Soumaya S?AANOUNI

Department of Mathematics,Faculty of Sciences of Tunis,University of Tunis El Manar, Campus University 2092 Tunis,Tunisia

Nihed TRABELSI

Higher Institut of Medicals Technologies of Tunis,University of Tunis El Manar, 9 Street Dr.Zouhair Essaf1006 Tunis,Tunisia

A BIFURCATION PROBLEM ASSOCIATED TO AN ASYMPTOTICALLY LINEAR FUNCTION?

Soumaya S?AANOUNI

Department of Mathematics,Faculty of Sciences of Tunis,University of Tunis El Manar, Campus University 2092 Tunis,Tunisia

E-mail:saanouni.soumaya@yahoo.com

Nihed TRABELSI

Higher Institut of Medicals Technologies of Tunis,University of Tunis El Manar, 9 Street Dr.Zouhair Essaf1006 Tunis,Tunisia

E-mail:nihed.trabelsi78@gmail.com

We study the existence of positive solutions to a two-order semilinear elliptic problem with Dirichlet boundary condition

extremal solution;regularity;bifurcation;stability

2010 MR Subject Classifcation35B65;34D35;70K50

1 Introduction and Statement of Main Results

The main interest of non-linear physics lies in its ability to explain the evolution of the problems:a phenomenon usually depends on a number of parameters,called control parameters that control the evolution of the system.By variation of the parameters and the result of nonlinearities,the system may undergo transitions.In math,they are bifurcations.

In biology;population dynamics,the equations of reactions-difusion were proposed by Turing(1952)for modeling morphogenesis phenomena.A model of interaction of species or chemicals is given by the diferential equations

where u represents the density,div(c?u)represents the substance of difusion through the system,and f models the interaction of substances.Moreover the broadcast functions c and the terms of reactions f may depend on(x,t)and the concentrations u it as a nonlinearity way.

In case where c and t are constants,various authors have studied the existence of weak solutions for the bifurcation problem

where ? is a bounded open subset of Rn,n≥2.

Mironescu and R?adulescu proved in[13,14]that there exists 0<λ?<∞,a critical value of the parameter λ,such that(Eλ)has a minimal,positive,classical solution uλfor 0<λ<λ?and does not have a weak solution for λ>λ?.

Our main interest here will be in the study of a bifurcation problem for λ>0

which is a generalization of(Eλ).Where ? be a smooth bounded domain in Rn(n≥2),c(x) is a smooth bounded positive function onis a positive,increasing and convex smooth function,such that

The value a was be crucial in the study of(Eλ?)and of the behavior of uλwhen λ tends to λ?.

Throughout the paper,we denote by k·k2,the L2(?)-norm,whereas we denote by k·k, the H10(?)-norm given by

We say that u∈H10(?)is a weak solution of the problem(Pλ),if f(u)∈L1(?)and

Such solutions are usually known as weak energy solutions.For short,we will refer to them simply as solutions wish is assuredly by the next lemma.

Lemma 1.1Since f(t)≤at+f(0),if u∈L1(?)is a weak solution of(Pλ),it is easily seen by a standard bootstrap argument that u is always a classical solution.

In the rest of this paper,we denote by a solution of problem(Pλ)any weak or classical solution.

Defnition 1.2We say that a solution uλof problem(Pλ)is minimal if uλ≤u in ? for any solution u of(Pλ).

Defnition 1.3We say that u∈H10(?)is a supersolution of(Pλ)if f(u)∈L1(?)and

Reversing the inequality one defnes the notion of subsolution.

Defnition 1.4A solution u of problem(Pλ)is stable if and only if the frst eigenvalue of the linearized operator

given by

is nonnegative.

If η1(λ,u)<0,the solution u is said to be unstable.

