Singular Third Order Multipoint Boundary Value Problem at Resonance

1. International Research Research and Development (IJTSRD) International Open Access Journal Third-Order Multipoint Boundary Value Problem at Resonance International Journal of Trend in Scientific Scientific (IJTSRD) International Open Access Journal ISSN No: 2456 ISSN No: 2456 - 6470 | www.ijtsrd.com | Volume 6470 | www.ijtsrd.com | Volume - 2 | Issue – 3 Singular Third Order Multipoint Boundary Value Problem G. Pushpalatha Department of Mathematics, Vivekanandha College of Arts and Sciences for Women, Tiruchengode, Namakkal, Tamilnadu, India S. K. Reka Department of Mathematics, Reka Assistant Professor, Department of Mathematics, Vivekanandha College of Arts and Sciences for Women, Tiruchengode, Namakkal, Tamilnadu, India Research Scholar, Department of Mathematics, Vivekanandha College of Arts and Science Women, Tiruchengode, Namakkal, Tamilnadu, India Women, Tiruchengode, Namakkal, Tamilnadu, India Vivekanandha College of Arts and Sciences for ABSTRACT The present paper is particularly exhibits about the derive results of a third-order singular multipoint boundary value problem at resonance using coindence degree arguments. The present paper is particularly exhibits about the Schauder continuation principle. These results correspond to the nonresonance case. The scope of this article is therefore to obtained the survive results (the resonance case) and when Letay-Schauder continuation principle. These results correspond to the nonresonance case. The scope of this article is therefore to obtained when ⋁ ????= 1 ?,??? (the resonance case) and when ? and ℎ have a singularity at ? order singular multipoint boundary value problem at resonance using coindence ??? ? = 1. Keywords: The present paper is particularly exhibits about the derive results of a third-order singular multipoint boundary value problem at resonance using coindence degree arguments. e present paper is particularly exhibits order singular Definition 1 multipoint boundary value problem at resonance Let ? and ? be real Banach spaces. One says that the linear operator ?:??? ? ⊂ ? mapping of index zero if Ker finite dimension, where ?? ? denotes the image of be real Banach spaces. One says that the ? → ? is a Fredholm if Ker ? and ?/?? ? are of denotes the image of ?. INTRODUCTION This paper derive the existence for the third singular multipoint boundary value problem at resonance of the form This paper derive the existence for the third-order multipoint boundary value problem at Note We will require the continuous projections ?:? → ?, such that Im P = Ker L,Ker Q = Im L, We will require the continuous projections ?:? → ? such that Im P = Ker L,Ker Q = Im L, ? =Ker L ⨁ Ker ?, ? ?|??? ? ⋂??? ? : dom L ⋂ isomorphism. ?′′′= ?(?),?(?),?′(?),?′′(?)) + ) + ℎ(?) ? =Im L ⨁ Im Q, ker ?→ Im ? Iis an ?′(0) = 0,?′′(0) = 0, ??? ? ?(1) = ? ???? ??????, ? Definition 2 ?,??? be a Fredholm mapping of index zero and Ω a bounded open subset of U such that dom Let ? be a Fredholm mapping of index zero and bounded open subset of U such that dom ?⋂Ω ≠ ?. The map M: ? → ? on ? ?, if the map ??(Ω ?) is bounded and compact, where one denotes by ??? ? ⋂??? ? the generalized inverse of addition ? is ?-completely continuous if it on every bounded Ω ⊂ ?. Where ? ∶ [0,1] × ℝ?