SOLUTIONS TO SECOND ORDER NON-HOMOGENEOUS MULTI-POINT BVPS USING A FIXED-POINT THEOREM

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1 Electronc Journal of Dfferental Equaton, Vol. 88, No. 96, pp. 5. ISSN: URL: or ftp ejde.math.txtate.edu logn: ftp SOLUTIONS TO SECOND ORDER NON-HOMOGENEOUS MULTI-POINT BVPS USING A FIXED-POINT THEOREM YUJI LIU Abtract. In th artcle, we tudy fve non-homogeneou mult-pont boundary-value problem BVP of econd order dfferental equaton wth the onedmenonal p-laplacan. Thee problem have a common equaton n dfferent functon doman and dfferent boundary condton. We fnd condton that guarantee the extence of at leat three potve oluton. The reult obtaned generalze everal known one and are llutrated by example. It alo hown that the approach for gettng three potve oluton by ung mult-fxed-pont theorem can be extended to nonhomogeneou BVP. The empha on the nonhomogeneou boundary condton and the nonlnear term nvolvng frt order dervatve of the unknown. Some open problem are alo propoed.. Introducton Mult-pont boundary-value problem BVP for econd order dfferental equaton wthout p-laplacan have receved a wde attenton becaue of ther potental applcaton and BVP are facnatng and challengng feld of tudy, one may ee the textbook by Ge [4]. There are four clae of uch BVP ncludng ther pecal cae tuded n known paper: x t ft, xt, x t =, t,, x = α xξ, x = β xξ, x t ft, xt, x t =, t,, x = α x ξ, x = β xξ,.. Mathematc Subject Clafcaton. 34B, 34B5, 35B. Key word and phrae. Second order; p-laplacan; potve oluton; mxed type mult-pont boundary-value problem. c 8 Texa State Unverty - San Marco. Submtted March 3, 8. Publhed July 5, 8. Supported by grant 6JJ58 from the Natural Scence Foundaton of Hunan provnce and grant from the Natural Scence Foundaton of Guangdong provnce, Chna.

2 Y. LIU EJDE-8/96 and x t ft, xt, x t =, t,, x = α xξ, x = β x ξ, x t ft, xt, x t =, t,, x = α x ξ, x = β x ξ,.3.4 where < ξ < < ξ m <, α, β R and f a contnuou functon. The man method to get oluton or multple potve oluton of thee BVP are a follow: fxed pont ndex theory [7, 8, 63, 64], or fxed pont theorem n cone n Banach pace [5, 7, 5, 35, 4, 4, 46], uch a the Guo-Kranoelk fxed-pont theorem [3, 44, 48, 5, 53, 57]; Leggett-Wllam theorem [33, 36, 5], the fvefunctonal fxed pont theorem []; the fxed pont theorem of Avery and Peteron [6], etc. Mawhn contnuaton theorem [, 9, 7, 8, 33, 34, 38, 39, 4, 4]; the hootng method [9, 5]; v upper and lower oluton method and monotone teratve technque [, 4, 67]; upper and lower oluton method and Leray-Schauder degree theory [3, 4, 5, 47], or the approach of a combnaton of nonlnear alternatve of Leray-Schauder wth the properte of the aocated vector feld at the x, x plane []; v the crtcal pont theory and varatonal method [3]; v topologcal degree theory [3, 4]; the Schauder fxed pont theorem n utable Banach pace [36, 4, 4, 43, 49]. In all the above paper, the boundary condton BC are homogeneou. However, n many applcaton, BVP cont of dfferental equaton coupled wth nonhomogeneou BC, for example y = λ y, ya = aα, t a, b, yb = β and y = y t yt α, ya = α, yb = β t a, b, whch are very well known and were propoed n 69 and 696, repectvely. In 964, the BVP tuded by Zhdkov and Shrkov [66] and by Lee [3] are alo nonhomogeneou. There are alo everal paper concerned wth the extence of potve oluton of BVP for dfferental equaton wth non-homogeneou BC. Ma Ruyun [45] tuded extence of potve oluton of the followng BVP contng of econd order dfferental equaton and three-pont BC x t atfxt =, t,, x =, x αxη = b,.5

3 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 3 Kwong and Kong [9] tuded the BVP y t = ft, yt, < t <, n θy co θy =, m y α yξ = b,.6 where ξ,, α, θ [, 3π/4], f a nonnegatve and contnuou functon. Under ome aumpton, t wa proved that there ext a contant b > uch that:.6 ha at leat two potve oluton f b, b ;.6 ha at leat one oluton f b = or b = b ;.6 ha no potve oluton f b > b. Palamde [[5]], under uperlnear and/or ublnear growth rate n f, proved the extence of potve oluton and monotone n ome cae of the boundary-value problem y t = ft, yt, y t, t, αy βy =, m y α yξ = b,.7 where α >, β >, the functon f contnuou, and ft, y, y, for all t [, ], y, y R. The approach baed on an analy of the correpondng vector feld on the face-plane and on Kneer property for the oluton funnel. Sun, Chen, Zhang and Wang [53] tuded the extence of potve oluton for the three-pont boundary-value problem u t atfut =, t, u =, m u α uξ = b,.8 where ξ,, α are gven. It wa proved that there ext b > uch that.8 ha at leat one potve oluton f b, b and no potve oluton f b > b. To tudy the extence of potve oluton of BVP.5,.6,.7,.8, the Green functon of the correpondng problem are etablhed and play an mportant role n the proof of the man reult. In recent paper, ung lower and upper oluton method, Kong and Kong [3, 4, 5] etablhed reult for oluton and potve oluton of the followng two problem and x t ft, xt, x t =, t,, x α x ξ = λ, x β xξ = λ, x t ft, xt, x t =, t,, x α xξ = λ, x β xξ = λ,.9. repectvely. We note that the boundary condton n [3, 4] are two-parameter non-homogeneou BC. There, the extence of lower and upper oluton wth certan relaton are aumed.

4 4 Y. LIU EJDE-8/96 In recent year, there have been many exctng reult concernng the extence of potve oluton of BVP of econd order dfferental equaton wth p-laplacan ubjected to dfferent mult-pont boundary condton: [x t] ft, xt, x t =, t,, x = α xξ, x = β xξ, [x t] ft, xt, x t =, t,, x = α x ξ, x = β xξ, [x t] ft, xt, x t =, t,, x = α xξ, x = β x ξ....3 Thee reult, can be found n [6, 8, 9,,, 3,,,, 37, 54, 55, 56, 58, 59, 6, 6, 6, 65]. In above mentoned paper, to obtan potve oluton, two knd of aumpton are uppoed. The frt one mpoed on α, β, the other one called growth condton mpoed on the nonlnearty f. To defne a cone P n Banach pace and to defne the nonlnear operator T : P P are mportant tep n the the proof of the reult. It eay to ee that x t =, t,, x = x = ha unque potve oluton xt = t t, but the BVP x t =, t,, x = A, x = B ha potve oluton xt = t B A t A f and only f A and B. It how u that the preence of nonhomogeneou BC can nduce nonextence of potve oluton of a BVP. Motvated by the fact mentoned above, th paper concerned wth the more generalzed BVP for econd order dfferental equaton wth p-laplacan coupled wth nonhomogeneou mult-pont BC;.e., [x t] ft, xt, x t =, t,, x = α x ξ A, x = β xξ B, [x t] ft, xt, x t =, t,, x = α xξ A, x = β x ξ B, [x t] ft, xt, x t =, t,, x = αx A, x = β xξ B,.4.5.6

