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Nonlinear Analysis

journal homepage:www.elsevier.com/locate/na

Self-similar solutions of the Euler equations with spherical symmetry

Chen-Chang Peng

a

, Wen-Ching Lien

b,

aDepartment of Applied Mathematics, National Chiayi University, No. 300, Xuefu Rd., Chiayi City, 60004, Taiwan

bDepartment of Mathematics, National Cheng Kung University, No. 1, Dasyue Rd., Tainan City, 70101, Taiwan

a r t i c l e i n f o

Article history:

Received 4 May 2012 Accepted 20 July 2012

Communicated by Enzo Mitidieri

MSC:

34B 35L 76N Keywords:

Euler equations Shock waves Self-similarity

a b s t r a c t

We consider self-similar flows arising from the uniform expansion of a spherical piston and preceded by a shock wave front. With appropriate boundary conditions imposed on the piston surface and the spherical shock, the isentropic compressible Euler system is transformed into a nonlinear ODE system. We formulate the problem in a simple form in order to present the analytic proof of the global existence of positive smooth solutions.

© 2012 Elsevier Ltd. All rights reserved.

1. Introduction

Spherically symmetric motion is important in the theory of explosion. The simplest way to simulate an explosion is to view it as a spherical piston motion pushing out undisturbed gas ahead of it; as a result, shock waves occur. In this paper, we study the spherical, self-similar flow which arises from the uniformly expanding piston and is preceded by a shock front moving with a constant speed. When the spherical piston expands at a constant speed, with the self-similar assumption, the problem can be reduced to that of solving two coupled nonlinear ODEs with appropriate boundary conditions imposed on the piston surface and the spherical shock. The main purpose here is to provide an analytic proof of the global existence of such solutions for the nonlinear system of ODEs.

We consider Euler’s equations with spherical symmetry:

 

 

∂ρ

t

+ ∂(ρ

u

)

r

= −

2

ρ

u

r

,

r

>

0

,

t

>

0

,

∂(ρ

u

)

t

+ ∂(ρ

u2

+

P

)

r

= −

2

ρ

u2 r

,

(1.1)

where

ρ,

u and P are the density, velocity and pressure of the gas respectively. Assume that P

(ρ) = ρ

γfor isentropic gases and

γ ∈ [

1

,

3

]

is the adiabatic constant. And the speed of sound is c

= √

P

(ρ)

. For the motions caused by the expansion of a spherical piston into still gas, the velocity of the flow is radial, and its magnitude, like those of the density, pressure, temperature and entropy, depends on the time t and the distance r from the center of the piston. Here, t is chosen to be

Corresponding author. Tel.: +886 6 2757575 65123; fax: +886 6 2743191.

E-mail addresses:ccpeng@mail.ncyu.edu.tw(C.-C. Peng),wlien@mail.ncku.edu.tw(W.-C. Lien).

0362-546X/$ – see front matter©2012 Elsevier Ltd. All rights reserved.

doi:10.1016/j.na.2012.07.019

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Fig. 1. Diagram of symbols.

zero when r

=

0, and the motion is supposed to be so small that only weak shocks are produced; therefore, the changes in entropy are ignored.

There are many studies of spherical shock waves. Self-similar solutions are well introduced and discussed in the books [1–4]. The nonlinear ODE problem was first proposed and solved numerically by Taylor [5]. He studied(1.1)for isentropic gases. Due to the self-similar motion, u and P are functions of x

=

r

/

t only. He introduced the following variables in order to express(1.1)in nondimensional form:

ξ =

u

/

x

, η =

c2

/

x2

,

z

=

ln x. The nonlinear ODEs are obtained to solve for

η(ξ)

and z

(ξ)

numerically. Taylor’s results show that if the radial velocity of the expanding piston is sufficiently large, the thickness of the layer of disturbed air is close to 6% of the radius of the piston. However, his numerical approach did not obtain suitable solutions for when the nondimensional expansion speed is small. Courant and Friedrich [6] investigated the vector field of the ODE system in a different form, to illustrate the solution curves. Lighthill [7] introduced the velocity potential in order to obtain a nonlinear second-order ODE. By ignoring certain terms in the equation, he obtained an approximate relation between the shock Mach number M and the nondimensional piston velocity

