The Erwin Schrodinger International Boltzmanngasse 9. Institute for Mathematical Physics A-1090 Wien, Austria

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1 ESI The Erwin Schrodinger International oltzmanngasse 9 Institute for Mathematical hysics -9 Wien, ustria Solutions of Finite Type of Sine{Gordon Equation Guosong Zhao Vienna, reprint ESI ) September, 997 Supported by Federal Ministry of Science and Research, ustria vailable via

2 Solutions of nite type of sine-gordon equation Guosong Zhao Department of Mathematics, Sichuan University, hengdu, 664, hina and Erwin Schrodinger International Institute for Mathematical hysics, Vienna, ustria October 2, 997 bstract In this paper we use the technique of dierential geometry to prove that the solutions of nite type of sine- Gordon equation x?' yy = sin ' cos ' can be obtained from a system of ordinary dierential equations 99 Mathematical Subject lassication: 532 Key words and phrases: surface of negative constant curvature, sine-gordon equation Introduction In [3] U incall and I Sterling discussed the solutions of the nite type of sinh? Gordon equation They transformed the solutions of type n of the equation x? ' yy = sinh ' into the solutions of some system of ordinary ddierential equations With this they obtained very excellent results in classifying constant mean curvature tori e inspired by [3] we consider the sine? Gordon equation x? ' yy = sin ' cos ': This equation can be derived from angular function of intersection of two asymptotic curves on the surface with negative constant Gaussian curvature in R 3 : Using the technique of dierential geometry, we transform the solutions of nite type of sine? Gordon equation into the solutions of some system of ordinary dierential equations This work is supported partly by National Science Foundation of hina

3 2 Solutions of nite type of sine-gordon equation Let be a simply connected domain in R 2 and let F :! R 3 be a immersed surface with constant Gaussian curvature? It is known that there are parometers ~x; ~y on such that the rst and the second fundamental forms of F are I = sin 2 'd~x 2 + cos 2 'd~y 2 II = sin ' cos 'd~x 2? d~y 2 ) here 2' is the angle of intersection of two asympototic curves and < 2' < : ' satises the sine? Gordon equation 2 ' ~x? 2 ' 2 ~y 2 =? sin ' cos ': From now on we assume that cos2') 6= We change the parometers and put ~x = x + y; ~y = x? y Then I = dx 2? 2 cos2')dxdy + dy 2 ; II = 2 sin2')dxdy; 2 ' xy + sin2') = : 2 Let N :! R 3 be a unit normal eld on F ) We denote the partial derivatives of ' and F with respect to x; y by ; ' y ; F x ; F y ; y ; F xy ; : ff x ; F y ; Ng form a frame eld on F ) Denote by <; > the inner product in R 3 Then < F x ; F x >= ; < F y ; F y >= ; < F x ; F y >=? cos2') 8 >< >: F xx = 2 cot2') F x + 2 sin2') y; F xy = sin2')n; F yy = 2' y sin2') x + 2 cot2')' y F y : Nx =? cot2')f x? sin2') y; N y =? sin2') x? cot2')f y : Let F t :! R 3 ;? < t < ; be a farmily of one-parameter surfaces in R 3 with F = F and each F t satises above conditions The calculations in the following is operated at t = : For the sake of convenience we write F := F t and _F := F t j t=; _F can be expressed as _ F = un + 2 F x + 2 F y ) 2) 2

