Existence results for a mixed boundary value problem

Existence results for a mixed boundary value problem

Armin Hadjian

and Stepan Tersian

1Department of Mathematics, Faculty of Basic Sciences, University of Bojnord, P.O. Box 1339, Bojnord 94531, Iran

2Department of Mathematics, University of Ruse, 7017-Ruse, Bulgaria

Received 27 October 2014, appeared 15 November 2015 Communicated by Gabriele Bonanno

Abstract. In the present paper, we obtain an existence result for a class of mixed bound- ary value problems for second-order differential equations. A critical point theorem is used, in order to prove the existence of a precise open interval of positive eigenvalues λ, for which the considered problem admits at least one non-trivial classical solution u_λ. It is proved that the norm ofu_λ tends to zero asλ→0.

Keywords: mixed boundary value problems, existence results, critical points theory.

2010 Mathematics Subject Classification: 34B15, 58E05.

1 Introduction

The aim of this paper is to study the following mixed boundary value problem (−(pu⁰)⁰+qu=λf(x,u) +g(u) in ]a,b[,

u(a) =u⁰(b) =_0, ^(1.1)

where p ∈ C¹([a,b]) and q ∈ C⁰([a,b]) are positive functions, λ is a positive parameter, f: [a,b]×_R→_Ris a continuous function such that

(f₁) |f(x,t)| ≤a₁+a₂|t|^r−1, a.e. x∈[a,b], t∈_R,

where a₁,a₂ ≥ 0 andr ∈ ]1,+_∞[, and g: R→ _Ris a Lipschitz continuous function with the Lipschitz constant L>0, i.e.,

|g(t₁)−g(t₂)| ≤ L|t₁−t₂| for every t₁,t₂∈_{R, and}g(0) =0.

Our goal here is to obtain some sufficient conditions which imply that the problem (1.1) has at least one classical solution (see Theorem 3.1). We use the variational method and a critical point theorem.

Motivated by the fact that such kind of problems are used to describe a large class of physical phenomena, many authors looked for existence and multiplicity of solutions for

BCorresponding author. Email: a.hadjian@ub.ac.ir, hadjian83@gmail.com

second-order ordinary differential nonlinear equations, with mixed conditions at the ends.

We cite the papers [1–3,6–10,14]. For instance, in [7], Bonanno and Tornatore established the existence of infinitely many weak solutions for the mixed boundary value problem

(−(pu⁰)⁰+qu=λf(x,u) in]a,b[, u(a) =u⁰(b) =0,

where p,q∈L^∞([a,b])are such that p₀ :=ess inf

x∈[a,b] p(x)>0, q₀:=ess inf

x∈[a,b] q(x)≥0, f: [a,b]×_R→_Ris a Carathéodory function andλis a positive parameter.

We also refer the reader to the paper [11] in which, by means of an abstract critical points result of Ricceri [13], existence of at least three solutions for the following two-point boundary value problem

(u⁰⁰+ (λf(t,u) +g(u))h(t,u⁰) =µp(t,u)h(t,u⁰) in ]a,b[, u(a) =u(b) =0,

whereλandµare positive parameters, f: [a,b]×_R→_Ris continuous,g: R→_Ris Lipschitz continuous withg(0) =0,h: [a,b]×_R→_Ris bounded, continuous, withm:=infh>0, and p: [a,b]×_R→_RisL¹-Carathéodory, are ensured.

The paper is organized as follows. In Section 2 we introduce our abstract framework and we give some notations. In Section 3 we prove the main result (Theorem3.1), while Section 4 is devoted to some consequences and remarks on the results of the paper. Here we give an application of the results (Example4.7).

2 Preliminaries

In order to prove our main result, that is Theorem3.1, we report here the result obtained in [5] (see [5, Theorem 3.1 and Remark 3.3]).

Theorem 2.1. Let X be a reflexive real Banach space, letΦ,Ψ: X →_Rbe two Gâteaux differentiable functionals such thatΦis strongly continuous, sequentially weakly lower semicontinuous and coercive in X and Ψ is sequentially weakly upper semicontinuous in X. Let I_λ be the functional defined as I_λ := _Φ−λΨ,λ∈ _R, and for any r>inf_XΦlet ϕbe the function defined as

ϕ(r):= inf

u∈_Φ⁻¹(]−_∞,r[)

sup

v∈_Φ⁻¹(]−_∞,r[)

Ψ(v)

!