α

. Also, local analytic solutions of the ODE system are pursued by many applied mathematicians, and we refer readers to Sachdev [8] for thorough discussions and [9,10] for related works. However, rigorous analysis of global existence is absent. In this paper, we introduce a suitable variable in terms of which to rewrite the system of nonlinear ODEs and present the global existence of smooth solutions for the first-order ODEs. Due to the effective form, we notice that the system is singular when the shock Mach number M

=

1.

In [6,7,5], they show (either by numerical solution or by means of the approximate equation) that M is a monotone function of the prescribed piston velocity

α

, and M approaches 1 when

α

is close to zero.

In this work, we assume that the spherical piston moves outward at a constant speed c0

α

and the gas flow is headed by a weak shock moving at a constant speed c0M. Here, M is the shock Mach number, M

>

1

, α

is the nondimensional piston velocity and

α >

0 is small. The constant c0

(= √

P

0

))

denotes the speed of sound for the undisturbed gas. We introduce a nondimensional variable

ξ

:

ξ =

r R

(

t

) ,

where R

(

t

)(=

c0Mt

)

is the shock radius at time t. Notice that

ξ =

1 represents the shock locus.Fig. 1shows the symbols that we use in this paper. Our goal is to seek the solutions of system(1.1)in the following self-similar form:

u

(

r

,

t

) =

c0Mf

(ξ),

ρ(

r

,

t

) = ρ

0h

(ξ).

(1.2)

By the above assumption of self-similarity, system(1.1)of partial differential equations is transformed into a system of nonlinear ordinary differential equations:

 

 

 

 

f

(ξ) =

2fhγ −1

[M

(ξ −

f

)

]2

ξ −

hγ −1

ξ ,

h

(ξ) =

2M2fh

(ξ −

f

)

[M

(ξ −

f

)

]2

ξ −

hγ −1

ξ .

(1.3)

For system(1.1), there are two families of elementary waves. Due to the Rankine–Hugoniot condition and the entropy condition [6], we can obtain the flow velocity and density immediately behind the two-shock wave as follows. We need to solve the following equation:

1

γ

Myγ +

1

1

γ

M

+

M

y

+

M

=

0

.

(1.4)

(4)

There is only one root greater than 1. We denote this root by y1and set the initial condition of system(1.3)as follows:

f

(

1

) =

1

1 y1

,

h

(

1

) =

y1

.

(1.5)

Furthermore, the kinematic condition at the piston requires that the flow velocity on the piston surface is the same as the piston velocity, which gives us

f

2

) = ξ

2

.

(1.6)

We note that

ξ

1

=

1 and

ξ

2

=

the radius of the piston the radius of the shock

= α

M

.

Now the problem can be treated as an initial value problem. For a given

γ ∈ [

1

,

3

]

and given M

>

1, we first solve system (1.3)with the initial condition(1.5). Then system(1.3)is integrated from

ξ =

1 backward until the value

ξ = ξ

2for which (1.6)holds is reached. We obtain the following main result.

Theorem 1.1. Consider the ODE system(1.3)satisfying the conditions(1.5)–(1.6). For

γ ≥

1 with any given M

>

1, there exists a unique positive smooth solution

(

f

(ξ),

h

(ξ))

for

ξ ∈ [ξ

2

,

1

]

. Moreover, f

(ξ)

and h

(ξ)

are decreasing functions for

ξ ∈ [ξ

2

,

1

]

. To prove the main theorem, we carefully investigate the common denominator of system(1.3)together with

ξ −

f

(ξ)

to show that the solutions are actually decreasing. The existence and uniqueness of the global solution for

γ =

1 is shown in Section2, and the proof for

γ >

1 is in Section3. The conclusions are finally given in Section4. We also present the numerical solutions for f

(ξ)

and h

(ξ)

of system(1.3)in Sections2and3to give a better understanding of the structure of the solution.

