Infinitely many solutions and Morse index for non-autonomous elliptic problems

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1 Infinitely many solutions and Mose index fo non-autonomous elliptic poblems Philip Koman Depatment of Mathematical Sciences Univesity of Cincinnati Cincinnati Ohio Abstact Ou fist esult is a change of vaiables which tansfoms adial k- Hessian equations into adial p-laplace equations. Then we genealize the classical esults of D.D. Joseph and T.S. Lundgen [9] by using the method we developed in [12] and [13]. We povide a consideably simple appoach, which yields additional infomation on the Mose index of solutions. Key wods: Global solution cuves, Infinitely many solutions, Mose index. AMS subject classification: 35J60, 35B40. 1 Intoduction J. Jacobsen and K. Schmitt [8] consideed a class of adial quasilinea elliptic equations on the unit ball 1.1) α u ) β u )) + γ fu) = 0, u 0) = u1) = 0, with paametes α, β and γ. This class of equations includes: the Laplacian, in case α = n 1, β = 0, γ = n 1, p-laplacian p > 1), fo α = n 1, β = p 2, γ = n 1, and k-hessian in case α = n k, β = k 1, γ = n 1, whee n is the dimension of the space, and 1 k n is an intege. Recall that the k-hessian is defined as the sum of all pincipal k k minos of the Hessian matix D 2 u. So that 1-Hessian is u, and n-hessian is det D 2 u, the Monge-Ampee opeato. The pape of J. Sánchez and V. Vegaa [20] has a detailed exposition of k-hessian equations, and the pevious contibutions 1

2 include L. Caffaelli et al [2], N.S. Tudinge and X.-J. Wang [21], J. Jacobsen [7], and J. Jacobsen and K. Schmitt [8]. Ou fist esult is a change of vaiables which tansfoms the poblem 1.2) into a Diichlet poblem fo a p-laplacian, with p = β + 2, in q+1 θ + 1 dimensions, on anothe ball aound the oigin, of adius q + 1) 1 q+1, whee θ and q ae defined below. In paticula, one can educe any k-hessian equation to a p-laplacian, and a class of non-autonomous p-laplace equations to autonomous p-laplace equations. This extends ou pevious esult in [14]. Then we pesent exhaustive esults on multiplicity of positive solutions, and global solution cuves fo p > 1, α 0) 1.2) u + n 1 u +λ α 1+u) p = 0, 0 < < 1, u 0) = 0, u1) = 0, genealizing the classical esults of D.D. Joseph and T.S. Lundgen [9], who consideed the case α = 0. The appoach in [9] was athe involved, and it elied on the old esults of E. Hopf [6], and S. Chandasekha [3] see p. 261 in [9]). We use a consideably simple method, involving geneating and guiding solutions, that we developed in [12] and [13]. In the pesent pape we futhe simplify and impove this method, paticulaly in the case when the solution cuve is monotone, see Theoem 3.2 below. Ou appoach uses the self-simila natue of the poblem 1.2), which allows paameteization of the entie solution cuve, see [12] fo the discussion. Singula solutions, aising as limits of the egula solutions, ae easy to undestand fo selfsimila poblems, while fo othe equations with polynomial nonlineaities this issue is athe involved, see e.g., F. Mele and L.A. Peletie [16]. In the last section we use the geneating and guiding solutions to show that all tuning points of the poblem ae non-degeneate, and that the Mose index of solutions inceases by one at each tuning point, when one follows the solution cuve of 1.2) in the diection of inceasing u0). Fo the Gelfand poblem, whee fu) = e u, such a esult was fist poved in K. Nagasaki and T. Suzuki [17], and then in P. Koman [13], by using the geneating and guiding solutions. 2 Reduction of k-hessian equation to p-laplacian Similaly to J. Jacobsen and K. Schmitt [8], we conside the poblem 2.1) α u ) β u )) + γ fu) = 0, u 0) = 0, 2

