A converse to the Ambrosetti-Prodi theorem

Transcription

1 A converse to the Ambrosetti-Prodi theorem Marta Calanchi, Università di Milano with Carlos Tomei and André Zaccur (PUC-Rio, Brazil) Varese, RISM, September 2015

2 The Ambrosetti-Prodi Theorem 1/14

3 The Ambrosetti-Prodi Theorem Let Ω R n be open, bounded, smooth, connected. 1/14

4 The Ambrosetti-Prodi Theorem Let Ω R n be open, bounded, smooth, connected. The Dirichlet Laplacian D in Ω has eigenvalues 1/14

5 The Ambrosetti-Prodi Theorem Let Ω R n be open, bounded, smooth, connected. The Dirichlet Laplacian D in Ω has eigenvalues σ D = {0 < µ 1,D < µ 2,D... } and first eigenvector ψ 1,D. 1/14

6 The Ambrosetti-Prodi Theorem Let Ω R n be open, bounded, smooth, connected. The Dirichlet Laplacian D in Ω has eigenvalues σ D = {0 < µ 1,D < µ 2,D... } and first eigenvector ψ 1,D. Take f : R R smooth such that 1/14

7 The Ambrosetti-Prodi Theorem Let Ω R n be open, bounded, smooth, connected. The Dirichlet Laplacian D in Ω has eigenvalues σ D = {0 < µ 1,D < µ 2,D... } and first eigenvector ψ 1,D. Take f : R R smooth such that f (t) > 0, t R, (convexity) 1/14

8 The Ambrosetti-Prodi Theorem Let Ω R n be open, bounded, smooth, connected. The Dirichlet Laplacian D in Ω has eigenvalues σ D = {0 < µ 1,D < µ 2,D... } and first eigenvector ψ 1,D. Take f : R R smooth such that f (t) > 0, t R, (convexity) 0 < lim f (t) < µ 1,D < lim f (t) < µ 2,D t t + (strict interaction with µ 1,D ; non resonance) 1/14

9 The Ambrosetti-Prodi Theorem Let Ω R n be open, bounded, smooth, connected. The Dirichlet Laplacian D in Ω has eigenvalues σ D = {0 < µ 1,D < µ 2,D... } and first eigenvector ψ 1,D. Take f : R R smooth such that f (t) > 0, t R, (convexity) 0 < lim f (t) < µ 1,D < lim f (t) < µ 2,D t t + (strict interaction with µ 1,D ; non resonance) Theorem (Ambrosetti-Prodi, Manes-Micheletti) For g C 0,α (Ω), there exists a critical height h c (g) R, F(u) = D u f (u) = g + h ψ 1,D has 0, 1, 2 solutions for h > h c (g), h = h c (g), h < h c (g). 1/14

10 Standard Boundary Conditions The AP-theorem still holds for standard boundary conditions. Take a domain of self-adjointness H 2 b H2 (Ω) of b : H 2 b H0. (For Dirichlet conditions, for example, H 2 b = H2 H 1 0.) By Kato-Rellich, for q L (Ω), T q : Hb 2 H0 H 0, v b v q v is also self-adjoint. Standard Boundary Conditions (SBC) 2/14

11 Standard Boundary Conditions The AP-theorem still holds for standard boundary conditions. Take a domain of self-adjointness H 2 b H2 (Ω) of b : H 2 b H0. (For Dirichlet conditions, for example, H 2 b = H2 H 1 0.) By Kato-Rellich, for q L (Ω), T q : Hb 2 H0 H 0, v b v q v is also self-adjoint. A boundary condition is standard if 1. The smallest eigenvalue λ q 1 of T q is simple. 2. There is a unique L 2 -normalized eigenfunction φ q 1 > 0 associated with λ q On bounded sets of potentials q, the sup norm of φ q 1 is uniformly bounded. Standard Boundary Conditions (SBC) 2/14

