On a superlinear periodic boundary value problem with vanishing Green’s function

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On a superlinear periodic boundary value problem with vanishing Green’s function

Dang Dinh Hai

^B

Department of Mathematics and Statistics, Mississippi State University Mississippi State, MS 39762, USA

Received 31 March 2016, appeared 24 July 2016 Communicated by Jeff R. L. Webb

Abstract. We prove the existence of positive solutions for the boundary value problem (y⁰⁰+a(t)y=λg(t)f(y), 0≤t≤2π,

y(0) =y(2π), y⁰(0) =y⁰(2π),

where λis a positive parameter, f is superlinear at ∞and could change sign, and the associated Green’s function may have zeros.

Keywords: superlinear, periodic, vanishing Green’s function.

2010 Mathematics Subject Classification: 34B15, 34B27.

1 Introduction

In this paper, we consider the existence of nonnegative solutions for the periodic boundary value problem

(y⁰⁰+a(t)y =λg(t)f(y), 0≤t≤2π,

y(₀) =y(_2π)_, y⁰(₀) =y⁰(_2π)_, ^(1.1) where the associated Green’s function is nonnegative and f is allowed to change sign. When a(t) = m², where m is a positive constant andm 6= 1, 2, . . . , the Green’s function for (1.1) is given by

G(t,s) = ^sin(m|t−s|) +sinm(2π−(|t−s|)

2m(1−cos 2mπ) ^, ^s,^t∈[_{0, 2π}]_.

Note that G(t,s)> _{0 on} [_{0, 2π}]×[_{0, 2π}]_iff m < _1/2 _andG(t,s) ≥ 0 = G(s,s)_on [_{0, 2π}]× [0, 2π]ifm=1/2. For a general nonnegative time-dependenta∈ L^p(0, 2π), 1≤ p≤_{∞, Torres} [14] showed that the Green’s function for (1.1) is positive (resp. nonnegative) provided thata>

BEmail: Dang@Math.Msstate.Edu

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0 on a set of positive measure,kak_p < K(2p^∗)(resp.kak_p ≤ K(2p^∗)), where p^∗ = p/(p−1) and

K(q) =







1 q(2π)^1/q

2 2+q

₁−2/q

Γ(¹_q) Γ(¹₂+¹_q)

2

if 1≤q<_∞,

1

2π ifq=_∞.

In particular, whena∈ L^∞(_{0, 2π}), the Green’s function is positive ifkak_∞<1/4 and nonnegative ifkak_∞ ≤1/4, which have been obtained in [12] whena is a constant. These conditions were extended to sign-changing a(t) with nonnegative average in [5]. Existence results for positive solutions of (1.1) when the associated Green’s function is positive have been obtained in [2,4,7,8,11,13,14,18] using Krasnosel’skii’s fixed point theorem on the cone

K=

u∈C[0, 2π]:u(t)≥ ^A

Bkuk_∞ ∀t

,

whereAandBdenote the minimum and maximum values ofG(t,s)on[0, 2π]×[0, 2π]respectively. WhenA=0, this cone becomes the cone of nonnegative functions and is not effective in obtaining the desired estimates. The case when the Green’s functionG(t,s)is nonnegative but β =min₀≤s≤2πR2π

0 G(t,s)dtis positive was studied by Graef et al. in [6]. Specifically, assume g is continuous with g(t) > 0 ∀t ∈ [0, 2π], they proved that (1.1) has a nonnegative solution for all λ > 0 when f is continuous, nonnegative with f₀ = _∞, f_∞ = 0 (sublinear), or when

f₀ =0, f_∞ = _∞(superlinear) and f is convex. Here f₀ =lim_u→0⁺ f(u)

u , f_∞ =limu→_∞ ^f(u) u . The method used in [6] is Krasnosel’skii’s fixed point theorem on the cone

K=

u∈C[0, 2π]:u≥0 on[0, 2π]and Z _2π

0 u(t)dt≥ ^β Bkuk_∞

.

