Multiple Parameter Function Approaches to the Equations of Dynamic Convection in a Sea

One of the most important topics in geophysics is to study convection in a sea. Based on the algebraic characteristics of the equations of dynamic convection in a sea, we introduce various schemes with multiple parameter functions to solve these equa…

Authors: Xiaoping Xu

Multiple P arameter F unction Approac hes t o the Equations of Dynamic Con v ecti on in a Sea 1 Xiaoping Xu Institute of Mathematics, Academy of Mathematics & System Sciences Chinese Academ y of Sciences, Beijing 100190, P .R. China 2 Abstract One of the most impo rtant topics in g eophysics is to study co nv ection in a sea. Based o n the algebraic characteristics of the equations of dynamic conv ection in a sea, we int r o duce v arious schemes with m ultiple parameter functions to solv e these equations and obta in families of new explicit exact solutions with multiple parameter functions. Moreover, symmetry tr ansformations are used to simplify our arg ument s. 1 In tro du ction Both the atmospheric and o ceanic flo ws are infl uenced by the rotation of the earth. In fact, the fast r otation and small asp ect ratio are t wo main c haracteristics of the large scale atmospheric and oceanic flo ws. The small asp ect rati o c haracteristic leads to t he primitiv e equations, and the fast rotation leads to th e qu asi-geost r op ic equations (cf. [2], [7], [8], [10]). A m ain ob jectiv e in climate dynamics and in geophysica l fluid dynamics is to un derstand and predict the p erio dic, quasi-p erio dic, ap erio d ic, and fu lly tu rbulent c h aracteristics of the large scale atmospheric and o ceanic flows (e.g., cf. [4], [6]). The general mo del of atmosph er ic and o ceanic flows is v ery complicated. V a rious s implified mo dels had b een established and studied. F o r instance, Boussinesq equations are simpler mo dels in atmospheric sciences (e.g., cf. [9]). Chae [1] pro ved the global regularit y , and Hou and Li [3] obtained the we ll-p osedness of the tw o-dimensional e qu ations. Hsia, Ma and W ang [4] studied the bifur cation and p erio dic solutions of th e three-dimensional equations. The follo wing equations in geoph ysics u x + v y + w z = 0 , ρ = p z , (1 . 1) ρ t + uρ x + v ρ y + wρ z = 0 , (1 . 2) u t + uu x + v u y + wu z + v = − 1 ρ p x , (1 . 3) 1 2000 Mathematical Sub ject Classification. Primary 35Q35, 35C05; S econdary 35L60. 2 Researc h supp orted by China NSF 10871193 1 v t + uv x + v v y + wv z − u = − 1 ρ p y , (1 . 