Stable-Range Approach to Short Wave and Khokhlov-Zabolotskaya Equations

Stable-Range Approac h to Short W a v e and Khokhlo v-Zab olotsk a y a Equations 1 Xiaopi ng Xu Institute of Mathematics, Academ y of Mathematics & System Scienc es Chinese A cadem y of Sciences, Beijing 10 0190, P .R. China 2 Abstract Short wa ve equations w ere introd u ce d in connection with the nonlinear reflection of w eak sho c k w a v es. They also relate to the modu la tion of a gas-fluid mixture. Khokhlov- Zab olotsk a ya equation are used to describ e the p ropaga tion of a diffraction sound b eam in a n o nlinear medium. W e give a new algebraic metho d of solving these equations b y using certain finite-dimensional stable r ange of the n onlinea r terms and obtain large families of n ew explicit exact solutions parameterized by s everal functions for them. These p arameter fu nctio ns enable one to find the solutions of some related practical mo dels a nd b oundary v alue problems. 1 In tro duction Khristiano vic h and R iz ho v [8] (1958) disco v ered the equations of short w av es in connection with the nonlinear reflec tion of weak sho c k w a v es. The equations are mathematically equiv alen t to the following equation of their p oten tial function u for the v elo city vec tor: 2 u tx − 2( x + u x ) u xx + u y y + 2 k u x = 0 , (1 . 1) where k is a real constant. F or conv enience, w e call the ab o v e equation “t he short w a v e equation”. The symmetry group and conserv ation laws of (1.1) w ere first studied by Kuc harczyk [14] (1965 ) and later b y Khamitov a [6] (1982). Bagdo ev and P etrosyan [2] (1985) sho w ed that the mo dulation equation of a gas-fluid mixture coincides in main orders with the corresp onding short-wa ve equation. Roy , R oy and D e [23] (1988) fo und a lo op algebra in the Lie symmetries for the short-w a v e equation. Kraenke l, Manna and Merle [13] (2 000) studied nonlinear short-w a v e propagatio n in ferrites and Ermak ov [3] (2006) inv estigated short-w a v e inte raction in film slic ks. 1 2000 Mathematica l Sub ject Classifica tion. Pr imary 35C05, 35Q35; Seconda r y 35C10, 35C15. 2 Research suppo rted b y China NS F 1 0871193 1 Khokhlo v and Zab olotsk a y a [7] (19 69) found t he equation 2 u tx + ( uu x ) x − u y y = 0 . (1 . 2) for quasi-plane wa v es in nonlinear acoustics of b ounded bundles. More sp ecifically , the equation describ es t he propagation of a diffraction sound b eam in a nonlinear medium (cf. [4], [20]). Kup ershmidt [15] (1994) constructed a geometric Hamiltonian form for the Khokhlo v-Zab olotsk a y a equation (1.2 ). Certain group-inv ar ia n t solutions o f (1.2) w ere found b y Korsunskii [1 2 ] (1991), and b y Lin and Zhang [16] (1995). The three-dimensional generalization 2 u tx + ( uu x ) x − u y y − u z z = 0 (1 . 3) and its symm etries w ere studied b y Krasil’shc hik, Lyc hagin and Vinogrado v [17] (1986) and b y Sc h w arz [25] (1987). Martinez-Moras and Ra mo s [18] (1993) sho w ed that the higher dimensional classical W-algebras a r e the P oisson s tructures associated w ith a higher dimens ional v ersion of the Kho k hlo v-Zab olotsk a y a hierarc hy . Kacdry a vtsev and Sap ozknik ov [9] (199 8) found the symmetries fo r a generalized Khokhlo v-Zab olotsk a y a equation. Sanc hez [24] (20 07) studied long w a v es in ferromagnetic media via K hokhlo v- Zab olotsk a y a equation. Morozo v [19] (200 8 ) deriv ed t w o non-equiv alen t co v erings fo r the mo dified Khokhlo v-Zab olotsk a y a equation fr om Maurer-Cartan forms of its s ym- metry pseudo-gr o up. Rozano v a [21, 22] (2007 , 2008) studied closely related Khokhlo v- Zab olotsk a y a-Kuzen tso v equation from a na ly tic p oin t of view. Ko stin a nd P a na s enk o [1 1 ] (2008) in v estigated nonlinear acoustics in heterogeneous media via Kho khlov- Zab olotsk ay a- Kuzen t s o v-t yp e equation. All the ab o v e equations are similar nonlinear algebraic partial differen tial equations. Observ e that the nonlinear terms in the a bov e equations k eep some finite-dimensional p olynomial space in x stable. In t his pap er, w e presen t a new algebraic method of solving these equations b y using this stabilit y . W e obtain a family of solutions o f the equation (1.1) with k = 1 / 2 , 2, whic h blo w up on a mov ing line y = f ( t ). They ma y reflect partial phenomena of g us t. Moreo v er, w e obtain another family of smo oth solutions parameterized b y six smo oth functions of t for any k . Similar results for the equation (1.2) are also giv en. F urthermore, we find a