Hyers – Ulam stability of linear difference equations of first order

We prove the Hyers – Ulam stability of linear difference equations of first order of the form ϕ(t)∆y(t)=y(t).

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Introduction

The theory of difference equations and their applications have been receiving intensive attention. See, for example [1-3] and the references cited therein. In this paper we consider the first order difference equation of the form ϕሺtሻ∆yሺtሻൌyሺtሻ, t∈I (1) where, I ൌNሺaሻൌሼa, a ൅1, a ൅2, … ሽ, (a is a fixed nonnegative integer), ∆ is the forward difference operator defined by ∆yሺtሻൌݕሺݐ൅1ሻെyሺtሻ. Assume further that ϕ: I → R is a given function.

By a solution of equation (1) we mean a sequence ሼݕሺݐሻሽ which is defined for ݐ∈ܫ, and satisfies equation (1). S.M Jung the author [4] who investigated the Hyers –Ulam stability of linear differential equation of first order ϕሺtሻݕᇱሺtሻൌyሺtሻ, which was the improvement of the papers [5, 6]. Indeed they dealt with the Hyers Ulam stability of differential equation ݕᇱሺtሻൌߣyሺtሻ, while Alsina and Ger investigated the differential equation ݕᇱሺtሻൌ yሺtሻ. The aim of this paper is to investigate the Hyers –ulam stability of the linear difference equation of first order (1).

More precisely, we prove that if ϕ(t) > 0 or ϕሺtሻ൏െ1 hold for all t ∈ I and further if the function y(t) satisfies |ϕሺtሻ∆yሺtሻെyሺtሻ|<ϵ ∀ t∈I then there exists a real number ܿ such that ቚy(t)-c ∑ 1+∅(s) ∅ሺ௦ሻ t-1 s=a ቚ൑߳ for any t ∈ I. 2. Preliminaries Following an idea of Scon – Mo Jung [4] we prove the following lemma Lemma:2.1 Assume that a function z: I → R is given, The inequality z(t) ≤ ∅(t)∆z(t) is true for t ∈ I, if and only if there exists a function α: I → R such that ∆α(t)∅(t) > 0 and z(t) = α(t) ∏ 1+∅(s) ∅ሺ௦ሻ t-1 s=a ∀ t ∈ I. Proof: Assume the inequality z(t) ≤ ∅(t)∆z(t) holds true for all t ∈ I. Let us define function α: I → R such that α(t)=z(t) ∏ ∅(s) ଵା∅ሺ௦ሻ t-1 s=a then, t-1 t ∆α(t)=∆൭ෑ ∅(s) 1 ൅∅ሺݏሻ s=a ൱z(t)+ ෑ ∅(s) 1 ൅∅ሺݏሻ s=a ∆z(t) ൌ∏ ∅(s) ଵା∅ሺ௦ሻ t-1 s=a ሾ∆zሺtሻሿ൅ zሺt ൅1ሻ ∅ሺ୲ሻ ଵା∅ሺ୲ሻ Maths 2016; 1(4): 34-36 M Vijaya Department of Mathematics Queen Mary’s College, Chennai, Tamil Nadu, India Correspondence M Vijaya Department of Mathematics Queen Mary’s College, Chennai, Tamil Nadu, India t-1 1 ൅∅ሺtሻ ∅(t)∆α(t)=∅(t)∆z(t) ෑ ∅(s) 1 ൅∅ሺݏሻ for every t ∈ I.