4) are used to describ e the d ynamic con ve ction in a sea, where u, v and w are comp onents of v elo cit y v ector of relativ e motion of fluid in Cartesian coordin ates ( x, y , z ), ρ = ρ ( x, y , z , t ) is the d ensit y of fluid and p is the pressure (e.g., cf. P age 203 in [5]). Ovsiann ik o v determined th e Lie p oint symm etries of the ab ov e equ ations and found t wo v ery sp ecial solutions (cf. [5]). In [11], we used the stable range of nonlinear term to solv e the equation of non s tationary transonic gas flo w. Moreo v er, w e [1 2] solv ed the th ree-dimensional Na vior-Stok es equations b y asymmetric tec hn iqu es and mo ving frames. Based on th e algebraic c haracteristics of the equations (1.1)-(1.4) of dynamic con vect ion in a sea, w e in tro du ce v arious schemes with multiple parameter functions to solve these equations and obtain families of new explicit exact solutions with multiple parameter fu nctions. Moreo ver, symmetry transformations are used to simplify our argument s. By sp ecifying these parameter functions, one can obtain the solutions of certain practical mo dels. F or con ve nience, we alwa ys assume that all the in vol ved partial deriv ative s of related func- tions alw a ys exist and we can c hange ord ers of taking p artial deriv ative s. W e also u se prime ′ to denote the deriv ativ e of any one-v ariable function. W e will use the follo wing symm etry transformations T 1 - T 4 due to Ovsianniko v ( cf. P age 204 of [5]) of the equation (1.1)-(1 .4) to simplify our solutions: T 1 ( u ) = u ( t, x + α, y , z + α ′ ′ x − α ′ y ) − α ′ , T 1 ( v ) = v ( t, x + α, y , z + α ′ ′ x − α ′ y ) , (1 . 5) T 1 ( w ) = w ( t, x + α, y , z + α ′ ′ x − α ′ y ) − α ′ ′ u + α ′ v − α ′ ′ ′ x + α ′ ′ y , (1 . 6) T 2 ( u ) = u ( t, x, y + α, z + α ′ x + α ′ ′ y ) , T 2 ( v ) = v ( t, x, y + α, z + α ′ x + α ′ ′ y ) − α ′ , (1 . 7 ) T 2 ( w ) = w ( t, x, y + α, z + α ′ x + α ′ ′ y ) − α ′ u − α ′ ′ v − α ′ ′ x − α ′ ′ ′ y , (1 . 8) T 1 ( p ) = p ( t, x + α, y , z + α ′ ′ x − α ′ y ) , T 2 ( p ) = p ( t, x, y + α, z + α ′ x + α ′ ′ y ) , (1 . 9) T 3 ( u ) = u ( t, x, y , z + α ) , T 3 ( v ) = v ( t, x, y , z + α ) , (1 . 10) T 3 ( w ) = w ( t, x, y , z + α ) − α ′ , T 3 ( p ) = p ( t, x, y , z + α ) , (1 . 11) T 4 ( p ) = p + α, T 4 ( F ) = F for F = u, v , w , (1 . 12) where α is an arbitrary fu nction of t . T he ab o ve transformations transform on e solution of the equations (1.1)-(1.4) into another solution. Applying th e ab o ve transform ations to an y solution found in this pap er will yield another solution with f our extra parameter functions. In Section 2, we u se a n ew v ariable of mo ving line to s olv e the equations (1.1)-(1.4) . A n approac h of using th e pro duct of cylindrical inv arian t function with z is introd uced in S ection 3. In Section 4 , we reduce the three-dimensional (spacial) equations (1.1)-(1.4) in to a tw o- dimensional pr oblem and then solv e it with three different ansatzes. 