Let be a real Banach space, and let be continuously differential, satisfying the P-S condition.
For the simplicity of our proof in Theorem 3.1, f ( t, x ) is a first-order continuously differentiable function satisfying Lipschitz condition.
Most of the known results on this problem are derived assuming only that the time-varying delay h ( t ) is a continuously differentiable function, satisfying some boundedness condition on its derivative: h ˙ ( t ) ≤ δ < 1.
Theorem 3.1 Let f : [ 0, π ] × X → X be a first-order continuously differentiable function satisfying the condition ∥ f ( t, x ) − f ( t, y ) ∥ ≤ L ∥ x − y ∥, ∀ t ∈ [ 0, π ], x, y ∈ X.
Assume that y : I → ℝ is a twice continuously differentiable function satisfying the differential inequality p 0 ( x ) y ″ + p 1 ( x ) y ′ + p 2 ( x ) y + f ( x ) ≤ φ ( x ) (3). for all x ∈ I and (2) is true.
Let (w(x)) be the weight function which is non-negative, twice continuously differentiable, and satisfying w(a)=w(b)=0,qquad w(a)=w(b)=0 (3.1) with (aleq xleq b).
In this paper, we shall solve this problem and explicitly give a new solution of the Neumann problem on (partialmathcal{T}_{n}). Define varepsilon_{0}=limsup_{rrightarrowinfty}tau^{-1}(r)r tau'(r)log r< 1, where (tau(r)) is a nondecreasing and continuously differentiable function satisfying (tau(r geq1) for any (rinmathbf{R}^cup{0}).
([6]) Let y : I → ℝ be a continuously differentiable function satisfying the differential inequality y ′ ( t ) + g ( t ) y ( t ) + h ( t ) ≤ φ ( t ). for all t ∈ I, where g, h : I → ℝ are continuous functions and φ : I → [0, ∞) is a function. Assume that.
Assume that y : I → ℝ is a twice continuously differentiable function satisfying the differential inequality ( μ ( x ) ) [ p 0 ( x ) y ″ + p 1 ( x ) y ′ + p 2 ( x ) y + f ( x ) ] ≤ φ ( x ) (9). for all x ∈ I and (8) is true.
Assume that y : I → ℝ is a twice continuously differentiable function satisfying the differential inequality y ″ + c y ′ + b y + f ( x ) ≤ φ ( x ) (12). for all x ∈ I. Then there exists a solution y 0 : I → ℝ of (10) such that y ( x ) - y 0 ( x ) ≤ exp { ( m - c ) ( x - a ) } ∫ x b exp ( - m v ) ∫ a v exp ( m t ) φ ( t ) exp { ( c - m ) ( x - a ) } d v. for all x ∈ I. Proof.
