# Oscillatory Solutions of Neutral Equations with Polynomial Nonlinearities

Authors: Vasil G. Angelov; Dafinka Tz. Angelova
Journal: International Journal of Differential Equations
 (2011)

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## Abstract

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


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## Body

## 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 1The voltage and currentu(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:(1.1)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)=R0i(0,t),t≥0,(1.2)C0du(Λ,t)dt=i(Λ,t)-f(u(Λ,t)),t≥0,(1.3)u(x,0)=u0(x),i(x,0)=i0(x),x∈[0,Λ],
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. [3–10]). Here, the V-I characteristic of the nonlinear resistive load is i=f(u)=∑n=1prnun, 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 equation(1.4)du(t)dt=2EC0(Z0+R0)-u(t)C0Z0-1C0∑n=1prn[u(t)]n-(Z0-R0)u(t-2T)Z0C0(Z0+R0)+Z0-R0C0(Z0+R0)∑n=1prn[u(t-2T)]n+Z0-R0Z0+R0du(t-2T)dt,t≥T,u(t)=υ0(t),du(t)dt=dυ0(t)dt,t∈[-T,T],
where (x,t)∈Π={(x,t)∈R2:(x,t)∈[0,Λ]×[0,∞)}, κ=|Z0-R0|/(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 x-νt=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.LetST={τk}k=0n,n∈N be the set of zeros of the initial function; that is, υ0(τk)=0 such that τ0=-T,τn=T≡t0.LetS={tk}k=0∞ be a strictly increasing sequence of real numbers satisfying the following conditions (C):(C1)
limk→∞tk=∞,(C2)
0<l0=inf{tk+1-tk:k=0,1,2,…}≤sup{tk+1-tk:k=0,1,2,…}=T0<∞,(C3)
for everyk there is s<k such that tk-T=ts where ts∈ST∪S.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(t-tk),t∈[tk,tk+1]}, where U0,μ are positive constants prescribed below.We assume that|υ0(t)|≤U0eμ(t-τk),t∈[τk,τk+1], (k=0,1,2,…,n-1).The setMSU turns out into a complete uniform space with respect to the family of pseudometrics ρμ(k)(f,g)=max{ρk(f,g),ρk(ḟ,ġ)}, (k=0,1,2,…), where ρk(f,g)=max{e-μ(t-tk)|f(t)-g(t)|:t∈[tk,tk+1]}, ρk(ḟ,ġ)=max{e-μ(t-tk)|ḟ(t)-ġ(t)|:t∈[tk,tk+1]}.One can verify thatMSU is closed subset of C1[t0,∞) with respect to the above metric.Remark 1.1.
The functions fromMS 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 operatorB:MSU→MSU by
(1.5)B(u)(t):=∫tktU(u)(s)ds-(t-tktk+1-tk)∫tktk+1U(u)(s)ds,t∈[tk,tk+1],(k=0,1,2,…),
where
(1.6)U(u)(t)=2EC0(Z0+R0)-u(t)C0Z0-1C0∑n=1prn[u(t)]n-κ(KTu)(t)Z0C0+κC0∑n=1prn[(KTu)(t)]n+κd(KTu)(t)dt,t≥T,
and (KTu)(t)=u(t-2T) is M. A. Krasnoselskii operator (cf. [14]).Remark 1.2.
The operatorKT 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 MSυ0={u(·)∈C1[-T,∞):u(t)=υ0(t),t∈[-T,T]}. Then, KT assigns to every function u(·)∈MS the function ũ(·)∈MSυ0 translated to the right on the interval [T,∞). So, the function (KTu)(t) coincides with υ0(t) on [t0,t0+2T]. Besides tk-2T=ts, and then
(1.7)(KTu)(tk)={u(tk-2T)=υ0(ts)=0,tk∈[T,3T],u(tk-2T)=u(tn)=0,t∈(3T,∞),
that is, (KTu)(·)∈MS.


