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Nonlinear Analysis
journal homepage:www.elsevier.com/locate/na
A Hamiltonian vorticity–dilatation formulation of the
compressible Euler equations
Mónika Polner
a,∗, J.J.W. van der Vegt
baBolyai Institute, University of Szeged, Aradi vértanúk tere 1, 6720 Szeged, Hungary
bDepartment of Applied Mathematics, University of Twente, P.O. Box 217, 7500 AE, Enschede, The Netherlands
a r t i c l e i n f o Article history:
Received 5 October 2013 Accepted 6 July 2014
Communicated by Enzo Mitidieri
MSC: 37K05 58A14 58J10 35Q31 76N15 93C20 65N30 Keywords:
Compressible Euler equations Hamiltonian formulation de Rham complex Hodge decomposition Stokes–Dirac structures Vorticity Dilatation
a b s t r a c t
Using the Hodge decomposition on bounded domains the compressible Euler equations of gas dynamics are reformulated using a density weighted vorticity and dilatation as pri-mary variables, together with the entropy and density. This formulation is an extension to compressible flows of the well-known vorticity–stream function formulation of the in-compressible Euler equations. The Hamiltonian and associated Poisson bracket for this new formulation of the compressible Euler equations are derived and extensive use is made of differential forms to highlight the mathematical structure of the equations. In order to deal with domains with boundaries also the Stokes–Dirac structure and the port-Hamiltonian formulation of the Euler equations in density weighted vorticity and dilatation variables are obtained.
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1. Introduction
The dynamics of an inviscid compressible gas is described by the compressible Euler equations and an equation of state. The compressible Euler equations have been extensively used to model many different types of compressible flows, since in many applications the effects of viscosity are small or can be neglected. This has motivated over the years extensive theoretical and numerical studies of the compressible Euler equations. The Euler equations for a compressible, inviscid and non-isentropic gas in a domainΩ
⊆
R3are defined asρ
t= −∇ ·
(ρ
u),
(1)ut
= −
u· ∇
u−
1
ρ
∇
p,
(2)st
= −
u· ∇
s,
(3)∗Corresponding author. Tel.: +36 304707243.
E-mail addresses:[email protected](M. Polner),[email protected](J.J.W. van der Vegt).
http://dx.doi.org/10.1016/j.na.2014.07.005
with u
=
u(
x,
t) ∈
R3the fluid velocity,ρ = ρ(
x,
t) ∈
R+the mass density and s(
x,
t) ∈
R the entropy of the fluid, which is conserved along streamlines. The spatial coordinates are x∈
Ωand time t and the subscript means differentiation with respect to time. The pressure p(
x,
t)
is given by an equation of statep
=
ρ
2∂
U∂ρ
(ρ,
s),
(4)where U
(ρ,
s)
is the internal energy function that depends on the densityρ
and the entropy s of the fluid. The compressible Euler equations have a rich mathematical structure [1] and can be represented as an infinite dimensional Hamiltonian system [2,3]. Depending on the field of interest, various types of variables have been used to define the Euler equations, e.g. conservative, primitive and entropy variables [1]. The conservative variable formulation is for instance a good starting point for numerical discretizations that can capture flow discontinuities [4], such as shocks and contact waves, whereas the primitive and entropy variables are frequently used in theoretical studies.In many flows vorticity is, however, the primary variable of interest. Historically, the Kelvin circulation theorem and Helmholtz theorems on vortex filaments have played an important role in describing incompressible flows, in particular the importance of vortical structures. This has motivated the use of vorticity as primary variable in theoretical studies of incompressible flows, see e.g. [5,2], and the development of vortex methods to compute incompressible vortex dominated flows [6].
The use of vorticity as primary variable is, however, not very common for compressible flows. This is partly due to the fact that the equations describing the evolution of vorticity in a compressible flow are considerably more complicated than those for incompressible flows. Nevertheless, vorticity is also very important in many compressible flows. A better insight into the role of vorticity, and also dilatation to account for compressibility effects, is not only of theoretical importance, but also relevant for the development of numerical discretizations that can compute these quantities with high accuracy.
In this article we will present a vorticity–dilatation formulation of the compressible Euler equations. Special attention will be given to the Hamiltonian formulation of the compressible Euler equations in terms of the density weighted vorticity and dilatation variables on domains with boundaries. This formulation is an extension to compressible flows of the well-known vorticity–stream function formulation of the incompressible Euler equations [5,2]. An important theoretical tool in this analysis is the Hodge decomposition on bounded domains [7]. Since bounded domains are crucial in many applications we also consider the Stokes–Dirac structure of the compressible Euler equations. This results in a port-Hamiltonian formulation [8] of the compressible Euler equations in terms of the vorticity–dilatation variables, which clearly identifies the flows and efforts entering and leaving the domain. An important feature of our presentation is that we extensively use the language of differential forms. Apart from being a natural way to describe the underlying mathematical structure it is also important for our long term objective, viz. the derivation of finite element discretizations that preserve the mathematical structure as much as possible. A nice way to achieve this is by using discrete differential forms and exterior calculus, as highlighted in [9–11].
The outline of this article is as follows. In the introductory Section2we summarize the main techniques that we will use in our analysis. A crucial element is the use of the Hodge decomposition on bounded domains, which we briefly discuss in Section2.2. This analysis is based on the concept of Hilbert complexes, which we summarize in Section2.1. The Hodge Laplacian problem is discussed in Section2.3. Here we show how to deal with inhomogeneous boundary conditions, which is of great importance for our applications. These results will be used in Section3to define via the Hodge decomposition a new set of variables, viz., the density weighted vorticity and dilatation, and to formulate the Euler equations in terms of these new variables. Section4deals with the Hamiltonian formulation of the Euler equations using the density weighted vorticity and dilatation, together with the density and entropy, as primary variables. The Poisson bracket for the Euler equations in these variables is derived in Section5. In order to account for bounded domains we extend the results obtained for the Hamiltonian formulation in Sections4and5to the port-Hamiltonian framework in Section6. First, we extend in Section6.1the Stokes–Dirac structure for the isentropic compressible Euler equations presented in [12] to the non-isentropic Euler equations. Next, we derive the Stokes–Dirac structure for the compressible Euler equations in the vorticity–dilatation formulation in Section6.3and use this in Section6.5to obtain a port-Hamiltonian formulation of the compressible Euler equations in vorticity–dilatation variables. Finally, in Section7we finish with some conclusions.
2. Preliminaries
This preliminary section is devoted to summarize the main concepts and techniques that we use throughout this paper in our analysis.
