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Contents lists available atScienceDirect

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

b

aBolyai 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.

© 2014 Elsevier Ltd. All rights reserved.

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 as

ρ

t

= −∇ ·

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

(2)

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

and time t and the subscript means differentiation with respect to time. The pressure p

(

x

,

t

)

is given by an equation of state

p

=

ρ

2

U

∂ρ

(ρ,

s

),

(4)

where U

(ρ,

s

)

is the internal energy function that depends on the density

ρ

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.

(3)

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+1and

dk+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, endowed

with the graph inner product

u

, v⟩

Vk

= ⟨

u

, v⟩

Wk

+

dku

,

dk

v

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 Hilbert

complex. 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 of

k-coboundaries is the image Bk

=

dk−1Vk−1, the kth harmonic space is the intersection Hk

=

Zk

BkW, 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 is

identical 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

)

:

Vk

Wk

Vk−1

Wk−1. The domain of d ∗ kis denoted by Vk, which is dense in Wk. Definition 5. We can define the k-cycles Z

k

=

ker d ∗ k

=

BkW and k-boundaries Bk

=

d ∗ k+1Vk.

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

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

(

)

the space of smooth differential k-forms on

,

d is the exterior

derivative 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 onas

ω, η⟩

L2Λk

=

ω ∧ ⋆η =

⟨⟨

ω, η⟩⟩v

=

⋆(ω ∧ ⋆η)v

,

(5) where

v

is the Riemannian volume form. The Hilbert space L2Λk

(

)

is the space of differential k-forms for which

ω∥

L2Λk

=

⟨

ω, ω⟩

L2Λk

< ∞

. WhenΩ is omitted from Lkin 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 by

HΛk

(

)

, is the space of differential forms in L2Λk

(

)

with the weak derivative in L2Λk+1

(

)

, that is

D

(

d

) =

HΛk

(

) = {ω ∈

Lk

(

) |

d

ω ∈

Lk+1

(

)},

which is a Hilbert space with the inner product

ω, η⟩

HΛk

= ⟨

ω, η⟩

L2Λk

+ ⟨

d

ω,

d

η⟩

L2Λk+1

.

For an oriented Riemannian manifoldΩ

R3, the L2de Rham complex is

0

L2Λ0

(

)

−→

d L2Λ1

(

)

−→

d L2Λ2

(

)

−→

d L2Λ3

(

) →

0

.

(6)

Note that d is a bounded map from HΛk

(

)

to L2Λk+1

(

)

andD

(

d

) =

HΛk

(

)

is densely-defined in L2Λk

(

)

. Since

HΛk

(

)

is complete with the graph norm, d is a closed operator (equivalent statement to the Closed Graph Theorem). Thus, the L2de Rham domain complex for

R3is

0

HΛ0

(

)

−→

d HΛ1

(

)

−→

d HΛ2

(

)

−→

d HΛ3

(

) →

0

.

(7)

The coderivative operator

δ :

Lk

(

) →

Lk−1

(

)

is defined as

(4)

Since we assumed thatΩhas Lipschitz boundary, the trace theorem holds and the trace operator tr

=

trmaps HΛk

(

)

boundedly into the Sobolev space H−1/2Λk

(∂

)

. Moreover, the trace operator extends to a bounded surjection from

H1Λk

(

)

onto H1/2Λ

(∂

)

, see [9]. We denote the space HΛk

(

)

with vanishing trace as

HΛk

(

) = {ω ∈

HΛk

(

) |

tr

ω =

0

}

.

(9) In analogy with HΛk

(

)

, we can define the space

H∗Λk

(

) = ω ∈

Lk

(

) | δω ∈

Lk−1

(

) .

(10) Since H∗Λk

(

) = ⋆

HΛnk

(

)

, for

ω ∈

H∗Λk

(

)

, the quantity tr

(⋆ω)

is well defined, and we can define

H∗Λk

(

) = ⋆

HΛnk

(

) = {ω ∈

H∗Λk

(

) |

tr

(⋆ω) =

0

}

.

(11) The adjoint d∗

=

dkof dk

−1has domainD

(

d

) =

H◦∗Λk

(

)

and coincides with the operator

δ

defined in(8), (see [10]).

