Regular solutions of a functional equation derived from the invariance problem of Matkowski means

Abstract

The main result of the present paper is about the solutions of the functional equation

$$\begin{aligned} F\Big (\frac{x+y}{2}\Big )+f_1(x)+f_2(y)=G(g_1(x)+g_2(y)),\quad x,y\in I, \end{aligned}$$

derived originally, in a natural way, from the invariance problem of generalized weighted quasi-arithmetic means, where $F,f_1,f_2,g_1,g_2:I\rightarrow {\mathbb {R}}$ and $G:g_1(I)+g_2(I)\rightarrow {\mathbb {R}}$ are the unknown functions assumed to be continuously differentiable with $0\notin g'_1(I)\cup g'_2(I)$, and the set I stands for a nonempty open subinterval of ${\mathbb {R}}$. In addition to these, we will also touch upon solutions not necessarily regular. More precisely, we are going to solve the above equation assuming first that F is affine on I and $g_1$ and $g_2$ are continuous functions strictly monotone in the same sense, and secondly that $g_1$ and $g_2$ are invertible affine functions with a common additive part.

1 Introduction

Before we define the key concepts and formulate the main problem, we will introduce some notions and conventions that will be indispensable later. In our main equation and the many related results, we will have functions, constants, etc. that are distinguished only by numbering. For brevity, whenever the index k appears as a subscript of a function, a constant, etc., it should be understood that the statement or condition in question is fulfilled for any index k, where k is running on the two-element set $\{1,2\}$.

Let $J\subseteq {\mathbb {R}}$ be a non-empty open subinterval. A function $f:J\rightarrow {\mathbb {R}}$ will be called affine on some subinterval $U\subseteq J$ if it fulfills Jensen’s Equation on U, that is, if

$$\begin{aligned} {f\Big (\frac{u+v}{2}\Big )=\frac{f(u)+f(v)}{2},\quad u,v\in U.} \end{aligned}$$

We shall say that f is locally affine on J or locally constant on J if f is affine or constant on some subinterval $U\subseteq J$ of positive length different from J. If f is neither affine (resp. constant) nor locally affine (resp. locally constant) on J, then it will be called nowhere affine (resp. nowhere constant) on J.

We say that a two-place function $M:J\times J\rightarrow {\mathbb {R}}$ is a two-variable mean on J or, shortly, a mean if

$$\begin{aligned} \min (u,v)\le M(u,v)\le \max (u,v),\quad u,v\in J. \end{aligned}$$

Elementary examples of means are the harmonic mean and the geometric mean over the set of positive reals, and the arithmetic mean. Rather than giving furter examples, we immediately introduce a general family of two-variable means, which covers the previous examples and which, incidentally, is also the object of our investigation.

A mean $M:J\times J\rightarrow {\mathbb {R}}$ will be called a generalized weighted quasi-arithmetic mean if one can find continuous functions $f,g:J\rightarrow {\mathbb {R}}$ strictly monotone in the same sense such that

$$\begin{aligned} {M(u,v)=(f+g)^{-1}(f(u)+g(v)),\quad u,v\in J,} \end{aligned}$$

(1.1)

where the functions f and g are called the generators of the mean in question. If the generators differ on J in an additive constant, then we come to the notion of quasi-arithmetic means. To see that our previous examples are indeed of such a type, set $f{:}{=}g{:}{=}\mathop {\text{ id }}\nolimits ^{-1}$, $f{:}{=}g{:}{=}\ln$, and $f{:}{=}g{:}{=}\mathop {\text{ id }}\nolimits$, respectively. (Obviously, the exponent $-1$ here stands for the multiplicative inverse.)

The class of generalized weighted quasi-arithmetic means was introduced in 2010 by Matkowski [27]. Motivated by this, means of the form (1.1) are often referred to as Matkowski means and, to indicate the generators too, denoted by ${\mathscr {M}}_{f,g}$. Note that, by the definition, any Matkowski mean is continuous and strictly monotone, furthermore is symmetric or balanced if and only if it is a quasi-arithmetic mean [19, 27].

A mean $M:J\times J\rightarrow {\mathbb {R}}$ will be called invariant with respect to the pair of means $(N,K):J\times J\rightarrow {\mathbb {R}}^2$ if the invariance equation

$$\begin{aligned} { M(N(u,v),K(u,v))=M(u,v),\quad u,v\in J} \end{aligned}$$

is fulfilled.

The most frequently cited (and an easy-to-check) example of this phenomenon in the literature is the invariance of the geometric mean with respect to the pair of arithmetic and harmonic means. A less trivial example is the arithmetic-geometric mean appearing already in Lagrange’s and Gauss’s works, which is invariant with respect to the pair of arithmetic and geometric means.

As more and more families of means have been introduced, over the last 20 years, the invariance problem has again become an area of active research. Without being exhaustive, we list some related papers. Interested readers can find a more detailed discussion of the invariance equation in [11, Daróczy–Páles] and in the survey [16, Jarczyk–Jarczyk].

The invariance of the arithmetic mean with respect to a pair of two quasi-arithmetic means under two times continuous differentiability of the generators was solved by Matkowski in 1999 [24]. As this regularity condition is not natural for the underlying problem, it was gradually weakened in the following years by Darôzy, Maksa, and Páles in the papers [8] and [10]. The final answer using only the necessary conditions was given in 2002 by Daróczy and Páles [11].

The invariance equation for three quasi-arithmetic means was investigated by Burai [5, 6], by Jarczyk and Matkowski [17], and by Jarczyk [15], where, in the last paper, the unnecessary regularity conditions were eliminated too.

A paper closely related to the present investigation is [3], where, under four times continuous differentiability of the generators, the authors solved the problem of invariance of the arithmetic mean with respect to a pair of Matkowski means.

More studies on the invariance problem concerning different classes of means can be found in [2, Baják–Páles], [25, 26, 28, 29, Matkowski], [7, Błasińska–Głazowska–Matkowski], [13, Głazowska–Jarczyk–Matkowski], [12, Domsta–Matkowski], and [31, Pàles–Zakaria].

2 The invariance equation and its reformulation

To make our underlying problem more manageable, relying only on continuity and strict monotonicity of the generators, we are going to give an equivalent formulation of the invariance equation of Matkowski means. This new equation will concern the composition of the initial generators and, like the original equation, will still contain six unknown functions. Then, under differentiability and a technical condition, we derive a system of two functional equations, in which the individual equations contain only 3–3 unknown functions, such that one of these will be common.

The following result states the relation between the invariance problem and our main equation mentioned in the abstract. Note that the domain of the latter is not necessarily a rectangle symmetric with respect to the diagonal.

Theorem 1

Let $(m_1,m_2),(n_1,n_2),(k_1,k_2):J\rightarrow {\mathbb {R}}^2$ be such that the coordinate-functions of each ordered pair are continuous and strictly monotone in the same sense. Then the invariance equation of Matkowski means

$$\begin{aligned} {\mathscr {M}}_{m_1,m_2}\big ({\mathscr {M}}_{n_1,n_2}(u,v),{\mathscr {M}}_{ k_1,k_2}(u,v)\big )={\mathscr {M}}_{m_1,m_2}(u,v),\quad u,v\in J \end{aligned}$$

(2.1)

holds if and only if for the composite functions

$$\begin{aligned} \begin{array}{lll} F{:}{=}-m_2\circ \Big (\dfrac{k_1+k_2}{2}\Big )^{-1}, &{}\quad f_1{:}{=}m_1\circ k_1^{-1}, &{}\quad f_2{:}{=}m_2\circ k_2^{-1},\\ G{:}{=}m_1\circ (n_1+n_2)^{-1}, &{}\quad g_1{:}{=}n_1\circ k_1^{-1}, &{}\quad g_2{:}{=}n_2\circ k_2^{-1} \end{array} \end{aligned}$$

(2.2)

we have

$$\begin{aligned} {F\Big (\frac{x+y}{2}\Big )+f_1(x)+f_2(y)=G(g_1(x)+g_2(y))} \end{aligned}$$

(2.3)

$for all x\in k_1(J) and y\in k_2(J)$.

Proof

Assume first that Eq. (2.1) holds on $J\times J$, and let $x\in k_1(J)$ and $y\in k_2(J)$ be arbitrary. The functions $k_1$ and $k_2$ are continuous and strictly monotone, therefore we uniquely have $u,v\in J$ with $k_1(u)=x$ and $k_2(v)=y$. Then $u=k_1^{-1}(x)$ and $v=k_2^{-1}(y)$. Applying Eq. (2.1) for the pair (u, v), expanding the Matkowski means included in it by definition, and, finally, applying the function $m_1+m_2$ on both sides, we get

$$\begin{aligned}&m_1\circ (n_1+n_2)^{-1}(n_1(u)+n_2(v))+ m_2\circ (k_1+k_2)^{-1}(k_1(u)+k_2(v))\\&\quad =m_1(u)+m_2(v). \end{aligned}$$

Using the definition of u and v, an obvious reformulation yields that

$$\begin{aligned}&-m_2\circ (k_1+k_2)^{-1}(x+y)+m_1\circ k_1^{-1}(x)\\&\quad +m_2\circ k_2^{-1}(y)= m_1\circ (n_1+n_2)^{-1}(n_1\circ k_1^{-1}(x)+n_2\circ k_2^{-1} (y)). \end{aligned}$$

Replacing $x+y$ in the argument of $\gamma _2\circ (\beta _1+\beta _2)$ by $2\cdot \frac{x+y}{2}$, and applying the definition of the functions in (2.2), we obtain that (2.3) is valid.

To prove that the validity of (2.3) implies the validity of (2.1), just reverse the above argumentation. $\square$

It is important to note that Eq. (2.3), considered on an interval of the form $]0,\alpha [$ with $\alpha >0$, has previously appeared in the paper [18] of Járai, Maksa, and Páles under the setting $-F\circ \frac{1}{2}\mathop {\text{ id }}\nolimits =f_1=f_2$ and $g{:}{=}g_1=g_2$. In [18] the authors solved the related equation assuming that G and g are continuous and strictly monotone.

Now we simplify our problem further. Unfortunately, to do this, in the rest of the paper we have to assume that $k_1(J)=k_2(J)=:I$ holds, where due to the properties of J, $k_1$, and $k_2$, the set I is a non-empty open subinterval of ${\mathbb {R}}$. Without this, by the way, though not a natural technical condition, we could not derive the following system of equations. On the other hand, let us point out that, to prove the next result, we do not yet require continuity of the derivatives.

Theorem 2

Let $(F,f_1,f_2,G,g_1,g_2)$ be a solution of (2.3) such that each coordinate-function is differentiable on its domain with $0\notin g_1'(I)\cup g_2'(I)$. Then the system of functional equations

$$\begin{aligned} {\left\{ \begin{array}{ll} \varphi \big (\frac{x+y}{2}\big )(\psi _1(x)+\psi _1(y)) =\Psi _1(x)+\Psi _1(y),\\ \varphi \big (\frac{x+y}{2}\big )(\psi _2(x)-\psi _2(y)) =\Psi _2(x)-\Psi _2(y), \end{array}\right. } \quad x,y\in I \end{aligned}$$

(2.4)

holds, where

$$\begin{aligned} \varphi {:}{=}\frac{1}{2}F',\quad \psi _k{:}{=}\frac{1}{g'_1}+(-1)^k\frac{1}{g'_2}, \quad \text {and}\quad \Psi _k{:}{=}-\frac{f_1'}{g_1'}+(-1)^{k-1}\frac{f_2'}{g_2'}. \end{aligned}$$

(2.5)

Proof

Differentiating (2.3) with respect to x and y separately, and putting $\varphi$ defined in (2.5), we obtain that

$$\begin{aligned}&\varphi \Big (\frac{x+y}{2}\Big )+f_1'(x) =G'(g_1(x)+g_2(y))g_1'(x)\;\text { and }\\&\varphi \Big (\frac{x+y}{2}\Big )+f_2'(y)=G'(g_1(x)+g_2(y))g_2'(y) \end{aligned}$$

hold for all $x,y\in I$, respectively. Multiplying the first and the second equation by $g_2'(y)$ and $-g_1'(x)$, respectively, and then adding up side by side the equations so obtained, we get

$$\begin{aligned} \varphi \Big (\frac{x+y}{2}\Big )\big (g_2'(y)-g_1'(x)\big )=f_2'(y)g_1'(x)-f_1'(x)g_2'(y),\quad (x, y)\in I\times I.\qquad \end{aligned}$$

(2.6)

This equation implies that

$$\begin{aligned} \varphi \Big (\frac{x+y}{2}\Big )\big (g_2'(x)-g_1'(y)\big ) =f_2'(x)g_1'(y)-f_1'(y)g_2'(x),\quad (x, y)\in I\times I.\qquad \end{aligned}$$

(2.7)

Dividing Eqs. (2.6) and (2.7) by $g_1'(x)g_2'(y)\ne 0$ and $g_1'(y)g_2'(x)\ne 0$, respectively, we obtain

$$\begin{aligned}&\varphi \Big (\frac{x+y}{2}\Big )\Big (\frac{1}{g_1'(x)}-\frac{1}{g_2'(y)}\Big ) =\frac{f_2'(y)}{g_2'(y)}-\frac{f_1'(x)}{g_1'(x)}\quad \text {and}\\&\varphi \Big (\frac{x+y}{2}\Big )\Big (\frac{1}{g_1'(y)}-\frac{1}{g_2'(x)}\Big ) =\frac{f_2'(x)}{g_2'(x)}-\frac{f_1'(y)}{g_1'(y)} \end{aligned}$$

for all $x,y\in I$. Taking the sum of the above equations and subtracting the second equation from the first, and then applying definition (2.5), we get the first and the second equation of (2.4), respectively. $\square$

Remark 3

Note that $\psi _2+\psi _1=\frac{2}{g'_1}\ne 0$ and $\psi _2-\psi _1=\frac{2}{g'_2}\ne 0$ hold on I. Thus, by (2.5) of Theorem 2, we have

$$\begin{aligned} F'=2\varphi ,\quad g'_j=\frac{2}{\psi _2+(-1)^{j-1}\psi _1},\quad \text {and}\quad f'_j=-\frac{\Psi _2+(-1)^{j-1}\Psi _1}{\psi _2+(-1)^{j-1}\psi _1}\quad \text {on }I. \end{aligned}$$

The individual functional equations of system (2.4) has a rich literature. In the rest of this section we recall the related results.

The upper equation of (2.4) Actually, this equation contains only two unknown functions, so, by substituting $x=y$, we immediately obtain that $\Psi _1$ must be of the form $\varphi \psi _1$. Thus it can be reformulated as

$$\begin{aligned} \varphi \Big (\frac{x+y}{2}\Big ) (\psi _1(x)+\psi _1(y))=\varphi (x)\psi _1(x)+\varphi (y)\psi _1(y),\quad x,y\in I. \end{aligned}$$

(2.8)

A functional equation of a similar form was studied by Lundberg in [23]. The functions involved were complex valued and, in essence, infinitely many times differentiable. Five years later, in order to solve the equality problem of two-variable functionally weighted quasi-arithmetic means (or shortly, Bajraktarević means) and quasi-arithmetic means, Daróczy, Maksa, and Páles [9] solved the above equation. Moreover, they determined the solutions under natural conditions needed to formulate their initial problem, to be more precise, $\varphi$ was supposed to be continuous and strictly monotone and $\psi _1$ was positive on its domain.

The best result known today concerning the above equation was obtained in [20]. In [20] Eq. (2.8) is totally solved under the continuity of $\varphi$ and assuming a regularity property of the zeros of the function $\psi _1$. More specifically, the validity of the inclusion $I\setminus {\mathscr {Z}}_{\psi _1}\subseteq \mathrm {cl}_I\circ \mathrm {conv}\big (I\setminus \mathrm {cl}_I\,{\mathscr {Z}}_{\psi _1}\big )$ is required, where ${\mathscr {Z}}_{\psi _1}$ denotes the zeros of $\psi _1$ in I and the operators $\mathrm {conv}$ and $\mathrm {cl}_I$ stand for the convex hull and the closure with respect to I, respectively. At first glance, this inculsion condition may seem artificial or technical, but notice that it is trivially satisfied provided that $\psi _1$ is continuous or injective. Hence, instead of the above general inclusion condition, we are going to formulate the corresponding theorem of [20] under the continuity of $\psi _1$. For our purposes, this will be enough.

Theorem 4

Let $\varphi ,\psi _1:I\rightarrow {\mathbb {R}}$ be continuous functions. Then $(\varphi ,\psi _1)$ solves Eq. (2.8) if and only if either

• (4.1) there is an interval $L\subseteq I$ such that $\psi _1=0$ on $I\setminus L$ and $\varphi$ is constant on $\frac{1}{2}(L+I)$, or

• (4.2) there exist constants $a,b,c,d,\gamma \in {\mathbb {R}}$ with $ad\ne bc$ such that exactly one of the conditions

  • (a) $\gamma <0$ and

    $$\begin{aligned} { \varphi (x)=\frac{c\sin (\kappa x)+d\cos (\kappa x)}{a\sin (\kappa x)+b\cos (\kappa x)} \quad \text {and}\quad \psi _1(x)=a\sin (\kappa x)+b\cos (\kappa x)\ne 0,\quad \text {or}} \end{aligned}$$

  • (b) $\gamma =0$ and

    $$\begin{aligned} { \varphi (x)=\frac{cx+d}{ax+b}\quad \text {and}\quad \psi _1(x)=ax+b\ne 0,\quad \text {or}} \end{aligned}$$

  • (c) $\gamma >0$ and

    $$\begin{aligned} { \varphi (x)=\frac{c\sinh (\kappa x)+d\cosh (\kappa x)}{a\sinh (\kappa x)+b\cosh (\kappa x)} \quad \text {and}\quad \psi _1(x)=a\sinh (\kappa x)+b\cosh (\kappa x)\ne 0} \end{aligned}$$

  holds for all $x\in I$, where $\kappa {:}{=}\sqrt{|\gamma |}$.

