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QUARTERLY OF APPLIED MATHEMATICS

VOLUME LXIII, NUMBER 2

JUNE 2005, PAGES 343–367

S 0033-569X(05)00964-6

Article electronically published on April 19, 2005

THE INCOMPLETENESS OF THE BORN-INFELD MODEL FORNON-LINEAR MULTI-D MAXWELL’S EQUATIONS

By

WLADIMIR NEVES (Instituto de Matematica, Universidade Federal do Rio de Janeiro, C. Postal68530, Rio de Janeiro, RJ 21945-970, Brazil)

and

DENIS SERRE (UMPA, Ecole Normale Superieure de Lyon, UMR 5669 CNRS, Lyon Cedex 07,France)

Abstract. We study the Born-Infeld system of conservation laws, which is the mostfamous model for non-linear Maxwell’s equations. This system is totally linear degen-erated and there exists a conjecture, see Y. Brenier, Hydrodynamic structure of the

augmented Born-Infeld equations, Arch. Rational Mech. Anal. 172 (2004), 65–91, thatshocks are not allowed to form. In fact, we show that this conjecture is false and thatthe Born-Infeld model is not complete by itself. It means that a further theory is neededto complete the model.

1. Introduction. We are concerned in this paper with the Born-Infeld system ofconservation laws, which has come into great discussion recently with the papers of Y.Brenier [4] and D. Serre [20]. This system is certainly the most famous model for non-linear Maxwell’s equations and, since it is totally linear degenerated (see Definition 1.2),there exists a conjecture that shock waves would not form. We show that this conjectureis false, by the appearance of shocks beyond the contact discontinuities in the resolutionof the Riemann Problem, at least when the initial-data are large. Moreover, when ararefaction shock (see Definition 1.3) is present, we show that the Riemann Problem isnot well-posed. In fact, we have a one parameter family of solutions for this problemand, in this way, the Born-Infeld model is not complete by itself. Therefore, for this classof Riemann Problems, we need a further local theory to complete the model.

It is well known that, when a system of conservation laws is endowed with a renormal-ized equation given by a uniformly convex entropy, the initial-value problem is locallywell-posed in the context of classical solutions; see Majda [18], Dafermos [10]. Further-more, we have uniqueness and continuous dependence on the initial-data for a broader

Received October 5, 2004.2000 Mathematics Subject Classification. Primary 35Q60.E-mail address: [email protected]

E-mail address: [email protected]

c©2005 Brown University

343

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344 WLADIMIR NEVES and DENIS SERRE

class of weak solutions that satisfies an admissibility inequality constraint; see [11, 12]and also Bressan et al. [5, 6, 7, 8] in the BV case. For instance, this is the case ofpiecewise smooth solutions in the class of Riemann Problems. On the other hand, theclassical problem of elastodynamics in nonlinear elasticity shows that the uniqueness ofsolution for the Riemann Problem is lost when there exists a lack of convexity on theentropy function; see for example [1, 10, 15, 17] and many other references therein. Thisdifficulty was resolved in [1, 17] by considering principally a further local theory given bya kinetic formulation. However, [10] overcame this problem of non-uniqueness for elasto-dynamics by considering extra conservation laws, called involutions, which compensatefor the lack of convexity. Analogously, it is the problem in fluid-dynamics for van derWaals fluid; see [15].

Here we have the same condition, that is, lack of convexity for the energy function,which is the entropy for the Born-Infeld model. As we shall see, the energy is strictlyconvex in a neighborhood of the origin, but is not far away from it. Hence, the Born-Infeld system is not hyperbolic in the large. Due to this lack of convexity, we are notallowed to discard some nonphysical types of jump discontinuities as non-classical shocks.Actually, they are a potential source of non-uniqueness, which is the case above and forthe Born-Infeld system of equations.

Finally, we mention the idea of an enlarged system of Born-Infeld equations introducedin [4, 20], following [10, 13] in the case of nonlinear elasticity. In [20], the polyconvexityof the energy (entropy) function was a sufficient condition to prove local well-posednessof the initial-value problem for classical solutions, that is, for initial-data in Hs, withs > 1 + d/2, where d is the dimension of the space. However, the Rankine-Hugoniotcondition for the enlarged system is not equivalent to the Born-Infeld one. Therefore,a couple of states, which satisfies the Rankine-Hugoniot condition for the Born-Infeldmodel, does not necessarily satisfy the enlarged one. Moreover, the enlarged systemcould suggest a local kinetic formulation in the class of Riemann Problems. We addressthese interesting questions in other works.

Let (t, x) ∈ R+ × R

3 be the points in the time-space domain. We consider the elec-tric intensity field E, the magnetic intensity field H, the electric induction D, and themagnetic induction B, all of them taking values in R

3. The models of electromagnetismcould be given in a rational continuum physics form through the stored energy functionh(D, B), see Coleman and Dill [9], by the following form:

∂tB + curlx(∂Dh) = 0, (1.1)

∂tD − curlx(∂Bh) = 0, (1.2)

divxD = 0, divxB = 0, (1.3)

where we have assumed for simplicity that there are neither charges nor currents. Fur-thermore, we have

∂Dih(D, B) =: Ei, ∂Bi

h(D, B) =: Hi (i = 1, . . . , 3).


BORN-INFELD EQUATIONS 345

The equations (1.1), (1.2) are respectively the Faraday and Ampere’s Law, and equation(1.3) are constraints, which are compatible with (1.1), (1.2). In the linear theory ofelectromagnetism, called Maxwell’s equations, we have

D = εE and B = µH,

where ε is the dielectric tensor and µ is the permeability tensor. However, there areseveral reasons to avoid the linear case. For instance, the electric field of a particle atrest decreases with the inverse square of the distance, which means that E grows withoutlimit as the distance tends to zero. To get rid of this fact, some non-linear models havebeen proposed; the most famous is due to M. Born and L. Infeld [3]. This model isobtained by taking

h(D, B) =√

1 + |D|2 + |B|2 + |P |2 (P := D × B). (1.4)

From (1.1), (1.2) and (1.4), we obtain

∂tD + curlx

(−B + D × P

h

)= 0, (1.5)

∂tB + curlx

(D + B × P

h

)= 0. (1.6)

Then, (1.3), (1.5) and (1.6) consist of the Born-Infeld system of conservation laws in 3-dimensions. Moreover, the stored energy function h given by (1.4) satisfies an additionalconservation law, that is

∂th + divxP = 0. (1.7)

Then, (h, P ) is an entropy pair (see Definition 1.1) for the Born-Infeld system. We notethat h(D, B) is a strictly convex function of D and B only in a neighborhood of theorigin, but fails to be convex far away from it. The entropy flux P is usually called thePoynting vector.

Now we focus on plane waves, which depend only on the time and one scalar spacevariable. We observe that the stored energy function which gives the Born-Infeld modeldescribes an isotropic medium. In fact, this behavior happens since h derives from aLorentz- and orientation-invariant Lagrangian, i.e.,

L = −√

1 + |B|2 − |E|2 − (E · B)2.

