We prove the existence of infinitely many solutions to a semilinear Dirichlet boundary value problem in a ball for a nonlinearity g(u) that grows subcritically for u positive and supercritically for u negative

Castro, Alfonso

Kwon, John

Tan, Chee Meng, \u2707

Scholarship@Claremont

Claremont CollegesScholarship @ ClaremontAll HMC Faculty Publications and Research HMC Faculty Scholarship8-1-2007Infinitely Many Radial Solutions for a Sub-SuperCritical Dirichlet Boundary Value Problem in a BallAlfonso CastroHarvey Mudd CollegeJohn KwonClaremont Graduate UniversityChee Meng Tan '07Harvey Mudd CollegeThis Article is brought to you for free and open access by the HMC Faculty Scholarship at Scholarship @ Claremont. It has been accepted for inclusionin All HMC Faculty Publications and Research by an authorized administrator of Scholarship @ Claremont. For more information, please contactscholarship@cuc.claremont.edu.Recommended CitationCastro, Alfonso; Kwon, John; and Tan, Chee Meng '07, "Infinitely Many Radial Solutions for a Sub-Super Critical Dirichlet BoundaryValue Problem in a Ball" (2007). All HMC Faculty Publications and Research. Paper 477.http://scholarship.claremont.edu/hmc_fac_pub/477Electronic Journal of Differential Equations, Vol. 2007(2007), No. 111, pp. 1–10.ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.eduftp ejde.math.txstate.edu (login: ftp)INFINITELY MANY RADIAL SOLUTIONS FOR A SUB-SUPERCRITICAL DIRICHLET BOUNDARY VALUE PROBLEM IN ABALLALFONSO CASTRO, JOHN KWON, CHEE MENG TANAbstract. We prove the existence of infinitely many solutions to a semilinearDirichlet boundary value problem in a ball for a nonlinearity g(u) that growssubcritically for u positive and supercritically for u negative.1. IntroductionIn this paper we consider the sub-super critical boundary-value problem∆u+ g(u(x)) = 0, x ∈ RN , ‖x‖ ≤ 1u(x) = 0 for ‖x‖ = 1, (1.1)whereg(u) ={up, u ≥ 0|u|q−1u, u < 0, (1.2)with1 < p <N + 2N − 2 < q <∞, (1.3)that is, g has subcritical growth for u > 0 and supercritical growth for u < 0. Ourresults hold for more general nonlinearities. For example, it is easy to see that (1.2)may be replaced by limu→+∞ g(r, u)/up ∈ (0,∞) and limu→−∞ g(r, u)/(|u|q−1u) ∈(0,∞), uniformly for r ∈ [0, 1].Our main result is as follows.Theorem 1.1. Problem (1.1) has infinitely many radial solutions.This theorem extends the results of [4] where it was established that if 1 < p <(N+1)/(N−1) and q > 1, or p, q ∈ (1, (N+2)/(N−2)), or p ∈ (1, (N+2)/(N−2))and q = (N+2)/(N−2), then (1.1) has infinitely many radial solutions. This resultis optimal in the sense that if p, q ∈ [(N + 2)/(N − 2),∞) then u = 0 is the onlysolution to (1.1) (see [12]). For related results for quasilinear equations the reader isreferred to [8] and [10]. Studies on positive solutions for sub-super critical problemsmay be found in [9]. For other studies on the critical case, p = q = (N +2)/(N −2)2000 Mathematics Subject Classification. 