Local mountain-pass for a class of elliptic problems in R N involving critical growth

Transcription

1 Nonlinear Analysis 46 (2001) Local mountain-pass for a class of elliptic problems in involving critical growth C.O. Alves a,joão Marcos do O b; ;1, M.A.S. Souto a;1 a Departamento de Matematica e Estatstica - Univ. Fed. Paraba, Campina Grande, Pb, Brazil b Departamento de Matematica, Univ. Fed. Paraba, João Pessoa, Pb, Brazil Received 8 April 1999; accepted 19 February 2000 Keywords: Elliptic problems; Critical growth; Variational methods; Mountain-pass theorem; Schrodinger equation; Standing wave solutions 1. Introduction In this paper we shall be concerned with the existence and the concentration behavior of positive bound-state solutions (solutions with bounded energy) for the problem 2 u + V (z)u = f(u)+u 2 1 ; in ; u C 2 ( ) H 1 ( );u(z) 0 for all z ; (P ) where 0; 2 =2N=(N 2); N 3; is the critical Sobolev exponent; V : R is a locally Holder continuous function satisfying and V (z) 0 for all z ; (V ) inf V V (V ) Corresponding author. address: jmbo@mat.ufpb.br (J.M. do O). 1 Partially supported by CNPq=Brazil X/01/$ - see front matter? 2001 Elsevier Science Ltd. All rights reserved. PII: S X(00)

2 496 C.O. Alves et al. / Nonlinear Analysis 46 (2001) for some bounded domain ; and the nonlinearity f : R + R is locally Lipschitz and such that (f 1 ) f(s)=o 1 (s) near the origin; (f 2 ) there are q 1 ;q 2 (1; 2 1); 0 such that f(s) f(s) s q1 for all s 0 and lim =0 s s q2 (when N =3; we need q 1 2; otherwise we require a suciently large ); (f 3 ) for some (2;q 2 + 1) we have 0 F(s) f(s)s for all s 0; where F(s)= s 0 f(t)dt; (f 4 ) the function f(s)=s is increasing for s 0. Since we are interested in positive solutions, we dene f(s) = 0 for s 0. Let us state our main result: Theorem 1. Suppose that the potential V satises (V ) (V ) and f satises (f 1 ) (f 4 ). Then there is an 0 0 such that problem (P ) possesses a positive bound state solution u ; for all 0 0. Moreover; u possesses at most one local (hence global) maximum z in ; which is inside ; such that lim V (z )=V 0 = inf V: 0 + Besides; there are Cand ; positive constants such that ( u (x) C exp x z ) for all x : We would like to remark that this kind of equation in (P ) arises from the problem of obtaining standing waves solutions of the nonlinear Schrodinger equation = 2 +(V (z)+e) 1 h( ) in (S where h(s) =f(s) +s 2 1. A standing wave solution to problem (S ) is one in the form (x; t) = exp( i 1 Et)u(x): In this case u is a solution of (P ). Some recent works have treated this problem in the subcritical case and we cite a couple of them. Floer and Weinstein [7] have studied the problem (S ) in the case N =1;h(s)=s 3 and bounded potentials with nondegenerate critical point. They show that for small, this problem has a solution which concentrates around each nondegenerate point (see also [18] for a related work). Roughly speaking, in this work, concentration behavior around the origin of a function means that it has the form (x) where is a C 2 function with exponential decay.

3 C.O. Alves et al. / Nonlinear Analysis 46 (2001) Under the potential condition inf V lim inf V (z); (1) z RN z Rabinowitz [14] has proved the existence of positive ground-state solutions (solutions with minimal energy) for (P ), in the case where h(s) behaves like s p ; 1 p 2 1, for small. The remaining concentration behavior result in this subcritical case was obtained by Wang [17]. Alves and Souto in [2] have established existence and concentration behavior of ground-state solutions when the nonlinearity has the critical form h(s)=s q + s 2 1 ; 1 q 2 1, under condition (1). In the very interesting article [5], del Pino and Felmer have obtained the complete treatment (existence and concentration behavior of solutions) with the potential under conditions (V ) and (V ). They have obtained bound-state solutions but not ground-state solutions, and this is reasonable, because some problems under condition (V ) do not admit any ground-state solution (see, for example, Theorem 4 of [2]). From this reason we cannot look for minimax critical points of the energy functional J : E R, J (u)= 1 ( 2 u 2 + V (z)u 2 )dz F(u + )dz 1 2 R 2 u 2 + dz N (where u + = max{u; 0}), dened on the Hilbert space { } E = u H 1 ( ): V (z)u 2 dz ; associated to (P ). A way to solve this problem is to modify the nonlinearity into one more convenient in order to apply the mountain-pass theorem. Namely, we will consider the following Caratheodory function: { (f(s)+s 2 1 )+ D f(s) if s 0; g(z; s)= 0 if s 0; where f(s)+s 2 1 if s a; f(s)= s if s a; k k ( 2) 1 1; a 0 is such that f(a) +a 2 1 = k 1 a; D = \ and A denotes the characteristic function of subset A of. Similar modied nonlinearity has been used by del Pino and Felmer in [5] to study the subcritical case. We have organized the paper as follows. In the second section, using the mountainpass theorem, we shall prove that the functional associated to the modied problem possesses a critical point. In the following section we shall prove Theorem 1 by a couple of lemmas which show that for small this critical point of the modied functional has a concentration behavior and it is also critical point of functional J.

