SOME ESTIMATES FOR THE MINIMAL EIGENVALUE OF THE STURM-LIOUVILLE PROBLEM WITH THIRD-TYPE BOUNDARY CONDITIONS

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1 MATHEMATICA BOHEMICA No. 4, SOME ESTIMATES FOR THE MINIMAL EIGENVALUE OF THE STURM-LIOUVILLE PROBLEM WITH THIRD-TYPE BOUNDARY CONDITIONS Elena Karulina, Moskva Received October 15, 29 Abstract. We consider the Sturm-Liouville problem with symmetric boundary conditions andanintegralcondition.weestimatethefirsteigenvalue λ 1 ofthisproblemfordifferent values of the parameters. Keywords: Sturm-Liouville problem, minimal eigenvalue MSC21:34B24,34L15 Consider the Sturm-Liouville problem 1. Introduction y x qxyx + λyx =, { y k 2 y =, y 1 + k 2 y1 =, where qx is a non-negative bounded summable function on [, 1] such that 1.3 q γ xdx = 1, γ. By A γ wedenotethesetofallsuchfunctions. Afunction yxiscalledasolutionofproblem ifitisdefinedon [, 1], satisfiesconditions 1.2,itsderivative y xisabsolutelycontinuous,andequation 1.1 holds almost everywhere on, 1. The author was partially supported by the Russian Foundation for Basic Researches Grant

2 Weestimatethefirsteigenvalue λ 1 qofthisproblemfordifferentvaluesof γ and k. Accordingtothevariationprinciple λ 1 q = inf Rq, y,where yx H 1,1\{} 1.4 Rq, y = y 2 xdx + qxy2 xdx + k 2 y 2 + y 2 1. y2 xdx Put m γ = inf λ 1 q, M γ = sup λ 1 q. qx A γ qx A γ Remark. Theproblemfortheequation y + λqxy =, qx A γ, with conditions y = y1 = wasconsideredin[1]. Theproblemforequation1.1, qx A γ,withconditions y = y1 = wasconsideredin[2],[3]. In[4]the problemfortheequation y + λqxy =, qx A γ,withconditions1.2was considered. 2. Results Theorem If γ,, 1,then M γ = +. 2 If γ 1,then M γ π 2 + 2; 3 if γ 1and k =,then M γ = 1. 4 If γ = 1and k,then M 1 =,where isthesolutiontotheequation arctan k2 = 1 2 ; M 1 1; 1 2 π π π 2 + 4forall k. Theorem If k =, γ > 1,then m γ = ; 2 if k =, γ 1,then m γ 1/4. 3 If < k 2 < 1 + 3/2,then m γ k 2 /2k 2 + 2forall γ ; 4 if k 2 [ 1 + 3/2; π/2,then m γ > k 4 forall γ ; 5 if k 2 = π/2,then m γ π 2 /4forall γ ; 6 if k 2 > π/2,then m γ > π 2 /4forall γ. 378

3 3. Proofs Proposition.If γ 1,then M γ 1 + 2k 2. Proof. Put y 1 x = ε,thenforany q A γ wehave λ 1 q = inf Rq, y Rq, y 1 yx H 1,1\{} = y 12 dx + qxy2 1 dx + k2 y1 2 + y y2 1 dx = ε2 qxdx + 2k2 ε 2 ε 2 = qxdx + 2k 2. If γ = 1,then qxdx = 1.For γ > 1,usingtheHölderinequality,weobtain qxdx Hence λ 1 q 1 + 2k 2,anditfollowsthat M γ = 1/γ 1 1/γ q γ xdx 1 dx γ/γ 1 = 1. sup λ 1 q sup 1 + 2k 2 = 1 + 2k 2. qx A γ qx A γ Proposition.If γ 1and k =,then M γ = 1. Proof. If qx 1,thenproblem hastheform y y + λy =, y = y 1 =. Notethat λ = 1isaneigenvalueofthisproblem. For λ < 1thesolutionto equation3.1is y = C 1 cosh 1 λx +C 2 sinh 1 λx.undercondition3.2 wehave C 2 =,and C 1 = or λ = 1.Thismeansthatproblem hasno eigenvalues λ < 1.So λ 1 = 1istheminimaleigenvalueofproblem with qx 1and k =. Itnowfollowsthat M γ = sup qx A γ λ 1 q 1. For γ 1wealreadygotthat M γ 1 + 2k 2,whichmeans M γ 1for k =. Combiningthese,wehavethe accurateestimate M γ =

4 Proposition. If γ = 1and k,then M 1 =,where isthesolutiontothe equation arctank 2 / = 1/2. Proof. 1.Considerthecontinuousfunction k 2 cos x + sin x, x < τ, y x = k 2 cos τ + sin τ, τ x < 1 τ, k 2 cos 1 x + sin 1 x, 1 τ x 1. If τ = 1 arctank 2 /,then y xiscontinuoustoo,and y xcanbeasolution to problem Now consider 3.3 Ly = y 2 dx + max x [,1] y2 x + k 2 y 2 + y 2 1. y2 xdx Since we have qxy 2 xdx max x [,1] y2 x λ 1 q = inf Rq, y inf y H 1,1\{} By denotethesolutiontotheequation Ly =. qxdx = max x [,1] y2 x, Ly. y H 1,1\{} Substituting y xinto3.3,weobtain 1 y = y 1 = /k 2, y x = + k 4 /k 2 for τ x < 1 τ; 2 since y xisincreasingfor x [, τ]anddecreasingfor x [1 τ, 1],wehave max x [,1] y2 x = + k4 /k 4 ; 3 38 y x 2 dx τ = + k 2 sin x + cos x 2 dx 1 τ k 2 sin 1 x cos 2 1 x dx

