Hamiltonian partial differential equations and Painlevé transcendents

The 6th TIMS-OCAMI-WASEDA Joint International Workshop on Integrable Systems and Mathematical Physics March 22-26, 2014 Hamiltonian partial differential equations and Painlevé transcendents Boris DUBROVIN SISSA, Trieste Lecture 2 1

Recall: the main goal is to compare the properties of solutions to the perturbed system u t = A(u)u x + A 2 (u; u x, u xx )+ 2 A 3 (u; u x, u xx, u xxx )+... u(x, 0; )= (x) with solutions to the dispersionless limit 0 v t = A(v)v x v(x, 0) = (x) 2

Recall: the main goal is to compare the properties of solutions to the perturbed system u t = A(u)u x + A 2 (u; u x, u xx )+ 2 A 3 (u; u x, u xx, u xxx )+... = 1 x H[u] u(x) u(x, 0; )= (x) with solutions to the dispersionless limit 0 v t = A(v)v x v(x, 0) = (x) 2

Of particular interest is the comparison near the point of gradient catastrophe of the dispersionless system v t = A(v)v x i.e., such a point (x 0,t 0 ) that the solution exists for t<t 0, there exists the limit lim v(x, t) t t 0 but, for some x 0 v x (x, t), v t (x, t) for (x, t) (x 0,t 0 ) The problem: to describe the asymptotic behaviour of the generic solution u(x, t; ), u(x, 0; )=u 0 (x) to the perturbed system for 0 in a neighborhood of the point of catastrophe (x 0,t 0 ) 3

Phase transition in KdV equation u t + uu x + ɛ 2 u xxx =0 4

Main Conjecture (B.D., 2005): a finite list of types of the critical behaviour 5

Main Conjecture (B.D., 2005): a finite list of types of the critical behaviour (Universality) 5

KdV asymptotics u t + uu x + 2 12 u xxx =0 u(x, t =0, )= (x) S(x, t) u(x, t, ) U ell ; r1 (x, t),r 2 (x, t),r 3 (x, t) U ell (z; r 1,r 2,r 3 ) Weierstrass elliptic function (U ell ) 2 + 4(U ell r 1 )(U ell r 2 )(U ell r 3 )=0 u(x, t, )=v(x, t)+o( ) where v t + vv x =0 v(x, 0) = (x) 6

KdV asymptotics u t + uu x + 2 12 u xxx =0 u(x, t =0, )= (x) S(x, t) u(x, t, ) U ell ; r1 (x, t),r 2 (x, t),r 3 (x, t) A.Gurevich, L.Pitayevski 73 U ell (z; r 1,r 2,r 3 ) Weierstrass elliptic function (U ell ) 2 + 4(U ell r 1 )(U ell r 2 )(U ell r 3 )=0 u(x, t, )=v(x, t)+o( ) where v t + vv x =0 v(x, 0) = (x) 6

KdV asymptotics u t + uu x + 2 12 u xxx =0 u(x, t =0, )= (x) S(x, t) u(x, t, ) U ell ; r1 (x, t),r 2 (x, t),r 3 (x, t) A.Gurevich, L.Pitayevski 73 P.Lax, D.Levermore 83 U ell (z; r 1,r 2,r 3 ) Weierstrass elliptic function (U ell ) 2 + 4(U ell r 1 )(U ell r 2 )(U ell r 3 )=0 u(x, t, )=v(x, t)+o( ) where v t + vv x =0 v(x, 0) = (x) 6

KdV asymptotics u t + uu x + 2 12 u xxx =0 u(x, t =0, )= (x) S(x, t) u(x, t, ) U ell ; r1 (x, t),r 2 (x, t),r 3 (x, t) A.Gurevich, L.Pitayevski 73 P.Lax, D.Levermore 83 U ell (z; r 1,r 2,r 3 ) Weierstrass elliptic function small oscillations Painlevé-II (U ell ) 2 + 4(U ell r 1 )(U ell r 2 )(U ell r 3 )=0 u(x, t, )=v(x, t)+o( ) where v t + vv x =0 v(x, 0) = (x) point of gradient catastrophe for the Hopf solution v(x, t) 6

