Pairs of commuting Hamiltonians quadratic in momenta
In the case of two degree system the pairs of quadratic in momenta Hamiltonians commuting according the standard Poisson bracket are considered. The new many-parametrical families of such pairs are founded. The universal method of constructing the fu…
Authors: V.G. Marikhin, V.V. Sokolov
P airs of omm uting Hamiltonians, quadrati in momen ta V.G. Marikhin 1 and V.V. Sok olo v 1 1 L.D. Landau Institute for Theoretial Ph ysis RAS, Moso w, Russia In the ase of t w o degree system the pairs of quadrati in momen ta Hamiltonians omm uting aording the standard P oisson bra k et are onsidered. The new man y-parametrial families of su h pairs are founded. The univ ersal metho d of onstruting the full solution of Hamilton - Jaobi equation in terms of in tegrals on some algebrai urv e is prop osed. F or some examples this urv e is non-h yp erellipti o v ering o v er the ellipti urv e. MSC n um b ers: 17B80, 17B63, 32L81, 14H70 A ddress : Landau Institute for Theoretial Ph ysis RAS, K osygina st.2 ,Moso w, Russia, 119334, E-mail : m vgitp.a.ru, sok olo vitp.a.ru 1 1 P airs of quadrati hamiltonians In pap ers [1, 2, 3, 4 , 5 , 6, 7℄ the problem of the omm uting pairs of Hamiltonians quadrati in momen ta w as onsidered. Consider pair of Hamiltonians in the form H = ap 2 1 + 2 bp 1 p 2 + cp 2 2 + dp 1 + ep 2 + f , (1.1) K = Ap 2 1 + 2 B p 1 p 2 + C p 2 2 + D p 1 + E p 2 + F , (1.2) omm uting with resp et to standart p oisson bra k et { p α , q β } = δ αβ . The o eien ts in form ulas (1.1 ),(1.2) - some (lo ally) analitial funtions of the v ariables q 1 , q 2 . Theorem 1. A ny p airs of ommuting Hamiltonians (1.1 )-(1.2 ) an b e anoni al ly tr ansforme d by ˆ P 1 = P 1 + ∂ F ( s 1 , s 2 ) ∂ s 1 , ˆ P 2 = P 2 + ∂ F ( s 1 , s 2 ) ∂ sq 2 to the p air of the form H = U 1 − U 2 s 1 − s 2 , K = s 2 U 1 − s 1 U 2 s 1 − s 2 , (1.3) wher e U 1 = S 1 ( s 1 ) P 2 1 + p S 1 ( s 1 ) S 2 ( s 2 ) Z s 1 ( s 1 − s 2 ) P 2 − S 1 ( s 1 ) Z 2 s 1 4( s 1 − s 2 ) 2 + V 1 ( s 1 , s 2 ) , U 2 = S 2 ( s 2 ) P 2 2 − p S 1 ( s 1 ) S 2 ( s 2 ) Z s 2 ( s 1 − s 2 ) P 1 − S 2 ( s 2 ) Z 2 s 2 4( s 2 − s 1 ) 2 + V 2 ( s 1 , s 2 ) , (1.4) V 1 = 1 2 p S 1 ( s 1 ) ∂ q 1 p S 1 ( s 1 )) Z 2 s 1 s 1 − s 2 + f 1 ( s 1 ) , V 2 = 1 2 p S 2 ( s 2 ) ∂ q 2 p S 2 ( s 2 )) Z 2 s 2 s 2 − s 1 + f 2 ( s 2 ) (1.5) for some funtions Z ( s 1 , s 2 ) , S i ( s i ) and f i ( s i ) . Poisson br aket { H , K } e quals to zer o if and only if Z s 1 ,s 2 = Z s 1 − Z s 2 2( s 2 − s 1 ) (1.6) and Z s 1 ∂ ∂ s 2 − Z s 2 ∂ ∂ s 1 V 1 − V 2 s 1 − s 2 = 0 . (1.7) Pro of W e in tro due new o ordinates s 1 , s 2 , su h that the quadrati parts of H , K (1.1,1.2 ) are diagonal: Let s 1 , s 2 b e the ro ots of equations Φ( s, q 1 , q 2 ) = ( B − bs ) 2 − ( A − as )( C − cs ) = 0 , (1.8) Then the anonial transformation ( q 1 , q 2 , p 1 , p 2 ) → ( s 1 , s 2 , P 1 , P 2 ) : p 1 = − ( Φ 1 q 1 Φ 1 s 1 P 1 + Φ 2 q 1 Φ 2 s 2 P 2 ) , p 2 = − ( Φ 1 q 2 Φ 1 s 1 P 1 + Φ 2 q 2 Φ 2 s 2 P 2 ) , (1.9) 2 where Φ i = Φ( s i , q 1 , q 2 ) under onditions { H , K } = 0 transforms pairs (1.1 ),(1.2) to the form H = U 1 − U 2 s 1 − s 2 , K = s 2 U 1 − s 1 U 2 s 1 − s 2 , (1.10) where U 1 = S 1 ( s 1 ) P 2 1 + ˜ dP 1 + ˜ eP 2 + ˜ f , U 2 = S 2 ( s 1 ) P 2 2 + ˜ D P 1 + ˜ E P 2 + ˜ F , (1.11) where S i ( s i ) = 1 (Φ i q i ) 2 (( as i − A )(Φ i q 1 ) 2 + 2( bs i − B )Φ i q 1 Φ i q 2 + ( cs i − C )(Φ i q 2 ) 2 ) (1.12) W e alulate a P oisson bra k et b et w een H and K. Then the o eien t of P 2 1 , P 2 2 , P 1 P 2 equal to zero i ˜ d = 2 S 1 ( s 1 ) ∂ F ( s 1 , s 2 ) ∂ s 1 , ˜ e = p S 1 ( s 1 ) S 2 ( s 2 ) Z s 1 ( s 1 − s 2 ) , ˜ D = − p S 1 ( s 1 ) S 2 ( s 2 ) Z s 2 ( s 1 − s 2 ) , ˜ E = 2 S 2 ( s 2 ) ∂ F ( s 1 , s 2 ) ∂ s 2 where Z ( s 1 , s 2 ) , F ( s 1 , s 2 ) - some funtions. and Z s 1 ,s 2 = Z s 1 − Z s 2 2( s 2 − s 1 ) (1.13) W e apply the anonial transformation ˆ P 1 = P 1 + ∂ F ( s 1 , s 2 ) ∂ s 1 , ˆ P 2 = P 2 + ∂ F ( s 1 , s 2 ) ∂ s 2 to equate ˜ d, ˜ E to zero. Then the o eien t of P 1 , P 2 equal to zero i U 1 , U 2 ha v e the form as in form ulation of Theorem 1. And nally the free o eien t in P oisson bra k et equals to zero i the equation (1.7) of the Theorem 1 is fullled just as exp eted. The general analytial solution of Euler - Darb oux equation ( 1.6) has near the line of singularities x = y the follo wing expansion: Z ( x, y ) = A + ln ( x − y ) B , A = ∞ X 0 a i ( x + y ) ( x − y ) 2 i , B = ∞ X 0 b i ( x + y ) ( x − y ) 2 i , where a 0 and a 1 - some funtion. The other o eien ts an b e expressed b y these t w o funtions and their deriv ativ es. F or example, b 0 = 1 2 a ′′ 0 . W e insert this expan tion in to (1.7) to obtain B = 0 . It is easy to he k that an y solution of the equation (1.6) with B = 0 has the form Z ( x, y ) = z 0 + δ ( x + y ) + ( x − y ) 2 ∞ X k =0 g (2 k ) ( x + y ) 2 (2 k ) k !( k + 1)! ( x − y ) 2 k , (1.14) where g ( x ) - some funtion and z 0 , δ - some onstan ts. W e all the funtion g ( x ) as gener ating funtion for (1.14). Without the loss of generalit y w e ho ose z 0 = 0 . The parameter δ , is v ery imp ortan t for lassiation of hamiltonians from Theorem 1. 3 W e nd all the funtions Z , orresp onding the rational generating funtions g . Cho osing g ( x ) = x n , w e obtain the innite set of p olynomial solutions Z ( n ) for (1.6). In partiular g ( x ) = 1 ⇐ ⇒ Z (0) ( x, y ) = ( x − y ) 2 g ( x ) = x ⇐ ⇒ Z (1) ( x, y ) = ( x + y )( x − y ) 2 , g ( x ) = x 2 ⇐ ⇒ Z (2) ( x, y ) = 1 4 ( x − y ) 2 + 4( x + y ) 2 ( x − y ) 2 . All set an b e obtained b y using 'reating' op erator