The results in [6] were improved by Webb [16], in which g is allowed to be 0 at some points and the existence of nonnegative nontrivial solutions were obtained when f ≥ 0 and either f_∞ < µ_1,λ < f₀ (sublinear) or f₀ <µ_1,λ, ^f⁽_R^R⁾ is large enough and f is convex on [0,T_λ]for a specificT_λ >0 (superlinear), whereµ_1,λ denote the principal characteristic value of the linear operator

L_λu=λ Z _2π

0 G(t,s)g(s)u(s)ds

onC[0, 2π]. The approach in [16] depends on fixed point theory on the modified cone K˜ =

u∈ C[0, 2π]:u≥0 on[0, 2π]and Z _2π

0 g(t)u(t)dt≥ B0kuk_∞

, whereB₀is a suitable positive constant. For results on the system

(y⁰⁰_i +a_i(t)y=λg_i(t)f_i(y), 0≤t ≤2π,

y_i(0) =y_i(2π), y⁰_i(0) =y⁰_i(2π), i=1, . . . ,n,

see [9], where both the sublinear and superlinear cases were discussed. Note that convexity is needed for one of the f_i in the superlinear case. Related results in the sublinear case when the Green’s function is nonnegative can be found in [4]. We refer to [10] for results in the case when the Green’s function may change sign. In this paper, motivated by the results in [6,16], we shall establish the existence of positive solutions to (1.1) when the Green’s function is nonnegative, and f is superlinear at ∞ without assuming convexity of f. We also allow

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the case when f can change sign. Note that nonnegative and convexity assumptions of f are essential for some of the proofs in [6,16]. Our approach depends on a Krasnosel’skii type fixed point theorem in a Banach space.

We shall make the following assumptions:

(A1) f :[0,∞)→_Ris continuous;

(A2) a:[0, 2π]→[0,∞)is continuous,a(t)≤1/4 for allt, anda6≡0;

(A3) g∈ L¹(_{0, 2π})_,g≥0 andg 6≡0 on any subinterval of(_{0, 2π})_. Our main result is the following.

Theorem 1.1. Let (A1)–(A3) hold. Then

(i) if f₀ =0, f_∞ =_∞, and f ≥₀then(1.1)has a positive solution for allλ>0;

(ii) if f_∞ = ∞, then there exists a constant λ^∗ > 0 such that (1.1) has a positive solution y_λ for λ<λ^∗. Furthermoreky_λk_∞ →_∞asλ→0⁺.

Example 1.2. Let c be a nonnegative constant, g satisfy (A3), and a satisfy (A2). Let f(y) = y^αcos² ¹_y

−c fory > 0, f(0) = −c, where α > 1. Then Theorem 1.1(i) gives the existence of a positive solution to (1.1) for c = 0 and λ > 0, while if c > 0, Theorem 1.1(ii) gives the existence of a large positive solution to (1.1) for λ > 0 small. Note that when α > 1,f is not convex on [0,T) for any T > 0 since it is easy to see that f ^y₂

6≤ ¹₂(f(y) + f(0) when y= ^π₂ +2nπ−1

,n∈N. Hence the results in [6,16] cannot be applied here.

2 Preliminary results

LetAC¹[0, 2π] ={u∈C¹[0, 2π]:u⁰is absolutely continuous on[0, 2π]}. We first recall the following fixed point result of Krasnosel’skii type in a Banach space (see e.g. [1, Theorem 12.3]).

Lemma A. Let X be a Banach space and T : X → X be a compact operator. Suppose there exist h∈X,h6=0and positive constants r,Rwithr6= Rsuch that

(a) Ify∈ Xsatisfiesy=θTy for someθ∈ (0, 1], thenkyk 6=r;

(b) Ify∈ Xsatisfiesy=Ty+ξhfor someξ ≥0, thenkyk 6= R.

ThenT has a fixed pointy ∈Xwith min(r,R)<kyk<max(r,R).

