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Page 1: Born-Infeld equations in the electrostatic caseAvenia.pdf · Born-Infeld equations in the electrostatic case Pietro d’Avenia Dipartimento di Meccanica, Matematica e Management Politecnico

Born-Infeld equations in the electrostatic case

Pietro d’Avenia

Dipartimento di Meccanica, Matematica e ManagementPolitecnico di Bari

Workshop in Nonlinear PDEs,Bruxelles, September 8, 2015

joint work with Denis Bonheure and Alessio Pomponio

September 8, 2015 1 / 33

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Introduction The infinity problem

Let us consider the Poisson equation

−∆φ = ρ in R3. (1)

In the classical Maxwell theory, φ is the electrostatic potentialgenerated by the charge density ρ.

If ρ = δ0, we get the

infinity problem associated with a point charge source:

the solution of (1) is φ(x) = 1/(4π|x|), but its energy is

H =1

2

∫R3

|E|2 dx =1

2

∫R3

|∇φ|2 dx = +∞.

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Introduction The infinity problem

When ρ ∈ L1(R3), which is another relevant physical case, we cannotsay, in general, that

−∆φ = ρ (1)

admits a solution with finite energy.Indeed

(i) by Gagliardo-Nirenberg-Sobolev inequality it is easy to see that ifρ ∈ L6/5(R3), then (1) has a unique and finite energy solution;

(ii) if, e.g.

ρ(x) =1

|x|5/2 + |x|7/2(∈ L1(R3) \ L6/5(R3))

then (1) has no radial solutions with H < +∞.

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Introduction Born solution

To avoid the violation of the principle of finiteness, Max Born in

M. Born, Modified field equations with a finite radius of theelectron, Nature 132 (1933), 282.

M. Born, On the quantum theory of the electromagnetic field, Proc.Roy. Soc. London Ser. A 143 (1934), 410–437.

proposed a nonlinear theory starting from a modification of Maxwell’sLagrangian density.

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Introduction Born solution

Newton’s mechanics → Einstein’s mechanicsLN = 1

2mv2 → LE = mc2(1−

√1− v2/c2)

(i) one of the simplest which is real only when v2 < c2;(ii) for small velocities LN ∼ LE.

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Introduction Born solution

By analogy, starting from Maxwell’s Lagrangian density in the vacuum

LM = −FµνFµν

4,

whereFµν = ∂µAν − ∂νAµ;(A0, A1, A2, A3) = (φ,−A) is the electromagnetic potential;(x0, x1, x2, x3) = (t, x);∂j denotes the partial derivative with respect to xj ;

and Born introduced the new Lagrangian density

LB = b2

(1−

√1 +

FµνFµν

2b2

)√−det(gµν),

whereb is a constant having the dimensions of e/r2

0 (e and r0 beingrespectively the charge and the radius of the electron);gµν is the Minkowski metric tensor with signature (+−−−).

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Introduction Born-Infeld action

Born’s action, as well as Maxwell’s action, is invariant only for theLorentz group of transformations (orthogonal transformations).

Some months later, Born and Infeld in

M. Born, L. Infeld, Foundations of the new field theory, Nature 132(1933), 1004.

M. Born, L. Infeld, Foundations of the new field theory, Proc. Roy.Soc. London Ser. A 144 (1934), 425–451.

introduced a modified version of the Lagrangian density

LBI = b2

(√−det(gµν)−

√−det

(gµν +

Fµνb

)),

whose integral is now invariant for general transformation.

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Introduction Born-Infeld action

Since the electromagnetic field (E,B) is given by

B = ∇×A and E = −∇φ− ∂tA,

we get

LM =|E|2 − |B|2

2, LB = b2

(1−

√1− |E|

2 − |B|2b2

)

and

LBI = b2

(1−

√1− |E|

2 − |B|2b2

− (E ·B)2

b4

).

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Introduction The electrostatic case

In the electrostatic case we infer that

LB = LBI = b2

(1−

√1− |E|

2

b2

)= b2

(1−

√1− |∇φ|

2

b2

).

In presence of a charge density ρ, we formally get the equation

−div

(∇φ√

1− |∇φ|2/b2

)= ρ,

which replaces the Poisson equation.

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Introduction Remarks

RemarkWhen ρ = δ0, one can easily explicitly compute the solution.

M.H.L. Pryce, On a Uniqueness Theorem, Math. Proc. CambridgePhilos. Soc. 31 (1935), 625–628.

φ′ρ(r) = − 1√1 + r2N−2

.

RemarkThe operator

Q−(φ) = −div

(∇φ√

1− |∇φ|2

),

also naturally appears in string theory and in classical relativity, whereQ− represents the mean curvature operator in Lorentz-Minkowskispace.