2. Existence for

γ =

1

In this section, we study the case of

γ =

1. The system of ODEs is written as follows:

 

 

f

(ξ) =

2f

[M

(ξ −

f

)

]2

ξ − ξ ,

h

(ξ) =

2M2fh

(ξ −

f

)

[M

(ξ −

f

)

]2

ξ − ξ ,

(2.1)

with the initial value

f

(

1

) =

1

M2

,

h

(

1

) =

M2

.

(2.2)

To prove the main theorem, we define a subsidiary function:

I1

(ξ) = (

M

(ξ −

f

))

2

1

,

which is the main ingredient of the denominator in the ODE system. We note that

I1

(

1

) =

M2

1

<

0

.

(2.3)

I1

(ξ) =

2M2

(ξ −

f

)(

1

f

),

(2.4)

provided that fexists. We also note that f

(

1

) >

0

,

f

(

1

) <

0

,

h

(

1

) >

0 and h

(

1

) <

0. Hence, in a small neighborhood of

ξ =

1

,

f

(ξ)

and h

(ξ)

are decreasing functions. By investigating the subsidiary function I1

(ξ)

together with

ξ −

f

(ξ)

, we can further show that f

(ξ)

is always negative, which will imply that f

(ξ)

is monotone decreasing.

Let the interval

(ˆξ,

1

]

denote the maximal interval of existence for the positive continuous solution of system(2.1)–(2.2), where

ξ ∈ [ ˆ

0

,

1

)

. We need the following technical lemma to prove that f

(ξ)

is always negative.

Lemma 2.1. Let

γ =

1 and M

>

1. Let f

(ξ)

and h

(ξ)

be the positive continuous solutions of system(2.1)satisfying the initial condition(2.2).

Let

ξ ∈ (ˆξ, ˜

1

)

be given. Suppose that

ξ >

f

(ξ)

for all

ξ ∈ [˜ξ,

1

]

; then f

(ξ) >

f

(

1

)

and I1

(ξ) <

0 for all

ξ ∈ [˜ξ,

1

)

.

Proof. We prove by contradiction. Since f

(

1

) <

0, we assume that there exists

ξ

∈ [ ˜ ξ,

1

)

such that f

) =

f

(

1

)

and f

(ξ) >

f

(

1

)

for all

ξ ∈ (ξ

,

1

)

. We discuss the following two cases.

(5)

Case 1: If f is differentiable on

,

1

)

, then by the Mean Value Theorem, there exists

ξ

m

∈ (ξ

,

1

)

such that f

m

) =

0.

However, f

(ξ) ̸=

0 for all

ξ ∈ (ξ

,

1

)

due to the first equation of system(2.1)and the fact that f

(ξ) >

f

(

1

) >

0.

We obtain a contradiction.

Case 2: Suppose that f is not differentiable at some point; that is, there exists at least one zero, say,

η

1

∈ [ ξ

,

1

)

such that I1

1

) =

0 and I1

(ξ) <

0 for all

ξ ∈ (η

1

,

1

]

.

Since f

(ξ) >

0 for all

ξ ∈ (ξ

,

1

]

, we obtain from(2.1)that lim

ξ→η+1

1

I1

(ξ) = −∞

and lim

ξ→η1+f

(ξ) = −∞.

From(2.4), we have lim

ξ→η+1 I1

(ξ) = +∞.

Hence there exists

ϵ >

0 such that I1

(ξ) >

0 for all

ξ ∈ (η

1

, η

1

+ ϵ)

.

However, I1

1

) =

0 and I1

(ξ) <

0 for all

ξ ∈ (η

1

,

1

]

. By the Mean Value Theorem, there exists

ξ

m

∈ (η

1

, η

1

+ ϵ)

such that I1

m

) <

0. This is a contradiction.

Therefore, we have shown f

(ξ) >

f

(

1

)

for all

ξ ∈ [˜ξ,

1

)

.