Therefore, we have a wave isotropy condition, see [16], and by symmetry considerationsthus far, only a single spatial variable is needed. Let us choose x = x1 as such a spatialcoordinate. Hence, all the fields involved in (1.3)–(1.6) depend on (t, x) ∈ R

+ × R.Moreover, it follows that

∂tD1 = 0, ∂xD1 = 0,

∂tB1 = 0, ∂xB1 = 0.



Thus, D1, B1 are constant functions, and for simplicity, we assume D1 = B1 = 0. Thenwe obtain the following system of conservation laws:

∂tD2 + ∂x(B3 + D2P

h) = 0, (1.8)

∂tD3 + ∂x(−B2 + D3P

h) = 0, (1.9)

∂tB2 + ∂x(−D3 + B2P

h) = 0, (1.10)

∂tB3 + ∂x(D2 + B3P

h) = 0, (1.11)

where P ≡ P1 = D2B3 − D3B2.

Next we present some mathematical considerations for systems of conservation laws.Set

u := (D2, D3, B2, B3), (1.12)

f(u) :=(

B3 + D2P

h,−B2 + D3P

h,−D3 + B2P

h,D2 + B3P

h

). (1.13)

So (1.8)–(1.11) could be written in the following simple form:

ut + f(u)x = 0 in R+ × R. (1.14)

The open set U ⊂ R4, such that u(t, x) ∈ U will be called the set of states and f the flux

function. We are concerned with the initial-value problem, that is, we seek a u(t, x) ∈ U

solution of (1.14) and satisfying an initial-data

u(0, x) = u0 x ∈ R, (1.15)

where u0 : R → U is a given bounded measurable function. As is well known, in generalfor conservation laws, there does not exist (global) solutions, even if the data is infinitelydifferentiable. Hence, we have to deal with the concept of weak solutions, which meansthat u ∈ L∞(R+ × R; R4) is a weak solution of (1.14), (1.15) if it satisfies∫ ∞

0

∫R

(u, f(u)) · ∇t,xφ(t, x) dxdt +∫

R

u0 φ(0, x) dx = 0, (1.16)

for any function φ ∈ C∞0 (R2). Moreover, if u ∈ L∞ is a C1 function outside a manifold

Γ (with codimension one), across which it has jump discontinuities, then it can be shownusing (1.16), see [10, 19], that u must satisfy the so-called Rankine-Hugoniot condition

nt[u] + nx[f(u)] = 0,

where n = (nt, nx) is the outward unit normal vector along the manifold Γ, [u] := u+−u−,[f(u)] := f(u+) − f(u−), and

u+ = limδ→0+

u((t, x) + δn), u− = limδ→0+

u((t, x) − δn).



Definition 1.1. A real Lipschitz function η is called an entropy for (1.14), withassociated entropy flux q ∈ W 1,∞(U), when for every open set Π ⊂ R

+×R and for everyu ∈ C1 that solves (1.14) pointwise, we have

∂tη(u) + ∂xq(u) = 0 in D′(Π).

If, in addition, η is a convex function, then we say that (η, q) is a convex entropy pair.Moreover, a weak solution of (1.14), (1.15) is called an entropy solution when ∂tη(u) +∂xq(u) ≤ 0 in the sense of distributions for every convex entropy pair.

We recall that (see [10, 19]) a system of conservation laws is said to be hyperbolic iffor any v ∈ U , the matrix

Ai,j(v) :=∂fi(v)∂vj

(i, j = 1, . . . , n),

has n real eigenvalues λ1(v) ≤ λ2(v) ≤ . . . ≤ λn(v) and is diagonalizable. Thus, thereexist ri(v), (i = 1, . . . , n) linearly independent (right) corresponding eigenvectors, and

A(v) ri(v) = λi(v) ri(v).

Here, n = 4 and this computation is hard to derive. Furthermore, since h is not convexin the large, it does not follow from the well-known result that the existence of a convexentropy pair for (1.14) implies the hyperbolicity. Although, taking account of the aug-mented theory (see [4]), the propagation speeds, i.e., λ’s, are easily calculated. We have

λ1 = λ2 =P − 1

h=: λ− < λ3 = λ4 =

P + 1h

=: λ+. (1.17)

Definition 1.2. For the system of conservation laws (1.14), a point v ∈ U is said tobe of linear degeneracy of the i-characteristic family when

∇vλi(v) · ri(v) = 0; (1.18)

otherwise, it is of genuine nonlinearity of the i-characteristic family. If (1.18) holds for ev-ery v ∈ U , then the i-characteristic family is called linear degenerated. Moreover, we saythat (1.14) is totally linear degenerated when every i-characteristic is linear degenerated.

Again, it is not easy to conclude from the above definition that the Born-Infeld modelis totally linear degenerated. However, from (1.17), it is an immediate application ofBoillat’s theorem; see [2].

In fact, for the study of shocks and completeness of the Born-Infeld model, we takethe Riemann Problem. Thus, we consider initial-data of the form

u0(x) =

{u�

0 if x < 0,

ur0 if x > 0,

(1.19)

where u�0, u

r0 are given constants. We seek self-similar solutions for

v(ξ) = u(t, x)(ξ =

x

t

)



in BV loc(R; R4) ∩ L∞(R; R4), satisfying in the sense of distributions the ordinary differ-ential equation

[ξ v(ξ) − f(v(ξ))]′ = v(ξ)(

′ ≡ d

dξ

)(1.20)

and the boundary conditions

v(−∞) = u�0, v(+∞) = ur

0. (1.21)

In fact, v is a Lipschitz function, and thus by Rademacher’s Theorem it is differentiableL1-a.e.; see [14]. Hence, (1.20) is satisfied by:

i) Constant states; for each Lebesgue point ξ, where

v′(ξ) = 0.

ii) Jump discontinuities; for each discontinuity point ξ, where the Rankine-Hugoniotjump condition must hold, i.e.,

ξ [v+ − v−] = f(v+) − f(v−) (v+ := v(ξ+), v− := v(ξ−)).

iii) Centered simple waves; for each Lebesgue point ξ, where v′(ξ) = 0. From (1.20), i.e.,[Df(v(ξ)) − ξ Id] v′(ξ) = 0, we must have

ξ = λi(v(ξ)) and v′(ξ) = c(ξ)ri(v(ξ)) (i = 1, . . . , 4). (1.22)

Moreover, if we set

C := {ξ ∈ R ; v′(ξ) = 0},

J := {ξ ∈ R ; the Rankine-Hugoniot condition holds},

W := {ξ ∈ R ; v′(ξ) = 0},

then R is the union of these pairwise disjoint sets. Therefore, the solutions v(ξ) of (1.20),(1.21) are given by a combination of (i)-(iii).

Remark 1.1. By differentiating the first relation in (1.22) and utilizing the second,we obtain

[Dλi(v(ξ)) · ri(v(ξ))]c(v(ξ)) = 1.

Since v′ is a locally finite Radon measure, we observe that the centered simple waves arepoints of genuine nonlinearity of the i-characteristic family. Moreover, from the aboveexpression, we determine the scalar function c. Therefore, for totally linear degeneratedsystems of conservation laws, we have W = ∅.

Usually, the jump v+−v− is called the amplitude and its size |v+−v−| is the strengthof the jump discontinuity. Moreover, when the strength of the jump discontinuity is lessthan a positive (sufficiently small) δ, we say that the jump discontinuity is weak.

