Lane-Emden problems: symmetries of low energy solutions

Lane-Emden roblems: symmetries of low energy solutions Ch. Grumiau Institut de Mathématique Université de Mons Mons, Belgium June 2012 Flagstaff, Arizona (USA) Joint work with M. Grossi and F. Pacella

The Problem Let N = 2, 1 < and a smooth oen bounded Ω R 2. We consider { u = u 1 u, in Ω, (P) u = 0, on Ω. Solutions are critical oints of the energy functional Energy functional E : H0 1(Ω) R : u 1 2 Ω u 2 1 +1 H0 1(Ω) := C 0 2(Ω) for u 2 := H0 1 Ω u 2 0 is always a solution Main goal : nodal solutions when + Remark : N = 2 no restriction on Ω u +1 Ch. Grumiau (UMons) Lane-Emden roblems 2 / 27

Outline 1 Symmetries of ground state solutions (non-trivial solution with minimum energy) 1 Existence 2 Symmetry results 2 Symmetries of least energy nodal solution (l.e.n.s. : sign-changing solution with minimum energy) Ch. Grumiau (UMons) Lane-Emden roblems 4 / 27

Existence of non-zero solutions for { u = u 1 u, in Ω u = 0, on Ω.. Mountain-Pass theorem (Z. Nehari A. Ambrosetti, P. H. Rabinowitz) There exists a non-zero solution. Sketch: The energy functional E ossesses a Mountain-Pass structure E (u) = 1 2 Ω u 2 1 +1 Ω u+1 N := {u 0 : (u, E(u)) To of the Mountain } and is called 0 λ u u u Nehari manifold Ch. Grumiau (UMons) Lane-Emden roblems 5 / 27

Remarks on N N is formed by all the non-zero functions u such that E(u)u = 0...you maximize energy in the direction of u Projection P : H 1 0 \ {0} N : u ( u 2 / u +1 +1 )1/( 1) u N is a manifold of codimension 1. So, Morse index of a ground state solution is 1. Ch. Grumiau (UMons) Lane-Emden roblems 6 / 27

Examles: Mountain-Pass algo of Y. Choi and J. McKenna, 93: Ch. Grumiau (UMons) Lane-Emden roblems 7 / 27

Symmetry result for ground state Theorem (B. Gidas,W. M. Ni, L. Nirenberg, 79) When Ω is convex, one and only one ground state (u to a multilicative factor by 1) it resects reflections that leaves Ω invariant in articular, ground state is radial on balls one and only one maximum (res. minimum) at the ositive (res. negative) ground state which is away from Ω ositive (res. negative) ground state is decreasing (res. increasing) from this maximum to the boundary Proosition (Adimurthi, M. Grossi, 04) When +, u e Idea : rescaling argument around the eaks (which are away from the boundary). Ch. Grumiau (UMons) Lane-Emden roblems 8 / 27

Outline 1 Symmetries of ground state solutions (non-trivial solution with minimum energy) 2 Symmetries of least energy nodal solution (l.e.n.s. : sign-changing solution with minimum energy) 1 Existence 2 Motivation : small 3 Partial results for + (in coll. with M. Grossi and F.Pacella) l.e.n.s. are of low energy; i.e. E (u ) 8πe For low energy solutions : rescaling around u ± Under some assumtions, u ± e Green and nodal line characterization : the nodal line intersects the boundary for large On balls, at the limit when +, the nodal line stays a diagonal. Ch. Grumiau (UMons) Lane-Emden roblems 9 / 27

Existence of nodal solution Theorem (A. Castro, J. Cossio, J. M. Neuberger, 97) There exists a nodal solution with minimum energy. N M := {u : u ± N } M u os u neg 0 λ u u u u 0 u 0 Projection: sh 1 0 \ {0} M : u P N (u + ) + P N (u ). Remark 1: You maximize energy in {tu + + su : t, s 0} Remark 2: Gidas, Ni and Nirenberg method fails. Ch. Grumiau (UMons) Lane-Emden roblems 10 / 27

Ex: radial domains ( u = u 3 ) Ch. Grumiau (UMons) Lane-Emden roblems 11 / 27

Symmetries? Theorem(T. Bartsch, T. Weth, M. Willem, 05) On radial domains, l.e.n.s. are foliated Schwarz symmetric. In articular, they are even with resect to N 1 directions. Theorem (G., C. Troestler, 08 ) For close to 1, on radial domains, l.e.n.s. is unique (u to rotations) and odd with resect to some direction. Moreover, on general domains, Theorem(D. Bonheure, V. Bouchez, G., J. Van Schaftingen, 08) When dim E 2 = 1 (second eigensace of ), l.e.n.s. resect the symmetries of function in E 2 when is close to 1 (ex : rectangle). For general domains, l.e.n.s. resect the symmetries of its orthogonal rojection on E 2 (ex : square). Question : large? Ch. Grumiau (UMons) Lane-Emden roblems 12 / 27

