Stability of Additive Functional Equation Originating from Characteristic Polynomial of Degree Three

Abstract

The authors use direct and fixed point methods to prove generalised Ulam–Hyers stability and a solution of the following new form of symmetric additive functional equation arising from characteristic polynomial of degree three in Banach spaces.

1. Introduction

Over the last eight decades, the stability analysis of functional equations has become a well-known topic. In 1940, S.M. Ulam [1] spoke at the University of Wisconsin’s Mathematical Colloquium, where he discussed a number of largely unsolved challenging problems. One of these, stability analysis, is the beginning point for a new type of research.

D.H. Hyers proved the stability of functional equations for the first time in 1941. He has completely answered Ulam’s question by assuming that G and H are Banach “spaces”. He proved the following famous theorem (see [2]).

The approach provided by Hyers that provides the additive function $a\left(x\right)$ is known as a direct method. The most important and powerful tool in this strategy is for studying the stability of different kinds of functional equations. A great number of studies have been written since the publication of Hyers’ theorem in connection with various generalisations of Ulam’s problem and Hyers theorem (see [3]). By reducing the criterion for the Cauchy difference for the addition of powers of norms, T. Aoki [4] modified the Hyers theorem for roughly linear transformations in Banach spaces in 1950. Th.M. Rassias [5] looked at a similar case in 1978 (see L. Maligranda [6]). For the “sum”, both proved the Hyers–Ulam–Aoki–Rassias theorem, J.M. Rassias [7] substituted the sum with the product of norm powers in 1982–1984. In reality, he proved the theorem. Ulam–Gavruta–Rassias stability is the stability indicated in the theorem by B. Bouikhalene, E. Elquorachi [8], Y.S. Jung [9], P. Nakmahachalasint [10], C. Park, A. Najati [11], A. Pietrzyk [12] and A. Sibaha et al. [13]. Th.M.Rassias queried whether the result was true for case $p\ge 1$ in 1989 at the 27th International Symposium on Functional Equations. Z. Gadja [14] in 1991 proved that the result was true for $p>1$ and untrue for $p=1$ by building a counter example using a small modification of Th.M.Rassias’ proof. When $p=\frac{1}{2}$ in the inequality is unique, the result is applied. In response to J.M. Rassias’ inquiry in [15] on the stability of the Cauchy equation, P. Gavruta provided a counter example in [16].

P. Gavruta [17] established the theorem in 1994 by generalising all of the above conclusions by taking the control function. The generalised Hyers–Ulam–Rassias stability of the functional equation is the name for this property. Several authors have studied the Hyers–Ulam–Rassias stability of various functional equations thoroughly during the last 35 years, and there are several noteworthy conclusions (see [18,19,20,21,22,23,24,25,26,27]).

Functional equations are a rapidly expanding branch of mathematics with far-reaching applications; they are increasingly employed to solve issues in mathematical analysis, combinatorics, biology, information theory, statistics, physical sciences, behavioural sciences and engineering. In order to discover solutions with minimum qualities, the symmetry types of functions used to take an equation or inequality can be analysed. In terms of concepts, understanding the symmetry features of significant mappings such as hypergeometric mappings and special polynomials may provide conditions for interesting outcomes. It is also possible to look at symmetry types for various sorts of operators connected with the concept of quantum calculus. The functional equation $h(x+y)+h(x-y)=2h\left(x\right)+2h\left(y\right)$ is related to symmetric bi-additive functions. Since $h\left(x\right)=c{x}^2$ is its solution, this equation is termed a quadratic functional equation for a reason.

The functional equations and its general solution and generalised Hyers–Ulam stability is described as follows:

$$\begin{array}{cc}\hfill & \sum _{i=1}^{m}h\left(m{x}_i+\sum _{j=1,j\ne i}^m{x}_j\right)+h\left(\sum _{i=1}^m{x}_i\right)=2h\left(\sum _{i=1}^{m}m{x}_i\right)\hfill \end{array}$$

(1)

$$\begin{array}{c}\\ \hfill & h\left(h\left(x\right)-h\left(y\right)\right)=h(x+y)+h(x-y)-h\left(x\right)-h\left(y\right)\hfill \end{array}$$

(2)

$$\begin{array}{c}\\ \hfill & \left({\lambda }^n+1\right)f\left(e^{\frac{in\pi }{3}}\vartheta +e^{\frac{-in\pi }{3}}\omega \right)+\left({\lambda }^n-1\right)f\left(e^{\frac{-in\pi }{3}}\vartheta +e^{\frac{in\pi }{3}}\omega \right)\hfill \\ \hfill & +e^{\frac{-in\pi }{3}}f(\vartheta -\omega )+e^{\frac{in\pi }{3}}f\left(\omega -\vartheta \right)=2{\lambda }^{n}cos\left(\frac{n\pi }{3}\right)\left[f\left(\vartheta \right)+f\left(\omega \right)\right]\hfill \end{array}$$

(3)

$$\begin{array}{c}\\ \hfill & h\left(K^{\mathcal{M}+\mathcal{N}}x+K^{\mathcal{L}}y\right)+h\left(K^{\mathcal{M}+\mathcal{N}}y+K^{\mathcal{L}}z\right)+K^{\mathcal{M}+\mathcal{N}}h(x-y)+K^{\mathcal{L}}h(y-z)\hfill \\ \hfill & =2\left(K^{\mathcal{M}+\mathcal{N}}h\left(x\right)+K^{\mathcal{L}}h\left(y\right)\right)\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill & \left(e^{2\left(\frac{n\pi }{3}\right)}+6e^{\left(\frac{n\pi }{3}\right)}\right)f\left(xsin\left(\frac{n\pi }{3}\right)+ycos\left(\frac{n\pi }{3}\right)\right)+\left(9-6e^{\left(\frac{n\pi }{3}\right)}\right)f\left(ysin\left(\frac{n\pi }{3}\right)+zcos\left(\frac{n\pi }{3}\right)\right)\hfill \\ \hfill & +6e^{\left(\frac{n\pi }{3}\right)}f\left(zsin\left(\frac{n\pi }{3}\right)+xcos\left(\frac{n\pi }{3}\right)\right)-6e^{\left(\frac{n\pi }{3}\right)}sin\left(\frac{n\pi }{3}\right)f(x-y)-6e^{\left(\frac{n\pi }{3}\right)}cos\left(\frac{n\pi }{3}\right)f(y-z)\hfill \\ \hfill & =\left(e^{2\left(\frac{n\pi }{3}\right)}sin\left(\frac{n\pi }{3}\right)+6e^{\left(\frac{n\pi }{3}\right)}cos\left(\frac{n\pi }{3}\right)\right)f\left(x\right)+\left(9sin\left(\frac{n\pi }{3}\right)+e^{2\left(\frac{n\pi }{3}\right)}cos\left(\frac{n\pi }{3}\right)\right)f\left(y\right)\hfill \\ \hfill & +\left(6e^{\left(\frac{n\pi }{3}\right)}sin\left(\frac{n\pi }{3}\right)+9cos\left(\frac{n\pi }{3}\right)\right)f\left(z\right)\hfill \end{array}$$

(5)

and they were discussed in (see [28,29,30,31,32,33]).

