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HyersUlamRassias stability of rst-order homogeneous linear dierence equations with

a small step size

Kanako Manabe ^∗ and Masakazu Onitsuka

Department of Applied Mathematics, Faculty of Science,

∗

Graduate School of Science, Okayama Uniervsity of Science,

1-1 Ridai-cho Kita-ku, Okayama 700-0005, Japan

Abstract. This paper deals with the stability in the sense of HyersUlamRassias of constant step size linear dierence equation ∆

x(t) − ax(t) = 0 , where a ∈ R. In particular, we establish stability results of this equation under the assumption that h > 0 is small step size: a ̸= 0 and a > −1/h.

Keywords: HyersUlamRassias stability; HyersUlam stability; linear dierence equation; constant step size;

small step size.

1 Introduction

The problem of HyersUlam stability (HUS) was posed by Ulam [25] in 1940. A year later, Hyers [10]

gave a partial answer to this problem. After that, it has been investigated and generalized by many researchers. It is well known that HyersUlamRassias stability (HURS) is a generalization of HUS.

These problems were rst addressed in the eld of functional equations. The reader see the books written by Brzd¦k, Popa, Ra³a and Xu [8], and Jung [11] for historical backgrounds on HyersUlam stability and HyersUlamRassias stability. For recent references on HURS of functional equations, see [12, 26]. In 1998, Alsina and Ger [1] studied HUS for the simple dierential equation

x

− x = 0.

This study has been improved and extended recently, and is ongoing (see, [18, 23]). In addition, this problem has spread to the other areas. For example, we can nd it in the eld of dierence equations, dynamic equations on time scales, delay dierential equations (see, [2, 3, 4, 6, 7, 15, 16, 17, 20, 21, 22]). In recent years, HyersUlamRassias stability has also been considered in the eld of dierential equations.

For example, see [5, 9, 13, 14, 19, 21, 24].

It is well known that the derivative x

can be approximated by the following dierence:

∆

x(t) := x(t + h) − x(t) h

for h > 0 . ∆

x(t) and h > 0 are so called the forward dierence operator and the step size. Dene h Z := { hk | k ∈ Z}

      (Received June 2, 2020; accepted December 11, 2020)

for h > 0 . In 2017, Onitsuka [16] studied HUS of the rst-order homogeneous linear dierence equation

∆

x(t) − ax(t) = 0, t ∈ h Z , (1.1)

where a ∈ R \ {− 1/h }. Note that we no longer have a dierence equation when a = − 1/h . For this reason, we assume a ̸ = − 1/h . Before giving a denition of the stability, we will give some sets. Let I be a nonempty open interval of R, and let T := h Z ∩ I . If the maximum of T exists, dene T

:= T \{ max T};

otherwise, T

:= T. Now we will give a denition of stability. (1.1) is HyersUlam stable (HUS) on T if and only if there exists a constant K > 0 such that the following holds:

Let ε > 0 be a given constant. If for any function ξ : T → R satisfying | ∆

ξ(t) − aξ (t) | ≤ ε for all t ∈ T

, there exists a solution x : T → R of (1.1) such that | ξ(t) − x(t) | ≤ Kε for all t ∈ T.

In [16], the following result was given.

Theorem A ([16], Theorem 1.5). Suppose that a ̸ = 0 and a > − 1/h . Let ε > 0 be a given constant.

Suppose also that a function ξ : T → R satises | ∆

ξ(t) − aξ(t) | ≤ ε for all t ∈ T

. Then the following hold:

(i) if a > 0 and t := max T exists, then any solution x(t) of (1.1) with | ξ(t) − x(t) | < ε/a satises that

| ξ(t) − x(t) | < ε/a for all t ∈ T;

(ii) if a > 0 and max T does not exist, then lim

ξ(t)(ah + 1)

exists, and there exists a unique solution

x(t) = n

lim

t→∞

ξ(t)(ah + 1)

o

(ah + 1)

of (1.1) such that | ξ(t) − x(t) | ≤ ε/a for all t ∈ T;

(iii) if − 1/h < a < 0 and t := min T exists, then any solution x(t) of (1.1) with | ξ(t) − x(t) | < ε/ | a | satises that | ξ(t) − x(t) | < ε/ | a | for all t ∈ T;

(iv) if − 1/h < a < 0 and min T does not exist, then lim

ξ(t)(ah + 1)

exists, and there exists a unique solution

x(t) =

lim

t→−∞

ξ(t)(ah + 1)

(ah + 1)

of (1.1) such that | ξ(t) − x(t) | ≤ ε/ | a | for all t ∈ T.

