حول االختزال املتدرج وهرميات بانليفة

The Islamic University of Gaza Deanship of Research and Graduate Studies Faculty of Science Master of Mathematical Science الجامعت اإلسالمي ت بغزة عمادة البحث العلمي والدراساث العليا كليت العلوم ماجستير العلوم الرياضيت On Scaling Reductions and Painlevé Hierarchies حول االختزال املتدرج وهرميات بانليفة By Asmaa S. Barakat Supervised by Prof. Ayman H. Sakka Professor of Mathematics A thesis submitted in partial fulfilment of the requirements for the degree of Master of Science in Mathematical Sciences October/2017

Abstract In this thesis we study scaling reduction of hierarchies of partial differential equations. We use scaling reduction of the Korteweg de Vries (KdV) to derive a thirty-fourth Painlevé hierarchy and we derive a fourth Painlevé hierarchy by scaling reduction of dispersive water wave (DWW). We also study generalized scaling reductions of Burgers hierarchy. Moreover we consider the mapping of nonisospectral KdV and DWW hierarchies onto isospectral KdV and DWW hierarchies. i

Acknowledgements Firstly, I thank Allah so much for this success and for finishing my studies. I would like to thank my father, mother,brothers and my sister who encouraged me to complete my studies and their continued support for my Masters thesis. Also, I would like to express my sincere gratitude to my advisor Prof. Ayman H. Sakka for the continuous support of my Master s study and related research, for his patience, motivation, and immense knowledge. His guidance helped me in all the time of research and writing of this thesis. I could not have imagined having a better advisor and mentor for my Master s study. Finally, I extend my thanks to all my colleagues and friends at the university for their endless love and support in good and bad circumstances. iii

Contents Abstract i Acknowledgements iii Contents iv Introduction 1 1 Scaling Reduction 5 1.1 Solving PDE by scaling method.................... 5 2 Scaling reductions and Painlevé hierarchies 11 2.1 The Korteweg-de Vries hierarchy................... 11 2.2 The dispersive water wave hierarchy................. 20 2.3 The Burgers hierarchy......................... 32 3 Nonisospectral scattering problems and scaling reductions 42 3.1 A nonisospectral Korteweg-de Vries hierarchy............ 42 3.2 Scaling reductions of the Korteweg-de Vries hierarchy....... 47 3.3 A nonisospectral dispersive water wave hierarchy.......... 49 3.4 Scaling reductions of the dispersive water wave hierarchy...... 54 Bibliography 57 iv

Introduction The Painlevé equations were first discovered at the end of the 19th century by Painlevé and his colleagues when they classified second order ordinary differential equations (ODEs) according to their singularities. Painlevé and his school found that forty-four of fifty equations are reducible in the sense that they can be solved in terms of previously known functions, and the remaining six equations are requiring the introduction of new special functions to solve them. These six non-linear differential equations are called Painlevé equations and their solutions are called Painlevé transcendents (Clarkson, 2003; Ince, 1956). The applications of Painlevé equations are very important in various areas of mathematics and physics. For example, in hydrodynamics and plasma physics, nonlinear waves, quantum gravity, quantum field theory, general relativity, nonlinear optics and fibre optics (Clarkson, 2003). A Painlevé hierarchy is an infinite sequence of non-linear ordinary differential equations whose first member is a Painlevé equation. Kudryashov was the first to derive a first Painlevé hierarchy and Airault was the first to derive a second Painlevé hierarchy (Sakka et al., 2009; Airault, 1979; Kudryashov, 1997). There are several ways to derive Painlevé hierarchies. Airault used higher order integrable partial differential equations (PDEs) to derive ODEs with Painlevé property to found a first member of the second Painlevé hierarchy, by similarity reduction of the Korteweg de Vries hierarchies (Gordoa and Pickering, 2011). (Gordoa et al., 2006) non-isospectral scattering problems is used to derive new second Painlevé 1

hierarchy and new fourth Painlevé hierarchy. After that a third Painlevé hierarchy was derived in (Sakka, 2008) by expansion of linear problems of the Painlevé equations in powers of the spectral variables. In 1977 (Ablowitz and Segur, 1977) noted that there exists a connection between completely integrable partial differential equation and Painlevé equation. They noted that the similarity reduction of completely integrable partial differential equations gave rises to ODEs with the Painlevé property (Gordoa et al., 2014). For example, the second Painlevé equation (Clarkson, 2003) d 2 w dz 2 = 2w3 + zw + α (0.1) can be obtained from the modified Korteweg-de Vries (mkdv) equation u t 6u 2 u x + u xxx = 0 (0.2) through the scaling reduction u(x, t) = w(z) (3t) 1 3, z = x. (0.3) (3t) 1 3 The goal of this thesis is to study scaling reduction of hierarchies of partial deferential equations and to derive thirty-fourth and fourth Painlevé hierarchies using scaling reduction. This thesis is organized as follows: In chapter one, we will give the basic definitions and theorems related to our study. In chapter two, we will derive a thirty-fourth Painlevé hierarchy from the KdV hierarchy using scaling reductions, and we will derive a fourth Painlevé hierarchy from scaling reductions of dispersive water wave (DWW). Then we will study scaling reductions of Burgers hierarchy. In chapter three, we show that the nonisospectral KdV hierarchy can be mapped onto isospectral KdV hierarchy, and its scattering problem is transformed into the nonisospectral scattering problem. Similarly, the nonisospectral DWW hierarchy can be mapped onto isospectral DWW hierarchy, and its scattering problem is 2

transformed into the nonisospectral scattering problem. Finally, we explain why the use of nonisospectral scattering problem and similarity reduction in KdV and DWW yields the same Painlevé hierarchy. 3

Chapter 1 Scaling Reduction

Chapter 1 Scaling Reduction There exist many techniques to find the solution to a linear partial differential equation (PDE) like separation of variables, integral transform and change of variables. But nonlinear partial differential equation can not be solved by these methods. Therefore we need other methods to solve these equations such as similarity method. A scaling reduction of a PDE is one that remains unchanged under scaling symmetries. The scaling reduction for a PDE in two variables can be found by solving an ordinary differential equation (ODE) and therefore make the solution much more simple. The main idea of a scaling reduction is that two or more independent variables can be linked by a transformation such that the PDE becomes ODE or a PDE with a reduced number of independent variables (Olver, 2014). 1.1 Solving PDE by scaling method Consider a PDE in two independent variable t and x: E(t, u, u t, u x..., ) = 0. (1.1) 5

A scaling reduction of equation (1.1) can be found by solving an ODE. To explain the similarity method in more details we consider the similarity variables τ = β a t, z = β b x, U(τ, z) = β c u(x, t), (1.2) where a, b, c are constants with a, b not both zero. Now, using the chain rule we find U τ = β c a u t, (1.3) U z = β c b u x, (1.4) U ττ = β c 2a u tt, (1.5) U τz = β 2c a b u tx, (1.6) and so on. Next we substitute these expressions in the given equation and we look for values of a, b, c, such that the equation for U is the same as that for u (Olver, 2014). Let us illustrate this by examples. Example 1.1. Consider the linear heat equation u t = u xx. (1.7) Assume By the chain rule we have τ = β a t, z = β b x, U(τ, z) = β c u(x, t). (1.8) U τ = β c a u t, U z = β c b u x, U zz = β c 2b u xx. (1.9) Equation (1.9) gives u t = β a c U τ, u xx = β 2b c U zz. (1.10) 6

Substituting u t and u xx from (1.10) into (1.7) we obtain β a c U τ β 2b c U zz = 0, that is U τ β 2b a U zz = 0. Thus the equation for U is the same as that for u if β 2b a = 1; that is a = 2b. As a result, if u(t, x) is any solution to (1.7), so is the function U(τ, z) = β c u(β a τ, β b z), where a = 2b. (1.11) The relation (1.11) can be written as u(t, x) = β c U(β a t, β b x). Let β = t 1/a. Then So, the scaling reduction has the form u(t, x) = t c a U(1, t 1 2 x). u(t, x) = t c a v(ξ), where v(ξ) = U(1, t 1 2 x) and ξ = t 1 2 x, where c is an arbitrary. By differentiating u with respect to t and x we obtain [ u t = t c 1 a 1 2 ξv (ξ) + c ] a v(ξ), u xx = t c a 1 v (ξ). Substituting u t and u xx into (1.7) yields [ t c 1 a 1 2 ξv (ξ) + c ] a v(ξ) or If c = 0, then the last equation becomes = t c a 1 v (ξ), v (ξ) + 1 2 ξv (ξ) c v(ξ) = 0. a v (ξ) + 1 2 ξv (ξ) = 0. (1.12) 7

Equation (1.12) is a linear second-order ordinary differential equation for v(ξ) and its solution is given by v(ξ) = c 1 erf( 1 2 ξ) + c 2, where c 1, c 2 are arbitrary constants and erf(x) = 2 x e ξ2 dξ is the error func- π tion. It follows that a scaling reduction of the heat equation (1.7) is given by ( ) x u(x, t) = c 2 + c 1 erf 2. t Example 1.2. Consider the modified Korteweg-de Vries equation (mkdv) 0 u t 6u 2 u x + u xxx = 0. (1.13) Assume τ = β a t, z = β b x, U = β c u(t, x), that is U(τ, z) = β c u(β a τ, β b z). By differentiation we have U τ = β c a u t, U z = β c b u x, U zzz = β c 3b u xxx. (1.14) From (1.14), we get u t = β a c U τ, u x = β b c U z, u xxx = β 3b c U zzz. (1.15) Substituting u, u t, u x, u xxx from (1.15) into equation (1.13) yields β a c U τ 6β b 3c U 2 U z + β 3b c U zzz = 0. Thus the conditions for invariance are a c = b 3c = 3b c; that is, a = 3b and c = b. As a result we get u(x, t) = β b U(β a t, β b x). Let β = (3t) 1/a. Then we have u(x, t) = (3t) 1 1 1 3 U(, (3t) 3 x). 3 8

