Normalized solutions for a Schrödinger-Bopp-Podolsky system

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(1)Normalized solutions for a Schrödinger-Bopp-Podolsky system Danilo Gregorin Afonso. Dissertation presented to the Institute of Mathematics and Statistics of the University of São Paulo to obtain the title of Master of Sciences Program: Mathematics Advisor: Prof. Dr. Gaetano Siciliano. During the development of this work the author was nancially supported by CNPq.. São Paulo, January 2020.

(2) Normalized solutions for a Schrödinger-Bopp-Podolsky system. This version of the dissertation contains the corrections and modications suggested by the Judging Committee during the defense of the original version of the work, which took place on 14/02/2020. A copy of the original version is available at the Institute of Mathematics and Statistics of the University of São Paulo.. Judging Committee:. •. Prof. Dr. Gaetano Siciliano (advisor) - IME-USP. •. Prof. Dr. Antoine Laurain - IME-USP. •. Prof. Dr. Francisco Odair Vieira de Paiva - UFSCar.

(3) Acknowledgements. To Gaetano, for all he has taught me and for the patience. To my parents, Ana Laura and Waldemar, for the constant support. To my mom, again, for her patience. To Hanna, for her companionship. To Elvis, Sereno, Bia, Vitor, Carol, Garapa, Felipe, Giovanni and Lissa, for being there. To all my friends at the Institute, for all the coees and teas and discussions. To the Wikipedia and MathStackExchange communities, for obvious reasons. To CNPq for the nancial support.. i.

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(5) Resumo. GREGORIN AFONSO, D.. Bopp-Podolsky.. Soluções normalizadas para um sistema de Schrödinger-. 2020. Dissertação (Mestrado) - Instituto de Matemática e Estatística,. Universidade de São Paulo, São Paulo, 2020.. Objetivamos estudar um sistema de tipo Schrödinger-Bopp-Podolsky, que consiste de duas equações diferenciais parciais não lineares. Apresentamos um resultado original de existência e multiplicidade de soluções fracas para o problema, ou seja, de existência de pontos críticos de um funcional restrito a uma subvariedade de um espaço de Hilbert. É desenvoldida a teoria do cálculo em espaços de Banach. Discutimos a teoria do gênero de Krasnoselskii e apresentamos o Lema de Deformação e noções correlatas. Discutimos subvariedades em um espaço de Banach e multiplicadores de Lagrange. Provamos a existência e multiplicidade de soluções fracas para o problema proposto.. Palavras-chave: sistema de Schrödinger-Bopp-Podolsky, gênero de Krasnoselskii, Lema de Deformação, multiplicadores de Lagrange, soluções fracas.. iii.

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(7) Abstract. GREGORIN AFONSO, D.. Normalized solutions for a Schrödinger-Bopp-Podolsky. system. 2020. Dissertation (Masters) - Instituto de Matemática e Estatística, Universidade de São Paulo, São Paulo, 2020.. The aim is to study a Schrödinger-Bopp-Podolsky system of partial dierential equations. We present an original result for the existence and multiplicity of weak solutions to the problem, which consists in the determination of critical points for a functional constrained to a submanifold of a Hilbert space. The calculus in Banach spaces is developed. Krasnoselskii's genus theory is discussed, after which the Deformation Lemma and some related notions are presented. Submanifolds of Banach spaces and Lagrange multipliers are discussed. The existence and multiplicity of solutions to the proposed problem is proved.. Keywords: Schrödinger-Bopp-Podolsky system, Krasnoselskii genus, Deformation Lemma, Lagrange multipliers, weak solutions.. v.

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(9) Contents. List of Figures. ix. 1 Introduction. 1. 2 Calculus in Banach spaces. 3. 2.1. 2.2. Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3. 2.1.1. Basic notions. 3. 2.1.2. Higher order derivatives. . . . . . . . . . . . . . . . . . . . . . . . . .. 9. 2.1.3. Partial derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 12. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Local Inversion Theorems. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 13. 2.2.1. The Inverse Function Theorem. . . . . . . . . . . . . . . . . . . . . .. 13. 2.2.2. The Implicit Function Theorem. . . . . . . . . . . . . . . . . . . . . .. 18. 3 Genus Theory. 21. 3.1. Denitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 21. 3.2. Properties of the genus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 22. 4 Deformations. 25. 4.1. Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 25. 4.2. Preliminary denitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 27. 4.3. The Palais-Smale condition. . . . . . . . . . . . . . . . . . . . . . . . . . . .. 28. 4.4. Tangent pseudo-gradient vector elds . . . . . . . . . . . . . . . . . . . . . .. 28. 4.5. The Deformation Lemma and its consequences . . . . . . . . . . . . . . . . .. 32. 4.6. Generalizations. 36. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 5 Submersions, Manifolds, Lagrange Multipliers. 37. 5.1. Submersions and submanifolds . . . . . . . . . . . . . . . . . . . . . . . . . .. 37. 5.2. The Theorem of Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . .. 40. 6 Proof of the main result. 43. 6.1. An auxiliary problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 6.2. The manifold. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 45. 6.3. Existence of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 50. M. vii. 43.

(10) viii. CONTENTS. 6.4. Multiplicity of solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. A Useful theorems. 56. 61. A.1. Neumann boundary value problems . . . . . . . . . . . . . . . . . . . . . . .. 61. A.2. Sobolev spaces. 61. A.3. The theorem of Borsuk-Ulam. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 62. Bibliography. 63. Index. 65.

(11) List of Figures. 4.1. Graph of. 4.2. Graph of. 4.3. Graph of. J(x) = x3 − 3x . . . . . . . . . . J(x, y) = (x2 + y 2 )2 − 2(x2 + y 2 ) J(x) = xe1−x . . . . . . . . . . .. ix. . . . . . . . . . . . . . . . . . . .. 26. . . . . . . . . . . . . . . . . . . .. 26. . . . . . . . . . . . . . . . . . . .. 27.

(12) x. LIST OF FIGURES.

(13) Chapter 1 Introduction. The Schrödinger-Poisson equation consists of a nonlinear coupling of the Schrödinger equation with a gravitational potential of newtonian form, representing the interaction of a particle with its own gravitational eld. In 1998, Benci and Fortunato (1998) treated a similar problem, where the coupling was with Maxwell's equations and represented the interaction of the particle with its own electromagnetic eld. This gives rise to a coupled system of PDEs. In their paper the authors studied the problem with Dirichlet boundary conditions on the unknowns. u. and. φ. and. employed variational methods and critical point theory to develop a procedure that would become standard to treat other similar problems. Later, Pisani and Siciliano (2013) treated a Schrödinger-Poisson system with Neumann boundary conditions on the scalar eld. φ. and considering the case in which the interaction. factor responsible for the coupling of the equations (the. q. below) is non-constant. This gives. rise to important and interesting considerations regarding the geometry of the manifold of possible solutions. In this dissertation we treat a modication of the problem dealt with by Pisani and Siciliano consisting in the addition of a biharmonic term in the second equation and imposition of appropriate boundary conditions. This can be interpreted as a coupling of Schrödinger equation with Bopp-Podolsky electrodynamics (for more information on this matter, see d'Avenia and Siciliano (2019) and the references therein), although here we focus on the mathematical aspects of the problem. The aim is to study the following system of partial dierential equations in a connected, Ω ⊂ R3 :. bounded, smooth open set. −∆u + qφu − κ|u|p−2 u = ωu ∆2 φ − ∆φ = qu2 in Ω where. ω ∈ R.. h1 , h2. continuous. The symbol. wards. Since. Ω. (1.1) (1.2). We assume the following boundary conditions:. u=0 ∂φ = h1 ∂n ∂∆φ = h2 ∂n with. in. u. on. ∂Ω. (1.3). on. ∂Ω. (1.4). on. ∂Ω. (1.5). n denotes the unit vector normal to ∂Ω pointing out-. represents the amplitude of the wave function of a particle conned in. 1. Ω,. we.

(14) 2. 1.0. INTRODUCTION. assume the following normalizing condition:. Z. u2 = 1.. (1.6). Ω We also assume that the coupling factor. q. is continuous on. Ω:. q ∈ C(Ω).. (1.7). Our main theorem is the following:. Theorem 1.0.1.. Let. Z. Z h2 ds −. α := ∂Ω. h1 ds. ∂Ω. Assume that inf Ω q < α < supΩ q and that |q −1 (α)| = 0. Then there exists a solution (u, ω, φ) ∈ H01 (Ω) × R ×H 2 (Ω) such that u ≥ 0. Moreover, there exists innitely many solutions (un , ωn , φn ) ∈ H01 (Ω) × R ×H 2 (Ω) to the problem (1.1) - (1.6), with Z |∇un |2 dx → +∞. Ω The motivation for the denition of. α and the importance of the hypothesis will be made. clear in Chapter 6.. κ|u|p−2 u, so we make appropriate values of p. See. In our computations, nothing important is lost if we drop the term this simplication. One only would have to take account of the Remark 6.4.3.. X, Y and eventually Z will denote U ⊂ X will be an open set. The symbol L(X, Y ) denotes the set of operators T : X −→ Y . The value of T at x ∈ X will be denoted by T [x].. It is now time to x some notation. In what follows, Banach spaces. Usually, continuous linear The letter. c. denotes a constant whose value can change from line to line..

(15) Chapter 2 Calculus in Banach spaces. 2.1 Derivatives 2.1.1. Basic notions. Recall from calculus in nite dimension the meaning of derivative of a function at some given point. The idea was to obtain for the given function a linear approximation for the function in some neighborhood of the point. In one dimension we had the angular coecient of the tangent line; in. N. dimensions we had the matrix of the linear transformation. The. concept extends immediately to innitely many dimensions:. Denition 2.1.1. Let X, Y be Banach spaces and U ⊂ X be an open set. We say that f : U −→ Y is Fréchet dierentiable(dierentiable for short) at the point u ∈ U if there exists. Lu ∈ L(X, Y ). such that. f (u + h) − f (u) = Lu [h] + R(h), with. ||R(h)|| = 0, ||h||→0 ||h|| lim. that is,. R(h) = o(h)1 .. Remark 2.1.2. X. and. Y. The map. Lu. is the. dierential of f. at. u.. The denition of Fréchet dierentiablity depends only on the topology of. and not on the specic norms these spaces are endowed with.. This denition was rst introduced by Fréchet. 2. in 1911.. Of course, all generalizations must reduce to the original cases. It is easy to see that the denition above reduces to the nite dimensional case if the spaces N be R or R . We remark that. the dierential is unique.. X. and. Y. are taken to. This is an immediate consequence of the. following. Lemma 2.1.3. Suppose L1 , L2 ∈ L(X, Y ) are such that L1 [h]−L2 [h] = o(h). Then L1 = L2 . Proof.. By hypothesis, we have that. ||L1 [h] − L2 [h]|| ||h||→0 −−−−→ 0. ||h|| 1 This is known as little-o notation. 2 Maurice Fréchet . 3.

