Parameter Dependence of Positive Solutions for Second-Order Singular Neumann Boundary Value Problems with Impulsive Effects

Abstract

The author considers the Neumann boundary value problem −𝑦^′′(𝑡)+𝐌𝑦(𝑡)=𝜆𝜔(𝑡)𝑓(𝑡,𝑦(𝑡)), 𝑡∈𝐽,𝑡≠𝑡_𝑘,−Δ𝑦^′|_{𝑡=𝑡𝑘} = 𝜆𝐼_𝑘(𝑡_𝑘,𝑦(𝑡_𝑘)),𝑘=1,2,…,𝑚,𝑦^′(0)=𝑦^′(1)=0 and establishes the dependence results of the solution on the parameter 𝜆, which cover equations without impulsive effects and are compared with some recent results by Nieto and O’Regan.

1. Introduction

Impulsive effects exist widely in many evolution processes in which their states are changed abruptly at certain moments of time. The theory and applications of impulsive differential equations have been emerging as an important area of investigation in recent years [1–6]. There is a vast literature on the existence of solutions by using different methods such as bifurcation theory [7, 8], fixed point theorems in cones [9–12], the method of lower and upper solutions [13, 14], and the theory of critical point theory and variational methods [15–19]. We remark that on second-order impulsive differential equations with parameter only a few results have been obtained; see, for instance, [9, 20–22]. To the best of our knowledge, these papers only consider the existence of positive solutions. However, the corresponding results for the dependence of the solution on the parameter 𝜆 for second-order impulsive differential equations are not investigated until now. In this paper, we try to solve this kind of problem.

Consider the Neumann boundary value problems −𝑦^′′(𝑡)+𝐌𝑦(𝑡)=𝜆𝜔(𝑡)𝑓𝑡,𝑦(𝑡),𝑡∈𝐽,𝑡≠𝑡_𝑘,−Δ𝑦^′_{𝑡=𝑡𝑘}=𝜆𝐼_𝑘𝑡_𝑘,𝑦𝑡_𝑘𝑦,𝑘=1,2,…,𝑚,_′(0)=𝑦^′(1)=0, (1) where 𝐌>0 is a constant, 𝜆>0 is a parameter, 𝐽=[0,1], 𝜔 is a nonnegative measurable function on (0,1), 𝜔≠0 on any open subinterval in (0,1), which may be singular at 𝑡=0 and/or 𝑡=1, 𝑡_𝑘(𝑘=1,2,…,𝑚) (where 𝑚 is fixed positive integer) are fixed points with 0=𝑡₀<𝑡₁<𝑡₂<⋯<𝑡_𝑘<⋯<𝑡_𝑚<𝑡_𝑚+1=1, and Δ𝑦^′|_{𝑡=𝑡𝑘}=𝑦^′(𝑡⁺_𝑘)−𝑥^′(𝑡⁻_𝑘), where 𝑦^′(𝑡⁺_𝑘) and 𝑦^′(𝑡⁻_𝑘) represent the right-hand limit and left-hand limit of 𝑦^′(𝑡) at 𝑡=𝑡_𝑘, respectively. In addition, 𝜔, 𝑓, and 𝐼_𝑘 satisfy(H₁)𝜔∈𝐿¹_loc(0,1);(H₂)𝑓∈𝐶(𝐽×ℛ⁺,ℛ⁺), 𝐼_𝑘∈𝐶(𝐽×ℛ⁺,ℛ⁺), where ℛ⁺=[0,+∞), 𝑘=1,2,…,𝑚.

Some special cases of (1) have been investigated. For example, Nieto and O’Regan [17] studied problem (1) with 𝜆=1 and 𝜔≡1 for 𝑡∈𝐽. By using variational methods and critical point theory, the authors proved the existence of solutions of problem (1).

