View
228
Download
0
Category
Preview:
Citation preview
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
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 = +∞.
September 8, 2015 2 / 33
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 < +∞.
September 8, 2015 3 / 33
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.
September 8, 2015 4 / 33
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.
September 8, 2015 5 / 33
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 (+−−−).
September 8, 2015 6 / 33
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.
September 8, 2015 7 / 33
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
).
September 8, 2015 8 / 33
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.
September 8, 2015 9 / 33
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.
September 8, 2015 10 / 33
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.
September 8, 2015 11 / 33
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, ...).
September 8, 2015 12 / 33
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
.
September 8, 2015 13 / 33
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.
September 8, 2015 14 / 33
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 ).
September 8, 2015 15 / 33
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
September 8, 2015 16 / 33
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.
September 8, 2015 17 / 33
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.
September 8, 2015 18 / 33
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.
September 8, 2015 19 / 33
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.
September 8, 2015 20 / 33
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〉.
September 8, 2015 21 / 33
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).
September 8, 2015 22 / 33
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).
September 8, 2015 23 / 33
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.
September 8, 2015 24 / 33
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〉.
September 8, 2015 25 / 33
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.
September 8, 2015 26 / 33
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).
September 8, 2015 27 / 33
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).
September 8, 2015 28 / 33
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.
September 8, 2015 29 / 33
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.
September 8, 2015 30 / 33
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.
September 8, 2015 31 / 33
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.
September 8, 2015 32 / 33
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.
September 8, 2015 33 / 33
Recommended