Hence, the dual complex of(7)is

0

H∗Λ0

(

)

←−

δ ◦ H∗Λ1

(

)

←−

δ ◦ H∗Λ2

(

)

←−

δ ◦ H∗Λ3

(

) ←

0

.

(12)

The integration by parts formula also holds

d

ω, η⟩ = ⟨ω, δη⟩ +

∂Ω tr

ω ∧

tr

(⋆η), ω ∈

Λk−1

(

), η ∈

Λk

(

),

(13) and we have

d

ω, η⟩ = ⟨ω, δη⟩ , ω ∈

HΛk−1

(

), η ∈

H∗Λk

(

).

(14)

Since the L2-de Rham complexes(6)and(7)are closed Hilbert complexes, the Hodge decomposition of L2Λkand HΛkare:

Lk

=

Bk

Hk

◦ B∗k

,

(15) HΛk

=

Bk

Hk

Zk

,

(16) where ◦ B∗k

= {

δω | ω ∈

H∗Λk+1

(

)}

, 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

(

) ∩

H∗Λk

(

) |

d

ω =

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 spaceHΛk

(

)

. The de Rham

complex with homogeneous boundary conditions onΩ

R3is

0

HΛ0

(

)

−→

d ◦ HΛ1

(

)

−→

d ◦ HΛ2

(

)

−→

d ◦ HΛ3

(

) →

0

.

(18)

From(13), we obtain that

d

ω, η⟩ = ⟨ω, δη⟩ , ω ∈

HΛk−1

(

), η ∈

H∗Λk

(

).

(19) Hence, the adjoint d∗of the exterior derivative with domain

HΛk

(

)

has domain HΛk

(

)

and coincides with the operator

δ

. Finally, the second Hodge decomposition of L2Λkand ofHΛkfollow immediately:

Lk

=

B◦k

◦ Hk

B∗k

,

(20) ◦ HΛk

=

B◦k

H◦k

Zk

,

(21) whereB◦k

=

d ◦ HΛk−1

(

),

Zk

=

HΛk

B

kand the space of harmonic forms is ◦

Hk

= {

ω ∈

(5)

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

(

L

k

) ⊂

Lk

Lk, 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

Lk, there exists a unique solution u

=

Kkf

D

(

Lk

)

satisfying

Lku

=

f

(

mod H

),

u

H

,

(24)

with Kkthe solution operator, see [10]. The solution u satisfies the Hodge Laplacian (homogeneous) boundary value problem

(

d

δ + δ

d

)

u

=

f

PHf inΩ

,

tr

(⋆

u

) =

0

,

tr

(⋆

du

) =

0 on

,

(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

Bkand

δ

du

B∗k, the differential equation in(25), or equivalently f

=

d

δ

u

+

δ

du

+

α, α ∈

Hk, is exactly the Hodge decomposition of f

L2Λk

(

)

. 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 d

δ

u

=

f

,

u

◦ Z∗k, where ◦ Z∗k

= {

ω ∈

H∗Λk

(

) | 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

=

d

δ

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

.

(27)

The B∗problem. If f

B

k, then u

=

Kkf satisfies

δ

du

=

f , with u

Zk. Similarly as for the B problem, it can be shown

that the solution u

B∗k. Consider u

D

(

Lk

)

and the Hodge decomposition u

=

uB

+

uH

+

u⊥. Then,

Lku

=

d

δ

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 Laplacian

boundary value problem

δ

du

=

f

,

δ

u

=

0

,

u

Zk inΩ (28)

tr

(⋆

du

) =

0

,

on

.

(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

=

d

δ

u

+

δ

du

=

f

(

mod ◦

H

),

inΩ (30)

tr

u

) =

0

,

tr u

=

0 on

,

(31)

with the condition u

H, which has a unique solution, u

=

Kkf . Both boundary conditions in(31)are essential in the mixed

variational 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

(6)

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 problem

d

δ

u

=

f

,

du

=

0

,

u

Z∗k inΩ (33)

tr

u

) =

0

,

on

.

(34)

The B∗problem. If f

B∗k, then u

=

Kkf satisfies

δ

du

=

f , with u

Zk. Similarly, u solves uniquely

δ

du

=

f

,

δ

u

=

0

,

u

Zk inΩ (35)

tr

(

u

) =

0

,

on

.