For ease of exposition, we set 𝑓⁰=limsup_𝑦→0+max_𝑡∈𝐽𝑓𝑡,𝑦𝑦,𝑓^∞=limsup_𝑦→∞max_𝑡∈𝐽𝑓𝑡,𝑦𝑦,𝑓₀=liminf_𝑦→0+min_𝑡∈𝐽𝑓𝑡,𝑦𝑦,𝑓_∞=liminf_𝑦→∞min_𝑡∈𝐽𝑓𝑡,𝑦𝑦,𝐼⁰(𝑘)=limsup_𝑦→0+max_𝑡∈𝐽𝐼_𝑘𝑡,𝑦𝑦,𝐼^∞(𝑘)=limsup_𝑦→∞max_𝑡∈𝐽𝐼_𝑘𝑡,𝑦𝑦,𝐼₀(𝑘)=liminf_𝑦→0+min_𝑡∈𝐽𝐼_𝑘𝑡,𝑦𝑦,𝐼_∞(𝑘)=liminf_𝑦→∞min_𝑡∈𝐽𝐼_𝑘𝑡,𝑦𝑦,𝑘=1,2,…,𝑚. (2)

Our main results are as follows.

Theorem 1. Assume that (𝐻₁) and (𝐻₂) hold. Then the following two conclusions hold:(H₃)if 𝑓⁰=0, 𝐼⁰(𝑘)=0, and 𝑓_∞=∞, 𝐼_∞(𝑘)=∞, 𝑘=1,2,…,𝑚, then for every 𝜆>0 problem (1) has a positive solution 𝑦_𝜆(𝑡) satisfying lim_𝜆→0⁺‖𝑦_𝜆‖_𝑃𝐶1=∞;(H₄)if 𝑓₀=∞, 𝐼₀(𝑘)=∞, and 𝑓^∞=0, 𝐼^∞(𝑘)=0, 𝑘=1,2,…,𝑚, then for every 𝜆>0 problem (1) has a positive solution 𝑦_𝜆(𝑡) satisfying lim_𝜆→0⁺‖𝑦_𝜆‖_𝑃𝐶1=0.

Remark 2. Assume that (H₁) and (H₂) hold. Furthermore, suppose that 𝑓_∞=∞ or 𝐼_∞(𝑘)=∞, 𝑘=1,2,…,𝑚, in (H₃) or 𝑓₀=∞ or 𝐼₀(𝑘)=∞, 𝑘=1,2,…,𝑚, in (H₄). Then the conclusions of Theorem 1 also hold.

Remark 3. It follows from the conditions of Theorem 1 that we develop some ideas of Guo and Lakshmikantham essentially; for detail, see Theorem 2.3.7 in [23].

Remark 4. For simplicity we only consider Neumann boundary conditions since all the results obtained in this paper can also be adapted with minor changes to the other boundary conditions.

2. Preliminaries

Let 𝐽^′=𝐽⧵{𝑡₁,𝑡₂,…,𝑡_𝑚}, 𝑘=1,2,…,𝑚, and PC¹[0,1]=𝑦∈𝐶[0,1]∶𝑦^′_{(𝑡𝑘,𝑡𝑘+1)}∈𝐶𝑡_𝑘,𝑡_𝑘+1,𝑦^′𝑡⁻_𝑘,𝑦^′𝑡⁺_𝑘.exists,𝑘=1,2,…,𝑚 (3) Then PC¹[0,1] is a real Banach space with norm 𝑦_PC1=max𝑦_∞,𝑦^′_∞, (4) where ‖𝑦‖_∞=sup_𝑡∈𝐽|𝑦(𝑡)|, ‖𝑦^′‖_∞=sup_𝑡∈𝐽|𝑦^′(𝑡)|.

A function 𝑦∈PC¹[0,1]∩𝐶²(𝐽^′) is called a solution of problem (1) if it satisfies (1).

In our main results, we will make use of the following lemmas.

