Note on the descriptions of the Euler-Poisson equations in various co-ordinate systems

0.37 MB·

by  Tetu Makino

Note on the descriptions of the Euler-Poisson
7
1 equations in various co-ordinate systems
0
2
r Tetu Makino ∗
a
M
March 17, 2017

6
1

] Abstract
P
In this note we derive thedescriptions of thesystem of Euler-Poisson
A
equations which governs the hydrodynamic evolution of gaseous stars in
.
h various co-ordinate systems. This note does not contain essentially new
t resultsforastrophysicists,butmathematicallyrigorousderivationscannot
a
be found in the literatures written by physicists so that it will be useful
m
to prepare details of rather stupidly honest derivations of the equations
[ in various flames when we are going to push forward the mathematical

4 research of theproblem.
v
0 Key words and phrases. Euler-Poisson equations, gaseous star, ax-
7 isymmetric solution, stellar oscillation, Lagrangian displacement.
3
6 2010MathematicsSubjectClassificationNumbers. 35L05,35L12,35L57,
0 76L10.
.
1
0
1 Euler-Poisson equations
7
1
: 1.1 Euler-Poisson equations and Newton potential
v
i
X The Euler-Poissonequations which governevolutions of a gaseous star are
r
a ∂ρ
+ (ρ~v)=0 (1.1a)
∂t ∇·
∂~v
ρ +(~v )~v + P = ρ Φ (1.1b)
∂t ·∇ ∇ − ∇
(cid:16)Φ=4πGρ. (cid:17) (1.1c)
△
The independent variable is (t,~x)=(t,x1,x2,x3) [0,T) R3. G is a positive
∈ ×
constant. The unknown functions are the density field ρ=ρ(t,x), the pressure
field P =P(t,x), the gravitational potential Φ=Φ(t,x), and the velocity field
~v =(v1,v2,v3)(t,x).
∗ProfessorEmeritusatYamaguchi University,Japan. e-mail: [email protected]
1
We are using the usual notations
3
∂
(ρ~v)= (ρvk),
∇· ∂xk
k=1
X
3 ∂vj
(~v )vj = vk , j =1,2,3,
·∇ ∂xk
k=1
X
∂Q ∂Q ∂Q
Q= , , for Q=P,Φ
∇ ∂x1 ∂x2 ∂x3
(cid:16)3 ∂2Φ (cid:17)
Φ= .
△ (∂xk)2
k=1
X
Also, we use the notation
3
D ∂ ∂ ∂
= +~v = + vk , (1.2)
Dt ∂t ·∇ ∂t ∂xk
k=1
X
which rewrite (1.1a), (1.1b) as
Dρ
+ρ ~v =0,
Dt ∇·
D~v
ρ + P = ρ Φ.
Dt ∇ − ∇
We assume that P is a given smooth function of ρ > 0 such that P >
0,dP/dρ>0 for ρ>0 and
P =Aργ(1+[ργ−1] ) (1.3)
1
as ρ +0, where A,γ are positive constants and 1<γ 2. Here [X] denotes
1
→ ≤
a convergent power series of the form a Xk. Of course we consider P = 0
k
k≥1
X
for ρ=0.
Definition 1 A solution ρ = ρ(t,~x),~v = ~v(t,~x),Φ = Φ(t,~x) will be called a
compactlysupportedclassicalsolutionifρ,~v,Φ C1([0,T) R3),Φ(t, )
C2(R3), ρ 0 everywhere, and the support of ρ(t, ) i∈s compact f×or t [0,·T)∈.
≥ · ∀ ∈
Without loss of generality we assume~v(t, ) is bounded on R3, since any modi-
