3. Scalar Conservation Law Equations- part II
We have seen now that scalar conservation laws have solutions which are not, in general,
global classical solutions. Discontinuous solutions for a Cauchy problem can arise either
spontaneously due to nonlinearities, or as the result of discontinuities in the initial
conditions. In any case, the singularities are resolved either as shock type solutions or as
fan type solutions and the selection of the physically relevant solution has, so far, been
based on the so called entropy condition that assert that a shock is formed only when the
characteristics carry information toward the shock. Then we have the following assertion,
(which we may prove later),
Admissible weak solutions to the equation / t uÝx, tÞ + / x FÝuÞ = 0, are unique. At this point,
we interpret admissible to mean that the solution satisfies the entropy condition. Other
interpretations of admissibility will be discussed later.
We suppose, F”ÝuÞ ® 0, and aÝuÞ = F v ÝuÞ then at a point where
lim uÝx 0 ? P, t 0 Þ = u L ® u R = lim uÝx 0 + P, t 0 Þ,
P¸0+
P¸0+
(a) a shock, x = xÝtÞ, forms if aÝu L Þ > aÝu R Þ
(Entropy condition)
the shock satisfies
ßßFÝuÞàà
x v ÝtÞ =
, xÝt 0 Þ = x 0 (Jump condition)
ßßuàà
(b) a fan forms if aÝu L Þ < aÝu R Þ. In this case,
where
uÝx, tÞ = G x ? x 0
t ? t0
in x 0 + aÝu L ÞÝt ? t 0 Þ < x < x 0 + aÝu R ÞÝt ? t 0 Þ
GÝ6Þ = a ?1 Ý6Þ
In this section we will consider more complicated examples of shock and fan solutions, and
note additional features about such weak solutions for scalar conservation laws.
3.1 More Difficult Examples
1. An Example With Secondary Shocks
Consider the following intial value problem for Burger’s equation
/ t uÝx, tÞ + u / x uÝx, tÞ = 0,
uÝx, 0Þ =
2 if | x| > 1
1 if | x| < 1
There are discontinuities in the initial data located at x = ±1. We shall have to determine
whether these discontinuities result in a shock or in a fan type solution. At x = ?1 we have
aÝu L Þ = 2 > aÝu R Þ = 1. Therefore at this discontinuity, the fast wave is behind a slow wave
and we have a shock. At x = +1 we have aÝu L Þ = 1 < aÝu R Þ = 2. Then at this discontinuity
a slow wave is behind a fast wave which means a rarefaction wave or fan will form.
As we have seen in past examples, the characteristics for the Burger’s equation are the
straight lines x ? u 0 t = x 0 , originating at the initial point Ýx 0 , 0Þ with u 0 = uÝx 0 , 0Þ. At x = ?1,
where the shock forms, we have
1
ßßFÝuÞàà
=
ßßuàà
x v ÝtÞ =
Then
xÝtÞ =
3
2
t ? 1,
1
2
u 2L ? 12 u 2R
=
ul ? uR
1
2
Ýu L + u R Þ =
3
2
,
xÝ0Þ = ?1.
t > 0,
and
uÝx, tÞ =
2 if x <
1 if x >
3
t ? 1,
2
3
t?1
2
At x = 1 there is a fan, bounded on the left by the characteristic x ? t = 1, and on the right
by x ? 2t = 1. Between these two rays we have
uÝx, tÞ = x ? 1 ,
t
t + 1 < x < 2t + 1,
t > 0.
Now observe that the characteristic that forms the left boundary of the fan intercepts the
shock line. That is 32 t ? 1 = t + 1 at t = 4, x = 5. At this point, the shock hits the fan (so to
say).
To the left of the shock line we have u L = 2, and to the right of the fan boundary we have
u R = x ? 1 , and since u L > u R , a secondary shock forms. This secondary shock satisfies
t
x v ÝtÞ = 12 Ýu L + u R Þ = 12 2 + x ? 1 ,
xÝ4Þ = 5.
t
The solution of this intial value problem is x 2 ÝtÞ = 2t + 1 ? 2 t . Then the solution to the
original Cauchy problem is,
2
uÝx, tÞ =
1
x?1
t
2
x<
if
if
3
2
3
2
t?1
t?1 < x < t+1
if t + 1 < x < 2t ? 1
if
for
0²t²4
x > 2t ? 1
2
if x < t + 1 or x > 2t ? 1
2
x?1
t
=
for t > 4
if t + 1 < x < 2t ? 1 ? 2 t
Clearly the secondary shock, x 2 ÝtÞ = 2t + 1 ? 2 t , never meets the far edge of the fan
which is the line, x ? 2t = 1.
