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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= −
2ρ
ur
,
r>
0,
t>
0,
∂(ρ
u)
∂
t+ ∂(ρ
u2+
P)
∂
r= −
2ρ
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 andγ ∈ [
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
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 forη(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 velocityα
. 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 whenα
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 andα >
0 is small. The constant c0(= √
P′
(ρ
0))
denotes the speed of sound for the undisturbed gas. We introduce a nondimensional variableξ
:ξ =
r R(
t) ,
where R
(
t)(=
c0Mt)
is the shock radius at time t. Notice thatξ =
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)
]2ξ −
hγ −1ξ ,
h′(ξ) =
2M2fh(ξ −
f)
[M
(ξ −
f)
]2ξ −
hγ −1ξ .
(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
−
1γ
M+
M
y
+
M=
0.
(1.4)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
(ξ
2) = ξ
2.
(1.6)We note that
ξ
1=
1 andξ
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 fromξ =
1 backward until the valueξ = 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))
forξ ∈ [ξ
2,
1]
. Moreover, f(ξ)
and h(ξ)
are decreasing functions forξ ∈ [ξ
2,
1]
. To prove the main theorem, we carefully investigate the common denominator of system(1.3)together withξ −
f(ξ)
to show that the solutions are actually decreasing. The existence and uniqueness of the global solution forγ =
1 is shown in Section2, and the proof forγ >
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 =
1In this section, we study the case of
γ =
1. The system of ODEs is written as follows:
f′
(ξ) =
2f[M
(ξ −
f)
]2ξ − ξ ,
h′(ξ) =
2M2fh(ξ −
f)
[M
(ξ −
f)
]2ξ − ξ ,
(2.1)
with the initial value
f(
1) =
1−
M−2,
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) =
M−2−
1<
0.
(2.3)I1′
(ξ) =
2M2(ξ −
f)(
1−
f′),
(2.4)provided that f′exists. We also note that f
(
1) >
0,
f′(
1) <
0,
h(
1) >
0 and h′(
1) <
0. Hence, in a small neighborhood ofξ =
1,
f(ξ)
and h(ξ)
are decreasing functions. By investigating the subsidiary function I1(ξ)
together withξ −
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), whereξ ∈ [ 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 thatξ >
f(ξ)
for allξ ∈ [˜ξ,
1]
; then f(ξ) >
f(
1)
and I1(ξ) <
0 for allξ ∈ [˜ξ,
1)
.Proof. We prove by contradiction. Since f′
(
1) <
0, we assume that there existsξ
∗∈ [ ˜ ξ,
1)
such that f(ξ
∗) =
f(
1)
and f(ξ) >
f(
1)
for allξ ∈ (ξ
∗,
1)
. We discuss the following two cases.Case 1: If f is differentiable on
(ξ
∗,
1)
, then by the Mean Value Theorem, there existsξ
m∈ (ξ
∗,
1)
such that f′(ξ
m) =
0.However, f′
(ξ) ̸=
0 for allξ ∈ (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(η
1) =
0 and I1(ξ) <
0 for allξ ∈ (η
1,
1]
.Since f
(ξ) >
0 for allξ ∈ (ξ
∗,
1]
, we obtain from(2.1)that limξ→η+1
1
I1
(ξ) = −∞
and limξ→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 allξ ∈ (η
1, η
1+ ϵ)
.However, I1
(η
1) =
0 and I1(ξ) <
0 for allξ ∈ (η
1,
1]
. By the Mean Value Theorem, there existsξ
m∈ (η
1, η
1+ ϵ)
such that I1′(ξ
m) <
0. This is a contradiction.Therefore, we have shown f
(ξ) >
f(
1)
for allξ ∈ [˜ξ,
1)
.We now prove that I1
(ξ) <
0 for allξ ∈ [˜ξ,
1)
. It is easy to see that I1(
1) =
M−2−
1<
0. Assume that there existsη
1∈ [ ˜ ξ,
1)
such that I1(η
1) =
0 and I1(ξ) <
0 for allξ ∈ (η
1,
1]
