AN ABSTRACT OF THE THESIS OF
JUrgen Gerlach for the degree of Doctor of Philosophy in
Mathematics presented on June 8, 1982
Title:
A Free Boundary Value Problem for the Gas Dynamic Equations
Abstract approved:
a.
1
Signature redacted for privacy.
2,0,.,,
R.B. Gu
ther
We discuss a mathematical model arising from the following
physical situation.
We consider a gas-filled cylinder with a piston
at one end whose motion is determined by the pressure of the gas
within the cylinder and by external forces.
Zero mass flux is
assumed at the piston while the mass flux is known at the fixed end.
This model leads to a free boundary value problem involving the gas
dynamic equations.
Existence and uniqueness theorems are presented
for the linearized problem and for a special case of the non-linear
problem.
In the non-linear case, the interaction between shocks and
the motion of the free boundary are studied.
Finally, numerical
algorithms for finding approximate solutions of the linear and
non-linear models are suggested and examples are performed.
A Free Boundary Value Problem for
the Gas Dynamic Equations
by
Jiirgen Gerlach
A THESIS
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Doctor of Philosophy
Completed June 8, 1982
Commencement June 1983
APPROVED:
Signature redacted for privacy.
ap,4af,t
charge of major
Professor of Mathematics
Signature redacted for privacy.
tkitt:
()$41-n,R,,
airman of Department of Mathematics
Signature redacted for privacy.
Dean of the
Gr7I
CI6L
41 ate School
Date thesis is presented
8 June 1982
Typed by Virginia Ruth Madsen for Jiirgen Gerlach
Acknowledgements
I want to thank Karl-Heinz Hoffmann, who suggested the topic
of the dissertation; Alan Coppola, for indispensible help with
graphics; and especially Ronald Guenther, for his time, patience,
and encouragement throughout the development of this dissertation.
Additionally, I wish to express my gratitude to the faculty, staff,
and graduate students at Oregon State University for their help,
interest, and support of my work.
TABLE OF CONTENTS
I. INTRODUCTION AND STATEMENT OF THE PROBLEM
Introduction
Derivation of the Equations
1
1
3
II. THE LINEAR PROBLEM
An Existence and Uniqueness Theorem
Discussion of the Results and Examples
Separation of Two Gases by a Piston
Numerical Computations for the Linear Problem
10
10
19
23
27
III. THE NON-LINEAR PROBLEM
An Existence and Uniqueness Theorem for the
Uniform Flow Case
Shocks
Free Boundary and Shocks
A Numerical Method for the Non-linear Problem
35
35
42
54
60
BIBLIOGRAPHY
70
APPENDIX
71
LIST OF FIGURES
Page
Figure
1
The physical situation
3
2
The non-linear model
7
3
Construction of the function
4
Motion of the piston between two gases
23
5
Grid point scheme for the linear model
28
6
Numerical solution for example 6.1
32
7
Numerical solution for example 6.2
33
8
Numerical solution for example 6.3
34
9
Illustration for the construction of a solution
in the non-linear case
38
Two possibilities arising from the mechanical
shock conditions
46
11
Regions of different entropy caused by a shock
52
12
Shocks caused by the motion of the free boundary
55
13
Reflection of a shock at the free boundary
56
14
Graph of the characteristics
61
15
Grid point scheme for the non-linear model
62
16
Grid points of the free boundary
65
17
Numerical solution for example 10.1
68
18
Numerical solution for example 10.2
69
10
T(t)
14
A FREE BOUNDARY VALUE PROBLEM FOR
THE GAS DYNAMIC EQUATIONS
INTRODUCTION AND STATEMENT OF THE PROBLEM
I.
1.
Introduction
This dissertation deals with a free boundary value problem for
the gas dynamic equations.
We consider a cylinder filled with a gas.
One end of the cylinder is assumed to be fixed, and there the mass
flux is assumed
known.
On the other end we have a piston which
moves due both to the influence of the internal pressure at the wall
of the piston and to external forces.
the piston.
No mass flux takes place at
The problem is to describe the motion of the piston and
the state of the gas at a given time.
A special case of this problem is known as "Lagrange's Ballistic
Problem" and was treated by Love and Pidduck [7] in 1922.
The governing equations for the physical situation described
above are derived in section 2, and a linearized version of the
problem is given there.
It should be noted that similar equations
arise in describing the filling of dry riverbeds (see [8]).
The second chapter (sections 3-6) deals with the linearized
problem.
In section 3 we give an existence and uniqueness theorem
which is discussed in section 4.
Section 5 treats the case in which
gases are on both sides of the piston, i.e., the external force is
caused by the motion of a gas on the other side of the piston.
This
problem was treated by Hill [4] for a cylinder of infinite length.
Section 6 describes an algorithm for the linear problem.
2
In the third chapter we consider the full non-linear problem.
Section 7 contains a local existence and uniqueness theorem for
uniform flow given initially.
Section 8 is concerned with shocks.
Special attention is paid to the mechanical shock conditions and the
necessity of increase of entropy across a shock.
Section 9 investi-
gates the interaction between shocks and the motion of the free
boundary, while the last section describes a numerical algorithm for
the approximation of the nonlinear problem.
Equations and formulas are referenced in numerical order in each
section.
Equations of other sections are refererred as follows:
(2.12) means formula 12 of section 2.
3
2.
Derivation of the Equations
We consider a one-dimensional cylinder filled with gas.
We
assume that one end of the cylinder is fixed while the other end is
equipped with a piston which moves.
forces acting on the piston.
The motion depends upon the
It will be further assumed that no gas
flows through the piston.
)
gas
1
moving piston
Figure 1
Let
u
the pressure.
denote the velocity of the gas,
pt
the density, and
p
We shall assume that the gas is an ideal gas and that
no body forces are acting.
be neglected.
p
In particular, gravitational forces may
The law governing the conservation of mass is given by
+ (up)
= 0
the so-called continuity equation, and the equation of motion takes
the form
P(ut + uux) + px = 0
.
4
In the case of an ideal gas and an adiabatic process, the
pressure is a function of the density alone and is given by
=A(-2-)'
p0
where
p0
y = c /c
p
is the quotient of the specific heats,
v
the specific heat obtained with constant pressure and
specific heat for constant volume.
gas we are dealing with.
200
po
C,
y = 1.4.
The value of
For all gases
In most cases
y
y
y > 1, e.g.
does not exceed
being
cp
cv
the
depends on the
for air at
5/3.
po
denote the average pressure and density, respectively, and
a constant which can be determined experimentally (see e.g.
Ell
and
A
or
[9])
Introducing the local sound speed
c2
clE
_A
dp
yp
0
p
c = c(p)
by the equation
y-1
p0
we can rewrite the system (1), (2) in the equivalent form as
P
(up)
UUU x
= 0
C2
p
px =0
The system (1), (2') governs the state in the interior of the
cylinder.
In order to derive boundary conditions, let us assume that the
fixed end of the cylinder is located at
the position of the piston at time
Denote by
so that
m(t)
t,
x = 0, and denote by
s(t)
s(0) = 2 > 0.
the mass present in the cylinder at time
t,
is
5
s(t)
m(t) =
f
p(x,t) dx
0
Differentiating (5) and using (1) we obtain for the rate of change of
mass
-d7 m(t) = P(s(t),t) ig(t) - u(s(t),01
P(0,t) u(0,t)
At the free boundary no mass is gained or lost, so the only change
occurring has to be given by the term
p(0,0 u(0,0 = r(t)
which
describes the rate at which mass is brought into or taken from the
From (6) we conclude that
system.
(t) = u(s(t),t)
which is a statement of the physically obvious fact that the velocity
of the gas at the piston is equal to the velocity of the piston.
The
flow of mass across the fixed boundary is assumed to be known so
that the boundary condition on the left hand side is
u(0,t) P(0,0 = r(t)
If
s(t)
.
is known, then (7) and (8) represent boundary conditions in
the ordinary sense.
The momentum of the entire process is given by
s(t)
M(t) =
f
pu dx + d t)
0
where
d
denotes the mass of the piston.
momentum we obtain using (1), (2) and (7)
For the rate of change of
6
AM(t)
= d .i(t) - p(s(t),t) + p(0,t) - p(0,t) u2(0,0
.
The rate of change of momentum on the free boundary equals the sum of
the forces acting on the piston.
Hence we obtain
d .(t) - p(s(t),t) = -k(t) + f(s(t),t)
where
describes an outer force acting on the piston at a
f(x,t)
given position and a given time.
k
is a coefficient of friction.
The equations we have derived so far all carry physical
However, we can introduce dimensionless variables so
dimensions.