Moreover, u u minimum of E : E 2 R : u Ω u2 log u 2 on N := {u 0 : E (u)u = 0}. Conjecture (D. Bonheure, V. Bouchez, G., J. Van Schaftingen, 08) On square, l.e.n.s. are odd with resect to a diagonal for small. Conjecture (D. Bonheure, V. Bouchez, G., J. Van Schaftingen, 08) There is a symmetry breaking on some rectangle close to the square. Using interval arithmetic, in collaboration with P. Hauweele and C. Troestler, the second conjecture has recently been roved. Question : what for other domains and large? Ch. Grumiau (UMons) Lane-Emden roblems 13 / 27

L.e.n.s. are of low energy Definition u is of low energy iff E (u ) 8πe. Proosition Least energy nodal solutions are of low energy. Ch. Grumiau (UMons) Lane-Emden roblems 15 / 27

Proof (lim su E (u ) 8πe) : test functions (1/2) 1 Let a, b Ω and r > 0 s.t. B(a, r), B(b, r) Ω and B(a, r) B(b, r) =. 2 test function : W defined on B(a, r) as 3 W (x) := ϕ(x) e (1 + where z(x) = 2 log(1 + x 2 8 ) and ε2 := 1 W +1 = W 2 = 8πe + o(1/) z( x a ε ) e 1. 4 Then, W = W on B(a, r) and the odd reflection on B(b, r) and roject W α W on M. 5 (α ) 1 = W ± 2 / W ± +1. So, α 1. ( ) ( ) 6 1 2 1 +1 u 2 2 1 2 1 +1 2(α ) 2 Ω W ± 2. ) Ch. Grumiau (UMons) Lane-Emden roblems 16 / 27

Proof (lim inf E (u ) 8πe) (2/2) 1 = (u ± ) +1 / u ± 2. for t > 1, u t D t t 1/2 u 2 where D t (8πe) 1/2 (Ren, Wei) ( ) 1 2 1 +1 ( + 1) +1 ( ) 1 Ω u± 2 1 2 1 +1 2 1 +1 D+1. Remark It doesn t deend on the fact that u is a least energy nodal solution. ( 1 2 1 +1 ) Ω u± 2 4πe and Ω u± 2 8πe. E (u ) 0, Ω u 2 0, Ω u± 2 0. Low energy solutions have 2 nodal domains. Ch. Grumiau (UMons) Lane-Emden roblems 17 / 27

Localization and characterization of x ± Let x ± be the maximum and minimum oint of u and λ 1 be the first eigenvalue of with DBC. Proosition u (x ± ) 1 λ 1. Sketch: 1 = Ω u± +1 Ω u± u ± (x ) 1 Ω (u± ) 2 2 Ω u± 2 u (x ± ) 1 λ 1 1 ( Ω ± ), where Ω ± are the nodal domains. We finish as Ω ± Ω. Remark: For ε > 0, u (x ± ) 1 ε. Goal : imrove it? Ch. Grumiau (UMons) Lane-Emden roblems 18 / 27

x + away from the boundary Assume that u (x + ) u (x ). Let ε 2 Proosition d(x +, Ω) ε +. := u (x + ) 1 +. Sketch : Else, d(x+, Ω) ε l 0 and x + x Ω (i.e. d(x+,x ) ε D: reflection of V V x l). u : odd reflection of u which is solution on D. z(x) := u (x+ ) (u (ε x + x + ) u(x + )) on Ω := D x+ ε R 2. z = 1 + z 1 ( ) 1 + z and 1 + z 1. Ch. Grumiau (UMons) Lane-Emden roblems 19 / 27

x + away from the boundary Assume that u (x + ) u (x ). Let ε 2 Proosition d(x +, Ω) ε +. := u (x + ) 1 +. Sketch : Else, d(x+, Ω) ε l 0 and x + x Ω (i.e. d(x+,x ) ε l). V x D D: reflection of V u : odd reflection of u which is solution on D. z(x) := u (x+ ) (u (ε x + x + ) u(x + )) on Ω := D x+ ε R 2. z = 1 + z 1 ( ) 1 + z and 1 + z 1. Ch. Grumiau (UMons) Lane-Emden roblems 19 / 27

x + away from the boundary V x D D: reflection of V u : odd reflection of u which is solution on D. z(x) := u (x+ ) (u (ε x + x + ) u(x + )) on Ω := D x+ ε R 2. z = 1 + z 1 ( ) 1 + z and 1 + z 1. Key oint: for large R, consider w = 1 + z w = 0, 1 (1 + z ), in B(0, R), on B(0, R). and use Harnack s inequalities on the harmonic function z w to get z uniformly bounded on B(0, R) (as z (0) = 0). Contradiction : y := x x+ ε B[0, l + 1] and z (y) =. Ch. Grumiau (UMons) Lane-Emden roblems 19 / 27

