Towards a unified field theory for classical
electrodynamics
Vieri Benci∗, Donato Fortunato∗∗∗Dipartimento di Matematica Applicata ”U. Dini”Universita degli Studi di Pisa Universita di Pisa
via Bonanno 25/b, 56126 Pisa, Italye-mail: [email protected]
∗∗Dipartimento di MatematicaUniversita di Bari,
Via Orabona 4, 70125 Bari, Italye-mail: [email protected]
January 5, 2005
Contents
1 Introduction 1
2 The semilinear Maxwell equations 42.1 Invariants of motion . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Static solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.3 Solitary waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.4 Analysis of static solutions . . . . . . . . . . . . . . . . . . . . . . 10
3 Existence of static solutions 133.1 Statement of the result . . . . . . . . . . . . . . . . . . . . . . . . 143.2 The functional setting . . . . . . . . . . . . . . . . . . . . . . . . 153.3 Proof of the existence theorem . . . . . . . . . . . . . . . . . . . 23
4 Conclusions 344.1 A formulation of classical physics . . . . . . . . . . . . . . . . . . 344.2 Final remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
1 Introduction
In a paper of 1934 Born and Infeld wrote [6]:
1
The relation of matter and the electromagnetic field can be in-terpreted from two opposite standpoints:
The first which may be called the unitarian standpoint assumesonly one physical entity, the electromagnetic field. The particlesof matter are considered as singularities of the field and mass isa derived notion to be expressed by field energy (electromagneticmass).
The second or dualistic standpoint takes field and particle as twoessentially different agencies. The particles are the sources of thefield, are acted on by the field but are not a part of the field; theircharacteristic property is inertia measured by a specific constant,the mass.
Then they report physical and phylosophical reasons to explain the advan-tages of the unitarian standpoint.
They present a unitary field theory based on a new formulation of theMaxwell equations; they replace the usual Lagrangian density of the electro-magnetic field E, H
L =12(E2 −H2
)(1)
with a modified Lagrangian
L = 1−√
1− (E2 −H2) =12(E2 −H2
)+ o
(E2 −H2
)(2)
They claim to have found nontrivial static solutions of the Euler-Lagrangeequation associated to the Lagrangian (2). Actually they made a mistake and,recently, it has been proved that there are not such solutions (for a proof see[11]).
However we think that the motivations for a new formulation of Maxwellequations are still valid. Nowadays, the methods of nonlinear analysis are suf-ficiently strong to analyse unitarian fields theories derived by perturbations ofLagrangian (1). In this paper we present a unitarian field theory based on aperturbation of (1). This perturbation produces a semilinear modification ofthe Maxwell equations.
Our theory presents the following peculiarities:
• matter particles are not singularities of the field but solitary waves due tothe presence of nonlinearities; thus particles have space extension;
• the equations are invariant for the Poincare group; then the basic prin-ciples of special relativity theory hold; in particular inertial mass equalsenergy;
• the energy of particles is finite: this point is very important, since it impliesfinite particle mass and makes electrodynamics consistent. In the usualapproach to classical electrodynamics, particles are points with infiniteinertial mass. This fact gives rise to well known difficulties (see for examplech. 17 of [8], sections 37, 75 of [9] or ch. 8 in [10]).
2
• the total charge is a constant of motion and its value depends on the initialconditions.
• any matter particle carries a magnetic moment; this is a kind of classicalanalogous of the spin.
• by the presence of the nonlinear term the equations of this theory are notgauge invariant. We point out that the gauge invariance is destroyed onlyin the region where the solitary wave is ”concentrated” ( the matter ) andit is preserved in the region where the nonlinear term is negligible so thatthere we have essentially the Maxwell equations in the empty space.
• this model deals with classical electrodynamics without touching any ques-tions related to quantum theory.
In the following we briefly sketch the content of the paper and the mainmathematical features.
In section 2, we introduce a semilinear perturbation of Maxwell equations(SME) and describe the general characteristics of the resulting equations. Wecannot exhibit explicit static solutions of SME and their existence is provedby using a suitable variational approach. In particular, in section 3, we studythe magnetostatic case (i.e. the case in which the electric field E = 0 and themagnetic field H does not depend on time). In this case SME are reduced tothe nonlinear elliptic degenerate equation
∇×∇×A = f ′(A) (3)
where ∇× denotes the curl operator, f ′ is the gradient of a strictly convexsmooth function f : R3 → R and A : R3 → R3 is the gauge potential related tothe magnetic field H (H = ∇×A).
The main difficulty in dealing with the above equation relies in the fact thatthe energy related to it
E(A) =∫ (
12|∇ ×A|2 − f (A)
)dx (4)
does not yield a priori bound on ‖∇A‖L2 . In particular the functional (4) isstrongly indefinite in the sense that it is not bounded from above or from belowand any possible critical point A has infinite Morse index; namely, the secondvariation of (4)
E ′′ (A) [v, v] =∫ (
|∇ × v|2 − f ′′ (A) [v, v])
dx
is negative definite on the infinite dimensional subspacev = ∇ϕ : ϕ ∈ C∞0
(R3,R
).
3
On the other hand, the nonlinearity f ′(A) destroys the gauge invariance of (3).So it is not possible to choose the Coulomb gauge (where ∇ ·A = 0) to avoidthis indefiniteness.
Another difficulty arises from the growth conditions of the nonlinearityf ′(A). In fact, physical considerations (see section 2) impose f ′(A) to be neg-ligeable when | A | is small (see W1), namely f ′′(0) = 0. This fact leads tostudy the problem in the funcion space
D :=
u ∈ L6 :∫|∇u|2 dx < +∞
and in the Orlicz space Lp + Lq (1 < p < 6 < q) in which D is continuouslyembedded (see (60)).
We will use a min-max argument. Strongly indefinite functional have beenlargely considered in relation to other problems arising in mathematical physicsand treated with min-max methods; we recall just few of them: periodic so-lutions of Hamiltonian system [3], Lorentzian geometry [1], Dirac equation [7].By its nature, (4) cannot be studied by a direct application of the theories pre-viuosly developed and a new approach is required. In particular we will use anew functional framework related to the Hodge splitting of the vector field A.
The last section is devoted to final considerations.AcknowledgmentsWe thank Marino Badiale and Charles Stuart for their useful suggestions.
2 The semilinear Maxwell equations
The Maxwell equations in the empty space are the Euler-Lagrange equations ofthe following action functional
S = −12
∫〈dξ, dξ〉ω (5)
Here
ξ =3∑
i=1
Aidxi + ϕdt
Ai, ϕ : R4 → R
is the gauge potential 1-form in the space-time R4, dξ denotes the exteriorderivative, ω is the volume form and
〈dξ, dξ〉 = ∗ (∗dξ ∧ dξ)
where ∗ is the Hodge operator with respect to the Minkowski metric in R4.More explicitly, S can be written as follows:
S =12
∫ ∫ [∣∣∣∣∂A∂t
+∇ϕ
∣∣∣∣2 − |∇ ×A|2]
dxdt (6)
4
where A = (A1, A2, A3).We modify the action (6) in the following way:
S =12
∫ ∫ [∣∣∣∣∂A∂t
+∇ϕ
∣∣∣∣2 − |∇ ×A|2 + W(|A|2 − ϕ2
)]dxdt (7)
where W : R → R.The argument of W is |A|2 − |ϕ|2 to make this expression invariant for the
Poincare group and the equations consistent with special relativity.Making the variation of S with respect to δA, δϕ respectively, we get the
equations
∂
∂t
(∂A∂t
+∇ϕ
)+∇× (∇×A) = W ′
(|A|2 − ϕ2
)A (8)
−∇ ·(
∂A∂t
+∇ϕ
)= W ′
(|A|2 − ϕ2
)ϕ (9)
If we setH = ∇×A (10)
E = −∂A∂t−∇ϕ (11)
ρ = W ′(|A|2 − ϕ2
)ϕ (12)
J = W ′(|A|2 − ϕ2
)A (13)
we get the equations:
∇×H− ∂E∂t
= J (A, ϕ) (14)
∇ ·E = ρ (A, ϕ) (15)
∇×E +∂H∂t
= 0 (16)
∇ ·H = 0 (17)
which, formally, are the Maxwell equations in the presence of matter if weinterpret ρ (A, ϕ) as the charge density and J (A, ϕ) as the current density.Notice that ρ and J are functions of the gauge potentials, so that we are in thepresence of a unified theory in the sense of Born-Infeld. Hereafter the system(8), (9) (or (14),...,(17)) will be called SME.