2 2 Mo ving-Line Approac h Let α and β b e giv en fu nctions of t . Denote  = α ′ x + β ′ y + z . (2 . 1) Supp ose that f , g , h are functions in t, x, y , z that are linear in x, y , z suc h that f x + g y + h z = 0 . (2 . 2) W e assume u = φ ( t,  ) + f , v = ψ ( t,  ) + g , (2 . 3) w = h − α ′ φ ( t,  ) − β ′ ψ ( t,  ) , p = ζ ( t,  ) , (2 . 4) where φ, ψ , ζ are t wo-v ariable fun ctions. Note th at the fi rst equation in (1.1) n aturally holds and ρ = p z = ζ  b y the second equation in (1.1). Moreo v er, (1.2)-(1.4 ) b ecome ζ  t + ζ   ( α ′ ′ x + β ′ ′ y + α ′ f + β ′ g + h ) = 0 , (2 . 5) f t + g + f f x + g f y + hf z + α ′ + φ t + ( f x − α ′ f z ) φ + ( f y − β ′ f z + 1) ψ + φ  ( α ′ ′ x + β ′ ′ y + α ′ f + β ′ g + h ) = 0 , (2 . 6) g t − f + f g x + g g y + hg z + β ′ + ψ t + ( g x − α ′ g z − 1) φ + ( g y − β ′ g z ) ψ + ψ  ( α ′ ′ x + β ′ ′ y + α ′ f + β ′ g + h ) = 0 . (2 . 7) In order to solve the ab o ve system of partial differentia l equations, we assu me α ′ ′ x + β ′ ′ y + α ′ f + β ′ g + h = − γ ′  = − γ ′ ( α ′ x + β ′ y + z ) (2 . 8) for some fun ction γ of t , and f t + g + f f x + g f y + hf z + α ′ = 0 , (2 . 9) g t − f + f g x + g g y + hg z + β ′ = 0 . (2 . 10) Then (2.5)-( 2.7) b ecome ζ  t − γ ′  ζ   = 0 , (2 . 11) φ t + ( f x − α ′ f z ) φ + ( f y − β ′ f z + 1) ψ − γ ′  φ  = 0 , (2 . 12) ψ t + ( g x − α ′ g z − 1) φ + ( g y − β ′ g z ) ψ − γ ′  ψ  = 0 . (2 . 13) According to (2.8), h = − α ′ ′ x − β ′ ′ y − α ′ f − β ′ g − γ ′  . (2 . 14) Substituting the ab ov e equation into (2.9) and (2.10) , we ha v e: f t + f ( f x − α ′ f z ) + g ( f y − β ′ f z + 1) − f z ( α ′ ′ x + β ′ ′ y + γ ′  ) + α ′ = 0 , (2 . 15) 3 g t + f ( g x − α ′ g z − 1) + g ( g y − β ′ g z ) − g z ( α ′ ′ x + β ′ ′ y + γ ′  ) + β ′ = 0 . (2 . 16) Our linearit y assumption implies that A =  f x − α ′ f z f y − β ′ f z + 1 g x − α ′ g z − 1 g y − β ′ g z  (2 . 17) is a matrix fu nction of t . In order to solv e th e system (2.12 ) and (2.13), and the system (2.15 ) and (2.16), we need the comm utativit y of A with dA/dt . F or simplicit y , we assume f y − β ′ f z + 1 = g x − α ′ g z − 1 = 0 . (2 . 18) So f y = β ′ f z − 1 , g x = α ′ g z + 1 . (2 . 19) Moreo ver, (2.15) and (2.16) b ecome f t + f ( f x − α ′ f z ) − f z ( α ′ ′ x + β ′ ′ y + γ ′  ) + α ′ = 0 , (2 . 20) g t + g ( g y − β ′ g z ) − g z ( α ′ ′ x + β ′ ′ y + γ ′  ) + β ′ = 0 . (2 . 21) W r ite f = α 1 x + ( β ′ α 2 − 1) y + α 2 z + α 3 , (2 . 22) g = ( α ′ β 2 + 1) x + β 1 y + β 2 z + β 3 (2 . 23) b y our linearity assumption and (2.19), wh er e α i and β j are functions of t . No w (2.20) is equiv alen t to the follo win g system of ord inary differen tial equations: α ′ 1 + α 1 ( α 1 − α ′ α 2 ) − α 2 ( α ′ ′ + γ ′ α ′ ) = 0 , (2 . 