N ov. 1 9
The S ingle C onservation L aw: Existence and Uniqueness
In this lecture we will prove the existence and uniqueness of entropy solutions to the single conservation
law
u t + f( u) x = 0,
(1)
u( x , 0) = u 0
where f > 0.
00
•
Many details of the proofs are omitted, because some of them are too hard while some others too
easy. Anyone interested in these details should read Chapter 1 6 of Smoller’ s book.
1 . Existence of entropy solution.
Theorem 1 . Let u 0 ∈ L ∞ ( R) , and let f ∈ C 2 R) with f 00 > 0 on
weak solution u with the following properties:
a ) | u( x , t) | 6 k u 0 k L ∞ ≡ M, ( x , t) ∈ R × R+ .
u: | u | 6 k u 0 k L ∞ . Then there exists a
00
u)
:
u
6
u
and A = max
b ) There
is
a
constant
E
>
0
,
depending
only
on
M,
µ
=
min
f
k
(
|
|
k
0
∞
L
0
| f ( u) | : | u | 6 k u 0 k L ∞ such that for every a > 0, t > 0 , and x ∈ R,
u( x + a , t) − u( x , t) E
< .
a
t
( 2)
c ) u is stable and depends continuously on u 0 in the following sense: If v 0 ∈ L ∞ ( R) with k v 0 k L ∞ 6
k u 0 k L ∞ , and v is the corresponding solution constructed from the process in the proof, then for
every x 1 , x 2 ∈ R with x 1 < x 2 , and every t > 0 ,
Z x2 + At
Zx2
( 3)
| u( x , t) − v( x , t) | dx 6
| u 0 ( x) − v 0 ( x) | dx.
x1 − At
x1
Remark 2. We first check that a) – c) are satisfied by every C 1 solution. In particular, recall that when
u is C 1 , we have
( 4)
u( x , t) = u 0 ( x 0 )
where x 0 + f 0 ( u 0 ( x 0 ) ) t = x. Thus
( 5)
u( x , t) = u 0 ( x − f 0 ( u( x , t) ) t) .
Differentiating we obtain
u x ( x , t) = u 00 ( x 0 ) [ 1 − f 00 ( u) u x t]
u x ( x , t) =
1+
u 00 ( x 0 )
0
u 0 ( x 0 ) f 00 ( u 0 ( x 0 ) )
t
6
( min f 00 )
t
−1
.
( 6)
Remark 3. Note that c) implies the uniqueness only for solutions obtained from the construction in the
proof. This does not exclude the existence that other entropy solutions not obtained from the particular
construction process.
Remark 4. The entropy condition implies some regularity in the solution. More specifically, recall that
the “total variation” of a function f over an interval [ a , b] is defined as
k f k TV ≡
sup
n
−1
X
n∈ N
i= 1
a = a1 < a2 < < an = b
| f( a i ) − f( a i − 1 ) | .
( 7)
For any t > 0, take c > E/ t, we see that u( x , t) = v( x) + c x where v( x) is decreasing. It is easy to verify
that
( 8)
k uk TV 6 k vk TV + k c xk TV < ∞
over any finite interval. Therefore although the initial data is only L ∞ , the solution becomes more regular
as soon as t > 0 in the sense that it has locally bounded total variation for each t > 0.
One can easily show that, a function with locally bounded total variation can have at most countably
many jump discontinuities, and is differentiable almost everywhere.
Now we sketch the proof of the theorem. The idea is to consider the following finite difference discretization of the equation
u kn + 1 − u kn + 1 + u kn − 1 / 2 f u kn + 1 − f u kn − 1
+
=0
( 9)
∆t
2 ∆x
with initial values
and show that
−
−
( 1 0)
u 0n ≡ u 0 ( n ∆x)
The solution satisfies the discrete versions of a) , b) , c) .
As ∆t, ∆x & 0, the discrete solutions ( or a subsequence of them) converge to a weak solution satisfying a) , b) , c) . The convergence would be a weak version of u kn → u( n ∆x , k ∆t) .
1 . The discrete
solutions satisfy a) – c) . Recall the notation M ≡ k u 0 k L ∞ , µ = min f 00 ( u) : | u | 6
k u 0 k L ∞ and A = max | f 0 ( u) | : | u | 6 k u 0 k L ∞ .
a) Show that u kn 6 M for all n ∈ Z, k ∈ Z + ∪ { 0} .