·
fication of~v outside the support of ρ is free.
2
For any compactly supported classical solution the Laplace equation (1.1c)
can be solved by the Newton potential
ρ(t,~x′)
Φ(t,~x)= G d (~x′). (1.4)
− ~x ~x′ V
Z | − |
Here d (~x) denotes the usual volume element dx1dx2dx3, and
V
~x = (x1)2+(x2)2+(x3)2
| |
for ~x=(x1,x2,x3)T. p
Inthisnote,weshallspecifythesolutionΦof(1.1c)bythisNewtonpotential
(1.4) for any compactly supported classical solution.
1.2 Conservation of mass, energy and angular momentum
It is well known that the total mass
M := ρ(t,~x)d (~x) (1.5)
V
Z
and the total energy
1 1
E := ρ~v 2+Ψ(ρ)+ ρΦ d (~x)
2 | | 2 V
Z (cid:16)1 (cid:17) G ρ(t,~x)ρ(t,~x′)
= ρ~v 2+Ψ(ρ) d (~x) d (~x)d (~x′) (1.6)
2 | | V − 2 ~x ~x′ V V
Z (cid:16) (cid:17) Z Z | − |
areconstantswithrespecttotalonganycompactlysupportedclassicalsolution.
Here the state quantities u=u(ρ) (enthalpy) and Ψ(ρ) is defined by
ρ dP ρ
u= , Ψ(ρ)= udρ. (1.7)
ρ
Z0 Z0
Remark 1 Note that
Aγ Aργ P
u= ργ−1, Ψ(ρ)= = ,
γ 1 γ 1 γ 1
− − −
when P =Aργ exactly.
Bywayofprecaution,letusverifytheconservationofthetotalmassM and
the total energy E.
First (1.1a) can be written as
∂ρ
= (ρ~v);
∂t −∇·
Hence
dM
= (ρ~v)d
dt − ∇· V
Z
= (ρ~v N~)d =0,
− | S
Z|~x|=R
3
where the support of ρ(t, ) is supposed to be included in = ~x < R
· D {| | }
and N~,d denote the outer normal vector and the surface area element of the
S
boundary ∂ = ~x =R ; This shows that dM/dt=0.
D {| | }
Next, (1.1a)(1.1b) imply
d 1 1 ∂
ρ~v 2d = ρ (vk)2 d
dt 2 | | V 2 ∂t V
Z Z (cid:16) Xk (cid:17)
1
= ρ vk∂ vk+ ∂ ρ (vk)2
t t
2 V
Z (cid:16) Xk Xk (cid:17)
= ρ vkvj∂ vk+ vk∂ P + ρvk∂ Φ+
j k k
−
Z (cid:16) Xj,k Xk Xk
1
+ ∂ (ρvj) (vk)2 d
j
2 V
Xk (cid:17)
1
= ρ vj∂ (vk)2+ vk∂ P +ρ vk∂ Φ+
j k k
− 2
Z (cid:16) Xj,k Xk Xk
1
+ ∂ (ρvj) (vk)2 d
j
2 V
Xj Xk (cid:17)
1
= ∂ ρvj(vk)2 (vk∂ P +ρvk∂ Φ)d
j k k
− 2 − V
Z Xj,k h i Z Xk
= ( ρvk∂ u+ ρvk∂ Φ)d
k k
− V
Z k k
X X
∂Ψ
= + ρvk∂ Φ d
k
− ∂t V
Z (cid:16) Xk (cid:17)
d
= Ψd + ∂ (ρvk)Φd
k
−dt V V
Z Z k
X
d ∂ρ
= Ψd Φd ;
−dt V − ∂t V
Z Z
However
∂ρ ρ(t,~x′)
Φd = G ∂ ρ(t,~x) d (~x)d (~x′)
∂t V − t ~x ~x′ V V
Z Z Z | − |
G d ρ(t,~x)ρ(t,~x′)
= d (~x)d (~x′)
−2 dt ~x ~x′ V V
Z | − |
1 d
= ρΦd ;
2dt V
Z
This shows dE/dt = 0. Here we put ~x = (x1,x2,x3) and ~v = (v1,v2,v3), and
∂ ∂
denote ∂ = ,∂ = , while k,j run 1,2,3.
t ∂t j ∂xj
4
Moreover the angular momentum
J~:= ~x (ρ~v)d (~x) (1.8)
× V
Z
is constant with respect to t along any compactly supported classical solution.
Here
x2v3 x3v2 v1
−
~x ~v = x3v1 x1v3 for ~v = v2 .
×  −   
x1v2 x2v1 v3
−
   
