ON THE ASYMPTOTIC BEHAVIOR OF THE PRINCIPAL EIGENVALUES OF SOME ELLIPTIC PROBLEMS

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1 ON THE ASYMPTOTIC BEHAVIOR OF THE PRINCIPAL EIGENVALUES OF SOME ELLIPTIC PROBLEMS TOMAS GODOY, JEAN-PIERRE GOSSE, AND SOFIA PACKA Abstract. This paper is concerned with nonselfadjoint elliptic problems involving inde nite weights and boundary conditions of the Dirichlet, Neumann or Robin type. We study the asymptotic behavior of the principal eigenvalues when the rst order term (drift term) becomes larger and larger. The basic results of [1] are extended to the present context. Moreover, answers are provided to some open problems raised in [1]. 1. Introduction This paper is mainly concerned with the study of the asymptotic behavior of the principal eigenvalues of the problem (1:1) L u := div (A (x) ru) + ha (x) ; rui + a 0 (x) u = m (x) u in ; u = 0 when the real parameter goes to in nity. Here is a bounded domain in R N ; A (x) is an uniformly elliptic matrix, a (x) is a divergence free vector eld, and m (x) is a (possibly inde nite) weight satisfying m + 6 0: Several works have been devoted to this type of question and we refer in particular to [1] for a recent survey as well as for interesting new results and applications. These applications concern for instance the behavior in nite time of the solution of a linear parabolic Cauchy problem with large advection, or the estimation of the speed of pulsating fronts in some nonlinear reaction di usion equation. See also [3] for another recent contribution with application to a competition model. The main result of [1], as far as the eigenvalue problem (1:1) is concerned, is the following. Assume the weight m (x) de nite, i.e. m (x) " > 0 in ; and let be the principal eigenvalue of (1:1). Then the limit of as! 1 always exists in R[f+1g, and this limit is nite if and only if the vector eld a (x) admits a rst integral w in H 1 0 () (i.e. w 2 H 1 0 () with w 6 0 and ha; rwi = 0 in ); moreover the limit of a ; when nite, is equal to the minimum of the Raleigh quotient of the associated selfadjoint problem (1:1) 0 over all such w 0 s; in addition this minimum for all : It is our purpose in this paper to extend this result of [1] to the case of an inde nite weight m (x) : In the process various complements will be obtained, which in particular answer three open problems raised in [1]. Our approach di ers from that in [1]; it is based on a minimax formula for the principal eigenvalue of (1:1). Date: December 3, Mathematics Subject Classi cation. 35J20, 35P15. Key words and phrases. Nonselfadjoint elliptic problems, principal eigenvalue, inde nite weight, minimax formula, asymptotic behavior, large drift. This work was carried out with the support of FNRS and SECyT. 1

2 2 TOMAS GODOY, JEAN-PIERRE GOSSE, AND SOFIA PACKA The rst di culty one faces when dealing with the inde nite case is that a principal eigenvalue may fail to exist (cf. e.g. [14], [4], [6], [9]). Denoting by the largest principal eigenvalue of (1:1) (provided a principal eigenvalue exists), we will show that if exists for one 0; then exists for all (cf. Proposition 3.2). We will also give examples showing that may fail to exist for all (cf. Example 3.7), or may exist for some s but not for others (cf. Examples 3.4 and 3.10). In the case where a (x) is a nontrivial gradient eld, we will see that always exists for large, with moreover! +1 as! +1 (cf. Proposition 3.6). Our main result concerning the function! in the general case, beside its continuity (cf. Proposition 4.1) and evenness (cf. Proposition 3.2), is that it is either constant on R or strictly increasing on [0; 1) (cf. Proposition 4.3). Note that the mere monotonicity of answers a question raised in Section 5 of [1]. The case constant can be characterized in term of the associated selfadjoint problem (1:1) 0 : is constant if and only if 0 exists and its associated eigenfunction is a rst integral (cf. Proposition 4.4). In the case non constant we have the following extension of the result of [1] described above: remains bounded as! +1 if and only if a (x) admits a rst integral w in H0 1 () with R mw2 > 0 (cf. Theorem 5.1). Moreover, when remains bounded with in addition either non constant, or a 0 0; or m 0; then the limit of is equal to the minimum of the Rayleigh quotient of (1:1) 0 over all rst integrals w 2 H0 1 () with R mw2 > 0 (cf. Theorem 5.4). When is constant with a and m changing sign, a rst integral w 2 H0 1 () with R mw2 > 0 may not exist (cf. Example 5.3). Another di culty which occurs in the inde nite case is that when m (x) changes sign, there may exist two principal eigenvalues. Distinguishing between the two is easy when a 0 0 (since then one is > 0 and the other is < 0), but requires more care when a 0 is not 0: We will use for this purpose the -function associated to (1:1), which we denote by L;m or brie y by : () is de ned for 2 R as the (unique) principal eigenvalue of (1:2) L u m (x) u = u in ; u = 0 The principal eigenvalues of (1:1) are thus exactly the zeros of the function : It turns out that the largest principal eigenvalue of (1:1) can be distinguished from the smallest by the property that the derivative of at is 0 (cf. Lemma 3.1). This observation of a technical nature will be used in some of our arguments, and more generally the function will play an important role in many of our proofs. As brie y indicated earlier, our approach is rather di erent from that in [1]. A basic ingredient for us is a minimax formula of variational type for the principal eigenvalues of a problem like to (1:1). Such a formula, for m 1 or m " > 0, goes back to [5] and [12], where it was derived by arguments involving respectively semigroups theory or stochastic di erential equations. The form we use in our present inde nite problems is that obtained recently in [9], whose derivation relies on the study of some degenerate elliptic equations in the context of weighted Sobolev spaces (cf. Proposition 2.2 below). In fact, we will also need here a variant of the formula from [9], whose derivation from the formula in [9] involves Ekeland variational principle (cf. Proposition 2.3 below). Note that the fact that Holland s minimax formula can be used to study the asymptotic behavior of provides an answer to another question raised in Section 5 of [1].

