Claudio Bombardelli

Jesus Peláez

On the Stability of Artificial Equilibrium Points in the Circular Restricted Three Body Problem

Keywords: Three Body Problem, Artficial Equilibrium Points, Stability, Solar Sails

The article analyses the stability properties of minimum-control artificial equilibrium points in the planar circular restricted three body problem. It is seen that when the masses of the two primaries are of different orders of magnitude, minimumcontrol equilibrium is obtained when the spacecraft is almost coorbiting with the second primary as long as their mutual distance is not too small. In addition, stability is found when the distance from the second primary exceeds a mnimum value which is a simple function of the mass ratio of the two primaries and their separation. Lyapunov stability under non-resonant conditions is demonstrated using Arnold s theorem. Among the most promising applications of the concept we find solar-sail-stabilized observatories coorbiting with the Earth, Mars, and Venus.

Introduction

The use of artificial equilibrium points (AEPs) have received considerable attention in the last decade as a means to stabilize a spacecraft in a region of space where equilibrium is not possible with gravitational forces alone but can be achieved by means of man-made propulsion devices providing continuous acceleration. Propellantless propulsion systems such as solar sails and electrodynamic tethers are ideal candidates for this concept as they open up the possibility of placing permanent or very long term outposts at different points of interest in space. These could serve the purpose of scientific observatories, communication stations or even habitable modules for astronauts in the far future.

The subject of artificial equilibrium points has been studied extensively in the literature. In 1990 Forward[Forward(1991)] first proposed the deployment of a stationary satellite hovering above the Earth poles with solar radiation pressure balancing gravitational forces. McInnes in 1994 [McInnes andMcDonald(1994)] derived continuous surfaces of unstable AEPs in the circular restricted three body problem (CRTBP) when a low-thrust control acceleration was added. Nevertheless, it was not until 2007 that Morimoto et al. [START_REF] Morimoto | Artificial equilibrium points in the low-thrust restricted three-body problem[END_REF]Morimoto, Yamakawa, and Uesugi] realized that linear (i.e. marginal) stability could be found in a region of space far enough from the second primary. Morimoto also noticed that linearly stable equilibrium positions belonging to the orbital plane of the two primaries required the smallest amount of control acceleration, which for the case of the Earth-Sun system could be smaller than 5 × 10 -5 m/s 2 . Although the issue of nonlinear stability was discussed by the previous author only a numerical simulation was provided to show a "qualitatively stable" behaviour in three dimensions. In the same year, and unaware of the previous study, Pelaez and Scheeres [START_REF] Pelaez | A permanent tethered observatory at jupiter. dynamical analysis[END_REF]] found a linearly stable region while studying the dynamics of an electrodynamic tether located in the proximity of the inner Jupiter moons. Using a Hill formulation of the circular three body problem they showed that a tethered system located in the orbital path of the moon at a distance greater than a given value enjoyed linear stability. The results obtained by Morimoto were confirmed by [START_REF] Baig | Artificial three-body equilibria for hybrid low-thrust propulsion[END_REF]] who proposed to control the AEPs with a hybrid solar sail and electric propulsion system.

In this article we analyze the stability of artificial equilibrium positions in the planar CRTBP focusing on the need to minimize the required control acceleration. First, we investigate the linear stability of the artificial equilibrium position with a series expansion of the system Hamiltonian which is later used to investigate non-linear (Lyapunov) stability. Next, we compute the locus of the points belonging to the plane of the two primaries for which equilibrium is obtained with minimum acceleration as a function of the distance between the spacecraft and the second primary. An approximate analytical solution is obtained with a straightforward perturbation technique in which the small parameter is the mass ratio µ of the two primaries. The solution is employed to compute the intersection between the minimum-control curve and the previously derived stability region. By exploiting the smallness of µ we derive simple and quite accurate analytical expressions to quantify the distance of the closest stable equilibrium point from the second primary and the magnitude of the required control acceleration. The location of resonance points in the stable region, for which stability cannot be granted on the base of a linear analysis, is also computed and simple analytical expressions are provided. Finally a survey of promising three-body "gravitational environments" in our solar system is performed highlighting the most favorable places to be considered for an application of the results obtained. Numerical simulations of the complete equations of motion are included in order to verify the validity of the proposed solution.

Equations of Motion

Under the framework of the circular restricted three body problem (CRTBP) we denote M 1 = (1 -µ)M and M 2 = µM to be the masses of the first and second primary, respectively. Here M is the total mass of the system and ℓ the distance between the primaries. The two primaries revolve around each other at angular velocity: where G is the gravitational constant. We introduce a synodic coordinate system < ux, uy, uz >with origin at the second primary and rotating with angular velocity ω = ωuz around an axis orthogonal to their orbital plane, with ux along the line going from the first to the second primary. We then consider a spacecraft (third body) of mass m placed in the xy plane at a distance ρ from the second primary and δ from the first, and undergoing a constant acceleration to maintain it in the neighborhood of its initial position (x 0 , y 0 )(figure 1).

ω = GM ℓ 3 , S/C G l x M 1 M 2 y 1 l 1-m d r w
From now on we will write all physical quantities in dimensionless coordinates taking as ℓ the unit of distance and as 1/ω the unit of time. The spacecraft position vector measured from the center of mass of the two primaries can then be written as:

r = (1 -µ + x, y) T ,
while the spacecraft distance from the first and second primary are, respectively:

δ = x 2 + y 2 + 2x + 1, (1)
ρ = x 2 + y 2 .
(2)

Eq.s (1,2) can be inverted for later use providing:

x = δ 2 -ρ 2 -1 2 , (3)
y = ± 1 2 4ρ 2 -(δ 2 -ρ 2 -1) 2 , (4)
with the contraint:

|1 -ρ| < δ < 1 + ρ. (5)
The inertial kinetic energy per unit mass is given by:

T = 1 2 ṙ + ω × r 2 = 1 2 ( ẋ2 + ẏ2 ) + ẏ(x + 1 -µ) -ẋy + 1 2 (x 2 + y 2 ) + x(1 -µ). (6)
The total potential due to the gravitational and the constant control acceleration is:

V = Vg + Va = - 1 -µ δ - µ ρ -axx -ayy. (7)
where ax, ay are the x and y components of the control acceleration (Figure 1).

The equations of motion can finally be derived through Lagrange equations and yield:

ẍ -2 ẏ = 1 -µ + x -(1 -µ) 1 + x δ 3 -µ x ρ 3 + ax, ÿ + 2 ẋ = y -(1 -µ) y δ 3 -µ y ρ 3 + ay.
The control acceleration for maintaining a desired equilibrium position (x 0 , y 0 ) is hence:

ax = (1 -µ) 1 + x 0 δ 3 0 + µ x 0 ρ 3 0 -x 0 + µ -1, (8)
ay = (1 -µ) y 0 δ 3 0 + µ y 0 ρ 3 0 -y 0 , (9)
where δ 0 , ρ 0 are the radial distances of the equilibrium point from the first and second primary, respectively.

Linear Stability Conditions

In this section we will study the linear stability of the system using a Hamiltonian approach which will be later extended to the investigation of non-linear stability.

From the kinetic and potential energy we obtain the Lagrangian function:

L = T -V,
which provides the canonical Hamiltonian variables:

px = ∂L ∂x = ẋ -y, py = ∂L ∂y = ẏ + x + 1 -µ,
so that the Hamiltonian can be derived as:

H = px ẋ + py ẏ -L, yielding: H = 1 2 (p 2 x + p 2 y ) + pxy -py(x + 1 -µ) - 1 -µ δ - µ ρ -axx -ayy.
In order to study the linear (local) equilibrium it is convenient to write the Hamiltonian with respect to o reference system with origin at the equilibrium point. This can be done by the following change of variables:

ξ = x -x 0 , η = y -y 0 , p ξ = px -p x0 = px + y 0 , pη = py -p y0 = py -x 0 -1 + µ,
which after eliminating constant terms yields the Hamiltonian:

H = 1 2 (p 2 ξ + p 2 η ) + p ξ η -pηξ + ξ(µ -1 -x 0 ) -ηy 0 - 1 -µ δ - µ ρ -axξ -ayη, (10)
where:

δ = (1 + ξ + x 0 ) 2 + (η + y 0 ) 2 , ρ = (ξ + x 0 ) 2 + (η + y 0 ) 2 .
Because the origin is an equilibrium point H can be expanded in series in which the linear term disappears:

H = H 2 + k>2 H k , (11)
where:

H 2 = 1 2 (p 2 ξ + p 2 η ) + ηp ξ -ξpη + aξ 2 + bξη + cη 2 , ( 12
)
and where a, b, c are function of the equilibrium point location and can be conveniently written in bipolar coordinates:

a = - µ 8ρ 5 3δ 4 -6δ 2 ρ 2 -6δ 2 + 3ρ 4 + 2ρ 2 + 3 - 1 -µ 8δ 5 3ρ 4 -6ρ 2 δ 2 -6ρ 2 + 3δ 4 + 2δ 2 + 3 , b = - 3 2ρ 2 (δ 2 + 1) -(δ 2 -1) 2 -ρ 4 4ρ 5 δ 5 ρ 5 δ 2 -ρ 7 , +ρ 5 + µ(-δ 5 + δ 7 -δ 5 ρ 2 -ρ 5 -ρ 5 δ 2 + ρ 7 ) c = µ 8ρ 5 3δ 4 -6δ 2 ρ 2 -6δ 2 + 3ρ 4 + 2ρ 2 + 3 + 1 -µ 8δ 5 3ρ 4 -6ρ 2 δ 2 -6ρ 2 + 3δ 4 -2δ 2 + 3 ,
The linearized equations of motion are finally:

ṗξ = - ∂H 2 ∂ξ = pη -2aξ -bη, ṗη = - ∂H 2 ∂η = -p ξ -2cη -bξ, ξ = ∂H 2 ∂p ξ = p ξ + η, η = ∂H 2 ∂pη = pη -ξ,
with the corresponding Jacobian matrix:

A =         0 1 -2a -b -1 0 -b -2c 1 0 0 1 0 1 -1 0        
and the characteristic polynomial:

P (λ) = λ 4 + 2f λ 2 + g (13)
where f, g can be written as:

f = a + c + 1 = 2δ 3 -1 2δ 3 + µ ρ 3 -δ 3 2ρ 3 δ 3 , (14)
g = -b 2 + 2a(c -1) -2c + 1 = δ 6 -2+δ 3 δ 6 + µ 4ρ 5 δ 6 -4ρ 5 δ 3 +16ρ 5 + 4ρ 2 δ 6 + 18ρ 2 δ + 18δ 3 -9δ 5 + 2ρ 2 δ 3 -9δρ 4 -9δ -µ 2 4ρ 6 δ 6 -9ρδ 5 + 18ρδ 3 -9ρ 5 δ + 18ρ 3 δ + 8δ 6 -9ρδ + 8ρ 6 + 2δ 3 ρ 3 , (15)
In order for the characteristic polynomial (13) to have pure imaginary roots the following conditions have to be satisfied: ) and for the Earth-Moon system (right) (µ = 1.2 × 10 -2 ). The light grey area delimits the stable zone while the dark grey solid line represents the minimum-control equilibrium points computed by numerically solving Eq. 27. The dark lines on the left figure represents equilibrium points with (2:1) and (3:1) resonance. Distances are measured with respect to the second primary in units of separation between the two primaries (ℓ).

a) f 2 -g > 0,
b) f > 0, c) g > 0.

(16)

The three conditions delimit a stable region in the orbital plane of the two primaries which is plotted in figures (2,3) for three different values of the contant µ. The stability region for the earth-sun system was already obtained by Morimoto et al. [START_REF] Morimoto | Artificial equilibrium points in the low-thrust restricted three-body problem[END_REF]Morimoto, Yamakawa, and Uesugi] although an explicit analytical representation of the curves was not provided.

We point out to the reader that, as an alternative to our Hamiltonian approach, it is possible and straightforward to study the system linear stability by just calculating the variational equations of the cartesian equations of motion. Fig. 3 Stability boundary for the Jupiter-Amalthea system (µ = 1.1 × 10 -9 ). The light grey area delimits the stable zone while the dark grey solid line represents the minimumcontrol equilibrium points computed by numerically solving Eq. 27. The dark lines represents equilibrium points with (2:1) and (3:1) resonance. Distances are measured with respect to the second primary in units of separation between the two primaries (ℓ).

Nonlinear Stability

As it is known, when a linearized system has all its eigenvalues in the left-half complex plane but at least one on the jω axis (i.e. the system is marginally stable) one cannot infer stability from the linear part alone, as higher order terms can drive the system towards instability. This is also the case of the triangular Lagrangian points in the CRTBP whose Lyapunov stability, for the two dimensional case, was first demonstrated by Leontovitch [Leontovich(1962)] in 1962 by investigating higher order terms of the system Hamiltonian and with the aid of Arnold s stability theorem.

Leontovitch showed that the system was stable for all values of the the mass ratio satisfying 0 < µ < (9 -√ 69)/18 ≃ 0.0385 . . . with the exclusion of a set of values of measure zero for which no conclusion about stability could be made. Later on, Deprit and Deprit-Bartholomé [START_REF] Deprit | Stability of the triangular lagrangian points[END_REF]] proved that the zero-measure set consisted of only three points:

µ 1 = (45 - √ 1833)/90 ≃ 0.0243, µ 2 = (15 - √ 213
)/30 ≃ 0.0135, µ 3 = 0.0109 the first two corresponding to 2:1 and 3:1 resonances between the two normal frequencies of oscillations around the La-grangian point, while the third having no straightforward physical meaning. Markeev, in 1972[Markeev(1972)], proved the instability of the two resonance values while stability was found for the third value.

Arnold s stability theorem can be applied to our case to guarantee Lyapunov stability along the whole linearly stable domain with the exception of a set of points with zero measure.

From the previous linearization we can derive the two oscillation frequencies around the AEP as:

iω 1,2 = -f ± f 2 -g, ω 1,2 = 2ρ 3 δ 3 -ρ 3 + µρ 3 -µδ 3 ± √ K 2ρ 3 δ 3 , ( 17
)
where

K = -8ρ 6 δ 3 + 9ρ 6 -ρ 18ρ 5 -8ρ 5 δ 3 + 8δ 6 ρ 2 -9δ -9δ 5 + 18δ 3 + 18δρ 2 -9δρ 4 µ + 9 2ρ 3 δ + 2ρδ 3 -ρδ 5 -δρ -ρ 5 δ + ρ 6 + δ 6 µ 2 . ( 18
)
According to Arnold s theorem the marginally stable equilibrium point (p ξ , pη, ξ, η) = (0, 0, 0, 0) is Lyapunov stable if two conditions are verified. The first, known as the nonresonance condition imposes that:

k 1 ω 1 + k 2 ω 2 = 0, (19)
for all pairs of rational integers such that:

|k 1 | + |k 1 | ≤ 4.
The second, known as the reversibility condition reads:

Aω 2 1 + Bω 1 ω 2 + Cω 2 2 = 0, (20)
where A, B, C are the coefficient of the fourth order part of the normalized Hamiltonian (see ref. [START_REF] Deprit | Stability of the triangular lagrangian points[END_REF]]):

H4 = 1 2 (AI 2 1 + BI 1 I 2 + CI 2 2 )
with:

I 1 = p2 ξ + ξ2 2 ; I 2 = p2 η + η2 2 ,
and where (p ξ , pη, ξ,η) are the new Hamiltonian variables after the normalization.

It can be readily verified that condition (19) leads to the exclusion of the two resonance cases:

ω 1 = 2ω 2 , (21)
ω 1 = 3ω 2 , (22)
Eq.s (21-22) represent two curves in the orbital plane which can be obtained in bipolar coordinates using Eq.s (17,18). After some algebra one obtains the 2:1 resonance curve as:

(-36ρ 6 -164µρ 3 + 216ν 2 )δ 6 + 225(-µ 2 ρ + µρ)δ 5 +(450µρ + 18µ 2 ρ 3 -18µρ 3 + 164µρ 6 -164ρ 6 + 450µ 2 ρ)δ 3 +225(µρ 5 -µ 2 ρ 5 + 2µ 2 ρ 3 -µ 2 ρ + µρ -2µρ 3 )δ +216(-2µρ 6 + µ 2 ρ 6 + ρ 6 ) = 0, (23)
while the 3:1 resonance curve yields:

(-64ρ 6 -136µρ 3 + 209µ 2 )δ 6 + 225(-µ 2 ρ + µρ)δ 5 +(-450µρ + 32µ 2 ρ 3 -32µρ 3 + 136µρ 6 -136ρ 6 + 450µ 2 ρ)δ 3 +225(µρ 5 -µ 2 ρ 5 + 2µ 2 ρ 3 -µ 2 ρ + µρ -2µρ 3 )δ +209(-2µρ 6 + µ 2 ρ 6 + ρ 6 ) = 0. ( 24
)
The curves cut through the stability domain as plotted in Figures (2- Similarily, condition (20) represents another curve in the orbital plane which will intersect the stable region. Along the resulting arcs stability or instability needs to be inferred by normalization of higher order terms in the Hamiltonian series expansion.

Given the complexity of the normalization procedure the derivation of curve (20) will not be dealt with in the present article.

As far as the three-dimensional problem is concerned "effective stability" can be assured given the analogy of the present problem with the triangular Lagrangian points in the CRTBP. The latter has been shown to be stable in three dimensions (although not in the sense of Lyapunov) by Markeev([Markeev(1972)]). [START_REF] Celletti | On the stability of the lagrangian points in the spatial restricted problem of three body[END_REF]]) also dealt with the problem of finding effective stability for the system (i.e. according to Nekhoroshev theory) and the interested reader can check the reference mentioned above.

Minimum Control Artificial Equilibrium Points

After having demonstrated the existence of a stable region in the planar perturbed CRTBP the next step is to identify stable points inside this region for which the required control acceleration is as small as possible. In this regard, our proposed approach is to fix an equilibrium distance ρ 0 from the second primary and to search for the spacecraft position that provides minimum control acceleration. This corresponds to minimize the objective function:

J = a 2 x + a 2 y , (25)
for a fixed value of ρ 0 . Using Eqs. (3,4) we can eliminate the variables x, y from Eqs. (8-25) and express J in bipolar coordinates. Omitting the subscript '0' from δ, ρ in order to simplify the notation we obtain:

J(ρ, δ) = µ(ρ 3 -1)(ρ -1)(µ + ρ 2 + ρ) ρ 4 - 1 -µ ρ 3 δ 4 × (µ -ρ 3 )δ 6 -(µ -2ρ 3 + µρ 3 )δ 3 + µ(ρ 3 -1)(ρ 2 -1)δ -(1 -µ)ρ 3 ,
subject to contraint (5).