(16) 4. 2.1. CALCULUS IN BANACH SPACES. Now, suppose the Lemma to be false. Then there is some ∗ Now, letting h = th we have. h∗. L1 [h∗ ] − L2 [h∗ ] 6= 0.. such that. ||L1 [th∗ ] − L2 [th∗ ]|| t||L1 [h∗ ] − L2 [h∗ ]|| = lim 6= 0, t→0 t→0 ||th∗ || t||h∗ ||. lim a contradiction.. The uniqueness of the dierential compels us to adopt the following. Notation. u. Let. f : X −→ Y. be a map dierentiable at. u ∈ X.. Then the dierential of. f. at. is the continuous linear transformation denoted by. f 0 (u). Remark 2.1.4.. If. f. df (u).. or sometimes by. is dierentiable at. u∈U. then it is continuous at. u.. Indeed,. h→0. f (u + h) − f (u) = f 0 (u)[h] + o(h) −−→ 0.. Denition 2.1.5. If a function f is dierentiable at all u ∈ U , we say that it is dierentiable in. U.. The map. f0 :. U −→ L(X, Y ) u 7−→ f 0 (u). derivative of f . If f 0 is continuous, we say that f is of class C 1 . Remark 2.1.6. Note the dierence between the concepts of dierential and derivative. The is the. dierential is a linear approximation for the function in the neighborhood of a given point, while the derivative is a map that associates to each point a linear map. We now see some examples.. Example 2.1.7. Example 2.1.8.. =⇒ f 0 (u) = 0. f. is constant. If. f ∈ L(X, Y ). then. for all. f 0 (u) = f. u ∈ U.. for all. u ∈ U.. Indeed, in this case. f (u + h) − f (u) = f (h), from which the claim immediately follows.. Example 2.1.9. Let b : X × Y at any. (u, v) ∈ X × Y. and. −→ Z be a continuous bilinear form. Then b is dierentiable db(u, v) is the linear map (h, k) 7→ b(u, k) + b(h, v). Indeed, note that. b(u + h, v + k) − b(u, v) = b(u, k) + b(h, v) + b(h, k) but by continuity we have (see Brezis (2010),. pp.. 138). ||b(h, k)|| ≤ c||h|| ||k|| and of course. c||h|| ||k|| = 0, (h,k)→0 ||h|| + ||k|| lim.

(17) 2.1. DERIVATIVES. 5. from which the claim follows.. Example 2.1.10.. Let. b : X × X −→ R :. f. be a continuous bilinear form. Dene. X −→ R u 7−→ b(u, u). Then. f 0 (u) = b(u, · ) + b( · , u). b is symmetric then f 0 (u) = 2b(u, · ). What if X is a Hilbert b(u, v) = (u|v), the scalar product? Then f (u) = ||u||2 and f 0 (u) = 2(u| · ). Of course, if. Example 2.1.11. matrix with. Ω aij ∈ L (Ω) Let ∞. be a smooth open set in R ∞ and let q ∈ L (Ω). Then. 1 J(u) = 2. N. ,. Z. A = A(x) Z. denes a functional on the Sobolev space. H 1 (Ω)3. whose Fréchet derivative is given by. Z (A∇u) · ∇v dx + 2. J (u)[v] = Ω Indeed, note that. J. quv dx. Ω. is the functional induced by the following bilinear form on. 1 b(u, v) = 2. symmetric. Ω. Z. 0. N ×N. qu2 dx. (A∇u) · ∇u dx + Ω. be an. space and. Z. H01 (Ω) ×H01 (Ω):. Z (A∇u) · ∇v dx + 2. Ω. quv dx. Ω. This completes the proof. The usual rules of calculus remain true for the Fréchet derivative.. Proposition 2.1.12. Let f, g be dierentiable functions at u ∈ X and let a, b ∈ R. Then af + bg is dierentiable at u and (af + bg)0 (u) = af 0 (u) + bg 0 (u). Proof.. Note that. (af + bg)(u + h) − (af + bg)(u) − af 0 (u)[h] − bg 0 (u)[h] = o(h), from which the claim immediately follows.. Proposition 2.1.13 (Chain Rule). Let X, Y. and Z be Banach spaces, u ∈ X and f : X −→ : Y , g Y −→ Z such that f is dierentiable at u and g is dierentiable at v = f (u). Then. g ◦ f : X −→ Z is dierentiable at u and 3 See Brezis (2010).. (g ◦ f )0 (u) = g 0 (f (u))f 0 (u)..

(18) 6. 2.1. CALCULUS IN BANACH SPACES. Proof.. We have. f (u + h) = f (u) + f 0 (u)[h] + o(h), g(v + k) = g(v) + g 0 (v)[k] + o(k). Then. g(f (u + h)) − g(f (u)) = g(f (u) + f 0 (u)[h] + o(h)) − g(f (u)) = g 0 (f (u)) f 0 (u)[h] + o(h) + o(f 0 (u)[h] + o(h)) = g 0 (f (u)) f 0 (u)[h] + g 0 (f (u))[o(h)] + o(f 0 (u)[h] + o(h)). We must show that the two last terms are. o(h).. That. g 0 (f (u))[o(h)] h→0 −−→ 0 ||h|| is clear by linearity and continuity of. g 0 (f (u)).. Now, note that. o(f 0 (u)[h] + o(h)) ||o(f 0 (u)[h] + o(h))|| o(f 0 (u)[h] + o(h)) = . ||h|| ||h|| ||o(f 0 (u)[h] + o(h))|| But. . ||o(f 0 (u)[h] + o(h))|| h o(h). = f 0 (u) +. ||h|| ||h|| ||h|| ≤ ||f 0 (u)|| +. o(h) ||h||. ≤C near. h = 0.. Since. h→0. f 0 (u)[h] + o(h) −−→ 0,. we have that. ||o(f 0 (u)[h] + o(h))|| h→0 −−→ 0, ||h|| which completes the proof. Just like in the case of by Gateaux. 4. RN , there is a concept of directional derivative. It was introduced. in 1913.. Denition 2.1.14.. f : X −→ Y . Then f is said to be Gateaux-dierentiable (Gat u ∈ X if there exists A ∈ L(X, Y ) such that for all h ∈ X there. Let. dierentiable for short) exists. lim+. t→0 The map. f (u + th) − f (u) := A[h]. t. A. is uniquely determined (this follows by the argument in Lemma 2.1.3) and 0 will be denoted by fG (u). It is immediate that Fréchet dierentiability implies Gateaux dierentiability. However, the converse is not true: Gateaux dierentiability doesn't even imply continuity. Consider. 4 René Gateaux ..

(19) 2.1. DERIVATIVES. the following example in. 7. R2 : x2 y x4 + y 2. F (x, y) =. !2 y 6= 0,. ,. F (x, 0) = 0. Notice that. √ F (( y, y)) = 1/2. but. F (0, 0) = 0.. We now prove the so-called Mean-Value Inequality. It plays a fundamental role in what follows.. Notation.. Given. u, v ∈ U. the segment between them is denoted by. . [u, v] := tu + (1 − t)v : t ∈ [0, 1] .. Theorem 2.1.15 (Mean Value Inequality). Let f : U −→ Y. be G-dierentiable at any point of the open set U ⊂ X . Given u, v such that [u, v] ⊂ U it holds that. ||f (u) − f (v)|| ≤ sup ||fG0 (w)|| ||u − v||.. (2.1). w∈[u,v]. Proof.. If. f (u) = f (v). the claim is trivial. Suppose then 5 ∗ Hahn-Banach theorem there exists ψ ∈ Y such that. f (u) 6= f (v).. By a corollary of the. ||ψ|| = 1 and. ψ f (u) − f (v) = ||f (u) − f (v)||. For. t ∈ [0, 1]. let. γ(t) = tu + (1 − t)v and consider the map. h : [0, 1] −→ R. dened by. h(t) = ψ f (γ(t) . Notice that. γ(t + τ ) = γ(t) + τ (u − v). From this it follows that. h(t + τ ) − h(t) =ψ τ Passing to the limit as. τ →0. . f (γ(t) + τ (u − v)) − f (γ(t)) τ. . we have. h0 (t) = ψ fG0 (γ(t))[u − v] . Now, by the classic Mean-Value Theorem we have that. 5 See Theorem 12.2 in Bachman and Narici (2000). h(1)−h(0) = h0 (θ) for some θ ∈ (0, 1)..

(20) 8. 2.1. CALCULUS IN BANACH SPACES. Now, note that. ||f (u) − f (v)|| = h(1) − h(0) = h0 (θ) = ψ fG0 (θu + (1 − θv))[u − v] ≤ ||ψ|| ||fG0 (θu + (1 − θ)v)|| ||u − v|| But recall that. ||ψ|| = 1. and that. θu + (1 − θ)v ∈ [u, v].. This completes the proof.. There is a well-known result in classic Calculus that says that if a function has continuous partial derivatives at a point then it is dierentiable at the given point. We have an analogous relation between the Gateaux and Fréchet derivatives.. Proposition 2.1.16.. Let f : U −→ Y be a G-dierentiable function in U and suppose fG0 to be continuous at u ∈ U . Then f is dierentiable at u and f 0 (u) = fG0 (u). Proof.. The proposition gives us a hint to the proof: it suces to show that. R(h) = f (u + h) − f (u) − fG0 (u)[h] is. o(h). Note that. R(h). is G-dierentiable for. h. small enough:. 1 R(h + tk) − R(h) = f (u + h + tk) − fG0 (u)[h + tk] − f (u + h) + fG0 (u)[h] t t f (u + h + tk) − f (u + h) − fG0 [k] = t → fG0 (u + h)[k] − fG0 (u)[k]. (h must be small enough so that Note that. R(0) = 0.. u + h ∈ U ).. We can then apply the Mean Value Inequality on the segment. [0, h]. to obtain. 0 ||R(h)|| ≤ sup ||RG (th)|| ||h||. 0≤t≤1. But, as we have seen,. 0 RG (h) = fG0 (u + h) − fG0 (u).. Thus. ||R(h)|| h→0 ≤ sup ||fG0 (u + th) − fG0 (u)|| −−→ 0, ||h|| 0≤t≤1 since. fG0. is continuous. This completes the proof.. Remark 2.1.17.. Computing Fréchet derivatives may be dicult, but Gateaux derivatives. are more treatable. In view of this, Proposition 2.1.16 provides a method to nd Fréchet derivatives: one computes the Gateaux derivative and then show that it is continuous. The next example will be useful afterwards.. Example 2.1.18.. Let. Ω. be a smooth, bounded open set and. u ∈ L6 (Ω) 7→ qu2 ∈ L6/5 (Ω). q ∈ C(Ω).. The map.