24) ( β ′ α 2 ) ′ + ( β ′ α 2 − 1)( α 1 − α ′ α 2 ) − α 2 ( β ′ ′ + γ ′ β ′ ) = 0 , (2 . 25) α ′ 2 + α 2 ( α 1 − α ′ α 2 − γ ′ ) = 0 , (2 . 26) α ′ 3 + α 3 ( α 1 − α ′ α 2 ) + α ′ = 0 . (2 . 27) Observe that (2 . 25) − β ′ × (2 . 26) b ecomes − α 1 + α ′ α 2 = 0 . (2 . 28) So (2.26) b ecomes α ′ 2 − γ ′ α 2 = 0 = ⇒ α 2 = b 1 e γ , b 1 ∈ R . (2 . 29) According to (2.28), α 1 = b 1 α ′ e γ . (2 . 30) With the data (2.29) and (2.30 ), (2.24) natur ally holds. By (2.27), w e take α 3 = − α. (2 . 31) 4 Note that (2.21) is equ iv alen t to the follo wing system of ordinary differential equations: α ′ β ′ 2 + ( α ′ β 2 + 1)( β 1 − β ′ β 2 ) − α ′ β 2 γ ′ = 0 , (2 . 32) β ′ 1 + β 1 ( β 1 − β ′ β 2 ) − β 2 ( β ′ ′ + β ′ γ ′ ) = 0 , (2 . 33) β ′ 2 + β 2 ( β 1 − β ′ β 2 − γ ′ ) = 0 , (2 . 34) β ′ 3 + β 3 ( β 1 − β ′ β 2 ) − β ′ = 0 . (2 . 35) Similarly , we ha v e: β 1 = b 2 β ′ e γ , β 2 = b 2 e γ , β 3 = β (2 . 36) with b 2 ∈ R . Moreo ver, (2.2) giv es γ ′ = 0 b y (2.14) , (2.28) and (2.36). W e ta ke γ = 0. Therefore, φ = ℑ (  ) and ψ = ι (  ) by (2.12) and (2.13) for some one-v ariable functions ℑ and ι . F u rthermore, w e tak e ζ = σ (  ) by (2.11) for another one-v ariable fu nction σ . In su mmary , w e h a v e: Theorem 2.1 . L et α, β b e functions of t and let b 1 , b 2 ∈ R . Supp ose that ℑ , ι and σ ar e arbitr ary one-variable functions. The fol lowing is a solution of the e quations (1.1)-(1.4) of dynamic c onve ction in a se a: u = b 1 α ′ x + ( b 1 β ′ − 1) y + b 1 z − α + ℑ ( α ′ x + β ′ y + z ) , (2 . 37) v = ( b 2 α ′ + 1) x + b 2 β ′ y + b 2 z + β + ι ( α ′ x + β ′ y + z ) , (2 . 38) w = − ( α ′ ′ + b 1 ( α ′ ) 2 + ( b 2 α ′ + 1) β ′ ) x − ( β ′ ′ + α ′ ( b 1 β ′ − 1) + b 2 ( β ′ ) 2 ) y − ( b 1 α ′ + b 2 β ′ ) z + αα ′ − β β ′ − α ′ ℑ ( α ′ x + β ′ y + z ) − β ′ ι ( α ′ x + β ′ y + z ) , (2 . 39) p = σ ( α ′ x + β ′ y + z ) , ρ = σ ′ ( α ′ x + β ′ y + z ) . (2 . 40) W e r emark that we hav e tried some other f orms of the matrix A in (2.17) suc h that A and dA/dt comm ute, but w e ha v e f ailed to get new solutions. 3 Approac h of Cylindrical Pro duct Let σ b e a fixed one-v ariable fu n ction and set  = σ ( x 2 + y 2 ) z . (3 . 1) Supp ose that f and g are functions in t, x, z that are linear homogeneous in x, y and h = γ σ − z ( f x + g y ) , (3 . 2) where γ is a fun ction of t . Ass u me u = f + y ψ ( t,  ) , v = g − xψ ( t,  ) , w = h, p = φ ( t,  ) (3 . 3) 5 where ψ and φ are t wo -v ariable functions. Note u t = f t + y ψ t , u x = f x + 2 xy z σ ′ ψ  , (3 . 4) u y = f y + ψ + 2 y 2 z σ ′ ψ  , u z = f z + y σ ψ  , (3 . 5) v t = g t − xψ t , v x = g x − ψ − 2 x 2 z σ ′ ψ  , (3 . 6) v y = g y − 2 xy z σ ′ ψ  , v z = g z − xσ ψ  . (3 . 