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International Journal of Differential Equations
Volume 2011 (2011), Article ID 949547, 12 pages
Research Article

Oscillatory Solutions of Neutral Equations with Polynomial Nonlinearities

1Department of Mathematics, University of Mining and Geology “St. I. Rilski”, 1700 Sofia, Bulgaria
2Department of Mathematics and Physics, Higher School of Civil Engineering “L. Karavelov”, 1373 Sofia, Bulgaria

Received 1 June 2011; Accepted 31 August 2011

Academic Editor: Elena Braverman

Copyright © 2011 Vasil G. Angelov and Dafinka Tz. Angelova. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


Existence uniqueness of an oscillatory solution for nonlinear neutral equations by fixed point method is proved.

1. Introduction

In [1, 2], we have considered a lossless transmission line terminated by a nonlinear resistive load and parallel connected capacitance (cf. Figure 1). The nonlinear boundary condition is caused by the polynomial type V-I characteristics of the nonlinear load at the second end of the transmission line (cf. Figure 1).

Figure 1

The voltage and current 𝑢(𝑥,𝑡), 𝑖(𝑥,𝑡) of the lossless transmission line can be found by solving the following mixed problem for the hyperbolic partial differential system:𝐶𝜕𝑢(𝑥,𝑡)+𝜕𝑡𝜕𝑖(𝑥,𝑡)𝜕𝑥=0,𝐿𝜕𝑖(𝑥,𝑡)+𝜕𝑡𝜕𝑢(𝑥,𝑡)𝜕𝑥=0,𝐸(𝑡)𝑢(0,𝑡)=𝑅0𝐶𝑖(0,𝑡),𝑡0,(1.1)0𝑑𝑢(Λ,𝑡)𝑑𝑡=𝑖(Λ,𝑡)𝑓(𝑢(Λ,𝑡)),𝑡0,(1.2)𝑢(𝑥,0)=𝑢0(𝑥),𝑖(𝑥,0)=𝑖0[],(𝑥),𝑥0,Λ(1.3) where 𝑢0(𝑥) and 𝑖0(𝑥) are prescribed initial functions, Λ is the length of the line, 𝐶 is the per-unit length capacitance, and 𝐿 is per-unit length inductance (cf. [310]). Here, the V-I characteristic of the nonlinear resistive load is 𝑖=𝑓(𝑢)=𝑝𝑛=1𝑟𝑛𝑢𝑛, where 𝑟𝑛 are real numbers, 𝐶0 is parallel connected capacitance, 𝐸 is the source voltage, 𝑅0 is the source resistance, and 𝑍0=𝐿/𝐶 is the line characteristic impedance.

The above formulated mixed problem can be reduced (cf. [1, 2, 11]) to an equivalent initial value problem for a neutral functional differential equation (cf. [12]). Here, we consider the problem of an existence uniqueness of oscillatory solutions of the equation𝑑𝑢(𝑡)=𝑑𝑡2𝐸𝐶0𝑍0+𝑅0𝑢(𝑡)𝐶0𝑍01𝐶0𝑝𝑛=1𝑟𝑛[]𝑢(𝑡)𝑛𝑍0𝑅0𝑢(𝑡2𝑇)𝑍0𝐶0𝑍0+𝑅0+𝑍0𝑅0𝐶0𝑍0+𝑅0𝑝𝑛=1𝑟𝑛[]𝑢(𝑡2𝑇)𝑛+𝑍0𝑅0𝑍0+𝑅0𝑑𝑢(𝑡2𝑇)𝑑𝑡,𝑡𝑇,𝑢(𝑡)=𝜐0(𝑡),𝑑𝑢(𝑡)=𝑑𝑡𝑑𝜐0(𝑡)[],𝑑𝑡,𝑡𝑇,𝑇(1.4) where (𝑥,𝑡)Π={(𝑥,𝑡)𝑅2(𝑥,𝑡)[0,Λ]×[0,)}, 𝜅=|𝑍0𝑅0|/(𝑍0+𝑅0)<1,𝑢(𝑡)=𝑢(Λ,𝑡). In fact, (1.4) is differential difference equation, and the initial function should be prescribed on an interval with length 2T. Let us note that the initial function 𝜐0(𝑡) can be obtained shifting the initial function 𝑢0(𝑥) from (1.3) along the characteristics 𝑥𝜈𝑡=const.,(𝜈=1/𝐿𝐶) on [0,𝑇] and along the characteristics 𝑥+𝜈𝑡=const. on [𝑇,0] (cf. [1, 2]). So, we obtain an initial function 𝜐0(𝑡) on [𝑇,𝑇].

Now, we are able to formulate the main problem: to find a solution of (1.4) with advanced prescribed zeros on the interval [𝑡0,),𝑇=𝑡0.

Let 𝑆𝑇={𝜏𝑘}𝑛𝑘=0,𝑛𝑁 be the set of zeros of the initial function; that is, 𝜐0(𝜏𝑘)=0 such that 𝜏0=𝑇,𝜏𝑛=𝑇𝑡0.

Let 𝑆={𝑡𝑘}𝑘=0 be a strictly increasing sequence of real numbers satisfying the following conditions (C):(C1)lim𝑘𝑡𝑘=,(C2)0<𝑙0=inf{𝑡𝑘+1𝑡𝑘𝑘=0,1,2,}sup{𝑡𝑘+1𝑡𝑘𝑘=0,1,2,}=𝑇0<,(C3)for every 𝑘 there is 𝑠<𝑘 such that 𝑡𝑘𝑇=𝑡𝑠 where 𝑡𝑠𝑆𝑇𝑆.

