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ON THE EXISTENCE OF STATIONARY STATES FOR
CLASSICAL NONLINEAR DIRAC FIELDS
Thierry Cazenave
Universite Pierre et Marie Curie & CNRS
Laboratoire Jacques-Louis Lions
B.C. 187, 4 place Jussieu75252 Paris Cedex 05, France
email address: [email protected]
URL: http://www.ljll.math.upmc.fr/cazenave/
1. Introduction
Our purpose in these notes is to describe and give a self-contained proof ofthe results of Cazenave and Vazquez [5] and Balabane, Cazenave, Douady andMerle [2]. We study the existence of stationary states for the following nonlinearDirac equation
i
3∑µ=0
γµ∂µψ −mψ + F (ψψ)ψ = 0. (1.1)
The notation is the following. ψ is defined on R4 with values in C4, ∂µ = ∂∂xµ
, m
is a positive constant ψψ = (γ0ψ,ψ), where (·, ·) is the usual scalar product on C4,and the γµ’s are the 4× 4 matrices of the Pauli-Dirac representation, given by
γ0 =
(I 00 −I
)and γk =
(0 σk
σk 0
)for k = 1, 2, 3,
where
σ1 =
(0 11 0
), σ2 =
(0 −ii 0
), σ3 =
(1 00 −1
).
Finally, F : R→ R models the nonlinear interaction.Nonlinear spinor fields giving rise to equations of the form (1.1) were considered
first by D. Ivanenko [15], H. Weyl [35] and by W. Heisenberg [13] in his unifiedtheory of elementary particles. Later, R. Finkelstein, C.F. Fronsdal and P. Kaus [11]considered the case of a spinor field with several types of fourth order self-couplings.But it was M. Soler [26] who was the first to investigate the stationary states ofthe nonlinear Dirac field with the scalar fourth order self-coupling (correspondingto F (x) = x in (1.1)) proposing them as a model of elementary extended fermions.Subsequently, the electromagnetic interaction was introduced [27, 23, 24] in orderto construct a model of extended charged fermion, which in spite of its simplicitydescribes with a reasonable accuracy the properties of the nucleons [22]. To improvethe model, the pseudoscalar fields were introduced in order to represent the cloudof pions[25, 12]. A summary of the above models, with the numerical computationsand further developments are described by Ranada [20, 21]. The case F (x) = xwas also considered by Rafelski [19], Takahashi [29] and Van der Merwe [30].
We are interested in stationary states, or localized solutions of (1.1), that issolutions ψ of the form ψ(t, x) = eiωϕ(x), where t = x0 and x = (x1, x2, x3). In
1
2 THIERRY CAZENAVE
addition, we seek finite energy solutions; and so we want ϕ to be at least squareintegrable. Clearly, the equation for ϕ : R3 → C4 is
i
3∑k=1
γk∂kϕ−mϕ+ ωγ0ϕ+ F (ϕϕ)ϕ = 0. (1.2)
In all the sequel, we assume that F satisfies the following hypotheses.
F ∈ C1(R,R),
F (0) = 0,
F (x) ≤ 0 for x ≤ 0,
F is increasing on (0,∞),
limx→+∞
F (x) > m+ ω,
F ′(F−1(m− ω)) > 0.
(H1)
We also assume that
0 < ω < m. (H2)
Then we have the following result.
Theorem 1.1. Assume (H1) and (H2) hold. Then (1.2) has infinitely many dif-ferent solutions. More precisely, for every integer n ≥ 0, there exists a solutionϕn ∈ C1(R3,C4) of (1.2) such that
(i) ϕn and ∇ϕn have exponential decay as |x| → ∞;(ii) the function x 7→ ϕnϕn is spherically symmetric and has n nodes as a function
of r.
Theorem 1.1 calls for a few comments. First, observe that it is conclusion (ii)that ensures that the solutions ϕn are nontrivial and distinct. Observe also that (i)implies that ϕn ∈ H1(R3,C4); and so ϕn has finite energy.
Concerning the assumption (H1), it is interesting to note that we did not imposeany restriction on the growth of F as x → ∞. In particular, we can take F (x) =|x|p−1x, for any p > 1. This is in striking contrast with the Klein-Gordon andSchrodinger equations where finite-energy stationary states exist only for p < 5 indimension 1 + 3 (compare [3]). Observe also that even though the argument of Ftakes negative values for the solution ϕn with n ≥ 1, the only assumption on Ffor x < 0 is F (x) ≤ 0. In particular, we do not impose any oddness condition.Assumption (H1) can be slightly improved with a few modifications in the proof.See the comments of Theorem 1.2 below and Section 6.
Equation (1.2) has a variational structure. More precisely, solutions of (1.2) are(formally) critical points of the Lagrangian L given by
L(ϕ) =
∫R3
{ 3∑k=1
(iγ0γk∂kγ, ϕ)−mϕϕ+ ω|ϕ|2 +G(ϕϕ)}dx,
where G is a primitive of F and (·, ·) is the scalar product in C4. Let us point outthat it would be extremely interesting to solve this variational problem since a betterknowledge of the structure of the Lagrangian might give relevant information on theinitial value problem for (1.1), and in particular on the stability of the stationarystates. However, this seems to be delicate, due to the defect of coerciveness of boththe term involving derivatives and the non-quadratic term in the Lagrangian.
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 3
Here, we do not attempt to solve the variational problem. Instead, and followingM. Wakano [33] and M. Soler [26], we seek solutions that are separable in sphericalcoordinates, of the form
ϕ(x) =
v(r)
(10
)iu(r)
(cos θ
sin θeiφ
) . (1.3)
Here, r = |x|, and (θ, φ) are the angular parameters. The Dirac equation then turnsto a nonautonomous planar differential system, in the r variable, which is u′ = −2u
r+ v[F (u2 − v2)− (m− ω)], (1.4)
v′ = u[F (u2 − v2)− (m− ω)]. (1.5)
In order to avoid solutions with singularity at the origin, due to the term 2ru in (1.4),
we impose
u(0) = 0; (1.6)
and since we are interested in finite energy solutions of (1.2), we seek solutionsof (1.4)-(1.5) that fulfill
|u(r)|+ |v(r)| −→r→∞
0. (1.7)
For every given x, there exists a local solution (u, v) of (1.4)-(1.6) with the initialcondition v(0) = x. The problem is to find x such that the corresponding solutionis global (i. e. defined for all r ≥ 0), and satisfies (1.7). We have the followingresult, from which Theorem 1.1 is an immediate consequence.
Theorem 1.2. Assume that (H1) and (H2) hold. There exists an increasing se-quence (xn)n≥0 of positive numbers with the following properties . For every n ≥ 0,
(i) the solution (un, vn) of (1.4)-(1.6) with vn(0) = xn is global;(ii) both un and vn have exactly n zeroes on (0,∞);
(iii) (un, vn) converges exponentially to (0, 0) as r →∞.(iv) Furthermore, if F (x) ≥ δ(log x)β for x large where δ > 0 and β > 2, the
sequence (xn)n≥0 is bounded.
Several remarks are in order, concerning Theorem 1.2. The first analytical studyof system (1.4)-(1.5) was done by L. Vazquez [31], who obtained sufficient conditionsfor the existence of solutions. The first existence result was obtained by T. Cazenaveand L. Vazquez [5]. They proved the existence of a solution without nodes (positiveu and v), which is essentially the solution (u0, v0) of Theorem 1.2. Later, that resultwas extended to a wider class of nonlinearities by F. Merle [18]. Theorem 1.2 inthe present form is due to M. Balabane, T. Cazenave, A. Douady and F. Merle [2].
We do not know whether or not the assumption (H2) is necessary in Theorem 1.1.However, it is almost necessary in Theorem 1.2. Indeed, it was shown in [31] thatwhen F (x) = x there is no solution of (1.4)-(1.6) such that ϕ is square-integrablewhen |ω| > m or ω = 0. The argument works as well when F satisfies (H1) . On theother hand, elementary calculations show that if −m ≤ ω < 0, no solution of (1.4)-(1.5) can converge to (0, 0) as r → ∞. In conclusion, the condition 0 < ω ≤ mis necessary in Theorem 1.2. Let us also mention that it does not restrict thegenerality to consider only the case v(0) > 0. Indeed, if (u, v) is a solution of (1.4)-(1.5), then (−u,−v) is also a solution. Our method of proof can be modified in
4 THIERRY CAZENAVE
order to weaken hypothesis (H1) . In particular, the conclusions of Theorem 1.2(and so these of Theorem 1.1) still hold if (H1) is replaced by (H1′) below (seeSection 6)
F ∈ C1(R,R),
F (0) = 0,
F (x) ≤ 0 for x ≤ 0,
F is increasing on (0,∞),
limx→+∞
F (x) > m− ω.
(H1′)
The numerical experiments performed on system (1.4)-(1.5) indicate the follow-ing (L. Vazquez [32], for F (x) = x). First, starting from x larger than somex?, the solutions blow-up (compare Proposition 3.2). It is also observed that theglobal solutions converge to one of the rest points of the system, which are (0, 0),
(0,√F−1(m− ω)) and (0,−
√F−1(m− ω)). The solutions wind around the origin
(in the plane (v, u)) before converging to a rest point. Note that the rest points
(0,±√F−1(m− ω)) are stable while (0, 0) is a saddle point. The set of positive
initial data for which the solution turns n/2 times (n being an integer) around
(0, 0) before converging to (0,±√F−1(m− ω)) seems to be an interval of the form
(xn−1,xn). The only solutions with positive initial data converging to (0, 0) appearto be those starting from xn, and they have n nodes. We prove these propertieshere, except uniqueness of the n-nodes solution with positive initial datum. TheMacMath of J.H. Hubbard and B.H. West [14] was also helpful to have qualitativeintuition for the dynamical system.
The proof of Theorem 1.2, that we give here is adapted from [5] and [2]. Weconsider (1.4)-(1.5) as a non-autonomous planar dynamical system (r being the timevariable), and we follow essentially the scheme suggested by the numerical results.For every n ≥ 0, we construct an open, non-empty set In of initial data for whichthe solution turns n/2 times around the origin and then remains trapped near oneof the stable rest points. Next, we show that the solutions with initial data in Inare bounded, uniformly in r ≥ 0 and in the initial datum. Finally, we show that thesolution with initial datum sup In is the expected n-node solution. The boundednessof the sequence (xn)n≥0 follows from a blow-up result (Proposition 3.2). It willappear to the reader that the proof of Theorem 1.2 given below is very long andtedious. However, this is the only proof available so far. Observe also that wedeal with system (1.4)-(1.5) without any sophisticated tool. All the arguments inthe proof are absolutely elementary; and since the conclusion is the existence ofinfinitely many solutions for a system of nonlinear partial differential equations, itis reasonable to expect that there is a lot of such argument to be put together! Weincluded a large number of figures in order to make the arguments clearer.
Theorem 1.2 raises some open questions, apart from the uniqueness problem.For example, the sequence (xn)n≥0 is bounded, but we do not know whether thecorresponding sequence (un, vn)n≥0 of solutions is bounded or not (the numericalexperiments indicate that it is unbounded). A related question is the following. Inthe case where (xn)n≥0 is bounded, consider the limit, say x∞ of xn. Does thesolution of (1.4)-(1.6) blow-up in a finite time when v(0) ≥ x∞?
Notice the importance of the term 2ru in (1.4). Indeed, it is the non-autonomous
term that allows the existence of infinitely many solutions (the same phenomenonappears in the semilinear elliptic problems, see [3]). More surprisingly, it is also the
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 5
non-autonomous term (even though it is linear) that makes some solutions blow-upin a finite time (when this term is removed, all the solutions are global solutions).
For completeness, let us indicate that for functions ϕ of the form (1.3), theLagrangian L defined above has the following form.
L(ϕ) =
∫ ∞0
{−u′v − 2uv
r+ v′u−m(v2 − u2) + ω(u2 + v2) +G(v2 − u2)
}r2dr.
Finally, observe that little seems to be known on the Cauchy problem (initialvalue problem) for (1.1) except global existence for small initial data (see Diasand Figueira [6, 7, 8]) and more recently global existence of weak solutions (seeDias and Figueira [9, 10]). In particular, we do not know whether stationary statesare stable or not. For instance, some authors claim that the stationary states areunstable [4, 34, 17], while others find regions of stable behaviour by using numeri-cal [1] or analytic [28] arguments.
The notes are organized as follows. In Section 2, we introduce the notationand we collect some basic properties of system (1.4)-(1.5). In Section 3, we studythe blowing-up of solutions and in Section 4, we establish the main boundednessproperty. Finally in Section 5, we complete the proof of Theorem 1.2, and Section 6is devoted to a few further comments.
2. Preliminary results on system (1.4)-(1.5)
We begin by introducing some notation. We consider a, b > 0 such that
F (a2) = m− ω, F (b2) = m+ ω. (2.1)
It is clear from (H1) that a and b are uniquely determined. We also define thefunctions G and H by
G(x) =
∫ x
0
F (s) ds x ∈ R (2.2)
H(u, v) =1
2{G(v2 − u2)−m(v2 − u2) + ω(v2 + u2)} (u, v) ∈ R2. (2.3)
It follows from (H1) thatG(x) ≥ 0 for x ≥ 0, (2.4)
and that there exist x0 > 0 and ε > 0 such that
G(x) ≥ (m− ω + ε)x for x ≥ x0. (2.5)
Putting together (2.4) and (2.5), it follows that there exists C such that
G(x) ≥ (m− ω + ε)x− C for x ∈ R. (2.6)
and so, for ε > 0 possibly smaller,
2H(u, v) ≥ ε(u2 + v2)− C for (u, v) ∈ R2. (2.7)
2.1. The associated Hamiltonian system. We will consider the Hamiltoniansystem associated with (1.4)-(1.5), which is{
u′ = v[F (u2 − v2)− (m− ω)], (2.8)
v′ = u[F (u2 − v2)− (m− ω)]. (2.9)
where ′ denotes the differentiation with respect to the r variable. The correspondingHamiltonian is H. The basic properties of H are summarized in the followinglemma.
