Transcendence of solutions of q-airy equation.

Transcription

1 Josai Mahemaical Monographs vol. 0 (207), pp Transcendence of soluions of q-airy equaion. Seiji NISHIOKA Absrac. In his paper, we prove ranscendence of soluions of he ieraed Riccai equaions associaed wih q-airy equaion when q is no a roo of uniy. The same resul is obained for a cerain q-bessel equaion. Previously, we sudied hem under a sronger assumpion ha q is a ranscendenal number.. Inroducion In his paper [3], he auhor sudied ranscendence of funcions which saisfy heraed Riccai equaions associaed wih q-airy equaion, y(q 2 )+qy(q) y() 0, when q is a ranscendenal number. In his paper, wnroduce a proof of ranscendence which requires only ha q is no a roo of uniy. Theraed Riccai equaions are obained in he following way. Seing z() y(q)/y(), we obain he following firs-order q-difference equaion, z(q) qz()+. z() We call his he (difference) Riccai equaion associaed wih q-airy equaion. By ieraions, we can epress z(q i ) in erms of z() such as z(q 2 ) (q3 2 + )z() q 2. qz()+ This is a q 2 -difference equaion of Riccai form. The resul of ranscendence menioned abovmplies unsolvabiliy of q-airy equaion in he Franke s Liouvillian sense (cf. S. Nishioka [3, 4]). A soluion of he above Riccai equaion saisfies q-painlevé II equaion of ype A () 6 (or (A + A ) () ), which is similar o he relaions beween Airy equaion and Painlevé II equaion. Moreover, each of he basic hypergeomeric soluions of q-airy equaion has a limi o he Airy funcion (see Hamamoo, Kajiwara and Wie [2]). 29

2 30 S. NISHIOKA The same resul of ranscendencs obained for a q-bessel equaion ( ) y(q 2 2 )+ 4 qν q ν y(q)+y() 0 in he very same way inroduced in his paper, where value of he parameer ν does no maer. This equaion is relaed o one of he q-bessel funcions, J ν (3) (; q). Here we se y(q) J ν (3) (q ν/2 ; q 2 ). For deails of his funcion, see he book [] by G. Gasper and M. Rahman. Noaion. Throughou he paper every field is of characerisic zero. When K is a field and τ is an isomorphism of K ino iself, namely an injecive endomorphism, he pair K (K, τ) is called a difference field. We call τ he (ransforming) operaor and K he underlying field. For a difference field K, K ofen denoes is underlying field. For a K, he elemen τ n a K (n Z), if i eiss, is called he n-h ransform of a and is someimes denoed by a n. If τk K, we say ha K is inversive. For an algebraic closure K of K, he ransforming operaor τ is eended o an isomorphism τ of K ino iself, no necessarily in a unique way. We call he difference field (K,τ) an algebraic closure of K. For p Z >0, K (p) denoes he difference field (K, τ p ). For difference fields K (K, τ) and K (K,τ ), K /K is called a difference field eension if K /K is a field eension and τ K τ. In his case, we say ha K is a difference overfield of K and ha K is a difference subfield of K. For breviy we someimes use (K, τ ) insead of (K, τ K ). We define a differencnermediae field in he proper way. Le K be a difference field, L (L, τ) a difference overfield of K and B a subse of L. The difference subfield K B L of L is defined o be he difference field (K(B,τB,τ 2 B,...),τ) and is denoed by K B for breviy. A soluion of a difference equaion over K is defined o be an elemen of some difference overfield of K which saisfies he equaion. We use he following lemma. Lemma. (Lemma 8 in S. Nishioka [3]). Le C be an algebraically closed field, q C no a roo of uniy, a ranscendenal elemen over C, F/C() a finie eension of degree n, and τ an isomorphism of F ino iself over C sending o q. Then F C(), n. 2. Noaion for difference Riccai equaion Le K (K, τ) be a difference field, and le ( ) ab A M 2 (K), cd ( ) a (i) b (i) A i c (i) d (i) (τ i A)(τ i 2 A) (τa)a (i, 2,...).