KdV asymptotics u t + uu x + 2 12 u xxx =0 u(x, t =0, )= (x) S(x, t) u(x, t, ) U ell ; r1 (x, t),r 2 (x, t),r 3 (x, t) A.Gurevich, L.Pitayevski 73 P.Lax, D.Levermore 83 U ell (z; r 1,r 2,r 3 ) Weierstrass elliptic function small oscillations Painlevé-II (U ell ) 2 + 4(U ell r 1 )(U ell r 2 )(U ell r 3 )=0 solitonic asymptotics u(x, t, )=v(x, t)+o( ) where v t + vv x =0 v(x, 0) = (x) point of gradient catastrophe for the Hopf solution v(x, t) 6

KdV asymptotics u t + uu x + 2 12 u xxx =0 u(x, t =0, )= (x) S(x, t) u(x, t, ) U ell ; r1 (x, t),r 2 (x, t),r 3 (x, t) A.Gurevich, L.Pitayevski 73 P.Lax, D.Levermore 83 U ell (z; r 1,r 2,r 3 ) Weierstrass elliptic function small oscillations Painlevé-II (U ell ) 2 + 4(U ell r 1 )(U ell r 2 )(U ell r 3 )=0 solitonic asymptotics T.Claeys, T.Grava u(x, t, )=v(x, t)+o( ) where v t + vv x =0 v(x, 0) = (x) point of gradient catastrophe for the Hopf solution v(x, t) 6

Universality Conjecture, case of scalar PDEs u t + a(u)u x + 2 c(u)u xxx + b(u)u xx u x + d(u)u 3 x + O 3 =0 u(x, 0; )= (x) Compare with solutions to the unperturbed equation with the same initial condition v t + a(v) v x =0 v(x, 0) = (x) Suppose v(x, t) has a generic gradient catastrophe at (x 0,t 0 ) v 0 := v(x 0,t 0 ) 7

Part 1. For t<t 0 u(x, t; ) v(x, t) as 0 Part 2. There exists = ( ) > 0 such that the solution u(x, t; ) can be extended till t = t 0 + Part 3. The limit lim 0 2/7 [u(x, t; ) v 0 ] x = x 0 + a 0 4/7 T + 6/7 X t = t 0 + 4/7 T exists and depends neither on the choice of generic solution nor on the choice of generic Hamiltonian perturbation = 2/7 1/7, = 1/7 3/7, = 3/7 2/7 = (a 0 t 0 + f 0 ) where f ( (x)) x, = 12c 0 a 0 8

Part 1. For t<t 0 u(x, t; ) v(x, t) as 0 Part 2. There exists = ( ) > 0 such that the solution u(x, t; ) can be extended till t = t 0 + Part 3. The limit lim 0 2/7 [u(x, t; ) v 0 ] =: U(X, T) x = x 0 + a 0 4/7 T + 6/7 X t = t 0 + 4/7 T exists and depends neither on the choice of generic solution nor on the choice of generic Hamiltonian perturbation = 2/7 1/7, = 1/7 3/7, = 3/7 2/7 = (a 0 t 0 + f 0 ) where f ( (x)) x, = 12c 0 a 0 8

Equivalently, for the the case of scalar Hamiltonian PDEs at the point of phase transition x = x 0,t= t 0,u= u 0 the generic solution has the following asymptotics u(x, t) =v 0 + 2/7 U x a0 (t t 0 ) x 0 6/7, t t 0 4/7 + O 4/7 where U(X, T) is a particular solution to the ODE X = TU [ 1 6 U 3 + 1 24 ( U 2 +2UU ) + 1 240 U IV ] P 2 I (a differential equation in X depending on the parameter T) 9

Proof of Part 1 (small time behaviour): D.Masoero and A.Raimondo u t = a(u)u x + k i=1 iu (2i+1) u(x, 0; )= (x) H s, s 2k +1 Thm. The solution is continuous in =( 1,..., k) R k 10

The smooth solution X = TU U(X, T) [ 1 6 U 3 + 1 24 to P 2 I ( U 2 +2UU ) + 1 240 U IV ] Proof of existence: T.Claeys, M.Vanlessen, 2007 11