7) Hence (1.3) b ecomes u t + uu x + v u y + wu z + v = f t + y ψ t + ( f + y ψ )( f x + 2 xy z σ ′ ψ  ) +( g − xψ )( f y + 1 + ψ + 2 y 2 z σ ′ ψ  ) + y σ hψ  = f t + f f x + g (1 + f y ) + x ( g x − f y − 1) ψ − xψ 2 + y [ ψ t + ( f x + g y ) ψ + (2( xf + y g ) σ ′ z + hσ ) ψ  ] = − 2 xz σ ′ σ (3 . 8) and (1.4) giv es v t + uv x + v v y + wv z − u = g t − xψ t + ( f + y ψ )( g x − 1 − ψ − 2 x 2 z σ ′ ψ  ) +( g − xψ )( g y − 2 xy z σ ′ ψ  ) − xσ hψ  = g t + f ( g x − 1) + g g y − y (1 + f y − g x ) ψ − y ψ 2 − x [ ψ t + ( f x + g y ) ψ + (2( xf + y g ) σ ′ z + hσ ) ψ  ] = − 2 y z σ ′ σ . (3 . 9) In order to solve the ab o ve system of different ial equations, we assum e f = α ′ x − y 2 , g = x 2 + α ′ y , σ ( x 2 + y 2 ) = 1 x 2 + y 2 (3 . 10) for some fun ction α of t . According to (3.2), h = γ σ − 2 α ′ z . (3 . 11) No w (3.8) b ecomes ( α ′ ′ + ( α ′ ) 2 + 4 − 1 − ψ 2 ) x + y [ ψ t + 2 α ′ ψ + ( γ − 4 α ′  ) ψ  ] = 2 x (3 . 12) and (3.9) yields ( α ′ ′ + ( α ′ ) 2 + 4 − 1 − ψ 2 ) y − x [ ψ t + 2 α ′ ψ + ( γ − 4 α ′  ) ψ  ] = 2 y  . (3 . 13) The ab o ve system is equiv alent to α ′ ′ + ( α ′ ) 2 + 4 − 1 − ψ 2 = 2  , (3 . 14) ψ t + 2 α ′ ψ + ( γ − 4 α ′  ) ψ  = 0 . (3 . 15) 6 By (3.14), we tak e ψ = q α ′ ′ + ( α ′ ) 2 + 4 − 1 − 2  , (3 . 16) due to the sk ew-symm etry of ( u, x ) and ( v , y ). Sub stituting (3.16) in to (3.15), we get α ′ ′ ′ + 2 α ′ α ′ ′ + 4 α ′ ( α ′ ′ + ( α ′ ) 2 + 4 − 1 − 2  ) − 2( γ − 4 α ′  ) = 0 , (3 . 17) equiv alen tly , γ = 2( α ′ ) 3 + 3 α ′ α ′ ′ + α ′ ′ ′ + α ′ 2 . (3 . 18) According to the second equation in (1.1), we h a v e ρ = σ φ  . Note ρ t = σ φ  t , ρ x = 2 xσ ′ ( φ  +  φ   ) , (3 . 19) ρ y = 2 y σ ′ ( φ  +  φ   ) , ρ z = σ 2 φ   . (3 . 20) So (1.2) b ecomes φ  t − 2 α ′ φ  + ( γ − 4 α ′  ) φ   = 0 . (3 . 21) Mo dulo T 4 in (1.12), the ab o v e equation is equiv alent to: φ t + 2 α ′ φ + ( γ − 4 α ′  ) φ  = 0 . (3 . 22) Set ˜ ψ = e 2 α ψ , ˜ φ = e 2 α φ. (3 . 23) Then (3.15) and (3.22) are equiv alen t to the equations: ˜ ψ t + ( γ − 4 α ′  ) ˜ ψ  = 0 , ˜ φ t + ( γ − 4 α ′  ) ˜ φ  = 0 , (3 . 24) resp ectiv ely . So w e h a v e the solution ˜ φ = ℑ ( ˜ ψ ) = ⇒ φ = e − 2 α ℑ  e 2 α q α ′ ′ + ( α ′ ) 2 + 4 − 1 − 2   (3 . 25) for some one-v ariable function ℑ . Thus w e hav e: Theorem 3.1 . L et α b e any function of t and let ℑ b e arbitr ary one-variable function. The fol lowing is a solution of the e q u ations (1.1)-(1.4) of dynamic c onve ction in a se a: u = α ′ x − y 2 + y r α ′ ′ + ( α ′ ) 2 + 1 4 − 2 z x 2 + y 2 , (3 . 26) v = α ′ y + x 2 − x r α ′ ′ + ( α ′ ) 2 + 1 4 − 2 z x 2 + y 2 , (3 . 27) w = 2( α ′ ) 3 + 3 α ′ α ′ ′ + α ′ ′ ′ + α ′ 2 ! ( x 2 + y 2 ) − 2 α ′ z , (3 . 28) p = e − 2 α ℑ  e 2 α r α ′ ′ + ( α ′ ) 2 + 1 4 − 2 z x 2 + y 2  , (3 . 29) ρ = − ℑ ′  e 2 α q α ′ ′ + ( α ′ ) 2 + 1 4 − 2 z x 2 + y 2  ( x 2 + y 2 ) q α ′ + α 2 + 1 4 − 2 z x 2 + y 2 . (3 . 