Applied Mathematical Sciences, Vol. 3, 009, no., 6-66 The Negative Neumann Eigenvalues of Second Order Differential Equation with Two Turning Points A. Neamaty and E. A. Sazgar Department of Mathematics, Faculty of Basic Sciences Mazandaran University, Babolsar, Iran neamaty@math.com Abstract In this work, we have studied the solution of the second order differential equation with two turning points case. It is well known that the solutions of the equation are obtained by the asymptotic solution. The aim of this article is to show the higher order distribution negative eigenvalues of Sturm-Liouville problem with Neumann boundary conditions in two turning points case. Mathematics Subject Classification: 3E05 Keywords: Turning point, Sturm-Liouville problem, Neumann conditions, Eigenvalue Introduction In the case of ordinary differential equations the book by Wasow [6] discusses several aspects with detail and rigour. One of the basic tools in the sturmliouville theory is the idea of asymptotic of solution, for example see []. The start of the systematic investigations of the asymptotic eigenvalues of sturm- Liouville problems has been made in Haupt [3] and Richardson[]. In [5] we obtained the positive asymptotic eigenvalues for sturm-liouville problems with boundary condition w (a) =w (b) = 0 with two turning points. The purpose of this paper is to derive asymptotic eigenvalue for differential equation d w dξ = {u ( ξ )+ϕ(ξ)}w () in which the independent variable ξ range over (a, b), and u is a large parameter and the ϕ(ξ) is continuous function on (a, b). We also saw the asymptotic eigenvalues for equation (), when the boundary conditions are w(a) = w(b) = 0
6 A. Neamaty and E. A. Sazgar with one turning point (see []). In this paper we study the negative asymptotic eigenvalues with Neumann boundary condition w (a) =w (b) = 0 with two turning points case. Relevant properties of parabolic functions In this paper we will consider the equation (). Any particular integral of the equation whereas the general solution of the corresponding homogeneous equation. This section we study the standard form of approximating differential equation d W dx =(l x )W () when l>0, the turning points are at x = ±l and for large positive l when l = μ and x = μy, by [] the solutions of () are in following form k ( μ )W ( μ,μy ) = π μ 6 ( η y ) Bi ( μ 3 η)( + O(u )) (3) k ( μ )W ( μ, μy ) = η π μ 6 ( y ) Ai ( μ 3 η)( + O(u )), () where η = 3 y (τ )dτ 3, k( μ = e πμ +O ( e πμ) and W ( μ,μy ), W ( μ, μy ) are two independent solutions of (). 3 Derivative of solutions For next section, in Neumann boundary condition, we need the derivative of solutions. For this purpose, let θ =( η ) ξ then dθ = dξ θ = (ξ ) η 5 (ξ ) 5 so the derivative of solutions (3) and () are in form of W( u, ξ u) ξ W( u, ξ u) ξ = k ( u)u π (θ B i + θb i ), (5) = k ( u)u π (θ A i + θa i). We consider differential equation (). By [] for ξ>0 have two solutions in following form w (u, ξ) =k u ( )W ( u, ξ u)+ɛ, w (u, ξ) =k u ( )W ( u, ξ u)+ɛ, (6)
Differential equation with two turning points 63 where ɛ and ɛ are error terms. So the derivative of this solution are w ξ = u π (θ B i + θb i)+ɛ, w ξ = u π (θ A i + θa i)+ɛ, (7) Lemma The asymptotic form of A i (u 3 η), A i(u 3 η), B i (u 3 η), B i(u 3 η) are given by (u,ξ >0),α= uη 3 3 b π A i ( u 3 ηb ) M cosα + M sinα, A i( u π u 3 ηb ) u b B i ( u 3 ηb ) M cosα M sinα, B i( u π u 3 ηb ) u b b π b π {M3 sinα M cosα}, (8) {M3 cosα + M sinα}. For ξ<0, a<, β = 3 uη 3 a π the Airy function are given by, A i ( u 3 η a ) M 5cosβ + M 6 sinβ, A i( u π u 3 η a ) u a B i ( u 3 η a ) M 6cosβ M 5 sinβ,b i( u π u 3 η a ) u a where the M i are given by aπ aπ (9) {M7 sinβ M 8 cosβ}, (0) {M7 cosβ + M 8 sinβ}, M = s=0 ( ) s u s ( 3, M = 3 uη b s=0 ( ) s u s+ ) s M 3 = s=0 ( ) s V s ( 3, M = 3 uη b s=0 ( ) s V s+ ) s ( 3, 3 uη b ) s+ M 5 = s=0 ( ) s u s ( 3, M 6 = 3 uη a s=0 ( ) s u s+ ) s ( 3, 3 uη a ) s+ M 7 = s=0 ( ) s V s ( 3, M 8 = 3 uη a s=0 ( ) s V s+ ) s ( 3. 3 uη a ) s+ We suppose ( 3 uη 3 b ) s+, () ψ = M sinα + M cosα, ψ = M 3 cosα + M sinα, ψ = α (b)( M cosα M sinα), ψ = α (b)( M 3 sinα+m cosα) ψ 3 = M 6 cosβ M 5 sinβ, ψ = M 5 cosβ + M 6 sinβ, ψ 3 = β (a)( M 5 cosβ M 6 sinβ), ψ = β (a)( M 5 sinβ+m 6 cosβ), θ =(b ), γ =(a ). For ξ< the solutions of equation () are in form of w (u, a) = π u (a ) η a π u a ( M 5 sin β + M 6 cos β),
