5. Systems of Conservation Law Equations
In the previous sections we have examined a number of issues relating to scalar valued
quasilinear partial differential equations of the conservation law form in one space variable
and one time variable. The most general problems of this type involve systems of equations
in m unknown function of n space variables and a time variable. Problems of this generality
are still poorly understood. However, we can generalize to problems involving m unknown
functions in one space variable and one time variable. We list now several examples of
such systems.
Example 5.1
1. The p-system- Consider the system of two equations for two unknown functions
u = uÝx, tÞ, v = vÝx, tÞ
/ t uÝx, tÞ ? / x vÝx, tÞ = 0
/ t vÝx, tÞ ? / x pÝuÞ = 0
2.
This can be written in vector notation as,
.
3 Ýu
3 Ýu
3
3 Þ = ß?v, ?pÝuÞà
3 Þ where
u = ßu, và,
F
u = /xF
/ t3
.
Note that
/ tt u = / tx v = / xt v = / xx pÝuÞ, hence this system is equivalent to a
single nonlinear wave equation of order two for the unknown function u.
.
Shallow Water Waves The following system for unknown functions
v = vÝx, tÞ and h = hÝx, tÞ can be used to describe waves propagating with
horizontal velocity v in a shallow pool of depth h.
/ t h + / x ÝvhÞ = 0
/ t v + / x Ýv 2 /2 + hÞ = 0
3.
.
The first equation expresses the conservation of mass and the second expresses
the conservation of momentum in the fluid.
.
Euler’s Formulation of the Equations of Gas Dynamics The equations for the
1-dimensional flow of a compressible gas are
.
Ýconservation of massÞ
/ t _ + / x Ý_vÞ = 0
2
/ t Ý_vÞ + / x Ý_v + pÞ = 0
Ýconservation of momentumÞ
/ts + v /xs = 0
Ýconservation of energyÞ
.
where the unknowns p, _, v and s are pressure, density, velocity and entropy.
If we add a state equation p = pÝ_, sÞ , then the second equation can be
replaced by
.
1
2
/ t v + v / x v + c_ / x _ + a / x s = 0
.
where c 2 = / _ pÞ s and a =
speed of propagation.
1
_
/ s pÞ _ . Then c has the interpretation of the local
Problem 5.1 Show how the state equation can be used to eliminate the pressure from the
momentum equation.
Problem 5.2 Suppose that the entropy is a constant, (i.e., isentropic flow) and that the
equation of state reduces to p = k_ L (a polytropic gas). What does the momentum equation
reduce to in this case?
Weak Solutions
As we did in the case of the scalar conservation law equations, we will define a weak
solution of the system,
3 Ýu
3Ýx, tÞÞ = 3
0,
uÝx, tÞ + / x F
/ t3
3
u 0 ÝxÞ
uÝx, 0Þ = 3
Ý5.1Þ
to be a function which is locally integrable in R 2+ with values in R n such that
X XR 2 3u 6 / t d3 + F3 Ýu3Þ 6 / x d3 dxdt + XR 3u 0 6 d3Ýx, 0Þ dx = 0
+
3.
for all vector valued test functions d
Rankine-Hugoniot Condition
If there is a curve x = xÝtÞ in the x-t plane across which 3
u=3
uÝx, tÞ experiences a jump
discontinuity, then we can use essentially the same argument that was employed in the
scalar case to show that
3 Ýu
3 Ýu
3L ? 3
3 R Þà
3L Þ ? F
x v ÝtÞ ßu
u R à= ßF
Ý5.2Þ
Note that unlike the scalar case, which was a scalar equation that could be solved for the
unknown function xÝtÞ, this is a system of n equations relating the 2n + 1 quantities,
3
u R and a = x v ÝtÞ. We will consider some shock problems later.
uL, 3
Travelling Waves and Strictly Hyperbolic Systems
Consider the quasilinear system of n equations in n unknown functions,
3Ýx, tÞÞ / x3
0.
/ t3
uÝx, tÞ + AÝu
uÝx, tÞ = 3
Ý5.3Þ
This is an equation of conservation law form if the n by n matrix A is equal to
2
/u1 F 1 /u2 F 1 ` /un F 1
/u1 F 2 `
3 Ýu
3Þ =
grad F
/un F 2
_
b _
/u1 F n
/un F n
A travelling wave solution to the equation (5.3) is a solution of the form 3
u = 3fÝx ? atÞ
3
where the wave form, f, and a, the wave speed, are to be found. We see that
/ t3
uÝx, tÞ = 3f v Ýx ? atÞ Ý?aÞ
3f v ÝSÞ Ý?aÞ + AÝu
3 Þ3f v ÝSÞ = 0, or
hence (5.1) becomes
3Þ 3
x = a3
x
AÝu
/ x3
uÝx, tÞ = 3f v Ýx ? atÞ Ý1Þ
and
where
3
x = 3f v ÝSÞ,
S = x ? at.
