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Hindawi Publishing Corporation Boundary Value Problems Volume 2009, Article ID 103867, 34 pages doi:10.1155/2009/103867 Research Article An Approximation Approach to Eigenvalue Intervals for Singular Boundary Value Problems with Sign Changing and Superlinear Nonlinearities ă Haishen Lu,1 Ravi P Agarwal,2, and Donal O’Regan4 Department of Applied Mathematics, Hohai University, Nanjing 210098, China Department of Mathematical Sciences, Florida Institute of Technology, Melbourne, FL 32901-6975, USA KFUPM Chair Professor, Mathematics and Statistics Department, King Fahd University of Petroleum and Minerals, Dhahran 31261, Saudi Arabia Department of Mathematics, National University of Ireland, Galway, Ireland Correspondence should be addressed to Haishen Lu, haishen2001@yahoo.com.cn ă Received 25 June 2009; Accepted October 2009 Recommended by Ivan T Kiguradze This paper studies the eigenvalue interval for the singular boundary value problem −u g t, u λh t, u , t ∈ 0, , u 0 u , where g h may be singular at u 0, t 0, 1, and may change sign and be superlinear at u ∞ The approach is based on an approximation method together with the theory of upper and lower solutions Copyright q 2009 Haishen Lu et al This is an open access article distributed under the Creative ă Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited Introduction The singular boundary value problems of the form −u f t, u , t ∈ 0, , 1.1 u 0 u occurs in several problems in applied mathematics, see 1–6 and their references In many papers, a critical condition is that f t, r ≥ for t, r ∈ 0, × 0, ∞ 1.2 Boundary Value Problems or there exists a constant L > such that for any compact set K ⊂ 0, , there is ε such that εK > f t, r ≥ L ∀t ∈ K, r ∈ 0, ε , 1.3 f t, r lim r →∞ r ∀t ∈ 0, We refer the reader to 1–4 In the case, when f t, r may change sign in a neighborhood of ∞ for t ∈ 0, , very few existence results are available r and lim supr → ∞ f t, r /r in literature In this paper we study positive solutions of the second boundary value problem −u g t, u λh t, u , t ∈ 0, , 1.4 u 0 u 1; here g : 0, × 0, ∞ → R and h : 0, × 0, ∞ → 0, ∞ are continuous, so as a result, our nonlinearity may be singular at t 0, and u Also our nonlinearity may change sign and be superlinear at u ∞ Our main existence results Theorems 1.1, 1.2 and 1.4 are new see Remark 1.5, Examples 3.1 and 3.2 A function u is a solution of the boundary value problem 1.4 if u : 0, → R, u satisfies the differential equation 1.4 on 0, and the stated boundary data Let C 0, denote the class of maps u continuous on 0, , with norm |u|∞ a ∧ b; max{a, b} a ∨ b Given α, β ∈ C 0, , α ≤ β, maxt∈ 0,1 |u t | We put min{a, b} let β Dα v | v ∈ C 0, , α ≤ v ≤ β 1.5 Let M h ∈ C 0, : |h s |ds < ∞ with lim t|h t | < ∞, lim− − t |h t | < ∞ t→0 t→1 1.6 In this paper, we suppose the following conditions hold: G1 suppose there exist gi : 0, × 0, ∞ → 0, ∞ that i 1, continuous functions such gi t, · is strictly decreasing for t ∈ 0, , g1 ·, rφ1 · , g2 ·, r ∈ M −g1 t, r ≤ g t, r ≤ g2 t, r where φ1 is defined in Lemma 2.1; ∀r > 