New Singular Standing Wave Solutions Of The Nonlinear Schrodinger Equation

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1 New Singula Standing Wave Solutions Of The Nonlinea Schodinge Equation W. C. Toy Abstact We pove existence, and asymptotic behavio as, of a family of singula solutions of 1) y + y +y y p 1 y = 0, 0 < <, < p 3, which satisfy lim 0 + y) = and lim y) = 0. We also pove that the limiting solution of this family is the gound state. Solutions of Eq. 1) ae adially symmetic standing waves of the nonlinea Schodinge equation. A cental component of ou investigation is the associated integal equation ) y k ) = k e t sinht )y kt) y k t) p 1 dt, 0 < <. Let < p 3. Fo each k > 0 thee is a unique solution y k ) of ), it satisfies 1), and y k ) k e as. We make use of ) to pove I) the existence of a bounded inteval 0,k ) such that fo each k 0,k ), y k ) < 0 > 0, y) as 0+, and y k ) k e as, II) when k = k, y k ) is a gound state solution satisfying 0 < y k 0) <, y k 0) = 0, y k ) < 0 > 0, and y k) k e as. We also state elated open poblems fo futue eseach. Declaations of inteest: none. AMS subject classifications: 34B40, 34C11, 45G10 Keywods: topological shooting, bound state, singula solution 1

2 1 Intoduction The nonlinea Schodinge NLS) equation iψ t + ψ + ψ ψ p 1 = 0, p > 1 1.1) is a canonical equation which aises in studies of lase beam self-focusing, in plasma physics, and in the fomation of Bose-Einstein condensates [7, 1, 0]. Coz [6] states in the theoy of nonlinea Schodinge equations, eithe solutions spead out o they concentate at one o moe points. In paticula, the fomation of singula solutions, and peak shaped solutions has played a cental ole in undestanding optical selffocusing, and Bose-Einstein condensate fomation. Coz [6] used vaiational techniques to pove existence of of standing wave solutions, i.e. solutions of 1.1) which have the fom ψ = e it φ, whee φ satisfies φ + φ φ p 1 φ = 0. 1.) Subsequently, Byeon and Wang [] investigated standing wave solutions when a singula potential tem is added onto 1.1). Goal. We focus on the ange < p 3, and the physically elevant dimension N = 3, and investigate adially symmetic solutions φ = y), of 1.), whee y satisfies y + y + y y p 1 y = 0, 0 < <. 1.3) Ou pimay goal is to pove existence, and asymptotic behavio as, of a family of singula solutions of 1.3) which satisfy see Figue 1, ight panel) y) > 0 > 0, lim y) = and lim 0 + y) = ) Section 3 descibes futue eseach into the full egime N > 1, 1 < p < N+ N. Pevious Results. Befoe stating ou main esult Theoem 1. below) we summaize esults of pevious studies. Fist, a bound state is a solution of 1.3) that satisfies y0) = y 0 0, y 0) = 0, 1.5) lim y), y )) = 0, 0). 1.6)

3 A gound state satisfies 1.3), 1.5), 1.6), and y) > 0 > 0. Methods used to pove existence of bound states include i) a vaiational appoach and ii) ODE methods. i) The Vaiational Appoach. In 1963 Nehai [16] esticted his attention to the physically elevant dimension N = 3, and used a vaiational method to pove His investigation of 1.3) was motivated by studies of spinos [], nucleons [15], and the nuclea coe [1]. Thus, Nehai investigated the poblem J = subject to the nomalizing condition 0 y t) + y t)) ) dt = minimum, 1.7) 0 y p+1 t) t p 1 dt = 1, 1.) whee y) 0 is continuous, y ) is piecewise continuous, y 0) = 0, and the integals 1.7), 1.) exist. Nehai poved that in this class of functions thee is at least one that minimizes J, and it is a gound state Figue 1). Nehai did not investigate singula solutions. Subsequently, Stauss [19] and Beestycki and Lions [1] used vaiational methods to pove existence of infinitely many adially symmetic solutions. ii) ODE Methods. A standad technique to detemine qualitative behavio of solutions of 1.3) is to investigate initial value poblem 1.3), 1.5). This appoach, combined with enegy estimates and compaison techniques has been used to pove existence and uniqueness of gound states and sign changing bound states [4, 5, 9, 11, 13, 14, 1]. A key obsevation is that the enegy functional satisfies E = y + y p+1 p + 1 y, 1.9) E0) = y 0 p+1 p + 1 y 0, and E = y An impotant consequence of 1.9)-1.10) is the following popety: I) Fo each y 0 0 the solution of 1.3), 1.5) is bounded on [0, ). 0 > ) Statement Of Main Results. Popety I) implies that the existence of unbounded singula solutions satisfying 1.3)-1.4) cannot be poved by analyzing initial value 3

