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

Authors: Ka Luen Cheung; Sen Wong
Journal: The Scientific World Journal
 (2016)

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## Abstract

The blowup phenomenon of solutions is investigated for the initial-boundary value problem (IBVP) of theN-dimensional Euler equations with spherical symmetry. We first show that there are only trivial solutions when the velocity is of the form c(t)xα-1x+b(t)(x/x) for any value of α≠1 or any positive integer N≠1. Then, we show that blowup phenomenon occurs when α=N=1 and c2(0)+c˙(0)<0. As a corollary, the blowup properties of solutions with velocity of the form (a˙t/at)x+b(t)(x/x) are obtained. Our analysis includes both the isentropic case (γ>1) and the isothermal case (γ=1).


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## Body

## 1. Introduction and Main Results

In this paper, we consider theN-dimensional Euler equations for compressible fluid:(1)ρt+∇·ρu=0,ρut+u·∇u+∇p=0,p=Kργ,γ≥1with boundary condition(2)u·nx=1=0,where ρ, u, and p represent the density, velocity, and pressure of the fluid, respectively. n is the unit normal vector on the unit sphere. The γ-law for p is given by (1)3. The fluid is called isentropic if γ>1 and is called isothermal if γ=1.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 [3–11].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 [10], the authors consider the solutions of (1) with velocity of the form(3)ut,x=ctxand show that, by using the standard argument of phase diagram, the solutions will be expanding if c(0) and c˙(0) satisfy some inequalities. It is natural to consider the more general velocity form:(4)ut,x=ctx+btxxfor the IBVP of system (1) in spherical symmetry, where b(t) is a time-dependent drifting function.For solutions in spherical symmetry, namely,ρ(t,x)=ρ(t,r) and u(t,x)=u(t,r)(x/r), system (1) together with (2) is transformed to(5)ρt+uρr+ρur+N-1rρu=0,t>0,r>1,ρut+uur+pr=0,t>0,r>1,p=Kργ,γ≥1,ut,1=0,where r=|x| is the length of the spatial variable x.Our main contributions in this paper are stated as follows.Theorem 1.
There are only trivial solutions to theN-dimensional Euler system (5) of the form u=c(t)rα+b(t) for any real α and integer N in the isentropic and isothermal cases except the case γ=α=N=1. For γ=α=N=1, one has the following two cases.(1)
Ifc2(0)+c˙(0)>0, then, for any t,c2(t)+c˙(t)>0, and c(t)→0, (ρ,u)→(0,0) as t→∞.(2)
Ifc2(0)+c˙(0)<0, then, for any t,c2(t)+c˙(t)<0. Moreover, in the region where c2(t)+c˙(t)<0, c(t)→-∞ and (ρ,u)→(∞,-∞) as t→∞.As a corollary, we also obtain the following.Corollary 2.
Let(ρ,u) be a solution for (5) with u=(a˙t/at)r+b(t), a0≔a(0)>0, and γ=N=1. Then a(t) satisfies(6)a¨=λafor some constant λ∈R. Furthermore, one has the following five cases.(1)
Ifλ<0, then the solution (ρ,u)→(∞,-∞) as t→T∗ for some finite T∗>0.(2)
Ifλ>0, then ρ is bounded above and the solution (ρ,u)→(0,0) as t→∞.(3)
Ifλ=0 and a1>0, ρ is bounded above and the solution (ρ,u)→(0,0) as t→∞.(4)
Ifλ=0 and a1=0, then the solution is trivial.(5)
Ifλ=0 and a1<0, then the solution (ρ,u)→(∞,-∞) as t→-a0/a1.