4 498 C.O. Alves et al. / Nonlinear Analysis 46 (2001) The modied functional In this section we will consider the energy functional J : E given by J (u)= 1 2 u 2 G(z; u)dz; where u 2 = ( u 2 + V (z)u 2 )dz: Here G(z; s)= (F(s)+ 1 2 s 2 )+ D F(s) and F(s)= s f(t)dt. 0 Notice that, using (f 1 ) (f 4 ), it is easy to check that (g 1 ) g(z; s)=f(s)+s 2 1 = o(s), near the origin, uniformly in z ; (g 2 ) g(z; s) f(s)+s 2 1 for all s 0; z ; (g 3 )0 G(z; s) g(z; s)s; for all z ; s 0orz D and s a; and 0 2G(z; s) g(z; s)s 1 k V (z)s2 for all z D; s 0; (g 4 ) the function s 1 g(z; s) is increasing in s 0 for each z xed. Lemma 2. J has the mountain-pass geometry. Proof. Let be a nonzero function in C0 (), such that 0. Then for all t 0 J (t) t2 2 2 t2 2 2 dz: Of course, J (t) as t. As the usual way, from (g 2 ) J (u)= 1 2 u 2 +o( u ), near origin. This completes the proof. Now in view of Lemma 2, we can apply a version of the mountain-pass theorem without (P:S:) condition (see [11]) to obtain a sequence {u n } such that J (u n )=c + o n (1) and J (u n ) = o n (1); (2) where c is the minimax level of functional J given by 0 c= inf max J (g(t)); (3) g 0 t 1 where = {g C 0 ([0; 1]; E) : such that g(0) = 0 and J (g(1)) 0}. Throughout this work o n (1) denotes a sequence converging to zero as n. Remark 1. As in [5,6], we shall use the equivalent characterization of c more adequate to our purpose, given by c = inf v E\{0} max t 0 J (tv):

5 C.O. Alves et al. / Nonlinear Analysis 46 (2001) Furthermore, it is easy to check that for each non-negative v E {0} there is a unique t 0 = t 0 (v) such that J (t 0 v) = max J (tv): t 0 We denote by S the best Sobolev constant for the embedding D 1;2 ( ), L 2 ( ). Lemma 3. There is a v E\{0} such that max J (tv) 1 t 0 N SN=2 ; therefore we have this estimate for the minimax level (3) of J: c 1 N SN=2 : Proof. For each h 0, consider the function 2)=4 [N (N 2)h](N h(z)= (h + z 2 ) (N 2)=2 : We recall that h satises the problem u = u 2 1 ; ; u(z) 0; u 2 dz and h 2 2 dz = h dz = S N=2 (see [16]). Now, consider v h (z)=( h (z)= h L 2 ) where C0 (RN ); 0 (z) 1 and { 1; z B1 ; (z)= 0; z B 2 ; where B 1 B 2 are concentric balls of radius and 2, respectively. From condition (f 2 ) we have J (tv h ) t2 ( v h 2 B2 2 + V L ()v 2h)dz tq1+1 v q1+1 h dz t2 q 1 +1 B 2 2 : (4) Using the same arguments explored in [12], there exists h 0 such that { [ ] } t 2 max t 0 B 2 2 ( v h 2 + V L ()vh) 2 tq1+1 q 1 +1 vq1+1 h dz t2 2 1 N SN=2 : (5) Therefore, from (4) and (5) we have that max t 0 J (tv h) 1 N SN=2 and the proof of the lemma is complete.