2.1. Review of Hilbert complexes
In this section we discuss the abstract framework of Hilbert complexes, which is the basis of the exterior calculus in Arnold, Falk and Winther [10] and to which we refer for a detailed presentation. We also refer to Brüning and Lesch [13] for a functional analytic treatment of Hilbert complexes.
Definition 1. A Hilbert complex
(
W,
d)
consists of a sequence of Hilbert spaces Wk, along with closed, densely-defined linear operators dk:
Wk→
Wk+1, possibly unbounded, such that the range of dkis contained in the domain of dk+1anddk+1
◦
dk=
0 for each k.A Hilbert complex is bounded if, for each k
,
dkis a bounded linear operator from Wkto Wk+1and it is closed if for each k, the range of dkis closed in Wk+1.Definition 2. Given a Hilbert complex
(
W,
d)
, a domain complex(
V,
d)
consists of domainsD(
dk) =
Vk⊂
Wk, endowedwith the graph inner product
⟨
u, v⟩
Vk= ⟨
u, v⟩
Wk+
dku
,
dkv
Wk+1.
Remark 1. Since dkis a closed map, each Vkis closed with respect to the norm induced by the graph inner product. From
the Closed Graph Theorem, it follows that dkis a bounded operator from Vkto Vk+1. Hence,
(
V,
d)
is a bounded Hilbertcomplex. The domain complex is closed if and only if the original complex
(
W,
d)
is.Definition 3. Given a Hilbert complex
(
W,
d)
, the space of k-cocycles is the null space Zk=
ker dk, the space ofk-coboundaries is the image Bk
=
dk−1Vk−1, the kth harmonic space is the intersection Hk=
Zk∩
Bk⊥W, and the kth reduced cohomology space is the quotient Zk/
Bk. When Bkis closed, Zk/
Bkis called the kth cohomology space.Remark 2. The harmonic space Hkis isomorphic to the reduced cohomology space Zk
/
Bk. For a closed complex, this isidentical to the homology space Zk
/
Bk, since Bkis closed for each k.Definition 4. Given a Hilbert complex
(
W,
d)
, the dual complex(
W∗,
d∗)
consists of the spaces W∗k
=
Wk, and adjoint operators d∗k=
(
dk−1)
∗:
V∗ k⊂
W ∗ k→
V ∗ k−1⊂
W ∗ k−1. The domain of d ∗ kis denoted by V ∗ k, which is dense in Wk. Definition 5. We can define the k-cycles Z∗k
=
ker d ∗ k=
Bk ⊥W and k-boundaries B∗ k=
d ∗ k+1V ∗ k.2.2. The L2-de Rham complex and Hodge decomposition
The basic example of a Hilbert complex is the L2-de Rham complex of differential forms. Let Ω
⊆
Rn be an
n-dimensional oriented manifold with Lipschitz boundary
∂
Ω, representing the space of spatial variables. Assume that there is a Riemannian metric⟨⟨
, ⟩⟩
onΩ. We denote byΛk(
Ω)
the space of smooth differential k-forms onΩ,
d is the exteriorderivative operator, taking differential k-forms on the domainΩto differential
(
k+
1)
-forms,δ
represents the coderivative operator and⋆
the Hodge star operator associated to the Riemannian metric⟨⟨
, ⟩⟩
. The operations grad,
curl,
div, ×, ·
from vector analysis can be identified with operations on differential forms, see e.g. [14].We can define the L2-inner product of any two differential k-forms onΩas
⟨
ω, η⟩
L2Λk=
Ωω ∧ ⋆η =
Ω⟨⟨
ω, η⟩⟩v
Ω=
Ω⋆(ω ∧ ⋆η)v
Ω,
(5) wherev
Ω is the Riemannian volume form. The Hilbert space L2Λk(
Ω)
is the space of differential k-forms for which∥
ω∥
L2Λk=
⟨
ω, ω⟩
L2Λk< ∞
. WhenΩ is omitted from L2Λkin the inner product(5), then the integral is always over Ω. The exterior derivative d=
dkmay be viewed as an unbounded operator from L2Λkto L2Λk+1. Its domain, denoted byHΛk
(
Ω)
, is the space of differential forms in L2Λk(
Ω)
with the weak derivative in L2Λk+1(
Ω)
, that isD
(
d) =
HΛk(
Ω) = {ω ∈
L2Λk(
Ω) |
dω ∈
L2Λk+1(
Ω)},
which is a Hilbert space with the inner product
⟨
ω, η⟩
HΛk= ⟨
ω, η⟩
L2Λk+ ⟨
dω,
dη⟩
L2Λk+1.
For an oriented Riemannian manifoldΩ
⊆
R3, the L2de Rham complex is0
→
L2Λ0(
Ω)
−→
d L2Λ1(
Ω)
−→
d L2Λ2(
Ω)
−→
d L2Λ3(
Ω) →
0.
(6)Note that d is a bounded map from HΛk
(
Ω)
to L2Λk+1(
Ω)
andD(
d) =
HΛk(
Ω)
is densely-defined in L2Λk(
Ω)
. SinceHΛk
(
Ω)
is complete with the graph norm, d is a closed operator (equivalent statement to the Closed Graph Theorem). Thus, the L2de Rham domain complex forΩ⊆
R3is
0
→
HΛ0(
Ω)
−→
d HΛ1(
Ω)
−→
d HΛ2(
Ω)
−→
d HΛ3(
Ω) →
0.
(7)The coderivative operator
δ :
L2Λk(
Ω) →
L2Λk−1(
Ω)
is defined asSince we assumed thatΩhas Lipschitz boundary, the trace theorem holds and the trace operator tr
=
tr∂Ωmaps HΛk(
Ω)
boundedly into the Sobolev space H−1/2Λk(∂
Ω)
. Moreover, the trace operator extends to a bounded surjection fromH1Λk
(
Ω)
onto H1/2Λ(∂
Ω)
, see [9]. We denote the space HΛk(
Ω)
with vanishing trace as ◦HΛk
(
Ω) = {ω ∈
HΛk(
Ω) |
trω =
0}
.
(9) In analogy with HΛk(
Ω)
, we can define the spaceH∗Λk
(
Ω) = ω ∈
L2Λk(
Ω) | δω ∈
L2Λk−1(
Ω) .
(10) Since H∗Λk(
Ω) = ⋆
HΛn−k(
Ω)
, forω ∈
H∗Λk(
Ω)
, the quantity tr(⋆ω)
is well defined, and we can define◦ H∗Λk
(
Ω) = ⋆
◦ HΛn−k(
Ω) = {ω ∈
H∗Λk(
Ω) |
tr(⋆ω) =
0}
.
