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HAL Id: hal-01312595

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Preprint submitted on 7 May 2016

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SINGULAR ELLIPTIC EQUATION INVOLVING THE GJMS OPERATOR ON COMPACT RIEMANNIAN

MANIFOLD

Mohammed Benalili, Ali Zouaoui

To cite this version:

Mohammed Benalili, Ali Zouaoui. SINGULAR ELLIPTIC EQUATION INVOLVING THE GJMS OPERATOR ON COMPACT RIEMANNIAN MANIFOLD. 2016. �hal-01312595�

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OPERATOR ON COMPACT RIEMANNIAN MANIFOLD

MOHAMMED BENALILI AND ALI ZOUAOUI

Abstract. In this paper we consider a singular elliptic equation involv- ing the GJMS (Graham-Jenne-Mason-Sparling) operator of orderk on n-dimensional compact Riemannian manifold with 2k < n. Mutiplicity and nonexistence results are established.

Let (M, g) be an n-dimensional Riemannian manifold. The k-th GJMS operator (Graham-Jenne-Mason-Sparling, see ([3])Pgis a differential opera- tor defined for any integerkif the dimension n is odd, and 2k≤notherwise.

In the following, we will consider the case 2k≤n. Pg is of the form Pg= ∆k+lot

where ∆ = −divg(∇) is the Laplacian-Beltrami operator and lot denotes the lower terms. One of the fundamental property of Pg is its behavior with respect to conformal change of metrics: for ϕ ∈C(M) , ϕ >0 and g=ϕn−2k4 g a conformal metric to g,

ϕn+2kn−2kP

egu=Pg(ϕu).

Pg is self-adjoint with respect to the L2-scalar product. To Pg is associ- ated a conformal invariant scalar function denoted Qg and is called the Q-curvature. For k = 1, the GJMS operator is ( up to a constant ) the conformal Laplacian and the corresponding Q-curvature function is simply the scalar curvature. Fork= 2, the GJMS operator is the Paneitz operator introduced in ([8]). For 2k < n, the Q-curvature is Qg = n−2k2 Pg(1). Many works was devoted theQ-curvature equation in the last two decades (see [1], [2], [5], [12]). Many authors investigated the interactions of conformal meth- ods with mathematical physic which led them to study the Einstein-scalar fields Lichnerowiz equations (see [4], [6], [9], [10], [11],). These methods have been extended to scalar fields Einstein-Licherowicz type equation in- volving the Paneitz operator, (see [7]). In this work we analyze an Einstein- Lichnerowicz scalar field equation containing thek-th order GJMS operator on a Riemannian n-dimensional manifold with 2k < n. Our work is or- ganized as follows: in a first step we show the existence of a solution to equation (1.1) obtained by means of the mountain-pass theorem. In the

1991Mathematics Subject Classification. 58J99-83C05.

Key words and phrases. GJMS operator,Critical Sobolev Growth.

1

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second step we prove, by the Ekeland lemma, the existence of a second so- lution to equation (1.1). In the last section we give a result of nonexistence of solution.

1. Existence of a first solution

Let (M, g) be an n-dimensional compact Riemannian manifold and k ∈ N, with 2k < n. Consider the following equation

(1.1)

(

Pg(u) =B(x)u2]−1+ A(x)

u2]+1 + C(x)up

u >0

where 2]= n−2k2n ,Pg is thek-th GJMS operator andp >1. In all the sequel of this paper we assume that the operator Pg is coercive which allows us ( see Proposition 2, [2]) to endow Hk2(M) with the following appropriated equivalent norm

kuk= v u u t Z

M

u.Pg(u)dvg.

So we deduce from the coercivity of Pg and the continuity of the inclusion Hk2(M)⊂L2](M), the existence of a constant S >0 such that

(1.2) kuk2]≤Skuk

where 2]= n−2k2n .

We assume also that the GJMS operator has a positive Green function.

Proposition 1. Suppose that the metric g is Einstein with positive scalar curvature of dimension n >2k, then the GJMS Pg admits a Green positive function.

Proof. Onn-dimensional Einstein manifold, the GJMS operator of order k (see [3]) is given by

Pg =

k

Y

l=1

(∆−clSc) wherecl= (n+2l−2)(n−2l)

4n(n−1) ,Scis the scalar curvature. If the scalar curvature is positive it is well known that the operator ∆−clSchas a positive Green function. Denote by Ll = ∆−clSc, l = 1, ..., k; by definition of the Green function of Ll we know that for allu∈C(M),

(Llu) (x) = Z

M

Gl+1((x, y) (Ll+1Llu) (y)dvg(y). So

u(x) = Z

M

Gl(x, z) (Llu) (z)dvg(z) + 1 V ol(M)

Z

M

u(x)dvg(x)

= Z

M

Z

M

Gl(x, z)Gl+1((z, y)dvg(z)

(Ll+1Llu) (y)dvg(y)+ 1 V ol(M)

Z

M

u(x)dvg(x)

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and letting

Gl,l+1(x, y) =Gl∗Gl+1(x, y) = Z

M

Gl(x, z)Gl+1((z, y)dvg(z).