0, for t, r ∈ 0, × 0, ∞ , 1.7 Boundary Value Problems H1 there exist hi : 0, × 0, ∞ → 0, ∞ i 1, continuous functions such that hi t, · is increasing for t ∈ 0, , h1 ·, r , h2 ·, r ∈ M h1 t, r ≤ h t, r ≤ h2 t, r 1.8 for r > 0, for t, r ∈ 0, × 0, ∞ ; H2 there exists r > such that h1 t, r > for t ∈ 0, The main results of the paper are the following Theorem 1.1 Suppose G1 , H1 , H2 and the following conditions hold: G2 for all r2 > r1 > 0, there exists γ · ∈ M such that g2 ·, r γ · r is increasing in r1 , r2 : H3 lim r →∞ h1 t, r r ∀t ∈ 0, ; H4 there exists a sequence {Rj }∞ such that limj → ∞ Rj j h2 s, Rj j →∞ Rj lim where a1 1 g a1 1.9 ∞ and 1.10 0, s, ds Then there exists λ∗ > such that for every λ ≥ λ∗ , 1.4 has at least one positive 1 solution u ∈ C 0, ∩ C1 0, and u > for t ∈ 0, Theorem 1.2 Suppose G1 , H1 , H2 and the following conditions hold: G3 for all r2 > r1 > there exists γ · ∈ M such that g t, r γ t r is increasing in r1 , r2 ; G4 there exists c1 > such that ≤ g t, r , t ∈ 0, , < r < c1 ; 1.11 G5 there exists c2 ∈ 0, c1 , < β < such that for all r ∈ 0, c2 t − t g t, rl t dt ≥ rπ, 1.12 where g m t, r and l t g t, r , min{t, − t} for t ∈ 0, m rβ for m ≥ 1, 1.13 Boundary Value Problems Then there exists λ∗ > such that i if < λ < λ∗ , 1.4 has at least one solution u ∈ C 0, ∩ C1 0, and u > for t ∈ 0, ; ii if λ > λ∗ , 1.4 has no solutions Remark 1.3 Notice that g m t, r satisfies G1 , G3 , G4 and for fixed m ≥ 1, t − t g m t, rl t dt ≥ rπ g t, r ≥ g m t, r ≥ g t, r for r ∈ 0, c2 , 1.14 for t ∈ 0, , r ∈ 0, ∞ Theorem 1.4 Suppose G1 , H1 , H2 and the following conditions hold: G6 there exists τ ≥ τ1 such that lim τr r →0 g − t, r h t, r 1.15 0, where τ1 is defined in Lemma 2.1 and g t, r −g t, r }; max{0, g t, r }, g − t, r H5 for all r2 > r1 > 0,there exists γ · ∈ M such that h t, r max{0, γ t r is increasing in r1 , r2 Then there exists λ∗ > such that i if < λ < λ∗ , 1.4 has at least one solution u ∈ C 0, ∩ C1 0, and u > for t ∈ 0, ; ii if λ > λ∗ , 1.4 has no solutions Remark 1.5 In 5, , the authors consider the boundary value problem 1.4 under the conditions lim r →∞ h2 t, r r 1.16 In Section 3, we give two examples see Examples 3.1 and 3.2 which satisfy the conditions in Theorem 1.1 or Theorem 1.2 but they not satisfy the conditions in 1–5 Proof of Main Results 2.1 Some Lemmas Lemma 2.1 Consider the following eigenvalue problem −u τu t , u u t ∈ 0, , 2.1 Boundary Value Problems Then the eigenvalues are mπ τm for m 1, 2, , 2.2 and the corresponding eigenfunctions are φm t sin mπt for m 1, 2, 2.3 Let G t, s be the Green’s function for the BVP: for t ∈ 0, , −u u u1 2.4 Then G t, s ⎧ ⎨s − t , ≤ s < t ≤ 1, ⎩t − s , ≤ t < s ≤ 2.5 Also for all t, s ∈ 0, × 0, , define N t, s ⎧ ⎪ G t, s ⎪ ⎪ ⎪ φ t ⎪ ⎪ ⎪ ⎪ ⎨ 1−s ⎪ ⎪ π ⎪ ⎪ ⎪ ⎪ ⎪ ⎪s ⎩ π if t / 0, 1, if t 0, if t 2.6 It follows easily that < G t, s ≤ t − t for t, s ∈ 0, × 0, , 2.7 s 1−s ≤ N t, s ≤ 2π for t, s ∈ 0, × 0, Define the operator A, B : M → C 0, by Ax t G t, s x s ds, 2.8 Bx t N t, s x s ds The following four results can be found in needed in the proofs there notice limr → ∞ h2 t, r /r is not Boundary Value Problems Lemma 