Differential Transcendence of Solutions of the Difference Equation $\Delta y=ay^2+by+c$

Abstract

By a purely algebraic way, we investigate a necessary and sufficient condition for the existence of differentially algebraic and transcendental solutions to certain difference equations whose form is related to Riccati equations. Our result roughly implies that all the solutions do not satisfy any algebraic differential equations if there is no algebraic solution.

1 Introduction

Differential transcendence of a function means that it does not satisfy any algebraic differential equations. This is also called hypertranscendence or transcendental transcendence. It is well-known that Gamma function has this property (see the paper [3] by O. Hölder). In his paper [1], P. Borwein studied this problem for the non-trivial solution of

$$\begin{aligned} y(x^2)=y(x)^2+cx \end{aligned}$$

or

$$\begin{aligned} z(x^2)=z(x)^2+c \end{aligned}$$

with a constant c. They are transformed into each other by $z(x)=y(x)/x^{1/2}$. When $c=2$, the former is solved by the generating function for counting the number of bracketing of n objects. The latter is related to Mandelbrot set (cf. the book [5]). In this paper, we shall study similar difference equations,

$$\begin{aligned} \tau (y(x))=a(x)y(x)^d+c(x),\quad d\in {\mathbb {Z}}_{>1}, \end{aligned}$$

with non-constant coefficients, where $\tau (y(x))$ can be $y(x^2)$, $y(x+1)$, y(2x), etc. This form is also derived from Riccati equation $y'=ay^2+by+c$ by the usual replacement of $y'$ with $\Delta y=\tau (y)-y$ in the following way,

$$\begin{aligned} \Delta y= & {} ay^2+by+c,\\ \tau (y)= & {} ay^2+(b+1)y+c,\\ \tau (z)= & {} az^2+\tau \left( \frac{b+1}{2a}\right) -\frac{(b+1)^2}{4a}+c, \end{aligned}$$

where

$$\begin{aligned} z=y+\frac{b+1}{2a}. \end{aligned}$$

Theorem 1 mentioned below in this section roughly implies that transcendental function solutions are differentially transcendental if the coefficients $a,c\in {\mathbb {C}}(x)$ do not satisfy

$$\begin{aligned} d\frac{c'}{c}=\frac{\tau (c/a)'}{\tau (c/a)}. \end{aligned}$$

It will be seen that this equality of the coefficients usually yields an algebraic function solution. Hence non-existence of algebraic function solutions leads to non-existence of differentially algebraic solutions, which is described in Theorem 2. Actually, each of the above equations Borwein studied satisfies the equality of the coefficients.

For an example of Mahler type, we introduce

$$\begin{aligned} p(x)y(x^d)=a(x)y(x)^d+c(x),\quad a,c,p\in {\mathbb {C}}[x], \end{aligned}$$

with $a(0)\ne 0$ and $(c(0),p(0))\ne (0,0)$. Choose $e\in {\mathbb {C}}^\times$ such that

$$\begin{aligned} a(0)\mathrm {e}^d-p(0)e+c(0)=0. \end{aligned}$$

By $z(x)=y(x)-e$, we obtain

$$\begin{aligned} p(x)(z(x^d)+e)=a(x)(z(x)+e)^d+c(x). \end{aligned}$$

Let

$$\begin{aligned} F(x,u,v)=p(x)(u+e)-a(x)(v+e)^d-c(x)\in {\mathbb {C}}[x,u,v]. \end{aligned}$$

Since it satisfies $F(0,0,0)=0$ and $F_v(0,0,0)\ne 0$, there exists a convergent power series $z=f(x)\in {\mathbb {C}}[[x]]$ which is effectively computable from the coefficients of F (see the book [5], Theorem 1.7.2). Hence $y=f(x)+e$ is a solution of the above equation in the field ${\mathbb {C}}((x))$ with the derivation $D=\frac{d}{dx}$ and the transforming operator $\tau :y(x)\mapsto y(x^d)$, which we call a DT field with $D\tau =dx^{d-1}\tau D$. We give one concrete example,

$$\begin{aligned} xy(x^d)=y(x)^d+1, \end{aligned}$$

whose coefficients do not satisfy the above equality. Since the solution $y=f(x)+e$ is transcendental, which is proved in an ordinary way (cf. the book [5], Theorem 1.3), it is differentially transcendental by Theorem 1.

