# Existence and Multiplicity of Solutions for a Periodic Hill's Equation with Parametric Dependence and Singularities

Authors: Alberto Cabada; José Ángel Cid
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We deal with the existence and multiplicity of solutions for the
periodic boundary value problemx″(t)+a(t)x(t)=λg(t)f(x)+c(t), x(0)=x(T), x′(0)=x′(T), where λ is a positive parameter. The function f:(0,∞)→(0,∞) is allowed to be singular, and the related Green's function is nonnegative
and can vanish at some points.


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## Body

## 1. Introduction and Preliminaries

In the recent paper [1], the authors obtain existence, multiplicity, and nonexistence results for the periodic problem(1.1)x′′(t)-k2x(t)=λg(t)f(x),x(0)=x(2π),x′(0)=x′(2π),
depending on the parameter λ>0. Although not explicitly mentioned in [1], we point out the important fact that the related Green’s function of (1.1) is strictly negative for all k>0.The aim of this paper is to give complementary results to those of [1] for the case of a nonnegative related Green's function. In particular, we will deal with problem(1.2)x′′(t)+a(t)x(t)=λg(t)f(x)+c(t),x(0)=x(T),x′(0)=x′(T),
assuming that its Green’s function is nonnegative (for instance, if a(t)=k2, this means 0<k≤π/T). Moreover, in order to give wider applicable results, we will also allow f(x) to be singular at x=0 (the reader may have in mind the model f(x)=1/xα, for some α>0).We note that analogous arguments have been developed in [2] for the fourth-order discrete equation (1.3)u(k+4)+Mu(k)=λg(k)f(u(k))+c(k),u(i)=u(T+i),i=0,…,3.The main tool used in this paper is Krasnoselskii's fixed point theorem in a cone, which is a classical tool extensively used in the related literature (see, for instance, [1, 3–5] and references therein). We will use cones of the form (1.4)K={x∈C([0,T],[0,∞)):φ(x)≥σ‖x‖},
where 0<σ≤1 is a fixed constant and φ:𝒞([0,T],[0,∞))→[0,∞) is a functional satisfying  (i)
φ(x+y)≥φ(x)+φ(y) for all x,y∈𝒞([0,T],[0,∞)),(ii)
φ(λx)=λφ(x) for all λ>0 and x∈𝒞([0,T],[0,∞)).In particular, in Section2, we use the standard choice φ(x)=mint∈[0,T]x(t), and in Section 3, we use φ(x)=∫0Tx(s)ds, which has been recently introduced in [3].We say that the linear problem(1.5)x′′+a(t)x=0,x(0)=x(T),x′(0)=x′(T)
is nonresonant when its unique solution is the trivial one. It is well known that if (1.5) is nonresonant, then the nonhomogeneous problem (1.6)x′′+a(t)x=h(t),a.e.t∈[0,T];x(0)=x(T),x′(0)=x′(T),
always has a unique solution which, moreover, can be written as (1.7)x(t)=∫0TG(t,s)h(s)ds,
where G(t,s) is Green's function related to (1.5). Thus, defining for each λ>0 the operator (1.8)Tλ:D(Tλ)≡{x∈C([0,T]):x(t)>0∀t∈[0,T]}⟶C([0,T]),
given by(1.9)Tλx(t)=λ∫0TG(t,s)g(s)f(x(s))ds+∫0TG(t,s)c(s)ds,t∈[0,T],
we have that x>0 is a solution of problem (1.2) if and only if x=𝒯λx.Throughout the paper, we will use the following notation:(1.10)γ(t)=∫0TG(t,s)c(s)ds,m=min⁡t,s∈[0,T]G(t,s),M=max⁡t,s∈[0,T]G(t,s),f0=lim⁡x→0+f(x)x,f∞=lim⁡x→∞f(x)x.Fora∈L1(0,T), we denote by a+=max{a,0} its positive part, and for 1≤p≤∞, we denote by p̂ its conjugate, that is 1/p+1/p̂=1. Moreover, for an essentially bounded function h:[0,T]→ℝ, we define (1.11)h*=inf⁡esst∈[0,T]⁡h(t),h*=sup⁡esst∈[0,T]⁡h(t),
and for x∈𝒞([0,T]), we will define (1.12)‖x‖=sup⁡t∈[0,T]|x(t)|.The following section is devoted to prove the existence, multiplicity, and nonexistence of solutions of problem (1.2) by assuming that the related Green’s function is strictly positive, whereas in Section 3, we turn out to the case, where the related Green's function is non negative. We point out that in the recent paper [6], the existence of solution for problem (1.2) with a sign-changing Green's function is studied, but only considering a regular f and c(t)≡0.