We now make the following assumption on W :
• (W1) there exsist two positive constants ε1, ε2 1
|W ′(t)| ≤ ε1 |t| for |t| ≤ 1; (18)
|W ′(t)| ≥ 1 for |t| ≥ 1 + ε2 (19)
5
We set
Ωt (A, ϕ) =
x ∈ R3 :∣∣∣|A (x, t)|2 − ϕ (x, t)2
∣∣∣ ≥ 1
Ωt represents the portion of space filled with matter at time t. Assumption (18)implies that ρ and J become negligible outside Ωt and the above equations canbe interpreted as the Maxwell equations in the empty space.
Assumption (19) implies that ρ and J become strong inside Ωt, at least inthe region where
∣∣∣|A (x, t)|2 − ϕ (x, t)2∣∣∣ ≥ 1 + ε2.
2.1 Invariants of motion
In this section, we will assume that SME have sufficiently smooth solutionsand we will analyse some of their properties. Also we will assume that thesesolutions are sufficienltly small at infinity in such a way that we can performintegrations by part with null ”boundary” terms. The main invariants of themotion of SME, namely the energy and the momentum can be calculated byusing Noether’s theorem. A direct calculation shows that they have the followingexpressions:
• Energy:
E (A, ϕ) =12
∫ (∣∣∣∣∂A∂t
∣∣∣∣2 − |∇ϕ|2 + |∇ ×A|2 −W(|A|2 − ϕ2
))dx
(20)
• Momentum:
P (A, ϕ) =∫ 3∑
i=1
(∂Ai
∂t+
∂ϕ
∂xi
)∇Aidx (21)
Another invariant is the
• Charge:
C (A, ϕ) =∫
ρ (A, ϕ) dx =∫
W ′(|A|2 − ϕ2
)ϕdx (22)
In fact, if we take the divergence in (14) and the derivative with respect to t in(15), we easily get the continuity equation
∂ρ
∂t+∇ · J =0
We can express the energy by a more meaningful expression which will beuseful later:
6
Proposition 1 The energy of the solutions of SME is
E (A, ϕ) =∫ (
12|E|2 +
12|H|2 −W ′ (σ) ϕ2 − 1
2W (σ)
)dx
=12
∫ (|E|2 + |H|2
)dx−
∫ (ρϕ +
12W (σ)
)dx
whereσ = |A|2 − ϕ2
Proof. If we multiply Eq. (15) by ϕ and integrate in x we get∫ (∂A∂t
· ∇ϕ + |∇ϕ|2)
dx−∫
W ′ (σ) ϕ2dx = 0;
We add this expression to E (A, ϕ) ; then
E (A, ϕ) =∫ (
12
∣∣∣∣∂A∂t
∣∣∣∣2 +12|∇ϕ|2 +∇ϕ · ∂A
∂t+
12|∇ ×A|2 −W ′ (σ)ϕ2 − 1
2W (σ)
)dx
=∫ (
12
∣∣∣∣∂A∂t
+∇ϕ
∣∣∣∣2 +12|∇ ×A|2 −W ′ (σ) ϕ2 − 1
2W (σ)
)dx
=∫ (
12|E|2 +
12|H|2 −W ′ (σ) ϕ2 − 1
2W (σ)
)dx
The second expression for the energy is obtained just using (12)The term
12
∫R3
(|E|2 + |H|2
)dx
represents the energy of the electromagnetic field, while
−∫R3
(12W (σ) + W ′(σ)ϕ2
)dx = −
∫R3
(ρϕ +
12W (σ)
)dx (23)
represents the energy of the matter (short range field such as the nuclearfields). It can be interpreted as bond energy and it is ”concentrated” essentiallyin Ωt
2.2 Static solutions
In this section we are interested in the static solutions of (8), (9), namely inthe solutions A, ϕ, depending only on the space variable x, of the followingequations:
∇× (∇×A) = W ′(|A|2 − ϕ2
)A (24)
−∆ϕ = W ′(|A|2 − ϕ2
)ϕ (25)
7
The static solutions are critical points of the energy functional:
E (A,ϕ) =12
∫ (|∇ ×A|2 − |∇ϕ|2 −W
(|A|2 − ϕ2
))dx (26)
Proposition 2 If (A,ϕ) is a finite energy, static solution of SME, then
E (A,ϕ) =13
∫ (|∇ ×A|2 − |∇ϕ|2
)dx
=∫
W(|A|2 − ϕ2
)dx
Proof. Let
ϕλ(x) = ϕ(λ−1x);
Aλ(x) = A(λ−1x);
then, setting y = λ−1x
E (Aλ,ϕλ) =12
∫ (|∇x ×Aλ (x)|2 − |∇xϕλ (x)|2
)dx− 1
2
∫W(|Aλ (x)|2 − ϕλ (x)2
)dx
=λ
2
∫ (|∇y ×A (y)|2 − |∇yϕ (y)|2
)dy − λ3
2
∫W(|A (y)|2 − ϕ (y)2
)dy
Since u is a critical point of E (Aλ,ϕλ) ,
d
dλE (Aλ,ϕλ)
∣∣∣∣λ=1
= 0 (27)
Let us compute this expression explicitly:
d
dλE (Aλ,ϕλ) =
12
∫|∇y ×A (y)|2−|∇yϕ(y)|2 dy−3
2λ2
∫W(|A (y)|2 − ϕ (y)2
)dy
For λ = 1, using (27), we get
13
∫ (|∇ ×A|2 − |∇ϕ|2
)dx =
∫W(|A|2 − ϕ2
)dx (28)
Then
E (A,ϕ) =12
∫ (|∇ ×A|2 − |∇ϕ|2
)dx− 1
2
∫W(|A|2 − ϕ2
)dx
=13
∫ (|∇ ×A|2 − |∇ϕ|2
)dx
And by (28) we have also
E (A,ϕ) =∫
W(|A|2 − ϕ2
)dx
8
2.3 Solitary waves
A solitary wave is a solution of a field equation whose energy travels as a localizedpacket. Since our equation is invariant for the action of the Poincare group, givena static solution
(A (x) ,ϕ (x))
it is possible to produce a solitary wave, moving with velocity v = (v, 0, 0), justmaking a Lorentz transformation:[
Av (x,t)ϕv (x, t)
]=[
A′ (x′ (x,t))ϕ′ (x′ (x,t))
](29)
where
x′ =
x1−vt√1−v2
x2
x3
; A′ =
A1+ϕt√1−v2
A2
A3
; ϕ′ =ϕ + A1t√
1− v2
Solitary waves have a particle-like behaviour. In particular, if (A, ϕ) is astatic solution with radial symmetry (whose existence will be proved in thenext section ), the region filled with matter
Ω =
x ∈ R3 :∣∣∣|A (x)|2 − ϕ (x)2
∣∣∣ ≥ 1
is a ball centered at the origin. Making the above Lorentz transformation
Ωt =
x ∈ R3 :∣∣∣|Av (x, t)|2 − ϕv (x, t)2
∣∣∣ ≥ 1
becomes a rotation ellipsoid moving in direction x1 with velocity v and havingthe shortest axis of lenght R
√1− v2 (where R is the radius of Ω). If we let the
nonlinear termW (t) = Wε(t) :=
1ε2
W1(t)
depend on a small parameter ε, then the radius of Ω becomes εR. Letting ε → 0,the particles approach a pointwise behaviour.
Moreover the solitary waves obtained by this method present the followingfeatures:
• It can be directly verified that the momentum P(Av, ϕv) in (21) of thesolitary wave (29) is proportional to the velocity v
P(Av, ϕv) = mv, m > 0
So the constant m defines the inertial mass of (29). Moreover, if wecalculate the energy E(Av, ϕv) (20) of (29), it can be seen that
m = E(Av, ϕv)
(Einstein equation; in our model we have set c = 1).
9
• the mass increases with velocity
m =m0√1− v2
where
m0 = E(A, ϕ) =13
∫ (|∇ ×A|2 − |∇ϕ|2
)dx =
∫W(|A|2 − ϕ2
)dx
is the rest mass.
