Throughout this paper, let $I\subseteq ℝ$ be a nonempty open interval. In the sequel, the classes of continuous strictly monotone and continuous positive real-valued functions defined on I will be denoted by $\mathcal{C}M\left(I\right)$ and $\mathcal{C}P\left(I\right)$, respectively.

Let $I\subset ℝ$ be an open interval. A two-variable function $M:I^2\to I$ is called a mean on the interval I if

$\min \left(x,y\right)\le M\left(x,y\right)\le \max \left(x,y\right),x,y\in I,$

holds. If for all $x,y\in I$, $x\ne y$, these inequalities are strict, M is called strict ( [1] ). Obviously, if M is a mean, then M is reflexive, i.e., $M\left(x,x\right)=x$ for all $x\in I$.

Let $M,N:I^2\to I$ be means. A mean $K:I^2\to I$ is called invariant with respect to the mean-type mappings $\left(M,N\right)$, shortly, $\left(M,N\right)$-invariant ( [2] ), if

$K\left(M\left(x,y\right),N\left(x,y\right)\right)=K\left(x,y\right),x,y\in I.$

The simplest example of when the invariance equation holds is the well-known identity

$G\left(\mathcal{A}\left(x,y\right),H\left(x,y\right)\right)=G\left(x,y\right),x,y>0,$

where

$A\left(x,y\right)=\frac{x+y}{2},G\left(x,y\right)=\sqrt{xy},H\left(x,y\right)=\frac{2xy}{x+y},$

are arithmetic mean, geometric mean and harmonic mean respectively.

The Cauchy mean $C_{f,g}:I^2\to I$ is defined by

$C_{f,g}\left(x,y\right):={\left(\frac{f}{g}\right)}^{-1}\left(\frac{{\displaystyle {\int }_0^{1}f}\left(tx+\left(1-t\right)y\right)\text{d}t}{{\displaystyle {\int }_0^{1}g}\left(tx+\left(1-t\right)y\right)\text{d}t}\right),x,y\in I,$

$f,g:I\to ℝ$ are two continuous functions such that $g\in \mathcal{C}P\left(I\right)$ and $\frac{f}{g}$ is bijective.

In 1984, Leach and Sholander [3] proposed the Cauchy mean theorem for difference quotients and the definition of Cauchy mean. Losonczi solved the comparison and equality problems of Cauchy means for more than two variables and the equality problem for the two-variable Cauchy means under the assumption of seven times differentiability in [4] - [6] . Then Matkowski reduced the regularity condition for the equality problem for the two-variable Cauchy means to first-order differentiability [7] . Other definitions for Cauchy have been in [8] [9] . The invariant equations of some special Cauchy mean-type mappings with geometric mean or arithmetic mean have been studied in [10] [11] . Other invariance of more generalized means generated by measure integral have been considered in [12] [13] . Recently, Zhang [14] gave a survey of results dealing with the problem of means.

In this paper, we will study the invariant equation of Cauchy mean with respect to the arithmetic mean, that is to solve

$\begin{array}{l}{\left(\frac{f}{g}\right)}^{-1}\left(\frac{{\displaystyle {\int }_0^{1}f}\left(tx+\left(1-t\right)y\right)\text{d}t}{{\displaystyle {\int }_0^{1}g}\left(tx+\left(1-t\right)y\right)\text{d}t}\right)+{\left(\frac{h}{k}\right)}^{-1}\left(\frac{{\displaystyle {\int }_0^{1}h}\left(tx+\left(1-t\right)y\right)\text{d}t}{{\displaystyle {\int }_0^{1}k}\left(tx+\left(1-t\right)y\right)\text{d}t}\right),\hfill \end{array}$(1.1)

where $x,y\in I$, $f,g,h,k:I\to ℝ$ are continuous functions such that $g,k\in \mathcal{C}P\left(I\right)$ and $\frac{f}{g},\frac{h}{k}$ are injective functions.

