Uniqueness of Positive Ground State Solutions of the Logarithmic Schrodinger Equation

Transcription

1 Achive fo Rational Mechanics and Analysis manuscipt No. will be inseted by the edito) William C. Toy Uniqueness of Positive Gound State Solutions of the Logaithmic Schodinge Equation Dedicated To The Memoy Of Byce McLeod Abstact We pove uniqueness of positive gound state solutions of the poblem d u d + n 1 du d + u ln u ) = 0, u) > 0 0, and u), u )) 0, 0) as. This equation is deived fom the logaithmic Schodinge equation iψ t = ψ + u ln u ), and also fom the classical equation u t = u+u u p 1) ) u. Fo each n 1 a positive gound state solution is u 0 ) = exp 4 + n, 0 <. We combine u 0 ) with enegy estimates, and associated Ricatti equation estimates, to pove that, fo each n [1, 9], u 0 ) is the only positive gound state. We also investigate the stability of u 0 ). Seveal open poblems ae stated. 1 Intoduction We investigate uniqueness of solutions of the gound state poblem u + n 1 u + u ln u ) = 0, 1) u ) = 0, u), u )) 0, 0) as, ) u) > 0 0, 3) Mathematics Depatment, Univesity of Pittsbugh, Pittsbugh, Pa toy@math.pitt.edu

2 William C. Toy Equation 1) is deived see the Appendix fo details) fom a escaling of the dimensionless logaithmic Schodinge equation iψ t = ψ + ψ ln ψ ), 4) whee ψ denotes the dimensionless wave function. The Appendix also shows how to deive 1), though a limiting pocess, as p 1 +, fom the classical equation u t = u + u u p 1 u, 5) and also fom the non-linea Klein-Godon equation u t = u + u u p 1 u. 6) A positive gound state solution of 1)-)-3) is given by u 0 ) = exp 4 + n ), 0 <, n 1. 7) This solution plays a cental ole in applications of the logaithmic Schodinge equation to quantum mechanics [4, 5], quantum optics [8], tanspot and diffusion phenomena [18], infomation theoy [7, 30], quantum gavity [31], and the theoy of Bose-Einstein condensation [1]. In these applications a physically impotant popety of u 0 ) is that it is the only positive gound state solution. Thus, ou goal is to pove Theoem 1 Uniqueness) Let 1 n 9. Then u 0 ) given in 7) is the only solution of 1)-)-3) Ou poof of Theoem 1 employs a new compaison method which combines u 0 ) with enegy estimates and associated Ricatti equation estimates. To undestand why ou appoach is new we need to descibe analytical techniques in pevious studies. In 1987 K. McLeod and J. Sein [] investigated existence and uniqueness of smooth solutions of the geneal equation u + n 1 u + fu) = 0, 8) whee n 1, u satisfies )-3), and f satisfies assumptions A 1 ) f C 1 [0, ), f0) = 0, f 0) < 0; A ) Thee is an α > 0 such that fu) < 0 fo u 0, α), fu) > 0, u > α; A 3 ) f α) > 0. Poblem 8)-)-3) aises in the study of solutions of the classical poblem u + fu) = 0, 9) ux) > 0 x R n, ux) 0 as x. 10)

3 Logaithmic Schodinge Equation 3 Unde suitable diffeentiability conditions on f, Gidas, Ni and Nienbeg [15] poved that solutions of 9)-10) must be adial and satisfy 1)-)-3). Geneal conditions on f have led to poofs of existence of positive gound states in othe impotant studies [, 6, 7]. In 1951 Finkelstein et al [16] analyzed u + u 3 u = 0 in the context of spino fields. In a 1973 classical pape, Coffman [9] poved uniqueness of a positive gound state solution of the initial value poblem u + u + u 3 u = 0, u0) = β, u 0) = 0, 11) Let u, β) denote the solution of 11), and let β 0 > 0 such that u 0 ) = u, β 0 ) satisfies )-3). To pove uniqueness of u 0 ), Coffman analyzed the behavio of w = βu, β), which solves the equation of fist vaiation w + n 1 w + 3u + 1)w = 0, w0) = 1, w 0) = 0. 1) He developed seveal functionals and inequalities involving w which help detemine the behavio of u, β) as β vaies. He used this infomation to pove that u, β 0 ) is the only positive gound state solution of 11). McLeod and Sein [, 3] extended Coffman s study and investigated the geneal, classical equation u + n 1 u + u p 1 u u = 0. 13) They made use of functionals in tems of, u and u, and technical compaison methods, to pove uniqueness of a positive gound state of 13) in the following paamete egimes: i) 1 < p < when 1 n <, n n ii) 1 < p when < n < 4, iii) 1 < p 8 n when 4 n 8.71 Kwong [0] poved uniqueness of positive gound state solutions of 13) when n > 1 and 1 < p < n+1 n 1. He followed Coffman s appoach and analyzed 13) by developing technical lemmas associated with the the equation of fist vaiation fo 13), namely d w d + n 1 dw d + p u p 1 1)w = 0, w0) = 1, w 0) = 0. 14) Uniqueness of positive solutions in annula domains has been poved by Coffman [11] and Kwong and Zhang [1]. In 000 Sein and Tang [6] extended the esults descibed above, and poved uniqueness of positive gound state solutions of the quasilinea equation u m u ) n 1 + u m= u + fu) = 0, > 0, n > m > 1, 15) whee n > m > 1 and fu) satisfies assumptions