For
that the essential features of the problem remain unchanged.
simplicity we want to assume from now on that all variables are
dimensionless, that the relation between pressure and density is
given by
y
1
(3')
P =
and that
s(0) = 1
P
(for the details of this reduction see Appendix).
The problem we want to solve now reads:
Let the constants
functions
r(t), t
be given.
uo(x),
> 0,
a > 0, y > 1, and i > 0, given initial
p(x)
on
satisfying
Find a function
[0,1]
and a boundary function
r(0) = u0(0) po(0), and a function f(x,t)
s(t)
and functions
such that the following equations are satisfied:
u(x,t) and
p(x,t)
7
(c)
pt
(M)
P(ut + uux) + px = 0
+ (pu)x = 0
where
p(x,t) =
for
t > 0
and
0 < x < s(t)
(P(x,t))
P(x,0) = Po(x)
(I)
u(x,0) = u(x)
P(0,0* u(0,t) = r(t)
(B)
u(s(t),t) = i(t)
s(t) +
= ap(s(t),t) + f(s(t),t)
(FB)
s(0) = 1,
6(0) = u0(1) =
z(t):=
Frequently we shall set
Pt
P(ut
6(t)
.
(uP)x =
uux.)
px =
((t)
+ ues = cp + f
=u
Figure 2
8
A number of complications arise in this non-linear problem,
e.g., shock waves might build up inside the cylinder.
Therefore,
we begin by considering first a linear model for the same process.
Let
be the average density, which is an appropriately
po
chosen numerical value, e.g.,
1
po
=I
p(x) dx
00
Define according to (4')
(12)
then
c
co
:= p
0
2
0
represents the average sound speed that corresponds to the
choice of
Define, furthermore,
.
Po
T
t =
:= ct
o
Co
:= ufc
a :=
u
0
P-Po
c
on
P = P0(1+a)
0
S := s
:=
=
c
F
R
z = coZ
c
o
a
:= -10
y o
o
f = c2F -
+
0
c2
r = c p
00
c p
00
We assume that
a
and
y0
R.
as well as their partial derivatives
n
are small, so that products of those terms can be neglected.
By the binomial theorem
p
can be approximated by
9
(13)P -- T
Y (1+0Y
1
and for
px
Po
=
1y pY
(1+ya)
0
we obtain the approximation
P
Pxy-2
p
P
(14)
Substituting
(13)
terms and writing
(c')
(N')
x
-p 0
and
(14)
P
+
nx
nt
+
ax
(x,0)
into (C) and (M), neglecting second order
instead of
t
at
ax
T, we obtain
=o
= 0
= no(x)
(I')
a(x,0) = ao(x)
n(0,0
= R(t)
(B')
11(S(0,0 = S(t)
+ N)
=
Ka + F
(FB')
S(0) = 1,
where K
=
apo
and
Z(0)
V=
= no(1)
-c0
10
II.
3.
THE LINEAR PROBLEM
An Existence and Uniqueness Theorem
We wish to solve the linearized problem as stated in section 2.
Assume that
is defined on the first quadrant of the
F
(x,t)-plane, that it is Lipschitz-continuous (L-continuous) in
IFl
is bounded by a constant
x and
In order to solve the equation
Fmax.
for the free boundary we observe that
Z = - V Z + Ka + F = -(V+K)Z + K(a+n) + F
.
Denoting the Riemann invariants by
(1)(x-0 := a(x,t) + 11(x,t)
T(x+t) := a(x,t) - n(x,t)
we know that
(1)
and
are well defined since
T
constant along the characteristic lines
x+t = const
dt
respectively.
Z(t) =
Assuming that
are
x-t = const, and
.
Fmax
v + K = A
F
M > 0
a-n
With this notation (1) becomes
M := 1
and obtain
and
Z(t) + K(13,(S(t)-t) + F(S(t),t)
X := (v+K)
where
a+.11
.
max-
,
v
we define a constant
M by
11
Theorem 1.
Let
ao,
no'
R
and
be given such that
is Lipschitz-
(I)
continuous and satisfies
-co < Y < y < 1
for
where the choice of
Y
is minimal, and let
< 1
1131
Then there exists a uniquely determined solution
defined on a maximal t-interval,
1Z(01
Proof:
on
< 1
and
I,
Z
of
(4)
satisfies
Z
I
We have to solve the system
=Z
(*)
Z = -AZ + K(I)(S-t) + F(S,t)
Its domain
Q
.
is restricted by the domain of
= {(x,t)
1
t >
o,
x > o,
and we get
(I)
Y < x - t < 1}
The right hand side of the system (*) is L-continuous, since both
and
F
are assumed to be L-continuous.
(1)
Hence, by the Picard-Lindel6f
Theorem there exists a unique solution of the initial value problem
and the solution approaches the boundary of
Q
Setting
H(S,t) := K(I)(S-t) + F(S,t)
we get the estimate
1H(S,01 < K,M
+
Fmax
= Ki-v= A
.
12
A solution of (*) yields
Z(t) =
e-At +
e-A(t-T) H(S(T),T) dT
and we obtain
lz(01
<e-At
+ A
11310
=
e-X(t-T) dT
e-At + (1-e At
1131
)
= 1 - (1-1131)e-Xt < 1
The proof of theorem 1 is similar to a proof given by Hill [4]
for the case of an infinite cylinder under very restrictive
assumptions.
Corollary.
If
Y
is infinite then either
I
is infinite and
0 < S(t) < 1 + t
Or
I = (0,T),
If
S(T) = 0
is finite then one of the statements is true:
I
is infinite
I = (0,T)
IZ(01
and
Y
I = (0,T),
Proof:
T > 1,
Y + t < S(t) < 1 + t
and
T > 1
is finite
S(T) = 0
and
and
S(T) = Y + T
Use the maximality of the interval of definition and
< 1
.
13
The crucial part of the further development of the existence
theory is to find conditions on
R(t)
such that
<m
1(1)(y)1
is satisfied.
Lemma 1.
Let
a
and
no
be L-continuous on
1110(x)1 , fa(x)( <
M
Then the assumptions of Theorem 1 are satisfied for some
Let
Proof:
0 < Y < 1
(I)
.
then
= no(Y)
1)(Y)
Hence
Y < 0
is L-continuous
and
a(y)
113.1
<N
.
Lemma 2.
Let
no
and
ao
be given as in Lemma 1,
let
R
be
L-continuous and satisfy
12R(t) -
no(01
<
1/2 M
for
0 < t < 1
;
then the assumptions of Theorem 1 are satisfied for some
Y < -1
.
14
Following the characteristics back, we obtain for
Proof:
4)(y)
expression
-1 < y < 0
2 R(-y) - no(-y) + ao(-y)
(4)
(1)(y) =
a (y) +
0 <y<1
no(y)
o
The compatability condition assures the continuity at
L-continuity and the estimate
14)1
y = 0, the
can be easily verified.
< M
The derivation of further conditions on
for
R(t)
t > I
involved the behaviour of the free boundary since the backward
characteristic from
at the position
meets the free boundary at a time
t1
1(t1)
as indicated in Figure 3.
S(T(ti))
Figure 3
Let us now assume that a solution
interval
[0,T]
TR := S(T) + T
(5)
satisfying
IS(t)1 = 12(01
and define for
h(T) :=
S(T)
S(t)
T-
1 < t <
t,
TR
0
is given on some
Let
< 1.
the function
<T < T
.
the
15
Lemma 3.
T=
There is a unique solution
The function
T(t)
(6)
dT
dt
Proof:
h(0) = 1 - t < 0
1
>0
h(T) = TR - t >
Hence
of
h
of the equation
h(T) = 0.
is differentiable with
Z(T(t))+1
dh
dT
T(t)
.
0
= Z(T) + 1 > 0
is strictly increasing and we obtain a unique solution
h(T) = 0.
depends on the choice of
T
This
t
and by the
implicit function theorem we obtain formula (6).
The function
is given, then
T(t)
characteristic from
in Figure 3.
that
T
has the following property.
T(t)
(0,t)
Hence,
T(1) = 0,
0
<T
_ _<
T
.
T(Ta) = T.
Using the function
T(t)
= T(-y)
y < -1,
T(t)
Formula (6) says
t, and this dependence is
h(T) = 0
S(T(t)) + T(t) = t
(7)
T=
meets the free boundary, as indicated
is an increasing function of
given in (4) for
t > 1
provides the time at which the backward
In particular
differentiable.
If a
yields the identity
.
we can extend the description of
and obtain with
t = -y
and
(1)(y)
16
(1)(y) = (1)(-y-2T) + 2(R(-y) - Z(T)),
alternatively
= 0(t-2T) + 2(R(t) - Z(-r))
.
