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ENTROPY AND KINETIC FORMULATIONS OF CONSERVATION LAWS
YUZHOU ZOU
Abstract. Entropy and kinetic formulations of conservation laws are introduced, in order to
create a well-posed theory. Existence and uniqueness results are proven for both formulations.
Convergence of approximations and compactness of solutions are also proven using the kineticformulation.
Contents
1. Introduction 12. Kruzkov’s entropy formulation 22.1. Distributional solutions 22.2. Entropy solutions and equivalent notions 42.3. Uniqueness 62.4. Existence 103. Kinetic Formulation 104. Existence and Uniqueness 124.1. Statement and Outline 124.2. Solving a linear approximation 134.3. Properties of approximate solutions and the approximate measure 154.4. Convergence of approximate solutions 184.5. Uniqueness 225. Convergence in the Diffusion Approximation 236. Compactness and Averaging Lemmas 24Acknowledgments 25References 25
1. Introduction
Conservation laws are prevalent in physics and are some of the most basic examples of nonlinearfirst-order PDEs. It is known that many conservation laws lack classical solutions for all time, evenif the initial data is smooth. A less restrictive notion of solution is thus needed for conservationlaws to be well-posed. In this paper, we will explore two alternative equivalent notions of solutions:entropy solutions and solutions to the kinetic formulation, or kinetic solutions. We will prove thatboth notions lead to conservation laws being well-posed, and using the kinetic formulation we willalso derive results concerning the convergence of solutions to approximations of conservation laws,as well as results concerning the compactness of a family of entropy solutions.
We assume the reader is familiar with Lebesgue integration and basic aspects of the theory ofdistributions.
Date: August 28, 2015.
1
2 YUZHOU ZOU
2. Kruzkov’s entropy formulation
2.1. Distributional solutions. It is well known (see [1]) that the problem ut+divx(A(u)) = 0 maynot have classical solutions for all positive time, even if A and the initial data u0 are both smooth.In order to provide a well-posed theory of conservation laws, we introduce alternative notions ofsolutions in order to obtain the existence and uniqueness of solutions.
Definition 2.1. Let T > 0, and denote πT = [0, T ] × Rd. Let A : R → Rd be C1, and let a = A′.We say that a bounded locally integrable function u : πT → R is a distributional solution to theproblem
(2.1) ut + divx(A(u)) = 0, u(0, x) = u0(x)
if it satisfies the above equations in the sense of distributions, i.e. if for all f ∈ C∞C ([0, T )×Rd) wehave ∫
πT
uft +A(u) · ∇xf dt dx+
∫Rd
u0(x)f(0, x) dx = 0.
For the rest of this section, we will use C∞C (πT ) to denote C∞ functions with compact supportstrictly inside (0, T )× Rd; in particular, these functions are zero at times t = 0 and t = T .
Note that all classical (i.e. C1) solutions to the PDE are distributional solutions, as can beseen from integrating the equation by parts against any smooth test function. Furthermore, anydistributional solution which is C1 is also a classical solution. In addition, the integral equality inthe definition of the distributional solution yields the following condition on the discontinuities ofu, known as the Rankine-Hugoniot condition:
Proposition 2.2. Suppose u is a distributional solution to (2.1) and is C1 in πT everywhere exceptalong a finite number of C1 surfaces, where no two surfaces intersect on a set of positive measure.Let (t, x) be a point belong to exactly one of the above curves. If n is the normal vector to the surfaceat (t, x), then
(u1 − u2, A(u1)−A(u2)) · n = 0
where u1 and u2 are the limits of u approaching (t, x) from either side of the discontinuity surface.In particular, if d = 1, and the discontinuity curve is parametrized by x = x(t), then
x(t)(u1 − u2) = A(u1)−A(u2).
Conversely, if u is a function which satisfies the PDE in (2.1) at every point in πT except alonga finite number of C1 surfaces, with no two surfaces intersecting on a set of positive measure, andeach non-intersection point of the surfaces satisfies the above relation, then u is a distributionalsolution to (2.1).
Proof. We prove this in one dimension for simplicity. Let u be a distributional solution, and assumefirst that we only have one curve of discontinuity C. Let L and R be the regions to the left andright, respectively, of C in the (t, x) plane, and let u1 and u2 be the respective limits. Since u issmooth in L and R, we know that ut+divx(A(u)) = 0 in the interior of L and R. By the divergencetheorem, we have∫L
uft +A(u)fx dt dx =
∫C
(u1f,A(u1)f) · nl ds−∫L
f(ut + divx(A(u))) dt dx =
∫C
(u1f,A(u1)f) · n1 ds
for all f ∈ C∞C (πT ). Similarly, we have∫R
uft +A(u)fx dt dx =
∫C
(u2f,A(u2)f) · n2 ds.
ENTROPY AND KINETIC FORMULATIONS OF CONSERVATION LAWS 3
We thus have
0 =
∫L
uft +A(u)fx +
∫R
uft +A(u)fx =
∫C
f(u1, A(u1)) · n1 + (u2, A(u2)) · n2 ds).
Letting n = (t, x) with x > 0, we have n1 = −n2 = n, and x(t) = − tx . We have∫
C
f((u1 − u2)t+ (A(u1)−A(u2))x) ds = 0.
Since this holds for all f , it follows that
(u1 − u2)t+ (A(u1)−A(u2))x = 0 =⇒ x(t)(u1 − u2) = − tx
(u1 − u2) = A(u1)−A(u2),
thus proving the relation in the case of one curve. For the case of multiple curves, we adapt theabove proof by testing with f whose support does not intersect any of the other curves.
The converse also follows easily from the above calculations, by noting that every point along a sin-gle curve of discontinuity satisfying the above relation will also satisfy the relation(u1 − u2)t+ (A(u1)−A(u2))x = 0. Denoting the curves by {Ci}, we thus have∑
i
∫Ci
f [(u1 − u2)t+ (A(u1)−A(u2))x] ds = 0.
If we denote the regions separated by the curves by {Rj}, we have∫πT
uft +A(u)fx dt dx =∑j
∫Rj
uft +A(u)fx dt dx
=∑j
∫∂Rj
(uf,A(u)f) · n ds−∫Rj
f(ut + divx(A(u))) dt dx
=∑i
∫Ci
f [(u1 − u2)t+ (A(u1)−A(u2))x] ds− 0 = 0,
thus implying that u is a distributional solution. �
While the distributional notion of a solution does not require the solution to be C1 and henceadmits a larger class of possible solutions, it has the disadvantage of admitting too many solutions,
in that solutions are not generally unique. For example, let A(u) = u2
2 on (0, T ) × R. (Thecorresponding PDE ut+uux = 0 is commonly known as Burgers’ equation.) The Rankine-Hugoniotconditions require the curves of discontinuity to satisfy
x(t)(u1 − u2) = A(u1)−A(u2) =(u1 − u2)(u1 + u2)
2=⇒ x(t) =
u1 + u22
.
For the initial data u0 ≡ 0, it is clear that u ≡ 0 is a distributional solution to the problem. However,consider the function
u1(t, x) =
1 0 < x < t2
−1 − t2 < x < 0
0 otherwise.
The initial data for u1 is also u0 ≡ 0. It has three curves of discontinuity (namely at x = 0 andx = ± t
2 ), and it is easy to check that the Rankine-Hugoniot condition is satisfied along each curve.
It follows that u1 is also a distributional solution to Burgers’ equation with u0 ≡ 0, showing thatthe problem does not admit a unique solution for u0 ≡ 0.
4 YUZHOU ZOU
2.2. Entropy solutions and equivalent notions. To avoid the issue of non-uniqueness, we in-troduce a different notion of solution more restrictive than simply satisfying the equation in thesense of distributions.
Definition 2.3. We say that a bounded locally integrable function u : πT → R is an entropysolution of the problem
ut + divx(A(u)) = 0, u(0, x) = u0(x)
on πT if, for all k ∈ R, we have the inequality
(2.2)∂
∂t(|u− k|) + sgn(u− k)divx(A(u)) ≤ 0
in the sense of distributions, i.e. for any nonnegative f ∈ C∞C (πT ) we have
(2.3)
∫πT
|u(t, x)− k|ft + sgn(u(t, x)− k)(A(u(t, x))−A(k)) · ∇xf dx dt ≥ 0,
and there exists E ⊂ [0, T ] such that [0, T ]\E has measure zero, the function u(t, ·) is defined almosteverywhere in Rd for all t ∈ E, and for any ball BR we have
(2.4) limt→0,t∈E
∫BR
|u(t, x)− u0(x)| dx = 0.
Remark 2.4. To see that (2.2) and (2.3) are equivalent, integrate (2.2) by parts against any non-negative f ∈ C∞C (πT ), noting that
divx(sgn(u− k)(A(u)−A(k))) = sgn(u− k)divx(A(u)−A(k)) +∇x(sgn(u− k)) · (A(u)−A(k))
= sgn(u− k)divx(A(u)) + 2δ(u− k)∇xu · (A(u)−A(k))
= sgn(u− k)divx(A(u)),
where δ is the Dirac delta. The last equality following from the fact that δ(u − k) = 0 wheneverA(u)−A(k) 6= 0.
We first show that all entropy solutions are distributional solutions, showing that the notion ofentropy solutions is more strict than that of distributional solutions.
Proposition 2.5. Let u ∈ L∞(πT ) satisfy (2.3). Then u is a distributional solution to the problem(2.1).
Proof. Let f ∈ C∞C (πT ) and k ∈ R. Choosing k > ‖u‖L∞ , we have |u− k| = k−u and sgn(u− k) =−1, and hence (2.3) becomes∫
πT
(k − u)ft − (A(u)−A(k)) · ∇xf dx dt ≥ 0.