For q-difference equations, there is a well-known paper [9] written by J. F. Ritt, who studied differentially algebraic meromorphic functions which satisfy Poincaré’s multiplication theorems of the form

$$\begin{aligned} y(qx)=R(y(x)), \quad R(Y)\in {\mathbb {C}}(Y). \end{aligned}$$

His conclusion was that if R is not linear, then the function can be represented by exponential functions and/or Weierstrass $\wp$-functions in some way. We introduce another kind of Poincaré’s multiplication theorems,

$$\begin{aligned} y(qx)=R(x,y(x)),\quad R\in {\mathbb {C}}(X,Y), \end{aligned}$$

where $|q| >1$, $R(0,0)=0$, and

$$\begin{aligned} B=\begin{pmatrix} q &{} 0 \\ \frac{\partial R}{\partial X}(0,0) &{} \frac{\partial R}{\partial Y}(0,0) \end{pmatrix} \end{aligned}$$

does not have eigenvalues $q^i$ $(i=2,3,\dots )$ but q. In his paper [7], H. Poincaré constructed meromorphic function solutions (cf. the paper [6]) with $y(0)=0$ and

$$\begin{aligned} \left( B-q\begin{pmatrix}1&{}0\\ 0&{}1\end{pmatrix}\right) \begin{pmatrix}1\\ y'(0)\end{pmatrix}=0. \end{aligned}$$

We give one concrete example again. The equation

$$\begin{aligned} y(qx)=y(x)^2+x \end{aligned}$$

does not satisfy the equality of the coefficients. It will be seen that the Poincaré’s solution is transcendental, and thus by Theorem 1, differentially transcendental in the field of meromorphic functions with the derivation $D=\frac{d}{dx}$ and the transforming operator $\tau :y(x)\mapsto y(qx)$, where $D\tau =q\tau D$.

As mentioned above, Borwein and Ritt studied difference equations with constant coefficients but both papers appear to have gaps. The footnote on page 219 in Borwein’s paper [1] notes there are cases omitted in his proof. In Ritt’s paper [9], on page 679 for example, it is not obvious why the coefficients of only the terms (21) and (22) are compared. In the eighth line, the case where $S_{n-1}=0$ looks missing.

Our results are written in terms of difference algebra and differential algebra.

Notation

Throughout the paper every field is of characteristic zero. The (transforming) operator $\tau$ of a difference field ${\mathcal {K}}=(K,\tau )$ is an isomorphism of the field K into itself which may not be surjective. For the calligraphic ${\mathcal {K}}$, the non-calligraphic K often denotes its underlying field.

For an algebraic closure $\overline{K}$ of K, the transforming operator $\tau$ is extended to an isomorphism $\overline{\tau }$ of $\overline{K}$ into itself, not necessarily in a unique way (cf. the book [10], Ch. II, §14, Theorem 33). We call the difference field $\overline{{\mathcal {K}}}=(\overline{K},\overline{\tau })$ an algebraic closure of ${\mathcal {K}}$.

For a difference field extension $(L,\tau )/(K,\tau |_K)$, we usually use $(K,\tau )$ instead of $(K,\tau |_K)$.

A solution of a difference equation $F(y,y_1,y_2,\dots )=0$ over ${\mathcal {K}}$ is defined to be an element f of some difference overfield ${\mathcal {L}}=(L,\tau )$ of ${\mathcal {K}}$ which satisfies the equation $F(f,\tau f,\tau ^2 f,\dots )=0$ (cf. the books [2, 4, 8]).

Let K be a field, D a derivation of K, which is a mapping $D:K\rightarrow K$ such that $D(a+b)=D(a)+D(b)$ and $D(ab)=D(a)b+aD(b)$, and $\tau$ a transforming operator of K. We simply call ${\mathcal {K}}=(K,D,\tau )$ a DT field, though some relation between the derivation and the transforming operator may be given. This term is short for a field with a derivation and a transforming operator. We say that a DT field ${\mathcal {K}}'=(K',D',\tau ')$ is a DT overfield of ${\mathcal {K}}$ when $K'/K$ is a field extension with $D'|_K=D$ and $\tau '|_K=\tau$.

Let ${\mathcal {L}}/{\mathcal {K}}$ be a differential field extension, and D the derivation of ${\mathcal {L}}$. When $f\in L$ satisfies $\mathop {\mathrm {tr.\, deg}}\nolimits K(f,Df,D^2f,\dots )/K<\infty$, we say that $f\in {\mathcal {L}}$ is differentially algebraic over K. It is equivalent that f satisfies a certain algebraic differential equation over ${\mathcal {K}}$ in ${\mathcal {L}}$.

Theorem 1

Let ${\mathcal {K}}=(K,D,\tau )$ be a DT field with $D\tau =s\tau D$ for a certain $s\in K^\times$. For the difference equation over ${\mathcal {K}}$,

$$\begin{aligned} y_1=ay^d+c,\quad a,c\in K^\times ,\ d\in {\mathbb {Z}}_{>1}, \end{aligned}$$

the following are equivalent:

• (i) There exists a solution f transcendental and differentially algebraic over K in a certain DT overfield ${\mathcal {U}}=(U,D,\tau )$ of ${\mathcal {K}}$ with $D\tau =s\tau D$.

• (ii) The coefficients satisfy

  $$\begin{aligned} d\frac{Dc}{c}=\frac{D\tau (c/a)}{\tau (c/a)}. \end{aligned}$$

In the usual situation such as ${\mathcal {K}}=({\mathbb {C}}(x),\frac{d}{dx},\tau )$ with

$$\begin{aligned} \{z\in {\mathbb {C}}(x)\;|\;\frac{d}{dx}z=0\} =\{z\in {\mathbb {C}}(x)\;|\;\tau z=z\} ={\mathbb {C}}, \end{aligned}$$

we will have the following simpler result.

Theorem 2

Let ${\mathcal {K}}=(K,D,\tau )$ be a DT field with $D\tau =s\tau D$ for a certain $s\in K^\times$, and suppose

$$\begin{aligned} \{z\in K\;|\;Dz=0\} =\{z\in K\;|\;\tau z=z\} =C, \end{aligned}$$

where C is an algebraically closed field. If the difference equation over ${\mathcal {K}}$,

$$\begin{aligned} y_1=ay^d+c,\quad a,c\in K^\times ,\ d\in {\mathbb {Z}}_{>1}, \end{aligned}$$

has a differentially algebraic solution over K in a certain DT overfield ${\mathcal {U}}=(U,D,\tau )$ of ${\mathcal {K}}$ with $D\tau =s\tau D$, then it has a solution in $\overline{{\mathcal {K}}}$, where $\overline{{\mathcal {K}}}$ is the algebraic closure of $(K,\tau )$ in an algebraic closure of $(U,\tau )$.

In Sect. 2, we prepare several lemmas. The proofs of the theorems are written in Sect. 3.

2 Lemmas

Throughout this section, ${\mathcal {U}}/{\mathcal {K}}$ is a difference field extension and $\overline{{\mathcal {U}}}=(\overline{U},\tau )$ an algebraic closure of ${\mathcal {U}}$. Moreover, $Y\in \overline{{\mathcal {U}}}$ is a transcendental element over K which satisfies

$$\begin{aligned} \tau Y=aY^d+c,\quad a,c\in K^\times ,\ d\in {\mathbb {Z}}_{>1}. \end{aligned}$$

Lemma 3

If $R\in K(Y)^\times$ satisfies

$$\begin{aligned} R=\alpha Y^e\tau (R),\quad \alpha \in K^\times ,\ e\in {\mathbb {Z}}_{>0}, \end{aligned}$$

then

$$\begin{aligned} d=2,\quad c^2=-2\tau \left( \frac{c}{a}\right) . \end{aligned}$$

Proof

By an expression,

$$\begin{aligned} R=\frac{M}{N},\quad M,N\in K[Y]\setminus \{0\}, \end{aligned}$$

where M and N are relatively prime and M is monic, we obtain

$$\begin{aligned} M\tau (N)=\alpha Y^e\tau (M)N. \end{aligned}$$

The polynomials $\tau (M)$ and $\tau (N)$ are also relatively prime, as there are $A,B\in K[Y]$ such that $AM+BN=1$, whose transform is $\tau (A)\tau (M)+\tau (B)\tau (N)=1$. This implies that $\tau (M)$ divides M. The degree of $\tau (M)$ is $d\cdot \deg M$, and is smaller than or equal to $\deg M$. Hence we see $M=1$ and

$$\begin{aligned} \tau (N)=\alpha Y^eN. \end{aligned}$$

(1)

Comparing the degrees, we obtain

$$\begin{aligned} d\cdot \deg N= & {} e+\deg N,\\ (d-1)\deg N= & {} e. \end{aligned}$$

By $\tau (N)\in K[Y^d]$, N has an expression,

$$\begin{aligned} N=\beta Y^m(Y^d-\lambda _1)\cdots (Y^d-\lambda _n),\quad \beta \in K^\times ,\ m\ge 0,\ \lambda _i\in \overline{K}^\times , \end{aligned}$$

where $\deg N=m+dn$. Then the Eq. (1) is

$$\begin{aligned} \begin{aligned}&\tau (\beta )(aY^d+c)^m((aY^d+c)^d-\tau \lambda _1)\cdots ((aY^d+c)^d-\tau \lambda _n)\\&=\alpha Y^e\cdot \beta Y^m(Y^d-\lambda _1)\cdots (Y^d-\lambda _n). \end{aligned} \end{aligned}$$

(2)

Each factor

$$\begin{aligned} (aY^d+c)^d-\tau \lambda _i=a^dY^{d^2}+\dots +dac^{d-1}Y^d+c^d-\tau \lambda _i \end{aligned}$$

can be divided by Y at most d times, and the right side of the Eq. (2) is divided by $Y^{e+m}$, where

$$\begin{aligned} e+m=(d-1)(m+dn)+m=dm+(d-1)dn\ge dn. \end{aligned}$$

Hence we see $\tau \lambda _i=c^d$ for all $i=1,\dots ,n$, which implies $\lambda _1=\dots =\lambda _n$. Comparing the numbers of divisibility by Y again, we obtain

$$\begin{aligned} dn= & {} e+m=dm+(d-1)dn,\\ m= & {} (2-d)n. \end{aligned}$$

Assume $d\ge 3$ to prove $d=2$. Since m and n are not negative, we see $m=n=0$ by the above equality, and thus $R=M/N=1/\beta \in K^\times$. This contradicts the assumption of this lemma. Hece we conclude $d=2$, and thus $m=0$ follows. If $n=0$, then we have $m=n=0$ again, which yields the same contradiction as above. Hence $n\ne 0$ also follows.

