Mathematical Continuum Mechanics - WebHomegw/alt-continuum.pdf · 5 Introduction Die Naturwissenschaft beschreibt und erkla¨rt die Natur nicht einfach, sie ist Teil des Wechselspiels

Technical University Munich 2018

Lectures on

MathematicalContinuum Mechanics

Prof. Dr. H.W. Alt

Version: 20180409 Last major change: 17.12.2017

Copyright 2011-2018 Prof. Dr. H.W. Alt

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This is the english version of the script, so far it is only partly translated.The script will be further developed parallel to the lecture. This version ispreliminary, it is subject to corrections.

To my parents

Contents

I Mass and momentum 81 Conservation laws . . . . . . . . . . . . . . . . . . . . . . . . 112 Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 Conservation of momentum . . . . . . . . . . . . . . . . . . . 444 Interfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 705 Change of coordinates . . . . . . . . . . . . . . . . . . . . . . 826 Reference coordinates . . . . . . . . . . . . . . . . . . . . . . 967 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

II Objectivity 1121 Classical observers transformations . . . . . . . . . . . . . . . 1152 Lorentz transformations . . . . . . . . . . . . . . . . . . . . . 1203 Objectivity of balance laws . . . . . . . . . . . . . . . . . . . 1324 Constitutive relations . . . . . . . . . . . . . . . . . . . . . . 1495 Objectivity in reference coordinates . . . . . . . . . . . . . . . 1656 Angular momentum . . . . . . . . . . . . . . . . . . . . . . . 1697 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

IIIEnergy and entropy 1821 Entropy inequality . . . . . . . . . . . . . . . . . . . . . . . . 1872 Energy equation . . . . . . . . . . . . . . . . . . . . . . . . . 1963 Distributional entropy . . . . . . . . . . . . . . . . . . . . . . 2044 Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2065 Lagrange multipliers . . . . . . . . . . . . . . . . . . . . . . . 2126 Dissipation inequality . . . . . . . . . . . . . . . . . . . . . . 2167 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220

IVVarious applications 2211 Tidal period . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2212 Fluids and gases . . . . . . . . . . . . . . . . . . . . . . . . . 2343 Navier-Stokes equation . . . . . . . . . . . . . . . . . . . . . . 2474 Eulers equation . . . . . . . . . . . . . . . . . . . . . . . . . 2585 Nonlinear elasticity . . . . . . . . . . . . . . . . . . . . . . . . 279

3

4

6 Tissue growth . . . . . . . . . . . . . . . . . . . . . . . . . . . 2877 Sound waves . . . . . . . . . . . . . . . . . . . . . . . . . . . 2928 vr-Vortices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3119 Fractionation . . . . . . . . . . . . . . . . . . . . . . . . . . . 32910 Reactive substances . . . . . . . . . . . . . . . . . . . . . . . 35011 Reaction-diffusion systems . . . . . . . . . . . . . . . . . . . . 35312 Combustion (Temperature dependent diffusion) . . . . . . . . 37313 Reactions in biology . . . . . . . . . . . . . . . . . . . . . . . 38714 Chemical reactions . . . . . . . . . . . . . . . . . . . . . . . . 39415 Prandtls boundary layer . . . . . . . . . . . . . . . . . . . . . 40716 Self-gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . 41417 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433

V Higher moments 4341 Cattaneos 8-Momente Gleichung . . . . . . . . . . . . . . . . 4352 Boltzmann Gleichung . . . . . . . . . . . . . . . . . . . . . . 4423 Die Chapman-Enskog Hierarchie . . . . . . . . . . . . . . . . 4504 Grads 13-Momente Gleichung . . . . . . . . . . . . . . . . . . 459

VI Speed of light 4641 Elektrodynamik . . . . . . . . . . . . . . . . . . . . . . . . . . 4662 Ubungen . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475

author: H.W. Alt title: Continuum Mechanics time: 2018 Apr 9

5

Introduction

Die Naturwissenschaft beschreibt und erklartdie Natur nicht einfach, sie ist Teil des Wechselspielszwischen der Natur und uns selbst.Werner Heisenberg (1901-1976)

Die mathematische Modellierung physikalischer Phanomene fuhrt zu Er-haltungsgleichungen, die von allen Beobachtern gleich formuliert werdenmussen. Daher stellen wir in der Vorlesung folgende Prinzipien auf:

die Formulierung mit Erhaltungssatzen,

die Objektivitat bei Beobachtertransformationen,

das Entropieprinzip bzw. die freie Energieungleichung,

wobei das letzte Prinzip ausdruckt, dass wir es mit irreversiblen Prozessenzu tun haben. Diese Prinzipien haben Auswirkungen auf die Behandlungphysikalischer Effekte, sie haben Konsequenzen was die mathematische Ex-istenztheorie betrifft, als auch fur die Entwicklung von numerischen Algo-rithmen. Es soll in dieser Vorlesung dargestellt werden, wie diese Prinzipienin Standardsituationen aussehen und welche Konsequenzen zu ziehen sind.Die abstrakten Formulierungen als partielle Differentialgleichung werden soin Zusammenhang mit alltaglichen Gleichungen gebracht. Die Idee zu dieserVorlesung ist aus meiner Veroffentlichung [18] entstanden und ich hoffe sehr,dass dieses Skript dazu beitragt zu verstehen, wie die physikalische Theorieauf ein einfaches System von Axiomen aufgebaut ist.

Es sei bemerkt, dass die allgemeinen Prinzipien in einem strengen Sinne zuverstehen sind, obwohl das im Text nicht immer so zum Ausdruck kommt.Das gilt in Standardsituationen als auch bei speziellen Theorien, sie sind all-gemeine physikalische Prinzipien. Dies bestimmt im wesentlichen den Auf-bau des Skriptes. Im ersten Abschnitt werden Erhaltungssatze vorgestellt,und zwar geben wir diese in der ublichen Differentialschreibweise an. EineFormulierung mit Hilfe von Testvolumina wird als Einfuhrung in das Kapi-tel I angegeben. Da viele physikalische Vorgange nichtklassische Losungenbeinhalten, wird danach, also moglichst fruh, der Begriff der Distribution


6

eingefuhrt. Nichtklassische Losungen sind etwa bei der Selbstgravitationund bei der Temperaturmessung der Standardfall. Es werden in dieser Vor-lesung jedoch nur solche Beweise uber Distributionslosungen gebracht, beidenen keine Groen auf der Flache auftreten, obwohl dies haufig der Fallware. Das heit, der Gausche Satz im Raum ist hinreichend fur die Be-weise, bei denen die Flachen von der Zeit nicht abhangen. Das Kapitel Ienthalt auch die Darstellung der Erhaltungssatze in Lagrange Koordinaten.Dazu wird eine allgemeine Transformationsformel bewiesen, die auch spaterbei der Beobachterunabhangigkeit sowohl im klassischen Newtonschen Fall,als auch bei den Lorentztransformationen benutzt wird. Damit sind indiesem Kapitel I alle mathematischen Hilfsmittel zusammengestellt.

Das Kapitel II enthalt alle Aussagen uber die Objektivitat, wobei bei diesemBegriff gemeint ist, dass physikalische Aussagen unabhangig vom Beobachtergetroffen werden mussen. Dies ist notwendig, da sonst eine Kommunikationzwischen beteiligten Wissenschaftlern unnotig verkompliziert wird, bzw. einephsikalische Beschreibung in Buchern bzw. elektronisch unmoglich wird.Groe Teile dieses Skripts basieren auf klassischen Newton Transformatio-nen, die in Abschnitt II.1 behandelt werden. Um die Abhangigkeit derTheorie von den Transformationen zu verdeutlichen, geben wir in diesemKapitel auch Lorentz Transformationen an, die allerdings erst im KapitelVI benotigt werden.

Das nachste Kapitel III handelt von der Energie und Entropie. Es ist einesder herausragenden Ergebnisse des 19. und 20. Jahrhunderts, die Irre-versibilitat von Prozessen mit einem Anstieg der Entropiedichte und desEntropieflusses in Verbindung zu setzen. Dabei wird hier der Standpunktvertreten, dass diese Groen an sich von vornherein unbekannt sind. Erstdurch die Anwendung des Prinzips wird deutlich, welche Bedingungen dasEntropieprinzip an die konstitutiven Funktionen stellt. Die Aufgabe bestehtalso darin, das Entropieprinzip mit zu berucksichtigen und so zu einemtragfahigen Modell zu kommen.

Das ist nun Aufgabe des Kapitels IV, in dem aus den verschiedensten Bere-ichen Modellgleichungen dargestellt werden, und zwar unter Benutzung desEntropieprinzips bzw. der Energieungleichung. Es wird klar, dass alle in denBeispielgleichungen gemachten Ungleichungen auf dieses Prinzip zuruckzu-fuhren sind.


7

Hinweise fur die LehrendenDie Anwendung der Distributionstheorie ist wesentlich fur dieses Skript, undwird gleich im zweiten Paragraphen eingefuhrt, wobei es zur Darstellung derPunktmechanik gebraucht wird. In den weiteren Kapiteln werden sie aufeindimensionalen Kurven und zweidimensionalen Flachen im Rn angewandt.Es wird, bei gleicher Definition, auch zwischen Distributionen in D(Rn) undDistributionen in D(Rn Rn) unterschieden, in dieser Vorlesung ist jedeDimension vertreten.

Die erste Vorlesung wurde ein Semester im Umfang von 4Std/Woche gehal-ten (im Wintersemester 2011). Dies umfasste die grundlegenden Kapitel I-III, und insgesamt funf Abschnitte aus Kapitel IV. Die Wahl der Abschnittekann nach der besonderen Situation der Universitat oder nach den speziellenWunschen des Lehrenden gewahlt werden.

Im Skript wurden oft mehrere Beweise gegeben, obwohl in der Vorlesungjeweils nur ein Beweis dargestellt wurde. Zum Teil sind auch Beweiseaufgeschrieben, die in der Vorlesung garnicht gebracht wurden. Dies istbei der Auswahl des Stoffes zu berucksichtigen.

Der Text ist z.Z. noch im Entwicklungsstadium und wird standig verbessertund erweitert. Die vorhandenen Paragraphen werden aber mit Sicherheitbleiben.


I Mass and momentum

The equations of continuum physics are based on systems of conservationlaws. In this chapter we focus on the simplest such system, namely theconservation of mass and momentum. Mathematically, we will introduceconservation laws and distributions. These are the main tools of this chapter.

Fig. 1: Gas and solid

For engineers, the conservation laws are introduced with the help of testvolumes V Rn. One writes the change of a physical quantity, whosedensity is u, as 1

d

dt

V

u(t, x) dx =

V

q(t, x)V (x) dHn1 +

V

r(t, x) dx .