Let v1be a positive eigenfunction associated with the frst eigenvalue λ1of the operator?div(c(x)?)with Dirichlet boundary conditions,namely,

Next,we let

We also let

The two values a and r0that we have already defned will be important in the bifurcation phenomena.More precisely,in the frame of the critical value λ?.

In this article,we want to show the following three theorems

Theorem 1.5There exists a critical value λ?∈(0,∞)such that the following properties hold true

(i)for any λ∈(0,λ?),problem(Pλ)has a minimal solution uλ,which is the unique stable solution of(Pλ);

(ii)for any λ∈(0,λ1/a),uλis the unique solution of problem(Pλ);

(iii)the mapping λ 7→uλis increasing;

An important role in our arguments will be played by

We distinguish two diferent situations strongly depending on the sign of l.

Theorem 1.6Assume that l≥0.Then

(i)λ?=λ1/a;

(ii)problem(Pλ?)has no solution;

Fig.1 Behavior of the minimal solution

Theorem 1.7Assume that l<0.Then the critical value λ?belongs to(λ1/a,λ1/r0) and(Pλ?)has a unique solution u?.In this case,problem(Pλ)has an unstable solution vλfor any λ∈(λ1/a,λ?)and the sequence(vλ)λhas the following properties:

Fig.2 Bifurcation branches in the case l<0

2 Proof of Theorem 1.5

The frst part of this article is conserved to prove Theorem 1.5,when the existence of the critical value λ?is a consequence of the following auxiliary result.

Lemma 2.1Problem(Pλ)has no solution for any λ>λ1/r0,but has at least one solution provided λ is positive and small enough.

ProofTo show that(Pλ)has a solution,we use the barrier method.To this aim,letwhich satisfes

The choice of w implies that w is a super-solution of(Pλ)for λ≤1/f(kwk∞).

Notice that for any λ>0,the function w≡0 is a sub-solution of(Pλ)since f(0)>0.

Next,we defne a sequence wn∈H2(?)by

The maximum principle(see[5])implies that

so that the sequence(wn)n≥0is increasing and bounded,then it converges.It follows that problem(Pλ)has a solution.

Assume now that u is a solution of(Pλ)for some λ>0.Using v1given in(1.1)as a test function and integrating by parts,we get

This yields

Since v1>0 and u>0,we conclude that the parameter λ should belong to(0,λ1/r0).This completes our proof.

Another useful result is stated in what follows.

Lemma 2.2Assume that problem(Pλ)has a solution for some λ∈(0,λ?).Then there exists a minimal solution denoted by uλ.Moreover,for any λ′∈(0,λ),problem(Pλ′)has a solution.

ProofFix λ∈(0,λ?)and let u be a solution of(Pλ).As above,we use the barrier method to obtain a minimal solution of(Pλ).The basic idea is to prove by induction that the sequence(wn)n≥0defned in(2.1)is increasing and bounded by u,so it converges to some solution uλ.Since uλis independent of the choice of u,then it is a minimal solution.

Now,if u is a solution of(Pλ),then u is a super-solution for problem(Pλ′)for any λ′in (0,λ)and 0 can be used always as a sub-solution.These complete the proof.

Remark 2.3Thanks to Lemmas 2.1 and 2.2,the set Λ is an interval not empty and bounded.

Proof of(i)of Theorem 1.5First,we claim that uλis stable.Indeed,arguing by contradiction,i.e.,the frst eigenvalue η1(λ,uλ)is negative.Then,there exists an eigenfunction ψ∈H2(?)such that

Consider uε:=uλ?εψ.Hence,by linearity,we have

Since η1(λ,uλ)<0,for ε>0 small enough,we have

Then,for ε>0 small enough,we use the strong maximum principle(Hopf’s lemma,see[9]) to deduce that uε≥0 is a super-solution of(Pλ).As before,we obtain a solution u such that u≤uεand since uε

Now,we show that(Pλ)has at most one stable solution.Assume the existence of another stable solution v 6=uλof problem(Pλ).Then the function w:=v?uλ>0 satisfes

Therefore

Thanks to the convexity of f,the term in the brackets is nonpositive,hence

which implies that f is afne over[uλ,v]in ?.So,there exists two real numbersˉa and b such that

Finally,since uλand v are two solutions to?div(c(x)?w)=λˉaw+λb,we obtain that

This is impossible since b=f(0)>0 and w=v?uλis positive in ?.

which shows that

Impossible for λ∈(0,λ1/a).So,η1(λ,u)≥0 and by(i),we obtain the uniqueness of u.

