# Journal of Nonlinear Mathematical Physics

Volume 26, Issue 4, July 2019, Pages 604 - 649

# Variational Operators, Symplectic Operators, and the Cohomology of Scalar Evolution Equations

Authors
M.E. Fels
Department of Mathematics and Statistics, Utah State University, Logan Utah, 84322, USA,mark.fels@usu.edu
E. Yaşar
Department of Mathematics, Uludag University, Bursa, Turkey,eyasar@uludag.edu.tr
Received 13 March 2019, Accepted 4 June 2019, Available Online 9 July 2019.
DOI
10.1080/14029251.2019.1640470How to use a DOI?
Keywords
Variational Bicomplex; Cohomology; Scalar Evolution Equation; Symplectic Operator; Hamiltonian Evolution Equation
Abstract

For a scalar evolution equation ut = K(t, x, u, ux, ..., u2m+1) with m ≥ 1, the cohomology space H1,2() is shown to be isomorphic to the space of variational operators and an explicit isomorphism is given. The space of symplectic operators for ut = K for which the equation is Hamiltonian is also shown to be isomorphic to the space H1,2() and subsequently can be naturally identified with the space of variational operators. Third order scalar evolution equations admitting a first order symplectic (or variational) operator are characterized. The variational operator (or symplectic) nature of the potential form of a bi-Hamiltonian evolution equation is also presented in order to generate examples of interest.

Open Access

## 1. Introduction

Given a scalar differential equation Δ = 0, the multiplier problem in the calculus of variations consists in determining whether there exists a smooth function m (the multiplier) and a smooth function L (the Lagrangian) such that

mΔ=E(L)(1.1)
where E is the Euler-Lagrange operator and E(L) is the Euler-Lagrange expression for L. The problem of determining whether m and L exists has a long history and is known as the inverse problem in the calculus of variations [4,7, 9,11,12,14,19].

The variational bicomplex [2, 3, 20] can be used to provide a solution to the inverse problem in the calculus of variations by utilizing the Helmholtz conditions. The result is that the existence of a solution to the inverse problem in equation (1.1) can be expressed in terms of the existence of special elements in the cohomology space Hn−1,2 where n is the number of independent variables. In some cases this in turn allows the solution to be expressed directly in terms of the invariants of the equation, see [7,12].

The main goal of this article is to give a description of the entire cohomology space H1,2() for scalar evolution equations ut = K(t, x, u, ux, ...) which extends the interpretation of the special elements which control the solution to the inverse problem. The result is a natural generalization of the inverse problem in equation (1.1), we call the variational operator problem, which we now state. Given a differential equation Δ = 0, does there exist a differential operator and Lagrangian L such that

(Δ)=E(L).(1.2)

A simple example is given by the potential cylindrical KdV equation, ut=uxxx+12ux2u2t which admits = tDx as a first order variational operator,

tDx(utuxxx12ux2+u2t)=E(12tuxut+12tuxuxxx+16tux3).(1.3)

The variational operator problem in equation (1.2) can be studied for either the case of scalar or systems of ordinary or partial differential equations. Here we restrict our attention to problem (1.2) in the case where Δ is a scalar evolution equation in order to relate this problem to the theory of symplectic and Hamiltonian operators for integrable systems.

In Section 2 we summarize the relevant facts about the variational bicomplex for the case of two independent and one dependent variable. Sections 3 and 4 provide normal forms for the cohomology spaces Hr,s() in the variational bicomplex associated with the equation Δ = 0. These normal forms are then used in Section 5 to show there exists a one to one correspondence between the solution to (1.2) and the cohomology space H1,2(). Even order evolution equations don’t admit non-zero variational operators (see Corollary 5.3) but we have the following theorem for odd order equations (the summation convention is assumed).

### Theorem 1.1.

Let =ri(t,x,u,ux,)Dxi, i = 0,...,k be a kth order differential operator and let the zero set of Δ = utK(t,x,u,ux,...,u2m+1), m ≥ 1 define an odd order evolution equation.

1. 1.

The operator ℰ is a variational operator for Δ if and only if ℰ is skew-adjoint and

ω=dxθ0εdtj=12m+1(a=1j(X)a1(Kujε)θja)(1.4)
is dH closed on, where ε=12riθi and θi are given in equation (2.11).

2. 2.

Let 𝒱op(Δ) be the vector space of variational operators for Δ. The function Φ : 𝒱op(Δ) → H1,2(ℛ) defined from equation (1.4) by

Φ()=[ω],(1.5)
is an isomorphism.

It follows immediately from Theorem 1.1 that a scalar evolution equation admits a (non-zero) variational operator if and only if H1,2() ≠ 0. Consequently the techniques developed for solving the multiplier inverse problem in terms of cohomology [4, 7, 12] can be used to solve the operator problem. The operator and the function L in (1.2) are easily determined from the cohomology class [ω] ∈ H1,2() (see Theorem 5.3).

The variational operator problem in equation (1.2) is related to the problem of whether a scalar evolution equation can be written in the form of a symplectic Hamiltonian evolution equation [10]. In the time independent case, a scalar evolution ut = K(x, u, ux, ..., un) equation is said to be Hamiltonian with respect to a time independent symplectic operator 𝒮=si(x,u,ux,)Dxi if there exists a function H such that,

𝒮(K)=E(H).(1.6)

For a time dependent equation and operator, the symplectic Hamiltonian condition is given in Definition 6.3 (see also Corollary 6.3). Symplectic Hamiltonian evolution equations are reviewed in Section 6 in terms of the variational bicomplex.

Symplectic operators exists on a different space than variational operators but there is a natural identification (see Remark 2.1) between symplectic operators and operators which can be variational operators. With this identification, the variational operator problems and the symplectic operator problem are shown to be the same in Section 7. This leads to the following theorem.

### Theorem 1.2.

Let 𝒮=si(t,x,u,ux,)Dxi be a differential operator and let Δ = utK(t, x, u, ux, ..., u2m+1). The operator 𝒮 is a symplectic operator and Δ = 0 is a symplectic Hamiltonian evolution equation for 𝒮 if and only if 𝒮 is a variational operator for Δ.

Theorem 1.2 shows that symplectic operators and variational operators for ut = K are the essentially the same so that Theorem 1.1 implies the following.

### Theorem 1.3.

The function Φ in equation (1.5) defines an isomorphism between the vector space of symplectic operators 𝒮==ri(t,x,u,ux,)Dxi for which Δ = utK is Hamiltonian, and the cohomology space H1,2().

With Theorem 1.3 in hand, the determination of a symplectic Hamiltonian formulation of ut = K is resolvable in terms of the cohomology H1,2() of the differential equation ut = K and subsequently the invariants of Δ. This characterization of symplectic Hamiltonian evolution equations in terms of H1,2() allows the techniques in [4,7,12] to be used in their study.

A key idea that directly explains the interplay between the symplectic Hamiltonian formulation for an evolution equation and the cohomology H1,2() is the fact that the equation manifold is canonically diffeomorphic to ℝ × J(ℝ,ℝ). The cohomology of the equation is expressed in terms of the geometric structure that arises from the embedding of the equation into J(ℝ2,ℝ) while the symplectic Hamiltonian formulation of an equation is expressed in terms of the contact structure on ℝ × J(ℝ,ℝ). Theorem 7.1 shows how these are related and this leads to Theorem 1.3. This idea also plays a role in the approach to geometric structures in the article [13].

In Section 8 the case of first order operators for third order equations is examined in detail and the following characterization is found.

### Theorem 1.4.

A third order scalar evolution equation ut = K(t, x, u, ux, uxx, uxxx) admits a first order symplectic operator (or variational operator) = 2RDx + DxR if and only if κ is a trivial conservation law, where

κ=K^2dx+(2K0+K1K^212(X(K3)K^22+K3K^23)+X(K3X(K^2)))dt(1.7)

and Ki = uiK, K^2=23K3(K2X(K3)), and X is the total x derivative on ℛ.

Furthermore, when κ = dH(logR) then ut = K admits the first order symplectic (or variational) operator ℰ = 2RDx + DxR.

In Section 8 we examine the relationship between the Hamiltonian form of an evolution equation and their potential form. In [15] it is shown that the (first order) potential form of a time independent Hamiltonian equation admits a variational operator. We examine this in more detail, as well as the role of bi-Hamiltonian systems as in [18]. In Example 9.1 the Krichever-Novikov equation (or Schwartzian KdV) is shown to be the potential form of the Harry-Dym equation. This demonstrates that the symplectic operators (or variational operators) for the Krichever-Novikov equation ([10]) arise as the lift of the Hamiltonian operators of the Harry-Dym equation as described in Section 8.1.

Theorem 1.4 should be contrasted to the problem of determining a Hamiltonian formulation of a scalar evolution equation in terms of a Hamiltonian operator. An evolution equation ut = K is Hamiltonian with respect to a Hamiltonian operator 𝒟 if there exists a Hamiltonian function H (see [1,10,17]) such that

ut=𝒟E(H).(1.8)

Conditions for the existence of 𝒟 and H in equation (1.8) in terms of the invariants of ut = K is unknown. We illustrate the difference in these problems with the cylindrical KdV and its potential form. The potential form of the cylindrical KdV is easily shown to admit at least two time dependent variational (or symplectic) operators. Section 8.1 then suggests that the cylindrical KdV is a time dependent bi-Hamiltonian system. See Example 9.3 where a bi-Hamiltonian formulation of the cylindrical KdV is proposed ([21] states that no Hamiltonian exists for the cylindrical KdV).

Lastly, in Appendix A we identify the elements of H1,1(), which don’t arise as the vertical differential of a conservation law, with a family of variational operators. This is demonstrated in Example 9.2.

The second author, E. Yaşar acknowledges the Scientific and Technological Research Council of Turkey (Tübitak) for financial support from the postdoctoral research program BIDEB 2219, and Utah State University for its hospitality.

## 2. Preliminaries

In this section we review some basic facts on the variational bicomplex associated with scalar evolution equations, see [6] for more details.

## 2.1. The Variational Bicomplex on J∞(ℝ2,ℝ)

The t and x total derivative vector fields on J(ℝ2,ℝ) with coordinates (t, x, u, ut, ux, utt, utx, uxx, ...) are given by

Dt=t+utu+uttut+utxux+Dx=x+uxu+utxut+uxxux+.

The contact forms on J(ℝ2,ℝ) are

ϑ0=duutdtuxdxϑi=Dxi(duutdtuxdx)=duiut,idtui+1dx,i1ζa,i=DxiDta(duutdtuxdx)=dua,iua+1,idtua,i+1dx,a1,i0(2.1)
where ui=Dxi(u) and ua,i=DxiDta(u)=utttt,xxx.

The variational bicomplex on J(ℝ2,ℝ) is denoted by Ωr,s(J(ℝ2,ℝ)) where ω ∈ Ωr,s(J(ℝ2,ℝ)) is a differential form of degree r + s which is horizontal of degree r = 0,1,2 and vertical of degree s = 0, 1, 2,... (see Section 2 in [6]). The forms dx and dt are horizontal, while the contact forms are vertical. For example if ω ∈ Ω1,2(J(ℝ2,ℝ)), then ω can be written

ω=dx(Aijϑiϑj+Biajϑiζa,j+Caibjζa,iζb,j)+dt(Fijϑiϑj+Giajϑiζa,j+Haibjζa,iζb,j)
where Aij, Biaj, Caibj, Fij, Giaj, HaibjC(J(ℝ2,ℝ)). The horizontal and vertical differentials
dH:Ωr,s(J(2,))Ωr+1,s(J(2,)),dV:Ωr,s(J(2,))Ωr,s+1(J(2,))
are anti-derivations which satisfy
dHω=dxDx(ω)+dtDt(ω),dVf=fuiϑi+fut,iζ1,i+,dHϑi=dxϑi+1+dtζ1,i,dVϑi=0,
where fC(J(ℝ2,ℝ)), ω ∈ Ωr,s(J(ℝ2,ℝ)) and Dx(ω),Dt(ω) are the Lie derivatives. Since d = dH + dV this implies,
dH2=0,dV2=0,anddHdV+dVdH=0.

The integration by parts operator I : Ω2,s(J(ℝ2,ℝ)) → Ω2,s(J(ℝ2,ℝ)) is defined by

and it has the following properties [2], [3],
I2=I,ω=I(ω)+dHη,forsomeηΩ1,s(J(2,)).(2.3)

If we let J : Ω2,s(J(ℝ2,ℝ) → Ω1,s(J(ℝ2,ℝ)) be

then I(κ)=1sϑ0J(κ). Both J and I satisfy,
KerJ=KerI=ImdH.(2.5)

The operator J is the interior Euler operator, see page 292 in [6] or page 43 in [3].

Let =riaDxiDta be a total differential operator. The formal adjoint * is the total differential operator characterized as follows. For any ρ ∈ Ω0,s(J(ℝ2,ℝ)) and ω ∈ Ω0,s′(J(ℝ2,ℝ)) there exists ζ ∈ Ω1,s+s′(J(ℝ2,ℝ)) depending on ρ and ω such that

(ρ(ω)*(ρ)ω)dtdx=dHζ.(2.6)

It follows from (2.6) that the formal adjoint satisfies (*)* = .

Let Δ be a smooth function on J(ℝ2,ℝ). The Fréchet derivative of Δ [17] is the total differential operator FΔ satisfying dV Δ = FΔ(ϑ0). If

Δ=utK(t,x,u,ux,,un),
then
dVΔ=ϑtKiϑiwhereKi=uiK(t,x,u,ux,,un),i=0,,n.(2.7)

The Fréchet derivative of Δ is determined from equation (2.7) to be the total differential operator

FΔ=Dti=0nKiDxi.(2.8)

The adjoint of the operator in (2.8) is,

FΔ*(ρ)=Dt(ρ)i=0n(Dx)i(Kiρ),ρΩ*(J(2,).