4 poblem 1.3), 1.5). Also, as pointed out above, Nehai did not pove existence of singula solutions. Thus, to pove the existence of solutions of 1.3)-1.4) we employ an entiely diffeent appoach, which we efe to as topological shooting fom infinity. A cental component of this appoach is the associated integal equation y k ) = k e t sinht )y kt) y k t) p 1 dt. 1.11) Lemma 1.1 Let < p 3 be fixed. Fo each k > 0 thee is a unique solution, y k ), of 1.11), it satisfies ODE 1.3), and y k ) k e as. Fo each > 0 whee y k ) continues to exist the solution and its deivatives ae continuous functions of and k. Equation 1.11) can be solved uniquely) by the method of successive appoximations, and this technique gives both the solution and its continuous dependence on k. This technique was used by Hastings and McLeod [9] and [10], pp. 4-49) in poving existence of a unique, monotonically deceasing solution of the Painleve type equation y = xy + y y α, α > 0, 1.1) yx) 1 x ) 1/α as x, and yx) Aix) as x, 1.13) whee Aix) is the Aiy function. Thus, although the method of poof of Lemma 1.1 is standad, we give a bief sketch of the details in the Appendix. We now define the topological shooting set { S = k > 0 if 0 < ˆk < k then y ˆk) < 0 > 0 and lim 0 + yˆk) = Theoem 1. Let < p 3. Thee exists k > 0 such that S = 0, k ). i) Singula Solutions Let k 0, k ). Then } 1.14) y k ) > 0 and y k ) < 0 > 0, 1.15) lim y k) = and y k ) k e 0 + as. 1.16) 4

5 ii) Gound State Let k = k. Thee exists y > 0 such that y k ) > 0 and y k ) < 0 > 0, 1.17) lim 0 +y k ), y k )) = y, 0) and y k ) k e as. 1.1) Remaks. Figue 1 and Figue illustate the stak diffeence between solutions of initial value poblem 1.3), 1.5) and solutions of the equivalent integal equation 1.11). By combining the esults of these divese analytical appoaches, i.e. 1) the investigation of behavio of solutions of initial value poblem 1.3), 1.5) as inceases fom = 0, and ) the investigation of behavio of solutions of integal equation 1.11) as deceases fom lage values, we have obtained new insights into the global stuctue of adially symmetic standing wave solutions of the nonlinea Schodinge equation 1.1). Section gives the poof of Theoem 1.. Section 3 descibes open poblems, and how a natual emegemce phenomenon is obseved when deceases fom lage values. Poof Of Theoem 1.. Thoughout the poof we focus on the lagest non-negative intevals whee solutions of 1.11) ae positive. Thus, let p, 3] be fixed, and fo each k > 0 define a k = inf{ˆ > 0 y k ) > 0 > ˆ}..1) It follows fom.1) that a k, ) is the lagest positive inteval whee y k ) > 0. Next, we give an outline of ou two pat pocedue to complete the poof. I) Fist, we pove thee auxiliay lemmas. a) Lemma.1 gives pactical uppe and lowe bounds on solutions y k of 1.11). b) Lemma. gives a citeion which guaantees that a k = 0, i.e. that a k, ) = 0, ). c) Lemma.3 gives a citeion fo a solution to satisfy lim 0 + y k ) =. 5

6 5.5 y 4.3 N=p=3 y k 4.3 N=p=3 3 k * =5.01 k= k=.1 k= Figue 1: Left) Solutions of initial value poblem 1.3), 1.5) when N = p = 3; y 0 = 3 gives a solution such that lim 0 + y) = 1, y 0 = 4.3 gives the gound state, y 0 = 5.5 gives a solution such that lim 0 + y) = 1. Right) Solutions of integal equation 1.11) when N = p = 3; k =.1, k = 1 give positive singula solutions which satisfy i) in Theoem 1., k = 5 gives the gound state which satisfies ii) in Theoem 1., and k = 10 gives a sign changing singula solution such that lim 0 + y 10 ) =. 6

7 II) We use Lemma.1, Lemma. and Lemma.3 to pove the following fou lemmas. d) Lemma.4 shows that S φ. e) Lemma.5 and Lemma.6 show that S is open. f) Lemma.7 shows that k 0 / S fo aome k 0 > 0. It follows fom Lemma 1.1, the definition of S given in 1.14), and d)-e)-f) above that S is a finite, connected, open inteval of the fom S = 0, k ), and that condition i) in Theoem 1. is satisfied. Finally, to pove that condition ii) holds we show, in Lemma., that y k ) is a gound state solution. Pat I) In Lemma.3 we deive uppe and lowe bounds on y k ) and y k) when a k, ). The poof uses the fact that 1.11) can be witten as y k ) = k e 1 t e t e t )) y k t) y k t) p 1 dt..) Lemma.1 Let < p 3 and k > 0. Then, fo all a k, ), k e 1 kp 1 e p 1) ) p 1) p 1 y k ) k e,.3) k e ) k y ) ke pkp 1 e p 1) p 1) kp 1 e p 1) )..4) p 1 p 1) p Poof. Fist, since sinht ) 0 when t, it follows fom.) that y k ) k e a k, )..5) Next, we substitute y k ) k e into.) and obtain y k ) k e 1 t e t e t )) k pe pt t dt a k, )..6) p Because 1 t 1 when t,.6) educes to y k ) k e kp p e t e t )) e pt dt a k, )..7) 7

