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ON DUAL EQUATION IN THEORY OF THE SECOND ORDER ODE’s Valerii Dryuma 7 0 0 Institute of Mathematics and Informatics, AS RM, 2 n 5 Academiei Street, 2028 Kishinev, Moldavia a J 2 2 Abstract ] I We study the relations between the second order nonlinear differential equations S . n y′′+a (x,y)y′3+3a (x,y)y′2 +3a (x,y)y′+a (x,y) = 0 1 2 3 4 i l n with arbitrary coefficients a (x,y) and dual the second order nonlinear differential equations [ i 1 ′′ ′ b = g(a,b,b ) v 7 ′ 4 with the function g(a,b,b = c) satisfying the nonlinear partial differential equation 0 1 g +2cg +2gg +c2g +2cgg aacc abcc accc bbcc bccc 0 7 0 +g2gcccc+(ga +cgb)gccc 4gabc 4cgbbc cgcgbcc − − − / n 3gg g g +4g g 3g g +6g = 0. bcc c acc c bc b cc bb i − − − l n : v 1 Introduction i X r The relation between the equations in form a y′′ +a (x,y)y′3+3a (x,y)y′2+3a (x,y)y′+a (x,y) = 0 (1) 1 2 3 4 and ′′ ′ b = g(a,b,b) (2) ′ with function g(a,b,b) satisfying the p.d.e g +2cg +2gg +c2g +2cgg + aacc abcc accc bbcc bccc g2g +(g +cg )g 4g 4cg cg g (3) cccc a b ccc abc bbc c bcc − − − − 3gg g g +4g g 3g g +6g = 0. bcc c acc c bc b cc bb − − from geometrical point of view was studied by E.Cartan [1]. 1 In fact, according to the expressions on curvature of the space of linear elements (x,y,y’) connected with equation (1) Ω1 = a[ω2 ω2], Ω0 = b[ω1 ω2], Ω0 = h[ω1 ω2]+k[ω2 ω2]. 2 ∧ 1 1 ∧ 2 ∧ ∧ 1 where: 1 ∂4f ∂b ∂µ 1 ∂2f ∂3f a = , h = , k = , −6∂y′4 ∂y′ −∂y′ − 6∂2y′∂3y′ and 6b = fxxy′y′ +2y′fxyy′y′ +2ffxy′y′y′ +y′2fyyy′y′ +2y′ffyy′y′y′ + f2fy′y′y′y′ +(fx +y′fy)fy′y′y′ 4fxyy′ 4y′fyyy′ y′fy′fyy′y′ − − − 3ffyy′y′ fy′fxy′y′ +4fy′fyy′ 3fyfy′y′ +6fyy . − − − two types of equations by a natural way are evolved: the first type from the condition a = 0 and second type from the condition b = 0. The first condition a = 0 lead to the the equation in form (1) and the second condition give ′ us the equations (2) where the function g(a,b,b) satisfies the above p.d.e. (3). From the elementary point of view the relation between both equations (1) and (2) is a result of the special properties of their General Integral F(x,y,a,b) = 0 which can be considered as the equation of some 3-dim orbifold. 2 Method of solution and reductions The equation (2) forming dual pair with some equation (1) can be find from the solutions of the p.d.e.(3). For solutions of this type of equation we use the method of solution of the p.d.e.’s described first in [3]. To integrate the partial nonlinear first order differential equation F(x,y,z,f ,f ,f ,f ,f ,f ,f ,f ,f ,f ,f ,..) = 0 (4) x y z xx xy xz yy yz xxx xyy xxy can be applied a following approach. We use the following parametric presentation of the functions and variables u t f(x,y,z) u(x,t,z), y v(x,t,z), f u v , x x x → → → − v t u u (ut) (u utv ) f u tv , f t, f vt t, f x − vt x t,... (5) z z z y yy xy → − v → v → v → v t t t t where variable t is considered as parameter. Remark that conditions of the type f = f , f = f ... xy yx xz zx are fulfilled at the such type of presentation. 