Thus the scaling reduction has the form u(x, t) = (3t) 1 3 v(ξ), v(ξ) = U(3 1, (3t) 1 3 x), ξ = (3t) 1 3 x. By differentiation u with respect to t and x, we have u t = (3t) 4 3 [ v(ξ) ξv (ξ)], u x = (3t) 2 3 v (ξ), u xxx = (3t) 4 3 v (ξ). (1.16) Substituting u and its derivatives from equation (1.16) into equation (1.13) yields (3t) 4 3 [ v(ξ) ξv (ξ)] 6(3t) 4 3 v 2 (ξ)v (ξ) + (3t) 4 3 v (ξ) = 0. Multiplying by (3t) 4 3 gives v (ξ) 6v 2 (ξ)v (ξ) ξv (ξ) v(ξ) = 0. By integration we obtain v (ξ) = 2v 3 (ξ) + ξv(ξ) + α, (1.17) where α is a constant. Equation (1.17) is the second Painlevé equation (Clarkson, 2003). 9

Chapter 2 Scaling reductions and Painlevé hierarchies

Chapter 2 Scaling reductions and Painlevé hierarchies In this chapter we will study scaling reductions of the Korteweg-de Vries, dispersive water wave and Burgers hierarchies. 2.1 The Korteweg-de Vries hierarchy The Korteweg-de Vries (KdV) hierarchy is given by (Gordoa et al., 2013) U t2n+1 = R n [U]U x + β i (t 2n+1 )R i [U]U x, where i=1 R[U] = 2 x + 4U + 2U x 1 x. (2.1) The recursion operator R[U] of the KdV hierarchy is the quotient R[U] = B 1 [U]B 1 0 [U] of the two Hamiltonian operators B 1 [U] = 3 x + 4U x + 2U x B 0 = x. 11

The KdV hierarchy can be written as U t2n+1 = R n [U]U x = B 0 [U]M n+1 = B 1 [U]M n [U], (2.2) where M n [U] are defined recursively by M 0 = 1 2, M n+1 [U] = B 1 0 [U]B 1 [U]M n [U]. (2.3) In this section, we will derive the thirty-fourth Painlevé hierarchy from a scaling reduction of the KdV hierarchy (2.2) (Gordoa et al., 2013). Lemma 2.1. (Gordoa and Pickering, 2011) The change of variables Ũ = U + C, where C is an arbitrary constant, in R n [Ũ]Ũx, yields R n [Ũ]Ũx = α n,j C n j R j [U]U x, (2.4) where R 0 [U] is the identity operator and the coefficients α n,j are determined recursively by α 0,0 = 1, and R 0 [U] = 1. α n,n = 1, (2.5) α n,j = 4α n 1,j + α n 1,j 1, (2.6) α n,0 = 4n + 2 α n 1,0, (2.7) n Proof. We must deal carefully with the operator 1 x which making the coefficient not binomial so that we added ( α n,0 Cn+1) as a constant of integration. For n = 1, n+1 x we have R[Ũ]Ũx = [ 2 x + 4Ũ ] 1 + 2Ũx x Ũx = Ũxxx + 6ŨŨx. 12

Using Ũ = U + C we obtain R[Ũ]Ũx = U xxx + 6(U + C)U x = U xxx + 6UU x + 6CU x = R[U]U x + 6CR 0 [U]U x. Therefore the relation (2.4) is true for n = 1. Suppose (2.4)-(2.5) are true for n. Now for n + 1 we have R n+1 [Ũ]Ũx = R[Ũ]Rn [Ũ]Ũx (2.8) Substituting Ũ = U + C in (2.8) and using the induction hypothesis we obtain R n+1 [Ũ]Ũx = ( x 2 + 4U + 2U x x 1 + 4C ) [ R n [U]U x + α n,j C n j R j [U]U x + Hence R n+1 [Ũ]Ũx = (R[U] + 4C) R n [U]U x + (R[U] + 4C) Simplifying we get [ n 1 ] α n,j C n j R j [U]U x + ( x 2 + 4U + 2U x x 1 + 4C ) [( ) α n,0 n + 1 Cn+1 x ]. ( ) αn,0 n + 1 Cn+1 x R n+1 [Ũ]Ũx = R n+1 [U]U x + 4CR n [U]U x + α n,j C n j R j+1 [U]U x + 4α n,j C n j+1 R j [U]U x It follows that + 2α n,0 n + 1 Cn+1 U x, = R n+1 [U]U x + α n,j 1 C n j+1 R j [U]U x + R n+1 [Ũ]Ũx = R n+1 [U]U x + = j=1 +4α n,0 C n+1 U x + 2α n,0 n + 1 Cn+1 U x. 4α n,j C n j+1 R j [U]U x j=1 (4α n,j + α n,j 1 ) C n+1 j R j [U]U x + j=1 n+1 α n+1,j C n+1 j R j [U]U x, 13 ( 4α n,0 + 2α ) n,0 C n+1 U x n + 1 ].

where α n+1,n+1 = 1, α n+1,j = 4α n,j + α n,j 1, α n+1,0 = 4n + 6 n + 1 α n,0. Thus the relation is true for n + 1 and the proof is finished. Proposition 2.1. (Gordoa et al., 2013) There exists a choice of coefficient functions β i (t 2n+1 ) and of the function c(t 2n+1 ) such that the substitution U = f(z) [2(2n + 1)g n 1 t 2n+1 ] 2/2n+1 + h, z = x [2(2n + 1)g n 1 t 2n+1 ] 1/2n+1 + c(t 2n+1), where g n 1 0 and h are arbitrary constants, into the KdV hierarchy (2.9) U t2n+1 = R n [U]U x + β i (t 2n+1 )R i [U]U x, (2.10) i=1 yields the generalized thirty-fourth Painlevé hierarchy K[f](K[f]) zz 1 2 (K[f] z) 2 + 2f(K[f]) 2 + 1 2 (g n 1 + α n ) 2 = 0, (2.11) where α n is an arbitrary constant and K[f] = M n [f] + B i M i [f] + g n 1 z. (2.12) i=0 In (2.12) the coefficients B i are arbitrary constants and M j [f] is as given in (2.3) with dependent variable f and independent variable z. Proof. Let U = f(z) T 2 +h, z = x T +c(t 2n+1), where T = [2(2n + 1)g n 1 t 2n+1 ] 1/2n+1. (2.13) Then we have U = T 2 f(z) + h, x = T (z c), 14

and By the chain rule, we have dz dt = 2xg n 1T 2n 2 + c. U t = T 2 f c 2(z c)g n 1 T (2n+3) f 4g n 1 ft (2n+3). and U x = T 3 f z. Since z = x T + c(t 2n+1), we have x = T 1 z, 1 x = T 1 z. Let w(x, t) = T 2 f(z). Then w x = T 3 f z and R[w]w x = ( x 2 + 4w + 2w x x 1 )T 3 f z = (T 2 2 z + 4T 2 f + 2T 2 f z 1 z )T 3 f z. So that R[w]w x = T 5 R[f]f z. Therefore R[w] = T 2 R[f]. By (2.9) we have U = w + h. Now using Lemma 2.1, we obtain R n [U]U x = α n,j h n j R j [w]w x. Using R[w] = T 2 R[f], we have R n [U]U x = = α n,j h n j R j [w]w x α n,j h n j (T 2j R j [f])t 3 f z. 15

Thus (2.10) becomes U t = i α n,j h n j T 2j 3 R j [f]f z + β i (t) α i,j h i j T 2j 3 R j [f]f z, i=1 Substituting equation (2.9) in equation (2.10) gives ( T 2 f c 2(z c)g n 1 T 2(n+3) f 4g n 1 ft (2n+3) α n,j h n j = R j [f]f T 2j+3 z + Or ( α n,j h n j R j [f]f T 2j+3 z + i=1 β i i α i,j T 2j+3 Rj [f]f z ) i=1 β i i α i,j T 2j+3 Rj [f]f z 1 T 2 f zc ( t 2n+1 )) 2 g n 1 T 2n+3 f zc+ g n 1 T 2n+3 (4f+2zf z) = 0. ) Then k=0 γ k R k [f]f z + g n 1 T 2n+3 (4f + 2zf z) = ( α n,j h n j R j [f]f T 2j+3 z + i=1 β i i ) α i,j h i j T 2j+3 Rj [f]f z 1 T 2 f zc ( t 2n+1 )) 2 g n 1 T 2n+3 f zc + g n 1 T 2n+3 (4f + 2zf z) = 0. We solve the equations γ k = B k /T 2n+3, k = n 1,..., 1 recursively for the coefficients β k and the equation γ 0 = B 0 /T 2n+3, for c, where all B k are constants. The resulting equation can be written as B 1 [f]k[f] = 0, where B 1 [f] = 3 z + 4f z + 2f z and K[f] is as given in (2.12). This last equation admits (2.11) as a first integral, where α n is an arbitrary constant of integration. Example 2.1. Consider the fifth order KdV equation U t5 = R 2 [U]U x + β 1 (t 5 )R[U]U x. 16