Definition 1.3. We say that the jump discontinuity (v−, v+; ξ) is an i-classical shock(or i-Lax shock, or i-compressive shock), when there exists an index i (1 ≤ i ≤ n) suchthat

λi(v+) < ξ < λi(v−),

λi−1(v−) < ξ < λi+1(v+).(1.23)



It means that at the point of discontinuity, there are n + 1 incoming characteristics, ofwhich the speeds are the eigenvalues

λ1(v+), . . . , λi(v+), λi(v−), . . . , λn(v−).

Moreover, (1.23) is called the Lax shock admissibility criterion. When the left or theright part of (1.23)1 is satisfied as equality, the jump discontinuity is called a left ora right i-contact discontinuity, and if both parts hold as equalities, then we have ani-contact discontinuity. When there are at least n+2 incoming characteristics, the jumpdiscontinuity (v−, v+; ξ) is called an i-overcompressive shock; that is, there exists anindex i such that

λi+1(v+) < ξ < λi(v−). (1.24)

When there are n incoming characteristics, the jump discontinuity (v−, v+; ξ) is called ani-undercompressive shock (or i-transitional shock); that is, there exists an index i suchthat

λi(v±) < ξ < λi+1(v±). (1.25)

When there are n−1 incoming characteristics, the jump discontinuity (v−, v+; ξ) is calledan i-rarefaction shock (or i-counter Lax shock); that is, there exists an index i such that

λi(v−) < ξ < λi(v+),

λi−1(v+) < ξ < λi+1(v−).(1.26)

In any (1.24)–(1.26) case, we say that the jump discontinuity is a non-classical shock.

2. Propagation speeds for the Born-Infeld model. The aim of this section isto study the Born-Infeld model, when given two constant states

u� = (D�2, D

�3, B

�2, B

�3), ur = (Dr

2, Dr3, B

r2 , Br

3),

not necessarily close, or small, how they could be connected. Since the Born-Infeldsystem of equations is totally linear degenerated, from Remark 1.1, we are not allowedto use centered simple waves. So, we have to connect u� ≡ u− and ur ≡ u+ by jumpdiscontinuities. Therefore, for any s := ξ ∈ J , we regard the Rankine-Hugoniot jumpcondition given from (1.8)–(1.11); that is

s(D+2 − D−

2 ) =B+

3 + D+2 P+

h+− B−

3 + D−2 P−

h− ,

s(D+3 − D−

3 ) =−B+

2 + D+3 P+

h+− −B−

2 + D−3 P−

h− ,

s(B+2 − B−

2 ) =−D+

3 + B+2 P+

h+− −D−

3 + B−2 P−

h− ,

s(B+3 − B−

3 ) =D+

2 + B+3 P+

h+− D−

2 + B−3 P−

h− ,

where h, P are given by (1.4) with the obvious notations for ±. Now, if we set

d±i :=D±

i

h± , b±i :=B±

i

h± (i = 2, 3),



then we can rewrite the above equations as

d+2 ζ+ − d−2 ζ− − b+

3 + b−3 = 0, (2.1)

d+3 ζ+ − d−3 ζ− + b+

2 − b−2 = 0, (2.2)

b+2 ζ+ − b−2 ζ− + d+

3 − d−3 = 0, (2.3)

b+3 ζ+ − b−3 ζ− − d+

2 + d−2 = 0, (2.4)

where ζ± = sh± − P±. So instead of s, we have two unknowns, i.e., ζ±. Hence, weobtain one more equation to be satisfied:

φ(ζ+, ζ−) :=P+ + ζ+

h+− P− + ζ−

h− = 0, (2.5)

which means that s must have the same value given by

ζ+ = s h+ − P+ or ζ− = s h− − P−.

Once we obtain ζ+, ζ− satisfying (2.1)–(2.5), the Rankine-Hugoniot condition is satisfied.Moreover, it follows from {(2.1) − (2.4)} and {(2.2) + (2.3)} that

(d+2 − b+

3 ) −(d−2 − b−3 )

(d+3 + b+

2 ) −(d−3 + b−2 )

(ζ+ + 1)

(ζ− + 1)

= 0. (2.6)

Analogously, from {(2.1) + (2.4)} and {(2.2) − (2.3)}, we have (d+

2 + b+3 ) −(d−2 + b−3 )

(d+3 − b+

2 ) −(d−3 − b−2 )

(ζ+ − 1)

(ζ− − 1)

= 0. (2.7)

Hence, we can equivalently search for ζ± satisfying (2.5)–(2.7). We begin regarding thevalues

ζ+ = ζ− = −1,

ζ+ = ζ− = +1.

The former trivially solves (2.6), and from (2.7) we must have

D+2 + B+

3

h+=

D−2 + B−

3

h− andD+

3 − B+2

h+=

D−3 − B−

2

h− . (2.8)

From (2.8), we obtain

(D+2 )2 + 2D+

2 B+3 + (B+

3 )2

(h+)2=

(D−2 )2 + 2D−

2 B−3 + (B−

3 )2

(h−)2

and(D+

3 )2 − 2D+3 B+

2 + (B+2 )2

(h+)2=

(D−3 )2 − 2D−

3 B−2 + (B−

2 )2

(h−)2.

Adding the two above equations, we have

|D+|2 + |B+|2 + 2P+

(h+)2=

|D−|2 + |B−|2 + 2P−

(h−)2,



where we have used the definition of P given by (1.4). Hence, after some algebra andthe definition of h given by (1.4), we obtain

1 − (P+ − 1)2

(h+)2= 1 − (P− − 1)2

(h−)2.

Therefore, it follows thatP+ − 1

h+− P− − 1

h− = 0

orP+ − 1

h++

P− − 1h− = 0.

In order to satisfy (2.5), we must have

P− − 1h− =

P+ − 1h+

(= s).

For the second case, that is ζ+ = ζ− = +1, we have (2.7) trivially satisfied, and from(2.6), we must have

D+2 − B+

3

h+=

D−2 − B−

3

h− andD+

3 + B+2

h+=

D−3 + B−

2

h− . (2.9)

Analogously, (2.9) and (1.4) imply

P− + 1h− =

P+ + 1h+

(= s).

Therefore, considering the BI system of conservation laws given by (1.8)–(1.11), we havethe following:

Lemma 2.1. Let u− = (D−2 , D−

3 , B−2 , B−

3 ), u+ = (D+2 , D+

3 , B+2 , B+

3 ) be two given con-stant states. Let (1.8), (1.11) be the system of conservation laws for u = (D, B). Thenu−, u+ could be connected by contact discontinuities in the following form:

i) When (u−, u+) satisfies (2.8), by a contact discontinuity of speed

s = λ−(u−) = λ−(u+).

ii) When (u−, u+) satisfies (2.9), by a contact discontinuity of speed

s = λ+(u−) = λ+(u+).

Remark 2.1. We recall the well-known result that any weak jump discontinuity as-sociated with a linear degenerated characteristic family is necessarily a contact discon-tinuity (see [10, 19]). Moreover, any two nearby states u− and u+ associated with alinear degenerated i-characteristic family could be connected to each other by a contactdiscontinuity of speed

s = λi(u−) = λi(u+).