Here q is the flux across the boundary of the test volume and r is the rateat which the quantity u in the volume is changed. The fact that no otherterms occur, is the characteristics of continuum physics. Another equivalentformulation of conservation laws is the version as differential equation forC1-functions

tu+ divq = r . (I0.1)

1Wir verwenden die Bezeichnung Hm fur das m-dimensionale Hausdorffma in jedemR

n mit n m und Ln fur das n-dimensionale Lebesguema im Rn.

8

I. Mass and momentum 9

This follows from the formulation for test volumes using the Gausss theo-rem, as one can see from the following calculation:

V

tu(t, x) dx =d

dt

V

u(t, x) dx

=

V

q(t, x)V (x) dHn1(x) +

V

r(t, x) dx

=

V

( divq(t, x) + r(t, x)) dx ,

consequently,

V

(tu(t, x) + divq(t, x) r(t, x)) dx = 0 .

Since the test domain V is arbitrary, we obtain the differential equation(I0.1). It should be mentioned that the formulation with test volumes followsfrom the strong differential equation just by reversing the above conclusions.

Es hat seinen besonderen Grund, dass in der Kontinuumsphysik die For-mulierung mit Differentialgleichungen gewahlt wird, und es beruhrt uber-haupt nicht die Struktur der Materie im Kleinen. So ist in Fig. 1 auf derlinken Seite dargestellt, wie sich die Atome irregular bewegen, so dass mannicht mehr wei, ob und wie die Atome im Moment zuvor angeordnet waren.Wahrenddessen ist auf der rechten Seite die Situation in einem Festkorperdargestellt. Hier bewegen sich die Atome nach denselben Gesetzen, abersie bleiben fast immer in derselben Anordnung. Das liegt daran, dass dieauf die Atome wirkenden Krafte ihr Vorzeichen andern, bevor sie selbst ihreOrdnung zu verlieren drohen. Also kommen wir zu dem folgenden Schluss:Wir mussen (t, x) als einen Punkt interpretieren, der viele Atome mitsamtihren lokalen Gesetzen beinhaltet, und die makroskopischen Erhaltungsgle-ichungen sind zu verstehen als eine Methode, diese lokalen Gesetze von Ortzu Ort zu vermitteln. Nichtsdestotrotz geben diese makroskopischen Gle-ichungen das Verhalten der Materie in der Natur wieder, wir werden dies beider Massen- und Impulserhaltung im einzelnen sehen. Die spater eingefuhrteTemperatur ist dann wie eine Verschlusselung der lokalen Bewegung derAtome.

However, many important functions are not classical solutions of the dif-ferential equation, for example, the earths gravitational field at the earthssurface. In this case, the formulation with test volumes becomes more com-plex. Therefore, we use test functions instead of test volumes. On the spaceof test functions

D(U) := { C(U) ; has compact support in U} ,


I. Mass and momentum 10

Fig. 2: Relevance of distributions (from [66])

where U R Rn is an open set, we consider linear forms U , Q, and R inthe dual space D (U) (see the section 2) so that

tU + divQ = R in D(U) , (I0.2)

where D (U) is called the space of distributions. This formulation insteadof (I0.1), see (I2.3), has the advantage that it is more general and mucheasier. This becomes particularly clear when one goes to descriptions ofconservation laws on surfaces. Both representations are very common inliterature, the representation of conservation laws using test volumes can befound usually in physics books. Both formulations are equivalent as one cansee if one replaces the characteristic functions XV (in the formulation withtest volumes) by smooth functions (in the formulation with test functions),which can be made rigorous by an approximation argument, that is, by aconvolution of the characteristic function.


I.1 Conservation laws 11

1 Conservation laws

We consider scalar conservation laws of the following form:

Conservation law:

tu+ div q = r

u physical quantity,

q associated flux,

r source term.

(I1.1)

So we have real-valued functions u, r, and qi for i = 1, . . . , n. Here n isthe space dimension. In physical reality this is 3, but it may also be 1 and2, when the quantities do not depend of the other space filling coordinates.Mathematically, n can be arbitrary. The functions depend on the time t Rand from the location x Rn, that is, (t, x) U R Rn and U is theconsidered region. This definition of a conservation law is only defined, if uand q are differentiable and r is continuous. For a continuously differentiablevector field q :R Rn Rn we write

q = (qi)i=1,...,n = (q1, . . . , qn) =

q1...qn

,

so we identify vectors with column matrices.

1.1 Remark. Die Erhaltungsgleichung tu + divq = r in t und x kannauch aufgefasst werden als Divergenzgleichung div(u, q) = r in (t, x), wobeidiv := (t, div).

For derivatives we have the following definitions. Note: We do not specifythe exact mathematical assumptions as differentiability or partial differen-tiability.

1.2 Definition of derivatives. For a function g :RN RM and a vectore RN the directional derivative in direction e is given by

eg(y) = limh0

1

h(g(y + he) g(y)) RM .

Important: The same definition holds if e is replaced by a map y 7 e(y),that is, the directional derivative depends on the variable.All other derivatives are based on this definition.

(1) If RN = R Rn, hence N = n+ 1, we write for the variables y = (t, x)and for e = (0, e) R Rn

eg(t, x) = (0,e)g(t, x) (as mapping on RN = R Rn)

= eg(t, x) (as mapping g(t, ) :Rn R) .



(2) On Rn we define for i = 1, . . . , n

ei := (0, . . . , 0, 1, 0, . . . , 0) Rn with a 1 on the i-th position. (I1.2)

Then {e1, . . . , en} is the standard orthonormal basis of Rn.

(3) The following formulas hold for g :R Rn R:

ig(t, x) := xig(t, x) :=

(0,ei)g(t, x) as mapping on R Rn,eig(t, x) as mapping g(t, ) :R

n R,

limh0

1

h(g(t, x+ hei) g(t, x)) ,

tg(t, x) := (1,0)g(t, x) = limh0

1

h(g(t+ h, x) g(t, x)) .

(4) For a mapping g :R Rn R the gradient of g is given by

g = (xig)i=1,...,n =

1g...ng

,

that is, (t, x) 7 g(t, x) Rn is a vector field. Important: The notation as well as the following notation involves only the space variables.

(5) The (space) derivative of a vector field q = (q1, . . . , qn) is

Dq = (xiqk)k,i=1,...,n =

1q1 nq1...

...1qn nqn

.

Remark: In literature sometimes the gradient q of the vector field q isused, and we define it as 2

q = (xiqk)i,k=1,...,n = (Dq)T =

1q1 1qn...

...nq1 nqn

.

For n = 1 this is in accordance with the gradient of a function.

(6) The divergence of q is given by the trace of Dq

div q :=n

i=1xiqi = traceDq .

2Throughout this book we use the following notation for matrices M : The transposedmatrix is MT, the symmetric part is MS = 1

2(M +MT), and the antisymmetric part or

skew symmetric part is MA = 12(M MT).



(7) Fur ein Vektorfeld q und eine Richtung e :Rn Rn gilt 3

eq = (e)q = Dq e fur alle e Rn.

Remark: It is div q = q where :=jejj in the world of the nablaoperator.

Please, keep these definitions in mind, we use them systematically in thisscript. Certain identities for derivatives can be found in exercise 7.2.

1.3 Representation of the divergence operator. For a differentiablevector field q and orthonormal bases {e1(t, x), . . . , en(t, x)} of the Euclideanspace Rn it holds

div q =n

i=1xiqi =

ni=1

eieiq . (I1.3)

Here the basis vectors can depend arbitrarily on (t, x).

It should be noted that in general

div q 6=n

i=1ei(eiq) =

ni=1

eieiq = div q

+( n

i=1eiei

i.A. 6= 0

)q ,

if ei are variable vectors. Remember that for the divergence operator prop-erty (I1.3) is true. This fact includes the isotropy of the empty space.

Proof. The orthonormality of {e1, . . . , en} means that

eiej = ij fur i, j = 1, . . . , n.

With 4

ei = (eik)k=1,...,n therefore eik = eiekthe orthonormality is

nk=1

eikejk = ij

orEET = Id if E = (eik)i,k=1,...,n . (I1.4)

Das besagt, dassET die Rechtsinverse von E ist, was aber gleich der Linksin-versen ist, eine Aussage fur endliche Matrizen, denn

(ETE Id)ET = ET (EET Id) = 0 ,3The Euclidean scalar product is given by xy := ni=1xiyi for x, y Rn. In analogy

we define the scalar product for matrices by RS := ni,j=1RijSij for R,S Rnn.Here Rnn stands for the set of real n n-matrices.

4for ek see (I1.2)



und da ET injektiv ist (folgt aus (I1.4)), somit surjektiv ist, schlieen wirETE Id = 0, also

ETE = Id

und damit

kl = (ETE)kl =

ni=1

eikeil .

Dann ist wegen 1.2(7)

ni=1

eieiq =n

i=1ei(Dq)ei

=n

k,l=1

ek(Dq)el = lqk

ni=1

( eiek = eik

)( eiel= eil

)

=n

k,l=1

lqkn

i=1eikeil =

nk,l=1

lqkkl =n

k=1

kqk = div q .

Now we give some examples for q in order to calculate divq.

1.4 Example. Let a matrix (t, r) 7 A(t, r) Rnn be given and

q(t, x) := A(t, |x|)x.

(1) If A depends only on time t, then

div q = traceA.

(2) For continously differentiable A we compute for x Rn \ {0}

div q(t, x) = traceA(t, |x|) +1

|x|

(xrA(t, |x|)

)x.

(3) Let n = 2 and a is continuous differentiable. If

A(t, r) = a(t, r)

[0 11 0

],

thenq(t, x) = a(t, |x|) ix with divq = 0.

Proof. Siehe die Ubung 7.6.

1.5 Plane polar coordinates. Let n = 2 and for x R2 \ {0} let

er(x) :=x

|x| , e(x) :=ix

|x| =(x2, x1)|x| . (I1.5)

Then {er(x), e(x)} is an orthonormal system of R2 and

erer(x) = 0 , eer(x) =1

|x|e(x) ,

ere(x) = 0 , ee(x) = 1

|x|er(x) .(I1.6)



0

er(x)e(x)

x

Fig. 3: The orthonormal system {er(x), e(x)} for x R2 \ {0}

Proof. On R2 polar coordinates are given by

x = (r, ) = rei

and thener = ei, e = iei .