The lower equation of (2.4) The first remarkable investigation concerning this equation is due to Aczél [1]. In 1963, he solved the equation under the assumption $\psi _2=\mathop {\text{ id }}\nolimits$. Keeping this condition, and even assuming that the right-hand side is the difference of two not necessarily equal functions, in 1979, Haruki [14] obtained the same result as Aczél.

In the early 2000s, the equation reappeared in the paper [11] of Daróczy and Páles,where, motivated by the underlying problem, the authors studied it assuming that $\Psi _2=\varphi \psi _2$.

Later, in 2016, by Balogh, Ibrogimov, and Mityagin [4], and then, in 2018, by Łukasik [22], literally

$$\begin{aligned} {\varphi \Big (\frac{x+y}{2}\Big )\big (\psi _2(x)-\psi _2(y)\big )=\Psi _2(x)-\Psi _2(y),\quad (x,y\in I) } \end{aligned}$$

(2.9)

was considered under the assumption that $\psi _2$ and $\Psi _2$ are three-times differentiable and continuously differentiable functions, respectively.

The result requiring the weakest regularity conditions known today can be found in [21], where the same solution was obtained as in [4, 22] but assuming only that $\varphi$ is continuous. To make this result easier to formulate, we introduce the following notation. For a subset $H\subseteq I$, define

$$\begin{aligned} { H_-(I){:}{=}\{x\in I\mid x<\inf H\}\quad \text {and}\quad H_+(I){:}{=}\{x\in I\mid \sup H<x\}. } \end{aligned}$$

Obviously, if H is empty, then $H_-(I)=H_+(I)=I$, furthermore, we have $\inf H=\inf I$ or $\sup H=\sup I$, if and only if $H_-(I)=\emptyset$ or $H_+(I)=\emptyset$, respectively.

Theorem 5

Let $\varphi ,\psi _2,\Psi _2:I\rightarrow {\mathbb {R}}$ be such that $\varphi$ is continuous. Then $(\varphi ,\psi _2,\Psi _2)$ solves Eq. (2.9) if and only if either

• (5.1) there exist $A^*\in \varphi (I)$, a closed interval $K\subseteq I$, and $\mu ^*\in {\mathbb {R}}$ such that $\Psi _2=A^*\psi _2+\mu ^*$ on I, the function $\psi _2$ is constant on $K_-(I)$ and $K_+(I)$, and $\varphi =A^*$ on the interval $\frac{1}{2}(K+I)$, or

• (5.2) there exist constants $a^*,b^*,c^*,d^*,\gamma ^*\in {\mathbb {R}}$ with $a^*d^*\ne b^*c^*$ and $\mu ^*,\lambda ^*\in {\mathbb {R}}$ such that exactly one of the conditions

  • (a) $\gamma ^*<0$ and

    $$\begin{aligned}&\varphi (x)=\frac{c^*\sin (\kappa ^*x)+d^*\cos (\kappa ^*x)}{a^*\sin (\kappa ^*x)+b^*\cos (\kappa ^*x)} \quad \text {and}\\&\begin{array}{l} \psi _2(x)=-a^*\cos (\kappa ^*x)+b^*\sin (\kappa ^*x)+\lambda ^*,\\ \!\Psi _2(x)=-c^*\cos (\kappa ^*x)+d^*\sin (\kappa ^*x)+\mu ^*, \end{array}\quad \text {or} \end{aligned}$$

  • (b) $\gamma ^*=0$ and

    $$\begin{aligned} \varphi (x)=\frac{c^*x+d^*}{a^*x+b^*}\quad \text {and}\quad \begin{array}{l} \psi _2(x)=\frac{1}{2}a^*x^2+b^*x+\lambda ^*,\\ \!\Psi _2(x)=\tfrac{1}{2}c^*x^2+d^*x+\mu ^*, \end{array}\quad \text {or} \end{aligned}$$

  • (c) $\gamma ^*>0$ and

    $$\begin{aligned}&\varphi (x)=\frac{c^*\sinh (\kappa ^*x)+d^*\cosh (\kappa ^*x)}{a^*\sinh (\kappa ^*x)+b^*\cosh (\kappa ^*x)} \quad \text {and}\\&\begin{array}{l} \psi _2(x)=a^*\cosh (\kappa ^*x)+b^*\sinh (\kappa ^*x)+\lambda ^*,\\ \!\Psi _2(x)=c^*\cosh (\kappa ^*x)+d^*\sinh (\kappa ^*x)+\mu ^* \end{array} \end{aligned}$$

  holds for all $x\in I$, where $\kappa ^*{:}{=}\sqrt{|\gamma ^*|}$.

The above results will be used to determine solutions that can be relevant to the invariance problem. But first, to handle the cases that arise, we need to discuss solutions that include affine functions.

3 Solutions with affine member

For the functions $g_1,g_2:I\rightarrow {\mathbb {R}}$, $U\subseteq I$, and $x\in I$, define the sets

$$\begin{aligned} U_1(x)&{:}{=}&\{y\in U\mid g_1(x)+g_2(y)\in g_1(U)+g_2(U)\} \quad \text {and}\\ U_2(x)&{:}{=}&\{y\in U\mid g_1(y)+g_2(x)\in g_1(U)+g_2(U)\}. \end{aligned}$$

Then, obviously, $U\subseteq U_1(x)=U_2(x)$ whenever $x\in U$.

Lemma 6

If $g_1,g_2:I\rightarrow {\mathbb {R}}$ are continuous functions strictly monotone in the same sense and $U{:}{=}\,]a,b[\,\subseteq I$ for some $a<b$, then the following statements hold.

• (1) If $a\in I$ (resp. $b\in I$), then $U\subseteq U_k(a)$ (resp. $U\subseteq U_k(b)$).

• (2) If $a\in I$ (resp. $b\in I$), then there exists $x\in I$ with $x<a$ (resp. with $b<x$) such that $U_k(x)\ne \emptyset$.

• (3) If $x\in I$ with $x<a$ (resp. $b<x$) and $U_k(x)\ne \emptyset$, then, for all $u\in [x,a]$ (resp. for all $u\in [b,x]$), we have $U_k(x)\subseteq U_k(u)$.

• (4) For all $x\in I$ with $x<a$ (resp. with $b<x$), we have $a<\inf U_k(x)$ (resp. $\sup U_k(x)<b$). Furthermore, if $U_k(x)\ne \emptyset$, then $\sup U_k(x)=b$ (resp. $a=\inf U_k(x)$).

Proof

For brevity, let $H{:}{=}g_1(U)+g_2(U)$. We perform the proof under the assumption $a\in I$ and for the index $k=1$. The proof of the other sub-cases can be done analogously. For consistency, in some steps, the strictly increasing and strictly decreasing cases will be treated in parallel.

Due to the facts that U is a non-empty open subinterval of I, and $g_1$ and $g_2$ are continuous, strictly monotone, $g_1(a)+g_2(U)$ and H are non-empty open intervals as well. On the other hand, if $g_1$ and $g_2$ are strictly increasing or strictly decreasing, then

$$\begin{aligned}&\inf g_1(a)+g_2(U)=\inf H=g_1(a)+g_2(a)\in {\mathbb {R}}\quad \text {and}\\&\sup g_1(a)+g_2(U)\le \sup H,\quad \text {or}\\&\sup g_1(a)+g_2(U)=\sup H=g_1(a)+g_2(a)\in {\mathbb {R}}\quad \text {and}\\&\quad \inf g_1(a)+g_2(U)\ge \inf H, \end{aligned}$$

respectively. These show that (1) holds.

To show (2), indirectly assume that, for all $x\in I$ with $x<a$, the sum $g_1(x)+g_2(y)$ is not contained in H for all $y\in U$. The two-place function $(x,y)\mapsto g_1(x)+g_2(y)$ is continuous on $]\inf I,a\,[\,\times \,U$, hence the image of this product by the function in question is a subinterval of ${\mathbb {R}}$. Moreover, by the strict monotonicity of $g_1$ and $g_2$, the image is of positive length. Therefore, in view of our indirect assumption, it follows that we have either $g_1(x)+g_2(y)\le \inf H$ or $\sup H\le g_1(x)+g_2(y)$ for all $x\in \,]\inf I,a\,[$ and for all $y\in U$.

If $g_1$ and $g_2$ are strictly increasing and the first inequality holds, then, taking $x\rightarrow a^-$ and using the continuity of $g_1$, we get that $g_2(y)\le g_2(a)$ holds for all $y\in U$. This contradicts the fact that $g_2$ is strictly increasing. If the second inequality is valid, then necessarily $\sup H\in {\mathbb {R}}$, and, particularly, for all $y\in U$ and for all $x\in \,]\inf I,a[\,$, we have $g_1(y)+g_2(y)\le \sup H\le g_1(x)+g_2(y)$. This reduces to $g_1(y)\le g_1(x)$, which is a contradiction again.

The proof of (2) for the strictly decreasing case can be easily constructed by modifying the preceding reasoning.

To prove (3), let $x\in I$ be such that $x<a$ and $U_1(x)\ne \emptyset$. For any $y\in U_1(x)$, the function $\xi \mapsto g_1(\xi )+g_2(y)$ is continuous on [x, a]. Then, by the choice of y, we have $g_1(x)+g_2(y)\in H$ and, by statement (1), $g_1(a)+g_2(y)\in H$. Hence, due to the Darboux Property of $\xi \mapsto g_1(\xi )+g_2(y)$, (3) follows.

To prove the first statement of (4), take $x\in I$ with $x<a$. If $U_1(x)=\emptyset$, then there is nothing to prove, hence we may assume that $U_1(x)\ne \emptyset$. Furthermore, assume indirectly that $\inf U_1(x)=a$ holds. Then, for some $r_0>0$, we have $a+r\in U_1(x)$ for all $0<r<r_0$. Hence $g_1(x)+g_2(a+r)\in H$. Consequently, if $g_1$ and $g_2$ are strictly increasing or strictly decreasing, by continuity, it follows that $g_1(a)\le g_1(x)$ or $g_1(x)\le g_1(a)$, respectively, contradicting that $x<a$.

The second statement of (4) is obvious. $\square$

Now, keeping the previous notation, let

$$\begin{aligned} {U^+{:}{=}\{x\in I\mid U_1(x)\cap U_2(x)\ne \emptyset \}. } \end{aligned}$$

By $U\subseteq U^+$, the set $U^+$ is non-empty. Furthermore, by (2) and (3) of Lemma 6, this inclusion is strict whenever $U\ne I$, and $U^+$ is a subinterval of I. Thus, roughly speaking, $U^+$ is a proper continuation of the interval U in I provided that $U\ne I$.

Proposition 7

Let $F:I\rightarrow {\mathbb {R}}$ be a function affine on some non-empty open subinterval $U\subseteq I$, and $g_1,g_2:I\rightarrow {\mathbb {R}}$ be continuous functions strictly monotone in the same sense. If $(F,f_1,f_2,G,g_1,g_2)$ solves functional Eq. (2.3) then there exist an additive function $B:{\mathbb {R}}\rightarrow {\mathbb {R}}$ and constants $\lambda _1,\lambda _2\in {\mathbb {R}}$, such that

$$\begin{aligned} { f_k=-\frac{1}{2}F+B\circ g_k+\lambda _k \quad \text {and}\quad G=B+\Lambda } \end{aligned}$$

(3.1)

hold on $U^+\cap (2U-U)\cap I$ and on $g_1(U)+g_2(U)$, respectively, where $\Lambda {:}{=}\lambda _1+\lambda _2$.

Proof

First, we are going to show that the formulas concerning the functions $f_k$ and G in (3.1) hold on U and on $g_1(U)+g_2(U)$, respectively.

Define $\ell _k{:}{=}F+2f_k$ on U. Expressing $f_k$ from here, substituting it back into (2.3), and using the fact that F is affine on U, we obtain that

$$\begin{aligned} G\big (g_1(x)+g_2(y)\big )=\frac{\ell _1(x)+\ell _2(y)}{2},\quad x,y\in U. \end{aligned}$$

Putting $u{:}{=}g_1(x)$ and $v{:}{=}g_2(y)$, and using that $g_1$ and $g_2$ are continuous and strictly monotone, it follows that

$$\begin{aligned} 2G(u+v)=\ell _1\circ g^{-1}(u)+\ell _2\circ g_2^{-1}(v),\quad (u,v)\in g_1(U)\times g_2(U). \end{aligned}$$

Then, in view of the celebrated Theorem 1 of [30, Radó–Baker], there exist $B^*:{\mathbb {R}}\rightarrow {\mathbb {R}}$ additive and $\lambda _1^*,\lambda _2^*\in {\mathbb {R}}$ such that

$$\begin{aligned} \ell _k=B^*\circ g_k+\lambda _k^* \quad \text {and}\quad G=\frac{1}{2}B^*+\frac{1}{2}(\lambda _1^*+\lambda _2^*) \end{aligned}$$

hold on U and on $g_1(U)+g_2(U)$, respectively. Define $B{:}{=}\frac{1}{2}B^*$, $\lambda _k{:}{=}\frac{1}{2}\lambda _k^*$, and let $\Lambda {:}{=}\lambda _1+\lambda _2$. Then, expressing $f_k$ in terms of $\ell _k$ and F, we obtain the desired decompositions listed in (3.1).

Now we are going to extend the form of $f_k$ to the subinterval $V{:}{=}U^+\cap (2U-U)\cap I$. If $U=I$, then $U^+=I$, hence, in this case, there is nothing to prove. Therefore assume that this is not the case and let $x\in V$ be any point. We may assume that $x\notin U$, moreover, without loss of generality, it can be assumed that $x<\inf U\in I$.

Then $U_1(x)\cap U_2(x)\ne \emptyset$ and there exists $v\in U$ such that $\frac{1}{2}(x+v)\in U$. Let $u\in U_1(x)\cap U_2(x)$ be any point and $y{:}{=}\max (u,v)\in U$. Then

$$\begin{aligned} \frac{x+y}{2}\in U\quad \text {and}\quad y\in U_1(x)\cap U_2(x). \end{aligned}$$

Indeed, if $u\le v$, then the first inclusion is trivially fulfilled and the second inclusion is implied by the fact $\sup U_k(x)=\sup U$. If $v\le u$, then the second inclusion holds automatically, furthermore, since U is an interval, the validity of the first inclusion is trivial.

Applying (2.3) for (x, y) and using that F is affine on U and the definition of y, the left hand side of (2.3) can be written as

$$\begin{aligned} F\Big (\frac{x+y}{2}\Big )+f_1(x)+f_2(y)= & {} \frac{1}{2}\big (F(x)+F(y)\big )+f_1(x)\\&-\frac{1}{2}F(y)+B\circ g_2(y)+\lambda _2\\= & {} \frac{1}{2}F(x)+f_1(x)+B\circ g_2(y)+\lambda _2. \end{aligned}$$

Thus

$$\begin{aligned} \frac{1}{2}F(x)+f_1(x)+B\circ g_2(y)+\lambda _2=G(g_1(x)+g_2(y))=B(g_1(x)+g_2(y))+\Lambda \end{aligned}$$

follows, which, after subtracting $B\circ g_2(y)$ from both sides and using that $\Lambda -\lambda _2=\lambda _1$, implies that

$$\begin{aligned} f_1(x)=-\frac{1}{2}F(x)+B\circ g_1(x)+\lambda _1. \end{aligned}$$

To get a similar conclusion for $f_2$, apply Eq. (2.3) for the pair (y, x) and perform the same reasoning. $\square$

Corollary 8

Let $F:I\rightarrow {\mathbb {R}}$ be affine and $g_1,g_2:I\rightarrow {\mathbb {R}}$ be continuous functions strictly monotone in the same sense. Then $(F,f_1,f_2,G,g_1,g_2)$ solves functional Eq. (2.3) if and only if there exist $B:{\mathbb {R}}\rightarrow {\mathbb {R}}$ additive and $\lambda _1,\lambda _2,\Lambda \in {\mathbb {R}}$ with $\Lambda =\lambda _1+\lambda _2$ such that (3.1) of Proposition 7 holds with $U{:}{=}I$.

Proof

The statement is a direct consequence of Proposition7. (At this point, it is worth noting that the proof of sufficiency does not require that $g_1$ and $g_2$ be continuous or strictly monotone.) $\square$

The following proposition is stated only for a subinterval of I, because we want to use it later in this form. The statement concerning the whole interval I is formulated as a corollary after the proposition.