30) 7 4 Dimensional Reduction Supp ose that u, v , ζ and η are functions in t, x, y . Assume w = ζ − ( u x + v y ) z , p = z + η , ρ = 1 . (4 . 1) Then the equations (1.1)-(1.4) are equiv alen t to the follo win g t wo -dimen sional problem: u t + uu x + v u y + v = − η x , (4 . 2) v t + uv x + v v y − u = − η y . (4 . 3) The compatibilit y η xy = η y x giv es ( u y − v x ) t + u ( u y − v x ) x + v ( u y − v x ) y + ( u x + v y )( u y − v x + 1) = 0 . (4 . 4) Supp ose that ϑ is a function in t, x , y suc h that ϑ xx + ϑ y y = 0 (4 . 5) (so ϑ is a time-dep end en t h armonic function). W e assume u = ϑ xx , v = ϑ xy . (4 . 6) Then (4.4) natur ally holds. Indeed, u t + uu x + vu y + v =  ϑ xt + 2 − 1 ( ϑ 2 xx + ϑ 2 xy ) + ϑ y  x , (4 . 7) v t + uv x + v v y − u =  ϑ xt + 2 − 1 ( ϑ 2 xx + ϑ 2 xy ) + ϑ y  y . (4 . 8) By (4.2) and (4.3), we tak e η = − ϑ xt − ϑ y − 1 2 ( ϑ 2 xx + ϑ 2 xy ) . (4 . 8) Hence w e h a v e the follo wing easy result: Prop osition 4.1 . L e t ϑ and ζ b e functions in t, x, y such that (4.5) holds. The fol lowing is a solution of the e q u ations (1.1)-(1.4) of dynamic c onve ction in a se a: u = ϑ xx , v = ϑ xy , w = ζ , (4 . 9) ρ = 1 , p = z − ϑ xt − ϑ y − 1 2 ( ϑ 2 xx + ϑ 2 xy ) . (4 . 10) The ab o v e appr oac h is the well- kn o wn rotation-free approac h. W e are more int erested in the approac hes that the rotatio n ma y not b e zero. Let f and g b e f unctions in t, x, y that are linear in x, y . Denote  = x 2 + y 2 . (4 . 11) 8 Consider u = f + y φ ( t,  ) , v = g − xφ ( t,  ) , (4 . 12) where φ is a tw o-v ariable fu n ction to b e determined. Then u x = f x + 2 xy φ  , u y = f y + φ + 2 y 2 φ  , (4 . 13) v x = g x − φ − 2 x 2 φ  , u y = g y − 2 xy φ  . (4 . 1 4) Th u s u x + v y = f x + g y , u y − v x = f y − g x + 2( φ )  . (4 . 15) F or simplicit y , w e assume f = − α ′ x 2 α − y 2 , g = x 2 x − α ′ y 2 α (4 . 16) for some fun ctions α and β of t . Th en (4.4) b ecomes (  φ )  t − α ′ α  (  φ )   − α ′ α (  φ )  = 0 . (4 . 17) Hence φ = γ + ℑ ( α )  (4 . 18) for some fun ction γ of t and one-v ariable function ℑ . No w (4.12), (4.16 ) and (4.18) imply u = − α ′ x 2 α − y 2 + ( γ + ℑ ( α )) y  , (4 . 19) v = x 2 − α ′ y 2 α − ( γ + ℑ ( α )) x  . (4 . 20) Moreo ver, (4.2) and (4.3) yield  ( α ′ ) 2 − 2 αα ′ ′ 4 α 2 + 1 4  x + γ ′ y  − xφ 2 = − η x , (4 . 21)  ( α ′ ) 2 − 2 αα ′ ′ 4 α 2 + 1 4  y − γ ′ x  − y φ 2 = − η y . (4 . 22) Th u s η = 1 2 Z ( γ + ℑ ( α )) 2 d  2 − 1 2  ( α ′ ) 2 − 2 αα ′ ′ 4 α 2 + 1 4   + γ ′ arctan y x . (4 . 23) Theorem 4.2 . L et α, γ b e any functions of t . Supp ose that ℑ is an arbitr ary one-variable function and ζ is any function in t , x, y . The fol lowing is a solution of the e quations (1.1)-(1.4) of dynamic c onve ction in a se a: u = − α ′ x 2 α − y 2 + ( γ + ℑ (( x 2 + y 2 ) α )) y x 2 + y 2 , (4 . 