Asymptotic analysis of solutions to elliptic and parabolic problems

Transcription

1 Linköping Sudies in Science and Technology. Disseraions No. 144 Asympoic analysis of soluions o ellipic and parabolic problems Peer Rand Maemaiska insiuionen Linköpings universie, SE Linköping, Sweden Linköping 26

2 ii Asympoic analysis of soluions o ellipic and parabolic problems c 26 Peer Rand Maemaiska insiuionen Linköpings universie SE Linköping, Sweden peran@mai.liu.se ISBN ISSN Prined by UniTryck, Linköping 26

3 iii Absrac In he hesis we consider wo ypes of problems. In Paper 1, we sudy small soluions o a ime-independen nonlinear ellipic parial differenial equaion of Emden-Fowler ype in a semi-infinie cylinder. The asympoic behaviour of hese soluions a infiniy is deermined. Firs, he equaion under he Neumann boundary condiion is sudied. We show ha any soluion small enough eiher vanishes a infiniy or ends o a nonzero periodic soluion o a nonlinear ordinary differenial equaion. Thereafer, he same equaion under he Dirichle boundary condiion is sudied, he nonlinear erm and righ-hand side now being slighly more general han in he Neumann problem. Here, an esimae of he soluion in erms of he righhand side of he equaion is given. If he equaion is homogeneous, hen every soluion small enough ends o zero. Moreover, if he cross-secion is sar-shaped and he nonlinear erm in he equaion is subjec o some addiional consrains, hen every bounded soluion o he homogeneous Dirichle problem vanishes a infiniy. In Paper 2, we sudy asympoics as of soluions o a linear, parabolic sysem of equaions wih ime-dependen coefficiens in,, where is a bounded domain. On, we prescribe he homogeneous Dirichle boundary condiion. For large values of, he coefficiens in he ellipic par are close o ime-independen coefficiens in an inegral sense which is described by a cerain funcion κ. This includes in paricular siuaions when he coefficiens may ake differen values on differen pars of and he boundaries beween hem can move wih bu sabilize as. The main resul is an asympoic represenaion of soluions for large. As a corollary, i is proved ha if κ L 1,, hen he soluion behaves asympoically as he soluion o a parabolic sysem wih ime-independen coefficiens. Acknowledgemens I would like o hank my supervisors Vladimir Kozlov and Mikael Langer for all heir suppor and invaluable hins during his work and Anders Björn and Jonna Gill for helping me wih L A TEX. Thanks also o everyone else who has helped me in some way.

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5 v Conens Inroducion 1 References 2 Paper 1: Asympoic analysis of a nonlinear parial differenial equaion in a semicylinder 7 1 Inroducion 7 2 The Neumann problem Noaion Problem formulaion and assumpions The main asympoic resul Corollaries of Theorem The corresponding problem in C An auxiliary ordinary differenial equaion The equaion for v Asympoics of small soluions of problem End of he proof of Theorem The Dirichle problem Problem formulaion and assumpions The main asympoic resul The corresponding problem in C End of he proof of Theorem The case of a sar-shaped cross-secion An esimae for soluions of a nonlinear ordinary differenial equaion A Some resuls from funcional analysis 4 A.1 Eigenvalues and eigenvecors of A.2 Exisence and uniqueness of bounded soluions of Poisson s equaion in C A.3 A local esimae for soluions of Poisson s equaion... 5 Paper 2: Asympoic analysis of soluions o parabolic sysems 57 1 Inroducion 57

6 vi 2 Problem formulaion and elemenary properies Noaion Spaces no involving ime Spaces involving ime Problem formulaion and assumpions An esimae for u Specral spliing of he soluion u 69 4 Esimaing he funcion v A general esimae Esimae for v Norm esimaes for R kl and g k w 84 6 Funcions h J+1,..., h M Definiion of funcions v, v 1 and v Inegro-differenial sysem for h J+1,..., h M A general esimae A paricular case of equaion Esimae for ȟ A represenaion for ȟ Funcions h 1,..., h J Equaion for ĥ; exisence and uniqueness resuls The homogeneous equaion A paricular soluion of Proof of Theorem Corollaries of Theorem A Eigenfuncions of a ime-independen operaor 124

7 1 Inroducion Mos differenial equaions and sysems are no possible o solve exacly. Hence, i is imporan o develop oher mehods of analyzing he properies of soluions. One of hese mehods is based on asympoic analysis. Alhough he soluions are unknown, i may be possible o find informaion abou heir behaviour as some variables end o some finie value or o infiniy. Asympoic analysis is used o sudy ime-dependen evoluion problems as well as ime-independen saionary problems. Frequenly, one is ineresed in behaviour of soluions as ime ends o infiniy, for example in quesions concerning sabiliy, periodiciy, rae of growh ec. A survey of evoluion problems and a general heory of analyzing hem, including asympoic analysis, can be found in Dauray, Lions [1], [2] or Lions, Magenes [8], [9], [1]. An imporan class of evoluion problems are reaciondiffusion problems. Such occur frequenly in biology and chemisry, see for example Fife [3] or Murray [11]. Imporan conribuions o asympoic mehods for evoluion problems can be found in Friedman [4], Pazy [12] and Vishik [13]. In his hesis, we use an approach developed in Kozlov, Maz ya, [6], [7]. Saring wih a linear or nonlinear equaion or sysem of equaions, he problem is reduced o firs order ordinary differenial equaions wih operaor coefficiens. Then, by use of a specral spliing, a finie dimensional sysem of firs order ordinary differenial equaions perurbed by a small inegro-differenial erm is obained for he leading erm. The main difficuly is o perform he above reducion and he sudy of he sysem of ordinary differenial equaions for he leading erm. The main resul is ha he asympoic behaviour of soluions of he iniial sysem of equaions is described by soluions of he above finie dimensional sysem. We use and exend his approach. In paper 1 we consider a nonlinear Emden-Fowler ype ime-independen parial differenial equaion in a semi-infinie cylinder and sudy he asympoics of soluions when he

8 2 unbounded coordinae ends o infiniy. The equaion is complemened by he Neumann or Dirichle boundary condiion. We analyze he asympoic behaviour of a given, small soluion of he problem. In he Neumann case, we obain a nonlinear ordinary differenial equaion for he leading erm in he asympoics of soluion. We also find an esimae for he remainder erm. From his asympoic formula i follows ha he soluion behaves asympoically like a periodic soluion. In he Dirichle case we show ha small soluions decrease exponenially. We also consider he case of a sarshaped cross secion and show ha if he nonlinear erm in he equaion is subjec o some addiional consrains, hen every bounded soluion of he homogeneous Dirichle problem vanishes a infiniy. The use of Pohožaev s ideniy is essenial in he proof. Paper 2 is devoed o he sudy of a linear parabolic sysem of equaions in a bounded domain under Dirichle boundary condiions and wih prescribed iniial values. We consider he asympoic behaviour of soluions as ime ends o infiniy. The ellipic par of he sysem is here considered as a perurbaion of ime-independen coefficiens. We consider a larger class of perurbaions han Kozlov, Maz ya [6]. Smallness of he perurbaions is assumed only in inegral sense. In paricular, we include such siuaions when he leading coefficiens may ake differen values on differen pars of and he boundaries beween hem can move wih bu sabilize as. Here we use anoher reducion han Kozlov, Maz ya [6] o obain he firs order sysem of ordinary differenial equaions perurbed by an inegro-differenial erm for he leading erms. Then an approach from Kozlov [5] is used o sudy he asympoic behaviour of soluions o his sysem. References [1] R. Dauray, J-L Lions, Mahemaical Analysis and Numerical Mehods for Science and Technology. Volume 5. Springer-Verlag, [2] R. Dauray, J-L Lions, Mahemaical Analysis and Numerical Mehods for Science and Technology. Volume 6. Springer-Verlag, [3] P. C. Fife, Mahemaical Aspecs of Reacing and Diffusing Sysems. Lecure Noes in Biomahemaics, 28. Springer-Verlag, [4] A. Friedman, Parial differenial equaions of parabolic ype. Prenice-Hall, Inc., Englewood Cliffs, N.J., [5] V. Kozlov, Asympoic represenaion of soluions o he Dirichle problem for ellipic sysems wih disconinuous coefficiens near he boundary. Elecron. J. Differenial Equaions 1 26, 46 pp. [6] V. Kozlov, V. Maz ya, Differenial Equaions wih Operaor Coefficiens. Springer-Verlag, 1999.