Proposition 9

Let $U\subseteq I$ be a nonempty open subinterval, $\mu _1,\mu _2\in {\mathbb {R}}$, $D:{\mathbb {R}}\rightarrow {\mathbb {R}}$ be an invertible additive function, and $g_1,g_2:I\rightarrow {\mathbb {R}}$ such that $g_k{:}{=}D+\mu _k$ on U. If $(F,f_1,f_2,G,g_1,g_2)$ solves (2.3) then there exist an additive function $C:{\mathbb {R}}\rightarrow {\mathbb {R}}$ and constants $\lambda _1,\lambda _2\in {\mathbb {R}}$, such that

$$\begin{aligned}&f_k(x)=Cx+\lambda _k \quad \text {and}\nonumber \\&G(u)=F\circ \frac{1}{2}D^{-1}(u-\mu )+C\circ D^{-1}(u-\mu )+\Lambda \end{aligned}$$

(3.2)

hold for all $x\in U$ and for all $u\in g_1(U)+g_2(U)$, where $\mu {:}{=}\mu _1+\mu _2$ and $\Lambda {:}{=}\lambda _1+\lambda _2$.

Proof

If $(F,f_1,f_2,G,g_1,g_2)$ solves (2.3), then

$$\begin{aligned} h(x+y)=f_1(x)+f_2(y),\quad x,y\in U \end{aligned}$$

with $h(v){:}{=}G(Dv+\mu )-F(\frac{1}{2}v)$, where $v\in 2U$. Hence, by Theorem 1 of [30, Radó–Baker], it follows that there exist an additive function $C:{\mathbb {R}}\rightarrow {\mathbb {R}}$ and constants $\lambda _1,\lambda _2\in {\mathbb {R}}$ such that

$$\begin{aligned} f_k(x)=Cx+\lambda _k \quad \text {and}\quad h(v)=Cv+\Lambda \end{aligned}$$

hold for all $x\in U$ and for all $v\in 2U$. Using the definition of h and putting $u{:}{=}Dv+\mu \in D(2U)+\mu =g_1(U)+g_2(U)$, we obtain the desired formula for G. $\square$

Corollary 10

Let $\mu _1,\mu _2\in {\mathbb {R}}$, $D:{\mathbb {R}}\rightarrow {\mathbb {R}}$ be an invertible additive function, and define $g_k{:}{=}D+\mu _k$ on I. Then $(F,f_1,f_2,G,g_1,g_2)$ solves (2.3) if and only if there exist $C:{\mathbb {R}}\rightarrow {\mathbb {R}}$ additive and $\lambda _1,\lambda _2\in {\mathbb {R}}$ such that (3.2) of Proposition 9 holds with $U{:}{=}I$.

Before we formulate and prove the main result of this section, let us recall a simple lemma, whose precise proof can be found in [20].

Lemma 11

For any subset $H\subseteq I$, the sum $H+I$ is an open subinterval of I such that

$$\begin{aligned} H+I={\left\{ \begin{array}{ll} \emptyset &{}\text {if }H=\emptyset ,\\ ]\inf H+\inf I,\sup H+\sup I[&{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

In the sequel, a 6-tuple of functions $(F,f_1,f_2,G,g_1,g_2)$ will be called regular if $F,f_k,g_k:I\rightarrow {\mathbb {R}}$ and $G:g_1(I)+g_2(I)\rightarrow {\mathbb {R}}$ are continuously differentiable with $0\notin g'_1(I)\cup g'_2(I)$.

In the next proof, $\mathrm{diam}\,H$ stands for the diameter of a set H, more precisely, $\mathrm{diam}\,H{:}{=}0$ if $H=\emptyset$ and $\mathrm{diam}\,H{:}{=}\sup \{x-y\mid x,y\in H\}$ otherwise.

Theorem 12

If $(F,f_1,f_2,G,g_1,g_2)$ is a regular solution of (2.3), then F is either affine or nowhere affine on I.

Proof

Let $(F,f_1,f_2,G,g_1,g_2)$ be a regular solution of Eq. (2.3) and, indirectly, assume that F is locally affine on I. Then, by Theorem 2, $(\varphi ,\psi _1,\Psi _1)$ and $(\varphi ,\psi _2,\Psi _2)$ solve the first and the second functional equations of system (2.4), where the coordinate-functions in question are defined in (2.5) of Theorem 2.

By our indirect assumption, $\varphi =\frac{1}{2}F'$ is locally constant on I. Thus, by Theorem 4 and Theorem 5, it follows that there exist nonempty subintervals $L,K\subseteq I$ different from I and constants $A^*,\mu ^*\in {\mathbb {R}}$ such that $\psi _1$ is identically zero on $L_-(I)\cup L_+(I)$, $\Psi _1=\varphi \psi _1$ holds on I, and $\varphi$ is constant on $L'{:}{=}\frac{1}{2}(L+I)$, furthermore that $\psi _2$ is constant on $K_-(I)$ and $K_+(I)$, $\Psi _2=A^*\psi _2+\mu ^*$ on I, and $\varphi =A^*$ on $K'{:}{=}\frac{1}{2}(K+I)$.

Then I cannot be ${\mathbb {R}}$, otherwise, by Lemma 11, $L'=K'=I$ follows, which contradicts our indirect assumption. If I is unbounded, then, regardless of the exact position of L relative to K, we have $L'\subseteq K'$ or $K'\subseteq L'$. Finally, if I is bounded, then $L'\cap K'$ is an interval of positive length. To see this, it is enough to observe that, by Lemma 11, $\min \{\mathrm{diam}\,L',\mathrm{diam}\,K'\}>\frac{1}{2}\mathrm{diam}\,I$ follows. Consequently, there exists a maximal closed subinterval $U\ne I$ of I such that $L'\cup K'\subseteq U$ and $\varphi$ is constant on U. We may and do assume that $\alpha {:}{=}\inf I<\min U$ holds. (The proof for the complementary case is analogous.) Then, obviously, we have the chain of inequalities

$$\begin{aligned} { -\infty<\alpha<\inf U<\min \{\inf L,\inf K\}=:\beta . } \end{aligned}$$

Define $W{:}{=}\,]\alpha ,\beta [$. Then, on the one hand, we have $W\subseteq L_-(I)\cap K_-(I)$, consequently, for all $x\in W$, we get

$$\begin{aligned}&\psi _1(x)=\frac{1}{g_1'(x)}-\frac{1}{g_2'(x)}=0\quad \text {and}\nonumber \\&\psi _2(x)=\frac{1}{g_1'(x)}+\frac{1}{g_2'(x)}=d \end{aligned}$$

(3.3)

for some $d\in {\mathbb {R}}$. Solving this system of differential equations, by $0\notin g'_1(I)+g'_2(I)$, we obtain that $d\ne 0$, and $g_1$ and $g_2$ are continuous affine functions on W with the common slope $\frac{2}{d}$, where condition $0\notin g'_1(I)+g'_2(I)$ provides that $d\ne 0$. Thus the conditions of Proposition 9 are fulfilled over the interval W, hence $f_1$ and $f_2$ are affine on W as well.

On the other hand, in view of Proposition 7, one can find $\varepsilon >0$ with $\inf U-\varepsilon >\alpha$ such that, for all points $x\in \,]\inf U-\varepsilon ,\beta \,[\,$, we have

$$\begin{aligned} F(x)=2A\circ g_k(x)-2f_k(x)+2\lambda _k \end{aligned}$$

for some additive function $A:{\mathbb {R}}\rightarrow {\mathbb {R}}$ and for some constant $\lambda _k\in {\mathbb {R}}$. The function G is continuously differentiable, hence, in view of (3.1), A must be continuous. Thus the above formula yields that F is affine, that is, $\varphi$ is constant on $]\inf U-\varepsilon ,\beta \,[\,$. This contradicts the maximality of U. $\square$

4 Regular solutions of (2.3)

In this section, we are going to determine the regular solutions of 2.3). In each cases, the proof of the sufficiency of the listed functions will be obvious as a matter of substitution. In contrast, for the necessity part, that is, to get some information about the shape of the functions in question, we are going to actively use our previous results. Let us detail the exact schedule below.

Having a regular solution $(F,f_1,f_2,G,g_1,g_2)$ of (2.3), by definition (2.5), consider the triplets of continuous functions $(\varphi ,\psi _1,\Psi _1)$ and $(\varphi ,\psi _2,\Psi _2)$, which, in view of Theorem 2, solve the first and the second equations of system (2.4), respectively. In view of Theorem 12, F is either affine or nowhere affine on I, which exactly means that $\varphi$ is either constant or nowhere constant on I. In the case when $\varphi$ is nowhere constant on I, the functions $\psi _1$ and $\psi _2$ can still be zero and constant on the interval I, respectively. Finally, the complementary case will be when $\varphi$ is nowhere constant and $\psi _1$ or $\psi _2$ is nowhere zero or constant on I, respectively.

Motivated by this, having a solution $(F,f_1,f_2,G,g_1,g_2)$, for the functions defined in (2.5), we are going to distinguish the following main cases: either

• (A) $\varphi$ is constant on I, or

• (B) $\varphi$ is nowhere constant on I and either

  • (B.1) (4.1) of Theorem 4 with $L=\emptyset$ and (5.1) of Theorem 5 with $K=\emptyset$ hold simultaneously, or

  • (B.2) at least one of the cases (4.2) of Theorem 4 or (5.2) of Theorem 5 is valid.

As we shall see, solutions satisfying conditions (A) or (B.1) will contain arbitrary members too. Motivated by this behavior, such solutions will be formulated within a theorem and case (B.2) will be discussed separately.

Theorem 13

If $(F,f_1,f_2,G,g_1,g_2)$ is a regular solution of (2.3) such that either (A) or (B.1) is met then either

• (1) $g_1$ and $g_2$ are arbitrary functions and there exist constants $A,B\in {\mathbb {R}}$ and $\lambda ,\lambda _1,\lambda _2\in {\mathbb {R}}$ such that

  $$\begin{aligned} F(x)=Ax+\lambda ,\quad f_k(x)=-\tfrac{1}{2}F(x)+Bg_k(x)+\lambda _k, \quad \text {and}\quad G(u)=Bu+\Lambda ,\text { or} \end{aligned}$$

• (2) F is an arbitrary function and there exist $C,D\in {\mathbb {R}}$ with $D\ne 0$ and $\lambda _1,\lambda _2,\mu _1,\mu _2\in {\mathbb {R}}$ such that

  $$\begin{aligned} f_k(x)=Cx+\lambda _k,\quad g_k(x)=Dx+\mu _k, \quad \text {and}\quad G(u)=F(\tfrac{u-\mu }{2D})+C\tfrac{u-\mu }{D}+\Lambda \end{aligned}$$

holds for all $x\in I$ and for all $u\in g_1(I)+g_2(I)$ with $\mu {:}{=}\mu _1+\mu _2$ and $\Lambda {:}{=}\lambda _1+\lambda _2$.

Conversely, $(F,f_1,f_2,G,g_1,g_2)$ with members defined as either in (1) or (2) solves (2.3).

Proof

If condition (A) holds, then, by (2.5), F is affine on I. Thus, applying Corollary 8, we obtain solution (1).

If condition (B.1) is valid, then the system (3.3) holds on I. Consequently, $g_1$ and $g_2$ are continuous affine functions with some common slope $D\in {\mathbb {R}}$ with $D\ne 0$. Thus, by Corollary 10, we obtain solution (2).

As we mentioned before, sufficiency of (1) and (2) is a matter of substitution. $\square$

To treat the sub-case of (B.2) when (4.2) of Theorem 4 and (5.2) of Theorem 5 are valid simultaneously, we need to formulate and prove two lemmas. First we recall the notion of Schwarzian derivative. Let $U\subseteq {\mathbb {R}}$ be a non-empty open subinterval and $f:U\rightarrow {\mathbb {R}}$ be an at least three-times differentiable function with non-vanishing first derivative. Then the Schwarzian derivative of f is defined by the formula

$$\begin{aligned} {\mathscr {S}}f{:}{=}\frac{f'''}{f'} -\frac{3}{2}\Big (\frac{f''}{f'}\Big )^2. \end{aligned}$$

As it is well known, the ,,Schwarzian lines” are exactly the Möbius transformations or linear fractions, that is, functions $g:U\rightarrow {\mathbb {R}}$ of the form

$$\begin{aligned} g(x)=\frac{cx+d}{ax+b},\quad x\in U, \end{aligned}$$

with $a,b,c,d\in {\mathbb {R}}$ such that $ad\ne bc$. More generally, it can be shown that if g is a linear fraction, then

$$\begin{aligned} {{\mathscr {S}}(g\circ f)={\mathscr {S}}f} \end{aligned}$$

(4.1)

holds on U for all functions f on U with appropriate properties. Roughly speaking, the Schwarzian derivative is invariant under linear fractions or, equivalently, linear fractions preserve Schwarzian derivatives.

We shall say that a function $f:U\rightarrow {\mathbb {R}}$ is a trigonometric fraction or a hyperbolic fraction if there exist constants $a,b,c,d,\kappa \in {\mathbb {R}}$ with $ad\ne bc$ and $\kappa >0$ such that

$$\begin{aligned}&f(x)=\frac{c\sin (\kappa x)+d\cos (\kappa x)}{a\sin (\kappa x)+b\cos (\kappa x)}\quad \text {or}\\&f(x) =\frac{c\sinh (\kappa x)+d\cosh (\kappa x)}{a\sinh (\kappa x)+b\cosh (\kappa x)},\quad x\in U, \end{aligned}$$

respectively. Obviously, on some subinterval of U, these functions can be rewritten as $g\circ \tan \circ (\kappa \mathop {\text{ id }}\nolimits )$ and $g\circ \tanh \circ (\kappa \mathop {\text{ id }}\nolimits )$, respectively, with some Möbius transformation g. Thus, in view of the property (4.1) the next lemma is suitable to give the Schwarzian derivative of trigonometric and hyperbolic fractions.

Lemma 14

For a given constant $\kappa \in {\mathbb {R}}$ with $\kappa \ne 0$, we have

$$\begin{aligned} { {\mathscr {S}}\tan (\kappa x)=2\kappa ^2\quad \text {and}\quad {\mathscr {S}}\tanh (\kappa y)=-2\kappa ^2 } \end{aligned}$$

for all $x\in {\mathbb {R}}\setminus \{\frac{(2\ell -1)\pi }{2\kappa }\mid \ell \in {\mathbb {Z}}\}$ and for all $y\in {\mathbb {R}}$. Furthermore, if $\kappa =0$, then the above formulas hold on ${\mathbb {R}}$.

Proof

Simple calculation. $\square$

Within case (B.2), at most one of the functions $\psi _1$ and $\psi _2$ can be constant on I, which, by Theorem 4, Theorem 5, and Lemma 14, yields that $\varphi =\frac{1}{2}F'$ is either (B.2.1) a trigonometric fraction, or (B.2.2) a linear fraction, or (B.2.3) a hyperbolic fraction on I. In addition, following from Lemma 14, we must have $\gamma =\gamma ^*$ provided that (4.2) of Theorem 4 and (5.2) of Theorem 5 are true simultaneously. Motivated by this, for simplicity and tractability, depending on the exact form of $F'=2\varphi$, case (B.2) will be discussed in three parts.

Before we turn to these results, we formulate and prove the following lemma which will help us to handle the different representations of the function $\varphi$.

Lemma 15

Let $U\subseteq {\mathbb {R}}$ be a nonempty open interval, $a,b,c,d\in {\mathbb {R}}$ and $a^*,b^*,c^*,d^*\in {\mathbb {R}}$ such that $ad\ne bc$ and $a^*d^*\ne b^*c^*$, and let $t\in \{\tan ,\mathop {\text{ id }}\nolimits ,\tanh \}$ be defined on U. Then we have

$$\begin{aligned} \frac{ct(x)+d}{at(x)+b}=\frac{c^*t(x)+d^*}{a^*t(x)+b^*},\quad x\in U \end{aligned}$$

(4.2)

if and only if there exists $p\ne 0$ such that $(a,b,c,d)=p(a^*,b^*,c^*,d^*)$.

Proof

The sufficiency is trivial. For the necessity, observe that (4.2) holds if and only if

$$\begin{aligned} \left( \begin{matrix} -c^*&{}\quad 0&{}\quad a^*&{}\quad 0\\ -d^*&{}\quad -c^*&{}\quad b^*&{}\quad a^*\\ 0&{}\quad -d^*&{}\quad 0&{}\quad b^* \end{matrix}\right) \left( \begin{matrix} a\\ b\\ c\\ d \end{matrix}\right) =\left( \begin{matrix} 0\\ 0\\ 0\\ 0 \end{matrix}\right) \end{aligned}$$

is valid. Solving this equation, we get that (a, b, c, d) must be of the form $p(a^*,b^*,c^*,d^*)$, where

$$\begin{aligned} { p{:}{=} {\left\{ \begin{array}{ll} \frac{b}{b^*}&{}\text {if }a^*=0,\\ \frac{a}{a^*}&{}\text {if }b^*=0,\\ \frac{a}{a^*}=\frac{b}{b^*}&{}\text {if }a^*b^*\ne 0. \end{array}\right. }} \end{aligned}$$

By condition $a^*d^*\ne b^*c^*$, the constants $a^*$ and $b^*$ cannot be zero simultaneously. Hence p is well-defined. On the other hand, by condition $ad\ne bc$, it follows that $a^*=0$ or $b^*=0$ if and only if $a=0$ or $b=0$, respectively. Consequently $p\ne 0$, which finishes the proof. $\square$

First we are dealing with the so-called Trigonometric Solutions, that is, with the case when $\gamma =\gamma ^*<0$.