## 2. Main Results

Lemma 2.1.
IfE≤U0, problem (1.4) has a solution u(·)∈MSU iff the operator B has a fixed point in MSU, that is,
(2.1)u(t)=B(u)(t).Proof.
Letu(·)∈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)=∫tktU(u)(s)ds⇔u(t)=∫tktU(u)(s)ds, and then,
(2.2)u(t)=∫tktU(u)(s)ds⟹0=u(tk+1)=∫tktk+1U(u)(s)ds⟹∫tktk+1U(u)(s)ds=0.
Therefore, u(t)satisfies
(2.3)u(t)=∫tktU(u)(s)ds⟺u(t)=∫tktU(u)(s)ds-(t-tktk+1-tk)∫tktk+1U(u)(s)ds,
that is, u(·)is a fixed point of B.
Conversely, letu(·)∈MSU be a solution of u=B(u); that is,
(2.4)u(t)=∫tktU(u)(s)ds-(t-tktk+1-tk)∫tktk+1U(u)(s)ds.
Then, introducingμ0=μT0, we obtain
(2.5)|∫tktk+1U(u)(s)ds|≤2EC0(Z0+R0)∫tktk+1eμ(t-tk)dt+1C0Z0∫tktk+1|u(t)|dt+1C0∑n=1p|rn|∫tktk+1|u(t)|ndt+κZ0C0∫tktk+1|u(t-2T)|dt+κC0∑n=1p|rn|∫tktk+1|u(t-2T)|ndt+κ|∫tktk+1u̇(t-2T)dt|≤2U0e-μTC0(Z0+R0)eμ(tk+1-tk)-1μ+U0C0Z0eμ(tk+1-tk)-1μ+1C0∑n=1p|rn|U0n∫tktk+1enμ(t-tk)dt+κU0e-2μTZ0C0eμ(tk+1-tk)-1μ+κC0∑n=1p|rn|U0ne-2nμT×∫tktk+1enμ(t-tk)dt+κ|u(tk+1-2T)-u(tk-2T)|≤2U0e-μTC0(Z0+R0)eμT0-1μ+U0C0Z0eμT0-1μ+1C0∑n=1p|rn|U0nenμT0-1nμ+U0κe-2μTC0Z0eμT0-1μ+κC0∑n=1p|rn|U0ne-2nμTenμT0-1nμ≤eμ0-1μC0(2U0e-μTZ0+R0+U0(1+κe-2μT)Z0)+1μC0∑n=1p|rn|U0n(1+κe-2nμT)(enμ0-1)n≡M(μ).
Let us assume that|∫tktk+1U(u)(t)dt|=β>0. We have just obtained that β≤M(μ). Then, for sufficiently large μ>0 (and sufficiently small T0>0), one can reach the inequality M(μ)<β. Consequently, ∫tktk+1U(u)(t)dt=0. It follows that u(t)=∫tktU(u)(s)ds and, after a differentiation, we obtain (1.4).
Lemma2.1 is thus proved.Theorem 2.2.
LetST={τk}k=0n,n∈N be the set of zeros of the initial function; that is, υ0(τk)=0 and υ0(·)∈C1[-T,T]. If E≤U0,|υ0(t)|≤U0eμ(t-τk), 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 thatB maps MSU into itself; that is, u∈MSU⇒B(u)∈MSU.
Indeed, for everyu(·)∈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(t-tk), t∈[tk,tk+1]. (The last inequalities imply that B(u)(t) is bounded because eμ(t-tk)≤eμT0,t∈[T,∞).)
We notice that|(t-tk)/(tk+1-tk)|≤1, t∈[tk,tk+1]. For sufficiently large μ, we obtain for t∈[tk,tk+1](2.6)|(Bu)(t)|≤|∫tktU(u)(s)ds|+|∫tktk+1U(u)(s)ds|≡B1+B2.
We have
