NOTES FOR THE COURSE

“SYSTEMS OF CONSERVATION LAWS IN ONE SPACE

VARIABLE”

LAURA V. SPINOLO

Abstract. These are my notes are for a course I am giving in March 2013 in SISSA,Trieste, Italy. As you can see, they are very rough and schematic. Nonetheless, I wouldbe very grateful to anybody pointing out mistakes and misprints.The tentative program of the course is the following.

• Classical solutions, the theory of characteristics.• Distributional solutions, Rankine-Hugoniot conditions.• Admissibility criteria for distributional solutions.• Existence of global in time, admissible distributional solutions. The wave-front tracking

algorithm.• Kruzkov’s theorem.• Lax’s solution of the Riemann problem and the Standard Riemann Semigroup.

These notes contain no new material and almost no references to original research papers.

Most of the material is taken from the books [1, 2, 6, 7], which also contain and extended

list of references.

1. Introduction

1.1. What is a system of conservation laws in one space variable?

1.1.1. The equation. A system of conservation laws in one space variable is a partialdifferential equation in the form

∂tU + ∂x[F (U)

]= 0, (1.1)

where

• (t, x) ∈]0,+∞[×R.• The unknown is U :]0,+∞[×R→ Rn.• The flux F : Rn → Rn is of class C2.

Throughout this course I will focus on the Cauchy problem obtained by coupling (1.1)with the Cauchy datum

U(0, x) = U0(x), (1.2)

where U0 is a given function U0 : R→ Rn.1

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1.1.2. Interpretation. To understand why equations in the form (1.1) are termed con-servation laws we can argue as follows. Assume that the solution of (1.1) is regular,then (1.1) implies that, given a, b ∈ R, a < b, we have

d

dt

∫ b

aU(t, x)dx =

∫ b

a∂tU(t, x)dx = −

∫ b

a∂x[F (U)

]dx = −F (U(b)) + F (U(a)).

The heuristic interpretation is that the rate of change in the amount of U “contained”in the interval ]a, b[ only depends on the inflow at a and on the outflow at b, and doesnot depend on dissipation phenomena that could in principle occur in between.

Some examples. The Euler equations of fluid dynamics are used to describe the mo-tion of particles in a fluid with zero viscosity. In the case when the space variable isone dimensional they take the form

∂tρ+ ∂x(ρv) = 0 conservation of mass

∂t(ρv) + ∂x(ρv2 + p) = 0 conservation of linear momentum

∂t(ρ[e+1

2v2]) + ∂x(ρv[e+

1

2v2] + pv) = 0 conservation of energy.

The unknowns u = (ρ, v, e) represent the density of the fluid, the velocity of theparticle of the fluids and the internal energy, respectively. The function p = p(ρ, e)is the pressure. The law p = p(ρ, e) is the equation of state and depends on the gasunder consideration.

Systems of conservation laws have several other applications (see the book byDafermos [2] for an extended discussion), for instance they are used to model bothvehicular and pedestrian traffic.

Finally, the simplest example of nonlinear conservation law is perhaps the so-calledBurgers’equation,

∂tu+ ∂x

(1

2u2

)= 0. (1.3)

1.2. Conservation laws in several space variables.

1.2.1. Scalar conservation law. A scalar conservation in several space variables is apartial differential equation in the form

∂tu+ divx[G(u)

]= 0, (1.4)

where u :]0,+∞[×Rd → R is the unknown and the flux G : R → Rd is of class C2.The symbol divx denotes the divergence with respect to the space variable only.

The interpretation is the same as before (provided we use the Gauss-Green formulainstead of the Fundamental Theorem of Calculus): if u is smooth, then for everyregular set Ω ⊆ Rd we have

d

dt

∫Ωu(t, x)dx = −

∫Ω

divx[G(u)

]= −

∫∂ΩG(u) · ~n dHd−1,

where ~n is the outwward pointing unit normal vector to ∂Ω. Namely, the rate ofchange in the amount of u “contained” in Ω only depends on the behavior of u at ∂Ω,

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and does not depend on dissipation phenomena that could in principle occur insideΩ.

1.2.2. Systems of conservation law in several space variables. Systems of conservationlaws in several space variables are equations in the form

∂tU +

d∑i=1

∂xi[F i(U)

]= 0, (1.5)

where U :]0,+∞[×Rd → Rn is the unknown and the flux functions F i : Rn → Rn ,i = 1, · · · , d are of class C2.

This is the most general case: the cases (1.1) and (1.4) can be obtained by takingd = 1 and n = 1, respectively. It is also the most interesting case, since for examplethe natural environment where to define the Euler equations is d = 3, n = 5.

1.2.3. Main results known so far. In the following I term “well-posedness theory” aresult establishing existence and uniqueness of a suitable notion of solution. Thatbeing said, the state of the art for conservation laws is, very roughly speaking, thefollowing.

• For scalar conservation laws in several state variables (1.4), there is a well-established and quite satisfactory well-posedness theory which dates back tothe pioneering work by Kruzkov [4].• For systems of conservation laws in one space variables (1.1), existence of

global in time solutions was first established by Glimm [3]. Uniqueness resultsare much more recent and are mainly due to Bressan and several collaborators(see the book by Bressan [1] for an overview). Note, however, that the well-posedness theory which is presently available is not completely satisfactoryin view of applications because it requires quite restrictive conditions on theinitial data.• The case of systems of conservation laws in several space variables (1.5) is, to

a large extent, still very poorly understood. No global existence or uniquenessresult is presently available for general systems. Also, recent developmentssuggest that the situation is much more complex than in the case of one spacevariable.