And by induction, we get u(x) =

Z

M

G1∗...∗Gk(x, y)Pg(y)dvg(y).

ThusPg admits a positive Green function.

In this section we establish the following theorem.

Theorem 1. Let(M, g)be a compact Riemannian manifold with dimension n > 2k and A > 0, B > 0, C > 0 are smooth functions on M. Suppose moreover that the operatorPg is coercive and have a positive Green function.

If there exists a constant C(n, p, k)>0 depending only onn, p, ksuch that (1.3) kϕk2]

2] Z

M

A(x)

ϕ2\ dvg ≤C(n, p, k)

Smaxx∈MB(x) 2+2

] 2−2]

and

(1.4) kϕkp−1 p−1

Z

M

C(x)

ϕp−1dvg ≤C(n, p, k)

Smaxx∈MB(x) p+1

2−2]

for some smooth function ϕ >0, then equation (1.1) admits a smooth solu- tion.

To prove the existence of solutions to (1.1), we consider the following -approximating equations

(1.5) Pg(u) =B(x) (u)2]−1+ A(x)u (ε+u2)2[+1

+ C(x)u (ε+u2)p+12

where 2[ = 22],p > 1.Which gives us a sequence (uε)ε of solutions to (1.5).

The solution of equation (1.1) is then obtained as the limiting of (uε)ε. Following Hebey-Pacard-Pollak ([4]), to get rid of negative exponents, we consider the energy functional associated to (1.5) defined by, for anyε >0

Iε(u) =I1(u) +Iε(2)(u) where I(1) :Hk2(M)→R is given by

I(1)(u) = 1 2

Z

M

uPg(u)dvg− 1 2]

Z

M

B(x) u+2]

dvg

and Iε(2):Hk2(M)−→Ris Iε(2)(u) = 1

2] Z

M

A(x)

ε+ (u+)2

2[dvg+ 1 p−1

Z

M

C(x)

ε+ (u+)2 p−1

2

dvg. It is easy to check the following inequality

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(1.6) Φ (kuk)≤I(1)(u)≤Ψ (kuk) with

Φ (t) = 1 2t2

Smax

M B(x) t2]

2] and

Ψ (t) = 1 2t2+

Smax

M B(x) t2]

2].

The Φ (t) is increasing on [0, t0] and decreasing on ]t0,+∞[, where

(1.7) t0 =

Smaxx∈MB(x) 2k−n4k

and

(1.8) Φ (t0) = 1

2 − 1

2] Smax

x∈MB(x) 2k−n

2k

= k nt0. Lemma 1. Let θ >0 such that

a 2

2

2\

< θ2 < 1

2 (n−k) =a22\ where

a= 1

(2 (n−k))2

\ 2

and put

t1 =θt0. Then we have the following inequality (1.9) Ψ (t1)≤θ22]+ 2

2]−2Φ (t0)< 1

2kΦ (t0). Proof. In fact

Ψ (t1) = 1 2t21+

Smaxx∈MB(x) t21]

2]

2 1

2t20+Smax

x∈MB(x)θ2]−2t20] 2]

! . Since

(1.10) t20]

Smaxx∈MB(x)

=t20

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we get

Ψ (t1) =θ2 1

2t202]−2 2] t20

!

2t20

"

1

2+ θ2]−2 2]

#

≤θ2t20 1

2 +n−2k 2n

≤θ2t202n−2k 2n

≤(n−k)θ2t20 n. And since

2]+ 2

2]−2 = n−k

k and Φ (t0) =kt20 n we infer that

Ψ (t1)≤θ22]+ 2

2]−2Φ (t0)< 1

2kΦ (t0) .

Now we check the Mountain-Pass lemma conditions for the functionalIε. Lemma 2. The functionalI satisfies the following conditions

i) There exit ρ >0 and t0 >0 such that I(u)≥ρ for everyu∈Hk2(M) with kuk=t0

ii)There exits isv∈Hk2(M) such that

kvk=t2> t0 and I(v)<0.

Proof. Letϕ∈C(M),ϕ >0 on M and without loss of generality we can assume kϕk= 1. Put

(1.11) C(n, p, k) = (2k−1)θ2]+p

4n ≤C1(n, k) = (2k−1)θ2]

4n .

