Positive solutions for singular boundary value problems involving integral conditions

Transcription

1 Hu Boundary Value Problems 22, 22:72 R E S E A R C H Open Access Positive solutions for singular boundary value problems involving integral conditions Liang-Gen Hu * * Correspondence: hulianggen@yahoo.cn Department of Mathematics, Ningbo University, Ningbo, 352, P.R. China Abstract We are interested in the following singular boundary value problem: u t+μwtfut =, <t <, u =, u = us das, where μ > is a parameter and us das is the Stieltjes integral. The function w C,,, + and w may be singular at t =and/ort =, f C[, +,, + and f =+. Some a priori estimates and the existence, multiplicity and nonexistence of positive solutions are obtained. Our proofs are based on the method of global continuous theorem, the lower-upper solutions methods and fixed point index theory. Furthermore, we also discuss the interval of parameter μ such that the problem has a positive solution. Keywords: singularity; global continuous theorem; solution of boundedness; fixed point index; positive solution Introduction We are concerned with the second order nonlocal boundary value problem: u t+μwtf ut =, <t <, u =, u = us das,. where μ > is a parameter and us das is a Stieltjes integral. The function w C,,, + and w may be singular at t = and/or t =,f C[, +,, + and f u f = lim u u =+. Integral boundary conditions and multi-point boundary conditions for differential equations come from many areas of applied mathematics and physics [ 7]. Recently, singular boundary value problems have been extensively considered in a lot of literature [, 2, 5, 8], since they model many physical phenomena including gas diffusion through porous media, nonlinear diffusion generated by nonlinear sources, chemically reacting systems as well as concentration in chemical or biological problems. In all these problems, positive solutions are very meaningful. 22 Hu; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

2 HuBoundary Value Problems 22, 22:72 Page 2 of 3 In [, 2], Webb and Infante considered the existence of positive solutions of nonlinear boundary value problem: u t+μhtg t, ut =, <t <, where h, g are nonnegative functions, subjected to the nonlocal boundary conditions u =, u = α[u]. Here α[u] is a linear functional on C[, ] given by α[u]= us das involving a Stieltjes integral with a signed measure, that is, A has bounded variation. They dealt with many boundary conditions given in the literature in a unified way by utilizing the fixed point index theory in cones. Recently, many researchers were interested in the global structure of positive solutions for the nonlinear boundary value problem see, e.g., [3, 6, 7]. In 29, Ma and An [3] considered the problem.. Assume that A A :[,] R is nondecreasing and At is not a constant on,, κ <with κ := tdat,and Gt, s dat for s [, ] for the definition of Gt, s,see 2.below. A w :, [, + is continuous and wt on any subinterval of [, ],and w L [, ] C,,where s=s s, s [, ]. A2 f C[, +, [, + and f s>for s >. A3 f = f =,wheref = lim u + f u u and f f u = lim u u. They obtained the following main result: Theorem. [3, Theorem.] Assume that A-A3 hold. Then there exists a component T in which joins, θ to,,and Proj R T =[ρ *,+ for some ρ * >. Moreover, there exists μ * ρ * >such that. has at least two positive solutions for μ μ *,.HereT joins, θ to, such that lim μ,u T, u μ u =, lim μ,u T, u > μ u =. Here, is the closure of the set of positive solutions of. on[, X, X = {u C [, ] : u =, u = us ds}, and the component of a set M is a maximal connected subset of M. A natural problem arises: How can we consider the global structure of positive solutions for the case f = f =? In this paper, we first obtain the global structure of positive solutions by the use of global continuous theorem, and some a priori estimates. Applying the analysis technique, we