(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

=

d

δ

u

+

δ

du

=

f inΩ (37)

tr u

=

rb

,

tr

u

) =

rN on

,

(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

(∂

),

tr

u

) =

rN

H1/2Λk−1

(∂

) .

(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

(∂

)

, using a bounded, linear trace lifting operator (see [15,9,7]), we can find

η ∈

Hk

(

)

, such that tr

η =

rb.

The Hodge decomposition of

η

η =

d

φ

η

+

δβ

η

+

α

η

,

d

φ

η

B◦k

, δβ

η

B∗k

, α

η

Hk means for the trace that

tr

η =

d

(

tr

φ

η

) +

tr

(δβ

η

) +

tr

α

η

=

tr

(δβ

η

),

viz. the components d

φ

ηand

α

ηof the extension

η

do not contribute to the tr

η

. Hence, we can construct

η = δβ

η. Then,

Lk

η = δ

d

δβ

ηand it can be easily shown that

Lk

η, α⟩ =

0 for all

α ∈

Hk.

On the other hand, given rN

H1/2Λk−1

(∂

)

, the extension result of Lemma 3.3.2 in [7], guarantees the existence of

¯

η ∈

H1Λk, such that

tr

η =

¯

0 and tr

(δ ¯η) =

rN

.

Takeu

¯

=

u

η − ¯η

. Then, Lku

¯

=

f

Lk

η −

Lk

η =: ¯

¯

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 conditions

Consider the Hodge Laplacian operator Lk

=

d

δ + δ

d

:

D

(

Lk

) ⊂

Lk

Lk, with domain DkN

=

u

HΛk

H∗Λk

|

du

H∗Λk+1

, δ

u

HΛk−1

,

tr

(⋆

u

) =

gb

H−1/2Λnk

(∂

),

tr

(⋆

du

) =

gN

H−1/2Λnk−1

(∂

) .

(41)

Our next step is to transform the inhomogeneous boundary value problem

(

d

δ + δ

d

)

u

=

f inΩ (42)

tr

(⋆

u

) =

gb

,

tr

(⋆

du

) =

gN on

,

(43)

with the condition

f

, α⟩ = −

∂Ω

(7)

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−1/2Λnk

(∂

)

we can find

τ ∈

H∗Λk

(

)

with tr

(⋆τ) =

gb. Note here that since

τ ∈

H∗Λk

(

)

, then

⋆τ ∈

HΛnk

(

)

, so tr

(⋆τ)

is well-defined. Moreover, using the Hodge decomposition

τ =

d

φ

τ

+

δβ

τ

+

α

τ

,

d

φ

τ

Bk

, δβ

τ

B∗k

, α

τ

Hk

,

we have tr

(⋆τ) =

tr

(⋆

d

φ

τ

)

, viz. the terms

δβ

τ and

α

τ do not contribute to the trace, hence we can take

τ =

d

φ

τ. Then,

Lk

τ =

d

δ

d

φ

τand

Lk

τ, α⟩ =

0 for all

α ∈

Hk. Similarly, for gN

H−1/2Λnk−1

(∂

)

, we can find

τ ∈

¯

H∗Λk

(

)

, with

⋆¯τ |

∂Ω

=

0

,

tr

(δ ⋆ ¯τ) =

gN

.

Takingu

¯

=

u

τ − ¯τ

, we obtain Lku

¯

=

f

Lk

τ −

Lk

τ =: ¯

¯

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

Lkis 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 function

β

.

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

(

)

,

the pressure p

Λ0

(

)

, and the velocity field u with a 1-form onΩ, viz., with an element ofΛ1

(

)

. 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

(

)

, the following relation between the interior product and Hodge star operator is valid, (see e.g. [16]),

iX

α = ⋆(

X

⋆α),

(45)

where Xis 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

(

)

, denoted by

ζ

,

ζ =

˜

ρ ∧

u

=

d

φ + δβ + α,

(49)

where

ρ = ⋆ρ

˜

. The choice of

ζ

will be motivated in the next section.

Definition 6. Using the Hodge decomposition(49), define the density weighted vorticity as

ω =

d

ζ

and the density weighted dilatation as

θ = −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) d

φ ∈

B◦1

, δβ ∈

B∗1

, α ∈

◦ H1.

