PHY411 Lecture Notes Part 8

PHY411 Lecture notes Part 8

Alice Quillen

November 26, 2018

Contents

1 Continuum Mechanics 3 1.1 Strings ...... 3

2 Inﬁnite dimensional Classical Hamiltonian systems 5 2.1 The functional derivative ...... 5

3 Examples with canonical coordinates 8 3.1 Wave equation ...... 8 3.2 Two-dimensional incompressible inviscid ﬂuid dynamics ...... 12 3.2.1 Dynamics of vortices in 2D ...... 13 3.2.2 Blinking Vortex model and time dependent models ...... 14 3.3 Conformal mapping ...... 14

4 Inviscid Barotropic Compressible Fluids 15 4.1 Eulerian and Lagrangian coordinates ...... 15 4.2 Inviscid Burger’s equation ...... 16 4.3 Working with the Determinant of the deformation matrix ...... 17 4.3.1 Functional derivatives involving the Jacobian matrix ...... 19 4.4 For Barotropic Compressible Fluids ...... 20 4.5 Lagrangian for Compressible Fluids ...... 21 4.6 Hamiltonian for Compressible Barotropic Fluids ...... 22 4.7 Changing coordinate ﬁelds and constructing a new Poisson bracket . . . . . 23 4.8 Another way to write Inviscid irrotational ﬂuid dynamics ...... 26

5 Poisson manifolds 27 5.1 Example: Inviscid Burgers’ equation ...... 29 5.1.1 The symplectic two form ...... 31 5.2 Conserved Quantities and Continuity Equations ...... 32

1 6 The KdV equation 33 6.1 Rescaling ...... 35 6.2 Soliton Solutions ...... 35 6.3 Infinite number of conserved quantities ...... 36 6.3.1 Integrability ...... 39 6.4 Lax pair ...... 39 6.5 Connection to Schrödingerequation ...... 41 6.6 Connection to Inverse Scattering method ...... 41 6.7 Zero Curvature Condition ...... 42 6.8 Two different Poisson brackets ...... 45

7 Symplectic Reduction 48

Figure 1: A string!

2 1 Continuum Mechanics

1.1 Strings Consider a string of linear mass µ(x) and under tension τ(x) where x is horizontal distance along the string when it is at rest. Its vertical displacement we describe with a function y(x, t) that is also a function of time. When the string is at rest y(x, t) = 0. A length of the string due to displacement is p ds = dx2 + dy2 s ∂y 2 = 1 + dx ∂x ! 1 ∂y 2 ∼ dx 1 + (1) 2 ∂x

The change in length locally is

dx ∂y 2 dl = ds − dx = 2 ∂x

The diﬀerential potential energy (force times distance)

1 ∂y 2 dU = τ(x)dl = τ(x) dx 2 ∂x

The diﬀerential kinetic energy

1 ∂y 2 dT = µ(x) dx 2 ∂t

We ﬁnd the total Lagrangian by integrating over x, Z L = dx L(y, y,˙ y0; x, t) and L we call the Lagrangian density

1 ∂y 2 1 ∂y 2 L(y, y,˙ y0; x, t) = µ(x) − τ(x) (2) 2 ∂t 2 ∂x

The action is a double integral

Z tb Z xb Z tb Z xb 1 S = dt dxL(y, y,˙ y0; x, t) = dt dx µy˙2 − τy02 ta xa ta xa 2

3 When we minimize the action we can consider minimizing it for paths y(x, t) → y(x, t)+ h(x, t).

Z tb Z xb h i ∂S = dt dx L(y + h, y˙ + h,˙ y0 + h0; x, t) − L(y, y,˙ y0; x, t) ta xa Z tb Z xb ∂L ∂L ˙ ∂L 0 = dt dx h + h + 0 h (3) ta xa ∂y ∂y˙ ∂y We integrate by parts " # Z xb Z tb Z xb Z tb tb ∂L d ∂L ∂L dx dt h˙ = dx − h + h ∂y˙ dt ∂y˙ ∂y˙ xa ta xa ta ta " # Z xb Z tb Z tb Z xb xb ∂L 0 d ∂L ∂L dx dt h = dt − h + h ∂y0 dx ∂y0 ∂y0 xa ta ta xa xa Inserting these into equation 3

Z tb Z xb Z xb tb Z tb xb ∂L d ∂L d ∂L ∂L ∂L ∂S = dt dx − − h + dx h + dt h ∂y dt ∂y˙ dx ∂y0 ∂y˙ ∂y0 ta xa xa ta ta xa With suitable boundary conditions ∂L d ∂L d ∂L − − = 0 ∂y dt ∂y˙ dx ∂y0 Evaluating the derivatives with τ, µ independent of x and t for the Lagrangian in equation 2 ∂L ∂L = µy˙ = −τy0 ∂y˙ ∂y0 giving ∂2y ∂2y µ − τ = 0 ∂t2 ∂x2 which is the wave equation. We can perform a Legendre transformation (at each value of x) ∂L π(x) = = µy˙(x) ∂y˙(x) Inverting this y˙(π(x)) = π(x)/µ and we ﬁnd a Hamiltonian or energy density π(x)2 y0(x)2 H(π(x), y(x), y0(x)) = π(x)y ˙(π) − L(y, y˙(π), y0, x) = + τ 2µ 2

4 The total energy is integrated over x Z E = dxH(π(x), y(x), y0(x)) and because the Hamiltonian does not depend on time, the total integrated energy should be a constant. It is confusing to see our functions depend on y0 here. It is helpful to think in the limit of a discrete system as the number of discrete items becomes large. The string can be considered as a large N limit of many coupled bodies, where the coupling depends on the differences between nearby bodies that is here represented by the spatial derivative. Each discrete body is at a different x, so we can think of N different bodies at N different x positions. In the large N limit the number of bodies becomes infinite. In this sense π(x) is an infinite dimensional vector of momenta conjugate to y(x), an infinite dimensional vector of coordinates. Our Hamiltonian is really depend on momenta π(x) and coordinates y(x) and y0(x) is a function of nearby coordinates.

2 Inﬁnite dimensional Classical Hamiltonian systems

2.1 The functional derivative Instead of just conjugate vectors p, q we can consider conjugate fields ρ(x) and φ(x). In the continuum limit the Hamiltonian is an integral over all space. Z H[ρ, φ] = dxf(ρ(x), φ(x)) and f(ρ, φ) a function that depends on the fields ρ(x), φ(x). The value of the fields at each position x serve as coordinates. These are functions of a continuous variable, effectively like an infinite number of coordinates. How do we take derivatives with respect to fields instead of with respect to coordinates? We use a functional derivative δF [u(x)] 1 ≡ lim (F [u(x) + δ(x − y)] − F [u(x)]) δu(y) →0

Here, as is common in physics, the delta function is used. Above F is a functional (operating on the function or ﬁeld u) and u is a function. More rigorously the functional derivative can be deﬁned Z δF 1 (x)g(x)dx = lim [F [ρ + g] − F [ρ]] δρ →0 d = [F [ρ + g]] (4) d =0

5 where F is a functional (function of functions) and ρ is a function and g is an arbitrary function. We can think of δF/δρ as the gradient of F at the point ρ. In analogy to

X ∂F dF = dρ ∂ρ i i i Z δF [ρ] δF = g(x)dx δρ(x) with g(x)dx serving as dρi. A function can be written as a functional using an integral and a delta function Z Fx[φ] = dy δ(y − x)φ(y) = φ(x)

For this point case the functional derivative δF [φ] 1 x = lim [φ(x) + δ(x − y) − φ(x)] δφ(y) →0 = δ(x − y)

This follows as the value of the derivative can only change at the location of x. Consider a slightly more diﬃcult but point case

Fx[φ] = ∇φ(x)

The functional derivative

δFx[φ] 1 = lim [∇x(φ(x) + δ(x − y)) − ∇xφ(x)] δφ(y) →0

= ∇xδ(x − y)

The functional derivative satisﬁes the product rule δ(FG)[ρ] δF [ρ] δG[ρ] = G[ρ] + F [ρ] δρ(x) δρ(x) δρ(x) If g is a function then the chain rule is δF [g(ρ)] δF [g(ρ)] dg(ρ) = δρ(y) δg(ρ(y)) dρ(y) If G is an operator or a functional then the chain rule is Z δ δF [G] δG[ρ] F [G[ρ]] = dx δρ(y) δG(x) G=G[ρ] δρ(y)

6 Suppose we have a function Z F [ρ] = f(x, ρ(x), ∇ρ(x))dx

The functional derivative is this δF [ρ] ∂f ∂f = − ∇ · . (5) δρ(x) ∂ρ ∂∇ρ

We now show this using the deﬁnition for the functional derivative in equation 4

Z ∂F [ρ] d Z g(x)dx = f(x, ρ(x) + g(x), ∇ρ + ∇g)dx δρ(x) d =0 1 Z ∂f ∂f = lim dx f(x) + g(x) + ∇g(x) − f(x) →0 ∂ρ ∂∇ρ Z ∂f ∂f = g + · ∇g dx ∂ρ ∂∇ρ Z ∂f ∂f ∂f = g + ∇ · g − ∇ · g dx ∂ρ ∂∇ρ ∂∇ρ Z ∂f ∂f = g − ∇ · g dx ∂ρ ∂∇ρ where we have assumed that φ or f are zero on the boundary of the integral. The last line shows that we have derived equation 5. For a Hamiltonian H[ρ, φ], with ρ acting like a momentum and φ acting like a coordinate, we can write Hamilton’s equations as