Evidently, the eigenvalues of A are the wave speeds and the associated eigenvector
determines the wave form that travels at that speed. Then the existence of travelling wave
solutions to (5.1) is dependent on the existence of real eigenvalues for the matrix A.
Therefore, we define the system (5.3) to be strictly hyperbolic if, for each 3z 5 R n , the n by
n matrix AÝz3Þ has n real, distinct eigenvalues. Although problems in which the matrix does
not have real distinct eigenvalues are of interest, they are more diificult to analyze and we
will therefore suppose throughout this section that our problems are strictly hyperbolic.
Example 5.2
1. The p-system- Consider the p-system for the unknown functions
u = uÝx, tÞ, v = vÝx, tÞ
.
/t
u
v
=
0
1
/x
p v ÝuÞ 0
u
v
Then
matrix eigenvalue V 1 ÝuÞ = ? p v ÝuÞ
eigenvector
3
x 1 = 1,
V 2 ÝuÞ =
p v ÝuÞ
3
x 2 = 1, ? p v ÝuÞ
p v ÝuÞ
It is evident that this system is strictly hyperbolic provided that p v ÝuÞ > 0 and in
that case, travelling wave solutions are of the form
3
x = 3f v ÝSÞ =
u v ÝSÞ
v v ÝSÞ
=
1
± p v ÝuÝSÞÞ
.
We find then
3
vÝSÞ = X
uÝSÞ = S ? S 0 ,
2.
S
p v ÝsÞ ds
S0
.
Shallow Water Waves Consider the equations
.
/th + v /xh + h /xv = 0
/ t v + v / x v + / x h = 0;
i.e.,
h
/t
v h
=
v
h
/x
1 v
.
v
Then
matrix eigenvalue V 1 = v ? h
3
eigenvector
x1 =
h , ?1
V2 = v + h
3
x2 =
h ,1
This example is also strictly hyperbolic, for h > 0, and simple wave solutions
have the form
h v ÝSÞ
3
x = 3f v ÝSÞ =
v v ÝSÞ
hÝSÞ
=
±1
.
i.e.,
hÝSÞ =
3.
1
4
ÝS ? S 0 Þ 2 + h 0 ,
vÝSÞ = ± ÝS ? v 0 Þ
.
Euler’s Formulation of the Equations of Gas Dynamics
The gas dynamics equations
/ t _ + / x Ý_vÞ = 0
2
/ t v + v / x v + c_ / x _ + a / x s = 0
/ts + v /xs = 0
can be written in matrix notation as follows,
_
/t
v
s
+
v
c2
_
_ 0
0
0 v
v a
_
/x
v
=3
0
s
Since this matrix has eigenvalues V = v, v ± c, it is clear that this system is also
strictly hyperbolic if c 2 = / _ pÞ s > 0. The eigens are,
4
matrix eigenvalue V 1 = v
3
x 1 = ß?a, 0, c 2 /_à
eigenvector
V2 = v + c
V3 = v ? c
3
x 2 = ß_, c, 0à
3
x 3 = ß?_, c, 0à
Riemann Invariants
3 Þ be arranged in
Suppose (5.3) is strictly hyperbolic and let the eigenvalues of A = AÝu
increasing order,
3 Þ < V 2 Ýu
3 Þ < ... < V n Ýu
3 Þ.
V 1 Ýu
3 Þr3m = V m 3r m .
3 Þ denote the associated right eigenvector; i.e., AÝu
For 1 ² m ² n, let 3r m Ýu
n
u. Since
Since the eigenvalues of A are distinct, the eigenvectors 3r m span R for every 3
3 Þ the eigenvector for A f Ýu
3Þ
A and A f share the same eigenvalues, we can denote by 3
q m Ýu
3
corresponding to the eigenvalue V m Ýu Þ; i.e.,
or
Note that
i.e.,
Then
3m = V m 3
3 Þq
A f Ýu
qm
1²m²n
3
3Þ = V m 3
q fm AÝu
q fm
3
3Þ
q fm is a left eigenvector for AÝu
3
q fj 3r k = 0
for
k ® j;
3 Þ 3r k = V j 3
Vk Ý 3
q fj 3r k Þ = 3
q fj V k 3r k = 3
q fj AÝu
q fj 3r k
(V j ? V k Þ 3
q fj 3r k = 0 and since V j ? V k ® 0 for
k ® j,
the result follows.