(2.7)B1≤[2C0(Z0+R0)∫tkt|E(s-T)|ds+1C0Z0∫tkt|u(s)|ds+1C0∑n=1p|rn|∫tkt|u(s)|nds+κZ0C0∫tkt|u(s-2T)|ds+κC0∑n=1p|rn|∫tkt|u(s-2T)|nds]+κ|∫tktu̇(s-2T)ds|≤[2U0e-μTC0(Z0+R0)eμ(t-tk)-1μ+U0C0Z0eμ(t-tk)-1μ+1C0∑n=1p|rn|U0n∫tktenμ(s-tk)ds+κU0e-2μTZ0C0eμ(t-tk)-1μ+κC0∑n=1p|rn|U0ne-2nμT∫tktenμ(s-tk)ds]+κ|u(t-2T)|≤eμ(t-tk)U0[1μC0(2e-μTZ0+R0+1+κe-2μTZ0+∑n=1p|rn|U0n-1(e(n-1)μT0-1)(1+κe-2nμT)n)+κe-2μT1μC0],B2≤[2U0e-μTC0(Z0+R0)eμ(tk+1-tk)-1μ+U0C0Z0eμ(tk+1-tk)-1μ+1C0∑n=1p|rn|U0n∫tktk+1enμ(s-T)ds+κU0e-2μTZ0C0eμ(tk+1-tk)-1μ+κC0∑n=1p|rn|U0ne-2nμT∫tktk+1enμ(s-T)ds]+κ|u(tk+1-2T)-u(tk-2T)|≤[2U0e-μTC0(Z0+R0)eμT0-1μ+U0C0Z0eμT0-1μ+1C0∑n=1p|rn|U0nenμT0-1nμ+κU0e-2μTC0Z0eμT0-1μ+κC0∑n=1p|rn|U0ne-2nμTenμT0-1nμ]≤eμ(t-tk)U0μC0(2e-μT(eμT0-1)Z0+R0+(eμT0-1)(1+κe-2μT)Z0+∑n=1p|rn|U0n-1(1+κe-2nμT)(enμT0-1)n).
Therefore, for sufficiently largeμ>0, we obtain
(2.8)|(Bu)(t)|≤eμ(t-tk)U0[1μC0(2e-μTZ0+R0+1+κe-2μTZ0+∑n=1p|rn|U0n-1(e(n-1)μT0-1)(1+κe-2nμT)n)+κe-2μT1μC0]+eμ(t-tk)U01μC0(2e-μT(eμT0-1)Z0+R0+(eμT0-1)(1+κe-2μT)Z0+∑n=1p|rn|U0n-1enμT0-1n(1+κe-2nμT))≤eμ(t-tk)U0[1μC0(2e-μTeμT0Z0+R0+eμT0(1+κe-2μT)Z0+∑n=1p|rn|U0n-1(enμT0+e(n-1)μT0-2)(1+κe-2nμT)n)+κe-2μT]≤eμ(t-tk)U0.
Consequently, the operatorB maps MSU into itself.
We show thatB is a contractive operator. Indeed,
(2.9)|B(u)(t)-B(u̅)(t)|≤|∫tkt[U(u)(s)-U(u̅)(s)]ds|+|∫tktk+1[U(u)(s)-U(u̅)(s)]ds|≡B1+B2,t∈[tk,tk+1].
We have(2.10)B1≤[1C0Z0∫tkt|u(s)-u̅(s)|ds+1C0∑n=1p|rn|∫tkt|un(s)-u̅n(s)|ds+κZ0C0∫tkt|u(s-2T)-u̅(s-2T)|ds+κC0∑n=1p|rn|∫tkt|un(s-2T)-u̅n(s-2T)|ds]+κ|∫tkt(u̇(s-2T)-u̅̇(s-2T))ds|≤[ρk(u,u̅)C0Z0eμ(t-tk)-1μ+1C0∑n=1pn|rn|esssup⁡{|un-1(s)|:s∈[tk,tk+1]}∫tkt|u(s)-u̅(s)|ds+κZ0C0ρk(u,u̅)e-2μTeμ(t-tk)-1μ+κC0∑n=1pn|rn|esssup⁡{un-1(s-2T):s∈[tk,tk+1]}∫tkt|u(s-2T)-u̅(s-2T)|ds]+κρk(u̇,u̅̇)e-2μTeμ(t-tk)-1μ≤eμ(t-tk)[ρk(u,u̅)μC0Z0+ρk(u,u̅)μC0∑n=1pn|rn|U0n-1e(n-1)μ(tk+1-tk)+κρk(u,u̅)e-2μTμZ0C0+κρk(u,u̅)e-2μTμC0∑n=1pn|rn|U0n-1e-2(n-1)μTe(n-1)μ(tk+1-tk)]+eμ(t-tk)κρk(u̇,u̅̇)e-2μTμ≤eμ(t-tk)ρk(u̇,u̅̇)[1μ2(1+κe-2μTC0Z0+1C0∑n=1pn|rn|U0n-1(1+κe-2nμT)e(n-1)μT0)+κe-2μTμ]≤eμ(t-tk)ρμ(k)(u,u̅)[1μ2(1+κe-2μTC0Z0+1C0∑n=1pn|rn|U0n-1(1+κe-2nμT)e(n-1)μT0)+κe-2μTμ],B2≤[1C0Z0∫tktk+1|u(s)-u̅(s)|ds+1C0∑n=1p|rn|∫tktk+1|un(s)-u̅n(s)|ds+κZ0C0∫tktk+1|u(s-2T)-u̅(s-2T)|ds+κC0∑n=1p|rn|∫tktk+1|un(s-2T)-u̅n(s-2T)|ds]+κ|∫tktk+1(u̇(s-2T)-u̇̇(s-2T))ds|≤[ρk(u,u̅)C0Z0eμ(tk+1-tk)-1μ+1C0∑n=1p|rn|n.esssup⁡{|un-1(s)|:s∈[tk,tk+1]}∫tktk+1|u(s)-u̅(s)|ds+κZ0C0ρk(u,u̅)e-2μTeμ(tk+1-tk)-1μ+κC0∑n=1p|rn|n.esssup⁡{un-1(s-2T):s∈[tk,tk+1]}∫tktk+