Asymmetric and Moving-Frame Approaches to the 2D and 3D Boussinesq Equations

Asymmetric and Mo ving-F rame Approac hes to the 2D and 3D B oussin esq Equations 1 Xiaopi ng Xu Institute of Mathematics, Academ y of Mathematics & System Sciences Chinese Academy of Sciences, Beijing 100 190, P .R. China 2 De dic ate d to 2008 Beijing O l ymp i c Games Abstract Boussinesq systems of nonlinear partial differential equations are fu ndament al equations in geoph ysical fluid dynamics. In this p ap er, we use asymmetric ideas an d mo ving f rames to solv e the t w o-dimensional Boussin esq equations with partial viscosit y terms stud ied by Chae ( A dv. Math. 203 (2006) , 497-513) and the three-dimensional stratified rotating Boussinesq equations studied b y Hsia, Ma and W a ng ( J. Math. Phys. 48 (2007), no. 6, 0656 0). W e obtain new families of explicit exact solutions w ith m ultiple parameter functions. Man y of them are the p erio dic, quasi-p erio d ic, ap erio dic solutions that may ha ve pr actica l significance. By F ourier expansion and some of our solutions, on e can obtain discon tin uous solutions. In add ition, Lie p oint symmetries are used to simp lify our arguments. 1 In tro duction Both the atmospheric and o ceanic flo ws are influenced by the rotat io n of the earth. In fact, the fast rotation and small a sp ect ra tio a r e t w o main characteristic s o f the large scale atmospheric and o ceanic flo ws. The small aspect ratio c haracteristic leads to the primitiv e equations, and the fast rotatio n leads to the quasi-geostropic equations (cf. [2], [6], [7], [9]). A main ob j ective in climate dynamics and in geoph ysical fluid dynamics is to understand and predict the p erio dic, quasi-p erio dic, ap erio dic, and fully turbulent c haracteristics of the lar g e scale atmospheric and o ceanic flows (e.g., cf. [4], [5]). 1 2000 Mathematical Sub ject Classificatio n. Pr imary 35C05, 35Q55 ; Secondar y 37K 10. 2 Research supp orted by China NSF 1 04310 40 1 The Bo ussines q sys tem for the incompressible fluid follows in R 2 is u t + uu x + v u y − ν ∆ u = − p x , v t + uv x + v v y − ν ∆ v − θ = − p y , (1 . 1) θ t + uθ x + v θ y − κ ∆ θ = 0 , u x + v y = 0 , (1 . 2) where ( u, v ) is the v elo city v ector field, p is the scalar pressure, θ is the scalar temp erature, ν ≥ 0 is the viscosit y and κ ≥ 0 is the t hermal diffusivit y . The ab ov e system is a simple mo del in atmospheric sciences (e.g., cf. [8]). Chae [1] prov ed the g lobal regularity , and Hou a nd Li [3] obta ined the w ell-p osedness of the ab o v e system. Aon ther slightly simplified version of the system of primitiv e equations is the three- dimensional stratified rot a ting Boussinesq system (e.g., cf. [7], [9]): u t + uu x + v u y + w u z − 1 R 0 v = σ (∆ u − p x ) , (1 . 3) v t + uv x + v v y + w v z + 1 R 0 u = σ (∆ v − p y ) , (1 . 4) w t + uw x + v w y + w w z − σ R T = σ (∆ w − p z ) , (1 . 5) T t + uT x + v T y + w T z = ∆ T + w , (1 . 6) u x + v y + w z = 0 , (1 . 7) where ( u, v , w ) is the v elo cit y v ector filed, T is the temp erature function, p is the pressure function, σ is the Prandtle n umber, R is the thermal Ra yleigh n um b er and R 0 is the Rossb y n um b er. Moreov er, t he vec tor (1 / R 0 )( − v , u, 0) represen ts the Coriolis force and the term w in (1.6) is deriv ed using s tratificatio n. So the abov e equations are the ex tensions of Navier-Stok es equations by adding the Coriolis f orce and the stratified temp erature equation. Due t o the Corio lis force, the t w o-dimensional system (1 .1 ) and (1.2) is not a sp ecial case of the abov e three-dimens ional syste m. Hsia, Ma and W ang [4] studied the bifurcation and p erio dic solutions of the ab ov e sy stem (1.3)- (1.7). In [1 0], w e used the stable r a nge o f nonlinear term to solv e the equation of nonsta- tionary transonic gas flow. Mor eov er, w e [11] solve d the three-dimensional Na vior-Stokes equations by asymmetric tec hniques and mo ving fra mes. Based on the algebraic charac- teristics of the ab ov e equations, w e use in this pap er asymmetric ideas and mo ving frames to solv e the a b o v e tw o Bo ussinesq systems o f partial differen tial equ ations. Ne w families of explicit exact solutions with m ultiple parameter functions are obta ined. Many of them are the perio dic, quasi-p erio dic, ap erio dic solutions that ma y hav e practical significance. Using F ourier expansion and some of our solutions, o ne can obtain discon tin uous solutions. The symmetry transformatio ns for these equations are used to simplify o ur ar g umen ts. 2 F or con v enience, w e a lwa ys assume that all the in v olv ed partial deriv atives of related functions alw a ys exist and w e can c hange orders of taking partial deriv ativ es. The pa- rameter f unctions are so chosen that the in volv ed expressions mak e sense . W e also use prime ′ to denote the deriv a tiv e of a ny one-v a riable function. Observ e that the tw o- dimensional Boussinesq system (1.1) and (1.2) is in v arian t under the action of the follow ing symmetry t r ansformation: T ( u ) = a − 1 ǫ 1 u ( a 2 ( t + b ) , aǫ 1 ( x + α ) , aǫ 2 ( y + β )) − α ′ , (1 . 8) T ( v ) = a − 1 ǫ 2 v ( a 2 ( t + b ) , a ( x + α ) , a ( y + β )) − β ′ , (1 . 9) T ( p ) = a − 2 p ( a 2 ( t + b ) , aǫ 1 ( x + α ) , aǫ 2 ( y + β )) + α ′ ′ x + β ′ ′ y + γ , (1 . 10) T ( θ ) = a − 3 ǫ 2 θ ( a 2 ( t + b ) , aǫ 1 ( x + α ) , aǫ 2 ( y + β )) , (1 . 