∂ρ(x) δH[φ, ρ] = − ∂t δφ(x) ∂φ(x) δH[φ, ρ] = ∂t δρ(x) in analogy to Hamilton’s equations for discrete or ﬁnite systems. What is the Poisson bracket? The equation of motion should be consistent with

φ˙(x) = {φ(x),H[φ, ρ]} ρ˙(x) = {ρ(x),H[φ, ρ]}

In the ﬁnite dimensional setting, the Poisson bracket is summed over derivatives with respect to all pairs of momenta and coordinates. We expect that in the continuum limit the Poisson bracket will involve an integral and instead of derivatives of functional H we will be taking functional derivatives of H. As ρ(x), φ(x) are both ﬁelds and canonical

7 coordinates the Poisson bracket with ρ will only involve an integral of a functional derivative of H with respect to φ and the Poisson bracket with φ would be the integral of a functional derivative of H with respect to ρ. We expect something like this Z δρ(y) δH {ρ(y),H[φ, ρ]} = dx(−) δρ(x) δφ(x) Z δH = − dx δ(y − x) δφ(x) Z δφ(y) δH {φ(y),H[φ, ρ]} = dx δφ(x) δρ(x) Z δH = dx δ(y − x) δρ(x) That the ﬁelds are like canonical coordinates is seen with the delta function in the integrals.

3 Examples with canonical coordinates

3.1 Wave equation We consider a ﬁeld φ(x). The time derivative φ˙(x). We can assume that the Lagrangian can be written as a sum of kinetic and potential energy terms

L(φ, φ˙) = T [φ˙] − W [φ]

The total kinetic energy is an integral over space of the kinetic term Z 1 T [φ˙] = φ˙2d3x 2 and the potential energy Z W [φ] = V (φ(x))d3x where V (φ) is a function (and it could depend on the gradient of the potential). We have a momentum δL δT π(x) = = = φ˙(x) δφ˙(x) δφ˙(x) Using the momentum π(x) we can construct a Hamiltonian Z 1 H[π, φ] = π(x)2 + V (φ(x)) dx 2 where π(x), φ(x). The potential energy can be something interesting that includes terms like 1 V (φ) = |∇φ|2 + U(φ) 2

8 Using the functional derivative we check δH 1 Z = lim dx[π(x) + δ(x − y)]2 − π(x)2] δπ(y) →0 2 1 Z = lim dx[π(x)2 + 2π(x)δ(x − y) − π(x)2] →0 2 1 Z = lim dx π(x)δ(x − y) →0 = π(y)

δH 1 Z 1 = lim dx [∇(φ(x) + δ(x − y))]2 + U(φ(x) + δ(x − y)) δφ(y) →0 2 Z 1 − dx[ [∇(φ(x)]2 + U(φ(x))] 2 Z = dx ∇φ(x) · ∇δ(x − y) + U 0(φ(x))δ(x − y)

= −∇2φ(y) + U 0(φ(y)) where the last step for the left term is done by parts. The equations of motion can be written as φ˙ = π π˙ = ∇2φ − U 0(φ) (6) and φ¨ = ∇2φ − U 0(φ) Neglecting U 0(φ) we recover the wave equation. For this system what is the Poisson bracket? The equation of motion should be consistent with φ˙(x) = {φ(x),H[φ, π]} π˙ (x) = {π(x),H[φ, π]} The Poisson bracket is summed over derivatives with respect to all pairs of momenta and coordinates. We expect that in the continuum limit the Poisson bracket will involve an integral and instead of derivatives we will be taking functional derivatives. As π(x), φ(x) are canonical coordinates the Poisson bracket with π will only involve an integral of a functional derivative of H with respect to φ and the Poisson bracket with φ would be the integral of a functional derivative of H with respect to π. In other words we expect something like this Z δπ(y) δH {π(y),H} = dx(−) δπ(x) δφ(x) Z = dx δ(y − x) ∇2φ(x) − U 0(φ(x))

= ∇2φ(y) − U 0(φ(y))

9 and this is consistent with our equations of motion (equation 6). Likewise

Z δφ(y) δH {φ(y),H} = dx δφ(x) δπ(x) Z = dx δ(y − x)π(x) = π(y)

This dynamical system is consistent with a Poisson bracket that gives

{φ(x), π(y)} = δ(x − y) {π(x), π(y)} = {φ(x), φ(y)} = 0 (7)

This is equivalent in the continuum limit with

{pi, qj} = δij

{pi, pj} = {qi, qj} = 0 where there are many possible coordinates. Our coordinates and momenta φ(x), π(x) are canonical because they obey the Poisson bracket that we expect in the continuum limit. In the ﬁnite dimensional setting we can use a single vector x = (q, p) for both coordinates and momenta. The Poisson bracket can be written

ij {xi, xj} = J with J the symplectic matrix 0 I J = −I 0 where I is the identity matrix. For functions ∂f ∂g {f, g} = J ij ∂xi ∂xj For the wave equation we have momenta and coordinate ﬁelds π and φ. We can describe this in terms of a single ﬁeld ρi with ρ0 = φ and ρ1 = π. With this notation

ij {ρi(x), ρj(y)} = J δ(x − y)

10 Figure 2: The velocity v = (−∂yψ, ∂xψ) depends on the stream function ψ. Because v · ∇ψ = 0, the velocity is perpendicular to the gradient of the stream function. This means that stream lines are level curves of the stream function. Here we show stream lines for ﬂow around a cylinder. The ﬂow is irrotational and so could also be described with a potential v = ∇φ = (∂xφ, ∂yφ). The velocity v is parallel to the gradient of the potential. This means that velocity potential level curves are everywhere perpendicular to streamlines.

Figure 3: This ﬂow has vorticity. If the ﬂow is 2d and incompressible we can describe it with a stream function.

11 3.2 Two-dimensional incompressible inviscid ﬂuid dynamics In a vorticity free setting we might use potential ﬂow (v = ∇φ). However in a two- dimensional incompressible setting we might use a stream function, v = ∇ × (ψzˆ). Let us use a stream function ψ(x, y) with velocity v = (u, v)

u = −∂yψ

v = ∂xψ and using short hand ∂x = ∂/∂x. Computing

∇ · v = ∂xu + ∂yv = −∂xyψ + ∂yxψ = 0

So by describing the system in terms of the stream function we automatically have assumed incompressible ﬂow. The vorticity (z component)

ωz = (∇ × v) · zˆ

= ∂xv − ∂yu 2 = ∂x∂xψ + ∂y∂yψ = ∇ ψ

So the function ∇2ψ is the vorticity. Using a Poisson bracket

{f, g} = ∂xf∂yg − ∂yf∂xg (8) we can compute

{x, ψ} = ∂yψ

{y, ψ} = −∂xψ

Here the coordinate x is conjugate to the coordinate y (which is acting like a momentum) and the coordinates are canonical coordinates. Now let us consider the motions of a tracer particle moving with the ﬂow with a trajectory x(t), y(t). Because the tracer particle is moving with the ﬂow equations of motion are

x˙(t) = u(x, y, t) = −∂yψ(x, y, t) = −{x, ψ}

y˙(t) = v(x, y, t) = ∂xψ(x, y, t) = −{y, ψ}

We can associate −ψ, minus the stream function, with the Hamiltonian and the Poisson bracket is canonical.

12 3.2.1 Dynamics of vortices in 2D The stream function acts like a potential and satisﬁes a Poisson equation, −∇2ψ = ω. The Greens’ function for a point vortex at position xi is Γ ψ(x) = − i log ||x − x || 2π i corresponding to a vorticity ﬁeld

ω(x) = Γiδ(x − xi) where Γi is a constant. We have assumed that we are working on the inﬁnite real plane. The associated velocity ﬁeld for this vortex is

Γi 1 u(x, y) = ∇ × (ψzˆ) = 2 [(y − yi)xˆ − (x − xi)yˆ] (9) 2π ||x − xi|| For N point vortices the vorticity ﬁeld

N X ω(x) = Γiδ(x − xi) i=1 Note: The above Greens function and associated velocity field is good when there is no boundary (or we are working on the plane). Solving Laplace’s equation given non-trivial boundaries (like a circular boundary) (and with point discontinuities) is more difficult! However because solutions are harmonic, analytical functions can be used to map the boundary to a nicer one. This can make it possible to solve the problem on a circle, for example, and then transfer back to real space. Since vorticity is advected by the flow, the motion of one vortex depends on the velocity fields caused by all the other vortices. Summing over the velocity fields (equation 9)

X Γj 1 x˙ i(t) = 2 [(yi − yj)xˆ − (xi − xj)yˆ] 2π ||xi − xj|| j6=i

These equations of motion can also be modeled with Hamiltonian and Poisson bracket

X ΓiΓj H(x , y , ....) = − log ||x − x || i i 4π i j i6=j

X 1 {f, g} = [∂ f∂ g − ∂ f∂ g] Γ xi yi yi xi i i The dynamics is that of an N-dimensional Hamiltonian system (2N phase space) so the inﬁnite dimensional system has been reduced to a ﬁnite dimensional one.

