Hindawi Publishing Corporation e Scientific World Journal Volume 2015, Article ID 494707, 6 pages http://dx.doi.org/10.1155/2015/494707

Research Article Stabilities for Nonisentropic Euler-Poisson Equations Ka Luen Cheung and Sen Wong Department of Mathematics and Information Technology, The Hong Kong Institute of Education, 10 Lo Ping Road, Tai Po, New Territories, Hong Kong Correspondence should be addressed to Sen Wong; [email protected] Received 6 December 2014; Revised 8 March 2015; Accepted 9 March 2015 Academic Editor: Carlo Bianca Copyright © 2015 K. L. Cheung and S. Wong. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 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.

1. Introduction The compressible nonisentropic Euler (𝛿 = 0) or EulerPoisson (𝛿 = ±1) system for fluids can be written as 𝜌𝑡 + ∇ ⋅ (𝜌𝑢) = 0, (𝜌𝑢)𝑡 + ∇ ⋅ (𝜌𝑢 ⊗ 𝑢) + ∇𝑃 + 𝛽𝜌𝑢 = −𝛿𝜌∇Φ, 𝑆𝑡 + 𝑢 ⋅ ∇𝑆 = 0,

Φ (𝑡, 𝑥) = 𝛼 (𝑁) ∫ (1)

ΔΦ (𝑡, 𝑥) = 𝛼 (𝑁) 𝜌, 𝑃 = 𝐾𝜌𝛾 𝑒𝑆 , where 𝛽 ≥ 0 is the frictional damping constant and 𝛼(𝑁) is a constant related to the unit ball in R𝑁. 𝛼(1) = 2, 𝛼(2) = 2𝜋 and 𝛼 (𝑁) = 𝑁 (𝑁 − 2) 𝑉 (𝑁) ,

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

(2)

where 𝑉(𝑁) is the volume of the unit ball in R𝑁. As usual, 𝜌 = 𝜌(𝑡, 𝑥) ≥ 0, 𝑢 = 𝑢(𝑡, 𝑥) ∈ R𝑁, 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. When 𝛿 = 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].

R𝑁

𝐺 (𝑥 − 𝑦) 𝜌 (𝑡, 𝑦) 𝑑𝑦,

(3)

where 𝐺 is the Green’s function for the Poisson equation in the 𝑁-dimensional spaces defined by |𝑥| , { { { { { 𝐺 (𝑥) := {log |𝑥| , { { −1 { { , { |𝑥|𝑁−2

𝑁 = 1; 𝑁 = 2;

(4)

𝑁 ≥ 3.

Notation. In the following discussion, classical solutions (𝜌, 𝑢, 𝑆) are 𝐶1 solutions with compact support Ω = Ω(𝑡) for each fixed time 𝑡. We also denote the total mass by 𝑀, where 𝑀 = ∫ 𝜌 𝑑𝑥 = ∫ Ω

Ω(0)

𝜌0 𝑑𝑥,

(5)

where 𝜌0 = 𝜌0 (𝑥) := 𝜌(0, 𝑥). Lastly, we will denote 𝑅 (𝑡) = the diameter of Ω (𝑡) .

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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 for 𝛾 > 1; namely, 1 1 𝛿 𝑃) 𝑑𝑥 + ∫ 𝜌Φ 𝑑𝑥 𝐸 (𝑡) = ∫ ( 𝜌 |𝑢|2 + 𝛾−1 2 Ω Ω 2

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Lemma 2. For the classical solution (𝜌, 𝑢, 𝑆) of system (1), we have 1 ∫ ( 𝜌 |𝑢|2 ) 𝑑𝑥 = − ∫ 𝑢 ⋅ ∇𝑃 − 𝛿 ∫ (∇Φ) ⋅ (𝜌𝑢) 𝑑𝑥 𝑡 Ω 2 Ω Ω (13) 2 − 𝛽 ∫ 𝜌 |𝑢| 𝑑𝑥, Ω

where 𝑃 is defined by (1)5 and Φ is the solution of (1)4 .

is conserved in time if the system (1) is not damped.

