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Technical University Munich 2017
Lectures on
MathematicalContinuum Mechanics
Prof. Dr. H.W. Alt
Version: 20171116 Last major change: 07.02.2017
Copyright 2011-2017 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.
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To my parents
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Contents
I Mass and momentum 81 Conservation laws . . . . . . . . . . . .
. . . . . . . . . . . . 112 Distributions . . . . . . . . . . . . .
. . . . . . . . . . . . . . 243 Conservation of momentum . . . . .
. . . . . . . . . . . . . . 444 Interfaces . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . 695 Change of coordinates . . . .
. . . . . . . . . . . . . . . . . . 816 Reference coordinates . . .
. . . . . . . . . . . . . . . . . . . 957 Exercises . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . 105
II Objectivity 1111 Classical observers transformations . . . .
. . . . . . . . . . . 1142 Lorentz transformations . . . . . . . .
. . . . . . . . . . . . . 1193 Objectivity of balance laws . . . .
. . . . . . . . . . . . . . . 1314 Constitutive relations . . . . .
. . . . . . . . . . . . . . . . . 1475 Angular momentum . . . . . .
. . . . . . . . . . . . . . . . . 1666 Exercises . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . 174
IIIEnergy and entropy 1791 Entropy inequality . . . . . . . . .
. . . . . . . . . . . . . . . 1842 Energy equation . . . . . . . .
. . . . . . . . . . . . . . . . . 1933 Mixtures . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . 2044 Lagrange multipliers .
. . . . . . . . . . . . . . . . . . . . . . 2095 Dissipation
inequality . . . . . . . . . . . . . . . . . . . . . . 2136
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
217
IVVarious applications 2181 Tidal period . . . . . . . . . . . .
. . . . . . . . . . . . . . . . 2182 Fluids and gases . . . . . . .
. . . . . . . . . . . . . . . . . . 2303 Navier-Stokes equation . .
. . . . . . . . . . . . . . . . . . . . 2424 Eulers equation . . .
. . . . . . . . . . . . . . . . . . . . . . 2535 Nonlinear
elasticity . . . . . . . . . . . . . . . . . . . . . . . . 2746
Tissue growth . . . . . . . . . . . . . . . . . . . . . . . . . . .
2817 Sound waves . . . . . . . . . . . . . . . . . . . . . . . . .
. . 287
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4
8 Vorticity . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . 3069 Fractionation . . . . . . . . . . . . . . . . . . . .
. . . . . . . 32410 Reaction-diffusion systems . . . . . . . . . .
. . . . . . . . . . 32711 Combustion (Temperature dependent
diffusion) . . . . . . . . 34912 Reactions in biology . . . . . . .
. . . . . . . . . . . . . . . . 36313 Chemical reactions . . . . .
. . . . . . . . . . . . . . . . . . . 37014 Prandtls boundary layer
. . . . . . . . . . . . . . . . . . . . . 38315 Self-gravitation .
. . . . . . . . . . . . . . . . . . . . . . . . . 39016 Exercises .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 409
V Higher moments 4101 Cattaneos 8-Momente Gleichung . . . . . .
. . . . . . . . . . 4112 Boltzmann Gleichung . . . . . . . . . . .
. . . . . . . . . . . 4183 Die Chapman-Enskog Hierarchie . . . . .
. . . . . . . . . . . 4264 Grads 13-Momente Gleichung . . . . . . .
. . . . . . . . . . . 435
VI Speed of light 4401 Elektrodynamik . . . . . . . . . . . . .
. . . . . . . . . . . . . 4422 Ubungen . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . 451
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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 [15] 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
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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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. Das wird klar insbesondere bei der
mathematischen Voraus-setzungen der Parabolizitat. Es zahlen aber
auch Gleichungen zu den Vo-raussetzungen, die keine
Entropieproduktion bewirken, aber durch das En-tropieprinzip
erzwungen werden.
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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.
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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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. To it 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 characteristic of continuum
physics.
We will first consider the conservation laws for C1-functions,
thus we write
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.
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I. Mass and momentum 9
This follows from the formulation for test volumes using the
Gausss theo-rem, as one sees 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 Gaussian domain V is arbitrary, we obtain the
differential equation(I0.1). By the way the formulation with test
volumes follows from the strongdifferential equation by reversing
the conclusions.
