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1 J. Math. Anal. Appl. 371 (010) Contents lists available at SieneDiret Journal of Mathematial Analysis an Appliations Singular Sturm omparison theorems Dov Aharonov, Uri Elias Department of Mathematis, Tehnion Israel Institute of Tehnology, Haifa 3000, Israel artile info abstrat Artile history: Reeive 8 February 010 Availableonline11June010 Submitte by J.S.W. Wong Keywors: Sturm omparison theorem Sturm s omparison theorem is isusse for ifferential equations whose oeffiients are ontinuous in an open, finite or infinite interval, but the oeffiients annot neessarily be extene ontinuously to the bounary points of the interval. 010 Elsevier In. All rights reserve. Sturm s omparison theorem is formulate usually as follows: Consier the two ifferential equations u p(x)u = 0, v P(x)v = 0, (1) () where P(x), p(x) are two ontinuous funtions in an interval [a, b], P(x) p(x) but P(x) p(x) there. If x 1, x are two zeros of a solution u of (1) then every solution v of () has at least one zero in (x 1, x ). This lassial theorem, whih was oneive in 1836, ontinues to raw attention. See, for example, [1]. The present work attempts to show one more point of view. What an be sai when the funtions P(x), p(x) are ontinuous only in an open interval but are unboune as its enpoints are approahe an the zeros of the solution u are loate at the singular enpoints? Suh is, for example, the ase for equation y 1 (1 x ) y = 0 (3) an its solution y(x) = (1 x ) 1/, whih playe a eisive role in the evelopment of the theory of isonjugate ifferential equations an univalent analyti funtions [5]. Quiet surprisingly the above formulation of Sturm s theorem is not neessarily vali, even if P(x)>p(x) an the ifferene between P(x) an p(x) grows near the bounary points. In [6, p. 703], Nehari applies Sturm s omparison theorem to Eq. (3) without giving any hint why it is permissible. The present artile is a result of our efforts to fill this gap. Several results about singular equations are summarize in [7, Chapter 1.4]. The aim of this isussion is to hek uner whih other singular situations the Sturm theorem ontinues to hol an to emonstrate by ounter-examples when it annot be extene. The following formulation treats two singular ases: a finite interval an unboune oeffiients an an infinite interval. * Corresponing author. aresses: ova@tx.tehnion.a.il (D. Aharonov), elias@tx.tehnion.a.il (U. Elias) X/$ see front matter 010 Elsevier In. All rights reserve. oi: /j.jmaa
2 760 D. Aharonov, U. Elias / J. Math. Anal. Appl. 371 (010) Theorem 1 (Singular Sturm theorem). Let P(x), p(x) be ontinuous funtions on the open, finite or infinite interval (a, b) (but not neessarily at its enpoints),anp(x) p(x), P(x) p(x) on (a, b). (i) Suppose that the ifferential equation u p(x)u = 0, a < x < b, has a solution u whih satisfies the bounary onitions a u (x) =, b Then every solution of the equation u (x) =. v P(x)v = 0, a < x < b, has a zero in (a, b). (ii) In partiular, if (a, b) is a finite interval an a positive solution u satisfies (4) (5) (6) u(x) M(x a) λ 1 near x = a, u(x) M(b x) λ near x = b, with λ 1,λ 1, then every solution of Eq. (6) has a zero in (a, b). Ifeither0 <λ 1 < 1 or if 0 <λ < 1, then the above laim is not neessarily true. In the lassial ase, when p(x) is ontinuous in the lose interval [a, b], the zeros of solutions are simple an are inlue in the ase λ 1 = λ = 1. (iii) If (a, b) = (0, ) an the ifferential equation u p(x)u = 0, 0 < x <, has a solution u whih is positive in (0, ) an satisfies u(x) Mx λ 1 near x = 0, u(x) Mx λ near x =, with λ 1 λ 1, then every solution of the equation (7) v P(x)v = 0, 0 < x <, (8) has a zero in (0, ).If0 <λ 1 < 1 or λ > 1, then this laim is not neessarily true. Proof. First