According to lemma 1.1 there exists a function α: I → R such that s=a ൅zሺt ൅1ሻ ∅ሺtሻଶ y(t)=ϵ+α(t) ∏ 1+∅(s) ∅ሺୱሻ t-1 s=a for all t ∈I. (6) Since by hypothesis ∅(t)∆z(t) ≥ z(t) ∅(t)∆α(t) ൒z(t) ∏ ∅(s) ଵା∅ሺ୲ሻ >0 and ଵା∅ሺ௦ሻ t-1 s=a ൅zሺt ൅1ሻ∅ሺ୲ሻమ αሺtሻ∏ ଵା∅ሺୱሻ ∅ሺ௦ሻ ୲ିଵ ୱୀୟ ൌz(t) ∏ ∅(s) ଵା∅ሺ௦ሻ t-1 s=a ∏ 1+∅(s) ∅ሺ௦ሻ t-1 s=a ൌz(t). where α additionally satisfies the condition, ∆α(t)∅(t) ൐0, for any t ∈ I (7) Analogously define z(t)=y(t)+ϵ, the inequality on the R.H.S of (5) implies that z(t)≥∅(t)∆y(t)=∅(t)∆z(t) holds for any t ∈I. According to lemma 1.2 there exists a function β: I → R such Conversely, assume that there exists a function α: I → R such that ∅(t)∆α(t) ≥ 0 for each t ∈ I. Let us define a function z: I → R by, that y(t)+ϵ=z(t)=βሺtሻ∏ 1+∅(s) ∅ሺୱሻ t-1 s=a (8) z(t)=α(t) ∏ 1+∅(s) ∅ሺୱሻ t-1 s=a , then, ∆z(t)=∆α(t) ∏ 1+∅(s) ∅ሺୱሻ t-1 s=a ൅ α(t+1)∆ቀ∏ 1+∅(s) ∅ሺୱሻ t-1 s=a ቁ and ∆β(t)∅(t) < 0 for all t ∈I. (9) From (6), Δyሺtሻൌα(t) ∏ 1+∅(s) Therefore ∅(t)∆z(t)=∅(t)∆α(t) ∏ 1+∅(s) ∅ሺୱሻ t-1 s=a ଵ ∅ሺୱሻ t s=a ∆α(t) (10) ∅(t) ൅∏ 1+∅(s) ∅ሺୱሻ t-1 s=a ቃ t-1 s=a ∅ሺୱሻ൅α(t+1) ቂ∏ 1+∅(s) Since by hypothesis ∅(t)∆α(t )≥ 0, we have From (8), ∆yሺtሻൌቀ∏ 1+∅(s) ∅ሺୱሻ t-1 s=a ቁ β(t) ∅ሺ௧ሻ+ ∏ 1+∅(s) ∅ሺୱሻ∆βሺtሻ t s=a (11) ∅(t)∆z(t) ≥ α(t+1) ∏ 1+∅(s) Also, from (8) and (6) ∅ሺୱሻ t-1 s=a ൒z(t), ∀ t ∈I βሺtሻ∏ 1+∅(s) ∅ሺୱሻ t-1 s=a ൌ2߳൅ α(t) ∏ 1+∅(s) ∅ሺ௦ሻ t-1 s=a substituting in (11) ∆yሺtሻൌ∆βሺtሻ∏ 1+∅(s) ∅ሺୱሻ t s=a ൅ ଶఢ ∅ሺୱሻ t-1 s=a ∅(t) ൅ α(t) ∅(t) ∏ 1+∅(s) which proves lemma. Lemma:2.2 Assume that a function z: I → R is given, The inequality z(t) ≥ ∅(t)∆z(t) holds true for any t ∈ I, if and only if there exists a function β: I → R such that ∆β(t)∅(t) < 0 and z(t)=β(t) ∏ 1+∅(s) ∅ሺ௦ሻ t-1 s=a , ∀ t ∈ I. substituting for ∆y(t) from (10) we get, ∆α(t) ∏ 1+∅(s) ∅ሺ௦ሻ t s=a ൌ∆β(t) ∏ 1+∅(s) ∅ሺ௦ሻ t s=a ൅ ଶఢ ∅(t) Proof: Assume the inequality zሺtሻ൒∅ሺtሻ∆zሺtሻ holds for all ∆α(t)= ∆β(t)+ ଶఢ ଵା∅ሺ௦ሻ t s=a ∅(t) ∏ ∅(s) t ∈I.

Let us define β: I → R to be β(t)= ∏ ∅(s) ଵା∅ሺ௦ሻ t-1 s=a z(t) then, ∅(t)∆β(t)=∅(t)∆α(t)-2ϵ ∏ ∅(s) ଵା∅ሺ௦ሻ t s=a ∆β(t)= ∏ ∅(s) ଵା∅ሺ௦ሻ t-1 s=a ∆z(t) ൅ zሺt ൅1ሻ ∆ቀ∏ ∅(s) ଵା∅ሺ௦ሻ t-1 s=a ቁ ∅(t)∆β(t)=∅(t)∆ z(t) ∏ ∅(s) ଵା∅ሺ௦ሻ t s=a ቃ t-1 s=a ଵା∅ሺ௦ሻെzሺt ൅1ሻቂ∏ ∅(s) from (7) and (9) we get, 0 ൑ ∆α(t)∅(t) ൑2ϵ ∏ ∅(s) ଵା∅ሺ௦ሻ t s=a , for any t ∈ I, which proves (4). Since, ∅ሺtሻ∆zሺtሻ൑zሺtሻ ∅(t)∆β(t)≤ z(t) ∏ ∅(s) ଵା∅ሺ௦ሻ t-1 s=a െ zሺt ൅1ሻቂ∏ ∅(s) ଵା∅ሺ௦ሻ t s=a ቃ < 0, and Conversely, assume that y: I → R is given by (3). β(t) ∏ 1+∅(s) y(t)=ϵ+α(t) ∏ 1+∅(s) ∅ሺ௦ሻ t-1 s=a ൌ z(t) ∏ ∅(s) ଵା∅ሺ௦ሻ t-1 s=a ∏ 1+∅(s) ∅ሺ௦ሻ t-1 s=a ൌzሺtሻ. ∅ሺୱሻ t-1 s=a ∀ t ∈ I and a function α: I → R satisfies Conversely, assume that ∆β(t)∅(t) ≤ 0 for each t ∈ I and let us define a function 0 ൑∆α(t)∅(t) ≤ 2ϵ ∏ 1+∅(s) t-1 s=a z: I → R by z(t)=β(t) ∏ 1+∅(s) ∅ሺୱሻ ∅ሺ௦ሻ t-1 s=a then, then from(10), t-1 t ∅(t)∆z(t) ൌ ∅(t)∆β(t) ∏ 1+∅(s) ∅ሺୱሻ t-1 s=a ൅ β(t+1) ∏ 1+∅(s) ∅ሺୱሻ t-1 s=a ∅(t)∆y(t)=α(t) ෑ1+∅(s) ∅ሺsሻ s=a +∅(t) ෑ1+∅(s) ∅ሺsሻ∆α(t) s=a since ∅(t)∆β(t) ≤ 0, ∅(t)∆z(t) ≤ β(t+1) ∏ 1+∅(s) ∅ሺୱሻ t-1 s=a Using (3) and (4) ∅(t)∆y(t)-y(t)= ∅(t)∆α(t) ∏ 1+∅(s) ∅ሺୱሻ-ϵ t s=a . Therefore ∅(t)∆z(t) ≤ z(t). Hence the proof.