8 An evaluation of the ight side of.7) gives y k ) k e 1 kp 1 e p 1) ) p 1) p 1 a k, )..) Next, we pove.4). Diffeentiating both sides of 1.11) with espect to gives y k) = k e ke + fo > a k. Substituting y k ) k e y k ) ke ke + kp cosht ) t + ) sinht ) y k t) y k t) p 1 dt.9) into.9) and using 1 1, we obtain t cosht ) p + fo all a k, ). An evaluation of the ight side of.10) gives y k ) ke Fom.9) we get y k ) k e pkp 1 e p 1) p 1) kp 1 e p 1) p 1 p 1) p ) sinht ) e pt dt.10) p+1 ) a k, )..11) ) ak, ). This completes the poof. In Lemma. we use Lemma.1 to pove a citeion which guantees that a k = 0. Lemma. Let < p 3 and k > 0. If then a k = 0. y k ) > 0 and y k ) < 0 a k, ),.1) Poof. Suppose, howeve, that p, 3] and k > 0 exist such that.1) holds, and a k > 0. Then.1) and.1) imply that y k ) ceases to exist as a + k, hence We conclude fom.1),.3) and.1) that lim y k ) =..13) a + k 0 < y k ) k e a k, )..14) It follows fom.14) that lim a + y k ) k e a k k a k conclude that a k = 0, and the lemma is poved. <, contadictiing.13). Thus, we

9 Next, in Lemma.3 we deive a citeion which guaantees that a positive solution of 1.11) satisfies y k ) as 0 +. The poof makes use of the function which satisfies H k ) = y k ), > a k,.15) H k) = y k) + y k ) and H k) = H k 1 y p 1 k )), > a k..16) Lemma.3 i) Let < p < 3 and k > 0. Assume that ) a k, 3 p p 1 exists such that Then a k = 0, y k ) > 1, y k ) < 0, y k ) > 0 and kp 1 3 p e p 1) p..17) y k ) > 0, y k) < 0 and y k) > 0 0, ), and lim y k) =..1) 0 + ii) Let p = 3 and k > 0. Assume that > 0 exists such that y k ) > 1, y k ) < 0, y k ) > 0, 0 < H k ) < Then a k = 0 and popety.1) holds 3 and H k ) > 0..19) Poof. Let < p 3 and k > 0. A diffeentiation of 1.3) show that y k satisfies y k + y k = p ) p y p 1 k y k + y k, > a k..0) We will also make use of the fact that.0) can be witten as ) y k) = p p y p 1 k y k + y k, > a k..1) Suppose, fo contadiction that thee is an ˆ a k, ) such that y k ) > 0 ˆ, ) and y ˆ) = 0..) k Then y k ˆ) 0..3) 9

10 It follows fom.) that y k ) < 0 and y k) > 1 [ˆ, )..4) To obtain a contadiction to.3) we need an uppe bound on the tem y p 1 k appeas in.0). Fo this we conside two cases, i) < p < 3 and ii) p = 3. i) Let < p < 3. It follows fom.3) that which y p 1 k ) k p 1 3 p e p 1) a k, )..5) It is easily veified that 3 p e p 1)) > 0 when ) a k, 3 p p 1. This popety, ou assumption that ˆ a k, ), assumption.17) and.5) imply that Fom.4) we get y k contadicting.3). ˆ y p 1 k ˆ) ˆ 3 p e p 1)ˆ < p..6) ˆ) < 0. This fact,.0) and.6) imply that y k ˆ) < 0, ii) Let p = 3. It follows fom.16) and.4) that H k = y k satisfies H k ) < 0 [ˆ, ). This popety and hypothesis.19) imply that H k ) > 0 and 0 < H k ) < 3 [ˆ, )..7) Thus, 0 < Hk ˆ) = ˆ yk ˆ) < and we conclude fom.0) and.4) that y 3 k ˆ) < 0, again contadicting.3). Thus, fo each p, 3], we conclude that y k) > 0 fo all a k, ). Fom this it follows that y k ) < 0 and 1 < y k) < k e fo all a k, ). These bounds and Lemma. imply that a k = 0, hence y k ) > 0, y k ) < 0 and 1 < y k) < k e 0, )..) It emains to pove that lim 0 + y k ) =. Suppose, fo contadiction, that p, 3] and M 1 > 1 exist such that 1 < y k ) M 1 0, )..9) Let y k ) = M. Since y k) > 0 0, ), we conclude that y k ) < M 0, ]..30) 10