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2. Classical solutions

From now on I will focus on the case of systems of conservation laws in one spacevariable (1.1). A classical solution of (1.1) is a C1 function that pointwise satisfiesthe equation.

A very powerful tool for the analysis of classical solutions relies is the method ofcharacteristics. For simplicity, we mostly focus on the scalar equation

∂tu+ ∂x[f(u)

]= 0, (2.1)

which is obtained from (1.1) in the case when n = 1.

2.1. The method of characteristics for the scalar equation (2.1). In this para-graph we always assume that u is a classical solution of (2.1). We denote by X(t, y)the solution of the Cauchy problem

dX

dt= f ′

(u(t,X)

)X(0, y) = y

(2.2)

and we term X(t, y) the characteristic curve starting from y.

2.2. Example. Consider the scalar, linear system

∂tu+ a ∂xu = 0 a ∈ R. (2.3)

Then the characteristic curve (2.2) is the line X(t, y) = at + y. Note that, if u is asolution of (2.3), then u is constant along X(t, y). Indeed,

d

dt

(u(t, at+ y)

)= ∂tu+ a∂xu = 0.

This result actually extends to the general nonlinear case.

2.3. Lemma. Let u0 ∈ C1c (R) and let u : [0, T [×R → R be a classical solution of the

Cauchy problem ∂tu+ ∂x

[f(u)

]= 0

u(0, x) = u0(x).(2.4)

Then u is constant along the characteristic curve X(t, y) for every y ∈ R. Thisimplies, in particular, that the characteristic curve is actually the line

X(t, y) = y + f ′(u0(y)

)t. (2.5)

Proof. We first show that u is constant along the characteristic curve X(t, y). Werewrite the first line of (2.4) as

∂tu+ f ′(u)∂xu = 0. (2.6)

By using the chain rule, we get

d

dt

[u(t,X(t)

)]= ∂tu+ ∂xu

dX

dt= ∂tu+ ∂xu f

′(u) = 0,

which allows us to conclude. Equality (2.5) immediately follows.

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2.4. Example. Consider the Burgers equation (1.3), then f ′(u) = u and hence theequation of the characteristic line X(t, y) is

X(t, y) = y + u0(y)t.

The following result ensures that the family of characteristic lines covers [0, T [×R,provided that T is sufficiently small.

2.5. Lemma. There is a constant T > 0, only depending on the datum u0 ∈ C1c (R)

and on the function f , such that the following holds. If t < T (u0, f), then for everyx ∈ R there is a unique y ∈ R such that X(t, y) = x.

Proof. We fix t > 0 and we consider the map X(t, ·) : R→ R, X(t, y) = y+f ′(u0(y))t.Note that the image of R through X(t, ·) is an interval by the continuity of X(t, ·).By using formula (2.5) and recalling that u0 is compactly supported, we obtain that

limy→−∞

X(t, y) = limy→−∞

y + f ′(u0)t = −∞ and limy→+∞

X(t, y) = +∞.

Hence, the image of R through X(t, ·) is the whole R. Also,

∂X

∂y= 1 + f ′′(u0)u′0(y)t

and hence ∂X/∂y > 0 if t is sufficiently small. We infer that X(t, ·) is invertible andthis concludes the proof of the lemma.

By combining Lemma 2.5 and 2.3, we obtain a very important property: equa-tion (2.1) has finite propagation speed. To understand what this means, we considerthe Cauchy problem (2.4) and we assume that suppu0 ⊆]−M,M [ and that f ′(0) = α.If u : [0, T [×R→ R is a classical solution of (2.4), then

u(t, x) ≡ 0 if x < −M + αt and x > M + αt.

Loosely speaking, we can say that the “information” u0 6= 0 travels with finite prop-agation speed and does not instantaneously propagate to the whole R.

Note that this behavior is dramatically different from that of other equations: forexample, consider the Cauchy problem for the heat equation

∂tv = ∂xxvv(0, x) = v0(x)

and assume that v0 ∈ C1c (R) and that v0(x) ≥ 0 for every x ∈ R. If v0 is not identically

zero, then the solution v satisfies v(t, x) > 0 for every x ∈ R and every t > 0. In thecase of the heat equation, the “information” v0 > 0 instantaneously propagates tothe whole R.

2.6. Local in time existence and uniqueness of classical solutions of thescalar equation (2.1). We can use the method of characteristics to establish localin time existence and uniqueness of classical solutions. We start with existence.

2.7. Proposition. Let u0 ∈ C1c (R), then there is T > 0, only depending on f and u0,

such that the Cauchy problem (2.4) admits a classical solution u : [0, T [×R→ R.

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As it is clear from the proof, the hypothesis that u0 is compactly supported canbe very much relaxed.

Proof. The heuristic idea is the following. To define the value u(t, x), we move back-wards along the characteristic line that at time t passes through the point x. We de-note by Y (t, x) the point we have reached at time 0 and we set u(t, x) := u0(Y (t, x)).We now give the formal proof.

Let T be as in the statement of Lemma 2.5 and for t ∈ [0, T [ denote by Y (t, ·) theinverse function of the function X(t, ·), namely

X(t, Y (t, x)

)= x. (2.7)

By Lemma 2.3, the only possible candidate for being a solution is u(t, x) :=u0

(Y (t, x)

). Since the function u is of class C1, we are only left to prove that it

satisfies equation (2.6). We compute

∂tu = u′0(Y )∂Y

∂tand ∂xu = u′0(Y )

∂Y

∂x. (2.8)

By using relation (2.7), we get

∂X

∂t+∂X

∂y

∂Y

∂t= 0

and, by recalling that ∂X/∂t = f ′(u0(Y )

), we finally get

∂Y

∂t= −

(∂X

∂y

)−1

f ′(u0)

By recalling (2.8), we finally get that (2.6) is satisfied.