x 2 ∂ ∂ x + y 2 ∂ ∂ y − 1 2 ( x + y ) , ating on Z (0) . The rational funtions g ( x ) = ( x − µ ) − n reate one more lass of exat solution of equation (1.6). F or example g µ ( x ) = 1 4 1 x − 2 µ ⇐ ⇒ Z µ ( x, y ) = p ( µ − x )( µ − y ) + 1 2 ( x + y ) − µ. The solution orresp onding the p oles of order n ≥ 2 , an b e obtained b y dieren tiating the last form ula b y parameter µ. Beause funtion Z is linear b y g w e obtained the solution Z with rational generating funtion g ( x ) = P i c i x i + P i,j d ij ( x − µ i ) − j . Hyp othesis 1. F or all Hamiltonians (1.3)-(1.7 ) generating funtion g is rational and has the form g ( x ) = P ( x ) S ( x ) , where P è S - some p olynomials with deg P < 5 , deg S < 6 . In pap ers [5, 6 ℄ the follo wing solution of the system ( 1.6), (1.7) w as onsidered: Z ( x, y ) = x + y , S 1 ( x ) = S 2 ( x ) = 6 X i =0 c i x i , f 1 ( x ) = f 2 ( x ) = − 3 4 c 6 x 4 − 1 2 c 5 x 3 + 2 X i =0 k i x i , where c i , k i - some onstan ts. A v ery imp ortan t fat is that Clebs h top and so (4) -S hottky- Manak o v top [8, 9, 10 ℄ are the partiular ases of this mo del [6℄. In pap er [6℄ a full solution of Hamilton - Jaobi equation of this mo del w as obtained in the form of some kind of separation of v ariables on a non-h yp erellipti urv e of gen us 4. 2 Univ ersal solution of Hamilton-Jaobi equation Let H and K ha v e the form (1.3)-(1.5 ). Consider system H = e 1 , K = e 2 , where e i - some onstan ts. Let p 1 = F 1 ( x, y ) , p 2 = F 2 ( x, y ) - b e its solution. W e use short notation x è y orresp onding q 1 è q 2 . Jaobi's lemma giv es that if { H , K } = 0 , then ∂ F 1 ∂ y = ∂ F 2 ∂ x . T o nd an ation S ( x, y , e 1 , e 2 ) , it is enough to solv e the follo wing system ∂ ∂ x S = F 1 , ∂ ∂ y S = F 2 . 4 W e rewrite the system H = e 1 , K = e 2 in the form p 2 1 + ap 2 + b = 0 , p 2 2 + Ap 1 + B = 0 , (2.15) where a = Z x x − y s S 2 ( y ) S 1 ( x ) , A = − Z y x − y s S 1 ( x ) S 2 ( y ) b = − Z 2 x 4( x − y ) 2 + V 1 − e 1 x + e 2 S 1 ( x ) , B = − Z 2 y 4( x − y ) 2 + V 2 − e 1 y + e 2 S 2 ( y ) . It easy to nd that 2 b y + Aa x + 2 aA x = 0 , 2 Aa y + aA y + 2 B x = 0 . (2.16) Using (1.6 ) and (1.7 ), it is easy to obtain the follo wing iden tit y Ab x − aB y + 2 A x b − 2 a y B = 0 . (2.17) Using a standard te hnique of Lagrange resolv en ts (see f.e. [11℄), w e rewrite system (2.15 ) to a system uv = 1 4 aA, (2.18) Au 3 + 4 b a u 2 v − 4 B A uv 2 − av 3 = 0 , (2.19) that is equiv alen t to the qubi equation on u 2 . Let ( u k , v k ) , k = 1 , 2 , 3 b e the solutions of (2.18 ), (2.19 ) su h that u 2 1 + u 2 2 + u 2 3 = − b, v 2 1 + v 2 2 + v 2 3 = − B u 1 u 2 u 3 = − 1 8 a 2 A, v 1 v 2 v 3 = − 1 8 A 2 a. Then, form ulas p 1 = u 1 + u 2 + u 3 , p 2 = v 1 + v 2 + v 3 ; p 1 = u 3 − u 1 − u 2 , p 2 = v 