## 2. Positive Green's Function

In this section, we assume the following hypotheses:(H0)
γ*>0  or  c(t)≡0,(H1)
problem (1.5) is nonresonant and the corresponding Green's function G(t,s) is strictly positive on [0,T]×[0,T],(H2)
g∈L1([0,T]), g(t)≥0 for a.e. t∈[0,T], and ∫0Tg(s)ds>0,(H3)
f:(0,∞)→(0,∞) is continuous,(H4)
c∈L1([0,T]).Notice that condition(H3) allows f to be singular at x=0. In particular, (H3) is satisfied when f(x)=1/xα, α>0 (the case 0<α<1 is called a weak singularity, while α≥1 is called an strong singularity).On the other hand, it is well known that for constanta(t)≡k2, condition (H1) is equivalent to 0<k<π/T. For a time-dependent and nonnegative potential a(t), Torres gave a sharp Lp-criterium [5] based on an antimaximum principle obtained in a previous work by  Torres and Zhang [7]. That criterium has been extended in [8] for sign-changing potentials with strictly positive average. The obtained result is the following.Proposition 2.1 (see [8, Theorem  3.2]).
Define(2.1)K(α,T)={2παT1+2/α(22+α)1-2/α(Γ(1/α)Γ(1/2+1/α))2,if1≤α<∞,4T,ifα=∞,
where Γ is the usual Gamma function.
Assume thata∈Lp(0,T) for some 1≤p≤∞, ∫0Ta(t)dt>0, and moreover,
(2.2)‖a+‖p<K(2p̂,T).
Then, G(t,s)>0 for all (t,s)∈[0,T]×[0,T].In [9], by studying antimaximum principles for the semilinear equation(2.3)(|u′|p-2u′)′+a(t)(|u|p-2u)=h(t),u(0)=u(T),u′(0)=u′(T),
the previous result has been extended to the potentials with nonnegative average as follows.Lemma 2.2 (see [9, Theorem  3.4 and Remark  3.7]).
Assume thata∈Lp(I) for some p≥1, ∫0Ta(t)dt≥0, and moreover,
(2.4)‖a+‖p≤K(2p̂,T).
Then, G(t,s)>0 for a. e. (t,s)∈I×I.Zhang constructs in [10] some examples of potentials a(t) for which the related Green's function is strictly positive, but (2.2) does not hold. In consequence, the best Sobolev constant K(2p̂,T) is not an optimal estimate to ensure the positiveness of Green's function. For optimal conditions in order to get maximum or antimaximum principles, expressed using eigenvalues, Green's functions, or rotation numbers, the reader is referred to the recent work of Zhang [11].Example 2.3.
By Proposition2.1, Hill's equation
(2.5)x′′+a(1+bcos⁡(t))x=0,
with the periodic boundary conditions
(2.6)x(0)=x(2π),x′(0)=x′(2π)
satisfies (H1), provided that a>0, and moreover,
(2.7)‖a(1+bcos⁡(t))+‖p<K(2p̂,2π),forsomep∈[1,∞],
where K is given by (2.1). So, for each b∈ℝ, the condition (H1) is fulfilled if
(2.8)0<a<M(b):=sup⁡p∈[1,∞]{K(2p̂,2π)‖(1+bcos⁡(t))+‖p}.
In particular, it is known thatM(0)=1/4 and M(1)=0.16448 (since the maximum of K(2p̂,2π)/∥1+cos(t)∥p is attained at p≈2.1941, see [10, Example  4.4]). The graphic of M(b) is showed in Figure 1.Figure 1
Graphic ofM(b).From(H1), it follows that m>0, and we define the cone(2.9)K:={x∈C([0,T],[0,∞)):min⁡t∈[0,T]x(t)≥σ‖x‖},
where(2.10)σ=min⁡{mM,γ*γ*},ifγ*>0orσ=mM,ifc(t)≡0.
In both cases, 0<σ<1, and for 0<r<R, we define(2.11)Kr,R:={x∈K:r≤‖x‖≤R}.Next, we give sufficient conditions for the solvability of problem (1.2).Theorem 2.4.
Assume that conditions(H0), (H1), (H2), (H3), and (H4) are fulfilled. Then, for each λ>0 and 0<r<R, the operator 𝒯λ:𝒦r,R→𝒦 given by (1.9) is well defined and completely continuous.
Moreover, if either(i)
∥𝒯λx∥≤∥x∥ for any x∈𝒦 with ∥x∥=r and ∥𝒯λx∥≥∥x∥ for any x∈𝒦 with ∥x∥=R, or(ii)
∥𝒯λx∥≥∥x∥ for any x∈𝒦 with ∥x∥=r and ∥𝒯λx∥≤∥x∥ for any x∈𝒦 with ∥x∥=R, then 𝒯λ has a fixed point in 𝒦r,R, which is a positive solution of problem (1.2).Proof.
Note that ifx∈𝒦r,R, then 0<σr≤x(t)≤R for all t∈[0,T], so 𝒦r,R⊂𝒟(𝒯λ), and then 𝒯λ:𝒦r,R→𝒞([0,T]) is well defined. Standard arguments show that 𝒯λ(𝒟(𝒯λ))⊂𝒦 and that 𝒯λ is completely continuous. Then, from Krasnoselskii's fixed point theorem (see [12, p. 148]), it follows the existence of a fixed point for 𝒯λ in 𝒦r,R which is, by the definition of 𝒯λ, a positive solution of problem (1.2).Before proving the existence and multiplicity results for problem (1.2), we need some technical lemmas proved in the next subsection.