let us reall that a solution u(x) whih satisfies b u (x) = is alle a prinipal solution at x = b. See [3, Chapter XI, Setion 6]. Thus (i) states that Sturm s theorem hols if the same solution u is a prinipal solution at both enpoints of the interval. (i) Suppose for a moment that u(x)>0in(a, b). Wetakeapoint in (a, b) so that P(x) p(x) both in (a, ) an in (, b) an onsier the solution v of (6) whih has at the same initial values as u has: v() = u()>0, v () = u (). It will be prove that in ase (i) this solution v(x) has a zero in (a, ) an a zero in (, b), i.e., at least two zeros in (a, b). Suppose, on the ontrary, that v(x) 0in(, b), an in fat, ue to the initial onitions, v(x)>0 there. From the ientity (vu uv ) = vu uv = (P p)uv it follows that ( vu uv ) x x = (P p)uv. By the initial value onitions (vu uv )() = 0. Sine u, v > 0 an sine P p, P(x) p(x) in [, b), the integral on the right-han sie inreases an is positive for < x < b. So for some suitable value, < < b, there exists a positive lower boun (vu uv ) (x) C > 0, x < b. (9)
3 D. Aharonov, U. Elias / J. Math. Anal. Appl. 371 (010) Sine u 0in[, b), ( v ) = vu uv C > 0, u u x < b. u Integration on [, x] yiels v(x) u(x) v() u() an by (5), [ lim x b v(x) u(x) ] =, C u (x) (10) ontraiting the assumption that u, v > 0on[, b). Thusv must vanish at least one in (, b). The proof that v(x) has another zero in (a, ) is analogous. One we know that a ertain solution v(x) of (6) has two zeros in (a, b), it follows by the lassial separation theorem that every solution of (6) has at least one zero there. Now suppose that u(x) vanishes at one point x = x 1 in (a, b). Then its zero at x 1 is a simple zero an so 1 /u =. Hene, we may repeat the above argument either for the interval (a, x 1 ) or for (x 1, b). Finally, if u(x) has two or more (even infinitely many) zeros in (a, b), we take two onseutive zeros x 1, x in (a, b) an apply the above argument (or Sturm s lassial result) for the interval [x 1, x ]. Note that this possibility annot be overlooke. Inee, as the referee pointe out, a prinipal solution may have infinitely many zeros. For example, the equation y ( 1 λ )y = 0, λ>1/4, 0 < x < 1, 4x x ln x has the solution y(x) = (x ln(1/x)) 1/ os( λ 1/4ln(ln(1/x))) whih is both prinipal an osillatory near x = 0 an also near x = 1. After the ompletion of this manusript we beame aware of []. In fat, our result an be eue also from the Relative onvexity lemma an Theorem 3 of []. However, our proof is more iret an entirely ifferent. (ii) By the assumption of (ii), u (x) K (b x) λ, x < b, where the onstant K is etermine by the ombination of two restritions: the inequality u(x) M(b x) λ neighborhoo of x = b an the bouneness of u away from x = b. Sine in ase (ii) we have λ 1, in some left u (x) 1 K (b x) λ (11) as x b, an the onlusion follows from (i). Fortunately, Nehari s equation (3) belongs to this lass with λ 1 = λ = 1. A onrete ounter-example for the ases 0 <λ 1 < 1 or 0 <λ < 1 is onstrute as follows: u(x) = (1 x)α (1 x) β is a positive solution of the ifferential equation u p αβ (x)u = 0, 1 < x < 1, with p αβ (x) = α(1 α) β(1 β) (1 x) (1 x) αβ 1 x. If 0 < α,β < 1, let us hoose A, B suh that 0 < α < A < 1,0<β<B < 1.Sineα(1 α)<a(1 A), et.,wehave p AB (x)>p αβ (x)>0 for 1 < x < 1. Nevertheless, the ifferential equation v p AB (x)v = 0 has the solution v(x) = (1 x) A (1 x) B whih has no zero in the interval ( 1, 1). Hene the Sturm omparison theorem annot be extene to this pair of singular equations. If only 0 < α < 1 while 1 β<1, the hoie 0 < α < A < 1, B = β leas to a similar ounter-example. (iii) When (a, b) = (0, ) an λ 1, inequality (11) is replae by u (x) 1 K x λ as x, an the onlusion follows. Note that in the ase b =, u(x) nee not ten to zero as x an the singular version of Sturm s theorem may be vali without two onseutive zeros. (1)