12 Standard Boundary Conditions The AP-theorem still holds for standard boundary conditions. Take a domain of self-adjointness H 2 b H2 (Ω) of b : H 2 b H0. (For Dirichlet conditions, for example, H 2 b = H2 H 1 0.) By Kato-Rellich, for q L (Ω), T q : Hb 2 H0 H 0, v b v q v is also self-adjoint. A boundary condition is standard if 1. The smallest eigenvalue λ q 1 of T q is simple. 2. There is a unique L 2 -normalized eigenfunction φ q 1 > 0 associated with λ q On bounded sets of potentials q, the sup norm of φ q 1 is uniformly bounded. Dirichlet, Neumann and periodic BC are standard. Standard Boundary Conditions (SBC) 2/14

13 A question raised by Dancer: a converse to AP The importance of being (non) convex: the Ambrosetti-Prodi theorem is unexpectedly tight 3/14

14 A question raised by Dancer: a converse to AP What happens if the convexity hypothesis fails? The importance of being (non) convex: the Ambrosetti-Prodi theorem is unexpectedly tight 3/14

15 A question raised by Dancer: a converse to AP What happens if the convexity hypothesis fails? For a general SBC, consider F(u) = b u f (u) = g. The importance of being (non) convex: the Ambrosetti-Prodi theorem is unexpectedly tight 3/14

16 A question raised by Dancer: a converse to AP What happens if the convexity hypothesis fails? For a general SBC, consider F(u) = b u f (u) = g. Let f : R R a smooth function such that The importance of being (non) convex: the Ambrosetti-Prodi theorem is unexpectedly tight 3/14

17 A question raised by Dancer: a converse to AP What happens if the convexity hypothesis fails? For a general SBC, consider F(u) = b u f (u) = g. Let f : R R a smooth function such that (1) (nonconvexity) f takes positive and negative values; The importance of being (non) convex: the Ambrosetti-Prodi theorem is unexpectedly tight 3/14

18 A question raised by Dancer: a converse to AP What happens if the convexity hypothesis fails? For a general SBC, consider F(u) = b u f (u) = g. Let f : R R a smooth function such that (1) (nonconvexity) f takes positive and negative values; (2) (strict interaction with µ 1,b ) There are m, M R for which f (R) = [m, M] and m < µ 1,b < M < µ 2,b ; The importance of being (non) convex: the Ambrosetti-Prodi theorem is unexpectedly tight 3/14

19 A question raised by Dancer: a converse to AP What happens if the convexity hypothesis fails? For a general SBC, consider F(u) = b u f (u) = g. Let f : R R a smooth function such that (1) (nonconvexity) f takes positive and negative values; (2) (strict interaction with µ 1,b ) There are m, M R for which f (R) = [m, M] and m < µ 1,b < M < µ 2,b ; (3) (nonresonance with µ 1,b ) There is an ɛ > 0 such that f (x) > µ 1,b + ɛ for x >> 0 and f (x) < µ 1,b ɛ for x << 0. The importance of being (non) convex: the Ambrosetti-Prodi theorem is unexpectedly tight 3/14

20 A question raised by Dancer: a converse to AP What happens if the convexity hypothesis fails? For a general SBC, consider F(u) = b u f (u) = g. Let f : R R a smooth function such that (1) (nonconvexity) f takes positive and negative values; (2) (strict interaction with µ 1,b ) There are m, M R for which f (R) = [m, M] and m < µ 1,b < M < µ 2,b ; (3) (nonresonance with µ 1,b ) There is an ɛ > 0 such that f (x) > µ 1,b + ɛ for x >> 0 and f (x) < µ 1,b ɛ for x << 0. The importance of being (non) convex: the Ambrosetti-Prodi theorem is unexpectedly tight 3/14

21 A question raised by Dancer: a converse to AP What happens if the convexity hypothesis fails? For a general SBC, consider F(u) = b u f (u) = g. Let f : R R a smooth function such that (1) (nonconvexity) f takes positive and negative values; (2) (strict interaction with µ 1,b ) There are m, M R for which f (R) = [m, M] and m < µ 1,b < M < µ 2,b ; (3) (nonresonance with µ 1,b ) There is an ɛ > 0 such that f (x) > µ 1,b + ɛ for x >> 0 and f (x) < µ 1,b ɛ for x << 0. Theorem (M.C., C. Tomei, A. Zaccur) Let F : B 2 b (= H2 b C2,α ) C 0,α. Then there is g C 0,α such that F(u) = g admits (at least) four solutions. The importance of being (non) convex: the Ambrosetti-Prodi theorem is unexpectedly tight 3/14