## 2. Lemmas

It is well-known thatρ is always positive if the initial datum ρ(0,r) is set to be positive. Thus, we suppose ρ(0,r)>0 in the following to avoid the trivial solutions ρ≡0.Lemma 3.
Forγ>1, one has(7)ργ-1γ-1t+ργ-1γ-1ru+ργ-1ur+N-1ru=0.Proof.
From (5)1, one has(8)ρt+ρru+ρur+N-1ru=0.Multiply both sides by ργ-2. Then, the result follows.Lemma 4.
Forγ>1, one has(9)ργ-1γ-1r=-1Kγut+uur,(10)ργ-1=ργ-1t,1-γ-1Kγ∫1rut+uurt,sds,(11)ργ-1γ-1t=ddtργ-1t,1γ-1-1Kγ∫1rut+uurtt,sds.Proof.
From (5)2, one has(12)ut+uur+Kργ-2ρr=0,ut+uur+Kργ-1γ-1r=0and the results follow.Similarly, we have the following two lemmas forγ=1.Lemma 5.
Forγ=1, one has(13)ln⁡ρt+ln⁡ρru+ur+N-1ru=0.Lemma 6.
Forγ=1, one has(14)ln⁡ρr=-1Kur+uur,(15)ln⁡ρ=ln⁡ρt,1-1K∫1rur+uurt,sds,(16)ln⁡ρt=ddtln⁡ρt,1-1K∫1rur+uurtt,sds.Lastly, one has the following lemma that will be used to prove that there are only trivial solutions when u(t,r)=c(t)r+b(t) and γ>1.Lemma 7.
Consider the following dynamical system(17)c3+A1cc˙+B1c¨=0,c3+A2cc˙+B2c¨=0with A1≠A2 or B1≠B2. If A≠0, B≠0, and(18)AA1B2-A2B1=2B2,then (17) is equivalent to(19)A2c2+Bc˙=0,where(20)A≔A2-A1,B≔B2-B1.Otherwise, the solution to (17) is trivial.Proof.
IfA1B2-A2B1=0, then it is clear that c=0 is the only solution. So we suppose A1B2-A2B1≠0. One has from (17) that(21)Acc˙+Bc¨=0.
IfB=0 and A≠0, then c=0 is the only solution.
IfA=0 and B≠0, then c=0 is the only solution.
So, we suppose bothA and B are not zero.
From (21), one has(22)A2c2+Bc˙=ξ,for some constant ξ.
From (21) and (22), one has(23)c¨=-ABcc˙,c˙=1Bξ-A2c2.Thus, from (17)1, one has(24)c1-AA1B2-A2B12B2c2+A1B2-A2B1B2ξ=0.If 1-AA1B2-A2B1/2B2≠0, then c=0 is the only solution. So we suppose it is zero; that is, (18) holds. Then, if ξ≠0, c=0 is the only solution. So we suppose ξ=0. Thus, we have(25)A2c2+Bc˙=0.Conversely, if one has (18) and (19), then system (17) is satisfied. The proof is complete.