Theorem 16

If $(F,f_1,f_2,G,g_1,g_2)$ is a regular solution of (2.3) such that, for the functions defined in (2.5), condition (B.2.1) holds, then there exist $A,B,C,D,T\in {\mathbb {R}}$ with $AD\ne 0$, $\alpha ,\beta ,\beta _1,\beta _2\in {\mathbb {R}}$ with $\alpha \ne 0$ and $\beta _1+\beta _2\in {\mathbb {Z}}\pi +\beta$, and $\lambda ,\lambda _1,\lambda _2,\mu _1,\mu _2\in {\mathbb {R}}$ such that F is of the form

$$\begin{aligned} F(x)= 2A\ln |\sin (2\alpha x+\beta )|+2Bx+\lambda ,\quad x\in I \end{aligned}$$

and exactly one of the following holds true:

• (T.1) $T^2<1$ and, for all $x\in I$ and $u\in g_1(I)+g_2(I)$,

  $$\begin{aligned}&f_k(x)=-A\ln |\cos (2\alpha x+2\beta _k)+T| -Bx+C\ln \big |\tfrac{\tau \tan (\alpha x+\beta _k)-1}{\tau \tan (\alpha x+\beta _k)+1}\big |+\lambda _k,\\&g_k(x)=D \ln \Big |\tfrac{\tau \tan (\alpha x+\beta _k)-1}{\tau \tan (\alpha x+\beta _k)+1}\Big |+\mu _k, G(u)=-2A\ln \big |T^*\sinh (\tfrac{u-\mu }{2D})\big |\\&\quad +C\tfrac{u-\mu }{D}+\Lambda \end{aligned}$$

• (T.2) $T^2=1$ and, for all $x\in I$ and $u\in g_1(I)+g_2(I)$,

  $$\begin{aligned}&f_k(x)=-A\ln |\cos (2\alpha x+2\beta _k)+T|-Bx+C\tan ^T(\alpha x+\beta _k)+\lambda _k,\\&g_k(x)=D\tan ^T(\alpha x+\beta _k)+\mu _k, G(u)\\&\quad =2A\ln |\tfrac{u-\mu }{2D}|+C\tfrac{u-\mu }{D}+\Lambda \end{aligned}$$

• (T.3) $T^2>1$ and, for all $x\in I$ and $u\in g_1(I)+g_2(I)$,

  $$\begin{aligned}&f_k(x)=-A\ln |\cos (2\alpha x+2\beta _k)+T|-Bx +C\arctan (\tau \tan (\alpha x+\beta _k))+\lambda _k,\\&g_k(x)=D\arctan (\tau \tan (\alpha x+\beta _k))+\mu _k,\\&G(u)=2A\ln \big |T^*\sin (\tfrac{u-\mu }{D})\big |+C\tfrac{u-\mu }{D}+\Lambda , \end{aligned}$$

where $\tau {:}{=}\big |\frac{T+1}{T-1}\big |^{1/2}$, $T^*{:}{=}|T^2-1|^{-1/2}$, $\mu {:}{=}\mu _1+\mu _2$, and $\Lambda {:}{=}\lambda +\lambda _1+\lambda _2$.

Conversely, in each of the above possibilities we obtain a regular solution of Eq. (2.3).

Proof

The proof of sufficiency of (T.1), or (T.2), or (T.3) is a simple calculation. Therefore we are going to focus on the necessity part.

Assume that $(F,f_1,f_2,G,g_1,g_2)$ is a regular solution of (2.3) such that, for the functions defined in (2.5), condition (B.2.1) holds. To measure the behavior of the functions $\psi _1$ and $\psi _2$ defined in (2.5), we introduce the following parameters. Let $p{:}{=}1$ and $q_k{:}{=}0$ if $\psi _1$ is zero on I and let $p\in {\mathbb {R}}$ and $q_k{:}{=}(-1)^{k-1}$ otherwise. In this terminology, $p=0$ stands for the case when $\psi _2$ is constant on I. Furthermore if $q_k\ne 0$ holds, then $\psi _1\ne 0$ and $\psi _2$ is not constant on I. Then, by our assumption, there exist constants $a,b,c,d\in {\mathbb {R}}$ with $ad\ne bc$, $\kappa >0$, and $\lambda ^*,\mu ^*\in {\mathbb {R}}$ such that

$$\begin{aligned}&F'(x)=\frac{c\sin (\kappa x)+d\cos (\kappa x)}{a\sin (\kappa x)+b\cos (\kappa x)},\\&g'_k(x)=\frac{2}{\omega _{2,k}\sin (\kappa x)+\omega _{1,k}\cos (\kappa x)+\omega _0},\quad \text {and} \\&f'_k(x)=-\frac{1}{2}\cdot \frac{\theta _{2,k}\sin (\kappa x)+\theta _{1,k}\cos (\kappa x)+\theta _0}{\omega _{2,k}\sin (\kappa x) +\omega _{1,k}\cos (\kappa x)+\omega _0} \end{aligned}$$

hold for all $x\in I$, where

$$\begin{aligned}&(\omega _{2,k},\omega _{1,k},\omega _0){:}{=} (pb+q_ka,-pa+q_kb,\lambda ^*)\\&\text {and}\quad (\theta _{2,k},\theta _{1,k},\theta _0){:}{=} (pd+q_kc,-pc+q_kd,2\mu ^*). \end{aligned}$$

To make it easier to handle the different sub-cases, first we are going to reformulate the right hand side of the above differential equations. Condition $ad\ne bc$ provides that

$$\begin{aligned}&t{:}{=}\omega _{1,k}^2+\omega _{2,k}^2=(p^2+1)(a^2+b^2)>0\quad \text {and}\\&s{:}{=}\theta _{1,k}^2+\theta _{2,k}^2=(p^2+1)(c^2+d^2)>0 , \end{aligned}$$

therefore there exist $\rho _k,\sigma _k\in [0,2\pi [$ such that

$$\begin{aligned} (\cos \rho _k,\sin \rho _k) =\tfrac{1}{\sqrt{t}}(\omega _{1,k},\omega _{2,k})\quad \text {and}\quad (\cos \sigma _k,\sin \sigma _k)= \tfrac{1}{\sqrt{s}}(\theta _{1,k},\theta _{2,k}). \end{aligned}$$

We note that, in general, t and s are independent from k, and $\sigma _1=\sigma _2=:\sigma$ and $\rho _1=\rho _2=:\rho$ provided that $\psi _1$ is zero on I. In light of this, we obtain that

$$\begin{aligned} g_k'(x)= & {} \frac{2}{\sqrt{t}}\cdot \frac{1}{\cos (\kappa x-\rho _k)+T}\quad \text {and}\nonumber \\ f_k'(x)= & {} -\frac{1}{2}\sqrt{\frac{s}{t}}\cdot \frac{\cos (\kappa x-\sigma _k)+S}{\cos (\kappa x-\rho _k)+T},\quad x\in I, \end{aligned}$$

(4.3)

where $T{:}{=}\frac{1}{\sqrt{t}}\omega _0$ and $S{:}{=}\frac{1}{\sqrt{s}}\theta _0$. Similarly, the function $F'$ can be written as

$$\begin{aligned} F'(x)= & {} \sqrt{\frac{s}{t}}\cos (\rho _0-\sigma _0)\nonumber \\&-\sqrt{\frac{s}{t}}\sin (\rho _0-\sigma _0)\tan (\kappa x-\rho _0),\quad x\in I \end{aligned}$$

(4.4)

with $\rho _0,\sigma _0\in [0,2\pi [$ satisfying the identities

$$\begin{aligned} (\sin \rho _0,\cos \rho _0)=\tfrac{1}{\sqrt{a^2+b^2}}(b,a) \quad \text {and}\quad (\sin \sigma _0,\cos \sigma _0)= \tfrac{1}{\sqrt{c^2+d^2}}(d,c). \end{aligned}$$

Equation (4.4) yields that there exists a constant $\lambda \in {\mathbb {R}}$ such that

$$\begin{aligned} F(x)= 2A\ln |\sin (2\alpha x+\beta )|+2Bx+\lambda ,\quad x\in I, \end{aligned}$$

(4.5)

where $\alpha {:}{=}\frac{\kappa }{2}>0$, $\beta =\rho _0$,

$$\begin{aligned} A{:}{=}\frac{ad-bc}{2\kappa (a^2+b^2)}\ne 0, \quad \text {and}\quad B{:}{=}\frac{ac+bd}{2(a^2+b^2)}. \end{aligned}$$

It turns out that the exact form of the solutions of the differential equations in (4.3) strongly depends on the value of T. Therefore, in the rest of the proof, we are going to distinguish three cases: $T^2<1$ or $T^2=1$ or $T^2>1$.

Case 1. Assume that $T^2<1$ holds. Then there exist constants $\lambda _1,\lambda _2,\mu _1,\mu _2\in {\mathbb {R}}$ such that

$$\begin{aligned} g_k(x)= & {} D \ln \Big | \tfrac{\tau \tan (\alpha x+\beta _k)-1}{\tau \tan (\alpha x+\beta _k)+1}\Big | +\mu _k\quad \text {and}\\ f_k(x)= & {} -A\ln |\cos (2\alpha x+2\beta _k)+T| -Bx +C\ln \big |\tfrac{\tau \tan (\alpha x+\beta _k)-1}{\tau \tan (\alpha x+\beta _k)+1}\big |+\lambda _k \end{aligned}$$

hold for all $x\in I$, where $\beta _k{:}{=}-\frac{\rho _k}{2}$, $\tau {:}{=}\sqrt{\frac{1-T}{1+T}}$,

$$\begin{aligned}&C{:}{=}-\frac{\mathop {\mathrm{sgn}}\nolimits (T-1)}{\kappa (a^2+b^2)\sqrt{1-T^2}}\big (S\sqrt{ (a^2+b^2)(c^2+d^2) } -\tfrac{1}{2}T(ac+bd)\big ),\quad \text {and}\\&D{:}{=}\frac{4\mathop {\mathrm{sgn}}\nolimits (T-1)}{\kappa \sqrt{t(1-T^2)}}\ne 0. \end{aligned}$$

Substituting $\xi {:}{=}\alpha x+\beta _1$, $\eta {:}{=}\alpha y+\beta _2$, $\mu {:}{=}\mu _1+\mu _2$, and $\Lambda {:}{=}\lambda +\lambda _1+\lambda _2$, Eq. (2.3) reduces to

$$\begin{aligned}&G\big (D\tfrac{\tau \tan \xi -1}{\tau \tan \xi +1} \tfrac{\tau \tan \eta -1}{\tau \tan \eta +1}+\mu \big ) =2A\ln |\sin (\alpha (x+y)+\beta )|+B(x+y)+\lambda \\&\qquad -A\ln |\cos (2\alpha x+2\beta _1)+T|-Bx +C\ln \big |\tfrac{\tau \tan (\alpha x+\beta _1)-1}{\tau \tan (\alpha x+\beta _1)+1}\big |+\lambda _1\\&\qquad -A\ln |\cos (2\alpha y+2\beta _2)+T|-By +C\ln \big |\tfrac{\tau \tan (\alpha y+\beta _2)-1}{\tau \tan (\alpha y+\beta _2)+1}\big |+\lambda _2\\&\quad =A\ln \big |\tfrac{\sin ^2(\xi +\eta )}{(\cos 2\xi +T)(\cos 2\eta +T)}\big | +C\ln \Big |\tfrac{\tau \tan \xi -1}{\tau \tan \xi +1} \tfrac{\tau \tan \eta -1}{\tau \tan \eta +1}\Big |+\Lambda , \end{aligned}$$

where in the last step, we used that $\beta _1+\beta _2-\beta$ is of the form $\ell \pi$ for some $\ell \in {\mathbb {Z}}$. (To see this, check that $\sin ^2(\beta _1+\beta _2-\beta )=0$ is valid.) In view of the identities

$$\begin{aligned} \frac{\tau \tan \xi -1}{\tau \tan \xi +1}\frac{\tau \tan \eta -1}{\tau \tan \eta +1} =\Big (1-\frac{ \tau \tan \xi +\tau \tan \eta }{1+\tau ^2\tan \xi \tan \eta }\Big ) \Big (1+\frac{\tau \tan \xi +\tau \tan \eta }{ 1+\tau ^2\tan \xi \tan \eta }\Big )^{-1} \end{aligned}$$

and

$$\begin{aligned} \frac{\sin ^2(\xi +\eta )}{(\cos 2\xi +T)(\cos 2\eta +T)}= & {} \frac{1}{1-T^2}\Big (\frac{\tau \tan \xi +\tau \tan \eta }{1+\tau ^2\tan \xi \tan \eta } \Big )^2\\&\times \Big (1-\Big (\frac{\tau \tan \xi +\tau \tan \eta }{1+\tau ^2\tan \xi \tan \eta }\Big )^2\Big )^{-1}, \end{aligned}$$

for $u{:}{=}D\ln \big |\frac{\tau \tan \xi -1}{\tau \tan \xi +1} \frac{\tau \tan \eta -1}{\tau \tan \eta +1} \big |+\mu \in g_1(I)+g_2(I)$, we get

$$\begin{aligned} G(u)=-2A\ln \big |\tfrac{1}{\sqrt{1-T^2}}\sinh (\tfrac{u-\mu }{2D})\big | +C\tfrac{u-\mu }{D}+\Lambda . \end{aligned}$$

Thus we obtained the solutions listed in (T.1).

Case 2. Assume that $T^2=1$ holds. Then there exist $\lambda _1,\lambda _2,\mu _1,\mu _2\in {\mathbb {R}}$ such that

$$\begin{aligned}&g_k(x)=D\tan ^T(\alpha x+\beta _k)+\mu _k\,\text { and }\\&f_k(x)=-A\ln |\cos (2\alpha x+2\beta _k)+T| -Bx+C\tan ^T(\alpha x+\beta _k)+\lambda _k \end{aligned}$$

hold for $x\in I$, where $\beta _k{:}{=}-\frac{\rho _k}{2}$,

$$\begin{aligned} C=-\frac{1}{2\kappa (a^2+b^2)} \big (S\sqrt{(a^2+b^2)(c^2+d^2)} +T(ac+bd)\big ), \quad \text {and}\quad D{:}{=}\frac{2}{\kappa \sqrt{t}}\ne 0. \end{aligned}$$

Particularly, $\beta _1+\beta _2={\mathbb {Z}}\pi +\beta$ holds. Substituting $\xi {:}{=}\alpha x+\beta _1$, $\eta {:}{=}\alpha y+\beta _2$, $\mu {:}{=}\mu _1+\mu _2$, and $\Lambda {:}{=}\lambda +\lambda _1+\lambda _2$, Eq. (2.3) reduces to

$$\begin{aligned}&G(D(\tan ^T\xi +\tan ^T\eta )+\mu )\\&\quad =2A\ln |\sin (\alpha (x+y)+\beta )|+B(x+y)+\lambda \\&\qquad -A\ln |\cos (2\alpha x+2\beta _1)+T|-Bx+C\tan ^T(\alpha x+\beta _1)+\lambda _1\\&\qquad -A\ln |\cos (2\alpha y+2\beta _2)+T|-By+C\tan ^T(\alpha y+\beta _2)+\lambda _2\\&\quad =A\ln \big |\tfrac{\sin ^2(\xi +\eta )}{(\cos 2\xi +T)(\cos 2\eta +T)}\big |\\&\qquad +C(\tan ^T\xi +\tan ^T\eta )+\Lambda . \end{aligned}$$

Using

$$\begin{aligned} \frac{\sin ^2(\xi +\eta )}{(\cos 2\xi +T)(\cos 2\eta +T)}=\tfrac{1}{4}(\tan ^T\xi +\tan ^T\eta )^2, \end{aligned}$$

for $u{:}{=}D(\tan ^T\xi +\tan ^T\eta )+\mu \in g_1(I)+g_2(I)$, we get

$$\begin{aligned} G(u)=2A\ln |\tfrac{u-\mu }{2D}|+C\tfrac{u-\mu }{D}+\Lambda . \end{aligned}$$

Thus we obtained the solutions listed in (T.2).