### 2.1. Auxiliary Results

Lemma 2.5.
Assume that conditions(H0), (H1), (H2), (H3), and (H4) are satisfied. Then, for each R>γ*, there exists λ0(R)>0 such that for every 0<λ≤λ0(R), we have
(2.12)‖Tλx‖≤‖x‖forx∈Kwith‖x‖=R.Proof.
FixR>γ*, and let x∈𝒦 with ∥x∥=R. If
(2.13)0<λ≤λ0(R):=R-γ*Mmax⁡u∈[σR,R]f(u)∫0Tg(s)ds,
then, for all t∈[0,T] the following inequalities hold:
(2.14)Tλx(t)=λ∫0TG(t,s)g(s)f(x(s))ds+γ(t)≤λMmax⁡u∈[σR,R]f(u)∫0Tg(s)ds+γ*≤R=‖x‖,
and thus ∥𝒯λx∥≤∥x∥.Lemma 2.6.
Assume that conditions(H0), (H1), (H2), (H3), and (H4) are fullfiled. Then, for each r>0, there exists λ0(r)>0 such that for every λ≥λ0(r), we have
(2.15)‖Tλx‖≥‖x‖,forx∈Kwith‖x‖=r.Proof.
Fixr>0, and let x∈𝒦 with ∥x∥=r. If
(2.16)λ≥λ0(r):=rmmin⁡u∈[σr,r]f(u)∫0Tg(s)ds,
then
(2.17)Tλx(t)=λ∫0TG(t,s)g(s)f(x(s))ds+γ(t)≥λmmin⁡u∈[σR,R]f(u)∫0Tg(s)ds+γ*≥r=‖x‖,
and thus ∥𝒯λx∥≥∥x∥.Lemma 2.7.
Suppose that conditions(H1), (H2), (H3), and (H4) are satisfied and c(t)≡0. Then, if f0=0, there exists r0(λ)>0 such that for every 0<r≤r0(λ), we have
(2.18)‖Tλx‖≤‖x‖,forx∈Kwith‖x‖=r.Proof.
Sincef0=0 for ε=ε(λ)=1/λM∫0Tg(s)ds, there exists r0(λ)>0 such that f(u)≤εu for each 0<u≤r0(λ).
Fix0<r≤r0(λ), and let x∈𝒦 with ∥x∥=r. Then,(2.19)Tλx(t)=λ∫0TG(t,s)g(s)f(x(s))ds≤λM∫0Tg(s)εx(s)ds≤λMε‖x‖∫0Tg(s)ds=‖x‖,
and thus ∥𝒯λx∥≤∥x∥.Lemma 2.8.
Assume that hypothesis(H0), (H1), (H2), (H3), and (H4) hold. Then, if f0=∞, there exists r0(λ)>0 such that for every 0<r≤r0(λ), we have
(2.20)‖Tλx‖≥‖x‖,forx∈Kwith‖x‖=r.Proof.
Sincef0=∞ for L=L(λ)=1/λmσ∫0Tg(s)ds, there exists r0(λ)>0 such that f(u)≥Lu for each 0<u≤r0(λ).
Fix0<r≤r0(λ), and let x∈𝒦 with ∥x∥=r. Then,(2.21)Tλx(t)=λ∫0TG(t,s)g(s)f(x(s))ds+γ(t)≥λm∫0Tg(s)Lx(s)ds+γ*≥λmLσ‖x‖∫0Tg(s)ds=‖x‖,
and thus ∥𝒯λx∥≥∥x∥.Lemma 2.9.
Suppose that conditions(H0), (H1), (H2), (H3), and (H4) are satisfied. Then, if f∞=0 then, there exists R0(λ)>0 such that for every R≥R0(λ), we have
(2.22)‖Tλx‖≤‖x‖,forx∈Kwith‖x‖=R.Proof.
Sincef∞=0 for ε(λ)=1/2λM∫0Tg(s)ds, there exists R1(λ)>0 such that f(u)≤εu for each u≥R1(λ). We define R0(λ):=max{R1(λ)/σ,2γ*}.
FixR≥R0(λ), and let x∈𝒦 with ∥x∥=R. Then,
(2.23)Tλx(t)=λ∫0TG(t,s)g(s)f(x(s))ds+γ(t)≤λM∫0Tg(s)εx(s)+γ*≤λMɛ‖x‖∫0Tg(s)ds+γ*=R2+γ*≤R2+R2=R=‖x‖,
and thus ∥𝒯λx∥≤∥x∥.Lemma 2.10.
Assume that(H0), (H1), (H2), (H3), and (H4) are fullfiled. Then, if f∞=∞, there exists R0(λ)>0 such that for every R≥R0(λ), we have
(2.24)‖Tλx‖≥‖x‖,forx∈Kwith‖x‖=R.Proof.
Sincef∞=∞ for L=L(λ)=1/λmσ∫0Tg(s)ds, there exists R1(λ)>0 such that f(u)≥Lu for each u≥R1(λ). We define R0(λ):=R1(λ)/σ.
FixR≥R0(λ), and let x∈𝒦 with ∥x∥=R. Then,
(2.25)Tλx(t)=λ∫0TG(t,s)g(s)f(x(s))ds+γ(t)≥λm∫0Tg(s)Lx(s)ds+γ*≥λmLσ‖x‖∫0Tg(s)ds=‖x‖,
and thus ∥𝒯λx∥≥∥x∥.In the sequel, we study separately the two different cases considered in condition(H0); that is, γ*>0 or c(t)≡0.


### 2.2. The Caseγ*>0

Theorem 2.11.
Assume that conditions(H1), (H2), (H3), and (H4) are fulfilled. If, moreover, γ*>0, the following results hold:(1)
there existsλ0>0 such that problem (1.2) has a positive solution if 0<λ<λ0,(2)
iff∞=0, then problem (1.2) has a positive solution for every λ>0,(3)
iff∞=∞, then there exists λ0>0 such that problem (1.2) has two positive solutions if 0<λ<λ0,(4)
iff0>0 and f∞>0, then there exists λ0>0 such that problem (1.2) has no positive solutions if λ>λ0.Proof.
Fix0<r<γ*. Then, for each λ>0 and x∈𝒦 with ∥x∥=r, we have
(2.26)‖Tλx‖≥Tλx(t)=λ∫0TG(t,s)g(s)f(x(s))ds+γ(t)≥γ*>r=‖x‖.Part 1.
FixR>γ*(≥γ*>r), and take λ0=λ0(R) given by Lemma 2.5. Then, from Theorem 2.4 (ii), it follows the existence of a positive solution for problem (1.2) if 0<λ<λ0.Part 2.
Fixλ>0, and take R>max{r,R0(λ)}, where R0(λ) is given by Lemma 2.9. Then, from Theorem 2.4 (ii), it follows the existence of a positive solution for problem (1.2).Part 3.
FixR2>R1>γ*(≥γ*>r), and take λ0=min{λ0(R1),λ0(R2)}, where λ0(R1) and λ0(R2) are the given by Lemma 2.5.
Now, fix0<λ<λ0, and take R>max{R2,R0(λ)}, where R0(λ) is given by Lemma 2.10. Therefore, from Theorem 2.4, it follows the existence of two positive solutions x1 and x2 for problem (1.2) such that
(2.27)r≤∥x1∥≤R1<R2≤∥x2∥≤R.Part 4.
Sincef0>0 and f∞>0, there exists L>0 such that f(u)≥Lu for all u>0. Define(2.28)λ0:=1mσL∫0Tg(s)ds.
If forλ>λ0, there exists a positive solution x of problem (1.2), we know that x∈𝒟(𝒯λ) and, as consequence, x=𝒯λ(x)∈𝒦. Therefore, we deduce the following inequalities:
(2.29)‖x‖=‖Tλx‖≥Tλx(t)=λ∫0TG(t,s)g(s)f(x(s))ds+γ(t)≥λm∫0Tg(s)Lx(s)ds+γ*≥λmLσ‖x‖∫0Tg(s)ds>‖x‖,
and we attain a contradiction.Example 2.12.
Let us consider the forced Mathieu-Duffing-type equation(2.30)x′′+a(1+bcos⁡(t))x-λx3=c(t),
which fits into expression (1.2) by defining a(t)=a(1+bcos(t)),g(t)=1 and f(x)=x3.
Equation (2.30), with c(t)≡0, was studied in [13], where a sufficient condition for the existence of a 2π-periodic solution is given. However, since the proof relies in the application of Schauder’s fixed point theorem in a ball centered at the origin, the trivial solution x(t)≡0 is not excluded. The existence of a nontrivial solution was later obtained by Torres in [5, Corollary 4.2]. More precisely, Torres proves that if function a(t)>0 for a.e. t∈[0,2π] and ∥a∥p<K(2p̂,2π), then the homogeneous problem (c(t)≡0) (2.30) has at least two nontrivial one-signed 2π-periodic solutions.
In this paper, as a consequence of Example2.3 and Theorem 2.11, Part 3, we arrive at the following multiplicity result for the inhomogeneous (c(t)≢0) equation (2.30) with a not necessarily constant sign function a(t).Corollary 2.13.
If condition (2.8) is satisfied and γ*>0, then there exists λ0>0 such that (2.30) has at least two positive 2π-periodic solutions, provided that 0<λ<λ0.