This result implies the following.

Corollary B ([16], Corollary 4.1). If a ̸ = 0 and a > − 1/h , then (1.1) is HUS on T.

If the step size h > 0 is suciently small, then a ̸ = 0 and a > − 1/h hold. We call this case small step size case. Note here that Theorem A and Corollary B were obtained under the assumption that h > 0 is small step size, but the other cases were also discussed in [16]. Recently, the results obtained in [16]

have already been extended to various general equations (see, [2, 3, 4, 17]). Moreover, (1.1) corresponds to the dierential equation x

− ax = 0 when h → 0 , so that, (1.1) is called an approximate equation of the dierential equation x

− ax = 0 . HUS of x

− ax = 0 was studied by Onitsuka and Shoji [18] in 2017. Namely, small step size case means that an approximation problem for dierential equations.

The purpose of this study is to extend the results obtained above to more general stability results.

(1.1) is said to be HyersUlamRassias stable (HURS) or AokiRassias stable on T if allowing ε > 0 and K > 0 in HUS to depend on t ∈ h Z, that is, if there exists a positive function ψ : T → R such that the following holds:

Let ϕ : T

→ R be a given positive function. If for any function ξ : T → R satisfying | ∆

ξ(t) − aξ(t) | ≤ ϕ(t) for all t ∈ T

, there exists a solution x : T → R of (1.1) such that | ξ(t) − x(t) | ≤ ψ(t) for all t ∈ T.

In the next section, we will discuss some properties for nonhomogeneous linear dierence equations.

In section 3, we will give the main theorem and its proof.

2 Nonhomogeneous linear dierence equations

In this section, we present some properties of the solutions of the rst-order nonhomogeneous linear dierence equation

∆

x(t) − ax(t) = f (t) (2.1)

on h Z, where a is a real number and f(t) is a real-valued function on h Z. Let

F (t; t

, Φ

, f ) :=

 

 

 

 

 



 

 

 

 

  Φ

+ h

(t−

X

f (t

+ (i − 1)h)(ah + 1)

, if t ≥ t

+ h,

Φ

, if t = t

,

Φ

− h

(t₀

X

f (t

− ih)(ah + 1)

, if t ≤ t

− h

(2.2)

for t ∈ h Z, where t

∈ h Z and Φ

∈ R are arbitrary constants. Then the following is known (see [17, Lemma 3.3]).

Lemma 2.1. Let t

∈ h Z and x

∈ R. If a ̸ = − 1/h then the solution of the initial-value problem (2.1) with x(t

) = x

is

x(t) = F

t; t

, x

(ah + 1)

, f

(ah + 1)

for t ∈ h Z, where F is the function given by (2.2).

Lemma 2.2. Suppose that a ̸ = 0 and a > − 1/h , and there exists an L > 0 such that 0 < f (t) ≤ L for all t ∈ h Z. Let t

∈ h Z and x

∈ R. Then F t; t

, x

(ah + 1)

, f

is an increasing function on h Z, where F is the function given by (2.2), and the following hold:

(i) if a > 0 and x

≤ − L/a , then lim

F t; t

, x

(ah + 1)

, f

exists and

F

t; t

, x

(ah + 1)

, f

< lim

t→∞

F

t; t

, x

(ah + 1)

, f ≤ 0

for all t ∈ h Z;

(ii) if − 1/h < a < 0 and x

≥ − L/a , then lim

F t; t

, x

(ah + 1)

, f

exists and F

t; t

, x

(ah + 1)

, f

> lim

t→−∞

F

t; t

, x

(ah + 1)

, f ≥ 0

for all t ∈ h Z.

Proof. For the simplicity, let G(t) := F t; t

, x

(ah + 1)

, f

on h Z. Since ah + 1 > 0 and f (t) is a positive function, G(t) is an increasing function on h Z.

First we prove the case a > 0 and x

≤ − L/a . Since G(t) is increasing on h Z, we have only to prove that G(t) < 0 for all t ≥ t

+ h . Using a > 0 and x

≤ − L/a , we obtain

G(t) ≤ x

(ah + 1)

+ hL

(t−

X

(ah + 1)

=

x

+ L a

(ah + 1)

− L

a (ah + 1)

< 0 for all t ≥ t

+ h , and thus,

G(t) < lim

t→∞

G(t) ≤ 0

for all t ∈ h Z.

