Multi-layer radial solutions for a supercritical Neumann problem

Transcription

1 Multi-layer radial solutions for a supercritical Neumann problem Massimo Grossi (Universitá di Roma I)..and

2 Multi-layer radial solutions for a supercritical Neumann problem Massimo Grossi (Universitá di Roma I)..and Denis Bonheure (ULB, Bruxelles)

3 Multi-layer radial solutions for a supercritical Neumann problem Massimo Grossi (Universitá di Roma I)..and Denis Bonheure (ULB, Bruxelles) Benedetta Noris (ULB, Bruxelles)

4 Multi-layer radial solutions for a supercritical Neumann problem Massimo Grossi (Universitá di Roma I)..and Denis Bonheure (ULB, Bruxelles) Benedetta Noris (ULB, Bruxelles) Susanna Terracini (Univ. di Torino)

5 Setting of the problem Let us consider the following problem u + u = u p in B 1 u > 0 in B 1, u ν = 0 on B 1 with N 2. This problem has a long history! Let us start by the Dirichlet case.

6 The Dirichlet case Some classical results Let us consider the following problem u + u = u p in B 1 u > 0 in B 1, u = 0 on B 1 with N 2.

7 The Dirichlet case Some classical results Let us consider the following problem u + u = u p in B 1 u > 0 in B 1, u = 0 on B 1 with N 2. All solutions are radial (Gidas, Ni and Nirenberg).

8 The Dirichlet case Some classical results Let us consider the following problem with N 2. u + u = u p in B 1 u > 0 in B 1, u = 0 on B 1 All solutions are radial (Gidas, Ni and Nirenberg). There is exactly one solution for 1 < p < N+2 N 2.

9 The Dirichlet case Some classical results Let us consider the following problem with N 2. u + u = u p in B 1 u > 0 in B 1, u = 0 on B 1 All solutions are radial (Gidas, Ni and Nirenberg). There is exactly one solution for 1 < p < N+2 N 2. There are no solutions for p N+2 N 2 (Pohozaev identity).

10 The Neumann case Let us consider the following problem u + u = u p in B 1 u > 0 in B 1, u = 0 on B 1 with N 2. The Neumann problem is much more involved...

11 The Neumann case Let us consider the following problem u + u = u p in B 1 u > 0 in B 1, u = 0 on B 1 with N 2. The Neumann problem is much more involved... First note that we have the trivial solution u 1. Then we are interested in the existence of noncostant solutions.

12 The Neumann case Let us consider the following problem u + u = u p in B 1 u > 0 in B 1, u = 0 on B 1 with N 2. The Neumann problem is much more involved... First note that we have the trivial solution u 1. Then we are interested in the existence of noncostant solutions. Moreover all previous results in the Dirichlet case are false! Indeed,

13 The Neumann case Let us consider the following problem u + u = u p in B 1 u > 0 in B 1, u = 0 on B 1 with N 2. The Neumann problem is much more involved... First note that we have the trivial solution u 1. Then we are interested in the existence of noncostant solutions. Moreover all previous results in the Dirichlet case are false! Indeed, Dirichlet problem (i) All solutions are radial (ii) Uniqueness for 1 < p < N+2 N 2 (iii) No solutions for p N+2 N 2 = Neumann problem (i) There exist nonradial solutions (ii) Nonuniqueness results (iii) Existence for any p > 1.

14 Some results the Neumann problem It is convenient to set our problem in a ball of radius R. So let us consider, u + u = u p in B R u > 0 in B R, u = 0 on B R

15 Some results the Neumann problem It is convenient to set our problem in a ball of radius R. So let us consider, u + u = u p in B R u > 0 in B R, u = 0 on B R Theorem, C.-S. Lin and W. M. Ni There exist radius R 0 and R 1 such that If 1 < p < N+2 N 2 there exists R 0 > 0 such that for any R R 0 the only solution is u 1. For any p > 1 there exists R 1 = R 1 (p) such that for any R R 1 there exists a nonconstant solution.

16 An interesting constraint In 2011 E. Serra and P. Tilli introduced the following interesting constraint, M = { u H 1 rad(b R ) u 0, u(r) u(s) for 0 r s 1 }.

17 An interesting constraint In 2011 E. Serra and P. Tilli introduced the following interesting constraint, M = { u H 1 rad(b R ) u 0, u(r) u(s) for 0 r s 1 }. Note that the function in M cannot concentrate, unless to have infinite H 1 norm.