• the solitary wave (29) in a given exterior field E,H experiences a force F(Lorentz force)
F = e (E + v ×H)
where e is the charge (22)
e =∫
W ′(|Av|2 − ϕ2
v
)ϕvdx.
Concluding, our solitary waves behave as relativistic particles except thatthey have space extension. These facts are a consequence of the invariance of theLagrangian density with respect to the Poincare group and simple computations.For more details on this matter, we refer to a paper in preparation ([2]). Inthe present work, we are interested to discuss the existence of solitary waveswhich is a problem involving techniques of calculus of variation and of nonlinearfunctional analysis.
2.4 Analysis of static solutions
In order to get the simplest static solutions, we make the following ansatz:
• ϕ 6= 0; A = 0;
• ϕ = 0; A 6= 0;
With these ansatz, we obtain the following equations:
• electrostatic equation:
−∆ϕ = W ′ (−ϕ2)ϕ (30)
• magnetostatic equation:
∇× (∇×A) = W ′(|A|2
)A (31)
10
They correspond to the critical points respectively of the functionals
E (ϕ) = −12
∫ (|∇ϕ|2 + W
(−ϕ2
))dx
E (A) =12
∫ (|∇ ×A|2 −W
(|A|2
))dx (32)
In order to get solutions we need the following technical assumptions:
• (W2) there exist positive constants c2, c3 > 0 such that
|W ′ (t)| ≤ c2 |t|p/2−1 ; p < 6 for |t| ≥ 1
|W ′ (t)| ≤ c3 |t|q/2−1 ; q > 6 for |t| ≤ 1
We have the following result:
Theorem 3 Assume that W satisfies (W2). Then (30) possesses a finite en-ergy, nontrivial solution if and only if there exists t0 such that
W (t0) < 0 (33)
Proof. Since W satisfies (W2) and (33), the if part follows from theorem 4in [4]. The only if part follows from the Pohozaev identity (see prop. 1 in [4]).
Unfortunately, by proposition 2, the energy (rest mass) of the solutions of
(30)
E (ϕ) = −13
∫|∇ϕ|2 dx =
∫W(−ϕ2
)dx
is negative; they are not physically acceptable for our program.Thus, if we want to avoid negative energy solutions, we are forced to assume
• (W+)W (t) ≥ 0 (34)
More exactly we have the following
Proposition 4 Assume that W satisfies (W+) and let (A, ϕ) be a finite energy,non trivial solution of the system (24,25) then:
• (i) A(x) 6= 0
• (ii) the total energy E (A,ϕ) in (26) is positive;
• (iii) the bond energy (23) is negative.
11
Proof. A(x) 6= 0 is a consequence of Theorem 3. In fact, arguing bycontradiction, assume that A = 0. Then ϕ solves (30). So, since W satisfies(34), by theorem 3 we get that also ϕ = 0. This contradicts the assumption that(A, ϕ) is nontrivial.
Since W (t) ≥ 0,by Prop. 2 we deduce that the total energy E (A,ϕ) ispositive . Consider now the bond energy
−∫ (
ρ(σ)ϕ +12W (σ)
)dx,
σ = |A|2 − ϕ2, ρ(σ) = W ′ (σ)ϕ
Now,∫
W (σ) dx is positive by (34); moreover by (25) we easily derive∫ρ(σ)ϕdx =
∫W ′(σ)ϕ2dx =
∫|∇ϕ|2 dx ≥ 0
Remark 5 The property (i) of the above proposition implies that any staticsolution carries a magnetic moment, even when ϕ = 0. Thus any ”particle”is sensitive to external magnetic field even if it has no charge. This can beinterpreted as the classical analogous of the spin.
Remark 6 The property (iii) of the above proposition suggests that solitarywaves must have some form of stability. It is an interesting open problem toinvestigate the orbital stability of them.
In the following we shall prove the existence of static solutions when W ≥ 0.In order to get solutions we need to make some other technical assumptions:
• (W3) There are constants M1 and M2 such that
W (t) ≥ M1 |t|p/2 ; 2 < p < 6 for |t| ≥ 1
W (t) ≥ M2 |t|q/2 ; q > 6 for |t| ≤ 1
• (W4) W ∈ C2 andW ′′(t) > 0 for t 6= 0
Clearly (W3) imply (W+). Moreover, given ε1 ε2 > 0, it is possible to choosesuitable constants in (W3) and (W2) such that (W1) holds.
We have the following result
Theorem 7 If (W2), (W3) and (W4) hold, then eq. (31) has a nontrivial,finite energy solution. This solutions has radial symmetry, namely
A(x) = g−1A(gx) ∀g ∈ O(3)
where O(3) is the orthogonal group in R3.
12
The proof of above theorem is quite involved and it is the main result ofsection 3. Here we give an heuristic argument of the proof. Any possible criticalpoint of (32) has infinite Morse index; namely, the second variation
E ′′ (A) [v]2 =∫ (
|∇ × v|2 −W ′(|A|2
)|v|2 − 2W ′′
(|A|2
)(A·v)2
)dx
is negative definite on the infinite dimensional subspace
v = ∇ϕ : ϕ ∈ C∞0 ;
so the usual tools of critical point theory cannot be used. On the other hand,it is not possible to work in the Coulomb gauge (∇ ·A = 0) since the nonlinearterm of the functional is not gauge invariant. To avoid this difficulty, we splitany radial the vector field A : R3 → R3 as follows
A = u + v = u +∇w (35)
where u : R3 → R3 is a divergence free vector field (∇·u = 0) and v : R3 → R3
is a potential vector field v = ∇w (w scalar field). Since W is strictly convex,for every u with ∇ · u = 0, we can find w0 which minimises the functional
w 7→∫
W(|u +∇w|2
)dx
and set w0 = Φ (u) . Replacing (35) in (32) with w = Φ (u) , we get a newfunctional
J(u) := E (u, Φ (u)) =12
∫ (|∇u|2 −W
(|u +∇Φ (u)|2
))dx
which depends only on u. This functional has the Mountain Pass geometry.Then, we expect the existence of a nontrivial critical point u0. Now, if J andthe map u → Φ (u) were sufficiently smooth in suitable function spaces, the field
A = u0 +∇ [Φ (u0)]
would solve equation (30). However, the lack of smoothness does not allow tocarry out a rigorous simple proof directly and the next section 3 will be devotedto the proof of the existence result.
Remark 8 It is still an open question to prove the existence of solutions of(24,25) with ϕ 6= 0 under the assumption W ≥ 0.
3 Existence of static solutions
This section is devoted to the proof of Th.7. The result will be proved in aslightly more general settings which will be stated in the following two subsec-tions.
13
3.1 Statement of the result
Let f : R3 → R be a C2 function satisfying the following assumptions:
f(0) = 0 and f strictly convex (36)
There are positive constants c1, c2, p, q with 2 < p < 6 < q such that
c1 |ξ|p ≤ f(ξ) for |ξ| ≥ 1 (37)
c1 |ξ|q ≤ f(ξ) for |ξ| ≤ 1 (38)
|f ′(ξ)| ≤ c2 |ξ|p−1 for |ξ| ≥ 1 (39)
|f ′(ξ)| ≤ c2 |ξ|q−1 for |ξ| ≤ 1 (40)We are interested in finding nontrivial, finite energy, weak solutions A : R3 →R3 of the following equation
∇×∇×A = f ′(A) (41)
where f ′ denotes the gradient of f.A weak solution of (41) means that both A and f ′(A) are in L1
loc and forall ϕ ∈ C∞0 (R3,R3)∫
(A |∇ × (∇× ϕ))dx =∫
(f ′(A) | ϕ) dx
where (· | ·) denotes the Euclidean inner product in R3.Moreover we recall that the energy of a weak solution A of (41) is (see (32))
E (A) =12
∫|∇ ×A|2 dx−
∫f(A)dx (42)
So A has finite energy if∫|∇ ×A|2 dx < ∞,
∫f(A)dx < ∞
In this case, by proposition 2, we have
E (A) =13
∫|∇ ×A|2 dx = 2
∫f(A)dx (43)
The following theorem holds
Theorem 9 If f satisfies assumptions (36), (37), (38), (39) and (40), equation(41) has at least a nontrivial, weak solution having finite and positive energy.Moreover this solution has radial symmetry, namely
A(x) =g−1A(gx) for all g ∈ O(3)
Cleary, if we set
f (ξ) =12W(|ξ|2)
theorem 7 follows from theorem 9.