Let t 5 = t. Then using R[U] = 2 x + 4U + 2U x 1 x U t = ( 2 x + 4U + 2U x 1 x )( 2 x + 4U + 2U x 1 x we obtain )U x + β 1 (t)( x 2 + 4U + 2U x x 1 )U x = U xxxxx + 10UU xxx + 20U x U xx + 30U 2 U x + β 1 (t)[u xxx + 6UU x ]. Equation (2.14) can be written as (2.14) U t = ( U xxxx + 10UU xx + 5U 2 x + 10U 3) x + β 1(t) ( U xx + 3U 2) x. (2.15) By Proposition 2.1, Equation (2.15) admits the scaling reduction From (2.16) we have U = f(z) T 2 + h, z = x T + c, T = [10g 1t] 1 5. (2.16) U = T 2 (f + ht 2 ), x = T (z c), and dt dt = 2g 1T 4, dz dx = T 1, By differentiating U with respect to t and x we obtain and z t = c 2g 1 xt 6. U t = T 7 [4g 1 f + 2g 1 (z c)f T 5 c f ] U x = T 3 f, U xx = T 4 f, U xxx = T 5 f, U xxxx = T 6 f (4), U xxxxx = T 7 f (5). Substituting into equation (2.15), we get [T 4 f + 3f 4 ( f + ht 2) 2 ] T [ 7 4g 1 f + 2g 1 (z c)f T 5 c f ] = β 1 (t)t 1 d dz +T 1 d [ T 6 f (4) + 10T 6 f ( f + ht 2) + 5T 6 (f ) 2 + 10T ( 6 f + ht 2) ] 3. dz Then d [ f4z + 10ff zz + 5fz 2 + 10hT 2 f zz + 10 ( f 3 + 3hT 2 f 2 + 3h 2 T 4 f + h 3 T 6)] dz +T 2 β 1 (t) d [ f + 3 ( f 2 + 2hT 2 f + h 2 T 4)] + 2g 1 (z c)f + 4g 1 f T 3 c f = 0. dz 17

Therefore d [ f4z + 10ff zz + 5fz 2 + 10f 3 + 10hT 2 (f zz + 3f 2 ) + 30h 2 T 4 f + 10h 3 T 6] + 2g 1 (zf + 2f) dz f (T 3 c + 2g 1 c) + T 2 β 1 (t) d [ fzz + 3f 2 6hT 2 f + 3h 2 T 4] = 0. dz Thus d [ fzzzz + 10ff zz + 5fz 2 + 10f 3] [ ] d + 2g 1 (zf z + 2f) + 10hT 2 dz dz (f zz + 3f 2 + 3hT 2 f z ) [ d ( +T 2 β 1 (t) fzz + 3f 2) ] ( + 6hT 2 f z f z T 5 c + 2g 1 c ) = 0. dz That is d [ f4z + 10ff zz + 5fz 2 + 10f 3] + 2g 1 (zf z + 2f) + T 2 [β 1 (t) + 10h] d ( fzz + 3f 2) dz dz + [ 30h 2 T 4 + 6hT 4 β 1 (t) T 5 c 2g 1 c ] f z = 0. (2.17) We set T 2 [β 1 (t) + 10h] = B 1 = constant, (2.18) and 6hT 4 (5h + β 1 (t)) T 5 c 2g 1 c = B 0 = constant. (2.19) Then, from equation (2.18) we have β 1 (t) = B 1 T 2 10h. Substituting β 1 (t) into equation (2.19) we have Since T 5 = 10g 1 t, we get ( T 5 c + 2g 1 c = B 0 + 6hT 4 5h + B ) 1 T 10h 2 = B 0 + 6hT 4 ( B1 T 2 5h ). 10g 1 tc + 2g 1 c = B 0 + 6hB 1 T 2 30h 2 T 4. (2.20) 18

By solving the differential equation (2.20) we have c = B 0 + hb 1 T 2 3h2 T 4 + c 0 2g 1 g 1 T, where c 0 = c 0 (10g 1 ) 4 5, and c 0 is a constant of integration. Equation (2.17) becomes or g 1 d [ f4z + 10ff zz + 5fz 2 + 10f 3] d ( + 2g 1 (zf z + 2f) + B 1 fzz + 3f 2) + B 0 f z = 0, dz dz (2.21) d [ f4z + 10ff zz + 5fz 2 + 10f 3 + B 1 f zz + 3B 1 f 2 + B 0 f ) + 2g 1 (zf z + 2f) = 0. dz (2.22) From equation (2.12), using n = 2 we find K[f] = f zz + 3f 2 + B 1 f + 1 2 B 0 + g 1 z. It follows that d dz K[f] = f zzz + 6ff z + B 1 f z + g 1, (2.23) d 2 dz K[f] 2 = f zzzz + 6fz 2 + 6ff zz + B 1 f zz. (2.24) Now will write equation (2.22) in terms of K[f]. Substituting f zzzz = d2 dz 2 K[f] 6f 2 z 6ff zz B 1 f zz in equation (2.22) yields or [ ] d d 2 dz dz K[f] f 2 2 z + 4ff zz + 10f 3 + 3B 1 f 2 + B 0 f + 2g 1 (zf z + 2f) = 0 d 3 dz 3 (K[f]) + 2f zf zz + 4ff zzz + 30f 2 f z + 6B 1 ff z + B 0 f z + 2g 1 zf z + 4g 1 f = 0. (2.25) Substituting f zzz = d dz K[f] 6ff z B 1 f z g 1 in equation (2.25) gives d 3 dz 3 (K[f]) + 4f d dz K[f] + 6f 2 f z + 2B 1 ff z + 2f z f zz + B 0 f z + 2g 1 zf z = 0 19

or d 3 dz 3 (K[f]) + 4f d dz K[f] z + 2f z K[f] = 0. (2.26) Multiplying both sides by K[f] and integrating with respect to z, we get K[f]K[f] zz 1 2 (K[f] z) 2 + 2f (K[f]) 2 + 1 2 (g 1 + α 2 ) 2 = 0, (2.27) where α 2 is an arbitrary constant. The equation (2.27) is called the thirty-fourth Painlevé equation (Ince, 1956). 2.2 The dispersive water wave hierarchy The dispersive water wave (DWW) hierarchy is a two-component hierarchy in u = (u, v) T given by (Gordoa et al., 2013) where u tn = R n [u]u x + γ i R i [u]u x, R[u] = 1 2 i=1 x(u x 1 ) x 2 2v + v x x 1 u + x. (2.28) The recursion operator R can be written as the quotient of two Hamiltonian operators R[u] = B 2 [u]b1 1 [u], where B 2 [u] = 1 2 2 x x u x 2 u x + x 2 v x + x v, and B 1 [u] = 0 x 0 The DWW hierarchy can be written as x. where the quantities L n [u] in (2.29) is defined by u tn = B 1 L n+1 [u] = B 2 L n [u], (2.29) L 0 = (0, 2) T, L n [u] = B 1 1 [u]b 2 [u]l n 1 [u]. 20

(2.30) In this section we will derive a fourth Painlevé hierarchy from a scaling reduction of DWW hierarchy (2.29). Lemma 2.2. (Gordoa and Pickering, 2011) The change of variables ũ = (u+c, v) T, where C is an arbitrary constant, in R n [ũ]ũ x, yields R n [ũ]ũ x = α n,j C n j R j [u]u x, (2.31) where where R 0 [u] is the identity operator and the coefficients α n,j are determined recursively by α n,n = 1, α 0,0 = 1 α n,j = 1 2 α n 1,j + α n 1,j 1, α n,0 = 1 2 n + 1 n α n 1,0, (2.32) Proof. For n = 1, we have R[ũ]ũ x = 1 2 = 1 2 xũ x 1 x 2 2v + v x x 1 ũ + x = 1 2 ũx v x ( xũ x 1 x )ũ x + 2v x (2v + v x x 1 )ũ x + (ũ + x )v x 2ũũ x ũ xx + 2v x. 2vũ x + v x ũ + v xx + v x ũ 21

Making the change of variables ũ = u + C we obtain R[ũ]ũ x = 1 2 = 1 2 = 1 2 2(u + C)u x + u xx + 2v x 2vu x + (u + C)v x + v xx + (u + C)v x 2uu x u xx + 2v x + 2Cu x 2vu x + v x u + v xx + uv x + 2Cv x 2uu x u xx + 2v x + 1 2Cu x 2vu x + v x u + v xx + uv 2 x 2Cv x = R[u]u x + CR 0 [u]u x Therefore, the relation (2.31) is true for n = 1. Suppose (2.31) is true for n. Before proving that (2.31) is true for n + 1 we will show that R[ũ] = R[u] + 1 CI, where I 2 is the identity matrix operator. R[ũ] = 1 2 x(ũ x 1 ) x 2 2v + v x x 1 ũ + x. Making the change of variables ũ = (u + C, v) T, we get R[ũ] = 1 2 xu x 1 x + x C 1 x 2 2v + v x 1 x u + x + C or R[ũ] = 1 2 xu x 1 x 2 2v + v x x 1 u + x + 1 C 0 2 0 C. Hence R[ũ] = R[u] + 1 CI. (2.33) 2 Now we will proof that (2.31) is true for n + 1. We must deal carefully with the operator 1 x which makes the coefficient not binomial so that we added ( α n,0 n+1 Cn+1) x as a constant of integration R n+1 [ũ]ũ x = R[ũ]R n [ũ]ũ x. 22

