1 Introduction

Set-valued functional equations in Banach spaces have received a lot of attention in the literature (see, for example, [1,2,3, 5, 6]). The pioneering papers by Aumann [3] and Debreu [5] were inspired by problems arising in Control Theory and Mathematical Economics. In 2009, Nikodem and Popa [17] considered the general solution of set-valued maps satisfying linear inclusion relation, which can be regarded as a generalization of the additive single-valued functional equation. By means of the inclusion relation, Lu and Park [13] investigated the approach of additive set-valued functional equations. Set-valued functional equations have been investigated by a number of authors and there are many interesting results concerning this problem (see [4, 10, 15, 16, 18]).

If X and Y are vector spaces and \(f:X\rightarrow Y\) is a mapping, then the operator \(\biguplus _{x_{2}}f(x_{1})\) is defined by the formula

$$\begin{aligned} \biguplus _{x_{2}}f(x_{1})=f(x_1+x_2)+f(x_1-x_2). \end{aligned}$$

The composite operator \(\biguplus _{x_{2},\ldots ,x_{n+1}}^nf(x_{1})\) is defined by

$$\begin{aligned} \biguplus _{x_{2},\ldots ,x_{n+1}}^nf(x_{1}) =\biguplus _{x_{n+1}}\left( \biguplus _{x_{2},\ldots ,x_{n}}^{n-1}f(x_{1})\right) \end{aligned}$$

for all \(n\in \mathbb N{\setminus }\{1\}\). Note that

$$\begin{aligned} 2^{3-n}\biguplus _{x_{2},x_{3},0,\ldots ,0}^{n-1}f(x_{1}) =\biguplus _{x_{2},x_{3}}^2f(x_{1})=\biguplus _{x_{3},x_{2}}^2f(x_{1}) \end{aligned}$$

and

$$\begin{aligned} \biguplus _{x_{2},\ldots ,x_{n+1}}^nf(x_{1})=\biguplus _{x_{2},\ldots ,x_{n}}^{n-1}f(x_{1}+x_{n+1}) +\biguplus _{x_{2},\ldots ,x_{n}}^{n-1}f(x_{1}-x_{n+1}). \end{aligned}$$

In [7, 8], Jun et al. proposed the following Euler–Lagrange type cubic functional equations:

$$\begin{aligned} \biguplus _{x_{2}}f(ax_{1})= a\biguplus _{x_{2}}f(x_{1})+ 2a(a^2 - 1)f (x_1) \end{aligned}$$
(1.1)

for a fixed integer a with \(a \ne 0,\pm 1\), and

$$\begin{aligned} \biguplus _{bx_{2}}f(ax_{1}) =ab^2\biguplus _{x_{2}}f(x_{1})+ 2a(a^2 - b^2)f (x_1) \end{aligned}$$
(1.2)

for fixed integers a, b with \(a \ne 0\), \(b \ne 0\), and \(a\pm b \ne 0\), and the equations being equivalent to the cubic functional equation. Also, Lee et al. [9, 12] proposed the following Euler–Lagrange type quartic functional equations:

$$\begin{aligned} \biguplus _{x_{2}}f(ax_{1})= a^2\biguplus _{x_{2}}f(x_{1})+ 2a^2(a^2 - 1)f (x_1)+ 2(1-a^2)f (x_2) \end{aligned}$$
(1.3)

for a fixed integer a with \(a \ne 0,\pm 1\), and

$$\begin{aligned} \biguplus _{bx_{2}}f(ax_{1})=a^2b^2\biguplus _{x_{2}}f(x_{1})+ 2a^2(a^2 - b^2)f (x_1)+ 2b^2(b^2-a^2)f (x_2) \end{aligned}$$
(1.4)

for fixed integers a, b with \(a \ne 0\), \(b \ne 0\), and \(a\pm b \ne 0\), and the equations being equivalent to the quartic functional equation. In [11], Kim extended the quadratic and quartic functional equations to the following generalized form:

$$\begin{aligned} \biguplus ^{n-1}_{x_{2},\ldots ,x_{n}}f(x_{1})=2^{n-2}\sum ^{}_{1\le i<j \le n}\biguplus _{x_{j}}f(x_{i}) +2^{n-1}(2-n)\sum ^{n}_{i=1}f(x_{i}). \end{aligned}$$
(1.5)