Next we prove the case − 1/h < a < 0 and x

≥ − L/a . We have only to prove that G(t) > 0 for all t ≤ t

− h . Using − 1/h < a < 0 and x

≥ − L/a , we obtain

G(t) ≥ x

(ah + 1)

− hL

(t₀

X

(ah + 1)

=

x

+ L a

(ah + 1)

− L

a (ah + 1)

> 0 for all t ≤ t

− h , and thus,

G(t) > lim

t→−∞

G(t) ≥ 0 for all t ∈ h Z.

Lemma 2.3. Suppose that a > − 1/h , and f (t) > 0 for all t ∈ h Z. Let x(t) be any solution of (2.1).

Then a real-valued function ξ : T → R satises | ∆

ξ(t) − aξ(t) | ≤ f (t) for all t ∈ T

if and only if 0 ≤ ∆

n

(ξ(t) + x(t))(ah + 1)

o ≤ 2f (t)(ah + 1)

= 2∆

x(t)(ah + 1)

for all t ∈ T

.

Proof. The inequalities in Lemma 2.3 is true since ah + 1 > 0 and the equality

∆

n

(ξ(t) + x(t))(ah + 1)

o

= 1

h { (ξ(t + h) + x(t + h)) − (ah + 1)(ξ(t) + x(t)) } (ah + 1)

= (∆

ξ(t) − aξ(t) + ∆

x(t) − ax(t))(ah + 1)

= (∆

ξ(t) − aξ(t) + f (t))(ah + 1)

holds for all t ∈ T

. If ξ(t) ≡ 0 then we have ∆

x(t)(ah + 1)

= f (t)(ah + 1)

from the equality above, and thus, this completes the proof.

Proposition 2.4. Suppose that a ̸ = 0 and a > − 1/h , and there exists an L > 0 such that 0 < f (t) ≤ L for all t ∈ h Z. Suppose also that a function ξ : T → R satises | ∆

ξ(t) − aξ (t) | ≤ f (t) for all t ∈ T

. Let G(t) := F(t; 0, − L/a, f ) , where F is the function given by (2.2). Then

G

=

 



t

lim

G(t), if a > 0,

t→−∞

lim G(t), if − 1

h < a < 0

exists, and there exist a nondecreasing function u : T → R and a nonincreasing function v : T → R such that

ξ(t) = (u(t) + G

− G(t))(ah + 1)

= (v(t) − G

+ G(t))(ah + 1)

(2.3) for all t ∈ T, and the following hold:

(i) if a > 0 and t := max T exists, then the inequality u(t) ≤ u(t) < v(t) ≤ v(t) holds for all t ∈ T;

(ii) if a > 0 and max T does not exist, then lim

u(t) and lim

v(t) exist, and u(t) ≤ lim

u(t) = lim

v(t) ≤ v(t) holds for all t ∈ T;

(iii) if − 1/h < a < 0 and t := min T exists, then the inequality v(t) ≤ v(t) < u(t) ≤ u(t) holds for all t ∈ T;

(iv) if − 1/h < a < 0 and min T does not exist, then lim

u(t) and lim

v(t) exist, and v(t) ≤

lim

v(t) = lim

u(t) ≤ u(t) holds for all t ∈ T.

Proof. By means of Lemma 2.2, we see that G(t) := F (t; 0, − L/a, f ) is an increasing function on h Z;

and lim

G(t) ≤ 0 exists if a > 0 ; lim

G(t) ≥ 0 exists if − 1/h < a < 0 ; and

G(t) − G

 



< 0, if a > 0,

> 0, if − 1

h < a < 0 (2.4)

for all t ∈ h Z. Let

x(t) = G(t)(ah + 1)

for all t ∈ h Z. Then x(t) is the solution of the initial-value problem (2.1) with x(0) = − L/a , by Lemma 2.1.