In this paper, We describe a novel type of symmetric additive functional equation derived from a characteristic polynomial of degree three of the following form:

$$\begin{array}{cc}\hfill & h\left[\left({\delta }^3+11\delta \right)u-6\left({\delta }^2+1\right)v\right]+h\left[\left(11\delta -6{\delta }^2\right)v+\left({\delta }^3-6\right)w\right]\hfill \\ \hfill & -h\left[\left({\delta }^3-6{\delta }^2\right)w+\left(11\delta -6\right)u\right]=\left({\delta }^3+6\right)h\left(u\right)\hfill \\ \hfill & +\left(11\delta -12{\delta }^2-6\right)h\left(v\right)+\left(6{\delta }^2-6\right)h\left(w\right)\hfill \end{array}$$

(6)

with $\delta \ne 0$, and using direct and fixed point methods, we study the general solution and generalised Ulam–Hyers stability in Banach spaces.

2. General Solution

We discuss the solution of the symmetric functional equation in this section.

Lemma 1.

Let the odd mapping $h:X\to Y$ fulfill the functional equation:

$$h(x+y)=h\left(x\right)+h\left(y\right)$$

(7)

if and only if $h:X\to Y$ fulfills the functional Equation (6) $\forall u,v,w\in X$.

Proof. 

An odd function $h:X\to Y$ fulfills functional Equation (7). Consider $x=y=0$ in (7), we obtain $f\left(0\right)=0$. Taking x by $-y$ in (7), we reach $h(-y)=-h\left(y\right)$ $\forall y\in X$. By “replacing” y by x in (7), we have the following:

$$h\left(2x\right)=2h\left(x\right)$$

(8)

for all $x\in X$. We arrive at n by induction.

$$h\left(nx\right)=nh\left(x\right)$$

(9)

Consider $u=\left({\delta }^3+11\delta \right)u,v=-6\left({\delta }^2+1\right)v$ in (7), and using (9), $h(-y)=-h\left(y\right)$, we obtain the following:

$$h\left(\left({\delta }^3+11\delta \right)u-6\left({\delta }^2+1\right)v\right)=\left({\delta }^3+11\delta \right)h\left(u\right)-6\left({\delta }^2+1\right)h\left(v\right)$$

(10)

$\forall u,v\in X$. Taking $v=\left(11\delta -6{\delta }^2\right)v,w=\left({\delta }^3-6\right)w$ in (7), and using (9), we arrive at the following:

$$h\left(\left(11\delta -6{\delta }^2\right)v+\left({\delta }^3-6\right)w\right)=\left(11\delta -6{\delta }^2\right)h\left(v\right)+\left({\delta }^3-6\right)h\left(w\right)$$

(11)

$\forall v,w\in X$. Taking $w=\left({\delta }^3-6{\delta }^2\right)w,u=\left(11\delta -6\right)u$ in (7), and “using” (9), we arrive at the following:

$$h\left(\left({\delta }^3-6{\delta }^2\right)w+\left(11\delta -6\right)u\right)=\left({\delta }^3-6{\delta }^2\right)h\left(w\right)+\left(11\delta -6\right)h\left(u\right)$$

(12)

$\forall u,w\in X$. Adding (10), (11), and “Subtracting” (12), we reach (6).

Conversely, setting $u=v=w=0$ in (6), we obtain $\left({\delta }^3-6{\delta }^2+11\delta -7\right)h\left(0\right)=0$ or $h\left(0\right)=0$ and substituting $(u,v,w)$ by $(u,u,u)$ in (6), we have the following:

$$h\left(\left({\delta }^3-6{\delta }^2+11\delta -6\right)\right)=\left({\delta }^3-6{\delta }^2+11\delta -6\right)h\left(u\right)$$

(13)

$\forall u\in X$. Taking $w=0$ in (6), we have the following:

$$\begin{array}{cc}\hfill & h\left[\left({\delta }^3+11\delta \right)u-6\left({\delta }^2+1\right)v\right]+h\left[\left(11\delta -6{\delta }^2\right)v\right]-h\left[\left(11\delta -6\right)u\right]\hfill \\ \hfill & =\left({\delta }^3+6\right)h\left(u\right)+\left(11\delta -12{\delta }^2-6\right)h\left(v\right)\hfill \end{array}$$

(14)

$\forall u,v\in X$. Using (13) in the above equation, we arrive at the following:

$$\begin{array}{c}\hfill h\left[\left({\delta }^3+11\delta \right)u-6\left({\delta }^2+1\right)v\right]=\left({\delta }^3+11\delta \right)h\left(u\right)-6\left({\delta }^2+1\right)h\left(v\right)\end{array}$$

(15)

$\forall u,v\in X$. Replacing $(u,v)$ by ${\displaystyle \left(\frac{x}{\left({\delta }^3+11\delta \right)},\frac{y}{-6\left({\delta }^2+1\right)}\right)}$ in (15) and using (13), we reach the following:

$$h(x+y)=h\left(x\right)+h\left(y\right)$$

(16)

$\forall x\in X$. Hence, the proof is complete. □

Here, we take $X,Y$ to be a normed linear space and Banach space, respectively:

$$\begin{array}{cc}\hfill & \mathsf{\Omega }(u,v,w)=h\left[\left({\delta }^3+11\delta \right)u-6\left({\delta }^2+1\right)v\right]+h\left[\left(11\delta -6{\delta }^2\right)v+\left({\delta }^3-6\right)w\right]\hfill \\ \hfill & -h\left[\left({\delta }^3-6{\delta }^2\right)w+\left(11\delta -6\right)u\right]-\left({\delta }^3+6\right)h\left(u\right)\hfill \\ \hfill & -\left(11\delta -12{\delta }^2-6\right)h\left(v\right)-\left(6{\delta }^2-6\right)h\left(w\right)\hfill \end{array}$$

with $\delta \ne 0$ $\forall u,v,w\in X.$

3. Direct Method of Stability Analysis

We analyse the stability of the symmetric functional equation in this section.

Theorem 1.