Note that
S(T) - T < (1+T) - T =
- 2T =
!S(T)I
since
and hence
< 1
1
is defined.
(1)(t-2T)
Lemma 4.
Let the assumptions of Theorem 1 hold for some finite
Let
Y < -1.
be the corresponding solution of the free boundary on the
S
maximal interval
If
R
and assume that
[0,T[
is L-continuous on
IR(t) -
[-Y,TR]
(z(T) - 1/2-4) (+t-2T))
S(T)
0
and
T
00
and satisfies
I
<
1'4
then the assumptions of Theorem 1 are satisfied at least for
Y
:=
Note that from the corollary, case c), we know that
Proof:
Y = S(T) - T = Y
Then
T - S(T) < T - S(T)
Thus the argument of
NO
< M
.
implies that
(I)
S(T) < S(T) + T - T.
Consequently,
S(T) - T > S(T) - T = Y
t - 2T =
and
ISI < 1
+ 2S(T).
is in the range where
.
(I)
According to (8) and (9) we obtain
is L-continuous
17
1.1)(y)1 = 14)(t-2T) + 2(R(t) -
To show the L-continuity of
of
R
(D
Z(T))I
<
M
we use formula (8), the L-continuity
,
and the differentiability of
and
Z
T(t) = T(-y).
Theorem 2.
Let
no,
Go
and
R
Suppose
be L-continuous.
no
and
ao
satisfy
In 0
R
Ila0 < k
[0
on
,1]
,
satisfies
12R(t) -
n0(01
M
<
on
[0,1]
and
IMO - (z(T) for
t > 1
as long as
T
1/24)(t-21-))1 <
and
S(T)
are defined.
Then there exists a uniquely determined solution
S
for the
free boundary satisfying
si
S
< 1
has a maximal interval of definition
I = (0,T),
T < co
,
and
one of the following cases holds:
Proof:
T
is finite,
T
is infinite
T > 1
and
and
S(T) = 0
.
0 < S(t) < 1 + t
for all
t > 0
.
From Lemmas 1 and 2 we know that the assumptions of Theorem 1
hold at least for
Y = -1.
some interval
The second condition on
I.
Hence, we can find a solution
R
that this solution can be extended as long as
S(t)
and Lemma 4 assure
S(t) 0 0.
on
18
So far little has been said about the solution of the gas dynamic
equations.
But once the free boundary is known, we know the Riemann
invariants and obtain
n(x,t) = ½((x-t) - T(x+0)
a(x,t) = 1/20(x-t) + T(x+t))
By the conditions imposed above,
so
p
and
a
(I)
and
.
T were only L-continuous,
are solutions of the partial differential equations
in a generalized sense.
S
is twice continuously differentiable and
has an L-continuous second derivative.
19
Discussion of the Results and Examples
4.
The main goal in the previous section was to find conditions
IZI < 1
such that
was guaranteed.
This implies that
does not
S
leave the domain of definition of the differential equation across
the line
It might be possible to achieve that goal by
x - t = 1.
other conditions which allow
S = Z
Co
IZI > 1.
However,
IZI > 1
implies
which means that the speed of the piston and hence the
speed of the gas near the free boundary is greater than the average
In this case the linearization itself becomes
sound speed.
questionable since it was based on the assumption of small velocities
compared to the speed of sound.
In Lemma 1
and
ao
no
were to be estimated independently.
The assumptions could be relaxed tolao(x)
+ no(x)I
< M
however,
:
for physical reasons the way chosen seems to be appropriate.
M
The constant
depends on the parameters of the problem.
no friction is present for the piston and if
More generally,
M
increases for increasing values of
decreases for increasing values of
Fmax.
In the limiting case
chosen so that
E 0
(I)
V,
it
If the parameters of the
,
M = 0,
no, ao,
and
R
have to be
according to Theorem 1.
Example 1:
Let
V
= 0
no(x) E ao(x) E
= 10(1) = 1/2
equation
.
and
.
1/2
Z
M = 1.
M < 0, the theory of the previous section does
problem are such that
not apply.
F E 0, we have
If
F E 0.
Then
Then
(1)(y) E 1
M = 1.
Let furthermore
for 0< y < 1
,
and
has to be determined from the differential
20
Z = -K (Z-1)
for
The corresponding solution is given by
S(t) > t.
S(t) = t +
Hence
1
(e
-Kt
- 1) + 1
.
217
is completely determined by the initial functions alone, if
S
K L1/2 , since then
According to Theorem 2,
S(t) > t.
R(t)
would
have to satisfy
0 < R(t) <
0 < t < 1
for
1/2
and
-1/2
where
R(t)
eKT < R(t) < 1 - eKT
is determined from
T(t)
t = 2T +
does not influence the path of
t > 1
for
1
,
te-KT - 1).
2K
However,
S.
Example 2:
Let
and for
v = 0,
F E0,
and
Let
K = 1.
choose
no
1
fl 0(x):=
<X<
1
2
< x < -3-
5
< x < 1
3
= -1/2
and
1
0
2
-3x +
Then
U(x) E 1/2,
R(t) E 0
21
2
-3y + 3
< y < 1
2
1
1
-3-
(1)(y)
3y +
=
1
1
1
-2-
2
0
Certainly,
and
ao
< y < - 23
-
-1
-3y- 2
The free boundary must be
satisfy Lemma 1.
no
determined from
= -3S + 3t+ 3
g + S = (1)(S(t) -
for
2
.< S(t) - t < 1.
We obtain
S(t) = t +
for
2
+
1-kt
(cos
-3- e
0 < t < t1 = 0.237,
where
t1
/TY t -
yields
8
yr
1/IT
sin k.
(S(t ) - t1 =
'Tr 0
2
Now
= a(s(t),t) - n(s(t),t)
w(s(0 +
= a(s(0,0 +
(s(0,0 -
2 2(0
= (1).(s(t) - t) - 2 Z(t)
= -
Obviously
T(1) = 1,
and
+ e
-1/2t
( 2 cos
t +
11
lin
sin 1/2
AT
)
22
dt
= T'(S(t) + t)(Z(t) + 1)
T(S(t) +
=
Hence
TIM
close to
some
1.
y < -1.
= 7,
Since
and
e -kt(-} cos
T(y) > 1
R(t) E 0
for
VD:
t
y > 1
it follows that
21
22
with
iff sin
y
1/2
/TT
.
sufficiently
(1)(y) = T(-y) > 1
This example shows that even the simple case
does not necessarily satisfy the conditions of Theorem 1.
for
R(t) E 0
23
5.
Separation of Two Gases by a Piston
We now want to consider a slightly different situation.
a cylinder of finite length filled with gas.
Assume
In the middle there is
a piston which separates the two sides and it is assumed that no gas
If we denote the states of the gas, the
flows through the piston.
initial and boundary functions on the left with subscript
i = 1, the
i = 2, the equations derived in
ones on the right with subscript
section 2 have to hold on each side.
Here we shall consider the
linear model.
alt +lx
n
it
= 0
+y lx =0
R2
a2t +2x =
01
G10'
2t
+
a2x
0
=0
Figure 4
G20' n20
10
According to the process of linearization we have to choose
average densities
assume that
p10
and
p10 = p20 =: po
p20
for each side.
and also
y1
For simplicity we
= y2 = y
.
The force
acting on the piston is given by the difference of the pressures on
either side and we obtain
Z(t) + vZ(t) =
after linearization.
K(a1 (S(t),t) -
a2(S(t),t))
Denoting the length of the cylinder by
L
24
we obtain the following problem (see Fig. 4):
+(n.)
ix
=o
it +(a.)
ix
= 0
(a.)
it
(Ti.)
0..(x,0) = a. (x)
1
10
(4)n.(x,0)
1
= n.10 (x)
1' 0 =
n (0
n2(L,t)
R1
(0
= R2(t)
.(S(0,0 = Z(t)
1
K(a1(8(0,0 -
Z(t) + vz(t) =
(8)
S(0) = 1,
(9)
n 10 (0) = R1 (0)
Z(0)
= n(l) =
n20(0
a2(S(0,0)
n2(1) =
= R2(0)
From equations (6) and (7) we can conclude
z(t) + (v+K)z(t) =
Ka1
(S(0,t) -
K(a2(S(0,0 - n2(S(0,0)
= Ka1 (s(t),t) - KT2(S(t) +
Hence (7) is equivalent to
z + vz =
where
v =V + K
Ka1 -
.
KT (S(t) +
2
,
.