6 THIERRY CAZENAVE
Lemma 2.1. H has the following properties.
(i) The critical points of H are (0, 0) and (0,±a),(ii) the minimum of H is negative and is achieved for (u, v) = (0,±a),
(iii) H(u, v)→∞ as |u|+ |v| → ∞,(iv) {H(u, v) ≤ H(0, b)} ⊂ {F (v2 − u2) ≤ m+ ω}, where b is defined by (2.1).
Proof. (i) is immediate, and (iii) follows from (2.7). Therefore, in order to prove(ii), it remains to verify that the minimum of H is negative. This is clear since
d
dvH(0, v) = v[F (v2)− (m− ω)] < 0 for 0 < v < a.
Note that in particularH(0, v) < 0 for 0 < v < a. (2.10)
Let (u, v) be such that F (v2 − u2) > m + ω. In particular, v2 − u2 > b2 andF (s)−m > ω for s ∈ [b2, v2 − u2]. It follows that
H(u, v) = H(0, b) +1
2
∫ v2−u2
b2(F (s)−m) ds+
ω
2(v2 + u2 − b2)
≥ H(0, b) +ω
2(v2 − u2 − b2) +
ω
2(v2 + u2 − b2)
= H(0, b) + ω(v2 − u2 − b2) + ωu2 > H(0, b).
Hence (iv). �
In order to study the solutions of (2.8)-(2.9), we define the curves ΓC for C ∈[minH,∞) by
ΓC = {(u, v) ∈ R2; H(u, v) = C}.It is clear from the definition of H that the curves ΓC are symmetric about boththe u and v axes. The properties of ΓC are described in the following lemma andin Figure 1.
Lemma 2.2. Let C ∈ [minH,∞). Then
(i) if C = minH, then ΓC = {0, a} ∪ {0,−a},(ii) if C ∈ [minH, 0), then ΓC is the union of two connected compact curves Γ+
C
and Γ−C with Γ+C ⊂ {v2 > u2; v > 0} and Γ+
C ⊂ {v2 > u2; v < 0},(iii) Γ0 is a connected compact curve and (0, 0) ∈ Γ0,(iv) if C > 0, then ΓC is a connected compact curve.
Proof. (i) follows from Lemma 2.1 (ii). Furthermore, it follows from (2.3) and (2.4)that if v2 ≤ u2, then H(u, v) ≥ 0. Therefore, if C ∈ [minH, 0), then ΓC ⊂ {v2 >u2}. It follows from the symmetries of H and Lemma 2.1 (iii) that ΓC is the unionof at least two connected closed curves. The uniqueness of the curves Γ+
C and Γ−Cfollows from the fact that F is increasing on (0,∞). The compactness propertyis a consequence of Lemma 2.1 (iii). Properties (iii) and (iv) follow from similarconsiderations. �
We can now describe the solutions of (2.8)-(2.9), which are illustrated by arrowsin Figure 1.
Lemma 2.3. Let (x, y) ∈ R2 and let (u, v) be the solution of (2.8)-(2.9) such thatu(0) = x and v(0) = y. Then
(i) if H(x, y) = 0 or H(x, y) = minH, then (u, v) ≡ (x, y),
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 7
-
v
6u
H = C < 0
H = 0
H = C > 0
Figure 1. The energy levels {H = C}
(ii) if H(x, y) = C ∈ (minH, 0) and (x, y) ∈ Γ±C then (u, v) is periodic and
(u, v)(R) = Γ±C ,(iii) if H(x, y) = C ∈ (0,∞), then (u, v) is periodic and (u, v)(R) = ΓC . Further-
more, if xy > 0, then the first zero of uv is a zero of v and the zeroes of uand v alternate.
Proof. (i) follows from Lemma 2.1 (i). Next, if ∇H(x, y) 6= 0 then ΓC is boundedand |∇H(u, v)| ≥ α > 0 for (u, v) ∈ ΓC ; and so the solution (u, v) must cover ina finite time the connected component of ΓC to which (x, y) belongs. Thereforeproperty (ii) and the first part of property (iii) follow from Lemma 2.2. For provingthe second part of property (iii) it is sufficient by symmetry to consider the casex > 0, y > 0. Since 2H(u, 0) = G(u2) + (m + ω)u2, it follows that H(u, 0) is anincreasing function of u that ranges from 0 to +∞. Thus ΓC intersects {v = 0, u >0}. Since (u, v)(R) = ΓC , uv has to vanish. Let ρ be the first zero of uv. Since(0, 0) 6∈ ΓC , we have either u(ρ) = 0, v(ρ) > 0 or u(ρ) > 0, v(ρ) = 0. If u(ρ) = 0,then u′(ρ) ≤ 0, which implies by (2.8) that F (v(ρ)2) < m−ω; and so 0 < v(ρ) < a.
8 THIERRY CAZENAVE
By (2.10) , we have then C = H(0, v(ρ)) < 0, which is a contradiction. Thus,v(ρ) = 0 and u(ρ) > 0. By a similar argument, one proves that the zeroes of u andv alternate, which completes the proof. �
2.2. Basic properties of the system (1.4)-(1.5). We now study the first prop-erties of system (1.4)-(1.5). The local existence properties are described in thefollowing Lemma.
Lemma 2.4. For every x ∈ R, there exists a unique, maximal solution (ux, vx)of (1.4)-(1.5) such that (ux(0), vx(0)) = (0, x). (ux, vx) is defined on the maximalinterval interval [0, Rx), (ux, vx) ∈ C1([0, Rx),R2), and if Rx < ∞ then |ux(r)| +|vx(r)| → ∞ as r ↑ Rx. In addition,
(ux, vx) depends continuously on x in C1([0, R],R2), for every R < Rx. (2.11)
Proof. We write system (1.4)-(1.6) in the formu(r) =
1
r
∫ r
0
s2v(s)[F (v2 − u2)− (m− ω)] ds,
v(r) =
∫ r
0
u(s)[F (v2 − u2)− (m+ ω)] ds,
Since the integrand is a locally Lipschitz-continuous function of (u, v), existenceof a maximal solution follows from the classical contraction mapping argument.The continuous dependence in C([0, Rx),R2) is easily seen on the above formula;then the continuous dependence in C1([0, Rx),R2) follows from the equations (1.4)-(1.5). �
We shall also need the following stability result.
Lemma 2.5. Let C > minH. For every T > 0 and ε > 0, there exists R > 0 withthe following property. Let (x, y) ∈ ΓC and let ρ > R. Let (u, v) be the solutionof (2.8)-(2.9) with initial data (x, y) and let (υ,$) ∈ C1([0, τ(ρ, x, y)),R2) be themaximal solution ofυ′ = − 2υ
ρ+ r+$[F ($2 − υ2)− (m− ω)],
$′ = υ[F ($2 − υ2)− (m+ ω)],(2.12)
such that (υ(0), $(0)) = (x, y). It follows that supr∈[0,T ]
|u(r)−υ(r)|+|v(r)−$(r)| ≤ ε.
Proof. First, observe that since ρ > 0, system (2.12) is not singular; and so theinitial value problem for (2.12) is well-posed. Let M = sup{|u| + |v|; (u, v) ∈ΓC} <∞. It follows easily from a contraction mapping argument that there existsµ > 0 such that for every ρ ≥ 1 and(x, y) ∈ R2 with |x| + |y| ≤ 2M the maximalsolution (υ,$) ∈ C1([0, τ ],R2) of (2.12) such that (υ,$)(0) = (x, y) satisfies τ > µand |υ(r)|+ |$(r)| ≤ 4M for every r ∈ [0, µ]. Let (υj , $j) ∈ C1([0, τj ],R2), j = 1, 2be two such solutions of (2.12) corresponding to ρj and (xj , yj), and let L be theLipschitz constant of the right hand side of (2.12) on the ball of R2 of radius 4M .It follows from (2.12) that
|υ2(r)− υ1(r)|+ |$2(r)−$1(r)| ≤ |x2 − x1|+ |y2 − y1|+ 8Mµ( 1
ρ2+
1
ρ1
)+ L
∫ r
0
[|υ2(s)− υ1(s)|+ |$2(s)−$1(s)|] ds,
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 9
for 0 ≤ r ≤ µ. On applying Gronwall’s lemma, it follows that there exists K ≥ 1,depending only on M , such that
|υ2(r)− υ1(r)|+ |$2(r)−$1(r)| ≤ K(|x2 − x1|+ |y2 − y1|+
1
ρ2+
1
ρ1
), (2.13)
for 0 ≤ r ≤ µ. Given T > 0 and 0 < ε < M , consider n such that nµ < T ≤ (n+1)µ,and let R be such that (n + 1)Kn + 1 ≤ εR. Let (x, y), (u, v) and (υ,$) be asin the statement of Lemma 2.5, and let ρ > R. On applying formula (2.13) with(x1, y1) = (x2, y2) = (x, y), ρ1 = ρ and ρ2 =∞, we obtain
|u(r)− υ(r)|+ |v(r)−$(r)| ≤ K
R≤M,
for 0 ≤ r ≤ µ. Therefore, we can iterate formula (2.13) with (x1, y1) = (υ(µ), $(µ)),(x2, y2) = (u(µ), v(µ)), ρ1 = ρ+ µ and ρ2 =∞. After n+ 1 iterations, we get
|u(r)− υ(r)|+ |v(r)−$(r)| ≤ 1
R
n+1∑j=1
Kj ≤ (n+ 1)Kn+1
R≤ ε,
for r ≤ (n+ 1)µ. Hence the result, since (n+ 1)µ ≥ T . �
The following corollary will be important in the sequel.
Corollary 2.6. Let C > 0 and consider an integer n ≥ 1. There exists T (n,C) <∞ with the following property. If x ∈ R is such that (ux(r), vx(r)) ∈ ΓC andux(r)vx(r) > 0 for some r > T (n,C), then there exists R ∈ (r,Rx) such that uxvxhas n zeroes on (r,R) and these zeroes are alternatively a zero of vx and a zero ofux, the first being a zero of vx.
Proof. This is an immediate consequence of Lemma 2.3 (iii) and of Lemma 2.5applied with (x, y) = (ux(r), vx(r)) and (υ(t), $(t)) = (ux(r + t), vx(r + t)). �
For convenience, we now define the function Hx, for x ∈ R, by
Hx(r) = H(ux(r), vx(r)) for every r ∈ (0, Rx).
It follows from straightforward calculations that
d
drHx(r) =
2
rux(r)2[F (vx(r)2 − ux(r)2)− (m+ ω)], (2.14)
for every x ∈ R and r ∈ (0, Rx); and so, the sign of the variation of Hx, at sometime r > 0, is determined only by the region of the plane where (ux, vx)(r) belongs.More precisely, the sign of the variation of Hx depends only of the position of(ux, vx)(r) with respect to the hyperbola {F (v2 − u2) = m+ ω} (see Figure 2).
We shall need the following identities, which are easily obtained from (1.4)-(1.5)and hold for x ∈ R and r ∈ (0, Rx) whenever the right hand side makes sense.
(v2x − u2
x)′ =4
ru2x − 4ωuxvx, (2.15)(ux
vx
)′=
1
v2x
[(v2x − u2
x)[F (v2x − u2
x)− (m− ω)] + 2ωu2x −
2
ruxvx
], (2.16)( vx
ux
)′=
1
u2x
[−(v2
x − u2x)[F (v2
x − u2x)− (m− ω)]− 2ωu2
x +2
ruxvx
]. (2.17)
It will also be useful to keep in mind the velocity field of the dynamical system (1.4)-(1.5) for various values of r, which is displayed in Figures 3 to 5.
10 THIERRY CAZENAVE
-
v
6u
F (v2 − u2) = m+ ω
H decreasing H increasingH increasing
Figure 2. The sign of H ′x
Next, in the following lemmas, we collect some basic properties of a geometricnature for solutions of (1.4)-(1.5).
Lemma 2.7. Let x 6= 0, x 6= ±a. If, for some r0 > 0 we have ux(r0) = 0, thenvx(r0) 6∈ {0,±a} and u′x(r0) 6= 0. If, for some r0 > 0, we have vx(r0) ∈ {0,±a},then ux(r0) 6= 0 and v′x(r0) 6= 0.
Proof. First, observe that the rest points of (1.4)-(1.5) are (0, 0) and (0,±a). Fur-thermore, for r0 > 0, the Cauchy problem for (1.4)-(1.5) is locally well-posed forany initial datum (u0, v0) ∈ R2, for both r ≥ r0 and r ≤ r0. Thus, a rest pointcannot be reached in a finite time. Hence the result. �
Lemma 2.8. Let x ∈ R. Assume Rx =∞, and (ux, vx)→ (0,±a) as r→∞. Thenux has infinitely many zeroes.