We now establish uniqueness.

2.8. Proposition. Let u0 ∈ C1c (R) and assume that u1, u2 : [0, τ [×R → R are two

classical solutions of the same Cauchy problem (2.4). Then u1(t, x) = u2(t, x) forevery (t, x) ∈ [0, τ [×R.

Again, by looking at the proof we realize that the assumption that u0 is compactlysupported can be very much relaxed.

Proof. The heuristic idea is that classical solutions must be constant along character-istic lines, whose equation only depends on the initial datum and on the flux. Hence,there is only “one possibility” and we cannot have more than one solution.

To achieve a formal proof, we set

t := supt ∈ [0, τ [: u1(t, x) = u2(t, x) for every x ∈ Rand observe that the set of which t is the supremum is nonempty since it contains 0.

By contradiction, we assume t < τ and we observe that this implies u1(t, x) =u2(t, x) for every x ∈ R. Indeed, if there were x ∈ R such that u1(t, x) 6= u2(t, x) thiswould violate the continuity of the map t 7→ u1(t, x)− u2(t, x).

We set u(x) := u1(t, x) = u2(t, x) and we observe that, due to Lemma 2.5, there isT > 0, only depending on u and f , such that t+T < τ and, for every (t, x) ∈ [t, t+T [,there is a unique characteristic line passing X through (t, x). By Lemma 2.3, both

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u1 and u2 are constant along this characteristic line and hence u1(t, x) = u2(t, x) forevery (t, x) ∈ [t, t + T [. This contradicts the definition of t and hence concludes theproof of the Lemma.

2.9. The method of characteristics for systems. Despite higher technical diffi-culties, the method of characteristics can be also applied to the analysis of a system ofconservation laws (1.1), provided that the system satisfies an hypothesis of so-calledstrict hyperbolicity.

2.10. Definition. System (1.1) is strictly hyperbolic if, for every U ∈ Rn, the Jacobianmatrix DF (U) admits n real and distinct eigenvalues

λ1(U) < λ2(U) < · · · < λn(U).

Note that strict hyperbolicity is trivially satisfied in the case of a single equa-tion (2.1) when n = 1.

We define the i-th characteristic curve Xi(t, y) by settingdXi

dt= λi(U(t,Xi))

Xi(0, y) = y .(2.9)

By using the method of characteristics combined with a suitable iteration scheme,one can establish local in time existence and uniqueness of classical solutions. See forinstance the book by Bressan [1, pp. 67-70].

2.11. Theorem. Assume that U0 ∈ C1c (R;Rn) and that the system (1.1) is strictly

hyperbolic. Then there is T > 0 such that the Cauchy problem (1.1)-(1.2) admits aclassical solution U : [0, T [×R→ Rn.

Note that, as in the scalar case, the hypothesis that U0 has compact support canbe very much relaxed.

2.12. Breakdown of classical solutions. We now want to show that, in general,classical solutions are only defined on a finite time interval.

To see this, we focus on the Cauchy problem for the scalar equation (2.4) and werecall that classical solutions are constant along the characteristic lines. In particular,let us consider the following Cauchy problem for the Burgers’ equation:

∂tu+ ∂x

(u2

2

)= 0

u(0, x) =1

1 + x2.

(2.10)

We recall that for the Burgers equation the equation of the characteristic line X(t, y)is

X(t, y) = y + u0(y)t.

and we infer the following considerations.

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• The characteristic line starting at y = 0 is

x = u(0)t+ 0 = t.

and u constantly attains the value 1 on this line.• The characteristic line starting at y = 1/2 is

x = u(1/2)t+ 1/2 =4

5t+

1

2.

and u constantly attains the value 4/5 on this line.• The two characteristic lines cross at (t, x) = (5/2, 5/2). This apparently gives

a contradiction: indeed, one should have that u(5/2, 5/2) is both 1 (becausethe point is on the first characteristic line) and 4/5 (because the point is onthe second characteristic line).• The contradiction is solved by observing that to prove that the solution is

constant along the characteristic line we had to a priori assume that thesolution is classical.

The previous example establishes the following theorem.

2.13. Theorem. Even if U0 ∈ C∞(R;Rn), classical solutions of the Cauchy prob-lem (1.1)-(1.2) are in general only defined on a finite time and breakdown in finitetime.

An heuristic interpretation of the classical solutions breakdown for (2.10) is thefollowing. If (t, x) is a point on the characteristic line (2.2), then

u(t, x) = u0

(x− f ′(u0(y))t

),

namely the value u0(y) “travels” in space-time with velocity f ′(u0(y)). In the caseof (2.10), the velocity is u0(y) and hence it is 1 at y = 0 and 4/5 at y = 1/2. Hence,the value u(0) is “faster” than the value u(1/2) and hence the graph of u(t, ·) getssteeper and steeper as t increases, until it develops a discontinuity and hence theclassical solution ceases to exist.

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3. Weak solutions

3.1. Definition of weak solution. Motivated by the fact that, in general, classicalsolutions are only defined on a finite time interval, we introduce a notion of weaksolution.

3.2. Definition. A weak solution of the system of conservation laws (1.1) is a locallyintegrable function U : [0,+∞[×R→ Rn satisfying the following two requirements.

i) The function U is a distributional solution of (1.1). Namely, for every testfunction ϕ ∈ C∞c (]0,+∞[×R), we have∫ T

0

∫RU∂tϕ+ F (U)∂xϕdxdt = 0. (3.1)

ii) The function t 7→ U(t, ·) is continuous from [0,+∞[ in L1loc(R) endowed with

the strong topology.