4 where u; ; are the dierentiable functions on : From ) and 2) we have 8 >< >: _F x = [u x + sin2')]n + [?u cot2') x + cot2') ]F x +[? u + sin2') 2 x + ' sin2') x]f y ; F_ y = [u y + sin2')]n + [? u + 2 sin2') 2 y + ' sin2') y]f x +[? cot2')u + 2 y + cot2')' y ]F y : 3) Dierentiating < F x ; F x > and < F y ; F y > with respect to t, we get x = x cos2') y = y cos2') 4) Dierentiating F xy and < F x ; F y >, we get _ N = 4 _' =?4u x sin2') + 2' y + 2 y sin2') 5) sin 2 2')?u x? 2 sin2')? 2u y cos2')? 2 sin2') cos2'))f x + sin 2 2')?u y? 2 sin2')? u x cos2')? 2 sin2') cos2'))f y: Dierentiating _ F x with respect to y and making inner product with N, we obtain < _ F xy ; N >= u xy?u cos2')+ cos2') + 2 y sin2')+ 2 x sin2')+ cos2')' y : On the other hand, the derivative of < F xy ; N > with respect to t gives 6) 7) < F_ xy ; N > =?2u cos2') + cos2') + 2 x sin2') cos2') +' y cos2') + 2 y sin2') cos2'): 8) omparing 7) with 8), we get u xy + u cos2') = 9) Making similar calculation for < _ F xx ; N > and < _ F yy ; N >, we have uxx + x sin2')? 2u x cot2')? 2u y sin2')? u = ; u yy + y sin2')? 2u x sin2') ' y? 2u y cot2')' y? u = : It follows from 4) and ) that x = cot2')?u xx + 2u x cot2') + 2u y ' sin2') x + u); y =?u sin2') yy + 2u x ' sin2') y + 2u y cot2')' y + u): x =?u sin2') xx + 2u x cot2') + 2u y ' sin2') x + u); y = cot2')?u yy + 2u x ' sin2') y + 2u y cot2')' y + u): ) ) 2) 3

5 straightforward calculation gives xy = yx ; xy = yx : Hence, and are determined by u up to a constant ut It is easy to check ~u = 2 + x sin2') = 2 + x tan2'): 3) ~u xy + ~u cos2') = 4) Formula 4) shows that any solution of 9) can generate a new solution of 9) It is easy to see that ~u is determined by u up to Note that is also a solution of 9) This arouses us construct a sequence fu l g of solutions from : u = ; u 2 = ~u ; ; u l+ = ~u l ; here ~u l is denied by 3) key step to construct the sequence is to nd l :! R which satises ) Set l =? cot2')u lx? sin2') u ly + l : 5) Then 8< : lx =? cot2')u lxx + 2u lx ' sin 2 2') x + 2 cos2') u sin 2 2') ly + cot2')u l + lx ; ly = 2 cos2') u sin 2 2') lx' y + cos2 2') u sin2') l + 2 cos2') u sin 2 2') ly' y? u sin2') lyy + ly : omparing it with ), we see lx =?2u lx ; ly =?2u l y = u l sin2'): 6) We get from 5) that u l+ = 2 l + lx sin2') =?u lxx + 2 l + u l : 7) It is easy to see that when l =, When l = 2, =?' 2 x =?u 2 ; u 2 =?u xx + u + 2 =?xx +? 2' 3 x : 2 = 2 xx + ' 2 xx + ' 2 x? 3' 4 x =?2u u 2 + u 2 x? u 2 + u 4 ; u 3 = xxxx + ' 2 x xx? 2xx + 4 ' 2 xx +? 6' 5 x : For l we have the following 4

6 roposition Let u = ; u 2 ; ; u l and ; 2 ; ; l? be given Then l := 8 >< >:?u 2 k? 2 k j;l?j if l = 2k? j=?u k u k+? kk? 2 k? j;l?j if l = 2k; j= ) where i;j = u i u j+ + u ix u jx + i j? u i u j : roof It is easy to get u l+;x = u lx cos2')? l sin2') + u ly ; i;jx = u i u j+;x? u i+ u jx ; i;jy = u iy u j+? u i+;y u j : Dierentiating *) with respect to x; y, we see that l satises 6) 2 straightforward calculation gives u lx =?u l?;xxx + u l?;x? 4u l?;x' 2 x + 2 l?x ; 8) u lxx =?u l?;xxxx + u l?;xx? 4u l?;xx' 2 x? 2u l?;x x + 2 l?xx ; 9) u ly = u l?;x cos2')? l? sin2') + u l?;y = u lyyy = Xl? u lyy = Xl? Xl? u ix cos2')? i sin2')); 2)?u i? 2u ix sin2')' y? 2 i cos2')' y ); 2) [?u iy? 4' 2 y u ix cos2')? i cos2'))? 2' yy u ix sin2') + i cos2'))]; here we have taken u = and = 2 Similarly, by setting 22) v = ' y ; v l+ =?v lyy + v l + l ' y ; l ; 23) we obtain another sequence fv l g of solutions 9) The l in 23) is given by l = 8 >< >:?vk 2? 2 k? j= j;l?j if l = 2k??v k v k+? k;k? 2 k? j;l?j if l = 2k; j= i;j = v i v i+ + v iy v jy + i j? v i v j ; 5