Theorem: 2.3 Given an ϵ ൐0, a function y: I → R is a solution of the following inequality ห∅(t)∆y(t)-y(t)ห൑߳ for all t ∈I (2) if and only if there exists a function α: I → R such that y(t)=ϵ+α(t) ∏ 1+∅(s) ∅ሺୱሻ t-1 s=a (3) and 0 ൑∆α(t)∅(t) ≤ 2ϵ ∏ ∅(s) ଵା∅ሺୱሻ t-1 s=a , ∀ t ∈ I. (4) By (4) and the last equation, we conclude that ห∅(t)∆z(t)-y(t)ห ൑ϵ. Which proves the theorem. 3. Hyers – Ulam Stability of difference equation (1) In the following theorem, we prove the Hyers-Ulam stability of the difference equation (1). Theorem: 3.1 If either ∅(t) > 0 holds for all t ∈ I, or 1 ൅∅(t) < 0 holds for all t ∈ I, and if a function y: I → R satisfies the inequality (2) then there exists a real number c such that ቚy(t)-c ∏ 1+∅(s) ∅ሺୱሻ t-1 s=a ቚ൑ϵ for any t ∈ I. (12) Proof: First we assume that y: I → R is a solution of the inequality (2) then y satisfies y(t)-ϵ ≤ ∅(t)∆y(t) ≤ y(t)+ϵ (5) for each t ∈ I.

Define z(t)=y(t)-ϵ then ∆z(t)=∆y(t), the inequality on the L.H.S of (5), becomes z(t) ≤ ∅(t)∆z(t) holds Proof: First we assume that ∅(t) > 0 holds for all t ∈I and a function y: I → R satisfies the inequality (2) for all t ∈ I. Let α: I → R is a function such that (3) and (4) hold, using Theorem 2.3 yሺtሻൌϵ ൅αሺtሻ∏ ଵା∅ሺୱሻ ∅ሺୱሻ ୲ିଵ ୱୀୟ and 0 ൑∆α(t)ሺ∅(t)ሻ ≤ 2ϵ ∏ ∅(s) ଵା∅ሺୱሻ t-1 s=a ∅ሺୱሻൌ1 ୟିଵ ୱୀୟ . Define c = lim t→a α(t). Also we use ∏ ଵା∅ሺୱሻ Now we can divide the above inequality by െሺ∅ሺtሻሻ we get, 0 ≥ -∆α(t) ≥ -2ϵ ሺ∅(t)ሻ∏ ∅(s) ଵା∅ሺୱሻ t-1 s=a Since 1+∅ሺtሻ > ∅ሺtሻ Taking the summation on both sides, we get 0 ≥ - ∑∆α(t) ≥ 2ϵ ∑ -1 ሺ1+∅(t)ሻ∏ ∅(s) ଵା∅ሺୱሻ t-1 s=a . 0 ≥ -α(t)+c ≥ 2ϵ ቀ∏ ∅(s) ଵା∅ሺୱሻ t-1 s=a ቁ -ϵ ≥ -α(t) ∏ 1+∅(s) ∅ሺୱሻ t-1 s=a +c ∏ 1+∅(s) ∅ሺୱሻ t-1 s=a െϵ ≥ ϵ -ϵ ≤ α(t) ∏ 1+∅(s) ∅ሺୱሻ t-1 s=a +ϵ-c ∏ 1+∅(s) ∅ሺୱሻ t-1 s=a ≤ ϵ -ϵ ≤ y(t)-c ∏ 1+∅(s) ∅ሺୱሻ t-1 s=a ≤ ϵ, ቚy(t)-c ∏ 1+∅(s) ∅ሺୱሻ t-1 s=a ቚ≤ ϵ for any t ∈I, which proves (12).

If we assume that 1 ൅∅(t) < 0 holds true for all t ∈I, then the proof is similar to the above procedure, hence we omit it. 4. Remark: Here, we notice that y(a) ∏ ∅(s) t-1 s=a is the general solution of the difference equation ∅(t)∆y(t)=y(t), where ∅(s) = ଵ+∅(s) ∅(s) .

Results

related to the exponential function, J. Inequal. Appl. 1998; 2:373-380. 6. Takahasi SE, Miura T, Miyajima S.

On the Hyers-Ulam stability of the Banach space- valued differential equation ݕᇱሺtሻൌߣyሺtሻ, Bull. Korean math. Soc. 2002; 39:309- 315.

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