(11) The adjoint d∗=
d∗ kof dk−1has domainD
(
d∗) =
H◦∗Λk(
Ω)
and coincides with the operatorδ
defined in(8), (see [10]).Hence, the dual complex of(7)is
0
←
◦ H∗Λ0(
Ω)
←−
δ ◦ H∗Λ1(
Ω)
←−
δ ◦ H∗Λ2(
Ω)
←−
δ ◦ H∗Λ3(
Ω) ←
0.
(12)The integration by parts formula also holds
⟨
dω, η⟩ = ⟨ω, δη⟩ +
∂Ω trω ∧
tr(⋆η), ω ∈
Λk−1(
Ω), η ∈
Λk(
Ω),
(13) and we have⟨
dω, η⟩ = ⟨ω, δη⟩ , ω ∈
HΛk−1(
Ω), η ∈
◦ H∗Λk(
Ω).
(14)Since the L2-de Rham complexes(6)and(7)are closed Hilbert complexes, the Hodge decomposition of L2Λkand HΛkare:
L2Λk
=
Bk⊕
Hk⊕
◦ B∗k,
(15) HΛk=
Bk⊕
Hk⊕
Zk⊥,
(16) where ◦ B∗k= {
δω | ω ∈
◦ H∗Λk+1(
Ω)}
, and Zk⊥=
HΛk∩
◦B∗kdenotes the orthogonal complement of Zkin HΛk. The space of harmonic forms, both for the original complex and the dual complex, is
Hk
= {
ω ∈
HΛk(
Ω) ∩
◦
H∗Λk
(
Ω) |
dω =
0, δω =
0}
.
(17) Problems with essential boundary conditions are important for applications. This is why we briefly review the de Rham complex with essential boundary conditions. Take as domain of the exterior derivative dkthe spaceH◦Λk(
Ω)
. The de Rhamcomplex with homogeneous boundary conditions onΩ
⊂
R3is0
→
◦ HΛ0(
Ω)
−→
d ◦ HΛ1(
Ω)
−→
d ◦ HΛ2(
Ω)
−→
d ◦ HΛ3(
Ω) →
0.
(18)From(13), we obtain that
⟨
dω, η⟩ = ⟨ω, δη⟩ , ω ∈
◦
HΛk−1
(
Ω), η ∈
H∗Λk(
Ω).
(19) Hence, the adjoint d∗of the exterior derivative with domain◦
HΛk
(
Ω)
has domain H∗Λk(
Ω)
and coincides with the operatorδ
. Finally, the second Hodge decomposition of L2Λkand ofH◦Λkfollow immediately:L2Λk
=
B◦k⊕
◦ Hk⊕
B∗k,
(20) ◦ HΛk=
B◦k⊕
H◦k⊕
Zk⊥,
(21) whereB◦k=
d ◦ HΛk−1(
Ω),
Zk⊥=
H◦Λk∩
B∗kand the space of harmonic forms is ◦
Hk
= {
ω ∈
◦2.3. The Hodge Laplacian problem
In this section we first review the Hodge Laplacian problem with homogeneous natural and essential boundary conditions. The main result of this section is to show how to deal with inhomogeneous boundary conditions, which are crucial for applications.
2.3.1. The Hodge Laplacian problem with homogeneous natural boundary conditions
Given the Hilbert complex(7) and its dual complex (12), the Hodge Laplacian operator applied to a k-form is an unbounded operator Lk
=
dk−1d∗k+
d∗ k+1d
k
:
D(
Lk
) ⊂
L2Λk→
L2Λk, with domain (see [10]) D(
Lk) = {
u∈
HΛk∩
◦
H∗Λk
|
dku∈
◦H∗Λk+1
,
d∗ku∈
HΛk−1}
.
(23) In the following, we will not use the subscripts and superscripts k of the operators when they are clear from the context and use d∗=
δ
.For any f
∈
L2Λk, there exists a unique solution u=
Kkf∈
D(
Lk)
satisfyingLku
=
f(
mod H),
u⊥
H,
(24)with Kkthe solution operator, see [10]. The solution u satisfies the Hodge Laplacian (homogeneous) boundary value problem
(
dδ + δ
d)
u=
f−
PHf inΩ,
tr(⋆
u) =
0,
tr(⋆
du) =
0 on∂
Ω,
(25) where PHf is the orthogonal projection of f into H, with the condition u⊥
Hrequired for uniqueness of the solution. The boundary conditions in(25)are both natural in the mixed variational formulation of the Hodge Laplacian problem, as discussed in [10].The Hodge Laplacian problem is closely related to the Hodge decomposition in the following way. Since d
δ
u∈
Bkandδ
du∈
B∗k, the differential equation in(25), or equivalently f=
dδ
u+
δ
du+
α, α ∈
Hk, is exactly the Hodge decomposition of f∈
L2Λk(
Ω)
. If we restrict f to an element of Bkor B∗k, we obtain two problems that are important for our applications
(see also [10]).
The B problem. If f
∈
Bk, then u=
Kkf satisfies dδ
u=
f,
u⊥
◦ Z∗k, where ◦ Z∗k= {
ω ∈
◦ H∗Λk(
Ω) | δω =
0}
. It also follows that the solution u∈
Bk. To see this, consider u∈
D(
Lk)
and the Hodge decomposition u=
uB+
uH+
u⊥, where uB∈
Bk,uH
∈
Hkand u⊥∈
B∗k∩
HΛk. Then,Lku
=
dδ
uB+
δ
du⊥=
f.
If f
∈
Bk, then u=
uB, hence u∈
Bk. Then, duB=
0 since d2=
0, and since Bk=
◦
Z∗k⊥, it follows that u
⊥
◦Z∗kis also satisfied and u solves uniquely the Hodge Laplacian boundary value problem, see [10],
d
δ
u=
f,
du=
0,
u⊥
◦Z∗k inΩ (26)
tr
(⋆
u) =
0,
on∂
Ω.
(27)The B∗problem. If f
∈
B∗k, then u
=
Kkf satisfiesδ
du=
f , with u⊥
Zk. Similarly as for the B problem, it can be shownthat the solution u
∈
B∗k. Consider u∈
D(
Lk)
and the Hodge decomposition u=
uB+
uH+
u⊥. Then,Lku
=
dδ
uB+
δ
du⊥=
f.
If f
∈
B∗k, then u=
u⊥, hence u∈
B∗k. Therefore,δ
u⊥=
0 and u⊥
Zk. Consequently, u solves uniquely the Hodge Laplacianboundary value problem
δ
du=
f,
δ
u=
0,
u⊥
Zk inΩ (28)tr
(⋆
du) =
0,
on∂
Ω.