It holds for functions g

(erg) = r(g) , (eg) =1

r(g) ,

due to

((erg))(r, ) = limh0

1

h(g(rei + hei) g(rei))

= limh0

1

h(g((r + h)ei) g(rei)) = r(g) ,

((eg))(r, ) = limh0

1

h(g(rei + hiei) g(rei))

= limh0

1

h(g(r (1 +

h

ri)

= eihr +O(h2)

ei) g(rei))

= limh0

1

rh(g(rei(+h) +O(h2)) g(rei)) = 1

r(g) .

Then

(erer) = r(er) = r(ei) = 0 ,

(eer) =1

r(er) =

1

r(e

i) =1

riei =

1

re ,

(ere) = r(e) = r(iei) = 0 ,

(ee) =1

r(e) =

1

r(ie

i) = 1rei = 1

rer .



We use this in order to calculate the divergence of a vector field q :R R2 R2.

1.6 Plain divergence. Each vector field q :R R2 R2 has a unique representation inR (R2 \ {0}):

q = s1er + s2e , where s1 = qer , s2 = qe ,

with s1, s2 :R (R2 \ {0}) R. Here the orthonormal system is chosen as in 1.5. Then

in R (R2 \ {0})

div q(t, x) = ers1(t, x) +s1(t, x)

|x|+ es2(t, x) .

If the vector field q is directed outward seen from the origin, then s1 0 and s2 = 0. Ifthe vector field q rotates around the origin, then s1 = 0 and s2 is arbitrary.

Proof 1.Version. Es ist nach 1.3

div q = ererq + eeq= erer (s1er + s2e) + ee (s1er + s2e)= ers1 + es2

+s1(ererer + eeer) + s2(erere + eee) .

Unter Benutzung der Regeln (I1.6) folgt mir r = |x|, dass dies

= ers1 + es2 +1

rs1 ,

also folgt die Behauptung.

Proof 2.Version. Es ist nach 1.3

div q = ererq + eeq= er (erq) + e (eq) (erer + ee)q= ers1 + es2 (erer + ee)q .

Unter Benutzung der Regeln (I1.6) folgt, dass dies

= ers1 + es2 +1

rerq

= ers1 + es2 +1

rs1 ,

also folgt die Behauptung.

The most famous example of a conservation law is the mass conservation,that is, we write u = , where > 0 is the mass density, which is themass per volume, and we set q = v + J, where v denotes the velocity ofthe mass and J the mass diffusion (we will derive this equation in detail



in II.3.4). Hence we get the 5

General mass conservation:

t+ divx(v + J) = r

> 0 mass density,

q = v + J mass flux,

v = (vi)i=1,...,n velocity,

J = (Ji)i=1,...,n mass diffusion,

r source term of the mass,

(I1.7)

what we can also write as

t+ div (v) transport

= r div J change of mass

.

The J-term has a twofold meaning. It can be written as divJ on the right-hand side of the equation, then it is an external term, or it can be writtenas J as part of the flux, then it is an internal term (for more informationon J specially in systems see section IV.14).

If one considers a system of Gases, that is, if one is confronted with a totalmass, which is the mixture of several constituents, an example is given inFig. 4, one has

=Mk=1

k , (I1.8)

where k are the M single masses and is the total mass.

1.7 Theorem. Let masses k as in (I1.8) be given satifying the generalpartial mass equation

tk + divx(kv + Jk) = rk fur k = 1, . . . ,M . (I1.9)

We introduce the concentration of the component k by

ck :=k

hence k = ck and > 0 .

Then if

J :=

kJk = 0 , r :=

krk = 0 ,

5 We write divx instead of div in order focus on the variables (t, x).



Fig. 4: From [Wikipedia: Atmosphere of earth] [103]

the system (I1.9) is equivalent to

t+ divx(v) = 0 ,

(tck + vck

)+ divxJk = rk for k = 1, . . . ,M .

(I1.10)

Attention: Since c1 + c2 + + cM = 1 the last M equations are linearlydependent.

Proof. Taking the sum of (I1.9) we get

t+ divx(v +

k

Jk

= 0

)=k

rk

= 0

,


https://en.wikipedia.org/wiki/Atmosphere_of_earth


hence t+ divx(v) = 0. The single equations of (I1.9) then become

rk divxJk = tk + divx(kv) = t(ck) + divx(ckv)

= ck(t+ divx(v)) + (tck +

ni=1

vixick)

= (tck + vck

).

The mass conservation of the total mass is usually valid without the J and rterms. Thus we assume that r = 0 and J = 0. Then the often used equationreads

Conservation of mass:

t+ divx(v) = 0

> 0 mass density,

q = v mass flux,

v = (vi)i=1,...,n velocity.

(I1.11)

We consider now this differential equation.

1.8 Relativity of velocity. Assume (, v) satisfies the mass conservation(I1.11). We move the mass density with a constant velocity v0 Rn, thatis, we define

(t, x) := (t, x+ tv0) .

Is there a v such that for (, v) the equation (I1.11) is satisfied? Yes, for

v(t, x) = v(t, x+ tv0) v0 .

Remark: This is the Doppler effect for constant v0. We will study thisphenomenon in detail in section II.3.

So (, v) and (, v) fulfill the same equation, thus, solutions of (I1.11)correspond to each other. The proof shows that this follows from a changeof coordinates.

Proof. We ask, what the property

t + divx(

v) = 0 (I1.12)

for v means. We consider the transformation

Y

([tx

])=

[t

x+ tv0

].



Then Y = and for any vector field qdivx(qY ) = ( divxq)Y , (I1.13)

hence we compute

t = t(Y ) = (t)Y + v0()Y

= (divx(v)

)Y + v0()Y (since t+ divx(v) = 0)

= (divx(v)

)Y +

(divx(v0)

)Y (since v0 is constant)

= (divx

((v v0)

))Y = divx

(((v v0))Y

)(nach (I1.13))

= divx(((v v0)Y )

).

Therefore the differential equation (I1.12) is satisfied, if

v = (v v0)Y = vY v0 ,which was the guess in the assertion.

In the following we consider a particle without mass in a fluid, or we thinkabout a flag which is assigned to a moving mass point.

1.9 Particle in a fluid. Let a fluid be modelled by a mass density satis-fying

t+ div (v) = 0 . (I1.14)

We are moving with the fluid, i.e., at time t we are somewhere, say, at thepoint (t) Rn, and we drift with the velocity v, i.e., is given by thedifferential equation 6

(t) = v(t, (t)) .

Define (t) := (t, (t)) the mass density at the position we are at time t.Then

(t) + a(t)(t) = 0 , a(t) := ( divv)(t, (t)) .

This means that the rate at which the mass density at our position changesis a(t). Therefore one writes (I1.14) as

+ div v = 0 ,

:= t+ v . (I1.15)

Proof. It ist+ div (v) = (t+ v) + div v ,

which implies (I1.15). We then compute

(t) =d

dt((t, (t))) = (t) (t, (t)) +

ni=1

(xi) (t, (t)) i(t)

= (t+ v())(t, (t))(da i(t) = vi(t, (t))

)

= ( divv)(t, (t)) = ( divv)(t, (t))(t) ,which is the statement.

6It is the time derivative of t 7 (t)



If one considers polar coordinates (r, ) for n = 2, it can be understood asthe case that the space functions do not depend on x3. We now consider thecase n = 3 and describe cylindrical coordinates (r, , z) with x3 = z, and weallow functions to depend on all variables.

1.10 Cylinder coordinates. In R R3 we consider the transformation

(t, x) = (t, x1, x2, x3) = (t, r, , z)

given byt = 0(t, r, , z) := t ,

x1 = 1(t, r, , z) := rcos ,

x2 = 2(t, r, , z) := rsin ,

x3 = 3(t, r, , z) := z .

We want to write the conservation law (I1.1)

tu+ div q = r

in cylindrical coordinates. To this we decompose the flux vector q withrespect to the cylindrical coordinates as

q = qrer + qe + qzez , (I1.16)

where er, e, ez are given by (compare (I1.6))7

er := (x21 + x

22)

12 (x1, x2, 0) , r = (0, cos , sin , 0) = (0, er) ,

e := (x21 + x

22)

12 (x2, x1, 0) ,

1

r = (0,sin , cos , 0) = (0, e) ,

ez := (0, 0, 1) , z = (0, 0, 0, 1) = (0, ez) ,{er(x), e(x), ez(x)} for x 6= 0 is an orthonormal basis of R3.

Further, if we define u = u , r = r , q = q (and therefore qr = qr ,q = q , qz = qz) it follows that

tu+ zqz + rqr +1

rqr

=

1

rr(rqr)

+1

rq = r .

(I1.17)

Multiplying the equation by r, we obtain

t(r u) + z(r qz) + r(r qr) + q = r r , (I1.18)

which is an equation also of divergence structure (compare the result in 5.1).

7 Notation for partial derivative: Wir bezeichnen partielle Ableitungen auch durchnachgestellte Ableitung, so z.B. in der Aussage r(t, r, , z) := r(t, r, , z) fur . Wirwerden diese neue Bezeichnung verwenden bei Koeffizientenfunktionen, um dadurch dieBeschreibung von partiellen Differentialgleichungen ubersichtlicher zu gestalten.



Proof (1.Version). We compute

divq = ererq + eeq + ezezq= er(erq) + e(eq) + ez(ezq)(erer + ee + ezez)q

= erqr + eq + ezqz +1

rqr

where r =x21 + x

22, since

erer = 0, ee = 1

rer, ezez = 0 (see (I1.6)).

Then, since for any function g

(erg) = r(g) ,

(eg) =1

r(g) ,

(ezg) = z(g) ,

we obtain

( divq) = r(qr) +1

r(q) + z(qz) +

1

rqr,

the assertion.

Proof (2.Version). We compute, since ez is constant,

divq = ererq + eeq + ezezq= erer(qrer + qe + qzez)

+ee(qrer + qe + qzez)+ezez(qrer + qe + qzez)

= erqr + eq + ezqz

+qr(ererer + eeer + ezezer)+q(erere + eee + ezeze)

= erqr + eq + ezqz +1

rqr ,

since

erer = 0, eer =1

re, ezer = 0,

ere = 0, ee = 1

rer, eze = 0.

(see (I1.6)).



Then, since for any function g

(erg) = r(g) ,

(eg) =1

r(g) ,

(ezg) = z(g) ,

we obtain

( divq) = r(qr) +1

r(q) + z(qz) +

1

rqr,

the assertion.


I.2 Distributions 24

2 Distributions

We multiply the scalar conservation law (I1.1) for C1-functions u, qi, r

tu+ div q = r in U R Rn (I2.1)

with a test function C0 (U) and obtain after integration by parts

0 =

U(tu div q + r) dLn+1

=

U(t u+q + r) dLn+1

where the last integral exists, if the functions u, qi und r are in L1loc(U).

