Born-Infeld equations in the electrostatic casealessiopomponio.altervista.org/Lavori/2015_Vienna -...
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Born-Infeld equations in the electrostatic case
Alessio Pomponio
Dipartimento di Meccanica, Matematica e Management, Politecnico di Bari
joint work with Denis Bonheure and Pietro d’Avenia
Quasilinear and nonlocal nonlinear Schrodinger equationsSeptember 28 - October 2, 2015,Wolfgang Pauli Institute, Vienna.
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 =12
∫R3|E|2 dx =
12
∫R3|∇φ|2 dx = +∞.
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 =12
∫R3|E|2 dx =
12
∫R3|∇φ|2 dx = +∞.
When ρ ∈ L1(R3), which is another relevant physical case, wecannot say, in general, that
− ∆φ = ρ (1)
admits a solution with finite energy.
Indeed
(i) by Gagliardo-Nirenberg-Sobolev inequality it is easy to seethat if ρ ∈ L6/5(R3), then (1) has a unique and finite energysolution;
(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 < +∞.
When ρ ∈ L1(R3), which is another relevant physical case, wecannot say, in general, that
− ∆φ = ρ (1)
admits a solution with finite energy.Indeed
(i) by Gagliardo-Nirenberg-Sobolev inequality it is easy to seethat if ρ ∈ L6/5(R3), then (1) has a unique and finite energysolution;
(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 < +∞.
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 ofMaxwell’s Lagrangian density.
Newton’s mechanics → Einstein’s mechanics
LN = 12 mv2 → LE = mc2(1−
√1− v2/c2)
Observe that
(i) for small velocities LN ∼ LE;
(ii) LE is real only when v2 6 c2 .
Newton’s mechanics → Einstein’s mechanics
LN = 12 mv2 → LE = mc2(1−
√1− v2/c2)
Observe that
(i) for small velocities LN ∼ LE;
(ii) LE is real only when v2 6 c2 .
Newton’s mechanics → Einstein’s mechanics
LN = 12 mv2 → LE = mc2(1−
√1− v2/c2)
Observe that
(i) for small velocities LN ∼ LE;
(ii) LE is real only when v2 6 c2 .
By analogy, starting from Maxwell’s Lagrangian density in thevacuum
LM = −FµνFµν
4,
where
Fµν = ∂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;
Born introduced the new Lagrangian density
LB = b2
(1−
√1 +
FµνFµν
2b2
)√−det(gµν),
where
b is a constant having the dimensions of e/r20 (e and r0 being
respectively the charge and the radius of the electron);
gµν is the Minkowski metric tensor with signature (+−−−).
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 +
Fµν
b
)),
whose integral is now invariant for general transformation.
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 +
Fµν
b
)),
whose integral is now invariant for general transformation.
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
).
The electrostatic case
In the electrostatic case we infer that
LB = LBI = b2
(1−
√1− |E|
2
b2
)= b2
(1−
√1− |∇φ|2
b2
).
Moreover when |E| ∼ 0 or b→ +∞, then LB = LBI ∼ LM.
In presence of a charge density ρ, we formally get the equation
−div
(∇φ√
1− |∇φ|2/b2
)= ρ,
which replaces the Poisson equation.
The electrostatic case
In the electrostatic case we infer that
LB = LBI = b2
(1−
√1− |E|
2
b2
)= b2
(1−
√1− |∇φ|2
b2
).
Moreover when |E| ∼ 0 or b→ +∞, then LB = LBI ∼ LM.
In presence of a charge density ρ, we formally get the equation
−div
(∇φ√
1− |∇φ|2/b2
)= ρ,
which replaces the Poisson equation.
The electrostatic case
In the electrostatic case we infer that
LB = LBI = b2
(1−
√1− |E|
2
b2
)= b2
(1−
√1− |∇φ|2
b2
).
Moreover when |E| ∼ 0 or b→ +∞, then LB = LBI ∼ LM.
In presence of a charge density ρ, we formally get the equation
−div
(∇φ√
1− |∇φ|2/b2
)= ρ,
which replaces the Poisson equation.
Remark
When ρ = δ0, one can easily explicitly compute the solution.
M.H.L. Pryce, On a Uniqueness Theorem, Math. Proc.Cambridge Philos. Soc. 31 (1935), 625–628.
The unique solution φρ is such that
φ′ρ(r) = −1√
1 + r2N−2.