Now, we study whether there exists ζ± = ±1 satisfying (2.5)–(2.7). We must have

(d+3 + b+

2 )(d−2 − b−3 ) − (d+2 − b+

3 )(d−3 + b−2 ) = 0, (2.10)

(d+3 − b+

2 )(d−2 + b−3 ) − (d+2 + b+

3 )(d−3 − b−2 ) = 0, (2.11)



which define a conic manifold M of dimension 6. Let σ = 0, θ = 0, γ±, α±, be realnumbers such that

ζ+ + 1 = γ+σ, ζ− + 1 = γ−σ,

ζ+ − 1 = α+θ, ζ− − 1 = α−θ.(2.12)

Once (2.10), (2.11) are satisfied, we may obtain γ±, α± = 0 from (2.6) and (2.7) respec-tively. Moreover, by eliminating ζ+ and ζ− in (2.12), we have

γ+ −α+

γ− −α−

σ

θ

=

2

2

.

Consequently, for α+γ− − α−γ+ = 0, it follows that

(α+γ− − α−γ+)σ = 2(α+ − α−),

(α+γ− − α−γ+)θ = 2(γ+ − γ−).

So γ+ = γ− and α+ = α−. Finally, we obtain from (2.12)

ζ+ =2α+γ+ − α+γ− − α−γ+

α+γ− − α−γ+, (2.13)

ζ− =α+γ− + α−γ+ − 2α−γ−

α+γ− − α−γ+. (2.14)

Since we seek ζ± = ±1, we are not allowed to assume that none of γ±, α± is zero.Although, we are also interested when

ζ+ = ±1 and ζ− = ±1,

ζ+ = ±1 and ζ− = ±1,

because this could be a solution for (2.5)–(2.7). It could be, when exactly one of theγ±, α± is zero. Indeed, if for instance γ+ = 0, then we must have γ− = 0, and α+ = 0,since

0 = α+γ− − α−γ+ ≡ α+γ−.

It follows from (2.13), (2.14) respectively that ζ+ = −1, ζ− = 1 − 2(α−/α+), and sinceα− = α+, ζ− = −1. Now, if we suppose α− = 0, then it is impossible to satisfy (2.5).Indeed, for γ+ = α− = 0, we must have

(d−2 − b−3 ) = 0, (d−3 + b−2 ) = 0,

(d+2 + b+

3 ) = 0, (d+3 − b+

2 ) = 0,

which implyP+ − 1

h+= −|D+|2 + |B+|2

h+− 1

h+< 0,

andP− + 1

h− =|D−|2 + |B−|2

h− +1

h− > 0.



Actually, the problem now is quite different; that is, we want to construct initial-datau−, u+ to be connected by a jump discontinuity that is not a contact discontinuity. As(2.10), (2.11) are 2 equations for 8 variables, we have a lot of solutions. Let

(δ2, δ3, β2, β3)± := (δ−2 , δ−3 , β−2 , β−

3 , δ+2 , δ+

3 , β+2 , β+

3 ) ∈ R8

be one of these solutions. We seek

(d2, d3, b2, b3)± = µ (δ2, δ3, β2, β3)±,

with µ ∈ R+ to be determined a posteriori. Moreover, for any µ, (d2, d3, b2, b3)± ∈ M.

We observe that (d, b)± must satisfy some conditions in order to yield the states u− andu+. For simplicity, we drop the superscript ±, i.e., + and −, whenever indifferent. Oncewe have (d, b) to obtain (D, B) from it, we need the value of h, which is not known. Infact, h = h(D, B), given by (1.4). However, if the following implicit equation for h, i.e.,

G(h) := h − H(d, b; h) = 0 (H(d, b; h) ≡ h(hd, hb)),

is resolved, then we have h = h(D, B), which means that D = hd and B = hb. SoX := h2 has to be a solution of

p2X2 + (|b|2 + |d|2 − 1)X + 1 = 0, (2.15)

where p = d2b3 − d3b2. We suppose temporarily that p = 0. For positive real roots, wemust have

(1 − |b|2 − |d|2)2 ≥ 4p2,

1 − |b|2 − |d|2 > 0.

Hence, we obtain the following constraint:

|b|2 + |d|2 + 2|p| ≤ 1,

which is the intersection of two orthogonal cylinders; that is,

(d3 + b2)2 + (b3 − d2)2 ≤ 1, (2.16)

(b2 − d3)2 + (b3 + d2)2 ≤ 1. (2.17)

Therefore, (2.16) and (2.17) define a convex set, and so a connected one. Moreover,(2.10), (2.11), (2.16), (2.17) are perfect compatible. Now, for |(d, b)| << 1, the roots of(2.15) are

X1 ∼ 1, X2 ∼ 1p2

≥ 1|b|2|d|2 .

On the other hand, the right-hand side of (2.15) is positive for X = 1. Consequently, thetwo roots are on the same side of 1. By connectedness, they remain in the same intervaland, moreover for |(d, b)| << 1, they are in (1,∞). Finally:

Lemma 2.2. If (d−2 , d−3 , b−2 , b−3 ), (d+2 , d+

3 , b+2 , b+

3 ) satisfy (2.16)–(2.17), then the respectiveroots of (2.15) belong to (1,∞), that is

X−1 , X−

2 ∈ (1,∞), X+1 , X+

2 ∈ (1,∞).



Set hi := X1/2i (i = 1, 2), Xi being the roots of (2.15), such that

1 ≤ X1 ≤ X2 < ∞.

Therefore, making Di = hid, Bi = hib, we obtain

h2i = 1 + |Bi|2 + |Di|2 + P 2

i (i = 1, 2),

where Pi := Di2B

i3 − Di

3Bi2. So, hi = h(Di, Bi) and moreover, h2

i ≥ 1, (i = 1, 2). If wecan take (d, b)± satisfying (2.10), (2.11), (2.16), (2.17), such that

P−i + ζ−

h−i

=P+

j + ζ+

h+j

(for some i, j = 1, 2), (2.18)

then we would make s this number, and the Rankine-Hugoniot condition is satisfied withζ± = ±1. In fact, we observe that the above equality is equation (2.5), and we can makethree different choices: i = j = 1, i = j = 2, and i = 1, j = 2. By symmetry, i = 1, j = 2and i = 2, j = 1 are equivalent.

Again, we suppose |(d, b)±| << 1. Hence, the conditions (2.16), (2.17) are triviallysatisfied. Furthermore, the function φ(ζ+, ζ−) with ζ+, ζ− given respectively by (2.13),(2.14), where by abuse of notation we shall write φ((d, b)±), could be approximated in asuitable manner. Since h±

1 ∼ 1, h±2 ∼ |p±|−1, it follows that

P±1

h±1

∼ p± andP±

2

h±2

∼ sgn(p±).

Thus, considering (2.18), we have the following approximations:

φ11((d, b)±) = p+ − p− + 2(γ+ − γ−)(α+ − α−)

α+γ− − α−γ+,

φ12((d, b)±) = p+ − p− + 2(γ+ − γ−)(α+ − p−α−)

α+γ− − α−γ+,

φ22((d, b)±) = sgn(p+) − sgn(p−)

+(γ+ − γ−)(α+|p+| − α−|p−|) + (α+ − α−)(γ+|p+| − γ−|p−|)

α+γ− − α−γ+,

where for φ12 we assumed, without loss of generality, that p− > 0. If one of the abovefunctions is null, in a non-trivial manner, i.e., |(d, b)±| > 0, over the manifold M, thenwe have a non-trivial, small solution of (2.5). Set φ11 = 0; then

{(p+ − p−)(α+γ− − α−γ+)} + 2{(γ+ − γ−)(α+ − α−)} = 0.

