Hindawi Publishing Corporation Boundary Value Problems Volume 2009, Article ID 970135, 20 pages doi:10.1155/2009/970135 Research Article Multiplicity Results Using Bifurcation Techniques for a Class of Fourth-Order m-Point Boundary Value Problems Yansheng Liu1 and Donal O’Regan2 Department of Mathematics, Shandong Normal University, Jinan 250014, China Department of Mathematics, National University of Ireland, Galway, Ireland Correspondence should be addressed to Yansheng Liu, yanshliu@gmail.com Received 13 March 2009; Accepted 12 April 2009 Recommended by Juan J Nieto By using bifurcation techniques, this paper investigates the existence of nodal solutions for a class of fourth-order m-point boundary value problems Our results improve those in the literature Copyright q 2009 Y Liu and D O’Regan This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited Introduction Consider the following fourth order m-point boundary value problem BVP, for short u4 t t ∈ 0, f u t ,u t , m−2 u 0, αi u ηi u i 1.1 m−2 u 0, αi u ηi , u i where f : R × R → R is a given sign-changing continuous function, m ≥ 3, ηi ∈ 0, , and αi > for i 1, , m − with m−2 αi < i 1.2 Boundary Value Problems Multi-point boundary value problems for ordinary differential equations arise in different areas of applied mathematics and physics The existence of solutions of the second order multi-point boundary value problems has been studied by many authors and the methods used are the nonlinear alternative of Leray-Schauder, coincidence degree theory, fixed point theorems in cones and global bifurcation techniques see 1–9 , and the references therein In , Ma investigated the existence and multiplicity of nodal solutions for u t f u t u 0, t ∈ 0, ; 0, 1.3 m−2 αi u ηi u i when ηi ∈ Q i 1, 2, , m − with < η1 < η2 < · · · < ηm−2 < 1, 1.4 and αi > for i 1, , m − satisfying 1.2 He obtained some results on the spectrum of the linear operator corresponding to 1.1 It should be pointed out that the main tool used in is results on bifurcation coming from the trivial solutions and we note no use was made of global results on bifurcation from infinity Recently 10 Wei and Pang studied the existence and multiplicity of nontrivial solutions for the fourth order m-point boundary value problems: u4 t t ∈ 0, f u t ,u t , m−2 u 0, u αi u ηi i 1.5 m−2 u 0, αi u ηi , u i where f : R × R → R is a given sign-changing continuous function, m ≥ 3, ηi ∈ 0, , and αi > for i 1, , m − satisfies 1.2 Motivated by 5, 10 , in this paper we consider the existence and multiplicity of nodal solutions for BVP 1.1 The method used here is Rabinowitz’s global bifurcation theorem To the best of our best knowledge, only 10 seems to have considered the existence of nontrivial or positive solutions of the nonlinear multi-point boundary value problems for fourth order differential equations As in 5, 10 we suppose 1.2 is satisfied throughout The paper is organized as follows Section gives some preliminaries Section is devoted to the existence of multiple solutions for BVP 1.1 To conclude this section we give some notation and state three lemmas, which will be used in Section Following the notation of Rabinowitz, let E be a real Banach space and L : E → E be a compact linear map If there exists μ ∈ R 0, ∞ and / v ∈ E such that v μLv, μ is said to be a real characteristic Boundary Value Problems value of L The set of real characteristic values of L will be denoted by σ L The multiplicity of μ ∈ σ L is ∞ dim N I − μL j , 1.6 j where N A denotes the null space of A Suppose that H : R × E → E is compact and H λ, u o u at u uniformly on bounded λ intervals Then u λLu H λ, u 1.7 possesses the line of trivial solutions Θ { λ, | λ ∈ R} It is well known that if μ ∈ R, a necessary condition for μ, to be a bifurcation point of 1.7 with respect to Θ is that μ ∈ σ L If μ is a simple characteristic value of L, let v denote the eigenvector of L corresponding to μ normalized so v By Σ we denote the closure of the set of nontrivial solutions of 1.7 A component of Σ is a maximal closed connected subset It was shown in Rabinowitz 11, Theorems 1.3, 1.25, 1.27 , the following Lemma 1.1 If μ ∈ σ L is simple, then Σ contains a component Cμ which can be decomposed into − two subcontinua Cμ , Cμ such that for some neighborhood B of μ, , − λ, u ∈ Cμ Cμ ∩ B, λ, u / μ, implies λ, u λ, αv w where α > α < and |λ − μ| − Moreover, each of Cμ , Cμ either i meets infinity in Σ, or ii meets μ, where μ / μ ∈ σ L , or iii contains a pair of points λ, u , λ, −u , u / o1, w 1.8 o |α| at α The following are global results for 1.7 on bifurcation from infinity see, Rabinowitz 9, Theorem 1.6 and Corollary 1.8 Lemma 1.2 Suppose L is compact and linear, H λ, u is continuous on R × E, H λ, u o u at u ∞ uniformly on bounded λ intervals, and u H λ, u/ u is compact If μ ∈ σ L is of odd multiplicity, then Σ possesses an unbounded component Dμ which meets μ, ∞ Moreover if Λ ⊂ R is an interval such that Λ ∩ σ L {μ} and ℘ is a neighborhood of μ, ∞ whose projection on R lies in Λ and whose projection on E is bounded away from 0, then either i Dμ \ ℘ is bounded in R × E in which case Dμ \ ℘ meets Θ { λ, | λ ∈ R} or ii Dμ \ ℘ is unbounded If (ii) occurs and Dμ \ ℘ has a bounded projection on R, then Dμ \ ℘ meets μ, ∞ where μ/μ ∈ σ L Lemma 1.3 Suppose the assumptions of Lemma 1.2 hold If μ ∈ σ L is simple, then Dμ can be − decomposed into two subcontinua Dμ , Dμ and there exists a neighborhood I ⊂ ℘ of μ, ∞ such that − λ, αv w where α > α < and λ, u ∈ Dμ Dμ ∩ I and λ, u / μ, ∞ implies λ, u |λ − μ| o , w o |α| at |α| ∞ 4 Boundary Value Problems Preliminaries C 0, with the norm u maxt∈ 0,1 |u t |, Y {u ∈ C1 0, : u 0, u αi u ηi } with the norm u max{ u , u }, Z {u ∈ C 0, : u 0, u αi u ηi } with the norm u max{ u , u , u } Then X, Y , and Z are Banach spaces 0, then t0 is a simple zero of u if u t0 / For any For any C1 function u, if u t0 ν integer k ∈ N and any ν ∈ {±}, as in , define sets Tk ⊂ Z consisting of the set of functions u ∈ Z satisfying the following conditions: Let X m−2 i m−2 i i u 0, νu > and u / 0; ii u has only simple zeros in 0, , and has exactly k − such zeros; iii u has a zero strictly between each two consecutive zeros of u − − − Note Tk −Tk and let Tk Tk ∪ Tk It is easy to see that the sets Tk and Tk are disjoint ν and open in Z Moreover, if u ∈ Tk , then u has at least k − zeros in 0, , and at most k − zeros in 0, Let E R×Y under the product topology As in 12 , we add the points { λ, ∞ : λ ∈ R} − to the space E Let Φk R × Tk , Φ− R × Tk , and Φk R × Tk k We first