14
3.2 The functional setting
The vector field A : R3 → R3 in (41) is regarded as sum
A = u + v = u +∇w (44)
where u : R3 → R3 is a divergence free vector field (∇·u = 0) and v : R3 → R3
is a potential vector field v = ∇w, (w scalar field).Let D(R3,R3) ( D(R3,R)) be the completion of C∞0 (R3,R3) (C∞0 (R3,R))
with respect to the norm
‖u‖D =(∫
R3|∇u|2 dx
) 12
Now (39), (40) with assumptions (37) and (38) imply that
c1 |ξ|p ≤ f(ξ) ≤ c3 |ξ|p ; for |ξ| ≥ 1 (45)
c1 |ξ|q ≤ f(ξ) ≤ c3 |ξ|q ; for |ξ| ≤ 1 (46)
So f (ξ) ' |ξ|p for |ξ| ≥ 1 and f (ξ) ' |ξ|q for |ξ| ≤ 1 and the choice of thefunction space for w is related to these growth properties of f. Consider firstthe space Lp + Lq made up by the vector fields v : R3 → R3 such that
v = v1 + v2, with v1 ∈ Lp, v2 ∈ Lq and 2 < p < 6 < q
Here Lp, Lq denote the usual Lebesgue spaces.Lp + Lq is a Banach space with the norm
‖v‖Lp+Lq = inf ‖v1‖Lp + ‖v2‖Lq : v1 + v2 = v (47)
It is well known that (see e. g. [5]) Lp + Lq coincides with the dual ofLp′ ∩ Lq′
, namely
Lp + Lq =(Lp′
∩ Lq′)′
, p′ =p
p− 1, q′ =
q
q − 1
and its norm is equivalent to the following one
‖v‖∗Lp+Lq = supϕ 6=0
∣∣∫ (v (x) | ϕ (x))dx∣∣
‖ϕ‖Lp′ + ‖ϕ‖Lq′(48)
.In the following we shall use the same symbol for (47) and (48).Now let Dp,q(R3,R) be the completion of C∞0 (R3,R) with respect to the
norm‖w‖Dp,q = ‖∇w‖Lp+Lq
First we prove some technical lemmas which give useful properties of thespace Lp + Lq.
If v ∈ Lp + Lq we set
Ω =x ∈ R3 : |v (x)| > 1
(49)
Clearly the measure |Ω| of Ω is finite.
15
Lemma 10 If v ∈ Lp + Lq, the following inequalities hold:max
(‖v‖Lq(R3−Ω) − 1, 1
1+|Ω|1/r ‖v‖Lp(Ω)
)≤ ‖v‖Lp+Lq ≤ max
(‖v‖Lq(R3−Ω) , ‖v‖Lp(Ω)
)where Ω is defined in (49) and
r =pq
q − p
Proof. First we prove the inequality
‖v‖Lp+Lq ≤ max(‖v‖Lq(R3−Ω) , ‖v‖Lp(Ω)
)(50)
Clearly∣∣∣∣∫ (v (x) | ϕ (x))dx
∣∣∣∣ = ∣∣∣∣∫R3−Ω
(v (x) | ϕ (x))dx +∫
Ω
(v (x) | ϕ (x))dx
∣∣∣∣ ≤≤ ‖v‖Lq(R3−Ω) ‖ϕ‖Lq′ (R3−Ω) + ‖v‖Lp(Ω) ‖ϕ‖Lp′ (Ω) ≤
≤ max‖v‖Lq(R3−Ω) , ‖v‖Lp(Ω)
(‖ϕ‖Lq′ (R3) + ‖ϕ‖Lp′ (R3)
)Then
‖v‖Lp+Lq = ‖v‖(Lp′∩Lq′)′ ≤ max(‖v‖Lq(R3−Ω) , ‖v‖Lp(Ω)
)Next we prove the inequality
max
(‖v‖Lq(R3−Ω) − 1,
1
1 + |Ω|1/r‖v‖Lp(Ω)
)≤ ‖v‖Lp+Lq (51)
Since p′ > q′ we have
‖ϕ‖Lq′ (Ω) ≤ |Ω|1/r · ‖ϕ‖Lp′ (Ω)
wherer =
pq
q − p
Then
16
‖v‖Lp+Lq = supϕ∈Lp′∩Lq′ ,ϕ 6=0
∣∣∫ (v (x) | ϕ (x))dx∣∣
‖ϕ‖Lp′ + ‖ϕ‖Lq′≥
≥ supϕ 6=0 , ϕ≡0 in R3−Ω
∣∣∫ (v (x) | ϕ (x))dx∣∣
‖ϕ‖Lp′ + ‖ϕ‖Lq′
= supϕ 6=0, ϕ∈Lp′ (Ω)
∣∣∫Ω(v (x) | ϕ (x))dx
∣∣‖ϕ‖Lp′ (Ω) + ‖ϕ‖Lq′ (Ω)
≥ supϕ 6=0, ϕ∈Lp′ (Ω)
∣∣∫Ω(v (x) | ϕ (x))dx
∣∣|Ω|1/r ‖ϕ‖Lp′ (Ω) + ‖ϕ‖Lp′ (Ω)
=1
1 + |Ω|1/rsup
ϕ 6=0, ϕ∈Lp′ (Ω)
∣∣∫Ω(v (x) | ϕ (x))dx
∣∣‖ϕ‖Lp′ (Ω)
=1
1 + |Ω|1/r‖v‖Lp(Ω)
Then1
1 + |Ω|1/r‖v‖Lp(Ω) ≤ ‖v‖Lp+Lq (52)
So, in order to complete the proof of (51), it remains to show that
‖v‖Lq(R3−Ω) − 1 ≤ ‖v‖Lp+Lq (53)
Let ε > 0. By (47) there exists v1 ∈ Lp such that
v − v1 ∈ Lq and ‖v‖Lp+Lq > ‖v1‖Lp + ‖v − v1‖Lq − ε ≥(∫Ω1
|v1|p dx +∫
Ωc−Ω1
|v1|p dx
) 1p
+(∫
Ω1
|v − v1|q dx +∫
Ωc−Ω1
|v − v1|q dx
) 1q
−ε
(54)where
Ωc = R3 − Ω =x ∈ R3 : |v(x)| ≤ 1
Ω1 = Ωc ∩
x ∈ R3 : |v1(x)| > 1
Now set
v(x) = v(x) for x ∈ Ω1 and v(x) = v1(x) for x /∈ Ω1
Clearly
for x ∈ Ω1 : |v(x)| = |v(x)| ≤ 1 < |v1(x)|for x ∈ Ωc − Ω1 : |v(x)| = |v1(x)| ≤ 1
Then, since p < q, we have∫Ω1
|v1|p dx +∫
Ωc−Ω1
|v1|p dx >∫
Ω1
|v|q dx +∫
Ωc−Ω1
|v|q dx = ‖v‖qLq(Ωc)
(55)
17
Moreover ∫Ωc−Ω1
|v − v1|q dx =∫
Ωc−Ω1
|v − v|q dx
Then, since v − v = 0 in Ω1, we get∫Ωc−Ω1
|v − v1|q dx =∫
Ωc−Ω1
|v − v|q dx+∫
Ω1
|v − v|q dx = ‖v − v‖qLq(Ωc)
(56)
By (54), (55) and (56) we easily deduce that
‖v‖Lp+Lq > ‖v‖qp
Lq(Ωc)+ ‖v − v‖Lq(Ωc)
− ε (57)
So, if‖v‖Lq(Ωc)
− 1 < 0,
(57) gives
‖v‖Lp+Lq > ‖v − v‖Lq(Ωc)+ ‖v‖Lq(Ωc)
− 1− ε > ‖v‖Lq(Ωc)− 1− ε (58)
We recall that qp > 1. Then, if
‖v‖Lq(Ωc)− 1 > 0,
(57) gives
‖v‖Lp+Lq > ‖v‖Lq(Ωc)+ ‖v − v‖Lq(Ωc)
− ε > ‖v‖Lq(Ωc)− ε (59)
Finally, since ε > 0 is arbitrary, (58) and (59) imply (53).Observe that the following embedding is continuous
D(R3,R3) ⊂ Lp + Lq (60)
In fact, by the Sobolev inequality, D(R3,R3) is continuously embedded in L6.Now let vn be a bounded sequence in L6 . Then, since p < 6 < q∫
R3−Ωn
|vn|q dx ≤∫R3−Ωn
|vn|6 dx
∫Ωn
|vn|p dx ≤∫
Ωn
|vn|6 dx
whereΩn =
x ∈ R3 : |vn(x)| > 1
So, by using (50), we deduce that vn is bounded also in Lp + Lq and (60)easily follows.