11 Let 1 0, ] satisfy M + Mp < 0..31) Fom 1.3) and.9) we get y k )) M p 1 0, ]..3) Integating.3) fom to 1, and using.30)-.31), we obtain y k ) 1 y k 1) + Mp ) 1 M Dividing both sides of.33) by, and integating fom to 1 gives y k ) y k 1 ) + M contadicting.9). This completes the poof. 0, 1 ]..33) 1 1 ) > M 1 when 0 < 1,.34) 1 Pat II) In this pat we pove Lemma.4, Lemma.5, Lemma.6 and Lemma.7. Fist, in Lemma.4 we pove that S φ. Lemma.4 i) Let p, 3). Then ii) Let p = 3. Then 0, ) 3 S. 0, p 1 ) ) 3 p p 3 p 1 p 1 e 3 p) S. Poof. i) Let p, 3). The key to the poof is to obtain an uppe bound on the tem y p 1 k which appeas on the ight side of.0). This bound.3) below) allows us to detemine the sign of y ). It follows fom.3) that We need an uppe bound on the function y p 1 k k p 1 3 p e p 1) > a k..35) f) = 3 p e p 1), 0..36) It is easily veified that f attains a positive maximum at p = 3 p, and that p 1 0 f) f p ) ) 3 p 3 p e p ) p 1 11

12 Combining.35),.36) and.37), we obtain if 0 < k < 0 < y p 1 k ) k p 1 3 p e p 1) k p 1 p ) 3 p 3 p e p 3 < p 1 p > a k.3) 1 ) 3 p p 3 p 1 p 1 e 3 p). Next, we conclude fom 1.3),.3),.4) that 0 < y k ) < 1, y k ) < 0 and y k ) > 0 when 1..39) Combining.1) with.3) and.39), we conclude that y k )) < 0 and y k ) > 0 a k, )..40) It follows fom.3),.39),.40) that if 0 < k < p 1 ) 3 p p 3 p 1 p 1 e 3 p), then 0 < y k ) k e, y k ) < 0 and y k ) > 0 a k, )..41) We conclude fom Lemma. and.41) that a k = 0 k 0, ) p 3 3 p e 3 p p p 1 1 p 1..4) Thus, the bounds in.41) hold fo all > 0. It emains to pove that lim y k) = k 0, ) 1 p 3 p 1 3 p e 3 p..43) 0 + p p 1 Suppose, howeve, that k 0, p 1 ) ) 3 p p 3 p 1 p 1 e 3 p) and M 1 > exist such that 0 < y k )) p M 1 > 0..44) The poof of i) is complete if we obtain a contadiction to.44). Fo this we focus on 0, 1] and let L = y k 1). Let 1 0, 1] satisfy L + M < 0..45) It follows fom 1.3), and the fact that y k ) > 0 > 0, that y k ) M 1 0, 1 ]. Integating fom to 1, we obtain y k) 1y k 1 ) + M ) 1y k 1 ) + M , 1 ]..46) 1

13 Next, obseve that y k 1) < L = y k 1) since.41) gives y k ) > 0 fo all > 0. Combining this inequality with.45) and.46), we obtain y k) L 1 0, 1 ]. Integating fom to 1 gives y k ) y k 1 ) + L ) > M 1 ) 1/p when 0 < 1,.47) 1 contadicting.44). ii) Let p = 3 and let k 0, ) 3. When p = 3, equation.1) becomes y k) = 3 ) 3 yk y k + y k, > a k..4) As above, we need to show that the tem 3 yk > 0 a k, ). It follows fom uppe bound.3), and the supposition that k 0, ) 3, that y k < k e < 3 > a k..49) Also, as above, we conclude fom 1.3),.3),.4) that 0 < y k ) < 1, y k ) < 0 and y k ) > 0 when 1..50) Combining.4) with.49) and.50), we conclude that y k )) < 0 and y k ) > 0 a k, )..51) It follows fom.3),.50) and.51) that if k 0, ) 3, then 0 < y k ) k e, y k ) < 0 and y k ) > 0 a k, )..5) We conclude fom Lemma. and.5) that a k = 0. Thus, the bounds in.5) hold fo all > 0. It emains to pove that lim y k) = when k 0 + Suppose, howeve, that k 0, ) 3 and M1 > exist such that 0,..53) 3 0 < y k )) 3 M 1 > 0..54) 13