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The problem Our equation

We consider the problem−div

(∇φ√

1− |∇φ|2

)= ρ, x ∈ RN ,

lim|x|→∞

φ(x) = 0,

(BI)

for general non-trivial charge distributions ρ.

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The problem References

This problem has motivated several publications in the past years.

R. Bartnik and L. Simon, Comm. Math. Phys. 87 (1982).(its ideas are fundamental in our arguments)

Moreover, the operator Q− has been studied in other situations bymany authors in the recent years (Azzollini, Bereanu, Bonheure,Brezis, Coelho, Corsato, Derlet, De Coster, Fortunato, Jebelean,Kiessling, Mawhin, Mugnai, Obersnel, Omari, Orsina, Pisani, Rivetti,Torres, Wang, Yu, ...).

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Functional setting The space

Assuming N > 3, we work on

X = D1,2(RN ) ∩ φ ∈ C0,1(RN ) | ‖∇φ‖∞ 6 1,

equipped with the norm defined by

‖φ‖X :=

(∫RN|∇φ|2 dx

)1/2

.

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Functional setting Properties of X

Lemma

(i) X is continuously embedded in W 1,p(RN ), for allp > 2∗ = 2N/(N − 2);

(ii) X is continuously embedded in L∞(RN );

(iii) if φ ∈ X , then lim|x|→∞ φ(x) = 0;

(iv) X is weakly closed.

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Functional setting Weak solutions

For a ρ ∈ X ∗, weak solutions are understood in the following sense.

Definition

A weak solution of (BI) is a function φρ ∈ X such that for all ψ ∈ X , wehave ∫

RN

∇φρ · ∇ψ√1− |∇φρ|2

dx = 〈ρ, ψ〉, (2)

where 〈 , 〉 denotes the duality pairing between X ∗ and X .

RemarkIf ρ is a distribution, the weak formulation of (2) extends to any testfunction ψ ∈ C∞c (RN ).

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Functional setting The functional

As Born-Infeld equation is formally the Euler equation of the actionfunctional

I(φ) =

∫RN

(1−

√1− |∇φ|2

)dx− 〈ρ, φ〉,

we expect that one can derive existence and uniqueness of thesolution from a variational principle.

Lemma

The functional I is bounded from below, coercive, continuous, strictlyconvex, weakly lower semi-continuous.

Thus one can look for the solution as the minimizer of I in X by thedirect methods of the Calculus of Variations.However, one needs to pay attention to the lack of regularity of thefunctional when ‖∇φ‖∞ = 1.Hence we use the following classical definitions

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Critical point in weak sense Definition

DefinitionLet X be a real Banach space and ψ : X → (−∞,+∞] be a convexlower semicontinuous function. Let D(ψ) = u ∈ X | ψ(u) < +∞ bethe effective domain of ψ. For u ∈ D(ψ), the set

∂ψ(u) = u∗ ∈ X∗ | ψ(v)− ψ(u) > 〈u∗, v − u〉, ∀v ∈ X

is called the subdifferential of ψ at u. If, moreover, we consider afunctional I = ψ + Φ, with ψ as above and Φ ∈ C1(X,R), thenu ∈ D(ψ) is said to be critical in weak sense if −Φ′(u) ∈ ∂ψ(u), that is

〈Φ′(u), v − u〉+ ψ(v)− ψ(u) > 0, ∀v ∈ X.

A. Szulkin, Ann. Inst. H. Poincaré Anal. Non Linéaire 3 (1986),77–109.

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Critical point in weak sense CP and minimum

Remark

Observe that, according to the previous definition, φρ is a critical pointin weak sense for the functional I if and only if, for any φ ∈ X we get∫RN

(1−

√1− |∇φ|2

)dx−

∫RN

(1−

√1− |∇φρ|2

)dx > 〈ρ, φ〉− 〈ρ, φρ〉,

which is simply equivalent to require that φρ is a minimum for I.

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Critical point in weak sense Existence and uniqueness

...and

Proposition

The infimum m = infφ∈X I(φ) is achieved by a unique φρ ∈ X \ 0.

easily follows from the properties of I.Thus we can conclude with

Theorem

For any ρ ∈ X ∗, there exists a unique critical point in weak sense φρ ofI.

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Critical point in weak sense Further properties

Proposition

Assume ρ ∈ X ∗. If φ ∈ X is a weak solution of (BI), then φ = φρ.

QuestionIs it true that the unique minimizer φρ is always a weak solution of(BI)?

We are not able to answer this question in its full generality but weconjecture a positive answer and the following statement goes in thatdirection.