(29)2.3.2. The Hodge Laplacian problem with homogeneous essential boundary conditions
Considering now the Hilbert complex with (homogeneous) boundary conditions(18)and its dual complex, the Hodge Laplacian problem with essential boundary conditions is
Lku
=
dδ
u+
δ
du=
f(
mod ◦H
),
inΩ (30)tr
(δ
u) =
0,
tr u=
0 on∂
Ω,
(31)with the condition u
⊥
H, which has a unique solution, u◦=
Kkf . Both boundary conditions in(31)are essential in the mixedvariational formulation of the Hodge Laplacian problem (see [10]). Here the domain of the Laplacian is D
(
Lk) = {
u∈
◦
HΛk
∩
H∗Λk|
du∈
H∗Λk+1, δ
u∈
◦We can briefly formulate the B and B∗problems as follows. The B problem. If f
∈
◦
Bk, then u
=
Kkf∈
◦Bksatisfies d
δ
u=
f,
u⊥
Z∗k. Then u solves uniquely the B problemd
δ
u=
f,
du=
0,
u⊥
Z∗k inΩ (33)tr
(δ
u) =
0,
on∂
Ω.
(34)The B∗problem. If f
∈
B∗k, then u=
Kkf satisfiesδ
du=
f , with u⊥
Zk. Similarly, u solves uniquelyδ
du=
f,
δ
u=
0,
u⊥
Zk inΩ (35)tr
(
u) =
0,
on∂
Ω.
(36)The next section shows how to transform the inhomogeneous Hodge Laplacian boundary value problem into a homoge-neous one.
2.3.3. The Hodge Laplacian with inhomogeneous essential boundary conditions
Consider the case when the essential boundary conditions are inhomogeneous, that is, the boundary value problem
Lku
=
dδ
u+
δ
du=
f inΩ (37)tr u
=
rb,
tr(δ
u) =
rN on∂
Ω,
(38)with the condition
⟨
f, α⟩ =
∂Ω rN∧
tr(⋆α), ∀α ∈
◦ Hk.
(39)Here the domain of the Hodge Laplacian operator is DkD
=
u
∈
HΛk∩
H∗Λk|
du∈
H∗Λk+1, δ
u∈
HΛk−1,
tr u
=
rb∈
H1/2Λk(∂
Ω),
tr(δ
u) =
rN∈
H1/2Λk−1(∂
Ω) .
(40)This problem has a solution, which is unique up to a harmonic form
α ∈
◦
Hk. Following the idea of Schwarz in [7], the inhomogeneous boundary value problem can be transformed to a homogeneous problem in the following way. For a given
rb
∈
H1/2Λk(∂
Ω)
, using a bounded, linear trace lifting operator (see [15,9,7]), we can findη ∈
H1Λk(
Ω)
, such that trη =
rb.The Hodge decomposition of
η
η =
dφ
η+
δβ
η+
α
η,
dφ
η∈
B◦k, δβ
η∈
B∗k, α
η∈
◦Hk means for the trace that
tr
η =
d(
trφ
η) +
tr(δβ
η) +
trα
η=
tr(δβ
η),
viz. the components d
φ
ηandα
ηof the extensionη
do not contribute to the trη
. Hence, we can constructη = δβ
η. Then,Lk
η = δ
dδβ
ηand it can be easily shown that⟨
Lkη, α⟩ =
0 for allα ∈
◦Hk.
On the other hand, given rN
∈
H1/2Λk−1(∂
Ω)
, the extension result of Lemma 3.3.2 in [7], guarantees the existence of¯
η ∈
H1ΛkΩ, such thattr
η =
¯
0 and tr(δ ¯η) =
rN.
Takeu
¯
=
u−
η − ¯η
. Then, Lku¯
=
f−
Lkη −
Lkη =: ¯
¯
f,
tru¯
=
0, tr(δ¯
u) =
0 and using the condition(39), we can show that¯
f
⊥
◦Hk. Hence,u is the solution of the homogeneous boundary value problem
¯
(30)–(31)with the right hand sidef .¯
2.3.4. The Hodge Laplacian with inhomogeneous natural boundary conditionsConsider the Hodge Laplacian operator Lk
=
dδ + δ
d:
D(
Lk) ⊂
L2Λk→
L2Λk, with domain DkN=
u
∈
HΛk∩
H∗Λk|
du∈
H∗Λk+1, δ
u∈
HΛk−1,
tr
(⋆
u) =
gb∈
H−1/2Λn−k(∂
Ω),
tr(⋆
du) =
gN∈
H−1/2Λn−k−1(∂
Ω) .
(41)Our next step is to transform the inhomogeneous boundary value problem
(
dδ + δ
d)
u=
f inΩ (42)tr
(⋆
u) =
gb,
tr(⋆
du) =
gN on∂
Ω,
(43)with the condition
⟨
f, α⟩ = −
∂Ω
and the side condition for uniqueness u
⊥
Hk, into the Hodge Laplacian homogeneous boundary value problem(25). This can be considered as the dual of the problem with inhomogeneous essential boundary conditions, treated in Section2.3.3. For completeness, we briefly summarize the steps of the construction.For gb
∈
H−1/2Λn−k(∂
Ω)
we can findτ ∈
H∗Λk(
Ω)
with tr(⋆τ) =
gb. Note here that sinceτ ∈
H∗Λk(
Ω)
, then⋆τ ∈
HΛn−k(
Ω)
, so tr(⋆τ)
is well-defined. Moreover, using the Hodge decompositionτ =
dφ
τ+
δβ
τ+
α
τ,
dφ
τ∈
Bk, δβ
τ∈
◦B∗k
, α
τ∈
Hk,
we have tr
(⋆τ) =
tr(⋆
dφ
τ)
, viz. the termsδβ
τ andα
τ do not contribute to the trace, hence we can takeτ =
dφ
τ. Then,Lk
τ =
dδ
dφ
τand⟨
Lkτ, α⟩ =
0 for allα ∈
Hk. Similarly, for gN∈
H−1/2Λn−k−1(∂
Ω)
, we can findτ ∈
¯
H∗Λk(
Ω)
, with⋆¯τ |
∂Ω=
0,
tr(δ ⋆ ¯τ) =
gN.
Takingu
¯
=
u−
τ − ¯τ
, we obtain Lku¯
=
f−
Lkτ −
Lkτ =: ¯
¯
f . Moreover, tr(⋆¯
u) =
0,
tr(⋆
du¯
) =
0 and by using(44), we obtain¯
f
⊥
Hk.Consequently, solving the inhomogeneous boundary value problem(42)–(44)for given f
∈
L2Λkis equivalent with solving the Hodge Laplacian problem with homogeneous boundary conditions(25), for given¯
f .Note that, since the B and B∗problems are special cases of the Hodge Laplacian problem, they can be solved also for inhomogeneous boundary conditions.