→ ℝ is caratheodory’s function (i.e., for each (?,?) ∈ ℝ? the function measurable on [0,1]; for almost everywhere the function ?(?,.,.) is continuous on ???∈ (0,1),?,? = 1,2,…,? − 3, and 1, where ? and ℎ have singularity at ?=1. is caratheodory’s function the function ?(.,?,?) is for almost everywhere ? ∈ [0,1], is continuous on ℝ?). Let and ⋁ =1. ? is called ?-compact is bounded and ??(? − ?) is compact, where one denotes by ??∶ Im ? → the generalized inverse of ?. In completely continuous if it ?-compact ??? ?,??? ????= In [1] Gupta et al. studied the above equation when and ℎ have no singularity and ⋁ obtained existence of a ??[0,1] solution by utilizing In [1] Gupta et al. studied the above equation when ? ??? ?,??? solution by utilizing ???≠ 1. They ?? @ IJTSRD | Available Online @ www.ijtsrd.com @ IJTSRD | Available Online @ www.ijtsrd.com | Volume – 2 | Issue – 3 | Mar-Apr 2018 Apr 2018 Page: 623

2. International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470 Where Theorem 1 Let ? be a Fredholm operator of index zero and let ? be ?-compact on Ω ?. Assume that the following conditions are satisfied : ??? ? = ?? ∈ ? ∶ ?′(0) = 0,?′′(0) = 0,?(1) ??? (i) ?? ≠ ??? for every (?,?) ∈ [(??? ? ∖ ??? ?) ∩ ??] × (0,1). ?? ∉ ?? ?, for every ? ∈ ??? ? ∩ ??. deg (??|??? ? ∩?? , ? ∩ ??? ?,0) ≠ 0, ? = ? ?????(?) ?,??? (ii) (iii) And ?:? → ? is defined by with ?:? → ? being a continuous projection such that ??? ? = ?? ?. then the equation ?? = ?? has at least one solution in ??? ? ∩ Ω ?. ?? = ? ??,?(?),?′(?),?′′(?)? + ℎ(?). (3) Then boundary value problem (1) can be written as Proof : ?? = ??. We shall make use of the following classical spaces, ?[0,1],??[0,1],??[0,1],??[0,1],??[0,1], Therefore ? = ?. and ?∞[0,1]. Let ??[0,1] denote the space of all absolute continuous functions on [0,1], ???[0,1] = {? ∈ ??[0,1] ∶ ?′′(?) ∈ ??[0,1]},???? ??:?|[?,?]∈ ??[0,1]? for every compact interval [0,?] ⊆ [0,1]. Lemma ?[0,1] = ??? ?,??? If ⋁ then ????= 1 (i) ??? ? = {? ∈ ??? ?:?(?) = ?,? ∈ ℝ,? ∈ [0,1]} ; ?????[0,1) = ??:?|[?,?]∈ ??[0,1]?. ? ∈ ?: ? ?? (ii) ? ? ? ?? ? = ? ??? ?,??? ????∫ ∫ ∫ ?(?)???? = 0 ? ?? ⋁ Let U be the Banach space defined by (iii) ?:??? ? ⊂ ? → ? is a Fredholm operator ?:? → ? can be defined by ?? =?? ℎ? ????? ? ? ?(?)???? ?,??? ?[0,1] ∶ (1 − ??)?(?) ∈ ??[0,1]}, ??? ? = {? ∈ ???? ? ? ? , With the norm ?? ?? ? ? (4) ‖?‖?= ? (1 − ??) ? |?(?)|??. Where Let ? be the Banach space ??? ?? + ??+ ??− ???− ???− 1? ? = {? ∈ ??[0,1):? ∈ ?[0,1],lim?→??(1 − ??)?′′ ??????} , ℎ = ? ???? ?,??? (iv) The linear operator ??:?? ?:→ ??? ? ⋂??? ? can be defined as ? ? With the norm ‖?‖?= max? ‖?‖∞ ,?(1 − ??)?′′(?)?∞?. (1) ? ??= ? ? ? ?(?)?? (5) ? ?? ?? ??????≤ ‖?‖? for all ? ∈ ?. (v) Where ‖?‖∞= sup?∈[?,?]|?(?)| . Proof : We denote the norm in ??[0,1] by ‖.‖?. we define the linear operator ?:??? ? ⊂ ? → ? by (i) It is obvious that ??? ? = {? ∈ ??? ?:?(?) = ?,? ∈ ℝ}. ?? = ?′′′(?), (2) @ IJTSRD | Available Online @ www.ijtsrd.com | Volume – 2 | Issue – 3 | Mar-Apr 2018 Page: 624