Given the complex structure of the function J we decided to first study the minimization problem numerically using a grid-sampling approach. Figures (4), dealing with the Earth-Moon and Sun-Earth systems respectively, represent the locus of the equilibrium points that, for a fixed value of the distance from the second primary, are characterized by minimum control acceleration. Taking for instance the Earth-Moon system as an example (Figure 4) we see that when ρ is large compared with the distances ρ L1 , ρ L2 between the second primary and the two Lagrangian points minimum acceleration is obtained when the spacecraft is almost coorbiting with the second primary. As ρ decreases the locus of the minimum acceleration points starts to deviate significantly from the trace of the primary orbit. When ρ drops to a critical value ρ * which is sligtly larger than ρ L2 a plateau is encountered in which the value of the minimum control acceleration is almost constant for δ increasing up to its boundary limit (ρ * + 1) which is reached when the locus crosses the y axis. This means the locus describes a portion of a circumference centered at the second primary and along which Fig. 4 Locus of the minimum control acceleration points (dark line) in the Earth-Moon system (left) and for the Sun-Earth system (right) for a given distance ρ from the second primary. A countour of the magnitude of the control acceleration is included (grey lines). The locus is computed numerically with a grid sampling approach.

the magnitude of the control aceeleration is virtually constant. As ρ further decreases the control acceleration, now directed away from the second primary, decreases until it becomes zero at L2. Reducing ρ even further will result in an increase in the magnitude of the control acceleration, now directed towards the second primary, until a jump occurs which brings the locus to the negative part of the x axis with ρ just slightly larger than ρ L1 and with negative radial acceleration. If ρ is further decreased we reach L1 with zero control acceleration. Finally, an increase in the required acceleration is encountered if one ones to obtain equilibrium at a distance smaller than ρ L1 in which case the spacecraft will be located on the negative x axis. Now that the structure of the minimization problem has been clarified we can approach the problem analitically.

Once ρ is fixed, the value of δ that minimizes J obeys:

∂J ∂δ = 0. ( 26
)
After performing the derivatives and with some algebraic simplifications Eq.( 26) yields a 6-degree polynomial equation in δ(ρ):

2 ρ 3 -µ δ 6 + 2ρ 3 -ρ 3 µ -µ δ 3 + 3µ ρ 2 -1 ρ 3 -1 δ -4ρ 3 (1 -µ) = 0 (27)
which can be numerically solved to obtain the optimum value δ opt . It is important to point out that Eq.27 only provides values of δ opt corresponding to stationary points for J(ρ, δ). On the other hand, as we have just seen, when ρ is small enough the locus of the minimum control points is located on the x axis where ρ = |δ -1|

and the minima are not necessarily stationary points (see boundary condition 5) and cannot be derived through Eq.27. Nevertheless Eq.27 is effective in the derivation of all minimum acceleration points that do not belong to the x axis, and, in particular, the ones inside the stable equilibrium region previously identified. As µ is usually small, an approximate analytical solution can be found with a straightforward perturbation method [Nayfeh(1981)]. To this purpose we set:

δ = δ 0 + µδ 1 + µ 2 δ 2 + O(µ 3 ). (28)
After substituting the latter into Eq.( 27), expanding in a Taylor series for small µ, and solving each term we eventually find:

δ opt ∼ = δ 0 + µδ 1 + µ 2 δ 2 = 1 + 1 -ρ 3 6ρ µ - ρ 8 + 2ρ 6 -2ρ 5 + 2ρ 3 + ρ 2 -4 36ρ 4 µ 2 (29)
which compared with the exact expression provided by Eq. ( 27) exhibits an error which increases exponentially with the inverse of the distance from the second primary.

It can be seen later that all along the stable equilibrium region previously computed Eq.( 29) provides an estimate of the optimal distance δ with a relative error on the second term:

e = δ opt -1 -µδ 1 δ opt -1
It can be verified that when µ < 0.1, which encompasses the totality of sun-planet and planet-satellite systems in our solar system, all terms of the type µρ k in the above expression can be neglected with a maximum loss of accuracy not exceeding 5%. In this way the three contraints simplify to:

ρ 6 -26ρ 3 µ + 9µ 2 > 0, (35)
ρ 3 -µ > 0, (36)
-2ρ 8 + 3ρ 6 -ρ 5 + 8ρ 3 -16µ > 0 (37)

From the first two conditions we obtain our stability boundary limit:

ρ > ρ min = (13µ + 4µ √ 10) 1/3 ≈ (26µ) 1/3 (38)
It can be easily checked that condition (37) is always verified for ρ > ρ lim .

The corresponding control acceleration computed at the boundary of the stable region is:

a(ρ min ) = (26µ) 1/3 |1 -26µ| 4 -(26µ) 2/3 52
In dimensional units the minimum distance from the second primary in order to obtain open-loop stability with minimum control acceleration is finally:

r min ≃ ℓ(26µ) 1/3 (39)
And the required dimensional control acceleration (Eq.( 30)) is:

â(r min ) ∼ = GM 2 µℓ 2   (26µ) 1/3 |1 -26µ| 4 -(26µ) 2/3 52   (40)
The intersections of the 2:1 and 3:1 resonance curves with the minimum control acceleration curve (Eq( 27)) can also be computed numerically. Nevertheless, similarily to what was done for the stability boundary, an approximate analytical expression that works very well for µ < 0.1 can be obtained by plugging Eq (29) int Eqs. (23,24) and neglecting all terms of the type µρ k . In this way we finally obtain the two intersections as:

ρ 21 ≃ 79 2 µ 1/3 ; ρ 31 ≃ 206 3 µ 1/3 6 Applications
The fact that a spacecraft can be made to hover in the vicinity of a planet or satellite and kept stable with just a constant force is very appealing from a general point of view. On the other hand the interest and applicability of the concept in a real mission scenario depends on the chosen three-body gravitational system and has to take into account the following aspects:

1. The minimum stable hovering distance (Eq.( 39)) should be as close as possible with the second primary in order to provide science and/or communication benefits 2. The minimum control acceleration (Eq.( 40)) should be as small as possible so that it can be provided by state of the art solar sail and/or other low-thrust propulsion systems ( [START_REF] Mengali | Non-Keplerian orbits for electric sails[END_REF]]).

Table (1) lists the equilibrium characteristics of a number of selected candidate sunplanet and planet-satellite systems. The four innermost planets enjoy favorable conditions in terms of minimum hovering distance, which is suitable for spacecraft communication purposes, and required acceleration, which could be easily supplied with state-of-the art solar sails ([McInnes(2003b)], [START_REF] Farres | Periodic and quasi-periodic motions of a solar sail close to SL 1 in the Earth-Sun system[END_REF]]) . Another important advantage is given by the fact that the sail orientation can be maintained with a fixed direction relative to the direction of both primaries and can be efficientlly repositioned at different azimuthal locations as proposed by Mcinnes [McInnes(2003a)]. (stable zone) and ρ = (24.7µ) 1/3 (unstable) on the minimum-control curve for the Sun-Earth system (µ = 3.0 × 10 -6 ). A disturbance velocity is applied with 1 m/s magnitude and 45 degrees orientation with respect to the x axis in the positive xy plane (Figure 1). The distance is measured in Earth radii (r E ) and the duration of the simulation is 50 years.

Conclusions

We have investigated the properties of minimum-control artificial equilibrium points in the planar circular restricted three body problem providing simple analytical expressions that characterize their location, required control acceleration and stability properties. It is seen that once the distance from the second primary is fixed and greater than the distance from the second collinear Lagrangian point minimum-control is obtained by having the spacecraft almost coorbiting with the second primary with an offset that is a simple function of the mass ratio µ of the two primaries. The required control acceleration would then decrease until becoming zero as the spacecraft location approaches the nearest triangular libration point. Lyapunov stability under minimum control conditions is seen to occur as the distance with the second primary reaches the value ρ min ≃ (26µ) 1/3 and taking the primaries' separation as unit of distance. As the spacecraft location moves away from the second primary along the minimum control curve, stability is always assured with the exclusion of 2:1 and 3:1 resonance conditions which appear at distances ρ 21 ≃ (79µ/2) 1/3 and ρ 31 ≃ (206µ/3) 1/3 , respectively. A

Fig. 1

 1 Fig. 1 Schematic of synodic plane and spacecraft position.

Fig. 2

 2 Fig.2Stability boundary for the Sun-Earth system (left) (µ = 3.0 × 10 -6 ) and for the Earth-Moon system (right) (µ = 1.2 × 10 -2 ). The light grey area delimits the stable zone while the dark grey solid line represents the minimum-control equilibrium points computed by numerically solving Eq. 27. The dark lines on the left figure represents equilibrium points with (2:1) and (3:1) resonance. Distances are measured with respect to the second primary in units of separation between the two primaries (ℓ).

