# Stabilities for Nonisentropic Euler-Poisson Equations

Authors: Ka Luen Cheung; Sen Wong
Journal: The Scientific World Journal
 (2015)

---

## Abstract

We establish the stabilities and blowup results for the nonisentropic Euler-Poisson equations by the energy method. By analysing the second inertia, we show that the classical solutions of the system with attractive forces blow up in finite time in some special dimensions when the energy is negative. Moreover, we obtain the stabilities results for the system in the cases of attractive and repulsive forces.


---

## Body

## 1. Introduction

The compressible nonisentropic Euler(
δ
=
0
) or Euler-Poisson (
δ
=
±
1
) system for fluids can be written as(1)
ρ
t
+
∇
·
ρ
u
=
0
,
ρ
u
t
+
∇
·
(
ρ
u
⊗
u
)
+
∇
P
+
β
ρ
u
=
-
δ
ρ
∇
Φ
,
S
t
+
u
·
∇
S
=
0
,
Δ
Φ
t
,
x
=
α
N
ρ
,
P
=
K
ρ
γ
e
S
,where β
≥
0 is the frictional damping constant and α
(
N
) is a constant related to the unit ball in R
N. α
(
1
)
=
2, α
(
2
)
=
2
π and (2)
α
(
N
)
=
N
(
N
-
2
)
V
(
N
)
,where V
(
N
) is the volume of the unit ball in R
N. As usual, ρ
=
ρ
(
t
,
x
)
≥
0, u
=
u
(
t
,
x
)
∈
R
N, and S
(
t
,
x
) are the density, the velocity, and the entropy, respectively. P is the pressure function, for which the constants K
≥
0 and γ
≥
1.Whenδ
=
1, the system is self-attractive. The system (1) is the Newtonian description of gaseous stars [1]. When δ
=
-
1, the system comprises the Euler-Poisson equations with repulsive forces and can be used as a semiconductor model [2, 3]. When δ
=
0, the system comprises the compressible Euler equations and can be applied as a classical model in fluid mechanics [3]. For more classical and recent results in these systems, readers can refer to [1, 4–10].It is well known that the solution for the Poisson equation (1)4 can be written as(3)
Φ
(
t
,
x
)
=
α
(
N
)
∫
R
N
‍
G
(
x
-
y
)
ρ
(
t
,
y
)
d
y
,where G is the Green’s function for the Poisson equation in the N-dimensional spaces defined by(4)
G
(
x
)
:
=
x
,
N
=
1
;
log
⁡
x
,
N
=
2
;
-
1
x
N
-
2
,
N
≥
3
.Notation. In the following discussion, classical solutions (
ρ
,
u
,
S
) are C
1 solutions with compact support Ω
=
Ω
(
t
) for each fixed time t. We also denote the total mass by M, where (5)
M
=
∫
Ω
‍
ρ
d
x
=
∫
Ω
(
0
)
‍
ρ
0
d
x
,where ρ
0
=
ρ
0
(
x
)
:
=
ρ
(
0
,
x
).Lastly, we will denote(6)
R
t
=
the
diameter
of
Ω
(
t
)
.