3. The Euler equations in density weighted vorticity and dilatation formulation
In this section we will derive, via the Hodge decomposition, a Hamiltonian formulation of the compressible Euler equations using a density weighted vorticity and dilatation as primary variables. This will provide an extension of the well known vorticity–stream function formulation of the incompressible Euler equations to compressible flows, see e.g. [2,3]. Special attention will be given to the proper boundary conditions for the potential
φ
and the vector stream functionβ
.The analysis of the Hamiltonian formulation and Stokes–Dirac structure of the compressible Euler equations is most easily performed using techniques from differential geometry. For this purpose we first reformulate(1)–(3)in terms of differential forms. We identify the mass density
ρ
and the entropy s with a 3-form onΩ, that is, with an element ofΛ3(
Ω)
,the pressure p
∈
Λ0(
Ω)
, and the velocity field u with a 1-form onΩ, viz., with an element ofΛ1(
Ω)
. Let u♯be the vector field corresponding to the 1-form u (using index raising, or sharp mapping), iu♯denotes the interior product by u♯.For an arbitrary vector field X and
α ∈
Λk(
Ω)
, the following relation between the interior product and Hodge star operator is valid, (see e.g. [16]),iX
α = ⋆(
X♭∧
⋆α),
(45)where X♭is the 1-form related to X by the flat mapping. Following the identifications of the variables suggested in [8] for the isentropic fluid using differential forms, and completing them with the entropy balance equation, the Euler equations of gas dynamics can then be formulated in differential forms as
ρ
t= −
d(
iu♯ρ),
(46) ut= −
d
1 2
u♯,
u♯
v
−
iu♯du−
1⋆ρ
d p (47) st= −
u∧
⋆
d(⋆
s) =
u∧
δ
s,
(48)where
⟨·
, ·⟩
vis the inner product of two vectors.Using the L2-de Rham complex described in Section2.2, let us start with the Hodge decomposition of the differential
1-form
ρ ∧
˜
u∈
L2Λ1(
Ω)
, denoted byζ
,ζ =
˜
ρ ∧
u=
dφ + δβ + α,
(49)where
ρ = ⋆ρ
˜
. The choice ofζ
will be motivated in the next section.Definition 6. Using the Hodge decomposition(49), define the density weighted vorticity as
ω =
dζ
and the density weighted dilatation asθ = −δζ
.There are two Hodge decompositions,(15)and(20), hence two sets of boundary conditions on the Hodge components (a) d
φ ∈
B1, δβ ∈
◦ B∗1, α ∈
H1, (b) dφ ∈
B◦1, δβ ∈
B∗1, α ∈
◦ H1.Remark 3. If the classical vorticity is
ξ =
du then, the density weighted vorticity isω =
dζ =
d(
ρ ∧
˜
u) =
d√
⋆ρ ∧
u+
√
⋆ρ ∧ ξ.
(50) When the flow is incompressible, thenω =
√⋆ρ ∧ξ
, i.e., the density weighted vorticity. Using the equation for the velocity (47), the vorticity evolution equation for constant density isξ
t= −
d
iu♯du
= −
d
iu♯ξ .
(51)On the other hand, the evolution equation for density weighted vorticity for incompressible flows is
ω
t=
√
⋆ρ ∧ ξ
t= −
d
iu♯
ω .
(52)Hence,
ω
andξ
satisfy the same evolution equations.The density weighted dilatation
θ
for an incompressible flow isθ = −δ
√
⋆ρ ∧
u = −
√
⋆ρ ∧ δ
u.
(53) The incompressibility constraint in differential forms isδ
u=
0, which impliesθ =
0.Lemma 1. The potential function
φ
and vector stream functionβ
in the Hodge decomposition(49)solve the following boundary value problems B-problem
dδβ = ω,
dβ =
0 inΩ tr(⋆β) =
0 on∂
Ω,
B ∗ -problem
δ
dφ = −θ
inΩ tr(⋆
dφ) =
0 on∂
Ω,
(54) whenζ ∈
HΛ1(
Ω) ∩
◦ H∗Λ1(
Ω) ∩
H1⊥and B-problem
dδβ = ω,
dβ =
0 inΩ tr(δβ) =
0 on∂
Ω,
B ∗ -problem
δ
dφ = −θ
inΩ tr(φ) =
0 on∂
Ω,
(55) whenζ ∈
◦ HΛ1(
Ω) ∩
H∗Λ1(
Ω) ∩
H◦1⊥.Proof. The proof of this lemma is constructive and we consider two cases. Case 1. The first approach is to choose
ζ ∈
HΛ1(
Ω) ∩
◦
H∗Λ1
(
Ω) ∩
H1⊥. Then, the Hodge decomposition (a) ofζ
is used to define two new variables.Since
ζ ∈
D(
d) =
HΛ1(
Ω)
, we can define the density weighted vorticityω ∈
B2⊂
L2Λ2(
Ω)
asω =
dζ =
d(
ρ ∧
˜
u) =
d(
dφ + δβ) =
dδβ,
(56) whereφ ∈
HΛ0(
Ω), β ∈
H∗Λ2(
Ω)
with tr(⋆β) =
0 andδβ ∈
D(
d) =
HΛ1(
Ω)
. Moreover, sinceω ∈
B2, it follows thatβ ∈
B2hence, dβ =
0. Observe here, that(56)is the B problem(26)–(27)with homogeneous natural boundary conditions.Consider now
ζ ∈
D(δ) =
◦
H∗Λ1
(
Ω)
. Define the density weighted dilatationθ ∈
B∗0⊂
L2Λ0(
Ω)
asθ = −δζ = −δ(
˜
ρ ∧
u) = −δ
dφ − δδβ = −δ
dφ,
(57) whereφ ∈
HΛ0(
Ω), β ∈
H◦∗Λ2(
Ω)
and dφ ∈
H∗Λ1(
Ω)
with 0=
tr(⋆ζ ) =
tr(⋆
dφ)
. Observe that(57)with this boundarycondition is the B∗problem(28)–(29)for
φ
, whereδφ =
0 is satisfied since HΛ−1is understood to be zero and tr(⋆φ) =
0since
⋆φ
is a 3-form.Case 2. Let us choose now
ζ ∈
◦HΛ1
(
Ω) ∩
H∗Λ1(
Ω) ∩
H◦1⊥. Then, we use the Hodge decomposition (b) ofζ
to define twonew variables. Since
ζ ∈
◦
HΛ1
(
Ω),
trζ =
0 and using the decomposition in (b), we obtain the weakly imposed essential boundary condition 0=
trζ =
tr(δβ)
. Define the density weighted vorticityω ∈
B◦2asω =
dζ =
dδβ,
(58)where
β ∈
H∗Λ2(
Ω)
. This is the B problem(33)–(34)forβ
with homogeneous essential boundary conditions. Similarly, sinceζ ∈
H∗Λ1(
Ω)
, we can define the density weighted dilatationθ ∈
B∗0, as
θ = −δ
dφ,
(59)where
φ ∈
◦
HΛ0
(
Ω),
dφ ∈
H∗Λ1(
Ω)
. The Hodge decomposition in (b) implies the strongly imposed boundary conditiontr
φ =
0. This is again the B∗problem(35)–(36)forφ
, with homogeneous essential boundary conditions.Remark 4. When the flow is incompressible, then 0
=
θ = −δζ = −δ
dφ
. This implies that 0= − ⟨
δ
dφ, φ⟩ = − ⟨
dφ,
dφ⟩ +
∂Ω
tr
φ ∧
tr(⋆
dφ) = − ⟨
dφ,
dφ⟩ .