Proof. We have

Lemma 1. For the classical solution (𝜌, 𝑢, 𝑆) of system (1), we have ∫ 𝑃𝑡 𝑑𝑥 = (𝛾 − 1) ∫ 𝑢 ⋅ ∇𝑃 𝑑𝑥, Ω

1 ( 𝜌 |𝑢|2 ) 2 𝑡 1 = (𝜌𝑢)𝑡 ⋅ 𝑢 − 𝜌𝑡 |𝑢|2 2

(8)

Ω

(14)

where 𝑃 is defined by (1)5 .

(by product rule)

(15)

= − (𝛿𝜌∇Φ + ∇ ⋅ (𝜌𝑢 ⊗ 𝑢) + ∇𝑃 + 𝛽𝜌𝑢) ⋅ 𝑢 (16) 1 (∇ ⋅ (𝜌𝑢)) |𝑢|2 by (1)1 and (1)2 . 2 One can check a detail proof of the following equality in the Appendix:

Proof. We have

+

𝑃𝑡 = 𝐾 (𝜌𝛾 𝑒𝑆 𝑆𝑡 + 𝑒𝑆 𝛾𝜌𝛾−1 𝜌𝑡 )

by (1)5

= 𝑃𝑆𝑡 + 𝐾𝑒𝑆 𝛾𝜌𝛾−1 [−∇ ⋅ (𝜌𝑢)]

by (1)5 and (1)1

= 𝑃 (−𝑢 ⋅ ∇𝑆) + 𝐾𝑒𝑆 𝛾𝜌𝛾−1 [−∇ ⋅ (𝜌𝑢)] 𝑁

𝑆 𝛾−1

= −𝑃 (∑𝑢𝑖 𝜕𝑖 𝑆) − 𝐾𝛾𝑒 𝜌 𝑖=1

𝑁

𝑁

𝑖=1

𝑖=1

𝑁

𝑁

𝑖=1

𝑖=1

1 ∫ ( |𝑢|2 ∇ ⋅ (𝜌𝑢) − 𝑢 ⋅ [∇ ⋅ (𝜌𝑢 ⊗ 𝑢)]) 𝑑𝑥 = 0. Ω 2

by (1)3

Thus,

𝑁

[∑ (𝜌𝜕𝑖 𝑢𝑖 + 𝑢𝑖 𝜕𝑖 𝜌)]

1 ∫ ( 𝜌 |𝑢|2 ) 𝑑𝑥 = − ∫ (𝛿𝜌∇Φ + ∇𝑃 + 𝛽𝜌𝑢) ⋅ 𝑢 𝑑𝑥 (18) 𝑡 Ω 2 Ω

𝑖=1

= −𝛾𝑃∑𝜕𝑖 𝑢𝑖 − ∑ [𝑃𝜕𝑖 𝑆 + 𝐾𝛾𝑒𝑆 𝜌𝛾−1 𝜕𝑖 𝜌] 𝑢𝑖

by (1)5

by (16) and (17). Thus, the proof is complete. Lemma 3. For the classical solution (𝜌, 𝑢, 𝑆) of system (1), we have 1 ∫ (∇Φ) ⋅ (𝜌𝑢) 𝑑𝑥 = ∫ Φ𝜌𝑡 𝑑𝑥 = ∫ [𝜌Φ]𝑡 𝑑𝑥, (19) 2 Ω Ω Ω

= −𝛾𝑃∑𝜕𝑖 𝑢𝑖 − ∑ (𝜕𝑖 𝑃) 𝑢𝑖 = −𝛾𝑃∇ ⋅ 𝑢 − 𝑢 ⋅ ∇𝑃. (9) Note that, by Divergence Theorem, − ∫ 𝑃∇ ⋅ 𝑢 𝑑𝑥 = ∫ 𝑢 ⋅ ∇𝑃 𝑑𝑥. Ω