13 3.2.2 Blinking Vortex model and time dependent models A time dependent low dimension dynamical system can be constructed by using point vortices that have time dependent vorticity Γi(t). Even with only 2 vortices this can be a chaotic system. I think one can construct a time dependent Hamiltonian for this system simply by replacing Γi with Γi(t) however this requires a time dependent Poisson bracket (and a procedure to handle when Γi(t) = 0). The vortices can be advected with the flow or held fixed. The blinking vortex model by Aref (H. Aref, Journal of Fluid Mechanics 143, 1 (1984)) with 2 fixed vortices that alternate in activity, each taking turns shifting between a fixed vorticity value and zero, became a paradigm for chaotic advection.

3.3 Conformal mapping We consider x, y to be the complex plane with z = x + iy. We deﬁne a complex potential function Φ(x, y) = φ(x, y) + iψ(x, y) (10) where φ and ψ are both real functions. The Cauchy-Reimann equations for Φ(x, y) to be diﬀerentiable with respect to z,

∂xφ = ∂yψ

∂yφ = −∂xψ. (11) We associate φ with the velocity potential

u = ∇φ

(ux, uy) = (∂xφ, ∂yφ). (12) We associate ψ with a stream function

(ux, uy) = (−∂yψ, ∂xψ). (13) Comparison between these relations show that the Cauchy-Reimann conditions (equations 11) are satisfied if derivatives of the velocity potential and stream function give the velocity vector as in equations 12 and 13. We can use a potential on the complex plane Φ that is differentiable. The velocity potential is harmonic (satisfies Laplace’s equation). In regions where there is no vorticity, the stream function is also harmonic. If the complex potential function Φ is analytic, (that means can be written as a power series in z and is infinitely differentiable) then both φ and ψ must be harmonic. This follows 2 2 because any function of f(x + iy) must satisfy Laplace’s equation; ∂xf = −∂y f. We can think of flow solutions as analytic functions Φ, that are harmonic everywhere except at boundaries where there might be vorticity or a point vortices. Then the real and imaginary parts of Φ are the velocity potential and stream function.

14 It is handy that complex functions provide a large supply of possible harmonic functions, or solutions to the the two-dimensional Laplace equation. A solution strategy for a boundary value problem on a complicated domain is to transform it into a solvable case by a change of variables that is realized by an analytical map. A handy map is z − 1 f(z) = z + 1 which maps the right half plane (Re z > 0) to the unit disk. If you can solve a problem on the unit disk, then you can also solve it on a half plane. Another handy map is the Joukowski map 1 1 f(z) = z + 2 z The Joukowski map squashes the unit circle down to the real line segment [-1,1] and is useful for solving ﬂows around airplane wings. A geometrical property enjoyed by analytic functions is that, at non-critical points (where f 0(z) 6= 0), they preserve angles, and therefore are conformal mappings.

4 Inviscid Barotropic Compressible Fluids

4.1 Eulerian and Lagrangian coordinates We refer to the background or Eulerian coordinates as x and those moving with the fluid or material as X and these are the Lagrangian coordinates. Quantities can be described as functions of x and time T or as functions of X and T . We refer to ∂t as a partial derivative with respect to time with x held fixed (the Eulerian derivative). We refer to ∂T as a partial derivative with respect to time with X held fixed (the Lagrangian derivative). We have an invertible map x(X, t). The deformation of space can be described in terms of the matrix ∂x ∂X ∂xi = F i (14) ∂Xj j with inverse ∂Xi = (F −1)i ∂xj j It will be useful later to have as short hand F = Dx with D a derivative, and j ∂ ∂x j ∂Xi = i = ∂xj = Fi ∂xj ∂X ∂Xi j ∂ ∂X −1 j ∂xi = i = ∂Xj = (F )i ∂Xj (15) ∂x ∂xi

15 The velocity of a material point

u(X,T ) = ∂T x(X,T ) and that expressed in terms of the background coordinates

u(x,T ) = u(x(X,T )) = U(X,T )

Using the chain rule, the material time derivative

∂xi ∂ ∂ = ∂ + T t ∂t ∂xi = ∂t + u · ∇x where ∇x refers to derivatives with respect to x. We can compute the acceleration of a material point using this time derivative ∂u ∂ u = + u · ∇ u T ∂t x We can describe a particle distribution (or density) as a function of either x or X. Let us describe the density in material coordinates as q(X,T ) and that in the background as ρ(x,T ). These two densities are related.

Z Z d3x q(X,T )d3X = q(X(x,T )) V V J with the Jacobian J = det F ∂xi i computed from the deformation matrix (equation 14; ∂Xj = Fj ). We identify the Eulerian density q(X(x,T ),T ) ρ(x,T ) = J Usually we consider material distributions that are not time dependent so q(X) and

q(X(x,T )) ρ(x,T ) = . J(x)

4.2 Inviscid Burger’s equation Ignoring the density and energy density for the moment, let us look at a Lagrangian containing only a kinetic energy term Z 1 L[∂ x, x] = (∂ x)2d3X. T 2 T

16 Here x(X) is a ﬁeld that is a function of X. To obtain a momentum δL = ∂T x(X) = π(X) = u(X) δ(∂T x(X)) Using this momentum ﬁeld Z π2 H[x, π] = d3X. 2 The equations of motion (and using a canonical Poisson bracket)

∂T x(X) = π(X) = u(X)

∂T π(X) = 0 Putting these together ∂T u = 0 and in 1 dimension this is equivalent to

u,t + uu,x = 0 which is known as the inviscid Burgers equation.

4.3 Working with the Determinant of the deformation matrix Below when considering a Hamiltonian or Lagrangian that contains terms dependent on density, we need to do a functional derivative involving the Jacobian matrix. The determinant in three dimensions for a matrix B can be written 1 det B = BiaBjbBkc 3! ijk abc Previously we deﬁned Jacobian matrix ∂xi F i = j ∂Xj and 1 J = det F = abcF iF jF k 3! ijk a b c Using this form for the Jacobian J (and with some index rearrangement)

∂J 1 abc i j k = ∂ l F F F l Fm ijk a b c ∂Fm 3! 1 h i = abcF iF jδkδc + abcF iF kδjδb + abcF jF kδiδa 3! ijk a b l m ijk a c l m ijk b c l m 1 = F iF j mab 2 a b lij

17 It may be useful for us to deﬁne a matrix A that looks like this

m ∂J 1 i j mab Al = l = FaFb lij (16) ∂Fm 2 If we multiply this by the matrix F 1 AmF l = F iF jF l mab l k 2 a b k lij 1 ∂xi ∂xj ∂xl = mab 2 ∂Xa ∂Xb ∂Xk lij Each term involves an antisymmetric sum of 3 derivatives. If m 6= k then two of the derivatives are the same and two terms cancel giving zero. If m = k then two terms are the same and the entire thing is equal to the Jacobian matrix. We can write

m l l Al Fk = Jδk (17)

This relation implies that A/J is the inverse of F and that means (see equation 15)

1 i ∂ k = A ∂ i (18) x J k X Using equation 16 we can show that

∂ 1 ∂ ∂xi ∂xj Am = mab ∂Xm j 2 ∂Xm ∂Xa ∂Xb lij ∂2xi ∂xj ∂xi ∂2xj = + mab ∂Xa∂Xm ∂Xb ∂Xa ∂Xb∂Xm lij ∂2xi ∂xj ∂xi ∂2xj = − mab ∂Xa∂Xm ∂Xb ∂Xb ∂Xa∂Xm lij = 0 (19)

The Jacobian J depends on the deformation F = Dx as J = det F. It might be useful to know that for matrix A

det(I + A) ∼ 1 + trA + O(2)...

This can be shown also using the above index notation.

18 We take the time derivative of J

∂J i ∂T J = ∂T Fj ∂ i Fj j i = Ai ∂Xj (∂T x ) j i = Ai ∂Xj (u ) j k i = Ai Fj ∂xk u k i = Jδi ∂xk u i = J∂xi u = J∇x · u

Using our relation between densities

ρJ = q

J∂T ρ + ρ∂T J = ∂T q ∂ q ∂ ρ + ρ∇ · u = T (20) T x J

We recover the mass continuity equation as long as we set ∂T q = 0. As q(X,T ) is mass distribution in material coordinates requiring it to be independent of T is probably rea- sonable.