22 Fibers and the height function Let ψ 1,b the first eigenfunction, ψ 1,b > 0, ψ 1,b 2 = 1 and V = ψ 1,b, H 1 b H2 = W 2 V, L 2 = W 0 V Main characters: fibers and height functions 4/14

23 Fibers and the height function Let ψ 1,b the first eigenfunction, ψ 1,b > 0, ψ 1,b 2 = 1 and V = ψ 1,b, Hb 1 H2 = W 2 V, L 2 = W 0 V Fibers are inverse of vertical lines {z + sψ 1,b, s R} under F. Main characters: fibers and height functions 4/14

24 Fibers and the height function Let ψ 1,b the first eigenfunction, ψ 1,b > 0, ψ 1,b 2 = 1 and V = ψ 1,b, H 1 b H2 = W 2 V, L 2 = W 0 V Fibers are inverse of vertical lines {z + sψ 1,b, s R} under F. For z H 0, the fiber is in H 2 b. If z C0,α, it is in C 2,α (as a consequence of regularity) Main characters: fibers and height functions 4/14

25 Fibers and the height function Let ψ 1,b the first eigenfunction, ψ 1,b > 0, ψ 1,b 2 = 1 and V = ψ 1,b, H 1 b H2 = W 2 V, L 2 = W 0 V Fibers are inverse of vertical lines {z + sψ 1,b, s R} under F. For z H 0, the fiber is in H 2 b. If z C0,α, it is in C 2,α (as a consequence of regularity) and may be parametrized by t, {u(z, t) = w(z, t) + tψ 1,b, t R}, Main characters: fibers and height functions 4/14

26 Fibers and the height function Let ψ 1,b the first eigenfunction, ψ 1,b > 0, ψ 1,b 2 = 1 and V = ψ 1,b, H 1 b H2 = W 2 V, L 2 = W 0 V Fibers are inverse of vertical lines {z + sψ 1,b, s R} under F. For z H 0, the fiber is in H 2 b. If z C0,α, it is in C 2,α (as a consequence of regularity) and may be parametrized by t, {u(z, t) = w(z, t) + tψ 1,b, t R}, Restrict F to a fiber u(z, t) = u(t) Bb 2 ( z B2 b is fixed, we drop it! ), F(u(t)) = z + h(u(t)) ψ 1,b for the height function h(u(t)) Main characters: fibers and height functions 4/14

27 Fibers and the height function Let ψ 1,b the first eigenfunction, ψ 1,b > 0, ψ 1,b 2 = 1 and V = ψ 1,b, H 1 b H2 = W 2 V, L 2 = W 0 V Fibers are inverse of vertical lines {z + sψ 1,b, s R} under F. For z H 0, the fiber is in H 2 b. If z C0,α, it is in C 2,α (as a consequence of regularity) and may be parametrized by t, {u(z, t) = w(z, t) + tψ 1,b, t R}, Restrict F to a fiber u(z, t) = u(t) Bb 2 ( z B2 b is fixed, we drop it! ), F(u(t)) = z + h(u(t)) ψ 1,b for the height function h(u(t)) The study of F restricted to a fiber is a calculus problem. Main characters: fibers and height functions 4/14

28 An example Here Ω = [0, 1] [0, 2], µ 1,D and µ 2,D Consider u f (u) = g, u = 0 in Ω, with 35 f (x) = µ 2 µ 1 π ( arctan( x 10 ) 2 5 x e (x/10)2 ) + µ1, f (x) 20 F x f u1 h On the right: the height function associated to the fiber obtained by inverting the vertical line through z = g(x, y) = 100 ( x(x 1)y 2 (y 2) ) 35 sin(πx) sin( πy 2 ) Numerics by José Cal Neto and Otavio Kaminski. 5/14

29 An example: four solutions The height value is attained by four preimages: A manifestation of the maximum principle: the four graphs sit one on top of the other. 6/14