Lemma 2.1. Letα,β∈_Rwithα<βand let y∈ AC¹[α,β]be a nonnegative solution of y⁰⁰+¹

4y≥0 a.e. on(α,β). (2.1)

Suppose one of the following conditions holds

(i) y⁰(α) =y(β) =0or y(α) =y⁰(β) =0andβ−α< π, (ii) y(α) =_y(β) =₀_andβ−α<_2π,

(iii) y(α) =y(β) =0, y⁰(α) =y⁰(β),andβ−α=2π.

Then y≡₀on[_α,β].

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Proof. (i) Supposey⁰(α) =y(β) =0. Multiplying (2.1) by sin ^π₂₍⁽_β^β₋⁻_α^t₎⁾

and integrating on[α,β], we obtain

0≥ ¹ 4−

π 2(β−α)

2! Z _β

α

y(t)sin

π(β−t) 2(β−α)

dt≥0,

which implies y ≡ 0 on [α,β]. On the other hand, if y(α) = y⁰(β) = 0 then the function

˜

y(t) =y(β+α−t)satisfies ˜y⁰(α) =y˜(β) =0 and (2.1). Hence ˜y ≡0 i.e.y≡0 on[α,β], which completes the proof.

(ii) Multiplying (2.1) by sin ^π⁽_β^β₋⁻_α^t⁾

and integrating on[α,β], we obtain 0≥ ¹

4 − π

β−α 2!

Z _β

α

y(t)sin

π(β−t) β−α

dt≥0, which impliesy ≡0 on[α,β].

(iii) Letτ∈[α,β]andh(t) =y⁰⁰(t) + ¹₄y(t). Multiplying the equation

y⁰⁰+¹

4y= h(t) (2.2)

by sin ^τ⁻₂^t

and integrating on[_α,τ] _gives 1

2y(τ)−y⁰(α)_sin

τ−α 2

=

Z _τ

α

h(t)_sin

τ−t 2

dt. (2.3)

Next, multiplying (2.2) by sin ^t⁻₂^τ

and integrating on[τ,β]gives 1

2y(τ) +y⁰(β)sin

β−τ 2

=

Z _β

τ

h(t)sin

t−τ 2

dt. (2.4)

Adding (2.3), (2.4) and usingy⁰(α) =y⁰(β)together withβ=α+2π, we obtain y(τ) =

Z _τ

α

h(t)sin

τ−t 2

dt+

Z _β

τ

h(t)sin

t−τ 2

dt. (2.5)

Since y(α) = 0 and h(t)sin ^t⁻₂^α

≥ 0 on (α,β), it follows that h(t)sin ^t⁻₂^α

= 0 for a.e.

t∈ (α,β). Henceh≡0 and therefore (2.5) impliesy(τ) =0 for allτ∈[α,β], which completes the proof.

As a consequence of Lemma2.1, we have the following result, which was obtained in [15]

(see also [12] when a is a constant). However, our proof is new and simple. We refer to [17]

for related results whena∈ L¹(_S,_R), whereSis the circle of length 1.

Corollary 2.2. Let y∈ AC¹[0, 2π]satisfy

(y⁰⁰+a(t)y≥0 a.e. on[0, 2π],

y(0) =y(2π), y⁰(0) =y⁰(2π). (2.6) Then either y>0on[0, 2π]or y≡0on[0, 2π]. In particular, if y_i, i=1, 2, satisfy

(y⁰⁰₁+a(t)y₁ ≥y₂⁰⁰+a(t)y₂ a.e. on[0, 2π], y_i(₀=y_i(_2π)_, y⁰_i(₀) =y⁰_i(_2π)_, i=_{1, 2,} then y₁≥ y₂on[_{0, 2π}].

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Proof. Extend y to be a 2π-periodic function on R. Then y ∈ C¹(_R) and y⁰ is absolutely continuous on R. Suppose y(τ) > 0 for some τ ∈ [0, 2π]. We claim that y > 0 on [0, 2π]. Suppose to the contrary that y(τ₀) ≤ 0 for some τ₀ ∈ [0, 2π]. Since y(τ₀) = y(τ₀±2π), there exists an interval (α,β) containingτ such that y > 0 on (α,β), y(α) = y(β) = 0, 0 <

β−α ≤ 2π, and (2.1) holds, which contradicts Lemma 2.1(ii) and (iii). Hence y > 0 on [0, 2π] as claimed. On the other hand, if y ≤ 0 on [0, 2π] then y⁰⁰ ≥ 0 a.e. on [0, 2π]. Let y(τ₁) = max_t_∈[_0,2π_]y(t). Then y⁰(τ₁) = 0, and hence y(t) = y(τ₁) for all t ∈ [0, 2π]. Hence (2.6) immediately gives y ≥ 0 on [0, 2π]. Consequentlyy ≡ 0, which completes the proof of the first part. The second part follows by using the first part with y=y₁−y₂.

