Critical Point Theory 0 and applications

Transcription

1 Critical Point Theory 0 and applications

2 Introduction This notes are the result of two Ph.D. courses I held at the University of Florence in Spring 2006 and in Spring 2010 on Critical Point Theory, as well as of the course Superior Analysis I have been delivering for the Master Degree in Mathematics at the University of Perugia since They are a basic introduction to the subject, and this is the reason for the title. The list of papers cited in the Bibliography reflects other subjects treated in the courses, such as Schroedinger-Poisson systems, Klein-Gordon-Maxwell systems, Born-Infeld-Maxwell type systems, reversed variational inequalities,... Many thanks to some friends and students for a careful reading of the manuscript; among them it is a pleasure to thank Matteo Rinaldi, and Alessio Fiscella. i

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4 Contents Introduction i 1 Some preliminary notions Variational tools Nemytskij operator Weak solutions Mountain pass theorems Applications: one solution Applications: infinitely many solutions Unbounded domains The linking theorem An application The saddle point theorem An application A new multiplicity abstract theorem An application iii

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6 List of Figures 2.1 The topological situation of a Mountain Pass The topological situation of a Linking The topological situation of Theorem v

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8 Chapter 1 Some preliminary notions In this chapter we recall some notions which will be used throughout these notes. 1.1 Variational tools Let us start with classical notions (see [110], [5]). In the following X are Y Banach spaces, A X is an open set and f : A Y is a given function. Here, and from now on, L (X, Y ) denotes the set of linear and continuous operator between X and Y. Definition We say that f is Gateaux differentiable at u 0 A along v X if f (u 0 )(v) := lim t 0 f(u 0 + tv) f(u 0 ) t exists. We say that f is Fréchet differentiable (or simply differentiable) at u 0 A if there exists df(u 0 ) L (X, Y ) such that f(u 0 + h) f(u 0 ) df(u 0 )(v) lim = 0. h 0 h If f is differentiable at any u A, then it is simply said differentiable in A. The element df(u 0 ), also denoted by f (u 0 ) or Df(u 0 ), is called the differential of f at u 0. As for the finite dimensional case, the following properties hold: if f is Fréchet differentiable at u 0, it is also Gateaux differentiable at u 0 along any vector; if f is differentiable at u 0, then it is continuous at u 0 ; if f is differentiable at u 0, then df(u 0 ) is unique. Particularly useful is the following classical result. Theorem (Total Differential Theorem). If f : A Y is such that f (u)(v) exists u A, v X, f (u) L (X, Y ) and the map u A f (u) L (X, Y ) is continuous in u 0 A, then f is differentiable at u 0 and df(u 0 ) = f (u 0 ). Example f(u) c R df(u) = 0 u; 2. f(u) = Lu, for some L L (X, Y ) df(u) = L u; 1

9 2 3. let B : X Y Z be a bilinear and continuous form with values in a Banach space Z, db(u, v)(h, k) = B(h, v) + B(u, k) (u, v), (h, k) X Y. In particular, if X = Y and B is symmetric, we get db(u, u)(v, v) = 2B(u, v); 4. X is a Hilbert space and f(u) = u 2 df(u) = 2u u. Definition Let f : A Y be a differentiable map. Denote by f : A L (X, Y ) its differential map, called the (Fréchet) differential map of f. If f is continuous, then f is said to be of class C 1. Notations. From now on, with no other specification, if H is a Hilbert space we denote by, its scalar product and by its norm. If H is a Hilbert space and f : H R is differentiable at u, we denote by f(u) or grad f(u) the gradient of f at u, i.e. the unique element D H such that df(u), v H,H = D, v v X, which exists by Riesz Representation Theorem. Remark Of course, if one changes the norms, the differential remains the same, but the gradient changes. 2. A scalar function f defined on an open set A of a Hilbert space H is of class C 1 if and only if the map f : A H is continuous. 3. In the case 2 here above, it is well known that (see [31]) f (u) H = max v =1 f (u), v = f(u), so that v = f(u) f(u) is the vector in H which maximizes the scalar product f (u), v on the unit sphere. Example Let L L (H, H) and consider the quadratic form f(u) = Lu, u. Then, for every u, v H we have f (u)(v) = Lu, v + Lv, u. By the Total Differential Theorem f is differentiable at any point and df(u)(v) = (L + L )u, v, where L is the adjoint of L, so that f(u) = (L + L )u. In particular, if L is self adjoint, then df = 2L and f(u) = 2Lu u H, a generalization of point 4 in Remark As we will see in the following pages, there is a strict relation between the set of solutions of a PDE with given boundary conditions and set of the critical points of a certain functional defined on a suitable Hilbert (or Banach) space. For this reason it is essential to investigate the existence of critical points. On the other hand there is no easy way to see if a given functional has critical points when working in infinite dimensional spaces. It turns out that a good way consists in investigating the topological properties of the sublevels of the functional itself. More precisely, the differences between a sublevel and a lower one will imply the existence of critical points in between. Let us start with some notations. For any a < b R we set f b a := {u X : a f(u) b}, f a := {u X : f(u) a} and if f is also differentiable, we also set K a = { u X f(u) = a and f (u) = 0 }. A fundamental concept is introduced in the following definition introduced in [32].

10 3 Definition Let X be a Banach space, f : X R be a C 1 function and c R. We say that a sequence (u n ) n in X is a Palais-Smale sequence at level c, or a (P S) c sequence, if lim f(u n) = c and lim f (u n ) = 0, n n We say that f satisfies the Palais-Smale condition at level c, or that (P S) c holds, if every Palais-Smale sequence at level c has a (strongly) converging subsequence. It is clear that, if the subsequence converges to u, then f(u) = c and f (u) = 0. Actually, a slightly more general condition was given before (P S) c was introduced and it is known as (P S) condition (so no level is considered). It states that if (u n ) n N in X is such that the sequence (f(u n )) n is bounded and lim n f (u n ) = 0, then there exists a converging subsequence. This version is actually a slight generalization of the compactness condition (C) originally introduced by Palais and Smale in [113], sounding as follows: if V is a subset of X on which f is bounded and on which df is not bounded away from zero, then there is a critical point of f in V. It is clear that (P S) implies (C), but they are not equivalent, since constant functions satisfy (C) but not (P S). As one can easily see, (P S) implies (P S) c at any level c, but not viceversa. On the other hand, in most standard cases both of them hold true. Let us also remark that this notion can be compared with that of an a priori estimate: indeed, it states that the set of (quasi) critical points (i.e. of (quasi) solutions of differential equations) is not too large. Example The function f(x) = e x satisfies (P S) c c 0 (since there are no (P S) c sequences), but it doesn t verify (P S) 0 : in fact x n = n is an unbounded (P S) 0 sequence. With this example it is clear that f doesn t satisfy (P S). We observe that the (P S) condition provides a sort of compactness for particular sequences of functions. In particular it is a sufficient condition for the existence of a minimizer for a differentiable functional which is bounded from below on X, thanks to Eckland s Variational Principle. Note also that (P S) implies that any set of critical points with uniformly bounded energy is relatively compact, as one can easily prove. We premise the following definition to the main ingredient of a classical critical point theory. Definition Let X be a topological space and Y, Z X. deformation retract of Y (in X) if We say that Z is a strong Z Y ; there exists a continuous map ϕ : [0, 1] Y X such that ϕ(0, u) = u u Y, ϕ(1, u) Z u Y and ϕ(t, u) = u (t, u) [0, 1] Z. Example The closed ball B(0, 1) = {x X : x 1} is a strong deformation retract of X by the application x if x B(0, 1), ϕ(t, x) = (1 t)x + t x if x > 1. x The basic tools for variational theorems are the following results.

