ; b ; 0: (4.19) b The case when ; (x) 2a can be treated similarly to the previous case, and in this case we have B(x; D v(x; )) 0. Now, we consider the case when a 2 ; (x) 6 2a. We want to show that B(x; D v(x; )) 0 if ; (x) 0; (4.20) B(x; D v(x; )) 6 0 if ; (x) 0: (4.21) We only deal with the case when ; (x) 0, i.e., the case when a 2 ; (x) 6 2a. The other case can be treated in a parallel way. Note that D f a = f a ; a a 3 ; D g a = g a ; a a 3 ; where f a = f ( ; a ) and g a = g ( ; a ). Therefore we have D p B(x; ˆp);uD f a u D p B(x; ˆp) f a a 3CM if a 6 ; (x) 6 2a a 0 if 1 a 6 ; (x) 6 a 2 Here and henceforth we have D p B(x; ˆp); and 3CM g a : a ˆp and C are those from (4.17) and (4.2), respectively. Also, ; 2 D g a = ; 2 g { a D p B(x; ˆp); M b ab ; 2 g a ab D p B(x; ˆp); 2g a ; g a ; b b 2 2aM 0; a g a : 2b } ;

16 1092 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) Fixing b = a 2 min{ (4.4) we obtain 6CM ; 1} and combining the above inequalities, in view of (4.17) and ( a B(x; D v(x; )) g a 2b 3CM a ) 0; which proves (4.20). Note that, since 6 a 4 on N, for (x; ) N Rn,if ; (x) a ( 4 resp:; ; (x) 6 a ) 4 ; then we have ; (x) a ( 2 resp:; ; (x) 6 a ) 2 ; and therefore, by (4.18) (4.20), B(x; D v(x; )) 0: This shows that (4.14) holds with a=4 in place of. Similar considerations to the above guarantee as well that (4.15) holds with a=4 in place of. Thus we conclude that the function v C 1;1 (N R n ) satises (4.12) (4.15), with replaced by a=4, for all (x; ) N R n. Step 2: We choose a function C 2 (R n ) so that (x) = 1 in a neighborhood the support of is contained in the interior of N, and inR n. Set ṽ(x; )=(x)v(x; )+(1 (x)) 2 for (x; ) R n R n : It is now obvious that the function ṽ has all the required properties. Theorem 4.4. Assume that (B1) (B3) hold. Then there are a function w C 1;1 ( ) and positive constants C and such that for all (x; y), (i) x y 4 6 w(x; y) 6 C x y 4, D x w(x; y) D y w(x; y) 6 C x y 3 ; (ii) B(x; D x w(x; y)) 0 if B(y; D y w(x; y)) 6 0 if (iii) D x w(x; y)+d y w(x; y) 6 C x y 4, (D x w(x; y); D y w(x; y)) 6 C x y if 0 x y 6, and for a.e. (x; y) ; (iv) D 2 w(x; y) 6 C{ x y 2 ( I I I I )+ x y 4 ( I 0 0 I )}. Remark 4.5. The second condition in (iii) of Theorem 4.4 assumes that if 0 x y 6, then D x w(x; y) 0 and D y w(x; y) 0. Proof of Theorem 4.4. According to Lemma 2.3, we can choose a function C having the properties: 0 on and D p B(x; p);d (x) 1 for (x; (R n \{0}):