The boundary integral is zero in either of the two Hodge decompositions. Then, d
φ =
0 hence, the Hodge decomposition isζ = δβ + α
and the B∗-problems inLemma 1cancel. The B-problems remain unchanged.Remark 5. Note, inLemma 1, we can consider inhomogeneous boundary conditions for all problems. More precisely, for Case 1, the B problem for
β
has a unique solutionβ ∈
B2, with the inhomogeneous boundary condition tr(⋆β) =
gb
∈
H−1/2Λ1(∂
Ω)
, by transforming it first to a homogeneous problem with modified right hand sideδ
dβ = ¯ω
, aswe discussed in Section2.3.4. The B∗problem for
φ
has a unique solution with the inhomogeneous boundary conditiontr
(⋆
dφ) =
gN∈
H−1/2Λ2(∂
Ω)
. Hence, solving the B∗problem(57)with inhomogeneous boundary conditions, is equivalentwith solving the B∗homogeneous problem with modified right hand side
δ
dφ = −¯θ
, withθ = θ+¯τ
¯
, as seen in Section2.3.4.In Case 2, the inhomogeneous essential boundary conditions for
φ
andβ
are trφ =
rb∈
H1/2Λ0(∂
Ω)
and tr(δβ) =
rN∈
H1/2Λ1
(∂
Ω)
, respectively. We can transform the equations into a homogeneous problem as in Section2.3.3. Corollary 1. The non-isentropic compressible Euler equations can be formulated in the variablesρ, ω, θ
and s asρ
t= −
d(
˜
ρ ∧ ⋆ζ ),
(60)ω
t=
dζ
t,
(61)θ
t= −
δζ
t,
(62) st=
1
˜
ρ
∧
ζ ∧ δ
s,
(63) withζ
given by(49).Proof. The statement of this corollary can easily be verified by introducing the Hodge decomposition(49)into the Euler equations(46)–(48).
Summarizing, we use the Hodge decomposition(49)to define the density weighted vorticity
ω
and dilatationθ
. In order to have them well defined, we choose the proper spaces forζ
. The potential functionφ
and vector stream functionβ
in the Hodge components are the solutions of the B∗and B problems, respectively, with natural or essential boundary conditions. 4. Functional derivatives of the Hamiltonian in density weighted vorticity–dilatation formulationIn this section we transform the Hamiltonian functional for the non-isentropic compressible Euler equations, into the new set of variables
ρ, θ, ω,
s, and calculate the variational derivatives with respect to these new variables.In order to simplify the discussion, we assume from here on that the domainΩis simply connected. Since the dimension of H1is equal to the first Betti number, i.e., the number of handles, of the domain, it follows that H1
=
0 if the domain issimply connected, see [10]. Then
α =
0 in the Hodge decomposition(49).Let us recall from van der Schaft and Maschke [8] the definition of the variational derivatives of the Hamiltonian functional when it depends on, for example, two energy variables. Consider a Hamiltonian density, i.e. energy per volume element,
H
:
Λp(
Ω) ×
Λq(
Ω) →
Λn(
Ω),
(64)whereΩis an n-dimensional manifold, resulting in the total energy H
[
α
p, α
q] =
Ω
H
(α
p, α
q) ∈
R,
(65)where square brackets are used to indicate thatHis a functional of the enclosed functions. Let
α
p, ∂α
p∈
Λp(
Ω)
, andα
q, ∂α
q∈
Λq(
Ω)
. Then under weak smoothness assumptions on H,H
[
α
p+
∂α
p, α
q+
∂α
q] =
Ω H(α
p+
∂α
p, α
q+
∂α
q)
=
Ω H(α
p, α
q) +
Ω
δ
pH∧
∂α
p+
δ
qH∧
∂α
q
+
higher order terms in∂α
p, ∂α
q,
(66)for certain uniquely defined differential forms
δ
pH∈
(
Λp(
Ω))
∗andδ
qH∈
(
Λq(
Ω))
∗, which can be regarded as thevariational derivatives ofH with respect to
α
p andα
q, respectively. The dual linear space(
Λp(
Ω))
∗ can be naturallyFor the non-isentropic compressible Euler equations, the energy density is given as the sum of the kinetic energy and internal energy densities. The Hamiltonian functional for the compressible Euler equations in differential forms is (see [8])
H
[
ρ,
u,
s] =
Ω
1 2
u♯,
u♯
vρ +
U( ˜ρ, ˜
s)ρ
,
(67)where
˜
s=
⋆
s. The Hamiltonian functional can further be written asH
[
ρ,
u,
s] =
Ω
1 2(˜ρ ∧
u)
♯, (
˜
ρ ∧
u)
♯
vv
Ω+
U( ˜ρ, ˜
s)ρ
=
Ω
1 2⟨⟨
˜
ρ ∧
u,
ρ ∧
˜
u⟩⟩
v
Ω+
U( ˜ρ, ˜
s)ρ
.
(68)The Hamiltonian written in form(68)motivates the choice of the variable
ζ
, defined in(49), and its Hodge decomposition. In the next lemma we compute the Hamiltonian functional and its variational derivatives with respect to the Hodge decomposition ofζ
.Remark 6. We have seen that the inhomogeneous B∗
problem for
φ
and the inhomogeneous B problem forβ
can be transformed into homogeneous boundary value problems with modified right hand side. Therefore, from here on we just use the standard de Rham theory for the bar variablesθ
¯
andω
¯
, with the corresponding homogeneous boundary conditions for theφ
andβ
variables.Lemma 2. The Hamiltonian density, when the variables
ρ, φ, β,
s are introduced, is a mapping H:
HΛ3×
D0×
D2×
HΛ3→
L2Λ3,
(ρ, φ, β,
s) →
H(ρ, φ, β,
s),
whereD0andD2are the domains of the Hodge Laplacian for 0-forms and 2-forms, respectively, with either essential or natural
inhomogeneous boundary conditions, as defined in(40)and(41). This results in the total energy
H
[
ρ, φ, β,
s] =
Ω
1 2(δ
dφ ∧ ⋆φ +
dδβ ∧ ⋆β) +
U( ˜ρ, ˜
s)ρ
.