11 4 Lemma (First Deformation Lemma). Let X be a real Banach space, f C 1 (X, R) and < a b, possibly b = if a R. If (1.1) inf{ f (u) : a f(u) b} > 0, then f a is a strong deformation retract of f b. Proof. We will not give the complete proof of this Lemma, which is a special case of the following Theorem below. We present here a simple proof in the special case that f is of class C 1,1 loc and X is a Hilbert space. Take u f b and consider the evolution problem U f(u (t)) (t) = (1.2) f(u (t)) 2, U (0) = u. Since f is locally Lipschitz continuous, by the Cauchy Uniqueness Theorem, there exists δ = δ(u) > 0 and a unique solution U : ( δ, δ) X of problem (1.2). Let us look for the largest domain of U in the future, i.e. for t > 0. First, let us note that (1.3) d dt f(u (t)) = f(u (t)), U (t) = 1 by (1.2), i.e. U is parameterized by the values of f. Let us prove that U can be extended to [0, f(u) a]. In fact, if the maximal interval is [0, T ) with T f(u) a, then from (1.3) we get f(u (t)) = f(u) t > a t [0, T ). But in the strip f b a by assumption we have f ε > 0 for a suitable ε, so that (1.2) implies U (t) = 1 f(u (t)) 1 ε, so that U is Lipschitz continuous in [0, T ), and thus uniformly continuous. Then, since X is complete, there exists u 0 X such that lim U (t) = u 0. But then we can solve the problem t T V f(v (t)) (t) = f(v (t)) 2, V (T ) = u 0, which provides a solution V that is an extension of U, and this is absurd. Let us denote by Ψ(t, u) the solution of (1.2), which depends continuously on the initial datum u and is defined in the set { (t, u) : a f(u) b, t [0, f(u) a] }. Finally, let us consider the function ( Ψ t ( f(u) a ) ), u if a f(u) b, ϕ(t, u) = u if f(u) a. It turns out that ϕ is the desired strong deformation. Of course, it is not clear how to get assumption (1.1) of Lemma The easiest case is when M is compact, so that it is sufficient to assume that f (u) 0 for all u such that a f(u) b. In the noncompact case we have the following

12 5 Proposition Let X be a real Banach space, f C 1 (X, R) and a b be real numbers. If f (u) 0 for every u such that a f(u) b and (P S) c holds for any c [a, b], then there exists ε > 0 such that inf { f (u) X : a ε f(u) b + ε } > 0. Proof. Assume by contradiction that there exists a sequence (u k ) k in X such that a 1 k f(u k ) b + 1 k and f (u k ) 0. Up to a subsequence, we can assume that f(u k ) c [a, b], so that we have a (P S) c sequence. Then we can extract a subsequence which converges to a point u with f(u) = c and f (u) = 0, which is absurd. The proof of the previous Lemma in general Banach spaces with f of class C 1 needs the notion of pseudo gradient introduced by Palais (see [112]). Moreover, more precise result can be proved, in which the presence of critical points in the strip f b a can be handled, and this is the content of the classical Deformation Lemma. Theorem (Deformation Lemma). Let X be a real Banach space and f C 1 (X, R). Let c R and assume that f satisfies (P S) c. Then for every ε > 0 and for every neighborhood O of K c there exist ε (0, ε) and a continuous one-parameter family of homeomorphisms η(t, ) of X onto X for every t [0, 1] such that 1. η(0, u) = u for all u in X; 2. η(t, u) = u for all t in [0, 1] and for all u such that f(u) [c ε, c + ε]; 3. η(t, u) u 1 for all t [0, 1] and u X; 4. f(η(t, u)) f(u) for all t [0, 1] and u X; 5. η(1, f c+ε \ O) f c ε and η(1, f c+ε ) f c ε O; 6. if K c =, η(1, f c+ε ) f c ε ; 7. η has the semigroup property: η(s, ) η(s, ) = η(s + t, ) for all s, t 0. See [124] for a proof. Remark The same result holds if X is replaced by a complete manifold M. Another difficulty one can meet is to check the validity of (P S). An easy way to prove it is with the help of the following theorem. Theorem Let H be a Hilbert space and f C 1 (H, R) such that f = T + K, where T : H H is a linear, continuous, invertible operator with T 1 continuous, and K : H H is a compact operator. If a bounded sequence (u n ) n in H is such that f(u n ) 0 as n, then (u n ) n is relatively compact. Proof. Since f(u n ) = T u n +K(u n ) 0 and T 1 is continuous, we get that u n +T 1 K(u n ) 0. Since K is compact, up to a subsequence, K(u n ) u in H, so that T 1 K(u n ) T 1 u. Then u n T 1 u. Remark It is clear that we can always consider T as the identity map on H, so that f = Id + K, i.e. f is a compact perturbation of the identity, a typical request in order to apply the Leray Schauder topological degree theory.