Blow-u on Ω to control x + Theorem z (x) := u (x + ) (u (ε x + x + ) u (x + )) on Ω x+ ε converges in R 2 to z = 2 log(1 + 1 8 x 2 )) solving z = e z with e z = 8π. Sketch: Same idea, Ω x+ ε w = 1 + z w = 0, R 2 and consider 1 (1 + z ), in B(0, R), on B(0, R). z bounded in L (B(0, R)) for any R. By regularity, z z in Cloc 2 (R2 ) and z = e z. As ( R e z µ < +, we are done (z(x) = log )). 2 (1+ µ 8 x 2 ) 2 Imrovement : it works on Ω + x+ ε where Ω + is the nodal domain of u. Ch. Grumiau (UMons) Lane-Emden roblems 20 / 27

Control on x We would like to work with defined on Ω x ε. z (x) := u (x + ) ( u (ε x + x ) u (x + )) We need d(x, Ω) ε +. Problem: z (0) bounded in the use of Harnack s inequalities. Assume (we conjecture it is true for low energy, at least l.e.n.s.) (B) ( u (x + ) + u (x ) ) K; Theorem z z solving the equation z = e z on R 2, z 0, R e z = 8π and ( 2 µ z(x) = log for some 0 < µ 1. When K = 0 in (1+ µ 8 x 2 ) 2 ) condition (B), we get µ = 1. Ch. Grumiau (UMons) Lane-Emden roblems 21 / 27

u e 1/2 Let u be a ground state solution : u e (see Adimurthi, M. Grossi). Working in the same way, we get Theorem u e. Under (B), u ± e. Sketch ( ): Ω + (ε ) = ( Ω + x + )/ε ( ) Ω 1 = u+ +1 u + (x + +1 ) u + +1 = u + ε 2 +1 1 + z +1 +1 Ω + (ε ) = u+ (x + ) 2 u + +1 1 + z +1 +1 Ω + (ε ) lim su + u + (x + ) 2 e z. 8πe R 2 Remark : For N > 3 and 2, it converges to + (M. Ben Ayed, K. El Medhi, F. Pacella). Ch. Grumiau (UMons) Lane-Emden roblems 22 / 27

Green s characterization Proosition Under (B), u 8πe 1/2 (G(, x + ) G(, x )) in C 1 loc (Ω \ {x, x + }) and x + x Ω. Sketch : 1 u (x) = Ω+ G(x, y) u (y) dy Ω G(x, y) u (y) dy. 2 u 8πe 1/2 (G(., x + ) G(., x )) u to a sub where x ± are the limit oints of x ± 3 alternatives Pohozaev s identities to get x + x Ω. 4 x ± / Ω x + x. Otherwise, u 0 and by Pohozaev s identities (times 2 ) 2 + 1 Ω u +1 = 1 (x ν)( ν (u )) 2. 4 Ω Ch. Grumiau (UMons) Lane-Emden roblems 23 / 27

Characterization of the nodal line. Proosition Under (B), the nodal line intersects the boundary for large. Sketch : Otherwise u is one-signed close to Ω. Then, ν u is one-signed on Ω. As x + x and Ω ν(g(, x + ) G(, x )) = 0, the normal derivative of the limit function changes its sign. It contradicts the C 1 -convergence of u to 8π e(g(, x + ) G(, x )) in a comact neighborhood of Ω. Ch. Grumiau (UMons) Lane-Emden roblems 24 / 27

Characterization of x ± Proosition Under (B), x + and x satisfy the system G (x +, x ) H (x +, x + ) = 0, x i x i G (x, x + ) H (x, x ) = 0, x i x i At the limit on balls, the nodal line of u goes to a diagonal. Sketch : 1 On balls, we know the Green and Robin functions. 2 The system imlies that x + and x are antiodal. Ch. Grumiau (UMons) Lane-Emden roblems 25 / 27

In rogress Same question for the least nodal radial solution : it seems that It seems that it is not of low energy (cometition between the nodal domain B(0, x ) and an annulus). Question : accurate localization of x. The rescaling argument is working for the domain with 0 and we get a eak at 0 of value e. The rescaling in the second domain gives some singularity in log t. Ch. Grumiau (UMons) Lane-Emden roblems 26 / 27

Thanks for your attention Ch. Grumiau (UMons) Lane-Emden roblems 27 / 27