25
(10) corresponds to equation (1) of section 4 if we focus our
role of the external force
< x < 1 - t}
{(x,t)10 < t < 1, 0 _
not defined on the triangle
overcome by a suitable
L-continuous extension of
If the initial functions
functions
(11)
Fl =
1112
are L-continuous, and given in a way such that
R1, R2
11 = 17110(1)
12 )
can be
T-120' G10' G20
=
KW2 <
for
,
z > 1
y < 1,
In20)1)I < 1
We obtain
we can apply Theorem 1 of section 3:
(
is
and the boundary
n
11)1(y) I, IT2(z)1 < 1 + - -2\)fc
and if
plays the
F(x,t) = -KT2(x+t)
The fact that
F.
-KT2
The expression
attention on the left side only.
=: F
K +
Fmax
from
max
Then M yields
Fmax -v
-
M = 1
=
and from (11) we have
for side
i = 1
V
1 - 1- -2-K- +
2K
(y)I
11,
V
+ V- + 1 = 1 + V2K
K
Applying Theorem 1 of section 3
< M.
we obtain existence and uniqueness of a solution
S(t) of the free boundary.
thefunctions(5.(x,t),
1
Once
11.(x,t)
1
S(t)
is known, one can derive
.
The crucial part again is to specify the conditions on
and
R2(0
such that (11) is satisfied.
For
i = 1
the estimates given in Theorem 2 of section 4; for
to use a similar technique.
R1(t)
one can apply
i = 2
one has
26
For the final time
T
one of the following two conditions
is true:
either
(A)
T < co
and
or
(B)
T = co
and 0< S(t) < L
either
S(T) = 0
or
S(T) = L
for all t >
27
Numerical Computations for the Linear Problem
6.
In this section we shall give an algorithm to approximate the
linear problem numerically.
step sizes
Ax
and
be denoted by
an
To settle the notation we choose fixed
The approximation for
At.
and likewise
,
a(m Ax, n At)
n(m Ax, n At) =
,
will
S(n At) = Sn
,
Z(n At) = Zn
For the numerical solution of the system
at
=0
+
flx
nt + ax
= 0
we choose the method of the characteristics.
the fact that the Riemann invariants
(I,
=a+
This method is based on
n
and
=
remain constant along the characteristic lines of slope
Ax = At =:
±1.
n
Choosing
h, we can approximate (1) by the equations
n+1
n
= %(a
+a+ a
=
+
n+1
m
1/2(nm-1
Hence, if the state on
time level
a-
t = (n+1) At
m+1
m+1
t = n At
+r m-1 -nm+1.1)
+
m-1
-am+11)
is known, we can proceed to the
in the interior.
At this stage it should be noted that in order to derive
convergence of the approximations to the true solutions,
have to be chosen such that
(At/Ax) < 1, for details see
satisfied this condition by the choice
Ax = At = h.
Ax
and
[3].
We
At
28
The values at the fixed boundary can be determined from
nn+1 = R((n+l)h)
0
,
an+1 = R((n+l)h) +
n
n
a1 - n1
.
On the free boundary we have to combine the solution in the
ODE
interior with an approximation of the second order
for
In particular we have to face the situation that in general
S(t).
Sn
does
not coincide with a grid point and that grid points may be gained or
An Euler method for the
lost due to the motion of the free boundary.
approximation of the free boundary can be obtained in the following
way.
Assume that the approximations
for some
and for
n ,
0 < in < M = M(n)
Then, according to (2)
m < (M-1)
.
nn , am ,
nn+1
The change in
am
,
Zn
where Mh < Sn < (M+1)h .
can be determined for
.
n known
a
+ n known
t = (n+l)h
t = nh
(M-2)h
are known
has to be determined from
Z = S
Z = - vZ + Ka + F
a,
n+1
and
Sn
Mh
(M-1)h
Figure 5
Sn
29
Since
not necessarily a grid point,
S'1
a(Sn,n h)
is not known.
However, we can approximate this value by a linear extrapolation
a
S
-= an + (Sn -M h)(an - an )/h
M
M
M-1
.
Then let
1n+1
+ F(Sn ,nh))
:= Zn + h(-\)Z'1 + Ka
Sn+1 := Sn + 1/21-1(Zn+1 + Zn)
For the new point
one of the following situations will occur:
Sn+1
n+1
(7a)
S
(7h)
M11 < Sn+1 < (M+1)h
(7c)
(M+1)h < Sn+1
< 14.11
In the case of (7a) we
lose a grid point and have all the
information needed to proceed in the algorithm and compute the state
for
t = (n+2)h.
n+1
.
In the case of (7h) it remains to find
an+1
and
Again we use a linear interpolation exploiting the
relationship
n
for the true solution:
Z(t) = n(S(t),t)
n+1
M
:=
set
n+1,,, n+1
n+1
n+1
+ h.(Z
- n
- (M-1)h)
)/tS
M-1
M-1
(8)
an+1
M
:=
n
GM
+
n
nM-1
n+1
--1
nM
In the case of (7c) we can use the same type of interpolation for
n+1
n M+1
and
n+1
am+1
,
just replacing
a second grid point is gained at
since
an
M+1
and
nn
M+1
M
by
M+1
x = (M+2) h
are not defined.
in (8).
However, if
our method fails,
Then we are in the
30
situation that the speed of the free boundary exceeds
1 -- a
situation which contradicts the theoretical discussion of the previous
sections, and we break off the computation.
The algorithm also fails as
approaches the fixed boundary.
Sn
In order to perform the interpolations in (5) and (8) we must have
M > 1
Sn > Ax
and thus
In the actual computations an improved Euler method was used to
compute the position of the free boundary--similar to the method
described above.
In the
Examples 1 and 2 correspond to those given in section 4.
first case we haveo(x) E ao(x) E
On the boundary we have chosen
S(t)
1/2,
K = 1,
and
F E 0,
V
= 0
.
R(t) E 0, but since the true solution
satisfies
S(t) = t + k(e-t + 1) > t +
(9)
the choice of
does not influence the motion of the free
R(t)
boundary.
In the second example we have chosen the data of example 2 of
section 4.
The numerical result shows that
--about 0.92--as
S
becomes fairly large
travels through the region where
S
(I)
> 1
however, it does not exceed the speed of sound.
In example 3, we let
friction term
by
S(0) = -0.8
= 1
V
.
no E ao E 0,
R(t) E 0 E F(x,t).
The
and the motion is caused by an impulse, given
What happens is that the piston moves into the
cylinder for a while, then changes its direction due to compression.
It oscillates and finally comes to rest due to the friction term. Since
no mass is gained or lost at the fixed end, and since entropy changes
are neglected, one would expect that the system comes to rest with
31
the piston being located at
about
x = 0.73
m(t) =
This paradox can be resolved
seems to be achieved.
by considering the mass
(10)
However, numerically a limit at
x = 1.
m(t)
present at time
S(t)
I
P(x,t) dx = p
0
Differentiation of
t.
We have
S(t)
(1 + a(x,t)) dx
f
00
(10) yields after substitution of the differential
equations and the boundary conditions
dm
dt =
po a(s(t),t) s(t)
.
This shows that in the process of linearization, the law of
conservation of mass has been modified, and our system is "leaking"
at the piston.
32
EXAMPLE 1
2.00 7
1.00 --
0.00
1
0.00
1
1
1
1
1
`1ba.
Figure 6
F
1
2.00
I
4
1
I
3.00
5.00 EXAMPLE 2
4.00 No
3.00 -
2.00 -
1.00 -
0.00
0.00
1.00
Figure 7
2.00
3.00
4.00
10.00
-
34
EXAMPLE 3
6,00 -
4.00 -
2.00
Figure 8
0.00
0.00
.50
1.00
35
THE NON-LINEAR PROBLEM
III.
An Existence and Uniqueness Theorem
7.
for the Uniform Flow Case
In this chapter we go back to the full non-linear problem in
dimensionless form as stated in the first chapter.
pt
+ (up)x = 0
for
t > 0
We have to solve
and
0 < x < s(t)
p(ut + uux) + px =0
p(x,0) = p(x)
0< x< 1
u(x,0) =
uo(x)
p(0,0-u(0,t) = r(t)
t > 0
u(s(t),t) = A(t)
u
= ap
f
= z
s(0) = 1
,
z(0) =
The initial functions are assumed to satisfy the compatability
conditions
r(0) = po(0)u0(0)
and
uo(1) =
[3
The relationship between pressure and density is given by
p
_1 py
and the local speed of sound is defined by
36
2
(12)
c
= p
y-1
clE
dp
The system (1) and (2) are the equations of one-dimensional,
isentropic flow, which have been extensively investigated.