Some remarks are here in order.

• Condition ii) rules out some “pathological” example, see Serre [6, Exercise2.11].• Also, note that, if U is only a locally integrable function, then the functionU(0, ·) is not, a priori, well-defined because the set (0, x), x ∈ R is negligiblein R2. However, condition ii) implies that the function U(0, ·) is well definedand hence we can prescribe U(0, ·) = U0 and obtain a “weak formulation” ofthe Cauchy problem (1.1)-(1.2). One could actually give a meaning to U(0, ·)without imposing condition ii), but this would require some more work.• Any classical solution is a weak solution: in particular, condition (3.1) is

verified because of the Integration by Parts Formula.

3.3. Rankine-Hugoniot conditions. As we have seen, heuristic considerations sug-gest that distributional solutions are, in general, discontinuous. To investigate thisissue, we focus on the simplest possible discontinuous function in the (t, x)−plane.Let λ ∈ R, U+, U− ∈ Rn and consider the function W : [0,+∞[×R → Rn which isdefined by setting

W (t, x) :=

U+ x > λtU− x < λt.

(3.2)

Then we have the following

3.4. Lemma. The function W : [0,+∞[×R → Rn is a distributional solution of thesystem of conservation laws (1.1) if and only if

F (U+)− F (U−) = λ(U+ − U−

)(3.3)

This system of n equations is usually referred to as Rankine-Hugoniot conditions.

Proof. For simplicity, we only give the proof in the case when n = 1, but the argumentstraightforwardly extends to the case of systems.

We decompose the set [0,+∞[×R as

[0,+∞[×R = Λ− ∪ Λ+ ∪ L,

10

whereΛ− = (t, x) : x < λt, Λ+ = (t, x) : x > λt

and L is the half lineL := (t, x) : x = λt.

The function W is a distributional solution of (1.1) if and only if it satisfies (3.1) forevery ϕ ∈ C∞c (]0,+∞[×R). Since the set L is negligible, then (3.1) can be rewrittenas ∫∫

Λ−

W∂tϕ+ F (W )∂xϕdxdt+

∫∫Λ+

W∂tϕ+ F (W )∂xϕdxdt = 0

and, by recalling the expression of W (3.2), this is equivalent to∫∫Λ−

U−∂tϕ+ F (U−)∂xϕdxdt+

∫∫Λ+

U+∂tϕ+ F (U+)∂xϕdxdt = 0. (3.4)

Next, we observe that the function

Υ : Λ− → R× R(t, x) 7→

(U−ϕ, F (U−)ϕ

)is of class C∞. Also,

divt,xΥ = U−∂tϕ+ F (U−)∂xϕ

and by applying the Gauss-Green formula and recalling that ϕ is compactly sup-ported, we get∫∫

Λ−

U−∂tϕ+ F (U−)∂xϕdxdt =

∫∫Λ−

divt,xΥ =

∫L

Υ · ~n dH1,

where ~n is the unit outward pointing normal vector to L. Since ~n = (−λ, 1)/√

1 + λ2,we eventually get∫∫

Λ−

U−∂tϕ+ F (U−)∂xϕdxdt =1√

1 + λ2

∫Lϕ(− λU− + F (U−)

)dH1. (3.5)

We now compute the integral over Λ+ by following exactly the same argument asbefore, the only difference is that the outward pointing unit normal vector is now(λ,−1)/

√1 + λ2. We eventually get∫∫

Λ−

U−∂tϕ+ F (U−)∂xϕdxdt =1√

1 + λ2

∫Lϕ(λU+ − F (U+)

)dH1. (3.6)

By plugging (3.5) and (3.6) into (3.4) we get∫Lϕ(λ(−U− + U+) + F (U−)− F (U+)

)dH1 = 0. (3.7)

If conditions (3.3) are satisfied, then (3.7) holds and hence W is a distributionalsolution of (1.1). Conversely, if W is a distributional solution then by (3.7) andthe arbitrariness of the function ϕ we deduce (3.3). This concludes the proof of thelemma.

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3.5. Non-uniqueness of distributional solutions. We now show that a givenCauchy problem has, in general, infinitely many distributional solutions. Considerthe Cauchy problem

∂tu+ ∂x

(u2

2

)= 0

u(0, x) =

1 if x > 00 if x < 0

(3.8)

We now fix α ∈]0, 1[ and consider the function uα which is defined by setting

uα(t, x) =

0 x <

α

2t

αα

2t < x <

1 + α

2t

1 x >1 + α

2t.

(3.9)

We have the following result.

3.6. Lemma. For every α ∈]0, 1[, the function uα defined as in (3.9) is a distributionalsolution of the Cauchy problem (3.8).

Proof. We proceed in 2 steps.Step 1: we observe that the map t 7→ uα(t, ·) is continuous in L1

loc(R) endowed withthe strong topology. Also, we prove that uα satisfies (3.1) if and only if the Rankine-Hugoniot conditions are verified at both discontinuities x = αt/2 and x = (1 +α)t/2.The details are left as an exercise.

Step 2: we check that the Rankine-Hugoniot conditions (3.3) are satisfied by bothshocks. For the discontinuity at x = αt/2, the Rankine-Hugoniot conditions reduceto

1

2(α2 − 02) =

α

2(α− 0),

which is satisfied. For the discontinuity at x = (1 + α)/2, the Rankine-Hugoniotconditions reduce to

1

2(1− α2) =

1 + α

2(1− α),

which is as well satisfied.