Ω

where Φ is the solution of (1)4 . Proof. We have

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∫ (∇Φ) ⋅ (𝜌𝑢) 𝑑𝑥 Ω

= − ∫ Φ∇ ⋅ (𝜌𝑢) 𝑑𝑥

Thus,

Ω

∫ 𝑃𝑡 𝑑𝑥 = (𝛾 − 1) ∫ 𝑢 ⋅ ∇𝑃 𝑑𝑥. Ω

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Ω

(11)

Next, the results of the following two lemmas will be used in the derivations of the energy functionals for both 𝛾 > 1 and 𝛾 = 1. It will be shown that in Section 3 the energy functional for 𝛾 = 1 is 1 𝛿 𝐸 (𝑡) = ∫ ( 𝜌 |𝑢|2 + 𝐾𝜌𝑒𝑆 (ln 𝜌 − 1)) 𝑑𝑥 + ∫ 𝜌Φ 𝑑𝑥 2 Ω Ω 2 (12) which is conversed in time if the system (1) is not damped.

= ∫ Φ𝜌𝑡 𝑑𝑥 Ω

by Divergence Theorem (20)

by (1)1 .

Thus, the first equality in (19) holds. Next, 1 ∫ ΦΔΦ𝑡 𝑑𝑥 by (1)4 ∫ Φ𝜌𝑡 𝑑𝑥 = 𝛼 (𝑁) Ω Ω =

1 ∫ ΔΦΦ𝑡 𝑑𝑥 𝛼 (𝑁) Ω

= ∫ 𝜌Φ𝑡 𝑑𝑥 Ω

by Green’s Formula (21)

by (1)4 .

Thus, the second equality in (19) holds.

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The lemma below is crucial to obtaining the energy functional for 𝛾 = 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 d ∫ 𝐾𝜌𝑒𝑆 (ln 𝜌 − 1) 𝑑𝑥 = ∫ 𝑢 ⋅ ∇𝑃 𝑑𝑥, d𝑡 Ω Ω

Proposition 5. For the classical solution (𝜌, 𝑢, 𝑆) of system (1) with 𝛾 > 1, let 1 1 𝛿 𝑃) 𝑑𝑥 + ∫ 𝜌Φ 𝑑𝑥. 𝐸 (𝑡) = ∫ ( 𝜌 |𝑢|2 + 2 𝛾 − 1 2 Ω Ω

(25)

Then, 𝐸̇ (𝑡) = −𝛽 ∫ 𝜌 |𝑢|2 𝑑𝑥, Ω

(26)

where 𝑃 is defined by (1)5 .

̇ is the devertive of 𝐸(𝑡) with respect to 𝑡. where 𝐸(𝑡) Thus, 𝐸(𝑡) is a decreasing function and is conserved if the system is not damped.

Proof. Note that

Proof. By Lemma 1,

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𝑆

1 ∫ 𝑃 𝑑𝑥 = ∫ 𝑢 ⋅ ∇𝑃 𝑑𝑥. 𝛾−1 Ω 𝑡 Ω

[𝜌𝑒 (ln 𝜌 − 1)]𝑡 𝑆

𝑆

= 𝑒 [𝜌 (ln 𝜌 − 1)]𝑡 + [𝜌 (ln 𝜌 − 1)] 𝑒 𝑆𝑡 𝑆

By Lemma 2,

𝑆

= 𝑒 [𝜌𝑡 + (ln 𝜌 − 1) 𝜌𝑡 ] + 𝑒 𝜌 (ln 𝜌 − 1) 𝑆𝑡 = 𝑒𝑆 (ln 𝜌) 𝜌𝑡 + 𝑒𝑆 𝜌 (ln 𝜌 − 1) 𝑆𝑡

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1 ∫ ( 𝜌 |𝑢|2 ) 𝑑𝑥 = − ∫ 𝑢 ⋅ ∇𝑃 − 𝛿 ∫ (∇Φ) ⋅ (𝜌𝑢) 𝑑𝑥 𝑡 Ω 2 Ω Ω − 𝛽 ∫ 𝜌 |𝑢| 𝑑𝑥. Ω

By Lemma 3,

by (1)1 and (1)3 .

