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Separable solutions of some quasilinear equations with source reaction

Marie-Françoise Bidaut-Veron, Mustapha Jazar, Laurent Veron

To cite this version:

Marie-Françoise Bidaut-Veron, Mustapha Jazar, Laurent Veron. Separable solutions of some quasilin- ear equations with source reaction. Journal of Differential Equations, Elsevier, 2008, 244, pp.274-308.

�hal-00282539�

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hal-00282539, version 1 - 27 May 2008

Separable solutions of some quasilinear equations with source reaction

Marie-Fran¸ coise Bidaut-V´ eron

Department of Mathematics, Universit´e Fran¸cois Rabelais, Tours, FRANCE

Mustapha Jazar

Department of Mathematics, Universit´e Libanaise, Beyrouth, LIBAN

Laurent V´ eron

Department of Mathematics, Universit´e Fran¸cois Rabelais, Tours, FRANCE

AbstractWe study the existence of singular solutions to the equation−div(|Du|p−2Du) =|u|q−1u under the formu(r, θ) =rβω(θ),r >0, θ∈SN−1. We prove the existence of an exponentqbelow which no positive solutions can exist. If the dimension is 2 we use a dynamical system approach to construct solutions.

1991 Mathematics Subject Classification. 35K60, 34.

Key words. p-Laplacian, Singularities, Phase-plane analysis, Poincar´e map, Painlev´e integral

1 Introduction

Letp >1,q >0 and Ω be an open subset ofRN. Ifu∈C(Ω\ {0})∩C1(Ω) is a solution of div

|Du|p2Du

+|u|q1u= 0 (1.1)

in Ω which vanishes on ∂Ω\ {0}, it may develop a boundary singularity at x = 0. The description of such boundary singularities should play a key role in the understanding of the boundary behaviour of solutions of these equations (see [3] for the equation with an absorption reaction term and [6] for the old results in the casep= 2). The interesting range for exponent q corresponds to a dominent reaction term, that is q > p−1, which will be always assumed. If Ω =RN

+ is a half space, or more generally a cone with vertex 0 (and this is most often the asymptotic form of the domain Ω at 0), a natural way to look for specific solutions of (1.1 ) is to look for solutions which can be written under the form

u(x) =u(r, σ) =rβω(σ) r >0, σ∈SN1. (1.2) For equations with an absorption nonlinearity

div

|Du|p2Du

− |u|q1u= 0, (1.3)

Supported by a grant from the Lebanese University

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such separable solutions are the key-stone elements for describing the boundary singularities of solutions of (1.3 ), as it is shown in [3]. It is expected that such will be the case for (1.1 ), even if the full theory will be much more difficult in particular due to the absence of comparison principle and simple a priori estimates of Keller-Osserman type. However, for such a solution, it is straightforward to check that

β = p

q+ 1−p:=βq, (1.4)

andω must be a solution of

−∇.

βq2ω2+|∇ω|2p/21

ω

− |ω|q1ω=λq,p

β2qω2+|∇ω|2p/21

ω, (1.5) inSN1, where∇ is the covariant gradient onSN1,∇.the divergence operator acting on vector fields onSN1 and

λq,pq(qβq−N).

Whenp= 2,βq = 2/(q−1) and (1.5 ) becomes

−∆ω− |ω|q1ω=λq,2ω, (1.6) where ∆ is the Laplace-Beltrami operator onSN1 and

λq,2= 2 q−1

2q q−1 −N

.

IfSis a subdomain ofSN1, equation (1.6 ), considered inS, is the Euler-Lagrange variation of the functional

I(ψ) = Z

S

1

2|∇ψ|2q,2

2 ψ2− 1

q+ 1|ψ|q+1

dσ. (1.7)

For any 1 < q < (N+ 1)/(N −3) (any q >1 if N = 2 or 3) this functional satisfies the Palais-Smale condition. Furthermore, if λq,2 < λS,2, (λS,2 is the first eigenvalue of−∆ in W01,2(S)), Ambrosetti-Rabinowitz theorem [1] or Pohozaev fibration method [12], [13] apply and yield to the existence of non-trivial positive solutions to (1.6 ) in S vanishing on ∂S;

while ifλq,2≥λS,2 no such solution exists.

Whenp6= 2, equation (1.5 ) cannot be associated to any functional defined on SN1, except if q = qc = (N(p−1) +p)/(N −p) (the critical Sobolev exponent −1 for W1,p, when N > p); therefore, finding functions satisfying it is not straitforward. Besides the constant solutions which exist as soon as qβq < N, it is not easy to find other solutions of this equation. However a classical method to construct signed solutions is to start from a fundamental simplicial domain S ⊂ SN1, corresponding to a tesselation of the sphere associated to a finite group of isometries generated by symmetries through hyperplanes, to look for one positive solution of (1.5 ) in S vanishing on ∂S, and then to extend it by reflexions accross∂S. It is natural to associate to S the sphericalp-harmonic spectral

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equation onSwhich is to find a couple (β, φ) whereβ >0,φ∈C1( ¯S) is a positive solution of





−∇.

β2φ2+|∇φ|2p/21

φ

β2φ2+|∇φ|2p/21

φ in S φ= 0 in∂S,

(1.8)

andλ=β(β(p−1) +p−N). Equivalently, this means that the functionv(r, σ) =rβφ(σ) is a positive p-harmonic function into the coneCS ={(r, σ) :r >0, σ∈S} which vanishes on∂S. Given a smooth subdomainS ⊂SN1, it is proved in [15], following the method of Tolksdorff [14], that there exists a couple (β, φ) = (βS, φS), whereβS is unique and φS up to an homothethy, such that (1.8 ) holds. Denoting

λSSS(p−1) +p−N),

the couple (φS, λS) is the natural generalization of the first eigenfunction and eigenvalue of the Laplace-Beltrami operator inW01,2(S) sinceλSS,2 whenp= 2. Our first theorem is a non-existence which extends the one already mentioned in the casep= 2.