Introduce the sets: 𝐶1[𝑡0,) consisting of all continuous and bounded functions differentiable with bounded derivatives on every interval (𝑡𝑘,𝑡𝑘+1) (the derivatives at 𝑡𝑘 do not necessary exist), 𝑀𝑆={𝑢()𝐶1[𝑡0,)𝑢(𝑡𝑘)=0(𝑘=0,1,2,)}, 𝑀𝑆𝑈={𝑢()𝑀𝑆|𝑢(𝑡)|𝑈0𝑒𝜇(𝑡𝑡𝑘),𝑡[𝑡𝑘,𝑡𝑘+1]}, where 𝑈0,𝜇 are positive constants prescribed below.

We assume that |𝜐0(𝑡)|𝑈0𝑒𝜇(𝑡𝜏𝑘),𝑡[𝜏𝑘,𝜏𝑘+1], (𝑘=0,1,2,,𝑛1).

The set 𝑀𝑆𝑈 turns out into a complete uniform space with respect to the family of pseudometrics 𝜌𝜇(𝑘)(𝑓,𝑔)=max{𝜌𝑘(𝑓,𝑔),𝜌𝑘(̇𝑓,̇𝑔)}, (𝑘=0,1,2,), where 𝜌𝑘(𝑓,𝑔)=max{𝑒𝜇(𝑡𝑡𝑘)|𝑓(𝑡)𝑔(𝑡)|𝑡[𝑡𝑘,𝑡𝑘+1]}, 𝜌𝑘(̇𝑓,̇𝑔)=max{𝑒𝜇(𝑡𝑡𝑘)|̇𝑓(𝑡)̇𝑔(𝑡)|𝑡[𝑡𝑘,𝑡𝑘+1]}.

One can verify that 𝑀𝑆𝑈 is closed subset of 𝐶1[𝑡0,) with respect to the above metric.

Remark 1.1. The functions from 𝑀𝑆 are not necessary differentiable at 𝑡𝑘 (𝑘=0,1,2,). That is why we consider a space with a countable family of pseudometrics, and then, we have to apply the fixed point theory from [13].
Define the operator 𝐵𝑀𝑆𝑈𝑀𝑆𝑈 by 𝐵(𝑢)(𝑡)=𝑡𝑡𝑘𝑈(𝑢)(𝑠)𝑑𝑠𝑡𝑡𝑘𝑡𝑘+1𝑡𝑘𝑡𝑘+1𝑡𝑘𝑡𝑈(𝑢)(𝑠)𝑑𝑠,𝑡𝑘,𝑡𝑘+1,(𝑘=0,1,2,),(1.5) where 𝑈(𝑢)(𝑡)=2𝐸𝐶0𝑍0+𝑅0𝑢(𝑡)𝐶0𝑍01𝐶0𝑝𝑛=1𝑟𝑛[]𝑢(𝑡)𝑛𝜅𝐾𝑇𝑢(𝑡)𝑍0𝐶0+𝜅𝐶0𝑝𝑛=1𝑟𝑛𝐾𝑇𝑢(𝑡)𝑛𝑑𝐾+𝜅𝑇𝑢(𝑡)𝑑𝑡,𝑡𝑇,(1.6) and (𝐾𝑇𝑢)(𝑡)=𝑢(𝑡2𝑇) is M. A. Krasnoselskii operator (cf. [14]).

Remark 1.2. The operator 𝐾𝑇 is well defined, because the initial function is defined on the interval [𝑇,𝑇]. We notice that 𝐾𝑇 maps 𝑀𝑆 into itself. Indeed, consider the set 𝐶1[𝑇,) consisting of all continuous and bounded functions differentiable with bounded derivatives on every interval (𝑡𝑘,𝑡𝑘+1). Introduce the set 𝑀𝜐0𝑆={𝑢()𝐶1[𝑇,)𝑢(𝑡)=𝜐0(𝑡),𝑡[𝑇,𝑇]}. Then, 𝐾𝑇 assigns to every function 𝑢()𝑀𝑆 the function ̃𝑢()𝑀𝜐0𝑆 translated to the right on the interval [𝑇,). So, the function (𝐾𝑇𝑢)(𝑡) coincides with 𝜐0(𝑡) on [𝑡0,𝑡0+2𝑇]. Besides 𝑡𝑘2𝑇=𝑡𝑠, and then 𝐾𝑇𝑢𝑡𝑘=𝑢𝑡𝑘2𝑇=𝜐0𝑡𝑠=0,𝑡𝑘[],𝑢𝑡𝑇,3𝑇𝑘𝑡2𝑇=𝑢𝑛=0,𝑡(3𝑇,),(1.7) that is, (𝐾𝑇𝑢)()𝑀𝑆.

2. Main Results

Lemma 2.1. If 𝐸𝑈0, problem (1.4) has a solution 𝑢()𝑀𝑆𝑈 iff the operator 𝐵 has a fixed point in 𝑀𝑆𝑈, that is, 𝑢(𝑡)=𝐵(𝑢)(𝑡).(2.1)