## 3. Proofs of Main Results

Proposition 8.
Assumeγ>1. Then there are only trivial solutions to the N-dimensional Euler system (5) of the form u=c(t)rα+b(t) with α≠1.Proof.
First, we set(26)ut,r=ctrα+bt.From (5)4, we have(27)u=ctrα-1.Then,(28)ut+uur=c˙rα+αc2r2α-1-αc2rα-1-c˙,(29)ut+uurt=c¨rα+2αcc˙r2α-1-2αcc˙rα-1-c¨,(30)ur+N-1ru=α+N-1crα-1-N-1cr-1.For N>1, if α=0, then u=0 from (27). It follows from (7) and (9) that ρ(t,r) is independent of t and r, respectively. Thus, ρ is a constant.
Forα≠0 and -1, after substituting (28), (29), and (30) into (9), (11), and (7), respectively, we see that (7) becomes(31)D1r3α-1+D2r2α+D3r2α-1+D4rα+1+D5rα+D6rα-1+D7r1+D8r0+D9r-1=0for all r≥1, where Dk are functions of t only. More precisely, one has(32)D1=-2α+γ-1α+N-12Kγc3,D2=-γ-1α+N-1+2α+2Kγα+1cc˙,D3=γ-12α+3N-3+4α2Kγc3,D4=-1Kγα+1c¨,D5=γ-1α+N+1+γ-1N-1α+1+41Kγcc˙,D6=-α+γ-1N-1Kγc3+α+N-1Fc,D7=1Kγc¨,D9=-N-1Fc,where(33)F≔ργ-1t,1-γ-1Kγαα+1c˙+c22.Note that we omitted D8 as it is irrelevant in the proof.
Ifα∉{0,1/3,1/2,2/3,1,2}, then the powers 3α-1 and 2α-1 are different and unique among the powers in (31). In this case, one has(34)D1=0,D3=0.As the two constants 2α+(γ-1)(α+N-1) and (γ-1)(2α+3N-3)+4α cannot be both zero for N≠1, we conclude that c=0. Hence, u=0 and ρ is a constant.
Forα∈{1/3,1/2,2/3,2}, we have Table 1.
Forα=1/3 or 2/3, as 2α-1 is unique among other powers, one has(35)D3=γ-12α+3N-3+4α2Kγc3=0.As (γ-1)(2α+3N-3)+4α≠0, we conclude that c=0. Thus, u=0 and ρ is a constant.
Forα=1/2, as α-1 and -1 are different and unique among other powers, one has(36)D6=0,D9=0,which is reduced to(37)-α+γ-1N-1Kγc3=0for N≠1. As α+(γ-1)(N-1)≠0, we conclude that c=0. Thus, u=0 and ρ is a constant.
Forα=2, as 3α-1 is unique among other powers, one has(38)D1=-2α+γ-1α+N-12Kγc3=0.As 2α+(γ-1)(α+N-1)≠0, we conclude that c=0. Thus, u=0 and ρ is a constant.
Next, we consider the caseα=-1.
Forα=-1, the corresponding equation of (31) is(39)E1ln⁡r+E2r-1ln⁡r+E3r-2ln⁡r+E4r+E5r0+E6r-1+E7r-2+E8r-3+E9r-4=0for all r≥1, where Ek are functions of t only. As ln⁡r is not a rational function, one has that all Ek=0. In particular, one has(40)E8=-4+γ-13N-52Kγc3=0,E9=2-N-2γ-12Kγc3=0.As -4+(γ-1)(3N-5) and 2-(N-2)(γ-1) cannot be both zero for N≠1, we conclude that c=0 and the solutions are trivial.
ForN=1 and α≠1, one can show that there are only trivial solutions with similar procedures. The proof is complete.Table 1
α
1/3
1/2
2
/
3
2
3
α
-
1 (D1)
0
1/2
1
5
2
α (D2)
2/3
1
4/3
4
2
α
-
1 (D3)
−1/3
0
1/3
3
α
+
1 (D4)
4/3
3/2
5/3
3
α (D5)
1/3
1/2
2/3
2
α
-
1 (D6)
−2/3
−1/2
−1/3
1
1 (D7)
1
1
1
1
0 (D8)
0
0
0
0
−1 (D9)
−1
−1
−1
−1Using similar analysis as that given for the caseγ>1 in Proposition 8, we obtain the following proposition for the case γ=1.Proposition 9.
Assumeγ=1. Then there are only trivial solutions to the N-dimensional Euler system (5) of the form u=c(t)rα+b(t) with α≠1.Next, the crucial caseα=1 will be analyzed as follows.Proposition 10.
Assumeγ>1. Then there are only trivial solutions to the N-dimensional Euler system (5) of the form u=c(t)rα+b(t) with α=1.Proof.
Forγ>1 and α=1, one has(41)D1+D2+D4=0,D3+D5+D7=0,D6+D8=0,D9=0.(41)1 and (41)2 are equivalent to(42)c3+A1cc˙+B1c¨=0,c3+A2cc˙+B2c¨=0,where(43)A1=γ-1N+4γ-1N+2,B1=1γ-1N+2,A2=3γ-1N+1+8γ-13N-1+4,B2=2γ-13N-1+4.Note that B1=B2 is equivalent to N=1 and A1=A2 is equivalent to(44)γ-1=N-52N,N≥6.Thus, we have either A1≠A2 or B1≠B2. Moreover, condition (18) is equivalent to(45)γ-1=-8NN+3,which is impossible for γ>1. Thus, we conclude by Lemma 7 that there are only trivial solutions.Next, we consider the remaining caseγ=α=1.Proposition 11.
