J Ér

Ôme Coville
email: jerome.coville@avignon.inra.fr

Juan D Ávila
email: jdavila@dim.uchile.cl

J Dávila

EXISTENCE OF RADIAL STATIONARY SOLUTIONS FOR A SYSTEM IN COMBUSTION THEORY

In this paper, we construct radially symmetric solutions of a nonlinear noncooperative elliptic system derived from a model for flame balls with radiation losses. This model is based on a one step kinetic reaction and our system is obtained by approximating the standard Arrehnius law by an ignition nonlinearity, and by simplifying the term that models radiation. We prove the existence of 2 solutions using degree theory.

Introduction

This paper deals with radial solutions of the system of equations

             ∆u -εu = -vf (u) in R 3 ∆v = vf (u) in R 3 u ≥ 0, v ≥ 0 in R 3 lim |x|→∞ u(x) = 0, lim |x|→∞ v(x) = 1, (1.1)
where ε ≥ 0 and f is an ignition type nonlinearity, that is, there exists 0 < θ < 1 such that

f (t) = 0 ∀t ≤ θ, f (t) > 0 ∀t > θ. (1.2)
System (1.1) arises as a model problem for some reaction diffusion systems in combustion theory. We will describe the connection with these models at the end of the section.

We call (u, v) = (0, 1) the trivial solution of (1.1). Note that for any solution (u, v) of (1.1) one has 0 ≤ u ≤ 1 and 0 ≤ v ≤ 1, as can be deduced from the maximum principle. We are mainly interested in existence and multiplicity of nontrivial radial solutions of (1.1). The first observation in this direction is that for ε > 0 large (1.1) has no nontrivial solutions. Indeed, by the maximum principle u + v ≤ 1 and then

-∆u ≤ (1 -u)f (u) -εu in R 3 .
If ε > 0 is large then (1 -u)f (u) -εu ≤ 0 for all u ∈ [0, 1] and we conclude that u ≡ 0. Better estimates for the quantity ε * = sup{ε > 0 : (1.1) has a nontrivial radial solution} are given in Section 6.

Date: January 26, 2011.

1 Suppose that u, v is a nontrivial radial solution. Then for some r > 0 we must have u(r) > θ. Otherwise f (u) ≡ 0 and then ∆v = 0, and by the Liouville theorem v ≡ 1. Since 0 ≤ u + v ≤ 1 this would imply u ≡ 0 and then u, v is trivial. So for a nontrivial solution u, v, since lim r→∞ u(r) = 0, there is β u > 0 such that u(β u ) = θ and u(r) < θ ∀r > β u .

The parameter β u , which we will write simply as β, then serves to distinguish different solutions.

Theorem 1.1. Assume f : [0, 1] → [0, +∞) is continuous, satisfies (1.2) and for some C > 0

f (t) ≤ Cf (u) ∀t ≤ u, t, u ∈ [θ, 1]. (1.3)
Then ε * > 0 and there exist 0 < ε 1 < ε * such that for 0 < ε < ε 1 there are at least 2 solutions of (1.1). One of them has bounded β as ε → 0 and the other has β in the range δ/ √ ε ≤ β ≤ 1/(θ √ ε) as ε → 0, where δ > 0 is fixed.

A very natural and interesting question is the stability of the solutions constructed in Theorem 1.1. Based on the works [START_REF] Buckmaster | The structure and stability of nonadiabatic flame balls[END_REF][START_REF] Buckmaster | The structure and stability of nonadiabatic flame balls: II. Effects of far-field losses[END_REF] we conjecture that the solution with bounded β is unstable and that in part of branch of solutions with large β the solution is stable, at least with respect to radial perturbations.

In some cases one may want to consider a discontinuous nonlinearity, such as the Heavisde function f (u) = χ [u>θ] . With this example in mind we introduce the following hypothesis f is continuous in (θ, 1] and lim u→θ+ f (u) exists. (1.4) Theorem 1.2. Assume f : [0, 1] → [0, +∞) satisfies (1.2), (1.3), and (1.4). Then ε * > 0 and there exist 0 < ε 1 < ε * such that for 0 < ε < ε 1 there are at least 2 solutions of (1.1). One of them has bounded β as ε → 0 and the other has β in the range δ/ √ ε ≤ β ≤ 1/(θ √ ε) as ε → 0, where δ > 0 is fixed.

The solution constructed in Theorem 1.2 is such that the set {r ∈ [0, ∞) : u(r) = θ} is finite, and hence the equation (1.1) holds a.e. in R 3 .

The motivation to consider a discontinuous nonlinearity is only mathematical. However the example f (u) = χ [u>θ] is interesting since it provides a situation where explicit calculations are possible. Theorem 1.2 shows that in part the conclusions obtained for f (u) = χ [u>θ] remain valid for more general non-linearities.

For f (u) = χ [u>θ] explicit calculations lead to an equation for β and ε in order for a radial solution to exist. In Figure 1 we show the numerical solution for this relation when θ = 0.5, with β in the vertical axis and √ ε in the horizontal axis. It shows that for 0 < ε < ε * there are 2 solutions. Solutions in the lower branch satisfy u > θ in [0, β) that is, the reaction takes place in the ball of radius β. The same happens for points in the upper branch which are to the right of the special point marked in the graph. To the left of that point the solution satisfies u > θ in an annulus of the form r ∈ (β -L β , β). Thanks to the explicit form of the relation bewteen β and ε we can compute the asymptotic behavior of the curve as β → ∞, and we find that

√ ε β u(r) > θ for r ∈ [0, β) u(r) > θ for r ∈ (β -L β , β) β = O(1/ √ ε) Figure 1. Bifurcation diagram for f = χ [u>θ]
with θ = 0.5; β is in the vertical axis and √ ε in the horizontal axis where a 0 , L 0 is the unique solution of the system of equations θ a tanh(a)

-

1 + θ(1 + a) = 1 1 - 1 L tanh(L) + 1 L sinh(L) = θ(1 + a).
Because of the information on the Heaviside nonlinearity one can conjecture that for general f there should be a similar relation for β and ε as β → +∞. We present in Section 2 nonexistence results for general ignition nonlinearities satisfying (1.2) and (1.3), that capture this relation, and roughly speaking say that no solution can exist if √ εβ is either too large or too small, provided β is taken large enough. Using these nonexistence results and degree theory we can give a proof of Theorem 1.1. This is done in Section 3. In Section 4 we give the proof of Theorem 1.2 by approximating the discontinuous nonlinearity by continuous ones. Section 5 is devoted to the explicit computations for the Heaviside function. Finally Section 6 contains a finer estimate of ε * .

As mentioned before, system (1.1) arises in connection with some models in combustion theory, more precisely, in the flame ball problem for a weakly premixed gas sensitive to radiative heat losses. In such a mixture, it is known that apparently stationary spherical structures appear, which are called flame balls [START_REF] Lewis | Combustion, Flames and Explosion of Gases[END_REF][START_REF] Ronney | Near-Limit Flame Structures at Low Lewis Number[END_REF][START_REF] Buckmaster | The structure and stability of nonadiabatic flame balls[END_REF][START_REF] Buckmaster | The structure and stability of nonadiabatic flame balls: II. Effects of far-field losses[END_REF].

In [START_REF] Lewis | Combustion, Flames and Explosion of Gases[END_REF][START_REF] Ronney | Near-Limit Flame Structures at Low Lewis Number[END_REF][START_REF] Buckmaster | The structure and stability of nonadiabatic flame balls[END_REF][START_REF] Buckmaster | The structure and stability of nonadiabatic flame balls: II. Effects of far-field losses[END_REF][START_REF] Kagan | Self-fragmentation of nonadiabatic cellular flames[END_REF][START_REF] Kagan | On self-drifting flame balls[END_REF][START_REF] Minaev | On self-drifting flame balls[END_REF] the following reaction diffusion system has been used to model a combustion process where flame balls can appear:

(1.5)            ρC p ∂T ∂t = ∇ • (λ∇T ) + QBY ρ m e -E RT -q(T ) ρ ∂Y ∂t = ∇ • (µ∇Y ) -BY ρe -E RT Y → Y ∞ , T → T ∞ as |x| → ∞
where T is the temperature, Y the reactant concentration, Y ∞ > 0, T ∞ > 0 are the reactant concentration and tempereature at infinity, and C p , R, Q and m are respectively the specific heat capacity at constant pressure, the perfect gas constant, the chemical heat release and the molecular mass of the reactant. The term q(T ) represents radiative losses. The reaction is characterized by the one-step Arrhenius kinetics ∼ Be -E

RT where B is a constant. Furthermore, the hydrodynamics effects are neglected, i.e the density ρ, the thermal conductivity λ and the diffusion coefficient µ are constant. See also numerical simulations in [START_REF] Bockhorn | An adaptive two-dimensional wavelet-vaguelette algorithm for the computation of flame balls[END_REF][START_REF] Kagan | Self-fragmentation of nonadiabatic cellular flames[END_REF][START_REF] Kagan | On self-drifting flame balls[END_REF].