5 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 5 and [x t] ft, xt, x t =, t,, x = α x ξ A, x = β x ξ B, [x t] ft, xt, x t =, t,, x = α xξ A, x = β xξ B,.7.8 where < ξ < < ξ m <, A, B R, α, α, β for all =,..., m, f contnuou and nonnegatve, called p-laplacan, x = x p x for x and = wth p >, t nvere functon denoted by x wth x = x q x for x and = wth /p /q =. The purpoe to etablh uffcent condton for the extence of at leat three potve oluton for.4.8. The reult n th paper are new nce there ext no paper concerned wth the extence of at leat three potve oluton of thee nonhomogeneou mult-pont BVP even when x = x. Maybe t the frt tme to ue the mult-fxed-pont theorem to olve thee knd of BVP. The remander of th paper organzed a follow: The man reult are preented n Secton Theorem.,.6,.,.5,.3. Some example to how the man reult are gven n Secton 3.. Man Reult In th ecton, we frt preent ome background defnton n Banach pace and tate an mportant three fxed pont theorem. Then the man reult are gven and proved. Defnton.. Let X be a em-ordered real Banach pace. The nonempty convex cloed ubet P of X called a cone n X f ax P for all x P and a and x X and x X mply x =. Defnton.. A map ψ : P [, a nonnegatve contnuou concave or convex functonal map provded ψ nonnegatve, contnuou and atfe ψtx ty tψx tψy, or ψtx ty tψx tψy, for all x, y P and t [, ]. Defnton.3. An operator T ; X X completely contnuou f t contnuou and map bounded et nto relatve compact et. Defnton.4. Let a, b, c, d, h > be potve contant, α, ψ be two nonnegatve contnuou concave functonal on the cone P, γ, β, θ be three nonnegatve contnuou convex functonal on the cone P. Defne the convex et: P c = {x P : x < c}, P γ, α; a, c = {x P : αx a, γx c}, P γ, θ, α; a, b, c = {x P : αx a, θx b, γx c}, Qγ, β;, d, c = {x P : βx d, γx c},

6 6 Y. LIU EJDE-8/96 Qγ, β, ψ; h, d, c = {x P : ψx h, βx d, γx c}. Lemma.5 []. Let X be a real Banach pace, P be a cone n X, α, ψ be two nonnegatve contnuou concave functonal on the cone P, γ, β, θ be three nonnegatve contnuou convex functonal on the cone P. There ext contant M > uch that αx βx, x Mγx for all x P. Furthermore, Suppoe that h, d, a, b, c > are contant wth d < a. Let T : P c P c be a completely contnuou operator. If C {y P γ, θ, α; a, b, c αx > a} = and αt x > a for every x n P γ, θ, α; a, b, c; C {y Qγ, θ, ψ; h, d, c βx < d} and βt x < d for every x n Qγ, θ, ψ; h, d, c; C3 αt y > a for y P γ, α; a, c wth θt y > b; C4 βt x < d for each x Qγ, β;, d, c wth ψt x < h, then T ha at leat three fxed pont y, y and y 3 uch that βy < d, αy > a, βy 3 > d, αy 3 < a. Chooe X = C [, ]. We call x y for x, y X f xt yt for all t [, ], defne the norm x = max{max t [,] xt, max t [,] x t }. It eay to ee that X a em-ordered real Banach pace. Chooe k, /, let σ = mn{k, k} = k. For a cone P X of the Banach pace X = C [, ], defne the functonal on P : P R by γy = max t [,] y t, y P, θy = βy = max yt, y P, t [,] max yt, y P, αy = mn yt, y P, t [k,k] t [k,k] ψy = mn yt, y P. t [k,k] It eay to ee that α, ψ are two nonnegatve contnuou concave functonal on the cone P, γ, β, θ are three nonnegatve contnuou convex functonal on the cone P and αy βy for all y P. Lemma.6. Suppoe that x X, xt for all t [, ] and x t decreang on [, ]. Then xt mn{t, t} max xt, t [, ].. t [,] Proof. Suppoe xt = max t [,] xt. If t, t, we get that there ext η t ξ t uch that Then xt x t xt x t = t[xt xt] t t[xt x] tt = tt tx ξ t ttx η tt tt tx η t ttx η tt =. xt t t xt t t x t t xt txt, t, t.

7 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 7 Smlarly xt txt, for t t,. It follow that xt mn{t, t} max t [,] xt for all t [, ]. The proof complete... Potve Soluton of.4. Frt, we etablh an extence reult for three decreang potve oluton of.4. We ue the followng aumpton: H f : [, ] [h,, ] [, contnuou wth ft, ch, B on each ub-nterval of [,], where h = P ; m β H A, B ; H3 α, β atfy m α m <, β < ; H4 h : [, ] [, a contnuou functon and ht on each ubnterval of [,]. Conder the boundary-value problem [y t] ht =, t,, y α y ξ = A, y β yξ =,. Lemma.7. Suppoe that H H4 hold. If y a oluton of., then y decreang and potve on,. Proof. Suppoe y atfe.. It follow from the aumpton that y decreang on [, ]. Then the BC n. and H3 mply y = α y ξ A α y A. It follow that y A α. We get y t for t [, ]. Then y = β yξ β y. So y. Then yt > y for t, nce y t on [, ]. The proof complete. Lemma.8. Suppoe that H H4 hold. If y a oluton of., then wth and yt = B h A h = t A h hudu d, α ξ A h hd A,.3 [ B h = β A h hudu d ξ β ] A h hudu d,

8 8 Y. LIU EJDE-8/96 where a = A P m α and m A b = α α Proof. It follow from. that yt = y t P m α P m y α hudu d, hd. and BC n. mply that y = α ξ y hd A,.4 and Let [ y = β y hudu d ξ β ] y hudu d. Gc = c It eay to ee that A Ga = G α A α On the other hand, one ee that m Gb b = α = ξ α = A P m α ξ c hd A. α hd b α P m α A P m α P m α A α α m α P m A α A b ξ P m α P m α P m α α hd A =. P m α P m α m α m α =. hd α m α hd

9 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 9 Gc c Hence Gb. It eay to know that contnuou and decreang on, and contnuou and ncreang on,, Hence Ga and Gb and m Gc lm c c =, lm Gc c c = α >, we get that there ext unque contant A h = y [b, a] uch that.3 hold. The proof completed. Note h = B/ β, and let xt h = yt. Then.4 tranformed nto the boundary-value problem Let P = [y t] f t, yt h, y t =, t,, y α y ξ = A, y β yξ =, { y X : yt for all t [, ], y t decreang on [, ], } yt mn{t, t} max yt for all t [, ]. t [,] Then P a cone n X. Snce yt = m β yξ m β y β m β ξ β max t [,] y t m = β ξ β γy, we obtan m max yt β ξ γy. t [,] β.5 It eay to ee that there ext a contant M > uch that y Mγy for all y P. Defne the nonlnear operator T : P X by T yt = B y where A y atfe A y = t A y fu, yu h, y udu d, y P, α ξ A y f, y h, y d A,.6 and B y atfe B y = β A y fu, yu h, y udu d ξ β A y fu, yu h, y udu d.