−_Ψ(u)

r−_Φ(u) ^{. (2.1)}

Then, for any r>infXΦand anyλ∈]0, 1/ϕ(r)[, the restriction of the functional I_λtoΦ⁻¹(]−_∞,r[) admits a global minimum, which is a critical point(precisely a local minimum)of I_λ in X.

Now, let f: [a,b]×_R →_Rbe a continuous function and g: R→_R be a Lipschitz contin- uous function with the Lipschitz constantL>_{0, i.e.,}

|g(t₁)−g(t₂)| ≤ L|t₁−t₂|

for every t₁,t₂∈_{R, and}g(0) =0.

Put

F(x,t):=

Z _t

0 f(x,ξ)dξ, G(t):=−

Z _t

0 g(ξ)dξ for all x∈[a,b]andt∈_{R. Denote}

X:= ⁿu∈W^1,2([a,b]):u(a) =0o

; the usual norm inXis defined by

kuk_X:= Z _b

a

(u(x))²dx+

Z _b

a

(u⁰(x))²dx 1/2

. For everyu,v∈X, we define

hu,vi:=

Z _b

a p(x)u⁰(x)v⁰(x)dx+

Z _b

a q(x)u(x)v(x)dx. (2.2) Clearly, (2.2) defines an inner product onXwhose corresponding norm is

kuk:= _{Z b}

a p(x)(u⁰(x))²dx+

Z _b

a q(x)(u(x))²dx 1/2

.

Then, it is easy to see that the normk · konX is equivalent tok · k_X. In fact, put p₀:= min

x∈[a,b]p(x)>0, q₀ := min

x∈[a,b]q(x)>0, m:=min{p₀,q₀}>0, and

p₁ := max

x∈[a,b]p(x), q₁:= max

x∈[a,b]q(x), M :=max{p₁,q₁}. Then, we have

m^1/2kuk_X ≤ kuk ≤ M^1/2kuk_X, ∀u∈X.

In the following, we will usek · kinstead ofk · k_Xon X. Note thatXis a reflexive real Banach space.

By standard regularity results, since f is a continuous function, p ∈ C¹([a,b]) _and q ∈ C⁰([a,b]), then weak solutions of problem (1.1) belong to C²([a,b]), thus they are classical solutions.

It is well known that the embeddingX,→C⁰([a,b])is compact and kuk_∞ ≤

s b−a

p₀ kuk (2.3)

for all u∈X(see, e.g., [15]).

Fixingr ∈ [1,+_∞[, from the Sobolev embedding theorem, there exists a positive constant cr such that

kuk_Lr([a,b])≤c_rkuk_X ≤ √^cr

mkuk, ∀u∈ X, (2.4)

and, in particular, the embedding X,→ L^r([a,b])is compact.

Suppose that the Lipschitz constantL>0 of the function gsatisfies L<_max

q₀, p₀ (b−a)²

. (2.5)

Consider the energy functional I_λ: X →_Rassociated to (1.1) defined as follows I_λ(u)_:= _Φ(u)−λΨ(u)_, ∀u∈ X,

where

Φ(_u)_:= ¹

2kuk²+

Z _b

a G(_u(_x))_dx and

Ψ(u):=

Z _b

a F(x,u(x))dx.

Lemma 2.2. Let the functionalΦbe defined as above. Then we have the following estimates for every u∈X:

q₀−L

2q₀ kuk²≤ _Φ(u)≤ ^q0+L

2q₀ kuk², (2.6)

p₀−L(b−a)²

2p₀ kuk²≤ _Φ(u)≤ ^p0+L(b−a)²

2p₀ kuk². (2.7)

Proof. Sincegis Lipschitz continuous and satisfies g(₀) =_{0, we have}

|g(t)| ≤ L|t|, ∀t∈_R, and so,

|G(t)| ≤L Z _t

0

|ξ|dξ = ^L

2t², ∀t∈_R.