9 3 [7] V. Kozlov, V. Maz ya, An asympoic heory of higher-order operaor differenial equaions wih nonsmooh nonlineariies. Journal of Funcional Analysis , [8] J. L. Lions, E. Magenes, Non-Homogeneous Boundary Value Problems and Applicaions. Volume I. Springer-Verlag, [9] J. L. Lions, E. Magenes, Non-Homogeneous Boundary Value Problems and Applicaions. Volume II. Springer-Verlag, [1] J. L. Lions, E. Magenes, Non-Homogeneous Boundary Value Problems and Applicaions. Volume III. Springer-Verlag, [11] J. D. Murray, Mahemaical Biology. Springer-Verlag, [12] A. Pazy, Semigroups of Linear Operaors and Applicaions o Parial Differenial Equaions. Springer-Verlag, [13] M. I. Vishik, Asympoic behaviour of soluions of evoluionary equaions. Cambridge Universiy Press, 1992.

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13 Asympoic analysis of a nonlinear parial differenial equaion in a semicylinder Peer Rand Absrac We sudy small soluions of a nonlinear parial differenial equaion in a semi-infinie cylinder. The asympoic behaviour of hese soluions a infiniy is deermined. Firs, he equaion under he Neumann boundary condiion is sudied. We show ha any soluion small enough eiher vanishes a infiniy or ends o a nonzero periodic soluion of a nonlinear ordinary differenial equaion. Thereafer, he same equaion under he Dirichle boundary condiion is sudied, bu now he nonlinear erm and righ-hand side are slighly more general han in he Neumann problem. Here, an esimae of he soluion in erms of he righ-hand side of he equaion is given. If he equaion is homogeneous, hen every soluion small enough ends o zero. Moreover, if he cross-secion is sar-shaped and he nonlinear erm in he equaion is subjec o some addiional consrains, hen every bounded soluion of he homogeneous Dirichle problem vanishes a infiniy. An esimae for he soluion is given. 1 Inroducion Le be a bounded domain in R n wih C 2 -boundary. We define he semi-infinie cylinder C + = {x = x, x n : x, x n > }. In Secion 2 we sudy bounded soluions of he equaion under he boundary condiion U + quu = H in C U ν = on,. 1.2 Our aim is o describe he asympoic behaviour as x n of soluions U of problem 1.1, 1.2 subjec o where Λ is a posiive consan. Ux Λ for x C +, 1.3 7

14 8 We assume ha qu > if u. Moreover, q is coninuous and s, Λ qss q C Λ s 1.4 wih C Λ < λ 1. Here, λ 1 is he firs posiive eigenvalue of he Neumann problem for he operaor n = 2 x 2 k=1 k in. We se C =, + 1 and define L r loc C +, 1 r, as he space of funcions which belong o L r C for every. We also suppose H L p loc C + and 1 + s H L p C s ds <, 1.5 where { p > n/2 if n 4 p = 2 if n = 2, The main resul of Secion 2 is Theorem 2.1, which saes ha one of wo alernaives is valid: 1. U admis he asympoic represenaion Ux = u h x n + wx as x n +, where u h is a nonzero periodic soluion of u h + qu h u h = and w as x n. An esimae for he remainder erm w is given. 2. U as x n. An esimae for U is given in he heorem. If, for example, H = and C Λ as Λ, hen Corollary 2.3 gives he following esimae for U in he second case: Ux, x n C ɛ e λ 1 ɛ x n, where ɛ is an arbirary small posiive number and C ɛ is a consan depending on ɛ. In Secions we sudy soluions U of 1.1 subjec o 1.3 under he Dirichle boundary condiion U = on,. 1.7

15 9 Now we suppose ha q is coninuous and ha qv C Λ if v Λ, where C Λ < λ D. Here, λ D is he firs eigenvalue of he Dirichle problem for in. We assume also ha H L p C, wih p as in 1.6, is a bounded funcion of,. The main resul is Theorem 3.1 which gives an explici bound for U L C in erms of he funcion H L p C. This implies in paricular ha U L C as if he same is valid for H L p C. If H = and q=, hen he esimae from Theorem 3.1 implies ha Ux, x n C ɛ e λ D ɛ x n, 1.8 where ɛ > is arbirary. In Secion 3.5 we sudy all bounded soluions of 1.1, 1.7. Here we suppose addiionally ha n 4, ha he domain is sar-shaped wih respec o he origin, ha he funcion q is coninuous wih q = and ha qu is posiive for u. We also assume ha n 3 quu 2 n 1 2 u qvv dv ɛ quu for some ɛ >. Then Theorem 3.4 saes ha every bounded soluion of 1.1, 1.7 wih H = saisfies 1.8. Some examples of funcions saisfying 1.9 are qu = u p, p > 4 n 3, qu = u p e u, p > 4 n 3, qu = u p e u 1, p > 7 n n 3, linear combinaions wih posiive coefficiens of he funcions above. A naural quesion: under which condiions on q is i possible o remove ɛ in he relaion 1.8? The aim of Secion 3.6 is o sudy a similar quesion for he ordinary differenial equaion u λu + quu =, 1.1 where λ >, q is coninuous wih q = bu qu > for u and 1 qu u du <. Theorem 3.5 saes ha every soluion u of 1.1 subjec o u as saisfies u + u = O e λ

16 1 for large posiive. The problem 1.1 under he boundary condiions 1.2 or 1.7 wih qu = U p, p > 1 has been sudied in Kozlov [14]. There i is shown ha he resricion 1.3 is essenial for Theorems 2.1 and 3.1. One of he goals of his hesis is o exend some resuls from [14] o he equaion 1.1. The equaion u a u q u = in C +, 1.11 where q > 1, a > and wih he boundary condiion 1.2 is considered in Kondraiev [11]. Furhermore, he problem Lu = in C + u ν + a u q u = on,, where L is an ellipic parial differenial operaor, a > and q > 1 are consans is sudied in Kondraiev [12]. In boh hese cases i is proved ha he soluions of hese problems have asympoics of he form ux, x n = Cx σ n wih σ >. This shows ha he minus sign in 1.11 essenially changes he asympoic behaviour of soluions a infiniy. There is a lo of research on posiive soluions of nonlinear problems in an infinie cylinder and oher unbounded domains. We direc he reader o Bandle and Essén [3], Beresycki [4], Beresycki, Caffarelli and Nirenberg [5], Beresycki, Larrouurou and Roquejoffre [6], Beresycki and Nirenberg [7] and Kondraiev [13] where also furher references can be found. Small global soluions of he equaion u + λu + fu, u x, u y = in a wo-dimensional srip wih homogeneous Dirichle boundary condiions are sudied in Amick, Toland [2] and Kirchgässner, Scheurle [1]. 2 The Neumann problem 2.1 Noaion The Laplace operaor and he gradien in R n are denoed by and, respecively. For he corresponding operaors in R n we inroduce = n 2 x 2 k=1 k and = x 1,..., x n.