(8)

Remark 3. If the classical vorticity is

ξ =

du then, the density weighted vorticity is

ω =

d

ζ =

d

(

ρ ∧

˜

u

) = 

d

⋆ρ ∧

u

+

⋆ρ ∧ ξ.

(50) When the flow is incompressible, then

ω =

√⋆ρ ∧ξ

, i.e., the density weighted vorticity. Using the equation for the velocity (47), the vorticity evolution equation for constant density is

ξ

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

and

ξ

satisfy the same evolution equations.

The density weighted dilatation

θ

for an incompressible flow is

θ = −δ 

⋆ρ ∧

u

 = −

⋆ρ ∧ δ

u

.

(53) The incompressibility constraint in differential forms is

δ

u

=

0, which implies

θ =

0.

Lemma 1. The potential function

φ

and vector stream function

β

in the Hodge decomposition(49)solve the following boundary value problems B-problem

d

δβ = ω,

d

β =

0 inΩ tr

(⋆β) =

0 on

,

B ∗ -problem

δ

d

φ = −θ

inΩ tr

(⋆

d

φ) =

0 on

,

(54) when

ζ ∈

HΛ1

(

) ∩

H∗Λ1

(

) ∩

H1⊥and B-problem

d

δβ = ω,

d

β =

0 inΩ tr

(δβ) =

0 on

,

B ∗ -problem

δ

d

φ = −θ

inΩ tr

(φ) =

0 on

,

(55) when

ζ ∈

HΛ1

(

) ∩

HΛ1

(

) ∩

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

(

) ∩

H∗Λ1

(

) ∩

H1⊥. Then, the Hodge decomposition (a) of

ζ

is used to define two new variables.

Since

ζ ∈

D

(

d

) =

HΛ1

(

)

, we can define the density weighted vorticity

ω ∈

B2

L2Λ2

(

)

as

ω =

d

ζ =

d

(

ρ ∧

˜

u

) =

d

(

d

φ + δβ) =

d

δβ,

(56) where

φ ∈

HΛ0

(

), β ∈

HΛ2

(

)

with tr

(⋆β) =

0 and

δβ ∈

D

(

d

) =

HΛ1

(

)

. Moreover, since

ω ∈

B2, it follows that

β ∈

B2hence, d

β =

0. Observe here, that(56)is the B problem(26)(27)with homogeneous natural boundary conditions.

Consider now

ζ ∈

D

(δ) =

H∗Λ1

(

)

. Define the density weighted dilatation

θ ∈

B∗0

L2Λ0

(

)

as

θ = −δζ = −δ(

˜

ρ ∧

u

) = −δ

d

φ − δδβ = −δ

d

φ,

(57) where

φ ∈

HΛ0

(

), β ∈

H◦∗Λ2

(

)

and d

φ ∈

HΛ1

(

)

with 0

=

tr

(⋆ζ ) =

tr

(⋆

d

φ)

. Observe that(57)with this boundary

condition is the B∗problem(28)(29)for

φ

, where

δφ =

0 is satisfied since HΛ−1is understood to be zero and tr

(⋆φ) =

0

since

⋆φ

is a 3-form.

Case 2. Let us choose now

ζ ∈

HΛ1

(

) ∩

HΛ1

(

) ∩

H◦1⊥. Then, we use the Hodge decomposition (b) of

ζ

to define two

new variables. Since

ζ ∈

HΛ1

(

),

tr

ζ =

0 and using the decomposition in (b), we obtain the weakly imposed essential boundary condition 0

=

tr

ζ =

tr

(δβ)

. Define the density weighted vorticity

ω ∈

B◦2as

ω =

d

ζ =

d

δβ,

(58)

where

β ∈

H∗Λ2

(

)

. This is the B problem(33)–(34)for

β

with homogeneous essential boundary conditions. Similarly, since

ζ ∈

HΛ1

(

)

, we can define the density weighted dilatation

θ ∈

B

0, as

θ = −δ

d

φ,

(59)

where

φ ∈

HΛ0

(

),

d

φ ∈

HΛ1

(

)

. The Hodge decomposition in (b) implies the strongly imposed boundary condition

tr

φ =

0. This is again the B∗problem(35)(36)for

φ

, with homogeneous essential boundary conditions. 