The Eq. (2) is therefore

$$\begin{aligned} \tau (\beta )(a^2Y^4+2acY^2)^n=\alpha Y^{2n}\cdot \beta (Y^2-\lambda _1)^n. \end{aligned}$$

Hence we have

$$\begin{aligned} \tau (\beta )a^{2n}\left( Y^2+\frac{2c}{a}\right) ^n=\alpha \beta (Y^2-\lambda _1)^n. \end{aligned}$$

The root of this polynomial in the transcendental element $Y^2$ is

$$\begin{aligned} -\dfrac{2c}{a}=\lambda _1, \end{aligned}$$

and its first transform is

$$\begin{aligned} -2\tau \left( \frac{c}{a}\right) =\tau \lambda _1=c^2. \end{aligned}$$

$\square$

Lemma 4

If $R\in K(Y)$ satisfies

$$\begin{aligned} R=\frac{\alpha }{Y}+daY^{d-1}\tau (R),\quad \alpha \in K^\times , \end{aligned}$$

then

$$\begin{aligned} d=2,\quad c^2=-2\tau \left( \frac{c}{a}\right) . \end{aligned}$$

Proof

By the given equation, we see $R\ne 0$. Hence R can be expressed as follows,

$$\begin{aligned} R=\frac{M}{N}, \quad M,N\in K[Y]\setminus \{0\}, \end{aligned}$$

where M and N are relatively prime and N is monic. Then we obtain

$$\begin{aligned} YM\tau (N)=\alpha N\tau (N)+daY^d\tau (M)N. \end{aligned}$$

(3)

Since $\tau (M)$ and $\tau (N)$ are relatively prime, $\tau (N)$ divides $Y^dN$. Their degrees satisfy $d\cdot \deg N\le d+\deg N$, and thus

$$\begin{aligned} \deg N\le \frac{d}{d-1}=1+\frac{1}{d-1}\le 2. \end{aligned}$$

In addition, from the Eq. (3), it follows that Y divides $N\tau (N)$. Hence the degree of N is 1 or 2.

In the case where $\deg N=2$, we see $d=2$ by the above inequality. We write

$$\begin{aligned} N=(Y-\lambda _1)(Y-\lambda _2),\quad \lambda _i\in \overline{K}. \end{aligned}$$

Then

$$\begin{aligned} \tau (N)=(aY^2+c-\tau \lambda _1)(aY^2+c-\tau \lambda _2). \end{aligned}$$

Thus $\tau (N)$ divides $Y^2N$ as mentioned above, and they both have the same degree 4, which imply $\tau (N)=a^2Y^2N$. Hence we have

$$\begin{aligned} (aY^2+c-\tau \lambda _1)(aY^2+c-\tau \lambda _2) =a^2Y^2(Y-\lambda _1)(Y-\lambda _2). \end{aligned}$$

We may suppose $\tau \lambda _1=c$, as Y divides the left-hand side. Then by

$$\begin{aligned} aY^2(aY^2+c-\tau \lambda _2)=a^2Y^2(Y-\lambda _1)(Y-\lambda _2), \end{aligned}$$

we obtain

$$\begin{aligned} aY^2+c-\tau \lambda _2=a(Y^2-(\lambda _1+\lambda _2)Y+\lambda _1\lambda _2). \end{aligned}$$

The coefficients satisfy

$$\begin{aligned} \lambda _1+\lambda _2=0,\quad a\lambda _1\lambda _2=c-\tau \lambda _2. \end{aligned}$$

Hence we see $\lambda _2=-\lambda _1$, and thus from the latter equation it follows that

$$\begin{aligned} -a\lambda _1^2= & {} c+\tau \lambda _1=2c,\\ \lambda _1^2= & {} -2\frac{c}{a}. \end{aligned}$$

Transforming this, we obtain

$$\begin{aligned} c^2=-2\tau \left( \frac{c}{a}\right) . \end{aligned}$$

The case where $\deg N=1$ has a contradiction. Indeed, the first transform of $N=Y-\lambda$, $\lambda \in K$, is

$$\begin{aligned} \tau (N)=aY^d+c-\tau \lambda , \end{aligned}$$

which divides $Y^dN=Y^d(Y-\lambda )$. This implies $\tau \lambda =c$. Moreover, from the Eq. (3), it follows that $N=Y-\lambda$ divides $Y\tau (N)=Y(aY^d)$. Hence we have $\lambda =0$, and thus $c=0$. This contradicts the assumption of this section. $\square$

Lemma 5

If $R\in K(Y)$ satisfies

$$\begin{aligned} daY^{d-1}R=a'Y^d+c'+s\tau (R),\quad a',c',s\in K,\ (a',c')\ne (0,0),\ s\ne 0, \end{aligned}$$

then $R=\alpha Y$, where

$$\begin{aligned} \alpha =\frac{1}{d}\left( \frac{a'}{a}-\frac{c'}{c}\right) ,\quad s\tau \alpha =-\frac{c'}{c}. \end{aligned}$$

Proof

We easily see $R\ne 0$. By an expression,

$$\begin{aligned} R=\frac{M}{N},\quad M,N\in K[Y]\setminus \{0\}, \end{aligned}$$

where M and N are relatively prime and N is monic, we obtain

$$\begin{aligned} daY^{d-1}M\tau (N)=(a'Y^d+c')N\tau (N)+s\tau (M)N. \end{aligned}$$

Hence $\tau (N)$ divides N, which implies $N=1$. Then the above equation is

$$\begin{aligned} daY^{d-1}M=(a'Y^d+c')+s\tau (M). \end{aligned}$$

We shall obtain $\deg M=1$ by comparing the degrees. Assuming $\deg M\ge 2$, we indeed have $d-1+\deg M=d\cdot \deg M$ which is impossible. If $\deg M=0$, then the degree of the left-hand side is $d-1$, though that of the right-hand side is d or 0.