3 Þ and note that
Consider now a scalar valued function Q j = Q j Ýu
3 Þ + V j / x Q j Ýu
3 Þ = 4Q j Ýu
3Þ 6 / t3
3Þ 6 / x3
/ t Q j Ýu
u + V j 4Q j Ýu
u
3Þ / x3
3 Þ 6 Ý/ t 3
3 Þ 6 Ý?AÝu
= 4Q j Ýu
u + V j / x3
uÞ = 4Q j Ýu
u + V j / x3
uÞ.
3Þ = 3
3 Þ, it follows that
q j Ýu
Now if 4Q j Ýu
3 Þ + V j / x Q j Ýu
3Þ = 0
/ t Q j Ýu
which, in turn implies that
5
3 Þ is constant along solution curves of x v ÝtÞ = V j Ýu
3 Þ.
Q j Ýu
3 Þ, 1 ² j ² n, can serve as alternative
Evidently, the scalar valued functions Q j = Q j Ýu
dependent variables in the equation (5.3). The advantage of this choice of dependent
variable is that the system becomes diagonal when written in terms of the Q vj s instead of the
u j ’s. This notion can be exploited in various ways.
Example 5.3
1. Consider the following linear system
/t
u
v
?1
0
+
?c
2
u
/x
0
=3
0.
v
The distinct real eigenvalues and the associated left eigenvectors are
3
q 1 = ßc, 1à
V 1 = ?c,
3
q 2 = ß?c, 1à
V2 = c ,
Then
3
q1 6 /t
u
v
+3
q1 6
?1
0
?c
2
0
/x
u
v
.=
.
c/ t u + / t v + V 1 Ýc/ x u + / x vÞ = 0
and
3
q2 6 /t
u
v
+3
q2 6
?1
0
?c
2
0
/x
u
v
.=
.
? c/ t u + / t v + V 2 Ý?c/ x u + / x vÞ = 0,
or
cÝ/ t u + V 1 / x uÞ + / t v + V 1 / x v = 0
? cÝ/ t u + V 2 / x uÞ + / t v + V 2 / x v = 0
Let C 1 , C 2 denote solution curves for the systems
C 1 : t v ÝJÞ = 1,
C 2 : t v ÝKÞ = 1,
x v ÝJÞ = V 1 = ?c
x v ÝKÞ = V 2 = c
i.e., xÝtÞ = ?ct + x 0
i.e., xÝtÞ = ct + x 0
These straight line solution curves are the characteristics for this constant
coefficient system of linear conservation law equations. Along characteristics, the
conservation laws reduce to odes as follows,
and
c d u+ d v = 0
dJ
dJ
?c d u+ d v = 0
dK
dK
on C 1
on C 2
6
This implies that R J Ýx, tÞ = vÝx, tÞ + cuÝx, tÞ is constant along C 1 characteristics and
R K Ýx, tÞ = vÝx, tÞ ? cuÝx, tÞ is constant along C 2 characteristics. This leads to,
R J Ýx, tÞ = vÝx, tÞ + cuÝx, tÞ = FÝx + ctÞ
R K Ýx, tÞ = vÝx, tÞ ? cuÝx, tÞ = GÝx ? ctÞ
2.
If we are given initial values for u and v, then these expressions permit the
computation of u and v at every point in the upper half plane.
Recall that for the system
.
/t
h
v h
+
v
1 v
/x
h
v
=3
0.
.
we have distinct real eigenvalues VÝuÞ = v ± h and the associated left
eigenvectors are
q 1 = ?1, h ,
V 1 ÝuÞ = v ? h corresponds to left eigenvector 3
V 2 ÝuÞ = v + h
corresponds to left eigenvector 3
q 2 = 1, h .
Then
3
q1 6 /t
h
v
+3
q1 6
v h
1 v
/x
h
v
.=
.
? / t h + h / t v + V 1 ?/ x h + h / x v
=0
.
3
q2 6 /t
h
v
+3
q2 6
v h
1 v
/x
h
v
.
.
= /th + h /tv + V2 /xh + h /xv
=0
or
/ t h + V 1 / x h ? h Ý/ t v + V 1 / t vÞ = 0
/ t h + V 2 / x h + h Ý/ t v + V 2 / t vÞ = 0
.
Let C 1 , C 2 denote solution curves for the systems
C 1 : t v ÝJÞ = 1,
C 2 : t v ÝKÞ = 1,
x v ÝJÞ = V 1
x v ÝKÞ = V 2
These solutions curves are the characteristics for this system of conservation law
equations. Along characteristics, the conservation laws reduce to odes as follows,
.
d h? h d v = 0
on C 1
dJ
dJ
and
7
d h+ h d v = 0
dK
dK
.