Let us show it. Note that (1.1b) can be written, under (1.1a), as
∂ ∂ ∂ ∂ ∂P ∂Φ
(ρv1)+ (ρ(v1)2)+ (ρv1v2)+ (ρv1v3)+ = ρ (1.9a)
∂t ∂x1 ∂x2 ∂x3 ∂x1 − ∂x1
∂ ∂ ∂ ∂ ∂P ∂Φ
(ρv2)+ (ρv2v1)+ (ρ(v2)2)+ (ρv2v3)+ = ρ (1.9b)
∂t ∂x1 ∂x2 ∂x3 ∂x2 − ∂x2
∂ ∂ ∂ ∂ ∂P ∂Φ
(ρv3)+ (ρv1v3)+ (ρv2v3)+ (ρ(v3)2)+ = ρ (1.9c)
∂t ∂x1 ∂x2 ∂x3 ∂x3 − ∂x3
Then we have
dJ1
= (~x Φ)xρd (~x),
dt − ×∇ V
Z
and so on, for J =(J1,J2,J3)T. Here
x2 ∂Φ x3 ∂Φ
∂x3 − ∂x2
(~x Φ)1
×∇  
~x Φ= (~x Φ)2 = x3 ∂Φ x1 ∂Φ
×∇  ×∇  ∂x1 − ∂x3
(~x Φ)3  
×∇  
  x1 ∂Φ x2 ∂Φ
 ∂x2 − ∂x1
 
On the other hand, the differentiation of the Newton potential (1.4) gives
∂Φ x3 (x3)′
=G − ρ(t,(x1)′,(x2)′,(x3)′)d (~x′),
∂x3 ~x ~x′ 3 V
Z | − |
and so on. Hence we see
dJ1 x2(x3 (x3)′)+x3(x2 (x2)′)
=G − − − ρ(t,~x)ρ(t,~x′)d (~x)d (~x′)
dt ~x ~x′ 3 V V
Z Z | − |
x2(x3)′
=G ρ(t,~x)ρ(t,~x′)d (~x)d (~x′)
~x ~x′ 3 V V
Z Z | − |
x3(x2)′
G ρ(t,~x)ρ(t,~x′)d (~x)d (~x′)
− ~x ~x′ 3 V V
Z Z | − |
=0.
By the same manner we can show dJ2/dt=dJ3/dt=0.(cid:4)
5
2 Axisymmetric solutions
2.1 Co-ordinate system (̟,φ,z)
Let (̟,φ,z) be the cylindrical coordinates defined by
x1 =̟cosφ, x2 =̟sinφ, x3 =z. (2.1)
(This somewhat clumsy notation with ̟ is historically standard for this prob-
lem.)
Note that, while the polar co-ordinates are
x1 =rsinθcosφ, x2 =rsinθsinφ, x3 =rcosθ,
we are taking ̟ =rsinθ =√r2 z2.
−
We have
∂ x1 ∂ x2 ∂
= ,
∂x1 ̟ ∂̟ − ̟2∂φ
∂ x2 ∂ x1 ∂
= + ,
∂x2 ̟ ∂̟ ̟2∂φ
∂ ∂
= ,
∂x3 ∂z
since
∂ x1 ∂ x2 ∂
= + ,
∂̟ ̟ ∂x1 ̟ ∂x2
∂ ∂
= ,
∂z ∂x3
∂ ∂ ∂
= x2 +x1 .
∂φ − ∂x2 ∂x2
Definition 2 A compactly supported solution ρ,~v,Φ will be said to be axisym-
metric if ∂ρ/∂φ = 0,∂Φ/∂φ = 0, that is, ρ = ρ(t,̟,z),Φ = Φ(t,̟,z) and if
the velocity field ~v is of the form
V x1 Ωx2
̟ −
 