24) v = x 2 − α ′ y 2 α − ( γ + ℑ (( x 2 + y 2 ) α )) x x 2 + y 2 , (4 . 25) 9 w = α ′ α z + ζ , ρ = 1 , (4 . 26) p = z + 1 2 Z ( γ + ℑ ( α )) 2 d  2 − 1 2  ( α ′ ) 2 − 2 αα ′ ′ 4 α 2 + 1 4  ( x 2 + y 2 ) + γ ′ arctan y x (4 . 27) with  = x 2 + y 2 . Next w e assu me u = ε ( t, x ) , v = φ ( t, x ) + ψ ( t, x ) y , (4 . 28) where ε, φ and ψ are fu nctions in t, x to b e determined. Su bstituting (4.28) into (4.4), we get φ tx + ψ tx y + ε ( φ xx + ψ xx y ) + ( φ + ψ y ) ψ x + ( ε x + ψ )( φ x + ψ x y − 1) = 0 , (4 . 29) equiv alen tly , ( φ t + εφ x + φψ − ε ) x − ψ = 0 , (4 . 30) ( ψ t + εψ x + ψ 2 ) x = 0 . (4 . 31) F or simplicit y , w e tak e ψ = − α ′ , (4 . 32) a function of t . Denote φ = ˆ φ + x. (4 . 33) Then (4.30) b ecomes ( ˆ φ t + ε ˆ φ x − α ′ ˆ φ ) x = 0 . (4 . 34) T o solv e the ab o ve equation, w e assume ε = β ˆ φ x − ϑ t ( t, x ) ϑ x ( t, x ) (4 . 35) for some fun ctions β of t , and ϑ of t and x . W e hav e th e follo wing solution of (4.34): ˆ φ = e α ℑ ( ϑ ) = ⇒ φ = e α ℑ ( ϑ ) + x = ⇒ v = e α ℑ ( ϑ ) + x − α ′ y (4 . 36) for another one-v ariable function ℑ . Moreo ver, ε = β e − α ϑ x ℑ ′ ( ϑ ) − ϑ t ϑ x . (4 . 37) Note u t + uu x + v u y + v = ( β e − α ) ′ ϑ x ℑ ′ ( ϑ ) − β e − α ( ϑ xt ℑ ′ ( ϑ ) + ϑ t ϑ x ℑ ′ ′ ( ϑ )) ( ϑ x ℑ ′ ( ϑ )) 2 − ϑ tt ϑ x − ϑ t ϑ xt ϑ 2 x − α ′ y +  β e − α ϑ x ℑ ′ ( ϑ ) − ϑ t ϑ x   β e − α ϑ x ℑ ′ ( ϑ ) − ϑ t ϑ x  x + e α ℑ ( ϑ ) + x, (4 . 38) v t + uv x + v v y − u = (( α ′ ) 2 − α ′ ′ ) y + β − α ′ x. (4 . 39) 10 By (4.2) and (4.3), η = Z  β e − α ( ϑ xt ℑ ′ ( ϑ ) + ϑ t ϑ x ℑ ′ ′ ( ϑ )) ( ϑ x ℑ ′ ( ϑ )) 2 + ϑ tt ϑ x − ϑ t ϑ xt ϑ 2 x − ( β e − α ) ′ ϑ x ℑ ′ ( ϑ ) − e α ℑ ( ϑ )  dx + α ′ xy − β y + ( α ′ ′ − ( α ′ ) 2 ) y 2 − x 2 2 − 1 2  β e − α ϑ x ℑ ′ ( ϑ ) − ϑ t ϑ x  2 . (4 . 40) Theorem 4.3 . L et α, β b e functions of t and let ℑ b e a one-variable function. Supp ose that ϑ and ζ ar e functions in t, x, y . The fol lowing i s a solution of the e quations (1.1)-(1.4) of dynamic c onve ction in a se a: u = β e − α ϑ x ℑ ′ ( ϑ ) − ϑ t ϑ x , v = e α ℑ ( ϑ ) + x − α ′ y , (4 . 41) w =  α ′ + β e − α ( ϑ xx ℑ ′ ( ϑ ) + ϑ 2 x ℑ ′ ′ ( ϑ )) ( ϑ x ℑ ′ ( ϑ )) 2 + ϑ xt ϑ x − ϑ t ϑ xx ϑ 2 x  z + ζ , ρ = 1 , (4 . 42) p = z + Z  β e − α ( ϑ xt ℑ ′ ( ϑ ) + ϑ t ϑ x ℑ ′ ′ ( ϑ )) ( ϑ x ℑ ′ ( ϑ )) 2 + ϑ tt ϑ x − ϑ t ϑ xt ϑ 2 x − ( β e − α ) ′ ϑ x ℑ ′ ( ϑ ) − e α ℑ ( ϑ )  dx + α ′ xy − β y + ( α ′ ′ − ( α ′ ) 2 ) y 2 − x 2 2 − 1 2  β e − α ϑ x ℑ ′ ( ϑ ) − ϑ t ϑ x  2 . (4 . 43) Finally , w e sup p ose that α, β are fun ctions of t and f , g are fun ctions of t, x, y that are linear homogeneous in x and y . Denote  = αx + β y . Assume u = f + β φ ( t,  ) , v = g − αφ ( t,  ) . (4 . 