Therefore, conditionq₀ >0 implies that

Z _b

a G(u(x))dx

≤ ^L 2

Z _b

a

(u(x))²dx

≤ ^L 2q₀

Z _b

a q(x)(u(x))²dx

≤ ^L 2q0

kuk², for everyu∈ X, and thus (2.6) follows.

On the other hand, the inequality (2.3) yields

Z _b

a G(u(x))dx

≤ ^L 2

Z _b

a

(u(x))²dx

≤ ^L(b−a)² 2p₀ kuk², for everyu∈ X. Therefore, we deduce (2.7). The proof is complete.

By the condition (2.5) and Lemma2.2we deduce thatΦis coercive.

By standard arguments, we have thatΦis Gâteaux differentiable and sequentially weakly lower semicontinuous and its Gâteaux derivative is the functionalΦ⁰(u)∈ X^∗, given by

Φ⁰(u)(v) =

Z _b

a p(x)u⁰(x)v⁰(x)dx+

Z _b

a q(x)u(x)v(x)dx−

Z _b

a g(u(x))v(x)dx

for everyv∈ X. Furthermore, the differentialΦ⁰: X→ X^∗ is a Lipschitzian operator. Indeed, for any u,v∈ X, there holds

k_Φ⁰(u)−_Φ⁰(v)k_X^∗ = sup

kwk≤1

|h_Φ⁰(u)−_Φ⁰(v),wi|

≤ sup

kwk≤1

|hu−v,wi|+ sup

kwk≤1

Z _b

a

|g(u(x))−g(v(x))| |w(x)|dx

≤ _sup

kwk≤1

ku−vk kwk

+ _sup

kwk≤1

Z _b

a

|g(u(x))−g(v(x))|²

1/2Z _b

a

|w(x)|² 1/2

.

Recalling that g is Lipschitz continuous and the embedding X ,→ L²([a,b]) is compact, the claim is true. In particular, we derive thatΦis continuously differentiable.

On the other hand, the fact that Xis compactly embedded into C⁰([a,b])implies that the functionalΨis well defined, continuously Gâteaux differentiable and with compact derivative, whose Gâteaux derivative is given by

Ψ⁰(u)(v) =

Z _b

a f(x,u(x))v(x)dx

for everyv∈ X. HenceΨis sequentially weakly (upper) continuous (see [16, Corollary 41.9]).

Fixing the real parameterλ, a functionu: [a,b]→_Ris said to be aweak solutionof problem (1.1) if u∈Xand

Z _b

a p(x)u⁰(x)v⁰(x)dx+

Z _b

a q(x)u(x)v(x)dx−λ Z _b

a f(x,u(x))v(x)dx−

Z _b

a g(u(x))v(x)dx=₀ for all v∈X.

Hence, the critical points of I_λ are exactly the weak (classical) solutions of (1.1).

In conclusion, we cite a recent monograph by Kristály, R˘adulescu and Varga [12] as a general reference on variational methods adopted here.

3 Main results

Put

α:=











2p₀

p₀−L(b−a)^{2, if q0} < ^p0 (b−a)^2, 2q₀

q₀−L, if q₀ ≥ ^p0 (b−a)^2. The main result in this paper is the following.

Theorem 3.1. Let f:[a,b]×_R→_Rbe a continuous function satisfying condition(f₁). In addition, if f(x, 0) =0for a.e. x∈[a,b], assume also that

(f₂) there exist a non-empty open set D⊆ ]a,b[and a set B ⊆ D of positive Lebesgue measure such that

lim sup

t→0⁺

inf_x∈BF(x,t)

t² = +_∞, and lim inf

t→0⁺

inf_x∈DF(x,t)

t² >−_∞.

Then, there exists a positive numberλ^? given by

λ^? :=rsup

γ>0

γ ra₁c_{1 m}^α1/2

+a₂c^r_{r m}^αr/2

γ^r−1

! ,

such that, for everyλ∈]0,λ^?[, problem(1.1)admits at least one non-trivial classical solution u_λ ∈X.

Moreover,

lim

λ→0⁺ku_λk=0,

and the functionλ7→ I_λ(u_λ)is negative and strictly decreasing in]0,λ^?[.

