A Generalized Bi-Quadratic–Drygas Functional System in Non-Archimedean Normed Spaces over p-Adic Numbers

Abstract

This work investigates the solution and the stability of a generalized system of bi-quadratic–Drygas functional equations in non-Archimedean normed spaces with unknown coefficients. The presence of asymmetric coefficients and reflection terms induces nontrivial coupling effects and symmetry-breaking phenomena, while simultaneously capturing additive, quadratic, and mixed additive–quadratic behaviors. An examination of the coefficients shows that the trivial solution is the only one satisfying the system of equations in asymmetric parameter configurations. For symmetric configurations, by exploiting the ultrametric structure of non-Archimedean norms and applying an iterative method combined with symmetry-based decomposition into even and odd parts, we establish the existence and uniqueness of an exact solution approximating a given mapping. Several known stability results for bi-Drygas functional equations are recovered with improvement as special cases.

1. Introduction

The study of the stability of functional equations originated with a question posed by Ulam about the stability of homomorphisms between algebraic structures. This question was first answered affirmatively by Hyers [1] in 1941, who established what is now known as Hyers–Ulam stability for additive functional equations. Subsequently, Rassias [2] introduced a significant generalization by allowing the stability control to depend on a power of the norm, leading to the well-known Hyers–Ulam–Rassias stability framework. Since then, the stability theory of functional equations has developed into a substantial research area with deep connections to analysis, algebra, and applied mathematics.

A fundamental structural insight was provided by Forti [3], who established a systematic framework for the Hyers–Ulam stability of functional equations in several variables. His work emphasized that stability phenomena in product spaces cannot, in general, be reduced to one-dimensional cases and that coupling effects among variables may fundamentally alter stability behavior. This perspective is particularly relevant when dealing with systems of functional equations rather than single identities. Jung [4] later developed a comprehensive framework for the Hyers–Ulam–Rassias stability of nonlinear functional equations, highlighting the role of iterative constructions and decomposition techniques in passing from functional inequalities to exact solutions.

Among various classes of functional equations, quadratic functional equations play a central role due to their close relationship with quadratic mappings and inner product structures. The classical quadratic functional equation

$$f(x+y)+f(x-y)=2f\left(x\right)+2f\left(y\right)$$

(1)

has been extensively studied from both exact and stability perspectives in normed spaces, Banach spaces, and non-Archimedean settings. The stability problem for (1) was first investigated by Škof [5], who showed that approximate quadratic mappings are close to exact quadratic mappings in Banach spaces. Czerwik [6] subsequently established global Hyers–Ulam stability via a direct iterative method, which became a standard tool in later developments. Numerous generalizations have since been proposed by modifying coefficients, introducing control functions, or extending domains to higher-dimensional and product spaces. In this context, Rassias and Šemrl [7] demonstrated that Hyers–Ulam stability does not hold universally and highlighted the decisive role of control functions in determining whether approximate solutions admit nearby exact solutions.

A notable extension of the quadratic equation is the Drygas functional equation

$$f(x+y)+f(x-y)=2f\left(x\right)+f\left(y\right)+f(-y),$$

which combines additive and quadratic behaviors into a single functional identity and thus provides a unified framework for studying mixed structural properties [8,9]. Drygas [8] introduced this equation in the context of quasi-inner products, providing a conceptual foundation for mixed-type functional equations. Cholewa [9] further emphasized that the stability of functional equations depends essentially on their structural properties, including symmetry and reflection, rather than solely on the validity of functional inequalities. Stability results for Drygas-type functional equations have been established under various assumptions on control functions and underlying spaces, revealing rich decomposition phenomena and structural constraints [10,11].

More recently, mixed additive–quadratic and additive–Drygas functional equations have been investigated to clarify the role of parity and symmetry. In particular, Choi and Lee [12] employed a Baire category approach to study the Hyers–Ulam stability of the mixed equations

$$\begin{array}{cc}\hfill 2f(x+y)+f(x-y)-3f\left(x\right)-3f\left(y\right)& =0,\hfill \\ \hfill 2f(x+y)+f(x-y)-3f\left(x\right)-2f\left(y\right)-f(-y)& =0,\hfill \end{array}$$

and showed that the odd and even parts of approximate solutions converge to additive and quadratic components, respectively, highlighting the essential role of reflection terms.

Parallel to these developments, increasing attention has been paid to multi-variable functional equations and systems of functional inequalities, where coupling among variables produces stability phenomena that cannot be reduced to one-dimensional arguments [13,14,15]. In this direction, Park, Bae, and their collaborators [16,17,18] initiated and developed stability results for multi-variable quadratic and cubic functional equations—motivated in part by structures related to elliptic-curve-type equations—within Banach and 2-Banach spaces, using decomposition and iterative techniques. Recent refinements further demonstrated that stability results for Drygas-type functional inequalities are highly sensitive to the choice of control functions and symmetry assumptions and that inappropriate parameter settings may lead to invalid conclusions [18,19].

Within the bi-variable setting, Park and Bae [20] investigated the solution and stability of a bi-quadratic functional equation in the form

$$\begin{array}{cc}\hfill f(x+y,z+w)& +f(x+y,z-w)+f(x-y,z+w)+f(x-y,z-w)\hfill \\ \hfill & =4\left[f(x,z)+f(x,w)+f(y,z)+f(y,w)\right],\hfill \end{array}$$

which provides a foundational model for bi-variable quadratic structures. In 2020, El-Fassi, El-Hady, and Nikodom [21] investigated set-valued solutions of a generalized bi-quadratic functional equation in the form

$$\begin{array}{cc}\hfill & f(x+y,z+w)+f(x+y,z-w)+f(x-y,z+w)+f(x-y,z-w)\hfill \\ \hfill & =af(x,z)+bf(x,w)+cf(y,z)+df(y,w),\hfill \end{array}$$

where $a,b,c$, and d are constants. More recently, Dehghanian, Izadi, and Sayyari [22] introduced and solved a system of bi-Drygas functional equations

$$\left\{\begin{array}{c}f(x+y,z)+f(x-y,z)=2f(x,z)+f(y,z)+f(-y,-z),\hfill \\ f(x,y+z)+f(x,y-z)=2f(x,y)+f(x,z)+f(-x,-z),\hfill \end{array}\right.$$

for which $x,y,z\in X$, and investigated the Hyers–Ulam stability of this system. In 2024, Sayyari, Dehghanian, and Park [23] further investigated the solutions and stability of a Pexider system of bi-additive and bi-quadratic functional equations

$$\left\{\begin{array}{c}f(x+y,z)+f(x-y,z)+h(x+y,z)=2f(x,z)+2f(y,z)+h(x,z)+h(y,z),\hfill \\ f(x,y+z)+f(x,y-z)+h(x,y+z)=2f(x,z)+2f(y,z)+h(x,z)+h(y,z),\hfill \end{array}\right.$$

highlighting the continuing development of bi-variable stability theory. Multi-quadratic mappings, which are quadratic in each coordinate variable, provide a natural multi-variable extension of quadratic functional equations and can be characterized by a single functional equation involving sign vectors ${\{-1,1\}}^n$. Their Hyers–Ulam stability has been studied using fixed-point methods in both normed and non-Archimedean spaces [24,25].

Motivated by these observations, the present work investigates a solution for and the stability of a generalized system of bi-quadratic–Drygas functional equations in the form

$$\left\{\begin{array}{c}f(x+y,z)+f(x-y,z)=af(x,z)+bf(y,z)+cf(-y,-z)\hfill \\ f(x,y+z)+f(x,y-z)=af(x,y)+bf(x,z)+cf(-x,-z),\hfill \end{array}\right.$$

(2)

where $a,b,c\in \mathbb{R}$ are fixed constants. This system constitutes a natural bi-variable generalization of quadratic and Drygas functional equations, enriched by asymmetric coefficients, coupling effects, and reflection terms.

The aim of this work is to establish existence, uniqueness, and stability results for the above system under appropriate assumptions of the control functions. Moreover, we provide solutions in solvable cases. By developing an iterative approach and exploiting symmetry-based structural properties of bi-quadratic and bi-additive mappings, we prove that approximate solutions are close to exact solutions, satisfying the associated functional equations. Several known stability results for bi-Drygas functional equations are recovered with improvement as special cases of the main theorems, demonstrating the generality and unifying nature of the proposed framework.

2. Preliminaries

We summarize the terminology and preliminary results in this section. Let S and G be abelian groups. Recall the definition of a bi-additive and a bi-quadratic mapping.

Definition 1.

A function $f:S^2\to G$ is said to be bi-additive if for any $x,y,z\in S$,

$$\left\{\begin{array}{c}f(x+y,z)=f(x,z)+f(y,z)\hfill \\ f(x,y+z)=f(x,y)+f(x,z).\hfill \end{array}\right.$$

Definition 2.