Lemma 5. If (𝐻₁) and (𝐻₂) hold, then problem (1) has a unique solution 𝑦 given by 𝑦(𝑡)=𝜆¹₀𝐺(𝑡,𝑠)𝜔(𝑠)𝑓𝑠,𝑦(𝑠)𝑑𝑠+𝜆^𝑚_𝑘=1𝐺𝑡,𝑡_𝑘𝐼_𝑘𝑡_𝑘,𝑦𝑡_𝑘, (5) where 𝐺()=√𝑡,𝑠cosh𝐌(1−𝑡)cosh√𝐌𝑠√𝐌sinh√𝐌√,0≤𝑠≤𝑡≤1,cosh𝐌𝑡cosh√𝐌(1−𝑠)√𝐌sinh√𝐌,0≤𝑡≤𝑠≤1. (6)

Proof. The proof is similar to that of Lemma 2.4 in [24].

By (6), we can prove that 𝐺(𝑡,𝑠) has the following property: 1√𝐌sinh√𝐌=𝛼≤𝐺(𝑡,𝑠)≤𝛽=cosh²√𝐌√𝐌sinh√𝐌,∀𝑡,𝑠∈𝐽. (7)

Define a cone in PC¹[0,1] by 𝐾=𝑦∈PC¹[0,1]∶𝑦≥0,𝑦(𝑡)≥𝛿𝑦_PC1,𝑡∈𝐽, (8) where 1𝛿=cosh²√𝐌. (9) It is easy to see 𝐾 is a closed convex cone of PC¹[0,1].

Define an operator 𝑇_𝜆∶𝐾→PC¹[0,1] by 𝑇_𝜆𝑦(𝑡)=𝜆¹₀𝐺(𝑡,𝑠)𝜔(𝑠)𝑓𝑠,𝑦(𝑠)𝑑𝑠+𝜆^𝑚_𝑘=1𝐺𝑡,𝑡_𝑘𝐼_𝑘𝑡_𝑘,𝑦𝑡_𝑘. (10)

From (10), we know that 𝑦∈PC¹[0,1] is a solution of problem (1) if and only if 𝑦 is a fixed point of operator 𝑇_𝜆.

Lemma 6. Suppose that (𝐻₁) and (𝐻₂) hold. Then 𝑇_𝜆(𝐾)⊂𝐾 and 𝑇_𝜆∶𝐾→𝐾 is completely continuous.

Proof. For 𝑦∈𝐾, it follows from (7) and (10) that 𝑇𝑦(𝑡)=𝜆¹₀𝐺(𝑡,𝑠)𝜔(𝑠)𝑓𝑠,𝑦(𝑠)𝑑𝑠+𝜆^𝑚_𝑘=1𝐺𝑡,𝑡_𝑘𝐼_𝑘𝑡_𝑘,𝑦𝑡_𝑘≤𝜆cosh²√𝐌√𝐌sinh√𝐌×¹₀𝜔(𝑠)𝑓𝑠,𝑦(𝑠)𝑑𝑠+^𝑚_𝑘=1𝐼_𝑘𝑡_𝑘,𝑦𝑡_𝑘,𝑡∈𝐽. (11) It follows from (7), (10), and (11) that 𝑇𝑦(𝑡)=𝜆¹₀𝐺(𝑡,𝑠)𝜔(𝑠)𝑓𝑠,𝑦(𝑠)𝑑𝑠+𝜆^𝑚_𝑘=1𝐺𝑡,𝑡_𝑘𝐼_𝑘𝑡_𝑘,𝑦𝑡_𝑘≥1√𝐌sinh√𝐌𝜆×¹₀𝜔(𝑠)𝑓𝑠,𝑦(𝑠)𝑑𝑠+^𝑚_𝑘=1𝐼_𝑘𝑡_𝑘,𝑦𝑡_𝑘≥𝛿cosh²√𝐌√𝐌sinh√𝐌𝜆×¹₀𝜔(𝑠)𝑓𝑠,𝑦(𝑠)𝑑𝑠+^𝑚_𝑘=1𝐼_𝑘𝑡_𝑘,𝑦𝑡_𝑘≥𝛿𝑇𝑦. (12) Thus, 𝑇(𝐾)⊂𝐾.
Next, by similar arguments of Lemmas 5 and 6 [12] one can prove that 𝑇∶𝐾→𝐾 is completely continuous. So it is omitted, and the theorem is proved.

To obtain positive solutions of problem (1), the following fixed point theorem in cones is fundamental, which can be found in [23, page 94].