In order to describe the regularity conditions related to the two unknown functions $f,g$ generating the mean $C_{f,g}$, we introduce some notations. The class ${\mathcal{C}}_0\left(I\right)$ consists of all those pairs $\left(f,g\right)$ of continuous functions $f,g:I\to ℝ$ such that $g\in \mathcal{C}\left(I\right)$ and $f/g\in \mathcal{C}ℳ\left(I\right)$. For $n\in \mathbb{N}$, we say that the pair $\left(f,g\right)$ is in the class ${\mathcal{C}}_n\left(I\right)$ if $\left(f,g\right)$ are n-times continuously differentiable functions such that and the function $f^{\prime }g-f{g}^{\prime }$ does not vanish anywhere on I. Obviously, the latter condition implies that $f/g$ is strictly monotone, i.e., $f/g\in \mathcal{C}ℳ\left(I\right)$.

For $\left(f,g\right)\in {\mathcal{C}}_2\left(I\right)$, we also introduce the notation

${\Phi }_{f,g}:=\frac{W_{f,g}^{2,0}}{W_{f,g}^{1,0}},{\Psi }_{f,g}:=-\frac{W_{f,g}^{2,1}}{W_{f,g}^{1,0}},$

where the $\left(i,j\right)$-order Wronskian operator $W_{i,j}$ is defined in terms of ith and jth derivatives by

$W_{f,g}^{i,j}:=\left|\begin{array}{ll}{f}^{\left(i\right)}\hfill & f^{\left(j\right)}\hfill \\ g^{\left(i\right)}\hfill & f^{\left(j\right)}\hfill \end{array}\right|.$

Lemma 1 (Lemma 5 in [15] ) Let $\left(f,g\right)\in {\mathcal{C}}_0\left(I\right)$. Then $f,g$ are solutions of the second-order differential equation

$y^{″}={\Phi }_{f,g}{y}^{\prime }+{\Psi }_{f,g}y.$

In what follows, we will give some formulae for the high-order directional derivatives of $C_{f,g}$ at the diagonal points of the Cartesian product $I\times I$. Given a pair $\left(f,g\right)\in {\mathcal{C}}_0\left(I\right)$ and a fixed element $x\in I$, define the function $m_x:=m_{x;f,g}$ in a neighborhood of origin by

$m_x\left(u\right)=m_{x;f,g}\left(u\right):=C_{f,g}\left(x+\frac{1}{2}u,x-\frac{1}{2}u\right).$(2.1)

Using Lemma 2.2 from [16] , we get the following

Lemma 2 Let $n\in \mathbb{N}$, $\left(f,g\right)\in {\mathcal{C}}_n\left(I\right)$. Then, for fixed $x\in I$, the function $m_x$ defined by (2.1) is n-times continuously differentiable at the origin and

$\underset{i=0}{\overset{n}{{\displaystyle \sum }}}\left(\begin{array}{l}n\hfill \\ i\hfill \end{array}\right)\left|\begin{array}{cc}{f}^{\left(i\right)}\left(x\right)& \left(f\circ m_x\right)f^{\left(n-i\right)}\left(0\right)\\ g^{\left(i\right)}\left(x\right)& \left(g\circ m_x\right)g^{\left(n-i\right)}\left(0\right)\end{array}\right|=0.$

Furthermore, $m_x^{\left(2k\right)}\left(0\right)=x$ for $k\in \mathbb{N}$ and for the cases $n=2,4$, we have

${m^{″}}_{x;f,g}\left(0\right)=\frac{1}{12}{\Phi }_{f,g}\left(x\right),$

$\begin{array}{c}{m}_{x;f,g}^{\left(4\right)}\left(0\right)=-\frac{1}{48}\left({\Phi }_{f,g}^3+2{\Phi }_{f,g}{\Psi }_{f,g}\right)\left(x\right)+\frac{1}{80}({{\Phi }^{″}}_{f,g}+3{{\Phi }^{\prime }}_{f,g}{\Phi }_{f,g}+{\Psi }_{f,g}^3\\ +2{\Phi }_{f,g}{\Psi }_{f,g}+2{{\Psi }^{\prime }}_{f,g})\left(x\right).\end{array}$

Lemma 3 If $\left(f,g\right),\left(h,k\right)\in {\mathcal{C}}_1\left(I\right)$, and Equation (1.1) holds, then

${\Phi }_{f,g}=-{\Phi }_{h,k}:=\Phi .$(2.2)

Proof. By Equation (2.1), the invariant Equation (1.1) can be rewritten by

$m_{x;f,g}\left(u\right)+m_{x;h,k}\left(u\right)=0.$(2.3)