Proof. Let 𝑢()𝑀𝑆𝑈 be a solution of (1.4). Then, integrating (1.4) on the interval [𝑡𝑘,𝑡][𝑡𝑘,𝑡𝑘+1] (𝑘=0,1,2), we obtain 𝑢(𝑡)𝑢(𝑡𝑘)=𝑡𝑡𝑘𝑈(𝑢)(𝑠)𝑑𝑠𝑢(𝑡)=𝑡𝑡𝑘𝑈(𝑢)(𝑠)𝑑𝑠, and then, 𝑢(𝑡)=𝑡𝑡𝑘𝑡𝑈(𝑢)(𝑠)𝑑𝑠0=𝑢𝑘+1=𝑡𝑘+1𝑡𝑘𝑈(𝑢)(𝑠)𝑑𝑠𝑡𝑘+1𝑡𝑘𝑈(𝑢)(𝑠)𝑑𝑠=0.(2.2) Therefore, 𝑢(𝑡)satisfies 𝑢(𝑡)=𝑡𝑡𝑘𝑈(𝑢)(𝑠)𝑑𝑠𝑢(𝑡)=𝑡𝑡𝑘𝑈(𝑢)(𝑠)𝑑𝑠𝑡𝑡𝑘𝑡𝑘+1𝑡𝑘𝑡𝑘+1𝑡𝑘𝑈(𝑢)(𝑠)𝑑𝑠,(2.3) that is, 𝑢()is a fixed point of B.
Conversely, let 𝑢()𝑀𝑆𝑈 be a solution of 𝑢=𝐵(𝑢); that is, 𝑢(𝑡)=𝑡𝑡𝑘𝑈(𝑢)(𝑠)𝑑𝑠𝑡𝑡𝑘𝑡𝑘+1𝑡𝑘𝑡𝑘+1𝑡𝑘𝑈(𝑢)(𝑠)𝑑𝑠.(2.4)
Then, introducing 𝜇0=𝜇𝑇0, we obtain ||||𝑡𝑘+1𝑡𝑘𝑈||||(𝑢)(𝑠)𝑑𝑠2𝐸𝐶0𝑍0+𝑅0𝑡𝑘+1𝑡𝑘𝑒𝜇(𝑡𝑡𝑘)1𝑑𝑡+𝐶0𝑍0𝑡𝑘+1𝑡𝑘||𝑢||+1(𝑡)𝑑𝑡𝐶0𝑝𝑛=1||𝑟𝑛||𝑡𝑘+1𝑡𝑘||||𝑢(𝑡)𝑛𝜅𝑑𝑡+𝑍0𝐶0𝑡𝑘+1𝑡𝑘||||+𝜅𝑢(𝑡2𝑇)𝑑𝑡𝐶0𝑝𝑛=1||𝑟𝑛||𝑡𝑘+1𝑡𝑘||||𝑢(𝑡2𝑇)𝑛||||𝑑𝑡+𝜅𝑡𝑘+1𝑡𝑘||||̇𝑢(𝑡2𝑇)𝑑𝑡2𝑈0𝑒𝜇𝑇𝐶0𝑍0+𝑅0𝑒𝜇(𝑡𝑘+1𝑡𝑘)1𝜇+𝑈0𝐶0𝑍0𝑒𝜇(𝑡𝑘+1𝑡𝑘)1𝜇+1𝐶0𝑝𝑛=1||𝑟𝑛||𝑈𝑛0𝑡𝑘+1𝑡𝑘𝑒𝑛𝜇(𝑡𝑡𝑘)+𝑑𝑡𝜅𝑈0𝑒2𝜇𝑇𝑍0𝐶0𝑒𝜇(𝑡𝑘+1𝑡𝑘)1𝜇+𝜅𝐶0𝑝𝑛=1||𝑟𝑛||𝑈𝑛0𝑒2𝑛𝜇𝑇×𝑡𝑘+1𝑡𝑘𝑒𝑛𝜇(𝑡𝑡𝑘)||𝑢𝑡𝑑𝑡+𝜅𝑘+1𝑡2𝑇𝑢𝑘||2𝑇2𝑈0𝑒𝜇𝑇𝐶0𝑍0+𝑅0𝑒𝜇𝑇01𝜇+𝑈0𝐶0𝑍0𝑒𝜇𝑇01𝜇+1𝐶0𝑝𝑛=1||𝑟𝑛||𝑈𝑛0𝑒𝑛𝜇𝑇01+𝑈𝑛𝜇0𝜅𝑒2𝜇𝑇𝐶0𝑍0𝑒𝜇𝑇01𝜇+𝜅𝐶0𝑝𝑛=1||𝑟𝑛||𝑈𝑛0𝑒2𝑛𝜇𝑇𝑒𝑛𝜇𝑇01𝑒𝑛𝜇𝜇01𝜇𝐶02𝑈0𝑒𝜇𝑇𝑍0+𝑅0+𝑈01+𝜅𝑒2𝜇𝑇𝑍0+1𝜇𝐶0𝑝𝑛=1||𝑟𝑛||𝑈𝑛01+𝜅𝑒2𝑛𝜇𝑇(𝑒𝑛𝜇01)𝑛𝑀(𝜇).(2.5)
Let us assume that |𝑡𝑘+1𝑡𝑘𝑈(𝑢)(𝑡)𝑑𝑡|=𝛽>0. We have just obtained that 𝛽𝑀(𝜇). Then, for sufficiently large 𝜇>0 (and sufficiently small 𝑇0>0), one can reach the inequality 𝑀(𝜇)<𝛽. Consequently, 𝑡𝑘+1𝑡𝑘𝑈(𝑢)(𝑡)𝑑𝑡=0. It follows that 𝑢(𝑡)=𝑡𝑡𝑘𝑈(𝑢)(𝑠)𝑑𝑠 and, after a differentiation, we obtain (1.4).
Lemma 2.1 is thus proved.

