1. A global compactness result and multiplicity of solutions for a class of critical exponent problems in the hyperbolic space.
- Author
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Bhakta, Mousomi, Ganguly, Debdip, Gupta, Diksha, and Sahoo, Alok Kumar
- Subjects
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GEODESIC distance , *THRESHOLD energy , *MULTIPLICITY (Mathematics) , *EQUATIONS , *ARGUMENT - Abstract
This paper deals with global compactness and the multiplicity of positive solutions to problems of the type −Δ픹Nu − λu = a(x)|u|2∗−2u + f(x)in 픹N, u ∈ H1(픹N), where 픹N denotes the ball model of the hyperbolic space of dimension N ≥ 4, 2∗ = 2N N−2, N(N−2) 4 < λ < (N−1)2 4 and f ∈ H−1(픹N) (f≢0) is a nonnegative functional in the dual space of H1(픹N). The potential a ∈ L∞(픹N) is assumed to be strictly positive, such that limd(x,0)→∞a(x) = 1, where d(x, 0) denotes the geodesic distance. We establish profile decomposition of the functional associated with the above equation. We show that concentration occurs along two different profiles, namely, hyperbolic bubbles and localized Aubin–Talenti bubbles. For f = 0 and a ≡ 1, profile decomposition was studied in Bhakta and Sandeep [Poincaré–Sobolev equations in the hyperbolic space,
Calc. Var. Partial Differential Equations 44 (2012) 247–269]. However, due to the presence of the potential a(.), an extension of profile decomposition to the present set-up is highly nontrivial and requires several delicate estimates and geometric arguments concerning the isometry group (Möbius group) of the hyperbolic space. Further, using the decomposition result, we derive various energy estimates involving the interacting hyperbolic bubbles and hyperbolic bubbles with localized Aubin–Talenti bubbles. Finally, combining these estimates with topological and variational arguments, we establish a multiplicity of positive solutions in the cases: a ≥ 1 and a < 1 separately. The equation studied in this paper can be thought of as a variant of a scalar-field equation with a critical exponent in the hyperbolic space, although such a critical exponent problem in the Euclidean space ℝN has only a trivial solution when f ≡ 0, a(x) ≡ 1 and λ < 0. [ABSTRACT FROM AUTHOR]- Published
- 2024
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