4 4 William C. Toy H1) Thee is a b > 0 such that f is continuous on 0, ), with fu) 0 on 0, b] and fu) > 0 fo u > b; H) f C 1 b, ), with gu) = uf u)/fu) non-inceasing on b, ). Gazzola, Sein and Tang [17] poved the existence of positive gound state solutions. Remak 1: The Case m=. Equation 15) educes to 8) when m =. Because of the constaint n > m > 1 in 15), it follows that, if m = then the Sein-Tang [6] theoem does not apply to 1) when 1 n To pove Theoem 1 we detemine the behavio of solutions of u + n 1 u + u ln u ) = 0, u0) = β > 0, u 0) = 0. 16) Equation 16) is fundamentally diffeent fom the investigations descibed above in two impotant ways: I) The nonlineaity fu) = u ln u ) is continuous on, ), with zeos at u = 0 and u = ±1, and satisfies A ) - A 3 ) stated above. Howeve, lim f u) = lim ln u ) + 1 =. 17) u 0 u 0 A key step in the McLeod-Sein analysis Lemma 3, p. 14 in []) makes use of the equiement f 0) = m < 0. Howeve, when fu) = u ln u ), popety 17) shows that f u) becomes unbounded as u 0. Thus, assumption A 1 ) is not satisfied and it is challenging to pove uniqueness of a gound state solution using the methods in []. The function fu) = u ln u ) does satisfy assumptions H1)-H) in [6]. Howeve, Remak 1 shows that the uniqueness esult in [6] does not apply to 1)-)-3) when 1 n. Thus, uniqueness of the positive gound state solution u 0 ) has not peviously been poved when 1 n. II) The uniqueness poofs in [9,11,0,1,9] make extensive use of the equation of fist vaiation fo the function w = βu, β). Fo 16) this equation is d w d + n 1 dw d + ln u ) + 1)w = 0, w0) = 1, w 0) = 0, 18) As Coffman [9] oiginally showed, a cucial step in poving uniqueness of gound states is to detemine the behavio of points R 1 = R 1 β) > 0 whee ur 1, β) = 0. The behavio of R 1 β) is detemined fom the equation dr 1 dβ = w R 1) u R 1 ). 19) Howeve, it is challenging to accuately detemine the behavio of u R 1 ) and w R 1 ) in 19) since the tem ln u ) in 16) and 18) becomes unbounded as u 0. The analytical difficulties descibed above have led us to develop a new appoach to pove uniqueness of the positive gound state u 0 ) given in 7).

