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On the Fredholm Alternative for the p ‐Laplacian in One Dimension
Author(s) -
Manásevich Raúl F.,
TakáČ Peter
Publication year - 2002
Publication title -
proceedings of the london mathematical society
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 1.899
H-Index - 65
eISSN - 1460-244X
pISSN - 0024-6115
DOI - 10.1112/plms/84.2.324
Subject(s) - mathematics , eigenfunction , combinatorics , bounded function , eigenvalues and eigenvectors , orthogonality , dimension (graph theory) , norm (philosophy) , p laplacian , laplace operator , mathematical analysis , boundary value problem , physics , geometry , quantum mechanics , political science , law
We investigate the existence of a weak solution u to the quasilinear two‐point boundary value problem (P) − ( | u ′| p − 2u ′ ) ′ = λ k | u | p − 2 u + f ( x ) , 0 < x < a ; u ( 0 ) = u ( a ) = 0.We assume that 1 < p < ∞ p ¬ = 2, 0 < a < ∞, and that f ∈ L 1 (0,a) is a given function. The number λ k stands for the k ‐th eigenvalue of the one‐dimensional p ‐Laplacian. Let ∈ p π p x/a) denote the eigenfunction associated with λ 1 ; then ∈ p (k π p x/a) is the eigenfunction associated with λ k . We show the existence of solutions to (P) in the following cases. (i) When k =1 and f satisfies the orthogonality condition∫ 0 a f ( x ) sin p ( π p x / a ) d x = 0 , the set of solutions is bounded. (ii) If k =1 and f t ∈ L 1 (0,a) is a continuous family parametrized by t ∈ [0,1], with∫ 0 a f 0 ( x ) sin p ( π p x / a ) d x < 0 < ∫ 0 a f 1 ( x ) sin p ( π p x / a ) d x , then there exists some t * ∈ [0,1] such that (P) has a solution for f = f t * . Moreover, an appropriate choice of t * yields a solution u with an arbitrarily large L 1 (0,a)‐norm which means that such f cannot be orthogonal to ∈ p π p x/a. (iii) When k ⩾ 2 and f satisfies a set of orthogonality conditions to ∈ p ( k π p x/a) and/or cos p ( k π p x / a ) on the subintervals [ m a / k , ( m + 1 ) a / k ] (with m = 0 , 1 , … , k − 1 ) and/or [ ( m − 1 2 ) a / k , ( m + 1 2 ) a / k ] (with m = 1 , 2 , … , k − 1 ), again, the set of solutions is bounded. (iv) If k ⩾ 2 and { f t ∈ L 1 ( 0 , a ) : t ∈ [ 0 , 1 ] } is a continuous family satisfying either∫ 0 a f 0 ( x ) sin p ( k π p x / a ) d x < 0 < ∫ 0 a f 1 ( x ) sin p ( k π p x / a ) d x or another related condition, then there exists some t * ∈ [0,1] such that (P) has a solution for f = f t* . Prüfer's transformation plays the key role in our proofs. 2000 Mathematical Subject Classification : primary 34B16, 47J10; secondary 34L40, 47H30.