(21) 2.1. DERIVATIVES. is of class. C 1.. Indeed:. lim+. t→0 Hence 6. fG0 (u)[h] = uh.. L (Ω).. 9. f (u + th) − f (u) = lim+ uh + th2 = uh. t→0 t. Let us show that. fG0. is continuous with respect to. u.. Let. un → u. in. Then. ||fG0 (u) − fG0 (un )|| = sup ||(u − un )h||6/5 ≤ ||u − un ||6/4 → 0 ||h||6 =1. by Hölder's inequality, which completes the proof. We close this section with an example that will be important in Section 2.1.3.. Example 2.1.19.. Let. f ∈ C([a, b], Y ). and let. 6. t. Z. f (ξ) dξ.. F (t) = a Then. F. is dierentiable and. F 0 (t) = f (t).. Indeed, we have. 1 F (t + h) − F (t) = h h But. t+h. Z. f (ξ) dξ. t. t+h. Z. f (t) dξ = hf (t). t Then we have. F (t + h) − F (t) 1 − f (t) = h h 1 = h. "Z. t+h. Z f (ξ) dξ −. f (t) dξ t. t. Z. #. t+h. t+h. (f (ξ) − f (t)) dξ. t. Hence,. F (t + h) − F (t). 1. − f (t) ≤ |h| sup ||f (ξ) − f (t)|| → 0.. h |h| [t,t+h] which completes the proof.. 2.1.2. Higher order derivatives. Denition 2.1.20.. Let. f ∈ C(U, Y ). (where, as always,. U ⊂ X. is an open subset) be a. dierentiable function. If the map. f 0 : U −→ L(X, Y ) is dierentiable at dierential of. f. at. u∈U u is. then. f. is said to be twice Fréchet dierentiable at. u.. The second. f 00 (u) = (f 0 )0 (u). 6 The theory of integration works just like for functions into. RN .. See Lang (1997) for details..

(22) 10. 2.1. CALCULUS IN BANACH SPACES. Higher order dierentials are dened in an analogous manner:. f (k) (u) = (f (k−1) )0 (u). Note that. f 00 (u) ∈ L(X, L(X, Y )). f0. L(X, Y ). 00 It is useful, however, to see f (u) as a bilinear map. Recall that we can identify L(X, L(X, Y )) with L2 (X, Y ), the space of continuous bilinear forms from X × X to Y . Indeed, given A ∈ L(X, L(X, Y )) let ΦA be the bilinear form Indeed,. takes values in. ΦA (u1 , u2 ) = [A(u1 )](u2 ). ΦA. Of course. L2 (X, Y ). is continuous, since both. A. and. A(u1 ). are continuous. Conversely, any. Φ∈. denes a map. φ :. X −→ L(X, Y ) . h 7−→ Φ(h, · ). Note that we have just dened an isomorphism between. L(X, L(X, Y )) and L2 (X, Y ). More-. over,. ||φ||L(X,L(X,Y )) = sup ||Φ(h)||L(X,Y ) ||h||≤1. = sup sup ||Φ(h, k)|| ||h||≤1 ||k||≤1. = ||Φ||L2 (X,Y ) and thus we actually have an isometry. 00 From now on, f (u) will be viewed as the bilinear form just dened. Its value at a pair. (h, k). will be denoted. f 00 (u)[h, k].. Denition 2.1.21. L2 (X, Y ). If. f. is twice dierentiable at every. is the second derivative of. f.. u ∈ U. we say that. f 00 : U −→. Higher order derivatives are dened in an analogous. manner.. Proposition 2.1.22. Proof.. Let. ϕ ∈ Y ∗.. Let f ∈ C 2 (U, Y ). Then f 00 (u) is symmetric for all u ∈ U .. Dene. g : (t, s) ∈ R2 7→ ϕ(f (u + tv + sw)) ∈ R . By the Chain Rule we have that. ∂g (t, s) = ϕ(f 0 (u + tv + sw)[v]), ∂t ∂ ∂g (t, s) = ϕ(f 00 (u + tv + sw)[v, w]). ∂s ∂t Analogously,. ∂ ∂t. . ∂g ∂s. . (t, s) = ϕ(f 00 (u + tv + sw)[w, v])..

(23) 2.1. 11. DERIVATIVES. Now, since. g. C2. is of class. (it is a composition of. C2. maps), it follows that. ϕ(f 00 (u + tv + sw)[v, w]) = ϕ(f 00 (u + tv + sw)[w, v]). The equality holds for all. ϕ ∈ Y ∗,. thus. f 00 (u + tv + sw)[v, w] = f 00 (u + tv + sw)[w, v] t, s.. for all. In particular, for. t=s=0. we get. f 00 (u)[v, w] = f 00 (u)[w, v], and the proof is complete. An analogous theorem holds for higher order derivatives. See Lang (1997) for details. Just like in classic Calculus, second derivatives are very useful for the study of extrema of functionals. For an extensive discussion, see Blanchard and Brüning (1992). (k) : If f U −→ Y is a continuous map we say that f ∈ C k (U, Y ). We can now generalize Taylor's formula.. Theorem 2.1.23. (Taylor's formula). .. open, convex neighborhood of u. Then. Let f : U −→ Y be a map of class C n dened in an. f (u + h) = f (u) + with. Z. 1. Rn = 0. n−1 X 1 (j) f (u)[h]j + Rn j! j=1. (2.2). (1 − t)n−1 (n) f (u + th)hn dt. (n − 1)!. Here, [h]j = [h, . . . , h] j times. Proof.. For. t ∈ [0, 1]. and. u, h ∈ U. let. φ :. Then. φ. is of class. Cn. γ(t) = u + th. and. [0, 1] −→ Y . t 7−→ f (γ(t)). and. φ0 (t) = f 0 (u + th)[h], φ00 (t) = f 00 (u + th)[h]2 , . . .. φ(n) (t) = f (n) (u + th)[h]n . But then. 1 1 φ(n) (0) + φ(1) = φ(0) + φ (0) + . . . + (n − 1)! (n − 1)! 0. Z. 1. (1 − t)n−1 φ(n) (t) dt,. 0. and hence the formula follows. For more details on higher order derivatives, see Lang (1997) and Ambrosetti and Prodi (1993)..

(24) 12. 2.1. CALCULUS IN BANACH SPACES. 2.1.3. Partial derivatives. The formulation of partial derivatives is as simple as in classic Calculus. Consider a product of Banach spaces:. X = X1 × . . . × Xn and let. Ui ⊂ Xi. be open subset. Denote. U = U1 × . . . × Un. and let. f : U −→ Y be a map. We can write elements. Denition 2.1.24.. Let. u∈U. in terms of their coordinates:. u1 , . . . , ui−1 , ui+1 , . . . un. u = (u1 , . . . , un ).. be xed. If the map. ui 7→ f (u1 , . . . , un ) U then its derivative is the partial derivative of f u ∈ U is denoted by Di f (u) or fu0 i (u).. is dierentiable in and its value at. Note that for each. u∈U. in the i-th coordinate,. we have. Di f (u) ∈ L(Xi , Y ). Just as in Calculus in nite dimension, smoothness is equivalent to smoothness of the partial derivatives:. Proposition 2.1.25.. Let Ui ⊂ Xi be open sets and let f : U = ΠN 1 Ui −→ Y . Then f is of class C if and only if each partial derivative k. Di f : U −→ L(Xi , Y ) is of class C k−1 . If this is the case, then 0. f (u)[h] =. n X. Di f (u)[hi ]. 1. for h = (h1 , . . . , hn ) ∈ Πn1 Xi Proof.. n = 2. For greater n, the argument is exactly the same. derivatives exist and are continuous. Let (u, v) ∈ U and h =. It suces to write a proof for. First, suppose the partial. (h1 , h2 ).. Then. f (u + h1 , v + h2 ) − f (u, v) = f (u + h1 , v + h2 ) − f (u + h1 , v) + f (u + h1 , v) − f (u, v) Z 1 Z 1 = D2 f (u + h1 , v + th2 )[h2 ] dt + D1 f (u + th1 , v)[h1 ] dt. 0. 0. Let. ψ(h1 , th2 ) := D2 f (u + h1 , v + th2 ) − D2 f (u, v).

(25) 2.2. LOCAL INVERSION THEOREMS. 13. The rst integral can be written as. 1. Z. 1. Z. Z. 1. ψ(h1 , th2 )[h2 ] dt D2 f (u, v)[h2 ] dt + 0 0 Z 1 ψ(h1 , th2 )[h2 ] dt. = D2 f (u, v)[h2 ] +. D2 f (u + h1 , v + th2 )[h2 ] dt = 0. 0 Now, note that.

(26) Z

(27)

(28) 1

(29)

(30)

(31) ψ(h1 , th2 )[h2 ] dt

(32) ≤ sup ||ψ(h1 , th2 )|| ||h2 ||

(33)

(34) 0

(35) t∈[0,1] ≤ ||h|| sup ||ψ(h1 , th2 )|| t∈[0,1]. = o(h) since, by continuity,. h→0. ψ(h1 , th2 ) = D2 f (u + h1 , v + th2 ) − D2 f (u, v) −−→ 0. Now let. φ(th1 , h2 ) = D1 f (u + th1 , v) − D1 f (u, v). We have. Z. 1. Z. 1. Z. 1. φ(th1 , h2 ) dt D1 f (u, v)[h1 ] dt + 0 0 Z = D1 f (u, v)[h1 ] + φ(th1 , h2 ) dt.. D1 f (u + th1 , v)[h1 ] dt = 0. Estimating the error as we have just done we conclude that. f (u + h1 , v + h2 ) − f (u, v) = D1 f (u, v)[h1 ] + D2 f (u, v)[h2 ] + o(h), f. hence. is dierentiable and the derivative is given as above.. To prove the converse statement one must only evaluate the derivative along the directions. (h1 , 0). and. (0, h2 ).. 2.2 Local Inversion Theorems One of the most important theorems in classic Calculus is the Inverse Function Theorem, together with its cousin Implicit Function Theorem. In this chapter we will see how they generalize to innite-dimensional spaces. We follow Ambrosetti and Prodi (1993) and Lang (1995).. 2.2.1. The Inverse Function Theorem. Denition 2.2.1. Let X and Y be Banach spaces. invertible if there exists A−1 ∈ L(Y, X) such that A−1 ◦ A = IX ,. A map. A ◦ A−1 = IY .. A ∈ L(X, Y ). is said to be.