Proof. We prove the result for the caseq₀< p₀/(b−a)². The proof for the caseq₀≥ p₀/(b−a)² is similar.

Fix λ ∈ ]0,λ^?[. Our aim is to apply Theorem 2.1 with the Sobolev space X and the functionals Φ and Ψ introduced in Section 2. As given in Section 2, Φ and Ψ satisfy the regularity assumptions of Theorem2.1. Clearly, infu∈XΦ(u) =0. Owing to(f₁), one has that

F(x,ξ)≤ a₁|ξ|+ ^a2

r |ξ|^r, (3.1)

for any(x,ξ)∈[a,b]×_R.

Since 0<λ<λ^?, there exists ¯γ>0 such that

λ< ^rγ¯

ra₁c₁

2p0

m(p0−L(b−a)²)

1/2

+a₂c^r_r

2p0

m(p0−L(b−a)²)

r/2

¯ γ^r−1

=:λ^?_γ¯. (3.2)

Now, setr∈]0,+_∞[and consider the function

χ(ρ):=

sup

v∈_Φ⁻¹(]−_∞,ρ[)

Ψ(v)

ρ .

Taking into account (3.1) it follows that Ψ(v) =

Z _b

a F(x,v(x))dx≤a₁kvk_L1([a,b])+ ^a2

r kvk^r_Lr([a,b]). Then, due to (2.7), we get

kuk<

2p₀ρ p₀−L(b−a)²

1/2

, (3.3)

for everyu∈ Xsuch thatΦ(u)<ρ.

Now, from (2.4) and by using (3.3), one has Ψ(v)<a₁c₁

2p₀

m(p₀−L(b−a)²) 1/2

ρ^1/2+a₂c^r_r r

2p₀

m(p₀−L(b−a)²) r/2

ρ^r/2,

for every v∈X such thatΦ(v)<ρ. Hence sup

v∈_Φ⁻¹(]−_∞,ρ[)

Ψ(v)≤ a₁c₁

2p₀

m(p0−L(b−a)²) 1/2

ρ^1/2+a₂c^r_r r

2p₀

m(p0−L(b−a)²) r/2

ρ^r/2. Then

χ(ρ)≤ a₁c₁

2p₀

m(p₀−L(b−a)²) 1/2

ρ^−1/2+a₂c^r_r r

2p₀

m(p₀−L(b−a)²) r/2

ρ^r/2−1. In particular, we deduce that

χ(γ¯²)≤a₁c₁

2p0

m(p₀−L(b−a)²) 1/2

¯

γ⁻¹+a₂c^r_r r

2p0

m(p₀−L(b−a)²) r/2

¯

γ^r−2. (3.4) At this point, observe that

ϕ(γ¯²) = inf

u∈_Φ⁻¹(]−_{∞, ¯}γ²[)

sup

v∈_Φ⁻¹(]−_{∞, ¯}γ²[)

Ψ(v)

!

−_Ψ(u)

γ¯²−_Φ(u) ≤χ(γ¯²), taking into account that 0_X ∈_Φ⁻¹(]−_{∞, ¯}γ²[)andΦ(0_X) =_Ψ(0_X) =0.

In conclusion, bearing in mind (3.2), the above inequality together with (3.4) yields ϕ(γ¯²)≤χ(γ¯²)

≤a₁c₁

2p₀

m(p₀−L(b−a)²) 1/2

¯

γ⁻¹+a₂c^r_r r

2p₀

m(p₀−L(b−a)²) r/2

¯ γ^r−2

< ¹ λ. In other words,

λ∈





0, rγ¯ ra₁c₁

2p0

m(p0−L(b−a)²)

1/2

+a₂c^r_r

2p0

m(p0−L(b−a)²)

r/2

¯ γ^r−1







⊆]0, 1/ϕ(γ¯²)[.

By Theorem2.1, there exists a functionuλ ∈_Φ⁻¹(]−_{∞, ¯}γ²[)such that I_λ⁰(u_λ) =_Φ⁰(u_λ)−_λΨ⁰(u_λ) =0,

and, in particular,u_λ is a global minimum of the restriction of I_λ toΦ⁻¹(]−_{∞, ¯}γ²[).