. By the same argument as in Case 2, we can obtain a contradiction. Hence, I1(ξ) <
0 for allξ ∈ [˜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 ofξ =
1,
f(ξ)
and h(ξ)
are decreasing andξ −
f(ξ) >
0. As the solution f(ξ)
is extended backward fromξ =
1, we want to find a fixed pointξ
2∈ (
0,
1)
such thatξ
2−
f(ξ
2) =
0.Let
(ˆξ,
1)
denote the maximal interval of existence. We prove thatξ
2exists by contradiction as follows. Assume that for allξ ∈ (ˆξ,
1] ,
f(ξ) ̸= ξ
. Since f(
1) <
1, we haveξ >
f(ξ)
for allξ ∈ (ˆξ,
1]
. ByLemma 2.1, we have f(ξ) >
f(
1)
and I1(ξ) <
0 for allξ ∈ (ˆξ,
1)
. This implies that f′(ξ) <
0 and h′(ξ) <
0 for allξ ∈ (ˆξ,
1]
and hence f is strictly decreasing on(ˆξ,
1]
. So we have that1
> ˆξ ≥
limξ→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 forξ ∈ (ˆξ − ϵ,
1]
, whereϵ
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(ξ
2) = ξ
2and, from(2.1), f′(ξ
2) = −
2. Thus, there existsη < ξ
2such thatη <
f(η)
. If there exists another fixed point,ξ
3∈ (ˆξ, ξ
2)
such that f(ξ
3) = ξ
3andξ <
f(ξ)
for allξ ∈ (ξ
3, ξ
2)
, then we must have that f′(ξ
3) ≥
1. However, f′(ξ
3) = −
2 due to(2.1). This is a contradiction.For
ξ ∈ (ξ
2,
1]
, we now haveξ >
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 1.1forγ =
1.Remark 2.1. Figs. 2and3show the numerical solutions of f
(ξ)
and h(ξ)
. Forγ =
1 and an assumed value of M, one may numerically solve system(1.3)with the initial values(1.5), starting fromξ =
1 and going backward to the valueξ
2for which f(ξ
2) = ξ
2. It is observed that both f(ξ)
and h(ξ)
are decreasing forξ ∈ [ξ
2,
1]
.3. Existence for
γ >
1In 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 valueξ
2such thatf
(ξ
2) = ξ
2.
We rewrite the system(1.3)as follows:
f′
(ξ) =
2fhγ −1(
M(ξ −
f))
2ξ −
hγ −1ξ ,
0< ξ ≤
1,
h′(ξ) =
2M2fh(ξ −
f)
(
M(ξ −
f))
2ξ −
hγ −1ξ ,
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
−
1γ
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))
2ξ −
gξ ,
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)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) = −
1y21
[
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 ofξ =
1,
f(ξ)
and h(ξ)
are decreasing functions. By investigating the subsidiary function I(ξ)
together withξ −
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) <
0and then solving L′
(
y) =
0 for y>
1, we obtain only one solution, denoted byy:ˆ
y
ˆ ≡
1+ γ
M2γ +
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 yˆ
γ +1−
M2>
0.
We rewritey
ˆ
γ +1−
M2as follows:y
ˆ
γ +1−
M2=
1+ γ
M2γ +
1
γ +γ1
−
M2=
1γ +
1+ γ γ +
1M2
γ +γ1−
M2.
Define a function l
(
c) = (
1−
c) +
cM2
1c−
M2, for c∈ [
12
,
1]
. Then l(
1) =
0 and l′(
c) =
e1cln(1−c+cM2)
′=
e1cln(1−c+cM2)
(−
1)
c−2ln(
1−
c+
cM2) +
c−1 −
1+
M2 1−
c+
cM2
=
1
−
c+
cM2
1c(−
1)(
c−2)
1−
c+
cM2
−1
1
−
c+
cM2
ln
(
1−
c+
cM2) − (
1−
c+
cM2) +
1
<
0due to the fact that x ln x
−
x+
1 for x>
0. Hence, l(
c) >
0,
for all c∈
1 2,
1 .
Set c
=
γγ +1; then we have
y
ˆ
γ +1−
M2=
1γ +
1+ γ γ +
1M2
γ +γ1−
M2=
l γ
γ +
1
>
0.
The proof is completed.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 pointξ
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 thatξ −
f(ξ) >
0 for allξ ∈ (˜ξ,
1]
. Then I(ξ) <
0 for allξ ∈ (˜ξ,
1]
.Proof. We prove by contradiction. Assume that there exists
η ∈ (˜ξ,
1]
such that I(η) =
0 and I(ξ) ̸=
0 for allξ ∈ (η,
1]
. Since I(
1) <
0,
I(ξ) <
0 for allξ ∈ (η,
1]
. This impliesξ→ηlim+ 1
I
(ξ) = −∞ .