1|u(s-2T)-u̅(s-2T)|ds]≤[ρk(u,u̅)C0Z0eμ(tk+1-tk)-1μ+ρk(u,u̅)C0eμ(tk+1-tk)-1μ∑n=1p|rn|nU0n-1e(n-1)μ(tk+1-tk)+κρk(u,u̅)e-2μTZ0C0eμ(tk+1-tk)-1μ+κρk(u,u̅)e-2μTC0eμ(tk+1-tk)-1μ∑n=1pn|rn|U0n-1e(n-1)(μT0-2μT)]≤ρk(u̇,u̅̇)eμT0-1μ2(1+κe-2μTC0Z0+1C0∑n=1p|rn|nU0n-1e(n-1)μT0(1+κe-2nμT))≤ρu(k)(u,u̅)eμT0-1μ2C0(1+κe-2μTZ0+∑n=1p|rn|nU0n-1e(n-1)μT0(1+κe-2nμT)).
Consequently,(2.11)|B(u)(t)-B(u̅)(t)|≤eμ(t-tk)ρμ(k)(u,u̅)[1μ2(1+κe-2μTC0Z0+1C0∑n=1pn|rn|U0n-1(1+κe-2nμT)e(n-1)μT0)+κe-2μTμ]+ρu(k)(u,u̅)eμT0-1μ2(1+κe-2μTC0Z0+1C0∑n=1p|rn|nU0n-1e(n-1)μT0(1+κe-2nμT))≤ρu(k)(u,u̅)[eμT0μ2(1+κe-2μTC0Z0+1C0∑n=1p|rn|nU0n-1e(n-1)μT0(1+κe-2nμT))+κe-2μTμ].≡eμ(t-tk)KUρμ(k)(u,u̅).
Therefore,ρk(Bu,Bu̅)≤KUρμ(k)(u,u̅).
It remains to estimate the derivative ofB.
We have(2.12)|Ḃ(u)(t)-Ḃ(u̅)(t)|≤|U(u)(s)-U(u̅)(s)|+1tk+1-tk|∫tktk+1[U(u)(s)-U(u̅)(s)]ds|≡Ḃ1+Ḃ2.
We have(2.13)Ḃ1≤1C0Z0|u(t)-u̅(t)|+1C0∑n=1p|rn||un(t)-u̅n(t)|+κC0Z0|u(t-2T)-u̅(t-2T)|+κC0∑n=1p|rn||un(t-2T)-u̅n(t-2T)|+κ|u̇(t-2T)-u̅̇(t-2T)|≤eμ(t-tk)ρk(u,u̅)C0Z0+1C0∑n=1p|rn|nesssup⁡{un-1(t):t∈[tk,tk+1]}|u(t)-u̅(t)|+κeμ(t-tk)ρk(u,u̅)e-2μTC0Z0+κC0∑n=1pn|rn|esssup⁡{un-1(t-2T):t∈[tk,tk+1]}|u(t-2T)-u̅(t-2T)|+κ|u̇(t-2T)-u̅̇(t-2T)|≤eμ(t-tk)ρk(u̇,u̅̇)μC0Z0+eμ(t-tk)ρk(u̇,u̅̇)μC0∑n=1p|rn|nU0n-1e(n-1)μ(tk+1-tk)+eμ(t-tk)κρk(u̇,u̅̇)e-2μTμC0Z0+eμ(t-tk)ρk(u̇,u̅̇)κe-2μTμC0∑n=1pn|rn|U0n-1e(n-1)μ(tk+1-tk)+eμ(t-tk)κρk(u̇,u̅̇)e-2μT≤eμ(t-tk)ρμ(k)(u,u̅)[1+κe-2μTμC0(1Z0+∑n=1p|rn|nU0n-1e(n-1)μT0)+κe-2μT],Ḃ2≤1tk+1-tk|∫tktk+1(U(u)(s)-U(u̅)(s))ds|≤1l0|∫tktk+1(U(u)(s)-U(u̅)(s))ds|≤ρu(k)(u,u̅)eμT0-1μ2C0l0(1+κe-2μTZ0+∑n=1p|rn|nU0n-1e(n-1)μT0(1+κe-2nμT)).
Therefore,(2.14)|Ḃ(u)(t)-Ḃ(u̅)(t)|≤eμ(t-tk)ρμ(k)(u,u̅)[1+κe-2μTμC0(1Z0+∑n=1p|rn|nU0n-1e(n-1)μT0)+κe-2μT]+ρu(k)(u,u̅)eμT0-1μ2C0l0(1+κe-2μTZ0+∑n=1p|rn|nU0n-1e(n-1)μT0(1+κe-2nμT))≤ρμ(k)(u,u̅)[(eμT0+μτ0-1)(1+κe-2μT)μ2C0l0(1Z0+∑n=1p|rn|nU0n-1e(n-1)μT0)+κe-2μT]≡eμ(t-tk)K̇Uρμ(k)(u,u̅).
It followsρk(Ḃ(u),Ḃ(u̅))≤eμ(t-tk)K̇Uρμ(k)(u,u̅).
Thenρμ(k)(B(u),B(u̅))≤max{KU,K̇U}ρμ(k)(u,u̅).
Consequently,(2.15)ρμ(k)(Bu,Bu̅)≤K̅ρμ(k)(u,u̅)(k=0,1,2,…),
where K̅=max{KU,K̇U}<1 does not depend on u and k.
We have to verify thatMSU is j-bounded. Indeed, since j is an identity mapping,(2.16)ρujn(k)(u,u̅)≤ρu(k)(u,u̅)<∞(n=0,1,2,…).
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).
Theorem2.2 is thus proved.