Now, we have to show that for any λ ∈ ]0,λ^?[ the solution u_λ is not the trivial zero function. If f(·, 0)6= 0, then it easily follows thatuλ 6≡0 inX, since the trivial function does not solve problem (1.1).

Let us consider the case when f(·, 0) =0 and let us fixλ∈]0,λ^?[. We will prove that the functionu_λ cannot be trivial inX. To this end, let us show that

lim sup

kuk→0⁺

Ψ(u)

Φ(u) = +_∞. (3.5)

For this, due to(f₂), we can fix a sequence{ξ_n} ⊂ _R⁺ converging to zero and two constants σandκ(withσ>0) such that

n→+lim_∞

inf_x∈BF(x,ξ_n)

ξ²_n = +_∞, and

xinf∈DF(x,ξ)≥κξ², for everyξ ∈[0,σ].

Now, fix a setC⊂ Bof positive measure and a functionv∈C₀^∞([a,b])⊂Xsuch that:

i) v(x)∈[0, 1], for every x∈ [a,b]; ii) v(x) =1, for every x∈C;

iii) v(x) =0, for every x∈]a,b[\D.

Finally, fixM>0 and consider a real positive numberηwith M< ^2p0

p0+L(b−a)²

ηmeas(C) +κR

D\Cv(x)²dx kvk^{2 .} Then, there isν∈_Nsuch thatξ_n<σand

xinf∈BF(x,ξ_n)≥ηξ²_n, for everyn>ν.

Now, for every n > ν, by the properties of the function v (that is, 0 ≤ ξ_nv(x) < σ for n sufficiently large), one has

Ψ(ξ_nv) Φ(ξnv) =

R

CF(x,ξ_n)dx+R

D\CF(x,ξ_nv(x))dx Φ(ξnv)

≥ ^2p0 p₀+L(b−a)²

ηmeas(C) +κR

D\Cv(x)²dx kvk² > M.

SinceM could be taken arbitrarily large, it follows that

n→+lim_∞

Ψ(ξnv)

Φ(ξ_nv) = +_∞,

from which (3.5) clearly follows. Hence, there exists a sequence{w_n} ⊂Xstrongly converging to zero, such that, for everynsufficiently large,wn ∈_Φ⁻¹(]−_{∞, ¯}γ²[)_{, and}

I_λ(w_n) =_Φ(w_n)−λΨ(w_n)<_0.

Sinceu_λ is a global minimum of the restriction of I_λ toΦ⁻¹(]−_∞, ¯γ²[), we conclude that I_λ(u_λ)<0= I_λ(0_X), (3.6) so thatu_λ is not trivial inX.

Moreover, from (3.6) we easily see that the map ]_0,λ^?[3λ7→ I_λ(u_λ)

is negative.

Now, we claim that

lim

λ→0⁺ku_λk=0.

Indeed, bearing in mind that Φ is a coercive functional and that u_λ ∈ _Φ⁻¹(]−_{∞, ¯}γ²[)_{, for} everyλ∈]0,λ^?_γ¯[, we obtain

ku_λk<

2p₀ p₀−L(b−a)²

1/2

¯ γ.

As a consequence, using the growth condition(f₁)together with the property (2.4), it follows that

Z _b

a f(x,u_λ(x))u_λ(x)dx

≤a₁ku_λk_L1([a,b])+a2ku_λk^r_Lr([a,b])

≤ ^a1c1

m^1/2ku_λk+ ^a2c

rr

m^r/2ku_λk^r (3.7)

<a₁c₁

2p₀

m(p₀−L(b−a)²) 1/2

+a2c^r_r

2p₀

m(p₀−L(b−a)²) r/2

=: M_γ¯, for every λ∈]0,λ^?_γ¯[.

Sinceu_λ is a critical point of I_λ, then I_λ⁰(u_λ)(v) =0, for anyv ∈ Xand everyλ ∈ ]0,λ^?_γ¯[. In particular, I_λ⁰(u_λ)(u_λ) =0, that is

Φ⁰(u_λ)(u_λ) =λ Z _b

a f(x,u_λ(x))u_λ(x)dx, (3.8) for every λ∈]0,λ^?_γ¯[. Hence, from (2.3), (3.7) and (3.8), it follows that

0≤ ^p0−L(b−a)² 2p0

ku_λk²≤_Φ⁰(u_λ)(u_λ)<λM_γ¯, for every λ∈]0,λ^?_γ¯[. Lettingλ→0⁺, we get lim_λ→0⁺ku_λk=0, as claimed.