11) where a, b ∈ R with a 6 = 0, ǫ 1 , ǫ 2 ∈ { 1 , − 1 } and α , β , γ are arbitrary functions of t . The ab ov e transformat io n transforms a solution of t he equation (1.1) and (1.2) in to another solution with additional three parameter functions. Denote ~ x = ( x, y ). The three-dimensional stratified rotating Boussinesq system is in v ar ia n t under the f o llo wing transformat io ns: T 1 [( u, v , w )] = (( u ( t + b, ~ xA, ǫz ) , v ( t + b, ~ xA ) , ǫz ) A, ǫw ) , (1 . 12) T 1 ( p ) = p ( t + b, ~ xA, ǫz ) , T 1 ( T ) = T ( t + b, ~ xA, ǫz ); (1 . 13) T 2 ( u ) = u ( t, x + α, y + β , z + γ ) − α ′ , T 2 ( v ) = v ( t, x + α , y + β , z + γ ) − β ′ , (1 . 14) T 2 ( w ) = w ( t, x + α, y + β , z + γ ) − γ ′ , T 2 ( T ) = T ( t, x + α , y + β , z + γ ) − γ , (1 . 15) T 2 ( p ) = p ( t, x + α, y + β , z + γ ) + σ − 1 ( α ′ ′ x + β ′ ′ y + γ ′ ′ z ) − Rγ z + µ ; (1 . 16) where ǫ = ± 1, b ∈ R , A ∈ O (2 , R ) , and α , β , γ , µ are arbitrary functions of t . T he abov e transformations transform a solution of the equation (1.3)- (1.7) into a nother solution. In particular, applying the transformatio n T 2 to an y solution in this pa p er yields another solution with extra four parameter functions. T o simplify problems, w e alwa ys solv e the Boussinesq systems mo dulo the ab ov e cor- resp onding symmetry transformatio ns, whic h is an idea that geometers and top ologists often use. The pap er is org anized as follows . In Section 2 , w e solv e the tw o-dimensional Bo ussi- nesq equations (1.1 ) -(1.2) a nd obtain fo ur fa milies of explicit exact solutions. In Section 3, w e presen t a n approach with u , v , w , T linear in x, y to the equations (1.3)-(1 .7 ), and obtain t w o families of explicit exact solutions. Assuming u z = v z = w z z = T z z = 0 in 3 Section 4, w e find another t w o families of explicit exact solutions of the equations (1.3)- (1.7). In Section 5, w e obtain a family of explicit exact solutions of ( 1 .3)-(1.7) that are indep enden t of x . The status can b e c hanged b y applying the transformation in (1.12) and (1 .13) to them. 2 Solutio ns of the 2D B oussin esq Equations In this section, w e solve the t w o-dimensional Boussinesq equations (1.1 ) - (1.2) b y an asym- metric metho d and by an moving frame. According to the second equation in ( 1 .2), w e take the p oten tial form: u = ξ y , v = − ξ x (2 . 1) for some functions ξ in t, x, y . Then the t w o-dimensional Boussinesq equations b ecome ξ y t + ξ y ξ xy − ξ x ξ y y − ν ∆ ξ y = − p x , ξ xt + ξ y ξ xx − ξ x ξ xy − ν ∆ ξ x + θ = p y , (2 . 2) θ t + ξ y θ x − ξ x θ y − κ ∆ θ = 0 . (2 . 3) By our assumption p xy = p y x , the compatible condition of the equations in (2.2) is (∆ ξ ) t + ξ y (∆ ξ ) x − ξ x (∆ ξ ) y − ν ∆ 2 ξ + θ x = 0 . (2 . 4) No w we first solv e the system (2.3) and (2.4). Our a symmetric approach is to a ssume θ = ε ( t, y ) , ξ = φ ( t, y ) + xψ ( t, y ) (2 . 5) for some functions ε, φ a nd ψ in t, y . Then (2.3) b ecomes ε t − ψ ε y − κε y y = 0 . (2 . 6) Moreo v er, (2.4) b ecomes φ y y t + xψ y y t + ( φ y + xψ y ) ψ y y − ψ ( φ y y y + xψ y y y ) − ν ( φ y y y y + xψ y y y y ) = 0 , (2 . 7) equiv alen tly , φ y y t + φ y ψ y y − ψ φ y y y − ν φ y y y y = 0 , (2 . 8) ψ y y t + ψ y ψ y y − ψ ψ y y y − ν ψ y y y y = 0 . (2 . 9) The a b o v e t w o equations are equiv alent to: φ y t + φ y ψ y − ψ φ y y − ν φ y y y = α 1 , (2 . 10) 4 ψ y t + ψ 2 y − ψ ψ y y − ν ψ y y y = α 2 (2 . 11) for some functions α 1 and α 2 of t to b e determined. Let c b e a fixed real constan t and let γ b e a fixed function of t . W e define ζ 1 ( s ) = e γ s − ce − γ s 2 , η 1 = e γ s + ce − γ s 2 , (2 . 12) ζ 0 ( s ) = sin γ s, η 0 ( s ) = cos γ s. (2 . 13) Then η 2 r ( s ) + ( − 1) r ζ 2 r ( s ) = c r (2 . 14) and ∂ s ( ζ r ( s )) = γ η r ( s ) , ∂ s ( η r ( s )) = − ( − 1) r γ ζ r ( s ) , (2 . 15) where we treat 0 0 = 1 when c = r = 0. First we assume ψ = β 1 y + β 2 ζ r ( y ) (2 . 16) for some functions β 1 and β 2 of t , where r = 0 , 1. Then (2.11) b ecomes β ′ 1 + c r β 2 2 γ 2 + β 2 1 + [( β 2 γ ) ′ + ( − 1) r ν β 2 γ 3 + 2 β 1 β 2 γ ] η r ( y ) +( − 1) r β 2 γ ( β 1 γ − γ ′ ) y ζ r ( y ) = α 2 , (2 . 17) whic h is implied by the follow ing equations: β ′ 1 + c r β 2 2 γ 2 + β 2 1 = α 2 , β 1 γ − γ ′ = 0 , (2 . 18) ( β 2 γ ) ′ + ( − 1) r ν β 2 γ 3 + 2 β 1 β 2 γ = 0 . (2 . 19) F or con v enience, we assume γ = √ α ′ for some function α of t . Th us w e ha v e β 1 = γ ′ γ = α ′ ′ 2 α ′ , β 2 = b 1 e − ( − 1) r ν α p ( α ′ ) 3 , b 1 ∈ R . (2 . 20) T o solv e (2.10), w e a ssume φ = β 3 η r ( y ) (2 . 21) for some function β 3 , mo dulo the transformation in (1.8)-(1.1 1 ). Now (2.10) b ecomes [( − 1) r (( β 3 γ ) ′ + β 1 β 3 γ ) + ν β 3 γ 3 ] ζ r ( y ) = − α 1 , (2 . 22) whic h is implied by ( − 1) r (( β 3 γ ) ′ + β 1 β 3 γ ) + ν β 3 γ 3 = 0 . (2 . 23) Th us β 3 = b 2 e − ( − 1) r ν α α ′ , (2 . 24) 5 where b 2 is a real constan t. In order to solv e (2.6), w e assume ε = be γ 1 η r ( y ) , (2 . 25) where b is a real constant and γ 1 is a function of t . Then (2.6) is implied by γ ′ 1 η r ( y ) + ( − 1) r β 2 γ γ 1 ζ 2 r ( y ) + κγ 2 γ 1 (( − 1) r η r ( y ) − γ 1 ζ 2 r ( y )) = 0 , (2 . 26) whic h is implied by γ ′ 1 + ( − 1) r κγ 2 γ 1 = 0 , ( − 1) r β 2 − κγ γ 1 = 0 . (2 . 27) Then the first equation implies γ 1 = b 3 e − ( − 1) r κα (2 . 28) for some constan t b 3 . By the second equations in (2.20) and (2.27) , w e hav e: ( − 1) r b 1 e − ( − 1) r ν α p ( α ′ ) 3 = b 3 κ √ α ′ e − ( − 1) r κα . (2 . 29) F or con v enience, we tak e b 1 = ( − 1) r b 2 0 κb 3 , b 0 ∈ R . (2 . 30) Then (2 .29) is implied by α ′ e ( − 1) r ( ν − κ ) α/ 2 = b 0 . (2 . 