Comparison of KdV with P 2 I 0 t=0.214 0 t=0.216 0 t=0.218 u!0.5 u!0.5 u!0.5!1!1.7!1.65!1.6!1.55!1.5 x t=0.220 0!1!1.7!1.65!1.6!1.55!1.5 x t=0.222 0!1!1.7!1.65!1.6!1.55!1.5 x t=0.224 0 u!0.5 u!0.5 u!0.5!1!1.7!1.65!1.6!1.55!1.5 x t=0.226 0!1!1.7!1.65!1.6!1.55!1.5 x t=0.228 0!1!1.7!1.65!1.6!1.55!1.5 x t=0.230 0 u!0.5 u!0.5 u!0.5!1!1.7!1.65!1.6!1.55!1.5 x!1!1.7!1.65!1.6!1.55!1.5 x!1!1.7!1.65!1.6!1.55!1.5 x 12

A non-integrable case: the Kawahara equation u t +6uu x + 2 u xxx 4 u xxxxx =0 versus P 2 I 13

Motivations (for a scalar Hamiltonian equation) u t + a(u)u x + 2 c(u)u xxx + b(u)u xx u x + d(u)u 3 x + O 3 =0 The solution to a Cauchy problem for the leading term v t + a(v) v x =0 v(x, 0) = (x) can be written in the implicit form x = a(v) t + f(v), f ( (x)) x Point of gradient x 0 = a(v 0 ) t 0 + f(v 0 ) catastrophe (x 0,t 0,v 0 ) such that 0=a (v 0 ) t 0 + f (v 0 ) 0=a (v 0 ) t 0 + f (v 0 ) 14

Local structure near the point of catastrophe v(x, t 0 ) v (x 0,v 0 ) x 15

Step 1: near the generic point of gradient catastrophe the solution to the dispersionless equation can be approximated by where x + a 0 v t 1 6 f 0 v 3 x = x x 0 a 0 (t t 0 ) t = t t 0 v = v v 0 a 0 = a(v 0 ), a 0 = a (v 0 ) etc. After rescaling x x t 2/3 t v 1/3 v one obtains the above cubic equation modulo O 1/3, 0 16

Step 2: replacing the original dispersive equation by KdV near the point (x 0,t 0,v 0 ) u t + a(u)u x + 2 c(u)u xxx + b(u)u xx u x + d(u)u 3 x + O 3 =0 Galilean transformation + rescaling yields x = x x 0 a 0 (t t 0 ) x t = t t 0 2/3 t ū = u v 0 1/3 ū 7/6 ū t + a 0 ū ū x + 2 c 0 ū x x x 0, c 0 = c(v 0 ) modulo O 1/3, 0 17

The calculation: after rescaling u t 1/3 u t A monomial in the rhs m u (i 1)...u (i k) i 1 + + i k = m +1 So m u (i 1)...u (i k) 7 6 m+ 1 3 k (i 1+ +i k ) m u (i 1)...u (i k) = m 6 + k 3 1 m u (i 1)...u (i k) Must have m 6 + k 3 1 1 3 or m +2k 4 Since m 2 m =2, k =1 Only one monomial m u (i 1)...u (i k) = 2 u xxx 18

Step 3: choosing a particular solution to KdV The trick: to replace PDE (the KdV) + initial condition by an ODE ( string equation ) u t + uu x + 2 12 u xxx =0 u(x, 0; )= (x) First, rewrite the solution x = vt+ f(v), f ( (x)) x to Hopf equation v t + vv x =0 in the form v v2 F (v)+t 2 xv =0, F (v) =f(v) 19

The idea: to determine the solution to the KdV equation u t + uu x + with the same, modulo 2 12 u xxx =0 u(x, 0; )=u 0 (x) O 2 initial data from the Euler-Lagrange equation u(x) H F [u; ]+ t u2 2 xu dx =0 ( string equation ) 20

Lemma. Given a first integral H F [u] of the KdV equation then the space of solutions to the string equation x = tu+ H F [u] u(x) is invariant wrt the KdV flow Recall: Lax-Novikov lemma (1974): Lax version: given a first integral H[u] then the set of stationary points is invariant wrt the KdV flow of KdV H[u] u(x) =0 21