13 6 1.2 Nemytskij operator In this section we denote by a measurable subset of R N and by u p the norm of a function u in L p (), p [1, ]. Definition We say that g : R R is a Carathéodory function if the function x g(x, s) is measurable s R; the function s g(x, s) is continuous for a.e. x. 2. The Nemytskij operator associated to g is the operator defined as N g (u)(x) := g(x, u(x)). For the ease of notation, for every function u : R we shall denote simply by g(x, u) the function x g(x, u(x)), i.e. the function N g (u)(x). Proposition Let g : R R be a Carathéodory function. We have: (a) if u k, u : R and u k u a.e. in, then g(x, u k ) g(x, u) a.e. in ; (b) if u : R is measurable, then g(x, u) is measurable; (c) if u, v : R coincide a.e. in, then g(x, u) = g(x, v) a.e. in. Proof. (a): Since the map s g(x, s) is continuous a.e. in and u k u a.e., it is immediate that g(x, u k (x)) g(x, u(x)) a.e., as claimed. (b): First assume that u is a simple function of the form u = k i=1 α 1χ Ai, where A i is a measurable subset of for any i = 1,..., k such that k i=1 A i = and A i A j = if i j. Then g(x, u) = k i=1 g(x, α i)χ Ai. But g(x, α i ) is measurable for every i = 1,..., k, χ Ai is measurable, and then g(x, u) is measurable as well. If u is a measurable function, there exists a sequence (u k ) k of step functions such that u k u a.e. in. By (a), g(x, u k ) g(x, u) a.e. in, so that g(x, u) is measurable, since it is limit of the sequence (g(x, u k ) k, which is measurable by the step above. (c) Straightforward. Remark Statement (b) of Proposition can be read as: the Nemytskij operator maps measurable functions in measurable functions. Theorem Let g : R be a Carathéodory function. Suppose there exist p 1, p 2 [1, ), a L p 2 () and b R such that (1.4) g(x, s) a(x) + b s p 1/p 2 s R, a.e. x. Then g(x, u) L p 2 () for any u L p 1 (), i.e. N g (L p 1 ()) L p 2 (). Moreover, the injection N g : L p 1 () L p 2 () is continuous. Proof. First note that if u L p 1, then g(x, u) is measurable by (b) of Proposition and by assumption g(x, u) a(x) + b u p 1/p 2 L p 2 (). Now consider a sequence (u k ) k such that u k u in L p 1 () and take any subsequence (still labelled by (u k ) k ). Up to a further subsequence, we can assume that u k u a.e. in. Then (a) of Proposition implies that g(x, u k ) g(x, u) a.e. in. Moreover, g(x, u k ) a(x) + b u k p 1/p 2 and u k p 1/p 2 converges to u p 1/p 2 in L p 2 (), so that the generalized Lebesgue Theorem of Dominated Convergence implies that g(x, u k ) g(x, u) in L p 2 (). Since this happens for any subsequence, the result follows.

14 7 There is a sort of viceversa of the previous result: Theorem ([7]). Let g : R be a Carathéodory function and p 1, p 2 [1, ). Suppose that N g (L p 1 ()) L p 2 (). Then there exist a L p 2 () and b R such that (1.4) holds. Note that the result is true without any continuity assumption on N, which is however re obtained by Theorem Proposition Let g : R be a linear Carathéodory function, p 1, p 2 [1, ] be such that p 2 > p 1 and N g : L p 1 () L p 2 () is defined. Then g is independent of s. Proof. By assumption there exist two measurable functions a, b : R such that g(x, s) = a(x) + b(x)s. Since N g (L p 1 ()) L p 2 (), we get that a = N g (0) L p 2 (), so that b(x)u L p 2 () for any u L p 1 (). Then b 0. Indeed, if there exist ε > 0 and with > 0 and b(x) ε a.e. x, we could take u L p 1 () \ L p 2 ( ) so that and a contradiction arises. L p 2 ( ) ε u b(x)u L p 2 () by assumption Remark The previous Proposition does not states that there are not linear continuous operators L p 1 L p 2 with p 2 > p 1 : of course there are many, but they are not Nemytskij operators. Consider, for example, the operator I : L 1 (0, 1) L (0, 1) defined as Iu(x) = x 0 u(t) dt. Up to now we only considered L p with p. Now we show a continuity result in L. Proposition Let g : R be a Carathéodory function and suppose that there exist a L p (), p 1, g 0 : R R locally bounded such that g(x, s) a(x)g 0 (s) s R, a.e. x. Then N g : L () L p () is continuous. Proof. By (b) of Proposition 1.2.2, g(x, u) is measurable, and from the growth assumption on g, we get that g(x, u) a(x)g 0 (u) L p () u L (). Moreover, if u k u in L (), then by (a) of Proposition 1.2.2, g(x, u k ) g(x, u) a.e. in and if u k M a.e. in n, then g(x, u k ) g(x, u) p 2 p 1 a(x) p max [ M,M] g p 0, and from the Lebesgue Theorem of Dominated Convergence we get that g(x, u k ) g(x, u) in L p (). It is immediate to prove the following result: Proposition Let g : R be a Carathéodory function and set G(x, s) := s 0 g(x, t) dt. Then G : R R is a Carathéodory function such that the map s G(x, s) is of class C 1 for every s R and a.e. x. From now on, though not stated all the times, we understand that g : R is a Carathéodory function and G(x, s) := s 0 g(x, t) dt. Theorem Let 1 p 1 < p 2 < and assume that there exist a L q (), q = p 1p 2 p 2 p 1 b R such that g(x, s) a(x) + b s p 2 1 p 1 s R a.e. in. Then N G : L p 2 () L p 1 () is of class C 1 and dn G (u)(v) = N g (u)(v) u, v L p 2 (). and

15 8 Proof. The growth condition on g implies that N G : L p 2 () L p 1 () is well defined, as well as N g : L p 2 () L q (), while Proposition ensures that the map s G(x, s) is of class C 1. Take u L p 2 (). The growth condition on g also implies that the map L p 2 () L p 1 (), v g(x, u)v is a well defined linear continuous operator. By the Total Differential Theorem we get that N G : L p 2 () L p 1 () is Fréchet differentiable at any point. The fact that dn G (u)(v) = N g (u)(v) u, v L p 2 () is a simple calculation. Finally, let us note that for any u 1, u 2, v L p 2 () by the Hölder inequality g(x, u 1 )v g(x, u 2 )v p1 g(x, u 1 ) g(x, u 2 ) q v p2, that is dn G (u 1 ) dn G (u 2 ) L (L p 2 (),L p 1 ()) g(x, u 1 ) g(x, u 2 ) q, so that (N G ) is continuous and N G is of class C 1. A special case of Theorem is when, setting p 2 = p > 2, we choose p 1 = p = so that we get the following particular version. Theorem Assume that there exist p > 2 and a L q (), q = g(x, s) a(x) + b s p 2 s R a.e. in. p p 2, such that p p 1 < 2, Then N G : L p () L p () is Fréchet differentiable and dn G (u)(v) = N g (u)(v) u, v L p (). Remark The extreme case p 1 = p 2 in Theorem , that is the case p = 2 in Theorem , are special ones. In this case, in general, N G is only Gateaux differentiable, but not Fréchet differentiable. An essential tool in our future investigation is the following result about the differentiability of the Nemytskij operator defined in Sobolev spaces. Theorem Let 1 p < N and g : R R be a Carathéodory function such that pn g(x, s) a(x) + b s q s R, a.e. x, where a L N(p 1)+p (), b R, q [1, p 1], and p = is the critical Sobolev exponent. Then the function G : W 1,p () R defined as G(u) = G(x, u) dx is a C 1 function such that G (u)(v) = g(x, u)v dx u, v W 1,p (). Proof. It is a consequence of the Theorems on the differentiability of the Nemytskij operator and of Lebesgue s Theorem. See also [124, Theorem C.1] for more general integral operators in the case p = 2. pn N p Of course a more general theory for the Nemytskij operator can be developed, but for the aims of this notes, the previous results are enough.