Lemma 7. Let 𝑃 be a cone in a real Banach space 𝐸. Assume Ω₁, Ω₂ are bounded open sets in 𝐸 with 0∈Ω₁, Ω₁⊂Ω₂. If 𝐴∶𝑃∩Ω₂⧵Ω₁⟶𝑃 (13) is completely continuous such that either (a)‖𝐴𝑥‖≤‖𝑥‖, ∀𝑥∈𝑃∩𝜕Ω₁, and ‖𝐴𝑥‖≥‖𝑥‖, ∀𝑥∈𝑃∩𝜕Ω₂, or(b)‖𝐴𝑥‖≥‖𝑥‖, ∀𝑥∈𝑃∩𝜕Ω₁, and ‖𝐴𝑥‖≤‖𝑥‖, ∀𝑥∈𝑃∩𝜕Ω₂,then 𝐴 has at least one fixed point in 𝑃∩(Ω₂⧵Ω₁).

3. Proofs of the Main Results

For convenience we introduce some notations 𝛾=¹₀𝜔(𝑠)𝑑𝑠,𝜌=max_{𝑡∈𝐽,0≤𝑦≤𝑐}𝑓,𝜌𝑡,𝑦_∗=max𝜌_𝑘,𝑘=1,2,…,𝑚, (14) where 𝜌_𝑘=max_{𝑡∈𝐽,0≤𝑦≤𝑐}𝐼_𝑘(𝑡,𝑦), 𝑘=1,2,…,𝑚, and 𝑐>0 is a constant.