convert BVP 1.1 into another form Suppose u t is a solution of BVP 1.1 Let v t −u t Notice that u t v t t ∈ I; 0, 2.1 m−2 u 0, αi u ηi u i Thus u t can be written as ut Lv t , 2.2 where the operator L is defined by H t, s v s ds, Lv t : ∀v ∈ Y, 2.3 where H t, s G t, s G t, s m−2 i 1− αi G ηi , s m−2 i αi ηi ⎧ ⎨1 − t, ≤ s ≤ t ≤ 1; ⎩1 − s, ≤ t ≤ s ≤ , 2.4 Boundary Value Problems Therefore we obtain the following equivalent form of 1.1 v t f Lv t , −v t 0, t ∈ 0, ; 2.5 m−2 v 0, αi v ηi v i For the rest of this paper we always suppose that the initial value problem v t f Lv t , −v t v t0 v t0 0, t ∈ 0, ; 2.6 has the unique trivial solution v ≡ on 0, for any t0 ∈ 0, ; in fact some suitable conditions such as a Lipschitz assumption or f ∈ C1 guarantee this Define two operators on Y by Av t : Fv t : f Lv t , −v t , LFv t , t ∈ I, v ∈ Y 2.7 Then it is easy to see the following lemma holds Lemma 2.1 The linear operator L and operator A are both completely continuous from Y to Y and Lv ≤ M v ≤ M v 1, ∀v ∈ Y, 2.8 m−2 m−2 where M max{1, 1/8 i αi / − i αi ηi } Moreover, u ∈ C 0, is a solution of BVP 1.1 if and only if v operator equation v Av −u is a solution of the Let the function Γ s be defined by Γs cos s − m−2 αi cos ηi s, s ∈ R 2.9 i Then we have the following lemma Lemma 2.2 i For each k ≥ 1, Γ s has exactly one zero sk ∈ Ik : s1 < s2 < · · · < sk −→ ∞ k − π, kπ , so k −→ ∞ ; 2.10 ii the characteristic value of L is exactly given by μk s2 , k 1, 2, , and the eigenfunction k cos sk t; φk corresponding to μk is φk t iii the algebraic multiplicity of each characteristic value μk of L is 1; iv φk ∈ Tk for k 1, 2, 3, , and φ1 is strictly positive on 0, Boundary Value Problems Proof From and by a similar analysis as in the proof of 6, Lemma 3.3 we obtain i and ii Now we assert iii holds Suppose, on the contrary, there exists y ∈ Y such that I − μk L y μ−1 φk Then y ∈ Z and k −y − s2 y k From y cos sk t 2.11 we know the general solution of this differential equation is y t sin sk t 2sk C cos sk t − 2.12 From i and ii of this lemma, C cos sk t satisfies the boundary condition Thus m−2 cos sk m−2 αi cos ηi sk , sin sk αi ηi sin ηi sk i 2.13 i Then, by 1.2 , m−2 αi cos ηi sk i ≤ m−2 αi ηi sin ηi sk i m−2 sin ηi sk sin ηj sk αi αj cos ηi sk cos ηj sk 2.14 i,j ≤ m−2 αi < 1, i a contradiction Thus the algebraic multiplicity of each characteristic value μk of L is Finally, from sk ∈ k − π, kπ and s1 ∈ 0, π/2 , it is easy to see that iv holds Lemma 2.3 For d d1 , d2 ∈ R × R \ { 0, }, define a linear operator Ld v t d1 L d2 L v t , ∀t ∈ I, v ∈ Y, 2.15 where L is defined as in 2.3 Then the generalized eigenvalues of Ld are simple and are given by < λ1 Ld < λ2 Ld < · · · < λk Ld −→ ∞ k −→ ∞ , 2.16 where λk Ld μ2 k d1 d2 μ k 2.17 Boundary Value Problems The generalized eigenfunction corresponding to λk Ld is φk t cos sk t, 2.18 where μk , sk , φk are as in Lemma 2.2 Proof Suppose there exist λ and v / such that v 2.7 and 2.15 it is easy to see that u / and λLd v Set u t t ∈ 0, ; λ d1 u t − d2 u t , u4 t Lv t Then from 2.2 – m−2 u 0, αi u ηi ; u 2.19 i u”’ m−2 0, αi u ηi u i Denote L−1 u −u for u ∈ Z Then there exist two complex numbers r1 and r2 such that L−1 − r2 I u t − λ d1 u t − d2 u t Now if there exists some ri i 2.20 1, such that L−1 − ri