2.2 Suppose G1 and H1 hold Let n0 ∈ N Assume that for every n > n0 , there exist an , δn , δ ∈ M such that ≤ an t , |δn t | ≤ δ t , lim δn t 0, n→∞ for t ∈ 0, 2.9 and there exist u, un , un , u ∈ C 0, such that < u t ≤ un t ≤ un t ≤ u t and u for t ∈ 0, , 2.10 If u − un t an t un t ≤ g t, − un t n v λh t, v δn t an t v t for t ∈ 0, , 2.11 an t un t ≥ g t, n v λh t, v δn t an t v t for t ∈ 0, , un where λ ≥ and v ∈ Dun , then 1.4 has a solution u ∈ C 0, ∩C1 0, such that u t ≤ u t ≤ u t for t ∈ 0, Lemma 2.3 Let ψ : 0, × 0, ∞ → 0, ∞ be a continuous function with ψ t, · is strictly decreasing, ψ ·, r ∈ M ∀r > 2.12 Then the problem −ω t n ψ t, ω t ω ω for t ∈ 0, , 2.13 has a solution ωn ∈ C 0, such that ωn t ≤ ωn t ≤1 ω1 t ≤ ψ s, ds for t ∈ 0, , n ∈ N 2.14 If we let ω t limn → ∞ ωn t for t ∈ 0, , then ω ∈ C 0, , −ω t ω t > for t ∈ 0, , ψ t, ω t ω ω for t ∈ 0, , 2.15 Boundary Value Problems Next we consider the boundary value problem −u a t ut t ∈ 0, , f t, u 0 u 1, 2.16 where a, f ∈ M, a t ≥ for t ∈ 0, Lemma 2.4 The following statements hold: i for any f ∈ M, 2.16 is uniquely solvable and u A au A f ; 2.17 ii if f t ≥ for t ∈ 0, , then the solution of 2.16 is nonnegative Corollary 2.5 Let Φ : M → C 0, ∩ C1 0, be the operator such that Φ f is the solution of 2.16 Then we have i if f1 t ≤ f2 t for t ∈ 0, , then Φ f1 t ≤ Φ f2 t for t ∈ 0, ; ii let E ⊂ M and β ∈ M If |f t | ≤ β t , t ∈ 0, for all f ∈ E, then Φ E is relatively compact with respect to the topology of C 0, Lemma 2.6 see Let f ∈ M, f ≥ 0, f / 0, u ∈ C 0, ∩ C1 0, satisfy ≡ −u u Then there exist m m f > 0, M f in 0, , u1 2.18 M f > such that ml t ≤ u t ≤ Ml t for t ∈ 0, 2.19 2.2 The Proof of Theorem 1.1 Claim see There exists λ∗ > 0, c > 0, independent of λ, such that for all λ ≥ λ∗ there 1 exist Rλ > c, u ∈ C 0, , with cφ1 t ≤ u t ≤ Rλ φ1 t and −u t −g1 t, u t u λh1 t, u t , u1 0, for t ∈ 0, , 2.20 Boundary Value Problems with g1 ·, u · , h1 ·, u · ∈ M 2.21 Let λ∗ > 0, c > and u ∈ C 0, be defined in Claim Define ψ t, r for t ∈ 0, g2 t, r 2.22 From G1 notice that ψ satisfies the assumptions of Lemma 2.3, so there exist ω, ωn ∈ C 0, , ωn t > 0, ω t > for t ∈ 0, such that −ωn t g2 t, n ωn ωn t ≤ ωn t ≤1 ω t −ω t for t ∈ 0, , ωn ωn 0, for t ∈ 0, , n ∈ N, ω1 t ≤ a1 n→∞ for t ∈ 0, , g2 t, ω t ω 2.23 for t ∈ 0, , lim ωn t ω 0, where a1 g s, ds Let λ ≥ λ∗ , n ∈ N be fixed We consider the following boundary value problem: −v t λh2 t, v ωn v λh1 t, u v h2 t, Rj j →∞ Rj a1 2.24 By H4 , there exist {Rj }∞ such that limj → ∞ Rj j lim for t ∈ 0, , ∞ and for t ∈ 0, , 2.25 so lim j →∞ λh2 t, Rj a1 λh1 t, u t Rj for t ∈ 0, 2.26 Boundary Value Problems There exists j0 ∈ N such that λh2 t, Rj0 a1 ≤ Rj0 λh1 t, u t 2.27 If v ∈ C 0, and ≤ v t ≤ Rj0 φ1 t for t ∈ 0, , then N t, s λh2 s, v s λh1 s, u ds ωn s ≤ N t, s λh2 s, v s λh1 s, u ds a1 ≤ N t, s λh2 s, Rj0 φ1 s a1 λh1 s, u ds N t, s λh2 s, Rj0 2.28 λh1 s, u ds ≤ a1 ≤ Rj0 , for t ∈ 0, , and so 0≤ G t, s λh2 s, v s ds ≤ Rj0 φ1 t λh1 s, u s ωn s for t ∈ 0, 2.29 Let Φ : C 0, → C 0, be the operator defined by Φv t : G t, s λh2 s, v s ωn s λh1 s, u s ds for v ∈ C 0, , t ∈ 0, 2.30 It is