Proof. It is equivalent to show that (ux, vx) cannot converge to (0,±a), while beingin one of the half-planes {u > 0} or {u < 0}. Let us first prove that (ux, vx) cannot
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 11
-
v
6u
a−a ?
6 66
??
��)
��)
��1
��1
Figure 3. The velocity field on the axes for r ≥ 0
converge to (0, a), while being in the half-plane {u > 0}. We argue by contradictionand we let (U, V ) = (ux, vx − a). By assumption (U, V ) → (0, 0) as r → ∞ andU > 0 for r large. The equations for U and V areU ′ = −2U
r+ (a+ V )[F (a2 + V 2 + 2aV − U2)− (m− ω)],
V ′ = U [F (a2 + V 2 + 2aV − U2)− (m+ ω)].
We have V 2 + 2aV −U2 → 0, and |V 2 + 2aV −U2| ≤ C(U + |V |). Therefore if weset
d = F ′(a2) > 0,
we have
F (a2 + V 2 + 2aV − U2) = (m− ω) + 2adV + o(U + |V |).
12 THIERRY CAZENAVE
-
v
6u
F (v2 − u2) = m+ ω
H = 0
H = C > 0
6
6
6
6
?
?
?
?
���
6
?
���
���@@I
@@R��*
���
���
���
���
���
PPi
PPq���
Figure 4. The velocity field for r > 1/ω
The equations for U and V become thenU ′ = −2U
r+ 2a2dV + o(U + |V |),
V ′ = −2ωU + Uo(U + |V |).(2.18)
Observe first that from the last equation in (2.18), it follows that V ′ < 0 for r large.Since V → 0, we get V > 0, for r large. It follows from (2.18) that
(V − U)′ = ω(V − U)− (ω − 2
r)U − (ω + 2a2d)V + o(|U |+ |V |) ≤ ω(V − U),
for r large. Therefore e−ωr(V −U) is nonincreasing for r large. Since e−ωr(V −U)→0, as r →∞, it follows that V −U ≥ 0 for r large. Thus we can replace o(|U |+ |V |)by o(V ). We get from the first equation in (2.18)
U ′ = a2dV − 2
rU + o(V ) ≥ a2d
2V ≥ 0,
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 13
-
v
6u
F (v2 − u2) = m+ ω
����
���
���
���
���
���
���
���
���
���
��
u = rωvH = 0
H = C > 0
6
6
6
6
?
?
?
?
���
���
��
��
Figure 5. The velocity field for r < 1/ω
for r large. Thus U is increasing for r large, which is a contradiction since U > 0and U → 0. By symmetry, (ux, vx) cannot converge to (0,−a), while being in thehalf-plane {u < 0}. The same proof applies to the two other situations. �
Lemma 2.9. Let x 6= 0, and let θ = min{Rx, 1ω}. Then v2
x(r) − u2x(r) ≥ e−4x2,
for 0 ≤ r < θ.
Proof. Let r = sup{r ∈ [0, θ); v2x ≥ u2
x on [0, r)}. Observe that on [0, ρ) we haveuxvx ≤ v2
x. Thus, for r ∈ [0, ρ) we have by (2.15)
(v2x − u2
x)′(r) =4
ru2x − 4ωuxvx ≥ 4ω(u2
x − uxvx) ≥ 4ω(u2x − v2
x).
It follows that on[0, ρ), e4ωr(v2x − u2
x) is nondecreasing. Therefore
(v2x − u2
x)(r) ≥ e−4ωrx2 ≥ e−4x2 > 0,
for r ∈ [0, ρ). Therefore ρ = θ. Hence the result. �
14 THIERRY CAZENAVE
Lemma 2.10. Let x 6= 0. Assume that (ux, vx) ∈ {F (v2 − u2) ≥ m+ ω} on someinterval [r0, r1) ⊂ [0, Rx). Then r1 − r0 <
32ω .
Proof. {F (v2−u2) ≥ m+ω} is the union of two disconnected components {F (v2−u2) ≥ m+ ω, v > 0} and {F (v2 − u2) ≥ m+ ω, v < 0}. Since vx is continuous, wemay assume, say, vx > 0 on [r0, r1). Thus, we have
− 1 <uxvx
< 1 (2.19)
on [r0, r1). From (2.16) we get(uxvx
)′≥ 1
v2x
[2ωv2
x −2
ruxvx
]on [r0, r1); and so (
r2uxvx
)′≥ 2ωr2. (2.20)
Integrating (2.20) between r0 and r1, and using (2.19), we get
r21 + r2
0 ≥2ω
3(r3
1 − r30) =
2ω
3(r1 − r0)(r2
1 + r1r0 + r20) ≥ 2ω
3(r1 − r0)(r2
1 + r20).
This completes the proof. �
Proposition 2.11. There exists a function Φ ∈ C(R,R) with the following prop-erty. If x ∈ R is such that Rx >
1ω and (ux, vx) ∈ {F (v2 − u2) ≥ m+ ω} on some
interval [r0, r1) ⊂ [ 1ω , Rx), then r1 < Rx and (|ux|+ |vx|)(r) ≤ Φ((|ux|+ |vx|)(r0)),
for every r ∈ [r0, r1).
Proof. By continuity and symmetry, we may assume, (ux, vx) ∈ {F (v2 − u2) ≥m+ ω, v > 0} for r ∈ [r0, r1). Let D+ = {F (v2 − u2) ≥ m+ ω, v ≥ 0, u ≥ 0}, andD− = {F (v2 − u2) ≥ m + ω, v > 0, u ≤ 0}. Observe that, from (1.5), u′x > 0 on[r0, r1); and so, if (ux, vx)(ρ) ∈ D+ for some ρ ∈ [r0, r1), then (ux, vx)(r) ∈ D+,for all r ∈ [ρ, r1). Thus, we may assume that (ux, vx)(r) ∈ D− on some interval[r0, r2], and (ux, vx)(r) ∈ D+ on [r2, r1). On (r0, r2], we have u′x > 0 and v′x < 0.Therefore,
(|ux|+ |vx|)(r) ≤ (|ux|+ |vx|)(r0), (2.21)
for every r ∈ [r0, r2]. Next, v2x − u2
x is nonincreasing on [r2, r1) by (2.15). Thus,by (1.4)-(1.5),
(ux + vx)′ ≤ (ux + vx)F ((v2x − u2
x)(r2)), (2.22)
on [r2, r1). Finally, by Lemma 2.10, we have r1 − r2 ≤ 32ω . Therefore, (2.21)
and (2.22) yield
(|ux|+ |vx|)(r) ≤ (|ux|+ |vx|)(r0)e32ωF ([(|ux|+|vx|)(r0)]2),
for every r ∈ [r0, r1). Hence the bound on (|ux| + |vx|); and so r1 < Rx, unlessr1 = Rx =∞. This is ruled out by Lemma 2.10. �
Proposition 2.12. (See Figure 6.) Let x 6= 0. Assume that for some r0 > 0, wehave vx(r0) ≤ ux(r0) and ux(r0) > 0 (respectively, ux(r0) ≤ vx(r0) and ux(r0) < 0).Then there exists r0 < r1 < Rx such that |ux| > 0 on (r0, r1), and ux(r1) = 0,|vx(r1)| > a. In addition , either vx(r0) > 0 (respectively , vx(r0) < 0), and thenvx has exactly one zero in (r0, r1), or else vx(r0) ≤ 0 (respectively vx(r0) ≥ 0), andthen |vx| > 0, on (r0, r1].
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 15
-
v
6u
F (v2 − u2) = m+ ω
%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%
eeeeeeeeeeeeeeeeee
eeeeeeeeeeeeeeeeee
?
Figure 6. Illustration for Proposition 2.12
Proof. Assume for example vx(r0) ≤ ux(r0) and ux(r0) > 0. Suppose first thatvx(r0) ≥ −ux(r0). Because the velocity field on {v = −u, u > 0} points towards{v < −u}, (ux, vx) can only enter {−u ≤ v ≤ u, u > 0} by crossing the diagonal{u = v}; and so, by Lemma 2.9, r0 >
1ω . For r ≥ r0, and while (ux, vx) belongs to
{−u ≤ v ≤ u, u > 0}, we have by (2.17)( vxux
)′≤ 1
u2x
( 2
r0uxvx − 2ωu2
x
)≤ −2
(ω − 1
r0
)< 0.
Thus, (ux, vx) exits {−u ≤ v ≤ u, u > 0} in a finite time, by crossing the u-axisonce, if vx(r0) > 0, and then, by crossing the diagonal {v = −u, u > 0} (notethat in the region {−u ≤ v ≤ u, u > 0}, the trajectory remains trapped in the set{H(u, v) ≤ Hx(r0)} due to (2.14)). Therefore, we may assume vx(r0) < −ux(r0).Suppose now that F (v2
x(r0) − u2x(r0)) < m + ω. While (ux, vx) belongs to the
region D = {F (v2 − u2) ≤ m + ω, u + v ≤ 0, u ≥ 0}, we have, by (2.5), H ′x ≤ 0;and so (ux, vx) is bounded. We claim that (ux, vx) must exit D in a finite time.
16 THIERRY CAZENAVE
Indeed, assume that (ux, vx) remains in D for r ≥ r0. We have v′x ≤ 0, thus vxhas a negative limit as r → ∞. It is not difficult to show that ux also has a limit(compare [5], proof of Lemma 2.10). The limit (k, h) of (ux, vx) is therefore a restpoint of (1.4)-(1.5) in the half-plane {v < 0}. Thus, (k, h) = (0,−a). This is ruledout by Lemma 2.8. Therefore, (ux, vx) must exit D in a finite time. Next, (ux, vx)cannot exit D by crossing the diagonal {u = −v, u > 0}, since the velocity fieldon {v = −u, u > 0} points towards {v < −u}. If (ux, vx) exits D by crossing thev-axis, then u′x < 0; and so vx < −a, which is the desired estimate. Therefore, wemay suppose that (ux, vx) exits D by crossing the hyperbola {F (v2−u2) = m+ω}.When (ux, vx) belongs to D′ = {F (v2 − u2) ≥ m + ω, v < 0, u ≥ 0}, we haveby (2.15)
(v2x − u2
x)′ ≤ 0;
and so, (ux, vx) cannot exit D′ by crossing again the hyperbola {F (v2 − u2) =m + ω}. By Lemma 2.10 and Proposition 2.11, we know that (ux, vx) must exitD′ in a finite time by crossing the v-axis; and when it does so, we have vx ≤ −a.Hence the result. �
Corollary 2.13. Let x 6= 0. Assume that, for some 0 < r0 < Rx, we havevx(r0) = 0. Then there exists r1 ∈ (r0, Rx), such that |ux| > 0, |vx| > 0 on (r0, r1),and ux(r1) = 0, |vx(r1)| > a.
Proof. If ux(r0) > 0, we have vx(r0) ≤ ux(r0), and if ux(r0) < 0, we have ux(r0) ≤vx(r0). Thus, we may apply Proposition 2.12. �
Lemma 2.14. Let x 6= 0 be such that Rx =∞. Assume that for some r0 > 0, wehave |ux| > 0 on [r0,∞). Then , we have the following .
(i) uxvx > 0 on (r0,∞),
(ii) there exists C such that 0 < |ux(r)| < |vx(r)| < Ce−m−ω
2 r, for r ∈ (r0,∞).
Proof. Assume, for example, ux > 0 on [r0,∞). Then, by Proposition 2.12, wemust have
0 < ux < vx,
on [r0,∞). This proves (i). By (2.15), and the above inequality, we have
(v2x − u2
x)′ < 0,
for r large. Therefore, by Lemma 2.10, we have
F (v2x − u2
x) < m+ ω,
for r large; and so, v′x < 0, for r large. Thus, vx has a limit as r → ∞. It isnot difficult to show that ux also has a limit (compare [5], proof of Lemma 2.10).The limit (k, h) of (ux, vx) is therefore a rest point of (1.4)-(1.5). By Lemmas 2.1and 2.8, we have (k, h) = (0, 0). Therefore, for r large, we have
0 < F (v2x − u2
x) <m− ω
2.
It follows from (1.4)-(1.5) that we have then
(ux + vx)′ ≤ −m− ω2
(ux + vx);
from which the exponential decay follows. �
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 17
Lemma 2.15. Let x ∈ R and let 0 ≤ r0 < Rx satisfy ux(r0) = 0 and F (vx(r0)2) >m+ ω. Let ρ ∈ (r0, Rx] be defined by
ρ = sup{r ∈ [r0, Rx), uxvx > 0 on (r0, r)}.
Then, there exists r0 < r1 ≤ Rx such that the function f(r) = F (v2x(r)− u2
x(r))−(m+ω) satisfies f(r) > 0 on [r0, r1)and f(r) < 0 on (r1, ρ). In addition, if r1 = ρ,then r1 = ρ = Rx ≤ 1
ω .