Proof of Theorem 1. We need only prove this theorem under condition (H₃) since the proof is similar when (H₄) holds, provided the proper adjustments are made.
If 𝑓⁰=0, 𝐼⁰(𝑘)=0, then there exist 𝑙>0 and 𝑟>0 such that 𝑓𝑡,𝑦<𝑙𝑦,𝐼_𝑘𝑡,𝑦<𝑙𝑦,∀𝑡∈𝐽,0≤𝑦≤𝑟,𝑘=1,2,…,𝑚, (15) where 𝑙 satisfies 𝜆max𝛽,1𝑙𝛾+𝑚≤1. (16) Then for 𝑦∈𝐾∩𝜕Ω_𝑟 we have 𝑇_𝜆𝑦(𝑡)=𝜆¹₀𝐺(𝑡,𝑠)𝜔(𝑠)𝑓𝑠,𝑦(𝑠)𝑑𝑠+𝜆^𝑚_𝑘=1𝐺𝑡,𝑡_𝑘𝐼_𝑘𝑡_𝑘,𝑦𝑡_𝑘≤𝜆𝛽¹₀𝜔(𝑠)𝑙𝑦(𝑠)𝑑𝑠+𝜆𝛽^𝑚_𝑘=1𝑙𝑦𝑡_𝑘≤𝜆𝛽𝑙𝑦_PC1¹₀𝜔(𝑠)𝑦𝑑𝑠+𝑚=𝜆𝛽𝑙_PC1𝛾+𝑚≤𝑦_PC1,𝑇_𝜆𝑦^′(𝑡)≤𝜆¹₀𝐺^′_𝑡(𝑡,𝑠)𝜔(𝑠)𝑓𝑠,𝑦(𝑠)𝑑𝑠+𝜆^𝑚_𝑘=1𝐺^′_𝑡𝑡,𝑡_𝑘𝐼_𝑘𝑡_𝑘,𝑦𝑡_𝑘≤𝜆¹₀𝜔(𝑠)𝑓𝑠,𝑦(𝑠)𝑑𝑠+𝜆^𝑚_𝑘=1𝐼_𝑘𝑡_𝑘,𝑦𝑡_𝑘≤𝜆𝑙𝛾+𝑚𝑦_PC1≤𝑦_PC1, (17) where 𝐺^′_𝑡()=√𝑡,𝑠−sinh𝐌(1−𝑡)cosh√𝐌𝑠sinh√𝐌√,0≤𝑠≤𝑡≤1,cosh𝐌𝑡sinh√𝐌(1−𝑠)sinh√𝐌,0≤𝑡≤𝑠≤1,max_{𝑡,𝑠∈𝐽,𝑡≠𝑠}𝐺^′_1𝑡(𝑡,𝑠)=1. (18) It follows from (17) that 𝑇_𝜆𝑦_PC1≤𝑦_PC1,∀𝑦∈𝐾∩𝜕Ω_𝑟. (19) If 𝑓_∞=∞, 𝐼_∞(𝑘)=∞, then there exist 𝐿>0 and 𝑅>𝑟>0 such that 𝑓𝑡,𝑦>𝐿𝑦,𝐼_𝑘𝑡,𝑦>𝐿𝑦,∀𝑡∈𝐽,𝑦≥𝑅,𝑘=1,2,…,𝑚, (20) where 𝐿 satisfies 𝜆𝛼𝐿𝛿𝛾+𝑚≥1. (21)
Let 𝜂=𝑅/𝛿. Thus, when 𝑦∈𝐾∩𝜕Ω_𝜂 we have 𝑦(𝑡)≥𝛿𝑦_PC1=𝛿𝜂=𝑅,𝑡∈𝐽, (22) and then we get 𝑇_𝜆𝑦(𝑡)=𝜆¹₀𝐺(𝑡,𝑠)𝜔(𝑠)𝑓𝑠,𝑦(𝑠)𝑑𝑠+𝜆^𝑚_𝑘=1𝐺𝑡,𝑡_𝑘𝐼_𝑘𝑡_𝑘,𝑦𝑡_𝑘≥𝜆𝛼¹₀𝜔(𝑠)𝐿𝑦(𝑠)𝑑𝑠+𝜆𝛼^𝑚_𝑘=1𝐿𝑦𝑡_𝑘≥𝜆𝛼𝐿𝛿𝑦_PC1¹₀𝜔(𝑠)𝑦𝑑𝑠+𝑚=𝜆𝛼𝐿𝛿_PC1≥𝑦𝛾+𝑚_PC1. (23) This yields 𝑇_𝜆𝑦_PC1≤𝑦_PC1,∀𝑦∈𝐾∩𝜕Ω_𝜂. (24)
Hence, for given 𝜆>0 condition (a) of Lemma 7 is satisfied of operator 𝑇_𝜆, which implies that 𝑇_𝜆 has a fixed point 𝑦_𝜆 in Ω_𝜂⧵Ω_𝑟.
It remains to prove ‖𝑦_𝜆‖_PC1=+∞ as 𝜆→0⁺. In fact, if not, there would exist a number 𝑐>0 and a sequence 𝜆_𝑛→0⁺ such that 𝑦_𝜆𝑛PC1≤𝑐(𝑛=1,2,3,…). (25) Furthermore, the sequence {‖𝑦_𝜆𝑛‖_PC1} contains a subsequence that converges into a number 𝑑, where 0≤𝑑≤𝑐. For simplicity, suppose that {‖𝑦_𝜆𝑛‖_PC1} itself converges into  𝑑.
If 𝑑>0, then ‖𝑦_𝜆𝑛‖_PC1>𝑑/2 for sufficiently large 𝑛 (𝑛>ℕ), and therefore 1𝜆_𝑛=¹₀𝐺(𝑡,𝑠)𝜔(𝑠)𝑓𝑠,𝑦_𝜆𝑛(𝑠+)𝑑𝑠_𝑚𝑘=1𝐺𝑡,𝑡_𝑘𝐼_𝑘𝑡_𝑘,𝑦_𝜆𝑛𝑡_𝑘PC1×𝑦_𝜆𝑛PC1⁻¹≤𝛽𝛾𝜌+𝑚𝜌^∗𝑦_𝜆𝑛PC1≤2𝛽𝛾𝜌+𝑚𝜌^∗𝑑(𝑛>ℕ), (26) which contradicts 𝜆_𝑛→0⁺.
If 𝑑=0, then ‖𝑦_𝜆𝑛‖_PC1→0 for sufficiently large 𝑛  (𝑛>ℕ), and therefore it follows from (H₃) that for any 𝜀>0 there exists 𝑏>0 such that 𝑓𝑡,𝑦_𝜆𝑛(𝑡)≤𝜀𝑦_𝜆𝑛,𝐼_𝑘𝑡,𝑦_𝜆𝑛(𝑡)≤𝜀𝑦_𝜆𝑛,∀𝑦_𝜆𝑛∶0≤𝑦_𝜆𝑛≤𝑏, (27) and hence it follows from (10) that 1𝜆_𝑛=¹₀𝐺(𝑡,𝑠)𝜔(𝑠)𝑓𝑠,𝑦_𝜆𝑛(𝑠+)𝑑𝑠_𝑚𝑘=1𝐺𝑡,𝑡_𝑘𝐼_𝑘𝑡_𝑘,𝑦_𝜆𝑛𝑡_𝑘PC1×𝑦_𝜆𝑛PC1⁻¹≤𝛽𝛾𝜀𝑦_𝜆𝑛PC1+𝑚𝜀𝑦_𝜆𝑛PC1𝑦_𝜆𝑛PC1=𝛽𝛾+𝑚𝜀. (28) Since 𝜀 is arbitrary, we have 𝜆_𝑛→+∞(𝑛→+∞) in contradiction to 𝜆_𝑛→0⁺. Therefore, ‖𝑦_𝜆‖→+∞ as 𝜆→0⁺ and the proof is complete.