In this article, we focus on the set-valued dynamics related to the theory of functional equations. We study the n-dimensional Euler–Lagrange cubic and quartic set-valued functional equations:

$$\begin{aligned} \biguplus ^{n-1}_{a_{2}x_{2},\ldots ,a_{n}x_{n}}f(a_{1}x_{1})=2^{n-2}\sum ^{n}_{i=2}a_{1}a^2_{i}\biguplus _{x_{i}}f(x_{1}) +2^{n-1}a_{1}\Big (a^2_{1}-\sum ^{n}_{i=2}a^2_{i}\Big )f(x_{1})\nonumber \\ \end{aligned}$$
(1.6)

and

$$\begin{aligned} \biguplus ^{n-1}_{a_{2}x_{2},\ldots ,a_{n}x_{n}}f(a_{1}x_{1})= & {} 2^{n-2}\sum ^{}_{1\le i<j \le n}a^2_{i}a^2_{j}\biguplus _{x_{j}}f(x_{i})\nonumber \\&+\,2^{n-1}\sum ^{n}_{i=1}a^2_{i}\Big (a^2_{i}-\sum ^{n}_{j=1, j\ne i}a^2_{j}\Big )f(x_{i}) \end{aligned}$$
(1.7)

for any fixed integers \(a_1,\ldots ,a_{n}\). We also establish some approaches of the above n-dimensional Euler–Lagrange cubic and quartic set-valued functional equations. More importantly, the corresponding single-valued functional equations acted as special cases will be included in our results. Note that, putting \(n=2\), \(a_1=a\) and \(a_2=1\) in (1.6) yields (1.1) and letting \(n=2\), \(a_1=a\) and \(a_2=b\) in (1.6) yields (1.2). Setting \(n=2\), \(a_1=a\) and \(a_2=1\) in (1.7) yields (1.3) and letting \(n=2\), \(a_1=a\) and \(a_2=b\) in (1.7) yields (1.4). Replacing each \(a_i\) in (1.7) with 1, we obtain (1.5).

2 Characterizations of set-valued dynamics

We first introduce some definitions and notations which are needed to prove the main theorems on this paper:

Let A and B be two nonempty subsets of a real vector space X and \(\lambda \) a real number. The (Minkowski) addition and scalar multiplication can be defined by

$$\begin{aligned} A + B = \left\{ a+b \in X :~ a \in A,~ b \in B \right\} ,~~~~~~~~~~~ \lambda A = \left\{ \lambda a \in X : ~a \in A \right\} . \end{aligned}$$

Lemma 2.1

([14]) Let \(\lambda \) and \(\mu \) be real numbers. If A and B are nonempty subsets of a real vector space X, then

$$\begin{aligned} \lambda (A+B)=\lambda A+\lambda B, \end{aligned}$$
$$\begin{aligned} (\lambda +\mu ) A\subseteq \lambda A+\mu A. \end{aligned}$$

In particular, if A is convex and \(\lambda \mu \ge 0\), then

$$\begin{aligned} (\lambda +\mu ) A= \lambda A+\mu A. \end{aligned}$$

A subset \(A \subseteq X\) is said to be a cone if \(A+ A \subseteq A\) and \(\lambda A \subseteq A\) for all \(\lambda > 0.\) If the zero vector in X belongs to A,  then we say that A is a cone with zero.

From now on, we assume that \(a_1,a_2,\ldots ,a_{n}\) are fixed positive integers with \(a_1\ne 1\) and \(a_{n}=1\), X is a real vector space, \(A \subseteq X\) is a cone with zero and Y is a Banach space. By CCZ(Y) we denote the family of all nonempty, convex and closed subsets, containing 0,  of Y.

