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1 Remarks on W 2;p -Solutions of Bilateral Obstacle Problems Srdjan Stojanovic Department of Mathematical Sciences University of Cincinnati Cincinnati, OH November 30, 995 Abstract The well known theorem on W 2;p -regularity of the solution of the bilateral obstacle problem is proved as a consequence of a necessary condition for an appropriate nonsmooth variational problem. Equivalence of the bilateral obstacle problem to a certain semilinear elliptic equation is proved, as well. Introduction Let be a bounded domain in R n. Let g 2 H () be a boundary value, and let ; 2 2H () be obstacles, given functions such that g 2. It is well known (see e.g. [5]) that the following variational inequality has the unique solution Find () z 2 K = fv 2 H (); v 2 ;v g 2H0()g; such that (2) rz r(v z)h u; v zi v 2 K In (2), u 2 H () is given. Moreover, it is well known that if data has more regularity (to be specied later in another context), then solution belongs to W 2;p () (see [5,,2]). In this paper we introduce another technique of proving W 2;p -regularity of solutions of bilateral obstacle problems. The technique is based on another variational characterization of the obstacle problem which takes full advantage of the regularity of the right-hand side, and of a variant of the maximum principle, and gives W 2;p -regularity of the solution simply, as a consequence of the necessary condition for the minimizer (Theorem ).

2 This method is introduced by the author in [4] in the case of the single obstacle problem (zero-obstacle). In that case it is enough to consider the variational functional (3) F (z) = jrzj 2 +2uz + dx; where z = z + z ;z 0, and where u is a given function such that (4) on an ane set in H () (5) u 2 L p () A = fz 2 H (); z (2 p<); g 2 H 0()g; where g 2 H ();g 0; and to consider the problem Find z 2 A such that (6) F (z) = min F (v) v2a The purpose of the present paper is to adapt these new ideas in the case of the bilateral obstacle problem. There are various formulations of the obstacle problem variational inequality, complementary problem, minimum problem (6), etc. But whatever the formulation of the problem is, in the end the solution obtained either satises the equation (7) z = u or it coincides with (one or) the (other) given obstacle i, in which case () z = i So, the most natural, or the most explicit, formulation of the (bilateral) obstacle problem seems to be elliptic equation (39), which happens to be semilinear, with discontinuous nonlinearity. The small problem is that equation (39) by itself, unless further assumptions are made (see (60)), is not complete description of the problem, since uniqueness does not hold. One has to add condition (4), i.e., (42), to get uniqueness, i.e., to identify the solution of (39) with minimal energy. Hence, the complete characterization of the obstacle problem is an equation (39) summarizing (7) and (), and an inequality (4), i.e., (42), granting the uniqueness. The key result here is the uniqueness (Lemma 4). Our approach leads naturally to such characterization of the bilateral obstacle problem. The semilinear elliptic equation obtained, can be thought of as an Euler equation for the minimum problem discussed in this paper, in a way in which the variational inequality (2) is an Euler \equation" of the corresponding minimum problem on K. In addition, this equation structure of the obstacle problem has been exploited recently by the author in deriving the perturbation formula for regular free boundaries [3]. 2

3 2 W 2;p -Regularity Via Nonsmooth Analysis We assume, in addition to (4), (9) (0) ; 2 2W 2;p () < 2 a.e. in g 2 We shall use the notation a _ b = max(a; b);a^b= min(a; b). Consider the variational functional G(z) = 2 jrzj2 + u( _ z ^ 2 )+( )(z ^ ) () +( 2 )( 2 _ z) for any z 2 A. Notice that because of (0), _ z ^ 2 is well dened. Also notice that G is locally Lipschitz continuous on A, but not dierentiable. Remark If z 2 K then G(z) = 2 jrzj2 + uz +( ) +( 2 ) 2 (2) = 2 jrzj2 + uz + const Consider the problem Find z 2 A such that (3) G(z) = min v2a G(v) Lemma Problem (3) has a solution. Proof We note only that, regarding last three terms in (), by passing to subsequences if necessary, weak convergence in H () implies strong convergence in L 2 (), and hence convergence a.e. in, so that we can pass the limit along a minimizing sequence. Lemma 2 Let z be a minimizer in Problem (3). Then (4) z 2 a.e. in, i.e., z 2 K. (5) Proof Since z ^ 2 2 A, wehave G(z)G(z^ 2 ) 3