3 ASYMPTOTIC BEHAVIOR OF PRINCIPAL EIGENVALUES 3 Finally it turns out that our approach can be easily adapted to the case of the Neumann or Robin boundary conditions (cf. Section 6). Dealing with the Robin boundary conditions provides an answer to another question raised in Section 5 of [1]. The plan of our paper is the following. Section 2, which is independent from the rest of the paper, is devoted to Holland s minimax formula. In Section 3, we mainly concentrate on the question of the existence of and in Section 4 on the properties of the function! : Section 5 deals with the asymptotic behavior of as! 1. Finally, in Section 6, we consider the Neumann-Robin boundary conditions. 2. Minimax formula Let us start by stating the assumptions to be imposed on the operator L and the weight m in (1:1) throughout the paper. is a bounded domain in R N of class C 2 and the coe cients of L satisfy: A is a symmetric uniformly positive de nite N N matrix, with A 2 C 0;1 ; a 2 C 0;1 ; R N and a 0 2 L 1 () : The weight m belongs to L 1 () and we will always assume m (the case m 0 can of course be treated by changing m into m). Since the parameter plays no speci c role in this section, we will assume it equal to 1 and write L and (1:1) instead of L 1 and (1:1) 1 : Note that we do not require in this section a (x) to be divergence free. Our purpose in this section is to present a variant of the minimax formula derived in [9] for the principal eigenvalues of (1:1) : By a principal eigenvalue we mean 2 R such that (1:1) admits a solution u 6 0 with u 0: The following result is well known (cf. e.g. [11], [4], [14], [6], [9]). Proposition 2.1. Assume rst a 0 0: If m 0 then (1:1) admits exactly one principal eigenvalue which is >0; if m changes sign, the (1:1) admits exactly two principal eigenvalues, one is > 0 and the other is < 0: Without the assumption a 0 0; (1:1) may have zero or one principal eigenvalues if m 0; and zero, one or two principal eigenvalues if m changes sign. All these eigenvalues are simple, and the associated positive eigenfunctions belong to W 2;p ()\D () for any 1 p < 1 (where D () is de ned below). We now recall the minimax formula from [9]. Let us de ne the weighted Sobolev space H 1 ; d 2 := u 2 Hloc 1 () : d 2 u 2 + jruj 2 < 1 where d (x) = dist ; as well as the set of functions comparable to d (x): D () := fu :! R measurable: c 1 d u c 2 d a. e. in ; for some positive constants c i = c i (u)g: Proposition 2.2. (cf. [9]) Assume the existence of a principal eigenvalue for (1:1) and let be the largest of these eigenvalues. Then (u) (2.1) = inf u2u D inf Q u (v) v2h 1 (;d 2 )

4 4 TOMAS GODOY, JEAN-PIERRE GOSSE, AND SOFIA PACKA where U D := u 2 H0 1 () \ D () : mu 2 = 1 ; (u) := haru; rui + ha; rui u + a 0 u 2 ; Q u (v) := u 2 (harv; rvi ha; rvi) : For any u 2 U D the in mum over v in (2:1) is achieved; moreover, when a 0 0 or when m (x) " > 0; the in mum over u in (2:1) is also achieved. In our study of the asymptotic behavior of the principal eigenvalue of (1:1), we will need a formula like (2.1) where u varies in the whole H 1 0 () instead of H 1 0 () \ D () : Proposition 2.3. Assume the existence of a principal eigenvalue for (1:1) and let be the largest of these eigenvalues. Then (u) (2.2) = inf u2u where U := u 2 H0 1 () : E u := v 2 H 1 () : inf Q u (v) v2e u mu 2 = 1 ; u 2 jrvj 2 < 1 : When a 0 0 or when m (x) " > 0; the in mum over u in (2.2) is achieved at some u 2 U D : The rest of this section is devoted to the proof of Proposition 2.3. A direct argument by approximation using Proposition 2.2 does not seem to work due to the lack of control on v in (2.2) when u lies only in H0 1 () : The proof below uses Proposition 2.2 as well as Ekeland variational principle. Let us start with some preliminaries concerning the space E u and the quadratic R expression Q u (v) : Fix u 2 H0 1 () with u 6 0: Then p u (v) := u2 jrvj is a seminorm on E u ; which becomes a norm on the quotient space G u of E u by F u := fv 2 E u : p u (v) = 0g : Call G u the completion of G u with respect to that norm and denote by p u the extended norm on G u : We will mainly deal below with G u ; which can be seen as the set of elements of E u de ned up to elements of F u : Our functional Q u (v) is made of a quadratic part R u2 harv; rvi and a linear part R u2 ha; rvi : It is clear that Q u (v) is well de ned for v 2 G u and extends by continuity into a functional Q u on G u : Moreover Q u is a C 1 functional on G u ; and its derivative at v 2 G u in the direction of h 2 G u is given by T u (v) h; where T u (v) h for v; h 2 G u denotes the extension by continuity of T u (v) h := u 2 (2 harv; rhi ha; rhi) where in this latter formula v; h 2 G u :

5 ASYMPTOTIC BEHAVIOR OF PRINCIPAL EIGENVALUES 5 Note the coercivity relation (2.3) Q u (v) c 1 p u (v) 2 c 2 8v 2 G u for some constants c 1 ; c 2 with c 1 > 0: It is this relation which will provide some control on v: One also clearly has (2.4) inf v2g u Q u (v) = inf v2e u Q u (v) : Moreover, if u 2 H 1 0 () \ D () ; then (2.5) inf Q u (v) = inf Q u (v) : v2e u v2h 1 (;d 2 ) Indeed, the inequality in (2.5) follows from the inclusion E u H 1 ; d 2 ; while the inequality follows from the density of C 1 (and so of E u ) in H 1 ; d 2 (cf. [15]). We are now ready to start the Proof of Proposition 2.3. We will rst prove that inequality holds in (2.2), i.e. (2.6) (u) inf v2e u Q u (v) for all u 2 U: Part of the argument will follow [9]. Let u be a positive eigenfunction associated to : Clearly it su ces to prove (2.6) for u 0: Take such a u and x " > 0: Multiplying both sides of the equation satis ed by u by u 2 = (u + ") and using the divergence theorem, one gets mu u 2 = (u (2.7) + ") Aru ; ru 2 = (u + ") h i = haru ; ru i = (u + ") 2 + ha; ru i = (u + ") u 2 + a 0 u u 2 = (u + ") : The use of the divergence theorem can be justi ed by observing that u 2 = (u + ") 2 W 1;s 0 () for some s > 1 and that div (Aru ) 2 L s0 () with s 0 the Hölder conjugate of s; it then su ces to approximate u 2 = (u + ") by Cc 1 () functions. Now, a simple calculation based on the idea of completing a square gives that for any x 2 and 2 R N ; h A (x) ; i ha (x) ; i = min h; i + 1 a (x) + ; A 1 (x) (a (x) + ) 2R N 4 (cf. (3.11) in [8]). We apply the inequality provided by the above relation with = ru = (u + ") to derive from (2.7) that for any measurable function :! R N with R u2 jj 2 < 1; one has mu u 2 = (u + ") Aru 2 (2.8) u 2 ; ru = (u + ") 1 a + ; A 1 (a + ) u 2 + a 0 u u 2 = (u + ") : 4 In order to select a suitable in (2.8) we recall the following consequence of Ekeland variational principle (cf. e.g. [18]): if a C 1 functional on a Banach space