4.3.1 Functional derivatives involving the Jacobian matrix

We can look at a functional Z 3 F1[Dx] = d XJ

Z δF1 1 3 ∂J ∂ i i i (Y) = lim d X i m x (X) + δ(X − Y) − x (X) δx →0 ∂Fm ∂X Z 3 ∂J ∂ = d X i m δ(X − Y) ∂Fm ∂X Z 3 ∂ ∂J = − d X m i δ(X − Y) ∂X ∂Fm ∂ = − Am(Y) ∂Xm i A slightly more complicated functional Z 3 F2[Dx] = d X f(J) (21)

19 and following the above computation δF ∂ 2 (X) = − f 0Am (X) (22) δxi ∂Xm i

Now consider Z 3 −1 F3[Dx] = d XJ and we follow the previous computation

δF ∂ 1 3 (X) = Am (X) δxi ∂Xm i J 2

Now consider Z 3 −1 F4(Dx) = d X f(J ) (23) and we follow the previous computation again

δF ∂ 1 4 (X) = f 0Am (X) (24) δxi ∂Xm i J 2

4.4 For Barotropic Compressible Fluids With e internal energy, v = 1/ρ the speciﬁc volume, p the pressure, T the temperature, s the speciﬁc entropy de = −pdv + T ds The enthalpy h = e + pv and dh = vdp + T ds With some arithmetic d(ρe) = hdρ + ρT ds Derivatives keeping s constant are p ∂ e = ρ ρ2 ∂ρ(ρe) = h ∂ p ∂ h = ρ (25) ρ ρ

20 4.5 Lagrangian for Compressible Fluids We consider a barotropic fluid with internal energy density e(ρ) a function of Eulerian density alone. With the wave equation the Lagrangian density depended on the time derivative of a field, and the gradient of the field. We expect the Lagrangian density here to depend on the local velocity, giving the kinetic term, and the energy density that depends on the deformation gradient. We integrate the Lagrangian density over the Lagrangian coordinates and our our field is x. The kinetic energy density 1 1 T [x](X) = q(∂ x)2 = qU2 2 T 2 The internal energy density U[x](X) = qe(ρ) = qe(q/J)

i The Jacobian depends on the derivative Dx which has two indices with components Fj = ∂xi ∂Xj . The Lagrangian density is the difference between the kinetic energy density and the internal energy density 1 L(∂ x, Dx, X) = q(∂ x)2 − qe(q/J(Dx)) T 2 T depending on the time and spatial derivatives of our field x and our Lagrangian coordinates X. The total Lagrangian is the integral of the Lagrangian density over the material coordinates Z 3 L[x] = L(∂T x, Dx, X)d X Z 1 = q(∂ x)2 − qe(q/J)d3X (26) 2 T Taking the functional derivative δL = q∂T x = q(X)U(X) (27) δ(∂T x(X)) The above equation is a vector equation and is satisfied for each index. The above equation also gives us a momentum π(X) = qU (28) field that we can use to form a Hamiltonian. We can use Lagrange’s equations to find the equations of motion or we can use the Hamiltonian and a canonical Poisson bracket that contains a delta function to generate the equations of motion. If we use Lagrange’s equa- δL tions, then we need to include a term to them that depends on ∂Xi i as the Lagrangian δFj contains gradient terms (through J). Lagrange’s equations are

∂T (∂∂T xL) + ∂X(∂DxL) − ∂XL = 0

21 or d ∂L d ∂L ∂L i + j i − i = 0 dt ∂(∂T x ) dX ∂(∂Xj x ) ∂X Further manipulation of this can be used to check that we get Euler’s equation. We will transfer this system into a Hamiltonian and check that we get Euler’s equation using that framework instead.

4.6 Hamiltonian for Compressible Barotropic Fluids Our Hamiltonian density corresponding to the Lagrangian in equation 26 is

π2 H[π, x](X) = + qe(q/J) 2q and the total Hamiltonian Z π2 H[π, x] = d3X + qe(q/J) (29) 2q

To ﬁnd the equations of motion we need to understand how to take the variational derivatives. The variational derivative with respect to π is straightforward giving

δH π(X) ∂ x(X) = = = U(X) T δπ(X) q and this is our ﬁrst equation of motion. For the second equation of motion we need to take the functional derivative of H with i i i respect to x . Writing the potential energy density as qe(q/J) = U(∂Xj x , ....) = U(Fj , ....) where these are functions of all possible i, j, Z δH 1 3 i (X) = lim d Y U(∂Xj (x + δ(X − Y), ...) − U(F) δxi →0 Z 1 3 X = lim d Y (∂∂ xi U(F))∂Xj δ(X − Y) →0 Xj j Z 3 0 q = − d Yqe (q/J) (∂F i J)∂Xj δ(X − Y) J 2 j h 0 2 i = ∂ j e (ρ)ρ (∂ i J) X Fj h ji = ∂Xj pAi and the second to last step we used equation 25 and the last step we have used equation 16 (also see equation 24) for functionals involving J −1). The Lagrangian or material density

22 q(X,T ) does not depend on x so its spatial derivatives don’t contribute to the variational derivative of H. We want the above expression in terms of ∂x = ∇x instead of ∂X = ∇X.

δH j j = p∂ j A + A ∂ j p δxi X i i X j k = 0 + Ai Fj (∂xk p) k = Jδi (∂xk p)

= J∂xi p δH = J∇ p (30) δx x where I have used equation 15 and equation 18 for the gradient and 19 for the zero diver- gence term. Hamilton’s equations will give δH ∂ π = ∂ (qU) = − T T δx Putting this together with equation 30

q∂T u + u∂T q = −J∇xp ∇ p u∂ q ∂ u = − x − T T ρ q The last term on the right must be zero in order to obey mass conservation, see equation 20. If we ignore it, then we have recovered Euler’s equations.

4.7 Changing coordinate ﬁelds and constructing a new Poisson bracket In a ﬁnite dimensional Hamiltonian system with coordinates x = q, p the Poisson bracket ∂f ∂g {f, g} = J ij ∂xi ∂xj dxi δH = J ij , dt δxj (here J is the symplectic matrix and nothing to do with the det F). If we have another set of coordinates that are not necessarily canonical, z dzi ∂zi dx = j dt ∂xj dt ∂zi δH = J jk ∂xj δxk ∂zi ∂zm δH = J jk ∂xj ∂xk δzm = {zi,H}

23 So we can change the coordinates and write the Poisson bracket as ∂f ∂zi ∂zm ∂g {f, g} = J jk ∂zi ∂xj ∂xk ∂zm ∂f ∂g = {zi, zm} ∂zi ∂zm Non-canonical coordinate transformations and the same Hamiltonian (but in new variables) can still give the correct equations of motion if the Poisson bracket is changed. If our original coordinates are canonical fields φ, π and F,G are functionals and we want to write the Poisson bracket in terms of a new fields q(φ, π), p(φ, π) then the Poisson bracket Z δF δG δG δF {F,G} = dx − δφ δπ δφ δπ Z δF δq δF δp δG δq δG δp = dx + + + δq δφ δp δφ δq δπ δp δπ δF δq δF δp δG δq δG δp − + + δq δπ δp δπ δq δφ δp δφ As an example we will work with the barotropic hydrodynamical system and change to new fields, finding a new Hamiltonian with respect to these new fields and a new and more complex Poisson bracket that is computed using these new fields. For this system the original fields x(X) and π(X) are functions of Lagrangian variable X and these are canonical variables. Suppose for this compressible fluid system we would rather not work with our deformation field x and associated momentum π = qU and instead would prefer to work with the variables ρ(x,T ), the Eulerian density, and the Eulerian momentum density m(x) = ρ(x,T )u(x,T ). In terms of our canonical variables and Lagrangian quantities q ρ(x) = J(x) q π(x) m(x) = ρ(x)u(x) = u(x) = . (31) J(x) J(x) We choose density in Lagrangian coordinates q = 1 to simplify the spatial derivatives (they are bad enough to work with even with q = 1). We will try to derive a new Poisson bracket integrated over Eulerian coordinate x instead of integrated over Lagrangian coordinate X. We start with any functional Z F [ρ, m] = d3xf(ρ(x), m(x)) (32)

Transferring the coordinates used to integrate to X Z F [ρ, m] = d3X Jf(ρ(x(X)), m(π(X), x(X))) (33)

24 Note the appearance of Jacobian J inside the integral. We will use equation 22 and 24 on our functional F () that depends on J and J −1 = ρ δF (setting q = 1) to compute δx k δF ∂f m 1 ∂f m π m = ∂ m A + ∂ m A − ∂ m (fA ) δxi X ∂ρ i J X ∂mk i J X i

i Equation 19 gives ∂Xm Am = 0 and we expand the last term k δF m ∂f m ∂f k m ∂f ∂ρ m ∂f ∂m = A ∂ m ρ + A ∂ m m − A − A δxi i X ∂ρ i X ∂mk i ∂ρ ∂Xm i ∂mk ∂Xm m ∂f m k ∂f = A ρ∂ m + A m ∂ m i X ∂ρ i X ∂mk

−1 i We can change from ∂X to ∂x using equation 18 (which is ∂xk = J Ak∂Xi ) δF ∂f k ∂f = J ρ∂ i + m ∂ i δxi x ∂ρ x ∂mk