2. An Example With Nonconstant Initial Conditions
Consider the following intial value problem for Burger’s equation
/ t uÝx, tÞ + u / x uÝx, tÞ = 0,
1 ? x if
uÝx, 0Þ =
0<x<1
if x < 0 or x > 1
0
The initial condition is discontinuous at x = 0 but at that point we have u L = 0 < u R = 1 so
there is no shock. Instead, a rarefaction wave originating at (0,0) develops and the solution
for this part of the problem is,
if x < 0, t > 0
0
x
t
uÝx, tÞ =
if 0 < x < t, t > 0
For characteristics originating in the interval 0 < x 0 < 1, we have
x ? u 0 t = x 0 , u 0 = fÝx 0 Þ = 1 ? x 0 and
u = fÝx ? utÞ =
1 ? Ýx ? utÞ
Solving this last equation for u leads to
u= t +
2
t 2 + 4Ý1 ? xÞ , t ?
2
1
2
1
2
t 2 + 4Ý1 ? xÞ
but only the first solution satisfies uÝx, 0Þ =
uÝx, tÞ = t +
2
1
2
t 2 + 4Ý1 ? xÞ ,
1 ? x for 0 < x < 1. Then we have
for t < x < xÝtÞ, t > 0,
where xÝtÞ is the equation for a shock which originates at (1,0) and is due to the following
collision of solutions,
uL = t +
2
1
2
t 2 + 4Ý1 ? xÞ > u R = 0.
Then the R-H condition asserts that
xvÝtÞ =
1
2
Ýu L + u R Þ =
1
2
t +
2
1
2
t 2 + 4Ý1 ? xÞ + 0
=
1
4
t + t 2 + 4Ý1 ? xÞ , xÝ0Þ = 1.
This is a nonlinear ode for x(t) which is found by inspection to have a solution of the form
xÝtÞ = k t 2 + 1
and substitution into the equation shows that k =
3
16
. We have then
3
uÝx, tÞ =
Note that at t =
t=
3 2
t
16
4
3
3 2
t
16
if x < 0 or x >
0
x
t
t +
2
+1
if 0 < x < t
1
2
t 2 + 4Ý1 ? xÞ if t < x <
3 2
t
16
0²t²
4
3
and x =
4
3
.
+1
, the right boundary of the fan, x = t, meets the shock, x =
+ 1, so t =
4
3
3 2
t
16
+ 1;
.
In view of the picture,
we have u L = x > 0 = u R so a secondary shock forms. This shock must satisfy
t
x 43 = 43 .
x v ÝtÞ = 12 Ýu L + u R Þ = 12 x + 0 ,
t
Then
x 2 ÝtÞ =
4t
3
and
0
uÝx, tÞ =
x
t
if x < 0 or x >
if 0 < x <
4t
3
4t
3
t>
4
3
.
It is apparent from this example that when a shock forms between two regions in which the
solution is constant, then the shock is a straight line. If the solution is nonconstant on either
side of the discontinuity, then the shock curve will necessarily be curved.
3. An Initial-Boundary Value Problem
Consider the following initial-boundary value problem
/ t uÝx, tÞ + 2u / x uÝx, tÞ = / t uÝx, tÞ + / x ÝuÝx, tÞÞ 2 = 0,
x > 0, t > 0,
4
uÝx, 0Þ =
1
2
,
x > 0,
and
1 if 0 < t < 1
uÝ0, tÞ =
1
2
if
t>1
Note that this is slightly different from Burger’s equation. We still have u = constant on
characteristics but the characteristic equations, x v ÝtÞ = 2u, u v ÝtÞ = 0, lead to straight line
characteristics given by, x ? 2u 0 t = C 0 . For a characteristic originating at a point Ýx 0 , 0Þ on
the line t = 0, this becomes x ? t = x 0 , since uÝx 0 , 0Þ = u 0 = 12 . For a characteristic that
originates at a point Ý0, t 0 Þ on the boundary x = 0, we get x ? 2t = C and C = ?2t 0 if
0 < t 0 < 1 and C = ?t 0 if t 0 > 1. Then x = 2Ýt ? t 0 Þ if 0 < t 0 < 1, but x = Ýt ? t 0 Þ if t 0 > 1.