If we let h = c 2 then
.
i.e.,
on C 2
d h = 2c d c etc and these equations reduce further to
dJ
dJ
2c d c ? c d v = 0
dJ
dJ
2c d c + c d v = 0;
dK
dK
and
d Ý2c ? vÞ = 0
dJ
d Ý2c + vÞ = 0
dK
and
.
Evidently, 2c ± v are constant along characteristics (which are curves, in
general).
2c-v, 2c+v constant on characteristics
3.
.
The eigenvalues and left eigenvectors for the system
.
_
/t
v
s
+
v
c2
_
_ 0
0
0 v
v a
_
/x
v
=3
0
s
are
matrix eigenvalue V 1 = v
eigenvector
Then
3
q 1 = ß0, 0, 1à
V2 = v + c
V3 = v ? c
3
q 2 = ßc 2 , _c, aà
3
q 3 = ßc 2 , ?_c, aà
3
q 1 6 Ý/ t 3
u + A / x3
uÞ = /ts + V1 /xs = 0
_
_
3
q 2 6 Ý/ t 3
u + A / x3
u Þ = / t _ + c / t v + V 2 Ý / x _ + c / x vÞ = 0
8
_
_
3
q 3 6 Ý/ t 3
u + A / x3
u Þ = / t _ ? c / t v + V 3 Ý / x _ ? c / x vÞ = 0.
.
If we now suppose that the gas is isentropic (i.e., s is constant) then / x p = c 2 / x _.
In addition, if the gas is a so called polytropic gas, then p = k _ L and c 2 = kL_ L?1 .
The equations now reduce to
.
c d_
dv
c
or
_ dJ + dJ = 0
_ Ý/ t _ + V 2 / x _Þ + / t v + V 2 / x v = 0
c d_
dv
c
or
_ dK ? dK = 0,
_ Ý/ t _ + V 3 / x _Þ ? Ý/ t v + V 3 / x vÞ = 0
.
where d/dJ, d/dK denote differentiation along the characteristics associated with
x v ÝtÞ = V 2 , x v ÝtÞ = V 3 , respectively. Since
.
X _c d_ = 2cÝ_Þ
L?1
.
2cÝ_Þ
2cÝ_Þ
+ v and r 2 =
? v, then
It follows that if r 1 =
L?1
L?1
r 1 = const along the characteristics associated with x v ÝtÞ = c + v
r 2 = const along the characteristics associated with x v ÝtÞ = c ? v
i.e., r 1 and r 2 are the Riemann invariants for this system.
We will now show how the Riemann invariants can be exploited in solving some problems.
6. Applications of Riemann Invariants
We will now show how Riemann invariants can be used to solve various problems for
systems of quasilinear pde’s.
Example 6.1 A Ripple Tank
A Ripple Tank of depth H
We consider the propagation through a tank of shallow water of a wave introduced at the
boundary of the tank. If h and v denote, respectively, the depth of the water and the fluid
velocity, then the governing equations are
9
/th + v /xh + h /xv = 0
/ t v + v / x v + / x h = 0.
Ý6.1Þ
hÝx, tÞ = H + wÝx, tÞ where w << 1, and vÝx, tÞ << 1,
Now we assume
wÝx, 0Þ = vÝx, 0Þ = 0,
and we suppose
and
vÝ0, tÞ = P sin It.
These conditions could be induced by moving the end of an initially quiet water tank back
and forth in a periodic fashion but with small amplitude. Then if we neglect all products of
small terms, the equations simplify to
/t
w
0 H
+
v
1 0
/x
w
v
=3
0.
Ý6.2Þ
These are the linearized shallow water equations. The eigenvalues and left eigenvectors of
the coefficient matrix are,
V1 =
H
3
q1 =
V2 = ? H 3
q2 =
Then
H ,H
H , ?H
3
q j 6 Ý/ t 3
u + A / x3
uÞ = 0, j = 1, 2
.
leads to,
H Ý/ t w + V 1 / x wÞ + HÝ/ t v + V 1 / x vÞ = 0
Ý6.3Þ
H Ý/ t w + V 2 / x wÞ ? HÝ/ t v + V 2 / x vÞ = 0.
The characteristics for this system are solution curves C k for x vk ÝtÞ = V k = ± H , k = 1, 2;
i.e.,
C1 :
C2 :
x 1 ÝtÞ = t H + x 0
x 2 ÝtÞ = ? t H + x 0 .