A function $f:S^2\to G$ is said to be bi-quadratic if for any $x,y,z\in S$,

$$\left\{\begin{array}{c}f(x+y,z)+f(x-y,z)=2f(x,z)+2f(y,z)\hfill \\ f(x,y+z)+f(x,y-z)=2f(x,y)+2f(x,z).\hfill \end{array}\right.$$

We denote ${\mathbb{N}}_0=\mathbb{N}\cup \left\{0\right\}$ and ${\left[k\right]}_0=\{0,1,\dots ,k\}$ for each $k\in {\mathbb{N}}_0$.

Lemma 1.

If f is bi-additive, then $f(kx,z)=kf(x,z)$ and $f(x,kz)=kf(x,z)$ for all $k\in {\mathbb{N}}_0$ and $x,z\in S$.

Proof. 

It is trivial by using mathematical induction. □

Lemma 2.

If f is bi-quadratic, then $f(kx,z)=k^{2}f(x,z)$ and $f(x,kz)=k^{2}f(x,z)$ for all $k\in {\mathbb{N}}_0$ and $x,z\in S$.

Proof. 

Let $x,z\in S$. Since f is bi-quadratic, $f(0,z)=0$ and $f(x,0)=0$. We show that $f(kx,z)=k^{2}f(x,z)$ by induction. The statement clearly holds for $k=0,1$. Moreover, for $k\ge 1$,

$$\begin{array}{cc}\hfill f\left(\right(k+1)x,z)& =2f(kx,z)+2f(x,z)-f\left(\right(k-1)x,z)\hfill \\ \hfill & =2k^{2}f(x,z)+2f(x,z)-{(k-1)}^{2}f(x,z)\hfill \\ \hfill & ={(k+1)}^{2}f(x,z).\hfill \end{array}$$

Similarly, it can be proved that $f(x,kz)=k^{2}f(x,z)$ for all $k\in {\mathbb{N}}_0$. □

Definition 3.

Let $f:S^2\to G$ be a function. We say that f is even if $f(-x,-y)=f(x,y)$ for all $x,y\in S$. We say that f is odd if $f(-x,-y)=-f(x,y)$ for all $x,y\in S$.

For any function $f:S^2\to G$, define

$$f_e(x,y)={\displaystyle \frac{f(x,y)+f(-x,-y)}{2}}$$

and

$$f_o(x,y)={\displaystyle \frac{f(x,y)-f(-x,-y)}{2}}$$

for all $x,y\in S$.

Definition 4.

A vector space V over a field $\mathbb{F}$ is said to be a non-Archimedean normed space if it is equipped with a map $\parallel ·\parallel :V\to [0,\infty )$, such that

1. 

$\parallel x\parallel =0$ if and only if $x=0$;

2. 

$\parallel \alpha x\parallel =\left|\alpha \right|\parallel x\parallel$;

3. 

$\parallel x+y\parallel \le max\{\parallel x\parallel ,\parallel y\parallel \}$ (ultrametric inequality);

for all $x,y\in V$ and $\alpha \in \mathbb{F}$.

Definition 5.

Let p be a prime number. Define the p-adic norm ${|·|}_p$ on $\mathbb{Q}$ by

$${\left|p^n{\displaystyle \frac{a}{b}}\right|}_p={\displaystyle \frac{1}{p^n}}$$

for any $a,b\in \mathbb{N}$, such that $p\nmid a$ and $p\nmid b$ and ${\left|0\right|}_p=0$.

Let ${\mathbb{Q}}_p$ denote the completion of $(\mathbb{Q},|·{|}_p)$.

Proposition 1.

The field ${\mathbb{Q}}_p$ is a non-Archimedean normed space. Moreover, any normed space over ${\mathbb{Q}}_p$ is non-Archimedean.

3. Main Results

Let S be an abelian group, X a normed space, and $f:S^2\to X$ a function. Consider the generalized bi-quadratic–Drygas functional system (2) where $a,b$ and c are constants with $a+b+c\ne 2$. We will investigate solutions to this system by examining the conditions on the constants a, b, and c.

Theorem 1.

Suppose that $a\ne 2-b-c$ and $a\ne 2$. Let $f:S^2\to X$ be a function satisfying (2). Then $f(x,y)=0$ for all $x,y\in S$.

Proof. 

By setting $x=y=z=0$ in (2), we obtain $2f(0,0)=(a+b+c)f(0,0)$, which implies that $f(0,0)=0$ since $a\ne 2-b-c$. Then letting $y=z=0$ in the first and $x=z=0$ in second equations of (2), we have

$$\left\{\begin{array}{c}2f(x,0)=af(x,0)\hfill \\ 2f(0,y)=af(0,y).\hfill \end{array}\right.$$

Since $a\ne 2$, we must have $f(x,0)=f(0,y)=0$ for all $x,y\in S$. Finally, using $y=0$ in the first equation of (2), we have

$$2f(x,z)=af(x,z)$$

which implies that $f(x,y)=0$ for all $x,y\in S$. □

The remaining case is when $a\ne 2-b-c$ and $a=2$; that is, $b\ne -c$. We consider two subcases depending on whether $b=c$, starting with $b\ne c$.

Theorem 2.

Suppose that $b\ne \pm c$. Let $f:S^2\to X$ be a function satisfying

$$\left\{\begin{array}{c}f(x+y,z)+f(x-y,z)=2f(x,z)+bf(y,z)+cf(-y,-z)\hfill \\ f(x,y+z)+f(x,y-z)=2f(x,y)+bf(x,z)+cf(-x,-z).\hfill \end{array}\right.$$

(3)

1. 

If $b\ne 2-c$, then $f(x,y)=0$.

2. 

If $b=2-c$, then the system of (3) becomes the system of bi-quadratic functional equations.

Proof. 

Suppose that $b\ne \pm c$. Setting $y=0$ and $z=0$ in the first and second equations of (3) respectively, we have

$$\left\{\begin{array}{c}0=bf(0,z)+cf(0,-z)\hfill \\ 0=bf(x,0)+cf(-x,0).\hfill \end{array}\right.$$

(4)

Then, replacing z with $-z$ and x with $-x$ in (4), we have

$$\left\{\begin{array}{c}0=bf(0,-z)+cf(0,z)\hfill \\ 0=bf(-x,0)+cf(x,0).\hfill \end{array}\right.$$

Adding the above systems obtains

$$\left\{\begin{array}{c}0=(b+c)\left[f\right(0,-z)+f(0,z\left)\right]\hfill \\ 0=(b+c)\left[f\right(-x,0)+f(x,0\left)\right].\hfill \end{array}\right.$$

Since $b\ne -c$, $f(0,-z)=-f(0,z)$ and $f(-x,0)=-f(x,0)$ for all $x,z\in S$. Hence, according to (4),

$$\left\{\begin{array}{c}0=(b-c)f(0,z)\hfill \\ 0=(b-c)f(x,0)\hfill \end{array}\right.$$

which implies that $f(x,0)=f(0,z)=0$ for all $x,z\in S$ (since $b\ne c$). Letting $x=0$ and replacing y with x in the first and letting $y=0$ in the second equations of (3), we have

$$\left\{\begin{array}{c}f(x,z)+f(-x,z)=bf(x,z)+cf(-x,-z)\hfill \\ f(x,z)+f(x,-z)=bf(x,z)+cf(-x,-z).\hfill \end{array}\right.$$

(5)

Hence, $f(-x,z)=f(x,-z)$. According to the first equation of (5), for all $x,z\in S$,

$$f(-x,z)=(b+c-1)f(x,z).$$

Thus,

$$f(x,z)=(b+c-1)f(-x,z)={(b+c-1)}^{2}f(x,z).$$

If $b\ne 2-c$, then $f(x,z)=0$. Assume that $b=2-c$. Then System (3) becomes

$$\left\{\begin{array}{c}f(x+y,z)+f(x-y,z)=2f(x,z)+bf(y,z)+(2-b)f(-y,-z)\hfill \\ f(x,y+z)+f(x,y-z)=2f(x,y)+bf(x,z)+(2-b)f(-x,-z);\hfill \end{array}\right.$$

that is,

$$\left\{\begin{array}{c}f(x+y,z)+f(x-y,z)=2f(x,z)+2f(y,z)\hfill \\ f(x,y+z)+f(x,y-z)=2f(x,y)+2f(x,z)\hfill \end{array}\right.$$

which is the system of bi-quadratic functional equations. □

Now, the case when $b\ne -c$ and $b=c$ must be considered. This particularly implies that $b\ne 0$ and System (3) becomes

$$\left\{\begin{array}{c}f(x+y,z)+f(x-y,z)=2f(x,z)+bf(y,z)+bf(-y,-z)\hfill \\ f(x,y+z)+f(x,y-z)=2f(x,y)+bf(x,z)+bf(-x,-z).\hfill \end{array}\right.$$

(6)

Note that, if $b=1$, the system becomes the system of bi-Drygas functional equations considered in [22]. So we investigate the case where $b=1$ and $b\ne 0,1$.