  3). Along those arcs Arnold s theorem cannot assure Lyapunov stability although the similarity with the case of the L 4 , L 5 Lagrangian points in the CRTBP would suggest the arcs are unstable. A stability analysis along the three resonant arcs is beyond the scope of the present article.

Fig. 5

 5 Fig.5Numerically-computed trajectories for a spacecraft located at a distance ρ = (30µ) 1/3 (stable zone) and ρ = (24.7µ) 1/3 (unstable) on the minimum-control curve for the Sun-Earth system (µ = 3.0 × 10 -6 ). A disturbance velocity is applied with 1 m/s magnitude and 45 degrees orientation with respect to the x axis in the positive xy plane (Figure1). The distance is measured in Earth radii (r E ) and the duration of the simulation is 50 years.

The work was supported by the DGI of the Spanish Ministry of Education and Science (research grant ESP2007-64068) Research Fellow, ETSI Aeronauticos, Plaza Cardenal Cisneros 3 Professor, ETSI Aeronauticos, Plaza Cardenal Cisneros 3 Address(es) of author(s) should be given

which is less than 3%.

The corresponding control acceleration is readily computed by substituting Eq.

(29) into Eqs. (8)(9)(10)(11)(12)(13)(14)(15)(16)(17)(18)(19)(20)(21)(22)(23)(24)(25). Neglecting terms of order µ 2 and higher we obtain: a = a 2

x + a 2 y ∼ = µ ρ 2 1 -ρ 2 + 8ρ 3 -2ρ 5 -4ρ 6 + ρ 8 4 (30)

Note that both Eq (29) and ( 30) are consistent with the location of the triangular Lagrangian points (ρ = 1) for which the control acceleration is zero.

The corresponding dimensional acceleration yields:

Let us now look for an approximate analytical estimation of the minimum distance from the second primary allowing stability. This can be done by substituding the value of δ opt (Eq. ( 29)) into conditions ( 16) and solving for ρ. After doing that and retaining terms up to µ 2 (for the first condition) and µ (for the latter two) the three conditions read:

For the different planet-satellite systems the case of Deimos seems to be the most favorable as an hovering spacecraft could be passively stabilized at about 100 km from the surface (i.e. less than 9 times the size of the satellite) with a relatively small acceleration. Finally we point out that the case of the inner Jupiter moons is also promising because the required control acceleration, probably too big for a solar sail, can be easily provided by electrodynamic tethers of modest size (see Pelaez and Scheeres [START_REF] Pelaez | A permanent tethered observatory at jupiter. dynamical analysis[END_REF]). Nevertheless the harsh radiation environment in the vicinity of Jupiter would make the last application prohibitive with current technology.

Before concluding the article we present numerical results supporting the analytical results obtained above. Figure 5 summarizes the stable/unstable behaviour of the 3body dynamics for the case of the Earth-Sun system showing a stable behavior for ρ > (26µ) 2.9 × 10 -3 Jupiter-Europa 2.53 × 10 -5 6.71 × 10 5 5.8 × 10 4 9.4 × 10 -4 Jupiter-Metis 6.31 × 10 -11 1.28 × 10 5 1.5 × 10 2 3.5 × 10 -4 Jupiter-Adrastea 3.95 × 10 -12 1.29 × 10 5 6.0 × 10 1 1.4 × 10 -4 Jupiter-Amalthea 1.09 × 10 -9 1.81 × 10 5 5.5 × 10 2 4.5 × 10 -4 Jupiter-Thebe 7.89 × 10 -10 2.22 × 10 5 6.1 × 10 2 2.7 × 10 -4

Table 1 Selected candidate gravitational systems with minimum required distance and acceleration for stable hovering quick inspection of promising applications in our solar system shows that stable solarsail-stabilized observatories or communication outposts coorbiting with Earth Venus and Mars are feasible from the point of view of the control acceleration required and advantageous given their vicinity with the planet of interest. Numerical simulations agree with the provided analytical formulas. Future work will be needed to assess the engineering feasibility of the concept.

Acknowledgments

The authors would like to thank Prof. Ryan Russell (Georgia Tech) for his constructive comments and suggestions in the review process.


J-L Fouquet

H Thuillier

J-M Vanherpe

A P Wojda

Al Mickiewicza

On (K q , k) vertex stable graphs with minimum size

Keywords:

A graph G is a (K q , k) vertex stable graph if it contains a K q after deleting any subset of k vertices. We give a characterization of (K q , k) vertex stable graphs with minimum size for q = 3, 4, 5.

Introduction

For terms not defined here we refer to [START_REF] Bondy | Graph theory[END_REF]. As usual, the order of a graph G is the number of its vertices (denoted by |G|) and the size of G is the number of its edges (denoted by e(G)). A complete subgraph of order q of G is called a q-clique of G. The complete graph of order q is denoted by K q . When a graph G contains a q-clique as subgraph, we say "G contains a K q ". The union of p mutually disjoint copies of K q is denoted by pK q .

When A is a set of vertices we denote by G -A the subgraph induced by V (G) -A.

In [START_REF] Horváth | Extremal P 4 -stable graphs[END_REF][START_REF] Katona | Extremal stable graphs[END_REF] Horvárth and G.Y Katona consider the notion of (H, k) stable graph: given a simple graph H, an integer k and a graph G containing H as subgraph, G is a a (H, k) stable graph whenever the deletion of any set of k edges does not lead to a H-free graph. These authors consider (P n , k) stable graphs and prove a conjecture stated in [START_REF] Frankl | Extremal k-edge hamiltonian hypergraphs[END_REF] on the minimum size of a (P 4 , k) stable graph. In [START_REF] Dudek | H, k) stable graphs with minimum size[END_REF], Dudek, Szymański and Zwonek are interested in a vertex version of this notion and introduce the (H, k) vertex stable graphs. 1 The research of APW was partially sponsored by polish Ministry of Science and Higher Education.

Definition 1.1. [START_REF] Dudek | H, k) stable graphs with minimum size[END_REF] Let H be a graph. A graph is a (H, k) vertex stable graph if it contains a graph isomorphic to H after deleting any subset of k vertices. By Q(H, k) we denote the minimum size of a (H, k) vertex stable graph. If G is (H, k) vertex stable of size Q(H, k) we call it minimum (H, k) vertex stable.

In this paper, we are only interested by (H, k) vertex stable graphs and, since no confusion will be possible, a (H, k) vertex stable shall be simply called a (H, k) stable graph.

In [START_REF] Dudek | H, k) stable graphs with minimum size[END_REF], the authors give values of Q(H, k) when H is isomorphic to C 3 , C 4 or K 4 and provide upper bounds for some other cases while in [START_REF] Dudek | On vertex stability with regard to complete bipartite subgraphs[END_REF][START_REF] Dudek | H, k) stable bipartite graphs with minimum size[END_REF] the bipartite case is considered.

It must be pointed out that in some cases the value of Q(H, k) can be obtained without the description of extremal graphs, that is (H, k) vertex stable graphs whose size is precisely Q(H, k). In this paper we describe the extremal (H, k) stable graphs when H is isomorphic to K q , for q ∈ {3, 4, 5} while in [START_REF] Fouquet | On (Kq, k) stable graphs with small k[END_REF] we describe the extremal (K q , k) stable graphs when k is small with respect to q.

By considering (H, k) stable graph with minimum size, it must be clear that we can add some isolated vertices, the resulting graph remains to be a (H, k) stable graph with minimum size. From now on, the graphs considered have no isolated vertices. Proposition 1.2. [START_REF] Dudek | H, k) stable graphs with minimum size[END_REF] If G is a (H, k) stable graph with minimum size then every vertex as well as every edge is contained in a subgraph isomorphic to H. Remark 1.3. Proposition 1.2 implies, in particular, that when H ≡ K q then the minimum degree of a (H, k) stable graph with minimum size is at least q -1.

Lemma 1.4. [2] Let k ≥ 1. If G is (H, k) stable then for any vertex v, G -{v} is (H, k -1) stable.
Definition 1.5. Let H be a non complete graph on q + p + 1 (p ≥ 0) vertices and u be one vertex. Let N be the neighbourhood of u and R = V (H)u -N . We shall say that H is a near complete graph (R,N,u) on q + p + 1 vertices (see Figure 1) when

• H -{u} is complete, • d H (u) = q + ǫ (ǫ ∈ {-1, 0, 1}). d H (u) = q. u N R K q+p Figure 1: A near complete graph (R,N,u) on q + p + 1 vertices

Preliminary results

Proposition 2.1. If G is a (K q , k) stable graph with minimum size (q ≥ 3) then G has no component isomorphic to a near complete graph (R,N,u) on q + p + 1 vertices.