We now prove that I1

(ξ) <

0 for all

ξ ∈ [˜ξ,

1

)

. It is easy to see that I1

(

1

) =

M2

1

<

0. Assume that there exists

η

1

∈ [ ˜ ξ,

1

)

such that I1

1

) =

0 and I1

(ξ) <

0 for all

ξ ∈ (η

1

,

1

]

. By the same argument as in Case 2, we can obtain a contradiction. Hence, I1

(ξ) <

0 for all

ξ ∈ [˜ξ,

1

]

. 

Proof of Theorem 1.1 for

γ =

1. By the local existence result for initial value problems [11], there exists a unique positive smooth solution

(

f

(ξ),

h

(ξ))

of system(2.1)satisfying the initial condition(2.2). Since f

(

1

) >

0

,

f

(

1

) <

0

,

h

(

1

) >

0 and h

(

1

) <

0, we know that in a small neighborhood of

ξ =

1

,

f

(ξ)

and h

(ξ)

are decreasing and

ξ −

f

(ξ) >

0. As the solution f

(ξ)

is extended backward from

ξ =

1, we want to find a fixed point

ξ

2

∈ (

0

,

1

)

such that

ξ

2

f

2

) =

0.

Let

(ˆξ,

1

)

denote the maximal interval of existence. We prove that

ξ

2exists by contradiction as follows. Assume that for all

ξ ∈ (ˆξ,

1

] ,

f

(ξ) ̸= ξ

. Since f

(

1

) <

1, we have

ξ >

f

(ξ)

for all

ξ ∈ (ˆξ,

1

]

. ByLemma 2.1, we have f

(ξ) >

f

(

1

)

and I1

(ξ) <

0 for all

ξ ∈ (ˆξ,

1

)

. This implies that f

(ξ) <

0 and h

(ξ) <

0 for all

ξ ∈ (ˆξ,

1

]

and hence f is strictly decreasing on

(ˆξ,

1

]

. So we have that

1

> ˆξ ≥

lim

ξ→ˆξ+

f

(ξ) ≥

f

(

1

) >

0

,

which also implies that limξ→ˆξ+h

(ξ)

is bounded due to the ODE system(2.1). By the continuation of solutions for initial value problems [11], the solution (f

(ξ),

h

(ξ)

) exists for

ξ ∈ (ˆξ − ϵ,

1

]

, where

ϵ

is a small number. This contradicts the fact that

(ˆξ,

1

]

is the maximal interval of existence of the solution.

Finally, we prove that

ξ

2is unique. f

2

) = ξ

2and, from(2.1), f

2

) = −

2. Thus, there exists

η < ξ

2such that

η <

f

(η)

. If there exists another fixed point,

ξ

3

∈ (ˆξ, ξ

2

)

such that f

3

) = ξ

3and

ξ <

f

(ξ)

for all

ξ ∈ (ξ

3

, ξ

2

)

, then we must have that f

3

) ≥

1. However, f

3

) = −

2 due to(2.1). This is a contradiction.

For

ξ ∈ (ξ

2

,

1

]

, we now have

ξ >

f

(ξ)

. Hence, byLemma 2.1, I1

(ξ) <

0 and f

(ξ)

and h

(ξ)

are strictly decreasing functions. We conclude that

(

f

(ξ),

h

(ξ)), ξ ∈ [ξ

2

,

1

]

, is the desired solution. This completes the proof ofTheorem 1.1for

γ =

1. 

Remark 2.1. Figs. 2and3show the numerical solutions of f

(ξ)

and h

(ξ)

. For

γ =

1 and an assumed value of M, one may numerically solve system(1.3)with the initial values(1.5), starting from

ξ =

1 and going backward to the value

ξ

2for which f

2

) = ξ

2. It is observed that both f

(ξ)

and h

(ξ)

are decreasing for

ξ ∈ [ξ

2

,

1

]

.

3. Existence for

γ >

1

In this section, we seek positive smooth solutions (f

(ξ),

h

(ξ)

) to the system(1.3), which satisfy the initial condition(1.5) and can be extended to the value

ξ

2such that

f

2

) = ξ

2

.