The first parenthesis is of O(|d±i d±j b±i b±j |), and the second is of O(|d±i d±j | + |d±i b±j | +|b±i b±j |) for (i, j = 2, 3). Then, it is not possible to obtain φ11 = 0, since we must have

γ+ = γ− or α+ = α−.

In fact, we already know that φ11 is not relevant. Since |(d, b)±| << 1 and h± ∼ 1, itfollows that |(D, B)±| << 1. Hence, from Remark 2.1, the jump discontinuity must be acontact discontinuity. Now, we set φ12 = 0; that is

{(p+ − p−)(α+γ− − α−γ+) − 2 p−α−(γ+ − γ−)} + 2{(γ+ − γ−)α+} = 0.



Again, the first parenthesis is of O(|d±i d±j b±i b±j |), and the second is of O(|d±i d±j |+|d±i b±j |+|b±i b±j |) for (i, j = 2, 3). Then, it is not possible to obtain φ12 = 0, since we must have

γ+ = γ− or α+ = 0.

The condition α+ = 0 could be acceptable as we have observed, but this implies acontradiction with the assumption that p− > 0. Indeed, if α+ = 0, then we must have

(d−2 + b−3 ) = 0 and (d−3 − b−2 ) = 0,

which implyp− = −|d−|2 = −|b−|2 < 0.

Finally, we set φ22 = 0; then

[ sgn(p+) − sgn(p−)](α+γ− − α−γ+)

+(γ+ − γ−)(α+|p+| − α−|p−|) + (α+ − α−)(γ+|p+| − γ−|p−|) = 0.

Now, if we prove that

(2α+γ+ − α+γ− − α−γ+)p+ + (2α−γ− − α−γ+ − α+γ−)p− = 0, (2.19)

and moreover we have

(2α+γ+ − α+γ− − α−γ+)(2α−γ− − α−γ+ − α+γ−) < 0, (2.20)

then we are done; that is, p+p− > 0 and thus φ22 = 0. Conveniently, we write

4p+ = R − S,

4p− = [(α+γ−)2R − (α−γ+)2S]/(α−γ−)2,(2.21)

where we have used (2.6), (2.7), and

R = [(d+2 + b+

3 )2 + (d+3 − b+

2 )2],

S = [(d+2 − b+

3 )2 + (d+3 + b+

2 )2].

From (2.19) and (2.21), we obtain

(α+γ− + α−γ+) [(γ−)2(α+ − α−)2 R − (α−)2(γ+ − γ−)2 S] = 0.

Consequently, we must have(α+γ− + α−γ+) = 0, (2.22)

or(γ−)2(α+ − α−)2 R = (α−)2(γ+ − γ−)2 S, (2.23)

and we observe that none of these imply a contradiction. Let m, n ∈ R∗, m = n, such

thatγ+ = m γ− and α+ = n α−.

It follows from (2.6) and (2.7) that

(d−2 − b−3 ) − m(d+2 − b+

3 ) = 0,

(d−3 + b−2 ) − m(d+3 + b+

2 ) = 0,

(d−2 + b−3 ) − n(d+2 + b+

3 ) = 0,

(d−3 − b−2 ) − n(d+3 − b+

2 ) = 0,

(2.24)



which define a sub-manifold N ⊂ M of dimension 4. Hence, for (d, b)± ∈ N , (2.20)implies

(2 n m − n − m)(2 − m − n) < 0. (2.25)

We seek if there exist m, n satisfying (2.25), which solves (2.22) or (2.23). The former,i.e., α−γ+ = −α+γ−, implies m = −n, and thus condition (2.25) is trivially satisfied,that is,

m = −n ⇒ (2 n m − n − m)(2 − m − n) = −4m2 < 0.

Consequently, over N with m = −n, we have φ22 = 0. Furthermore, from (2.21) we get

p− = m2p+,

and from equations (2.13), (2.14) we obtain respectively

ζ+ = m and ζ− = 1/m.

Therefore, we have two solutions:

ζ+ = ±

√p−

p+, ζ− = ±

√p+

p−, (2.26)

which yield a difference of ±1 when p+ = p−. Moreover, for u− sufficiently close to u+,and thus from (1.4) h− ∼ h+ and P− ∼ P+, we have ζ± ∼ ±1. Therefore, we recoverthe contact discontinuity solution in the limit case of weak jump discontinuities. Thesecond condition, i.e., (2.23), implies

(n − 1)2 R = (m − 1)2 S, (2.27)

i.e., one equation for two variables, which means that we have many solutions for m, n

and one of them could satisfy (2.25). Although, for u− sufficiently close to u+, from(2.21) we have

(1 − n2) R = (1 − m2) S. (2.28)

Then, from (2.27) and (2.28) we must have m = n, which is a contradiction. There-fore, for all m, n satisfying (2.27), we do not recover the contact discontinuity solution.Consequently, from now on, we consider m = −n. We remember that we have supposedp+p− = 0. From (2.21),

p+ = 0 ⇔ p− = 0,

which implies ζ± = ±1. Consequently, for ζ± = ±1, we have p+p− > 0.

In fact, we have shown that φ22 = 0, but not φ yet. Before we do this, we shall showsome qualitative properties of the jump discontinuities given by (2.26). Let ζ+, ζ− begiven by (2.26); when p− > p+, we have

ζ+ < −1 ⇔ ζ− > −1 (ζ± < 0),

ζ+ > 1 ⇔ ζ− < 1 (ζ± > 0).



In the first case, we obtain

λ−(u−) =P− − 1

h− <P− + ζ−

h− = s,

λ+(u−) =P− + 1

h− >P− + ζ−

h− = s,

λ−(u+) =P+ − 1

h+>

P+ + ζ+

h+= s.

Consequently, from Definition 1.3 we have a Rarefaction Shock, since

λ−(u−) < s < λ−(u+),

s < λ+(u−).

In the second case, we obtain

λ+(u−) =P− + 1

h− >P− + ζ−

h− = s,

λ−(u−) =P− − 1

h− <P− + ζ−

h− = s,

λ+(u+) =P+ + 1

h+<

P+ + ζ+

h+= s.

Consequently, from Definition 1.3 we have a Lax Shock, since

λ+(u+) < s < λ+(u−),

λ−(u−) < s.

Analogously, when p− < p+, we have

ζ+ > −1 ⇔ ζ− < −1 (ζ± < 0),

ζ+ < 1 ⇔ ζ− > 1 (ζ± > 0).

From the former, we obtain

λ−(u−) =P− − 1

h− >P− + ζ−

h− = s,

λ−(u+) =P+ − 1

h+<

P+ + ζ+

h+= s,

λ+(u+) =P+ + 1

h+>

P+ + ζ+

h+= s.

Consequently, from Definition 1.3 we have a Lax Shock, since

λ−(u+) < s < λ−(u−),

s < λ+(u+).