(36) 14. 2.2. CALCULUS IN BANACH SPACES. We set. Inv(X, Y ) := A ∈ L(X, Y ) : A. is invertible. .. In order to prove our next proposition we need a preliminary lemma:. Lemma 2.2.2. Let A ∈ L(X, X) such that ||A|| < 1. Then I −A is invertible and its inverse. is given by the convergent series. (I − A)−1 =. ∞ X. An. 0. Proof.. Note that. ||AA|| ≤ ||A|| ||A||,. thus. ||An || ≤ ||A||n .. Then. ||I + A + . . . + An || ≤ 1 + ||A|| + . . . + ||A||n which converges, since. ||A|| < 1 (it is the geometric series). Thus the series really converges.. Now, note that. (I − A)(I + A + . . . + An ) = I − An+1 = (I + A + . . . + An )(I − A) and that. An → 0.. Taking the limit completes the proof.. The above lemma shows that there is a ball of radius one around the identity of invertible operators. Indeed, if. ||I − A|| < 1. Proposition 2.2.3.. If A ∈ Inv(X, X) then every T ∈ L(X, X) such that. then. A = I − (I − A). ||T − A|| <. is invertible.. 1 ||A−1 ||. is invertible. Hence, Inv(X, X) is a open subset of L(X, X). Proof.. Note that. ||T A−1 − I|| = ||(T − A)A−1 || ≤ ||A−1 || ||T − A|| ≤ 1. Thus. T A−1. is invertible. Hence,. Proposition 2.2.4.. T = T A−1 A. is invertible.. The map. ϕ : A ∈ Inv(X, X) 7→ A−1 ∈ L(X, X) is innitely dierentiable, with. ϕ0 (A)[B] = −A−1 ◦ B ◦ A−1 . Proof.. For. B. suciently small we have that. (A + B)−1 − A−1 = (A(I + A−1 B))−1 − A−1 = [(I + A−1 B)−1 − I]A−1 . By the above lemma, if. B. is so small that. ||A−1 B|| < 1. then. (I + A−1 B)−1 = I − A−1 B + (A−1 B)2 g(B).

(37) 2.2. 15. LOCAL INVERSION THEOREMS. where. g. is a convergent power series. Hence. (A + B)−1 − A−1 = −A−1 BA−1 + (A−1 B)2 g(B)A−1 . Now, notice that the series is bounded, because it is convergent. Then the second term above can be estimated:. ||(A−1 B)2 g(B)A−1 || ≤ C||B||2 . Thus. ϕ0 (A)[B] = −A−1 ◦ B ◦ A−1 . That. ϕ. is innitely dierentiable follows from the fact that. ϕ0. is a composition of inverses. and continuous bilinear maps (the composition of linear operators).. Remark 2.2.5.. By Proposition 2.2.3 the set of invertible linear maps is open, and thus the. derivative above really makes sense. Above we dened the set of invertible. linear. maps. Let us now treat general maps.. Denition 2.2.6. Let f ∈ C(X, Y ) be a continuous map and let u ∈ X . Then f is said to be locally invertible at u if there exist neighborhoods U of u and V of f (u) and a map f −1 : V −→ U. such that. f ◦ f −1 = IV , We say that. f −1 ◦ f = IU .. f ∈ Hom(U, V ).. Proposition 2.2.7. The following properties hold: (i) (Transitivity) Let f ∈ C(X, Y ) be locally invertible at u ∈ X. invertible at v = f (u). Then g ◦ f is locally invertible at u.. (ii). and g ∈ C(Y, Z) be locally. (Stability) Let f be locally invertible at u. Then f is locally invertible at any point in some neighborhood of u.. Proof.. (i). We have. f ∈ Hom(U, V 0 ). and. g ∈ Hom(V 00 , W ).. Take. V = V 0 ∩ V 00 .. Then. g ◦ f : f −1 (V ) −→ g(V ) is well dened and is clearly a bijection, thus is an invertible map.. (ii). f ∈ Hom(U, V ). Let u0 ∈ U and take some ball B around u0 contained f |B : B −→ f (B) is a bijection, thus f is locally invertible at u0 .. in. U.. is dierentiable at. u0. We have Then. be a locally invertible map at u0 ∈ U and suppose that −1 and that f is dierentiable at v0 = f (u0 ). We have that Let. f. f −1 ◦ f (u) = u ∀u ∈ U,. f ◦ f −1 (v) = v. f. ∀v ∈ V.. Dierentiating and applying the chain rule we get. (f −1 )0 (v0 ) ◦ f 0 (u0 ) = IX , Hence,. f 0 (u0 ) ∈ Inv(X, Y ). and. f 0 (u0 ) ◦ (f −1 )0 (v0 ) = IY .. (f 0 (u0 ))−1 = (f −1 )0 (v0 ).. We have just proved that invertibility of maps implies the invertibility of derivatives. Now, derivatives are linear maps, for which it is much easier to study invertibility. It would.

(38) 16. 2.2. CALCULUS IN BANACH SPACES. then be interesting to try and obtain information about the invertibility of maps from the invertibility of derivatives. This is our next theorem.. Theorem 2.2.8. .. Let f ∈ C 1 (X, Y ) and suppose f 0 (u0 ) ∈ Inv(X, Y ). Then f is locally invertible at u0 with C 1 inverse. In other words, there exist neighborhoods U of u0 and V of v0 = f (u0 ) such that. (i). (Inverse Function Theorem). f ∈ Hom(U, V ).. (ii). f −1 ∈ C 1 (V, X) and. (iii). (f −1 )0 (v0 ) = f 0 (u0 )−1. If f ∈ C k (X, Y ) then f −1 ∈ C k (V, X).. Proof.. We begin by making some simplications to our problem.. A ◦ f , where A is any invertible linear map, then the theorem −1 know that A is locally invertible. Then, by the transitivity. If we prove the theorem for is proved for. f.. Indeed, we. property, the map. A−1 ◦ [A ◦ f ] = f 0 −1 is locally invertible. Choosing A = [f (0)] we reduce the problem to proving the theorem 0 −1 for F : X −→ X, F (u) = [f (0)] ◦ f (u).. u0 = 0. We consider the case where. and. v0 = F (0) = 0.. The general case follows imme-. diately from this one by considering compositions with translations (which are, of course, locally invertible).. g(u) = u − F (u). Note that g ∈ C 1 (X, X) ball of radius r > 0 such that for p ∈ Br (0). Let some. ||g 0 (p)|| <. and that. g 0 (0) = 0.. By continuity, there is. we have. 1 . 2. Now, by the Mean Value Inequality (Theorem 2.1.15) we have that. ||g(p) − g(q)|| ≤ sup ||g 0 (w)|| ||p − q|| w∈[p,q]. ≤ for all. p, q ∈ Br (0).. Hence,. g. is a contraction on. ||g(p)|| ≤ For. v ∈ X,. 1 ||p − q|| 2 Br (0).. In particular,. 1 ||p|| ∀p ∈ Br (0). 2. dene. φv (u) = v − g(u). Note that. φv. is a contraction in. Br (0).. Moreover, if for. u ∈ Br (0). and. v ∈ Br/2 (0). we have. that. ||φv (u)|| ≤ ||v|| + ||g(u)|| ≤ r Then, if we choose. v ∈ Br/2 (0). then. φv. Br (0) into itself. u ∈ Br (0) such that. is a contraction that maps. by the Banach Fixed Point Theorem it has a unique xed point. u = v − g(u).. Then,.

(39) 2.2. LOCAL INVERSION THEOREMS. 17. Thereof,. v = F (u). F. Hence. is bijective from. Br (0). into. Br/2 (0). and thus we can dene an inverse. F −1 : Br/2 (0) −→ Br (0). F −1. To show that. is continuous let. u = F −1 (v). and. w = F −1 (z),. that is,. u + g(u) = v w + g(w) = z Then. ||u − w|| ≤ ||v − z|| + ||g(u) − g(w)|| 1 ≤ ||v − z|| + ||u − w|| 2 and thus. ||F −1 (v) − F −1 (z)|| ≤ 2||v − z||. We have thus proved that. F |U ∈ Hom(U, V ) where. V = Br/2 (0). For at. u,. and. (ii), suppose r. (i). U = Br (0) ∩ F −1 (V ).. Hence is proved. 0 small enough such that F (u) is invertible. Since. F. is dierentiable. we may write. F (u) − F (w) = F 0 (u)(u − w) + ||u − w||ψ(u − w) where. ψ. is some function such that. lim ψ(u − w) = 0.. w→u Now, note that. ||F −1 (v) − F −1 (z) − F 0 (u)−1 (v − z)|| = ||u − w − F 0 (u)−1 (F (u) − F (w))|| = ||u − w − u + w + F 0 (u)−1 ||u − w||ψ(u − w)|| ≤ C||v − z|| ||ψ(F −1 (v) − F −1 (z))|| → 0. Hence,. F −1. is dierentiable and. (F −1 )0 (v) = F 0 (u)−1 . To see that. (F −1 )0 : v 7→ F 0 (F −1 (v))−1. is continuous, note that it is a composition of. continuous maps:. F −1. F0. ϕ. v −−→ u −→ F 0 (u) − → (F 0 (u))−1 . Item that. (iii) follows by induction, using the same arguments presented above and noting. ϕ ∈ C ∞.. Remark 2.2.9.. It can be show that the continuity of the derivative cannot be dropped in.