We rewrite the system(1.3)as follows:

 

 

 

 

f

(ξ) =

2fhγ −1

(

M

(ξ −

f

))

2

ξ −

hγ −1

ξ ,

0

< ξ ≤

1

,

h

(ξ) =

2M2fh

(ξ −

f

)

(

M

(ξ −

f

))

2

ξ −

hγ −1

ξ ,

(6)

Fig. 2. The numerical solution forγ =1 and M=1.1.

Fig. 3. The numerical solution forγ =1 and M=1.01.

with the same initial data as(1.5):

f

(

1

) =

1

1 y1

,

h

(

1

) =

y1

,

where y1

>

1 and satisfies Eq.(1.4):

1

γ

Myγ +

1

1

γ

M

+

M

y

+

M

=

0

.

We introduce a transformation:

g

(ξ) =

h

(ξ)

γ −1

.

(Note that

γ >

1.) The system(1.3)is simplified as follows:

f

(ξ) =

2fg

(

M

(ξ −

f

))

2

ξ −

g

ξ ,

g

(ξ) = (γ −

1

)

M2

(ξ −

f

)

f

,

(3.1)

with its initial data

f

(

1

) =

1

1 y1

,

g

(

1

) =

y1γ −1

.

(3.2)

We note that f

(

1

) >

0

,

g

(

1

) >

0.

We define a subsidiary function I

(ξ)

:

I

(ξ) ≡ (

M

(ξ −

f

(ξ)))

2

g

(ξ),

(3.3)

(7)

which is the denominator in the ODE system. By direct calculations, I

(ξ) =

2M2

(ξ −

f

(ξ)) 

1

f

(ξ) −

g

(ξ)

=

M2

(ξ −

f

(ξ)) 

2

− (γ +

1

)

f

(ξ)

(3.4)

and

I

(

1

) = −

1

y21

[

yγ +1 1

M2

] .

We first establish a technical lemma to prove that

I

(

1

) <

0

,

for M

∈ (

1

, ∞).

(3.5)

From(3.5), we know that f

(

1

) >

0

,

f

(

1

) <

0, h

(

1

) >

0 and h

(

1

) <

0. Hence, in a small neighborhood of

ξ =

1

,

f

(ξ)

and h

(ξ)

are decreasing functions. By investigating the subsidiary function I

(ξ)

together with

ξ −

f

(ξ)

, we can further show that f

(ξ)

is always negative, which implies that f

(ξ)

is monotone decreasing.

Lemma 3.1. For any

γ >

1 with any given M

>

1, we have I

(

1

) <

0

.

Proof. Let L

(

y

) =

yγ +1

− (

1

+ γ

M2

)

y

+ γ

M2and L

(

1

) =

L

(

y1

) =

0. By straightforward calculations, we have L

(

y

) = (γ +

1

)

yγ

− (

1

+ γ

M2

)

and

L

(

1

) = (γ +

1

) − (

1

+ γ

M2

) <

0

and then solving L

(

y

) =

0 for y

>

1, we obtain only one solution, denoted byy:

ˆ

y

ˆ ≡

1

+ γ

M2

γ +

1

γ1

.

We knowy

ˆ <

y1by observing the graph of L

(

y

)

for y

>

1. Therefore, to show I

(

1

) <

0, we only need to prove that y

ˆ

γ +1

M2

>

0

.

We rewritey

ˆ

γ +1

M2as follows:

y

ˆ

γ +1

M2

=

1

+ γ

M2

γ +

1

γ +

γ1

M2

=

1

γ +

1

+ γ γ +

1M

2

γ +γ1

M2

.

Define a function l

(

c

) = (

1

c

) +

cM2

1c

M2, for c

∈ [

1

2

,

1

]

. Then l

(

1

) =

0 and l

(

c

) = 

e1cln(1c+cM2)

=

e1cln(1c+cM2)

(−

1

)

c2ln

(

1

c

+

cM2

) +

c1

 −

1

+

M2 1

c

+

cM2



= 

1

c

+

cM2

1c

(−

1

)(

c2

) 

1

c

+

cM2

1



1

c

+

cM2

ln

(

1

c

+

cM2

) − (

1

c

+

cM2

) +

1

<

0

due to the fact that x ln x

x

+

1 for x

>

0. Hence, l

(

c

) >

0

,

for all c

1 2

,

1

 .