16 9 1.3 Weak solutions Let us consider the following Dirichlet problem in presence of a nonlinear Poisson equation (1.5) { u = g(x, u) in, u = 0 on, where is a domain (open and connected set) of R N, N 1, and g : R is a Carathéodory function. If u H 2 () H0 1() satisfies (1.5) a.e. in, take any v H1 0 () and multiply both sides of (1.5) by v. By the Gauss Green formula we finally end up with (1.6) Du Dv dx = Going the other way, we give the following g(x, u)v dx v H 1 0 (). Definition We say that u H0 1 () is a weak solution of problem (1.5) if (1.6) holds. Remark One can define weak solutions requiring that (1.6) holds for any v C0 1 () or for any v C0 (), but this is equivalent to our choice since both C1 0 () and C 0 () are dense in H0 1(). It is clear that if g has non supercritical growth in the sense of the Sobolev exponent, then problem (1.5) is variational: denoting by f : H0 1 () R the functional defined as (1.7) f(u) = 1 Du 2 dx G(x, u) dx, 2 where as usual G(x, s) = s 0 g(x, t) dt, then f is differentiable by Theorem and u is a weak solution of problem (1.5) if and only if u is a critical point of f. Indeed, u is a critical point for f if and only if f (u)(v) = 0 for any v H0 1 (), that is (1.6). Then, in order to find solutions of problem (1.5), we will look for critical points of f. Historically, problem (1.5) is called the strong Euler Lagrange equation associate to f, while (1.6) is called the weak Euler Lagrange equation associate to f. We have seen that (strong) solutions of the strong Euler Lagrange equation (1.5) are also (weak) solutions of the weak Euler Lagrange equation (1.6). Viceversa, by regularity results it is possible to show that in most essential cases weak solutions are also strong solutions; this is the case, for example, when g has subcritical growth in the sense of the Sobolev exponent. Theorem Let be of class C 2 and let u H0 1 () be a weak solution of (1.5), where g : R is a Carathédory function such that there exist C > 0 and p > 0 for which g(x, s) C(1 + s p 1 ) s R a.e. x. Moreover, assume that p 2N N 2 if N 3. Then u C 1,α () for some α (0, 1). If in addition g is Hölder continuous, then u is of class C 2 (). Finally, if is of class C k+2 and g is of class C k,α ( R), then u W k+2,2 () and if is of class C and g is of class C ( R), then u C ().

17 10 We underline the fact that such a regularity result requires that g is not supercritical, but it lets g be sublinear. The proof of this theorem in based on Moser s iteration technique (see [104]) and its development due to Brezis and Kato (see [33]): first one shows that u belongs to W 2,q () H 1 0 () q <, so that by the Morrey Theorem u C 1,α () (see [124], Lemma B.3 and the following statements). If in addition g is Hölder continuous, the C 2 () regularity is proved with the help of Schauder techniques, see [73, Chapter 6]. See also [73, Theorem 8.13] for some estimates of the solution in terms of the data and [105] for some L H 1 0 estimates. Throughout these notes we will consider only elliptic problems where the governing operator is the Laplacian, but everything can be immediately extended to other problems where the principal part is replaced by a more general uniformly elliptic operator in divergence form and the boundary condition is not 0. Let us see how: consider the problem (1.8) N x i,j=1 i u = u 0 ( a ij (x) u ) = g(x, u) in, x j on, where is a domain of R N, N 1, g : R is a Carathéodory function and the principal part is uniformly elliptic according to the following Definition The operator L = N i,j=1 x i ( a ij (x) ) x j is called uniformly elliptic if a ij = a ij L () i, j = 1,..., N, and there exists λ 0 > 0 (the ellipticity constant) such that N a ij (x)ξ i ξ j λ 0 ξ 2 ξ R N a.e. x. i,j=1 First, let us note that in problem (1.8) we can assume u 0 = 0, provided u 0 is sufficiently regular, say u 0 is the restriction to of a function ũ 0 H 1 (). In fact, consider the solution v H 1 () to the problem (1.9) N x i,j=1 i v = u 0 ( a ij (x) v ) = 0 in, x j on, (which exists, for example, by [73, Theorem 8.3]); then the function z := u v solves (both weakly and strongly) the problem N ( a ij (x) z ) = g(x, u) = g(x, z + v) := g(x, z) in, (1.10) x i,j=1 i x j z = 0 on, and g satisfies all the same growth condition satisfied by g in the s variable (if there are any). Moreover, we can apply Theorem to the function v, so that higher regularity on and g imply higher regularity on v itself.

18 11 Now let us show that problem (1.8) with u 0 = 0 is variational and then it can be treated as (1.5). With an obvious generalization of Definition 1.3.1, we say that u H0 1 () is a weak solution of N ( a ij (x) u ) = g(x, u) in, (1.11) x i,j=1 i x j u = 0 on, if N i,j=1 u v a ij dx = g(x, u)v dx x j x i v H 1 0 (), so that weak solutions of (1.11) are critical points of the functional F : H0 1 () R defined as F (u) = 1 2 N i,j=1 u u a ij dx G(x, u) dx. x j x i We remark that the previous statement is not true any more if we drop the symmetry assumption a ij = a ji i, j = 1,..., N. Then we can proceed in a standard way: the operator L = N i,j=1 x i ( a ij (x) ) x j has a diverging sequence of eigenvalues 0 < µ 1 < µ 2... on H0 1 () which play the role of the eigenvalues 0 < µ 1 < µ 2... of on H0 1(). Moreover it provides H1 0 () with the new scalar product N u v u, v = a ij dx + uv dx, x j x i i,j=1 which is equivalent to the usual one, and which induces the new norm u 2 = N i,j=1 u u a ij dx + u 2 dx. x j x i As usual, the second integral in the two definitions above can be removed if is bounded. In this last case the gradient of F with respect to the new scalar product is again given by F (u) = u + L 1 (g(x, u)), in the same way as the gradient of the functional f defined in (1.7) with respect the standard scalar product is f(u) = u + 1 (g(x, u)), as we will se in Section 2.1. Remark Concerning the definition of ellipticity, there are other cases considered in literature, which are slightly different from Definition For example we can say that the operator N ( L = a ij (x) ) x i x j i,j=1

19 12 is uniformly elliptic if a i,j is a measurable function i, j = 1,..., N, and there exist Λ 0 > λ 0 > 0 such that (1.12) λ 0 ξ 2 N a ij (x)ξ i ξ j Λ 0 ξ 2 ξ R N a.e. x. i,j=1 Surprisingly, this new definition and the one above coincide only under the symmetry assumption a ij = a ji i, j = 1,..., N. Indeed, if L satisfies Definition 1.3.4, then N i,j=1 a ij (x)ξ i ξ j max i,j { a ij } N ξ i ξ j M 2 i,j=1 N ( ξ i 2 + ξ j 2 ) = MN ξ 2, i,j=1 where we have set M = max i,j { a ij }. Conversely, if L satisfies this new definition of ellipticity, then a ij L () i, j = 1,..., N. Otherwise if a ī j L () and for example a ī j > 0 and a ī j L ( ) with, take ξ such that ξ ī = ξ j = 1 and ξ i = 0 i ī, j, so that 2Λ 0 = Λ 0 ξ 2 N a ij (x)ξ i ξ j = 2a ī j L (). i,j=1 Note that we used the symmetry a ij = a ji i, j = 1,..., N precisely in this last calculation. If this condition is removed, the equivalence fails: it is enough to consider a 12 = log x, a 21 = log x + 1 and a ij = δ ij in the other cases. Then N a ij (x)ξ i ξ j = ξ 1 ξ 2 + ξ 2, i,j=1 so that (1.12) is satisfied, but of course a 12 and a 21 are not bounded.