Case 3. Finally, assume that $T^2>1$ holds. Then there exist $\lambda _1,\lambda _2,\mu _1,\mu _2\in {\mathbb {R}}$ such that

$$\begin{aligned} g_k(x)= & {} D\arctan (\tau \tan (\alpha x+\beta _k))+\mu _k\quad \text {and}\\ f_k(x)= & {} -A\ln |\cos (2\alpha x+2\beta _k)+T| -Bx +C\arctan (\tau \tan (\alpha x+\beta _k))+\lambda _k \end{aligned}$$

hold for all $x\in I$, where $\beta _k{:}{=}-\frac{\rho _k}{2}$, $\tau {:}{=}\sqrt{\frac{T-1}{T+1}}$,

$$\begin{aligned}&C{:}{=}-\frac{\mathop {\mathrm{sgn}}\nolimits (T+1)}{\kappa (a^2+b^2)\sqrt{T^2-1}}\big (S\sqrt{(a^2+b^2)(c^2+d^2)} -\tfrac{1}{2}T(ac+bd)\big ),\quad \text {and}\\&D{:}{=}\frac{4\mathop {\mathrm{sgn}}\nolimits (T+1)}{\kappa \sqrt{t(T^2-1)}}\ne 0. \end{aligned}$$

Particularly, $\beta _1+\beta _2={\mathbb {Z}}\pi +\beta$ holds. Substituting $\xi {:}{=}\alpha x+\beta _1$, $\eta {:}{=}\alpha y+\beta _2$, $\mu {:}{=}\mu _1+\mu _2$, and $\Lambda {:}{=}\lambda +\lambda _1+\lambda _2$, Eq. (2.3) reduces to

$$\begin{aligned}&G\big (D \arctan \big (\tfrac{\tau \tan \xi +\tau \tan \eta }{1-\tau ^2\tan \xi \tan \eta }\big )+\mu \big )= 2A\ln |\sin (\alpha (x+y)+\beta )|+B(x+y)+\lambda \\&\qquad -A\ln |\cos (2\alpha x+2\beta _1)+T|-Bx +C\arctan (\tau \tan (\alpha x+\beta _1))+\lambda _1\\&\qquad -A\ln |\cos (2\alpha y+2\beta _2)+T| -By +C\arctan (\tau \tan (\alpha y+\beta _2))+\lambda _2\\&\quad =A\ln \big |\tfrac{\sin ^2(\xi +\eta )}{(\cos 2\xi +T)(\cos 2\eta +T)}\big | +C\arctan \big (\tfrac{\tau \tan \xi +\tau \tan \eta }{1-\tau ^2\tan \xi \tan \eta }\big )+\Lambda . \end{aligned}$$

In view of the identity

$$\begin{aligned}&\frac{\sin ^2(\xi +\eta )}{(\cos 2\xi +T)(\cos 2\eta +T)}\\&\quad =\frac{1}{T^2-1}\Big (\frac{\tau \tan \xi +\tau \tan \eta }{1-\tau ^2\tan \xi \tan \eta }\Big )^2 \Big (1+\Big (\frac{\tau \tan \xi +\tau \tan \eta }{1-\tau ^2\tan \xi \tan \eta }\Big )^2\Big )^{-1}, \end{aligned}$$

for $u=D \arctan \big (\tfrac{\tau \tan \xi +\tau \tan \eta }{1-\tau ^2\tan \xi \tan \eta }\big )+\mu \in g_1(I)+g_2(I)$, we get

$$\begin{aligned} G(u)=2A\ln \big |\tfrac{1}{\sqrt{T^2-1}}\sin (\tfrac{u-\mu }{D})\big |+C(\tfrac{u-\mu }{D})+\Lambda . \end{aligned}$$

Thus we obtained the solutions listed in (T.3). $\square$

Now we turn to the case of Polynomial Solutions of Eq. (2.3), that is, when $\gamma =\gamma ^*=0$. As we will see, in terms of F, solutions can be classified into two groups. In one case, F is a linear combination of an affine function and the logarithm of an affine function. This group will include four solutions. In the other case, F is a second order polynomial, which gives two further solutions. Since the behavior of the latter two solutions is fundamentally different, they will be formulated separately within the theorem.

Theorem 17

If $(F,f_1,f_2,G,g_1,g_2)$ is a regular solution of (2.3) such that, for the functions defined in (2.5), condition (B.2.2) holds, then there exist $A,B,C,D\in {\mathbb {R}}$ with $AD\ne 0$, $\alpha ,\beta ,\beta _1,\beta _2\in {\mathbb {R}}$ with $\alpha \ne 0$ and $\beta _1+\beta _2=\beta$, and $\lambda ,\lambda _1,\lambda _2,\mu _1,\mu _2\in {\mathbb {R}}$ such that either F is of the form

$$\begin{aligned} F(x)=2A\ln |2\alpha x+\beta |+2Bx+\lambda ,\quad x\in I \end{aligned}$$

and either

$$\begin{aligned} \begin{array}{l} f_k(x)=-A\ln |(\alpha x+\beta _k)^2+1|-Bx+C\arctan (\alpha x+\beta _k)+\lambda _k,\\ g_k(x)=D\arctan (\alpha x+\beta _k)+\mu _k,\quad G(u){:}{=}2A\ln |\sin (\tfrac{u-\mu }{D})|+C\tfrac{u-\mu }{D}+\Lambda , \end{array} \end{aligned}$$

(P1.1)

or

$$\begin{aligned} \begin{array}{l} f_k(x)=-2A\ln |\alpha x+\beta _k|-Bx+C(\alpha x+\beta _k)^{-1}+\lambda _k,\\ g_k(x)=D(\alpha x+\beta _k)^{-1}+\mu _k,\quad G(u)=2A\ln |\tfrac{u-\mu }{D}|+C\tfrac{u-\mu }{D}+\Lambda , \end{array} \end{aligned}$$

(P1.2)

or

$$\begin{aligned} \begin{array}{l} f_k(x)=-A\ln |(\alpha x+\beta _k)^2-1|-Bx +C\ln |\tfrac{\alpha x+\beta _k-1}{\alpha x+\beta _k+1}| +\lambda _k,\\ g_k(x)=D\ln |\tfrac{\alpha x+\beta _k-1}{\alpha x+\beta _k+1}|+\mu _k,\quad G(u)=2A\ln |\sinh (\tfrac{u-\mu }{2D})|+C\tfrac{u-\mu }{D}+\Lambda , \end{array} \end{aligned}$$

(P1.3)

or there exist $A_1,A_2\in {\mathbb {R}}$ with $\frac{1}{2}(A_1+A_2)=A$ such that

$$\begin{aligned} \begin{array}{l} f_k(x)=-A_k\ln |\alpha x+\beta _k|-Bx+\lambda _k,\\ g_k(x)=(-1)^{k-1}D\ln |\alpha x+\beta _k|+\mu _k,\\ G(u)=2A\ln |\exp (\tfrac{u-\mu }{D})+1|-A_1\tfrac{u-\mu }{D}+\Lambda , \end{array} \end{aligned}$$

(P1.4)

hold for all $x\in I$ and for all $u\in g_1(I)+g_2(I)$ or there exist $A_k,B_k,D_k\in {\mathbb {R}}$ with $D_1D_2(A+4A_k)=D_k^2A$ and $B+2B_k=2D_kC$ such that

$$\begin{aligned} \begin{array}{ll} F(x)=Ax^2+2Bx+\lambda , &{}\quad f_k(x)=A_kx^2+B_kx+\lambda _k,\\ g_k(x)=D_kx+\mu _k, &{}\quad G(u)=\tfrac{1}{4D_1D_2}A(u-\mu )^2+C(u-\mu )+\Lambda , \end{array} \end{aligned}$$

(P2.1)

or

$$\begin{aligned} \begin{array}{ll} \,F(x)=A(2x+\beta )^2+2B+\lambda ,&{} \quad f_k(x)=-A(x+\beta _k)^2-Bx+C\ln |x+\beta _k|+\lambda _k,\\ g_k(x)=D\ln |x+\beta _k|+\mu _k, &{} \quad G(u)=2A\exp (\tfrac{u-\mu }{D})+C\tfrac{u-\mu }{D}+\Lambda \end{array} \end{aligned}$$

(P2.2)

hold for all $x\in I$ and for all $u\in g_1(I)+g_2(I)$, where $\mu {:}{=}\mu _1+\mu _2$ and $\Lambda {:}{=}\lambda +\lambda _1+\lambda _2$.

Conversely, in each of the above possibilities we obtain a regular solution of Eq. (2.3)

Proof

The proof of sufficiency is a simple calculation, therefore we will focus only on the necessity.

Similarly to the previous proof, to indicate the behavior of $\psi _1$ and $\psi _2$, let $p{:}{=}1$ and $q_k{:}{=}0$ if $\psi _1$ is zero on I and let $p\in {\mathbb {R}}$ and $q_k{:}{=}(-1)^{k-1}$ otherwise. Again, $p=0$ corresponds to the case when $\psi _2$ is constant on I. Furthermore, if $q_k\ne 0$, then we get the case when $\psi _1$ and $\psi _2$ are not constant on I. Then, in view of our assumption concerning the derivative of F, there exist $a,b,c,d\in {\mathbb {R}}$ with $ad\ne bc$ such that

$$\begin{aligned}&F'(x)=\frac{cx+d}{ax+b},\quad g'_k(x)=\frac{4}{\omega _2x^2+\omega _{1,k}x+\omega _{0,k}},\quad \text {and}\\&f'_k(x)=-\frac{1}{2}\cdot \frac{\theta _{2}x^2+\theta _{1,k}x+\theta _{0,k}}{\omega _2x^2+\omega _{1,k}x+\omega _{0,k}} \end{aligned}$$

hold for all $x\in I$, where

$$\begin{aligned}&(\omega _2,\omega _{1,k},\omega _{0,k}){:}{=}(pa,2pb+2q_ka,2q_kb+2\lambda ^*)\quad \text {and}\\&(\theta _2,\theta _{1,k},\theta _{0,k}){:}{=} (pc,2pd+2q_kc,2q_kd+4\mu ^*). \end{aligned}$$

Then, an elementary calculation yields that there exists a constant $\lambda _0\in {\mathbb {R}}$, such that

$$\begin{aligned} F(x)= {\left\{ \begin{array}{ll} A_0x^2+B_0x+\lambda _0&{}\text {if }a=0,\\ 2A_0\ln |ax+b|+2B_0+\lambda _0&{}\text {if }a\ne 0, \end{array}\right. }\quad x\in I, \end{aligned}$$

(4.6)

where

$$\begin{aligned} 0\ne A_0{:}{=} {\left\{ \begin{array}{ll} \frac{c}{2b}&{}\text {if }a=0,\\ \frac{ad-bc}{2a^2}&{}\text {if }a\ne 0, \end{array}\right. }\quad \text {and}\quad B_0{:}{=} {\left\{ \begin{array}{ll} \tfrac{d}{b}&{}\text {if }a=0,\\ \tfrac{c}{2a}&{}\text {if }a\ne 0. \end{array}\right. } \end{aligned}$$

(4.7)

The exact form of the functions $g_k$ and $f_k$ strongly depends on the degree of the polynomial in their denominators. Therefore we are going to distinguish two main cases: either $\omega _2=0$ or $\omega _2\ne 0$.

Case 1. Assume that $\omega _2=0$ holds. Then $a=0$ or $p=0$ holds such that whenever $p=0$ is valid then $g_1-g_2$ cannot be constant on I.

(1.1) Suppose that $a=0$ and $p\ne 0$. Then condition $ad\ne bc$ implies that $bc\ne 0$ and there exist constants $\lambda _1,\lambda _2,\mu _1,\mu _2\in {\mathbb {R}}$ such that

$$\begin{aligned}&g_k(x)=D\ln |x+\beta _k|+\mu _k\quad \text {and}\\&f_k(x)=-A(x+\beta _k)^2-Bx+C\ln |x+\beta _k|+\lambda _k \end{aligned}$$

hold for all $x\in I$, where $A{:}{=}\tfrac{1}{4}A_0$, $\beta _k{:}{=}\frac{\omega _{0,k}}{2pb}$, $B{:}{=}\frac{d}{2b}-\lambda ^*\frac{c}{2pb^2}$,

$$\begin{aligned}&C{:}{=}-\frac{1}{4b^3}\cdot {\left\{ \begin{array}{ll} 4\mu ^*b^2-\lambda ^*(2bd-c\lambda ^*)&{}\text {if }g_1-g_2\text { is constant on }I,\\ p^{-2}(4p\mu ^*b^2-\lambda ^*(2pbd-c\lambda ^*)-cb^2)&{}\text {otherwise}, \end{array}\right. }\\&\quad \text {and}\quad D{:}{=}\frac{2}{pb}\ne 0. \end{aligned}$$

Observe that F in (4.6) can be reformulated as

$$\begin{aligned} F(x)=A(2x+\beta )^2+2Bx+\lambda ,\quad x\in I, \end{aligned}$$

with $\beta {:}{=}\beta _1+\beta _2$ and $\lambda {:}{=}\lambda _0-A\beta ^2$. Substituting $\xi {:}{=}x+\beta _1$, $\eta {:}{=}y+\beta _2$, $\mu {:}{=}\mu _1+\mu _2$, and $\Lambda {:}{=}\lambda +\lambda _1+\lambda _2$, Eq. (2.3) reduces to

$$\begin{aligned}&G(D\ln |\xi \eta |+\mu ) =A\big (x+y+\beta \big )^2+B(x+y)+\lambda \\&\qquad -A(x+\beta _1)^2-Bx+C\ln |x+\beta _1|+\lambda _1\\&\qquad -A(y+\beta _2)^2-By+C\ln |y+\beta _2|+\lambda _2\\&\quad =A(\xi +\eta )^2-A(\xi ^2+\eta ^2)+C\ln |\xi \eta |+\Lambda =2A\xi \eta +C\ln |\xi \eta |+\Lambda . \end{aligned}$$

Consequently, for $u{:}{=}D\ln |\xi \eta |+\mu \in g_1(I)+g_2(I)$, we get

$$\begin{aligned} G(u)=2A\exp (\tfrac{u-\mu }{D})+C\tfrac{u-\mu }{D}+\Lambda . \end{aligned}$$

Thus we obtained the solutions listed in (P2.2).

(1.2) Suppose that $a\ne 0$ and $p=0$. Then there exist constants $\lambda _1,\lambda _2,\mu _1,\mu _2\in {\mathbb {R}}$ such that

$$\begin{aligned}&g_k(x)=(-1)^{k-1}D\ln |\alpha x+\beta _k| +\mu _k \quad \text {and}\\&f_k(x)=-A_k\ln |\alpha x+\beta _k|-Bx+\lambda _k \end{aligned}$$

for all $x\in I$, where $\alpha {:}{=}a\ne 0$, $B{:}{=}B_0$, $D{:}{=}\frac{2}{a}\ne 0$, $\beta _k{:}{=}b+(-1)^{k-1}\lambda ^*$, and

$$\begin{aligned} A_k{:}{=}\tfrac{1}{2a^2}(ad-bc+(-1)^{k-1}(2a\mu ^*-c\lambda ^*)). \end{aligned}$$

Observe that $\frac{1}{2}(A_1+A_2)=A_0=:A$ and that the function F in (4.6) can be written as

$$\begin{aligned} F(x)=2A\ln |2\alpha x+\beta |+2Bx+\lambda , \end{aligned}$$

with $\beta {:}{=}\beta _1+\beta _2$, and $\lambda {:}{=}\lambda _0-\ln 4 A$. Hence, substituting $\xi {:}{=}\alpha x+\beta _1$, $\eta {:}{=}\alpha y+\beta _2$, $\mu {:}{=}\mu _1+\mu _2$, and $\Lambda {:}{=}\lambda +\lambda _1+\lambda _2$, Eq. (2.3) reduces to

$$\begin{aligned} G(D\ln |\tfrac{\xi }{\eta }|+\mu )= & {} 2A\ln |\alpha (x+y)+\beta |+B(x+y)+\lambda \\&-A_1\ln |\alpha x+\beta _1|-Bx+\lambda _1\\&-A_2\ln |\alpha y+\beta _2|-By+\lambda _2\\= & {} 2A\ln |\tfrac{\xi }{\eta }+1|-A_1\ln |\tfrac{\xi }{\eta }|+\Lambda . \end{aligned}$$

Consequently, for $u{:}{=}D\ln |\tfrac{\xi }{\eta }|+\mu \in g_1(I)+g_2(I)$, we get

$$\begin{aligned} G(u)=2A\ln |\exp (\tfrac{u-\mu }{D})+1|-A_1\tfrac{u-\mu }{D}+\Lambda . \end{aligned}$$

Thus we obtained the solutions listed in (P1.4).

(1.3) Finally, suppose that $a=p=0$. Then, on the one hand, by $ad\ne bc$, we must have $c\ne 0$. On the other hand, in view of

Remark 18

$\omega _{0,k}\ne 0$ follows. Thus there exist constants $\lambda _1,\lambda _2, \mu _1,\mu _2\in {\mathbb {R}}$ such that

$$\begin{aligned} g_k(x)=D_kx+\mu _k\quad \text {and}\quad f_k(x)=A_kx^2+B_kx+\lambda _k \end{aligned}$$

hold for all $x\in I$, where

$$\begin{aligned}&A_k{:}{=}\tfrac{(-1)^k}{2}\omega _{0,k}^{-1}c\ne 0,\quad B_k{:}{=}(-1)^k\omega _{0,k}^{-1}(d+(-1)^{k-1}2\mu ^*),\\&\quad \text {and}\quad D_k{:}{=}4\omega _{0,k}^{-1}\ne 0. \end{aligned}$$

Observe that the identity $D_1D_2(A+4A_k)=D_k^2A$ holds. Note further that the constants $\frac{1}{2D_1}(B+2B_1)$ and $\frac{1}{2D_2}(B+2B_2)$ are equal to each other. Denote their common value by C. Substituting $\xi {:}{=}2\alpha _1x$, $\eta {:}{=}2\alpha _2y$, $\mu {:}{=}\mu _1+\mu _2$, and $\Lambda {:}{=}\lambda +\lambda _1+\lambda _2$, Eq. (2.3) reduces to

$$\begin{aligned} G(\xi +\eta +\mu )= & {} \tfrac{1}{4}A(x+y)^2+\tfrac{1}{2}B(x+y)+\lambda +A_1x^2+B_1x+\lambda _1\\&+ A_2y^2+B_2y+\lambda _2\\= & {} \tfrac{1}{4D_1D_2}A(\xi +\eta )^2 +C(\xi +\eta )+\Lambda . \end{aligned}$$

Consequently, for $u{:}{=}\xi +\eta +\mu \in g_1(I)+g_2(I)$, we get

$$\begin{aligned} G(u)=\tfrac{1}{4D_1D_2}A(u-\mu )^2+C(u-\mu )+\Lambda . \end{aligned}$$

Thus we obtained the solutions listed in (P2.1).