To solve (6), we will obtain stability theorems for this system in the setting of a non-Archimedean normed space. Let p be prime, S a $2p$-divisible abelian group, and X a normed space over the field ${\mathbb{Q}}_p$ with the completion $\overline{X}$. Consider any function $\phi ,\psi :S^3\to [0,\infty )$. First, we obtain two preparatory theorems for the stability. For each $k\in \mathbb{N}$, we define

$$\begin{array}{cc}\hfill {\phi }_{max}(x,y,z)& =max\left\{\phi \right(\alpha x,\alpha y,\gamma z):\alpha ,\gamma \in \{-1,0,1\left\}\right\},\hfill \\ \hfill {\psi }_{max}(x,y,z)& =max\left\{\psi \right(\alpha x,\gamma y,\gamma z):\alpha ,\gamma \in \{-1,0,1\left\}\right\},\hfill \\ \hfill {\phi }_k(x,z)& =max\{\phi (nx,\alpha x,\gamma z):|n|\in {[k-1]}_0,\alpha ,\gamma \in \{-1,0,1\}\},\hfill \\ \hfill {\psi }_k(x,z)& =max\{\psi (\alpha x,nz,\gamma z):|n|\in {[k-1]}_0,\alpha ,\gamma \in \{-1,0,1\}\}.\hfill \end{array}$$

Denote

$${\Phi }_A(x,z)=\underset{n\in {\mathbb{N}}_0}{sup}{\displaystyle \frac{1}{p^n}}{\phi }_p\left({\displaystyle \frac{x}{p^{n+1}}},z\right)\mathrm{and}{\Psi }_A(x,z)=\underset{n\in {\mathbb{N}}_0}{sup}{\displaystyle \frac{1}{p^n}}{\psi }_p\left(x,{\displaystyle \frac{z}{p^{n+1}}}\right).$$

Consider the following condition:

$$\left\{\begin{array}{l}\underset{n\to \infty }{lim}\frac{1}{p^n}\left[{\phi }_{max}\left(\frac{x}{p^n},\frac{y}{p^n},z\right)+{\phi }_{max}\left(x,y,\frac{z}{p^n}\right)\right]=0,\hfill \\ \underset{n\to \infty }{lim}\frac{1}{p^n}\left[{\psi }_{max}\left(\frac{x}{p^n},y,z\right)+{\psi }_{max}\left(x,\frac{y}{p^n},\frac{z}{p^n}\right)\right]=0,\\ \underset{n\to \infty }{lim}\frac{1}{p^n}{\phi }_p\left(\frac{x}{p^{n+1}},z\right)=0,\\ \underset{n\to \infty }{lim}\frac{1}{p^n}{\psi }_p\left(x,\frac{z}{p^{n+1}}\right)=0.\end{array}\right.$$

(7)

Theorem 3.

Suppose that Condition (7) holds and a function $f:S^2\to X$ satisfies

$$\left\{\begin{array}{c}\parallel f(x+y,z)+f(x-y,z)-2f(x,z)\parallel \le {\phi }_{max}(x,y,z)\hfill \\ \parallel f(x,y+z)+f(x,y-z)-2f(x,y)\parallel \le {\psi }_{max}(x,y,z).\hfill \end{array}\right.$$

(8)

with $f(x,0)=f(0,z)=0$ for all $x,z\in S$. Then there are two bi-additive mappings, $A_1,A_2:S^2\to \overline{X}$, such that

$$\left\{\begin{array}{c}\parallel f(x,z)-A_1(x,z)\parallel \le {\Phi }_A(x,z)\hfill \\ \parallel f(x,z)-A_2(x,z)\parallel \le {\Psi }_A(x,z).\hfill \end{array}\right.$$

Moreover, if there are bi-additive mappings, $A_1^{\prime },A_2^{\prime }:S^2\to \overline{X}$, such that

$$\left\{\begin{array}{c}\parallel f(x,z)-A_1^{\prime }(x,z)\parallel \le M_1{\Phi }_A(x,z)\hfill \\ \parallel f(x,z)-A_2^{\prime }(x,z)\parallel \le M_2{\Psi }_A(x,z)\hfill \end{array}\right.$$

for some $M_1,M_2>0$, then $A_1^{\prime }=A_1$ and $A_2^{\prime }=A_2$.

Proof. 

We first prove by induction that $\parallel f(kx,z)-kf(x,z)\parallel \le {\phi }_k(x,z)$ for all $k\in {\mathbb{N}}_0$, which is clear for $k=0,1$. For $k\ge 1$, according to the first equation of (8) and the induction hypothesis, we have

$$\begin{array}{cc}\hfill \parallel f\left(\right(k+1)x,z)-& (k+1)f(x,z)\parallel \hfill \\ \hfill & =\parallel \left[f\right((k+1)x,z)+f((k-1)x,z)-2f(kx,z\left)\right]\hfill \\ \hfill & -\left[f\right((k-1)x,z)-(k-1\left)f\right(x,z\left)\right]+2\left[f\right(kx,z)-kf(x,z\left)\right]\parallel \hfill \\ \hfill & \le max\{{\phi }_{max}(kx,x,z),{\phi }_{k-1}(x,z),|2{|}_p{\phi }_k(x,z)\}\hfill \\ \hfill & \le {\phi }_{k+1}(x,z),\hfill \end{array}$$

which proves the claim. Hence, for any $m\in \mathbb{N}$ and $n\in {\mathbb{N}}_0$,

$$\begin{array}{cc}\hfill \parallel p^n& f\left({\displaystyle \frac{x}{p^n}},z\right)-p^{n+m}\left({\displaystyle \frac{x}{p^{n+m}}},z\right)\parallel ={\left|p^n\right|}_p∥f\left({\displaystyle \frac{x}{p^n}},z\right)-p^m\left({\displaystyle \frac{x}{p^{n+m}}},z\right)∥\hfill \\ \hfill & ={\displaystyle \frac{1}{p^n}}∥\left[f\left({\displaystyle \frac{x}{p^n}},z\right)-pf\left({\displaystyle \frac{x}{p^{n+1}}},z\right)\right]+\dots +p^{m-1}\left[f\left({\displaystyle \frac{x}{p^{n+m-1}}},z\right)-pf\left({\displaystyle \frac{x}{p^{n+m}}},z\right)\right]∥\hfill \\ \hfill & \le {\displaystyle \frac{1}{p^n}}\underset{0\le i\le m-1}{max}\left\{{\displaystyle \frac{1}{p^i}}{\phi }_p\left({\displaystyle \frac{x}{p^{n+i+1}}},z\right)\right\}\hfill \\ \hfill & =\underset{0\le i\le m-1}{max}\left\{{\displaystyle \frac{1}{p^{n+i}}}{\phi }_p\left({\displaystyle \frac{x}{p^{n+i+1}}},z\right)\right\}.\hfill \end{array}$$

(9)

According to the assumption, the last term converges to 0 as $n\to \infty$, implying that $\left(p^{n}f\left({\displaystyle \frac{x}{p^n}},z\right)\right)$ is a Cauchy sequence for each $(x,z)\in S^2$. Define $A_1:(x,z)\mapsto {lim}_{n\to \infty }{p}^{n}f\left({\displaystyle \frac{x}{p^n}},z\right)$. Then,

$$∥p^{n}f\left({\displaystyle \frac{x+y}{p^n}},z\right)-p^{n}f\left({\displaystyle \frac{x}{p^n}},z\right)-p^{n}f\left({\displaystyle \frac{y}{p^n}},z\right)∥\le {\displaystyle \frac{1}{p^n}}{\phi }_{max}\left({\displaystyle \frac{x}{p^n}},{\displaystyle \frac{y}{p^n}},z\right)$$

and

$$∥p^{n}f\left({\displaystyle \frac{x}{p^n}},y+z\right)-p^{n}f\left({\displaystyle \frac{x}{p^n}},y\right)-p^{n}f\left({\displaystyle \frac{x}{p^n}},z\right)∥\le {\displaystyle \frac{1}{p^n}}{\psi }_{max}\left({\displaystyle \frac{x}{p^n}},y,z\right).$$