3 with paametes α, β and γ. We make a change of vaiables 2.2) t = q+1 q + 1, with q to be chosen shotly. Clealy, u = u t q, and we have α+qβ+1) u t β u t ) + γ q fu) = 0. We now choose q to equalize the powes of : α + qβ + 1) = γ q θ, 2.3) q = γ α β + 2. The common powe θ is then 2.4) θ = β + 1)γ + α β + 2, and we have 2.5) θ u t β u t ) + θ fu) = 0. Theoem 2.1. Assume that fu0)) > 0, β + 1 > 0, and 2.6) γ > α β 2. The change of vaiables 2.2), with q given by 2.3), tansfoms the poblem 2.1) into 2.7) ) t θ q+1 u t) β u θ t) + t q+1 fu) = 0, u 0) = 0, i.e., into β + 2)-Laplacian in θ q dimensions. Poof: Replacing in 2.5), = q +1) 1 q+1t 1 q+1, and using pimes to denote the deivatives in t, we obtain the equation in 2.7). Since u ) < 0 fo small > 0, and β + 1 > 0, we expess fom 2.1) u ) ) β+1 = 1 α 3 0 z γ fuz)) dz.

4 Using again that β + 1 > 0, we have obseve that 0 as t 0, since q + 1 > 0, by 2.6)) and [ du u ) 0) = lim dt 0 q = lim 0 u )) β+1 lim 0 qβ+1) = lim 0 1 α+qβ+1) since γ > α + qβ + 1) 1, by 2.6). u )) β+1 0 qβ+1) ] 1 β+1 z γ fuz)) dz = 0, In paticula the k-hessian equation is tansfomed into 2.7), with β = k 1, and q+1 θ = k)+n 2k, i.e., into a p-laplace equation, with p = β + 2 = k + 1, in θ q = kn+n 2k dimensions. Radial solutions of hee s > 1 is a eal paamete) 2.8) u + s fu) = 0, u 0) = 0 in n dimensions satisfy, 2.9) n 1 u ) + n+s 1 fu) = 0, u 0) = 0. The change of vaiables 2.2) becomes t = s/2+1 s/2+1, and it tansfoms 2.9) into 2.10) t m 1 u ) + t m 1 fu) = 0, du dt 0) = 0. with m = n+s s/2+1. This coesponds to u + fu) = 0 in m dimensions. So that the non-autonomous tem s in 2.8) got emoved with the dimension changing to m). We fist poved this esult in [14]. 3 A genealization of a esult of Joseph and Lundgen By the classical theoem of B. Gidas, W.-M. Ni and L. Nienbeg [5], all positive solutions of the Diichlet poblem hee u = ux), x R n ) u + λ 1 + u) p = 0 fo x < 1, u = 0 when x = 1 4

5 ae adially symmetic, i.e., u = u), = x, and they satisfy 3.1) u + n 1 u +λ 1+u) p = 0 fo 0 < < 1, u 0) = 0, u1) = 0. Hee λ is a positive paamete, p > 1 a constant. Similaly to J.A. Pelesko [19], we set v = 1 + u, followed by v = aw, and t = b, whee a = v0) = 1 + u0). The constants a and b ae assumed to satisfy 3.2) λ = b2 a p 1. Then 3.1) becomes 3.3) w + n 1 w + w p = 0, w0) = 1, w 0) = 0. t The solution of this poblem is easily seen to be a deceasing function, going to zeo. In case of sub-citical p, 1 < p < n+2, it is known that wt) vanishes at some t 0 > 0, since the Diichlet poblem fo the equation in 3.3) has a unique) solution on any ball as follows by the mountain pass lemma). If p n+2, then wt) has no oots on 0, ), which follows by Pohozhaev s identity also a well known fact). Once we compute wt) fom 3.3), u) = 1 + awb), and since u1) = 0, we have 1 = awb), so that a = 1 wb), and then λ = b2 w p 1 b). The global solution cuve is λ, u0)) = b 2 w p 1 b), ), wb) paameteized by b 0, t 0 ) in the sub-citical case, and b 0, ) fo the supe-citical and citical cases. The solution of 3.1) at the paamete value of b is u) = awb) 1 = wb) wb) 1 It will be convenient to use the lette t instead of b as the paamete. So that the global solution cuve fo p n+2 3.4) λ, u0)) = is t 2 w p 1 t), ) wt), t [0, ), and the solution of 3.1) at the paamete value of t is u) = wt) wt) 1, whee wt) is the solution of 3.3). 5