Remark
The operator
Q−(φ) = −div
(∇φ√
1− |∇φ|2
),
also naturally appears in string theory and in classical relativity,where Q− represents the mean curvature operator inLorentz-Minkowski space.
References
The operator Q− has been studied in other situations by manyauthors 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, ...).
Let me mention in particular
R. Bartnik and L. Simon, Comm. Math. Phys. 87 (1982),whose ideas are fundamental in our arguments.
References
The operator Q− has been studied in other situations by manyauthors 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, ...).
Let me mention in particular
R. Bartnik and L. Simon, Comm. Math. Phys. 87 (1982),whose ideas are fundamental in our arguments.
The Born-Infeld 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 and with N > 3.
The functional framework
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
.
Properties of X
Lemma
(i) X is continuously embedded in W1,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.
Weak solutions
Definition
For a ρ ∈ X ∗, a weak solution of (BI) is a function φρ ∈ X suchthat for all ψ ∈ X , we have∫
RN
∇φρ · ∇ψ√1− |∇φρ|2
dx = 〈ρ, ψ〉,
where 〈 , 〉 denotes the duality pairing between X ∗ and X .
The functional I
The Born-Infeld equation is formally the Euler-Lagrange equationof the action functional
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,strictly convex, weakly lower semi-continuous.
Thus one can look for the solution as the minimizer of I in X bythe direct methods of the Calculus of Variations.
However, one needs to pay attention to the lack of regularity of thefunctional when ‖∇φ‖∞ = 1. In this case we cannot makevariations.
Lemma
The functional I is bounded from below, coercive, continuous,strictly convex, weakly lower semi-continuous.
Thus one can look for the solution as the minimizer of I in X bythe direct methods of the Calculus of Variations.
However, one needs to pay attention to the lack of regularity of thefunctional when ‖∇φ‖∞ = 1. In this case we cannot makevariations.
Lemma
The functional I is bounded from below, coercive, continuous,strictly convex, weakly lower semi-continuous.
Thus one can look for the solution as the minimizer of I in X bythe direct methods of the Calculus of Variations.
However, one needs to pay attention to the lack of regularity of thefunctional when ‖∇φ‖∞ = 1. In this case we cannot makevariations.
Definition
Let X be a real Banach space and ψ : X→ (−∞,+∞] be aconvex lower semicontinuous function. LetD(ψ) = u ∈ X | ψ(u) < +∞ be the effective domain of ψ. Foru ∈ 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 for I if−Φ′(u) ∈ ∂ψ(u), that is
〈Φ′(u), v− u〉+ ψ(v)− ψ(u) > 0, ∀v ∈ X.
A. Szulkin, Ann. Inst. H. Poincare Anal. Non Lineaire 3(1986), 77–109.
Remark
Observe that, according to the previous definition, φρ is a criticalpoint in weak sense for the functional I if and only if, for anyφ ∈ 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 minimizer for I.
Proposition
The infimum m = infφ∈X I(φ) is achieved by a uniqueφρ ∈ X \ 0.
Proposition
For any ρ ∈ X ∗, there exists a unique critical point in weak senseφρ of I.
Proposition
Assume ρ ∈ X ∗. If φ ∈ X is a weak solution of (BI), thenφ = φρ.
Proposition
The infimum m = infφ∈X I(φ) is achieved by a uniqueφρ ∈ X \ 0.
Proposition
For any ρ ∈ X ∗, there exists a unique critical point in weak senseφρ of I.
Proposition
Assume ρ ∈ X ∗. If φ ∈ X is a weak solution of (BI), thenφ = φρ.
Proposition
The infimum m = infφ∈X I(φ) is achieved by a uniqueφρ ∈ X \ 0.
Proposition
For any ρ ∈ X ∗, there exists a unique critical point in weak senseφρ of I.
Proposition
Assume ρ ∈ X ∗. If φ ∈ X is a weak solution of (BI), thenφ = φρ.
Question
Is 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 inthat direction.
Question
Is 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 inthat direction.
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 functionφρ satisfies ∫
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〉.
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).
The radial case
For τ ∈ O(N), φ ∈ X and ρ ∈ X ∗, we define φτ ∈ X asφτ(x) = φ(τx), for all x ∈ RN, and ρτ ∈ X ∗ as 〈ρτ, ψ〉 = 〈ρ, ψτ〉,for all ψ ∈ X .
Definition
We say that ρ ∈ X ∗ is radially distributed if ρτ = ρ, for anyτ ∈ 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)weak solution φρ ∈ X of (BI).