Theorem 1. Let S ⊂SN1 be a smooth subdomain. If βq ≥ βS there exists no positive solution of (1.5 ) inS which vanishes on ∂S.

Apart the casep= 2, the existence counterpart of this theorem is not known in arbitrary dimension, except if q=qc where (1.5 ) is the Euler-Lagrange equation of the functional

J(ψ) = Z

S

1 p

βq2cψ2+|∇ψ|2p/2

− 1

qc+ 1|ψ|qc+1

dσ. (1.9)

An application of the already mentioned variational methods leads to an existence result.

IfN = 2 the problem of finding solutions of (1.1 ) under the form (1.2 ) can be completely solved using dynamical systems methods. In order to point out a richer class of phenomena, we shall imbed this problem into the more general class of quasilinear equations with a potential, authorizing the valuep= 1.

div

|Du|p2Du

+|u|q1u− c

|x|p |u|p2u= 0 (1.10) in R2\ {0}, withq > p−1≥0 andc∈R. Looking again for solutions under the form (1.2 ), we see clearly that β must be equal toβq, whileω is any 2π-periodic solution of

d dσ

 β2qω2+ dω

2!p2)/2

dω dσ

+λq

"

βq2ω2+ dω

2#(p2)/2

ω +|ω|q1ω−c|ω|p2ω= 0,

(1.11)

where

λqq(qβq−2) =βq(p−2 + (p−1)βq). (1.12)

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Set

cqqp2λq =pp1(p−2)q+ 2(p−1)

(q+ 1−p)p , (1.13)

If c ≤cq, the only constant solution is the zero function, while if c > cq, there exist two other constant solutions±(c−cq)1/(q+1p).Then we can describe the set

F=±E+∪ E

of nonzero solutions of (1.11 ) on S1, where E+ is the set of nonnegative solutions, andE the set of sign changing solutions. Our main result is the following:

Theorem 2. Assumep >1, q > p−1.

(i)

E= [

kN, k=kq

ωk(.+ψ) :ψ∈S1 , (1.14)

where ωk is a function with least period 2π/k,where kq = 1 if c≥cq; ifc < cq, then kq is the smallest positive integer such that kq > Mq,where

Mq = πβ1qp

2 Z π/2

0

1 + (p−1) tan2θ

βqp(p−1) tan2θ+cq−ccosp2θdθ

. (1.15)

(ii) If c ≤ cq, then E+ = ∅. If 0 < c−cq ≤ βqp1/p, then E+ =

(c−cq)1/(q+1p) . If c−cq> βpq1/p,thenE+ contains a set of the form

n(c−cq)1/(q+1p)o

k+q

[

kN, k=1

ωk+(.+ψ) :ψ∈S1 , (1.16)

whereω+k is a positive function with least period2π/k,andk+q is the largest integer smaller than(pβq1p(c−cq))1/2.

As a consequence we can prove the existence counterpart of Theorem 1 in dimension 2 Corollary 1. Let N = 2 and S be any angular sector ofS1. Then there exists a positive solution of (1.5 ) vanishing at the two end points of S if and only if βq < βS. Further- more this solution is unique. In particular, existence holds for any sector if p < 2 and q≥2(p−1)/(2−p).

The casep= 1 appears as a limiting case of the preceding one. In that case we observe thatuis a positive solution of (1.10 ) if and only ifv=uq is a solution of the same equation relative toq= 1,

div

|Dv|1Dv

+v− c

|x| = 0. (1.17)

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The initial case c = 0 is easily treated while the case c 6= 0, that we shall analyse in full generality is much richer and delicate.

Theorem 3. Assumep= 1andq >0.

(i) Ifc6= 0,or c= 0 andq >1,thenE =∅. Ifc= 0and q≤1 then E =

ω0(.+ψ) :ψ∈S1 ,

where σ7→ω0(σ) := 21/q|sinσ|1/q1sinσis aC1 solution of (1.11 ).

(ii) If c≤ −1, thenE+ =∅. If−1< c <0,then E+=

(c+ 1)1/q . Ifc >0,then E+=n

(c+ 1)1/qo

k2

[

kN, k=k1

ωk+(.+ψ) :ψ∈S1 ,

where ω+k is a positive function with least period 2π/k,and k2 is the largest integer strictly smaller than (c+ 1)1/2, andk1 is the smallest integer greater thanπ/2

Z π/2 0

q cosθ cosθ+2cdθ.If c= 0, then

E+={1} ∪ [

K(0,1)

ωK+(.+ψ) :ψ∈S1

( ∅ ifq≥1 ω+0(.+ψ) :ψ∈S1 ifq <1 where

ωK+ =p

1−K2sin2σ−Kcosσ1/q

, and ω0+= (2|sinσ|)1/q ∀σ∈S1.

A striking phenomenon is the existence of a 2-parameter family of solutions whenc= 0.

Our paper is organized as follows: 1- Introduction. 2- The N-dimensional case. 3- The 2-dim dynamical system. 4- The casep >1. 5-The casep= 1.

2 The N-dimensional case

2.1 The spherical p-harmonic spectral problem

Letq > p−1>0 andS be a smooth connected domain onSN1 andCS be the cone with vertex 0 generated byS. We look for positive solutions of thep-Laplace equation

−div(|Du|p2Du) =uq, (2.1)

in CS\ {(0)}vanishing on∂CS\ {(0)}, under the form

u(r, σ) =rβω(σ), (2.2)

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thenβ =p/(q+ 1−p) =βq andω solves

−∇.

q2ω2+|∇ω|2)(p2)/2ω

−ωqq,p2qω2+|∇ω|2)(p2)/2ω in S ω= 0 on ∂S.