The inequality (1.3) becomes

(1.12) 1

2] Z

M

A(x)

(t1ϕ)2]dvg ≤ 2k−1 4k Φ (t0) . Indeed, we have

(1.13) Φ(to) = k

nt2o. and by (1.11), we get

1 2\

Z

M

A(x)

(t1ϕ)2]dvg ≤ C1(n, k) t2o\θ2\

S.max

M B(x) 2+2

] 2−2]

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= 2k−1 4k Φ(to).

Analogously and by putting

C2(n, p, k) = (2k−1)θp−1 4n we obtain

(1.14) 1

p−1 Z

M

C(x)

(t1ϕ)p−1dvg ≤ 2k−1 4k Φ(to).

Let us prove know (i). By relations (1.6), (1.9), (1.12) and (1.14), we infer that

I(t1ϕ)≤Ψ (kt1ϕk)+1 2]

Z

M

A(x)

ε+ (t1ϕ)2

2[dvg+ 1 p−1

Z

M

C(x)

ε+ (t1ϕ)2p−12 dvg

(1.15)

≤Ψ (t1)+1 2]

Z

M

A(x)

ε+ (t1ϕ)2

2[dvg+ 1 p−1

Z

M

C(x)

ε+ (t1ϕ)2p−12

dvg ≤Φ (t0). Again from (1.6), we deduce that

I(t0ϕ)≥Φ (t0) + 1 2]

Z

M

A(x)

ε+ (t0ϕ)2

2[dvg+ 1 p−1

Z

M

C(x)

ε+ (toϕ)2 p−1

2

dvg

and since A andC are assumed with positive values, we obtain (1.16) I(t0ϕ)≥Φ (t0) .

Finally from (1.15) and (1.16), we get

I(t1ϕ)<Φ (t0)≤I(t0ϕ) Now to have (i) just take

u=t0ϕand ρ= Φ (t0) . it remains to prove (ii); to do so we remark that

t→+∞lim I(tϕ) = lim

t→+∞

 1

2ktϕk2Pg − 1 2]

Z

M

B(x) (tϕ)2]dvg− A(x)

ε+ (tϕ)22[

dvg

− lim

t→+∞

1 p−1

Z

M

C(x)

ε+ (tϕ)2p−12 dvg

= lim

t→+∞t2] 1

2t2]−2 − 1 2]

Z

M

B(x)ϕ2]dv(g)

. Since by assumptionB >0, we have

t→+∞lim I(tϕ) =−∞

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consequently there ist2 such that

t2 > t0 and I(t2ϕ)<0

Hence, puttingu2=t2ϕ,we get the assumption (ii) of Lemma 2.

Lemma 1 allows us to apply the Mountain-Pass Lemma to the functional I. Let

C= inf

γ∈Γmax

u∈γI(u)

where Γ denotes the set of paths in Hk2(M) joining the functions u1 =t1ϕ and u2 =t2ϕ.

So Cε is a critical value of Iε and moreover C>Φ (t0)

and by puttingγ(t) =tϕ, for t∈[t1, t2], we see thatCε is uniformly when εgoes to 0, so we have

(1.17) 0<Φ (t0)< C≤C forεsufficiently small and C >0 not depending onε.

Consequently there exists a sequence (um)m of functions in Hk2(M) such that

(1.18) I(um) →

m→+∞C and DI(uk) →

m→+∞0

By Lemma 2 the sequence (um)m∈N ofHk2(M) is a Palais-Smale sequence (P-S) for the functional I.

Theorem 2. The Palais-Smale sequence (um)m∈N is bounded in Hk2(M) and converges weakly to nontrivial smooth solutionuεof equation (1.5).

Proof. By (1.18) we get for anyϕ∈Hk2(M) DI(um)ϕ=o(1) i.e. for any ϕ∈Hk2(M) one has

(1.19)

Z

M

ϕPgumdvg = Z

M

B(x) u+m2]−1

ϕdvg

+ Z

M

A(x)u+mϕ

ε+ u+m22[+1dvg+ Z

M

C(x)u+mϕ

ε+ u+m2p2+1dvg+o(1) in particular, forϕ=um we have

Z

M

umPgumdvg− Z

M

B(x) u+m2]

dvg = Z

M

A(x) (u+m)2

ε+ u+m

22[+1dvg

+ Z

M

C(x) (u+m)2

ε+ (um)2 p

2+1dvg+o(1) .