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Critical point in weak sense Further properties

Proposition

Assume ρ ∈ X ∗ and let φρ be the unique minimizer of I in X . Then

E = x ∈ RN | |∇φρ| = 1

is a null set (with respect to Lebesgue measure) and the function φρsatisfies ∫

RN

|∇φρ|2√1− |∇φρ|2

dx 6 〈ρ, φρ〉.

Moreover, for all ψ ∈ X , we have the variational inequality∫RN

|∇φρ|2√1− |∇φρ|2

dx−∫RN

∇φρ · ∇ψ√1− |∇φρ|2

dx 6 〈ρ, φρ〉 − 〈ρ, ψ〉.

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Critical point in weak sense Further properties

Remark

If φρ satisfies further∫RN

|∇φρ|2√1− |∇φρ|2

dx = 〈ρ, φρ〉,

then it is easy to see that φρ is a weak solution of (BI).

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Radially distributed charge densities Definition and Theorem

For τ ∈ O(N), φ ∈ X and ρ ∈ X ∗, we define φτ ∈ X as φτ (x) = φ(τx),for all x ∈ RN , and ρτ ∈ X ∗ as 〈ρτ , ψ〉 = 〈ρ, ψτ 〉, for all ψ ∈ X .

DefinitionWe say that ρ ∈ X ∗ is radially distributed if ρτ = ρ, for any τ ∈ O(N).

We next define

Xrad = φ ∈ X | φτ = φ for every τ ∈ O(N).

Theorem

If ρ ∈ X ∗ is radially distributed, then there exists a unique (radial) weaksolution φρ ∈ X of (BI).

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Radially distributed charge densities Proof

The argument is borrowed from

E. Serra, P. Tilli, Ann. Inst. H. Poincaré Anal. Non Linéaire 28(2011).

• φρ ∈ Xrad;• Define, for k ∈ N∗, the sets

Ek =

r > 0 |φ′ρ(r)| > 1− 1

k

and

|r > 0 | |φ′ρ(r)| = 1| = 0⇒

∣∣∣∣∣∣⋂k>1

Ek

∣∣∣∣∣∣ = 0

• Take ψ ∈ Xrad ∩ C∞c (RN ) with suppψ ⊂ [0, R] and let

ψk(r) = −∫ +∞

rψ′(s)[1− χEk(s)]ds.

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Radially distributed charge densities Proof

• We have suppψk ⊂ [0, R], for any k > 1 and, if |t| is sufficientlysmall, then φρ + tψk ∈ X .• Since φρ is the minimizer of I

0 = limt→0

I(φρ + tψk)− I(φρ)

t

= ωN

∫ +∞

0

φ′ρψ′√

1− |φ′ρ|2[1− χEk ]rN−1dr − 〈ρ, ψk〉

• Since χEk → 0 a.e. in RN and by Lebesgue’s DominatedConvergence Theorem, we have∫ +∞

0

φ′ρψ′√

1− |φ′ρ|2[1− χEk ]rN−1dr →

∫ +∞

0

φ′ρψ′√

1− |φ′ρ|2rN−1dr.

• Since ψk → ψ in X , we have

〈ρ, ψk〉 → 〈ρ, ψ〉.

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Radially distributed charge densities Proof

• Thus for any ψ ∈ Xrad ∩ C∞c (RN ), we conclude that∫RN

∇φρ · ∇ψ√1− |∇φρ|2

dx = 〈ρ, ψ〉. (3)

• To show that (3) holds for any ψ ∈ Xrad, we construct(ψn)n ⊂ C∞c (RN ), ψn radially symmetric such that ψn → ψ inD1,2(RN ) and with ‖∇ψn‖∞ 6 C. Then we apply

Lemma

Assume ρ ∈ X ∗ and let φρ be the unique minimizer of I in X . If(ψn)n ⊂ D1,2(RN ) is such that ‖∇ψn‖∞ 6 C for some C > 0 andψn → ψ in D1,2(RN ) then, up to a subsequence,

limn→∞

∫RN

∇φρ · ∇ψn√1− |∇φρ|2

dx =

∫RN

∇φρ · ∇ψ√1− |∇φρ|2

dx.

• To show that (3) holds for any ψ ∈ X , we take ψ = φρ in (3) (φρ isradially symmetric) and we conclude.

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Radially distributed charge densities Proof

Assuming further hypotheses on ρ, we can prove

Theorem

Assume that ρ is a radially symmetric function such thatρ ∈ Ls(RN ) ∩ Lσ(Bδ(0)), for some s > 1, σ > N and δ > 0. Then theweak solution φρ of (BI) is C1(RN ;R).