Since f ∈ C∞C (πT ), we have∫πT
ft = 0 and∫πT
∇xf = 0. Thus we have
∫πT
kft +A(k) · ∇xf dx dt = 0
and hence ∫πT
−uft −A(u) · ∇xf dx dt ≥ 0 =⇒∫πT
uft +A(u) · ∇xf dx dt ≤ 0.
ENTROPY AND KINETIC FORMULATIONS OF CONSERVATION LAWS 5
Similarly, choosing k < −‖u‖L∞ yields∫πT
(u− k)ft + (A(u)−A(k)) · ∇xf dx dt ≥ 0 =⇒∫πT
uft +A(u) · ∇xf dx dt ≥ 0.
Thus, we have ∫πT
uft +A(u) · ∇xf dx dt = 0
for all nonnegative f ∈ C∞C (πT ), which is enough to conclude that u satisfies (2.1) in the sense ofdistributions. �
Remark 2.6. If u ∈ W 1,1(πT ) and u is a distributional solution, then u is an entropy solution aswell, since we may apply the chain rule to obtain
∂
∂t(|u− k|) + sgn(u− k)divx(A(u)) = sgn(u− k)(ut + divx(A(u))) = 0.
In general, distributional solutions need not be entropy solutions as well. Consider the function u1listed above in the Burgers’ equation example. It is not an entropy solution, since it is easy to verifythat for f ∈ C∞C (πT ) we have∫
πT
|u1|ft + sgn(u1)u212fx dt dx = −
T∫0
f(t, 0) dt.
Hence, if f ≥ 0 in πT , and f∣∣∣x=0
is not identically zero, then the right-hand side is negative, showing
that u1 is not an entropy solution. Of course, u1 is not in W 1,1(πT ). One way to see this is to notethat ∫
πT
u1ft dt dx = −
T2∫
−T2
f(2|x|, x) dx
which implies that the right-hand side is 0 if f ∈ C∞C (πT \{(t, x) : t = 2|x|}). This shows that thedistributional time derivative of u1 is zero almost everywhere outside of {(t, x) : t = 2|x|} and hencezero almost everywhere in πT , implying that it cannot be in L1(πT ) as u1 is not constant in time.
Finally, we prove a property of entropy solutions which allows an alternative characterization ofentropy solutions that, while appearing more restrictive at first, is logically equivalent.
Proposition 2.7. Let u be an entropy solution to (2.1). Then the inequality
∂
∂t(S(u)) + divx(η(u)) ≤ 0
holds for all convex S, where η satisfies η′ = S′A′.
Note that the definition is equivalent to the above inequality holding for all S of the form S(u) =|u − k|, k ∈ R. Hence, entropy solutions can be defined by the solutions satisfying the aboveinequality for all convex S, and not just those of the form S(u) = |u− k|.
Proof. Notice that F (k) = |k|/2 is the fundamental solution to the Laplace equation in 1 dimension.This implies that S′′(u) = (S′′ ∗ | · |/2)′′(u), and hence
S(u) =
∫R
S′′(k)
2|u− k| dk + au+ b
6 YUZHOU ZOU
for some constants a and b in the sense of distributions. From Proposition 2.5, we know that u is adistributional solution, and so if S is linear (say S(u) = au+ b), then
∂
∂t(S(u)) + divx(η(u)) = a(ut + divx(A(u))) = 0.
Hence, assume that S(u) =∫R
S′′(k)2 |u− k| dk. Then η′(u) = S′(u)A′(u) =
∫R
S′′(k)2 sgn(u− k)A′(u) dk,
and hence η(u) =∫R
S′′(k)2 sgn(u− k)(A(u)−A(k)) dk, up to an additive constant. For any f ∈
C∞C (πT ) we have∫πT
S(u)ft + η(u) · ∇xf dt dx =
∫R
∫πT
S′′(k)
2|u− k|ft +
S′′(k)
2sgn(u− k)(A(u)−A(k)) · ∇xf dt dx dk
=
∫R
S′′(k)
2
∫πT
|u− k|ft + sgn(u− k)(A(u)−A(k)) · ∇xf dt dx dk ≥ 0,
since S convex implies S′′ ≥ 0. �
2.3. Uniqueness. In defining the notion of entropy solutions, we aimed to create a notion of solu-tion where existence held in order to fix the main drawback to the notion of distributional solutions.We will now show that entropy solutions are indeed unique.
Theorem 2.8. Let u, v be entropy solutions with corresponding initial data u0 and v0. Let Msatisfy ‖u‖L∞ , ‖v‖L∞ ≤M , and let N = max
|u|≤M|a(u)|. Then, for any R > 0, we have∫
Sτ
|u(τ, x)− v(τ, x)| dx ≤∫BR
|u0(x)− v0(x)| dx
for almost every 0 < τ < T0 = min(T,R/N), where
Sτ = {x : |x| ≤ R−Nτ}is the cross-section of the plane t = τ of the cone
C = {(t, x) : |x| ≤ R−Nt, 0 ≤ t ≤ T0}.
Note that if we take R→∞, we obtain the L1 contraction property
‖u(t, ·)− v(t, ·)‖L1(Rd) ≤ ‖u0 − v0‖L1(Rd).
The proof will follows Kruzkov’s “doubling variables” proof in [3]. We proceed by proving twolemmas:
Lemma 2.9. Let g ∈ C∞C (πT × πT ) be nonnegative. Then
(2.5)
∫πT×πT
|u(t, x)− v(τ, y)|(gt + gτ )+
sgn(u(t, x)− v(τ, y))(A(u(t, x))−A(v(τ, y))) · (∇xg +∇yg) dt dx dτ dy ≥ 0.
Proof. We wish to apply the inequality (2.3) by by integrating over the variables t, x, τ, y, integratingover dt dx and using v(τ, y) as a constant, and then integrating over dτ dy, treating u(t, x) as aconstant.
We thus have∫πT
∫πT
|u(t, x)− v(τ, y)|gt + sgn(u(t, x)− v(τ, y))(A(u(t, x))−A(v(τ, y))) · ∇xg dt dx
dτ dy ≥ 0
ENTROPY AND KINETIC FORMULATIONS OF CONSERVATION LAWS 7
and∫πT
∫πT
|v(τ, y)− u(t, x)|gτ + sgn(v(τ, y)− u(t, x))(A(v(τ, y))−A(u(t, x))) · ∇yg dτ dy
dt dx ≥ 0
Adding the two inequalities yields the desired lemma. �
Lemma 2.10. Let f ∈ C∞C (πT ) be nonnegative. Then∫πT
|u− v|ft + sgn(u− v)(A(u)−A(v)) · ∇f dt dx ≥ 0.
Proof. Let supp f ⊂ K for some compact K. We apply Lemma 2.9 with
g(t, x, τ, y) = f
(t+ τ
2,x+ y
2
)ρε
(t− τ
2,x− y
2
)where ρε is the standard mollifier on Rd+1, supp ρε ⊂ Bε, and ε is chosen small enough for g to bewell-defined. For notational purposes, let t = t+τ
2 , x = x+y2 , ∆t = t−τ
2 , and ∆x = x−y2 . We have
(gt + gτ )(t, x, τ, y) = ft (t, x) ρε (∆t,∆x)
and
(∇xg +∇yg)(t, x, τ, y) = (∇f (t, x)) ρε (∆t,∆x)
I now claim that
∫πT×πT
|u(t, x)− v(τ, y)|ft (t, x) ρε (∆t,∆x) dt dx dτ dyε→0−−−→ 2d+1
∫πT
|u(t, x)− v(t, x)|ft(t, x) dt dx
(2.6)
and ∫πT×πT
sgn(u(t, x)− v(τ, y))(A(u(t, x))−A(u(τ, y))) · (∇f (t, x) ρε (∆t,∆x)) dt dx dτ dy
ε→0−−−→ 2d+1
∫πT
sgn(u(t, x)− v(t, x))(A(u(t, x))− v(t, x)) · ∇f(t, x) dt dx.(2.7)
For simplicity, we prove (2.6). We re-write the integral on the left hand side as∫πT×πT
[|u(t, x)− v(τ, y)|ft (t, x)− |u(t, x)− v(t, x)|ft(t, x)] ρε (∆t,∆x) dt dx dτ dy
+
∫πT×πT
|u(t, x)− v(t, x)|ft(t, x)ρε (∆t,∆x) dt dx dτ dy.(2.8)
After a change-of-variables, the second integral becomes
2d+1
∫πT×πT
|u(t, x)− v(t, x)|ft(t, x)ρε(τ − t, y − x) dt dx dτ dy
= 2d+1
∫πT
((|u− v|ft) ∗ ρε) (τ, y) dτ dyε→0−−−→ 2d+1
∫πT
|u(τ, y)− v(τ, y)|ft(τ, y) dτ dy.
8 YUZHOU ZOU
It suffices to show that the first integral in (2.8) vanishes as ε→ 0. To do so, we make the followingestimate:
||u(t, x)− v(τ, y)|ft (t, x)− |u(t, x)− v(t, x)|ft(t, x)|≤ ||u(t, x)− v(τ, y)| − |u(t, x)− v(t, x)||ft (t, x) + |u(t, x)− v(t, x)| |ft (t, x)− ft(t, x)|≤ |v(t, x)− v(τ, y)|‖ft‖L∞ + |u(t, x)− v(t, x)|‖∇ft‖L∞ |(∆t,∆x)|≤ ‖ft‖L∞ + ε‖∇ft‖L∞ |u(t, x)− v(t, x)|.