~v = V x2+Ωx1 (2.2)
̟
 
 
 W 
 
 
with V =V(t,̟,z),W =W(t,̟,z),Ω=Ω(t,̟,z).
6
Note that if ∂ρ/∂φ= 0, then the Newton potential Φ given by (1.4) neces-
sarily satisfies ∂Φ/∂φ=0.
Of course a spherically symmetric solution, for which
V v(t,r) v(t,r)
ρ=ρ(t,r), = , Ω=0, W = z
̟ r r
with r =√̟2+z2, is axisymmetric in this sense.
Let us derive the equations which governaxisymmetric solutions.
First we nte that the following formula is easily verified:
D ∂ ∂ ∂ ∂
= +V +Ω +W . (2.3)
Dt ∂t ∂̟ ∂φ ∂z
. In fact, we have
∂ ∂ ∂ ∂ ∂ ∂
x1 +x2 =̟ , x2 +x1 = .
∂x1 ∂x2 ∂̟ − ∂x1 ∂x2 ∂φ
Then the equation (1.1a) reads
Dρ ∂V V ∂W
+ρ + + =0. (2.4)
Dt ∂̟ ̟ ∂z
(cid:16) (cid:17)
Using the calculations
Dx1 V
= x1 Ωx2,
Dt ̟ −
Dx2 V
= x2+Ωx1,
Dt ̟
D̟
=V,
Dt
we see that the equation (1.1b) reads
x1DV DΩ x2 ∂P ∂Φ
ρ x2 2 VΩ x1Ω2 + = ρ , (2.5a)
̟ Dt − Dt − ̟ − ∂x1 − ∂x1
hx2DV DΩ x1 i ∂P ∂Φ
ρ +x1 +2 VΩ x2Ω2 + = ρ , (2.5b)
̟ Dt Dt ̟ − ∂x2 − ∂x2
hDW ∂P ∂Φ i
ρ + = ρ . (2.5c)
Dt ∂x3 − ∂x3
7
1
Taking (x1 (2.5a)+x2 (2.5b)) and x2 (2.5a)+x1 (2.5b)), wesee that
̟ · · − · ·
(2.5a) (2.5b) is equivalent to
∧
DV ∂P ∂Φ
ρ ̟Ω2 + = ρ , (2.6a)
Dt − ∂̟ − ∂̟
hD i
ρ (̟2Ω)=0. (2.6b)
Dt
On the other hand te Laplace equation (1.1c) reads
1 ∂ ∂Φ ∂2Φ
̟ + =4πGρ, (2.7)
̟∂̟ ∂̟ ∂z2
(cid:16) (cid:17)
and the Newton potential (1.4) reads
+∞ ∞
Φ(̟,z)= G K (̟,̟′,z z′)ρ(̟′,z′)̟′d̟′dz′, (2.8)
I
− −
Z−∞ Z0
where
π/2 dα
K (̟,̟′,z z′)=4 (2.9)
I
− Z0 (̟ ̟′)2+(z z′)2+4̟̟′sin2α
− −
q
Summing up, the full system which governs axisymmetric solutions is
(2.4)(2.6a)(2.6b)(2.5c)(2.7), that is,
Dρ ∂V V ∂W
+ρ + + =0, (2.10a)
Dt ∂̟ ̟ ∂z
DV (cid:16) ∂P (cid:17) ∂Φ
ρ ̟Ω2 + = ρ , (2.10b)
Dt − ∂̟ − ∂̟
hDW ∂P i ∂Φ
ρ + = ρ . (2.10c)
Dt ∂z − ∂z
D
ρ (̟2Ω)=0. (2.10d)
Dt
HerewenotetheoperatorD/Dtactingonfunctionswhichareaxisymmetric
reduces to
D ∂ ∂ ∂
= +V +W .
Dt ∂t ∂̟ ∂z
Note that (2.10d) is a linear first order partial differential equation of Ω,
provided that ρ=0 and the components of the velocity fields V,W are known.
6
Therefore, given V,W, the equation (2.10d) can be solved explicitly as follows.
Let
Ω0(̟,z)=Ω (2.11)
t=0
|
8
be the initial data. For t [0,T),̟ > 0, z < , we consider the solution
∈ | | ∞
τ (ϕ(τ ; t,̟,z),ψ(τ ; t,̟,z)) of the ordinary differential equations
7→
dϕ dψ
=V(τ,ϕ,ψ), =W(τ,ϕ,ψ)
dτ dτ
satisfying the initial conditions
ϕ(t ; t,̟,z)=̟, ψ(t ; t,̟,z)=z.
Then the solution exists on the time interval [0,t], provided that V,W are
bounded, and Ω is given by
φ(0 ; t,̟,z)2
Ω(t,̟,z)= Ω0(ϕ(0 ; t,̟,z),ψ(0 ; t,̟,z)). (2.12)
̟2
Especially let us note that, if C is an arbitrary constant, then
C
Ω(t,̟,z)= (2.13)
̟2
satisfiestheequation(2.10b),whateverρ,V,W maybe,when (2.10a),(2.10b),
{
(2.10c), (2.7) turns out to be a closed system for only ρ,V,WΦ. However, if
}
C =0, then Ωx2,Ωx1 are unbounded at the axis ̟ =0.
6 −
Let us calculate the angular momentum for the axisymmetric solution.
We see
x1 V x1 Ωx2
̟ −
   