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or

−1 2

Z

M

umPgumdvg+1 2

Z

M

B(x) u+m2

dvg

(1.20) +1 2

Z

M

A(x) (u+m)2

ε+ u+m22[+1dvg+ Z

M

C(x) (u+m)2

ε+ u+m

2p2+1dvg =o(1) . On the other hand it comes from (1.18) that

1 2

Z

M

umPgumdvg− 1 2]

Z

M

B(x) u+m2]

dvg (1.21)

+1 2]

Z

M

A(x) (u+m)2

ε+ u+m

22[+1dvg+1 p

Z

M

Z

M

C(x) (u+m)2

ε+ u+m

2p

2+1dvg =C+o(1) . So by adding (1.20) and (1.21) we get

(1.22) 2 n

Z

M

B(x) u+m2]

dvg+1 2

Z

M

A(x) (u+m)2

ε+ u+m

22[+1dvg+1 2]

Z

M

A(x) (u+m)2

ε+ u+m

22[+1dvg

1 2

Z

M

C(x) (u+m)2

ε+ u+m

2p2+1dvg+ 1 2]

Z

M

C(x) (u+m)2

ε+ u+m

2p2+1dvg =C+o(1) . For sufficiently largem we deduce that

2 n

Z

M

B(x) u+m2]

dvg ≤2C+o(1)

or 1

2] Z

M

B(x) u+m2]

dvgdv(g)≤ n

2]C+o(1) and plugging this last inequality with in (1.21) we obtain

1 2

Z

M

umPgumdv(g)≤C+ n

2]C+o(1)

≤nC+n(n−4)

2n C+o(1)≤2 (n−2)C+o(1) . Hence for m large enough

Z

M

umPgumdv(g)≤4nC+o(1)≤4nC+ 1 i.e.

(1.23) kumk2≤4nC+ 1

Thus we prove the sequence (um)mis bounded inHk2(M) so we can extract a subsequence, still denoted (um)m which verifies:

1. um →uε weakly inHk2(M).

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2. um →uε strongly in Lp(M),∀p < n−2k2n 3. um →uε a.e. in M.

4. (um)2]−1→u2ε]−1 weakly inL

2] 2]−1 (M).

Furthermore, putting g(x) = ε1q, where ε > 0 and q > 0, we get by Lebesgue’s dominated convergence theorem that

∀k∈N:

u+m2

−q

< ε−q and ε−q ∈Lp(M) ∀p≥1 thus

(u+m)2−q

→ u2

−q

strongly in Lp(M) ∀p≥1 and with (ii) we infer that u+m

(u+m)2q u+

ε+(u+ε)2q strongly inL2(M). So if we let m go to +∞ in (1.19) we obtain that u is a weak solution to the equation (1.24) Pguε=B(x) u+ε2]−1

+ A(x)u+ε

ε+ u+ε

22[+1 + C(x)u+ε

ε+ u+ε2p+12 where 2b= 22\ and p >1.

Our solution u is not identically zero; indeed by (1.21) we have

−1 2]

Z

M

B(x) u+m2]

dv(g)≤Cε+o(1)≤2Cε+o(1) or

−2 n

Z

M

B(x) u+m2]

dv(g)≤ 4

n.2]Cε+o(1) . In addition by (1.22), we get

1 2]

Z

M

A(x)

ε+ (um)2

2[dv(g)≤Cε− 2 n

Z

M

B(x) u+m2]

dv(g)

≤Cε+ 4

n2]Cε+o(1)≤ 2]−1

Cε+o(1) . Letting k→+∞ and taking into account of (1.17) we infer that

(1.25) 1

2] Z

M

A(x)

ε+ u+ε22[dv(g)≤

2]−1 C

whereC is the upper bound ofCε. Now if for a sequence εj →0

j→+∞

( withj >0,∀j ∈N); uεj goes to 0, then it follows

(1.26) 1

2](2]−1)ε2j[ Z

M

A(x)dv(g)≤C.

So if j →+∞,it leads to a contradiction since by assumptionA >0.

Finally, forεsufficiently small,uε is a solution not identically zero of the equation (1.5).

By writing the equation (1.24) in the form

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kgu+

k−1

X

l=1

Al(x)

lu,∇lu +

A0(x)− A(x)

ε+ u+22b+1 − C(x)

ε+ u+

2p+1

2

 u+ε

=B u+ε2\−1

where Al is a smooth tensor field of type T02l and symmetric in the sense thatAl(X, Y) =Al(Y, X) for tensorsX,Y of type Tl0 on M (see [12]).

Noticing that

A0(x)− A(x)

ε+ u+

22b+1 − C(x)

ε+ u+2p+12

∈L(M)

we deduce by standard regularity theory that u∈C2k(M).

Now we are in position to prove Theorem 1.