Note that we may assume |(∆t,∆x)| < ε since supp ρε ⊂ Bε. Furthermore, under this assumptionwe have supp ft (t, x) ⊂ K+Bε = {x+y|x ∈ K, |y| ≤ ε}. Hence, we can take the limits of integrationin the integral to be (K +Bε)× (K +Bε). We thus have∣∣∣∣∣∣∫
πT×πT
[|u(t, x)− v(τ, y)|ft (t, x)− |u(t, x)− v(t, x)|ft(t, x)] ρε (∆t,∆x) dt dx dτ dy
∣∣∣∣∣∣≤
∫(K+Bε)×(K+Bε)
(‖ft‖L∞ |v(t, x)− v(τ, y)|+ ε‖∇ft‖L∞ |u(t, x)− v(t, x)|)ρε (∆t,∆x) dt dx dτ dy.
(2.9)
Since ∫(K+Bε)×(K+Bε)
|u(t, x)− v(t, x)|ρε (∆t,∆x) dt dx dτ dy
= 2d+1
∫(K+Bε)×(K+Bε)
|u(t, x)− v(t, x)|ρε(τ − t, y − x) dt dx dτ dyε→0−−−→ 2d+1
∫K
|u(t, x)− v(t, x)| dt dx
it follows that
ε‖∇ft‖L∞
∫(K+Bε)×(K+Bε)
|u(t, x)− v(t, x)|ρε (∆t,∆x) dt dx dτ dyε→0−−−→ 0.
Furthermore, we have∫(K+Bε)×(K+Bε)
‖ft‖L∞ |v(t, x)− v(τ, y)|ρε (∆t,∆x) dt dx dτ dy
≤ ‖ft‖L∞‖ρ‖L∞
∫K+Bε
1
εd+1
∫(t,x)+Bε
|v(t, x)− v(τ, y)| dτ dy
dt dx.(2.10)
By the Lebesgue Differentiation Theorem, almost every (t, x) ∈ πT is a Lebesgue point, i.e.
1
εd+1
∫(t,x)+Bε
|v(t, x)− v(τ, y)| dτ dyε→0−−−→ 0
for almost every (t, x) ∈ πT . Since
1
εd+1
∫(t,x)+Bε
|v(t, x)− v(τ, y)| dτ dy ≤ 1
εd+1
∫(t,x)+Bε
2M dτ dy = 2M |B1|
and the outer integral in (2.10) is taken over a bounded set, we can apply the Dominated ConvergenceTheorem to conclude that the integral in (2.10) vanishes as ε→ 0. Hence, the first integral in (2.8)vanishes, so we arrive at (2.6). The statement in (2.7) can be proven similarly.
ENTROPY AND KINETIC FORMULATIONS OF CONSERVATION LAWS 9
Hence, applying Lemma 2.9 with our choice of g, and letting ε→ 0, we have
2d+1
∫πT
|u(t, x)− v(t, x)|ft(t, x) + sgn(u(t, x)− v(t, x))(A(u(t, x))−A(v(t, x))) · ∇f(t, x) dt dx ≥ 0,
which proves the desired lemma. �
With these two lemmas proven, we may now proceed to prove the main theorem.
Proof of Theorem 2.8. Formally, Lemma 2.10 implies that
∂
∂t(|u− v|) + divx(sgn(u− v)(A(u)−A(v))) ≤ 0
in the sense of distributions. Furthermore, we have |A(u) − A(v)| ≤ N |u − v| by the definition ofN , and hence −N |u− v| ≤ sgn(u− v)(A(u)−A(v)) · ν for any unit vector ν. It follows that
d
dt
∫St
|u− v| dx
=
∫St
∂
∂t(|u− v|) dx−N
∫∂St
|u− v| dS
≤∫St
∂
∂t(|u− v|) dx+
∫∂St
sgn(u− v)(A(u)−A(v)) · n dS
≤∫St
∂
∂t(|u− v|) + divx(sgn(u− v)(A(u)−A(v))) dx ≤ 0.
We can formalize the argument as follows: let µ(t) =∫St
|u(t, x)− v(t, x)| dx, let Eµ be the set of
Lebesgue points of µ, and let Eu and Ev be the subsets of [0, T ] involved in the definition of entropysolutions. Let E = Eµ ∩ Eu ∩ Ev. Then [0, T ]\E has measure zero. It suffices to show that µ isdecreasing on E, and that µ(t) approaches µ(0) =
∫BR
|u0(x)− v0(x)| dx for t ∈ E approaching 0.
To show that µ(t)→ µ(0), note that∫St
|u(t, x)− v(t, x)| dx ≤∫BR
|u(t, x)− u0(x)| dx+
∫BR
|v(t, x)− v0(x)| dx+
∫BR
|u0(x)− v0(x)| dx.
For t ∈ E approaching 0, the first and second terms on the right-hand side vanish, so we obtain thedesired result.
We now show that µ is decreasing on E. Let ρε be the standard mollifier in R, and let χε(x) =x∫−∞
ρε(y) dy. Note that χε(x) = 0 for x < −ε and χε(x) = 1 for x > ε. For t1 < t2 ∈ E, let
f(t, x) = (χε(t− t1)− χε(t− t2))(1− χε′(|x|+Nt−R+ ε′))
with ε′ chosen so that 2ε′ < R − Nt2. Notice that supp f ⊂ [t1 − ε, t2 + ε] × C, and that f ≥ 0.Furthermore, f is clearly infinitely differentiable wherever x 6= 0, and at x = 0, either t ≤ t2, inwhich case Nt2 − R + ε′ < −ε′, and hence χε′(|x| + Nt − R + ε′) = 0 in a neighborhood of (t, x),or t > t2, in which case f = 0 in a neighborhood of (t, x). Hence, f ∈ C∞C (πT ), so we may applyLemma 2.10. We have
ft(t, x) = (ρε(t−t1)−ρε(t−t2))(1−χε′(|x|+Nt−R+ε′))+(χε(t−t1)−χε(t−t2))(−Nρε′(|x|+Nt−R+ε′))
and
∇f(t, x) = (χε(t− t1)− χε(t− t2))
(ρε′(|x|+Nt−R+ ε′)
x
|x|
).
10 YUZHOU ZOU
Hence, we have
sgn(u− v)(A(u)−A(v)) · ∇f(t, x) ≤ N |u− v|(χε(t− t1)− χε(t− t2))ρε′(|x|+Nt−R+ ε′)
= |u− v|((ρε(t− t1)− ρε(t− t2))(1− χε′(|x|+Nt−R+ ε′))− ft).(2.11)
Applying Lemma 2.10 to f and combining with inequality (2.11), we obtain
(2.12)
∫πT
|u(t, x)− v(t, x)|(ρε(t− t1)− ρε(t− t2))(1− χε′(|x|+Nt−R+ ε′)) dt dx ≥ 0.
As ε′ → 0, we have
1− χε′(|x|+Nt−R+ ε′)→{
1 if |x| < R−Nt0 if |x| > R−Nt
and hence we have(2.13)
T∫0
µ(t)(ρε(t− t1)− ρε(t− t2)) dt =
T∫0
∫St
|u(t, x)− v(t, x)|(ρε(t− t1)− ρε(t− t2)) dx dt ≥ 0.
Since t1 is a Lebesgue point of µ, we have∣∣∣∣∣∣ T∫
0
µ(t)ρε(t− t1) dt
− µ(t1)
∣∣∣∣∣∣ =
∣∣∣∣∣∣T∫
0
(µ(t)− µ(t1))ρε(t− t1) dt
∣∣∣∣∣∣≤
t1+ε∫t1−ε
|µ(t)− µ(t1)| ‖ρ‖L∞
εdt
=‖ρ‖L∞
ε
∫|t−t1|<ε
|µ(t)− µ(t1)| dtε→0−−−→ 0.
A similar result holds for t2. Hence, the left-hand side of (2.13) converges to µ(t1)−µ(t2) as ε→ 0,so (2.13) implies that µ(t1) ≥ µ(t2) for t1 < t2, thus proving the theorem. �
2.4. Existence. We can also establish a result regarding the existence of entropy solutions. Theidea is to consider, for ε > 0, the solution to uε of the problem
ut + divx(A(u)) = ε∆u, u|t=0 = u0
It is well known[2] that this equation admits a unique classical solution uε if u0 is bounded and hassufficient bounded derivatives. Hence, for regular enough u0, it suffices to show that the family {uε}is compact, in order to extract a subsequence εn and a limit u such that uεn → u, with u satisfyingthe desired entropy inequalities, while for u0 ∈ L∞ we can approximate by smooth initial data u0h,thus getting a family {uε,h} which converges to some u as h→ 0 and ε→ 0.
The proof involves finding equicontinuity estimates on uε and is similar to the correspondingproof for kinetic solutions described later in this paper, so the proof, which can be found in [3], willbe omitted. In fact, we will investigate the rate of convergence of the parabolic approximation uεto the entropy solution u later in this paper.
3. Kinetic Formulation
We now turn our attention to a reformulation of conservation laws which generalizes the notion ofentropy solutions. In the kinetic formulation, a function χ(ξ, u) is introduced to turn the nonlinearconservation law into a linear equation on the nonlinear quantity χ(ξ, u). This structure providesa method to construct solutions by approximating with solutions of a linear equation, as well as
ENTROPY AND KINETIC FORMULATIONS OF CONSERVATION LAWS 11
nice estimates on such approximations and compactness results on solutions without requiring com-pactness of initial data. We will follow the approach of Perthame[5] in introducing the kineticformulation, existence results, convergence estimates, and compactness results.