~x ~v = x2 V x2+Ωx1 =
× × ̟
   
   
x3  W 
   
 Vx3x2 Ωx3x1+Wx2
−̟ −
 
= V x3x1 Ωx3x2 Wx1 =
̟ − −
 
 
 Ω((x1)2+(x2)2) 
 
 
Vzsinφ Ωz̟cosφ+W̟sinφ
− −
= Vzcosφ Ωz̟sinφ W̟cosφ .
 − − 
Ω̟2
 
Integrating this,we get
0
J~= 0 , (2.14)
 
J
 
where
+∞ ∞
J =2π ρ(t,̟,z)Ω(t,̟,z)̟3d̟dz. (2.15)
Z−∞ Z0
9
In fact, we note that
+∞ 2π ∞
ρVzsinφ ̟d̟dφdz =0,
·
Z−∞ Z0 Z0
2π
since sinφdφ=0 and so on.
Z0
We know that J is constant with respect to t .
V V
Definition 3 Let ρ = ρ(t,̟,z),~v = ( x1 Ωx2, x2+Ωx1,W)T with V =
̟ − ̟
V(t,̟,z),Ω = Ω(t,̟,z),W = W(t,̟,z) be an axisymmetric solution. This
solution is said to be equatorially symmetric if
ρ(t,̟, z)=ρ(t,̟,z), V(t,̟, z)=V(t,̟,z),
− −
Ω(t,̟, z)=Ω(t,̟, z), W(t,̟, z)= W(t,̟,z)
− − − −
for z.
∀
Then the potential Φ given by (1.4) necessarily satisfies
Φ(t,̟, z)=Φ(t,̟,z)
−
for z.
∀
2.2 Co-ordinate system (r,φ,ζ)
Sometimes instead the co-ordinates (̟,z) one uses the co-ordinates (r,ζ) de-
fined by
z z
r = ̟2+z2, ζ = = . (2.16)
r √̟2+z2
p
Note that, while the polar co-ordinates are
x1 =rsinθcosφ, x2 =rsinθsinφ, x3 =rcosθ,
we are taking ζ =cosθ so that
x1 =r 1 ζ2cosφ, x2 =r 1 ζ2sinφ, x3 =rζ, (2.17)
− −
p p
while
∂ x1 ∂ x1ζ ∂ x2 ∂
= , (2.18a)
∂x1 r ∂r − r2 ∂ζ − r2(1 ζ2)∂φ
−
∂ x2 ∂ x2ζ ∂ x1 ∂
= , (2.18b)
∂x2 r ∂r − r2 ∂ζ − r2(1 ζ2)∂φ
−
∂ ∂ 1 ζ2 ∂
=ζ + − , (2.18c)
∂x3 ∂r r ∂ζ
10