∫ (∇Φ) ⋅ (𝜌𝑢) 𝑑𝑥 =

Thus,

Ω

d ∫ 𝐾𝜌𝑒𝑆 (ln 𝜌 − 1) 𝑑𝑥 d𝑡 Ω

(29)

Proposition 6. For the classical solution (𝜌, 𝑢, 𝑆) of system (1) with 𝛾 = 1, let

Ω

1 𝛿 𝐸 (𝑡) = ∫ ( 𝜌 |𝑢|2 + 𝐾𝜌𝑒𝑆 (ln 𝜌 − 1)) 𝑑𝑥 + ∫ 𝜌Φ 𝑑𝑥. 2 Ω Ω 2 (30)

− ∫ 𝐾𝑒𝑆 𝜌 (ln 𝜌 − 1) (𝑢 ⋅ ∇𝑆) 𝑑𝑥 Ω

= ∫ (𝜌𝑢) ⋅ ∇ [𝐾𝑒𝑆 (ln 𝜌)] 𝑑𝑥

Then,

Ω

𝐸̇ (𝑡) = −𝛽 ∫ 𝜌 |𝑢|2 𝑑𝑥,

(24)

Ω

by Divergence Theorem = ∫ 𝑢 ⋅ [𝐾𝜌∇ (𝑒𝑆 ln 𝜌) − 𝐾𝑒𝑆 𝜌 (ln 𝜌 − 1) ∇𝑆] 𝑑𝑥 Ω

= ∫ 𝑢 ⋅ [𝐾𝑒𝑆 ∇𝜌 + 𝐾𝑒𝑆 𝜌∇𝑆] 𝑑𝑥 Ω

Ω

1 ∫ [𝜌Φ]𝑡 𝑑𝑥. 2 Ω

Thus, the proof is complete.

= − ∫ 𝐾𝑒𝑆 (ln 𝜌) [∇ ⋅ (𝜌𝑢)] 𝑑𝑥

= ∫ 𝑢 ⋅ ∇𝑃 𝑑𝑥

(28)

2

= −𝑒𝑆 (ln 𝜌) (∇ ⋅ (𝜌𝑢)) − 𝑒𝑆 𝜌 (ln 𝜌 − 1) (𝑢 ⋅ ∇𝑆)

− ∫ 𝐾𝑒𝑆 𝜌 (ln 𝜌 − 1) (𝑢 ⋅ ∇𝑆) 𝑑𝑥

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Ω

̇ is the derivative of 𝐸(𝑡) with respect to 𝑡. where 𝐸(𝑡) Thus, 𝐸(𝑡) is a decreasing function and is conserved if the system is not damped. Proof. By Lemma 2, 1 ∫ ( 𝜌 |𝑢|2 ) 𝑑𝑥 = − ∫ 𝑢 ⋅ ∇𝑃 − 𝛿 ∫ (∇Φ) ⋅ (𝜌𝑢) 𝑑𝑥 𝑡 Ω 2 Ω Ω

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).

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2

− 𝛽 ∫ 𝜌 |𝑢| 𝑑𝑥. Ω

By Lemma 3, ∫ (∇Φ) ⋅ (𝜌𝑢) 𝑑𝑥 = Ω

1 ∫ [𝜌Φ]𝑡 𝑑𝑥. 2 Ω

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By Lemma 4, d ∫ 𝐾𝜌𝑒𝑆 (ln 𝜌 − 1) 𝑑𝑥 = ∫ 𝑢 ⋅ ∇𝑃 𝑑𝑥. d𝑡 Ω Ω Thus, the proof is complete.