Let the mapping $h:\mathbb{R}\to \mathbb{R}$ satisfy the following inquality:

$$\begin{array}{cc}\hfill & ∥h\left[\left({\delta }^3+11\delta \right)u-6\left({\delta }^2+1\right)v\right]+h\left[\left(11\delta -6{\delta }^2\right)v+\left({\delta }^3-6\right)w\right]\hfill \\ & -h\left[\left({\delta }^3-6{\delta }^2\right)w+\left(11\delta -6\right)u\right]-\left({\delta }^3+6\right)h\left(u\right)\hfill \\ & \left(11\delta -12{\delta }^2-6\right)h\left(v\right)-\left(6{\delta }^2-6\right)h\left(w\right)∥ \le \mathsf{\Theta }\left(u,v,w\right)\hfill \end{array}$$

(17)

with the following condition:

$$\underset{n\to \infty }{lim}\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{n}u,{\mathsf{\Delta }}^{n}v,{\mathsf{\Delta }}^{n}w\right)}{{\mathsf{\Delta }}^n}=0$$

$\forall u,v,w\in \mathbb{R}$, then $\exists$ is a unique symmetric additive function $A:\mathbb{R}\to \mathbb{R}$ and $\mathsf{\Delta }=\left({\delta }^3-6{\delta }^2+11\delta -6\right)$ such that the following is the case:

$$\left|\right|A\left(u\right)-h\left(u\right)\left|\right| \le \frac{1}{\mathsf{\Delta }}\sum _{r=1}^n\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u\right)}{{\mathsf{\Delta }}^r}$$

(18)

$\forall u\in \mathbb{R}$.

Proof. 

Let $h:\mathbb{R}\to \mathbb{R}$ be a real function satisfying the following:

$$∥\mathsf{\Omega }(u,v,w)∥ \le \mathsf{\Theta }\left(u,v,w\right)$$

(19)

for all $u,v,w\in \mathbb{R}$ and for some $\mathsf{\Theta }\ge 0$. Instead of $(u,v,w)$ by $(0,0,0)$ in (19), we have $∥\left({\delta }^3-6{\delta }^2+11\delta -7\right)h\left(u\right)∥ = 0$ or $h\left(u\right)=0$. Letting $u=v=w=u$ in (19), we obtain the following:

$$∥h\left(\mathsf{\Delta }u\right)-\mathsf{\Delta }h\left(u\right)∥ \le \mathsf{\Theta }\left(u,u,u\right)$$

(20)

$\forall u\in \mathbb{R}$, where $\mathsf{\Delta }=\left({\delta }^3-6{\delta }^2+11\delta -6\right)$. We replace u by ${\mathsf{\Delta }}^{r-1}u$ (for $r\in \mathbb{N}$ and $r\ge 1$), and we obtain the following:

$$∥h\left({\mathsf{\Delta }}^{r}u\right)-\mathsf{\Delta }h\left({\mathsf{\Delta }}^{r-1}u\right)∥ \le \mathsf{\Theta }\left({\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u\right)$$

$\forall u\in \mathbb{R}$. Multiplying both sides of the above inequality by ${\displaystyle \frac{1}{{\mathsf{\Delta }}^r}}$, by combining the n inequalities that occur, we arrive at the following.

$$\sum _{r=1}^n\frac{1}{{\mathsf{\Delta }}^r}∥h\left({\mathsf{\Delta }}^{s}u\right)-\mathsf{\Delta }h\left({\mathsf{\Delta }}^{r-1}u\right)∥ \le \sum _{r=1}^n\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u\right)}{{\mathsf{\Delta }}^r}$$

We utilise triangle inequality

$$\left|a+b\right| \le \left|a\right|+\left|b\right|$$

and the following.

$$∥\frac{1}{{\mathsf{\Delta }}^n}h\left({\mathsf{\Delta }}^{n}u\right)-h\left(u\right)∥ \le \sum _{r=1}^n\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u\right)}{{\mathsf{\Delta }}^r}$$

(21)

Since the following is the case:

$$\sum _{r=1}^n\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u\right)}{{\mathsf{\Delta }}^r} \le \sum _{r=1}^{\infty }\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u\right)}{{\mathsf{\Delta }}^r}$$

inequality (21) yields the following:

$$∥\frac{1}{{\mathsf{\Delta }}^n}h\left({\mathsf{\Delta }}^{n}u\right)-h\left(u\right)∥ \le \sum _{r=1}^{\infty }\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u\right)}{{\mathsf{\Delta }}^r}$$

$\forall u\in \mathbb{R}$. Through induction (21), it is true that $\forall n\in N$. If $m>n>0$, then $m-n\in N$ and taking n by $m-n$ in (21), we reach the following:

$$∥\frac{1}{{\mathsf{\Delta }}^{m-n}}h\left({\mathsf{\Delta }}^{m-n}u\right)-h\left(u\right)∥ \le \sum _{r=1}^{\infty }\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u\right)}{{\mathsf{\Delta }}^r}$$

(22)

which is stated as follows:

$$∥\frac{1}{{\mathsf{\Delta }}^m}h\left({\mathsf{\Delta }}^{m-n}u\right)-\frac{1}{{\mathsf{\Delta }}^n}h\left(u\right)∥ \le \frac{1}{{\mathsf{\Delta }}^n}\sum _{r=1}^{\infty }\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u\right)}{{\mathsf{\Delta }}^r}$$

(23)

for all $u\in \mathbb{R}$. Replacing u by ${\mathsf{\Delta }}^{n}u$ in (23), we obtain the following.

$$∥\frac{1}{{\mathsf{\Delta }}^m}h\left({\mathsf{\Delta }}^{m}u\right)-\frac{1}{{\mathsf{\Delta }}^n}h\left({\mathsf{\Delta }}^{n}u\right)∥ \le \frac{1}{{\mathsf{\Delta }}^n}\sum _{r=1}^{\infty }\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r+n-1}u,{\mathsf{\Delta }}^{r+n-1}u,{\mathsf{\Delta }}^{r+n-1}u\right)}{{\mathsf{\Delta }}^r}$$

(24)

Since the following is the case:

$$\underset{n\to \infty }{lim}\frac{1}{{\mathsf{\Delta }}^n}=0$$

from (24), we obtain the following.

$$\underset{n\to \infty }{lim}∥\frac{1}{{\mathsf{\Delta }}^m}h\left({\mathsf{\Delta }}^{m}u\right)-\frac{1}{{\mathsf{\Delta }}^n}h\left({\mathsf{\Delta }}^{n}u\right)∥=0$$

Therefore, the following is the case:

$${\displaystyle {\left\{\frac{h\left({\mathsf{\Delta }}^{n}u\right)}{{\mathsf{\Delta }}^n}\right\}}_{n=1}^{\infty }}$$

and it is a Cauchy sequence. The limit of the sequence is $\mathbb{R}$. Then, we obtain the following:

$$A\left(u\right)=\underset{n\to \infty }{lim}\frac{h\left({\mathsf{\Delta }}^{n}u\right)}{{\mathsf{\Delta }}^n}$$

$\forall u\in \mathbb{R}$. Hence, we prove that $A:\mathbb{R}\to \mathbb{R}$ is additive.