Proof. From what precedes u is a C2k(M) nontrivial solution to equation (1.5), moreover u is a weak limit of the sequence (uk)kwhich allows us by the lower semicontinuity of the norm to write

kuk ≤ lim

k→+∞infkukk.

And by the inequalities (1.17), (1.23) we deduce that the sequence (uε)ε of the ε-approximating solutions is bounded in Hk2(M) for sufficiently small >0 i.e.

(1.27) kuk2≤4nC+ 1

thus we can extract a subsequence still labelled (uk)k satisfying:

i)uk−→u weakly inHk2(M)

ii) uk−→u strongly in Lp(M) for p <2] iii)uk−→u a.e. in M.

vi)u2k]−1−→u2]−1weakly in L

2] 2]−1.

Furthermore the sequence (uk)kis bounded below: indeed as the functions uk are continuous, denote by xk their respective maximums on M and put xo = limxk ( a subsequence of (xk)k still labelled (xk)k ). Since by assumption the operator Pg admits a positive Green function, then we can write

uk(xk) = Z

M

G(xk, y)

B(y) u+k (y)2]−1

+ A(y)u+k (y)

ε+ u+k (y)22[+1 + C(y)u+k(y)

ε+ u+k (y)2p+1

2

 dvg

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and by the Fatou’s lemma, we get lim inf

k uk(xk)≥ Z

M

lim inf

k G(xk, y)

B(y) u+k (y)2]−1

+ A(y)u+k (y)

ε+ u+k (y)22[+1 + C(y)u+k (y)

ε+ u+k (y)2p+12

 dvg

= Z

M

lim inf

k G(xk, y)

B(y) u+(y)2]−1

+ A(y)u+(y)

ε+ (u+(y))2

2[+1 + C(y)u+(y)

ε+ (u+(y))2 p+12

 dvg.

And since the functions A, B, C are positive, then lim infkuk(xk) = 0 implies that u+ = 0. This contradicts relation (1.26). Thus, there exists δ > 0, such that uk ≥ δ. We can once again use Lebesgue’s dominated convergence theorem to get

1

εk+ (uk)2q → 1

(u)2q strongly in Lp(M) ,∀p≥1,∀q ≥1.

Since for klarge enough uk>0 there is ˜ε >0 such that 1

εk+u2kq ≤ 1

˜

εq withq >0.

Thus by the Lebesgue’s dominated convergence theorem, we infer that 1

εk+ (uk)2

q −→ 1

u2q strongly inLp(M) , ∀p≥1,∀q >0.

Finally, with ii), it follows that uk

εk+ (uk)2

2[+1 −→ 1

u2]+1 strongly in L2(M) .

with u >0. Letting εk → 0 in (1.24) as k→ +∞,we get that u is a weak solution of equation (1.1) and since the regularity of u is the one of the functionγ(x) =|x|n−2k4k and since u >0, thenu∈C(M).

2. Existence of a second solution

According to the previous section our functional admits a local maximum Cε , this means the following inequalities

Iε(t1ϕ)<Φ(t0)< Iε(t0ϕ)≤Cε

where t0, t1 are real numbers satisfying 0< t1 < t0 and ϕ∈C(M) with ϕ >0 andkϕk= 1.

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On the other hand and as Iε(tϕ) tends to −∞ astgoes to +∞, there is t2 >> t0 such thatIε(t2ϕ)<0. Now if we lett and εtend both to 0+, the functional laIε tends to +∞. Indeed,

t→0lim+

ε→0lim+Iε(t.ϕ)

= lim

t→0+

h

I(1)(t.ϕ) +I0(2)(t.ϕ) i

= lim

t→0+

Z

M

(t.ϕ)Pg(t.ϕ)− 2

2]B(x)(t.ϕ)2]

dv(g)

+ lim

t→0+

 1 2]

Z

M

A(x)

(t.ϕ)2]dv(g) + 1 p−1

Z

M

C(x)

(t.ϕ)p−1dv(g).

= +∞.

and it follows that for ε small enough, there is near 0 a real number 0 <

t0 << t1 such that Iε(t0ϕ) > Φ(t0 > Iε(t1ϕ). What let us see, from this effect, that our function has a local lower bound. We will give the necessary conditions for this lower bound to exist, then we show by Ekeland’s lemma that this lower bound is reached.

We will need the following version of the Ekeland’s lemma

Lemma 3. Let V be a Banach space,J be aC1lower bounded function on a closed subsetF ofV andc= infF J. Letuε ∈F such thatc≤J(uε)≤c+ε.