10 Y. LIU EJDE-8/96 Then for y P, T yt = t A y β β fu, yu h, y udu d ξ A y fu, yu h, y udu d. Lemma.9. Suppoe that H H3 hold. It eay to how that the followng equalte hold: [T y t] f t, yt h, y t =, t,, T y α T y ξ = A, T y β T yξ = ; T y P for each y P ; x a oluton of.4 f and only f x = y h and y a oluton of the operator equaton y = T y n P ; v T : P P completely contnuou. Proof. The proof of, and are mple. To prove v, t uffce to prove that T contnuou and T compact. We dvde the proof nto two tep: Step. Prove that T contnuou about y. Suppoe y n X and y n y X. Let A yn be decded by.3 correpondng to y n for n =,,,.... We wll prove that A yn A y a n tend to nfnty. Snce y n y unformly on [,] and f contnuou, we have that for ɛ =, there ext N, when n > N, for each t [, ], uch that ft, y n t h, y nt ft, y t h, y t max t [,] ft, y t h, y t. Hence Lemma.8 mple that A yn an element n the nterval [ m A α α P m α P m f, y n h, y nd, α A ] α m A α [ α A α ]. P m α P m α max ft, y t h, y t, t [,] It follow that {A yn } bounded. If {A yn } doe not converge to A y, then there ext two ubequence {A y nk } and {A y nk } of {A yn } wth A y nk C, A y nk C, k, C C. By the contructon of A yn, A y nk = α A y nk ξ f, y nk h, y n k d A.

11 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS Snce ft, y n t h, y nt unformly bounded, by Lebegue domnated convergence theorem, lettng k, we get C = α ξ C f, y h, y d A. Snce A y atfe A y = α ξ A y f, y h, y d A, Lemma.8 mple that A y = C. Smlarly, we can prove that A y = C. Th contradct to C C. Therefore for each y n y, we have A yn A y. It follow that A y contnuou about y. So the contnuty of T obvou. Step. Prove that T compact. Let Ω P bet a bounded et. Suppoe that Ω {y P : y M}. For y Ω, we have f, y h, y d It follow from the defnton of T and Lemma.8 that T yt = A y t β β max ft, u, v =: D. t [,],u [h,mh],v [M,M] fu, yu h, y udu d ξ m A α α A y fu, yu h, y udu d β P m α β ξ fu, yu h, y udu d P m α P m α m A α α P m fu, yu h, y udu α m A α D m α P m α m β ξ β m A α D m α P m m = E β ξ β E, T y t = A y t fu, yu h, y udu d fu, yu h, y udu fu, yu h, y udu d α

12 Y. LIU EJDE-8/96 m A α α P m α P m fu, yu h, y udu α m A α D m α P m α = E, where m A E = α α D P m fu, yu h, y udu d α For the unform contnuty of on the nterval [E, E], for each ɛ >, there ext a ρ > uch that Put Y = A y Y Y < ɛ, Y, Y [E, E], Y Y < ρ. t t fu, yuh, y udu, Y = A y fu, yuh, y udu. Snce Y Y = t t fu, yu h, y udu D t t, t eay to ee that there ext δ > ndependent of ɛ uch that Y Y < ρ for all t, t wth t t < δ. Hence there δ > ndependent of ɛ uch that T y t T y t = Y Y < ɛ, whenever t, t [, ] and t t < δ. Th how that T yt equ-contnuou on [,]. The Arzela-Akol theorem guarantee that T Ω relatve compact, whch mean that T compact. Hence the contnuty and the compactne of T mply that T completely contnuou. Theorem.. Suppoe that H H3 hold and there ext potve contant e, e, c, L = m α m α d β β m α ξ m α d, Q = mn { c c }, L ; P m α e W = k ; E = e. σ k k d L uch that If c e σ > e > e σ > e >, m A Q α α., Q > W.

13 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 3 A ft, u, v < Q for all t [, ], u [h, c h], v [c, c]; A ft, u, v > W for all t [k, k], u [e h, e /σ h], v [c, c]; A3 ft, u, v < E for all t [, ], u [h, e /σ h], v [c, c]; then.4 ha at leat three decreang potve oluton x, x, x 3 uch that x < e h, x k > e h, x 3 > e h, x 3 k < e h. Remark.. In paper [3], uffcent condton are found for the extence of oluton of.9 baed on the extence of lower and upper oluton wth certan relaton. Ung the obtaned reult, under ome other aumpton, the explct range of value of λ and λ are preented wth whch.9 ha a oluton, ha a potve oluton, and ha no oluton, repectvely. Furthermore, t proved that the whole plane for λ and λ can be dvded nto two djont connected regon E and N uch that.9 ha a oluton for λ, λ E and ha no oluton for λ, λ N. When applyng Theorem. to.9, t how u that.9 ha at leat three decreang potve oluton under the aumpton λ, λ and ome other aumpton. Remark.. Conder the cae A and B <, when H3 and H4 hold, we can prove mlarly that Lemma.7 and Lemma.8 are vald. Defne the ame operator T on the cone P. Theorem. how that.5 ha at leat three decreang and potve oluton y, y, y 3. Hence.4 ha at leat three decreang oluton x = y h, x = y h and x 3 = y 3 h, whch need not be B potve nce h = P m <. In cae A >, B and A >, B <, the β author could not get the uffcent condton guaranteeng the extence of multple potve oluton of.4. Proof of Theorem.. To apply Lemma.5, we prove that t hypothee are atfed. By the defnton, t eay to ee that α, ψ are two nonnegatve contnuou concave functonal on the cone P, γ, β, θ are three nonnegatve contnuou convex functonal on the cone P and αy βy for all y P, there ext contant M > uch that y Mγy for all y P. Lemma.8 mple that x = xt a potve oluton of.4 f and only f xt = yt h and yt a oluton of the operator equaton y = T y and T : P P completely contnuou. Correpondng to Lemma.5, c = c, h = σ e, d = e, a = e, b = e σ. Now, we prove that all condton of Lemma.5 hold. One ee that < d < a. The remander dvded nto four tep. Step. Prove that T : P c P c ; For y P c, we have y c. Then yt c for t [, ] and c y t c for all t [, ]. So A mple that ft, yt h, y t Q, t [, ]. It follow from Lemma.9 that T y P. Then Lemma.8 mple T yt = t A y β β fu, yu h, y udu d ξ A y fu, yu h, y udu d

14 4 Y. LIU EJDE-8/96 A m α α P m α P m α fu, yu h, y udu fu, yu h, y udu d β P m α P m β α ξ m A α α β P m α P m β α Q β ξ m A α α fu, yu h, y udu P m α Q Q d P m α P m β ξ α P m α m A α α Q Q d [ = Q m c. β β ξ Q P m α P m α m α α d m α fu, yu h, y udu d Q Q d Q Q d m α ] d Smlarly to the above dcuon, we have from Lemma.7 that m A A y α m α P m fu, yu h, y udu. α Then T y t T y = A y m A α α m A α α P m P m α α Q fu, yu h, y udu

15 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 5 Q m α Q c. It follow that T y = max{max t [,] T yt, max t [,] T y t } c. Then T : P c P c. Step. Prove that {y P γ, θ, α; a, b, c αy > a} = {y P γ, θ, α; e, e σ, c αy > e } and αt y > e for every y P γ, θ, α; e, e σ, c. Chooe yt = e σ for all t [, ]. Then y P and αy = e σ > e, θy = e σ e σ, γy = < c. It follow that {y P γ, θ, α; a, b, c : αy > a}. For y P γ, θ, α; a, b, c, one ha αy = Then mn yt e, θy = max yt e, t [k,k] t [k,k] σ Thu A mple Snce we get αt y σ [ σ [ e yt e σ, t [k, k], y t c. ft, yt h, y t W, αt y = A y β n [k, k]. mn T yt σ max T yt, t [k,k] t [,] β A y ξ [ σ A [ k σ k fu, yu h, y udu d A y ] fu, yu h, y udu d α A α k σ W k d = e. k γy = max t [,] y t c. ] fu, yu h, y udu d ] fu, yu h, y udu d k ] fu, yu h, y udu d Th complete Step. Step 3. Prove that {y Qγ, θ, ψ; h, d, c : βy < d} whch equal to {y Qγ, θ, ψ; σ e, e, c : βy < e } not empty, and βt y < e for every y Qγ, θ, ψ; h, d, c = Q γ, θ, ψ; σ e, e, c ;

