One of the results can be formulated as follows [6]: if there exists a positive constant ε such that τ ( t ) > ε and p ( t ) > ε, then there exist unbounded solutions to (1.7).
If (2.2). then there exist unbounded solutions of (1.4).
There exist unbounded solutions of corresponding homogeneous equation x ″ ( t ) + a ( t ) x ( t − τ ) = 0, t ∈ [ 0, + ∞ ) (see, for example, [[4], Chapter III, Section 16, p.106] in the case of constant a, and noted above in Section 1 sufficient conditions for existence of unbounded solutions from the paper [6]).
Does there exist an unbounded closed convex subset of a Banach space which has the fixed point property for nonexpansive mappings?
If (H1) holds, then there exists an unbounded closed connected component (C=C_{lambda}) of solutions for (4.1) λ.
(3) ⟺ (2) Suppose that, for some x ∈ C, there exists an unbounded subsequence { T j n i x } of { T j n x }.
Similar to the proof of Theorem 5.1, there exists an unbounded continuum (mathscr{D}^{nu[n]}) of solutions of problem (5.8) emanating from ((frac{lambda^{nu}}{rf_{0}},0)), such that (mathscr{D}^{nu [n]}subset((mathbb{R}times P^{nu} cup{(frac{lambda^{nu }}{rf_{0}},0)})) and (mathscr{D}^{nu[n]}) joins ((frac{lambda^{nu }}{rf_{0}},0)) to ((frac{lambda^{nu}}{rn},infty)).
Similar to the proof of Theorem 5.1, there exists an unbounded continuum (mathscr{D}^{nu[n]}) of solutions of problem (5.7) emanating from ((frac{lambda^{nu}}{rn},0)), such that (mathscr{D}^{nu[n]}subset ((mathbb{R}times P^{nu} cup{(frac{lambda^{nu}}{rn},0)})), and (mathscr{D}^{nu[n]}) joins ((frac{lambda^{nu}}{rn},0)) to ((frac {lambda^{nu}}{rf_{infty}},infty)).
Assume that is unbounded, then there exist a ( can be chosen large arbitrarily) and such that (2.40).
Then, for each k ∈ N, there exist two unbounded sub-continua C k ± in bifurcating from ( λ k / f 0, 0,).
There exist two unbounded sub-continua (mathscr{D}^) and (mathscr{D}^) of solutions of (1.4) in (mathbb{R}times E), bifurcating from (Itimes{0}), and (mathscr{D}^{nu}subset(mathbb{R}times P^{nu} cup(Itimes{0})) for (nu=+) and (nu=-).
