International Journal of Differential Equations

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Recent Advances in Oscillation Theory 2011

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Volume 2011 |Article ID 949547 | https://doi.org/10.1155/2011/949547

Vasil G. Angelov, Dafinka Tz. Angelova, "Oscillatory Solutions of Neutral Equations with Polynomial Nonlinearities", International Journal of Differential Equations, vol. 2011, Article ID 949547, 12 pages, 2011. https://doi.org/10.1155/2011/949547

Oscillatory Solutions of Neutral Equations with Polynomial Nonlinearities

Academic Editor: Elena Braverman
Received01 Jun 2011
Accepted31 Aug 2011
Published20 Nov 2011

Abstract

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).

The voltage and current 𝑒(π‘₯,𝑑), 𝑖(π‘₯,𝑑) of the lossless transmission line can be found by solving the following mixed problem for the hyperbolic partial differential system:πΆπœ•π‘’(π‘₯,𝑑)+πœ•π‘‘πœ•π‘–(π‘₯,𝑑)πœ•π‘₯=0,πΏπœ•π‘–(π‘₯,𝑑)+πœ•π‘‘πœ•π‘’(π‘₯,𝑑)πœ•π‘₯=0,𝐸(𝑑)βˆ’π‘’(0,𝑑)=𝑅0𝐢𝑖(0,𝑑),𝑑β‰₯0,(1.1)0𝑑𝑒(Ξ›,𝑑)𝑑𝑑=𝑖(Ξ›,𝑑)βˆ’π‘“(𝑒(Ξ›,𝑑)),𝑑β‰₯0,(1.2)𝑒(π‘₯,0)=𝑒0(π‘₯),𝑖(π‘₯,0)=𝑖0[],(π‘₯),π‘₯∈0,Ξ›(1.3) where 𝑒0(π‘₯) and 𝑖0(π‘₯) are prescribed initial functions, Ξ› is the length of the line, 𝐢 is the per-unit length capacitance, and 𝐿 is per-unit length inductance (cf. [3–10]). Here, the V-I characteristic of the nonlinear resistive load is βˆ‘π‘–=𝑓(𝑒)=𝑝𝑛=1π‘Ÿπ‘›π‘’π‘›, where π‘Ÿπ‘› are real numbers, 𝐢0 is parallel connected capacitance, 𝐸 is the source voltage, 𝑅0 is the source resistance, and 𝑍0=√𝐿/𝐢 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 equation𝑑𝑒(𝑑)=𝑑𝑑2𝐸𝐢0𝑍0+𝑅0ξ€Έβˆ’π‘’(𝑑)𝐢0𝑍0βˆ’1𝐢0𝑝𝑛=1π‘Ÿπ‘›[]𝑒(𝑑)π‘›βˆ’ξ€·π‘0βˆ’π‘…0𝑒(π‘‘βˆ’2𝑇)𝑍0𝐢0𝑍0+𝑅0ξ€Έ+𝑍0βˆ’π‘…0𝐢0𝑍0+𝑅0𝑝𝑛=1π‘Ÿπ‘›[]𝑒(π‘‘βˆ’2𝑇)𝑛+𝑍0βˆ’π‘…0𝑍0+𝑅0𝑑𝑒(π‘‘βˆ’2𝑇)𝑑𝑑,𝑑β‰₯𝑇,𝑒(𝑑)=𝜐0(𝑑),𝑑𝑒(𝑑)=π‘‘π‘‘π‘‘πœ0(𝑑)[],𝑑𝑑,π‘‘βˆˆβˆ’π‘‡,𝑇(1.4) where (π‘₯,𝑑)∈Π={(π‘₯,𝑑)βˆˆπ‘…2∢(π‘₯,𝑑)∈[0,Ξ›]Γ—[0,∞)}, πœ…=|𝑍0βˆ’π‘…0|/(𝑍0+𝑅0)<1,𝑒(𝑑)=𝑒(Ξ›,𝑑). 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(𝑑) can be obtained shifting the initial function 𝑒0(π‘₯) from (1.3) along the characteristics √π‘₯βˆ’πœˆπ‘‘=const.,(𝜈=1/𝐿𝐢) on [0,𝑇] and along the characteristics π‘₯+πœˆπ‘‘=const. on [βˆ’π‘‡,0] (cf. [1, 2]). So, we obtain an initial function 𝜐0(𝑑) on [βˆ’π‘‡,𝑇].

Now, we are able to formulate the main problem: to find a solution of (1.4) with advanced prescribed zeros on the interval [𝑑0,∞),𝑇=𝑑0.

Let 𝑆𝑇={πœπ‘˜}π‘›π‘˜=0,π‘›βˆˆπ‘ be the set of zeros of the initial function; that is, 𝜐0(πœπ‘˜)=0 such that 𝜏0=βˆ’π‘‡,πœπ‘›=𝑇≑𝑑0.

Let 𝑆={π‘‘π‘˜}βˆžπ‘˜=0 be a strictly increasing sequence of real numbers satisfying the following conditions (C):(C1)limπ‘˜β†’βˆžπ‘‘π‘˜=∞,(C2)0<𝑙0=inf{π‘‘π‘˜+1βˆ’π‘‘π‘˜βˆΆπ‘˜=0,1,2,…}≀sup{π‘‘π‘˜+1βˆ’π‘‘π‘˜βˆΆπ‘˜=0,1,2,…}=𝑇0<∞,(C3)for every π‘˜ there is 𝑠<π‘˜ such that π‘‘π‘˜βˆ’π‘‡=𝑑𝑠 where π‘‘π‘ βˆˆπ‘†π‘‡βˆͺ𝑆.

Introduce the sets: 𝐢1[𝑑0,∞) consisting of all continuous and bounded functions differentiable with bounded derivatives on every interval (π‘‘π‘˜,π‘‘π‘˜+1) (the derivatives at π‘‘π‘˜ do not necessary exist), 𝑀𝑆={𝑒(β‹…)∈𝐢1[𝑑0,∞)βˆΆπ‘’(π‘‘π‘˜)=0(π‘˜=0,1,2,…)}, π‘€π‘†π‘ˆ={𝑒(β‹…)βˆˆπ‘€π‘†βˆΆ|𝑒(𝑑)|β‰€π‘ˆ0π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜),π‘‘βˆˆ[π‘‘π‘˜,π‘‘π‘˜+1]}, where π‘ˆ0,πœ‡ are positive constants prescribed below.

We assume that |𝜐0(𝑑)|β‰€π‘ˆ0π‘’πœ‡(π‘‘βˆ’πœπ‘˜),π‘‘βˆˆ[πœπ‘˜,πœπ‘˜+1], (π‘˜=0,1,2,…,π‘›βˆ’1).