Case 2. Assume now that $\omega _2\ne 0$. Then the polynomials appearing in the denominator in the formulas concerning $g_k$ and $f_k$ are of degree two and have a common discriminant

$$\begin{aligned} \Delta {:}{=} {\left\{ \begin{array}{ll} 4b^2-8a\lambda ^*&{} \text {if }\,g_1-g_2\,\text { is constant on }\,I,\\ 4p^2b^2-8pa\lambda ^*+4a^2&{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

Accordingly, depending on the sign of $\Delta$, within Case 2., we are going to distinguish three further sub-cases.

(2.1) Suppose that $\Delta <0$. Then there exist constants $\lambda _1,\lambda _2,\mu _1,\mu _2\in {\mathbb {R}}$ such that

$$\begin{aligned}&g_k(x)=D\arctan (\alpha x+\beta _k)+\mu _k\quad \text {and}\\&f_k(x)=-A\ln |(\alpha x+\beta _k)^2+1|-Bx+C\arctan (\alpha x+\beta _k)+\lambda _k \end{aligned}$$

hold for all $x\in I$, where $A{:}{=}A_0$, $B{:}{=}B_0$, $\alpha {:}{=}\frac{2}{\sqrt{-\Delta }}\omega _2\ne 0$, $\beta _k{:}{=}\frac{1}{\sqrt{-\Delta }}\omega _{1,k}$, $D{:}{=}\frac{4}{\sqrt{-\Delta }}\ne 0$, and

$$\begin{aligned} C{:}{=}\frac{1}{a\sqrt{-\Delta }}(c\lambda ^*-2a\mu ^*)+\frac{pb}{a^2\sqrt{-\Delta }}(ad-bc). \end{aligned}$$

Note that F can be reformulated as

$$\begin{aligned} F(x)=2A\ln |2\alpha x+\beta |+2Bx+\lambda ,\quad x\in I \end{aligned}$$

with $\beta {:}{=}\beta _1+\beta _2$ and $\lambda {:}{=}\lambda _0+2A\ln |\frac{\sqrt{-\Delta }}{4p}|$. Hence, substituting $\xi {:}{=}\alpha x+\beta _1$, $\eta {:}{=}\alpha y+\beta _2$, $\mu {:}{=}\mu _1+\mu _2$, and $\Lambda {:}{=}\lambda +\lambda _1+\lambda _2$, Eq. (2.3) reduces to

$$\begin{aligned}&G(D\arctan (\tfrac{\xi +\eta }{1-\xi \eta })+\mu )= 2A\ln |\alpha (x+y)+\beta |+B(x+y)+\lambda \\&\qquad -A\ln |(\alpha x+\beta _1)^2+1|-Bx+C\arctan (\alpha x+\beta _1)+\lambda _1\\&\qquad -A\ln |(\alpha y+\beta _2)^2+1|-By+C\arctan (\alpha y+\beta _2)+\lambda _2\\&\quad =A\ln \big |\tfrac{(\xi +\eta )^2}{(\xi ^2+1)(\eta ^2+1)}\big |+C\arctan (\tfrac{\xi +\eta }{1-\xi \eta } )+\Lambda . \end{aligned}$$

Consequently, for $u{:}{=}D\arctan (\frac{\xi +\eta }{1-\xi \eta })+\mu \in g_1(I)+g_2(I)$, we get

$$\begin{aligned} \frac{(\xi +\eta )^2}{(\xi ^2+1)(\eta ^2+1)} =\frac{\big (\frac{\xi +\eta }{1-\xi \eta }\big )^2}{1+\big (\frac{\xi +\eta }{1-\xi \eta }\big )^2}=\frac{\tan ^2\big (\frac{u-\mu }{D}\big )}{1+\tan ^2\big (\frac{u-\mu }{D}\big )}=\sin ^2\Big (\frac{u-\mu }{D}\Big ), \end{aligned}$$

and hence

$$\begin{aligned} G(u){:}{=}2A\ln |\sin (\tfrac{u-\mu }{D})|+C\tfrac{u-\mu }{D}+\Lambda . \end{aligned}$$

Thus we obtained the solutions listed in (P1.1).

(2.2) Suppose that $\Delta =0$. Then there exist constants $\lambda _1,\lambda _2,\mu _1,\mu _2\in {\mathbb {R}}$ such that

$$\begin{aligned}&g_k(x)=D(\alpha x+\beta _k)^{-1}+\mu _k\quad \text {and}\\&f_k(x)=-2A\ln |\alpha x +\beta _k|-Bx+C(\alpha x+\beta _k)^{-1}+\lambda _k \end{aligned}$$

hold for all $x\in I$, where $A{:}{=}A_0$, $B{:}{=}B_0$, $\alpha {:}{=}\omega _2\ne 0$, $\beta _k{:}{=}\frac{1}{2}\omega _{1,k}$, $D{:}{=}-4\ne 0$, and

$$\begin{aligned} C{:}{=}-\frac{1}{a}(c\lambda ^*-2a\mu ^*)-\frac{pb}{a^2}(ad-bc). \end{aligned}$$

Observe that F can be written as

$$\begin{aligned} F(x)=2A\ln |2\alpha x+\beta |+2Bx+\lambda ,\quad x\in I \end{aligned}$$

with $\beta {:}{=}\beta _1+\beta _2$ and $\lambda {:}{=}\lambda _0-\ln (4p^2)A$. Hence, substituting $\xi {:}{=}\alpha x+\beta _1$, $\eta {:}{=}\alpha y+\beta _2$, $\mu {:}{=}\mu _1+\mu _2$, and $\Lambda {:}{=}\lambda +\lambda _1+\lambda _2$, Eq. (2.3) reduces to

$$\begin{aligned} G\left( D\tfrac{\xi +\eta }{\xi \eta }+\mu \right)= & {} 2A\ln |\alpha (x+y)+\beta |+B(x+y)+\lambda \\&-2A\ln |\alpha x+\beta _1|-Bx+C(\alpha x+\beta _1)^{-1}+\lambda _1\\&-2A\ln |\alpha y+\beta _2|-By+C(\alpha y+\beta _2)^{-1}+\lambda _2=2A_0\ln |\tfrac{\xi +\eta }{\xi \eta }|\\&+C\tfrac{\xi +\eta }{\xi \eta }+\Lambda . \end{aligned}$$

Consequently, for $u{:}{=}D\frac{\xi +\eta }{\xi \eta }+\mu \in g_1(I)+g_2(I)$, we get

$$\begin{aligned} G(u)=2A\ln |\tfrac{u-\mu }{D}|+C\tfrac{u-\mu }{D}+\Lambda . \end{aligned}$$

Thus we obtained the solutions listed in (P1.2).

(2.3) Finally, suppose that $\Delta >0$. Then there exist constants $\lambda _1,\lambda _2,\mu _1,\mu _2\in {\mathbb {R}}$ such that

$$\begin{aligned} g_k(x)= & {} D\ln |\tfrac{\alpha x+\beta _k-1}{\alpha x+\beta _k+1}| +\mu _k \quad \text {and}\\ f_k(x)= & {} -A\ln |(\alpha x+\beta _k)^2-1|-Bx +C\ln |\tfrac{\alpha x+\beta _k-1}{\alpha x+\beta _k+1}| +\lambda _k \end{aligned}$$

for all $x\in I$, where $A{:}{=}A_0$, $B{:}{=}B_0$, $\alpha {:}{=}\frac{2}{\sqrt{\Delta }}\omega _2\ne 0$, $\beta _k{:}{=}\frac{1}{\sqrt{\Delta }}\omega _{1,k}$, $D{:}{=}\frac{4}{\sqrt{\Delta }}\ne 0$, and

$$\begin{aligned} C{:}{=} \frac{1}{a\sqrt{\Delta }}(c\lambda ^*-2a\mu ^*)+\frac{pb}{a^2\sqrt{\Delta }}(ad-bc) \end{aligned}$$

Observe that F can be written as

$$\begin{aligned} F(x)=2A\ln |2\alpha x+\beta |+2Bx+\lambda ,\quad x\in I, \end{aligned}$$

with $\beta {:}{=}\beta _1+\beta _2$ and $\lambda {:}{=}\lambda _0+2A_0\ln |\frac{\sqrt{\Delta }}{4p}|$. Hence, substituting $\xi {:}{=}\alpha x+\beta _1$, $\eta {:}{=}\alpha y+\beta _2$, $\mu {:}{=}\mu _1+\mu _2$, and $\Lambda {:}{=}\lambda +\lambda _1+\lambda _2$, Eq. (2.3) reduces to

$$\begin{aligned}&G\big (D\ln \big |\tfrac{(\xi -1)(\eta -1)}{(\xi +1)(\eta +1)}\big |+\mu \big ) =2A\ln |\alpha (x+y)+\beta |+B(x+y)+\lambda \\&\qquad -A\ln |(\alpha x+\beta _1)^2-1|-Bx +C\ln |\tfrac{\alpha x+\beta _1-1}{\alpha x+\beta _1+1}|+\lambda _1\\&\qquad -A\ln |(\alpha y+\beta _2)^2-1|-By +C\ln |\tfrac{\alpha y+\beta _2-1}{\alpha y+\beta _2+1}|+\lambda _2\\&\quad =A\ln \big |\tfrac{(\xi +\eta )^2}{(\xi ^2-1)(\eta ^2-1)}\big |+C\ln \big |\tfrac{(\xi -1)(\eta -1)}{ (\xi +1)(\eta +1)}\big |+\Lambda . \end{aligned}$$

Putting $u{:}{=}D\ln \big |\frac{(\xi -1)(\eta -1)}{(\xi +1)(\eta +1)}\big |+\mu \in g_1(I)+g_2(I)$, we obtain that

$$\begin{aligned} \frac{(\xi +\eta )^2}{(\xi ^2-1)(\eta ^2-1)} =\frac{\big (1-\frac{(\xi -1)(\eta -1)}{(\xi +1)(\eta +1)}\big )^2}{4\cdot \frac{(\xi -1)(\eta -1)}{ (\xi +1)(\eta +1)}} =\frac{(1-\exp (\frac{u-\mu }{D}))^2}{4\exp (\frac{u-\mu }{D})} =\sinh ^2\Big (\frac{u-\mu }{2D}\Big ), \end{aligned}$$

consequently

$$\begin{aligned} G(u)=2A\ln |\sinh (\tfrac{u-\mu }{2D})|+C\tfrac{u-\mu }{D}+\Lambda \end{aligned}$$

holds.

Thus we obtained the solutions listed in (P1.3). $\square$

Finally, consider the so-called Hyperbolic Solutions, that is, the case when $\gamma =\gamma ^*>0$ holds.

Theorem 19

If $(F,f_1,f_2,G,g_1,g_2)$ is a regular solution of (2.3) such that, for the functions defined in (2.5), condition (B.2.3) holds, then there exist $A,B,C,D\in {\mathbb {R}}$ with $AD\ne 0$, $\kappa >0$, and $\lambda ,\lambda _k,\mu _k\in {\mathbb {R}}$ such that either F is of the form

$$\begin{aligned} F(x)=Ae^{-2q\kappa x}+2Bx+\lambda ,\quad x\in I \end{aligned}$$

for some $|q|=1$ and exactly one of the following holds true:

• (H1.1) there exist $\alpha ,\beta _k,A_k\in {\mathbb {R}}$ with $\alpha \beta _kA_k\ne 0$, $\beta _2A=\alpha qA_1$, and $\beta _1A=\alpha qA_2$ such that, for all $x\in I$ and $u\in g_1(I)+g_2(I)$, we have

  $$\begin{aligned} \begin{array}{l} f_k(x)=qA_ke^{-q\kappa x}-Bx+C\ln |\alpha e^{-q\kappa x}+\beta _k|+\lambda _k,\\ g_k(x)=D\ln |\alpha e^{-q\kappa x}+\beta _k|+\mu _k,\\ G(u)=\alpha ^{-2}A(\exp (\tfrac{u-\mu }{D})-\beta _1\beta _2)+C\tfrac{u-\mu }{D}+\Lambda \end{array} \end{aligned}$$

• (H1.2) there exist $|p|\notin \{0,1\}$, $A_k,C_k,D_k\in {\mathbb {R}}$ with $-\frac{1}{2}p^*A=q_k^2A_k$, $p^*C=q_kC_k$, and $p^*D=q_kD_k$ such that, for all $x\in I$ and $u\in g_1(I)+g_2(I)$, we have

  $$\begin{aligned} \begin{array}{l} f_k(x)=-A_ke^{-2q\kappa x}-Bx+qC_k e^{-q\kappa x}+\lambda _k,\\ g_k(x)=qD_ke^{-q\kappa x}+\mu _k,\quad G(u)=\tfrac{1}{2p^*}A(\tfrac{u-\mu }{D})^2+C\tfrac{u-\mu }{D}+\Lambda , \end{array} \end{aligned}$$

  where $(p^*,q_k){:}{=}(1,1)$ if $\psi _1=0$ on I and $(p^*,q_k){:}{=}(1-p^2,q+(-1)^{k-1}p)$ otherwise,

or F is of the form

$$\begin{aligned} F(x)=2A\ln |\alpha e^{2\kappa x}-\beta |+2Bx+\lambda ,\quad x\in I \end{aligned}$$

for some $\alpha ,\beta \in {\mathbb {R}}$ with $\alpha \beta \ne 0$ and exactly one of the following holds true:

• (H2.1) there exist $\alpha _k,\gamma \in {\mathbb {R}}$ with $\alpha _1\alpha _2=\alpha$ and $\gamma ^2=\beta$ such that, for all $x\in I$ and $u\in g_1(I)+g_2(I)$, we have

  $$\begin{aligned} \begin{array}{l} f_k(x)=-2A\ln |\alpha _ke^{\kappa x}+\gamma | -Bx+C(\alpha _ke^{\kappa x}+\gamma )^{-1}+\lambda _k,\\ g_k(x)=D(\alpha _ke^{\kappa x}+\gamma )^{-1}+\mu _k,\quad G(u)=2A\ln |1-\gamma \tfrac{u-\mu }{D}|+C\tfrac{u-\mu }{D}+\Lambda \end{array} \end{aligned}$$

• (H2.2) there exist $\alpha _k,\gamma \in {\mathbb {R}}$ with $\alpha _1\alpha _2=\alpha$, $\gamma \ne 0$, and $\gamma ^2+1=\beta$, such that, for all $x\in I$ and $u\in g_1(I)+g_2(I)$, we have

  $$\begin{aligned} \begin{array}{l} f_k(x)=-A\ln |(\alpha _k e^{\kappa x}+\gamma )^2+1| -Bx +C\arctan (\alpha _ke^{\kappa x}+\gamma )+\lambda _k,\\ g_k(x)=D\arctan (\alpha _ke^{\kappa x}+\gamma )+\mu _k,\\ G(u)=2A\ln |\gamma \sin (\tfrac{u-\mu }{D})+\cos (\tfrac{u-\mu }{D})|+C\tfrac{u-\mu }{D}+\Lambda \end{array} \end{aligned}$$

• (H2.3) there exist $\alpha _k,\gamma _k,A_k\in {\mathbb {R}}$ with $\alpha _1\alpha _2=\alpha$, $\gamma _1\gamma _2=\beta$, and $\frac{1}{2}(A_1+A_2)=A$ such that, for all $x\in I$ and $u\in g_1(I)+g_2(I)$, we have

  $$\begin{aligned} \begin{array}{l} f_k(x)=-A_1\ln |\alpha _ke^{\kappa x}+\gamma _1|-A_2\ln |\alpha _ke^{\kappa x}+\gamma _2|-Bx+\lambda _k,\\ g_k(x)=D\ln \big |\tfrac{\alpha _ke^{\kappa x}+\gamma _1}{\alpha _ke^{\kappa x}+\gamma _2}\big |+\mu _k,\\ G(u)=2A\ln |\tfrac{\gamma _2}{\gamma _2-\gamma _1}\exp (\tfrac{u-\mu }{D})-\tfrac{\gamma _1}{ \gamma _2-\gamma _1} |-A_1\tfrac{u-\mu }{D}+\Lambda \end{array} \end{aligned}$$

• (H2.4) there exist $|p|=1$ and $\gamma ,\beta _k,A_k,B_k\in {\mathbb {R}}$ with $\frac{1}{2}(A_1+A_2)=A$ such that, for all $x\in I$ and $u\in g_1(I)+g_2(I)$, we have

  $$\begin{aligned}&f_k(x)=-A_k\ln |\gamma e^{(-1)^kp\kappa x}+\beta _k|-B_kx+\lambda _k,\\&g_k(x)=(-1)^kD\ln |\gamma e^{(-1)^kp\kappa x}+\beta _k|+\mu _k,\\&G(u)=2A\ln |\beta _1\exp (\tfrac{u-\mu }{D})-\beta _2\Big |-A_2\tfrac{u-\mu }{D}+\Lambda \end{aligned}$$

  with $(B_1,B_2)=\frac{1+p}{2}(2\kappa A+B,B)+\frac{1-p}{2}(B,2\kappa A+B)$ and $\gamma (\beta _1,\beta _2)=\frac{1+p}{2}(\alpha ,-\beta )+\frac{1-p}{2}(-\beta ,\alpha )$

• (H2.5) there exist $|p|=1$ and $B_k,C_k,D_k\in {\mathbb {R}}$ with $-\frac{p}{2\kappa }(B_2-B_1)=A$ such that, for all $x\in I$ and $u\in g_1(I)+g_2(I)$, we have

  $$\begin{aligned} \begin{array}{l} f_k(x)=C_ke^{(-1)^kp\kappa x}-B_k x+\lambda _k,\\ g_k(x)=D_ke^{(-1)^{k}p\kappa x}+\mu _k,\quad G(u)=2A\ln |\tfrac{u-\mu }{D}|+C\tfrac{u-\mu }{D}+\Lambda , \end{array} \end{aligned}$$

  where $B=\frac{1+p}{2}B_2+\frac{1-p}{2}B_1$ and

  $$\begin{aligned} \alpha \beta (C,D) =-\tfrac{1+p}{2}\alpha (C_1,D_1)-\tfrac{1-p}{2}\alpha (C_2,D_2) =\tfrac{1+p}{2}\beta (C_2,D_2)+\tfrac{1-p}{2}\beta (C_1,D_1). \end{aligned}$$

Proof

For brevity, in this proof, whenever we write $\psi _1\equiv 0$, we mean that $\psi _1=0$ on the whole interval I. The sufficiency of the listed functions is a matter of substitution.