Write $M=\alpha Y+\beta$, where $\alpha ,\beta \in K$ and $\alpha \ne 0$. Then we have

$$\begin{aligned} daY^{d-1}(\alpha Y+\beta )=(a'Y^d+c')+s(\tau (\alpha )(aY^d+c)+\tau (\beta )), \end{aligned}$$

and thus

$$\begin{aligned} da\alpha Y^d+da\beta Y^{d-1}=(a'+s\tau (\alpha ) a)Y^d+c'+s(\tau (\alpha ) c+\tau (\beta )). \end{aligned}$$

Comparing the coefficients, we first have $\beta =0$, by which $R=M=\alpha Y$ and

$$\begin{aligned} \left\{ \begin{aligned} da\alpha&=a'+s\tau (\alpha ) a,\\ 0&=c'+s\tau (\alpha ) c. \end{aligned} \right. \end{aligned}$$

From the latter, $s\tau (\alpha )=-c'/c$ follows. Hence we obtain

$$\begin{aligned} \alpha =\frac{1}{d}\left( \frac{a'}{a}+s\tau (\alpha )\right) =\frac{1}{d}\left( \frac{a'}{a}-\frac{c'}{c}\right) . \end{aligned}$$

$\square$

3 Proofs of Theorems

Proof of Theorem 1

((i)$\Rightarrow$(ii)). We may suppose $(Da,Dc)\ne (0,0)$ because $Da=Dc=0$ clearly implies (ii). For brevity let $f',f'',\dots ,f^{(k)}$ denote $Df,D^2f,\dots ,D^kf$ respectively. From $\tau f=af^d+c$, their first transforms can be calculated by repeated differentiation. For example,

$$\begin{aligned} s\tau f'=D\tau f=c'+a'f^d+daf^{d-1}f', \end{aligned}$$

and for $f''$,

$$\begin{aligned} \begin{aligned} s'\tau f'+s^2\tau f''=&c''+a''f^d+da'f^{d-1}f'\\&+da'f^{d-1}f'+da((d-1)f^{d-2}f'^2+f^{d-1}f''),\\ s^2\tau f''=&-\frac{s'}{s}(c'+a'f^d+daf^{d-1}f')\\&+c''+a''f^d+da'f^{d-1}f'\\&+da'f^{d-1}f'+d(d-1)af^{d-2}f'^2+daf^{d-1}f''\\ =&(\hbox { an element of}\ K(f)) +\left( -\frac{s'}{s}\cdot da+2da'\right) f^{d-1}f'\\&+d(d-1)af^{d-2}f'^2+daf^{d-1}f''. \end{aligned} \end{aligned}$$

The first transforms of $f^{(k)}$ for $k\ge 3$ are as follows,

$$\begin{aligned} \begin{aligned} s^k\tau f^{(k)}=&(\text {a polynomial in }f',\dots ,f^{(k-2)} \text { over }K(f))\\&+(\hbox { an element of}\ K)f^{d-1}f^{(k-1)}\\&+kd(d-1)af^{d-2}f'f^{(k-1)}+daf^{d-1}f^{(k)}. \end{aligned} \end{aligned}$$

Let $n\ge 1$ be the minimum number such that $f',\dots ,f^{(n)}$ are algebraically dependent over K(f). If $n\ge 2$, $f',\dots ,f^{(n-1)}$ are algebraically independent over K(f). Hence we can uniquely define polynomials $Z_1,\dots ,Z_n\in K(f)[Y_1,\dots ,Y_{n-1}]$ $(n\ge 1)$ by

$$\begin{aligned} s^k\tau f^{(k)}-daf^{d-1}f^{(k)}=Z_k(f',\dots ,f^{(n-1)}). \end{aligned}$$

It is seen that $Z_k\in K(f)[Y_1,\dots ,Y_{k-1}]$. Let $G\in K(f)[Y_1,\dots ,Y_n]\setminus K(f)$ be an irreducible polynomial such that $G(f',\dots ,f^{(n)})=0$. We may choose G of the form

$$\begin{aligned} G=\sum _{i=(i_1,\dots ,i_n)}R_iY_1^{i_1}\cdots Y_n^{i_n}, \quad R_i\in K(f),\ R_p=1, \end{aligned}$$

where p is the maximum of i with $R_i\ne 0$ in the sense of

$$\begin{aligned} (i_1,\dots ,i_n)<(j_1,\dots ,j_n) \Longleftrightarrow i_n=j_n,\dots ,i_{m+1}=j_{m+1},\ i_m<j_m. \end{aligned}$$