Let I₁ = [^π₂,^3π₄ ], I₂ = [π,^5π₄ ], I₃ = [^3π₂ ,^7π₄ ], I₄ = [^5π₄ ,^3π₂ ] and J₁ = [0^π₂], J₂ = [^π₂,π], J₃ = [_π,^3π₂ ]_, J₄ = [^3π₂ _{, 2π}]. The next result plays an important role in the proof of the main results.

Lemma 2.3. There exists a positive constant m such that all solutions y∈ AC¹[0, 2π]of (2.6)satisfy y(t)≥mkyk

for t ∈ I_i for some i∈ {1, 2, 3, 4}.

Proof. Let y ∈ AC¹[0, 2π] be a solution of (2.6). Theny ≥ 0 on [0, 2π] by Corollary 2.2. Let kyk=y(τ)for someτ∈[0, 2π]. Theny⁰(τ) =0. Let z_τ satisfy

(z⁰⁰_τ+a(t)zτ =0 on [0, 2π],

z_τ(τ) =_1, z⁰_τ(τ) =_0. ^(2.7)

Note that the existence of a unique solution z_τ ∈ C²[0, 2π] follows from the basic theory for linear differential equations (see e.g. [3, Theorem 3.7.1]). We shall verify that zτ is bounded in C²[0, 2π] by a constant independent ofτ ∈ [0, 2π]. Indeed, by integrating the equation in (2.7), we get

z_τ(t) =1−

Z _t

τ

(t−s)a(s)z_τ(s)ds fort ∈[0, 2π], which, together with (A2), implies

|z_τ(t)| ≤1+^π 2

Z _t

τ

|z_τ(s)|ds fort ≥τ, and

|zτ(t)| ≤1+ ^π 2

Z _τ

t

|zτ(s)|ds fort ≤τ.

Hence Gronwall’s inequality gives

|zτ(t)| ≤e⁽^π/2^)|^t⁻^τ^| ≤e^π² (2.8) fort∈ [0, 2π]. Sincez⁰_τ(t) =−Rt

τa(s)zτ(s)dsandz⁰⁰_τ = −a(t)zτ on[0, 2π], it follows from (2.8) that zτ is bounded inC²[0, 2π]by a constant independent of τ∈ [0, 2π].

Claim 1:There exists a constant m>0such that zτ(t)≥ m for allτ∈ J_i and t∈ I_i, i∈ {1, 2, 3, 4}. Suppose to the contrary that there exists i ∈ {1, 2, 3, 4}and sequences (τ_n) ⊂ J_i, (t_n) ⊂ I_i, (zn)⊂C²[0, 2π]such thatzn(tn)≤ ¹_n for all nand

(z⁰⁰_n+a(t)z_n=₀ _on[_{0, 2π}]_, z_n(τ_n) =1, z⁰_n(τ_n) =0.

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Since (z_n) is bounded in C²[0, 2π] by the above discussion, and (τ_n),(t_n) are bounded in J_i,I_i respectively, by passing to a subsequence if necessary, we can assume that there exist τ_i ∈ J_i,t_i ∈ I_i, andz ∈ C¹[0, 2π]such thatτ_n →τ_i,t_n →t_i, andz_n →z inC¹[0, 2π]. Note that t_n≥τ_nfori<4 andn∈_{N, and so}t_i ≥τ_i fori<4. Since

zn(t) =1−

Z _t

τn

(t−s)a(s)zn(s)ds, by passing to the limit asn →∞, we obtain

z(t) =1−

Z _t

τi

(t−s)a(s)z(s)ds, i.e.z satisfies

(z⁰⁰+a(t)z=0 on [0, 2π], z(τ_i) =1, z⁰(τ_i) =0.