16 6 Y. LIU EJDE-8/96 Chooe yt = σ e. Then y P, and ψy = σ e h, βy = θy = σ e < e = d, γy = c. It follow that {y Qγ, θ, ψ; h, d, c βy < d}. For y Qγ, θ, ψ; h, d, c, one ha ψy = Hence yt e σ So βt y = max T yt t [,] = A y β mn yt h = e σ, θy = max yt d = e, t [k,k] t [k,k] γy = max t [,] y t c. and c y t c, for t [, ]. Then A3 mple ft, yt h, y t E, t [, ]. β fu, yu h, y udu d ξ m A α α A y fu, yu h, y udu d β P m α P m β α ξ m A α α β P m α P m β α ξ P m α P m fu, yu h, y udu d α m A α α fu, yu h, y udu E E d P m α E P m β β ξ α P m α P m α m A α α E E d E P m α P m α fu, yu h, y udu fu, yu h, y udu d E E d E E d

17 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 7 [ = E m α m α d β β m α ] ξ m α d = e = d. Th complete Step 3. Step 4. Prove that αt y > a for y P γ, α; a, c wth θt y > b; For y P γ, α; a, c = P γ, α; e, c wth θt y > b = e σ, we have that αy = mn t [k,k] yt e and γy = max t [,] yt c and max t [k,k] T yt > e σ. Then αt y = mn T e yt σ βt y > σ = e = a. t [k,k] σ Th complete Step 4. Step 5. Prove that βt y < d for each y Qγ, β; d, c wth ψt y < h. For y Qγ, β; d, c wth ψt y < d, we have γy = max t [,] yt c and βy = max t [,] yt d = e and ψt y = mn t [k,k] T yt < h = e σ. Then βt y = max T yt mn t [,] σ T yt < e σ = e = d. t [k,k] σ Th complete the Step 5. Then Lemma.5 mple that T ha at leat three fxed pont y, y and y 3 uch that βy < e, αy > e, βy 3 > e, αy 3 < e. Hence.4 ha three decreang potve oluton x, x and x 3 uch that max x t < e h, t [,] max x 3t > e h, t [,] mn x t > e h, t [k,k] mn x 3t < e h. t [k,k] Hence x < e h, x k > e h, x 3 > e h, x 3 k < e h... Potve Soluton of.5. Now we prove the extence of three potve ncreang oluton of.5. We ue the aumpton: H5 f : [, ] [, [, [, contnuou wth ft, ch, on each ub-nterval of [,] for all c, where h = H6 A, B ; H7 α, β atfy m α, m β < ; Conder the boundary-value problem A P m α ; [y t] ht =, t,, y α yξ =, y β y ξ = B..7 Lemma.3. Suppoe that H4, H6, H7 hold. If y a oluton of.7, then y ncreang and potve on,.

18 8 Y. LIU EJDE-8/96 Proof. Suppoe y atfe.7. It follow from the aumpton that y decreang on [, ]. Then the BC n.7 and H4 mply y = β y ξ B β y. It follow that y. We get that y t for t [, ]. Then y = α yξ α y. It follow that y. Then yt > y for t, nce y t for all t [, ]. The proof complete. Lemma.4. Suppoe that H4, H6, H7 hold. If y a oluton of.7, then t yt = B h A h hudu d, wth and where B h = A h = β A h α ξ ξ α A h a = m B b = β β Proof. It follow from.7 that yt = y t B β, P m β P m y and the BC n.7 mply y = β y Let y = α ξ hd B,.8 β ξ α y Gc = c It eay to ee that B Ga = G β c B β β β ξ hudu d, hd. hudu d, hd B, hd B. B β hudu d. B

19 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 9 =. On the other hand, one ee that Gb m b = β = β B P m ξ hd b β P m β B P m B b β P m β B β P m P m β P m β P m β β Gc c ξ hd P m β P m β hd hd β m β =. Hence Gb. It eay to know that contnuou and decreang on, and contnuou and ncreang on,, Hence Ga and Gb and m Gc lm c c =, lm Gc c c = β >. Then there ext unque contant A h = y [a, b] uch that.8 hold. The proof complete. A Note h = P m, and xt h = yt. Then.5 tranformed nto the α boundary-value problem Let [y t] f t, yt h, y t =, t,, y α yξ =, y β y ξ = B, P = { y X : yt for all t [, ], y t decreang on [, ], yt mn{t, t} max t [,] yt for all t [, ]}. Then P a cone n X. Snce yt = m α yξ m α y α m α ξ α max t [,] y t = m α ξ α γy,.9 we have m max yt α ξ t [,] α γy. It eay to ee that there ext a contant M > uch that y Mγy for all y P.

20 Y. LIU EJDE-8/96 Defne the nonlnear operator T : P X by t T yt = B y A y fu, yu h, y udu d, y P, where A y atfe A y = and B y atfe B y = Then for y P, T yt = β A y α t ξ ξ α A y A y α f, y h, y d B,. fu, yu h, y udu d ξ α A y fu, yu h, y ud d. Lemma.5. Suppoe that H5, H6, H7 hold. Then the followng equalte hold: fu, yu h, y ud d. [T y t] f t, yt h, y t =, t,, T y α T yξ =, T y β T y ξ = B; T y P for each y P ; x a oluton of.5 f and only f x = y h and y a oluton of the operator equaton y = T y n cone P ; v T : P P completely contnuou. The proof of the above lemma mlar to that of Lemma.9 and t omtted. Theorem.6. Suppoe that H5 H7 hold and there ext potve contant e, e, c, P L = m β P m d β P ξ α α m β P m d, β Q = mn { c c } e, L ; W = k ; P m β σ k k d uch that c e σ > e > e σ > e >, E = e L. m B Q β β, Q > W.

21 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS If A, A and A3 from n Theorem. hold, then.5 ha at leat three ncreang potve oluton x, x, x 3 uch that x < e h, x k > e h, x 3 > e h, x 3 k < e h. Proof. To apply Lemma.5, we prove that all t condton are atfed. By the defnton, t eay to ee that α, ψ are two nonnegatve contnuou concave functonal on the cone P, γ, β, θ are three nonnegatve contnuou convex functonal on the cone P and αy βy for all y P, there ext contant M > uch that y Mγy for all y P. Lemma.5 mple that x = xt a potve oluton of.5 f and only f xt = yt h and yt a oluton of the operator equaton y = T y and T : P P completely contnuou. Correpondng to Lemma.5, c = c, h = σ e, d = e, a = e, b = e σ. Now, we prove that all condton of Lemma.5 hold. One ee that < d < a. The remander dvded n four tep. Step. Prove that T : P c P c ; For y P c, we have y c. Then yt c for t [, ] and c y t c for all t [, ]. So A mple that ft, yt h, y t Q, t [, ]. It follow from Lemma.5 that T y P. Then Lemma.4 mple T yt A y α fu, yu h, y udu d ξ α A y m B β β fu, yu h, y udu d α P m β P m ξ α β m B β α P m β P m β ξ α β P m β P m β m B β fu, yu h, y udu d β fu, yu h, y udu Q Q d P m β P m β m B β β fu, yu h, y udu fu, yu h, y udu d Q Q d

22 Y. LIU EJDE-8/96 P Q m β P m Q Q d β P ξ α α Q m β P m Q Q d β [ P = Q m β P m d β P ξ α α m β ] P m d β c. Smlarly to above dcuon, from Lemma.4, we have T y t T y = A y m B β β fu, yu h, y udu P m β P m β fu, yu h, y udu fu, yu h, y udu m B β m β P m Q β Q m β Q c. It follow that T y = max{max t [,] T yt, max t [,] T y t } c. Then T : P c P c. Step. Prove that {y P γ, θ, α; a, b, c αy > a} = {y P γ, θ, α; e, e σ, c : αy > e } and αt y > e for every y P γ, θ, α; e, e σ, c ; Chooe yt = t [, ]. Then y P and αy = e σ > e, θy = e σ e σ, γy = < c. It follow that {y P γ, θ, α; a, b, c αy > a}. For y P γ, θ, α; a, b, c, one ha αy = mn yt e, θy = max yt e, t [k,k] t [k,k] σ Then e yt e σ, t [k, k], y t c. Thu A mple ft, yt h, y t W, t [k, k]. Snce αt y = mn t [k,k] T yt σ max t [,] T yt, e σ for all γy = max t [,] y t c.