Proof. Note first that |vx(r0)| > a, and so (see Figure 3) uxvx > 0 on (r0, r0 + ε),for some ε > 0. Therefore ρ > r0. Furthermore, since f(r0) > 0, we can define thenumber r1 ∈ (r0, ρ] by
r1 = sup{r ∈ [r0, ρ]; f(r) > 0 on (r0, r)},
and we may assume by symmetry that vx(r0) > 0. Suppose first that r1 = ρ. Thenwe have r1 = ρ = Rx ≤ 1
ω . To see this, observe that on [r0, r1], vx is nondecreasing;and so if we had ρ < Rx, we would have
vx(ρ) ≥ vx(r0) > a. (2.23)
By definition of ρ this would imply ux(ρ) = 0; and so we would get u′x(ρ) ≤ 0,which is incompatible with (2.23) (see Figure 3). Therefore, r1 = ρ = Rx, and thenby Proposition 2.11 we get Rx ≤ 1
ω . Suppose now that r1 < ρ. Then we have
f(r1) = 0, f ′(r1) ≤ 0. (2.24)
From (2.24) and (1.5), we get
v′x(r1) = 0. (2.25)
Furthermore, v2x − u2
x cannot be increasing in a neighborhood of r1 since otherwisef also would be increasing, which is impossible by definition of r1. Thus we have
(v2x − u2
x)′(r1) ≤ 0. (2.26)
It follows from (2.25) and (2.26) that
u′x(r1) ≥ 0. (2.27)
Next, we claim that there exists ε > 0 such that f(r) < 0 on (r1, r1 + ε). Indeedif u′x(r1) > 0, then it follows from (2.26) that (v2
x − u2x)′(r1) < 0, which implies
the claim. On the other hand, if u′x(r1) = 0, then it follows from (2.25) that(v2x − u2
x)′(r1) = 0. These formulas, together with (1.4) and (1.5) imply thatu′′x(r1) = 2
r1ux(r1) > 0 and v′′x(r1) = 0; and so
(v2x − u2
x)′′(r1) = 2vxv′x(r1)− 2uxu
′x(r1) = −4ux(r1)
r1< 0,
which again implies the claim. It remains to prove that f(r) < 0 on (r1, ρ). Weargue by contradiction and we assume that there exists r2 ∈ (r1, ρ) such thatf(r) < 0 on (r1, r2) and f(r2) = 0 (see Figure 7). From (2.15) and (2.26) we get
r1 ≥ux(r1)
2ωvx(r1). (2.28)
As well, we have f(r2) = 0 and f ′(r2) ≥ 0; and so
r2 ≤ux(r2)
2ωvx(r2). (2.29)
18 THIERRY CAZENAVE
-v
6u
�������������������������������������
u = v
F (v2 − u2) = m+ ω
?
(u(r1), v(r1))
(u(r2), v(r2))
Figure 7. Notation for Lemma 2.15
Note that on (r1, r2) we have v′x < 0; and so vx(r2) < vx(r1). Furthermore we have
ux(r1) > 0 and ux(r2) > 0, and f(r1) = f(r2) = 0. This implies ux(r2)2ωvx(r2) <
ux(r1)2ωvx(r1) .
Applying (2.28) and (2.29) we get r2 < r1, which is a contradiction. This completesthe proof of the lemma. �
Corollary 2.16. (See Figure 8.) Let x 6= 0, x 6= ±a. Assume that for somer0 ∈ [0, Rx) we have ux(r0) = 0. Then, one of the following properties holds .
(i) |vx(r0)| < a, and there exists r1 ∈ (r0, Rx) such that uxvx < 0 on (r0, r1),0 < |ux| < |vx| on (r0, r1), |vx(r1)| > a, and ux(r1) = 0;
(ii) |vx(r0)| > a, and there exists r1 ∈ (r0, Rx) such that uxvx > 0 on (r0, r1),0 < |ux| < |vx| on (r0, r1), |vx(r1)| < a, and ux(r1) = 0;
(iii) |vx(r0)| > a, Rx =∞, uxvx > 0 on (r0,∞), and 0 < |ux| < |vx| < Ce−m−ω
2 r,on (r0,∞);
(iv) |vx(r0)| > a, and there exists r1 ∈ (r0, Rx) such that uxvx > 0 on (r0, r1), andvx(r1) = 0;
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 19
-v
6u
�������������������������������������
u = v
a
���
(v)
Y
(iv)
(iii)N (ii)O
(i)
Figure 8. Illustration for Corollary 2.16
(v) |vx(r0)| > a, Rx ≤ 1ω , uxvx > 0 on (r0, Rx) and F (v2
x − u2x) > m + ω on
(r0, Rx).
Proof. Assume for example vx(r0) > 0. By Lemma 2.7, we have vx(r0) 6= 0, a. Ifvx(r0) < a, then u′x(r0) < 0. Applying Proposition 2.12, we get (i), except theproperty 0 < |ux| < |vx| on (r0, r1). Observe that by (2.10) we have Hx(r0) < 0,and that on (r0, r1), we have H ′x ≤ 0 and v′x ≥ 0, until possibly (ux, vx) crosses thehyperbola H = {F (v2 − u2) = m + ω}; and so, until then , we have Hx < 0, fromwhich it follows (cf. Lemma 2.2 (ii)) that 0 < |ux| < |vx|. Now, the velocity fieldon H ∩ {u < 0} points towards {F (v2 − u2) > m + ω}. Thus, if (ux, vx) crossesH, it cannot come back in the region {F (v2 − u2) ≤ m+ ω} on (r0, r1). Hence (i).Suppose now vx(r0) > a, and let ρ = sup{r ∈ (r0, Rx); uxvx > 0 on (r0, r)}. Ifρ < Rx, we have either vx(ρ) = 0, in which case we get (iv), or else ux(ρ) = 0.In the last case, we must have 0 < |ux| < |vx| on (r0, ρ), by Proposition 2.12;hence (ii). Assume now ρ = Rx. By Lemma 2.15 we have either (v) or else there
20 THIERRY CAZENAVE
exists r0 < r1 < ρ such that the function f(r) = F (v2x(r)−u2
x(r))−(m+ω) satisfiesf > 0 on (r0, r1) and f < 0 on (r1, ρ). It follows from (2.14) that Hx(r) ≤ Hx(r1)for r1 ≤ r ≤ ρ = Rx. By Lemma 2.1 (ii) and Lemma 2.4, this implies Rx = ∞.Applying Lemma 2.14, we obtain (iii). This completes the proof. �
Lemma 2.17. Let x 6= 0. Assume that vx has at least two zeroes. Then, betweentwo consecutive zeroes of vx, ux has an odd number of zeroes .
Proof. Assume for example that vx > 0 on (r0, r1), and vx(r0) = vx(r1) = 0.Because v′x < 0 when vx = 0, ux > 0, and v′x > 0 when vx = 0, ux < 0, we haveux(r0)ux(r1) < 0. Hence the result. �
Lemma 2.18. Let x 6= 0. Assume that Rx ≥ 1ω , and that ux has a finite number
of zeroes . Then Rx =∞, and |ux|+ |vx| → 0 as r →∞.
Proof. In the region {F (v2 − u2) ≤ m+ ω}, Hx is nonincreasing, by (2.15); and so(ux, vx) cannot blow up. Now, the region{F (v2− u2) > m+ω} is the union of twoconnected, open components D1 and D2, where D1 = {F (v2−u2) ≤ m+ω, v > 0}and D2 = {F (v2 − u2) ≤ m + ω, v < 0}. Considering the velocity field on theboundary of D1, and for r ≥ 1
ω , it is not difficult to show that (ux, vx) can enterD1 only in the half-plane {u < 0}, and can exit D1 only in the half-plane {u > 0}.Thus, when the solution crosses D1 after r = 1
ω , ux has at least one zero. Bysymmetry, the same holds for D2; and so, (ux, vx) can only cross D1 or D2 a finitenumber of times. Thus, by Proposition 2.11, we have Rx = ∞. Note that vxcannot vanish after the last zero of ux, by Corollary 2.13. Therefore, we may applyLemma 2.14, from which the result follows. �
Lemma 2.19. Assume that F (x2) ≤ m+ω. Then Rx =∞. Furthermore F (v2x(r)−
u2x(r)) ≤ m+ ω and Hx(r) ≤ H(0, x), for every r ≥ 0.
Proof. It follows from Lemma 2.1 (iv) that {H(u, v) ≤ H(0, x)} ⊂ {F (v2 − u2) ≤m + ω}. Therefore (2.14), together with an obvious continuity argument, impliesthat Hx(r) ≤ H(0, x) for every r ∈ [0, Rx); and so F (v2
x(r) − u2x(r)) ≤ m + ω, for
every 0 ≤ r < Rx. It follows from Lemma 2.1 (iii) and Lemma 2.4 that Rx =∞. �
3. A blow-up result
In this section, we prove that under some additional assumption on F and for|x| large enough, the solution (ux, vx) blows up in a finite time, and remains inthe region {F (v2 − u2) > m + ω}. This shows the existence of an a priori boundon admissible initial data for the solutions of (1.4)-(1.7). Let us introduce somenotation. We define the function Φ on [0,∞) by
Φ(x) = F (F−1(m+ ω) + 4ωe−4x)− (m+ ω) ≥ 0.
By assumption (H1), Φ is increasing. We set
Ψ(x) =
∫ x
0
sΦ(s) ds,
for x ≥ 0. Ψ is also increasing on [0,∞) and we assume in this section that∫ ∞0
ds√1 + Ψ(s)
<∞. (3.1)
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 21
Remark 3.1. Assumption (3.1) requires that F goes to ∞ fast enough at infinity.For example, it is easy to check that (3.1) is satisfied if there exists α > 0 and β > 2such that
F (x) ≥ α(log x)β ,
for x large,
Our main result is the following.
Proposition 3.2. For every τ > 0, there exists B(τ) such that if |x| ≥ B(τ), thenRx < τ and F (v2
x − u2x) > m+ ω for r ∈ (0, Rx).
Proof. We use a slightly simpler argument than that of [2]. Let τ ∈ (0, 12ω ) and let
B > 0 be large enough so that
F (e−4B2) > m+ ω, (3.2)
and ∫ ∞0
ds√B2 + 2ω
3 Ψ(s)< τ. (3.3)
It is clear from (3.1) that (3.2) and (3.3) can be achieved for B large enough.By symmetry, it is sufficient to consider the case x > 0. Let x ≥ B and letθ = min{Rx, τ}. From (2.15), we get for r ∈ [0, θ)
(v2x − u2
x)′(r) =4
ru2x − 4ωuxvx ≥ 8ωu2
x − 4ωuxvx = 8ω(u2x − v2
x) + 4ω(2v2x − uxvx).
By Lemma 2.9, we have uxvx ≤ v2x on [0, θ); and so
(v2x − u2
x)′(r) ≥ 8ω(u2x − v2
x) + 4ωv2x,
for r ∈ [0, θ). Therefore,
[e8ωr(v2x − u2
x)]′(r) ≥ 4ωe8ωrv2x ≥ 4ωv2
x, (3.4)
for r ∈ [0, θ). Integrating (3.4) between 0 and r < θ < 12ω , and applying (3.2), we
get
(v2x − u2
x)(r) ≥ F−1(m+ ω) + 4ωe−4
∫ r
0
vx(s)2ds,
for r ∈ [0, θ). Therefore,
(v2x − u2
x)− (m+ ω) ≥ Φ(f(r)), (3.5)
for r ∈ [0, θ), with
f(r) =
∫ t
0
vx(s)2ds,
for r ∈ [0, θ). Observe that by (3.5) and (1.5), vx is increasing on [0, θ), and so f isconvex. On the other hand, it follows from (3.5) that we can apply formula (2.20),which yields
ux(r) ≥ 2ω
3rvx(r), (3.6)
for r ∈ [0, θ). It follows from (1.5), (3.5) and (3.6) that
v′x(r) ≥ 2ω
3rvx(r)Φ(f(r)), (3.7)
for r ∈ [0, θ). On multiplying (3.7) by vx we get
f ′′(r) ≥ 2ω
3rf ′(r)Φ(f(r)), (3.8)
22 THIERRY CAZENAVE
for r ∈ [0, θ). Since f is convex, positive and increasing, we have rf ′(r) ≥ f(r);and so (3.8) yields
f ′′(r) ≥ 2ω
3f(r)Φ(f(r)), (3.9)
for r ∈ [0, θ). We now multiply (3.9) by f ′ to get
(f ′2)′(r) ≥ 4ω
3f ′(r)f(r)Φ(f(r)) =
4ω
3Ψ(f(r))′, (3.10)
for r ∈ [0, θ). Integrating (3.10) we obtain
(f ′2)(r) ≥ x2 +4ω
3Ψ(f(r)), (3.11)
for r ∈ [0, θ). Taking the square root of (3.11) we get
f ′(r) ≥√x2 +
4ω
3Ψ(f(r)), (3.12)
for r ∈ [0, θ). It follows from (3.12) and (3.3) that
r ≤∫ f(r)
0
ds√x2 + 4ω
3 Ψ(s)≤∫ ∞
0
ds√x2 + 4ω
3 Ψ(s)< τ, (3.13)
for r ∈ [0, θ). By definition of θ, this implies that θ = Rx; and so Rx < τ . Thiscompletes the proof. �
4. The main estimate
We introduce the sets In, An, and En (see figure 9) defined for n ∈ N by
In = {x > a,∃rx,n ∈ (0, Rx) s.t. both ux and vx
have exactly n zeroes on (0, rx,n) and ux(rx,n) = 0},
An = {x > a,Rx = +∞ and both ux and vx
have exactly n zeroes on (0,∞) and |ux|+ |vx| → 0 as r →∞},
En = In ∪An.For x ∈ An, we set
rx,n = +∞.Finally, we define the sets I, A and E by
I = ∪n≥0
In, A = ∪n≥0
An, E = ∪n≥0
En.
In this section, we shall establish a boundedness property for solutions of (1.4)-(1.5),with initial data in En . Our main result is the following.