We define $H\in K(f)[Y_1,\dots ,Y_n]$ by

$$\begin{aligned} \begin{aligned} H=&G^\tau (s^{-1}(Z_1+daf^{d-1}Y_1),\dots ,s^{-n}(Z_n+daf^{d-1}Y_n))\\ =&\sum _i\tau (R_i)\{s^{-1}(Z_1+daf^{d-1}Y_1)\}^{i_1} \cdots \{s^{-n}(Z_n+daf^{d-1}Y_n)\}^{i_n}, \end{aligned} \end{aligned}$$

which satisfies

$$\begin{aligned} H(f',\dots ,f^{(n)}) =G^\tau (\tau f',\dots ,\tau f^{(n)}) =\tau (G(f',\dots ,f^{(n)}))=0. \end{aligned}$$

Then G divides H because of the irreducibility (cf. the book [10], Ch. II, §13, Lemma 2).

Using $\Vert i\Vert =i_1+i_2+\dots +i_n$ and $[i]=i_1+2i_2+\dots +ni_n$, we obtain

$$\begin{aligned} \begin{aligned} H=&\sum _i\tau (R_i)s^{-[i]}((daf^{d-1})^{\Vert i\Vert }Y_1^{i_1}\cdots Y_n^{i_n}+\cdots )\\ =&s^{-[p]}(daf^{d-1})^{\Vert p\Vert }Y_1^{p_1}\cdots Y_n^{p_n}+\cdots , \end{aligned} \end{aligned}$$

and thus

$$\begin{aligned} H=s^{-[p]}(daf^{d-1})^{\Vert p\Vert }G. \end{aligned}$$

(4)

Let $p=(0,\dots ,0,p_m,\dots ,p_n)$ with $p_m\ne 0$ and $m\ge 1$. We shall consider separately the cases where (1) $m\ge 3$, (2) $m=2$ or (3) $m=1$.

Case 1. When $m\ge 3$,

$$\begin{aligned} \begin{aligned}&\{s^{-1}(Z_1+daf^{d-1}Y_1)\}^{p_1}\cdots \{s^{-n}(Z_n+daf^{d-1}Y_n)\}^{p_n}\\ =&s^{-[p]}(Z_m+daf^{d-1}Y_m)^{p_m}\cdots (Z_n+daf^{d-1}Y_n)^{p_n}\\ =&s^{-[p]}\{(daf^{d-1})^{\Vert p\Vert }Y_m^{p_m}\cdots Y_n^{p_n}\\&+p_m\cdot md(d-1)af^{d-2}\cdot (daf^{d-1})^{\Vert p\Vert -1} Y_1Y_{m-1}Y_m^{p_m-1}Y_{m+1}^{p_m+1}\cdots Y_n^{p_n}\\&+(\text {terms of smaller indices})\}. \end{aligned} \end{aligned}$$

The index related to the second term is

$$\begin{aligned} q=(1,0,\dots ,0,1,p_m-1,p_{m+1},\dots ,p_n), \end{aligned}$$

which has values $[q]=[p]$ and $\Vert q\Vert =\Vert p\Vert +1$.

If there exists $q<r<p$ such that $R_r\ne 0$, let r be the maximum. This r has the form,

$$\begin{aligned} r=(r_1,\dots ,r_{m-1}\ge 1,p_m-1,p_{m+1},\dots ,p_n), \end{aligned}$$

where $r_1+\dots +r_{m-1}\ge 2$. Hence we see

$$\begin{aligned} \Vert r\Vert -\Vert p\Vert \ge 2+(\Vert p\Vert -1)-\Vert p\Vert =1 \end{aligned}$$

and

$$\begin{aligned}{}[r]>[q]=[p]. \end{aligned}$$

Since $R_i=0$ for any $r<i<p$, the coefficients of $Y_1^{r_1}\dots Y_n^{r_n}$ in the Eq. (4) give

$$\begin{aligned} s^{-[p]}(daf^{d-1})^{\Vert p\Vert }R_r=\tau (R_r)s^{-[r]}(daf^{d-1})^{\Vert r\Vert }, \end{aligned}$$

and thus

$$\begin{aligned} R_r=s^{[p]-[r]}(daf^{d-1})^{\Vert r\Vert -\Vert p\Vert }\tau (R_r). \end{aligned}$$

By Lemma 3, we obtain $d=2$ and $c^2=-2\tau (c/a)$ whose logarithmic derivative is the equality required in (ii).

If there is no such r, the coefficients of terms of index q in the Eq. (4) give

$$\begin{aligned} \begin{aligned}&s^{-[p]}(daf^{d-1})^{\Vert p\Vert }R_q\\ =&s^{-[p]}p_m\cdot md(d-1)af^{d-2}\cdot (daf^{d-1})^{\Vert p\Vert -1} +\tau (R_q)s^{-[q]}(daf^{d-1})^{\Vert q\Vert }, \end{aligned} \end{aligned}$$

and thus

$$\begin{aligned} R_q=\frac{m(d-1)p_m}{f}+daf^{d-1}\tau (R_q). \end{aligned}$$

By Lemma 4, we obtain the same result as above.

