An isoperimetric inequality for a nonlinear eigenvalue problem - PowerPoint PPT Presentation

An isoperimetric inequality for a nonlinear eigenvalue problem Gisella Croce joint work with A. Henrot and G. Pisante PICOF ’12 - Ecole Polytechnique, Palaiseau

Let Ω be an open bounded subset of R N .   �∇ v � L p (Ω) �   : v � = 0 , v ∈ W 1 , p λ p , q (Ω) := inf | v | q − 2 v = 0 (Ω) , 0 � v � L q (Ω)   Ω

Let Ω be an open bounded subset of R N .   �∇ v � L p (Ω) �   : v � = 0 , v ∈ W 1 , p λ p , q (Ω) := inf | v | q − 2 v = 0 (Ω) , 0 � v � L q (Ω)   Ω Among the sets of given volume, which one minimizes λ p , q (Ω)?

Let Ω be an open bounded subset of R N .   �∇ v � L p (Ω) �   : v � = 0 , v ∈ W 1 , p λ p , q (Ω) := inf | v | q − 2 v = 0 (Ω) , 0 � v � L q (Ω)   Ω Among the sets of given volume, which one minimizes λ p , q (Ω)? THEOREM (C., Henrot and Pisante, Annales de l’IHP) � 1 < q < p ∗ = Np N − p , if 1 < p < N Assume 1 < p < ∞ and . 1 < q < ∞ , if p ≥ N λ p , q (Ω) ≥ λ p , q ( B 1 ∪ B 2 ) , where B 1 and B 2 are two disjoint balls of volume | Ω | / 2.

Freitas-Henrot, On the first twisted Dirichlet eigenvalue : λ 2 , 2 (Ω) Motivation: Barbosa-B´ erard, Eigenvalue and twisted eigenvalue problems, applications to cmc surfaces : second variation of constant mean curvature immersions.

Freitas-Henrot, On the first twisted Dirichlet eigenvalue : λ 2 , 2 (Ω) Motivation: Barbosa-B´ erard, Eigenvalue and twisted eigenvalue problems, applications to cmc surfaces : second variation of constant mean curvature immersions. Dacorogna-Gangbo-Sub´ ıa, Sur une g´ en´ eralisation de l’in´ egalit´ e de Wirtinger : the minimizers of λ p , q (( − 1 , 1)) are odd.

Freitas-Henrot, On the first twisted Dirichlet eigenvalue : λ 2 , 2 (Ω) Motivation: Barbosa-B´ erard, Eigenvalue and twisted eigenvalue problems, applications to cmc surfaces : second variation of constant mean curvature immersions. Dacorogna-Gangbo-Sub´ ıa, Sur une g´ en´ eralisation de l’in´ egalit´ e de Wirtinger : the minimizers of λ p , q (( − 1 , 1)) are odd.  1  � v ′ � L p (( − 1 , 1)) �   λ p , q : v ∈ W 1 , p | v | q − 2 v = 0 per (( − 1 , 1)):=inf per (( − 1 , 1)) , � v � L q (( − 1 , 1))   − 1 = λ p , q (( − 1 , 1))

Freitas-Henrot, On the first twisted Dirichlet eigenvalue : λ 2 , 2 (Ω) Motivation: Barbosa-B´ erard, Eigenvalue and twisted eigenvalue problems, applications to cmc surfaces : second variation of constant mean curvature immersions. Dacorogna-Gangbo-Sub´ ıa, Sur une g´ en´ eralisation de l’in´ egalit´ e de Wirtinger : the minimizers of λ p , q (( − 1 , 1)) are odd.  1  � v ′ � L p (( − 1 , 1)) �   λ p , q : v ∈ W 1 , p | v | q − 2 v = 0 per (( − 1 , 1)):=inf per (( − 1 , 1)) , � v � L q (( − 1 , 1))   − 1 = λ p , q (( − 1 , 1)) [ L ( ∂ A )] 2 ≥ 4 λ p , p ′ ∀ A ⊆ R 2 per (( − 1 , 1)) | A | 1 � ( x ′ ( t ) , y ′ ( t )) � l p , if ∂ A = { ( x ( t ) , y ( t )) : t ∈ [ − 1 , 1] } � L ( ∂ A ) = − 1

� | u | q − 2 u = 0 makes the problem difficult. Why? The constraint Ω

� | u | q − 2 u = 0 makes the problem difficult. Why? The constraint Ω A classical example: the first eigenvalue of the laplacian Among the sets of given volume, which one minimizes � � �∇ v � L 2 (Ω) : v � = 0 , v ∈ H 1 λ (Ω) = inf 0 (Ω) ? � v � L 2 (Ω)

� | u | q − 2 u = 0 makes the problem difficult. Why? The constraint Ω A classical example: the first eigenvalue of the laplacian Among the sets of given volume, which one minimizes � � �∇ v � L 2 (Ω) : v � = 0 , v ∈ H 1 λ (Ω) = inf 0 (Ω) ? � v � L 2 (Ω) The ball!