## 2. Lemmas

In this section, we establish some lemmas for the proof of the main results. The following lemma will be used to derive the energy functional forγ
>
1; namely, (7)
E
(
t
)
=
∫
Ω
‍
1
2
ρ
u
2
+
1
γ
-
1
P
d
x
+
δ
2
∫
Ω
‍
ρ
Φ
d
xis conserved in time if the system (1) is not damped.Lemma 1.
For the classical solution(
ρ
,
u
,
S
) of system (1), we have (8)
∫
Ω
‍
P
t
d
x
=
(
γ
-
1
)
∫
Ω
‍
u
·
∇
P
d
x
,where P is defined by (1)5.Proof.
We have(9)
P
t
=
K
ρ
γ
e
S
S
t
+
e
S
γ
ρ
γ
-
1
ρ
t
by
1
5
=
P
S
t
+
K
e
S
γ
ρ
γ
-
1
-
∇
·
ρ
u
by
1
5
and
1
1
=
P
-
u
·
∇
S
+
K
e
S
γ
ρ
γ
-
1
-
∇
·
ρ
u
by
1
3
=
-
P
∑
i
=
1
N
‍
u
i
∂
i
S
-
K
γ
e
S
ρ
γ
-
1
∑
i
=
1
N
‍
ρ
∂
i
u
i
+
u
i
∂
i
ρ
=
-
γ
P
∑
i
=
1
N
‍
∂
i
u
i
-
∑
i
=
1
N
‍
P
∂
i
S
+
K
γ
e
S
ρ
γ
-
1
∂
i
ρ
u
i
by
1
5
=
-
γ
P
∑
i
=
1
N
‍
∂
i
u
i
-
∑
i
=
1
N
‍
∂
i
P
u
i
=
-
γ
P
∇
·
u
-
u
·
∇
P
.Note that, by Divergence Theorem, (10)
-
∫
Ω
‍
P
∇
·
u
d
x
=
∫
Ω
‍
u
·
∇
P
d
x
.Thus, (11)
∫
Ω
‍
P
t
d
x
=
(
γ
-
1
)
∫
Ω
‍
u
·
∇
P
d
x
.Next, the results of the following two lemmas will be used in the derivations of the energy functionals for bothγ
>
1 and γ
=
1. It will be shown that in Section 3 the energy functional for γ
=
1 is (12)
E
(
t
)
=
∫
Ω
‍
1
2
ρ
u
2
+
K
ρ
e
S
ln
⁡
ρ
-
1
d
x
+
δ
2
∫
Ω
‍
ρ
Φ
d
xwhich is conversed in time if the system (1) is not damped.Lemma 2.
For the classical solution(
ρ
,
u
,
S
) of system (1), we have(13)
∫
Ω
‍
1
2
ρ
u
2
t
d
x
=
-
∫
Ω
‍
u
·
∇
P
-
δ
∫
Ω
‍
(
∇
Φ
)
·
(
ρ
u
)
d
x
-
β
∫
Ω
‍
ρ
u
2
d
x
,where P is defined by (1)5 and Φ is the solution of (1)4.Proof.
We have(14)
1
2
ρ
u
2
t
(15)
=
ρ
u
t
·
u
-
1
2
ρ
t
u
2
(
by
product
rule
)
(16)
=
-
δ
ρ
∇
Φ
+
∇
·
ρ
u
⊗
u
+
∇
P
+
β
ρ
u
·
u
+
1
2
∇
·
ρ
u
u
2
by
1
1
and
1
2
.One can check a detail proof of the following equality in the Appendix:(17)
