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International Journal of Differential Equations
Volume 2011 (2011), Article ID 949547, 12 pages
doi:10.1155/2011/949547
Research Article

Oscillatory Solutions of Neutral Equations with Polynomial Nonlinearities

Vasil G. Angelov1 and Dafinka Tz. Angelova2

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.

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).

949547.fig.001
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 ) = 𝑅 0 𝐶 𝑖 ( 0 , 𝑡 ) , 𝑡 ≥ 0 , ( 1 . 1 ) 0 𝑑 𝑢 ( Λ , 𝑡 ) 𝑑 𝑡 = 𝑖 ( Λ , 𝑡 ) − 𝑓 ( 𝑢 ( Λ , 𝑡 ) ) , 𝑡 ≥ 0 , ( 1 . 2 ) 𝑢 ( 𝑥 , 0 ) = 𝑢 0 ( 𝑥 ) , 𝑖 ( 𝑥 , 0 ) = 𝑖 0 [ ] , ( 𝑥 ) , 𝑥 ∈ 0 , Λ ( 1 . 3 ) where 𝑢 0 ( 𝑥 ) and 𝑖 0 ( 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. [3–10]). Here, the V-I characteristic of the nonlinear resistive load is ∑ 𝑖 = 𝑓 ( 𝑢 ) = 𝑝 𝑛 = 1 𝑟 𝑛 𝑢 𝑛 , where 𝑟 𝑛 are real numbers, 𝐶 0 is parallel connected capacitance, 𝐸 is the source voltage, 𝑅 0 is the source resistance, and 𝑍 0 = √ 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 equation 𝑑 𝑢 ( 𝑡 ) = 𝑑 𝑡 2 𝐸 𝐶 0 𝑍 0 + 𝑅 0 − 𝑢 ( 𝑡 ) 𝐶 0 𝑍 0 − 1 𝐶 0 𝑝 𝑛 = 1 𝑟 𝑛 [ ] 𝑢 ( 𝑡 ) 𝑛 − 𝑍 0 − 𝑅 0 𝑢 ( 𝑡 − 2 𝑇 ) 𝑍 0 𝐶 0 𝑍 0 + 𝑅 0 + 𝑍 0 − 𝑅 0 𝐶 0 𝑍 0 + 𝑅 0 𝑝 𝑛 = 1 𝑟 𝑛 [ ] 𝑢 ( 𝑡 − 2 𝑇 ) 𝑛 + 𝑍 0 − 𝑅 0 𝑍 0 + 𝑅 0 𝑑 𝑢 ( 𝑡 − 2 𝑇 ) 𝑑 𝑡 , 𝑡 ≥ 𝑇 , 𝑢 ( 𝑡 ) = 𝜐 0 ( 𝑡 ) , 𝑑 𝑢 ( 𝑡 ) = 𝑑 𝑡 𝑑 𝜐 0 ( 𝑡 ) [ ] , 𝑑 𝑡 , 𝑡 ∈ − 𝑇 , 𝑇 ( 1 . 4 ) where ( 𝑥 , 𝑡 ) ∈ Π = { ( 𝑥 , 𝑡 ) ∈ 𝑅 2 ∶ ( 𝑥 , 𝑡 ) ∈ [ 0 , Λ ] × [ 0 , ∞ ) } , 𝜅 = | 𝑍 0 − 𝑅 0 | / ( 𝑍 0 + 𝑅 0 ) < 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 𝜐 0 ( 𝑡 ) can be obtained shifting the initial function 𝑢 0 ( 𝑥 ) from (1.3) along the characteristics √ 𝑥 − 𝜈 𝑡 = c o n s t . , ( 𝜈 = 1 / 𝐿 𝐶 ) on [ 0 , 𝑇 ] and along the characteristics 𝑥 + 𝜈 𝑡 = c o n s t . on [ − 𝑇 , 0 ] (cf. [1, 2]). So, we obtain an initial function 𝜐 0 ( 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 [ 𝑡 0 , ∞ ) , 𝑇 = 𝑡 0 .

Let 𝑆 𝑇 = { 𝜏 𝑘 } 𝑛 𝑘 = 0 , 𝑛 ∈ 𝑁 be the set of zeros of the initial function; that is, 𝜐 0 ( 𝜏 𝑘 ) = 0 such that 𝜏 0 = − 𝑇 , 𝜏 𝑛 = 𝑇 ≡ 𝑡 0 .

Let 𝑆 = { 𝑡 𝑘 } ∞ 𝑘 = 0 be a strictly increasing sequence of real numbers satisfying the following conditions (C):(C1) l i m 𝑘 → ∞ 𝑡 𝑘 = ∞ ,(C2) 0 < 𝑙 0 = i n f { 𝑡 𝑘 + 1 − 𝑡 𝑘 ∶ 𝑘 = 0 , 1 , 2 , … } ≤ s u p { 𝑡 𝑘 + 1 − 𝑡 𝑘 ∶ 𝑘 = 0 , 1 , 2 , … } = 𝑇 0 < ∞ ,(C3)for every 𝑘 there is 𝑠 < 𝑘 such that 𝑡 𝑘 − 𝑇 = 𝑡 𝑠 where 𝑡 𝑠 ∈ 𝑆 𝑇 ∪ 𝑆 .

Introduce the sets: 𝐶 1 [ 𝑡 0 , ∞ ) consisting of all continuous and bounded functions differentiable with bounded derivatives on every interval ( 𝑡 𝑘 , 𝑡 𝑘 + 1 ) (the derivatives at 𝑡 𝑘 do not necessary exist), 𝑀 𝑆 = { 𝑢 ( ⋅ ) ∈ 𝐶 1 [ 𝑡 0 , ∞ ) ∶ 𝑢 ( 𝑡 𝑘 ) = 0 ( 𝑘 = 0 , 1 , 2 , … ) } , 𝑀 𝑆 𝑈 = { 𝑢 ( ⋅ ) ∈ 𝑀 𝑆 ∶ | 𝑢 ( 𝑡 ) | ≤ 𝑈 0 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) , 𝑡 ∈ [ 𝑡 𝑘 , 𝑡 𝑘 + 1 ] } , where 𝑈 0 , 𝜇 are positive constants prescribed below.

We assume that | 𝜐 0 ( 𝑡 ) | ≤ 𝑈 0 𝑒 𝜇 ( 𝑡 − 𝜏 𝑘 ) , 𝑡 ∈ [ 𝜏 𝑘 , 𝜏 𝑘 + 1 ] , ( 𝑘 = 0 , 1 , 2 , … , 𝑛 − 1 ).