Now we consider the functions

u(t) = (ξ(t) + x(t))(ah + 1)

− G

and v(t) = (ξ(t) − x(t))(ah + 1)

+ G

on T, where ξ : T → R satises | ∆

ξ(t) − aξ(t) | ≤ f (t) for all t ∈ T

. Then it is clear that (2.3) holds on T. By (2.3), we have

u(t) − v(t) = 2x(t)(ah + 1)

− 2G

= 2(G(t) − G

) (2.5) for all t ∈ T. From this and (2.4), we obtain

u(t)

 



< v(t), if a > 0,

> v(t), if − 1

h < a < 0 (2.6)

for all t ∈ T. Note that u : T → R is a nondecreasing function and v : T → R is a nonincreasing function on T by using Lemma 2.3.

First, we will prove (i). From the facts above with t = max T, u(t) and v(t) become the maximum of u(t) and the minimum of v(t) on T, respectively. Hence, together with (2.6), we see that the claim of (i) is true. The argument for (iii) is similar to that given above for (i). Therefore we omit the proof of (iii).

Next, we prove (ii). Fix the constant s ∈ T. Since v : T → R is a nonincreasing function on T and (2.6) holds on T, we have

u(t) < v(t) ≤ v(s) < ∞

for t ≥ s and t ∈ T. That is, u(t) is bounded above for t ≥ s . Since u : T → R is a nondecreasing function, we see that lim

u(t) exists. By (2.5), we obtain the inequality in the claim of (ii). The argument for (iv) is similar to that given above for (ii). Therefore we omit the proof of (iv). The proof is now complete.

3 Main result

We will present the main theorem and its proof.

Theorem 3.1. Suppose that a ̸ = 0 and a > − 1/h , and there exists an ε > 0 such that 0 < ϕ(t) ≤ ε for all t ∈ h Z. Suppose also that a function ξ : T → R satises | ∆

ξ(t) − aξ(t) | ≤ ϕ(t) for all t ∈ T

. Let G(t) := F (t; 0, − ε/a, ϕ) , where F is the function given by (2.2). Then

G

=

 



t

lim

G(t), if a > 0,

t→−∞

lim G(t), if − 1

h < a < 0

exists, and the following hold:

(i) if a > 0 and t := max T exists, then any solution x(t) of (1.1) with | ξ(t) − x(t) | < (G

− G(t))(ah+

1)

satises that | ξ(t) − x(t) | < (G

− G(t))(ah + 1)

for all t ∈ T;

(ii) if a > 0 and max T does not exist, then lim

ξ(t)(ah + 1)

exists, and there exists a unique solution

x(t) = n

t

lim

ξ(t)(ah + 1)

o

(ah + 1)

of (1.1) such that | ξ(t) − x(t) | ≤ (G

− G(t))(ah + 1)

for all t ∈ T;

(iii) if − 1/h < a < 0 and t := min T exists, then any solution x(t) of (1.1) with | ξ(t) − x(t) | <

(G(t) − G

)(ah + 1)

satises that | ξ(t) − x(t) | < (G(t) − G

)(ah + 1)

for all t ∈ T;

(iv) if − 1/h < a < 0 and min T does not exist, then lim

ξ(t)(ah + 1)

exists, and there exists a unique solution

x(t) =

t→−∞

lim ξ(t)(ah + 1)

(ah + 1)

of (1.1) such that | ξ(t) − x(t) | ≤ (G(t) − G

)(ah + 1)

for all t ∈ T.

Proof. From Lemma 2.2, we see that G

exists and

G(t) − G

= F

t; 0, − ε a , ϕ

− G

 



< 0, if a > 0,

> 0, if − 1

h < a < 0 (3.1) for all t ∈ h Z. By Proposition 2.4, we can nd a nondecreasing function u : T → R and a nonincreasing function v : T → R such that

ξ(t) = (u(t) + G

− G(t))(ah + 1)

= (v(t) − G

+ G(t))(ah + 1)

(3.2) for all t ∈ T.

First we prove case (i). We consider any solution x

(t) of (1.1) with

| ξ(t) − x

(t) | < (G

− G(t))(ah + 1)

, (3.3) where a > 0 and t := max T. Note here that G

− G(t) is positive by (3.1), and x

(t) is written as

x

(t) = x

(t)(ah + 1)

for t ∈ T, so that this together with Proposition 2.4 (i), (3.2) and (3.3) implies that u(t) ≤ u(t) < x

(t)(ah + 1)

< v(t) ≤ v(t)

for t ∈ T. From this and (3.2), we obtain

(ξ(t) − x

(t))(ah + 1)

= u(t) + G

− G(t) − x

(t)(ah + 1)

< G

− G(t) and

(ξ(t) − x

(t))(ah + 1)

= v(t) − G

+ G(t) − x

(t)(ah + 1)

> − (G

− G(t)) for t ∈ T. Thus, the claim of (i) is true.