6 500 C.O. Alves et al. / Nonlinear Analysis 46 (2001) Lemma 4. Every sequence {u n } satisfying (2) is bounded in E. Proof. To check this, observe that using (g 3 ) we have J (u n ) 1 ( 1 J (u n )= 2 1 ) u n [u n g(z; u n ) G(z; u n )] dz ( ) u n [u n g(z; u n ) G(z; u n )] dz D ( ( ) 1 2 ) u n 2 + G(z; u n )dz D 2 1 ( ) ( ) 2 u n 2 + V (z)un 2 dz 2k D ( ) [ ( 2 = u n R k N Applying (2) in inequality (6) we have the upper bound of u n. ) ] V (z)un 2 dz: (6) Lemma 5. There is a sequence {z n } un 2 dz : B R(z n) and R 0; 0 such that Proof. Suppose by contradiction that the lemma does not hold. Then by Lion s result (see [10] or [4]) it follows that u n q dz = o n (1); as n for all 2 q 2 and then F(u n )dz = u n f(u n )dz = o n (1): This implies that G(z; u n )dz 1 R 2 (u n + ) 2 dz + un 2 dz + o n (1) (7) N {u n a} 2k D {u n a} and u n g(z; u n )dz = n {u n a}(u + ) 2 dz + un 2 dz + o n (1): (8) k D {u n a} From equality (8) and J (u n ) u n = o n (1), we conclude that u n 2 un 2 dz + o n (1) = (u n + ) 2 dz: (9) k D {u n a} {u n a}

7 C.O. Alves et al. / Nonlinear Analysis 46 (2001) Let 0 be such that u n 2 un 2 dz : k D {u n a} It is easy to check that 0, otherwise we have u n 0 which contradicts c 0. From (9) (u n + ) 2 dz : {u n a} From inequality (7) and J (u n )=c + o n (1) we have Nc and hence 0. Now, using the denition of the constant S, we have u n 2 ( ) 2=2 un 2 dz S un 2 dz : k D {u n a} {u n a} Taking the limit in the above inequality, as n, we achieve that S 2=2 ; which, together with (10), implies c 1 N SN=2 ; which contradicts Lemma 3. Lemma 6. The sequence {z n } is bounded in. Proof. For each 0 consider a dierentiable function such that { 0 if z ; (z)= 1 if z 2 and (z) C 1, for all z. Use J (u n )( u n )=o n (1) to obtain ( 1 1 ) [ ( un 2 u n 2 + V (z) ) ] un 2 dz k R k N = u n u n dz [ + g(z; u n )u n ] k u2 n dz + o n (1): If is large enough, we have B (0). Furthermore, from (g 3 )weget ( 1 1 ) un 2 dz C k R u n 2 H + o n(1): (11) 1 N The inequality (11) completes the proof. (10)

8 502 C.O. Alves et al. / Nonlinear Analysis 46 (2001) Using standard arguments, upto subsequence, we may assume that there is u H such that u n *uin H; u n u in L q loc (RN ) and u n (x) u(x) a:e: in : From Lemmas 5 and 6, u is nontrivial and using the weak convergence of u n we can see that u is a critical point of J. Finally, from Remark 1, (g 3 ) and Fatou s lemma c J (u)=j (u) 1 2 J (u)u = 1 [ug(z; u) 2G(z; u)] dz 2 { } 1 lim inf [u n g(z; u n ) 2G(z; u n )] dz n 2 = lim inf {J (u n ) 12 } J (u n )u n = c n and then u is a solution with minimal energy J (u)=c. We have proved up this moment the following result: Proposition 7. For all 0; there is a positive critical point u E associated to the functional J (u)= 1 ( 2 u 2 + V (z)u 2 dz) G(z; u)dz 2 at the level c = inf J (tv): t 0 v E\{0} max 3. Proof of Theorem 1 In order to proof Theorem 1, let us x some notations. First we suppose, without loss of generality is smooth and 0. Furthermore, V (0) = V 0 = inf V: We will denote by I 0 : H 1 ( ) R the functional given by I 0 (u)= 1 ( u 2 + V 0 u 2 )dx [F(u)+ 12 ] 2 R (u+ ) 2 dx; N associated to the autonomous problem u + V 0 u = f(u)+ u 2 2 u; in : (12) It is known that under assumptions (f 1 ) (f 4 ), (12) possesses a ground-state solution! at the level c 0 = I 0 (!) = inf v H 1 \{0} max t 0 I 0(tv) (13)

