Elliptic Kirchhoff equations - PDF Free Download

Elliptic Kirchhoff equations David ARCOYA Universidad de Granada Sevilla, 8-IX-2015 Workshop on Recent Advances in PDEs: Analysis, Numerics and Control In honor of Enrique Fernández-Cara for his 60th birthday January 25th - 27th, 2017 Supported by FEDER-MINECO MTM2015-68210-P and Junta de Andalucía FQM-116

Joint works with Antonio Ambrosetti

Joint works with Antonio Ambrosetti Kirchhoff (1883) introduced the hyperbolic equation ( 2 u L ) t 2 1 + u 2 2 u x dx = f, u(0, t) = u(l, t) = 0, x 2 0 to describe the small transversal oscillations of an elastic clamped string which is not homogeneous. Motivated by this many authors have studied the stationary states of this equation or more generally, of the elliptic Kirchhoff equation in a bounded smooth domain Ω R N, (N 3) M ( Ω u 2 dx ) u = λf (x, u), x Ω (K) u(x) = 0, x Ω under appropriate hypotheses on λ R and the functions f and M.

Previous works See e.g. C. Alves, F. Corrêa, T. Ma 2005 M. Chipot, F.J.S.A. Corrêa, 2009 G. Figueiredo, A. Suárez, 2015 G. Figueiredo, C. Morales-Rodrigo, J. Santos, A. Suárez, 2014 Z. Liang, F. Li, J. Shi, 2014 D. Naimen, 2015 which contain further references.

A non-local problem via nonlinear functional analysis Notice that the fact that the evaluation of M is depending on the all values of u(x) in the whole domain Ω M ( Ω u 2 dx ) u = λf (x, u), x Ω (K) u(x) = 0, x Ω. This means that the associated problem to the Kirchhoff equations is a non-local problem. We will show that, in spite of the presence of the non-local coefficient M( Ω u 2 dx), the Kirchhoff equations do not require special tools, but can be handled by standard arguments of nonlinear functional analysis.

Asymptotically linear nonlinearities f First, we use global bifurcation theory to prove the existence of branches of positive solutions when we deal with asymptotically linear f, i.e. satisfying (f 1 ) f (x, 0) = 0, x, a(x) := D u f (x, 0) > 0. (f 2 ) f (x, u) = b(x)u + g(x, u), where b(x) > 0 and g(x, u) lim = 0. u + u The results we find depends mainly on the interplay between the behavior of f and M at zero and infinity. More precisely, we consider three cases: M(t) > 0 for t 0 and lim M(t) = M > 0 t + or else M(t) = +. lim t + the coefficient M vanishes at zero and/or at infinity.

An equivalent fixed point problem If we assume: (M 1 ) m 0 > 0 such that M(t) m 0 for all t 0. the problem can be seen as a fixed point problem in E = H 1 0 (Ω) (with u 2 = Ω u 2 dx).

An equivalent fixed point problem If we assume: (M 1 ) m 0 > 0 such that M(t) m 0 for all t 0. the problem can be seen as a fixed point problem in E = H0 1 (Ω) (with u 2 = Ω u 2 dx). Indeed, since M( u 2 ) > 0, M ( Ω u 2 dx ) u = λf (x, u), x Ω u(x) = 0, x Ω. u = λf (x,u) M( u 2 ), x Ω u(x) = 0, x Ω. Let L = ( ) 1 : E L 2 (Ω). We see that the fixed points u = T (u) := λlf (u) M( u 2 ), u E (1) are just the weak and, by regularity, classical solutions of the Kirchhoff equation (K).

Bifurcation from zero The linearized equation of u = T (u), u E at u = 0 gives rise to M(0) v = λa(x)v, in Ω, v Ω = 0. (2) The eigenvalues of (2) are M(0)λ m [a] and have the same multiplicity of λ m [a] and the same eigefunctions. In particular Λ 1 := M(0)λ 1 [a] is simple with eigenfunctions that do not change sign. In the sequel we will denote by φ the eigenfunction associated to (λ 1 [a] and) Λ 1 such that φ > 0, normalized by φ = 1. Lemma If (f 1 ), (f 2 ) and (M 1 ) hold, then Λ 1 = M(0)λ 1 [a] is a bifurcation point of positive solution for (K), and the only one. Furthermore, from (Λ 1, 0) emanates an unbounded branch of positive solutions Γ (0, + ) E.

