International Journal of Trend in Scientific Research and Development (IJTSRD) International Open Access Journal ISSN No: 2456 - 6470 | www.ijtsrd.com | Volume - 2 | Issue – 3 Singular Third Third-Order Order Multipoint Boundary Value Problem at Resonance G. Pushpalatha Assistant Professor, Department of Mathematics, Vivekanandha College of Arts and Sciences for Women, Tiruchengode, Namakkal, Tamilnadu, India S. K. Reka Research Scholar, Department of Mathematics, Vivekanandha College of Arts and Sciences Science for Women, Tiruchengode, Namakkal, Tamilnadu, India ABSTRACT The present paper is particularly exhibits about the derive results of a third-order order singular multipoint boundary value problem at resonance using coindence degree arguments. Keywords: Thee present paper is particularly exhibits about the derive results of a third-order order singular multipoint boundary value problem at resonance using coindence degree arguments. INTRODUCTION This paper derive the existence for the third third-order singular multipoint boundary value problem at resonance of the form π’′′′ = π(π‘), π’(π‘), π’′ (π‘), π’′′ (π‘))) + β(π‘) π’′ (0) = 0, π’′′ (0) = 0, π’(1) = ππ π’ π , , Where π βΆ [0,1] × β → β is caratheodory’s function (i.e., for each (π’, π£) ∈ β the function π(. , π’, π£) is measurable on [0,1]; for almost everywhere π‘ ∈ [0,1], the function π(π‘, . , . ) is continuous on β ). Let π ∈ (0,1), π, π = 1,2, … , π − 3, and β , π π = 1, where π and β have singularity at π‘=1. =1. In [1] Gupta et al. studied the above equation when π and β have no singularity and β , π π ≠ 1. They obtained existence of a πΆ [0,1] solution by utilizing Letay-Schauder Schauder continuation principle. These results correspond to the nonresonance case. The scope of this article is therefore to obtained the survive results when β , π π = 1 (the resonance case) and when π and β have a singularity at π‘ = 1. Definition 1 Let π and π be real Banach spaces. One says that the linear operator πΏ: πππ πΏ ⊂ π → π is a Fredholm mapping of index zero if Ker πΏ and π/πΌπ πΏ are of finite dimension, where πΌπ πΏ denotes the image of πΏ. Note We will require the continuous projections π: π → π, π: π → π such that Im P = Ker L,Ker Q = Im L, π =Ker L β¨ Ker π, π =Im L β¨ Im Q, πΏ| : dom L β ker π → Im πΏ Iis an β isomorphism. Definition 2 Let πΏ be a Fredholm mapping of index zero and Ω a bounded open subset of U such that dom πΏβΩ ≠ π. The map M: π → π is called π³-compact on π, if the map ππ(Ω) is bounded and π (πΌ − π) is compact, where one denotes by π βΆ Im πΏ → πππ πΏ β πΎππ π the generalized inverse of πΏ. In addition π is πΏ-completely completely continuous if it πΏ-compact on every bounded Ω ⊂ π. @ IJTSRD | Available Online @ www.ijtsrd.com | Volume – 2 | Issue – 3 | Mar-Apr Apr 2018 Page: 623 International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470 Where Theorem 1 Let πΏ be a Fredholm operator of index zero and let π be πΏ-compact on Ω. Assume that the following conditions are satisfied : (i) (ii) (iii) πΏπ’ ≠ π ππ’ for every (π’, π ) ∈ [(πππ πΏ β πΎππ πΏ) ∩ ππ ] × (0,1). ππ’ ∉ πΌπ πΏ, for every π’ ∈ πΎππ πΏ ∩ ππ . deg (ππ| ∩ , π ∩ πΎππ πΏ, 0) ≠ 0, πππ πΏ = π’ ∈ π βΆ π’′ (0) = 0, π’′′ (0) = 0, π’(1) = π π π’(π) , And π: π → π is defined by with π: π → π being a continuous projection such that πΎππ π = πΌπ πΏ. then the equation πΏπ’ = ππ’ has at least one solution in πππ πΏ ∩ Ω. πΏπ’ = π π‘, π’(π‘), π’′ (π‘), π’′′ (π‘) + β(π‘). (3) Then boundary value problem (1) can be written as Proof : We shall make use of the following classical spaces, πΆ[0,1], πΆ [0,1], πΆ [0,1], πΏ [0,1], πΏ [0,1], ∞ [0,1]. and πΏ Let π΄πΆ[0,1] denote the space of all absolute continuous functions on [0,1], π΄πΆ [0,1] = {π’ ∈ πΆ [0,1] βΆ π’′′ (π‘) ∈ π΄πΆ[0,1]}, πΏ [0,1] = π’: π’|[ , ] ∈ πΏ [0,1] for every compact interval [0, π] ⊆ [0,1]. π΄πΆ [0,1) = π’: π’|[ , ] ∈ π΄πΆ[0,1] . Therefore πΏ = π. Lemma If β , (i) π π = 1 then πΎππ πΏ = {π’ ∈ πππ πΏ: π’(π‘) = π, π ∈ β, π‘ ∈ [0,1]} ; π£ ∈ π§: πΌπ πΏ = β π π ∫ ∫ ∫ π£(π)ππππ = 0 , (ii) Let U be the Banach space defined by π = {π’ ∈ πΏ πΏπ’ = ππ’. (iii) πΏ: πππ πΏ ⊂ π → π is a Fredholm operator π: π → π can be defined by [0,1] βΆ (1 − π‘ )π’(π‘) ∈ πΏ [0,1]}, ππ£ = With the norm βπ£β = π β ππ π£(π)ππππ , , (4) (1 − π‘ ) |π£(π‘)|ππ‘. Where Let π be the Banach space π = {π’ ∈ πΆ [0,1): π’ ∈ πΆ[0,1], lim π‘ ) π’′′ ππ₯ππ π‘π } , → β= (1 − , (iv) With the norm βπ’β = max βπ’β∞ , (1 − π‘ )π’′′ (π‘) (1) Where βπ’β∞ = sup ∈[ , ] |π’(π‘)| ∞ . (2) The linear operator π : πΌπ πΏ: → πππ πΏ β πΎππ π can be defined as π = (v) . We denote the norm in πΏ [0,1] by β. β . we define the linear operator πΏ: πππ πΏ ⊂ π → π by πΏπ’ = π’′′′ (π‘), π π π+π +π −π −π −1 π π£ ≤ βπ£β π£(π)ππ (5) for all π£ ∈ π. Proof : (i) It is obvious that πΎππ πΏ = {π’ ∈ πππ πΏ: π’(π‘) = π, π ∈ β}. @ IJTSRD | Available Online @ www.ijtsrd.com | Volume – 2 | Issue – 3 | Mar-Apr 2018 Page: 624 International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470 (ii) π€(π‘) = π€(0) + We show that πΌπ πΏ = π£ ∈ π: π π ∫ ∫ ∫ π£(π)ππππ = 0 . β, Where π£ ∈ π (7) π€ ′′′ (π‘) = π£(π‘) (iii) To do this, we consider the problem π€ ′′′ (π‘) = π£(π‘) ππ£ = And we show that (5) has a solution π€(π‘) satisfying πππ€ ππ (12) For π£ ∈ π, we define the projection ππ£ as (8) π€ ′′ (0) = 0, π€ ′ (0) = 0, π€(π‘) = π£(π)ππππ , π β ππ π£(π)ππππ , , π‘ ∈ [0,1], (13) Where , β= If and only if π π π + π + π − π − π − 1 ≠ 0. , ππ We show that π: π → π is well defined and bounded. π£(π)ππππ = 0 , (9) |ππ£(π‘)| ≤ Suppose (3) has a solmution π€(π‘) satisfying π€ ′′ (0) = 0, π€ ′ (0) = 0, π€(π‘) = |π | |β| = πππ€ ππ 1 |β| , Then we obtain from (5) that π€(π‘) = π€(0) + βππ£β π£(π)ππππ , ≤ (10) 1 |β| , , |π | π βπ£β |π | , (1 − π‘) |ππ£(π‘)|ππ‘ ≤ , = π£(π)ππππ (1 − π ) |π£(π )|ππ |π | π βπ£β |π | (1 − π ) ππ And applying the boundary conditions we get ππ |π | π 1 |β| |π | π βπ£β βπ β . , In addition it is easily verified that = Since β , π£(π)ππππ , (11) π π = 1, and using (i) and we get ππ π£(π)ππππ = 0 π π£ = ππ£, π£ ∈ π. (14) We therefore conclude that π: π → π is a projection. If π£ ∈ πΌπ πΏ, then from (6) ππ£(π‘) = 0. Hence πΌπ πΏ ⊆ πΎππ π. Let π£ = π£ − ππ£ ; that is, π£ ∈ πΎππ π. Then , On the other hand if (6) holds, let π’ ∈ β; then @ IJTSRD | Available Online @ www.ijtsrd.com | Volume – 2 | Issue – 3 | Mar-Apr 2018 Page: 625 International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470 (πΏπ )π£(π‘) = π£ (π)ππππ ππ ′′ π π£ (π‘) = π£(π‘) (17) And for π’ ∈ πππ πΏ ∩ πΎππ π we know that , = ππ π£(π)ππππ − 1 β π’′′ (π)ππππ (π πΏ) π’(π‘) = , π£(π)ππππ (π‘ − π )π’′′ ππ = , (18) Here β = 0 we get = π’(π‘) − π’′ (0) π‘ − π’(0) = π’(π‘) π£ (π)ππππ ππ Since π’ ∈ πππ πΏ ∩ πΎππ π, π’(0) = 0, and ππ’ = 0. , = ππ π£(π)ππππ This shows that π = (πΏ| (v) , π π£ Therefore π£ = π£. ∞ ≤ max ≤ Thus π£ ∈ πΌπ πΏ and therefore πΎππ π ⊆ πΌπ πΏ and hence π = πΌπ πΏ + πΌπ π = πΌπ πΏ + β. ∩ ∈[ , ] ∫ ) . (π‘ − π ) |π£(π )|ππ (π‘ − π ) |π£(π )|ππ ≤ βπ£β . We conclude that It follows that since πΌπ πΏ ∩ β = {0}, then π = πΌπ πΏ β¨ πΌπ π. π π£ ≤ βπ£β . Therefore, References dim πΎππ πΏ = dim πΌπ π = dim β = πππππ πΌπ πΏ = 1. [1] Gupta C.P, Ntouyas S.K, and Tsamatos P.C, “Solvability of an π-point boundary value problem for second order ordinary differential equations,” Journal of Mathematical Analysis and Applications, vol. 189, no. 2, pp. 575-584, 1995. This implies that πΏ is Fredholm mapping of index zero. (iv) We define π βΆ π → π by ππ’ = π’(0), (15) And clearly π is continuous and linear and π π’ = π(ππ’) = ππ’(0) = π’(0) = ππ’ and πΎππ π = {π’ ∈ π βΆ π’(0) = 0}. We now show that the generalized inverse πΎ = πΌπ πΏ → πππ πΏ ∩ πΎππ π of πΏ is given by π π£= For π£ ∈ πΌπ πΏ we have π£(π)ππ (16) [2] Ma R and O’Regan D, “Solvability of singular second order π-point boundary value problem,” Journal of Mathematical Analysis and Applications, vol. 301, no. 1, pp. 124-134, 2005. [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. [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