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Additional info for A Cauchy Problem for an Ultrahyperbolic Equation
1u) γ W (2) ∂S ∩ ∂γS ⊂ ΛS . Generally, we say S is admissible if S is ε-admissible for some ε > 0. A proper rectangle R is (ε-) admissible if we can write R = ΓS, for some (ε-) admissible proper D-rectangle S. If S is an ε-admissible proper D-rectangle, we let S(ε, s) (resp S(ε, u)) be the set of points x ∈ S for which (1s) (resp (1u)) is satisfied. We similarly define R(ε, s), R(ε, u) for admissible proper rectangles. Obviously, R(ε, s) ∩ R(ε, u) ⊃ R◦ . 6. It may be shown that the definitions of R(ε, s), R(ε, u) are independent of choice of proper D-rectangle or slice D.
Let Λ be transversally hyperbolic. There exist an open Γ-invariant neighborhood U of Λ in M and constants β, ρ > 0 such that whenever x ∈ Λ, y ∈ U and there exists a sequence (τn ) ⊂ Γ such that for all n ∈ Z we have (a) dΓ (τn , τn+1 ) < ρ, (b) d(f n (x), τn f n (y)) < β, then y ∈ Γx. Proof. Our condition on y implies that the f -orbit of y stays close to Λ. 6], we may choose an open Γ-invariant neighborhood U of Λ such that if Λ = ∩n∈Z f n (U¯ ), then TΛ has a partially hyperbolic splitting. 6 that Λ is transversally hyperbolic.
The map π : Σ→Λ defined by −k (Rxk ) π(x) = ∩∞ k=−∞ f is well-defined. Furthermore, (a) π is H¨older continuous and surjective. (b) π is 1:1 on a residual subset of Σ. (c) #π −1 (x) ≤ n2 for all x ∈ Λ. (d) f π = πσ. (e) π is Γ-equivariant. (f) Every isotropy group in Λ occurs as the isotropy group of a point in Σ. Proof. Everything except (e,f) is proved in [6, 7]. Statement (e) is immediate from the fact that Γ acts on R. It remains to prove (f). Suppose H ∈ I(Λ). Since R is a Γ-invariant Markov partition, it follows that RH is a Markov partition for f : ΛH →ΛH .
A Cauchy Problem for an Ultrahyperbolic Equation by Kostomarov