We first introduce a simple, yet important function whose properties are critical in forming thekinetic formulation:
Definition 3.1. The function χ : R× R→ R is defined by
χ(ξ, u) =
1 if 0 < ξ < u−1 if u < ξ < 00 otherwise
We prove a few basic properties:
Proposition 3.2. We have
(1)∫RS′(ξ)χ(ξ, u) dξ = S(u) − S(0) for S locally Lipschitz, i.e. S′ ∈ L∞loc, and in particular∫
Rχ(ξ, u) dξ = u,
(2)∫R|χ(ξ, u)− χ(ξ, v)| dξ = |u− v|,
(3) ∂∂ξ (χ(ξ, u)) = δ(ξ)− δ(ξ − u), and
(4) ∂∂u (χ(ξ, u)) = δ(ξ − u) for ξ 6= 0.
The last two statements are made in the sense of distributions.
The proof of these properties is an easy exercise to verify.We now consider an entropy solution u to ut+div(A(u)) = 0, and define the distribution m(t, x, ξ)
by
m(t, x, ξ) =∂
∂t
ξ∫0
χ(ζ, u(t, x)) dζ
+ divx
ξ∫0
a(ζ)χ(ζ, u(t, x)) dζ
.
It turns out that m has some interesting properties. For example, by differentiating both sides inξ, we have the distributional equation
∂m
∂ξ=
∂
∂t(χ(ξ, u)) + a(ξ) · ∇x(χ(ξ, u)).
Furthermore, we can show that m is nonnegative. Multiplying m by ϕ(ξ), where ϕ ∈ C∞C (R), andintegrating by parts yields
−∫R
ϕ′(ξ)m(t, x, ξ) dξ =∂
∂t
∫R
ϕ(ξ)χ(ξ, u(t, x)) dξ
+ divx
∫R
ϕ(ξ)a(ξ)χ(ξ, u(t, x)) dξ
In particular, if we choose ϕ approaching S′ for S convex, we obtain, using the properties above,that
−∫R
S′′(ξ)m(t, x, ξ) dξ =∂
∂t
∫R
S′(ξ)χ(ξ, u(t, x)) dξ
+ divx
∫R
η′(ξ)χ(ξ, u(t, x)) dξ
=
∂
∂t(S(u(t, x))− S(0)) + divx(η(u(t, x))− η(0))
=∂
∂t(S(u(t, x))) + divx(η(u(t, x))) ≤ 0.
Since this holds for all S convex (i.e. for all S′′ ≥ 0), it follows that m is nonnegative.
12 YUZHOU ZOU
Inspired by these results, we define the kinetic formulation of the conservation law as follows:
Definition 3.3. Let u ∈ C(R+;L1(Rd)). We say that u is a kinetic formulation to the equation
ut + divx(A(u)) = 0, u(0, x) = u0(x)
if there exists a nonnegative bounded measure m ∈ C0(R, w −M1(R+ × Rd)) such that
∂
∂t(χ(ξ, u(t, x))) + a(ξ) · ∇x (χ(ξ, u(t, x))) =
∂m
∂ξ(t, x, ξ), χ(ξ, u(0, x)) = χ(ξ, u0(x))
in the sense of distributions.
Note that this definition does not require a L∞ bound on u, and as such can be applied to initialdata which is not L∞.
Remark 3.4. If u is regular enough (say W 1,1), then applying the chain rule just as we did inProposition 2.5 yields
∂
∂t(χ(ξ, u)) + a(ξ) · ∇x(χ(ξ, u)) = δ(ξ − u)(ut + a(ξ) · ∇xu) = δ(ξ − u)(ut + divx(A(u))) = 0.
Hence, for u ∈W 1,1, the corresponding measure is identically zero.
Remark 3.5. If u is in L∞, then the corresponding measure m is compactly supported in ξ, withsupport contained in |ξ| ≤ ‖u‖L∞ . In particular, the measure associated with entropy solutions arecompactly supported in ξ. To see this, note that for |ξ| > ‖u‖L∞ we have
m(t, x, ξ) =∂
∂t
ξ∫0
χ(ζ, u(t, x)) dζ
+ divx
ξ∫0
a(ζ)χ(ζ, u(t, x)) dζ
=
∂
∂t
∞∫−∞
χ(ζ, u(t, x)) dζ
+ divx
∞∫−∞
a(ζ)χ(ζ, u(t, x)) dζ
=
∂
∂t(u(t, x)) + divx(A(u(t, x))−A(0)) = 0.
The second equality follows from the fact that χ(ζ, u) = 0 for ζ not between 0 and u, while the lastequality follows from the fact that u satisfies the conservation law in the sense of distributions.
4. Existence and Uniqueness
4.1. Statement and Outline. We have the following existence/uniqueness theorem:
Theorem 4.1. Let u0 ∈ L1(Rd), and let A be locally Lipschitz, i.e. a ∈ (L∞loc(R))d. Then thereexists a unique distributional distribution u ∈ C(R+;L1(Rd)) to the kinetic formulation.
The proof will follow the following steps: We first solve the equation
∂fλ∂t
+ a(ξ) · ∇xfλ = λ(χ(ξ, uλ)− fλ), fλ
∣∣∣t=0
= χ(ξ, u0)
for every λ > 0 through a fixed point argument, where uλ =∫fλ dξ. For every λ > 0, we find a
(nonnegative bounded) measure mλ such that
∂mλ
∂ξ= λ(χ(ξ, uλ)− fλ)
and hence∂fλ∂t
+ a(ξ) · ∇xfλ =∂mλ
∂ξ.
ENTROPY AND KINETIC FORMULATIONS OF CONSERVATION LAWS 13
This is similar to the desired equation in the kinetic formulation, except that we do not know if fλhas the structure of χ(ξ, u) for some u. Hence, we will show that {mλ} is uniformly bounded for
all λ > 0, and that the initial condition fλ
∣∣∣t=0
= χ(ξ, u0) imposes extra conditions on the sign and
bounds of fλ, in the hope that
fλ − χ(ξ, uλ) =1
λ
∂mλ
∂ξ
λ→∞−−−−→ 0
in some way. We then argue that as λ→∞, the sequences {uλ}, {fλ}, and {mλ} converge to u, f ,and m, so that
∂f
∂t+ a(ξ) · ∇xf =
∂m
∂ξ.
and that fλ → χ(ξ, u), i.e. f = χ(ξ, u), thus proving the existence of the desired solution.
4.2. Solving a linear approximation. We first investigate properties of the solutions of theproblem
(4.1) ft + a(ξ) · ∇xf + λf = g, f∣∣∣t=0
= f0
in R+ × Rd × Rξ, where λ is a fixed positive parameter.
Theorem 4.2. Let f0 ∈ L1(Rd × Rξ), g ∈ L1((0, T ) × Rdx × Rξ), and a ∈ (L∞loc(R))d. Then thereexists a distributional solution f ∈ C(R+;L1(Rd × Rξ)) to (4.1) given by the formula
f(t, x, ξ) = f0(x− a(ξ)t, ξ)e−λt +
t∫0
e−λsg(t− s, x− a(ξ)s, ξ) ds.
Furthermore, this solution satisfies the properties
d
dt
∫Rd×Rξ
f(t, x, ξ) dx dξ
+ λ
∫Rd×Rξ
f(t, x, ξ) dx dξ =
∫Rd×Rξ
g(t, x, ξ) dx dξ
and
d
dt
∫Rd×Rξ
|f(t, x, ξ)| dx dξ
+ λ
∫Rd×Rξ
|f(t, x, ξ)| dx dξ ≤∫
Rd×Rξ
|g(t, x, ξ)| dx dξ.
Proof. For f0 and g smooth, we have
d
dt
(eλtf(t, x+ a(ξ)t, ξ)
)= eλt(ft + a(ξ) · ∇xf + λf)(t, x+ a(ξ)t, ξ) = eλtg(t, x+ a(ξ)t, ξ).
Integrating from s = 0 to s = t gives the desired formula. For f0 and g in L1, let {f0n} and {gn}be sequences of smooth functions such that f0n → f0 in L1(Rd+1) and gn → g in L1((0, T )×Rd+1).Letting {fn} denote the corresponding solutions, we note that the preceding formula gives
fn(t, x, ξ)− fm(t, x, ξ) = (f0n − f0m)(x− a(ξ)t, ξ)e−λt +
t∫0
e−λs(gn − gm)(t− s, x− a(ξ)s, ξ) ds.
Hence, integrating along x and ξ, and making an appropriate change of variables, we obtain
‖fn(t, ·)− fm(t, ·)‖L1(Rd+1) ≤ e−λt‖f0n − f0m‖L1 +
t∫0
e−λs‖gn(s, ·)− gm(s, ·)‖L1 ds.
14 YUZHOU ZOU
Thus, {fn} is a Cauchy sequence in C((0, T );L1(Rd+1)), and thus converges to some f satisfyingthe same distributional formula.
To derive the integral equation, we multiply (4.1) by ϕR(x) = ϕ(x/R), where ϕ is a compactlysupported cutoff function, and integrate with respect to x and ξ. As R → ∞, we have
∫ftϕR →
ddt
(∫f),∫λfϕR → λ
∫f ,∫gϕR →
∫g, and∫
a(ξ) · ϕR∇xf =
∫a(ξ) ·
(−∫f∇xϕR dx
)dξ =
1
R
∫a(ξ) ·
∫f(x)∇xϕ(
x
R) dx dξ → 0.