family o f solutions of the equation (1.3) blo wing up on a rotat ing and t r a ns lating plane cos α ( t ) y + sin α ( t ) z = f ( t ), whic h ma y reflect partial phenomena of sound sho c k, and a family of solutions parameterized b y time-dep ende n t harmonic functions in y and z , whose sp ecial cases are smo oth solutions. Since our solutions con tain parameter functions, they can b e used to solv e certain related practical mo dels and b oundary-v alue problems for these equations. On the list of the Lie po int symmetries of the equation (1.1) in the w orks o f Kuc harczyk [14] and of Khamito v a [6] (e.g. cf. P age 301 in [5]), the most sophistic ated o nes are those with r e sp ect to the following v ector fields: X 1 = − α ′ y ∂ x + α∂ y +  xy ( α ′ ′ + α ′ ) − y 3 3 ( α ′ ′ ′ + ( k + 1) α ′ ′ + k α ′ )  ∂ u , (1 . 4) X 2 = β ∂ x + [ y 2 ( β ′ ′ + ( k + 1) β ′ + k β ) − x ( β ′ + β )] ∂ u , (1 . 5) 2 where α and β are arbitrary functions of t . Among the know n Lie p o in t symmetries of the Khokhlo v-Zab olotsk a y a equation (1.2 ) in t he w o r ks of Vinogradov and V orob’ev [26], and of Sch w arz [25] (e.g. cf. P age 299 in [5]), the most in teresting o ne s are those with resp e ct to the follow ing v ector fields: X 3 = 1 2 α ′ y ∂ x + α∂ y − 1 2 α ′ ′ y ∂ u , (1 . 6) X 4 = β ∂ t + 2 β ′ x + β ′ ′ y 2 6 ∂ x + 2 3 β ′ y ∂ y − 4 β ′ u + 2 β ′ ′ x + β ′ ′ ′ y 2 6 ∂ u . (1 . 7) The symmetries of the three-dimensional Khokhlo v-Zab olotsk a y a equation (1.3) causing our at ten tion are those with respect to the v ector fields (e.g. cf. P age 301 in [5]): X 5 = 10 t 2 ∂ t + (4 tx + 3 y 2 + 3 z 2 ) ∂ x + 12 ty ∂ y + 12 tz ∂ z − (4 x + 16 tu ) ∂ u , (1 . 7) X 6 = 1 2 α ′ y ∂ x + α∂ z − 1 2 α ′ ′ y ∂ u , (1 . 8) X 7 = 1 2 β ′ z ∂ x + β ∂ y − 1 2 β ′ ′ z ∂ u . (1 . 9) W e find that the group-inv a r ia n t solutions with respect to the ab o ve v ector fields X 1 - X 7 are p olynomial in x . This motiv ates us to find more exact solutions o f t he equations (1.1)-(1.3) p olynomial in x . In Section 2, we solv e t he short-w a v e equation (1.1). Although the equation (1.2) can b e view ed as a sp ecial case of the equation (1.3), we first solve (1.2) in Section 3 f or simplicit y b ecause our appro a c h to (1 .3) in v o lv es time-dep enden t harmonic functions and sophisticated in tegrals. The exact solutions of the equation (1.3 ) will b e g iven Section 4. 2 Short W a v e Equation In this section, w e study solutions p olynomial in x for the short w a v e equation (1.1 ). By comparing the terms of hig hest degree in x , w e find that suc h a solution m ust b e of the form: u = f ( t, y ) + g ( t, y ) x + h ( t, y ) x 2 + ξ ( t, y ) x 3 , (2 . 1) where f ( t, y ) , g ( t, y ) , h ( t, y ) and ξ ( t, y ) are suitably-differentiable functions to b e deter- mined. Note u x = g + 2 hx + 3 ξ x 2 , u xx = 2 h + 6 ξ x, (2 . 2) u tx = g t + 2 h t x + 3 ξ t x 2 , u y y = f y y + g y y x + h y y x 2 + ξ y y x 3 , (2 . 3) No w (1 .1) b ecomes 2( g t + 2 h t x + 3 ξ t x 2 ) − 2( g + (2 h + 1) x + 3 ξ x 2 )(2 h + 6 ξ x ) + f y y + g y y x + h y y x 2 + ξ y y x 3 + 2 k ( g + 2 hx + 3 ξ x 2 ) = 0 , (2 . 4) whic h is equiv alen t t o the following systems of partial differential equations: ξ y y = 36 ξ 2 , (2 . 5) 3 h y y = 6 ξ (6 h + 2 − k ) − 6 ξ t , (2 . 6) g y y = 8 h 2 + 4(1 − k ) h + 12 ξ g − 4 h t , (2 . 7) f y y = 4 g h − 2 g t − 2 k g . (2 . 8) First w e observ e that ξ = 1 ( √ 6 y + β ( t )) 2 (2 . 9) is a solution of the equation (2.5) for any differentiable function β of t . Substituting (2.9) in to (2.6), w e get h y y = 12 β ′ ( t ) ( √ 6 y + β ( t )) 3 + 6(6 h + 2 − k ) ( √ 6 y + β ( t )) 2 . (2 . 10) Denote b y Z the ring of in tegers. W rite h ( t, y ) = X i ∈ Z a i ( t )( √ 6 y + β ( t )) i . (2 . 11) Then h y y = X i ∈ Z 6( i + 2 )( i + 1) a i +2 ( t )( √ 6 y + β ( t )) i . (2 . 12) Substituting (2.11) and (2.12) in to (2.10), w e obtain X i ∈ Z 6[( i + 2 )( i + 1) − 6] a i +2 ( t )( √ 6 y + β ( t )) i = 12 β ′ ( t ) ( √ 6 y + β ( t )) 3 + 6(2 − k ) ( √ 6 y + β ( t )) 2 . (2 . 1 3 ) So − 24 a − 1 ( t ) = 1 2 β ′ ( t ) , − 36 a 0 ( t ) = 6 (2 − k ) (2 . 14) and 6( i + 4 )( i − 1) a i +2 ( t ) = 0 , i 6 = − 2 , − 3 . (2 . 15) Th us h = α ( √ 6 y + β ) 2 − β ′ 2( √ 6 y + β ) + k − 2 6 + γ ( √ 6 y + β ) 3 , (2 . 16) where α and γ a re arbitrary differen tiable functions of t . Note h t = − 2 αβ ′ ( √ 6 y + β ) 3 + 2 α ′ + ( β ′ ) 2 2( √ 6 y + β ) 2 − β ′ ′ 2( √ 6 y + β ) + 3 γ β ′ ( √ 6 y + β ) 2 + γ ′ ( √ 6 y + β ) 3 (2 . 