Sincez(t_i) =lim_n→_∞z_n(t_n)≤0, we obtain fori<4 thatt_i >τ_i (sincet_i 6=τ_i), and there exists t˜_i ∈(τ_i,t_i]such thatz>0 on(τ_i, ˜t_i)andz(^˜t_i) =0. Since ˜t_i−τ_i ≤ ^3π₄ , Lemma2.1(i) givesz=0 on(τ_i, ˜t_i), a contradiction. On the other hand, ifi=4 thent₄<τ₄and there exists ˜t₄ ∈[t₄,τ₄) such that z > 0 on (t^˜₄,τ₄)and z(^˜t₄) = 0. Since τ₄−t^˜₄ ≤ ^3π₄ , we obtain a contradiction with Lemma2.1(i). This proves the claim.

Letu=y− kykz_τ. Thenu satisfies

(u⁰⁰+a(t)u≥0 a.e. on[_{0, 2π}]_, u(τ) =0, u⁰(τ) =0.

Claim 2: u≥0 on[0, 2π].

Indeed, suppose u(τ˜) < 0 for some ˜τ ∈ [0, 2π]with ˜τ < τ. Then there exists ˜τ₀ ∈ (τ,˜ τ] such thatu<0 on (τ, ˜˜ τ₀)andu(τ˜₀) =0. Hence

u⁰⁰≥ −a(t)u≥0 a.e. on(τ, ˜˜ τ₀]. (2.9) If u⁰(τ˜0) ≤ 0, then (2.9) implies u⁰ ≤ 0 on (τ, ˜˜ τ0] and so u(t) ≥ u(τ˜0) = 0 on (τ, ˜˜ τ0], a contradiction. On the other hand, ifu⁰(τ˜₀) > 0 then there exists ˜τ₁ ∈ (τ˜₀,τ] such thatu > 0 on (τ˜₀, ˜τ₁) and u(τ˜₁) = 0. Since ˜τ₁−τ˜₀ < 2π, Lemma 2.1(ii) implies u ≡ 0 on (τ˜₀, ˜τ₁), a contradiction. Similarly, we reach a contradiction in the case ˜τ>τ, which proves claim 2.

Sinceτ∈ ∪⁴_i₌₁J_i, it follows from claims 1 and 2 that there existsi∈ {1, 2, 3, 4}such that y(t)≥ kykzτ(t)≥mkyk

for allt∈ I_i, which completes the proof of Lemma2.3.

By Lemma2.6below, there existsz∈ AC¹[0, 2π]satisfying (z⁰⁰+a(t)z =g(t) a.e. on[0, 2π],

z(0) =z(2π), z⁰(0) =z⁰(2π). (2.10) Sinceg6≡0, Corollary 2.2givesz>_{0 on} [_{0, 2π}]_.

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Corollary 2.4. Let k be a positive constant and y∈ AC¹[0, 2π]satisfy (y⁰⁰+a(t)y≥ −λkg(t) a.e. on[_{0, 2π}]_,

y(0) =y(2π), y⁰(0) =y⁰(2π). (2.11) Then

(i) y≥ −λkz on[_{0, 2π}]

(ii) Ifkyk ≥2λkkzk(m+1)m⁻¹ then

y(t)≥ m₀kyk (2.12)

for t ∈ I_i for some i∈ {1, 2, 3, 4},where m₀= m/2and m is given by Lemma2.3.

Proof. Letu=y+λkz. Thenusatisfies

u⁰⁰+a(t)u≥0 a.e. on[0, 2π], from which Corollary2.2and Lemma2.3give u≥0 on[0, 2π]and

y(t) +λkz(t) =u(t)≥ kukm=ky+λkzkm fort ∈ I_i for somei∈ {1, 2, 3, 4}. Thus y≥ −λkzon[0, 2π]and

y(t)≥ kykm−λkkzk(m+1), from which (2.12) follows ifkyk ≥2λkkzk(m+₁)m⁻¹. Lemma 2.5. Let U,V ∈C²[0, 2π]be the solutions of

(U⁰⁰+a(t)U=0 on[0, 2π], U(0) =1, U⁰(0) =0,

and (

V⁰⁰+a(t)V =0 on [0, 2π], V(0) =0, V⁰(0) =1.