The radial case
For τ ∈ O(N), φ ∈ X and ρ ∈ X ∗, we define φτ ∈ X asφτ(x) = φ(τx), for all x ∈ RN, and ρτ ∈ X ∗ as 〈ρτ, ψ〉 = 〈ρ, ψτ〉,for all ψ ∈ X .
Definition
We say that ρ ∈ X ∗ is radially distributed if ρτ = ρ, for anyτ ∈ 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)weak solution φρ ∈ X of (BI).
Sketch of the proof
The argument is borrowed from
E. Serra, P. Tilli, Ann. Inst. H. Poincare Anal. Non Lineaire 28(2011).
• φρ ∈ Xrad;
• for k ∈N∗, we consider
Ek =
r > 0 |φ′ρ(r)| > 1− 1
k
;
• since |r > 0 | |φ′ρ(r)| = 1| = 0 then
|Ek| → 0 as k→ +∞;
• take ψ ∈ Xrad ∩ C∞c (RN) with supp ψ ⊂ [0, R] and let
ψk(r) = −∫ +∞
rψ′(s)[1− χEk(s)]ds.
Sketch of the proof
The argument is borrowed from
E. Serra, P. Tilli, Ann. Inst. H. Poincare Anal. Non Lineaire 28(2011).
• φρ ∈ Xrad;
• for k ∈N∗, we consider
Ek =
r > 0 |φ′ρ(r)| > 1− 1
k
;
• since |r > 0 | |φ′ρ(r)| = 1| = 0 then
|Ek| → 0 as k→ +∞;
• take ψ ∈ Xrad ∩ C∞c (RN) with supp ψ ⊂ [0, R] and let
ψk(r) = −∫ +∞
rψ′(s)[1− χEk(s)]ds.
Sketch of the proof
The argument is borrowed from
E. Serra, P. Tilli, Ann. Inst. H. Poincare Anal. Non Lineaire 28(2011).
• φρ ∈ Xrad;
• for k ∈N∗, we consider
Ek =
r > 0 |φ′ρ(r)| > 1− 1
k
;
• since |r > 0 | |φ′ρ(r)| = 1| = 0 then
|Ek| → 0 as k→ +∞;
• take ψ ∈ Xrad ∩ C∞c (RN) with supp ψ ⊂ [0, R] and let
ψk(r) = −∫ +∞
rψ′(s)[1− χEk(s)]ds.
Sketch of the proof
The argument is borrowed from
E. Serra, P. Tilli, Ann. Inst. H. Poincare Anal. Non Lineaire 28(2011).
• φρ ∈ Xrad;
• for k ∈N∗, we consider
Ek =
r > 0 |φ′ρ(r)| > 1− 1
k
;
• since |r > 0 | |φ′ρ(r)| = 1| = 0 then
|Ek| → 0 as k→ +∞;
• take ψ ∈ Xrad ∩ C∞c (RN) with supp ψ ⊂ [0, R] and let
ψk(r) = −∫ +∞
rψ′(s)[1− χEk(s)]ds.
Sketch of the proof
The argument is borrowed from
E. Serra, P. Tilli, Ann. Inst. H. Poincare Anal. Non Lineaire 28(2011).
• φρ ∈ Xrad;
• for k ∈N∗, we consider
Ek =
r > 0 |φ′ρ(r)| > 1− 1
k
;
• since |r > 0 | |φ′ρ(r)| = 1| = 0 then
|Ek| → 0 as k→ +∞;
• take ψ ∈ Xrad ∩ C∞c (RN) with supp ψ ⊂ [0, R] and let
ψk(r) = −∫ +∞
rψ′(s)[1− χEk(s)]ds.
Sketch of the proof
• supp ψk ⊂ [0, R], for any k > 1 and, if |t| 1, thenφρ + 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, we have∫ +∞
0
φ′ρψ′√1− |φ′ρ|2
[1−χEk ]rN−1dr→
∫ +∞
0
φ′ρψ′√1− |φ′ρ|2
rN−1dr;
• since ψk → ψ in X , we have
〈ρ, ψk〉 → 〈ρ, ψ〉.
Sketch of the proof
• supp ψk ⊂ [0, R], for any k > 1 and, if |t| 1, thenφρ + 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, we have∫ +∞
0
φ′ρψ′√1− |φ′ρ|2
[1−χEk ]rN−1dr→
∫ +∞
0
φ′ρψ′√1− |φ′ρ|2
rN−1dr;
• since ψk → ψ in X , we have
〈ρ, ψk〉 → 〈ρ, ψ〉.