Consider the following.

$$\begin{array}{cc}\hfill & ∥h\left[\left({\delta }^3+11\delta \right)u-6\left({\delta }^2+1\right)v\right]+h\left[\left(11\delta -6{\delta }^2\right)v+\left({\delta }^3-6\right)w\right]\hfill \\ & -h\left[\left({\delta }^3-6{\delta }^2\right)w+\left(11\delta -6\right)u\right]-\left({\delta }^3+6\right)h\left(u\right)\hfill \\ & \left(11\delta -12{\delta }^2-6\right)h\left(v\right)-\left(6{\delta }^2-6\right)h\left(w\right)∥ \le \mathsf{\Theta }\left(u,v,w\right)\hfill \\ \hfill & \underset{n\to \infty }{lim}\frac{1}{{\mathsf{\Delta }}^n}∥h\left[{\mathsf{\Delta }}^n\left({\delta }^3+11\delta \right)u-{\mathsf{\Delta }}^{n}6\left({\delta }^2+1\right)v\right]+h\left[{\mathsf{\Delta }}^n\left(11\delta -6{\delta }^2\right)v+{\mathsf{\Delta }}^n\left({\delta }^3-6\right)w\right]\hfill \\ & -h\left[{\mathsf{\Delta }}^n\left({\delta }^3-6{\delta }^2\right)w+{\mathsf{\Delta }}^n\left(11\delta -6\right)u\right]-\left({\delta }^3+6\right)h\left({\mathsf{\Delta }}^{n}u\right)\hfill \\ & \left(11\delta -12{\delta }^2-6\right)h\left({\mathsf{\Delta }}^{n}v\right)-\left(6{\delta }^2-6\right)h\left({\mathsf{\Delta }}^{n}w\right)∥ \le \underset{n\to \infty }{lim}\frac{1}{{\mathsf{\Delta }}^n}\mathsf{\Theta }\left({\mathsf{\Delta }}^{n}u,{\mathsf{\Delta }}^{n}v,{\mathsf{\Delta }}^{n}w\right)=0\hfill \end{array}$$

Hence, we obtain the following:

$$\begin{array}{cc}\hfill & A\left[\left({\delta }^3+11\delta \right)u-6\left({\delta }^2+1\right)v\right]+A\left[\left(11\delta -6{\delta }^2\right)v+\left({\delta }^3-6\right)w\right]\hfill \\ \hfill & -A\left[\left({\delta }^3-6{\delta }^2\right)w+\left(11\delta -6\right)u\right]=\left({\delta }^3+6\right)A\left(u\right)\hfill \\ \hfill & +\left(11\delta -12{\delta }^2-6\right)A\left(v\right)+\left(6{\delta }^2-6\right)A\left(w\right)\hfill \end{array}$$

where $\forall u\in \mathbb{R}$. Next, we have the following.

$$\begin{array}{cc}\hfill \left|\right|A\left(u\right)-h\left(u\right)\left|\right|& =\left|\right|\underset{n\to \infty }{lim}\frac{h\left({\mathsf{\Delta }}^{n}u\right)}{{\mathsf{\Delta }}^n}-h\left(u\right)\left|\right|\hfill \\ \hfill & =\underset{n\to \infty }{lim}\left|\right|\frac{h\left({\mathsf{\Delta }}^{n}u\right)}{{\mathsf{\Delta }}^n}-h\left(u\right)\left|\right|\hfill \\ \hfill & \le \underset{n\to \infty }{lim}\sum _{r=1}^n\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u\right)}{{\mathsf{\Delta }}^r}\hfill \end{array}$$

Hence, we obtain the following.

$$\left|\right|A\left(u\right)-h\left(u\right)\left|\right| \le \sum _{r=1}^{\infty }\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u\right)}{{\mathsf{\Delta }}^r}$$

(25)

for all $u\in \mathbb{R}$.

Next, we show that A is unique. Let us consider another additive function $B:\mathbb{R}\to \mathbb{R}$ such that the following is the case.

$$\left|\right|B\left(u\right)-h\left(u\right)\left|\right| \le \sum _{r=1}^n\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u\right)}{{\mathsf{\Delta }}^r}$$

Hence, the following is obtained.

$$\begin{array}{cc}\hfill \left|\right|B\left(u\right)-A\left(u\right)\left|\right|& \le \left|\right|B\left(u\right)-h\left(u\right)\left|\right|+\left|\right|A\left(u\right)-h\left(u\right)\left|\right|\hfill \\ \hfill & \le \sum _{r=1}^n\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u\right)}{{\mathsf{\Delta }}^r}+\sum _{r=1}^n\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u\right)}{{\mathsf{\Delta }}^r}\hfill \\ \hfill & =2\sum _{r=1}^n\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u\right)}{{\mathsf{\Delta }}^r}\hfill \end{array}$$

Here, A and B are symmetric additives, and we reach the following.

$$\begin{array}{cc}\left|\right|A\left(u\right)-B\left(u\right)\left|\right|& =\frac{1}{{\mathsf{\Delta }}^n}\left|\right|A_t\left({\mathsf{\Delta }}^{n}x\right)-B_t\left({\mathsf{\Delta }}^{n}x\right)\left|\right|\hfill \\ \hfill & \le \frac{2}{{\mathsf{\Delta }}^n}\sum _{r=1}^n\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r+n-1}u,{\mathsf{\Delta }}^{r+n-1}u,{\mathsf{\Delta }}^{r+n-1}u\right)}{{\mathsf{\Delta }}^r}\hfill \end{array}$$

(26)

By considering the limit as $n\to \infty$, we reach the following from (26).

$$\underset{n\to \infty }{lim}\left|\right|A\left(u\right)-B\left(u\right)\left|\right| \le \underset{n\to \infty }{lim}\frac{2}{{\mathsf{\Delta }}^n}\sum _{r=1}^n\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r+n-1}u,{\mathsf{\Delta }}^{r+n-1}u,{\mathsf{\Delta }}^{r+n-1}u\right)}{{\mathsf{\Delta }}^r}$$

At the end, we obtain the following.

$$\left|\right|A\left(u\right)-B\left(u\right)\left|\right| \le 0$$

Finally, $A\left(u\right)=B\left(u\right)$ for all $u\in \mathbb{R}$. Hence, A is unique. □

Corollary 1.

Let the mapping $f:X\to Y$ fulfill the inequality and let $\mathsf{\Lambda },s\in R-0$ and the following be fulfilled:

$$\begin{array}{cc}\hfill & ∥\mathsf{\Theta }(u,v,w)∥ \le \left\{\begin{array}{cc}\mathsf{\Lambda },\hfill & \\ \mathsf{\Lambda }\left\{{\left|\right|u\left|\right|}^s+{\left|\right|v\left|\right|}^s+{\left|\right|w\left|\right|}^s\right\},\hfill & s\ne 1;\hfill \\ {\mathsf{\Lambda }\left|\right|u\left|\right|}^s{\left|\right|v\left|\right|}^s{\left|\right|w\left|\right|}^s,\hfill & 3s\ne 1;\hfill \\ \mathsf{\Lambda }\left\{{\left|\right|u\left|\right|}^s{\left|\right|v\left|\right|}^s{\left|\right|w\left|\right|}^s+\left\{{\left|\right|u\left|\right|}^{3s}+{\left|\right|v\left|\right|}^{3s}+{\left|\right|w\left|\right|}^{3s}\right\}\right\},\hfill & 3s\ne 1;\hfill \end{array}\right.\hfill \end{array}$$