17) and h 2 = α 2 ( √ 6 y + β ) 4 − αβ ′ ( √ 6 y + β ) 3 + 3( β ′ ) 2 + 4( k − 2) α 12( √ 6 y + β ) 2 + (2 − k ) β ′ 6( √ 6 y + β ) + ( k − 2) 2 36 +2 αγ ( √ 6 y + β ) − β ′ γ ( √ 6 y + β ) 2 + ( k − 2) γ 3 ( √ 6 y + β ) 3 + γ 2 ( √ 6 y + β ) 6 . (2 . 18) Substituting the ab o v e tw o equations into (2.7), w e ha v e: g y y − 12 g ( √ 6 y + β ) 2 = 8 α 2 ( √ 6 y + β ) 4 − 4[( k + 1) α + 3 α ′ ] 3( √ 6 y + β ) 2 + 2(( k + 1) β ′ + 3 β ′ ′ ) 3( √ 6 y + β ) 4 + 2( k − 2)(1 − 2 k ) 9 + 16 αγ ( √ 6 y + β ) − 20 β ′ γ ( √ 6 y + β ) 2 − 4[( k + 1) γ + 3 γ ′ ] 3 ( √ 6 y + β ) 3 + 8 γ 2 ( √ 6 y + β ) 6 . (2 . 19) W rite g ( t, y ) = X i ∈ Z b i ( t )( √ 6 y + β ) i . (2 . 20) Then X i ∈ Z 6[( i + 2)( i + 1) − 2] b i +2 ( t )( √ 6 y + β ) i = 8 α 2 ( √ 6 y + β ) 4 − 4[( k + 1) α + 3 α ′ ] 3( √ 6 y + β ) 2 + 2[( k + 1) β ′ + 3 β ′ ′ ] 3( √ 6 y + β ) + 2( k − 2)(1 − 2 k ) 3 + 16 αγ ( √ 6 y + β ) − 20 β ′ γ ( √ 6 y + β ) 2 − 4[( k + 1) γ + 3 γ ′ ] 3 ( √ 6 y + β ) 3 + 8 γ 2 ( √ 6 y + β ) 6 . (2 . 21) Comparing the constan t terms, we get k = 1 / 2 , 2. Moreo ver, the coefficien ts of the other terms giv e b − 2 = α 2 3 , b 0 = ( k + 1) α + 3 α ′ 9 , b 1 = − ( k + 1) β ′ + 3 β ′ ′ 18 , (2 . 22) b 3 = 2 αγ 3 , b 4 = − β ′ γ 3 , b 5 = − ( k + 1) γ + 3 γ ′ 81 , b 8 = 2 γ 2 81 (2 . 23) and ( i + 3) ib i +2 = 0 for i 6 = − 4 , − 2 , − 1 , 1 , 2 , 3 , 6 . (2 . 24) Therefore g = α 2 3( √ 6 y + β ) 2 + σ √ 6 y + β + ( k + 1) α + 3 α ′ 9 − ( k + 1) β ′ + 3 β ′ ′ 18 ( √ 6 y + β ) + ρ ( √ 6 y + β ) 2 + 2 αγ 3 ( √ 6 y + β ) 3 − β ′ γ 3 ( √ 6 y + β ) 4 − ( k + 1) γ + 3 γ ′ 81 ( √ 6 y + β ) 5 + 2 γ 2 81 ( √ 6 y + β ) 8 , (2 . 25) where σ and ρ are arbitra r y differen tiable functions of t . Observ e that g t = − 2 α 2 β ′ 3( √ 6 y + β ) 3 + (2 αα ′ − 3 σ β ′ ) 3( √ 6 y + β ) 2 + σ ′ √ 6 y + β + ( k + 1)(2 α ′ − ( β ′ ) 2 ) 18 + 2 α ′ ′ − β ′ β ′ ′ 6 + 36 β ′ ρ − ( k + 1) β ′ ′ − 3 β ′ ′ ′ 18 ( √ 6 y + β ) + ( ρ ′ + 2 αβ ′ γ )( √ 6 y + β ) 2 + 2 αγ ′ + 2 α ′ γ − 4( β ′ ) 2 γ 3 ( √ 6 y + β ) 3 − 42 β ′ γ ′ + 5( k + 1) β ′ γ + 27 β ′ ′ γ 81 ( √ 6 y + β ) 4 − ( k + 1) γ ′ + 3 γ ′ ′ 81 ( √ 6 y + β ) 5 + 16 β ′ γ 2 81 ( √ 6 y + β ) 7 + 4 γ γ ′ 81 ( √ 6 y + β ) 8 , (2 . 26) 5 g h = α 3 3( √ 6 y + β ) 4 + 6 ασ − α 2 β ′ 6( √ 6 y + β ) 3 + k α 2 + 2 αα ′ − 3 β ′ σ 6( √ 6 y + β ) 2 + 3( k − 2) σ − 3 αβ ′ ′ − 2( k + 1) αβ ′ − 3 α ′ β ′ 18( √ 6 y + β ) + αρ + ( k + 1)( β ′ ) 2 + 3 β ′ β ′ ′ 36 + ( k − 2)(( k + 1) α + 3 α ′ ) 54 + [ α 2 γ − β ′ ρ 2 − ( k − 2)(( k + 1) β ′ + 3 β ′ ′ ) 108 ]( √ 6 y + β ) + ( k − 2) ρ + 6 γ σ − 4 α β ′ γ 6 ( √ 6 y + β ) 2 + [ ( β ′ ) 2 γ 6 + (17 k − 1 0) αγ − 3 αγ ′ 81 + α ′ γ 3 ]( √ 6 y + β ) 3 + (10 − 1 7 k ) β ′ γ + 3 β ′ γ ′ − 27 β ′ ′ γ 162 ( √ 6 y + β ) 4 +[ γ ρ − ( k − 2)(( k + 1) γ + 3 γ ′ ) 486 ]( √ 6 y + β ) 5 + 56 αγ 2 81 ( √ 6 y + β ) 6 − 28 β ′ γ 2 81 ( √ 6 y + β ) 7 + ( k − 2) γ 2 243 ( √ 6 y + β ) 8 + 2 γ 3 81 ( √ 6 y + β ) 11 . (2 . 27) Substituting (2.25), (2.26) and (2.27) in to (2.8), we obta in f y y = 4 α 3 3( √ 6 y + β ) 4 − 2[6( k + 1) σ + 3 α β ′ ′ + 2( k + 1) α β ′ + 3 α ′ β ′ + 9 σ ′ ] 9( √ 6 y + β ) + 12 ασ + 2 α 2 β ′ 3( √ 6 y + β ) 3 + 4 αρ + 2[ β ′ β ′ ′ − ( k + 1) α ′ − α ′ ′ ] 3 + 2( k + 1)( β ′ ) 2 9 − 4( k + 1) 2 α 27 + [4 α 2 γ − 6 β ′ ρ + β ′ ′ ′ + ( k + 1) β ′ ′ 3 + 2( k + 1) 2 β ′ 27 ]( √ 6 y + β ) +[4 γ σ − 2 ρ ′ − 4( k + 1) ρ + 20 α β ′ γ 3 ]( √ 6 y + β ) 2 + [ − 40(( k + 1) αγ + 3 α γ ′ ) 81 + 10( β ′ ) 2 γ 3 ]( √ 6 y + β ) 3 + 10( k + 1) β ′ γ + 30 β ′ γ ′ 27 ( √ 6 y + β ) 4 + [4 γ ρ + 4( k + 1) 2 γ 243 + 2( k + 1) γ ′ + 2 γ ′ ′ 27 ]( √ 6 y + β ) 5 + 224 αγ 2 81 ( √ 6 y + β ) 6 − 16 β ′ γ 2 9 ( √ 6 y + β ) 7 − 8 γ (( k + 1 ) γ + 3 γ ′ ) 243 ( √ 6 y + β ) 8 + 8 γ 3 81 ( √ 6 y + β ) 11 . (2 . 28) Th us f = α 3 27( √ 6 y + β ) 2 + 6 ασ + α 2 β ′ 18( √ 6 y + β ) + θ + ϑy + 2 αρy 2 + ( k + 1)( β ′ ) 2 9 y 2 − 6( k + 1) σ + 3 αβ ′ ′ + 2( k + 1) α β ′ + 3 α ′ β ′ + 9 σ ′ 27 ( √ 6 y + β )[ln( √ 6 y + β ) − 1] + β ′ β ′ ′ − ( k + 1) α ′ − α ′ ′ 3 y 2 − 2( k + 1) 2 α 27 y 2 + [ α 2 γ 9 − β ′ ρ 6 + β ′ ′ ′ + ( k + 1) β ′ ′ 108 + ( k + 1) 2 β ′ 486 ]( √ 6 y + β ) 3 + [ 2 γ σ − ρ ′ 36 − ( k + 1) ρ + 5 αβ ′ γ 54 ]( √ 6 y + β ) 4 +[ ( β ′ ) 2 γ 36 − ( k + 1) α γ + 3 αγ ′ 243 ]( √ 6 y + β ) 5 + ( k + 1) β ′ γ + 3 β ′ γ ′ 486 ( √ 6 y + β ) 6 +[ γ ρ 63 + ( k + 1) 2 γ 15309 + ( k + 1) γ ′ + γ ′ ′ 3402 ]( √ 6 y + β ) 7 + 2 αγ 2 243 ( √ 6 y + β ) 8 − β ′ γ 2 243 ( √ 6 y + β ) 9 − 2 γ (( k + 1) γ + 3 γ ′ ) 32805 ( √ 6 y + β ) 10 + γ 3 9477 ( √ 6 y + β ) 13 , (2 . 29) where θ and ϑ are arbitrary functions of t . 