Proposition 4.1. Let n ∈ N and assume En 6= ∅. It follows that
supx∈En
supr∈[0,rx,n)
|ux(r)|+ |vx(r)| <∞.
Before proceeding to the proof of Proposition 4.1, we need some preliminaryresults, where we set topological features of the sets En and properties of trajectorieswith initial data in En. We begin with
Lemma 4.2. For every n ≥ 0, In is an open subset of (a,+∞).
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 23
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%%%%%%%%%%%%%%%%%%
eeeeeeeeeeeeeeeeee
eeeeeeeeeeeeeeeeee
6
> x1 y1
Figure 9. Trajectories with x1 ∈ I1 and y1 ∈ A1
Proof. Observe that if vx = 0 (respectively ux = 0), then by Lemma 2.7, we havev′x 6= 0 (respectively u′x 6= 0). Therefore, the result follows from (2.11). �
Next, we consider the case n = 0, for which the proof is different from the generalcase. We begin with the following result.
Lemma 4.3. Let x > a and let ρ = sup{r ∈ (0, Rx), 0 < ux < vx on (0, r)}. Thereexists θ ∈ [0, ρ] such that
(i) Hx and vx are increasing and F (v2x(r)− u2
x(r)) > m+ ω on (0, θ);(ii) Hx and vx are decreasing and F (v2
x(r)− u2x(r)) < m+ ω on (θ, ρ).
Proof. This follows from Lemma 2.15 and (1.5). �
Corollary 4.4. Assume E0 6= ∅. It follows that
supx∈E0
supr∈[0,rx,0)
|ux(r)|+ |vx(r)| <∞.
24 THIERRY CAZENAVE
Proof. Consider C > 0 and set
M = sup{|u|+ |v|; H(u, v) = C}.If x ∈ E0 is such that Hx(r) ≤ C on (0, rx,0), then |ux| + |vx| ≤ M on (0, rx,0).Consider now x ∈ E0 such that there exists r ∈ (0, rx,0) with Hx(r) > C. Observethat if x ∈ I0 then vx(rx,0) ∈ (0, a) (Corollary 2.16, case (ii)); and so Hx(rx,0) < 0(by (2.10)). On the other hand, if x ∈ A0 then Hx(r) → 0, as r → rx,0. Thusin both cases there exists τ ∈ (0, rx,0) such that Hx(τ) = C and Hx(r) < C forr ∈ (τ, rx,0). By Corollary 2.6 and the definition of E0, it follows that there existsΘ < ∞ independent of x ∈ E0 such that τ ≤ Θ. Let us now apply Corollary 2.16with r0 = 0. By definition of E0 we are either in case (ii) if x ∈ I0 or in case (iii)if x ∈ A0. It follows that 0 < ux < vx on (0, rx,0). Therefore we may applyLemma 4.3 with ρ = rx,0. It follows that
0 ≤ ux ≤ vx ≤ vx(θ) on [0, rx,0); (4.1)
0 ≤ F (v2x − u2
x) ≤ m+ ω on [θ, τ ]. (4.2)
It follows from (4.2) and (1.5) that |v′x| ≤ (m+ ω)vx on (θ, τ). This implies that
vx(θ) ≤ vx(τ)e(m+ω)(τ−θ) ≤Me(m+ω)τ ≤Me(m+ω)Θ. (4.3)
The result follows from (4.1) and (4.3). �
We now consider the case n ≥ 1. In the following Lemma, we describe sometopological properties of (ux, vx), when x ∈ In.
Lemma 4.5. Let n ≥ 1 and x ∈ In. Then ,
(i) rx,n >1ω ,
(ii) the first zero of uxvx in (0, rx,n) is a zero of vx,(iii) the 2n zeroes of uxvx in (0, rx,n) are alternatively one zero of vx and one zero
of ux,(iv) the last two zeroes of uxvx in (0, rx,n] are zeros of ux,(v) we have |vx(rx,n)| < a.
Proof. Let r0 be the first zero of vx. By Corollary 2.13, ux has at least n zeroesin (r0, rx,n]. If ux has a first zero, say, ρ0 ∈ (0, r0), we have u′x(ρ0) < 0, and so0 < vx(ρ0) < a. By corollary 2.16, ux must have another zero in (0, r0); and so uxhas at least n + 2 zeroes in (0, rx,n], which is impossible by definition of En. Thisproves (ii). Therefore, (ux, vx) must cross the diagonal {u = v} in (0, r0). Hence (i),by Lemma 2.9. Next, note that, by Corollary 2.13, a zero of ux follows every zeroof vx; hence (iii) and (iv). Property (v) follows from (iii), (iv), and Corollary 2.16,part (ii). �
A similar result holds for solutions with initial data in An. More precisely, wehave.
Lemma 4.6. Let n ≥ 1, and x ∈ An. Then , the first zero of uxvx in (0, rx,n) is azero of vx. The 2n zeroes of uxvx in (0, rx,n) are alternatively one zero of vx andone zero of ux. Furthermore , for some r0 > 0, we have uxvx > 0 on (r0,∞), andthere exists C such that
(|ux|+ |vx|)(r) ≤ Ce−m−ω
2 r,
for r≥ 0.
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 25
Proof. The alternate character of the zeroes of ux and vx is proved with the argu-ment of the proof of Lemma 4.5. The other properties follow from Lemma 2.14. �
The following corollary is an immediate consequence of Lemmas 4.5 and 4.6.
Corollary 4.7. We have I ∩A = ∅. Furthermore, for k 6= j, we have Ik ∩ Ij = ∅.
Finally, we will need the following two results.
Lemma 4.8. For any K > 0 and n ∈ N, there exists C(K,n) with the followingproperty. Assume x > 0 is such that Rx >
1ω . Assume furthermore that there exists
r0, r1 ∈ [ 1ω , Rx] such that Hx(r0) ≤ K and ux has at most n zeroes on (r0, r1).
Then(|ux|+ |vx|)(r) ≤ C(K,n),
for r ∈ (r0, r1).
Proof. Considering the velocity field on {F (v2 − u2) = m + ω} for r > 1ω (see
Figure 4), we observe that (ux, vx) can enter D1 = {F (v2 − u2) > m + ω, v > 0}only in the half-plane {u < 0}. Similarly, (ux, vx) can exit D1 only in the half-plane{u > 0}. Therefore, when (ux, vx) crosses D1, ux has at least one zero. The sameholds for D2 = {F (v2 − u2) > m + ω, v < 0}; and so (ux, vx) crosses D1 or D2,at most n times on (r0, r1). When (ux, vx) crosses the set {F (v2 − u2) ≥ m + ω},we have H ′x ≤ 0, and so (ux, vx) remains bounded by (2.7). Therefore, the resultfollows from Proposition 2.11. �
Lemma 4.9. For every n ≥ 1, and K > 0, there exists τ(K,n) > 0, with thefollowing property . Suppose x ∈ En and Hx(ρ) = K, for some ρ ∈ [0, rx,n).Thenρ ≤ τ(K,n).
Proof. This follows from the definition of En and Corollary 2.6. �
As a consequence of the previous results, we have the following.
Corollary 4.10. For every x > a, x 6∈ I, one of the following properties is satisfied.
(i) Rx ≤ 1ω , and ux, vx and [F (v2
x − u2x)− (m+ ω)] are positive on (0, Rx),
(ii) Rx >1ω , both ux and vx have infinitely many zeroes on (0, Rx), and they are
alternate,(iii) x ∈ An, for some n ∈ N.
Proof. Let x be as above. We have u′x(0) > 0, and so ux > 0 and vx > 0, for sometime. Let
ρ = sup{r ∈ (0, Rx), vx > ux > 0 on (0, r)}.Suppose first that ρ = Rx. Then, either vx is bounded, and therefore Rx =∞. ByLemma 2.14, we have then x ∈ A0. Otherwise, vx is unbounded. In that case, v′xis positive somewhere. Thus,
J = {r ∈ [0, Rx); F (v2x(r)− u2
x(r)) > m+ ω} 6= ∅.Therefore, by Lemma 2.15, J is either the interval [0, Rx), or else some interval[0, R), with R < Rx. In the latter case, v′x is negative on [R,Rx), and so vx isbounded, which is a contradiction. Thus, J = [0, Rx). Applying Proposition 2.11,we must have Rx ≤ 1
ω , which implies property (i). Suppose now that ρ < Rx.Then, we have either ux(ρ) = 0, or else ux(ρ) = vx(ρ). In the first case, we have0 < vx(ρ) < a (see Figure 3). This implies x ∈ I0, (Corollary 2.16 (i)) which is a
26 THIERRY CAZENAVE
contradiction. Therefore ux(ρ) = vx(ρ); and so, Proposition 2.12 implies that vxhas at least one zero, which is the first zero of uxvx. Furthermore, by Corollary 2.13,after every zero of vx, the next zero of uxvx is a zero of ux. A zero of vx cannotbe followed by two zeroes of ux, since we would have x ∈ I, and this is ruled outby Corollary 4.7. Therefore, if vx has infinitely many zeroes, then (ii) is satisfied,and if vx has a finite number of zeroes, then (iii) holds, by Lemma 2.18. Hence theresult. �
Proof of Proposition 4.1. From Corollary 4.4, we need only prove the property forn ≥ 1. Therefore, by Lemma 4.5, we have rx,n >
1ω . Now, we fix K > 0. Consider
x ∈ En. By Lemma 4.5 (v) and (2.10), we have Hx(rx,n) < 0, if x ∈ In. Further-more, Hx(r) → 0 as r → ∞, if x ∈ An. Therefore, there exists r ∈ ( 1
ω , rx,n), withHx(r) < K. We define the number ρ by
ρ =
{1ω if Hx( 1
ω ) ≤ K,inf{r ∈ ( 1
ω , rx,n); Hx(r) < K} if Hx( 1ω ) > K.
By Lemma 4.8, we have
(|ux|+ |vx|)(r) ≤ C(K,n) for r ∈ [ρ, rx,n).
Therefore, it remains to bound the solution on (0, ρ). Note that, from Lemma 4.9,there exists Θ > 0, depending only on n, such that ρ ≤ Θ. We shall bound thesolution separately on [0, 1
ω ] and on [ 1ω , ρ].
Step 1. A bound on [0, 1ω ]. Let us establish the following estimate.
(|ux|+ |vx|)(r) ≤2
ωem+ωω (|ux|+ |vx|)
( 1
ω
), r ∈
[0,
1
ω
]. (4.4)
We first apply Corollary 2.16 with r0 = 0. By definition of En we are in case (iv). Itfollows that 0 < ux < vx untill the solution crosses the line {u = v}. By Lemma 2.9,this happens after 1
ω . Therefore we may apply Lemma 4.3 with ρ > 1ω . It follows
that
0 ≤ ux ≤ vx ≤ vx(θ) on[0,
1
ω
]; (4.5)
0 ≤ F (v2x − u2
x) ≤ m+ ω on(θ,
1
ω
); (4.6)
for some θ ∈ [0, 1ω ]. If θ = 1
ω , (4.4) follows from (4.5). If θ < 1ω , it follows from (4.6)
and (1.5) that |v′x| ≤ (m+ ω)vx on (θ, 1ω ). This implies that
vx(θ) ≤ e(m+ω)( 1ω−θ)vx
( 1
ω
)≤ e
m+ωω vx
( 1
ω
). (4.7)
(4.4) is then a consequence of (4.5) and (4.7). Consequently, in order to estimatethe solution on [0, ρ], it is sufficient to estimate it on [ 1
ω , ρ]. This is the object ofStep 2 below.Step 2. A bound on [ 1
ω , ρ]. Let us consider the regions ∆1, ∆2, ∆3, and ∆4,(see figure 10), defined by
∆1 = {F (v2 − u2) > m+ ω; H(u, v) > K, v > 0},∆2 = {F (v2 − u2) < m+ ω; H(u, v) > K,u > 0},∆3 = {F (v2 − u2) > m+ ω; H(u, v) > K, v < 0},∆4 = {F (v2 − u2) < m+ ω; H(u, v) > K,u < 0}.
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 27
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v
6u
F (v2 − u2) = m+ ω
H = K > 0
∆1
∆2
∆3
∆4
Figure 10. The sets ∆i, i = 1, 2, 3, 4
Assume that (ux, vx) enters ∆1 at some time t ∈ [1/ω, ρ]. Considering the velocityfield on the hyperbola {F (v2 − u2) = m + ω} and on the set {H(u, v) = K} (seeFigure 4), we obtain that ux(t) < 0. By Lemma 2.10, (ux, vx) must exit ∆1 in afinite time. The velocity field on the hyperbola {F (v2 − u2) = m+ ω} and on theset {H(u, v) = K} forces (ux, vx) to exit ∆1 in the half-plane {u > 0}. Therefore,ux has at least one zero in ∆1. The same holds for ∆3, by symmetry. A similarargument shows that, when (ux, vx) crosses ∆2 or ∆4, vx has at least one zero (seeFigures 3 and 4). Therefore, (ux, vx) crosses at most n times each of the sets ∆i.Thus, the proof of Proposition 4.1 will be complete, provided we show the followinglemma. �
Lemma 4.11. For every i ∈ {1, 2, 3, 4}, and for every Ξ > 0, A > 0, there existsB > 0, with the following property. Let x ∈ R be such that Rx > Ξ. If on someinterval [r0, r1] ⊂ [ 1
ω ,Ξ], we have (ux, vx) ∈ ∆i, and if (|ux| + |vx|)(r1) ≤ A, then(|ux|+ |vx|)(r) ≤ B, for every r ∈ [r0, r1].