## 2.2. The Variational Bicomplex on ℛ∞ and Hr,s(ℛ∞)

An nth order scalar evolution equation is given by ut = K(t, x, u, ux, ..., un) with KC(J(ℝ2,ℝ)) and Kn = unK nowhere vanishing. Let Δ = utK(t,x,u,ux,...,un) and let be the infinite dimensional manifold which is the zero set of the prolongation of Δ = 0 in J(ℝ2,ℝ). With coordinates (t, x, u, ux, uxx, ...) on the embedding ı : J(ℝ2,ℝ) is given by

ι=[t=t,x=x,u=u,ut=K,ux=ux,utt=T(K),utx=X(K),uxx=uxx,],(2.9)
where the vector fields T and X are the restriction of Dt and Dx to given by,
X=x+uxu+uxxux+T=t+Ku+X(K)ux+(2.10)
and these satisfy [X,T] = 0. The Pfaffian system = {θi}i≥0 on is generated by the pullback of the 1-forms ϑi in equation (2.1)
θi=ι*ϑi=duiXi(K)dtui+1dx.(2.11)

The forms

{dt,dx,θi=duiXi(K)dtui+1dx}i=0,1,(2.12)
form a coframe on , and give rise to a vertical and horizontal splitting in the complex of differential forms leading to the bicomplex Ωr,s(), r = 0, 1, 2 and s = 0, 1,.... For example if ω ∈ Ω1,2() then
ω=dx(aijθiθj)+dt(bijθiθj)
where aij, bijC(). The bicomplex Ωr,s() is the pullback of the unconstrained bicomplex Ωr,s(J(ℝ2,ℝ)) by the embedding ı : J(ℝ2,ℝ).

The horizontal exterior derivative dH : Ωr,s() → Ωr+1,s() and vertical exterior derivative dV : Ωr,s() → Ωr,s+1() are anti-derivations computed from the equations,

dH(ω)=dxX(ω)+dtT(ω),dV=ddH.(2.13)

The horizontal and vertical differentials satisfy

dH2=0,dV2=0,anddHdV=dVdH.(2.14)

The structure equations of are computed using (2.11) to be

dHθi=dxθi+1+dtXi(dVK)anddVθi=0.(2.15)

Since dH2=0, the complex dH : Ωr,s() → Ωr+1,s() is a differential complex and Hr,s() is defined to be its cohomology,

Hr,s()=Ker{dH:Ωr,s()Ωr+1,s()}Im{dH:Ωr1,s()Ωr,s()}.

The conservation laws of Δ are the dH closed forms in Ω1,0() while H1,0() is the space of equivalence classes of conservation laws modulo the horizontal derivative of a function dH f, fC().

The vertical complex dV : Ωr,s() → Ωr,s+1() is a differential complex whose cohomology is trivial [3], [6]. Specifically, dV is the ordinary exterior derivative in the variables ui, and the DeRham homotopy formula (in ui variables with parameter) applies. The property dHdV = −dV dH make dV : Hr,s() → Hr,s+1() a co-chain map up to sign, see Appendix A.

### Remark 2.1.

Every function of the form Q(t, x, u, ux, uxx, ..., uk) on J(ℝ2,ℝ) factors through π : J(ℝ2,ℝ) → , π(t, x, u, ut, ux, utt, utx, uxx,...) = (t, x, u, ux, uxx, ...), where π is a left inverse of ı in equation (2.9). Therefore by an abuse of notation, we view a function of the form Q(t, x, u, ux, uxx, ..., uk) either on J(ℝ2,ℝ) or where the context will determine which. For example,

FQ=QiDxi
is a differential operator on J(ℝ2,ℝ) while
LQ=QiXi
is a differential operator on which satisfies π* (FQ) = LQ . Given a differential operator ¯=ri(t,x,u,ux,)Xi on the differential operator =riDxi satisfies π*=¯ and we’ll call the (canonical) lift. The formal adjoint of ¯ acting on a form ω is (−Xi)i(riω). The operator ¯ is skew-adjoint if and only if is skew-adjoint.

## 3. Normal Forms for H1,s(ℛ∞) and Characteristic Forms

The universal linearization (see [6]) of Δ = utK(t,x,u,ux,...,un) on is the differential operator (on ),

LΔ=Ti=0nKiXi(3.1)
where Ki = uiK, and the vector fields T and X are defined in equation (2.10). The operator LΔ is the restriction of the Fréchet derivative of Δ to . The adjoint of LΔ is the differential operator defined by
LΔ*(ρ)=T(ρ)i=0n(X)i(Kiρ),ρΩ*().

This next theorem provides a normal form for a representative of the cohomology classes in H1,s() and is analogous to Theorem 5.1 in [6].

### Theorem 3.1.

Let Δ = utK(t, x, u, ux, ..., un) define an nth order evolution equation Δ = 0 and let Hr,s() be its cohomology with s ≥ 1. For any [ω] ∈ H1,s() there exists a representative,

ω=dxθ0ρdtβ,(3.2)
where ρ ∈ Ω0,s−1(), β ∈ Ω0,s() and LΔ*(ρ)=0.

### Proof.

The proof follows Theorem 5.1 of [6]. Choose ω˜0Ω1,s(J(2,)) such that

ι*(ω˜0)[ω].(3.3)
where ı : J(ℝ2,ℝ) is given in equation (2.9). Since ι*(dHω˜0)=0, it there exists ζ˜abΩ0,s(J(2,)), μ˜abΩ0,s1(J(2,)) such that (see Lemma 5.2 in [6])

Applying the identical integration by parts argument on page 292 [6] to (3.4), implies there exists ζ˜Ω0,s(J(2,)), ρ˜Ω0,s1(J(2,)) and ω˜Ω1,s(J(2,)) such that ω˜=ω˜0+d˜Hη˜ and ι*η˜=0 (hence ι*ω˜=ω) and where

dHω˜=dtdx(Δζ˜+dVΔρ˜).(3.5)

We now apply ı* ∘ J to equation (3.5), where J is defined in equation (2.4). For the first term in right hand side of equation (3.5) we find

ι*J(Δζ˜)=ι*(Δu(ζ˜)Dt(Δuζ˜)Dx(Δuxζ˜)+)=0(3.6)
since each term contains a total derivative of Δ, and these vanish under pullback to .

We now apply ı*J to the second term in the right hand side of (3.5),

u1,0dVΔ=1,u0,idVΔ=Ki
with all other ua,idV Δ = 0, so that equation (3.7) becomes,
ι*J(dVΔρ˜)=ι*(Dt(ρ˜)+i=0n(1)iDxi(Kiρ˜))=LΔ*(ι*ρ˜).(3.8)

By equation (2.5) J(dHω˜)=0, so that applying ı* ∘ J to equation (3.5) implies ι*J(dVΔρ˜)=0, and so equation (3.8) gives LΔ*(ι*ρ˜)=0.

We now turn to showing that equation (3.2) holds using the horizontal homotopy operator (equations (5.15), (5.16) and below (5.16) in [6]), see also proposition 4.12 page 117 of [3] or equation (5.133) in [17]. Using the notation hHr,s from [3], this operator satisfies

ω˜=hH2,s(dHω˜)+dH(hH1,sω˜),ω˜Ω1,s(J(2,)).

Applying the pull back by ı to this formula gives the representative for [ω],

ω=ι*hH2,s(dHω˜)(3.9)
with dHω˜ in (3.5).

To utilize the formula in [3] for hH2,s let (x1 = t, x2 = x) and so for example ϑ1122 = DxDxDtDtϑ0 and let k be the max of |I| (number of derivatives) of ϑI terms in (ϑtKiϑi) ∧ ρ. Then by definition 4.13 on page 117 in [3] (or 5.134 in [17])

hH2,s(dHω˜)=1s|I|=0k1DI(ϑ0JIj((dHω˜)j))=|I|=0k1DI(ϑ0(1)jdx˜jJIj(Δζ˜+dVΔρ˜))(3.10)
where (dHω˜)j=DxjdHω˜=(1)j+1dx^j(Δζ˜+dVΔρ˜), (x^1=x,x^2=t), I = (i1,...,il), |I|= l, and
JIj(Δζ˜+dVΔρ˜)=|L|=0k|I|1(|I|+|L|+1|L|)(D)L(u(jIL)(Δζ˜+dVΔρ˜)).(3.11)

Applying ı* to equation (3.10) we have the ι*hH2,s(dtdx(Δζ˜))=0 because all terms in (3.11) on Δζ˜ involve total derivatives of Δ. Therefore using equation (3.10), equation (3.9) becomes,

ω=1sι*(dx|I|=0k1DI[ϑ0|L|=0k|I|1(|I|+|L|+1|L|)(D)L((u(1IL)dVΔ)ρ˜)]dt|I|=0k1DI[ϑ0|L|=0k|I|1(|I|+|L|+1|L|)(D)L((u(2IL)dVΔ)ρ˜)])(3.12)

Consider the first term in equation (3.12). The only non-zero interior product is (with u(1) = ut, u(2) = ux etc.)

u(1)dVΔ=1,
since Δ does not depend on derivatives such as utx. Therefore the only non-zero terms have |I| = 0, |L| = 0 in the first term of (3.12) giving,
dx|I|=0k1DI[ϑ0|L|=0k|I|1(|I|+|L|+1|L|)(D)L((u(1IL)dVΔ)ρ˜)]=dxϑ0ρ˜(3.13)

Combining equation (3.13) with (3.12) we have

ω=1sι*(dxϑ0ρ˜dt|I|=0k1DI[ϑ0|L|=0k|I|1(|I|+|L|+1|L|)(D)L((u(2IL)dVΔ)ρ˜])(3.14)
which produces equation (3.2) with ρ=ιsι*ρ˜. Equations (3.14) and (3.8) shows that ρ=ιsι*ρ˜ satisfies LΔ*(ρ)=0.

If s = 1 in Theorem 3.1. then ρC() is the characteristic function for the cohomology class [ω] ∈ H1,1(), see Theorem 3.3 and Theorem A.1. In general ρ in equation (3.2) is called a characteristic form for [ω] see [6]. The form β in (3.2) is given in terms of ρ by formula (3.14) which is simplified in Corollary 3.2 for H1,1() and H1,2(). The term dxθ0ρ in equation (3.2) generalizes the conserved density of a conservation law, and plays a critical role in Section 7.

### Theorem 3.2.

Let ut = K(t, x, u, ux, ..., un) be an nth order evolution equation where n ≥ 2. The cohomology satisfies H1,s() = 0 for all s ≥ 3.

This is essentially Theorem 1 in [13] and we give a different proof.

### Proof.

Suppose ω is a representative for an element of H1,s+1(), (s ≥ 2), in the form (3.2), where ρ ∈ Ω0,s() is given by

ρ=Ai1isθi1θis.(3.15)
and satisfies LΔ*(ρ)=0.

Suppose for ρ in (3.15) that the highest form order is (no sum) Am1 ... msθm1θm2···θms where we use lexicographic order so that max of (m1 > m2 > ··· > ms) determines the highest order. We first claim that in LΔ*(ρ) the coefficient of θm1+nθm2···θms is

[LΔ*(ρ)]m1+n,m2,,ms=(1(1)n)KnAm1ms,(3.16)
and that the coefficient of θm1+n−1θm1···θms when m1 > m2 + 1 is
[LΔ*(ρ)]m1+n1,m2+1,,ms=(1)nnKnAm1ms(3.17)
and when m1 = m2 + 1 and n is odd, the coefficient of θm1+n−1θm1···θms
[LΔ*(ρ)]m1+n1,m1,,ms=(1)nnKnAm1ms.(3.18)

Therefore LΔ*(ρ)=0 implies Am1...ms = 0, ρ = 0 and ω = dtβ. The condition dHω = 0 gives X(β) = 0. This implies β = 0 since β ∈ Ω0,s+1() and so ω = 0.

We compute LΔ*(ρ)=T(ρ)(X)i(Kiρ) to find equations (3.16), (3.17), (3.18),

LΔ*(ρ)=T(Ai1is)θ1iθisAi1isT(θi1)θisAi1isθi1T(θi2)θis+(1)nXn(KnAi1isθi1θis)(1)n1Xn1(Kn1Ai1isθi1θis)(3.19)
where from equation (2.15) we have T(θi) = Knθi+n +lower order. Consider also the highest order terms in expanding Xn(KnAi1...isθi1···θis) (no sum),
Xn(KnAm1msθm1θm2θms)=Xn(KnAm1ms)θm1θm2θms+KnAm1msθm1+nθm2θms+nKnAm1msθm1+n1θm2+1θms++KnAm1msθm1θm2+nθms+=KnAm1msθm1+nθm2θms+nKnAm1msθm1+n1θm2+1θmsKnAm1msθm2+nθm1θms+lowerorder(3.20)

The coefficient of θm1+nθm2···θms (which is the highest order) occurring in equation (3.19) comes from the second term on the right hand side in (3.19) and the first term on the last right hand side in equation (3.20) to give (3.16).

We consider the next highest order term in (3.19). From equation (3.19), the only possible term that can contain θm1+n−1θm2+1θm3 ···θms when m1 > m2 + 1 is from second term on the last right hand side of equation (3.20). Therefore 3.20 produces (3.17).