The set π‘€π‘†π‘ˆ turns out into a complete uniform space with respect to the family of pseudometrics πœŒπœ‡(π‘˜)(𝑓,𝑔)=max{πœŒπ‘˜(𝑓,𝑔),πœŒπ‘˜(̇𝑓,̇𝑔)}, (π‘˜=0,1,2,…), where πœŒπ‘˜(𝑓,𝑔)=max{π‘’βˆ’πœ‡(π‘‘βˆ’π‘‘π‘˜)|𝑓(𝑑)βˆ’π‘”(𝑑)|βˆΆπ‘‘βˆˆ[π‘‘π‘˜,π‘‘π‘˜+1]}, πœŒπ‘˜(̇𝑓,̇𝑔)=max{π‘’βˆ’πœ‡(π‘‘βˆ’π‘‘π‘˜)|̇𝑓(𝑑)βˆ’Μ‡π‘”(𝑑)|βˆΆπ‘‘βˆˆ[π‘‘π‘˜,π‘‘π‘˜+1]}.

One can verify that π‘€π‘†π‘ˆ is closed subset of 𝐢1[𝑑0,∞) with respect to the above metric.

Remark 1.1. The functions from 𝑀𝑆 are not necessary differentiable at π‘‘π‘˜ (π‘˜=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 π΅βˆΆπ‘€π‘†π‘ˆβ†’π‘€π‘†π‘ˆ by ξ€œπ΅(𝑒)(𝑑)∢=π‘‘π‘‘π‘˜ξ‚΅π‘ˆ(𝑒)(𝑠)π‘‘π‘ βˆ’π‘‘βˆ’π‘‘π‘˜π‘‘π‘˜+1βˆ’π‘‘π‘˜ξ‚Άξ€œπ‘‘π‘˜+1π‘‘π‘˜ξ€Ίπ‘‘π‘ˆ(𝑒)(𝑠)𝑑𝑠,π‘‘βˆˆπ‘˜,π‘‘π‘˜+1ξ€»,(π‘˜=0,1,2,…),(1.5) where π‘ˆ(𝑒)(𝑑)=2𝐸𝐢0𝑍0+𝑅0ξ€Έβˆ’π‘’(𝑑)𝐢0𝑍0βˆ’1𝐢0𝑝𝑛=1π‘Ÿπ‘›[]𝑒(𝑑)π‘›βˆ’πœ…ξ€·πΎπ‘‡π‘’ξ€Έ(𝑑)𝑍0𝐢0+πœ…πΆ0𝑝𝑛=1π‘Ÿπ‘›πΎξ€Ίξ€·π‘‡π‘’ξ€Έξ€»(𝑑)𝑛𝑑𝐾+πœ…π‘‡π‘’ξ€Έ(𝑑)𝑑𝑑,𝑑β‰₯𝑇,(1.6) and (𝐾𝑇𝑒)(𝑑)=𝑒(π‘‘βˆ’2𝑇) 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 [βˆ’π‘‡,𝑇]. We notice that 𝐾𝑇 maps 𝑀𝑆 into itself. Indeed, consider the set 𝐢1[βˆ’π‘‡,∞) consisting of all continuous and bounded functions differentiable with bounded derivatives on every interval (π‘‘π‘˜,π‘‘π‘˜+1). Introduce the set π‘€πœ0𝑆={𝑒(β‹…)∈𝐢1[βˆ’π‘‡,∞)βˆΆπ‘’(𝑑)=𝜐0(𝑑),π‘‘βˆˆ[βˆ’π‘‡,𝑇]}. Then, 𝐾𝑇 assigns to every function 𝑒(β‹…)βˆˆπ‘€π‘† the function ̃𝑒(β‹…)βˆˆπ‘€πœ0𝑆 translated to the right on the interval [𝑇,∞). So, the function (𝐾𝑇𝑒)(𝑑) coincides with 𝜐0(𝑑) on [𝑑0,𝑑0+2𝑇]. Besides π‘‘π‘˜βˆ’2𝑇=𝑑𝑠, and then ξ€·πΎπ‘‡π‘’π‘‘ξ€Έξ€·π‘˜ξ€Έ=ξ‚»π‘’ξ€·π‘‘π‘˜ξ€Έβˆ’2𝑇=𝜐0𝑑𝑠=0,π‘‘π‘˜βˆˆ[],𝑒𝑑𝑇,3π‘‡π‘˜ξ€Έξ€·π‘‘βˆ’2𝑇=𝑒𝑛=0,π‘‘βˆˆ(3𝑇,∞),(1.7) that is, (𝐾𝑇𝑒)(β‹…)βˆˆπ‘€π‘†.

2. Main Results

Lemma 2.1. If πΈβ‰€π‘ˆ0, problem (1.4) has a solution 𝑒(β‹…)βˆˆπ‘€π‘†π‘ˆ iff the operator 𝐡 has a fixed point in π‘€π‘†π‘ˆ, that is, 𝑒(𝑑)=𝐡(𝑒)(𝑑).(2.1)