Bifurcation from infinity We say that λ R is a bifurcation point from infinity for (K) if there exists a sequence (λ k, u k ) R C0 2 (Ω) with λ k λ, u k +, such that u k a solution of M ( Ω u k 2 dx ) u k = λ k f (x, u k ), x Ω u k (x) = 0, x Ω. Lemma Suppose that (f 2 ) and (M ) M > 0 such that lim t + M(t) = M. hold. Then Λ = M λ 1 [b] is a bifurcation point from infinity of positive solutions to (K). Furthermore, from (Λ, ) emanates a branch of positive solutions, and the only one.

Global Bifurcation Γ meets (µ, + ) (λ k, u k ) Γ such that λ k µ and u k +.

Global Bifurcation Γ meets (µ, + ) (λ k, u k ) Γ such that λ k µ and u k +. Theorem Let (f 1 ), (f 2 ) and (M 1 ) hold and consider the branch Γ of positive solutions of (K) bifurcating from (Λ 1, 0). (i) If condition (M ) is also satisfied and (f 3 ) κ > 0 such that f (x, u) κu, (x, u) Ω R +, then Γ meets (Λ, + ). In particular, there exists positive sol. for every λ between Λ 1 and Λ. (ii) If lim t M(t) =, then the projection of Γ on the λ axis is an unbounded interval [l, + ) with l Λ 1. In particular, there exists positive sol. for every λ > Λ 1.

Global Bifurcation.. 0 Λ1 Λ λ 0 Λ1 λ Figura: Bifurcation diagrams

Degenerate coefficients M Theorem Suppose (f 1 ), (f 2 ), (f 3 ) and (M 2 ) M(t) > 0, for every t 0 and lim t + M(t) = 0 hold. Then there is an unbounded branch Γ of positive solutions of (K) bifurcating from (Λ 1, 0) that meets (0, + ). In particular, (K) has at least one positive solution for all 0 < λ < Λ 1. Proof. For each integer k > 0, let us consider the perturbed equation ( 1 k + M ( u 2) ) u = λf (x, u), in Ω, u Ω = 0. (3) Since 1 k + M ( u 2) 1 k > 0, previous theorem applies yielding a branch of positive solutions Γ k of (3) bifurcating from (Λ 1,k, 0), where Λ 1,k = ( 1 k + M(0))λ 1[a], that meets ( 1 k, ). By using Whyburn lemma we can pass to the limit and to conclude the proof.

Degenerate coefficients M Similarly we can handle the case in which M(0) = 0.

Degenerate coefficients M Similarly we can handle the case in which M(0) = 0. Theorem Suppose (f 1 ), (f 2 ), (f 3 ) hold. Assume also that M(0) = 0, M(t) > 0, t > 0, and that M 0 such that lim M(t) = M. t + Then there is a branch Γ of positive solutions of (K) emanating from (0, 0) which meets (M λ 1 [b], ). In particular: 1 if M > 0 then (K) has at least one positive solution for all 0 < λ < M λ 1 [b]; 2 if M = 0 then there exists λ > 0 such that (K) has at least two positive solutions for all 0 < λ < λ.

Variational methods Setting M(t) = 1 + γg (t), the Kirchhoff problem becomes variational.

Variational methods Setting M(t) = 1 + γg (t), the Kirchhoff problem becomes variational. Specifically, we deal with the non-homogeneous Kirchhoff boundary value problem { ( 1 + γ G ( u 2 ) ) u = F (u) + h(x), x Ω, (VK) u(x) = 0, x Ω, where h L 2 (Ω), F C 1 (R) and G C 1 (R + ) satisfy (F 1) a 1, a 2 > 0 such that F (u) a 1 + a 2 u p with 2n, if n 3, 2 < p < 2 = n 2 +, if n = 1, 2; (G0) G(0) = G (0) = 0 and G (t) 0, t 0; (G1) there exist k > 0 and q 1 such that G(t) kt q, t 0.