14 The poof of i) is complete if we obtain a contadiction to.54). Such contadiction is obtained exactly as in pat i), and we omit the details fo the sake of bevity. Ou next goal is to pove that S is open Lemma.6). Fo this we fist need to pove the following auxiliay esult. Lemma.5 i) Let < p < 3 and k S. Then y p 1 k ) 0 as ) ii) Let p = 3 and k S. Then y k ) 0 as ) Poof. i) Let < p < 3 and k S. It follows fom.3) that 0 < y p 1 k ) k p 1 3 p e p 1) 0, )..57) Since < p < 3, we obseve that 3 p e p 1) 0 as 0 +. Combining this popety with.57) gives popety.55). ii) Let p = 3 and k S. Since k S, thee is an > 0 such that y k ) > 1, y k ) < 0 0, ), and lim y k) =..5) 0 + Suppose that.56) doesn t hold. Then eithe L > 0 and L 0, ) exist such that y k ) L 0, L ],.59) o else 0 = lim inf 0 + y k) < lim sup 0 + y k )..60) Fist, assume that.59) holds. Let L 0, ) be chosen so small that < L y k) 0, L ]..61) it follows fom 1.3) and.61) that ) ) y k = y k y3 k y3 k y3 k L3 0, L]..6) 14

15 An integation fom to L gives y k ) L y k L) + L3 ln L) ln)) > 0 when 0 < L,.63) contadictiing.5). Thus,.59) cannot hold. Finally, suppose that.60) holds. Then thee exists > 0 such that y k ) and H k ) = y k ) satisfy y k ) > 1, H k ) = 0 and H k ) 0..64) Howeve, it follows fom the fact that y k ) > 1 and.16) that H k ) = H k )1 )) < 0, contadicting.64). Thus,.60) cannot hold. Theefoe, we conclude y k that popety.56) holds and the poof is complete. We make use of Lemma.5 to pove Lemma.6 Let < p 3. Then S is open. Poof. Let k S. The definition of S implies that > 0 exists such that y k) > 1, y k) < 0 0, ], and lim y k) =..65) 0 + To complete the poof we conside two sepaate cases, i) < p < 3 and ii) p = 3. i) < p < 3. It follows fom citeion.17) in Lemma.3 that we only need to pove y kˆ) > 1, y kˆ) < 0, y k ˆ) > 0 and kp 1ˆ 3 p e p 1)ˆ < p,.66) fo some ˆ ) 0, 3 p p 1 when 0 < k k 1. Thus, let be chosen to also satisfy 0, 3 p ) and p 1 k p 1 3 p e p 1) < p..67) Obseve that 3 p e p 1)) ) > 0 0, 3 p p 1. This,.66) and.67) imply that y p 1 k ) k p 1 p 1 e p 1) < p 0, ]..6) Finally, since y k) as 0 +, thee is an ˆ 0, ) such that y k ˆ) > 0. Othewise, if y k ) < 0 fo all 0, ], then two integations fom to give y k) y k ) + y k ) ) < [0, ],.69) 15

16 contadicting.65). Thus,.66) holds when k = k. It follows fom continuity that.66) holds when 0 < k k 1. Thus, citeion.17) fo blowup of solutions in Lemma.17 is satisfied when 0 k k 1. This completes the poof of i) ii) p = 3. By Lemma.3, we only need to pove that citeion.19) holds fo some > 0 when 0 k k 1. It follows fom.16),.5),.65) that H k = y k satisfies H k) > 0, H k ) < 0 0, ], and lim H k) = 0..70) 0 We conclude fom.70) that can be chosen to satisfy the additional popety 0 < H k ) < 3, and H k) > 0 0, ]..71) Next, we claim that thee is an ˆ 0, ] such that y k ˆ) > 0..7) If not then y k ) < 0 0, ], and two integations give.69), again contadicting.65). Thus,.7) holds at some ˆ 0, ], and we conclude fom.65),.71) that y kˆ) > 1, y kˆ) < 0, 0 < H kˆ) < 3, and H kˆ) > 0..73) Continuity implies that popeties.7) and.73) hold if 0 k k 1. Thus, when p = 3, citeion.19) fo blowup of solutions in Lemma.3 is satisfied when 0 k k 1. This completes the poof of ii). We have now poved that S is non-empty and open. Finally, we pove Lemma.7 Let < p 3. Thee exists k 0 > 0 such that k 0 / sup S. Poof. It follows fom.3) that fo each k 0 > 0, the solution y k0 satisfies e k 0 1 kp 1 0 e p 1) ) e y p 1) p 1 k0 ) k 0 a k0, )..74) To poceed futhe we conside two cases, i) < p < 3 and ii) p = 3. i) Let < p < 3, and define 0 and coesponding k 0 > 0 by 1 0 = ) 1/p 1) 1 + π p ) 1/p 1) 1, k 79 7 p 0 = 0 e 0.75) 1 p 1 16