Since u is positive, one has �∇ u ∗ � L 2 ( B ) �∇ u � L 2 (Ω) λ (Ω) = ≥ ≥ λ ( B ) � u ∗ � L 2 ( B ) � u � L 2 (Ω) where u ∗ is the Schwarz rearrangement of u (rearrangement of the level sets of u in balls of same volume) and B is a ball with | B | = | Ω | .

Since u is positive, one has �∇ u ∗ � L 2 ( B ) �∇ u � L 2 (Ω) λ (Ω) = ≥ ≥ λ ( B ) � u ∗ � L 2 ( B ) � u � L 2 (Ω) where u ∗ is the Schwarz rearrangement of u (rearrangement of the level sets of u in balls of same volume) and B is a ball with | B | = | Ω | . For   �∇ v � L p (Ω) �   : v � = 0 , v ∈ W 1 , p | v | q − 2 v = 0 λ p , q (Ω) = inf (Ω) , 0 � v � L q (Ω)   Ω the minimizers are forced to change sign, because of the constraint!

IDEA OF THE PROOF OF OUR RESULT: 1. Let u be a minimizer and Ω ± = { u ≷ 0 } . Let B ± be two balls such that | B ± | = | Ω ± | . Then λ p , q (Ω) ≥ λ p , q ( B + ∪ B − ) We reduce to Ω = B 1 ∪ B 2 , | B 1 ∪ B 2 | fixed; u radial on B 1 and B 2

IDEA OF THE PROOF OF OUR RESULT: 1. Let u be a minimizer and Ω ± = { u ≷ 0 } . Let B ± be two balls such that | B ± | = | Ω ± | . Then λ p , q (Ω) ≥ λ p , q ( B + ∪ B − ) We reduce to Ω = B 1 ∪ B 2 , | B 1 ∪ B 2 | fixed; u radial on B 1 and B 2 2. u solves − div ( |∇ u | p − 2 ∇ u ) = [ λ p , q (Ω)] p � u � p − q L q (Ω) | u | q − 2 u p − 1 p − 1 � � � � ∂ u 1 ∂ u 2 � � � � = ⇒ | ∂ B 1 | = | ∂ B 2 | � � � � ∂ν 1 ∂ν 2 � � � �

IDEA OF THE PROOF OF OUR RESULT: 1. Let u be a minimizer and Ω ± = { u ≷ 0 } . Let B ± be two balls such that | B ± | = | Ω ± | . Then λ p , q (Ω) ≥ λ p , q ( B + ∪ B − ) We reduce to Ω = B 1 ∪ B 2 , | B 1 ∪ B 2 | fixed; u radial on B 1 and B 2 2. u solves − div ( |∇ u | p − 2 ∇ u ) = [ λ p , q (Ω)] p � u � p − q L q (Ω) | u | q − 2 u p − 1 p − 1 � � � � ∂ u 1 ∂ u 2 � � � � = ⇒ | ∂ B 1 | = | ∂ B 2 | � � � � ∂ν 1 ∂ν 2 � � � � � � � � ∂ u 1 ∂ u 2 3. ˙ λ p , q ( B 1 ∪ B 2 ) = 0 ⇔ � � � � � = � . � � � � ∂ν 1 ∂ν 2 � � ⇓ | B 1 | = | B 2 |

CONCLUSION: 1. λ p , q (Ω) ≥ λ p , q ( B + ∪ B − )

CONCLUSION: 1. λ p , q (Ω) ≥ λ p , q ( B + ∪ B − ) 2 | = | B + ∪ B − | 2. λ p , q ( B + ∪ B − ) ≥ λ p , q ( B ′ 1 ∪ B ′ 2 ) for | B ′ 1 | = | B ′ 2