31) If ν = κ , then w e hav e α = b 0 t + c 0 . Modulo the transformation in (1.8)-(1.1 1 ), we tak e b 0 = 1 and c 0 = 0, that is, α = t . When ν 6 = κ , we similarly tak e b 0 = 1 and α = 2( − 1) r ν − κ ln[( − 1) r ( ν − κ ) t/ 2 + c 0 ] , c 0 ∈ R . (2 . 32) Supp ose ν = κ . Then γ = 1 and φ = b 2 e − ( − 1) r ν t η r ( y ) , ψ = ( − 1) r b 3 ν e − ( − 1) r ν t ζ r ( y ) . (2 . 33) Moreo v er, θ = b exp( b 3 e − ( − 1) r ν t η r ( y )) , (2 . 34) ξ = b 2 e − ( − 1) r ν t η r ( y ) + ( − 1) r b 3 ν e − ( − 1) r ν t xζ r ( y ) (2 . 35) b y (2 .5). According to (2 .1), u = ξ y = ( − 1) r [ − b 2 e − ( − 1) r ν t ζ r ( y ) + b 3 ν e − ( − 1) r ν t xη r ( y )] , (2 . 36) v = − ξ x = − ( − 1) r b 3 ν e − ( − 1) r ν t ζ r ( y ) . (2 . 37) 6 Note u t + uu x + v u y − ν ∆ u = b 2 3 ν 2 c r e − ( − 1) r 2 ν t x, (2 . 38) v t + uv x + v v y − ν ∆ v − θ = v v y − b exp( b 3 e − ( − 1) r ν t η r ( y )) . (2 . 39) By (1.1) , w e hav e p = b Z exp( b 3 e − ( − 1) r ν t η r ( y )) dy − 1 2 b 2 3 ν 2 e − ( − 1) r 2 ν t ( c r x 2 + ζ 2 r ( y )) (2 . 40) mo dulo the t r ansformation in (1.8)- (1.11). Consider the case ν 6 = κ . Then γ = √ α ′ = 1 p ( − 1) r ( ν − κ ) t/ 2 + c 0 (2 . 41) b y (2 .32). Moreo v er, φ = b 2 [( − 1) r ( ν − κ ) t/ 2 + c 0 ] 2 ν / ( κ − ν )+1 η r ( y ) (2 . 42) b y (2 .21) and (2.24 ) . F urthermore, ψ = ( − 1) r ( κ − ν ) y 4[( − 1) r ( ν − κ ) t/ 2 + c 0 ] + ( − 1) r b 3 κ [( − 1) r ( ν − κ ) t/ 2 + c 0 ] 2 ν / ( κ − ν )+3 / 2 ζ r ( y ) (2 . 43) b y (2 .16), (2.20 ) and (2.30). According to (2.25), (2.28) and ( 2.32), θ = be b 3 [( − 1) r ( ν − κ ) t/ 2+ c 0 ] 2 κ/ ( κ − ν ) η r ( y ) . (2 . 44) Similarly , w e hav e u t + uu x + v u y − ν ∆ u = b 2 3 c r κ 2 [( − 1) r ( ν − κ ) t/ 2 + c 0 ] 4 ν / ( κ − ν )+2 x + 3( ν − κ ) 2 x 16[( − 1) r ( ν − κ ) t/ 2 + c 0 ] 2 , (2 . 45) v t + uv x + v v y − ν ∆ − θ = − ψ t + ψ ψ y + ν ψ y y − θ = − be b 3 [( − 1) r ( ν − κ ) t/ 2+ c 0 ] 2 κ/ ( κ − ν ) η r ( y ) + 3 4 b 3 κ ( κ − ν )[( − 1) r ( ν − κ ) t/ 2 + c 0 ] 2 ν / ( κ − ν )+1 / 2 ζ r ( y ) + 3( ν − κ ) 2 y 16[( − 1) r ( ν − κ ) t/ 2 + c 0 ] 2 + b 2 3 2 κ 2 [( − 1) r ( ν − κ ) t/ 2 + c 0 ] 4 ν / ( κ − ν )+3 ∂ y ζ 2 r ( y ) . (2 . 46) According (1.1 ), w e hav e p = b Z e b 3 [( − 1) r ( ν − κ ) t/ 2+ c 0 ] 2 κ/ ( κ − ν ) η r ( y ) dy − b 2 3 2 c r κ 2 [( − 1) r ( ν − κ ) t/ 2 + c 0 ] 4 ν / ( κ − ν )+2 x 2 − 3( ν − κ ) 2 ( x 2 + y 2 ) 32[( − 1) r ( ν − κ ) t/ 2 + c 0 ] 2 − b 2 3 2 κ 2 [( − 1) r ( ν − κ ) t/ 2 + c 0 ] 4 ν / ( κ − ν )+3 ζ 2 r ( y ) + 3 4 ( − 1) r b 3 κ ( κ − ν )[( − 1) r ( ν − κ ) t/ 2 + c 0 ] 2 ν / ( κ − ν )+1 η r ( y ) (2 . 47) 7 mo dulo the t r ansformation in (1.8)- (1.11). Theorem 2.1 . L et b, b 2 , b 3 , c, c 0 ∈ R and let r = 0 , 1 . If ν = κ , we ha ve the solution (2.34), (2.36 ), (2.37) and (2.40) of the two-dimensional Boussinesq e quations (1.1) -(1.2), wher e ζ r ( y ) an d η r ( y ) ar e define d in (2 .12)-(2.13) with γ = 1 . When ν 6 = κ , we have the fol lowing solutions of the two-dimensional Boussine s q e quations (1.1)-(1.2): u = ( − 1) r ( κ − ν ) x 4[( − 1) r ( ν − κ ) t/ 2 + c 0 ] + ( − 1) r b 3 κ [( − 1) r ( ν − κ ) t/ 2 + c 0 ] 2 ν / ( κ − ν )+1 xη r ( y ) − ( − 1) r b 2 [( − 1) r ( ν − κ ) t/ 2 + c 0 ] 2 ν / ( κ − ν )+1 / 2 ζ r ( y ) , (2 . 48) v = ( − 1) r ( ν − κ ) y 4[( − 1) r ( ν − κ ) t/ 2 + c 0 ] − ( − 1) r b 3 κ [( − 1) r ( ν − κ ) t/ 2 + c 0 ] 2 ν / ( κ − ν )+3 / 2 ζ r ( y ) , (2 . 49) θ is given in (2.44) and p is given in (2.47), w her e ζ r ( y ) a n d η r ( y ) a r e defin e d in (2.1 2)- (2.13) with γ = [( − 1) r ( ν − κ ) t/ 2 + c 0 ] − 1 / 2 . Observ e that ψ = 6 ν y − 1 (2 . 50) is another solution of (2.11). In order t o solve (2.10), we assume φ = ∞ X i =1 γ i y i (2 . 51) mo dulo the transformation in (1.8) - (1.11), where γ i are functions of t to b e determined. No w ( 2 .10) b ecomes − 6 ν γ 1 y − 2 − 18 ν γ 2 y − 1 + ∞ X i =1 [ iγ ′ i − ν ( i + 2)( i + 3)( i + 4) γ i +2 ] y i − 1 = α 1 , (2 . 52) equiv alen tly , γ 1 = γ 2 = 0 , α 1 = − 60 ν γ 3 , (2 . 53) iγ ′ i − ν ( i + 2)( i + 3)( i + 4) γ i +2 = 0 , i > 1 . (2 . 54) Th us γ 2 i +2 = 2 iγ ′ 2 i ν (2 i + 2)(2 i + 3)( 2 i + 4) = 0 , i ≥ 1 , (2 . 55) γ 2 i +3 = (2 i + 1) γ ′ 2 i +1 ν (2 i + 3)(2 i + 4)( 2 i + 5) = 360 γ ( i ) 3 ν i (2 i + 2)(2 i + 5)! , i ≥ 1 . (2 . 56) Hence φ = 36 0 ∞ X i =0 α ( i ) y 2 i +3 ν i (2 i + 3)(2 i + 5)! , (2 . 57) where α is an arbitr ary function of t suc h that the series conv erges, say , a p olynomial in t . 8 T o solve (2.6), w e also assume ε = ∞ X i =0 β i y i , (2 . 58) where β i are functions of t . Then (2.6) b ecomes 6 ν β 1 y − 1 + ∞ X i =0 [ β ′ i + ( i + 2)(6 ν − ( i + 1) κ ) β i +2 ] y i = 0 , (2 . 58) that is, β 1 = 0 a nd β ′ i − ( i + 2)(6 ν + ( i + 1) κ ) β i +2 = 0 , i ≥ 0 . (2 . 59) Hence θ = β + ∞ X i =1 β ( i ) y 2 i 2 i i ! Q i r =1 (6 ν + (2 r − 1) κ ) , (2 . 60) where β is an arbitrary function o f t suc h that the series conv erg es, sa y , a p olynomial in t . In this case, u t + uu x + v u y − ν ∆ u = − 60 ν α , (2 . 61) v t + uv x + v v y − ν ∆ − θ = − 36 ν 2 y − 3 − β − ∞ X i =1 β ( i ) y 2 i 2 i i ! Q i r =1 (6 ν + (2 r − 1) κ ) . (2 . 62) According (1.1 ), w e hav e p = 60 ν αx − 18 ν 2 y − 2 + β y + ∞ X i =1 β ( i ) y 2 i +1 2 i i !(2 i + 1) Q i r =1 (6 ν + (2 r − 1) κ ) (2 . 63) mo dulo the t r ansformation in (1.8)- (1.11). Theorem 2.2 . We have the fol lowing solutions o f the two-dime n sional Boussinesq e q uations (1.1)-(1.2): u = 36 0 ∞ X i =0 α ( i ) y 2 i +2 ν i (2 i + 5)! − 6 ν xy − 2 , v = − 6 ν y − 1 , (2 . 