Then there is uε ∈F such that

c≤J(uε)≤c+ε kuε−uεkV ≤2√

ε

∀u∈F, u6=uε, J(u)−J(uε) +√

εku−uεkV >0.

If moreover, uε is in the interior of F, then kDJ(uε)kV0 ≤√

ε.

We can consider the sequence ( uε )ε in the interior of F. Indeed if uε is on the border of F then by the continuity of J there is uε belonging to interior of F such that |J(uε)−J(uε)|< ε. Which gives, for ε sufficiently, c−ε < J(uε)< c+ 2ε and J(u)−J(uε) +√

εku−uεkV =J(u)−J(uε) + J(uε)−J(uε) +√

εku−uε+uε−uεkV

≥ J(u)−J(uε)−ε+√

εku−uεkV −√

εkuε−uεkV > J(u)−J(uε) +

√εku−uεkV −2ε >0. So we can speak about the differential DJ(uε).

Before stating the main result of this section we will establish some pre- liminary lemmas.

Lemma 4. Let θ >0 such that (2.1)

a 2

2

2]

< θ2 < a

2 2\

where

a= 1

(2 (n−k))2

\ 2

(14)

and put

t3 = a

2 1

2]

t0

then we have the following inequality

(2.2) Φ(t3)> a

2Φ(t0).

Proof. Since (t1 being defined as in Lemma 1),

(2.3) t3 =

a 2

1

2]

t0 < θt0 =t1. and

(2.4)

a 2

2

2]

> a 2. by (1.7), we get

Φ(t3) = 1 2t23

S.max

M B(x) t23]

2]

= 1 2

a 2

1

2]

.t0 2

− 1 2]t20a

2

= 2n k

1 2

a 2

2

2]

−a 2

n−2k 2n

k 2nt20. Knowing by (1.13) that

k

2nt20= 1 2Φ(t0) we deduce

Φ(t3) = n

k a

2 2

2]

−n a 2k +a

1 2Φ(t0)

= n

k a

2 2

2]

− a 2

+a

1 2Φ(t0)

> a 2Φ(t0).

Where we used the inequality (2.4) in the last line.

Lemma 5. Given a Riemannian compact manifold (M, g ) of dimension n >2k, k∈N? and 3< p <2]+ 1.

So there is a constantλ>0such that: ∀ε∈]0, λ?[the following inequalities take place

(2.5) Z

M

A(x) (ε+ (t3ϕ)2)2

] 2

dvg ≥ 2 a.

 Z

M

pA(x)dvg

2

1 t0a1

2]

(15)

and

(2.6) Z

M

C(x) (ε+ (t3ϕ)2)

p−1 2

dvg≥ 2

a p−1

2]

 Z

M

pC(x)dvg

2

1 t0a2

p−1

where a1, a2 are positive constants andt3, a are chosen as in Lemma 4.

Proof. Letϕ∈C(M), ϕ >0 in M with kϕk= 1. Put

(2.7) β1=

a 2V (M)

2

2] t0 a1

√2 2

and

(2.8) β2 =

1 V (M)

p−12 a

2 2

2]

t0a2

√2 2

whereV(M) denotes the volume ofM. Let λ? = min (β1, β2) . By H¨older’s inequality, we get

 Z

M

pA(x)dvg

2

=

 Z

M

pA(x) [ε+ (t3ϕ)2]2

] 4

ε+ (t3ϕ)22

] 4 dvg

2

 Z

M

A(x) (ε+ (t3ϕ)2)2

] 2

dvg

.

 Z

M

ε+ (t3ϕ)22

] 2 dvg

(2.9) ≤I

kε+ (t3ϕ)2k2] 2

2

] 2

where

I = Z

M

A(x) (ε+ (t3ϕ)2)2

] 2

dvg.

Independently, the Minkowski’s inequality can be written kε+ (t3ϕ)2k2]

2

≤ kεk2] 2

+t232k2] 2

consequently (2.10)

kε+ (t3ϕ)2k2] 2

2

] 2

kεk2] 2

+t232k2] 2

2

] 2

. Notice that

kεk2] 2

=ε.[Vg(M)]

2 2]

and

2k2] 2

=kϕk22]

(16)

By continuity of inclusion Hk2(M) ⊂L2](M), we deduce the existence of a positive constant S1 such that

kϕk2]≤S1kϕkH2 k

and as the operator Pg is coercive, the norms k.kH2

k and k.k are equivalent (this results from Proposition2, reference [12]) and the exists an another constantC >0 such that

kϕkH2

k ≤CkϕkPg ∀ϕ∈Hk2(M) and therefore (2.10) becomes

kε+ (t3ϕ)2k2] 2

2

] 2

ε(V(M))

2

2] +t23(S1C)2 2

] 2

where we have used the fact thatkϕkPg = 1.