For the second, we obtain

λ+(u−) =P− + 1

h− <P− + ζ−

h− = s,

λ−(u+) =P+ − 1

h+<

P+ + ζ+

h+= s,

λ+(u+) =P+ + 1

h+>

P+ + ζ+

h+= s.

Consequently, from Definition 1.3 we have a Rarefaction Shock, since

λ+(u−) < s < λ+(u+),

λ−(u+) < s.

Then it remains to show that φ = 0. The following lemma will be used in the proofof it.

Lemma 2.3. Let U be the manifold defined as

U := {A ∈ Mn(R) : detA = 0, ‖A‖ = 1}.Then U is a connected set.

Proof. 1. Let a ∈ Rn, ‖a‖ = 1, and we set

U(a) := {A ∈ Mn(R); A a = 0, ‖A‖ = 1}.

Then, we haveU =

⋃‖a‖=1

U(a).

Furthermore, we defineU1 := {A ∈ Mn(R); AT e1 = 0},

where {ei}ni=1 is the standard basis of R

n. Then,

U(a) ∩ U1 = ∅,

for any vector a. Indeed, if we take M = e2 ⊗ b, where b ∈ Rn, ‖b‖ = 1, and b · a = 0,

then ‖M‖ = ‖e2‖‖b‖ = 1, and

Ma = (e2 ⊗ b)a = (b · a)e2 = 0,

MT e1 = (e2 ⊗ b)T e1 = (b ⊗ e2)e1 = (e2 · e1)b = 0,

that is, M ∈ U(a) ∩ U1.

2. Now, we observe that U(a) is equivalent to the unit sphere of Mn×(n−1)(R), andthus pathwise-connected. Indeed, upon rotating and relabelling the coordinate axes ifnecessary, we may assume a = e1. Therefore,

U(a) ∼ U(e1) = {A ∈ Mn(R); Ai1 = 0, Ai(j+1) = Bij ,

B ∈ Mn×(n−1)(R), ‖B‖ = 1}.It follows easily that U(e1) is pathwise-connected. Analogously, we have that U1 ispathwise-connected.



3. Finally, let A1, A2 ∈ U . Then, there exist a1, a2 ∈ Rn, ‖a1‖ = ‖a2‖ = 1, such that

A1 ∈ U(a1), A2 ∈ U(a2).

Since U1, U(a1), and U(a2) are pathwise-connected and U(a) ∩ U1 = ∅ for any a, we canalways make a path in U with endpoints A1, A2. Therefore, U is pathwise-connected,and thus a connected set. �

Theorem 2.1. Let (1.8)–(1.11) be the system of Born-Infeld equations, and let us nowconsider the Riemann Problem. Then there exist large states

u− = (D−2 , D−

3 , B−2 , B−

3 ), u+ = (D+2 , D+

3 , B+2 , B+

3 )

that can be connected by jump discontinuities, which are not contact discontinuities.Moreover, these jump discontinuities could be either Lax Shocks or Rarefaction Shocks.

Proof. 1. Let M be the manifold defined by (2.10), (2.11). Since M is conic, for all0 = ω ∈ M, we may write ω = (µ, v) with µ ∈ R

+ and v ∈ V , that is

M = (0,∞) × V ,

where V is a manifold of dimension 5. Moreover, we may assume without loss of generalitythat

V = M∩ S7,

and thus ‖ω‖ = µ. Further, V is globally compact. Analogously, for N , the manifolddefined by (2.24) with m = −n, we have

N = (0,∞) ×W ,

where W is a manifold of dimension 3. As N ⊂ M, it follows that W ⊂ V .

2. For α+γ− − α−γ+ = 0, φ : M → R, given by (2.5), (2.13), and (2.14), is well definedand, moreover, a C∞(M)-function. Let δ > 0 be sufficiently small. We set φ : M → R,

φ(µ, v) = φ22(µ, v)

for any µ < δ and v ∈ V . Since φ is not a continuous function, for each � ∈ Z+, we define

φ� : M → R,

φ�(µ, v) := H ′�(p

+) − H ′�(p

−)

+(γ+−γ−)(α+H�(p+)−α−H�(p−)) + (α+−α−)(γ+H�(p+)−γ−H�(p−))

α+γ− − α−γ+

for all µ < δ and v ∈ V , where

H�(z) = (z2 + 1/�2)1/2 − 1/� (z ∈ R, � ∈ Z+).

Consequently, φ� ∈ C∞(M) and φ� → φ uniformly as � → ∞; that is, given ε > 0, thereexists L ∈ Z

+ such that, for any µ < δ and v ∈ V ,

φ(µ, v) − ε < φ�(µ, v) < φ(µ, v) + ε,

for all � ≥ L. Furthermore, since φ(µ, w) = 0 for all (µ, w) ∈ (0, δ) × W , and ε > 0 isarbitrary, we have for any � ≥ L

φ�(µ, w) = 0.



Therefore, there exist v1, v2 ∈ V such that, for all � ≥ L,

φ�(µ, v1) < 0 < φ�(µ, v2).

3. Finally, let us show that there exists (µ0, v0) ∈ M such that φ(µ0, v0) = 0. Forδ > 0 above, we have φ(µ, v) ∼ φ(µ, v) for all (µ, v) ∈ (0, δ) × V ; then for � ∈ Z

+

sufficiently large, it follows that φ(µ, v) ∼ φ�(µ, v). In fact, since φ, φ� are continuousfunctions and V is compact, given ε > 0, there exist 0 < δ1 < δ2 < δ such that, for any(µ, v) ∈ [δ1, δ2] × V ,

φ(µ, v) − ε < φ�(µ, v) < φ(µ, v) + ε.

Now, from item 2 there exist v1, v2 ∈ V such that, for any µ ∈ [δ1, δ2],

φ(µ, v1) − ε < φ�(µ, v1) < 0 < φ�(µ, v2) < φ(µ, v2) + ε.

Letting ε → 0+, there exists µ0 ∈ [δ1, δ2] such that

φ(µ0, v1) < 0 < φ(µ0, v2).

Thus from the Intermediate Value Theorem, there exists v0 ∈ V such that

φ(µ0, v0) = 0,

where we have used that V is connected, which follows easy from Lemma 2.3. Indeed, itis enough to observe that V = U × U . �

3. The ill-posed Riemann problem. Since shocks could be formed, we shall showin this section that a further microscopic, i.e., local, theory is needed to complete theBorn-Infeld model. In fact, as the following theorem shows, this could happen in thepresence of rarefaction shocks. In this case, the Riemann Problem for the system ofBorn-Infeld equations can have many solutions, and thus the Born-Infeld model is notcomplete by itself.

Theorem 3.1. Let u� = (D�2, D

�3, B

�2, B

�3), ur = (Dr

2, Dr3, B

r2 , Br

3) be two states satisfyingthe Rankine-Hugoniot condition for the system of Born-Infeld equations. If u�, ur areconnected by a rarefaction shock, then the Riemann Problem of (1.8)–(1.11) in R

+ × R,with initial-data

(D2, D3, B2, B3)(0, x) =

{(D�

2, D�3, B

�2, B

�3) if x < 0,

(Dr2, D

r3, B

r2 , Br

3) if x > 0,(3.1)

is not well-posed. Moreover, the Born-Infeld model is not complete.