Fig. 2: Bedeutung der Distributionen (aus [61])
However, many specific functions are not classical solutions of
the differ-ential equation, for example, the earths gravitational
field at the (earths)surface. In this case, the formulation with
test volumes becomes more com-plex. Therefore, we use test
functions instead of test volumes. On the space
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I. Mass and momentum 10
of test functions
D(U) := { C(U) ; has compact support in U} ,
where U RRn is an open set, we consider distributions U , Q, and
R inD (U) (see the section 2), so that
tU + divQ = R in D(U) . (I0.2)
This formulation of (I0.1) with test functions has the advantage
that it ismore general and much easier. This becomes particularly
clear when onegoes to descriptions of conservation laws on
surfaces. Both representationsare common, the representation of a
conservation law using test volumescan be found usually in physics
books. And both 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.
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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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 specify the exact mathematical assumptions as
differentia-bility or partial differentiability.
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) .
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I.1 Conservation laws 12
(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).
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.1 Conservation laws 13
(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)
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I.1 Conservation laws 14
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)
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.1 Conservation laws 15
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 .
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I.1 Conservation laws 16
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
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I.1 Conservation laws 17
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.13).
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).
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.1 Conservation laws 18
Fig. 4: From [Wikipedia: Atmosphere of earth] [90]
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
,
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.1 Conservation laws 19
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
].
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.1 Conservation laws 20
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)
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I.1 Conservation laws 21
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.
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I.1 Conservation laws 22
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)).
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.1 Conservation laws 23
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.
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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.
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.2 Distributions 25
Fig. 5: Functions as functionals (see [61])
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 .
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I.2 Distributions 26
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 [62].
MathematischeEinfuhrungen werden fur N = 1 in [61], fur beliebiges
N in [56, in Ab-schnitt 3], [58, Kapitel I-II], [59, Kapitel 1-9],
[60, Kapitel 1-2] gegeben. Ichhabe auch ein eigenes Skript [55]
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 [56, 3.19] and[55, section 6] where also the
completeness of D (U) is discussed.
Proof of equivalence: The statements (1) and (2) are equivalent.
For example see [56, 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 [55, 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 .
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.2 Distributions 27
(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 [55, 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 [55, 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.
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I.2 Distributions 28
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)
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I.2 Distributions 29
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.Then it follows from the distributional mass conservation that
this mass hasto be constant. This proves that the distributional
conservation law is theright 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
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.2 Distributions 30
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)
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I.2 Distributions 31
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 [16, 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([]) = [] ,
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I.2 Distributions 32
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).
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.2 Distributions 33
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)
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
16
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I.2 Distributions 34
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,
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
16
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I.2 Distributions 35
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 ,
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.2 Distributions 36
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
-
I.2 Distributions 37
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).
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.2 Distributions 38
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
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
16
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I.2 Distributions 39
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, 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).
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.2 Distributions 40
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
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
16
http://www-m6.ma.tum.de/~alt/QUELLEN/GRACE_globe_animation.gif
-
I.2 Distributions 41
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.
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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https://en.wikipedia.org/wiki/Gravity_of_Earth
-
I.2 Distributions 42
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 coincides with the
solution obtainedif one sets the star as a point mass with the same
total mass. In thenext section 3 we will consider the conservation
of momentum and we willshow that in the stationary incompressible
case homogeneous stars producea gravitational field like the one
here (see 4.5). In the compressible case, werefer to section IV.15,
where radially symmetric mass distributions of starsare
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
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.2 Distributions 43
for D(R Rn). Further, let
s :=m
Hn1(BR(0))
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)))
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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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.
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.3 Conservation of momentum 45
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 .
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
16
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I.3 Conservation of momentum 46
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. The first differential equation in (I3.4), that is the
mass conservation,is treated as in 2.8. This results in the
equations (I3.5). Further, for thesecond equation then it applies
for all D(R Rn;Rn) that
, t(mv) divx(mvvT) + f
=k
k , t(mvk) divx(mvkv) + fk
=k
(tk , mvk
+k , mvkv
+k , fk
)
=k
R
m vk(t, (t)) = k(t)
(tk + vk)(t, (t)) =
d
dtk(t, (t))
dt
+k
R
(kfk)(t, (t)) dt
=k
R
( ddt
(mk(t)) + fk(t, (t))
)k(t, (t)) dt .
Then the assertion follows.
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.3 Conservation of momentum 47
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
(I3.7)
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 asthe 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),
(I3.8)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. It canalso happen that there are several
particles after the collision (as in Fig. 9),which leads to
corresponding formulas.
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 to
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.3 Conservation of momentum 48
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)
consider 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 (I3.7), 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.
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
16
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I.3 Conservation of momentum 49
We now choose test functions D(R Rn;R) which have a support in
aneighborhood 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 (I3.8).