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Proposition 7. Let

Thirdly, 𝐻 (𝑡) = ∫ 𝜌 |𝑥|2 𝑑𝑥.

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Ω

− 𝛿 ∫ 𝜌𝑥 ⋅ ∇Φ 𝑑𝑥 Ω

= −𝛿𝛼 (𝑁) ∫ ∫ 𝜌 (𝑥) 𝜌 (𝑦) [∇𝑥 𝐺 (𝑥 − 𝑦) ⋅ 𝑥] 𝑑𝑦 𝑑𝑥

We have

Ω Ω

by (3)

𝐻̈ (𝑡) = −𝛽𝐻̇ (𝑡) + 2 ∫ (𝜌 |𝑢|2 + 2𝑃) 𝑑𝑥 − 2𝜋𝛿𝑀2 , Ω

=: −𝛿𝛼 (𝑁) 𝐼,

for 𝑁 = 2,

(40) (36)

𝐻̈ (𝑡) = −𝛽𝐻̇ (𝑡) + 2 ∫ (𝜌 |𝑢|2 + 𝑁𝑃) 𝑑𝑥 Ω

+ (𝑁 − 2) 𝛿 ∫ 𝜌Φ 𝑑𝑥, Ω

for 𝑁 ≥ 3,

̇ ̈ ̇ where 𝐻(𝑡) = (d/d𝑡)𝐻(𝑡), 𝐻(𝑡) = (d/d𝑡)𝐻(𝑡), (𝜌, 𝑢, 𝑆) is a classical solution of system (1), and 𝑀 is defined by (5).

where ∇𝑥 is the gradient operator with respect to the spatial variable 𝑥. Note that 𝐼 = ∫ ∫ 𝜌 (𝑥) 𝜌 (𝑦) [∇𝑥 𝐺 (𝑥 − 𝑦) ⋅ 𝑥] 𝑑𝑦 𝑑𝑥 Ω Ω

= ∫ ∫ 𝜌 (𝑥) 𝜌 (𝑦) [∇𝑥 𝐺 (𝑥 − 𝑦) ⋅ (𝑥 − 𝑦)] 𝑑𝑦 𝑑𝑥 Ω Ω

Proof.

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+ ∫ ∫ 𝜌 (𝑥) 𝜌 (𝑦) [∇𝑥 𝐺 (𝑥 − 𝑦) ⋅ 𝑦] 𝑑𝑦 𝑑𝑥.

𝐻̇ (𝑡) = − ∫ ∇ ⋅ (𝜌𝑢) |𝑥| 𝑑𝑥 2

Ω

= ∫ ∇ |𝑥|2 ⋅ (𝜌𝑢) 𝑑𝑥 Ω

Ω Ω

by (1)1

For 𝑁 = 2, ∇𝑥 𝐺 (𝑥) = ∇𝑥 log |𝑥| =

by Divergence Theorem

= ∫ 2𝑥 ⋅ (𝜌𝑢) 𝑑𝑥,

1 𝑥. |𝑥|2

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Thus,

Ω

𝐼 = ∫ ∫ 𝜌 (𝑥) 𝜌 (𝑦) 𝑑𝑦 𝑑𝑥 − 𝐼,

𝐻̈ (𝑡) = 2 ∫ 𝑥 ⋅ (−∇ ⋅ (𝜌𝑢 ⊗ 𝑢) − ∇𝑃 − 𝛿𝜌∇Φ − 𝛽𝜌𝑢) 𝑑𝑥

Ω Ω

Ω

1 𝑀2 𝐼 = ∫ ∫ 𝜌 (𝑥) 𝜌 (𝑦) 𝑑𝑦 𝑑𝑥 = . 2 Ω Ω 2

by (1)2 = −𝛽𝐻̇ (𝑡) + 2 ∫ 𝑥 ⋅ (−∇ ⋅ (𝜌𝑢 ⊗ 𝑢) − ∇𝑃 − 𝛿𝜌∇Φ) 𝑑𝑥. Ω

(37)