2 In result instead of equation (4) one get the relation between the new variables u(x,t,z) and v(x,t,z) and their partial derivatives Φ(u,v,u ,u ,u ,v ,v ,v ...) = 0. (6) x z t x z t In some cases the solution of such type of indefinite equation is more simple problem than solution of the equation (4). The equation (3) has many types of reductions and the simplest of them are g = cαω[acα−1], g = cαω[bcα−2], g = cαω[acα−1,bcα−2], g = a−αω[caα−1], g = b1−2αω[cbα−1], g = a−1ω(c b/a), g = a−3ω[b/a,b ac], g = aβ/α−2ω[bα/aβ,cα/aβ−α]. − − For every type of reduction we can write corresponding equation (3) and then integrate it. Remark that the first examples of solutions of equation (3) were obtained in [2-8]. Proposition 1 Equation (3) can be represent in form g +gg g2/2+cg 2g = h(a,b,c), (7) ac cc − c bc − b h +gh g h +ch 3h = 0. ac cc c c bc b − − So in standard name of variables we get f +ff f2/2+yf 2f = h(x,z,y), xy yy − y yz − z h +fh f h +yh 3h = 0. (8) xy yy y y xz z − − 3 Two-dimensional short-cut (x,y) equation At the condition h(x,z,y) = 0 we get the equation f +ff f2/2+yf 2f = 0. (9) xy yy − y yz − z In particular case f(x,z,y) = f(x,y) the equation takes the form f +ff f2/2 = 0. (10) xy yy − y It was integrated in the [4] by the Legendre transformation andanother methods in more latest publications. We consider a new approach to integration of this equation. The change from the equation (10) to the (6) lead to the relation between the functions u(x,t) and v(x,t) and their derivatives ∂2 ∂ 2 ∂2 ∂ ∂ 2 u(x,t) v(x,t) 2 u(x,t) v(x,t) v(x,t)+ ∂t∂x ! ∂t ! − ∂t2 ! ∂x ! ∂t ∂ ∂ ∂2 ∂ ∂2 ∂ +2 u(x,t) v(x,t) v(x,t) 2 u(x,t) v(x,t) v(x,t)+ ∂t ! ∂x ! ∂t2 − ∂t ! ∂t∂x ! ∂t ∂2 ∂ ∂ ∂2 ∂ 2 ∂ +2u(x,t) u(x,t) v(x,t) 2u(x,t) u(x,t) v(x,t) u(x,t) v(x,t) = 0 ∂t2 ! ∂t − ∂t ! ∂t2 − ∂t ! ∂t 3 The substitutions here of the form u(x,t) = t ∂ ω(x,t) ω(x,t), t − v(x,t) = ∂ ω(x,t) t give us the linear p.d.e. equation ∂2 ∂ ∂2 2 ω(x,t)+2t ω(x,t) 2ω(x,t) t2 ω(x,t) = 0. (11) − ∂t∂x ∂t − − ∂t2 The equation (11) is transformed into the form ∂2 ∂ ω(ξ,η) ω(ξ,η) ∂ξ ω(ξ,η)+4 2 = 0 ∂η∂ξ ξ +η − ( ξ +η)2 − − with the change of variables ξ = x+2/t, η = x and can be integrated by the Laplace-method. In particular case the equation (11) admits the solution ω(x,t) = C1 +4 C2 ln(t)t+4 C2 t+ C3 t+ C4 t2 +x C1 t+ C2 t2 − (cid:16) (cid:17) with arbitrary parameters Ci and elimination of the parameter t from the relations ∂ f(x,y) t