Using R n+1 [ũ] = RR n [ũ], equations (2.31) and (2.33), we obtain [ R n+1 [ũ]ũ x = R[u] + 1 ] ( 2 CI R n [u]u x + α n,j C n j R j [u]u x + Thus R n+1 [ũ]ũ x = R n+1 [u]u x + 1 2 CRn [u]u x + α n,j C n j R j+1 [u]u x + + 1 2 ( ) αn,0 C n+1 u x. n + 1 Collecting similar terms, we obtain R n+1 [ũ]ũ x = R n+1 [u]u x + 1 2 CRn [u]u x + + 1 2 As a result, we get R n+1 [ũ]ũ x = R n+1 [u]u x + where = ( α n,0 + α ) n,0 C n+1 u x. n + 1 ( ) ) αn,0 n + 1 Cn+1 x α n,j 1 C n j+1 R j [u]u x + j=1 ( 1 2 α n,j + α n,j 1 )C n j R j [u]u x + 1 2 j=1 n+1 α n+1,j C n+1 j R j [u]u x, α n+1,n+1 = 1, α n+1,j = 1 2 α n,j + α n,j 1, α n+1,0 = 1 ( ) n + 2 α n,0. 2 n + 1 Therefore the relation is true for n + 1 and the proof is completed. 1 2 α n,jc n+1 j R j [u]u x j=1 1 2 α n,jc n+1 j R j [u]u x ( ) n + 2 α n,0 C n+1 u x n + 1 Proposition 2.2. (Gordoa et al., 2013) There exists a choice of coefficient functions γ j (t n ) and of the function c(t n ) such that the substitution u = f(z) [ 1(n + 1)g 2 nt n ] +h, v = g(z) 1/(n+1) [ 1(n + 1)g 2 nt n ], z = x 2/(n+1) [ 1(n + 1)g 2 nt n ] +c(t n), 1/(n+1) 23 (2.34)

where g n 0 and h are arbitrary constants, into the DWW hierarchy u tn = R n [u]u x + γ i R i [u]u x (2.35) i=1 yields the fourth Painlevé hierarchy in f = (f, g) T 0 = 2K + fl + g n 2α n L z, (2.36) ( 0 = K + 1 ) 2 2 g n α n 1 4 β2 n gl 2 K z L. (2.37) Here α n and β n are arbitrary constants and K and L are the components of K[f], K = (K, L) T, and K[f] = L n + B i L i [f] + g n 0 i=0 z. (2.38) In (2.38), coefficients B i are arbitrary constants and L j [f] is as given in (2.30) with dependent variables f and g and independent variable z. Proof. Let u = f(z) T Then + h, v = g(z) T 2 z = x T + c(t n), where T = u = T 1 f(z)+h, v = T 2 g, x = T (z c), [ ] 1/(n+1) 1 2 (n + 1)g nt n. (2.39) and Differentiating u with respect to t yields u t v t = dz dt = 1 2 xg nt (n+2) + c. 1 2 g 2T (n+2) + T 1 f c 1 2 T 1 g n T (n+2) + 1 2 g nt (n+2) cf T 2 g c 1 2 g ng T (2n+1) z + 1 2 g ng T (2n+1) c. (2.40) Differentiating u with respect to x we get u x = T 2 (t)f z T 3 (t)g z 24.

Thus u x can be written as Let u x = T 2 0 0 T 3 f z. w(x, t) = (w 1, w 2 ) T, where w 1 = T 1 f(z), w 2 = T 2 (t)g(z), z = T 1 x + c(t). Hence x = T 1 z, 1 x = T 1 z, w x = u x and R[w]w x = 1 2 xw 1 1 x x 2 2w 2 + (w 2 ) x 1 x T 2 0 f z. w 1 + x 0 T 3 Substituting w 1 = T 1 f(z) and w 2 = T 2 g(z) and using x = T 1 z, 1 x have R[w]w x = 1 2 = 1 2 T 1 z T 1 ft z 1 T 1 z 2 2T 2 g + T 3 g z T z 1 T 1 f + T 1 z T 1 [ z f z 1 z ] 2 T 2 [2g + g z z 1 ] T 1 [f + z ] = T we T 2 0 f z. 0 T 3 T 2 0 f z. 0 T 3 It follows that and hence R[w]w x = T 3 0 R[f]f z, 0 T 4 R[w] = T 1 0 R[f], 0 T 1 u = (w 1 + h, w 2 ) T 25

Now using Lemma 2.2, we obtain R n [u]u x = = α n,j h n j R j [w]w x α n,j h n j T 1 0 0 T 1 j R j [f] T 2 0 0 T 3 f z. Then where Thus u t = R n [u]u x = α n,j h n j T j+2 R j [f]f z, T j = T j (t) 0. 0 T (j+1) α n,j h n j T j+2 R j [f]f z + i=1 γ i i α i,j h i j T j+2 R j [f]f z. Using u t as in equation (2.40) we obtain 1g 2 nt (n+2) + T 1 f c 1T 1 g 2 n T (n+2) + 1g 2 nt (n+2) cf T 2 g c 1g 2 ng T (2n+1) z + 1g 2 ng T (2n+1) c ( i ) + γ i α i,j h i j T j+2 R j [f]f z, i=1 or = ( i ) α n,j h n j T j+2 R j [f]f z + γ i α i,j h i j T j+2 R j [f]f z i=1 1g 2 nt (n+2) + T 1 f c 1T 1 g 2 n T (n+2) + 1g 2 nt (n+2) cf = 0. T 2 g c 1g 2 ng T (2n+1) z + 1g 2 ng T (2n+1) c 0 α n,j h n j T j+2 R j [f]f z Then ( i ) α n,j h n j T j+2 R j [f]f z + γ i α i,j h i j T j+2 R j [f]f z i=1 + 1 2 g T (n+2) 0 (zf) z n = 0. 0 T (2n+1) 2g + zg z 0 T 1 0 0 T 2 f g c 26

Therefore ( i ) α n,j h n j T j+2 R j [f]f z + γ i α i,j h i j T j+2 R j [f]f z T 1 f z c tn 1 2 g nt j+2 f z c i=1 + 1 2 g nt j+2 (zf) z = 0. 2g + zg z 0 That is Γ k R k [f]f z + 1 2 g nt j+2 k=0 (zf) z 2g + zg z = n 1 α n,j h n j T j+2 R j [f]f z + i=1 γ i i α i,j h i j T j+2 R j [f]f z T 1 f z c tn 1 2 g nt j+2 f z c + 1 2 g nt n+2 (zf) z = 0, 2g + zg z 0 where R[f] is obtained from R[u] by replacing u by f and x by z. We recall that each α i,i = 1, and so in particular Γ n = T n+2. We solve the equations Γ k = B k /T n+2, k = n 1,..., 1 recursively for the coefficients γ k and the equation Γ 0 = B 0 /T n+2 for c, where all B k are constants. The resulting ODE equation can be written B 2 [f]k[f] = 0, where B 2 [f] = 3 z + 4f z + 2f z and K[f] is as given in (2.38). The last equation admits (2.54) as a first integral, where α n and β n are arbitrary constants of integration. Example 2.2. The second nontrivial dispersive water wave. 27

Let t 2 = t. Then the hierarchy (2.35) for n = 2 gives u = R 2 [u]u x + γ(t)r[u]u x v t = 1 u xxx 3uu xx + 6uv x + 6vu x + 3u 2 u x + 3u 2 x 4 v xxx + 6vu x + 3uv xx + 3u x v x + 6uu x v + 3u 2 v x + 1 2 γ(t) 2uu x + 2v x u xx 2vu x + 2v x + v xx. (2.41) Equation (2.41) can be written as u = 1 u xx 3uu x + u 3 + 6uv v 4 v xx + 3v 2 + 3uv x + 3u 2 v t x + 1 2 γ(t) 2v + u2 u x 2vu x + v x By Proposition 2.2 equation (2.42) admits the scaling reduction. (2.42) x u = f(z) T From (2.43), we have + h, v = g(z) T 2, z = x T + c, T = ( 3 2 g 2t ) 1/3. (2.43) dt dt = 1 2 g 2T 2, dz dx = T 1, x = T 1 d dz, dz dt = c 1 2 g 2xT 4. By differentiating u with respect to x we obtain u x = T 2 f, u xx = T 3 f, u xxx = T 4 f. By differentiating v with respect to x we have v x = T 3 g, v xx = T 4 g, v xxx = T 5 g. Differentiating u and v with respect to t gives u t = T 4 [ 1 2 g 2f + 1 2 g 2f T 3 c f ], and v t = T 5 [ g 2 g + 1 2 g 2(z c)g T 3 g c ]. 28