I u t then by Lemma 2.2 we know ri L−1 − r1 I u t s2 k 0, 2.21 μk for some k ∈ N, and consequently u t cos sk t 2.22 is a nontrivial solution Substituting 2.22 into 2.19 , we have λ μ2 k d1 d2 μk 2.23 On the other hand, suppose, for example, L−1 − r1 I u t / 0, L−1 − r2 I L−1 − r1 I u t 2.24 Boundary Value Problems Reasoning as previously Let w t : L−1 − r1 I u t Then L−1 − r2 I w t s2 for some k ∈ N, and consequently w t a cos sk t a / is a mentioned, we have r2 k nontrivial solution Therefore, L−1 − r1 I u t If r1 0, is a cos sk t 2.25 s2 , then the general solution of the differential equation 2.25 , satisfying u k C cos sk t − u t a t sin sk t, 2sk 2.26 which is similar to 2.12 Reasoning as in the proof of Lemma 2.2 we can get a contradiction Thus r1 / s2 and the general solution of 2.25 , satisfying u 0, is k u t u t a cos sk t , s2 − r1 k 2.27 where u t is the general solution of homogeneous differential equation corresponding to 2.25 L−1 − r1 I u t t 2.28 Notice the term a cos sk t/ s2 − r1 in 2.27 satisfies the boundary condition of 1.1 at k 1, so u t also satisfies m−2 u 0, αi u ηi u 2.29 i C cos sj t for some j ∈ N, and consequently Therefore, by Lemma 2.2 we know u t r1 s2 / s2 , j k ut C cos sj t a cos sk t s2 − s2 j k 2.30 By substituting this into 2.19 , we have aλ d1 d2 μ k aμ2 , k Cλ d1 d2 μ j Cμ2 j 2.31 Since μj / μk , if there exists some λ such that 2.31 holds, then d1 d1 d2 μ k d2 μj μ2 k μ2 j , 2.32 Boundary Value Problems which implies d1 d2 / 0, μk d1 μj −d2 , 2.33 a contradiction with d1 > and d2 > Consequently, 2.24 does not hold This together with 2.20 – 2.23 and Lemma 2.2 guarantee that the generalized eigenvalues of Ld are given by < λ1 Ld < λ2 Ld < · · · < λk Ld −→ ∞ k −→ ∞ , 2.34 μ2 / d1 d2 μk The generalized eigenfunction corresponding to λk Ld is where λk Ld k cos sk t φk t Now we are in a position to show the generalized eigenvalues of Ld are simple Clearly, from above we know for λk : λk Ld , I−λk Ld φk and dimN I−λk Ld Suppose there exists an v ∈ C2 such that φk t μk I − λk L d v 2.35 This together with 2.3 and 2.15 guarantee that v ∈ Y If we let u t then we have u t − λk d1 u t − d2 u t m−2 u 0, t ∈ 0, , 2.36 m−2 u”’ αi u ηi ; u cos sk t, Lv t as above, 0, αi u ηi u i 2.37 i Consider the following homogeneous equation corresponding to 2.36 : u4 t − μ2 k d1 d2 μ k d1 u t − d2 u t 2.38 The characteristic equation associated with 2.38 is λ4 − μ2 k d1 d2 μ k d1 − d2 λ2 2.39 Then there exists a real number η such that λ2 Notice that −ημk μk −d1 μ2 / d1 k λ2 − η λ4 − μ2 k d1 d2 μ k d1 − d2 λ2 d2 μk < if d1 > So η > if d1 > 0, and η 2.40 if d1 10 Boundary Value Problems First we consider the case d1 > In this case the general solution of 2.38 is c1 e √ ηt c2 e − √ ηt c3 cos sk t c4 sin sk t 2.41 After computation we obtain that the general solution of 2.36 is u t c1 e √ ηt c2 e − √ ηt where a − d1 d2 μk /2sk 2d1 μk 2.37 it follows that c3 cos sk t c4 sin sk t d2 μ2 From boundary condition u k η c1 − c2 sk c4 By η > and μk > 0, we know c1 − c2 c1 e and c4 √ ηt e− √ ηt 2.42 u”’ 0 in 0; 2.43 η η c − c − s3 c k ut at sin sk t, 0 Then 2.42 can be rewritten as c3 cos sk t at sin sk t 2.44 Notice that the term c3 cos sk t satisfies 2.37 From the boundary condition m−2 m−2 αi u ηi , u1 αi u ηi , u