Proof. Suppose, by contradiction, that there exists such a component H = (R, N, u) on q + p + 1 vertices with d H (u) = q + ǫ (ǫ ∈ {-1, 0, 1}). Since G is a (K q , k)-stable graph with minimum size, G -{u} is not (K q , k)-stable. There exists a set S with at most k vertices such that S intersects every K q of G -{u}. There exists a K q in G -S and clearly such a K q contains u. Since N is a K q+ǫ and N -S contains no K q , |S ∩N | ≥ ǫ+1

(trivial for ǫ = -1). If |S ∩ N | ≥ ǫ + 2 then |N -S| ≤ q -2
, and hence S intersects every

K q containing u, a contradiction. Thus, |S ∩ N | = ǫ + 1 and |N -S| = q -1. If there exists v in R -S then (N -S) + {v} is a K q in G -{u}, a contradiction. Thus, R ⊂ S. Let a ∈ R and b ∈ N -S, and set S ′ = S -{a} + {b}. We have | S ′ |≤ k and G -S ′ contains no K q , a contradiction.
Lemma 2.2. Let q ≥ 3 and k ≥ 1 and let G be a minimum (K q , k) stable graph.If u is a vertex of degree q -1 then one of the following statements is true

• ∀v ∈ N (u) d(v) ≥ q + 1, • Q(K q , k -1) + 3(q -2) ≤ Q(K q , k).
Proof. Since d(u) = q -1, {u} + N (u) induces a complete graph on q vertices. Assume that some vertex w ∈ N (u) has degree q + a (a = -1 or a = 0) and let v ∈ N (u) distinct from w. Then Gv is a (K q , k -1) stable graph (Lemma 1.4). Since the degree of u in G -{v} is q -2, no edge incident with u can be contained in a K q . We can thus delete these q -2 edges and the resulting graph (say G ′ ) is still a (K q , k -1) stable graph.

In G ′ , the degree of w is now q + a -2. Hence, no edge incident with w in G ′ can be contained in a K q . Deleting these q + a -2 edges from G ′ leads to a graph G " which remains to be a (K q , k -1) stable graph.

By deleting v, we have e(G -{v}) ≤ e(G) -(q -1) and hence

e(G ′ ) ≤ e(G) -(q -1) -(q -2)
We get thus

Q(K q , k -1) ≤ e(G ′′ ) ≤ e(G) -(q -1) -(q -2) -(q + a -2)
Since e(G) = Q(K q , k), the result follows.

Proposition 2.3. If G is a minimum (K q , 1) stable graph (q ≥ 4) then it is isomorphic to K q+1 .

Proof.

Let G be a minimum (K q , 1) stable graph. Since K q+1 is (K q , 1) stable, clearly e(G) ≤ q+1 2 . We can assume that G is connected. Otherwise, each component contains a K q , but q+1 2 < 2 q 2 as soon as q ≥ 4, a contradiction. Let u be a vertex of G and Q be a subgraph of G -{u} isomorphic to K q . Assume that there exists a vertex v outside Q and distinct from u. Note that v can be a neighbour of u. Since

d(u) ≥ q -1 and d(v) ≥ q -1, e(G) ≥ e(Q) + 2(q -1) -1 = q 2 + 2q -3 = q+1 2 + q -3.
Thus, e(G) > e(K q+1 ), a contradiction. Hence,

V (G) = V (Q) ∪ {u} with d(u) ≥ q -1.
Since for any edge e K q+1 -{e} is not (K q , 1) stable, we see that

d(u) = q, that is G is isomorphic to K q+1 .
Remark 2.4. It is easy to see that the minimum (K 3 , 1) stable graphs are 2K 3 and K 4 .

Proposition 2.5. If G is a minimum (K q , 2) stable graph (q ≥ 4) then it is isomorphic to K q+2 .

Proof.

Since K q+2 is a (K q , 2) stable graph, we can suppose that G has at most q+2 2

edges. We can suppose, moreover, that G is not complete, otherwise G is obviously reduced to K q+2 . Let u be a vertex of minimum degree (recall that the minimum degree is at least q -1) and let v be one of its neighbours.

Assume that d G (u) = q -1. G -{v} is a (K q , 1) stable graph, but it is not minimum, since none of the remaining edge incident with u can be contained in a complete graph on q vertices. By deleting the q -2 edges incident with u, we get thus a (K q , 1) stable graph.

If d(v) ≥ q + 1, this graph has at most q+2 2 -(2q -1) edges. Since this number of edges must be greater than q+1 2 by Proposition 2.3, we have (q + 2)(q + 1) -4q + 2 ≥ (q + 1)q That leads to q ≤ 2, a contradiction. If d(v) ≤ q, by Lemma 2.2, we have Q(K q , 1) + 3q -6 ≤ Q(K q , 2) and hence q(q + 1) + 6q -12 ≤ (q + 1)(q + 2)

Which gives q ≤ 3, a contradiction.

We can thus assume that the minimum degree of G is at least q. Let u and v be two non adjacent vertices of G. Since G -{u, v} contains a K q (say Q), let a and b be two distinct vertices of Q. Since G -{a, b} must contain also a K q , there is certainly a vertex w distinct from v and u, outside Q, inducing with q -1 other vertices of G -{a, b} a K q .

Hence G contains three vertices (u, v and w) at least in G -Q and we have:

q + 2 2 ≥ e(G) ≥ q 2 + 3q -2
Which gives q < 3, a contradiction. Hence G is complete and the proposition follows.

Lemma 2.6. Let G be a minimum (K q , 3) stable graph, q ≥ 5. Let u be a vertex of minimum degree in G and suppose that d G (u) = q + l, where -1 ≤ l ≤ 1. Then for every

neighbour v of u we have d G (v) ≥ q + l + 2.
Proof. Suppose, contrary to our claim, that d G (v) ≤ q + l + 1 for a neighbour v of u.

Since, by Proposition 1.2, the edge uv is contained in a clique of order q and q ≥ 5, there is a set A of vertices of G such that |A| = l + 2 and the vertices of the set A ∪ {u, v} are mutually adjacent. The graph

G ′ = G -A is (K q , 3 -(l + 2)) stable. We have d G ′ (u) = q + l -(l + 2) = q -2, hence also G ′′ = G ′ -{u} is (K q , 1 -l) stable. But in G ′′ the degree of the vertex v is at most q -2 and therefore G ′′′ = G ′′ -{v} is (K q , 1 -l)
stable. Since every vertex of the set A ∪ {u, v} has at least q -3 neighbours outside this set, we have

q + 1 -l 2 ≤ e(G ′′′ ) ≤ q + 3 2 -(l + 4)(q -3) - l + 4 2 which contradicts q ≥ 5. Proposition 2.7. If G is a minimum (K q , 3) stable graph (q ≥ 5) then it is isomorphic to K q+3 .
Proof.

Note first that to prove the proposition it is sufficient to prove that every vertex of G has the degree at least q + 2.

Let u be a vertex of the minimum degree in G and suppose, contrary to our claim, that

d G (u) ≤ q + l, where -1 ≤ l ≤ 1. Let v 1 , v 2 , ..., v l+2 be such vertices of G that the set {u, v 1 , v 2 , ..., v l+2 } induce a clique
in G (such vertices exist since u is contained in a clique of order q by Proposition 1.2 and q ≥ 5). By Lemma 2.6, we have

d G (v i ) ≥ q + l + 2 for i = 1, 2, ..., l + 2. Set G ′ = G -{v 1 , v 2 , ..., v l+2 }. The graph G ′ is clearly (K q , 1 -l) stable. Moreover, since d G ′ (u) = q -2, the graph G ′′ = G ′ -{u} is also (K q , 1 -l) stable and we have q + 1 -l 2 ≤ e(G ′′ ) ≤ q + 3 2 -(l + 2)(q -1) -(q -2) - l + 3 2
which contradicts q ≥ 5.

3. A characterization of (K 3 , k) stable graph with minimum size

Dudek, Szymański and Zwonek in [START_REF] Dudek | H, k) stable graphs with minimum size[END_REF] have shown that Q(K 3 , k) = 3k + 3 for every nonnegative integer k. In this section we characterize all that (K 3 , k) stable graphs with minimum size.

Clearly, K 3 is the unique minimum (K 3 , 0) stable graph.

The following theorem characterize all graphs which are (K 3 , k) stable with minimum size.

Theorem 3.1. Let G = (V, E) be a (K 3 , k) stable graph with minimum size. G is isomorphic to pK 4 +qK 3 , where p and q are such nonnegative integers that 2p+q = k +1.

Proof.

By Remark 2.4, K 3 is the unique minimum (K 3 , 0) stable graph, and the minimum (K 3 , 1) stable graphs are 2K 3 and K 4 . Clearly, the graph (k + 1)K 3 is a (K 3 , k) stable graph and has 3k + 3 edges. Let k 0 ≥ 1 and suppose that for every k < k 0 every minimum (K 3 , k) stable graph is a union of p copies of K 4 and q copies K 3 with

2p + q = k + 1. Let G be a (K 3 , k 0 ) stable graph of minimum size. Since G -{v} is (K 3 , k 0 -1) stable for every vertex v, we have 3k 0 ≤ e(G -{v}) ≤ e(G) -d G (v) ≤ 3k 0 + 3 -d G (v), that is d G (v) ≤ 3.
If every vertex of G has degree equal to 2, then G is a union of k 0 + 1 copies of K 3 , and the theorem is proved. So we may suppose that there is a vertex v 0 of degree

3. But then G -{v 0 } is (K 3 , k 0 -1) stable and e(G -{v 0 }) = 3k o , that is G -{v 0 } is minimum (K 3 , k 0 -1) stable. By the induction hypothesis, G -{v 0 } is isomorphic to p ′ K 4 + q ′ K 3 , where 2p ′ + q ′ = k 0 .
It is clear that all the neighbours of v 0 are in the same component of G, (otherwise one of the edges incident with v 0 is not contained in any triangle, contrary to Proposition 1.2). Now it is easy to see that G is isomorphic to

(p ′ + 1)K 4 + (q ′ -1)K 3 and 2(p ′ + 1) + (q ′ -1) = k 0 + 1 (otherwise there is a set A of cardinality k 0 which is transversal of all cliques of order 3 in G).
4. A characterization of (K 4 , k) stable graph with minimum size

In [START_REF] Dudek | H, k) stable graphs with minimum size[END_REF] the minimum number of edges of a (K 4 , k) stable graph is given. H is not isomorphic to a K 5 minus one edge, hence there must exist some new vertex w ′ adjacent to w. Since each vertex in {u 1 , u 2 , u 3 , w} has degree 4, we cannot find a K 4 using the edge ww ′ , a contradiction with Proposition 1.2.