(27)

$\forall u,v,w\in X$. Then, $\exists$ is a unique symmetric additive function $A:X\to Y$ such that the following is the case:

$$\begin{array}{c}\hfill ∥h\left(u\right)-A\left(u\right)∥ \le \left\{\begin{array}{c}{\displaystyle \frac{\mathsf{\Lambda }}{|\mathsf{\Delta }-1|},}\hfill \\ {\displaystyle \frac{{3\mathsf{\Lambda }\left|\right|u\left|\right|}^s}{|\mathsf{\Delta }-{\mathsf{\Delta }}^s|},}\hfill \\ {\displaystyle \frac{{\mathsf{\Lambda }\left|\right|u\left|\right|}^{3s}}{|\mathsf{\Delta }-{\mathsf{\Delta }}^{3s}|},}\hfill \\ {\displaystyle \frac{{4\mathsf{\Lambda }\left|\right|u\left|\right|}^{3s}}{|\mathsf{\Delta }-{\mathsf{\Delta }}^{3s}|}}\hfill \end{array}\right.\end{array}$$

(28)

for all $u\in X.$

Proof. 

Define a mapping $h:\mathbb{R}\to \mathbb{R}$ by (27). Hence, from (25), we obtain the following.

$$\begin{array}{cc}\hfill & \sum _{r=1}^{\infty }\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u\right)}{{\mathsf{\Delta }}^r}=\sum _{r=1}^{\infty }\frac{\mathsf{\Lambda }}{{\mathsf{\Delta }}^r}\hfill \\ \hfill & \sum _{r=1}^{\infty }\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u\right)}{{\mathsf{\Delta }}^r}=\sum _{r=1}^{\infty }\frac{\mathsf{\Lambda }\left\{\left|\right|{\mathsf{\Delta }}^{r-1}{u\left|\right|}^s+\left|\right|{\mathsf{\Delta }}^{r-1}{u\left|\right|}^s+\left|\right|{\mathsf{\Delta }}^{r-1}u{\left|\right|}^s\right\}}{{\mathsf{\Delta }}^r}\hfill \\ \hfill & \sum _{r=1}^{\infty }\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u\right)}{{\mathsf{\Delta }}^r}=\sum _{r=1}^{\infty }\frac{\mathsf{\Lambda }\left\{\left|\right|{\mathsf{\Delta }}^{r-1}{u\left|\right|}^s\left|\right|{\mathsf{\Delta }}^{r-1}{u\left|\right|}^s\left|\right|{\mathsf{\Delta }}^{r-1}u{\left|\right|}^s\right\}}{{\mathsf{\Delta }}^r}\hfill \\ \hfill & \sum _{r=1}^{\infty }\frac{\mathsf{\Theta }\left({\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u,{\mathsf{\Delta }}^{r-1}u\right)}{{\mathsf{\Delta }}^r}\hfill \\ \hfill & =\sum _{r=1}^{\infty }\frac{\mathsf{\Lambda }\left\{\left|\right|{\mathsf{\Delta }}^{r-1}{u\left|\right|}^s\left|\right|{\mathsf{\Delta }}^{r-1}{u\left|\right|}^s\left|\right|{\mathsf{\Delta }}^{r-1}u{\left|\right|}^s+\left\{\left|\right|{\mathsf{\Delta }}^{r-1}{u\left|\right|}^{3s}+\left|\right|{\mathsf{\Delta }}^{r-1}{u\left|\right|}^{3s}+\left|\right|{\mathsf{\Delta }}^{r-1}u{\left|\right|}^{3s}\right\}\right\}}{{\mathsf{\Delta }}^r}.\hfill \end{array}$$

(29)

By using the theorem (1), we reach the corollary result. □

4. Fixed Point Method of Stability Analysis

We derive the stability results in Banach spaces by approaching the fixed point technique (see [34]) in this part.

Theorem 2.

Let the mapping $h:V\to B$ for which $\exists$ functions $\mathsf{\Theta },\beta ,\gamma :V^3\to [0,\infty )$ with the following condition:

$$\underset{t\to \infty }{lim}\frac{\mathsf{\Theta }({\mu }_i^{t}u,{\mu }_i^{t}v,{\mu }_i^{t}w)}{{\mu }_i^t}=0,$$

(30)

where the following is the case:

$$\begin{array}{c}\hfill {\mu }_i=\left\{\begin{array}{cc}\mathsf{\Delta },\hfill & i=0,\hfill \\ \frac{1}{\mathsf{\Delta }},\hfill & i=1\hfill \end{array}\right.\end{array}$$

which satisfy the following functional inequality:

$$∥\mathsf{\Omega }(u,v,w)∥ \le \mathsf{\Theta }\left(u,v,w\right)$$

(31)

$\forall u,v,w\in V$. If the function $\exists$ is $L=L\left(i\right)<1$:

$$u\to \gamma \left(u\right)=\frac{1}{\mathsf{\Delta }}\mathsf{\Theta }\left(\frac{u}{\mathsf{\Delta }},\frac{u}{\mathsf{\Delta }},\frac{u}{\mathsf{\Delta }}\right)$$

such that it has the following property:

$$\gamma \left(u\right)=L{\mu }_i\gamma \left(\frac{u}{{\mu }_i}\right)$$

(32)

$\forall u\in V$, then function $A:V\to B$ is a symmetric mapping and the following:

$$∥h\left(u\right)-A\left(u\right)∥ \le \frac{L^{1-i}}{1-L}\gamma \left(u\right)$$

(33)

holds $\forall u\in V$.

Proof. 

Let us take $X=\{P/P:V\to B,p(0)=0\}$ and the generalised metric introduced on X.

$$d(P,Q)=inf\{K\in (0,\infty ):\Vert P(u)-Q(u)\Vert \le K\gamma (u),u\in V\}.$$

Hence, we knoew $(X,d)$ is complete.

Define the mapping $T:X\to X$ as follows.

$$TP\left(u\right)=\frac{1}{{\mu }_i}P\left({\mu }_{i}u\right),\forall u\in V.$$

Now, $P,Q\in X$:

$$\begin{array}{cc}\hfill d(P,Q) \le K& \Rightarrow ∥P\left(u\right)-Q\left(u\right)∥ \le K\gamma \left(u\right),u\in V.\hfill \\ & \Rightarrow ∥\frac{1}{{\mu }_i}P\left({\mu }_{i}u\right)-\frac{1}{{\mu }_i}Q\left({\mu }_{i}u\right)∥ \le \frac{1}{{\mu }_i}K\gamma \left({\mu }_{i}u\right),u\in V,\hfill \\ & \Rightarrow ∥\frac{1}{{\mu }_i}P\left({\mu }_{i}u\right)-\frac{1}{{\mu }_i}Q\left({\mu }_{i}u\right)∥ \le LK\gamma \left(u\right),u\in V,\hfill \\ & \Rightarrow ∥TP\left(u\right)-TQ\left(u\right)∥ \le LK\gamma \left(u\right),u\in V,\hfill \\ & \Rightarrow d(TP,TQ) \le LK.\hfill \end{array}$$

$$⟹d(TP,TQ) \le Ld(P,Q),$$

$\forall P,Q\in X$, i.e., T is the strictly contractive mapping on X and is L Lipschitz constant.