64) θ is given in (2.60) and p is given in (2.63), wher e α and β ar e arbitr ary functions of t such that the r elate d series c onver ge, say, p o l ynom ials in t . Let γ b e a function of t . Denote the mov ing f r a me ˜  = x cos γ + y sin γ , ˆ  = y cos γ − x sin γ . (2 . 65) Then ∂ t ( ˜  ) = γ ′ ˆ  , ∂ t ( ˆ  ) = − γ ′ ˜  . (2 . 66) 9 Moreo v er, ∂ ˜  = cos γ ∂ x + sin γ ∂ y , ∂ ˆ  = − sin γ ∂ x + cos γ ∂ y . (2 . 67) In particular, ∆ = ∂ 2 x + ∂ 2 y = ∂ 2 ˜  + ∂ 2 ˆ  . (2 . 68) W e assume ξ = φ ( t, ˜  ) − γ ′ 2 ( x 2 + y 2 ) , θ = ψ ( t, ˜  ) , (2 . 69) where φ and ψ are functions in t, ˜  . Then (2 .3 ) b ecomes ψ t − κψ ˜  ˜  = 0 (2 . 70) and (2 .4) b ecomes − 2 γ ′ ′ + φ t ˜  ˜  − ν φ ˜  ˜  ˜  ˜  + ψ ˜  cos γ = 0 . (2 . 71) Mo dulo the tr a nsformation in (1.8)- (1.11), the ab ov e equation is equiv alent to − 2 γ ′ ′ ˜  + φ t ˜  − ν φ ˜  ˜  ˜  + ψ cos γ = 0 . (2 . 72) Assume ν = κ . W e take the following solution of (2 .70): ψ = m X i =1 a i d i e a 2 i κt cos 2 b i + a i ˜  cos b i sin( a 2 i κt sin 2 b i + a i ˜  sin b i + b i + c i ) (2 . 73) where a i , b i , c i , d i are real n umbers. Moreov er, (2.72 ) is equiv alen t to solving the fo llo wing equation: 2 ν γ ′ − γ ′ ′ ˜  2 + φ t − ν φ ˜  ˜  + [ m X i =1 d i e a 2 i κt cos 2 b i + a i ˜  cos b i × sin ( a 2 i κt sin 2 b i + a i ˜  sin b i + c i )] cos γ = 0 (2 . 74) b y (2 .1). Th us w e hav e the following solution of (2.7 4): φ = − [ m X i =1 d i e a 2 i κt cos 2 b i + a i ˜  cos b i sin( a 2 i κt sin 2 b i + a i ˜  sin b i + c i )] Z cos γ dt + γ ′ ˜  2 + n X s =1 ˆ d s e ˆ a 2 s κt cos 2 ˆ b s +ˆ a s ˜  cos ˆ b s sin(ˆ a 2 s κt sin 2 ˆ b s + ˆ a s ˜  sin ˆ b s + ˆ c s ) , (2 . 75) where ˆ a s , ˆ b s , ˆ c s , ˆ d s are real n um b ers. Supp ose ν 6 = κ . T o mak e (2.7 2) solv able, we c ho ose the follow ing solution of (2.70): ψ = m X i =1 a i d i e a 2 i κt + a i ˜  . (2 . 76) 10 No w ( 2 .72) is equiv alent to solving the f ollo wing equation: ν γ ′ − γ ′ ′ ˜  2 + φ t − ν φ ˜  ˜  + m X i =1 d i e a 2 i κt + a i ˜  cos γ = 0 (2 . 77) b y (2 .1). W e obtain the f ollo wing solution of (2.7 7 ): φ = γ ′ ˜  2 + n X s =1 ˆ d s e ˆ a 2 s κt cos 2 ˆ b s +ˆ a s ˜  cos ˆ b s sin(ˆ a 2 s κt sin 2 ˆ b s + ˆ a s ˜  sin ˆ b s + ˆ c s ) − m X i =1 d i e a 2 i ν t + a i ˜  Z e a 2 i ( κ − ν ) t cos γ dt. (2 . 78) Note u = φ  sin γ − γ ′ y , v = γ ′ x − φ  cos γ . (2 . 79) By (2.72 ), u t + uu x + v u y − ν ∆ u = ( φ  t − ν φ    ) sin γ + 2 γ ′ φ  cos γ − γ ′ 2 x − γ ′ ′ y = (2 γ ′ ′ ˜  − ψ cos γ ) sin γ + 2 γ ′ φ  cos γ − γ ′ 2 x − γ ′ ′ y , = γ ′ ′ ( x sin 2 γ − y cos 2 γ ) + ( 2 γ ′ φ  − ψ sin γ ) cos γ − γ ′ 2 x, (2 . 80) v t + uv x + v v y − ν ∆ v − θ = ( ν φ    − φ  t ) cos γ + 2 γ ′ φ  sin γ − γ ′ 2 y + γ ′ ′ x − ψ = ( ψ cos γ − 2 γ ′ ′ ˜  ) cos γ + 2 γ ′ φ  sin γ − γ ′ 2 y + γ ′ ′ x − ψ = − γ ′ ′ ( x cos 2 γ + y sin 2 γ ) + (2 γ ′ φ  − ψ sin γ ) sin γ − γ ′ 2 y . (2 . 81) According to (1 .1 ), p = γ ′ 2 − γ ′ ′ sin 2 γ 2 x 2 + γ ′ 2 + γ ′ ′ sin 2 γ 2 y 2 + γ ′ ′ xy cos 2 γ + Z ψ d ˜  sin γ − 2 γ ′ φ (2 . 82) mo dulo the t r ansformation in (1.8)- (1.11). Theorem 2.3 . L et γ b e any function of t and denote ˜  = x cos γ + y sin γ . T ake { a i , b i , c i , d i , ˆ a s , ˆ b s , ˆ c s , ˆ d s | i = 1 , ..., m ; s = 1 , ..., n } ⊂ R . (2 . 83) If ν = κ , we have the fol lo wing solutions of the two-d imensional B oussinesq e q uations (1.1)-(1.2): u = − γ ′ y + sin γ { 2 γ ′ ˜  + n X s =1 ˆ a s ˆ d s e ˆ a 2 s κt cos 2 ˆ b s +ˆ a s ˜  cos ˆ b s sin(ˆ a 2 s κt sin 2 ˆ b s + ˆ a s ˜  sin ˆ b s + ˆ b s + ˆ c s ) − [ m X i =1 a i d i e a 2 i κt cos 2 b i + a i ˜  cos b i sin( a 2 i κt sin 2 b i + b i + a i ˜  sin b i + c i )] Z cos γ dt } , (2 . 84) 11 v = γ ′ x − cos γ { 2 γ ′ ˜  + n X s =1 ˆ a s ˆ d s e ˆ a 2 s κt cos 2 ˆ b s +ˆ a s ˜  cos ˆ b s sin(ˆ a 2 s κt sin 2 ˆ b s + ˆ a s ˜  sin ˆ b s + ˆ b s + ˆ c s ) − [ m X i =1 a i d i e a 2 i κt cos 2 b i + a i ˜  cos b i sin( a 2 i κt sin 2 b i + a i ˜  sin b i + b i + c i )] Z cos γ dt } , (2 . 85) θ = ψ in (2.73), and p = (sin γ + 2 γ ′ Z cos γ )[ m X i =1 d i e a 2 i κt cos 2 b i + a i ˜  cos b i sin( a 2 i κt sin 2 b i + a i ˜  sin b i + c i )] + γ ′ 2 − γ ′ ′ sin 2 γ 2 x 2 + γ ′ 2 + γ ′ ′ sin 2 γ 2 y 2 + γ ′ ′ xy cos 2 γ − 2 γ ′ 2 ˜  2 − 2 γ ′ n X s =1 ˆ d s e ˆ a 2 s κt cos 2 ˆ b s +ˆ a s ˜  cos ˆ b s sin(ˆ a 2 s κt sin 2 ˆ b s + ˆ a s ˜  sin ˆ b s + ˆ c s ) . (2 . 86) When ν 6 = κ , we have the fol lowing so l utions of the two-dimensio n al Bo ussi n esq e qua- tions ( 1 .1)-(1.2): u = { n X s =1 ˆ a s ˆ d s e ˆ a 2 s κt cos 2 ˆ b s +ˆ a s ˜  cos ˆ b s sin(ˆ a 2 s κt sin 2 ˆ b s + ˆ a s ˜  sin ˆ b s + ˆ b s + ˆ c s ) +2 γ ′ ˜  − m X i =1 a i d i e a 2 i ν t + a i ˜  Z e a 2 i ( κ − ν ) t cos γ dt } sin γ − γ ′ y , (2 . 87) v = −{ n X s =1 ˆ a s ˆ d s e ˆ a 2 s κt cos 2 ˆ b s +ˆ a s ˜  cos ˆ b s sin(ˆ a 2 s κt sin 2 ˆ b s + ˆ a s ˜  sin ˆ b s + ˆ b s + ˆ c s ) +2 γ ′ ˜  − m X i =1 a i d i e a 2 i ν t + a i ˜  Z e a 2 i ( κ − ν ) t cos γ dt } cos γ + γ ′ x, (2 . 88) θ = ψ in (2.76), and p = γ ′ 2 − γ ′ ′ sin 2 γ 2 x 2 + γ ′ 2 + γ ′ ′ sin 2 γ 2 y 2 + γ ′ ′ xy cos 2 γ − 2 γ ′ 2 ˜  2 − 2 γ ′ n X s =1 ˆ d s e ˆ a 2 s κt cos 2 ˆ b s +ˆ a s ˜  cos ˆ b s sin(ˆ a 2 s κt sin 2 ˆ b s + ˆ a s ˜  sin ˆ b s + ˆ c s ) + m X i =1 d i e a 2 i ν t + a i ˜  (2 γ ′ + sin γ ) Z e a 2 i ( κ − ν ) t cos γ dt ) . (2 . 89) Remark 2.4 . By F ourier expansion, w e can use the ab ov e solution to obtain the one dep ending on tw o piecew ise con tin uous f unctions of ˜  . 3 Asymmetric Approac h I to t he 3 D Equati ons Starting from this section, w e use asymmetric approaches dev elop ed in [11] to solv e the stratified rot a ting Boussinesq equations (1.3)- ( 1 .7). 