easy to see that Φ is a continuous and completely continuous operator Also if ≤ v t ≤ Rj0 φ1 t for t ∈ 0, , then ≤ Φ v t ≤ Rj0 φ1 t for t ∈ 0, , so Schauder’s fixed point v, that is, theorem guarantees that there exists v ∈ 0, Rj0 φ1 such that Φ v −v t λh2 t, v t v λh1 t, u t , ωn s v 2.31 Let un t ωn t t for t ∈ 0, 2.32 10 Boundary Value Problems Then un ∈ C 0, , un −un t un 0, and −ωn t − t g2 t, n ωn λh2 t, ωn ≥ g2 t, n un λh1 t, u λh2 t, un ω t Rj0 φ1 t for t ∈ 0, , λh1 t, u 2.33 for t ∈ 0, Let u t 2.34 so ≤ un t ≤ u t for t ∈ 0, 2.35 From Claim 1, we obtain −u t −g1 t, u λh1 t, u ≤ λh1 t, u ≤ λh1 t, u g2 t, n 2.36 un λh2 t, un ≤ −un t for t ∈ 0, , − u − un t ≤ for t ∈ 0, that is, 2.37 A standard argument yields u t ≤ un t for t ∈ 0, From G2 , there exists γ ∈ M such that r → g2 t, 1/n 0, |u|∞ Let un u From 2.35 and 2.38 , we have < u t ≤ un t ≤ un t ≤ u t 2.38 r for t ∈ 0, γ t r is increasing on 2.39 20 Boundary Value Problems Let ⎧ ⎪ ⎨ ⎛ ⎞ ⎜ π sup B⎝ r m > max M ⎪ ⎩ m β u∧χ t∈ 0,1 uβ ⎟ e⎠ t , r Note u ≤ u since r m > r Let v ≥ u and notice note g − ·, r A g m ·, n v∧χ δn ≤ G t, s g m s, ≤ 1 n v ∧ χ − g s, v∧χ 1/n β ⎡ G t, s ⎣ ⎢ G t, s ⎣ v∧χ uβ β g − s, n n u∧χ u g s, u ∧ χ g s, u ds ds 2.93 ⎤ m u∧χ β uβ ⎥ e⎦ds ⎞⎤ m u∧χ β uβ ⎟⎥ e⎠⎦ t ⎞⎤ ⎡ ⎛ ⎢ ⎜ ≤ π ⎣B⎝ if < r < c1 from G4 e⎦ds ⎡ ⎛ ⎢ ⎜ φ1 t ⎣B⎝ 2.92 ⎤ m ⎡ ≤ ⎪ ⎭ t m G t, s ⎫ ⎪ ⎬ m u∧χ β uβ ⎟⎥ e⎠⎦ t · l t for t ∈ 0, On the other hand, A μh ·, v ∧ χ t μ G t, s h s, v ∧ χ ds ≤λ G t, s h2 s, χ ds v t ≤ Ml t for t ∈ 0, , 2.94 Boundary Value Problems 21 so A g m ·, n v∧χ ≤ A g m ·, n δn μh ·, v ∧ χ v∧χ δn ⎡ ⎛ ⎢ ⎜ ≤ π ⎣B⎝ t A μh ·, v ∧ χ t t ⎞⎤ m uβ β u∧χ 2.95 ⎟⎥ e⎠⎦ t · l t Ml t for t ∈ 0, , v ∈ u, um , n ≥ ≤ um t Step Let < μ < λ Let n, m ≥ be fixed There exists βn,m ∈ C 0, such that u t ≤ βn,m t ≤ um t , −βn,m t g m t, n βn,m ∧ χ μh t, βn,m ∧ χ βn,m βn,m for t ∈ 0, , δn t 2.96 Let n, m > be fixed From Remark 1.3, there exist γn ∈ M, γn ≥ such that g m t, r γn t r is increasing in 1/n, 1/n r m /2 We easily prove that g m t, r ∧ χ γn t r is increasing in 1 , n n rm 2.97 Let γ t γn We have g m t, 1/n r ∧ χ γ t r is increasing in 0, r m /2 From 2.83 and 2.89 , we have for fixed v ∈ C 0, , u t ≤ v t ≤ um t that u t A γu t ≤ A g m ·, n u∧χ δn t A γu t ≤ A g m ·, n u∧χ γu δn μh ·, v ∧ χ n v∧χ ≤ A g m ·, ≤ um t t 2.98 A γ um t γv δn μh ·, v ∧ χ t 22 Boundary Value Problems Fix v ∈ C 0, with u t ≤ v t ≤ um t From Lemma 2.4, there exists Ψ v ∈ C 0, such that γ tΨ v t −Ψ v t g m t, Ψ v n v∧χ Ψ v γ t v t μh t, v ∧ χ δn t for t ∈ 0, 2.99 Then Ψ v t A γΨ v A g m ·, t n v∧χ γv μh ·, v ∧ χ δn for t ∈ 0, , t 2.100 so 2.98 implies that A γu t ≤ Ψ v t u t A γΨ v t 2.101 ≤ um t A γ um for t ∈ 0, t From Corollary 2.5, we have u t ≤ Ψ v t ≤ um t for t ∈ 0, 2.102 Also, g m t, n v∧χ ≤ g1 t, φ1 t n ≡β t ∈M γv μh t, v ∧ χ δn g2 t, n γ um ∞ |δn t | λh2 t, χ ∞ 2.103 for t ∈ 0, u u Now Ψ : Dum → Dum is compact, so Schauder’s fixed point theorem implies that there exists βn,m ∈ C 0, such that u t ≤ βn,m t ≤ um t and Ψ βn,m t −βn,m t g m t, n βn,m ∧ χ βn,m g m t, n βn,m ∧ χ μh t, βn,m ∧ χ βn,m μh t, βn,m ∧ χ δn t δn t βn,m t for t ∈ 0, : for t ∈ 0, , 