6 A. Neamaty and E. A. Sazgar similarly w (u, a) = 3 u (a ) (M6 cos β M 5 sin β), () w (u, a) = 3 u (a ) (M5 cos β + M 6 sin β). (3) The derivative of w (u, a) and w (u, a) are in form of w (u, a) = 3 u (γ ψ 3 + γψ 3 ) w (u, a) = 3 u (γ ψ + γψ ), () where β = 3 u(a )η a = β u, γ =(a ), ψ 3 = β (a) ψ, ψ = β (a) ψ 3. (5) Asymptotic of the eigenvalues Now we will study distribution of eigenvalue of equation () with Neumann condition w (a) =w (b) = 0. The eigenvalue of equation () are the zero of Δ(u) = 0 where Δ(u) is defined as following form therefore we get Δ(u) = w (u, b) w (u, b) w (u, a) w (u, a) w (u, b)w (u, a) w (u, a)w (u, b) =0, (6) Δ(u) =[γ ψ 3 γβ (a)ψ ](θ ψ +θψ ) (γ ψ +γβ (a)ψ 3 )(θ ψ +θψ ) = 0 (7) Δ(u) = [γ (M 6 cosβ M 5 sinβ) γβ (M 5 cosβ + M 6 sinβ)] [ θ (M 3 cosα + M sinα)+θα (b) ( M 3sinα + M cosα) ] [γ (M 5 cosβ + M 6 sinβ)+γβ ( M 5 sinβ + M cosβ)] [θ ( M sinα + M cosα)+θα ( M cosα M sinα)] = 0. Here we have M = M 5 = u 0 + O(u ) and M 3 = u 0 + O(u ) M 6 cos β M 5 sin β = ( ) s u s+ cos β ( ) s u s sin β s=0 ( uη 3 3 a ) s+ s=0 ( uη 3 3 a ) s u u 3 =( +...) cos β (u uη 3 3 a ( uη 0 u +...) sin β = (u 3 3 a ) uη 0 +O(u )) sin β. 3 3 a (8)
Differential equation with two turning points 65 Similarly we will have M 5 cos β M 6 sin β = (u 0 + O(u )) cos β M cos α + M sin α = O(u ) M 3 sin α + M cos α = (v 0 + O(u )) sin α M 5 cos β + M 6 sin β =(u 0 + O(u )) cos β M 5 sin β + M 6 cos β = (u 0 + O(u )) sin β M sin α + M cos α = (u 0 + O(u )) sin α M cos α M sin α = (u 0 + O(u )) cos α. (9) Therefore by substituting (9) in Δ(u) then we will have Δ(u) = [ ( γ sin β γβ u cos β)(u 0 + O(u )) ][ O(u ) θα u(v 0 + O(u )) sin α ] [ (γ cos β γβ u sin β)(u 0 + O(u )) ][ ( θ sin α θα u cos α)(u 0 + O(u )) ] =0 (0) where β = β u and α = α u. By deviation (0) to u then [O(u ) γβ (u 0 + O(u )) cos β][o(u ) θα (v 0 + O(u )) sin α] [O(u ) γβ (u 0 + O(u )) sin β][o(u ) θα (u 0 + O(u )) cos α] = 0 () O(u )+γθβ α (u 0 + O(u ))(v 0 + O(u )) sin α cos β (O(u )+γθβ α (u 0 + O(u ))(u 0 + O(u )) sin β cos α) =0. () By deviation from α β γθ in () then we will have [O(u )+(u 0 +O(u ))(v 0 +O(u )) sin α cos β (u 0 +O(u ))(u 0 +O(u )) sin β cos α =0 For u we have since tan β = β + β3 + β5 3 5 tan α = M 5 M 3 tan α. cot β = u 0 + O(u ) v 0 + O(u ) = M 5 M 3, +... then we see that [ tan β = M ] 5 β + β3 M 3 3 + β5 5 +..., we suppose tan α = x and we get arctan of two side, therefore α = nπ + arctan x = nπ + π x + 3x +... x > 3 5x5
66 A. Neamaty and E. A. Sazgar Hence we get at last 3 uη 3 b π = nπ + 3π ( +O(x ) ), x u n = nπ + 3π η 3 3 b 3 η 3 b x ( +O(x ) ). Theorem Under the conditions of the given Neumann problem, the eigenvalues of Neumann at the origin are References u n = nπ + 3π η 3 3 b 3 η 3 b x ( +O(x ) ). [] Olver.F.W.J., Second-order linear differential equations with two turning points. Phil. Trans. R. Soc. Lon. A.78 37-7(975) [] Jodayree Akbarfam A., Higher-order asymptotice approximatios to the eigenvalues of the Sturm -liouville problems in one turning point case, Bulletin Iranian Math. society, vol. 3, No..(997)37-53. [3] O. Haupt, Uber eine Methode zum Beweise von oszillations theoreme, Mat. Ann., 76 (95) 67-0. [] R.G.D. Richardson, Contributions to the study of oscillation properties of the solutions of linear differential equations of the second order, Amer.J. Math., 0, (98) 83-36. [5] A. Neamaty and A. Sazgar, The Neumann conditions for Sturm-Liouville problems with turning points, Int. J. Contemp. Math. Sciences,(To appear). [6] W. Wosow, Asymptotic expansions for ordinary differential equations,interscience Publishers, New York (965). Received: September 6, 008