6 In paticula, λ = λt) = t 2 w p 1 t), and 3.5) λ t) = tw p 2 [ 2w + p 1)tw ], so that the solution cuve tavels to the ight left) in the λ, u0)) plane if 2w + p 1)tw > 0 < 0). The tuning points coespond to the oots of the function 2w + p 1)tw. If we set this function to zeo 2w + p 1)tw = 0, then the geneal solution of this equation is If we choose wt) = ct β, with β = 2 p 1. c = c 0 = [βn 1) ββ + 1)] 1 p 1, then w 0 t) = c 0 t β is also a solution of the equation in 3.3), and the issue tuns out to be how many times wt) and w 0 t) coss as t, as the following lemma shows. Lemma 3.1. Assume that wt) and w 0 t) intesect infinitely many times. Then the solution cuve of 3.1) makes infinitely many tuns. Poof: Let {t n } denote the points of intesection. At {t n } s, wt) and w 0 t) have diffeent slopes by uniqueness fo initial value poblems). Since 2w 0 t n ) + p 1)t n w 0 t n) = 0, it follows that 2wt n ) + p 1)t n w t n ) > 0 < 0) if wt) intesects w 0 t) fom below above) at t n. Hence, on any inteval t n, t n+1 ) thee is a point t 0, whee 2wt 0 )+p 1)t 0 w t 0 ) = 0, i.e., λ t 0 ) = 0, and t 0 is a citical point. Since λ t n ) and λ t n+1 ) have diffeent signs, the solution cuve changes its diection ove t n, t n+1 ). Using the teminology fom [12], wt) is called the geneating solution, while w 0 t) is the guiding solution. The lineaized equation fo 3.3) is z + n 1 z + pw p 1 z = 0. t At w = w 0 t), the lineaized equation becomes 3.6) z + n 1 z + pγ 1 t t2z = 0, with γ = βn 1) ββ + 1). 6

7 This is Eule s equation. The oots of its chaacteistic equation ae = n 2) ± n 2) 2 4pγ 2 If n 2) 2 4pγ < 0, the oots ae complex, the fundamental solution set of 3.6) consists of t 2 cosω ln t) and t 2 sinω ln t), with ) 2 4pγ 4ω 2, and it is natual to expect that wt) w 0 t) changes sign infinitely many times, and then the solution cuve makes infinitely many tuns. By the esult of P. Koman [10], this may happen only if p > n+2. Lemma 3.2. Assume that p > n+2. Then n 2)2 4pγ < 0 if and only if. 3.7) 4p p p p 1 > n 2. Poof: We have n 2) 2 4pγ = n 2) 2 4pβn 2) + 4pβ 2 < 0, povided that n 2 lies between the oots of this quadatic, i.e., 4p p p 1 4 p 1 < n 2 < 4p p p p 1. The condition p > n+2 completing the poof. implies that n > 2p+2 p 1. Then n 2 > 4 p 1 > 4p p p 1 4 p 1, Obseve that the left hand side of 3.7) is a deceasing function, tending to 8 as p. It follows that in dimensions 2 < n 10 the condition 3.7) holds fo all p > 1. The following esult is known as Bihai s inequality. Lemma 3.3. Assume that the functions at) 0, and ut) 0 ae continuous fo t t 0, and we have t 3.8) ut) C + as)u m s) ds, fo t t 0, t 0 7

8 with some constants C > 0 and m > 1. Assume also that Then Poof: ut) C 1 m m 1) t t 0 as) ds > 0, fo t t 0. 1 [ C 1 m m 1) t t 0 as) ds ] 1 m 1, fo t t 0. Denote the ight hand side of 3.8) by wt). Then wt 0 ) = C, and w = at)u m at)w m. Divide by w m, and integate ove t 0, t). Lemma 3.4. Assume that p > n+2, n 3, and in case n > 10 assume additionally that 3.7) holds. The geneal solution of is y + n 1 y + pγ t t 2 y = ft) yt) = Ct 2 sinω ln t + D) + 1 t t 2 sin ω ln t ) s n 2 fs) ds, ω t 0 s with ) 2 4pγ 4ω 2, fo any constant t 0 fixed, and abitay constants C and D. Poof: By Lemma 3.2, the fundamental set of the coesponding homogeneous equation consists of y 1 t) = t 2 cosω ln t) and y 2 t) = t 2 sin ω lnt). Compute thei Wonskian Wy 1, y 2 ) = ωt 1 n, and apply the method of vaiation of paametes. The following esult is known, see e.g., J. Dávila et al [4]. Ou poof is a little diffeent fom the usual one, in that we avoid the language of heteoclinic connections. Lemma 3.5. Assume that p > n+2, and let w) and w 0) be as above. Then lim w) w 0 ) = 1. Poof: In the initial value poblem detemining w): 3.9) y + n 1 y + y p = 0, y0) = 1, y 0) = 0, > 0 8