17 11 By we denoe a bounded domain in R n wih C 2 -boundary and n- dimensional Lebesgue measure. We inroduce he cylinder and he semicylinder C = {x, x n : x and x n R} C + = {x, x n : x and x n > }. We le ν and ν denoe he ouward uni normals o C and, respecively. Thus ν R n while ν R n and ν = ν,. Afer inroducing C =, + 1, we say ha a funcion u : R n R belongs o L r loc C or W k,r loc C, 1 r and k =, 1,..., if i belongs o Lr C or W k,r C for every R. 2.2 Problem formulaion and assumpions Assume ha p is subjec o 1.6. We sudy he asympoic behaviour as x n of soluions U W 2,p loc C + of he problem U + quu = H in C + U ν = on, 2.1 saisfying 1.3. We assume ha q is coninuous and posiive for u and saisfies 1.4. We suppose furher ha H L p loc C + is subjec o 1.5 In order o moivae 1.6, le us consider a bounded soluion U W 1,2 loc C + of 2.1. By Lemma A.16 in Secion A.3 we ge ha U W 2,p loc C +. Furhermore, i follows from well-known Sobolev inequaliies, see for example Theorem 5.6 in Evans [8], ha, since p > n/2, here exiss a posiive γ such ha eiher U C,γ C or U C 1,γ C for every >. Hence i is meaningful o assume ha he sudied soluion belongs o W 2,p loc C + and is bounded. 2.3 The main asympoic resul The aim of Secion 2 is o prove he following heorem concerning he asympoic behaviour of soluions of 2.1 subjec o 1.3: Theorem 2.1 Suppose ha U W 2,p loc C +, where p saisfies 1.6, is a soluion of 2.1 subjec o 1.3. Suppose also ha q is coninuous, qu > if u and ha he Lipschiz condiion 1.4 is fulfilled. Finally, assume ha H L p loc C + saisfies 1.5. Then one of he following alernaives is valid:

18 12 1. Ux = u h x n + wx, where u h is a nonzero periodic soluion of and w L C C + u h + qu h u h = s H L p C s ds e λ 1 C Λ s H L p C s ds + e λ 1 C Λ for 1. The righ-hand side ends o as. 2. U, x n L as x n. Furhermore, Ux = u x n + wx, where u 2 u + qvv dv C H L 2 p C s ds e λ 1 C Λ s H L p C s ds + e λ 1 C Λ and w L C C for 1. e λ 1 C Λ s H L p C s ds + e λ 1 C Λ 2.3 The proof of his heorem is conained in Secions Theorem 2.1 is a generalizaion of Theorem 3 in Kozlov [14], where he case qu = U p, p > 1 is sudied. We use he same approach in his hesis. Since mos of he proofs in [14] are brief or absen, we presen here complee proofs of all asserions. Our resricion 1.5 is differen from he corresponding resricion in Theorem 3 [14]. This rigorous analysis of he proofs indicaes ha possibly 1.5 is he righ assumpion also in [14]. In he nex secion we give some corollaries of Theorem Corollaries of Theorem 2.1 Corollary 2.2 Suppose, in addiion o he condiions in Theorem 2.1, ha he nonlinear erm q has he propery ha he consan C Λ in 1.4 ends o as Λ ends o. Then he esimaes 2.2 and 2.3 can be improved, namely, he consan λ 1 C Λ can be replaced by λ 1 ɛ where ɛ > is arbirary. In his case, he consan C appearing in 2.2 and 2.3 is dependen of ɛ. Proof. Since U, x n L as x n we can apply he heorem for he semicylinder T, where T is sufficienly large and Λ small enough.

19 13 Corollary 2.3 Suppose, in addiion o he condiions in Corollary 2.2, ha H =. If case 2 in Theorem 2.1 occurs, hen, for every ɛ, λ 1, here exiss a consan C ɛ such ha Ux, x n C ɛ e λ 1 ɛ x n for x n Proof. We begin wih proving ha here exiss a consan A ɛ such ha From 2.2 i follows ha and Since we ge Furhermore, 2.3 gives Ux, x n A ɛ e 1 2 λ1 ɛ x n for x n u Ce 1 2 λ1 C Λ u as. u = where C does no depend on. Since we ge from 2.6 and 2.7 ha u s ds u Ce 1 2 λ1 C Λ. 2.6 w L C Ce λ 1 C Λ, 2.7 Ux = u x n + wx Ux, x n Ce 1 2 λ1 C Λ x n. 2.8 Using ha C Λ as Λ and considering problem 2.1 in a semicylinder,, where is sufficienly large, we can suppose ha C Λ < ɛ. Then he esimae 2.5 follows from 2.8. We se C = + 1/4, + 3/4. Lemma A.16 in Secion A.3 implies ha so from 1.4 i follows ha U W 2,2 C C U L 2 C + quu L 2 C U W 2,2 C C U L 2 C

20 14 and by using 2.5 we ge he esimae U W 2,2 C Ce 1 2 λ1 ɛ. Corollary in Kozlov and Maz ya [15], wih he parameers k =, k + = λ 1, m = 2 and m + = 1 ogeher wih he fac ha R can be made suiably small, implies ha U W 2,2 C Ce λ 1 ɛ. 2.9 We can now use local esimaes and an ieraion procedure as in he proof of Lemma A.16 in Secion A.3 o obain U L C C U W 2,2 C, where C = + 3/8, + 5/8. Combinaion of he las esimae and 2.9 gives if 3/8. This implies 2.4. U L C Ce λ 1 ɛ 2.5 The corresponding problem in C We now begin proving Theorem 2.1. Before urning o equaion 2.1, we sudy a soluion u W 2,p loc C of he problem u + quu = h in C u ν = on C 2.1 saisfying sup ux Λ, 2.11 x C where Λ is he same consan as in 1.3 and p is subjec o 1.6. By he Sobolev embedding heorem he soluion is coninuous. Thus we do no need o use essenial supremum. As before, we assume ha q is coninuous and posiive for u and saisfies 1.4. We also suppose ha h L p loc C and ha 1 + s h L p C s ds < Theorem A.8 in Secion A.1 saes ha here exiss an ON-basis of L 2 consising of eigenfuncions of he operaor for he Neumann problem in. Le φ, x n denoe he eigenfuncion wih L 2 -norm equal o 1 corresponding o he eigenvalue λ =, i.e. φ = /2. Se u

21 15 o he orhogonal projecion of u ono he subspace of L 2 spanned by φ, ha is ux n = 1 ux, x n dx, and define vx by he equaliy ux = ux n + vx Insering 2.13 in 2.1 and inegraing over, we obain u x n + 1 vx, x n dx + 1 fux, x n dx = hx n, 2.14 where and hx n = 1 hx, x n dx f = q Due o he homogeneous boundary condiion in 2.1, Greens formula gives u dx =. Therefore 1 v dx = 1 = 1 v dx + d2 1 dx 2 v dx n u dx =. Equaion 2.14 can now be wrien as u x n + 1 fux, x n dx = hx n 2.16 and by defining we ge Ku, vx n = fux n fu + vx n 2.17 u x n + fux n = hx n + Ku, vx n We will ofen wrie Kx n or K insead of Ku, vx n. Using 2.1, he equaion v = h fu + v u is obained which, ogeher wih 2.16, implies ha v = Mu, v + h 1 in C v ν = on C, 2.19