5 Logaithmic Schodinge Equation 5 Ou appoach is to combine u 0 ) with enegy based estimates and associated Ricatti equation estimates to detemine the behavio of solutions of the initial value poblem 16) fo each β > 0. Thus, to pove Theoem 1 we only need to pove the equivalent esult Theoem Uniqueness) Let 1 n 9 and β > 0. i) If β = e n/ then the solution of 16) is the positive gound state solution 7). ii) If β e n/ then the solution of 16) is not a positive gound state. Remak. We pove uniqueness when 1 n 9, appoximately the same ange whee the McLeod-Sein [] theoem holds. Also, Theoem applies to the peviously unesolved paamete egime 1 n see Remak 1), in paticula to the physically impotant value n =. Poof of Theoem. The fist step is to conside the case n = 1 and obseve that 16) has the fist integal u ) + u lnu) 1 ) = E, 0) whee E is constant. Substituting u), u )) 0, 0) as into 0) gives E = 0, and it easily follows fom u ) + u lnu) 1 ) ) = 0 that the only positive gound state is u 0 ) = exp Outline of poof when n > 1. In Section we pove Theoem when 1 < n 9. Thee, we detemine behavio of solutions in the distinct paamete egimes 0 < β < e n/ and β > e n/. i) When 0 < β < e n/ we show that u) > 0 on its maximal inteval of existence, and that u) cannot satisfy both the positivity condition u) > 0 0, and the limiting condition lim u), u )) = 0, 0). 1) ii) When β > e n/ we pove that thee is a fist R 1 = R 1 β) > 0 such that u R 1 ) = 0 and u R 1 ) < 0. ) Thus, the positivity condition u) > 0 0 is violated and the solution cannot be a positive gound state. Uniqueness poofs in pevious studies descibed above ae necessaily vey technical. In Section ou poof, which is also somewhat technical, is completed with the help of auxiliay lemmas. The ole of each lemma is explained as we poceed. Section 3 contains conclusions and a statement of open poblems. Section 4 is the Appendix. Futue Study: Sign Changing Solutions. The existence of multi-zeo gound state solutions of 13) was poved by McLeod, Toy and Weissle [4], and Jones and Kuppe [19]. Toy [8] poved existence of multi-zeo gound states fo 1)-) when fu) = u p 1 u u q. Toy [9] also poved uniqueness of the gound state solution of 8)-) with exactly one positive zeo when

6 6 William C. Toy fu) is piecewise linea. Fo moe geneal geneal fu), existence and uniqueness of multi-zeo gound state solutions was poved by Cotaza, Gacia- Huidboo and Yau [1 14]. Thei esults pove uniqueness of multi-zeo gound states of 16) when n = 3 o n = 4. Uniqueness emains an open poblem when n / {3, 4}. The poof of uniqueness of multi-zeo gound states of the classical cubic equation 11) is also unesolved. It is hoped that a combination of techniques in [1 14] and the enegy and Ricatti based estimates developed in this pape may give new insights into poving uniqueness of sign-changing gound state solutions of 16) and 11). Uniqueness In this section we keep 1 < n 9 fixed and pove that the solution of initial value poblem 16) is not a positive gound state when β > 0 and β e n/. We conside two sepaate cases: 0 < β < e n/ and β > e n/. Case I. 0 < β < e n/. Let u denote the solution of 16). The enegy functional satisfies Q = n 1 Q = u ) + u lnu) 1 ) u ), Q 0) = 0 and Q0) = β 3) lnβ) 1 ). 4) The following esult gives conditions that a positive gound state solution must satisfy. Lemma 1 Let β 0, e n/ ). A positive gound state solution of 16) satisfies u0) = β e 1/ and Q0) = β lnβ) 1 ) 0, 5) u ) < 0, Q) > 0 and Q ) < 0 > 0, and lim Q) = 0. 6) Poof. A positive gound state solution satisfies conditions )-3). Thus, These popeties imply that Q ) 0 > 0 and lim Q) = 0. 7) Q) 0 > 0. 8) We conclude fom 16), 3) and 8) that Q0) = β lnβ) ) 1 0, hence β e 1/. Next, it follows fom 16) that u 0) = β lnβ) < 0 since β e 1/. This implies that u) > 0 and u ) < 0 on an inteval 0, ǫ). If thee is a fist > 0 such that u ) > 0 and u ) = 0 then u ) = u )lnu )) 0. Thus, lnu )) 0 and Q ) = u ) lnu )) ) 1 < 0, contadicting 8). We conclude that u ) < 0 > 0, hence Q ) = n 1)u ) < 0 > 0. This popety and 7) imply that Q) > 0 0. This completes the poof. We use the esults of Lemma 1 to pove