Substituting u t, v t into equation (2.42) we get, T [ 4 1 2 g 2f + 1 2 g 2f T 3 c f ] T [ 5 g 2 g + 1 2 g 2(z c)g T 3 g c ] = 1 4 T 1 d T [ 3 f zz 3f z (f + ht ) + (f + ht ) 3 + 6(f + ht )g ] dz T [ 4 g zz + 3g 2 + 3g z (f + ht ) + 3g(f + ht ) 2] + 1 2 γ 1(t)T 1 d T [ ] 2 2g + (f + ht ) 2 f z. dz T 3 [2g(f + ht )] Multiplying both sides by T 4 0, gives 0 T 5 1 2 g 2f + 1 2 g 2f T 3 c f g 2 g + 1 2 g 2(z c)g T 3 g c = 1 d [f zz 3f z (f + ht ) + (f + ht ) 3 + 6(f + ht )g] 4 dz [g zz + 3g 2 + 3g z (f + ht ) + 3g(f + ht ) 2 ] + 1 2 γ 1(t)T 1 d T 2 [2g + (f + ht ) 2 f z ]. dz T 3 [2g(f + ht )] Therefore 1 d f zz 3f z f + f 3 + 6gf + ht [3f 2 3f z + 6g + 3hT f] + h 3 T 3 + 1 4 dz g zz + 3g 2 + 3g z f + 3gf 2 + ht [3g z + 6fg + 3hT g] 2 g zf z + f 2 zg z + 2g + 1 2 γ 1(t)T d 2g + f 2 + 2fhT f z + h 2 T 2 [ T 3 c + 1 ] dz 2gf + 2hT g + g 2 g 2c f = 0. z g 0 Thus + 1 2 γ 1(t)T 1 d f zz 3f z f + f 3 + 6gf 4 dz g zz + 3g 2 + 3g z f + 3gf 2 d [f 2 f dz z + 2g] + 2hT f z [g z + 2fg] + 2hT g z + 3 4 ht + 1 2 g zf z + f 2 zg z + 2g d dz [f 2 f z + 2g] + ht f z d [g dz z + 2fg] + ht g z [ T 3 c + 1 ] 2 g 2c f = 0 g 0. So 1 d f zz 3f z f + 3f 3 + 6gf + 1 [ 4 dz g zz + 3g 2 + 3g z f + 3gf 2 2 T γ 1 (t) + 3 ] d 2 h f 2 f z + 2g dz g z + 2fg [ 3 + 4 h2 T 2 + ht 2 γ 1 (t) T 3 c 1 ] 2 g 2c f z + 1 g 2 g zf z + f 2 = 0. z zg z + 2g 0 29

(2.44) We set T [γ 1 (t) + 32 ] h = B 1 = constant. (2.45) ( ) 3 ht 2 4 h + γ 1(t) T 3 c 1 2 g 2c = B 0 = constant. (2.46) Then, from equation (2.45) we have γ 1 (t 2 ) = B 1 T 3 2 h. Substituting γ 1 (t) in equation (2.46) we have T 3 c + 1 ( 3 2 g 2c = B 0 + ht 2 4 h + B 1 T 3 ) 2 h ( = B 0 + ht 2 B1 T 3 ) 4 h. Since T 3 = 3 2 g 2t, we get 3 2 g 2tc + 1 2 g 2c = B 0 + hb 1 T 3 4 T 2 h 2. (2.47) By solving the differential equation (2.47) we have c = 2B 0 g 2 + ht g 2 B 1 h2 2g 2 T 2 + c 0 T, where c 0 = c 0 ( 3 2 g 2) 1 3, and c 0 is the constant of integration. Equation (2.44) becomes 1 d f zz 3f z f + 3f 3 + 6gf + 1 4 dz g zz + 3g 2 + 3g z f + 3gf 2 2 B d f 2 f z + 2g 1 dz g z + 2fg + 1 2 g zf z + f 2 = 0, zg z + 2g 0 + B 0 f z g z or f zz 3f z f + 3f 3 + 6gf + 2B 1 f 2 2B 1 f z + 4B 1 g + 4B 0 f g zz + 3g 2 + 3g z f + 3gf 2 + 2B 1 g z + 4B 1 fg + 4B 0 g 30 +2g 2 zf z + f = 0 zg z + 2g 0 z (2.48).

From equation (2.38), using n = 2 we find K = 1 2fg + g z + B 1 g L 2 2g + f 2 f z f It follows that K L z + B 0 0 2 = f zg + g z f + 1g 2 zz + B 1 g z. g z + ff z 1f 2 zz + B 1 f z + g 2 + g 2 0 z Now will write equation (2.48) in terms of (K, L) T. Substituting g zz = 2K z 2f zg 2g z f 2B 1 g z in equation (2.48) f zz 2L z + 2g z + 2ff z + 2B 1 f z + 2g 2 yields 2 L z + 2g [ zz + f z fz + 3f 2 + 6g + 4B 1 f + 4B 0 + 2g 2 z ] ff zz + 6g z f + 4B 1 g z + 2g 2 f [ K z g zz f + g z fz + 3f 2 + 6g + 4B 1 f + 4B 0 + 2g 2 z ] 2gf zz + 6gff z f + 4B 1 gf z + fg 2 5 z + 2g 2 0 z = 0, 0 or 2 L z + 2f zl + f [ f zz + 2g z + 2ff z + 2B 1 f z + 2g 2 ] + [2g zz + 4B 1 g z + 4g z f + 4gf z ] K z 2g z L + g [ 2f zz + 4g z + 4ff z + 4B 1 f z + 4g 2 ] + f [g zz + 2g z f + 2B 1 g z + 2gf z ] z + 2g 2 0 z = 0. 0 It follows that 2 L z K z + 2f zl + fl z + 4K z 2g z L + 4gL z + 2Kf z + 2g 2 0 z z = 0. 0 Multiplying both sides by 1 and integrating with respect to z, we get 2 2K + fl + g 2 2α 2 L z ( K + 1 g ) = 0 (2.49) 2 2 2 α 2 1 4 β2 2 gl 2 K z L 0 where α 2 and β 2 are arbitrary constants. The equation (2.49) is called the fourth Painlevé equation. 31.

2.3 The Burgers hierarchy The Burgers hierarchy is given by (Gordoa et al., 2013) U tn+1 = R n [U]U x, R[U] = x ( x + 1 2 ) 1 x, (2.50) or alternatively U tn+1 = x L n [U] = x T n [U]U, T [U] = x + 1 U, (2.51) 2 where L n [U] are defined by L 1 [U] = 2, L 0 [U] = T 0 [U]U = U, L 1 [U] = T [U]U = U x + 1 2 U 2, L n [U] = T n [U]U. In this section we will study a scaling reduction of Burgers hierarchy (2.51) and this reduction leads to a hierarchy of ODEs. Lemma 2.3. (Gordoa and Pickering, 2011) The change of variables Ũ = U + C, where C is an arbitrary constant, in L n [Ũ], yields L n [Ũ] = n + 1 ( ) n j 1 j= 1 j + 1 2 C L j [U], (2.52) where we define L 1 [U] = 2. Proof. We will prove this Lemma by induction. For n = 1, we have L 1 [Ũ] = T [Ũ]Ũ = ( x + 1 2Ũ)Ũ = Ũx + 1 2Ũ2. 32

Since Ũ = U + C, we have L 1 [Ũ] = (U + C) x + 1 (U + C)2 2 = U x + 1 2 U 2 + UC + 1 2 C2 = L 1 [U] + L 0 [U]C + 1 2 C2. This show that (2.52) is true for n = 1. Suppose (2.52) is true for n, which can be written as and L n [Ũ] = n + 1 j= 1 j + 1 ( ) n j 1 2 C T j [U]U, T [Ũ] = x + 1 2Ũ Making the change of variables Ũ = U + C, we obtain T [Ũ] = x + 1 (U + C) 2 = T [U] + 1 2 C. Now for n + 1, we have n L n+1 [Ũ] = T [Ũ]T [Ũ]Ũ. Making the change of variables Ũ = U + C, we obtain L n+1 [Ũ] = [ T [U] + 1 ] 2 C n + 1 ( ) n j 1 j= 1 j + 1 2 C T j [U]U = T [U] n + 1 ( ) n j 1 j= 1 j + 1 2 C T j [U]U + 1 2 C n + 1 ( ) n j 1 j= 1 j + 1 2 C T j [U]U = n + 1 ( ) n j 1 j= 1 j + 1 2 C T j+1 [U]U + n + 1 ( ) n j+1 1 j= 1 j + 1 2 C T j [U]U. Hence, from (2.52) we obtain n+1 L n+1 [Ũ] = n + 1 ( ) n (j 1) 1 j 2 C L j [U] + n + 1 j= 1 j + 1 ( ) n j+1 1 2 C L j [U]. 33