∂f δF Using equation 32 we can write ∂ρ as δρ . Wring the above equation in vector notation δF δF δF = J ρ∇ + ∇ · m (34) δx x δρ x δm It is more straightforward to compute the other functional derivative that we need. Using equation 33 δF δF = (35) δπ δm Because π(X) and x(X) are canonical coordinates they satisfy a canonical Poisson bracket. For functionals F,G Z δF δG δG δF {F,G} = d3X − δx δπ δx δπ Inserting equation 34 and 35 into this for both F and G we ﬁnd Z δF δF δG {F,G} = d3X J ρ∇ + ∇ · m · x δρ x δm δm δG δG δF −J ρ∇ + ∇ · m · x δρ x δm δm Z δF δG δF δG = ρ ∇ · − · ∇ x δρ δm δm x δρ δF δG δF δG + ∇ · m · − · ∇ · m d3x (36) x δm δm δm x δm

25 and the last step we have changed the integral to be in x rather than X. This is our new Poisson bracket! Remaining is to rewrite our Hamiltonian (equation 29) in terms of our new variables. Setting q = 1

Z π2 H[x, π] = d3X + e(1/J) 2 Z π2 = d3x J + e(1/J) 2 Z m2 = d3x + ρe(ρ) 2ρ so our new Hamiltonian is Z m2 H[ρ, m] = d3x + ρe(ρ) 2ρ

4.8 Another way to write Inviscid irrotational fluid dynamics Inviscid irrotational fluid dynamics (but not incompressible) can be described with a Hamil- tonian Z 1 H[ρ, φ] = d3x ρ(x)(∇φ(x))2 + e(ρ(x)) 2 and a Poisson bracket {ρ(x), φ(y)} = δ3(x − y) (we note that these coordinates are canonical). The choice of a velocity potential φ means we have chosen a vorticity free fluid. Both fields ρ, φ are functions of x. In the Hamiltonian the function e(ρ) is an internal energy density. For barotropic flow

∂2e 1 ∂p ∂e p = = ∂ρ2 ρ ∂ρ ∂ρ ρ2 Since we have two fields we get two equations of motion dρ δH = = −ρ∇2φ (37) dt δφ dφ δH 1 ∂e = − = − (∇φ)2 − e(ρ) − ρ (38) dt δρ 2 ∂ρ Let us define a vector field that we call the velocity

u = ∇φ

26 so φ is the velocity potential ﬁeld and there can be no vorticity in this model as

ω = ∇ × u = ∇ × ∇φ = 0

Inserting the expression for velocity into our equations of motion dρ = −ρ∇ · u (39) dt dφ u2 ∂e u2 p = + e(ρ) + ρ = + e(ρ) + (40) dt 2 ∂ρ 2 ρ The first equation can be recognized as conservation of mass. The second equation can be recognized as Bernoulli’s equation. Both require the variables to be moving with the fluid! Taking the gradient of equation 40 d ∂e p ∇p ∇φ = −(u · ∇)u − ∇ρ + ∇ρ − dt ∂ρ ρ2 ρ du p p ∇p = −(u · ∇)u − ∇ρ + ∇ρ − dt ρ2 ρ2 ρ ∇p = −(u · ∇)u − . ρ This is Euler’s equation. Why is it that ρ is a Lagrangian variable (moving with the fluid) and not φ?

5 Poisson manifolds

For infinite dimensional systems instead of equations of motions for a single particle that is two differential equations, we would expect partial differential equations, like the Euler equation for fluid mechanics or the KdV equation. A choice of Hamiltonian and Poisson bracket is not necessarily unique given equations of motion (in this case partial differential equations). When we work with fields, we can relax the requirement that we have a canonical set of coordinates in which to define the Poisson bracket. For finite dimensional systems we often had a canonical basis in which the Poisson brackets of the elements gave ±1 or 0. Here we can assume instead that we know the Poisson brackets in a particular coordinate basis (but that this coordinate basis is not canonical). The system is symplectic if we ensure that the Poisson brackets satisfy the properties of Poisson brackets. For example we can write Poisson brackets for two functions that depend on coordinates

27 x ∂A(x) ∂B(x) ∂A(x) ∂B(x) {A(x),B(x)} = − ∂qi ∂pi ∂pi ∂qi ∂A ∂xj ∂B ∂xk ∂A ∂xj ∂B ∂xk = − ∂xj ∂qi ∂xk ∂pi ∂xj ∂pi ∂xk ∂qi ∂A ∂xj ∂xk ∂xj ∂xk ∂B = − ∂xj ∂qi ∂pi ∂pi ∂qi ∂xk j k = ∂jA{x , x }∂kB where we have assumed in the middle that we know how to compute Poisson brackets for the coordinates x (that are not necessarily canonical) and we have used summation notation and the notation ∂ ∂i ≡ ∂xi This we write again i j {A(x),B(x)} = ∂iA{x , x }∂jB (41) Let us check that the Poisson bracket is consistent for the wave equation which has canonical coordinates (and {xi, xj} = J ij). Using equation 41 but with both functions inside the Poisson bracket a function of φ, π ∂π(y) δH {π(y),H[π, φ]} = {xi, xj} ∂xi δxj Z δπ(y) δH = dxdz {π(x), φ(z)} δπ(x) δφ(z) Z δH = dxdzδ(y − x){π(x), φ(z)} δφ(z) Z δH = dz{π(y), φ(z)} δφ(z) Z δH = dzδ(y − z) δφ(z) For the wave equation Z δH φ˙(y) = {φ(y),H[π, φ]} = δ(y − x) dx δπ(x) Z δH π˙ (y) = {π(y),H[π, φ]} = − δ(y − x) dx δφ(x) The equations of motion for the wave equation are consistent with the Hamiltonian Z 1 H[π, φ] = π(x)2 + V (φ(x)) dx 2

28 and the Poisson bracket shown in equation 7 that has {xi, xj} = J ij. In a more general case we might not be able to ﬁnd coordinates that are everywhere canonical, but we might still have a way of computing something like a Poisson bracket. In this case we consider a Poisson structure on a smooth manifold M. This structure is in the algebra of smooth functions on the manifold. For smooth functions f, g, h the Poisson bracket satisﬁes

1. Skew symmetry {f, g} = −{g, f}

2. Leibnitz rule {fg, h} = f{g, h} + g{f, h}

3. Jacobi identity {f, {g, h}} + {g, {h, f}} + {h, {f, g}} = 0

For canonical coordinates x = (q, p) we can write the Poisson bracket as

∂f ∂g {f, g} = J µν (42) ∂xµ ∂xν

Now consider the possibility that J ij is replaced by something that depends on the coordinates or ∂f ∂g {f, g} = F µν ∂xµ ∂xν We require F µν = −F νµ so that that is antisymmetric. The Leibnitz rule is satisfied by any Poisson bracket in this form. Inserting the bracket into the Jacobi identity we find that the coefficients must satisfy

∂J jk ∂J ij ∂J ki F im + F km + F jm = 0 ∂xm ∂xm ∂xm

5.1 Example: Inviscid Burgers’ equation Using the inviscid Burgers’ equation to explore an inﬁnite dimensional (continuous) Hamil- tonian system with a non-trivial Poisson structure. This system has a Poisson structure that is also used with the KdV equation. A pressure-less ﬂuid with constant density ∂u + u · ∇u = 0 ∂t In one dimension u,t + uu,x = 0

29 also known as the Inviscid Burger’s equation. We can ﬁnd the equations of motion using a Hamiltonian Z 1 H(u) = − u(x)3 dx (43) 3! that is only a function of one ﬁeld and a Poisson bracket that is ∂ {u(x), u(y)} = δ(x − y) (44) ∂x This is antisymmetric, but how do we use it? For two functions of u Z δf δg {f(u), g(u)} = dx dy {u(x), u(y)} δu(x) δu(y) Z δf ∂ δg = dx dy δ(x − y) δu(x) ∂x δu(y) Z δf ∂ Z δg = dx dyδ(x − y) δu(x) ∂x δu(y) Z δf ∂ δg = dx δu(x) ∂x δu(x) With suitably well behaved boundaries we could have integrated by parts in x giving Z ∂ δf δg {f(u), g(u)} = − dy ∂y δu(y) ∂u(y) Adding these two we can construct an expression that makes the asymmetry of the bracket clearer Z 1 δf ∂ δg ∂ δf δg {f(u), g(u)} = dx − 2 δu(x) ∂x δu(x) ∂x δu(x) ∂u(x) µν ∂ We can associate F from equation 42 with ∂x δ(x − y) from equation 44. Let’s use the Poisson bracket in equation 44 and the Hamiltonian in equation 43 to compute the equations of motion using du(x) = {u(x),H} dt We compute the functional derivative using the Hamiltonian in equation 43 (with energy density −u3/3!) δH[u] 1 Z = lim dy [H(u(y) + δ(y − x)) − H(y)] δu(x) →0 1 Z 1 = lim − dy (u(y) + δ(y − x))3 − u(y)3 →0 3! 1 Z 1 = lim − dy 3u(y)2δ(y − x) →0 3! u(x)2 = − 2

30 The equations of motion Z δu(x) ∂ δH {u(x),H} = dy dz δ(y − z) δu(y) ∂y δu(z) Z ∂ u(y)2 = − dyδ(x − y) ∂y 2 ∂ u(x)2 = − = −uu ∂x 2 ,x And this is consistent with our equation of motion for the inviscid Burger’s equation.

u,t = −uu,x

Burger’s equation has one field u but previously we derived it using two fields. I think that the Hamiltonian Poisson manifolds with one field cannot be derived from a Lagrangian. There could a way to derive single field Poisson brackets with a projection to a submanifold.