Theorem 2.2. Let 𝑆𝑇={𝜏𝑘}𝑛𝑘=0,𝑛𝑁 be the set of zeros of the initial function; that is, 𝜐0(𝜏𝑘)=0 and 𝜐0()𝐶1[𝑇,𝑇]. If 𝐸𝑈0,|𝜐0(𝑡)|𝑈0𝑒𝜇(𝑡𝜏𝑘), 𝑡[𝜏𝑘,𝜏𝑘+1],𝜐0(𝑡0)=0, then, there exists a unique oscillatory solution of the initial value problem (1.4), belonging to 𝑀𝑆𝑈.

Proof. We show that 𝐵 maps 𝑀𝑆𝑈 into itself; that is, 𝑢𝑀𝑆𝑈𝐵(𝑢)𝑀𝑆𝑈.
Indeed, for every 𝑢()𝑀𝑆𝑈, the function 𝐵(𝑢)(𝑡) is continuous on [𝑡0,) and differentiable on every (𝑡𝑘,𝑡𝑘+1). We have also 𝐵(𝑢)(𝑡𝑘)=0 and 𝐵(𝑢)(𝑡𝑘+1)=0.
We show that |(𝐵𝑢)(𝑡)|𝑈0𝑒𝜇(𝑡𝑡𝑘), 𝑡[𝑡𝑘,𝑡𝑘+1]. (The last inequalities imply that 𝐵(𝑢)(𝑡) is bounded because 𝑒𝜇(𝑡𝑡𝑘)𝑒𝜇𝑇0,𝑡[𝑇,).)
We notice that |(𝑡𝑡𝑘)/(𝑡𝑘+1𝑡𝑘)|1, 𝑡[𝑡𝑘,𝑡𝑘+1]. For sufficiently large 𝜇, we obtain for 𝑡[𝑡𝑘,𝑡𝑘+1]||||||||(𝐵𝑢)(𝑡)𝑡𝑡𝑘||||+||||𝑈(𝑢)(𝑠)𝑑𝑠𝑡𝑘+1𝑡𝑘𝑈||||(𝑢)(𝑠)𝑑𝑠𝐵1+𝐵2.(2.6) We have 𝐵12𝐶0𝑍0+𝑅0𝑡𝑡𝑘||||1𝐸(𝑠𝑇)𝑑𝑠+𝐶0𝑍0𝑡𝑡𝑘||||1𝑢(𝑠)𝑑𝑠+𝐶0𝑝𝑛=1||𝑟𝑛||𝑡𝑡𝑘||||𝑢(𝑠)𝑛+𝜅𝑑𝑠𝑍0𝐶0𝑡𝑡𝑘||||𝜅𝑢(𝑠2𝑇)𝑑𝑠+𝐶0𝑝𝑛=1||𝑟𝑛||𝑡𝑡𝑘||||𝑢(𝑠2𝑇)𝑛||||𝑑𝑠+𝜅𝑡𝑡𝑘||||̇𝑢(𝑠2𝑇)𝑑𝑠2𝑈0𝑒𝜇𝑇𝐶0𝑍0+𝑅0𝑒𝜇(𝑡𝑡𝑘)1𝜇+𝑈0𝐶0𝑍0𝑒𝜇(𝑡𝑡𝑘)1𝜇+1𝐶0𝑝𝑛=1||𝑟𝑛||𝑈𝑛0𝑡𝑡𝑘𝑒𝑛𝜇(𝑠𝑡𝑘)+𝑑𝑠𝜅𝑈0𝑒2𝜇𝑇𝑍0𝐶0𝑒𝜇(𝑡𝑡𝑘)1𝜇+𝜅𝐶0𝑝𝑛=1||𝑟𝑛||𝑈𝑛0𝑒2𝑛𝜇𝑇𝑡𝑡𝑘𝑒𝑛𝜇(𝑠𝑡𝑘)||𝑢||𝑑𝑠+𝜅(𝑡2𝑇)𝑒𝜇(𝑡𝑡𝑘)𝑈01𝜇𝐶02𝑒𝜇𝑇𝑍0+𝑅0+1+𝜅𝑒2𝜇𝑇𝑍0+𝑝𝑛=1||𝑟𝑛||𝑈0𝑛1𝑒(𝑛1)𝜇𝑇011+𝜅𝑒2𝑛𝜇𝑇𝑛+𝜅𝑒2𝜇𝑇,𝐵22𝑈0𝑒𝜇𝑇𝐶0𝑍0+𝑅0𝑒𝜇(𝑡𝑘+1𝑡𝑘)1𝜇+𝑈0𝐶0𝑍0𝑒𝜇(𝑡𝑘+1𝑡𝑘)1𝜇+1𝐶0𝑝𝑛=1||𝑟𝑛||𝑈𝑛0𝑡𝑘+1𝑡𝑘𝑒𝑛𝜇(𝑠𝑇)+𝑑𝑠𝜅𝑈0𝑒2𝜇𝑇𝑍0𝐶0𝑒𝜇(𝑡𝑘+1𝑡𝑘)1𝜇+𝜅𝐶0𝑝𝑛=1||𝑟𝑛||𝑈𝑛0𝑒2𝑛𝜇𝑇𝑡𝑘+1𝑡𝑘