Lemma 11 Let vn be a sequence in Lp+Lq and set Ωn =x ∈ R3 : |vn (x)| > 1
.
Then vn is bounded in Lp + Lq if and only if the sequences |Ωn| and‖vn‖Lq(R3−Ωn) + ‖vn‖Lp(Ωn)
are bounded.
18
Proof. The if part clearly follows from the second inequality in lemma 10.Now we prove the only if part, so we assume that vn is a bounded sequencein Lp + Lq. Then, by the first inequality in lemma 10, there exists c > 0 suchthat for any positive integer n(∫
Ωn|vn(x)|p dx
) 1p
1 + |Ωn|1/r≤ c
So, since |vn(x)| > 1 for x ∈ Ωn, we have
|Ωn| ≤∫
Ωn
|vn(x)|p ≤ cp(1 + |Ωn|1/r
)p
(61)
On the other handp
r= 1− p
q< 1 (62)
then from (61) and (62) we deduce that
|Ωn| is bounded
Then, from the first inequality in lemma 10 we get that‖v‖Lq(R3−Ωn) + ‖v‖Lp(Ωn)
is bounded.
The energy functional is invariant under the action of the space translations,
namely for any vector field A and x ∈ R3
E(A) =E(Ax)
where Ax(x) = A(x + x).This invariance causes a lack of compactness. To overcome this difficulty
we take u and w in (44) belonging to suitable subspaces of D(R3,R3) andDp,q(R3,R) respectively. To this end we need some preliminaries.
Let Tg, T′
g be the actions of the orthogonal group O(3) on the fields u : R3 →R3 and w : R3 → R defined respectively as follows:
for all g ∈ O(3) and x ∈ R3
Tgu(x) = g−1u(gx)T ′gw(x) = w(gx)
Moreover let F and F ′ be the sets of the fixed points for the actions Tg and T ′grespectively, namely
F =u : R3 → R3 : Tgu = u for all g ∈ O(3)
F ′ =
w : R3 → R : T ′gw = w for all g ∈ O(3)
19
It can be easily shown that for any g ∈ O(3), u : R3 → R3 and w : R3 → Rsufficiently smooth we have
T ′g(∇ · u) = ∇ · (Tgu) (63)
∇(T ′gw) = Tg(∇w) (64)
where ∇· denotes the divergence operator. As a consequence we have:
(u ∈ F ) =⇒ (∇ · u ∈ F ′) (65)
(w ∈ F ′) =⇒ (∇w ∈ F ) (66)
Finally we set
V =u ∈ D(R3,R3) : ∇ · u = 0
∩ F (67)
W= Dp,q(R3,R) ∩ F ′ (68)
Observe that V 6= 0.In fact, let z ∈ F a non potential field, decaying to zero at infinity. By (66)
∇ · z ∈ F ′. Then the equation
∆ϕ = ∇ · z (69)
has solution ϕ ∈ F ′. So, by (66), ∇ϕ ∈ F and consequently
u = z −∇ϕ ∈ F
and∇ · u = ∇ · z −∆ϕ = 0
Moreover, since z is not potential, u = z −∇ϕ 6= 0.The following compactness result can be proved.
Proposition 12 V is compactly embedded into Lp + Lq
In order to prove the above proposition we need some lemmas.
Lemma 13 D(R3,R) ∩ F ′ is compactly embedded into the space Lp(R3,R) +Lq(R3,R).
Proof. Let wn be a sequence in D(R3,R) ∩ F ′ such that
wn → 0 weakly in D(R3,R) (70)
We shall prove that
wn → 0 strongly in Lp + Lq (71)
20
By a well known radial lemma [4], there exists c > 0 such that
|wn(x)| ≤ c
|x|12
for all n (72)
Let ε > 0. By (72) and since q > 6, for R > 0 sufficiently large we have∫|x|≥R
|wn(x)|q dx ≤ c
∫|x|≥R
dx
|x|q2
< ε for all n (73)
On the other hand, since p < 6, we have
D(R3,R) ⊂ H1,2loc ⊂⊂ Lp
loc
where the first embedding is continuous and the second is compact. Then, by(70), for n large enough we have
‖wn‖pLp(BR) < ε (74)
whereBR =
x ∈ R3 : |x| ≤ R
By (73) and (74) we get for n large enough
‖wn‖qLq(R38BR) + ‖wn‖p
Lp(BR) < 2ε (75)
Now if R is large enough, (72) gives for all n
|wn(x)| ≤ 1 for |x| > R
Then for all n we have
Ωn =x ∈ R3 : |wn(x)| > 1
⊂ BR
Therefore
‖wn‖qLq(R38Ωn) + ‖wn‖p
Lp(Ωn) = ‖wn‖qLq(R38BR) + ‖wn‖q
Lq(BR8Ωn) +
+ ‖wn‖pLp(BR) − ‖wn‖p
Lp(BR8Ωn) (76)
Since p < q and|wn(x)| ≤ 1 for x ∈ BR8Ωn
we have‖wn‖q
Lq(BR8Ωn) ≤ ‖wn‖pLp(BR8Ωn) (77)
So, by (76) and (77) and (75), we get for n large enough
‖wn‖qLq(R38Ωn) + ‖wn‖p
Lp(Ωn) ≤ ‖wn‖qLq(R38BR) + ‖wn‖p
Lp(BR) < 2ε (78)
Then‖wn‖Lq(R38Ωn) < (2ε)
1q and ‖wn‖Lp(Ωn) < (2ε)
1p (79)
21
Finally, by lemma 10 and (79), we have for n large
‖wn‖Lp+Lq ≤ max(‖wn‖Lq(R38Ωn) , ‖wn‖Lp(Ωn)
)< max
((2ε)
1q , (2ε)
1p
)from which we clearly get the conclusion.
Lemma 14 Let un ⊂ D(R3,R3) be a bounded sequence with |un| ∈ F ′, then(up to a subsequence)
un → u strongly in Lp + Lq
Proof. Since|un| is bounded in D(R3,R), we can assume
|un| → χ weakly in D(R3,R)
So, by lemma 13, we have
|un| → χ strongly in Lp(R3,R) + Lq(R3,R) (80)
then passing to a subsequence
|un| → χ a.e. in R3 (81)
Now, since un ⊂ D(R3,R3) is bounded, (up to a subsequence) we have
un → u in L2loc and un → u a. e. in R3 (82)
So, from (81) and (82), we get
χ = |u| (83)
By (80) and (83) we have
‖un‖Lp+Lq → ‖u‖Lp+Lq (84)
On the other hand by (60) we have
un → u weakly in D(R3,R3) =⇒ un → u weakly in Lp + Lq (85)
By (84) and (85) we get
un → u strongly in Lp + Lq
Finally the proof of Proposition 12 easily follows; in fact:Let un ⊂ D(R3,R3) ∩ F such that
un → u weakly in D(R3,R3)
Since |un| ∈ F ′, by lemma 14 we have
un → u strongly in Lp + Lq
22
3.3 Proof of the existence theorem
Lemma 15 f ′ is a bounded map from Lp + Lq into Lp′ ∩ Lq′
Proof. Let v ∈ Lp + Lq and set
Ω =x ∈ R3 : |v (x)| > 1
Then, since q = p + α, α > 0, by (39) and (40) we have for all ϕ ∈ Lp
(R3)
∣∣∣∣∫ (f ′(v) | ϕ
)dx
∣∣∣∣ ≤ c2
∫R3−Ω
|v|q−1 |ϕ| dx + c2
∫Ω
|v|p−1 |ϕ| dx
= c2
∫R3−Ω
|v|p−1 |ϕ| |v|α dx + c2
∫Ω
|v|p−1 |ϕ| dx
≤ c2
(∫R3−Ω
|v|p |v|α dx
)1/p′
·(∫
R3−Ω
|ϕ|p |v|α dx
)1/p
+ c2 ‖v‖p−1Lp(Ω) · ‖ϕ‖Lp(Ω)
≤ c2
(∫R3−Ω
|v|q dx
)1/p′
·(∫
R3−Ω
|ϕ|p dx
)1/p
+ c2 ‖v‖p−1Lp(Ω) · ‖ϕ‖Lp(Ω)
≤ c2 ‖v‖q/p′
Lq(R3−Ω) · ‖ϕ‖Lp(R3) + c2 ‖v‖p−1Lp(Ω) · ‖ϕ‖Lp(R3)
(86)
Then, by lemma 10, we have
f′(v) ∈ Lp′
. (87)
Now let ϕ ∈ Lq(R3); then, since Ω is bounded, we have
∣∣∣∣∫ (f ′(v) | ϕ
)∣∣∣∣ ≤ c2
∫R3−Ω
|v|q−1 |ϕ|+ c2
∫Ω
|v|p−1 |ϕ|
≤ c2 ‖v‖q−1Lq(R3−Ω) · ‖ϕ‖Lq(R3−Ω)
+ c2 |Ω|1/r
(∫Ω
|v|p dx
)1/p′
‖ϕ‖Lq(Ω)
≤ c2 ‖v‖q−1Lq(R3−Ω) · ‖ϕ‖Lq(R3)
+ c2 |Ω|1/r ‖v‖p/p′
Lp(Ω) ‖ϕ‖Lq(R3) (88)
where r = pqq−p .