The
Riemann invariants are defined by
R :=
k(u +
2
Y-1
c)
and
(13)
2
S :(u
y-1
and they satisfy the equations
Rt + (u + c)R
= 0
- c)Sx = 0
St +
From (14) it follows that
R
.
and
characteristics, i.e. the curves
dxdt
-
remain constant along the
S
and
x+(t)
x(t)
satisfying
u± c
However, in contrast to the linear case, the characteristics-formerly lines of slope
the solution.
x+
and
x
±1--are not known a priori; they depend on
In other words:
u
and
c
have to be known before
can be determined from (15).
In our discussion we shall restrict ourselves to the case that
initially a uniform flow is given, i.e. the functions
and
po(x) E po
are constant.
u(x) E
Then the forward characteristics
impinging on the free boundary will carry the constant value
Ro := 1/2(uo +
co).
Hence, on
uo
s
we must have
x+
37
y-1
V o(s(t),t)
=
2
+
2
(s(t),t))
p
Using (6) and (11) we can solve for
.
and obtain
p
2y
p =
1
y-1
(
y
(2R
o
2
6))Y-1
and substitution into (7) furnishes the equation
2y,
=
+ 1(6
y
(1=1
(2R 0
2
MY-1
The domain of (18) is restricted by
initially--unless
f(x,t)
co
< 0
gO) = uo
<. 2Ro, which is satisfied
If
which is physically unreasonable.
is L-continuous in
s(0) = 1,
+ f
x, then the initial value problem
and (18) has a unique solution
s(t)
on the
domain of definition of (18).
It is a well known fact in the theory of gas dynamics that in
a region adjacent to a uniform flow region one set of characteristics
is
straight lines (simple waves) (see e.g.
11)).
illustrate this, take a line of a given slope
prescribe a value
So
on that line.
characteristics carry the same value
solutions
G,
m
In order to
in region II and
All intersecting
Ro,
x+
and we obtain a pair of
given by
= Ro +S o
e
azi
2
If
fl -
(R
0
-s)
o
e = m, then our line was a characteristic.
Thus, in order
to construct a solution in region II we have to find m
and
So
in
38
a suitable way.
Along the free boundary, both
c =
y-1
2
points
s(T) + (u(s(T),T) -
x =
S = S(T)
Given a point
on which
9,(T)
T = T(x,t).
are known:
u =
and
c(s(T),T))(t-T)
on those lines by
S(T) = Vu(s(T),T) -
line
c
by
2,(T):
and prescribe
and
We can define a set of lines emanating from the
(2Ro - u).
(s(T),T)
u
(x,t)
(x,t)
2
c(s(T),T)) = i(T) - R 0
near the free boundary we want to find the
lies, i.e., we have to solve (20)for
In order to do so, consider the function
H(x,t,T) := s(T) - x + (u(s(T),T) - c(s(T),T))(t-T)
Figure 9
.
39
and moreover
H(s(T),T,T,) = 0
Obviously
(s(T),T,T)
= c(s(T),T)
> 0
Hence, for any point on the free boundary we can find a function
T(x,t), defined in a neighborhood of
which satisfies
Consequently, there exists a function
H(x,t,T(x,t)) E O.
which is defined on a neighborhood
H(x,t,T(x,t)) E 0
satisfies
(s(T),T)
and which
(s(t),t)
there.
(x,t) 6 U1
For any point
of
U1
T(x,t)
we set
u(x,t) := Ro - S(T(x,t))
(23)
c(x,t) :- Y-1 (R
2
o
+ S(T(x,t)))
and according to (13) we obtain for the Riemann invariants on
R(x,t) = Ro
and therefore
and
u
Those functions satisfy (14)
S(x,t) = S(T(x,t)).
and
c
U1
as defined in (23) provide a solution to
the system (1), (2).
On the fixed boundary we apply a similar procedure.
The
incoming characteristics all carry the constant value
So := 1/2(u-4--o
y=1 co)
.
2So = 11(0,0
and
c(0,t)
Using (5) we can determine
2
r(t)
)1-1
y-1 (11(0,0
2
from
2
2So = r(t) c(0,t) Y-1
on a certain t-interval.
2
c(0
y-1*
u(0,t)
from
40
Again we define a set of lines
(0,T)
with slope
R(T)
invariants
R(T)
and define the Riemann
m(T) := u(0,T) + c(0,T)
along those lines by
:=
2
(u(0, T) +
For each point
y-1 c(0'
T))
.
in a neighborhood
(x,t)
boundary we can find a
line
emanating from the points
2,(T)
T(x,t)
such that
2,(T(x,t)), and a solution of (1) and
(x,t)
(2)
of the fixed
U2
belongs to the
U2
on
is given by
u(x,t) := R(T(x,t)) - So
c(x,t)
(R(T(x,t)) +
2
Differentiability of
So)
.
needs to be assumed in order to obtain
r(t)
differentiable solutions to the system (1) and (2).
This completes the construction of solutions in
U1
and
U2
and we can summarize our analysis in a theorem.
Theorem 3.
Suppose
suppose
uo(x) E 110
r(t)
co(x) E co > 0
is differentiable.
of the initial line
(1) - (10)
and
Then there exists a neighborhood
in which the free boundary problem
t = 0
has a unique solution.
solution of (18),
u(x,t) = uo
region, otherwise
u
and
are constant, and
c
and
The free boundary is given as a
c(x,t) = co
in the steady flow
are given by (23) and (27).
Some remarks about Theorem 3 seem to be in order.
It could be stated in the form:
"There exists a
T > 0
such
that the free boundary has a unique solution."
It should be well understood that (18) represents the ordinary
differential equation for the free boundary on a first interval
41
only, i.e. for small
t
,
where we can take advantage of the
fact that the incoming Riemann invariants all have the same
value.
(c)
The construction of the solution in the regions (II) and (III)
breaks down
if the lines emanating from the boundary intersect.
Then the equations for
u
and
c
are overdetermined, and we
In particular,
have to consider discontinuous solutions (shocks).
if
i
increases, it follows from
decreases and therefore
u - c
Ro
= k(u
2
+
y-1
also increases.
c)
that
c
This means that
the lines emanating from the free boundary will not intersect
and we obtain a rarefaction wave.
decreases, the slopes
becomes likely.
u - c
On the other hand, if
of the lines decrease and a shock
42
8.
Shocks
It is well known that for hyperbolic systems, continuous
solutions for large time intervals may cease to exist.
For example,
-
whenever two characteristics of the same type, i.e.
intersect, the equations for
u
and
p
x - or
are overdetermined.
x -curves,
Hence,
we encounter "curves of discontinuities" which separate the plane
into regions where the solution is continuous.
Such curves could,
mathematically, be chosen at random; however, we want to determine
those curves which are physically reasonable.
First, we are going to consider the equations of conservation of
mass and conservation of momentum in integral form.
From here we
will derive the mechanical shock conditions, and we will study some
conclusions that can be drawn from the mechanical shock conditions
alone.
In the second part we will add a thermodynamical condition,
which is derived from the law of conservation of energy across a
shock.
As it turns out, the relation
shock condition.
p=
1
py
violates the third
Hence, in the second part we will consider
p
as
a function of density and entropy.
8.1.
Mechanical shock conditions
Conservation of mass and balance of linear momentum in integral
form are given by
ar
dt
p dx = 0
and
al(t)
ar
dt
a1
f
pU dx = p(al(t),t) - p(ar(t),t)
43
where
1(0, ar(0
are the paths of two particles.
of discontinuity is given by
E(t)
I
dt
If a curve
we require that
E(t)
a2(t)
pdx
=
dt
al(t)
pdx = 0
E(t)
and
a
(t)
d
pu dx +
dt
a1(t)
uL
1
hold.
a1
r
f
r
pu dx = p(a
1
- p(ar(t),t)
If we carry out the differentiation, and if we let
E(t), we obtain
(t), ar(t)
U(pi - pr) = ui pi - ur pr
and
2
2
U(u1 pl - ur pr) = pl - pr + ui p
where
'
E(t)
dE
dt
U =
- ur pr
denotes the shock velocity, and the subscripts
indicate the state on the left or on the right of the shock.
The
equations (5) and (6) are called the mechanical shock conditions.
If we denote the velocities relative to the shock velocity by
i.e.
v := u - U, then the mechanical shock conditions can be
written in the form
vl
pl
= p
v
r
=: m
m vi + pi = m vr + pr =: P
m
.
is called the mass flux across the shock, and
P
is the total
and m always have the
momentum flux.
(7) implies that
same sign.
in > 0, then particles cross the shock from the left
If
to the right and vice versa.
For
v1, vr
m > 0
we call the state on the
v,
44
left the front side of the shock and the state on the right the back
of the shock, and vice versa.