Taking $n\to \infty$ in the two inequalities above, we conclude, according to the assumption, that $A_1$ is bi-additive. Moreover, according to (9),

$$∥f(x,z)-A_1(x,z)∥=∥f(x,z)-\underset{n\to \infty }{lim}{p}^{n}f\left({\displaystyle \frac{x}{p^n}},z\right)∥\le {\Phi }_A(x,z).$$

For the uniqueness, according to Lemma 1, observe that for each $m\in \mathbb{N}$,

$$\begin{array}{cc}\hfill \parallel A_1(x,z)-A_1^{\prime }(x,z)\parallel & ={\left|p^m\right|}_p∥A_1\left({\displaystyle \frac{x}{p^m}},z\right)-A_1^{\prime }\left({\displaystyle \frac{x}{p^m}},z\right)∥\hfill \\ \hfill & \le {\displaystyle \frac{1}{p^m}}max\left\{∥f\left({\displaystyle \frac{x}{p^m}},z\right)-A_1\left({\displaystyle \frac{x}{p^m}},z\right)∥,∥f\left({\displaystyle \frac{x}{p^m}},z\right)-A_1^{\prime }\left({\displaystyle \frac{x}{p^m}},z\right)∥\right\}\hfill \\ \hfill & \le {\displaystyle \frac{1}{p^m}}max\left\{{\Phi }_A\left({\displaystyle \frac{x}{p^m}},z\right),M_1{\Phi }_A\left({\displaystyle \frac{x}{p^m}},z\right)\right\}\hfill \\ \hfill & \le {\displaystyle \frac{1}{p^m}}(M_1+1){\Phi }_A\left({\displaystyle \frac{x}{p^m}},z\right)\hfill \\ \hfill & =(M_1+1)\underset{n\ge m}{sup}{\displaystyle \frac{1}{p^n}}{\phi }_p\left({\displaystyle \frac{x}{p^{n+1}}},z\right).\hfill \end{array}$$

Since

$$\underset{m\to \infty }{lim\; sup}{\displaystyle \frac{1}{p^m}}{\phi }_p\left({\displaystyle \frac{x}{p^{m+1}}},z\right)=\underset{n\to \infty }{lim}{\displaystyle \frac{1}{p^n}}{\phi }_p\left({\displaystyle \frac{x}{p^{n+1}}},z\right)=0,$$

we have $A_1=A_1^{\prime }$.

For the remaining part, one can show that $\left(p^{n}f\left(x,{\displaystyle \frac{z}{p^n}}\right)\right)$ is a Cauchy sequence for each $(x,z)\in S^2$; then we can define $A_2:(x,z)\mapsto {lim}_{n\to \infty }{p}^{n}f\left(x,{\displaystyle \frac{z}{p^n}}\right)$, which is bi-additive. The details are similar to the first part of the proof. □

Remark 1.

This theorem provides the stability theorem for System (6) when $b=0$.

Corollary 1.

Let ${\epsilon }_1,{\epsilon }_2>0$. Suppose that a function $f:S^2\to X$ satisfies

$$\left\{\begin{array}{c}\parallel f(x+y,z)+f(x-y,z)-2f(x,z)\parallel \le {\epsilon }_1\hfill \\ \parallel f(x,y+z)+f(x,y-z)-2f(x,y)\parallel \le {\epsilon }_2.\hfill \end{array}\right.$$

with $f(x,0)=f(0,z)=0$ for all $x,z\in S$. Then there are two bi-additive mappings, $A_1,A_2:S^2\to \overline{X}$, such that

$$\left\{\begin{array}{c}\parallel f(x,z)-A_1(x,z)\parallel \le {\epsilon }_1\hfill \\ \parallel f(x,z)-A_2(x,z)\parallel \le {\epsilon }_2\hfill \end{array}\right.$$

which are unique regardless of a constant multiple on ${\epsilon }_1$ and ${\epsilon }_2$.

Proof. 

By taking ${\phi }_{max}={\epsilon }_1$ and ${\psi }_{max}={\epsilon }_2$, Condition (7) holds, ${\Phi }_A={\epsilon }_1$, and ${\Psi }_A={\epsilon }_2$. Hence, the result follows immediately from Theorem 3. □

For the next theorem, denote

$${\Phi }_Q(x,z)=\underset{n\in {\mathbb{N}}_0}{sup}{\displaystyle \frac{1}{p^{2n}}}{\phi }_p\left({\displaystyle \frac{x}{p^{n+1}}},z\right)\mathrm{and}{\Psi }_Q(x,z)=\underset{n\in {\mathbb{N}}_0}{sup}{\displaystyle \frac{1}{p^{2n}}}{\psi }_p\left(x,{\displaystyle \frac{z}{p^{n+1}}}\right)$$

and consider the condition

$${\displaystyle {\displaystyle \left\{{\displaystyle \begin{array}{l}{\displaystyle \underset{n\to \infty }{lim}\frac{1}{p^{2n}}\left[{\phi }_{max}\left(\frac{x}{p^n},\frac{y}{p^n},z\right)+{\phi }_{max}\left(x,y,\frac{z}{p^n}\right)\right]=0,}\hfill \\ \underset{n\to \infty }{lim}\frac{1}{p^{2n}}\left[{\psi }_{max}\left(\frac{x}{p^n},y,z\right)+{\psi }_{max}\left(x,\frac{y}{p^n},\frac{z}{p^n}\right)\right]=0,\\ \underset{n\to \infty }{lim}\frac{1}{p^{2n}}{\phi }_p\left(\frac{x}{p^{n+1}},z\right)=0,\\ \underset{n\to \infty }{lim}\frac{1}{p^{2n}}{\psi }_p\left(x,\frac{z}{p^{n+1}}\right)=0.\end{array}}\right.}}$$

(10)

Theorem 4.

Suppose that Condition (10) holds and a function $f:S^2\to X$ satisfies

$$\left\{\begin{array}{c}\parallel f(x+y,z)+f(x-y,z)-2f(x,z)-2f(y,z)\parallel \le {\phi }_{max}(x,y,z)\hfill \\ \parallel f(x,y+z)+f(x,y-z)-2f(x,y)-2f(x,z)\parallel \le {\psi }_{max}(x,y,z).\hfill \end{array}\right.$$

(11)

with $f(x,0)=f(0,z)=0$ for all $x,z\in S$. Then there are two bi-quadratic mappings, $Q_1,Q_2:S^2\to \overline{X}$, such that

$$\left\{\begin{array}{c}\parallel f(x,z)-Q_1(x,z)\parallel \le {\Phi }_Q(x,z)\hfill \\ \parallel f(x,z)-Q_2(x,z)\parallel \le {\Psi }_Q(x,z)\hfill \end{array}\right.$$

which are unique regardless of a constant multiple on Φ and Ψ.

Proof. 

We first prove by induction that $\parallel f(kx,z)-k^{2}f(x,z)\parallel \le {\phi }_k(x,z)$ for all $k\in {\mathbb{N}}_0$, which is clear for $k=0,1$. For $k\ge 1$, by substituting x with $kx$ and y with x in the first equation of (11), we have

$$\begin{array}{cc}\hfill {\parallel f((k+1)x,z)-(k+1)}^{2}f(x,z)\parallel & =\parallel \left[f\right((k+1)x,z)+f((k-1)x,z)-2f(kx,z)-2f(x,z\left)\right]\hfill \\ & -[f((k-1)x,z)-{(k-1)}^{2}f(x,z)]+2[f(kx,z)-k^{2}f(x,z)]\parallel \hfill \\ & \le max\{{\phi }_{max}(kx,x,z),{\phi }_{k-1}(x,z),|2{|}_p{\phi }_k(x,z)\}\hfill \\ & \le {\phi }_{k+1}(x,z),\hfill \end{array}$$

which proves the claim.