After the seminal work [START_REF] Zeldovich | A theory of the limit of slow flame propagation[END_REF], the traveling front problem for systems like (1.5) has been investigated by several authors, for example, [START_REF] Giovangigli | Nonadiabatic plane laminar flames and their singular limits[END_REF][START_REF] Glangetas | Rigorous derivation of the dispersion relation in a combustion model with heat losses[END_REF][START_REF] Roques | Study of the premixed flame model with heat losses. The existence of two solutions[END_REF].

In absence of radiation, i.e. when q = 0, there are many works dealing with (1.5), see for instance [START_REF] Berestycki | Quelques aspects mathmatiques de la propagation des flammes prmlanges. Nonlinear partial differential equations and their applications[END_REF][START_REF] Berestycki | Traveling wave solutions to combustion models and their singular limits[END_REF][START_REF] Lederman | Mathematical justification of a nonlinear integrodifferential equation for the propagation of spherical flames[END_REF][START_REF] Marion | Qualitative properties of a nonlinear system for laminar flames without ignition temperature[END_REF][START_REF] Sagon | Steady fronts solutions of semilinear elliptic equations in exterior domains arising in flame propagation[END_REF] and the references therein. Also we remark that the stationary version of (1.5) without radiation, leads to system (1.1) with ε = 0 which reduces to a scalar equation, since v = 1 -u. There is a huge amount of literature concerning existence of radial ground states for semilinear equations, so we mention here only some classical references [START_REF] Berestycki | Nonlinear scalar field equations. I. Existence of a ground state[END_REF][START_REF] Kwong | Uniqueness of positive solutions of ∆u -u + u p = 0 in R n[END_REF] on the problem in entire space. When the problem is treated in a bounded domain see the book [START_REF] Aris | The Mathematical Theory of Diffusion and Reaction in Permeable Catalysts[END_REF] and [START_REF] Du | Proof of a conjecture for the perturbed Gelfand equation from combustion theory[END_REF][START_REF] Kapila | Reactive-diffusive system with Arrhenius kinetics: peculiarities of the spherical geometry[END_REF] for multiplicity results in the case of Arrehnius non-linearity. The paper [START_REF] Brindley | Stationary-state solutions for coupled reaction-diffusion and temperature-conduction equations. II. Spherical geometry with Dirichlet boundary conditions[END_REF] contains interesting numerical computations of the bifurcation diagram in the case of the full coupled system in a interior of a sphere.

A common simplification of (1.5) under the assumption of large activation energy, that is E >> 1, is to assume that the source term for the reaction is concentrated on a very thin layer, typically a sphere. This approach is taken for example in [START_REF] Buckmaster | The structure and stability of nonadiabatic flame balls[END_REF][START_REF] Buckmaster | The structure and stability of nonadiabatic flame balls: II. Effects of far-field losses[END_REF][START_REF] Lee | The structure and stability of flame balls: a near-equidiffusional flame analysis[END_REF][START_REF] Minaev | On self-drifting flame balls[END_REF] and leads to the free boundary problem

(1.6)                      ∂T ∂t = ∆T + Be -E 2T * δ(r -R(t)) -q(T ) in R 3 ∂Y ∂t = 1 Le ∆Y -Be -E 2T * δ(r -R(t)) in R 3 Y ≡ 0 in B(0, R(t)) q ≡ 0 in R 3 \ B(0, R(t)) Y → Y ∞ , T → T ∞ as |x| → ∞
where R(t) is the radius of the front where the reaction takes place, δ is the Dirac measure and T * is the front temperature. In [START_REF] Buckmaster | The structure and stability of nonadiabatic flame balls[END_REF][START_REF] Buckmaster | The structure and stability of nonadiabatic flame balls: II. Effects of far-field losses[END_REF][START_REF] Lee | The structure and stability of flame balls: a near-equidiffusional flame analysis[END_REF] the authors analyze the stability of stationary solutions of (1.6). In a similar framework, existence and stability of flame balls and travelling flame balls have also been studied in [START_REF] Minaev | On self-drifting flame balls[END_REF][START_REF] Guyonne | Instability in a flame ball problem[END_REF][START_REF] Guyonne | On a model of flame ball with radiative transfer[END_REF][START_REF] Shah | Stability of a spherical flame ball in a porous medium[END_REF][START_REF] Van Den Berg | Flame balls for a free boundary combustion model with radiative transfer[END_REF]. We arrive at (1.1) by introducing the following simplifications: a) assume the radiative loss to be linear, i.e., q(T ) = a(T -T ∞ ) where a > 0, b) approximate e -E RT by η(T -T 0 )BY e -E RT where T 0 is an activation temperature and η is a cut-off function satisfying η > 0 in R + and η ≡ 0 in R -. As in [START_REF] Kagan | On self-drifting flame balls[END_REF][START_REF] Shah | Stability of a spherical flame ball in a porous medium[END_REF] one can model radiative heat losses using Stefan's law q(T ) = ε(T 4 -T 4 ∞ ) for some constant ε > 0. When T is close to T ∞ we can write q(T ) ≈ 4εT 3 ∞ (T -T ∞ ). As a step towars understanding more general situations, we assume that this linear relation holds for all T , that is, we assume a). Other linear or piecewise linear approximations have been used before, for instance in [START_REF] Brailovsky | On stationay and travelling flameballs[END_REF][START_REF] Minaev | On self-drifting flame balls[END_REF][START_REF] Shah | Stability of a spherical flame ball in a porous medium[END_REF]. Assumption b) corresponds the a standard approximation in combustion theory to avoid the cold boundary difficulty, see [START_REF] Berestycki | Mathematical investigation of the cold boundary difficulty in flame propagation theory[END_REF].

After introducing dimensionless variables u, v, corresponding to temperature and reactant concentration, the stationary version of (1.5) becomes

(1.7)        ∆u + vg(u) -cu = 0 in R 3 1 Le ∆v -vg(u) = 0 in R 3 u → 0, v → v ∞ as |x| → +∞
where Le > 0 is the Lewis number, c > 0, v ∞ > 0 and g is an ignition type function, that is, there is θ > 0 such that g(u) = 0 if u ≤ θ.

We stress that our results are valid for any value of Le > 0. Indeed, since we are considering stationary solutions, the following change of variables will allow us to assume that Le

= 1. Letting v(x) = v ∞ ṽ( √ Le x), u(x) = v∞ Le ũ( √ Le x) transforms system (1.7) into (1.1), where ε = c
Le > 0 and f (u) = g( v∞ Le u). We observe that f is still is an ignition type nonlinearity.

Apriori estimates

The purpose in this section is to establish nonexistence results in some ranges of the parameters.

Given a nontrivial solution u, v of (1.1) let β > 0 be such that u(β) = θ and u(r) < θ ∀r > β.

Setting ũ(r) = u(βr), ṽ(r) = v(βr) these new functions satisfy

             ∂ ∂r r 2 ∂u ∂r -εr 2 β 2 u = -r 2 β 2 vf (u) in R + ∂ ∂r r 2 ∂v ∂r = r 2 β 2 vf (u) in R + lim r→∞ u(r) = 0, lim r→∞ v(r) = 1 (2.1)
In the sequel we will study (2.1) in the following set of functions

S = {(u, v) : u, v are C 1 ([0, ∞)
) and u(1) = θ and u(r) < θ for r > 1}.

Let h 0 : R → R be such that

h 0 (t) = 0 ∀t ≤ θ and h 0 (t) > 0 ∀t > θ (2.2) h 0 is nondecreasing. (2.3)
We consider now functions f : R → R such that

f (t) = 0 ∀t ≤ θ, (2.4) f ≥ h 0 in R (2.5) f (t) ≤ C 0 f (u) ∀θ ≤ t ≤ u ≤ 1. (2.6) Lemma 2.1. There is no solution in S to (2.1) if β √ ε > 1 θ . Proof. Since ∆u -εβ 2 u = 0 for r > 1, u(1) = θ, lim r→+∞ u(r) = 0,
we have an explicit formula

u(r) = θ e -(r-1)β √ ε r for all r ≥ 1.
¿From this we find

u ′ (1) = -θ(1 + β √ ε). (2.7)
Similarly, since ∆v = 0 for r ≥ 1 and lim r→∞ v(r) = 1 we have

v(r) = 1 - γ r ∀r ≥ 1
where 0 < γ < 1. This yields v ′ (1) = γ ∈ (0, 1).

Integrating the equation for v in (0,1) implies

β 2 1 0 s 2 vf (u) ds = v ′ (1) = γ. (2.8)
Let r 0 ∈ [0, 1) be such that u ′ (r 0 ) = 0. Then integrating the equation for u in (r 0 , 1) yields

u ′ (1) = εβ 2 1 r0 s 2 u ds -β 2 1 r0 s 2 vf (u) ds.
This formula together with (2.7) and (2.8) gives

v ′ (1) ≥ β 2 1 r0 s 2 vf (u) ds = εβ 2 1 r0 s 2 u ds + θ(1 + β √ ε)
and it follows that

θβ √ ε ≤ 1.
Lemma 2.2. There is β 0 > 0, δ > 0 depending only on h 0 , θ, C 0 such that for all β ≥ β 0 , β √ ε ≤ δ and all f satisfying (2.4), (2.5), (2.6) there is no solution in S to the system (2.1).