The set 𝑀 𝑆 𝑈 turns out into a complete uniform space with respect to the family of pseudometrics 𝜌 𝜇 ( 𝑘 ) ( 𝑓 , 𝑔 ) = m a x { 𝜌 𝑘 ( 𝑓 , 𝑔 ) , 𝜌 𝑘 ( ̇ 𝑓 , ̇ 𝑔 ) } , ( 𝑘 = 0 , 1 , 2 , … ), where 𝜌 𝑘 ( 𝑓 , 𝑔 ) = m a x { 𝑒 − 𝜇 ( 𝑡 − 𝑡 𝑘 ) | 𝑓 ( 𝑡 ) − 𝑔 ( 𝑡 ) | ∶ 𝑡 ∈ [ 𝑡 𝑘 , 𝑡 𝑘 + 1 ] } , 𝜌 𝑘 ( ̇ 𝑓 , ̇ 𝑔 ) = m a x { 𝑒 − 𝜇 ( 𝑡 − 𝑡 𝑘 ) | ̇ 𝑓 ( 𝑡 ) − ̇ 𝑔 ( 𝑡 ) | ∶ 𝑡 ∈ [ 𝑡 𝑘 , 𝑡 𝑘 + 1 ] } .

One can verify that 𝑀 𝑆 𝑈 is closed subset of 𝐶 1 [ 𝑡 0 , ∞ ) with respect to the above metric.

Remark 1.1. The functions from 𝑀 𝑆 are not necessary differentiable at 𝑡 𝑘 ( 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 𝐵 ∶ 𝑀 𝑆 𝑈 → 𝑀 𝑆 𝑈 by 𝐵 ( 𝑢 ) ( 𝑡 ) ∶ = 𝑡 𝑡 𝑘 𝑈 ( 𝑢 ) ( 𝑠 ) 𝑑 𝑠 − 𝑡 − 𝑡 𝑘 𝑡 𝑘 + 1 − 𝑡 𝑘 𝑡 𝑘 + 1 𝑡 𝑘 𝑡 𝑈 ( 𝑢 ) ( 𝑠 ) 𝑑 𝑠 , 𝑡 ∈ 𝑘 , 𝑡 𝑘 + 1 , ( 𝑘 = 0 , 1 , 2 , … ) , ( 1 . 5 ) where 𝑈 ( 𝑢 ) ( 𝑡 ) = 2 𝐸 𝐶 0 𝑍 0 + 𝑅 0 − 𝑢 ( 𝑡 ) 𝐶 0 𝑍 0 − 1 𝐶 0 𝑝 𝑛 = 1 𝑟 𝑛 [ ] 𝑢 ( 𝑡 ) 𝑛 − 𝜅 𝐾 𝑇 𝑢 ( 𝑡 ) 𝑍 0 𝐶 0 + 𝜅 𝐶 0 𝑝 𝑛 = 1 𝑟 𝑛 𝐾 𝑇 𝑢 ( 𝑡 ) 𝑛 𝑑 𝐾 + 𝜅 𝑇 𝑢 ( 𝑡 ) 𝑑 𝑡 , 𝑡 ≥ 𝑇 , ( 1 . 6 ) and ( 𝐾 𝑇 𝑢 ) ( 𝑡 ) = 𝑢 ( 𝑡 − 2 𝑇 ) is M. A. Krasnoselskii operator (cf. [14]).

Remark 1.2. The operator 𝐾 𝑇 is well defined, because the initial function is defined on the interval [ − 𝑇 , 𝑇 ] . We notice that 𝐾 𝑇 maps 𝑀 𝑆 into itself. Indeed, consider the set 𝐶 1 [ − 𝑇 , ∞ ) consisting of all continuous and bounded functions differentiable with bounded derivatives on every interval ( 𝑡 𝑘 , 𝑡 𝑘 + 1 ) . Introduce the set 𝑀 𝜐 0 𝑆 = { 𝑢 ( ⋅ ) ∈ 𝐶 1 [ − 𝑇 , ∞ ) ∶ 𝑢 ( 𝑡 ) = 𝜐 0 ( 𝑡 ) , 𝑡 ∈ [ − 𝑇 , 𝑇 ] } . Then, 𝐾 𝑇 assigns to every function 𝑢 ( ⋅ ) ∈ 𝑀 𝑆 the function ̃ 𝑢 ( ⋅ ) ∈ 𝑀 𝜐 0 𝑆 translated to the right on the interval [ 𝑇 , ∞ ) . So, the function ( 𝐾 𝑇 𝑢 ) ( 𝑡 ) coincides with 𝜐 0 ( 𝑡 ) on [ 𝑡 0 , 𝑡 0 + 2 𝑇 ] . Besides 𝑡 𝑘 − 2 𝑇 = 𝑡 𝑠 , and then 𝐾 𝑇 𝑢 𝑡 𝑘 = 𝑢 𝑡 𝑘 − 2 𝑇 = 𝜐 0 𝑡 𝑠 = 0 , 𝑡 𝑘 ∈ [ ] , 𝑢 𝑡 𝑇 , 3 𝑇 𝑘 𝑡 − 2 𝑇 = 𝑢 𝑛 = 0 , 𝑡 ∈ ( 3 𝑇 , ∞ ) , ( 1 . 7 ) that is, ( 𝐾 𝑇 𝑢 ) ( ⋅ ) ∈ 𝑀 𝑆 .

2. Main Results

Lemma 2.1. If 𝐸 ≤ 𝑈 0 , problem (1.4) has a solution 𝑢 ( ⋅ ) ∈ 𝑀 𝑆 𝑈 iff the operator 𝐵 has a fixed point in 𝑀 𝑆 𝑈 , that is, 𝑢 ( 𝑡 ) = 𝐵 ( 𝑢 ) ( 𝑡 ) . ( 2 . 1 )