Sketch of the proof
• supp ψk ⊂ [0, R], for any k > 1 and, if |t| 1, thenφρ + 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, we have∫ +∞
0
φ′ρψ′√1− |φ′ρ|2
[1−χEk ]rN−1dr→
∫ +∞
0
φ′ρψ′√1− |φ′ρ|2
rN−1dr;
• since ψk → ψ in X , we have
〈ρ, ψk〉 → 〈ρ, ψ〉.
• Thus for any ψ ∈ Xrad ∩ C∞c (RN), we conclude that∫
RN
∇φρ · ∇ψ√1− |∇φρ|2
dx = 〈ρ, ψ〉. (2)
• We show that (2) holds also for any ψ ∈ Xrad.
• Since (2) holds for any ψ ∈ X , we take ψ = φρ in (2) (φρ isradially symmetric) and we conclude.
• Thus for any ψ ∈ Xrad ∩ C∞c (RN), we conclude that∫
RN
∇φρ · ∇ψ√1− |∇φρ|2
dx = 〈ρ, ψ〉. (2)
• We show that (2) holds also for any ψ ∈ Xrad.
• Since (2) holds for any ψ ∈ X , we take ψ = φρ in (2) (φρ isradially symmetric) and we conclude.
• Thus for any ψ ∈ Xrad ∩ C∞c (RN), we conclude that∫
RN
∇φρ · ∇ψ√1− |∇φρ|2
dx = 〈ρ, ψ〉. (2)
• We show that (2) holds also for any ψ ∈ Xrad.
• Since (2) holds for any ψ ∈ X , we take ψ = φρ in (2) (φρ isradially symmetric) and we conclude.
A regularity result
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. Thenthe weak solution φρ of (BI) is C1(RN; R).
Bounded charge densities
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= yand the line segment xy ⊂ Ω;
• strictly spacelike if φ is spacelike, φ ∈ C1(Ω) and |∇φ| < 1 inΩ.
Moreover a line of slope 1 is called a light ray.
Theorem
If ρ ∈ L∞(RN) ∩ X ∗, then φρ is a strictly spacelike weak solutionof (BI).
Bounded charge densities
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= yand the line segment xy ⊂ Ω;
• strictly spacelike if φ is spacelike, φ ∈ C1(Ω) and |∇φ| < 1 inΩ.
Moreover a line of slope 1 is called a light ray.
Theorem
If ρ ∈ L∞(RN) ∩ X ∗, then φρ is a strictly spacelike weak solutionof (BI).
Bounded charge densities
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= yand the line segment xy ⊂ Ω;
• strictly spacelike if φ is spacelike, φ ∈ C1(Ω) and |∇φ| < 1 inΩ.
Moreover a line of slope 1 is called a light ray.
Theorem
If ρ ∈ L∞(RN) ∩ X ∗, then φρ is a strictly spacelike weak solutionof (BI).
Sketch of the proof
Let Ω be an arbitrary bounded domain with smooth boundary inRN. 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−
∫Ω
ρφ dx.
It is easy to see that φρ|Ω is a minimizer for IΩ in C0,1φρ(Ω).
Sketch of the proof
By [Bartink & Simon, CMP 1982] we have that φρ is strictlyspacelike weak solution 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| andxy ⊂ Ω.
If K = ∅, then φρ is strictly spacelike weak solution in Ω.
Sketch of the proof
By [Bartink & Simon, CMP 1982] we have that φρ is strictlyspacelike weak solution 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| andxy ⊂ Ω.
If K = ∅, then φρ is strictly spacelike weak solution in Ω.
Sketch of the proof
Assume by contradiction that K 6= ∅. Hence there exist x, y ∈ ∂Ωsuch that x 6= y, xy ⊂ Ω and |φρ(x)− φρ(y)| = |x− y|.Assume for example that φρ(x) > φρ(y). Then for all t ∈ (0, 1)
φρ(tx + (1− t)y) = φρ(y) + t|x− y|. (3)
Since, for any R > 0 such that Ω ⊂ BR, φρ|BR is a minimizer of
IBR in C0,1φρ(BR), then, by [Bartink & Simon, CMP 1982], we have
that (3) holds for all t ∈ R such that tx + (1− t)y ∈ BR. Now wereach a contradiction with the boundedness of φρ, for an Rsufficiently large.