9 we make a change of vaiables y) = w 0 )v), followed by = e t. Then vt) satisfies 3.10) v +av γv v p ) = 0, v ) = v ) = 0, t, ), with a = 2β + n 2 > 0, and as above γ = β n β 2) > 0. Since v = 1 c 0 β y, we have dv dt t= = dv d d dt t= = c 1 [ 0 β β 1 y) + β y ) ] =0 = 0.) We need to show that lim t vt) = 1. Since the enegy Et) = 1 ) 2 v 2 v 2 γ 2 vp+1 p + 1 is deceasing along the solution of 3.10), it follows that vt) is bounded, and it cannot tend to zeo as t. Obseve that E ) = 0, and hence Et) < 0 fo all t.) Fom the equation 3.10), vt) > 0 can have local minimums only if 0 < v < 1, and local maximums only if v > 1, and it can tend only to 1, as t. Hence, eithe vt) tends to 1 monotonously, o it oscillates infinitely often aound 1. We show next that in the latte case vt) also tends to 1. Let v 1 = z. By ou assumption, zt) has infinitely many oots, 1 < z < z 0, fo some z 0 > 0, and it satisfies z + az + γfz) = 0, whee fz) behaves like z + z 2 on 1, z 0 ), to which fz) is equal in case p = 2. The enegy Ēt) = 1 2 z 2 +γfz), whee Fz) = z 0 ft) dt, is positive fo z > 1, and deceasing, since 3.11) Ē t) = az 2. Let E 0 = lim t Ēt) 0. Assume that E 0 > 0. If t k ae the oots of zt), then z t k ) get abitaily close to 2E 0, fo k lage. Fom the equation 3.10), we get a bound on z t k ). Hence, z t) cannot change fast, so that we can find an inteval of fixed length t k, t k + θ) on which z t) > 1 2 2E0, independently of k. Integating 3.11), we conclude that Ēt) dops by at least θ 2 2E0 fo each k, and eventually Et) dops below E 0, a contadiction. Hence, E 0 = 0. Let τ k be the point of maximum minimum) of zt) on t k, t k+1 ). Then Fzτ k )) = Ēτ k ) 0, poving that zt) 0 as t. Remaks 9

10 1. Fo p = n+2, the lemma does not hold the poof beaks down since a = 0 in 3.10)). In that case, the geneating solution, i.e., the solution 1 of 3.3) is well-known to be w) =. In standad 1+ 1 n) 2 2 teminology, w) has slow decay fo p > n+2, and fast decay at p = n The tansfomation y = w 0 v is equivalent to the classical Fowle tansfomation w = t p 1v. 2 So that the Fowle tansfomation can be seen as the extension of the elementay eduction of ode method, when one of the solutions hee w 0 t)) is known. Theoem 3.1. Assume that p > n+2, n 3, and in case n > 10 assume additionally that 3.7) holds. Then the solution cuve of 3.1) makes infinitely many tuns. Moeove, the poblem 3.1) has infinitely many solutions at λ = γ = βn 1) ββ + 1), with β = 2 p 1, and at most finitely many solutions at othe λ s. As u0), the solutions of 3.1) tend to β 1, fo 0, which is a singula solution of 3.1). Poof: By Lemma 3.2, the oots of the chaacteistic equation of 3.6) ae 2 ±iω, with )2 4pγ = 4ω 2. We will show that yt) wt) w 0 t) has infinitely many oots. This will follow fom the following asymptotic fomula 3.12) t 2 yt) = C sinω ln t + D) + Ot n/2+3 βp 2) ), as t, with some constants C 0 and D, and with n/2 + 3 βp 2) < 0. o The function yt) satisfies y + n 1 y + pw p 1 0 y = ft), t 3.13) y + n 1 y + pγ t t 2 y = ft), [ whee ft) = yt) + w 0 t)) p w p 0 t) pwp 1 the geneal solution of 3.13) is 3.14) yt) = Ct 2 sinω lnt + D) + 1 t 2 ω 10 ] 0 t)yt). Using Lemma 3.4, t t 0 sin ω ln t ) s n 2 fs) ds. s