Finally, we have to show that the mapλ7→ I_λ(u_λ)is strictly decreasing in ]0,λ^?[. For this, we observe that for anyu∈ X, one has

I_λ(u) =λ

Φ(u)

λ −_Ψ(u)

. (3.9)

Now, let us fix 0 < λ₁ < λ₂ < λ^?_γ¯ and let u_λi be the global minimum of the functional I_λi restricted toΦ⁻¹(]−_{∞, ¯}γ²[)_fori=1, 2. Also, let

m_λi :=

Φ(u_λi)

λ_i −_Ψ(u_λi)

= inf

v∈_Φ⁻¹(]−_{∞, ¯}γ²[)

Φ(v)

λ_i −_Ψ(v)

, for every i=1, 2.

Clearly, the negativity of the mapλ7→ I_λ(u_λ)in]0,λ^?[together with (3.9) and the positiv- ity of λimply that

m_λi <_{0, for}i=_{1, 2. (3.10)}

Moreover,

mλ₂ ≤mλ₁, (3.11)

thanks to 0< λ₁ < λ₂ and Φ≥ 0 by Lemma 2.2. Then, by (3.9)–(3.11) and again by the fact that 0<λ₁< λ₂, we get that

Iλ₂(uλ₂) =λ₂mλ₂ ≤λ₂mλ₁ < λ₁mλ₁ = Iλ₁(uλ₁),

so that the mapλ7→ Iλ(uλ)is strictly decreasing in]0,λ^?[, which completes the proof.

Remark 3.2. Theorem 3.1 can be also obtained applying Theorem 2.3 of [4], which directly ensures that the local minimum is non-zero.

4 Additional results and comments

In this section we give some consequences, remarks and an example.

Remark 4.1. By direct computation, it follows that the parameter λ^? in Theorem 3.1 can be expressed as

λ^? =



















+_∞, if 1<r<2,

2m

αa₂c²₂, ifr=2, rγe_max

ra₁c_{1 m}^α1/2

+a₂c^r_{r m}^αr/2

γe^r_max⁻¹

, if 2<r<+_∞, where

γe_max:=^m α

1/2

ra₁c₁ a₂c^r_r(r−2)

1/(r−1)

.

Remark 4.2. From the above expressions, it follows that if the term f is sublinear at infinity (i.e., r ∈ ]1, 2[ in (f₁)), Theorem3.1 ensures that, for all λ > 0, problem (1.1) admits at least one nontrivial classical solution.

Remark 4.3. We observe that if f(x, 0) = 0 for a.e. x ∈ [a,b], Theorem 3.1 is a bifurcation result. Indeed, in this setting, it follows that the trivial solution solves problem (1.1) for every parameterλ. Hence,λ = 0 is a bifurcation point for problem (1.1), in the sense that the pair (0, 0)belongs to the closure of the set

(u_λ,λ)∈X×]0,+_∞[: u_λ is a nontrivial classical solution of (1.1) in the spaceX×_R.

Indeed, by Theorem3.1we have that

ku_λk →0 as λ→0⁺.

Hence, there exist two sequences{u_j}_j∈Nin Xand{λ_j}_j∈NinR⁺(hereu_j := u_λj) such that λ_j →0⁺ and ku_jk →0,

asj→+_∞.

Moreover, for any λ₁,λ₂ ∈ ]0,λ^?[, with λ₁ 6= λ₂, the solutions u_λ1 and u_λ2 given by Theorem3.1 are different, thanks to the fact that the map

]_0,λ^?[3λ7→ I_λ(u_λ) is strictly decreasing.

The next result is an immediate consequence of Remark4.1.