Sketch of the proof
Assume by contradiction that K 6= ∅. Hence there exist x, y ∈ ∂Ωsuch that x 6= y, xy ⊂ Ω and |φρ(x)− φρ(y)| = |x− y|.Assume for example that φρ(x) > φρ(y). Then for all t ∈ (0, 1)
φρ(tx + (1− t)y) = φρ(y) + t|x− y|. (3)
Since, for any R > 0 such that Ω ⊂ BR, φρ|BR is a minimizer of
IBR in C0,1φρ(BR), then, by [Bartink & Simon, CMP 1982], we have
that (3) holds for all t ∈ R such that tx + (1− t)y ∈ BR. Now wereach a contradiction with the boundedness of φρ, for an Rsufficiently large.
An approximation result
Theorem
Let ρ ∈ X ∗ and suppose that there exist (ρn)n ⊂ X ∗ and ρ ∈ X ∗such that ρ > 0, ρn → ρ in X ∗ and −ρ 6 ρn 6 ρ. Then φρn (theminimizer associated to ρn) converges to φρ weakly in X anduniformly in RN.
Corollary
If ρ ∈ Lp(RN) ∩ X ∗, with 1 6 p < +∞ and (ρn)n is a standardsequence of mollifications of ρ, then φρ is the limit (weakly in Xand uniformly in RN) of a sequence (φρn)n of smooth strictlyspacelike solutions of (BI) with the data ρn.
An approximation result
Theorem
Let ρ ∈ X ∗ and suppose that there exist (ρn)n ⊂ X ∗ and ρ ∈ X ∗such that ρ > 0, ρn → ρ in X ∗ and −ρ 6 ρn 6 ρ. Then φρn (theminimizer associated to ρn) converges to φρ weakly in X anduniformly in RN.
Corollary
If ρ ∈ Lp(RN) ∩ X ∗, with 1 6 p < +∞ and (ρn)n is a standardsequence of mollifications of ρ, then φρ is the limit (weakly in Xand uniformly in RN) of a sequence (φρn)n of smooth strictlyspacelike solutions of (BI) with the data ρn.
The k point charges case
Let
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→ ∞.(4)
Let Γ =⋃
i 6=j xixj.
The k point charges case
Let
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→ ∞.(4)
Let Γ =⋃
i 6=j xixj.
Theorem (Part I)
The minimizer φρ is a classical solution of the equation in RN \ Γ,namely
−div
(∇φ√
1− |∇φ|2
)= 0
in the classical sense in RN \ Γ.
Furthermore, we have that
(i) φρ ∈ C∞(RN \ Γ) ∩ C(RN);
(ii) φρ is strictly spacelike on RN \ Γ;
(iii) for i 6= j, either φρ is a classical solution on xixj, or
φρ(txi + (1− t)xj) = tφρ(xi) + (1− t)φρ(xj), 0 < t < 1.
Theorem (Part I)
The minimizer φρ is a classical solution of the equation in RN \ Γ,namely
−div
(∇φ√
1− |∇φ|2
)= 0
in the classical sense in RN \ Γ. Furthermore, we have that
(i) φρ ∈ C∞(RN \ Γ) ∩ C(RN);
(ii) φρ is strictly spacelike on RN \ Γ;
(iii) for i 6= j, either φρ is a classical solution on xixj, or
φρ(txi + (1− t)xj) = tφρ(xi) + (1− t)φρ(xj), 0 < t < 1.
Sketch of the proof
Let Ω be an arbitrary bounded open domain with smoothboundary in RN \ x1, . . . , xk. Here ρ = 0.
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−
k
∑i=1
aiφ(xi).
It is easy to see that φρ|Ω is a minimizer for IΩ in C0,1φρ(Ω).
Sketch of the proof
Let Ω be an arbitrary bounded open domain with smoothboundary in RN \ x1, . . . , xk. Here ρ = 0. 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−
k
∑i=1
aiφ(xi).
It is easy to see that φρ|Ω is a minimizer for IΩ in C0,1φρ(Ω).
Sketch of the proof
By [Bartnik & Simon, CMP 1982] we have that φρ is strictlyspacelike weak solution 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| andxy ⊂ Ω.
If K = ∅, then φρ is strictly spacelike weak solution in Ω.Hence we will assume K 6= ∅ and we now show that K contains atmost Γ.
Sketch of the proof
By [Bartnik & Simon, CMP 1982] we have that φρ is strictlyspacelike weak solution 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| andxy ⊂ Ω.