The following theorem is an immediate consequence of Lemma 3.6.

3.7. Theorem. A given Cauchy problem (1.1)-(1.2) admits, in general, infinitelymany weak solutions.

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4. Admissibility criteria

4.1. Introduction. If we sum up what we have obtained so far by investigating thewell-posedness of the Cauchy problem (1.1)-(1.2), we obtain the following.

• If we only consider classical solutions, then we cannot hope for global existenceresults because in general classical solutions breakdown in finite time.• On the other side, if we consider weak solutions then we cannot hope for

uniqueness results because a given Cauchy problem admits, in general, infin-itely many weak solutions.

In the attempt at establishing a well-posedness (existence and uniqueness) theory,various admissibility criteria have been introduced in the literature (see in particu-lar Dafermos [2] for an extended discussion). Introducing an admissibility criterionamounts to augmenting the definition of weak solution with additional conditions inan attempt at selecting uniqueness. In this section we go over two of the most popularadmissibility criteria. Note that the fact that these criteria actually work (namely,that they actually succeeds at selecting a unique solution) is far from being trivialand it is the object of current investigation.

4.2. Entropy admissibility criterion. We first introduce the definition of entropyadmissible solution and we then make some related remark.

4.3. Definition. An entropy-entropy flux pair for system (1.1) is a couple of C1

functions (η, q), η : Rn → R, q : Rn → R, such that

∇η(U) ·DF (U) = ∇q(U) ∀U ∈ Rn. (4.1)

Note that (4.1) is a system of n equations in 2 unknowns (η and q). Hence,when n > 2 it is overdetermined and admits no solution. However, systems withphysical applications typically admit entropy-entropy flux pairs. One usually definesthe mathematical entropy η as the opposite of the physical entropy.

The following results states that, for classical solutions, the total entropy is con-served.

4.4. Lemma. Let U be a classical solution of (1.1). Then

∂tη(U) + ∂xq(U) = 0. (4.2)

Proof. Equality (4.2) is an immediate consequence of the chain rule. Indeed, byusing (4.1) and the regularity of the function U we obtain

∂tη(U) + ∂xq(U) = ∇η(U) · ∂tU +∇q(U)∂xU = ∇η(U)(∂tU +DF (U)∂xU

)= 0.

Here is the definition of entropy admissible solution of (1.1).

4.5. Definition. A locally integrable function U : [0,+∞[×R → Rn is an entropyadmissible solution of (1.1) if the following conditions are all satisfied.

(1) System (1.1) admits at least an entropy-entropy flux pair (η, q) with η convex.(2) U is a weak solution of (1.1).

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(3) For every entropy-entropy flux pair (η, q) with η convex, the inequality

∂tη(U) + ∂xq(U) ≤ 0 (4.3)

is satisfied in the sense of distributions. Namely,∫ +∞

0

∫Rη(U)∂tϕ+ q(U)∂xq(U) ≥ 0 (4.4)

for every ϕ ∈ C∞c (]0,+∞[×R) such that ϕ(t, x) ≥ 0 for every (t, x).

As mentioned above, the mathematical entropy is usually defined as the oppositeof the physical entropy. Hence, (4.3) is consistent with the Second Law of Thermody-namics since it prescribes that the total (mathematical) entropy should not increase,or equivalently that the total physical entropy should not decrease.

4.6. Entropy admissible discontinuities. Lemma 4.4 ensures that classical solu-tions are always entropy admissible. It is not, in general, immediate to determinewhether or not a given weak solution is entropy admissible. The following lemmaestablishes a simple criterium valid in the case when the weak solution is in theform (3.2).

4.7. Lemma. Given λ ∈ R, U+, U− ∈ Rn, let W be the function defined as in (3.2).Then W is an entropy admissible solution of (1.1) if and only if the following condi-tions are both satisfied.

1. The Rankine-Hugoniot conditions are satisfied, namely (3.3) holds.2. There is at least an entropy-entropy flux pair (η, q) with η convex. Also, for

any such pair we have

λ[η(U+)− η(U−)

]≥ q(U+)− q(U−). (4.5)

The proof follows the same argument we used in the proof of Lemma 3.4. Byrelying on Lemma 4.7 we obtain the following.

4.8. Lemma. Let uα be the distributional solution of (3.8) defined as in (3.9). Forevery α ∈]0, 1[, uα is not an entropy admissible solution.

Proof. We set

η(u) :=1

2u2, q(u) :=

1

3u3

and we verify that (η, q) is an entropy-entropy flux pair:

η′(u)f ′(u) = u · u = q′(u).

We recall that the Rankine-Hugoniot conditions are verified at both discontinuitylines. One easily infers that the weak solution uα is entropy admissible if and only ifthe discontinuities at x = αt/2 and x = (1 + α)t/2 both satisfy (4.5). Let us verifythat condition (4.5) is violated by the discontinuity at x = αt/2:

α2

4=α

2

(1

2α2 − 1

202)

1

3α3 − 1

303 =

α3

3.

One can actually verify that also the discontinuity at x = (1 + α)t/2 violates condi-tion (4.5).

14

4.9. Lax admissibility criterion. We now introduce an admissibility criterion in-troduced by P. Lax in [5]. This criterion only applies to solutions that are (locally)in the form (3.2). On the other hand, it applies to all systems, namely no existenceof entropy-entropy flux pairs with convex entropies is required. Also, testing it isparticularly simple.