Euler functional Solutions of { ( 1 + γ G ( u 2 ) ) u = F (u) + h(x), x Ω, u(x) = 0, x Ω, are the stationary points u E of the C 1 functional J(u) = 1 2 u 2 + 1 2 γ G( u 2 ) F(u) Ω hu, where F(u) = F (u). Ω

Euler functional Solutions of { ( 1 + γ G ( u 2 ) ) u = F (u) + h(x), x Ω, u(x) = 0, x Ω, are the stationary points u E of the C 1 functional J(u) = 1 2 u 2 + 1 2 γ G( u 2 ) F(u) where F(u) = Ω F (u). In order to study J we will consider separately the cases p < 2q, p > 2q and p = 2q. Ω hu,

The case p < 2q Theorem Suppose that (F 1) a 1, a 2 > 0 such that F (u) a 1 + a 2 u p with 2n, if n 3, 2 < p < 2 = n 2 +, if n = 1, 2; (G0) G(0) = G (0) = 0 and G (t) 0, t 0; (G1) there exist k > 0 and q 1 such that G(t) kt q, t 0; hold and let p < 2q. Then (VK) has a solution for all h L 2 (Ω) and all γ > 0.

The case p < 2q Proof Assumptions (F 1), (G 1) and Sobolev embedding imply J(u) 1 2 γ k u 2q c 1 c 2 u p c 3 u. (4) p < 2q and γ > 0 = J is coercive. p < 2 = F is weakly continuous. G (t) 0 = G( u 2 ) is w.l.s.c. In conclusion, J is coercive and w.l.s.c. and thus has a global minimum that gives rise to a solution of (VK).

The case p < 2q Example Take γ = 1, G(t) = 1 q tq, F (u) = 1 p u p and consider the problem (1 + u 2q 2 ) u = u p 2 u + h(x), u Ω = 0. = a sol. for all p (2, 2 ) with 2q > p and all h L 2 (Ω).

The case p < 2q Example Take γ = 1, G(t) = 1 q tq, F (u) = 1 p u p and consider the problem (1 + u 2q 2 ) u = u p 2 u + h(x), u Ω = 0. = a sol. for all p (2, 2 ) with 2q > p and all h L 2 (Ω). Rmk Existence of solutions to the problem u = u p 2 u + h(x), u Ω = 0 is known provided 2 < p < 2 /2, (see A. Bahri and H. Berestycki, 1981 and A. Bahri and P.L. Lions, 1988). Moreover, we recall also that if 2 < p < 2 then a solution exists for h in a dense subset of L 2 (Ω), (see A. Bahri, 1981).

The case p < 2q Example When b(u) be a bounded continuous function, it is well known (Landesman-Lazer) that if the problem at resonance has a solution, then u = λ k u + b(u) + h(x), u Ω = 0 (5) inf b(u) u R Ω φ Ω hφ sup b(u) φ, u R Ω for every eigenfunction φ associated to λ k. Conversely if, for example, b(u) is monotone and inf b(u) φ < hφ < sup b(u) φ, u R Ω Ω u R Ω then (5) has a solution. Notice that F (u) = 1 2 λ ku 2 + u 0 b(s)ds satisfies (F 1) with p = 2, and hence, the equation (1 + G ( u 2 )) u = λ k u + b(u) + h(x), u Ω = 0, possesses a solution for all h L 2 (Ω), provided G(t) t q and q > 1.