20 Chapter 2 Mountain pass theorems In [6] a fundamental result in critical point theory was proved. Due to the geometrical structure involved therein, it is called the Mountain Pass Theorem. We present the following version of the theorem, where the geometrical situation is referred to the point 0, but if the situation is translated somewhere else, of course the result still holds. Theorem (Mountain Pass Theorem, [6]). Let X be a real Banach space and f C 1 (X, R) be such that f(0) = 0. Assume that there exist ρ, α > 0 such that inf f(u) = α u =ρ and there exists e in X \ B ρ such that f(e) 0. Set Γ := {γ C([0, 1], X) : γ(0) = 0, γ(1) = e} and β = inf γ Γ max t [0,1] f(γ(t)). Finally, assume that (P S) β holds. Then β is a critical value. Proof. First, take γ Γ, so that γ([0, 1]) S ρ, where S ρ = {u X : u = ρ}. In this way max f(γ(t)) α, t [0,1] and then β α. Moreover, taking γ(t) = te, the function (γ(t)) is continuous on [0, 1], so that it admits a maximum M <. Then β M. Assume by contradiction that β is not a critical value. Then by Proposition , there exists ε > 0 such that inf { f (u) : β ε f(u) β + ε } > 0, and of course we can choose ε such that β ε > 0. By the First Deformation Lemma (Theorem ), there exists strong deformation U (t, ) of f β+ε on f β ε. By definition of β there exists γ Γ such that max f(γ) < β + ε. [0,1] Now consider the function h(t) := U (1, γ(t)), which is of class C ([0, 1], X). First note that h(0) = U (1, γ(0)) = U (1, 0) = 0, since 0 f 0 f β ε, and f β ε is kept fixed by U. In the same way h(1) = e, so that h Γ and so β max t [o,1] f(h(t)). 13

21 14 On the other hand, γ([0, 1]) f β+ε, so that U (1, γ([0, 1])) f β ε, that is and a contradiction arises. max f(h(t)) β ε, t [0,1] 0 e Figure 2.1: The topological situation of a Mountain Pass Definition (See [75]). Let X be a real Banach space, f C 1 (X, R) and u X be such that f (u) = 0. We say that u is a mountain pass point in the sense of Katriel if for any neighborhood U of u the set U {x X : f(x) < f(u)} is non empty and disconnected; in the sense of Hofer if for any neighborhood U of u the set U {x X : f(x) < f(u)} is non empty and not path-connected. It may happen that critical points found with the help of the Mountain Pass Theorem are not mountain pass points: Example Consider the function f(x) = x 2 x 4 on X. Then all the critical points found with the Mountain Pass Theorem are maxima points. As it is natural, symmetries in the functional lead to the existence of several critical points: if f is even and u is a critical point at level c, then also u is a critical point at level c. We recall some classical theorems from Ljusternik and Schnirelmann theory. Theorem (Ljusternik Schnirelmann, [89]). Let f C 1 (R N, R) be a even function. Then f S N 1 admits at least N couples (u, u) of critical points. In infinite dimension the compactness property of the sphere is lost, so that an additional assumption is needed. Again, the (P S) condition turns out to be a good one, so that we get the following extension of Theorem

22 15 Theorem (Theorem 8.10, [116]). Let f be a even C 1 function defined on an infinite dimensional Banach space X. If there exists α R such that f Sα satisfies (P S) c c R and f Sα is bounded from below, then f Sα has infinitely many couples (u, u) of distinct critical points. Verifying (P S) for f constrained on spheres is not so immediate, and it is more natural to work with unconstrained functionals. In this setting the following result is extremely useful (see also [16, Theorem 9.12]). Theorem (Bartolo Benci Fortunato, [13]). Let f be a even C 1 function defined on an infinite dimensional Banach space X such that f(0) = 0. Assume that X is decomposable as direct sum of two closed subspaces X = X 1 X 2 with dimx 1 <. Suppose that (i) there exist α, ρ > 0 such that inf f(s ρ X 2 ) α; (ii) for any finite dimensional subspace Y X there exists R = R(Y ) > 0 such that f(u) 0 u Y with u R; (iii) (P S) c holds c α. Then f has an unbounded sequence of positive critical values. Remark If X 1 = {0}, condition (i) of Theorem is nothing but the first condition in Theorem 2.0.6, and, indeed, a first version of Theorem was proved in [6] in the case X = X Applications: one solution Let us consider the problem { u λu = g(x, u) in, (P ) u = 0 on, where is a bounded domain of R N, N 1, λ R and g : R R. First of all we assume the standard assumptions for a superlinear and subcritical nonlinearity, whose prototype is the function g(x, s) = s p 2 s (such assumptions are quite natural and appear quite often while studying nonlinear subcritical problems, even in less general forms, see [6], [81], [116]): (g 1 ) g : R is a Carathéodory function; (g 2 ) there exits constants a 1, a 2 > 0 and p > 2 such that s R and for a.e. x (2.1) g(x, s) a 1 + a 2 s p 1, where p < 2N N 2 if N 3; (g 3 ) g(x, s) = o( s ) as s 0 uniformly in ; (g 4 ) µ > 2 and R 0 such that s with s > R and for a.e. x (2.2) 0 < µg(x, s) g(x, s)s, and there exist a 3 > 0 and a positive function a 4 L 1 () such that s R and for a.e. x, G(x, s) a 3 s µ a 4 (x).