This proves the integral equation. Finally, to derive the L1 inequality, we multiply (4.1) by
ηε(f)ϕR(x) and integrate, where ηε(f) = 2f∫−∞
ρε(y) dy − 1 and ρε is the standard mollifier on
R. By similar arguments as above, as R→∞ we have∫a(ξ) · ηε(f)ϕR∇xf → 0, while the ϕR term
drops out in the other terms, leading to the equation∫ftηε(f) + λ
∫fηε(f) =
∫gηε(f).
Letting ε→ 0 yieldsd
dt
(∫|f |)
+ λ
∫|f | =
∫gsgn(f) ≤
∫|g|
as desired. �
We use the results of this theorem to find solutions to the problem
(4.2)∂fλ∂t
+ a(ξ) · ∇xfλ = λ(χ(ξ, uλ)− fλ), fλ
∣∣∣t=0
= χ(ξ, u0)
To do this, we first fix v ∈ C((0, T );L1(Rd)), and consider the solution f to the equation
(4.3)∂f
∂t+ a(ξ) · ∇xf + λf = λχ(ξ, v), f
∣∣∣t=0
= χ(ξ, u0)
Let Φ : C((0, T ), L1(Rd)) → C((0, T ), L1(Rd)) be the operator sending v to the integral of thecorresponding f , i.e.
Φ(v) : (t, x) 7→ u(t, x) =
∫R
f(t, x, ξ) dξ.
We aim to show that Φ is a strict contraction on the Banach space C((0, T ), L1(Rd)), with the norm
‖u‖C((0,T );L1(Rd)) = supt∈(0,T )
‖u(t, ·)‖L1(Rd)
in order to apply the Banach fixed point theorem and obtain a solution to (4.2). To do this,let v1, v2 ∈ C((0, T );L1(Rd)). Letting f = f1 − f2, where f1 and f2 are the solutions to (4.3)corresponding to v1 and v2, we see that f is a solution to (4.3) with g = χ(ξ, v1) − χ(ξ, v2) andf0 ≡ 0. We thus have
d
dt
∫Rd×Rξ
|f(t, x, ξ)| dx dξ
+ λ
∫Rd×Rξ
|f(t, x, ξ)| dx dξ ≤ λ∫
Rd×Rξ
|χ(ξ, v1(t, x))− χ(ξ, v2(t, x))| dx dξ
(4.4)
= λ
∫Rd
|v1(t, x)− v2(t, x)| dx
≤ λ‖v1 − v2‖C((0,T );L1(Rd)).
ENTROPY AND KINETIC FORMULATIONS OF CONSERVATION LAWS 15
It follows that
d
dt
eλt ∫Rd×Rξ
|f(t, x, ξ)| dx dξ
≤ d
dt
(eλt‖v1 − v2‖C((0,T );L1(Rd))
)and, using the fact that
∫|f(0, x, ξ)| dx dξ =
∫|f0| dx dξ = 0, we have∫
Rd×Rξ
|f(t, x, ξ)| dx dξ ≤ (1− e−λt)‖v1 − v2‖C((0,T );L1(Rd)).
Hence, we have
‖Φ(v1)− Φ(v2)‖C((0,T );L1(Rd)) = supt∈(0,T )
∫Rd
∣∣∣∣∣∣∫R
f1(t, x, ξ)− f2(t, x, ξ) dξ
∣∣∣∣∣∣ dx
≤ supt∈(0,T )
∫Rd×Rξ
|f(t, x, ξ)| dx dξ
≤ (1− e−λT )‖v1 − v2‖C((0,T );L1(Rd))
showing that Φ is a strict contraction on C((0, T );L1(Rd)). Hence, for all T > 0, there exists a fixedpoint of Φ, i.e. there exists uλ ∈ C((0, T );L1(Rd)) and fλ ∈ C((0, T );L1(Rd × Rξ)) such that
∂fλ∂t
+ a(ξ) · ∇xfλ = λ(χ(ξ, uλ)− fλ), fλ
∣∣∣t=0
= χ(ξ, u0)
and
uλ(t, x) =
∫R
fλ(t, x, ξ) dξ.
Moreover, uλ and fλ are defined on (0, T ) for all T > 0 and hence can be extended to be definedfor all positive time.
4.3. Properties of approximate solutions and the approximate measure. We now provesome properties of the solutions fλ and uλ obtained above. Since these properties do not dependon the specific value of λ, we drop it in the subscript for ease of notation.
Theorem 4.3. Let u and f be as above. Then we have the representational formula
f(t, x, ξ) = χ(ξ, u0(x− a(ξ)t))e−λt + λ
t∫0
e−λsχ(ξ, u(t− s, x− a(ξ)s)) ds.
Furthermore, we have the following properties:
(1) The total mass is conserved, i.e.
d
dt
∫Rd
u(t, x) dx
= 0.
(2) For any initial data u01 and u02, with corresponding u1, f1 and u2, f2, we have an L1 con-traction property
‖u1(t, ·)− u2(t, ·)‖L1(Rd) ≤ ‖f1(t, ·)− f2(t, ·)‖L1(Rd+1) ≤ ‖u01 − u02‖L1(Rd).
16 YUZHOU ZOU
Hence, for any initial data u0, setting u01 = u0(·+h) and u02 = u0 yields the space-oscillationcontraction property
‖u(·+ h, t)− u(·, t)‖L1(Rd) ≤ ‖u0(·+ h)− u0(·)‖L1(Rd).
(3) |f(t, x, ξ)| = sgn(ξ)f(t, x, ξ) ≤ 1. In other words, the sign of f matches the sign of ξ, and|f | ≤ 1.
(4) If u0 ∈ L∞(Rd), then ‖u(t, ·)‖L∞ ≤ ‖u0‖L∞ , and f(t, x, ξ) = 0 for |ξ| > ‖u0‖L∞ .
Proof. Since u and f satisfy the equation
∂f
∂t+ a(ξ) · ∇xf + λf = g, f
∣∣∣t=0
= χ(ξ, u0),
with g = λχ(ξ, u), the representation formula holds, as well as the integral equality
d
dt
∫Rd×Rξ
f dx dξ
+ λ
∫Rd×Rξ
f dx dξ =
∫Rd×Rξ
λχ(ξ, u) dx dξ = λ
∫Rd
u dx.
Since∫
Rd×Rξf dx dξ =
∫Rdu dx, it follows that
d
dt
∫Rd
u(t, x) dx
+ λ
∫Rd
u(t, x) dx = λ
∫Rd
u(t, x) dx,
thus proving mass conservation. The L1 contraction property follows similarly, since f1−f2 satisfies(4.3) with g = χ(ξ, u1)− χ(ξ, u2), and hence
d
dt
(∫|f1 − f2| dx dξ
)+ λ
∫|f1 − f2| dx dξ ≤ λ
∫|χ(ξ, u1)− χ(ξ, u2)| dx dξ
= λ
∫|u1 − u2| dx
≤ λ∫|f1 − f2| dx dξ
Hence, ddt
(∫|f1 − f2| dx dξ
)≤ 0, and since (f1 − f2)
∣∣∣t=0
= u01 − u02, the second inequality in (2)
follows. The first inequality follows easily from the fact that u =∫f dξ.
To prove the sign property on f , we note |χ| ≤ 1, and the sign of χ(ξ, u) matches the sign of ξ.Using the representational formula, we see that the sign of fλ also matches the sign of ξ, while
|fλ(t, x, ξ)| ≤ e−λt + λ
t∫0
e−λs ds = 1
as desired.To show the L∞ bounds on u, we aim to show that the set
{v ∈ C((0, T );L1(Rd)) : ‖v‖L∞((0,T )×Rd) ≤ ‖u0‖L∞(Rd)},
a closed subset of C((0, T );L1(Rd)), is invariant under Φ, and hence the fixed point (i.e. u) mustlie in that subset. For v with ‖v‖L∞ ≤ ‖u0‖L∞ , consider the associated solution fv. We have
fv(t, x, ξ) = χ(ξ, u0(x− a(ξ)t))e−λt + λ
t∫0
e−λsχ(ξ, v(t− s, x− a(ξ)s, ξ)) ds.
ENTROPY AND KINETIC FORMULATIONS OF CONSERVATION LAWS 17
From this formula, we see immediately that fv = 0 whenever |ξ| ≥ ‖u0‖L∞ , and, using similararguments as those used to prove (3), we obtain the sign property |fv| = sgn(ξ)fv ≤ 1. We can thuswrite
Φ(v)(t, x) = uv(t, x) =
∫R
fv(t, x, ξ) dξ =
‖u0‖L∞∫0
fv(t, x, ξ) dξ −0∫
−‖u0‖L∞
|fv(t, x, ξ)| dξ.
Notice that both integrals on the right-hand side are positive, and since |fv| ≤ 1, both integralsare bounded by ‖u0‖L∞ . Thus uv, as a difference of two positive terms, must have absolute valuebounded by the maximum of the two terms, and hence |Φ(v)(t, x)| ≤ ‖u0‖L∞ for all t and x. Thisshows that ‖u‖L∞((0,T )×Rd) ≤ ‖u0‖L∞(Rd). Finally, from the representational formula for f , we
clearly see that f(t, x, ξ) = 0 for |ξ| > ‖u0‖L∞ .�
We now show that we can find a function mλ such that
∂mλ
∂ξ= λ(χ(ξ, uλ)− fλ) =
∂fλ∂t
+ a(ξ) · ∇xfλ.
This brings the equation to a form similar to that in the kinetic formulation. We obtain mλ throughthe formula
mλ(t, x, ξ) = λ
ξ∫−∞
χ(ζ, uλ(t, x))− fλ(t, x, ζ) dζ.