In the case when m1 = m2 + 1 we have from the second term in right side of (3.19) at highest order giving (no sum)

Am1msθm1T(θms)θms=Am1msθm1(Knθm2+n)θms+lowerorder,=KnAm1msθm11+nθm1θms+lowerorder.(3.21)

The second and third term on the last right hand side in equation (3.20) are (no sum)

nKnAm1msθm1+n1θm2+1θmsKnAm1msθm1+n1θm2+1θms.(3.22)

Using m1 = m2 + 1, equations (3.21), (3.22) and that n is odd in equation (3.19), gives equation (3.18).

We also have as a corollary of Theorem 3.1.

### Corollary 3.1.

If ut = K(t,x,u,...,u2m), m ≥ 1 is an even order evolution equation, then H1,2() = 0.

### Proof.

We show that the only ρ ∈ Ω0,1() satisfying LΔ*(ρ)=0 is ρ = 0. Suppose ρ = riθi, i = 0,...,k then by direct computation and the fact 2m > 0

LΔ*(ρ)=T(ρ)(X)i(Kiρ)2rkK2mθ2m+kmodθ0,,θ2m+k1
which is non-zero unless rk = 0. Therefore ρ = 0.

This next theorem is a partial converse to Theorem 3.1 which will be used in Corollary 3.2 below to provide a formula for β in equation (3.2).

### Theorem 3.3.

Let Δ = utK(t, x, u, ux, ..., un) define an nth order evolution equation Δ = 0. Let ρ ∈ Ω0,s−1() (s = 1, 2) satisfies θ0LΔ*(ρ)=0, then

ω=dxθ0ρdtβ(ρ),β(ρ)=i=1n(a=1i(X)a1(Kiρ)θia),(3.23)
satisfies dHω = 0.

### Proof.

First suppose ρ ∈ Ω0,s−1() and satisfies θ0LΔ*(ρ)=0, and let ω ∈ Ω1,s() be as in equation (3.23). We compute dHω,

dHω=dtdx[T(θ0ρ)+X(β(ρ))].(3.24)

To compute X(β (ρ)) we need the telescoping identity,

X(a=1i(X)a1(Kiρ)θia)=(X)i(Kiρ)θ0+Kiρθi(nosumoni).(3.25)

Using equation (3.25) in the formula for β (ρ) in equation (3.23) gives

X(β(ρ))=i=1nX[(a=1i(X)a1(Kiρ)θia)]=i=1n(X)i(Kiρ)θ0+Kiρθi.(3.26)

We then use T(θ0ρ) = T(θ0) ∧ ρ + θ0T(ρ) so that together with equation (3.26), equation (3.24) becomes (adding and subtracting K0θ0ρ)

dHω=dtdx(T(θ0)i=0nKiθi)ρ+dtdxθ0(T(ρ)+i=0n(X)i(Kiρ))(3.27)

Now by equation (2.7), ι*dV(Δ)=T(θ0)+i=0nKiθi=0, and so equation (3.27) becomes,

dHω=dtdxθ0(T(ρ)+i=0n(X)i(Kiρ))=dtdxθ0LΔ*(ρ)=0.

### Corollary 3.2.

For any [ω] ∈ H1,s(), s = 1,2 the representative, ω ∈ Ω1,s(), s = 1,2 in equation (3.2) is given by

ω=dxθ0ρdtβ(ρ),β(ρ)=i=1n(a=1i(X)a1(Kiρ)θia),(3.28)
where ρ ∈ Ω1,s−1(). If s = 1, then the representative (3.28) is unique.

### Proof.

Starting with the representative in equation (3.2) of Theorem 3.1 we have ω = dxθ0ρdtβ where L* (ρ) = 0. Let ω^ be the form in (3.28) using this ρ which satisfies LΔ*(ρ)=0 and so by Theorem 3.3. dHω^=0. The form ω=ωω^ is then dH closed, leading to

0=dHω=dH(dt(ββ(ρ))=dxdtX(ββ(ρ)).(3.29)

This implies X(ββ (ρ)) = 0, where ββ (ρ) ∈ Ω0,s(),s = 1,2. However, the only contact form satisfying this condition is the zero form. So β = β (ρ). This proves equation (3.28).

For the final statement in the theorem, suppose ωa = dxθ0 ·Qadtβa, a = 1,2 where QaC() and βa ∈ Ω0,1() satisfy [ω1] = [ω2] ∈ H1,1(). This implies there exists ξ = gjθj such that ω1ω2 = dHξ so that

dxθ0(Q1Q2)=dxX(gjθj).(3.30)

Since X(θi) = θi+1, equation (3.30) can only be satisfied when Q1 = Q2 and gj = 0. Therefore ω1 = ω2 and the form ω in equation (3.28) when s = 1 is unique.

The form ω in equation (3.28) can be derived by a rather lengthy calculation from the second term in equation (3.14).

A form ρ ∈ Ω0,1() can be written ρ = riXi(θ0). We define the adjoint of ρ by ρ* = (−X)i(riθ0) while (ρ*)* = ρ because the operator riXi has this property, see Remark 2.1.

### Theorem 3.4.

Suppose [ω] ∈ H1,2() admits a representative

ω=dxθ0εdtβ,εΩ0,1(),βΩ0,2()
where ɛ* = −ɛ. Then LΔ*(ε)=0 and
β=β(ε)=i=1n(a=1i(X)a1(Kiε)θia).

### Proof.

Write β = Babθ aθb, and choose in equation (3.3), ω˜0=dxϑ0ε˜0dtβ˜, where ε˜0=riϑi, β˜=Babϑaϑb(see Remark 2.1). Then there exists sabC(J(ℝ2,ℝ)) such that,

dH(ω˜0)=dtdx(ϑtε˜0+ϑ0Dt(ε˜0)+Dx(β˜))=dtdx((dVΔ+Kmϑm)ε˜0+ϑ0Dt(ε˜0)+Dx(β˜))=dtdx((dVΔ+Kmϑm)ε˜0+ϑ0(ri,tϑi+riϑti)+Dx(β˜))=dtdx(dVΔε˜0+ϑ0(riDxi(dVΔ))+sabϑaϑb)(3.31)
since Dt(ϑ0) = dV Δ+Kmϑm. Now using equations (3.25) and (3.26) with X = Dx, ρ = ϑ0, Ki = ri, and θ0 = dV α while adding and subtracting r0ϑ0dV Δ we have
dtdxϑ0(riDxi(dVΔ))=dtdx(Dx)i(riϑ0)dVΔdHη˜(3.32)
where
η˜0=dti=1kj=1i(Dx)j1(riϑ0)Dxij(dVΔ)(3.33)
and η˜0 satisfies ι*η˜=0. Since ɛ* = −ɛ, we have ε˜0*=ε˜0, and combining this with equation (3.31) and (3.32) we have
dH(ω˜0)=dtdx(dVΔ2ε˜0+sabϑaϑb)dHη˜0(3.34)

Therefore comparing equations (3.34) with equation (3.5) we have ρ˜=2ε˜0. By equations (3.14) in the proof of Theorem 3.1, ρ=12ι*ρ˜=ε satisfies LΔ*(ε)=0. Finally Theorem 3.3 implies β = β(ɛ).

### Corollary 3.3.

Let ɛ ∈ Ω0,1() satisfy ɛ* = −ɛ. Then θ0LΔ*(ε)=0 if and only if LΔ*(ε)=0.

### Proof.

Let ɛ be as stated and satisfy θ0L* (ɛ) = 0. The form ω with ρ = ɛ in equation (3.23) of Theorem 3.3 satisfies dHω = 0. By Theorem 3.4 LΔ*(ε)=0. The if part of the statement is trivial.

## 4. A Canonical Form for H1,2(ℛ∞) and the Snake Lemma

We now refine Theorem 3.1 to produce a canonical form for elements of H1,2() by determining a unique representative for any [ω] ∈ H1,2().

### Theorem 4.1.

Let [ω] ∈ H1,2(). There exists a unique representative for [ω] of the form

ω=dxθ0εdtβ(ε),β(ε)=i=1n(a=1i(X)a1(Kiε)θia),(4.1)
where ɛ ∈ Ω0,1(), ɛ* = −ɛ, and LΔ*(ε)=0.

### Proof.

We begin by utilizing equation (3.26) and make the substitution ρ = θ0, Ki = ri giving the identity,

X(i=1k(a=1i(X)a1(riθ0)θia))=i=1k((X)i(riθ0)θ0+riθ0θi).(4.2)

If we now write ρ=i=1kriθi and let

η=i=1k(a=1i(X)a1(riθ0)θia),(4.3)
the identity (4.2) gives
X(η)=θ0(ρ*+ρ)(4.4)

Suppose now [ω] ∈ H1,2() with representative ω = dxθ0ρdtβ (ρ) with ρ = riθi from Theorem 3.2. Let ω^=ω12dH(η) where η is given in equation (4.3), so that [ω^]=[ω]. We then use equation (4.4) to replace X(η) in the following,

ω^=dxθ0ρdtβ(ρ)12dxX(η)12dtT(η)=dx(θ0ρ12X(η))dtβ(ρ)+12T(η))=dx(θ0ρ12(θ0ρ+θ0ρ*))dt(β(ρ))+12T(η)=dxθ012(ρρ*)dt(β(ρ)+12T(η)).(4.5)

The representative ω^ in (4.5) satisfies the skew-adjoint condition in the theorem with

ε=12(ρρ*),
while Theorem 3.4 shows LΔ*(ε)=0 and β(ρ)+12T(η)=β(ε).

We now show the representative (4.1) unique. Suppose that

ωα=dxθ0εαdtβˇ(εa),α=1,2(4.6)
where εα*=εα and [ω1] = [ω2]. This implies there exists ξ = ξabθ aθb ∈ Ω0,2() such that
dxθ0(ε1ε2)=dxX(ξabθaθb).(4.7)

Now let ε˜α=ri,αϑiΩ0,1(J(2,)) and ξ˜ab=ξabϑaϑbΩ0,2(J(2,)) so that ι*ε˜α=εα, ι*ξ˜=ξ. Equation (4.7) implies

dtdxϑ0(ε˜1ε˜2)=dt[Dx(ξ˜)]=dH(dtξ˜).(4.8)

Applying the integration by parts operator I (using (2.5)) to equation (4.8) and that εα*=εα gives

dtdxθ0(ε˜1ε˜2)=0.

Since ε˜1ε˜2 is skew-adjoint, this implies ε˜1ε˜2 and hence ɛ1 = ɛ2. Therefore β (ɛ1) = β (ɛ2) and ω1 = ω2.

### Corollary 4.1.

If [ω] ∈ H1,2() with representative ω = dxθ0ρdtβ (ρ) then the unique representative in Theorem 4.1 has

ε=12(ρρ*).(4.9)

The second part of Corollary 3.3 provides an isomorphism between H1,1() and the solution space LΔ*(Q)=0, QC(). We now extend this to H1,2().

### Corollary 4.2.

Let S = {ɛ ∈ Ω0,1() | ɛ* = −ɛ, and θ0L* (ɛ) = 0}. The linear map χ : SH1,2() given by χ(ɛ) = [dxθ0ɛdtβ(ɛ)] where β(ɛ) is given in equation (4.1), is an isomorphism.

### Proof.

Given ɛ satisfying the conditions of the corollary, Theorem 3.3 shows χ(ɛ) ∈ H1,2(). Theorem 4.1 shows directly that χ is onto, while the uniqueness of the representative in Theorem 4.1 shows that χ is one-to-one.

See Section 8.1 for an application of Corollary 4.2.

We now refine Theorem 3.1 and provide a third (non-unique) normal form.

### Theorem 4.2.

Given [ω] ∈ H1,2(), there exists a representative ω such that

ω=dxθ0dVQdtdVγ=dV(dxθ0Qdtγ)(4.10)
where Q is a smooth function on ℛ and γ ∈ Ω0,1().

### Proof.

We start with equation (3.2) in Theorem 3.1 where a representative for [ω] can be written

ω=dxθρdtβ,
where w.l.o.g. ρ = raθ a, a = 1,...,m. Now dV ωH1,3(), and so by Theorem 3.2 there exists ξ ∈ Ω0,3() such that
dVω=dHξ.

Writing ξ = Aijkθiθjθk, this gives

dxX(Aijkθiθjθk)=dxθ0(dVρ).(4.11)

We now show ξ has the form

ξ=Aiθiθ1θ0,i=0,,m1(4.12)

Suppose there is a term in ξ with θM1θM2θM3, 1 ≤ M1 < M2 < M3, and assume we have the one with the highest M3. On the left side of (4.11) there will be

dxX(θM1θM2θM3)
which contains dxθM1θM2θM3+1, which can’t occur on the right side since there is no θ0. Suppose now that there are terms in ξ of the form θ0θM2θM3 with 1 < M2 < M3. Consider the maximal M3, and again
dxX(θ0θM2θM3),
will contain a term dxθ0θM2θM3+1 which can’t occur on the right hand side of equation (4.11). This shows equation (4.12).

Now

dHdVξ=0
but since ξ ∈ Ω0,3(), this implies dV ξ = 0. We apply vertical exactness and let ζ ∈ Ω0,2() be such that dV ζ = ξ. By the vertical homotopy on ξ we may assume ζ = 0θ1. Finally, we let
ω˜=ω+dHζ=dxθ0ρ˜dtβ˜,(4.13)
where ρ˜=ρ+X(Aθ1θ0) and β˜=βT(ζ). Therefore,
dVω˜=dVω+dVdHζ=dVω˜dHdVζ=dVω˜dHξ=0.

This proves there is a representative ω˜ for [ω] with dVω˜=0.

Again we use dV exactness to find η ∈ Ω1,1() such that,

ω˜=dVη(4.14)
where
η=dxαdtγ,α,γΩ0,1().(4.15)

Writing α = ajθj, j = 0,...,m, equation (4.14) and (4.15) give

θ0ρ˜=dV(ajθj).