Proof. We treat the case ε > 0 since the situation ε = 0 is similar.

As before

u(r) = θ e -(r-1)β √ ε r for all r ≥ 1.
¿From this we find

u ′ (1) = -θ(1 + β √ ε). (2.9) Similarly v(r) = 1 - γ r ∀r ≥ 1
where 0 < γ < 1. This yields

v ′ (1) = γ ∈ (0, 1).
Integrating the equation for u we see that

r 2 u ′ (r) = εβ 2 r 0 s 2 u ds -β 2 r 0 s 2 vf (u) ds. (2.10)
¿From this and (2.9) it follows that

β 2 1 0 s 2 vf (u) ds = θ(1 + β √ ε) + εβ 2 1 0 s 2 u ds ≤ θ + δ(θ + δ). (2.11)
Integrating the equation for v in (0,1) implies

β 2 1 0 s 2 vf (u) ds = v ′ (1) = γ (2.12)
and combined with (2.11) yields

γ ≤ θ + δ(θ + δ). (2.13)
Integrating the equation for v in (0, r) we obtain

v ′ (r) = β 2 r 0 (s/r) 2 v(s)f (u(s)) ds. (2.14) Integrating this on (0, R) with 0 ≤ R ≤ 1 yields v(R) -v(0) = β 2 R 0 r 0 (s/r) 2 v(s)f (u(s)) ds dr = β 2 R 0 R s r -2 drs 2 v(s)f (u(s)) ds = β 2 R 0 (s -1 -R -1 )s 2 v(s)f (u(s)) ds.
In particular, with R = 1

β 2 1 0 (s -s 2 )v(s)f (u(s)) ds = v(1) -v(0) ≤ 1
This and (2.12) give

β 2 1 0 sv(s)f (u(s)) ds ≤ 2. Now going back to (2.14) we obtain v ′ (r) = β 2 r 0 (s/r) 2 v(s)f (u(s)) ds ≤ β 2 r 0 (s/r)v(s)f (u(s)) ds ≤ 2 r 0 < r ≤ 1.
(2.15)

Now consider the function z = u + v which satisfies ∆z = εβ 2 u in R 3 . Integrating the relation (s 2 z ′ ) ′ = εβ 2 s 2 u in (0, r) z ′ (r) = εβ 2 r 0 (s/r) 2 u(s) ds so that 0 ≤ z ′ (r) ≤ δ 2 r r ≥ 0. (2.16) It follows that - 2 r ≤ u ′ (r) ≤ δ 2 r 0 < r ≤ 1. (2.17)
Observe that integrating (2.16) on (r, 1) we find

0 ≤ z(1) -z(r) ≤ δ 2 and since z(1) = θ + 1 -γ 1 + θ -γ -δ 2 ≤ z(r) ≤ 1 + θ -γ 0 ≤ r ≤ 1.
In particular, using (2.13),

v = z -u ≥ 1 + θ -γ -δ 2 -u ≥ 1 -δ(θ + 2δ) -u 0 ≤ r ≤ 1. (2.18)
Step 1. For any r 0 > 0 there exists β 1 (r 0 ) depending on r 0 , h 0 and θ only such that max

[r0,1] u < 1 + θ 2 ∀β ≥ β 1 . (2.19)
To prove this, suppose that max [r0,1] u ≥ 1+θ 2 and let r ∈ [r 0 , 1] be such that u(r) = 1+θ

2 . Let M = 2/r 0 so that from (2.17)

|u ′ (s)| ≤ M r 0 ≤ s ≤ 1. Then 1 + θ 2 - 1 -θ 4 ≤ u(s) ≤ 1 + θ 2 + 1 -θ 4 ∀s ∈ [r 0 , 1] such that |s -r| ≤ 1 -θ 4M .
(2.20)

Using (2.12) and since f ≥ h 0

1 ≥ β 2 1 r0 s 2 v(s)f (u(s)) ds ≥ β 2 1 r0 s 2 v(s)h 0 (u(s)) ds. If r ≤ 1+r0 2 let I = [r -1-θ 4M , r] and if r ≥ 1+r0 2 let I = [r, r + 1-θ 4M ]. Using (2.18) and (2.20): 1 ≥ β 2 I s 2 v(s)f (u(s)) ds ≥ β 2 r 2 0 1 -θ 4M h 0 1 + θ 2 - 1 -θ 4 1 -δ(θ + 2δ) - 3 + θ 4 .
We take δ > 0 small only depending on θ such that

1 -δ(θ + 2δ) - 3 + θ 4 ≥ 1 -θ 8 . (2.21) Then 1 ≥ β 2 r 3 0 (1 -θ) 2 64 h 0 1 + θ 2 - 1 -θ 4
and the claim follows.

Step 2. For any r 0 > 0 there exists C > 0 such that (max

[r0,1] u -θ)h 0 max [r0,1] u -θ 2 + θ ≤ C β 2 .
The constant C depends only on r 0 and θ. The conclusion from this is that lim sup

β→∞ max [r0,1] (u -θ) ≤ 0, (2.22)
and this is uniform with respect to u and f . As before, setting M = 2/r 0 we have |u

′ (s)| ≤ M for r 0 ≤ s ≤ 1 by (2.17). Let r m ∈ [r 0 , 1] be a point such that u(r m ) = max [r0,1] u and let us write u m = u(r m ) = max [r0,1] u. Then u(r) ≥ u m -θ 2 + θ ∀r ∈ [r 0 , 1] such that |r -r m | ≤ u m -θ 2M .
Since f ≥ h 0 , using (2.12), (2.18), (2.19), (2.21) we obtain

1 ≥ β 2 1 0 s 2 v(s)f (u(s)) ds ≥ β 2 1 0 s 2 v(s)h 0 (u(s)) ds ≥ β 2 r 2 0 u m -θ 2M h 0 u m -θ 2 + θ 1 -δ(θ + 2δ) - 1 + θ 2 ≥ β 2 r 2 0 u m -θ 2M h 0 u m -θ 2 + θ 1 -θ 8 .
Thus there exists C depending only on r 0 , θ such that for β ≥ β 1

(u m -θ)h 0 u m -θ 2 + θ ≤ C β 2 . This proves the claim. Step 3. For 0 ≤ a ≤ 1 1 a r 2 (u ′ ) 2 dr ≤ 2a + δ 2 + max [a,1] (u -θ). (2.23)
Indeed multiplying the equation

-(r 2 u ′ ) ′ = -εβ 2 r 2 u + β 2 r 2 vf (u) (2.24)
by (u -θ) and integrating over (a, 1) we get

1 a r 2 (u ′ ) 2 dr = r 2 u ′ (u -θ) 1 a -εβ 2 1 a r 2 u(u -θ) dr + β 2 1 a r 2 vf (u)(u -θ) dr.
By (2.17)

r 2 u ′ (u -θ) 1 a = -a 2 u ′ (a)(u(a) -θ) ≤ 2a. Since u ≤ 1 we have εβ 2 1 a r 2 u(u -θ) dr ≤ δ 2
Finally using (2.12)

β 2 1 a r 2 vf (u)(u -θ) dr ≤ max [a,1] (u -θ)β 2 1 a r 2 vf (u) dr ≤ max [a,1] (u -θ).
Step 4. We finish the proof of the Lemma using a modification of Pohozaev's identity. Let 0 < a < 1 be a fixed number. Multiplying (2.24) by (r -a)u ′ and integrating over (a, 1)

- 1 a (r 2 u ′ ) ′ (r -a)u ′ dr = εβ 2 1 a r 2 (r -a)uu ′ dr + β 2 1 a r 2 (r -a)vf (u)u ′ dr.
(2.25)

A computation shows that the left hand side is given by

1 a (r 2 u ′ ) ′ (r -a)u ′ dr = 1 2 r 2 (r -a)(u ′ ) 2 1 a + 1 2 1 a r 2 (u ′ ) 2 dr -a 1 a r(u ′ ) 2 dr = 1 2 (1 -a)u ′ (1) 2 + 1 2 1 a r 2 (u ′ ) 2 dr -a 1 a r(u ′ ) 2 dr. (2.26) Thanks to (2.23) we find a 1 a r(u ′ ) 2 dr ≤ 1 a r 2 (u ′ ) 2 dr ≤ 2a + δ 2 + max [a,1] (u -θ). Since u ′ (1) = -θ(1 +
√ εβ) we obtain from (2.26), (2.23) and the previous estimate

1 a (r 2 u ′ ) ′ (r -a)u ′ dr - 1 2 (1 -a)θ 2 (1 + √ εβ) 2 ≤ 4a + 2δ 2 + 2 max [a,1]
(u -θ).