30 An example: four solutions The height value is attained by four preimages: A manifestation of the maximum principle: the four graphs sit one on top of the other. 6/14

31 The height function in the non convex context (Smoothness) F(u(t)) = z + h(u(t)) ψ 1,b, is a smooth map. The asymptotic behavior is a key ingredient in Berger and Podolak s approach to AP. 7/14

32 The height function in the non convex context (Smoothness) F(u(t)) = z + h(u(t)) ψ 1,b, is a smooth map. (Asymptotic behavior) By (3) (f near ± ), on each fiber u(t) B 2 b, lim h(u(t)) =. t ± The asymptotic behavior is a key ingredient in Berger and Podolak s approach to AP. 7/14

33 The height function in the non convex context (Smoothness) F(u(t)) = z + h(u(t)) ψ 1,b, is a smooth map. (Asymptotic behavior) By (3) (f near ± ), on each fiber u(t) B 2 b, lim h(u(t)) =. t ± (a local minimum along the fiber) To do: The asymptotic behavior is a key ingredient in Berger and Podolak s approach to AP. 7/14

34 The height function in the non convex context (Smoothness) F(u(t)) = z + h(u(t)) ψ 1,b, is a smooth map. (Asymptotic behavior) By (3) (f near ± ), on each fiber u(t) B 2 b, lim h(u(t)) =. t ± (a local minimum along the fiber) To do: there is a fiber u(z, t) whose height function has positive second derivative at a stationary point The asymptotic behavior is a key ingredient in Berger and Podolak s approach to AP. 7/14

35 The height function in the non convex context (Smoothness) F(u(t)) = z + h(u(t)) ψ 1,b, is a smooth map. (Asymptotic behavior) By (3) (f near ± ), on each fiber u(t) B 2 b, lim h(u(t)) =. t ± (a local minimum along the fiber) To do: there is a fiber u(z, t) whose height function has positive second derivative at a stationary point The asymptotic behavior is a key ingredient in Berger and Podolak s approach to AP. 7/14

36 The height function in the non convex context (Smoothness) F(u(t)) = z + h(u(t)) ψ 1,b, is a smooth map. (Asymptotic behavior) By (3) (f near ± ), on each fiber u(t) B 2 b, lim h(u(t)) =. t ± (a local minimum along the fiber) To do: there is a fiber u(z, t) whose height function has positive second derivative at a stationary point convex case non convex case The asymptotic behavior is a key ingredient in Berger and Podolak s approach to AP. 7/14

37 Basic spectral theory Denote by λ 1 (u) the first eigenvalue of DF (u) = b + f (u) and φ 1 (u) the associated (positive, normalized) eigenfunction. Notice that Z -convergence is not induced by a metric. 8/14

38 Basic spectral theory Denote by λ 1 (u) the first eigenvalue of DF (u) = b + f (u) and φ 1 (u) the associated (positive, normalized) eigenfunction. DF(u) is not invertible iff λ 1 (u) = 0. Notice that Z -convergence is not induced by a metric. 8/14

39 Basic spectral theory Denote by λ 1 (u) the first eigenvalue of DF (u) = b + f (u) and φ 1 (u) the associated (positive, normalized) eigenfunction. DF(u) is not invertible iff λ 1 (u) = 0. λ 1 : B 2 b (:= H2 b C2,α ) R and φ 1 : B 2 b H0 are smooth. Notice that Z -convergence is not induced by a metric. 8/14

40 Basic spectral theory Denote by λ 1 (u) the first eigenvalue of DF (u) = b + f (u) and φ 1 (u) the associated (positive, normalized) eigenfunction. DF(u) is not invertible iff λ 1 (u) = 0. λ 1 : B 2 b (:= H2 b C2,α ) R and φ 1 : B 2 b H0 are smooth. Equip Z = L with the L 2 -norm, and define Z -convergence, u n Z u if u n u in L 2 and {u n } n is bounded in L. Notice that Z -convergence is not induced by a metric. 8/14