If \(\left( E_m\right) _{m\ge 0}\) and \(\left( K_m\right) _{m\ge 0}\) are decreasing sequences of closed sets in Y and \(K_0\) is compact we have (see [14])

$$\begin{aligned} \bigcap ^{\infty }_{n=0}\left( E_m+K_m\right) =\bigcap ^{\infty }_{n=0}E_m +\bigcap ^{\infty }_{n=0}K_m. \end{aligned}$$
(2.1)

Lemma 2.2

([8], Theorem 2.2]) A mapping \(f:X\rightarrow Y\) satisfies the functional equation (1.1) if and only if there exists a cubic mapping \(\mathcal {C}:X\rightarrow Y\) such that \(f (x) =\mathcal {C}(x)\) for all \(x \in X\).

Theorem 2.3

Suppose \(\mathcal {S}: A+(-1)A \rightarrow CCZ(Y)\) is a set-valued mapping satisfying

$$\begin{aligned}&\biguplus ^{n-1}_{a_{2}x_{2},\ldots ,a_{n}x_{n}}\mathcal {S}(a_{1}x_{1}) +2^{n-1}a_{1}\sum ^{n}_{i=2}a^2_{i}\mathcal {S}(x_{1})\nonumber \\&\qquad \subseteq 2^{n-2}a_{1} \sum ^{n}_{i=2}a^2_{i}\biguplus _{x_{i}}\mathcal {S}(x_{1})+2^{n-1}a^{3}_{1}\mathcal {S}(x_{1}) \end{aligned}$$
(2.2)

for all \(x_1,\ldots ,x_n \in A\) and \( \sup \left\{ \mathrm{{diam}}\left( \mathcal {S}(x)\right) : x \in A\right\} < +\infty \). Then there exists a unique cubic mapping \(\mathcal {C}:A+(-1)A \rightarrow Y\) such that \(\mathcal {C}(x) \in \mathcal {S}(x)\) for all \(x \in A.\)

Proof

Putting \(x_1=x\) and replacing each \(x_2,\ldots ,x_n\) in (2.2) with 0 yield that

$$\begin{aligned} \mathcal {S}(a_1 x)\subseteq a_1^{3}\mathcal {S}(x) \end{aligned}$$

for all \(x \in A.\) Replacing x by \(a_1^kx,\) \(k\in \mathbb {N},\) in the above and dividing by \(a_1^{3(k+1)}\), we observe that

$$\begin{aligned} \frac{1}{a_1^{3(k+1)}}\mathcal {S}(a_1^{k+1} x)\subseteq \frac{1}{a_1^{3k}}\mathcal {S}(a_1^kx) \end{aligned}$$

for all \(x \in A.\) Let \(\mathcal {S}_k(x) = \frac{1}{a_1^{3k}}\mathcal {S}(a_1^kx)\), \(x \in A\), \(k\in \mathbb {N}\), we obtain that \((\mathcal {S}_k(x))_{k\ge 0}\) is a decreasing sequence of closed subsets of the Banach space Y. We also get

$$\begin{aligned} \mathrm{{diam}}\left( \mathcal {S}_k(x)\right) = \frac{1}{a_1^{3k}}\mathrm{{diam}}\left( \mathcal {S}(a_1^kx)\right) \end{aligned}$$

for all \(x \in A.\) Taking into account that \(\sup \left\{ \mathrm{{diam}}\left( \mathcal {S}(x)\right) : x \in A\right\} < +\infty \), we obtain for all \(x \in A\),

$$\begin{aligned} \lim _{k\rightarrow +\infty } \mathrm{{diam}}(\mathcal {S}_k(x)) = 0. \end{aligned}$$

Using the Cantor theorem for the sequence \((\mathcal {S}_k(x))_{k\ge 0}\), we obtain that the intersection \(\bigcap _{k\ge 0}\mathcal {S}_k(x)\) is a singleton set and we denote this intersection by \(\mathcal {C}(x)\) for all \(x \in A.\) Thus we obtain a mapping \(\mathcal {C}:A+(-1)A \rightarrow Y\) and \(\mathcal {C}(x) \in \mathcal {S}_0(x)=\mathcal {S}(x)\) for all \(x \in A.\)