5.1.1 The symplectic two form Our Poisson bracket (equation 44) that we write again here ∂ {u(x), u(y)} = δ(x − y) (45) ∂x has a kind of inverse. If you integrate over a delta function you would get a step function so we expect the inverse would involve integrating with a step function. It is possible to find a symplectic structure consistent with the Poisson bracket of equation 44. The symplectic two form takes two vectors and returns a real number. Here we have infinite dimensional vectors which we can describe with u(x) with all x values. So our symplectic structure should take two different u functions and return a single number. An antisymmetric function that does this is 1 Z ∞ Ω(u1, u2) = [û1(x)u2(x) − uˆ2(x)u1(x)]dx (46) 2 −∞ and where Z x uˆ(x) ≡ dy u(y) −∞ If u1(x) = ∂v/∂x for some v(x) then Z x ∂v uˆ1 = dy = v(x) −∞ ∂y

31 and 1 Z ∞ Z x ∂v Ω(u1, u2) = v(x)u2(x) − dy u2(y) dx 2 −∞ −∞ ∂x 1 Z ∞ Z x ∞ = dx [v(x)u2(x) + u2(x)v(x)] − u2(y)dy v(x) 2 −∞ −∞ −∞ Z = v(x)u2(x)dx (47)

(where we have integrated by parts) and have assumed that u2, v vanish at the boundaries. δH ∂ δH Let us use v(x) = δu(x) giving u1(x) = ∂x δu(x) and so equation 46 gives

∂ δH Z δH Ω , u = dx u (x) (48) ∂x δu(x) 2 δu(x) 2

The symplectic two-form can be used to generate Hamiltonian flows from a Hamiltonian function H. Previously we did this by finding a vector V such that the two form ω(V, ?) = i dH. The vector V gives x˙ and generates the flow. Previously dH = ∂xi Hdx . Now our variables are u(x) so by analogy we use the functional derivative

δH[u] dH = du(x) δu(x) and we are implicitly taking into account all u(x). The vector ﬁeld V = XH (u) givingu ˙ . Z δH[u] Ω(X (u), u ) = dx u (x) H 2 δu(x) 2

∂ δH But comparing this we equation 47 implies that XH (u) = ∂x δu(x) and du ∂ δH = dt ∂x δu(x) and this is consistent with what we found using the Poisson bracket of equation 44.

5.2 Conserved Quantities and Continuity Equations An integrated quantity Z Q(u, t) = f(u(x, t))dx that is conserved would satisfy dQ d Z = f(u(x, t))dx = 0 dt dt

32 Here I have written the local quantity f(u) depending on a ﬁeld u(x, t) that is dependent on x, t. An example of a continuity equation is mass continuity in one dimension

∂tρ + ∂x(ρv) = 0 with ρ the local mass density. This equation implies that the total mass R dx ρ is conserved. Here ρv is the mass flux and v can be any velocity field that moves mass around. If a function locally (but everywhere) satisfies a continuity equation then its integral is a conserved quantity. For example, suppose that

∂t (f(u(x, t))) + ∂x (g(u(x, t))) = 0 for some functions f(u) and g(u). The above relation is called a continuity equation or in conservation law form. We can integrate the continuity equation Z ∞ Z ∞ dx∂t (f(u(x, t))) = − dx∂x (g(u(x, t))) −∞ −∞ d Z dx f(u(x, t)) = −g(u(x, t))|∞ dt −∞ dQ = 0 dt We have assumed that g is zero on the boundary. We say that g(u(x, t)) is the ﬂux or current of f(u(x, t)).

6 The KdV equation

Another interesting non-linear inﬁnite dimension Hamiltonian system is based on the KdV equation. To put the KdV equation into Hamiltonian form is more challenging because we don’t obviously have conjugate momenta. The KdV diﬀerential equation

u,t = uu,x + u,xxx or ∂u ∂u ∂3u = u + ∂t ∂x ∂x3 that can also be written in conservation law form

2 u,t = ∂x(u /2 + u,xx) (49) The KdV equation can be derived from a Hamiltonian Z u3 1 H[u] = dx − (u )2 (50) 3! 2 ,x

33 Using our deﬁnition of a functional derivative Z 3 δH 1 (u(y) + δ(y − x)) 1 2 = lim dy − [∂y(u(y) + δ(y − x))] δu(x) →0 3! 2 Z 1 h 2 i = lim dy u(y) δ(y − x) − ∂yu(y)∂yδ(y − x) →0 2 u2(x) ∂2u(x) = + 2 ∂x2 where the last step is done by integrating by parts. We see that this is contained within the derivative inside the equation of motion in conservation law from (equation 48) The equations of motion can be written as ∂ δH u = ,t ∂x δu(x) What type of Poisson bracket gives this equation of motion? We desire ∂ δH {u(x),H[u]} = ∂x δu(x) Using equation 41 Z δu(x) δH {u(x),H[u]} = dydz {u(z), u(y)} δu(z) δu(y) Z δH = dy{u(x), u(y)} δu(y) If we use a Poisson bracket ∂ {u(x), u(y)} = δ(x − y) (51) ∂x then Z ∂ δH {u(x),H[u]} = dy δ(x − y) ∂x δu(y) ∂ Z δH = dyδ(x − y) ∂x δu(y) ∂ δH = ∂x δu(x) as desired. Thus the KdV equation arises from the Hamiltonian in equation 49 with Poisson bracket in equation 50.

Remark It is possible to check that this Poisson bracket obeys the Jacobi identity and so is well behaved. Instead of coordinates and momenta here we only have u. This is why the Poisson bracket has a derivative of a delta function rather than just a delta function for p, q analogs. How is possible that we can use one function for both canonical coordinates?

34 6.1 Rescaling We show here that is it not important what coefficients are in front of each term. If there are coefficients in front the terms in the differential equation, it is possible to rescale u, x, t so that it resembles the form we have used above. For

u,t = auu,x + bu,xxx we can rescale with x → b1/3x u → a−1b1/3u to get the form used above for the KdV equation.

6.2 Soliton Solutions Soliton solutions can be found by searching for a solution in the form of a traveling wave u(x, t) = f(x + vt) with constant traveling speed v. This has

0 0 000 u,t = vf u,x = f uxxx = f If we insert this in to the KdV equation

vf 0 = ff 0 + f 000

Treating these as x derivatives only (like setting time to zero) d vf = (f 2/2 + f ) ,x dx xx Integrating d2f 1 + f 2 − vf = constant dx 2 Requiring a zero on the inﬁnite boundaries, implies that the constant is zero. Multiplying by fx d2f 1 f + f f 2 − vf f = 0 x dx2 2 x x 1 d 1 v (f )2 + (f 3) − (f 2) = 0 2 dx x 6 x 2 x Integrating and again setting a constant to zero 1 1 v (f )2 + f 3 − f 2 = 0 2 x 6 2

Looking for a solution with fx = 0 at the origin, √ vx u(x) = 3v sech2 2

35 and restoring time √ v(x + vt) u(x, t) = 3v sech2 2 This is a traveling wave with amplitude or height directly proportional to its travel speed. The solution is non-dispersive; the wave maintains its shape as a function of time and only travels to the left. Higher amplitude solitons travel faster than (and catch up to) lower amplitude solitons. Solitons are a characteristic of integrable systems. Two solitons can pass through one another, but delayed in time by the interaction. By considering asymptotic limits and using inverse scattering theory, it might be possible to compute the delay?