We now modify η in equation (4.15) and the representative ω˜ for [ω] in equation (4.13) by

η^=ηdH(amθm1),ω^=ω˜+dHdV(amθm1),nosummation
so that ω^=dVη^. In particular we note
η^=dx(a^jθj)dtγ^,j=0,m1.

Continuing by induction, there exists a representative ω¯ for [ω] and an η¯Ω1,1(), where ω¯=dVη¯ and

η¯=dxθ0Qdtγ¯(4.16)
where Q is a smooth function on . Therefore
ω¯=dxθ0dVQdtdVγ¯.

Combining Theorem 4.2 and Corollary 4.1 gives the following.

### Corollary 4.3.

If [ω] ∈ H1,2() with representative ω = dV (dxθ0·Qdtγ) from Theorem 4.2 then the unique representative in Theorem 4.1 is determined by

ε=12(LQLQ*)θ0(4.17)
where LQ = QiXi.

The snake lemma from the variational bicomplex is the following.

### Lemma 4.1.

Let ω ∈ Ω1,2() satisfy dHω = 0, dV ω = 0. Let η ∈ Ω1,1() such that dV η = ω. Then there exists λ = Ldtdx ∈ Ω2,0() such that dHη = dV λ.

### Proof.

We have

dVdHη=dHdVη=dHω=0,
and the vertical exactness of the variational bicomplex implies the lemma.

The relationship between ω, λ and η in Lemma 4.1 is represented by the diagram,

### Corollary 4.4.

Let [ω] ∈ H1,2() and let ω be a dV closed representative as in equation (4.10), and let λ ∈ Ω2,0() be as in Lemma 4.1, so that dHη = dV λ. The linear map Λ : H1,2() → H2,0() given by

Λ([ω])=[λ](4.18)
is well defined.

### Proof.

Suppose ωa ∈ Ω1,2(), a = 1, 2 where [ω1] = [ω2] and that ωa = dV ηa, a = 1, 2 are dV closed representatives. Let λa ∈ Ω2,0() satisfy dHηa = dV λa, a = 1, 2. To demonstrate [λ1] = [λ2] ∈ H2,0 we show there exists κ ∈ Ω1,0() such that λ1λ2 = dHκ.

Since [ω1] = [ω2], there exists ξ ∈ Ω0,2() such that,

dV(η1η2)=dHξ.(4.19)

Taking dV of equation (4.19) gives dV dHξ = −dHdV ξ = 0, which implies dV ξ = 0 since H0,2() = 0 (or dHμ = 0, μ ∈ Ω0,s() implies μ = 0). Therefore there exists ϕ ∈ Ω0,1() such that ξ = dV ϕ. Substituting ξ = dV ϕ in equation (4.19) gives

dV(η1η2+dHφ)=0.(4.20)

The vertical exactness of the variational bicomplex applied to equation (4.20) implies that there exists κ ∈ Ω1,1() such that

η1η2+dHφ=dVκ.(4.21)

Taking dH of equation (4.21) and using dHηa = dV λa gives

dVλ1dVλ2=dHdVκ
so that
dV(λ1λ2+dHκ)=0.(4.22)

Again by vertical exactness of the augmented variational bicomplex applied to equation (4.22), there exists μ ∈ Ω2,0(ℝ) such that

λ1λ2=dHκ+μ(4.23)
where μ ∈ Ω2(ℝ2). The deRham cohomology of ℝ2 is trivial so μ = = dHα,α ∈ Ω1(ℝ2). Therefore equation (4.23) becomes
λ1λ2=dH(κ+α)
and [λ1] = [λ2] ∈ H2,0().

The relevance of the kernel of Λ is given in Theorem A.2.

## 5. Variational Operators and H1,2(ℛ∞)

A scalar evolution equation defined by the zero set of Δ = utK(t, x, u, ux, ..., un) is said to admit a variational operator of order k if there exists a differential operator

=i=0kri(t,x,u,ux,uxx,)Dxi
with rk nowhere vanishing, and a function LC(ℝ2,ℝ) such that,
(Δ)=E(L(t,x,u,ut,ux,utt,utx,uxx,))(5.1)
where E(L) is the Euler-Lagrange expression of LC(J(ℝ2,ℝ)). Theorem 2.6 in [7] relates the existence of a multiplier (or zero order operator) to the cohomology of Δ. In this section we will prove a generalization of this result and ultimately prove Theorem 1.1. We start with the following.

### Theorem 5.1.

Let =riDxi be a variational operator for Δ = utK(t, x, u, ux, ..., un) with Lagrangian L satisfying (5.1). Then there exists η ∈ Ω1,1(J(ℝ2,ℝ)) such that

dV(Ldtdx)=dtdxϑ0(Δ)+dHη.(5.2)

With ı : J(ℝ2,ℝ) in equation (2.9) let,

ω=dV(ι*η)Ω1,2().

Then dHω = 0.

### Proof.

Suppose and L are given satisfying (5.1), then using the standard formula in the calculus of variations (for example equation (3.2) in [2]), we have on account of (5.1)

dV(Ldtdx)=dtdxϑ0E(L)+dHη=dtdxϑ0(Δ)+dHη(5.3)
which shows (5.2).

By applying ı* to equation (5.2) we have

dVι*λ=dHι*η.(5.4)

Letting ω = dV ı* η, we compute dHω using equation (5.4) and get

dHω=dHdVι*η=dVdHι*η=dV(ι*dVλ)=dV2(ι*λ)=0.

Therefore dHω = 0.

A formula for ω in terms of in Theorem 5.1 is given in Theorem 5.2 below. Before giving Theorem 5.2 we note the following property of variational operators for evolutions equations.

### Lemma 5.1.

If Δ = utK(t, x, u, ux, ..., un) admits the kth order variational operator =ri(t,x,u,ux,)Dxi, then ℰ is skew-adjoint.

### Proof.

Suppose (utK) = E(L) then applying IdV to equation (5.2) using dV2=0 along with the property (2.5) for I, we have

IdV(dtdxϑ0(Δ))=0.(5.5)

With Δ = utK let

κ=dV((Δ))=riDxiDt(ϑ0)+ut,idVridV(riDxi(K)).(5.6)
so that condition (5.5) gives

In the term ua,jκ where κ is given in equation (5.6) we note that ua,j (rj) = 0, ua,j (K) = 0, a ≥ 1, j ≥ 0. Therefore the only possible non-zero terms in the summation term in equation (5.7) with ua,j with a ≥ 1, j ≥ 0 satisfy

Dt(utκ)=Dt(r0ϑ0)r0ϑtmod{ϑj}j0DxDt(ut,1κ)=DxDt(r1ϑ0)Dx(r1θt)mod{ϑj}j0(1)kDxtDt(ut,1κ)=(1)kDxtDt(rkϑ0)(1)kDxk(rkθt)mod{ϑj}j0.(5.8)

Writing the condition I(dtdxϑ0κ) mod {ϑ j}j≥0 using equation (5.7) and (5.8) gives

2I(dtdxϑ0κ)dtdxϑ0(riDxiϑt+i=0k(Dx)i(riϑi))mod{ϑj}j0dtdxϑ0((ϑt)+*(ϑt))mod{ϑi}i0.(5.9)

In order for the right side of equation (5.9) to be zero we must have * = −.

### Theorem 5.2.

Let =ri(t,x,u,ux,)Dxi, i = 0,...,k be a kth order variational operator for Δ = utK(t, x, u, ux, ..., un) and let [dV ı*η] ∈ H1,2() from Theorem 5.1. Then the unique representative for [dV ı*η] in Theorem 4.1 is

ω=dxθ0εdtβ(ε),ε=12ι*(ϑ0)=12riθi.(5.10)

### Proof.

Let ω˜0=dVη˜0 where η˜0 satisfies equation (5.3). We have from equations (5.3) and (3.32)

dHω˜0=dVdH(η˜0)=dtdxdV(ϑ0(Δ))=dtdx(Dxi(Δ)dVriϑ0+riDxi(dVΔ)ϑ0)=dtdx(Dxi(Δ)dVriϑ0+dVΔ(Dx)i(riϑ0))+dHη˜(5.11)
where η˜ is given in equation (3.33) and satisfies ι*η˜=0. As remarked in the proof of Theorem 3.1. the term Dxi(Δ)dVriϑ0 in (5.11) does not contribute to the form ρ˜ in equation (3.5). Therefore we have from equation (5.11) that ρ˜ in equation (3.5) is,
ρ˜=(Dx)i(riϑ)=*(ϑ0).

Since ρ=12ι*ρ˜ and is skew-adjoint we get equation (5.10).

We now come to the last main theorem in this section which proves the converse to Theorem 5.1. The proof is again a generalization of the argument given in Theorem 2.6 of [7] for the multiplier problem.

### Theorem 5.3.

Let [ω] ∈ H1,2() with representative ω as in equation (4.10) of Theorem 4.2,

ω=dVη,whereη=dxθQdtγ.(5.12)

Let λ = Ldtdx satisfying dHη = dV λ from Lemma 4.1. Then =FQ*FQ is a variational operator and,

(Δ)=(FQ*FQ)Δ=E(QΔ+L).(5.13)

The proof requires considerable care whether we are working on or J(ℝ2,ℝ), see Remark 2.1.

### Proof.

We start by writing γ = gjθj in equation (5.12) and define the form η˜Ω1,1(J(2,)) given by

η˜=dxϑ0Qdt[gjϑj](5.14)
which satisfies ι*η˜=η in equation (5.12), and where the forms ϑ j are defined in (2.1). Now define the vector fields on J(ℝ2,ℝ),
T˜=t+Ku+Dx(K)ux+Dx2(K)uxx+,V=Δu+Dx(Δ)ux+,X˜=x+uxu+uxxux+Dx(K)ut+,W=Dx(Δ)ut+Dx2(Δ)utx+(5.15)
so that Dt=T˜+V, Dx=X˜+W. Then with η˜ in equation (5.14) we have,
dHη˜=dtdx[Dt(ϑ0Q)+Dx(gjϑj)]=dtdx[T˜(ϑ0Q)+X˜(gjϑj)]+dtdx[V(ϑ0Q)+W(gjϑj)].(5.16)

Since gj = gj(t, x, u, ux, ...) then W(gj) = 0, while dtdxDx(ϑj)=dtdxX˜(ϑj) and V(ϑ0) = dV Δ. Equation (5.16) then can be written,

dHη˜=dtdx[T˜(Qϑ0)+X˜(gjϑj)]+dtdx[ϑ0V(Q)+dVΔQ].(5.17)

The condition dHη = dV λ (on ) is

dtdx[T(θQ)X(gjθj)]=dtdxθaLa.(5.18)

Now π*dtdxθj = dtdxϑj (see Remark 2.1), and using the vector fields in (5.15) we have

π*(dVλ)=dV(π*λ)=dV(Ldtdx),π*(dtdx[T(Qθ0)])=dtdx[T˜(Qϑ0)]π*(dtdx[X(gjθj)])=dtdx[X˜(gjϑj)](5.19)

Therefore applying π* to (5.18) and using (5.19) we have

dtdx[T˜(Qϑ0)+X˜(gjϑj)]=dV(Ldtdx),(5.20)

The first variational formula for dV (Ldtdx) on J(ℝ2,ℝ) applied to the right side of (5.20) gives

dtdx[T˜(Qϑ0)+X˜(gjϑj)]=dtdxϑ0E(L)+dHζ˜1(5.21)
where ζ˜1Ω1,1(J(2,)) ∈ Ω1,1(J(ℝ2,ℝ)). Inserting equation (5.21) into (5.17) we have,
dHη˜=dtdxϑ0E(L)+dtdx[ϑ0V(Q)+dVΔQ]+dHζ˜1(5.22)

The terms dV Δ · Q in equation (5.22) can be written as

dtdx[QdVΔ]=dtdx[dV(QΔ)ΔdVQ](5.23)

We now apply the integration by parts operator (see equation 2.8 in [2]) and use the first variational formula for dV (QΔdtdx), in equation (5.23) and get

dtdx[QdVΔ]=dtdxϑ0[E(QΔ)(Dx)i(Q,iΔ)]+dHζ˜2=dtdxϑ0[E(QΔ)FQ*(Δ)]+dHζ˜2(5.24)

Next we expand the term V(Q) in equation (5.22) using V in (5.15) and

V(Q)=Dxi(Δ)Q,i=FQ(Δ)(5.25)

Inserting (5.24), and (5.25) into (5.22) and letting ζ˜1=ζ˜1+ζ˜2 gives,

dHη˜=dtdxϑ0[E(L)+FQ(Δ)FQ*(Δ)+E(QΔ)]+dHζ˜.

This implies dH(η˜ζ˜) is a source-form. This is only possible if dH(η˜ζ˜)=0, and so

(FQ*FQ)Δ=E(QΔ+L),
which is equation (5.13) as required.

### Remark 5.1.

In general three applications of the vertical homotopy operator are required to determine λ ∈ Ω2,0() from [ω] ∈ H1,2(). The first is to find a representative ωH1,2() with dV ω = 0 (Theorem 4.2). The second is to find η such that dV η = ω, and the third is to find λ such that dV λ = dHη.

We now have the following corollaries.

### Corollary 5.1.

Let [ω] ∈ H1,2() with unique representative ω = dxθ0ɛdtβ(ɛ), ɛ = riθi as in Theorem 4.1. Then Δ admits =2riDxi as a variational operator.

### Proof.

Starting with equation (5.12), Corollary 4.3 implies

ε=12(LQLQ*)θ0=riθi.(5.26)

Equation (5.26) together with the fact FQ=QiDxi gives ε=ι*12(FQFQ*)(ϑ0)=riθi., we have =FQ*FQ=2riDxi is a variational operator by Theorem 5.3.