In other words, particles always
cross the shock from the front to the back.
Note that this
definition does not refer to the shock velocity
The case m = 0
implies
=
v1
vr
= 0,
U.
particles travel
i.e.
with the shock curve, and from (8) we obtain
However,
pi = pr.
jumps in the density are possible.
Equation (8) implies
and using (7)
m(vi - vr ) = pr - pi
we can derive
v
2
1
- v
2
r
1
1
+
=
pr
pl
r
- p
)
.
1
From (9) we conclude that the absolute values of the velocities and
The same is true for
the pressures change in the opposite sense.
the absolute values of the velocities and the densities, according
to (7).
In the remaining part dealing with the mechanical shock
conditions, we shall assume that the relation
p=
is valid
1- pY
on both sides of the shock.
Assume that the state on the left
as well as the shock velocity
m and
V1
P.
ui
,
are known.
U
and
pl
pi -
1piy
Then we can compute
In order to determine the state on the right, we
'
v
eliminate
P
from (8)
m2.1
Pr
2
and obtain
py
r
)
Pr
Equation (10) has at least one solution
and no jumps occur.
pr = pi
Differentiation provides
.
Then
vr
=
v1
45
2
dpr
f(Pr) =c r
- v2
and
r
d2
f(p ) = 2m2 p-r3 + (y - 1) p7-2 > 0
2
dpr
Hence
cl
pi = pr
= 11,11
,
U = u - cl
is the only solution of (10) if and only if
which is equivalent to either
Or
In this case the shock has a characteristic direction.
.
Iv11 0
If
U = u1 + cl
cl
,
then
f(pr) = P
has a second solution
pr
0
p1
and we obtain
Ivr1
Hence
1v1
cr
iff
>
Iv11
c1
is subsonic with respect to the state on one side and
supersonic with respect to the state on the other side.
It should
be noted that (13) can be derived without the assumption
p = -
py
7
(see [1]).
As far as the mechanical shock conditions are concerned, it is
not possible to conclude that e.g. the pressure is higher or lower
on the front side, compared to the back.
Let m > 0, i.e. the front
is located on the left, and consider the cases
(a)
and
P1 > Pr
In case (a) we must have
we obtain from (11) that
0 <
(b)
<
v1
2
2
vr
cl - v1 > 0
according to (13), and we obtain
P1
and
Pr
p/ >
Then
v1
Since
pr.
<
c1
and
pr
vr
<
>
p1
c1
46
ul
U <
-
cl
ur
< U <
-c r
+
ul
<
ur
cl
+
cr
By similar reasoning we obtain for case
U < u1 - C1 < u1 +
ur
-c r
< U <
ur
+
(b)
c1
cr
Case (b) is the physically reasonable case.
It also allows us to
compute the state on the left, and the value of the Riemann invariant
along
xr
together with the two mechanical shock conditions provides
shock
'
(a)
shock
(b)
u
+
-cr
Ur +C r
-cr
Figure 10
47
a system of three equations for the three unknowns
Lax
[51
ur
and
pr
for example, includes (14b) in the definition of a
[61
,
U,
shock--in addition to (5) and (6).
8.2.
The Thermodynamical Shock Condition
For polytropic gases the law of conservation of energy takes
the form
(0
a
d
dt aj (t)
(15)
(a (t),t)
2
rt.
P) dx = - pu
(2121
1
(a'
from which we derive the third shock condition
2
2
Plul
1
+
2
y-1
p u
r r
p)v - (---- +
1
2
1
1
-15 iv=pu
rry-lrr
pl ul
Using the notation of the first part of this section we obtain
v2
m( 1
T :=
for
v2
m( r
y-1 131 Tl
2
where
)
.4.
1/p
2
_a__
y-1 pr Tr)
is the specific volume.
m
If
we can solve
0
2
vi - v2r, and combining this expression with (9) we obtain
.21(:(1
Let
82
(T
(pr Tr - pl T1) = (Tr + T1)(pr - p1)
:= (y - 1)/(y + 1).
-
62
T1) pr
Then (18) becomes
= (T1 - 62
Tr) p1
which is equivalent to
(Pi - 62 Pr) Pr = (P
- 62 Pi)
pi
.
.
.
.
48
Assume
p =
y
1
p
x := pr
and let
,
and
y := pl
.
Multiplying
y, we have to solve
by
wx,y)
0
Since
H(x,y) = -H(y,x)
(20).
If we fix
x = y
,
62 yy+1
will always provide a solution of
and consider
y > 0
yyx
yxy
_62 xy+1
h(x) := H(x,y)
on
10,y1,
we find that
11.1(x) = -(y-1,x
y-1
y
+ y yx
- y
and
h-(x) = y(y-i)J-2 (y-x).
Since
hl(y) = 0
h(x) > 0
on
and
ht(x) > 0
0 < x < y ,
for
is the only
x = y
and that the diagonal
(0,x)
we conclude that
solution of (21).
Hence the assumption
p =
1
y
violates the law of conservation
of energy, as soon as discontinuities occur.
Equation (20) implies
Pr
Pr
p1
p1
and it follows that
cannot exceed
1/62
62 P1
+ 62
p1
62 < pr/p1 < 1/62
.
Investigating the relation between
(20)
Pr
Y
Pr
i.e. the compression
,
P1
= K(pr,p)
P1
p
and
pi
we derive from
49
where
A
(25)
r
- 62 P
2p)
K(pr,P1)
pl - 6
Assume that
62
<
pl
pl
k(pr) := K(pr,p1)
for
We find that
.
k(pr) =
d
Thus,
r
Pr
is fixed and consider
pr < p1/62
(26)
P
r_l)y
1
k(pr)
is increasing with
k(pi) = 1
and
1(1(y =
0.
In the theory of gas dynamics the pressure is considered to be
a function of the density
and of the specific entropy
p
S.
For polytropic gases this function becomes
p = A(S)
where
pY,
S-S
A(S) = (y-1) exp
where
i.e.
So
(7=2)
is a suitable constant.
In particular we have A'(S) > 0,
at constant density, pressure, and entropy increase in the
same sense.
the state on the left is given, then we obtain
Assume
from (24) and (27)
A(Sr) = k(pr) A(S1)
Then
A(Sr) < A(Si)
,
if
pr < pl
.
,
which implies
Sr < S1
.
By a
consequence of the second law of thermodynamics, the entropy of a
particle is nondecreasing.
right.
Hence the shock front has to be on the
See also Dafermos [2].
Collecting our results we obtain
50
pressure and density are less on the front side than on the
back of a shock
particles pass a shock from the front side to the back.
Their relative speed with respect to the shock is supersonic on the front side and subsonic on the back (compare
equation (13)).
The entropy of a particle increases when passing a shock.
As it turns out,
Hence
A(Sr) = A(S1
expressed by
kl(pi) = 0,
k"(pi) = 0,
but
k'"(pi)
0.
up to third order terms of the shock strength--
)
Thus for "small" jumps, the concept developed
pr - pl.
via mechanical shock conditions alone provides an appropriate
description of the situation.
The increase of entropy across the
shock stresses "case (b)" mentioned in that section.
The third shock condition allows us to derive some more properties
of the shock transition.
c2
* -= 62 v2
'
c*
1
From (17) we derive for
+ (1-6) c2
= 62 v2
+ (1-6)
m
c2
0
.
1
is called the critical speed and it yields Frandtl's theorem
2
c* = vivr
For a proof of (31) see, e.g.
(1).
The Mach numbers
are defined by
M1
and
Mr
=Mr :Iv11
(32)N1
or
:
Ivr1
cr
1
and the pressure ratio can be expressed by
(33)
= (1+62) M2 - 62
pl
1
51
Assume that the state on the left is given and that the shock
velocity
U
is known.
Then the state on the right is completely
determined:
First compute
vi := ui - U,
M1 :=
and
v1 /c1
c2 :=2
m := vi pi,
v21 +
(1-62) c2
and
pr =
((1+82)14
P
:=
.
1
Then for the state on the right we obtain
pr := m/vr
C1
vr
:= c2/v
*
1
- 82) pi
If we take entropy into account, we have to extend our former
system of equations describing the state of the gas.
Assume that initially the gas has constant entropy
case
So
in (28) can be chosen such that
A(S) =
1
S.
In this
and the former
system of two equations, derived from conservation of mass and
balance of linear momentum, is appropriate until shocks occur.
The
model that particles maintain their entropy unless they cross shocks
can be described by
St + u Sx
=0
and the jump condition
m Sr
S1
Changes of entropy along the shock curve can be computed from
52
(36)
Sr - S1 = c
ln
Pr
(P1)y I
P1
Pr
12
2
Note that by definition the sound speed was given
by.c2
which is independent of possible entropy change.