2.104 0, ≤ 3g2 t, u ∧ χ λh2 t, χ Boundary Value Problems 23 Let m ≥ be fixed We consider the sequence {βn,m }∞ Fix n0 ∈ {2, 3, } Let us n look at the interval 1/2n0 , − 1/2n0 The mean value theorem implies that there exists τ ∈ 1/2m0 , − 1/2m0 with |βn,m τ | ≤ 8/3 supt∈ 0,1 um t As a result βn,m t ∞ n n0 1 ,1 − n 2n0 is bounded, equicontinuous family on 2.105 The Arzela-Ascoli theorem guarantees the existence of subsequence Nn0 of integers and a function zn0 ,m ∈ 1/2n0 , − 1/2n0 with βn,m converging uniformly to zn0 ,m on 1/2n0 , − 1/2n0 as n → ∞ through Nn0 Similarly, βn,m ∞ n n0 is bounded, equicontinuous family on 1 ,1 − n 2n0 2 , 2.106 so there is a subsequence Nn0 of Nn0 and a function zn0 1,m ∈ C 1/2n0 , − 1/2n0 with βn,m converging uniformly to zn0 1,m on 1/2n0 , − 1/2n0 as n → ∞ through Nn0 Note zn0 ,m on 1/2n0 , − 1/2n0 since Nn0 ⊆ Nn0 Proceed inductively to obtain zn0 1,m subsequences of integers Nn0 ⊇ Nn0 ⊇ · · · ⊇ Nk ⊇ · · · and functions zk,m ∈ C 1/2k , − 1/2k with βn,m converging uniformly to zk,m on 1/2k , − 1/2k as n → ∞ through Nk , and zk,m zk−1,m on 1/2k , − 1/2k zk,m t on 1/2k , − 1/2k and Define a function um : 0, → 0, ∞ by um t um Notice um is well defined and u t ≤ um t ≤ um t for t ∈ 0, Next, fix um t ∈ 0, without loss of generality assume t / 1/2 and let n∗ ∈ {n0 , n0 1, } be such that ∗ ∗ ∗ ∗ 1/2n < t < − 1/2n Let Nn∗ {i ∈ Nn : i ≥ n∗ } Now βn,m , n ∈ Nn∗ satisfies the integral equation βn,m t βn,m t 1/2 βn,m t− 2.107 s−t gm s, n βn,m ∧ χ μh s, βn,m ∧ χ δn s ds, ∗ for t ∈ 1/2n , − 1/2n Notice take t 2/3 say that {βn,m 1/2 }, n ∈ Nn∗ , is a bounded sequence since u t ≤ βn,m t ≤ um t for t ∈ 0, Thus {βn,m 1/2 }n∈N ∗∗ has a convergent n subsequence; for convenience we will let {βn,m 1/2 }n∈N ∗∗ denote this subsequence also, and n ∗ let τ ∈ R be its limit Now for the above fixed t, and let n → ∞ through Nn∗ to obtain gm t, n βn,m ∧ χ −→ gm t, zk,m ∧ χ , h t, βn,m ∧ χ −→ h t, zk,m ∧ χ , δn −→ 2.108 24 Boundary Value Problems As a result, zk,m t zk,m τ t− τ t− t t 1/2 s − t g m s, zk,m ∧ χ μh s, zk,m ∧ χ ds, 2.109 that is, um t um 1/2 s − t g m s, um ∧ χ μh s, um ∧ χ ds 2.110 We can this argument for each t ∈ 0, and so −um t g m t, um ∧ χ μh t, um ∧ χ for t ∈ 0, 2.111 It remains to show that um is continuous at and Let ε > be given Since um ∈ C 0, there exists δ > with um t < ε/2 for t ∈ 0, δ As a result u t ≤ βn,m t ≤ um t < ε/2 for t ∈ 0, δ Consequently, u t ≤ um t ≤ ε/2 < ε for t ∈ 0, δ and so um is continuous at Similarly, um is continuous at As a result um ∈ C 0, and −um t g m t, um ∧ χ um μh t, um ∧ χ um for t ∈ 0, , 2.112 Next we prove um t ≤ χ t for t ∈ 0, 2.113 Suppose 2.113 is not true Let y t um t − χ t and σ ∈ 0, be the point where y t attains its maximum over 0, We have y σ > 0, y σ ≤ 2.114 On the other hand, since um σ > χ σ , we have y σ um σ − χ σ −g m σ, um ∧ χ − μh σ, um ∧ χ −g m σ, χ σ − μh σ, χ σ ≥ λ − σ h σ, χ σ This is a contradiction, so 2.113 is true > g σ, χ g σ, χ σ λh σ, χ 2.115 λh σ, χ σ Boundary Value Problems 25 Thus we have −um gm t, um um μh t, um , um u t ≤ um t ≤ χ t 0, 2.116 for t ∈ 0, By the same reason as