By differentiating the above equation twice with respect to variable u, it can be obtained that

${m^{″}}_{x;f,g}\left(u\right)+{m^{″}}_{x;h,k}\left(u\right)=0,$

letting $u=0$

${m^{″}}_{x;f,g}\left(0\right)+{m^{″}}_{x;h,k}\left(0\right)=0.$

And because

${m^{″}}_{x;f,g}\left(0\right)=\frac{1}{12}{\Phi }_{f,g}\left(x\right),{m^{″}}_{x;h,k}\left(0\right)=\frac{1}{12}{\Phi }_{h,k}\left(x\right),$

then

${m^{″}}_{x;f,g}\left(0\right)+{m^{″}}_{x;h,k}\left(0\right)=\frac{1}{12}\left({\Phi }_{h,k}+{\Phi }_{h,k}\right)\left(x\right)=0,$

that is

${\Phi }_{f,g}=-{\Phi }_{h,k}=\Phi .$

$\square$

Lemma 4 If $\left(f,g\right),\left(h,k\right)\in {\mathcal{C}}_4\left(I\right)$, and Equation (1.1) holds, then

$\frac{3}{80}\Phi {\Phi }^{\prime }-\frac{1}{120}\Phi \left({\Psi }_{f,g}-{\Psi }_{f,g}\right)+\frac{1}{80}\left({{\Psi }^{\prime }}_{f,g}+{{\Psi }^{\prime }}_{h,k}\right)=0.$(2.4)

Proof. By differentiating Equation (2.3) fourth with respect to variable u, it can be obtained that

$m_{x;f,g}^{\left(4\right)}\left(u\right)+m_{x;h,k}^{\left(4\right)}\left(u\right)=0,$

letting $u=0$, we have

$m_{x;f,g}^{\left(4\right)}\left(0\right)+m_{x;h,k}^{\left(4\right)}\left(0\right)=0.$

Since

$\begin{array}{l}{m}_{x;f,g}^{\left(4\right)}\left(0\right)=-\frac{1}{48}\left({\Phi }_{f,g}^3+2{\Phi }_{f,g}{\Psi }_{f,g}\right)\left(x\right)\\ +\frac{1}{80}\left({{\Phi }^{″}}_{f,g}+3{{\Phi }^{\prime }}_{f,g}{\Phi }_{f,g}+{\Psi }_{f,g}^3+2{\Phi }_{f,g}{\Psi }_{f,g}+2{{\Psi }^{\prime }}_{f,g}\right)\left(x\right),\end{array}$

$\begin{array}{l}{m}_{x;h,k}^{\left(4\right)}\left(0\right)=-\frac{1}{48}\left({\Phi }_{h,k}^3+2{\Phi }_{h,k}{\Psi }_{h,k}\right)\left(x\right)\\ +\frac{1}{80}\left({{\Phi }^{″}}_{h,k}+3{{\Phi }^{\prime }}_{h,k}{\Phi }_{h,k}+{\Psi }_{h,k}^3+2{\Phi }_{h,k}{\Psi }_{h,k}+2{{\Psi }^{\prime }}_{h,k}\right)\left(x\right),\end{array}$

then we get

$\frac{3}{80}\Phi {\Phi }^{\prime }-\frac{1}{120}\Phi \left({\Psi }_{f,g}-{\Psi }_{f,g}\right)+\frac{1}{80}\left({{\Psi }^{\prime }}_{f,g}+{{\Psi }^{\prime }}_{h,k}\right)=0.$

$\square$

Although it is difficult to directly solve the Cauchy mean invariant equation with respect to the arithmetic mean, some types of invariant equations can also be solved with certain prerequisites. Next, we will solve the Cauchy mean invariant equation with respect to the arithmetic mean when the denominator functions $g,k$ are equal and satisfy the harmonic oscillator equation

$y^{″}=py.$(3.1)

for some $p\in ℝ$. Then, we introduce the sine and cosine type functions $S_p,C_p:ℝ\to ℝ$ by

$S_p\left(x\right):=\{\begin{array}{ll}\sin \sqrt{-p}x,\hfill & p<0,\hfill \\ x,\hfill & p=0,\hfill \\ \sinh \left(\sqrt{p}x\right),\hfill & p>0,\hfill \end{array}{C}_p\left(x\right):=\{\begin{array}{ll}\cos \sqrt{-p}x,\hfill & p<0,\hfill \\ x,\hfill & p=0,\hfill \\ \cosh \left(\sqrt{p}x\right),\hfill & p>0.\hfill \end{array}$

Due to basic results on the second-order linear differential equations, the functions $S_p$ and $C_p$ given above form a fundamental system of solutions for the differential Equation (3.1).