41 Basic spectral theory Denote by λ 1 (u) the first eigenvalue of DF (u) = b + f (u) and φ 1 (u) the associated (positive, normalized) eigenfunction. DF(u) is not invertible iff λ 1 (u) = 0. λ 1 : B 2 b (:= H2 b C2,α ) R and φ 1 : B 2 b H0 are smooth. Equip Z = L with the L 2 -norm, and define Z -convergence, u n Z u if u n u in L 2 and {u n } n is bounded in L. The extensions λ 1 : Z R and φ 1 : Z H 0 are Z-sequentially continuous. Notice that Z -convergence is not induced by a metric. 8/14

42 Critical points of the height function Differentiate the formula for h, F(u(t)) = z + h(u(t))ψ 1,b : 9/14

43 Critical points of the height function Differentiate the formula for h, F(u(t)) = z + h(u(t))ψ 1,b : ( ) d b u (t) f (u)u (t) = DF(u(t))u (t) = dt h(u(t)) ψ 1,b. 9/14

44 Critical points of the height function Differentiate the formula for h, F(u(t)) = z + h(u(t))ψ 1,b : ( ) d b u (t) f (u)u (t) = DF(u(t))u (t) = dt h(u(t)) ψ 1,b. The derivative of h(u(t)) is zero exactly at critical points u(t 0 ) of F. 9/14

45 Critical points of the height function Differentiate the formula for h, F(u(t)) = z + h(u(t))ψ 1,b : ( ) d b u (t) f (u)u (t) = DF(u(t))u (t) = dt h(u(t)) ψ 1,b. The derivative of h(u(t)) is zero exactly at critical points u(t 0 ) of F. The eigenfunction φ 1 (u(t 0 )) is a positive multiple of u (t 0 ), the tangent vector to the fiber at u(t 0 ). 9/14

46 Critical points of the height function Differentiate the formula for h, F(u(t)) = z + h(u(t))ψ 1,b : ( ) d b u (t) f (u)u (t) = DF(u(t))u (t) = dt h(u(t)) ψ 1,b. The derivative of h(u(t)) is zero exactly at critical points u(t 0 ) of F. The eigenfunction φ 1 (u(t 0 )) is a positive multiple of u (t 0 ), the tangent vector to the fiber at u(t 0 ). This is slightly harder, h is essentially λ 1 : If λ 1 (u(t 0 )) 0, there is p : Bb 2 R, p > 0, for which d dt h(u(t)) = Dh(u(t)) u (t) = p(u(t)) λ 1 (u(t)) near t 0. 9/14

47 Critical points of the height function Differentiate the formula for h, F(u(t)) = z + h(u(t))ψ 1,b : ( ) d b u (t) f (u)u (t) = DF(u(t))u (t) = dt h(u(t)) ψ 1,b. The derivative of h(u(t)) is zero exactly at critical points u(t 0 ) of F. The eigenfunction φ 1 (u(t 0 )) is a positive multiple of u (t 0 ), the tangent vector to the fiber at u(t 0 ). This is slightly harder, h is essentially λ 1 : If λ 1 (u(t 0 )) 0, there is p : Bb 2 R, p > 0, for which d dt h(u(t)) = Dh(u(t)) u (t) = p(u(t)) λ 1 (u(t)) near t 0. We search for u = u(t 0 ) B 2 b such that h (u) = 0 h (u) > 0 (a local minimum of h along the fiber). 9/14

48 A transversality check: δ 1 From the previous formula, to get a second derivative of h, we need the derivative of λ 1, a standard computation: Dλ 1 (u) v = λ 1 (u), v = f (u)φ 2 1 (u) v, λ 1 (u) = f (u)φ 2 1 (u). ( λ 1 admits a (Z -sequentially continuous) extension to Z = L.) Ω δ 1 = 0 φ 1 (u) λ 1 10/14

49 A transversality check: δ 1 From the previous formula, to get a second derivative of h, we need the derivative of λ 1, a standard computation: Dλ 1 (u) v = λ 1 (u), v = f (u)φ 2 1 (u) v, λ 1 (u) = f (u)φ 2 1 (u). ( λ 1 admits a (Z -sequentially continuous) extension to Z = L.) Define δ 1 (u) := λ 1 (u), φ 1 (u) = Ω f (u)φ 3 1 (u). Ω δ 1 = 0 φ 1 (u) λ 1 10/14