7 Logaithmic Schodinge Equation 7 Theoem 3 Let 1 < n 9 and β 0, e n/ ). Then the solution of 16) is not a positive gound state solution. Poof. We assume that thee is a β 0, e n/ ) such that the solution of 16) is a positive gound state, and obtain a contadiction. By Lemma 1, the solution must satisfy conditions 5)-6). Thus, if we pove that one of these conditions does not hold, then we have a contadiction of the assumption that the solution is a positive gound state. Fist, when 0 < β < e 1/ it is easily veified that Q0) = β lnβ) 1 ) < 0, hence 5) does not hold. Next, when e 1/ β < e n/ we claim that 6) does not hold. To pove this claim we make use of the Ricatti function ρ = u u, which satisfies ρ + ρ + n 1) ρ + lnu) = 0, ρ0) = 0 and ρ 0) = lnβ) < 0, 9) n Below we show that ρ) deceases ) until it eaches a negative minimum, then inceases until ρˆ) 1, 0 and 0 < uˆ) < 1 at some ˆ > 0, hence Qˆ) = u ˆ) u ˆ) uˆ) ) ) + ln uˆ)) 1 < 0. 30) Thus, 6) does not hold and Lemma 1 implies that u) is not a gound state. A diffeentiation of 9) gives ρ + n 1) ρ n 1) ρ + ρρ + ρ = 0, ρ 0) = 0. 31) ) 4 + n is the We also use the function ρ 0 = u 0 u 0, whee u 0 ) = exp positive gound state given in 7). Then ρ 0 satsifies ρ 0 ) = and ρ 0) = ) Impotant popeties ae and ρ 0) = lnβ) n < 0 β [ e 1/, e n/) 33) ρ 0) ρ 0 0) = lnβ) n + 1 [ > 0 β e 1/, e n/). 34) Next, let 0, max ) denote the lagest inteval whee < ρ) < 0. That is, max = sup {ˆ > 0 if 0 < < ˆ then < ρ) < 0} 35)

8 8 William C. Toy To pove that condition 6) does not hold we need two auxiliay esults Lemma and Lemma 3 below) which detemine the behavio of ρ 0, ρ and ρ. Lemma gives pactical lowe bounds fo ρ ) and ρ ) ρ 0 ) ove the lagest inteval 0, ) 0, max ) whee ρ 0) < ρ ) < 0. Thus, define = sup {ˆ 0, max ) if 0 < < ˆ then ρ 0 ) < ρ ) < 0} 36) It follows fom 33)-34) and continuity that > 0 β [ e 1/, e /). Lemma Let e 1/ β < e n/. Then ρ ) > 0 and ρ ) ρ 0 ) lnβ) n + 1 > 0 0, ]. 37) Poof. It follows fom 31) and 3) that ρ ) + n 1 We conclude fom 38) that ) + ρ ρ = ρ ρ 0 ) ρ + ρ ). 38) n+1 e ) ρx)dx 0 ρ ) = n 1 e ρx)dx 0 ρ ρ 0 ) ρ + ρ ). 39) The definitions of and max imply that the ight side of 39) is positive fo all 0, ). Also, ecall fom 31) that ρ 0) = 0. Thus, an integation of 39) gives ρ ) > 0 0, ]. 40) It follows fom 3) and 40) that d d ρ ) ρ 0 )) = ρ ) > 0 0, ]. An integation gives ρ ) ρ 0 ) lnβ) n + 1 > 0 0, ]. 41) This completes the poof of Lemma. Ou second auxiliay esult Lemma 3) shows that ρ has at most one zeo on 0, max ), and that the second inequality in 37) extends to the entie inteval 0, max ). Lemma 3 Let e 1/ β < e n/. Suppose that ρ ) = 0 at some 0, max ). Then and ρ ) < 0 0, ), ρ ) = 0, and ρ ) > 0, max ), 4) ρ ) ρ 0 ) lnβ) n + 1 > 0 [, max). 43)

9 Logaithmic Schodinge Equation 9 Poof. Wite equation 31) as ) n 1 ρ + + ρ ρ = ) n 1 1 ρ. 44) Let 0, max ) denote the fist positive zeo of ρ. Then ρ ) 0, and 44) gives ) n 1 ρ ) = 1 ρ ) 0. 45) Lemma and the definitions of and max imply that ρ ) > 0. Also, ρ ) < 0 since 0 < < max. Combining these popeties with 45), we conclude that > n 1, and that ρ ) > 0 on an inteval, + ǫ). If thee is a next zeo of ρ at some, max ) then ρ ) 0. 46) Howeve, ρ ) < 0 and > > n 1, and theefoe 44) gives ρ ) > 0, contadicting 46). This poves popety 4). Finally, 4) shows that ρ ) 0 [, max ). Thus, we conclude that ρ ) ρ 0 ) lnβ) n + 1 > 0 [, max). 47) This completes the poof of the Lemma 3. We now complete the poof of Theoem 3. The assumption that u is a positive gound state implies that u) > 0 and u ) < 0 fo all > 0, and u) 0 as. Thus, max = and ρ) = u ) u) and thee is an > 0 such that < 0 > 0, 48) 0 < u) 1 and ln u)) 0. 49) Also, it follows fom 3) and Lemma 1 that u Q) = u ) ) ) ) + ln u)) 1 u) > ) Combining 48), 49) and 50), we conclude that ρ ) > 1. Thus, ρ) < 1. 51) Ou goal is to pove that ρ) > 1 when, contadicting 51). It follows fom 31) and identities ρ 0) = 1 and ρ = ρ ρ) that ρ ρ) + n 1 ρ ρ) = ρ ρ ρ 0) > 0, 5)