∫
Ω
‍
1
2
u
2
∇
·
(
ρ
u
)
-
u
·
∇
·
ρ
u
⊗
u
d
x
=
0
.Thus, (18)
∫
Ω
‍
1
2
ρ
u
2
t
d
x
=
-
∫
Ω
‍
δ
ρ
∇
Φ
+
∇
P
+
β
ρ
u
·
u
d
xby (16) and (17).
Thus, the proof is complete.Lemma 3.
For the classical solution(
ρ
,
u
,
S
) of system (1), we have (19)
∫
Ω
‍
∇
Φ
·
(
ρ
u
)
d
x
=
∫
Ω
‍
Φ
ρ
t
d
x
=
1
2
∫
Ω
‍
ρ
Φ
t
d
x
,where Φ is the solution of (1)4.Proof.
We have(20)
∫
Ω
‍
∇
Φ
·
ρ
u
d
x
=
-
∫
Ω
‍
Φ
∇
·
ρ
u
d
x
by
Divergence
Theorem
=
∫
Ω
‍
Φ
ρ
t
d
x
by
1
1
.Thus, the first equality in (19) holds.
Next,(21)
∫
Ω
‍
Φ
ρ
t
d
x
=
1
α
N
∫
Ω
‍
Φ
Δ
Φ
t
d
x
by
1
4
=
1
α
N
∫
Ω
‍
Δ
Φ
Φ
t
d
x
by
Green’s
Formula
=
∫
Ω
‍
ρ
Φ
t
d
x
by
1
4
.Thus, the second equality in (19) holds.The lemma below is crucial to obtaining the energy functional forγ
=
1. Comparing the left hand sides of (8) and (22), we note that the left hand side of (22) (given in the next lemma), which contains the term ln
⁡
ρ
-
1, is nontrivial to be found.Lemma 4.
For the classical solution(
ρ
,
u
,
S
) of system (1) with γ
=
1, we have (22)
d
d
t
∫
Ω
‍
K
ρ
e
S
ln
⁡
ρ
-
1
d
x
=
∫
Ω
‍
u
·
∇
P
d
x
,where P is defined by (1)5.Proof.
Note that(23)
ρ
e
S
ln
⁡
ρ
-
1
t
=
e
S
ρ
ln
⁡
ρ
-
1
t
+
ρ
ln
⁡
ρ
-
1
e
S
S
t
=
e
S
ρ
t
+
ln
⁡
ρ
-
1
ρ
t
+
e
S
ρ
ln
⁡
ρ
-
1
S
t
=
e
S
(
ln
⁡
ρ
)
ρ
t
+
e
S
ρ
(
ln
⁡
ρ
-
1
)
S
t
=
-
e
S
ln
⁡
ρ
∇
·
ρ
u
-
e
S
ρ
ln
⁡
ρ
-
1
u
·
∇
S
mmmmmmmmmmmmmmmmm
by
1
1
and
1
3
.Thus,(24)
d
d
t
∫
Ω
‍
K
ρ
e
S
(
ln
⁡
ρ
-
1
)
d
x
=
-
∫
Ω
‍
K
e
S
ln
⁡
ρ
∇
·
ρ
u
d
x
-
∫
Ω
‍
K
e
S
ρ
ln
⁡
ρ
-
1
u
·
∇
S
d
x
=
∫
Ω
‍
(
ρ
u
)
·
∇
K
e
S
ln
⁡
ρ
d
x
-
∫
Ω
‍
K
e
S
ρ
ln
⁡
ρ
-
1
u
·
∇
S
d
x
mmmmmmlm
by
Divergence
Theorem
=
∫
Ω
‍
u
·
K
ρ
∇
e
S
ln
⁡
ρ
-
K
e
S
ρ
ln
⁡
ρ
-
1
∇
S
d
x
=
∫
Ω
‍
u
·
K
e
S
∇
ρ
+
K
e
S
ρ
∇
S
d
x
=
∫
Ω
‍
u
·
∇
P
d
x
by
1
5
.The proof is complete.