𝑒𝑛𝜇(𝑠𝑇)||𝑢𝑡𝑑𝑠+𝜅𝑘+1𝑡2𝑇𝑢𝑘||2𝑇2𝑈0𝑒𝜇𝑇𝐶0𝑍0+𝑅0𝑒𝜇𝑇01𝜇+𝑈0𝐶0𝑍0𝑒𝜇𝑇01𝜇+1𝐶0𝑝𝑛=1||𝑟𝑛||𝑈𝑛0𝑒𝑛𝜇𝑇01+𝑛𝜇𝜅𝑈0𝑒2𝜇𝑇𝐶0𝑍0𝑒𝜇𝑇01𝜇+𝜅𝐶0𝑝𝑛=1||𝑟𝑛||𝑈𝑛0𝑒2𝑛𝜇𝑇𝑒𝑛𝜇𝑇01𝑛𝜇𝑒𝜇(𝑡𝑡𝑘)𝑈0𝜇𝐶02𝑒𝜇𝑇𝑒𝜇𝑇01𝑍0+𝑅0+𝑒𝜇𝑇011+𝜅𝑒2𝜇𝑇𝑍0+𝑝𝑛=1||𝑟𝑛||𝑈0𝑛11+𝜅𝑒2𝑛𝜇𝑇𝑒𝑛𝜇𝑇01𝑛.(2.7)
Therefore, for sufficiently large 𝜇>0, we obtain ||||(𝐵𝑢)(𝑡)𝑒𝜇(𝑡𝑡𝑘)𝑈01𝜇𝐶02𝑒𝜇𝑇𝑍0+𝑅0+1+𝜅𝑒2𝜇𝑇𝑍0+𝑝𝑛=1||𝑟𝑛||𝑈0𝑛1𝑒(𝑛1)𝜇𝑇011+𝜅𝑒2𝑛𝜇𝑇𝑛+𝜅𝑒2𝜇𝑇+𝑒𝜇(𝑡𝑡𝑘)𝑈01𝜇𝐶02𝑒𝜇𝑇𝑒𝜇𝑇01𝑍0+𝑅0+𝑒𝜇𝑇011+𝜅𝑒2𝜇𝑇𝑍0+𝑝𝑛=1||𝑟𝑛||𝑈0𝑛1𝑒𝑛𝜇𝑇01𝑛1+𝜅𝑒2𝑛𝜇𝑇𝑒𝜇(𝑡𝑡𝑘)𝑈01𝜇𝐶02𝑒𝜇𝑇𝑒𝜇𝑇0𝑍0+𝑅0+𝑒𝜇𝑇01+𝜅𝑒2𝜇𝑇𝑍0+𝑝𝑛=1||𝑟𝑛||𝑈0𝑛1𝑒𝑛𝜇𝑇0+𝑒(𝑛1)𝜇𝑇021+𝜅𝑒2𝑛𝜇𝑇𝑛+𝜅𝑒2𝜇𝑇𝑒𝜇(𝑡𝑡𝑘)𝑈0.(2.8)
Consequently, the operator 𝐵 maps 𝑀𝑆𝑈 into itself.
We show that B is a contractive operator. Indeed, ||𝐵(𝑢)(𝑡)𝐵𝑢||||||(𝑡)𝑡𝑡𝑘𝑈(𝑢)(𝑠)𝑈𝑢||||+||||(𝑠)𝑑𝑠𝑡𝑘+1𝑡𝑘𝑈(𝑢)(𝑠)𝑈𝑢||||(𝑠)𝑑𝑠𝐵1+𝐵2𝑡,𝑡𝑘,𝑡𝑘+1.(2.9)
We have 𝐵11𝐶0𝑍0𝑡𝑡𝑘||𝑢(𝑠)||1𝑢(𝑠)𝑑𝑠+𝐶0𝑝𝑛=1||𝑟𝑛||𝑡𝑡𝑘||𝑢𝑛(𝑠)𝑢𝑛||+𝜅(𝑠)𝑑𝑠𝑍0𝐶0𝑡𝑡𝑘||𝑢(𝑠2𝑇)||𝜅𝑢(𝑠2𝑇)𝑑𝑠+𝐶0𝑝𝑛=1||𝑟𝑛||𝑡𝑡𝑘||𝑢𝑛(𝑠2𝑇)𝑢𝑛||||||(𝑠2𝑇)𝑑𝑠+𝜅𝑡𝑡𝑘̇̇𝑢(𝑠2𝑇)||||𝜌𝑢(𝑠2𝑇)𝑑𝑠𝑘𝑢,𝑢𝐶0𝑍0𝑒𝜇(𝑡𝑡𝑘)1𝜇+1𝐶0𝑝𝑛=1𝑛||𝑟𝑛||||𝑢esssup𝑛1||𝑡(𝑠)𝑠𝑘,𝑡𝑘+1𝑡𝑡𝑘||𝑢(𝑠)||+𝜅𝑢(𝑠)𝑑𝑠𝑍0𝐶0𝜌𝑘𝑢,𝑢𝑒2𝜇𝑇𝑒𝜇(𝑡𝑡𝑘)1𝜇+𝜅𝐶0𝑝𝑛=1𝑛||𝑟𝑛||𝑢esssup𝑛1𝑡(𝑠2𝑇)𝑠𝑘,𝑡𝑘+1𝑡𝑡𝑘||𝑢(𝑠2𝑇)||𝑢(𝑠2𝑇)𝑑𝑠+𝜅𝜌𝑘̇̇𝑢,𝑢𝑒2𝜇𝑇𝑒𝜇(𝑡𝑡𝑘)1𝜇𝑒𝜇(𝑡𝑡𝑘)𝜌𝑘𝑢,𝑢𝜇𝐶0𝑍0+𝜌𝑘𝑢,𝑢𝜇𝐶0𝑝𝑛=1𝑛||𝑟𝑛||𝑈0𝑛1𝑒(𝑛1)𝜇(𝑡𝑘+1𝑡𝑘)+𝜅𝜌𝑘𝑢,𝑢𝑒2𝜇𝑇𝜇𝑍0𝐶0+𝜅𝜌𝑘𝑢,𝑢𝑒2𝜇𝑇𝜇𝐶0𝑝𝑛=1𝑛||𝑟𝑛||𝑈0𝑛1𝑒2(𝑛1)𝜇𝑇𝑒(𝑛1)𝜇(𝑡𝑘+1𝑡𝑘)+𝑒𝜇(𝑡𝑡𝑘)𝜅𝜌𝑘̇̇𝑢,𝑢𝑒2𝜇𝑇𝜇𝑒𝜇(𝑡𝑡𝑘)𝜌𝑘̇̇𝑢,𝑢1𝜇21+𝜅𝑒2𝜇𝑇𝐶0𝑍0+1𝐶0𝑝𝑛=1𝑛||𝑟𝑛||𝑈0𝑛11+𝜅𝑒2𝑛𝜇𝑇𝑒(𝑛1)𝜇𝑇0+𝜅𝑒2𝜇𝑇𝜇𝑒𝜇