Taking account of

t3 = a

2 1

2]

t0

and letting a1 =√

2S1C, (2.9) becomes

 Z

M

pA(x)dvg

2

≤I ε[Vg(M)]

2 2] +

aε 2

2

2]

t20 a1

√ 2

2!2

] 2

.

And since 0< ε < λ? ≤β1,we get

 Z

M

pA(x)dvg

2

≤I 2aε 2

2

2] t0α1

√2 2!2

] 2

.

Finally we deduce I ≥ 2

aε

 Z

M

pA(x)dvg

2

1 t01

2]

.

Let us now prove the second inequality; again with the H¨older’s inequality, we have

 Z

M

pC(x)dvg

2

=

 Z

M

pC(x) [ε+ (t3ϕ)2]p−14

. ε+ (t3ϕ)2p−14 dvg

2

(2.11)

 Z

M

C(x) (ε+ (t3ϕ)2)

p−1 2

dvg

.

 Z

M

ε+ (t3ϕ)2p−12 dvg

≤J.

h

kε+ (t3ϕ)2kp−1 2

ip−1 (2.12) 2

(17)

where

J = Z

M

C(x) (ε+ (t3ϕ)2)p−12

dvg. Again with Minkowski’s inequality, we have

kε+ (t3ϕ)2kp−1 2

≤ kεkp−1

2 +t232kp−1 2

(2.13)

≤ε.[Vg(M)]p−12 +t23kϕk2p−1 (2.14)

and since the embeddingHk2(M)⊂Lp−1(M) is continuous, there is a posi- tive constant S2 such that

kϕkp−1≤S2kϕkH2 k. The norms k.kH2

k,k.kbeing equivalent, there is a positive constantC1 such that

kϕkH2

k ≤C1kϕk, ∀ϕ∈Hk2(M).

consequently (2.11) becomes

 Z

M

pC(x)dvg

2

≤J

εVg(M)p−12 +t32S22C2p−12 .

Replacing t3 by his expression and posing a2 =√

2S2C1 it comes that

 Z

M

pC(x)dvg

2

≤εJ.V(M)p−12 + a

2 2

2]

t0

a2

√ 2

2

now since 0< ε < λ? whereλ?≤β2 , then it follows that

 Z

M

pC(x)dvg

2

≤J 2a 2

2

2] t0 a2

√2

2!p−12

i.e.:

Z

M

C(x) (ε+ (t3ϕ)2)

p−1 2

dvg

 Z

M

pC(x)dvg

2

2 a

p−1

2] 1 t0a2

p−1

. Now we are able to prove the existence of a second solution to equation (1.1)

Theorem 3. Let (M, g) be a compact Riemannian manifold of dimension n >2k, (k∈N?). Suppose the operator Pg is coercive, has a Green positive

(18)

function and there is a constant C(n, p, k)>0which depends only on n, p , k such that:

kϕk2] 2]

Z

M

A(x)

ϕ2\ dvg ≤C(n, p, k)

Smaxx∈MB(x) 2+2

] 2−2]

and

kϕkp−1 p−1

Z

M

C(x)

ϕp−1dvg ≤C(n, p, k)

Smaxx∈MB(x) p+1

2−2]

for some smooth function ϕ > 0 on M. If moreover for every ε∈ ]0, λ?[ , where λ? is a positive constant, the following two conditions occur

(2.15) 2

a

 Z

M

pA(x)dvg

2

1 t0a1

2]

>2]kt20

4n(2−a) and

(2.16)

2 a

p−1

2]

 Z

M

pC(x)dvg

2

1 t0a2

p−1

>(p−1)kt20

4n(2−a) where a1, a2 are positive constants, 2] = n−2k2n , 3 < p < 2]+ 1. Then the equation (1.1) admits a second smooth solution.

Proof. The proof will be done in three steps

1th-step. The functionalIε has a local lower bound.

This consists to find a strictly positive real number λ? such ∀ε∈ ]0, λ?[ one has the following inequality

Iε(t3ϕ)>Φ(t0) ∀ϕ∈C(M), kϕk= 1, with t3 < t1. Indeed, according to Lemma (4) inequality (2.2), one has

Iε(t3ϕ) =I(1)(t3ϕ) +I(2)(t3ϕ) (2.17)

> a

2Φ(t0) + 1 2]

Z

M

A(x) (ε+ (t3ϕ)2)2b

dvg

+ 1

p−1 Z

M

C(x)

(ε+ (t3ϕ)2)p−1dvg. and as by assumption

2 a.