We will now prove that \(\mathcal {C}\) is cubic. We have (note Lemma 2.1)

$$\begin{aligned}&\biguplus ^{n-1}_{a_{2}x_{2},\ldots ,a_{n}x_{n}}\mathcal {S}_k(a_{1}x_{1}) +2^{n-1}a_{1}\sum ^{n}_{i=2}a^2_{i}\mathcal {S}_k(x_{1})\nonumber \\&\quad =\frac{1}{a^{3k}_1}\biguplus ^{n-1}_{a_{1}^ka_{2}x_{2},\ldots ,a_{1}^ka_{n}x_{n}}\mathcal {S}(a_{1}^{k+1}x_{1})\nonumber \\&\qquad +\,\frac{2^{n-1}a _{1}}{a^{3k}_1}\sum ^{n}_{i=2}a^2_{i}\mathcal {S}(a_{1}^kx_{1}) \nonumber \\&\quad \subseteq \frac{2^{n-2}a _{1}}{a^{3k}_1} \sum ^{n}_{i=2}a^2_{i}\biguplus _{a_{1}^kx_{i}}\mathcal {S}(a_{1}^kx_{1})+\,\frac{2^{n-1}a^{3}_{1}}{a^{3k}_1} \mathcal {S}(a_{1}^kx_{1})\nonumber \\&\quad =2^{n-2}a _{1} \sum ^{n}_{i=2}a^2_{i}\biguplus _{x_{i}}\mathcal {S}_k(x_{1})+\,2^{n-1}a^{3}_{1}\mathcal {S}_k(x_{1}) \end{aligned}$$
(2.3)

for all \(x_1,\ldots ,x_n \in A.\) It follows from (2.1), (2.3) and the definition of \(\mathcal {C}\) that

$$\begin{aligned}&\biguplus ^{n-1}_{a_{2}x_{2},\ldots ,a_{n}x_{n}}\mathcal {C}(a_{1}x_{1}) +2^{n-1}a _{1}\sum ^{n}_{i=2}a^2_{i}\mathcal {C}(x_{1})\\&\qquad =\biguplus ^{n-1}_{a_{2}x_{2},\ldots ,a_{n}x_{n}}\bigcap _{k=0}^{\infty }\mathcal {S}_k(a_{1}x_{1}) +2^{n-1}a _{1}\sum ^{n}_{i=2}a^2_{i}\bigcap _{k=0}^{\infty }\mathcal {S}_k(x_{1})\\&\qquad =\bigcap _{k=0}^{\infty }\left( \biguplus ^{n-1}_{a_{2}x_{2},\ldots ,a_{n}x_{n}}\mathcal {S}_k(a_{1}x_{1}) +2^{n-1}a _{1}\sum ^{n}_{i=2}a^2_{i}\mathcal {S}_k(x_{1})\right) \\&\qquad \subseteq \bigcap _{k=0}^{\infty }\left( 2^{n-2}a _{1} \sum ^{n}_{i=2}a^2_{i}\biguplus _{x_{i}}\mathcal {S}_k(x_{1})+2^{n-1}a^{3}_{1}\mathcal {S}_k(x_{1})\right) \end{aligned}$$

and

$$\begin{aligned}&2^{n-2}a _{1} \sum ^{n}_{i=2}a^2_{i}\biguplus _{x_{i}}\bigcap _{k=0}^{\infty }\mathcal {S}_k(x_{1}) +2^{n-1}a^{3}_{1}\bigcap _{k=0}^{\infty }\mathcal {S}_k(x_{1})\\&\quad =2^{n-2}a _{1} \sum ^{n}_{i=2}a^2_{i}\biguplus _{x_{i}}\mathcal {C}(x_{1}) +2^{n-1}a^{3}_{1}\mathcal {C}(x_{1}) \end{aligned}$$

for all \(x_1,\ldots ,x_n \in A.\) Thus we obtain

$$\begin{aligned} \biguplus ^{n-1}_{a_{2}x_{2},\ldots ,a_{n}x_{n}}\mathcal {C}(a_{1}x_{1})=2^{n-2}\sum ^{n}_{i=2}a_{1}a^2_{i}\biguplus _{x_{i}}\mathcal {C}(x_{1}) +2^{n-1}a_{1}\Bigg (a^2_{1}-\sum ^{n}_{i=2}a^2_{i}\Bigg )\mathcal {C}(x_{1}) \end{aligned}$$
(2.4)