3 Transcendence of q-airy 3 In his paper, Eq(A, i)/k denoes he equaion over K, We easily see he following. y i (c (i) y + d (i) )a (i) y + b (i). Lemma 2.. If f is a soluion of Eq(A, k)/k in a difference field eension L/K, f Lis also a soluion of Eq(A, ki)/k (i, 2,...). Lemma 2.2. Le B A k and B i (τ k(i ) B)(τ k(i 2) B) B (i, 2,...). Then B i A ki. Lemma 2.3. For any k, l, m Z >0, f Lis a soluion of Eq(A k, lm)/k (k) f L (l) is a soluion of Eq(A kl,m)/k (kl), where L is a difference overfield of K (k). 3. Proof of ranscendence Le C be an algebraically closed field and a ranscendenal elemen over C. Le q C and K (C(),τ q : q). I is easy o prove ha he Riccai equaion associaed wih q-airy equaion has no raional funcion soluion, and ha is one of he keys o ranscendence. Lemma 3.. The equaion over K, y y qy +, has no soluion in C(). Proof. We prove his by conradicion. Assume ha here eiss a soluion f C(). Le f P/Q, where P, Q C[] \{0} are relaively prime. Then we obain and so P P Q Q qp Q +, P P qp Q + Q Q. This implies P Q and Q P. Hence, we find deg P deg Q. However, he above equaion yields 2 deg P 2 deg P +, a conradicion. Le A ( ) q GL 2 (C()) 0

4 32 S. NISHIOKA and Then ( ) a (i) b (i) A i c (i) d (i) (τq i A)(τq i 2 A) (τ q A)A (i, 2,...). A 2 (τ q A)A ( ) q q 2, q and for i 2, ( ) ( q a (i ) b (i ) ) A i (τ q A i )A c (i ) d (i ) 0 ( qa (i ) + b (i ) a (i ) qc (i ) + d (i ) c (i ) ) and ( )( ) q A i (τq i i a (i ) b (i ) A)A i 0 c (i ) d (i ) ( ) q i a (i ) + c (i ) q i b (i ) + d (i ). a (i ) b (i ) Hence we find and so for i 3, By inducion, we easily see for all i. This implies b (i) a (i ), c (i) a (i ), d (i) b (i ) c (i ), a (i) q i a (i ) + c (i ) q i a (i ) + a (i 2). a (i) ( ) i q i(i+)/2 i + ( erms of deg i 2) () c (i) ( ) i q (i )i/2 i + ( erms of deg i 3) (i ), (2) and b (i) ( ) i q (i )(i+2)/2 i + ( erms of deg i 3) (i ) (3) { ( ) i 2 d (i) q (i 2)(i+)/2 i 2 + ( erms of deg i 4) (i 2), 0 (i ). (4)

5 Transcendence of q-airy 33 Lemma 3.2. Eq(A k, )/K (k) has a unique soluion f (k) of he form ( ) i, C, e 0, in (C((/)),τ k :/ q k (/)). Moreover, f () f (2) f (3) holds. Proof. (Uniqueness) Suppose here eiss a soluion f of Eq(A k, )/K (k) (C((/)),τ k ) which is epressed as in ( ) i f, C, e 0. Then f saisfies τ k (f)(c (k) f + d (k) )a (k) f + b (k). The lef sids τ k (f)(c (k) f + d (k) ) ( q ki ( ) )( i c (k) ) i + d (k) ) (5) and he righ sids a (k) f + b (k) a (k) Comparing he coefficiens of (/) k+, we obain ) i + b (k). (6) 0( ) k q k(k+)/2 e +( ) k q (k )(k+2)/2, and so e q. For j 2, he coefficien of (/) k+j of he formula (6) is ( ) k q k(k+)/2 e j + P j, where P j is deermined by e,...,e j. On he oher hand, for j 2, he coefficien of (/) k+j of he formula (5) is equal o he coefficien of (/) k+j of ( j q ki ( ) i j )(c (k) ) i + d (k) ) which is denoed by Q j and also deermined by e,...,e j. Hence we find ( ) k q k(k+)/2 e j Q j P j,,

6 34 S. NISHIOKA and so e j ( ) k q k(k+)/2 (Q j P j ), which implies e j is deermined by e,...,e j. Therefore we conclude ha f is unique. (Eisence) Define (i, 2,...) as e q, ( ) k q k(k+)/2 (Q i P i ) (i 2). By he above discussion, i follows ha f (k) is a soluion of Eq(A k, )/K (k). (Ideniy) Fi k. Since f () is a soluion of Eq(A, )/K in (C((/)),τ ), i is a soluion of Eq(A, k)/k in (C((/)),τ ). Hence f () is a soluion of Eq(A k, )/K (k) in (C((/)),τ k τ k ). By he uniqueness, we find f (k) f (). Theorem 3.3. Suppose q is no a roo of uniy. Then for any k, Eq(A k, )/K (k) has no soluion algebraic over C(). Proof. We prove his by conradicion. Assume here eiss k such ha Eq(A k, )/K (k) has a soluion f algebraic over C(). Le L (L, τ) K (k) f. Then f saisfies ) i τ(f)(c (k) f + d (k) )a (k) f + b (k). (7) We obain de A k 0 from de A. Hence c (k) f + d (k) 0, and so τ(f) a(k) f + b (k) C(, f). c (k) f + d (k) This means L C(, f). Le n [L : C()] be he degree of he eension. By Lemma., we find L C(), n. I follows ha is ranscendenal over C. By he calculaion, ( τ ) n τ( n ) n τ τ q k q k, we obain τ/ C. Le r C deno. Then τ r holds. Noe f C() and A k M 2 (C[ n ]). Epressing f as f P/Q, where P, Q C[] are relaively