6 6 TOMAS GODOY, JEAN-PIERRE GOSSE, AND SOFIA PACKA is bounded from below, then there exists a minimizing sequence satisfying the (PS) condition; moreover the elements of this sequence can be taken in any pregiven dense subset of the Banach space. Applying this result to our functional Q u on G u ; with G u as a dense subset, we obtain a sequence v k 2 E u such that for k! 1; (2.9) Q u (v k )! inf v2e u Q u (v) ; (2.10) u 2 h2arv k a; rvi = o (p u (v)) for v 2 E u where the o (p u (v)) in (2.10) is uniform with respect to v 2 E u as k! 1: We then take = a + 2A (rv k + ru= (u + ")) in (2.8), which gives, after some simple calculations, mu u 2 = (u + ") 2 haru; ru i u (1 u= (u + ")) = (u + ") + u 2 h2arv k a; r log (u + ")i haru; rui u 2 = (u + ") 2 + u 2 h2arv k a; r log (u + ")i + u 2 ha; rui = (u + ") + u 2 harv k ; rv k i + a 0 u u 2 = (u + ") : Since log (u + ") and log (u + ") belong to E u ; we deduce from (2.10) that when k! 1; the third integral of the left-hand side and the second integral of the Rright -hand side are o (1) as k! 1; moreover one also derives from (2.10) that u2 harv k ; rv k i di ers from Q u (v k ) by a o (p u (v k )) as k! 1: It thus follows that mu u 2 = (u + ") 2 haru; ru i u (1 u= (u + ")) = (u + ") haru; rui + ha; rui u 2 = (u + ") + a 0 u u 2 = (u + ") Q u (v k ) + o (1) + o (p u (v k )) : Since by (2.3), v k remains bounded in G u ; one has that o (p u (v k ))! 0 as k! 1: Going to the limit as k! 1 in the above inequality thus leads to mu u 2 = (u + ") 2 haru; ru i u (1 u= (u + ")) = (u + ") haru; rui + ha; rui u 2 = (u + ") + a 0 u u 2 = (u + ") inf Q u (v) : v2e u Finally, letting "! 0 and using dominated convergence, we get (2.6). It remains to prove that inequality holds in (2.2). Let " > 0: By Proposition 2.2 there exists u 2 U D such that + " (u) inf Q u (v) : v2h 1 (;d 2 )

7 ASYMPTOTIC BEHAVIOR OF PRINCIPAL EIGENVALUES 7 Since (2.5) holds for such a u (in fact only the trivial inequality in (2.5) is used here) and since U D U; we deduce that + " inf (u) u2u inf Q u (v) : v2e u Letting "! 0 leads to the conclusion. Finally, the result relative to the achievement of the in mum over u in (2.2) follows from the corresponding result in (2.1) using (2.5). Remark 2.4. In the applications of the minimax formula treated in this paper, any of the two versions above can be used, except in the proof of Proposition 4.4 (resp. Theorem 5.1) where the old (resp. new) version is needed. More comments on the need of this new version are given after the statement of Theorem Existence of The assumptions in this section as well as in Sections 4 and 5 are those stated at the beginning of Section 2, with in addition (3.1) div a = 0 in : As observed in Proposition 2.1, problem (1:1) may have zero, one or two principal eigenvalues. In case of existence, we denote by the largest principal eigenvalue. Our purpose in this section is mainly to investigate various situations of existence or nonexistence of : Existence for any certainly holds when a 0 (x) 0 (by Proposition 2.1), or when m (x) " > 0 (by writing L u = mu as L u + lmu = ( + l) mu; which allows to reduce for l large to the preceding case of a nonnegative zero order coe cient). We will use repeatedly in this paper the -function () de ned in (1:2). Clearly is a principal eigenvalue for (1:1) if and only if vanishes at : The following lemma (which does not depend on assumption (3.1)) collects some properties of this function. References include [13], [10], [2], [16]; the proof of most of the properties below can for instance be adapted from the proof of Lemma 2.5 in [8]. Lemma 3.1. (i) If a 0 0; then (0) > 0: (ii) If m + 6 0; then ()! 1 as! 1; if m 6 0; then ()! 1 as! 1: (iii) The function! () is real analytic; moreover a solution u of (1:2) can be chosen so that! u 2 C0 1 is real analytic. (iv) The function! () is concave. (v) If m 0; then! () is strictly decreasing. Proposition 2.3 can be used in (1:2) to get the following expression for () : (3.2) () = inf (u) u mu 2 inf v Q u (v)

8 8 TOMAS GODOY, JEAN-PIERRE GOSSE, AND SOFIA PACKA where u varies in H0 1 () with R u2 = 1; v varies in E u ; and where (u) and Q u (v) denote (u) := haru; rui + ha; rui u + a0 u 2 ; Q u (v) = u 2 (harv; rvi ha; rvi) : Note that as this stage, Proposition 2.2 could be used as well (see Remark 2.4). A simple calculation based on the divergence theorem and hypothesis (3.1) shows that (3.3) (u) = 0 (u) 8u 2 H 1 0 () ; 8 2 R; moreover, replacing v by v in Q u (v) gives (3.4) inf v Q u (v) = 2 inf v Q1 u (v) 8 2 R; consequently, (3.5) () = inf u 0 (u) mu 2 2 inf v Q1 u (v) where u and v vary as above. These formulas (3.3), (3.4), (3.5) will be used repeatedly. We are now ready to start the study of the existence of : Proposition 3.2. (i) exists if and only if exists, and then = : (ii) If exists for some 0; then exists for all and : Proof. (i) By (3.5), () = () ; and the conclusion follows. (ii) We know that is the largest zero of : Using (3.5) and the fact that inf v Q 1 u (v) 0; one has that : Since by Lemma 3.1, ()! 1 as! +1; one deduces that has a zero which is : The conclusion follows. Corollary 3.3. If 0 exists, then exists for all ; and 0 for all. As will be seen in Example 3.10, the existence of for all 6= 0 does not imply the existence of 0 : Because of (i) from Proposition 3.2, we will from now on mostly restrict ourselves to the study of for 0: The rest of this section is mainly devoted to various examples related to the existence or nonexistence of : Our starting point will be the observation that if m changes sign, or if m 0 but vanishes on a ball B ; then, for l su ciently large, the problem (3.6) u lu = mu in ; u = 0 has no principal eigenvalue (cf. e.g. [9]). Our rst example shows that may exist for some but not for others. Example 3.4. Take m such that (3.6) has no principal eigenvalues for l large. Consider the problem (3.7) u + ha; rui lu = mu in ; u = 0

9 ASYMPTOTIC BEHAVIOR OF PRINCIPAL EIGENVALUES 9 with a (x) = a a nonzero constant vector eld. Writing u = e ha;xi v; one easily sees that (3.7) is equivalent to (3.8) v + ( 2) ha; rvi + ( 1) jaj 2 l v = mv in ; v = 0 For = 2 we deduce from (3.6) that (3.8) (and so (3.7)) has no principal eigenvalue for l su ciently large. Fixing such a l; we then see that for su ciently large, the zero order coe cient of (3.8) is nonnegative, and consequently (3.8) (and so (3.7)) has a principal eigenvalue. Remark 3.5. By a change of unknown function as in the above example, one can see that if A (x) I and a (x) = a is a nonzero constant vector eld, then (1:1) always has a principal eigenvalue for su ciently large. This is in fact a particular case of the proposition below, which concerns gradient elds. Note that the proof of this proposition relies on some unique continuation property. Proposition 3.6. If a (x) 6 0 satis es (3.1) and is of the form a = 2Arg for some g 2 C 1;1 () ; then exists for su ciently large, and moreover! +1 as! 1: that Proof. It follows from (3.5), after completing the square in Q 1 u (v) = u 2 ha (rv) 2rg; rvi ; (3.9) () = inf u 0 (u) + 2 harg; rgi u 2 mu 2 : This latter expression is the -function associated to the selfadjoint operator L 0 u+ 2 harg; rgi u and the weight m: By Proposition 3.2,! () is nondecreasing on 0: We look at (0) : If (0) goes to +1 as! 1; then Lemma 3.1 implies that has a zero for large, and the existence of follows. If on the contrary (0) remains bounded as! 1; then we consider (3.10) L 0 u + 2 harg; rgi u = () u where u is an eigenfunction associated to (0) and normalized by ku k L 2 = 1: Taking u as test function in the weak form of (3.10), one sees that u remains bounded in H0 1 () : Going to a weakly convergent subsequence and dividing by 2 ; one deduces the existence of u 2 H0 1 () with kuk L 2 = 1 and (3.11) harg; rgi u 2 = 0: But since by (3.1), div (Arg) = 0 in ; one has that fx 2 : rg = 0g has measure zero (cf. [7]). This contradicts (3.11). We have thus proved the existence of for large. We now prove that! +1 as! 1: Since! is nondecreasing, we can assume by contradiction that remains bounded. We go back to (3.9), however now with = ; and obtain L o u + 2 harg; rgi u = mu