4.10. Definition. Given λ ∈ R, U+, U− ∈ Rn, consider a function W in theform (3.2) and assume that the Rankine-Hugoniot conditions (3.3) are satisfied. Also,assume that the system (1.1) is strictly hyperbolic in the sense of Definition 2.10. Thefunction W is a Lax admissible discontinuity for (1.1) if

λi(U+) ≤ λ ≤ λi(U−) (4.6)

for some i = 1, · · · , n. In the previous expression, λi(U) is the i−th eigenvalue of theJacobian matrix DF (U).

An heuristic justification of the Lax admissibility criterion is the following. Letus for simplicity focus on the scalar case (2.1), then the Lax admissibility criterionreduces to

f ′(u+) ≤ λ ≤ f ′(u−). (4.7)

We recall that the characteristic lines have slope f ′(u) and are lines along whichthe solution is constant. In other words, the “information” (i.e., the solution u) istransported along the characteristic lines. Note that requiring Lax condition (4.7)amounts to impose that the characteristic lines are running into the discontinuityline x = λt. Hence, heuristically speaking Lax condition expresses the fact that, inphysical systems, information can be only destroyed and cannot be created out ofnowhere.

Finally, let us go back to the example of the Cauchy problem (3.8) and observethat, for every α ∈]0, 1[, the solution uα defined as in (3.9) is not Lax admissible.Indeed, for the discontinuity at x = αt/2 we have

λ(u+) = u+ = α α

2 λ(u−) = u− = 0

and similarly for the discontinuity at x = (1 + α)t/2 we have

λ(u+) = 1 1 + α

2 λ(u−) = u− =

α

2.

15

5. Weak, admissible solutions of a scalar conservation law

This section is devoted to the analysis of weak, admissible solutions of the scalarconservation law (2.1).

5.1. Rankine-Hugoniot conditions. In the case of the scalar equation (2.1), theRankine-Hugoniot conditions can be reformulated by imposing

λ =f(u+)− f(u−)

u+ − u−if u− 6= u+. (5.1)

This condition has a simple geometric interpretation: λ must be equal to the angularcoefficient of the line connecting the points (u−, f(u−)) and (u+, f(u+)) in R2. Thisline is in general secant to the graph of f .

5.2. Entropy admissible solutions of the scalar conservation law (2.1). First,we observe that in the case n = 1, condition (4.1) defining entropy-entropy flux pairsreduces to

η′(u)f ′(u) = q′(u) ∀u ∈ R.Hence, any C1 function is an entropy for (2.1) and the associated flux is

q(u) =

∫ u

0η′(θ)f ′(θ)dθ.

We thus have plenty of entropy-entropy flux pairs (η, q) with η convex.We now introduce a family of entropy-entropy flux pairs first considered by

Kruzkov [4].

5.3. Definition. Given k ∈ R, a Kruzkov entropy-entropy flux pair is a couple offunctions (η, q), η, q : R→ R in the form

ηk(u) := |u− k| qk(u) := sign(u− k) ·(f(u)− f(k)

)k ∈ R. (5.2)

Note that an Kruzkov entropy-entropy flux pair is not, strictly speaking, anentropy-entropy flux pair in the sense of Definition 4.3 because the functions arenot of class C1. However, we have the following characterization.

5.4. Lemma. Let u : [0,+∞[×R → R be a measurable bounded function. Then thefollowing two conditions are equivalent.

i) u is an entropy admissible solution of (2.1).ii) The map t 7→ u(t, ·) is continuous from [0,+∞[ to L1

loc(R) and, for everyk ∈ R and every ϕ ∈ C∞c (]0,+∞[×R) satisfying ϕ ≥ 0, we have∫ +∞

0

∫R|u− k|∂tϕ+ sign(u− k) ·

(f(u)− f(k)

)∂xϕdxdt ≥ 0. (5.3)

Note that, in particular, Lemma 5.4 implies that from inequality (5.3) we inferthat u is a distributional solution of (2.1), namely∫ T

0

∫Ru∂tϕ+ f(u)∂xϕdxdt = 0 ∀ϕ ∈ C∞c (]0,+∞[×R).

The proof of Lemma 5.4 relies on an approximation argument, see for instance thebook by Serre [6, pp.34-35].

16

We now focus on single-discontinuity functions in the form

w(t, x) =

u+ x > λtu− x < λt,

(5.4)

where λ, u+ and u− are given real numbers. By using Lemma 5.4 and mimicking theproof of Lemma 3.4 (see also Lemma 4.7), we obtain the following characterization.

5.5. Lemma. Let w be as in (5.4), then w is an entropy admissible solution of (2.1)if and only if

λ(|u+−k|− |u−−k|

)≥ sign(u+−k) ·

(f(u+)−f(k)

)− sign(u−−k) ·

(f(u−)−f(k)

)(5.5)

for every k ∈ R.

Note that if we choose k > maxu+, u−, then (5.5) implies

λ(u+ − u−) ≥ f(u+)− f(u−).

Conversely, if k < minu+, u−, then (5.5) implies

λ(−u+ + u−) ≥ −f(u+) + f(u−)

and hence from (5.5) we recover the Rankine-Hugoniot conditions (5.1).

5.6. Entropy admissible solutions of a scalar conservation law with convexflux. We now want to investigate the admissibility criterion in the case of a scalarconservation law with strictly convex flux. Note that the weak solution uα defined asin (3.9) is neither entropy admissible nor Lax admissible. This is not a case, indeedwe have the following simple characterization.

5.7. Lemma. Consider the scalar conservation law (2.1) and assume that f ′′ > 0.Given λ, u+, u− ∈ R, u+ 6= u−, consider the scalar consideration law (2.1) and thefunction w defined as in (5.4). Then the following conditions are equivalent.

i) The Rankine-Hugoniot condition (5.1) is satisfied and w is Lax admissible.ii) The function w is an entropy admissible solution of (2.1), namely (5.5) holds.iii) The Rankine-Hugoniot condition (5.1) is satisfied and u+ < u−.