18 An interesting constraint In 2011 E. Serra and P. Tilli introduced the following interesting constraint, M = { u H 1 rad(b R ) u 0, u(r) u(s) for 0 r s 1 }. Note that the function in M cannot concentrate, unless to have infinite H 1 norm. They proved the following result, Theorem, E. Serra and P. Tilli (2011) Assume that a(r) L 1 (B 1 ) is a radial, increasing, positive and nonconstant function. Then for any p > 1 the problem u + u = a( x )u p in B 1 u > 0 in B 1, u = 0 on B 1 admits at least one radially increasing solution.

19 A generalization to the autonomous case The previous result was generalized to the case a 1 by, and T. Weth (2012) Assume that p > λ2 (B1 ). Then the problem p in B1 u + u = u u>0 in B1, u=0 on B1 admits at least one radially increasing solution.

20 Extension to non-spherical domains The previous results lead to the following conjecture,

21 Extension to non-spherical domains The previous results lead to the following conjecture, Open problem Let us consider the problem, u + u = u p in Ω u > 0 in Ω, u = 0 on Ω, where Ω is a convex domain.

22 Extension to non-spherical domains The previous results lead to the following conjecture, Open problem Let us consider the problem, u + u = u p in Ω u > 0 in Ω, u = 0 on Ω, where Ω is a convex domain. Then, for any p > 1 (not too small), there exists a solution u p satisfying, (x x p ) u p (x) 0 for some x p Ω and for any x Ω.

23 The case where p + (Dirichlet problem) If the exponent p goes to + we have additional information. Let us recall the following result in the Dirichlet case,

24 The case where p + (Dirichlet problem) If the exponent p goes to + we have additional information. Let us recall the following result in the Dirichlet case, (2005) Let u p be the radial solution of the problem the problem u = u p in A = { x R N a x b } u > 0 in A = { x R N a x b } u = 0 on x = a and x = b. Then, as p + we have that,

25 The case where p + (Dirichlet problem) If the exponent p goes to + we have additional information. Let us recall the following result in the Dirichlet case, (2005) Let u p be the radial solution of the problem the problem u = u p in A = { x R N a x b } u > 0 in A = { x R N a x b } u = 0 on x = a and x = b. Then, as p + we have that, (i) u p 1,

26 The case where p + (Dirichlet problem) If the exponent p goes to + we have additional information. Let us recall the following result in the Dirichlet case, (2005) Let u p be the radial solution of the problem the problem u = u p in A = { x R N a x b } u > 0 in A = { x R N a x b } u = 0 on x = a and x = b. Then, as p + we have that, (i) u p 1, (ii) u p (r) G(r,r 0) G(r 0,r 0 ) where G(r, s) is the Green function of the operator u N 1 r u and r 0 is given by ( a 2 N + b 2 N r 0 = 2 ) 1 2 N

27 Ideas of the proof of the previous result

28 Ideas of the proof of the previous result (i) Why is the maximum of u p going to 1?

29 Ideas of the proof of the previous result (i) Why is the maximum of u p going to 1? Let us recall that u p can be characterized as I p = inf u H 1 0 (A) ( B 1 u 2 B 1 u p+1 ) 2 p+1

30 Ideas of the proof of the previous result (i) Why is the maximum of u p going to 1? Let us recall that u p can be characterized as I p = inf u H 1 0 (A) ( B 1 u 2 B 1 u p+1 ) 2 p+1 It is not difficult to see that 0 < C 0 I p C 1. Then,

31 Ideas of the proof of the previous result (i) Why is the maximum of u p going to 1? Let us recall that u p can be characterized as I p = inf u H 1 0 (A) ( B 1 u 2 B 1 u p+1 ) 2 p+1 It is not difficult to see that 0 < C 0 I p C 1. Then, (a) If u p M > 1 we get that u p + in a set of positive measure and so I p +.

32 Ideas of the proof of the previous result (i) Why is the maximum of u p going to 1? Let us recall that u p can be characterized as I p = inf u H 1 0 (A) ( B 1 u 2 B 1 u p+1 ) 2 p+1 It is not difficult to see that 0 < C 0 I p C 1. Then, (a) If u p M > 1 we get that u p + in a set of positive measure and so I p +. (b) On the other hand if u p M < 1 we get that u p 0 uniformly and this is not possible since 0 is an isolated solution.