Case 2. In the case where $m=2$,

$$\begin{aligned} \begin{aligned}&\{s^{-1}(Z_1+daf^{d-1}Y_1)\}^{p_1}\cdots \{s^{-n}(Z_n+daf^{d-1}Y_n)\}^{p_n}\\ =&s^{-[p]}(Z_2+daf^{d-1}Y_2)^{p_2}\cdots (Z_n+daf^{d-1}Y_n)^{p_n}\\ =&s^{-[p]}\{(daf^{d-1})^{\Vert p\Vert }Y_2^{p_2}\cdots Y_n^{p_n}\\&+p_2\cdot d(d-1)af^{d-2}\cdot (daf^{d-1})^{\Vert p\Vert -1}Y_1^2Y_2^{p_2-1}Y_3^{p_3} \cdots Y_n^{p_n}\\&+(\text {terms of smaller indices}) \}. \end{aligned} \end{aligned}$$

The index related to the second term is

$$\begin{aligned} q=(2,p_2-1,p_3,\dots ,p_n), \end{aligned}$$

which has values $[q]=[p]$ and $\Vert q\Vert =\Vert p\Vert +1$.

If there exists $q<r<p$ such that $R_r\ne 0$, let r be the maximum. This r has the form,

$$\begin{aligned} r=(r_1\ge 3,p_2-1,p_3,\dots ,p_n). \end{aligned}$$

Hence we see $\Vert r\Vert>\Vert q\Vert >\Vert p\Vert$ and $[r]>[q]=[p]$. The coefficients of terms of index r in the Eq. (4) give

$$\begin{aligned} s^{-[p]}(daf^{d-1})^{\Vert p\Vert }R_r=\tau (R_r)s^{-[r]}(daf^{d-1})^{\Vert r\Vert }, \end{aligned}$$

and thus

$$\begin{aligned} R_r=s^{[p]-[r]}(daf^{d-1})^{\Vert r\Vert -\Vert p\Vert }\tau (R_r). \end{aligned}$$

We find (ii) by Lemma 3.

If there is no such r, it follows that

$$\begin{aligned} \begin{aligned}&s^{-[p]}(daf^{d-1})^{\Vert p\Vert }R_q\\ =&s^{-[p]}p_2\cdot d(d-1)af^{d-2}\cdot (daf^{d-1})^{\Vert p\Vert -1} +\tau (R_q)s^{-[q]}(daf^{d-1})^{\Vert q\Vert }. \end{aligned} \end{aligned}$$

Hence we have

$$\begin{aligned} R_q=\frac{p_2(d-1)}{f}+daf^{d-1}\tau (R_q), \end{aligned}$$

which implies (ii) by Lemma 4.

Case 3. Lastly, we consider the case where $m=1$. In this case,

$$\begin{aligned} \begin{aligned}&\{s^{-1}(Z_1+daf^{d-1}Y_1)\}^{p_1}\cdots \{s^{-n}(Z_n+daf^{d-1}Y_n)\}^{p_n}\\ =&s^{-[p]}(Z_1+daf^{d-1}Y_1)^{p_1}\cdots (Z_n+daf^{d-1}Y_n)^{p_n}\\ =&s^{-[p]}\{(daf^{d-1})^{\Vert p\Vert }Y_1^{p_1}\cdots Y_n^{p_n}\\&+p_1(a'f^d+c')(daf^{d-1})^{\Vert p\Vert -1}Y_1^{p_1-1}Y_2^{p_2}\cdots Y_n^{p_n}\\&+(\text {terms of smaller indices})\}. \end{aligned} \end{aligned}$$

The index related to the second term is

$$\begin{aligned} q=(p_1-1,p_2,\dots ,p_n), \end{aligned}$$

which has values $\Vert q\Vert =\Vert p\Vert -1$ and $[q]=[p]-1$. From the Eq. (4), it follows that

$$\begin{aligned} \begin{aligned}&s^{-[p]}(daf^{d-1})^{\Vert p\Vert }R_q\\ =&s^{-[p]}p_1(a'f^d+c')(daf^{d-1})^{\Vert p\Vert -1} +\tau (R_q)s^{-[q]}(daf^{d-1})^{\Vert q\Vert }, \end{aligned} \end{aligned}$$

and thus we have

$$\begin{aligned} daf^{d-1}\left( \frac{R_q}{p_1}\right) =a'f^d+c'+s\tau \left( \frac{R_q}{p_1}\right) . \end{aligned}$$

Lemma 5 says that

$$\begin{aligned} \alpha =\frac{1}{d}\left( \frac{a'}{a}-\frac{c'}{c}\right) \end{aligned}$$

satisfies

$$\begin{aligned} s\tau \alpha =-\frac{c'}{c}. \end{aligned}$$

Hence we see

$$\begin{aligned} d\frac{Dc}{c}=-ds\tau \alpha =s\tau \left( \frac{c'}{c}-\frac{a'}{a}\right) =s\frac{\tau D(c/a)}{\tau (c/a)} =\frac{D\tau (c/a)}{\tau (c/a)}, \end{aligned}$$

which is the equality required in (ii).