17 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) Let v and be a function on R n and a positive constant, respectively, for which the conditions of Lemma 4.3 hold. We set g = v 2 and choose a constant C 1 0so that for a.e. (x; ) R n, g(x; ) D x g(x; ) 6 C 1 4 and D g(x; ) D x D g(x; ) 6 C 1 3 : We x a Lipschitz constant M 1 of the function B on, and set (x)=mc 1 (x) for x. Then we have D p B(x; p);d (x) MC 1 for (x; (R n \{0}): (4.22) We dene u(x; y)=e (x)+ (y) g(x; x y) for (x; y) : It is easy to see that inequality (i), with w replaced by u, holds for some constant C. It is a standard observation (see, e.g., the proof of Theorem 4.1 in [8]) that (iv), with w replaced by u, holds for some constant C. Since is a bounded C 1 domain, we can nd d 1 0 so that if y, and x y 6 d 1, then (x);y x 6 y x. Let 0 d6d 1. We intend to show that if d is small enough, then u satises condition (iii), with replaced by d. To do this, writing ê =e (x)+ (y), we calculate that for x; y, D x u(x; y)=ê(g(x; x y)d (x)+d x g(x; x y)+d g(x; x y)); D y u(x; y)=ê(g(x; x y)d (y) D g(x; x y)): Then we observe that for x; y, D x u(x; y)+d y u(x; y) 6 ê( D (x)+d (y) g(x; x y)+ D x g(x; x y) ) 6 (2C 2 +1)C 1 ê x y 4 ; (4.23) where C 2 = max{ D (x) : x }. Observing by Euler s identity that 4g(x; ) = D g(x; );, we nd that D g(x; ) 4 3. Using this, we get g(x; )D (x)+d x g(x; )+D g(x; ) 4 3 C 1 (C 2 +1) 4 = 3 (4 C 1 (C 2 +1) ): Similarly, we have g(x; )D (y) D g(x; ) 3 (4 C 1 C 2 ): We assume henceforth that C 1 (C 2 +1)d63. Then we have g(x; )D (x)+d x g(x; )+D g(x; ) 3 ; g(x; )D (y) D g(x; ) 3 : Thus we see that if x; y and 0 x y 6 d, then D x u(x; y)+d y u(x; y) D x u(x; y) D y u(x; y) 6 (2C 2 +1)C 1 x y ; and conclude that (iii), with u in place of w, holds for some constant C.

18 1094 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) We want to show that (ii), with u in place of w, holds if x y 6 d. Let and y satisfy x y 6 d. It suces to show that B(x; D 1 u(x; y)) 0 and B(x; D 2 u(y; x)) 6 0; (4.24) where D 1 and D 2 denote, respectively, the dierentiation of any function f(x; y) on R n R n with respect to the rst variable x R n and the second variable y R n.if x = y, then D x u(x; y)=d y u(x; y) = 0 and hence, B(x; D 1 u(x; y)) = B(x; D 2 u(y; x))=0: Henceforth, we may assume that x y. Since (x);x y x y, by virtue of (4.15) we have B(x; D g(x; x y)) 0: Using this and writing ê =e (x)+ (y), we calculate that B(x; D 1 u(x; y)) êb(x; D g(x; x y)) for some ˆp R n \{0}, and that + D p B(x; ˆp);u(x; y)d (x)+êd x g(x; x y) MC 1 u(x; y) MC 1 u(x; y)=0 B(x; D 2 u(y; x)) = B(x; u(y; x)d (x)+êd g(y; y x)) 6 êb(x; D g(y; y x)) D p B(x; ˆq);u(y; x)d (x) 6 ê(b(x; D g(x; y x)) + M D g(y; y x) D g(x; y x) ) MC 1 u(y; x) 6 MC 1 ê x y 4 MC 1 u(y; x) 6 0 for some ˆq R n \{0}. Thus, by choosing d = min{d 1 ;C 1 (C 2 +1)=3}, we conclude that (iii), with u and d in place of w and, holds. We choose a function C 2 (R) such that 0 6 (r) 6 1 for r R, (r) =r for r 6 d 4 =2, and (r)=0 for r d 4. We x a constant C 1 so that u(x; y) 6 C x y 4 for x; y, and choose a constant 0 so that C 4 6 d 4 =2. Notice that d. Select a constant A 1 so that A( x y 4 ) x y 4 for all x; y, and dene w C 1;1 ( ) by setting w(x; y)=a(u(x; y)). Noting that Dw = A (u)du, we see easily that condition (i) holds. To check (ii), let x; y. We observe that if Dw(x; y) 0, then u(x; y) d 4, hence x y d, and therefore B(x; D x w(x; y)) = A (u(x; y))b(x; D x u(x; y)) 0 if B(y; D y w(x; y)) = A (u(x; y))b(y; D y u(x; y)) 6 0 We now easily conclude that (ii) holds. if