i 2.45 i 2sk cos sk t − s2 t sin sk t k t sin sk t we have c1 e √ η e− m−2 √ η αi c1 e a sin sk √ ηηi e− √ ηηi aηi sin sk ηi , 2.46 − aηi s2 sin sk ηi k 2.47 i c1 η e √ η e− √ η m−2 − as2 sin sk k αi c1 η e √ ηηi e− √ ηηi i Multiply 2.46 by s2 and then add to 2.47 to obtain k c1 η s2 k e √ η e− √ η c1 η s2 k m−2 αi e √ ηηi e− √ ηηi 2.48 i On the other hand, from 1.2 it can be seen that e √ η e− √ η m−2 > αi e i √ ηηi e− √ ηηi 2.49 Boundary Value Problems 11 This together with 2.48 guarantee that c1 Therefore, 2.42 reduces to c3 cos sk t ut at sin sk t 2.50 Similar to 2.12 , a contradiction can be derived Next consider the case d1 Then η from above In this case the general solution of 2.38 is c1 c2 t c3 cos sk t c4 sin sk t 2.51 By a similar process, one can easily get a contradiction To sum up, the generalized eigenvalues of Ld are simple, and the proof of this lemma is complete Main Results We now list the following hypotheses for convenience H1 There exists a a1 , a2 ∈ R × R \ { 0, } such that a1 x − a2 y f x, y o x, y , as x, y −→ 0, 3.1 −→ ∞ 3.2 where x, y ∈ R × R, and | x, y | : max{|x|, |y|} H2 There exists b b1 , b2 ∈ R × R \ { 0, } such that b1 x − b2 y f x, y o x, y , as x, y H3 There exists R > such that f x, y < R , M for x, y ∈ x, y : |x| ≤ MR, y ≤ R , 3.3 where M is defined as in Lemma 2.1 H4 There exist two constants r1 < < r2 such that f x, −r1 ≥ and f x, −r2 ≤ for x ∈ −Mr, Mr , and f x, −y satisfies a Lipschitz condition in y for x, y ∈ −Mr, Mr × r1 , r2 , where r max{|r1 |, r2 } Now we are ready to give our main results Theorem 3.1 Suppose (H1)-(H2) hold Suppose there exists two integers i0 ≥ and k > such that either μ20 i a1 k a2 μi0 k α < and |λ − λk | o , w o |α| at α ν By 3.7 and the continuity of the operator A : Y → Z, the set Ck lies in R × Z and the ν ν injection Ck → R × Z is continuous Thus, Ck is also a continuum in R × Z, and the above properties hold in R × Z Since Tk is open in Z and φk ∈ Tk , we know v α w ∈ Tk α φk 3.11 for / α sufficiently small Then there exists ε0 > such that for ε ∈ 0, ε0 , we have Ck { λk , } λ, v ∈ Φk , ∩ Bε ⊂ Φk , 3.12 where Bε is an open ball in R × Z of radius ε centered at λk , It follows from the proof of 6, Proposition 4.1 that λ, v ∈ Ck ∩ R × ∂Tk ⇒u 0, 3.13 which means Ck \ { λk , } ∩ ∂Φk ∅ Consequently, Ck lies in Φk ∪ { λk , } ν or − Similarly we have that Ck lies in Φν ∪ { λk , } ν k Next we show alternative ii of Lemma 1.1 is impossible If not, without loss of generality, assume that Ck meets λi , with λk / λi ∈ σ La Then there exists a sequence ξm , zm ∈ Ck with ξm → λi and zm → as m → ∞ Notice that zm ξm La zm H ξm , zm 3.14 o zm as m → Dividing this equation by zm and using Lemma 2.1 and H ξm , zm ∞, we may assume without loss of generality that zm / zm → z as m → ∞ Thus from 3.14 it follows that z λi La z 3.15 Since z / 0, by Lemmas 2.2 and 2.3, z belongs to Ti or Ti− By 3.1λ and the continuity of the operator A : Y → Z, from zm − z → it follows that zm − z → Notice that Ti and 14 Boundary Value Problems Ti− are open in Z Therefore, zm ∈ Ti or Ti− for m sufficiently large, which is a contradiction with zm ∈ Tk m ≥ , i / k Hence alternative ii of Lemma 1.1 is not possible Finally it remains to show alternative iii of Lemma 1.1 is impossible In fact, notice − − − ν −Tk , and Tk ∩ Tk ∅ If u ∈ Tk , then −u ∈ Tk This guarantees that Ck does not that Tk contain a pair of points λ, v , λ, −v , v / Therefore alternative i of Lemma 1.1 holds This implies that for each integer k ∈ N ν and each ν , or −, there exists a continua Ck of solutions of 3.1λ in Φν ∪ { λk , }, which k meets { λk , } and ∞ in Σ Under the condition H2 , 3.1λ can be rewritten as v λLb v Hb λ, v , 3.16 λAv − λLb v, Lb is defined as in 2.12 replacing d with b here Hb λ, v Let h x, y : f x, y − b1 x b2 y Then from H2 it follows that lim| x,y | x, y | Define a function h r : max h x, y : x, y ≤r | → ∞h x, y / 3.17 Then h r is nondecreasing and lim r →∞ h r r 3.18 Obviously, by 3.18 and Lemma 2.1, it can be seen that Hb λ, v is o v for v near ∞ uniformly on bounded λ intervals and Lb is a compact linear map on Y Similar to 3.8 , by Lemma 2.3, Lb possesses an increasing sequence of simple eigenvalues < λ1 < λ2 < · · · < λk < · · · , Note φk with λk μ2 k b1 b2 μk as k −→ ∞ 3.19 cos sk t is an eigenfunction corresponding to λk Obviously, it is in Tk Lemma 3.5 Assume that (H1)-(H2) holds Then for each integer k > and each ν , or −, there ν ν exists a continua Dk of Σ in Φk ∪ { λk , ∞ } coming from { λk , ∞ }, which meets λk , or has an unbounded projection on R Proof From 2.7 , 3.16 , and 3.18 it follows that Hb λ, v is continuous on E, Hb λ, v o v at v ∞ uniformly on bounded λ intervals Moreover, as in the proof of 12, Theorem 2.4 , one can see that v Hb λ, v/ v is compact From Lemma 2.3 we know λk is a simple 1 characteristic value of Lb for each integer k ∈ N Thus by Lemmas 1.2 and 1.3, Σ contains a − component Dk which can be decomposed into two subcontinua Dk , Dk which meet { λk , ∞ } − Now we show that for a smaller neighborhood I ⊂ ℘ of λk , ∞ , λ, v ∈ Dk Dk ∩ I − and λ, v / λk , ∞ imply that v ∈ Tk Tk In fact, by Lemma 1.3 we already know that there Boundary Value Problems 15 − exists a neighborhood I ⊂ ℘ of λk , ∞ satisfying λ, v ∈ Dk Dk ∩ I and λ, v / λk , ∞ imply λ, v λ, αvk w where α > α < and |λ − λk | o , w o |α| at |α| ∞ ν As in the proof of Lemma 3.4, Dk is also a continuum in R×Z, and the above properties ν ∞, hold in R × Z Since Tk is open in Z and w/α is smaller compared to φk ∈ Tk near α φk w/α and therefore v αφk w ∈ Tk for α near ∞ Here and in the following the same argument works if is replaced by − Therefore, Dk ∩I ⊂ R×Tk ∪ λk , ∞ Now we have two cases to consider, that is, Dk \I is bounded or unbounded First suppose Dk \ I is bounded Then there exists λ, v ∈ ∂Dk with v ∈ ∂Tk If v / 0, by Lemma 3.3 we know v ∈ Tjν for some positive integer j / k and ν ∈ { , −} As in the proof of Lemma 3.4, we get a contradiction, which means v Thus there exists a sequence ξm , zm ∈ Dk with zm → v ≡ as m → ∞ This together with H1 guarantee that ξm , zm satisfies 3.14 As in the proof of Lemma 3.4, we may assume without loss of generality that zm / zm → z and ξm → ξ as m → ∞ Then we have z ξLa z 3.20 Since z / 0, ξ / is an eigenvalue of operator La From this, 3.9 , and 3.19 it follows that ξ λj for some positive integer j Then by Lemma 2.3, z belongs to Tj or Tj− Notice that zm − z → and so zm − z → as in the proof of Lemma 3.4 Thus zm ∈ Tj or Tj− for m sufficiently large since Tj and Tj− are open This