Therefore the conservation law contains the following three contributions

7

Ut u dLn+1 ,

7

Uq dLn+1 ,

7

U r dLn+1 ,

(I2.2)

which are all linear in the test function . These linear functions are, as weshall see, distributions with N = n+ 1.

Definition of Distributions

We start with the essential property of distributions.

2.1 Distributions. Let U RN be an open set. We denote by

D(U) := C0 (U)

the space of test functions. We consider mappings

T :D(U) R linear

and call them distributions with the notation T D (U), if they satisfythe estimate 2.4(1). We introduce the notation

, T D(U) := T () ,

which is motivated by the integral in (I2.2). Often we simply write , T = , T

D(U), if the domain U is fixed.

There are two things which are important for a distribution, taking thederivative and multiplying with a function.



Fig. 5: Functions as functionals (see [66])

2.2 Operations on distributions.

(1) Derivative. For j {1, . . . , N} a linear map jT :D(U) R is definedby

, jT D(U) := j , T D(U) .General: For higher derivatives see 2.5(1).

Definition in spacetime: Let N = n+ 1 with n 1. Then U R Rn andj runs from 0 to n. We then have 0 = t and i = xi for i = 1, . . . , n.

(2) Multiplication. For a Cloc(U) a linear map aT :D(U) R is definedby

, aT D(U) := a , T D(U) .

Both, jT and aT are again distributions, since they still satisfy 2.4(1).

These are all definitions for distributions we need, and for our three terms(I2.2) in the conservation law we have to define

2.3 Functions as distribution. Let us consider special mappings T = [g]where g L1loc(U), defined for test functions D(U) by

, [g] D(U) :=

U g dLN .



Remark: The Lebesgue-measurable function g can be reconstructed from itsdistribution [g] (see exercise 7.9).

The remark says that g can be recovered from its distribution almost every-where (see also the text in Fig. 5). Similarly this follows for the derivativei[g], provided this distribution is represented by a function. For example,if g is a Lipschitz continuous function, it is i[g] = [gi] with a boundedmeasurable function gi (see the definition in 2.5(2))

References: Zur Geschichte der Distributionen siehe [67]. MathematischeEinfuhrungen werden fur N = 1 in [66], fur beliebiges N in [61, in Ab-schnitt 3], [63, Kapitel I-II], [64, Kapitel 1-9], [65, Kapitel 1-2] gegeben. Ichhabe auch ein eigenes Skript [60] dazu angefertigt. Siehe auch die sehr guteDarstellung in [Wikipedia: Distribution (Mathematik)].

We have yet to specify the full definition of distributions.

2.4 Estimate satified by distributions. Let U RN be an open set and consider thespace D(U) = C0 (U). A distribution satifies by definition one of the following equivalentproperties:

(1) A linear mapping T :D(U) R satisfies 8

U U : kU N {0} and CU 0 :

D(U) with supp U : , T

D(U)

CUCkU (U) .

(2) D (U) is the set of linear continuous mappings, in fact the dual space of D(U), ifwe assign D(U) with the following topology T :

T := {V C0 (U) ; V : : + V V } .

Thereby = (j)jN and

V := conv(

jN

{ C0 (U) ; supp () Uj and p() < j}),

(Uj)jN is an open covering of U with Uj U compact,

p() :=

k=0

2kCk(U)

1 + Ck(U)for supp () U , U compact in U .

Result: Hence D(U) becomes a locally convex topoplogical vector space, see [61, 3.19] and[60, section 6] where also the completeness of D (U) is discussed.

Proof of equivalence: The statements (1) and (2) are equivalent. For example see [61, 3.21Der Dualraum von D(U)], but you can visit any book involving distributions.

The mathematical definition of distributions essentially show that functions as distribu-tions are dense in the set of distributions (siehe [60, End of section 2]). However, wewill not use this estimate (except in 2.9). Here some of the important properties whichdistributions have.

2.5 Some properties of distributions.

8U U means that U U and U is compact in U , in words: U is relative compactin U .


https://de.wikipedia.org/wiki/Distribution_(Mathematik)


(1) Higher derivatives. For all multi-indices s the distributional derivative sT isthe linear map sT : D(U) R defined by

, sT D(U) = (1)

| s | s , T D(U) for D(U).

Es giltsT = r1(r2T ) for all r1, r2 with r1 + r2 = s .

(2) Partial derivative. Fur g C1(U) gilt j [g] = [jg] wegen der Regel der partiellenIntegration. Man definiert daher in Analogie dazu

W1,ploc (U) := {g L

ploc(U) ; i : gi L

ploc(U) : i[g] = [gi]} .

Hierbei ist 1 p . (Entsprechend ist W k,ploc (U) definiert.)

(3) Vector valued distributions. Analog ist die Definition von [g] fur vektorwertigesg :U RM , es ist dann 9 fur D(U ;RM ) = C0 (U ;R

M )

, [g] D(U) :=

U

g dLN .

Wir schreiben dann [g] D (U ;RM ) (siehe auch [60, 5.4]).

(4) Order of a distribution. A distribution T is of order k, if 2.4(1) is satisfied alwayswith the same kU = k. It holds: If T is a distribution of order k, then

sT is a distributionof order k + | s |.

(5) Extended distributions. Ist T eine Distribution der Ordnung k, so kann T eindeutigfortgesetzt werden zu einer linearen Abbildung auf Ck0 (U). Es ist also , T := T () fur Ck0 (U) als Fortsetzung definiert. Es folgt, dass aT als Distribution definiert ist fura Ck(U), es ist (siehe auch [60, 4.1]).

, aT D(U) := a , T Ck

0(U) fur D(U).

What does it mean for our conservation law?

Back to the conservation law

We will now write conservation laws in the context of distributions, wherewe set N = n + 1, i.e. it is U R Rn and the distributions, we consider,live in spacetime:

2.6 Distributions in spacetime. Let N = n + 1 with n 1 and U R Rn. Then for g L1loc(U) the distribution [g] D(U) satisfies

, [g] D(U) =

U g dLn+1 =

R

Ut

(t, x)g(t, x) dx dt ,

where Ut := {x Rn ; (t, x) U}.

9 With we denote the scalar product of the Euclidic space.



With these definitions we obtain for the law tu+ div q = r

0 =

U(tu div q + r) dLn+1

=

U(t u+q + r) dLn+1

=

Ut u dLn+1 +

Uq dLn+1 +

U r dLn+1

= t , [u] D(U) + , [q] D(U) + , [r] D(U)= , t[u] div [q] + [r] D(U) ,

where [u], [r], [qj ] (j = 1, . . . , n) are defined as in 2.3. Consequently theconservation law (I2.1) now is for functions u, r, qj L1loc(U)

t[u] + div[q] = [r] in D(U) , (I2.3)

and for general distributions U,Qj , R :D(U) R the equation becomes

Distributional conservation law:

tU + divQ = R in D(U),

U,Qj , R D (U) for j = 1, . . . , n.

(I2.4)

This definition means that for D(U)

0 = , tU divQ+R D(U)= t , U D(U) + , Q D(U) + , R D(U)= t , U D(U) +

j

j , Qj D(U) + , R D(U) .

Mass points

The first example shows that the motion of a mass point is a solution of thedistributional mass conservation. In the context of momentum conservationin the next section 3 we come back to this example.

2.7 Moving mass point. We are thinking about a moving mass pointwith mass m > 0 which moves through the space in virtue of a continuouslydifferentiable map t 7 (t) Rn, i.e. in time and space

t 7 (t, (t)) R Rn

is the trajectory. On the trajectory the velocity

v(t, x) := (t) fur x = (t) (I2.5)



of the mass point is continuous.

Assertion: The distributional mass conservation

t(m) + div (mv) = 0 (I2.6)

in D (R Rn) is satisfied, where the distribution D (R Rn) is givenby

,

D(RRn)

:=

R

(t, (t)) dt fur D(R Rn)

=

R

(t, ) , (t)

D(Rn)

dt .

(I2.7)

Dirac Distribution: Fur x0 Rn ist x0 D (Rn) definiert durch

, x0 D(Rn) = (x0) fur D(Rn). (I2.8)

Consequently definition (I2.5) for the velocity v is equivalent to the dis-tributional mass conservation (I2.6), where m is the mass distribution.That is, at the time t the mass m is concentrated at the point (t), so thedistribution has the trajectory of the movement as support.

Proof. Since is continuously differentiable, and vi are distributions.It is

, t(m) + div (mv)

= mt ,

+m

ni=1

xi , vi

= m

R

((t)(t, (t)) +

ni=1

(xi)(t, (t)) vi(t, (t)) = i(t)

)dt

= m

R

d

dt

((t, (t))

)dt = 0 ,

because has compact support.

We now suppose a-priori that the mass of the particle depends on time, wewrite t 7 m(t) or t 7 m(t, (t)) if the mass belongs to the moving point. Then it follows from the distributional mass conservation that this masshas to be constant. This proves that the distributional conservation law isthe right thing to consider.

2.8 Lemma. Let be as in 2.7, that is, is continuously differentiable,and

:= {(t, x) ; x = (t)} ,m : R continuous and positive,v : Rn continuous



witht(m) + div (mv) = 0 . (I2.9)

Thenm is a constant,

v(t, (t)) = (t) .

So the total mass of the point is constant.

Proof. Let D(R Rn). Then

0 = , t(m) div (mv)

=t , m

+ , mv

=

R

(mt +mv)(t, (t)) dt .

By looking at the velocity of ,

v(t, x) := (t) for x = (t) , that is, (t, x) ,

we can write the integral by partial integration as

=

R

(m(t + v))(t, (t)) dt+

R

(m(v v))(t, (t)) dt

=

R

m(t, (t))d

dt((t, (t))) dt+

R

(m(v v))(t, (t)) dt

=

R

d

dt(m(t, (t)))(t, (t)) dt+

R

(m(v v))(t, (t)) dt .

This is true for all C0 (R Rn), and therefore by approximation alsofor all C10 (R Rn) (see 2.5(5)). Now use

(t, x) = (t)(t, x) ,

C10 (R) , C1(R Rn) ,

such that for some > 0

(t, x) = 1 for dist ((t, x),) < ,

(t, x) = 0 for dist ((t, x),) 2 .

Then (t, (t)) = (t)(t, (t)) = 0 and the above integrals become

=

R

d

dt(m(t, (t)))(t) dt .

This implies, since is arbitrary,

d

dt(m(t, (t))) = 0 , (I2.10)



and therefore m is constant. Inserting this in the above integral gives for alltest functions

0 =

R

(m(v v))(t, (t))(t, (t)) dt .

Now we choose

(t, x) = (x (t))w(t) withw C10 (R;Rn) , C0 (Rn;Rn) ,(z) = z for small |z|.

