Addendum to the paper “an iterative method for a functional-differential equation of second order with mixed type argument“

Diana Otrocol
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Abstract

Authors

Diana Otrocol,
Tiberiu Popoviciu Institute of Numerical Analysis of Romanian Academy, Cluj-Napoca, Romania

Diana Otrocol

Veronica Ilea,
Department of Applied Mathematics, Babe¸s-Bolyai University, Cluj-Napoca, Romania

Cornelia Revnic,
Department of Mathematics and Computer Science University of Medicine and Pharmacy “Iuliu Hatieganu” Cluj-Napoca, Romania

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D. Otrocol, V. Ilea, C. Revnic, Addendum to the paper “an iterative method for a functional-differential equation of second order with mixed type argument“, Fixed Point Theory, 14(2013) no. 2, 427-434, Fixed Point Theory, 16 (2015) no. 1, pp. 191-192

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Fixed Point Theory

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University Babes-Bolyai, Cluj-Napoca, Department of Mathematics

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Fixed Point Theory, 16(2015), No.1, …-…

http://www.math.ubbcluj.ro/∼nodeacj/sfptcj.html

Addendum to the paper “An iterative method for a functional-differential equation of second order with mixed type argument“, Fixed Point Theory, 14(2013), No. 2, 427-434

Diana Otrocol∗, Veronica Ilea∗∗ and Cornelia Revnic∗∗∗

∗ Tiberiu Popoviciu Institute of Numerical Analysis of Romanian Academy,
Cluj-Napoca, Romania
E-mail: dotrocol@ictp.acad.ro
∗∗Department of Applied Mathematics, Babeş-Bolyai University,
Cluj-Napoca, Romania
E-mail: vdarzu@math.ubbcluj.ro
∗∗∗Department of Mathematics and Computer Science
University of Medicine and Pharmacy “Iuliu Haţieganu”
Cluj-Napoca, Romania
E-mail: cornelia.revnic@umfcluj.ro

In the paper “An iterative method for a functional-differential equation of second order with mixed type argument“, Fixed Point Theory, 14(2013), No. 2, 427-434, we study the problem

x′′​(t)=f​(t,x​(t),x′​(t),x​(t−h),x​(t+h)),t∈[−h,T],x^{\prime\prime}(t)=f(t,x(t),x^{\prime}(t),x(t-h),x(t+h)),\ t\in[-h,T], (1)
x​(t)=φ​(t),t∈[−h,h]x(t)=\varphi(t),\ t\in[-h,h] (2)

in the following conditions

  • (C1)

    f∈Ck​([−T,T]×ℝ4,ℝ),φ∈Ck​([−h,h],ℝ),k=[Th]+1;f\in C^{k}([-T,T]\times\mathbb{R}^{4},\mathbb{R}),\varphi\in C^{k}([-h,h],\mathbb{R}),k=[\frac{T}{h}]+1;

  • (C2)

    ∂f​(t,u,v,w,z)∂z∈ℝ∗,\frac{\partial f(t,u,v,w,z)}{\partial z}\in\mathbb{R}^{\ast}, ∀t∈[−T,T]\forall t\in[-T,T], ∀u,v,w,z∈ℝ;\forall u,v,w,z\in\mathbb{R};

  • (C3)

    |∂f​(t,u,v,w,z)∂z|≤M1,∀t∈[−T,T],∀u,v,w,z∈ℝ\left|\frac{\partial f(t,u,v,w,z)}{\partial z}\right|\leq M_{1},\forall t\in[-T,T],\ \forall u,v,w,z\in\mathbb{R};

  • (C4)

    ∀t∈[−T,T],u,v,w,z,η∈ℝ,\forall t\in[-T,T],u,v,w,z,\eta\in\mathbb{R}, the equation f​(t,u,v,w,z)−η=0f(t,u,v,w,z)-\eta=0 has a unique solution.

In this addendum we add the condition

  • (C5)(C_{5})

    dkd​tk​φ′′​(t)|t=0=dkd​tk​f​(t,φ​(t),φ′​(t),φ​(t−h),φ​(t+h))|t=0\left.\frac{d^{k}}{dt^{k}}\varphi^{\prime\prime}(t)\right|_{t=0}=\left.\frac{d^{k}}{dt^{k}}f\Big(t,\varphi(t),\varphi^{\prime}(t),\varphi(t-h),\varphi(t+h)\Big)\right|_{t=0}.