Then, by Lemma 10, we have
f′(v) ∈ Lq′
. (89)
23
Finally (87) and (89) imply that f ′ maps Lp + Lq into Lp′ ∩ Lq′.
Now we show that f ′ is bounded.Let vn a bounded sequence in Lp +Lq. Then, by lemma 11, the sequences
|Ωn| and‖vn‖Lq(R3−Ωn) + ‖vn‖Lp(Ωn)
are bounded
Then (86) and (88) imply that
f ′(vn) is bounded in Lp′∩ Lq′
Let
v ∈ Lp + Lq
then ∫f(v(x))dx < ∞ (90)
In fact, by (45) and (46)
f(v(x)) ' |v(x)|p in Ω and f(v(x)) ' |v(x)|q in R3 \ Ω (91)
whereΩ =
x ∈ R3 : |v (x)| > 1
Since v ∈ Lp + Lq, by Lemma 10, we have∫
Ω
|v(x)|p dx,
∫R3\Ω
|v(x)|q dx < ∞ (92)
then by (91) and (92) we deduce (90).Now observe that, if u ∈ V and w ∈ W, by (60) v = u + ∇w belongs to
Lp + Lq. So (90) implies that∫f(u +∇w)dx < ∞
Lemma 16 Consider the functionals
v ∈ Lp + Lq → F (v) =∫
f(v(x))dx (93)
andw ∈ W → F (u, w) = F (u +∇w) =
∫f(u(x) +∇w(x))dx (94)
where u is given in V. The functionals (93) and (94) are continuous and strictlyconvex.
24
Proof. Letvn → v in Lp + Lq =
(Lp′
∩ Lq′)′
(95)
Clearly we can write∫(f(vn(x))− f(v(x)))dx =
∫(f ′(θn(x)) | vn(x)− v(x)) dx
where θn(x) is a suitable convex combination of vn(x) and v(x) . So, since vnis bounded in Lp + Lq, also θn is bounded in Lp + Lq. Then, by lemma 15,f ′(θn) is bounded in Lp′ ∩ Lq′
and, using (95), we deduce that∫(f(vn(x))− f(v(x)))dx =
∫(f ′(θn(x)) | vn(x)− v(x)) dx → 0
So the functional
v ∈ Lp + Lq → F (v) =∫
f(v(x))dx
is continuous.Now let wn be a sequence in W such that
wn → w in W
then∇wn → ∇w in Lp + Lq
Clearly
F (u, wn)− F (u, w) =∫
(f ′(u(x) + θn(x)) | ∇wn(x)−∇w(x)) dx
where θn(x) is a suitable convex combination of ∇wn(x) and ∇w(x). Then,arguing as before, we deduce that
F (u, wn)− F (u, w) → 0
Now the strict convexity of f implies the strict convexity of the functionals (93)and (94).
Theorem 17 For any u ∈ V, there exists a unique w ∈ W which minimizesthe functional F (u, ·)
Proof. Let u ∈ V. Since F (u, ·), is continuous and convex, it is also weaklylower semicontinuous. Now we prove that F (u, ·) is coercive on W. For anyw ∈ W set
Ω =x ∈ R3 : |u(x) +∇w (x)| > 1
25
then
F (u, w) =∫
f(u +∇w)dx
=∫R3−Ω
f(u +∇w)dx +∫
Ω
f(u +∇w)dx ≥ (by (37) and (38))
≥ c1
∫R3−Ω
|u +∇w|q dx + c1
∫Ω
|u +∇w|p dx
So, by lemma 10, it can be deduced that
F (u, w) →∞ as ‖w‖W = ‖∇w‖Lp+Lq →∞
Then F (u, ·) is coercive and it possesses a minimizer which is unique, since f isstrictly convex.
For any u ∈ V we denote by Φ(u) the minimizer of F (u, ·), so
F (u, Φ(u)) ≤ F (u, w) ∀w ∈ W (96)
Now consider the functional on V
u → F (u, Φ(u)) =∫
f(u +∇Φ(u))dx (97)
Theorem 18 The functional u →∫
f(u +∇Φ(u))dx is weakly continuous onV .
Proof. Letun → u weakly in V
By Proposition 12
un → u strongly in Lp + Lq
Then by the continuity of the functional F (u) (see lemma 16)
F (un) =∫
f(un)dx →∫
f(u)dx (98)
Moreover by (96)
0 ≤ F (un,Φ(un)) ≤ F (un, 0) = F (un) (99)
Then from (98) and (99) we deduce that F (un,Φ(un)) is bounded, so wecan assume that (up to a subsequence)
F (un,Φ(un)) =∫
f(un +∇Φ(un))dx → c ∈ R (100)
26
We want to show that
c =∫
f(u +∇Φ(u))dx (101)
Now the sequence F (un,Φ(un)) is bounded and F is coercive on Lp + Lq
(see the proof of Theorem 17), then
un +∇Φ(un) is bounded in Lp + Lq (102)
Then, since un converges strongly in Lp + Lq, (102) implies that Φ(un) isbounded in W, so (up to a subsequence)
∇Φ(un) → ∇w weakly in Lp + Lq
and thenun +∇Φ(un) → u +∇w weakly in Lp + Lq (103)
F is convex and continuous on Lp +Lq, then it is weakly lower semicontinuous.So, by (100), (103) and (96) we get
c = limn→∞
F (un +∇Φ(un))
≥ F (u +∇w) = F (u, w)
≥ F (u, Φ(u)) =∫
f(u +∇Φ(u))dx
then ∫f(u +∇Φ(u))dx ≤ c (104)
Now, since f is C2, we have∫f(u +∇Φ(u))dx =
=∫
f(un +∇Φ(un) + (u +∇Φ(u))− (un +∇Φ(un)))dx
=∫
f(un +∇Φ(un))dx
+∫
(f ′(un +∇Φ(un)) | (u +∇Φ(u))− (un +∇Φ(un))) dx
+12
∫f ′′ (θn(x)) [(un +∇Φ(un))− (u +∇Φ(u))]2 dx
=∫
f(un +∇Φ(un))dx
+∫
(f ′(un +∇Φ(un)) | u− un) dx
+∫
(f ′(un +∇Φ(un)) | ∇(Φ(u)− Φ(un)))dx
+12
∫f ′′ (ϑn(x)) [(un +∇Φ(un))− (u +∇Φ(u))]2 dx (105)
27
where ϑn(x) is a suitable convex combination of un(x) + ∇Φ(un)(x) andu(x) +∇Φ(u)(x).