33 Ideas of the proof of the previous result (i) Why is the maximum of u p going to 1? Let us recall that u p can be characterized as I p = inf u H 1 0 (A) ( B 1 u 2 B 1 u p+1 ) 2 p+1 It is not difficult to see that 0 < C 0 I p C 1. Then, (a) If u p M > 1 we get that u p + in a set of positive measure and so I p +. (b) On the other hand if u p M < 1 we get that u p 0 uniformly and this is not possible since 0 is an isolated solution. (ii) Why is u p converging to G(r,r 0) G(r 0,r 0 )?

34 Ideas of the proof of the previous result (i) Why is the maximum of u p going to 1? Let us recall that u p can be characterized as I p = inf u H 1 0 (A) ( B 1 u 2 B 1 u p+1 ) 2 p+1 It is not difficult to see that 0 < C 0 I p C 1. Then, (a) If u p M > 1 we get that u p + in a set of positive measure and so I p +. (b) On the other hand if u p M < 1 we get that u p 0 uniformly and this is not possible since 0 is an isolated solution. (ii) Why is u p converging to G(r,r 0) G(r 0,r 0 )? By the previous step we derive that u p (r) u < 1 far away from the maximum point. Passing to the limit we get that u N 1 r u = 0 in (a, b) and the claim follows.

35 The case where p + (Neumann problem) In the case of Neumann boundary condition we have that the maximum point is located at the boundary,

36 The case where p + (Neumann problem) In the case of Neumann boundary condition we have that the maximum point is located at the boundary, and (2012) Let up be the solution of the problem the problem p in B1 u + u = u u>0 in B1, u=0 on B1

37 The case where p + (Neumann problem) In the case of Neumann boundary condition we have that the maximum point is located at the boundary, and (2012) Let up be the solution of the problem the problem p in B1 u + u = u u>0 in B1, u=0 on B1 Then, as p +, there exists a solution up such that G(r, 1) up (r ), G(1, 1) where G(r, s) is the Green function of u 00 N 1 + u with r u 0 (0) = u 0 (1) = 0.

38 The case where p + (Neumann problem) In the case of Neumann boundary condition we have that the maximum point is located at the boundary, and (2012) Let up be the solution of the problem the problem p in B1 u + u = u u>0 in B1, u=0 on B1 Then, as p +, there exists a solution up such that G(r, 1) up (r ), G(1, 1) where G(r, s) is the Green function of u 00 N 1 + u with r u 0 (0) = u 0 (1) = 0. Finally u minimizes the following infimum, R u 2 + u 2 B1 inf such that u c in B \ B ρ 1 2! p+1 1 (B ) u Hrad 1 R p+1 u B1

39 Some remarks on the Green function Here we recall some properties of the Green function of the radial laplacian operator.

40 Some remarks on the Green function Here we recall some properties of the Green function of the radial laplacian operator. Properties of the Green function The Green function G(r, s) of the operator u N 1 r u + u with Neumann boundary condition is defined as { G rr (r, s) N 1 r G r (r, s) + G(r, s) = δ s (r) in (0, 1) G r (0, s) = G r (1, s) = 0 for any s (0, 1).

41 Some remarks on the Green function Here we recall some properties of the Green function of the radial laplacian operator. Properties of the Green function The Green function G(r, s) of the operator u N 1 r u + u with Neumann boundary condition is defined as { G rr (r, s) N 1 r G r (r, s) + G(r, s) = δ s (r) in (0, 1) G r (0, s) = G r (1, s) = 0 for any s (0, 1). Note that G(r, s) is bounded in L (0, 1)

42 Some remarks on the Green function Here we recall some properties of the Green function of the radial laplacian operator. Properties of the Green function The Green function G(r, s) of the operator u N 1 r u + u with Neumann boundary condition is defined as { G rr (r, s) N 1 r G r (r, s) + G(r, s) = δ s (r) in (0, 1) G r (0, s) = G r (1, s) = 0 for any s (0, 1). Note that G(r, s) is bounded in L (0, 1) and its graph is given by

43 Some remarks on the Green function Here we recall some properties of the Green function of the radial laplacian operator. Properties of the Green function The Green function G(r, s) of the operator u N 1 r u + u with Neumann boundary condition is defined as { G rr (r, s) N 1 r G r (r, s) + G(r, s) = δ s (r) in (0, 1) G r (0, s) = G r (1, s) = 0 for any s (0, 1). Note that G(r, s) is bounded in L (0, 1) and its graph is given by If N = 3 the Green function can be explicitly computed, { e r e r G(r, s) = 2r e s s for r s e s e s 2r e r s for r > s.