9 (see [1] for instance). Furthermore, c 0 1 N SN=2 : C.O. Alves et al. / Nonlinear Analysis 46 (2001) Remark 2. The dependence of the mountain-pass level c 0 on the potential V 0 is continuous and increasing (see [14]). From a result due to Gidas et al. [8], any solution of problem (12) must be spherically symmetric about some point in 0 for all r 0, where r is the radial coordinate. Let I denote the energy functional I (u)= 1 ( u 2 + V (x)u 2 )dx G(x; u)dx; 2 dened in { } E = u H 1 ( ): V (x)u 2 dx ; associated to the problem u + V (x)u = g(x; u) in : (MP ) From Proposition 7, the family of nonnegative functions v (x)=u (z)=u (x); z = x is such that each v is a critical point of I at the level b = I (v ) = inf max I (tv): v E \{0} t 0 It is easy to check that b = N c. Furthermore, from Lemma 3, for each 0we have b 1 N SN=2 : (14) Moreover, Lemma 8. lim sup 0 b c 0, the mountain-pass minimax level of I 0. Proof. Fix! dened in (13) and consider! (x)= (x)!(x), where is the function dened in Lemma 3. Here we assume that B 1 = B (0); B 2 = B 2 (0). It is easy to see that!! in H 1 ( );I 0 (! ) I 0 (!), as 0 +, and the support of! is in = {x ; x }. By denition of b we have b max t 0 I (t! )=I (t! ) = t2 [! 2 + V (x)! 2 ]dx 2 [F(t! )+ t2 2!2 ] dx; (15)

10 504 C.O. Alves et al. / Nonlinear Analysis 46 (2001) for some t 0. It is easy to verify that t 1as 0 +. On the other hand, I (t! )=I 0 (t! )+ t2 (V (x) V 0 )! 2 dx: (16) 2 Since V (x) is bounded on the support of!, by the Lebesgue Dominated Convergence Theorem and (15) and (16), we conclude the proof. Notice that I (v ) c 0 + o (1), where o (1) goes to zero as 0. From (V ) and (g 2 ), we have I (u) I(u) = : 1 ( u 2 + u 2 )dx [F(u)+ 12 ] 2 R (u+ ) 2 dx: N Hence, b is bounded from below by c 0; the minimax level of functional I. Now, using similar arguments as of Lemma 5, we have the following result. Lemma 9. There are 0 0; a family {y } {0 0} such that v 2 dx ; for all 0 0 : B R(y ) and positive constants R; Lemma 10. y is bounded in. Moreover; dist(y ; ) R Proof. For 0, we dene K = {x : dist(x; ) }. Weset (x) =(x), where C ( ; [0; 1]) is such that { 1; x K ; (x)= 0; x and C 1. Taking v as test function, using property (g 3 ) and the fact that the support of does not intercept, we obtain ( 1 1 ) [ ( v 2 dx v 2 + V (x) ) ] v 2 dx k R k N [ = v v dx + v f(v ) ] R k v2 dx N v v dx: So we have ( 1 1 ) v 2 dx C 1 v 2 H 1: (17) k If for some sequence n 0 we have B R (y n ) {x ; n x K } = (18)