10 10 TOMAS GODOY, JEAN-PIERRE GOSSE, AND SOFIA PACKA where here u is an eigenfunction associated to and normalized by ku k L 2 = 1: Arguing as above, one reaches again (3.11) for some u with kuk L 2 = 1; and the same contradiction follows. In opposition with the previous situation, in the following example fails to exist for all : Example 3.7. Take the unit disk in R 2 ; m (x) as in (3.6) with in addition m (x) radial, and x l such that (3.6) has no principal eigenvalues. Consider the - function l;m () corresponding to (3.6), which we abbreviate into () ; and let w be the positive normalized (by kw k L 2 = 1) eigenfunction associated to () : (3.12) w lw mw = () w in ; w = 0 Simplicity implies that w is radial; moreover the nonexistence of a principal eigenvalue for (3.6) implies () < 0 for all ; nally, multiplying (3.12) by w and integrating give (3.13) jrw j 2 lw 2 mw 2 = () < 0: We now take the vector eld a (x) = ( x 2 ; x 1 ) : Clearly (3.1) holds, and since w is radial, we have ha; rw i = 0 in : Thus w 2 H0 1 () is a rst integral for a; and so, by Lemma 3.8 below, (3.14) inf v2e w Q 1 w (v) = 0 where the matrix A (x) in Q 1 w is in the present example the identity matrix. Consider now the function () associated to the problem (3.15) u + ha; rui lu = mu in ; u = 0 Taking u = w in formula (3.5), using (3.14) and (3.13), one has () jrw j 2 lw 2 mw 2 2 inf v Q1 w (v) < R; 8 2 R. This implies that for any ; (3.15) has no principal eigenvalue. Lemma 3.8. Let w 2 H 1 0 () be a rst integral for a: Then inf Q 1 w (v) = 0: v2e w Proof. Using (3.1) and the fact that w is a rst integral, one obtains w 2 ha; rvi = 0 for any v 2 E w : The derivation of this equality involves truncating w and using the divergence theorem, which can be justi ed by standard approximation arguments. It follows that inf Q 1 w (v) = inf w 2 harv; rvi = 0: v2e w v2e w

11 ASYMPTOTIC BEHAVIOR OF PRINCIPAL EIGENVALUES 11 We conclude this section with the study of the following question: when exists for some but not for others, it seems natural to look at 0 := inf f 0 : existsg and to ask wether 0 exists or not. The following two examples show that no general answer can be given. We will use below the easily veri ed fact that 0 exists if and only if remains bounded when! 0 ; > 0 : Example 3.9. If m changes sign, then 0 exists. Indeed, picking u 2 H0 1 () such that R mu2 > 0; one deduces from (3.5) that for 2 ] 0 ; 0 + 1[ ; () c 1 for a constant c 1 > 0 and for all > some 1 > 0; with c 1 and 1 independent of : Picking u 2 H0 1 () such that R mu2 < 0 one similarly deduces that () c 2 for some c 2 > 0 and all < some 2 < 0; This implies that remains between 2 and 1 when! 0 ; > 0 ; and the conclusion follows. Example If m 0 vanishes on some ball B ; then 0 may not exist. To build an example, we start from (3.6) with l = l 0 ; where l 0 is chosen such that (3.6) with l = l 0 has no principal eigenvalue but (3.6) with any l < l 0 has a principal eigenvalue. The existence of such a l 0 easily follows by looking at the shape of the corresponding function () (cf. Lemma 3.1). We then consider (3.16) u + ha; rui l 0 u = mu in ; u = 0 where a is a nonzero constant vector eld. Writing u = e ha;xi v; one easily veri es that (3.16) is equivalent to v + ( 2) ha; rvi l jaj 2 jaj 2 v = mv in ; v = 0 For = =2; this latter problem becomes a problem of the form (3.6): v l jaj2 v = mv in ; v = 0 We can then conclude that (3.16) has a principal eigenvalue for > 0 but does not have a principal eigenvalue for = 0: 4. Function We will limit ourselves to the study of the function! on 0; which, by Proposition 3.2, is no real restriction. The domain A R + of this function is thus either empty or of the form [r; +1[ or ]r; +1[ for some r 0: We start with a continuity result which, for m 1; is a particular case of Proposition 5.1 of [2]. Proposition 4.1. Suppose A non empty. Then is continuous in A. Proof. We rst show that if 0 exists for some 0 0; then is continuous from the right at 0 : Let " > 0 and consider 0 : By Proposition 2.3, there

12 12 TOMAS GODOY, JEAN-PIERRE GOSSE, AND SOFIA PACKA exists u = u ("; 0 ) 2 H 1 0 () with R mu2 = 1 such that 0 + " 0 (u) inf v Q0 u (v) = (u) inf v Q u (v) inf v inf v Q1 u (v) ; Q1 u (v) where we have used (3.3), (3.4), and where the in ma are taken for v varying in E u : So " inf v Q1 u (v) ; and the continuity from the right follows. We will now show that if exists in a neighborhood of some 0 ; then is continuous from the left at 0 : Consider a sequence k! 0 ; with k < 0 : By monotonicity, the limit of k exists and is 0 : Continuity from the left will follow if we show that for any given such that k for all k; one has 0 : Fix such a and call u k the positive eigenfunction associated to k ; normalized by ku k k C 1 0 () = 1: Since k and k is the largest principal eigenvalue of (1:1) k ; one has k 0: Moreover by Lemma 4.2 below, k remains bounded as k! 1; and consequently, taking l su ciently large in the equation satis ed by u k : (L 0 + l) u k = mu k + k u k k ha; ru k i + lu k ; one deduces that u k remains bounded in W 2;p () for any nite p: It follows that for a subsequence, k! 0; u k! u in C 1 0 with kuk C 1 0 () = 1 and u 0; with moreover, from the equation above, L 0 u = mu + u: This implies that = 0 ; and consequently 0 0: To go on we will now use repeatedly the properties of the function 0 () as described in Lemma 3.1. If m 0; we immediately deduce from 0 0 that 0 ; and we are nished. If m changes sign and (1:1) 0 has two principal eigenvalues, then we observe that the preceding argument can be repeated for any 0 and gives 0 ( 0 ) 0; this now implies 0 : Finally if m changes sign and (1:1) 0 has only one principal eigenvalue 0 ; then we argue di erently: since k 0 implies k 0 (cf. the proof of Proposition 3.2), one necessarily has in this situation k = 0 for all k; and the conclusion a0 follows. Lemma 4.2. () remains bounded when and remain bounded. Proof. For any u 2 H0 1 () with R mu2 = 1; one has, by (3.5), () 0 (u) mu 2 2 inf v Q1 u (v) const (u) when and remain bounded. On the other hand, for any u as above, 0 (u) mu 2 2 inf v Q1 u (v) 0 (u) mu 2 L0;m () and consequently () L0;m () : The conclusion follows.