23 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 3 from Lemma.4, we have αt y σ max t [,] T yt [ = σ A y fu, yu h, y udu d ξ α α ] A h fu, yu h, y ud d [ σ B β fu, yu h, y udu d ξ α α B β ] fu, yu h, y ud d σ B β σ fu, yu h, y udu d k σ k k fu, yu h, y udu d k σ W k d = e. k fu, yu h, y udu d Th complete Step. Step 3. Prove that the et {y Qγ, θ, ψ; h, d, c βy < d} whch equal to {y Q γ, θ, ψ; σ e, e, c : βy < e } not empty, and βt y < e for every y Qγ, θ, ψ; h, d, c = Q γ, θ, ψ; σ e, e, c. Chooe yt = σ e. Then y P, and ψy = σ e h, βy = θy = σ e < e = d, γy = c. It follow that {y Qγ, θ, ψ; h, d, c : βy < d}. For y Qγ, θ, ψ; h, d, c, one ha ψy = mn yt h = e σ, θy = max yt d = e, t [k,k] t [k,k] γy = max t [,] y t c. Hence we have yt e σ and c y t c for t [, ]. Then A3 mple ft, yt h, y t E, t [, ]. So that βt y = max T yt t [,] = A y α fu, yu h, y udu d ξ α A y fu, yu h, y udu d

24 4 Y. LIU EJDE-8/96 m B β β P m β P m fu, yu h, y udu β fu, yu h, y udu d α P m β P m β ξ α fu, yu h, y udu d m B β β α P m β m B β β fu, yu h, y udu P m β P m β ξ α m B β β P m E E d β P E m β P m E E d β P ξ α α E m β P m β [ P = E m β P m d β P ξ α α m β P m β = e = d. E E d E E d ] d Th complete Step 3. Step 4. Prove that αt y > a for y P γ, α; a, c wth θt y > b; For y P γ, α; a, c = P γ, α; e, c wth θt y > b = e σ, we have that αy = mn t [k,k] yt e and γy = max t [,] yt c and max t [k,k] T yt > e σ. Then αt y = mn T e yt σ βt y > σ = e = a. t [k,k] σ Th complete Step 4. Step 5. Prove that βt y < d for each y Qγ, β; d, c wth ψt y < h. For y Qγ, β; d, c wth ψt y < d, we have γy = max t [,] yt c and

25 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 5 βy = max t [,] yt d = e and ψt y = mn t [k,k] T yt < h = e σ. Then βt y = max T yt mn t [,] σ T yt < e σ = e = d. t [k,k] σ Th complete the Step 5. Then Lemma.5 mple that T ha at leat three fxed pont y, y and y 3 n P uch that βy < e, αy > e, βy 3 > e, αy 3 < e. Hence.5 ha three ncreang potve oluton x, x and x 3 uch that Hence max x t < e h, t [,] max x 3t > e h, t [,] mn x t > e h, t [k,k] mn x 3t < e h. t [k,k] x < e h, x k > e h, x 3 > e h, x 3 k < e h. The proof complete. Remark.7. For.5, when A <, B, we can alo get the extence reult for three ncreang oluton of.5 mlarly, but the oluton need not be potve. By the way, t nteretng to etablh uffcent condton guarantee the extence of potve oluton of.5 when B <..3. Potve Soluton of.6. Now we prove an extence reult for three potve oluton of.6. We ue the followng there condton: H8 f : [, ] [h, R [, contnuou wth ft, c h, on each ub-nterval of [,] for all c, where h = H9 A, B ; H α, β atfy m β < and A B Conder the followng boundary-value problem P ; m β B P m β ; [y t] ht =, t,, y αy = D, y β yξ =,. Lemma.8. Suppoe that H4, H hold and D. If y a oluton of., then y potve on,. Proof. Suppoe y atfe.. It follow from aumpton H4 that y decreang on [, ]. If y >, the BC n. and H4 mply y t > for all t [, ]. Then y = β yξ β y. Then y. On the other hand, y = αy D. Th a contradcton nce y t >.

26 6 Y. LIU EJDE-8/96 If y and y, then we get that y t for all t [, ]. The BC n. and H4 mply y = β yξ β y. It follow that y. Then yt y for all t [, ]. H4 mple yt > for all t,. If y and y >, then y = αy D. It follow from. and H4 that y = β yξ β mn{y, y}. If y y, then y m β y mple that y. If y > y, then y > nce y. It follow from H4 that yt mn{y, y}. Then y potve on,. The proof complete. Lemma.9. Aume H4, H hold and D. If y a oluton of., then t yt = B h A h hudu, t [, ], where A h [b, ], [ α A h = ξ ] hudu D β A h β ξ A h B h = α A h b = Proof. From. t follow that y ξ D hudu β hudu d, hudu d hudu D, a = α β yt = y β = t y y β a β ξ. ξ β y, udu d, hudu d hudu d,

27 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 7 Thu Let y = α y [ α y = Gc =. y hudu D. ] hudu D ξ β y = [ α c c β [ α c ξ hudu d β hudu d. ] hudu D c hudu d hudu β c β ξ c hudu d ] D hudu d β β hudu d It eay to ee that Gc ncreang on,, G. Snce a = α β β ξ, we get D Gb = G = [ α D m β a β β ξ β a D D hudu ] D β a β a β hudu d hudu d

28 8 Y. LIU EJDE-8/96 [ α D β a β β ξ D we get y hudu D Let D ] D β a β a P m β a β d d =, = b. The proof complete. P 3 = { y X : yt for all t [, ], y t decreang on [, ], yt mn{t, t} max t [,] yt for all t [, ]}. Then P 3 a cone n X. Snce, for y P 3, we have we get yt = yt y y y θ t y m max t [,] y t β yξ m β y β m β ξ β max t [,] y t m = β ξ β γy, max yt t [,] m β ξ β γy. It eay to ee that there ext a contant M > uch that y Mγy for all y P 3. Suppoe that H hold. Let xt h = yt. Then.6 tranformed nto [y t] ft, yt h, y t =, t,, y αy B = A β, y β yξ =, Defne the nonlnear operator T 3 : P 3 X by t T 3 yt = B y A y fu, yu h, y udu d, y P 3, where [ α A y fu, yu h, y B ] udu A β β

29 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 9 = β A y β ξ A y fu, yu h, y udu d fu, yu h, y udu d, and B h = α A y fu, yu h, y B uduu A β. For y P 3, the defnton of A y mple T 3 yt = α A y = t t A y A y β fu, yu h, y udu A β fu, yu h, y udu d fu, yu h, y udu d ξ A y B β fu, yu h, y udu d. Lemma.. Suppoe that H8 H hold. Then T 3 y P 3 for each y P 3 ; x a oluton of.7 f and only f x = y h and y a oluton of the operator equaton T 3 y = y n P 3 ; T 3 : P 3 P 3 completely contnuou; v the followng equalte hold: [T 3 y t] f t, yt h, y t =, t,, T 3 y αt 3 y B ξ = A β, T 3 y β T 3 yξ = ; The proof of the above lemma mlar to that of Lemma.9, o t omtted. Theorem.. Suppoe that H8 H hold and that there ext potve contant e, e, c, L = β β ξ, Q = mn { c c }, ; L e W =. uch that c e σ k σ k kd > e > e σ > e >, ; E = e L B A P m m Q β β α m β, Q > W. ξ