Proof. Let 𝑒(β‹…)βˆˆπ‘€π‘†π‘ˆ be a solution of (1.4). Then, integrating (1.4) on the interval [π‘‘π‘˜,𝑑]βŠ‚[π‘‘π‘˜,π‘‘π‘˜+1] (π‘˜=0,1,2…), we obtain 𝑒(𝑑)βˆ’π‘’(π‘‘π‘˜βˆ«)=π‘‘π‘‘π‘˜βˆ«π‘ˆ(𝑒)(𝑠)𝑑𝑠⇔𝑒(𝑑)=π‘‘π‘‘π‘˜π‘ˆ(𝑒)(𝑠)𝑑𝑠, and then, ξ€œπ‘’(𝑑)=π‘‘π‘‘π‘˜ξ€·π‘‘π‘ˆ(𝑒)(𝑠)π‘‘π‘ βŸΉ0=π‘’π‘˜+1ξ€Έ=ξ€œπ‘‘π‘˜+1π‘‘π‘˜ξ€œπ‘ˆ(𝑒)(𝑠)π‘‘π‘ βŸΉπ‘‘π‘˜+1π‘‘π‘˜π‘ˆ(𝑒)(𝑠)𝑑𝑠=0.(2.2) Therefore, 𝑒(𝑑)satisfies ξ€œπ‘’(𝑑)=π‘‘π‘‘π‘˜ξ€œπ‘ˆ(𝑒)(𝑠)π‘‘π‘ βŸΊπ‘’(𝑑)=π‘‘π‘‘π‘˜ξ‚΅π‘ˆ(𝑒)(𝑠)π‘‘π‘ βˆ’π‘‘βˆ’π‘‘π‘˜π‘‘π‘˜+1βˆ’π‘‘π‘˜ξ‚Άξ€œπ‘‘π‘˜+1π‘‘π‘˜π‘ˆ(𝑒)(𝑠)𝑑𝑠,(2.3) that is, 𝑒(β‹…)is a fixed point of B.
Conversely, let 𝑒(β‹…)βˆˆπ‘€π‘†π‘ˆ be a solution of 𝑒=𝐡(𝑒); that is, ξ€œπ‘’(𝑑)=π‘‘π‘‘π‘˜ξ‚΅π‘ˆ(𝑒)(𝑠)π‘‘π‘ βˆ’π‘‘βˆ’π‘‘π‘˜π‘‘π‘˜+1βˆ’π‘‘π‘˜ξ‚Άξ€œπ‘‘π‘˜+1π‘‘π‘˜π‘ˆ(𝑒)(𝑠)𝑑𝑠.(2.4)
Then, introducing πœ‡0=πœ‡π‘‡0, we obtain ||||ξ€œπ‘‘π‘˜+1π‘‘π‘˜π‘ˆ||||≀(𝑒)(𝑠)𝑑𝑠2𝐸𝐢0𝑍0+𝑅0ξ€Έξ€œπ‘‘π‘˜+1π‘‘π‘˜π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)1𝑑𝑑+𝐢0𝑍0ξ€œπ‘‘π‘˜+1π‘‘π‘˜||𝑒||+1(𝑑)𝑑𝑑𝐢0𝑝𝑛=1||π‘Ÿπ‘›||ξ€œπ‘‘π‘˜+1π‘‘π‘˜||||𝑒(𝑑)π‘›πœ…π‘‘π‘‘+𝑍0𝐢0ξ€œπ‘‘π‘˜+1π‘‘π‘˜||||+πœ…π‘’(π‘‘βˆ’2𝑇)𝑑𝑑𝐢0𝑝𝑛=1||π‘Ÿπ‘›||ξ€œπ‘‘π‘˜+1π‘‘π‘˜||||𝑒(π‘‘βˆ’2𝑇)𝑛||||ξ€œπ‘‘π‘‘+πœ…π‘‘π‘˜+1π‘‘π‘˜||||≀̇𝑒(π‘‘βˆ’2𝑇)𝑑𝑑2π‘ˆ0π‘’βˆ’πœ‡π‘‡πΆ0𝑍0+𝑅0ξ€Έπ‘’πœ‡(π‘‘π‘˜+1βˆ’π‘‘π‘˜)βˆ’1πœ‡+π‘ˆ0𝐢0𝑍0π‘’πœ‡(π‘‘π‘˜+1βˆ’π‘‘π‘˜)βˆ’1πœ‡+1𝐢0𝑝𝑛=1||π‘Ÿπ‘›||π‘ˆπ‘›0ξ€œπ‘‘π‘˜+1π‘‘π‘˜π‘’π‘›πœ‡(π‘‘βˆ’π‘‘π‘˜)+π‘‘π‘‘πœ…π‘ˆ0π‘’βˆ’2πœ‡π‘‡π‘0𝐢0π‘’πœ‡(π‘‘π‘˜+1βˆ’π‘‘π‘˜)βˆ’1πœ‡+πœ…πΆ0𝑝𝑛=1||π‘Ÿπ‘›||π‘ˆπ‘›0π‘’βˆ’2π‘›πœ‡π‘‡Γ—ξ€œπ‘‘π‘˜+1π‘‘π‘˜π‘’π‘›πœ‡(π‘‘βˆ’π‘‘π‘˜)||𝑒𝑑𝑑𝑑+πœ…π‘˜+1ξ€Έξ€·π‘‘βˆ’2π‘‡βˆ’π‘’π‘˜ξ€Έ||β‰€βˆ’2𝑇2π‘ˆ0π‘’βˆ’πœ‡π‘‡πΆ0𝑍0+𝑅0ξ€Έπ‘’πœ‡π‘‡0βˆ’1πœ‡+π‘ˆ0𝐢0𝑍0π‘’πœ‡π‘‡0βˆ’1πœ‡+1𝐢0𝑝𝑛=1||π‘Ÿπ‘›||π‘ˆπ‘›0π‘’π‘›πœ‡π‘‡0βˆ’1+π‘ˆπ‘›πœ‡0πœ…π‘’βˆ’2πœ‡π‘‡πΆ0𝑍0π‘’πœ‡π‘‡0βˆ’1πœ‡+πœ…πΆ0𝑝𝑛=1||π‘Ÿπ‘›||π‘ˆπ‘›0π‘’βˆ’2π‘›πœ‡π‘‡π‘’π‘›πœ‡π‘‡0βˆ’1β‰€π‘’π‘›πœ‡πœ‡0βˆ’1πœ‡πΆ02π‘ˆ0π‘’βˆ’πœ‡π‘‡π‘0+𝑅0+π‘ˆ0ξ€·1+πœ…π‘’βˆ’2πœ‡π‘‡ξ€Έπ‘0ξƒͺ+1πœ‡πΆ0𝑝𝑛=1||π‘Ÿπ‘›||π‘ˆπ‘›0ξ€·1+πœ…π‘’βˆ’2π‘›πœ‡π‘‡ξ€Έ(π‘’π‘›πœ‡0βˆ’1)𝑛≑𝑀(πœ‡).(2.5)
Let us assume that |βˆ«π‘‘π‘˜+1π‘‘π‘˜π‘ˆ(𝑒)(𝑑)𝑑𝑑|=𝛽>0. We have just obtained that 𝛽≀𝑀(πœ‡). Then, for sufficiently large πœ‡>0 (and sufficiently small 𝑇0>0), one can reach the inequality 𝑀(πœ‡)<𝛽. Consequently, βˆ«π‘‘π‘˜+1π‘‘π‘˜π‘ˆ(𝑒)(𝑑)𝑑𝑑=0. It follows that βˆ«π‘’(𝑑)=π‘‘π‘‘π‘˜π‘ˆ(𝑒)(𝑠)𝑑𝑠 and, after a differentiation, we obtain (1.4).
Lemma 2.1 is thus proved.

Theorem 2.2. Let 𝑆𝑇={πœπ‘˜}π‘›π‘˜=0,π‘›βˆˆπ‘ be the set of zeros of the initial function; that is, 𝜐0(πœπ‘˜)=0 and 𝜐0(β‹…)∈𝐢1[βˆ’π‘‡,𝑇]. If πΈβ‰€π‘ˆ0,|𝜐0(𝑑)|β‰€π‘ˆ0π‘’πœ‡(π‘‘βˆ’πœπ‘˜), π‘‘βˆˆ[πœπ‘˜,πœπ‘˜+1],𝜐0(𝑑0)=0, then, there exists a unique oscillatory solution of the initial value problem (1.4), belonging to π‘€π‘†π‘ˆ.