28 THIERRY CAZENAVE
-
v
6u
%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%
eeeeeeeeeeeeeeeeee
eeeeeeeeeeeeeeeeee
δ1
δ2δ3
δ4
Figure 11. The sets δi, i = 1, 2, 3, 4
Proof. If i = 1, or i = 3, Hx is nondecreasing on [r0, r1]; and the result followsfrom (2.7). Now, assume for example i = 2, the case i = 4 being symmetric. Wesplit ∆2 in four subdomains δ1, δ2, δ3, and δ4, defined by (see figure 11)
δ1 = {(u, v) ∈ ∆2; u ≤ v},δ2 = {(u, v) ∈ ∆2; 0 ≤ v ≤ u},δ3 = {(u, v) ∈ ∆2; 0 ≤ −v ≤ u},δ4 = {(u, v) ∈ ∆2; u ≤ −v}.
Considering the velocity field on the axes {u = v}, {v = 0} and {u = −v}, forr > 1
ω , we obtain easily that (ux, vx) can only move from δ1 to δ2, from δ2 toδ3, and from δ3 to δ4. Therefore, we may assume, without loss of generality, that(ux, vx)(r0) ∈ δ1, and (ux, vx)(r1) ∈ δ4. We denote by τi, for i = 1, 2, 3, the timewhen (ux, vx) moves from δi to δi+1. On [τ3, r1], we have vx < 0, v′x < 0, and
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 29
ux ≤ −vx. Thus,
(|ux|+ |vx|)(r) ≤ 2A on [τ3, r1]. (4.8)
Next, in δ3 we set
g(r) = ux(r)− vx(r) = |ux(r)|+ |vx(r)| for r ∈ [τ2, τ3].
Considering (1.4)-(1.5), we obtain
g′ = −2
rg − 2
rvx − ω(vx + ux) +mg − gF (v2
x − u2x) on [τ2, τ3];
Observe that in δ4 we have − 2r g ≥ −2ωg, − 2
rvx ≥ 0, −ω(vx + ux) ≥ −ωg and
−gF (v2x − u2
x) ≥ 0. Therefore
g′ ≥ −2ωg + (m− ω)g ≥ −2ωg. (4.9)
Integrating (4.9) between r ∈ [τ2, τ3] and τ3 yields
g(r) ≤ g(τ3)e2ωΞ for every r ∈ [τ2, τ3]. (4.10)
Putting together (4.8) and (4.10) , we obtain
(|ux|+ |vx|)(r) ≤ 2Ae2ωΞ on [τ2, r1]. (4.11)
Then, in δ2, for r ∈ [τ1, τ2], we set
h(r) = u2x(r)− v2
x(r) ≥ 0.
Applying (2.6), we obtain
h′ ≥ 4ω(uxvx − u2x) ≥ −4ωh on [τ1, τ2]. (4.12)
Integrating (4.12) between r ∈ [τ1, τ2] and τ2, and applying (4.11), we obtain
h(r) ≤ h(τ2)e4ωΞ ≤ (2Ae4ωΞ)2 for every r ∈ [τ1, τ2]. (4.13)
Considering (1.4) and (4.13), we obtain
u′x ≥ −2
rux −Mvx ≥ −(M + 2ω)ux for every r ∈ [τ1, τ2], (4.14)
where M = sup{−F (−x); 0 ≤ x ≤ (2Ae4ωΞ)2}. Since ux ≥ vx on [τ1, τ2], integrat-ing (4.14) between r ∈ [τ1, τ2] and τ2, and applying (4.11), gives
(|ux|+ |vx|)(r) ≤ 4Ae2ωΞe(M+ω)Ξ on [τ1, r1]. (4.15)
Finally, in δ1 observe that 0 ≤ F (v2x − u2
x) ≤ m+ ω; and so by (1.5)
|v′x| ≤ (m+ ω)vx on [r0, τ1].
It follows that for r ∈ [r0, τ1] we have
vx(r) ≤ vx(τ1)e(m+ω)Ξ. (4.16)
Since in δ1 we have 0 < ux < vx, it follows from (4.15) and (4.16) that
(|ux|+ |vx|)(r) ≤ 4Ae2ωΞe(M+ω)Ξe(m+ω)Ξ. (4.17)
If we define B as the right hand side of (4.17), the result follows. �
30 THIERRY CAZENAVE
5. Proof of Theorem 1.2
We begin with two preliminary observations.
Lemma 5.1. Let n ≥ 0. Assume An 6= ∅, and let x belong to the closure of An.Then x ∈ ∪
0≤j≤nAj.
Proof. By Corollary 4.7, we have I ∩ A = ∅. Thus, since I is open (Lemma 4.2),x 6∈ I. By (2.11) and Proposition 4.1, (ux, vx) is bounded. This proves Rx = ∞.If ux and vx have more than n zeroes, then by (2.11) it is the same for x′ close tox. This is impossible, since x belongs to the closure of An. Therefore, the resultfollows from Corollary 4.10. �
Lemma 5.2. Let n ≥ 0. Assume In 6= ∅, and let x > a belong to the boundary ofIn. Then x ∈ ∪
0≤j≤nAj.
Proof. By Proposition 4.1, In 6= (a,∞), so its boundary is nonempty. By Corol-lary 4.7, and since In is open, x 6∈ In. Assume x ∈ Ij for some j 6= n. Thenby Lemma 4.2), we would have In ∩ Ij 6= ∅. This is ruled out by Corollary 4.7.Therefore, x 6∈ I. Now, we apply Corollary 4.10. Property (i) is ruled out byProposition 4.1, while property (ii) is ruled out by (2.11). Thus, x belong to someAj . By continuous dependence, again, we must have j ≤ n. �
In order to show that In is nonempty, we need the following lemmas.
Lemma 5.3. For every C > 0, there exists T > 0 with the following property. Letx 6= 0 be such that Rx ≥ T . Assume that for some ρ ≥ T , we have vx(ρ) = 0 and
|ux(ρ)| ≤ Ce−m−ω2 ρ. Then, there exists θ ∈ (ρ,Rx) such that |ux| > 0 and |vx| ≤ a
on (ρ, θ), |vx(θ)| = a and Hx(θ) < 0.
Proof. (See figure 12) Let x and ρ be such that ρ ≤ Rx, vx(ρ) = 0, and
|ux(ρ)| ≤ Ce−m−ω
2 ρ. (5.1)
Assume for example ux(ρ) > 0. Then, vx will become negative before ux vanishes(see Figure 3). Let µ ∈ (0, a) be such that F (µ2) = m−ω
2 . If ρ is large enough, wehave ux(ρ)) ≤ µ
2 . We define the number R > 0, by
R = sup{r ∈ (ρ,Rx), ux + |vx| ≤ µ on [ρ, r]}. (5.2)
By Corollary 2.13, we have R < Rx. Thus,
(ux + |vx|)(R) = µ. (5.3)
Furthermore, still by Corollary 2.13 we have ux > 0 and vx < 0 on (ρ,R]; and so
(ux − vx)′ = −(ux − vx)F (v2x − u2
x) + (m+ ω)(ux − vx)− 2
rux + 2ωvx
≤ (ux − vx)(m+ ω − F (v2x − u2
x)).
Note further that on [ρ,R] we have v2x − u2
x ≥ −µ2. Thus, if we set
c = −min{F (x),−µ2 ≤ x ≤ 0},
we obtain
(ux − vx)′ ≤ (m+ ω + c)(ux − vx), on [ρ,R]. (5.4)
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 31
-v
6u
@@@
@@@
@@
@@@
@@@
@@@
@@@
@@
@@@
@@@
@@@
@@@@
−a
H = 0
Y
(ux(ρ), vx(ρ))
(ux(θ), vx(θ))
Figure 12. Notation for Lemma 5.3
Integrating (5.4) between ρ and R, we obtain
(ux + |vx|)(R) ≤ ux(ρ)e(m+ω+c)(R−ρ). (5.5)
Putting together (5.1), (5.3), and (5.5), we obtain
(m+ ω + c)(R− ρ) ≥ log( µC
)+m− ω
2ρ.
Hence, for some δ > 0 and ρ large enough
R ≥ (1 + δ)ρ. (5.6)
We claim that, if ρ is large enough, then Hx(R) < 0. To see this, we first makethe following observation. Consider x, y such that x2 ≤ µ2 and y2 ≤ m−ω
2(m+ω)x2. It
follows from (2.3) that
2H(y, x) = G(x2 − y2)− (m− ω)x2 + (m+ ω)y2 ≤ G(x2 − y2)− m− ω2
x2.
32 THIERRY CAZENAVE
Furthermore x2 − y2 ≥ 0; and so
G(x2 − y2) ≤ G(x2) ≤ x2F (x2) ≤ m− ω2
x2.
Therefore, H(y, x) ≤ 0. It follows that
u2 ≥ m− ω2(m+ ω)
v2, for all u, v such that H(u, v) ≥ 0 and |u|+ |v| ≤ µ. (5.7)
Let now
τ = sup{r ∈ [ρ,R]; Hx ≥ 0 on [ρ, r]}. (5.8)
From (5.7) and (5.8), we get
Λux(r) ≥ −vx(r), for r ∈ [ρ, τ ], (5.9)
where Λ > 0 is given by Λ2 = 2(m+ω)m−ω . Next, on [ρ,R], we have
F (v2x − u2
x)− (m+ ω) ≤ −m+ ω
2; F (v2
x − u2x)− (m− ω) ≤ −m− ω
2. (5.10)
Therefore, we have, by (1.4) and (1.5),
(ux − vx)′ ≥ −2
rux +
m− ω2
(ux − vx). (5.11)
Since −ux ≥ −(ux − vx), and if ρ > 8m−ω , we get from (5.11)
(ux − vx)(r) ≥ ux(ρ)e(m−ω)(r−ρ)
4 for r ∈ [ρ,R]. (5.12)
Putting together (5.9) and (5.12), we obtain
u2x(r) ≥ k2u2
x(ρ)e(m−ω)(r−ρ)
4 for r ∈ [ρ, τ ], (5.13)
where k = 1/(1 + Λ). Now, from (5.8), (5.10) and (2.5), we have
0 ≤ Hx(τ) ≤ Hx(ρ)− (m+ ω)
∫ τ
ρ
1
rux(r)2dr. (5.14)
Next, observe that if ρ is large enough, we have by (2.3) and (5.1)
Hx(ρ) = H(ux(ρ), 0) ≤ 1
2G(−u2
x(ρ)) +m+ ω
2u2x(ρ) ≤ (m+ ω)u2
x(ρ). (5.15)
Therefore, putting together (5.13), (5.14) and (5.15), we obtain
0 ≤ 1− k2
∫ τ
ρ
1
re
(m−ω)(r−ρ)2 dr. (5.16)
We deduce easily from (5.16) that
k2
τ[e
(m−ω)(τ−ρ)2 − 1] ≤ m− ω.
Therefore, for any α > 0, we get for ρ large enough
τ ≤ (1 + α)ρ. (5.17)
Putting together (5.6) and (5.17), we obtain τ < R. Since by (2.14) we haveH ′x < 0, on [τ,R], we get
Hx(R) < 0, if ρ is large enough. (5.18)
To conclude, observe that by Corollary 2.13, vx must decrease to some value lessthan −a, before ux vanishes. Note also that when |vx| ≤ a, we have H ′x < 0; and
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 33
-v
6u
−a
H = 0
�
(ux(ρ), vx(ρ))
(ux(τ), vx(τ))
Figure 13. Notation for Lemma 5.4
so by (2.14), at the time when vx = −a, we have Hx < 0. Thus, the proof ofLemma 5.3 is complete. �
Lemma 5.4. There exists R with the following property. Assume x 6= 0 is suchthat Rx ≥ R. Assume further that for some ρ ≥ R, we have vx(ρ) = −a, ux(ρ) > 0(respectively vx(ρ) = a, ux(ρ) < 0), and Hx(ρ) < 0. Then , there exists τ > ρ suchthat |vx| > 0 on (ρ, τ) and such that ux has two zeroes on (ρ, τ).
Proof. (See figure 13) Let x be as above, and assume, for example, ux(ρ) > 0. ByProposition 2.12, and Corollary 2.16, there exists ρ1 ∈ (ρ,Rx), such that vx(ρ1) =−a, vx < −a on (ρ, ρ1), and ux has exactly one zero on (ρ, ρ1). Assume thatHx(ρ1) < 0 and let
Σ = {H(u, v) < Hx(ρ1), 0 > v > −a, u < 0}.Observe that H ′x(ρ1) < 0, and v′x(ρ1) > 0. Therefore, (ux, vx) will belong to Σ, forr ∈ (ρ1, ρ1 + ε], where ε is some postive number. Since Hx(ρ1) < 0, Σ is bounded
34 THIERRY CAZENAVE
away from the u-axis by Lemma 2.2 (ii). Furthermore, when (ux, vx) ∈ Σ, we haveH ′x < 0 and v′x > 0, by (1.5) and (2.14). Thus, (ux, vx) cannot exit Σ by crossingthe line {v = −a} or the curve {H(u, v) = Hx(ρ1)}. Finally, since Σ is boundedaway from the u-axis, it follows from Corollary 2.16, that (ux, vx) exits Σ in a finitetime, by crossing the v-axis. This is the desired result.