6.3 Inﬁnite number of conserved quantities We start with a transformation 2 u = v + i∂ v + v2 (52) x 6 This transformation takes a function v, generates a function u, and the transformation depends on the constant . We compute some derivatives

2 u = v + iv + vv t t xt 3 t 2 u = v + iv + vv x x xx 3 x vv u = v + iv + 2 v v + xxx xxx xxx xxxx x xx 3 Putting these expressions into the KdV equation

ut = uux + uxxx and rearranging terms

2v 2 1 − i∂ − v − vv − v2v − v = 0 x 3 t x 6 x xxx

If v satisﬁes a modiﬁed equation

2 v = vv + v2v + v (53) t x 6 x xxx then using the transformation 51 we obtain a u that is a solution to the KdV equation. In the limit of → 0 the modiﬁed equation is the KdV equation. In the limit of → ∞

36 √ and after a rescaling v/ 6 → v we obtain what is known as the modified KdV or MKdV equation 2 vt = v vx + vxxx A transformation that takes a solution of the MKdV to a solution of the KdV is known as a Miura transformation (and there are various different forms of it depending on the signs and coefficients used). We can write our modified equation (equation 52) as a continuity equation

v2 v3 v = ∂ ( + + v ) t x 2 18 xx If v satisﬁes our modiﬁed equation then Z ∞ dx v(x) −∞ is a conserved quantity. Now write ∞ X n v = vn n=0 and put this into equation 51

∞ X 2 X X u = (nv + in+1∂ v ) + n+mv v n x n 6 n m n=0 n m ∞ n−2 ! X 1 X = n v + i∂ v + v v n x n−1 6 m n−2−m n=0 m=0 with the convention that vi is zero for i < 0. This is a way of inverting v → u. If u satisfies the KdV equation, then v satisfies a modified equation and this is true for all possible . This means that each vn should satisfy the modified equation also. Since each vn satisfy an equation in conservation law form, each one should be a conserved quantity. To the first few orders of

u = v0

0 = v1 + i∂xv0 1 0 = v + i∂ v + (v )2 2 x 1 6 0 1 0 = v + i∂ v + v v 3 x 2 3 0 1 1 1 0 = v + i∂ v + v2 + v v 4 x 3 6 1 3 0 2 0 = ....

37 This is a recursive relation! At each step we see vi in terms of vj with j < i. To each order in n−2 1 X v = −i∂ v − v v n x n−1 6 m n−2−m m=0 In the other direction

v0 = u

v1 = −i∂xv0 = −i∂xu 1 v = −i∂ v − (v )2 2 x 1 6 0 1 = −u − u2 xx 6 1 1 v = −i∂ v − (v )2 − v v 3 x 2 6 1 3 0 1 2i u2 = iu + uu = i∂ u + xxx 3 x x xx 3 1 1 v = −i∂ v − v2 − v v 4 x 3 6 1 3 0 2 1 u3 u2 u2 = − x + ∂ + u 3 6 2 xx 2 xx

If we assert that u is a solution of the KdV equation, then each vn must individually must satisfy a continuity equation and each vn must be conserved quantity. The iterative nature of the series implies that there are an inﬁnite number of them. The odd ones give rise to imaginary terms that are also derivatives. If they are total derivatives then then integrating by parts with the continuity equation gives a trivial relation. The even quantities are not derivatives so are non-trivial conserved quantities. Conserved quantities are not necessarily in involution. However if we use Z ∞ n Hn = 3(−1) dx v2n −∞ it is possible to show that 1 δH δH D3 + (D u(x) + u(x)D ) m = D m+1 (54) x 3 x x δu(x) x δu(x) and this can be used to show that

{Hn,Hm} = 0 Each of the conserved quantities can be thought of as its own Hamiltonian, each giving rise to its own evolution equation, and these are known as higher order equations of the KdV hierarchy.

38 As we will see below, the operators on either side of equation 53 are two diﬀerent Poisson brackets. Thus the Lax equation is related to having a dual Poisson structure.

6.3.1 Integrability One way to define an infinite dimensional system as integrable is to require that there are an infinite number of conserved quantities that are in involution.

6.4 Lax pair A Lax pair is a pair of time dependent diﬀerential operators L, B such that dL = [B,L] (55) dt Here the commutator [L, B] = LB − BL should not be confused with the Poisson bracket. If the Lax relation is satisﬁed then eigenvalues of L are not time dependent. Suppose Lφ = λφ with λ an eigenvalue and φ,t = Bφ gives a time derivative of φ. We can show this

∂t(Lφ) = ∂t(λφ) = λ,tφ + λφ,t

= λ,tφ + λBφ

= λ,tφ + BLφ and

∂t(Lφ) = (∂tL)φ + Lφ,t

= (∂tL)φ + LBφ Putting these together gives (∂tL)φ = [B,L]φ + λ,tφ dλ If the Lax equation (equation 58) is satisﬁed then λ,t = dt = 0 and the eigenvalues of φ are time independent. It should be noticed that the time derivative of L refers to the derivative of the operator before it is applied to a function. The Lax relation implies that −iB acts like a Hamiltonian function for a Shr¨odinger equation. In analogy L(0) = U †(t)L(t)U(t) (56) with dU(t) = B(t)U(t) = e−iHt dt Taking the time derivative of equation 55 † † † 0 = iHU LU + U ∂tLU − U LiHU † 0 = U (iHL + ∂tL − iLH) U

0 = ∂tL + [iH, L]

39 And the eigenvalues (and spectrum) of L are independent of time (L is isospectral as time varies). As a consequence the solution of the equation Lψ = λψ can be solved at time zero and then propagated to another time with the operator B ∂ψ = Bψ (57) ∂t (again −iB is acting like a Hamiltonian function). It was noted by Peter Lax that using the differential operators 1 L = D + u xx 6 1 B = 4D + (D u + uD ) (58) xxx 2 x x the equation (known as the Lax equation) satisfy the following relation dL + [L, B] = 0 (59) dt and that this equation is equivalent to the KdV equation. Let us check that this is true. 1 L = u t 6 t 5 5 u2 1 1 LB = 4D5 + uD3 + u D2 + D + 2u D + uu + u x 3 x 2 x x 6 x xx x 12 x 2 xxx 5 5 u2 1 2 BL = 4D5 + uD3 + u D2 + D + 2u D + uu + u x 3 x 2 x x 6 x xx x 4 x 3 xxx Hence 1 L + [L, B] = (u − uu − u ) t 6 t x xxx Terms without Dx or its powers are considered operators using an identity operator. Here the operators L, B can be written as sums of powers of the operator Dx = ∂x. Lax noted that an equation that can be cast into the framework of a Lax equation with two operators L, B can display many of the characteristics of the KdV equation including an infinite number of conserved quantities. The operator L to any power, Ln, gives conserved quantities (and if L is a matrix take the trace). A Lax pair is not unique, as new pairs can be generated from a single pair. If L, B are a Lax pair and g an invertible transformation then L0 = gLg−1 B0 = gBg−1 +gg ˙ −1 (60) also defines a Lax pair. L to any power gives a Lax pair with B. Any operator B˜ with [L, B − B˜] = 0 gives a Lax pair with L. Here g an operator, can it be time dependent? Check this xxxx

40 6.5 Connection to Schr¨odingerequation Consider a time independent Schr¨odingerequation 1 D ψ + ( u(x) + λ)ψ = 0 (61) xx 6 for u(x, t) is a solution to the KdV equation and the variable t changes the equation (so we are not solving a time dependent thing, just using u to set a potential for the equation). This expression should look like it contains the operator L in it (see equation 57). We can invert the equation for u giving

ψ u(x, t) = −6 λ + xx (62) ψ

Taking spatial and time derivatives of this relation we find that u satisfies the KdV equation if the eigenvalue λ is independent of time. If u satisfies the KdV equation then eigenvalues of the associated Schrödingerequation are time independent.

6.6 Connection to Inverse Scattering method Given u(x, t = 0), compute the spectrum of L(t = 0). Choose an eigenvalue and compute the associated wavefunction ψ(x, t = 0) . Then predict the wavefunction at all later times using the evolution operator B and using equation 56. From this compute L(t) and u(x, t) at later times. It may be useful to consider asymptotic limits of u and the wavefunction. If the asymptotic limits at x → ±∞ are wave-like then the problem can be described in terms of reﬂection and transmission coeﬃcients. If the asymptotic limits are exponentially dropping then the scattering problem is discussed in terms of bound states. This sounds simple and can be done for any initial u so is a general technique. As an example let’s look at the soliton solution for the KdV equation.

u(x, t = 0) = 12 sech2x

The associated Shr¨odingerequation (plugging u into equation 60)

2 ψxx + 2sech x ψ = −λψ

A solution exists with λ = −1 and 1 ψ(x) = sechx 2

41 Let us make sure that this is a solution to the associated Shr¨odingerequation. 1 sinhx ψx = 2 cosh2x 1 ψ = − sechx 2sech2x − 1 xx 2 1 5sinhx 6sinh3x ψxxx = − − 2 cosh2x cosh4x

1 ψ + 2sech2x ψ = − sechx(2sech2x − 1) + sech3x xx 2 = ψ and so ψ is an eigenfunction with λ = −1. To take into account time evolution asymptotic behavior must be taken into account. It is not as simple as just applying B to the wave function. This is the soliton solution so the eigenvalue is the wave speed. More generally one discusses bound and free states separately and computes reﬂection and transmission coeﬃcients.