above, we obtain subsequences of integers Nm0 ⊇ Nm0 ⊇ · · · ⊇ Nk ⊇ · · · and functions zm ∈ C 1/2k , − 1/2k with um converging uniformly to zk on 1/2k , − 1/2k as m → ∞ through Nk , and zk zk−1 on 1/2k , − 1/2k Define a function u : 0, → 0, ∞ by u t zk t on 1/2k , − 1/2k and u u Notice u is well defined and u t ≤ u t ≤ χ t for t ∈ 0, Next fix t ∈ 0, ∗ without loss of generality assume t / 1/2 and let m∗ ∈ {m0 , m0 1, } be such that 1/2m < ∗ ∗ ∗ t < 1−1/2m Let Nm∗ {k ∈ Nm∗ : k ≥ m∗ } Now um , m ∈ Nm∗ satisfies the integral equation um t um um ∗ t− t 1/2 s − t g m s, um μh s, um ds 2.117 ∗ ∗ for t ∈ 1/2m , − 1/2m Notice take t 2/3 say that {um 1/2 }, m ∈ Nm∗ is a bounded sequence since u t ≤ um t ≤ χ t for t ∈ 0, Thus {um 1/2 }m∈N ∗ ∗ has a convergent m subsequence; for convenience we will let {um 1/2 }m∈N ∗ ∗ denote this subsequence also, and m ∗ let τ ∈ R be its limit Now for the above fixed t, and letting m → ∞ through Nk to obtain u t u τ t− t s − t g s, u μh s, u ds 2.118 1/2 we can this argument for each t ∈ 0, and so −u t g t, u μh t, u for t ∈ 0, 2.119 Also reasoning as before we have that u is continuous at and Thus we have −u g t, u u Now let λ∗ u1 μh t, u , 2.120 sup Λ > Then i if < λ < λ∗ , 1.4 has at least one solution u ∈ C 0, ∩ C1 0, and u > for t ∈ 0, ; ii if λ > λ∗ , 1.4 has no solutions 26 Boundary Value Problems 2.4 The Proof of Theorem 1.4 Claim Let λ∗ N maxt∈ 0,1 t, s h2 s, a3 φ1 ds > 0; 2.121 here a3 1 g2 s, h2 s, φ1 s φ1 ∞ φ1 s ds 2.122 Then 0, λ∗ ∈ Λ Proof of Claim Let λ ∈ 0, λ∗ be fixed From assumption G6 , it follows that there is τ ≥ τ1 and c3 ∈ 0, , such that if n > 2/c3 , < k < c3 /2 < 1, we have < k φ1 τ 1/n c3 , < ∞ 0< n g − t, 1/n kφ1 t h t, 1/n kφ1 t < c3 , 2.123 kφ1 t ≤ λ kφ1 t Thus, g − t, 1/n τkφ1 t h t, 1/n kφ1 t kφ1 t ≤ λ 2.124 Then, for n > 2/c3 , g − t, τkφ1 t n ≤ λh t, kφ1 t n kφ1 t , 2.125 and we have τkφ1 t ≤ λh t, ≤g t, g t, n kφ1 t − g − t, n kφ1 t n kφ1 t − g − t, n kφ1 t n kφ1 t − λh t, kφ1 t g t, n kφ1 t λh t, n kφ1 t λh t, kφ1 t λh t, kφ1 t δn t , λh t, n kφ1 t 2.126 Boundary Value Problems 27 where δn t Let u t λh t, n − λh t, kφ1 t kφ1 t 2.127 kφ1 t We have −u t n τ1 kφ1 t ≤ τkφ1 t ≤ g t, u t λh t, u t for t ∈ 0, δn t 2.128 Let ψ t, s g2 t, s λh2 t, φ1 t φ1 ∞ φ1 t 2.129 From G1 notice that ψ satisfies the assumptions of Lemma 2.3, so there exist ω, ωn ∈ C 0, such that −ωn t g2 t, n ωn λh2 t, ωn ω t ωn t ≤ 1 g2 s, h2 s, ωn for t ∈ 0, , 0, for t ∈ 0, , lim ωn t n→∞ φ1 t φ1 ∞ φ1 t φ1 s φ1 ∞ φ1 s ds a3 for t ∈ 0, , n ∈ N 2.130 Consider the boundary value problem −v t λh2 t, ωn v v v for t ∈ 0, , 2.131 Let Φ : C 0, → C 0, be the operator defined by Φv t : λ G t, s h2 s, ωn v ds for v ∈ C 0, , t ∈ 0, 2.132 28 Boundary Value Problems It is easy to see that Φ is a continuous and completely continuous operator Also if ≤ v t ≤ φ1 t for t ∈ 0, , then 0≤Φ v t λ G t, s h2 s, ωn v ds ≤ λ∗ G t, s h2 s, a3 φ1 ds 2.133 φ1 t N t, s h2 s, a3 φ1 ds maxt∈ 0,1 N t, s h2 s, a3 φ1 ds ≤ φ1 t for t ∈ 0, Thus Schauder’s fixed point theorem guarantees that there exists ∈ , that is, Φ −vn t , λh2 t, ωn vn 0, φ1 such that 2.134 Let un t ωn t Then un , u ∈ C 0, , un −un t t , un ut 0, u ω t u φ1 t for t ∈ 0, 2.135 0, ≤ un t ≤ u t for t ∈ 0, , and −ωn t − t g2 t, n ωn λh2 t, φ1 t φ1 t φ1 λh2 t, ωn ≥ g2 t, n un λh2 t, φ1 t φ1 t φ1 λh2 t, un for t ∈ 0, 2.136 We next prove that u t ≤ un t for t ∈ 0, 2.137 u t − un t and σ ∈ 0, be the point where Suppose 2.137 is not true Let y t y t attains its maximum over 0, We have y σ > 0, y σ ≤ 2.138 Boundary Value Problems 29 On the other hand, since u σ > un σ , we have −u σ ≤ g σ, n uσ λh σ, u σ g σ, n uσ λh σ, ≤ g2 σ, n u σ < g2 σ, n un σ δn σ n λh2 σ, u σ λh2 σ, 2.139 φ1 σ φ1 σ φ1 φ1 σ λh2 σ, un σ ≤ −un σ Thus y σ u σ − un σ > 0, and this is a contradiction As a result, 2.137 is true On the other hand, we have |δn t | ≤ λ h t, ≤ 2λh2 n kφ1 t t, − h t, kφ1 t 2.140 φ1 for n > 2/c3 Consequently, for t ∈ 0, , δn → and n → ∞ From assumptions G2 and H5 , there exists a γ, τ ∈ M, n > 2/c3 , so that g t, 1/n un γ t τ t Let un u t For v ∈ Dun , r h t, r a t r is increasing in 0, |u|∞ , where a t we have − un t a t un t ≤ g t, n un t λh t, un t δn t ≤ g t, n un t λh t, un t a t un t λh t, n ≤ g t, n ≤ g2 t, n λh2 t, ≤ −un t − λh t, un t un t un t un t a t un t λh t, un t λh2 t, un t φ1 t an t un t a t un t a t un t φ1 t φ1 ∞ λh t, n 2.141 un t 30 Boundary Value Problems Reasoning as in the proof of Theorem 1.1, Lemma 2.2 guarantees that 1.4 has a solution u ∈ C 0, ∩ C1 0, Thus 1.4 has a solution for λ ∈ 0, λ∗ so Claim holds In particular, Λ / ∅ and sup Λ > Claim If λ ∈ Λ, then 0, λ ∈ Λ Proof of Claim We may assume that λ > Let χ be a positive solution of 1.4 , that is, −χ g t, χ λh t, χ , χ 0 t ∈ 0, , χ1 2.142 We first prove that there exists ρ > such that χ t ≥ ρl t for t ∈ 0, 2.143 By G6 , there exists σ > such that for all r ∈ 0, σ , we have τr g − t, r ≤ λ, h t, r 2.144 that is, τr ≤ λh t, r − g − t, r From the continuity of χ and χ for t ∈ 0, , r ∈ 0, σ 2.145 χ , it follows that there is < δ < 1/2 such that for t ∈ 0, δ ∪ − δ, χ t such that χ t ≥ ρl t for t ∈ 0, 2.148 Boundary Value Problems 31 We consider the boundary value problem g t, u ∧ χ −u u μh t, u ∧ χ , u1 2.149 0, g1 t, u ∧ χ , g2 t, u g2 t, u ∧ χ , h1 t, u h1 t, u ∧ χ , and where μ ∈ 0, λ Let g1 t, u h2 t, u ∧ χ We easily prove that the conditions of 6, Theorem 1.2 are satisfied so h2 t, u 2.149 has a positive solution u ∈ C1 0, ∩ C 0, We next prove that u t ≤χ t Suppose 2.150 is not true Let y t attains its maximum over 0, We have for t ∈ 0, 2.150 u t − χ t and σ ∈ 0, be the point where y t y σ ≤ y σ > 0, 2.151 On the other hand, since u σ > χ σ , we have y σ u σ −χ σ −g σ, u ∧ χ − μh σ, u ∧ χ −g σ, χ σ − μh σ, χ σ g σ, χ g σ, χ σ λh σ, χ λh σ, χ σ 2.152 λ − μ h σ, χ σ > This is a contradiction, so u t ≤χ t for t ∈ 0, 2.153 Thus we have −u g t, u u Let λ∗ u1 μh t, u , 2.154 sup Λ > Then i if < λ < λ∗ , 1.4 has at least one solution u ∈ C 0, ∩ C1 0, and u > for t ∈ 0, ; ii if λ > λ∗ , 1.4 has no solutions 32 Boundary Value Problems Example Example 3.1 Consider the boundary value problem −√ u where λ > Define {xn }∞ as x1 n q r λq u u −u u1 2, x2n ∀0 < t < 1, 3.1 x2n−1 , x2n ⎧ ⎪r , ⎪ ⎪ ⎪ ⎪ ⎨ x2n−1 , ⎪ √ ⎪ ⎪x ⎪ 2n − x2n ⎪ ⎩ r − x2n x2n − x2n 0, x2n 1, and if r ∈ 0, , if r ∈ x2n−1 , x2n , √ x2n , if r ∈ x2n , x2n 3.2 Then, Theorem 1.1 implies that there exists λ∗ > such that for every λ ≥ λ∗ , 3.1 has at least 1 one positive solution u ∈ C 0, ∩ C1 0, and u > for t ∈ 0, To see this, let g1 t, r g2 t, r h2 t, r h1 t, r √ r for t, r ∈ 0, × 0, ∞ , for t, r ∈ 0, × 0, ∞ , q r ⎧√ ⎨ r for t, r ∈ 0, × 16, ∞ , ⎩q r for t, r ∈ 0, × 0, 16 It is easy to see that G1 , H1 , H2 , and H3 are satisfied √ For all r2 > r1 > 0, let γ t 1/2r1 r1 Then g2 t, r r1 , r2 √ 1/ s ds and let Rj On the other hand, a1 h2 s, Rj j →∞ Rj lim 3.3 a1 √ 1/2r1 r1 r is increasing in x2j − 3, so we have x2j h2 s, x2j · j →∞ x2j x2j − √ x2j x2j lim · j → ∞ x2j x2j − lim 3.4 Thus G2 and H4 are satisfied Then Theorem 1.1 implies that there exists λ∗ > such that for every λ ≥ λ∗ , 3.1 has at least one positive solution u ∈ C 0, ∩ C1 0, and u > for t ∈ 0, Boundary Value Problems 33 Example 3.2 Consider the boundary value problem −u g t, u t ∈ 0, , λh t, u , u 0 3.5 u 1, where g t, r ⎧ ⎪ sin , ⎪ α ⎨r r ⎪ ⎪ ⎩− sin , rα r h t, r 0 Then Theorem 1.2 guarantees that there exists λ∗ > such that i if < λ < λ∗ , 3.5 has at least one solution u ∈ C 0, ∩ C1 0, and u > for t ∈ 0, ; ii if λ > λ∗ , 3.5 has no solutions 1/r β π α , and g2 t, r 1/r α , for t, r ∈ To see this, let β min{1/2, α/2}, g1 t, r h2 t, r r , for t, r ∈ 0, × 0, ∞ Notice that G1 , H1 , and 0, × 0, ∞ , and h1 t, r H2 are satisfied For all r2 > r1 > 0, let γ t ≡ sup r∈Λ ∂g ∂r where Λ r1 , r2 \ {nπ | n ∈ N}, so we have g t, r Let c1 1/π and we have ≤ g t, r , < ∞, 3.7 γ t r is increasing in r1 , r2 t ∈ 0, , < r < c1 3.8 Let n0 be fixed such that 21/ α−β < n0 π π 3.9 Let c2 ∈ 0, c1 be such that c2 < ∞ 2 − β π 1−β n n0 6n 2−β − 6n 2−β , 3.10 34 Boundary Value Problems and we have for n ≥ n0 , r ∈ 0, c2 , rt α sin 1 ≥ rt rt for t ∈ β r nπ 5π/6 , r nπ π/6 3.11 Also we have 1/2 t − t g m t, rl t dt ≥ ≥ ≥ ≥ t − t g m t, rl t dt 1/2 ∞ n n0 tg m t, rl t dt 1/r nπ π/6 t 1/r nπ 5π/6 ∞ 2r β n n0 1/r nπ π/6 rt β dt 3.12 t1−β dt 1/r nπ 5π/6 ∞ 2r 2 − β π 2−β n n0 6n 2−β − 6n 2−β ≥ rπ Thus G5 is satisfied Acknowledgment This research is supported by NNSF of China 10871059 References R P Agarwal and D O’Regan, Singular Differential and Integral Equations with Applications, Kluwer Academic Publishers, Dordrecht, The Netherlands, 2003 R P Agarwal and D O’Regan, “Twin solutions to singular Dirichlet problems,” Journal of Mathematical Analysis and Applications, vol 240, no 2, pp 433–445, 1999 D O’Regan, Theory of Singular Boundary Value Problems, World Scientific, River Edge, NJ, USA, 1994 P Habets and F Zanolin, “Upper and lower solutions for a generalized Emden-Fowler equation,” Journal of Mathematical Analysis and Applications, vol 181, no 3, pp 684–700, 1994 H Lu, D O’Regan, and R P Agarwal, “An approximation approach to eigenvalue intervals for ă singular boundary value problems with sign changing nonlinearities, Mathematical Inequalities and Applications, vol 11, no 1, pp 81–98, 2007 H Lu, D O’Regan, and R P Agarwal, Existence to singular 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