ω(x,t)+ω(x,t) = 0 − ∂t and ∂ y ω(x,t) = 0 − ∂t give us the solution of the equation (10) taking of of the form at the condition C4 = 0 f(x,y) = 4 C2 LambertW(1/2xe−1/4−y+8 C2+C2C3+ C1x) 2x−1+ (cid:16) (cid:17) +8 C2 LambertW(1/2xe−1/4−y+8 C2+C2C3+ C1x)x−1 + C1. So the equation d2b = 4 C2 LambertW(1/2ae−1/4−ddab(a)+8 CC22+ C3+ C1a) 2a−1+ da2 ! +8 C2 LambertW(1/2ae−1/4−ddab(a)+8 CC22+ C3+ C1a)a−1 + C1 (12) is dual for the some equation in form (1). In fact General solution of the equation (12) is defined by the relation b 8 C2 a+8x C2 C3 a+ C3 x 1/2 C1 a2 +1/2 C1 x2 − − − − 4 C2 aln(2)+4 C2 xln(2) 4 C2 aln(( a+x)−1) y(x) = 0, − − − − 4 and elimination of the variables a, b from here give us the equation in form (1) d2 y(x)+ dx2 2 d y(x) ( 8 C2 ln(2) 2 C3 2 C1 x 12 C2) d y(x) +1/4 dx +1/4 − − − − dx + (cid:16) x C2(cid:17) x C2 +1/4 C12x+2 C1 C2 ln(2)+1/2 C3 C1 +2 C2 C1+ 32 C22 +12 C2 C3 + C32 +8 C3 C2 ln(2)+48 C22ln(2)+16 C22(ln(2))2 +1/4 = 0. x (13) Remark that the equation (13) is the Rikkati equation with respect of variable z(x) = dy. dx 4 Two-dimensional short-cut (z,y) equation The next example is the equation (9) at the conditions h(x,z,y) = 0 and f(x,z,y) = f(z,y) ff f2/2+yf 2f = 0. (14) yy − y yz − z For a such equation we get the relation ∂2 ∂ ∂ ∂2 ∂ 2 ∂ 2u(z,t) u(z,t) v(z,t) 2u(z,t) u(z,t) v(z,t) u(z,t) v(z,t)+ ∂t2 ! ∂t − ∂t ! ∂t2 − ∂t ! ∂t ∂2 ∂ 2 ∂2 ∂ ∂ +2v(z,t) u(z,t) v(z,t) 2v(z,t) u(z,t) v(z,t) v(z,t)+ ∂t∂z ! ∂t ! − ∂t2 ! ∂z ! ∂t ∂ ∂ ∂2 ∂ ∂2 ∂ +2v(z,t) u(z,t) v(z,t) v(z,t) 2v(z,t) u(z,t) v(z,t) v(z,t) ∂t ! ∂z ! ∂t2 − ∂t ! ∂t∂z ! ∂t − 3 2 ∂ ∂ ∂ ∂ ∂ 4 u(z,t) v(z,t) +4 u(z,t) v(z,t) v(z,t) = 0 − ∂z ! ∂t ! ∂t ! ∂z ! ∂t ! which is equivalent the p.d.e ∂ ∂2 ∂ ∂2 2ω(z,t)+2t ω(z,t) t2 ω(z,t) 2 ω(z,t) ω(z,t)+ − ∂t − ∂t2 − ∂t ! ∂t∂z ∂2 ∂ +4 ω(z,t) ω(z,t) = 0 (15) ∂t2 ! ∂z after the substitution u(z,t) = t ∂ ω(z,t) ω(z,t), t − v(z,t) = ∂ ω(z,t). t A simplest solution of the equation (15) can be find in form ω(z,t) = A(t)+zt2 5 where the function A(t) satisfies the linear equation d d2 2A(t) 2t A(t)+3t2 A(t) = 0 − − dt dt2 having the solution C1 A(t) = + C2 t2. √3 t After inverse transformationin thecase C2 = 0 we find corresponding solution ofthe equation (14) in implicit form 351918y3z2 C13(f(z,y))2 +84672y5z C13f(z,y) 34992y2(f(z,y))6z2+ − − +8748y4(f(z,y))5z +46656 (f(z,y))7z3 729y6(f(z,y))4 +518616yz3 C13(f(z,y))3+ − +823543z4 C16 6912y7 C13 = 0. − So the second order ODE d 3 d2 2 d 5 d2 351918 b(a) (b(a))2 C13 b(a) +84672 b(a) b(a) C13 b(a) − da ! da2 ! da ! da2 − d 2 d2 6 d 4 d2 5 34992 b(a) b(a) (b(a))2 +8748 b(a) b(a) b(a)+ − da ! da2 ! da ! da2 ! d2 7 d 6 d2 4 +46656 b(a) (b(a))3 729 b(a) b(a) + da2 ! − da ! da2 ! d d2 3 +518616 b(a) (b(a))3 C13 b(a) +823543 (b(a))4 C16 da ! da2 ! − 7 d 6912 b(a) C13 = 0 − da ! is dual equation for the some equation of the form (1). It can be reduced to the first order ODE 2 d d 351918 (h(b))5b2 C13 h(b) +84672 (h(b))6b C13 h(b) − db ! db − 6 5 d d 34992 (h(b))8 h(b) b2 +8748 (h(b))9 h(b) b+ − db ! db ! 