Then U(2π),V⁰(2π)<1.

Proof. Suppose U(2π) ≥ 1. If there exists τ ∈ (0, 2π) such that U(τ) < 0 then, since U(0) > 0, there exists an interval [α,β] ⊂ (0, 2π) such that U < 0 on (α,β) and U(α) = U(β) = 0. Since a(t) ≤ 1/4, it follows from Lemma 2.1(ii) with y = −U that U = 0 on (α,β), a contradiction. On the other hand, if U ≥ 0 on (0, 2π) then U⁰⁰ ≤ 0 on (0, 2π) i.e.

U⁰ is nonincreasing on [_{0, 2π}]_{. Hence} U⁰ ≤ _{0 on} [_{0, 2π}], which implies U(_2π) ≤ U(₀) = _1.

Thus U(2π) = 1 = U(0) and since U is nonincreasing, we deduce that U = 1 on [0, 2π]. Consequently, the equation in U gives a(t) = 0 for all t ∈ [0, 2π], a contradiction. Hence U(_2π) < 1. Next, we show that V⁰(_2π) < _{1. Since} V(₀) = _{0 and} V⁰(₀) > 0, it follows that V(t)>0 fort >0 near 0. Hence ifV(τ₀)<0 for someτ₀∈ (0, 2π)then there existsβ∈(0,τ₀) such that V > 0 on(0,β)and V(β) = 0 = V(0), a contradiction with Lemma 2.1(ii). Hence V ≥ _{0 on} (_{0, 2π}), which implies V⁰⁰ ≤ _{0 on} (_{0, 2π}). Consequently, V⁰(_2π) ≤ V⁰(₀) = _{1. If} V⁰(2π) =1 thenV⁰ =1 on[0, 2π], which impliesV(t) =t fort ∈[0, 2π]. Using the equation in V, we see that a(t) = 0 for all t ∈ [0, 2π], a contradiction. Hence V⁰(2π) < 1, which completes the proof.

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Lemma 2.6. Let h∈ L¹(0, 2π). Then the problem

(y⁰⁰+a(t)y=h(t) a.e. on[0, 2π],

y(0) =y(2π), y⁰(0) =y⁰(2π) ^(2.13) has a unique solution y∈ AC¹[0, 2π],which is given by

y(t) =

Z _2π

0 G(t,s)h(s)ds, (2.14)

where

G(t,s) =c₁V(t)V(s)−c₂U(t)U(s) +

(c₃U(s)V(t)−c₄U(t)V(s), 0≤s≤t ≤2π, c3U(t)V(s)−c₄U(s)V(t), 0≤t≤s ≤2π, c₁ = ^U⁰⁽_D^2π⁾,c₂ = ^V⁽_D^2π⁾,c₃ = ^U⁽^2π_D⁾⁻¹,c₄ = ^V⁰⁽^2π_D⁾⁻¹, D = U(2π) +V⁰(2π)−2, and U,V are defined in Lemma2.5.

Proof. By Corollary2.2, the only solution of

(y⁰⁰+a(t)y=0 a.e. on[0, 2π], y(0) =y(2π), y⁰(0) =y⁰(2π),

is the trivial one. Hence Fredholm’s alternative theorem implies that the inhomogeneous problem (2.13) has a unique solution, which is given by (2.14) (see [2, Theorem 2.4]). Note that G(t,s)is defined sinceD<0 in view of Lemma2.5. From (2.14), a calculation shows that

y⁰(t) =c₁ _Z _2π

0 V(s)h(s)ds

V⁰(t)−c₂ _Z _2π

0 U(s)h(s)ds

U⁰(t) +c₃

_Z _t

0 U(s)h(s)ds

V⁰(t)−c₄ _Z _t

0 V(s)h(s)ds

U⁰(t) +c3

Z _2π

t V(s)h(s)ds

U⁰(t)−c₄ Z _2π

t U(s)h(s)ds

V⁰(t), from which we see thaty∈ AC¹[_{0, 2π}]and satisfies (2.13).