To show the necessity, let $(F,f_1,f_2,G,g_1,g_2)$ be a regular solution of (2.3). Then there exist constants $a,b,c,d\in {\mathbb {R}}$ with $ad\ne bc$, $\kappa >0$, and $\lambda ^*,\mu ^*\in {\mathbb {R}}$ such that

$$\begin{aligned}&F'(x)=\frac{(d+c)e^{2\kappa x}+d-c}{(b+a)e^{2\kappa x}+b-a},\\&g'_k(x)=\frac{4e^{\kappa x}}{(\omega _{1,k} +\omega _{2,k})e^{2\kappa x} +2\omega _{0,k}e^{\kappa x}+\omega _{1,k}-\omega _{2,k }}, \end{aligned}$$

and

$$\begin{aligned} f'_k(x)=-\frac{1}{2}\frac{(\theta _{1,k}+\theta _{2,k})e^{2\kappa x}+2\theta _{0,k}e^{\kappa x}+\theta _{1,k} -\theta _{2,k}}{(\omega _{1,k}+\omega _{2,k})e^{2\kappa x} +2\omega _{0,k}e^{\kappa x}+\omega _{1,k}-\omega _{2,k}} \end{aligned}$$

hold for all $x\in I$, where

$$\begin{aligned} (\omega _{2,k},\omega _{1,k},\omega _0){:}{=} {\left\{ \begin{array}{ll} (b,a,\lambda ^*)&{}\text {if }\psi _1\equiv 0,\\ (pb+(-1)^{k-1}a,pa+(-1)^{k-1}b,\lambda ^*)&{}\text {otherwise} \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} (\theta _{2,k},\theta _{1,k},\theta _0){:}{=} {\left\{ \begin{array}{ll} (d,c,2\mu ^*)&{}\text {if }\psi _1\equiv 0,\\ (pd+(-1)^{k-1}c,pc+(-1)^{k-1}d,2\mu ^*)&{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

Note that $p=0$ corresponds to the case when the function $\psi _2$ is constant on I. On the other hand, if $\psi _1$ is different from zero and $\psi _2$ is not constant on I, by Lemma 15, parameter p is different from zero.

From the differential equation concerning F we directly get that there exist constants $\lambda ,\lambda _0\in {\mathbb {R}}$ such that

$$\begin{aligned} F(x)={\left\{ \begin{array}{ll} A\exp (-2q\kappa x)+2Bx+\lambda &{}\text {if }|a|=|b|,\\ 2A\ln |\alpha _0\exp (2\kappa x)+\beta _0|+2Bx+\lambda _0&{}\text {if } |a|\ne |b|, \end{array}\right. }\quad x\in I,\quad \end{aligned}$$

(4.8)

where $q{:}{=}\mathop {\mathrm{sgn}}\nolimits (ab)\ne 0$, $\alpha _0{:}{=}a+b\ne 0$, $\beta _0{:}{=}b-a\ne 0$,

$$\begin{aligned} { 0\ne A{:}{=} {\left\{ \begin{array}{ll} \frac{c-qd}{2\kappa (b+qa)}&{}\text {if }|a|=|b|,\\ \frac{bc-ad}{2\kappa (b^2-a^2)}&{}\text {if }|a|\ne |b|, \end{array}\right. } \quad \text {and}\quad B{:}{=} {\left\{ \begin{array}{ll} \frac{d+qc}{2(b+qa)}&{}\text {if }|a|=|b|,\\ \frac{d-c}{2(b-a)}&{}\text {if }|a|\ne |b|. \end{array}\right. }}\nonumber \\ \end{aligned}$$

(4.9)

Correspondingly, we are going to distinguish the following main cases: either $|a|=|b|$ or $|a|\ne |b|$.

Case 1. Assume first that $|a|=|b|$ is valid. In this case, the form of $g_k$ and $f_k$ strongly depends on the value of the coefficient $\omega _0$, therefore, within Case 1., we will distinguish two sub-cases: either $\omega _0=0$ or $\omega _0\ne 0$.

(1.1) Assume that $\omega _0=0$ holds. Then we must have $|p|\ne 1$whenever $\psi _1$ is not zero on I. Indeed, our assumption $|a|=|b|$ and condition $ad\ne bc$ imply that either $a=b$ or $a=-b$ holds. Consequently, for a given $k=1,2$, at least one of $\omega _{1,k}+\omega _{2,k}=(p+(-1)^{k-1})(a+b)$ and $\omega _{1,k}-\omega _{2,k}=(p-(-1)^{k-1})(a-b)$ must be zero. If $|p|=1$ were true, then, it is easy to see, that there would exist $k\in \{1,2\}$, for which both of the previous coefficients are zero. It would then follow that exactly one of the functions $\psi _2-\psi _1$ or $\psi _2+\psi _1$ is identically zero on I, which, in view of Remark 18, is impossible.

Thus there exist $\lambda _1,\lambda _2,\mu _1,\mu _2\in {\mathbb {R}}$ such that

$$\begin{aligned} g_k(x){:}{=}qD_ke^{-q\kappa x}+\mu _k\quad \text {and}\quad f_k(x){:}{=}-A_ke^{-2q\kappa x}-Bx-qC_ke^{-q\kappa x}+\lambda _k, \end{aligned}$$

hold for all $x\in I$, where

$$\begin{aligned} (C,D){:}{=} {\left\{ \begin{array}{ll} \frac{1}{\kappa (a+qb)}\big (2\mu ^*,-4\big )&{}\text {if }\psi _1\equiv 0,\\ \frac{1}{\kappa (a+qb)p(p^2-1)}\big (2\mu ^*,4p\big )&{}\text {otherwise,} \end{array}\right. } \end{aligned}$$

and $-\frac{1}{2}p^*A=q_k^2A_k$, $p^*C=q_kC_k$, and $p^*D=q_kD_k$, with

$$\begin{aligned} (p^*,q_k){:}{=} {\left\{ \begin{array}{ll} (1,1)&{}\text {if }\psi _1\equiv 0\\ (1-p^2,q+(-1)^{k-1}p)&{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

(4.10)

Observe that we have $0\notin \{q-p,q+p\}$. Now, substituting $(\xi ,\eta ):= (e^{-q\kappa x},e^{-q\kappa y})$ if $\psi _1\equiv 0$ holds or $(\xi ,\eta ){:}{=}((q-p)e^{-q\kappa x},(q+p)e^{-q\kappa y})$ otherwise, furthermore putting $\mu {:}{=}\mu _1+\mu _2$ and $\Lambda {:}{=}\lambda +\lambda _1+\lambda _2$, we get that

$$\begin{aligned} G(qD(\xi +\eta )+\mu )= & {} Ae^{-q\kappa (x+y)}+B(x+y)+\lambda \\&-A_1e^{-2q\kappa x}-Bx-qC_1e^{-q\kappa x}+\lambda _1\\&-A_2e^{-2q\kappa y}-By-qC_2e^{-q\kappa y}+\lambda _2\\= & {} {\left\{ \begin{array}{ll} \tfrac{1}{2}A(\xi +\eta )^2+qC(\xi +\eta )+\Lambda &{}\text {if }\psi _1\equiv 0,\\ \tfrac{1}{2(1-p^2)}A(\xi +\eta )^2+qC(\xi +\eta )+\Lambda &{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

Consequently, for $u{:}{=}qD(\xi +\eta )+\mu \in g_1(I)+g_2(I)$, we get

$$\begin{aligned} G(u)=\tfrac{1}{2p^*}A(\tfrac{u-\mu }{D})^2+C\tfrac{u-\mu }{D}+\Lambda . \end{aligned}$$

Thus we obtained the solutions listed in (H1.2).

(1.2) Now assume that $\omega _0\ne 0$ holds. By Case 1. and condition $ad\ne bc$, we have $a\ne -qb$ and $d\ne qc$. Then there exist constants $\lambda _1,\lambda _2,\mu _1,\mu _2\in {\mathbb {R}}$ such that

$$\begin{aligned}&g_k(x)=D\ln |\alpha e^{-q\kappa x}+\beta _k|+\mu _k\quad \text {and}\\&f_k(x)=qA_ke^{-q\kappa x} -Bx +C\ln |\alpha e^{-q\kappa x}+\beta _k|+\lambda _k \end{aligned}$$

hold for all $x\in I$, where $\alpha {:}{=}2\lambda ^*\ne 0$, $D{:}{=}-q\frac{2}{\kappa \lambda ^*}\ne 0$,

$$\begin{aligned} C{:}{=}\frac{q}{4\kappa \lambda ^*}\cdot {\left\{ \begin{array}{ll} 4\mu ^*-4\lambda ^*B-\frac{1}{2\lambda ^*}(a+qb)(c-qd)&{}\text {if }\psi _1\equiv 0,\\ 4\mu ^*-4\lambda ^*B-\tfrac{1}{2\lambda ^*}(p^2-1)(b+qa)(d-qc)&{}\text {otherwise}, \end{array}\right. }\\ { 0\ne A_k{:}{=}\frac{1}{4\kappa \lambda ^*}\cdot {\left\{ \begin{array}{ll} c-qd&{}\text {if }\psi _1\equiv 0,\\ ((-1)^{k-1}p-q)(d-qc)&{}\text {otherwise}, \end{array}\right. }} \end{aligned}$$

and

$$\begin{aligned} 0\ne \beta _k{:}{=} {\left\{ \begin{array}{ll} a+qb&{}\text {if }\psi _1\equiv 0,\\ ((-1)^{k-1}p+q)(b+qa)\ne 0&{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

Note that, regardless of the behavior of the function $\psi _1$, the identities $\beta _2A=q\alpha A_1$ and $\beta _1A=q\alpha A_2$ hold. Hence, substituting $\xi {:}{=}\alpha e^{-q\kappa x}+\beta _1$, $\eta {:}{=}\alpha e^{-q\kappa y}+\beta _2$, $\mu {:}{=}\mu _1+\mu _2$, and $\Lambda {:}{=}\lambda +\lambda _1+\lambda _2$, Eq. (2.3) reduces to

$$\begin{aligned} G(D\ln |\xi \eta |+\mu )= & {} Ae^{-q\kappa (x+y)}+B(x+y)+\lambda \\&+qA_1e^{-q\kappa x}-Bx+C\ln |\alpha e^{-q\kappa x}+\beta _1|+\lambda _1\\&+qA_2e^{-q\kappa y}-By+C\ln |\alpha e^{-q\kappa y}+\beta _2|+\lambda _2\\= & {} \tfrac{1}{\alpha ^2}A(\xi \eta -\beta _1\beta _2) +C\ln |\xi \eta |+\Lambda . \end{aligned}$$

Consequently, for $u{:}{=}D\ln |\xi \eta |+\mu \in g_1(I)+g_2(I)$, we get

$$\begin{aligned}&G(u)=\alpha ^{-2}A(\exp (\tfrac{u-\mu }{D})-\beta _1\beta _2) +C\tfrac{u-\mu }{D}+\Lambda . \end{aligned}$$

Thus we obtained the solutions listed in (H1.1).

Case 2. Assume now that $|a|\ne |b|$ holds. Here we will distinguish two further cases: either $|p|=1$ or $|p|\ne 1$.

(2.1) Suppose that $|p|=1$ holds. Then, it is easy to see, that the shape of the corresponding solutions still depends on the value of $\omega _0$.

Within sub-case (2.1), suppose that $\omega _0=0$. Then there exist constants $\lambda _1,\lambda _2,\mu _1,\mu _2\in {\mathbb {R}}$ such that

$$\begin{aligned} g_k(x)=D_ke^{(-1)^{k}p\kappa x}+\mu _k \quad \text {and}\quad f_k(x)=C_ke^{(-1)^kp\kappa x}-B_k x+\lambda _k \end{aligned}$$

hold for all $x\in I$, with

$$\begin{aligned}&B_k{:}{=}\frac{d+(-1)^{k-1}pc}{2(b+(-1)^{k-1}pa)}, \quad \text {and}\quad \\&\alpha \beta (C,D)= {\left\{ \begin{array}{ll} \beta (C_1,D_1)=\alpha (C_2,D_2)&{}\text {if }p=-1\\ \alpha (C_1,D_1)=\beta (C_2,D_2)&{}\text {if }p=1, \end{array}\right. } \end{aligned}$$

where $\alpha {:}{=}\alpha _0$, $\beta {:}{=}-\beta _0\in {\mathbb {R}}$, $C{:}{=}\frac{p\mu ^*}{\kappa (b^2-a^2)}$, and $D{:}{=}-\frac{2}{\kappa (b^2-a^2)}\ne 0$.

Observe that $-\frac{p}{2\kappa }(B_2-B_1)=A\ne 0$, and either $B_1=B$ or $B_2=B$ if either $p=-1$ or $p=1$, respectively, where $A,B\in {\mathbb {R}}$ are defined in (4.9). Substituting either $(\xi ,\eta ){:}{=}(\alpha e^{\kappa x},\beta e^{-\kappa y})$ or $(\xi ,\eta ){:}{=}(\beta e^{-\kappa x},\alpha e^{\kappa y})$ if either $p=-1$ or $p=1$, respectively, furthermore putting $\mu {:}{=}\mu _1+\mu _2$ and $\Lambda {:}{=}\lambda +\lambda _1+\lambda _2$, Eq. (2.3) reduces to

$$\begin{aligned} G(D(\xi +\eta )+\mu )= & {} 2A\ln |\alpha e^{\kappa (x+y)} -\beta |+B(x+y)+\lambda \\&+C_1e^{-p\kappa x}-B_1 x+\lambda _1+C_2e^{p\kappa y}-B_2 y+\lambda _2\\= & {} 2A\ln |\alpha e^{\kappa (x+y)}-\beta |\\&+(B-B_1)x +(B-B_2)y+C(\xi +\eta )+\Lambda . \end{aligned}$$

Thus, depending on the exact value of p, we have

$$\begin{aligned} G(D(\xi +\eta )+\mu )= & {} {\left\{ \begin{array}{ll} 2A\ln |\alpha e^{\kappa (x+y)}-\beta |-2A\kappa y+C(\xi +\eta )+\Lambda &{}\text {if }p=-1\\ 2A\ln |\alpha e^{\kappa (x+y)}-\beta |-2A\kappa x+C(\xi +\eta )+\Lambda &{}\text {if }p=1 \end{array}\right. }\\= & {} 2A\ln |\xi +\eta |+C(\xi +\eta )+\Lambda . \end{aligned}$$

This, for $u{:}{=}D(\xi +\eta )+\mu \in g_1(I)+g_2(I)$, yields that

$$\begin{aligned} G(u)=2A\ln |\tfrac{u-\mu }{D}|+C\tfrac{u-\mu }{D}+\Lambda . \end{aligned}$$

Thus we obtained the solutions listed in (H2.5).