## 3. Main Results

In this section, we find out the energy functionals for the system (1) in the case of γ
>
1 (Proposition 5) and γ
=
1 (Proposition 6). Moreover, we establish the stabilities results (Proposition 8) and a blowup result (Proposition 9) for system (1).Proposition 5.
For the classical solution(
ρ
,
u
,
S
) of system (1) with γ
>
1, let (25)
E
(
t
)
=
∫
Ω
‍
1
2
ρ
u
2
+
1
γ
-
1
P
d
x
+
δ
2
∫
Ω
‍
ρ
Φ
d
x
.Then, (26)
E
˙
(
t
)
=
-
β
∫
Ω
‍
ρ
u
2
d
x
,where E
˙
(
t
) is the devertive of E
(
t
) with respect to t.
Thus,E
(
t
) is a decreasing function and is conserved if the system is not damped.Proof.
By Lemma1,(27)
1
γ
-
1
∫
Ω
‍
P
t
d
x
=
∫
Ω
‍
u
·
∇
P
d
x
.By Lemma 2,(28)
∫
Ω
‍
1
2
ρ
u
2
t
d
x
=
-
∫
Ω
‍
u
·
∇
P
-
δ
∫
Ω
‍
(
∇
Φ
)
·
(
ρ
u
)
d
x
-
β
∫
Ω
‍
ρ
u
2
d
x
.By Lemma 3, (29)
∫
Ω
‍
(
∇
Φ
)
·
(
ρ
u
)
d
x
=
1
2
∫
Ω
‍
ρ
Φ
t
d
x
.Thus, the proof is complete.Proposition 6.
For the classical solution(
ρ
,
u
,
S
) of system (1) with γ
=
1, let (30)
E
(
t
)
=
∫
Ω
‍
1
2
ρ
u
2
+
K
ρ
e
S
ln
⁡
ρ
-
1
d
x
+
δ
2
∫
Ω
‍
ρ
Φ
d
x
.Then, (31)
E
˙
(
t
)
=
-
β
∫
Ω
‍
ρ
u
2
d
x
,where E
˙
(
t
) is the derivative of E
(
t
) with respect to t.
Thus,E
(
t
) is a decreasing function and is conserved if the system is not damped.Proof.
By Lemma2,(32)
∫
Ω
‍
1
2
ρ
u
2
t
d
x
=
-
∫
Ω
‍
u
·
∇
P
-
δ
∫
Ω
‍
(
∇
Φ
)
·
(
ρ
u
)
d
x
-
β
∫
Ω
‍
ρ
u
2
d
x
.By Lemma 3, (33)
∫
Ω
‍
(
∇
Φ
)
·
(
ρ
u
)
d
x
=
1
2
∫
Ω
‍
ρ
Φ
t
d
x
.By Lemma 4, (34)
d
d
t
∫
Ω
‍
K
ρ
e
S
ln
⁡
ρ
-
1
d
x
=
∫
Ω
‍
u
·
∇
P
d
x
.Thus, the proof is complete.Proposition 7.
Let(35)
H
(
t
)
=
∫
Ω
‍
ρ
x
2
d
x
.We have(36)
H
¨
t
=
-
β
H
˙
t
+
2
∫
Ω
‍
ρ
u
2
+
2
P
d
x
-
2
π
δ
M
2
,
mmmmmmmmmmmmmmmmm
f
o
r
N
=
2
,
H
¨
t
=
-
β
H
˙
(
t
)
+
2
∫
Ω
‍
ρ
u
2
+
N
P
d
x
+
N
-
2
δ
∫
Ω
‍
ρ
Φ
d
x
,
f
o
r
N
≥
3
,where H
˙
(
t
)
=
(
d
/
d
t
)
H
(
t
), H
¨
(
t
)
=
(
d
/
d
t
)
H
˙
(
t
), (
ρ
,
u
,
S
) is a classical solution of system (1), and M is defined by (5).Proof.
(37)
H
˙
t
=
-
∫
Ω
‍
∇
·
(
ρ
u
)
x
2
d
x
by
1
1
=
∫
Ω
‍
∇
x
2
·
ρ
u
d
x
by
Divergence
Theorem
=
∫
Ω
‍
2
x
·
ρ
u
d
x
,
H
¨
t
=
2
∫
Ω
‍
x
·
-
∇
·
ρ
u
⊗
u
-
∇
P
-
δ
ρ
∇
Φ
-
β
ρ
u
d
x
mmmmmmmmmmmmmmmmmmmmm
by
1
2
=
-
β
H
˙
(
t
)