For the existence,we consider the minimization problem

where

with

If λ∈(0,λ1/a),there exist ε>0 and A>0 depending on λ such that

Standard arguments imply that J(u)is coercive,bounded from below and weakly lower semi-continuous in H10(?).Hence,the minimum of J is attained by some function u∈H10(?) and also by u+since J(u+)≤J(u).So,the critical point u of J gives a solution of(Pλ).

Proof of(iii)of Theorem 1.5By sub-and super-solution method,see Lemma 2.2 we obtain that the mapping λ 7?→uλis increasing and this proves(iii).

Proof of(iv)of Theorem 1.5Consider the nonlinear operator

where α∈(0,1)and E is the function space defned by

Remark 2.4Thanks to Lemma 2.1 and(ii)of Theorem 1.5,the critical value λ?satisfes

3 Proof of Theorem 1.6

To prove this theorem,we can assume that c(x)≥1,we show that the three assertions are equivalent.And fnally,we prove that one is true.

We frst recall the following result which is due to H¨ormander[10].

Lemma 3.1Let ? be an open bounded subset of Rn,n≥2 with smooth boundary.Let (un)be a sequence of super-harmonic nonnegative functions defned on ?.Then the following alternative holds:

(ii)or(un)contains a subsequence which converges into some function u.

Remark 3.2The result of H¨ormander is also true if(un)is a sequence of a superbiharmonic nonnegative functions.

First,we assume that λ?=λ1/a.If(Pλ?)has a solution u?,then as we have already observed in(iv)of Theorem 1.5 η1(λ?,u?)=0.Thus,there exists ψ∈H2(?)satisfying

Using v1,given in(1.1),as a test function and integrating by parts,we obtain

therefore

Since λ1?λ?f′(u?)≥0,the above equation forces λ1?λ?f′(u?)=0.Hence

If not,Using v1and integrating by parts,we have

then

i.e.,

Hence,problem(Pλ?)has no solution and(i)implies(ii).

with

Since f(t)≤at+f(0)and c(x)≥1,we have

where cλis a positive constant independent on λ.

Moreover,by the trace theorem,

We deduce that

Moreover,it simply shows that(ii)and(iii)are equivalent.

Then,up to a subsequence,we obtain

Moreover,

and

Then,

Multiplying by v1,which is defned in(1.1),we obtain

This proves(i).

To fnish the proof of Theorem 1.6,we need only to show that(Pλ1/a)has no solution. Indeed,assume that u is a solution of(Pλ1/a).Since f(t)?at≥0,we have

Multiplying the previous equation by v1and integrating by parts,we get f(u)=au in ?,which contradicts f(0)>0.This concludes the proof of Theorem 1.6.

Remark 3.3Observe that the equivalence of the assertions of Theorem 1.6 does not depend on the sign of l.

4 Proof of Theorem 1.7

For the frst part of Theorem 1.7,we have already seen in Remark 2.4 that λ1/a≤λ?≤λ1/r0.Hence it sufces to prove that λ?6=λ1/a and λ?6=λ1/r0.

First,assume that λ?=λ1/a.Let uλbe the minimal solution to(Pλ).Then,multiplying (Pλ)by v1given in(1.1)and integrating,we obtain

Passing to the limit in the last inequality as λ tends to λ?,we fnd

which is impossible.