Proof. Step 1: we prove that i) ⇐⇒ iii). For a scalar equation, Lax admissibilitycondition reduces to

f ′(u+) ≤ f(u+)− f(u−)

u+ − u−≤ f ′(u−). (5.6)

Since f ′′ > 0, if condition (5.6) is satisfied then u+ < u−. Also, if u+ < u− thencondition (5.6) is satisfied because the different quotients of a convex function arenondecreasing.

Step 2: we prove that ii) ⇐⇒ iii). We recall that w is an entropy admissiblesolution of (2.1) if and only if (5.5) holds. Also, we recall the discussion after thestatement of Lemma 5.5 and we observe that the Rankine-Hugoniot condition (5.1)is equivalent to the validity of (5.5) for both k < minu+, u− and k > maxu+, u−.

Hence, we are left to investigate the implications of (5.5) when k = θu+ +(1−θ)u−for θ ∈]0, 1[.

17

If u+ < u−, then (5.5) reduces to

f(u−)− f(u+)

u− − u+

(2k − u+ − u−

)≥(2f(k)− f(u+)− f(u−)

)and by multiplying both members by the positive quantity u− − u+ we get[

f(u−)− f(u+)][

2k − u+ − u−]≥[2f(k)− f(u+)− f(u−)

](u− − u+).

We now write k as k = θu− + (1− θ)u+ and by using

f(u−)(2k − u+ − u− + u− − u+

)= 2f(u−)(1− θ)(u− − u+)

andf(u+)

(u+ + u− − 2k + u− − u+

)= 2f(u+)θ(u− − u+)

we get(1− θ)f(u−) + θf(u+) ≥ f(θu+ + (1− θ)f(u−)),

which is satisfied for every α ∈]0, 1[ since f is convex.If u− < u+ then (5.5) reduces to(

f(u+)− f(u−))(u− + u+ − 2k

)≥(f(u−) + f(u+)− 2f(k)

)(u+ − u−

)and by arguing as before this is equivalent to

2(u− − u+)(θf(u+) + (1− θ)f(u−)

)≥ 2f(k)(u− − u+),

namelyθf(u+) + (1− θ)f(u−) ≤ f(θu+ + (1− θ)u−)

for every θ ∈]0, 1[. This contradicts the assumption f ′′ > 0 and hence (5.5) holds ifand only if u+ < u−.

5.8. Admissible solution of the Riemann problem for a single equation withstrictly convex flux. The Riemann problem is a particular Cauchy problem posedby coupling the system of conservation laws (1.1) with an initial datum in the form

U0(x) =

U− x < 0U+ x > 0,

(5.7)

where U− and U+ are given values U−, U+ ∈ Rn.Why is the Riemann problem relevant?

• It is the simplest possible example of discontinuous initial datum and, as weare going to see, its solution can be explicitly computed. Hence, it is simplebut it provides information on how to tackle one of the greatest challengesposed by the analysis of systems of conservation laws, namely the finite timediscontinuity formation.• As we will see, it provides information on the local in space-time structure of

admissible solutions of general Cauchy problems.• As we will see, its analysis is the building block for the construction of ap-

proximation algorithms (Glimm scheme, wave front-tracking algorithm) thatare used to establish global existence and uniqueness results.• It is the building block for the construction of numerical schemes and hence

it is also very relevant in view of applications.

18

We now focus on the case of a scalar equation with convex flux. We start by consid-ering the example of the Burger equation.

5.9. Example. We focus on the Riemann problem ∂tu+ ∂x(u2/2) = 0

u(0, x) =

u− x < 0u+ x > 0.

(5.8)

where u−, u+ ∈ R, u+ 6= u− are given. Our goal is constructing an entropy admissiblesolution (5.8). We distinguish between two cases.

CASE 1: u+ < u−. Let λ be as in (5.1), then by applying Lemma 5.7 we have thatthe function w defined as in (5.4) is an entropy admissible solution of (2.1), whichmoreover satisfies the same initial condition as in (5.8).

CASE 2: u+ < u−. We start by considering the case when u+ = 1, u− = 0. Notethat in this case the Riemann problem (5.8) reduces to (3.8). Note that Lemma 5.7tells us that solutions in the form (5.4) cannot be entropy admissible.

Let us consider the function u defined by setting

u(t, x) =

0 x < 0x/t x = θt, θ ∈]0, 1[1 x > t.

First, we observe that the function t 7→ u(t, ·) is continuous from [0,+∞[ in L1loc(R)

endowed with the strong topology. In particular,

u(0, x) =

0 x < 01 x > 0.

Also,

∂tu =

0 x < 0−x/t2 x = θt, θ ∈]0, 1[0 x > t.

and

∂xu =

0 x < 01/t x = θt, θ ∈]0, 1[0 x > t.

and hence the equation ∂tu + u ∂xu = 0 is pointwise satisfied for L2-a.e. (t, x) ∈]0,+∞[×R. By using that u is locally Lipschitz continuous in ]0,+∞[×R, we finallyget that u is a solution in the sense of distributions, namely (3.1) holds.

Next, we observe that u is of class C∞ away from the lines x = 0 and x = t. Byrecalling Lemma 4.4 we infer that away from these line the equation ∂tη(u)+∂xq(u) =0 is pointwise satisfied for every entropy-entropy flux pair (η, q). By again using thatu is locally Lipschitz continuous in ]0,+∞[×R, we finally get that∫ +∞

0

∫Rη(u) ∂tϕ+ q(u) ∂xϕdxdt = 0

for every ϕ ∈ C∞c (]0,+∞[×R) and every entropy-entropy flux pair (η, q). This impliesthat u is an entropy admissible solution of (2.1).