44 The main result Let k > 0 be an integer.

45 The main result Let k > 0 be an integer. Then there exists p(k ) such that for any p > p(k )

46 The main result Let k > 0 be an integer. Then there exists p(k ) such that for any p > p(k ) there exists a radial solution up (r ) having exactly k maximum points α1,p,..., αk,p.

47 The main result Let k > 0 be an integer. Then there exists p(k ) such that for any p > p(k ) there exists a radial solution up (r ) having exactly k maximum points α1,p,..., αk,p. Furthermore we have that Pk (i) up (r ) converges pointwise to j=1 Aj G(r, αj ),

48 The main result Let k > 0 be an integer. Then there exists p(k ) such that for any p > p(k ) there exists a radial solution up (r ) having exactly k maximum points α1,p,..., αk,p. Furthermore we have that Pk (i) up (r ) converges pointwise to j=1 Aj G(r, αj ), where (A1,..., Ak ) is a solution of the system k X j=1 Aj G(αi, αj ) = 1, i = 1,.., k.

49 The main result Let k > 0 be an integer. Then there exists p(k ) such that for any p > p(k ) there exists a radial solution up (r ) having exactly k maximum points α1,p,..., αk,p. Furthermore we have that Pk (i) up (r ) converges pointwise to j=1 Aj G(r, αj ), where (A1,..., Ak ) is a solution of the system k X Aj G(αi, αj ) = 1, i = 1,.., k. j=1 (ii) (α1,p,..., αk,p ) (α1,..., αk ) as p

50 The main result Let k > 0 be an integer. Then there exists p(k ) such that for any p > p(k ) there exists a radial solution up (r ) having exactly k maximum points α1,p,..., αk,p. Furthermore we have that Pk (i) up (r ) converges pointwise to j=1 Aj G(r, αj ), where (A1,..., Ak ) is a solution of the system k X Aj G(αi, αj ) = 1, i = 1,.., k. j=1 (ii) (α1,p,..., αk,p ) (α1,..., αk ) as p and (α1,..., αk ) is a critical point of the function ϕ(α1,., αk ) = inf{kuk2h 1 rad (B1 ) in the set 0 < α1 <... < αk < 1. : u(α1 ) =.. = u(αk ) = 1},

51 The main result Let k > 0 be an integer. Then there exists p(k ) such that for any p > p(k ) there exists a radial solution up (r ) having exactly k maximum points α1,p,..., αk,p. Furthermore we have that Pk (i) up (r ) converges pointwise to j=1 Aj G(r, αj ), where (A1,..., Ak ) is a solution of the system k X Aj G(αi, αj ) = 1, i = 1,.., k. j=1 (ii) (α1,p,..., αk,p ) (α1,..., αk ) as p and (α1,..., αk ) is a critical point of the function ϕ(α1,., αk ) = inf{kuk2h 1 rad (B1 ) : u(α1 ) =.. = u(αk ) = 1}, in the set 0 < α1 <... < αk < 1. k P We will see that ϕ(α1,., αk ) = Ai αjn 1 i=1

52 The case of one peak

53 The case of one peak There exists p such that for any p > p there exists a radial solution up (r ) having exactly 1 maximum point α1,p such that

54 The case of one peak There exists p such that for any p > p there exists a radial solution up (r ) having exactly 1 maximum point α1,p such that (i) up (r ) A1 G(r, α1 ) = G(r,α1 ) G(α1,α1 ). A1 = So 1, G(α1, α1 )

55 The case of one peak There exists p such that for any p > p there exists a radial solution up (r ) having exactly 1 maximum point α1,p such that (i) up (r ) A1 G(r, α1 ) = G(r,α1 ) G(α1,α1 ). A1 = So 1, G(α1, α1 ) (ii) α1,p α1 as p and α1 is a critical point of the function ϕ(α1 ) = inf{kuk2h 1 rad in the set 0 < α1 < 1. (B1 ) : u(α1 ) = 1},