Proof. Let 𝑢 ( ⋅ ) ∈ 𝑀 𝑆 𝑈 be a solution of (1.4). Then, integrating (1.4) on the interval [ 𝑡 𝑘 , 𝑡 ] ⊂ [ 𝑡 𝑘 , 𝑡 𝑘 + 1 ] ( 𝑘 = 0 , 1 , 2 … ), we obtain 𝑢 ( 𝑡 ) − 𝑢 ( 𝑡 𝑘 ∫ ) = 𝑡 𝑡 𝑘 ∫ 𝑈 ( 𝑢 ) ( 𝑠 ) 𝑑 𝑠 ⇔ 𝑢 ( 𝑡 ) = 𝑡 𝑡 𝑘 𝑈 ( 𝑢 ) ( 𝑠 ) 𝑑 𝑠 , and then, 𝑢 ( 𝑡 ) = 𝑡 𝑡 𝑘 𝑡 𝑈 ( 𝑢 ) ( 𝑠 ) 𝑑 𝑠 ⟹ 0 = 𝑢 𝑘 + 1 = 𝑡 𝑘 + 1 𝑡 𝑘 𝑈 ( 𝑢 ) ( 𝑠 ) 𝑑 𝑠 ⟹ 𝑡 𝑘 + 1 𝑡 𝑘 𝑈 ( 𝑢 ) ( 𝑠 ) 𝑑 𝑠 = 0 . ( 2 . 2 ) Therefore, 𝑢 ( 𝑡 ) satisfies 𝑢 ( 𝑡 ) = 𝑡 𝑡 𝑘 𝑈 ( 𝑢 ) ( 𝑠 ) 𝑑 𝑠 ⟺ 𝑢 ( 𝑡 ) = 𝑡 𝑡 𝑘 𝑈 ( 𝑢 ) ( 𝑠 ) 𝑑 𝑠 − 𝑡 − 𝑡 𝑘 𝑡 𝑘 + 1 − 𝑡 𝑘 𝑡 𝑘 + 1 𝑡 𝑘 𝑈 ( 𝑢 ) ( 𝑠 ) 𝑑 𝑠 , ( 2 . 3 ) that is, 𝑢 ( ⋅ ) is a fixed point of B.
Conversely, let 𝑢 ( ⋅ ) ∈ 𝑀 𝑆 𝑈 be a solution of 𝑢 = 𝐵 ( 𝑢 ) ; that is, 𝑢 ( 𝑡 ) = 𝑡 𝑡 𝑘 𝑈 ( 𝑢 ) ( 𝑠 ) 𝑑 𝑠 − 𝑡 − 𝑡 𝑘 𝑡 𝑘 + 1 − 𝑡 𝑘 𝑡 𝑘 + 1 𝑡 𝑘 𝑈 ( 𝑢 ) ( 𝑠 ) 𝑑 𝑠 . ( 2 . 4 )
Then, introducing 𝜇 0 = 𝜇 𝑇 0 , we obtain | | | | 𝑡 𝑘 + 1 𝑡 𝑘 𝑈 | | | | ≤ ( 𝑢 ) ( 𝑠 ) 𝑑 𝑠 2 𝐸 𝐶 0 𝑍 0 + 𝑅 0 𝑡 𝑘 + 1 𝑡 𝑘 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 1 𝑑 𝑡 + 𝐶 0 𝑍 0 𝑡 𝑘 + 1 𝑡 𝑘 | | 𝑢 | | + 1 ( 𝑡 ) 𝑑 𝑡 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑡 𝑘 + 1 𝑡 𝑘 | | | | 𝑢 ( 𝑡 ) 𝑛 𝜅 𝑑 𝑡 + 𝑍 0 𝐶 0 𝑡 𝑘 + 1 𝑡 𝑘 | | | | + 𝜅 𝑢 ( 𝑡 − 2 𝑇 ) 𝑑 𝑡 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑡 𝑘 + 1 𝑡 𝑘 | | | | 𝑢 ( 𝑡 − 2 𝑇 ) 𝑛 | | | | 𝑑 𝑡 + 𝜅 𝑡 𝑘 + 1 𝑡 𝑘 | | | | ≤ ̇ 𝑢 ( 𝑡 − 2 𝑇 ) 𝑑 𝑡 2 𝑈 0 𝑒 − 𝜇 𝑇 𝐶 0 𝑍 0 + 𝑅 0 𝑒 𝜇 ( 𝑡 𝑘 + 1 − 𝑡 𝑘 ) − 1 𝜇 + 𝑈 0 𝐶 0 𝑍 0 𝑒 𝜇 ( 𝑡 𝑘 + 1 − 𝑡 𝑘 ) − 1 𝜇 + 1 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑈 𝑛 0 𝑡 𝑘 + 1 𝑡 𝑘 𝑒 𝑛 𝜇 ( 𝑡 − 𝑡 𝑘 ) + 𝑑 𝑡 𝜅 𝑈 0 𝑒 − 2 𝜇 𝑇 𝑍 0 𝐶 0 𝑒 𝜇 ( 𝑡 𝑘 + 1 − 𝑡 𝑘 ) − 1 𝜇 + 𝜅 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑈 𝑛 0 𝑒 − 2 𝑛 𝜇 𝑇 × 𝑡 𝑘 + 1 𝑡 𝑘 𝑒 𝑛 𝜇 ( 𝑡 − 𝑡 𝑘 ) | | 𝑢 𝑡 𝑑 𝑡 + 𝜅 𝑘 + 1 𝑡 − 2 𝑇 − 𝑢 𝑘 | | ≤ − 2 𝑇 2 𝑈 0 𝑒 − 𝜇 𝑇 𝐶 0 𝑍 0 + 𝑅 0 𝑒 𝜇 𝑇 0 − 1 𝜇 + 𝑈 0 𝐶 0 𝑍 0 𝑒 𝜇 𝑇 0 − 1 𝜇 + 1 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑈 𝑛 0 𝑒 𝑛 𝜇 𝑇 0 − 1 + 𝑈 𝑛 𝜇 0 𝜅 𝑒 − 2 𝜇 𝑇 𝐶 0 𝑍 0 𝑒 𝜇 𝑇 0 − 1 𝜇 + 𝜅 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑈 𝑛 0 𝑒 − 2 𝑛 𝜇 𝑇 𝑒 𝑛 𝜇 𝑇 0 − 1 ≤ 𝑒 𝑛 𝜇 𝜇 0 − 1 𝜇 𝐶 0 2 𝑈 0 𝑒 − 𝜇 𝑇 𝑍 0 + 𝑅 0 + 𝑈 0 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝑍 0 + 1 𝜇 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑈 𝑛 0 1 + 𝜅 𝑒 − 2 𝑛 𝜇 𝑇 ( 𝑒 𝑛 𝜇 0 − 1 ) 𝑛 ≡ 𝑀 ( 𝜇 ) . ( 2 . 5 )
Let us assume that | ∫ 𝑡 𝑘 + 1 𝑡 𝑘 𝑈 ( 𝑢 ) ( 𝑡 ) 𝑑 𝑡 | = 𝛽 > 0 . We have just obtained that 𝛽 ≤ 𝑀 ( 𝜇 ) . Then, for sufficiently large 𝜇 > 0 (and sufficiently small 𝑇 0 > 0 ) , one can reach the inequality 𝑀 ( 𝜇 ) < 𝛽 . Consequently, ∫ 𝑡 𝑘 + 1 𝑡 𝑘 𝑈 ( 𝑢 ) ( 𝑡 ) 𝑑 𝑡 = 0 . It follows that ∫ 𝑢 ( 𝑡 ) = 𝑡 𝑡 𝑘 𝑈 ( 𝑢 ) ( 𝑠 ) 𝑑 𝑠 and, after a differentiation, we obtain (1.4).
Lemma 2.1 is thus proved.

Theorem 2.2. Let 𝑆 𝑇 = { 𝜏 𝑘 } 𝑛 𝑘 = 0 , 𝑛 ∈ 𝑁 be the set of zeros of the initial function; that is, 𝜐 0 ( 𝜏 𝑘 ) = 0 and 𝜐 0 ( ⋅ ) ∈ 𝐶 1 [ − 𝑇 , 𝑇 ] . If 𝐸 ≤ 𝑈 0 , | 𝜐 0 ( 𝑡 ) | ≤ 𝑈 0 𝑒 𝜇 ( 𝑡 − 𝜏 𝑘 ) , 𝑡 ∈ [ 𝜏 𝑘 , 𝜏 𝑘 + 1 ] , 𝜐 0 ( 𝑡 0 ) = 0 , then, there exists a unique oscillatory solution of the initial value problem (1.4), belonging to 𝑀 𝑆 𝑈 .