4 76 D. Aharonov, U. Elias / J. Math. Anal. Appl. 371 (010) Fig. 1. For ounter-examples to ase (iii) we utilize u(x) = x α (1 x) β α. This funtion is a positive solution of the ifferential equation u p αβ (x)u = 0, 0 < x <, with p αb (x) = β(1 β)x α(1 β)x α(1 α) x (1 x) (13) an it satisfies 0 < u(x) Mx α near x = 0,0< u(x) Mx β near x =. If 0 < α < 1 an 0 β 1, we hoose A, B suh that α < A < 1, B = β. Then p AB (x)>p αβ (x)>0 for 0< x <, but the ifferential equation v p AB (x)v = 0 has the solution v(x) = x A (1 x) B A, whih has no zero in (0, ). Hene the Sturm omparison theorem annot be extene to this pair of singular equations. If 0 <β< 1 while 0 α 1, the hoie A = α, β<b < 1 leas to a similar ounter-example. Example. u(x) = (1 x ) λ is a solution of the equation [ ] u 4λ(1 λ) λ(λ 1) (1 x ) 1 x u = 0, while v(x) = (1 x ) λ C (λ 1/) n [ v 4λ(1 λ) (x) (where C (μ) n (λ n)(λ 1 n) (1 x ) 1 x (x) enote the Gegenbauer polynomials) satisfies the equation ] v = 0. It is known that for λ 1 = μ > 1 (λ 1/), i.e., for λ>0, the Gegenbauer polynomials C n (x) are orthogonal in ( 1, 1) an all their n zeros are in ( 1, 1). See [4, p. 41]. Hene for these two equations the onsequenes of Sturm s theorem hol when λ>0. Thus the onitions λ 1,λ 1 of (ii) are suffiient but not neessary. The utilization of the initial values (9) enables another twist in the formulation of Sturm s theorem even in the lassial ase of ontinuous oeffiients. The initial value onitions (9) mean that the graphs of u(x) an v(x) are tangent. The following ontains an aitional geometri variant of Sturm s theorem. Theorem. Let P(x), p(x) be ontinuous, P(x)>p(x) on an interval [a, b] an let u(x) be a solution of (4) with two zeros, x 1,x. Then every solution v(x) of (6) whih is tangent to the solution u(x) of (4) at some point x =, x 1 < < x, has at least two zeros in (x 1, x ), one on eah sie of the tangeny point x =. Between an the two ajaent zeros z 1,z of v(x) on the two sies of, the graph of v(x) lies stritly on one sie of the graph of u(x) (Fig. 1). Proof. Suppose, for simpliity, that u(x)>0forx 1 < x < x. In aition to (9), v() = u()>0, we have, by P(x)>p(x), that v () = u (), v () = P()v()< p()u() = u (). Hene 0 < v(x)<u(x) on some small intervals on both sies of x =. It is known from Theorem 1(i) that v(x) has a zero in (x 1, ) an a zero in (, x ). Suppose that v(x) meets u(x) again at some point x =, < < x, without vanishing first, i.e., 0 < v(x)<u(x), < x <. Asx from the left sie, v(x) approahes u(x) from below, so the slope u ( ) annot be bigger than the slope v ( ). Consequently v( ) = u( )>0, v ( ) u ( ). (14)
5 D. Aharonov, U. Elias / J. Math. Anal. Appl. 371 (010) But then the ientity vu uv = (P p)uv together with the initial values (9), (14), yiel 0 u( ) [ u ( ) v ( ) ] 0 = (P p)uv> 0, a ontraition. So the graph of v(x) must be below the graph of u(x) for (z 1, z ),asinfig.1. An analogous laim may be formulate also for a singular equation on an open interval (a, b). Referenes [1] W.O. Amreim, A.M. Hinz, D.B. Pearson (Es.), Sturm Liouville Theory, Birkhäuser Verlag, Basel, 005. [] M. Chuaqui, P. Duren, B. Osgoo, D. Stowe, Osillation of solutions of linear ifferential equations, Bull. Austral. Math. So. 79 (009) [3] P. Hartman, Orinary Differential Equations, John Wiley an Sons, NY, [4] H. Hohstat, The Funtions of Mathematial Physis, Wiley Intersiene, NY, [5] Z. Nehari, The Shwarzian erivative an shliht funtions, Bull. Amer. Math. So. 55 (1949) [6] Z. Nehari, Some riteria of univalene, Pro. Amer. Math. So. 5 (1954) [7] C.A. Swanson, Comparison an Osillation Theory of Linear Differential Equations, Aaemi Press, NY, 1968.
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