A BIFURCATION PROBLEM ASSOCIATED TO AN ASYMPTOTICALLY LINEAR FUNCTION∗

Soumaya SˆAANOUNI

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 SˆAANOUNI

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 Ω 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 Ω 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(Ω)-norm,whereas we denote by k·k, the H10(Ω)-norm given by

We say that u∈H10(Ω)is a weak solution of the problem(Pλ),if f(u)∈L1(Ω)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(Ω)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 Ω for any solution u of(Pλ).

Defnition 1.3We say that u∈H10(Ω)is a supersolution of(Pλ)if f(u)∈L1(Ω)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(Ω)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(Ω)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 Ω.So,there exists two real numbers¯a and b such that

Finally,since uλand v are two solutions to−div(c(x)∇w)=λ¯aw+λb,we obtain that

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

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(Ω).Hence,the minimum of J is attained by some function u∈H10(Ω) 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 Ω be an open bounded subset of Rn,n≥2 with smooth boundary.Let (un)be a sequence of super-harmonic nonnegative functions defned on Ω.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(Ω)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 Ω,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 Ω,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(Ω).Now(4.2)gives that

Note that f(un)−→f(u)in L2(Ω)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(Ω)-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(Ω).Indeed,the L2(Ω)-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(Ω).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∈Ω: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(Ω)and,thus,up to a subsequence,wnis dominated in L2(Ω)(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∂Ω,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(Ω)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 ϕn∈H10(Ω)such that Z

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

Let ϕ∈E be such that,up to a subsequence,ϕn⇀ϕ in H10(Ω).Then

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

fnally,since kϕk2=1,we get

Obviously,the fact that the function v belongs to C2(¯Ω)∩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 Ω.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(Ω),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(Ω)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(Ω)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 Ω⊂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 Ω.We also prove the existence of critical value and claim the uniqueness of extremal solutions.