Proof. We show that 𝐵 maps 𝑀 𝑆 𝑈 into itself; that is, 𝑢 ∈ 𝑀 𝑆 𝑈 ⇒ 𝐵 ( 𝑢 ) ∈ 𝑀 𝑆 𝑈 .
Indeed, for every 𝑢 ( ⋅ ) ∈ 𝑀 𝑆 𝑈 , the function 𝐵 ( 𝑢 ) ( 𝑡 ) is continuous on [ 𝑡 0 , ∞ ) and differentiable on every ( 𝑡 𝑘 , 𝑡 𝑘 + 1 ) . We have also 𝐵 ( 𝑢 ) ( 𝑡 𝑘 ) = 0 and 𝐵 ( 𝑢 ) ( 𝑡 𝑘 + 1 ) = 0 .
We show that | ( 𝐵 𝑢 ) ( 𝑡 ) | ≤ 𝑈 0 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) , 𝑡 ∈ [ 𝑡 𝑘 , 𝑡 𝑘 + 1 ] . (The last inequalities imply that 𝐵 ( 𝑢 ) ( 𝑡 ) is bounded because 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) ≤ 𝑒 𝜇 𝑇 0 , 𝑡 ∈ [ 𝑇 , ∞ ) .)
We notice that | ( 𝑡 − 𝑡 𝑘 ) / ( 𝑡 𝑘 + 1 − 𝑡 𝑘 ) | ≤ 1 , 𝑡 ∈ [ 𝑡 𝑘 , 𝑡 𝑘 + 1 ] . For sufficiently large 𝜇 , we obtain for 𝑡 ∈ [ 𝑡 𝑘 , 𝑡 𝑘 + 1 ] | | | | ≤ | | | | ( 𝐵 𝑢 ) ( 𝑡 ) 𝑡 𝑡 𝑘 | | | | + | | | | 𝑈 ( 𝑢 ) ( 𝑠 ) 𝑑 𝑠 𝑡 𝑘 + 1 𝑡 𝑘 𝑈 | | | | ( 𝑢 ) ( 𝑠 ) 𝑑 𝑠 ≡ 𝐵 1 + 𝐵 2 . ( 2 . 6 ) We have 𝐵 1 ≤ 2 𝐶 0 𝑍 0 + 𝑅 0 𝑡 𝑡 𝑘 | | | | 1 𝐸 ( 𝑠 − 𝑇 ) 𝑑 𝑠 + 𝐶 0 𝑍 0 𝑡 𝑡 𝑘 | | | | 1 𝑢 ( 𝑠 ) 𝑑 𝑠 + 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑡 𝑡 𝑘 | | | | 𝑢 ( 𝑠 ) 𝑛 + 𝜅 𝑑 𝑠 𝑍 0 𝐶 0 𝑡 𝑡 𝑘 | | | | 𝜅 𝑢 ( 𝑠 − 2 𝑇 ) 𝑑 𝑠 + 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑡 𝑡 𝑘 | | | | 𝑢 ( 𝑠 − 2 𝑇 ) 𝑛 | | | | 𝑑 𝑠 + 𝜅 𝑡 𝑡 𝑘 | | | | ≤ ̇ 𝑢 ( 𝑠 − 2 𝑇 ) 𝑑 𝑠 2 𝑈 0 𝑒 − 𝜇 𝑇 𝐶 0 𝑍 0 + 𝑅 0 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) − 1 𝜇 + 𝑈 0 𝐶 0 𝑍 0 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) − 1 𝜇 + 1 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑈 𝑛 0 𝑡 𝑡 𝑘 𝑒 𝑛 𝜇 ( 𝑠 − 𝑡 𝑘 ) + 𝑑 𝑠 𝜅 𝑈 0 𝑒 − 2 𝜇 𝑇 𝑍 0 𝐶 0 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) − 1 𝜇 + 𝜅 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑈 𝑛 0 𝑒 − 2 𝑛 𝜇 𝑇 𝑡 𝑡 𝑘 𝑒 𝑛 𝜇 ( 𝑠 − 𝑡 𝑘 ) | | 𝑢 | | 𝑑 𝑠 + 𝜅 ( 𝑡 − 2 𝑇 ) ≤ 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝑈 0 1 𝜇 𝐶 0 2 𝑒 − 𝜇 𝑇 𝑍 0 + 𝑅 0 + 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝑍 0 + 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑈 0 𝑛 − 1 𝑒 ( 𝑛 − 1 ) 𝜇 𝑇 0 − 1 1 + 𝜅 𝑒 − 2 𝑛 𝜇 𝑇 𝑛 + 𝜅 𝑒 − 2 𝜇 𝑇 , 𝐵 2 ≤ 2 𝑈 0 𝑒 − 𝜇 𝑇 𝐶 0 𝑍 0 + 𝑅 0 𝑒 𝜇 ( 𝑡 𝑘 + 1 − 𝑡 𝑘 ) − 1 𝜇 + 𝑈 0 𝐶 0 𝑍 0 𝑒 𝜇 ( 𝑡 𝑘 + 1 − 𝑡 𝑘 ) − 1 𝜇 + 1 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑈 𝑛 0 𝑡 𝑘 + 1 𝑡 𝑘 𝑒 𝑛 𝜇 ( 𝑠 − 𝑇 ) + 𝑑 𝑠 𝜅 𝑈 0 𝑒 − 2 𝜇 𝑇 𝑍 0 𝐶 0 𝑒 𝜇 ( 𝑡 𝑘 + 1 − 𝑡 𝑘 ) − 1 𝜇 + 𝜅 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑈 𝑛 0 𝑒 − 2 𝑛 𝜇 𝑇 𝑡 𝑘 + 1 𝑡 𝑘 𝑒 𝑛 𝜇 ( 𝑠 − 𝑇 ) | | 𝑢 𝑡 𝑑 𝑠 + 𝜅 𝑘 + 1 𝑡 − 2 𝑇 − 𝑢 𝑘 | | ≤ − 2 𝑇 2 𝑈 0 𝑒 − 𝜇 𝑇 𝐶 0 𝑍 0 + 𝑅 0 𝑒 𝜇 𝑇 0 − 1 𝜇 + 𝑈 0 𝐶 0 𝑍 0 𝑒 𝜇 𝑇 0 − 1 𝜇 + 1 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑈 𝑛 0 𝑒 𝑛 𝜇 𝑇 0 − 1 + 𝑛 𝜇 𝜅 𝑈 0 𝑒 − 2 𝜇 𝑇 𝐶 0 𝑍 0 𝑒 𝜇 𝑇 0 − 1 𝜇 + 𝜅 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑈 𝑛 0 𝑒 − 2 𝑛 𝜇 𝑇 𝑒 𝑛 𝜇 𝑇 0 − 1 𝑛 𝜇 ≤ 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝑈 0 𝜇 𝐶 0 2 𝑒 − 𝜇 𝑇 𝑒 𝜇 𝑇 0 − 1 𝑍 0 + 𝑅 0 + 𝑒 𝜇 𝑇 0 − 1 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝑍 0 + 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑈 0 𝑛 − 1 1 + 𝜅 𝑒 − 2 𝑛 𝜇 𝑇 𝑒 𝑛 𝜇 𝑇 0 − 1 𝑛 . ( 2 . 7 )
Therefore, for sufficiently large 𝜇 > 0 , we obtain | | | | ( 𝐵 𝑢 ) ( 𝑡 ) ≤ 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝑈 0 1 𝜇 𝐶 0 2 𝑒 − 𝜇 𝑇 𝑍 0 + 𝑅 0 + 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝑍 0 + 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑈 0 𝑛 − 1 𝑒 ( 𝑛 − 1 ) 𝜇 𝑇 0 − 1 1 + 𝜅 𝑒 − 2 𝑛 𝜇 𝑇 𝑛 + 𝜅 𝑒 − 2 𝜇 𝑇 + 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝑈 0 1 𝜇 𝐶 0 2 𝑒 − 𝜇 𝑇 𝑒 𝜇 𝑇 0 − 1 𝑍 0 + 𝑅 0 + 𝑒 𝜇 𝑇 0 − 1 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝑍 0 + 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑈 0 𝑛 − 1 𝑒 𝑛 𝜇 𝑇 0 − 1 𝑛 1 + 𝜅 𝑒 − 2 𝑛 𝜇 𝑇 ≤ 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝑈 0 1 𝜇 𝐶 0 2 𝑒 − 𝜇 𝑇 𝑒 𝜇 𝑇 0 𝑍 0 + 𝑅 0 + 𝑒 𝜇 𝑇 0 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝑍 0 + 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑈 0 𝑛 − 1 𝑒 𝑛 𝜇 𝑇 0 + 𝑒 ( 𝑛 − 1 ) 𝜇 𝑇 0 − 2 1 + 𝜅 𝑒 − 2 𝑛 𝜇 𝑇 𝑛 + 𝜅 𝑒 − 2 𝜇 𝑇 ≤ 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝑈 0 . ( 2 . 8 )
Consequently, the operator 𝐵 maps 𝑀 𝑆 𝑈 into itself.
We show that B is a contractive operator. Indeed, | | 𝐵 ( 𝑢 ) ( 𝑡 ) − 𝐵 𝑢 | | ≤ | | | | ( 𝑡 ) 𝑡 𝑡 𝑘 𝑈 ( 𝑢 ) ( 𝑠 ) − 𝑈 𝑢 | | | | + | | | | ( 𝑠 ) 𝑑 𝑠 𝑡 𝑘 + 1 𝑡 𝑘 𝑈 ( 𝑢 ) ( 𝑠 ) − 𝑈 𝑢 | | | | ( 𝑠 ) 𝑑 𝑠 ≡ 𝐵 1 + 𝐵 2 𝑡 , 𝑡 ∈ 𝑘 , 𝑡 𝑘 + 1 . ( 2 . 9 )