(43)

The result for 𝑁 = 2 is established. For 𝑁 ≥ 3, ∇𝑥 𝐺 (𝑥) = −∇𝑥

We split the last term of the above equality into three parts. Firstly,

1 1 = (𝑁 − 2) 𝑁 𝑥. 𝑁−2 |𝑥| |𝑥|

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Thus, − ∫ 𝑥 ⋅ [∇ ⋅ (𝜌𝑢 ⊗ 𝑢)] 𝑑𝑥

𝐼 = ∫ ∫ 𝜌 (𝑥) 𝜌 (𝑦) [∇𝑥 𝐺 (𝑥 − 𝑦) ⋅ (𝑥 − 𝑦)] 𝑑𝑦 𝑑𝑥

Ω

𝑁

Ω Ω

𝑁

= − ∑ ∫ 𝑥𝑗 ∑𝜕𝑖 (𝜌𝑢𝑖 𝑢𝑗 ) 𝑑𝑥 𝑗=1 Ω

by definitions

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𝑖=1

= ∫ 𝜌 |𝑢|2 𝑑𝑥 Ω

Ω Ω

=−

by integration by parts.

(𝑁 − 2) ∫ 𝜌Φ 𝑑𝑥 − 𝐼, 𝛼 (𝑁) Ω 𝐼=−

Secondly, − ∫ 𝑥 ⋅ ∇𝑃 𝑑𝑥 = ∫ 𝑃∇ ⋅ 𝑥 𝑑𝑥 Ω

+ ∫ ∫ 𝜌 (𝑥) 𝜌 (𝑦) [∇𝑥 𝐺 (𝑥 − 𝑦) ⋅ 𝑦] 𝑑𝑦 𝑑𝑥

Ω

by Divergence Theorem

𝑁−2 ∫ 𝜌Φ 𝑑𝑥. 2𝛼 (𝑁) Ω

The results for 𝑁 ≥ 3 are also established. Now, we are ready to present the stability results.

= ∫ 𝑁𝑃 𝑑𝑥. Ω

(45)

(39)

Proposition 8. Considering the classical solutions of system (1), we have the following.

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Case 1. For 𝛿 = 1, 𝛽 = 0, 𝑁 = 3 or 4, 𝛾 ≥ 2(𝑁 − 1)/𝑁, and 𝐸(0) ≥ 0, we have lim inf

𝑡→∞

𝑅 (𝑡) √ (𝑁 − 2) 𝐸 (0) ≥ . 𝑡 𝑀

Case 2. By Propositions 5 and 7, we have 𝐻̈ (𝑡) = 2 ∫ (𝜌 |𝑢|2 + 𝑁𝑃) 𝑑𝑥 + (𝑁 − 2) ∫ 𝜌 (−Φ) 𝑑𝑥 Ω

(46)

Ω

= 2 [2𝐸 (𝑡) −

Case 2. For 𝛿 = −1, 𝛽 = 0, 𝑁 ≥ 4, and 𝛾 ≥ (𝑁 + 2)/𝑁, we have lim inf

𝑛→∞

𝑅 (𝑡) √ 2𝐸 (0) ≥ . 𝑡 𝑀

+ 2𝑁 ∫ 𝑃 𝑑𝑥 + (𝑁 − 2) ∫ 𝜌 (−Φ) 𝑑𝑥 Ω

(47)

𝑛→∞

𝑅 (𝑡) √ ≥ 𝜋𝑀. 𝑡

Ω(𝑡)

(48)

𝜌 |𝑥|2 𝑑𝑥 ≤ 𝑅2 (𝑡) 𝑀.