Due to the ODE system, f
(ξ) >
0 and g(ξ) >
0 forξ ∈ (ˆξ,
1]
and we have thatξ→ηlim+f′
(ξ) = −∞.
By the assumption that
ξ −
f(ξ) >
0 forξ ∈ [η,
1]
and(3.4), we get thatξ→ηlim+I′
(ξ) = +∞,
which implies that there exists
η
1∈ (η,
1]
such that I′(ξ) >
0 for allξ ∈ (η, η
1)
. Since I(ξ) <
0 for allξ ∈ (η,
1]
and I(η) =
0, by the Mean Value Theorem, there exists some pointξ ∈ (η, η
1)
such that I′(ξ) <
0. This is a contradiction because I′(ξ) >
0 for allξ ∈ (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. Ifξ >
f(ξ)
for allξ ∈ (ξ
∗,
1]
, then f(ξ) >
f(
1)
and g(ξ) >
g(
1)
for allξ ∈ (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 allξ ∈ (˜ξ,
1)
.Since f
(
1) >
0 and f′(
1) <
0, there existsη ∈ (˜ξ,
1)
such that f′(η) >
0. But byLemma 3.2, f(ξ) >
0 and g(ξ) >
0 for allξ ∈ (˜ξ,
1]
and I(ξ) <
0 for allξ ∈ (˜ξ,
1]
. Due to the ODE system(3.1), this implies f′(ξ) <
0 for allξ ∈ (˜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 allξ ∈ (˜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 allξ ∈ (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(ξ
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 allξ ∈ (ˆξ,
1] ,
f(ξ) ̸= ξ
. That isξ >
f(ξ)
for allξ ∈ (ˆξ,
1]
.ByLemma 3.3, f
(ξ) >
f(
1),
g(ξ) >
g(
1)
for allξ ∈ (ˆξ,
1)
. Since I(
1) <
0 and byLemma 3.2, I(ξ) <
0 for allξ ∈ (ˆξ,
1]
. Thus, from(3.1), we have that f′(ξ) <
0 and g′(ξ) <
0 for allξ ∈ (ˆξ,
1]
. That is, f and g are strictly decreasing on(ˆξ,
1]
. FromLemma 3.3, 1≥ ξ >
f(ξ) >
f(
1)
for anyξ ∈ (ˆξ,
1]
, and hence1
> 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 limξ→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) ≥
M1and
g′
=
2(γ −
1)
M2f(ξ −
f)
g (
M(ξ −
f))
2−
g
ξ ≥ −
cgFig. 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(p−ξ)for allξ ∈ (ˆξ,
p]
and thus limξ→ˆξ+g
(ξ) ≤
g(
p)
ec(p− ˆξ),
which contradicts(3.6).We now prove that
ξ
2is unique. Since f(ξ
2) = ξ
2,
g(ξ
2) >
0 and by(3.1), we obtain f′(ξ
2) = −
2. Hence, there existsη < ξ
2such thatη <
f(η)
. If there existsξ
3∈ (ˆξ, ξ
2)
such that f(ξ
3) = ξ
3andξ <
f(ξ)
for allξ ∈ (ξ
3, ξ
2)
, then f′(ξ
3) >
1.This is a contradiction since f′
(ξ
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 ofξ =
1,
f(ξ)
and g(ξ)
are decreasing andξ −
f(ξ) >
0. As the solution f(ξ)
is extended backward fromξ =
1, we want to find a fixed pointξ
2∈ (
0,
1)
such thatξ
2−
f(ξ
2) =
0.ByProposition 3.1, we know that
ξ
2exists uniquely and f(ξ)
satisfies condition(1.6). Forξ ∈ (ξ
2,
1]
, we haveξ >
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)
. Forγ =
1.
4 and an assumed value of M, one may numerically solve system(1.3)with the initial values(1.5), starting fromξ =
1 and going backward to the valueξ
2for which f(ξ
2) = ξ
2. It is observed that both f(ξ)
and h(ξ)
are decreasing forξ ∈ [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
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.
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