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.3 Conservation of momentum 50
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
(I3.9)
wirkt. Es ist dies die Newtonsche Kraft(dichte), die fur f
bedeutet
Newtons force density:
f = g
f force density,
(I3.10)
wobei hier angenommen wird, dass es die alleinige Kraft ist, im
Allgemeinenkonnen noch andere Krafte wirksam sein. Die zugehorige
Beschleunigungist
a = g . (I3.11)
Bemerkung: Let n=3. It is g = 4G with
G = 6.67384 1011 m3
kg s2(I3.12)
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
Soweit eine gemeinsame Massendichte. Wir denken uns nun die
gesamteMassendichte aus disjunkten Teilmassen zusammengesetzt. Fur
dasGravitationspotential gilt dann wegen der Linearitat des
Gravitationsgeset-zes
= , =
, div([]) = [] .
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.3 Conservation of momentum 51
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 (I3.9) fur
jedes
div([]) = [] ,t + div(v + u) = 0 ,
t(v + u) + div(v vT + u v
T + vuT)
= g
div(uuT +) .
(I3.13)
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.
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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https://de.wikipedia.org/wiki/Sonnensystem
-
I.3 Conservation of momentum 52
Weiter konvergiert dann
[u] 0 punktweise in D (R Rn).
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). (I3.14)
Siehe dazu 4.5 fur Kugeln und IV.15.1 fur rotierende
kompressible Planeten.Setzen wir nun diese Resultate in (I3.12)
ein, so lauten die Gleichungen imLimes
div([]) = m ,t(m) + div(mv) = 0 ,
t(mv) + div(mv vT) = gm(
: 6=
) .(I3.15)
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)) . (I3.16)
wobei wir die letzte Gleichung noch durch m dividiert haben. In
diesemZusammenhang sei auf [16, N -body problem] verwiesen, wo der
Einflussvon Planeten auf die Perihelbewegung mit der allgemeinen
Formel (I3.15)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 (I3.10). Now the sun
takes about 99.86%of the mass of the whole sun system (see Fig.
11), hence m
-
I.3 Conservation of momentum 53
Fig. 11: Fotomontage zum Groenvergleich zwischen Erde (links)
undSonne. Das Kerngebiet (Umbra) des groen Sonnenflecks hat etwa
5-fachenErddurchmesser aus [Wikipedia: Sonne]
In the following statement the potential of the sun 0 and the
position ofthe planet have no index. The planet moves with t 7 (t)
in a centralgravitational field.
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)) , (I3.18)
where is the gravitational potential of the sun, given by
div([]) = m00 . (I3.19)
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 (I3.16). The first
differential equationis (I3.18), and with the boundary condition
(t, x) 0 as |x| it hasthe solution
(t, x) =m04|x| hence (t, x) =
m04
x
|x|3 .
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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https://de.wikipedia.org/wiki/Sonne
-
I.3 Conservation of momentum 54
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 (I3.17)
(t) = g(t, (t)) ,
i.e. with g = 4G
(t) = Gm0(t)
|(t)|3 . (I3.20)
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
.
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.3 Conservation of momentum 55
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 (I3.19) 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.
(I3.21)
3. Step. Solution of the first equation in (I3.20).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. (I3.22)
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.3 Conservation of momentum 56
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 (I3.20).We get from
(I3.21)
= 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 (I3.21)
c2
d2r2 = r
2r2r r . (I3.23)
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 , (I3.24)
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
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.3 Conservation of momentum 57
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 (I3.22) to
c2
d2r2 = r
2r2r r = r
2
p.
This is equivalent to the condition
p =d2
c2. (I3.25)
The equations (I3.21), (I3.23) and (I3.24) are the Kepler motion
(siehe [16,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 [16, N -body prob-lem]). Die Bestimmung der Position
der Sterne von Aufnahmen derselbenist nichttrivial, die Schritte,
die dabei gebraucht werden, sind in Gillessen
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.3 Conservation of momentum 58
et. al. [34] dargestellt, see also [35], 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 [12, 2.2 and 2.4] and [15,
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) (I3.26)
in t, where mappings F :Rn \ {0} Rn for 6= are given with
F(z) = F(z) . (I3.27)
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,
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.3 Conservation of momentum 59
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, .
(I3.28)
To prove this, we let F := 0 and F := F( ) for 6= , so that
by assumption (I3.26)
F( ) = F( ) for 6= ,
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.3 Conservation of momentum 60
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 (I3.28) 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:
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.3 Conservation of momentum 61
(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.