3 HuBoundary Value Problems 22, 22:72 Page 3 of 3 construct the lower and upper solutions. Those combined with the fixed point index theory, the existence, multiplicity and nonexistence of positive solutions to. inthecase f = f = areinvestigated.finally,wediscusstheintervalofparameterμ such that the problem. has positive solutions. The proof of the method which is based on the construction of some bounds of the solution together with global continuous theorem and fixed point index is of independent interest, and is different from the other papers. This paper is arranged as follows. We will give some hypotheses and lemmas in Section 2.InSection3, new criteria of the existence, multiplicity and nonexistence of a positive solution are obtained. Moreover, an example is given to illustrate our result. 2 Preliminaries and lemmas Let X denote the Banach space C[, ] with the maximumnorm u = max ut. t [,] Define K := {u C[, ] : u is concave in [, ] and ut },thenk is a cone. Let ts, s t, Gt, s= st, t s. 2. Denote κ = Gs:= tdat, G t, s=gt, s+ Gt, s dat, for s [, ], t Gs, κ. κ 2.2 Throughout this paper, we suppose that the following conditions hold: H A :[,] R is nondecreasing, dat on, and κ := tdat with κ <. H w C,,, + and w may be singular at t =and/or, satisfying < G t, sws ds <, t,. 2.3 H2 f C[,,, f > obviously holds. f u H3 f = lim u u =+. Remark 2. It is easy to see from H that Gs [A A]Gs, s. Therefore, if we assume that < Gs, sws ds <,then2.3 obviously holds. Define an operator T : K X as follows: Tut =Tμ, ut:=μ G t, swsf us ds.

4 HuBoundary Value Problems 22, 22:72 Page of 3 Assume that the conditions H-H2 hold, then it is easy to verify that T : K K is well defined and completely continuous. Lemma 2. [9] Global continuation theorem Let X be a Banach space and let K be an order cone in X. Consider the equation u = Tμ, u, 2. where μ R + =[, and u X. Suppose that T : R + K K is completely continuous and T, u =θ for all u K, then L + K, the component of the solution set of 2. containing, is unbounded. Remark 2.2 We note that u is a positive solution of the problem.ifandonlyifu = Tμ, u on K. 2 If Tμ, θ θ for μ and T, u=θ for all u K, then we get from Lemma 2. that there exists an unbounded continuum L emanating from,θ in the closure of the set of positive solutions.inr + K. Lemma 2.2 [] Let X be a Banach space, K an order cone in X and S an open bounded set in X with θ S. SupposethatT: S K K is a completely continuous operator. If Ty λy, for all y S Kandallλ,theniT, S K, K=. 3 Main results Lemma 3. Let H-H3 hold and let J =[μ, with μ >. Then there exists a constant Q J >such that for all μ J and all possible positive solutions u μ of., the inequality u μ < Q J holds. Proof Suppose on the contrary that there exist a sequence {μ n } J and a sequence {u μn } of the positive solutions of. corresponding to μ n such that u μn, asn. Denote u n := u μn.fromtheconcavityofu n, it follows that u n t u n [ >, fort, 3 ]. 3. Choose η := π2 μ w,wherew = min t [, 3 ] wt. Then we find from H3 that there exists a constant R >suchthat f u>ηu, for all u > R. 3.2 Since lim n u n =,weget u N >R, for enough large N.

5 HuBoundary Value Problems 22, 22:72 Page 5 of 3 This, together with 3. and3.2, implies f u N t > ηu N t, for t, Put ψt:= cos 2πt,fort [, 3 ]. Hence, we know from 3.3that 3 μ N wtf u N t 3 ψt dt ημ N wtu N tψt dt 3 ηwμ N u N tψt dt. On the other hand, multiplying.byψ and integrating by parts, we obtain that leads to μ N 3 wtf u N t 3 ψt dt = u N tψt dt 3 = u N tψt + 3 = 3 u N tψ t dt u N tψ t dt + u N tψ t 3 π 2 u N tψt dt 3 i.e., 3 3 ηwμ N u N tψt dt π 2 u N tψt] dt, π 2 μ w = η π 2 μ N w < π 2 μ w. This is a contradiction. Lemma 3.2 Assumethat thehypotheses H-H3 hold. Then there exists ρ >such that if the problem. has a positive solution for parameter μ,thenμ ρ. Proof Let u be a positive solution of. corresponding to μ. From the hypotheses H2 and H3, it follows that there exists ϱ >suchthatf u ϱu for all u >.Consequently, we have μϱwtut μwtf ut = u t, t,. 3.