Note that since the system is linear with constant coefficients, the characteristics are two
families of parallel straight lines. The equations (6.3) can be written as
H dw + H dv = 0
dJ
dJ
H dw ? H dv = 0
dK
dK
where d/dJ and d/dK denote differentiation along characteristics C 1 and C 2 respectively. It
10
follows that the Riemann invariants are
r 1 Ýx, tÞ =
H wÝx, tÞ + H vÝx, tÞ on characteristics C 1
r 2 Ýx, tÞ =
H wÝx, tÞ ? H vÝx, tÞ on characteristics C 2
For C 1 characteristics originating at points Ýx 0 , 0Þ with x 0 ³ 0, we have
r 1 Ýx, tÞ = r 1 Ýx 0 , 0Þ
But
hence
r 1 Ýx 0 , 0Þ =
H wÝx 0 , 0Þ + H vÝx 0 , 0Þ = 0
r 1 Ýx, tÞ = 0
Similarly,
and
hence
on x ? t H = x 0 .
on x ? t H = x 0 ³ 0, t > 0.
r 2 Ýx, tÞ = r 2 Ýx 0 , 0Þ
r 2 Ýx 0 , 0Þ =
r 2 Ýx, tÞ = 0
on x + t H = x 0 ³ 0,
H wÝx 0 , 0Þ ? H vÝx 0 , 0Þ = 0,
on x + t H = x 0 ³ 0, t > 0
Then r 1 Ýx, tÞ = r 2 Ýx, tÞ = 0 for t > 0, x > t H and this leads to wÝx, tÞ = vÝx, tÞ = 0 in
t > 0, x > t H . This region is referred to as the ”zone of quiet”. Now consider the region
t > 0, 0 < x < t H .
Since r 1 Ýx, tÞ is constant on C 1 characteristics and r 2 Ýx, tÞ is constant on C 2 , we have, for
any Ýx D , t D Þ 5 t > 0, 0 < x < t H there exist x 2 > x 1 > 0, and t 2 > 0 such that
and,
iÞ r 1 Ý0, t 2 Þ = r 1 Ýx D , t D Þ,
iiÞ r 2 Ýx 2 , 0Þ = r 2 Ýx D , t D Þ,
iiiÞ r 2 Ýx 1 , 0Þ = r 2 Ý0, t 2 Þ.
11
Then iii) implies
H wÝ0, t 2 Þ ? H vÝ0, t 2 Þ, =
i.e.,
wÝ0, t 2 Þ =
H wÝx 1 , 0Þ ? H vÝx 1 , 0Þ = 0
H vÝ0, t 2 Þ = P H sin It 2 .
Also, the C 1 characteristic originating at Ý0, t 2 Þ is given by x 1 ÝtÞ = Ýt ? t 2 Þ H and since
D
Ýx D , t D Þ lies on this line, it follows that t 2 = t D ? x .
H
D D
D D
Now ii) implies r 2 Ýx , t Þ = 0 so wÝx , t Þ = H vÝx D , t D Þ, and finally, iÞ implies
wÝx D , t D Þ +
H vÝx D , t D Þ = wÝ0, t 2 Þ +
H vÝ0, t 2 Þ.
and we can combine all these equalities to conclude that at any point
Ýx D , t D Þ 5 t > 0, 0 < x < t H
D
vÝx D , t D Þ = vÝ0, t 2 Þ = P sin I t D ? x
H
D
H vÝx D , t D Þ = P H sin I t D ? x
H
D
hÝx D , t D Þ = H + H P sin I t D ? x
,
H
wÝx D , t D Þ =
from which we see that the response to the moving boundary is a travelling wave of
amplitude H P, having the same frequency as the input but with a delay due to the finite
speed of propagation.
Now compare this result to what we find from solving the quasilinear system (6.1). We
know from example 5.3.1 that
r 1 Ýx, tÞ = 2cÝx, tÞ + vÝx, tÞ is constant along solution curves of x v1 ÝtÞ = v + h
r 2 Ýx, tÞ = 2cÝx, tÞ ? vÝx, tÞ is constant along solution curves of x v2 ÝtÞ = v ? h ,
where c =
h.
Then for characteristics C 1 .C 2 originating at arbitrary points x 2 > x 1 > 0 we have
12
r 1 Ýx, tÞ = r 1 Ýx 1 , 0Þ = 2cÝx 1 , 0Þ + vÝx 1 , 0Þ = 2 H , on
C 1 : x v1 ÝtÞ = v + h , x 1 Ý0Þ = x 1
r 2 Ýx, tÞ = r 2 Ýx 2 , 0Þ = 2cÝx 2 , 0Þ ? vÝx 2 , 0Þ = 2 H , on
C 2 : x v2 ÝtÞ = v ? h , x 2 Ý0Þ = x 2 .