22 16 where and Mu, vx = fux n + vx fu + vx n 2.2 h 1 = h h The equaions 2.18 and 2.19 will play a cenral role in he sequel. 2.6 An auxiliary ordinary differenial equaion In his secion we sudy he equaion ξ + qξξ = g,, 2.22 where 1 is given. We assume ha qv is posiive for every v R, possibly excep for v =, and coninuous. We also assume ha qvv is Lipschiz coninuous on every finie inerval and ha sgs ds < Since soluions of 2.22 wih g = will play an imporan role in he asympoic represenaion of ξ as, we will now describe hem. We have hus he equaion ξ + qξξ =, Muliplying 2.24 by ξ and inegraing, we obain 1 2 ξ 2 + qξξξ ds = 1 2 ξ 2. Using ha where dgξ d Gv = v = qξξξ, 2.25 qww dw, we ge 1 2 ξ 2 + Gξ = c, where c is consan. If { } c < min qss ds, qss ds, 2.26 hen ξ and ξ are bounded and ξ is periodic. Conversely, if we know in advance ha ξ is a bounded soluion, hen c saisfies This fac and an even more general resul will be deduced in he proof of Theorem 2.4.

23 17 We also noe ha if he righ-hand side of 2.26 is infinie, hen every soluion of 2.24 is periodic. In he following heorem we describe he asympoic behaviour of bounded soluions of Theorem 2.4 Le ξ be a bounded soluion of Then one of he wo following alernaives occurs: 1. ξ = ξ h +w, where ξ h is a nonzero periodic soluion of 2.24 and w + w = O sgs ds 2.27 as. 2. Boh ξ and ξ end o as and ξ 2 ξ + qvv dv = O gs ds The remaining par of his secion is devoed o he proof of his heorem. We sar wih he following lemma: Lemma 2.5 Le ξ be a bounded soluion of Then 1 2 ξ 2 + Gξ = c + O gs ds as, where c is a nonnegaive consan depending on and ξ Proof. We begin wih proving ha ξ is bounded. Muliplying 2.22 by ξ and inegraing, we obain 1 2 ξ 2 Now 2.25 implies ha gsξ s ds = 1 2 ξ 2 qξsξsξ s ds = Gξ Gξ, qξsξsξ s ds. 2.3 where Gξ is uniformly bounded in due o he boundedness of ξ. This means ha also he lef hand side of 2.3 is bounded in. Se C T = sup T ξ. We have for T gξ ds C T g ds,

24 18 which is finie because of Therefore, from 2.3 i follows ha 1 2 ξ 2 C T g ds C wih a consan C independen of and T. Taking supremum over T we ge 1 2 C2 T C T g ds C which gives an upper bound for C T independen of T. Thus ξ is bounded for all. The equaion 2.3 is equivalen o 1 2 ξ 2 + Gξ = 1 2 ξ 2 + Gξ + From 2.23 and he boundedness of ξ, we ge gξ ds = C 1 gξ ds = C 1 + O gξ ds g ds for some consan C 1. This equaliy applied o 2.31 finally implies By leing i follows ha c. Proof of Theorem 2.4: Since ξ is bounded, here exiss a number L such ha ξ L, We rewrie 2.22 as he sysem of firs order equaions { y 1 = y 2 y 2 = g qy 1 y 1, where y 1 = ξ and y 2 = ξ. In polar coordinaes, y 1 = r cos φ, y 2 = r sin φ, he above sysem akes he form { r cos φ rφ sin φ = r sin φ r sin φ + rφ cos φ = g qr cos φr cos φ. This implies ha Define he funcion ρ as r = g sin φ + 1 qr cos φr sin φ cos φ φ = g cos φ r qr cos φ cos 2 φ sin 2 φ ρ = r2 sin 2 φ 2 + r cos φ qvv dv 2.34

25 19 and observe ha ρ. Since ρ = 1 2 ξ 2 + ξ i follows from Lemma 2.5 ha ρ = c + O qvv dv, gs ds 2.35 wih c. If c =, hen he second alernaive in he heorem is valid. Suppose now ha c >. I is more convenien o use he variables ρ, φ insead of r, φ. The second equaion of 2.33 becomes where φ = g cos φ F ρ, φ, 2.36 rρ, φ F ρ, φ = qrρ, φ cos φ cos 2 φ + sin 2 φ For readabiliy we will make an abuse of noaion and someimes consider r as a funcion of and someimes of ρ and φ. We now show ha c saisfies If boh inegrals in 2.26 are infinie, his is obvious. Suppose ha one of hem is finie. Firs we conclude ha here exiss a posiive number r and a 1 such ha r r 2.38 for 1. Indeed, if r j for some sequence { j } j=2 ρ j which conradics Thus 2.38 follows. Inegraing 2.36 from 1 o, where 1, we obain hen also φ = φ 1 + gr cos φ ds 1 q cos 2 φ ds 1 sin 2 φ ds From 2.23 and 2.38 i follows ha he firs inegral in 2.39 has a finie limi as. Le us show ha one of he las wo inegrals ends o infiniy as. Suppose ha 1 sin 2 φ ds <. Then l{ 1 : sin 2 φ > 1/2} <, where ld denoes he Lebesgue measure of D. Thus le = for E = { 1 : cos 2 φ 1/2}. From 2.38 and 2.32 i hen follows ha r / 2 r cos φ L

26 2 on E. Therefore, here exiss a posiive consan q such ha qr cos φ q on E, which implies ha This proves ha 1 q cos 2 φ ds q 2 le =. φ 2.4 as. Because of 2.38 and 2.4, we can find sequences {τ j }, {s j } such ha τ j, s j as j and rτ j sin φτ j = rs j sin φs j =, rτ j cos φτ j < bu rs j cos φs j >. Equaions 2.34 and 2.32 hen imply ha ρs j = Analogously, rsj cos φs j ρτ j L qvv dv L qvv dv < Since ρ c as, we have { L c min qvv dv, { < min qvv dv, and 2.26 is proved. By 2.26 here exiss an ɛ such ha { < ɛ < min qvv dv, qvv dv < L qvv dv. } qvv dv } qvv dv } qvv dv c. qvv dv. Nex, we prove ha here exiss posiive consans A 1, A 2, B 1 and B 2 such ha A 1 rρ, φ A and B 1 F ρ, φ B if c ɛ ρ c + ɛ. By he same argumen as in he proof of he exisence of r in 2.38, we can show ha here exiss a posiive A 1 such ha r A 1. To prove he righ inequaliy in 2.41 we inroduce r 1 φ as he unique soluion of he equaion ρr, φ = c + ɛ. Tha his equaion has one soluion r = r 1 φ for every φ is a consequence from hese hree facs which all follow from 2.34:

27 21 1. ρ, φ is sricly increasing. 2. ρr, φ as r for every φ. 3. We have for φ nπ ρ as r and for φ = nπ ρ ± qvv dv, where boh inegrals are larger han c + ɛ. Since ρ is coninuous, so is r 1. We can hus define A 2 = max r 1φ φ [,2π] and conclude ha he consan A 2 saisfies he righ inequaliy in The exisence of B 2 follows from 2.37 ogeher wih he coninuiy of q and he righ inequaliy in To show he exisence of B 1 we proceed as follows. Suppose ha here exiss sequences {ρ j } j=1 and {φ j} j=1 wih such ha ρ j [c ɛ, c + ɛ], j = 1, 2,... F ρ j, φ j = qρ j, φ j rρ j, φ j 2 cos 2 φ j + rρ j, φ j 2 sin 2 φ j rρ j, φ j 2 as j. Here we use he abbreviaion qρ, φ for qrρ, φ cos φ. Due o 2.41, his implies ha { qρj, φ j cos 2 φ j sin 2 φ j. This can happen only if cos φ j and sin φ j which is impossible so he lef inequaliy in 2.42 is proved. Le us show ha F ρ, φ = F c, φ + O ρ c 2.43 for ρ c < ɛ. clearer, we se In order o make he compuaions somewha visually r = rρ, φ r = rc, φ q = qr cos φ q = qr cos φ