11 We claim that fo lage t 3.15) ft) c 1 ωt βp 2) y 2, with some constant c 1 > 0. Indeed, by the two-tem Taylo s fomula with a emainde tem ft) = pp 1) 2 [θwt) + 1 θ)w 0 t)] p 2 y 2 ĉ w p 2 0 y 2, with some θ 0, 1) and ĉ > 0, giving 3.15). We estimated wt) by a multiple of w 0 fom above below), in case p 2 p < 2), using Lemma 3.5.) Setting vt) t 2 yt), we estimate fom 3.14) 3.16) vt) C + c 1 t Since p > n+2, it follows that By Bihai s inequality Lemma 3.3) t 0 s n/2+2 βp 2) v 2 s) ds. n/2 + 2 βp 2) < ) vt) C 1 t c 1 t 0 s n/2+2 βp 2) ds 1, c 2 with t 0 chosen lage enough, so that C 1 t c 1 t 0 s n/2+2 βp 2) ds > c 2 > 0 fo all t > t 0, and some constant c 2. By 3.14), yt) = Ct 2 sinω ln t + D) 1 t 2 ω t1 t sin ω ln t ) s n 2 fs) ds s gives fo t < t 1, with t 1 lage and fixed, the unique solution of 3.13), which satisfies the appopiate initial conditions at t 1 equal to the ight hand side and its deivative evaluated at t 1 ). This solution is witten using an integal of itself. Then t 2 yt) = C sinω ln t + D) 1 ω t t1 t t sin ω ln t ) s n 2 fs) ds, s and using 3.17) we estimate, fo t lage, t1 sin ω ln t ) t1 s n 2 fs) ds c1 s n/2+2 βp 2) v 2 s) ds. s 11

12 c t1 1 c 2 s n/2+2 βp 2) ds = Ot n/2+3 βp 2) ) = o1), 2 t poving 3.12). Hence, wt) and w 0 t) have infinitely many points of intesection, and by Lemma 3.1, the solution cuve makes infinitely many tuns. Tuning to the othe claims, fo lage t, we have, by Lemma 3.5, u) = wt) wt) 1 w 0t) w 0 t) 1 = β 1, which is the singula solution of 3.1), and λ = t 2 w p 1 t 2 w p 1 0 = γ, gives the vetical asymptote of the solution cuve of 3.1). Remak In case n 11, define p 0 as the solution of fp) 4p p p p 1 = n 2. The function fp) is deceasing, f n+2 theoem equies that n + 2 n 2 < p < p 0. ) > n 2, so that n+2 < p 0, and the To teat the case when the condition 3.7) fails, we need the following elementay lemmas. Lemma 3.6. Let the functions y) and z) of class C 2 [0, ) satisfy 3.18) y + n 1 y + a)y = 0, 3.19) z + n 1 z + a)z > 0, 3.20) 0 < y 0 ) = z 0 ), y 0 ) = z 0 ) at some 0 > 0, fo some a) C[0, ). Assume that y) > 0 on 0, ). Then z) > 0 on 0, ). Poof: Assuming the contay, let 1 > 0 be the fist oot of z). Fom 3.18) and 3.19) we get [ n 1 z y zy )] > 0 on 0, 1 ). Integating ove 0, 1 ), we get which is a contadiction. n 1 1 z 1 )y 1 ) > 0, 12