(2.3)

where

λq,pq(qβq−N).

We denote by βS the exponent corresponding to the first spherical singular p-harmonic function and by φS the corresponding function. This means that βS > 0 and u(r, σ) = rβSφS(σ) isp-harmonic inCS\{(0)}and vanishes on∂CS\{(0)}. Furthermoreφ=φS >0 and satisfies

−∇.

S2φ2+|∇φ|2)(p2)/2φ

SS2φ2+|∇φ|2)(p2)/2φ in S φ= 0 on ∂S.

(2.4)

where

λSSS(p−1) +p−N).

Ifϕ∈C1(S), we denote T(ϕ) =−∇.

2qϕ2+|∇ϕ|2)(p2)/2ϕ

−λq,p2qϕ2+|∇ϕ|2)(p2)/2ϕ (2.5) We recall that (βS, φS) is unique up to an homothety uponφ. FurthermoreφS is positive in S,∂φS/∂ν <0 on∂S and

S⊂S, S6=S=⇒βS > βS.

2.2 Non-existence

Proof of Theorem 1. We put

θ= βq

βS

and η=φθS. Thenθ≥1 and

η=θφθS1φS,

βq2η2+|∇η|22φ2(θS 1)2Sφ2S+|∇φS|2),

q2η2+|∇η|2)(p2)/2p2φ(pS2)(θ1)S2φ2S+|∇φS|2)(p2)/2,

.(βq2η2+|∇η|2)(p2)/2η=θp1φ(pS1)(θ1).(β2Sφ2S+|∇φS|2)(p2)/2φS

+ (p−1)(θ−1)θp2φ(pS1)(θ1)1S2φ2S+|∇φS|2)(p2)/2|∇φS|2 Using (2.4 ) withφ=φS, we derive

T(η) =−(p−1)θp1(θ−1)φ(pS1)(θ1)1S2φ2S+|∇φS|2)p/2in S. (2.6)

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Becauseω is a nonnegative nontrivial solution of (2.3 ), it is nonpositive inS. Furthermore

∂ω/∂ν <0 on ∂S. Therefore we can chooseφS as the maximal positive solution of (2.4 ) such thatη ≤ω. Ifθ >1 there existsσ∈S such that

ω(σ) =η(σ)>0 and ω(σ)≥η(σ) ∀σ∈S.¯ (2.7) Ifθ= 1, the graphs ofω andη could be tangent only on ∂S. This means that either (2.7 ) holds, or there exists ¯σ∈∂Ssuch that

∂ω(¯σ)/∂ν=∂η(¯σ)/∂ν <0 and ω(σ)< η(σ) ∀σ∈S. (2.8) Letψ=ω−η and we first consider the case where (2.7 ) holds. Letg= (gij) be the metric tensor onSN1. We recall the following expressions in local coordinatesσj aroundσ,

|∇ϕ|2=X

j,k

gjk∂ϕ

∂σj

∂ϕ

∂σk

,

for anyϕ∈C1(S), and

.X= 1 p|g|

X

∂σ

p|g|X

= 1

p|g| X

ℓ,i

∂σ

p|g|gℓiXi

,

for any vector fieldX ∈C1(T SN1), if we lower the indices by settingX=X

i

gℓiXi. We derive from the mean value theorem

2qω2+|∇ω|2)(p2)/2∂ω

∂σi −(βq2η2+|∇η|2)(p2)/2∂η

∂σi

=X

j

αij∂(ω−η)

∂σj

+bi(ω−η), where

bi= (p−2)

βq2(η+t(ω−η))2+|∇(η+t(ω−η))|2(p4)/2

×(η+t(ω−η))∂(η+t(ω−η))

∂σi

, and

αij= (p−2)

β2q(η+t(ω−η))2+|∇(η+t(ω−η))|2(p4)/2

×∂(η+t(ω−η))

∂σi

X

k

gjk∂(η+t(ω−η))

∂σk

ij

β2q(η+t(ω−η))2+|∇(η+t(ω−η))|2(p2)/2

. Since the graph ofη andω are tangent atσ,

η(σ) =ω(σ) =P0>0 and∇η(σ) =∇ω(σ) =Q.

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Thus

bi) = (p−2)

βq2P02+|Q|2(p4)/2

P0Qi, and

αij) =

β2qP02+|Q|2(p4)/2

δijq2P02+|Q|2) + (p−2)Qi

X

k

gjkQk

! . Now

T(ω)− T(η) =ωq+ (p−1)θp1(θ−1)φ(pS1)(θ1)1S2φ2S+|∇φS|2)p/2

= −1 p|g|

X

ℓ,i

∂σ

p|g|gℓi

q2ω2+|∇ω|2)p21∂ω

∂σi −(β2qη2+|∇η|2)p21∂η

∂σi

−λq,p

2qω2+|∇ω|2)p21ω−(βq2η2+|∇η|2)p21η ,

=− 1 p|g|

X

ℓ,i

∂σ

 p|g|gℓi

 X

j

αij∂(ω−η)

∂σj

+bi(ω−η)

+X

i

Ci

∂(ω−η)

∂σi

=− 1 p|g|

X

ℓ,j

∂σ

aj∂(ω−η)

∂σj

+X

i

Ci

∂(ω−η)

∂σi

, where theCi are continuous functions and

aj =p

|g|X

i

gℓiαij. The matrix αij0)

is symmetric, definite and positive since it is the Hessian of the strictly convex function

X = (X1, . . . , Xn1)7→ 1 p

P02+|X|2p/2

=1 p

P02+X

j,k

gjkXjXk

p/2

.