23 16 Note that (g 3 ) implies that G(x, s ) = o( s 2 ) as s 0 uniformly in and that u = 0 is a solution of problem (P ). Condition (g 4 ) is known as the Ambrosetti Rabinowitz condition. Remark Condition (g 2 ) implies that G(x, s) a 1 s + a 2 p s p s R and for a.e. x, so that it is a subcritical assumption on G, since p is strictly less than 2 = 2N N 2, the Sobolev embedding exponent. Remark As seen in (g 2 ), no upper bound on the exponent p is given when N = 1, 2, since H0 1() L () if N = 1 and H0 1() Lq () for any q < if N = 2. Actually, by the Truedinger Moser inequality (see [73, Theorem 7.15]) in the case N = 2 we could replace (2.1) with g(x, s) a 1 e ϕ(s) ϕ(s) + a 2, where lim s s 2 = 0. Remark Condition (g 4 ) says that G is superquadratic, or that g is superlinear. We also note that the final request in (g 4 ) is useless if is smooth and g : R is continuous. In fact, when R > 0 integrating (2.2) gives G(x, s) a 3 s µ a 4 (x), where and { a 3 = min min x G(x, R) R µ, min x { a 4 (x) = max max G(x, σ), max σ [0,R] } G(x, R) R µ σ [ R,0] } G(x, σ). If R = 0 the value R in the definitions above must be replaced by limits. Let us denote by 0 < λ 1 < λ 2... the diverging sequence of the eigenvalues of in H 1 0 (). If i 1 set H i = Span(e 1,..., e i ), where e j is the eigenfunction associated to λ j, and define H i as the orthogonal complement of H i in H 1 0 (). We want to prove the following Theorem Assume (g 1 ) (g 4 ) and λ < λ 1. Then problem (P ) admits a nontrivial solution. Proof. The proof will be made with the help of the Mountain Pass Theorem, for which the case λ λ 1 cannot be handled (for this case, see Chapter 3). First of all let us recall that H0 1 (), as usual, is endowed with the scalar product u, v = Du Dv, which induces the usual norm u 2 = Du 2. Now consider the functional f : H0 1 () R defined as f(u) = 1 Du 2 dx λ u 2 dx G(x, u) dx. 2 2 It is easy to see that f is a C 1 functional on H0 1 () whose critical points solve problem (P ). Thus Theorem will be proved if we show the existence of a nontrivial critical point for f. To this aim, we want to apply Theorem to f. First, f(0) = 0 and by (g 3 ), fixed ε > 0, there exists δ > 0 such that G(x, s) ε 2 s 2 for any s such that s δ and for a.e. x. Moreover, if s δ, by Remark we get G(x, s) a 1 s + a ( 2 a1 p s p δ p 1 + a ) 2 s p = c δ s p, p

24 17 so that (2.3) G(x, s) ε 2 s 2 + c δ s p s R and for a.e. x. By the Poincaré, Hölder and Sobolev inequalities there exist c 1 = c 1 (δ, ) > 0 such that (2.4) G(x, s) dx ε 1 u 2 + c 1 u p. 2λ 1 In this way (we can assume λ > 0, the case λ 0 being easier) [ ( 1 f(u) 1 λ + ε ) ] c 1 u p 2 u 2. 2 λ 1 Let us choose ε so small that λ + ε < λ 1. Now take { 1 (2.5) u = ρ < 2c 1 ( 1 λ + ε )} 1/(p 2), λ 1 and obtain inf f > 0. S ρ Now take u 0 in H 1 0 () and t > 0. Then by (g 4) so that f(tu) t2 2 Du 2 dx λt2 2 u 2 dx a 3 t µ lim f(tu) =, t u µ dx + a 4 (x) dx, and the last geometrical condition of the Theorem follows. Now, let us prove the Palais Smale condition. By Theorem it is enough to prove that for every c R (more than required, actually) any (P S) c sequence is bounded and that f has the form Id + K, where K is a compact operator. First step: f = Id + K. Indeed, we have f(u) = u + 1 (λu + g(x, u)), where 1 : H 1 () H0 1() is the operator which associates to any element h H 1 () = (H0 1()) the unique solution in H0 1 () of the problem { u = h in, u = 0 on, which exists by the Lax Milgram Theorem. By (g 2 ), for any u H0 1(), the function λu + g(x, u) belongs to Lp () and since p > 2, L p () H 1 (), so that 1 (λu + g(x, u)) is well defined. Claim: 1 : H 1 () H0 1 () is compact. Indeed, consider the Nemytzskij operator N g : H0 1() Lp () L p () H 1 () u u g(, u) g(, u);

25 18 since the embedding H 1 0 () Lp () is compact, we get that N g is compact as well. Second step: any (P S) c sequence is bounded. Take c R and a (P S) c sequence (u n ) n, that is f(u n ) 0 and f(u n ) c as n. Then there exist M, N > 0 such that (2.6) µf(u n ) f(u n ), u n M + N u n. On the other hand, (2.7) µf(u n ) f(u n ), u n = Now, let us write ( µ ) ( µ ) 2 1 u n 2 λ 2 1 u 2 n dx + [g(x, u n )u n µg(x, u n )] dx = {x : u n(x) R} +, {x : u n(x) >R} so that, by (g 2 ) and (g 4 ) there exists c R > 0 such that ( µ ) ( µ ) 2 1 u n 2 λ 2 1 u 2 n dx + [g(x, u n )u n µg(x, u n )] dx (2.8) ( µ ) ( λ ) u n 2 c R. λ 1 Of course, here we used the Poincaré inequality and again we only considered the case λ > 0. By (2.6), (2.7) and (2.8) we finally get (1 λλ1 ) (µ2 1 ) u n 2 c R M + N u n, that is (u n ) n is bounded. Since all the assumptions of Theorem are fulfilled, we can conclude that problem (P ) has a solution u, which is not the trivial one, since f(u) > 0, and so Theorem is proved. 2.2 Applications: infinitely many solutions In the assumptions of the previous section, if g is odd in the s variable, we expect many solutions. Indeed we have: Theorem Assume (g 1 ) (g 4 ), g(x, s) = g(x, s) s R and for a.e. x, and λ < λ 1. Then problem (P ) admits an unbounded sequence (u n, u n ) n of couples of solutions. Proof. We apply Theorem to the already introduced functional f(u) = 1 Du 2 dx λ u 2 dx G(x, u) dx, 2 2 and we only need to verify the geometrical conditions (i) and (ii), since all the other request are verified, because this is a special case in which in addition f is even. Condition (i). Proceeding as in the proof of Theorem 2.1.4, we obtain (2.4), so that with the same choice of (2.5), we prove (i) of Theorem with the choice X 1 = {0} and X 2 = H0 1(). Condition (ii). Let Y H0 1 () be a finite dimensional subspace. We already observed that f(u) 1 Du 2 dx λ u 2 dx a 3 u µ dx + a 4 (x) dx, 2 2