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Bounded charge densities Theorem

Definition

Let φ ∈ C0,1(Ω), with Ω ⊂ RN . We say that φ is• weakly spacelike if |∇φ| 6 1 a.e. in Ω;• spacelike |φ(x)− φ(y)| < |x− y| whenever x, y ∈ Ω, x 6= y and the

line segment xy ⊂ Ω;• strictly spacelike if φ is spacelike, φ ∈ C1(Ω) and |∇φ| < 1 in Ω.

Theorem

If ρ ∈ L∞loc(RN ) ∩ X ∗, then φρ is a (locally strictly) space-like weaksolution of (BI).

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Bounded charge densities Proof

Let Ω be an arbitrary bounded domain with smooth boundary in RN .We set

C0,1φρ

(Ω) =φ ∈ C0,1(Ω) | φ|∂Ω = φρ|∂Ω, |∇φ| 6 1

,

K = xy ⊂ Ω | x, y ∈ ∂Ω, x 6= y, |φρ(x)− φρ(y)| = |x− y| ,

and define IΩ : C0,1φρ

(Ω)→ R by

IΩ(φ) =

∫Ω

(1−

√1− |∇φ|2

)dx−

∫Ωρφ dx.

It is easy to see that φρ|Ω is a minimizer for IΩ in C0,1φρ

(Ω).By [BS82, Corollary 4.2] we have that φρ is strictly spacelike in Ω \Kand Q−(φρ) = ρ in Ω \K. Furthermore,

φρ(tx+ (1− t)y) = tφρ(x) + (1− t)φρ(y), 0 < t < 1

for every x, y ∈ ∂Ω such that |φρ(x)− φρ(y)| = |x− y| and xy ⊂ Ω.If K = ∅, then φρ is strictly spacelike in Ω.

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Bounded charge densities Proof

Assume by contradiction that K 6= ∅. Then there exist x, y ∈ ∂Ω suchthat x 6= y, xy ⊂ Ω and |φρ(x)− φρ(y)| = |x− y|.Without loss of generality we can assume that φρ(x) > φρ(y). It is easyto see that for all t ∈ (0, 1)

φρ(tx+ (1− t)y) = φρ(y) + t|x− y|. (4)

Since, for any R > 0 such that Ω ⊂ BR, φρ|BR is a minimizer of IBR inC0,1φρ

(BR), then, by [BS82,Theorem 3.2], we have that (4) holds for allt ∈ R such that tx+ (1− t)y ∈ BR. Now we reach a contradiction withthe boundedness of φρ, for an R sufficiently large.

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The k point charges case Definitions

ρ =

k∑i=1

aiδxi ,

where ai ∈ R and xi ∈ RN , for i = 1, . . . , k, k ∈ N∗.We consider the problem −div

(∇φ√

1− |∇φ|2

)=

k∑i=1

aiδxi , in RN ,

φ(x)→ 0, as x→∞.(5)

The existence of a unique minimizer φρ of the associated energyfunctional

I(φ) =

∫RN

(1−

√1− |∇φ|2

)dx−

k∑i=1

aiφ(xi),

is done before.We want to prove that this minimizer solves (5) in a weak or a strongsense. We are able to prove this fact in some particular cases only.

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The k point charges case Theorem

Let Γ =⋃i 6=j xixj .

Theorem

The minimum φρ is a distributional solution of the Euler-Lagrangeequation in RN \ x1, . . . , xk. Namely, for everyψ ∈ C∞c (RN \ x1, . . . , xk), we have∫

RN

∇φρ · ∇ψ√1− |∇φρ|2

dx = 〈ρ, ψ〉.

It is a classical solution of the equation in RN \ Γ, namelyφρ ∈ C∞(RN \ Γ) and

−div

(∇φ√

1− |∇φ|2

)= 0

in the classical sense in RN \ Γ.

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The k point charges case Theorem

Moreover,

(i) for any fixed xi ∈ RN , i = 1, . . . , k, there existsσ = σ(x1, . . . , xk) > 0 such that if

maxi=1,...,k

|ai| < σ,

then φρ is a classical solution in RN \ x1, . . . , xk;

(ii) for any ai ∈ R, i = 1, . . . , k, there exists τ = τ(a1, . . . , ak) > 0 suchthat if

mini,j=1,...,k, i 6=j

|xi − xj | > τ,

then φρ is a classical solution in RN \ x1, . . . , xk.In these last cases, φρ ∈ C∞(RN \ x1, . . . , xk), it is strictly spacelikeon RN \ x1, . . . , xk and

limx→xi

|∇φρ(x)| = 1.

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