6 HuBoundary Value Problems 22, 22:72 Page 6 of 3 Let λ be the first eigenvalue of ϕ t+λwtϕt=, t,, ϕ = ϕ =, and let ϕ be the positive eigenfunction corresponding to λ see [8]. It is easy to see that ϕ. Multiplying 3.byϕ and integrating by parts, we get that μϱ implies wtutϕ t dt u tϕ t dt = ϕ tu t + u tϕ t dt = uϕ uϕ utϕ t dt = λ wtutϕ t dt utϕ t dt μ λ ϱ ρ. This completes the proof. Lemma 3.3 Assume that H-H3 hold. Then we have lim u μ =. μ,u μ L uμ,μ Proof Suppose this fails, that is, there exists {μ, u μ } L such that μ and u μ M, where M is a positive constant. Since μ, u μ L,wegetthatforallt [, ] u μ t =μ = μ μ M max t [,] G t, swsf u μ s ds ds u μ G t, sws f u μs u μ G t, sws ds u μ, where { } f uμ t M = max u μ : u μ M, t [, ],.

7 HuBoundary Value Problems 22, 22:72 Page 7 of 3 Hence, we find that { μ M max t [,] G t, sws ds} >. Thus, it implies that μ. This is contradiction. Theorem 3. Assume that the conditions H-H3 hold and dat <. Then there exists a constant μ >such that the problem. has at least two positive solutions for <μ < μ, and at least one positive solution for μ = μ, and no positive solution for μ > μ. Proof Define μ := sup { μ * > : for all μ, μ *, there exist at least two positive solutions of. }. 3.5 From Lemma 2. and Remark 2.2, we can find that there exists an unbounded continuum L emanating from, θ in the closure of the set of positive solutions in R + K and T, u= θ for all u K. Meanwhile, Lemma 3. and Lemma 3.3 respectively imply that u μ is bounded μ >,μ and unbounded μ >,μ and u μ. Therefore, we conclude that the set of 3.5 is nonempty. Those combined with Lemma 3.2 follows that μ >iswelldefinedandμ, ρ]. From the definition of μ, it is easy to see that the problem. has at least two positive solutions for μ, μ. Again, since the continuum is a compact connected set and T is a completely continuous operator, the problem. has at least one positive solution at μ = μ. Next, we only show that the problem. has no positive solution for any μ > μ.suppose on the contrary that there exists some μ 2 > μ such that the problem. hasa positive solution u 2 corresponding to μ 2. Then we will prove that the problem.hasat least two positive solutions for any μ [μ, μ 2 which contradicts the definition of 3.5. For the sake of obtaining the contradiction, we divide the proof into four steps. Step. Constructing a modified boundary value problem. Choose arbitrarily a constant μ [μ, μ 2. Since f is uniformly continuous on [, u 2 + ], there exists a constant σ >suchthat f u + σ <f u+ɛ, foru [, u 2 + ], 3.6 where ɛ := μ 2 μ min u [, u2 +] f u 2μ >. Denote ζ t:=u 2 t+σ,fort [, ]. Then we claim that ζ t+μwtf ζ t <, t,, 3.7 and ζ = σ, ζ = u 2 s das+σ. 3.8

8 HuBoundary Value Problems 22, 22:72 Page 8 of 3 Indeed, using 3.6, we have ζ t+μwtf ζ t = u 2 t+μwtf u 2 t+σ = μ 2 wtf u 2 t + μwtf u 2 t+σ < wt [ μ 2 f u 2 t + μf u 2 t + μɛ ] μ 2 μ wtf u 2 t 2 <. Define a set S := {u C[, ] <ut<ζ t, t, }. Then the set S is bounded and open in C[, ]. Now, we construct the modified second-order boundary value problem: u t+μwtf ξ ut =, t,, 3.9 u =, u = us das, where ξ : R R + is defined by ξ ut ζ t, if ut>ζ t, = ut, if ut ζ t,, if ut<. Step 2. We will show that if u is a positive solution of 3.9, then u S K. Let u be a positive solution of 3.9, then we claim that u S K. 3. Suppose this fails, that is, u / S K. Clearly, we only show that ut ζ t, for t, ]. Comparing the boundary conditions 3.8 and3.9, the only following three cases need to be considered. Case I. There exists t, such that ut =ζ t and<ut<ζ t, for t t σ, t t, t + σ andsomeσ > ; Case II. There exists t 2, ] such that ut 2 =ζ t 2 andut ζ t fort t 2, ]. Case III. There exists [t 3, t ], ] such that ut ζ t, t [t 3, t ], ut 3 =ζ t 3 andut =ζ t. See the three Figures, 2 and 3. Case I. From 3.7, it implies that there exists a constant ɛ >suchthat max { ζ t+μwtf ζ t } = ɛ <. 3. t [t σ,t +σ ] Figure Case I.