(2.27)

Let us compute

εβ 2 1 a r 2 (r -a)uu ′ dr = 1 2 εβ 2 r 2 (r -a)u 2 1 a - 1 2 εβ 2 1 a (3r 2 -2ar)u 2 dr.
Recalling that √ εβ ≤ δ this shows that

εβ 2 1 a r 2 (r -a)uu ′ dr ≤ 3δ 2 . (2.28) Define F (u) = u θ f (t) dt. Then β 2 1 a r 2 (r -a)vf (u)u ′ dr = β 2 r 2 (r -a)vF (u) 1 a -β 2 1 a (3r 2 -2ar)vF (u) dr -β 2 1 a (r 3 -ar 2 )v ′ F (u) dr = β 2 1 a (3r 2 -2ar)vF (u) dr -β 2 1 a (r 3 -ar 2 )v ′ F (u) dr.
These terms can be estimated using (2.6):

F (u) ≤ C 0 f (u)(u -θ) ∀θ ≤ u ≤ 1
and hence by (2.12)

β 2 1 a r 2 vF (u) dr ≤ C 0 1 a r 2 vf (u)(u -θ) dr ≤ C 0 max [a,1] (u -θ) Similarly aβ 2 1 a rvF (u) dr ≤ β 2 1 a r 2 vF (u) dr ≤ C 0 max [a,1] (u -θ).
To estimate the remaining terms we use (2.15)

β 2 1 a r 3 v ′ F (u) dr ≤ 2β 2 1 a r 2 F (u) dr ≤ 16 1 -θ 1 a r 2 vF (u) dr
where the last inequality is consequence of (2.18), (2. [START_REF] Guyonne | On a model of flame ball with radiative transfer[END_REF]) and (2.21) provided β ≥ β 1 (a) (a ∈ (0, 1) is fixed). Hence

β 2 1 a r 3 v ′ F (u) dr ≤ 16C 0 1 -θ max [a,1] (u -θ),
and this implies also

β 2 1 a ar 2 v ′ F (u) dr ≤ β 2 1 a r 3 v ′ F (u) dr ≤ 16C 0 1 -θ max [a,1] (u -θ).
Therefore

β 2 1 a r 2 (r -a)vf (u)u ′ dr ≤ K 1 max [a,1] (u -θ) (2.29)
where K 1 = 5C 0 + 32C0 1-θ . ¿From (2.25), (2.27), (2.28), (2.29) we arrive at

1 2 (1 -a)θ 2 (1 + √ εβ) 2 ≤ 4a + 5δ 2 + (K 1 + 2) max [a,1] (u -θ) ∀β ≥ β 1 (a). (2.30)
We fix δ > 0 even smaller if necessary so that 5δ 2 ≤ 1 3

θ 2 4 .
Then fix 0 < a < 1 2 such that 4a ≤ 1 3

θ 2 4 .
Then for (2.30) yields 1 3

θ 2 4 ≤ (K 1 + 2) max [a,1] (u -θ) ∀β ≥ β 1 (a)
which is not possible for β large enough by (2.22).

Existence of solutions

In this section we prove Theorem 1.1, and throughout it we assume that f is continuous and satisfies (1.2) and (1.3).

We work in the Banach space

X = {(u, β) : u ∈ C 1,α ([0, 1]), u(1) = 0, β ∈ R} endowed with its natural norm (u, β) X = u C 1,α ([0,1]) + |β|, where 0 < α < 1 is fixed. Define for t ∈ [0, 1] f t (u) = tf (u) + (1 -t)(u -θ) + t ∈ [0, 1],
where s + = max(s, 0). Let us consider (2.1) with nonlinearity f t that is

       ∆u -εβ 2 u = -β 2 vf t (u) in R 3 ∆v = β 2 vf t (u) in R 3 lim |x|→∞ u(x) = 0, lim |x|→∞ v(x) = 1 (3.1)
To apply the non-existence results of the previous section we need to exhibit a function h 0 satisfying (2.2) and (2.3). For this purpose define

g(t) = inf {r ∈ [0, 1] : f ≥ t in [r, 1]} for all t ≥ 0, and h(r) = inf {t ≥ 0 : g(t) ≥ r} for all r ∈ [0, 1].
The following properties then follow from these definitions: a) g is strictly increasing, continuous from the left and satisfies lim t→0

+ g(t) = θ. b) h(r) = for r ∈ [0, θ], h(r) > 0 for r ∈ (θ, 1], h is nondecreasing and continuous in [0, 1]. c) f ≥ h in [0, 1]. The function h 0 = min(h, (u -θ) + ) satisfies (2.
2) and (2.3) and the nonlinearity f t satisfies (2.4), (2.5), (2.6) for all t ∈ [0, 1]. Since any solution to (3.1) is bounded above by 1 we may redefine f t (u) as a constant for u ≥ 1. Thus we may assume that for some constant M > 0 we have

|f t (u)| ≤ M ∀u ∈ R, ∀t ∈ [0, 1].
We define a nonlinear map T ε,t : X → X as follows. Let (ũ, β) ∈ X. Then solve

∆v = β2 vf t (ũ + θ) in B 1 , v(1) + v ′ (1) = 1 on ∂B 1 . (3.2)
This problem has a unique solution which can be found for example by minimizing

1 2 B1 |∇v| 2 + β2 f t (ũ + θ)v 2 + 1 2 ∂B1 v 2 - ∂B1 v with v ∈ H 1 (B 1 ). Since f t (ũ + θ) ∈ L ∞ (B 1 ) by standard elliptic regularity v ∈ C 1,α ([0, 1]
). Then find β > 0 and u such that

-∆u + ε β2 u = β 2 vf t (ũ + θ) -ε β2 θ + β 2 -β2 in B 1 u(1) = 0, u ′ (1) = -θ(1 + β√ ε) (3.3)
admits a solution. This problem has a unique solution (u, β) which furthermore is C 1,α ([0, 1]) (this assertion is verified in the proof of Lemma 3.1 below). We define

T ε,t (ũ, β) = (u, β). Observe that (u, β) is a fixed point of T ε,t if and only if (u + θ, v) is a solution of (3.1). Lemma 3.1. Write T (u, β, ε, t) = T ε,t (u, β). Then T : X × [0, ∞) × [0, 1] → X is compact. Moreover T (u, β, ε, t) X ≤ C(1 + sinh( √ εβ) + β 2 ) ∀(u, β) ∈ X, ∀ε ≥ 0, ∀t ∈ [0, 1]. (3.4)
Proof. First we remark that the solution v to (3.2) satisfies

0 ≤ v ≤ 1 in B 1
because 0 is a subsolution and 1 is a supersolution. Given a ≥ 0 let Z a be the solution to

-∆Z a + aZ a = 0 in B 1 , Z a (1) = 1. Note that Z a is explicit: Z a (r) = sinh (
-u ′ (1) = β 2 B1 (vf (ũ + θ) + 1) -β2 B1 (εθ + 1)Z a .
¿From the boundary condition in (3.3) it follows that β is explicitly given by

β 2 = θ(1 + √ ε β) + β2 B1 (εθ + 1)Z a B1 (vf (ũ + θ) + 1)Z a , (3.6)
where a = ε β2 . By (3.5)

B1 (vf (ũ + θ) + 1)Z a ≥ √ ε β sinh( √ ε β ) (3.7)
¿From formulas (3.6), (3.7) we see that

β 2 ≤ C(1 + sinh( √ ε β) + β2 ).
Using this inequality, standard elliptic estimates and the facts that v ≤ 1 and f t is uniformly bounded we obtain (3.4). From here we deduce that T is continuous and compact. Indeed, for the latter assertion, note that if B be a bounded set of

X × [0, ∞) × [0, 1] then T (B) is a bounded set in C 1,µ ([0, 1]) × R for any µ ∈ (0, 1)
and taking µ > α it follows that this set is precompact in C 1,α ([0, 1]) × R.

Lemma 3.2. The operator T 0,0 has a unique fixed point (u 0 , β 0 ) in X.

Proof. In this situation u + v ≡ 1 and hence the system reduces to

-∆u = β 2 (1 -u)(u -θ) + in B 1 , u| ∂B1 = θ
with the additional requirement that

- ∂u ∂ν = θ.
Let w = u -θ. Then the equation for w becomes

-∆w = β 2 (1 -θ -w)w + in B 1 , w| ∂B1 = 0. (3.8)
This equation is of logistic type and many properties are well known (see [START_REF] Cantrell | Spatial ecology via reaction-diffusion equations[END_REF][START_REF] Hutson | Limit behaviour for a competing species problem with diffusion, Dynamical systems and applications[END_REF] It follows that there is a unique β > β * such that

- ∂w β ∂ν = θ.
We call this value β 0 and the let u 0 = w β0 . Then (u 0 , β 0 ) ∈ X is the fixed point of T 0,0 .

Lemma 3.3. Let (u 0 , β 0 ) denote the unique fixed point of T 0,0 found in Lemma 3.2. Then (u 0 , β 0 ) is nondegenerate.