10 10 William C. Toy Combining 37), 43), 51) and 5) gives ρ ρ) + n 1 ρ ρ) A, 53) whee A = ) lnβ) n + 1 > 0 since β [e 1/, e n/ ). Integating 53) gives n 1 ρ ρ) An+1 n B, 54) whee B = A ) n+1 n+1 + ) n 1 ρ ) ρ )). Next, divide 54) by n+1 and get ρ An integation of 55) gives ρ) ) A n ρ ) + A n + 1 ) + B n B n+1. 55) 1 ) n 1 ) n. 56) Because A > 0, the ight side of 56) is positive when 1. Thus, ρ) > 1 when 1, contadicting 51). This completes the poof of Theoem 3. Case II) β > e n/. In this egime we pove that the solution of 16) is not a gound state solution by showing that thee is an 1 > 0 such that u) > 0 and u ) < 0 0, 1 ), u 1 ) = 0 and u 1 ) < 0. 57) To pove 57) we make use of the tansfomation f) = u)exp 4 n ). 58) Define α = βe n/. Then β > e n/ α > 1, and f satisfies ) n 1 f + f + f lnf) = 0, f0) = α > 1, f 0) = 0. 59) It follows fom 59) that f 0) = α lnα) n < 0 α > 1. 60) We need to detemine the behavio of f) fo each α 1. When α = 1, uniqueness of the constant solution f = 1 implies that f) = 1 0. The coesponding solution of 16) is gound state solution 7) since u) = f)exp 4 + n ) = exp 4 + n ) 0. 61) Goal. When α > 1 we show that f) eaches f = 0 at a finite value. Define 1 = sup{ˆ > 0 if 0 < < ˆ then f) > 0}. 6)

11 Logaithmic Schodinge Equation 11 Theoem 4 Let α > 1 and 1 < n 9. The solution of 59) satisfies 1 <, f) > 0 and f ) < 0 0, 1 ), f 1 ) = 0 and f 1 ) < 0. 63) Implications of Theoem 4. Let β > e n/ so that α = βe n/ > 1, and let f denote the solution of 59). It follows fom 58) and Theoem 4 that the coesponding solution of 16) satisfies u0) = β > e n/ and u 0) = 0, u) = f)exp 4 + n ) > 0 0, 1 ), 64) u ) = u 1 ) = f 1 )e f ) ) f) exp 4 + n ) < 0 0, 1 ), 65) n ) ) = 0 and u 1 ) = f 1 )e 4 + n < 0. 66) Thus, popety 57) is veified and the poof of Theoem is complete Poof of Theoem 4 The fist step is to show that f) deceases until f 0 ) = 1 at a finite 0 > 0. Thus, define 0 = sup{ˆ 0, 1 ) if 0 < < ˆ then f) > 1} 67) Lemma 4 Let α > 1. Then 0 <, f) > 1 and f ) < 0 0, 0 ), f 0 ) = 1 and f 0 ) < 0. 68) Poof. It follows fom 59) and definition 67) that ) ) n 1 e f )) = n 1 e f lnf) < 0 0, 0 ). 69) Combining 59) with69) gives f) > 1 and f ) < 0 0, 0 ). This poves the fist pat of 68). It emains to show that 0 is finite, and that f 0 ) = 1 and f 0 ) < 0. 70) Suppose, howeve, that 0 = fo some α > 1. Then 59) implies that f ) < 0, f) > 1 and f ) = ) n 1 f f lnf) < 0 > n 1. These popeties imply that f) = 1 at a finite > n 1, contadicting the supposition 0 =. Thus, 0 <, f 0 ) = 1 and f 0 ) 0. Uniqueness of the constant solution f 1 implies that f 0 ) < 0. This poves 70). Remak. It follows fom Lemma 4 that f) > 0 and f ) < 0 on an inteval 0, 0 + ǫ). This fact and the definitions of 0 and 1 imply that 0 < 0 < 1 α > 1. The next step in the poof of Theoem 4 is to develop a citeion which guaantees that 1 <, f 1 ) = 0 and f 1 ) < 0. We do this in