The case p > 2q Theorem Suppose that F, G C 1 satisfy (F 1), (F 2), (G0), (G1 ) K > 0, q 1 such that G(t) K t q, t 0. (G2) pg(t) 2G (t)t, for all t 0. and let p > 2q. Then there exists ɛ > 0 such that (VK) has at least two solutions provided h L 2 < ɛ, for all γ 0. Proof Step 1. (F 2) and (G2) = (PS). Step 2. Let J 0 denote the functional J with h = 0. G 0, γ 0 = J 0 (u) = 1 2 u 2 + 1 2 γ G( u 2 ) F(u) 1 2 u 2 F(u). = δ, r > 0 such that J 0 (v) δ, v = r. = ɛ > 0 s.t. J(v) δ/2 v = r, h L 2 (Ω) < ɛ. As a consequence, since (PS) holds, J(u) has a strict local minimum u 1 in the interior of the ball centered at zero with radius r, (1 st sol. of (VK)).

The case p > 2q, cont. Step 3 Fixed v E, v = 1, v > 0, (G1 ) = J(tv) t2 2 + γ Kt2q F (tv) t Ω Ω (F 2) t2 2 + γ Kt2q c 2 t p v p t Ω hv. Ω hv.

The case p > 2q, cont. Step 3 Fixed v E, v = 1, v > 0, (G1 ) = J(tv) t2 2 + γ Kt2q F (tv) t Ω Ω (F 2) t2 2 + γ Kt2q c 2 t p v p t Ω hv. Ω hv. p > 2q = lim J(u) = = z E s.t. J(z) < J(u 1) t + (MP th.) = second sol. u 2 u 1.

The case p = 2q Theorem Let F C 1 (R) and G C 1 (R + ). (a) If (F 1), (G0) and (G1) hold, with p = 2q, then (VK) has a solution for all h L 2 (Ω) and all γ 1; (b) if (F 1), (F 2), (G0), (G1 ) and (G2) hold, with p = 2q, then there exists ɛ > 0 such that (VK) has at least two solutions provided h L 2 < ɛ and 0 γ 1.

Discontinuous nonlinearities We suppose that F C(R) is convex (by simplicity), satisfies the growth condition (F 1) and F has an upward jump discontinuity at a point u = a: (F 3) a R such that F C 1 (R {a}) and the limits F (a±) = lim u a± F (u) exist, are finite, with F (a ) < F (a+). A function F satisfying the preceding conditions has a subdifferential F (u) = {ξ R : F (v) F (u) ξ (v u), v R} {F (u)} if u a, = [F (a ), F (a+)] if u = a.

Discontinuous nonlinearities. Example A model nonlinearity that satisfies (F 3) with a = 0 is the function αu if u 0, F (u) = βu if u > 0, with α < β.

Discontinuous nonlinearities. Example A model nonlinearity that satisfies (F 3) with a = 0 is the function αu if u 0, F (u) = βu if u > 0, with α < β. In this case, F (u) = H(u) = α if u < 0, β if u > 0, [α, β] if u = 0, is a Heaviside like function.

Discontinuous nonlinearities Rmk ((F 1) and) (F 3) = F(u) = Ω F (u) is not Fréchet differentiable!

Discontinuous nonlinearities Rmk ((F 1) and) (F 3) = F(u) = Ω F (u) is not Fréchet differentiable! But it is locally Lipschitz continuous with subdifferential given by F(u) = F (u), (Chang, 1981).

Discontinuous nonlinearities Rmk ((F 1) and) (F 3) = F(u) = Ω F (u) is not Fréchet differentiable! But it is locally Lipschitz continuous with subdifferential given by F(u) = F (u), (Chang, 1981). As a consequence, also the functional (with γ = 1) J(u) = 1 2 u 2 + 1 2 G( u 2 ) F(u) hu, u E, is only locally Lipschitz continuous with J(u) = A(u) F(u) h = A(u) F (u) h, where A is defined by setting A(u), v = ( 1 + G ( u 2 ) ) u, v. Ω

Discontinuous nonlinearities Theorem Suppose that F C(R) is convex and satisfies (F 1), (F 3) and let G C 1 (R + ) satisfies (G0) and (G1), with 2q > p. Then for all h L 2 (Ω), problem (1 + G ( u 2 ) u(x) h(x) F (u(x)), a.e. in Ω. (dvk) has a solution u E. Moreover, the set Ω a = {x Ω : u(x) = a} has null Lebesgue measure and hence u solves (dvk) almost everywhere in Ω.