(9)

Remark 4. When the flow is incompressible, then 0

=

θ = −δζ = −δ

d

φ

. This implies that 0

= − ⟨

δ

d

φ, φ⟩ = − ⟨

d

φ,

d

φ⟩ +

∂Ω

tr

φ ∧

tr

(⋆

d

φ) = − ⟨

d

φ,

d

φ⟩ .

The boundary integral is zero in either of the two Hodge decompositions. Then, d

φ =

0 hence, the Hodge decomposition is

ζ = 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 solution

β ∈

B2, with the inhomogeneous boundary condition tr

(⋆β) =

gb

H−1/2Λ1

(∂

)

, by transforming it first to a homogeneous problem with modified right hand side

δ

d

β = ¯ω

, as

we discussed in Section2.3.4. The B∗problem for

φ

has a unique solution with the inhomogeneous boundary condition

tr

(⋆

d

φ) =

gN

H−1/2Λ2

(∂

)

. Hence, solving the B∗problem(57)with inhomogeneous boundary conditions, is equivalent

with solving the B∗homogeneous problem with modified right hand side

δ

d

φ = −¯θ

, with

θ = θ+¯τ

¯

, as seen in Section2.3.4.

In Case 2, the inhomogeneous essential boundary conditions for

φ

and

β

are tr

φ =

rb

H1/2Λ0

(∂

)

and tr

(δβ) =

rN

H1/2Λ1

(∂

)

, 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 variables

ρ, ω, θ

and s as

ρ

t

= −

d

(

˜

ρ ∧ ⋆ζ ),

(60)

ω

t

=

d

ζ

t

,

(61)

θ

t

= −

δζ

t

,

(62) st

=

1

˜

ρ

ζ ∧ δ

s

,

(63) with

ζ

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 dilatation

θ

. In order to have them well defined, we choose the proper spaces for

ζ

. The potential function

φ

and vector stream function

β

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 formulation

In 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 is

simply 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

(

) ×

Λq

(

) →

Λn

(

),

(64)

whereΩis an n-dimensional manifold, resulting in the total energy H

[

α

p

, α

q

] =

H

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

(

)

, and

α

q

, ∂α

q

Λq

(

)

. Then under weak smoothness assumptions on H,

H

[

α

p

+

∂α

p

, α

q

+

∂α

q

] =

H

p

+

∂α

p

, α

q

+

∂α

q

)

=

H

p

, α

q

) +

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

(

))

∗and

δ

qH

(

Λq

(

))

∗, which can be regarded as the

variational derivatives ofH with respect to

α

p and

α

q, respectively. The dual linear space

(

Λp

(

))

∗ can be naturally

(10)

For 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

] =

1 2

u

,

u

v

ρ +

U

( ˜ρ, ˜

s

,

(67)

where

˜

s

=

s. The Hamiltonian functional can further be written as

H

[

ρ,

u

,

s

] =

1 2

(˜ρ ∧

u

)

, (

˜

ρ ∧

u

)

v

v

+

U

( ˜ρ, ˜

s

=

1 2

⟨⟨

˜

ρ ∧

u

,

ρ ∧

˜

u

⟩⟩

v

+

U

( ˜ρ, ˜

s

.

(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 of

ζ

.

Remark 6. We have seen that the inhomogeneous B

problem for

φ

and the inhomogeneous B problem for

β

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 variables

θ

¯

and

ω

¯

, with the corresponding homogeneous boundary conditions for the

φ

and

β

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

] =

1 2

d

φ ∧ ⋆φ +

d

δβ ∧ ⋆β) +

U

( ˜ρ, ˜

s

.

(69)

The variational derivatives of the Hamiltonian are

δ

H

δρ

=

∂ ˜ρ

( ˜ρ

U

( ˜ρ, ˜

s

)),

δ

H

δφ

= −

⋆ ¯θ,

δ

H

δβ

=

⋆ ¯ω,

δ

H

δ

s

=

U

( ˜ρ, ˜

s

)

∂˜

s

ρ.