28 22 and obain F ρ, φ F c, φ = q cos 2 φ q cos 2 φ qr r cos φ q r r cos φ rr From 2.34 we ge ha r qr cos φ q r cos φ + r r q r cos φ rr. r = rf ρ 2.44 and by 2.41 and 2.42, he righ-hand side is bounded uniformly in φ. Therefore r r C ρ c, 2.45 where C is a consan independen of r, ρ and φ. Using he Lipschiz coninuiy of he funcion qvv and 2.45, we derive from 2.44 he relaion If ρ and φ are considered as funcions of, he represenaion 2.35 shows ha F ρ, φ = F c, φ h, where Now choose 2 such ha h = O g ds ρ c ɛ if 2 and se g 1 = r g cos φ The equaion 2.36 can now be rewrien as This implies ha φ φ 2 dψ F c, ψ = φ = F c, φ + g 1 + h. 2 + = g 1 s F c, φs + hs ds F c, φs 2 g 1 s F c, φs ds + 2 hs F c, φs ds. 2.48

29 23 Using 2.41, 2.42 and 2.47, we ge g 1 s 2 F c, φs ds = C g cos φ 1 rf c, φ ds = C 1 O gs ds Furhermore, he relaions 2.42 and 2.46 imply ha hs 2 F c, φs ds = C 2 + O hs ds = C 2 + O sgs ds. 2.5 By using 2.49 and 2.5, we derive from 2.48 he relaion φ dψ F c, ψ = c 1 + O sgs ds, Le us rewrie he equaion 2.24 in he variables ρ and φ. The above calculaions can be used wih g =. Equaions 2.35 and 2.51 become ρ h = a φh dψ F c, ψ = a ,. We show ha he homogeneous equaion 2.24 has a soluion which saisfies 2.52 wih a = c and a 1 = c 1. Firs choose ρ h and φ h such ha ρ h = c φh dψ + F c, ψ = c 1. The las equaion is solvable because of We can now reconsruc ξ h and ξ h and ake hese as he Cauchy daa for The equaliy 2.35 and he firs equaliy in 2.52 wih a = c imply ha ρ = ρ h + O gs ds 2.53 and by using 2.42 we obain φ dψ φ φ h B 2 F c, ψ. φ h Furhermore, 2.51 ogeher wih he second equaion in 2.52 wih a 1 = c 1 gives φ φ h dψ F c, ψ = O sgs ds.

30 24 Therefore we have φ = φ h + O sgs ds The nex sep is o sudy he relaion beween r and r h = rρ h, φ h. Expanding rρ, φ near ρ h, φ h and using 2.53, 2.54, we obain r = r h + O sgs ds. This, ogeher wih he fac ha ξ = r cos φ, ξ = r sin φ, finally gives ξ = ξ h + O sgs ds ξ = ξ h + O sgs ds and he heorem follows. 2.7 The equaion for v We now sudy soluions v of In Secion 2.5 we inroduced M and h 1 by 2.2 and I is sraighforward o check ha h 1 L p loc C, and h 1 x, x n dx =, 2.55 M dx =, 2.56 vx, x n dx = M = w w, 2.57 where w = fu fu 2.58 and f is given by Furhermore, by noing ha h 1 L p C 2 h L p C, we ge from 2.12 ha 1 + s h 1 L p C s ds < Clearly, he funcion v belongs o L C and we have he following resul:

31 25 Lemma 2.6 The funcion v in 2.19 saisfies he esimae v L C C where C depends on p, n,, Λ and C Λ. Proof. We sar by proving he inequaliy e λ 1 C Λ s h L p C s ds, v, x n L 2 2 e λ 1 C Λ x n s h 1, s L λ 1 C 2 ds. Λ 2.61 Defining w as in 2.58, inequaliy 1.4 implies ha w C Λ v 2ΛC Λ. I follows from 2.57 ha M 4ΛC Λ, which ogeher wih 2.59 gives ha e λ 1 s M + h 1 L2 C s ds <. Also, he equaions 2.55 and 2.56 imply ha Mu, vx, x n + h 1 x, x n dx =. Because of he orhogonaliy beween w w and w in L 2, we ge from 2.57 and he Pyhagorean heorem ha Mu, v, x n L 2 w, x n L 2. The condiion 1.4 hen gives Mu, v, x n L 2 C Λ v, x n L 2. This and Lemma A.12 in Secion A.2 show ha 2.19 has a soluion v fulfilling v, x n L e λ 1 x n s h 1, s L λ 2 + C Λ v, s L 2 ds Insering he righ-hand side of his expression in he las occurrence of v, s L 2 and ieraing, we obain v, x n L 2 k= T k, where C k Λ T k = 2 λ 1 k+1 e λ1 x k n + R k+1 j= j j+1 h 1, k L 2 d d 1 d k. The variable ransformaion j = s + s j+1 for j =,..., k 1, k = s applied o 2.62, gives v, x n L2 Gx n s h 1, s L2 ds, 2.63

32 26 where G = 1 2 e λ1 + λ 1 R k e k=1 C k Λ 2 λ 1 k+1 λ1 s 1 + s 1 s s k s k + s k ds 1 ds 2... ds k In order o calculae he funcion G we consider he wo differenial operaors d2 d +λ 2 1 C Λ and d2 d +λ 2 1. Clearly, heir fundamenal soluions are 1 g = 2 e λ1 C Λ λ 1 C Λ and h = 1 2 e λ1. λ 1 Therefore dg2 d 2 + λ 1g = δ + C Λ g, which implies ha g = h δ + C Λ g, i.e. g = h + C Λ h sgs ds. Insering his expression for g ino he righ-hand side and repeaing his procedure, we obain g = 1 2 e λ1 + λ 1 e R k k=1 C k Λ 2 λ 1 k+1 λ1 s 1 + s 1 s s k s k + s k ds 1 ds 2... ds k, where he righ-hand side coincides wih he righ-hand side of Hence, 1 G = 2 e λ1 C Λ λ 1 C Λ and 2.61 now follows from In he remaining par of his secion, le C denoe a generic consan depending only on p, n,, Λ and C Λ. We define C = + 1/4, + 3/4 and C = + 1/8, + 7/8. From Corollary A.18 in Secion A.3 i follows ha v L C C v L p C + M L p C + h L p C. Furhermore, since w C Λ v and w L p C /p w L p C, i follows from 2.57 ha M L p C C Λ 1 + /p v L p C.