7 and l satises Similarly, v lxxx = Xl? v lxx = v lx = Xl? Xl? lx = v l sin2') ly =?2v ly ' y : 24) v iy cos2')? i sin2')) 25)?v i? 2v iy sin2')? 2 i cos2') ) 26) [?v ix? 4' 2 xv iy cos2')? i sin2'))? 2x v iy sin2') + i cos2'))] 27) v ly =?v l?;yyy + v l?;y? 4v l?;y' 2 y + 2 l?' yy 28) v lyy =?v l?;yyyy + v l?;yy? 4v l?;yy' 2 y? 2v l?;y' y ' yy + 2 l?' yyy : 29) Here we have taken v = and = For the sake of convenience, we put 2 roposition 2 i): ii): ' 2 x u l+?u l ' 2 x l? i= u l = v l = l = l = ; l < : ) y = 2 sin2 2') u lx sin2') ) x; v i ) y =? 2 sin2 2') sin2') v lx) x ; iii): cos2')' 2 x u l+?u l ) x =? cos 2 2')' 2 u lxx x cos2') ) x? 4' 4 x u l+?u l ) y ; iv): cos2')' 2 x l? v i ) x =? cos 2 2')' 2 v lxx x cos2') ) x? 4' 4 x l? v i ) y : i= roof i) It follows from 7) and 2) that ' 2 x u l+?u l ) y = u lx cos2')? l sin2')) + 2 u l+? u l ) sin2') = u lx cos2')? 2 u lxx sin2') =? 2 sin2 2') sin2') u lx) x : ii) It follows from 25) and 26) that? 2 sin2 2') v sin2') lx) x = l? v 2 i + 2v iy sin2') + 2 i cos2') ) sin2') + l? = l? = ' 2 x l? i= v iy cos2')? i sin2')) cos2') 2 v i sin2')? v iy ) i= v i ) y : 6

8 iii) It follows from 7), 8) and i) that cos2')' 2 x u l+? u l )) x + 4' 4 x u l+? u l ))y = cos2')?u lxxx? 4u lx ' 2 x + 2 l x )? cos2')x?u lxx + 2 l x ) +2' 2 x? sin2')u lxx + 2 cos2')u lx ) =? cos2') u lxxx + cos2')x u lxx? 2 sin2')' 2 x u lxx =? cos 2 2')' 2 x u lxx cos2') ) x : iv) It follows from 25), 26) and 27) that? cos 2 v 2') lxx cos2') ) x =? cos2')v lxxx + x v lxx cos2')? 2' 2 x sin2 2') v sin2') lx) x? 4' 3 x cos2')v lx = 4' 4 x l? v i ) y + cos2')? v lxxx + x v lxx? 4' 3 x v lx) 2 i= = 4' 4 x l? = 4' 4 x l? v i ) y + cos2') l? i= i= i= v ix? v i x ) v i ) y + cos2')' 2 x l? v i ) x : i= We can use roposition 2 to prove roposition 3 ssume that there are two sets of constants, a ; a 2 ; ; a n 6= and b ; b 2 ; ; b m 6= ; such that := a l u l + mx b l v l = : Then there exists constants and ~ such that n ) a l u l+ + m b l l v i ) = u = ; i= n 2) a l l u i ) + m b l v l+ = v ~ = ' ~ y : i= roof We need only to prove that ~ := a l u l+? u l ) + mx Xl? b l v i ) is a constant Dierentiating ~ with respect to x; y and using roposition 2, we see that ' 2 ~ x y = n b l ' 2 x l? a l ' 2 x u l+?u l ) y + m i= v i ) y i= = n a l [? sin2') u 2 sin2') lx) x ] + m =? 2 sin2 2')[ n a sin2') l u lx + m b l v lx )] x =? 2 sin2 2') sin2') x) x = : 7 b l [? 2 sin2 2') sin2') v lx) x ]