(40) 18. 2.2. CALCULUS IN BANACH SPACES. the theorem. Indeed, consider the following elementary counterexample:. ( x + 2x2 sin(1/x), f (x) = 0, x=0. x 6= 0. 0 0 We have that f (x) = 1 + 4x sin(x) − 2 cos(1/x), so the limit limx→0 f (x) does not exist, 0 although f (0) = 1. Hence it is discontinuous at x = 0. Moreover, it oscillates up and down, so it is not invertible around the origin.. 2.2.2. The Implicit Function Theorem. Suppose we know a given point. (u0 , v0 ). is a solution to the equation. f (u, v) = 0. It may be of interest to nd a path of solutions near the given point, that is, to nd a. u 7→ v(u) such that v(u0 ) = v0 and f (u, v(u)) = 0. In other words, we want to nd a v = v(u) implicitly dened by f . But does such a function always exist? If so, what smoothness properties does it have? Does it depend on f ? How so?. map. function. The Implicit Function Theorem answers all these questions.. Theorem 2.2.10. (Implicit Function Theorem). Let U ⊂ X × Y be an open neighborhood of (u0 , v0 ) and f : U −→ Z , where X, Y, Z are Banach spaces. Suppose that. f (u0 , v0 ) = 0. Also, assume that fv (u0 , v0 ) exists and is bijective and that both f and fv are continuous at (u0 , v0 ). Then. (i). There are r0 , r > 0 such that for every u ∈ Br0 (u0 ) there is exactly one v(u) ∈ Br (v0 ) ⊂ Y and f (u, v(u)) = 0.. (ii). The sequence. vn+1 = vn − fv0 (u0 , v0 )−1 f (u, vn (u)). converges to v(u) for u ∈ Br0 (u0 ).. (iii). If f is continuous in a neighborhood of (u0 , v0 ) then v( · ) is continuous in a neighborhood of u0 .. (iv). If f is of class C k then v( · ) is of class C k .. Proof.. Without loss of generality, we can suppose. (u0 , v0 ) = 0. (it suces to compose with. translations). Set. g(u, v) := fv0 (0, 0)[v] − f (u, v). Then the equation. f (u, v) = 0. is equivalent to. v = fv0 (0, 0)−1 [g(u, v)] =: Tu (v) Here. u. is an index to the function. T,. not a partial derivative. We must show that there. exists a unique solution to this equation..

(41) 2.2. LOCAL INVERSION THEOREMS. Let. ||u||, ||v||, ||w|| ≤ r.. 19. Note that. gv0 (u, v) = fv0 (0, 0) − fv0 (u, v), hence. gv0. is continuous at. (0, 0). (because. fv0. is) and. gv0 (0, 0) = 0.. Then Taylor's Theorem. implies that. r→0. ||g(u, v) − g(u, w)|| ≤ sup ||gv0 (u, t(v − w)|| ||v − w|| −−→ 0. 0<t<1. Moreover, since. g. is continuous at. (0, 0). and. g(0, 0) = 0, r→0. ||g(u, v)|| −−→ 0. Let. . M = v ∈ Y : ||v|| ≤ r .. Then if. ||Tu v|| ≤ r,. are small enough we have that. ||Tu (v) − Tu (w)|| ≤. 1 ||v − w|| 2. u ∈ X such that ||u|| ≤ r0 . It then follows that Tu : M −→ M is a contraction. Thus for each xed u there is a unique xed point, v = v(u), that is the solution to the equation f (u, v) = 0. This proves. for all. v, w ∈ M. r, r0 > 0. (i).. Item. (ii). and xed. follows from the construction of the xed point in the proof of the Banach. Fixed Point Theorem. The continuity of. v. follows from the continuity of the xed point on a parameter (see. Zeidler (1986), Proposition 1.2). This proves It remains to prove. (iii).. (iv). First, let k = 1. We will compute the Gateaux derivative of v. and show that it is continuous. From the existence and continuity of. v(u). we have that. f (u + th, v(u + th)) = f (u, v). But since. f. is dierentiable we have. 0 = f (u + th, v(u + th)) − f (u, v(u)) = fu0 (u, v(u))[th] + fv0 (u, v(u))[v(u + th) − v(u)] + o(t). Then. v(u + th) − v(u) = −tfv0 (u, v(u))−1 fu0 (u, v(u))[h] + o(t) and hence the Gateaux derivative of. v. exists:. 0 vG (u)[h] = −fv0 (u, v(u))−1 fu0 (u, v(u))[h]. fv0 (u, v(u)) is invertible in a neighborhood of the origin by Proposition 2.2.3. Since f is of class C 1 then both fu0 and fv0 are continuous, then vG ! is continuous. Thus follows 1 that v is of class C . Let k = 2. Note that Note that. G(u) = fv0 (u, v(u))v 0 (u)[h] + fu0 (u, v(u))[h] = 0 for all. x. in a neighborhood of the origin. Therefore. G(u + tk) = 0 for small t. Repeating the v exists and is continuous.. argument above we can show that the Gateaux derivative of.

(42) 20. 2.2. CALCULUS IN BANACH SPACES. For higher k , the argument is analogous. The key fact in the argument is the existence −1 of fv (u, u(v)) . The proof is now complete.. Remark 2.2.11.. Note that the proof above gives us an expression for. v 0 (u) = −fv0 (u, v(u))−1 fu0 (u, v(u)).. v 0 (u): (2.3).

(43) Chapter 3 Genus Theory. In this chapter we study a topological tool that is of fundamental importance in the proof of our main theorem: the Krasnoselskii genus. It was rst introduced, as the reader may guess, by Krasnoselskii in the study of nonlinear integral equations. The original denition can be found, for example, in Krasnoselskii (1964). We use an equivalent denition by Coman (1969). The proof of the equivalence can be found in Rabinowitz (1973). For the properties of the genus we follow Kavian (1993). Another interesting reference is Ambrosetti and Malchiodi (2007), which presents some properties of the genus and a general framework for the application of the theory to the existence and multiplicity of critical points of even unbounded functionals.. 3.1 Denitions We begin by dening the domain of genus theory.. Denition 3.1.1.. Let. symmetric subsets of. X. X. be a Banach space. Then. s(X). denotes the family of closed,. that do not contain the origin:. o n /A s(X) := A ⊂ X : A = A, A = −A, 0 ∈ We now dene the genus of a set in. s(X).. Denition 3.1.2. For A ∈ s(X), the genus of A is the least integer n such that there exist a continuous, odd function between. A. and. Rn \{0}:. γ(A) = inf n ≥ 1 : ∃ϕ : A −→ Rn \{0} For convenience, we set. γ(∅) = 0.. continuous and odd. If no such function exists, we set. .. γ(A) = ∞.. Example 3.1.3. Let x0 ∈ X and R < ||x0 ||. Set A = BR (x0 )∪BR (−x0 ). Of course A ∈ s(X). γ(A) = 1. Indeed, ϕ(x) = −1 otherwise.. Then we can compute. x ∈ BR (x0 ). and. it suces to take. We now present an important class of sets of genus. Example 3.1.4. Then. Let. X = RN. and let. Ω. ϕ : A −→ R. with. ϕ(x) = 1. if. N.. be an open, symmetric set containing the origin.. γ(∂Ω) = N .. It is immediately seen that. ∂Ω ∈ s(RN ).. Of course we have that γ(∂Ω) ≤ N , since ∂Ω onto RN \{0}. Now, suppose. the identity is a trivial odd, continuous function from. 21.

(44) 22. 3.2. GENUS THEORY. γ(∂Ω) ≤ N − 1.. Then there exists and odd, continuous function. ϕ : ∂Ω −→ RN −1 \{0}.. Then, by the theorem of Borsuk-Ulam (see Kavian (1993), or the Appendix) there is some. x ∈ ∂Ω. such that. ϕ(x) = 0,. a contradiction. Hence. γ(∂Ω) = N .. 3.2 Properties of the genus We now present some of the most useful properties of the genus.. Theorem 3.2.1. Let X be a Banach space and let A, B ∈ s(X). (i) If there exist and odd, continuous function f : A −→ B then γ(A) ≤ γ(B). (ii). If A ⊂ B then γ(A) ≤ γ(B).. (iii). If there is and odd homeomorphism f : A −→ B then γ(A) = γ(B).. (iv). γ(A ∪ B) ≤ γ(A) + γ(B).. (v). If A is compact then γ(A) < ∞.. (vi). If A is compact, it has a closed neighborhood of the same genus. More precisely, there exists an ε > 0 such that if Aε = x ∈ X : d(x, A) ≤ ε then γ(Aε ) = γ(A).. (vii). If γ(B) < ∞ then. γ(A \ B) ≥ γ(A) − γ(B).. Proof.. (i). If. γ(B) = ∞. there is nothing to show. Assume then that γ(B) = n < ∞. Then ϕ : B −→ Rn \{0}. Now, the composition. there exists and odd, continuous function. ϕ ◦ f : A −→ Rn \{0} is an odd, continuous function. Hence. (ii). γ(A) ≤ n.. The canonical inclusion is an odd, continuous function from follows from. (i).. (iii). By. (iv). If the genus of either. A. to. B,. hence the result. (i) we have that γ(A) ≤ γ(B) and also the converse inequality, hence γ(A) = γ(B).. A or B is ∞, then there is nothing to prove. Suppose then γ(A) = m < ∞, γ(B) = n < ∞. Then there exist two odd, continuous functions. that. ϕ : A −→ Rm \{0} and. ψ : B −→ Rn \{0}. By Tietze's theorem we can extend both functions to odd, continuous functions dened in. X: ϕ˜ : X −→ Rm. odd, continuous such that. ϕ| ˜A=ϕ. and. ψ˜ : X −→ Rn. odd, continuous such that. ˜ B = ψ. ψ|. Now, set. f. :. A ∪ B −→ Rm+n \{0} . ˜ x 7−→ (ϕ(x), ˜ ψ(x)).