19 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) To see (iii), we observe from (4.23) that D x w + D y w = A (u) D x u + D y u 6 A(2C 2 +1)C 1 u; and that for x; y, if x y 6, then u(x; y) 6 C x y 4 6 C 4 6 d 4 =2 and hence w(x; y)=au(x; y). Therefore, for x; y, if0 x y 6, then D x w(x; y)=ad x u(x; y) 0; D y w(x; y)=ad y u(x; y) 0; (D x w(x; y); D y w(x; y)) = (D x u(x; y); D y u(x; y)) 6 (2C 2 +1)C 1 x y : To check (iv), we note rst that we may assume, by reselecting C if necessary, that for a.e. x; y, I I I 0 D 2 u(x; y) 6 C x y 2 + C x y 4 : I I 0 I Note also that, since w is a C 1;1 function, I 0 D 2 w(x; y) 6 C 3 0 I for a.e. x; y and for some constant C 3 0. This readily yields I I I 0 D 2 w(x; y) 6 AC x y 2 + max{ac; 4 C 3 } x y 4 I I 0 I for a.e. x; y. 5. Examples In this section we discuss typical examples of F and B to which Theorems 2.1 and 3.1 apply. First we treat examples of F. Let A : (R n \{0}) M n m, where M n m denotes the space of real n m matrices, be a continuous function which is homogeneous of degree zero, i.e. A(x; p)=a(x; p) for all (x; p; ) (R n \{0}) (0; ) (5.1) and which satises A(x; p) A(y; q) 6 C 1 ( x y + p q ) (5.2) for all x; y and p; q S n 1, where C 1 0 is a constant and S n 1 denotes the unit sphere { R n : =1}. It follows that for all x; y and p; q R n \{0}, A(x; p) A(y; q) 6 C 1 ( x y + p p q ) ( q 6 C 1 x y + 6 C 1 ( x y +2(p; q)): ) 2 p q p q

20 1096 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) Let b C( ; R n ) satisfy b(x) b(y) 6 C 2 x y for all x; y : (5.3) Furthermore, let c; f C( ; R) be given. Dene the function F C( R (R n \{0}) S n )by F(x; u; p; X )= tr[a(x; p)a(x; p) X ]+ b(x);p + c(x)u + f(x): (5.4) As is observed in [4], if X; Y S n and 1 ; 2 [0; ) satisfy X 0 I I I Y I I 0 I; then for any x; y and p; q R n \{0}, tr[a(x; p)a(x; p) X ]+tr[a(y; q)a(y; q) Y ] 6 1 tr(a(x; p) A(y; q)) (A(x; p) A(y; q)) + 2 tr(a(x; p) A(x; p)+a(y; q) A(y; q)) 6 n 1 A(x; p) A(y; q) 2 + n 2 ( A(x; p) 2 + A(y; q) 2 ) 6 C 3 1 ( x y 2 + (p; q) 2 )+C 3 C 4 2 ; where C 3 0 is a constant depending only on n and C 1 and C 4 = su p A(x; p) 2 : (x;p) S n 1 It is now easy to see that F satises condition (F3). Also, it is immediate to see that condition (F2) is satised with 6 min c. To check (F4), we note that for any (t; x; r; p; X ) [0; T] R (R n \{0}) S n, F(t; x; r; p; X ) c(x)r f(x) 6 nc 4 X + C 5 p ; where C 5 =max{ b(x) : x }, and nd that F (t; x; r; 0; 0)=F (t; x; r; 0; 0)=c(x)r+f(x) for all (t; x) [0; T]. Thus F satises (F1) (F4). If A(x; p)=i p 2 (p p), b = 0, and c = f = 0, then it is the case of the mean curvature ow equation and the above conditions on A, b, c, and f are valid. A standard way of generalizing the above example is to take the max min of a family of functions F having the form of (5.1). More precisely, let A and B be two non-empty index sets, and let A C( (R n \{0});M n m ), b C( ; R n ), c C, and f C, with (; ) A B, be given. Assume that these sets of functions are uniformly bounded, that {c } and {f } are equi-continuous, that {A } satises (5.1) and (5.2) with a uniform constant C 1, and that {b } is equi-lipschitz