Proof. 1. By hypothesis, the states u�, ur satisfy the Rankine-Hugoniot jump condi-tion of a rarefaction shock; we denote by s its speed. So from Theorem 2.1,

s =P � + ζ�

h�=

P r + ζr

hr,

with ζ�, ζr = ±1, P �P r ≥ 0, where h and P are given by (1.4) with the obvious notations.Let u�, ur be two states to be determined, and we assume that (u�, ur) ∈ R

8 is closeenough to (u�, ur) ∈ R

8; that is,

u� ∼ u� and ur ∼ ur.



Therefore, u�, u� and ur, ur could be connected by two different contact discontinuities.For instance, we set

s1 = λ−(u�) = λ−(u�),

s2 = λ+(ur) = λ+(ur),(3.2)

with s1 < s2, the respective speeds of their contact discontinuities. Now, we seek if thereexists a jump discontinuity such that u�, ur are connected by it. Let us denote by s itsspeed. Moreover, we must have

s1 < s < s2,

and this jump discontinuity must be a rarefaction shock. Indeed, it is not possible tohave s < s1 or s > s2, since we have assumed (u�, ur) ∼ (u�, ur). Furthermore, we have

s1 =P � − 1

h�<

P � + ζ−

h�= s (ζ− > −1 ⇔ ζ+ < −1),

i.e., s is a rarefaction shock. Analogously,

s2 =P r + 1

hr>

P r + ζ+

hr= s (ζ+ < 1 ⇔ ζ− > 1).

2. The states u�, u�, ur, and ur of item 1 have to satisfy the following Rankine-Hugoniotjump conditions:

s1(D�2 − D�

2) = (B�3 + D�

2P�)(h�)−1 − (B�

3 + D�2P

�)(h�)−1,

s1(D�3 − D�

3) = (−B�2 + D�

3P�)(h�)−1 − (−B�

2 + D�3P

�)(h�)−1,

s1(B�2 − B�

2) = (−D�3 + B�

2P�)(h�)−1 − (−D�

3 + B�2P

�)(h�)−1,

s1(B�3 − B�

3) = (D�2 + B�

3P�)(h�)−1 − (D�

2 + B�3P

�)(h�)−1,

s(Dr2 − D�

2) = (Br3 + Dr

2Pr)(hr)−1 − (B�

3 + D�2P

�)(h�)−1,

s(Dr3 − D�

3) = (−Br2 + Dr

3Pr)(hr)−1 − (−B�

2 + D�3P

�)(h�)−1,

s(Br2 − B�

2) = (−Dr3 + Br

2P r)(hr)−1 − (−D�3 + B�

2P�)(h�)−1,

s(Br3 − B�

3) = (Dr2 + Br

3P r)(hr)−1 − (D�2 + B�

3P�)(h�)−1,

s2(Dr2 − Dr

2) = (Br3 + Dr

2Pr)(hr)−1 − (Br

3 + Dr2P

r)(hr)−1,

s2(Dr3 − Dr

3) = (−Br2 + Dr

3Pr)(hr)−1 − (−Br

2 + Dr3P

r)(hr)−1,

s2(Br2 − Br

2) = (−Dr3 + Br

2P r)(hr)−1 − (−Dr3 + Br

2P r)(hr)−1,

s2(Br3 − Br

3) = (Dr2 + Br

3P r)(hr)−1 − (Dr2 + Br

3P r)(hr)−1.



From the above equations and s1, s2 given by (3.2), we obtain

ψ1 := [D�2 + B�

3](h�)−1 − [D�

2 + B�3](h

�)−1 = 0,

ψ2 := [D�3 − B�

2](h�)−1 − [D�3 − B�

2](h�)−1 = 0,

ψ3 := [Br3 − Dr

2(shr − P r)](hr)−1 − [B�

3 − D�2(sh

� − P �)](h�)−1 = 0,

ψ4 := [−Br2 − Dr

3(shr − P r)](hr)−1 − [−B�2 − D�

3(sh� − P �)](h�)−1 = 0,

ψ5 := [−Dr3 − Br

2(shr − P r)](hr)−1 − [−D�3 − B�

2(sh� − P �)](h�)−1 = 0,

ψ6 := [Dr2 − Br

3(shr − P r)](hr)−1 − [D�2 − B�

3(sh� − P �)](h�)−1 = 0,

ψ7 := [Dr2 − Br

3 ](hr)−1 − [Dr2 − Br

3 ](hr)−1 = 0,

ψ8 := [Dr3 + Br

2 ](hr)−1 − [Dr3 + Br

2 ](hr)−1 = 0.

(3.3)

Hence, we have 8 equations to be satisfied by 9 variables, that is, ψi(s, u�, ur) = 0,(i = 1, . . . , 8). If there exists (s0, u

�0, u

r0) such that

ψ(s0, u�0, u

r0) = 0,

then we have a solution of the Riemann problem.

3. By definition, ψ : R × R8 → R

8 is a C∞ function; we set

[A]ij := ∂(j+1)ψi(s, u�, ur) (i, j = 1, . . . , 8).



After a straightforward calculation, we obtain the matrix A; for convenience we writeA = B + C, with B and C given respectively by

1−a�1f�

4√

h�

h�

a�1f�

3√

h�

h�

a�1f�

2√

h�

h�

1−a�1f�

1√

h�

h� 0 0 0 0

−a�2f�

4√

h�

h�

1+a�2f�

3√

h�

h�

−1+a�2f�

2√

h�

h�

−a�2f�

1√

h�

h� 0 0 0 0

a�31+f�

1f�4√

h�

h�

a�32−f�

1f�3√

h�

h�

a�33−f�

1f�2√

h�

h�

a�34+(f�

1)2√

h�

h� 0 0 0 0

a�41+f�

2f�4√

h�

h�

a�42−f�

2f�3√

h�

h�

a�43−(f�

2)2√

h�

h�

a�44+f�

2f�1√

h�

h� 0 0 0 0

a�51+f�

3f�4√

h�

h�

a�52−(f�

3)2√

h�

h�

a�53−f�

3f�2√

h�

h�

a�54+f�

3f�1√

h�

h� 0 0 0 0

a�61+(f�

4)2√

h�

h�

a�62−f�

4f�3√

h�

h�

a�63−f�

4f�2√

h�

h�

a�64+f�

4f�1√

h�

h� 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

,

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0ar35−fr

1 fr4√

hr

hrar36+fr

1 fr3√

hr

hrar37+fr

1 fr2√

hr

hrar38−(fr

1 )2√

hr

hr

0 0 0 0ar45−fr

2 fr4√

hr

hrar46+fr

2 fr3√

hr

hrar47+(fr

2 )2√

hr

hrar48−fr

2 fr1√

hr

hr

0 0 0 0ar55−fr

3 fr4√

hr

hrar56+(fr

3 )2√

hr

hrar57+fr

3 fr2√

hr

hrar58−fr

3 fr1√

hr

hr

0 0 0 0ar65−(fr

4 )2√

hr

hrar66+fr

4 fr3√

hr

hrar67+fr

4 fr2√

hr

hrar68−fr

4 fr1√

hr

hr

0 0 0 0−1+ar

7fr4√

hr

hr−ar

7fr3√

hr

hr−ar

7fr2√

hr

hr1+ar

7fr1√

hr

hr

0 0 0 0ar8f4

√hr

h�

−1−ar8f3

√hr

h�

−1−ar8f2

√hr

h�

ar8f1

√hr

h�

,

where f � = f(u�), fr = f(ur), f is the flux function of the Born-Infeld system, and

a�1 =

D�2 + B�

3

h�, a�

2 =D�

3 − B�2

h�, ar

7 =Dr

2 − Br3

hr, ar

8 =Dr

3 + Br2

hr.