30 3 Y. LIU EJDE-8/96 If A A3 n Theorem. hold, then.6 ha at leat three potve oluton x, x, x 3 uch that max x t < e h, t [,] mn x 3t > e h, t [k,k] mn x t > e h, t [k,k] mn x 3t < e h. t [k,k] The proof of the above theorem mlar to that of Theorem., o t omtted. Aume the followng three condton: H f : [, ] [h, R [, contnuou wth ft, c h, on each ub-nterval of [,] for all c, where h = A; H A, B ; H3 α, β atfy m β < and B A β ; Conder the boundary-value problem [y t] ht =, t,, y αy =, y β yξ = D,. Lemma.. Aume H4, H3 and D. If y a oluton of., then y potve on,. Proof. Suppoe y atfe.. It follow from the aumpton that y decreang on [, ]. If y <, the BC n. and H4 mply that y t < for all t [, ] and y. Then y = β yξ D β y D. D Then y P m. It follow from D that y y, a contradcton to β y t < for all t [, ]. If y and y >, we get that y t > for all t [, ]. The BC n. and H4 mply that y = αy. Then yt > y for all t [, ]. If y and y, then y = αy. It follow from. that y = β yξ D β mn{y, y}. If y y, then y m β y mple that y. If y > y, then y > nce y. It follow that yt mn{y, y}. Then y potve on,. The proof complete. Lemma.3. Aume H4, H3, D. If y a oluton of., then yt = B h A h hudu, t [, ], where A h [a, b], α A h t ξ hudu β

31 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 3 and a = D c β A h β, b = D c ξ Proof. It follow from. that Thu y = A h B h = α y yt = y hudu, β = β y t hudu d hudu d D =, ξ y ξ y hudu, c = α y β y β ξ y m β β ξ. hudu d, hudu d = αy. hudu d D, hudu d αy hudu d D. It follow from the proof of Lemma. that y. Let Gc = β c hudu d α β c β ξ c hudu d D. It eay to ee that Gc ncreang on, and Ga = G D a. Snce Gb = G D hudu c β α D c β D c β ξ D c hudu d hudu hudu d D

32 3 Y. LIU EJDE-8/96 D m D β c α β c β ξ D c D =, we get a y D c hudu. The proof complete. Let Suppoe that H3 hold. Let xt h = yt. Then.6 tranformed nto [y t] ft, yt h, y t =, t,, y y αy =, A β yξ = B β. P 3 = {y X : yt for all t [, ], y t decreang on [, ], yt mn{t, t} max yt for all t [, ]}. t [,] Then P 3 a cone n X. Snce, for y P 3, we have yt yt y y max t [,] y t α max t [,] y t = αγy, we get max t [,] yt αγy. It eay to ee that there ext a contant M > uch that y Mγy for all y P 3. Defne the nonlnear operator T 3 : P 3 X by T 3yt = B y A y fu, yu h, y udu d, y P 3, where and Then = β A y β B y = T 3yt = = ξ t t t A y A y A y A y A y fu, yu h, y udu d α A y fu, yu h, y A udu d B β. fu, yu h, y udu d α A y. fu, yu h, y udu d α A y fu, yu h, y udu d fu, yu h, y udu d α A y, y P 3. Lemma.4. Aume H H3. Then the followng hold: T 3y P 3 for each y P 3; x a oluton of.6 f and only f x = y h and y a oluton of the operator equaton T 3y = y n P 3;

33 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 33 T 3 : P 3 P 3 completely contnuou; v the followng equalte hold: [T 3y t] f t, yt h, y t =, t,, T 3y αt 3yξ =, T 3y β T 3yξ = B β A; The proof of the above lemma mlar to that of Lemma.9, o t omtted. Theorem.5. Suppoe that H H3 hold and there ext potve contant e, e, c, uch that c e σ and Q = mn { c, d α W = σ mn { k k E = > e > e σ > e > e d α c } ; k d, k k kd } e. α β α m B Q β β β ξ, Q > W. B ; A β, If A A3 defned n Theorem. hold, then.6 ha at leat three potve oluton x, x, x 3 uch that max x t < e h, t [,] mn x 3t > e h, t [k,k] mn x t > e h, t [k,k] mn x 3t < e h. t [k,k] The proof of the above theroem mlar to that Theorem.6 and t omtted. Remark.6. Kwong and Wong [9], Palamde [5] tuded the extence of potve oluton of.6 and.7 the man reult may be een n Secton. When applyng Theorem. to.6, we get three potve oluton f θ, π/] and the other aumpton n Theorem. hold..4. Potve Soluton of.7. We prove extence reult for three potve oluton of.7. The followng aumpton are ued n th ub-ecton. H4 f : [, ] [, [, [, contnuou wth ft, ch, on each ub-nterval of [,] for all c, where h = A; H5 α, β atfy m α <, m β < ; H6 A, B wth B A P m β.

34 34 Y. LIU EJDE-8/96 Suppoe that H4 H6 hold and conder the boundary-value problem [y t] ht =, t,, y α y ξ =, y β y ξ = D,.3 Lemma.7. Aume H4, H5, H6 and D. If y a oluton of.3, then y potve on,. Proof. Suppoe y atfe.3. It follow from the aumpton that y decreang on [, ]. Then the BC n.3 and H4 mply y = β y ξ D β y. It follow that y. We get y t for t [, ]. Then y = α yξ α y. It follow that y. Then yt > y for t, nce y t for all t [, ]. The proof complete. Lemma.8. Aume H4 H6. If y a oluton of.3, then t yt = B h A h hudu d and there ext unque A h [a, b] uch that A h = β A h and B h = α A h a = m D b = β β Proof. It follow from.3 that yt = y The BC n.3 mply y = y = t ξ ξ hd D,.4 hudu, D β, P m β P m y β y α y β ξ hudu d hd D, ξ hudu. hd.

35 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 35 Let It eay to ee that Ga = On the other hand, Gc = c D β D β Gb m b = β = β c β β β D P m ξ D β hd D. D β D =. ξ hd b β P m β D P m D b β P m β D β P m P m β P m β P m β β Gc c ξ hd ξ P m β hd D P m β hd β m β =. hd Hence Gb. It eay to know that contnuou and decreang on, and contnuou and ncreang on,, Hence Ga and Gb and m Gc lm c c =, lm Gc c c = β >. Then there ext unque contant A h = y [a, b] uch that.4 hold. The proof complete. Suppoe H6 hold. Let xt A = yt. Then.7 tranformed nto Let [y t] f t, yt h, y t =, t,, y α y ξ =, y β y A ξ = B β, P 4 = {y X : yt for all t [, ], y t decreang on [, ], yt ty for all t [, ]}..5

36 36 Y. LIU EJDE-8/96 Then P 4 a cone n X. For y P 4, nce yt = yt y y α max t [,] y t = we get max yt t [,] α γy. α γy, It eay to ee that there ext a contant M > uch that y Mγy for all y P. Defne the operator T 4 : P 4 X by t T 4 yt = B y A y fu, yu h, y udu d, y P 4, where A h = and Then β A h B y = ξ α A y T 4 yt = α A y t It follow from Lemma.8 that B A P m β β fu, yuh, y A udu B β,.6 A y A y B ξ ξ fu, yu h, y udu. A P m fu, yu h, y udu fu, yu h, y udu d. β P m β P m β Lemma.9. Suppoe that H4 H6. Then the followng equalte hold: m β β fu, yu h, y udu. [T 4 y t] f t, yt h, y t =, t,, T 4 y α T 4 y ξ =, T 4 y T 4 y P 4 for each y P 4 ; β T 4 y A ξ = B β ;