If K = ∅, then φρ is strictly spacelike weak solution in Ω.
Hence we will assume K 6= ∅ and we now show that K contains atmost Γ.
Sketch of the proof
By [Bartnik & Simon, CMP 1982] we have that φρ is strictlyspacelike weak solution 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| andxy ⊂ Ω.
If K = ∅, then φρ is strictly spacelike weak solution in Ω.Hence we will assume K 6= ∅ and we now show that K contains atmost Γ.
Sketch of the proof
Assume by contradiction there exist x, y ∈ ∂Ω such that x 6= y,xy ⊂ Ω and |φρ(x)− φρ(y)| = |x− y| and such that the straightline spanned by xy intersects Γ at a finite number of points,possibly zero, at least in one direction.
Assume for example that φρ(x) > φρ(y). Then for all t ∈ (0, 1)
φρ(tx + (1− t)y) = φρ(y) + t|x− y|. (5)
Sketch of the proof
Assume by contradiction there exist x, y ∈ ∂Ω such that x 6= y,xy ⊂ Ω and |φρ(x)− φρ(y)| = |x− y| and such that the straightline spanned by xy intersects Γ at a finite number of points,possibly zero, at least in one direction.Assume for example that φρ(x) > φρ(y). Then for all t ∈ (0, 1)
φρ(tx + (1− t)y) = φρ(y) + t|x− y|. (5)
We can arbitrarily stretch Ω in the direction xy to build new opensets Ω′ ⊂ RN \ x1, . . . , xk with smooth boundaries and suchthat Ω ⊂ Ω′.
Since φρ|Ω′ is a minimizer for IΩ′ in C0,1φρ(Ω′), by [Bartnik & Simon,
CMP 1982], (5) holds for all t ∈ R such that tx + (1− t)y ∈ Ω′.We reach a contradiction with the boundedness of φρ by choosingΩ′ long enough in the direction xy.
Moreover, since ρ = 0 in RN \ Γ, by bootstrap arguments we inferthat φρ ∈ C∞(RN \ Γ).
We can arbitrarily stretch Ω in the direction xy to build new opensets Ω′ ⊂ RN \ x1, . . . , xk with smooth boundaries and suchthat Ω ⊂ Ω′.Since φρ|Ω′ is a minimizer for IΩ′ in C0,1
φρ(Ω′), by [Bartnik & Simon,
CMP 1982], (5) holds for all t ∈ R such that tx + (1− t)y ∈ Ω′.We reach a contradiction with the boundedness of φρ by choosingΩ′ long enough in the direction xy.
Moreover, since ρ = 0 in RN \ Γ, by bootstrap arguments we inferthat φρ ∈ C∞(RN \ Γ).
We can arbitrarily stretch Ω in the direction xy to build new opensets Ω′ ⊂ RN \ x1, . . . , xk with smooth boundaries and suchthat Ω ⊂ Ω′.Since φρ|Ω′ is a minimizer for IΩ′ in C0,1
φρ(Ω′), by [Bartnik & Simon,
CMP 1982], (5) holds for all t ∈ R such that tx + (1− t)y ∈ Ω′.We reach a contradiction with the boundedness of φρ by choosingΩ′ long enough in the direction xy.
Moreover, since ρ = 0 in RN \ Γ, by bootstrap arguments we inferthat φρ ∈ C∞(RN \ Γ).
Theorem (Part II)
The minimizer 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〉.
Sketch of the proof
As consequence of a classical Trudinger result, for every boundeddomain Ω such that Ω ⊂ RN \ x1, . . . , xk, there exists a uniquedistributional solution φρ of the problem
−div
∇φ√1− |∇φρ|2
= 0, in Ω,
φ = φρ, on ∂Ω.
By Theorem - Part I and the uniqueness of the Trudinger result,we infer that φρ = φρ.
Sketch of the proof
As consequence of a classical Trudinger result, for every boundeddomain Ω such that Ω ⊂ RN \ x1, . . . , xk, there exists a uniquedistributional solution φρ of the problem
−div
∇φ√1− |∇φρ|2
= 0, in Ω,
φ = φρ, on ∂Ω.
By Theorem - Part I and the uniqueness of the Trudinger result,we infer that φρ = φρ.
Theorem (Part III)
Let φρ the minimizer, as before. Then
(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 spacelike on
RN \ x1, . . . , xk andlimx→xi|∇φρ(x)| = 1.