So the complete Theorem 2.1 should be:

Theorem 2.1.

In the conditions (C1)−(C3)(C_{1})-(C_{3}) we have

  • a)

    The problem (1.1)-(1.2) has in C2​[−T,T]C^{2}[-T,T] (which is in fact in Ck​[−T,T]C^{k}[-T,T]) a unique solution

    x∗​(t)={φ​(t),t∈[−h,h]x1∗​(t),t∈[h,2​h]⋮xn∗​(t),t∈[n​h,T].x^{\ast}(t)=\begin{cases}\varphi(t),&t\in[-h,h]\\ x_{1}^{\ast}(t),&t\in[h,2h]\\ \vdots&\\ x_{n}^{\ast}(t),&t\in[nh,T].\end{cases}
  • b)

    We suppose that the conditions (C1)−(C5)(C_{1})-(C_{5}) are satisfied. Then the sequence defined by

    (p0)​x​(t)=φ​(t)={x−1​(t),t∈[−h,0],x0​(t),t∈[0,h];\displaystyle(p_{0})\ x(t)=\varphi(t)=\left\{\begin{array}[c]{l}x_{-1}(t),t\in[-h,0],\\ x_{0}(t),t\in[0,h];\end{array}\right.
    (p1)​x1​m​(t)=x1,m−1​(t)−G​(t,x1∗​(t))​F​(t,x1,m−1​(t)),t∈[h,2​h];\displaystyle(p_{1})\ x_{1m}(t)=x_{1,m-1}(t)-G(t,x_{1}^{\ast}(t))F(t,x_{1,m-1}(t)),t\in[h,2h];
    (p2)​x2​m​(t)=x2,m−1​(t)−G​(t,x2∗​(t))​F​(t,x2,m−1​(t)),t∈[2​h,3​h];\displaystyle(p_{2})\ x_{2m}(t)=x_{2,m-1}(t)-G(t,x_{2}^{\ast}(t))F(t,x_{2,m-1}(t)),t\in[2h,3h];
    (p3)​x3​m​(t)=x3,m−1​(t)−G​(t,x3∗​(t))​F​(t,x3,m−1​(t)),t∈[3​h,4​h];\displaystyle(p_{3})\ x_{3m}(t)=x_{3,m-1}(t)-G(t,x_{3}^{\ast}(t))F(t,x_{3,m-1}(t)),t\in[3h,4h];
    ⋮\displaystyle\vdots
    (pn)​xn​m​(t)=xn,m−1​(t)−G​(t,xn∗​(t))​F​(t,xn,m−1​(t)),t∈[n​h,T].\displaystyle(p_{n})\ x_{nm}(t)=x_{n,m-1}(t)-G(t,x_{n}^{\ast}(t))F(t,x_{n,m-1}(t)),t\in[nh,T].

    is convergent and limm→∞​xi​m=xi∗,i=1,n¯;\underset{m\rightarrow\infty}{\lim}x_{im}=x_{i}^{\ast},\ i=\overline{1,n};

Proof.

On each interval of the form: [k​h,(k+1)​h]∪[(k+1)​h,T][kh,(k+1)h]\cup[(k+1)h,T], k∈ℤk\in\mathbb{Z}, from condition (C​5)(C5) we have xk​(k​h)=xk−1​(k​h),xk′​(k​h)=xk−1′​(k​h)x_{k}(kh)=x_{k-1}(kh),\ x^{{}^{\prime}}_{k}(kh)=x^{{}^{\prime}}_{k-1}(kh) and xk′′​(k​h)=xk−1′′​(k​h)x^{{}^{\prime\prime}}_{k}(kh)=x^{{}^{\prime\prime}}_{k-1}(kh). We choose a start function xk,0​(t)x_{k,0}(t) such that xk,0​(k​h)=xk−1,0​(k​h)x_{k,0}(kh)=x_{k-1,0}(kh). So we obtain xk,m​(k​h)=xk−1,m​(k​h)x_{k,m}(kh)=x_{k-1,m}(kh).

Acknowledgement. The authors would like to express their gratitude to Prof. Dirk-André Deckert for careful reading of the manuscript and insightful comments.

Received: ; Accepted:

2015

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