44) Then u y − v x = f y − g x + ( α 2 + β 2 ) φ  , u x + v y = f x + g y . (4 . 45) No w (4.4) b ecomes f y t − g xt + ( α 2 + β 2 ) ′ φ  + ( α 2 + β 2 )( φ  t + ( α ′ x + β ′ y + αf + β g ) φ   ) +( f x + g y )( f y − g x + 1 + ( α 2 + β 2 ) φ  ) = 0 . (4 . 46) In order to solve the ab o ve equation, w e assume g x = ϕ, f y = ϕ − 1 , (4 . 47) α ′ x + β ′ y + αf + β g = 0 (4 . 48) for some fun ction ϕ of t . The equation (4.48) is equiv alen t to: α ′ + αf x + ϕβ = 0 = ⇒ f x = − α ′ + ϕβ α , (4 . 49) β ′ + β g y + α ( ϕ − 1) = 0 = ⇒ g y = − β ′ + α ( ϕ − 1) β . (4 . 50 ) 11 No w (4.46) b ecomes φ  t −  α ′ + ϕβ α + β ′ + α ( ϕ − 1) β − ( α 2 + β 2 ) ′ α 2 + β 2  φ  = 0 . (4 . 51) Th u s w e hav e the follo w ing s olution: φ = α + β α 2 + β 2 e R ( αβ − 1 ( ϕ − 1)+ α − 1 β ϕ ) dt ℑ ′ (  ) , (4 . 52) where ℑ is an arb itrary one-v ariable fun ction. Note u t + uu x + v u y + v = α ( α + β ) α 2 + β 2  αβ ′ − α ′ β α 2 + b 2 − 1  e R ( αβ − 1 ( ϕ − 1)+ α − 1 β ϕ ) dt ℑ ′ (  ) +  2( α ′ ) 2 + ( ϕβ ) 2 + 3 α ′ β ϕ − α ( ϕβ ) ′ − αα ′ ′ α 2 + ϕ 2  x +  ϕ ′ − ( ϕ − 1)( α ′ + ϕβ ) α − ϕ ( β ′ + α ( ϕ − 1)) β  y , (4 . 53) v t + uv x + v v y − u = β ( α + β ) α 2 + β 2  αβ ′ − α ′ β α 2 + b 2 − 1  e R ( αβ − 1 ( ϕ − 1)+ α − 1 β ϕ ) dt ℑ ′ (  ) + [( ϕ − 1) 2 − β (( ϕ − 1) α ) ′ + β β ′ ′ − 2( β ′ ) 2 − (( ϕ − 1) α ) 2 − 3 αβ ′ ( ϕ − 1) β 2 ] y +  ϕ ′ − ( ϕ − 1)( α ′ + ϕβ ) α − ϕ ( β ′ + α ( ϕ − 1)) β  x. (4 . 54) By (4.2) and (4.3), η = y 2 2  β (( ϕ − 1) α ) ′ + β β ′ ′ − 2( β ′ ) 2 − (( ϕ − 1) α ) 2 − 3 αβ ′ ( ϕ − 1) β 2 − ( ϕ − 1) 2  − x 2 2  2( α ′ ) 2 + ( ϕβ ) 2 + 3 α ′ β ϕ − α ( ϕβ ) ′ − αα ′ ′ α 2 + ϕ 2  + [ ( ϕ − 1)( α ′ + ϕβ ) α − ϕ ′ + ϕ ( β ′ + α ( ϕ − 1)) β ] xy + α + β α 2 + β 2  1 − αβ ′ − α ′ β α 2 + b 2  e R ( αβ − 1 ( ϕ − 1)+ α − 1 β ϕ ) dt ℑ (  ) . (4 . 55) Theorem 4.4 . L et α, β , ϕ b e func tions of t and let ℑ b e a one-variable function. Supp ose that ζ is func tions in t, x, y . The fol lowing is a solution of the e quations (1.1)-(1.4) of dynamic c onve ction in a se a: u = ( ϕ − 1) y − ( α ′ + ϕβ ) x α + β ( α + β ) α 2 + β 2 e R ( αβ − 1 ( ϕ − 1)+ α − 1 β ϕ ) dt ℑ ′ ( αx + β y ) , (4 . 56) v = ϕx − ( β ′ + ( ϕ − 1) α ) y β − α ( α + β ) α 2 + β 2 e R ( αβ − 1 ( ϕ − 1)+ α − 1 β ϕ ) dt ℑ ′ ( αx + β y ) , (4 . 57) w =  α ′ + ϕβ α + β ′ + ( ϕ − 1) α β  z + ζ , ρ = 1 , (4 . 58) p = z + y 2 2  β (( ϕ − 1) α ) ′ + β β ′ ′ − 2( β ′ ) 2 − (( ϕ − 1) α ) 2 − 3 αβ ′ ( ϕ − 1) β 2 − ( ϕ − 1) 2  − x 2 2  2( α ′ ) 2 + ( ϕβ ) 2 + 3 α ′ β ϕ − α ( ϕβ ) ′ − αα ′ ′ α 2 + ϕ 2  + xy [ ( ϕ − 1)( α ′ + ϕβ ) α − ϕ ′ + ϕ ( β ′ + α ( ϕ − 1)) β ] + α + β α 2 + β 2  1 − αβ ′ − α ′ β α 2 + b 2  e R ( αβ − 1 ( ϕ − 1)+ α − 1 β ϕ ) dt ℑ ( αx + β y ) . (4 . 59) 12 References [1] D. 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