4 Hence, it is easy to see that (6) jrzj 2 + 2( 2 )(z _ 2 ) jr(z ^ 2 )j 2 + 2( 2 ) 2 ; which implies, since z _ H (), 0 r(z z ^ 2 ) r(z+z^ 2 ) (7) 2 r 2 r(z_ 2 2 ) This implies () jr(z z ^ 2 )j 2 +2 r(z z^ 2 )r(z^ 2 ) 2 r 2 r(z_ 2 2 ); and hence, since r(z z ^ 2 ) r(z^ 2 )=r 2 r(z_ 2 2 ), (9) jr(z z ^ 2 )j 2 0; i.e., z = z ^ 2, i.e., z 2. Similarly, we can prove that z. Remark 2 No explicit obstacles are introduced, but a variant of a maximum principle forces a minimizer to stay above and below 2. Remark 3 Lemma 2 will follow also from the necessary condition of minimality (more precisely, from Lemma 3). Corollary A minimizer in Problem (3) is unique. Proof By Lemma 2, z 2 K. But by Remark, G is strictly convex on K, which implies that a minimizer is unique. Corollary 2 Problems (3) and (2) are equivalent. (20) For D R n let x2d I D (x) = 0 x62 D be the characteristic function of the set D. In the next Theorem we shall not use the knowledge obtained in Lemma 2, i.e., we shall do computations on sets fz < gand fz > 2 g,aswell. That should help understanding the relationship between necessary and sucient conditions of minimality to follow. 4

5 Theorem The necessary and sucient condition for z to be a minimizer for Problem (3) is ui f z< 2g ( )I fz< g ( 2 )I fz 2g h >< rz rh (2) ui f <z 2g ( )I fz g ( 2 )I fz> 2g h > z g 2H 0 () h 2 H 0();h0 Proof We will rst prove the necessity. The suciency will follow from the consideration in the next Section. Suppose rst that h 2 C 0 (). Since ( + h) _ 0; fh <0g (22)! = h h; fh 0g + as 0, a.e. in, we have < h; fz > (z+h) _ z _ g (23)! h + ; fz= g 0; fz< g as 0, a.e. in. Similarly, < (z + h) ^ 2 z ^ 2 (24)! Hence, (25) as 0, a.e. in. Also, (26) and (27) h; fz < 2 g h ; fz= 2 g 0; fz> 2 g ((z + h) _ ) ^ 2 (z _ ) ^ 2! h; f <z< 2 g >< h! + ; fz= g h ; fz= 2 g > 0; fz< g[fz> 2 g (z + h) ^ z ^ <! 0; fz > g h ; fz= g h; fz < g < 0; fz < (z + h) _ 2 z _ 2 g 2! h + ; fz= 2 g h; fz > 2 g 5

6 as 0, a.e. in. Notice also, for example, that ((z + h) _ ) ^ 2 (z _ ) ^ 2 (2) khk L 2 () ; L 2 () and similar inequalities hold for other relevant quantities. Hence, if z is a minimizer, using uniform integrability, we can compute the one-sided directional derivative, and (29) for all h 2 C 0 (30) and (3) 0 G 0 G(z + h) G(z) (z; h) = lim 0 = rz rh+u I f <z< 2gh + I fz= gh + + I fz= 2gh +( ) I fz= g( h )+I fz< gh for all h 2 C 0 +( 2 ) I fz= 2gh + + I fz> 2gh ; (). This implies, by rst taking h 0, and then also h, rz rh ui f z< 2g ( )I fz< g ( 2 )I fz 2g h; rz rh ui f <z 2g ( )I fz g ( 2 )I fz> 2g h; ();h0, from which (2) follows. Corollary 3 Let z be a minimizer in Problem (3). Then (32) z 2 W 2;p loc () If g 2 W 2;p (), then z 2 W 2;p (), and the following apriori estimate holds kzk W 2;p () c kuk L p () + k k L p () + k 2 k L p () + kgk W 2;p () (33) Proof Consider the linear functional (34) H0() 3 h 7! rz rh By Theorem, this functional can be extended to L 2 (), i.e., there exists f 2 L 2 () such that (35) rz rh= fh h 2 H0(); 6

7 and then (36) which implies ui f z< 2g ( )I fz< g ( 2 )I fz 2g h fh ui f <z 2g ( )I fz g ( 2 )I fz> 2g h h 2 L 2 ();h0 (37) ui f z< 2g ( )I fz< g ( 2 )I fz 2g f ui f <z 2g ( )I fz g ( 2 )I fz> 2g a.e. in. From (35) and (37), Elliptic L p -estimates imply (32) and (33). Corollary 4 Let z be a minimizer in Problem (3). Then (3) ui f z< 2g ( )I fz< g ( 2 )I fz 2g z ui f <z 2g ( )I fz g ( 2 )I fz> 2g Proof Estimate (3) follows from (35) and (37). 3 Bilateral Obstacle Problem as a Semilinear Equation Consider following equations and inequalities (39) (40) (4) z 2 W 2;p loc () \ H () z + ui f <z< 2g +( )I fz g+( 2 )I fz 2g=0 z g 2 H0(); z 2 W 2;p loc () \ H () z + ui f <z< 2g +( )I fz= g+( 2 )I fz= 2g=0 z g 2 H (); 0 (u ) I fz= g =0 (u 2 ) + I fz= 2g=0 Remark 4 More explicitly, (4) can be written as (42) u in \fz= g u 2 in \fz= 2 g a.e. in a.e. in a.e. in 7