By [34] with (20), we arrive at the following:

$$\begin{array}{c}\hfill ∥h\left(u\right)-\frac{h\left(\mathsf{\Delta }u\right)}{\mathsf{\Delta }}∥ \le \frac{\mathsf{\Theta }\left(u,u,u\right)}{\mathsf{\Delta }}\end{array}$$

(34)

where the following is the case:

$$\beta \left(u\right)=\frac{\mathsf{\Theta }\left(u,u,u\right)}{\mathsf{\Delta }}$$

$\forall u\in V$. For the condition $i=0$, using (32), we reach the following:

$$∥\frac{1}{\mathsf{\Delta }}h\left(\mathsf{\Delta }u\right)-h\left(u\right)∥ \le \frac{1}{\mathsf{\Delta }}\gamma \left(u\right)$$

$\forall u\in V$.

$$\mathrm{i}.\mathrm{e}.,d(Th,h) \le \frac{1}{\mathsf{\Delta }}=L=L^{1-0}=L^{1-i}<\infty .$$

Let ${\displaystyle u=\frac{u}{\mathsf{\Delta }}}$ in (34), and we obtain the following:

$$∥h\left(u\right)-\mathsf{\Delta }h\left(\frac{u}{\mathsf{\Delta }}\right)∥ \le \frac{1}{\mathsf{\Delta }}\mathsf{\Theta }\left(\frac{u}{\mathsf{\Delta }},\frac{u}{\mathsf{\Delta }},\frac{u}{\mathsf{\Delta }}\right).$$

for all $u\in V$. For condition $i=1$, using (32) $i=1$, we have the following:

$$∥h\left(u\right)-\mathsf{\Delta }h\left(\frac{u}{\mathsf{\Delta }}\right)∥ \le \gamma \left(u\right)$$

$\forall u\in V$.

$$\mathrm{i}.\mathrm{e}.,d(h,Th) \le 1=L^0=L^{1-1}=L^{1-i}<\infty .$$

In the above cases, we arrive at the following.

$$d(h,Th) \le L^{1-i}.$$

Therefore by [34] with $\left(B_2\left(i\right)\right)$, the result holds.

By [34] with $\left(B_2\left(ii\right)\right)$, then $\exists$ is a fixed point A of T in X such that the following is the case.

$$A\left(u\right)=\underset{t\to \infty }{lim}\frac{h\left({\mu }_i^{t}u\right)}{{\mu }_i^t},\forall x\in V.$$

(35)

To derive $A:V\to B$ is to use symmetric additives. Taking $(u,v,w)$ by $\left({\mu }_i^{t}u,{\mu }_i^{t}y,{\mu }_i^{t}w\right)$ in (31) and ÷${\mu }_i^t,$ followed by (30) and (35), A satisfies (6) for all $u,v,w\in X$.

By [34] with $\left(B_2\left(iii\right)\right)$, in the set $Y=\{h\in X:d(Th,A)<\infty \},$ A is the unique fixed point of T with the unique function such that the following is the case:

$$∥h\left(u\right)-A\left(u\right)∥ \le K\gamma \left(u\right)$$

$\forall u\in V$ and $K>0.$ Hence, by [34] with $\left(B_2\left(iv\right)\right)$, we reach the following.

$$d(h,A) \le \frac{1}{1-L}d(h,Th)$$

$$⟹d(h,A) \le \frac{L^{1-i}}{1-L}.$$

At the end, we obtain the following:

$$\Vert h\left(u\right)-A\left(u\right)\Vert \le \frac{L^{1-i}}{1-L}\gamma \left(u\right).$$

$\forall u\in V$. □

Corollary 2.

Let the mapping $h:V\to B$ and $\exists$ $\mathsf{\Lambda },s\in R$ such that (27) $\forall u,v,w\in V$, then $\exists$ is a unique symmetric additive function $A:V\to B$:

$$∥h\left(u\right)-A\left(u\right)∥ \le \left\{\begin{array}{c}{\displaystyle \frac{\mathsf{\Lambda }}{|\mathsf{\Delta }-1|},}\hfill \\ {\displaystyle \frac{{3\mathsf{\Lambda }\left|\right|u\left|\right|}^s}{|\mathsf{\Delta }-{\mathsf{\Delta }}^s|},}\hfill \\ {\displaystyle \frac{{\mathsf{\Lambda }\left|\right|u\left|\right|}^{3s}}{|\mathsf{\Delta }-{\mathsf{\Delta }}^{3s}|},}\hfill \\ {\displaystyle \frac{{4\mathsf{\Lambda }\left|\right|u\left|\right|}^{3s}}{|\mathsf{\Delta }-{\mathsf{\Delta }}^{3s}|}}\hfill \end{array}\right.$$

(36)

for all $u\in V$.

Proof. 

Let us set in (27) $\forall u,v,w\in V$. Now, we have the following.

$$\begin{array}{cc}\hfill \frac{\mathsf{\Theta }({\mu }_i^{t}u,{\mu }_i^{t}v,{\mu }_i^{t}w)}{{\mu }_i^t}& =\left\{\begin{array}{c}{\displaystyle \frac{\mathsf{\Lambda }}{{\mu }_i^t},}\hfill \\ {\displaystyle \frac{\mathsf{\Lambda }}{{\mu }_i^t}\left\{\left|\right|{\mu }_i^t{u\left|\right|}^s+\left|\right|{\mu }_i^t{v\left|\right|}^s+\left|\right|{\mu }_i^{t}w{\left|\right|}^s\right\},}\hfill \\ {\displaystyle \frac{\mathsf{\Lambda }}{{\mu }_i^n}\left|\right|{\mu }_i^n{u\left|\right|}^s\left|\right|{\mu }_i^n{v\left|\right|}^s\left|\right|{\mu }_i^{n}w{\left|\right|}^s,}\hfill \\ {\displaystyle \frac{\mathsf{\Lambda }}{{\mu }_i^t}\left\{\left|\right|{\mu }_i^t{u\left|\right|}^s\left|\right|{\mu }_i^t{v\left|\right|}^s\left|\right|{\mu }_i^{t}w{\left|\right|}^s\left\{\left|\right|{\mu }_i^t{u\left|\right|}^{3s}+\left|\right|{\mu }_i^t{v\left|\right|}^{3s}+\left|\right|{\mu }_i^{t}w{\left|\right|}^{3s}\right\}\right\}}\hfill \end{array}\right.\hfill \\ \hfill & =\left\{\begin{array}{c}\to 0\mathrm{as}t\to \infty ,\hfill \\ \to 0\mathrm{as}t\to \infty ,\hfill \\ \to 0\mathrm{as}t\to \infty ,\hfill \\ \to 0\mathrm{as}t\to \infty .\hfill \end{array}\right.\hfill \end{array}$$