It is not hard to see that since r 1 and r 2 are both constant along characteristics
2c P + v P = 2 H + 0 = 2 H
2c P ? v P = 2 H ? 0 = 2 H
hence
cP =
H
and
v P = 0;
i.e., cÝx, tÞ and vÝx, tÞ are separately constant along C 1 characteristics so that
cÝx, tÞ = H , and vÝx, tÞ = 0 at all points in the zone of quiet, t > 0, x > t H . Both
families of characteristics throughout that region are families of straight lines
x 1 ÝtÞ = t H + x 0
and
x 2 ÝtÞ = ?t H + x 0
for x 0 ³ 0.
This part of the solution agrees with what we found for the linearized problem.
Now consider the region t > 0, 0 < x < t H . We can show that the C 1
characteristics in this region are straight lines although the C 2 characteristics are, in
general, not straight lines in this region. Note first in the following figure that we have
cÝPÞ = cÝQÞ and vÝPÞ = vÝQÞ since v and c are separately constant along C 1 characteristics
in the zone of quiet. In addition, we have
r 1 = 2cÝRÞ + vÝRÞ = 2cÝSÞ + vÝSÞ,
and
r 2 = 2cÝPÞ ? vÝPÞ = 2cÝRÞ ? vÝRÞ,
r 2 = 2cÝQÞ ? vÝQÞ = 2cÝSÞ ? vÝSÞ.
Since cÝPÞ = cÝQÞ and vÝPÞ = vÝQÞ, the left sides of these last two equations are identical,
leading to
2cÝRÞ ? vÝRÞ = 2cÝSÞ ? vÝSÞ.
In combination with the first equation for r 1 this implies cÝRÞ = cÝSÞ and vÝRÞ = vÝSÞ.
Then the equation for C 1 characteristics in this region is
13
on
C 1 : x v1 ÝRÞ = vÝRÞ + cÝRÞ = vÝSÞ + cÝSÞ =
H , constant
Then the C 1 characteristics in this region are straight lines. Note that this same argument
does not extend to the C 2 characteristics.
Now note that for arbitrary Ýx D , t D Þ 5 t > 0, 0 < x < t H there exist x 1 > x 2 > 0 and
t 2 > 0 such that
r 2 Ý0, t 2 Þ = r 2 Ýx 2 , 0Þ
r 1 Ýx D , t D Þ = r 1 Ý0, t 2 Þ.
r 2 Ýx D , t D Þ = r 2 Ýx 1 , 0Þ.
Then
r 2 Ýx 2 , 0Þ = 2 H ,
cÝ0, t 2 Þ =
leads to
r 2 Ý0, t 2 Þ = 2cÝ0, t 2 Þ ? vÝ0, t 2 Þ = 2 H ? P sin It 2 ,
H + P sin It 2 .
2
and
r 2 Ýx D , t D Þ = r 2 Ýx 1 , 0Þ = 2 H , and r 1 Ý0, t 2 Þ = r 1 Ýx D , t D Þ;
In addition,
i.e.,
2 H = 2cÝx D , t D Þ ? vÝx D , t D Þ
2 H + P sin It 2 + P sin It 2 = 2cÝx D , t D Þ + vÝx D , t D Þ.
This leads to
cÝx D , t D Þ =
D
H + P sin I t D ? x
2
H
D
vÝx D , t D Þ = P sin I t D ? x
H
,
and, since h = c 2
D
hÝx D , t D Þ = H + H P sin I t D ? x
H
D
= H + H P sin I t D ? x
H
2
D
+ P sin 2 I t D ? x
4
H
2
+ P
8
D
1 ? cos 2 I t D ? x
H
Evidently, the nonlinear solution consists of the linearized solution plus an additional term
which, although of small amplitude, contains a travelling wave with double the frequency of
the input.
Example 6.2 A Wave in a Polytropic Gas
Consider a polytropic gas in a long tube containing a piston. Suppose the gas is initially at
rest in a uniform state (i.e., constant pressure and density) and at t = 0 the piston begins to
move. If X = XÝtÞ denotes the position of the piston at time t, then we suppose
XÝ0Þ = X v Ý0Þ = 0,
and
X v ÝtÞ < 0,
X”ÝtÞ < 0, for t > 0.