12 1 William C. Toy Lemma 5 Let n > 1 and α > 1. If 0 n 1 then 1 <, 0 < f) 1, f ) < 0 [ 0, 1 ), f 1 ) = 0 and f 1 ) < 0. 71) Poof. It follows fom 59) and 70) that f 0 ) < 0 and f 0 ) 0. 7) A diffeentiation of 59) gives ) n 1 f + f + lnf) n 1 ) f = 0. 73) We conclude fom 70), 7) and 73) that n 1 exp ) f )) = n 1 exp ) ) n 1 lnf) f ) < 74) 0 fo all in an inteval 0, 0 + ǫ). It follows fom 7) and 74) that 0 < f) < 1, f ) < 0 and f ) < 0 0, 1 ). In tun this implies that f ) < f 0 ) < 0 0, 1 ), and we conclude that 1 <, f 1 ) = 0 and f 1 ) f 0 ) < 0. This completes the poof of Lemma 5. We now use Lemmas 4 and 5 to detemine the behavio of f and f fo each α > 1. Lemma 6 Let 1 < α e. Then n 1 < 0 < 1 <, f ) < 0 0, 1 ] and f 1 ) = 0. 75) Poof. Fist, Lemma 4 and the fact that f 0) = 0 imply that f ) < 0 and 1 f) e 0, 0 ). 76) Next, we show that 0 > n 1 α 1, e]. Suppose, fo contadiction, that thee is an α 1, e] fo which 0 < 0 n 1. This and popety 76) imply that n 1 lnf)) n 1 1 > 0 0, 0 ). 77) Recall fom 60) that f 0) < 0. Then Lemma 4, 74) and 77) imply that f ) < 0 0, 0 ]. 78) Since we assume that 0 < 0 n 1, we conclude fom 76) and 59) that f 0 ) 0, contadicting 78). Theefoe, it must be the case that 0 > n 1. Thus, since 0 > n 1, Lemma 5 implies that 1 <, f 1 ) = 0 and f 1 ) < 0. This completes the poof of Lemma 6. To complete the poof of Theoem 4 we need to show that popety 63) holds when α > e. Fist, if 0 n 1, then Lemma 5 guaantees that 63) holds. It emains to pove that 63) also holds when 0 < 0 < n 1. This is done in

13 Logaithmic Schodinge Equation 13 Lemma 7 Let 1 < n 9 and α > e. If 0 < 0 < n 1 then 0 < 1 <, f) > 0 and f ) < 0 [ 0, 1 ), f 1 ) = 0 and f 1 ) < 0. 79) Poof. It follows fom 59), the assumption 0 < 0 < n 1, and Lemma 4 that f 0 ) = 1, f 0 ) < 0 and f 0 ) = 0 n 1 ) f 0 ) > 0. 80) 0 We claim that 80) implies that thee is an A 0, 0 ) such that ) ) n 1 f A ) = exp A and f n 1) n 1 A ) A 3 exp A. 81) Suppose, howeve, that thee is an α > e such that ) n 1 1 < f) < exp 0, 0 ). 8) Then lnf)) < n 1 0, 0 ), and it follows fom 59), 60) and 73) that ) ) ) n 1 e f )) = n 1 e n 1 lnf) f ) < 0 83) when 0, 0 ). An integation gives f ) < 0 0, 0 ], hence f 0 ) < 0, contadicting 80). ) We conclude that thee is an A 0, 0 ) such that ) n 1 n 1 f A ) = exp. This popety, and the fact that f A 0 ) = 1 < exp, 0 imply that we can choose A such that ) ) n 1 n 1 f A ) = exp and 1 < f) < exp A, 0 ). 84) A exp It follows fom 84) that f A ) n 1) n 1 A 3 A To complete the poof of Lemma 7, we make use of 81) and enegy functional S = f ) + f ), and 81) is poved. lnf) 1 ), 85) which satisfies S = n 1 ) f ). 86) It follows fom 85) and 86) that S 0) = 0 and S0) = α and that S also satisfies n 1) S + lnα) 1 ) > 0, ) S = n 1 ) 1 )f lnf). 87)