6 Theorem 2.1 . When k = 1 / 2 , 2 , we ha v e the fol lowing solution of the e quation ( 1 .1) blowing up on the surfac e √ 6 y + β ( t ) = 0 : u = x 3 ( √ 6 y + β ) 2 + [ α ( √ 6 y + β ) 2 − β ′ 2( √ 6 y + β ) + k − 2 6 + γ ( √ 6 y + β ) 3 ] x 2 +[ α 2 3( √ 6 y + β ) 2 + σ √ 6 y + β + ( k + 1) α + 3 α ′ 9 − ( k + 1) β ′ + 3 β ′ ′ 18 ( √ 6 y + β ) + ρ ( √ 6 y + β ) 2 + 2 αγ 3 ( √ 6 y + β ) 3 − β ′ γ 3 ( √ 6 y + β ) 4 − ( k + 1) γ + 3 γ ′ 81 ( √ 6 y + β ) 5 + 2 γ 2 81 ( √ 6 y + β ) 8 ] x + α 3 27( √ 6 y + β ) 2 + 6 ασ + α 2 β ′ 18( √ 6 y + β ) + 2 αρy 2 + ( k + 1)( β ′ ) 2 9 y 2 − 6( k + 1) σ + 3 αβ ′ ′ + 2( k + 1) α β ′ + 3 α ′ β ′ + 9 σ ′ 27 ( √ 6 y + β )[ln( √ 6 y + β ) − 1] + β ′ β ′ ′ − ( k + 1) α ′ − α ′ ′ 3 y 2 − 2( k + 1) 2 α 27 y 2 + [ α 2 γ 9 − β ′ ρ 6 + β ′ ′ ′ + ( k + 1) β ′ ′ 108 + ( k + 1) 2 β ′ 486 ]( √ 6 y + β ) 3 + [ 2 γ σ − ρ ′ 36 − ( k + 1) ρ + 5 α β ′ γ 54 ]( √ 6 y + β ) 4 + θ + ϑy +[ ( β ′ ) 2 γ 36 − ( k + 1) α γ + 3 αγ ′ 243 ]( √ 6 y + β ) 5 + ( k + 1) β ′ γ + 3 β ′ γ ′ 486 ( √ 6 y + β ) 6 +[ γ ρ 63 + ( k + 1) 2 γ 15309 + ( k + 1) γ ′ + γ ′ ′ 3402 ]( √ 6 y + β ) 7 + 2 αγ 2 243 ( √ 6 y + β ) 8 − β ′ γ 2 243 ( √ 6 y + β ) 9 − 2 γ (( k + 1) γ + 3 γ ′ ) 32805 ( √ 6 y + β ) 10 + γ 3 9477 ( √ 6 y + β ) 13 , (2 . 30) wher e α , β , γ , σ, ρ, θ a nd ϑ ar e arbitr ary function s of t , whose derivative s a p p e ar e d in the ab ove exist in a c ertain op en set of R . When α = γ = σ = ρ = θ = ϑ = 0, the ab o ve solution b ecomes u = x 3 ( √ 6 y + β ) 2 + [ k − 2 6 − β ′ 2( √ 6 y + β ) ] x 2 − ( k + 1) β ′ + 3 β ′ ′ 18 ( √ 6 y + β ) x + ( k + 1)( β ′ ) 2 9 y 2 + β ′ β ′ ′ 3 y 2 + [ β ′ ′ ′ + ( k + 1) β ′ ′ 108 + ( k + 1) 2 β ′ 486 ]( √ 6 y + β ) 3 . (2 . 31) T ak e the t rivial solution ξ = 0 of (2.5), whic h is the only solution p olynomial in y . Then (2.6) and (2.7) b ecome h y y = 0 , g y y = 8 h 2 + 4(1 − k ) h − 4 h t , (2 . 32) Th us h = α ( t ) + β ( t ) y . (2 . 33) Hence g y y = 4(2 α 2 + (1 − k ) α − α ′ ) + 4 (4 αβ + (1 − k ) β − β ′ ) y + 8 β 2 y 2 . (2 . 34) So g = γ + σ y + 2(2 α 2 + (1 − k ) α − α ′ ) y 2 + 2 3 (4 αβ + (1 − k ) β − β ′ ) y 3 + 2 3 β 2 y 4 , (2 . 35) 7 where γ a nd σ a re arbitrary functions o f t . Now (2.8) yields f y y = 4( α + β y )[ γ + σ y + 2(2 α 2 + (1 − k ) α − α ′ ) y 2 + 2 3 (4 αβ + (1 − k ) β − β ′ ) y 3 + 2 3 β 2 y 4 ] − 2[ γ ′ + σ ′ y + 2(4 α α ′ + (1 − k ) α ′ − α ′ ′ ) y 2 + 2 3 (4 α ′ β + 4 α β ′ +(1 − k ) β ′ − β ′ ′ ) y 3 + 4 3 β β ′ y 4 ] − 2 k [ γ + σ y + 2(2 α 2 + (1 − k ) α − α ′ ) y 2 + 2 3 (4 αβ + (1 − k ) β − β ′ ) y 3 + 2 3 β 2 y 4 ] = 4 αγ − 2 γ ′ − 2 k γ + 2(2 ασ + 2 β γ − σ ′ − k γ ) y + 8((1 − k ) α − α ′ ) β y 3 +4(4 α 3 + 2(1 − 2 k ) α 2 − 6 αα ′ + k ( k − 1) α + (2 k − 1 ) α ′ + α ′ ′ + β σ ) y 2 + 4 3 (20 α 2 β + 2(1 − 3 k ) αβ − 6 α β ′ − 4 α ′ β + (2 k − 1) β ′ + β ′ ′ − k (1 − k ) β ) y 3 + 4 3 (10 αβ 2 + (2 − 3 k ) β 2 − 4 β β ′ ) y 4 + 8 3 β 3 y 5 . (2 . 36) Therefore, f = (2 αγ − γ ′ − k γ ) y 2 + 2 ασ + 2 β γ − σ ′ − k γ 3 y 3 + 2((1 − k ) α − α ′ ) 5 y 5 + τ + ρy + 1 3 (4 α 3 + 2(1 − 2 k ) α 2 − 6 αα ′ + k ( k − 1) α + (2 k − 1 ) α ′ + α ′ ′ + β σ ) y 4 + 1 15 (20 α 2 β + 2(1 − 3 k ) αβ − 6 α β ′ − 4 α ′ β + (2 k − 1) β ′ + β ′ ′ − k (1 − k ) β ) y 5 + 2 45 (10 αβ 2 + (2 − 3 k ) β 2 − 4 β β ′ ) y 6 + 4 β 3 63 y 7 . (2 . 37) Theorem 2.2 . The fol lo wing is a solution of the e quation (1 . 1 ): u = ( α + β y ) x 2 + [ γ + σ y + 2(2 α 2 + (1 − k ) α − α ′ ) y 2 + 2 3 (4 αβ + (1 − k ) β − β ′ ) y 3 + 2 3 β 2 y 4 ] x + (2 α γ − γ ′ − k γ ) y 2 + 2 ασ + 2 β γ − σ ′ − k γ 3 y 3 + 2((1 − k ) α − α ′ ) 5 y 5 + τ + ρy + 1 3 (4 α 3 + 2(1 − 2 k ) α 2 − 6 α α ′ + k ( k − 1) α + (2 k − 1 ) α ′ + α ′ ′ + β σ ) y 4 + 1 15 (20 α 2 β + 2(1 − 3 k ) αβ − 6 αβ ′ − 4 α ′ β + (2 k − 1) β ′ + β ′ ′ − k (1 − k ) β ) y 5 + 2 45 (10 αβ 2 + (2 − 3 k ) β 2 − 4 β β ′ ) y 6 + 4 β 3 63 y 7 , (2 . 38) wher e α, β , γ , σ, ρ and τ ar e arbitr ary functions of t , whose deri v a t ives app e ar e d in the ab ove exist in a c ertain o p en s e t of R . Mor e over, any solution p olynomia l in x and y of (1.1) must b e of the ab ove form. Th e ab ove solution is smo oth (an alyt ic) if a l l α, β , γ , σ, ρ and τ ar e smo oth (an a lyt ic) functions of t . Remark 2.3 . In addition to the nonzero solution (2.9) of the equation (2.5), the other nonzero solutions are of the form ξ = ℘ ι ( √ 6 y + β ( t )) , (2 . 39) 8 where ℘ ι ( w ) is the W eierstrass’s elliptic function suc h that ℘ ′ ι ( w ) 2 = 4( ℘ ι ( w ) 3 − ι ) , (2 . 40) and ι is a nonzero constan t and β is any function of t . When β is not a constan t, the solutions of ( 2 .6)-(2.8) are extremely complicated. If β is constan t, w e can ta k e β = 0 b y adjusting ι . An y solution of (2.6)- (2.8) with h 6 = 0 is also v ery complicated. Th us the only simple solutio n of the equation (1.1) in this case is u = ℘ ι ( √ 6 y ) x 3 . (2 . 