(69)The variational derivatives of the Hamiltonian are
δ
Hδρ
=
∂
∂ ˜ρ
( ˜ρ
U( ˜ρ, ˜
s)),
δ
Hδφ
= −
⋆ ¯θ,
δ
Hδβ
=
⋆ ¯ω,
δ
Hδ
s=
∂
U( ˜ρ, ˜
s)
∂˜
sρ.
˜
(70)Proof. Introducing the Hodge decomposition(49)into the Hamiltonian in(68)and using the inner product(5), we obtain the following Hamiltonian when the variables
ρ, φ, β,
s are introducedH
[
ρ, φ, β,
s] =
1 2
Ω⟨⟨
dφ + δβ,
dφ + δβ⟩⟩v
Ω+
Ω U( ˜ρ, ˜
s)ρ
=
1 2⟨
dφ + δβ,
dφ + δβ⟩ +
Ω U( ˜ρ, ˜
s)ρ.
(71)Using the definition of the density weighted vorticity and dilatation, after partial integration, the inner product in the Hamiltonian in(71)reduces to
⟨
dφ + δβ,
dφ + δβ⟩ = ⟨
dφ,
dφ⟩ + ⟨
dφ, δβ⟩ + ⟨δβ, δβ⟩ + ⟨δβ,
dφ⟩
= ⟨
φ, δ
dφ⟩ + ⟨β,
dδβ⟩ = φ, −¯θ + ⟨β, ¯ω⟩ ,
where
⟨
dφ, δβ⟩ =
0 and⟨
δβ,
dφ⟩ =
0 since(49)is an orthogonal decomposition. Note that this is valid for both types of boundary conditions, since in either case the boundary integrals are zero. Introducing this inner product into(71)we obtain(69).Consider(71)in the form H
[
ρ, φ, β,
s] =
12
(⟨
dφ,
dφ⟩ + ⟨δβ, δβ⟩) +
Ω
U
( ˜ρ, ˜
s)ρ,
(72)where the inner products are in the space L2Λ1
(
Ω)
. Let us calculateδ
φHfirst. For
φ, ∂φ ∈
D(
L0)
andβ ∈
D(
L2)
, witheither essential or natural homogeneous boundary conditions on
φ
and∂φ
, we have H[
ρ, φ + ∂φ, β,
s] =
Ω H(ρ, φ, β,
s) + ⟨
dφ,
d(∂φ)⟩ + {
h.o.t. in∂φ}
=
Ω H(ρ, φ, β,
s) + ⟨δ
dφ, ∂φ⟩ + {
h.o.t. in∂φ}.
Therefore,
δ
Hδφ
=
⋆δ
dφ = −
d(⋆
dφ) = − ⋆ ¯θ.
(73)Similarly, let us calculate the variational derivative
δ
βH. Forβ, ∂β ∈
D(
L2)
, we have for either boundary conditions,H
[
ρ, φ, β + ∂β,
s] =
Ω
H
(ρ, φ, β,
s) + ⟨∂β,
dδβ⟩ + {
h. o. t. in∂β}.
(74) Hence, we obtain thatδ
Hδβ
=
⋆
dδβ = ⋆ ¯ω,
(75)which completes the proof of this lemma. The variational derivatives of the Hamiltonian with respect to the variables
ρ
ands, given in(70), can easily be calculated, see [8].
Remark 7. We have defined the problem in the bar variables to account for inhomogeneous boundary conditions, see Section2.3. From here on we drop the bar to make the notation simpler.
Our aim is now to formulate the Hamiltonian as a functional of
ρ, θ, ω,
s and calculate the variational derivatives withrespect to these new variables.
Lemma 3. The Hamiltonian density in the variables
ρ, θ, ω,
s is a mapping H:
HΛ3×
B∗0×
B2×
HΛ3→
L2Λ3,
(ρ, θ, ω,
s) →
H(ρ, θ, ω,
s),
(76)which results in the total energy
H
[
ρ, θ, ω,
s] =
Ω
1 2(θ ∧ ⋆
K0θ + ω ∧ ⋆
K2ω) +
U( ˜ρ, ˜
s)ρ
,
(77)where Kkis the solution operator of the Hodge Laplacian operator Lk
,
k=
0,
2. The variational derivatives of the Hamiltonianfunctional are:
δ
Hδρ
=
∂
∂ ˜ρ
( ˜ρ
U( ˜ρ, ˜
s)),
δ
Hδω
=
⋆
K2ω = ⋆β,
(78)δ
Hδθ
=
⋆
K0θ = − ⋆ φ,
δ
Hδ
s=
∂
U( ˜ρ, ˜
s)
∂˜
sρ.
˜
(79)Proof. UsingLemma 2, and that
θ
andω
are in the domain of the solution operators K0 and K2of the Hodge Laplacianproblems for
φ
andβ
, respectively, the Hamiltonian can be written as H[
ρ, θ, ω,
s] =
Ω
1 2(θ ∧ ⋆
K0θ + ω ∧ ⋆
K2ω) +
U( ˜ρ, ˜
s)ρ
.
(80)Let
θ, ∂θ ∈
B∗0andω, ∂ω ∈
B2. The variational derivatives of the Hamiltonian with respect toθ
andω
can be obtained from H[
ρ, θ + ∂θ, ω + ∂ω,
s] =
1 2
Ω(θ + ∂θ) ∧ ⋆
K0(θ + ∂θ) +
Ω U( ˜ρ, ˜
s)ρ +
1 2
Ω(ω + ∂ω) ∧ ⋆
K2(ω + ∂ω)
=
Ω H(ρ, θ, ω,
s) +
1 2
Ω(θ ∧ ⋆
K0(∂θ) + ∂θ ∧ ⋆
K0θ)
+
1 2
Ω(ω ∧ ⋆
K2(∂ω) + ∂ω ∧ ⋆
K2ω) + {
h.o.t. in∂θ, ∂ω}.