9 and cos2')' 2 x ~ x = n a l cos2')' 2 x[ u l+? u l )] x + m =? cos 2 n 2')? cos 2 2')' 2 x m n l+ u a l lxx cos2') ) x? 4' 4 x v b l lxx cos2') ) x? 4' 4 x a l u lxx + b l cos2')' 2 x l? n m =? cos 2 2')' 2 x cos2') ) x? 4' 4 ~ x y =? cos 2 2')' 2 xx x cos2') ) x = : m b l v lxx This proves ) The proof of 2) is similar to that of ) 2 i= v i ) x a l [ u l+? u l )] y b l l? i= v i ) y From roposition 3 we see that if u n+ can be linearly expressed by u ; ; u n ; v ; ; v m, then when l > n and k > m, u l and v k can linearly expressed by u ; ; u n ; v ; ; v m In fact that if then v m+ = u n+ =? where a n+ =? and b is a constant a l u l + a i+ b m )u l + mx mx b l v l ; b l? b l? )v l Denition If there exists positive integer n such that when l > n, u l and v l can be linearly expressed by u ; ; u n ; v ; ; v n ; then ' is called a solution of nite type of sine-gordon equation y + sin2') = If such n is the smallest, 2 ' is called a solution of type n b m 3 The proof of Theorem Let ' :! R be a solution of type n and let u n+ = a l u l + b l v l ); v n+ = >From roposition 3 we see that a l ; b l and c l ; d l satisfy and c l =? c l u l + d l v l ): a i+ b n ; d l = b l? b l? b n ; a n+ =?) 3) a l = c l? c l? c n ; b l =? 8 d i+ c n ; d n+ =?): 3)

10 It is easy to obtain from 7) and 23) that ul = 2l? + l ; ; 2l?2); v l = ' y 2l? + ~ l ' y ; ; ' y 2l?2) 32) here = ; = ; k = k ' x k ; ' y k = k ' y k and l is a polynomial in ; 2; ; 2l?2 and ~ l is a polynomial in ' y ; ; ' y 2l?2: Therefore 2n+ and ' y 2n+ are polynomials in ; ; 2n; ' y ; ; ' y 2n : 2n+ ' y 2n+ = Q ; ; 2n; ' y ; ; ' y 2n); = Q' ~ x ; ; 2n; ' y ; ; ' y 2n): It shows from y =? sin2') that any partial derivatives of ' can be calculated 2 from ; ; 2n; ' y ; ; ' y 2n: Now we consider the set of 2n? jets of ' From above discussion we see that the set is a 4n+)?dimensional vector space and we can take '; ; ; 2n; ' y ; ; ' y 2n as its coordinates Thus the space of 2n?jets of ' is dieomorphic to R 4n+ : Denote the jet?mapping by W : W : x; y) 7! W x; y) = It is easy to see from 32) and 33) that partial derivatives with respect to x; y of W are functions of W : W x = W y = ' 2n ' y ' y 2n ' 2n ' y ' y 2n x y = = 2 ' 2n ' y ' y 2n Q ; ; ' y 2n)? 2 sin2')? 2 sin2')) y 2n? ' y? 2 sin2')? 2 sin2')) x 2n? ' y 2 ~Q ; :' y 2n) : : 33) 34) 9

11 We see that the 2n?jet mapping W of a solution of type n of sine-gordon equation is a solution of 34) conversly, the rst component of a solution W of 34) is a solution of type n of sine-gordon equation From 32) and 33) we can take '; u ; u x ; u 2 ; ; u n ; u nx ; v ; v y ; v 2 ; ; v n ; v ny as the coordinates of the space of 2n?jets of ' Write U = '; u ; u x ; u 2 ; ; u n ; u nx ; v ; v y ; v 2 ; ; v n ; v ny ):

12 With the U?coordinate we can get system of ordinary dierential equations: U x = ' u u x u l u lx u n u nx v v y v l v ly v n v ny x = u u x u? u u u lx u l? u l+ + 2 l u u nx u n? n a l u l + b l v l ) + 2 n u? 2 sin2')?v cos2') l? v iy cos2')? i sin2'))?v l cos2') n? v iy cos2')? i sin2'))?v n cos2') ; U y = ' u u x u l u lx u n u nx v v y v l v ly v n v ny y = v? 2 sin2')?u cos2') l? u ix cos2')? i sin2'))?u l cos2') n u ix cos2')? i sin2'))?u n cos2') v y v? v v v ly v l? v l+ + 2 l v v ny v n? n c l u l + d l v l ) + 2 n v : 35)