13 Lemma 3.7. Conside the function y) = α s 1 +β s 2, with constants s 1 < s 2 < 0 and α, β R. Assume that at some 0 > 0 we have y 0 ) = A > 0, y 0 ) = B < 0, and 3.21) B A s 1 0 > 0. Then y) > 0 fo > 0. Poof: Wite ) s1 ) s2 ) s1 ) ] s2 s 1 y) = c 1 + c 2 = [c 1 + c 2, with c 1 = α s 1 0 and c 2 = β s 2 0. We have s 2 s 1 > 0, and the poof will follow if c 2 > 0. We have y 0 ) > 0, and if y) has a oot, it lies to the left of 0.) Since 0 it follows that in view of 3.21). c 2 = y 0 ) = c 1 + c 2 = A y 0 ) = s 1 0 c 1 + s 2 0 c 2 = B, B A s ) 1 0 > 0, 0 s 2 s 1 Theoem 3.2. Assume that p > n+2, n 11, and assume that 3.7) fails, i.e., 4p p 3.22) p p 1 n 2. Then the solution cuve of 3.1) is monotone inceasing in the λ, u0)) plane, i.e., λ t) > 0 fo all t > 0. Moeove, lim t λt) = γ = βn 1) ββ + 1), with β = 2 p 1. It follows that that the poblem 3.1) has a unique positive solution fo λ < γ, and no positive solution fo λ γ. As u0), the solutions of 3.1) tend to β 1, fo 0, which is a singula solution of 3.1). Poof: Assume that the inequality in 3.22) is stict, the case of equality is simila. The oots of the chaacteistic equation of 3.6) ae now eal and negative 3.23) s 1 = n 2) n 2) 2 4pγ < s 2 = n 2) + n 2) 2 4pγ

14 We claim that the geneating and guiding solutions satisfy 3.24) w) < w 0 ) fo all > 0. Let z) = w 0 ) w). We wish to show that z) > 0 fo all > 0. We have z + n 1 z + pw p 1 0 z = w p w p 0 pw 0 p 1 w w 0 ) > 0. Denote A = A 0 ) = w 0 0 ) w 0 ), and B = B 0 ) = w 0 0) w 0 ). Let y) be the solution of y + n 1 y + pw p 1 0 y = 0, y 0 ) = A, y 0 ) = B, so that y) = α s 1 + β s 2, with some constants α, β, and s 1 < s 2 < 0 as in 3.23). To apply Lemma 3.7, we shall find 0 > 0 small, so that A > 0, B < 0, and 3.25) 0 B 0 ) A 0 ) > s 1. Recalling that w 0 ) = c 0 β, with β = 2 p 1, and c 0 as above, we ewite 3.25) as c 0 β β+1 0 w 0 ) c 0 β 0 w > s 1. 0) This will hold fo sufficiently small 0, povided that c 0 β n 2) n 2) > s 1 = 2 4pγ. c 0 2 The last inequality follows fom β = p 1 2 < 2, which is in tun equivalent to p > n+2. By Lemma 3.7, y) > 0 fo > 0, and then by Lemma 3.6, w) < w 0 ) fo > 0, and since 0 can be chosen abitaily small, 3.24) is justified. 3.26) By 3.5) we have λ t) > 0 fo all t > 0, povided that 2 p 1 + tw t) > 0, fo all t > 0. wt) Recall that wt) and w 0 t) ae both solutions of w + n 1 t w + w p = 0. Witing this equation at wt) and at w 0 t), we conclude, in view of 3.24), [ t n 1 w 0 w w 0w )] ) = w0 w w p 1 w p 1 0 < 0, fo all t > 0. 14

15 The function qt) t n 1 w 0 w w 0w ) satisfies q0) = 0 obseve that n 1 > β + 1), and q t) < 0. It follows that qt) < 0, o that w t) wt) > w 0 t) w 0 t) fo all t > 0. Then 2 p 1 + tw t) wt) > 2 p 1 + tw 0 t) w 0 t) = 0, justifying 3.26), and so the solution cuve is monotone. Tuning to the othe claims, fo lage t, we have, by Lemma 3.5, u) = wt) wt) 1 w 0t) w 0 t) 1 = β 1, and λ = t 2 w p 1 t 2 w p 1 0 = γ, while using 3.24), λt) = t 2 w p 1 < t 2 w p 1 0 = γ, fo all t > 0. We now conside the poblem 3.1) fo the citical and sub-citical cases, p n+2 n+2. Fo p =, the geneating solution, i.e., the solution of 3.3) is well-known: 1 wt) = ), n) t2 2 see e.g., [14] fo the efeences. In view of 3.4), the global solution cuve of u + n 1 u + λ 1 + u) n+2 = 0 fo 0 < < 1, u 0) = 0, u1) = 0 is given explicitly by 3.27) λ, u0)) = t n) t2 ) 2, 1 + fo t [0, ) obseve that hee p 1 = 4 ). Theoem 3.3. Conside the poblem ) 1 2 nn 2) t2 1, u + n 1 u + λ 1 + u) p = 0 fo 0 < < 1, u 0) = 0, u1) = 0, with p n+2. All positive solutions lie on a single smooth solution cuve, which begins at λ = 0, u = 0), and makes exactly one tun to the left at some λ 0 > 0, tending to infinity as λ 0. Poof: In case p < n+2 n+2, this is Theoem 2.20 in [11]. Fo p =, the esult follows fom the epesentation 3.27), paticulaly fom [ ] 2 t λt) = 1 + 1, n) t2 15

16 which shows that λt) has a unique point of maximum, and tends to zeo as t. One can explicitly compute λ 0 = n) 4. We now conside the poblem 3.28) u + n 1 u +λ s 1+u) p = 0, 0 < < 1, u 0) = 0, u1) = 0. We shall assume that the constant p > 0, while the case p < 0 we consideed in [12] and [13], in the context of so called MEMS poblems. Ou esult will allow s < 0, unde some conditions, i.e., singula poblems. The following is the main esult of this section, which fo s = 0 was poved in D.D. Joseph and T.S. Lundgen [9]. Theoem 3.4. Assume that n 1. Define m = n+s 4 p p 1, and assume that m 1. s/2+1, fp) = 4p p 1 + i) If 0 < p < 1, the poblem 3.28) has a unique positive solution fo any 0 < λ <. ii) If 1 < p m+2 m 2, then thee is a citical λ 0 > 0 so that the poblem 3.28) has exactly two positive solutions fo λ 0, λ 0 ), exactly one positive solution at λ = λ 0, and no solutions fo λ > λ 0. iii) If p > m+2 m 2 and fp) > m 2, then the solution cuve of 3.28) makes infinitely many tuns. Moeove, the poblem 3.28) has infinitely many solutions at λ = γ s/2 + 1) 2 [βm 1) ββ + 1)], with β = 2 p 1, and at most finitely many solutions at othe λ s. iv) If p > m+2 m 2 and fp) m 2, then the solution cuve of 3.28) is monotone inceasing in the λ, u0)) plane. Moeove, lim t λt) = γ, lim t u0) =. It follows that the poblem 3.28) has a unique positive solution fo λ < γ, and no positive solution fo λ γ. Poof: By above, see 2.9) and 2.10), the change of vaiables t = s/2+1 s/2 + 1, followed by τ = s/2 + 1)t, and µ = s/2 + 1) 2 λ, tansfoms 3.28) into u + m 1 u + µ 1 + u) p = 0, 0 < τ < 1, u 0) = 0, u1) = 0, τ with m = n+s s/2+1. Then the esult follows by the Theoems 3.1, 3.2 and 3.3, while the case 0 < p < 1 is well-known. 16

17 4 Mose index of solutions We now use the geneating and guiding solutions to show that all tuning points of the poblem 3.1) ae non-degeneate, and that the Mose index of solutions inceases by one at each tuning point. We addess the nondegeneacy fist. Theoem 4.1. Let ut n ) be a singula solution of 3.1), i.e., λ t n ) = 0. Then ut n ) is non-degeneate, i.e., λ t n ) 0. Poof: Recall that Since λ t n ) = 0, we have λ t) = tw p 2 [ 2w + p 1)tw ], 4.1) 2wt n ) + p 1)t n w t n ) = 0. Then 4.2) λ t n ) = t n w p 2 t n ) [ p + 1)w t n ) + p 1)t n w t n ) ], and we need to show that S p + 1)w t n ) + p 1)t n w t n ) 0, to conclude that λ t n ) 0. Using the equation 3.3), and then 4.1), we expess S = [p + 1 n 1)p 1)]w t n ) p 1)t n w p t n ) [ ] 2 [p + 1 n 1)p 1)] 4.3) = wt n ) p 1)t n w p 1 t n ). p 1)t n Fo the guiding solution w 0 t) = c 0 t β, with β = 2 p 1 and c 0 = [βn 1) ββ + 1)] p 1, 1 we have by a diect computation 2 [p + 1 n 1)p 1)] p 1)t n w p 1 0 t n ) = 0. p 1)t n Obseving that the quantity in the squae backet in 4.3) is a deceasing function of wt n ), we conclude that S 0, once we show that 4.4) wt n ) w 0 t n ). 17