11 C.O. Alves et al. / Nonlinear Analysis 46 (2001) then ( 1 1 ) v 2 k n dx C 1 n v n 2 H 1; B R(y n ) which contradicts Lemma 9. Thus (18) does not hold, that is, for all there is an x such that x K and x y R. It is easy to verify that this implies dist(y ; ) R+. From this fact we conclude the proof. From Lemma 10 we can suppose that the family {y }, dened in Lemma 9, can be taken such that y for all (0; 0 ]. Indeed, if not, we replace y by 1 x, where x comes from Lemma 10, so that y x R. Observing that y (x =) R, we can replace R by 2R in Lemma 9. Let us consider the following subset of : E = {x : v (x) a and x }: Observe that, in fact, we want to show that E is empty if is small. Let F be the following translation of E : F = {x : v (x + y ) a and x + y }: It is easy to see that E = F ( A indicates the Lebesgue measure of the subset A). Lemma 11. The following limits hold: (i) lim 0 V (y )=V 0 ; (ii) lim 0 b = c 0 ; (iii) lim 0 E =0. Proof. If n 0 and y n =y n, are such that n y n x 0, we must prove that V (x 0 )=V 0. We already know that x 0, that is, V (x 0 ) V 0. Let us set v n (x)=v n (x);! n (x)= v n (x + y n );E n = E n and F n = F n. From Lemma 9 we have! n 2 dx 0; for all n; B R(0)! n + V ( n x + n y n )! n = g( n x + n y n ;! n ) in and! n H 1 = v n H 1 is bounded. Let! H 1 ( ) such that! n *!in H 1 ( ). We have! 0;! 0 and! + V (x 0 )! = (x)[f(!)+! 2 1 ]+(1 (x)) f(!)= g(x;!) in ; (19) where (x) = lim n ( n x + n y n ) almost everywhere in. It is easy to verify that if x 0 we have (x) = 1 for all x and therefore! is a solution of the problem! + V (x 0 )! = f(!)+! 2 1 in : (20) On the other hand, if x without loss of generality, we may suppose that the outer normal vector in x 0 is (1; 0;:::;0): Let P = {x : x 1 0}: Observe that on P; 1: In fact, for each x P; n x + n y n ; for all large n:

12 506 C.O. Alves et al. / Nonlinear Analysis 46 (2001) Then in both cases, g(x; s)=f(s)+s 2 1 for all x P: This implies that the functional energy Ĩ associated to problem (19) has the same minimax level c of the functional Ĩ x0 associated to problem (20). In eect, for all u H 1 ( ) we have Ĩ x0 (u) Ĩ(u) and then C x0 c (where C x0 is the minimax level of Ĩ x0 ): On the other hand Ĩ x0 (u)=ĩ(u); for all u with support inside P: From the autonomous case (see Remark 2) theory, c 0 Ĩ(!): It is easy to check, from Fatou s Lemma and Lemma 8, that c 0 Ĩ(!)= 1 [! g(x;!) 2 G(x;!)] dx 2 { } 1 lim inf [! n g( n x + n y n ;! n ) 2G( n x + n y n ;! n )] dx n 2 \F n lim inf n = lim inf n { } 1 [v n g( n x; v n ) 2G( n x; v n )] dx 2 \E n {I n (v n ) 12 } I n (v n )v n c 0 ; (21) where G denotes the primitive of g. Thus, (ii) follows from (21). Suppose that limit (i) does not hold, that is V (x 0 ) V 0 : It comes from Remark 2, c 0 c Ĩ(!)=c 0 which is a contradiction, then V (x 0 )=V 0 : To show the part (iii) we have from (21) that 1 lim [v n g( n x; v n ) 2G( n x; v n )] dx = c 0 : (22) n 2 \E n The same approach used in the whole instead of \E n in inequality (21) also shows that 1 lim [v n g( n x; v n ) 2G( n x; v n )] dx = c 0 : (23) n 2 Since (22) and (23) hold, we have 1 lim [v n g( n x; v n ) 2G( n x; v n )] dx =0: n 2 E n But, from denition of F we have [ [v n g( n x; v n ) 2G( n x; v n )] dx = E n k a2 2F(a) 2 a2 and this implies the limit (iii). 2 ] E n 0 From the proof of Lemma 11,! n converges in the weak sense to a! that is a solution of problem (12) satisfying I 0 (!)=c 0 ; that is,! is a ground state solution of

13 C.O. Alves et al. / Nonlinear Analysis 46 (2001) (12). From that proof we also have that b n c 0 and that g( n x + n y n ;s) converges uniformly over compacts to f(s)+s 2 1. Moreover, from (g 3 )! g(x + y ;! )dx! 2 dx F k F k! 2 L F (2 2)=2 = o 2 (1): (24) In the same way we have G(x + y ;! )dx = o (1): (25) F Also from (22) and the denition of g we have 2c 0 + o n (1) = [! n g( n x + n y n ;! n ) 2G( n x + n y n ;! n )] dx \F n = [! n f(! n ) 2F(! n )] dx + (1 22 ) \F! 2 n dx; n \F n which implies that dx dx; (26)! n 2 \F n! 2 and [! n f(! n ) 2F(! n )] dx \F n [!f(!) 2F(!)] dx: From limit (26) we have proved the following result. Lemma 12.! \F converges to! in L 2 ( ). In order to prove the concentration of the solution, we state the following two fundamental propositions. The rst one is an adequate version of a result due to Brezis and Kato [3] (see also [15] for the details). Proposition 13. Let v H0 1 (); RN open; satisfying v +(b (x) q(x))v = f(x; v) in ; where f: R R + is a Caratheodory function such that 0 f(x; s) Cf (sr + s); for all s 0; x ; b C(; [0; + )); q L N=2 (); 1 r (N +2)=(N 2): Then v L p () for all 2 p ; and there is a positive constant C p depending on p; q and Cf such that v L p () C p v H 1 (): Moreover; the dependence on q of C p can be given uniformly on Cauchy sequences q k in L N=2. The next proposition is a very particular version of Theorem 8:17 in [9].