Then (t, (t)) = w(t) and therefore

0 =

R

(m(v v))(t, (t))w(t) dt .

Since w is arbitrary, it follows (m(vv))(t, (t)) = 0, and sincem is positive

v(t, (t)) = v(t, (t)) = (t) . (I2.11)

The statements (I2.10) and (I2.11) are the assertion.

In [19, 1 Flug eines Asteroiden] we extend this to a decreasing total mass.

Gravitational law

As another example consider the gravity, the corresponding field equationhas a distributional solution, so it is not a smooth function in the generalcase, because the characteristic function for the mass density has a jump.The field equation is

Newtons gravitation:

div([]) = []in entire R Rn (physically n = 3), i.e. in D (R Rn) total mass density (as a function),

gravitational field (is a function),

(t, x) 0 for |x| (if n = 3),G = 6.67384 1011 m3

kg s2gravitational constant,

literature = 4G potential in the literature (n = 3).

(I2.12)

One can imagine this equation also as conservation law

t0 + div([]) = [] ,



thus is seems to be a general mass conservation without any mass. (Butthis is misleading, since the 0 arises if the speed of light goes to .) Inthe literature the gravitational field is literature and therefore the equationreads literature = 4G. One can also write div[] = [], hence[] = []. In general the gravitational field and the mass may bedistributions and R satisfying the equation

General gravitational law:

div() = R in D (R Rn)

R the total mass as distribution,

the gravity field as distribution.

(I2.13)

In the law of gravitation the time t occurs only as a parameter. There is noexplicit time derivative in the Newtonian physics considered here. Therefore,the general gravity law is related to the distributional Poisson equation,which is:

Distributional Poisson equation:

= R in D (Rn)

R the source term as a distribution,

the solution as a distribution.

the Laplace operator in Rn.

(I2.14)

In the following we apply the Poisson equation where n is the space dimension and wherethe time is a parameter. We compare it with Newtons law in spacetime R Rn withdimension n+ 1.

2.9 Remark. We assume that Ut, Rt D(Rn) for t R are distributions of order k, that

is,| , Ut D(Rn) | + | , Rt D(Rn) | C(t)Ck(Rn)

with an integrable function C L1(R). If they satisfy the Poisson equation

divUt = Rt in D(Rn) for almost all t,

then (under the assumption of measurability on t 7 Rt, t 7 Ut)

, R D(RRn) :=

R

(t, ) , Rt D(Rn) dt ,

, U D(RRn) :=

R

(t, ) , Ut D(Rn) dt

define distributions U,R D (R Rn) and they fulfill the general law of gravitation

divU = R in D (R Rn).

Attention: Not each distribution R D (R Rn) can be represented as shown (see,e.g. Exercise 7.13, but keep 2.10 in mind).



Proof. Both U and R are distributions, and it is

, U +R D(RRn) = , U D(RRn) + , R D(RRn)

=

R

((t, ) , Ut D(Rn) + (t, ) , Rt D(Rn)

)dt

=

R

(t, ) , Ut +Rt D(Rn) dt .

As an example we choose a moving mass point.

2.10 Example. If R D (R Rn) is a distribution that belongs to a mass point withthe trajectory {(t, (t)) ; t R}, then the definition R = m implies that

, R D(RRn) = m

R

(t, (t)) dt =

R

(t, ) , Rt D(Rn) dt ,

where Rt = m(t), i.e.

, Rt D(Rn) := m ((t)) for D(Rn) ,

hence Rt is given by the Dirac distribution.

We are now focusing first on the Poisson equation. Here we can consider in(I2.14) as a special case R = x0 for x0 Rn, see (I2.8). Then the solution = [] to R = 0 with L1loc(Rn) is the fundamental solution for thenegative Laplace operator:

2.11 Fundamental solution for the Laplace operator. Let n 3. Thesolution L1loc(Rn) of the equation

[] = 0 in D (Rn) ,

with the boundary condition (x) 0 as |x| , is given by

(x) :=1

n(n 2)|x|2n for |x| > 0 . (I2.15)

Remark: It is the fundamental solution for , that is, the negativeLaplace operator. Definition: It is n := H

n1(B1(0)) = nn the surfaceof the unit sphere in Rn, and n := L

n(B1(0)) the volume of the unitball in Rn.

n 1 2 3 arbitrary

n 2 43 L

n(B1(0))

n 2 2 4 Hn1(B1(0)) = nn

(I2.16)



Proof. It is for D(Rn)

, i[] = i , [] =1

n(n 2)

Rn

i(x)dx

|x|n2

=1

n(n 2)lim0

Rn\B(0)xi(x)

dx

|x|n2

= lim0

1

n(n 2)

B(0)(x)eiB(0)(x)

1

n2dHn1(x)

lim0

1

n(n 2)

Rn\B(0)(x)xi

1

|x|n2 dx

=1

n

Rn

(x)xi|x|n dx

= , [Fi]

with

F (x) :=1

n

x

|x|n ,

hence [] = [F ] in D (Rn;Rn). Now

, div[F ] = , [F ] = 1n

Rn

(x) x|x|n dx

= 1n

lim0

Rn\B(0)(x) x|x|n dx

=1

nlim0

B(0)(x) B(0)(x)

x

|x|n =

1

n1

dHn1(x)

+ lim0

1

n

Rn\B(0)(x) div

x

|x|n = 0

dx (see 7.12)

= (0) = , 0 ,

that is, F is the fundamental solution of the divergence operator.

In the case N = 1, 2 there are also fundamental solutions of the Laplace op-erator, however they are physically only of interest in finite neighbourhoodsof the singularity. They are

(x) =

12

log |x| if N = 2,

12|x| if N = 1,



and it is[] = [F ] in D (RN ;RN )

F (x) :=1

N

x

|x|N

for all N 1.

In the case N = n = 3 the fundamental solution is exactly the solution ofthe general gravity law, modulo the statement 2.9.

2.12 Gravitational potential of a point-shaped star. Let n = 3. Ifm > 0 and R := m is the density of a mass point t 7 (t), then thesolution, i.e. the distribution , of the general law of gravitation

div() = R := m in D (R Rn)

is given by = [],

(t, x) :=m

4|x (t)| for x 6= (t) . (I2.17)

The solution is uniquely determined by the condition that as |x| thepotential (t, x) 0.

Proof. This follows essentially in the same way as the proof of 2.11, thedifference is that one deals with integrals over R Rn. The uniqueness isderived from the following. We mention that (t, ) L1loc(Rn).

2.13 Uniqueness. Let n 3 and R D (R Rn). Then there exists atmost one L1loc(R Rn) with

div([]) = R in D (R Rn),(t, x) 0 for |x| for almost all t.

Proof. Because 1 and 2 are solutions to R, it follows with := 1 2that

[] = 0 in D (R Rn) ,(t, x) 0 fur |x| for almost all t.

It follows for (t, x) = 0(t)1(x)

0 = , [] = , [] =

R

0(t)

Rn

1(x)(t, x) dx dt

Since this holds for all 0, it follows for almost all t

0 =

Rn

1(x)(t, x) dx ,



hence (t, ) or better [(t, )] is a harmonic distribution, that is,

[(t, )] = 0 in D (Rn) ,

and for such functions the mean value property of spheres applies (see[PDE]), i.e.

(t, x0) =1

nrn1

Br(x0)(t, x) dHn1(x) 0 as r .

Consequently, it is = 0.

We will now calculate the gravitational force of a planet. The solution isa distribution because it models the boundary between a solid body andvacuum, that is, the density makes a jump. (Hence it is L, and thebest of what could be shown by the regularity theory is that C1,1. Thisis because there is the sharp statement that Lp implies W 2,p forp


Note: For the result 2.14 the requirement R = [] with L(D) isessential. For example if we have 10

R = +[LNxD+] + [L

NxD] + 0[H

N1x] ,

then is only continuous with

+ = + 0

on . The case 0 6= 0 is treated in 2.17.Proof. For a test function D(D) it holds

, [] = , R .

The right-hand side is

, R =

D

dLN =

D+

+ dLN

D

dLN .

The left-hand side is

, [] = , [] =

D

dLN

=

D+

+ dLN +

D

cot dLN

=

D+

+ dLN

D

dLN

+

(+D+ + D

)dHN1

=

D+

+ dLN +

D

dLN

+

(

(+D+ + D

)

(+D+ +D

))dHN1 .

We now choose in D(D+) we conclude

D+

+ dLN =

D+

+ dLN

for all such test functions, hence

+ = + in D+ .

Accordingly in the same way, it means chosing in D(D), it follows

= in D .10 Ist ein Ma, so ist das Ma xS definiert durch (xS)(E) := (S E).



By plugging these identities into the above equation we obtain for arbitrarytest functions

0 =

(

(+D+ + D

)

(+D+ +D

))dHN1 .

Now we extend this argument from C0 (D) to C10 (D) by an approx-imation. Let C1(D) be a function which vanishes on , and for which 6= 0 on applies (e.g. (x) = signdist (x,)). Set

= 0 C10 (D) ,

where 0 C0 (D) so that on

= 0 , = 0

is satisfied. It follows

0 =

0

(+D+ + D

)dHN1

and, since 0 is arbitrary,

0 = (+D+ + D

)on

and therefore+ = on ,

i.e. is continuous across . Thus the identity for arbitrary test functionsis now

0 =

(+D+ +D

)dHN1 .

It follows since is an arbitrary test function

+D+ +D = 0 on ,

i.e. the differentiability of .

Proof der Bemerkung. Wir betrachten nur den Fall R = 0[HN1x]. Der

Beweis ist derselbe bis auf die Tatsache, dass nun

, R =

0 dH

N1

und der Term auf

0 dH

N1 =

(

(+D+ + D

)

(+D+ +D

) )dHN1



ist. Mit dergleichen Argumentation wie oben ist dann = + auf unddaher

0 =

(0 +D+ D

)dHN1

fur alle , weswegen +D+ +D = 0.

The situation in 2.14 occurs for example for the gravity when the body hasa smooth surface. This is true for the spherical case.

2.15 Gravitational potential of a globe. Let be n 3 (physically n = 3),m > 0 the mass and t 7 (t) the motion of the center of the planet. Then

(t, x) =m

Ln(BR((t)))XBR((t))(x)

is the mass distribution of the planet idealized as a homogeneous mass den-sity on a sphere of radius R (see also 4.5). The total mass of the planetis

m =

Rn

(t, x) dx .

We are seeking a solution of the differential equation

div([]) = [] in D (R Rn)with (t, x) 0 if |x| ,

(I2.18)

which is of order C1 in the space variables (see statement 2.14).