12 F or con v enience o f computation, w e denote Φ 1 = u t + uu x + v u y + w u z − 1 R 0 v − σ ( u xx + u y y + u z z ) , (3 . 1) Φ 2 = v t + uv x + v v y + w v z + 1 R 0 u − σ ( v xx + v y y + v z z ) , (3 . 2) Φ 3 = w t + uw x + v w y + w w z − σ R T − σ ( w xx + w y y + w z z ) . (3 . 3) Then the equations (1.3) - (1.5) b ecome Φ 1 + σ p x = 0 , Φ 2 + σ p y = 0 , Φ 3 + σ p z = 0 . (3 . 4) Our stra t egy is to solve the follo wing compatibilit y conditions: ∂ y (Φ 1 ) = ∂ x (Φ 2 ) , ∂ z (Φ 1 ) = ∂ x (Φ 3 ) , ∂ z (Φ 2 ) = ∂ y (Φ 3 ) . (3 . 5) First we assume u = φ z ( t, z ) x + ς ( t, z ) y + µ ( t, z ) , v = τ ( t, z ) x + ψ z ( t, z ) y + ε ( t, z ) , (3 . 6) w = − φ ( t, z ) − ψ ( t, z ) , T = ϑ ( t, z ) + z , (3 . 7) where φ, ϑ, ς , µ, τ , and ε are f unctions of t, z to b e determined. Then Φ 1 = φ tz x + ς t y + µ t + φ z ( φ z x + ς y + µ ) + ( ς − 1 /R 0 )( τ x + ψ z y + ε ) − ( φ + ψ )( φ z z x + ς z y + µ z ) − σ ( φ z z z x + ς z z y + µ z z ) = [ φ tz + φ 2 z + τ ( ς − 1 /R 0 ) − φ z z ( φ + ψ ) − σ φ z z z ] x +[ ς t + ς φ z + ψ z ( ς − 1 / R 0 ) − ς z ( φ + ψ ) − σ ς z z ] y + µ t + µφ z + ( ς − 1 / R 0 ) ε − µ z ( φ + ψ ) − σ µ z z , (3 . 8) Φ 2 = τ t x + ψ tz y + ε t + ψ z ( τ x + ψ z y + ε ) + ( τ + 1 /R 0 )( φ z x + ς y + µ ) − ( φ + ψ )( τ z x + ψ z z y + ε z ) − σ ( τ z z x + ψ z z z y + ε z z ) = [ ψ tz + ψ 2 z + ς ( τ + 1 /R 0 ) − ( φ + ψ ) ψ z z − σ ψ z z z ] y +[ τ t + τ ψ z + ( τ + 1 / R 0 ) φ z − ( φ + ψ ) τ z − σ τ z z ] x + ε t + εψ z + ( τ + 1 /R 0 ) µ − ( φ + ψ ) ε z − σ ε z z , (3 . 9) Φ 3 = − φ t − ψ t + ( φ + ψ )( φ z + ψ z ) − σ R ( ϑ + z ) + σ ( φ z z + ψ z z ) . (3 . 10) Th us (3.5) is equiv alent to the following system of partial differen tial equations: φ tz + φ 2 z + τ ( ς − 1 /R 0 ) − φ z z ( φ + ψ ) − σ φ z z z = α 1 , (3 . 11) 13 ς t + ς φ z + ψ z ( ς − 1 /R 0 ) − ς z ( φ + ψ ) − σ ς z z = α, (3 . 12) µ t + µφ z + ( ς − 1 /R 0 ) ε − µ z ( φ + ψ ) − σ µ z z = α 2 , (3 . 13) ψ tz + ψ 2 z + ς ( τ + 1 /R 0 ) − ( φ + ψ ) ψ z z − σ ψ z z z = β 1 , (3 . 1 4) τ t + τ ψ z + ( τ + 1 /R 0 ) φ z − ( φ + ψ ) τ z − σ τ z z = α, (3 . 15) ε t + εψ z + ( τ + 1 /R 0 ) µ − ( φ + ψ ) ε z − σ ε z z = β 2 (3 . 16) for some α, α 1 , α 2 , β 1 , β 2 are f unctions of t . Let 0 6 = b and c b e fixed real constan ts. Recall the not io ns in (2.12) a nd (2.13) with γ = b . W e a ssume φ = b − 1 γ 1 ζ r ( z ) , ψ = b − 1 ( γ 2 ζ r ( z ) + γ 3 η r ( z )) , (3 . 17) ς = γ 4 ( γ 2 η r ( z ) − ( − 1) r γ 3 ζ r ( z )) , τ = γ 5 γ 1 η r ( z ) , γ 4 γ 5 = 1 , (3 . 18) where γ i are f unctions of t t o b e determined. Moreov er, (3.11) b ecomes ( γ ′ 1 + ( − 1) r b 2 σ γ 1 − γ 1 γ 5 /R 0 ) η r ( z ) + ( γ 1 + γ 2 ) γ 1 c r = α 1 , (3 . 19) whic h is implied by α 1 = ( γ 1 + γ 2 ) γ 1 c r , (3 . 20) γ ′ 1 + ( − 1) r b 2 σ γ 1 − γ 1 γ 5 /R 0 = 0 . (3 . 21) On the other hand, ( 3 .15) b ecomes [( γ 1 γ 5 ) ′ + γ 1 /R 0 + ( − 1) r b 2 σ γ 1 γ 5 ] η r + γ 1 γ 5 ( γ 1 + γ 2 ) c r = α, (3 . 22) whic h giv es α = γ 1 γ 5 ( γ 1 + γ 2 ) c r , (3 . 23) ( γ 1 γ 5 ) ′ + ( − 1) r b 2 σ γ 1 γ 5 + γ 1 /R 0 = 0 . (3 . 24) Solving (3 .21) and (3.24 ) f or γ 1 and γ 1 γ 5 , w e get γ 1 = b 1 e − ( − 1) r b 2 σt sin t R 0 , γ 1 γ 5 = b 1 e − ( − 1) r b 2 σt cos t R 0 , (3 . 25) where b 1 is a real constan t. In par t icular, we tak e γ 5 = cot t R 0 . (3 . 26) Observ e that (3.12) b ecomes [( γ 2 γ 4 ) ′ + ( − 1) r b 2 σ γ 2 γ 4 − γ 2 /R 0 ] η r ( z ) + γ 4 ( γ 1 γ 2 + γ 2 2 + ( − 1) r γ 2 3 ) c r − ( − 1) r [( γ 3 γ 4 ) ′ + ( − 1) r b 2 σ γ 2 γ 4 − γ 3 /R 0 ] ζ r ( z ) = α (3 . 27) 14 and (3 .14) b ecomes [ γ ′ 2 + ( − 1) r b 2 σ γ 2 + γ 2 γ 4 /R 0 ] η r ( z ) + ( γ 1 γ 2 + γ 2 2 + ( − 1) r γ 2 3 ) c r − ( − 1) r [ γ ′ 3 + ( − 1) r b 2 σ γ 3 + γ 3 γ 4 /R 0 ] ζ r ( z ) = β 1 , (3 . 28) equiv alen tly , α = γ 4 ( γ 1 γ 2 + γ 2 2 + ( − 1) r γ 2 3 ) c r , ( 3 . 29) β 1 = ( γ 1 γ 2 + γ 2 2 + ( − 1) r γ 2 3 ) c r , (3 . 30) ( γ 2 γ 4 ) ′ + ( − 1) r b 2 σ γ 2 γ 4 − γ 2 /R 0 = 0 , (3 . 31) γ ′ 2 + ( − 1) r b 2 σ γ 2 + γ 2 γ 4 /R 0 = 0 , (3 . 32) ( γ 3 γ 4 ) ′ + ( − 1) r b 2 σ γ 2 γ 4 − γ 3 /R 0 = 0 , (3 . 33) γ ′ 3 + ( − 1) r b 2 σ γ 3 + γ 3 γ 4 /R 0 = 0 . (3 . 34) Solving (3 .31)-(3.34) under the assumption γ 4 γ 5 = 1 , we obtain γ 2 γ 4 = b 2 e − ( − 1) r b 2 σt sin t R 0 , γ 2 = b 2 e − ( − 1) r b 2 σt cos t R 0 , (3 . 35) γ 3 γ 4 = b 3 e − ( − 1) r b 2 σt sin t R 0 , γ 3 = b 3 e − ( − 1) r b 2 σt cos t R 0 . (3 . 36) In particular, we ha v e: γ 4 = ta n t R 0 . (3 . 37) According to (3 .2 3) and ( 3.29), γ 1 γ 5 ( γ 1 + γ 2 ) c r = γ 4 ( γ 1 γ 2 + γ 2 2 + ( − 1) r γ 2 3 ) c r , ( 3 . 38) equiv alen tly − 2 b 1 b 2 cos 2 t R 0 + ( b 2 2 − b 2 1 + ( − 1) r b 2 3 ) sin 2 t R 0 = 0 . (3 . 39 ) Th us b 1 b 2 = 0 , b 2 2 − b 2 1 + ( − 1) r b 2 3 = 0 . (3 . 40) So r = 0 , b 2 = 0 , b 1 = b 3 (3 . 41) or r = 1 , b 1 = 0 , b 2 = b 3 . (3 . 42) Assume r = 0 and b 1 6 = 0. Then φ = b − 1 b 1 e − b 2 σt sin bz sin t R 0 , ψ = b − 1 b 1 e − b 2 σt cos bz cos t R 0 , (3 . 43) 15 ς = − b 1 e − b 2 σt sin bz sin t R 0 , τ = b 1 e − b 2 σt cos bz cos t R 0 . (3 . 44) Moreo v er, w e take µ = ε = ϑ = 0. Then Φ 1 = γ 2 1 ( x + γ 5 y ) = b 2 1 e − 2 b 2 σt sin t R 0  x sin t R 0 + y cos t R 0  (3 . 45) b y (3 .8), ( 3 .11)-(3.12 ) , (3.20 ) and (3.23). Similarly Φ 2 = b 2 1 e − 2 b 2 σt cos t R 0  x sin t R 0 + y cos t R 0  . (3 . 46) According to (3 .1 0) Φ 3 =  b − 1 R − 1 0 b 1 e − b 2 σt − b − 1 b 2 1 e − 2 b 2 σt cos  bz − t R 0  sin  bz − t R 0  − Rσ z . (3 . 47) By (3.4) , w e hav e p = Rz 2 2 + b 1 e − b 2 σt b 2 σ R 0 cos  bz − t R 0  − b 2 1 e − 2 b 2 σt 2 σ b 2 cos 2  bz − t R 0  − b 2 1 e − 2 b 2 σt 2 σ  y 2 cos 2 t R 0 + x 2 sin 2 t R 0 + xy sin 2 t R 0  (3 . 