(45) 3.2. PROPERTIES OF THE GENUS. 23. f is an odd, continuous function and that indeed f (x) 6= 0 for all x ∈ A ∪ B . γ(A ∪ B) ≤ m + n.. Note that Hence. (v). Let. x ∈ A.. Take. R(x) = ||x||/2. and set. ω(x) = BR(x) (x) ∪ BR(x) (−x), A(x) = ω(x) . We know from Example 3.1.3 that γ(A(x)) = 1. family ω(x) is an open cover of A, hence, by compactness, x∈A set of points x1 , . . . , xn such that [ A⊂ ω(xi ). Now, of course the there is some nite. 1≤i≤n and hence. [. A⊂. A(xi ).. 1≤i≤n. (iv)) we have that γ(A) ≤ n.. By the subaditivity property of the genus (. (vi). (v) we know that γ(A) = n < ∞, so there exists an odd, continuous function : ϕ A −→ Rn \{0}. By Tietze's theorem, such a function admits an odd, continuous extension ϕ ˜ : X −→ Rn .. By. A,. By the compactness of there is no such. ε.. ε > 0 such that 0 ∈ / ϕ(A ˜ ε ). Indeed, sequences (xk ) ⊂ X , (yk ) ⊂ A such that. there exists an. Then there are. ϕ(x ˜ k ) = 0, A,. By the compactness of. yk → y ∈ A.. Consequently,. d(xk , yk ) = ||xk − yk || ≤. the sequence. (yk ). suppose. 1 . k. is convergent (up to a subsequence),. xk → y :. ||xk − y|| ≤ ||xk − yk || + ||yk − y|| → 0. But then, since some. ε>0. ϕ. ϕ(y) ˜ = ϕ(y) = 0, n ϕ(A ˜ ε ) ⊂ R \{0}. Hence. is continuous,. we have that. a contradiction. Consequently, for. n = γ(A) ≤ γ(Aε ) ≤ n.. (vii). Of course we have that. A ⊂ B ∪ A \ B.. Then. γ(A) ≤ γ(B) + γ(A \ B). Since. γ(B) < ∞, γ(A \ B) ≥ γ(A) − γ(B),. which completes the proof..

(46) 24. GENUS THEORY. 3.2.

(47) Chapter 4 Deformations. The present chapter is devoted to the Deformation Lemma. It basically says that the topological structure of the sublevels of a functional change when we pass through a critical value, or rather that such structure remains the same in a neighborhood of a regular value. Such a change in topology is useful to characterize the critical levels, and thus provides us with a tool for nding critical points. We present here a fairly general version, for functionals constrained on a manifold of codi1,1 mension 1 and such that the map dening the constraint is of class Cloc . It can be extended 1 in a number of ways, for example to the case of C manifolds of any nite codimension, see 1,1 Bonnet (1993), or to the case of complete C -Finsler manifolds, see Palais (1970). After some motivations to look at the topology of sublevels when dealing with critical points, we present the Palais-Smale condition, the construction of tangent pseudo-gradient vector elds, state and prove the Deformation Lemma and present some of its applications. We also present, briey, some generalizations that will be useful in Chapter 6.. Notation. In what follows, let X be a Banach Space and J : X −→ R a C 1 functional. The J 0 . Given b ∈ R, J b denotes the sublevel {x ∈ X; J(x) ≤ b}. The sets Jb and so on are dened analogously. Kb := {u ∈ X; J(u) = b, J 0 (u) = 0} denotes the set of critical points of J at level b.. Fréchet derivative of. J. is denoted by. 4.1 Motivation The topology of the sublevels of a functional is intrinsically related to the existence of critical points for the given functional.. Example 4.1.1.. J(x) = x3 − 3x. It is easily seen that J 0 (x) = x = ±1, with critical values c1 := J(1) = −2 and. Consider the real functional. 2. 3x − 3 and so the critical points c2 := J(1) = 2. Now, note that •. for. a1 < c1. •. for. c1 < a 2 < c 2. we have. are. J a1 = [−∞, J −1 (a1 )]. we have. (one connected component).. J a2 = [−∞, α] ∪ [β, γ]. for some. α<β<γ. (two connected. components).. •. for. a3 > c2. we have. J a3 = [−∞, J −1 (a3 )]. (again, one connected component).. J ai and J ai +ε have the same topological structure, i.e. are homeomorphic, and so can be continuously deformed into one another. For the values ci , however, it is impossible to nd such ε. Indeed, for small Note that for each. ai. there is some. ε>0. small enough so that. 25.

(48) 26. 4.1. DEFORMATIONS. Figure 4.1:. values of. ε>0. the sets. J ci −ε. and. J ci +ε. Graph of J(x) = x3 − 3x have a dierent number of connected components. and thus are not homeomorphic. However, a change in the number of connected components is not enough as a criterion for the existence of critical points.. Example 4.1.2.. Consider in. R2. Figure 4.2:. It is clear that. J. the functional. Graph of J(x, y) = (x2 + y2)2 − 2(x2 + y2). has critical values. •. for. a1 < c1. •. for. c1 < a2 < c2. •. for. a3 > c2. we have. J(x, y) = (x2 + y 2 )2 − 2(x2 + y 2 ).. c1 := −1. and. c2 := 0.. Now, note that. J a1 = ∅. the sublevel is a ring, with the inner radius shrinking as. the sublevel is some ball. a2. goes to. c2. BR (0).. Now, in the last two cases the presence of a critical level did not cause a change in the number of connected sets of the sublevels. Nonetheless, there was a (more subtle) change of a a topology: while J 3 is simply connected, J 2 is not.. a change in the topology of sublevels, alone, is not enough to assert the existence of critical points Now,.

(49) 4.2. PRELIMINARY DEFINITIONS. Figure 4.3:. 27. Graph of J(x) = xe1−x. Example 4.1.3.. 1−x Consider the real functional J(x) = xe (see Figure 4.3). We have that 1−x x)e , so J has a (maximum) critical point at x = 1, with critical value. J 0 (x) = (1 − c = 1. We also have that limx→+∞ J(x) = 0. Then, although 0 is not a critical level, for any 0 < a < c the sublevel J a has two connected components, while for a < 0 the sublevel J a is an interval. In the previous example, it is intuitive that some compactness is missing. To overcome. such an issue, the notion of Palais-Smale condition is introduced.. 4.2 Preliminary denitions The formal denition of a constraint is as follows:. Denition 4.2.1.. Let. X. be a Banach space. A. constraint is a set. M = {v ∈ X; G(v) = 0} where. G ∈ C 1 (X, R). is such that. G0 (v) 6= 0. for all. v ∈ M.. It can be shown (see Ambrosetti and Malchiodi (2007)) that codimension 1 of. M. is a submanifold of. X.. Denition 4.2.2. Let M be a constraint dened by the map G and let p ∈ M . The tangent space to M at the point p is the set Tp M := {v ∈ X; G0 (p)[v] = 0}. Note that the derivative of the function that denes the restriction is "orthogonal" to X = Tp M ⊕ hG0 (p)i.. the restriction at each point. Indeed,. Denition 4.2.3. Let J ∈ C 1 (X, R) be a functional and let M G. A constrained critical point is a point u ∈ M such that J 0 (u)[v] = 0 ∀v ∈ Tu M. Equivalently,. u. is such that. J 0 (u) = λG0 (u).. be a constraint dened by.

(50) 28. 4.4. DEFORMATIONS. λ. The real number. is called. Lagrange multiplier.. We provide more details on Lagrange Multipliers in the next chapter. Note that when dealing with functional restricted to manifolds the critical points are those such that the derivative in the direction of the manifold (that is, in the tangent space) is identically zero.. 4.3 The Palais-Smale condition As we have seen in Example 4.1.3, some notion of compactness is necessary if we wish to deal with critical points via topology. The Palais-Smale condition is a notion that emulates the necessary compactness and allows us to construct the tangent pseudo-gradient vector eld later on.. Denition 4.3.1. the map. G.. Then. Let. J|M. X. be a Banach space,. and. M. a constraint dened by. satises the Palais-Smale condition if every sequence. (J(un )) has a converging subsequence. If at level. J ∈ C 1 (X, R). is bounded,. (un ). such that. J|0M (un ) → 0. J(un ) → b we say that J. satises the Palais-Smale condition. b.. Remark 4.3.2.. Let. (λn ). is a sequence. Remark 4.3.3.. (un ). be a sequence satisfying the Palais-Smale condition. Then there. such that. λn → λ,. the Lagrange multiplier.. It is not strictly necessary that the functional be dened in all of. may ask for a functional dened in a neighborhood of. M. as well as in. M. X.. One. only, if taking care. of the denition of the derivative on manifolds.. Remark 4.3.4. With the notation as above, if J|M satises the Palais-Smale condition then the following sets are compact:. {(u, λ) ∈ M × R; J(u) = b, J 0 (u) = λG0 (u)}, Kb := {u ∈ M ; J(u) = b and ∃λ ∈ R such that J 0 (u) = λG0 (u)}, ∪a≤b≤c Kb for all. a, c ∈ R.. 4.4 Tangent pseudo-gradient vector elds The idea of the proof of the Deformation Lemma in nite dimensions (or in Hilbert spaces in general) is to construct the deformations as ows in the direction of the gradient. Since in general Banach spaces there is no notion of gradient at our disposal, we emulate the necessary properties with the notion of a pseudo-gradient vector eld. In the case of constrained functionals, we talk about. Notation. C 1 (X, R). pseudo-gradient vector elds.. X be a Banach space and let M be a constraint dened by G. u ∈ M denote . ||J 0 (u)||∗ := sup |J 0 (u)[y]|; y ∈ X, ||y|| = 1, G0 (u)[y] = 0 .. Let. and. tangent. For. J ∈.