8 We have Theorem 2 (a) Necessary conditions of minimality for the variational Problem (3) are (4), (39), (40), and (4). (b) A function z is a solution of (39) if and only if it is a solution of (40) and (4). (c) Sucient conditions of minimality for the variational Problem (3) are (39) and (4), or alternatively (4), (40), and (4). In particular, Problem (39) and (4) admits a unique solution. Proof From (3) we can see on fz < gwehave z, i.e., (43) z = a.e. on fz < g; on fz = g we have u, i.e., (44) u ; z = a.e. on fz = g; on f <z< 2 gwehave u z u, i.e., (45) z = u a.e. on f <z< 2 g; on fz = 2 g we have 2 2 u, i.e., (46) u 2 ; z = 2 a.e. on fz = 2 g; and on fz > 2 gwehave 2 z 2, i.e., (47) z = 2 a.e. on fz > 2 g So, we see that (39) and (4) hold. Lemma 3 If z is a solution of (39) then z 2. (4) Proof If z is solution of (39) then rz r' + ui f <z< 2g +( )I fz g+( 2 )I fz 2g '=0 Set ' =(z 2 ) + in (4). Since (49) rz r(z 2 ) + = '2H 0 () jr(z 2 ) + j 2 + r 2 r(z 2 ) + ;

9 and (50) 2 I fz 2g(z 2 ) + = r 2 r(z 2 ) + ; we get (5) jr(z 2 ) + j 2 =0 and hence, z 2. Similarly we can deduce z, and Lemma follows. This proves (40) and reproves (4). So, (a) holds. Also, we have proved (b). To prove (c), it suces to prove the uniqueness, for example, for Problem (4), (40), (4). Lemma 4 Solution of Problem (4), (40), (4) is unique. Proof (52) Let z and z 2 be two solutions. Then (z z 2 )+u I f <z < 2g I f <z 2< 2g +( ) I fz= g I fz2= g +( 2 ) I fz= 2g I fz2= 2g =0 We need to use some kind of monotonicity of the nonlinear term. To this end, notice (53) u =(u ) + (u ) + From (52) we have (54) (z z 2 )+(u ) + I f <z 2g I f <z 2 2g u I fz= 2g I fz2= 2g +( 2 ) I fz= 2g I fz2= 2g =(u ) I fz2= g I fz= g ( ) I fz2= g I fz= g In (54) we have used also ( ) I fz= g I fz2= g (55) I f <z 2g I f <z 2 2g = I fz2= g I fz= g From (54) we get (56) (z z 2 )+(u ) + I f <z 2g I f <z 2 2g +( 2 u) + I fz= 2g I fz2= 2g =(u ) I fz2= g I fz= g +( 2 u) I fz= 2g I fz2= 2g =0; 9

10 because of (4), and since ( 2 u) = (u 2 ) +. Finally, notice the monotonicity (on K) (57) and (5) (u ) + I f <z 2g I f <z 2 2g (z z 2 ) 0; So, multiplying (56) by (z (59) ( 2 u) + I fz= 2g I fz2= 2g (z z 2 ) 0 z 2 ) and integrating, we get jr(z z 2 )j 2 0 This completes the proof of the Lemma and of the Theorem. Corollary 5 If (60) then a solution of (39) is unique. u 2 Example In general, solution of (39) is not unique. Let =( 2; 2) R ;g=0; = ; 2 =;u(x)= sin x. Then (60), obviously, does not hold. Let (6) z (x) = sin x (62) x2( z 2 (x) = 2 ;3) 2 sin x x 2 n ( 2 2 (63) x2( 3 z 3 (x) = 2 ; 2 ) sin x x 2 n ( 3 2 ; 2 ) Then all three z ;z 2 ; and z 3 are solutions of (39), while only z is the solution of (39), (4). Example 2 In general, (40) does not imply (39). Let =( 5 4 ; 5 4 )R ;g = 0; = 2; 2 (x)=x 2 ;u=0. Then (60), obviously, holds. Let (64) (65) x 2 x2( z (x) = 5 jxj x2n( 4 z 2 (x) =0 2 ; 2 ) 2 ; 2 ) Then both z and z 2 are solutions of (40), while only z is the solution of (39). 0

11 References [] H. Brezis and G. Stampacchia, Sur la regularite de la solution d'inequations elliptiques, Bull. Soc. Math. France 96, 53-0 (96). [2] A. Friedman, Variational Principles and Free-Boundary Problems, John Wiley & Sons, New York, 92. [3] S. Stojanovic, Perturbation formula and optimal control of regular free boundaries in obstacle problems, preprint. [4] S. Stojanovic, New proof of W 2;p -regularity for the obstacle problem, preprint. [5] G. M. Troianiello, Elliptic Dierential Equations and Obstacle Problems, Plenum Press, New York, 97.

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