Collecting similar terms, we obtain L n+1 [Ũ] = n + 1 ( ) n+2 1 0 2 C L 1[U] + n + 1 n + 1 + n + 1 ( ) n j+1) 1 j 2 C L j [U] + n + 1 j + 1 ( ) n+2 ( ) 0 1 1 = 2 C L 1[U] + 2 C L n+1[u] + + n + 1 ( ) n j+1 1 j + 1 j 2 C L j [U]. ( ) n (n+1 1) 1 2 C L n+1[u] ( 1 2 C ) n j+1 L j [U]. So L n+1 [Ũ] = n + 2 0 + n + 2 j + 1 ( ) n+2 1 2 C L 1[U] + n + 2 n + 2 ( ) n j+1 1 2 C L j [U]. ( ) 0 1 2 C L n+1[u] That is n+1 L n+1 [Ũ] = j= 1 n + 2 j + 1 ( ) n j+1 1 2 C L j [U]. Therefore the relation is true for n + 1 and the proof is complete. Proposition 2.3. (Gordoa et al., 2013) There exists a choice of coefficient functions β i (t n+1 ) and of the function c(t n+1 ) such that the substitution U = f(z) [(n + 1)g n 1 t n+1 ] + h, z = x 1/(n+1) [(n + 1)g n 1 t n+1 ] + c(t n+1), (2.53) 1/(n+1) where g n 1 0 and d are arbitrary constants, into the hierarchy yields the hierarchy of ODEs U tn+1 = R n [U]U x + β i (t n+1 )R i [U]U x, (2.54) L n [f] + i= 1 i=1 B i L i [f] + g n 1 zf = 0, (2.55) 34

where L n [f] = n + 1 j= 1 j + 1 ( ) n j 1 2 h L j [U], with dependent variable f and independent variable z, and where the coefficient B i are arbitrary constant. Proof. Let U = f(z) T + h, z = x T + c(t n+1), where T = [(n + 1)g n 1 t n+1 ] 1/n+1. (2.56) Then and It follows that U = T 1 f(z) + h, x = T (z c), z t = xg n 1T (n+2) + c. U x = T 1 f z. dz dx = 1 T, U t = g n 1 ft (n+2) g n 1 f zt (n+2) + g n 1 f T (n+2) + T 1 f c t n+1. (2.57) Since z = x T + c(t n+1), we have x = T 1 z, 1 x = T 1 z. Let w(x, t) = T 1 f(z). Then w x = T 2 f z and R[w]w x = x L n [w]w x = x [ x + 1 2 w] 1 x T 2 f z 35

So that R[w]w x = T 3 R[f]f z. Therefore R[w] = T 1 R[f]. By (2.53) we have U = w + h. Now using Lemma 2.3, we obtain R n [U]U x = x L n [w]w x, = x n + 1 j + 1 = n + 1 j + 1 ( ) n j 1 2 h L j [w]w x, ( ) n j 1 2 h T (j+2) R j [f]f z Thus (2.54) becomes U t = n + 1 ( ) n j 1 j + 1 2 h T (j+2) R j [f]f z +n 1 β i (t) i=1 i n + 1 j + 1 ( ) i j 1 2 h T (j+2) R j [f]f z. Substituting equation (2.53) into equation (2.54) gives 1 T f zc tn+1 + g n 1 T n+2 f zc g n 1 T n+2 zf z = n + 1 j + 1 i + β i i + 1 i=0 j + 1 ( ) n j 1 2 h 1 T R j[f]f j+2 z ( ) i j 1 2 h 1 T R j[f]f j+2 z. Then n + 1 j + 1 ( ) n j 1 2 h 1 T R j[f]f j+2 z + i=0 β i i i + 1 j + 1 ( ) i j 1 2 h 1 T R j[f]f j+2 z 1 T f zc t n+1 g n 1 T n+2 f zc + g n 1 T n+2 (zf) z = 0. That is k=0 γ k R k [f]f z + g n 1 T n+2 f zc + g n 1 T n+2 (zf) z = n + 1 j + 1 i + β i i + 1 i=0 j + 1 ( ) n j 1 2 h 1 T R j[f]f j+2 z ( ) i j 1 2 h 1 T R j[f]f j+2 z 1 T f zc t n+1 g n 1 T n+2 f zc + g n 1 T n+2 (zf) z = 0, 36

where L 1 is constant and γ n = 1/T n+2. We solve the equations γ k = B k /T n+2, k = n 1,..., 1, recursively for the coefficients γ k and the equation γ 0 = B 0 /T n+2 for c, where all B k are constants. Integrating the resulting ODE then yields (2.55), where we include a constant of integration as the term B 1 L 1 [f] = 2B 1. Proposition 2.4. (Gordoa et al., 2013) The hierarchy (2.55) is linearizable by the Cole-Hopf transformation f = 2ϕz to the hierarchy of ODEs ϕ z n+1 ϕ + B i z i+1 ϕ + g n 1 zϕ = 0. (2.58) Proof. To prove equation (2.58) we want to verify that i=1 We will use induction to prove this. First of all we have L n [f] = 2 ϕ n+1 z ϕ. (2.59) L n [f] = T n [f]f = ( z + 1 2 f)n f. Let f = 2ϕ z /ϕ. Now for n = 1 we obtain L 1 [f] = T[f]f = ( z + 1 2 f)f = f z + 1 2 f 2. Using f = 2ϕ z /ϕ we have L 1 [f] = 2 ϕ ( ) 2 zz ϕ 2 ϕz + 1 ( 2 ϕ ) 2 z. ϕ 2 ϕ Thus L 1 [f] = 2 ϕ 2 zϕ, (2.60) 37

and we have the relation (2.59) is true for n = 1. Suppose the relation (2.59) is true for n. For n + 1 we have Hence we have L n+1 [f] = T[f]T n [f]f = ( z + 1 2 f)( 2 ϕ n+1 z ϕ) = z ( 2 ϕ n+1 z = 2 ϕ n+2 z ϕ 2 ϕ z ϕ) + 1 ϕ f n+1 z ϕ ϕ 2 n+1 z ϕ + 2 ϕ z ϕ 2 n+1 z ϕ. L n+1 [f] = 2 ϕ n+2 z ϕ. (2.61) That is the relation (2.59) is true for n + 1. Substituting L n, L i and f in hierarchy (2.55) gives 2 ϕ n+1 z ϕ + Multiplying both side by ϕ 2 i=1 ( ) 2 B i ϕ i+1 z ϕ + g n 1 z we obtain i=1 ( ) 2ϕz = 0. ϕ z n+1 ϕ + B i z i+1 ϕ + g n 1 zϕ z = 0. Example 2.3. Consider the second nontrivial member of the Burgers hierarchy U t3 = R 2 [U]U x + β 1 R[U]U x. (2.62) We have and R 2 [U]U x = x L 2 [U] = x ( U xx + 3 2 UU x + 1 4 U 3 ), (2.63) R[U]U x = x L 1 [U] = x ( U + 1 2 U 2 ). (2.64) Let t 3 = t. From equations (2.63)-(2.64) we have U t = (U xx + 32 UU x + 14 ) U 3 + (U + 12 ) U 2. (2.65) x x 38

By Proposition 2.3 equation (2.65) admits the scaling reduction It follows that U = f(z) T + h, z = x T + c(t), T = [3g 1t] 1 3. (2.66) U = T 1 (f + ht ), x = T (z c), dz dx = T 1, and dt dt = g 1T 2, x = T 1 d dz, By differentiating U with respect to x and t we obtain and z t = xg 1T 4 + c. U x = T 2 f, U xx = T 3 f, U xxx = T 4 f, U t = T 4 [ g 1 (f + f z) f (g 1 c + T 3 c ) ]. Substituting into equation (2.65) gives [T 3 f + 32 T 3 f (f + ht ) 14 T 3 (f + ht ) 3 ] T [ 4 g 1 (f + f z) f (g 1 c + T 3 c ) ] = T 1 d dz +β 1 T 1 d [T 2 f + 12 ] dz T 2 (f + ht ) 2. Then d [f zz + 32 dz f zf + 32 f zht + 14 f 3 + 34 f 2 ht + 34 fh2 T 2 + 14 ] h3 T 3 +β 1 (t)t d [f z + 12 dz f 2 + fht + 12 ] h2 T 2 + g 1 (f + zf z ) f z (g 1 c + T 3 c ) = 0. Therefore d [f zz + 32 dz f zf + 14 ] f 3 + g 1 (f + zf z ) + 3 2 ht d [f z + 12 ] dz f 2 + 3 4 h2 T 2 f z +β 1 (t)t d [f z + 12 ] dz f 2 + β 1 (t)t 2 hf z f z (g 1 c + T 3 c ) = 0. That is d [f zz + 32 dz f zf + 14 ] [ ] 3 d f 3 + g 1 (f + zf z ) + 2 ht + β 1(t)T dz (f z + 1 2 f 2 ) [ ] 3 + 4 h2 T 2 + β 1 (t)t 2 h T 3 c g 1 c f z = 0. 39

(2.67) We set T [β 1 (t) + 32 ] h = B 1 = constant, (2.68) and Then, from equation (2.68), we get ht 2 ( 3 4 h + β 1(t)) T 3 c g 1 c = B 0 = constant. (2.69) Substituting for β 1 (t) from (2.70) in (2.69), we have β 1 (t) = B 1 T 3 h. (2.70) 2 Since T 3 = 3g 1 t, equation (2.71) becomes T 3 c + g 1 c = B 0 + ht 2 ( B 1 T 3 h). (2.71) 4 c + 1 3t c = 1 3g 1 t B 0 + B 1 t 2 3 (3g1 ) 2 3 h 3 4 h2 t 1 3 (3g1 ) 1 3. (2.72) By solving the ODE (2.72), we have c = B 0 g 1 + B 1h 2g 1 T h2 4g 1 T 2 + c T, where c = c 0 (3g 1 ) 1 3, and c 0 is a constant of integration. Equation (2.67) becomes d [f zz + 32 dz f zf + 14 ] f 3 d + g 1 (f + zf z ) + B 1 dz (f z + 1 2 f 2 ) + B 0 f z = 0. Integration with respect to z gives f zz + 3 2 f zf + 1 4 f 3 + g 1 zf + B 1 (f z + 1 2 f 2 ) + B 0 f + 2B 1 = 0, (2.73) where 2B 1 is a constant of integration. Equation (2.73) is linearizable by the Cole- Hopf transformation f = 2ϕz ϕ to ϕ zzz + B 1 ϕ zz + B 0 ϕ z + B 1 ϕ + g 1 zϕ z = 0. 40