3 − v 1 − v 2 ; p 1 = u 2 − u 1 − u 3 , p 2 = v 2 − v 1 − v 3 ; p 1 = u 1 − u 2 − u 3 , p 2 = v 1 − v 2 − v 3 dene four solutions of (2.15). Consider the rst of them. Lemma 1. F or i = 1 , 2 , 3 fol lowing e quations ar e ful lle d ∂ u i ∂ y = ∂ v i ∂ x . Pro v e. Dieren tiating equations (2.18 ) and (2.19 ) on x and y , w e nd u y and v x as the funtions on u and v . Then expressing v through u, w e obtain that u y = v x is equiv alen t to iden tities (2.16) and (2.17 ). Lemma 1 means, that in v ariables u 1 , u 2 , u 3 w e nd partiular separation v ariables. Really S = S 1 + S 2 + S 3 , where S is the ation, and funtions S i dened from a system ∂ ∂ x S i = u i , ∂ ∂ y S i = v i . 5 Let's u = 1 2 Z x x − y s y − ξ x − ξ , v = − 1 2 Z y x − y s x − ξ y − ξ . It easy to see that pair ( u, v ) for all ξ are a solution of (2.18 ). If Z is a solution of (1.6), then ∂ u ∂ y = ∂ v ∂ x . Using this fat w e in tro due a funtion σ ( x, y , ξ ) so that ∂ σ ∂ x = u, ∂ σ ∂ y = v . In a ase of rational funtion g , orresp onding funtion Z is expressed through quadrati radials and the funtion σ an b e obtained. Let's Y = ∂ σ ∂ ξ . After m ultipliation of expression (2.19) b y expression − 2 p S 1 ( x ) p S 2 ( y ) √ x − ξ √ y − ξ ( x − y ) Z x Z y , left side of (2.19 ) an b e written in the form − e 2 + e 1 ξ + y − ξ x − y V 1 − S 1 ( x ) Z 2 x 4( x − ξ )( x − y ) − x − ξ x − y V 2 + S 2 ( y ) Z 2 y 4( y − ξ )( x − y ) . (2.20) Prop osition 1 . L et the expr ession (1.6 ), (1.7 ) b e full le d . Then the expr ession ( 2.20 ) is a funtion of Y and ξ v ariables only . Pro v e . W e assign the funtion (2.20) as Ψ( x, y , ξ ) . Consider Jaobian J = ∂ Ψ ∂ x ∂ Y ∂ y − ∂ Ψ ∂ y ∂ Y ∂ x . W e hange ∂ Y ∂ y and ∂ Y ∂ x to ∂ v ∂ ξ and ∂ u ∂ ξ , resp etevily , then Jaobian J equals to zero iden tially taking in to aoun t (1.6 ), (1.7 ). Due to Prop osition 1, the relation Ψ( x, y , ξ ) = 0 an b e rewritten in the form φ ( ξ , Y ) = 0 . One an nd the funtion φ b y assuming y = x . Equation φ ( ξ , Y ) = 0 denes a urv e, and the dieren tials of this urv e dene the funtion of ation S. W e note ξ k ( x, y ) , where k = 1 , 2 , 3 , the ro ots of ubi equation Ψ( x, y , ξ ) = 0 . Theorem 2. The funtion of ation S has the form S ( x, y ) = 3 X k =1 σ ( x, y , ξ k ) − ξ k Z Y ( ξ ) dξ , (2.21) wher e Y ( ξ ) - alebr ai funtion on the urve φ ( ξ , Y ) = 0 . Pro v e. W e obtain ∂ ∂ x S ( x, y ) = 3 X k =1 σ x ( x, y , ξ k ) + 3 X k =1 { σ ξ ( x, y , ξ k ) − Y ( ξ k ) } ξ k ,x = 3 X k =1 u k = p 1 . 6 Analogously ∂ ∂ y S ( x, y ) = p 2 . 