Proof. Let us write (u 0 , β 0 ) ∈ X the solution of I -T 0,0 = 0 of Lemma 3.2. We have to verify that the linearization of I -T 0,0 around this solution is an invertible operator. Note that the operator T 0,0 involves only the nonlinearity f 0 (s) = (s-θ) + and that the fixed point (u 0 , β 0 from Lemma 3.2 satisfies u 0 > 0 in [0, 1) and u ′ 0 (1) < 0. Therefore for ϕ in a neighborhood of u 0 in the topology of C 1,α ([0, 1]) we have f 0 (u 0 + ϕ+ θ) = (u 0 + ϕ) + = u 0 + ϕ. Hence T 0,0 is differentiable at (u 0 , β 0 ).

We next compute the derivative of T 0,0 at (u 0 , β 0 ) in the direction of (ϕ, σ) ∈ X, which we write as (ψ, γ) = DT 0,0 (u 0 , β 0 )(ϕ, σ).

Then

ψ = ψ 2 + σψ 1 , γ = γ 2 + σγ 1 (3.9)
where (ψ 2 , γ 2 ) and (ψ 1 , γ 1 ) are computed as follows.

To compute ψ 2 we have linearize (3.3) with respect to ũ and then set ũ = u = u 0 , β = β = β 0 . Since for t = 0, we have

f 0 (u 0 + θ) = u 0 and f ′ 0 (u 0 + θ) = 1, the function ψ 2 satisfies    -∆ψ 2 = 2β 0 γ 2 v 0 u 0 + β 2 0 ∂v ∂ ũ u 0 + β 2 0 v 0 ϕ + 2β 0 γ 2 in B 1 ψ 2 (1) = 0, ψ ′ 2 (1) = 0, (3.10)
where γ 2 is adjusted so that ψ 2 satisfies both boundary conditions, ∂v ∂ ũ is found solving

     ∆ ∂v ∂ = β 2 0 ∂v ∂ ũ u 0 + β 2 0 v 0 ϕ in B 1 ∂v ∂ ũ (1) + ∂ ∂r ∂v ∂ ũ (1) = 0, (3.11)
and v 0 is the solution of (3.2) with β = β 0 , ũ = u 0 and t = 0. As explained in the proof of Lemma 3.2 then v 0 = 1 -θ -u 0 .

To compute ψ 1 we linearize (3.3) with respect to β and then set ũ = u = u 0 , β = β = β 0 . Therefore ψ 1 satisfies

     -∆ψ 1 = 2β 0 γ 1 v 0 u 0 + β 2 0 ∂v ∂ β u 0 + 2β 0 γ 1 -2β 2 in B 1 ψ 1 (1) = 0, ψ ′ 1 (1) = 0, (3.12)
where γ 1 is chosen so that both boundary conditions are satisfied and ∂v ∂ β is computed from the equation

       ∆ ∂v ∂ β = 2β 0 v 0 u 0 + β 2 0 ∂v ∂ β u 0 in B 1 ∂v ∂ β (1) + ∂ ∂r ∂v ∂ β (1) = 0. (3.13)
We need to find the kernel of I -D ũ, β T 0,0 (u 0 , β 0 ), that is solutions (ϕ, σ) ∈ X of:

D ũ, β T 0,0 (u 0 , β 0 )(ϕ, σ) = (ϕ, σ) (3.
-∆ϕ = β 2 0 (1 -θ -2u 0 ) ϕ + 2σβ 0 (1 -θ -u 0 )u 0 But also ϕ satisfies ϕ(1) = 0, ϕ ′ (1) = 0. The linear elliptic operator -∆-β 2 0 (1 -θ -2u 0 )
with Dirichlet boundary condition arises as the linearization of (3.8) around the solution u 0 , and it is well known that it has a positive first eigenvalue. Since β 0 (1 -θ -u 0 )u 0 > 0, for positive σ we would have ϕ > 0 and ϕ ′ (1) < 0 and for negative σ we would have ϕ < 0 and ϕ ′ (1) > 0, which are both impossible since ϕ ′ (1) = 0. Hence σ = 0 and also ϕ = 0. This shows that the linear equation D ũ, β T 0,0 (u 0 , β 0 )(ϕ, σ) = (ϕ, σ) admits only the trivial solution. Since D ũ, β T 0,0 (u 0 , β 0 ) is compact we see that I -D ũ, β T 0,0 (u 0 , β 0 ) is an isomorphism.

Proof of Theorem 1.1. By Lemma 2.2 there exists δ > 0 and β 1 > 0 such that for all β ≥ β 1 , β √ ε ≤ δ and all t ∈ [0, 1] there is no solution in S to the system (2.1). This means that T ε,t has no fixed point (u,

β) if β ≥ β 1 and β √ ε ≤ δ. Let R 1 = C(1 + sinh(δ) + β 2 1 ) and define Ω 1 = {(u, β) ∈ X : u C 1,α ([0,1]) < R 1 , 0 < β < β 1 }. Let ε 1 = (δ/β 1 ) 2 .
Then Ω 1 is a bounded open set of X and for 0 ≤ ε ≤ ε 1 and 0 ≤ t ≤ 1 the operator T ε,t has no fixed point in ∂Ω 1 . Indeed, suppose (u, β) ∈ ∂Ω 1 is a fixed point of T ε,t . It is not possible that β = β 1 by Lemma 2.2. The case β = 0 is also impossible. This means that 0 < β < β 1 and u C 1,α ([0,1]) = R 1 . But by inequality (3.4) we would have

R 1 = u C 1,α ([0,1]) ≤ C(1 + sinh( √ εβ) + β 2 ) < C(1 + sinh(δ) + β 2 1 ) = R 1 which is impossible. Thus the Leray-Schauder degree deg(I -T ε,t , Ω 1 , 0) is well defined for 0 ≤ ε ≤ ε 1 , t ∈ [0, 1]
. By Lemmas 3.2 and 3.3 we have

deg(I -T 0,0 , Ω 1 , 0) = ±1
and by homotopy invariance

deg(I -T ε,t , Ω 1 , 0) = ±1 ∀0 ≤ ε ≤ ε 1 , ∀t ∈ [0, 1].
This shows T ε,t has at least one fixed point in Ω 1 for 0 ≤ ε ≤ ε 1 and t ∈ [0, 1]. In particular the system (2.1) has a solution with bounded β for any 0 ≤ ε ≤ ε 1 .

We know there exists ε * > 0 such that for ε > ε * the system (2.1) has no solution, and hence T ε,1 has no fixed points in X for such ε. Let 0 < ε 0 < ε 1 and define

Ω 2 = {(u, β) ∈ X : u C 1,α ([0,1]) < R 2 , 0 < β < β 2 } where R 2 = C(1 + sinh( √ 2ε * β 2 ) + β 2 2 ) and β 2 = θ/ √ ε 0 . Then T ε,t has no fixed points on ∂Ω 2 for ε 0 ≤ ε ≤ 2ε * and t ∈ [0, 1]. Then using a homotopy along ε ∈ [ε 0 , 2ε * ] we find deg(I -T ε,1 , Ω 2 , 0) = 0 ∀ε 0 ≤ ε ≤ 2ε * .
This implies that for ε 0 ≤ ε ≤ ε 1 the system has at least another solution and that this solution has β in the range δ/ √ ε ≤ β ≤ θ/ √ ε. Since ε 0 is arbitrary we obtain the same conclusion for 0 < ε ≤ ε 1 .

Existence when f is discontinuous

In this section we prove Theorem 1.2. We assume that f satisfies (1.2), (1.4), (1.3) and that

η = lim u→θ + f (u) > 0. For n ≥ 1 define f n (t) = min(f (t), n(t -θ) +
). Then f n is continuous and satisfies (1.2) and (1.3) with a fixed constant. Moreover there is h 0 satisfying (2.2) and (2.3) and such that f n ≥ h 0 for all n. Such h 0 can be taken for instance as h 0 (t) = min((t -θ) + , σ) for some σ > 0 small enough.

By Theorem 1.1 there is ε 1 > 0 such that the system (2.1) admits 2 solutions (u

1 n , v 1 n ), (u 2 n , v 2
n ) for all ε ∈ (0, ε 1 ). Moreover there is a fixed number β 1 such that

β 1 n ≤ β 1 and β 2 n ∈ (δ/ √ ε, 1/(θ √ δ)
) for all n ≥ 1 and ε ∈ (0, ε 1 ). From now on we fix ε ∈ (0, ε 1 ) and study the limit as n → ∞ of any of the 2 solutions which we call just (u n , v n ) with parameter β n . Since β n is bounded and by standard elliptic estimates we may assume that Setting ϕ(r) = 0 for r > 1 we see that u, v satisfy

β n → β, u n → u and v n → v in C 1,α ([0, 1]). Since f n (u n ) is uniformly bounded we may extract a further subsequence such that f n (u n ) → ϕ weakly-* in L ∞ (0, 1). The function ϕ then satisfies 0 ≤ ϕ ≤ max [0,1] f in [0, 1]. Since
       ∆u -εβ 2 u = -β 2 vϕ in R 3 ∆v = β 2 vϕ in R 3 lim |x|→∞ u(x) = 0, lim |x|→∞ v(x) = 1 The set D = {r ∈ [0, ∞) : u(r) = θ} is open (relative to [0, ∞)) and for any r ∈ D we have f n (u n (r)) → f (u(r)) as n → ∞. It follows that ϕ = f (u) in D. Outside D we have ϕ ≤ η a.e. Indeed, let ψ ∈ C ∞ 0 ((0, 1)), ψ ≥ 0. Then 1 0 f n (u n )ψ = D∩[0,1] f n (u n )ψ + D c ∩[0,1] f n (u n )ψ. Since f n (u n ) → f (u) in D by dominated convergence we have D∩[0,1] f n (u n )ψ → D∩[0,1] f (u)ψ. On the other hand, if r ∈ D c lim sup n→+∞ f n (u n (r)) ≤ η so that (f n (u n )-η) + → 0 on D c . Then D c ∩[0,1] (f n (u n ) -η)ψ = D c ∩[0,1] (f n (u n ) -η) + ψ - D c ∩[0,1] (f n (u n ) -η) -ψ ≤ o(1)
where o(1) denotes a sequence converging to 0 as n → +∞. It follows that

D c ∩[0,1] f n (u n )ψ ≤ D c ∩[0,1] ηψ + o(1)
and hence

1 0 ϕψ = lim n→+∞ 1 0 f n (u n ) ≤ D∩[0,1] f (u)ψ + D c ∩[0,1]
ηψ.