13 ASYMPTOTIC BEHAVIOR OF PRINCIPAL EIGENVALUES 13 We now turn to the study of the strict monotonicity of the function! on A R + : Proposition 4.3. Suppose A nonempty. Then either is strictly increasing on A; or is de ned and constant for all 0: Proposition 4.3 follows from Proposition 4.4. below, which in particular characterizes the constant case in terms of the associated selfadjoint problem (1:1) 0 : is constant if and only if 0 exists and its eigenfunction is a rst integral. It will convenient from now on to call I 0 the set of rst integrals of a in H 1 0 () ; i.e. I 0 := w 2 H 1 0 () : w 6 0 and ha; rwi = 0 a.e. in : Proposition 4.4. (i) If = for some 0 < ; then 0 exists and is equal to = ; moreover the eigenfunction u associated to 0 belongs to I 0 : (ii) Conversely if 0 exists and if its eigenfunction belongs to I 0 ; then exists and = 0 for all 0: The rest of this section is mostly devoted to the proof of Proposition 4.4. Proof of Proposition 4.4. We rst deal with (i) in the particular case where a 0 0: Let 0 < and assume = : Our starting point will be the minimax formula (2.1) for ; where, since a 0 0; the in mum over u 2 U D is achieved at some eu 2 H0 1 () \ D () with R meu2 = 1: Using (3.3) and (3.4), formula (2.1) for gives (eu ) inf v Q eu (v) = (eu ) inf v Q eu (v) 2 2 inf v Q1 eu (v) = 2 2 inf v Q1 eu (v) where the in ma are taken for v varying in H 1 0; the above implies (4.1) inf v Q1 eu (v) = 0: ; d 2 : Since = and inf v Q1 eu (v) We now apply Lemma 3.2 from [9]: the in mum in (4.1) is achieved at some W eu 2 H 1 ; d 2 ; which satis es (4.2) 2ArWeu a; r' = 0 8' 2 H 1 ; d 2 ; eu 2 R and the value of this in mum is eu2 ArWeu ; rw eu : Consequently, by (4.1), this last integral has value zero, and so W eu is constant on : Replacing in (4.2) gives (4.3) eu 2 ha; r'i = 0 8' 2 H 1 ; d 2 : We now apply another result from [9]: denoting by u the positive eigenfunction q of (1:1) associated to ; one can write eu as u G for some function G 2

14 14 TOMAS GODOY, JEAN-PIERRE GOSSE, AND SOFIA PACKA ; d 2 \ L 1 () which is some " > 0 and which satis es (4.4) u 2 ArG + ag ; r' = 0 8' 2 H 1 ; d 2 H 1 (cf. Lemma 3.3 and the second part of the proof of Theorem 3.1 in [9]). Using (4.3) in (4.4) yields u 2 ArG ; r' = 0 8' 2 H 1 ; d 2 which implies, taking ' = G ; that G is a constant on : Consequently (4.3) becomes u 2 ha; r'i = 0 8' 2 H 1 ; d 2 : Applying the divergence theorem as well as (3.1) to this last relation gives a; ru ' = 0 8' 2 H 1 ; d 2 u D E and so a; ru = 0 in ; i.e. u belongs to I 0: The equation in (1:1) thus becomes L 0 u = mu ; which shows that is a principal eigenvalue for (1:1) 0 ; with u as associated positive eigenfunction. Since 0 and 0 is the largest principal eigenvalue of (1:1) 0 ; we conclude that = 0 : The proof of (i) in the case a 0 0 is thus completed. We now turn to the proof of (i) in the general case where a 0 is not necessarily 0: Let as before 0 < with = and call this common value = : Suppose rst > 0 and take l su ciently large so that the zero order coe cient of L +l is 0 and m+l is 0: Denoting as before by u the positive eigenfunction of (1:1) associated to ; one has L + l u = (m + l) u ; and it follows from Proposition 2.1 that is the largest principal eigenvalue of L +l for the weight m+l: Of course a similar conclusion holds for L + l u = (m + l) u : is the largest principal eigenvalue for L + l for the weight m + l. By the part of (i) which is already proved, we thus conclude that L 0 + l with the weight m+l admits as the largest principal eigenvalue and that the corresponding positive eigenfunction u belongs to I 0 : Clearly is then a principal eigenvalue for (1:1) 0 with u as positive eigenfuinction. Moreover = 0 ; and we conclude that = 0 : The proof of (i) is thus completed in the case under consideration. Suppose now < 0: The preceding argument can be easily adapted by writing L u = mu as L l u = ( m + l) u and by taking l su ciently large. Suppose nally = 0: Again the preceding argument can be easily adapted by writing L u = 0 as (L + l) u = lu and by taking l su ciently large. We now turn to the proof of (ii). By assumption 0 exists and its positive L 2 normalized eigenfunction u satis es ha; ru i = 0: One thus has L u = L 0 u = 0 mu ; which shows that for any 0; 0 is a principal eigenvalue for (1:1) : The proof of (ii) will be completed if we show that 0 is the largest principal eigenvalue of (1:1) :

15 ASYMPTOTIC BEHAVIOR OF PRINCIPAL EIGENVALUES 15 This is easily veri ed in the cases a 0 0 or m 0; by using Proposition 2.1. Indeed, if a 0 0; then 0 is > 0 and consequently is the largest principal eigenvalue of (1:1) ; and if m 0; then (1:1) has only one principal eigenvalue. Our purpose in the general case is to prove that (4.5) 0 ( 0 ) 0: The properties of the function () (cf. Lemma 3.1) then clearly imply the desired conclusion that 0 is the largest principal eigenvalue of (1:1). To prove (4.5) we apply Lemma 4.5 below with = 0 : Here u ; = u ; moreover, since u is a rst integral for a; one has L u = L u and so u ; = u : It follows that 0 ( 0 ) = R m (u ) 2 R (u ) 2 ; which in particular implies that 0 ( 0 ) does not depend on : Since 0 is the largest principal eigenvalue of (1:1) 0 ; one has 0 0 ( 0 ) 0; and consequently 0 ( 0 ) 0. The following lemma was used in the proof above, and will be used again later. Consider (1:2) and let u ; be a positive eigenfunction associated to () : L u ; mu ; = () u ; : We will also consider (1:2) ; () and u ; : Recall that by Proposition 3.2, () = () Note in connection with the lemma below that by (3.1), the adjoint of L (with zero Dirichlet boundary condition) is L (with zero Dirichlet boundary condition). Lemma 4.5. Let be a principal eigenvalue of (1:1) ; i.e. 0 R = R mu ; u ; u ; u : ; = 0: Then m and let v be an asso- Proof. Call () the -function associated to L ciated positive eigenfunction: L m v mv = () v : Clearly () = + so that (0) = 0 and 0 (0) = 0 ; moreover, up to a multiple, v 0 = u ; : We recall that v can be chosen so that () 2 R and v 2 C0 1 are analytic functions of ; moreover the above equation for v can be rewritten, for l su ciently large, as v = T [mv + () v + lv ] where T is the continuous operator T := L m + l 1 : L 1 ()! C0 1 : Taking the derivative with respect to and making = 0 = T (mv 0 ) + (0) T (v 0 ) + lt : =0 Now a simple calculation using (3.1) and the divergence theorem shows that R R ut v = (T u) v for u; v 2 L 1 () : Multiplying (4.6) by u ; ; integrating and using