3 Case of ubis Consider a ase when the urv e(2.20 ) an b e written in the form ˜ φ ( ξ , η ) = 0 ⇔ φ ( ξ , Y ) , so that p oin ts ( ξ 1 , η 1 ) , ( ξ 2 , η 2 ) , ( ξ 3 , η 3 ) lie on a straigh t line, that equiv alen t to denition η = ξ a ( x, y ) + b ( x, y ) , its substitution in to ˜ φ, giv es a urv e Ψ( x, y , ξ ) = 0 . F orm ula (2.20) giv es a urv e in a new v ariables ξ , η − e 2 + e 1 ξ + C 2 ( ξ ,η ) C 1 ( ξ ,η ) = 0 , where C 1 ( ξ , η ) → 0 at x → 0 or y → 0 . Using rev ersible urv e equation C 1 ( ξ , η ) = 0 → η = f ( ξ ) using η , w e nd the expressions for a ( x, y ) , b ( x, y ) a ( x, y ) = f ( x ) − f ( y ) x − y , b ( x, y ) = y f ( x ) − xf ( y ) x − y On the other hand the equiv alene of the urv e φ ( ξ , Y ) = 0 è ˜ φ ( ξ , η ) = 0 giv es Y x η y = Y y η x ⇔ u ξ η y = v ξ η x ⇔ ( ξ − y ) Z x η y = ( x − ξ ) Z y η x , or Z = Z ( a ) , b x = − y a x , b y = − x a y . 4 Examples In this Setion w e onsider all the pairs of Hamiltonians kno wn at the momen t (1.3 )-(1.7 ). 4.1 Class 1 F or the mo dels of this lass S 1 = S 2 = S, f 1 = f 2 = f . (4.22) Theorem 3. L et g = ˜ G S , ˜ G = G − δ 10 S ′ , f = − 4 ˜ G 2 S − 4 δ 3 ˜ G ′ − δ 2 12 S ′′ , wher e S ( x ) = s 5 x 5 + s 4 x 4 + s 3 x 3 + s 2 x 2 + s 1 x + s 0 , G ( x ) = g 3 x 3 + g 2 x 2 + g 1 x + g 0 , wher e s i , g i , δ - some onstants. Then funtions S, f and funtion Z , orr esp onding (see. 1) gener ation funtion g , full l the systems ( 1.6), (1.7). 7 Remark. P arameter δ from Theorem 3 oinsides with parameter δ from (1.14 ). Consider the ase δ = 0 in the form ula (1.14 ), Then all pairs of Hamiltonians (1.3)-(1.7 ), (4.22 ), that fulll this ondition are desrib ed b y Theorem 3. Consider a general ase S ( x ) = s 5 ( x − µ 1 )( x − µ 2 )( x − µ 3 )( x − µ 4 )( x − µ 5 ) , where s 5 6 = 0 and all ro ots µ i of p olynomial S are distint. then the funtion Z has the form Z ( x, y ) = 5 X i =1 ν i p ( µ i − x )( µ i − y ) , (4.23) where ν i - some onstan ts. Co etien ts g i and δ are expressed through onstan ts ν j from (1.14 ). F or example, 2 δ = − P ν i . F untion f is dened b y f ( x ) = − 1 16 5 X i =1 ν 2 i S ′ ( µ i ) x − µ i + k 1 x + k 0 , where k 1 , k 0 - some onstan ts. Calulation for a funtion (4.23 ) giv es σ ( x, y , ξ ) = − 1 2 5 X i =1 ν i log √ x − ξ √ y − µ i + √ y − ξ √ x − µ i √ x − y √ µ i − ξ , (4.24) Y = 1 4 N X i =1 ν i p ( x − µ i )( y − µ i ) ( ξ − µ i ) p ( x − ξ )( y − ξ ) . Algebrai urv e has the form of h yp erellipti urv e of gen us = 2 φ ( Y , ξ ) = S ( ξ ) Y 2 + f ( ξ ) − ξ e 1 + e 2 = 0 Steklo v top on so (4) [12℄ is a partiular ase of Theorem 3. 4.2 Class 2 F untions Z for the mo dels of this lass are the sp eial ases of the funtions Z of Class 1. But this Class on tains m u h more parameters them Theorem 3. Su h funtions Z an b e dened as the solutions of system Z xy = Z x − Z y 2( y − x ) = 1 3 U ( Z ) Z x Z y , (4.25) where U - some funtions of one v ariable. Remark. It easy to see that this lass of solutions of Euler - Darb oux equation Z xy = Z x − Z y 2( y − x ) oinide with the lass of solutions of the form Z = F h ( x ) − h ( y ) x − y , 8 where F and h - some funtions of one v ariable and U = F ′′ /F ′ 2 . Lemma. The system (4.25 ) is omp atible if and only if U = 3 2 B ′ B , B ( Z ) = b 2 Z 2 + b 1 Z + b 0 , wher e b i - some onstants . In a ase deg B = 2 Z ( x, y ) = p ( x − µ 1 )( y − µ 1 ) + p ( x − µ 2 )( y − µ 2 ) , (4.26) where b 2 = 1 , b 1 = 0 , b 0 = − ( µ 1 − µ 2 ) 2 . If deg B = 1 , then Z ( x, y ) = √ x y + 1 2 ( x + y ) , (4.27) b 1 = 1 , b 2 = b 0 = 0 . If deg B = 0 , then Z ( x, y ) = x + y . (4.28) 1. Consider funtion Z of the form (4.26). Then S ( x ) = ( x − µ 1 )( x − µ 2 ) P ( x ) + ( x − µ 1 ) 3 / 2 ( x − µ 2 ) 3 / 2 Q ( x ) , deg P ≤ 3 , deg Q ≤ 2 , è f ( x ) = f 0 + f 1 x + k 2 ( x − µ 1 ) 1 / 2 ( x − µ 2 ) 1 / 2 + ( µ 2 − µ 1 ) 16 n P ( µ 1 ) x − µ 1 − P ( µ 2 ) x − µ 2 o + ( µ 2 − µ 1 ) 32 ( x − µ 1 ) 1 / 2 ( x − µ 2 ) 1 / 2 n Q ( µ 1 ) x − µ 1 − Q ( µ 2 ) x − µ 2 o . In a ase when Q = 0 , k 2 = 0 , These form ulas oinside with orresp onding form ulas of Class 1. The funtions σ , Y are dened the same form ula (4.24) as for Class 1 : σ ( x, y , ξ ) = − 1 2 2 X i =1 log √ x − ξ √ y − µ i + √ y − ξ √ x − µ i √ x − y √ µ i − ξ , Y = 1 4 2 X i =1 p ( x − µ i )( y − µ i ) ( ξ − µ i ) p ( x − ξ )( y − ξ ) . Algebrai urv e in this ase has the form [ S R ( ξ ) + η S I ( ξ )] Y 2 − [ k R ( ξ ) + η k I ( ξ )] = 0 , (4.29) where S R ( x ) = ( x − µ 1 )( x − µ 2 ) P ( x ) , S I ( x ) = ( x − µ 1 )( x − µ 2 ) Q ( x ) , k R ( x ) = − e 2 + e 1 x − f 0 − f 1 x − ( µ 2 − µ 1 ) 16 n P ( µ 1 ) x − µ 1 − P ( µ 2 ) x − µ 2 o , k I ( x ) = k 2 − 1 32 ( µ 1 − µ 2 ) 2 − 1 16 ( µ 1 − µ 2 ) n Q ( µ 1 ) x − µ 1 − Q ( µ 2 ) x − µ 2 o , 1 η = 1 √ ξ − µ 1 √ ξ − µ 2 s 1 − ( µ 1 − µ 2 ) 2 16( ξ − µ 1 ) 2 ( ξ − µ 2 ) 2 Y 2 . 9 Expressing Y as a funtion of ( ξ , η ) and substituting to (4.29 ), w e obtain 10-parameter ubi in ( ξ , η ) , v ariables. So in a general ase the urv e φ ( Y , ξ ) = 0 , is a o v ering o v er an ellipti urv e W e obtain η = ξ − µ 1 √ x − µ 1 √ x − µ 2 + √ y − µ 1 √ y − µ 2 + ξ − µ 2 √ x − µ 2 √ x − µ 1 + √ y − µ 2 √ y − µ 1 , therefore p oin ts ( ξ 1 , η 1 ) , ( ξ 2 , η 2 ) , ( ξ 3 , η 3 ) lie on a straigh t line. 2. F or the funtion Z of the form (4.27 ) w e ha v e S ( x ) = xP ( x ) + x 3 / 2 Q ( x ) , deg P ≤ 3 , deg Q ≤ 2 , f ( x ) = − 1 16 x P ( x ) − 1 32 √ x Q ( x ) + f 1 x + f q √ x + f 0 . The funtion Y is dened b y Y = ξ + √ x √ y 4 ξ √ x − ξ √ y − ξ . The urv e in this ase an b e written in the form (4.29 ), where S R ( x ) = xP ( x ) , S I ( x ) = xQ ( x ) , k R ( x ) = − e 2 + e 1 x − f 0 − f 1 x + 1 16 x P ( x ) , k I ( x ) = 1 16 x Q ( x ) − f q , η = 4 Y ξ 3 / 2 p 16 Y 2 ξ 2 − 1 . In ( ξ , η ) v ariables it also has the form of arbitrary ubi. F orm ula η = ξ + √ xy √ x + √ y giv es the fat that p oin ts ( ξ 1 , η 1 ) , ( ξ 2 , η 2 ) , ( ξ 3 , η 3 ) lie on a straigh t line 3. F or the funtion Z , giv en b y (4.28 ), w e obtain S ( x ) = s 6 x 6 + s 5 x 5 + s 4 x 4 + s 3 x 3 + s 2 x 2 + s 1 x + s 0 , f ( x ) = − 1 40 S ′′ ( x ) − 1 32 √ x Q ( x ) + f 2 x 2 + f 1 x + f 0 . In this ase Y = 1 2 √ x − ξ √ y − ξ . Algebrai urv e S ( ξ ) Y 6 − F ( ξ ) Y 4 − 1 8 F ′′ ( ξ ) + 7 1920 S I V ( ξ ) − k 2 2 Y 2 − s 6 64 = 0 , F ( ξ ) = − e 2 + e 1 ξ − f ( ξ ) and in ( ξ , η ) , v ariables where η = ξ 2 − 1 4 Y 2 , has the form of arbitrary ubi. Beause η = ξ ( x + y ) − xy , p oin ts ( ξ 1 , η 1 ) , ( ξ 2 , η 2 ) , ( ξ 3 , η 3 ) b elong to a straigh t line. 10 4.3 Class 3 W e in tro due 'non-symmetrial Hamiltonian ( 1.3)-(1.7 ) if S 1 ( x ) 6 = S 2 ( x ) , or f 1 ( x ) 6 = f 2 ( x ) . Theorem 4. [7℄ In non-symmetri al ase the funtions Z , S i , f i is the solutions of (1.6 ), (1.7 ) if and only if δ = 0 , g = 1 H , S 1 , 2 = W H ± M H 3 / 2 , f 1 , 2 = − 4 W H ∓ 2 M H − 1 / 2 ± aH 1 / 2 , wher e g - gener ation funtion of Z , W ( x ) = w 3 x 3 + w 2 x 2 + w 1 x + w 0 , H ( x ) = h 2 x 2 + h 1 x + h 0 , M ( x ) = m 2 x 2 + m 1 x + m 0 . Her e w i , h i , m i , a - some onstants . Consider the general ase H ( x ) = ( x − µ 1 )( x − µ 2 ) . Algebrai urv e in this ase is dened b y Ψ( ξ , Y ) = − e 2 + e 1 ξ − R W ( ξ ) 2( ξ − µ 1 )( ξ − µ 2 )( µ 2 − µ 1 ) + 4 M ( ξ ) √ 2 Y √ ξ − µ 1 √ ξ − µ 2 ( µ 2 − µ 1 ) 3 / 2 √ R + 8 b √ 2 Y ( ξ − µ 1 ) 3 / 2 ( ξ − µ 2 ) 3 / 2 √ R √ µ 2 − µ 1 = 0 , (4.30) where Y = p ( x − µ 1 )( y − µ 1 ) ( ξ − µ 1 ) p ( x − ξ )( y − ξ ) − p ( x − µ 2 )( y − µ 2 ) ( ξ − µ 2 ) p ( x − ξ )( y − ξ ) , R = 1 6 ( ξ − µ 1 ) 2 ( ξ − µ 2 ) 2 Y 2 − ( µ 1 − µ 2 ) 2 . Substituting Y = 1 4 ( µ 1 − µ 2 ) 3 2 η ( ξ − µ 2 )( ξ − µ 1 ) p η 2 ( µ 2 − µ 1 ) − 8( ξ − µ 1 )( ξ − µ 2 ) in to (4.30 ), W e obtain the ubi in v ariables ( ξ , η ) with a full set of ten indep enden t parameters. It easy to see that η = a ( x, y ) ξ + b ( x, y ) , where a, b - some funtions. Therefore in the ases of Class 2 and 3 the algebrai urv e is non-h yp erellipti o v ering o v er the ellipti urv e. The dynamis of the p oin ts ( ξ 1 , Y 1 ) , ( ξ 2 , Y 2 ) , ( ξ 3 , Y 3 ) on this urv e (see. theorem 2) denes the follo wing ondition: their pro jetion on the ellipti base ( ξ 1 , η 1 ) , ( ξ 2 , η 2 ) , ( ξ 3 , η 3 ) lies on the straigh t line. Hyp othesis 2. An y pair of the Hamiltonians (1.3)-(1.7 ) b elongs to one of ab o v e three lasses. 