### Corollary 5.2.

Let [ω] ∈ H1,2() with ω = dxθ0 ∧ (riθi)−dtβ(ρ) as in Theorem 3.1. Then Δ admits

=(riDxi)*riDxi(5.27)
as a variational operator.

### Proof.

By Corollary 4.1 the unique representative ω^=dxθ0εdtβ(ε) has ε=12(ρρ*)=12(riθi(Xi)(riθ0)). Therefore by Corollary 5.1, in equation (5.27) is a variational operator.

### Corollary 5.3.

If an even order evolution equation ut = K(t, x, u, ..., u2m) m ≥ 1 admits a varia- tional operator ℰ, then ℰ = 0.

### Proof.

Let ω ∈ Ω2,1() be the dH closed form from Theorem 5.2. On the other hand using the representative for [ω] from Theorem 3.1 combined with Corollaries 3.1 and 5.2 implies ω = 0. Hence = 0.

Finally we may also restate Theorem 5.3 without reference to the equation manifold as follows.

### Corollary 5.4.

The operator =ri(t,x,u,ux,)Dxi, i = 0,...,k is a variational operator for ut = K if and only if there exists Q(t, x, u, ux, uxx, ...) and L(t, x, u, ux, uxx, ...) such that

=FQ*FQand(utK)=E(Q(utK)+L).(5.28)

Lastly we combine the results of Theorems 5.1 and 5.3 to prove Theorem 1.1.

### Proof.

(Theorem 1.1) We define a linear transformation Φ^:H1,2()𝒱op(Δ) by using the unique representative in Theorem 4.1 to be

Φ^([ω])=Φ^[(dxθ0εdtβ(ε))]=2riDxi(5.29)
where ɛ = riθi and is skew-adjoint. Then Corollary 5.1 shows Φ([ω]) ∈ 𝒱op(Δ).

We check Φ^=Φ1. With =2riDxi a variational operator, let ɛ = riθi. We have from equation (1.5) (or Theorem 5.2. and equation (5.29),

Φ^Φ()=Φ^([dxθ0εdtβ(ε)])=
and
Φ^Φ^([dxθ0εdtβ(ε)])=Φ(2riDxi)=[dxθ0εdtβ(ε)].

Therefore Φ in equation (1.5) is invertible with Φ^ in equation (5.29) as the inverse.

## 6. Functional 2-Forms, Symplectic Forms and Hamiltonian Vector Fields

In this section we review the space of functional forms on J(ℝ,ℝ) as in [3], [2] and relate these to symplectic forms, symplectic operators and Hamiltonian vector fields.

## 6.1. Functional Forms

On the space J(ℝ,ℝ) with coordinates (x,u,ux,...,ui,...) the contact forms are θi = duiui+1dx and Dx = x +uxu +...ui+1ui +... is the total x derivative operator. Again Ωr,s(J(ℝ,ℝ)) denotes the r = 0,1 horizontal, s ≥ 0 vertical forms on J(ℝ,ℝ). The horizontal and vertical differentials dH : Ωr,s(J(ℝ,ℝ)) → Ωr+1,s(J(ℝ,ℝ)), dV : Ωr,s(J(ℝ,ℝ)) → Ωr,s+1(J(ℝ,ℝ)), are anti-derivations which satisfy

dHω=dxDx(ω)dx,dVf=fuiθi=fiθi,dHθi=dxθi+1,dVθi=0,
where fC(J(ℝ,ℝ)), ω ∈ Ωr,s(J(ℝ,ℝ)) and Dx(ω) is the Lie derivative. Since d = dH + dV this implies,
dH2=0,dV2=0,anddHdV+dVdH=0.

The integration by parts operator I : Ω1,s(J(ℝ,ℝ)) → Ω1,s(J(ℝ,ℝ)) is

I(Σ)=1sθ0i=0(1)i(Dx)i(uiΣ),ΣΩ1,s(J(,))(6.1)
and I satisfies the same properties as in (2.3),
Σ=I(Σ)+dHη,I2=I,KerI=ImagedH.(6.2)

The space of functional s forms (s ≥ 1) on J(ℝ,ℝ), s(J(ℝ,ℝ)) ⊂ Ω1,s(J(ℝ,ℝ)), is defined to be the image of Ω1,s(E) under I,

s(J(,))=I(Ω1,s(J(,))).(6.3)

Equation (6.1) applied to Definition 6.3 shows that if Σ ∈ s(E) then there exists α ∈ Ω0,s−1(J(ℝ,ℝ)) such that

Σ=dxθ0α.(6.4)

However, not every differential form Σ ∈ Ω1,s(J(ℝ,ℝ)) written in the form (6.4) is in the space s(J(ℝ,ℝ)). In the case of 2(J(ℝ,ℝ)) the following is easy to show using the definition of I in (6.1), see also Proposition 3.6 and 3.7 in [3].

### Lemma 6.1.

Let Σ ∈ 2(J(ℝ,ℝ)), then there exists a unique skew-adjoint differential operator, 𝒮=siDxi such that,

Σ=dxθ0𝒮(θ0).(6.5)

The differential δV : s(J(ℝ,ℝ)) → s+1(J(ℝ,ℝ)) is defined by

δV=IdV:s(J(,))s+1(J(,)),s=0,
where we let 0(J(ℝ,ℝ)) = Ω1,0(J(ℝ,ℝ)). This leads to the differential complex
C(J(,))dH0(J(,))δV1(J(,))δV2(J(,)),(6.6)
which is exact and is known as the Euler complex, see Theorem 2.7 [2].

## 6.2. Symplectic Forms, Symplectic Operators, and Hamiltonian Vector Fields

Let Γ be the Lie algebra of prolonged evolutionary vector fields on J(ℝ,ℝ). We begin by recalling the appropriate definitions (see Section 2.5 [10]).

### Definition 6.1.

An element Σ ∈ 2(J(ℝ,ℝ)) is a symplectic form on Γ if Σ ≠ 0 and δV (Σ) = 0. A skew-adjoint differential operator 𝒮=siDxi is symplectic if dxθ0𝒮(θ0) is a symplectic form.

Definition 6.1 combined with Lemma 6.1 shows there is a one-to-one correspondence between symplectic forms and symplectic operators. We now define Hamiltonian vector fields.

### Definition 6.2.

Let Σ be a symplectic form. A vector field Y ∈ Γ is Hamiltonian if

δVI(YΣ)=0.(6.7)

The vector field Y is a degenerate direction if I(Y ⌋ Σ) = 0.

Definition 6.2 is equivalent to Σ being invariant under the flow of Y on 2(J(ℝ,ℝ)) as shown in the following theorem.

### Theorem 6.1.

Let Σ be a symplectic form. An evolutionary vector field Y ∈ Γ is Hamiltonian with respect to Σ if and only if

YΣ=Iπ1/2YΣ=0,(6.8)
where ℒ♮ = Iπ1,2ℒ is the projected Lie derivative on ℱ2(J(ℝ,ℝ)), see Theorem 3.21 in [3].

### Proof.

Using Lemma 3.24 in [3] and the fact that δV Σ = 0, we have

YΣ=IdV(YΣ)+I(YδVΣ)=IdV(YΣ).(6.9)

By the first property in equation (6.2), IdVI = IdV, so conditions (6.7) and (6.8) are equivalent through equation (6.9).

We now write out definition 6.2 in a more familiar form. The exactness of the Euler complex and the condition δVI(Y ⌋ Σ) = 0 implies there exists λ = 2Hdx 0(J(ℝ,ℝ)) such that

I(YΣ)=δVλ=dxθ0E(2H).(6.10)

Writing Y = pr(K∂u) and Σ = dxθ0𝒮(θ0) where 𝒮=siDxi is a skew-adjoint differential operator, the left side of equation (6.10) is then

I(YΣ)=I(dx(siDxi(K)θ0Ksiθi))=dxθ0(siDxi(K)(Dx)i(Ksi))=dxθ02𝒮(K).(6.11)

Using this computation in (6.10) shows that condition (6.7) (or (6.8)) is then equivalent to the following.

### Corollary 6.1.

Let Σ be a symplectic form with corresponding symplectic operator 𝒮. The evolutionary vector field Y = pr(K∂u) ∈ Γ is Hamiltonian if and only if there exists HC(J(ℝ,ℝ)) such that

𝒮(K)=E(H).(6.12)

Corollary 6.1 just shows that Definition 6.2 agrees with the standard symplectic Hamiltonian formulation for time independent evolution equations [10].

## 6.2.1. Symplectic Potential

If Σ is a symplectic form, the exactness of the δV complex implies there exists ψ 1(J(ℝ,ℝ)) such that Σ = δV (ψ). The functional form ψ is a symplectic potential for Σ.

### Lemma 6.2.

Let Σ ∈ 2(J(ℝ,ℝ)) be symplectic (and so δV closed), then there exists a smooth function PC(J(ℝ,ℝ)) such that

Σ=dxθ0𝒮(θ0),where𝒮=12(FPFP*)(6.13)
where FP=PiDxi is the Fréchet derivative of P.

### Proof.

A symplectic potential ψ1(J(ℝ,ℝ)) for Σ can be written using (6.4) as

ψ=dxθ0P,PC(J(,)).(6.14)

Writing Σ = δV ψ and using equation (6.14) produces (6.13).

The Hamiltonian condition on Y ∈ Γ in terms of a symplectic potential ψ is the following.

### Lemma 6.3.

The evolutionary vector field Y ∈ Γ is Hamiltonian for the symplectic form Σ = δV ψ if and only if there exists λ 0(J(ℝ,ℝ)) such that

Yψ=δVλ.(6.15)

### Proof.

Using the exactness of the δV complex we show δVYψ=0 which is equivalent to equation (6.15). By Theorem 6.1. Y is Hamiltonian if and only if

0=YδVψ=δVYψ(6.16)
where we have used YδV=δVY (Lemma 3.24 [3]). This proves the lemma.

Using either Lemma 6.3 or equations (6.13) and (6.1) we have the following simple corollary.

### Corollary 6.2.

Let Σ be a symplectic form with symplectic potential ψ = dxθ0 ·P. The evolutionary vector field V = pr(K∂u) ∈ Γ is Hamiltonian if and only if there exists HC(J(ℝ,ℝ)) such that

12(FPFP*)(K)=𝒮(K)=E(H)(6.17)
where FP is the Fréchet-derivative of P on J(ℝ,ℝ).

A straight forward computation writing Σ = δV ψ classifies the first order symplectic operators, see also Theorem 6.2 in [10]

### Lemma 6.4.

An element Σ ∈ 2(J(ℝ,ℝ)) of the first order form,

Σ=dxθ0θ1AAC(J(,))
is symplectic, if and only if there exists P(x, u, ux, uxx) ∈ C(J(ℝ,ℝ)) depending on up to the second order derivative, such that
A=PuxDx(Puxx).(6.18)

## 6.3. Time Dependent Systems

Most of the definitions and results from Sections 6.1 and 6.2 extend immediately to the case of time dependent systems. Let E = ℝ × J(ℝ,ℝ), and label the extra ℝ with the parameter t. The contact forms are

θEi=duiui+1dx(6.19)
and we let Ωtsbr,s(E) be the bicomplex of t semi-basic forms on E,
Ωtsbr,s(E)={ωΩr,s(E)|tω=0,r=0,1;s=0}.

A generic form ωΩtsb1,2(E) is given by

ω=dx(ξijθEiθEi),ξijC(E).

The anti-derivations dHE:Ωtsbr,s(E)Ωtsbr+1,s(E) and dVE:Ωtsbr,s(E)Ωtsbr,s+1(E) are determined by

dHE(ω)=dxDx(ω)dx,dEV(f)=fiθEi,dVEθEi=0,(6.20)
and satisfy (dHE)2=0, (dVE)2=0, dHEdVE+dVEdHE=0. However ddHE+dVE. The integration by parts operator I induces a map IE:Ωtsbr,s(E)Ωtsbr,s(E) having the formula (6.1) and properties (6.2). We let
tsbr,s(E)=IE(Ωtsbr,s(E)).(6.21)

The mapping δVE=IEdVE gives rise to the exact sequence as in (6.6). A form Σtsb2(E) is symplectic if δVEΣ=0 and Lemma 6.2 becomes the following.

### Lemma 6.5.

An element Σ=dxθE0𝒮(θE0)tsb2(E) where 𝒮=siDxi, and siC(E) is symplectic if and only if there exists PC(E) such that

𝒮=12(LPLP*)(6.22)
where LP=PiDxi.

We use Theorem 6.1 to define Hamiltonian vector fields in this case.

### Definition 6.3.

An evolutionary vector field Y = pr(K∂u) where KC(E) is Hamiltonian with respect to the symplectic form Σtsb2(E) if

TΣ=IEπ1/2TΣ=0(6.23)
where T = t +Y and T=IEπ1/2T is the projected Lie derivative.

Note that T in Definition 6.3 agrees with T in equation (2.10). We can also write condition (6.23) as follows.

### Lemma 6.6

An evolutionary vector field Y = pr(K∂u) is a Hamiltonian vector field for the symplectic form Σ=dxθE0(siθEi) if and only if there exists ξ ∈ Ω0,2(E) such that

π1,2T(dxθE0(siθEi))=dxDx(ξ)=dHEξ.(6.24)

### Proof.

We have kernel IE=ImagedHE, therefore equation (6.23) can be written as equation (6.24).

A formula for ξ in equation (6.24) in terms of ρ=siθEi is given in the proof of Theorem 7.1.