33 27 Thus, v L C C v L p C + h L p C and, afer ieraing as in he proof of Lemma A.16 in Secion A.3, we ge v L C C v L 2 C + h L p C We are now in posiion o perform he las sep of he proof of 2.6. Begin by looking a h L p C and se C,1 =, + 1/2. For τ 1/2, we have h L p C,1 h L p C τ and inegraing from 1/2 o we ge h L p C,1 2e λ 1 C Λ /2 /2 e λ 1 C Λ τ h L p C τ dτ. Wih a similar esimae of h L p C,2, where C,2 = + 1/2, + 1, we ge h L p C C e λ 1 C Λ τ h L p C τ dτ To find an esimae for v L 2 C, we use 2.61 ogeher wih Minkowski s inequaliy yielding v L 2 C = C +1 v, τ 2 L 2 dτ 1/2 +1 e 2 λ 1 C Λ τ s dτ 1/2 h, s L 2 ds. Using he inequaliy +1 1/2 e 2 λ 1 C Λ τ s dτ Ce λ 1 C Λ s and making he subsiuion s s + τ, we arrive a v L 2 C C From he inequaliy we now obain 1 1 v L 2 C C e λ 1 C Λ s τ h, s + τ L 2 dτds. h, s + τ L 2 dτ h L 2 C s, e λ 1 C Λ s h L 2 C s ds. This, ogeher wih 2.65 and 2.66, gives he esimae 2.6 wih v L C replaced by v L C. Since v L C v L C /2 + v L C + v L C +1/2, he inequaliy for v L C implies 2.6 and he proof is complee.

34 Asympoics of small soluions of problem 2.1 In his secion we coninue o sudy soluions of problem 2.1. Our goal is o find asympoics of u as x n. Lemma 2.7 Le u W 2,p loc C be a soluion of 2.1 subjec o Then eiher 1. ux = u h x n + wx, where u h is a nonzero periodic soluion of u h + qu h u h = 2.67 or and for 1 w L C C s h L p C s ds + e λ 1 C Λ s h L p C s ds 2. u, x n L as x n. If u = u + v as before, hen 2.68 u u 2 + qvv dv 2 C h L p C s ds + and e λ 1 C Λ s h L p C s ds 2.69 v L C C e λ 1 C Λ s h L p C s ds. 2.7 The righ-hand sides of 2.68, 2.69 and 2.7 end o as. Proof. We se α = λ 1 C Λ and, as before, represen u as u = u + v. Le us firs sudy he erm u which saisfies he equaion We denoe he righ-hand side of 2.18 by g, i.e. gx n = hx n + Ku, vx n wih K given by Since K = 1 quu quu dx, i follows from 1.4 ha Kx n C v, x n L 1. From 2.61 we ge gx n C hx n + e α xn s h 1, s L2 ds. 2.71

35 29 Le us find a bound for sgs ds for 1. Obviously, shs ds 1 s h, s L 1 ds C s h, s L 2 ds, 2.72 and s e α s τ h 1, τ L 2 dτ ds C I 1, τ h, τ L 2 dτ, 2.73 where I 1, τ = For τ, a direc compuaion yields I 1 = e ατ e α α + e α α 2 se α s τ ds Ce α τ If τ, we make he subsiuion s τ s in 2.74 and obain I 1, τ s e α s ds + τ e α s ds 2α 2 + α τ The combinaion of 2.71, 2.72, 2.73 and he use of he esimaes 2.75 and 2.76 for I 1 give sgs ds C s h, s L 2 ds + e α s h, s L 2 ds Nex we show ha i is possible o replace h, s L 2 by h L 2 C s in Denoe he righ hand side of 2.77 by G and fix a τ [, 1]. Making he variable subsiuion s s + τ we ge G C s + τ h, s + τ L 2 ds τ τ + e α s h, s + τ L 2 ds. Since τ [, 1], his implies ha G C e α s h, s + τ L 2 ds + s h, s + τ L 2 ds.

36 3 Therefore where sgs ds C k, s h, s + τ L2 ds, 2.78 k, s = { e α s if s < s if s. Inegraing 2.78 wih respec o τ over [, 1] and using he inequaliy 1 we obain sgs ds C h, s + τ L 2 dτ C h L p C s, s h L p C s ds + e α s h L p C s ds The righ-hand side is bounded because of Therefore he assumpion 2.23 is verified and we can apply Theorem 2.4 on he funcion u. Hence eiher or 1. ux n = u h x n + w 1 x n, where u h is a nonzero periodic soluion of 2.67 and w 1 saisfies ux n as x n and 2.28 is valid. Le us esimae he righ-hand sides in 2.27 and From 2.79 we ge ha w 1 C s h L p C s ds + e α s h L p C s ds and o obain he esimae 2.69, we can use ha gs ds C h L p C s ds + e α s h L p C s ds. This follows from 2.71 and calculaions similar o hose done o obain he esimae Seing wx = w 1 x n + vx in he firs case and using Lemma 2.6 o esimae v L C, we arrive a The esimae 2.7 follows direcly from Lemma 2.6. Le us show ha he righ-hand sides of 2.68, 2.69 and 2.7 end o as. In fac, i is enough o show his for Due o 2.12, he firs erm in he righ-hand side of 2.68 ends o as. The second erm is esimaed by

37 31 + M e α s h Lp C s ds C M e α 2 s 1 + s h L p C s ds M C e α 2 s 1 + s h L p C s ds + e α 2 s 1 + s h L p C s ds M s h L p C s ds for any M. Given ɛ >, we can choose M so large ha he las inegral is less han ɛ/2. The inegral M e α 2 s 1 + s h L p C s ds is hen majorized by C exp α/2 where C depends on M and h and he expression in he righ-hand side of 2.8 is less han ɛ if is large enough. Hence he righ-hand side of 2.68 ends o as and he proof is complee. 2.9 End of he proof of Theorem 2.1 We are now in posiion o complee he proof of Theorem 2.1. Proof of Theorem 2.1: Le η be a smooh funcion wih η 1, η = if 1 and η = 1 if 2. We se ux = ηx n Ux. Then { u + quu = h in C ν u = on C, where h = ηh + quu ηquu + 2η U xn + η U. Obviously, ηh L p loc C. The funcions η U and χ = quu ηquu are bounded and equal o for x n 1 and x n 2. Furhermore, we have ha he funcion η U xn belongs o L p C and is also equal o for x n 1 and x n 2. Thus, he inequaliy 2.12 follows from 1.5. Now we can apply Lemma 2.7 on u. In he firs case we have he represenaion u = u h + w 1 where u h and w 1 are subjec o 2.67 and 2.68, respecively. We se w 1 if x n 2 w = U u h if 1 x n < 2 oherwise.

38 32 Then w L C + and for > 2, w L C C + 2 s H L p C s ds + e α s h L p C s ds C s H L p C s ds + 2 e α s H L p C s ds e α s H L p C s ds + e α. I is easy o see ha his esimae is valid also for 1 2 so he firs case in Theorem 2.1 is proved. In he second case of Lemma 2.7, we have u = u + v, where u and v are subjec o 2.69 and 2.7, respecively. We se u = u and v if x n 2 w = U u if 1 x n < 2 oherwise. The inequaliies 2.2 and 2.3 now follow from 2.69 and 2.7 and he proof is complee. 3 The Dirichle problem 3.1 Problem formulaion and assumpions In Secion 3 we sudy bounded soluions of he Dirichle problem { U + quu = H in C+ U = on,. 3.1 We assume ha U fulfills sup Ux Λ 3.2 x C + and ha q is coninuous. Le λ D be he firs eigenvalue of he Dirichle problem for in. We suppose ha here is a consan C Λ < λ D such ha v Λ qv C Λ. 3.3 Finally, we assume ha H L p loc C +, where p saisfies 1.6, and ha H L p C is a bounded funcion of,.