Assertion: The solution, which disappears at infinity (for n 3), is 11

(t, x) :=

m

2n

1

Rn

(R2

n 2 |x (t)|2

n

)if |x (t)| R,

m

nn(n 2)1

|x (t)|n2 if |x (t)| R.(I2.19)

Proof. Without restrictions let (t) = 0. Since nRn = Ln(BR(0)), one

computes

(t, x) =m

nRnXBR(0)(x) = mRXBR(0)(x) if mR :=

m

nRn.

Further let

(t, x) :=

{(t, x) if |x| < R,+(t, x) if |x| > R,

where(t, x) = mR if |x| < R,+(t, x) = 0 if |x| > R .

11 It is n := Ln(B1(0)), see (I2.16).



R 0 +R

Quadratic in r

Mass 14r

r: Distance from center

R: Radius of sphere

Fig. 7: Gravitational field of an incompressible ball (n = 3)

This is satisfied if(t, x) = c0

mR2n|x|2,

+(t, x) = c1

|x|n2 .

Then is continuous in space, if (t, x) = +(t, x) for x BR(0), i.e.

c0 mR2n

R2 = cR2n . (I2.20)

Then is continuous differentiable in space if and only if (t, x) =+(t, x) for x BR(0), i.e.

mRnR = (n 2)cR1n . (I2.21)

From (I2.20) and (I2.21) it follows that is continuously differentiable inspacetime, and it is

c =mR

n(n 2)Rn =

m

nn(n 2),

c0 = mR

(1

2n+

1

n(n 2)

)R2 =

m

2n(n 2)R2n .

(I2.22)

The conditions in (I2.22) yield (I2.19).

Usually it is only an approximation if we consider a planet to be a ball. Thereason is that there are mountains on the surface, or inside the planet isnot constant, see Fig. 8 and [GRACE globe animation.gif]. But whatever is, in any case (I2.18) says what the gravitation potential has to be. Wenow treat the case that the planet degenerates to a point of mass m.

2.16 Convergence to a mass point. Let n 3. As fixed mass m > 0and as R 0 the gravity solution of 2.15 converges in L1loc(R Rn) to a


http://www-m6.ma.tum.de/~alt/QUELLEN/GRACE_globe_animation.gif


solution of div([]) = m in D (R Rn) ,

(t, x) 0 as |x| .This solution is given by 12

(t, x) :=m

n(n 2)|x (t)|2n if |x (t)| > 0 . (I2.23)

Fig. 8: Earths gravity measured by NASA GRACE mission, showingdeviations from the theoretical gravity of an idealized smooth Earth, theso-called earth ellipsoid. Red shows the areas where gravity is stronger thanthe smooth, standard value, and blue reveals areas where gravity is weaker.[Wikipedia: Gravity of Earth].

Proof of convergence. The gravity solution R and R of 2.15 fulfills

div([R]) = [R] in D (R Rn) ,

or with test functions C0 (R Rn)R dL

n+1 =

R dLn+1 =

R dLn+1 .

Now it holds R in L1loc(R Rn). This is because of Lebesgues con-vergence theorem and the estimate

R(t, x) = (t, x) if |x (t)| R ,0 R(t, x) (t, x) if |x (t)| R .

12 We define n := Hn1(B1(0)), B1(0)) R

n, so that n = nn.


https://en.wikipedia.org/wiki/Gravity_of_Earth


The first identity follows from the definition of R. The second inequalityreads R(t, x) (t, x) for 0 r = |x (t)| R This holds if and only if

m

2n

1

Rn

(R2

n 2 r2

n

) mnn(n 2)

r2n

R2

(n 2)Rn 2

n(n 2)r2n +

r2

nRn

sn2

n 2 2

n(n 2) +sn

nfor s =

r

R 1

sn2 2n+n 2n

sn for s =r

R 1,

which is true by Youngs inequality. (For the L1-convergence it is enoughto show that R(t, x) C|x (t)|2n for all R, where C is independent ofR.) Also RL

n+1 m as R 0, which follows fromR dL

n+1 =

R

BR((t))(t, x)

m

nRndx dt

=

R

B1(0)(t, (t) +Ry)

m

ndy dt

R

(t, (t))mdt = , m

.

Therefore altogether

, m

= , ,

qed.

Therefore, the solution outside a star (if the star has a constant massdensity) coincides with the solution obtained if one sets the star as a pointmass with the same total mass. In the next section 3 we will considerthe conservation of momentum and we will show that in the stationaryincompressible case homogeneous stars produce a gravitational field like theone here (see 4.5). In the compressible case we refer to section IV.16, whereradially symmetric mass distributions of stars are considered.

That the solution of the gravity equation is C1, is not true if the mass densityis supported on a surface. As it turns out the solution is only Lipschitzcontinuous. The following example is for a homogeneous mass distribution.

2.17 Hollow sphere. Let n 3, m > 0 be constant and t 7 (t) themovement of the center of a shell. Its support is supposed to ly on BR((t)).Then let D (R Rn) be given by

, :=

R

BR((t))(t, x) dHn1(x) dt



for D(R Rn). Further, let

s(t, x) :=m

Hn1(BR(0))XBR((t))(x)

the constant mass density on BR((t)). Then the solution of equation

div([]) = s in D (R Rn) ,(t, x) 0 as |x| ,

is given by

(t, x) =

m

n(n 2)|x (t)|2n if |x (t)| R ,

mR2n

n(n 2)if |x (t)| R .

The solution is thus only of class C0.

Proof. If is as in the formula, we get

(t, x) =

mn

x (t)|x (t)|n if x R

n \ BR((t)) ,

0 if x BR((t)) ,

where = div = 0 in Rn \ BR((t)). Hence for D(R Rn;Rn)

, [] = div , [] =

R

Rn

div(t, x) (t, x) dx dt

=

R

Rn\BR((t))(t, x)(t, x) dx dt ,

because is continuous. Therefore it holds for D(R Rn;R) , div([]) = , []

=

R

Rn\BR((t))(t, x)(t, x) dx dt

=

R

Rn\BR((t)))(t, x) (t, x)

= 0

dx dt

+

R

BR((t)))(t, x) (t, x)

Rn\BR((t)) =

m

n|x (t)|n1

dHn1(x) dt

= , s ,where in the last integral (t, x) is taken from outside, i.e.

(t, x) = limh0(t, x+ hBR((t)))


I.3 Conservation of momentum 44

3 Conservation of momentum

The momentum conservation needs for its formulation a mass conservation,which results from the observer transformations in Section II.3. Hence, thissystem of mass-momentum balance reads 13

General mass-momentum equation:

t+ div(v + J) = r ,

t(v) + div(vvT + vJT +) = f

where besides the quantities in (I1.7)

= (ij)i,j=1,...,n pressure tensor,

f =(fi

)i=1,...,n

general force density.

(I3.1)

Here at first (,J, r) and (v,, f) are arbitrary terms, so we have writtendown the general version of the conservation equations. The f -term includesboth external forces and internal forces such as the self-gravity. Strictlyspeaking f is a force density since it is a function of (t, x). There is acorrespondence between the pressure term and the force term f , in fact itis similar as between J and r (see the remark following (I1.7)). Thus partsof the forces can be written under the divergence term, that is, as part ofthe pressure tensor. Such terms will be denoted as internal force. Thev-terms in the fluxes result from objectivity reasons (see Section II.3, wealso refer to this section if you want a precise definition of f). In general,the divergence is defined by the fact that it acts on the last index, for amatrix see the following definition.

Definition: If

M = (Mij)i,j=1,...,n =

M11 . . . M1n...

...Mn1 . . . Mnn

is a matrix-valued function, the divergence of it is defined by

divM :=

(n

j=1xjMij

)

i=1,...,n

.

13 While (x, y) 7 xy = xT y denotes the scalar product, the tensor product is ex-pressed by (x, y) 7 x yT = xy.



Further, in the above equation it is

v vT =

v1...vn

[ v1 . . . vn ] =

v1v1 . . . v1vn...

...vnv1 . . . vnvn

= (vivj)i,j=1,...,n ,

vJT =

v1...vn

[J1 . . . Jn ] =

v1J1 . . . v1Jn...

...vnJ1 . . . vnJn

= (viJj)i,j=1,...,n .

ThusvvT + vJT + = (vivj + viJj +ij)i,j=1,...,n

and therefore the system of differential equations can be written as a systemof n+ 1 equations

t+n

j=1j(vj + Jj) = r ,

t(vk) +n

j=1j(vkvj + vkJj +kj) = fk for k = 1, . . . , n.

(I3.2)

If is the total mass again, it is usually J = 0 and r = 0, the term f ,which we called general force density, now becomes the force density f(for explanation see the mass-momentum balance in section II.4). We thenobtain the

Mass-momentum conservation:

t+ div(v) = 0 ,

t(v) + div(vvT +) = f

mass density, v velocity,

= (ij)i,j=1,...,n pressure tensor,

f = (fi)i=1,...,n force density.

(I3.3)

First we treat the case that the pressure tensor equals 0, but the force termis arbitrary.

Momentum of mass points

This is the case for the motion of a mass point. Here the trajectory is againdenoted by t 7 (t) Rn (as in 2.7) and the mass-momentum conservationhas a distributional formulation. In fact, one has to think about the masspoint at (t) as a limit of a body with small diameter, for which the mass-momentum equations are satisfied. We show that this is equivalent to anordinary differential equation of second order for .



3.1 Mass point. We consider the mass point introduced in 2.8 which moveswith t 7 (t) Rn and whose total mass is given by t 7 m(t, (t)) > 0.The following is equivalent

(1) The distributional equations

t(m) + div (mv) = 0 ,

t(mv) + div (mvvT) = f

(I3.4)

are fulfilled. Here the distribution is given by (I2.7).

(2) It ism constant,

v(t, (t)) = (t) velocity,(I3.5)

and the ordinary differential equation

m = f (I3.6)

is satisfied.

We mention that here f is the force density (in the distributional mo-mentum equation), whereas f , the right-hand side of the ODE (I3.6), iscalled force.

Proof (1)(2). We consider the second equation of (I3.4). For test functions D(R Rn;Rn) we compute

0 = , t(mv) divx(mvvT) + f

=k

k , t(mvk) divx(mvkv) + fk

=k

(tk , mvk

+k , mvkv

+k , fk

)

=k

R

(mvk)(t, (t)) (tk + vk)(t, (t)) =

d

dtk(t, (t))

dt

+k

R

(kfk)(t, (t)) dt

=k

R

( ddt

((mvk)(t, (t))

)+ fk(t, (t))

)k(t, (t)) dt .