19

In the general case when u−, u+ are two given values satisfying u− < u+, we defineu by setting

u(t, x) :=

u− x < u−tx/t x = θt, θ ∈]u−, u+[u+ x > u+t.

and by repeating the same computations as before we obtain that u is an entropyadmissible solution of (5.8).

By mimicking the construction discussed in Example 5.9, we obtain the followingcharacterization.

5.10. Lemma. Let f be a C2 function satisfying f ′′ > 0 and let u−, u+ be given realvalues, u− 6= u+. Then an entropy admissible solution of the Riemann problem ∂tu+ ∂x

[f(u)

]= 0

u(0, x) =

u− x < 0u+ x > 0,

(5.9)

is obtained as follows.

i) If u+ < u−, then we define λ as in (5.1) and u as in (5.4).ii) If u+ > u−, then we define u as follows:

u(t, x) =

u− x < f ′(u−)t(f ′)−1(x/t) f ′(u−) < x/t < f ′(u+)u+ x > f ′(u+)t.

(5.10)

Note that the function f ′ is invertible since f ′′ > 0.

Proof. We proceed by following the same argument as in Example 5.9. The func-tion (5.4) is entropy admissible by Lemma 5.7.

Also,

∂tu =

u− x < f ′(u−)t

1

f ′′(

(f ′)−1(x/t)) · x−t2 , f ′(u−) < x/t < f ′(u+)

u+ x > f ′(u+)t.

and

∂xu =

u− x < f ′(u−)t

1

f ′′(

(f ′)−1(x/t)) · 1

t , f ′(u−) < x/t < f ′(u+)

u+ x > f ′(u+)t.

Hence, by exactly repeating the same argument as in Example 5.9 we obtain that uis an entropy admissible solution of (5.9).

20

6. Kruzkov uniqueness result

We now introduce a fundamental uniqueness result that was originally establishedby Kruzkov [4]. Although this result actually applies to a scalar conservation laws inseveral space variables (1.4), here we only focus on the case when the space variableis one dimensional. Also, note that the proof of Theorem 6.1 only works in the caseof a scalar equation. Uniqueness results for systems of conservation laws in one spacevariable are more technical and will be discussed later. There are presently no generaluniqueness results for systems of conservation laws in several space variables.

Here is Kruzkov’s uniqueness result.

6.1. Theorem. Let f ∈ C2(R) and let u, v : [0,+∞[×R→ R satisfy

‖u‖L∞(]0,+∞[×R), ‖v‖L∞(]0,+∞[×R) ≤M.

Assume moreover that u and v are two entropy admissible solutions of the scalarconservation law (2.1). Then∫ R

−R|u(t, x)− v(t, x)|dx ≤

∫ R+Lt

−R−Lt|u(0, x)− v(0, x)|dx (6.1)

for every R > 0 and t > 0. In the previous expression, the constant L is given by theinequality

|f(w)− f(z)| ≤ L|w − z| for every w, z ∈ [−M,M ].

As an immediate consequence of Theorem 6.1, we get uniqueness of bounded en-tropy admissible solutions of a given Cauchy problem.

6.2. Corollary. Assume that f ∈ C2(R), then the Cauchy problem (2.4) admits atmost one weak, entropy admissible solution u that satisfies ‖u‖L∞ < +∞.

Proof. Assume by contradiction that the Cauchy problem (2.4) admits two differentsolutions u1 and u2, then (6.1) gives∫ R

−R|u1(t, x)− u2(t, x)|dx ≤ 0

for every R > 0, t ≥ 0. Hence, u1(t, x) = u2(t, x) for a.e.-(t, x) ∈ [0,+∞[×R.

Also, note that, if u(0, ·) and v(0, ·) both elong to L1, then by letting R → +∞in (6.1) we get ∫

R|u(t, x)− v(t, x)|dx ≤

∫R|u(0, x)− v(0, x)|dx (6.2)

For the proof of Theorem 6.1 we follow [1, pp. 114-117].

References

[1] A. Bressan. Hyperbolic systems of conservation laws. The one-dimensional Cauchy problem, vol-ume 20 of Oxford Lecture Series in Mathematics and its Applications. Oxford University Press,Oxford, 2000.

[2] Constantine M. Dafermos. Hyperbolic conservation laws in continuum physics, volume 325 ofGrundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sci-ences]. Springer-Verlag, Berlin, third edition, 2010.

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[3] J. Glimm. Solutions in the large for nonlinear hyperbolic systems of equations. Comm. Pure Appl.Math., 18:697–715, 1965.

[4] S. N. Kruzkov. First order quasilinear equations with several independent variables. Mat. Sb.(N.S.), 81 (123):228–255, 1970.

[5] P. D. Lax. Hyperbolic systems of conservation laws. II. Comm. Pure Appl. Math., 10:537–566,1957.

[6] D. Serre. Systems of conservation laws. 1. Cambridge University Press, Cambridge, 1999. Hyper-bolicity, entropies, shock waves, Translated from the 1996 French original by I. N. Sneddon.

[7] D. Serre. Systems of conservation laws. 2. Cambridge University Press, Cambridge, 2000. Geo-metric structures, oscillations, and initial-boundary value problems, Translated from the 1996French original by I. N. Sneddon.

L.V.S. IMATI-CNR, via Ferrata 1, I-27100 Pavia, Italy.E-mail address: spinolo@imati.cnr.it