The analogue to Lemma 6.3 also holds in this case where Y is replaced by T. In order to prove this we now show the commutation formula in equation (6.16) holds where Y is replaced by T.

### Lemma 6.7.

If ψ=dxθE0·P where PC(E), then TδVEψ=δVETψ.

### Proof.

Since T = t +Y and YδVEψ=δVEYψ (Lemma 3.24 [3]), we need to check

tδVEψ=δVEtψ

We write out both side of this equation. The left side is

IEπ1/2(12dxθE0[Pi,tθEi(Dx)i(Pi,tθE0)])=12dxθE0[Pi,tθEi(Dx)i(Pi,tθE0)].(6.25)

The right side is

δV(dxθE0Pt)=12dxθE0[Pt,iθEi(Dx)i(Pt,iθE0)].(6.26)

Since the mixed partials are equal Pt,i = Pi,t, equations (6.25) and (6.26) are equal, which proves the lemma.

### Lemma 6.8.

The evolutionary vector field Y ∈ Γ is Hamiltonian for the symplectic form Σ=δVEψ if and only if there exists λ0(E) such that

Tψ=δVEλ.(6.27)

### Proof.

Using the exactness of the δVE complex we show δVETψ=0 which is equivalent to equation (6.27). By Definition 6.3, Y is Hamiltonian if and only if

0=TΣ=TδVEψ=δVETψ
where we have used Lemma 6.7. This proves the lemma.

Using Lemma 6.8 we have the following corollary which is the t-dependent version of Corollary 6.2.

### Corollary 6.3.

Let Σ=dxθE0𝒮(θE0) be a symplectic form with symplectic potential ψ = dxθ0 · P. The evolutionary vector field Y = pr(K∂u) ∈ Γ is Hamiltonian if and only if there exists HC(E) such that

12(Pt+(LPLP*)(K))=12Pt+𝒮(K)=E(H).(6.28)

### Proof.

We just need to compute

T(dxθE0P)=dxθE0Pt+dxθE0(LPKLP*K)=dxθE0(Pt+LPKLP*K)

Using this computation in equation (6.27) with λ = 2Hdx gives equation (6.28).

The function H in (6.28) is the Hamiltonian.

### Remark 6.1.

A symplectic form Σ is t-invariant if tΣ = 0. In this case Σ determines a well defined symplectic form Σ¯ on the quotient of E by the flow of t, q : EE/t = J(ℝ,ℝ) such that q*Σ¯=Σ. Definition 6.3 where Y and Σ are t-invariant implies Definition 6.2 for Σ¯ and Y¯=q*Y and equation (6.28) becomes equation (6.17).

## 7. Variational and Symplectic Operator Equivalence

A time independent evolution equation ut = K(x,u,ux,...,un) is a symplectic Hamiltonian evolution equation [10] if there exists a symplectic operator 𝒮 and a function H called the Hamiltonian such that equation (1.6) holds. With this definition, the determination of the possible symplectic Hamiltonian evolution equations is typically approached in two ways. The first way consists of determining the possible symplectic operators of a certain order [10]. Then for a given class of symplectic operators 𝒮, determine K which satisfy equation (1.6). The second approach starts with a given K and then determines if there exists a symplectic operator 𝒮 such that equation (1.6) holds.

By comparison Theorem 1.3. whose proof is given in this section, combines these two questions and resolves the characterization of symplectic Hamiltonian evolution equations by the invariants H1,2(). This can simultaneously solve the existence of 𝒮 and the existence of the Hamiltonian function H in equation (1.6) as done for the special case given in Theorem 1.4.

## 7.1. H1,2(ℛ∞) and Symplectic Hamiltonian Evolution Equations

Given a scalar evolution equation ut = K(t, x, u, ux, ...), we identify the manifolds and E = ℝ × J(ℝ,ℝ) by identifying their coordinates which in turn induces an identification of smooth functions. Define the projection map Π : T*() → T*(E) by

Π(θi)=θEi,Π(dx)=dx,Π(dt)=0,(7.1)
where θi are given in equation (2.11) and θEi are in equation (6.19). Also denote by Π the induced projection map Π:Ωr,s()Ωtsbr,s(E) where for example
Π(dxθ0(siθi)+dtβ)=dxθE0(siθEi).(7.2)

### Lemma 7.1.

The map Π:Ωr,s()Ωtsbr,s(E) is a bicomplex co-chain map,

Π(dHω)=dHEΠ(ω),Π(dVω)=dVEΠ(ω).(7.3)

### Proof.

Equation (7.3) follows for the case ω = θi directly from equations (2.15) and (6.20), and generically from the anti-derivation property of the operators.

### Lemma 7.2.

The function Π:Ω1,2()Ωtsb1,2(E) induces a well defined injective linear map Π^:H1,2()KerδVEtsb2(E) defined by

Π^([ω])=IEΠ(ω)(7.4)
where ω is a representative of [ω].

### Proof.

To show Π^ is well defined, suppose ω′ = ω + dHξ. Then by equation (7.3) and property 3 in equation (6.2) applied to IE gives

IEΠ(ω)=IE(Π(ω)+Π(dHξ))=IE(Π(ω)+dHE(Πξ))=IEΠ(ω).

Therefore Π^ is well defined.

We now show Π^(ω) is δEV closed. We use equation (7.3) and compute

IEdVEIEΠ(ω)=IEdVEΠ(ω)=IEΠ(dVω).(7.5)

Since dV ωH1,3(), Theorem 3.2 implies there exists ξ ∈ Ω0,3() such that dV ω = dHξ so equation (7.5) becomes

IEdVEIEΠ(ω)=IEΠ(dHξ)=IEdHEΠ(ξ)=0.

Therefore Π^(ω) is δV closed.

We now show Π^ is injective. Let [ω] ∈ H1,2() and let ω = dxθ0ɛdtβ (ɛ) be the unique representative from Theorem 4.1. where ɛ = riθi and ɛ* = −ɛ. Then Π^([ω])=dxθE0(siθEi), and (siθEi)*=(siθEi) since X = Dx. If Π^([ω])=0, then siθEi=0 and ω = 0. This shows Π^([ω])tsb2(E) and that Π^ is injective.

In particular we have

### Corollary 7.1.

If [ω] ≠ 0 then Π^([ω])tsb2(E) is a symplectic form.

We now set out to prove the fact that Π^ in Lemma 7.2 is in fact a bijection which will imply Theorem 1.2 in the Introduction. We will use the following Lemma.

### Lemma 7.3.

Let si,ξijC() then

1]dt(π1/2T(dxθE0siθEi))=dtT(dxθ0siθi),2]dtdxDx(ξijθEiθEi)=dtdxX(ξijθiθj).(7.6)

### Proof.

Since dtθEi=dtθi and X = Dx these identities follow.

We now have the main theorem.

### Theorem 7.1.

Let 𝒮=siDxi be a skew-adjoint differential operator on E. The form Σ=dxθE0(siθEi) is symplectic, and Y = pr(K∂u) is a Hamiltonian vector-field for Σ if and only if

ω=dxθ0εdtβ(ε)(7.7)
satisfies dHω = 0, where ɛ = 𝒮(θ0) and β (ɛ) is given in equation (4.1).

### Proof.

Supposed Σ is symplectic and Y is Hamiltonian, then Lemma 6.6 produces ξ=ξabθEaθEb satisfying equation (6.24). Let

ω=dxθ0(siθi)+dt(ξabθaθb).(7.8)

Equations (7.6) and (6.24) give

dHω=dtT(dxθ0(siθi))+dxX(dtξabθaθb)=dtT(dxθE0(siθEi)+dxdtDx(ξabθEaθbE)=dt(π1/2T(dxθE0(siθEi))dxDx(ξabθEaθEb))=0.

Therefore [ω] ∈ H1,2(). Now Theorem 3.4 implies ξabθ aθb = −β (siθi) so that ω in equation (7.8) and equation (7.7) are the same.

Suppose now that ω in equation (7.7) is dH closed. By Lemma 7.2, Σ=Π^(ω) is a symplectic form. So we need only show that Y is Hamiltonian. Again we refer to Lemma 6.6 and show the existence of ξ=ξabθEaθEb in equation (6.24).

Writing β (siθi) = Babθ aθb and using equations (7.6) we have

dHω=dtT(dxθ0(siθi))dxX(dtBabθaθb)=dt(π1/2T(dxθE0(siθEi))+dxDx(BabθEaθEb)).(7.9)

This will vanish if and only if

π1/2T(dxθE0(siθEi))+dxDx(BabθEaθEb)=0(7.10)
because this term is t semi-basic. Equation (7.10) produces ξ=Π(β(siθi))=BabθEaθbE in equation (6.24) and therefore Y is a Hamiltonian vector field for Σ.

We now summarize the results of Lemma 7.2 and Theorem 7.1 by the following corollary.

### Corollary 7.2.

Let ut = K be an evolution equation, and let Y = pr(K∂u) be the corresponding evolutionary vector field on E and let 𝒵Y(E)tsb2(E) be the subset of symplectic forms for which Y is a Hamiltonian vector field. Define the function Ψ : 𝒵Y (E) → H1,2() given by

Ψ(dxθE0εE)=[dxθ0εdtβ(ε)],(7.11)
where dxθEi=εE𝒵Y(E) with εE=𝒮(θE0) and 𝒮=siDxi is the corresponding symplectic operator, and ɛ = 𝒮(θ0) = siXi(θ0). The function Ψ : 𝒵Y (E) → H1,2() is an isomorphism and Ψ^=Ψ1 where Ψ^ is defined in equation (7.4).

With Theorem 1.1 and Corollary 7.2 in hand the proof of Theorem 1.2 and 1.3 are now easily given.

### Proof.

(Theorems 1.2 and 1.3) We start with Theorem 1.2 and suppose that 𝒮=siDxi𝒵Y(E) is a symplectic operator for the scalar evolution equation Δ = utK and that Y = pr(K∂u) is a Hamiltonian vector field for Σ. Then with Ψ from equation (7.11) in Corollary 7.2 and Φ in equation (1.5) in Theorem 1.1 we have,

Φ1Ψ(𝒮)=12siDxi
is a variational operator and so 𝒮 is a variational operator for Δ (by the abuse of notation in Remark 2.1). The fact that Φ−1 ∘ Ψ is an isomorphism then proves Theorem 1.2.

To prove Theorem 1.3 we identify as above, a symplectic operator 𝒮 on E as an operator on J(ℝ2,ℝ) (see Remark 2.1). The function Φ in equation (1.5) defines an isomorphism between symplectic operators for Δ and H1,2(). This proves Theorem 1.3.

As our final Lemma we show for completeness how formula (5.13) can be determined from the symplectic potential.

### Lemma 7.4.

Let 𝒮=siDxi be a symplectic operator and let ψ=dxθE0PC(E) be a symplectic potential. The unique representative for Ψ(𝒮) ∈ H1,2() in Theorem 5.2 has ɛ = 𝒮(θ0). Furthermore there exists a representative ω for Ψ(𝒮) where ω in equation (5.12) can be written ω = dV η where

η=dxθ0Pdtγ.(7.12)

### Proof.

By equation (7.11) of Corollary 7.2 we have the unique representative as stated in the lemma.

We prove the second part of the lemma by first using Theorem 4.2 to construct a representative ω0 for Ψ(𝒮) such that ω0 = dV η0 with η0 = dxθ0 ·Qdtγ0. By equation (4.17) of Corollary 4.3 and equation (6.22) for the operator in the form ɛ = 𝒮(θ0) gives

ε=12(LQLQ*)θ0=12(LPLP*)θ0.(7.13)

Lemma 6.5 and equation (7.13) show ψ=dxθE0Q is a symplectic potential for 𝒮 and that δVEψ0=δVEψ. Therefore using equation (6.2) (for IE) and the exactness of the dVE complex,

for some AC(E) and ξ ∈ Ω0,1(E). We then let
where we are computing dH and dV on . Note that by equation (7.3) and (7.14) we have Π(η) = ψ so that η has the form in equation (7.12). We then compute using equation (7.15) and ω0 = dV η0 that
dVη=dV(η0+dHξ)=ω0+dVdHξ=ω,
which proves the lemma.

## 7.2. Time Independent Operators

Equation (1.6) defines when the time independent evolution equation ut = K(x, u, ux, ...) is a Hamiltonian evolution equation with symplectic operator 𝒮. This is precisely the same definition that the ordinary differential equation K(x, u, ux, ...) = 0 admits 𝒮 as a variational operator. The following simple lemma is the key to decoupling the variational operator problem for time independent scalar evolution equations.

### Lemma 7.5.

Let 𝒮 = siDxi be a time independent symplectic operator with symplectic potential PC(J(ℝ,ℝ)) (equation (6.13) in Lemma 6.2). Then

𝒮(ut)=E(12Put).(7.16)

### Proof.

By the product formula in the calculus of variations (equation 5.80 in [17]) the left side of equation (7.16) is

E(12Put)=12(FP*ut+Fut*P)=12(FP*utDtP)=12(FP*utPiDxiut)=12(FP*utFPut).(7.17)

Equation (7.17) together with the fact from equation (6.13) that 2𝒮(ut)=FPutFP*ut show that the two sides of equation (7.16) agree.

We then have the following.

### Theorem 7.2.

Let 𝒮 be a t-independent symplectic operator. The following are equivalent,

1. (1)

ut = K(x, u, ux, ..., u2m+1), m ≥ 1, satisfies 𝒮(K) = E(H).

2. (2)

𝒮 is a symplectic variational operator for the ODE K = 0,

3. (3)

𝒮 is a variational operator for ut = K (see Remark 2.1).

This converts the symplectic Hamiltonian question for the evolution equation into a variational operator problem for the ODE K = 0.