for all \(x_1,\ldots ,x_n \in A.\) Letting \(x_1=\cdots =x_n=0\) in (2.4), we obtain \(\mathcal {C}(0)=0\) since \(a_1>1\). Putting \(x_2=\cdots =x_{n-1}=0\) in (2.4) and using \(\mathcal {C}(0)=0\), we gain

$$\begin{aligned} 2^{n-2}\biguplus _{a_{n}x_{n}}\mathcal {C}(a_1x_{1})= 2^{n-2}a_{1}a^2_{n}\biguplus _{x_{n}}\mathcal {C}(x_{1})+ 2^{n-1} a_{1}(a^2_{1}-a^2_{n})\mathcal {C} (x_1) \end{aligned}$$

for all \(x_{1},x_{n}\in A.\) Since \(a_n=1\), we can conclude that

$$\begin{aligned} \biguplus _{x_{n}}\mathcal {C}(a_1x_{1})= a_1\biguplus _{x_{n}}\mathcal {C}(x_{1})+ 2a_1(a_1^2 - 1)\mathcal {C} (x_1) \end{aligned}$$

for all \(x_1,x_n \in A\). Then it follows from Lemma 2.2 that the mapping \(\mathcal {C}\) is cubic. Therefore, we conclude that there exists a cubic mapping \(\mathcal {C}:A+(-1)A \rightarrow Y\) such that \(\mathcal {C}(x) \in \mathcal {S}(x)\) for all \(x \in A.\)

Next, let us prove the uniqueness property of \(\mathcal {C}\).

Suppose that \(\mathcal {S}\) have two cubic selections \(\mathcal {C}, \mathcal {D}: A + (-1)A \rightarrow Y\). We have

$$\begin{aligned} (3k)^3 \mathcal {C}(x) =\mathcal {C}(3kx) \in \mathcal {S}(3k x),~~~(3k)^3 \mathcal {D}(x) =\mathcal {D}(3kx) \in \mathcal {S}(3k x) \end{aligned}$$

for all \(x \in A\) and \(k \in \mathbb {N}\). Then we get

$$\begin{aligned}&(3k)^3\left\| \mathcal {C}(x)-\mathcal {D}(x)\right\| = \left\| (3k)^3\mathcal {C}(x) -(3k)^3\mathcal {D}(x)\right\| \\&\quad = \left\| \mathcal {C}(3kx)-\mathcal {D}(3kx)\right\| \le 2 \cdot \mathrm{{diam}}\left( \mathcal {S}(3kx)\right) \end{aligned}$$

for all \(x \in A\) and \(k \in \mathbb {N}\). It follows from \(\sup \left\{ \mathrm{{diam}}\left( \mathcal {S}(x)\right) : x \in A\right\} < +\infty \) that \(\mathcal {C}(x) = \mathcal {D}(x)\) for all \(x \in A\), as desired. \(\square \)

Lemma 2.4

([12], Theorem 2]). A mapping \(f:X\rightarrow Y\) satisfies the functional equation (1.3) if and only if there exists a quartic mapping \(\mathcal {Q}:X\rightarrow Y\) such that \(f (x) =\mathcal {Q}(x)\) for all \(x \in X\).

Theorem 2.5

Suppose \(\mathcal {S}: A+(-1)A \rightarrow CCZ(Y)\) is a set-valued mapping satisfying \(\mathcal {S}(0)=\{0\}\), \(\sup \left\{ \mathrm{{diam}}\left( \mathcal {S}(x)\right) : x \in A\right\} < +\infty \) and

$$\begin{aligned}&\biguplus ^{n-1}_{a_{2}x_{2},\ldots ,a_{n}x_{n}}\mathcal {S}(a_{1}x_{1}) +2^{n-1}\sum ^{n}_{i=1}a^2_{i}\Bigg (\sum ^{n}_{j=1, j\ne i}a^2_{j}\Bigg )\mathcal {S}(x_{i})\nonumber \\&\quad \subseteq 2^{n-2}\sum ^{}_{1\le i<j \le n}a^2_{i}a^2_{j}\biguplus _{x_{j}}\mathcal {S}(x_{i}) +2^{n-1}\sum ^{n}_{i=1}a^4_{i}\mathcal {S}(x_{i}) \end{aligned}$$
(2.5)

for all \(x_1,\ldots ,x_n \in A\). Then there exists a unique quartic mapping \(\mathcal {Q}:A+(-1)A \rightarrow Y\) such that \(\mathcal {Q}(x) \in \mathcal {S}(x)\) for all \(x \in A.\)