Since this is true for arbitrary test function we obtain

d

dt

((mv)(t, (t))

)= f(t, (t)) . (I3.7)



So far the momentum equation. The first differential equation in (I3.4), thatis the mass conservation, is treated as in 2.8. This results in the equations(I3.5). So we get the differental equation m = f .

Die eingerahmte Gleichung (I3.7) besagt also, wie Newton in seinen Principiaschreibt, siehe Newton [95, Axiomata sive Leges Motus: Lex.II] oder Newton[96, Axioms, or the Laws of Motion: Law 2]14,

Change in momentum = force

und das ist bei sich andernder Masse richtig, also wenn die erste Massengle-ichung lautet t(m) + div (mv) = r. Die Newtonsche Physik ist alsoin den distributionellen Masse-Impuls Gleichungen enthalten. Wenn sich dieMasse nicht andert, so lautet die Aussage Masse Beschleunigung = Kraft.

3.2 Collision of mass points. Let be given two mass points, as in 2.7,

t 7 (t) Rn continuous, = 1, 2,

whose trajectories meet exactly in the spacetime point (t0, x0),

x0 = 1(t0) =

2(t0) .

We denote the distributions as in (I2.7). The masses are given bybounded continuous functions

t 7 m(t, (t)) > 0 for t 6= t0 .

Thus, the distributional total mass of the system is given by

=1,2

m D (R Rn) .

Assertion: Let the distributional equations

t

(m

)+ div

(mv

)= 0 ,

t

(mv

)+ div

(mv vT

)=f

(I3.8)

be satisfied, where v(t, (t)) := (t) are the velocities, and let the deriva-tives (t) for t 6= t0 be piecewise continuous up to the point t0, as well as

14I. Bernard Cohen writes there in A Guide to Newtons Principia: For example, inlaw 2, Newton writes that a change in motion is proportional to the motive force.Here he means change in the quantity of motion or, in our terminology, change inmomentum.



the vector fields f. It follows

m = f ,

m locally constant in t

}for t 6= t0and = 1, 2,

m1 +m2 = m

1+ +m

2+ (mass conservation in t0),

m1v1 +m

2v

2 = m

1+v

1+ +m

2+v

2+ (momentum conservation in t0),

(I3.9)where

m := limtt0

m(t, (t)) , m+ := limtt0

m(t, (t))

and similarly v and v+.

Thus, it is not described what happenes to the particles when colliding, butit is set up a total mass balance and a total momentum balance. This isdone under the assumption that after the collision the only thing which is leftare again two particles. It can also happen that there are several particlesafter the collision (as in Fig. 9), which leads to corresponding formulas,or during the collision a light flash is emitted, which changes the formulasdramatically. We refer to III.?? where we present also an energy balance.

Fig. 9: Particle tracks from the collision of an accelerated nucleus of aniobium atom with another niobium nucleus. The single line on the leftis the track of the incoming projectile nucleus, and the other tracks arefragments from the collision. (Courtesy of the Department of Physics andAstronomy, Michigan State University)



Proof. Outside the point (t0, x0) the two trajectories are apart from each

other. Let t 6= t0. In a neighbourhood of the point (t, (t)) we have toconsider only the -phase, this means that in this neighbourhood we haveto consider

t(m

)+ div

(mv

)= 0 ,

t(mv

)+ div

(mv vT

)= f .

Due to 3.1 and 2.8, it follows that m is constant in this region, thatv(t, (t)) = (t), and that m(t) = f(t, (t)) for t 6= t0.Therefore we have to compute the mass and momentum contribution nearthe point (t0, x0). We write the mass conservation and the components ofthe momentum conservation, see (I3.8), in one equation

t

(g

)+ div

(gv

)=f ,

where

g := m , f := 0 for the mass conservation,

g := mvk , f := fk , k = 1, . . . , n for the momentum conservation.

We now choose test functions D(R Rn;R) which have a support in a



neighborhood of (t0, x0). We calculate

0 =

, t

(g

) div

(gv

)+f

=

t ,

g

+

,

gv

+

,f

=

(

R\{t0}(t)(t,

(t))g(t, (t)) dt

+

R\{t0}()(t, (t)) (gv)(t, (t))

= g(t, (t))(t)

dt

+

R\{t0}(t, (t))f(t, (t)) dt

)

=

R\{t0}

( ddt

((t, (t))

)g(t, (t)) + (t, (t))f(t, (t))

)dt

(now we integrate by parts)

=

R\{t0}(t, (t))

( d

dt

(g(t, (t))

)+ f(t, (t))

)dt

+

R\{t0}

d

dt

((t, (t))g(t, (t))

)dt

= (t0,

(t0))(g g+)

= (t0, x0)(g g+)

=

R\{t0}(t, (t))

( d

dt

(g(t, (t))

)+ f(t, (t))

)dt

+(t0, x0)(g g+) ,

whereg := lim

tt0g(t, (t)) , g+ := lim

tt0g(t, (t)) .

Since the test function is arbitrarily, it follows

d

dt

(g(t, (t))

)= f(t, (t)) for t 6= t0 and = 1, 2,g =

g+ .

This gives all the equations in (I3.9).

Gravity applied to space objects



Wir betrachten nun das Newtonsche Gravitationsgesetz (I2.12)

div([]) = []

fur die gesamte Masse . Wir stellen uns die Frage, wie als Kraft auf dieImpulserhaltung (I3.3)

t+ div(v) = 0 ,

t(v) + div(vvT +) = f

(I3.10)

wirkt. Es ist dies die Newtonsche Kraft(dichte), die fur f bedeutet

Newtons force density:

f = g

f force density,

(I3.11)

wobei hier angenommen wird, dass es die alleinige Kraft ist, im Allgemeinenkonnen noch andere Krafte wirksam sein. Die zugehorige Beschleunigungist

a = g . (I3.12)

Bemerkung: Let n=3. It is g = 4G with

G = 6.67384 1011 m3

kg s2(I3.13)

being the gravitational constant. Here is a list of some dimensions:

kg

mdiv , kg

m3

kgm2

tkg

m3s

a = g ms2

a , f , g kgm2s2

Wir denken uns nun die gesamte Massendichte aus disjunkten Teilmassen zusammengesetzt. Fur das Gravitationspotential gilt dann wegen derLinearitat des Gravitationsgesetzes

= , =

, div([]) = [] .

Da wir annehmen, dass die Teilmassen alle verschiedenen Trager haben,sagen wir disjunkte D R Rn fur den -Trager, konnen wir definieren

v = v + u und = in D,



wobei v die Bewegung des Himmelkorpers als Ganzes und u z.B. die lokaleRotationsbewegung ist. Es gilt damit nach (I3.10) fur jedes

div([]) = [] ,t + div(v + u) = 0 ,

t(v + u) + div(v vT + u v

T + vuT)

= g

div(uuT +) .

(I3.14)

Fig. 10: Die Umlaufbahnen der Objekte des Sonnensystems im Mastabaus [Wikipedia: Sonnensystem] (2-dimensionale Projektion)

Wir lassen nun die Teilkorper gegen Punktmassen konvergieren, also kon-vergiert fur alle

[] m punktweise in D (R Rn),v gleichmaig in Raum und Zeit R Rn.

Weiter konvergiert dann

[u] 0 punktweise in D (R Rn).


https://de.wikipedia.org/wiki/Sonnensystem


Aber es gibt Probleme mit dem Term , da hier beide Faktoren en-tarten, geht gegen einen Punkt und dort geht gegen unendlich, esgibt also keinen einfachen Limes. Jedoch gilt in der hier gegebenen Situa-tion, dass

[] div[uuT +] 0 punktweise in D (R Rn). (I3.15)

Siehe dazu 4.5 fur Kugeln und IV.16.1 fur rotierende kompressible Planeten.Setzen wir nun diese Resultate in (I3.14) ein, so lauten die Gleichungen imLimes

div([]) = m ,t(m) + div(mv) = 0 ,

t(mv) + div(mv vT) = gm(

: 6=

) .(I3.16)

Nach 3.1 sind diese Gleichungen aquivalent dazu, dass fur alle die Massem konstant ist, dass v(t, (t)) = (t) ist, und dass gilt

div([]) = m ,(t) = g

: 6=

(t, (t)) . (I3.17)

wobei wir die letzte Gleichung noch durch m dividiert haben. In diesemZusammenhang sei auf [19, N -body problem] verwiesen, wo der Einflussvon Planeten auf die Perihelbewegung mit der allgemeinen Formel (I3.17)numerisch gezeigt wird.

Momentum of a single planet

We consider now the sun system and assume that we are in the center ofgravity, hence

m(t) = 0 ,

and we orientate ourselves on stars in the surroundings. This is the reasonwhy we took only one f -term in (I3.11). Now the sun takes about 99.86%of the mass of the whole sun system (see Fig. 11), hence m


Fig. 11: Fotomontage zum Groenvergleich zwischen Erde (links) undSonne. Das Kerngebiet (Umbra) des groen Sonnenflecks hat etwa 5-fachenErddurchmesser aus [Wikipedia: Sonne]

Therefore the planet moves with t 7 (t) in a central gravitational field.In the following statement the potential of the sun 0 and the position ofthe planet have no index.

3.3 Keplers laws of planetary motion. The mass of the sun is concen-trated on the point {0}. The planet is modeled as a mass point {(t)} attime t with mass m and satisfies

m = f , f(t) = mg(t, (t)) , (I3.19)

where is the gravitational potential of the sun, given by

div([]) = m00 . (I3.20)

It is assumed that (t) 6= 0. Then (with some exceptions of one dimensionalmovement in the positive or negative direction to the sun) the equations ofKeplerian motion apply, that is, the movement is in a plane spanned by anorthonormal system {e1, e2} with the representation

(t) = r((t))(cos(t) e1 + sin(t) e2

)

andr() =

p

1 + e cos ,

=d

r()2, d2 = pGm0 > 0 .

The independent quantities are p > 0 and e. If |e| < 1 the planet makes aperiodic movement. (See also exercise 7.17.)

Proof. We have to solve the system (I3.18). The first differential equationis (I3.20), and with the boundary condition (t, x) 0 as |x| it hasthe solution

(t, x) =m04|x| hence (t, x) =

m04

x

|x|3 .


https://de.wikipedia.org/wiki/Sonne


Fig. 12: Elliptical orbits of stars at the galactic center. The massive blackhole is at coordinate (0,0). Star S2 has an orbital period of about 15 yearsfrom Department of Physics and Astronomy (California State L.A.). Seealso Fig. 14.