Now,assume that λ?=λ1/r0and let u be a solution of problem(Pλ?).Multiplying(Pλ?) by v1and integrating by parts,we have

Recall that η1(λ?,u?)=0,so let ψ be the corresponding eigenfunction.Multiplying the last inequality by ψ and integrating by parts,we fnd

Therefore,we must have equalityin ?,which implies that f is linear inand this leads a contradiction as in the proof of Theorem 1.6.

The second part of Theorem 1.7 concerning the existence of a non stable solution vλof (Pλ)will be proved by using the mountain pass theorem of Ambrosetti and Rabinowitz[2]in the following form.

Theorem 4.1Let E be a real Banach space and J∈C1(E,R).Assume that J satisfes the Palais-Smale condition and the following geometric assumptions:

(?)there exist positive constants R and ρ such that

(??)there exists v0∈E such that kv0?u0k>R and J(v0)≤J(u0).

Then the functional J possesses at least a critical point.The critical value is characterized by

where

and satisfes

In our case,

where E is the function space defned in(2.2)and

We take u0as the stable solution uλfor each λ∈(λ1/a,λ?).

Remark 4.2The energy functional J belongs to C1(E,R)and

Since η1(λ,uλ)≥0,the function uλis a local minimum for J,and in order to transform it into a local strict minimum,we apply the mountain pass theorem not for J,but to the perturbed functional Jεdefned by

for all ε∈[0,ε0],where

We observe that Jεis also in C1(E,R)and

Using the same arguments of Mironescu and R?adulescu in[14,Lemma 9],we show in the next lemma that Jεsatisfes the Palais-Smale compactness condition.

Lemma 4.3Let(un)?E be a Palais-Smale sequence,that is,

Then(un)is relatively compact in E.

ProofSince any subsequence of(un)verifes(4.1)and(4.2)it is enough to prove that (un)contains a convergent subsequence.It sufces to prove that(un)contains a bounded subsequence in E.Indeed,suppose we have proved this.Then,up to a subsequence,un?→u weakly in E,strongly in L2(?).Now(4.2)gives that

Note that f(un)?→f(u)in L2(?)because|f(un)?f(u)|≤a|un?u|.This shows that

That is

The above equality multiplied by u gives

Now(4.2)multiplied by(un)gives

in view of the boundedness of(un)and the L2(?)-convergence of unand f(un),we have

and

Hence,(4.3)and(4.4)gives

which insures us that un?→ u in E.Actually,it is enough to prove that(un)is(up to a subsequence)bounded in L2(?).Indeed,the L2(?)-boundedness of(un)implies that E-boundedness of(un)as it can be seen by examining(4.1).

We shall conclude the proof obtaining a contradiction from the supposition that kunk2→∞.Let un=knwnwith kn>0,kn?→∞and kwnk2=1.Then

which can still write as follows

However,since|f(t)|≤a|t|+b,we have

This shows that

We claim that

Indeed,(4.2)divided by kngives

for each v∈E.Now

Hence(4.5)can be concluded from(4.6)if we show that 1/knf(un)converges(up to a subsequence)to a w+in L2(?).Now 1/knf(un)=1/knf(knwn)and it is easy to see that the required limit is equal to a w in the set{x∈?:wn(x)?→w(x)6=0}.

If w(x)=0 and wn(x)?→w(x),let ε>0 and n0be such that|wn(x)|<ε for n≥n0. Then

that is the required limit is 0.Thus,f(un)/kn→a w+a.e.Here b=f(0).Now wn→w in L2(?)and,thus,up to a subsequence,wnis dominated in L2(?)(see theorem IV.9 in[5]).

Since 1/knf(un)≤a|wn|+1/knb,it follows that 1/knf(un)is also dominated.Hence(4.5) is now obtained.Now(4.5)and the maximum principle imply w≥0 and(4.5)becomes

Since u?uλis not harmonic,we can choose

and this makes uλbecomes a strict local minimal for Jε,which proves(?).

