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## Qualitative Theory of Differential Equations, Difference Equations, and Dynamic Equations on Time Scales

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Research Article | Open Access

Volume 2016 |Article ID 3781760 | 6 pages | https://doi.org/10.1155/2016/3781760

# Blowup Phenomenon of Solutions for the IBVP of the Compressible Euler Equations in Spherical Symmetry

Accepted19 Jan 2016
Published03 Feb 2016

#### Abstract

The blowup phenomenon of solutions is investigated for the initial-boundary value problem (IBVP) of the -dimensional Euler equations with spherical symmetry. We first show that there are only trivial solutions when the velocity is of the form for any value of or any positive integer . Then, we show that blowup phenomenon occurs when and . As a corollary, the blowup properties of solutions with velocity of the form are obtained. Our analysis includes both the isentropic case and the isothermal case .

#### 1. Introduction and Main Results

In this paper, we consider the -dimensional Euler equations for compressible fluid:with boundary conditionwhere , , and represent the density, velocity, and pressure of the fluid, respectively. is the unit normal vector on the unit sphere. The -law for is given by (1)3. The fluid is called isentropic if and is called isothermal if .

Euler equation (1) is one of the most important fundamental equations in inviscid fluid dynamics. Many interesting fluid dynamic phenomena can be described by system (1) [1, 2]. The Euler equations are also the special case of the noted Navier-Stokes equations, whose problem of whether there is a formation of singularity is still open and long-standing. Thus, the singularity formation in fluid mechanics has been attracting the attention of a number of researchers .

In particular, in [3, 4], the authors obtain blowup results for the IBVP of the Euler equations, namely, system (1) with boundary condition (2). By making use of the finite propagation speed property [5, 6], they are able to apply the integration method to derive differential inequalities and show that if the initial weighted functionals of velocity or momentum are large enough, then blowup occurs.

In , the authors consider the solutions of (1) with velocity of the formand show that, by using the standard argument of phase diagram, the solutions will be expanding if and satisfy some inequalities. It is natural to consider the more general velocity form:for the IBVP of system (1) in spherical symmetry, where is a time-dependent drifting function.

For solutions in spherical symmetry, namely, and , system (1) together with (2) is transformed towhere is the length of the spatial variable .

Our main contributions in this paper are stated as follows.

Theorem 1. There are only trivial solutions to the -dimensional Euler system (5) of the form for any real and integer in the isentropic and isothermal cases except the case . For , one has the following two cases.(1)If , then, for any ,, and , as .(2)If , then, for any ,. Moreover, in the region where , and as .

As a corollary, we also obtain the following.

Corollary 2. Let be a solution for (5) with , , and . Then satisfiesfor some constant . Furthermore, one has the following five cases.(1)If , then the solution as for some finite .(2)If , then is bounded above and the solution as .(3)If and , is bounded above and the solution as .(4)If and , then the solution is trivial.(5)If and , then the solution as .

#### 2. Lemmas

It is well-known that is always positive if the initial datum is set to be positive. Thus, we suppose in the following to avoid the trivial solutions .

Lemma 3. For , one has

Proof. From (5)1, one hasMultiply both sides by . Then, the result follows.

Lemma 4. For , one has

Proof. From (5)2, one hasand the results follow.

Similarly, we have the following two lemmas for .

Lemma 5. For , one has

Lemma 6. For , one hasLastly, one has the following lemma that will be used to prove that there are only trivial solutions when and .

Lemma 7. Consider the following dynamical systemwith or . If , , andthen (17) is equivalent towhereOtherwise, the solution to (17) is trivial.

Proof. If , then it is clear that is the only solution. So we suppose . One has from (17) thatIf and , then is the only solution.
If and , then is the only solution.
So, we suppose both and are not zero.
From (21), one hasfor some constant .
From (21) and (22), one hasThus, from (17)1, one hasIf , then is the only solution. So we suppose it is zero; that is, (18) holds. Then, if , is the only solution. So we suppose . Thus, we haveConversely, if one has (18) and (19), then system (17) is satisfied. The proof is complete.