### 2.3. The Casec(t)≡0

Theorem 2.14.
Assume that conditions(H1), (H2), (H3), and (H4) hold. If moreover c(t)≡0 the following results hold: (1)
iff0=∞ or f∞=∞, then there exists λ0>0 such that problem (1.2) has a positive solution if 0<λ<λ0,(2)
iff∞=0 then there exists λ0>0 such that problem (1.2) has a positive solution for every λ>λ0,(3)
iff0=∞ and f∞=0 then problem (1.2) has a positive solution for every λ>0,(4)
iff0=∞ and f∞=∞ then there exists λ0>0 such that problem (1.2) has two positive solutions if 0<λ<λ0,(5)
iff0=0 and f∞=∞, then problem (1.2) has a positive solution for every λ>0,(6)
iff0=0 and f∞=0 then there exists λ0>0 such that problem (1.2) has two positive solutions if λ>λ0,(7)
iff0>0 and f∞>0, then there exists λ0>0 such that problem (1.2) has no positive solutions if λ>λ0.Proof of Part 1.
FixR>γ*=0 and take λ0=λ0(R)>0 given by Lemma 2.5. In consequence, for all 0<λ≤λ0(R) we have
(2.31)‖Tλx‖≤‖x‖,forx∈Kwith‖x‖=R.
Now, let0<λ<λ0 be fixed, and choose 0<r<min{R,r0(λ)}, where r0(λ) is given by Lemma 2.8 when f0=∞. In case of f∞=∞, we get r>max{R,R0(λ)}>R, with R0(λ) given by Lemma 2.10. In both situations, we arrive at
(2.32)‖Tλx‖≥‖x‖,forx∈Kwith‖x‖=r.
Thus, Theorem2.4 implies the existence of a positive solution for problem (1.2). Part 2.
Fixr>0, and take λ0=λ0(r)>0 given by Lemma 2.6. Now, for each λ>λ0, take R>max{r,R0(λ)}, with R0(λ) given by Lemma 2.9, and apply Theorem 2.4.Part 3.
For eachλ>0, take r0(λ)<R0(λ) given by Lemmas 2.8 and 2.9, respectively, and apply Theorem 2.4.Part 4.
FixR2>R1>γ*=0, and take λ0=min{λ0(R1),λ0(R2)} given by Lemma 2.5. Now, for each 0<λ<λ0, take r<min{R1,r0(λ)} given by Lemma 2.8 and R>max{R2,R0(λ)} given by Lemma 2.10. Then, Theorem 2.4 implies the existence of two positive solutions x1 and x2 for problem (1.2) such that(2.33)r≤‖x1‖≤R1<R2≤‖x2‖≤R.Part 5.
Use Lemmas2.7 and 2.10 and Theorem 2.4.Part 6.
Use Lemmas2.6, 2.7, and 2.9 and Theorem 2.4 twice.Part 7.
The proof follows the same steps as Part 4 in Theorem2.11.Remark 2.15.
Theorem2.14 complements [1, Theorem  2.1], since it provides similar results for the problem x′′+k2x=λg(t)f(x), with 0<k<π/T.Example 2.16.
Consider as a model the problem(2.34)x′′+a(1+bcos⁡(t))x=λ(1xα+μxβ),x(0)=x(2π),x′(0)=x′(2π),
where a>0, b∈ℝ and α,β,μ≥0. When b=1 and μ=0,  (2.34) is the Brillouin-beam focusing equation which has been widely studied in the literature (see [5, 10, 14] and references therein). Now, we have the following:Corollary 2.17.
Assume condition (2.8). Then, the following results are satisfied:(i)
if0≤β<1, then problem (2.34) has a positive solution for every λ>0.(ii)
ifβ=1 and μ>0, then there exists λ0>0 such that problem (2.34) has a positive solution for every 0<λ<λ0, and there exists λ1>0 such that the problem has no positive solution for λ>λ1.(iii)
ifβ>1 and μ>0, then there exists λ0>0 such that problem (2.34) has two positive solutions for every 0<λ<λ0.Proof.
Condition (2.8) implies that condition (H1) is satisfied. Now, to prove (i), (ii) or (iii) it is enough to apply Theorem 2.14 Part 3, Part 1 and Part 7 or Part 4, respectively.