17 Ou goal is to pove that k 0 / S. Fo this we need uppe and lowe bounds on y k0 0 ), and an uppe bound on y k 0 0 ). Fist, obseve that fo any 0 > 0, 1 kp 1 0 e p 1) p 1) = 1 1 ) 0 p 1) 7 p 1 e 0 ) 0..76) Thus, fo each 0 > 0, if k 0 = ) p 1 1/p 1) 0 e 0 then.74) and.76) give 0 < 7 p 1 ) 1/p 1) 0 ) e 0 y k0 ) p 1 fo all 0. Fom.77) we conclude that 0 > a k0, and ) 1/p 1) 0 ) e 0 < 1.77) p ) 1/p 1) 1 p ) 1/p 1) 1 y k0 0 ) < 1..7) 0 < 7 We also need an uppe bound on y k 0 0 ). Fo this we make use of the functional whee E = E = 1 y k0 ) + 1 p + 1 y k 0 ) p+1 1 y k 0 ),.79) y k0 ) 0 > ak0 and E ) = 0. Thus, E) > 0 0 and 1 y k0 0 ) ) 1 > y k 0 0 )) 1 p + 1 y k 0 0 )) p+1..0) Fom.7) and.0), and using the fact that 1 < p 3, we obtain the uppe bound y k 0 0 ) p ) 1/p 1) 1..1) We conclude fom 1.3),.7),.1) that y k 0 ) = y k 0 )+y k0 ) y k0 y p 1 ) > 0 and y k 0 ) y k 0 0 ) fo < 0 as long as y k0 ) < 1. Integating y k 0 ) y k 0 0 ) fom to 0, and making use of.7) and.1), we conclude that y k0 ) 7 p 1 ) 1/p 1) fo < 0 as long as y k0 ) < 1. Next, obseve that 7 p 1 ) 1/p 1) p ) 1/p 1) 1 0 ),.) p ) 1/p 1) 1 0 ) = 1.3) 17

18 when = 0 + It follows fom.75) and.4) that 7 p = ) 1/p 1) 1 ) p 1 1/p 1)..4) π p 1..5) We conclude fom.1),.),.3),.4) and.5) that thee is an 1 [ π p 1, 0 ) such that yk0 0 ) y k0 ) < 1 fo all 1, 0 ], y k0 1 ) = 1 and y k 0 1 ) y k 0 0 ) < 0..6) Next, we detemine the behavio of y k0 ) when < 1. Fist, we solve v + v + p 1)v = 0, v 1 ) = 0 and v 1 ) = y k 0 1 )..7) Thee exist unique A and B such that v) = A sin p 1) peiod π p 1, and it follows that [ 1 π p 1, 1 ) exists such that + B cos p 1). Thus, v) has v) > 0, 1 ), v ) = 0 and v ) > 0..) We claim that y k0 ) = 0 fo some [, 1 ). Suppose, howeve, that y k0 ) > 1 [, 1 )..9) It is easily veified that the wonskian vy k 0 y k0 1)v satisfies vy k 0 y k0 1)v )) = v y p k 0 y k0 p 1)y k0 1) ) < 0.90) fo all, 1 ). It follows fom.6),.),.9) and.90) that v)y k 0 ) y k0 ) 1)v ) ) > 0 [, 1 )..91) When =,.91) educes to y k0 ) 1)v ) < 0, a contadiction since v 1 ) > 0 and ou supposition in.9) that y k0 ) > 1. Thus, [, 1 ) exists with y k0 ) = 0. In tun, since y k0 0 ) = 0 and y k 0 0 ) < 0, we conclude that ˆ, 0 ) exists such that y k 0 ˆ) = 0. Thus, k 0 / S, and the poof is complete, 1

19 Lemma. Let < p 3 and k = sup S. Then y k ) is a gound state solution. Poof. We need to pove that y k ) > 0 and y k ) < 0 > 0, lim y 0 + k ) <, lim 0 y + k ) = 0..9) Suppose, fist of all, that thee exists > 0 such that y k ) > 0 and y k ) < 0 > and y k ) = 0..93) Then y k ) 0. If y k ) = 0 then 1.3) implies that y k ) = 0 and y k ) = 1, contadicting uniqueness of the constant solution y 1. Thus, y k ) < 0. This and continuity imply that if 0 < k k 1, then k > 0 exists such that y k ) < 0 > k and y k k) = 0. Theefoe, k / S if 0 k k 1, contadicting k = sup S. Thus, It emains to pove that y k ) > 0 and y k ) < 0 > 0..94) lim y 0 + k ) < and lim 0 y + k ) = 0..95) Suppose, fist of all, that lim 0 + y k ) =. Then k S, and it follows fom the fact that S is open that k sup S, a contadiction. Thus it must be the case that 0 < lim 0 + y k ) <. Finally, we assume fo contadiction that lim 0 + y k ) 0. Then eithe L > 0 and 3 > 0 exist such that o else y k ) L 0, 3),.96) lim inf y k ) < lim sup y k ) = 0..97) Assume that.96) holds, let M = lim 0 + y k )) p, and let 3 > 0 also satisfy L + M 3 3 < 0..9) It follows fom 1.3) that y k )) M 0, 3 ). An integation, combined with.9), gives y k ) L 3 y k ) y k 3 ) + L 3 0, 3 ). A futhe integation gives ) 0, 3 ),.99) 19