Within sub-case (2.1), assume that $\omega _0=\lambda ^*\ne 0$. Then there exist $\lambda _1,\lambda _2,\mu _1,\mu _2\in {\mathbb {R}}$ such that

$$\begin{aligned}&g_k(x)=(-1)^kD\ln |\gamma e^{(-1)^kp\kappa x} +\beta _k|+\mu _k \quad \text {and}\\&f_k(x)=-A_k\ln |\gamma e^{(-1)^kp\kappa x} +\beta _k|-B_kx+\lambda _k \end{aligned}$$

hold for all $x\in I$, where $D{:}{=}\frac{2p}{\kappa \lambda ^*}\ne 0$, $\gamma {:}{=}2\lambda ^*\ne 0$,

$$\begin{aligned}&A_k{:}{=}\tfrac{(-1)^kp}{2\kappa \lambda ^*}\big (2\mu ^*-\lambda ^*\tfrac{d+(-1)^{k-1}pc}{b+(-1)^{k-1}pa }\big ),\\&B_k{:}{=}\tfrac{d+(-1)^{k-1}pc}{2(b+(-1)^{k-1}pa)},\quad \text {and}\quad \beta _k{:}{=}2((-1)^{k-1}pb+a)\ne 0. \end{aligned}$$

Note that $\frac{1}{2}(A_1+A_2)=A$, and that the function F defined in (4.8) can be written as

$$\begin{aligned} F(x)=2A\ln |\alpha e^{2\kappa x}-\beta |+2Bx+\lambda , \end{aligned}$$

where $\lambda {:}{=}\lambda _0-2A\ln |4\lambda ^*|$, and either $(\alpha ,\beta )=\gamma (\beta _2,-\beta _1)$ or $(\alpha ,\beta )=\gamma (\beta _1,-\beta _2)$ if either $p=-1$ or $p=1$, respectively. Substituting $\xi {:}{=}\gamma e^{-p\kappa x}+\beta _1$, $\eta {:}{=}\gamma e^{p\kappa y}+\beta _2$, $\mu {:}{=}\mu _1+\mu _2$, and $\Lambda {:}{=}\lambda +\lambda _1+\lambda _2$, Eq. (2.3) reduces to

$$\begin{aligned}&G(D\ln |\tfrac{\eta }{\xi }|+\mu ) ={\left\{ \begin{array}{ll} 2A\ln |\alpha e^{\kappa (x+y)}-\beta |+B(x+y)\\ -A_1\ln |\xi |-B_1x-A_2\ln |\eta |-B_2y+\Lambda &{}\text {if }p=-1,\\ 2A\ln |\alpha e^{\kappa (x+y)}-\beta |+B(x+y)\\ -A_1\ln |\xi |-B_1x-A_1\ln |\eta |-B_2x+\Lambda &{} \text {if }p=1, \end{array}\right. }\\&\quad ={\left\{ \begin{array}{ll} 2A\ln |\beta _1\tfrac{\eta }{\xi }-\beta _2|-A_2\ln |\tfrac{\eta }{\xi }|+(B-B_1)x+(2\kappa A+B-B_2)y+\Lambda &{}\text {if }p=-1,\\ 2A\ln |\beta _1\tfrac{\eta }{\xi }-\beta _2|-A_2\ln |\tfrac{\eta }{\xi }|+(B-B_2)y+(2\kappa A+B-B_1)x+\Lambda &{}\text {if }p=1. \end{array}\right. } \end{aligned}$$

A short calculation shows that we have either $B_1=B$ and $B_2=2\kappa A+B$ or $B_1=2\kappa A+B$ and $B_2=B$ if either $p=-1$ or $p=1$, respectively. Consequently, for $u{:}{=}D\ln |\tfrac{\eta }{\xi }|+\mu \in g_1(I)+g_2(I)$, we get

$$\begin{aligned} G(u)=2A\ln |\beta _1\exp (\tfrac{u-\mu }{D})-\beta _2|-A_2\tfrac{u-\mu }{D}+\Lambda . \end{aligned}$$

Thus we obtained the solution listed in (H2.4).

(2.2) Suppose that $|p|\ne 1$ holds. Then we will distinguish three sub-cases depending on the sign of the discriminant

$$\begin{aligned} \Delta {:}{=}4(\lambda ^*)^2-4\cdot {\left\{ \begin{array}{ll} a^2-b^2&{}\text {if }\psi _1\equiv 0,\\ (p^2-1)(a^2-b^2)&{}\text {otherwise} \end{array}\right. } \end{aligned}$$

of the second degree polynomial in the denominator of $g_k$.

Within sub-case (2.2), suppose that $\Delta <0$ holds. Then there exist constants $\lambda _1,\lambda _2,\mu _1,\mu _2\in {\mathbb {R}}$ such that

$$\begin{aligned} g_{k}(x)= & {} D\arctan (\alpha _ke^{\kappa x}+\gamma ) +\mu _k\quad \text {and}\\ f_k(x)= & {} -A\ln |(\alpha _k e^{\kappa x}+\gamma )^2+1| -Bx +C\arctan (\alpha _ke^{\kappa x}+\gamma )+\lambda _k \end{aligned}$$

hold for all $x\in I$, where $\alpha _k{:}{=}\frac{2}{\sqrt{-\Delta }}(a+b)\ne 0$ if $\psi _1\equiv 0$ and $\alpha _k{:}{=}\frac{2}{\sqrt{-\Delta }}(p+(-1)^{k-1})(a+b)\ne 0$ otherwise, furthermore $\gamma {:}{=}\frac{2\lambda ^*}{\sqrt{-\Delta }}\ne 0$,

$$\begin{aligned}&C{:}{=}-\frac{1}{\kappa \sqrt{-\Delta }}\Big (2\mu ^*-\lambda ^*\frac{ac-bd}{a^2-b^2}\Big ), \quad \text {and}\\&D{:}{=}\frac{8}{\kappa \sqrt{-\Delta }}\ne 0. \end{aligned}$$

Observe that F can be reformulated as

$$\begin{aligned} F(x)=2A\ln |\alpha e^{2\kappa x}-\beta |+2Bx+\lambda , \end{aligned}$$

where $\alpha {:}{=}\alpha _1\alpha _2$, $\beta {:}{=}\gamma ^2+1$, and $\lambda {:}{=}\lambda _0+2A\ln \big |\frac{\Delta }{4(a+b)}\big |$ or $\lambda {:}{=}\lambda _0+2A\ln \big |\frac{\Delta }{4(p^2-1)(a+b)}\big |$ depending on whether $\psi _1\equiv 0$ or not, respectively. Hence, substituting $\xi {:}{=}\alpha _1e^{\kappa x}+\gamma$, $\eta {:}{=}\alpha _2e^{\kappa y}+\gamma$, $\mu {:}{=}\mu _1+\mu _2$, and $\Lambda {:}{=}\lambda +\lambda _1+\lambda _2$, Eq. (2.3) reduces to

$$\begin{aligned}&G(D\arctan (\tfrac{\xi +\eta }{1-\xi \eta })+\mu )\\&\quad =2A\ln |\alpha _1\alpha _2e^{\kappa (x+y)}-\gamma ^2-1|+B(x+y)+\lambda \\&\qquad -A\ln |(\alpha _1 e^{\kappa x}+\gamma )^2+1| -Bx +C\arctan (\alpha _1e^{\kappa x}+\gamma )+\lambda _1\\&\qquad -A\ln |(\alpha _2 e^{\kappa y}+\gamma )^2+1| -By +C\arctan (\alpha _2e^{\kappa y}+\gamma )+\lambda _2\\&\quad =A\ln |(\gamma \tfrac{\xi +\eta }{1-\xi \eta }+1)^2((\tfrac{\xi +\eta }{1-\xi \eta })^2+1)^{-1} | +C\arctan (\tfrac{\xi +\eta }{1-\xi \eta } )+\Lambda , \end{aligned}$$

where in the last step, we used the identity

$$\begin{aligned}&\frac{(\alpha _1\alpha _2e^{\kappa (x+y)}-\gamma ^2-1)^2}{(\xi ^2+1)(\eta ^2+1)}\\&\quad =\frac{(1-\xi \eta +\gamma (\xi +\eta ))^2}{(1-\xi \eta )^2+(\xi +\eta )^2}=\Big (\gamma \frac{\xi +\eta }{1-\xi \eta }+1\Big )^2 \Big (\Big (\frac{\xi +\eta }{1-\xi \eta }\Big )^2+1\Big )^{-1}. \end{aligned}$$

Consequently, for $u{:}{=}D\arctan (\frac{\xi +\eta }{1-\xi \eta })+\mu \in g_1(I)+g_2(I)$, we get

$$\begin{aligned} G(u)=2A\ln |\gamma \sin (\tfrac{u-\mu }{D})+\cos (\tfrac{u-\mu }{D})|+C\tfrac{u-\mu }{D}+\Lambda . \end{aligned}$$

Thus we obtained the solutions listed in (H2.2).

Within sub-case (2.2), suppose that $\Delta =0$ holds. Then there exist $\lambda _1,\lambda _2,\mu _1,\mu _2\in {\mathbb {R}}$ such that

$$\begin{aligned}&g_k(x)=D(\alpha _ke^{\kappa x}+\gamma )^{-1}+\mu _k \quad \text {and}\\&f_k(x)=-2A\ln |\alpha _ke^{\kappa x}+\gamma |-Bx+C(\alpha _ke^{\kappa x}+\gamma )^{-1}+\lambda _k \end{aligned}$$

hold for all $x\in I$, where $A,B\in {\mathbb {R}}$ are defined in (4.9),

$$\begin{aligned} C{:}{=}\frac{1}{\kappa }\Big (2\mu ^*-\lambda ^*\frac{ac-bd}{a^2-b^2}\Big ),\quad D{:}{=}-\frac{4}{\kappa }\ne 0,\quad \gamma {:}{=}\lambda ^*\ne 0, \end{aligned}$$

furthermore either $\alpha _k{:}{=}a+b\ne 0$ or $\alpha _k{:}{=}(p+(-1)^{k-1})(a+b)\ne 0$ if $\psi _1\equiv 0$ or not, respectively.

Observe that, due to our assumption $\Delta =0$, the function F can be written as

$$\begin{aligned} F(x)=2A\ln |\alpha e^{2\kappa x}-\beta |+2Bx+\lambda , \end{aligned}$$

where $\alpha {:}{=}\alpha _1\alpha _2$, $\beta {:}{=}\gamma ^2$, and either $\lambda {:}{=}\lambda _0-2A\ln |a+b|$ or $\lambda {:}{=}\lambda _0-A\ln |(p^2-1)(a+b)|$ if $\psi _1\equiv 0$ or not, respectively. Hence, substituting $\xi {:}{=}\alpha _1e^{\kappa x}+\gamma$ and $\eta {:}{=}\alpha _2e^{\kappa y}+\gamma$, $\mu {:}{=}\mu _1+\mu _2$, and $\Lambda {:}{=}\lambda +\lambda _1+\lambda _2$, Eq. (2.3) reduces to

$$\begin{aligned} G\big (D\tfrac{\xi +\eta }{\xi \eta }+\mu \big )= & {} 2A\ln |\alpha _1\alpha _2e^{\kappa (x+y)}-\gamma ^2|+B(x+y)+\lambda \\&-2A\ln |\alpha _1e^{\kappa x}+\gamma |-Bx+C(\alpha _1e^{\kappa x}+\gamma )^{-1}+\lambda _1\\&-2A\ln |\alpha _2e^{\kappa y}+\gamma |-By+C(\alpha _2e^{\kappa y}+\gamma )^{-1}+\lambda _2\\= & {} 2A\ln |1-\gamma \tfrac{\xi +\eta }{\xi \eta }| +C\tfrac{\xi +\eta }{\xi \eta } +\Lambda . \end{aligned}$$

Consequently, for $u{:}{=}D\tfrac{\xi +\eta }{\xi \eta }+\mu \in g_1(I)+g_2(I)$, we get

$$\begin{aligned} G(u)=2A\ln |1-\gamma \tfrac{u-\mu }{D}|+C\tfrac{u-\mu }{D}+\Lambda . \end{aligned}$$

Thus we obtained the solutions listed in (H2.1).

Finally, suppose that $\Delta >0$ holds. Then there exist constants $\lambda _1,\lambda _2,\mu _1,\mu _2\in {\mathbb {R}}$ such that

$$\begin{aligned} g_k(x)= & {} D\ln \big | \tfrac{\alpha _ke^{\kappa x}+\gamma _1}{\alpha _ke^{\kappa x}+\gamma _2}\big | +\mu _k \quad \text {and}\\ f_k(x)= & {} -A_1\ln |\alpha _ke^{\kappa x} +\gamma _1| -A_2\ln |\alpha _ke^{\kappa x}+\gamma _2|-Bx+\lambda _k \end{aligned}$$

hold for all $x\in I$, where either $\alpha _k=2(a+b)$ or $\alpha _k{:}{=}2(p+(-1)^{k-1})(a+b)$ if $\psi _1\equiv 0$ or not, respectively, furthermore $\gamma _k{:}{=}2\lambda ^*+(-1)^k\sqrt{\Delta }$,

$$\begin{aligned} A_k{:}{=}A+\frac{(-1)^{k-1}}{\kappa \sqrt{\Delta }} \Big (2\mu ^* -\lambda ^*\frac{ac-bd}{a^2-b^2}\Big ),\quad \text {and}\quad D{:}{=}\frac{4}{\kappa \sqrt{\Delta }}\ne 0. \end{aligned}$$

Observe that $\gamma _1\gamma _2$ cannot be zero, $\frac{1}{2}(A_1+A_2)=A\ne 0$, and that the function F can be written as

$$\begin{aligned} F(x)=2A\ln |\alpha e^{2\kappa x}-\beta |+2Bx+\lambda ,\quad x\in I, \end{aligned}$$

where $\alpha {:}{=}\alpha _1\alpha _2$, $\beta {:}{=}\gamma _1\gamma _2$, and either $\lambda {:}{=}\lambda _0-2A\ln |4(a+b)|$ or $\lambda {:}{=}\lambda _0-2A\ln |4(p^2-1)(a+b)|$ if $\psi _1\equiv 0$ or not, respectively. Hence, substituting $\xi _k{:}{=}\alpha _1e^{\kappa x}+\gamma _k$, $\eta _k{:}{=}\alpha _2e^{\kappa y}+\gamma _k$, $\mu {:}{=}\mu _1+\mu _2$, and $\Lambda {:}{=}\lambda +\lambda _1+\lambda _2$, and using that $\gamma _2-\gamma _1=2\sqrt{\Delta }\ne 0$, Eq. (2.3) reduces to

$$\begin{aligned}&G(D\ln |\tfrac{\xi _1\eta _1}{\xi _2\eta _2}|+\mu )\\&\quad =2A\ln |\alpha _1\alpha _2e^{\kappa (x+y)}-\gamma _1\gamma _2|+B(x+y)+\lambda \\&\qquad -A_1\ln |\alpha _1e^{\kappa x}+\gamma _1|-A_2\ln |\alpha _1e^{\kappa x}+\gamma _2|-Bx+\lambda _1\\&\qquad -A_1\ln |\alpha _2e^{\kappa y}+\gamma _1|-A_2\ln |\alpha _2e^{\kappa y}+\gamma _2|-By+\lambda _2\\&\quad =2A\ln |\alpha _1\alpha _2e^{\kappa (x+y)}-\gamma _1\gamma _2|\\&\qquad -A_1\ln |\xi _1|+(A_1-2A)\ln |\xi _2|-A_1\ln |\eta _1|+(A_1-2A)\ln |\eta _2|+\Lambda \\&\quad =2A\ln \big | \tfrac{\alpha _1\alpha _2e^{\kappa (x+y)}-\gamma _1\gamma _2}{\xi _2\eta _2}\big |-A_1\ln \big |\tfrac{\xi _1\eta _1}{\xi _2\eta _2}\big |+\Lambda \\&\quad =2A\ln \big |\tfrac{ \gamma _2}{\gamma _2-\gamma _1} \tfrac{\xi _1\eta _1}{\xi _2\eta _2}-\tfrac{\gamma _1}{\gamma _2-\gamma _1}\big |-A_1\ln \big |\tfrac{ \xi _1\eta _1}{ \xi _2\eta _2}\big |+\Lambda , \end{aligned}$$

where, in the last step, we used the identity

$$\begin{aligned} \alpha _1\alpha _2e^{\kappa (x+y)}-\gamma _1\gamma _2 =\tfrac{\gamma _2}{\gamma _2-\gamma _1} \xi _1\eta _1-\tfrac{\gamma _1}{\gamma _2-\gamma _1}\xi _2\eta _2. \end{aligned}$$

Consequently, for $u{:}{=}D\ln |\tfrac{\xi _1\eta _1}{\xi _2\eta _2}|+\mu \in g_1(I)+g_2(I)$, we get

$$\begin{aligned} G(u)=2A\ln |\tfrac{\gamma _2}{\gamma _2-\gamma _1}\exp (\tfrac{u-\mu }{D}) -\tfrac{\gamma _1}{ \gamma _2-\gamma _1}|-A_1\tfrac{u-\mu }{D}+\Lambda . \end{aligned}$$

Thus we obtained the solutions listed in (H2.3). $\square$