48) mo dulo the t r ansformation in (1.14) - (1.16). Supp ose r = 1 and b 2 6 = 0. Then φ = τ = µ = ε = ϑ = 0 , ψ = b − 1 b 2 e bz + b 2 σt cos t R 0 , ς = b 2 e bz + b 2 σt sin t R 0 . (3 . 4 9) Moreo v er, Φ 1 = Φ 2 = 0 , Φ 3 = b − 1 b 2 R − 1 0 e bz + b 2 σt sin t R 0 + b − 1 b 2 2 e 2( bz + b 2 σt ) cos 2 t R 0 − Rσ z . (3 . 50) According to (3 .4 ), p = Rz 2 2 − b 2 e bz + b 2 σt b 2 σ R 0 sin t R 0 − b 2 2 e 2( bz + b 2 σt ) 2 b 2 σ cos 2 t R 0 (3 . 51) mo dulo the t r ansformation (1.1 4 )-(1.16). Theorem 3.1 . L et b, b 1 , b 2 ∈ R with b 6 = 0 . We have the fol l o wing sol utions of the thr e e-dimension a l str atifie d r otating B oussinesq e quations (1.3 ) -(1.7): (1) u = b 1 e − b 2 σt ( x cos bz − y sin bz ) sin t R 0 , v = b 1 e − b 2 σt ( x cos bz − y sin bz ) cos t R 0 , (3 . 52) w = − b − 1 b 1 e − b 2 σt cos  bz − t R 0  , T = z (3 . 53 ) and p is give n in (3.48); (2) u = b 2 e bz + b 2 σt y sin t R 0 , v = b 2 e bz + b 2 σt y cos t R 0 , (3 . 54) 16 w = − b − 1 b 2 e bz + b 2 σt cos t R 0 T = z (3 . 55) and p is give n in (3.51). Next w e assume φ = ς = ψ = τ = 0. Then µ t − 1 R 0 ε − σ µ z z = α 2 , ε t + 1 R 0 ν − σ ε z z = β 2 , ϑ t − ϑ z z = 0 . (3 . 56) Solving them, we get: Theorem 3.2 . L e t a i , b i , c i , d i , ˆ a r , ˆ b r , ˆ c r , ˆ d r , ˜ a s , ˜ b s , ˜ c s , ˜ d s b e r e al numb ers. We have the fol lowing solutions of the thr e e-dim ensional str atifie d r otating Boussinesq e quations (1.3)- (1.7): u = cos t R 0 m X i =1 d i e a 2 i σt cos 2 b i + a i z cos b i sin( a 2 i σ t sin 2 b i + a i z sin b i + c i ) + sin t R 0 n X r =1 ˆ d r e ˆ a 2 r σt cos 2 ˆ b r + a r z cos ˆ b r sin(ˆ a 2 r σ t sin 2 ˆ b r + ˆ a r z sin ˆ b r + ˆ c r ) , (3 . 57) v = − sin t R 0 m X i =1 d i e a 2 i σt cos 2 b i + a i z cos b i sin( a 2 i σ t sin 2 b i + a i z sin b i + c i ) + cos t R 0 n X r =1 ˆ d r e ˆ a 2 r σt cos 2 ˆ b r + a r z cos ˆ b r sin(ˆ a 2 r σ t sin 2 ˆ b r + ˆ a r z sin ˆ b r + ˆ c r ) , (3 . 58) w = 0 , T = z + k X s =1 ˜ a s ˜ d s e ˜ a 2 s t cos 2 ˜ b s +˜ a s z cos ˜ b s sin(˜ a 2 s t sin 2 ˜ b s + ˜ a s z sin ˜ b s + ˜ b s + ˜ c s ) , (3 . 59) p = Rz 2 2 + R m 3 X s =1 ˜ d s e ˜ a 2 s t cos 2 ˜ b s +˜ a s z cos ˜ b s sin(˜ a 2 s t sin 2 ˜ b s + ˜ a s z sin ˜ b s + ˜ c s ) . (3 . 60) Remark 3.3 . By F ourier expansion, w e can use the ab ov e solution to obtain the one dep ending on three ar bitr ary piecewise con tin uous functions of z . 4 Asymmetric Approac h I I to the 3D Equation s In this section, we solv e the stratified rotating Boussinesq equations (1.4)-(1.7 ) under the assumption u z = v z = w z z = T z z = 0 . (4 . 1) Let γ b e a function of t and w e use the mov ing f rame ˜  in (2.65 ). Assume u = f ( t, ˜  ) sin γ − γ ′ y , v = − f ( t, ˜  ) cos γ + γ ′ x, (4 . 2 ) 17 According to (4 .3 ), w e assume w = φ ( t,  ) , T = ψ ( t,  ) + z , (4 . 3) for some functions f , φ a nd ψ in t and ˜  . Using (2.6 6 )-(2.68), we get Φ 1 = − ( γ ′ 2 + γ ′ /R 0 ) x − γ ′ ′ y + f t sin γ + (2 γ ′ + 1 /R 0 ) f cos γ − σ f ˜  ˜  sin γ , (4 . 4) Φ 2 = − ( γ ′ 2 + γ ′ /R 0 ) y + γ ′ ′ x − f t cos γ + (2 γ ′ + 1 /R 0 ) f sin γ + σ f ˜  ˜  cos γ , (4 . 5) Φ 3 = φ t − σ φ ˜  ˜  − σ R ( ψ + z ) . (4 . 6) By (3.5) , w e hav e − 2 γ ′ ′ + f ˜  t − σ f ˜  ˜  ˜  = 0 , (4 . 7) φ t − σ φ ˜  ˜  − σ Rψ = 0 . (4 . 8) Moreo v er, (1.6) b ecomes ψ t − ψ ˜  ˜  = 0 . (4 . 9) Solving (4 .7), we ha v e: f = 2 γ ′ ˜  + m X i =1 a i d i e a 2 i κt cos 2 b i + a i ˜  cos b i sin( a 2 i κt sin 2 b i + a i ˜  sin b i + b i + c i ) , (4 . 10) where a i , b i , c i , d i are a rbitrary r eal n um b ers. Moreov er, (4.8 ) and (4.9) yield φ = n X r =1 ˆ d r e ˆ a 2 r t cos 2 ˆ b r +ˆ a r ˜  cos ˆ b r sin(ˆ a 2 r t sin 2 ˆ b i + ˆ a r ˜  sin ˆ b r + ˆ c r ) + σ Rtψ , ( 4 . 11) ψ = k X s =1 ˜ d s e ˜ a 2 s t cos 2 ˜ b s +˜ a s ˜  cos ˜ b s sin(˜ a 2 s t sin 2 ˜ b s + ˜ a s ˜  sin ˜ b s + ˜ c s ) (4 . 12) if σ = 1, and φ = n X r =1 ˆ d r e ˆ a 2 r σt cos 2 ˆ b r +ˆ a r ˜  cos ˆ b r sin(ˆ a 2 r σ t sin 2 ˆ b i + ˆ a r ˜  sin ˆ b r + ˆ c r ) + σ R 1 − σ k X s =1 ˜ d s e ˜ a 2 s t cos 2 ˜ b s +˜ a s ˜  cos ˜ b s sin(˜ a 2 s t sin 2 ˜ b s + ˜ a s ˜  sin ˜ b s + ˜ c s ) , (4 . 13) ψ = k X s =1 ˜ a 2 s ˜ d s e ˜ a 2 s t cos 2 ˜ b s +˜ a s ˜  cos ˜ b s sin(˜ a 2 s t sin 2 ˜ b s + ˜ a s ˜  sin ˜ b s + 2 ˜ b s + ˜ c s ) (4 . 14) when σ 6 = 1, where ˆ a r , ˆ b r , ˆ c r , ˆ d r , ˜ a s , ˜ b s , ˜ c s , ˜ d s are a rbitrary real n um b ers. No w Φ 1 = ( γ ′ ′ sin 2 γ − γ ′ 2 − γ ′ /R 0 ) x − γ ′ ′ y cos 2 γ + (2 γ ′ + 1 /R 0 ) f cos γ , (4 . 15) 18 Φ 2 = − ( γ ′ ′ sin 2 γ + γ ′ 2 + γ ′ /R 0 ) y − γ ′ ′ x cos 2 γ + (2 γ ′ + 1 /R 0 ) f sin γ (4 . 16 ) and Φ 3 = − σ Rz . According (3.4 ) , w e hav e p = − 2 γ ′ + 1 /R 0 σ [ γ ′ ˜  2 + m X i =1 d i e a 2 i κt cos 2 b i + a i ˜  cos b i sin( a 2 i κt sin 2 b i + a i ˜  sin b i + c i )] + R 2 z 2 + ( γ ′ 2 + γ ′ /R 0 )( x 2 + y 2 ) + γ ′ ′ ( y 2 − x 2 ) sin 2 γ 2 σ + γ ′ ′ σ xy cos 2 γ (4 . 17) mo dulo the t r ansformation in (1.14) - (1.16). Theorem 4.1 . L et a i , b i , c i , d i , ˆ a r , ˆ b r , ˆ c r , ˆ d r , ˜ a s , ˜ b s , ˜ c s , ˜ d s b e r e al numb ers and l e t γ b e any function of t . Deno te ˜  = x cos γ + y sin γ . We have the fol lowing solutions of the thr e e-dimension a l str atifie d r otating B oussinesq e quations (1.3 ) -(1.7): u = [ m X i =1 a i d i e a 2 i κt cos 2 b i + a i ˜  cos b i sin( a 2 i κt sin 2 b i + a i ˜  sin b i + b i + c i ) +2 γ ′ ˜  ] sin γ − γ ′ y , (4 . 18) v = [ − m X i =1 a i d i e a 2 i κt cos 2 b i + a i ˜  cos b i sin( a 2 i κt sin 2 b i + a i ˜  sin b i + b i + c i ) +2 γ ′ ˜  ] cos γ + γ ′ x, (4 . 