18 If, on the contay, wt n ) = w 0 t n ), we conclude fom 4.1) and the identity 2w 0 t n ) + p 1)t n w 0t n ) = 0, that w t n ) = w 0 t n), and then by the uniqueness fo initial value poblems wt) = w 0 t), a contadiction. By C.S. Lin and W.-M. Ni [15], any solution of the lineaized poblem fo 3.1) is adially symmetic, and hence it satisfies 4.5) ω + n 1 ω +λp1+u) p 1 ω = 0, 0 < < 1, ω 0) = ω1) = 0. We call u) a singula solution of 3.1) if the poblem 4.5) has a non-tivial solution. Diffeentiating 3.1) in t, and setting t = t n, it is easy to see that a solution is singula if and only if λ t n ) = 0.) The following lemma gives explicitly any non-tivial solution of 4.5). Lemma 4.1. Let u) be a singula solution of 3.1). Then gives a solution of 4.5). ω) = u ) + 2 p 1 u) + 2 p 1. Poof: The function v) u )+ p 1 2 u)+ p 1 2 solves the equation in 4.5), and we have v 0) = ω 0) = 0, v0) > 0. By scaling of ω), we may assume that ω0) = v0), and then by the uniqueness esult fo this type of initial value poblems see [18]), it follows that ω) v). We now pesent the main esult of this section fo positive solutions of the poblem 4.6) u + λ 1 + u) p = 0, fo x < 1, u = 0, when x = 1. Theoem 4.2. Let ut n ) > 0 be a singula solution of 4.6), which means that the coesponding lineaized poblem 4.5) has non-tivial solutions, o that λ t n ) = 0. Then ut n ) is non-degeneate, i.e., λ t n ) 0. Moeove, when one follows the solution cuve of 4.6) in the diection of inceasing u0), the Mose index of solution inceases by one at each tun. Poof: By [5] positive solutions of 4.6) ae adially symmetic, and hence they satisfy 3.1). The Mose index of solution is the numbe of negative eigenvalues µ of ω + λp 1 + u) p 1 ω + µω = 0, fo x < 1, ω = 0, when x = 1. 18

19 By [15] solutions of this poblem ae adially symmetic. At a singula solution µ = 0, and then ω) = u ) + 2 p 1 u) + 2 p 1 by Lemma 4.1. Assume that at a singula solution ut n ), µt n ) = 0 is the k-th eigenvalue. Following [17], we will show that µ t n ) < 0, which means that fo t < t n t > t n ) the k-th eigenvalue is positive negative), i.e., the Mose index inceases by one though t = t n. We shall show that the sign of µ t n ) is the same as that of λ t n )) 2, which is negative by the Theoem 4.1. Denoting fu) = 1 + u) p, ecall the following known fomulas, which also hold fo geneal fu) hee u = ut n ), ω is a solution of 4.5), and B is the unit ball aound the oigin in R n ): µ t n ) ω 2 dx = λt n ) f u)ω 3 dx p. 11 in [11]), λt n ) B B B B f u)ω 3 dx = λ t n ) B fu)ω dx p. 3 in [11]), fu)ω dx = 1 2λt n ) u 1)ω 1) p. 5 in [11]). Putting them togethe, we conclude 4.7) µ t n ) ω 2 dx = λ t n ) 2λt n ) u 1)ω 1). By Lemma 4.1, B ω 1) = p + 1 p 1 u 1) + u 1). Recall that u) = wtn) wt 1, so that n) u 1) = w t n)t n wt n) Using these expessions and 4.2), we have and finally, fom 4.7), ω 1) = µ t n ) ω 2 dx = B λ t n ) p 1)t n w p 1 t n ), λ t n )) 2 u 1) 2λt n )p 1)t n w p 1 t n ) < 0, and u 1) = w t n)t 2 n wt n). because u 1) < 0 and λ t n ) 0. We conclude that µ t n ) < 0, completing the poof. 19

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