## 2.1. Auxiliary Results

Lemma 2.5.
Assume that conditions(H0), (H1), (H2), (H3), and (H4) are satisfied. Then, for each R>γ*, there exists λ0(R)>0 such that for every 0<λ≤λ0(R), we have
(2.12)‖Tλx‖≤‖x‖forx∈Kwith‖x‖=R.Proof.
FixR>γ*, and let x∈𝒦 with ∥x∥=R. If
(2.13)0<λ≤λ0(R):=R-γ*Mmax⁡u∈[σR,R]f(u)∫0Tg(s)ds,
then, for all t∈[0,T] the following inequalities hold:
(2.14)Tλx(t)=λ∫0TG(t,s)g(s)f(x(s))ds+γ(t)≤λMmax⁡u∈[σR,R]f(u)∫0Tg(s)ds+γ*≤R=‖x‖,
and thus ∥𝒯λx∥≤∥x∥.Lemma 2.6.
Assume that conditions(H0), (H1), (H2), (H3), and (H4) are fullfiled. Then, for each r>0, there exists λ0(r)>0 such that for every λ≥λ0(r), we have
(2.15)‖Tλx‖≥‖x‖,forx∈Kwith‖x‖=r.Proof.
Fixr>0, and let x∈𝒦 with ∥x∥=r. If
(2.16)λ≥λ0(r):=rmmin⁡u∈[σr,r]f(u)∫0Tg(s)ds,
then
(2.17)Tλx(t)=λ∫0TG(t,s)g(s)f(x(s))ds+γ(t)≥λmmin⁡u∈[σR,R]f(u)∫0Tg(s)ds+γ*≥r=‖x‖,
and thus ∥𝒯λx∥≥∥x∥.Lemma 2.7.
Suppose that conditions(H1), (H2), (H3), and (H4) are satisfied and c(t)≡0. Then, if f0=0, there exists r0(λ)>0 such that for every 0<r≤r0(λ), we have
(2.18)‖Tλx‖≤‖x‖,forx∈Kwith‖x‖=r.Proof.
Sincef0=0 for ε=ε(λ)=1/λM∫0Tg(s)ds, there exists r0(λ)>0 such that f(u)≤εu for each 0<u≤r0(λ).
Fix0<r≤r0(λ), and let x∈𝒦 with ∥x∥=r. Then,(2.19)Tλx(t)=λ∫0TG(t,s)g(s)f(x(s))ds≤λM∫0Tg(s)εx(s)ds≤λMε‖x‖∫0Tg(s)ds=‖x‖,
and thus ∥𝒯λx∥≤∥x∥.Lemma 2.8.
Assume that hypothesis(H0), (H1), (H2), (H3), and (H4) hold. Then, if f0=∞, there exists r0(λ)>0 such that for every 0<r≤r0(λ), we have
(2.20)‖Tλx‖≥‖x‖,forx∈Kwith‖x‖=r.Proof.
Sincef0=∞ for L=L(λ)=1/λmσ∫0Tg(s)ds, there exists r0(λ)>0 such that f(u)≥Lu for each 0<u≤r0(λ).
Fix0<r≤r0(λ), and let x∈𝒦 with ∥x∥=r. Then,(2.21)Tλx(t)=λ∫0TG(t,s)g(s)f(x(s))ds+γ(t)≥λm∫0Tg(s)Lx(s)ds+γ*≥λmLσ‖x‖∫0Tg(s)ds=‖x‖,
and thus ∥𝒯λx∥≥∥x∥.Lemma 2.9.
Suppose that conditions(H0), (H1), (H2), (H3), and (H4) are satisfied. Then, if f∞=0 then, there exists R0(λ)>0 such that for every R≥R0(λ), we have
(2.22)‖Tλx‖≤‖x‖,forx∈Kwith‖x‖=R.Proof.
Sincef∞=0 for ε(λ)=1/2λM∫0Tg(s)ds, there exists R1(λ)>0 such that f(u)≤εu for each u≥R1(λ). We define R0(λ):=max{R1(λ)/σ,2γ*}.
FixR≥R0(λ), and let x∈𝒦 with ∥x∥=R. Then,
(2.23)Tλx(t)=λ∫0TG(t,s)g(s)f(x(s))ds+γ(t)≤λM∫0Tg(s)εx(s)+γ*≤λMɛ‖x‖∫0Tg(s)ds+γ*=R2+γ*≤R2+R2=R=‖x‖,
and thus ∥𝒯λx∥≤∥x∥.Lemma 2.10.
Assume that(H0), (H1), (H2), (H3), and (H4) are fullfiled. Then, if f∞=∞, there exists R0(λ)>0 such that for every R≥R0(λ), we have
(2.24)‖Tλx‖≥‖x‖,forx∈Kwith‖x‖=R.Proof.
Sincef∞=∞ for L=L(λ)=1/λmσ∫0Tg(s)ds, there exists R1(λ)>0 such that f(u)≥Lu for each u≥R1(λ). We define R0(λ):=R1(λ)/σ.
FixR≥R0(λ), and let x∈𝒦 with ∥x∥=R. Then,
(2.25)Tλx(t)=λ∫0TG(t,s)g(s)f(x(s))ds+γ(t)≥λm∫0Tg(s)Lx(s)ds+γ*≥λmLσ‖x‖∫0Tg(s)ds=‖x‖,
and thus ∥𝒯λx∥≥∥x∥.In the sequel, we study separately the two different cases considered in condition(H0); that is, γ*>0 or c(t)≡0.


## 2.2. The Caseγ*>0

Theorem 2.11.
Assume that conditions(H1), (H2), (H3), and (H4) are fulfilled. If, moreover, γ*>0, the following results hold:(1)
there existsλ0>0 such that problem (1.2) has a positive solution if 0<λ<λ0,(2)
iff∞=0, then problem (1.2) has a positive solution for every λ>0,(3)
iff∞=∞, then there exists λ0>0 such that problem (1.2) has two positive solutions if 0<λ<λ0,(4)
iff0>0 and f∞>0, then there exists λ0>0 such that problem (1.2) has no positive solutions if λ>λ0.Proof.
Fix0<r<γ*. Then, for each λ>0 and x∈𝒦 with ∥x∥=r, we have
(2.26)‖Tλx‖≥Tλx(t)=λ∫0TG(t,s)g(s)f(x(s))ds+γ(t)≥γ*>r=‖x‖.Part 1.
FixR>γ*(≥γ*>r), and take λ0=λ0(R) given by Lemma 2.5. Then, from Theorem 2.4 (ii), it follows the existence of a positive solution for problem (1.2) if 0<λ<λ0.Part 2.
Fixλ>0, and take R>max{r,R0(λ)}, where R0(λ) is given by Lemma 2.9. Then, from Theorem 2.4 (ii), it follows the existence of a positive solution for problem (1.2).Part 3.
FixR2>R1>γ*(≥γ*>r), and take λ0=min{λ0(R1),λ0(R2)}, where λ0(R1) and λ0(R2) are the given by Lemma 2.5.
Now, fix0<λ<λ0, and take R>max{R2,R0(λ)}, where R0(λ) is given by Lemma 2.10. Therefore, from Theorem 2.4, it follows the existence of two positive solutions x1 and x2 for problem (1.2) such that
(2.27)r≤∥x1∥≤R1<R2≤∥x2∥≤R.Part 4.
Sincef0>0 and f∞>0, there exists L>0 such that f(u)≥Lu for all u>0. Define(2.28)λ0:=1mσL∫0Tg(s)ds.
If forλ>λ0, there exists a positive solution x of problem (1.2), we know that x∈𝒟(𝒯λ) and, as consequence, x=𝒯λ(x)∈𝒦. Therefore, we deduce the following inequalities:
(2.29)‖x‖=‖Tλx‖≥Tλx(t)=λ∫0TG(t,s)g(s)f(x(s))ds+γ(t)≥λm∫0Tg(s)Lx(s)ds+γ*≥λmLσ‖x‖∫0Tg(s)ds>‖x‖,
and we attain a contradiction.Example 2.12.
Let us consider the forced Mathieu-Duffing-type equation(2.30)x′′+a(1+bcos⁡(t))x-λx3=c(t),
which fits into expression (1.2) by defining a(t)=a(1+bcos(t)),g(t)=1 and f(x)=x3.
Equation (2.30), with c(t)≡0, was studied in [13], where a sufficient condition for the existence of a 2π-periodic solution is given. However, since the proof relies in the application of Schauder’s fixed point theorem in a ball centered at the origin, the trivial solution x(t)≡0 is not excluded. The existence of a nontrivial solution was later obtained by Torres in [5, Corollary 4.2]. More precisely, Torres proves that if function a(t)>0 for a.e. t∈[0,2π] and ∥a∥p<K(2p̂,2π), then the homogeneous problem (c(t)≡0) (2.30) has at least two nontrivial one-signed 2π-periodic solutions.
In this paper, as a consequence of Example2.3 and Theorem 2.11, Part 3, we arrive at the following multiplicity result for the inhomogeneous (c(t)≢0) equation (2.30) with a not necessarily constant sign function a(t).Corollary 2.13.
If condition (2.8) is satisfied and γ*>0, then there exists λ0>0 such that (2.30) has at least two positive 2π-periodic solutions, provided that 0<λ<λ0.


