On the Equations of Nonstationary Transonic Gas Flows

On the Equations of Nonstationary T ransonic Gas Flo ws V aleri i Dryuma ∗ Institute of Mathem atics and Informatics, AS RM, 5 A c ademiei Str e et, 20 28 Kishinev, Moldova , e-mail: v alery@dryuma.c o m; c a inar@mail.md Abstract The examples of solutio ns of the Equations of No nstationary T ransonic Gas Flo ws are considered. Their pr op erties are discussed. 1 In tro duction Tw o-dimensional equation of Nonstationary T ransonic Gas F lo w ha s the form 2 ∂ 2 ∂ x∂ z f ( x, y , z ) + ∂ ∂ x f ( x, y , z ) ! ∂ 2 ∂ x 2 f ( x, y , z ) − ∂ 2 ∂ y 2 f ( x, y , z ) = 0 , (1) where v a riable z is considered as the time-v ariable . Three-dimensional generalization of this equation is defined b y the equation 2 ∂ 2 ∂ x∂ z f ( x, y , z , s ) + ∂ ∂ x f ( x, y , z , s ) ! ∂ 2 ∂ x 2 f ( x, y , z , s ) − ∂ 2 ∂ y 2 f ( x, y , z , s ) − − ∂ 2 ∂ s 2 f ( x, y , z , s ) = 0 . (2) where v a riable z here is considered as the time-v ariable. The solutions of these equations and a corresp onding bibliograph y ha v e b een considered re- cen tly in [1] In this article we apply the metho d o f solution of the p.d.e.’s describ ed first in [2] and dev elop ed then in [3], [4]. This metho d allo w us to c onstruct particular solutions of the partial nonlinear differen tia l equation F ( x, y , z , f x , f y , f z , f xx , f xy , f xz , f y y , f y z , f xxx , f xy y , f xxy , .. ) = 0 . (3) ∗ W ork supp orted in part by Grant RFFI, Rus s ia-Moldov a 1 with the help of transformation of the function and v ariables. Essence of metho d consists in a follow ing presen tation of t he functions and v aria bles f ( x, y , z , s ) → u ( x, t, z , s ) , y → v ( x, t, z , s ) , f x → u x − u t v t v x , f s → u s − u t v t v s , f z → u z − u t v t v z , f y → u t v t , f y y → ( u t v t ) t v t , f xy → ( u x − u t v t v x ) t v t , ... (4) where v a riable t is considered as parameter. Remark that conditions of the type f xy = f y x , f xz = f z x , f xs = f sx ... are fulfilled at the suc h type of presen tat io n. In result instead of equation (3) one g et the relation b et we en the new v a riables u ( x, t, z ) and v ( x, t, z ) a nd their partial deriv atives Ψ( u, v , u x , u z , u t , u s , v x , v z , v t , v s ... ) = 0 . (5) This relation coincides with init ia l p.d.e at t he condition v ( x, t, z , s ) = t a nd ta kes more general form after presen tation of the functions u, v in fo r m u ( x, t, z , , s ) = F ( ω , ω t ... ) and v ( x, t, z , s ) = Φ( ω , ω t ... ) with some function ω ( x, t, z , s ) . Example. The equation of Riemann w a ve ∂ ∂ x f ( x, y ) + f ( x, y ) ∂ ∂ y f ( x, y ) = 0 after ( u, v )-transformatio n tak es the form ∂ ∂ x u ( x, t ) ! ∂ ∂ t v ( x, t ) − ∂ ∂ t u ( x, t ) ! ∂ ∂ x v ( x, t ) + u ( x, t ) ∂ ∂ t u ( x, t ) = 0 . The substitution here o f the expressions v ( x, t ) = t ∂ ∂ t ω ( x, t ) − ω ( x, t ) , u ( x, t ) = ∂ ∂ t ω ( x, t ) giv e us the linear equation ∂ ∂ x ω ( x, t ) + ∂ ∂ t ω ( x, t ) = 0 with general solution ω ( x, t ) = F1 ( t − x ) , where F1 ( t − x ) is arbitrary function. Choice of the function F1 ( t − x ) and elimination of the parameter t from the relations y − t D( F1 )( t − x ) + F1 ( t − x ) = 0 , f ( x, y ) − D( F1 )( t − x ) = 0 lead to the function f ( x, y ) satisfying the R iemann w av e equation. 