The second equation is (I3.19)

(t) = g(t, (t)) ,

i.e. with g = 4G

(t) = Gm0(t)

|(t)|3 . (I3.21)

By assumption (t) is non-zero. From the differential equation it followsthat (t) has at most finitely many zeros. So we can assume that (0) 6= 0and (0) 6= 0.1. Step. We show that we only need to treat the two-dimensional case.We denote with H that subspace which contains 0, (0), und (0). Wedecompose

(t) = x(t) H

+ y(t)H

.



Then y satisfies the differential equation

y(t) = Gm0y(t)

|(t)|3 , y(0) = 0, y(0) = 0

(|(t)|3 is in the denominator). Because of the homogeneous initial conditionit follows from the differential equation that y = 0. Consequently (t) H.Then H is a hyperplane provided (0) and (0) are linearly independent. Ifnot, then H is one-dimensional and (t) goes to 0 or infinity.2. Step. We assume that (I3.21) applies and that the motion is two-dimensional, hence without loss of generality t 7 (t) R2. Then we canintroduce locally in time polar coordinates

(t) = r((t))ei(t)

that means, r is a function of . Then with

c :=Gm0

one computes

c2

r2ei = c2 (t)|(t)|3 = =

d2

dt2(rei) =

d

dt((r + ir)e

i)

= ((r + ir) + 2(r r + 2ir ))ei

and therefore

(r + ir) + 2(r r + 2ir ) =

c2

r2.

Real part and imaginary part result in the two equations

r + 22r = 0 ,

r + 2(r r) =

c2

r2.

(I3.22)

3. Step. Solution of the first equation in (I3.22).If 6= 0, the first equation can be written as

+ 2

r

r= 0 ,

thusd

dt(log ||+ 2log r()) = 0 ,

therefore with a constant d 6= 0

=d

r()2. (I3.23)



Fig. 13: All planets move in elliptical orbits, with the sun at one focusfrom hyperphysics.phy-astr.gsu.edu/hbase/kepler.html

4. Step. Solution of the second equation in (I3.22).We get from (I3.23)

= 2dr()3

r () = 2r r2

If we plug this into the second equation, we obtain

c2

r2= r +

2(r r) = 2(

2r2r

+ r r),

that means with (I3.23)

c2

d2r2 = r

2r2r r . (I3.24)

This is a ordinary differential equation of second order in 7 r(). If wenow set, with p 6= 0 and a given e R,

r() =p

1 + e cos , (I3.25)

then it is

r =p e sin

(1 + e cos)2=e sin

pr2

r =e cos

pr2 +

2e sin

pr r

=(pr 1)r2p

+2e2 sin 2

p2r3

= r r2

p+

2e2 sin 2

p2r3 = r r

2

p+

2r2r



Fig. 14: Left: An image of Sgr A and S2 from the 8.2m VLT YEPUNtelescope at the ESO Paranal Observatory. Right: The orbit of S2 aroundSgr A, highlighting the last close encounter, in 2002 from ESO. The nextencounter of S2 with Sgr A will occur in 2018 from www.chandra.si.eduChandra X-Ray Observatory. See also Fig. 12.

and thus becomes (I3.24) to

c2

d2r2 = r

2r2r r = r

2

p.

This is equivalent to the condition

p =d2

c2. (I3.26)

The equations (I3.23), (I3.25) and (I3.26) are the Kepler motion (siehe [19,Keplers laws]).

Sogar im Zentrum der Milchstrasse ist die Bewegung der Sterne um das Zen-trum (Schwarzes Loch, SgrA) nahe einer Kepler Bewegung, wie Fig. 12und Fig. 14 zeigt. Eine Abweichung davon ist wie bei den Planeten des Son-nensystems (insbesondere die Bewegung des Merkur) eine Perihelbewegung,verursacht durch die Gravitation der ubrigen Sterne (siehe [19, N -body prob-lem]). Die Bestimmung der Position der Sterne von Aufnahmen derselbenist nichttrivial, die Schritte, die dabei gebraucht werden, sind in Gillessen



et. al. [38] dargestellt, see also [39], wo auch eine Abschatzung der Massevon SgrA gegeben wird.

Collection of mass points

We now consider a collection of mass points (or particles) with interactingforces. We obtain in the momentum conservation a special matrix , thatmeans the interacting forces have in part a form which allows them to bereformulated, so that they are expressed as a term under the divergenceoperator (see [15, 2.2 and 2.4] and [18, section 7]).

3.4 Multiple mass points. We consider N mass points with mass m atthe position t 7 (t) for = 1, . . . , N . They should satisfy the ordinarydifferential equation

m(t) =

: 6=

F((t) (t)

)+ f(t) (I3.27)

in t, where mappings F :Rn \ {0} Rn for 6= are given with

F(z) = F(z) . (I3.28)

We consider a t-region in which the mass points are disjoint, i.e. they do notmeet (see REFenergy.???). With velocities v(t,

(t)) = (t) the followingmass-momentum equations hold

t

(m

)+ divx

(mv

)= 0 ,

t

(mv

)+ divx

(mv v

T

12

,: 6=

F( )

(

)T,

)=f .

Here the distributions and , are given for test functions by

,

:=

R

(t, (t)) dt (as in (I2.7)),

, ,

:=

R

1

0(t, (1 s)(t) + s(t)) ds dt .

We see that in the ordinary differential equation m = f the force term

f := : 6=

F(

)+ f

consists of two terms, where the first one can be written as internal term

12

,: 6=

F( )

(

)T,



in the flux of the momentum equation. Therefore we call

: 6=

F(

)

an internal force density.

Proof. For a single mass point, the mass m is constant and it holds

m = f :=

: 6=

F(

)+ f .

Therefore follows as in 3.1 that this is equivalent to

t(m) + divx(mv) = 0 ,

t(mv) + divx(mv vT) = f

.

By forming the sum, of course, it follows

t

(m

)+ divx

(mv

)= 0 ,

t

(mv

)+ divx

(mv v

T)=f .

It remains to rewrite the force term. (This manipulation is not so apparentin the existing literature.) Now

f =

F +

f

whereF :=

: 6=

F( ) .

We show that the F-expressions can be written as term under the diver-gence, hence we call it an internal force term. Thus we have to show thatfor D(R Rn;Rn)

, F

= D , M = , divxM ,

M :=1

2

,: 6=

F,( )

(

)T, .

(I3.29)

To prove this, we let F := 0 and F := F( ) for 6= , so that

by assumption (I3.28)

F( ) = F( ) for 6= ,



or F = F for all and . Thus we obtain

, F

=,

, F,

=,

R

(t, (t))F,(t) dt

=,

R

1

2

((t, (t))F,(t) + (t, (t))F,(t)

)dt

=,

R

1

2

((t, (t)) (t, (t))

)F,(t) dt

=1

2

,

R

1

0

(D(t, (1 s)(t) + s(t))((t) (t))

)F,(t) ds dt

(since (Dz1)z2 = D(z2 z1T), D matrix, z1, z2 vectors)

=1

2

,

R

1

0D(t, (1 s)(t) + s(t))

(F,(t)(

(t) (t))T)ds dt

=1

2

,

D , F, (

)T,= D , M

=j,k

jk , Mkj = j,k

k , jMkj = , divxM .

Consequently the assertion follows.

Hence we have seen, that the mass and momentum conservation plays anessential role even for models with particles. We have also seen how New-tons mechanics, if it is interpreted using distributions, is part of continuummechanics.

Navier-Stokes equation

Now we come back to the general systen of conservation laws (I3.3). Thissystem models the motion of fluids, where is given by = pIdS with apressure p and a stress tensor S. The stress tensor is assumed to be linear inDv. The force term f is now usually an external force. (These properties arederived in Chapter II, and the inequalities in (I3.30) follow from the entropyprinciple, see Chapter III.) For a single fluid we dont need a distributionalmass-momentum law, and with J = 0, r = 0, we obtain the following:



(Compressible) Navier-Stokes equations:

t+ div(v) = 0 ,

t(v) + div(vvT +) = f ,

where: = pId S pressure tensor, p pressure,S = (Dv +(Dv)T) + ( divv) Id

= 2(Dv)S + ( divv) Id tension tensor,

> 0 and +2

n 0 Lame coefficients,

and: f force density.

(I3.30)

It should be noted15 that the representation of the pressure tensor andthe tension S is a consequence of II.4.12. It is

S = 2(Dv)S + ( divv) Id

= 2 ( (Dv)S 1ndiv(v)Id

trace free

) + (+2

n) div(v)Id ,

hence the positivity of the coefficients becomes clear. This positivity wewill later in the connection with the entropy principle study in detail (seee.g. III.2.5). The pressure p is a function of , among other things, if weregard a compressible fluid (see Section IV.2). Another version of the mo-mentum balance comes from the following computations, which are true ingeneral,

t(v) + div(vvT)

= ( t+ div(v) = 0

)v + (tv + vv) ,

anddiv = div(pId S) = p divS .

15 Lame introduces his coefficients in the theory of elasticity, to use his name here is byanalogy of the physical terms, see [Wikipedia: Lame parameters].


https://en.wikipedia.org/wiki/Lam\unhbox \voidb@x \bgroup \let \unhbox \voidb@x \setbox \@tempboxa \hbox {e\global \mathchardef \accent@spacefactor \spacefactor }\accent 19 e\egroup \spacefactor \accent@spacefactor _parameters


Thus the (compressible) Navier-Stokes equations are equivalent to

(Compressible) Navier-Stokes equations:

t+ div(v) = 0 ,

(tv + vv) +p divS = fp pressure (e.g. a function of ),

S = (Dv +(Dv)T) + ( divv) Id ,

f force density, for the other quantities see (I3.30).

(I3.31)

We test the equations by looking at the example of a centrifuge.

3.5 Centrifuge. We model a centrifuge by an infinitely long pipe in Rn,n = 3, that is {x R3 ; |(x1, x2)| < R}, and we denote by r the distancefrom the axis r =

x21 + x

22. We consider stationary solutions, that

means solutions that do not depend on the time variable.16 We make theansatz

= (r) , p = p(r) , f = 0 ,

v(x) = (x2, x1, 0) .(I3.32)

Then the compressible Navier-Stokes equation (see (I3.31)) in the stationarycase is equivalent to the differential equation

rp = 2r . (I3.33)

That is, the pressure increases with the radius r.

Proof. The stationary equations (see (I3.31)) are

div(v) = 0 ,

vv +p divS = 0 .(I3.34)

It is

v =

x2x10

also Dv =

0 1 01 0 00 0 0

,

so (Dv)S = 0 and divv = 0, hence S = 0. Moreover

(p(r)) = rpr =rp

r(x1, x2, 0) .

It followsdiv(v) = divv + ()v

= r

r(x1, x2, 0)(x2, x1, 0) = 0

16This is the general definition of stationary.


I.3 Conservation of momentum

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