39 The main asympoic resul The main resul of Secion 3 is he following heorem concerning he asympoic behaviour of soluions U of 3.1. Theorem 3.1 Assume ha U W 2,p loc C +, where p saisfies 1.6, is a soluion of 3.1 subjec o 3.2. Assume furher ha q is coninuous and ha 3.3 is saisfied. Also, assume ha H L p loc C + and ha H L p C is a bounded funcion of for. Then U L C C e λ D C Λ s H L p C s ds + e λ D C Λ, 3.4 where C is independen of. In paricular, if H = and q =, hen U L C = O e λ D ɛ, 3.5 where ɛ > is arbirary. Remark 3.2 If H L p C as, hen, by he same reasoning as in he end of he proof of Lemma 2.7, we see ha he righ-hand side of 3.4 ends o as. The proof of Theorem 3.1 is similar o he proof of Theorem 2.1 bu shorer. I is conained in Secion 3.3 and 3.4. Theorem 3.1 is a generalizaion of Theorem 2i in Kozlov [14]. As in he Neumann problem, he case qu = U p is sudied in [14]. 3.3 The corresponding problem in C As in he Neumann problem in Secion 2, we firs consider he problem in he whole cylinder. Suppose ha u W 2,p loc C is a soluion of he problem { u + quu = h in C u = on C 3.6 saisfying wih he same Λ as in 3.2. We have he following: sup ux Λ 3.7 x C Lemma 3.3 Le u be a soluion of 3.6 fulfilling 3.7 and suppose ha h L p loc C wih h L p C bounded on R. Then we have he esimae u L C C e λ D C Λ s h L p C s ds. 3.8

40 34 Proof. We consider he boundary value problem { v = g in C v = on C, where e λ 1 s g L 2 C s ds <. From Secion A.2 i follows ha i has a unique bounded soluion v W 1,2 loc C wih v, x n L λ D e λ D x n s g, s L 2 ds. Applying his esimae o he soluion u of he problem 3.6, we obain u, x n L e λ D x n s h, s L λ 2 + C Λ u, s L 2 ds. D Ieraing his esimae in he same way as in he proof of Lemma 2.6 in Secion 2.7, we ge 1 u, x n L 2 2 λ D C Λ Now use he local esimae e λ D C Λ x n s h, s L 2 ds. u L C C u L 2 C + h L p C 3.9 compare wih 2.65 and esimae he erm u L 2 C by 3.9. Doing he same calculaions as in Secion 2.7 we arrive a End of he proof of Theorem 3.1 We now complee he proof of Theorem 3.1. Proof of Theorem 3.1: We use he same smooh funcion η as in he proof of Theorem 2.1 page 31 and ge, for u = ηu, he problem { u + quu = h in C u = on C where h = ηh + quu ηquu + 2η U xn + η U. Clearly, h L p C is bounded and he esimae 3.4 follows from 3.8.

41 35 Suppose now ha H = and q =. From 3.4 i follows ha U L C = O e λ D C Λ. By choosing T large enough and considering 3.1 in T, insead of C +, we can make Λ and, since qu as u, also C Λ arbirary small. From his 3.5 follows. 3.5 The case of a sar-shaped cross-secion In his secion we show ha under some special assumpions on and q, every bounded soluion of 3.1 wih H = will saisfy 3.5. This is a generalizaion of Theorem 2iii in [14] where he case qu = U p is sudied. Theorem 3.4 Suppose ha n 4 and is sar-shaped wih respec o he origin and has C 2 -boundary. Also assume ha q is coninuous wih q =, and qu > oherwise, and ha n 3 quu 2 n 1 2 u qvv dv ɛ quu for some ɛ >. Then every bounded soluion of 3.1 wih H = is subjec o 3.5. Before proving his heorem we give some examples of funcions q saisfying 3.1. Le us check ha all funcions of he form qu = f u u a+δ 3.11 wih a = 4/n 3, δ > and f being a nondecreasing funcion saisfy 3.1. Obviously, he funcion q in 3.11 is even. Therefore also boh sides of he inequaliy 3.1 are even and we can assume ha u. We have u u qvv dv fu v a+1+δ dv = fuua+2+δ a δ and by using his inequaliy we obain n 3 quu 2 n 1 2 n 3 fuu a+2+δ 2 bfuu a+2+δ, u qvv dv n 1fuua+2+δ a δ = where b = δn 3 2a δ.

42 36 Obviously, b > for every δ > since n 4. Choosing ɛ = b, we see ha 3.1 is fulfilled. Here are some examples of funcions saisfying 3.11: i qu = u p, p > 4 n 3. ii qu = u p e u, p > 4 n 3. iii qu = u p e u 1, p > 7 n n 3. iv Linear combinaions wih posiive coefficiens of funcions from i - iii. Proof of Theorem 3.4: The funcion u is a soluion of he problem { u + quu = in C u = on,. As before, we se Gu = u qvv dv. Using Pohožaev s ideniy wih respec o he x -variables, compare wih Secion 3.2, Kozlov [14], we ge = u + fux u + u xn x n x u = div ux u x u 2 + x Gu 2 n 1Gu + u xn x n x u. + n 3 u 2 2 This implies ha = div ux u + div x u 2 + n 3 u 2 n 1Gu x Gu + u 2 x n Se C N = 1, N and Γ N = 1, N and observe ha 3.13 u = u ν ν on Γ N. This follows by represening he vecor u a a cerain poin on Γ N as he sum of a normal and a angen vecor. The angen vecor is hen zero because of he homogeneous Dirichle boundary condiion. Using ha Gu = on Γ N and inegraing 3.13 over C N, we obain n 3 = u 2 + u 2 x n n 1Gu dx C N 2 Γ N x ν 2 u ds + F 1 N F 1 1, ν 3.14

43 37 where F 1 = u xn x u dx. xn = Muliplying 3.12 by u and using Green s formula, we have = u 2 + quu 2 dx + F 2 N F 2 1, 3.15 C N where F 2 = u xn u dx. xn = A linear combinaion of 3.14 and 3.15 gives 2 u + n 3 quu 2 n 1Gu dx C N x n u x ν ds = F 1 F N, 2 Γ N ν 3.16 where F = F 1 + n 3 F 2. 2 Le us show ha F is bounded on [1,. For given >, choose r > n. Using he Sobolev embedding heorem, we obain u L C C u W 1,r C C u W 2,r C. Now, Lemma A.16 in Secion A.3 gives u W 2,r C C quu L r C + u L 2 C C 1. Therefore, u L C is a bounded funcion. This implies ha F 1 and F 2 are bounded funcions on [1,. Since is sar-shaped, i follows ha x ν. This and 3.1 imply ha every erm in 3.16 is non-negaive and, from he previous analysis, bounded in N. Thus quu 2 dx < 3.17 C + and from 3.15 i herefore follows ha Corollary A.18 in Secion A.3 gives C u 2 dx as u L C C quu L p C + u L 2 C. 3.19

44 38 From Poincaré s inequaliy i follows ha and using Hölder s inequaliy, we ge u L 2 C C u L 2 C quu p L p C qu1/2 u L 2 C qu p/2 u p L 2 C. Since u is bounded and q coninuous, he righ-hand side is also bounded. Applying he las wo inequaliies for esimaing he righ-hand side in 3.19, we obain u L C C quu 2 1/2p L 1 C + u L 2 C. Using 3.17 and 3.18 we see ha u L C as. Applying Theorem 3.1 on vx, x n = ux, x n + N for N large enough complees he proof. 3.6 An esimae for soluions of a nonlinear ordinary differenial equaion The esimae 3.5 conains an arbirary small parameer ɛ. Is i possible o remove ɛ from his esimae? In order o see wha kind of requiremens we need on he funcion q, we consider here he model equaion u λu + quu =. 3.2 Theorem 3.5 Le u be a soluion of 3.2 subjec o he condiion u as Suppose ha λ >, ha he funcion fu = quu is Lipschiz coninuous wih f = and ha qu > if u. Suppose also ha 1 qu u du < Then u + u = O e λ. Proof. We may assume ha u is no idenically zero. Using 3.21, we see from 3.2 ha u as. Since u + 1 = u + u +1 s 1u s ds,