Chapter 3 Nonisospectral scattering problems and scaling reductions

Chapter 3 Nonisospectral scattering problems and scaling reductions Two operators are called isospectral if they have the same spectrum this means they have the same sets of eigenvalues. In this chapter we will show that nonisospectral hierarchies and their scattering problems can be mapped onto standard PDE hierarchies and their isospectral scattering problems. 3.1 A nonisospectral Korteweg-de Vries hierarchy The KdV hierarchy can be solved by the Inverse Scattering Transform (IST), having the corresponding scattering problem ψ xx + (U λ)ψ = 0, (3.1) where λ is the constant spectral parameter. Consider the nonisospectral KdV hierarchy u τ = R n [u]u ξ + B j (τ)r j [u]u ξ + g n 1 (τ)(4u + 2ξu ξ ) + g n (τ), (3.2) 42

where R[u] is given in (2.1) with U replaced by u and x by ξ. Equation (3.2) is important since its stationary reduction when g n 1 (τ), g n (τ) and all B j (τ) are constants yields the equation R n [u]u ξ + B j R j [u]u ξ + g n 1 (4u + 2ξu ξ ) + g n = 0, (3.3) which is a generalized thirty-fourth Painlevé hierarchy if g n = 0. In this section we will show that nonisospectral KdV hierarchy can be mapped to an isospectral KdV hierarchy. Proposition 3.1. (Gordoa et al., 2014) The nonisospectral KdV hierarchy (3.2) is equivalent to the isospectral KdV hierarchy U t = R n [U]U x + β i (t)r i [U]U x, (3.4) under a change of variables of the form i=1 U = g(t) 2 u(ξ, τ) + m(t), ξ = g(t)x + h(t), τ = k(t). (3.5) Proof. By differentiating U with respect to t and x we get Hence In a similar way we can found U t = g 2 u t + 2gg u + m (3.6) [ ] = g 2 τ u τ dt + u ξ ξ + 2gg u + m. (3.7) dt U t = g 2 [u τ k + (g x + h )u ξ ] + 2gg u + m. U x = g 2 [gu ξ ] = g 3 u ξ, u ξ = g 3 U x. Since ξ = gx + h, we have x = g ξ, 1 x = g 1 1 ξ. 43

Let w(x, t) = g 2 u(ξ, τ). Then w x = g 3 u ξ, and R[w]w x = ( x 2 + 4w + 2w x x 1 )g 3 u ξ = (g 2 2 ξ + 4g 2 u + 2g 2 u ξ 1 ξ )g 3 u ξ. So that R[w]w x = g 5 R[u]u ξ. Therefore R[w] = g 2 R[u]. By (3.5) we have U = w + m. Now using Lemma 2.1, we obtain R n [U]U x = α n,j m n j R j [w]w x. Using R[w] = g 2 R[u], we have R n [U]U x = = α n,j m n j R j [w]w x α n,j m n j (g 2j R j [u])g 3 u ξ. Thus (3.4) becomes U t = i α n,j m n j g 2j+3 R j [u]u ξ + β i (t) α i,j m i j g 2j+3 R j [u]u ξ. i=1 Using U t = g 2 k u τ + g 2 (g x + h )u ξ + 2gg u + m, we get g 2 k u τ + [g 2 h + gg (ξ h)]u ξ + 2gg u + m = α n,j m n j g 2j+3 R j [u]u ξ + β i (t) i=1 i α i,j m i j g 2j+3 R j [u]u ξ. Which can be written as g 2 k u τ = g 2n+3 R n [u]u ξ + B i k(t)g 2n+3 R i [u]u ξ gg u ξ (2u + ξu ξ ) m. (3.8) i=0 44

Let k (t) = g 2n+1, m (t) = g n (k(t))g 2n+3, g (t) = 2g n 1 (k(t))g 2n+2. (3.9) Then substituting k, m, and g in equation (3.8), yields g 2n+3 u τ = g 2n+3 R n [u]u ξ + B i k(t)g 2n+3 R i [u]u ξ +2g n 1 (k(t))g 2n+3 u ξ (2u+ξu ξ )+g n (k(t))g 2n+3. i=0 Multiplying both sides by g 2n 3 gives u τ = R n [u]u ξ + B i k(t)r i [u]u ξ + 2g n 1 (k(t))u ξ (2u + ξu ξ ) + g n (k(t)). i=0 Witting the coefficient as a function of τ gives u τ = R n [u]u ξ + B i (τ)r i [u]u ξ + g n 1 (τ)(4u + 2ξu ξ ) + g n (τ). i=0 Proposition 3.2. (Gordoa et al., 2014) Extending the change of variables (3.5) with ψ(x, t) = ϕ(ξ, τ), µ(k(t)) = λ m(t) g(t) 2, (3.10) the scattering problem (3.1) is transformed into the nonisospectral scattering problem ϕ ξξ + [u µ(τ)]ϕ = 0, (3.11) where µ(τ) satisfies the nonisospectral condition Proof. Since dµ dτ = 4g n 1(τ)µ(τ) + g n (τ). (3.12) ψ(x, t) = ϕ(ξ, τ), λ = g 2 (t)µ(k(t)) + m(t), we have and ψ x = ϕ ξ dξ dx = g(t)ϕ ξ, ψ xx = g 2 (t)ϕ ξξ. 45

Substituting ψ x and ψ xx in equation (3.1) we have g(t) 2 ϕ ξξ + [ g(t) 2 u(ξ, τ) + m(t) µ(k(t))g(t) 2 m(t) ] ϕ = 0. It follows that or g(t) 2 [ϕ ξξ + (u(ξ, τ) µ(k(t)))ϕ] = 0, ϕ ξξ + [u µ(τ)] ϕ = 0. To proof that µ(τ) satisfies the nonisospectral condition (3.12) we calculate Then It follows that d dt µ = µ τ τ dt = m (t)g(t) 2 2g(t)g (t)(λ m(t)) g(t) 4. µ τ k (t) = 2g(t)g (t)[λ m(t)] m (t)g(t) 2. g(t) 4 g(t) 4 µ τ k (t) = 2 ( λ m(t) g(t) 2 ) ( ) g (t) m (t) g(t) g(t). (3.13) 2 Substituting g (t), m (t) and k (t) from (3.9) into equation (3.13) we have ( µ τ g2n+1 (t) = 2 λ m(t) ) ( ) 2gn 1 (k(t))g(t) 2n+2 + g n(k(t))g(t) 2n+3 g(t) 2 g(t) g(t) 2 µ τ g2n+1 (t) = ( 2 λ m(t) ) ( ) 2gn 1 (k(t))g(t) 2n+2 + g n(k(t))g(t) 2n+3 g(t) 2 g(t) g(t) 2 = 4 [λ m(t)] g n 1 (k(t))g 2n 1 (t) + g n (k(t))g 2n+1 (t). Substituting λ = µ(k(t))g 2 (t) + m(t) gives µ τ g2n+1 (t) = g(t) 2n+1 [4µ(k(t))g n 1 (k(t)) + g n (k(t))]. Multiplying by g 2n 1 (t) and writing the coefficients as functions of τ. we obtain µ τ = 4g n 1(τ)µ(τ) + g n (τ). 46

3.2 Scaling reductions of the Korteweg-de Vries hierarchy We have studied in Chapter 2 the scaling reduction of KdV. In this section we will explain the relationship between the derivation of Painlevé hierarchies using nonisospectral scattering problems and scaling reductions and show that the same Painlevé hierarchies are found using the two method. Corollary 3.1. (Gordoa et al., 2014) For the special case of equation (3.4) and transformation (3.5) having g(t) = 1, m(t) = g n t (g n constant), k(t) = t and h(t) and all β k (t) such that all B k (t), k = 0, 1,... n 1 are constants, the corresponding equivalent nonisospectral equation is u t = R n [u]u ξ + B j (t)r j [u]u ξ + g n. (3.14) Proof. If g(t) = 1, m(t) = g n t and k(t) = t, the transformation (3.5) becomes U = u(ξ, τ) g n t, ξ = x + h, τ = t. (3.15) It follows that U t = u τ g n, U x = u ξ. By using Proposition 3.1, substituting U t and U x in equation (3.4) we have u τ g n = R n [U]u ξ + β i (t)r i [U]u ξ, Since τ = k(t) = t, we obtain u t = R n [U]u ξ + β i (t)r i [U]u ξ + g n. i=1 i=1 Corollary 3.2. (Gordoa et al., 2014) For the special case of equation (3.4) and transformation (3.5) having 47