Hence

This yields,using the defnition of v1mentioned in(1.1),

since kv1k2=1,then we have

which implies that,

Therefore

So,there exists v0∈E such that

and(??)is proved.Finally,for all ε∈[0,ε0],let vε(respectively,cε)be the critical point (respectively,critical value)of Jε.

Remark 4.4The fact that Jεincreases with ε implies that for all ε∈[0,ε0],cε∈[c0,cε0[. Then,cεis uniformly bounded.Thus,for all ε∈[0,ε0],the critical point vεsatisfes kvε?uλk≥R.

Recall that for any ε∈[0,ε0],the function vεbelongs to E and satisfes

and

Thanks to Lemma 4.3,Remark 4.4,(4.8)and(4.9),there exists v∈E such that

satisfying

From Remark 4.4,we see that v 6=uλ,which can be also proved using the same arguments of Mironescu and R?adulescu in[14].

Indeed,note that vεis a solution to(4.8)which is diferent from the unique stable solution uλ.Then,vεis unstable,that is,

since(4.8)can be written as

where gεis convex,positive and hεis positive.Thus,if(4.10)has solutions satisfying vε= 0 in??,then it has a minimal one,say wε,which is stable.Now,thanks to Theorem 1.5,all other solutions vεof(4.10)are unstable.

The next lemma states that the limit of a sequence of unstable solutions is also unstable (the proof is similar to that of Lemma 11 in[14]).

Lemma 4.5Let un? u in H10(?)andμn→ μbe such that η1(μn,un)<0.Then, η1(μ,u)<0.

ProofThe fact that η1(μn,un)<0 is equivalent to the existence of a ?n∈H10(?)such that Z

Since f′≤a,(4.11)shows that(?n)is bounded in H10(?).

Let ?∈E be such that,up to a subsequence,?n?? in H10(?).Then

This can be seen by extracting from(?n)a subsequence dominated in L2(?)as in Theorem IV.9 in[5].Now we have Z

fnally,since k?k2=1,we get

Obviously,the fact that the function v belongs to C2(ˉ?)∩E follows from a bootstrap argument.

Proof of(i)of Theorem 1.7Thanks to Lemma 3.1,if(i)does not occur,then there is a sequence of positives scalars(μn)and a sequence(vn)of unstable solutions to(Pμn)such that vn→v infor some function v.

We frst claim that(vn)cannot be bounded in E.Otherwise,let w∈E be such that,up to a subsequence,

Therefore,

and

which implies that?div(c(x)?w)=λ1af(w)in ?.It follows that w∈E and solves(Pλ1/a). From Lemma 4.5,we deduce that

Relation(4.12)shows that w 6=uλ1/awhich contradicts the fact that(Pλ1/a)has a unique solution.Now,since?div(c(x)?vn)=μnf(vn),the unboundedness of(vn)in E implies that this sequence is unbounded in L2(?),too.To see this,let

Then

So,we have convergence also in the sense of distributions and(wn)is seen to be bounded in E with standard arguments.We obtain

The desired contradiction is obtained since w∈E.

Proof of(ii)of Theorem 1.7As before,it is enough to prove the L2(?)boundedness of vλnear λ?and to use the uniqueness property of u?.Assume that kvnk2→∞asμn→λ?, where vnis a solution to(Pμn).We write again vn=lnwn.Then,

The fact that the right-hand side of(4.13)is bounded in L2(?)implies that(wn)is bounded in E.Let(wn)be such that(up to a subsequence)

A computation already done shows that

which forces λ?to be λ1/a.This contradiction concludes the proof.

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?Received May 27,2015;revised April 25,2016.

where ??Rn;n≥2 is a smooth bounded domain;f is a positive,increasing and convex source term and c(x)is a smooth bounded positive function on ?.We also prove the existence of critical value and claim the uniqueness of extremal solutions.