This shows that ϕ ≤ η a.e. in D c . Our main objective is to show that the complement of D is finite. If u ′ (r) = 0 whenever u(r) = θ then D c is discrete and since it is contained in [0, 1] it is finite.

Let us analyze the case where for some r ∈ [0, 1] we have u(r) = θ and u ′ (r) = 0. Let r 0 = sup {r > 0 : u(r) = θ, u ′ (r) = 0}. Then r 0 < 1 and u(r 0 ) = θ, u ′ (r 0 ) = 0. We assert that there is a small interval (r 0 , r 0 + σ), σ > 0 such that u > θ in (r 0 , r 0 + σ). To prove this we start ruling out the possibility that u(r n ) = θ for some infinite sequence r n ց r 0 . We actually may assume that if n is even then u > θ on (r n+1 , r n ) and if n is odd u < θ on (r n+1 , r n ). Let us see that the following holds

u ′ (r n+1 ) = -u ′ (r n ) + O(ℓ 2 n ) (4.1)
where ℓ n = r n+1 -r n and O(ℓ 2 n ) denotes a sequence bounded by Cℓ 2 n with C independent of n as n → ∞. Suppose first that u < θ on (r n+1 , r n ) and define ũn by

u(r) = θ + ℓ 2 n ũn ((r -r n+1 )/ℓ n ) r ∈ [r n+1 , r n ]. Then ũ′′ n + 2ℓ n rℓ n + r n+1 ũ′ n = εβ 2 (ℓ 2 n ũn + θ) for r ∈ [0, 1].
ũn (0) = ũn (1) = 0. This equation implies that ũn and ũ′ n are uniformly bounded on [0, 1]. Therefore

ũ′′ n = εβ 2 θ + O(ℓ n ) for r ∈ [0, 1] hence integrating ũn (r) = 1 2 εβ 2 (1 + O(ℓ n ))r(r -1) for r ∈ [0, 1] and ũ′ n (r) = 1 2 εβ 2 (1 + O(ℓ n ))(2r -1) for r ∈ [0, 1]. (4.2)
This proves (4.1) in this case. Now suppose that u > θ on (r n+1 , r n ). Defining ũn as before we now obtain the following equation

ũ′′ n + 2ℓ n rℓ n + r n+1 ũ′ n = εβ 2 (ℓ 2 n ũn + θ) -β 2 ṽn f (θ + ℓ 2 n ũn ) for r ∈ [0, 1]. ũn (0) = ũn (1) = 0 where ṽn (r) = v(r n+1 + ℓ n r) r ∈ [0, 1].
This equation implies that ũn , ũ′ n are uniformly bounded on [0, 1]. Thus

ũ′′ n = εβ 2 θ -β 2 ṽn (0)f (θ + ℓ 2 n ũn ) + O(ℓ n ) for r ∈ [0, 1]
. Multiplying this equation by ũn and integrating in [0, 1] yields

1 2 (ũ n (1) ′ ) 2 - 1 2 (ũ n (0) ′ ) 2 = O(ℓ n ) which shows that ũn (1) ′ = -ũ n (0) ′ + O(ℓ n )
and proves (4.1) in this case. Using (4.1) inductively we find

u ′ (r n ) = (-1) k u ′ (r n+k ) + n+k-1 j=n O(ℓ 2 j ) ∀n ≥ 1, k ≥ 1.
Letting k → +∞, and using that u ′ (r 0 ) = 0 yields

u ′ (r n ) = ∞ j=n O(ℓ 2 j ) ∀n ≥ 1. Choose a subsequence n i → ∞ as i → ∞ such that ℓ ni ≥ ℓ j for all j ≥ n i . Then |u ′ (r ni )| ≤ Cℓ ni ∞ j=ni ℓ j . But (4.2) implies that |u ′ (r n )| = ℓ n εβ 2 (1 + O(ℓ n ))/2. It follows that ℓ ni εβ 2 (1 + O(ℓ ni ))/2 ≤ Cℓ ni ∞ j=ni ℓ j
which is impossible as i → ∞. This establishes the assertion that u cannot oscillate infinitely many times around θ to the right of r 0 . It follows that for some σ > 0 either u > θ in (r 0 , r 0 + σ) or u < θ in (r 0 , r 0 + σ). The latter can in fact not occur by the Hopf lemma.

Since u satisfies the ODE u ′′ + 2 r u ′ = εβ 2 u -β 2 vf (u) in (r 0 , r 0 + σ) and in this interval u > θ we see that u is C 2 in [r 0 , r 0 + σ). Since r 0 is a minimum of u restricted to [r 0 , r 0 + σ) it follows that lim r→r + 0 u ′′ ≥ 0. Which yields the following inequality

εβ 2 θ -β 2 v(r 0 )η ≥ 0. (4.3) Now, if v is constant in [0, r 0 ]
this means that ϕ = 0 a.e. in [0, r 0 ] and u ≤ θ in [0, r 0 ]. Actually u < θ in [0, r 0 ) by the strong maximum principle and this finishes the proof in this case. If v is not constant, then v(r) < v(r 0 ) for all r ∈ [0, r 0 ). Let us verify that D c is discrete. Suppose that for some r 1 ∈ [0, r 0 ) we have u(r 1 ) = θ. If u ′ (r 1 ) = 0 then r 1 is isolated. If u ′ (r 1 ) = 0 then integrating the equation for u we have

r 2 u ′ (r) = r r1 t 2 (εβ 2 u -β 2 vϕ) dt. For r near r 1 we have if u(r) < θ then εβ 2 u(r) -β 2 v(r)ϕ(r) > 0 if u(r) = θ then εβ 2 u(r) -β 2 v(r)ϕ(r) > εβ 2 θ -β 2 v(r 0 )η ≥ 0 if u(r) > θ then εβ 2 u(r) -β 2 v(r)ϕ(r) = εβ 2 u(r) -β 2 v(r)f (u(r)).
In the second line above we have used that v(r) < v(r 0 ), ϕ ≤ η and (4.3). In the third line above we may say that εβ 2 u(r) -β 2 v(r)f (u(r)) > 0 if r is sufficiently close to r 1 by continuity, v(r) < v(r 0 ) and (4.3) (we may regard f as continuous here since we are working with values above θ). This shows that u is strictly convex in a neighborhood of r 1 and hence there are no other points in D c close to r 1 . This shows that D c is discrete, hence finite, and finishes the proof of the theorem.

The Heaviside ignition

In this section we perform explicit calculations for the ignition nonlinearity f (u) = χ [u>θ] , where 0 < θ < 1. We first reduce the differential equations to a finite number of equations in some parameters. In a second part we obtain rigorously the bifurcation diagram of Figure 1 for small ε.

We rewrite (2.1) in the form

       ∆u -β 2 εu = 0 in (1, +∞) ∆v = 0 in (1, +∞) u(1) = θ, v(1) = γ u(+∞) = 0 v(+∞) = 1 (5.1) and 
          ∆u -β 2 εu = -β 2 vf (u) in (0, 1) ∆v = β 2 vf (u) in (0, 1) u(1) = θ v(1) = γ u ′ (0) = 0 v ′ (0) = 0 u ′ (1 -) = u ′ (1 + ) v ′ (1 -) = v ′ (1 + ) (5.2)
where γ is a parameter to be adjusted.

