Local invertible analytic solutions for an iterative differential equation related to a discrete derivatives sequence

The Exist Local Analytic Solutions of an Iterative Functional Differential Equation

This paper is concerned with an iterative functional differential equation x′ x r z = c0z+ c1x z + c2x x z + · · · + cmx m z , where r and m are nonnegative integers, x 0 z = z x 1 z = x z x 3 z = x x x z , etc. are the iterates of the function x z , and ∑mj=0 cj = 0. By constructing a convergent power series solution y z of a companion equation of the form αy ′ αr+1z = y ′ αz ∑mj=0 cjy αz , an...

Local Analytic Solutions of an Iterative Functional Differential Equation Depend on the State Derivative

An analytic solution for a non-local initial-boundary value problem including a partial differential equation with variable coefficients

‎This paper considers a non-local initial-boundary value problem containing a first order partial differential equation with variable coefficients‎. ‎At first‎, ‎the non-self-adjoint spectral problem is derived‎. ‎Then its adjoint problem is calculated‎. ‎After that‎, ‎for the adjoint problem the associated eigenvalues and the subsequent eigenfunctions are determined‎. ‎Finally the convergence ...

Analytic Solutions for Iterative Functional Differential Equations

Because of its technical difficulties the existence of analytic solutions to the iterative differential equation x′(z) = x(az + bx(z) + cx′(z)) is a source of open problems. In this article we obtain analytic solutions, using Schauder’s fixed point theorem. Also we present a unique solution which is a nonconstant polynomial in the complex field.

Analytic Solutions for a Functional Differential Equation Related to a Traffic Flow Model

and Applied Analysis 3 Proof. If we let h s ∑∞ n 1 ans n and substituting into 1.3 , we have α2 ∞ ∑ n 0 n 1 λ − 1 an 1s 2α ∞ ∑ n 1 ( n−1 ∑ i 0 i 1 ( λn−i − 1 )( λ − 1 ) an−iai 1 ) s ∞ ∑ n 2 ( n−2 ∑ i 0 n−i−1 ∑ k 1 i 1 ( λn−k−i − 1 )( λ − 1 )( λ − 1 ) an−k−iakai 1 ) s β ∞ ∑ n 1 ( n−1 ∑ i 0 i 1 ( λn−i − 1 )2 an−iai 1 ) s. 2.3 Comparing coefficients we obtain α2 ( λ0 − 1 ) a1 0, 2.4 α2 n 1 λ − 1 a...