3 Proof of the main results

Let X be the Banach space C[0, 2π] equipped with the norm kuk = sup_t_∈[_0,2π_]|u(t)|. For u∈X, define

Tu(t) =λ Z _2π

0 G(t,s)g(s)f(|u(s)|)ds

fort ∈[_{0, 2π}]_{, where} G(t,s)is the Green’s function ofy⁰⁰+a(t)y with the periodic boundary conditions in (1.1) given by Lemma2.6. Theny =Tu∈ AC¹[0, 1]satisfies

(y⁰⁰+a(t)y= λg(t)f(|u|) _{a.e. on}[_{0, 2π}]_, y(0) =y(2π), y⁰(0) =y⁰(2π).

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It is easy to see that T : X → X is continuous and since T maps bounded sets in X into bounded sets in C¹[0, 2π], Tis a compact operator. For the rest of the paper, we shall use the following notations:

f^0,z = sup

0≤t≤z

|f(t)| and f_z,∞ =inf

t≥zf(t) forz≥0.

Note that f^0,z and fz,∞ are nondecreasing on[0,∞).

Proof of Theorem1.1. (i) By Corollary2.2, Tu≥0 for allu. Let 0<ε< ¹

λkzk, wherezis defined by (2.10). Since f0 =0, there exists a constantr >0 such that

f(z)<εz forz ∈(0,r].

We shall verify that the conditions of Lemma A with h≡1 are satisfied.

(a)Let y∈ X satisfy y= θTy for someθ ∈(0, 1]. Thenkyk 6=r.

Indeed, suppose to the contrary thatkyk=r. Then

y⁰⁰+a(t)y=λθg(t)f(|y|)≤ λεg(t)kyk a.e. on[0, 2π], from which Corollary2.2implies

y≤λεzkyk on [0, 2π]. Henceλεkzk ≥1, a contradiction with the choice of ε.

(b)Let y∈ X satisfy y= Ty+ξ for someξ ≥0. Thenkyk< R for R>>1.

Note thatysatisfies

y⁰⁰+a(t)y=a(t)ξ+λg(t)f(|y|) a.e. on[0, 2π]. Let M be a constant such that λMmc>π/2, wherec= min₁≤i≤4

R

Ii g(t)dt andmis given by Lemma2.3. Since f_∞ =∞, there exists a constantA>0 such that

f(z)> Mz forz≥ A.

We claim thatkyk< RforR> A/m. Indeed, supposekyk ≥R> A/m. By Lemma2.3, there existsi∈ {1, 2, 3, 4}such that

y(t)≥ kykm≥Rm > A fort ∈ I_i, which implies

f(y(t))> My(t)≥ Mmkyk fort ∈ I_i. Thus

y⁰⁰+a(t)y≥

(λMmkykg(t), t ∈ I_i,

0 t ∈/ I_i a.e. on[0, 2π], and upon integrating on[_{0, 2π}]_{, we get}

Z _2π

0 a(t)y(t)dt≥λMmkyk

Z

Ii

g(t)dt≥λMmckyk. Sincea ≤1/4 on[0, 2π], this implies

π

2kyk ≥λMmckyk,

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i.e.π/2≥ λMmc, a contradiction with the choice ofM. Hencekyk< Ras claimed.