We have 𝐵 1 ≤ 1 𝐶 0 𝑍 0 𝑡 𝑡 𝑘 | | 𝑢 ( 𝑠 ) − | | 1 𝑢 ( 𝑠 ) 𝑑 𝑠 + 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑡 𝑡 𝑘 | | 𝑢 𝑛 ( 𝑠 ) − 𝑢 𝑛 | | + 𝜅 ( 𝑠 ) 𝑑 𝑠 𝑍 0 𝐶 0 𝑡 𝑡 𝑘 | | 𝑢 ( 𝑠 − 2 𝑇 ) − | | 𝜅 𝑢 ( 𝑠 − 2 𝑇 ) 𝑑 𝑠 + 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑡 𝑡 𝑘 | | 𝑢 𝑛 ( 𝑠 − 2 𝑇 ) − 𝑢 𝑛 | | | | | | ( 𝑠 − 2 𝑇 ) 𝑑 𝑠 + 𝜅 𝑡 𝑡 𝑘 ̇ ̇ 𝑢 ( 𝑠 − 2 𝑇 ) − | | | | ≤ 𝜌 𝑢 ( 𝑠 − 2 𝑇 ) 𝑑 𝑠 𝑘 𝑢 , 𝑢 𝐶 0 𝑍 0 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) − 1 𝜇 + 1 𝐶 0 𝑝 𝑛 = 1 𝑛 | | 𝑟 𝑛 | | | | 𝑢 e s s s u p 𝑛 − 1 | | 𝑡 ( 𝑠 ) ∶ 𝑠 ∈ 𝑘 , 𝑡 𝑘 + 1 𝑡 𝑡 𝑘 | | 𝑢 ( 𝑠 ) − | | + 𝜅 𝑢 ( 𝑠 ) 𝑑 𝑠 𝑍 0 𝐶 0 𝜌 𝑘 𝑢 , 𝑢 𝑒 − 2 𝜇 𝑇 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) − 1 𝜇 + 𝜅 𝐶 0 𝑝 𝑛 = 1 𝑛 | | 𝑟 𝑛 | | 𝑢 e s s s u p 𝑛 − 1 𝑡 ( 𝑠 − 2 𝑇 ) ∶ 𝑠 ∈ 𝑘 , 𝑡 𝑘 + 1 𝑡 𝑡 𝑘 | | 𝑢 ( 𝑠 − 2 𝑇 ) − | | 𝑢 ( 𝑠 − 2 𝑇 ) 𝑑 𝑠 + 𝜅 𝜌 𝑘 ̇ ̇ 𝑢 , 𝑢 𝑒 − 2 𝜇 𝑇 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) − 1 𝜇 ≤ 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝜌 𝑘 𝑢 , 𝑢 𝜇 𝐶 0 𝑍 0 + 𝜌 𝑘 𝑢 , 𝑢 𝜇 𝐶 0 𝑝 𝑛 = 1 𝑛 | | 𝑟 𝑛 | | 𝑈 0 𝑛 − 1 𝑒 ( 𝑛 − 1 ) 𝜇 ( 𝑡 𝑘 + 1 − 𝑡 𝑘 ) + 𝜅 𝜌 𝑘 𝑢 , 𝑢 𝑒 − 2 𝜇 𝑇 𝜇 𝑍 0 𝐶 0 + 𝜅 𝜌 𝑘 𝑢 , 𝑢 𝑒 − 2 𝜇 𝑇 𝜇 𝐶 0 𝑝 𝑛 = 1 𝑛 | | 𝑟 𝑛 | | 𝑈 0 𝑛 − 1 𝑒 − 2 ( 𝑛 − 1 ) 𝜇 𝑇 𝑒 ( 𝑛 − 1 ) 𝜇 ( 𝑡 𝑘 + 1 − 𝑡 𝑘 ) + 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝜅 𝜌 𝑘 ̇ ̇ 𝑢 , 𝑢 𝑒 − 2 𝜇 𝑇 𝜇 ≤ 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝜌 𝑘 ̇ ̇ 𝑢 , 𝑢 1 𝜇 2 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝐶 0 𝑍 0 + 1 𝐶 0 𝑝 𝑛 = 1 𝑛 | | 𝑟 𝑛 | | 𝑈 0 𝑛 − 1 1 + 𝜅 𝑒 − 2 𝑛 𝜇 𝑇 𝑒 ( 𝑛 − 1 ) 𝜇 𝑇 0 + 𝜅 𝑒 − 2 𝜇 𝑇 𝜇 ≤ 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝜌 𝜇 ( 𝑘 ) 𝑢 , 𝑢 1 𝜇 2 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝐶 0 𝑍 0 + 1 𝐶 0 𝑝 𝑛 = 1 𝑛 | | 𝑟 𝑛 | | 𝑈 0 𝑛 − 1 1 + 𝜅 𝑒 − 2 𝑛 𝜇 𝑇 𝑒 ( 𝑛 − 1 ) 𝜇 𝑇 0 + 𝜅 𝑒 − 2 𝜇 𝑇 𝜇 , 𝐵 2 ≤ 1 𝐶 0 𝑍 0 𝑡 𝑘 + 1 𝑡 𝑘 | | 𝑢 ( 𝑠 ) − | | 1 𝑢 ( 𝑠 ) 𝑑 𝑠 + 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑡 𝑘 + 1 𝑡 𝑘 | | 𝑢 𝑛 ( 𝑠 ) − 𝑢 𝑛 | | + 𝜅 ( 𝑠 ) 𝑑 𝑠 𝑍 0 𝐶 0 𝑡 𝑘 + 1 𝑡 𝑘 | | 𝑢 ( 𝑠 − 2 𝑇 ) − | | 𝜅 𝑢 ( 𝑠 − 2 𝑇 ) 𝑑 𝑠 + 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑡 𝑘 + 1 𝑡 𝑘 | | 𝑢 𝑛 ( 𝑠 − 2 𝑇 ) − 𝑢 𝑛 | | | | | | ( 𝑠 − 2 𝑇 ) 𝑑 𝑠 + 𝜅 𝑡 𝑘 + 1 𝑡 𝑘 ̇ | | | | ≤ 𝜌 ̇ 𝑢 ( 𝑠 − 2 𝑇 ) − ̇ 𝑢 ( 𝑠 − 2 𝑇 ) 𝑑 𝑠 𝑘 𝑢 , 𝑢 𝐶 0 𝑍 0 𝑒 𝜇 ( 𝑡 𝑘 + 1 − 𝑡 𝑘 ) − 1 𝜇 + 1 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | | | 𝑢 𝑛 . e s s s u p 𝑛 − 1 | | 𝑡 ( 𝑠 ) ∶ 𝑠 ∈ 𝑘 , 𝑡 𝑘 + 1 𝑡 𝑘 + 1 𝑡 𝑘 | | 𝑢 ( 𝑠 ) − | | + 𝜅 𝑢 ( 𝑠 ) 𝑑 𝑠 𝑍 0 𝐶 0 𝜌 𝑘 𝑢 , 𝑢 𝑒 − 2 𝜇 𝑇 𝑒 𝜇 ( 𝑡 𝑘 + 1 − 𝑡 𝑘 ) − 1 𝜇 + 𝜅 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑢 𝑛 . e s s s u p 𝑛 − 1 𝑡 ( 𝑠 − 2 𝑇 ) ∶ 𝑠 ∈ 𝑘 , 𝑡 𝑘 + 1 𝑡 𝑘 + 1 𝑡 𝑘 | | 𝑢 ( 𝑠 − 2 𝑇 ) − | | ≤ 𝜌 𝑢 ( 𝑠 − 2 𝑇 ) 𝑑 𝑠 𝑘 𝑢 , 𝑢 𝐶 0 𝑍 0 𝑒 𝜇 ( 𝑡 𝑘 + 1 − 𝑡 𝑘 ) − 1 𝜇 + 𝜌 𝑘 𝑢 , 𝑢 𝐶 0 𝑒 𝜇 ( 𝑡 𝑘 + 1 − 𝑡 𝑘 ) − 1 𝜇 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑛 𝑈 0 𝑛 − 1 𝑒 ( 𝑛 − 1 ) 𝜇 ( 𝑡 𝑘 + 1 − 𝑡 𝑘 ) + 𝜅 𝜌 𝑘 𝑢 , 𝑢 𝑒 − 2 𝜇 𝑇 𝑍 0 𝐶 0 𝑒 𝜇 ( 𝑡 𝑘 + 1 − 𝑡 𝑘 ) − 1 𝜇 + 𝜅 𝜌 𝑘 𝑢 , 𝑢 𝑒 − 2 𝜇 𝑇 𝐶 0 𝑒 𝜇 ( 𝑡 𝑘 + 1 − 𝑡 𝑘 ) − 1 𝜇 𝑝 𝑛 = 1 𝑛 | | 𝑟 𝑛 | | 𝑈 0 𝑛 − 1 𝑒 ( 𝑛 − 1 ) ( 𝜇 𝑇 0 − 2 𝜇 𝑇 ) ≤ 𝜌 𝑘 ̇ ̇ 𝑢 , 𝑢 𝑒 𝜇 𝑇 0 − 1 𝜇 2 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝐶 0 𝑍 0 + 1 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑛 𝑈 0 𝑛 − 1 𝑒 ( 𝑛 − 1 ) 𝜇 𝑇 0 1 + 𝜅 𝑒 − 2 𝑛 𝜇 𝑇 ≤ 𝜌 𝑢 ( 𝑘 ) 𝑢 , 𝑢 𝑒 𝜇 𝑇 0 − 1 𝜇 2 𝐶 0 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝑍 0 + 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑛 𝑈 0 𝑛 − 1 𝑒 ( 𝑛 − 1 ) 𝜇 𝑇 0 1 + 𝜅 𝑒 − 2 𝑛 𝜇 𝑇 . ( 2 . 1 0 )
Consequently, | | 𝐵 ( 𝑢 ) ( 𝑡 ) − 𝐵 𝑢 | | ( 𝑡 ) ≤ 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝜌 𝜇 ( 𝑘 ) 𝑢 , 𝑢 1 𝜇 2 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝐶 0 𝑍 0 + 1 𝐶 0 𝑝 𝑛 = 1 𝑛 | | 𝑟 𝑛 | | 𝑈 0 𝑛 − 1 1 + 𝜅 𝑒 − 2 𝑛 𝜇 𝑇 𝑒 ( 𝑛 − 1 ) 𝜇 𝑇 0 + 𝜅 𝑒 − 2 𝜇 𝑇 𝜇 + 𝜌 𝑢 ( 𝑘 ) 𝑢 , 𝑢 𝑒 𝜇 𝑇 0 − 1 𝜇 2 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝐶 0 𝑍 0 + 1 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑛 𝑈 0 𝑛 − 1 𝑒 ( 𝑛 − 1 ) 𝜇 𝑇 0 1 + 𝜅 𝑒 − 2 𝑛 𝜇 𝑇 ≤ 𝜌 𝑢 ( 𝑘 ) 𝑢 , 𝑢 𝑒 𝜇 𝑇 0 𝜇 2 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝐶 0 𝑍 0 + 1 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑛 𝑈 0 𝑛 − 1 𝑒 ( 𝑛 − 1 ) 𝜇 𝑇 0 1 + 𝜅 𝑒 − 2 𝑛 𝜇 𝑇 + 𝜅 𝑒 − 2 𝜇 𝑇 𝜇 . ≡ 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝐾 𝑈 𝜌 𝜇 ( 𝑘 ) 𝑢 , 𝑢 . ( 2 . 1 1 )