14 14 William C. Toy Obseve that f 1 lnf)) 1 f > 0. Then 87) educes to ) n 1) S + S 1 n 1 ) 88) when 0, min{ n 1, 1 } ). It follows fom 88) that ) n 1 e S n 1 n 1) n 3) e 89) when A, min{ n 1, 1 } ). Integation of both sides of 89) fom A to gives n e S) n A e A SA ) + 1 n A e A n e ), 90) 4 whee A, min{ n 1, 1 } ). It follows fom 81), 85), and the fact that 0 < A n 1, that a pactical lowe bound on S A) is S A ) n 1) n 1) A 6 e A n 1) + e A n 1 1 ) A n 1) n 1) A 6 e A 91) To complete the poof of Lemma 7 we also need pactical lowe bounds on the poduct n S). We conside two cases: Case A: 1 < n 6 and Case B: 6 < n 9. Case A: 1 < n 6. Combine 90) and 91), multiply by e, and get n 1) n S) n 1) n 8 A e A n 4 9) n 1) e A when A min{ n 1, 1 }. The tem n 8 A is deceasing in A when 1 < A n 1 and 1 < n 6. Thus, we substitute A = n 1 and the uppe bound = n 1 into 9) and conclude that, if 1 < n 6, then n S) when A min{ n 1, 1 }. n 1)n 1 4 Case B: 6 < n 9. The tem n 8 n 1) A [ ] 8 n 1 e 1 > 0 93) n 1) e A attains its positive elative n 1) n 4 minimum at A = n 4 < n 1 when n > 6. Substitute A = and the uppe bound = n 1 into 9) and conclude that, if 6 < n 9, then ) n 4 n n 1)n 1 8 e S) 1) > 0 94) 4 n 1 n 4

15 Logaithmic Schodinge Equation 15 when A min{ n 1, 1 }. We now complete the poof of Lemma 7. Fist, define K n = K n = n 1)n 1 4 n 1)n 1 4 [ 8 n 1 [ ] 8e n 1 1 > 0 if 1 < n 6, 95) ) n 4 e 1] > 0 if 6 < n 9, 96) n 4 Next, let n 1, 9] be fixed. It follows fom 93), 94), 95), and 96) that n S) K n, A min{ n 1, 1 } 97) We combine 85) with 97) and conclude that f )) + f)) lnf) 1 ) K n, 98) n 1) n 1 when A min{ n 1, 1 }. Suppose that 0 < 1 n 1. It follows fom 98) and initial conditions f 0 ) = 1 and f 0 ) < 0 that f K n ) < n 1) n 1, 0 < f) < 1 and lnf)) < 0 0, 1 ). 99) We conclude fom 99), the assumption that 1 < n 1, and continuity, that f 1 ) = 0 and f K n 1 ) n 1) n ) Finally, suppose that 1 > n 1. Again, it follows fom 98) and initial conditions f 0 ) = 1 and f 0 ) < 0 that f K n ) < n 1) n 1 and 0 < f) < 1 0, n 1). 101) At = n 1, we conclude fom 85), 98), 101) and continuity that S n 1 ) K n, 10) n 1) n 1 f n 1 ) K n < n 1) n 1 and 0 < f n 1) < )

16 16 William C. Toy It follows fom 86) and 10) that S ) > 0 and S) K n n 1) n 1 when > n 1. Combining these popeties with 85), we conclude that inequality 98) extends to f )) + f)) lnf) 1 ) K n n 1) n 1 [ ) n 1, 1.104) It easily follows fom 103) and 104) that101) holds on [ ) n 1, 1. That is, f K n ) < n 1) n 1 and 0 < f) < 1 [ ) n 1, ) An integation of f ) < Kn n 1) n 1 implies that 1 <, f 1 ) = 0 and f 1 ) Kn n 1) n 1. This completes the poof of Lemma 7. 3 Conclusions In this pape) we pove uniqueness of the positive gound stat e u 0 ) = exp 4 + n, which satisfies 1)-)-3). Ou main theoetical advance is to develop an appoach to poving uniqueness which is diffeent fom taditional methods. Ou method combines u 0 ) with estimates deived fom associated enegy functionals and Ricatti equations. It is hoped that futue extensions of ou techniques can be combined with methods in pevious studies to esolve open poblems such as the following: Poblem 1. When 1 < n < 9 detemine whethe u 0 ) is the only positive solution of such that u + u ln u )) = 0, 106) u x 1, x,.., x N ) 0 as x 1, x,.., x N ). 107) Poblem. When n > 1 ae sign changing solutions of 1)-) with pescibed numbes of zeos unique? What is the physical ole of these solutions fo the logaithmic Schodinge equation 4)? Do they epesent highe enegy states? Ae they stable? Poblem 3. Real vaiable models. Detemine stability of the positive gound state solution u 0 ) of the eal vaiable patial diffeential equations and u t = u + u ln u )), 108) u = u + u ln u )). 109) t