26 19 and by the Poincaré inequality (again we consider only λ > 0) f(u) 1 ) (1 λλ1 u 2 a 3 u µ dx + 2 a 4 (x) dx. Since all norms are equivalent in Y, then f(u) when Y and u. Then by Theorem there exists a sequence (u n ) n such that f(u n ). But by (g 4 ) u n 2 2 = f(u n ) + G(x, u n ) dx f(u n ) c R, where c R is the same constant appearing in (2.8). The theorem is now completely proved. 2.3 Unbounded domains In the previous section we saw that an essential tool in proving (P S) is that the embedding i : H 1 0 () Lp () is compact for any p < 2 when is bounded, so that the adjoint i is compact as well. When is not bounded, say for example = R N, this compactness cannot hold any more, since it is possible to translate functions in every direction. For example, take u 0 in H 1 (R N ), then u h ( + h) has the same norm of u, but it is impossible to extract any converging sequence from (u h ) h. In some cases, however, some compactness can still be recovered, like in the case of radial functions. Let us denote by H 1 r (R N ) the set of radial Sobolev functions in R N, i.e. u H 1 r if and only if u H 1 (R N ) and u(x) = u( x ). It holds: Theorem ([27], [121]). The immersions H 1 r (R N ) L p r(r N ) is compact for any p (2, 2 ). Of course, by duality, L p r (R N ) is compactly embedded in (H 1 r (R N )), so that it suggests to follow the approach of Section 2.1 in order to show solutions for superlinear and subcritical problems in R N. The first result in this spirit is the one of [27], where the authors prove the following. Theorem If g satisfies (g 1 ) (g 4 ), then there exists a radial nontrivial solution of problem { u + u = g(x, u) in R N, u H 1 (R N ). Actually g was supposed to be a bit more regular, but the proof is essentially the same in this case and it is modelled on the proof of Theorem 2.1.4: denote by f : Hr 1 (R N ) R the functional f(u) = 1 Du 2 dx + 1 u 2 dx λ u 2 dx G(x, u) dx, 2 R N 2 R N 2 R N R N and then prove that f satisfies the geometrical conditions of Theorem Finally, in order to prove (P S), again it is enough to show that every (P S) sequence is bounded, since f(u) = u + 1 (g(x, u)),

27 20 where 1 H 1 r : Hr 1 (R N ) = (Hr 1 (R N )) the unique solution in H 1 (R N ) of the problem (R N ) H 1 r (R N ) is the operator which associates to any element h { u u = h in, u H 1 r (R N ), which exists by the Lax Milgram Theorem. We leave the details to the reader, while in the following chapters we present three systems of differential equations arising in quantum mechanics which can be studied with the tools presented above.

28 Chapter 3 The linking theorem Theorem (Linking Theorem, [26]). Let X be a Banach space, with X = X 1 X 2, where X 1 and X 2 are closed subspaces with dimx 1 <. Let f : X R be of class C 1 and assume that (i) there exist ρ, α > 0 such that ) inf f (S ρ X 2 = α; (ii) there exists e in X 2 \{0} and R > ρ such that sup f(σ) 0, where, denoting (e) = Span(e), we have set = ( B R X 1 ) { te : t [0, R] } and Σ = X1 (e) ; (iii) setting H = {h C(, X) : h = Id on Σ} and assume that (P S) β holds. Then β is a critical value. β = inf h H max u f(h(u)), Proof. The proof follows the one given in [116]. First, let us note that α β <. Indeed, take h = Id H, so that β sup f( ) < since is compact. In order to prove rigorously that β α we would need an intersection theory, but it goes beyond air purposes, so that we only underline that intuitively it is clear that (S ρ X 2 ) h( ) h H, so that there exists u h (S ρ X 2 ) h( ) h H. Of course f(u h ) α h H and then β α. Now assume by contradiction that β is not critical for f. Then by Proposition there exists ε > 0 such that inf { f (u) X : a ε f(u) a + ε } > 0. By Lemma there exists U C([0, 1], X) which deforms f β+ε on f β ε strongly, and we can assume ε so small that β ε > 0. By definition of β there exists h H such that (3.1) sup f(h(u)) < β + ε. u 21

29 22 Since f(σ) 0, for our choice of ε we get Σ f β ε, so that U(t, h(u)) = u for all t [0, 1] and for all u Σ. In particular, if u, U(1, h(u)) H, so that β sup f(u(1, h(u))). On the other hand, f(h(u)) β + ε u. Since U is a strong deformation, we get U(1, h(u)) f β ε and then sup f(u(1, h(u))) β ε. Absurd. 2 e Figure 3.1: The topological situation of a Linking Remark If X 1 = { 0}, then Theorem is nothing but Theorem If Σ and are replaced by homeomorphic manifolds not centered at 0 and the topological situation is translated in another point, the theorem still holds. For a more complete version of this theorem, where the existence of another critical point is proved, see [92]. 3.1 An application Let us go back to problem (P ) { u λu = g(x, u) in, u = 0 on, where is a bounded domain of R N, N 1. In Theorem we showed that, under conditions (g 1 ) (g 4 ) introduced in Section 2.1, problem (P ) has a nontrivial solution when λ < λ 1. With the help of Theorem we will remove this bound on λ. For simplicity, we will make an extra assumption on G.

30 23 Theorem Assume (g 1 ) (g 4 ) of Section 2.1 and if R > 0 in (g 4 ), assume also that G(x, s) 0 s R and for a.e. x. Then problem (P ) has a nontrivial solution for any λ R. Proof. Of course, we consider only the case λ λ 1, the other case being already considered in Theorem Therefore we assume that there exists a natural number i 1 such that λ i λ < λ i. Set H i = Span(e 1,..., e i ) and denote by Hi the orthogonal complement of H i in H0 1 (), so that H0 1() = H i Hi. We apply Theorem with X = H1 0 (), X 1 = H i and X 2 = Hi. First of all, let us recall the following form of the Poincaré inequality, which can be easily proved by a spectral decomposition in H0 1(), (3.2) λ i+1 u 2 dx Du 2 dx u Hi, and its counterpart (3.3) Du 2 dx λ i u 2 dx u H i. Take u Hi. Proceeding as in the in the proof of Theorem we get (2.3), so that by (3.2) we get (after relabelling ε) f(u) = 1 Du 2 dx λ u 2 dx G(x, u) dx 1 ( 1 λ ) ε u 2 C u p, λ i+1 where C > 0 is a constant depending only on and ε. Of course we choose ε so small that 1 λ λ i+1 ε > 0. Now take u Hi such that u = ρ < [ ( 1 1 λ )] 1/(p 2) ε. 2C λ i+1 In this way condition (i) of Theorem holds. By (3.3) and by the positivity of G it is readily seen that f(h i ) 0. Moreover, take u H i, t > 0 and consider f(u + te i+1 ) = 1 2 u 2 λ u 2 dx + t2 ( ) e i+1 1 λ e 2 i+1 dx G(x, u + te i+1 ) dx, 2 2 since u and e i+1 are orthogonal. By (3.3) and (g 4 ) we get f(u + te i+1 ) t2 2 (λ i+1 λ) a 3 t µ u t µ + e i+1 µ dx + a 4 1 so that condition (ii) of Theorem is satisfied with e = e i+1. We now prove that (P S) c holds c R. First we note that f(u) = u + 1 (λu + g(x, u)), t, so that we only need to prove that (P S) sequences are bounded, like in the proof of Theorem 2.1.4, since 1 is compact and the problem is subcritical. So, let (u n ) n be such that f(u n ) c and f(u n ) 0. Take k (2, µ) and note that (3.4) kf(u n ) f(u n ), u n M + N u n