Set c

=

γ

γ +1; then we have

y

ˆ

γ +1

M2

=

1

γ +

1

+ γ γ +

1M

2

γ +γ1

M2

=

l

 γ

γ +

1

>

0

.

The proof is completed. 

(8)

The interval

(ˆξ,

1

]

still denotes the maximal interval of existence in

(

0

,

1

]

for the positive continuous solution of system (3.1)–(3.2). To find the fixed point

ξ

2for the solution f , we need the following lemmas.

Lemma 3.2. Let

ξ ∈ ( ˜

0

,

1

)

be given. f

(ξ)

and g

(ξ)

are the positive continuous solutions of system(3.1)satisfying the initial condition(3.2). Suppose that

ξ −

f

(ξ) >

0 for all

ξ ∈ (˜ξ,

1

]

. Then I

(ξ) <

0 for all

ξ ∈ (˜ξ,

1

]

.

Proof. We prove by contradiction. Assume that there exists

η ∈ (˜ξ,

1

]

such that I

(η) =

0 and I

(ξ) ̸=

0 for all

ξ ∈ (η,

1

]

. Since I

(

1

) <

0

,

I

(ξ) <

0 for all

ξ ∈ (η,

1

]

. This implies

ξ→ηlim+ 1

I

(ξ) = −∞ .

Due to the ODE system, f

(ξ) >

0 and g

(ξ) >

0 for

ξ ∈ (ˆξ,

1

]

and we have that

ξ→ηlim+f

(ξ) = −∞.

By the assumption that

ξ −

f

(ξ) >

0 for

ξ ∈ [η,

1

]

and(3.4), we get that

ξ→ηlim+I

(ξ) = +∞,

which implies that there exists

η

1

∈ (η,

1

]

such that I

(ξ) >

0 for all

ξ ∈ (η, η

1

)

. Since I

(ξ) <

0 for all

ξ ∈ (η,

1

]

and I

(η) =

0, by the Mean Value Theorem, there exists some point

ξ ∈ (η, η

1

)

such that I

(ξ) <

0. This is a contradiction because I

(ξ) >

0 for all

ξ ∈ (η, η

1

)

.

Hence, we conclude that I

(ξ) <

0. 

Lemma 3.3.

γ >

1 and M

>

1. Let f

(ξ)

and g

(ξ)

be the positive continuous solutions of system(3.1)satisfying the initial condition(3.2).

Let

ξ

∈ [ ˆ ξ,

1

)

be given. If

ξ >

f

(ξ)

for all

ξ ∈ (ξ

,

1

]

, then f

(ξ) >

f

(

1

)

and g

(ξ) >

g

(

1

)

for all

ξ ∈ (ξ

,

1

)

. Proof. We consider the following two cases.

Case 1: Assume that there exists

ξ ∈ (ξ ˜

,

1

)

such that f

(˜ξ) =

f

(

1

),

g

(˜ξ) ≥

g

(

1

)

, and that f

(ξ) >

f

(

1

)

and g

(ξ) >

g

(

1

)

for all

ξ ∈ (˜ξ,

1

)

.

Since f

(

1

) >

0 and f

(

1

) <

0, there exists

η ∈ (˜ξ,

1

)

such that f

(η) >

0. But byLemma 3.2, f

(ξ) >

0 and g

(ξ) >

0 for all

ξ ∈ (˜ξ,

1

]

and I

(ξ) <

0 for all

ξ ∈ (˜ξ,

1

]

. Due to the ODE system(3.1), this implies f

(ξ) <

0 for all

ξ ∈ (˜ξ,

1

]

. We obtain a contradiction.