50 A transversality check: δ 1 From the previous formula, to get a second derivative of h, we need the derivative of λ 1, a standard computation: Dλ 1 (u) v = λ 1 (u), v = f (u)φ 2 1 (u) v, λ 1 (u) = f (u)φ 2 1 (u). ( λ 1 admits a (Z -sequentially continuous) extension to Z = L.) Define δ 1 (u) := λ 1 (u), φ 1 (u) = Ω f (u)φ 3 1 (u). Combining with the formula for h, At a critical point u with λ 1 (u) 0, h (u) = c δ 1 (u), c > 0. So we search for a critical point u with δ 1 (u) > 0. Ω δ 1 = 0 φ 1 (u) λ 1 10/14

51 Simple two valued functions in Ω L s min Ω s s R c Ω s s max Let e R n. Hyperplanes in R n orthogonal to e split Ω in two sets, Ω s = {x Ω, x, e s}, Ω c s = {x Ω, x, e > s} It looks like we forgot completely boundary conditions 11/14

52 Simple two valued functions in Ω L s min Ω s s R c Ω s s max Let e R n. Hyperplanes in R n orthogonal to e split Ω in two sets, Ω s = {x Ω, x, e s}, Ω c s = {x Ω, x, e > s} with characteristic functions χ s and χ c s, s [s min, s max ]. It looks like we forgot completely boundary conditions 11/14

53 Simple two valued functions in Ω L s min Ω s s R c Ω s s max Let e R n. Hyperplanes in R n orthogonal to e split Ω in two sets, Ω s = {x Ω, x, e s}, Ω c s = {x Ω, x, e > s} with characteristic functions χ s and χ c s, s [s min, s max ]. The set of two-valued functions is V = {u(l, R, s) = L χ s + R χ c s, L, R, s R} Z. We look for u V so that λ 1 (u) = 0 and δ 1 (u) > 0. It looks like we forgot completely boundary conditions 11/14

54 Simple two valued functions in Ω L s min Ω s s R c Ω s s max Let e R n. Hyperplanes in R n orthogonal to e split Ω in two sets, Ω s = {x Ω, x, e s}, Ω c s = {x Ω, x, e > s} with characteristic functions χ s and χ c s, s [s min, s max ]. The set of two-valued functions is V = {u(l, R, s) = L χ s + R χ c s, L, R, s R} Z. We look for u V so that λ 1 (u) = 0 and δ 1 (u) > 0. Mollification of u then yields the required u B 2 b. It looks like we forgot completely boundary conditions 11/14

55 Looking for u with δ 1 (u) > 0 & λ 1 (u) = 0, I We first find a family u s, s [s min, s max ], for which δ 1 (u s ) > 0. 12/14

56 Looking for u with δ 1 (u) > 0 & λ 1 (u) = 0, I We first find a family u s, s [s min, s max ], for which δ 1 (u s ) > 0. Take x s.t. f (x ) < 0, with f (x ) < µ 1,b (other cases are similar). M 0 x f µ 1 - x m 12/14

57 Looking for u with δ 1 (u) > 0 & λ 1 (u) = 0, I We first find a family u s, s [s min, s max ], for which δ 1 (u s ) > 0. Take x s.t. f (x ) < 0, with f (x ) < µ 1,b (other cases are similar). M 0 x f µ 1 - x m Lucky case: there exists x 0 such that f (x 0 ) = 0 and f (x 0 ) > µ 1,b. 12/14

58 Looking for u with δ 1 (u) > 0 & λ 1 (u) = 0, I We first find a family u s, s [s min, s max ], for which δ 1 (u s ) > 0. Take x s.t. f (x ) < 0, with f (x ) < µ 1,b (other cases are similar). M 0 x f µ 1 - x m Lucky case: there exists x 0 such that f (x 0 ) = 0 and f (x 0 ) > µ 1,b. For u s = x χ s + x 0 χ c s, we have f (u) = f (x )χ s + f (x 0 )χ c s and δ 1 (u s ) = Ω s f (x )φ 1 (u s ) 3 f (x 0 )φ Ω c 1 (u s ) 3 > 0. s 12/14