Therefore αij

has the same property in some neighborhood ofσ, and the same holds true with aj

. Finally the functionψ=ω−η is nonnegative, vanishes at σand satifies

− 1 p|g|

X

ℓ,j

∂σ

aj∂ψ

∂σj

+X

i

Ci

∂ψ

∂σi ≥0. (2.9)

Then ψ = 0 in a neighborhood of S. Since S is connected, ψ is identically 0 which a contradiction.

If (2.8 ) holds, then θ = 1 and the graphs of η and ω are tangent at ¯σ. Proceeding as above and using the fact that∂η/∂ν exists and never vanishes on the boundary, we see that ψ=η−ω satisfies (2.9 ) with a strongly elliptic operator in a neighborhoodN of ¯σ.

Moreoverψ >0 inN,ψ(¯σ) = 0 and∂ψ/∂ν(¯σ) = 0. This is a contradiction, which ends the

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proof.

Remark. Ifp= 2, the proof of non-existence is straightforward by multiplying the equation in ωby the first eigenfunction φS and get

Z

S

((λS−λq,2)ω−ωqSdσ= 0, a contradiction sinceλS ≤λq,2.

2.3 Existence results

Let us consider the case q = qc = (N(p−1) +p)/(N −p) (N > p > 1), and let S be any smooth subdomain of SN1. The research of solution of (1.1 ) under the form (1.2 ) vanishing on∂CS leads to





−∇.

βq2cω2+|∇ω|2p/21

ω

− |ω|qc1ω=−β2qc

βq2cω2+|∇ω|2p/21

ω inS ω= 0 in ∂S,

(2.10) where βqc = N/p−1. This equation is the Euler-Lagrange variation of the functional J defined onW01,p(S) by

J(ψ) = Z

S

1 p

βq2cψ2+|∇ψ|2p/2

− 1

qc+ 1|ψ|qc+1

dσ. (2.11)

Theorem 2.1 Problem (2.10 ) admits a positive solution.

Proof. Clearly the functional is well defined onW01,p(S) sinceqcis smaller than the Sobolev exponent pN1 for W1,p in dimension N-1. For anyψ ∈ W01,p(S), limt→∞J(tψ) = −∞. Furthermore there existδ >0 andǫ >0 such thatJ(ψ)≥ǫfor anyψ∈W01,p(S) such that kψkW1,p =δ. Assume now that {ψn} is a sequence of W01,p(S) such that J(ψn)→αand kDJ(ψn)kW−1,p′ →0 asn→ ∞. Then

−∇.

β2qcψn2+|∇ψn|2p/21

ψn

−|ψn|qc1ψn2qc

βq2cψ2n+|∇ψn|2p/21

ψnn →0.

Then Z

S

βq2cψ2n+|∇ψn|2p/2

− |ψn|qc+1

dσ=hǫn, ψni. SinceJ(ψn)→αit follows

Z

S

βq2cψn2+|∇ψn|2p/2

dσ→p(qc+ 1)α/(qc+ 1−p).

Therefore{ψn}remains bounded inLqc+1(S), and relatively compact inLr(S), for any 1<

r < qc+ 1. Multiplying the equationDJ(ψn)−ǫn byTk,θn) whereθ∈(1,(pN1−1)/qc),

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k >0 andTk,θ(r) = sgn min{|r|, k} and using standard bootstrap arguments yields to the boundedness of{ψn} inL(S). Combining this fact with the compactness of inLr(S), we derive the compactness in any anyLs, fors <∞. Therefore{ψn} is relatively compact in W01,p(S). This means thatJ satisfies the Palais-Smale condition.

3 The 2-dim dynamical system

3.1 Extension of the data

We study more generally the existence of 2π-periodic solutions of equation d

 β2ω2+ dω

2!p/21

dω dσ

+λ

"

β2ω2+ dω

2#p/21

ω+g(ω)−c|ω|p2ω= 0, (3.1) where λ, β, c are real arbitrary parameters, withβ > 0, and g ∈C0(R)∩ C1(R\ {0}), g is odd,

slim0+g(s)/sq= 1, lim

s→∞g(s)/sp1=∞, d

ds(g(s)/|s|p1)>0 on (0,∞), (3.2) with q > p−1 ≥ 0. In fact we can easily reduce the problem to a simpler form, and particularly in the casep= 1,where the equation has a remarkable homogeneity property.

The next statement is straightforward.

Lemma 3.1 (i) Letp >1.Setting

τ =βσ, andω(σ) =βp/(q+1p)w(τ); (3.3) equation (3.1 ) takes the form, wherew =dw/dτ:

d dτ

w2+w2p/21

w

−b w2+w2p/21

w+f(w)−d|w|p2w= 0, (3.4) where

b=−λ

β2, d= c

βp, f(s) =βpq/(q+1p)g(βp/(q+1p)s), (3.5) thusf satisfies the same assumptions (3.2 ) asg.

(ii) Let p= 1.At any point where ω(σ)6= 0,setting

τ =βq σ, andω(σ) = (βq)1/q|w(τ)|1/q1w(τ), (3.6) then wsatisfies an equation of type (3.4 ) with

b=−λ/β2q, d=c/βq, f1(s) =βqg((βqs)1/q), (3.7) andf1 satisfies the assumptions (3.2 ) withq= 1 :

slim0+f1(s)/s= 1, lim

s→∞f1(s) =∞, f1(s)>0 on (0,∞), (3.8) From above, the changes of variables (3.3 ) and (3.6 ) reduce the problem to study the existence of periodic solutions of equation (3.4 ) and the range of values of the period function, for anyq > p−1 ifp >1,and forq >0 ifp= 1.