(51) 4.4. 0 is the norm of the projection of J (u) onto the hyperplane tangent 0 0 0 at u. Also note that ||J (u)||∗ = 0 means precisely that J (u) = λG (u), i.e., u is a. Note that to. 29. TANGENT PSEUDO-GRADIENT VECTOR FIELDS. M. ||J 0 (u)||∗. critical points of. Notation.. J|M .. With the notation as above, denote by. ˜ M. the set of regular points of. J|M ,. that. is,. ˜ := {u ∈ M ; ∀λ ∈ R J 0 (u) − λG0 (u) 6= 0}. M We now dene tangent pseudo-gradient vectors and vector elds.. Denition 4.4.1. G.. Then. v. is a. X be a Banach space, J ∈ C 1 (X, R) and M a constraint tangent pseudo-gradient vector of J with respect to S at u if ( ||v(|| ≤ 2||J 0 (u)||∗ ; J 0 (u)[v] ≥ ||J 0 (u)||2∗ if G0 (u)[v] = 0. Let. ˜ −→ X is a tangent pseudo-gradient vector eld of J if V V :M ˜ and V (u) is a tangent pseudo-gradient vector for every u ∈ M ˜. M. A map in. Remark 4.4.2.. Let. gradient vectors at. u ∈ S.. u. dened by. is locally Lipschitz. It is immediately seen that the the set of tangent pseudo-. is convex. It then follows that convex combinations of two or more. tangent pseudo-gradient vector elds is also a tangent pseudo-gradient vector eld. Tangent pseudo-gradient vector elds do not always exist, but under some conditions on C 1 regularity.. the constraint we can prove they exist. We require a little more than. Notation. onto. X∗. The set of. is denoted by. C 1 functionals 1,1 Cloc (X, R).. such that the derivative is locally Lipschitz from. Proposition 4.4.3.. X. 1,1 Let X be a Banach space, J ∈ C 1 (X, R), G ∈ Cloc (X, R) and M a constraint dened by G. Suppose that J is not constant on M . Then there exists a map V ˜ and whose restriction to M ˜ is a that is locally Lipschitz in an open neighborhood NM˜ of M tangent pseudo-gradient vector eld. Moreover, if J and G are even then NM˜ can be chosen symmetric with respect to the origin and V is odd.. Proof.. Let. ˜. u0 ∈ M. Then, by the denition of. ||y0 || = 1, Set. ||J(u0 )||∗. G0 (u0 )[y0 ] = 0,. there exists. y0 ∈ X. such that. 2 J 0 (u0 )[y0 ] ≥ ||J 0 (u0 )||∗ . 3. 5 v0 = ||J 0 (u0 )||∗ y0 . 3. Then we have. Now, since.   ||v0 || = 5 ||J 0 (u0 )||∗ 3 10  0 J (u0 )[v0 ] ≥ ||J 0 (u0 )||2∗ , 9 u0 ∈ M , then G0 (u0 ) 6= 0 and so there. G0 (u0 )[v0 ] = 0. exists. z0 ∈ X. 2 G0 (u0 )[z0 ] ≥ ||G0 (u0 )||. 3 Setting. x0 :=. z0 0 G (u0 )[z0 ]. such that. ||z0 || = 1. and.

(52) 30. 4.4. DEFORMATIONS. we have that. 3 ||x0 || ||G0 (u0 )|| ≥ . 2 0 Now, follows form the continuity of G that there exists R1 > 0 such that for all u ∈ BR1 (u0 ) G0 (u0 )[x0 ] = 1,. we have. 1 G0 (u)[x0 ] ≥ , 2. Then, taking. R < R1. ||x0 || ||G0 (u)|| ≤ 2.. small enough we have that. v := v(u, u0 ) = v0 −. G0 (u)[v0 ] x0 G0 (u)[x0 ]. ˜. u ∈ BR (u0 ) ∩ M

(53) 0

(54)

(55) G (u)[v0 ]

(56) . kvk ≤ kv0 k + 2 0 G (u0 )[z0 ]. is a tangent pseudo-gradient vector of. By continuity, there is. R. J. at. u. such that for some. for all. 0<α<2. Indeed, rst note that. we have. ||v0 || ≤ α||J 0 (u)||∗ and. |G0 (u)[v0 ]| ≤. 2−α 0 G (u0 )[z0 ]||J 0 (u)||∗ 2. from what follows that. ||v|| ≤ 2||J 0 (u)||∗ Now, note that we can nd an. R>0. small enough such that for all. ˜ u ∈ BR (u0 ) ∩ M. we. have. G0 (u)[v0 ] 0 J (u)[x0 ] G0 (u)[x0 ] G0 (u)[v0 ] 0 = J 0 (u)[v0 ] − 0 J (u)[z0 ] G (u)[z0 ] 10 |G0 (u)[v0 ]| 0 ≥ − ε ||J 0 (u)||2∗ − 0 ||J (u)||∗ 9 |G (u)[z0 ]| 10 3 0 0 2 0 − ε ||J (u)||∗ − ||G (u0 )|| + ε |G0 (u)[v0 ]| ||J 0 (u)||∗ ≥ 9 2 10 3 0 0 2 0 ≥ − ε ||J (u)||∗ − ||G (u0 )|| + ε |G0 (u)[v0 ]| sup ||J 0 (u)||∗ 9 2 BR (u0 ). J(u)[v] = J 0 (u)[v0 ] −. ≥ ||J 0 (u)||2∗ Various. R. where considered, it suces to take the smallest one.. Let us show now that. u 7→ v(u, u0 ) is a locally Lipschitz map in. BR (u0 ). Let K. be some compact subset. Then, since. G0 is locally.

(57) 4.4. TANGENT PSEUDO-GRADIENT VECTOR FIELDS. Lipschitz, for. u, w ∈ K. 31. we have:. 0. G0 (u)[v0 ]. G (w)[v ] 0 v(u, u0 ) − v(w, u0 ) = . G0 (u)[x0 ] x0 − G0 (w)[x0 ] x0 0. 0 G (u)[v0 ]. 1 G (w)[v ] 0.

(58) ≤

(59)

(60) 0 −. G (u0 )[z0 ]

(61) G0 (u)[x0 ] G0 (w)[x0 ] . G0 (u)[v ] G0 (w)[v ] G0 (w)[v ] G0 (w)[v ] . 0 0 0 0 − 0 − − 0 = c 0. 0 G (u)[x0 ] G (u)[x0 ] G (w)[x0 ] G (u)[x0 ] 0. 0. G (w)[v0 ] G0 (w)[v0 ] 0. ≤ c G (u) − G (w) kv0 k + 0 − G (w)[x0 ] G0 (u)[x0 ] . 0 G (w)[v0 ] G0 (w)[v0 ] . − = cku − wk + 0 G (w)[x0 ] G0 (u)[x0 ] Now,. 0. G (w)[v0 ] G0 (w)[v0 ]

(62) 0

(63) . 1 1.

(64) G (w)[v0 ]

(65) . − G0 (w)[x0 ] − G0 (u)[x0 ] ≤ sup. G0 (w)[x0 ] G0 (u)[x0 ] K

(66)

(67) . ≤ 4 sup

(68) G0 (w)[v0 ]

(69) G0 (u)[x0 ] − G0 (w)[x0 ] K. ≤ cku − vk and thus. u 7→ v(u, u0 ). is locally Lipschitz.. Since. NM˜ :=. [. BR (u). ˜ M is paracompact (because it is a subset of a metric space) there is a locally nite renement. (ωj )j∈I and a partition of unity (θj )j∈I u0j such that ωj ⊂ BR (u0j ). Then. subordinated to. V (u) =. X. (ωj )j∈I . For each j ∈ I. there is some. = θj (u)v(u, u0j ). j is a locally Lipschitz tangent pseudo-gradient vector eld for Suppose now that. G. and. J. are even. Then, taking. that. [. R(u). J as. ˜. M min{R(u), R(−u)}. over. we have. BR (u). ˜ M is symmetric with respect to the origin. Then. 1 V1 (u) = (V (u) − V (−u)) 2 V− := −V (−u) is a by the symmetry of G and J −u0 . Thus, if u ∈ BR (u0 ) and. is an odd tangent pseudo-gradient vector eld. It suces to show that tangent pseudo-gradient vector for the construction of. y0 , z0. and. x0. J. at. ˜. u∈M. Note that,. is the same for both. u0. and. −u ∈ BR (−u0 ), v(u, u0 ) = v(−u, −u0 ). It then follows that. V (−u). is a convex combination of tangent pseudo-gradient vectors at.

(70) 32. u,. 4.5. DEFORMATIONS. and thus the claim follows.. Remark 4.4.4.. Note that for all. u ∈ BR (u0 ). G0 (u)[v] = G0 (u)[v0 ] − This implies that. G0 (u)[V (u)] = 0. for all. we have. G0 (u)[v0 ] 0 G (u)[x0 ] = 0. G0 (u)[x0 ]. u ∈ NM˜ .. 4.5 The Deformation Lemma and its consequences We are now in position to prove the Deformation Lemma for constrained functionals. We remark that it would suce to consider a functional dened in a neighborhood of the manifold. M.. Theorem 4.5.1 (Deformation Lemma).. 1,1 (X, R) dening Let X be a Banach space, G ∈ Cloc the constraint M , J ∈ C 1 (X, R) such that J|M satises the Palais-Smale condition, J|M is not constant and b ∈ R is not a critical value for J|M . Then there exists ε0 > 0 such that for all 0 < ε < ε0 there exists a map η ∈ C(R ×M, M ) such that. (i). η(0, u) = u for all u ∈ M ;. (ii) (iii) (iv) (v). b+ε0 If u ∈ / J|M b−ε then η(t, u) = u for all t ∈ R; 0. For each t ∈ R η(t, · ) is a homeomorphism from M onto M ; For all u ∈ M the function t 7→ J|M η(t, u) is decreasing; b+ε η(1, J|M ) ⊂ J|b−ε M ;. (vi). If J|M and G are even then η(t, · ) odd.. Proof.. The idea is to build a ow as the solution of a convenient ordinary dierential equation. and then to check that it veries conditions. (i) to (vi).. We know by Proposition 4.4.3 that there exists a map V dened in an open neighborhood ˜ , locally Lipschitz over N ˜ such that V | ˜ is a tangent pseudo-gradient vector eld. NM˜ of M M M If J|M and G are even then NM ˜ is symmetric with respect to the origin. Let us note that there exists ε1 > 0 and δ > 0 such that. ||J|0M (u)||∗ ≥ δ. 1 ∀u ∈ J|M b+ε b−ε1 .. δ ≤ 1. Indeed, suppose that εn → 0, δn → 0 such that for all n ∈ N. It is convenient to choose are sequences. n ∃un ∈ J|M b−ε b−εn. But then. (un ). Dene. ||J|0M (un )|| < δn .. such that. is a Palais-Smale sequence at level. subsequence, which implies, by continuity, that. (. b. the assertion is not true. Then there. b. for. J.. Hence there is a converging. is a critical level, a contradiction.. δ2 ε0 := min ε1 , 8. ) ..