2 2 Tw o-dimensi onal case The equation (1) after applying ( u, v )- transfor ma t ion with conditions u ( x, t, z ) = t ∂ ∂ t ω ( x, t, z ) − ω ( x, t, z ) and v ( x, t, z ) = ∂ ∂ t ω ( x, t, z ) tak es the form ∂ 2 ∂ t 2 ω ( x, t, z ) ! ∂ ∂ x ω ( x, t, z ) ! ∂ 2 ∂ x 2 ω ( x, t, z ) − 1 − 2 ∂ 2 ∂ t 2 ω ( x, t, z ) ! ∂ 2 ∂ x∂ z ω ( x, t, z )+ +2 ∂ 2 ∂ t∂ x ω ( x, t, z ) ! ∂ 2 ∂ t∂ z ω ( x, t, z ) − ∂ 2 ∂ t∂ x ω ( x, t, z ) ! 2 ∂ ∂ x ω ( x, t, z ) = 0 . (6) Its solution of the form ω ( x, t, z ) = A ( t, z ) − xB ( t ) lead to the equation on the function A ( t, z ) − 1 − 2 d dt B ( t ) ! ∂ 2 ∂ t∂ z A ( t, z ) + d dt B ( t ) ! 2 B ( t ) = 0 with solution A ( t, z ) = F2 ( t ) + F1 ( z ) +   − 1 / 2 Z d dt B ( t ) ! − 1 dt + 1 / 4 ( B ( t )) 2   z where F2 ( t ) , B ( t ) , F1 ( z ) are a rbitrary functions. In result w e find that the function ω ( x, t, z ) = F2 ( t ) + F1 ( z ) +   − 1 / 2 Z d dt B ( t ) ! − 1 dt + 1 / 4 ( B ( t )) 2   z − xB ( t ) is the solution of the equation (6). After the c hoice of the functions F2 ( t ) , B ( t ) and elimination of the par a meter t fro m the relations f − t ∂ ∂ t ω ( x, t, z ) − ω ( x, t, z ) ! = 0 y − ∂ ∂ t ω ( x, t, z ) = 0 one gets the solution of the equation (1). Let us consider some examples. In the case F2 ( t ) = 0 , B ( t ) = t − 1 , w e find the relations 4 f ( x, y , z ) t 2 − 4 / 3 z t 5 − 8 xt + 3 z + 4 F1 ( z ) t 2 = 0 , 3 and 2 y t 3 − z t 5 − 2 xt + z = 0 . Elimination of the pa r a meter t from these relat io ns in the case F1 ( z ) = 0 lead to the solutio n f ( x, y , z ) of the equation (1) satisfying the a lg ebraic equation 248832 ( f ( x, y , z )) 5 z 2 − 248 832 x 2 ( f ( x, y , z )) 4 z +  − 1451520 z 2 xy − 221184 y 3 z  ( f ( x, y , z )) 3 + +  − 216000 z 4 x + 475200 z 3 y 2 + 1327 104 y z x 3 + 2211 84 y 3 x 2  ( f ( x, y , z )) 2 + +  90000 y z 5 + 9953 28 y 4 z x + 6 14400 z 3 x 3 + 1290 240 y 2 z 2 x 2  f ( x, y , z ) − − 518400 y 3 z 3 x − 3 93216 x 5 z 2 − 117 9648 x 4 y 2 z + 31 25 z 7 − 373 248 y 5 z 2 − − 192000 z 4 x 2 y − 88473 6 y 4 x 3 = 0 . 3 Three-dimen sional generalizati on In a thr ee dimensional case the equation of Nonstationary T ransonic Ga s F lo w tak es the form 2 ∂ 2 ∂ x∂ z f ( x, y , z , s ) + ∂ ∂ x f ( x, y , z , s ) ! ∂ 2 ∂ x 2 f ( x, y , z , s ) − − ∂ 2 ∂ y 2 f ( x, y , z , s ) − ∂ 2 ∂ s 2 f ( x, y , z , s ) = 0 . (7) Recall that the v ariable z in this equation play the role o f a time-v a r ia ble. After application of u , v - transformation of the form u ( x, t, z , s ) = t ∂ ∂ t ω ( x, t, z , s ) − ω ( x, t, z , s ) , v ( x, t, z , s ) = ∂ ∂ t ω ( x, t, z , s ) w e find from ( 7) the equation ∂ 2 ∂ t 2 ω ( x, t, z , s ) ! ∂ 2 ∂ s 2 ω ( x, t, z , s ) − 2 ∂ 2 ∂ t 2 ω ( x, t, z , s ) ! ∂ 2 ∂ x∂ z ω ( x, t, z , s )+ + ∂ 2 ∂ t 2 ω ( x, t, z , s ) ! ∂ ∂ x ω ( x, t, z , s ) ! ∂ 2 ∂ x 2 ω ( x, t, z , s ) − − ∂ 2 ∂ t∂ x ω ( x, t, z , s ) ! 