Some manipulation of the difference equation gives
1
∆t 0 k 1
∆t 0 k k+1
k
+
f θn un − 1 +
−
f θ n u nk + 1 .
un =
(11)
2 2 ∆x
2 2 ∆x
This implies
k+1
u n 6 max u nk − 1 , u kn + 1 ( 1 2)
as long as the CFL condition
A ∆t
61
∆x
( 1 3)
E ≡ min ( µ/ 2, A/ 4 M) − 1 .
( 1 4)
u kn − u kn − 2
E
6
.
2 ∆x
k ∆t
( 1 5)
u kn − u kn − 2 m
E
6
2 m ∆x
k ∆t
( 1 6)
is satisfied.
b) Let
We will show that
Note that this implies
which is the discrete version of the entropy condition.
The proof is too long to be included here. Details can be found in J. Smoller’ s book, pp.
269 – 272.
c) Let u kn and v nk be solutions corresponding to the initial values u 0n and v n0 , respec tively, with sup u 0n , sup v n0 6 M. Then for k > 0 we have
X X u kn − v nk ∆x 6
u 0k − v k0 ∆x.
( 1 7)
| n| 6 N
| n| 6 N+ k
The proof of this is very similar to that of a) after setting w nk ≡ u kn − v nk and representing
w nk + 1 by w nk − 1 and w nk + 1 .
2. Convergence to entropy solutions
•
Preparations – uniform bounds.
To be able to pass to the limit, we need the following bounds ( proofs are omitted. See
Smoller’ s book for details) :
a. L ∞ bound. This has already been established in the first step;
b. TV bound. We have
X
| n| < X/∆x
k
u n + 2 − u kn 6 c
( 1 8)
for any X > 0 and k ∆t > α > 0. The RHS c depends on X and α.
c. L 1 local Lipschitz continuity in t. We have
Lemma. If ∆t/ ∆x > δ > 0 , and ∆t, ∆x 6 1 , then there exists an L > 0 , independent
of ∆t, ∆x, such that if k > p, k − p is even and p ∆t > α > 0 ,
X
k
u n − u np ∆x 6 L ( k − p) ∆t.
( 1 9)
| n | 6 X/ ∆x
A similar estimate for k − p odd also holds.
Note that the above is the discrete version of the L 1 local Lipschitz continuity in
t:
k u( x , t) − u( x , t 0 ) k L 1 ( − X , X ) 6 C | t − t 0 | .
( 20)
The reason why k − p is even or odd matters is that u kn + 1 depends on u kn − 1 but not
u kn .
•
The limit exists.
The first thing we need
to settle is in what sense are we talking about the convergence,
as for each ( ∆t, ∆x) u kn is just a bunch of numbers,
while the alleged limit is a function.
To fix this, we define a function U∆ t , ∆ x out of each u kn as follows:
for all n ∆x 6 x < ( n + 1 ) ∆x , k ∆t 6 t < ( k + 1 ) ∆t.
U∆ t , ∆ x ( x , t) = u kn
( 21 )
We will now show that there is a subsequence Ui ( x , t) which is a Cauchy sequence in L 1 .
More specifically, we show
Lemma. There exists a subsequence { Ui } ⊂ { U∆ t , ∆ x } which converges to a measurable
function u( x , t) in the sense that for any X > 0 , t > 0 , and T > 0 , both
Z
0
( 22)
| Ui ( x , t) − u( x , t) | dx
and
| x| 6 X
Z
0 6 t6 T
Z
| x| 6 X
| Ui ( x , t) − u( x , t) | dx dt
0.
( 23)
Proof. The main steps are the following.
a. Fix any k > 0. Since U∆ t , ∆ x ( · , k ∆t) ’ s are uniformly bounded and each U∆ t , ∆ x has
locally finite total variation and therefore can be written as the difference of two
monotone functions, we can find a subsequence converging pointwise according to a
theorem obtained by E. Helley in 1 91 2. 1
1 . T he “H elley S election T heorem” says, every sequence of uniformly bounded sequence of monotone functions contains a
subsequence which converges at every point.