Remark 8. Comparing with Nieto and O’Regan [17], the main features of this paper are as follows.(i)The parameter 𝜆>0 is considered.(ii)The parameter dependence of the solution is available.(iii)𝜔∈𝐿¹_loc(0,1), not 𝜔(𝑡)≡1 for 𝑡∈𝐽.

4. An Example

To illustrate how our main results can be used in practice we present an example.

Example 9. Consider the following boundary value problems −𝑦^′′(𝑡)(𝑡)1+𝑦=𝜆√𝑡³√1+𝑡²𝑦²(𝑡),𝑡∈𝐽,𝑡≠12,−Δ𝑦^′_𝑡1=1/2=1𝑦1+𝑡³𝑦1+𝑦,𝑘=1,_′(0)=𝑦^′(1)=0. (29) Evidently, 𝑦(𝑡)≡0 is the trivial solution of problem (29).

Conclusion. Problem (29) has at least one positive solution for any 𝜆>0.

Proof. Problem (29) can be regarded as a problem of the form (1), where 𝐌=1,𝑡₁=12,𝜔(𝑡)=1√𝑡,𝑓𝑡,𝑦=√³1+𝑡²𝑦²(𝑡),𝐼₁=()𝑦𝑡,𝑦1+𝑡³.1+𝑦 (30)
It follows from the definition of 𝜔, 𝑓, and 𝐼 that (H₁) and (H₂) hold, and 𝜔(𝑡) is singular at 𝑡=0 and 𝑡=1. By calculating, we have 1𝛿=cosh²(1),𝛾=¹₀1√𝑡1𝑑𝑡=2,𝛼=sinh(1),𝛽=cosh²(1)(1),𝑓sinh₀=limsup_𝑦→0max_𝑡∈𝐽³√1+𝑡²𝑦²(𝑡)𝑦𝐼=0,₀(𝑘)=limsup_𝑦→0max_𝑡∈𝐽(1+𝑡)𝑦³/1+𝑦𝑦𝑓=0,_∞=liminf_𝑦→∞min_𝑡∈𝐽³√1+𝑡²𝑦²(𝑡)𝑦𝐼=∞,_∞(𝑘)=limsup_𝑦→∞min_𝑡∈𝐽(1+𝑡)𝑦³/1+𝑦𝑦=∞. (31) Then, the condition (H₃) of Theorem 1 holds. Hence, by Theorem 1, the conclusion follows, and the proof is complete.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors are indebted to the referee’s suggestions. These have greatly improved this paper. This work is sponsored by the Project NSFC (11301178) and the Fundamental Research Funds for the Central Universities (2014MS58).