Therefore, 𝜌 𝑘 ( 𝐵 𝑢 , 𝐵 𝑢 ) ≤ 𝐾 𝑈 𝜌 𝜇 ( 𝑘 ) ( 𝑢 , 𝑢 ) .
It remains to estimate the derivative of B.
We have | | ̇ 𝐵 ̇ 𝐵 ( 𝑢 ) ( 𝑡 ) − 𝑢 | | ≤ | | 𝑈 ( 𝑡 ) ( 𝑢 ) ( 𝑠 ) − 𝑈 𝑢 | | + 1 ( 𝑠 ) 𝑡 𝑘 + 1 − 𝑡 𝑘 | | | | 𝑡 𝑘 + 1 𝑡 𝑘 𝑈 ( 𝑢 ) ( 𝑠 ) − 𝑈 𝑢 ( | | | | ≡ ̇ 𝐵 𝑠 ) 𝑑 𝑠 1 + ̇ 𝐵 2 . ( 2 . 1 2 )
We have ̇ 𝐵 1 ≤ 1 𝐶 0 𝑍 0 | | 𝑢 ( 𝑡 ) − | | + 1 𝑢 ( 𝑡 ) 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | | | 𝑢 𝑛 ( 𝑡 ) − 𝑢 𝑛 | | + 𝜅 ( 𝑡 ) 𝐶 0 𝑍 0 | | 𝑢 ( 𝑡 − 2 𝑇 ) − | | + 𝜅 𝑢 ( 𝑡 − 2 𝑇 ) 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | | | 𝑢 𝑛 ( 𝑡 − 2 𝑇 ) − 𝑢 𝑛 | | | | ̇ ( 𝑡 − 2 𝑇 ) + 𝜅 ̇ 𝑢 ( 𝑡 − 2 𝑇 ) − 𝑢 | | ≤ 𝑒 ( 𝑡 − 2 𝑇 ) 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝜌 𝑘 𝑢 , 𝑢 𝐶 0 𝑍 0 + 1 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑢 𝑛 e s s s u p 𝑛 − 1 𝑡 ( 𝑡 ) ∶ 𝑡 ∈ 𝑘 , 𝑡 𝑘 + 1 | | 𝑢 ( 𝑡 ) − 𝑢 | | + ( 𝑡 ) 𝜅 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝜌 𝑘 𝑢 , 𝑢 𝑒 − 2 𝜇 𝑇 𝐶 0 𝑍 0 + 𝜅 𝐶 0 𝑝 𝑛 = 1 𝑛 | | 𝑟 𝑛 | | 𝑢 e s s s u p 𝑛 − 1 𝑡 ( 𝑡 − 2 𝑇 ) ∶ 𝑡 ∈ 𝑘 , 𝑡 𝑘 + 1 | | 𝑢 ( 𝑡 − 2 𝑇 ) − | | | | ̇ 𝑢 ( 𝑡 − 2 𝑇 ) + 𝜅 ̇ 𝑢 ( 𝑡 − 2 𝑇 ) − | | ≤ 𝑒 𝑢 ( 𝑡 − 2 𝑇 ) 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝜌 𝑘 ̇ ̇ 𝑢 , 𝑢 𝜇 𝐶 0 𝑍 0 + 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝜌 𝑘 ̇ ̇ 𝑢 , 𝑢 𝜇 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑛 𝑈 0 𝑛 − 1 𝑒 ( 𝑛 − 1 ) 𝜇 ( 𝑡 𝑘 + 1 − 𝑡 𝑘 ) + 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝜅 𝜌 𝑘 ̇ ̇ 𝑢 , 𝑢 𝑒 − 2 𝜇 𝑇 𝜇 𝐶 0 𝑍 0 + 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝜌 𝑘 ̇ ̇ 𝑢 , 𝑢 𝜅 𝑒 − 2 𝜇 𝑇 𝜇 𝐶 0 𝑝 𝑛 = 1 𝑛 | | 𝑟 𝑛 | | 𝑈 0 𝑛 − 1 𝑒 ( 𝑛 − 1 ) 𝜇 ( 𝑡 𝑘 + 1 − 𝑡 𝑘 ) + 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝜅 𝜌 𝑘 ̇ ̇ 𝑢 , 𝑢 𝑒 − 2 𝜇 𝑇 ≤ 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝜌 𝜇 ( 𝑘 ) 𝑢 , 𝑢 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝜇 𝐶 0 1 𝑍 0 + 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑛 𝑈 0 𝑛 − 1 𝑒 ( 𝑛 − 1 ) 𝜇 𝑇 0 + 𝜅 𝑒 − 2 𝜇 𝑇 , ̇ 𝐵 2 ≤ 1 𝑡 𝑘 + 1 − 𝑡 𝑘 | | | | 𝑡 𝑘 + 1 𝑡 𝑘 𝑈 ( 𝑢 ) ( 𝑠 ) − 𝑈 𝑢 | | | | ≤ 1 ( 𝑠 ) 𝑑 𝑠 𝑙 0 | | | | 𝑡 𝑘 + 1 𝑡 𝑘 𝑈 ( 𝑢 ) ( 𝑠 ) − 𝑈 𝑢 | | | | ( 𝑠 ) 𝑑 𝑠 ≤ 𝜌 𝑢 ( 𝑘 ) 𝑢 , 𝑢 𝑒 𝜇 𝑇 0 − 1 𝜇 2 𝐶 0 𝑙 0 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝑍 0 + 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑛 𝑈 0 𝑛 − 1 𝑒 ( 𝑛 − 1 ) 𝜇 𝑇 0 1 + 𝜅 𝑒 − 2 𝑛 𝜇 𝑇 . ( 2 . 1 3 )
Therefore, | | ̇ 𝐵 ̇ 𝐵 ( 𝑢 ) ( 𝑡 ) − 𝑢 | | ( 𝑡 ) ≤ 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) 𝜌 𝜇 ( 𝑘 ) 𝑢 , 𝑢 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝜇 𝐶 0 1 𝑍 0 + 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑛 𝑈 0 𝑛 − 1 𝑒 ( 𝑛 − 1 ) 𝜇 𝑇 0 + 𝜅 𝑒 − 2 𝜇 𝑇 + 𝜌 𝑢 ( 𝑘 ) 𝑢 , 𝑢 𝑒 𝜇 𝑇 0 − 1 𝜇 2 𝐶 0 𝑙 0 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝑍 0 + 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑛 𝑈 0 𝑛 − 1 𝑒 ( 𝑛 − 1 ) 𝜇 𝑇 0 1 + 𝜅 𝑒 − 2 𝑛 𝜇 𝑇 ≤ 𝜌 𝜇 ( 𝑘 ) 𝑢 , 𝑢 𝑒 𝜇 𝑇 0 + 𝜇 𝜏 0 − 1 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝜇 2 𝐶 0 𝑙 0 1 𝑍 0 + 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑛 𝑈 0 𝑛 − 1 𝑒 ( 𝑛 − 1 ) 𝜇 𝑇 0 + 𝜅 𝑒 − 2 𝜇 𝑇 ≡ 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) ̇ 𝐾 𝑈 𝜌 𝜇 ( 𝑘 ) 𝑢 , 𝑢 . ( 2 . 1 4 )
It follows 𝜌 𝑘 ( ̇ ̇ 𝐵 ( 𝑢 ) , 𝐵 ( 𝑢 ) ) ≤ 𝑒 𝜇 ( 𝑡 − 𝑡 𝑘 ) ̇ 𝐾 𝑈 𝜌 𝜇 ( 𝑘 ) ( 𝑢 , 𝑢 ) .
Then 𝜌 𝜇 ( 𝑘 ) ( 𝐵 ( 𝑢 ) , 𝐵 ( 𝑢 ) ) ≤ m a x { 𝐾 𝑈 , ̇ 𝐾 𝑈 } 𝜌 𝜇 ( 𝑘 ) ( 𝑢 , 𝑢 ) .
Consequently, 𝜌 𝜇 ( 𝑘 ) 𝐵 𝑢 , 𝐵 𝑢 ≤ 𝐾 𝜌 𝜇 ( 𝑘 ) 𝑢 , 𝑢 ( 𝑘 = 0 , 1 , 2 , … ) , ( 2 . 1 5 ) where 𝐾 = m a x { 𝐾 𝑈 , ̇ 𝐾 𝑈 } < 1 does not depend on 𝑢 and 𝑘 .
We have to verify that 𝑀 𝑆 𝑈 is j-bounded. Indeed, since j is an identity mapping, 𝜌 𝑗 𝑛 𝑢 ( 𝑘 ) 𝑢 , 𝑢 ≤ 𝜌 𝑢 ( 𝑘 ) 𝑢 , 𝑢 < ∞ ( 𝑛 = 0 , 1 , 2 , … ) . ( 2 . 1 6 )
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: 1 𝜇 𝐶 0 2 𝑒 − 𝜇 𝑇 𝑒 𝜇 𝑇 0 𝑍 0 + 𝑅 0 + 𝑒 𝜇 𝑇 0 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝑍 0 + 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑈 0 𝑛 − 1 𝑒 𝑛 𝜇 𝑇 0 + 𝑒 ( 𝑛 − 1 ) 𝜇 𝑇 0 − 2 1 + 𝜅 𝑒 − 2 𝑛 𝜇 𝑇 𝑛 + 𝜅 𝑒 − 2 𝜇 𝑇 𝐾 ≤ 1 , 𝑈 = 𝑒 𝜇 0 𝜇 2 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝐶 0 𝑍 0 + 1 𝐶 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑛 𝑈 0 𝑛 − 1 𝑒 ( 𝑛 − 1 ) 𝜇 0 1 + 𝜅 𝑒 − 2 𝑛 𝜇 𝑇 + 𝜅 𝑒 − 2 𝜇 𝑇 𝜇 ̇ 𝐾 < 1 , 𝑈 = 𝑒 𝜇 0 + 𝜇 𝜏 0 − 1 1 + 𝜅 𝑒 − 2 𝜇 𝑇 𝜇 2 𝐶 0 𝑙 0 1 𝑍 0 + 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑛 𝑈 0 𝑛 − 1 𝑒 ( 𝑛 − 1 ) 𝜇 0 + 𝜅 𝑒 − 2 𝜇 𝑇 < 1 . ( 3 . 1 )