Evidently there is a discontinuity in the data located at Ý0, 0Þ where we have
aÝu L Þ = 2 u L = 2 > 1 = 2 u R = aÝu R Þ
Therefore, a shock forms, satisfying the equation
x v ÝtÞ =
u2 ? u2
ßßFÝuÞàà
= u LL ? u RR = u L + u R =
ßßuàà
3
2
,
xÝ0Þ = 0.
Then x 1 ÝtÞ = 32 t is the equation for the straight line shock originating at the origin. There is
also a discontinuity in the data located at Ý0, 1Þ where
aÝu L Þ = 2 u L = 1 < 2 = 2 u R = aÝu R Þ
so a fan type solution must form in the wedge shaped region W = át ? 1 < x < 2t ? 2â.
Since aÝuÞ = 2u, the inverse function is GÝuÞ = a ?1 ÝuÞ = 12 u and so
uÝx, tÞ = G
x
t?1
=
1
2
x
t?1
in W.
Now it can be seen from the picture that the right boundary of the wedge, x = 2t ? 2, meets
the shock x = 32 t, at t = 4, x = 6. At this point the solution to the left is greater than the
solution value on the right so a secondary shock will form which satisfies
5
x v ÝtÞ = u L + u R =
Then
1
2
x +
t?1
1
2
,
xÝ4Þ = 6.
x 2 ÝtÞ = t ? 1 + 3t ? 3 , t > 4,
describes the (curved) secondary shock and the global weak solution is,
1
2
uÝx, tÞ =
1/2
x
t?1
1
if
0 < x < t ? 1, or 3t/2 < x
if
t ? 1 < x < 2t ? 2
if
0 < x < 3t/2
2t ? 2 < x < 3t/2
1/2
x
t?1
1
2
for 1 < t < 4
if 0 < x < t ? 1, or x 2 ÝtÞ < x
if
0<t<4
for 0 < t < 1
t ? 1 < x < x 2 ÝtÞ
t>4
3.2 Equivalence of Conservation Law Equations
If u = uÝx, tÞ is a smooth solution of the Burger’s equation
/ t uÝx, tÞ + u / x uÝx, tÞ = 0,
then u is also a smooth solution of each of the following equations
u / t uÝx, tÞ + u 2 / x uÝx, tÞ = / t Ýu 2 /2Þ + / x Ýu 3 /3Þ = 0,
u 2 / t uÝx, tÞ + u 3 / x uÝx, tÞ = / t Ýu 3 /3Þ + / x Ýu 4 /4Þ = 0,
_
n
n+1
= / t ÝPÝuÞÞ + / x ÝFÝuÞÞ = 0.
/ t un + / x u
n+1
These equations all have the form of a conservation law corresponding to density field PÝuÞ
and flux field given by FÝuÞ. Then the R-H condition for the shocks for these equations
reads,
x v ÝtÞ =
n+1
u n+1
FÝu L Þ ? FÝu R Þ
L ? uR
= n
,
n
n
n + 1 uL ? uR
PÝu L Þ ? PÝu R Þ
Ýn = 1Þ
x v ÝtÞ =
1
2
Ýn = 2Þ
x v ÝtÞ =
2
3
Ýu L + u R Þ,
u 2L + u L u R + u 2R
,
uL + uR
etc
Evidently the equations for various values of n have the same smooth solutions but each
has its own shock solution different from all of the others. Therefore, when studying the
weak solutions of a given conservation law equation, it is not permissable to change the
conservation law to a form which, for smooth solutions is equivalent since the form does
not, in general, have the same weak solutions.
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3.3 Transformations and Conservation Laws
Consider the conservation law equation,
/ t uÝx, tÞ + / x FÝuÝx, tÞÞ = 0
Ý3.1Þ
where F”ÝuÞ ® 0 so the equation is genuinely nonlinear. If we let vÝx, tÞ = F v ÝuÝx, tÞÞ then
/ t vÝx, tÞ = F”ÝuÞ / t uÝx, tÞ,
and
/ x vÝx, tÞ = F”ÝuÞ / x uÝx, tÞ,
hence
F”ÝuÞÝ/ t u + F v ÝuÞ / x uÞ = / t vÝx, tÞ + v / x vÝx, tÞ = 0.