Therefore, it remains to prove that Hx(ρ1) < 0. Note that for a fixed value ofHx(ρ), and for ρ large enough, this is a consequence of the fact that the trajectoryremains close to a trajectory of the Hamiltonian motion (Lemma 2.5). Actually,we need an estimate on ρ that does not depend on the value of Hx(ρ). To provethis, we consider two cases.
If H(0, b) ≥ 0 (b given by (2.1)), then the sets {F (v2 − u2) ≥ m − ω} and{H(u, v) < 0} are disconnected by Lemma 2.1 (iv); and so, by (2.14), the set{H(u, v) < 0} is a trapping region for the solutions of (1.4)-(1.5). Thus, we haveHx(ρ1) < 0.
Suppose now H(0, b) < 0. By Lemma 2.19, the set {H(u, v) ≤ H(0, b)} is atrapping region. Thus, if for some r ∈ [ρ, ρ1], we have Hx(r) ≤ H(0, b), we getimmediately Hx(ρ1) < 0. Therefore, it only remains to consider the case where xis such that Hx(ρ) < 0, and
H(0, b) < Hx(r) for every r ∈ [ρ, ρ1]. (5.19)
Let Λ be the set of such x’s. Consider the set
J = {z > 0, H(0, b) < H(z,−a) < 0}.
Clearly, we have
ux(ρ) ∈ J for every x ∈ Λ. (5.20)
Note that, since H(0, b) < 0, there exists δ ∈ (0, a) such that H(0, δ) = H(0, b),since H(0, 0) = 0 and H(0, a) < H(0, b). It is not too difficult to show that forz ∈ J , there exists Tz > 0, such that the solution (uz, vz) of (2.8)-(2.9) withuz(ρ) = z, vz(ρ) = −a, will satsify
vz(r) < −δ for t ∈ (ρ, ρ+ Tz), uz(ρ+ Tz) < 0, and vz(ρ+ Tz) = −δ. (5.21)
The set {0 ≥ H(u, v) ≥ H(0, b), v < −δ} is bounded away from the critical pointsof H, and so, there exists M > 0 such that
Tz ≤M for every z ∈ J. (5.22)
Note also that, since J is bounded,
(uz(r), vz(r)) is uniformly bounded, with respect tor ∈ R and z ∈ J. (5.23)
Next, by Lemma 2.5 we know that for every x ∈ Λ, there exists rx > 0, such thatif ρ ≥ rx, then
|(ux, vx)− (uz, vz)| ≤a− δ
2on (ρ, ρ+M); (5.24)
where z = ux(ρ). Note that, since J is bounded, rx can be chosen to be bounded,uniformly in x ∈ Λ. Thus, if R is large enough, we have (5.24), for every x ∈ Λ.Therefore, from (5.21), (5.22) and (5.24), we get
ρ1 − ρ ≤M. (5.25)
On the other hand, by (5.23), (5.24), (5.25) and (1.4)-(1.5) we have
(ux, vx) is bounded in C1([ρ, ρ1],R2),uniformly in x ∈ Λ. (5.26)
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 35
Let now σ be the area of the set {H(u, v) ≤ H(0, b), v ≤ −a}. It follows from (5.25)and (5.26), that if R is large enough, we have∣∣∣∫ ρ1
ρ
(1− ρ
r
)ux(r)v′x(r) dr
∣∣∣ ≤ σ
2for every x ∈ Λ. (5.27)
Next, observe that the curve ∪r∈[ρ,ρ1]{ux(r), vx(r)} has no multiple points, since on
(ρ, ρ1) we have (v2x − u2
x)′ > 0 while ux > 0, and (v2x − u2
x)′ < 0 while ux < 0. Thiscomes from (2.15), assuming R > 1/ω. Let Σx be the region contained betweenthe curve and the line {v = −a}, and let δx be its area. By (5.19), we haveΣx 3 {H(u, v) ≤ H(0, (m+ ω)1/2, v ≤ −a}; and so
δx ≥ σ. (5.28)
Now, applying Green’s formula, we have
δx = −∫ ρ1
ρ
ux(r)v′x(r) dr. (5.29)
Putting together (5.27), (5.28) and (5.29), we get∫ ρ1
ρ
1
rux(r)v′x(r) dr ≤ − σ
2ρ. (5.30)
Integrating (2.14), between ρ and ρ1, and using (1.5) and (5.30), we obtain theinequality
Hx(ρ1)−Hx(ρ) ≤ −σ/ρ < 0;
from which the result follows. �
Corollary 5.5. Let n ≥ 0 be such that An 6= ∅, and let xn = supAn. Assume thatxn ∈ An and that xn ≥ sup In. Then , there exists ε > 0 such that (xn, xn + ε) ⊂In+1.
Proof. Let us set (U, V ) = (uxn , vxn) and let R be the last zero of U . Assume forexample that U > 0 on (R,∞). By Lemma 4.6, there exists τ > R such that
0 < U(r) < V (r) ≤ a
4for r ≥ τ. (5.31)
Next, observe that for x close to xn, we have the following. Both ux and vx haven zeroes on (0, τ), which are alternate, and
0 < ux(τ) < vx(τ) ≤ a
2. (5.32)
This follows from (2.10) and Lemma 4.6. Now, assume in addition that x > xn.Let
ρx = sup{r ∈ (τ,Rx), vx > 0 and ux > 0 on (τ, r)}.Observe that, from (5.32) and (1.4)-(1.5) it follows easily that u′x < 0 and v′x < 0on (τ, ρx), and so
0 < ux ≤a
2and 0 < vx ≤
a
2on (τ, ρx). (5.33)
Next, x 6∈ An by definition of xn; and so, by Lemma 2.14, ρx must be finite.Therefore, ux(ρx) = 0 or vx(ρx) = 0. Furthermore, x 6∈ In, and so ux(ρx) 6= 0.Thus vx(ρx) = 0. Note that, from (5.31) and (2.10), we have
ρx −→x→xn
+∞. (5.34)
36 THIERRY CAZENAVE
Finally, from (1.4)-(1.5) and (5.33), we get
(ux + vx)′ ≤ −1
2(m− ω)(ux + vx) on (τ, ρx). (5.35)
Integrating (5.35), we obtain
ux(ρx) ≤ a
2em−ω
2 τe−m−ω
2 ρx . (5.36)
Putting together (5.34), (5.36), and applying Lemmas 5.3 and 5.4, we obtain thatif x is close enough to xn, then uxvx has at least two zeroes after ρx, and thefirst two zeroes are zeroes of ux; and so x ∈ In+1. This completes the proof ofCorollary 5.5. �
End of the proof of Theorem 1.2. Let us first show that I0 6= ∅. To see this, con-sider x > a such that F (x2) < m + ω and H(0, x) < 0 (such an x exists bydefinition of a and by Lemma 2.1 (ii)). By Lemma 2.19 we have Rx = ∞ andHx(r) ≤ H(0, x) for r≥ 0. By Lemma 2.2 (ii) we have then vx(r) ≥ α for r ≥ 0,for some α > 0. Applying Corollary 2.16 with r0 = 0, it follows that we are incase (ii); and so x ∈ I0. Consider now y0 = sup I0. y0 is finite by Proposition 4.1.By Lemma 5.2, y0 ∈ A0. Let now x0 = supA0. x0 is also finite by Proposition 4.1.Applying Lemma 5.1, we get x0 ∈ A0. Therefore, by Corollary 5.5, there existsε0 > 0, such that (x0, x0 + ε0) ⊂ I1. Thus, I1 6= ∅. Let y1 = sup I1. y1 is finiteby Proposition 4.1. We have y1 > x0 ≥ y0; and so, by Lemma 5.2, y1 ∈ A1; thenby Lemma 5.1, x1 := supA1 ∈ A1 (x1 is also finite by Proposition 4.1). Iterat-ing this argument, we construct an increasing sequence (xn)n ≥ 0, with xn ∈ An.The exponential decay follows from Lemma 4.6. This proves Theorem 1.2 (i), (ii)and (iii). Finally, (iv) follows from Proposition 3.2 and Remark 3.1. �
6. Further results
Our goal in this section is to show how to modify the proof of Theorem 1.2 inorder to weaken the assumptions on F , and to establish the following result.
Theorem 6.1. Assume that (H1′) and (H2) hold. There exists an increasingsequence (xn)n≥0 of positive numbers with the following properties . For everyn ≥ 0,
(i) the solution (un, vn) of (1.4)-(1.6) with vn(0) = xn is a global solution ,(ii) both un and vn have exactly n zeroes on (0,+∞),(iii) (un, vn) converges exponentially to (0, 0), as r → +∞.(iv) Furthermore, if F (x) ≥ δ(log x)β for x large, where δ > 0 and β > 2, the
sequence (xn)n≥0 is bounded.
Remark 6.2. It follows from Theorem 6.1 that the conclusions of Theorem 1.1hold if we replace assumption (H1) by assumption (H1′).
The differences between (H1) and (H1′) are the following. Instead of assumingthat F (x) > m+ω for x large, we assume only that F (x) > m−ω for x large, andwe remove the assumption F ′(F−1(m−ω)) > 0. However, note that F is increasingon (0,∞), so that F ′(F−1(m − ω)) ≥ 0. Therefore, with respect to Theorem 1.2,the only new situations are the cases where
F (x) < m+ ω for all x ∈ R, (6.1)
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 37
and/or
F ′(F−1(m− ω)) = 0. (6.2)
We shall explain how to modify the proof of Theorem 1.2 when either (6.1) or (6.2)hold. The case where both (6.1) and (6.2) hold requires both series of modifications.
Modifications under assumption (6.1). The proof of Theorem 1.2 applies as itis to this case. However, it can be simplified a lot, due to the following reason. Theset {F (v2 − u2) ≥ m + ω} is empty. In particular, it follows from (2.14) that Hx
is always nonincreasing; and so for every x ∈ R we have Rx =∞ and the solution(ux, vx) is uniformly bounded. Let us now indicate the main simplifications thatarise in the proof (the rest of the proof can be leaved unchanged). In Section 2, thenumber b defined in (2.1) does not exist anymore. Furthermore, Lemma 2.1 (iv),Lemma 2.10, Proposition 2.11, Lemma 2.15, Corollary 2.16 (v) and Lemma 2.19can be removed. Section 3 can be removed entierely, since F never satisfies (3.1).In Section 4, Lemma 4.3, one always has θ = 0. The proof of Lemma 4.8 is trivialsince Hx is always nonincreasing, and property (i) of Corollary 4.10 can be removed.Finally, step 2 of the proof of Proposition 4.1 is simplified since ∆1 = ∆3 = ∅. InSection 5, the proof of Lemma 5.4 is trivial since Hx is nonincreasing (we areformally in the case where H(0, b) ≥ 0).
Modifications under assumption (6.2). In this case, the proof does not applyas it is and must be adapted. Indeed, the assumption F ′(F − 1(m − ω)) > 0 wasused in Lemma 2.8. Therefore, the conclusion of Lemma 2.8 may not hold anymoreand we must admit the possibility of solutions converging to (0,±a) in one of thehalf-planes {u > 0} or {u < 0}. Lemma 2.8 was applied several times in the sequel;and so this induces the following modifications. In Section 2, Lemma 2.8 does nothold anymore and must be removed. Therefore, Proposition 2.12 must be replacedby the following result, whose proof is essentially the same.
Proposition 2.12′. (See Figure 14) Let x 6= 0. Assume that for some r0 > 0, wehave vx(r0) ≤ ux(r0) and ux(r0) > 0 (respectively , ux(r0) ≤ vx(r0) and ux(r0) <0). Then one of the following properties hold.
(i) There exists r0 < r1 < Rx such that |ux| > 0 on (r0, r1), and ux(r1) = 0,|vx(r1)| > a. In addition, either vx(r0) > 0 (respectively , vx(r0) < 0),and then vx has exactly one zero in (r0, r1), or else vx(r0) ≤ 0 (respectivelyvx(r0) ≥ 0), and then |vx| > 0 on (r0, r1];
(ii) Rx = ∞, |ux| > 0 on (r0,∞), ux → 0 and vx ↓ −a as r → ∞ (respectivelyvx ↑ a as r →∞).
This implies modifications of Corollary 2.13 and Lemma 2.14 as follows.
Corollary 2.13′. Let x 6= 0. Assume that, for some 0 < r0 < Rx, we havevx(r0) = 0. Then one of the following properties hold.
(i) There exists r1 ∈ (r0, Rx) such that |ux| > 0, |vx| > 0 on (r0, r1), and ux(r1) =0, |vx(r1)| > a.
(ii) Rx =∞, |ux| > 0 on (r0,∞), ux → 0 and |vx| ↑ a as r →∞.
Lemma 2.14′. Let x 6= 0 be such that Rx =∞. Assume that for some r0 > 0, wehave |ux| > 0 on [r0,∞). Then , one of the following properties hold.
(i) uxvx > 0 on (r0,∞), and there exists C such that 0 < |ux(r)| < |vx(r)| <Ce−(1/2)(m−ω)r for r ∈ (r0,∞).
38 THIERRY CAZENAVE
-
v
6u
%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%
eeeeeeeeeeeeeeeeee
eeeeeeeeeeeeeeeeee
?
�
Figure 14. Illustration for Proposition 2.12′
(ii) ux → 0 and vx → ±a as r →∞.
The necessary changes in the proofs are obvious. Corollary 2.16 also has to bemodified as follows (with little change in the proof).