## 2.3. The Casec(t)≡0

Theorem 2.14.
Assume that conditions(H1), (H2), (H3), and (H4) hold. If moreover c(t)≡0 the following results hold: (1)
iff0=∞ or f∞=∞, then there exists λ0>0 such that problem (1.2) has a positive solution if 0<λ<λ0,(2)
iff∞=0 then there exists λ0>0 such that problem (1.2) has a positive solution for every λ>λ0,(3)
iff0=∞ and f∞=0 then problem (1.2) has a positive solution for every λ>0,(4)
iff0=∞ and f∞=∞ then there exists λ0>0 such that problem (1.2) has two positive solutions if 0<λ<λ0,(5)
iff0=0 and f∞=∞, then problem (1.2) has a positive solution for every λ>0,(6)
iff0=0 and f∞=0 then there exists λ0>0 such that problem (1.2) has two positive solutions if λ>λ0,(7)
iff0>0 and f∞>0, then there exists λ0>0 such that problem (1.2) has no positive solutions if λ>λ0.Proof of Part 1.
FixR>γ*=0 and take λ0=λ0(R)>0 given by Lemma 2.5. In consequence, for all 0<λ≤λ0(R) we have
(2.31)‖Tλx‖≤‖x‖,forx∈Kwith‖x‖=R.
Now, let0<λ<λ0 be fixed, and choose 0<r<min{R,r0(λ)}, where r0(λ) is given by Lemma 2.8 when f0=∞. In case of f∞=∞, we get r>max{R,R0(λ)}>R, with R0(λ) given by Lemma 2.10. In both situations, we arrive at
(2.32)‖Tλx‖≥‖x‖,forx∈Kwith‖x‖=r.
Thus, Theorem2.4 implies the existence of a positive solution for problem (1.2). Part 2.
Fixr>0, and take λ0=λ0(r)>0 given by Lemma 2.6. Now, for each λ>λ0, take R>max{r,R0(λ)}, with R0(λ) given by Lemma 2.9, and apply Theorem 2.4.Part 3.
For eachλ>0, take r0(λ)<R0(λ) given by Lemmas 2.8 and 2.9, respectively, and apply Theorem 2.4.Part 4.
FixR2>R1>γ*=0, and take λ0=min{λ0(R1),λ0(R2)} given by Lemma 2.5. Now, for each 0<λ<λ0, take r<min{R1,r0(λ)} given by Lemma 2.8 and R>max{R2,R0(λ)} given by Lemma 2.10. Then, Theorem 2.4 implies the existence of two positive solutions x1 and x2 for problem (1.2) such that(2.33)r≤‖x1‖≤R1<R2≤‖x2‖≤R.Part 5.
Use Lemmas2.7 and 2.10 and Theorem 2.4.Part 6.
Use Lemmas2.6, 2.7, and 2.9 and Theorem 2.4 twice.Part 7.
The proof follows the same steps as Part 4 in Theorem2.11.Remark 2.15.
Theorem2.14 complements [1, Theorem  2.1], since it provides similar results for the problem x′′+k2x=λg(t)f(x), with 0<k<π/T.Example 2.16.
Consider as a model the problem(2.34)x′′+a(1+bcos⁡(t))x=λ(1xα+μxβ),x(0)=x(2π),x′(0)=x′(2π),
where a>0, b∈ℝ and α,β,μ≥0. When b=1 and μ=0,  (2.34) is the Brillouin-beam focusing equation which has been widely studied in the literature (see [5, 10, 14] and references therein). Now, we have the following:Corollary 2.17.
Assume condition (2.8). Then, the following results are satisfied:(i)
if0≤β<1, then problem (2.34) has a positive solution for every λ>0.(ii)
ifβ=1 and μ>0, then there exists λ0>0 such that problem (2.34) has a positive solution for every 0<λ<λ0, and there exists λ1>0 such that the problem has no positive solution for λ>λ1.(iii)
ifβ>1 and μ>0, then there exists λ0>0 such that problem (2.34) has two positive solutions for every 0<λ<λ0.Proof.
Condition (2.8) implies that condition (H1) is satisfied. Now, to prove (i), (ii) or (iii) it is enough to apply Theorem 2.14 Part 3, Part 1 and Part 7 or Part 4, respectively.