Consider a line with the following specific parameters: 1 Λ = 1 m , 𝐿 = 0 , 2 𝜇 H / m , 𝐶 = 8 0 p F / m , 𝜈 = √ = 1 𝐿 𝐶 √ 0 , 2 ⋅ 1 0 − 6 ⋅ 8 0 ⋅ 1 0 − 1 2 = 1 4 ⋅ 1 0 − 9 = 2 , 5 ⋅ 1 0 8 , 𝑍 0 = 𝐿 𝐶 = 0 , 2 ⋅ 1 0 − 6 8 0 ⋅ 1 0 − 1 2 = 5 0 Ω , 𝑅 0 = 4 5 Ω , 𝐶 0 = 8 p F = 8 ⋅ 1 0 − 1 2 F . ( 3 . 2 )

Then, √ 𝑇 = Λ 𝐿 𝐶 = 4 . 1 0 − 9 s ; 𝜅 = ( 𝑍 0 − 𝑅 0 ) / ( 𝑍 0 + 𝑅 0 ) = 1 / 1 9 = 0 , 0 5 2 6 .

Let us check the propagation of millimeter waves 𝜆 0 = 1 0 − 3 m . We have 𝑓 0 = 1 𝜆 0 √ = 1 𝐿 𝐶 1 0 − 3 ⋅ 4 ⋅ 1 0 − 9 = 2 , 5 ⋅ 1 0 1 1 H z ⟹ 𝑇 0 = 1 𝑓 0 = 1 2 , 5 ⋅ 1 0 1 1 = 4 ⋅ 1 0 − 1 2 s e c . ; 𝑙 = 2 ⋅ 1 0 − 1 2 s e c . ( 3 . 3 ) If we choose 𝜇 = ( 1 / 4 ) 1 0 1 2 , then 𝜇 𝑇 0 = 𝜇 0 = 1 , 𝜇 𝜏 0 = ( 1 / 2 ) , and 𝑇 = 4 ⋅ 1 0 − 9 ⋅ ( 1 / 4 ) ⋅ 1 0 1 2 𝑇 0 = 1 0 0 0 ⋅ 𝑇 0 .