45 39 we have u u + u sup u s. [,+1] Therefore also u as. As before, we use he noaion Gu = u qvv dv. We muliply 3.2 by u and obain u 2 λu 2 + 2Gu =. Therefore u 2 λu 2 + 2Gu is a consan. Since he funcions u and u vanish a we conclude ha his consan is. Thus u = ± λu 2 2Gu Le us show ha he funcion λu 2 2Gu has no zeros for large enough. Firs, suppose ha u = for some R. Then 3.23 implies ha u = and we ge u by he uniqueness of soluions o he Cauchy problem of equaion 3.2. Furhermore, define Since Qu = 1 λu 2 Qu 1 λ u u we have, because of 3.21 and 3.22, ha qvv dv. qv v Qu as. This implies ha λu 2 2Gu is posiive for large. From 3.23 i follows ha u has no zeros for large. Since u as, we mee here one of wo possibiliies: Eiher u > and u < or else u < and u >. We consider he firs possibiliy. The second one is considered analogously. We have dv, u = λ u 1 2Qu Power series expansion yields ha here exiss an ɛ > such ha x 2 x 3.25 if x ɛ. Furhermore, since Qv as v, here exiss a δ > such ha Qv ɛ if v δ. Choose so large ha u δ. Inegraing 3.24, we obain 1 δ dv λ v 1 2Qv = C u

46 4 The lef-hand side can be wrien as 1 δ λ u dv v + 1 λ δ Furhermore, we ge from 3.25 ha δ dv v 1 2Qv 2 u u 1 v 1 1 2Qv 1 1 Qv v dv dv <, 3.28 where he las inequaliy follows from From 3.26, 3.27 and 3.28 i follows ha ln u = B λ, where B is a bounded funcion. Therefore u = O we also obain ha u = O e λ. e λ. Using 3.24, A Some resuls from funcional analysis A.1 Eigenvalues and eigenvecors of Here we show ha if is smooh enough here exiss an ON-basis of L 2 consising of eigenvecors of he operaor. For he Dirichle problem his resul is ofen proved in exbooks in parial differenial equaions so we focus on he Neumann problem and prove i since in his case he proof is no easily found. On our way we need some lemmas which will be saed wihou proof. Throughou his appendix, le H denoe a Hilber space and, H he inner produc in H. If A is a bounded linear operaor on H, we le σa denoe he specrum of A and σ p A he se of eigenvalues of A, i.e. he poin specrum. The following lemmas are well-known resuls from funcional analysis. Lemma A.1 Le be an open se in R. Then C is dense in L 2. Definiion A.2 If X is a normed space we say ha a subse M X is oal in X if he span of M is dense in X. Lemma A.3 If M is oal in X and for some x X we have ha x M, hen x =. Conversely, if X is complee and x M implies x =, hen M is oal in X. Lemma A.4 Suppose K is a compac linear operaor on H. Then NI K is finie dimensional. If dim H = hen also

47 41 i σk. ii σk {} = σ p K {}. iii σk {} is eiher finie or counable wih he unique limi poin. Lemma A.5 Suppose A is a linear, bounded and symmeric operaor on H and define Then m, M σa [m, M]. m = inf{ Au, u H : u H, u = 1} M = sup{ Au, u H : u H, u = 1}. Remark A.6 Since A is symmeric, Au, u H is always real. Lemma A.7 Suppose H is separable and A : H H is linear, bounded, symmeric and compac. Then here exiss a counable orhonormal basis of H consising of eigenvecors of A. Theorem A.8 Le be an open, bounded region in R n wih C 1 -boundary. Then here exiss an ON-basis {φ k } k= of L2 consising of eigenfuncions of he operaor for he Neumann problem, i.e. φ k v dx = λ k φ k v dx, v H 1, A.1 and he eigenvalues λ k are subjec o = λ < λ 1 λ 2 λ 3..., λ k as k, A.2 where each eigenvalue is repeaed according o is mulipliciy. Moreover, {φ k } k= is also an orhogonal basis of H1. Proof. Given u H 1, he funcional v vu dx is linear and bounded on H 1. According o Riesz represenaion heorem, here exiss a unique T u H 1 such ha vu dx = v, T u H 1, v H 1. A.3 We herefore define he operaor T : H 1 H 1 by he relaion A.3. I is easy o see ha T is linear and from he closed graph heorem i follows ha T is bounded. Namely, le Γ = {x, T x : x H 1 } be he graph of T and suppose ha x n, y n Γ, x n, y n x, y in H 1 H 1. From A.3 i follows ha y n T x, y n T x H 1 = y n T xx n x which, ogeher wih Hölder s inequaliy, shows ha y n T x and hus y = T x. Hence he graph is closed.

48 42 T is also symmeric, because he equaliy T u, v H 1 = u, T v H 1 follows by combining he fac ha T u, v H1 = v, T u H1 wih he relaion A.3. Finally, T is compac: le i : H 1 L 2 be he inclusion of H 1 in L 2 and T 1 : L 2 H 1 be he exension of T o L 2 as defined by he lef-hand side of A.3. Since H 1 is compacly embedded in L 2 and T 1, as T, is bounded, i follows ha T = T 1 i is compac. Lemma A.7 now gives ha here exiss a counable orhonormal basis {ψ j } j= of H1 consising of eigenfuncions of T and we will now prove ha, afer normalizaion, he same se is an ON-basis for L 2. Le µ j be he eigenvalue corresponding o ψ j. From A.3 we ge ha ψ jψ k dx = µ k ψ j, ψ k H 1 so he eigenfuncions are orhogonal o each oher also in L 2. By seing ψ j φ j =, ψ j L 2 we hus ge an orhonormal sequence {φ j } j= in L2, each φ j also an eigenfuncion of T wih he same eigenvalue as ψ j. I remains o see ha each funcion in L 2 can be wrien as a finie or infinie linear combinaion of hem. Se M = {φ j } j=. The fac ha M is a basis of H1 implies ha M is oal in H 1. Since H 1 is dense in L 2 according o Lemma A.1, M is also oal in L 2. For f L 2 we consruc g = f, φ k L 2 φ k, k= which is convergen in L 2. We ge immediaely f g, φ j L 2 = for every j so Lemma A.3 gives ha f = g. Thus {φ j } j= is an ON-basis for L 2 and he coordinaes of f are is usual Fourier coefficiens. The final sep is o invesigae he eigenvalues of T. From A.3 i follows ha T u, u H 1 = u 2 L 2 A.4 so Lemma A.5 gives ha σt [, 1] wih, 1 σt. A.4 also shows ha is no an eigenvalue. Furhermore, if λ is an eigenvalue, Lemma A.4 shows ha NλI T = NI λ T is finie dimensional. Bu since dim H 1 = and he eigenfuncions form a basis, we conclude ha T has infiniely many eigenvalues. Lemma A.4 also gives ha 1 really is an eigenvalue wih consans as eigenfuncions and ha we can label he eigenvalues in decreasing order as We see ha 1 = µ µ 1... >, µ k as k. A.5 u, v H 1 = fv dx, v H 1