together with zm ∈ Tk m ≥ guarantee that k j This means Dk meets λk , if Dk \ I is bounded Next suppose Dk \ I is unbounded In this case we show Dk \ I has an unbounded projection on R If not, then there exists a sequence ζm , ym ∈ Dk \ I with ζm → ζ and ym → ∞ as m → ∞ Let xm : ym / ym , m ≥ From the fact that ym ζm Lb ym xm ζm Lb xm Hb ζm , ym 3.21 it follows that Hb ζm , ym ym 3.22 Notice that Lb : Y → Y is completely continuous We may assume that there exists w ∈ Y with w 1 such that xm − w → as m → ∞ Letting m → ∞ in 3.22 and noticing Hb ζm , ym / ym → as m → ∞ one obtains w ζLb w 3.23 Since w / 0, ζ / is an eigenvalue of operator Lb , that is, ζ λk0 for some positive integer − k0 / k Then by Lemma 2.3 w belongs to Tk0 or Tk0 Notice the fact that xm − w → and − so xm − w → as in the proof of Lemma 3.4 Thus xm ∈ Tk0 or Tk0 for m sufficiently large − since Tk0 and Tk0 are open This is a contradiction with xm ∈ Tk m ≥ Thus Dk \ I has an unbounded projection on R 16 Boundary Value Problems Proof of Theorem 3.1 Suppose first that μ20 i a1 k a2 μi0 k Thus 3.27 is not true, which means 3.26 holds Next suppose that μ20 i b1 k b2 μi0 k < · · · < λi0 k Boundary Value Problems 17 guarantees that 3.30 is satisfied On the other hand, if Diν0 j has an unbounded projection on R, notice that λ, v 0, is the unique solution of 3.1λ in which λ in E, so 3.30 also holds Proof of Theorem 3.2 First suppose that H3 holds Then there exists ε > such that < ε f x, y R , M x, y : |x| ≤ MR, y ≤ R for x, y ∈ Let λ, v be a solution of 3.1λ such that ≤ λ < 3.1λ , 3.31 and Lemma 2.1 it is easy to see v λ Av λ LFv ε and v M max λf Lv t , −v t ≤ λM Fv t∈ 0,1 3.31 ≤ R Then by 2.7 , r2 , t ∈ 0, 3.38 v t < r1 , t ∈ 0, 3.39 t∈ 0,1 or t∈ 0,1 In fact, like in 13 , suppose on the contrary that there exists λ, v ∈ Ciν ν Dj such that either max{v t : t ∈ 0, } r2 3.40 min{v t : t ∈ 0, } r1 3.41 or for some i, j First consider the case max{v t : t ∈ 0, } r2 Let v t τ0 : inf t ∈ 0, t : v s ≥ r2 Then there exists t ∈ 0, such that for s ∈ t, t , 3.42 τ1 : sup t ∈ t, : v s ≥ for s ∈ t, t Then max{v t : t ∈ τ0 , τ1 } ≤ v t ≤ r2 , r2 , 3.43 t ∈ τ0 , τ1 3.44 Therefore, v t is a solution of the following equation −v t λf Lv t , −v t , t ∈ τ0 , τ1 3.45 v τ1 if < τ0 < τ1 < and v τ0 if τ0 with v τ0 My is strictly increasing in y for By H4 , there exists M ≥ such that f x, −y x, y ∈ r1 , r2 × −Mr, Mr , where r max{|r1 |, r2 } Then −v λMv λ f Lv t , −v t Mv , t ∈ τ0 , τ1 3.46 Boundary Value Problems 19 Using H4 and Lemma 2.1 again, we can obtain − r2 − v t λM r2 − v t −λ f Lv t , −v t Mv t − Mr2 3.47 −λ f Lv t , −v t ≥ 0, and if τ1 Mv t − f Lv t , −r2 Mr2 − λf Lv t , −r2 t ∈ τ0 , τ1 1, by 1.2 we know v < r2 Therefore, r2 − v τ0 > 0, r2 − v τ1 > if < τ0 < τ1 < 1; r2 − v τ0 if τ0 r2 − v τ1 > if τ1 0; 3.48 This together with 3.47 and the maximum principle 14 imply that r2 − v t > in τ0 , τ1 , which contradicts 3.43 The proof in the case min{v t : t ∈ 0, } r1 is similar, so we omit it Acknowledgments The Project Supported by NNSF of P R China 10871120 , the Key Project of Chinese Ministry of Education No: 209072 , and the Science & Technology 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in different areas of applied mathematics and physics The existence of solutions of the