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3.2 Reduction to dynamical systems

Assume p≥1.Equation (3.4 ) can be re-written as the system,

w =F(w, y) =y

y =G(w, y) = bw3+ (b+ 2−p)w y2−(f(w)−d|w|p2w)(w2+y2)2p/2

w2+ (p−1)y2 . (3.9)

We denote byhthe odd function defined on Rby h(s) =

( f(s)/|s|p2s ifs6= 0

0 ifs= 0. (3.10)

If b+d≤ 0, it has no non-trivial stationary point, while if b+d > 0, it admits the two stationary points±P0,with

P0= (a,0), a=h1(b+d).

FurthermoreP0is a center since the linearized system atP0is given by the matrix

0 1

−ah(a) 0

. System (3.9 ) is singular at (0,0),and,a priori, along the linew= 0 ifp <2,andq <1 or d 6= 0, or if p = 1. In fact for p > 1 it is not singular at the points (0, σ) for σ 6= 0.

Indeed the problemw′′=G(w, w) with initial dataw(0) = 0, w(0) =σhas a unique local solution: indeed near (0, σ), Gis continuous with respect to w and C1 with respect toy, thus w is C2; locally t can be expressed in terms ofw, and setting w(t) =p(w), pis C1 near 0, p(0) = 1 anddp/dw =G(w, p)/p,and J(w, p) =G(w, p)/p isC1 with respect to p and continuous with respect to w, thus one gets local uniqueness ofp; and then the local uniqueness of problemw(t) =p(w(t)), w(0) = 1, since pis of classC1.

The phase plane of system (3.9 ) is invariant by symmetries with respect to the two axes of coordinates, becauseF is even with respect towand odd with respect toy,andGis odd with respect to wand even with respect to y.Thus from now we can restrict the study to the first quadrant

Q\ {(0,0)}, whereQ= (0,∞)×(0,∞) ;

in particularw≥0. In casep >1,due to the symmetries, any trajectory which meets the two axes in finite timesτ, τ+T is a closed orbit of period 4T.

Remark. It is sometimes useful to introduce the slopeξ=w/w,(or a function of the slope) as a new variable. This was first used forp >1 in [11] for the homogeneous problem

d dτ

w2+w2p/21

w

−b w2+w2p/21

w= 0;

in that case it be written under the form d

1 +ξ2p/21

ξ

=−((p−1)ξ2−b) 1 +ξ2p/21

,

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forw >0 , and this equation is completely integrable in terms of u= 1 +ξ2p/21

ξ.

We can transform (3.9 ) by using polar coordinates inQ

(w, y) = (ρcosθ, ρsinθ), ρ >0, θ∈(0, π/2).

It gives

θ =b−(p−1) tan2θ+ (d−h(ρcosθ)) cosp2θ

1 + (p−1) tan2θ , ρ=ρ(1 +θ) tanθ. (3.11) Equivalently, we introduce the slopeξ= tanθ∈(0,∞) instead of the angle, and set

u=φ(ξ) = cos1pθsinθ, φ(ξ) = (1 +ξ2)(p2)/2ξ,

the function φis strictly increasing, from (0,∞) into (0,∞) when p > 1, from (0,∞) into (0,1) whenp= 1.Indeedφ(ξ) = (1 +ξ2)(p4)/2(1 + (p−1)ξ2).Defining

ϕ=φ1, and E(ξ) = (p−1)ξ2−b

(1 +ξ2)p/21, (3.12)

we obtain (

w =wϕ(u),

u=−E(ϕ(u))−h(w) +d. (3.13)

This system is still singular on the linew= 0 ifh6∈C1([0,∞)) near 0.In the sequel we set Ψ(u) =

u

Z

0

ϕ(s)ds. (3.14)

Noticing that

E(ξ) = p(p−1)ξ2+ 2(p−1)−(p−2)b

(1 +ξ2)(p2)/2ξ, (3.15) we derive thatEis increasing on (0,∞) when (p−2)b≤2(p−1). When (p−2)b >2(p−1), E is decreasing on (0, η) and then increasing, where

p(p−1)η2= (p−2)b−2(p−1), (3.16) and

minE=E(η) =− 2 p−2

(p−2)(b+p−1) p(p−1)

p/2

. (3.17)

In the case of initial problem (1.11 ),E is increasing.

Remark. Ifp >1, system (3.9 ) is singular at (0,0).If we replace the assumption lims0+f(s)/sq = 1,by the stronger one

slim0+f(s)/sq1=q, (3.18)

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we can transform system (3.13 ) in (0,∞)×R in a system of the same type, but without singularity, by performing the substitution

v=wq+1p. Then

v = (q+ 1−p)vϕ(u), u=−E(ϕ(u))−˜h(v) +d. (3.19) wherev7→˜h(v) =h(v1/(q+1p)∈C1([0,1).In particular, iff(w) =|w|q1w, we find

v= (q+ 1−p)vϕ(u), u=−E(ϕ(u))−v+d. (3.20) Remark. In the case f(w) =|w|q1w, we can differentiate the equation relative tou and obtain a second order equation that satisfiesu,

u′′=B(ϕ(u))u+ (q+ 1−p)(E(ϕ(u))−d)ϕ(u) (3.21) whereE is given above, and

B(ξ) = (p−2)b+q−3(p−1) + (q+ 1−2p)(p−1)ξ2

1 + (p−1)ξ2 ξ. (3.22)

Notice that equation (3.21 ) has no singularity forp >1.

4 The case p > 1

4.1 Existence of a first integral

A natural question is to see if equation (3.4 ) admits a variational structure.