41) 3 2-D Khokhlo v-Zab olots k a y a Equ a tion The solution of the equation (1.2) p olynomial in x mus t b e of the form u = f ( t, y ) + g ( t, y ) x + ξ ( t, y ) x 2 . (3 . 1) Then u x = g + 2 ξ x, u tx = g t + 2 ξ t x, u y y = f y y + g y y x + ξ y y x 2 , (3 . 2) ( uu x ) x = ∂ x ( f g + ( g 2 + 2 f ξ ) x + 3 g ξ x 2 + 2 ξ 2 x 3 ) = g 2 + 2 f ξ + 6 g ξ x + 6 ξ 2 x 2 . (3 . 3) Substituting them in to (1.2), w e get 2( g t + 2 ξ t x ) + g 2 + 2 f ξ + 6 g ξ x + 6 ξ 2 − f y y − g y y x − ξ y y x 2 = 0 , (3 . 4) equiv alen t ly , ξ y y = 6 ξ 2 , (3 . 5) g y y − 6 g ξ = 4 ξ t , (3 . 6) f y y − 2 f ξ = 2 g t + g 2 . (3 . 7) First w e observ e that ξ = 1 ( y + β ( t )) 2 (3 . 8) is a solution of the equation (3.5) for any differentiable function β of t . Substituting (3.8) in to (3.6), w e obtain g y y − 6 g y + β ( t )) 2 = − 8 β ′ ( t ) ( y + β ( t )) 3 . (3 . 9) W rite g ( t, y ) = X i ∈ Z a i ( t )( y + β ( t )) i . (3 . 10) Then (3.9) b ec omes X i ∈ Z [( i + 2)( i + 1) − 6] a i +2 ( t )( y + β ( t )) i = − 8 β ′ ( t ) ( y + β ( t )) 3 . (3 . 11) Th us a − 1 = 2 β ′ , ( i + 4)( i − 1) a i +2 = 0 for i 6 = − 3 . (3 . 12) 9 Hence g = α ( t ) ( y + β ( t )) 2 + 2 β ′ ( t ) y + β ( t ) + γ ( t )( y + β ( y )) 3 , (3 . 13) where α and γ a re arbitrary differen tiable functions of t . Note g t = − 2 αβ ′ ( y + β ) 3 + α ′ − 2( β ′ ) 2 ( y + β ) 2 + 2 β ′ ′ y + β + 3 γ β ′ ( y + β ) 2 + γ ′ ( √ 3 y + β ) 3 (3 . 14) and g 2 = α 2 ( y + β ) 4 + 4 αβ ′ ( y + β ) 3 + 4( β ′ ) 2 ( y + β ) 2 + 2 αγ ( y + β ) + 4 γ β ′ ( y + β ) 2 + γ 2 ( y + β ) 6 . (3 . 15 ) Substituting the ab o v e tw o equations into (3.7), w e ha v e: f y y − 2 f ( y + β ) 2 = α 2 ( y + β ) 4 + 2 α ′ ( y + β ) 2 + 4 β ′ ′ y + β + 2 αγ ( y + β ) +10 γ β ′ ( y + β ) 2 + 2 γ ′ ( y + β ) 3 + γ 2 ( y + β ) 6 . (3 . 16) W rite f ( t, y ) = X i ∈ Z b i ( t )( y + β ) i . (3 . 17) Then (3.16) b ecome s X i ∈ Z [( i + 2) ( i + 1) − 2] b i +2 ( y + β ) i = α 2 ( y + β ) 4 + 2 α ′ ( y + β ) 2 + 4 β ′ ′ y + β +2 αγ ( y + β ) + 10 β ′ γ ( y + β ) 2 + 2 γ ′ ( y + β ) 3 + γ 2 ( y + β ) 6 . (3 . 18) Th us b − 2 = α 2 4 , b 0 = − α ′ , b 1 = − 2 β ′ ′ , b 3 = αγ 2 , (3 . 19) b 4 = β ′ γ , b 5 = γ ′ 9 , b 8 = γ 2 54 , ( 3 . 20) ( i + 3) ib i +2 = 0 for i 6 = − 4 , − 2 , − 1 , 1 , 2 , 3 , 6 . (3 . 21) Therefore, f = α 2 4( y + β ) 2 + σ y + β − α ′ − 2 β ′ ′ ( y + β ) + ρ ( y + be ) 2 + αγ 2 ( y + β ) 3 + β ′ γ ( y + β ) 4 + γ ′ 9 ( y + β ) 5 + γ 2 54 ( y + β ) 8 , (3 . 22) where σ and ρ are arbitra r y functions of t . Theorem 3.1 . We have the fol lowing so lut ion of the e quation (1.1) b l o wing up on the surfac e y + β ( t ) = 0 : u = x 2 ( y + β ) 2 + αx ( y + β ) 2 + 2 β ′ x y + β + γ ( y + β ) 3 x + α 2 4( y + β ) 2 + σ y + β − α ′ − 2 β ′ ′ ( y + β ) + ρ ( y + be ) 2 + αγ 2 ( y + β ) 3 + β ′ γ ( y + β ) 4 + γ ′ 9 ( y + β ) 5 + γ 2 54 ( y + β ) 8 , (3 . 23) 10 wher e α, β , γ , σ and ρ ar e arbitr ary functions of t , who s e derivatives app e ar e d in the ab o v e exist in a c e rt ain op en set of R . When α = γ = σ = ρ = 0, the ab o v e solution b ecomes u = x 2 ( y + β ) 2 + 2 β ′ x y + β − 2 β ′ ′ ( y + β ) . (3 . 24) T ak e the t rivial solution ξ = 0 of (3.5), whic h is the only solution p olynomial in y . Then (3.6) and (3.7) b ecome g y y = 0 , f y y = 2 g t + g 2 . (3 . 25) Th us g = α ( t ) + β ( t ) y . (3 . 26) Hence f y y = α 2 + 2 α ′ + 2( β ′ + αβ ) y + β 2 y 2 . (3 . 27) So f = γ + σ y + α 2 + 2 α ′ 2 y 2 + β ′ + αβ 3 y 3 + β 2 12 y 4 , (3 . 28) where γ a nd σ a re arbitrary functions o f t . Theorem 3.2 . The fol lo wing is a solution of the e quation (1 . 2 ): u = ( α + β y ) x + γ + σ y + α 2 + 2 α ′ 2 y 2 + β ′ + αβ 3 y 3 + β 2 12 y 4 , ( 3 . 29) wher e α, β , γ and σ ar e arbitr ary functions of t , whose de riv atives app e ar e d in the ab ove exist in a c ertain op en set of R . Mor e over, any so lut ion p olynomi a l in x and y of ( 1.2) must b e of the ab ove form. Th e ab ove solution is smo oth (analytic) if al l α, β , γ and σ ar e smo oth (analytic) functions of t . Remark 3.3 . In addition to the solutions in Theorems 3.1 and 3.2, the equation (1.2) has the fo llo wing simple solution: u = ℘ ι ( y ) x 2 , (3 . 30) where ℘ ι ( w ) is the W eierstrass’s elliptic function satisfying (2.40). 4 3-D Khokhlo v-Zab olots k a y a Equ a tion By comparing the terms of hig he st degree, w e find that a solution p olynomial in x of the equation (1.3) must b e of the fo rm: u = f ( t, y , z ) + g ( t, y , z ) x + ξ ( t, y , z ) x 2 , (4 . 