Corollary 4.4. Let f: [a,b]×_R → _Rbe a continuous function with f(x, 0) =0 for a.e. x ∈ [a,b], satisfying the following subcritical growth condition

|f(x,t)| ≤a₁+a₂|t|^r−1, a.e.x ∈[a,b], t∈_R, (4.1) where a₁,a₂ ≥ ₀and r ∈ ]_2,+_∞[. Further, assume that there exists a non-empty open set B⊆ ]a,b[ such that

tlim→0⁺

inf_x∈BF(x,t)

t² = +_{∞. (4.2)}

Then, there exists a positive numberλ^? given by

λ^? := ^m(r−2) αa₁c₁(r−1)

ra₁c₁ a₂c^r_r(r−2)

1/(r−1)

,

such that, for everyλ∈]0,λ^?[, problem(1.1)admits at least one nontrivial classical solution u_λ ∈X.

Moreover,

lim

λ→0⁺ku_λk=0,

and the functionλ→ I_λ(u_λ)is negative and strictly decreasing in]0,λ^?[. We state an example on the following special case of our results.

Theorem 4.5. Let f: R→_Rbe a continuous function such that f(0) =0, and

tlim→0⁺

f(t)

t = +_{∞, lim}

|t|→+_∞

f(t)

|t|^s = _0,

for some 0 ≤ s < +∞. Then, there exists λ^? > 0 such that, for every λ ∈ ]0,λ^?[, the following autonomous mixed problem

(−(pu⁰)⁰+qu=λf(u) +g(u) in]a,b[, u(a) =u⁰(b) =_0,

admits at least one nontrivial classical solution u_λ ∈X. Moreover, lim

λ→0⁺ku_λk=0, and the mapping

λ7→_Φ(u_λ)−λ Z _b

a

Z _u

λ(x)

0 f(x,t)dt

dx is negative and strictly decreasing in]0,λ^?[.

Proof. The conclusion follows immediately from Theorem3.1. Indeed, if

tlim→0⁺

f(t)

t = +_∞, then, we have

tlim→0⁺

F(t)

t² = +_∞, and condition(_f2)holds true. Moreover, hypothesis

|t|→+lim_∞ f(t)

|t|^s =0, where 0≤s <+∞, implies the growth condition(_f1)_.

Remark 4.6. We observe that if f is a non-negative function, our results guarantee that the attained solution is non-negative. To this end, letu0be a solution of problem (1.1). Arguing by contradiction, assume that the set A:= {x ∈ [a,b] :u₀(x)< 0}is non-empty and of positive Lebesgue measure. Put ¯v(x):=min{0,u₀(x)}for allx∈ [a,b]. Clearly, ¯v∈ Xand, taking into account thatu0 is a weak solution and by choosingv=v, one has¯

Z _b

a p(x)u⁰₀(x)v¯⁰(x)dx+

Z _b

a q(x)u₀(x)v¯(x)dx−λ Z _b

a f(x,u₀(x))v¯(x)dx−

Z _b

a g(u₀(x))v¯(x)dx =_0, that is,

Z

Ap(x)|u⁰₀(x)|²dx+

Z

Aq(x)|u0(x)|²dx−

Z

Ag(u0(x))u0(x)dx≤0.

On the other hand, ifq₀ < p₀/(b−a)², then p₀−L(m(A))²

p₀ ku₀k²_W1,2(A) ≤

Z

Ap(x)|u⁰₀(x)|²dx+

Z

Aq(x)|u₀(x)|²dx−

Z

Ag(u₀(x))u₀(x)dx, wherem(A)is the Lebesgue measure of the setA, and ifq0≥ p0/(b−a)², then

q₀−L q0

ku₀k²_W1,2(A) ≤

Z

Ap(x)|u⁰₀(x)|²dx+

Z

Aq(x)|u₀(x)|²dx−

Z

Ag(u₀(x))u₀(x)dx.

Hence,u0 ≡0 on Awhich is absurd. So, it follows thatu0is non-negative.

The next example deals with a nonlinearity f has vanishes at zero. The existence of one nontrivial solution for the mixed problem involving the map f is achieved by using Corol- lary4.4.