Proof. We show that 𝐡 maps π‘€π‘†π‘ˆ into itself; that is, π‘’βˆˆπ‘€π‘†π‘ˆβ‡’π΅(𝑒)βˆˆπ‘€π‘†π‘ˆ.
Indeed, for every 𝑒(β‹…)βˆˆπ‘€π‘†π‘ˆ, the function 𝐡(𝑒)(𝑑) is continuous on [𝑑0,∞) and differentiable on every (π‘‘π‘˜,π‘‘π‘˜+1). We have also 𝐡(𝑒)(π‘‘π‘˜)=0 and 𝐡(𝑒)(π‘‘π‘˜+1)=0.
We show that |(𝐡𝑒)(𝑑)|β‰€π‘ˆ0π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜), π‘‘βˆˆ[π‘‘π‘˜,π‘‘π‘˜+1]. (The last inequalities imply that 𝐡(𝑒)(𝑑) is bounded because π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)β‰€π‘’πœ‡π‘‡0,π‘‘βˆˆ[𝑇,∞).)
We notice that |(π‘‘βˆ’π‘‘π‘˜)/(π‘‘π‘˜+1βˆ’π‘‘π‘˜)|≀1, π‘‘βˆˆ[π‘‘π‘˜,π‘‘π‘˜+1]. For sufficiently large πœ‡, we obtain for π‘‘βˆˆ[π‘‘π‘˜,π‘‘π‘˜+1]||||≀||||ξ€œ(𝐡𝑒)(𝑑)π‘‘π‘‘π‘˜||||+||||ξ€œπ‘ˆ(𝑒)(𝑠)π‘‘π‘ π‘‘π‘˜+1π‘‘π‘˜π‘ˆ||||(𝑒)(𝑠)𝑑𝑠≑𝐡1+𝐡2.(2.6) We have 𝐡1≀2𝐢0𝑍0+𝑅0ξ€Έξ€œπ‘‘π‘‘π‘˜||||1𝐸(π‘ βˆ’π‘‡)𝑑𝑠+𝐢0𝑍0ξ€œπ‘‘π‘‘π‘˜||||1𝑒(𝑠)𝑑𝑠+𝐢0𝑝𝑛=1||π‘Ÿπ‘›||ξ€œπ‘‘π‘‘π‘˜||||𝑒(𝑠)𝑛+πœ…π‘‘π‘ π‘0𝐢0ξ€œπ‘‘π‘‘π‘˜||||πœ…π‘’(π‘ βˆ’2𝑇)𝑑𝑠+𝐢0𝑝𝑛=1||π‘Ÿπ‘›||ξ€œπ‘‘π‘‘π‘˜||||𝑒(π‘ βˆ’2𝑇)𝑛||||ξ€œπ‘‘π‘ +πœ…π‘‘π‘‘π‘˜||||≀̇𝑒(π‘ βˆ’2𝑇)𝑑𝑠2π‘ˆ0π‘’βˆ’πœ‡π‘‡πΆ0𝑍0+𝑅0ξ€Έπ‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)βˆ’1πœ‡+π‘ˆ0𝐢0𝑍0π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)βˆ’1πœ‡+1𝐢0𝑝𝑛=1||π‘Ÿπ‘›||π‘ˆπ‘›0ξ€œπ‘‘π‘‘π‘˜π‘’π‘›πœ‡(π‘ βˆ’π‘‘π‘˜)+π‘‘π‘ πœ…π‘ˆ0π‘’βˆ’2πœ‡π‘‡π‘0𝐢0π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)βˆ’1πœ‡+πœ…πΆ0𝑝𝑛=1||π‘Ÿπ‘›||π‘ˆπ‘›0π‘’βˆ’2π‘›πœ‡π‘‡ξ€œπ‘‘π‘‘π‘˜π‘’π‘›πœ‡(π‘ βˆ’π‘‘π‘˜)ξƒ­||𝑒||𝑑𝑠+πœ…(π‘‘βˆ’2𝑇)β‰€π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)π‘ˆ01πœ‡πΆ02π‘’βˆ’πœ‡π‘‡π‘0+𝑅0+1+πœ…π‘’βˆ’2πœ‡π‘‡π‘0+𝑝𝑛=1||π‘Ÿπ‘›||π‘ˆ0π‘›βˆ’1𝑒(π‘›βˆ’1)πœ‡π‘‡0βˆ’1ξ€Έξ€·1+πœ…π‘’βˆ’2π‘›πœ‡π‘‡ξ€Έπ‘›ξƒͺ+πœ…π‘’βˆ’2πœ‡π‘‡ξ‚Ή,𝐡2≀2π‘ˆ0π‘’βˆ’πœ‡π‘‡πΆ0𝑍0+𝑅0ξ€Έπ‘’πœ‡(π‘‘π‘˜+1βˆ’π‘‘π‘˜)βˆ’1πœ‡+π‘ˆ0𝐢0𝑍0π‘’πœ‡(π‘‘π‘˜+1βˆ’π‘‘π‘˜)βˆ’1πœ‡+1𝐢0𝑝𝑛=1||π‘Ÿπ‘›||π‘ˆπ‘›0ξ€œπ‘‘π‘˜+1π‘‘π‘˜π‘’π‘›πœ‡(π‘ βˆ’π‘‡)+π‘‘π‘ πœ…π‘ˆ0π‘’βˆ’2πœ‡π‘‡π‘0𝐢0π‘’πœ‡(π‘‘π‘˜+1βˆ’π‘‘π‘˜)βˆ’1πœ‡+πœ…πΆ0𝑝𝑛=1||π‘Ÿπ‘›||π‘ˆπ‘›0π‘’βˆ’2π‘›πœ‡π‘‡ξ€œπ‘‘π‘˜+1π‘‘π‘˜π‘’π‘›πœ‡(π‘ βˆ’π‘‡)ξƒ­||𝑒𝑑𝑑𝑠+πœ…π‘˜+1ξ€Έξ€·π‘‘βˆ’2π‘‡βˆ’π‘’π‘˜ξ€Έ||β‰€ξƒ¬βˆ’2𝑇2π‘ˆ0π‘’βˆ’πœ‡π‘‡πΆ0𝑍0+𝑅0ξ€Έπ‘’πœ‡π‘‡0βˆ’1πœ‡+π‘ˆ0𝐢0𝑍0π‘’πœ‡π‘‡0βˆ’1πœ‡+1𝐢0𝑝𝑛=1||π‘Ÿπ‘›||π‘ˆπ‘›0π‘’π‘›πœ‡π‘‡0βˆ’1+π‘›πœ‡πœ…π‘ˆ0π‘’βˆ’2πœ‡π‘‡πΆ0𝑍0π‘’πœ‡π‘‡0βˆ’1πœ‡+πœ…πΆ0𝑝𝑛=1||π‘Ÿπ‘›||π‘ˆπ‘›0π‘’βˆ’2π‘›πœ‡π‘‡π‘’π‘›πœ‡π‘‡0βˆ’1ξƒ­π‘›πœ‡β‰€π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)π‘ˆ0πœ‡πΆ02π‘’βˆ’πœ‡π‘‡ξ€·π‘’πœ‡π‘‡0ξ€Έβˆ’1𝑍0+𝑅0+ξ€·π‘’πœ‡π‘‡0βˆ’1ξ€Έξ€·1+πœ…π‘’βˆ’2πœ‡π‘‡ξ€Έπ‘0+𝑝𝑛=1||π‘Ÿπ‘›||π‘ˆ0π‘›βˆ’1ξ€·1+πœ…π‘’βˆ’2π‘›πœ‡π‘‡π‘’ξ€Έξ€·π‘›πœ‡π‘‡0ξ€Έβˆ’1𝑛ξƒͺ.(2.7)