(𝑡𝑡𝑘)𝜌𝜇(𝑘)𝑢,𝑢1𝜇21+𝜅𝑒2𝜇𝑇𝐶0𝑍0+1𝐶0𝑝𝑛=1𝑛||𝑟𝑛||𝑈0𝑛11+𝜅𝑒2𝑛𝜇𝑇𝑒(𝑛1)𝜇𝑇0+𝜅𝑒2𝜇𝑇𝜇,𝐵21𝐶0𝑍0𝑡𝑘+1𝑡𝑘||𝑢(𝑠)||1𝑢(𝑠)𝑑𝑠+𝐶0𝑝𝑛=1||𝑟𝑛||𝑡𝑘+1𝑡𝑘||𝑢𝑛(𝑠)𝑢𝑛||+𝜅(𝑠)𝑑𝑠𝑍0𝐶0𝑡𝑘+1𝑡𝑘||𝑢(𝑠2𝑇)||𝜅𝑢(𝑠2𝑇)𝑑𝑠+𝐶0𝑝𝑛=1||𝑟𝑛||𝑡𝑘+1𝑡𝑘||𝑢𝑛(𝑠2𝑇)𝑢𝑛||||||(𝑠2𝑇)𝑑𝑠+𝜅𝑡𝑘+1𝑡𝑘̇||||𝜌̇𝑢(𝑠2𝑇)̇𝑢(𝑠2𝑇)𝑑𝑠𝑘𝑢,𝑢𝐶0𝑍0𝑒𝜇(𝑡𝑘+1𝑡𝑘)1𝜇+1𝐶0𝑝𝑛=1||𝑟𝑛||||𝑢𝑛.esssup𝑛1||𝑡(𝑠)𝑠𝑘,𝑡𝑘+1𝑡𝑘+1𝑡𝑘||𝑢(𝑠)||+𝜅𝑢(𝑠)𝑑𝑠𝑍0𝐶0𝜌𝑘𝑢,𝑢𝑒2𝜇𝑇𝑒𝜇(𝑡𝑘+1𝑡𝑘)1𝜇+𝜅𝐶0𝑝𝑛=1||𝑟𝑛||𝑢𝑛.esssup𝑛1𝑡(𝑠2𝑇)𝑠𝑘,𝑡𝑘+1𝑡𝑘+1𝑡𝑘||𝑢(𝑠2𝑇)||𝜌𝑢(𝑠2𝑇)𝑑𝑠𝑘𝑢,𝑢𝐶0𝑍0𝑒𝜇(𝑡𝑘+1𝑡𝑘)1𝜇+𝜌𝑘𝑢,𝑢𝐶0𝑒𝜇(𝑡𝑘+1𝑡𝑘)1𝜇𝑝𝑛=1||𝑟𝑛||𝑛𝑈0𝑛1𝑒(𝑛1)𝜇(𝑡𝑘+1𝑡𝑘)+𝜅𝜌𝑘𝑢,𝑢𝑒2𝜇𝑇𝑍0𝐶0𝑒𝜇(𝑡𝑘+1𝑡𝑘)1𝜇+𝜅𝜌𝑘𝑢,𝑢𝑒2𝜇𝑇𝐶0𝑒𝜇(𝑡𝑘+1𝑡𝑘)1𝜇𝑝𝑛=1𝑛||𝑟𝑛||𝑈0𝑛1𝑒(𝑛1)(𝜇𝑇02𝜇𝑇)𝜌𝑘̇̇𝑢,𝑢𝑒𝜇𝑇01𝜇21+𝜅𝑒2𝜇𝑇𝐶0𝑍0+1𝐶0𝑝𝑛=1||𝑟𝑛||𝑛𝑈0𝑛1𝑒(𝑛1)𝜇𝑇01+𝜅𝑒2𝑛𝜇𝑇𝜌𝑢(𝑘)𝑢,𝑢𝑒𝜇𝑇01𝜇2𝐶01+𝜅𝑒2𝜇𝑇𝑍0+𝑝𝑛=1||𝑟𝑛||𝑛𝑈0𝑛1𝑒(𝑛1)𝜇𝑇01+𝜅𝑒2𝑛𝜇𝑇.(2.10)
Consequently, ||𝐵(𝑢)(𝑡)𝐵𝑢||(𝑡)𝑒𝜇(𝑡𝑡𝑘)𝜌𝜇(𝑘)𝑢,𝑢1𝜇21+𝜅𝑒2𝜇𝑇𝐶0𝑍0+1𝐶0𝑝𝑛=1𝑛||𝑟𝑛||𝑈0𝑛11+𝜅𝑒2𝑛𝜇𝑇𝑒(𝑛1)𝜇𝑇0+𝜅𝑒2𝜇𝑇𝜇+𝜌𝑢(𝑘)𝑢,𝑢𝑒𝜇𝑇01𝜇21+𝜅𝑒2𝜇𝑇𝐶0𝑍0+1𝐶0𝑝𝑛=1||𝑟𝑛||𝑛𝑈0𝑛1𝑒(𝑛1)𝜇𝑇01+𝜅𝑒2𝑛𝜇𝑇𝜌𝑢(𝑘)𝑢,𝑢𝑒𝜇𝑇0𝜇21+𝜅𝑒2𝜇𝑇𝐶0𝑍0+1𝐶0𝑝𝑛=1||𝑟𝑛||𝑛𝑈0𝑛1𝑒(𝑛1)𝜇𝑇01+𝜅𝑒2𝑛𝜇𝑇+𝜅𝑒2𝜇𝑇𝜇.𝑒𝜇(𝑡𝑡𝑘)𝐾𝑈𝜌𝜇(𝑘)𝑢,𝑢.(2.11)
Therefore, 𝜌𝑘(𝐵𝑢,𝐵𝑢)𝐾𝑈𝜌𝜇(𝑘)(𝑢,𝑢).
It remains to estimate the derivative of B.
We have||̇𝐵̇𝐵(𝑢)(𝑡)𝑢||||𝑈(𝑡)(𝑢)(𝑠)𝑈𝑢||+1(𝑠)𝑡𝑘+1𝑡𝑘||||𝑡𝑘+1𝑡𝑘𝑈(𝑢)(𝑠)𝑈𝑢(||||̇𝐵𝑠)𝑑𝑠1+̇𝐵2.(2.12)