Theorem 1 If $\left(f,g\right),\left(h,k\right)\in {\mathcal{C}}_4\left(I\right)$. Assume $k=g$ satisfying Equation (3.1) and Equation (1.1) holds, then exists $c\in ℝ$ such that

$f=g{\displaystyle \int \frac{1}{g^2}\exp }\left({\displaystyle \int c{g}^{\frac{4}{9}}}\right),h=g{\displaystyle \int \frac{1}{g^2}\exp }\left(-{\displaystyle \int c{g}^{\frac{4}{9}}}\right).$(3.2)

Proof. Since Equation (3.1) holds, by the definitions of ${\Phi }_{f,g}$ and ${\Psi }_{f,g}$ we get

${\Psi }_{f,g}=-\frac{\left|\begin{array}{cc}{f}^{″}& f^{\prime }\\ g^{″}& g^{\prime }\end{array}\right|}{\left|\begin{array}{cc}{f}^{\prime }& f\\ g^{\prime }& g\end{array}\right|}=-\frac{\left|\begin{array}{cc}{f}^{″}& f^{\prime }\\ pg& g^{\prime }\end{array}\right|}{\left|\begin{array}{cc}{f}^{\prime }& f\\ g^{\prime }& g\end{array}\right|}=-\frac{g^{\prime }}{g}{\Phi }_{f,g}+p.$

Similarly, we have

${\Psi }_{h,k}=-\frac{k^{\prime }}{k}{\Phi }_{h,k}+p.$

Using (2.2) and the above two equalities, we have

${\Psi }_{f,g}+{\Psi }_{h,k}=2p,{\Psi }_{f,g}-{\Psi }_{h,k}=-2\frac{g^{\prime }}{g}\Phi ,$

and thus ${{\Psi }^{\prime }}_{f,g}+{{\Psi }^{\prime }}_{h,k}=0$. Consequently, (2.4) becomes

$\frac{3}{80}\Phi {\Phi }^{\prime }+\frac{g^{\prime }}{60g}{\Phi }^2=0,$

which implies that there exists some $c\in ℝ$ such that

$\Phi =c{g}^{\frac{4}{9}}.$

Hence, we obtain

${\Phi }_{f,g}=\frac{{\left(W_{f,g}^{1,0}\right)}^{\prime }}{W_{f,g}^{1,0}}=c{g}^{\frac{4}{9}},{\Phi }_{h,k}=\frac{{\left(W_{h,k}^{1,0}\right)}^{\prime }}{W_{h,k}^{1,0}}=-c{g}^{\frac{4}{9}},$

Since $W_{f,g}^{1,0}=\left(\frac{f}{g}\right)g^2$ and $W_{h,k}^{1,0}=\left(\frac{h}{k}\right)k^2$, integrating the above equations, we get (3.2). $\square$

In what follows, we will restrict to the basic solutions to the equation of harmonic oscillator, i.e. consider the functions

(H1) $g\left(x\right)=x$ for $x\in I\subseteq \left(0,+\infty \right)$;

(H2) $g\left(x\right)={\text{e}}^x$ for $x\in I\subseteq ℝ$;

(H3) $g\left(x\right)=\sin x$ for $x\in I\subseteq \left(0,\frac{\pi }{2}\right)$.