3. International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470 We show that ? ? ? ?(?) = ?(0) + ? ? ? ?(?)????, ?? (ii) ?? ? ?? ? = Where ? ∈ ? ? ∈ ?: ? ?? ?. (7) ? ? ? ??? ?,??? ????∫ ∫ ∫ ?(?)???? = 0 ? ?? ⋁ ?′′′(?) = ?(?) (12) (iii) For ? ∈ ?, we define the projection ?? as ?? =?? ℎ? ????? ? ? ?(?)???? ?,??? ? ∈ [0,1], (13) To do this, we consider the problem ??? ? ? ? ?′′′(?) = ?(?) (8) , ?? ?? ? And we show that (5) has a solution ?(?) satisfying ??? Where ?′′(0) = 0, ?′(0) = 0,?(?) = ? ??????????? ??? ?,??? ?? + ??+ ??− ???− ???− 1? ≠ 0. ℎ = ? ???? If and only if ?,??? ??? ? ? ? We show that ?:? → ? is well defined and bounded. ? ????? ? ? ?(?)???? = 0 ?? ?? ?,??? ? ? ??? |??(?)| ≤|??| (9) ?(1 − ?)? |?(?)|?? |ℎ|?|??|???? ?,??? Suppose (3) has a solmution ?(?) satisfying ? ??? ??? 1 ‖?‖?|??| |ℎ|?|??|???? ?,??? = ?′′(0) = 0, ?′(0) = 0,?(?) = ? ??????????? ?,??? ? Then we obtain from (5) that ‖??‖?≤ ?(1 − ?)? |??(?)|?? ? ? ? ? ?(?) = ?(0) + ? ? ? ?(?)????, ?? (10) ? ??? 1 ?? ? ‖?‖?|??|?(1 − ?)? ?? |ℎ|?|??|???? ?,??? ≤ ? And applying the boundary conditions we get ??? 1 ‖?‖?‖??‖? . |ℎ|?|??|???? ?,??? = ??? ? ? ? ? ????? ? ? ?(?)???? ?? ?? ?,??? ? ? In addition it is easily verified that ? ? (11) =? ? ? ?(?)????, ? ?? ?? ??? = ??,? ∈ ?. (14) We therefore conclude that ?:? → ? is a projection. If ? ∈ ?? ?, then from (6) ??(?) = 0. Hence ?? ? ⊆ ??? ?. Let ??= ? − ?? ; that is, ??∈ ??? ?. Then ??? ?,??? Since ⋁ , and using (i) and we get ????= 1 ??? ? ? ? ? ????? ? ? ?(?)???? = 0 ?? ?? ?,??? ? On the other hand if (6) holds, let ??∈ ℝ; then @ IJTSRD | Available Online @ www.ijtsrd.com | Volume – 2 | Issue – 3 | Mar-Apr 2018 Page: 625

4. International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470 ′′= ?(?) (17) ??? ? ? ? (???)?(?) = ??????(?)? ? ????? ? ? ??(?)???? ?? ?? ?,??? ? ??? And for ? ∈ ??? ? ∩ ??? ? we know that ? ? ? ? ? ? = ? ????? ? ? ?(?)???? ?? ?,??? ??? (?? ?) ?(?) = ? ? ? ?′′(?)???? ?? ?? ? ?? ? ? ? ? −1 ℎ? ? ? ? ?(?)???? ?? ?? ?,??? ? = ?(? − ?)?′′ ? ?? (18) ? Here ℎ = 0 we get = ?(?) − ?′(0) ? − ?(0) = ?(?) ??? ? ? ? ? ????? ? ? ??(?)???? ?? ?? ?,??? Since ? ∈ ??? ? ∩ ??? ?,?(0) = 0, and ?? = 0. ? ??? This shows that ??= (?|??? ?∩??? ?)??. ? ? ? = ? ????? ? ? ?(?)???? ?? ?,??? ? ?? ? ?????∞≤ max?∈[?,?]∫ (? − ?)?|?(?)|?? ? (v) ? Therefore ??= ?. ≤ ?(? − ?)?|?(?)|?? Thus ??∈ ?? ? and therefore ??? ? ⊆ ?? ? and hence ? ≤ ‖?‖?. ? = ?? ? + ?? ? = ?? ? + ℝ. We conclude that It follows that since ?? ? ∩ ℝ = {0}, then ? = ?? ? ⨁ ?? ?. ??????≤ ‖?‖?. Therefore, References [1] Gupta C.P, Ntouyas S.K, and Tsamatos P.C, “Solvability of an ?-point boundary value problem for second order ordinary differential equations,” Journal of Mathematical Analysis and Applications, vol. 189, no. 2, pp. 575-584, 1995. dim??? ? = dim?? ? = dimℝ = ????? ?? ? = 1. This implies that ? is Fredholm mapping of index zero. (iv) We define ? ∶ ? → ? by ?? = ?(0), (15) [2] Ma R and O’Regan D, “Solvability of singular second order ?-point boundary value problem,” Journal of Mathematical Analysis and Applications, vol. 301, no. 1, pp. 124-134, 2005. And clearly ? is continuous and linear and ??? = ?(??) = ??(0) = ?(0) = ?? {? ∈ ? ∶ ?(0) = 0}. We now show that the generalized inverse ??= ?? ? → ??? ? ∩ ??? ? of ? is given by and ??? ? = [3] Infante G, and Zima M.A, “Positive solutions of multi-point boundary value problems at resonance,” Nonlinear Analysis: Theory, Methods and Applications, vol. 69, no. 8, pp. 2458-2465, 2008. ? ? ? ??? = ? ? ? ?(?)?? (16) ? ?? ?? For ? ∈ ?? ? we have [4] Kosmatov N, “A singular non-local problem at resonance,” Journal of Mathematical Analysis and Applications, vol. 394, no. 1, pp. 425-431, 2012. @ IJTSRD | Available Online @ www.ijtsrd.com | Volume – 2 | Issue – 3 | Mar-Apr 2018 Page: 626