From this, we can prove several properties:
Proposition 4.4. The function mλ satisfies the following properties:
(1) mλ is nonnegative.(2) For any convex S with S′(0) = 0, we have
∞∫0
∫Rd+1
S′′(ξ)mλ(t, x, ξ) dx dξ dt ≤∫Rd
S(u0(x))− S(0) dx.
In particular, for S(ξ) = (ξ ∓ ξ0)± (ξ0 ≥ 0) we have
∞∫0
∫Rd
mλ(t, x,±ξ0) dx dt ≤∫Rd
(u0(x)∓ ξ0)± dx ≤ µ(±ξ0)
where µ(ξ) ≤ ‖u0‖L1 and lim|ξ|→∞
µ(ξ) = 0, and if u0 ∈ L2, then for S(ξ) = ξ2
2 we have
‖mλ‖L1 ≤ 1
2‖u0‖L2
.(3) If u0 ∈ L∞, then mλ(t, x, ξ) = 0 for |ξ| > ‖u0‖L∞ .
Proof. Note that if uλ ≥ 0, then by the sign property of fλ we have
χ(ζ, uλ)− fλ(t, x, ζ) = −fλ(t, x, ζ) ≥ 0
for ζ ≤ 0, Similarly, for 0 < ζ < uλ, the integrand is 1 − fλ(t, x, ζ) ≥ 0, while for ζ > uλ theintegrand is −fλ(t, x, ζ) ≤ 0. Hence, χ(ζ, uλ)− fλ(t, x, ζ) is nonnegative for ζ < uλ and nonpositive
18 YUZHOU ZOU
for ζ ≥ uλ, assuming uλ ≥ 0. A similar statement also holds for uλ < 0. It follows that mλ(t, x, ξ)is increasing in ξ for ξ < uλ and ξ > uλ, and furthermore
limξ→∞
mλ(t, x, ξ) =
∞∫−∞
χ(ξ, uλ(t, x))− fλ(t, x, ξ) dξ = uλ(t, x)−∞∫−∞
fλ(t, x, ξ) dξ = 0.
Hence, mλ is nonnegative.To prove (2), we multiply the equation ∂mλ
∂ξ = ∂fλ∂t + a(ξ) · ∇xfλ by S′(ξ) and integrate in x
and ξ. By applying the arguments in Theorem 4.2, we may assume that the term∫S′(ξ)a(ξ) · ∇xf
vanishes. We thus have∫∂fλ∂t
S′(ξ) dx dξ =
∫S′(ξ)
∂mλ
∂ξdx dξ = −
∫S′′(ξ)mλ dx dξ
and hence ∫S′′(ξ)mλ dx dξ = − d
dt
(∫S′(ξ)fλ dx dξ
).
Integrating both sides with respect to time gives
T∫0
∫Rd+1
S′′(ξ)mλ dx dξ dt =
∫S′(ξ)χ(ξ, u0(x)) dx dξ −
∫S′(ξ)fλ(T, x, ξ) dx dξ
≤∫S′(ξ)χ(ξ, u0(x)) dx dξ
=
∫S(u0(x)) dx.
Note that S convex and S′(0) = 0 implies that the sign of S′(ξ) matches that of ξ, and henceS′(ξ)fλ(t, x, ξ) ≥ 0 for all ξ. Letting T →∞ yields the general estimate. For S(ξ) = (ξ ∓ ξ0)±, wehave S′′(ξ) = δ(ξ ∓ ξ0), so
∞∫0
∫Rd+1
S′′(ξ)m(t, x, ξ) dx dξ dt =
∞∫0
∫Rd
m(t, x,±ξ0) dx dt,
yielding the first inequality. We note that∫Rd
(u0(x)∓ ξ0)± dx is clearly bounded by ‖u0‖L1 , while
the integral goes to 0 as ξ0 →∞ from the Dominated Converge Theorem. Finally, the last inequality
follows from noting that S(ξ) = ξ2
2 =⇒ S′′(ξ) = 1.
To prove (3), we simply note that χ(ξ, uλ) = fλ = 0 for |ξ| > ‖u0‖L∞ from Theorem 4.3, so bydefinition mλ = 0 for |ξ| > ‖u0‖L∞ . �
4.4. Convergence of approximate solutions. To obtain a kinetic solution from a family ofapproximate solutions, we seek to show that the family of approximate solutions is compact, inorder to extract subsequences {fλ}, {uλ}, and {mλ} converging to f , u, and m as λ→∞. We alsoaim to show that fλ − χ(ξ, uλ)→ 0 as λ→∞, which will imply that f = χ(ξ, u), as desired.
To show compactness, we need uniform boundedness, equicontinuity, and uniform integrability.Uniform boundedness follows from the results of Theorem 4.3, while equicontinuity can be shownusing the space-oscillation contraction in Theorem 4.3, combined with the following time continuityestimate:
Proposition 4.5. There exists a modulus of continuity ω, independent of λ, such that
‖uλ(k, ·)− u0(·)‖L1(Rd) ≤ ‖fλ(k, ·)− χ(·, u0)‖L1(Rd×Rξ) ≤ ω(k).
ENTROPY AND KINETIC FORMULATIONS OF CONSERVATION LAWS 19
Consequently, setting u0 = u(t, x) for some t > 0 yields the time continuity estimate
‖uλ(t+ k, ·)− uλ(t, ·)‖L1 ≤ ‖fλ(t+ k, ·)− fλ(t, ·)‖L1 ≤ ω(k).
Proof. Set u0ε = u0 ∗ ρε, where ρε is the standard mollifier in Rd. Then ‖u0ε‖L∞ ≤ ‖ρ‖L∞
ε ‖u0‖L1 .
Set χ0ε = χ(ξ, u0ε), and let Nε = sup
|ξ|≤‖u0ε‖L∞
|a(ξ)|. We have
∂
∂t(f − χ0
ε) + a(ξ) · ∇x(f − χ0ε) + λ(f − χ0
ε) = (ft + a(ξ) · ∇xf + λf)− λχ0ε − a(ξ) · ∇xχ0
ε
= λ(χ(ξ, u)− χ0ε)− a(ξ) · ∇xχ0
ε .
Hence, f − χ0ε satisfies equation (4.1), with g = λ(χ(ξ, u) − χ0
ε) − a(ξ) · ∇xχ0ε and initial data
χ(ξ, u0)− χ0ε . We note that
‖g‖L1 ≤ λ‖χ(ξ, u)− χ(ξ, u0ε)‖L1 + ‖∇xχ0ε‖M1 sup
|ξ|≤‖u0ε‖L∞
|a(ξ)|
and
‖χ(ξ, u)− χ(ξ, u0ε)‖L1 = ‖u− u0ε‖L1 ≤ ‖f − χ0ε‖L1 .
From Theorem 4.2, we have
d
dt(‖f(t)− χ0
ε‖L1) + λ‖f − χ0ε‖L1 ≤ λ‖f − χ0
ε‖L1 +Nε‖∇xu0ε‖M1 .
Integrating in time thus gives
‖f(k)− χ(ξ, u0)‖L1 ≤ ‖χ(ξ, u0)− χ(ξ, u0ε)‖L1 + kNε‖∇xu0ε‖L1 = ‖u0 − u0ε‖L1 + kNε‖∇xu0ε‖M1 .
Letting ω1(k) = sup|h|≤k
‖u0(·+ h)− u0‖L1 , we have
‖u0 − u0ε‖L1 ≤∫ ∫
|u0(x− y)− u0(x)|ρε(y) dx dy ≤ ω1(ε).
By a similar argument, we obtain
‖∇xχ0ε‖M1 = ‖∇xu0ε‖L1 ≤ ‖∇ρε‖L∞ω1(ε) ≤ ‖∇ρ‖L
∞
εω1(ε).
Hence, we have
‖f(k)− χ(ξ, u0)‖L1 ≤(
1 +kNεε
)ω1(ε).
Setting ω(k) equal to the infimum of the right-hand side over all ε > 0 gives the desired modulus ofcontinuity. �
We now show the uniform integrability of uλ, which shows the compactness of the family.
Proposition 4.6. Let u0 ∈ L∞, and let N = sup|ξ|≤‖u0‖L∞
|a(ξ)|. Then
∫|x|≥R
|uλ(t, x)| dx ≤∫
|x|≥R2
|u0(x)| dx+CNt‖u0‖L1
R.
Note that we only have to consider u0 ∈ L∞ since, given u0 ∈ L1, we can always regularize byconvolution with some mollifiers such that the resulting convolution approximates u0.
20 YUZHOU ZOU
Proof. Let ϕ : Rd → R be a nonnegative smooth function satisfying ϕ(x) = 1 for |x| ≥ 1, ϕ(x) = 0for |x| ≤ 1
2 , and ‖ϕ‖L∞ = 1. Set ϕR(x) = ϕ(x/R). We have
∂
∂t(fλϕR) + a(ξ) · ∇x(fλϕR) + λfλϕR = ϕR(ft + a(ξ) · ∇xf + λfλ) + fλa(ξ) · ∇xϕR
= λχ(ξ, uλ)ϕR + fλa(ξ) · ∇xϕR.
From Theorem 4.2, we have
d
dt
(∫|fλ|ϕR dx dξ
)+ λ
∫|fλ|ϕR dx dξ ≤ λ
∫|χ(ξ, uλ)|ϕR dx dξ +
∫|fλ||a(ξ)||∇xϕR| dx dξ.