˜

(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 introduced

H

[

ρ, φ, β,

s

] =

1 2

⟨⟨

d

φ + δβ,

d

φ + δβ⟩⟩v

+

U

( ˜ρ, ˜

s

=

1 2

d

φ + δβ,

d

φ + δβ⟩ +

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

d

φ + δβ,

d

φ + δβ⟩ = ⟨

d

φ,

d

φ⟩ + ⟨

d

φ, δβ⟩ + ⟨δβ, δβ⟩ + ⟨δβ,

d

φ⟩

= ⟨

φ, δ

d

φ⟩ + ⟨β,

d

δβ⟩ = φ, −¯θ + ⟨β, ¯ω⟩ ,

where

d

φ, δβ⟩ =

0 and

δβ,

d

φ⟩ =

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

] =

1

2

(⟨

d

φ,

d

φ⟩ + ⟨δβ, δβ⟩) +

U

( ˜ρ, ˜

s

)ρ,

(72)

where the inner products are in the space L2Λ1

(

)

. Let us calculate

δ

φHfirst. For

φ, ∂φ ∈

D

(

L0

)

and

β ∈

D

(

L2

)

, with

either essential or natural homogeneous boundary conditions on

φ

and

∂φ

, we have H

[

ρ, φ + ∂φ, β,

s

] =

H

(ρ, φ, β,

s

) + ⟨

d

φ,

d

(∂φ)⟩ + {

h.o.t. in

∂φ}

=

H

(ρ, φ, β,

s

) + ⟨δ

d

φ, ∂φ⟩ + {

h.o.t. in

∂φ}.

(11)

Therefore,

δ

H

δφ

=

⋆δ

d

φ = −

d

(⋆

d

φ) = − ⋆ ¯θ.

(73)

Similarly, let us calculate the variational derivative

δ

βH. For

β, ∂β ∈

D

(

L2

)

, we have for either boundary conditions,

H

[

ρ, φ, β + ∂β,

s

] =

H

(ρ, φ, β,

s

) + ⟨∂β,

d

δβ⟩ + {

h. o. t. in

∂β}.

(74) Hence, we obtain that

δ

H

δβ

=

d

δβ = ⋆ ¯ω,

(75)

which completes the proof of this lemma. The variational derivatives of the Hamiltonian with respect to the variables

ρ

and

s, 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 with

respect 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

] =

1 2

(θ ∧ ⋆

K0

θ + ω ∧ ⋆

K2

ω) +

U

( ˜ρ, ˜

s

,

(77)

where Kkis the solution operator of the Hodge Laplacian operator Lk

,

k

=

0

,

2. The variational derivatives of the Hamiltonian

functional are:

δ

H

δρ

=

∂ ˜ρ

( ˜ρ

U

( ˜ρ, ˜

s

)),

δ

H

δω

=

K2

ω = ⋆β,

(78)

δ

H

δθ

=

K0

θ = − ⋆ φ,

δ

H

δ

s

=

U

( ˜ρ, ˜

s

)

∂˜

s

ρ.

˜

(79)

Proof. UsingLemma 2, and that

θ

and

ω

are in the domain of the solution operators K0 and K2of the Hodge Laplacian

problems for

φ

and

β

, respectively, the Hamiltonian can be written as H

[

ρ, θ, ω,

s

] =

1 2

(θ ∧ ⋆

K0

θ + ω ∧ ⋆

K2

ω) +

U

( ˜ρ, ˜

s

.

(80)

Let

θ, ∂θ ∈

B∗0and

ω, ∂ω ∈

B2. The variational derivatives of the Hamiltonian with respect to

θ

and

ω

can be obtained from H

[

ρ, θ + ∂θ, ω + ∂ω,

s

] =

1 2

(θ + ∂θ) ∧ ⋆

K0

(θ + ∂θ) +

U

( ˜ρ, ˜

s

)ρ +

1 2

(ω + ∂ω) ∧ ⋆

K2

(ω + ∂ω)

=

H

(ρ, θ, ω,

s

) +

1 2

(θ ∧ ⋆

K0

(∂θ) + ∂θ ∧ ⋆

K0

θ)

+

1 2

(ω ∧ ⋆

K2

(∂ω) + ∂ω ∧ ⋆

K2

ω) + {

h.o.t. in

∂θ, ∂ω}.

(81)

Here

∂θ

and

∂ω

denote the variation of

θ

and

ω

, respectively, to avoid confusion with the coderivative operator

δ

. In order to further investigate the last two integrals in(81), we useLemma 2, where the variational derivatives of the Hamiltonian with respect to

φ

and

β

are given, then apply the variational chain rule to obtain

δ

θHand

δ

ωH.