(g(t) = 1/[2(2n + 1)g n 1 t] 1/(2n+1) ) (g n 1 constant), m(t) = d (constant), k(t) = log(t)/[2(2n + 1)g n 1 ] and h(t) and all β k (t) such that all B k (υ), k = 0, 1,... n 1 are constants, the corresponding equivalent nonisospectral equation is u τ = R n [u]u ξ + B j (t)r j [u]u ξ + g n 1 (4u + 2ξu ξ ). (3.16) Proof. Let 2(2n + 1)g n 1 t = T By differentiating g(t), k(t), m(t) and h(t) with respect to t we obtain g (t) = 2g n 1 [2(2n + 1)g n 1 t] 2n 2/(2n+1) = 2g n 1 T 2n 2/(2n+1), and k (t) = [2(2n + 1)g n 1 t] 1 = T 1, h (t) = 0, m (t) = 0 Substituting k, g, m and h in equation (3.8) gives T 2n 3/(2n+1) u τ = T 2n 3/(2n+1) R n [u]u ξ + B i k(t)t 2n 3/(2n+1) R i [u]u ξ i=0 +2T 2n 3/(2n+1) u ξ (2u + ξu ξ ). Multiplying both side by T 2n+3/(2n+1) gives u τ = R n [u]u ξ + B i k(t)r i [u]u ξ + 2u ξ (2u + ξu ξ ). i=0 It follows that u τ = R n [u]u ξ + B i R i [u]u ξ + g n 1 (4u + 2ξu ξ ), i=0 The stationary reduction of (3.16) yields (3.3) with g n = 0, R n [u]u ξ + B i R i [u]u ξ + g n 1 (4u + 2ξu ξ ) = 0. i=0 The scaling reduction of (3.4) yields a P XXXIV hierarchy. That is, the change of variables of Corollary 3.2 explains why the stationary reduction of the nonisospectral 48

hierarchy (3.16) can also be obtained as a scaling reduction of the corresponding case of the standard hierarchy (3.4). Both techniques then yield the same Painlevè hierarchy. 3.3 A nonisospectral dispersive water wave hierarchy The DWW hierarchy has an energy-dependent scattering problem, [ ψ xx = (λ 1 2 U)2 + 1 ] 2 U x v ψ, (3.17) where λ is the constant spectral parameter. Consider the nonisospectral DWW hierarchy in u = (u, v) T u τ = R n [u]u ξ + B j (τ)r j [u]u ξ + 1 2 g n(τ) (ξu) ξ 2v + ξv ξ +g n+1 (τ) 1 0, (3.18) where R[u] is given in (2.28) with U replaced by u and x by ξ. Equation (3.18) when g n+1 (τ), g n (τ) and all B j (τ) are constants yields the equation R n [u]u ξ + B j R j [u]u ξ + 1 2 g (ξu) ξ n + g n+1 1 = 0, (3.19) 2v + ξv ξ 0 which is a generalized fourth Painlevé hierarchy if g n+1 = 0. In this section we will show that nonisospectral DWW hierarchy can be mapped to an isospectral DWW hierarchy. Proposition 3.3. (Gordoa et al., 2014) The nonisospectral hierarchy (3.18) is equivalent to the isospectral DWW hierarchy under a change of variables of the form U t = R n [U]U x + γ i (t)r i [U]U x, (3.20) i=1 U = g(t)u(ξ, τ)+m(t), V = g(t) 2 v(ξ, τ), ξ = g(t)x+h(t), τ = k(t). (3.21) 49

Proof. By differentiating U with respect to t we obtain U t = g(t)[u ξ(g x + h ) + u τ k ] + g u + m g 2 (t)[v ξ (g x + h ) + v τ k ] + 2gg v That is U t = g (t)(ξu) ξ g(t)g (t)(2v + ξv ξ ) + [g(t)h (t) g (t)h(t)]u ξ g(t)[g(t)h (t) g (t)h(t)]v ξ + g(t)k (t)u τ g 2 (t)k (t)v τ +m (t) 1 0. Differentiating U with respect to x we get U x = g2 (t)u ξ g 3 (t)v ξ Thus U x can be written as U x = g2 (t) 0 u ξ. 0 g 3 (t) Let w(x, t) = (w 1, w 2 ) T, where w 1 = g(t)u(ξ, τ), w 2 = g 2 (t)v(ξ, τ), ξ = g(t)x + h(t). Then x = g ξ, 1 x = g 1 1 ξ, and R[w]w x = 1 2 xw 1 1 x x 2 2w 2 + (w 2 ) x 1 x g2 (t) 0 u ξ. w 1 + x 0 g 3 (t) Substituting w 1 = g(t)u(ξ, τ) and w 2 = g 2 (t)v(ξ, τ) we have R[w]w x = 1 g ξgug 1 1 ξ g ξ 2 g2 (t) 0 u ξ. 2 2g 2 v + g 3 v ξ g 1 1 ξ gu + g ξ 0 g 3 (t) = 1 g[ ξu 1 ξ ξ ] 2 g2 (t) 0 u ξ. 2 g 2 [2v + v ξ 1 ξ ] g[u + ξ ] 0 g 3 (t) 50

where It follows that R[w]w x = g3 (t) 0 R[u]u ξ, 0 g 4 (t) R[w] = g 0 R[u], 0 g U = (w 1 + m, w 2 ) T By (3.21) we have U = (w 1 + m, w 2 ). Now using Lemma 2.2, we obtain R n [U]U x = = α n,j m n j R j [w]w x α n,j m n j g 0 0 g j R j [u] g2 (t) 0 0 g 3 (t) u ξ. Then where Thus R n [U]U x = α n,j m n j G j+2 R j [u]u ξ, G j = gj (t) 0. 0 g j+1 (t) U t = α n,j m n j G j+2 R j [u]u ξ + i=1 γ i i α i,j m i j G j+2 R j [u]u ξ. Using U t as in equation (??) we obtain g(t)k (t)u τ g 2 (t)k (t)v τ = α n,j m n j G j+2 R j [u]u ξ + i=1 γ i i α i,j m i j G j+2 R j [u]u ξ g (t)(ξu) ξ g(t)g (t)(2v + ξv ξ ) [g(t)h (t) g (t)h(t)]u ξ g(t)[g(t)h (t) g (t)h(t)]v ξ m (t) 1 0. 51

Which can be written as g(t)k (t)u τ = G n+2 R n [u]u ξ + B i k(t)g n+2 R i [u]u ξ g 2 (t)k (t)v τ i=1 g (t)(ξu) ξ m (t) 1 g(t)g (t)(2v + ξv ξ ) 0. (3.22) Let k (t) = g n+1 (t), g (t) = 1 2 g n(k(t))g n+2 (t), m (t) = g n+1 (k(t))g n+2 (t) (3.23) Substituting k, m, and g in equation (3.22), and multiplying both side by G 1 n+2 gives u τ v τ = R n [u]u ξ + B i k(t)r i [u]u ξ i=1 + 1 2 g nk(t) (ξu) ξ + g n+1 1. 2v + ξv ξ 0 Witting the coefficient as a function of τ gives u τ = R n [u]u ξ + B i (τ)r i [u]u ξ + 1 2 g n(τ) i=0 (ξu) ξ 2v + ξv ξ + g n+1 1 0. Proposition 3.4. (Gordoa et al., 2014) Extending the change of variables (3.21) with ψ(x, t) = ϕ(ξ, τ), µ(k(t)) = 2λ m(t), (3.24) 2g(t) the scattering problem (3.17) is transformed into the nonisospectral scattering problem ϕ ξξ = where µ(τ) satisfies the nonisospectral condition [ ( µ(τ) 1 ) ] 2 2 u + 1 2 u ξ v ϕ (3.25) dµ dτ = 1 2 g n(τ)µ(τ) + 1 2 g n+1(τ). (3.26) 52

Proof. Since ψ(x, t) = ϕ(ξ, τ), λ = 1 [2g(t)µ(k(t)) + m(t)], 2 we have and ψ x = ϕ ξ dξ dx = g(t)ϕ ξ, ψ xx = g(t) 2 ϕ ξξ. Substituting ψ xx in equation (3.17) we have [ (2g(t)µ(τ) + m(t) g(t)u(ξ, τ) g(t) 2 ϕ ξξ = 2 ) ] 2 m(t) + 1 2 g(t)2 u ξ g(t) 2 v(ξ, τ) ϕ = 0. Then That is g(t) 2 ϕ ξξ = g(t) [(µ(τ) 2 1 2 u)2 + 1 ] 2 u ξ v(ξ, τ) ϕ. ϕ ξξ = [ (µ(τ) 1 2 u)2 + 1 ] 2 u ξ v(ξ, τ) ϕ. To proof that µ(τ) satisfies the nonisospectral condition, we calculate Then That is d dt µ = µ τ τ dt = 2m (t)g(t) 2g (t)(2λ m(t)) 4g(t) 2. µ τ k (t) = 2g (t)(2λ m(t)) 2m (t)g(t) 4g(t) 2 4g(t) 2 µ τ k (t) = ( 2λ m(t) 2g(t) ) ( ) g (t) 1 m (t) g(t) 2 g(t). (3.27) Substituting g (t), m (t) and k (t) from (3.23) into equation (3.27) we have µ 2λ m(t) 1 τ gn+1 (t) = g 2 n(k(t))g(t) n+2 + 1 g n+1 (k(t))g(t) n+2 2g(t) g(t) 2 g(t) = 1 4 g n(k(t))g(t) n (2λ m(t)) + 1 2 g n+1(k(t))g(t) n+1. 53