Observe that problem (5.1) can explicitly be solved. Indeed, the following functions solve (5.1)

u = θe -β √ ε(r-1) r v = 1 - 1 -γ r .
Hence the last condition in (5.2) becomes

u ′ (1 + ) = -θ(1 + β √ ε) v ′ (1 + ) = 1 -γ.
5.1. Solving (5.2) with the assumption u > θ in B(0, 1). Assuming that u > θ, in B(0, 1) the subsystem (5.2) is reduced to the following linear problem ∆u -

β 2 εu = -β 2 v in B(0, 1) (5.3) ∆v = β 2 v in B(0, 1) (5.4) u(1) = θ v(1) = γ (5.5) v ′ (1 -) = 1 -γ. (5.6) u ′ (1 -) = -θ(1 + β √ ε) (5.7)
For ε = 1, observe that the two following functions v := 1 βch(β) sh(βr) r r ∈ (0, 1) (5.8)

u := - v 1 -ε + θ + th(β) β(1-ε) sh(β √ ε)
sh(β √ εr) r r ∈ (0, 1) (5.9) solve equations (5.3)- (5.6). ¿From the formula for v we can see that

γ = th(β) β .
Now to obtain a solution of subsystem (5.2), it then remains to adjust β and ε to get (5.7). This equation is

-θ(1 + β √ ε) = u ′ (1 -) = v ′ (1) 1 -ε + θ -th(β) β(1-ε) sh(β √ ε) (β √ εch(β √ ε) -sh(β √ ε))
which after simplification yields (5.10)

θ(1 -ε)(1 + th(β √ ε)) + th(β) β = th(β √ ε) β √ ε .
Let us define

g(ε, β) := θ(1 -ε)(1 + th(β √ ε)) + th(β) β - th(β √ ε) β √ ε . (5.11)
It can be shown that there is ε * > 0 that for ε ∈ (0, ε * ), g(ε, •) has 2 zeros, β - ε < β + ε . The function g(0, •) has a unique zero β 0 = lim ε→0 β - ε . A calculation (see Lemma 5.1) shows that for ε > 0 small and β = β - ε the solution u obtained by the formula (5.9) satisfies u > θ in [0, 1). Along the lower branch β - ε , 0 ≤ ε ≤ ε * we observe numerically that u > θ in [0, 1). After reaching ε * we continue the curve along the upper branch β + ε now decreasing ε from ε * to 0 until we reach a critical value ε = ε 0 where the corresponding solution given by (5.9) satisfies u > θ in (0, 1) but u(0) = θ. 5.2. Solving the subsystem (5.2) with the assumption u > θ in B(0, 1) \ B(0, η) and u ≤ θ in B(0, η) for some η > 0.

Under this assumption the problem (5.2) decouples again into two subsystems

           ∆u -β 2 εu = -β 2 v in B(0, 1) \ B(0, η) ∆v = β 2 v in B(0, 1) \ B(0, η) u(1) = θ = u(η) v(1) = γ v(η) = δ u ′ (1 -) = u ′ (1 + ) v ′ (1 -) = v ′ (1 + ) u ′ (η -) = u ′ (η + ) v ′ (η -) = v ′ (η + ) (5.12) and    ∆u -β 2 εu = 0 in B(0, η) ∆v = 0 in B(0, η) u(η) = θ v(η) = δ (5.13)
where δ, γ, η are parameters to be found. Using the conditions at r = η and r = 1 we will be able to reduce the parameters to only β and η which will be implicitly defined as functions of ε by 2 equations.

Observe that since v is harmonic in B(0, η) with the Dirichlet condition v(η) = δ, we deduce that v ≡ δ and therefore v ′ (η-) = 0.

Let us now solve (5.12) explicitly. First, observe that the function v = A sh(βr) r + B ch(βr) r satisfies ∆v -β 2 v = 0 in B(0, 1) \ B(0, η).

Therefore, choosing A and B such that

γ = A sh(β) + B ch(β) 1 -γ = A[βch(β) -sh(β)] + B[βsh(β) -ch(β)],
it follows that v satisfies the right boundary conditions on ∂B(0, 1). Solving for A and B we find

A = ch(β) -γβsh(β) β and B = sh(β) -γβch(β) β .
To obtain a full solution it remains to adjust γ in such way that v ′ (η) = 0. But ¿From (5.13) we know that in B(0, η), we have the following

v ′ = A[ βch(βr) r - sh (
u = θη sh(ηβ √ ε) sh(rβ √ ε) r .
Therefore,

u ′ (η) = θ β √ εch(ηβ √ ε) sh(ηβ √ ε) - 1 η = θch(ηβ √ ε) ηsh(ηβ √ ε) [ηβ √ ε -th(ηβ √ ε)]. (5.15)
Let us now find the solution u in B(0, 1) \ B(0, η). As for the construction of v, let us observe that for ε = 1,

u := 1 ε -1 v + E sh(rβ √ ε) r + F ch(rβ √ ε) r (5.16) solves the following equation ∆u -β 2 εu = -β 2 v
By taking

E(β, η, γ) := ch(β √ ε) -γβ √ εsh(β √ ε) (1 -ε)β √ ε -θ ch(β √ ε) + sh(β √ ε) (5.17) F (β, η, γ) := - sh(β √ ε) -γβ √ εch(β √ ε) (1 -ε)β √ ε + θ ch(β √ ε) + sh(β √ ε) (5.18)
we easily verify that u(1) = θ and u ′ (1 -) = -θ(1 + √ εβ). It remains now to impose the boundary condition on ∂B(0, η). From the formula for u (5.16) we have

u(η) = δ ε -1 + E sh(ηβ √ ε) η + F ch(ηβ √ ε) η u ′ (η) = Eβ √ ε ch(β √ εη) η + F β √ ε sh(β √ εη) η -E sh(β √ εη) η 2 -F ch(β √ εη) η 2 (5.19) since v ′ (η) = 0. Thus the equation u(η) = θ becomes δ ε -1 + E(β, η, γ(β, η)) sh(ηβ √ ε) η + F (β, η, γ(β, η)) ch(ηβ √ ε) η = θ (5.20)
and (5.15) combined with (5. [START_REF] Guyonne | On a model of flame ball with radiative transfer[END_REF]) becomes

E(β, η, γ(β, η))ηβ √ εch(β √ εη) + F (β, η, γ(β, η))ηβ √ εsh(β √ εη) -E(β, η, γ(β, η))sh(β √ εη) -F (β, η, γ(β, η))ch(β √ εη) = θ β √ εcoth(ηβ √ ε) - 1 η . (5.21)
We solve numerically equations (5.20) and (5.21) with β and η as unknowns that depend on ε, for 0 < ε < ε 0 , where ε 0 is the critical value of ε described in Subsection 5.1. The result from this numerical computation is shown in Figure 1.

5.3.

Results for small ε. Lemma 5.1. For ε > 0 suitably small β - ε is a smooth function of ε and the solution given by (5.9) corresponding to β = β - ε satisfies u > θ in [0, 1).

Proof. Let β 1 the unique solution of the equation

th(β) β = (1 -θ) 2 . Then g(ε, β 1 , θ) = θ(1 -ε)(1 + th(β 1 √ ε)) + 1 -θ 2 - th(β 1 √ ε) β 1 √ ε = θ(1 -ε)th(β 1 √ ε)) + 1 + θ -2θε 2 - th(β 1 √ ε) β 1 √ ε = θ(1 -ε)th(β 1 √ ε)) - 1 -θ 2 -θε + [1 - th(β 1 √ ε) β 1 √ ε ] For β 1 √ ε < 1, using that β 1 √ ε -(β1 √ ε) 3 3 ≤ th(β 1 √ ε) ≤ β 1 √ ε, we end up with g(ε, β 1 , θ) ≤ θ(1 -ε)β 1 √ ε - 1 -θ 2 -θε + (β 1 √ ε) 2 3 ≤ (β 1 √ ε) 2 3 + θβ 1 √ ε - 1 -θ 2 -θε.
Computing now the positive roots of

X 2 3 + θX -1-θ 2 , yields g(ε, β 1 , θ) < 0 for √ εβ 1 ≤ 3θ 1+ 2(1-θ) 3θ 2 -1 2
. Therefore, we achieve g(ε, β 1 , θ) < 0, for ε in (0, min{

  3θ 1+ 2(1-θ) 3θ 2 -1 2β1   2 , 1 β 2 1 }).

Let us denote

ε 1 :=     3θ 1 + 2(1-θ) 3θ 2 -1 2β 1     2 .
Hence, for any (ε, θ) in (0, min{ε 1 , 1 β 2 1 }) × (0, 1) fixed there exists two possible solution to g(ε, β, θ) = 0. Moreover, β -< β 1 < β + .

To obtain a solution, to problem (5.2), we still need to show that the constructed solutions effectively satisfies the conditions u > θ in B(0, 1). Let us observe that u -θ satisfies:

∆(u -θ) -εβ 2 (u -θ) = -β 2 (v -εθ).
If v(0) -εθ > 0 then using the maximum principle, it follows that (u -θ) > 0.

Since v(r) = 1 ch(β)β sh(βr) r

, it follows that v(0) = 2 ch(β) . Hence, we end up with the condition 2 ch(β) ≥ εθ.

We can conclude with a final estimate on ε, namely since β -< β 1 , and 1 ch(β) is a decreasing function, we have the following uniform estimates on ε ε ≤

(1 -θ)β 1 θsh(β 1 ) .