To see this, note that first by the standard diagonal argument we can find a subsequence ( all subsequences will still be
denoted { fn } ) which converges on a countable dense subset of R, say Q . Denote the limit by f . N ow we extend the definition of f from Q to the whole R by setting f ( x) = sup r ∈ Q , r 6 x f ( r) ( wlog we can assume all fn are non-decreasing, thus f is
non-decreasing over Q ) . S uch f ( x) is non-decreasing and therefore can have only countably many discontinuities. We choose
a subsequence once more to obtain fn such that { fn ( x) } is C auchy for all x ∈ Q and for all x which is a discontinuity of f .
F inally, at any continuous point of f , fn ( x)
f ( x) by the standard ε/ 3 argument.
b. For this subsequence, Lebesgue’ s dominated convergence theorem implies Ui ( x , t) converges in L 1 for this particular t = k ∆t.
c. By a diagonal argument, we can obtain a new subsequence Ui ( x , t) which converges in
L 1 for all t ∈ Q + .
d. By the standard ε/ 3-type argument, we can show that Ui ( x , t) converges in L 1 ( R) for
every t. Denote the limit by u( x , t) .
e. Since Ui is uniformly bounded, so is u, and therefore k Ui − u k L 1 ( − X , X ) is a bounded
function in t. Application of Lebesgue’ s dominated convergence theorem then gives
Z
Z
0.
( 24)
| Ui ( x , t) − u( x , t) | dx dt
0 6 t6 T
•
| x| 6 X
The limit satisfies the entropy condition.
This follows from straightforward computation of
Ui ( x , t) − Ui ( x 0 , t)
x − x0
combined with the pointwise convergence.
•
( 25)
The limit is a weak solution.
a. First as u 0 ∈ L ∞ and therefore locally L 1 , Ui ( x , 0) → u 0 in L 1 .
b. In this last step, we take any φ ∈ C03 , multiply the difference equation by φ kn ∆x ∆t
where φ kn = φ( n ∆x , k ∆t) , and then sum over n, k. It is easy to obtain
Z
ZZ
Ui φ = 0
( 26)
lim
[ Ui φ t + f( Ui ) φ x ] +
n, k %∞
t= 0
which immediately gives the same equality for u.
2. Uniqueness of entropy solution.
To show uniqueness, we need to show that if u , v are two entropy solutions, then necessarily u − v = 0
1
x
for almost all ( x , t) . Since u, v ∈ L 1 , we only need to show that u ε − v ε = 0 for all ε where u ε = ε ρ ε ∗ u
is the mollified function from u. As u ε and v ε are smooth, we finally see that the only thing need to be
shown is
Z
for all φ in, say,
•
( 27)
( u − v) φ = 0
C01 .
Main idea of the proof.
What we have is the weak formulation which is satisfied by both u and v:
ZZ
Z
ZZ
Z
u ψ t + f( u) ψ x +
u 0 ψ = 0,
v ψ t + f( v) ψ x +
v0 ψ = 0
t= 0
where ψ is any C01 function.
Subtracting the two equations, and remembering u 0 = v 0 , we have
ZZ
f( u) − f( v)
ψ x = 0.
( u − v) ψ t +
u− v
Now setting F( x , t) ≡
C01 such that
f ( u) − f ( v )
u− v
( 28)
t= 0
( 29)
, all we need to do is to show that for any φ ∈ C01 , we can find ψ ∈
ψ t + F( x , t) ψ x = φ.
( 30)
T he fn constructed converges everywhere, the limit is f at all rational p oints as well as all points where f is continuous,
at the remaining p oints, the limit may b e other values than f ( x) .
For initial conditions, we assume φ = 0 for t > T and take ψ = 0 along t = T.
•
Where’ s the catch.
The above transport equation can be solved ( formally) by the method of characteristics. Let
x( t) solves
dx
= F( x( t) , t) ,
( 31 )
dt
we obtain
dψ
ψ t= T = 0
( 32)
( x( t) , t) = φ( x( t) , t) ,
dt
which leads to
Zt
ψ( x( t) , t) =
φ( x( s) , s) ds.
( 33)
T
Remember that we want ψ ∈ C 1 and at the same time has compact support.
•
−
Does ψ have compact support?
Recall first that we solve ψ by setting ψ = 0 for t > T. Next notice that ψ can be nonzero only along those characteristics which passes the support of φ. Now since F is uniformly bounded, the slope of the characteristics are uniformly bounded and the boundedness
of ψ’ s support follows.