Novikov version: given a first integral H[u] of the KdV then the space of stationary points of the Hamiltonian flow u s = x is invariant wrt the KdV flow H[u] u(x) u s =0 u t = x H KdV [u] u(x) Indeed, from commutativity of Hamiltonians {H KdV,H} =0 it follows commutativity of the flows (u t ) s =(u s ) t So, if the KdV initial data satisfies u s =0, then so does the solution. Observe u s =0 where H[u] =H[u] const udx H[u] u(x) = const H[u] u(x) =0 22

Galilean symmetry for KdV equation x x ct t t u u + c Infinitesimal form u =1 tu x, (u ) t =(u t ) So, the string equation x = tu+ H F [u] u(x) describes the stationary points of the linear combination s u =0 Here u s = x H F [u] u(x) 23

Construction of the functional H F [u; ] uses the theory of deformations of the conservation laws (see the previous lecture) For the Hopf equation v t + vv x =0 the functional H 0 F = F (v) dx is a conservation law for an arbitrary function F 24

Theorem. For any function F there exists a deformed functional H F = F (u) 2 24 F (u)u2 x + 4 1 480 F (4) u 2 xx 1 3456 F (6) u 4 x +... dx being a conservation law for the KdV equation u t + uu x + 2 12 u xxx =0 25

An explicit formula in terms of Lax operator L = 2 2 d 2 dx 2 + u(x) Then H F = h F dx where h F = res F (1/2) (L) 26

So, one arrives at studying solutions to the Euler-Lagrange equation u(x) H F [u; ]+ t u2 2 xu dx =0 where H F = F (u) 2 24 F (u)u2 x + 4 1 480 F (4) u 2 xx 1 3456 F (6) u 4 x +... dx Explicitly x = ut+ f(u)+ 2 24 2f (u)u xx + f (u)u 2 x + 4 240 f (u)u xxxx +... 27

The last step: to apply to the string equation x = ut+ f(u)+ 2 24 2f (u)u xx + f (u)u 2 x + 4 240 f (u)u xxxx +... a rescaling near the point of phase transition x = x x 0 v 0 (t t 0 ) x t = t t 0 2/3 t ū = u v 0 1/3 ū 7/6 to arrive at x =ū t + 1 6 f 0 ū 3 + 2 4 2ū ū x x +ū 2 x 4 + 40ū x x x x + O 1/3 Choosing = 6/7 one obtains P 2 I 28

Solutions to P 2 I X = TU [ 1 6 U 3 + 1 24 ( U 2 +2UU ) + 1 240 U IV ] satisfy KdV U T + UU X + 1 12 U XXX =0 Matching condition: for large X U(X, T) (unique) root of cubic equation X = UT 1 6 U 3 Matching + smoothness choice of a particular solution 29

Proof of the Universality Conjecture for analytic solutions to KdV in T.Claeys, T.Grava, 2008 Uses: Riemann-Hilbert formulation of inverse scattering asymptotics of the scattering data of Lax operator L = 2 2 Deift-Zhou asymptotic analysis of the Riemann- Hilbert problem d 2 dx 2 + u 30

Second order systems u t = A(u)u x + A 2 (u; u x, u xx )+ 2 A 3 (u; u x, u xx, u xxx )+... 1) Hyperbolic case: eigenvalues of A(u) are real and distinct. Riemann invariants diagonalize the leading term r t + + + (r)r x + + O( )=0 r t + (r)r x + O( )=0 For the leading order system only one of Riemann invariants breaks down. Critical behaviour described by the same solution to P 2 I 2) Elliptic case. The Riemann invariants are complex conjugate. They break down simultaneously 31

Dispersionless case, characteristic directions x +, x r + x =0, r =0 x + At the critical point: Whitney fold catastrophe (hyperbolic case) x + = r + x = r + r 1 6 r3 In the elliptic case x = x +, hence the Cauchy-Riemann eqs. r + x + =0 At the critical point x + = r 2 + (elliptic umbilic singularity) 32

How to solve dispersionless equations? Example of (focusing) NLS H 0 = 1 Z (u 2 uv 2 ) dx 2 u t +(uv) x =0 v t + vv x u x =0 u t = @ x H 0 v(x) 9 >= First integrals F 0 = Z f(u, v) dx v t = @ x H 0 u(x) >; 0={H 0,F 0 } = Z apple @h0 @u @ x @f 0 @v + @h 0 @v @ x @f 0 @u dx, f vv + uf uu =0 33