7 Transcendence of q-airy 35 prime, we obain he following equaion from he equaion (7), τp τq a(k) P Q + b(k) c (k) P Q + a(k) P + b(k) Q d(k) c (k) P + d (k) Q. Since τp,τq are relaively prime, here eiss R C[] such ha { Rτ(P )a (k) P + b (k) Q, Rτ(Q) c (k) P + d (k) Q. (8) Noing de A k ( ) k, we can calculae as follows, ( ) ( )( ) τp a (k) b (k) P R τq c (k) d (k), Q ( )( ) ( ) d ( ) k (k) b (k) τp P R c (k) a (k). τq Q Since P, Q are relaively prime, we find R C. Comparing he degrees of he equaion (8), we obain Since deg a (k) kn, which means deg (a (k) P + b (k) Q) deg (Rτ(P )) deg P. deg a (k) P deg b (k) Q, deg Q deg P deg a (k) deg b (k) kn (k )n n. By his resul, epress f as f ( i, C, e n 0, ) in and eend hsomorphism τ : C(/) C(/) sending / o r (/) o he isomorphism τ : C((/)) C((/)) sending / o r (/). We will show f C(). We prove ha n i implies 0(i n) by conradicion. Assume here eiss i n such ha n i and 0. Le ln + m (0 <m<n) be he

8 36 S. NISHIOKA minimum of such numbers. The firs erm of a (k) f + b (k) a (k) ( e n ) n + + e ln ( ) ln + e ln+m whose eponen is no divisible by n has he eponen On he oher hand, he firs erm of τ(f)(c (k) f + d (k) ) { ( e n r n ( {c (k) ) n + + e ln r ln e n ) n + + e ln kn +(ln + m). ( ) ln + e ln+m r ln+m ( ) ln + e ln+m ( ) ln+m + ) + b (k) ( ) ln+m + } ( ) } ln+m + ) + d (k) whose eponen is no divisible by n has he eponen greaer han or equal o Hence we obain (2 k)n +(ln + m). kn +(ln + m) (2 k)n +(ln + m), a conradicion. We proved ha n i implies 0(i n), which means f C(((/) n )) C(/) C((/) n )C(/) C(). I follows from he above resul ha L C(, f) C() and n [L : C()]. Hence we find, r q k and f ( ) i C(), C, e 0. Since f is a soluion of Eq(A k, )/K (k) in (C((/)),τ:/ q k (/)), f is a soluion of Eq(A, )/K by Lemma 3.2. However, Lemma 3. says ha Eq(A, )/K has no soluion in C(). Remark 3.4. Considering he proofs in he auhor s paper [3] or paper [4], we

9 Transcendence of q-airy 37 easily obain he same heorem for q-bessel equaion, ( ) y(q 2 2 )+ 4 qν q ν y(q)+y() 0, in he very same way. This resul is independen of value of he parameer ν. Acknowledgmens. This work was parially suppored by JSPS KAKENHI Gran Number References [] Gasper, G., Rahman, M., Basic Hypergeomeric Series 2nd edn., Cambridge Universiy Press, [2] Hamamoo, T., Kajiwara, K., Wie, N., Hypergeomeric soluions o he q-painlevé equaion of ype (A + A )(), In. Mah. Res. No. 2006, Ar. ID [3] Nishioka, S., Solvabiliy of Difference Riccai Equaions by Elemenary Operaions, J. Mah. Sci. Univ. Tokyo, 7 (200), [4] Nishioka, S., Proof of unsolvabiliy of q-bessel equaion using valuaions. To appear in J. Mah. Sci. Univ. Tokyo. Seiji Nishioka Faculy of Science, Yamagaa Universiy Kojirakawa-machi -4-2, Yamagaa-shi, , Japan nishioka@sci.kj.yamagaa-u.ac.jp