17 Logaithmic Schodinge Equation 17 The Appendix shows how 108) and 109) aise fom classical models though a limiting pocess as p 1 +. A fist step in poving stability is to lineaize 108) aound u 0 ) and set u = u 0 +ǫe λt v, whee ǫ 1. To fist ode in ǫ, v satisfies v + ln u 0 ) )) + 1 λ)v = ) The bounded, positive function v = u 0 ) satisfies 110) when λ = 1. This suggests that u 0 ) is linealy unstable. It emains to esolve the following: i) Is u 0 ) also unstable as a solution of the nonlinea equation 108)? ii) Ae thee solutions of of 108) which blow up in finite time, o as t? iii) Investigates the same issues fo equation 109). 4 Appendix Hee we have thee goals. In pat I below we give a standad deivation of u + n 1 u + u ln u ) = 0 111) fom the dimensionless logaithmic Schodinge equation iψ t = ψ + ψ ln ψ ). 11) In pats II and III we apply a limiting pocess as p 1 + ) to deive 111) fom the classical equation and the non-linea Klein-Godon equation I. Recall that u = N i=1 u x i Then 11) educes to u t = u + u u p 1 u, 113) u t = u + u u p 1 u. 114) and set the wavefunction ψ = exp iωt + ω ) ux 1,.., x N ). N i=1 u x i + u ln u ) = ) Define x i = x i, i = 1,.., N and tansfom 115) into N i=1 u x i + u ln u ) = )

18 18 William C. Toy Substitute = N i=1 x i into 116) and get 111). II. Next, we deive 111) fom 113). Set t = p 1)t and x i = p 1x i, i = 1,.., N, and ecast 113) in tems of these new coodinates. Divide the esulting equation by p 1 and get u N t = u x i i=1 + u u p 1 ) u ) p ) A fomal application of L Hopital s ule gives lim p 1 + Combining this esult with 117) gives u u p 1 ) u) p 1 = u ln u ). u N t = u x i i=1 + u ln u ). 118) Time independent, adially symmetic solutions of 118) satisfy 111). III. Finally, conside the nonlinea Klein-Godon equation 114). The same pocess descibed above in pat II with t = p 1)t eplaced by t = p 1t) educes 114) to u t = N u x i i=1 + u ln u ). 119) Time independent, adially symmetic solutions of 119) satisfy 111). Poblem 4. Beestycki and Lions [] poved that 114) has positive gound state solutions when N 1. Beestycki and Cazenave [3] see also [5]) poved stong instability of the gound state when n 3 and 1 < p < n, and that petubations fom the gound state blow up in finite time. a) Detemine stability popeties of the positive gound state solution u 0 ) of 111) as a solution of the time dependent equation 119). b) Fo equation 119) detemine whethe petubations fom the gound state can blow up in finite time, o as t. Refeences 1. Avdeenkov, A. V., Zloshchastiev, K.G.: Quantum Bose liquids with logaithmic nonlineaity: Self-sustainability and emegence of spatial extent. J. Phys. B: At. Mol. Opt. Phys ). Beestycki, H., Lions, P.: Non-linea scala field equations. I, existence of a gound state; II, Existence of infinitely many solutions. Ach. Rat. Mech. Anal. 8, ) 3. Beestycki, H., Cazenave, T.: Instabilite des etats stationnaies dans les equations de Schodinge et de Klein-Godon non lineaes. C. R. Acad. Sci Pais 93, ) 4. Bialynicki-Biula, I. & Mycielski, J.: Nonlinea wave mechanics. Annals of Physics 100, ) 5. I. Bialynicki-Biula, I., Mycielski, J.: Gaussons: Solitons of the Logaithmic Schodinge equation. Physica Scipta 0, )

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