Consequently, 𝜇 𝑇 = ( 1 / 4 ) 1 0 1 2 ⋅ 2 ⋅ 1 0 − 8 = ( 1 / 2 ) 1 0 4 , 𝜇 𝐶 0 = ( 1 / 4 ) 1 0 1 2 ⋅ 8 ⋅ 1 0 4 1 0 − 1 2 = 2 , and 𝜇 2 𝐶 0 = ( 1 / 2 ) ⋅ 1 0 1 2 .

Since 𝑒 − 𝜇 𝑇 = 𝑒 − 5 0 0 0 = 0 , then the above inequalities (omitting the second one) become 𝑒 + 1 0 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑈 0 𝑛 − 1 𝑒 𝑛 + 𝑒 𝑛 − 1 − 2 ̇ 𝐾 2 𝑛 ≤ 1 , 𝑈 1 = 2 𝑒 − 2 1 + 5 0 𝑝 𝑛 = 1 | | 𝑟 𝑛 | | 𝑛 𝑈 0 𝑛 − 1 𝑒 𝑛 − 1 < 1 . ( 3 . 4 )

If the V-I characteristic of the nonlinear resistive element is 𝑓 ( 𝑢 ) = − 0 , 1 2 𝑢 + 0 , 8 𝑢 3 , then 𝑈 0 ≤ 0 , 4 1 ; ̇ 𝐾 𝑈 = 𝑈 0 < 0 , 0 6 . It follows that 𝑈 0 < 0 , 0 6 .

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