Theorem 4.1. [2] If G is a (K 4 , k) stable graph with minimum size (k ≥ 1) then • Q(K 4 , 0) = 6, • Q(K 4 , k) = 5k + 5 when k ≥ 1. Proposition 4.2. If G is a (K 4 ,
Theorem 4.5. If G is (K 4 , k) stable (k ≥ 1)
with minimum size then it is isomorphic to pK 5 + qK 6 , where p and q are nonnegative integers such that 2p + 3q = k + 1.

Proof. The proof is by induction on k. By Proposition 2.3, the only minimum (K 4 , 1)

stable graph with minimum size is K 5 . Let k 0 ≥ 2 and suppose that for every integer k, such that 1 ≤ k < k 0 every (K 4 , k) stable graph with minimum size is isomorphic to pK 5 + qK 6 , where p and q are nonnegative integers such that 2p + 3q = k + 1.

Let G be a (K 4 , k 0 ) stable graph with minimum size. By Theorem 4.

A characterization of (K 5 , k) stable graph with minimum size

In this section we provide the value of Q(K 5 , k) for k ≥ 5, as well as a description of the corresponding minimum stable graphs.

Lemma 5.1. Let G be a (K 5 , k) stable graph containing a component isomorphic to K p with p ≥ 9. Then the graph G ′ obtained from G by deleting two vertices v and v ′ in this K p and adding a disjoint K 6 is a (K 5 , k) stable graph such that

• if p ≥ 10 then e(G ′ ) < e(G),
• if p = 9 then e(G ′ ) = e(G).

Proof.

Let A be the set of vertices created by the adjunction of the new K 6 . Let S be a set of vertices with

|S| ≤ k in G ′ . If |S ∩ A| ≤ 1, G -S obviously contains a K 5 . If |S ∩ A| ≥ 2 then S ′ = S -A + {v, v ′ } is a subset of G with at most k vertices. Hence G -S ′ contains a K 5 which still exists in G ′ -S.
If p ≥ 10 then at least 17 edges are deleted and 15 edges are created, thus e(G ′ ) < e(G)

. If p = 9, 15 edges are deleted while 15 edges are created so e(G) = e(G ′ ).

Lemma 5.2. Let G be a (K 5 , k) stable graph with minimum size. Then G does not contain 2 components isomorphic to a K p with 5 ≤ p ≤ 6.

Proof.

If we have two components (say K and L) isomorphic to a complete graph with 5 vertices then the graph G ′ obtained from G by deleting these two components and adding a complete graph on 6 vertices is still a (K 5 , k) stable graph. Indeed, let S ′ be any subset of

V (G ′ ) with |S ′ | ≤ k. If G ′ -S ′
does not contain any K 5 then S ′ must contain at least 2 vertices v and w of the new K 6 . Let S = S ′ -{v, w} + {a, b}, where a ∈ K and b ∈ L, then G -S does not contain any K 5 , a contradiction.

When we have a K 5 and a K 6 , we get the same kind of contradiction when replacing these two complete graphs with a K 7 as well as when we have two K 6 s replaced by a K 8 .

Lemma Therefore, we have to consider only the case when G is the vertex disjoint union of complete graphs isomorphic to K 7 or K 8 and at most one K 5 or one K 6 . Replacing a K 5 and a K 7 by one K 8 leads to a (K 5 , k) stable graph with a number of edges less than the number of edges of G, a contradiction. Replacing a K 6 and a K 8 by two K 7 leads to a (K 5 , k) stable graph with a number of edges less than the number of edges of G, a contradiction.

It remains to consider the case where the components are all isomorphic to a K 7 with the exception of one K 6 or all isomorphic to a K 8 with the exception of one K 5 . When we have at least two K 7 and a K 6 , these three complete graphs can be replaced by two K 8 , the resulting graph is still a (K 5 , k) stable graph, but the number of edges is less than the number of edges of G, a contradiction. When we have at least two K 8 and a K 5 , these three complete graphs can be replaced by three K 7 , the resulting graph is still a (K 5 , k) stable graph, but the number of edges is less than the number of edges of G, a contradiction.

When G is reduced to a K 8 and a K 5 or to a K 7 and a K 6 , we must have k ≤ 4, which is impossible.

Lemma 5.4. Let G be a (K 5 , k) stable graph with minimum size and maximum degree 6. Assume that some component contains a K 6 . Then either the component is equal to this K 6 or to K 7 .

Proof.

Let A = {v 1 . . . v 6 } be the set of vertices of the K 6 . If d(v i ) = 5 for each vertex in A the proof is complete. Assume that the vertex v 1 has degree 6 and let w be its neighbour outside A. Since v 1 w must be contained in a K 5 by 1.2, w must be adjacent to 3 other vertices in V (say v 2 , v 3 and v 4 ). In the same way, if v 5 or v 6 has a neighbour outside A, this vertex must be adjacent to 4 vertices of A, which is impossible if this vertex is distinct from w.

Let w ′ ∈ A be a neighbour of w (if any). Since ww ′ must be contained in a K Proof. Let H be a connected component containing vertices of degree 5 or 6 only. By Proposition 1.2, every edge is contained in a K 5 . Let U = {u 1 , u 2 , u 3 , u 4 , u 5 } be a set of vertices inducing a K 5 in H.

case 1 : ∃ i 1 ≤ i ≤ 5 d H (u i ) = 6.
Without loss of generality we may assume that i = 1. Let w and w ′ the two neighbours of u 1 outside U . Since u 1 w must be contained in a K 5 and since this K 5 contains 4 neighbours of u 1 , w must be adjacent to at least two vertices in U -{u 1 }. Without loss of generality, assume that wu 2 ∈ E(G) and wu 3 ∈ E(G). Let us remark that w is not joined to the two vertices u 4 and u 5 , otherwise, H contains a K 6 and H is thus isomorphic to a complete graph by Lemma 5.4. For the same reason, w ′ is not joined to all the vertices in U .

subcase 1.1 : If w or w ′ has no other neighbour in U , say w, we must have

ww ′ ∈ E(G), w ′ u 2 ∈ E(G) and w ′ u 5 ∈ E(G).
One of u 4 or u 5 , say u 4 , is not adjacent to w ′ , and there must be a vertex w ′′ adjacent to u 4 (d H (u 4 ) ≥ 5), but the edge u 4 w ′′ cannot be on any K 5 , which is impossible.

subcase 1.2 : If w has an other neighbour (say u 5 ) in U . When w ′ is not adjacent to w, w ′ must be adjacent to precisely 3 vertices in {u 2 , u 3 , u 4 , u 5 }. If u 4 w ′ is an edge, there must be an edge incident with w ′ (d H (w ′ ) ≥ 5), but this edge cannot be contained in any K 5 , a contradiction. If u 4 w ′ is not an edge, there must be an edge incident with u 4 and this edge cannot be contained in any K 5 , which is impossible. Thus, w and w ′ are adjacent and there must be 2 vertices in {u 2 , u 3 , u 5 } adjacent to w ′ , say u 2 and u 3 . But now, there is an additionnal edge incident with u 4 and this edge is u 4 w ′ otherwise it is not contained in any K 5 . It is a routine matter to check that there is no additionnal vertex nor edge in H. Hence H is isomorphic to the graph depicted in Figure 2, a contradiction with Lemma 5.5.

case 2 : ∀ i 1 ≤ i ≤ 5 d H (u i ) = 5.
Let w be the last neighbour of u 1 outside U . Since wu 1 must be contained in a K 5 , w must be adjacent to u 2 , u 3 and u 4 , without loss of generality. Hence, wu 5 ∈ E(G) or H is complete. Since d H (u 5 ) = 5, let w ′ = w be the last neighbour of u 5 outside U . Then u 5 w ′ is not contained in a K 5 , which is impossible.

Lemma 5.7. Q(K 5 , 4) = 36.

Proof. Since K 9 and K 6 + K 7 are (K 5 , 4) stable graphs, we certainly have Q(K 5 , 4) ≤

36.

Assume that some graph G with e(G) ≤ 35 is a (K We can thus assume that the maximum degree of G is at most 6. If some vertex u has degree 4, let v be one of its neighbours. We know, by Lemma 2.6 that d(v) = 6.