Evidently this change of variable transforms the general conservation law (3.1) for
u = uÝx, tÞ, into the Burger’s equation for v = vÝx, tÞ. If the solutions to these equations are
smooth, then solving one of them leads to the solution of the other. However, there is no
such equivalence for shock solutions to these equations since
/ t uÝx, tÞ + / x FÝuÝx, tÞÞ = 0
leads to
x v ÝtÞ =
FÝu L Þ ? FÝu R Þ
uL ? uR
while
/ t vÝx, tÞ + v / x vÝx, tÞ = 0
leads to
x v ÝtÞ =
1
2
Ýv L + v R Þ =
1
2
ÝF v Ýu L Þ + F v Ýu R ÞÞ
and, in general, these are not the same.
More generally, consider the initial value problem for Burger’s equation in example
2.1.1. The shock solution here is given by
1 if x <
uÝx, tÞ =
0 if x >
1
2
1
2
t
Ý3.2Þ.
t
On the other hand, the transformation, v = e u , leads to / t v = e u / t u, / x v = e u / x u, and it
follows that
e u ß / t u + u / x uà= / t vÝx, tÞ + u / x vÝx, tÞ = / t vÝx, tÞ + Ýln vÞ / x vÝx, tÞ = 0
Ý3.3Þ
vÝx, 0Þ =
e if x < 0
1 if x > 0
.
Since F v ÝvÞ = ln v, the shock relation for (3.3) is
x v ÝtÞ =
ßßv ln v ? vàà
Ýe ln e ? eÞ ? Ý1 ln 1 ? 1Þ
=
= 1
e?1
e?1
ßßvàà
and this is not the shock that is obtained by subjecting (3.2) to the change of variable,
v = eu.
The implication of these examples is that smooth transformations do not preserve weak
solutions, even though smooth solutions are preserved by such smooth transformations.
3.4 Time Reversals for Conservation Laws
We recall that the solution to an initial value problem for a linear first order equation can be
7
written as uÝx, tÞ = fÝdÝx, tÞÞ where f denotes a function determined by the initial data and
d = const is an implicit representation for the base characteristics of the equation. Then is
clearly possible to recover the initial state from the specification of the solution at any time
after the initial time. On the other hand, for 0 < c < 1, consider the initial value problem
/ t uÝx, tÞ + u / x uÝx, tÞ = 0,
uÝx, 0Þ =
1
x ? c/2
?c
0
if
x < ?c/2
if ?c/2 < x < c/2
if
x > c/2
The solution to this initial value problem is easily found to be
uÝx, tÞ =
=
1
x ? c/2
t?c
0
if x < t ? c/2
if t ? c/2 < x < c/2
for 0 ² t ² c
if x > c/2
1 if x < t/2
0 if x > t/2
for t > c
Solutions for various choices of c are shown in the figure,
Since the solutions for all choices of c, 0 ² c ² 1, are identical for t > 1, we observe that it
is not possible, in general, to recover earlier states from later states when the solutions
involved are weak solutions to nonlinear equations
We can summarize some of the differences we have noted between linear and
nonlinear initial value problems:
Linear Problems
¾ for all smooth initial data, there is a smooth global solution
¾ the smooth solution is uniquely determined by the equation and the data and the
solution is reversible in time,
i.e,
uÝx, tÞ = SÝtÞßuÝx, 0Þà,
uÝx, 0Þ = SÝ?tÞßuÝx, tÞà
¾ discontinuities in the initial data propagate along characteristics
¾ solutions are invariant under smooth transformations and equivalent equations have
the same smooth solutions
8
.
Nonlinear Problems
¾ no global classical solution need exist, even for smooth initial data; global weak
solutions exist but are not reversible
¾ weak solutions are not uniquely determined by the equation and the initial data, an
additional selection principle is needed.
¾ discontinuities can arise spontaneously and are propagated along noncharacteristic
curves called shocks
¾ weak solutions are not invariant under smooth transformations and equations which
are algebraically equivalent need not have the same weak solutions
9