Since ∫|χ(ξ, uλ)| dx dξ =
∫|uλ| dx ≤
∫|fλ| dx dξ
and∫|fλ||a(ξ)||∇xϕR| dx dξ ≤ N‖∇ϕR‖L∞
∫|fλ| dx dξ ≤ N‖∇ϕ‖L∞
R‖uλ(t, ·)‖L1 ≤ CN
R‖u0‖L1
we have
d
dt
(∫|fλ|ϕR dx dξ
)≤ CN
R‖u0‖L1
and hence∫|x|≥R
|uλ(t, x)| dx ≤∫|fλ|ϕR dx dξ ≤
∫|χ(ξ, u0)|ϕR dx dξ+
CNt
R‖u0‖L1 ≤
∫|x|≥R2
|u0| dx+CNt
R‖u0‖L1 ,
as desired. �
We thus obtain a sequence λn and u ∈ L1 such that uλn → u in L1([0, T ]× Rd). The next stepis to show the convergence of {fλn}.
Proposition 4.7. fλn → χ(ξ, u) in L1([0, T ]× Rd × Rξ) for the sequence {λn} obtained above.
Proof. Assume first that u0 ∈ L∞. We use the representational formula to have
fλ(t, x, ξ)− χ(ξ, u(t, x)) = χ(ξ, u0(x− a(ξ)t))e−λt + λ
t∫0
e−λsχ(ξ, uλ(t− s, x− a(ξ)s)) ds
− χ(ξ, u(t, x))(e−λt + (1− e−λt))
= e−λt(χ(ξ, u0(x− a(ξ)t))− χ(ξ, u(t, x)))(4.4)
+ λ
t∫0
e−λs(χ(ξ, uλ(t− s, x− a(ξ)s))− χ(ξ, u(t, x))) ds.
ENTROPY AND KINETIC FORMULATIONS OF CONSERVATION LAWS 21
If we integrate the LHS over x and ξ, the first term in the RHS is bounded by 2‖f0‖L1e−λt, whilethe second term is bounded by
t∫0
λe−λs∫|χ(ξ, uλ(t− s, x− a(ξ)s))− χ(ξ, u(t− s, x− a(ξ)s))| dx dξ ds
+
t∫0
λe−λs∫|χ(ξ, u(t− s, x− a(ξ)s))− χ(ξ, u(t, x− a(ξ)s))| dx dξ ds
+
t∫0
λe−λs∫|χ(ξ, u(t, x− a(ξ)s))− χ(ξ, u(t, x))| dx dξ ds.
Substituting λ = λn and letting n→∞, we note that the first term is, after a change of variables,bounded by
t∫0
λne−λns
∫|χ(ξ, uλn(t− s, y))− χ(ξ, u(t− s, y))| dy dξ ds =
t∫0
λne−λns‖uλn(t− s, ·)− u(t− s, ·)‖L1 ds,
which vanishes as n → ∞ since uλn → u. The second term also vanishes as n → ∞ by a similarargument. To control the third term, we note that if u is locally Lipschitz and compactly supported,with N = sup
|ξ|≤‖u‖L∞
|a(ξ)| and supp u ⊂ [0, T ]×K, then
∫|χ(ξ, u(t, x− a(ξ)s))− χ(ξ, u(t, x))| dx dξ ≤
∫Rd
∫R
1ξ∈(u(t,x−a(ξ)s),u(t,x)) dx dξ
≤∫Rd
sup|y|≤Ns
|u(t, x− y)− u(t, x)| dx
≤ Ns∫K
‖∇xu‖L∞ dx = Ns|K|‖∇xu‖L∞
and hence
t∫0
λe−λs∫|χ(ξ, u(t, x− a(ξ)s))− χ(ξ, u(t, x))| dx dξ ds ≤ N |K|‖∇xu‖L∞
t∫0
λnse−λns ds
n→∞−−−−→ 0.
Otherwise, we can regularize u by space convolution and truncation to obtain a locally Lipschitzand compactly supported uδ such that ‖u(t, ·)− uδ(t, ·)‖L1 ≤ δ and supp uδ ⊂ B1/δ for each t, andthe integral with u can thus be controlled by controlling the integral with uδ. Hence, we see thatfλn converges to χ(ξ, u) if u0 is L∞.
For u0 not in L∞, we can simply regularize by convolution to obtain L∞ initial data {u0δ} such
that ‖u0δ−u0‖L1δ→0−−−→ 0, and similarly with u and {fλ}. The conclusion follows from the contraction
‖fλn,δ − fλ‖L1 ≤ ‖u0δ − u0‖L1 . �
We conclude that fλn → χ(ξ, u). To conclude the proof of existence, we note that the functionsmλ satisfy the uniform local bound
∞∫0
∫Rd×(−R,R)
mλ(t, x, ξ) dx dξ dt ≤ 2R max|ξ|≤R
µ(ξ) ≤ 2R‖u0‖L1 .
22 YUZHOU ZOU
Hence, we can extract a subsequence from {mλn} which converges weakly to some measure m, thusproving the existence of a solution to the kinetic formulation.
4.5. Uniqueness. To finish showing that the kinetic formulation of conservation laws is indeedwell-posed, we must show that kinetic solutions are unique and depend continuously (in L1) on theinitial data. Both can be shown by showing the following contraction principle:
Theorem 4.8. Let u1 and u2 be two kinetic solutions with corresponding initial data u01 and u02.Then ∫
Rd
|u(t, x)− v(t, x)| dx ≤∫Rd
|u01(x)− u02(x)| dx.
The proof will be sketched below without regard to regularity or rigor. A completely rigorousproof can be found in [5] and involves regularizing the χ functions by convolution in time and space.
Proof. Let m1 and m2 be the corresponding measures. Note that, for fixed (t, x), we have
|u1 − u2| =∫|χ(ξ, u1)− χ(ξ, u2)| dξ =
∫|χ(ξ, u1)− χ(ξ, u2)|2 dξ
=
∫|χ(ξ, u1)|2 + |χ(ξ, u2)|2 − 2χ(ξ, u1)χ(ξ, u2) dξ
=
∫|χ(ξ, u1)|+ |χ(ξ, u2)| − 2χ(ξ, u1)χ(ξ, u2) dξ
since |χ|2 = |χ| and |χ(ξ, u1)−χ(ξ, u2)| can only take the values 0 or 1. It thus suffices to show that
d
dt
(∫|χ(ξ, u1)|+ |χ(ξ, u2)| − 2χ(ξ, u1)χ(ξ, u2) dx dξ
)≤ 0.
If we multiply the equation ∂∂t (χ(ξ, u)) + a(ξ) · ∇xu = ∂m
∂ξ by sgn(ξ) and integrate, we obtain
d
dt
(∫|χ(ξ, u)| dx dξ
)=
∫sgn(ξ)
∂m
∂ξdx dξ = −2
∫m∣∣∣ξ=0
dx.
(As before, the term containing a(ξ) drops out after integration). It follows that
(4.5)d
dt
(∫|χ(ξ, u1)|+ |χ(ξ, u2)| dx dξ
)= −2
∫(m1 +m2)
∣∣∣ξ=0
dx.
Similarly, if we multiply the equation ∂∂t (χ(ξ, u1)) + a(ξ) · ∇xu1 = ∂m1
∂ξ by χ(ξ, u2), switch the roles
of u1 and u2, and add, we obtain
∂
∂t(χ(ξ, u1)χ(ξ, u2)) + a(ξ) · ∇x(χ(ξ, u1)χ(ξ, u2)) = χ(xi, u2)
∂m1
∂ξ+ χ(ξ, u1)
∂m2
∂ξ
and hence integration yields
d
dt
(∫2χ(ξ, u1)χ(ξ, u2) dx dξ
)= −2
∫m1
∂
∂ξ(χ(ξ, u2)) +m2
∂
∂ξ(χ(ξ, u1)) dx dξ
= −2
∫m1(δ(ξ)− δ(ξ − u2)) +m2(δ(ξ)− δ(ξ − u1)) dx dξ
= −2
∫(m1 +m2)δ(ξ) dx dξ + 2
∫m1δ(ξ − u2) +m2δ(ξ − u1) dx dξ
≥ −2
∫(m1 +m2)δ(ξ) dx dξ = −2
∫(m1 +m2)
∣∣∣ξ=0
dx.
Subtracting the above inequality from equation (4.5) gives the desired result. �
ENTROPY AND KINETIC FORMULATIONS OF CONSERVATION LAWS 23
5. Convergence in the Diffusion Approximation
We now prove convergence estimates for solutions of the conservation laws approximationut + divx(A(u)) = ε∆u, using the kinetic formulation. In this section, let u be a kinetic solution tout + divx(A(u)) = 0 with initial data u0, and let ω0 and ω1 be the time modulus of continuity of uand initial modulus of continuity of u0, respectively, i.e.
ω0(k) = sup0≤s≤k
‖u(s, ·)− u0‖L1
and
ω1(k) = sup|h|≤k
‖u0(·+ h)− u0‖L1 .
We have the following estimate:
Theorem 5.1. Suppose v ∈ L∞((0, T );L1(Rd)) satisfies the equation
∂
∂t(χ(ξ, v)) + a(ξ) · ∇x(χ(ξ, v)) =
∂m
∂ξ+ α+
∂β0∂t
+ divx(β) +∑i,j
∂2
∂xi∂xjγij
for some nonnegative measure m, with initial data v0, where α,β0,β, and γ satisfy
α, β, γ ∈M1((0, T )× Rd), β0 ∈ L∞((0, T );M1(Rd))
where α(t, x) = ‖α(t, x, ·)‖L1(R), and similarly for the other terms. Then, for T > 0 and anyε1, ε2 > 0, we have
‖v(T )− u(T )‖L1 ≤ ‖v0 − u0‖L1 + ω1(ε2) + ω0(ε1)
+ ‖α‖M1 +C
ε2‖β‖M1 +
C
ε22‖γ‖M1 +
(2 +
CT
ε1
)supt∈[0,T ]
∫¯β0(t, x) dx.