31 24 for some M, N > 0. On the other hand kf(u ( n ) f(u ) n ), u ( n ) k k = 2 1 u n λ ( ) ( ) k k 2 1 u n λ u 2 n dx + [g(x, u n )u n kg(x, u n )] dx u 2 n dx + (µ k) G(x, u n ) dx C R for some constant C R 0, as already found in (2.8) in the proof of Theorem By (g 4 ) we finally get (3.5) kf(u ( n ) f(u ) n ), u ( n k k 2 1 u n ) λ u 2 n dx + (µ k)a 3 u n µ dx C for some constant C 0. By the Hölder and Young inequalities we get that for any u and for any ε > 0 where C ε as ε 0. Then by (3.5) we get u 2 2 C() u 2 µ ε u µ µ + C ε, (3.6) kf(u ( n ) f(u ) n ), u n k 2 1 u n 2 + [ (µ k)a 3 ε ( ) ] k 2 1 λ u n µ dx C ε, where C ε is independent on n. Let us choose ε < (µ k)a 3 ( k 2 1) λ, so that (3.6) gives kf(u n ) f(u n ), u n ( ) k 2 1 u n 2 C ε. Together with (3.4) this gives that (u n ) n is bounded, as claimed. Remark An existence theorem for the Dirichlet problem { u = h(x, u) in, u = 0 on, continues to hold if h is such that h(u) λu + g(x, u), where g is as above, and the proof is based on Theorem or Theorem according to the cases λ < λ 1 or λ λ 1, respectively. With the aim of Theorem one can prove the following multiplicity result. Theorem Assume (g 1 ) (g 4 ) of Section 2.1 and if R > 0 in (g 4 ), assume also that G(x, s) 0 s R and for a.e. x. Suppose that g(x, s) = g(x, s) s R and for a.e. x, and λ < λ 1. Then problem (P ) admits infinitely many couples (u n, u n ) n of solutions. The proof is left to the reader.

32 Chapter 4 The saddle point theorem As usual, though not stated explicitly, X is a Banach space and f C 1 (X, R). Definition A point u 0 X is called a saddle point if f (u 0 ) = 0 and for any neighborhood U of u 0, the function f f(u 0 ) is not definite positive either negative in U. Theorem (Saddle Point Theorem). Suppose that there exist two closed subspaces X 1 and X 2 of X with X 1 {0} and dimx 1 <. Moreover, assume that R > 0 such that ) (4.1) sup f (S(0, R) X 1 < inf f(x 2 ). Set and Γ = {h C (B R X 1, X) : h = Id on B R } β = inf sup f(h(u)) h Γ u B R X 1 and assume that (P S) β holds. Then β inf f(x 2 ) is a critical value for f. See [92] or [116] for a proof. Let us only remark that again the Palais Smale condition is essential. Indeed, the function (x, y) = arctan y x 2 defined in R 2 is of class C, f(0, y) π 2 y R, f(x, 0) as x, but it has no critical points, since f(x, y) = ( 2x, 1/(1 + y 2 )). 4.1 An application While the Mountain Pass Theorem and the Linking Theorem are useful in presence of superlinear problems, the natural way to face problems with linear growth is the Saddle Point Theorem. In this section we assume that is a bounded domain of R N, N 1 and g : R is a Catathéodory function such that there exist a L 2 () and b R such that (4.2) g(x, s) a(x) + b s s R and a.e. x. We have the following. Theorem Assume that g(x, s) g(x, s) lim inf := α(x) lim sup := α(x) s s s s 25

33 26 are measurable functions. If α(x) < λ 1 or there exists i N such that λ i < α(x) α(x) < λ i+1, then problem (Q) { u = g(x, u) + h in, u = 0 on has a solution for any h L 2 (). Remark Without loss of generality, we can assume h 0, since the function g(x, s) = g(x, s) + h(x) satisfies the same assumptions of g. Proof of Theorem We look for critical points of the C 1 functional f : H0 1 () R given as usual by f(u) = 1 Du 2 dx λ u 2 dx G(x, u) dx. 2 2 I: α(x) < λ 1. For ε > 0 there exists K > 0 such that Integrating we get g(x, s) s α(x) < ε s > K. (4.3) G(x, s) α(x) + ε (s 2 R 2 ) s > K, 2 while (4.2) implies (4.4) G(x, s) In this way we prove that (a(x)r + b2 R ) R s K. (4.5) lim sup s G(x, s s 2 α(x) 2. Indeed by (4.3) and (4.4) there exists C K > 0 such that so that R G(x, s) 0 s 2 = + s R g(x, σ) dσ s 2 < lim sup s for every ε > 0, and then (4.5) follows. Now let us show G(x, s s 2 C K + α(x) + ε (s 2 R 2 ) 2 s 2, α(x) + ε 2 f(u) (4.6) lim u u 2 > 0. Take (u n ) such that u n. Up to a subsequence we can assume that v n := u n u n converges to a function u weakly in H0 1(), strongly in L2 () and a.e. in. Moreover u 1. Then G(x, u n ) u n 2 a(x) u n + b u n 2 /2 u n 2 b 2 u2 in L 2 ()

34 27 an by the generalized Fatou Lemma G(x, u n ) (4.7) lim sup n u n 2 But dx lim sup n G(x, u n ) u n 2 dx. = {x : u n (x) is bounded} {x : u n (x) }, and G(x,u n) u n 2 0 in {x : u n (x) is bounded}, while in {x : u n (x) } lim sup n by (4.5). Therefore (4.7) gives G(x, u n ) u n 2 = lim sup n G(x, u n ) α(x) lim sup n u n 2 G(x, u n ) u 2 n 2 u2 u 2 n u n 2 α(x) 2 u2 (x) < λ 1 u 2 dx if u 0, 2 = 0 if u = 0, so that 1 f(u n ) lim n u n 2 > 2 λ 1 u 2 dx if u 0, 2 1 if u = 0. 2 By the Poincaré inequality and the fact that u 1, we get 1 f(u n ) lim n u n 2 > 2 1 Du 2 dx 0 if u 0, 2 1 if u = 0, 2 and (4.6) follows. Finally, let us note that the map u u 2 is lower semicontinuous in the weak topology of H0 1(), while the map u G(x, u) is continuous, so that f is lower semicontinuous and by the Weierstrass Theorem there exists a minimum of f on H0 1 (), which is clearly a solution of problem (Q). II: i N such that λ i < α(x) α(x) < λ i+1. As usual, let us set H i =Span(e 1,..., e i ) and H i = Span(e i+1,...), so that H 1 0 () = H 1 H i and dimh i = i <. Reasoning as above we can prove that lim inf s G(x, s) s 2 α(x) s 2, so that, by using (3.3) which implies f(u) lim sup u H i, u u 2 < 0, (4.8) lim f(u) =. u H i, u Moreover, in an analogous way one can prove lim inf u H i, u f(u) u 2 > 0.