9 HuBoundary Value Problems 22, 22:72 Page 9 of 3 Figure 2 Case II. Figure 3 Case III. Since f is uniformly continuous on [, u 2 +],thereexistsaσ >suchthatifu, v [, ζ ]and u v < σ,thenweget f u f v < ɛ W, 3.2 where W := μ max t [t σ,t +σ ] wt >. From the assumption of Case I, it follows that there exists a subinterval [r, s] t σ, t + σ witht r, ssuchthat σ < ut ζ t, for t [r, s] and u ζ r> and u ζ s<. This together with 3.and3.2leadsto >u ζ s u ζ r= = = s r s r s r s [ = ε r s [ μwtf ut + ζ t ] dt r [ u t ζ t ] dt [ μwt f ut f ζ t + ζ t+μwtf ζ t ] dt [ μwtɛ W ɛ ] dt μwt W This is a contradiction. ] dt.

10 Hu Boundary Value Problems 22, 22:72 Page of 3 Case II. Let xt:=ut ζ t, for t [, ]. Then we have that for t t 2,] xt 2 =, xt, 3.3 x t= ut ζ t = μwtf ζ t ζ t>. 3. Obviously, we have x t 2. From 3.3, it implies that x t>,foranyt t 2,].Hence, the function xt isstrictlyincreasingin[t 2,]. From 3.8 and the boundary condition 3.9, we find x = u ζ = us das u 2 s das σ = us ζ s das+σ das = xs das σ das < x. We obtain a contradiction. In particular, if t 2 =,thenweobtain x = u ζ = xs das +σ das <. This contradicts 3.3. Case III. Since ξut = ζ t, for t t 3, t, we have that ut ζt >,fort t 3, t. Again since u ζ t 3 =u ζ t =, we know from the maximum principle that ut< ζ t, for t t 3, t. This contradicts the assumption of Case III. Therefore, we conclude that the claim 3.holds. Step 3. i T μ, S K, K = the definition of T μ see below. Using 2.2, we define an operator T μ : K X by T μ ut=μ G t, swsf ξ us ds, t [, ]. Then T μ : K K is completely continuous and u is a positive solution of 3.9ifandonly if u = T μ u onk. From the definition of ξut, it implies that there exists R >such that T μ u < R, for all u K. Consequently, we get from Lemma 2.2 that i T μ, B R K, K=, where B R = {u X : u < R }. Applying the conclusion of Step 2 and the excision property of fixed point index, we find that i T μ, S K, K=i T μ, B R K, K=. 3.5 Step. We conclude that the problem. has at least two positive solutions corresponding to μ.

11 Hu Boundary Value Problems 22, 22:72 Page of 3 Since the problem. is equivalent to the problem 3.9 ons K, we get that the problem. has a positive solution in S K. Without loss of generality, we may suppose that T has no fixed point on S K otherwise the proof is completed. Then it, S K, K is well defined and 3.5implies it, S K, K=. 3.6 On the other hand, from Lemma 3.2, wechooseμ * > ρ such that the problem. has no positive solution in K.Byaprioriestimatein J =[μ, μ * ], there exists R 2 > R >such that for all possible positive solutions u λ of.withλ [μ, μ * ], we know that u λ < R 2. Define G :[,] B R2 K K by Gν, u=t νμ + νμ *, u. Then it is easy to verify that G is completely continuous on [, ] K and Gν, u u for all ν, u [, ] B R2 K. From the property of homotopy invariance, it follows that it, B R2 K, K=i Tμ,, B R2 K, K = i Tμ *,, B R2 K, K =. Hence, by the additivity property and 3.6, we have i T,B R2 \S K, K =. Then we conclude that the problem. has at least two positive solutions corresponding to μ. Remark 3. i From the hypotheses H2 and H3, it implies that there exists L >such that D = f L L f u = min u> u. 3.7 Let f attain its maximum at the point L * of [, L].If G t, sws ds <,then adopting the similar method as in [, Theorem ], we get that for G t, sws ds, the problem. has at least two positive solutions <μ < f L* L u t and u 2 t such that < u < L < u 2 by the use of compression of conical shells in [2, Corollary 2.]. Consequently, we know that μ f L* L G t, sws ds. ii If u is a positive solution of the equation. corresponding to μ,thenwehave u = max ut = μ max t [,] i.e., from 3.7, =μ max t [,] t [,] G t, sws f us u G t, swsf us ds ds μd max t [,] Therefore, we get that μ * D max t [,] G t, sws ds. G t, sws ds.