16 16 TOMAS GODOY, JEAN-PIERRE GOSSE, AND SOFIA PACKA the fact that T u ; = 1 l u ; ; one obtains mv 0 u ; + 0 (0) u ; v 0 = 0: This is the desired relation since v 0 = u ; : Here is a consequence of Lemma 4.5 which has independent interest. recall that we systematically assume m + 6 0: Let us Corollary 4.6. Consider the selfadjoint problem (1:1) 0 and assume it admits a principal eigenvalue. Call 0 its largest principal eigenvalue and let u be an associated positive eigenfunction. Then R m (u ) 2 0: More precisely R m (u ) 2 = 0 in the case when m changes sign and has a unique principal eigenvalue, and R m (u ) 2 > 0 in all the other cases (i.e. m changes sign and (1:1) 0 has two principal eigenvalues, or m 0). Proof. One has by Lemma 4.5 m (u ) 2 = 0 0 ( 0 ) (u ) 2 ; and the conclusion follows from the shape of the function 0 () (cf. Lemma 3.1). We conclude this section with an example illustrating Propositions 4.3 and 4.4. Example 4.7. Let be the unit disc in R 2 ; m 1; and let 0 be the (unique) principal eigenvalue of on H0 1 () ; with u the associated positive eigenfunction. By simplicity, u is radial. So if we take a (x) = ( x 2 ; x 1 ) ; we see that (3.1) holds and that u belongs to I 0 : Moreover, by Proposition 2.1, the problem u + ha; rui = u in ; u = 0 has a unique principal eigenvalue, which is clearly 0 : Thus = 0 for all : This conclusion of course also follows from part (ii) of Proposition Asymptotic behavior of The results in this section provide a complete extension to the inde nite case of the main result of [1] recalled in the introduction. The new feature is the condition R mw2 > 0 which involves the weight. Theorem 5.1. Assume A nonempty and nonconstant. Then remains bounded as! 1 if and only if the vector eld a (x) admits a rst integral w in H 1 0 () with R mw2 > 0: Proof. We start with the necessary condition. So we assume bounded as! 1: Call u the L 2 normalized positive principal eigenfunction associated to in (1:1) : Multiplying both sides of L u = mu by u and integrating, using (3.1), one sees that u remains bounded in H 1 0 () : Consequently, for a subsequence, u converges to some w 2 H 1 0 () ; weakly in H 1 0 () and strongly in L 2 () ; and one has w 0; kwk L 2 = 1: Dividing both sides of L u = mu by ; one

17 ASYMPTOTIC BEHAVIOR OF PRINCIPAL EIGENVALUES 17 obtains that ha; ru i! 0 in D 0 () : Since on the other hand ha; ru i! ha; rwi in D 0 () ; one concludes that w 2 I 0 : The proof of the necessary condition would be completed if we could show that (5.1) mw 2 > 0: Inequality (5.1) is easily veri ed when m (x) " > 0 or when a 0 0: Indeed, the rst case is trivial. In the second case, multiplying as above the equation by u ; one obtains (5.2) harw; rwi + a 0 w 2 lim mw 2 ; since a 0 0; the left-hand side is > 0; as well lim (because! is nondecreasing and each is > 0); (5.1) thus follows. In the general case, using the assumption nonconstant, we will construct another rst integral which will satisfy an inequality like (5.1). For this purpose we consider the -function () de ned in (1:2) ; we will denote below by u ; the associated L 2 normalized positive eigenfunction. Using the fact that remains bounded as! 1 and is nonconstant, one has the following Claim. There exist 1 and a constant c > 0 such that (5.3) 0 ( ) c for 1 : Accepting this claim for a moment, one deduces from the concavity of the function! () that (5.4) () c ( ) for 1 and : It then follows from the equation satis ed by u ; that (5.5) haru ; ; ru ; i + [a 0 + c ( )] u 2 ; mu 2 ;: Since remains bounded, one can select with > 0 and 1 := sup in such a way that the bracket [a 0 + c ( )] in (5.5) is 0 for all : Call this value of : By (5.5) u ; remains bounded in H0 1 () as! 1; and so, for a subsequence, u ; converges to some w 2 H0 1 () ; weakly in H0 1 () and strongly in L 2 () ; and one has kwk L 2 R = 1: One then deduces from (5.5) that harw; rwi R mw2 ; which implies R mw2 > 0: We will now show that w is a rst integral. For this purpose observe that remains bounded as! 1 since ( ) = 0 and is non decreasing by Proposition 3.2. It then easily follows, as before, by dividing by the equation that R ha; rwi = 0; i.e. mw2 > 0: L u ; mu ; = u ; ; w 2 I 0 : So w provides a rst integral in H 1 0 () with Proof of the Claim. The proof goes in 3 steps: (i) 0 ( ) 0 for all 2 A; (ii) 0 ( ) < 0 for 2 Int (A) ; (iii) inequality (5.3) holds for su ciently large.