Note that −Φ is a positive function for 𝑁 ≥ 3 by (3) and (4). Thus, 𝐻 (𝑡) ≥ 2𝐸 (0) 𝑡2 + 𝐻̇ (0) 𝑡 + 𝐻 (0) .

lim inf

𝑛→∞

𝑅 (𝑡) √ 2𝐸 (0) ≥ . 𝑡 𝑀

(56)

Case 3. By Proposition 7, we have 𝐻̈ (𝑡) = 2 ∫ (𝜌 |𝑢|2 + 2𝑃) 𝑑𝑥 + 2𝜋𝑀2

𝐻̈ (𝑡) = 2 ∫ (𝜌 |𝑢|2 + 𝑁𝑃) 𝑑𝑥 + (𝑁 − 2) ∫ 𝜌Φ 𝑑𝑥

Ω

Ω

(57) 2

≥ 2𝜋𝑀 .

= 2 ∫ 𝜌 |𝑢|2 𝑑𝑥 + 2𝑁 ∫ 𝑃 𝑑𝑥 Ω

(55)

It follows that

(49)

Case 1. By Propositions 5 and 7, we have

Ω

Ω

= 4𝐸 (0) .

Proof. First of all, by definitions (35), (6), and (5), we always have 𝐻 (𝑡) = ∫

(54)

≥ 4𝐸 (𝑡)

Case 3. For 𝛿 = −1, 𝛽 = 0, 𝑁 = 2, and 𝛾 > 1, we have lim inf

2 ∫ 𝑃 𝑑𝑥 − ∫ 𝜌 (−Φ) 𝑑𝑥] 𝛾−1 Ω Ω

Ω

2 + (𝑁 − 2) [2𝐸 (𝑡) − ∫ 𝑃 𝑑𝑥 − ∫ 𝜌 |𝑢|2 𝑑𝑥] 𝛾−1 Ω Ω

It follows that lim inf

𝑛→∞

𝑅 (𝑡) √ ≥ 𝜋𝑀. 𝑡

(58)

= [2 − (𝑁 − 2)] ∫ 𝜌 |𝑢|2 𝑑𝑥 Ω

+ [2𝑁 −

2 (𝑁 − 2) ] ∫ 𝑃 𝑑𝑥 + 2 (𝑁 − 2) 𝐸 (𝑡) 𝛾−1 Ω

Finally, we can give the blowup result. Proposition 9. If 𝛿 = 1, 𝑁 ≥ 4, 1 < 𝛾 ≤ 2(𝑁 − 1)/𝑁, and 𝐸(0) < 0, then the classical solutions of (1) blow up in finite time.

≥ 2 (𝑁 − 2) 𝐸 (𝑡) = 2 (𝑁 − 2) 𝐸 (0) . (50)

Proof. Case 1 (𝛽 = 0). As before, we have, from Propositions 5 and 7, that

Thus, 𝐻 (𝑡) ≥ (𝑁 − 2) 𝐸 (0) 𝑡2 + 𝐻̇ (0) 𝑡 + 𝐻 (0) .

(51)

In view of inequality (49), we have 𝑅 (𝑡) √ (𝑁 − 2) 𝐸 (0) 𝐻̇ (0) 𝐻 (0) . ≥ + + 𝑡 𝑀 𝑀𝑡 𝑀𝑡2

2 (𝑁 − 2) ] ∫ 𝑃 𝑑𝑥 𝐻̈ (𝑡) = (4 − 𝑁) ∫ 𝜌 |𝑢|2 𝑑𝑥 + [2𝑁 − 𝛾−1 Ω Ω + 2 (𝑁 − 2) 𝐸 (𝑡)

(52)

≤ 0 + 0 + 2 (𝑁 − 2) 𝐸 (𝑡) = 2 (𝑁 − 2) 𝐸 (0) . (59)

Thus, 𝑅 (𝑡) √ (𝑁 − 2) 𝐸 (0) ≥ . lim inf 𝑡→∞ 𝑡 𝑀

It follows that (53)

𝐻 (𝑡) ≤ (𝑁 − 2) 𝐸 (0) 𝑡2 + 𝐻̇ (0) 𝑡 + 𝐻 (0) .