⟹ It holds for (30), and we have the following.

$$\gamma \left(u\right)=\frac{1}{\mathsf{\Delta }}\left[\mathsf{\Theta }\left(\frac{u}{\mathsf{\Delta }},\frac{u}{\mathsf{\Delta }},\frac{u}{\mathsf{\Delta }}\right)\right].$$

Hence, the following is the case.

$$\begin{array}{c}\hfill \gamma \left(u\right)=\frac{1}{\mathsf{\Delta }}\left[\mathsf{\Theta }\left(\frac{u}{\mathsf{\Delta }},\frac{u}{\mathsf{\Delta }},\frac{u}{\mathsf{\Delta }}\right)\right]=\left\{\begin{array}{c}{\displaystyle \frac{\mathsf{\Lambda }}{\mathsf{\Delta }},}\hfill \\ {\displaystyle \frac{3\mathsf{\Lambda }}{{\mathsf{\Delta }}^s}{\left|\right|u\left|\right|}^s,}\hfill \\ {\displaystyle \frac{\mathsf{\Lambda }}{{\mathsf{\Delta }}^{3s}}{\left|\right|u\left|\right|}^{3s},}\hfill \\ {\displaystyle \frac{4\mathsf{\Lambda }}{{\mathsf{\Delta }}^{3s}}{\left|\right|u\left|\right|}^{3s}.}\hfill \end{array}\right.\end{array}$$

Moreover, we have the following.

$$\begin{array}{cc}\hfill \frac{1}{{\mu }_i}\gamma \left({\mu }_{i}u\right)& =\left\{\begin{array}{c}{\displaystyle \frac{\mathsf{\Lambda }}{{\mu }_i·\mathsf{\Delta }},}\hfill \\ {\displaystyle \frac{3\mathsf{\Lambda }}{{\mu }_i·{\mathsf{\Delta }}^s}\left|\right|{\mu }_{i}u{\left|\right|}^s,}\hfill \\ {\displaystyle \frac{\mathsf{\Lambda }}{{\mu }_i·{\mathsf{\Delta }}^{3s}}\left|\right|{\mu }_{i}u{\left|\right|}^{3s},}\hfill \\ {\displaystyle \frac{\mathsf{\Lambda }}{{\mu }_i·{\mathsf{\Delta }}^{3s}}\left|\right|{\mu }_{i}u{\left|\right|}^{3s}.}\hfill \end{array}\right.\hfill \\ \hfill & =\left\{\begin{array}{c}{\displaystyle {\mu }_i^{-1}\frac{\mathsf{\Lambda }}{},}\hfill \\ {\displaystyle {\mu }_i^{s-1}\frac{3\mathsf{\Lambda }}{{\mathsf{\Delta }}^s}{\left|\right|u\left|\right|}^s,}\hfill \\ {\displaystyle {\mu }_i^{3s-1}\frac{\mathsf{\Lambda }}{{\mathsf{\Delta }}^{3s}}{\left|\right|u\left|\right|}^{3s},}\hfill \\ {\displaystyle {\mu }_i^{3s-1}\frac{4\mathsf{\Lambda }}{{\mathsf{\Delta }}^{3s}}{\left|\right|u\left|\right|}^{3s}.}\hfill \end{array}\right.\hfill \\ \hfill & =\left\{\begin{array}{c}{\displaystyle {\mu }_i^{-1}\gamma \left(u\right),}\hfill \\ {\displaystyle {\mu }_i^{s-1}\gamma \left(u\right),}\hfill \\ {\displaystyle {\mu }_i^{3s-1}\gamma \left(u\right),}\hfill \\ {\displaystyle {\mu }_i^{3s-1}\gamma \left(u\right).}\hfill \end{array}\right.\hfill \end{array}$$

The results are obtained by using (33).

Criteria 1:$L={\mathsf{\Delta }}^{-1}$ if $i=0$.

$$\begin{array}{cc}\hfill ∥h\left(u\right)-A\left(u\right)∥ \le \frac{L^{1-i}}{1-L}\gamma \left(u\right)& =\frac{{\left({\mathsf{\Delta }}^{-1}\right)}^{1-0}}{1-{\left(\mathsf{\Delta }\right)}^{-1}}·\frac{\mathsf{\Lambda }}{\mathsf{\Delta }}\hfill \\ \hfill & =\frac{\mathsf{\Lambda }}{\left(\mathsf{\Delta }-1\right)}.\hfill \end{array}$$

Criteria 2:$L=\mathsf{\Delta }$ if $i=1$.

$$\begin{array}{cc}\hfill ∥h\left(u\right)-A\left(u\right)∥ \le \frac{L^{1-i}}{1-L}\gamma \left(u\right)& =\frac{{\left(\mathsf{\Delta }\right)}^{1-1}}{1-\mathsf{\Delta }}·\frac{\mathsf{\Theta }}{\mathsf{\Delta }}\hfill \\ \hfill & =\frac{\mathsf{\Theta }}{\left(1-\mathsf{\Delta }\right)}.\hfill \end{array}$$

Criteria 1:$L={\mathsf{\Delta }}^{r-1}$ if $i=0$.

$$\begin{array}{cc}\hfill ∥h\left(u\right)-A\left(u\right)∥ \le \frac{L^{1-i}}{1-L}\gamma \left(u\right)& =\frac{{\left({\mathsf{\Delta }}^{r-1}\right)}^{1-0}}{1-{\mathsf{\Delta }}^{r-1}}\frac{3\mathsf{\Lambda }}{{\mathsf{\Delta }}^s}{\left|\right|u\left|\right|}^s\hfill \\ \hfill & =\frac{{\mathsf{\Delta }}^r}{\mathsf{\Delta }-{\mathsf{\Delta }}^r}\frac{3\mathsf{\Lambda }}{{\mathsf{\Delta }}^s}{\left|\right|u\left|\right|}^s\hfill \\ \hfill & =\frac{{3\mathsf{\Lambda }\left|\right|u\left|\right|}^s}{\left(\mathsf{\Delta }-{\mathsf{\Delta }}^s\right)}.\hfill \end{array}$$

Criteria 2:$L=\frac{1}{{\mathsf{\Delta }}^{r-1}}$ if $i=1$.