18 18 TOMAS GODOY, JEAN-PIERRE GOSSE, AND SOFIA PACKA Step (i) is a direct consequence of the fact that is the largest principal eigenvalue of (1:1) : To prove step (ii), let us x any 0 2 A and consider, for 2 R, ( 0 ) ; with v its associated L 2 normalized positive eigenfunction: (5.6) L v 0 mv = ( 0 ) v : By Proposition 3.2, the function! ( 0 ) is nondecreasing in R + ; moreover, by Proposition 4.4, it is strictly increasing if ha; rv 0 i 6 0: Let us rst consider this case. We thus have, for > 0 ; ( 0 ) > 0 ( 0 ) = 0; the shape of the function (cf. Lemma 3.1) then implies the conclusion 0 ( ) < 0: To complete the proof of step (ii), we will show that the situation where (5.7) ha; rv 0 i 0 is impossible. Indeed, if (5.7) holds, then, by Proposition 4.4, ( 0 ) as a function of is constant on R; and so ( 0 ) = 0 ( 0 ) = R: It follows that (5.6) becomes (5.8) L v = 0 mv ; which shows that 0 is a principal eigenvalue of L for the weight m: We will prove that it is the largest principal eigenvalue for the weight m; i.e. that 0 = 8 2 R, contradicting the assumption that the function! is nonconstant. That 0 is the largest principal eigenvalue of L for the weight m follows if we show (5.9) 0 ( 0 ) 0: To prove (5.9) we rst deduce from (5.8) and (5.7) that L v 0 = L 0 v 0 = 0 mv 0 ; comparing with (5.8) implies that v = v R: Applying now Lemma 4.5 to (1:1) with = 0 and observing that with the notations of Lemma 4.5, u 0 ; = v and u 0 ; = v ; we obtain R 0 ( 0 ) = R mv v v v R = R mv 0 v 0 v = 0 0 v 0 ( 0 ) 0; 0 where the latter inequality holds by de nition of 0 : Thus (5.9) holds and the proof of step (ii) is complete. To prove step (iii), x 0 2 Int (A) : Since by step (ii) 0 0 ( 0 ) < 0; one can nd < 0 such that 0 > 0: Let now > 0 : By Proposition 3.2, the concavity of the function! () and the fact that 0 ( ) 0; one has Consequently 0 < 0 0 ( ) 0 ( ) 1 : 0 ( ) 0 1 ; and the conclusion of the step (iii) follows. This concludes the proof of the claim and of the necessary part of Theorem 5.1.

19 ASYMPTOTIC BEHAVIOR OF PRINCIPAL EIGENVALUES 19 It remains to prove the su cient part in Theorem 5.1. So suppose the existence of w 2 I 0 with R R mw2 > 0: Normalizing we can without loss of generality assume mw2 = 1: By Proposition 2.3, (3.3) and (3.4), one has (5.10) = inf u2u 0 (u) 2 inf v2e u Q 1 u (v) 0 (w) ; where the last inequality follows from Lemma 3.8. This shows that remains bounded as! 1: The proof of Theorem 5.1 is now complete. Remark 5.2. When is constant and either a 0 0 or m 0; then there still exists a rst integral w in H0 1 () satisfying R mw2 > 0: Indeed, by Proposition 4.4, the positive eigenfunction u 0 associated to 0 belongs to I 0 : Moreover one derives from the equation for u 0 that (5.11) haru 0 ; ru 0 i + a 0 u 2 0 = 0 mu 2 0: When m 0; then R mu2 0 > 0 since u 0 (x) > 0 in : When a 0 0; then 0 > 0 and the left-hand side of (5.11) is > 0; which again imply R mu2 0 > 0: The example below shows that a rst integral w in H0 1 () satisfying R mw2 > 0 may fail to exist when is constant, a and m changes sign. Example 5.3. Let be the unit disc in R 2 and denote by 1 the principal eigenvalue of on H0 1 () with ' 1 a corresponding positive eigenfunction. For a (x) = ( x 2 ; x 1 ) ; we consider the problem (5:12) u + ha; rui + a 0 u = mu in ; u = 0 where m (x) := 1 + a 0 (x) and a 0 is chosen so that m (x) = 1 and the half-disk x 1 < 0 and m (x) = 1 on the half-disk x 1 > 0: Clearly div (a) = 0 in ; and since ' 1 is radial one also has ha; r' 1 i = 0 in ; so that ' 1 2 H0 1 () is a rst integral. Moreover, for = 0; 1 is clearly a principal eigenvalue of (5:12) 0, with ' 1 as associated eigenfunction. We now show that 1 is the largest principal eigenvalue of R (5:12) 0 : For that purpose we look at 0 0 (1) which, by Lemma 4.5, is equal to m'2 1= R '2 1; by construction of the weight, (5.13) m = 0 jxj=r for any 0 r 1; which implies, since ' 1 is radial, that R m'2 1 = 0; consequently 0 0 (1) = 0 and we see that, in fact, (5:12) 0 has a unique principal eigenvalue. Proposition 4.4 thus applies and yields that is constant. We now claim that for any rst integral w 2 H0 1 () ; one has (5.14) mw 2 = 0: Take a rst integral w: Since hrw; ai = 0; we see that for a.e. x 2 ; rw (x) is a multiple of x; i.e. of the form f (x) x; going to polar coordinates and writing w (r cos ; r sin ) = ew (r; ) ; one easily veri es ew=@ = 0; and so w is radial. Using (5.13), one concludes that (5.14) holds.

20 20 TOMAS GODOY, JEAN-PIERRE GOSSE, AND SOFIA PACKA We nally turn to the study of the limiting value of as! 1 and investigate in particular the validity of (5.15) lim = inf harw; rwi + a 0 w 2 : w 2 I 0 and mw 2 = 1 : Theorem 5.4. Assume A nonempty and bounded as! 1: Then (5.15) holds if either is nonconstant, or a 0 0; or m 0: Moreover, in each of these cases, the in mum in (5.15) is achieved. Remark 5.5. Note that Example 5.3 shows that (5.15) may fail to hold when is constant, a and m changes sign. Note also that Theorem 5.1 implies that (5.15) still holds, in the trivial form 1 = 1; when is unbounded as! 1: Proof of Theorem 5.4. Theorem 5.1 (when is nonconstant) and Remark 5.2 R (when is constant with a 0 0 or m 0) imply the existence of w 2 I 0 with mw2 = 1: Moreover (5.10) shows that inequality always holds in (5.15). When is constant with either a 0 0 or m 0; then the argument in Remark 5.2 yields equality in (5.15) as well as the achievement of the in mum. We now consider the case where is nonconstant. We start by observing the general fact that (5.15) holds with the achievement of the minimum whenever remains bounded and m (x) " > 0: This is a simple consequence of (5.2). We apply this observation for each 2 R to () (for which the weight is 1). This is possible since, as observed at the beginning of the proof, there exists w 2 I 0 with R mw2 = 1; and so (3.5) and Lemma 3.8 imply that for each 2 R, () remains bounded as! 1: Consequently, writing 1 () := lim () where the limit exists by Proposition 3.2, one has (5.16) 1 () = inf harw; rwi + a 0 w 2 mw 2 : w 2 I 0 ; kwk L 2 = 1 ; and the in mum is achieved at some w : The idea of the proof of Theorem 5.4 in the case nonconstant is now the following. We will approximate 1 := lim from the right by and show that for a subsequence, w converges to some ew satisfying R ew 2 I (5.17) 0 ; k ewk L 2 = 1; m ew2 > 0; 1 R m ew2 = R har ew; r ewi + a 0 ew 2 : The existence of such a ew clearly completes the proof of Theorem 5.4. Claim 1. Let! 1 with > 1 : Then, for a subsequence, w! ew weakly in H0 1 () ; strongly in L 2 () ; and one has ew 2 I 0 ; k ewk L 2 = 1; R m ew2 > 0 and (5.18) 1 ( 1 ) = har ew; r ewi + a 0 ew 2 1 m ew 2 : Note that the existence of ew satisfying the conditions of Claim 1, with the exception of R m ew2 > 0; follows directly by taking = 1 in (5.16). The key point is that the approximation procedure from Claim 1 yields a ew which also satis es R m ew2 > 0: Claim 2. 1 ( 1 ) = 0:

21 ASYMPTOTIC BEHAVIOR OF PRINCIPAL EIGENVALUES 21 Claims 1 and 2 clearly yield a function ew with the properties stated in (5.17). Proof of Claim 1. The function 1 : R! R is concave and so continuous, and consequently 1 ()! 1 ( 1 ) as! 1 : It then follows from (5.16) that w remains bounded in H0 1 () and so, for a subsequence, w converges to some ew 2 H0 1 () ; weakly in H0 1 () and strongly in L 2 () : One has k ewk L 2 = 1 and ew 2 I 0 (this latter point can be seen by going to the limit in R ' ha; rw i = 0 for ' 2 Cc 1 ()). One also obtains by going to the limit in (5.16) that 1 ( 1 ) har ew; r ewi + a 0 ew 2 1 m ew 2 : But one also has the inequality in the above relation by applying (5.16) with = 1 : So (5.18) follows. It remains to show that R m ew2 > 0: For this purpose we use (5.16) twice to get, for > 1 ; 1 () = harw ; rw i + a 0 w 2 mw 2 ; 1 ( 1 ) harw ; rw i + a 0 w 2 1 mw 2 : Substracting yields 1 () 1 ( 1 ) mw : 2 1 On the other hand, applying successively the mean value theorem, the concavity of and the claim from the proof of Theorem 5.1 (which can be used since we are dealing here with the case bounded and nonconstant), one obtains, for some with < 1 < < ; () ( 1 ) 1 = 0 () 0 ( ) c for su ciently large, with c a constant > 0. Going to the limit as! 1 and comparing with the previous inequality give mw 2 c: The conclusion R m ew2 > 0 now follows by letting! 1 : This completes the proof of Claim 1. Proof of Claim 2. We rst prove the existence of a constant c 0 > 0 such that (5.19) 0 ( 1 ) c 0 for su ciently large. Indeed, for some with < 1 < < 1 + 1; and consequently when! 1; since 1 ( 1 + 1) ( 1 + 1) ( 1 ) = 0 () 0 ( 1 ) ; lim inf 0 ( 1 ) 1 ( 1 + 1) 1 ( 1 ) ; 1 ( 1 ) is a xed number, (5.19) follows.

22 22 TOMAS GODOY, JEAN-PIERRE GOSSE, AND SOFIA PACKA We now look at 1 ( 1 ) and rst observe that since < 1 ; one has ( 1 ) < ( ) = 0; and so 1 ( 1 ) 0: Assume by contradiction 1 ( 1 ) = M < 0: Since is concave, one has 0 = ( ) ( 1 ) + 0 ( 1 ) ( 1 ) : This implies, for su ciently large, using (5.19), 0 M 2 + c0 ( 1 ) ; i.e. 1 M=2c 0 ; which contradicts the fact that! 1 : This completes the proof of Claim 2 and of Theorem 5.4. We conclude this section with several examples illustrating various situations which can occur in relation with the asymptotic behavior of as! 1: When a is a gradient eld,! +1 as! 1 (cf. Proposition 3.6). The case constant together with the existence (resp. nonexistence) of w 2 I 0 satisfying R mw2 > 0 is illustrated by Example 4.7 (resp. Example 5.3). The following example describes a situation where is nonconstant and has a nite limit as! 1: Example 5.5. Let be the unit disk in R 2 ; a (x) = ( problem (5.20) u + ha; rui + a 0 u = m (x) u in ; u = 0 x 2 ; x 1 ) and consider the By the choice of a; any nontrivial radial function w 2 H0 1 () belongs to I 0 : Fix Rsuch a w: We then take a 0 2 L 1 () with a 0 0; m 2 L 1 () with m 0 and mw2 > 0; with moreover either m radial and a 0 nonradial, or m nonradial and a 0 radial. With these choices, (5.20) has a unique principal eigenvalue for each ; is nonconstant (because the eigenfunction associated to 0 is nonradial and so does not belong to I 0 ; which allows the application of Proposition 4.4), and remains bounded (by Theorem 5.1). 6. Neumann-Robin boundary conditions Our purpose in this last section is to indicate how the preceding results can be adapted to deal with the problem L u = m (x) u in ; (6:1) A + b 0 (x) u = 0 Here A := ha; rui denotes the conormal derivative associated to the operator L ; with the exterior unit normal The assumptions on ; the operator L and the weight m are the same as those stated at the beginning of Section 2 (in particular m + 6 0); moreover b 0 2 C 0;1 (@) with b 0 0 It is well known that (6:1) admits 0, 1 or 2 principal eigenvalues, which are simple (cf. e.g. [10], [17], [8]). In case of existence we will again denote by the largest principal eigenvalue of (6:1). Proposition 6.1. Assume the existence of : Then (6.2) = inf (u) + b 0 u 2 inf u2u 0 v2h 1 () Q u

23 ASYMPTOTIC BEHAVIOR OF PRINCIPAL EIGENVALUES 23 where ; Q u have the same meaning as in Section 2 and where U 0 := u 2 H 1 () \ L 1 () : essinf (u) > 0 and mu 2 = 1 : For any u 2 U 0 the in mum over v in (6.2) is achieved; moreover when a 0 0 with either a or b 0 6 0; or when m (x) " > 0; the in mum over u in (6.2) is also achieved. Proposition 6.1 is proved in [8] when a 0 0 (cf. Theorems 3.1 and 3.6). Its derivation in the general case can be carried out by adapting to the present boundary conditions the arguments of the proof of Theorem 3.5 from [9]. The -function associated to (6:1) is de ned for 2 R as the unique principal eigenvalue = () of (6:3) L u mu = u in ; Bu = 0 This function enjoys properties similar to those stated in Lemma 3.1. The only changes occur in (i) where now (0) is > 0 if a 0 0 and either a or b 0 6 0; and in (iii) where now u should be considered as varying in C (cf. Lemma 2.5 in [8]). From now on we will assume, in addition to the conditions stated at the beginning of this section, that the vector eld a (x) satis es (6.4) div (a) = 0 in and ha; i = 0 It then follows from (6.2) that (6.5) () = inf u 0 (u) b 0 u 2 mu 2 2 inf v2h 1 () Q1 u (v) where u varies in U 0 0 where U 0 0 is de ned as U 0 with however now m 1. This formula for () implies the following Proposition 6.2. (i) exists if and only if exists, and then = : (ii) If exists for some 0; then exists for all ; and : As in the Dirichlet case, we will limit ourselves to the study of the function! for 0; and denote by A R + its domain. Arguing as in Section 4, we obtain Proposition 6.3. Assume A nonempty. Then is continuous on A. Moreover either is strictly increasing on A, or is de ned and constant for all 0: In addition the constant case occurs if and only if 0 exists and its eigenfunction belongs to I := w 2 H 1 () : w 6 0 and ha (x) ; rw (x)i = 0 a.e. in ; the set of rst integrals of a in H 1 () : The proof of Proposition 6.3 is similar to that of Propositions 4.1 and 4.4. The main di erence consists at the beginning of the proof of (i) from Proposition 4.4 to replace the condition a 0 0 by the condition a 0 0; a 0 6 0: This guarantees that the in mum over u in the minimax formula is achieved. Of course references should