We have ̇𝐵11𝐶0𝑍0||𝑢(𝑡)||+1𝑢(𝑡)𝐶0𝑝𝑛=1||𝑟𝑛||||𝑢𝑛(𝑡)𝑢𝑛||+𝜅(𝑡)𝐶0𝑍0||𝑢(𝑡2𝑇)||+𝜅𝑢(𝑡2𝑇)𝐶0𝑝𝑛=1||𝑟𝑛||||𝑢𝑛(𝑡2𝑇)𝑢𝑛||||̇(𝑡2𝑇)+𝜅̇𝑢(𝑡2𝑇)𝑢||𝑒(𝑡2𝑇)𝜇(𝑡𝑡𝑘)𝜌𝑘𝑢,𝑢𝐶0𝑍0+1𝐶0𝑝𝑛=1||𝑟𝑛||𝑢𝑛esssup𝑛1𝑡(𝑡)𝑡𝑘,𝑡𝑘+1||𝑢(𝑡)𝑢||+(𝑡)𝜅𝑒𝜇(𝑡𝑡𝑘)𝜌𝑘𝑢,𝑢𝑒2𝜇𝑇𝐶0𝑍0+𝜅𝐶0𝑝𝑛=1𝑛||𝑟𝑛||𝑢esssup𝑛1𝑡(𝑡2𝑇)𝑡𝑘,𝑡𝑘+1||𝑢(𝑡2𝑇)||||̇𝑢(𝑡2𝑇)+𝜅̇𝑢(𝑡2𝑇)||𝑒𝑢(𝑡2𝑇)𝜇(𝑡𝑡𝑘)𝜌𝑘̇̇𝑢,𝑢𝜇𝐶0𝑍0+𝑒𝜇(𝑡𝑡𝑘)𝜌𝑘̇̇𝑢,𝑢𝜇𝐶0𝑝𝑛=1||𝑟𝑛||𝑛𝑈0𝑛1𝑒(𝑛1)𝜇(𝑡𝑘+1𝑡𝑘)+𝑒𝜇(𝑡𝑡𝑘)𝜅𝜌𝑘̇̇𝑢,𝑢𝑒2𝜇𝑇𝜇𝐶0𝑍0+𝑒𝜇(𝑡𝑡𝑘)𝜌𝑘̇̇𝑢,𝑢𝜅𝑒2𝜇𝑇𝜇𝐶0𝑝𝑛=1𝑛||𝑟𝑛||𝑈0𝑛1𝑒(𝑛1)𝜇(𝑡𝑘+1𝑡𝑘)+𝑒𝜇(𝑡𝑡𝑘)𝜅𝜌𝑘̇̇𝑢,𝑢𝑒2𝜇𝑇𝑒𝜇(𝑡𝑡𝑘)𝜌𝜇(𝑘)𝑢,𝑢1+𝜅𝑒2𝜇𝑇𝜇𝐶01𝑍0+𝑝𝑛=1||𝑟𝑛||𝑛𝑈0𝑛1𝑒(𝑛1)𝜇𝑇0+𝜅𝑒2𝜇𝑇,̇𝐵21𝑡𝑘+1𝑡𝑘||||𝑡𝑘+1𝑡𝑘𝑈(𝑢)(𝑠)𝑈𝑢||||1(𝑠)𝑑𝑠𝑙0||||𝑡𝑘+1𝑡𝑘𝑈(𝑢)(𝑠)𝑈𝑢||||(𝑠)𝑑𝑠𝜌𝑢(𝑘)𝑢,𝑢𝑒𝜇𝑇01𝜇2𝐶0𝑙01+𝜅𝑒2𝜇𝑇𝑍0+𝑝𝑛=1||𝑟𝑛||𝑛𝑈0𝑛1𝑒(𝑛1)𝜇𝑇01+𝜅𝑒2𝑛𝜇𝑇.(2.13)
Therefore, ||̇𝐵̇𝐵(𝑢)(𝑡)𝑢||(𝑡)𝑒𝜇(𝑡𝑡𝑘)𝜌𝜇(𝑘)𝑢,𝑢1+𝜅𝑒2𝜇𝑇𝜇𝐶01𝑍0+𝑝𝑛=1||𝑟𝑛||𝑛𝑈0𝑛1𝑒(𝑛1)𝜇𝑇0+𝜅𝑒2𝜇𝑇+𝜌𝑢(𝑘)𝑢,𝑢𝑒𝜇𝑇01𝜇2𝐶0𝑙01+𝜅𝑒2𝜇𝑇𝑍0+𝑝𝑛=1||𝑟𝑛||𝑛𝑈0𝑛1𝑒(𝑛1)𝜇𝑇01+𝜅𝑒2𝑛𝜇𝑇𝜌𝜇(𝑘)𝑢,𝑢𝑒𝜇𝑇0+𝜇𝜏011+𝜅𝑒2𝜇𝑇𝜇2𝐶0𝑙01𝑍0+𝑝𝑛=1||𝑟𝑛||𝑛𝑈0𝑛1𝑒(𝑛1)𝜇𝑇0+𝜅𝑒2𝜇𝑇𝑒𝜇(𝑡𝑡𝑘)̇𝐾𝑈𝜌𝜇(𝑘)𝑢,𝑢.(2.14)
It follows 𝜌𝑘(̇̇𝐵(𝑢),𝐵(𝑢))𝑒𝜇(𝑡𝑡𝑘)̇𝐾𝑈𝜌𝜇(𝑘)(𝑢,𝑢).
Then 𝜌𝜇(𝑘)(𝐵(𝑢),𝐵(𝑢))max{𝐾𝑈,̇𝐾𝑈}𝜌𝜇(𝑘)(𝑢,𝑢).
Consequently, 𝜌𝜇(𝑘)𝐵𝑢,𝐵𝑢𝐾𝜌𝜇(𝑘)𝑢,𝑢(𝑘=0,1,2,),(2.15) where 𝐾=max{𝐾𝑈,̇𝐾𝑈}<1 does not depend on 𝑢 and 𝑘.
We have to verify that 𝑀𝑆𝑈 is j-bounded. Indeed, since j is an identity mapping,𝜌𝑗𝑛𝑢(𝑘)𝑢,𝑢𝜌𝑢(𝑘)𝑢,𝑢<(𝑛=0,1,2,).(2.16)
Therefore, in view of the fixed point theorem for contractive mappings in uniform spaces (cf. [13]), the operator B has a unique fixed point, and it is an oscillatory solution of (1.4).
Theorem 2.2 is thus proved.