(12)

To obtain the variational derivative of the Hamiltonian with respect to

θ

, apply the variational chain rule as follows

δ

H

δφ

∂φ =

δ

H

δθ

∂θ = −

δ

H

δθ

δ

d

(∂φ) = −

δ

δθ

H

, δ

d

(∂φ)

= −

Ω d

δ

δ

H

δθ

∂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 of

φ

and

θ

, to avoid confusion with the coderivative operator

δ

. Analogously, to obtain the variational derivative of the Hamiltonian with respect to

ω

, consider the variational chain rule

δ

H

δβ

∂β =

δ

H

δω

∂ω =

δ

H

δω

d

δ(∂β) =

δ

δω

H

,

d

δ(∂β)

=

δ

d

δ

H

δω

∂β +

∂Ω tr

(δ(∂β)) ∧

tr

δ

H

δω

∂Ω tr

(⋆∂β) ∧

tr

d

δ

H

δω

.

(83)

We now have two choices.

First, when Case 1 applies inLemma 1, then tr

(⋆

d

(∂φ)) =

0, hence the variational equation(82)becomes

δ

H

δφ

∂φ = −

Ω d

δ

δ

H

δθ

∂φ +

∂Ω tr

(∂φ) ∧

tr

δ

δ

δθ

H

,

(84)

∂φ ∈

HΛ0

(

)

, with tr

(⋆

d

(∂φ)) =

0. Choose

∂φ

such that the boundary integral in(84)is zero. We thus obtain thatδH

δθ

solves the differential equation d

δ

δ

H

δθ

= −

δ

H

δφ

=

⋆θ

inΩ

.

(85)

Consider now the variation

∂φ ∈

D

(

L0

)

arbitrary. Inserting(85)into the variational equation(84), we obtain that

tr

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

H

δθ

= −

⋆ φ +

h

,

(87)

with h

Z∗3, the null space of

δ

. The B problem forδδθH has however, a unique solution δδθH

B◦3, hence the side condition

δH

δθ

Z∗3is satisfied. Consequently,

δ

H

δθ

= −

⋆ 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 condition

δ

d

δ

H

δω

=

δ

H

δβ

inΩ

,

tr

δ

H

δω

=

0 on

.

(89)

Combined with(75), we obtain the following equation

δ

d

δ

H

δω

=

d

δβ

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

H

δω

=

⋆β,

(91)

(13)

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

H

δθ

=

⋆θ

inΩ

,

tr

δ

H

δθ

=

0 on

,

(92)

and the B∗problem with natural boundary condition forδδωH

δ

d

δ

H

δω

=

⋆ω

inΩ

,

tr

d

δ

H

δω

=

0 on

.

(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

and

β

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

} = −

δ

F

δρ

d

δ

G

δ

u

δ

G

δρ

d

δ

F

δ

u



T1

+

iX

δ

δ

G u

δ

F

δ

u



T2

+

1

˜

ρ

d

˜

s

δ

F

δ

s

δ

G

δ

u

δ

G

δ

s

δ

F

δ

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 variables

ρ, θ, ω,

s and to properly account for the boundary conditions. We derive the bracket in a well-chosen functional space

using 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 that

ζ ∈

H◦Λ1

(

) ∩

H∗Λ1

(

) ∩

H◦1⊥

,

and tr

δ

δθ

F

=

0

,

tr

δ

δθ

G

=

0 (95) or

ζ ∈

HΛ1

(

) ∩

H∗Λ1

(

) ∩

H1⊥

,

and tr

δ

F

δω

=

0

,

tr

δ

G

δω

=

0

.

(96)

Then, the bracket(94)in terms of the variables

ρ, θ, ω,

s has the form

{

F

,

G

} = −

δ

F

δρ

d



˜

ρ ∧ α(

G

)

δ

G

δρ

d



˜

ρ ∧ α(

F

)

+

ζ

2

ρ

˜

α(

F

)

d



ρ ∧ α(

˜

G

)

ζ

2

ρ

˜

α(

G

)

d



ρ ∧ α(

˜

F

)

+

d

˜

s

˜

ρ

δ

F

δ

s

α(

G

) −

δ

G

δ

s

α(

F

)

,

(97)

where

α(·)

is the following operator on functionals

α(·) =

d

δ ·

δω

+

δ

δ ·

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