(81)Here
∂θ
and∂ω
denote the variation ofθ
andω
, respectively, to avoid confusion with the coderivative operatorδ
. In order to further investigate the last two integrals in(81), we useLemma 2, where the variational derivatives of the Hamiltonian with respect toφ
andβ
are given, then apply the variational chain rule to obtainδ
θHandδ
ωH.To obtain the variational derivative of the Hamiltonian with respect to
θ
, apply the variational chain rule as follows
Ωδ
Hδφ
∧
∂φ =
Ωδ
Hδθ
∧
∂θ = −
Ωδ
Hδθ
∧
δ
d(∂φ) = −
⋆
δ
δθ
H, δ
d(∂φ)
= −
Ω dδ
δ
Hδθ
∧
∂φ +
∂Ω tr(∂φ) ∧
tr
δ
δ
δθ
H
+
∂Ω tr(⋆
d(∂φ)) ∧
tr
⋆
δ
δθ
H
,
(82) where∂φ, ∂θ = −δ
d(∂φ)
denote the variations ofφ
andθ
, to avoid confusion with the coderivative operatorδ
. Analogously, to obtain the variational derivative of the Hamiltonian with respect toω
, consider the variational chain rule
Ωδ
Hδβ
∧
∂β =
Ωδ
Hδω
∧
∂ω =
Ωδ
Hδω
∧
dδ(∂β) =
⋆
δ
δω
H,
dδ(∂β)
=
Ωδ
d
δ
Hδω
∧
∂β +
∂Ω tr(δ(∂β)) ∧
trδ
Hδω
−
∂Ω tr(⋆∂β) ∧
tr
⋆
dδ
Hδω
.
(83)We now have two choices.
First, when Case 1 applies inLemma 1, then tr
(⋆
d(∂φ)) =
0, hence the variational equation(82)becomes
Ωδ
Hδφ
∧
∂φ = −
Ω dδ
δ
Hδθ
∧
∂φ +
∂Ω tr(∂φ) ∧
tr
δ
δ
δθ
H
,
(84)∀
∂φ ∈
HΛ0(
Ω)
, with tr(⋆
d(∂φ)) =
0. Choose∂φ
such that the boundary integral in(84)is zero. We thus obtain thatδHδθ
solves the differential equation d
δ
δ
Hδθ
= −
δ
Hδφ
=
⋆θ
inΩ.
(85)Consider now the variation
∂φ ∈
D(
L0)
arbitrary. Inserting(85)into the variational equation(84), we obtain thattr
δ
δ
δθ
H
=
0,
(86)which together with(85)is precisely the B problem with essential boundary conditions(33)–(34)for δδθH, with weakly imposed boundary condition(86). On the other hand, combining(85)with(73)leads to
δ
Hδθ
= −
⋆ φ +
h,
(87)with h
∈
Z∗3, the null space ofδ
. The B problem forδδθH has however, a unique solution δδθH∈
B◦3, hence the side conditionδH
δθ
⊥
Z∗3is satisfied. Consequently,δ
Hδθ
= −
⋆ φ,
(88)where
φ
is the unique solution of the B∗problem(28)–(29).We still need to calculate
δ
ωH, when Case 1 applies. Since tr(⋆∂β) =
0, the last integral in(83)cancels. Using the same arguments as before, we obtain the B∗problem with essential boundary conditionδ
d
δ
Hδω
=
δ
Hδβ
inΩ,
tr
δ
Hδω
=
0 on∂
Ω.
(89)Combined with(75), we obtain the following equation
δ
d
δ
Hδω
=
⋆
dδβ
inΩ,
(90) which leads toδH δω=
⋆β +
h, with h∈
◦Z1, the null space of d. On the other hand, the B∗problem(89)has a unique solution δH δω
∈
B∗1=
◦ Z1⊥. Therefore,δ
Hδω
=
⋆β,
(91)When Case 2 applies, the first boundary integral in both variational equations(82)and(83)will be zero. In a completely analogous way as in Case 1, we obtain the following B problem forδδθH when natural boundary conditions hold
d
δ
δ
Hδθ
=
⋆θ
inΩ,
tr
⋆
δ
Hδθ
=
0 on∂
Ω,
(92)and the B∗problem with natural boundary condition forδδωH
δ
d
δ
Hδω
=
⋆ω
inΩ,
tr
⋆
dδ
Hδω
=
0 on∂
Ω.
(93)Analyzing the solution of these problems leads to the same variational derivatives(88)and(91).
The variational derivatives of the Hamiltonian with respect to the variables
ρ
and s, given in(78)and(79), respectively, can easily be calculated, see [8].Summarizing, when
φ
andβ
solve a B∗ and B problem, respectively, with (in)homogeneous natural or essential boundary conditions, the variational derivativesδδθH andδδωH will solve a dual problem, viz. a B and B∗problem, respectively,with the corresponding (dual) boundary conditions.
5. Poisson bracket
The nonlinear system(46)–(48)has a Hamiltonian formulation with the Poisson bracket of Morrison and Green [3,2] with the Hamiltonian given by(67)in the
ρ,
u,
s variables, (see also [8]). The Poisson bracket has the form{
F,
G} = −
Ω
δ
Fδρ
∧
dδ
Gδ
u−
δ
Gδρ
∧
dδ
Fδ
u
T1+
⋆
iX⋆
⋆
δ
δ
G u∧
⋆
δ
Fδ
u
T2+
1˜
ρ
∧
d˜
s∧
δ
Fδ
s∧
δ
Gδ
u−
δ
Gδ
s∧
δ
Fδ
u
T3,
(94)with the vector field X
=
(
⋆du ˜ρ
)
♯. The aim of this section is to transform the Poisson bracket into the new set of variablesρ, θ, ω,
s and to properly account for the boundary conditions. We derive the bracket in a well-chosen functional spaceusing the functional chain rule.
Lemma 4. Consider the Hodge decomposition of
ζ
in(49)and let F[
ρ, θ, ω,
s]
andG[
ρ, θ, ω,
s]
be arbitrary functionals. Assume thatζ ∈
H◦Λ1(
Ω) ∩
H∗Λ1(
Ω) ∩
H◦1⊥,
and tr
⋆
δ
δθ
F
=
0,
tr
⋆
δ
δθ
G
=
0 (95) orζ ∈
HΛ1(
Ω) ∩
◦ H∗Λ1(
Ω) ∩
H1⊥,
and tr
δ
Fδω
=
0,
tr
δ
Gδω
=
0.
(96)Then, the bracket(94)in terms of the variables
ρ, θ, ω,
s has the form{
F,
G} = −
Ω
δ
Fδρ
∧
d
˜
ρ ∧ α(
G)
−
δ
Gδρ
∧
d
˜
ρ ∧ α(
F)
+
⋆
ζ
2ρ
˜
∧
α(
F)
∧
d
ρ ∧ α(
˜
G)
−
⋆
ζ
2ρ
˜
∧
α(
G)
∧
d
ρ ∧ α(
˜
F)
+
d˜
s
˜
ρ
∧
δ
Fδ
s∧
α(
G) −
δ
Gδ
s∧
α(
F)
,
(97)where