Lemma. Given a first integral f, the system x = f u (u, v) t = f v (u, v), @(xu+tv @(xu+tv f(u,v)) @u = 0 f(u,v)) @v = 0 defines a solution to dispersionles NLS. Any generic solution can be obtained in this way. 34

Dispersive deformations of first integrals (NLS case) 2 h f = D NLS f = f f uuu + 3 12 2u f uu u 2 x +2f uuv u x v x uf uuu vx 2 1 + 4 f uuuu + 5 120 2u f uuu u 2 xx +2f uuuv u xx v xx uf uuuu vxx 2 1 80 f uuuuu xx vx 2 1 48u f uuuvv xx u 2 1 x 3456u 3 30f uuu 9uf uuuu + 12u 2 f 5u +4u 3 f 6u u 4 x 1 432u 2 3f uuuv +6uf uuuuv +2u 2 f 5uv u 3 x v x + 1 + 1 2160 (9f uuuuv + 10uf 5uv ) u x v 3 x 288u 9f uuuu +9uf 5u +2u 2 f 6u u 2 x v 2 x u 4320 (18f 5u +5uf 6u ) v 4 x + O( 6 ) where f vv + uf uu =0 Remark. Using dispersionless limit of the standard NLS conservation laws one obtains only half of solutions f(u,v). Another half from construction of Extended Toda/ Extended NLS hierarchy, G.Carlet, B.D., Youjin Zhang 2004 35

Adding dispersive corrections Hyperbolic case r + = r+ 0 + cx + + + 4/7 U a 6/7 x ; b 4/7 x + + O 6/7 r = r 0 + 2/7 U a 6/7 x ; b 4/7 x + + O 4/7 where U(X,T) is the solution to P 2 I 36

Elliptic case r + = r+ 0 + 2/5 w 4 5 (x+ x 0 +) + O 4/5 where w = w(z) is the tritronquée solution to the Painlevé-I eq. w =6w 2 z 37

Thm. (O.Costin et al. 2012) analyticity of the tritronquée solution in the sector arg z < 4 5 38

NLS, dispersionless NLS, tritronquée solution to PI 8 7 6 5 u 4 3 2 1 0.5 0.4 0.3 0.2 0.1 0 0.1 0.2 0.3 0.4 0.5 x u = 2 39

NLS, dispersionless NLS, tritronquée solution to PI 1 0.5 0!0.5 u!1!1.5!2!2.5 0.2 0.3 0.4 0.5 0.6 0.7 x v = Im(log ) x 40

About tritronquée solution to Painlevé-1 equation w =6w 2 z Any solution is meromorphic on the entire complex plane Distribution of poles: Theorem of Boutroux (1913) Poles of a generic solution to P-1 accumulate along the rays arg z = 2πn 5, n =0, ±1, ±2 41

Boutroux ansatz (1913): solutions to P-1 w(z) z ( ) 4 5 z5/4 for large z 42

Boutroux ansatz (1913): solutions to P-1 w(z) z ( ) 4 5 z5/4 for large z Hint: do a substitution 42

Boutroux ansatz (1913): solutions to P-1 w(z) z ( ) 4 5 z5/4 for large z Hint: do a substitution w = z 1/2 W Z = 4 5 z5/4 42

Boutroux ansatz (1913): solutions to P-1 w(z) z ( ) 4 5 z5/4 for large z Hint: do a substitution w = z 1/2 W Z = 4 5 z5/4 to arrive at 42

Boutroux ansatz (1913): solutions to P-1 w(z) z ( ) 4 5 z5/4 for large z Hint: do a substitution w = z 1/2 W Z = 4 5 z5/4 to arrive at W =6W 2 1 1 Z W + 4 25 W Z 2 42

Boutroux ansatz (1913): solutions to P-1 w(z) z ( ) 4 5 z5/4 for large z Hint: do a substitution w = z 1/2 W Z = 4 5 z5/4 to arrive at W =6W 2 1 1 Z W + 4 25 W Z 2 6 W 2 1 42

Z-plane z-plane 43

2) For any three consecutive rays there exists a unique tritronquée solution w(z) such that the lines of poles truncate along these three rays for large z Denote w 0 (z) the tritronquée solution with lines of poles truncated along the green rays w 0 (z) z 6, z, arg z < 4 5 44