Next we will prove case (ii). Using Proposition 2.4 (ii), we see that lim

u(t) and lim

v(t) exist, and

u(t) ≤ lim

t→∞

u(t) = lim

t→∞

v(t) ≤ v(t) holds for all t ∈ T. From this and (3.2), we conclude that

t

lim

ξ(t)(ah + 1)

= lim

t→∞

(u(t) + G

− G(t)) = lim

t→∞

u(t)

exists and

u(t) ≤ lim

t→∞

ξ(t)(ah + 1)

≤ v(t) for all t ∈ T. Now we consider the solution

x

(t) = n

t

lim

ξ(t)(ah + 1)

o

(ah + 1)

of (1.1) on h Z. Using (3.2) and the inequality above, we have

(ξ(t) − x

(t))(ah + 1)

= u(t) + G

− G(t) − lim

t→∞

ξ(t)(ah + 1)

≤ G

− G(t) and

(ξ(t) − x

(t))(ah + 1)

= v(t) − G

+ G(t) − lim

t→∞

ξ(t)(ah + 1)

≥ − (G

− G(t)) for t ∈ T, and thus, | ξ(t) − x

(t) | ≤ (G

− G(t))(ah + 1)

for all t ∈ T.

We next prove the uniqueness of x

(t) . By way of contradiction, we consider a solution y(t) of (1.1) such that y(t) ̸≡ x

(t) and | ξ(t) − y(t) | ≤ (G

− G(t))(ah + 1)

for all t ∈ T. Since the uniqueness of solutions of (1.1) are guaranteed for the initial value problem, we can rewrite y(t) as

y(t) = c(ah + 1)

for t ∈ h Z, where c ̸ = lim

ξ(t)(ah + 1)

. Therefore, we obtain 0 ̸ = c − lim

t→∞

ξ(t)(ah + 1)

= | y(t) − x

(t) | (ah + 1)

≤ ( | y(t) − ξ(t) | + | ξ(t) − x

(t) | )(ah + 1)

≤ 2(G

− G(t))

for all t ∈ T, however, this contradicts the fact that lim

G(t) = G

. Thus, the claim of (ii) is true.

Using the same arguments in (i) and (ii), we can conclude that the claims of (iii) and (iv) are also true. The proof of Theorem 3.1 is now complete.

Theorem 3.1 implies the following result.

Corollary 3.2. Suppose that a ̸ = 0 and a > − 1/h , and there exists an ε > 0 such that 0 < ϕ(t) ≤ ε for all t ∈ h Z. Suppose also that a function ξ : T → R satises | ∆

ξ(t) − aξ(t) | ≤ ϕ(t) for all t ∈ T

. Then there exist a positive function ψ : T → R and a solution x : T → R of (1.1) such that | ξ(t) − x(t) | ≤ ψ(t) for all t ∈ T.

Remark 3.1. The statement of Corollary 3.2 do not exactly match HURS because of the restriction of the function ϕ(t) . Therefore, the future subject is to consider whether the boundedness of ϕ(t) is necessary. However, it can be seen that the above Theorem 3.1 and Corollary 3.2 include Theorem A and Corollary B, respectively.

Now we will show that this claim. Consider the case where ϕ(t) ≡ ε and x

= ε/a . Since

G(t) = F

t; 0, − ε a , ε

= − ε a + hε

X

(ah + 1)

= − ε

a (ah + 1)

holds if t ≥ t

+ h , and

G(t) = F

t; 0, − ε a , ε

= − ε a − hε

−

X

(ah + 1)

= − ε

a (ah + 1)

holds if t ≤ t

− h , we see that

G(t) = − ε

a (ah + 1)

for all t ∈ h Z. Then we have G

= 0 , so that

| G

− G(t) | (ah + 1)

= ε

| a |

for all t ∈ h Z. Therefore, we can conclude that Theorem 3.1 includes Theorem A. Obviously, Corollary B

implies Corollary 3.2.

Acknowledgements

The second author was supported by JSPS KAKENHI Grant Number JP20K03668.

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