Hence, for any $m\in \mathbb{N}$ and $n\in {\mathbb{N}}_0$,

$$\begin{array}{cc}\hfill \parallel & p^{2n}f\left({\displaystyle \frac{x}{p^n}},z\right)-p^{2n+2m}\left({\displaystyle \frac{x}{p^{n+m}}},z\right)\parallel ={\left|p^{2n}\right|}_p∥f\left({\displaystyle \frac{x}{p^n}},z\right)-p^{2m}\left({\displaystyle \frac{x}{p^{n+m}}},z\right)∥\hfill \\ \hfill & ={\displaystyle \frac{1}{p^{2n}}}∥\left[f\left({\displaystyle \frac{x}{p^n}},z\right)-p^{2}f\left({\displaystyle \frac{x}{p^{n+1}}},z\right)\right]+\cdots +p^{2m-2}\left[f\left({\displaystyle \frac{x}{p^{n+m-1}}},z\right)-p^{2}f\left({\displaystyle \frac{x}{p^{n+m}}},z\right)\right]∥\hfill \\ \hfill & \le {\displaystyle \frac{1}{p^{2n}}}\underset{0\le i\le m-1}{max}\left\{{\displaystyle \frac{1}{p^{2i}}}{\phi }_p\left({\displaystyle \frac{x}{p^{n+i+1}}},z\right)\right\}\hfill \\ \hfill & =\underset{0\le i\le m-1}{max}\left\{{\displaystyle \frac{1}{p^{2(n+i)}}}{\phi }_p\left({\displaystyle \frac{x}{p^{n+i+1}}},z\right)\right\}.\hfill \end{array}$$

(12)

According to the assumption, the last term converges to 0 as $n\to \infty$, implying that $\left(p^{2n}f\left({\displaystyle \frac{x}{p^{2n}}},z\right)\right)$ is a Cauchy sequence for each $(x,z)\in S^2$. Define $Q_1:(x,z)\mapsto {lim}_{n\to \infty }{p}^{2n}f\left({\displaystyle \frac{x}{p^n}},z\right)$. Then,

$$∥p^{2n}f\left({\displaystyle \frac{x+y}{p^n}},z\right)+p^{2n}f\left({\displaystyle \frac{x-y}{p^n}},z\right)-2p^{2n}f\left({\displaystyle \frac{x}{p^n}},z\right)-2p^{2n}f\left({\displaystyle \frac{y}{p^n}},z\right)∥\le {\displaystyle \frac{1}{p^{2n}}}{\phi }_{max}\left({\displaystyle \frac{x}{p^n}},{\displaystyle \frac{y}{p^n}},z\right)$$

and

$$∥p^{2n}f\left({\displaystyle \frac{x}{p^n}},y+z\right)+p^{2n}f\left({\displaystyle \frac{x}{p^n}},y-z\right)-2p^{2n}f\left({\displaystyle \frac{x}{p^n}},y\right)-2p^{2n}f\left({\displaystyle \frac{x}{p^n}},z\right)∥\le {\displaystyle \frac{1}{p^{2n}}}{\psi }_{max}\left({\displaystyle \frac{x}{p^n}},y,z\right).$$

Taking $n\to \infty$ in the two inequalities above, we conclude, according to the assumption, that $Q_1$ is bi-quadratic. Moreover, according to (12),

$$∥f(x,z)-Q_1(x,z)∥=∥f(x,z)-\underset{n\to \infty }{lim}{p}^{2n}f\left({\displaystyle \frac{x}{p^n}},z\right)∥\le {\Phi }_Q(x,z).$$

For the uniqueness, according to Lemma 2, observe that for each $m\in \mathbb{N}$,

$$\begin{array}{cc}\hfill \parallel Q_1(x,z)-Q_1^{\prime }(x,z)\parallel & ={\left|p^{2m}\right|}_p∥Q_1\left({\displaystyle \frac{x}{p^m}},z\right)-Q_1^{\prime }\left({\displaystyle \frac{x}{p^m}},z\right)∥\hfill \\ \hfill & \le {\displaystyle \frac{1}{p^{2m}}}max\left\{∥f\left({\displaystyle \frac{x}{p^m}},z\right)-Q_1\left({\displaystyle \frac{x}{p^m}},z\right)∥,∥f\left({\displaystyle \frac{x}{p^m}},z\right)-Q_1^{\prime }\left({\displaystyle \frac{x}{p^m}},z\right)∥\right\}\hfill \\ \hfill & \le {\displaystyle \frac{1}{p^{2m}}}max\left\{{\Phi }_Q\left({\displaystyle \frac{x}{p^m}},z\right),M_1{\Phi }_Q\left({\displaystyle \frac{x}{p^m}},z\right)\right\}\hfill \\ \hfill & \le {\displaystyle \frac{1}{p^{2m}}}(M_1+1){\Phi }_Q\left({\displaystyle \frac{x}{p^m}},z\right)\hfill \\ \hfill & =(M_1+1)\underset{n\ge m}{sup}{\displaystyle \frac{1}{p^{2n}}}{\phi }_p\left({\displaystyle \frac{x}{p^{n+1}}},z\right).\hfill \end{array}$$

Since

$$\underset{m\to \infty }{lim\; sup}{\displaystyle \frac{1}{p^{2m}}}{\phi }_p\left({\displaystyle \frac{x}{p^{m+1}}},z\right)=\underset{n\to \infty }{lim}{\displaystyle \frac{1}{p^{2n}}}{\phi }_p\left({\displaystyle \frac{x}{p^{n+1}}},z\right)=0,$$

we have $Q_1=Q_1^{\prime }$.

For the remaining part, one can show that $\left(p^{2n}f\left(x,{\displaystyle \frac{z}{p^{2n}}}\right)\right)$ is a Cauchy sequence for each $(x,z)\in S^2$; then we can define $Q_2:(x,z)\mapsto {lim}_{n\to \infty }{p}^{2n}f\left(x,{\displaystyle \frac{z}{p^n}}\right)$, which is bi-quadratic. The details are similar to the first part of the proof. □

Corollary 2.

Let ${\epsilon }_1,{\epsilon }_2>0$. Suppose that a function $f:S^2\to X$ satisfies

$$\left\{\begin{array}{c}\parallel f(x+y,z)+f(x-y,z)-2f(x,z)-2f(y,z)\parallel \le {\epsilon }_1\hfill \\ \parallel f(x,y+z)+f(x,y-z)-2f(x,y)-2f(x,z)\parallel \le {\epsilon }_2.\hfill \end{array}\right.$$

with $f(x,0)=f(0,z)=0$ for all $x,z\in S$. Then there are two bi-quadratic mappings, $Q_1,Q_2:S^2\to \overline{X}$, such that

$$\left\{\begin{array}{c}\parallel f(x,z)-Q_1(x,z)\parallel \le {\epsilon }_1\hfill \\ \parallel f(x,z)-Q_2(x,z)\parallel \le {\epsilon }_2\hfill \end{array}\right.$$

which are unique regardless of a constant multiple on ${\epsilon }_1$ and ${\epsilon }_2$.

We are now ready to establish the stability theorem for System (6) when $b=1$. Combining Condition (7) and Condition (10) gives

$$\left\{\begin{array}{c}{\displaystyle \underset{n\to \infty }{lim}\frac{1}{p^n}{\phi }_{max}\left(\frac{x}{p^n},\frac{y}{p^n},z\right)+\frac{1}{p^n}{\phi }_{max}\left(x,y,\frac{z}{p^n}\right)=0}\hfill \\ {\displaystyle \underset{n\to \infty }{lim}\frac{1}{p^n}{\psi }_{max}\left(\frac{x}{p^n},y,z\right)+\frac{1}{p^n}{\psi }_{max}\left(x,\frac{y}{p^n},\frac{z}{p^n}\right)=0}\hfill \\ {\displaystyle \underset{n\to \infty }{lim}\frac{1}{p^n}{\phi }_p\left(\frac{x}{p^{n+1}},z\right)=0}\hfill \\ {\displaystyle \underset{n\to \infty }{lim}\frac{1}{p^n}{\psi }_p\left(x,\frac{z}{p^{n+1}}\right)=0.}\hfill \end{array}\right.$$

(13)

Note that ${\Phi }_Q(x,z)\le {\Phi }_A(x,z)$ and ${\Psi }_Q(x,z)\le {\Psi }_A(x,z)$. We define

$${\Phi }_{max}(x,z)=max\left\{{\displaystyle \frac{1}{{\left|4\right|}_p}}{\Phi }_A(x,z),{\displaystyle \frac{1}{{\left|2\right|}_p}}{\Phi }_Q(x,z)\right\}={\displaystyle \frac{1}{{\left|4\right|}_p}}{\Phi }_A(x,z)$$

and

$${\Psi }_{max}(x,z)=max\left\{{\displaystyle \frac{1}{{\left|4\right|}_p}}{\Psi }_A(x,z),{\displaystyle \frac{1}{{\left|2\right|}_p}}{\Psi }_Q(x,z)\right\}={\displaystyle \frac{1}{{\left|4\right|}_p}}{\Psi }_A(x,z).$$

Theorem 5.

Suppose that Condition (13) holds and a function $f:S^2\to X$ satisfies

$$\left\{\begin{array}{c}\parallel f(x+y,z)+f(x-y,z)-2f(x,z)-f(y,z)-f(-y,-z)\parallel \le \phi (x,y,z)\hfill \\ \parallel f(x,y+z)+f(x,y-z)-2f(x,y)-f(x,z)-f(-x,-z)\parallel \le \psi (x,y,z).\hfill \end{array}\right.$$

(14)

Then there are two bi-quadratic functions, $Q_1,Q_2:S^2\to \overline{X}$, such that

$$\left\{\begin{array}{c}\parallel [f(x,z)-f(x,0)-f(0,z)+f(0,0)]-Q_1(x,z)\parallel \le {\Phi }_{max}(x,z)\hfill \\ \parallel [f(x,z)-f(x,0)-f(0,z)+f(0,0)]-Q_2(x,z)\parallel \le {\Psi }_{max}(x,z)\hfill \end{array}\right.$$

which are unique regardless of a constant multiple on ${\Phi }_{max}$ and ${\Psi }_{max}$.