3. Numerical Example

Finally, we summarize all inequalities needed for the applications:1𝜇𝐶02𝑒𝜇𝑇𝑒𝜇𝑇0𝑍0+𝑅0+𝑒𝜇𝑇01+𝜅𝑒2𝜇𝑇𝑍0+𝑝𝑛=1||𝑟𝑛||𝑈0𝑛1𝑒𝑛𝜇𝑇0+𝑒(𝑛1)𝜇𝑇021+𝜅𝑒2𝑛𝜇𝑇𝑛+𝜅𝑒2𝜇𝑇𝐾1,𝑈=𝑒𝜇0𝜇21+𝜅𝑒2𝜇𝑇𝐶0𝑍0+1𝐶0𝑝𝑛=1||𝑟𝑛||𝑛𝑈0𝑛1𝑒(𝑛1)𝜇01+𝜅𝑒2𝑛𝜇𝑇+𝜅𝑒2𝜇𝑇𝜇̇𝐾<1,𝑈=𝑒𝜇0+𝜇𝜏011+𝜅𝑒2𝜇𝑇𝜇2𝐶0𝑙01𝑍0+𝑝𝑛=1||𝑟𝑛||𝑛𝑈0𝑛1𝑒(𝑛1)𝜇0+𝜅𝑒2𝜇𝑇<1.(3.1)

Consider a line with the following specific parameters:1Λ=1m,𝐿=0,2𝜇H/m,𝐶=80pF/m,𝜈==1𝐿𝐶0,2106801012=14109=2,5108,𝑍0=𝐿𝐶=0,2106801012=50Ω,𝑅0=45Ω,𝐶0=8pF=81012F.(3.2)

Then, 𝑇=Λ𝐿𝐶=4.109s;𝜅=(𝑍0𝑅0)/(𝑍0+𝑅0)=1/19=0,0526.

Let us check the propagation of millimeter waves 𝜆0=103m. We have𝑓0=1𝜆0=1𝐿𝐶1034109=2,51011Hz𝑇0=1𝑓0=12,51011=41012sec.;𝑙=21012sec.(3.3) If we choose 𝜇=(1/4)1012, then 𝜇𝑇0=𝜇0=1,𝜇𝜏0=(1/2), and 𝑇=4109(1/4)1012𝑇0=1000𝑇0.

Consequently, 𝜇𝑇=(1/4)10122108=(1/2)104,𝜇𝐶0=(1/4)101281041012=2, and 𝜇2𝐶0=(1/2)1012.

Since 𝑒𝜇𝑇=𝑒5000=0, then the above inequalities (omitting the second one) become𝑒+100𝑝𝑛=1||𝑟𝑛||𝑈0𝑛1𝑒𝑛+𝑒𝑛12̇𝐾2𝑛1,𝑈1=2𝑒21+50𝑝𝑛=1||𝑟𝑛||𝑛𝑈0𝑛1𝑒𝑛1<1.(3.4)

If the V-I characteristic of the nonlinear resistive element is 𝑓(𝑢)=0,12𝑢+0,8𝑢3, then 𝑈00,41; ̇𝐾𝑈=𝑈0<0,06. It follows that 𝑈0<0,06.


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