Corollary 16′. (see Figures 8 and 15) Let x 6= 0, x 6= ±a. Assume that for somer0 ∈ [0, Rx) we have ux(r0) = 0. Then, one of the following properties holds .
(i) |vx(r0)| < a, and there exists r1 ∈ (r0, Rx) such that uxvx < 0 on (r0, r1),0 < |ux| < |vx| on (r0, r1), |vx(r1)| > a, and ux(r1) = 0;
(ii) |vx(r0)| > a, and there exists r1 ∈ (r0, Rx) such that uxvx > 0 on (r0, r1),0 < |ux| < |vx| on (r0, r1), |vx(r1)| < a, and ux(r1) = 0;
(iii) |vx(r0)| > a, Rx = ∞, uxvx > 0 on (r0,∞), and 0 < |ux| < |vx| <C exp (−(1/2)(m− ω)r) on (r0,∞);
(iv) |vx(r0)| > a, and there exists r1 ∈ (r0, Rx) such that uxvx > 0 on (r0, r1), andvx(r1) = 0;
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 39
-v
6u
�������������������������������������
u = v
�
a (vi)
Figure 15. Illustration for Corollary 16′ (vi)
(v) |vx(r0)| > a, Rx ≤ 1/ω, uxvx > 0 on (r0, Rx), and F (v2x − u2
x) > m + ω on(r0, Rx).
(vi) |vx(r0)| > a, Rx =∞, uxvx > 0 on (r0,∞), ux → 0 and vx → a as r →∞.
Finally, the following result is easily obtained.
Lemma 2.18′. Let x 6= 0. Assume that Rx ≥ 1/ω, and that ux has a finite numberof zeroes . Then Rx =∞, and either |ux|+ |vx| → 0 as r →∞, or else ux → 0 andvx → ±a as r →∞.
Section 3 in unchanged, since everything takes place in the set {F (v2 − u2) ≥m+ ω}, which does not contain the points (0,±a).
Major changes occur in Section 4, concerning the definition of the sets In. Onehas to define
I0 = J0 ∪U0,
40 THIERRY CAZENAVE
where
J0 = {x > a;∃rx,0 ∈ (0, Rx) s.t. ux > 0 and vx > 0 on (0, rx,0), and ux(rx,0) = 0},and
U0 = {x > a;Rx =∞, ux > 0 and vx > 0 on (0, Rx),
and ux → 0, vx → a as r →∞}.For x ∈ U0, we set rx,0 =∞. For n ≥ 1, one must define
In = Jn ∪Un ∪On,where (see Figure 16)
Jn = {x > a;∃rx,n ∈ (0, Rx) s.t. ux and vx
have exactly n zeroes on (0, rx,n), and ux(rx,n) = 0},Un = {x > a;Rx =∞, both ux and vx have exactly
n zeroes on (0, Rx), and ux → 0, ux → ±a as r →∞},On = {x > a;Rx =∞, vx has exactly n zeroes on (0, Rx), ux has exactly
n− 1 zeroes on (0, Rx), and ux → 0, vx → ±a as r →∞}.
With these new definitions of the sets In , the rest of Section 4 is essentiallyunchanged. Only the proof of Lemma 4.2) is a little bit more delicate, but is stilleasy due to the fact that the points (0,±a) are dynamically stable for system (1.4)-(1.5) (indeed, Hx is nonincreasing in a neighborhood of (0,±a)).
In Section 5, Lemmas 5.3 and 5.4 have to be modified in the following way, withan obvious adaptation of the proof.
Lemma 5.3′. For every C > 0, there exists T > 0 with the following property .Let x 6= 0 be such that Rx ≥ T . Assume that for some ρ ≥ T , we have vx(ρ) = 0and |ux(ρ)| ≤ C exp(−(1/2)(m− ω)ρ). Then,
(i) either there exists θ ∈ (r,Rx) such that |ux| > 0 and |vx| ≤ a on (ρ, θ),|vx(θ)| = a, and Hx(θ) < 0 (see Figure 12);
(ii) or else Rx = ∞, |ux| > 0 and |vx] ≤ a on (ρ,∞), ux → 0 and |vx| ↑ a, asr →∞ (see Figure 17).
Lemma 5.4′. There exists R with the following property. Assume x 6= 0 is suchthat Rx ≥ R. Assume further that for some ρ ≥ R, we have vx(ρ) = −a, ux(ρ) > 0(respectively vx(ρ) = a, ux(ρ) < 0), and Hx(ρ) < 0. Then,
(i) either there exists τ > ρ such that |vx| > 0 on (ρ, τ) and such that ux has twozeroes on (ρ, τ) (see Figure 13).
(ii) or else Rx = ∞, |vx| > 0 on (ρ,∞), ux has one zero on (ρ,∞) and ux → 0,|vx| → a as r →∞(see Figure 18).
The rest of the proof is essentially unchanged.
References
[1] Alvarez A. and Soler M. Energetic stability criterion for a nonlinear spinorial model, Phys.Rev. Lett. 50 (1983), no. 17 1230–1233. (link: http://prl.aps.org/pdf/PRL/v50/i17/p1230_
1)
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 41
-
v
6u
%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%
eeeeeeeeeeeeeeeeee
eeeeeeeeeeeeeeeeee
� � z1 y1 x1
Figure 16. Trajectories with x1 ∈ J1, y1 ∈ U1 and z1 ∈ O1
[2] Balabane M., Cazenave T., Douady A. and Merle F. Existence of excited states for anonlinear Dirac field, Comm. Math. Phys. 119 (1988), 153–176. (968485) (link: http:
//projecteuclid.org/euclid.cmp/1104162275)
[3] Berestycki H. and Lions P.-L. and Peletier L.A. An O.D.E. approach to the existence ofpositive solutions for semilinear problems in Rn, Indiana Univ. Math. J. 30 (1981), no. 1, 141–157. (MR0600039) (link: http://www.iumj.indiana.edu/IUMJ/FTDLOAD/1981/30/30012/pdf)
[4] Bogolubsky I.L. On spinor soliton stability, Phys. Lett. A 73 (1979), no. 2, 87–90.(MR0592382) (doi: 10.1016/0375-9601(79)90442-0)
[5] Cazenave T. and Vazquez L. Existence of localized solutions for a nonlinear classical Dirac
field, Comm. Math. Phys. 105 (1986), 35–47. (847126) (link: http://projecteuclid.org/
euclid.cmp/1104115255)[6] Dias J.-P. and Figueira M. Global existence of solutions with small initial data in Hs for the
massive nonlinear Dirac equations in three space dimensions, Boll. Un. Mat. Ital. B (7) 1(1987), no. 3, 861–874. (MR0916298)
[7] Dias J.-P. and Figueira M. Remarque sur le probleme de Cauchy pour une equation de Diracnon lineaire avec masse nulle, Portugal. Math. 45 (1988), no. 4, 327–335. (MR0982901) (link:http://purl.pt/3144/1/)
42 THIERRY CAZENAVE
-v
6u
@@@
@@@
@@
@@@
@@@
@@@
@@@
@@
@@@
@@@
@@@
@@@@
−a
H = 0
�
(ux(ρ), vx(ρ))
Figure 17. Illustration for Lemma 5.3′ (ii)
[8] Dias J.-P. and Figueira M. Sur l’existence d’une solution globale pour une equation de Dirac
non lineaire avec masse nulle. C. R. Acad. Sci. Paris Ser. I Math. 305 (1987), no. 11, 469–472.(MR0916312)
[9] Dias J.-P. and Figueira M. On the existence of weak solutions for a nonlinear time depen-
dent Dirac equation, Proc. Royal Soc. Edinburgh Sect. A 113 (1989), no. 1-2, 149–158.(MR1025460)
[10] Dias J.-P. and Figueira M. Solutions faibles du probleme de Cauchy pour certaines equations
de Dirac non lineaires, Portugal. Math. 46 (1989), suppl., 475–484. (MR1080767) (link: http:
//purl.pt/3771)
[11] Finkelstein R., Fronsdal C.F. and Kaus P. Nonlinear spinor fields, Phys. Rev. 103 (1956),
no. 5, 1571–1579. (doi: 10.1103/PhysRev.103.1571)[12] Garcia L. and Ranada A.F. A classical model of the nucleon, Progr. Theoret. Phys. 64 (1980),
no. 2, 671–693. (MR0588482) (doi: 10.1143/PTP.64.671)
[13] Heisenberg W. Doubts and hopes in quantumelectrodynamics, Physica 19 (1953), 897–908.(MR0058472) (doi: 10.1016/S0031-8914(53)80100-X)
[14] Hubbard J. H. and West B. H. MacMath, c© Cornell University (1985).[15] Iwanenko D. Bemerkungen zur theorie der wechselwirkung, Physikalische Zeitschrift der Sow-
jetunion 13 (1938), 141-150. (link: http://www.g-sardanashvily.ru/Nonlinear1.pdf)
STATIONARY STATES FOR CLASSICAL NONLINEAR DIRAC FIELDS 43
-v
6u
−a
H = 0
�
(ux(ρ), vx(ρ))
Figure 18. Illustration for Lemma 5.4′ (ii)
[16] Jones C. and Kupper T. On the infinitely many solutions of a semilinear elliptic equation,
SIAM J. Math. Anal. 17 (1986), no. 4, 803–836. (MR0846391) (doi: 10.1137/0517059)[17] Mathieu P. and Morris T.F. Instability of stationary states for nonlinear spinor models
with quartic self-interaction, Phys. Lett. B 126 (1983), no. 1-2, 74–76. (doi: 10.1016/
0370-2693(83)90019-9)
[18] Merle F. Existence of stationary states for nonlinear Dirac equations, J. Differential Equations74 (1988), no. 1, 50–68. (MR0949625) (doi: 10.1016/0022-0396(88)90018-6)
[19] Rafelski J. Soliton solutions of a selfinteracting Dirac field in three space dimensions, Phys.Lett. B 66 (1977), no. 3, 262–266. (doi: 10.1016/0370-2693(77)90876-0)
[20] Ranada A.F. Classical nonlinear Dirac field models of extended particles, in Quantum theory,
groups, fields and particles, A.O. Barut (ed.), Reidel, Amsterdam, 1982.
[21] Ranada A.F. and Ranada M.F. Nonlinear model of c-number confined Dirac quarks, Phys.Rev. D 29 (1984), no. 5, 985–993. (doi: 10.1103/PhysRevD.29.985)
[22] Ranada A.F., Ranada M.F., Soler M. and Vazquez L. Classical electrodynamics of a nonlinearDirac field with anomalous magnetic moment, Phys. Rev. D 10 (1976), no. 2, 517–525. (doi:
10.1103/PhysRevD.10.517)
44 THIERRY CAZENAVE
[23] Ranada A.F. and Soler M. Perturbation theory for an exactly soluble spinor model in in-
teraction with its electromagnetic field, Phys. Rev. D 8 (1973), no. 10, 3430–3433. (doi:
10.1103/PhysRevD.8.3430)[24] Ranada A.F., Uson J. and Vazquez L. Born-Infeld effects in the electromagnetic mass of an
extended Dirac particle, Phys. Rev. D 22 (1980), no. 10, 2422–2424. (doi: 10.1103/PhysRevD.
22.2422)[25] Ranada A.F. and Vazquez L. Classical system of nonlinear Dirac and Klein-Gordon fields,
Progr. Theoret. Phys. 56 (1976), no. 1, 311–323. (doi: 10.1143/PTP.56.311)
[26] Soler M. Classical, stable, nonlinear spinor field with positive rest energy, Phys. Rev. D 1(1970), no. 10, 2766–2769. (doi: 10.1103/PhysRevD.1.2766)
[27] Soler M. Classical electrodynamics for a nonlinear spinor field: Perturbative and exact ap-
proaches, Phys. Rev. D 8 (1973), no. 10, 3424–3429. (doi: 10.1103/PhysRevD.8.3424)[28] Strauss W.A. and Vazquez L. Stability under dilations of nonlinear spinor fields. Phys. Rev.
D (3) 34 (1986), no. 2, 641–643. (MR0848095) (doi: 10.1103/PhysRevD.34.641)[29] Takahashi K. Soliton solutions of nonlinear Dirac equations, J. Math. Phys. 20 (1979), no.
6, 1232–1238. (doi: 10.1063/1.524176)
[30] Van Der Merwe, P. du T. Space-time symmetries and nonlinear field theory, Nuovo CimentoA (11) 60 (1980), 247–264. (doi: 10.1007/BF02902461)
[31] Vazquez L. Localized solutions of a nonlinear scalar field with a scalar potential, J. Mathe-
matical Phys. 18 (1977), no. 7, 1341–1342. (MR0449276) (doi: 10.1063/1.523426)[32] Vazquez L. Personal communication.
[33] Wakano M. Intensely localized solutions of the classical Dirac-Maxwell field equations, Progr.
Theoret. Phys. 35 (1966), no. 6, 1117–1141. (doi: 10.1143/PTP.35.1117)[34] Werle J. Stability of particle-like solutions of nonlinear Klein-Gordon and Dirac equations,
Acta Phys. Polon. B 12 (1981), no. 6, 601–616. (link: http://th-www.if.uj.edu.pl/acta/
vol12/pdf/v12p0601.pdf)[35] Weyl H. A remark on the coupling of gravitation and electron, Physical Rev. (2) 77 (1950),
no. 5, 699–701. (MR0033250) (doi: 10.1103/PhysRev.77.699)