Since f is convex
12
∫f ′′ (ϑn(x)) [(un +∇Φ(un))− (u +∇Φ(u))]2 dx ≥ 0 (106)
On the other hand, by (96), Φ(un) is a minimizer of F (un, ·) in W, then∫(f ′(un +∇Φ(un)) | ∇w) dx = 0 for all w ∈ W
In particular, taking w = Φ(u)− Φ(un) ∈ W, we have∫(f ′(un +∇Φ(un)) | ∇(Φ(u)− Φ(un))) dx = 0 (107)
Now, by (102) and lemma 15, we deduce that f ′(un +∇Φ(un)) is boundedin Lp′ ∩ Lq′
. Then, since unconverges to u strongly in Lp + Lq, we deduce∫(f ′(un +∇Φ(un)) | u− un)dx → 0 (108)
Then, taking the limit in (105) for n →∞, and using (108), (107) and (100),we get ∫
f(u +∇Φ(u))dx ≥ c (109)
Clearly we get the conclusion from (104) and (109).Now consider the functional
J(u) := E (u, Φ (u)) =12
∫|∇u|2 dx− F (u, Φ(u)), u ∈ V
We do not know if J is C1and we cannot use directly the usual tools of criticalpoint theory. So we argue as follows. Let u0 ∈ V , u0 6= 0. Then
u0 +∇Φ(u0) 6= 0 (110)
In fact, arguing by contraddiction, assume that
u0 = −∇Φ(u0) (111)
then∇ · u0 = −∆Φ(u0)
So, since u0 ∈ V, we have∆Φ(u0) = 0
Then, by (111) we get u0 = 0, contraddicting u0 6= 0.Then t0 = 1
2F (u0,Φ (u0)) > 0.
28
Now consider the functional
Jβ(u) =12
∫|∇u|2 dx− β (F (u, Φ (u)))
where F (u, Φ (u)) is defined in (97) and β : R → R is a function such that
• β (t) ≥ 0; β (t) = 0 for t ≤ 0
• β′ (t) ≥ 0 ∀t
• β (F (u0,Φ (u0))) ≥ 1 + 12 ‖u0‖2
• β is bounded.
Lemma 19 Jβ has a minimizer u in V such that Jβ(u) ≤ −1
Proof. Jβ is bounded from below and coercive in V. Moreover, by lemma18, β (F (u, Φ (u))) is weakly continuous on V. Then, since Jβ (u0) ≤ −1, we getthe existence of a minimizer u in V such that Jβ(u) ≤ Jβ(u0) ≤ −1.
In the following we denote by (∇u | ∇h) the Hilbert product of the Jacobian
matrices ∇u,∇h of the vector fields u, h, namely we set
(∇u | ∇h) = Tr[(∇u) (∇h)T
]where (∇h)T is transposed of ∇h and Tr denotes the trace.
Lemma 20 Assume that there exists u ∈ V and µ > 0 such that∫(∇u | ∇h) dx− µ2
∫(f ′(u +∇Φ(u)) | h) dx = 0 for all h ∈ V. (112)
Then uλ(x) = u(λx), λ = 1µ , solves the equation∫
(∇uλ | ∇h) dx−∫
(f ′(uλ +∇Φ(uλ)) | h) dx = 0 for all h ∈ V (113)
Proof. Setw = Φ(u) and wλ(x) = w(λx)
First of all we show that
Φ(uλ)(x) =1λ
Φ(u)(λx) =1λ
wλ(x). (114)
In fact:w = Φ(u) is minimizer of F (u, ·), and f ′ is strictly monotone. Then w is the
unique solution of the equation∫(f ′(u(x) +∇w(x)) | ∇v) dx = 0 for all v ∈ W (115)
29
Since v ∈ W is arbitrary, in (115) we take v 1λ(x) = v(x
λ ) and then changethe variable y = x
λ :∫ (f ′(u(x) +∇w(x)) | ∇v 1
λ
(x))
dx = 0 for all v ∈ W∫ (f ′(u(λy) +∇w(λy)) | 1
λ∇v(y)
)λ3dy = 0 for all v ∈ W∫
(f ′(u(λy) +∇w(λy)) | ∇v(y)) dy = 0 for all v ∈ W
Then ∫ (f ′(uλ(y) +
1λ
(∇wλ) (y)) | ∇v(y))
dy = 0 for all v ∈ W
and this implies that wλ
λ minimizes F (uλ.·) i.e.
Φ(uλ)(x) =wλ(x)
λ=
w(λx)λ
=Φ(u)(λx)
λ
Now we verify that uλ satisfies (113). So, inserting (114) in (113), we have forall h ∈ V∫
(∇uλ | ∇h) dx−∫
(f ′(uλ +∇Φ(uλ)) | h) dx
=∫
λ (∇u(λx) | ∇h(x)) dx−∫ (
f ′(u(λx) +1λ∇wλ(x)) | h(x)
)dx = (y = λx)
= λ
∫1λ3
(∇u(y) | λ∇h 1
λ(y))
dy −∫
1λ3
(f ′(u(y) +∇w(y)) | h 1
λ(y))
dy
=1λ
(∫ (∇u(y) | ∇h 1
λ(y))
dy − 1λ2
∫ (f ′(u(y) +∇w(y)) | h 1
λ(y))
dy
)= µ
(∫ (∇u(y) | ∇h 1
λ(y))
dy − µ2
∫ (f ′(u(y) +∇w(y)) | h 1
λ(y))
dy
)=
= ( u solves (112)) = 0
Theorem 9 will immediately follow from lemma 19 and the following result
Theorem 21 Let u be a minimizer of Jβ in V and uλ(x) = u(λx) with λ =1
(β′(ϑ))12, θ = F (u, Φ (u)). Then A = uλ + ∇Φ(uλ) is a weak, nontrivial solu-
tion of equation (41) and it has positive, finite energy. Moreover it has radialsymmetry, namely
A(x) =g−1A(gx) for all g ∈ O(3)
30
Proof. Let u a minimizer u of Jβ in V. Then ∀h ∈ V,
0 ≤ Jβ (u + εh)− Jβ (u) (116)
=∫
(∇u | ε∇h) dx +12‖εh‖2D − β (F (u + εh, Φ (u + εh))) + β (F (u, Φ (u)))
=∫
(∇u | ε∇h) dx +12‖εh‖2D − β′ (ϑε) (F (u + εh, Φ (u + εh))− F (u, Φ (u)))
=∫
(∇u | ε∇h) dx +12‖εh‖2D
− β′ (ϑε) ·∫
[f(u + εh +∇Φ(u + εh))− f(u +∇Φ(u))]
=∫
(∇u | ε∇h) dx +12‖εh‖2D
− β′ (ϑε)∫
(f ′(u +∇Φ(u)) | εh +∇ [Φ(u + εh)− Φ(u)])dx
− β′ (ϑε)∫
f ′′ (v(x)) [εh +∇ [Φ(u + εh)− Φ(u)]]2 dx (117)
where
ϑε = (F (u, Φ (u)) + θ [F (u + εh, Φ (u + εh))− F (u, Φ (u))]); θ ∈ (0, 1) (118)
and v(x) is a convex combination of v1(x) = u(x) +∇Φ(u)(x) and v2(x) =u(x) + εh(x) +∇Φ(u + εh)(x).
Since Φ (u) is minimizer for F (u, ·), we have∫(f ′(u +∇Φ(u)) | ∇w) dx = 0 for all w ∈ W (119)
So, taking in (119) w = Φ(u + εh)− Φ(u), we get∫(f ′(u +∇Φ(u)) | ∇(Φ(u + εh)− Φ(u))) dx = 0 (120)
Since f is convex and β′ (ϑε) ≥ 0, we get, inserting (120) in (117),
0 ≤ Jβ (u + εh)− Jβ (u)
≤∫
(∇u | ε∇h) dx +12‖εh‖2D − β′ (ϑε)
∫(f ′(u +∇Φ(u)) | εh) dx
Then for any h ∈ V, ε > 0
1ε
[∫(∇u | ε∇h) dx +
12‖εh‖2D1,2 − β′ (ϑε)
∫(f ′(u +∇Φ(u)) | εh) dx
]≥ 0
(121)Now the functional u → F (u, Φ (u) is continuous on V by Theorem 18. Then
F (u + εh, Φ (u + εh)) → F (u, Φ (u)) as ε → 0.