Assumeγ=1. Then there are only trivial solutions to the N-dimensional Euler system (5) of the form u=c(t)rα+b(t) with α=1 and N>1. For N=α=1, one has the following two cases.(1)
Ifc2(0)+c˙(0)>0, then, for any t, c2(t)+c˙(t)>0, and c(t)→0, (ρ,u)→(0,0) as t→∞.(2)
Ifc2(0)+c˙(0)<0, then, for any t, c2(t)+c˙(t)<0. Moreover, in the region where c2(t)+c˙(t)<0, c(t)→-∞ and (ρ,u)→(∞,-∞) as t→∞.Proof.
Forγ=α=1, the corresponding system of (41) is(46)G1+G2+G4=0,G3+G5+G7=0,G6+G8=0,G9=0,where(47)G1=-1Kc3,G2=-2Kcc˙,G3=2Kc3,G4=-12Kc¨,G5=4Kcc˙,G6=-1Kc3+Nc,G7=1Kc¨,G8=-1Kcc˙+ddtln⁡ρt,1-1K12c¨+cc˙,(48)G9=-N-1c.It is clear that from (46)4 and (48) we have only the trivial solutions if N≠1. So we suppose N=1. Then (46) is equivalent to(49)2c3+4cc˙+c¨=0,ddtln⁡ρt,1=-c.Note that (49)1 is a special case of equation (7) in [10] when we set the parameter N in (7) to be zero. Thus, by Theorem  2.1 in [10], the results (1) and (2) in the proposition follow.Remark 12.
From (49)2 and (15), the density function ρ is given by(50)ρt,r=ρ0,1e-∫0tcsdse-c2+c˙/K1/2r2-r+1/2.Thus, the total mass is finite if c2(0)+c˙(0)>0 and is infinite if c2(0)+c˙(0)<0. From (1) and (2) in the proposition, we see that blowup can occur only when the total mass is infinite.Proof of Theorem1.
Theorem1 is followed from Propositions 8–11.Finally, we are ready to present the proof of Corollary2.Proof of Corollary2.
Letc=a˙/a in (49)1. One has(51)a⃛a+a˙a¨a2=0.It follows that(52)ddtaa¨=0,aa¨=λ,where λ≔a0a¨(0). Thus, a satisfies (6). Consider (6) with initial conditions(53)a0=a0>0,a1≔a˙0.Set(54)T∗≔sup⁡t≥0:a>0on0,t>0.First, note that if T∗ is finite, then the one-sided limit limt→T∗a(t) must be zero. More precisely, if T∗ is finite and limt→T∗a(t)>0, then we can extend the solution by solving (6) with initial condition a(T∗)≔limt→T∗a(t)>0. This contradicts the definition of T∗.
Next, supposeλ<0, and then(55)a¨=λa,(56)a˙a¨=λa˙a,ddt12a˙2-λln⁡a=0,12a˙2-λln⁡a=μ,where μ≔(1/2)a12-λln⁡a0. It follows that(57)μ+λln⁡a=12a˙2≥0,ln⁡a≤-μλ,a≤e-μ/λ.From (55), one has(58)a¨≤λeμ/λ,at≤12λeμ/λt2+a1t+a0.As the coefficient of t2 is negative, we see that a(t) will be negative if t is sufficiently large. This implies that T∗<(a1+a12-2a0λeμ/λ)/-λeμ/λ is finite.
On the other hand, from (55), one has(59)a˙=a1+∫0tλasds≤a1+λeμ/λt,limt→T∗⁡a˙≤a1+λeμ/λT∗<-a12-2a0λeμ/λ<0.Thus,(60)limt→T∗⁡a˙a=-∞.Thus, u→-∞ and ρ→∞ as t→T∗.
Forλ>0, one has(i)
a≥e-μ/λ>0,(ii)
T∗=∞,(iii)
(61)ρt,r=a0ρ0,1ate-λ/2Ka2tr-12.From (i), (ii), and (iii) above, we see that ρ is bounded above by(62)ρ0,1a0eμ/λ.Moreover, we have limt→∞a(t)=∞. This is because if limt→∞a(t) is finite, then a(t) is bounded by some positive number M>0. But, from (6), one has(63)a¨≥λM,at≥λ2Mt2+ta1+a0,which implies that a is unbounded as the coefficient of t2 is positive.
Next, we show that(64)limt→∞⁡a˙a=0.If limt→∞a˙ is finite, then (64) is clearly held. If limt→∞a˙ is not finite, then (65)limt→∞⁡a˙a=limt→∞⁡a¨a˙=limt→∞⁡λaa˙=0.Thus, for λ>0, (ρ,u)→(0,0) as t→∞.
As the cases forλ=0 can be verified trivially, the proof is complete.



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