Towards rigorous proof: the focusing NLS case: inverse spectral transform for the Zakharov - Shabat operator i d dx + semiclassical asymptotics for scattering data Deift - Zhou nonlinear steepest descent analysis of the Riemann - Hilbert problem 45

Towards rigorous proof: the focusing NLS case: inverse spectral transform for the Zakharov - Shabat operator i d dx + semiclassical asymptotics for scattering data??? Deift - Zhou nonlinear steepest descent analysis of the Riemann - Hilbert problem Bertola & Tovbis, 2010 45

Remark. Universality and random matrices. Z N (t; ) = 1 Vol N Z N N e 1 Tr V (M) dm Integral over the space of N N Hermitean matricesm = M V (M) = 1 2 M 2 X k t k M k+1 (k + 1)! Vol N = Vol (U(N)/U(1) U(1)) N = x, x = t Hooft parameter ) Z N (t; )! Z(x, t; ) 46

Claim. = Z(x, t; ) is tau function of a solution to Toda hierarchy i.e., the functions u = log (x+ ) (x ) 2 (x) v = @ log (x+ ) @t 0 (x) satisfy @L @t k = 1 (k + 1)! (L k+1 ) +,L where L = + v(x)+e u(x) 1, = e @ x difference Lax operator. 47

Proof uses representation Z N = h 0 h 1...h N 1 in terms of monic orthogonal polynomials p n (x) =x n +..., Z 1 1 p n (x)p m (x)e 1 V (x) dx = h n mn For t =0 p n (x) H n (x) (Hermite polynomials) ) initial conditions u(x, t =0, ) = log x, v(x, t =0, ) =0 Gradient catastrophies=various critical regimes 48

(from J.Jurkiewicz, Phys. Lett. B, 1991) 49

References B.Dubrovin, Hamiltonian perturbations of hyperbolic systems of conservation laws, II: universality of critical behaviour, Comm. Math. Phys. 267 (2006) 117-139 B.Dubrovin, T.Grava. C.Klein, On universality of critical behavior in the focusing nonlinear Schrödinger equation, elliptic umbilic catastrophe and the tritronquée solution to the Painlevé-I equation, J. Nonlinear Sci. 19 (2009) 57-94. B.Dubrovin, On universality of critical behaviour in Hamiltonian PDEs, Amer. Math. Soc. Transl. 224 (2008) 59-109 T.Claeys, M.Vanlessen, The existence of a real pole-free solution of the fourth order analogue of the Painlevé-I equation, Nonlinearity 20 (2007) 1163-1184 T.Claeys, T.Grava, Universality of the break-up profile for the KdV equation in the small dispersion limit using the Riemann - Hilbert problem, Comm. Math. Phys. 286 (2009) 979-1009. D.Masoero, Poles of integrales tritronquée and anharmonic oscillators. A WKB approach, J. Phys. A 43 (2010) 5201 B.Dubrovin, T.Grava, C.Klein, Numerical Study of breakup in generalized Korteweg-de Vries and Kawahara equations, SIAM J. Appl. Math. 71 (2011) 983-1008. D.Masoero, A.Raimondo, Semiclassical limit for generalized KdV equation before the time of gradient catastrophe, arxiv:1107.0461 O.Costin, M.Huang, S.Tanveer, Proof of the Dubrovin conjecture and analysis of tritronquée solutions of PI, arxiv:1209.1009 B.Dubrovin, M.Elaeva, On critical behaviour in nonlinear evolutionary PDEs with small viscosity, Russ. J. Math. Phys. 19 (2012) 13-22. 50

Thank you! 51

About viscous case u t + a(u)u x = [b 1 (u)u xx + b 2 (u)u 2 x]+ 2 [c 1 (u)u xxx + c 2 (u)u xx u x + c 3 (u)u 3 x]+... Critical behaviour (shock wave) u(x, t; ) =u 0 + 1/4 x x0 a 0 (t t 0 ) 3/4, t t 0 1/2 + O 1/2 Here (, ) is the logarithmic derivative of the Pearcey function (, ) = 2 log e 1 8(z 4 2z 2 +4z ) dz B.D., M.Elaeva, 2012 Particular case u t + a(u)u x = (A.Il in 1985) u xx 52