7 4 d d +46656 h(b) (h(b))7b3 729 (h(b))10 h(b) + db ! − db ! 3 d +518616 (h(b))4b3 C13 h(b) +823543b4 C16 db ! − 6912 (h(b))7 C13 = 0 − having singular solution 7 h(b) = 1085/7 7 b4 C13. 108 − q 6 In general case C1 = 0, C2 = 0 we also get the first order ODE having singular solution 6 6 in the form 7 d 823543 823543 b(a) + C13 C24 + C13 C23b+ da ! 11664 2916 823543 823543 823543 + C13 C22b2 + C13 C2 b3 + C13b4 = 0. 1944 2916 11664 which corresponds the function determined from the equation 1 a+4 ( 1/12b(a) 1/12 C2) 7 11664 + C3 = 0. − − v− C13( C2 +b(a))4 u u t In more general case the solution of the equation (15) has the form ω(z,t) = t−1+k4k 4 C1 t(−1+4k)−1 −2t−1 + C2t2 +kzt2 (cid:16) (cid:17) (cid:16) (cid:17) where k is essential parameter. With help of the function ω(z,t) a large class of solutions of the equation (14) radically de- pending from the choice of parameter k can be produced. 5 Two-dimensional full (x,y)- equation In the case f(x,z,y) = f(x,y), h(x,z,y) = h(x,y) = 0 from the system (8) we find full (x,y) - 6 equation ∂4 ∂ ∂3 ∂4 ∂4 f(x,y)+ f(x,y) f(x,y)+2f(x,y) f(x,y)+(f(x,y))2 f(x,y) ∂y∂x2∂y ∂x ! ∂y3 ∂y2∂x∂y ∂y4 − ∂ ∂3 f(x,y) f(x,y) = 0. (16) − ∂y ! ∂y∂x∂y This equation can be transformed into the form ([5]) Ω 1 x Ω + Ω 1 = 0 (17) (xfff) (xxff) Ω Ω − f f with the help of presentation y Ω(f(x,y),x) = 0. − From the equation (17) we find that the function Ω (f,x) defined by the relation x ∂Λ(x,f) Ω(f,x) = ∂x satisfies the equation ∂2 ∂ 3 ∂ 2 ∂ Λ(x,f) = 1/6 Λ(x,f) +α(f) Λ(x,f) +β(f) Λ(x,f)+µ(f), ∂f2 ∂f ! ∂f ! ∂f with arbitrary coefficients. 7 From solutions of this equation we can find the function g(a,c) from the relation c Ω(g(a,c),a) = 0 − and the equations forming dual pair with the equations y′′ +a (x)y′3 +3a (x)y′2 +3a (x)y′ +a (x) = 0 (18) 1 2 3 4 can be obtained by such a way. Remark that the equation (18) has the form of Abel’s equation with respect the variable ′ z(x) = y . Inthecaseofitssolvability(Bernoulliandothers)therearealotpossibilitiestogetanexamples of of dual equation in explicit form. Let us consider the solutions of the equation (16) in form Ay′′2 +(By′3+Cy′2 +Ey′ +F)y′′+Hy′6 +Ky′5 +Ly′4 +My′3 +Ny′2 +Py′+Q = 0 were the coefficients depend from the variable x A = A(x), B = B(x).... Joint consideration of the relation Af(x,y)2 +(By3 +Cy2+Ey +F)f(x,y)+Hy6 +Ky5+Ly4 +My3 +Ny2 +Py +Q = 0 and (16) lead to the conditions for determination of the coefficients. As example we find 26244A2(f(x,y))2 + 18x3y3 +2916Axy2 