−
Is ψ ∈ C 1 ?
This is where the catch is. As u , v are only in L 1 , F( x , t) is in general not Lipschitz and
therefore the characteristics may collide with one another. When that happens, ψ is not in
C 1 anymore. In fact, as F( x , t) can only be expected to be in L 1 ∩ L ∞ , even the existence of
the solution is questionable!
Fixing the problem.
We have seen that the obstacle is that F is not Lipschitz. To overcome this, we replace F by a
smooth approximation F ε such that F ε → F locally in L 1 , and call the corresponding solution ψ ε .
Then we have
ZZ
ZZ
( 34)
( u − v) [ ψ tε + F ε ( x , t) ψ xε ] dx dt.
( u − v) φ =
Comparing with the definition of weak solutions, we obtain
ZZ
ZZ
( u − v) [ F( x , t) − F ε ( x , t) ] ψ xε dx dt.
( u − v) φ =
As soon as we have shown the uniform boundedness of ψ xε , we can take ε & 0 and obtain
ZZ
( u − v) φ = 0
( 35)
( 36)
and finish the proof.
•
Uniform boundedness of ψ xε .
If we naïvely mollify F, there is no way we can obtain this bound as in general ψ xε grows as the
Lipschitz constant of F ε grows, and the latter grows like ε − 1 . On the other hand, the subtle situation here is that we can make F smooth in a more sophisticated manner, which allows us to take
advantage of the entropy condition ( which hasn’ t been used so far! ) .
Instead of mollifying F directly, we mollify u, v and define
Z1
f( u ε ) − f( v ε )
F ε ( x , t) =
=
f 0 ( θ u ε + ( 1 − θ) v ε ) dθ.
( 37)
uε − v ε
0
This gives
Z1
∂u ε
∂F ε
∂v ε
00
ε
ε
=
f ( θ u + ( 1 − θ) v ) θ
+ ( 1 − θ)
dθ
( 38)
∂x
∂x
∂x
0
which is uniformly bounded from above if
the entropy condition.
∂u ε
∂x
and
∂v ε
∂x
are so – and this is indeed the case due to
Thus we obtain
∂F ε C
6
∂x
t
( 39)
| ψ xε ( x , t) | 6 C log t − 1
( 40)
for some positive constant C. 2
From this bound one can show that
using the argument presented on p. 287 of J. Smoller’ s book.
Finally, recall that we want to prove
ZZ
( u − v) [ F − F ε ] | ψ xε | dx dt
t> 0
0
( 41 )
as ε → 0. It is easy see that F − F ε → 0 in L 1 . Combine this with the fact tat F − F ε is uniformly
bounded, we can show that
Fε
Now the desired limit holds as |
Further readings.
Missing details can be found in
•
F
ψ xε |
in L p for any 1 6 p < ∞ .
( 42)
is uniformly bounded for any 1 6 q < ∞ . 3
J. Smoller, Shock Waves and Reaction-Diffusion Equations, § A, § B of Chapter 1 6.
Exercises.
E xercise 1 . L et f : R
finite. P rove that
R b e of locally bounded total variation, that is the total variation over any finite interval is
1 . f can have at most countably many jump discontinuities. ( H int: C onsider the numb er of jumps of size > 1 / n over
the interval [ − n , n] and then sum up. )
2 . f is differentiable almost everywhere. ( H int: Define g( x) = T he total variation of f over [ − x , x] , show that both g
and f − g are monotone. T he fact that monotone functions are differentiable almost everywhere can b e taken for
granted. )
E xercise 2 . T he most “natural” finite difference discretization of the equation is obviously
k
k
u kn + 1 − u kn f u n + 1 − f u n
+
= 0.
∆t
∆x
( 43)
Try to carry out the existence proof using this scheme. At which step does it break down?
2 . N ote that if we solve a transp ort equation forward, u t + a( x , t) u x = φ, then when a is increasing, that is a x > 0, the
characteristics are m oving away from one another, which means u x remain b ounded; In the current situation, we are solving
∂F ε
backwards from t = T to t = 0, thus a x 6 0 is “good” and a x > 0 is “bad”. T his is why we do not need a lower bound of ∂x .
3 . For an elementary – and therefore much more tricky – argument, see pp. 2 87 – 2 9 0 of J . S moller’ s b ook.