By LemmaA, Thas a fixed pointywithr<kyk< R. By Corollary2.2,y>0 on [0, 2π]. (ii) Let k be a positive constant such that f(z) ≥ −k for allz ≥ 0. By Lemma 2.6, there existz_i, ˜z_i ∈ AC¹[0, 2π]satisfying

z⁰⁰_i +a(t)z_i =

(g(t) t ∈ I_i,

0, t 6∈I_i z_i(0) =z_i(2π), z⁰_i(0) =z⁰_i(2π), and

˜

z⁰⁰_i +a(t)z˜_i =

(0, t∈ I_i,

kg(t)_, t6∈ I_i, z˜_i(0) =z˜_i(2π), ˜z⁰_i(0) =z˜⁰_i(2π),

fori∈ {_{1, 2, 3, 4}}. Note thatz_i >_{0 on} [_{0, 2π}]_{for all}iby Corollary2.2. Chooser >_{0 so that} fm₀r,∞ min

1≤i≤4,t∈[0,2π]z_i(t)> max

1≤i≤4kz˜_ik, (3.1) wherem₀is given by Corollary2.4. Letλ>0 be such that

λmax{f^0,rkzk, 2kkzk(m+1)m⁻¹}<r. (3.2) We shall verify that

(a)Let y∈X satisfy y=θTy for someθ∈ (0, 1]. Thenkyk 6= r.

Suppose to the contrary thatkyk=r. Then

−λf^0,rg(t)≤y⁰⁰+a(t)y≤λf^0,rg(t) _{a.e. on}(_{0, 2π})_, from which it follows that

|y(t)| ≤λf^0,rz(t), fort ∈[_{0, 2π}]_{, where}z is defined in (2.10). Hence

r =kyk ≤λf^0,rkzk, a contradiction with (3.2), which proves (a).

(b)There exists a constant R_λ > r such that any solution y ∈ X of y = Ty+ξ for someξ ≥ 0 satisfieskyk 6= R_λ.

Lety∈ Xsatisfyy= Ty+ξ for someξ ≥0. Since limz→_∞ ^f^z,∞_z =∞, there exists a constant R_λ >r be such that

λ

f_m₀_R_λ_,∞ min

1≤i≤4,t∈[0,2π]z_i(t)− max

1≤i≤4kz˜_ik

>R_λ. (3.3)

Supposekyk= R_λ. Sincekyk ≥2λkkzk(m+1)m⁻¹ and

y⁰⁰+a(t)y≥ λg(t)f(|y|)≥ −λkg(t) a.e. on [0, 2π],

it follows from Corollary 2.4 that y ≥ −λkz on [0, 2π] andy(t) ≥ m0kykfor t ∈ I_i for some i∈ {1, 2, 3, 4}. Hence

y⁰⁰+a(t)y≥ λg(t)f(|y|)≥ λg(t)f_|_y_|_,∞

≥ λ f_m₀_k_y_k_,∞

(g(t), t∈ I_i, 0, t∈/ I_i,−

(0, t ∈ I_i kg(t), t ∈/ I_i

!

a.e. on(0, 2π).

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By Corollary2.2,

y≥ λ(f_m₀_k_y_k_,∞z_i−z˜_i) on [0, 2π], (3.4) which implies by (3.3) that

Rλ =kyk ≥λ

fm₀R_λ,∞ min

1≤i≤4,t∈[0,2π]z_i(t)−max

1≤i≤4kz˜_ik

>Rλ, a contradiction. Hence kyk 6= R_λ, which proves (b).

By LemmaA, T has a fixed pointy_λ ∈ Xwith r< ky_λk< R. Since (3.4) holds, we obtain from (3.1) that

y_λ≥ λ

f_m₀_r,∞ min

1≤i≤4,t∈[0,2π]z_i(t)− max

1≤i≤4kz˜_ik

>0 on [0, 2π]. It remains to show thatky_λk →_∞asλ→0⁺. Since

y⁰⁰_λ+a(t)y_λ =λg(t)f(y_λ)≤λg(t)f^0,^k^y^λ^k a.e. on(0, 2π), it follows that

y_λ≤ λf^0,^k^y^λ^kz on [_{0, 2π}]_, which implies

f^0,^k^y^λ^k ky_λk ≥ ¹

λkzk^.

Since ky_λk > r, it follows that ky_λk → _∞ as λ → 0⁺, which completes the proof of Theo- rem1.1.

Acknowledgement

The author thanks the referee for carefully reading the manuscript and providing helpful suggestions.

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