## 3. Numerical Example

Finally, we summarize all inequalities needed for the applications:(3.1)1μC0(2e-μTeμT0Z0+R0+eμT0(1+κe-2μT)Z0+∑n=1p|rn|U0n-1(enμT0+e(n-1)μT0-2)(1+κe-2nμT)n)+κe-2μT≤1,KU=eμ0μ2(1+κe-2μTC0Z0+1C0∑n=1p|rn|nU0n-1e(n-1)μ0(1+κe-2nμT))+κe-2μTμ<1,K̇U=(eμ0+μτ0-1)(1+κe-2μT)μ2C0l0(1Z0+∑n=1p|rn|nU0n-1e(n-1)μ0)+κe-2μT<1.Consider a line with the following specific parameters:(3.2)Λ=1m,L=0,2μH/m,C=80pF/m,ν=1LC=10,2⋅10-6⋅80⋅10-12=14⋅10-9=2,5⋅108,Z0=LC=0,2⋅10-680⋅10-12=50Ω,R0=45Ω,C0=8pF=8⋅10-12F.Then,T=ΛLC=4.10-9s;κ=(Z0-R0)/(Z0+R0)=1/19=0,0526.Let us check the propagation of millimeter wavesλ0=10-3m. We have(3.3)f0=1λ0LC=110-3⋅4⋅10-9=2,5⋅1011Hz⟹T0=1f0=12,5⋅1011=4⋅10-12sec.;l=2⋅10-12sec.
If we choose μ=(1/4)1012, then μT0=μ0=1,μτ0=(1/2), and T=4·10-9·(1/4)·1012T0=1000·T0.Consequently,μT=(1/4)1012·2·10-8=(1/2)104,μC0=(1/4)1012·8·10410-12=2, and μ2C0=(1/2)·1012.Sincee-μT=e-5000=0, then the above inequalities (omitting the second one) become(3.4)e100+∑n=1p|rn|U0n-1en+en-1-22n≤1,K̇U=2(e-12)(150+∑n=1p|rn|nU0n-1en-1)<1.If theV-I characteristic of the nonlinear resistive element is f(u)=-0,12u+0,8u3, then U0≤0,41; K̇U=U0<0,06. It follows that U0<0,06.



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*Source: 949547-2011-11-20.xml*