20 hence lim 0 + y k ) =, contadicting the fact that y k ) M 1/p suppose that.97) holds. Since y k ) < M 1/p thee is an 4 > 0 such that > 0. Finally, p y p 1 k ) > 0 0, 4)..100) Popety.97) implies that y k ) changes sign infinitely often as 0+. Thus, thee exists 5 0, 4 ) such that y k 5) = 0, and y k 5) 0, Howeve, since y k 5) < 0, it follows fom a diffeentiation of 1.3), and popety.100) that y k 5) < 0, a contadiction. This completes the poof of the lemma. 3 Conclusions It is hoped that the shooting fom infinity technique developed in this pape can be adapted to povide new insights into a wide vaiety of futue eseach poblems. Fo example, in I)-III) below we conjectue how Theoem 1. might be extended to a wide ange of p and dimension N values. In Figue ight panel) we descibe how a natual emegence type phenomenon is obseved as deceases fom +. I) Extend the Theoem 1. to hold fo all N > 1 and < p < N+ N. II) Singula solutions and bound states with one o moe zeos. Let N > 1 and < p < N+ N. Pove that a positive inteval k N 1, k N ] exists such that i) if k k N 1, k N ) then y k ) has exactly N positive zeos, lim y k) = if N is odd, y k ) 0 as if N is even. 3.1) 0 + ii) if k = k N then y kn ) is a bound state solution with exactly N positive zeos. Figue illustates singula solutions and a bound state with one zeo when p = 3. III) Uniqueness. A long standing unesolved poblem is to pove uniqueness of bound states with moe than one zeo. 4 Appendix The method of successive appoximations is a standad technique e.g. see Coddington and Levinson [3], Chapte 13)) to pove existence and uniqueness of solu- 0

21 y k 0 k=10 y k 4.3 k * =5.01 N=p=3 7 k 1 =17.1 N=p=3 0 7 k 1 = Figue : Solutions of integal equation 1.11) when N = p = 3; Left) k = 10, k = 14 give singula solutions with one zeo, k 1 = 17.1 gives a bound state with one zeo. Right) When 0 < k k 1 = 17.1 an emegence type phenomenon is obseved; fist, when 1 all solutions ae small and behave linealy, i.e. y k k e ; second, as deceases fom lage values nonlinea inteactions cause solutions to ultimately begin gowing in amplitude and emege into fou distinct types, 1) positive singula solutions when 0 < k < k 5, ) the gound state when k = k, 3) singula solutions with one zeo when k < k < k , 4) a bound state with one zeo when k = k 1. 1

22 tions of integal equations. In the intoduction we pointed out that Hastings and McLeod [10], pp. 4-49) give complete details of the application of this method to pove existence of a unique solution of the integal equation associated with the Painleve type poblem 1.1)-1.13). The method of successive appoximations applies equally well to pove Lemma 1.1, i.e. to pove, fo each fixed k > 0 and < p 3 that thee exist X 0 > 0 and a unique solution of the integal equation y k ) = k e t sinht )y kt) y k t) p 1 dt, X 0, 4.1) which satisfies y k ) k e as, 4.) and is also continuous in x and k. Although the poof is standad, ou goal hee is to give a bief sketch of the details of the application of the successive appoximations method to pove existence. Fo this we follow pecisely the same steps in the poof given by Hastings and McLeod [10], pp. 4-49) fo the Painleve poblem. Step 1. The fist step is to deive two fundamental integal estimates, 4.6) and 4.7) below. Let k > 0 and < p 3 be fixed, let X > 0, and wite 4.1) as y k ) = k e 1 Next, define y 0 ) = k e Theefoe t e t e t )) y k t) y k t) p 1 dt. 4.3) > 0, and obseve that 0 < te t y 0 t) ) p k p e p 1)t = t X. 4.4) t p 1) te t y 0 t)) p dt conveges fo X, and 0 < e te t y 0 t) ) p e k p e p 1)t dt = dt X. 4.5) t p 1) We conclude fom 4.5) that X 0 > 0 can be chosen such that 0 < e te t y 0 t) ) p k e dt p+1, X ) Similaly, it follows fom the fact that e t e when t > 0, that 0 < e te t y 0 t) ) p dt = e e k p e t) e p 1)t dt t p 1) k p e p 1)t dt t p 1) k e p+1, X )