1) 11 where f ( t, y , z ) , g ( t, y , z ) a nd ξ ( t, y , z ) a re suitably-differentiable functions to b e deter- mined. As (3.2)- ( 3 .7), the equation (1.3) is equiv alen t to: ξ y y + ξ z z = 6 ξ 2 , (4 . 2) g y y + g z z − 6 g ξ = 4 ξ t , (4 . 3) f y y + f z z − 2 f ξ = 2 g t + g 2 . (4 . 4) First w e observ e that ξ = 1 ( y cos α ( t ) + z sin α ( t ) + β ( t )) 2 (4 . 5) is a solution of the equation (4.2), where α and β are suitable differen tiable functions of t . With the ab o v e ξ , (4.3) b ecomes g y y + g z z − 6 g ( y cos α ( t ) + z sin α ( t ) + β ( t )) 2 = − 8( α ′ ( − y sin α + z cos α ) + β ′ ) ( y cos α + z sin α + β ) 3 . (4 . 6) In order to solve (4.6), w e c hange v ariables: ζ = cos α y + sin α z + β , η = − sin α y + cos α z . (4 . 7) Then ∂ y = cos α ∂ ζ − sin α ∂ η , ∂ z = sin α ∂ ζ + cos α ∂ η . ( 4 . 8) Th us ∂ 2 y + ∂ 2 z = (cos α ∂ ζ − sin α ∂ η ) 2 + (sin α ∂ ζ + cos α ∂ η ) 2 = ∂ 2 ζ + ∂ 2 η . (4 . 9) Note ∂ t ( ζ ) = α ′ η + β ′ , ∂ t ( η ) = α ′ ( β − ζ ) . (4 . 10) The equation (4.6) can b e rewritten as: g ζ ζ + g ηη − 6 ζ − 2 g = − 8( α ′ η + β ′ ) ζ − 3 . (4 . 11) In order to solve the ab o v e equation, we assume g = X i ∈ Z a i ( t, η ) ζ i . (4 . 12) No w (4 .11) b ecomes X i ∈ Z [(( i + 2 ) ( i + 1) − 6) a i +2 + a iηη ] = − 8( α ′ η + β ′ ) ζ − 3 , (4 . 13) whic h is equiv alen t t o − 4 a − 1 + a − 3 ηη = − 8( α ′ η + β ′ ) , ( i + 4 )( i − 1) a i +2 + a iηη = 0 for − 3 6 = i ∈ Z . (4 . 14) Hence a − 1 = 1 4 a − 3 ηη + 2( α ′ η + β ′ ) , ( i + 4)( i − 1) a i +2 = − a iηη for − 3 6 = i ∈ Z . (4 . 15) 12 When i = − 4 and i = 1, w e get a − 4 ηη = a 1 ηη = 0. Moreo v er, a − 2 and a 3 can b e an y functions. T ak e a 3 = σ, a − 2 = ρ, a − 1 = 2( α ′ η + β ′ ) , (4 . 16) a 1 = a − 1 − 2 i = a − 2 − 2 i = 0 for 0 < i ∈ Z (4 . 17) in order to a v oid infinite num b er of negativ e p o wers of ζ in (4.12), where σ a nd ρ a re arbitrary functions of t and η differen tiable in a certain domain. By (4.1 5 ), a 3+2 k = ( − 1) k ∂ 2 k η ( σ ) Q k i =1 (2 i + 5 )(2 i ) = ( − 1) k 15 ∂ 2 k η ( σ ) (2 k + 5)(2 k + 3)(2 k + 1)! , (4 . 18) a − 2+2 k = ( − 1) k ∂ 2 k η ( ρ ) Q k i =1 (2 i )(2 i − 5) = ( − 1) k (2 k − 1)(2 k − 3 ) ∂ 2 k η ( ρ ) 3(2 k )! . (4 . 19) Therefore, g = 2( α ′ η + β ′ ) ζ − 1 + ∞ X k =0 ( − 1) k [ 15 ∂ 2 k η ( σ ) ζ 3 (2 k + 5)(2 k + 3)(2 k + 1)! + (2 k − 1)(2 k − 3) ∂ 2 k η ( ρ ) ζ − 2 3(2 k )! ] ζ 2 k (4 . 20) is a solution of (4.1 1). By ( 4 .9), (4.4) is equiv alen t to f ζ ζ + f ηη − 2 ζ − 2 f = 2 g t + g 2 . (4 . 21) Note g t = 2( α ′ ′ η + β ′ ′ + ( α ′ ) 2 β ) ζ − 1 − 2( α ′ ) 2 − 2( α ′ η + β ′ ) 2 ζ − 2 + ∞ X k =0 ( − 1) k ζ 2 k { × 15 ∂ 2 k η ( σ t + α ′ ( β − ζ ) σ η ) ζ 3 (2 k + 5)(2 k + 3)(2 k + 1)! + (2 k − 1) (2 k − 3) ∂ 2 k η ( ρ t + α ′ ( β − ζ ) ρ η ) ζ − 2 3(2 k )! ! +( α ′ η + β ′ )[ 15 ∂ 2 k η ( σ ) ζ 2 (2 k + 5)(2 k + 1)! + (2 k − 1) (2 k − 2)(2 k − 3) ∂ 2 k η ( ρ ) ζ − 3 3(2 k )! ] } . (4 . 22) F or con v enience of solving the equation (4 .21), w e denote 2 g t + g 2 = ∞ X i = − 4 b i ( t, η ) ζ i (4 . 23) b y (4.2 0) and (4.22). In particular, b − 4 = ρ 2 , b − 3 = 0 , (4 . 24) b − 2 = 2( ρ t + α ′ β ρ η ) + ρ ηη ρ 3 , (4 . 25) b − 1 = 4[ α ′ ′ η + β ′ ′ + ( α ′ ) 2 β ] − 2 α ′ ρ η + 2 3 ( α ′ η + β ′ ) ρ ηη , (4 . 26) 13 b 0 = − 4( α ′ ) 2 + 1 3 ( ρ tηη + α ′ β ρ ηη η ) + 1 12 ∂ 4 η ( ρ ) ρ + 1 36 ρ 2 ηη . (4 . 27) Supp ose that f = X i ∈ Z c i ( t, η ) ζ i (4 . 29) is a solution (4.21 ) . Then X i ∈ Z [(( i + 2)( i + 1) − 2) c i +2 + c iηη ] ζ i = ∞ X r = − 4 b r ζ r , (4 . 30) equiv alen t ly ( i + 3) ic i +2 = b i − c iηη , ( r + 3) r c r +2 = − c r ηη , r < − 4 ≤ i. (4 . 31) By t he ab ov e second equation , we take c r = 0 for r < − 4 (4 . 32) to av oid infinite n um b er of negativ e p o w ers of ζ in (4.2 9 ). L etting i = − 3 , 0, w e get b − 3 = c − 3 ηη , b 0 = c 0 ηη . (4 . 33) The first equation is naturally satisfied b ecause c − 3 = − c − 5 ηη / 10 = 0. T aking i = − 2 , − 4 and r = − 6 in (4.31), w e o btain c 0 = 1 2 c − 2 ηη − 1 2 b − 2 , c − 2 = 1 4 b − 4 . ( 4 . 34) So c 0 = 1 8 ∂ 2 η ( b − 4 ) − 1 2 b − 2 . (4 . 35) Th us we get a constrain t: b 0 = 1 8 ∂ 4 η ( b − 4 ) − 1 2 ∂ 2 η ( b − 2 ) , (4 . 36) equiv alen t ly , − 4( α ′ ) 2 + 1 3 ( ρ tηη + α ′ β ρ ηη η ) + 1 12 ∂ 4 η ( ρ ) ρ + 1 36 ρ 2 ηη = 1 8 ∂ 4 η ( ρ 2 ) − ρ tηη − α ′ β ρ ηη η − ∂ 2 η ( ρ ηη ρ ) 6 . (4 . 37) Th us 96( ρ tηη + α ′ β ρ ηη η ) + 6 ∂ 4 η ( ρ ) ρ + 2 ρ 2 ηη − 9 ∂ 4 η ( ρ 2 ) + 12 ∂ 2 η ( ρ ηη ρ ) = 288( α ′ ) 2 . (4 . 