19) p is given in (4.17); w = n X r =1 ˆ d r e ˆ a 2 r t cos 2 ˆ b r +ˆ a r ˜  cos ˆ b r sin(ˆ a 2 r t sin 2 ˆ b i + ˆ a r ˜  sin ˆ b r + ˆ c r ) + σ R t k X s =1 ˜ d s e ˜ a 2 s t cos 2 ˜ b s +˜ a s ˜  cos ˜ b s sin(˜ a 2 s t sin 2 ˜ b s + ˜ a s ˜  sin ˜ b s + ˜ c s ) , (4 . 20) T = z + k X s =1 ˜ d s e ˜ a 2 s t cos 2 ˜ b s +˜ a s ˜  cos ˜ b s sin(˜ a 2 s t sin 2 ˜ b s + ˜ a s ˜  sin ˜ b s + ˜ c s ) (4 . 21) if σ = 1 , a nd w = n X r =1 ˆ d r e ˆ a 2 r σt cos 2 ˆ b r +ˆ a r ˜  cos ˆ b r sin(ˆ a 2 r σ t sin 2 ˆ b i + ˆ a r ˜  sin ˆ b r + ˆ c r ) + σ R 1 − σ k X s =1 ˜ d s e ˜ a 2 s t cos 2 ˜ b s +˜ a s ˜  cos ˜ b s sin(˜ a 2 s t sin 2 ˜ b s + ˜ a s ˜  sin ˜ b s + ˜ c s ) , (4 . 22) T = z + k X s =1 ˜ a 2 s ˜ d s e ˜ a 2 s t cos 2 ˜ b s +˜ a s ˜  cos ˜ b s sin(˜ a 2 s t sin 2 ˜ b s + ˜ a s ˜  sin ˜ b s + 2 ˜ b s + ˜ c s ) (4 . 23) when σ 6 = 1 . 19 Remark 4.2 . By F ourier expansion, w e can use the ab ov e solution to obtain the one dep ending on three ar bitr ary piecewise con tin uous functions of ˜  . Next w e let α b e an y fixed function of t and set  = α ( x 2 + y 2 ) . (4 . 24) W e assume u = y φ ( t,  ) − α ′ 2 α x, v = − xφ ( t,  ) − α ′ 2 α y , (4 . 25) w = ψ ( t,  ) + α ′ α z , T = ϑ ( t,  ) + z (4 . 26) where φ, ψ and ϑ are functions in t,  . Note Φ 1 = − α ′ 2 + 2 αα ′ ′ 4 α 2 x + α ′ 2 R 0 α y + y φ t +  x R 0 − α ′ α y  φ − xφ 2 − 4 σ αy (  φ )   , (4 . 27) Φ 2 = − α ′ 2 + 2 αα ′ ′ 4 α 2 y − α ′ 2 R 0 α x − xφ t +  y R 0 + α ′ α x  φ − y φ 2 + 4 σ α x (  φ )   . (4 . 28) According to the first equation in (3 .5),    φ t − α ′ α φ − 4 σ α (  φ )     + α ′ 2 R 0 α = 0 , (4 . 29) equiv alen tly , (  φ ) t − α ′ α  φ − 4 σ α (  φ )   + α ′  2 R 0 α = α β ′ (4 . 30) for some function β of t . W rite ˆ φ =  φ α +  2 R 0 α − β . (4 . 31) Then (4 .30) b ecomes ˆ φ t − 4 σ α ˆ φ   = 0 . (4 . 32) Supp ose ˆ φ = ∞ X i =1 γ i  i , (4 . 33) where γ i are f unctions of t t o b e determined. Equation (4.32) yields ( γ i ) t = 4 i ( i + 1 ) σ α γ i +1 . (4 . 34) Hence γ i +1 = ( α − 1 ∂ t ) i ( γ ) i !( i + 1)!(4 σ ) i (4 . 35) for some function γ o f t . Thus ˆ φ = ∞ X i =0 ( α − 1 ∂ t ) i ( γ )  i +1 i !( i + 1)!(4 σ ) i . (4 . 36) 20 By (4.31 ), w e get φ = αβ  − 1 2 R 0 + α ∞ X i =0 ( α − 1 ∂ t ) i ( γ )  i i !( i + 1)!(4 σ ) i . (4 . 37) Note Φ 3 = ψ t + α ′ α ψ − 4 σ (  ψ  )  − σ R ( ϑ + z ) . (4 . 38 ) By the last t w o equations in (3.5), ψ t + α ′ α ψ − 4 σ (  ψ  )  − σ R ϑ = 0 (4 . 39) mo dulo the t r ansformation in (1.14) - (1.16). On the other hand, (1.6) b ecomes ϑ t − 4(  ϑ  )  = 0 . (4 . 40) Hence ϑ = ∞ X i =0 θ ( i ) 1  i +1 4 i (( i + 1)!) 2 (4 . 41) mo dulo the transformation in (1 .1 4)-(1.16), where θ 1 is an arbitrary function of t . Sub- stituting (4 .41) into (4.39), w e o btain ψ = α − 1 θ 2  + α − 1 ∞ X i =1 θ ( i ) 2 + R P i − 1 r =0 σ i − r ( αθ ( i − s − 1) 1 ) ( s ) (4 σ ) i (( i + 1)!) 2  i +1 , (4 . 42) where θ 2 is another arbitra ry function of t . No w Φ 1 = − α ′ 2 + 2 αα ′ ′ 4 α 2 x + αβ ′ y  + x R 0 φ − xφ 2 , (4 . 43) Φ 2 = − α ′ 2 + 2 αα ′ ′ 4 α 2 y − αβ ′ x  + y R 0 φ − y φ 2 (4 . 44) b y (4 .27) and (4.28 ) , and Φ 3 = ( α − 1 α ′ − σ R ) z (4 . 45) b y (4 .38). According to (3.4), we obtain p =  α ′ 2 + 2 αα ′ ′ 4 σ α 2 + 3 8 σ R 2 0  ( x 2 + y 2 ) + β ′ σ arctan y x + ( R 0 αγ − 1) β σ R 0 ln α ( x 2 + y 2 ) − σ − 1 β 2 2( x 2 + y 2 ) + σ R − α − 1 α ′ R 2 σ z 2 − 1 σ R 0 ∞ X i =0 ( α − 1 ∂ t ) i ( γ ) α i +1 ( x 2 + y 2 ) i +1 (( i + 1)!) 2 (4 σ ) i + α 2 σ ∞ X i,j =0 ( α − 1 ∂ t ) i ( γ )( α − 1 ∂ t ) j ( γ )( α ( x 2 + y 2 )) i + j +1 i ! j !( i + 1)!( j + 1)!( i + j + 1)(4 σ ) i + j + αβ 2 σ ∞ X i =1 ( α − 1 ∂ t ) i ( γ )( α ( x 2 + y 2 )) i i !( i + 1)! i (4 σ ) i (4 . 46) 21 mo dulo the transformation in (1.14 ) - (1.16). By (4.25), (4.26 ), (4.37), (4.4 1 ) and ( 4.42), w e ha v e: Theorem 4.3 L et α, β , γ , θ 1 , θ 2 b e any function of t such that the fol lowing involv e d series c onver ge. We have the fol lowing solutions of the thr e e-dimensional str atifie d r otating Boussinesq e quations (1 .3)-(1.7): u = β y x 2 + y 2 − y 2 R 0 − α ′ 2 α x + αy ∞ X i =0 ( α − 1 ∂ t ) i ( γ ) α i ( x 2 + y 2 ) i i !( i + 1)!(4 σ ) i , (4 . 47) v = x 2 R 0 − α ′ 2 α y − β x x 2 + y 2 + αx ∞ X i =0 ( α − 1 ∂ t ) i ( γ ) α i ( x 2 + y 2 ) i i !( i + 1)!(4 σ ) i , (4 . 48) w = θ 2 ( x 2 + y 2 ) + α ′ α z + 1 α ∞ X i =1 θ ( i ) 2 + R P i − 1 r =0 σ i − r ( αθ ( i − s − 1) 1 ) ( s ) (4 σ ) i (( i + 1)!) 2 α i +1 ( x 2 + y 2 ) i +1 , (4 . 4 9) T = z + ∞ X i =0 θ ( i ) 1 α i +1 ( x 2 + y 2 ) i +1 4 i (( i + 1)!) 2 (4 . 50) and p is give n in (4.46). 5 Asymmetric Approac h I I I to t he 3 D Equati ons In this section, w e solv e (1.3 ) -(1.7) with v x = w x = T x = 0 . Let c b e a real constant. Set  = y cos c + z sin c. (5 . 1) Supp ose u = f ( t,  ) , v = φ ( t,  ) sin c, ( 5 . 2) w = − φ ( t,  ) cos c, T = ψ ( t,  ) + z , (5 . 3) where f , φ and ψ are functions in t a nd  . Then Φ 1 = f t − σ f   − sin c R 0 φ, (5 . 4) Φ 2 = ( φ t − σ φ   ) sin c + 1 R 0 f , (5 . 5) Φ 3 = ( σ φ   − φ t ) cos c − σ R ( ψ + z ) . (5 . 6) By (3.5) , f  t − σ f    − sin c R 0 φ  = 0 , (5 . 7) ( φ t − σ φ   )  + sin c R 0 f  + σ Rψ  cos c = 0 . (5 . 8) 22 Mo dulo (1.14)-( 1 .16), we hav e f t − σ f   − sin c R 0 φ = 0 , (5 . 9) φ t − σ φ   + sin c R 0 f + σ Rψ cos c = 0 . (5 . 10) Denote  ˆ f ˆ φ  =  cos t si n c R 0 − sin t si n c R 0 sin t sin c R 0 cos t si n c R 0   f φ  . (5 . 11) Then (5 .9) and ( 5 .10) b ecome ˆ f t − σ ˆ f   − σ Rψ cos c sin t sin c R 0 = 0 , (5 . 