2 ∂ ∂ x ω ( x, t, z , s ) + 2 ∂ 2 ∂ t∂ x ω ( x, t, z , s ) ! ∂ 2 ∂ t∂ z ω ( x, t, z , s ) − − 1 − ∂ 2 ∂ s∂ t ω ( x, t, z , s ) ! 2 = 0 . (8) F rom here in the case ω ( x, t, z , s ) = A ( t, s ) + k ( x + z ) t one gets the Monge-Amp ere equation on the function A ( t, s ) 1 − ∂ 2 ∂ t 2 A ( t, s ) ! ∂ 2 ∂ s 2 A ( t, s ) + ∂ 2 ∂ s∂ t A ( t, s ) ! 2 = 0 . (9) 4 It is p ossible to sho w that the equation (9) can b e in tegrated with the help of corresp onding ( u, v )-transformation. Its solutions are dep enden t from solutions of the linear Laplace equation. In fact, the equation ∂ 2 ∂ x 2 f ( x, y ) ! ∂ 2 ∂ y 2 f ( x, y ) − ∂ 2 ∂ x∂ y f ( x, y ) ! 2 − 1 = 0 after ( u, v )-transformatio n with u ( x, t ) = tω t − ω , v ( x, t ) = ω t tak es the form o f linear Laplace equation ∂ 2 ∂ x 2 ω ( x, t ) + ∂ 2 ∂ t 2 ω ( x, t ) = 0 and its particular solutions after elimination of para meter t giv e us the solutions of the Monge- Amp ere equation (9). The substitution of a nother form ω ( x, t, z , s ) = A ( t, z ) + ( s + x ) t in to the equation (8) lead to the equation o n the function A ( t, z ) − 2 − t + 2 ∂ 2 ∂ t∂ z A ( t, z ) = 0 ha ving the general solution A ( t, z ) = F2 ( t ) + F1 ( z ) + tz + 1 / 4 t 2 z where F2 ( t ) , F1 ( z ) are arbitrary functions. The choice of the functions F2 ( t ) , F1 ( z ) allo w us to construct solutions of initia l equation. F or example in t he case F2 ( t ) = t − 1 , F1 ( z ) = 0 elimination of the parameter t fro m the r elat io ns 4 f ( x, y , z , s ) t − t 3 z + 8 = 0 , 2 y t 2 − 2 t 2 z − t 3 z − 2 t 2 s − 2 t 2 x + 2 = 0 giv e us the solution of the equation (7) satisfying the alg ebraic equation 16 ( f ( x, y , z , s )) 3 z + +  32 z y − 32 z x − 32 z s − 32 sx + 32 sy − 16 y 2 − 16 z 2 − 16 s 2 − 16 x 2 + 32 y x  ( f ( x, y , z , s )) 2 + +  72 z y − 72 z x − 72 z 2 − 72 z s  f ( x, y , z , s ) + 192 z s 2 + 64 s 3 − 64 y 3 − − 384 z y x + 384 z sx − 384 z y s + 27 z 2 + 192 sx 2 + 64 x 3 + 64 z 3 + 192 z 2 s + 192 z 2 x + 192 z y 2 + +192 sy 2 + 192 y 2 x − 1 92 y z 2 − 192 y s 2 − 192 y x 2 + 192 z x 2 + 192 s 2 x − 3 84 y sx = 0 . 5 4 The case of axisymmetri c e quation The equation (7) 2 ∂ 2 ∂ x∂ y f ( x, y , z , s ) + ∂ ∂ x f ( x, y , z , s ) ! ∂ 2 ∂ x 2 f ( x, y , z , s ) − ∆ f ( x, y , z , s ) = 0 , where ∆ = ∂ ∂ z 2 + ∂ ∂ s 2 , in p o lar co ordinat es s = r cos( φ ) , z = r sin ( φ ) tak es the form 2 ∂ 2 ∂ x∂ y f ( x, y , z ) + ∂ ∂ x f ( x, y , z ) ! ∂ 2 ∂ x 2 f ( x, y , z ) − ∂ 2 ∂ z 2 f ( x, y , z ) − ∂ ∂ z f ( x, y , z ) z = 0 (10) where we r eplace v ariable r on v ariable z After applying ( u, v )-transformation with v ( x, t, z ) = t ∂ ∂ t ω ( x, t, z ) − ω ( x, t, z ) , u ( x, t, z ) = ∂ ∂ t ω ( x, t, z ) the equation (10) takes the form − ∂ 2 ∂ t 2 ω ( x, t, z ) ! z t 2 ∂ ∂ x ω ( x, t, z ) ! ∂ 2 ∂ x 2 ω ( x, t, z ) + t 3 ∂ ∂ z ω ( x, t, z ) ! ∂ 2 ∂ t 2 ω ( x, t, z )+ + ∂ 2 ∂ t 2 ω ( x, t, z ) ! t 3 z ∂ 2 ∂ z 2 ω ( x, t, z ) − t 3 z ∂ 2 ∂ t∂ z ω ( x, t, z ) ! ∂ 2 ∂ z 2 ω ( x, t, z )+ + t 2 z ∂ ∂ z ω ( x, t, z ) ! ∂ 2 ∂ z 2 ω ( x, t, z ) + z t 2 ∂ 2 ∂ t∂ x ω ( x, t, z ) ! 