13 roposition 4 [U x ; U y ] vanishes on if and only if 8 n >< n a i+ )u lx cos2')? l sin2')) + n b l v ly = l= n n a i+ )u lx + n b l v ly cos2')? l sin2')) = >: or 8 > < >: n n l= l= l= c l u lx cos2')? l sin2')) + n n d i+ )v ly = c l u lx + n n d i+ )v ly cos2')? l sin2')) = l= where [; ] denote the Lie bracket of vector elds roof Let [U x ; U y ] acts on coordinate functions Then 36) 37) [U x ; U y ] ' = y? ' yx = u ) y? v ) x = ; [U x ; U y ] u = u xy? u yx =?u cos2')??u cos2')) = ; Here derivatives with respect to x and y are calculated by using 35)The actions of [U x ; U y ] on coordinate functions, except on u nx and on v ny, are zero It follows from u nx ) xy = u n? n a l u l + b l v l ) + 2 n u ) y and = 2u u n sin2')? n sin2') + n? u ix cos2')? i sin2'))? n l? [a l u ix cos2')? i sin2')) + b l v ly ] u nx ) yx =?u n cos2')) x =?u nx cos2') + 2u u n sin2') that [U x ; U y ]u nx = if and only if l= a i+ )u lx cos2')? l sin2')) + Similarly, [U x ; U y ]v ny = if and only if l= c l u lx + l= b l v ly = : d i+ )v ly cos2')? l sin2')) = : It is easy to check from 3) and 3) that 36) and 37) are equivalent2 by To remove the restraint of 36) and 37), we replace u nx and v ny respectively f := sin2') a i+ )u lx + cot2') b l v ly + 2v n X b l v l ; a n+ =?; 2

14 Write g := cot2') Then 35) becomes where u nx = V x = V y = v ny = sin2') c l u lx + 2 n X c l u l + sin2') V := '; u ; u x ; u 2 ; ; u n ; f; v ; v y ; v 2 ; ; v n ; g): ' u u x u n f v v n g ' u u x u n f v v n g x y = = [?f + n? sin2') sin2') u u x u? u u u nx 2v n l? b l? 2 sin2') n? d i+ )v ly ; d n+ =?: v iy cos2')? i sin2')) v ly cos2')? l sin2')) l= n 2u x c l u l v? 2 sin2')?u cos2') n u lx cos2')? l sin2')) l= n 2v y v y v ny 2u b l v l n l? c l n +b n cot2')[?g + cot2') n? n? [?g + cot2') +c n cot2')[?f + sin2') u ix cos2')? i sin2')) a i+ )u lx + cot2') n? c l u lx + sin2') c l u lx + n? sin2') n? n n? b l v ly + 2v n n a i+ )u lx + cot2') n? 3 n ; b l v l ] d i+ )v ly + 2u d i+ )v ly + 2u n c l u l ] b l v ly + 2v n n 38) c l u l ]; b l v l ]:

15 >From 38), it is easy to check n f yx =?2 cos2')v b l v l + 2v y f xy = 2v y n +2v n l? b l l? b l l? b l n l? b l v iy cos2')? i sin2')) [v i? v i+ + 2v i ) cos2') v iy cos2')? i sin2')); +2v?v iy sin2') + v iy sin2')? i cos2'))] n l? n l? = 2v y b l v iy cos2')? i sin2')) + 2v b l v i? v i+ ) cos2') n = 2v y v iy cos2')? i sin2'))? 2v cos2') n b l v l = f yx Similarly, g xy = g yx We see that the vector elds V x and V y given by 38) satisfy [V x ; V y ] = on Therefore for any initial vector V ) = V 2 R 4n+, 38) has unique solution V :! R 4n+, where is a simply connected domain containing the origin in R 2 We have proved the following Theorem solution of type n of sine-gordon equation determines a system of ordinary dierential equations 38) For any initial value the system 38) has unique solution V The rst component ' of V is a solution of type m n of sine-gordon equation obtained in this way2 ll solutions of nite type of sine-gordon equation are cknowledgement uthor is grateful to rofessoe eter W Michor for his kind invitation to Erwin Schrodinger International Institute for Mathematical hysics and for his many helps uthor would like to express his gratitude to rofessor -M Li for his constant encouragement References [] SS hern, Moving frame methods, Mathematical Notes of eijing University, 978 [2] -M Li, U Simon and G Zhao, Global ne Dierential Geometry of Hypersurfaces, de Gruyter Expositions in Mathematics, Walter de Gruyter, erlin-new York,993 [3] U inkall and I Sterling: On the classication of constant mean curvature tori, nnals of Math, 3989),

16 [4] Su, H Hu, Dierential Geometry, dvanced Education ress, eijing, 978 [5] J Wolf, Spaces of constant curvature, McGraw-Hill, New York, 967 5

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