Whenp= 2, it is the case, for any realsb, d: indeed (3.4 ) takes the form w′′−(b +d)w+f(w) = 0,

hence denotingF(w) = Z w

0

f(s)ds,it is the Euler equation of the functional H2(w, w) = w2

2 + (b+d)w2

2 − F(w) and thus it has a first integralw2= (b +d)w2−2F(w) +C.

When p6= 2, p >1, we find a first integral only in the caseb= 1; in this case equation (3.4 ) is the Euler equation of the functional

H(w, w) = w2+w2p/2

p +d|w|p

p − F(w), from which follows the Painlev´e integral:

1

p w2+w2p/21

(p−1)w2−w2

=d|w|p

p − F(w)−K. (4.1)

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Using the function Eintroduced at (3.12 ), then (4.1 ) is equivalent forw >0 to E

w w

=E(ϕ(u)) =d−pK+F(w)

wp (4.2)

and hereE is increasing on (0,∞) from−b=−1 to +∞.

Therefore, in the general case, we cannot use a first integral for studying the periodicity properties of the solutions, while it was the main tool in [2] forp= 2. We are lead to use phase plane techniques. For the initial problem (1.11 ), the valueb= 1 corresponds to the casep <2 andq= (3p−2)/(2−p)).

4.2 Description of the solutions

Next we describe in more details the trajectories of system (3.9 ) in the phase plane (w, y).

Recall that the system can be singular on the axisw= 0.

Proposition 4.1 Assume p > 1. Then all the orbits of system (3.9 ) are bounded. Any trajectoryT[P] issued of any pointP inQis either a closed orbit surrounding(0,0), or (only if b+d >0) a closed orbit surrounding P0 but not (0,0), or an homoclinic orbit, starting from (0,0) with the slope m such that E(m) = d, ending at (0,0) with the slope −m and with Rfor interval of existence.

Proof. First look at the vector field on the boundary ofQ. At any point (0, σ) withσ >0, it is given by (σ,0), thus it is transverse and inward. At any point ( ¯w,0) with ¯w >0, it is given by (0,w(b¯ +d−h( ¯w)). Thus it is transverse and outward whenever b+d ≤0 or b+d >0 and ¯w > a,and inward wheneverb+d >0 and ¯w < a.

Consider any solution (w, y) of the system, such that P = (w(0), y(0)) ∈ Q , and let (τ1, τ2) be its maximal interval existence inQ.At each pointτ whereu(τ) = 0,andu >0, u′′(τ) =−h(w)wϕ(u)<0 from (3.13 ). Thus if τ exists, it is unique, and it is a maximum foru.

Sincew =y >0,then whas a limit ℓ2 ∈(0,∞] at τ2 and ℓ1 ∈[0,∞) atτ1. Anduis strictly monotonous nearτ1, τ2,thus it has limitsu1, u2∈[0,∞],in other wordsθhas limits θ1, θ2∈[0, π/2]

(i) Let us go forward in time. On any interval whereuis increasing, one hasE(ϕ(u))≤d, thusuis bounded. Thenu2 is finite. Ifℓ2=∞, thenθ tends to−∞,thusρis decreasing thus it is bounded, which is contradictory; thus ℓ2 is finite. Ifu2>0 then (ℓ2, ℓ2ϕ(u2)) is stationary, which is impossible. Thusuis decreasing to 0,and the trajectory converges to (ℓ2,0).Ifb+d >0 andℓ2=a, thenu tends to 0 from (3.13 ), and

u′′=−(E◦ϕ)(u)u−h(w)wϕ(u) =−h(a)aϕ(u)(1 +o(1))

thus u′′ <0 nearτ2, which is impossible. Then either b+d≤0, orb+d >0 and ¯w > a, andτ2 is finite, the trajectory leavesQ transversally atτ2.

(ii) Next let us go backward in time.

•Suppose thatu1= 0,then the trajectory converges to (ℓ1,0); then necessarilyb+d >0 andℓ1≤a,andℓ1< aas above . The trajectory entersQtransversally atτ2,and from the symmetries it is a closed orbit surrounding only the stationnary pointP0.

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• Suppose thatu1 =∞, that meansθ tends toπ/2. Then from (3.11 ), θ tends to 1, thusτ1is finite and π/2−θ= (τ−τ1)(1 +o(1)),tanθ= (τ−τ1)1(1 +o(1)),and

(p−1)(τ−τ1)1ρ

ρ = (b+ 1 + (d−h(ρcosθ)) cosp2θ)(1 +o(1)

Ifp≥2,thenρ/ρ=O((τ−τ1)); ifp <2 thenρ/ρ=O((τ−τ1)p1).In any case, lnρcase is integrable, thus ρhas a finite limit ¯y >0. Then the trajectory entersQtransversally at τ1 and from the symmeries it is a closed orbit surrounding (0,0). From the considerations in §3-2, for any ¯y >0 there exist such an orbit, and it is unique. Moreover inQ the slope ξ=ϕ(u) is decreasing from∞to 0; indeed it decreases nearτ1 andτ2 and can only have a maximal point.

• Suppose that 0 < u1 < ∞. If ℓ1 > 0, then (ℓ1, ℓ1ϕ(u1)) is stationary, which is im- possible. Thus (y, w) converges to (0,0). Andw/w tends to ϕ(u1), thus τ1 =−∞. And u converges to d−E(ϕ(u1)), thus tanθ =ϕ(u) has a limit m ≥0 such that E(m) = d.

From the symmetries the trajectory is homoclinic and the solutionwis defined onR. The next precises the behaviour of solutions according to the sign ofb+d.

Theorem 4.2 Consider system (3.9 ) withp >1.