23 Step. We now use 4.6) and 4.7) in ou deivation of the appopiate unifomly bounded sequence of functions, 4.13) below. If y 1 ) = k e 1 t e t e t )) y 0 t) ) p dt, 4.) then y 1 ) is defined fo x X 0, and it follows fom 4.6)-4.7)-4.) that y 1 ) k + k p+1 ) e ke, X ) Next, let j 1 and suppose that y j ) has been defined on [X 0, ), and that Define y j ) k e, X ) y j+1 ) = k e 1 t e t e t )) y j t) ) p dt, X ) It follows fom 4.6), 4.7), 4.10) and 4.11) that y j+1 ) k e, X ) We conclude fom 4.), 4.9), 4.10), 4.11) and 4.1) that y j ) k e X 0 and j ) Step 3. We use 4.13) in the completion of the poof. Fo each j 1 define y j+1 ) y j t ) = e t e t )) y j t) ) p y j 1 t) ) p ) dt, 4.14) whee X 0. Fo each j 1 and t it follows fom the mean value theoem that ζ j t) exists which satisfies y j t) ) p y j 1 t) ) p = p ζ j t) ) p 1 y j t) y j 1 t) ), 4.15) whee y j t) ζ j t) y j+1 t) if y j t) y j+1 t), 4.16) 3

24 y j+1 t) ζ j t) y j t) if y j+1 t) y j t). 4.17) Let M = p p k p 1 and let f denote the sup nom of f. Fom 4.13), 4.14), 4.15), 4.16), 4.17) and the fact that p >, we conclude that X 0 can be futhe chosen such that if X 0 then Thus, y j+1 ) y j ) pt e t e t )) ζ j t) p 1 y j t) y j 1 dt Mt e t e t )) e p 1)t y j t) y j 1 t) dt t p 1 M e p 1)t et y j t) y j 1 t) dt t p y j y j 1 y j y j 1 M et e p 1)t t p Me e p )t dt t p dt < 1 y j y j ) y j+1 y j 1 y j y j 1 j 1, 4.19) and we conclude that y 0 + j=0 y j+1 y j ) conveges unifomy on [X 0, ) to a continuous function y k ) which solves 4.3). As pointed out by Hastings and McLeod [10], pp. 4-49), uniqueness and continuity with espect to k follow in the usual way fo poofs using successive appoximations Coddingtion [3], Chapte 13). Refeences [1] H. Beestycki and P. L. Lions, Nonlinea scala field equations I, existence of a gound state, existence of infinitely many solutions, Achives Fo Rational Mechanics And Analysis 193), [] J. Byeon and Z. Wang, Standing waves fo nonlinea Schodinge equations with singula potentials, Annales de l, Institut Heni Poincae C) Non Linea Analysis 6,3 009), [3] E. Coddington and N. Levinson, Theoy Of Odinay Diffeential Equations, McGaw-Hill 1955) 4

25 [4] C. Coffman, Uniqueness of the gound state solution fo u u = u 3 and a vaiational chaacteization of othe solutions, Achives Fo Rational Mechanics And Analysis 46, 197), 1-95 [5] C. Cotaza, M. Gacia-Huidobo and P. Heeos, Uniqueness of sign changing bound state solutions of a semilinea equation with weights, axiv pepint axiv: ) [6] S. Le Coz, Standing waves in nonlinea Schodinge equations, Analytical and Numeical Aspects of Patial Diffeential Equations, eds. Emmich and Wittbold eds., de Guyle, Belin 00), 1-41 [7] G. Fibich, N. Gavish and X. P. Wang New singula solutions of the nonlinea Schodinge equation, Physica D ), [] R. Finklstein, R. Levelie and M. Rudeman, Non-linea Spino Fields, Physics Review 3 197), 36 [9] S. P. Hastings and J. B. McLeod, A Bounday Value Poblem Associated with the Second Painleve Tanscendent and the Koteweg-de Vies Equation, Achives Fo Rational Mechanics And Analysis ), [10] S. P. Hastings and J. B. McLeod,j Classical methods in odinay diffeential equations: with applications to bounday value poblems, Ame. Math. Soc. 011) [11] M. Kwong, Uniqueness of the positive solution of u u = u p = 0 in R N, Achives Fo Rational Mechanics And Analysis 105, 3 199), [1] B. LeMesuie, G. Papanicolau, C. Sulem and and P. Sulem, Local stuctue of the self-focusing nonlineaity of the nonlinea Schodinge equation, Physica D 3 19), 10-0 [13] K. McLeod and J. Sein, Uniqueness of positive adial solutions of u+fu) = 0 in R N, Achives Fo Rational Mechanics And Analysis ), [14] K. McLeod, W. C. Toy and F. Weissle, Radial solutions of u + fu) = 0 with pescibed numbes of zeos, J. Diff. Eqs ),

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