### Proof.

Suppose ut = K(x, u, ..., u2m+1) is Hamiltonian for the t-independent symplectic operator 𝒮, so that 𝒮(K) = E(H) on J(ℝ,ℝ). By definition 𝒮 is a variational operator for the ODE K = 0. So (1) and (2) are trivially equivalent.

We show (1) implies (3). Suppose that 𝒮(K) = E(H). Using equation (7.17) in Lemma 7.5 we have

𝒮(utK)=𝒮(ut)E(H)=E(12PutH),
where P is a symplectic potential for 𝒮 (see equation (6.13)). Therefore 𝒮 is a variational operator for utK.

Finally we show (3) implies (1). Lemma 7.4 which allows Q in equation (5.13) to be replaced with the symplectic potential P so that hypothesis (3) implies,

(FP*FP)(utK)=E(ΔP+L)=E(PutPK+L).(7.18)

Substituting from equation (7.17) into equation (7.18) we get

(FP*FP)(K)=E(PK+L).

Therefore

𝒮(K)=E(2(LPK)),
and ut = K is a time independent Hamiltonian evolution equation for the symplectic operator 𝒮.

It is worth noting that if [ω] ∈ H1,2() then [tω]K=0H1,2(K = 0) [5]. That is, in the time independent case, the form β in Lemma 3.2 (restricted to the ODE K = 0) defines an H0,2 cohomology class for the ODE K = 0.

## 8.1. First Order Operators for Third Order Equations

For a third order evolution equation

ut=K(t,x,u,ux,uxx,uxxx)(8.1)
we write the conditions for when = 2RDx +X(R), RC() is a variational operator. This will prove Theorem 1.4 in the Introduction.

### Proof.

(Theorem 1.4) By Theorem 1.1 and Corollary 4.2 the skew-adjoint operator = 2RDx + X(R) is a variational operator for (8.1) if and only if the skew-adjoint form ε=Rθ112R0θ0 solution to

LΔ*(ε)θ0=(T(ε)X3(K3ε)+X2(K2ε)X(K1ε)+K0ε)θ0=0(8.2)
where Ki = uiK. Using T(θ0) = dV K = Kiθi and T(θ1) = X(dV K) = X(Kiθi) we have
T(ε)=T(R)θ112T(X(R)))θ0RX(Kiθi)12X(R)Kiθi.(8.3)

The highest possible θiθ0 term in equation (8.2) using (8.3) is θ4. We find from equation (8.2)

[θ4θ0]=RK3+RK3=0.

While for θ3θ0, θ2θ0 and θ1θ0 we have from equations (8.3) and (8.2),

[θ3θ0]=2X(K3)R2K2R+3K3X(R),[θ2θ0]=3X(K2R)+3X2(RK3)+32X(K3X(R)),[θ1θ0]=T(R)2K0R+K1X(R)+32X2(K3X(R))+X3(K3R)X2(K2R)X(K2X(R)).(8.4)

For the coefficient of θ3θ0 to be zero we have from equation (8.4),

X(R)=23K3(K2X(K3))R=K^2R(8.5)
where K^2=23K3(K2X(K3)) . The coefficient of θ2θ0 in equation (8.4) is zero on account of (8.5). For the coefficient of θ1θ0 in (8.4) to be zero gives
T(R)=2K0R+K1X(R)+32X3(K3X(R))+X3(K3R)X2(K2R)X(K2X(R)).(8.6)

Simplifying equation (8.6) using equation (8.5) we get

T(R)=(2K0+K1K^212(X(K3)K^22+K3K^23)+X(K3X(K^2)))R(8.7)

It follows that a non-vanishing R (which we may assume to be positive) satisfying equations (8.5) and (8.7), which is necessary and sufficient for the existence of a first order variational operator for Δ = utK in equation (8.1), is equivalent to A = X(logR) and B = T(logR) satisfying the conditions in Theorem 1.4. This proves Theorem 1.4.

The form κ in equation (1.7) for the KdV equation ut = uxxx + uux satisfies

κ=uxdt,dHκ=uxxdxdt.

Therefore according to Theorem 1.4 there is no first order formulation for the KdV equation as a symplectic Hamiltonian evolution equation.

## 8.2. First Order Hamiltonians Operators and their Potential Form

Let vt = 𝒟E(H(x,v,vx,...)) be a time independent Hamiltonian evolution equation where 𝒟 is a first order Hamiltonian operator. According to [16] or [1] we may choose coordinates (using a contact transformation) such that 𝒟 = Dx. The following is Theorem 1 in [15] in the context of scalar evolution equations.

### Lemma 8.1.

The potential form of the Hamiltonian evolution equation,

vt=DxE(H1(x,v,vx,)).(8.8)
is given by the equation
ut=E(H1)|v=ux.(8.9)

Equation (8.9) admits ℰ = Dx as a first order variational operator, and satisfies

Dx(ut=E(H1)|v=ux)=E(12uxut+H1|v=ux).(8.10)

There is an abuse of notation in this lemma where Dx is used as the total x derivative operator in either variable u or v depending on context.

### Proof.

Starting with equation (8.8), let v = ux so that (8.8) becomes

utx=(DxE(H1)|v=ux=DxE(H1)|v=ux).(8.11)

Integrating equation (8.11) with respect to x gives the potential form (8.9).

To prove equation (8.10) holds, we simply need the change of variables formula, see exercise 5.49 in [17],

E(H1|v=ux)=(Dx)*E(H1|v=ux))=DxE(H1|v=ux)).(8.12)

Equation (8.12) together with the simple fact −2E(utux) = utx proves equation (8.10).

The second term in the right hand side of equation (8.10) is just the pullback of the Hamiltonian function in (8.8). We also note the following simple corollary.

### Corollary 8.1.

Every Hamiltonian evolution equation vt = 𝒟(E(H1(x, v, vx, ...))) with first order Hamiltonian operator 𝒟 is the symmetry reduction of an equation ut = K(x, u, ux, ...), of the same order, which admits an invariant first order variational operator.

## 8.3. Bi-Hamiltonian Evolution Equations with a First Order Hamiltonian Operator

We now present sufficient conditions when the potential form of a compatible bi-Hamiltonian system admits another variational operator.

### Theorem 8.1.

Let vt = K(x, v, vx, ...) = Dx (E(H1(x,v,vx,...))) be a Hamiltonian evolution equation with potential form

ut=E(H1)|v=ux.(8.13)

Let 𝒟0 be second time independent Hamiltonian operator with Hamiltonian H0(x, v, vx, ...) satisfying,

vt=Dx(E(H1))=𝒟0(E(H0)).

Assume 𝒟0 also satisfies the compatibility condition (equation 7.29 in [17])

𝒟0(E(H1))=DxE(H2).(8.14)

Then the right hand side of the potential form satisfies

(E(H1)|v=ux)=E(H2)|v=ux)(8.15)
where a, = 𝒟0|v=ux. Furthermore if ℰ = 𝒟0|v=ux is symplectic, then ℰ is a variational operator for the evolution equation (8.13) and
(utK)=E(Qut+H2|v=ux)(8.16)
where Q is defined in equation (5.28) where =FQ*FQ.

### Proof.

First we apply = 𝒟0|v=ux to the right hand side of equation (8.13), and use condition (8.14) to get

(E(H1)|v=ux)=(𝒟0(E(H1))|v=ux=(Dx(E(H2))|v=ux=E(H2)|v=ux).(8.17)

Again the last line follows from the change of variables formula in the calculus variations (exercise 5.49 in [17]). This verifies equation (8.15). Then by part (1) of Theorem 7.2 equation (8.15) shows that is a variational operator for equation (8.13). If Q is the function from equation (5.28) we then have (ut)=(FQ*FQ)(ut)=E(Qut) and equation (8.17) that

(utE(H1)|v=ux)=E(Qut+H2|v=ux).(8.18)

Theorem 8.1 makes the hypothesis that = 𝒟0|v=ux is a symplectic operator. This holds in the case of the Hamiltonian operators given by Theorem 5.3 in [10],

𝒟0=h(v)(c1+c20v1h(y)dyDxc1+c20v1h(y)dy+Dx3)h(v)(8.19)
satisfy the compatibility conditions with Dx in Corollary 3.2 of [8] when h(v) = (k1v + k2)−1. This gives
=𝒟0|v=ux=1k1ux+k2(c1+12k1ux2+k2uxDxc1+12k1ux2+k2ux+Dx3)1k1ux+k2(8.20)
which are symplectic [10].

## 9. Examples

### Example 9.1.

The Harry-Dym equation zt = z3zxxx [10,21] is a compatible bi-Hamiltonian system,

zt=𝒟^1E(H^1)=𝒟^0E(H^0)(9.1)
where
𝒟^1=z2Dxz2,H^1=12zx2z,and𝒟^0=z3Dx3z3,H^0=1z.(9.2)

The change of variable z = v−1 maps the Hamiltonian operator 𝒟^1 to canonical form [1,16], and the Hamilonian operators and the associated Hamiltonians in equation (9.2) become

𝒟1=Dx,H1=12vx2v3,and𝒟0=v1Dx3v1,H0=v.(9.3)

The Harry-Dym equation (9.1) in these coordinates is then,

vt=DxE(H1)=𝒟0(E(H0))=vxxxv36vxvxxv4+6vx3v5.(9.4)

The potential form of equation (9.4) is found by letting v = wx and integrating to get (see also (8.9))

wt=E(H1)|v=wx=wxxxwx33wxx22wx4.(9.5)

Equation (8.10) of Lemma 8.1 as it applies to the potential Harry-Dym equation (9.5) produces the following variational operator equation for Dx,

Dx(wtwxxxwx3+3wxx22wx4)=E(12wxwt12wxx2wx3).

We now apply Theorem 8.1 to obtain a second variational operator. The compatibility condition in equation (8.14) is satisfied with the operators from equation (9.3) with

𝒟0(E(H1))=𝒟x(E(H2)),whereH2=12vxx2v5158vx4z7.(9.6)

The operator 𝒟0 in equation (9.3) is of the form (8.19) so that by equation (8.20)

=𝒟0|v=wx=wx1Dx3wx1(9.7)
is a symplectic or variational operator. Since the compatibility condition in equation (8.14) is satisfied and is a symplectic operator Theorem 8.1 applies. Therefore the operator in equation (9.7) is a variational operator for the potential Harry-Dym equation in (9.5). The function Q in equation (5.28) is easily determined for (using the fact that −2 is a symplectic operator) to be
Q=wxx2wxwxxx2wx3.(9.8)

Equation (8.16) with Q in equation (9.8) and H2 in equation (9.6) (with v = wx) gives the variational operator equation for the potential Harry-Dym equation (9.5),

(wtwxxxwx3+3wxx22wx4)=E(wxx2wxwxxx2wx3wt+12wxxx2wx5158wxx5wx7).

If we return to the original coordinates for the Harry-Dym equation and make the change of variable given by x = u, w = x, wx=ux1, to the potential form in equation (9.5) we get the Krichever-Novikov equation (or Schwarzian KdV), pg. 120 in [10],

ut=uxxx32uxx2ux.(9.9)

In particular the Krichever-Novikov in equation (9.9) is the potential form of the Harry-Dym equation (9.1). These different coordinate representations of the Harry-Dym equation and the Krichever-Novikov equation is summarized by the diagram,

(9.10)

The variational or symplectic operators for the Krichever-Novikov equation are obtained by applying the change of variables x = u,w = x, wx=ux1, to Dx and equation (9.7) giving the well known symplectic or variational operators for the Krichever-Novikov equation [10],

1=ux1Dxux1=1ux2Dxuxxux3,0=1ux2Dx33uxxux3Dx2+(3uxx2ux4uxxxux3)Dx.(9.11)

With quotient map q(t,x,u,ux,uxx,)=(t=t,x=u,z=ux,zx=uxxux1,), the operators from (9.11) project q*i=𝒟^i to the Hamiltonian operators in equation (9.2).

We now compute the explicit unique representative for the H1,2() cohomology class for the Krichever-Novikov equation (9.9) corresponding to the first operator in (9.11) (Theorem 4.1). This is computed using formula (1.4) in Theorem 1.1 to be,

ω1=12ux2dxθ0θ1+dt[θ0(4uxxxux3uxx24ux4θ1+uxx2ux3θ212ux2θ3)+1ux2θ1θ2].(9.12)

We have dV ω1 = 0 and for the forms η and λ in Theorem 5.3 we may choose

η1=12uxdxθ0+dt(uxx22uxxxux4ux3θ0+uxx2ux2θ1+12uxθ2)λ1=3uxx24ux2dtdx.(9.13)

Likewise formula (1.4) for the second operator in (9.11) gives the unique cohomology representative (Theorem 4.1),

ω^0=dxθ0(uxuxxx3uxx22ux4θ1+3uxx2ux3θ212ux2θ3)12ux2dtθ2θ3+dtθ1(12ux2θ4uxxux3θ35uxuxxx6uxx22ux4θ2)12ux2dtθ0θ5+2ux3uxxxxx18ux2uxxuxxxx12ux2uxxx2+69uxuxx2uxxx39uxx44u6xdtθ0θ1+dtθ0(10ux2uxxxx48uxuxxuxxx+39uxx34ux5θ2+3(4uxuxxx7uxx2)4ux4θ3+2uxxux3θ4).(9.14)

In this case dVω^00, but [ω^0]=[ω0] where

ω0=ω^0+dH(uxx2ux3θ0θ1)(9.15)
and dV ω0 = 0. Furthermore with ω0 in equation (9.15) the forms η and λ in Theorem 5.3 can be chosen to be
η0=2uxx2uxuxxxx2ux3dxθ0+2uxx2uxuxxxx2ux3dtθ2+uxxxxux23uxuxxuxxx+uxx32ux4dtθ12ux3uxxxxx10ux2uxxuxxxx6ux2uxxx2+27uxuxx2uxxx12uxx44ux5dtθ0λ0=uxx48ux4dtdx.(9.16)

For λi in equations (9.13) and (9.16), it is difficult to determine whether [λi] ∈ H2,0() is trivial or not (see Theorem A.2). However, it is possible but not easy to show λii where κi is t-invariant by using the infinite sequence of conservation laws [10] for the Krichever-Novikov (Schwarzian KdV) equation (9.9). The forms λi define a non-trivial cohomology class in the t-invariant variational bi-complex for (9.9).