(i) Assumeb+d >0.Then there exists a unique homoclinic trajectoryHstarting from(0,0) in Q with the slope md =E1(d) ( m0 =p

b/(p−1) if d = 0), and ending at (0,0) with the slope−md,and surroundingP0.Up to the stationary points, the other orbits are closed, and either they surround only one of the points P0 or −P0, in the domain delimitated by H, corresponding to solutionsw of constant sign, or they are exterior to±H and surround (0,0) and±P0,corresponding to sign changing solutionsw.

(ii) Assume b+d≤0.Then

• if (p−2)b ≤2(p−1), or [(p−2)b >2(p−1) andd < E(η)], there is no homoclinic trajectory.

• if [(p−2)b >2(p−1)andE(η)< d≤ −b] ; then denoting by m1,d < m2,d the two roots of equation E(m) =d, there exists an infinity of homoclinic trajectories H1 starting from (0,0) inQ with the slopem1 and ending at(0,0) with the slope−m1,d,and a unique homoclinic trajectory H2 starting from (0,0) in Q with the slope m2 and ending at (0,0) with the slope −m2.

Proof. (i)Case b+d >0.Then the equationE(m) =dhas a unique solutionm=E1(d);

and w/w tends to m; thus the trajectory starts from (0,0) with a slope m.Then for any P ∈ Q, the trajectory T[P] passing throughP meets the axisy = 0 afterP at some point (µ,0) withµ > a.Let

U=

P ∈ Q:T[P]∩ {(0, σ) :σ >0} 6=∅ , V=

P ∈ Q:T[P]∩ {(µ,0) : 0< µ < a} 6=∅ . Then eitherP ∈ U and the trajectory is a closed orbit surrounding (0,0) and±P0,and inQ. OrP ∈ V and the trajectory is a closed orbit surrounding onlyP0. OrT[P]is an homoclinic orbit Hstarting from (0,0) with the slope m, wherem is the unique solution of equation E(m) =d (, such thatm > η ifE is not monotone, see (3.15 ). Now U andV are open,

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since the vector field on the axes is transverse, thusU ∪ V 6=Q.This shows the existence of such an orbitH.

(ii)Case b+d≤0.

• Eitherb+d < 0 andE is increasing, or E has a minimum atη andd < E(η).Then equation E(m) = d has no solution: there is no homoclinic orbit. OrE is increasing and b+d= 0; thenE(ϕ(u))>−b=d,thusu<0,thusucannot tend to 0, the same conclusion holds.

• 0rE has a minimum atη andE(η)< d≤ −b; then the equationE(m) = dhas two rootsm1, m2such that 0≤m1< η < m2≤mb, whereE(mb) =−b.Consider again the set U. Any trajectoryT[P] such thatP ∈ U satisfies u <0, that means h(w)> d−E(ϕ(u)) andudescribes (0,∞),thus there existsτ such thatϕ(u)(τ) =η,thush(w(τ))> d−E(η) and y(τ) = ηw(τ). Next consider any trajectory T[P˜] going through P˜ = ( ˜w, ηw) such˜ that h( ˜w) ≤ d−E(η) at time 0. It cannot be a trajectory of the preceeding type, thus (y, w)→(0,0) asτ →τ1,and θ tends toθ1, with tanθ1 =m1 or m2; moreoveru(0)≥0 , and u <0 near τ2, thus there exists a unique τ ≥0 such thatu(τ) = 0; then u >0 in (τ1, τ), thus tanθ1 < η, thus tanθ1 =m1.Thus there exist an infinity of such trajectories H1,starting with the slopem1. Next fix one trajectoryT[P˜0] such thath( ˜w0)≤d−E(η).

LetRbe the subdomain ofQdelimitated byT[P˜0] andT[(0,1)].Let V=

P ∈ R:T[P]∩ {(w, ηw) : 0< w <w˜0} 6=∅ .

Then alsoV is open. Indeed the intersection with the liney=ηwforw <w˜0 is transverse:

at the intersection point, h( ˜w)< d−E(η), thusu>0,andy/w=ϕ(u) =η,and y

y =ϕ(u) +ϕ(u)

ϕ(u)u> η=w w

Then (U ∩ R)∪ V 6=R.Then there exists at least a trajectoryH1,starting from (0,0) with the slopem2.

(iii)Uniqueness of HandH2. Let m=m0 or m2,d. Suppose that system (3.9 ) has two solutions (w1, y1), (w2, y2) near −∞, such that wi >0 and wi tends to 0 and yi/wi

tend tom.Then the system (3.13 ) has two local solutions (w1, u1),(w2, u2) such thatϕ(ui) tends to mat −∞.Then wi >0 locally, thus one can expressui as a function ofwi.Then at the same pointw,

wd(Ψ(ui)

dw =wϕ(ui)dui

dw =−E(ϕ(ui))−h(w) +d, wd(Ψ(u2)−Ψ(u1))

dw =−(E(ϕ(u2))−E(ϕ(u1)) =E(ϕ(u))ϕ(u)(u2−u1)

for someu betweenu1andu2,andE(ϕ(u)) =E(m)(1 +o(1)); andE(m)>0.Then for smallw

d(Ψ(u2)−Ψ(u1))

dw (Ψ(u2)−Ψ(u1))<0

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thus (Ψ(u2)−Ψ(u1))2is decreasing; and its limit at 0 is 0, thus Ψ(u2) = Ψ(u1),thusu2≡u1

near−∞then from (3.13 ),h(w1) =h(w2),andhis injective, thusw1≡w2 near−∞. The global uniqueness follows, since the system is regular except at (0,0).And all the trajectories are described.