Ý1Þ
14
A long tube with a moveable piston
In example 5.3.2 we showed that the polytropic gas equations have the following Riemann
invariants,
r1 =
2cÝ_Þ
+v
L?1
r2 =
and
2cÝ_Þ
? v,
L?1
Ý2Þ
constant along solution curves
C 1 Ýx 0 Þ : x v1 ÝtÞ = v + c,
C 2 Ýx 0 Þ : x v2 ÝtÞ = v ? c,
and
x 1 Ý0Þ = x 0
x 2 Ý0Þ = x 0 ,
respectively. Since we have the initial conditions
vÝx 0 , 0Þ = 0,
_Ýx 0 , 0Þ = _ 0 ,
cÝx 0 , 0Þ = c 0 ,
pÝx 0 , 0Þ = p 0 , for x 0 ³ 0,
Ý3Þ
which means that this
It is clear that
x v2 Ý0Þ = vÝx 0 , 0Þ ? cÝx 0 , 0Þ = 0 ? c 0 < 0
characteristic curve will go initially to the left.
Now right at the surface of the piston, the fluid velocity is equal to the speed of the piston,
so we have
vÝXÝtÞ, tÞ = X v ÝtÞ < 0.
Ý4Þ
But this implies that the C 2 Ýx 0 Þ characteristic will meet the piston curve at some time
t 0 > 0 since at this point we will have
x v2 ÝtÞ = X v Ýt 0 Þ ? cÝXÝt 0 Þ, t 0 Þ < X v Ýt 0 Þ.
Now
r 2 Ýx 0 , 0Þ = 2c 0 ? 0 = r 2 Ýx 2 Ýt; x 0 Þ, tÞ = constant along C 2 Ýx 0 Þ.
L?1
Similarly,
15
r 2 Ýx 1 , 0Þ = 2c 0 ? 0 = r 2 Ýx 2 Ýt; x 1 Þ, tÞ = constant along C 2 Ýx 1 Þ
L?1
r 2 Ýx 2 , 0Þ = 2c 0 ? 0 = r 2 Ýx 2 Ýt; x 2 Þ, tÞ = constant along C 2 Ýx 2 Þ.
L?1
Evidently, not only is r 2 Ýx, tÞ constant along every C 2 Ýx p Þ, it equals the same constant on
all the curves in this family that originate at a point Ýx p , 0Þ, x p ³ 0. That is,
r 2 Ýx, tÞ = 2c 0
L?1
for all Ýx, tÞ 5 S =
t > 0, x ³ XÝtÞ .
Ý5Þ
Now consider the solution curves, C 1 , for x v1 ÝtÞ = v + c. If the curve originates at a point
Ýx p , 0Þ for x p ³ 0, then in just the same way as with r 2 Ýx, tÞ, we can show that not only is
r 1 Ýx, tÞ constant along every C 1 Ýx p Þ originating at x p ³ 0, it is the case that r 1 Ýx, tÞ equals
the same constant on every C 1 Ýx p Þ originating at x p ³ 0. In particular,
r 1 Ýx, tÞ = 2c 0
L?1
along every C 1 Ýx p Þ originating at x p ³ 0
Ý6Þ
Using (5) and (6) together, we conclude that v = 0, and c = c 0 throughout the ”zone of
quiet” x ³ c 0 t, t > 0 . Then all the C 1 characteristics in the region are straight lines given
by x 1 Ýt; x p Þ = c 0 t + x p , x p ³ 0. Now consider the C 1 characteristics which originate at a
point on the piston curve. Although v and c are not constants in the region
XÝtÞ ² x ² c 0 t, t > 0 , these curves also turn out to be straight lines. To see this, write
r 1 Ýx, tÞ =
2c + v = K p (constant) on C 1 Ýx p Þ originating at Ýx p , t p Þ = ÝXÝt p Þ, t p Þ,
L?1
r 2 Ýx, tÞ =
2c ? v = 2c 0 , for all Ýx, tÞ 5 S =
L?1
L?1
t > 0, x ³ XÝtÞ .
Then, in particular, r 2 Ýx, tÞ is constant on all the C 1 Ýx p Þ ? characteristics so we can add and
subtract these two equations to conclude that along the curve C 1 Ýx p Þ,
c=
L?1
K p + 2c 0
L?1
4
v = 1 K p ? 2c 0
L?1
2
L?1
= 1 c0 +
Kp
2
2
:= c p
:= v p .
Then c and v are constant along C 1 Ýx p Þ characteristics, but they are equal to different
constants on distinct characteristics. Since C 1 Ýx p Þ is the solution curve for
x v1 ÝtÞ = v + c = v p + c p , t > t p ,
x 1 Ýt p Þ = x p ,
it follows that the C 1 characteristic that originates at Ýx p , t p Þ is the straight line whose
equation is, x 1 ÝtÞ = Ýv p + c p Þ Ýt ? t p Þ + x p .