 Z

M

pA(x)dvg

2

1 t0a1

2]

>2]kt20

4n(2−a) and

λ? = min (β1, β2)

(19)

where β1, β2 are given by (2.7) and (2.8) and knowing that Φ(t0) = k

nt20 it follows by Lemma5 that,∀ε∈]0, λ?[

(2.18) 1

2] Z

M

A(x) (ε+ (t3ϕ)2)2

] 2

dvg >

1 2 −a

4

Φ(t0).

Similarly, one has 2

a p−1

2]

 Z

M

pC(x)dvg

2

1 t0.a2

p−1

>(p−1)kt20

4n(2−a) which implies with Lemma (5) that ∀ε∈]0, λ?[

(2.19) 1

(p−1) Z

M

C(x) (ε+ (t3ϕ)2)p−12

dvg>

1 2 −a

4

.Φ(t0).

Finally, by combination of (2.17), (2.18) and (2.19) we get

Iε(t3ϕ)> a

2Φ(t0) + 2 1

2 −a 4

Φ(t0) (2.20)

>Φ(t0).

Hence our result.

2th− step. The infimum of the functionalIε is reached.

Denote by B(0, t1) =

u∈Hk2(M) :kuk ≤t1 the closed ball centred at the origin 0 of radius t1 in Hk2(M). In this section we will show that cε = infB(0,t1)Iε ( cε < Φ(t0) ) is reached. By Ekeland’s Lemma, there exists a sequence (um)m∈N in B(0, t1) such that Iε(u) → cε =InfB(0,t1)Iε

and DIε(um) → 0 strongly in the dual space of Hk2(M). That is to say (um) is a Palais-Smale sequence, so by the same arguments as in Theorem2 and Theorem1, we get that equation (1.1) has a smooth solution v. Since the ε-approximating solutions are obtained as weak limit of sequences of functions from B(0, t1), it follows by the weak lower semi-continuity of the norm that these ε-approximating solutions are in B(0, t1). As in turn v is obtained as a limit of a sequence ofε-approximating solutions,v∈B(0, t1).

3th-step. The two solutions are distinct. Indeed, we have on the one other hand the energy of the solution v given by c = limε→0+cε ≤ Φ(t0) and secondly the energy of the solution u obtained in the first section is C = limε→0+Cε where C = inf

γ∈Γmax

u∈γI(u) and Γ denotes the set of paths joining the functionsu1=t1ϕand u2 =t2ϕ. Now since any path joiningu1 andu2 intersects the sphere centred at the origin and of radiust0 inHk2(M)

(20)

then C = inf

γ∈Γmax

u∈γI(u)≥(Iε(t0ϕ)) =I(1)(t0ϕ) +Iε(2)(t0ϕ)≥Φ (t0) +Iε(2)(t0ϕ)

≥Φ (t0) + 1 2]

Z

M

A(x)

ε+ (t0ϕ)22[dvg+ 1 p−1

Z

M

C(x)

ε+ (t0ϕ)2 p−12

dvg

Thus the Lebesgue’s dominated theorem, we get C= lim

ε→0+Cε≥Φ (t0) + 1 2]

Z

M

A(x)

(t0ϕ)2\dvg+ 1 p−1

Z

M

C(x) (t0ϕ)p−1dvg

>Φ (t0).

So the solutionsu and v are of different energies and therefore are distinct.

3. Nonexistence of solution

In this section we will be placed in a closed ballB(0, R) ofHk2(M) centered at the origin 0 and of radius R > 0, we prove that under some condition (inequality 3.1) that the equation (1.1) admits no solution.

Theorem 4. Given (M, g) a compact Riemannian manifold of dimension n > 2k, (k ∈ N?) and A, B, C are positive smooth functions on M and 2< p <2]+ 1. Assume that

(3.1) C(n, p, k)

 R

M

√B.Cdvg

R

M

Bdvg

2. 2]

p−1+2]

Z

M

Bdvg >(SR)2 where S, R are positive constants and

C(n, p, k) = 2]+p−1 p−1

p−1 2]

2

] 2]+p−1

.

Then the equation (1.1) has no smooth positive solutionuwith energykukH2 k(M) ≤ R.

Proof. Suppose that there exists a smooth positive solutionu∈Hk2(M) such that kukH2

k(M) ≤R. By multiplying both sides of equation (1.1) by u end integrating overM, we get

Z

M

uPg(u)dvg= Z

M

B(x)u2]+A(x)

u2] +C(x) up−1

dvg.

And since kukpg =rR

M

uPg(u)dvg is a norm equivalent to kukH2

k(M), there exists a constant S >0 such that

kuk ≤SkukH2 k(M).

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