Theorem (Part III)
Let φρ the minimizer, as before. Then
(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 spacelike on
RN \ x1, . . . , xk andlimx→xi|∇φρ(x)| = 1.
Theorem (Part III)
Let φρ the minimizer, as before. Then
(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 spacelike on
RN \ x1, . . . , xk andlimx→xi|∇φρ(x)| = 1.
Sketch of the Proof
Proof of (i)
Arguing as before, we aim to prove that K = ∅.
Assume by contradiction that K 6= ∅. Then there exist x, y ∈ ∂Ωsuch that x 6= y, xy ⊂ Ω and |φρ(x)− φρ(y)| = |x− y|. We canassume that φρ(x) > φρ(y) and so for all t ∈ (0, 1)
φρ(tx + (1− t)y) = φρ(y) + t|x− y|.
Two possibilities occur: either xy intersects Γ in a finite number ofpoints (possibly zero), or xy intersects Γ in an infinite number ofpoints.
In the first case, we conclude as before.
Sketch of the Proof
Proof of (i)
Arguing as before, we aim to prove that K = ∅.Assume by contradiction that K 6= ∅. Then there exist x, y ∈ ∂Ωsuch that x 6= y, xy ⊂ Ω and |φρ(x)− φρ(y)| = |x− y|. We canassume that φρ(x) > φρ(y) and so for all t ∈ (0, 1)
φρ(tx + (1− t)y) = φρ(y) + t|x− y|.
Two possibilities occur: either xy intersects Γ in a finite number ofpoints (possibly zero), or xy intersects Γ in an infinite number ofpoints.
In the first case, we conclude as before.
Sketch of the Proof
Proof of (i)
Arguing as before, we aim to prove that K = ∅.Assume by contradiction that K 6= ∅. Then there exist x, y ∈ ∂Ωsuch that x 6= y, xy ⊂ Ω and |φρ(x)− φρ(y)| = |x− y|. We canassume that φρ(x) > φρ(y) and so for all t ∈ (0, 1)
φρ(tx + (1− t)y) = φρ(y) + t|x− y|.
Two possibilities occur: either xy intersects Γ in a finite number ofpoints (possibly zero), or xy intersects Γ in an infinite number ofpoints.
In the first case, we conclude as before.
Sketch of the Proof
In the second case we can assume that xy coincides with a piece ofx1x2 and we can consider any extension of xy in x1x2.
Lemma
For any ε > 0 there exists σ > 0 such that, if maxi=1,...,k |ai| < σ,then ‖φρ‖∞ < ε.
Fixing ε > 0 such that 2ε < mini,j=1,...,k, i 6=j |xi − xj|, there existsσ > 0 such that, if maxi=1,...,k |ai| < σ, then ‖φρ‖∞ < ε. Since we
can find x′, y′ ∈ x1x2 with |x′ − y′| > 2ε and x′y′ 6= x1x2, we reacha contradiction:
2ε < |x′ − y′| = |φρ(x′)− φρ(y′)| < 2ε.
Sketch of the Proof
In the second case we can assume that xy coincides with a piece ofx1x2 and we can consider any extension of xy in x1x2.
Lemma
For any ε > 0 there exists σ > 0 such that, if maxi=1,...,k |ai| < σ,then ‖φρ‖∞ < ε.
Fixing ε > 0 such that 2ε < mini,j=1,...,k, i 6=j |xi − xj|, there existsσ > 0 such that, if maxi=1,...,k |ai| < σ, then ‖φρ‖∞ < ε. Since we
can find x′, y′ ∈ x1x2 with |x′ − y′| > 2ε and x′y′ 6= x1x2, we reacha contradiction:
2ε < |x′ − y′| = |φρ(x′)− φρ(y′)| < 2ε.
Sketch of the Proof
In the second case we can assume that xy coincides with a piece ofx1x2 and we can consider any extension of xy in x1x2.
Lemma
For any ε > 0 there exists σ > 0 such that, if maxi=1,...,k |ai| < σ,then ‖φρ‖∞ < ε.
Fixing ε > 0 such that 2ε < mini,j=1,...,k, i 6=j |xi − xj|, there existsσ > 0 such that, if maxi=1,...,k |ai| < σ, then ‖φρ‖∞ < ε. Since we
can find x′, y′ ∈ x1x2 with |x′ − y′| > 2ε and x′y′ 6= x1x2, we reacha contradiction:
2ε < |x′ − y′| = |φρ(x′)− φρ(y′)| < 2ε.