#### 3. Proofs of Main Results

Proposition 8. Assume . Then there are only trivial solutions to the -dimensional Euler system (5) of the form with .

Proof. First, we setFrom (5)4, we haveThen,For , if , then from (27). It follows from (7) and (9) that is independent of and , respectively. Thus, is a constant.
For and , after substituting (28), (29), and (30) into (9), (11), and (7), respectively, we see that (7) becomesfor all , where are functions of only. More precisely, one haswhereNote that we omitted as it is irrelevant in the proof.
If , then the powers and are different and unique among the powers in (31). In this case, one hasAs the two constants and cannot be both zero for , we conclude that . Hence, and is a constant.
For , we have Table 1.
For or , as is unique among other powers, one hasAs , we conclude that . Thus, and is a constant.
For , as and are different and unique among other powers, one haswhich is reduced tofor . As , we conclude that . Thus, and is a constant.
For , as is unique among other powers, one hasAs , we conclude that . Thus, and is a constant.
Next, we consider the case .
For , the corresponding equation of (31) isfor all , where are functions of only. As is not a rational function, one has that all . In particular, one hasAs and cannot be both zero for , we conclude that and the solutions are trivial.
For and , one can show that there are only trivial solutions with similar procedures. The proof is complete.

 1/3 1/2 () 0 1/2 1 5 () 2/3 1 4/3 4 () −1/3 0 1/3 3 () 4/3 3/2 5/3 3 () 1/3 1/2 2/3 2 () −2/3 −1/2 −1/3 1 1 () 1 1 1 1 0 () 0 0 0 −1 () −1 −1 −1 −1

Using similar analysis as that given for the case in Proposition 8, we obtain the following proposition for the case .

Proposition 9. Assume . Then there are only trivial solutions to the -dimensional Euler system (5) of the form with .

Next, the crucial case will be analyzed as follows.

Proposition 10. Assume . Then there are only trivial solutions to the -dimensional Euler system (5) of the form with .

Proof. For and , one has(41)1 and (41)2 are equivalent towhereNote that is equivalent to and is equivalent toThus, we have either or . Moreover, condition (18) is equivalent towhich is impossible for . Thus, we conclude by Lemma 7 that there are only trivial solutions.

Next, we consider the remaining case .

Proposition 11. Assume . Then there are only trivial solutions to the -dimensional Euler system (5) of the form with and . For , one has the following two cases.(1)If , then, for any , , and , as .(2)If , then, for any , . Moreover, in the region where , and as .

Proof. For , the corresponding system of (41) iswhereIt is clear that from (46)4 and (48) we have only the trivial solutions if . So we suppose . Then (46) is equivalent toNote that (49)1 is a special case of equation in  when we set the parameter in (7) to be zero. Thus, by Theorem  2.1 in , the results and in the proposition follow.

Remark 12. From (49)2 and (15), the density function is given byThus, the total mass is finite if and is infinite if . From and in the proposition, we see that blowup can occur only when the total mass is infinite.

Proof of Theorem 1. Theorem 1 is followed from Propositions 811.

Finally, we are ready to present the proof of Corollary 2.

Proof of Corollary 2. Let in (49)1. One hasIt follows thatwhere . Thus, satisfies (6). Consider (6) with initial conditionsSetFirst, note that if is finite, then the one-sided limit must be zero. More precisely, if is finite and , then we can extend the solution by solving (6) with initial condition . This contradicts the definition of .
Next, suppose , and thenwhere . It follows thatFrom (55), one hasAs the coefficient of is negative, we see that will be negative if is sufficiently large. This implies that is finite.
On the other hand, from (55), one hasThus,Thus, and as .
For , one has(i),(ii),(iii)From (i), (ii), and (iii) above, we see that is bounded above byMoreover, we have . This is because if is finite, then is bounded by some positive number . But, from (6), one haswhich implies that is unbounded as the coefficient of is positive.
Next, we show thatIf is finite, then (64) is clearly held. If is not finite, then Thus, for , as .
As the cases for can be verified trivially, the proof is complete.

#### 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 no. MIT/SRG02/15-16 from the Department of Mathematics and Information Technology of the Hong Kong Institute of Education.

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