Proof. 

Define $F(x,z)=f(x,z)-f(x,0)-f(0,z)+f(0,0)$. Then $F(x,0)=F(0,z)=0$ for all $x,z\in S$. Setting $z=0$ in the first and $y=z=0$ in the second equations of (14), we have

$$\left\{\begin{array}{c}\parallel f(x+y,0)+f(x-y,0)-2f(x,0)-f(y,0)-f(-y,0)\parallel \le \phi (x,y,0)\hfill \\ \parallel f(x,0)+f(x,0)-2f(x,0)-f(x,0)-f(-x,0)\parallel \le \psi (x,y,0).\hfill \end{array}\right.$$

(15)

Similarly, setting $x=y=0$ in the first and $x=0$ in the second equations of (14), we have

$$\left\{\begin{array}{c}\parallel f(0,z)+f(0,z)-2f(0,z)-f(0,z)-f(0,-z)\parallel \le \phi (0,0,z)\hfill \\ \parallel f(0,y+z)+f(0,y-z)-2f(0,y)-f(0,z)-f(0,-z)\parallel \le \psi (0,y,z).\hfill \end{array}\right.$$

(16)

Finally, setting $x=y=z=0$ in (14), then combining it with (14)–(16), we obtain, using the ultrametric inequality,

$$\left\{\begin{array}{c}\parallel F(x+y,z)+F(x-y,z)-2F(x,z)-F(y,z)-F(-y,-z)\parallel \le max\left\{\phi \right(x,y,z),\phi (x,y,0),\phi (0,0,z),\phi (0,0,0\left)\right\}\hfill \\ \parallel F(x,y+z)+F(x,y-z)-2F(x,y)-F(x,z)-F(-x,-z)\parallel \le max\left\{\psi \right(x,y,z),\psi (x,0,0),\psi (0,y,z),\psi (0,0,0\left)\right\}.\hfill \end{array}\right.$$

Then, by replacing x, y and z with $-x$, $-y$ and $-z$, respectively, we have

$$\left\{\begin{array}{c}\parallel F(-(x+y),-z)+F(-(x-y),-z)-2F(-x,-z)-F(-y,-z)-F(y,z)\parallel \hfill \\ \le max\left\{\phi \right(-x,-y,-z),\phi (-x,-y,0),\phi (0,0,-z),\phi (0,0,0\left)\right\}\hfill \\ \parallel F(-x,-(y+z\left)\right)+F(-x,-(y-z\left)\right)-2F(-x,-y)-F(-x,-z)-F(x,z)\parallel \hfill \\ \le max\left\{\psi \right(-x,-y,-z),\psi (-x,0,0),\psi (0,-y,-z),\psi (0,0,0\left)\right\}.\hfill \end{array}\right.$$

Again, using the ultrametric inequality with the two systems of equations above, we obtain

$$\left\{\begin{array}{c}\parallel F_o(x+y,z)+F_o(x-y,z)-2F_o(x,z)\parallel \le {\displaystyle \frac{1}{{\left|2\right|}_p}}max\{\phi (\alpha x,\alpha y,\gamma z):\alpha ,\gamma \in \{-1,0,1\}\}\hfill \\ \parallel F_o(x,y+z)+F_o(x,y-z)-2F_o(x,y)\parallel \le {\displaystyle \frac{1}{{\left|2\right|}_p}}max\{\psi (\alpha x,\gamma y,\gamma z):\alpha ,\gamma \in \{-1,0,1\}\}.\hfill \end{array}\right.$$

and

$$\left\{\begin{array}{c}\parallel F_e(x+y,z)+F_e(x-y,z)-2F_e(x,z)-2F_e(y,z)\parallel \le {\displaystyle \frac{1}{{\left|2\right|}_p}}max\{\phi (\alpha x,\alpha y,\gamma z):\alpha ,\gamma \in \{-1,0,1\}\}\hfill \\ \parallel F_e(x,y+z)+F_e(x,y-z)-2F_e(x,y)-2F_e(x,z)\parallel \le {\displaystyle \frac{1}{{\left|2\right|}_p}}max\{\psi (\alpha x,\gamma y,\gamma z):\alpha ,\gamma \in \{-1,0,1\}\}.\hfill \end{array}\right.$$

According to Theorems 3 and 4, there are two bi-additive mappings, $A_1,A_2:S^2\to \overline{X}$ such that,

$$\left\{\begin{array}{c}{\left|2\right|}_p\parallel F_o(x,z)-A_1(x,z)\parallel \le {\Phi }_A(x,z)\hfill \\ {\left|2\right|}_p\parallel F_o(x,z)-A_2(x,z)\parallel \le {\Psi }_A(x,z)\hfill \end{array}\right.$$

(17)

and two bi-quadratic mappings, $Q_1,Q_2:S^2\to \overline{X}$, such that

$$\left\{\begin{array}{c}{\left|2\right|}_p\parallel F_e(x,z)-Q_1(x,z)\parallel \le {\Phi }_Q(x,z)\hfill \\ {\left|2\right|}_p\parallel F_e(x,z)-Q_2(x,z)\parallel \le {\Psi }_Q(x,z).\hfill \end{array}\right.$$

(18)

Note that $F_o$ is odd, while $A_1$, $A_2$, $\Phi$, and $\Psi$ are even. By replacing x with $-x$ and z with $-z$ in (17), we have

$$\left\{\begin{array}{c}{\left|2\right|}_p\parallel F_o(x,z)+A_1(x,z)\parallel \le {\Phi }_A(x,z)\hfill \\ {\left|2\right|}_p\parallel F_o(x,z)+A_2(x,z)\parallel \le {\Psi }_A(x,z).\hfill \end{array}\right.$$

Then

$$\left\{\begin{array}{c}{\left|4\right|}_p\parallel F_o(x,z)\parallel \le {\Phi }_A(x,z)\hfill \\ {\left|4\right|}_p\parallel F_o(x,z)\parallel \le {\Psi }_A(x,z).\hfill \end{array}\right.$$

(19)

By combining (18) and (19), we obtain

$$\left\{\begin{array}{c}\parallel F(x,z)-Q_1(x,z)\parallel \le {\Phi }_{max}(x,z)\hfill \\ \parallel F(x,z)-Q_2(x,z)\parallel \le {\Psi }_{max}(x,z)\hfill \end{array}\right.$$

as desired.

For the uniqueness, suppose that there are two bi-quadratic functions, $Q_1$ and $Q_2$, such that

$$\left\{\begin{array}{c}\parallel F(x,z)-Q_1^{\prime }(x,z)\parallel \le M_1{\Phi }_{max}(x,z)\hfill \\ \parallel F(x,z)-Q_2^{\prime }(x,z)\parallel \le M_2{\Psi }_{max}(x,z),\hfill \end{array}\right.$$

for some constants $M_1,M_2>0$. Since $Q_1^{\prime }$, $Q_2^{\prime }$, ${\Phi }_{max}$, and ${\Psi }_{max}$ are even, decomposing F into $F_e$ yields

$$\left\{\begin{array}{c}\parallel F_e(x,z)-Q_1^{\prime }(x,z)\parallel \le {\displaystyle \frac{1}{{\left|2\right|}_p}}{M}_1{\Phi }_{max}(x,z)\hfill \\ \parallel F_e(x,z)-Q_2^{\prime }(x,z)\parallel \le {\displaystyle \frac{1}{{\left|2\right|}_p}}{M}_2{\Psi }_{max}(x,z),\hfill \end{array}\right.$$