(71) 4.5. 33. THE DEFORMATION LEMMA AND ITS CONSEQUENCES. For each. ˜ v∈M. set. Rv = 21 d v, X \ NM˜ Ω :=. . . Dene. [. BRv (v).. ˜ M Now, for. 0 < ε < ε0. dene. B := J|M b+ε b−ε. A := J|M b−ε0 ∪ J|M b+ε0 ∪ X \ Ω , and. α(u) = Of course, case. J|M. d(u, A) . d(u, A) + d(u, B). α(B) = {1} and α(A) = {0}. We also know that α is locally Lipschitz G are even then A and B are symmetric with respect to the origin. and. in. X . In α is. and. even. Let. W (u) :=.  n α(u) min 1, 0. Then. u∈X. W. 1 ||V (u)||. o. V (u),. u∈ /A. otherwise. is locally Lipschitz and. ||W (u)|| ≤ 1. for all. u ∈ X.. It then follows that given. there exists a unique solution to the problem.  d η(t, u) = −W (η(t, u)) (∗) dt η(0, u) = u η( · , u) ∈ C 1 (R, X), that η(t, η(s, u)) = homeomorphism of X onto X for all t ∈ R. It remains to. dened on the whole real line. We know that. η(t + s, u). and that. prove that. η. η(t, · ). is an. satises properties (i) to (vi). (i) is trivial, by the initial condition of the problem (∗). Property (ii) is trivial, since for u ∈ / J|M b+ε b−ε the right hand side of the equation in (∗). Property. 0 0. is zero and thus the solution is constant. To prove property. (iii) it suces to show that if u ∈ M. then. η(t, u) ∈ M. for all. t ∈ R.. η(−t, u) and continuity of both follows from continuity of 0 the solution to (∗) with respect to initial conditions. The claim is trivial if u ∈ / J|M b+ε b−ε0 . Let 0 ˜ u ∈ J|M b+ε ˜ for t b−ε0 . Then, u ∈ M . Since t 7→ η(t, u) is continuous we have that η(t, u) ∈ NM small enough. But then we have that. Indeed, the inverse is given by. d d 0 G(η(t, u)) = G (η(t, u)) η(t, u) dt dt 1 = −α(η(t, u)) min 1, G0 (η(t, u)) V (η(t, u)) ||V (η(t, u))|| = 0. Thus. t 7→ G(η(t, u)). is constant, which implies that. η(t, u) ∈ M. for all. t ∈ R..

(72) 34. 4.5. DEFORMATIONS. To prove. (iv), note that. d d 0 J|M (η(t, u)) = J|M (η(t, u)) η(t, u) dt dt 1 J|0M (η(t, u)) V (η(t, u)) = −α(η(t, u)) min 1, ||V (η(t, u))|| 1 ≤ −α(η(t, u)) min 1, ||J|0M (η(t, u))||2∗ ||V (η(t, u))|| ≤ 0. For. (v), since we have (iv) it suces to prove that J|M (η(1, u)) ≤ b − ε for u ∈ J|M b+ε b−ε .. First, note that. ||J|0M (u)||2∗ ≤ J|0M (u)[V (u)] ≤ ||J|0M (u)||∗ ||V (u)||, thus. ||J|0M (u)||∗ ≤ ||V (u)|| ≤ 2||J|0M (u)||∗ .. Then we have that. 1 d J|M (η(t, u)) ≤ −α(η(t, u)) min 1, ||J|0M (η(t, u))||2∗ dt ||V (η(t, u))|| 1 1 ≤ − min 1, ||V (η(t, u))||2 4 ||V (η(t, u))||  − 1  if ||V (η(t, u))|| ≥ 1 42 ≤  − δ if ||V (η(t, u))|| < 1 4 Integrating, we conclude that. J|M (η(1, u)) ≤ −. V , and consequently W , can be taken to be odd. In this case, it is easy to see that −η(t, u) is a solution to the Cauchy problem (∗) with initial condition −η(0, u) = −u. By the uniqueness of the solution, η(t, · ) is odd for all t ∈ R. At last, note that if. J|M. δ2 δ2 + J|M (η(0, u)) ≤ − + b + ε ≤ b − ε. 4 4. and. G. are even, then. Remark 4.5.2.. There are plenty versions of the Deformation Lemma. One of them does. not require that. b. be a regular value for. J|M .. In this case, the deformation is takes the. complement of a neighborhood of the critical points to a lower sublevel. The proof is very similar, and the key ideas are the same. See Struwe (2008) for details. We now analyze some important consequences of the Deformation Lemma.. Theorem 4.5.3 (Morse Theorem).. Let X be a Banach space, M be a constraint dened by G∈ J ∈ C (X, R) such that J|M is not constant and b a regular value. Then there exists ε0 > 0 such that for all 0 < ε < ε0 the sublevel J|b−ε M is a deformation retract of . J|b+ε M 1,1 Cloc (X, R),. Proof.. 1. With the notation as in the proof of Theorem 4.5.1, dene. ϕ(t, u) = η. 4t + (J|M (u) − b + ε) , u δ2.

(73) 4.5. THE DEFORMATION LEMMA AND ITS CONSEQUENCES. We show that. ϕ denes a deformation of J. b+ε. into. J. b−ε. . That is,. . ϕ ∈ C [0, 1] ×. 35. b−ε J|b+ε M , J|M. . is such that. ϕ(0, · ) = Id,. ϕ (t, · ) = Id. ϕ 1, J|c+ε = J|c−ε M M .. ∀t ∈ [0, 1],. ϕ(0, · ) = η(0, · ) = Id. If u ∈ J|b−ε then J|M (u) − b + ε ≤ −2ε < 0 so M ϕ(t, u) = η(0, u) = u as well. b+ε b+ε Now, let u ∈ J|M b−ε and τ > 0 such that η(t, u) ∈ J|M b−ε for 0 ≤ t ≤ τ . Recall from the. Of course,. proof of Theorem 4.5.1 that. b − ε < J|M (η(τ, u)) ≤ J|M (u) − that is,. τ<. δ2 τ, 4. 4 (J|M (u) − b + ε) =: t0 . δ2. Then. ϕ(1, u) = η(t0 , u) ∈ J|b−ε M . Indeed, note that. δ2 J|M (η(t0 , u)) ≤ J|M (u) − t0 = b − ε, 4 which completes the proof.. Theorem 4.5.4 (Minimax Principle).. Let X be a Banach space, M be a constraint dened by G ∈ J ∈ C (X, R) such that J|M is not constant and satises the PalaisSmale condition, and let B a nonempty family of nonempty subsets of M . Suppose that for each b ∈ R and ε > 0 the ow η constructed in the proof of the Deformation Lemma is such that η(1, A) ∈ B ∀A ∈ B. 1,1 Cloc (X, R),. 1. Let. ˜b := inf sup J|M (v). A∈B v∈A. Then if ˜b ∈ R we have that ˜b is a critical value of J|M . Proof. Suppose ˜b ∈ R and is not a critical point of J|M . ε > 0 small enough such that for some A ∈ B we have. By the denition of. ˜b,. there is an. c˜ ≤ sup J|M (v) ≤ ˜b + ε. v∈A But then. ˜. b−ε η(1, A) ∈ B ⊂ J|M ,. The families. B. a contradiction.. of sets used in the Minimax Principle are dened with the help of the. Ljusternik-Schnirelman category (see Ambrosetti and Malchiodi (2007)) or of the genus. For example, we have the following theorem:. Theorem 4.5.5.. 1,1 Let X be a Banach space, G ∈ Cloc (X, R), M a constraint dened by G and J ∈ C 1 (X, R). Suppose that G and J|M are even, that J|M is not constant and satises the Palais-Smale condition on M and that 0 ∈ / M . For k ∈ N, set . Bk = A ∈ s(X) : A ⊂ M, γ(A) ≥ k.

(74) 36. 4.6. DEFORMATIONS. and. bk := inf sup J|M (v). A∈Bk v∈A. Then. (i). For all k ∈ N such that Bk 6= ∅ and bk ∈ R, bk is a critical value of J|M . Moreover,. bk ≤ bk+1 and if for some j ∈ N we have Bj+k 6= ∅ and bk = bj+k ∈ R then γ(Kbk ) ≥ j + 1.. (ii). If for all k ∈ N we have Bk 6= ∅ and bk ∈ R, then. lim bk = +∞. n→∞. 4.6 Generalizations As already mentioned, the theorems of this chapter can be modied to more general situations. Of remarkable interest is the following theorem, which gives a Minimax Principle for 1 manifolds that are only C :. Theorem 4.6.1.. Let M be a closed symmetric submanifold of a Banach space X such that 0∈ / M . Let J ∈ C (M, R) be even and bounded from below. Dene 1. bk = inf sup J(v) A∈Bk v∈A. . where Bk = A ∈ s(X) : A ⊂ M, γ(A) ≥ k, A is compact . If Bk 6= ∅ for some k and if J satises the Palais-Smale condition then J has at least k distinct pairs of critical points. The proof relies heavily on the category of Ljusternik-Schnirelman and on the Finsler structure of the manifold. See Szulkin (1988). There is also a generalization of the category and of the genus and for the results presented here to variational problems that admit some kind of the references therein.. index. theory. See Struwe (2008) and.

(75) Chapter 5 Submersions, Manifolds, Lagrange Multipliers. In the proof of Theorem 1.0.1, we will consider the restriction of a functional to a subset of our Hilbert space. The fact that this subset is a manifold has a fundamental role in the theorems we will apply. In this chapter, we recall some denitions and prove two important theorems: Ljusternik's theorem on the construction of manifolds via submersions and the Theorem of Lagrange Multipliers.. 5.1 Submersions and submanifolds Denition 5.1.1.. X be a Banach space. A subset A of X is a (smooth) submanifold of X if for every u ∈ A there exists a neighborhood Uu of u in X and a (smooth) map ξu : Uu −→ Vu , where Vu is an open subset of X , and a closed complemented subspace S of X such that ξu (Uu ∩ A) = Vu ∩ S. Let. Denition 5.1.2. Let X, Y be two Banach spaces. A map submersion at u0 ∈ X if (i) G is continuously dierentiable in a neighborhood of u0 ; (ii) (iii). G0 (u0 ). G : X −→ Y. is said to be a. is surjective;. there exists a continuous projection operator. P : X −→ ker G0 (u0 ).. Hence,. X = ker G0 (u0 ) ⊕ (I − P )(X).. Remark 5.1.3.. X is a Hilbert space and Y = Rk then condition (iii) is immediately 0 0 satised and the map G = (G1 , . . . , Gk ) is a submersion at u0 if and only if G1 (u0 ), . . . , Gk (u0 ) If. are linearly independent.. Denition 5.1.4. Let M ⊂ X and u0 ∈ M be xed. (i) An admissible curve in M through u0 is a map u : t ∈ (−ε, ε) 7→ u(t) ∈ M u(0) = u0. (ii). and the derivative. h is called a tangent 0 that h = u (0).. 0. u (0). such that. exists.. vector to M at u0 if and only if there is an admissible curve such 37.