By deleting v, we get a graph Gv which is a (K 5 , 3) stable graph. In that graph, the edges incident with u are not contained in a K 5 since the degree of u is now 3. We can thus delete these edges and we obtain a (K 5 , 3) stable graph with at most 27 edges, a contradiction with Proposition 2.7.

Hence every vertex must have degree 5 or 6. By Lemma 5. We can thus assume that the maximum degree of G is at most 7. If some vertex u has degree 4, let v be one of its neighbours. We know, by Lemma 2.2 that d(v) ≥ 6.

By deleting v, we get a graph G -{v} which is a (K 5 , 3) stable graph. In that graph, the edges incident with u are not contained in a K 5 since the degree of u is now 3.

We can thus delete these edges and we obtain a (K 5 , 3) stable graph with 27 edges, a contradiction with Proposition 2.7.

Hence the degree of each vertex is 5, 6 or 7.

In the following Claims Q 1 and Q 2 denote any two induced K 5 of G.

Claim 5.8.1. such that x ∈ V (Q 1 ) and y ∈ V (Q 2 ). Then we can find a vertex x ′ ∈ V (Q 1 ) and a vertex

|V (Q 1 ) ∩ V (Q 2 )| = 1. Proof Assume that |V (Q 1 ) ∩ V (Q 2 )| =
y ′ ∈ V (Q 2 ) such that {x, x ′ , y, y ′ } is contained in an induced K 5 of G. Moreover the 5 th vertex of this K 5 must be contained in V (Q 1 ) ∪ V (Q 2 ).

Proof

Since G is a minimum (K 5 , 4) stable graph, the edge xy must be contained in a K 5 (say Q). By Claim 5.8.1 Q contains at least one vertex more in Q 1 (say x ′ ) and one vertex more in Q 2 (say y ′ ). Let a be the 5 th vertex of Q and assume that a ∈

V (Q 1 ) ∪ V (Q 2 ).
G -{a} is a (K 5 , 3) stable graph but it is not minimum since the edges between {x, x ′ } and {y, y ′ } cannot be contained in a K 5 . By deleting these 4 edges in G -{a} we get a Claim 5.8.4.

|V (Q 1 ) ∩ V (Q 2 )| = 3 or G is isomorphic to K 7 + K 6 . Proof Assume that |V (Q 1 ) ∩ V (Q 2 )| = {x,
= G - (V (Q 1 ) ∪ V (Q 2 )) is a (K 5 , 1) stable graph. If w ∈ V (Q 1 ) ∪ V (Q 2 )
is adjacent to some vertex w ′ in G 2 then a K 5 using that edge forces 4 more edges more between these two subgraphs, a contradiction since G would have at least 37 edges (by Proposition 2.3 G 2 has at least 15 edges).

If some edge is missing between {u 1 , u 2 } and {v 1 , v 2 } (say

u 1 v 1 ∈ E(G)), then G 3 = G -{u 2 , v 2 } is a (K 5 , 2
) stable graph where x, y, z, u 1 , v 1 are not contained in any K 5 .

The graph G 3 is still (K 5 , 2) stable. Hence, by Proposition 2.5 G must have at least 38 edges, a contradiction.

We can thus suppose that

V (Q 1 ) ∪ V (Q 2 ) induces a K 7 .
The remaining part of G is the (K 5 , 1) stable graph G 2 described above. This graph must have exactly 15 edges.

Hence, G 2 is isomorphic to K 6 by Proposition 2.3. That means that G is isomorphic to

K 7 + K 6 . Claim 5.8.5. |V (Q 1 ) ∩ V (Q 2 )| = 4 or G is isomorphic to K 7 + K 6 .
Proof Assume on the contrary that |V (Q 1 ) ∩ V (Q 2 )| = {x, y, z, t} and G is not isomorphic to K 7 + K 6 . Let u and v be the remaining vertices of Q 1 and Q 2 respectively.

Let r be a neighbour of u, if any, outside Proof. We can check that the property holds for k = 5 (G is the vertex disjoint union of two K 7 s by Lemma 5.10). Assume that the property holds for any k (5 ≤ k < k 0 ) and let us consider a (K 5 , k 0 ) stable graph G with minimum size. Assume that G has at most 7k 0 +6 edges and let v be a vertex of maximum degree. Since G-v is a (K 5 , k 0 -1) stable graph, it must have 7k 0 edges, which means that d(v) ≤ 6. Moreover, by Proposition 1.2, we certainly have d(v) ≥ 4.

V (Q 1 ) ∪ V (Q 2 ). Let Q 3 be a K 5 containing the edge ur. Then V (Q 1 ) ∩ V (Q 3 )
Let z be a vertex of degree 4 in some component of G. If z has a neighbour v whose degree is 6 then Gv has exactly 7k 0 edges. Hence Gv is a (K 5 , k 0 -1) stable graph with minimum size. Since the degree of z is 3 in Gv, any edge incident with z in Gv is not contained in a K 5 , a contradiction.

If z has a neighbour v whose degree is 5 then Gv has exactly 7k 0 + 1 edges. Gv is a (K 5 , k 0 -1) stable graph. This graph does not have minimum size since the 3 remaining edges incident with z are not contained in a K 5 . If we delete these 3 edges, we still have a (K 5 , k 0 -1) stable graph, but the number of edges is then 7k 0 -2, which is impossible by the induction hypothesis.

Hence the neighbours of z have also degree 4, that means that the component containing a vertex of degree 4 is a 4 regular graph containing a K 5 . That is, this component is a K 5 .

Since each component containing only vertices of degree 5 or 6 are complete by Lemma 5.6, we have thus that all the connected components of G are complete. By Lemma 5.3, each component has 7 vertices or 8 vertices (recall that k 0 ≥ 5). Assume that we have p components isomorphic to a K 7 and q isomorphic to a K 8 , then k 0 ≤ 3p + 4q -1 and G

  [START_REF] Bondy | Graph theory[END_REF] we have e(G) = 5k 0 +5. Note that it is sufficient to prove that every component of G is isomorphic either to K 5 or to K 6 .ByProposition 4.3, we have 3 ≤ d G (v) ≤ 5 for every vertex v of G. Since by Proposition 1.2, every edge of G is contained in a K 4 , all the neighbours of a vertex v are in the same component of G -{v}.Suppose first that there is a vertexv in G such that d G (v) = 5. Then G -{v} is (K 4 , k 0 -1) stable and moreover, since e(Gv) = 5k 0 , G -{v} is minimum (K 4 , k 0 -1)stable. Hence every component of G -{v} is either isomorphic to K 5 or to K 6 . If v is connected in G to a K 6 , then the component of G which contains v is a near complete graph, contradicting Proposition 2.1. So v is connected to a K 5 and G is a union of graphs isomorphic to K 5 or K 6 , as desired.Assume now that no component has a vertex of degree 5. Then, by Proposition 4.4, each component is a 4-regular subgraph.Let v be any vertex and let N (v) = {u 1 , u 2 , u 3 , u 4 } be its neighbourhood. Since v is contained in a K 4 by Proposition 1.2, we can suppose, without restriction of generality, that u 1 u 2 , u 1 u 3 and u 2 u 3 are edges of G. Since vu 4 must be contained in a K 4 by Proposition 1.2, u 4 must be adjacent to at least 2 vertices of N (say, without loss of generality, u 2 and u 3 ).

case 1 :

 1 u 1 u 4 ∈ E(G). Then the component containing v is a K 5 . case 2: u 1 u 4 ∈ E(G).Let w be a new vertex adjacent to u 1 (this new vertex must exist since the component of v is 4-regular). Then u 1 w cannot be contained in a K 4 , a contradiction.

5 by Proposition 1 . 2 , 4 and v 5 ,Figure 2 :Lemma 5 . 5 .

 1245255 Figure 2: Forbidden component of a (K 5 , k) stable graph with minimum size.

case 1 :

 1 contains 4 vertices (Claims 5.8.1 and 5.8.3) butV (Q 2 ) ∩ V (Q 3 ) contains 3 vertices, a contradiction. Since d(u) ≥ 5, we must have uv ∈ E(G) (and, moreover, d(u) = d(v) = 5). There are neighbours of {x, y, z, t} outside V (Q 1 ) ∪ V (Q 2 ).Let s be such a neighbour of x,. The edge xs being contained in a K 5 , this K 5 must have 4 common vertices with Q 1 and 4 common vertices with Q 2 (Claims 5.8.1, 5.8.3 and 5.8.4). Hence, s must be adjacent to the 4 vertices of V (Q 1 ) ∩ V (Q 2 ) and {x, y, z, t, s} induces a K 5 with 4 common vertices with Q 1 and 4 common vertices with Q 2 . By the If G has a vertex u of degree 4, let v be one of its neighbours. By deleting v we get a (K 5 , 4) stable graph where the edges incident with the vertex u are not contained in any K 5 since the degree of u in that graph is 3. By deleting these edges we get a (K 5 , 4) stable graph with at most 35 edges, a contradiction with Lemma 5.7.Hence every vertex has 5 or 6. By Lemma 5.6, the connected components of G are complete. It is an easy task to see that the only convenient graph G is isomorphic to K 7 + K 7 , as claimed. Theorem 5.11. If G is (K 5 , k) stable (k ≥ 5) with minimum size then |E(G)| = 7k + 7.