Now consider the characteristic C 1 Ýt p Þ that originates at a point ÝXÝt p Þ, t p Þ, t p ³ 0, on the
piston curve. This line has the equation
x 1 ÝtÞ = Ýv + cÞ Ýt ? t p Þ + XÝt p Þ
for t ³ t p .
To determine the values for v and c on this line, C 1 Ýt p Þ, note that vÝXÝt p Þ, t p Þ = X v Ýt p Þ. But
then for some x p > 0, there is a curve C 2 Ýx p Þ which meets the piston curve at
ÝXÝt p Þ, t p Þ and then
16
r2 =
2cÝx, tÞ
? X v Ýt p Þ = 2c 0 .
L?1
L?1
L?1 v
X Ýt p Þ. which leads to
2
L?1 v
L+1 v
X Ýt p Þ = c 0 +
X Ýt p Þ
v + c = X v Ýt p Þ + c 0 +
2
2
This allows us to solve for cÝx, tÞ = c 0 +
and,
x 1 ÝtÞ =
c0 +
L+1 v
X Ýt p Þ Ýt ? t p Þ + XÝt p Þ
2
for t ³ t p .
Then the region S\Q is covered by C 1 characteristics. In order to determine v at some point
x! , !t 5 S\Q we find the C 1 ?characteristic passing through the point, x! , !t , and find the
point ÝXÝt 0 Þ, t 0 Þ where this characteristic meets the piston curve. Then v x! , !t = X v Ýt 0 Þ and
L?1
c x! , !t = c 0 +
v x! , !t . Then _, p are obtained from the equation of state and the
2
definition c 2 = p v Ý_Þ.
L+1 v
Note that the C 1 characteristics have slopes m p = c 0 +
X Ýt p Þ. Then for t q > t p , it
2
follows from X”ÝtÞ < 0 that m q < m p which implies that the lines x 1 Ýt; t p Þ and x 1 Ýt; t q Þ don’t
cross. Instead, we have an expansion wave in the region S\Q. As long as X”ÝtÞ < 0, the
expansion wave persists and the solution is a classical solution. On the other hand,
suppose X”ÝtÞ > 0 for t 1 < t < t 2 . Then x 1 Ýt; t 1 Þ has a steeper slope than x 1 Ýt; t 2 Þ and these
lines will meet, leading to the formation of a shock.
In a situation where, instead of a gradual withdrawl, the piston is suddenly withdrawn,
we would have
XÝ0Þ = 0,
X v Ý0Þ = ?V < 0.
Then v = 0 and c = c 0 in Q = t > 0, x > c 0 t . In addition, by the same reasoning used
previously, we see that the C 1 characteristics are straight lines and at each point of S,
r 2 Ýx, tÞ =
2c ? v = 2c 0 .
L?1
L?1
Now it follows that there is a wedge shaped region, bounded on the left by the piston curve,
where v = ?V and c = constant. In this region, we have
2c ? Ý?VÞ = 2c 0
L?1
L?1
i.e.,
c = c0 ?
L?1
V.
2
Then the C 1 characteristic that meets the piston curve at the origin forms the right boundary
of this wedge. This C 1 characteristic is the solution curve for
x v1 ÝtÞ = c + v = c 0 ?
i.e.,
x 1 ÝtÞ = ßc 0 ?
L?1
V ? V = c0 ?
2
L+1
V,
2
x 1 Ý0Þ = 0
L+1
Và t.
2
17
Evidently, the zone of quiet, where v = 0 and c = c 0 , is a wedge whose left boundary is the
C 1 characteristic, x 1 ÝtÞ = c 0 t. Also there is the uniform region where v = ?V and c =
constant, and this wedge has as its right boundary, the C 1 characteristic we just computed.
In between these two C 1 characteristics, there is a wedge shaped region containing no
characteristics. This is the region
L+1
Và t < x < c 0 t,
2
ßc 0 ?
1?
i.e.,
L+1
2
t > 0.
V
x
c 0 < c 0 t < 1.
If we let
vÝx, tÞ = 2c 0
L+1
x ?1
c0t
in this wedge shaped region
then, since
r 2 Ýx, tÞ =
2cÝx, tÞ
? vÝx, tÞ = 2c 0
L?1
L?1
at all Ýx, tÞ, it then follows that in the characteristic-free wedge,
cÝx, tÞ =
L?1
vÝx, tÞ + c 0 = c 0
2
L?1 x
+ 2
.
L + 1 c0t
L+1
This solution joins the two uniform regions with an expansion fan. This, is consistent with
the entropy condition discussed previously since
ßc 0 ?
L+1
Và t < c 0 t
2
i.e., the velocity of the wave on the left is less than the velocity of the wave on the right,
leading to the formation of an expansion fan.
18