+
2
∫
Ω
‍
x
·
-
∇
·
ρ
u
⊗
u
-
∇
P
-
δ
ρ
∇
Φ
d
x
.We split the last term of the above equality into three parts.
Firstly,(38)
-
∫
Ω
‍
x
·
∇
·
ρ
u
⊗
u
d
x
=
-
∑
j
=
1
N
‍
∫
Ω
‍
x
j
∑
i
=
1
N
‍
∂
i
ρ
u
i
u
j
d
x
by
definitions
=
∫
Ω
‍
ρ
u
2
d
x
.
by
integration
by
parts
.Secondly,(39)
-
∫
Ω
‍
x
·
∇
P
d
x
=
∫
Ω
‍
P
∇
·
x
d
x
by
Divergence
Theorem
=
∫
Ω
‍
N
P
d
x
.Thirdly,(40)
-
δ
∫
Ω
‍
ρ
x
·
∇
Φ
d
x
=
-
δ
α
N
∫
Ω
‍
∫
Ω
‍
ρ
x
ρ
y
∇
x
G
x
-
y
·
x
d
y
d
x
mmmmmmmmmmmmmmmmmmmmmmmmm
by
(
3
)
=
:
-
δ
α
(
N
)
I
,where ∇
x is the gradient operator with respect to the spatial variable x.
Note that(41)
I
=
∫
Ω
‍
∫
Ω
‍
ρ
x
ρ
y
∇
x
G
x
-
y
·
x
d
y
d
x
=
∫
Ω
‍
∫
Ω
‍
ρ
(
x
)
ρ
(
y
)
[
∇
x
G
(
x
-
y
)
·
(
x
-
y
)
]
d
y
d
x
+
∫
Ω
‍
∫
Ω
‍
ρ
x
ρ
y
∇
x
G
x
-
y
·
y
d
y
d
x
.For N
=
2, (42)
∇
x
G
(
x
)
=
∇
x
log
⁡
x
=
1
x
2
x
.Thus, (43)
I
=
∫
Ω
‍
∫
Ω
‍
ρ
x
ρ
y
d
y
d
x
-
I
,
I
=
1
2
∫
Ω
‍
∫
Ω
‍
ρ
(
x
)
ρ
(
y
)
d
y
d
x
=
M
2
2
.The result for N
=
2 is established.
ForN
≥
3, (44)
∇
x
G
(
x
)
=
-
∇
x
1
x
N
-
2
=
(
N
-
2
)
1
x
N
x
.Thus,(45)
I
=
∫
Ω
‍
∫
Ω
‍
ρ
x
ρ
y
∇
x
G
x
-
y
·
x
-
y
d
y
d
x
+
∫
Ω
‍
∫
Ω
‍
ρ
(
x
)
ρ
(
y
)
[
∇
x
G
(
x
-
y
)
·
y
]
d
y
d
x
=
-
N
-
2
α
N
∫
Ω
‍
ρ
Φ
d
x
-
I
,
I
=
-
N
-
2
2
α
(
N
)
∫
Ω
‍
ρ
Φ
d
x
.The results for N
≥
3 are also established.Now, we are ready to present the stability results.Proposition 8.
Considering the classical solutions of system (1), we have the following.
Case
1. For δ
=
1, β
=
0, N
=
3 or 4, γ
≥
2
(
N
-
1
)
/
N, and E
(
0
)
≥
0, we have (46)
lim
t
→
∞
inf
⁡
R
(
t
)
t
≥
(
N
-
2
)
E
(
0
)
M
.
Case
2. For δ
=
-
1, β
=
0, N
≥
4, and γ
≥
(
N
+
2
)
/
N, we have(47)
lim
n
→
∞
inf
⁡
R
(
t
)
t
≥
2
E
(
0
)
M
.
Case
3. For δ
=
-
1, β
=
0, N
=
2, and γ
>
1, we have (48)
lim
n
→
∞
inf
⁡
R
(
t
)
t
≥
π
M
.Proof.
First of all, by definitions (35), (6), and (5), we always have(49)
H
(
t
)
=
∫
Ω
(
t
)
‍
ρ
x
2
d
x
≤
R
2
(
t
)
M
.
Case
1. By Propositions 5 and 7, we have(50)
H
¨
t
=
2
∫
Ω
‍
ρ
u
2
+
N
P
d
x
+
(
N
-
2
)
∫
Ω
‍
ρ
Φ
d
x
=
2
∫
Ω
‍
ρ
u
2
d
x
+
2
N
∫
Ω
‍
P
d
x
+
(
N
-
2
)
2
E
(
t
)
-
2
γ
-
1
∫
Ω
‍
P
d
x
-
∫
Ω
‍
ρ
u
2
d
x
=
2
-
(
N
-
2
)
∫
Ω
‍
ρ
u
2
d
x
+
2
N
-
2
(
N
-
2
)
γ
-
1
∫
Ω
‍
P
d
x
+
2
(