59 Looking for u with δ 1 (u) > 0 & λ 1 (u) = 0, I We first find a family u s, s [s min, s max ], for which δ 1 (u s ) > 0. Take x s.t. f (x ) < 0, with f (x ) < µ 1,b (other cases are similar). M 0 x f µ 1 - x m Lucky case: there exists x 0 such that f (x 0 ) = 0 and f (x 0 ) > µ 1,b. For u s = x χ s + x 0 χ c s, we have f (u) = f (x )χ s + f (x 0 )χ c s and δ 1 (u s ) = Ω s f (x )φ 1 (u s ) 3 f (x 0 )φ Ω c 1 (u s ) 3 > 0. s This holds for any s. In the extremal cases, f (u s ) µ 1,b is a constant which is either smaller or larger than zero for an intermediate value of s, u s is a critical point! 12/14

60 Looking for u with δ 1 (u) > 0 & λ 1 (u) = 0, II If you are out of luck, x 0 such that f (x 0 ) = 0 and f (x 0 ) > µ 1. x + m! 1 f x - Search for almost critical points of f : a sequence x + m such that R m = f (x + m ) R + > µ 1,b with f (x + m ) 0. For R m = f (u + m) R +, the (Z -convergent) potentials f (x )χ s + R m χ c s Z f (x )χ s + R + χ c s yield (uniformly bounded, by SBC!) eigenfunctions φ 1 (u + m) φ 1. Since again δ 1 (x χ s + R + χ c s) > 0, by continuity δ 1 (u K = x χ s + R K χ c s) > 0 for some large K. 13/14

61 Looking for u with δ 1 (u) > 0 & λ 1 (u) = 0, II If you are out of luck, x 0 such that f (x 0 ) = 0 and f (x 0 ) > µ 1. x + m! 1 f x - Search for almost critical points of f : a sequence x + m such that R m = f (x + m ) R + > µ 1,b with f (x + m ) 0. For R m = f (u + m) R +, the (Z -convergent) potentials f (x )χ s + R m χ c s Z f (x )χ s + R + χ c s yield (uniformly bounded, by SBC!) eigenfunctions φ 1 (u + m) φ 1. Since again δ 1 (x χ s + R + χ c s) > 0, by continuity δ 1 (u K = x χ s + R K χ c s) > 0 for some large K. Balance s as before to get u K, λ 1 (u k ) = 0, δ 1 (u K ) > 0. 13/14

62 Smoothing: u D with λ 1 (u) = 0, δ 1 (u) > 0 Take a ball U Z centered in u K in which δ 1 is positive. The choice of D reintroduces the boundary conditions. 14/14

63 Smoothing: u D with λ 1 (u) = 0, δ 1 (u) > 0 Take a ball U Z centered in u K in which δ 1 is positive. For small τ, along the segment u(τ) = u(x + τ, x + K, s K ) H 0 f (x ) < 0, λ 1 (u(τ)) at τ = 0 is strictly monotonic. Take D dense in H 2 b δ 1> 0 U u k o The choice of D reintroduces the boundary conditions. 14/14

64 Smoothing: u D with λ 1 (u) = 0, δ 1 (u) > 0 Take a ball U Z centered in u K in which δ 1 is positive. For small τ, along the segment u(τ) = u(x + τ, x + K, s K ) H 0 f (x ) < 0, λ 1 (u(τ)) at τ = 0 is strictly monotonic. Take D dense in H 2 b U δ 1> 0 o u k o o The choice of D reintroduces the boundary conditions. 14/14

65 Smoothing: u D with λ 1 (u) = 0, δ 1 (u) > 0 Take a ball U Z centered in u K in which δ 1 is positive. For small τ, along the segment u(τ) = u(x + τ, x + K, s K ) H 0 f (x ) < 0, λ 1 (u(τ)) at τ = 0 is strictly monotonic. Take D dense in H 2 b U λ 1>0 u + o δ 1> 0 u k o λ1<0 u - o The choice of D reintroduces the boundary conditions. 14/14