## 3. Nonnegative Green's Function

In this section instead of conditions(H1) and (H3), we assume  (H1)̃
Problem (1.5) is nonresonant, the corresponding Green's function G(t,s) is nonnegative on [0,T]×[0,T], and β=mint∈[0,T]∫0TG(t,s)ds>0,(H3)̃
f:[0,∞)→[0,∞) is continuous, and f(u)>0 for all u>0.Notice thatx(t)=∫0TG(t,s)ds is the unique solution of the problem (3.1)x′′(t)+a(t)x(t)=1,x(0)=x(T),x′(0)=x′(T),
and then (H1)̃ asks for this solution to be strictly positive. On the other hand, assumption (H3)̃ allows us to consider only regular problems.  We will discuss to singular problems in Section 3.1 by means of a truncation technique.For constanta(t)≡k2, condition (H1)̃ is equivalent to 0<k≤π/T. For nonconstant a(t), condition (H1)̃ is satisfied provided that Lemma 2.2 holds.On the other hand, under condition(H1)̃, it is allowed that(3.2)m=min⁡t,s∈[0,T]G(t,s)=0,
so σ=m/M can be equal to 0, and thus, the arguments used in the previous section do not work. So, by assuming that γ*≥0, let us define (3.3)K̃:={x∈C([0,T],[0,∞)):∫0Tx(s)ds≥σ̃‖x‖},
where σ̃=min{β/TM,∫0Tγ(s)ds/T∥γ∥} if ∥γ∥>0 or σ̃=β/TM if ∥γ∥=0. As far as we know, the cone 𝒦̃ was introduced in [3].Clearly,0<σ̃≤1, and for 0<r<R, we define (3.4)K̃r,R:={x∈K̃:r≤‖x‖≤R}.Next, we prove the following result similar to Theorem2.4.Theorem 3.1.
Assume that(H0),(H1)̃,(H2),(H3)̃, and (H4) hold. Then, for each λ>0 and 0<r<R, the operator 𝒯λ:𝒦̃r,R→𝒦̃ given by (1.9) is well defined and completely continuous.
Moreover, if either(i)
∥𝒯λx∥≤∥x∥ for any x∈𝒦̃ with ∥x∥=r and ∥𝒯λx∥≥∥x∥ for any x∈𝒦̃ with ∥x∥=R, or(ii)
∥𝒯λx∥≥∥x∥ for any x∈𝒦̃ with ∥x∥=r and ∥𝒯λx∥≤∥x∥ for any x∈𝒦̃ with ∥x∥=R,then 𝒯λ has a fixed point in 𝒦̃r,R, which is a nonnegative solution of problem (1.2).Proof.
Ifx∈𝒞([0,T],[0,∞)) and assuming ∥γ∥>0 (the case ∥γ∥=0 being analogous), we obtain
(3.5)∫0TTλx(t)dt=λ∬0TG(t,s)g(s)f(x(s))dsdt+∫0Tγ(t)dt=λ∫0Tg(s)f(x(s))(∫0TG(t,s)dt)ds+∫0Tγ(s)ds≥λβ∫0Tg(s)f(x(s))ds+∫0Tγ(s)ds=λβTMTM∫0Tg(s)f(x(s))ds+∫0Tγ(s)dsT‖γ‖T‖γ‖≥σ̃(λTM∫0Tg(s)f(x(s))ds+T‖γ‖)≥σ̃‖Tλx‖.
Thus, 𝒯λ(𝒞([0,T],[0,∞)))⊂𝒦̃, and it is standard to show that 𝒯λ is completely continuous. In consequence, from Krasnoselskii's fixed point theorem (see [12, p.148]), it follows the existence of a fixed point for 𝒯λ in 𝒦̃r,R which it is, by the definition of 𝒯λ, a non negative solution of problem (1.2).Now, we are going to give sufficient conditions to obtain∥𝒯λx∥≤∥x∥ or ∥𝒯λx∥≥∥x∥. The combination of the next lemmas with Theorem 3.1 will allow us to prove existence and multiplicity results for problem (1.2).Lemma 3.2.
Suppose that the conditions(H0),(H1)̃,(H2),(H3)̃, and (H4) are satisfied. Then, for each R>γ*, there exists λ0(R)>0 such that for every 0<λ≤λ0(R), we have
(3.6)‖Tλx‖≤‖x‖,forx∈K̃with‖x‖=R.Proof.
FixR>γ*, and let x∈𝒦̃ with ∥x∥=R. If
(3.7)0<λ≤λ0(R):=R-γ*Mmax⁡u∈[0,R]f(u)∫0Tg(s)ds,
then
(3.8)Tλx(t)=λ∫0TG(t,s)g(s)f(x(s))ds+γ(t)≤λMmax⁡u∈[0,R]f(u)∫0Tg(s)ds+γ*≤R=‖x‖,
and thus ∥𝒯λx∥≤∥x∥.Lemma 3.3.
Assume that(H1̃),(H2),(H3̃), (H4), and ∫0Tγ(s)ds>0 are satisfied. Then, there exists r0>0 such that for each 0<r<r0, we have
(3.9)‖Tλx‖≥‖x‖,forx∈K̃with‖x‖=r.Proof.
Fix0<r<r0:=1/T∫0Tγ(s)ds, and let x∈𝒦̃ with ∥x∥=r. Then,
(3.10)‖Tλx‖≥1T∫0TTλx(t)dt=λT∬0TG(t,s)g(s)f(x(s))dsdt+1T∫0Tγ(t)dt≥r0>r=‖x‖,
and thus ∥𝒯λx∥≥∥x∥.Lemma 3.4.
Let(H0),(H1̃),(H2),(H3̃), and (H4) be fulfilled. Then, if f∞=0, there exists R0(λ)>0 such that for every R≥R0(λ), we have
(3.11)‖Tλx‖≤‖x‖,forx∈K̃with‖x‖=R.Proof.
Definef̃(u)=max0≤z≤uf(z). Clearly, f̃ is a nondecreasing function on [0,∞); moreover, since f∞=0, it is obvious that
(3.12)lim⁡u→∞f̃(u)u=0.
Therefore, we have that forɛ(λ)=1/2λM∫0Tg(s)ds, there exists R1(λ)>0 such that f̃(u)≤εu for each u≥R1(λ).
DefineR0(λ):=max{R1(λ),2γ*}, fix R≥R0(λ), and let x∈𝒦̃ with ∥x∥=R. Then
(3.13)Tλx(t)=λ∫0TG(t,s)g(s)f(x(s))ds+γ(t)≤λ∫0TG(t,s)g(s)f̃(‖x‖)ds+γ(t)≤λMε‖x‖∫0Tg(s)ds+γ*=R2+γ*≤R2+R2=R=‖x‖,
and thus ∥𝒯λx∥≤∥x∥.Theorem 3.5.
Assume(H0),(H1̃),(H2),(H3̃), and (H4). The following results hold:(1)
if∫0Tγ(s)ds>0. then there exists λ0>0 such that problem (1.2) has a nonnegative solution if 0<λ<λ0,(2)
if∫0Tγ(s)ds>0 and f∞=0, then problem (1.2) has a nonnegative solution for every λ>0.Proof.
The first assertion is a direct consequence of Lemmas3.2 and 3.3. The second part follows from Lemmas 3.3 and 3.4Now, we will impose a strong condition on functiong by assuming that g is strictly positive on the whole interval.  (H2)̃
g∈L1([0,T]), g(t)≥g*>0 for a.e. t∈[0,T].Lemma 3.6.
Assume that conditions(H0),(H1)̃,(H2)̃,(H3)̃, and (H4) are satisfied. Then, if f0=∞, there exists r0(λ)>0 such that for every 0<r≤r0(λ), we have
(3.14)‖Tλx‖≥‖x‖,forx∈K̃with‖x‖=r.Proof.
Sincef0=∞ for L=L(λ)=T/λβσ̃g*, there exists r0(λ)>0 such that f(u)≥Lu for each 0≤u≤r0(λ).
Fix0<r≤r0(λ), and let x∈𝒦̃ with ∥x∥=r. Then,
(3.15)‖Tλx‖≥1T∫0TTλx(t)dt=λT∬0TG(t,s)g(s)f(x(s))dsdt+1T∫0Tγ(t)dt≥λT∬0TG(t,s)g(s)f(x(s))dtds≥λTβg*L∫0Tx(s)ds≥λTβg*Lσ̃‖x‖=‖x‖,
and thus ∥𝒯λx∥≥∥x∥.Now, we are in a position to present the main result of this section.Theorem 3.7.
Suppose that conditions(H0),(H1̃),(H2̃),(H3̃), and (H4) are fulfilled. The following assertions are satisfied:(1)
iff0=∞, then there exists λ0>0 such that problem (1.2) has a nonnegative solution if 0<λ<λ0,(2)
iff0=∞ and f∞=0, then problem (1.2) has a nonnegative solution for every λ>0,(3)
iff0>0 and f∞>0, then there exists λ0>0 such that problem (1.2) has no nonnegative solutions if λ>λ0.Proof.
The first assertion is a direct consequence of Lemmas3.2 and 3.6. The second part follows from Lemmas 3.4 and 3.6.
To prove Part 3, by using thatf0>0 and f∞>0, we know that there exists L>0 such that f(u)≥Lu for all u≥0.
By defining(3.16)λ0:=Tβσ̃Lg*,
we have that if there is any λ>λ0 for which there exists a nonnegative solution x of problem (1.2), then x=𝒯λx∈𝒦. So, we arrive at the following contradiction:
(3.17)‖Tλx‖≥1T∫0TTλx(t)dt=λT∬0TG(t,s)g(s)f(x(s))dsdt+1T∫0Tγ(t)dt≥λT∬0TG(t,s)g(s)f(x(s))dtds≥λTβg*L∫0Tx(s)ds≥λTβg*Lσ̃‖x‖>‖x‖.