Hence, the state of
=
dp
= y
,
the gas is given by the system
pt
+ (up)x = 0
p(ut + uux) + c2 px = 0
St + u
S
=0
The set of characteristics now is
u
c
and
u
and the value of
the entropy becomes a third Riemann invariant which remains constant
along the corresponding characteristic--the path of the particle.
shock
;particle
path
II
Figure 11
53
In region I indicated in figure 11, (37) and (38) together with
A(S) = 1-
suffice to describe the state of the gas.
contains particles with "shock experience," i.e.
entropy.
Region II
with increased
The regions are separated by the shock curve and a contact
discontinuity, i.e.
a shock curve with m = 0.
This curve has
characteristic direction with respect to both regions, since it is
the path of a particle.
m = 0
implies that
u
and
p
are
continuous along this curve whereas the density in II is less than
in I due to the increase of entropy.
54
Free Boundary and Shocks
9.
In this section we shall study the relationship between shocks
and the motion of the free boundary.
First we discuss the question
of how the motion of the piston can generate shocks.
In the second
part we study the problem of how the piston reacts on impinging
shocks.
If the piston moves towards the fixed end, in general we will
obtain a compression depending on the initial state of the gas.
Increasing pressure goes along with an increase of the density and
the speed of sound.
Hence, for a compression the characteristic
curves of slope
c
arise.
u
have a tendency to intersect and shocks
However, such a shock would occur somewhere in the interior
of the cylinder and not directly at the piston.
If the piston moves away from the fixed end, then we obtain a
rarefaction wave.
In general, rarefaction waves are shock-free.
Shocks can be generated at the free boundary
occur.
if jumps in
Since the motion of the free boundary is governed by
(1)
= a p(s(t),t)
g+
we see that jumps in
continuous.
If
f
f
+ f
will cause jumps in
N; however,
remains
is a 6-distribution, which corresponds to a
sudden impulse at the free boundary, we will have a jump in
some time
t*.
at
If the impulse goes in the positive x-direction, we
obtain a centered rarefaction wave with center at
(s(t*),t*).
the impulse is directed towards the interior, we obtain a shock
If
55
emanating from
u(s(0,t)
and
The limits of
(s(t*),t*).
as
t
p(s(t),t),
p(s(t),t)
determine the state on the left of the
t t*
.
shock,
in particular we have
u1
=
.
Together with
have enough information to determine the shock velocity
.
Pr
ur
pr
U,
and
In particular, the state Dn the left becomes the front side,
the state on the right is in the back of the shock,
A different situation arises as a shock curve
the free boundary at some instant
extend the solution for
t > t*.
t*.
etc.
in > 0,
(t)
intersects
There are various ways to
One could assume that
remains
continuous and the shock is reflected, or one assumes a jump in
e.g.
we
= s4-
u1
= i-
and
ur
= i+
combination of both cases.
and the shock is absorbed, or a
Our aim is to find the physically
reasonable model.
rarefaction wave
Figure 12
56
The linear momentum of the entire process is the sum of the momentum
of the gas and the momentum of the piston and can be expressed by
s(t)
f
M(t) =
Pu dx + de
0
(see (2.9), and note that in the dimensionless version of the
problem
d = 1).
The shock conditions along
along a reflected shock
(t)
A
has those properties.
external force occurs,
Then
M(t)
is continuous or
C1
as
Unless an impulse caused by the
has to be continuous at
M+(t* = lim M(t) = lim M(t) =
t+t*
and possibly
take care of the conservation laws
in the interior of the cylinder and M(t)
long as
(t)
t
t*.
and we derive
-(t*)
ttt*
M+(t*) - M-(t*) = d(i+ - A-) = 0
since the integral terms cancel out due to the shock conditions in
the interior.
Hence
A
is continuous at
of the piston is negligible.
Figure 13
t = t*, unless the mass
57
In our notation we shall indicate the states on the different
sides of the shock
E(t)
by subscripts
simplify the notation, the values
as limits as
approaches
t
The region between
E(t)
t*, i.e.
and
etc. are understood
m
u
ur'
= lim
u1
< v
r
A
pi > pr,
P1 > Pr,
6(t*) = u.
r
1
Since
etc.
necessarily belongs to the
s(t)
Moreover, we have
< O.
u(E(t),t)
t+ t*
front side of the shock and we obtain m < 0,
v
1, and in order to
and
r
ui =
ur
t*, the inequality
is continuous at
with the front side on the
generates a reflected shock curve
E(t)
left, the back side on the right.
The states on either side of
t's
R
will be indicated by subscripts
E(t)
understand
approaching
uL, uR
etc. as the corresponding limits for
t
t*.
We then have
pi = pL,
The continuity of
determine
L, and again we
and
requires
A
uR, pR,
ui = uL,
pi =
but
vi
vL.
Our aim is to
ur = A = uR.
from the information about the initial
pR
shock.
First we determine
vL*
,2
2
Let
c
- = o
*
'
2
,
,2,
+ kl-o
vL
2
)
cL.
Then
from Prandtl's theorem we obtain
c2
*
= v v
L
R
= v2 + (u -u_ ) v
R
L
L
.
L
Combining both expressions we obtain a quadratic equation for
(4)
2
vL
1+1
2
(uR
uL
)
vL
cL2 = 0
.
vL:
58
coefficients in (4) are known, and since
side we must have
vL
> O.
Once
L
vL
4
belongs to the front
v
Hence,
y+1
(5)
2
1/((+1)2
CUR
All
1/(1+82) = (y+1)/2).
82 = (y-1)/(y+1), hence
(Recall that
UL)
U
is known, we can compute
UL)
CUR
4
= UL - vL
and proceeding as
in the previous section we can completely determine the state on the
right of
E
.
For the motion of the piston the jump of the pressure at
is of interest.
In order to describe
pR/pr
in terms of
t*
pl/pr
consider first the numbers
A
u1-U
vi
A
M =
and
M :=
(6)
c1
c1
u1-U _ vL
Cl
Cl
For the initial shock we can derive a quadratic equation for
in the same way as we did for
v1
is a solution of (4) as well.
satisfy
v1
< 0
and since
'
UL =
u1
and
CL =
This time, however, we have to
and pick the negative square root term.
Multiplication then yields
(7)
vL
vl
vL *IT1
M M = -1
=
2
-c1
and we obtain
.
Using the relationship between the Mach numbers and the pressure
ratios (8.32) and (8.33) we can write
c1
59
Pr
= (1+62) M2 -
cS2
P1
and
PR
"2
2
2
= (1+8 ) M
.
P1
From (5) and (6) we derive
9
(1+2(5-)
PR
(9)
pi
' - 62
Pr
Pr
l+ S
.
2 P
P1
Pr
Pr
Considering
pR
as a function of
Pr
p1
P1
, we obtain that for
Pr
is the arithmetic mean of
pR
whereas when
and
'jr'
p1
p1
=
,
pr
÷ co
Pr
we have the approximation
PR
1
P1
= (2 + --)
(S2
pr
Hence after the impact of the shock we have to encounter a
tremendous increase of pressure, depending on the strength of the
initial shock which pushes the piston in the positive x-direction.
In particular, we have a jump in
t = t* .
and a jump in
M(t)
at
60
A Numerical Method for the
10.
Non-linear Problem
In this section we shall find approximate solutions to the
non-linear problem, restricting ourselves to the case that no shocks
occur.
In order to solve the non-linear hyperbolic system
Pt
'
(up)x =
p(ut + uux) + px =0
we again apply the method of characteristics.
express
p
in terms of
p
by setting
p :=
First of all, we
1-
py
,
as done before.
Since the slopes of the characteristics are given by
the Riemann invariants are functions of
prefer to work with approximations of
The relation between
c
and
p
u
c
and
c
and
u ± c
as well, we
rather than those of
p
is given by
Y-1
c
p
2
In the non-linear case the characteristics are no longer
straight lines.
Taking care of this situation, we have to work with
a variable grid point net.
We introduce new variables
and consider
c
x, t, u
and
as functions of
a
a
and
system (1) can then be written in canonical form (see e.g.
and
.
The
(31])
61
x
a
= (u+c) t
a
xa = (u-c) ta
2
(3)
a
y_1 ca
2
y-1
a
and
are parameters for the characteristics.
we have for example
R = 1/2(u +
a
2
y-1
dx
= u + c,
dt
remains constant.
Figure 14
Along
and the Riemann invariant
= const,
62
In our grid point system we want to find approximations
cn,
m
un
m
,
0 _<m <_ M
xn
m
or
and
tn
m
(M-1)
where
indicates the iteration step and
n
'
is the number of grid points for each
n.