21 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) continuous (i.e., satises (5.3) with a uniform constant C 2 ). Dene F (x; u; p; X )= tr[a (x; p)a (x; p)x ] + b (x);p + c (x)u + f (x); and F(x; u; p; X ) = sup inf F (x; u; p; X ): A B Then the function F satises (F1) (F4). Next we deal with the boundary condition. Consider the function B of the form B(x; p)= (x); p C(x)p ; where : R n R n is a C 1;1 vector eld over R n and C : R n M n n is a C 1;1 function satisfying det C(x) 0 in a neighborhood It is clear that (B2) is satised. We can modify the denition of B so that the resulting function B satises (B1) and B(x; )=B(x; ) for all x in a neighborhood As before let (x) denote the unit outer normal of at By calculation, we have D p B(x; p)=(x) C(x) C(x)p if p 0; C(x)p and we see that (B3) is equivalent to the condition (x);(x) ; C(x)(x) for all (x; S n 1 : A particular case is when = and C(x) =a(x)i for some a C 1;1 (R n ) such that 0 a(x) 1 for which corresponds to the capillary condition. In this case the boundary regularity of should be of class C 2;1 in order that = C 1;1 (R n ) is satised, which is one of requirements of Theorems 2.1 and 3.1. It is interesting to nd that the results in [2] need the same C 2;1 regularity of the boundary. References [1] G. Barles, Fully nonlinear Neumann type boundary conditions for second-order elliptic and parabolic equations, J. Dierential Equations 106 (1) (1993) [2] G. Barles, Nonlinear Neumann boundary conditions for quasilinear degenerate elliptic equations and applications, J. Dierential Equations 154 (1) (1999) [3] Y.-G. Chen, Y. Giga, S. Goto, Uniqueness and existence of viscosity solutions of generalized mean curvature ow equations, J. Dierential Geom. 33 (3) (1991) [4] M. Crandall, H. Ishii, P.-L. Lions, User s guide to viscosity solutions of second order partial dierential equations, Bull. Am. Math. Soc. (N.S.) 27 (1) (1992) [5] S.-I. Ei, M.-H. Sato, E. Yanagida, Stability of stationary interfaces with contact angle in a generalized mean curvature ow, Am. J. Math. 118 (3) (1996) [6] L.C. Evans, J. Spruck, Motion of level sets by mean curvature. I, J. Dierential Geom. 33 (3) (1991) [7] Y. Giga, M.-H. Sato, Neumann problem for singular degenerate parabolic equations, Dierential Integral Equations 6 (6) (1993) [8] H. Ishii, Fully nonlinear oblique derivative problems for nonlinear second-order elliptic PDEs, Duke Math. J. 62 (3) (1991)

22 1098 H. Ishii, M.-H. Sato / Nonlinear Analysis 57 (2004) [9] H. Ishii, P.E. Souganidis, Generalized motion of noncompact hypersurfaces with velocity having arbitrary growth on the curvature tensor, Tohoku Math. J. (2) 47 (2) (1995) [10] P.-L. Lions, Neumann type boundary conditions for Hamilton Jacobi equations, Duke Math. J. 52 (4) (1985) [11] M.-H. Sato, Interface evolution with Neumann boundary condition, Adv. Math. Sci. Appl. 4 (1) (1994) [12] M.-H. Sato, Capillary problem for singular degenerate parabolic equations on a half space, Dierential Integral Equations 9 (6) (1996)