(I3.29)
It should be noted that the representation of the pressure
tensor and thetension S is a consequence of II.4.15. 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 (see
III.2.5). The pres-sure p is a function of among other things, if
we regard a compressiblefluid (see Section IV.2). Another version
of the momentum balance comesfrom the following computations, which
are true in general,
t(v) + div(vvT)
= ( t+ div(v) = 0
)v + (tv + vv) ,
anddiv = div(pId S) = p divS .
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 (I3.28).
(I3.30)
We test the equations by looking at the example of a
centrifuge.
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.3 Conservation of momentum 62
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.14 We
make theansatz
= (r) , p = p(r) , f = 0 ,
v(x) = (x2, x1, 0) .(I3.31)
Then the compressible Navier-Stokes equation (see (I3.29)) in
the stationarycase is equivalent to the differential equation
rp = 2r . (I3.32)
That is, the pressure increases with the radius r.
Proof. The stationary equations (see (I3.29)) are
div(v) = 0 ,
vv +p divS = 0 .(I3.33)
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
and the momentum equation becomes
vv +p = 0 .
We compute 15
vv =3
i=1vixiv
= 2(x2e2 x1e1) = 2(x1, x2, 0) ,
hence it follows that
0 = vv +p = (2 + rpr
)(x1, x2, 0) ,
14This is the general definition of stationary.15It is ei =
(ij)j=1,...,n the i-th basis vector of R
n.
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.3 Conservation of momentum 63
that is1
rrp =
2 .
3.6 Different materials. We discuss different constitutive
relations for pand its consequences for the identity (I3.31).
(1) Let = const > 0 and let p be an arbitrary free variable.
Then it followsfrom (I3.31) that
p =2
2r2 + const.
(2) Let p = c with > 1. Then (I3.31) is
=
(( 1)2
2cr2 + const
) 11
(3) Let p = c. Then (I3.31) is
= exp
(2
2cr2 + const
).
(4) Let p = f () f() with a given function f . Then (I3.31)
is
f () =2
2r2 + const.
(5) Let p = c1+ c2 mit > 1. Then p is as in (4) if
f() = c1(log 1) +c2
1 + const c0 .
Hence (I3.31) is equivalent to the formula in (4), where f is a
convex functionand f monotone increasing.
Remark: The function f is the internal free energy, see Section
III.5.Constitutive functions you also find in [31, 1.3.2 and
1.4].
Proof (1). This follows by integrating (I3.31).
Proof (2). It is
2r =c
r(
) = c2r = r(h()) ,
ifh () = c
2 hence h() = c
11 + const .
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.3 Conservation of momentum 64
Therefore
r
(h()
2
2r2)
= 0 ,
that meansc
11 + const = h() =
2
2r2 + const.
From this it follows the result.
Proof (3). It is
2r =c
r = c r(log )
and thus
r
(c log
2
2r2)
= 0 ,
that is,
log =2
2cr2 + const.
Proof (4). It is p = f , and thus
rp = p r = f ()r .
Since rp = 2r, we have
f ()r = 2r .
That means
r
(f ()
2
2r2)
= 0 ,
and from this follows the assertion.
Proof (5). It is
f () = c1log +c2
11 + const,
hencef () f() = c1log +
c2
11 + const.
In many applications it is assumed that the fluid is
incompressible, that is, = 0 = const > 0. In this case, the mass
conservation reduces to
0 = t+ div(v) = 0 divv ,
author: H.W. Alt title: Continuum Mechanics time: 2017 Nov
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I.3 Conservation of momentum 65
hence it is divv = 0 and, with (Dv)S = 12(Dv +(Dv)T
),
S = 2(Dv)S + divv= 0
Id = 2(Dv)S .
Thus we obtain from (I3.29) the following for the incompressible
Navier-Stokes equations
Incompressible Navier-Stokes equation:
divv = 0 ,
0(tv + vv) +p divS = f
0 > 0 constant, v velocity, p pressure,
S = (Dv +(Dv)T) = 2(Dv)S tension tensor,
> 0 Lame coefficient, f force density.
(I3.34)
If in addition = const, then
2 div((Dv)S) =
(n
j=1j(jvk + kvj)
)
k=1,...,n
= v +
(n
j=1kjvj
)
k=1,...,n
= v +( divv) ,
thereforedivS = div(2(Dv)S) = 2 div((Dv)S) = v .
Hence, if = const, then (I3.33) is equivalent to
Special Navier-Stokes e