37 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 37 x a oluton of.7 f and only f x = y h and y a oluton of the operator equaton y = T 4 y n P 4 ; v T 4 : P 4 P 4 completely contnuou. The proof of the above lemma mlar to that of Lemma.9; o we omt t. Theorem.3. Suppoe that H4 H6 hold, and there ext potve contant e, e, c, and P L = m β P m d β P α m β P m ξ, β uch that c e σ W = Q = mn { c L, e c k σ k k d > e > e σ > e >, P m β } ; ; E = e L A B P m m Q β β β, Q > W. If A A3 n Theorem. hold, then.7 ha at leat three ncreang potve oluton x, x, x 3 uch that max x t < e h, t [,] mn x 3t > e h, t [k,k] mn x t > e h, t [k,k] mn x 3t < e h. t [k,k] The proof of the above theorem mlar to that of Theorem.6; Therefore, t omtted..5. Potve Soluton of.8. Fnally, we prove an extence reult for three potve oluton of.8. The followng aumpton wll be ued n the proof of all reult n th ubecton. H7 f : [, ] [h, R [, contnuou wth ft, c h, on A each ub-nterval of [,] for all c, where h = P ; m α H8 α, β atfy m α m <, β < ; H9 A, B wth B P m β P m Conder the boundary-value problem α A. [y t] ht =, t,, y α yξ =, y β yξ = B,..7 Lemma.3. Aume H4, H8, H9. If y a oluton of.7, then y potve on,.

38 38 Y. LIU EJDE-8/96 Proof. Suppoe y atfe.7. It follow from the aumpton that y decreang on [, ]. Then the BC n.7 and H4 mply that yt mn{y, y} for all t [, ]. Then y α mn{y, y}..8 Smlarly, we get y β mn{y, y}..9 It follow from.8 and.9 that mn{y, y} mn { m α, } β mn{y, y}. Then H8 mple that mn{y, y}. So yt mn{y, y} for all t [, ]. The proof complete. Lemma.3. Aume H4, H8, H9. If y a oluton of.7, then there ext unque A h [, b] uch that β ξ α A h hudu d where b = α A h ξ β A h hudu B a Proof. From.7 t follow that yt = y, a = m t hudu d hudu d B =. α ξ y β β ξ. The BC n.7 mple ξ y = y α α y and Then y y hudu d ξ = y β β y β α ξ α y hudu d. hudu d hudu d B. hudu d

39 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 39 Let y ξ β y Gc = β α c ξ β c hudu d ξ α c hudu d = β ξ α α c β c β ξ c hudu d B =. hudu d B hudu d hudu d hudu d hudu d B. It eay to ee that Gc ncreang on, and and G = β ξ α α hudu d β hudu d β ξ hudu d B <, Gb = G hudu B a = β ξ α α B hudu d a β B hudu d a β ξ β α B a hudu d B ξ α B d a

40 4 Y. LIU EJDE-8/96 β B d a β B d B =. a ξ Then G <, Gb and that Gc ncreang on, mply that A h = y [, b] and A h atfe β ξ α A h hudu d Let α A h ξ β A h hudu d hudu d B =. Let xt h = yt. Then.8 tranformed nto [y t] f t, yt h, y t =, t,, y α yξ =, y β yξ = B β α A, P 5 = { y X : yt for all t [, ], y t decreang on [, ], yt mn{t, t} max yt for all t [, ]} t [,] Then P 5 a cone n X. For y P 5, nce yt = yt y y. y θ t y max t [,] y t m α yξ m α y α m α ξ α max t [,] y t m = α ξ α γy, we get m max yt α ξ t [,] α γy. It eay to ee that there ext a contant M > uch that y Mγy for all y P 5.

41 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 4 Defne the operator T 5 : P 5 X by where and T 5 yt = B y β B y = t α A y A y ξ α A y ξ β A y α fu, yu h, y udu d, y P 5, fu, yu h, y udu d ξ α A y fu, yu h, y udu d fu, yu h, y udu d B =. Lemma.33. Aume H7, H8, H9. Then the followng equalte hold: fu, yu h, y udu d. [T 5 y t] f t, yt h, y t =, t,, T 5 y α T 5 yξ =, T 5 y β T 5 yξ = B β α A ; T 5 y P 5 for each y P 5 ; x a oluton of.8 f and only f x = y h and y a oluton of the operator equaton y = T 5 y; v T 5 : P 5 P 5 completely contnuou. The proof mlar to that of Lemma.9; therefore, t omtted. Theorem.34. Suppoe that H7, H8, H9 hold, and that there ext potve contant e, e, c, L = m β m β d α d, α W = ξ Q = mn { c L m β m β } ; c, e k σ k k d P m β P m β ; E = e L.

42 4 Y. LIU EJDE-8/96 uch that c e σ > e > e σ > e >, m B Q β β, Q > W. If A A3 n Theorem. hold, then.8 ha at leat three potve oluton x, x, x 3 uch that max x t < e h, t [,] mn x 3t > e h, t [k,k] mn x t > e h, t [k,k] mn x 3t < e h. t [k,k] The proof of the above theorem mlar to that of Theorem.; t omtted. For.8, we have the followng aumpton: H9 f : [, ] [h, R [, contnuou wth ft, c h, on B each ub-nterval of [,] for all c, where h = P ; m β H α, β atfy m α <, m β < ; H A, B wth B P m β P m Conder the boundary-value problem α A. [y t] ht =, t,, y α yξ = A, y β yξ =,. Lemma.35. Aume H4, H, H. If y a oluton of., then y potve on,. The proof of the above lemma mlar to that of Lemma.4; t omtted. Lemma.36. Aume H4, H, H. If y a oluton of., then yt = B h A h hudu d. where α t β β A h α ξ A h ξ A h hudu d hudu d hudu d A =, b = hudu A, a a = α β β ξ α α ξ, B h = β β y hudu d. ξ

43 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 43 Proof. It follow from. that yt = y The BC n. mple y = y β β and Then Let y t y = y α α α β y ξ ξ β y α ξ ξ y Gc = α β c α ξ y hudu d y hudu d. y hudu d β ξ hudu d hudu d A. hudu d hudu d A =. c hudu d c = α β α β ξ α c ξ c hudu d A c hudu d hudu d A. hudu d hudu d It eay to ee that Gc decreang on, and G = α β β hudu d ξ α hudu d

44 44 Y. LIU EJDE-8/96 and Gb = G ξ α = α β hudu A a β ξ α ξ α α β ξ α β α ξ hudu d A <, hudu A d a hudu A d a hudu A d A a A a A d a d A d A =. a Then G <, Gb and that Gc decreang on, mply that A h = y [b, ] and A h atfe α β β A h α ξ A h ξ A h hudu d hudu d hudu d A =. Let xt h = yt. Then.8 tranformed nto Let [y t] f t, yt h, y t =, t,, y α yξ = A α β B, y β yξ =, P 6 = { y X : yt for all t [, ], y t decreang on [, ], yt mn{t, t} max ytfor all t [, ]} t [,].