CONCLUSION: 1. λ p , q (Ω) ≥ λ p , q ( B + ∪ B − ) 2 | = | B + ∪ B − | 2. λ p , q ( B + ∪ B − ) ≥ λ p , q ( B ′ 1 ∪ B ′ 2 ) for | B ′ 1 | = | B ′ 2 3. λ p , q is decreasing: if Ω 1 ⊂ Ω 2 , then λ p , q (Ω 1 ) ≥ λ p , q (Ω 2 )

CONCLUSION: 1. λ p , q (Ω) ≥ λ p , q ( B + ∪ B − ) 2 | = | B + ∪ B − | 2. λ p , q ( B + ∪ B − ) ≥ λ p , q ( B ′ 1 ∪ B ′ 2 ) for | B ′ 1 | = | B ′ 2 3. λ p , q is decreasing: if Ω 1 ⊂ Ω 2 , then λ p , q (Ω 1 ) ≥ λ p , q (Ω 2 ) 2 ) ≥ λ p , q ( B 1 ∪ B 2 ) for | B 1 | = | B 2 | = | Ω | ⇒ λ p , q ( B ′ 1 ∪ B ′ = 2

CONCLUSION: 1. λ p , q (Ω) ≥ λ p , q ( B + ∪ B − ) 2 | = | B + ∪ B − | 2. λ p , q ( B + ∪ B − ) ≥ λ p , q ( B ′ 1 ∪ B ′ 2 ) for | B ′ 1 | = | B ′ 2 3. λ p , q is decreasing: if Ω 1 ⊂ Ω 2 , then λ p , q (Ω 1 ) ≥ λ p , q (Ω 2 ) 2 ) ≥ λ p , q ( B 1 ∪ B 2 ) for | B 1 | = | B 2 | = | Ω | ⇒ λ p , q ( B ′ 1 ∪ B ′ = 2 Summarizing we get | B 1 | = | B 2 | = | Ω | λ p , q (Ω) ≥ λ p , q ( B 1 ∪ B 2 ) , 2

Technique of Freitas-Henrot ( p = q = 2): 1. Reduction to pairs of balls

Technique of Freitas-Henrot ( p = q = 2): 1. Reduction to pairs of balls 2. A minimizer u = u ( r ) solves u ′′ + N − 1 u ′ + [ λ 2 , 2 (Ω)] 2 u = µ 0 r on [0 , R 1 ] and [0 , R 2 ]. N.B.: explicit expression of the solution (Bessel functions)

Current projects: p → 1 λ p , q (Ω): 1. lim

Current projects: p → 1 λ p , q (Ω): we think that at least lim p → 1 λ p , p (Ω) equals either 1. lim � � | ∂ E 1 | | E 1 | , | ∂ E 2 | � � inf max , | E 1 | = | E 2 | , E 1 ∩ E 2 = ∅ , E 1 , E 2 ⊂ Ω | E 2 | or

Current projects: p → 1 λ p , q (Ω): we think that at least lim p → 1 λ p , p (Ω) equals either 1. lim � � | ∂ E 1 | | E 1 | , | ∂ E 2 | � � inf max , | E 1 | = | E 2 | , E 1 ∩ E 2 = ∅ , E 1 , E 2 ⊂ Ω | E 2 | or inf { max { h ( E 1 ) , h ( E 2 ) } , | E ∗ 1 | = | E ∗ 2 | , E 1 ∩ E 2 = ∅ , E 1 , E 2 ⊂ Ω } where, for any set E , h denotes its Cheeger constant: | ∂ D | | D | = h ( E ∗ ) h ( E ) = inf D ⊂ E

Current projects: p → 1 λ p , q (Ω): we think that at least lim p → 1 λ p , p (Ω) equals either 1. lim � � | ∂ E 1 | | E 1 | , | ∂ E 2 | � � inf max , | E 1 | = | E 2 | , E 1 ∩ E 2 = ∅ , E 1 , E 2 ⊂ Ω | E 2 | or inf { max { h ( E 1 ) , h ( E 2 ) } , | E ∗ 1 | = | E ∗ 2 | , E 1 ∩ E 2 = ∅ , E 1 , E 2 ⊂ Ω } where, for any set E , h denotes its Cheeger constant: | ∂ D | | D | = h ( E ∗ ) h ( E ) = inf D ⊂ E p →∞ λ p , q (Ω) 2. lim

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