We therefore have construct a solution to the problem (1), when (ε, θ) belongs to (0, ε 0 (θ)) × (0, 1) where

ε 0 (θ) := min{ 1 β 2 1 ; (1 -θ)β 1 θsh(β 1 ) ; ε 1 ; (1 -θ) 2 2 }
Lemma 5.2. For ε > 0 small the system (5.20), (5.21) has a unique solution β + ε , η with η > 0, β + ε is a smooth function of ε and u defined by (5.16) satisfies u > θ in (η, 1).

Proof. The 2 equations (5.20) and (5.21) can be written in the form

E sinh(aη) + F cosh(aη) = ηθ + ηδ 1 -ε E (aη cosh(aη) -sinh(aη)) + F (aη sinh(aη) -cosh(aη)) = ηθ aη tanh(aη) -1 where a = β √ ε.
Instead of ε, β, η consider the variables t, a, x defined by the following relations

t = √ ε, a = β √ ε, η = 1 - x β .
Consider also the function G(t, a, x) = (G 1 (t, a, x), G 2 (t, a, x)) where

G 1 (t, a, x) = β E sinh(aη) + F cosh(aη) -ηθ - ηδ 1 -ε G 2 (t, a, x) = E (aη cosh(aη) -sinh(aη)) + F (aη sinh(aη) -cosh(aη)) -ηθ aη tanh(aη) - 1
where δ, γ, E, F are as before. We see that we have a solution to the system if and only if G(t, x, a) = 0. Using the fact that lim which is seen to have a unique solution a 0 ∈ (0, 1/θ -1) since the left hand side defines a strictly increasing function of a which goes to zero as a → 0 and goes to +∞ as a ր 1/θ -1. Now the equation G 1 (0, a 0 , x) = 0 can be written as 1 -1 x tanh(x) + 1 x sinh(x) = θ(1 + a 0 ) ∈ (0, 1). (5.22) The above left hand side is strictly increasing with limits 1/2 as x → 0 and 1 as x → ∞. Using the inequality a tanh(a)

-1 ≤ a ∀a ≥ 0 we see that 1 -θ -θa 0 ≤ θa 0 which yields a 0 ≥ 1-θ 2θ . Hence

θ(1 + a 0 ) ≥ 1 2 + θ 2 > 1 2 .
Thus (5.22) has a unique solution x 0 ∈ (0, ∞).

Estimate on ε *

As mentioned before the largest value of ε such that (1.1) has a nontrivial solution satisfies ε * ≤ ε 0 where ε 0 = inf{ε > 0 : g ε (s) ≤ 0 ∀s ∈ [0, 1]}, and g ε (s) := (1 -s)f (s) -εs.

In the particular case of f (u) = χ [u>θ] , 0 < θ < 1 we obtain the value ε 0 = 1 θ . This estimate can be sharpened.

For this, given ε > 0 let us introduce For 0 < ε < ε 0 define also a(ε) as the smallest zero of g ε in the interval (0, 1). We b(ε) be the largest b in [0, 1] such that g ε is positive on (a(ε), b(ε)). There is a vast literature concerning the existence of positive solutions of the above equation. We will just mention a necessary and sufficient condition for the existence of positive solutions. Proof of Proposition 6.1. Fix ε > ε 1 . From our assumption, we have G(b(ε), ε) < 0. Let g a smooth function such that gε ≥ g ε and b(ε) 0 gε (s) ds < 0. This is always possible since G(b(ε), ε) < 0. Observe now that from the necessary and sufficient condition given in theorem 6.2, the following scalar problem (6.3) ∆w = -g ε (w) in Ω w = 0 on ∂Ω does not have any positive solution for any bounded domain Ω. Now, we argue by contradiction, assume there exists a non trivial couple (u, v) solution of the system (1.1). An easy computation shows that u is a sub-solution of the problem (6.3) with Ω = R n . Take δ > 0 small, and define g ε,δ = gε (s + δ).

So for δ small enough,(i.e δ < δ 0 ),we have: Therefore g ε,δ does not satisfy the necessary existence condition of Theorem 6.2. Now, take Ω = B(0, R). Since u goes uniformly to 0 when |x| goes to infinity, we have sup ∂Ω u → 0 as R → +∞. Thus, with δ = sup ∂Ω u < δ 0 , u δ = u -δ is a sub-solution of the following scalar problem: (6.4) ∆w = -g δ (w) in B(0, R) w = 0 on ∂B(0, R)

Observe that the constant b(ε) -δ is a super-solution of (6.4) and u δ < b(ε) -δ, then we can apply the monotone iterative scheme to obtain at least one positive solution which contradicts theorem 6.2.

In the case f (u) = χ [u>θ] where 0 < θ < 1 with a similar argument we can show that for ε > (1-θ) 2

2

the system (1.1) has no nontrivial solution. Indeed, given any σ > 0 choose a smooth function f ≥ f such that f (u) = 1 for u ≥ θ and f (u) = 0 for u ≤ θ -σ. With the same argument as before, (1.1) has no nontrivial solution for ε > ε1 where ε1 is given by (6.1) with f replaced by f . A computation then shows that as σ → 0, ε1 → (1-θ) 2 2 .

  , β) = T (ũ, β, ε, t). Then choosing a = ε β2 , multiplying (3.3) by Z a and integrating in B 1 we find

  ): a) Any solution w to (3.8) satisfies 0 ≤ w < 1 -θ, and either w ≡ 0 or w > 0 in B 1 . b) (3.8) has a nontrivial solution if and only if β > β * where β * is such that (β * ) 2 (1 -θ) = λ 1 and λ 1 denotes the first Dirichlet eigenvalue of -∆ in the unit ball. c) For β > β * there is a unique non-trivial solution, which we write as w β . d) w β is monotone increasing with respect to β and lim β→∞ w β = 1 -θ uniformly on compact sets of B 1 and lim β→β * w β = 0. ¿From the above properties it follows that lim β→∞

14 )

 14 Using the notation above we may writeϕ = ψ 2 + σψ 1 , σ = γ 2 + σγ 1 . We claim that To prove this, let z = ϕ + ∂v ∂ ũ + σ ∂v ∂ β .Then a calculation shows that ∆z = 0 in B 1 and z(1) + z ′ (1) = 0. This implies z = 0 and (3.15) follows. ¿From the formulas (3.9)-(3.13) we get

  u n satisfies a linear equation for r > 1 and vanishes at infinity it has an explicit formula u n (r) = θ e -(r-1)βn √ ε r for all r ≥ 1.

  βr) r 2 ] + B[ βsh(βr) r -ch(βr) r 2 ] and setting v ′ (η) = 0 in the above equation yields βη) -sh(βη) + B ηβsh(βη) -ch(βη) = 0 Substituting now the values of A and B in the above equation, it follows ch(β) -γβsh(β) C -sh(β) -γβch(β) D = 0, where C := ηβch(βη) -sh(βη) D := ηβsh(βη) -ch(βη) Therefore, we have (5.14) γ = C -Dth(β) β Cth(β) -D which after simplification reads γ = γ(β, η) = ηβ -th(ηβ) -th(β) ηβth(βη) -1 β ηβ -th(ηβ) th(β) + 1 -ηβth(ηβ).Thus we may compute now δ = v(η):

1

 1 difficult to verify that G maybe extended in a C 1 manner for t = 0, a, x ∈ (0, ∞) with the valuesG 1 (0, a, x) = -x + θax + 1 tanh(x) + xθ -1 sinh(x) G 2 (0, a, x) = 1 -θ(1 + a) -θ a tanh(a) -Note that the equation G 2 (0, a, x) = 0 is equivalent to a tanh(a) -1 1 -θ(1 + a) = 1 θ

G ε (b) = b 0 g

 0 ε (s) ds.

Proposition 6 . 1 .

 61 Assume that f satisfies (1.2) and is continuous. Then (1.1) has no nontrivial solution if ε > ε 1 whereε 1 = sup{ε > 0 | G ε (b(ε)) > 0}. (6.1) Before proving the above proposition, let us recall a result on the scalar problem on a bounded domain Ω ⊂ R N : (6.2) -∆w = h(w) in Ω w = 0 on ∂Ω.

Theorem 6 . 2 .

 62 (Clément and Sweers [14]) Let Ω ⊂ R N be a bounded smooth domain and h ∈ C 1 (R). Assume there exists b 1 < b 2 with b 2 > 0 such that h(b 1 ) = h(b 2 ) = 0 and h > 0 on (b 1 , b 2 ). Then there exists positive solution w to (6.2) with max w ∈ (b 1 , b 2 ) if and only if b2 b h(s)ds > 0 for all b ∈ [0, b 2 ).

G

  δ ( b(ε) -δ) = b(ε)-δ 0 g ε,δ (s)ds = b(ε) δ gε (s)ds = G( b(ε), ε) -δ 0 gε (s)ds ≤ 0

Acknowledgment. J. Coville warmly thanks Professor Sivashinsky for proposing the formulation of this problem and enlightening discussions. He is also indebted to Professor Dolbeault for useful conversations on this problem.

J. Coville was partially supported Tel Aviv University, Université Paris Dauphine and Max Planck Institute for Mathematics in the Science. J. Dávila was supported by Fondecyt 1057025, 1090167. Both authors acknowledge the support of Ecos-Conicyt project C05E04.