Therefore, for sufficiently large πœ‡>0, we obtain ||||(𝐡𝑒)(𝑑)β‰€π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)π‘ˆ01πœ‡πΆ02π‘’βˆ’πœ‡π‘‡π‘0+𝑅0+1+πœ…π‘’βˆ’2πœ‡π‘‡π‘0+𝑝𝑛=1||π‘Ÿπ‘›||π‘ˆ0π‘›βˆ’1𝑒(π‘›βˆ’1)πœ‡π‘‡0βˆ’1ξ€Έξ€·1+πœ…π‘’βˆ’2π‘›πœ‡π‘‡ξ€Έπ‘›ξƒͺ+πœ…π‘’βˆ’2πœ‡π‘‡ξ‚Ή+π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)π‘ˆ01πœ‡πΆ02π‘’βˆ’πœ‡π‘‡ξ€·π‘’πœ‡π‘‡0ξ€Έβˆ’1𝑍0+𝑅0+ξ€·π‘’πœ‡π‘‡0βˆ’1ξ€Έξ€·1+πœ…π‘’βˆ’2πœ‡π‘‡ξ€Έπ‘0+𝑝𝑛=1||π‘Ÿπ‘›||π‘ˆ0π‘›βˆ’1π‘’π‘›πœ‡π‘‡0βˆ’1𝑛1+πœ…π‘’βˆ’2π‘›πœ‡π‘‡ξ€Έξƒͺβ‰€π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)π‘ˆ01πœ‡πΆ02π‘’βˆ’πœ‡π‘‡π‘’πœ‡π‘‡0𝑍0+𝑅0+π‘’πœ‡π‘‡0ξ€·1+πœ…π‘’βˆ’2πœ‡π‘‡ξ€Έπ‘0+𝑝𝑛=1||π‘Ÿπ‘›||π‘ˆ0π‘›βˆ’1ξ€·π‘’π‘›πœ‡π‘‡0+𝑒(π‘›βˆ’1)πœ‡π‘‡0βˆ’2ξ€Έξ€·1+πœ…π‘’βˆ’2π‘›πœ‡π‘‡ξ€Έπ‘›ξƒͺ+πœ…π‘’βˆ’2πœ‡π‘‡ξƒ­β‰€π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)π‘ˆ0.(2.8)
Consequently, the operator 𝐡 maps π‘€π‘†π‘ˆ into itself.
We show that B is a contractive operator. Indeed, ||𝐡(𝑒)(𝑑)βˆ’π΅π‘’ξ€Έ||≀||||ξ€œ(𝑑)π‘‘π‘‘π‘˜ξ€Ίξ€·π‘ˆ(𝑒)(𝑠)βˆ’π‘ˆπ‘’ξ€Έξ€»||||+||||ξ€œ(𝑠)π‘‘π‘ π‘‘π‘˜+1π‘‘π‘˜ξ€Ίπ‘ˆξ€·(𝑒)(𝑠)βˆ’π‘ˆπ‘’ξ€Έξ€»||||(𝑠)𝑑𝑠≑𝐡1+𝐡2𝑑,π‘‘βˆˆπ‘˜,π‘‘π‘˜+1ξ€».(2.9)
We have 𝐡1≀1𝐢0𝑍0ξ€œπ‘‘π‘‘π‘˜||𝑒(𝑠)βˆ’||1𝑒(𝑠)𝑑𝑠+𝐢0𝑝𝑛=1||π‘Ÿπ‘›||ξ€œπ‘‘π‘‘π‘˜||𝑒𝑛(𝑠)βˆ’π‘’π‘›||+πœ…(𝑠)𝑑𝑠𝑍0𝐢0ξ€œπ‘‘π‘‘π‘˜||𝑒(π‘ βˆ’2𝑇)βˆ’||πœ…π‘’(π‘ βˆ’2𝑇)𝑑𝑠+𝐢0𝑝𝑛=1||π‘Ÿπ‘›||ξ€œπ‘‘π‘‘π‘˜||𝑒𝑛(π‘ βˆ’2𝑇)βˆ’π‘’π‘›||ξƒ­||||ξ€œ(π‘ βˆ’2𝑇)𝑑𝑠+πœ…π‘‘π‘‘π‘˜ξ‚€Μ‡Μ‡π‘’(π‘ βˆ’2𝑇)βˆ’ξ‚||||β‰€ξƒ¬πœŒπ‘’(π‘ βˆ’2𝑇)π‘‘π‘ π‘˜ξ€·π‘’,𝑒𝐢0𝑍0π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)βˆ’1πœ‡+1𝐢0𝑝𝑛=1𝑛||π‘Ÿπ‘›||ξ€½||𝑒esssupπ‘›βˆ’1||𝑑(𝑠)βˆΆπ‘ βˆˆπ‘˜,π‘‘π‘˜+1ξ€œξ€»ξ€Ύπ‘‘π‘‘π‘˜||𝑒(𝑠)βˆ’||+πœ…π‘’(𝑠)𝑑𝑠𝑍0𝐢0πœŒπ‘˜ξ€·π‘’,π‘’ξ€Έπ‘’βˆ’2πœ‡π‘‡π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)βˆ’1πœ‡+πœ…πΆ0𝑝𝑛=1𝑛||π‘Ÿπ‘›||𝑒esssupπ‘›βˆ’1𝑑(π‘ βˆ’2𝑇)βˆΆπ‘ βˆˆπ‘˜,π‘‘π‘˜+1ξ€œξ€»ξ€Ύπ‘‘π‘‘π‘˜||𝑒(π‘ βˆ’2𝑇)βˆ’||𝑒(π‘ βˆ’2𝑇)𝑑𝑠+πœ…πœŒπ‘˜ξ‚€Μ‡Μ‡π‘’,π‘’ξ‚π‘’βˆ’2πœ‡π‘‡π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)βˆ’1πœ‡β‰€π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)ξƒ¬πœŒπ‘˜ξ€·π‘’,π‘’ξ€Έπœ‡πΆ0𝑍0+πœŒπ‘˜ξ€·π‘’,π‘’ξ€Έπœ‡πΆ0𝑝𝑛=1𝑛||π‘Ÿπ‘›||π‘ˆ0π‘›βˆ’1𝑒(π‘›βˆ’1)πœ‡(π‘‘π‘˜+1βˆ’π‘‘π‘˜)+πœ…πœŒπ‘˜ξ€·π‘’,π‘’ξ€Έπ‘’βˆ’2πœ‡π‘‡πœ‡π‘0𝐢0+πœ…πœŒπ‘˜ξ€·π‘’,π‘’ξ€Έπ‘’βˆ’2πœ‡π‘‡πœ‡πΆ0𝑝𝑛=1𝑛||π‘Ÿπ‘›||π‘ˆ0π‘›βˆ’1π‘’βˆ’2(π‘›βˆ’1)πœ‡π‘‡π‘’(π‘›βˆ’1)πœ‡(π‘‘π‘˜+1βˆ’π‘‘π‘˜)ξƒ­+π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)πœ…πœŒπ‘˜ξ‚€Μ‡Μ‡π‘’,π‘’ξ‚π‘’βˆ’2πœ‡π‘‡πœ‡β‰€π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)πœŒπ‘˜ξ‚€Μ‡Μ‡π‘’,𝑒1πœ‡21+πœ…π‘’βˆ’2πœ‡π‘‡πΆ0𝑍0+1𝐢0𝑝𝑛=1𝑛||π‘Ÿπ‘›||π‘ˆ0π‘›βˆ’1ξ€·1+πœ…π‘’βˆ’2π‘›πœ‡π‘‡ξ€Έπ‘’(π‘›βˆ’1)πœ‡π‘‡0ξƒͺ+πœ…π‘’βˆ’2πœ‡π‘‡πœ‡ξƒ­β‰€π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)πœŒπœ‡(π‘˜)𝑒,𝑒1πœ‡21+πœ…π‘’βˆ’2πœ‡π‘‡πΆ0𝑍0+1𝐢0𝑝𝑛=1𝑛||π‘Ÿπ‘›||π‘ˆ0π‘›βˆ’1ξ€·1+πœ…π‘’βˆ’2π‘›πœ‡π‘‡ξ€Έπ‘’(π‘›βˆ’1)πœ‡π‘‡0ξƒͺ+πœ…π‘’βˆ’2πœ‡π‘‡πœ‡ξƒ­,𝐡2≀1𝐢0𝑍0ξ€œπ‘‘π‘˜+1π‘‘π‘˜||𝑒(𝑠)βˆ’||1𝑒(𝑠)𝑑𝑠+𝐢0𝑝𝑛=1||π‘Ÿπ‘›||ξ€œπ‘‘π‘˜+1π‘‘π‘˜||𝑒𝑛(𝑠)βˆ’π‘’π‘›||+πœ…(𝑠)𝑑𝑠𝑍0𝐢0ξ€œπ‘‘π‘˜+1π‘‘π‘˜||𝑒(π‘ βˆ’2𝑇)βˆ’||πœ…π‘’(π‘ βˆ’2𝑇)𝑑𝑠+𝐢0𝑝𝑛=1||π‘Ÿπ‘›||ξ€œπ‘‘π‘˜+1π‘‘π‘˜||𝑒𝑛(π‘ βˆ’2𝑇)βˆ’π‘’π‘›||ξƒ­||||ξ€œ(π‘ βˆ’2𝑇)𝑑𝑠+πœ…π‘‘π‘˜+1π‘‘π‘˜ξ€·Μ‡ξ€Έ||||β‰€ξƒ¬πœŒΜ‡π‘’(π‘ βˆ’2𝑇)βˆ’Μ‡π‘’(π‘ βˆ’2𝑇)π‘‘π‘ π‘˜ξ€·π‘’,𝑒𝐢0𝑍0π‘’πœ‡(π‘‘π‘˜+1βˆ’π‘‘π‘˜)βˆ’1πœ‡+1𝐢0𝑝𝑛=1||π‘Ÿπ‘›||ξ€½||𝑒𝑛.esssupπ‘›βˆ’1||𝑑(𝑠)βˆΆπ‘ βˆˆπ‘˜,π‘‘π‘˜+1ξ€œξ€»ξ€Ύπ‘‘π‘˜+1π‘‘π‘˜||𝑒(𝑠)βˆ’||+πœ…π‘’(𝑠)𝑑𝑠𝑍0𝐢0πœŒπ‘˜ξ€·π‘’,π‘’ξ€Έπ‘’βˆ’2πœ‡π‘‡π‘’πœ‡(π‘‘π‘˜+1βˆ’π‘‘π‘˜)βˆ’1πœ‡+πœ…πΆ0𝑝𝑛=1||π‘Ÿπ‘›||𝑒𝑛.esssupπ‘›βˆ’1𝑑(π‘ βˆ’2𝑇)βˆΆπ‘ βˆˆπ‘˜,π‘‘π‘˜+1ξ€œξ€»ξ€Ύπ‘‘π‘˜+1π‘‘π‘˜||𝑒(π‘ βˆ’2𝑇)βˆ’||ξƒ­β‰€ξƒ¬πœŒπ‘’(π‘ βˆ’2𝑇)π‘‘π‘ π‘˜ξ€·π‘’,𝑒𝐢0𝑍0π‘’πœ‡(π‘‘π‘˜+1βˆ’π‘‘π‘˜)βˆ’1πœ‡+πœŒπ‘˜ξ€·π‘’,𝑒𝐢0π‘’πœ‡(π‘‘π‘˜+1βˆ’π‘‘π‘˜)βˆ’1πœ‡π‘ξ“π‘›=1||π‘Ÿπ‘›||π‘›π‘ˆ0π‘›βˆ’1𝑒(π‘›βˆ’1)πœ‡(π‘‘π‘˜+1βˆ’π‘‘π‘˜)+πœ…πœŒπ‘˜ξ€·π‘’,π‘’ξ€Έπ‘’βˆ’2πœ‡π‘‡π‘0𝐢0π‘’πœ‡(π‘‘π‘˜+1βˆ’π‘‘π‘˜)βˆ’1πœ‡+πœ…πœŒπ‘˜ξ€·π‘’,π‘’ξ€Έπ‘’βˆ’2πœ‡π‘‡πΆ0π‘’πœ‡(π‘‘π‘˜+1βˆ’π‘‘π‘˜)βˆ’1πœ‡π‘ξ“π‘›=1𝑛||π‘Ÿπ‘›||π‘ˆ0π‘›βˆ’1𝑒(π‘›βˆ’1)(πœ‡π‘‡0βˆ’2πœ‡π‘‡)ξƒ­β‰€πœŒπ‘˜ξ‚€Μ‡Μ‡π‘’,π‘’ξ‚π‘’πœ‡π‘‡0βˆ’1πœ‡21+πœ…π‘’βˆ’2πœ‡π‘‡πΆ0𝑍0+1𝐢0𝑝𝑛=1||π‘Ÿπ‘›||π‘›π‘ˆ0π‘›βˆ’1𝑒(π‘›βˆ’1)πœ‡π‘‡0ξ€·1+πœ…π‘’βˆ’2π‘›πœ‡π‘‡ξ€Έξƒͺβ‰€πœŒπ‘’(π‘˜)𝑒,π‘’ξ€Έπ‘’πœ‡π‘‡0βˆ’1πœ‡2𝐢01+πœ…π‘’βˆ’2πœ‡π‘‡π‘0+𝑝𝑛=1||π‘Ÿπ‘›||π‘›π‘ˆ0π‘›βˆ’1𝑒(π‘›βˆ’1)πœ‡π‘‡0ξ€·1+πœ…π‘’βˆ’2π‘›πœ‡π‘‡ξ€Έξƒͺ.(2.10)
Consequently, ||𝐡(𝑒)(𝑑)βˆ’π΅π‘’ξ€Έ||(𝑑)β‰€π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)πœŒπœ‡(π‘˜)𝑒,𝑒1πœ‡21+πœ…π‘’βˆ’2πœ‡π‘‡πΆ0𝑍0+1𝐢0𝑝𝑛=1𝑛||π‘Ÿπ‘›||π‘ˆ0π‘›βˆ’1ξ€·1+πœ…π‘’βˆ’2π‘›πœ‡π‘‡ξ€Έπ‘’(π‘›βˆ’1)πœ‡π‘‡0ξƒͺ+πœ…π‘’βˆ’2πœ‡π‘‡πœ‡ξƒ­+πœŒπ‘’(π‘˜)𝑒,π‘’ξ€Έπ‘’πœ‡π‘‡0βˆ’1πœ‡21+πœ…π‘’βˆ’2πœ‡π‘‡πΆ0𝑍0+1𝐢0𝑝𝑛=1||π‘Ÿπ‘›||π‘›π‘ˆ0π‘›βˆ’1𝑒(π‘›βˆ’1)πœ‡π‘‡0ξ€·1+πœ…π‘’βˆ’2π‘›πœ‡π‘‡ξ€Έξƒͺβ‰€πœŒπ‘’(π‘˜)𝑒,π‘’ξ€Έξƒ¬π‘’πœ‡π‘‡0πœ‡21+πœ…π‘’βˆ’2πœ‡π‘‡πΆ0𝑍0+1𝐢0𝑝𝑛=1||π‘Ÿπ‘›||π‘›π‘ˆ0π‘›βˆ’1𝑒(π‘›βˆ’1)πœ‡π‘‡0ξ€·1+πœ…π‘’βˆ’2π‘›πœ‡π‘‡ξ€Έξƒͺ+πœ…π‘’βˆ’2πœ‡π‘‡πœ‡ξƒ­.β‰‘π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)πΎπ‘ˆπœŒπœ‡(π‘˜)𝑒,𝑒.(2.11)
Therefore, πœŒπ‘˜(𝐡𝑒,𝐡𝑒)β‰€πΎπ‘ˆπœŒπœ‡(π‘˜)(𝑒,𝑒).
It remains to estimate the derivative of B.
We have||̇𝐡̇𝐡(𝑒)(𝑑)βˆ’π‘’ξ€Έ||≀||π‘ˆξ€·(𝑑)(𝑒)(𝑠)βˆ’π‘ˆπ‘’ξ€Έ||+1(𝑠)π‘‘π‘˜+1βˆ’π‘‘π‘˜||||ξ€œπ‘‘π‘˜+1π‘‘π‘˜ξ€Ίξ€·π‘ˆ(𝑒)(𝑠)βˆ’π‘ˆπ‘’ξ€Έ(ξ€»||||≑̇𝐡𝑠)𝑑𝑠1+̇𝐡2.(2.12)
We have ̇𝐡1≀1𝐢0𝑍0||𝑒(𝑑)βˆ’||+1𝑒(𝑑)𝐢0𝑝𝑛=1||π‘Ÿπ‘›||||𝑒𝑛(𝑑)βˆ’π‘’π‘›||+πœ…(𝑑)𝐢0𝑍0||𝑒(π‘‘βˆ’2𝑇)βˆ’||+πœ…π‘’(π‘‘βˆ’2𝑇)𝐢0𝑝𝑛=1||π‘Ÿπ‘›||||𝑒𝑛(π‘‘βˆ’2𝑇)βˆ’π‘’π‘›||||Μ‡(π‘‘βˆ’2𝑇)+πœ…Μ‡π‘’(π‘‘βˆ’2𝑇)βˆ’π‘’||≀𝑒(π‘‘βˆ’2𝑇)πœ‡(π‘‘βˆ’π‘‘π‘˜)πœŒπ‘˜ξ€·π‘’,𝑒𝐢0𝑍0+1𝐢0𝑝𝑛=1||π‘Ÿπ‘›||𝑒𝑛esssupπ‘›βˆ’1𝑑(𝑑)βˆΆπ‘‘βˆˆπ‘˜,π‘‘π‘˜+1||𝑒(𝑑)βˆ’π‘’||+(𝑑)πœ…π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)πœŒπ‘˜ξ€·π‘’,π‘’ξ€Έπ‘’βˆ’2πœ‡π‘‡πΆ0𝑍0+πœ…πΆ0𝑝𝑛=1𝑛||π‘Ÿπ‘›||𝑒esssupπ‘›βˆ’1𝑑(π‘‘βˆ’2𝑇)βˆΆπ‘‘βˆˆπ‘˜,π‘‘π‘˜+1||𝑒(π‘‘βˆ’2𝑇)βˆ’||||̇𝑒(π‘‘βˆ’2𝑇)+πœ…Μ‡π‘’(π‘‘βˆ’2𝑇)βˆ’||≀𝑒𝑒(π‘‘βˆ’2𝑇)πœ‡(π‘‘βˆ’π‘‘π‘˜)πœŒπ‘˜ξ‚€Μ‡Μ‡π‘’,π‘’ξ‚πœ‡πΆ0𝑍0+π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)πœŒπ‘˜ξ‚€Μ‡Μ‡π‘’,π‘’ξ‚πœ‡πΆ0𝑝𝑛=1||π‘Ÿπ‘›||π‘›π‘ˆ0π‘›βˆ’1𝑒(π‘›βˆ’1)πœ‡(π‘‘π‘˜+1βˆ’π‘‘π‘˜)+π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)πœ…πœŒπ‘˜ξ‚€Μ‡Μ‡π‘’,π‘’ξ‚π‘’βˆ’2πœ‡π‘‡πœ‡πΆ0𝑍0+π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)πœŒπ‘˜ξ‚€Μ‡Μ‡π‘’,π‘’ξ‚πœ…π‘’βˆ’2πœ‡π‘‡πœ‡πΆ0𝑝𝑛=1𝑛||π‘Ÿπ‘›||π‘ˆ0π‘›βˆ’1𝑒(π‘›βˆ’1)πœ‡(π‘‘π‘˜+1βˆ’π‘‘π‘˜)+π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)πœ…πœŒπ‘˜ξ‚€Μ‡Μ‡π‘’,π‘’ξ‚π‘’βˆ’2πœ‡π‘‡β‰€π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)πœŒπœ‡(π‘˜)𝑒,𝑒1+πœ…π‘’βˆ’2πœ‡π‘‡πœ‡πΆ01𝑍0+𝑝𝑛=1||π‘Ÿπ‘›||π‘›π‘ˆ0π‘›βˆ’1𝑒(π‘›βˆ’1)πœ‡π‘‡0ξƒͺ+πœ…π‘’βˆ’2πœ‡π‘‡ξƒ­,̇𝐡2≀1π‘‘π‘˜+1βˆ’π‘‘π‘˜||||ξ€œπ‘‘π‘˜+1π‘‘π‘˜ξ€·ξ€·π‘ˆ(𝑒)(𝑠)βˆ’π‘ˆπ‘’ξ€Έξ€Έ||||≀1(𝑠)𝑑𝑠𝑙0||||ξ€œπ‘‘π‘˜+1π‘‘π‘˜ξ€·ξ€·π‘ˆ(𝑒)(𝑠)βˆ’π‘ˆπ‘’ξ€Έξ€Έ||||(𝑠)π‘‘π‘ β‰€πœŒπ‘’(π‘˜)𝑒,π‘’ξ€Έπ‘’πœ‡π‘‡0βˆ’1πœ‡2𝐢0𝑙01+πœ…π‘’βˆ’2πœ‡π‘‡π‘0+𝑝𝑛=1||π‘Ÿπ‘›||π‘›π‘ˆ0π‘›βˆ’1𝑒(π‘›βˆ’1)πœ‡π‘‡0ξ€·1+πœ…π‘’βˆ’2π‘›πœ‡π‘‡ξ€Έξƒͺ.(2.13)
Therefore, ||̇𝐡̇𝐡(𝑒)(𝑑)βˆ’π‘’ξ€Έ||(𝑑)β‰€π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)πœŒπœ‡(π‘˜)𝑒,𝑒1+πœ…π‘’βˆ’2πœ‡π‘‡πœ‡πΆ01𝑍0+𝑝𝑛=1||π‘Ÿπ‘›||π‘›π‘ˆ0π‘›βˆ’1𝑒(π‘›βˆ’1)πœ‡π‘‡0ξƒͺ+πœ…π‘’βˆ’2πœ‡π‘‡ξƒ­+πœŒπ‘’(π‘˜)𝑒,π‘’ξ€Έπ‘’πœ‡π‘‡0βˆ’1πœ‡2𝐢0𝑙01+πœ…π‘’βˆ’2πœ‡π‘‡π‘0+𝑝𝑛=1||π‘Ÿπ‘›||π‘›π‘ˆ0π‘›βˆ’1𝑒(π‘›βˆ’1)πœ‡π‘‡0ξ€·1+πœ…π‘’βˆ’2π‘›πœ‡π‘‡ξ€Έξƒͺβ‰€πœŒπœ‡(π‘˜)𝑒,π‘’ξ€Έξƒ¬ξ€·π‘’πœ‡π‘‡0+πœ‡πœ0βˆ’1ξ€Έξ€·1+πœ…π‘’βˆ’2πœ‡π‘‡ξ€Έπœ‡2𝐢0𝑙01𝑍0+𝑝𝑛=1||π‘Ÿπ‘›||π‘›π‘ˆ0π‘›βˆ’1𝑒(π‘›βˆ’1)πœ‡π‘‡0ξƒͺ+πœ…π‘’βˆ’2πœ‡π‘‡ξƒ­β‰‘π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)Μ‡πΎπ‘ˆπœŒπœ‡(π‘˜)𝑒,𝑒.(2.14)
It follows πœŒπ‘˜(̇̇𝐡(𝑒),𝐡(𝑒))β‰€π‘’πœ‡(π‘‘βˆ’π‘‘π‘˜)Μ‡πΎπ‘ˆπœŒπœ‡(π‘˜)(𝑒,𝑒).
Then πœŒπœ‡(π‘˜)(𝐡(𝑒),𝐡(𝑒))≀max{πΎπ‘ˆ,Μ‡πΎπ‘ˆ}πœŒπœ‡(π‘˜)(𝑒,𝑒).
Consequently, πœŒπœ‡(π‘˜)𝐡𝑒,π΅π‘’ξ€Έβ‰€πΎπœŒπœ‡(π‘˜)𝑒,𝑒(π‘˜=0,1,2,…),(2.15) where 𝐾=max{πΎπ‘ˆ,Μ‡πΎπ‘ˆ}<1 does not depend on 𝑒 and π‘˜.
We have to verify that π‘€π‘†π‘ˆ is j-bounded. Indeed, since j is an identity mapping,πœŒπ‘—π‘›π‘’(π‘˜)𝑒,π‘’ξ€Έβ‰€πœŒπ‘’(π‘˜)𝑒,𝑒<∞(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:1πœ‡πΆ02π‘’βˆ’πœ‡π‘‡π‘’πœ‡π‘‡0𝑍0+𝑅0+π‘’πœ‡π‘‡0ξ€·1+πœ…π‘’βˆ’2πœ‡π‘‡ξ€Έπ‘0+𝑝𝑛=1||π‘Ÿπ‘›||π‘ˆ0π‘›βˆ’1ξ€·π‘’π‘›πœ‡π‘‡0+𝑒(π‘›βˆ’1)πœ‡π‘‡0βˆ’2ξ€Έξ€·1+πœ…π‘’βˆ’2π‘›πœ‡π‘‡ξ€Έπ‘›ξƒͺ+πœ…π‘’βˆ’2πœ‡π‘‡πΎβ‰€1,π‘ˆ=π‘’πœ‡0πœ‡21+πœ…π‘’βˆ’2πœ‡π‘‡πΆ0𝑍0+1𝐢0𝑝𝑛=1||π‘Ÿπ‘›||π‘›π‘ˆ0π‘›βˆ’1𝑒(π‘›βˆ’1)πœ‡0ξ€·1+πœ…π‘’βˆ’2π‘›πœ‡π‘‡ξ€Έξƒͺ+πœ…π‘’βˆ’2πœ‡π‘‡πœ‡Μ‡πΎ<1,π‘ˆ=ξ€·π‘’πœ‡0+πœ‡πœ0βˆ’1ξ€Έξ€·1+πœ…π‘’βˆ’2πœ‡π‘‡ξ€Έπœ‡2𝐢0𝑙01𝑍0+𝑝𝑛=1||π‘Ÿπ‘›||π‘›π‘ˆ0π‘›βˆ’1𝑒(π‘›βˆ’1)πœ‡0ξƒͺ+πœ…π‘’βˆ’2πœ‡π‘‡<1.(3.1)