$$\begin{array}{cc}\hfill ∥h\left(u\right)-A\left(u\right)∥ \le \frac{L^{1-i}}{1-L}\gamma \left(u\right)& =\frac{{\left(\frac{1}{{\mathsf{\Delta }}^{r-1}}\right)}^{1-1}}{1-\frac{1}{{\mathsf{\Delta }}^{r-1}}}\frac{3\mathsf{\Lambda }}{{\mathsf{\Delta }}^s}{\left|\right|u\left|\right|}^s\hfill \\ \hfill & =\frac{{\mathsf{\Delta }}^s}{{\mathsf{\Delta }}^s-\mathsf{\Delta }}\frac{3\mathsf{\Lambda }}{{\mathsf{\Delta }}^s}{\left|\right|u\left|\right|}^s\hfill \\ \hfill & =\frac{{3\mathsf{\Lambda }\left|\right|u\left|\right|}^s}{\left({\mathsf{\Delta }}^s-\mathsf{\Delta }\right)}.\hfill \end{array}$$

Criteria 1:$L={\mathsf{\Delta }}^{3s-1}$ if $i=0$.

$$\begin{array}{cc}\hfill ∥h\left(u\right)-A\left(u\right)∥ \le \frac{L^{1-i}}{1-L}\gamma \left(u\right)& =\frac{{\left({\mathsf{\Delta }}^{3s-1}\right)}^{1-0}}{1-{\mathsf{\Delta }}^{3s-1}}\frac{\mathsf{\Lambda }}{{\mathsf{\Delta }}^{3s}}{\left|\right|u\left|\right|}^{3s}\hfill \\ \hfill & =\frac{{\mathsf{\Delta }}^{3s}}{\mathsf{\Delta }-{\mathsf{\Delta }}^{3s}}\frac{\mathsf{\Lambda }}{{\mathsf{\Delta }}^{3s}}{\left|\right|u\left|\right|}^{3s}\hfill \\ \hfill & =\frac{{\mathsf{\Lambda }\left|\right|u\left|\right|}^{3s}}{\left(\mathsf{\Delta }-{\mathsf{\Delta }}^{3s}\right)}.\hfill \end{array}$$

Criteria 2:$L=\frac{1}{{\mathsf{\Delta }}^{3s-1}}$ if $i=1$.

$$\begin{array}{cc}\hfill ∥h\left(u\right)-A\left(u\right)∥ \le \frac{L^{1-i}}{1-L}\gamma \left(u\right)& =\frac{{\left(\frac{1}{{\mathsf{\Delta }}^{3s-1}}\right)}^{1-1}}{1-\frac{1}{{\mathsf{\Delta }}^{3s-1}}}\frac{\mathsf{\Lambda }}{{\mathsf{\Delta }}^{3s}}{\left|\right|u\left|\right|}^{3s}\hfill \\ \hfill & =\frac{{\mathsf{\Delta }}^{3s}}{{\mathsf{\Delta }}^{3s}-\mathsf{\Delta }}\frac{\mathsf{\Lambda }}{{\mathsf{\Delta }}^{3s}}{\left|\right|u\left|\right|}^{3s}\hfill \\ \hfill & =\frac{{\mathsf{\Lambda }\left|\right|u\left|\right|}^{3s}}{\left({\mathsf{\Delta }}^{3s}-\mathsf{\Delta }\right)}.\hfill \end{array}$$

Criteria 1:$L={\mathsf{\Delta }}^{3s-1}$ if $i=0$.

$$\begin{array}{cc}\hfill ∥h\left(u\right)-A\left(u\right)∥\le \frac{L^{1-i}}{1-L}\gamma \left(u\right)& =\frac{{\left({\mathsf{\Delta }}^{3s-1}\right)}^{1-0}}{1-{\mathsf{\Delta }}^{3s-1}}\frac{4\mathsf{\Lambda }}{{\mathsf{\Delta }}^{3s}}{\left|\right|u\left|\right|}^{3s}\hfill \\ \hfill & =\frac{{\mathsf{\Delta }}^{3s}}{\mathsf{\Delta }-{\mathsf{\Delta }}^{3s}}\frac{4\mathsf{\Lambda }}{{\mathsf{\Delta }}^{3s}}{\left|\right|u\left|\right|}^{3s}\hfill \\ \hfill & =\frac{{4\mathsf{\Lambda }\left|\right|u\left|\right|}^{3s}}{\left(\mathsf{\Delta }-{\mathsf{\Delta }}^{3s}\right)}.\hfill \end{array}$$

Criteria 2:$L=\frac{1}{{\mathsf{\Delta }}^{3s-1}}$ if $i=1$.

$$\begin{array}{cc}\hfill ∥h\left(u\right)-A\left(u\right)∥ \le \frac{L^{1-i}}{1-L}\gamma \left(u\right)& =\frac{{\left(\frac{1}{{\mathsf{\Delta }}^{3s-1}}\right)}^{1-1}}{1-\frac{1}{{\mathsf{\Delta }}^{3s-1}}}\frac{4\mathsf{\Lambda }}{{\mathsf{\Delta }}^{3s}}{\left|\right|u\left|\right|}^{3s}\hfill \\ \hfill & =\frac{{\mathsf{\Delta }}^{3s}}{{\mathsf{\Delta }}^{3s}-\mathsf{\Delta }}\frac{4\mathsf{\Lambda }}{{\mathsf{\Delta }}^{3s}}{\left|\right|u\left|\right|}^{3s}\hfill \\ \hfill & =\frac{{4\mathsf{\Lambda }\left|\right|u\left|\right|}^{3s}}{\left({\mathsf{\Delta }}^{3s}-\mathsf{\Delta }\right)}.\hfill \end{array}$$

□

5. Applications

The matrix of the quadratic form $2x^2+2y^2+2z^2+2xy$ is stated as follows: $A=\left(\begin{array}{ccc}2& 1& 0\\ 1& 2& 0\\ 0& 0& 2\end{array}\right)$ and the characteristic equation of the matrix A is ${\lambda }^3-6{\lambda }^2+11\lambda -6=0$ and the eigenvalues are $\lambda =1,2,3$. By using this characteristic polynomial, we frame the new form of additive functional equation.

$$\begin{array}{cc}\hfill & h\left[\left({\delta }^3+11\delta \right)u-6\left({\delta }^2+1\right)v\right]+h\left[\left(11\delta -6{\delta }^2\right)v+\left({\delta }^3-6\right)w\right]\hfill \\ \hfill & -h\left[\left({\delta }^3-6{\delta }^2\right)w+\left(11\delta -6\right)u\right]=\left({\delta }^3+6\right)h\left(u\right)\hfill \\ \hfill & +\left(11\delta -12{\delta }^2-6\right)h\left(v\right)+\left(6{\delta }^2-6\right)h\left(w\right)\hfill \end{array}$$

The solution of the above functional equation is $h\left(x\right)=x$. Therefore, the functional equation is related with eigenvalue and eigen vectors.

6. Conclusions

We demonstrated the generalised Hyers–Ulam stability results of the new symmetric additive functional equation in Banach spaces using the direct and fixed point technique. We also presented a number of findings that can be derived by using the appropriate mapping $\mathsf{\Theta }$, such as the sum or multiplication of the power of norms. Furthermore, our functional equation is derived from a characteristic polynomial of degree three and is related to the matrix theory.