66 Smoothing: u D with λ 1 (u) = 0, δ 1 (u) > 0 Take a ball U Z centered in u K in which δ 1 is positive. For small τ, along the segment u(τ) = u(x + τ, x + K, s K ) H 0 f (x ) < 0, λ 1 (u(τ)) at τ = 0 is strictly monotonic. Take D dense in H 2 b U λ 1>0 u + o δ 1> 0 u k o λ1<0 u - o The choice of D reintroduces the boundary conditions. 14/14

67 Smoothing: u D with λ 1 (u) = 0, δ 1 (u) > 0 Take a ball U Z centered in u K in which δ 1 is positive. For small τ, along the segment u(τ) = u(x + τ, x + K, s K ) H 0 f (x ) < 0, λ 1 (u(τ)) at τ = 0 is strictly monotonic. Take D dense in H 2 b δ 1> 0 U u + o λ 1>0 + ou u k o λ1<0 u - o o - u The choice of D reintroduces the boundary conditions. 14/14

68 Smoothing: u D with λ 1 (u) = 0, δ 1 (u) > 0 Take a ball U Z centered in u K in which δ 1 is positive. For small τ, along the segment u(τ) = u(x + τ, x + K, s K ) H 0 f (x ) < 0, λ 1 (u(τ)) at τ = 0 is strictly monotonic. Take D dense in H 2 b δ 1> 0 U u + o λ 1>0 + ou u k o λ1<0 u - o o - u The choice of D reintroduces the boundary conditions. 14/14

69 Smoothing: u D with λ 1 (u) = 0, δ 1 (u) > 0 Take a ball U Z centered in u K in which δ 1 is positive. For small τ, along the segment u(τ) = u(x + τ, x + K, s K ) H 0 f (x ) < 0, λ 1 (u(τ)) at τ = 0 is strictly monotonic. Take D dense in H 2 b δ 1> 0 U u + o λ 1>0 + ou u k o o u λ1<0 u - o o - u The choice of D reintroduces the boundary conditions. 14/14

70 Smoothing: u D with λ 1 (u) = 0, δ 1 (u) > 0 Take a ball U Z centered in u K in which δ 1 is positive. For small τ, along the segment u(τ) = u(x + τ, x + K, s K ) H 0 f (x ) < 0, λ 1 (u(τ)) at τ = 0 is strictly monotonic. Take D dense in H 2 b δ 1> 0 U u + o λ 1>0 + ou u k o o u λ1<0 u - o o - u Final check: Because δ 1 (u) > 0, λ 1 (u) 0 and thus also near u B 2 b the functions h and δ 1 are essentially the same. The choice of D reintroduces the boundary conditions. 14/14

71 Smoothing: u D with λ 1 (u) = 0, δ 1 (u) > 0 Take a ball U Z centered in u K in which δ 1 is positive. For small τ, along the segment u(τ) = u(x + τ, x + K, s K ) H 0 f (x ) < 0, λ 1 (u(τ)) at τ = 0 is strictly monotonic. Take D dense in H 2 b δ 1> 0 U u + o λ 1>0 + ou u k o o u λ1<0 u - o o - u Final check: Because δ 1 (u) > 0, λ 1 (u) 0 and thus also near u B 2 b the functions h and δ 1 are essentially the same. For a different point of view come to Carlos talk. The choice of D reintroduces the boundary conditions. 14/14

72 Have a nice time in Varese!.../14

73 A. Ambrosetti and G. Prodi, On the inversion of some differentiable mappings with singularities between Banach spaces, Ann. Mat. Pura Appl. 93 (1972) M. Berger and P.T. Church, Complete integrability and perturbation of a nonlinear Dirichlet problem. I, Indiana Univ. Math. J. 28 (1979) M.S. Berger and E. Podolak, On the solutions of a nonlinear Dirichlet problem, Indiana Univ. Math. J. 24 (1974) M. Calanchi, C. Tomei and A. Zaccur, Fibers and the global geometry of functions, N. Dancer, On the ranges of certain weakly nonlinear elliptic partial differential equations, J. Math. Pures Appl. 57 (1978) A. Manes and A.M. Micheletti, Un estensione della teoria variazionale classica degli autovalori per operatori ellittici del secondo ordine, Boll. U. Mat. Ital. 7 (1973) Some references..../14