### 3.1. Applications to Singular Equations

Despite the fact that in the previous results we deal with regular functions, it is possible to apply some of them to the singular equation(3.18)x′′(t)+a(t)x(t)=λg(t)f(x(t))+c(t),x(0)=x(T),x′(0)=x′(T),
by means of a truncation technique.To this end, we will consider a functionf that satisfies  (H5)
f:(0,∞)→(0,∞) is a continuous function such that f∞=0.Theorem 3.8.
Assume thatγ*>0 and conditions (H1̃), (H2), (H4), and (H5) hold.
Then, problem (3.18) has a positive solution for every λ>0.Proof.
Letr=γ*>0, and define the function
(3.19)fr(u)={f(r),if0≤u<r,f(u),ifu≥r.
From(H5), it follows that fr satisfies condition (H3̃), and (fr)∞=0. Moreover, γ*>0 implies that ∫0Tγ(s)ds>0. As consequence, Theorem 3.5, Part 2, implies that the modified problem
(3.20)x′′(t)+a(t)x(t)=λg(t)fr(x(t))+c(t),x(0)=x(T),x′(0)=x′(T)
has a nonnegative solution xr for all λ>0. Such function is given by the expression
(3.21)xr(t)=λ∫0TG(t,s)g(s)fr(xr(s))ds+γ(t).
The nonnegativeness of functions G, g and fr implies that the solution xr(t)≥γ*=r for all t∈[0,T]. Therefore, xr is a positive solution of problem (3.18).Remark 3.9.
Theorem3.8 is an alternative result to those obtained in [15, 16] by means of Schauder's fixed point theorem.Example 3.10.
Let us consider the repulsive singular differential equation(3.22)x′′+a(t)x=λg(t)(x1/2-ln⁡(x))+c(t).
Since f(x)=x1/2-ln(x) satisfies (H5), we can apply Theorem 3.8 to obtain the following.Corollary 3.11.
Assume that(H1̃), (H2), and (H4) hold. If γ*>0 then, (3.22) has a positive T-periodic solution for every λ>0.


## 3.1. Applications to Singular Equations

Despite the fact that in the previous results we deal with regular functions, it is possible to apply some of them to the singular equation(3.18)x′′(t)+a(t)x(t)=λg(t)f(x(t))+c(t),x(0)=x(T),x′(0)=x′(T),
by means of a truncation technique.To this end, we will consider a functionf that satisfies  (H5)
f:(0,∞)→(0,∞) is a continuous function such that f∞=0.Theorem 3.8.
Assume thatγ*>0 and conditions (H1̃), (H2), (H4), and (H5) hold.
Then, problem (3.18) has a positive solution for every λ>0.Proof.
Letr=γ*>0, and define the function
(3.19)fr(u)={f(r),if0≤u<r,f(u),ifu≥r.
From(H5), it follows that fr satisfies condition (H3̃), and (fr)∞=0. Moreover, γ*>0 implies that ∫0Tγ(s)ds>0. As consequence, Theorem 3.5, Part 2, implies that the modified problem
(3.20)x′′(t)+a(t)x(t)=λg(t)fr(x(t))+c(t),x(0)=x(T),x′(0)=x′(T)
has a nonnegative solution xr for all λ>0. Such function is given by the expression
(3.21)xr(t)=λ∫0TG(t,s)g(s)fr(xr(s))ds+γ(t).
The nonnegativeness of functions G, g and fr implies that the solution xr(t)≥γ*=r for all t∈[0,T]. Therefore, xr is a positive solution of problem (3.18).Remark 3.9.
Theorem3.8 is an alternative result to those obtained in [15, 16] by means of Schauder's fixed point theorem.Example 3.10.
Let us consider the repulsive singular differential equation(3.22)x′′+a(t)x=λg(t)(x1/2-ln⁡(x))+c(t).
Since f(x)=x1/2-ln(x) satisfies (H5), we can apply Theorem 3.8 to obtain the following.Corollary 3.11.
Assume that(H1̃), (H2), and (H4) hold. If γ*>0 then, (3.22) has a positive T-periodic solution for every λ>0.



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