31
So, taking the limit in (118), we have
ϑε → ϑ = F (u, Φ (u)) as ε → 0 (122)
Then, taking the limit as ε → 0+ in (121), we get∫(∇u | ∇h) dx− β′ (ϑ)
∫(f ′(u +∇Φ(u)) | h) dx ≥ 0
Since h ∈ V is arbitrary, we get
∫(∇u | ∇h) dx− β′ (ϑ)
∫(f ′(u +∇Φ(u)) | h) dx = 0 for all h ∈ V (123)
Now β′ (ϑ) > 0. In fact, arguing by contradiction, assume that β′ (ϑ) =0. Then by (123)
∫(∇u | ∇u) dx = 0 and consequently u = 0. So Jβ (u) = 0.
This contradicts Jβ (u) ≤ −1.Now, by lemma 20, the rescaled function
uλ(x) = u(λx) with λ =1
(β′ (ϑ))12
solves the equation∫(∇uλ | ∇h) dx−
∫(f ′(uλ +∇Φ(uλ)) | h) dx = 0 for all h ∈ V
and since ∇ · u = 0 and ∇ · h = 0, we have∫(∇× uλ | ∇ × h) dx−
∫(f ′(uλ +∇Φ(uλ)) | h) dx = 0 for all h ∈ V (124)
Now we want to show that
A = uλ +∇Φ(uλ)
is a weak solution of (41), namely that for all ϕ ∈ C∞0 (R3,R3) we have∫(A |∇ × (∇× ϕ))dx−
∫(f ′(A) | ϕ) dx = 0 (125)
To this end define the distribution M as follows: for all ϕ ∈ C∞0 (R3,R3)
〈M,ϕ〉 :=∫
(∇× uλ | ∇ × ϕ) dx−∫
(f ′(A) | ϕ) dx (126)
Any vector field ϕ ∈ C∞0 (R3,R3) can be written as follows
ϕ = ϕr + ϕ∗ = h +∇w + ϕ∗
32
where ϕr ∈ D(R3,R3) ∩ F , ϕ∗ ∈(D(R3,R3) ∩ F
)⊥, h ∈ V and w ∈ W.
So〈M,ϕ〉 = 〈M,h〉+ 〈M,∇w〉+ 〈M,ϕ∗〉 (127)
By (124)〈M,h〉 = 0 (128)
Since Φ(uλ) is the minimizer of the functional F (uλ, ·) on W, we have,∫(f ′(A) | ∇w) dx = 0
Then, since ∇× (∇w) = 0, we have
〈M,∇w〉 =∫
(∇× uλ | ∇ × (∇w))dx−∫
(f ′(A) | ∇w) dx = 0 (129)
Now the following implications hold(ϕ∗ ∈
(D(R3,R3) ∩ F
)⊥)=⇒
(∇× ϕ∗ ∈
(L2(R3,R3) ∩ F
)⊥)(uλ ∈ D(R3,R3) ∩ F
)=⇒
(∇× uλ ∈ L2(R3,R3) ∩ F
)Then
〈M,ϕ∗〉 =∫
(∇× uλ | ∇ × ϕ∗) dx−∫
(f ′(A) | ϕ∗) dx = −∫
(f ′(A) | ϕ∗) dx
(130)By (78) A ∈F ∩ (Lp + Lq) , then, using (102), we have
f ′(A) ∈ F ∩ Lp′∩ Lq′
So, for a suitable ξ ∈ D(R3,R3) ∩ F, we have∫(f ′(A) | ϕ∗) dx = (ξ | ϕ∗)D = 0 (131)
where (· | ·)D denotes the inner product in D(R3,R3).Replacing (131) in (130) we get
〈M,ϕ∗〉 = 0 (132)
Inserting (132), (129), (128) in (127) we have
〈M,ϕ〉 =∫
(uλ | ∇ ×∇× ϕ) dx−∫
(f ′(A) | ϕ) dx = 0 (133)
Moreover∫(∇Φ(uλ) | ∇ ×∇× ϕ) dx = −
∫Φ(uλ)(∇ · (∇×∇× ϕ))dx = 0 (134)
33
By (133) and (134) we get (125). Let us finally show that A 6=0, A haspositive, finite energy and it has radial symmetry:
1) A is nontrivial. In fact, since uλ 6= 0 and ∇ · uλ = 0, we have
A = uλ +∇Φ(uλ) 6= 0
2) A has finite and positive energy. In fact, since uλ ∈ D(R3,R3) andA ∈Lp + Lq, we have ∫
|∇ ×A|2 dx =∫|∇uλ|2 dx < ∞∫
f(A)dx < ∞.
Then, by (42) and (43), A has finite and positive energy.3) A has radial symmetry. In fact uλ ∈ V ⊂F (see ( 67) ) and ∇Φ(uλ) ∈ F,
since Φ(uλ) ∈ F ′ (see (66)).
4 Conclusions
4.1 A formulation of classical physics
As we have discussed in section 2, the equations of classical electrodynamicscan be derived from (7); it is possible to make a slight modification of (7) in thespirit of general relativity in order to derive the equations of classical physics(electrodynamics and gravitation theory).
Let (M, g) be a Lorentzian manifold. Using the formalism of (5), the action(7) takes the following expression:
S(ξ) = −12
∫M
[〈dξ, dξ〉+ W (−〈ξ, ξ〉)]ω (135)
whereξ = A1dx1 + A2dx2 + A3dx3 + ϕdt
is the gauge potential form, ω is the volume form relative to g and
〈α, α〉 = ∗ (∗α ∧ α)
where ∗ is the Hodge operator for the Lorentzian metric g. In this way theEuler-Lagrange equations relative to (135) provide the equations of classicalphysics in a assigned gravitational field g. In order to consider also the actionof matter-energy on g, it is sufficient to take the Hilbert functional
12
∫M
R (g)ω
34
where R (g) is the scalar curvature and add this them to (135):
S(ξ, g) =12
∫M
[R (g)− 〈dξ, dξ〉 −W (−〈ξ, ξ〉)]ω
The first term describes the gravitational force, the second term describes theelecromagnetic force and the last term produces short range forces and it isresponsable of the presence of matter; it reminds the weak force.
4.2 Final remarks
It is well known that classical electrodynamics is not a completely mathemat-ically consistent theory. In fact, if the particles are supposed pointwise, theenergy becomes infinite. Moreover
a completely satisfacory treatment of the reactive effects of ra-diation does not exists. The difficulties presented by this problemtouch one of the most fundamental aspects of physics, the nature ofan elementary particle. Although partial solutions, workable withinlimited areas can be given, the basic problems remain unsolved ([8],section 17.1 pag. 579).
In the last fourty years, the nonlinear functional analysis and variationalmethods have been largely developed and fundational problems in classicalphysics can be faced with new powerful tools. The present paper representsa step in this direction.
References
[1] Benci V., Fortunato D., On the existence of infinitely many geodesicson space-time manifolds. Adv. in Math., 105 (1994),1-25.
[2] Benci V., Fortunato D., Solitary waves of the semilinear Maxwell equa-tions and their dynamical properties, in preparation.
[3] Benci V., Rabinowitz P.H., Critical points theorems for indefinite func-tionals, Invent. Math., 52 (1979), 241-273.
[4] Berestycki H., Lions P.L., Nonlinear Scalar Field Equations, I - Ex-istence of a Ground State, Arch. Rat. Mech. Anal., 82 (4) (1983), 313-345.
[5] Berg J., Lofstrom J., Interpolation Spaces, Springer-Verlag, Berlin,Heidelberg, New York, (1976).
[6] Born M., L.Infeld L., Foundations of the new field theory, Proc. R. Soc.Lon. A, 144 (1934), 425-451.
35
[7] Esteban M., Sere E., Stationary states of the nonlinear Dirac equation:a variational approach. Comm. Math. Phys. , 171 (1995), 323-350.
[8] Jackson J.D., Classical electrodynamics, John Wiley & Sons., New York,London, (1962).
[9] Landau L., Lifchitz E., Theorie des Champs, Editions Mir, Moscou,(1970).
[10] Thirring W., Classical Mathematical Physics, Springer, New York, Vi-enna, (1997).
[11] Yang Y., Classical solutions in the Born-Infeld theory, Proc. R. Soc. Lon.A (2000), 456, 615-640.
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