Case 2: Assume that there exists

ξ ∈ (ξ ˜

,

1

)

such that f

(˜ξ) ≥

f

(

1

),

g

(˜ξ) =

g

(

1

)

, and that f

(ξ) >

f

(

1

)

and g

(ξ) >

g

(

1

)

for all

ξ ∈ (˜ξ,

1

)

.

By arguments similar to those used in Case 1, we can obtain a contradiction.

We thus conclude that f

(ξ) >

f

(

1

)

and g

(ξ) >

g

(

1

)

for all

ξ ∈ (ξ

,

1

)

. 

Proposition 3.1. Let

γ >

1 and M

>

1. Let f

(ξ)

and g

(ξ)

be the positive continuous solutions of system(3.1)satisfying the initial condition of (3.2),

ξ ∈ (ˆξ,

1

]

.

There exists a unique

ξ

2

∈ (ˆξ,

1

)

such that f

2

) = ξ

2. Proof. It is easy to see 1

>

f

(

1

) >

0 and g

(

1

) >

1.

We first prove the existence of

ξ

2by contradiction. Assume that for all

ξ ∈ (ˆξ,

1

] ,

f

(ξ) ̸= ξ

. That is

ξ >

f

(ξ)

for all

ξ ∈ (ˆξ,

1

]

.

ByLemma 3.3, f

(ξ) >

f

(

1

),

g

(ξ) >

g

(

1

)

for all

ξ ∈ (ˆξ,

1

)

. Since I

(

1

) <

0 and byLemma 3.2, I

(ξ) <

0 for all

ξ ∈ (ˆξ,

1

]

. Thus, from(3.1), we have that f

(ξ) <

0 and g

(ξ) <

0 for all

ξ ∈ (ˆξ,

1

]

. That is, f and g are strictly decreasing on

(ˆξ,

1

]

. FromLemma 3.3, 1

≥ ξ >

f

(ξ) >

f

(

1

)

for any

ξ ∈ (ˆξ,

1

]

, and hence

1

> ˆξ ≥

f

(

1

).

By the continuation of solutions for initial value problems [11] and the fact that f is bounded on

(ˆξ,

1

]

, we must have lim

ξ→ˆξ+

g

(ξ) = +∞.

(3.6)

Note that(3.6)implies that there exists some point p

∈ (ˆξ,

1

)

, and two constants M1

>

1 and c

>

0 such that g

(ξ) ≥

M1

and

g

=

2

(γ −

1

)

M2f

(ξ −

f

)

g

 (

M

(ξ −

f

))

2

g

ξ ≥ −

cg

(9)

Fig. 4. The numerical solution forγ =1.4 and M=1.1.

Fig. 5. The numerical solution forγ =1.4 and M=1.01.

for all

ξ ∈ (ˆξ,

p

]

. The above inequality leads to g

(ξ) ≤

g

(

p

)

ec(pξ)for all

ξ ∈ (ˆξ,

p

]

and thus lim

ξ→ˆξ+g

(ξ) ≤

g

(

p

)

ec(p− ˆξ)

,

which contradicts(3.6).

We now prove that

ξ

2is unique. Since f

2

) = ξ

2

,

g

2

) >

0 and by(3.1), we obtain f

2

) = −

2. Hence, there exists

η < ξ

2such that

η <

f

(η)

. If there exists

ξ

3

∈ (ˆξ, ξ

2

)

such that f

3

) = ξ

3and

ξ <

f

(ξ)

for all

ξ ∈ (ξ

3

, ξ

2

)

, then f

3

) >

1.

This is a contradiction since f

3

) = −

2 due to(3.1). 

Proof of Theorem 1.1 for

γ >

1. By the local existence result for initial value problems [11], there exists a unique positive smooth solution

(

f

(ξ),

g

(ξ))

of system(3.1)satisfying the initial condition(3.2). Since f

(

1

) >

0

,

f

(

1

) <

0

,

g

(

1

) >

0 and g

(

1

) <

0, we know that in a small neighborhood of

ξ =

1

,

f

(ξ)

and g

(ξ)

are decreasing and

ξ −

f

(ξ) >

0. As the solution f

(ξ)

is extended backward from

ξ =

1, we want to find a fixed point

ξ

2

∈ (

0

,

1

)

such that

ξ

2

f

2

) =

0.