N
-
2
)
E
(
t
)
≥
2
(
N
-
2
)
E
(
t
)
=
2
(
N
-
2
)
E
(
0
)
.Thus, (51)
H
(
t
)
≥
(
N
-
2
)
E
(
0
)
t
2
+
H
˙
(
0
)
t
+
H
(
0
)
.In view of inequality (49), we have (52)
R
(
t
)
t
≥
(
N
-
2
)
E
(
0
)
M
+
H
˙
(
0
)
M
t
+
H
(
0
)
M
t
2
.Thus, (53)
lim
t
→
∞
inf
⁡
R
(
t
)
t
≥
(
N
-
2
)
E
(
0
)
M
.
Case
2. By Propositions 5 and 7, we have (54)
H
¨
t
=
2
∫
Ω
‍
ρ
u
2
+
N
P
d
x
+
N
-
2
∫
Ω
‍
ρ
-
Φ
d
x
=
2
2
E
(
t
)
-
2
γ
-
1
∫
Ω
‍
P
d
x
-
∫
Ω
‍
ρ
(
-
Φ
)
d
x
+
2
N
∫
Ω
‍
P
d
x
+
(
N
-
2
)
∫
Ω
‍
ρ
(
-
Φ
)
d
x
≥
4
E
(
t
)
=
4
E
(
0
)
.Note that -
Φ is a positive function for N
≥
3 by (3) and (4). Thus,(55)
H
(
t
)
≥
2
E
(
0
)
t
2
+
H
˙
(
0
)
t
+
H
(
0
)
.It follows that (56)
lim
n
→
∞
inf
⁡
R
(
t
)
t
≥
2
E
(
0
)
M
.
Case
3. By Proposition 7, we have(57)
H
¨
t
=
2
∫
Ω
‍
ρ
u
2
+
2
P
d
x
+
2
π
M
2
≥
2
π
M
2
.It follows that (58)
lim
n
→
∞
⁡
inf
⁡
⁡
R
(
t
)
t
≥
π
M
.Finally, we can give the blowup result.Proposition 9.
Ifδ
=
1, N
≥
4, 1
<
γ
≤
2
(
N
-
1
)
/
N, and E
(
0
)
<
0, then the classical solutions of (1) blow up in finite time.Proof.
Case
1 (
β
=
0
). As before, we have, from Propositions 5 and 7, that(59)
H
¨
t
=
4
-
N
∫
Ω
‍
ρ
u
2
d
x
+
2
N
-
2
N
-
2
γ
-
1
∫
Ω
‍
P
d
x
+
2
(
N
-
2
)
E
(
t
)
≤
0
+
0
+
2
(
N
-
2
)
E
(
t
)
=
2
(
N
-
2
)
E
(
0
)
.It follows that (60)
H
(
t
)
≤
(
N
-
2
)
E
(
0
)
t
2
+
H
˙
(
0
)
t
+
H
(
0
)
.Suppose the solutions exist globally; then for sufficient large t, we see that H
(
t
) is negative as the leading coefficient of the right hand side of (60) is negative. However, H
(
t
) is nonnegative by definition (35). This is a contradiction. As a result, the solutions blow up in finite time.
Case
2 (
β
≠
0
). Now (59) becomes(61)
H
¨
(
t
)
+
β
H
˙
(
t
)
≤
2
(
N
-
2
)
E
(
0
)
.It follows by multipying an integral factor e
β
t on both sides and taking integration that (62)
H
(
t
)
≤
A
1
+
A
2
e
-
β
t
+
2
(
N
-
2
)
E
(
0
)
β
tfor some constants A
1 and A
2. Note that this implies that H
(
t
) is negative for sufficient large t as β
>
0 and (
N
-
2
)
E
(
0
)
<
0. Therefore, the solutions blow up in finite time.



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*Source: 494707-2015-03-16.xml*