(ii$\Rightarrow$(i)). Let

$$\begin{aligned} \alpha =\frac{1}{d}\left( \frac{a'}{a}-\frac{c'}{c}\right) =-\frac{1}{d}\cdot \frac{D(c/a)}{c/a}\in K. \end{aligned}$$

It follows that

$$\begin{aligned} s\tau \alpha =-\frac{1}{d}\cdot \frac{s\tau D(c/a)}{\tau (c/a)} =-\frac{1}{d}\cdot \frac{D\tau (c/a)}{\tau (c/a)} =-\frac{c'}{c}. \end{aligned}$$

Let f be an element transcendental over K, and extend the transforming operator $\tau$ and the derivation D to those for K(f) by $\tau f=af^d+c$ and $Df=-\alpha f$. Then we find $D\tau f=s\tau Df$ by

$$\begin{aligned} s\tau Df=-(s\tau \alpha )\tau f=\frac{c'}{c}\cdot af^d+c' \end{aligned}$$

and

$$\begin{aligned} D\tau f=a'f^d+adf^{d-1}(-\alpha f)+c' =(a'-da\alpha )f^d+c' =\frac{c'}{c}\cdot af^d+c'. \end{aligned}$$

Using notations $(\ )_X$ and $(\ )^\tau$ defined by

$$\begin{aligned} (a_nX^n+a_{n-1}X^{n-1}+\dots +a_0)_X =na_nX^{n-1}+(n-1)a_{n-1}X^{n-2}+\dots +a_1,\\ (a_nX^n+a_{n-1}X^{n-1}+\dots +a_0)^\tau =\tau (a_n)X^n+\tau (a_{n-1})X^{n-1}+\dots +\tau (a_0), \end{aligned}$$

we shall examine $D\tau =s\tau D$ for $P(f)/Q(f)\in K(f)$ ($P(X),Q(X)\in K[X]$). Firstly, for P(f),

$$\begin{aligned} \begin{aligned} D\tau P(f)=&D(P^\tau (\tau f))\\ =&(P^\tau )^D(\tau f)+(D\tau f)(P^\tau )_X(\tau f)\\ =&s(P^D)^\tau (\tau f)+s(\tau Df)(P_X)^\tau (\tau f)\\ =&s\tau (P^D(f)+(Df)P_X(f))\\ =&s\tau D(P(f)). \end{aligned} \end{aligned}$$

Secondly,

$$\begin{aligned} \begin{aligned} D\tau \frac{P(f)}{Q(f)}=&D\left( \frac{\tau P(f)}{\tau Q(f)}\right) \\ =&\frac{(D\tau P(f))(\tau Q(f))-(\tau P(f))(D\tau Q(f))}{(\tau Q(f))^2}\\ =&\frac{(s\tau D(P(f)))(\tau Q(f))-(\tau P(f))(s\tau D(Q(f)))}{\tau (Q(f)^2)}\\ =&s\tau D\frac{P(f)}{Q(f)}. \end{aligned} \end{aligned}$$

Hence ${\mathcal {U}}=(K(f),D,\tau )$ is a DT overfield of ${\mathcal {K}}$ with $D\tau =s\tau D$. By the definitions above, $f\in {\mathcal {U}}$ is a solution of our difference equation and the first-order linear differential equation $y'=-\alpha y$. $\square$

Proof of Theorem 2

Let f be such a differentially algebraic solution. We may assume that f is transcendental over K. By Theorem 1,

$$\begin{aligned} d\frac{Dc}{c}=\frac{D\tau (c/a)}{\tau (c/a)}. \end{aligned}$$

From this equality, it follows that the logarithmic derivative of $\tau (c/a)/c^d$ is 0. Since the field of constants of ${\mathcal {K}}$ is C, it belongs to C.

Let $(\overline{U},\overline{\tau })$ be an algebraic closure of $(U,\tau )$, and $(\overline{K},\overline{\tau })$ the algebraic closure of $(K,\tau )$ in it. Then an algebraic element $\varepsilon \in \overline{K}$ defined by

$$\begin{aligned} \varepsilon =\frac{\overline{\tau } g}{c},\quad g=\left( \frac{c}{a}\right) ^\frac{1}{d}\in \overline{K}, \end{aligned}$$

is in C, as

$$\begin{aligned} \varepsilon ^d=\frac{\overline{\tau }(g^d)}{c^d}=\frac{\tau (c/a)}{c^d}\in C. \end{aligned}$$

Choose $\gamma \in C$ such that $\gamma \varepsilon =\gamma ^d+1$, and let $h=\gamma g$. This h belongs to $\overline{K}$ and satisfies

$$\begin{aligned} \overline{\tau }h=\gamma \overline{\tau }g=\gamma \varepsilon c=(\gamma ^d+1)c =\gamma ^d(ag^d)+c=ah^d+c. \end{aligned}$$

$\square$