12) ˆ φ t − σ ˆ φ   + σ Rψ cos c cos t sin c R 0 = 0 . (5 . 13) On the other hand, ( 1 .6) b ecomes ψ t − ψ   = 0 . (5 . 14) Assume σ = 1. W e hav e the following solution: ψ = m X i =1 a i d i e a 2 i t cos 2 b i + a i  cos b i sin( a 2 i t sin 2 b i + a i  sin b i + b i + c i ) , (5 . 15) ˆ f = − RR 0 cot c cos t sin c R 0 m X i =1 a i d i e a 2 i t cos 2 b i + a i  cos b i sin( a 2 i t sin 2 b i + a i  sin b i + b i + c i ) + n X r =1 ˆ a r ˆ d r e ˆ a 2 r t cos 2 ˆ b r +ˆ a r  cos ˆ b r sin(ˆ a 2 r t sin 2 ˆ b i + ˆ a r  sin ˆ b r + ˆ b r + ˆ c r ) , (5 . 16 ) ˆ φ = − RR 0 cot c sin t sin c R 0 m X i =1 a i d i e a 2 i t cos 2 b i + a i  cos b i sin( a 2 i t sin 2 b i + a i  sin b i + b i + c i ) + k X s =1 ˜ a s ˜ d s e ˜ a 2 s t cos 2 ˜ b s +˜ a s  cos ˜ b s sin(˜ a 2 s t sin 2 ˜ b s + ˜ a s  sin ˜ b s + ˜ b s + ˜ c s ) , (5 . 17) where a i , b i , c i , ˆ a r , ˆ b r , ˆ c r , ˆ d r , ˜ a s , ˜ b s , ˜ c s , ˜ d s are a rbitrary real n um b ers. According to (5.11), f = − RR 0 cot c cos 2 t sin c R 0 m X i =1 a i d i e a 2 i t cos 2 b i + a i  cos b i sin( a 2 i t sin 2 b i + a i  sin b i + b i + c i ) + cos t sin c R 0 n X r =1 ˆ a r ˆ d r e ˆ a 2 r t cos 2 ˆ b r +ˆ a r  cos ˆ b r sin(ˆ a 2 r t sin 2 ˆ b i + ˆ a r  sin ˆ b r + ˆ b r + ˆ c r ) + sin t sin c R 0 k X s =1 ˜ a s ˜ d s e ˜ a 2 s t cos 2 ˜ b s +˜ a s  cos ˜ b s sin(˜ a 2 s t sin 2 ˜ b s + ˜ a s  sin ˜ b s + ˜ b s + ˜ c s ) , (5 . 18) 23 φ = − sin t sin c R 0 n X r =1 ˆ a r ˆ d r e ˆ a 2 r t cos 2 ˆ b r +ˆ a r  cos ˆ b r sin(ˆ a 2 r t sin 2 ˆ b i + ˆ a r  sin ˆ b r + ˆ b r + ˆ c r ) + cos t sin c R 0 k X s =1 ˜ a s ˜ d s e ˜ a 2 s t cos 2 ˜ b s +˜ a s  cos ˜ b s sin(˜ a 2 s t sin 2 ˜ b s + ˜ a s  sin ˜ b s + ˜ b s + ˜ c s ) . (5 . 19) Supp ose σ 6 = 1. W e take the f ollo wing solution of (5.1 1 )-(5.14): ψ = m X i =1 a i d i e a 2 i t + a i  , (5 . 20) ˆ f = σ R m X i =1 a i d i e a 2 i t + a i  cos c h a 2 i (1 − σ ) sin t si n c R 0 − R − 1 0 sin c cos t si n c R 0 i a 4 i (1 − σ ) 2 + R − 2 0 sin 2 c + n X r =1 ˆ a r ˆ d r e ˆ a 2 r σt cos 2 ˆ b r +ˆ a r  cos ˆ b r sin(ˆ a 2 r σ t sin 2 ˆ b i + ˆ a r  sin ˆ b r + ˆ b r + ˆ c r ) , (5 . 21) ˆ φ = σ R m X i =1 a i d i e a 2 i t + a i  cos c h a 2 i ( σ − 1) cos t sin c R 0 − R − 1 0 sin c sin t si n c R 0 i a 4 i (1 − σ ) 2 + R − 2 0 sin 2 c + k X s =1 ˜ a s ˜ d s e ˜ a 2 s σt cos 2 ˜ b s +˜ a s  cos ˜ b s sin(˜ a 2 s σ t sin 2 ˜ b s + ˜ a s  sin ˜ b s + ˜ b s + ˜ c s ) , (5 . 22) where a i , b i , c i , ˆ a r , ˆ b r , ˆ c r , ˆ d r , ˜ a s , ˜ b s , ˜ c s , ˜ d s are a rbitrary real n um b ers. According to (5.11), f = cos t sin c R 0 n X r =1 ˆ a r ˆ d r e ˆ a 2 r σt cos 2 ˆ b r +ˆ a r  cos ˆ b r sin(ˆ a 2 r σ t sin 2 ˆ b i + ˆ a r  sin ˆ b r + ˆ b r + ˆ c r ) + sin t sin c R 0 k X s =1 ˜ a s ˜ d s e ˜ a 2 s σt cos 2 ˜ b s +˜ a s  cos ˜ b s sin(˜ a 2 s σ t sin 2 ˜ b s + ˜ a s  sin ˜ b s + ˜ b s + ˜ c s ) − σ R m X i =1 a i d i e a 2 i t + a i  sin 2 c 2 R 0 ( a 4 i (1 − σ ) 2 + R − 2 0 sin 2 c ) , (5 . 23) φ = − sin t sin c R 0 n X r =1 ˆ a r ˆ d r e ˆ a 2 r σt cos 2 ˆ b r +ˆ a r  cos ˆ b r sin(ˆ a 2 r σ t sin 2 ˆ b i + ˆ a r  sin ˆ b r + ˆ b r + ˆ c r ) + cos t sin c R 0 k X s =1 ˜ a s ˜ d s e ˜ a 2 s σt cos 2 ˜ b s +˜ a s  cos ˜ b s sin(˜ a 2 s σ t sin 2 ˜ b s + ˜ a s  sin ˜ b s + ˜ b s + ˜ c s ) − σ R m X i =1 a 3 i d i ( σ − 1) e a 2 i t + a i  cos c a 4 i (1 − σ ) 2 + R − 2 0 sin 2 c . (5 . 24) By (5.4) -(5.6), ( 5 .9) and (5.10), Φ 1 = 0 , Φ 2 =  cos c R 0 f − σ Rψ sin c  cos c, (5 . 25) 24 Φ 3 =  cos c R 0 f − σ Rψ sin c  sin c − σ Rz . (5 . 26) According to (3 .4 ), p = R cos 2 c sin c cos 2 t sin c R 0 m X i =1 d i e a 2 i t cos 2 b i + a i  cos b i sin( a 2 i t sin 2 b i + a i  sin b i + c i ) − cos c R 0 cos t sin c R 0 n X r =1 ˆ d r e ˆ a 2 r t cos 2 ˆ b r +ˆ a r  cos ˆ b r sin(ˆ a 2 r t sin 2 ˆ b i + ˆ a r  sin ˆ b r + ˆ c r ) − cos c R 0 sin t sin c R 0 k X s =1 ˜ d s e ˜ a 2 s t cos 2 ˜ b s +˜ a s  cos ˜ b s sin(˜ a 2 s t sin 2 ˜ b s + ˜ a s  sin ˜ b s + ˜ c s ) + R sin c m X i =1 d i e a 2 i t cos 2 b i + a i  cos b i sin( a 2 i t sin 2 b i + a i  sin b i + c i ) + R 2 z 2 (5 . 27) mo dulo if σ = 1, and p = − cos c σ R 0 cos t sin c R 0 n X r =1 ˆ d r e ˆ a 2 r σt cos 2 ˆ b r +ˆ a r  cos ˆ b r sin(ˆ a 2 r σ t sin 2 ˆ b i + ˆ a r  sin ˆ b r + ˆ c r ) − cos c σ R 0 sin t sin c R 0 k X s =1 ˜ d s e ˜ a 2 s σt cos 2 ˜ b s +˜ a s  cos ˜ b s sin(˜ a 2 s σ t sin 2 ˜ b s + ˜ a s  sin ˜ b s + ˜ c s ) + m X i =1 d i Re a 2 i t + a i  sin 2 c cos c 2 R 2 0 ( a 4 i (1 − σ ) 2 + R − 2 0 sin 2 c ) + R sin c m X i =1 d i e a 2 i t + a i  + R 2 z 2 , (5 . 28) mo dulo the t r ansformation in (1.14) - (1.16). In summary , w e ha v e: Theorem 5.1 . L et a i , b i , c i , ˆ a r , ˆ b r , ˆ c r , ˆ d r , ˜ a s , ˜ b s , ˜ c s , ˜ d s , c b e arbitr ary r e al numb ers. D e - note  = y cos x + z sin c . We have the fol lowing solutions of the thr e e-dimensional str atifie d r otating Boussinesq e quations (1.3)-(1.7): u = f , v = φ sin c, w = − φ cos c, T = ψ + z , (5 . 29) wher e (1) f is given in (5 .18), φ is given i n ( 5 .19), ψ is given in (5 .15) and p is given i n (5.27) if σ = 1 ; (2) f is given in (5.23 ) , φ is given in (5.24), ψ is given in (5.20) and p is g i v en in ( 5 .28) when σ 6 = 1 . Remark 5.2 . By F ourier expansion, w e can use t he ab ov e solution to obtain the one dep ending on three arbitrary pie cewise con tin uous func tions of  . Applying the transformation T 1 in (1.12) -(1.13) to the ab ov e solution, w e get a solution in v olving all the v ariables t, x, y , z . 25 References [1] D. 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