Sketch of the Proof
Proof of (ii)
We repeat the previous arguments and use the following
Lemma
There exists C = C(a1, . . . , ak) > 0 such that, for all xi ∈ RN,i = 1, . . . , k, ‖φρ‖∞ < C.
The behavior of the gradient of φρ near the singularities xi is aconsequence of a result contained in
K. Ecker, Area maximizing hypersurfaces in Minkowski spacehaving an isolated singularity, Manuscr. Math. 56 (1986),375–397.
Sketch of the Proof
Proof of (ii)
We repeat the previous arguments and use the following
Lemma
There exists C = C(a1, . . . , ak) > 0 such that, for all xi ∈ RN,i = 1, . . . , k, ‖φρ‖∞ < C.
The behavior of the gradient of φρ near the singularities xi is aconsequence of a result contained in
K. Ecker, Area maximizing hypersurfaces in Minkowski spacehaving an isolated singularity, Manuscr. Math. 56 (1986),375–397.
Approximations of the minimizer
Formally, the operator Q− can be expended as a sum of2h-Laplacian, namely
Q−(φ) = −∞
∑h=1
αh∆2hφ,
where for all h > 1, αh > 0 and ∆2hφ := div(|∇φ|2h−2∇φ).
The curvature operator is formally the Gateaux derivative of thefunctional∫
RN
(1−
√1− |∇φ|2
)dx =
∫RN
∞
∑h=1
αh
2h|∇φ|2h dx,
where the power series in the right hand side converges pointwisewhen |∇φ(x)| 6 1.
Approximations of the minimizer
Formally, the operator Q− can be expended as a sum of2h-Laplacian, namely
Q−(φ) = −∞
∑h=1
αh∆2hφ,
where for all h > 1, αh > 0 and ∆2hφ := div(|∇φ|2h−2∇φ).
The curvature operator is formally the Gateaux derivative of thefunctional∫
RN
(1−
√1− |∇φ|2
)dx =
∫RN
∞
∑h=1
αh
2h|∇φ|2h dx,
where the power series in the right hand side converges pointwisewhen |∇φ(x)| 6 1.
For n > 1, we define X2n as the completion of C∞c (RN) with
respect to the norm
‖φ‖2X2n
:=∫
RN|∇φ|2 +
(∫RN|∇φ|2ndx
)1/n
.
Assuming ρ ∈ X ∗2n, let us denote the nth approximation of thefunctional I by
In := φ ∈ X2n 7→n
∑h=1
αh
2h
∫RN|∇φ|2h dx− 〈ρ, φ〉X2n .
This functional is C1.
For n > 1, we define X2n as the completion of C∞c (RN) with
respect to the norm
‖φ‖2X2n
:=∫
RN|∇φ|2 +
(∫RN|∇φ|2ndx
)1/n
.
Assuming ρ ∈ X ∗2n, let us denote the nth approximation of thefunctional I by
In := φ ∈ X2n 7→n
∑h=1
αh
2h
∫RN|∇φ|2h dx− 〈ρ, φ〉X2n .
This functional is C1.
Proposition
Given n0 > 1 and ρ ∈ X ∗2n0, then, for all n > n0, the functional
In : X2n → R has one and only one critical point φn.
We can cover the following case
• ρ ∈ L1(RN);
• ρ = ∑ki=1 aiδxi ;
• ρ is Radon measure.
Theorem
If ρ ∈ X ∗2n0for some n0 > 1, then φn tends to φρ weakly in X2m
for all m > n0.
Proposition
Given n0 > 1 and ρ ∈ X ∗2n0, then, for all n > n0, the functional
In : X2n → R has one and only one critical point φn.
We can cover the following case
• ρ ∈ L1(RN);
• ρ = ∑ki=1 aiδxi ;
• ρ is Radon measure.
Theorem
If ρ ∈ X ∗2n0for some n0 > 1, then φn tends to φρ weakly in X2m
for all m > n0.
Proposition
Given n0 > 1 and ρ ∈ X ∗2n0, then, for all n > n0, the functional
In : X2n → R has one and only one critical point φn.
We can cover the following case
• ρ ∈ L1(RN);
• ρ = ∑ki=1 aiδxi ;
• ρ is Radon measure.
Theorem
If ρ ∈ X ∗2n0for some n0 > 1, then φn tends to φρ weakly in X2m
for all m > n0.
Thank you
for your attention!!!