Initially we start with a discretization of the interval
to obtain the points
points we have
to = 0
xo
,
and
where
co
xo
= 0
uo
and
and
xm = 1.
[0,11
For those
can be determined from
the initial functions.
Next we approximate the characteristics emanating from
by lines of slope
uo ± co.
Then the new grid points are given by
the points of intersection of two neighboring
0
x°
xo
rn
Figure 15
±
characteristics.
63
We obtain the formulas
t
1_
x)/(u
=
1
xm
1
cm
u1
,
(xo
m
m+i
=
o,+ o
xm
kUm
o
o
o
Ut
Ut
m+1
o,
1
M
M
+C
)*t
co
m
)
m+1
can be derived from the constancy of the Riemann invariants
along the characteristics.
From (4) we see that we loose a grid point in the first step,
since
t1
cannot be determined.
We will make up for that loss by
including the boundary conditions in the next step, as we will see
later.
After the first iteration step the grid points
not necessarily lie on a straight line with constant
1
,
k
X
m
t
,
1
t
m
)
do
values.
Thus, the scheme (4) has to be modified to obtain a general formula.
We set
an := un +c
n
bn
*= un
- c
m+i
m+1
m+I
,
and obtain
xn
- x: + an tn - bn
tn
m+1
m+I m+1
m in
n+1
tm
n+1
xm
an - bn
m+1
=
n
xm
n+1
n
+a n,
kt
-t)
m
m
The approximations for
u
and
c
are given by
64
n+1
um
n
+um+i
y-1 , n
n+1
c
n
=
(u
4
in
kn
))
,cn -c m+1
2
+
y-i
-u-
, n
2
+
u+
in
y-1 kCm
This algorithm becomes unreasonable
n+1
< tn
tm
m+1
if
+c
tn+1 < tn
-M
M
or
which indicates that characteristics of the same
'
We do not pursue this question
type intersect and a shock builds up.
any further.
In order to describe the algorithm on the boundary, we shall
assume that
n > 2
un-1
n-1
cm
and
n-1
is even, and that the data
0 _
< m <
_ M-1
are given for
n-1
,
xm
tm
In the interior
.
we apply the schemes (6) and (7) with a shift of subscripts to
tn
determine
xn,
,
M
111
un
and
cn
for
M
1 _
< m _
< M-1.
The additional grid point on the fixed boundary is computed
n-1
n-1
co
uo
by the intersection of the characteristic of slope
emanating from the point
n-1
(xo
n-1
)
,
to
x = 0
and the line
obtain
xo
t
o
The value
co
=0
= t
n-1
o
- xn-1/(un-
.
is to be determined from
2
r(tn) (cn)-2/(Y-1
o
0
y-1
is known we can find
Once
- cn-1 )
un
en
0
from
con
uo
= r(tn).(0n)-2/(y-1)
0
0
2
0
y-1
en-1
0
.
We
65
For the free boundary we use a similar idea.
The new grid point
is given by the intersection of the characteristic with slope
n-1
n-1
+ cm...1
n-1
emanating from the point
n-1
t)
and the line
approximating the motion of the free boundary.
slope
zn-2 =
u;11-2
,
passing through
n-2
(xm
The latter is of
n-2
tm
).
,
then has the coordinates
n-2
tn
M
n
xM
n-2
n-1
uM -1 (tM-1
n-1n-1
cm..1
M-1
n-2
n
n-2
(tm - tm )
um
The rate of change of
(12)
n-1
xM-1
tn -1 4. xM
z=a
n-2
tM )
n-2
- um
xit.11 -2
was given by
= -pz + ap + f
t)
free boundary
n-2
n-2
(xm
Figure 16
,
n-2
tm )
-
The grid point
66
thus, we can approximate
(recall t) = z(t) = u(s(t),t))
u
at
the new grid point by
11.;
where
Az
=
.Az
tr141-2)
is given by
Az = -pu;
and
(t;
11;11-2
-2
+
we can compute
n
CM =
n-1
cm_i
+
a ,kcmn-2,2y/(y-1)
)
cn
y-1
f(x;-2
n-2 ,
tm
)
by
n
n-1
Finally, we illustrate this algorithm in two examples.
Example 1
Here we assume that the same initial data are given as in
example 1 of section 6, now in the setting of the non-linear model.
We obtain
a = 1, u(x) E
p = 0,
f(x,t) E -1/y
.
0.5,
p(x)
E 1.5
and
The boundary conditions have been modified into
r(t) = u(0,t) E 0.5.
Comparison shows that the motion of the free
boundary is almost identical in the linear and the non-linear case.
In the plot (Figure 17) the characteristics and the free boundary
are graphed.
Example 2
We assume that the gas is initially at rest and that the
cylinder is closed at the fixed end.
The motion is caused by a
higher pressure outside, which is twice the internal pressure.
we have
u(x) E 0,
co(x) E
1,
r(t) E 0
assume that friction is present and choose
and
f(x,t) = -2/1
p = 1
and
a=
1.
Thus
.
We
The
67
numerical result (see Figure 18) shows a motion of the piston
towards the fixed end, and the computation breaks down in the 13th
iteration step when a shock builds up.
68
1
0
0
I
0.5
1.0
Figure 17
1.5
2.0
if
=
69
4
1
I
1.i
0
0
04
0
0 0
.
Figure 18
70
BIBLIOGRAPHY
R. Courant and K.O. Friedrichs, Supersonic Flow and Shock Waves,
Interscience, New York, 1948.
C.M. Dafermos, The Equations of Elasticity are Special, (1980)
Pitman Publications, London.
G.E. Forsythe, W. Wasow, Finite Difference Methods for Partial
Differential Equations, Wiley, New York, 1960.
C.D. Hill, A Hyperbolic Free Boundary Problem, J. Math. Anal.
Appl. 31 (1970), 117-129.
P.D. Lax, Hyperbolic Systems of Conservation Laws II, Comm.
Pure Appl. Math 10 (1957), 537-566.
P.D. Lax, The Formation and Decay of Shock Waves, Math. Monthly
79 (1972), 227-241.
A.E.H. Love and F.B. Pidduck, Lagrange's Ballistic Problem,
Trans. Roy. Soc. London 222 (1922), 167-226.
B. Sherman, A Free Boundary Problem Arising in a Kinematic
Wave Model of Channel Flow with Infiltration, Quart. Appl.
Math. 39 (1981), 87-96.
A.N. Tychonow und A.A. Samarski, Differentialgleichungen der
mathematischen Physik, VEB Deutscher Verlag der Wissenschaften,
Berlin, 1959.
APPENDIX
71
We consider the free boundary problem as stated in section 2,
given by the equations (1) - (4), (7), (8), (11), where all
quantities carry physical dimensions.
For an ideal gas in an adiabatic process the relationship
between pressure and density is given by
2- = (-2-)Y
o
o
where
po
depends on the gas we are dealing with.
y
and
for
po
remain to be chosen appropriately.
po
,
y,
co
to obtain
po
po.
po,
po
are known, let
c
Then
E.g., given a choice
determine the pressure of the gas at density
the value for
Once
The quantities
o
:=
if17-0
y
p
0
has the dimensions of a velocity.
In fact, it is the
average sound speed corresponding to the density
po
and pressure
po.
In order to rewrite the equations in a dimensionless form we
introduce the dimensionless variables
72
x=EZ
T
:=
t =
t
:= 11.
Po
U = Ti
C
0
= C c
C
s = S Z
Z :=
z
c
A = z = Z
=
o
p
F :=
R :=
7 d po
f
c p
00
d po
f = 7 F z
r=Rc 0 p 0
Next we express all quantities of the original problem in terms
of the dimensionless variables and we obtain the equivalent
problem
co
73
(PT
40
(0)
+
= o
E
+
T
nng)
+P =0
= (1)0(E)
11(,0)
{(1)(E,O)
= no(E)
= R(T)
n(s(T),T) = z(T) =
Igo,T)n(0,T)
dT
z + pz = al) +
s(T)
F
(FB)
s(0) = 1,
where
p :=
k
z(0) = 710(1)
a .-
and
d---
po
co
This is exactly the system of equations as stated in the end
of the second section, if we replace
=
P
(I)
,
u=
n
,
E = x, etc.
Finally, we want to remark that at the first glance the choice
of
ypo
as a reference-pressure seems to be unnatural.
p = P po
However,
provides
2
n
k
and equation
= Po YPo p
= Po p
(M)
E
y kpo
E
1 Po
y
o
2PE
would be instead
(n-r +nn)P +
Y
= 0,
E
which differs from the original form of the equation of motion by
the factor
.