14 508 C.O. Alves et al. / Nonlinear Analysis 46 (2001) Proposition 14. Suppose that t N; g L t=2 () and u H 1 () satises in the weak sense u g(x) in where is an open subset of : Then for any ball B 2R (y) ; sup u C( u + L 2 (B 2R(y)) + g L t=2 (B 2R(y))) B R(y) where Cdepends on N; t and R: From Lemma 12, the sequence! 2 2 n is a Cauchy sequence in L N=2. Using Proposition 13, with q(x) =! 2 2 n ( n x + n y n ) L N=2 ; b(x) =V ( n x + n y n ) and f(x; s)=g( n x + n y n ;s)! 2 1 n ( n x + n y n ), we have! n L t for all t 2 and! n L t C t! n H 1 (27) where C t does not depend on n: Still from Lemma 12, lim R x R! 2 n dx + x R! 2 n dx = 0 uniformly on n: (28) We will apply Proposition 14 to the following inequality:! n g n (x)=g( n x + n y n ;! n ) in : (29) Combining (27) with the fact that! n H 1 bounded, we have a t N such that g n L t C; for all n: Using Proposition 14 in inequality (29), for all y sup! n C(! n L 2 (B 2(y)) + g n Lt (B 2(y))); (30) B 1(y) which implies an uniform bound for! n L and consequently, we have an uniform bound for! L for 0 0 : Moreover, combining the limit (28) with inequality (30) we reach lim n(x) = 0 x uniformly on n (31) and we have for! (x)=v (x + y ) lim (x) = 0 x uniformly on (0; 0 ]: (32) From limit (32) there is a 0 such that! (x) a for all x for all n, that is! + V (x + y )! = f(! )+! 2 1 in x : On the other hand, if x ; g(x + y ;s)=f(s)+s 2 1 when B (0), then! + V (x + y )! = f(! )+! 2 1 in for all (0; 0 ]: Notice that, upthis moment, we have obtained the existence of solutions u for problem (P ): In order to prove the concentration behavior of these solutions, we shall prove:

15 C.O. Alves et al. / Nonlinear Analysis 46 (2001) Lemma 15. If 0 is suciently small! possesses at most one local (hence global) maximum x. Proof. If y 1 is a local maximum of! n we must have! n (y 1 ) q1 1 +! n (y 1 ) 2 2 : (33) From limit (32), it is sucient to consider the problem in a xed ball B R (0) of : Since! n L is uniformly bounded, elliptic regularity theory implies that! n converges in Cloc 2 (RN )to!: Let x n B R (0) a global maximum of! n. The translation of! n ;! n (x)=! n (x + x n ), attains its global maximum at the origin. Proceeding with! n as we have proceeded with! n, it is easy to see that! n converges to! in Cloc 2 (RN ). Now, by Lemma 4:2 in [13], if n is sucient large,! n possesses no critical points other than the origin. At this moment we can consider the sequence {y } in Lemma 11, as the sequence of global maximum points of v : For suciently small, u attains its global maximum at an unique z which must satisfy u (z )=v (y ); y = z : (34) Then from Lemma 11 lim V (z )=V 0 : 0 Finally! has an exponential decay: Lemma 16. There are C 0 and 0 such that! (x) Ce x for all x : Proof. From limit (31) and (f 1 ) there is a R 0 0 such that f(! (x)) +! (x) 2 1 V 0 2! (x) for all x R 0 : (35) Fix (x)=me x with 2 V 0 =2 and Me R0! (x) for all x = R 0 : It is easy to verify that 2 for all x 0: (36) Dene =! : Using (35) and (36) we have + V in x R 0 ; 0 on x = R 0 ; lim (x)=0: x The classical maximum principle implies that 0in x R 0 and we conclude! (x) Me x ; for all x R 0 and (0; 0 ]: The proof is complete.

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