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.
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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
or
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,
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,
where
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
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
with \(a(0)\ne 0\) and \((c(0),p(0))\ne (0,0)\). Choose \(e\in {\mathbb {C}}^\times \) such that
By \(z(x)=y(x)-e\), we obtain
Let
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,
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
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,
where \(|q| >1\), \(R(0,0)=0\), and
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
We give one concrete example again. The equation
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 \({\mathcal {K}}\),
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
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
where C is an algebraically closed field. If the difference equation over \({\mathcal {K}}\),
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
Lemma 3
If \(R\in K(Y)^\times \) satisfies
then
Proof
By an expression,
where M and N are relatively prime and M is monic, we obtain
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
Comparing the degrees, we obtain
By \(\tau (N)\in K[Y^d]\), N has an expression,
where \(\deg N=m+dn\). Then the Eq. (1) is
Each factor
can be divided by Y at most d times, and the right side of the Eq. (2) is divided by \(Y^{e+m}\), where
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
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
Hence we have
The root of this polynomial in the transcendental element \(Y^2\) is
and its first transform is
\(\square \)
Lemma 4
If \(R\in K(Y)\) satisfies
then
Proof
By the given equation, we see \(R\ne 0\). Hence R can be expressed as follows,
where M and N are relatively prime and N is monic. Then we obtain
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
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
Then
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
We may suppose \(\tau \lambda _1=c\), as Y divides the left-hand side. Then by
we obtain
The coefficients satisfy
Hence we see \(\lambda _2=-\lambda _1\), and thus from the latter equation it follows that
Transforming this, we obtain
The case where \(\deg N=1\) has a contradiction. Indeed, the first transform of \(N=Y-\lambda \), \(\lambda \in K\), is
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
then \(R=\alpha Y\), where
Proof
We easily see \(R\ne 0\). By an expression,
where M and N are relatively prime and N is monic, we obtain
Hence \(\tau (N)\) divides N, which implies \(N=1\). Then the above equation is
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
and thus
Comparing the coefficients, we first have \(\beta =0\), by which \(R=M=\alpha Y\) and
From the latter, \(s\tau (\alpha )=-c'/c\) follows. Hence we obtain
\(\square \)
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,
and for \(f''\),
The first transforms of \(f^{(k)}\) for \(k\ge 3\) are as follows,
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
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
where p is the maximum of i with \(R_i\ne 0\) in the sense of
We define \(H\in K(f)[Y_1,\dots ,Y_n]\) by
which satisfies
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
and thus
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 \(m\ge 3\),
The index related to the second term is
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,
where \(r_1+\dots +r_{m-1}\ge 2\). Hence we see
and
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
and thus
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
and thus
By Lemma 4, we obtain the same result as above.
Case 2. In the case where \(m=2\),
The index related to the second term is
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,
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
and thus
We find (ii) by Lemma 3.
If there is no such r, it follows that
Hence we have
which implies (ii) by Lemma 4.
Case 3. Lastly, we consider the case where \(m=1\). In this case,
The index related to the second term is
which has values \(\Vert q\Vert =\Vert p\Vert -1\) and \([q]=[p]-1\). From the Eq. (4), it follows that
and thus we have
Lemma 5 says that
satisfies
Hence we see
which is the equality required in (ii).
(ii\(\Rightarrow \)(i)). Let
It follows that
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
and
Using notations \((\ )_X\) and \((\ )^\tau \) defined by
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),
Secondly,
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,
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
is in C, as
Choose \(\gamma \in C\) such that \(\gamma \varepsilon =\gamma ^d+1\), and let \(h=\gamma g\). This h belongs to \(\overline{K}\) and satisfies
\(\square \)
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Acknowledgements
The author appreciates the reviewer’s thoughtful suggestions, by which the section of Introduction greatly improved. This work was partially supported by Japan Society for the Promotion of Science (JSPS) KAKENHI Grant Number 18K03318.
Funding
This work was partially supported by Japan Society for the Promotion of Science (JSPS) KAKENHI Grant Number 18K03318.
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Nishioka, S. Differential Transcendence of Solutions of the Difference Equation \(\Delta y=ay^2+by+c\). Results Math 77, 197 (2022). https://doi.org/10.1007/s00025-022-01731-3
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DOI: https://doi.org/10.1007/s00025-022-01731-3
Keywords
- Differential transcendence
- transcendental transcendence
- hypertranscendence
- difference equation
- difference algebra
- differential algebra