f(x,y)+27x2y4+3888y3A = 0 − (cid:16) (cid:17) or d2 2 d2 d 2916A2 y(x) 324A y(x) y(x) y(x)+ − dx2 ! − dx2 ! dx ! d 2 d2 d 3 +3 (y(x))2 y(x) 2 (y(x))3 y(x)+432A y(x) = 0. dx ! − dx2 dx ! From General Integral y C1 x2 +2y C1 x C2 +y C1 C22 +108A 108A C1 x 108A C1 C2 = 0, − − we find dual equation d d 3 d2 3 b(a)+ b(a) a4 +2 b(a) a = 0. da da ! da2 ! The equation b′′ = A/b3 has General Integral A b2 x(a y)2 = 0 − − − x Corresponding dual equation looks as d2 2 d d 3 d2 Ax6 y(x) + 6 y(x) x5A 2 y(x) x9 y(x)+A2 − dx2 ! − dx ! − dx !  dx2 −   4 2 d d 3x8 y(x) 6x4 y(x) A = 0 − dx ! − dx ! 8 6 Short-cut (x,y,z) equation The equation has the form ∂2 ∂2 ∂ 2 ∂2 ∂ f(x,y,z)+f(x,y,z) f(x,y,z) 1/2 f(x,y,z) +y f(x,y,z) 2 f(x,y,z) = 0 ∂x∂y ∂y2 − ∂y ! ∂y∂z − ∂z On rearrangement we find the relation ∂2 ∂ 2 ∂ ∂2 ∂ 2 u(x,t,z) v(x,t,z) 2 u(x,t,z) v(x,t,z) v(x,t,z) ∂t∂x ! ∂t ! − ∂t ! ∂t∂x ! ∂t − ∂ ∂2 ∂ ∂ ∂ ∂2 2 v(x,t,z) u(x,t,z) v(x,t,z)+2 v(x,t,z) u(x,t,z) v(x,t,z)+ − ∂x ! ∂t2 ! ∂t ∂x ! ∂t ! ∂t2 ∂2 ∂ ∂ ∂2 +2u(x,t,z) u(x,t,z) v(x,t,z) 2u(x,t,z) u(x,t,z) v(x,t,z) ∂t2 ! ∂t − ∂t ! ∂t2 − ∂ 2 ∂ ∂2 ∂ 2 u(x,t,z) v(x,t,z)+2v(x,t,z) u(x,t,z) v(x,t,z) − ∂t ! ∂t ∂t∂z ! ∂t ! − ∂ ∂2 ∂ 2v(x,t,z) v(x,t,z) u(x,t,z) v(x,t,z) − ∂t ! ∂t2 ! ∂z − ∂ ∂ ∂2 2v(x,t,z) v(x,t,z) u(x,t,z) v(x,t,z)+ − ∂t ! ∂t ! ∂t∂z ∂2 ∂ ∂ +2v(x,t,z) v(x,t,z) u(x,t,z) v(x,t,z) ∂t2 ! ∂t ! ∂z − 3 2 ∂ ∂ ∂ ∂ ∂ 4 v(x,t,z) u(x,t,z)+4 v(x,t,z) u(x,t,z) v(x,t,z) = 0. − ∂t ! ∂z ∂t ! ∂t ! ∂z From here with the help of substitution u(x,t,z) = t ∂ ω(z,t) ω(z,t), t − v(x,t,z) = ∂ ω(z,t) t we get the equation ∂2 ∂ ∂2 2 ω(x,t,z)+2t ω(x,t,z) 2ω(x,t,z) t2 ω(x,t,z) − ∂t∂x ∂t − − ∂t2 − ∂ ∂2 ∂2 ∂ 2 ω(x,t,z) ω(x,t,z)+4 ω(x,t,z) ω(x,t,z) = 0. (19) − ∂t ! ∂t∂z ∂t2 ! ∂z We consider the solutions of the equation (19) in form ω(x,t,z) = A(x,t)+kzt2 where k is parameter. 9 In result we get linear equation for the function A(x,t) ∂ ∂2 ∂2 (2t 4kt) A(x,t)+ t2 +4kt2 A(x,t) 2 A(x,t) 2A(x,t) = 0. − ∂t − ∂t2 − ∂t∂x − (cid:16) (cid:17) It can be transformed into the form ∂ ∂2 4 (3k 1)( ξ +η) A(ξ,η) A(ξ,η) ( ξ +η)2( 1+4k) 2A(ξ,η) = 0 (20) − − ∂η − ∂η∂ξ ! − − − with help of the substitutions x xk 1 ξ = x,η = +4 2 . ( − 1+4k 1+4k − ( 1+4k)t) − − − In result the Laplace-equation ∂2 (12k 4) ∂ A(ξ,η) A(ξ,η) A(ξ,η) − ∂η +2 = 0 (21) ∂η∂ξ − ( 1+4k)( ξ +η) ( ξ +η)2( 1+4k) − − − − with the invariants 1 H = 2 − ( ξ +η)2( 1+4k) − − and 2k 1 K = 6 − ( ξ +η)2( 1+4k) − − has been obtained. For a given case we have K p = = 6k +3 H − and ∂ ∂ (lnH) ξ η q = = 1+4k H − i.e. all invariants of the Laplace-sequence of the equation (19) are in a fixed ratio ([9]). In particular the condition HN/H = 1+(1 p)N 1/2qN (N +1) = − − = 1+(6k 2)N 1/2 ( 1+4k)N (N +1) − − − is fulfilled. From here we have N = 2, N = ( 1+4k)−1. − − 10