5 App endix 1. Steklo v top W e sho w that the ase of Steklo v top on so (4) is a partiular ase of Class 1 after restrition on the sympleti leafs. Hamiltonian and the additional in tegral in this ase ha v e the form H = ( ~ S 1 , A ~ S 1 ) + ( ~ S 1 , B ~ S 2 ) , K = ( ~ S 1 , ¯ A ~ S 1 ) + ( ~ S 1 , ¯ B ~ S 2 ) , 11 where A = − α 2 diag ( 1 α 2 1 , 1 α 2 2 , 1 α 2 3 ) , B = α diag ( α 1 , α 2 , α 3 ) , ¯ A = − diag ( α 2 1 , α 2 2 , α 2 3 ) , ¯ B = α d iag ( 1 α 1 , 1 α 2 , 1 α 3 ) , α = α 1 α 2 α 3 . Here ~ S i - three-dimensional v etors with omp onen ts S α i . It is easy to see that, H and K omm ute under a spin P oisson bra k et { S α i , S β j } = κ ε αβ γ δ ij S γ i . It is on v enien t to hose κ = − 2 i. W e x the Casimirs for spin bra k et: ( ~ S k , ~ S k ) = j 2 k , k = 1 , 2 . Then the form ulas ~ S k = π k ~ K ( Q k ) + j k 2 ~ K ′ ( Q k ) , ã äå ~ K ( Q ) = (( Q 2 − 1) , i ( Q 2 + 1) , 2 Q ) , dene the Darb oux o ordinate π 1 , π 2 , Q 1 , Q 2 for the simpleti leaf of P oisson manifold with o ordinates ~ S k , k = 1 , 2 . As this transformation is linear b y momen ta π k , as a result, w e obtain a pair of omm uting Hamiltonians quadrati in momen ta under the bra k et { π α , Q β } = δ αβ . Consider the anonial transformation that transforms their pair to the form (1.3)-(1.5 ), (4.22 ). W e apply the anonial transformation P 1 = π 1 p r ( Q 1 ) , P 2 = π 2 p R ( Q 2 ) , dX = dQ 1 p r ( Q 1 ) , d Y = dQ 2 p R ( Q 2 ) , where r ( Q 1 ) = ( ~ K ( Q 1 ) , A ~ K ( Q 1 )) , R ( Q 2 ) = ( ~ K ( Q 2 ) , ¯ A ~ K ( Q 2 )) , to obtain H = P 2 1 + 2 P 1 P 2 V + j 2 P 1 V Y + j 1 P 2 V X + 1 2 j 1 j 2 V X,Y + j 2 1 6 g ′′ 1 ( X ) g 1 ( X ) − 3 2 ( g ′ 1 ( X ) g 1 ( X ) ) 2 , K = P 2 2 + 2 P 1 P 2 W + j 2 P 1 W Y + j 1 P 2 W X + 1 2 j 1 j 2 W X,Y + j 2 2 6 g ′′ 2 ( Y ) g 2 ( Y ) − 3 2 ( g ′ 2 ( Y ) g 2 ( Y ) ) 2 . where V ( X , Y ) = ( ~ K ( Q 1 ) , B ~ K ( Q 2 )) p r ( q 1 ) p R ( Q 2 ) , W ( X , Y ) = ( ~ K ( Q 1 ) , ¯ B ~ K ( Q 2 )) p r ( Q 1 ) p R ( Q 2 ) , g 1 ( X ) = p r ( Q 1 ) , g 2 ( Y ) = p R ( Q 2 ) . W e apply the anonial transformation ( P 1 , P 2 , X , Y ) → ( p 1 , p 2 , x, y ) of the form dX = 1 2 dx p S ( x ) + dy p y S ( y ) , d Y = − 1 2 dx p xS ( x ) + dy p S ( y ) , P 1 = 2 √ x − √ y h p 1 − j 1 + j 2 4( x − y ) r y x p S ( x ) − p 2 + j 1 + j 2 4( x − y ) r x y p S ( y ) i , 12 P 2 = 2 √ x − √ y h p 1 − j 1 + j 2 4( x − y ) r y x p y S ( x ) − p 2 + j 1 + j 2 4( x − y ) r x y p x S ( y ) i , to obtain (1.3)-(1.5 ), (4.22), where S ( x ) = − 4 x (1 + α 2 1 x )(1 + α 2 2 x )(1 + α 2 3 x ) , Z ( x, y ) = − 1 2 j 1 ( x + y ) − j 2 √ xy , f ( x ) = 1 4 j 2 1 α 2 x 2 + j 2 2 x − j 2 2 1 4 α 2 1 α 2 1 + 1 α 2 2 + 1 α 2 3 x. A kno wledgemen ts. The author are grateful to E V F erap on to v for useful disussion. The resear h w as partially supp orted b y RFBR gran t 05-01-00189 and NSh 6358.2006.2. Ñïèñîê ëèòåðàòóðû [1℄ B. Dorizzi, B. Grammatios, A. Ramani and P . Win ternitz, In tegrable Hamiltonian systems with v elo it y dep enden t p oten tials, J. Math. 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