We will deduce that $Q_1=Q_1^{\prime }$ and $Q_2=Q_2^{\prime }$ by using a similar argument as in Theorem 4. According to Lemma 2, observe that, for each $m\in \mathbb{N}$,

$$\begin{array}{cc}\hfill \parallel Q_1(x,z)-Q_1^{\prime }(x,z)\parallel & =|p^{2m}{|}_p∥Q_1\left({\displaystyle \frac{x}{p^m}},z\right)-Q_1^{\prime }\left({\displaystyle \frac{x}{p^m}},z\right)∥\hfill \\ \hfill & \le {\displaystyle \frac{1}{p^{2m}}}max\left\{∥F_e\left({\displaystyle \frac{x}{p^m}},z\right)-Q_1\left({\displaystyle \frac{x}{p^m}},z\right)∥,∥F_e\left({\displaystyle \frac{x}{p^m}},z\right)-Q_1^{\prime }\left({\displaystyle \frac{x}{p^m}},z\right)∥\right\}\hfill \\ \hfill & \le {\displaystyle \frac{1}{p^{2m}}}max\left\{{\Phi }_{max}\left({\displaystyle \frac{x}{p^m}},z\right),{\displaystyle \frac{1}{{\left|2\right|}_p}}{M}_1{\Phi }_{max}\left({\displaystyle \frac{x}{p^m}},z\right)\right\}\hfill \\ \hfill & \le {\displaystyle \frac{1}{p^m}}\left({\displaystyle \frac{1}{{\left|2\right|}_p}}{M}_1+1\right){\Phi }_{max}\left({\displaystyle \frac{x}{p^m}},z\right)\hfill \\ \hfill & =\left({\displaystyle \frac{1}{{\left|2\right|}_p}}{M}_1+1\right)\underset{n\ge m}{sup}{\displaystyle \frac{1}{p^n}}{\phi }_p\left({\displaystyle \frac{x}{p^{n+1}}},z\right),\hfill \end{array}$$

Since the last term converges to 0 as $m\to \infty$, we conclude that $Q_1=Q_1^{\prime }$. Apply a similar argument to $Q_2$ and $Q_2^{\prime }$ to get $Q_2=Q_2^{\prime }$. □

If $\phi$ and $\psi$ are constant functions, our result improves that of [22] but in a non-Archimedean normed space setting.

Corollary 3.

Let ${\epsilon }_1,{\epsilon }_2>0$. Suppose that a function $f:S^2\to X$ satisfies

$$\left\{\begin{array}{c}\parallel f(x+y,z)+f(x-y,z)-2f(x,z)-f(y,z)-f(-y,-z)\parallel \le {\epsilon }_1\hfill \\ \parallel f(x,y+z)+f(x,y-z)-2f(x,y)-f(x,z)-f(-x,-z)\parallel \le {\epsilon }_2.\hfill \end{array}\right.$$

Then there is a bi-quadratic function, $Q:S^2\to \overline{X}$, such that

$$\parallel [f(x,z)-f(x,0)-f(0,z)+f(0,0)]-Q(x,z)\parallel \le min\{{\epsilon }_1,{\epsilon }_2\}$$

which is unique regardless of a constant multiple on ${\epsilon }_1$ and ${\epsilon }_2$.

Remark 2.

Modified results for Theorems 3 and 4, as well as Corollaries 1 and 2, can be obtained without assuming $f(x,0)=f(0,z)=0$ for all $x,z\in S$. To establish this, we consider the function F defined by $F(x,y)=f(x,y)-f(x,0)-f(0,y)+f(0,0)$; see Theorem 5.

Example 1.

Let X be a field extension of ${\mathbb{Q}}_p$ and $a,b,c,d,q,r,s,t\in {\mathbb{Q}}_p$. Let $f(x,y)=a{x}^2{y}^2+bx{y}^2+c{x}^{2}y+d{x}^2+e{y}^2+qxy+rx+sy+t$. Suppose that $f:X^2\to X$ satisfies

$$\left\{\begin{array}{c}\parallel f(x+y,z)+f(x-y,z)-2f(x,z)-f(y,z)-f(-y,-z)\parallel \le \epsilon \hfill \\ \parallel f(x,y+z)+f(x,y-z)-2f(x,y)-f(x,z)-f(-x,-z)\parallel \le \epsilon .\hfill \end{array}\right.$$

for some $\epsilon >0$. Then $b=c=q=0$; that is, $f(x,y)=a{x}^2{y}^2+d{x}^2+e{y}^2+rx+sy+t$.

Proof. 

Note that $f(x,y)-f(x,0)-f(0,y)+f(0,0)=a{x}^2{y}^2+bx{y}^2+c{x}^{2}y+qxy$. According to Corollary 3, there is a bi-quadratic function, $Q:X^2\to \overline{X}$, such that

$$\parallel (a{x}^2{y}^2+bx{y}^2+c{x}^{2}y+qxy)-Q(x,y)\parallel \le \epsilon .$$

Replacing x with $-x$, we have

$$\parallel (a{x}^2{y}^2-bx{y}^2+c{x}^{2}y-qxy)-Q(x,y)\parallel \le \epsilon .$$

So

$$\parallel bx{y}^2+qxy\parallel \le {\displaystyle \frac{1}{{\left|2\right|}_p}}\epsilon .$$

Similarly, one can obtain

$$\parallel c{x}^{2}y+qxy\parallel \le {\displaystyle \frac{1}{{\left|2\right|}_p}}\epsilon$$

and

$$\parallel bx{y}^2+c{x}^{2}y\parallel \le {\displaystyle \frac{1}{{\left|2\right|}_p}}\epsilon .$$

Using these three inequalities, we deduce that

$$\parallel bx{y}^2\parallel \le {\displaystyle \frac{1}{{\left|4\right|}_p}}\epsilon ,\parallel c{x}^{2}y\parallel \le {\displaystyle \frac{1}{{\left|4\right|}_p}}\epsilon ,\mathrm{and}\parallel qxy\parallel \le {\displaystyle \frac{1}{{\left|4\right|}_p}}\epsilon .$$

Since x and y are arbitrary, we must have $b=c=q=0$, as desired. □

Remark 3.

This example indicates that f must contain only bi-quadratic parts $a{x}^2{y}^2$, $d{x}^2$, and $e{y}^2$ with the exception of excess terms $rx$, $sy$, and t, making $f(x,0)\ne 0$ and $f(0,y)\ne 0$.

Finally, for System (6) with $b\ne 0,1$, we consider it when f is an even function.

Theorem 6.

Suppose that $b\ne 0,1$ and a function $f:S^2\to X$ satisfies

$$\left\{\begin{array}{c}f(x+y,z)+f(x-y,z)=2f(x,z)+bf(y,z)+bf(-y,-z)\hfill \\ f(x,y+z)+f(x,y-z)=2f(x,y)+bf(x,z)+bf(-x,-z).\hfill \end{array}\right.$$

If f is even and $f(x,0)=f(0,y)=0$ for all $x,y\in S$, then $f(x,y)=0$ for all $x,y\in S$.

Proof. 

Suppose that f is even with $f(x,0)=f(0,y)=0$ for all $x,y\in S$. Then the system of equations reduces to

$$\left\{\begin{array}{c}f(x+y,z)+f(x-y,z)=2f(x,z)+2bf(y,z)\hfill \\ f(x,y+z)+f(x,y-z)=2f(x,y)+2bf(x,z).\hfill \end{array}\right.$$

(20)

We define

$$f_{ee}(x,y)={\displaystyle \frac{1}{2}}[f(x,y)+f(-x,y)]\mathrm{a}\mathrm{n}\mathrm{d}{f}_{oo}(x,y)={\displaystyle \frac{1}{2}}[f(x,y)-f(-x,y)].$$

By using the fact that f is an even function, it can be easily seen that $f_{ee}$ is even in both arguments; that is,

$$f_{ee}(-x,y)=f_{ee}(x,-y)=f_{ee}(x,y),$$

$f_{oo}$ is odd in both arguments; that is,

$$f_{oo}(-x,y)=f_{oo}(x,-y)=-f_{oo}(x,y),$$

and $f=f_{ee}+f_{oo}$.

We decompose f into $f_{ee}$ and $f_{oo}$ as follows. Replacing x with $-x$ and y with $-y$ in the first and x with $-x$ in the second equations of (20), we have

$$\left\{\begin{array}{c}f(-(x+y),z)+f(-(x-y),z)=2f(-x,z)+2bf(-y,z)\hfill \\ f(-x,y+z)+f(-x,y-z)=2f(-x,y)+2bf(-x,z).\hfill \end{array}\right.$$

Together with (20), we obtain

$$\left\{\begin{array}{c}{f}_{ee}(x+y,z)+f_{ee}(x-y,z)=2f_{ee}(x,z)+2b{f}_{ee}(y,z)\hfill \\ f_{ee}(x,y+z)+f_{ee}(x,y-z)=2f_{ee}(x,y)+2b{f}_{ee}(x,z)\hfill \end{array}\right.$$

(21)

and

$$\left\{\begin{array}{c}{f}_{oo}(x+y,z)+f_{oo}(x-y,z)=2f_{oo}(x,z)+2b{f}_{oo}(y,z)\hfill \\ f_{oo}(x,y+z)+f_{oo}(x,y-z)=2f_{oo}(x,y)+2b{f}_{oo}(x,z).\hfill \end{array}\right.$$

(22)

To solve for $f_{ee}$, letting $y=0$ in (21), we have

$$\left\{\begin{array}{c}{f}_{ee}(x,z)+f_{ee}(x,z)=2f_{ee}(x,z)\hfill \\ f_{ee}(x,z)+f_{ee}(x,z)=2b{f}_{ee}(x,z).\hfill \end{array}\right.$$

Hence $2(b-1)f_{ee}(x,z)=0$, but then $b\ne 1$. We conclude that $f_{ee}(x,z)=0$. To solve for $f_{oo}$, letting $y=0$ in the second equation of (22), we have $2b{f}_{oo}(x,z)=0$, which implies that $f_{oo}(x,z)=0$ since $b\ne 0$. Therefore, $f=f_{ee}+f_{oo}=0$. □

Corollary 4.