The estimate is proven in a similar fashion to that of the uniqueness theorem.We now wish to consider a solution uε to the diffusion problem
(uε)t + divx(A(uε)) = ε∆uε, uε(0, x) = u0(x)
and estimate the rate of convergence of uε to u as ε→ 0. We have
∂
∂t(χ(ξ, uε)) + a(ξ) · ∇x(χ(ξ, uε)) = δ(ξ − uε)((uε)t + a(ξ) · ∇xuε)
= δ(ξ − uε)((uε)t + divx(A(uε))) = δ(ξ − uε)ε∆uε.
We can apply Theorem 5.1 either with γij = εδijδ(ξ − uε)uε (δij being the Kronecker delta) and allother terms equal to zero, or with β = εδ(ξ − uε)∇xuε and all other terms zero. These lead to thefollowing results:
Theorem 5.2. For any ε′ > 0, we have
‖uε(T )− u(T )‖L1 ≤ ω1(ε′) +C
(ε′)2(Tε‖u0‖L1).
If, in addition, u0 ∈ BV , we have
‖uε(T )− u(T )‖L1 ≤ C‖u0‖TV√Tε.
Proof. If we let γij = εδijδ(ξ − uε)uε, then in Theorem 5.1 we can take ε1 = 0 (since all termsassociated with ε1 vanish) and ε2 = ε′. We thus have
‖uε(T )− u(T )‖L1 ≤ ω1(ε′) +C
(ε′)2‖γ‖M1 .
24 YUZHOU ZOU
Since ‖γij‖M1 = εδij‖uε‖M1 = εδijT∫0
‖uε(T )‖L1(Rd) dt ≤ Tεδij‖u0‖L1 , it follows that
‖γ‖M1 =d∑
i,j=1
‖γij‖M1 ≤ Tεd‖u0‖L1 , and the first result follows. If we assume u0 ∈ BV and take
β = εδ(ξ − uε)∇xuε, we can again take ε1 = 0 and ε2 = ε′, this time yielding
‖uε(T )− u(T )‖L1 ≤ ω1(ε′) +C
ε′‖β‖M1 ≤ ε′‖u0‖TV +
C
ε′‖β‖M1 .
Since ‖β‖M1 = ε∫δ(ξ − uε)|∇xuε| dξ dx dt = ε
T∫0
∫|∇xuε(t, x)| dx dx ≤ Tε‖u0‖TV , we obtain
‖uε(T )− u(T )‖L1 ≤ ‖u0‖TV(ε′ +
CTε
ε′
).
Minimizing the RHS with respect to ε′ yields the desired result. �
6. Compactness and Averaging Lemmas
We conclude by proving a result on the compactness of entropy solutions. To do so, we will usethe following averaging compactness theorem:
Theorem 6.1. Let R > 0, and let a satisfy the non-degeneracy condition
|{ξ : |ξ| < R, a(ξ) · ζ + α = 0}| = 0 ∀α ∈ R, ζ ∈ Sd−1.Consider the transport equation
ft + a(ξ) · ∇xf =
m∑k=0
∂k
∂ξk(divt,x,ξgk),
and let {fn}n and {gk,n}n be sequences of functions satisfying this equation. Let ψ ∈ C∞C (R),and let ρn(t, x) =
∫ψ(ξ)fn(t, x, ξ) dξ. If, for some 1 < q < ∞, the sequence {fn} is bounded in
Lq(R+ × Rd), and {gk,n} is relatively compact in Lq(R+ × Rd × Rξ,Rd+2), then the averages {ρn}are relatively compact in Lq(R+ × Rd).
The proof can be found in [4] or [5]. The idea is to take the Fourier transform with respect totime and space of the transport equation to obtain
(τ + a(ξ) · η)f =
m∑k=0
∂k
∂ξk
((τ, η,
∂
∂ξ
)· gk)
and hence
f =1
τ + a(ξ) · η
m∑k=0
∂k
∂ξk
((τ, η,
∂
∂ξ
)· gk).
where τ and η are the time and space Fourier variables, respectively. It follows that f decaysquickly (and hence ensures regularity) whenever the quantity τ + a(ξ) · η is far from zero, and thenon-degeneracy conditions ensure that this is indeed the case almost everywhere.
We can now prove the following compactness result on entropy solutions:
Theorem 6.2. Assume a satisfies the non-degeneracy condition in Theorem 6.1. Let {un} be asequence of entropy solutions to ut + divx(A(u)) = 0 with uniform L1 and L∞ bounds. Then {un}is locally relatively compact in Lp(R+ × Rd) for all 1 ≤ p <∞.
Remark 6.3. Notice that no compactness requirement is needed for the initial data, only that thesolutions be uniformly bounded. Furthermore, the non-degeneracy condition is needed for the result,to rule out the cases of transport equations of the form ut + divx(au) = 0, a ∈ Rd.
ENTROPY AND KINETIC FORMULATIONS OF CONSERVATION LAWS 25
Proof. Let χn = χ(ξ, un). From the kinetic formulation, we have ∂χn∂t + a(ξ) · ∇xχn = ∂mn
∂ξ . We
first localize the functions χn so that they are uniformly supported in t and x for every ξ in orderto attain convergence more easily, since we are only seek local compactness results on un.
More specifically, let 0 < t1 < t2, let K ⊂ Rd be compact, let φ1 ∈ C∞C (R+) satisfy φ1(t) = 1 fort1 ≤ t ≤ t2, let φ2 ∈ C∞C (Rd) satisfy φ2(x) = 1 for x ∈ K, and set fn(t, x, ξ) = φ1(t)φ2(x)χn. Then{fn} is uniformly supported in t and in x, and since {un} has a uniform L∞ bound, it follows that{χn} and hence {fn} are uniformly supported in ξ as well. We have
∂fn∂t
+ a(ξ) · ∇xfn = φ′1φ2χn + φ1φ2∂χn∂t
+ φ1(a(ξ) · ∇xφ2)χn + φ1φ2a(ξ) · χn
= φ1φ2∂mn
∂ξ+ (φ′1φ2 + φ1a(ξ) · ∇xφ2)χn
:=∂m1
n
∂ξ+m2
n
where m1n = φ1φ2mn and m2
n = (φ′1φ2 + φ1a(ξ) · ∇xφ2)χn are uniformly bounded measures withuniform compact support.
We now claim that there exist M1n and M2
n in W 1,1(R+ × Rd × Rξ,Rd+2) such that min =
divt,x,ξMin. Indeed, if we solve the equations ∆vin = mi
n, and set M in = ∆vin, then mi
n = divt,x,ξMin,
and the uniform bound on {min} implies that {vin} uniformly bounded in W 2,1, and hence {M I
n} isuniformly bounded in W 1,1. Since {M i
n} are also compactly supported, we can apply the Rellich-Kondrachov theorem to conclude that {M i
n}n is compact in Lq for 1 ≤ q < d+2d+1 .
Hence, if we fix some 1 < q < d+2d+1 , we can apply Theorem 6.1 to conclude that {ρn} is com-
pact in Lq(Rd+1), where ρn =∫ψ(ξ)fn(t, x, ξ) dξ, for any ψ ∈ C∞C (R). Since fn(t, x, ξ) = 0 for
|ξ| > sup ‖un‖L∞ , we can choose ψ so that ψ(ξ) = 1 for |ξ| ≤ sup ‖un‖L∞ , in which case
ρn(t, x) =
∫φ1(t)φ2(x)χ(ξ, un) dξ = φ1(t)φ2(x)un(t, x).
It follows that {φ1φ2un} is compact in Lq(Rd+1), and in particular {un} is compact in Lq(K ′),where K ′ = [t1, t2] × K. Hence, there exists u ∈ Lq(K ′) such that un → u, up to subsequence.Since {un} is uniformly bounded in L∞(K ′), it follows that u ∈ L∞(K ′). Note that we have thecontinuous injection Lp(K ′) ↪→ Lq(K ′)∩L∞(K ′) for all q < p <∞, and since K ′ has finite measure,we also have the continuous injection Lp(K ′) ↪→ Lq(K ′) for all 1 ≤ p < q. It follows that u ∈ Lp(K ′)and un → u in Lp(K ′) for all 1 ≤ p <∞. This shows that {un} is relatively compact in Lp(K ′) forall 1 ≤ p < ∞. Since every compact K ′ ∈ R+ × Rd is contained in [t1, t2] ×K for some 0 < t1 <2
and K ⊂ Rd compact, it follows that {un} is locally relatively compact in Lp for all 1 ≤ p <∞.�
Acknowledgments. I would like to thank Professor Takis Souganidis for his guidance as my pri-mary mentor in the study of conservation laws. I would also like to Casey Rodriguez for answeringmany of my questions regarding integration and distribution theory, and to Professor Peter May fororganizing the University of Chicago Mathematics REU.
References
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Sbornik. 1970. Vol. 10. No.2
[4] P.-L. Lions, B. Perthame, and E. Tadmor. A kinetic formulation of multidimensional scalar conservation laws andrelated equations. J. Amer. Math. Soc. 1994. Vol.7
[5] B. Perthame. Kinetic Formulation of Conservation Laws. Oxford University Press. 2002.