Consider a line with the following specific parameters:1Ξ›=1m,𝐿=0,2πœ‡H/m,𝐢=80pF/m,𝜈=√=1𝐿𝐢√0,2β‹…10βˆ’6β‹…80β‹…10βˆ’12=14β‹…10βˆ’9=2,5β‹…108,𝑍0=𝐿𝐢=ξƒŽ0,2β‹…10βˆ’680β‹…10βˆ’12=50Ξ©,𝑅0=45Ξ©,𝐢0=8pF=8β‹…10βˆ’12F.(3.2)

Then, βˆšπ‘‡=Λ𝐿𝐢=4.10βˆ’9s;πœ…=(𝑍0βˆ’π‘…0)/(𝑍0+𝑅0)=1/19=0,0526.

Let us check the propagation of millimeter waves πœ†0=10βˆ’3m. We have𝑓0=1πœ†0√=1𝐿𝐢10βˆ’3β‹…4β‹…10βˆ’9=2,5β‹…1011HzβŸΉπ‘‡0=1𝑓0=12,5β‹…1011=4β‹…10βˆ’12sec.;𝑙=2β‹…10βˆ’12sec.(3.3) If we choose πœ‡=(1/4)1012, then πœ‡π‘‡0=πœ‡0=1,πœ‡πœ0=(1/2), and 𝑇=4β‹…10βˆ’9β‹…(1/4)β‹…1012𝑇0=1000⋅𝑇0.

Consequently, πœ‡π‘‡=(1/4)1012β‹…2β‹…10βˆ’8=(1/2)104,πœ‡πΆ0=(1/4)1012β‹…8β‹…10410βˆ’12=2, and πœ‡2𝐢0=(1/2)β‹…1012.

Since π‘’βˆ’πœ‡π‘‡=π‘’βˆ’5000=0, then the above inequalities (omitting the second one) become𝑒+100𝑝𝑛=1||π‘Ÿπ‘›||π‘ˆ0π‘›βˆ’1𝑒𝑛+π‘’π‘›βˆ’1βˆ’2̇𝐾2𝑛≀1,π‘ˆξ‚€1=2π‘’βˆ’21+50𝑝𝑛=1||π‘Ÿπ‘›||π‘›π‘ˆ0π‘›βˆ’1π‘’π‘›βˆ’1ξƒͺ<1.(3.4)

If the V-I characteristic of the nonlinear resistive element is 𝑓(𝑒)=βˆ’0,12𝑒+0,8𝑒3, then π‘ˆ0≀0,41; Μ‡πΎπ‘ˆ=π‘ˆ0<0,06. It follows that π‘ˆ0<0,06.

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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.


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