5 τ 2 1 cos2 x = 2 k cos2 x 2 2 k 2 sin2 x = 2 k 4 x sin2 x 2 τ + x + sin2 x 2 τ + k 2 cos2 x = 2 k 4 τ k2 + k 4 + τ + k2 + k 4 = 1 arctan k2 2 k 4 + k 2 ; + k 2 k 4 + k 4 1 τ dx 4 yxdx 2 = τ + = k 4 + k 2 cos x + sin 2 τ + k 4 x dx + τ k 4 dx 1 τ k 2 cos 1 x + sin 2 1 x dx x + sin2 x 2 τ + x sin2 x 2 τ 1 k 2 cos2 x τ 1 k τ = 1 arctan k2 k k 2 + k Finally,wehavethat isasolutiontotheequation arctank 2 / = 1 2 1/. Put t = > andconsidertheequation arctank 2 /t = 1 2 t2 1/tfor t, +. Thefunction arctank 2 /tisdecreasingfor t >,tendsto π/2as t +,to as t + seefig.1. Thefunction 1 2 t2 1/tisincreasingfor t >,tendsto as t +,to + as t +,isequalto for t = 1. Itfollowsthatthis equationhasauniquepositivesolution t,and t > 1. π/2 1 t Figure

6 Besides,thoughthesolutiondependson k 2,itispossibletoindicatetheinterval which t belongsto,wheretheboundsdonotdependon k 2,andtoestimate t on thesebounds.accordingtothebehaviourof arctank 2 /t,weget: 1 if k 2,then t 1 + ; 2 if k 2 +,then arctank 2 /t π/2,and t tendstothepositivesolutionof theequation 1 2 t2 1/t = π/2,whichmeans t π + π 2 + 4/2; 3 t 1, π + π 2 + 4/2forall k. For = t 2 weobtain: 1 if k 2,then 1 + ; 2 if k 2 +,then 1 2 π π π 2 + 4; 3 1, 1 2 π π π 2 + 4forall k. 3.Consider y x = y x.thisfunctionisasolutiontotheproblems y + λy =, y k 2 y = for x < τ, y y + λy =, for τ x < 1 τ, y + λy =, y 1 + k 2 y1 = for 1 τ x 1 where λ =.Itfollowsthat y xisasolutiontoproblem ,where, x < τ, qx = q x =, τ x < 1 τ,, 1 τ x < 1 notethat q xsatisfiescondition1.3. Since y x > on, 1,itisthefirst eigenfunctionofproblem ,and isthefirsteigenvalueofthisproblem. Finally, the following conditions hold: = λ 1 q M 1 = sup q A γ inf Rq, y inf y H 1,1\{} Ly Ly =. y H 1,1\{} Therefore M 1 =. Proposition.If k =, γ > 1,then m γ =. Proof. Substituting k = in1.2,wehave y = y 1 = ;similarly,from 1.4weget Rq, y = y 2 xdx + qxy2 xdx. y2 xdx Put y 1 = 1, q ε x = { ε 1/γ, < x < ε,, ε < x <

7 Then,since γ > 1,wehave m γ = inf q A γ inf Rq, y Rq ε, y 1 = ε 1 1/γ as ε. y H 1,1\{} Thusweconcludethat m γ =. Proposition.If k =, γ 1,then m γ 1/4. Proof. Put = {yx: yx H 1, 1 \ {}, y2 xdx = 1, yx }. Notethat λ 1 = inf Rq, y = inf Rq, y. y H 1,1\{} y Put α = y 2 xdx, β = min y = y,where [, 1]. y [,1] Using yx = y + x y sdsandthehölderinequality,weobtain x 2 y 2 x 2β y sds 2β x y 2 sds 2β 2 + 2α. For yx weget 2β 2 + 2α 1. Iffollowsthatoneofthefollowingcasestakes place:a 2α 1/2;b 2β 2 1/2. asuppose α 1/4.Hencefor yx and qx A γ weget Rq, y = α + qxy2 dx bsuppose β 1/2. Since yx β forall yx [, 1],for yx and qx A γ weget Rq, y = y 2 xdx + qxy2 dx 1 qxy 2 dx 1 4 qx dx. and Using the Hölder inequality, we have 1 = q γ/γ 1 q γ/1 γ dx q γ xdx whence qxdx 1. γ/γ 1 1/1 γ qx dx q γ dx γ/γ 1 = qxdx for γ <, γ 1 γ dx qx dx 1 1/1 γ for γ, 1], 383

8 Hence, Rq, y 1/4inbothcases,and m γ = inf q A γ inf Rq, y = inf y H 1,1\{} q A γ inf Rq, y 1 y 4. References [1] Yu. Egorov, V. Kondratiev: On Spectral Theory of Elliptic Operators. Birkhäuser, Basel, zbl [2] O. V. Muryshkina: On estimates for the first eigenvalue of the Sturm-Liouville problem with symmetric boundary conditions. Vestnik Molodyh Uchenyh Series: Applied Mathematics and Mechanics. 1 25, [3] V.A.Vinokurov, V.A.Sadovnichii:Ontherangeofvariationofaneigenvaluewhen the potential is varied. Dokl. Math. 6823, ; Translation from Dokl. Akad. Nauk, Ross. Akad. Nauk 39223, zbl [4] S. S. Ezhak: On the estimates for the minimum eigenvalue of the Sturm-Liouville problem with integral condition. J. Math. Sci., New York 14527, In English.; Translation from Sovrem. Mat. Prilozh. 3625, zbl Author s address: Elena Karulina, Moscow State University of Economics, Statistics and Informatics, Moskva, 11951, Russia, KarulinaES@yandex.ru. 384