56 The case of one peak There exists p such that for any p > p there exists a radial solution up (r ) having exactly 1 maximum point α1,p such that (i) up (r ) A1 G(r, α1 ) = G(r,α1 ) G(α1,α1 ). A1 = So 1, G(α1, α1 ) (ii) α1,p α1 as p and α1 is a critical point of the function ϕ(α1 ) = inf{kuk2h 1 rad (B1 ) : u(α1 ) = 1}, in the set 0 < α1 < 1. Moreover we have that ϕ(α1 ) = A1 α1n 1 = αn 1 1 G(α1,α1 )

57 The case of two peaks

58 The case of two peaks There exists p such that for any p > p there exists a radial solution up (r ) having exactly 2 maximum points α1,p and α2,p such that

59 The case of two peaks There exists p such that for any p > p there exists a radial solution up (r ) having exactly 2 maximum points α1,p and α2,p such that (i) up (r ) A1 G(r, α1 ) + A2 G(r, α2 ) with G(α2, α2 ) G(α1, α2 ) A1 =, G(α1, α1 )G(α2, α2 ) G(α1, α2 )G(α2, α1 ) A2 = G(α1, α1 ) G(α2, α1 ). G(α1, α1 )G(α2, α2 ) G(α1, α2 )G(α2, α1 )

60 The case of two peaks There exists p such that for any p > p there exists a radial solution up (r ) having exactly 2 maximum points α1,p and α2,p such that (i) up (r ) A1 G(r, α1 ) + A2 G(r, α2 ) with G(α2, α2 ) G(α1, α2 ) A1 =, G(α1, α1 )G(α2, α2 ) G(α1, α2 )G(α2, α1 ) A2 = G(α1, α1 ) G(α2, α1 ). G(α1, α1 )G(α2, α2 ) G(α1, α2 )G(α2, α1 ) (ii) (α1,p, α2,p ) (α1, α2 ) as p and (α1, α2 ) is a critical point of the function ϕ(α1, α2 ) = inf{kuk2h 1 rad in the set 0 < α1 < α2 < 1. (B1 ) : u(α1 ) = u(α2 ) = 1},

61 The case of two peaks There exists p such that for any p > p there exists a radial solution up (r ) having exactly 2 maximum points α1,p and α2,p such that (i) up (r ) A1 G(r, α1 ) + A2 G(r, α2 ) with G(α2, α2 ) G(α1, α2 ) A1 =, G(α1, α1 )G(α2, α2 ) G(α1, α2 )G(α2, α1 ) A2 = G(α1, α1 ) G(α2, α1 ). G(α1, α1 )G(α2, α2 ) G(α1, α2 )G(α2, α1 ) (ii) (α1,p, α2,p ) (α1, α2 ) as p and (α1, α2 ) is a critical point of the function ϕ(α1, α2 ) = inf{kuk2h 1 rad (B1 ) : u(α1 ) = u(α2 ) = 1}, in the set 0 < α1 < α2 < 1. Moreover we have that ϕ(α1, α2 ) = A1 α1n 1 + A2 α2n 1 = (G(α2,α2 ) G(α1,α2 ))αn 1 1 G(α1,α1 )G(α2,α2 ) G(α1,α2 )G(α2,α1 ) + (G(α1,α1 ) G(α2,α1 ))αn 1 2 G(α1,α1 )G(α2,α2 ) G(α1,α2 )G(α2,α1 )

62 The graph of a two peak-solution Here there is the graph of the solution,

63 Sketch of the proof

64 Sketch of the proof Construction of the 1-peak solution gluing an increasing solution in (0, α) and a decreasing solution in (α, 1).

65 Sketch of the proof Construction of the 1-peak solution gluing an increasing solution in (0, α) and a decreasing solution in (α, 1). Construction of the 2-peak solution gluing the 1- peak solutions in (0, α) and (α, 1) respectively.

66 Sketch of the proof Construction of the 1-peak solution gluing an increasing solution in (0, α) and a decreasing solution in (α, 1). Construction of the 2-peak solution gluing the 1- peak solutions in (0, α) and (α, 1) respectively. The general case follows by a degree argument.

67 Comparison with perturbative methods Question Is it possible to get the same result using the Lyapounov-Schmidt reduction? Namely, u p (r) = k PU λp.α p (r) + φ p (r) i=1 where PU λ.α is the projection of a function U λ.α solving some limit problem and φ p 0 as p +? The answer (very likely) is YES, but if needs many computations! In our opinion this approach is simpler.