Moreover, we have with the correspondent �, r superscripts

a31 = a64 = −a35 = −a68 = sh − (D2B3 + P ),

a42 = a53 = −a46 = −a57 = sh − (−D3B2 + P ),

a34 = −a38 = −[1 + (D2)2], a43 = −a47 = [1 + (D3)2],

a52 = −a56 = [1 + (B2)2], a61 = −a65 = −[1 + (B3)2],

a32 = −a54 = −a36 = a58 = D2B2, a51 = −a62 = −a55 = a66 = −B2B3,

a41 = −a63 = −a45 = a67 = −B3D3, a33 = −a44 = −a37 = a48 = D3D2.



Now, we study whether A is non-singular. To do this, we verify whether some row orcolumn is a linear combination of the others. We begin regarding the columns and, inthis case from the particular structure of A, it is enough to verify for B or C. Let usstudy B, and in order to simplify the notation, we drop the superscript �. Then, makinga linear combination with the columns of B, we obtain

α1f4 − α2f3 − α3f2 + α4f1 = 0,

α1 + α4 = 0, α2 − α3 = 0.(3.4)

4∑i=1

αiaji = 0 (j = 3, 4, 5, 6). (3.5)

The former equation of (3.4) implies for all (D2, D3, B2, B3) = 0,

(α1 + α4)(P + 1) = 0, (α2 − α3)(P + 1) = 0,

(α1 − α4)(P − 1) = 0, (α2 + α3)(P − 1) = 0.

Hence, when P = ±1, we have αi = 0, (i = 1, . . . , 4). However, we shall show that thecolumns of A are linear independent in any case. Indeed, for P = −1, we have α1 = α4

and α2 = −α3. Then, from the second equation of (3.4),

αi = 0 (i = 1, . . . , 4).

For P = 1, we have α1 = −α4 =: α and α2 = α3 =: α; thus, the second equation of (3.4)is satisfied, and from (3.5),

α [sh + D2(D2 − B3)] + α D2[B2 + D3] = 0,

α [sh − B3(D2 − B3)] − α B3[B2 + D3] = 0,

α D3[D2 − B3] + α [sh + D3(B2 + D3)] = 0,

α B2[D2 − B3] + α [sh + B2(B2 + D3)] = 0.

That is, we have four linear homogeneous equations for the two unknowns α and α. Itfollows that we could get a solution (α, α) = 0 if and only if D2 = −B3, D3 = B2 orD2 = B3, D3 = −B2, but both of them imply contradictions. In the first case, we have

P = −|D|2 = −|B|2 < 0.

For the second, we obtain

α sh = 0, α sh = 0,

and since h > 0, s must be zero, which implies from item 1 ζ = −1. Consequently, thecolumns of A are linear independent. Now, making a linear combination with the rows



of A, we obtain

α1 a�1 + α2 a�

2 −6∑

i=3

αif�(i−2) = 0, (3.6)

6∑i=3

αifr(i−2) − α7 ar

7 − α8 ar2 = 0, (3.7)

α1 +∑6

i=3 αia�i1 = 0, α2 +

∑6i=3 αia

�i2 = 0,

α2 −∑6

i=3 αia�i3 = 0, α1 +

∑6i=3 αia

�i4 = 0,

(3.8)

∑6i=3 αia

ri5 − αr

7 = 0,∑6

i=3 αiari6 − αr

8 = 0,∑6i=3 αia

ri7 − αr

8 = 0,∑6

i=3 αiari8 + αr

7 = 0.(3.9)

From (3.6), (3.7), we get for all (D�2, D

�3, B

�2, B

�3), (Dr

2, Dr3, B

r2 , Br

3) =0,

2α1 = (α3 + α6)(P � + 1), (α3 − α6)(P � − 1) = 0,

2α2 = (α4 − α5)(P � + 1), (α4 + α5)(P � − 1) = 0,

2α7 = (α3 − α6)(P r − 1), (α3 + α6)(P r + 1) = 0,

2α8 = (α4 + α5)(P r − 1), (α4 − α5)(P r + 1) = 0.

Then, when P �, P r = ±1, we have

αi = 0 (i = 1, . . . , 8).

Moreover, since P �P r ≥ 0 , we obtain respectively for P �, P r = −1 and P �, P r = 1

α1 = α2 = α7 = α8 = 0,

α3 = α6 = α, α4 = −α5 = α,(3.10)

α1 = α2 = α7 = α8 = 0,

α3 = −α6 = α, α4 = α5 = α.(3.11)

Now, we obtain from (3.8), (3.9) and (3.10) with the correspondent �, r superscripts

α [sh − B3(D2 + B3)] + α B3[B2 − D3] = 0,

α [sh − D2(D2 + B3)] + α D2[B2 − D3] = 0,

α B2[D2 + B3] − α [sh − B2(B2 − D3)] = 0,

α D3[D2 + B3] − α [sh + D3(B2 − D3)] = 0.

Thus, a similar argument as above for the columns implies α = α = 0. Analogously, weobtain for P �, P r = 1. Consequently, the matrix A is non-singular.

4. Finally, we show that for any s sufficiently close to s, that is, (s, u�, ur) in a neighbor-hood of (s, u�, ur), we have a solution for the Riemann problem. First, from item 2, weobserve that (s, u�, ur) satisfies

ψ(s, u�, ur) = 0.



Moreover, we recall from item 1 that s must be a rarefaction shock, which is the case fors. Now, from item 3, the matrix A defined as

[A]ij = ∂(j+1)ψi(s, u�, ur) (i, j = 1, . . . , 8)

is non-singular. Then by the Implicit Function Theorem, there exist a neighborhood U

of (s, u�, ur), an open set S ⊂ R, with s ∈ S, and φ ∈ C∞(S; R8), such that

{(s0, u�0, u

r0) ∈ U : ψ(s0, u

�0, u

r0) = 0} = {(s0, φ(s0)) : s0 ∈ S}.

So (1.8)–(1.11) are not complete by themselves; they must be augmented by some selec-tion criteria. �

Acknowledgments. Wladimir Neves’ research was supported in part by CNPq-Brazil, proc. 202351/02-5, FAPERJ, proc. 170439/03, FUJB, proc. 10643-7, and In-ternational Cooperation Agreement Brazil-France. The first author would like to thankthe staff of UMPA of the ENS de Lyon, and he is grateful for their warm hospitality.

Denis Serre’s research is partially funded by the European IHP project HYKE underthe contract HPRN-CT-2002-00282.

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€¦ · 366 wladimir neves and denis serre moreover, we recall from item 1 that ˜s must be a...

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