Suppose that $b\ne 0,1$ and a function $f:S^2\to X$ satisfies

$$\left\{\begin{array}{c}f(x+y,z)+f(x-y,z)=2f(x,z)+bf(y,z)+bf(-y,-z)\hfill \\ f(x,y+z)+f(x,y-z)=2f(x,y)+bf(x,z)+bf(-x,-z).\hfill \end{array}\right.$$

If f is even, then $f(x,y)=f(x,0)+f(0,y)$ for all $x,y\in S$.

Proof. 

Note that $f(0,0)=0$. By considering the function F given by $F(x,y)=f(x,y)-f(x,0)-f(0,y)$, we are done. □

Note that any function f can be decomposed as $f=f_e+f_o$. Theorem 6 suggests that if the function f satisfies System (6), where $b\ne 0,1$, the even part $f_e$ must vanish, so the function must be considered when f is odd.

Theorem 7.

Suppose that $b\ne 0,1$, Condition (7) holds, and a function $f:S^2\to X$ satisfies

$$\left\{\begin{array}{c}\parallel f(x+y,z)+f(x-y,z)-2f(x,z)-bf(y,z)-bf(-y,-z)\parallel \le \phi (x,y,z)\hfill \\ \parallel f(x,y+z)+f(x,y-z)-2f(x,y)-bf(x,z)-bf(-x,-z)\parallel \le \psi (x,y,z).\hfill \end{array}\right.$$

(23)

If f is odd, then

$$\left\{\begin{array}{c}\parallel f(x,z)-f(x,0)-f(0,z)+f(0,0)\parallel \le {\displaystyle \frac{1}{{\left|2\right|}_p}}{\Phi }_A(x,z)\hfill \\ \parallel f(x,z)-f(x,0)-f(0,z)+f(0,0)\parallel \le {\displaystyle \frac{1}{{\left|2\right|}_p}}{\Psi }_A(x,z).\hfill \end{array}\right.$$

Proof. 

Since f is odd, (23) reduces to

$$\left\{\begin{array}{c}\parallel f(x+y,z)+f(x-y,z)-2f(x,z)\parallel \le \phi (x,y,z)\hfill \\ \parallel f(x,y+z)+f(x,y-z)-2f(x,y)\parallel \le \psi (x,y,z).\hfill \end{array}\right.$$

Define $F(x,z)=f(x,z)-f(x,0)-f(0,z)+f(0,0)$. We obtain

$$\left\{\begin{array}{c}\parallel F(x+y,z)+F(x-y,z)-2F(x,z)\parallel \le {\phi }_{max}(x,y,z)\hfill \\ \parallel F(x,y+z)+F(x,y-z)-2F(x,y)\parallel \le {\psi }_{max}(x,y,z).\hfill \end{array}\right.$$

By applying Theorem 3 with the same argument used for the odd part in Theorem 5, we obtain

$$\left\{\begin{array}{c}\parallel F(x,z)\parallel \le {\displaystyle \frac{1}{{\left|2\right|}_p}}{\Phi }_A(x,z)\hfill \\ \parallel F(x,z)\parallel \le {\displaystyle \frac{1}{{\left|2\right|}_p}}{\Psi }_A(x,z).\hfill \end{array}\right.$$

□

Corollary 5.

Let ${\epsilon }_1,{\epsilon }_2>0$. Suppose that $b\ne 0,1$ and a function $f:S^2\to X$ satisfies

$$\left\{\begin{array}{c}\parallel f(x+y,z)+f(x-y,z)-2f(x,z)-bf(y,z)-bf(-y,-z)\parallel \le {\epsilon }_1\hfill \\ \parallel f(x,y+z)+f(x,y-z)-2f(x,y)-bf(x,z)-bf(-x,-z)\parallel \le {\epsilon }_2.\hfill \end{array}\right.$$

(24)

If f is odd, then

$$\parallel f(x,z)-f(x,0)-f(0,z)+f(0,0)\parallel \le {\displaystyle \frac{1}{{\left|2\right|}_p}}min\{{\epsilon }_1,{\epsilon }_2\}.$$

Corollary 5 indicates that if an odd function f satisfies $f(x,0)=f(0,z)=0$ and (5), then $f=0$.

Example 2.

Let X be a field extension of ${\mathbb{Q}}_p$ and $f(x,y)=ax{y}^2+b{x}^{2}y$, where $a,b\in {\mathbb{Q}}_p$. Suppose that $f:X^2\to X$ satisfies

$$\left\{\begin{array}{c}\parallel f(x+y,z)+f(x-y,z)-2f(x,z)-bf(y,z)-bf(-y,-z)\parallel \le \epsilon \hfill \\ \parallel f(x,y+z)+f(x,y-z)-2f(x,y)-bf(x,z)-bf(-x,-z)\parallel \le \epsilon .\hfill \end{array}\right.$$

for some $\epsilon >0$. According to Corollary 5,

$$\parallel ax{y}^2+b{x}^{2}y\parallel \le {\displaystyle \frac{1}{{\left|2\right|}_p}}\epsilon .$$

Since x and y are arbitrary, we have $a=b=0$, which implies that $f=0$.

4. Discussion and Conclusions

The results obtained in this work demonstrate that coupled bi-variable functional systems of the bi-quadratic–Drygas type exhibit structural behaviors that differ fundamentally from those of single or uncoupled functional equations. The presence of asymmetric coefficients together with reflection terms generates intrinsic coupling effects. As shown by the main theorems, in the non-Archimedean setting, the stability of the system of equations requires simultaneous control of both variables.

A crucial aspect of the analysis is the use of symmetry-based decomposition. The separation of mappings into components reflecting additive and quadratic behavior enables a precise characterization of approximate solutions. In non-Archimedean normed spaces, the ultrametric inequality plays a decisive role in stabilizing the iterative process and yields uniqueness of the exact solution under suitable control functions. The reflection terms appearing in the system are shown to govern the interaction between symmetry and coupling, rather than acting as negligible perturbations.

The main theorems also unify several known stability results. By appropriately choosing parameters and control functions, bi-quadratic and bi-Drygas stability results are recovered as special cases. This confirms that the proposed framework provides a coherent extension of existing theories rather than an isolated generalization.

In summary, this work establishes Hyers–Ulam–Rassias stability for a generalized coupled bi-quadratic–Drygas functional system in non-Archimedean normed spaces. The results clarify the roles of symmetry, reflection, and ultrametric structure in determining stability behavior and provide a unified approach to analyzing coupled functional equations.

Further extensions to higher-dimensional systems or weaker control conditions may offer additional insights into stability phenomena in non-classical settings. In such a framework, for $n>2$, one considers a mapping $f:S^n\to X$ satisfying quadratic–Drygas-type identities in each coordinate separately, namely,

$$f(\dots ,x_i+y_i,\dots )+f(\dots ,x_i-y_i,\dots )=af(\dots ,x_i,\dots )+bf(\dots ,y_i,\dots )+cf(\dots ,-y_i,\dots )$$

for each $i=1,\dots ,n$. However, the normalization procedure becomes substantially more delicate. In the 2-tuple setting, the mapping $F(x,y)=f(x,y)-f(x,0)-f(0,y)+f(0,0)$ removes boundary traces in a canonical way. For an n-tuple, one would need to construct a higher-dimensional normalization eliminating all lower-dimensional boundary components. Consequently, the main difficulty in extending the theory lies not only in handling coordinate-wise quadratic behavior but also in defining an appropriate mapping that addresses lower-dimensional boundary conditions. Nevertheless, the method developed in this paper is expected to ensure convergence of iterative sequences of the form $p^{2m}f(\dots ,x_i/p^m,\dots ),$ leading to the existence and uniqueness of n-variable quadratic mappings. Such an extension would establish a higher-dimensional stability theory clarifying how reflection symmetry and coupling constraints propagate across multiple interacting variables in non-Archimedean normed spaces.