38) It can b e pro v ed b y considering the terms o f highest degree that an y solution of (4.38) p olynomial in η must b e of the fo rm ρ = γ 0 ( t ) + γ 1 ( t ) η + γ 2 ( t ) η 2 . (4 . 39) Then (4.38) b ecome s 6 γ ′ 2 − 5 γ 2 2 = 9( α ′ ) 2 . ( 4 . 40) 14 So α ′ = ǫ 3 q 6 γ ′ 2 − 5 γ 2 2 ⇒ α = ǫ 3 Z q 6 γ ′ 2 − 5 γ 2 2 dt, (4 . 41) where ǫ = ± 1. R e place β b y − β if necessary , w e can tak e ǫ = 1. Under the assumption (4.39), g = ρζ − 2 + 2( α ′ η + β ′ ) ζ − 1 + γ 2 6 + ∞ X k =0 ( − 1) k 15 ∂ 2 k η ( σ ) ζ 3+2 k (2 k + 5)(2 k + 3)(2 k + 1)! (4 . 42) and b − 2 = 2( ρ t + α ′ β ρ η ) + 2 3 γ 2 ρ, (4 . 43) b − 1 = 4[ α ′ ′ η + β ′ ′ + ( α ′ ) 2 β ] − 2 α ′ ρ η + 4 3 ( α ′ η + β ′ ) γ 2 , (4 . 44) b 0 = − 4( α ′ ) 2 + 2 3 γ ′ 2 + γ 2 2 9 . (4 . 45) Denote Ψ h β ,ρ,σ i ( t, η , ζ ) = ∞ X i =1 b i ζ i . (4 . 46) F or an y real f unction F ( t, η ) analytic at η = η 0 , w e define F ( t, η 0 + √ − 1 ζ ) = ∞ X r =0 ∂ r η ( F )( t, η 0 ) r ! ( √ − 1 ζ ) r . (4 . 47) Note ∞ X k =0 ( − 1) k 15 ∂ 2 k η ( σ ) ζ 3+2 k (2 k + 5)(2 k + 3)(2 k + 1)! = 15 ζ 2 Z ζ 0 ∞ X k =0 ( − 1) k ∂ 2 k η ( σ ) τ 2 k 1 (2 k + 5)(2 k + 3)(2 k )! ! dτ 1 = 15 Z ζ 0 τ 2 Z τ 2 0 ∞ X k =0 ( − 1) k ∂ 2 k η ( σ ) τ 2 k 1 (2 k + 5)(2 k )! ! dτ 1 dτ 2 = 15 ζ − 2 Z ζ 0 τ 3 Z τ 3 0 τ 2 Z τ 2 0 ∞ X k =0 ( − 1) k ∂ 2 k η ( σ ) τ 2 k 1 (2 k )! ! dτ 1 dτ 2 dτ 3 = 15 2 ζ − 2 Z ζ 0 τ 3 Z τ 3 0 τ 2 Z τ 2 0 [ σ ( t, η + √ − 1 τ 1 ) + σ ( t, η − √ − 1 τ 1 )] dτ 1 dτ 2 dτ 3 , (4 . 48) ∂ t " ∞ X k =0 ( − 1) k 15 ∂ 2 k η ( σ ) ζ 3+2 k (2 k + 5)(2 k + 3)(2 k + 1)! # = ∞ X k =0 ( − 1) k 15 ∂ 2 k η ( σ t ) ζ 3+2 k (2 k + 5)(2 k + 3)(2 k + 1)! + α ′ β ∞ X k =0 ( − 1) k 15 ∂ 2 k + 1 η ( σ ) ζ 3+2 k (2 k + 5)(2 k + 3)(2 k + 1)! − α ′ ∞ X k =0 ( − 1) k 15 ∂ 2 k + 1 η ( σ ) ζ 4+2 k (2 k + 5)(2 k + 3)(2 k + 1)! + ( α ′ η + β ′ ) ∞ X k =0 ( − 1) k 15 ∂ 2 k η ( σ ) ζ 2+2 k (2 k + 5)(2 k + 1)! = 15 2 ζ − 2 Z ζ 0 τ 3 Z τ 3 0 τ 2 Z τ 2 0 [ σ t ( t, η + √ − 1 τ 1 ) + σ t ( t, η − √ − 1 τ 1 )] dτ 1 dτ 2 dτ 3 + 15 α ′ ( ζ − β ) 2 ζ 2 √ − 1 ζ − 2 Z ζ 0 τ 2 Z τ 2 0 τ 1 [ σ ( t, η + √ − 1 τ 1 ) − σ ( t, η − √ − 1 τ 1 )] dτ 1 dτ 2 + 15 2 ( α ′ η + β ′ ) ζ − 3 Z ζ 0 τ 3 2 Z τ 2 0 [ σ ( t, η + √ − 1 τ 1 ) + σ ( t, η − √ − 1 τ 1 )] dτ 1 dτ 2 . (4 . 49) 15 Hence g = ρζ − 2 + 2( α ′ η + β ′ ) ζ − 1 + γ 2 6 + 15 2 ζ − 2 × Z ζ 0 τ 3 Z τ 3 0 τ 2 Z τ 2 0 [ σ ( t, η + √ − 1 τ 1 ) + σ ( t, η − √ − 1 τ 1 )] dτ 1 dτ 2 dτ 3 , (4 . 50) b y (4.4 2) and (4.48). According to (4.23) and (4.46), w e hav e Ψ h β ,ρ,σ i ( t, η , ζ ) = 225 4 ζ − 4  Z ζ 0 τ 3 Z τ 3 0 τ 2 Z τ 2 0 [ σ ( t, η + √ − 1 τ 1 ) + σ ( t, η − √ − 1 τ 1 )] dτ 1 dτ 2 dτ 3  2 +15 ζ − 2 Z ζ 0 τ 3 Z τ 3 0 τ 2 Z τ 2 0 [ σ t ( t, η + √ − 1 τ 1 ) + σ t ( t, η − √ − 1 τ 1 )] dτ 1 dτ 2 dτ 3 + 15 α ′ ( ζ − β ) ζ 2 √ − 1 Z ζ 0 τ 2 Z τ 2 0 τ 1 [ σ ( t, η + √ − 1 τ 1 ) − σ ( t, η − √ − 1 τ 1 )] dτ 1 dτ 2 +15( α ′ η + β ′ ) ζ − 3 Z ζ 0 τ 3 2 Z τ 2 0 [ σ ( t, η + √ − 1 τ 1 ) + σ ( t, η − √ − 1 τ 1 )] dτ 1 dτ 2 +15  Z ζ 0 τ 3 Z τ 3 0 τ 2 Z τ 2 0 [ σ ( t, η + √ − 1 τ 1 ) + σ ( t, η − √ − 1 τ 1 )] dτ 1 dτ 2 dτ 3  × ζ − 2  ρζ − 2 + ( α ′ η + β ′ ) ζ − 1 + γ 2 6  . (4 . 51) No w c − 2 = ρ 2 4 (4 . 52) b y (4.24) and (4.34). According to (4.31) with i = − 3 , 0, c − 1 and c 2 can b e arbitrary . F o r con v enience, w e redenote c − 1 = κ ( t, η ) , c 2 = ω ( t, η ) . (4 . 53) Moreo v er, (4.24), (4.35) and (4.43) imply c 0 = ρ 2 η 4 − ρ t − α ′ β ρ η + γ 2 ρ 6 . (4 . 54) F urthermore, (4.31) and (4.44) yield c 1 = κ ηη 2 − 2( α ′ ′ η + β ′ ′ + ( α ′ ) 2 β ) + α ′ ρ η − 2 3 ( α ′ η + β ′ ) γ 2 . (4 . 55) In addition, (4.31) and (4.5 3 ) ga v e c 2 k + 3 = ( − 1) k +1 ∂ 2 k + 4 η ( κ ) 2( k + 2)(2 k + 2)! + k X i =0 ( − 1) k − i ( i + 1) ( 2 i )! ( k + 2)(2 k + 2)! ∂ 2( k − i ) η ( b 2 i +1 ) , (4 . 56) c 2 k + 4 = ( − 1) k +1 3 ∂ 2 k + 2 η ( ω ) (2 k + 5)(2 k + 3)! + k X i =0 ( − 1) k − i (2 i + 3 )(2 i + 1)! (2 k + 5)(2 k + 3)! ∂ 2( k − i ) η ( b 2 i +2 ) (4 . 57) for 0 ≤ k ∈ Z . 16 Set Φ h β ,ρ,σ ,κ,ω i ( t, η , ζ ) = κζ − 1 + κ ηη ζ 2 + ω ζ 2 + ∞ X i =3 c i ζ i = − ζ ∂ ζ ζ − 1 " ∞ X k =0 ( − 1) k ∂ 2 k η ( κ ) ζ 2 k (2 k )! # + ζ 2 ∞ X k =0 ( − 1) k 3 ∂ 2 k η ( ω ) ζ 2 k (2 k + 3)(2 k + 1)! + ∞ X k =0 k X i =0 ( − 1) k − i ( i + 1) ( 2 i )! ( k + 2)(2 k + 2)! ∂ 2( k − i ) η ( b 2 i +1 ) ζ 2 k + 3 + ∞ X k =0 k X i =0 ( − 1) k − i (2 i + 3 )(2 i + 1)! (2 k + 5)(2 k + 3)! ∂ 2( k − i ) η ( b 2 i +2 ) ζ 2 k + 4 . (4 . 58) Note ζ 2 ∞ X k =0 ( − 1) k 3 ∂ 2 k η ( ω ) ζ 2 k (2 k + 3)(2 k + 1)! = 3 2 ζ − 1 Z ζ 0 τ 2 Z τ 2 0 [ ω ( t, η + √ − 1 τ 1 ) + ω ( t, η − √ − 1 τ 1 )] dτ 1 dτ 2 . (4 . 59) Moreo v er, Ψ h β ,ρ,σ i ( t, η , 0) = 0 , b i = ∂ i ζ (Ψ h β ,ρ,σ i )( t, η , 0) i ! for 0 < i ∈ Z . (4 . 