Proof

Putting \(x_1=x\), replacing each \(x_2,\ldots ,x_n\) in (2.5) with 0 and using \(\mathcal {S}(0)=\{0\}\) yield that

$$\begin{aligned} \mathcal {S}(a_1 x)\subseteq a_1^{4}\mathcal {S}(x) \end{aligned}$$

for all \(x \in A.\) Replacing x by \(a_1^kx,\) \(k\in \mathbb {N},\) in the above and dividing by \(a_1^{4k+4}\), we see that

$$\begin{aligned} \frac{1}{a_1^{4k+4}}\mathcal {S}(a_1^{k+1} x)\subseteq \frac{1}{a_1^{4k}}\mathcal {S}(a_1^kx) \end{aligned}$$

for all \(x \in A.\) Let \(\mathcal {S}_k(x) = \frac{1}{a_1^{4k}}\mathcal {S}(a_1^kx)\), \(x \in A\), \(k\in \mathbb {N}\), we obtain that \((\mathcal {S}_k(x))_{k\ge 0}\) is a decreasing sequence of closed subsets of the Banach space Y. We also get \(\mathrm{{diam}}\left( \mathcal {S}_k(x)\right) = \frac{1}{a_1^{4k}}\mathrm{{diam}}\left( \mathcal {S}(a_1^kx)\right) \) for all \(x \in A.\) Taking into account that \(\sup \left\{ \mathrm{{diam}}\left( \mathcal {S}(x)\right) : x \in A\right\} < +\infty \), we conclude that \(\lim _{k\rightarrow +\infty } \mathrm{{diam}}(\mathcal {S}_k(x)) = 0\) for all \(x \in A\). Using the Cantor theorem for the sequence \((\mathcal {S}_k(x))_{k\ge 0}\), we obtain that the intersection \(\bigcap _{k\ge 0}\mathcal {S}_k(x)\) is a singleton set and we denote this intersection by \(\mathcal {Q}(x)\) for all \(x \in A.\) Thus we obtain a mapping \(\mathcal {Q}:A+(-1)A \rightarrow Y\) and \(\mathcal {Q}(x) \in \mathcal {S}_0(x)=\mathcal {S}(x)\) for all \(x \in A.\)

In the same way as in Theorem 2.3, we obtain that \(\mathcal {Q}:A+(-1)A \rightarrow Y\) satisfies

$$\begin{aligned} \biguplus ^{n-1}_{a_{2}x_{2},\ldots ,a_{n}x_{n}}\mathcal {Q}(a_{1}x_{1})= & {} 2^{n-2}\sum ^{}_{1\le i<j \le n}a^2_{i}a^2_{j}\biguplus _{x_{j}}\mathcal {Q}(x_{i})\\&+\, 2^{n-1}\sum ^{n}_{i=1}a^2_{i}\Bigg (a^2_{i}-\sum ^{n}_{j=1, j\ne i}a^2_{j}\Bigg )\mathcal {Q}(x_{i}) \end{aligned}$$

for all \(x_1,\ldots ,x_n \in A\). Putting \(x_2=\cdots =x_{n-1}=0\) in the above and using \(\mathcal {C}(0)=0\) and \(a_n=1\), we gain

$$\begin{aligned} \biguplus _{x_{n}}\mathcal {Q}(a_1x_{1})= a_1^2\biguplus _{x_{n}}\mathcal {Q}(x_{1})+ 2a_1^2(a_1^2 - 1)\mathcal {Q} (x_1)+ 2(1-a_1^2)f (x_n) \end{aligned}$$

for all \(x_1,x_n \in A\). Then it follows from Lemma 2.4 that the mapping \(\mathcal {Q}\) is quartic.

The rest of the proof is similar to the proof of Theorem 2.3. \(\square \)