### Example 9.2.

The Harry Dym equation can be written in the form

vt=Dx3(1v)=𝒟i(E(Hi)),i=0,1(9.17)
where the Hamilonian operators and their Hamiltonians are
𝒟0=Dx3,H0=2vand𝒟1=2vDx+vx,H1=18v52vx2.

Equation (9.17) is obtained from equation (9.4) by substituting v=212v^.

Another potential form (or integrable extension) for the Harry-Dym equation (9.17) can be obtained by letting v = uxxx in equation (9.17) so that

utxxx=(Dx3E(H0))|v=uxxx,
which after integrating three times gives,
ut=E(H0)|v=uxxx=1uxxx.(9.18)

We show that Dx3 is a variational operator. First using the change of variables formula in the calculus of variation for v = uxxx (exercise 5.49 [17]) we have

(Dx3E(H0))|v=uxxx=E(H0|v=uxxx).(9.19)

The operator Dx3 is symplectic which together with equation (9.19) shows that Dx3 is a variational operator for equation (9.18) and giving,

Dx3(utE(H0)|v=uxxx)=utxxx+E(H0|v=uxxx)=E(12utuxxx+2uxxx).

In equation (8.14) compatibility was used to show the second Hamiltonian operator for a bi-Hamiltonian equation became a variational operator for the potential form. In order to use a similar argument in this case we need to show 𝒟1E(H0) = 𝒟0E(H−1). We find

𝒟1(E(H0))=(2vDx+vx)(1v)=0=𝒟0(0).(9.20)

In analogy to equation (8.14), this gives rise with H−1 = 0 to the variational operator

=𝒟1|v=uxxx=2uxxxDx+uxxxx.(9.21)

Using the fact that operator in equation (9.21) is a symplectic operator, the compatibility condition (9.20) gives

(utE(H0)|v=uxxx)=2uxxxutx+uxxxxut(E(H0))|v=uxxx=E(12uxx2ut).(9.22)

Equation (9.22) shows directly that in (9.21) is a variational operator for equation (9.18).

It is worth noting that (K) = 0 in this example and that [ω] = dV [η] where [η] ∈ H1,1(). The representative

ω=dxθ0εdtβ(ε)+dH(θ0θ1uxx)+13dHdV(uuxxθ1(uxuxx+uuxxx)θ0)
with ε=12(θ0)=uxxxθ112uxxxxθ0 satisfies ω = dV η where
η=dxθ0(23uxuxxx13uuxxxx)dtβ(23uxuxxx13uuxxxx).

Since dHη = 0, [η] ∈ H1,1(). This also produces an example where

Q=23uxuxxx13uuxxxx
satisfies LΔ*(Q)=0, as well as equation (A.6). By Theorem A.1, Corollary A.1 or Corollary A.3, Q is not the characteristic of a classical conservation law

### Example 9.3.

The cylindrical KdV equation is (see [21])

vt=vxxx+vvxv2t(9.23)
while it’s potential form is
ut=uxxx+12ux2u2t.(9.24)

The form κ in equation (1.7) in Theorem 1.4 is κ = t−1dt = dH(logt) and so equation (9.24) admits 1 = tDx as a variational operator. In equation (5.13) we have Q1=12tux leading to

1(utuxxx12ux2+u2t)=E(12tux(utuxxx12ux2+u2t)112tux3).

Note that the Lagrangian on the right side of this equation differs from that in equation (1.3) by a total divergence.

By solving the equation θ0LΔ*(ε)=0 from (4.1) for third order forms ɛ we find that equation (9.24) admits a third order symplectic or variational operator,

0=t2Dx3+13(2t2ux+tx)Dx+16(2t2uxx+t).

For 0 we have

Q0=16(t2ux2+txux+3uxxxt2)
0(utuxxx12ux2+u2t)=E(Q0(utuxxx12ux2+u2t)172(t2ux4+2txux3))

If we now compute the reduction of the potential cylindrical KdV by substituting w=tux into the x-derivative of equation (9.24) we get

wt=wxxx+1twwx=𝒟1(E(H1))=𝒟0(E(H0))(9.25)
where
𝒟1=Dx,H1=12wx2+16tw3,𝒟0=Dx3+2w3tDx+wx3t,H0=12w2.(9.26)

Equation (9.25) can of course be obtained from the cylindrical KdV equation (9.23) by the change of variables w=tv. It is unclear (to the authors) if the cylindrical KdV in equation (9.24) is a bi-Hamiltonian evolution equation for which 𝒟1 and 𝒟0 in equation (9.26) are Hamiltonian operators. Reference [21] states there are no Hamiltonians for the cylindrical KdV. It is straightforward to work out the symplectic or variational operators for the potential cylindrical KdV in these new variables from equation (9.25) by following Theorem 8.1.

More generally any evolution equation of the form

ut=uxxx+a(t)ux2.(9.27)
admits Dx as a first order variational operator. We find after a long computation that equation (9.27) admits a third order variational operator in the case where a(t) (t) ≠ 0 only when
a(t)=±1c1t+c2.(9.28)

For the + sign in equation (9.28), the change of variables t=c11(t^c2), x=c113x^, u=12c113u^, takes equation (9.27) with a(t) in (9.28) to the potential form of the cylindrical KdV obtained from equation (9.25). The same result holds in the other cases with slightly different changes of variable.

## 10. Conclusions

The determination of a variational or symplectic operator for a scalar evolution equation has been shown to be equivalent to the non-vanishing of a cohomology class in H1,2(). The arguments used to prove this clearly extend to other types of differential equations including systems. For example Theorem 5.1 holds independently of Δ being a evolution equation and so the variational operators for Δ always determine an element of the cohomology Hn−1,2() as in Theorem 5.1.

There remain many open theoretical questions such as how the compatibility condition for symplectic operators appears in the cohomology. Another interesting problem is to determine under what conditions the symmetry reduction of a variational operator equation is a Hamiltonian system (the converse of Lemma 8.1).

Many difficult computational questions have also not been resolved. We were unable to compute the dimension of H1,2() in our examples. Preliminary computations using equation θ0LΔ*(ρ)=0 from Theorem 3.1 suggests that dimH1,2() = 2 for the Krichever-Novikov equation in Example 1. However we were not able to give a full proof of this fact. We have also not explored in any detail the obvious generalization of Noether’s Theorem which arises from the existence of a variational operator or equivalently by utilizing a non-trivial element of H1,2(). This would provide an alternate derivation for identifying symmetries and conservation laws for symplectic Hamiltonian systems, see Theorem 7.1. in [17] and [13].

## A. The Vertical Differential

The vertical differential induces a mapping dV : Hr,s() → Hr,s+1() defined by dV [ω] = [dV ω]. Let ut = K be a scalar evolution equation with equation manifold . We now examine when [ω] ∈ ImagedV .

### Theorem A.1.

Let [ζ] ∈ H1,1(). There exists [κ] ∈ H1,0() such that [ζ] = dV [κ] if and only if δV ∘ Π(ζ) = 0 where Π:Ω1,1()Ωtsb1,1(E) is the induced map from equation (7.1) and ζ is any representative of [ζ].

This answers the question of when [ζ] is the image of a classical conservation law [κ]. To relate Theorem A.1 to the theory of characteristics for a conservation law, suppose [ζ] ∈ H1,1() with (unique) canonical representative given in Theorem 3.3 by

ζ=dxθ0Qdtβ(Q)
where the function Q satisfies LΔ*(Q)=0. Theorem A.1 states that the function Q is the characteristic of a classical conservation law for Δ if and only if Q = E(L). The test for this condition is the Helmholtz condition δVE(dxθE0Q)=0.

### Proof.

Suppose [ζ] ∈ H1,1() where ζ = dx ∧ (aiθi) − dtβ is a representative, then

IEdVEΠ(dx(aiθi)dtβ)=IEdVE(dx(aiθEi))=0.

This implies by equation (6.2),

dx(aiθEi)=dVE(gdx)+dHE(miθEi).(A.1)

Let μ = miθi ∈ Ω1,0() and

ζ^=ζdHμ.(A.2)
so that [ζ^]=[ζ]. Now by equation (7.3), the definition of Π in equation (7.1), and equation (A.1),
Π(ζ^)=Π(ζ)dHEΠ(μ)=dx(aiθEi)dHE(miθEi)=dVE(gdx)=giθEi.

Therefore there exists β^Ω0,1() such that,

ζ^=giθidxdtβ^=dV(gdx)dtβ^.(A.3)

Now dHdVζ^=dVdHζ=0. and therefore from equation (A.3),

dHdVζ^=dtdxX(dVβ)=0.(A.4)

However dV β ∈ Ω0,2() and the only way the contact two form dV β satisfies equation (A.4) is if dV β = 0. This implies from equation (A.3) that dVη^=0.

Using the vertical exactness of Ω1,1() we conclude there exists κ ∈ Ω1,0() such that ζ^=dVκ. Now

dVdHκ=dHdVκ=dHζ^=0.

Again by vertical exactness of the (augmented) variational bicomplex for dV : Ω2,0() → Ω2,1() applied to dHκ we have,

dHκ=a(t,x)dtdx.

Since ℝ2 is simply connected we may write

dHκ=a(t,x)dtdx=d(g(t,x)dx+h(t,x)dt).(A.5)

Finally let

κ^=κg(t,x)dxh(t,x)dt,
so that dVκ^=dVκ=ζ^, and equation (A.5) gives
dHκ^=dHκd(g(t,x)dx+h(t,x)dt)=0.

Therefore [ζ]=dV[κ^] and [κ^]H1,0().

### Corollary A.1.

Let [ζ] ∈ H1,1() with canonical representative given by

ζ=dxθ0Qdtβ
where LΔ*(Q)=0 (see Theorem 3.1). Then [ζ] = dV [κ] where [κ] ∈ H1,0() if and only if the function Q is in the image of the Euler operator. That is if and only if there exists A(t, x, u, ux, uxx, ...) ∈ C(E) such that Q = E(A).

### Corollary A.2.

If ut = K(t, x, u, ..., u2m), m ≥ 1 is even order, then every solution Q to LΔ*(Q)=0 is the characteristic of a conservation law.

As is well known, the characteristic of a conservation law is a solution to LΔ*(Q)=0 but the converse is not necessarily true. Corollary A.1 identifies the characteristics which come from conservation laws. See Example 9.2 for a solution to LΔ*(Q)=0 which is not the characteristic of a conservation law.

We now examine the case of H1,2().

### Theorem A.2.

Let [ω] ∈ H1,2(). Then [ω] = dV [η] where [η] ∈ H1,1() if and only if [ω] ∈ KerΛ where Λ : H1,2() → H2,0() is defined in equation (4.18).

### Proof.

Let [ω] ∈ H1,2() with representative ω satisfying ω = dV η and λ be as in Lemma 4.1. That is

[ω]=[dVη]dHη=dVλ.

Suppose now that λ = dHκ so that [ω] ∈ KerΛ. Let η^=η+dVκ. Then

[dVη^]=[dVη],dHη^=dHη+dHdVκ=dVλdVdHκ=0.

Therefore [ω]=dV[η^] where [η^]H1,1(). This proves sufficiency of the condition.

Suppose now that [ω] = dV [η] where [η] ∈ H1,1(). Let ω be the representative such that ω = dV η. By hypothesis dHη = 0 and so for λ in Lemma 4.1 we have

dHη=dVλ=0.

The same argument as in the second part of the proof of Corollary 4.4 implies that there exists κ ∈ Ω1,0(ℝ2) such that λ = dHκ. Therefore Λ([ω]) = [λ] = [dHκ] = 0.

Theorem A.2 is demonstrated in Example 9.2. As a simple corollary to Theorem A.2 we can identify the elements of H1,1() which are not the image of a conservation law as follows.

### Corollary A.3.

The map dV : H1,1()/dV (H1,0()) → KerΛ is an isomorphism. Moreover, we can identify ηH1,1()/dV (H1,0()) with the space of functions QC() such that

(FQ*FQ)(utK)=E(Q(utK))δVE(dxθEQ)0.(A.6)

## Footnotes

a

𝒟0 is the push-forward of by the quotient map q : (t,x,u,ux,...) → (t, x, v, vx, ...).

Journal
Journal of Nonlinear Mathematical Physics
Volume-Issue
26 - 4
Pages
604 - 649
Publication Date
2019/07/09
ISSN (Online)
1776-0852
ISSN (Print)
1402-9251
DOI
10.1080/14029251.2019.1640470How to use a DOI?
Open Access

TY  - JOUR
AU  - M.E. Fels
AU  - E. Yaşar
PY  - 2019
DA  - 2019/07/09
TI  - Variational Operators, Symplectic Operators, and the Cohomology of Scalar Evolution Equations
JO  - Journal of Nonlinear Mathematical Physics
SP  - 604
EP  - 649
VL  - 26
IS  - 4
SN  - 1776-0852
UR  - https://doi.org/10.1080/14029251.2019.1640470
DO  - 10.1080/14029251.2019.1640470
ID  - Fels2019
ER  -