12 Hu Boundary Value Problems 22, 22:72 Page 2 of 3 Corollary 3. Assume that H-H3 hold. Consider the following m-point boundary value problem u t+μwtf ut =, <t <, m 2 u =, u = α i uη i, i= 3.8 where μ is a positive parameter, α i [,, i=,2,...,m 2, <η < < η m 2 <and m 2 i= α i η i <. Then there exists a constant μ >such that the problem 3.8 hasat least two positive solutions for <μ < μ, and at least one positive solution for μ = μ, and no positive solution for μ > μ. Proof In the boundary condition of., if we let m 2 As= α i χs η i, i= where χs is the characteristic function on [,, i.e.,, if s, χs=, if s <. Then the boundary condition of. reduces to the m-point boundary condition of 3.8. Applying the method of Theorem 3.,wegettheconclusion. Example 3. We consider the following singular boundary value problem: u t+μ 3 2+sin ut+e ut =, t,, t u =, u = us das, 3.9 where As=s 2 2 s. Computing yields κ = Gs= tdat= = s Gt, s dat td 2t 2 t 3 = 7 2, st t 3t 2 dt + = 5 2 s 2 3 s3 + s, G t, s=gt, s+ t 7 5s 8s 3 +3s. s t t 3t 2 sdt

13 Hu Boundary Value Problems 22, 22:72 Page 3 of 3 We find that for any t, ], < G t, sws ds Gt, s = 3 ds + t s s 2 +3s 3 ds s 7 2,988 = t 2,695 9 t2 <. Thus, it implies that 2.3 holds. It is easy to verify that the conditions H2 and H3 hold. Therefore, by Theorem 3., we obtain that there exists a constant μ > such that the problem 3.9 has at least two positive solutions for < μ < μ, and at least one positive solution for μ = μ, and no positive solution for μ > μ. Competing interests The author declares that they have no competing interests. Author s contributions The author typed, read and approved the final manuscript. Acknowledgement The author would like to thank the anonymous referees very much for helpful comments and suggestions which led to the improvement of presentation and quality of the work. The work was supported partly by NSCF of Tianyuan Youth Foundation No. 2625, K. C. Wong Magna Fund of Ningbo University, Subject Foundation of Ningbo University No. xkl and Hulan s Excellent Doctor Foundation of Ningbo University. Received: 2 December 2 Accepted: June 22 Published: 5 July 22 References. Webb, JRL, Infante, G: Positive solutions of nonlocal boundary value problems: a unified approach. J. Lond. Math. Soc. 7, Webb, JRL, Infante, G: Positive solutions of nonlocal boundary value problems involving integral conditions. Nonlinear Differ. Equ. Appl. 5, Ma, RY, An, YL: Global structure of positive solutions for nonlocal boundary value problems involving integral conditions. Nonlinear Anal. TMA 7, Anderson, DR, Smyrlis, G: Solvability for a third-order three-point boundary value problem on time scales. Math. Comput. Model. 9, Lan, KQ, Webb, JRL: Positive solutions of semilinear differential equations with singularities. J. Differ. Equ. 8, Li, HY, Sun, JX: Positive solutions of superlinear semipositone nonlinear boundary value problems. Comput. Math. Appl. 6, Rynne, B: Spectral properties and nodal solutions for second-order, m-point, boundary value problems. Nonlinear Anal. TMA 67, Asakawa, H: On nonresonant singular two-point boundary value problems. Nonlinear Anal. 7, Zeidler, E: Nonlinear Functional Analysis and Its Applications, I. Fixed-Point Theorems. Springer, New York 986. Guo, DJ, Lakshmikantham, V: Nonlinear Problems in Abstract Cones. Academic Press, New York 988. Sánchez, J: Multiple positive solutions of singular eigenvalue type problems involving the one-dimensional p-laplacian. J. Math. Anal. Appl. 292, Deimling, K: Nonlinear Functional Analysis. Springer, Berlin 985 doi:.86/ Cite this article as: Hu: Positive solutions for singular boundary value problems involving integral conditions. Boundary Value Problems 22 22:72.