  k) stable graph with minimum size (k ≥ 1) then it has no connected component isomorphic to K 4 . contains some vertex v with degree 3. Let N (v) = {u 1 , u 2 , u 3 } be its neighbourhood. By Proposition 1.2, N (v) is complete. Since H is not isomorphic to K 4 by Proposition 4.2, assume that, without loss of generality, u 1 is joined to some new vertex w. Since u 1 w must be contained in a K 4 by Proposition 1.2, the vertices of this K 4 are in {u 1 , u 2 , u 3 , w, v}, thus w must be adjacent to u 2 and u 3 . By Proposition 2.1,

	Proof.	By Proposition 4.3 the vertices of G have degree 3 or 4. Assume to the
	contrary that H

Proof. Let us consider k ≥ 2. Assume that some component H of G is isomorphic to a K 4 with the vertices of H being a, b, c, d. Then G -H has 5k -1 edges. Since G -H is not a (K 4 , k -1) stable graph, there is a set S with at most k -1 vertices intersecting each K 4 of G -H. Then S + {a} intersects each K 4 of G while S has at most k -1 vertices, a contradiction. When k = 1, G must have 10 edges by Theorem 4.1. Since for each vertex v the graph Gv contains a K 4 , v is joined to this K 4 by 4 edges. Hence G is a K 5 and the result holds. Proposition 4.3. If G is a (K 4 , k) stable graph with minimum size (k ≥ 1) then every vertex of G has degree 3, 4 or 5.

Proof.

By Proposition 1.2 every vertex is contained in a K 4 , hence its degree is at least 3. Assume that G has a vertex v with d(v) ≥ 6. Then, by Lemma 1.4, Gv is a (K 4 , k -1) stable graph and therefore has at least 5k edges, which is impossible since G has exactly 5k + 5 edges, by Theorem 4.1.

Proposition 4.4. Let G = (V, E) be a (K 4 , k) stable graph with minimum size (k ≥ 1).

If H is a component containing no vertex of degree 5, then each vertex of H has degree 4.

  5.3. Let k ≥ 5 and let G be a (K 5 , k) stable graph with minimum size which is the vertex disjoint union of complete graphs. Then each component is a K 7 or a K 8 .

Proof. By Lemma 5.1, we can consider that each component is a K p with 5 ≤ p ≤ 9. By Lemma 5.2, at most one component is a K 5 or a K 6 . If some component is isomorphic to a K 9 then let us replace this component by a K 6 and a K 7 By Lemma 5.1 the resulting graph is still a (K 5 , k) stable graph with minimum size. It is clear that no component is isomorphic to a K 9 now. Indeed, applying once more the operation described above leads to a (K 5 , k) stable graph with minimum size having two K 6 s, a contradiction with Lemma 5.2.

  5 , 4) stable graph with minimum size. Let v be a vertex with maximum degree. If d(v) ≥ 8 then Gv is a (K 5 , 3) stable graph with at most 27 edges, a contradiction with Proposition 2.7. If d(v) = 7then G -{v} is a (K 5 , 3) stable graph with at most 28 edges. Hence we must have e(G -{v}) = 28 and G is a (K 5 , 3) stable graph with minimum size. By Proposition 2.7, G -{v} is a K 8 and G is a K 9 minus one edge, a contradiction with Proposition 2.1.

  6, the components of G are complete graphs. It can be easily checked that the only convenient graphs are K 9 and K 6 + K 7 , a contradiction with e(G) ≤ 35.Lemma 5.8. K 6 + K 7 and K 9 are the unique (K 5 , 4) stable graph with minimum size.Proof. By Lemma 5.7, let G be a (K 5 , 4) stable graph with 36 edges.If G has a vertex of degree at least 8 then G -{v} is a (K 5 , 4) stable graph with at most 28 edges. Hence G -{v} must have exactly 28 edges and d(v) = 8. Since, by Proposition 2.7 G -{v} is a K 8 , G itself is a K 9 .

  1 then the vertex in the intersection must have degree at least 8, which is impossible.

Claim 5.8.2. Assume that Q 1 and Q 2 are vertex disjoint and let xy ∈ E(G) (if any)

  y, z}. Let {u 1 , u 2 } and {v 1 , v 2 } be the sets of remaining vertices of Q 1 and Q 2 respectively.Then G 1 = G -{x, y, z} is a (K 5 , 1) stable graph in which the vertices u 1 , u 2 , v 1 , v 2 are not contained in any K 5 by Claims 5.8.1 and 5.8.3. That means that G 2

Acknowledgement. APW wishes to thank Laboratoire d'Informatique Fondamentale de l'Université d'Orléans, where a part of this paper was written, for the invitation and hospitality.

Claim 5.8.3.

Proof Assume that V (Q 1 ) ∩ V (Q 2 ) = {x, y}. Let us remark that these two vertices have degree 7. Let {u 1 , u 2 , u 3 } and {v 1 , v 2 , v 3 } be the sets of remaining vertices of Q 1 and Q 2 respectively.

Assume that some edge is missing between {u 1 , u 2 , u 3 } and {v 1 , v 2 , v 3 } (say u 1 v 1 ∈ E(G)). Then G 1 = G -{u 2 , v 2 , v 3 } is a (K 5 , 1) stable graph in which the vertices x and y are not contained in any K 5 . Hence G 2 = G 1 -{x, y} is a (K 5 , 1) stable graph. Since

contains 19 edges, we need to find two more edges. By Claim 5.8.2 no edge can connect V (Q 1 ) ∪ V (Q 2 ) to G 4 . Whatever is the place of these edges, G -{x, y} is a (K 5 , 2) stable graph, where no vertex in {u 1 , u 2 , u 3 } nor That means that u 1 and u 3 have 3 neighbours outside V (Q 1 ) ∪ V (Q 2 ). In the same way, we can consider that u 2 has also three such neighbours (take G 1 = G-{u 3 , v 2 , v 3 }) as well as v 1 , v 2 and v 3 by symmetry. Hence G 3 contains the 19 edges of V (Q 1 )∪V (Q 2 ) and 18 edges connecting {u 1 , u 2 , u 3 } and {v 1 , v 2 , v 3 } to the vertices outside, a contradiction.

We can thus suppose that every vertex in {u 1 , u 2 , u 3 } is joined to every vertex in

in this graph are not contained in any K 5 , we can delete them and the (K 5 , 2) stable graph obtained in this way must have 21 edges exactly. This component is a K 7 by Proposition 2.5, a contradiction.

To end our proof, it is sufficient to say that any two induced K 5 of G must be disjoint Claims 5.8.1, 5.8.3, 5.8.4 and 5.8.5. That means that each component of G is a K 5 , which is impossible since G must have 36 edges. Lemma 5.9. Q(K 5 , 5) = 42.

Proof.

Since K 7 + K 7 is a (K 5 , 5) stable graphs, we certainly have Q(K 5 , 5) ≤ 42.

Let G be a (K 5 , 5) stable graph with minimum size and assume that e(G) ≤ 41. Let us remark that the size of G is certainly greater than Q(K 5 , 4).

If G has a vertex of degree at least 6 then G-v is a (K 5 , 4) stable graph with at most 35 edges, a contradiction with Lemma 5.7. If G has a vertex of degree 4 then, since the degree of every neighbour is at most 5, we must have, by Lemma 2.2, Q(K 5 , 5) ≥ Q(K 5 , 4) + 9, a contradiction.

Hence, every vertex must have degree 5 and by Lemma 5.6, the connected component of G are isomorphic to K 6 . It is easy to see that no such graph can exist.

Lemma 5.10. K 7 + K 7 is the unique (K 5 , 5) stable graph with minimum size.

Proof. By Lemma 5.9, let G be a (K 5 , 5) stable graph with 42 edges.

If G has a vertex of degree at least 7 then Gv is a (K 5 , 4) stable graph with at most 35 edges, a contradiction with Lemma 5.7.

has 21p + 28q edges. If k 0 = 3p + 4q -1, we have 21p + 28q = 7k 0 + 7, a contradiction. If k 0 < 3p + 4q -1 then deleting one vertex in some component leaves the graph (K 5 , k 0 ) stable, which is impossible.

Dudek, Szymański and Zwonek proposed the following conjecture.

Conjecture 5.12. [START_REF] Dudek | H, k) stable graphs with minimum size[END_REF] For every integer q ≥ 5 there is an integer k(q) such that

Theorem 5.11 proves this conjecture for q = 5 with k(q) = 5.

• each connected component is isomorphic to a complete graph with 7 or 8 vertices,

• there are p components isomorphic to K 7 and q components isomorphic to K 8 for any choice of p and q with 3p + 4q = k + 1.

Proof. By Theorem 5.11, the first claim is true. We can check that the property of the second claim holds for k = 5 (G is the vertex disjoint union of two K 7 s). Assume that the property holds for any k (5 ≤ k < k 0 ) and let us consider a (K 5 , k 0 ) stable graph G with minimum size.

If G has a vertex v of degree at least 8, then Gv has at most 7k 0 -1 edges and cannot be a (K 5 , k 0 -1) stable graph, a contradiction. Thus the maximum degree of G is at most 7.

In that case, It is now a routine matter to check that the third claim holds.