ByProposition 3.1, we know that

ξ

2exists uniquely and f

(ξ)

satisfies condition(1.6). For

ξ ∈ (ξ

2

,

1

]

, we have

ξ >

f

(ξ)

. Hence, byLemma 3.2, I

(ξ) <

0 and f

(ξ)

and g

(ξ)

are strictly decreasing functions. Set h

(ξ) =

g

(ξ)

γ −11. Then

(

f

(ξ),

h

(ξ))

is the desired solution. 

Remark 3.1. Figs. 4and5show the numerical solutions of f

(ξ)

and h

(ξ)

. For

γ =

1

.

4 and an assumed value of M, one may numerically solve system(1.3)with the initial values(1.5), starting from

ξ =

1 and going backward to the value

ξ

2for which f

2

) = ξ

2. It is observed that both f

(ξ)

and h

(ξ)

are decreasing for

ξ ∈ [ξ

2

,

1

]

. We also refer the reader to [12] for more numerical results.

4. Conclusions

For a spherical piston motion pushing out undisturbed gas ahead of it, as modeled by(1.1), shock waves occur. Assuming that the spherical piston moves outward at a constant speed c0

α

and the gas flow is headed by a weak shock moving at a constant speed c0M, the nonlinear ODEs are introduced. We have presented an analytic proof of the global existence of positive solutions for the nonlinear ODE system(1.3)satisfying the conditions(1.5)and(1.6).

In this simple model, the motion is supposed to be so small that only weak shocks are produced; hence, the changes in entropy are ignored. Besides, to maintain constant speed of the piston in three-dimensional space, an energy supply at

(10)

an increasing rate is required. However, in the actual situation of explosion, the total energy is given (usually by a given mass of explosive), and the strength of the shock and the change of entropy rapidly decrease; therefore, the flow is no longer isentropic. Even with such limitations, one can still see the present model as a proper formulation of the problem in intermediate time and space intervals.

Acknowledgments

The authors would like to thank Prof. M-H. Chen for helpful discussions and the programming of the numerical solutions.

Lien was supported in part by NSC Grant 97-2115-M-006-012.

References

[1] G.I. Barenblatt, Scaling, Self-Similarity and Intermediate Asymptotics, Cambridge University Press, New York, 1996.

[2] G.I. Barenblatt, Scaling, Cambridge University Press, New York, 2003.

[3] M.H. Giga, Y. Giga, Nonlinear Partial Differential Equations, Asymptotic Behavior of Solutions and Self-Similar Solutions, Birkhäuser, 2010.

[4] G.B. Whitham, Linear and Nonlinear Waves, Wiley, New York, 1974.

[5] G.I. Taylor, The air wave surrounding an expanding sphere, Proc. R. Soc. Lond. Ser. A 186 (1946) 273–292.

[6] R. Courant, K.O. Friedrich, Supersonic Flow and Shock Waves, Interscience, New York, 1948.

[7] M.J. Lighthill, The position of the shock-wave in certain aerodynamic problems, Quart. J. Mech. Appl. Math. 1 (1948) 309–318.

[8] P.L. Sachdev, Shock Waves and Explosions, in: Chapman and Hall/CRC Monographs and Surveys in Pure and Applied Mathematics, vol. 132, 2004.

[9] M.H. Rogers, Similarity flows behind strong shock waves, Quart. J. Mech. Appl. Math. 11 (1958) 411–422.

[10] P.L. Sachdev, et al., Exact solutions of compressible flow equations with spherical symmetry, Stud. Appl. Math. 114 (2005) 325–342.

[11] J.K. Hale, Ordinary Differential Equations, Wiley, New York, 1969.

[12] K.T. Lin, The self-similar flow of the spherical piston problem, Master’s Thesis, Department of Mathematics, National Cheng Kung University, Taiwan, 2008.

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