2 ∂ ∂ x ω ( x, t, z ) − − 2 z t ∂ 2 ∂ t∂ x ω ( x, t, z ) ! ∂ ∂ x ω ( x, t, z ) ! 2 + z ∂ ∂ x ω ( x, t, z ) ! 3 + 2 z t ∂ ∂ x ω ( x, t, z ) − − 2 z t 2 ∂ 2 ∂ t∂ x ω ( x, t, z ) = 0 . This equation has solution of the form ω ( x, t, z ) = A ( x, z ) t + B ( t ) where function B ( t ) is arbitrary and the function A ( x, z ) satisfies the equation z ∂ ∂ x A ( x, z ) ! ∂ 2 ∂ x 2 A ( x, z ) − ∂ ∂ z A ( x, z ) − z ∂ 2 ∂ z 2 A ( x, z ) = 0 . (11) The solutions of the equation (11) can b e obtained with t he help of ( u, v )-transformatio n and a simplest of them has the form A ( x, z ) = C2 − C1 − C2 ln( C2 z ) − x. 6 Using the express ion ω ( x, t, z ) = A ( x, z ) t + B ( t ) with a giv en function A ( x, z ) and arbitrary function B ( t ) the solution of the equation (10) can be constructed. As example, in the case B ( t ) = ln ( t ) the elimination of the pa r ameter t from the r elat io ns f ( x, y , z ) t − t C2 + t C1 + t C2 ln( C2 z ) + tx − 1 = 0 , y − 1 + ln( t ) = 0 lead to the solution of the equation (1 0) f ( x, y , z ) = −  − e − y +1 C2 + e − y +1 C1 + e − y +1 C2 ln( C2 z ) + e − y +1 x − 1   e − y +1  − 1 . In the case B ( t ) = te t b y ana lo gy wa y w e find the solution f ( x, y , z ) = − C1 − C2 ln( C2 z ) − x + C2 + 1 / 2 y L amb ertW (1 / 2 √ y ) + +1 / 4 y  L amb ertW (1 / 2 √ y )  2 . The solution of the equation (11 ) A ( x, z ) = 2 / 9 x 3 z 2 lead to the function ω ( x, t, z ) = 2 / 9 x 3 t z 2 + B ( t ) where B ( t ) is arbitra ry function. In the case B ( t ) = ln( t ) w e find f ( x, y , z ) = 2 / 9 x 3 z 2 +  e − y +1  − 1 . In the case B ( t ) = t 2 + t − 2 w e get f ( x, y , z ) = 2 / 9  q 2 y + 2 √ y 2 + 12 √ y 2 + 12 + q 2 y + 2 √ y 2 + 12 y  x 3  y + √ y 2 + 12  q 2 y + 2 √ y 2 + 12 z 2 + +2 / 9 72 z 2 + 18 z 2 y 2 + 18 z 2 y √ y 2 + 12  y + √ y 2 + 12  q 2 y + 2 √ y 2 + 12 z 2 . 7 5 Ac kno wled gemen t This researc h w as partially supp orted b y the GRant 0 6 .01 CRF of HCSTD ASM and the RFBR Gran t References: 1. Xiaoping Xu, Stable-R ange Appr o ach to the Equation of Nonstationar y T r ansonic Gas flows , ArXiv: 0706.4 189 v1 , physics.flu-dyn, 28 Jun 2007, p. 1-19. 2. V. Dryuma, The Riema nn and E i nsten-Weyl ge ometries in the ory of differ ential e quations, their applic ations and al l that . A.B.Shabat et all.(eds.), New T rends in Inte grabilit y an d P artial S olv abilit y , Klu wer Academic Publishers, P r in ted in the Netherlands , 2004, p.115–156 . 3. Dryuma V.S., On solutions of t he he aven ly e quations and their gener alizations , ArXiv:gr- qc/0611001 v1 31 Oct 2006, p.1- 14. 3. Dryuma V.S., On dual e quation in the ory of the se c ond o r der OD E’s , ArXiv:nlin/0001047 v1 22 Jan 2007, p.1-17. 8