Remark. Under the assumption (3.18 ), existence and uniqueness of H and H2 can be obtained more directly whenever d 6= E(η). Indeed the system (3.19 ) relative to (v, u) is regular, with stationary points (0,0), (0,±ϕ1(m)), where m = m0, m1 or m2 and also (±a,0) if b+d > 0. The linearized system at (0, ϕ1(m)) is given by the matrix m(q+ 1−p) 0

0 K(m)

, with K(m) = p(p−1)(η2−m2)/(1 + (p−1)m2). If m= m1,d, then it is a source, and we find again the existence of an infinity of solutions. Ifm=md or m=m2,d,thenK(m)<0,thus this point is a saddle point. Then in the phase plane (v, u), there exists precisely one trajectory defined near −∞,such that v > 0 and converging to

(0, m) at−∞,andu/v converges to 0.

Remark. Supposef(w) =|w|q1w, then we can study the critical case (p−2)b >2(p−1) and E(η) =d: there exists an infinity of homoclinic trajectoriesH1 starting from (0,0) in Qwith an infinite slope and ending at (0,0) with an infinite slope, and a unique homoclinic trajectoryH2starting from (0,0) inQwith the slopeη and ending at (0,0) with the slope

−η.Indeed using system (3.20 ) and setting u=ϕ1(η) +z,and ζ= (q+ 1−p)ηz+v, it can be written under the form

ζ=P(ζ, v), v = (q+ 1−p)ηv+Q(ζ, v),

where P and Qboth start with quadratic terms. Moreover the quadratic part ofP(ζ, v) is given byp2,0ζ2+p1,1ζv+p0,2v2,where by computation,

p2,0=−p(p−1)

q+ 1−pηϕ2(ϕ(η))(1 +η2)(p2)/2<0;

then the results follow from the description of sadle-node behaviour given in [8, Theorem 9.1.7].

Remark. In the caseb= 1>−d,we have a representation of the homoclinic trajectory : it corresponds toK= 0 in (4.2 ).In the casef(w) =|w|q1w, in terms ofuwe obtain

u =q+ 1−p

p (E(ϕ(u))−d), which allows to computeuby a quadrature.

4.3 Period of the solutions

First we consider the sign changing solutions

Theorem 4.3 Consider system (3.9 ) with p >1. For any ν > 0, consider the trajectory T[(0,ν)] which goes through (0, ν). Let T(ν) be its least period . Then T is decreasing on (0,∞).

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(i) Ifb+d≤0, andE is increasing ord <minE,it decreases from Td to0, where Td= 4

Z

0

du

E(ϕ(u))−d = 4 Z π/2

0

1 + (p−1) tan2θ

(p−1) tan2θ−b−dcosp2θdθ, (4.3) andTd is finite if and only if b+d <0. Ifb <0 =d,then

T0= 2π (p−1)γ+ 1

(p−1)γ(γ+ 1), γ=p

|b|/(p−1)). (4.4) (ii) If b+d >0 orb+d≤0andd≥minE,it decreases from∞ to0.

Proof. (i) Monotonicity of T : consider the part of the trajectoriesT[(0,ν)]located inQ, given by (wν, yν).We have shown thatuis decreasing with respect toτ from ∞to 0, then E(ϕ(u)) +h(wν(u))−d >0 andwν can be expressed in terms ofu,and

T(ν) = 4 Z

0

du

E(ϕ(u)) +h(wν(u))−d.

Let λ > 1. Since the trajectories T[(0,ν)]and T[(0,λν)]. have no intersection point, then wλν(u) > wν(u) for any u > 0, and h is nondecreasing, thus T(λν) < T(ν), thus T is decreasing.

(ii) Behaviour near ∞: Let νn≥1, such that limνn=∞.Observe that for fixedu,for any integern ≥1, there exists a unique ˜νn >0 (depending onu), such that w˜νn(u) =n;

let ˆνn = max(˜νn, n). Then h(wνˆn(u)) ≥ h(n) thus h(wνˆn(u)) converges to ∞; since ν 7→

h(wν(u)) is nondecreasing thenh(wνn(u)) converges to∞,and T(νn) converges to 0, from the Beppo-Levi theorem.

(iii) Behaviour near 0 :

•First assumeb+d≤0,andE is increasing, ord < E(η).Then all the orbits are of the typeT[(0,ν)].Letσn ∈(0,1),such that limνn= 0.For fixedu,for any integern≥1,there exists a unique ¯νn >0 (depending onu), such thatwν¯n(u) = 1/n; let ˇνn = min(¯νn,1/n).

Thenh(wνˇn(u)≤h(1/n), thush(wˇνn(u)) converges to 0,thus againh(wνn(u)) converges to 0.ThenT(νn) converges to Td given by (4.3 ), from the Beppo-Levi theorem. Ifb+d <0, then Td is finite: indeed near ∞, E(ϕ(u)) = (p−1)up/(p1)(1 +o(1)); if E is increasing, thenE(ϕ(u))−d >−(b+d)>0; if d < E(η),thenE(ϕ(u))−d≥E(η)−d >0.

Ifb+d= 0 andEis increasing, thenTd=∞: indeed near 0, E(ϕ(u))−d=u2(E′′(0)/2+

o(1)) and E′′(0) = 2(p−1)−(p−2)b; if (p−2)b = 2(p−1), then E(ϕ(u))−d= (p(p− 1)/4)u4(1 +o(1)) ; in any case the integral is divergent.

Whenb <0 =d,one can computeT0: T0

4 = Z

0

du E(ϕ(u)) =

Z 0

φ(ξ)dξ E(ξ) =

Z 0

1 + (p−1)ξ2

(|b|+ (p−1)ξ2)(1 +ξ2)dξ

= π 2 + ( 1

p−1 −γ2) Z

0

ds

2+s2)(1 +s2) = π

2(1 + 1−(p−1)γ2 (p−1)γ(γ+ 1)).

Hence (4.4 ) holds.

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