(60)

6

The Scientific World Journal

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.

𝐻̈ (𝑡) + 𝛽𝐻̇ (𝑡) ≤ 2 (𝑁 − 2) 𝐸 (0) .

(61)

It follows by multipying an integral factor 𝑒𝛽𝑡 on both sides and taking integration that 2 (𝑁 − 2) 𝐸 (0) 𝑡 𝛽

(62)

for 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.

Appendix

1 ∫ ( |𝑢|2 ∇ ⋅ (𝜌𝑢) − 𝑢 ⋅ [∇ ⋅ (𝜌𝑢 ⊗ 𝑢)]) 𝑑𝑥 = 0. Ω 2

(A.1)

Proof. Firstly, by divergence theorem,

Ω

Ω

= ∫ (|𝑢|2 ∇ ⋅ (𝜌𝑢) + =∫

Ω

1 (𝜌𝑢) ⋅ ∇ |𝑢|2 ) 𝑑𝑥 2

1 2 |𝑢| ∇ ⋅ (𝜌𝑢) 𝑑𝑥. 2

(A.4)

by (A.2) .

Thus, equality (A.1) is established.

Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment This research paper is partially supported by Grant (MIT/SRG02/14-15) from the Department of Mathematics and Information Technology of the Hong Kong Institute of Education.

References

We here complement the proof of Lemma 2 by proving the equality (17); namely,

∫

∫ 𝑢 ⋅ [∇ ⋅ (𝜌𝑢 ⊗ 𝑢)] 𝑑𝑥

Ω

Case 2 (𝛽 ≠ 0). Now (59) becomes

𝐻 (𝑡) ≤ 𝐴 1 + 𝐴 2 𝑒−𝛽𝑡 +

Thus,

1 2 1 |𝑢| ∇ ⋅ (𝜌𝑢) 𝑑𝑥 = − ∫ (𝜌𝑢) ⋅ ∇ |𝑢|2 𝑑𝑥. 2 2 Ω

(A.2)

Secondly, by definitions of the operations, 𝑁

𝑁

𝑗=1

𝑖=1

𝑁

𝑁

𝑗=1

𝑖=1

𝑢 ⋅ [∇ ⋅ (𝜌𝑢 ⊗ 𝑢)] = ∑ 𝑢𝑗 (∑𝜕𝑖 (𝜌𝑢𝑖 𝑢𝑗 ))

= ∑ 𝑢𝑗 (∑ (𝑢𝑗 𝜕𝑖 (𝜌𝑢𝑖 ) + 𝜌𝑢𝑖 𝜕𝑖 𝑢𝑗 )) 𝑁

𝑁

𝑁

𝑁

𝑗=1

𝑖=1

𝑗=1

𝑖=1

= ∑ 𝑢𝑗2 ∑𝜕𝑖 (𝜌𝑢𝑖 ) + ∑𝑢𝑗 ∑𝜌𝑢𝑖 𝜕𝑖 𝑢𝑗 𝑁 𝑁 1 = |𝑢|2 ∇ ⋅ (𝜌𝑢) + ∑𝜌𝑢𝑖 ∑ 𝜕𝑖 𝑢𝑗2 𝑖=1 𝑗=1 2

1𝑁 = |𝑢|2 ∇ ⋅ (𝜌𝑢) + ∑𝜌𝑢𝑖 𝜕𝑖 |𝑢|2 2 𝑖=1 = |𝑢|2 ∇ ⋅ (𝜌𝑢) +

1 (𝜌𝑢) ⋅ ∇ |𝑢|2 . 2

(A.3)

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