60) Th us Φ h β ,ρ,σ ,κ,ω i ( t, η , ζ ) = 3 2 ζ − 1 Z ζ 0 τ 2 Z τ 2 0 [ ω ( t, η + √ − 1 τ 1 ) + ω ( t, η − √ − 1 τ 1 )] dτ 1 dτ 2 − 1 2 ζ ∂ ζ ζ − 1 [ κ ( t, η + √ − 1 τ 1 ) + κ ( t, η − √ − 1 τ 1 )] + ∞ X k =0 k X i =0 ( − 1) k − i ( i + 1) ∂ 2( k − i ) η ∂ 2 i +1 ζ (Ψ h β ,ρ,σ i )( t, η , 0) (2 i + 1 )( k + 2)(2 k + 2)! ζ 2 k + 3 + ∞ X k =0 k X i =0 ( − 1) k − i (2 i + 3 ) ∂ 2( k − i ) η ∂ 2 i +2 ζ (Ψ h β ,ρ,σ i )( t, η , 0) (2 i + 2 )(2 k + 5)(2 k + 3)! ζ 2 k + 4 , (3 . 61) in whic h the summations a re finite if σ ( t, η ) is po lynomial in η . Acc ording to (4.52)-(4.58) and (4.61), f = Φ h β ,ρ,σ ,κ,ω i ( t, η , ζ ) + ρ 2 4 ζ − 2 + ρ 2 η 4 − ρ t − α ′ β ρ η + γ 2 ρ 6 − [2( α ′ ′ η + β ′ ′ + ( α ′ ) 2 β ) − α ′ ρ η + 2 3 ( α ′ η + β ′ ) γ 2 ] ζ . (4 . 62) Theorem 4.1 . In terms o f the notions in (4.7), we have the fol lowin g solution of the 17 e quation (1.3) blowing up on the hyp ers urfac e cos α ( t ) y + sin α ( t ) z + β ( t ) = 0 ( ζ = 0 ): u = x 2 ζ − 2 + [ ρζ − 2 + 2( α ′ η + β ′ ) ζ − 1 + γ 2 6 + 15 2 ζ − 2 × Z ζ 0 τ 3 Z τ 3 0 τ 2 Z τ 2 0 [ σ ( t, η + √ − 1 τ 1 ) + σ ( t, η − √ − 1 τ 1 )] dτ 1 dτ 2 dτ 3 ] x +Φ h β ,ρ,σ ,κ,ω i ( t, η , ζ ) + ρ 2 4 ζ − 2 + ρ 2 η 4 − ρ t − α ′ β ρ η + γ 2 ρ 6 − [2( α ′ ′ η + β ′ ′ + ( α ′ ) 2 β ) − α ′ ρ η + 2 3 ( α ′ η + β ′ ) γ 2 ] ζ , (4 . 63) wher e the involve d p ar ametric functions ρ is given in (4.39), α is given in (4.41) and β is any function of t . Mor e ov e r, σ , κ, ω ar e r e al functions in r e al variable t and η , and Φ h β ,ρ,σ ,κ,ω i ( t, η , ζ ) is given i n (4.61) via (4.51). When σ = κ = ω = 0 , the ab o v e solution b ecomes : u = x 2 ζ − 2 + [ ρζ − 2 + 2( α ′ η + β ′ ) ζ − 1 + γ 2 6 ] x + ρ 2 4 ζ − 2 + ρ 2 η 4 − ρ t − α ′ β ρ η + γ 2 ρ 6 − [2( α ′ ′ η + β ′ ′ + ( α ′ ) 2 β ) − α ′ ρ η + 2 3 ( α ′ η + β ′ ) γ 2 ] ζ , (4 . 64) Next w e consider ξ = 0, whic h is the only solution p olynomial in y and z of (4.2 ) . In this case, ( 4.3) and (4.4) b ecomes : g y y + g z z = 0 , f y y + f z z = 2 g t + g 2 . (4 . 65) The ab o v e first equation is classical tw o-dimensional Laplace equation, whose solutions are called harmonic functions . In order to find simpler expressions of the solutions of the ab o v e equations, w e in tro duce a new notion. A complex function G ( µ ) is called b ar-homomorphic if G ( µ ) = G ( µ ) . (4 . 66) F or instance, trigonometric functions, p olynomials with real co efficien ts and elliptic f unc- tions with bar-inv a r ia n t p erio ds are bar-ho momorphic functions. The extended function F ( t, µ ) in (4 .47) is bar- homomorphic in µ . As (4.20), it can be prov ed by p o w er series that the general solution of the first e quation in (4.65) is: g = ( σ + √ − 1 ρ )( t, y + √ − 1 z ) + ( σ − √ − 1 ρ )( t, y − √ − 1 z ) , (4 . 67) where σ ( t, µ ) and ρ ( t, µ ) are complex functions in real v ariable t and bar-homomorphic in complex v ariable µ . Set w = y + √ − 1 z , w = y − √ − 1 z . (4 . 68) Then the L aplace op erator ∂ 2 y + ∂ 2 z = 4 ∂ w ∂ w . (4 . 69) 18 The second equation in (4 .65) is equiv alen t to: ∂ w ∂ w ( f ) = g t 2 + g 2 4 = 1 2 ( σ t + √ − 1 ρ t )( t, w ) + ( σ t − √ − 1 ρ t )( t, w ) + 1 4 [( σ + √ − 1 ρ )( t, w ) + ( σ − √ − 1 ρ )( t, w )] 2 . (4 . 70) Hence the g eneral solution of the second equation in (4.65) is: f = Z w w 1 Z w w 1 { 1 2 [( σ t + √ − 1 ρ t )( t, µ 1 ) + ( σ t − √ − 1 ρ t )( t, µ 1 )] + 1 4 [( σ + √ − 1 ρ )( t, µ 1 ) + ( σ − √ − 1 ρ )( t, µ 1 )] 2 } dµ 1 dµ 1 +( κ + √ − 1 ω )( t, w ) + ( κ − √ − 1 ω )( t, w ) , (4 . 71) where κ ( t, µ ) and ω ( t, µ ) are complex functions in real v ariable t and bar-homomorphic in complex v ariable µ , a nd w 1 is a complex constan t. Theorem 4.3 . In terms of the notions in (4.67), the fol lowing is a solution p olynomia l in x of the e quation (1.3): u = [( σ + √ − 1 ρ )( t, w ) + ( σ − √ − 1 ρ )( t, w )] x + Z w w 1 Z w w 1 { 1 2 [( σ t + √ − 1 ρ t )( t, µ 1 ) +( σ t − √ − 1 ρ t )( t, µ 1 )] + 1 4 [( σ + √ − 1 ρ )( t, µ 1 ) + ( σ − √ − 1 ρ )( t, µ 1 )] 2 } dµ 1 dµ 1 +( κ + √ − 1 ω )( t, w ) + ( κ − √ − 1 ω )( t, w ) , (4 . 72) wher e σ ( t, µ ) , ρ ( t, µ ) , κ ( t, µ ) and ω ( t, µ ) ar e c o m plex functions in r e al variable t and b ar- homomorphic in c omplex variable µ (cf. (4.66). Mor e over, the ab ove solution is smo oth (analytic) if al l σ , ρ, κ and ω ar e smo oth (analytic) functions. In p articular, any solution of the e quation (1.3) p o l ynom ial in x, y , z must b e of the form (4.72 ) in whi c h σ, ρ, κ and ω ar e p olynom ial in µ . Remark 4.4 . 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