45 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 45 Then P 6 a cone n X. For y P 6, nce yt = m β yξ m β y β m β ξ β max t [,] y t m = β ξ β βy, we get m max yt β ξ t [,] β γy. It eay to ee that there ext a contant M > uch that y Mγy for all y P 6. Defne the operator T 6 : P 6 X by T 6 yt = B y A y fu, yu h, y udu d, y P 6, where A h [b, ] atfe t α β β ξ A h α ξ A h and B h atfe B h = β β ξ A h hudu d hudu d hudu d A =, y Lemma.37. Suppoe that H9 H hold. Then the followng equalte hold: hudu d. [T 6 y t] f t, yt h, y t =, t,, T 6 y α T 6 yξ = A α β B, T 6 y β T 6 yξ = ; T 6 y P 6 for each y P 6 ; x a oluton of.8 f and only f x = y h and y a oluton of the operator equaton y = T 6 y; v T 6 : P 6 P 6 completely contnuou. The proof of the above lemma mlar to that of Lemma.9; we omt t.

46 46 Y. LIU EJDE-8/96 Theorem.38. Suppoe that H9 H hold and that there ext potve contant e, e, c, L = m β m β d α d, uch that c e σ α W = ξ Q = mn { c L m β m β } ;, c P m β e k σ k k d > e > e σ > e >, m B Q β β P m β ; E = e L, Q > W. If A A3 n Theorem. hold, then.8 ha at leat three potve oluton x, x, x 3 uch that max x t < e h, t [,] mn x 3t > e h, t [k,k] mn x t > e h, t [k,k] mn x 3t < e h. t [k,k] The proof of the above theorem mlar to that of Theorem.; t omtted. Remark.39. In paper [3, 5], uffcent condton are found for the extence of oluton of.. It wa proved that the whole plane dvded by a contnuou decreang curve Γ nto two djont connected regon E and N uch that. ha at leat one oluton for λ, λ Γ, ha at leat two oluton for λ, λ E \ Γ, and ha no oluton for λ, λ N. The explct ubregon of E where. ha at leat two oluton and two potve oluton, repectvely. When applyng Theorem.3 to., t how u that. ha at leat three potve oluton under the aumpton λ P m β P m α λ and ome other aumpton. When applyng Theorem.34 to., t how u that. ha at leat three potve oluton under the aumpton λ P m α P m β λ and ome other aumpton. Remark.4. In paper [58], the author tuded the extence of multple potve oluton of. under the aumpton α β for all =,..., m and other aumpton, when we apply Theorem.34 and Theorem.38 to., the aumpton α β for all =,..., m are deleted. So Theorem.34 and Theorem.38 generalze and mprove the theorem n [58]. Remark.4. The extence problem on multple potve oluton of.8 olved n the cae A, B, but uch problem reman unolved n the cae A, B <, A <, B and A <, B <..

47 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS Example Now, we preent three example, whoe three potve oluton can not be obtaned by theorem n known paper, to llutrate the man reult. Example 3.. Conder the boundary-value problem x t ft, xt, x t =, t,, x = 4 x /4, x = 4 x/. 3. Correpondng to.4, one ee that x = x = x, α =, ξ = /4, ξ = /, α = /4, α =, β =, β = /4, A =, B =. Chooe k = /4, then σ = 4, chooe e =, e = 5, c =. Then L = m α m α d β β m α ξ m α d = 99, Q = mn { c c, L W = uch that c e σ If and e k σ k k d > e > e σ > e >, P m α m A Q α α } = ; 99 = 6; E = e L, Q > W. 5 99x, x [, 4], 6 99, x [4, 44], 6 4 f u = 99 x , x [44, 54], 4, x [54, 4], x 64, x 4, = 99. ft, u, v = f u n t u v, t eay to ee that c e > e σ > e σ > e >, m A Q α α, Q > W and A ft, u, v < 99 for all t [, ], u [4, 4], v [, ]; A ft, u, v > 6 for all t [/4, 3/4], u [54, 84], v [, ]; A3 ft, u, v 99 for all t [, ], u [4, 44], v [, ]; then Theorem. mple that 3. ha at leat three decreang and potve oluton x, x, x 3 uch that x < 4, x 3/4 > 54, x 3 > 4, x 3 3/4 < 54.

48 48 Y. LIU EJDE-8/96 Example 3.. Conder the boundary-value problem x t ft, xt, x t =, t,, x = x /4, x = 4 x /4 4 x / Correpondng to.7, one ee that x = x = x, α =, ξ = /4, ξ = /, α = /, α =, β = /4 = β, A =, B = 5, h =. Chooe k = /3, then σ = /3, e =, e = 8, c = 3 and L = W = uch that c e σ If α Q = mn { c L, e P m β P m β P m β P m β k σ k k d > e > e σ > e >, c d P m β ξ = 55 8, } = 48 ; = 43; E = e L A B P m m Q β β β, Q > W. = 3. A ft, u, v < 48 for all t [, ], u [, 3], v [3, 3]; A ft, u, v > 43 for all t [/3, /3], u [8, 7], v [3, 3]; A3 ft, u, v 3 for all t [, ], u [, 6], v [3, 3]; then Theorem.3 mple that 3. ha at leat three potve oluton x, x, x 3 uch that max x t <, t [,] max x 3t >, t [,] mn x t > 8, t [/3,/3] mn x 3t < 8. t [k,k] Example 3.3. Conder the boundary-value problem x t ft, xt, x t =, t,, x = x/4 x/, x = 4 x/4 x/ 8. 4 Correpondng to.8, one ee that x = x = x, ξ = /4, ξ = /, α = /, α = /3, β = /4 = β, A =, B = 8, h P m = 6. I= β

49 EJDE-8/96 SOLUTIONS TO SECOND ORDER BVPS 49 Chooe k = /4, then σ = /4. Chooe e = 5, e = 5, c = 4 and L = m β m β d α α m β ξ m β d = 48 64, W = uch that c e σ If Q = mn { c L, e k σ k k d c P m β P m β } 56 = ; 48 = 8; E = e L > e > e σ > e >, m B Q β β, Q > W. = A ft, u, v < for all t [, ], u [4, 44], v [4, 4]; A ft, u, v > 8 for all t [/4, 3/4], u [54, 44], v [4, 4]; A3 ft, u, v 3 48 for all t [, ], u [4, 4], v [4, 4]; then Theorem.34 mple that 3.3 ha at leat three potve oluton x, x, x 3 uch that max x t < 54, t [,] max x 3t > 54, t [,] mn x t > 54, t [k,k] mn x 3t < 54. t [k,k] Acknowledgement. The author grateful to the anonymou referee for h/her detaled readng and contructve comment whch mprove the preentaton of th artcle. The author alo want to thank the managng edtor of th journal for h uggeton. Reference [] R. Agarwal, D. O Regan, P. Palamde; The generalzed Thoma-Ferm ngular boundaryvalue problem for neutral atom. Math. Method n the Appl. Sc. 96, pp [] D. Anderon, R. Avery, J. Henderon; Corollary to the fve functonal fxed pont theorem. Journal of Nonlnear Stude. 8, pp [3] S. Chen, J. Hu, L. Chen, and C. Wang; Extence reult for n-pont boundary-value problem of econd order ordnary dfferental equaton. J. Comput. Appl. Math. 85, pp [4] W. Cheung, J. Ren; Potve oluton for m-pont boundary-value problem. J. Math. Anal. Appl. 335, pp [5] K. Demlng; Nonlnear Functonal Analy, Sprnger, Berln, Germany, 985. [6] H. Feng, W. Ge; Trple ymmetrc potve oluton for mult-pont boundary-value problem wth one-dmenonal p-laplacan. Math. Comput. Modellng. 478, pp [7] M. Feng, W. Ge; Potve oluton for a cla of m-pont ngular boundary-value problem. Math. Comput. Modellng. 467, pp [8] H. Feng, W. Ge; Extence of three potve oluton for m-pont boundary-value problem wth one-dmenonal p-laplacan. Nonl. Anal. 688, pp 7-6.

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