Example 4.7. Consider the following problem

(−(2e^xu⁰)⁰+u(e^x−₁) =λf(x,u) _in]_{0, 1}[_,

u(0) =u⁰(1) =0, (4.3)

where f(x,u) := α(x)|u|^h−2u+β(x)|u|^l−2u andα,β: [0, 1] → _R are two continuous positive and bounded functions, and 1<h<₂<l. Then, for everyλ∈]_0,λ^?[_{, where}

λ^? := ^l−2

8(l−1)max{kαk_∞,kβk_∞}

lc₁ c^l_l(l−2)

!1/(l−1)

, problem (4.3) admits at least one nontrivial classical solution

u_λ ∈Y:= u ∈W^1,2([0, 1]):u(0) =0 . Moreover, by

ku_λk:= Z ₁

0 2e^x(u⁰_λ(x))²dx+

Z ₁

0 e^x(u_λ(x))²dx 1/2

, we have

lim

λ→0⁺ku_λk=_0, and the function

λ7→ ¹

2ku_λk²+

Z ₁

0 u_λ(x)dx−λ Z ₁

0

_{Z u}

λ(x)

0 f(x,t)dt

dx

is negative and strictly decreasing in ]0,λ^?[. To prove this, we can apply Corollary4.4with

f(x,t):=α(x)|t|^h−2t+β(x)|t|^l−2t, p(x)_:=2e^x, q(x)_:=e^x, g(t)_:= t,

for every (x,t)∈[_{0, 1}]×R. In fact, f(x, 0) =_{0 for a.e.} x∈ [_{0, 1}]and it is easy to verify that

|f(x,t)| ≤2 max{kαk_∞,kβk_∞}1+|t|^l−1, a.e.x∈ [0, 1], t∈_R.

Then, condition (4.1) holds. Moreover, a direct computation shows that

tlim→0⁺

infx∈BF(x,t)

t² ≥ ^infx∈Bα(x) h

tlim→0⁺

1 t^2−h

= +_∞,

where B ⊆ ]0, 1[is an arbitrary non-empty open set. Hence, assumption (4.2) is verified and the conclusion follows.

Remark 4.8. We point out that the energy functional I_λ associated to problem (4.3) is un- bounded from below. Indeed, fixu∈ X\ {0}and let τ∈_{R. We have}

I_λ(τu) =_Φ(τu)−λ Z ₁

0

_{Z τu(x)}

0 f(x,t)dt

dx

≤ ³

4kuk²−λ

τ^hinf_x∈[0,1]α(x)

h kuk^h_Lh([0,1])−λ

τ^linf_x∈[0,1]β(x)

l kuk^l_Ll([0,1])→ −_∞, asτ→+∞, bearing in mind that h<2<l.

Hence, since the functionalI_λ is not coercive, the classical direct method result cannot be applied to the case treated in Example4.7.

Remark 4.9. We note that, applying Theorem2.1, we have the relevant result of Theorem3.1 for the following mixed boundary value problem with a complete equation

(−(pu¯ ⁰)⁰+nu¯ ⁰+qu¯ = λf(x,u) +g(u) in ]a,b[,

u(a) =u⁰(b) =0, (4.4)

where ¯p ∈ C¹([a,b]) and ¯q, ¯n ∈ C⁰([a,b]) such that ¯p and ¯q are positive functions, λ is a positive parameter, f: [a,b]×_R→_Ris a continuous function such that

|f(_x,t)| ≤e^R(x)

a1+_a2|t|^r−1, a.e. x∈[_a,b]_{, t}∈_R,

where a₁,a₂ ≥0 and r ∈ ]1,+_∞[andR is a primitive of ¯n/ ¯p, while g:R → _R is a Lipschitz continuous function with the Lipschitz constant L>0 satisfying

L<max (

xmin∈[a,b]e^−R(x)q¯(x),min_x∈[a,b]e^−R(x)p¯(x) (b−a)²

) , andg(₀) =_0.

In fact, since the solutions of problem (4.4) are solutions of the problem (−(e^−Rpu¯ ⁰)⁰+_e^−Rqu¯ = (λf(_x,u) +_g(_u))_e^−R _in]_a,b[_,

u(a) =u⁰(b) =0,

we can state and prove a result for problem (4.4) similar to Theorem3.1.

Acknowledgements

The authors express their gratitude to the anonymous referee for useful comments and re- marks.

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