6.7 Zero Curvature Condition The idea here is that we can look at integrable non-linear problems as a compatibility condition of two linear problems and these can involve a spectral parameter λ. We change from linear operators to matrices introducing a vector wavefunction

ψ ψ = ψx

Consider two matrices X, T that can be a function of a time independent parameter λ. Suppose that our wavefunction satisﬁes

ψx = Xψ

ψt = Tψ (63)

Note that these two equations are linear. We need to ﬁgure out if they are consistent with each other. Compute

ψxt = Xtψ + Xψt

= Xtψ + XTψ

ψtx = Txψ + Tψx

= Txψ + TXψ

42 Subtracting these two (Xt − Tx + [X, T]) ψ = 0 A pair X, T satisﬁes the condition

Xt − Tx + [X, T] = 0 (64) iff the accompanying differential equation is satisfied. Such a pair of matrix operators can be called a Lax pair. For the KdV equation we have (equation 57) 1 L = D + u xx 6 1 B = 4D3 + (uD + Du) 2 Using Lψ = λψ u ψ = λ − ψ xx 6 and 0 1 ψ ∂xψ = = Xψ λ − u/6 0 ψx with 0 1 X = (65) λ − u/6 0

Computing ψt using the B and L operators (equations 57) 1 ψ = Bψ = 4Dψ + u ψ + uψ t xx 2 x x u 1 = 4λ + ψ − u ψ 3 x 6 x

u 1 1 ψ = 4λ + ψ − u ψ + u ψ xt 3 xx 6 xx 6 x x u2 1 1 1 = 4λ2 − − uλ − u ψ + u ψ 18 3 6 xx 6 x x with ∂tψ = Tψ we ﬁnd 1 u − 6 ux 4λ + 3 T = 2 u2 uλ 1 1 (66) 4λ − 18 − 3 − 6 uxx 6 ux

43 Using the two matrices we can compute the zero curvature condition and we find 0 0 Xt − Tx + [X, T] = 1 − 6 (ut − uux − uxxx) 0 and this is zero if u satisfies the KdV equation. Matrices X, T can be constructed from a Lax operator pair of operators (if there are higher order derivatives we could use a larger dimension vector for the wavefunction). The name zero curvature condition comes from the following geometrical interpretation. The equations (∂x − X)ψ = 0 and (∂t − T)ψ = 0 define a connection on a two-dimensional vector bundle over the (x, t)-plane. The first equation describes how to parallel-translate a vector ψ in the x-direction, and the second equation describes how to parallel- translate a vector ψ in the t-direction. The matrices X and T are the connection coefficients. A connection is said to have zero curvature if parallel translation of a vector ψ along a path from a point (x0, t0) to another point (x1, t1) gives the same result independent of path. This is the same thing as asserting the existence of a full two-dimensional basis of simultaneous solutions of the equations (∂x − X)ψ = 0 and (∂t − T)ψ = 0 that must be satisfied by the connection coefficients and the zero curvature condition. Every solution of the KdV equation defines a connection with zero curvature. The zero curvature condition is important when considering evolution or scattering operators and it makes possible to reorder integrals. Let us see if we start with a zero-curvature condition and derive an operator equation.

Xt − Tx + [X, T] = 0 we can write as [∂t − T, ∂x − X] = 0 (67) To check this work with the operators

0 = [∂t − T, ∂x − X]

= [∂t, ∂x] − [∂t, X] − [T, ∂x] + [T, X]

= 0 − ∂t(Xφ) + X∂tφ − T∂xφ + ∂x(Tφ) + [T, X]φ

= −∂tX + ∂xT + [T, X] Now let us deﬁne B = T (68)

L = ∂x − X (69) and rewrite equation 66

0 = [∂t − B, L]

= [∂t, L] − [B, L]

= ∂t(Lφ) − L∂tφ − [B, L]φ

= ∂tL − [B, L]

44 and this is the operator form of the Lax equation. It seems that is possible to construct an operator Lax pair from a matrix Lax pair and vice versa. However I am not sure if a one-dimensional operator Lax pair can always be constructed from a matrix Lax pair. The zero curvature condition looks remarkably similar to the Cartan-Maurer equation. If we associate

−1 X = g ∂xg −1 T = g ∂tg and have g be an element of a Lie group. Then the zero curvature equation (or Cartan- Maurer equation) determines the structure constants of the group. T, X = A0,A1 are a gauge potential or a connection. Apparently for SL(2,R) the Cartan-Maurer equation gives the KdV equation. This implies that every Lie group gives rise to a integrable system (and hierachy of diﬀerential equations). Is this true in the opposite direction? Does every integrable system with a Lax Pair have a Lie group associated with it? Since Lie groups are classiﬁed, does this mean all integrable systems are too?

6.8 Two diﬀerent Poisson brackets Above we found that the KdV equation is the equations of motion for

{u(x), u(y)}1 = ∂xδ(x − y) ! Z ∞ u3(x) ∂u2 H[u]1 = dx − −∞ 6 ∂x

The equations of motion

∂u(x, t) = {u(x),H[u] } → u = uu + u ∂t 1 1 t x xxx There is a second pair of Poisson bracket and Hamiltonian that gives the same equations of motion ∂3 1 ∂ ∂ {u(x), u(y)} = + u(x) + u(x) δ(x − y) 2 ∂x3 3 ∂x ∂x Z ∞ u2(x) H[u]2 = = dx −∞ 2

45 We can check the equation of motion Z ∞ ∂u δH2 = dy {u(x), u(y)}2 ∂t −∞ δu(y) Z ∞ ∂3 1 ∂ ∂ = dyu(y) 3 + u(x) + u(x) δ(x − y) −∞ ∂x 3 ∂x ∂x ∂3 1 ∂ ∂ = + u(x) + u(x) u(x) ∂x3 3 ∂x ∂x 2 1 = u + uu + uu xxx 3 x 3 x = uxxx + uux So ∂u(x, t) = {u(x),H[u] } → u = uu + u ∂t 2 2 t x xxx A conserved quantity can be operated on by either Poisson bracket. A conserved quantity can be used to generate its own flow (a Hamiltonian flow) and using either Poisson bracket. Each conserved quantity can be used to generate a new time evolution equation (a new differential equation). It might be useful to write each Poisson bracket as an operator

p1 = Dx (70) 1 p = D + (D u + uD ) (71) 2 xxx 3 x x Reexamine equation 53 and notice the relation between the diﬀerent conserved quantities! The existence of two Poisson brackets is related to the fact that the equation is integrable. Suppose we have Hamiltonians H0 and H1 and Poisson brackets for each that can be written

0,µν {f(y), g(y)}0 = f ∂µf∂νg 1,µν {f(y), g(y)}1 = f ∂µf∂νg

0,µν Here f can be a function of coordinates y. Time evolution can be computed using H0 and the ﬁrst Poisson bracket or H1 and the second Poisson bracket. Inverting f1 we deﬁne a tensor µ 0,λν Sν = f1,µλf and a tensor ν ν 0,µλ 1,µλ Uµ (y) = ∂µy˙ = ∂µ(f ∂λH0) = ∂µ(f ∂λH1) and after computing some derivatives dSν µ = SλU ν − U λSν dt µ λ µ λ

46 which can be written as dS = [S, U] dt and this is a Lax equation. 1 n Quantities Kn = n traceS (for n > 0) and K0 = log |detS| are conserved quantities. We should go on to prove that conserved quantities are independent and can be involution. For the KdV equation the two Poisson structures give

f0(x, y) = Dxδ(x − y) 1 f (x, y) = D3 + (uD + D u) δ(x − y) 1 x 3 x x

−1 The inverse of f0 is D and from this we get L and we can check that the Lax equation gives the KdV equation. The inverse of Dx is −1 f0 (x, y) = (x − y) which is a step function (equal to −1/2 for x − y < 0 and equal to +1/2 for x − y > 0).

∂x(x − y) = δ(x − y)

Z Z 1 S(x − z) = dyf −1(x − y)f (y − z) = dy(x − y) D3 + (uD + D u) δ(y − z) 0 1 x 3 x x In operator language 1 S = D3 + (uD + Du) D−1 3 The KdV equation itself we can write

3 u˙ = uxxx + uux = D u + u(Du)

δu˙ U = = D3 + Du δu (though this is not obvious to me).

1 SU = D3 + (uD + Du) D−1(D3 + Du) 3 2 2 1 1 = D5 + D3u + uD3 + DuD2 + uDu + DuD2 + u(Du) 3 3 3 3 2 1 2 1 US = D5 + D3u + D3(Du)D−1 + DuD2 + Du2 + Du(Du)D−1 3 3 3 3

47 and combining we get the KdV equation. More here to be more careful with computation and deﬁnitions! To summarize, the existence of two Hamiltonians (each with its own Poisson bracket) that are consistent with the equations of motion implies that there must be a Lax pair and inﬁnite number of conserved quantities. Likely the existence of a Lax pair implies the dual Poisson structure.

7 Symplectic Reduction

A goal here is to use a symmetry of the problem to reduce inﬁnite dimensional systems to ﬁnite dimensional ones. Or to just simplify the Hamiltonian at the expense of complicating the Poisson bracket. Someday a nice exploration of how to do this here, hopefully with examples! Perhaps look at the recent paper by Podvin and Sergent.

Notes: Much of this is following Ashok Das’s book on Integrable Models. The hydrodynamics examples are primarily taken from P. J. Morrison’s review Hamiltonian description of the ideal ﬂuid, 1998, Rev Mod Physics, 70, 467 or a nice review entitled Hamiltonian Fluids by John Hunter.