Theorem 2 If $\left(f,g\right),\left(h,k\right)\in {\mathcal{C}}_4\left(I\right)$. Assume $k=g$ satisfying one of the conditions (H1) - (H3) and Equation (1.1) holds, then

1) For case (H1), there exists $b_1,b_2\in {ℝ}_{+},c\in ℝ$,

$f\left(x\right)=b_1{\displaystyle \int \frac{1}{x^2}\exp }\left(c{x}^{\frac{13}{9}}\right),h\left(x\right)=b_2{\displaystyle \int \frac{1}{x^2}\exp }\left(c{x}^{\frac{13}{9}}\right),x\in I.$

2) For case (H2), there exists $b_1,b_2\in {ℝ}_{+},c\in ℝ$,

$f\left(x\right)=b_1{\text{e}}^x{\displaystyle \int {\text{e}}^{-2x}\exp }\left(c{\text{e}}^{\frac{4x}{9}}\right),h\left(x\right)=b_2{b}_1{\text{e}}^x{\displaystyle \int {\text{e}}^{-2x}\exp }\left(-c{\text{e}}^{\frac{4x}{9}}\right),x\in I.$

3) For case (H3), there exists $c\in ℝ$,

$\begin{array}{l}f\left(x\right)=\cos x{\displaystyle \int \frac{1}{{\cos }^{2}x}\exp }\left({\displaystyle \int c{\cos }^{\frac{4}{9}}x}\right),\\ h\left(x\right)=\cos x{\displaystyle \int \frac{1}{{\cos }^{2}x}\exp }\left({\displaystyle \int -c{\cos }^{\frac{4}{9}}x}\right),x\in I.\end{array}$

Next, we will consider the solution of the invariant equation for (1.1) when the denominator functions are power functions, that is $g\left(x\right)=x^p$, $k\left(x\right)=x^r$.

Theorem 3 Let $g\left(x\right)=x^p,k\left(x\right)=x^r,p=r$ and the invariant Equation (1.1) holds, then there exists $c_1,c_2\in ℝ$ such that

$f=g{\displaystyle \int \frac{1}{g^2}\exp }\left(\pm {\displaystyle \int \sqrt{c_1\beta +c_2}}\right),h=g{\displaystyle \int \frac{1}{g^2}\exp }\left(\pm {\displaystyle \int \sqrt{c_1\beta +c_2}}\right),$

where $\beta =\frac{p^3\left(1-p\right)}{x^2}$.

Proof. When $g=k=x^p$we have $g^{\prime }\left(x\right)=k^{\prime }\left(x\right)=p{x}^{p-1}$ and $g^{″}\left(x\right)=k^{″}\left(x\right)=p\left(p-1\right)x^{p-2}$, the relationship between ${\Phi }_{f,g}$ and ${\Psi }_{f,g}$ is

${\Psi }_{f,g}\left(x\right)=-\frac{p}{x}{\Phi }_{f,g}\left(x\right)+\frac{p\left(p-1\right)}{x^2}.$

Similarly

${\Psi }_{h,k}\left(x\right)=-\frac{p}{x}{\Phi }_{h,k}\left(x\right)+\frac{p\left(p-1\right)}{x^2}.$

Using (2.2) and the above two equalities, we have

${\Psi }_{f,g}\left(x\right)+{\Psi }_{h,k}\left(x\right)=\frac{2p\left(p-1\right)}{x^2},{\Psi }_{f,g}\left(x\right)-{\Psi }_{h,k}\left(x\right)=-\frac{2p\Phi }{x},$

and thus

${{\Psi }^{\prime }}_{f,g}\left(x\right)+{{\Psi }^{\prime }}_{h,k}\left(x\right)=-\frac{4p\left(p-1\right)}{x^3}.$

Consequently, (2.4) becomes

$\frac{3}{80}\Phi \left(x\right){\Phi }^{\prime }\left(x\right)+\frac{p}{60x}{\Phi }^2\left(x\right)-\frac{p\left(p-1\right)}{20x^3}=0,$

which implies that there exist $c_1,c_2\in ℝ$ such that

${\Phi }^2\left(x\right)=c_1\frac{p^3\left(1-p\right)}{x^2}+c_2.$

Hence, we obtain

${\Phi }_{f,g}=\frac{{\left(W_{f,g}^{1,0}\right)}^{\prime }}{W_{f,g}^{1,0}}=c{g}^{\frac{4}{9}},{\Phi }_{h,k}=\frac{{\left(W_{h,k}^{1,0}\right)}^{\prime }}{W_{h,k}^{1,0}}=-c{g}^{\frac{4}{9}}.$

By $W_{f,g}^{1,0}=\left(\frac{f}{g}\right)g^2$ and $W_{h,k}^{1,0}=\left(\frac{h}{k}\right)k^2$, integrating the above equations, we get the result. $\square$