35J65, 34B16.Key words and phrases. Sub-super critical; radial solutions; nonlinear elliptic equation;Pohozaev identity.c©2007 Texas State University - San Marcos.Submitted February 4, 2007. Published August 14, 2007.12 A. CASTRO, J. KWON, C. M. TAN EJDE-2007/111and lim|u|→∞ /(u|u|p−1) ∈ R, see [1, 3, 4, 6, 7]. In [2] the reader can find a completeclassification of the radial solutions to (1.1) for 1 < p = q < (N +2)/(N −2). For arecent survey of radial solutions for elliptic boundary-value problems that includesthe case where the Laplacian operator is replaced by the more general k-Hessianoperator, see [11].Radial solutions to (1.1) are the solutions to the singular ordinary differentialequationu′′ +ntu′ + g(u(t)) = 0u′(0) = u(1) = 0,(1.4)where, and henceforth, n = N − 1.For d > 0 let u(t, d) be the solution to the initial-value problemu′′ +ntu′ + g(u(t)) = 0u(0) = d, u′(0) = 0.(1.5)We define the energy functionE(t, d) ≡ (u′(t, d))22+G(u(t, d)), (1.6)where G(u) =∫ u0g(s)ds. For future reference we note thatdEdt(t) = −nt(u′(t))2 ≤ 0. (1.7)The proof of Theorem 1.1 is based on the properties of the energy and the argumentfunction defined below (see (1.9)).Theorem 1.2. There exists D > 0 such that if d ≥ D, thentN−1(tE(t) +N − 22u(t)u′(t))≥ cdξ for all t ≥√Nd(1−p)/2, (1.8)where ξ = N+2−p(N−2)2 . Also u(t) ≥ d/2 for t ∈ [0,√Nd(1−p)/2].As a consequence of Theorem 1.2 we see that, for d ≥ D, ρ(t, d) ≡ u2(t, d) +(u′(t, d))2 > 0 for all t ∈ [0, 1]. Hence there exists a continuous function θ :[0, 1]× [D,∞)→ R such thatu(t, d) = ρ(t, d) cos(θ(t, d)) and u′(t, d) = −ρ(t, d) sin(θ(t, d)). (1.9)In section 7 we prove thatlimd→∞θ(1, d) = +∞, (1.10)see (7.7) below.2. First zeroLet d > 0 and t0 > 0 be such that u(t0) = d/2, and u(t) > d/2 for t ∈ (0, t0).Following the arguments in [4], based on−u′(t) = t−n∫ t0sng(u(s))ds, (2.1)it is easily seen that √Nd(1−p)/2 ≤ t0 ≤√2pNd(1−p)/2. (2.2)EJDE-2007/111 A SUB-SUPER CRITICAL DIRICHLET PROBLEM 3Multiplying (1.5) by rN−1u and integrating on [s, t], then multiplying the sameequation by rNu′ and integrating also on [s, t] one has the following identity, knownas Pohozaev’s identity,tnH(t) = snH(s) +∫ tsrn(NG(u(r))− N − 22u(r)g(u(r)))dr, (2.3)where H(x) ≡ xE(x) + N−22 u′(x)u(x). In particular, taking s = 0 and t = t0 wehavetn0H(t0) ≥tN0 γdp+12p+1N≥ N(N−2)/2γ2p+1dξ ≡ c1dξ, (2.4)where γ = N/(p+ 1)− (N − 2)/2 and ξ is as in (1.8). Also, from (2.3), if u(r) ≥ 0for all r ∈ [0, t] we havetN (u′(t))2 = −(N − 2)u · tnu′ − 2tN up+1p+ 1+ 2∫ t0γrnup+1 dr. (2.5)Thus from (2.5) and the fact that t−n∫ t0snup+1 ds ≥ −u(t)u′(t) we have(−tu′u)′ = (−tu′′ − u′)u+ t(u′)2u2=−t(−nt u′ − up)u− uu′ + t(u′)2u2=(N−2︷ ︸︸ ︷n− 1)uu′ + tup+1 + t(u′)2u2=2t−n∫ t0snγup+1 ds− 2 tup+1p+1 + tup+1u2=2t−n∫ t0snγup+1 ds+ t(p−1p+1)up+1u2≥ 2γt(−tu′(t)u(t)),(2.6)provided u(s) > 0 for s ∈ (0, t). Integrating (2.6) on [t0, t) we haveln( −tu′(t)/u(t)−t0u′(t0)/u(t0))≥ ln ( tt0)γ.Letting Γ = −t0u′(t0)/u(t0) we conclude that−tu′(t)u(t)≥ Γ( tt0)γ.For future reference we note thatΓ ≥ 21−p, (2.7)where we have used (2.2), and −u′(t0) ≥ t0dp/(2pN) (see (2.1)). Integrating againin [t0, t] yieldsln(u(t0)u(t)) ≥ Γγtγ0[tγ − tγ0 ].Assuming that u(t) ≥ 0 for all t ∈ [t0, t0 ln1/γ(d) ≡ T ] we haveu(T ) ≤ u(t0)(ed−1)Γ/γ = eΓ/γ2d1−Γ/γ .4 A. CASTRO, J. KWON, C. M. TAN EJDE-2007/111Now we estimate E for t ≥ t0 with u(s) ≥ 0 for s ∈ (t0, t]. Since E′(t) ≥ −2nE(t)/t,E(t) ≥ E(s)(s/t)2n for any 0 ≤ s ≤ t ≤ 1. (2.8)Thus(u′(T ))22≥ E(t0)( t0T)2n − up+1(T )p+ 1≥ dp+1(p+ 1)2p+1 ln2n/γ(d)− 1p+ 1(eΓ/γ2d1−Γ/γ)p+1≥ dp+1(p+ 1)2p+2 ln2n/γ(d),(2.9)for d sufficiently large.Let us suppose now that that u(t) > 0, for any t ∈ [T, 2T ]. Arguing as in (2.9)we have(u′(t))22≥ E(T )(Tt)2n− up+1(T )p+ 1≥ dp+1(p+ 1)2p+1+2n ln2n/γ(d)− 1p+ 1(eΓ/γ2d1−Γ/γ)p+1≥ dp+1(p+ 1)2p+2+2n ln2n/γ(d),(2.10)for d large. Integrating on [T, t] we have0 ≤ u(t) = u(T ) +∫ tTu′(s)ds≤ eΓ/γ2d1−Γ/γ − (t− T )√2d(p+1)/221+n+(p/2) lnn/γ(d)√p+ 1.(2.11)Hence u has a zero in [d(1−p)/2, T + eΓ/γd(1−p)/2−Γ/γ2n+(p/2) lnn/γ(d)√p+ 1]. Wesummarize the above in the following lemma.Lemma 2.1. For d > 0 sufficiently large, there existst1 ∈ (√Nd(1−p)/2, 2√Nd(1−p)/2 ln1/γ(d)) (2.12)such that u(t1) = 0, u(s) > 0 for s ∈ [0, t1), anddp+1(p+ 1)2p+2+2n ln2n(d)≤ E(t1) ≤ dp+1p+ 1(2.13)EJDE-2007/111 A SUB-SUPER CRITICAL DIRICHLET PROBLEM 53. First local minimumLet t ∈ (t1, t1 + (1/2)(2/(q + 1))q/(q+1)|u′(t1)|(1−q)/(1+q) ≡ t1 + τ). From (2.12)and (2.13),t1t≥ 1− τt1 + τ≥ 1− τt1≥ 1− (1/2)(2/(q + 1))q/(q+1)|u′(t1)|(1−q)/(1+q)√Nd(1−p)/2≥ 1− (1/2)(2/(q + 1))q/(q+1)(d(p+1)/2/√p+ 1)(1−q)/(1+q)√Nd(1−p)/2≡ 1−md(p−q)/(1+q) ≥ 0.91/n.(3.1)for d large. Hence for d positive and largeu′(t) = t−n[tn1u′(t1)−∫ tt1sn|u(s)|q−1u(s)ds]≤ 0.9u′(t1) + (t− t1)(q + 12)q/(q+1)|u′(t1)|(2q)/(q+1)≤ 0.4u′(t1),(3.2)where we have used that, since E′ ≤ 0, |u(t)|q+1 ≤ (q + 1)(u′(t1))2/2 for t ≥ t1with u(t) ≤ 0. This and (3.2) yieldu(t1 + τ) ≤ 0.4u′(t1)τ ≤ −0.2(2/(q + 1))q/(q+1)|u′(t1)|2/(1+q). (3.3)Now for t ≥ t1 + τ with u(s) ≤ −0.2(2/(q + 1))q/(q+1)|u′(t1)|2/(1+q) for all s ∈(t1 + τ, t) we haveu′(t)= t−n[tn1u′(t1)−∫ tt1sn|u(s)|q−1u(s)ds]≥ u′(t1) + t−n(0.2(2/(q + 1))q/(q+1))q|u′(t1)|2q/(1+q)∫ tt1+τsnds≥ −u′(t1)[− 1 + t−n(0.2(2/(q + 1))q/(q+1))q|u′(t1)|(q−1)/(1+q) tN − (t1 + τ)NN]≥ −u′(t1)[− 1 + (0.2(2/(q + 1))q/(q+1))qN|u′(t1)|(q−1)/(1+q)(t− (t1 + τ))].(3.4)This and the definition of τ imply the following lemma.Lemma 3.1. There exists τ1 in(t1, t1 +{(1/2)(2/(q + 1))q/(q+1) +N(.2(2/(q + 1))q/(q+1))q}|u′(t1)|(1−q)/(1+q)]≡ (t1, t1 + κ1|u′(t1)|(1−q)/(1+q)]such that u′(τ1) = 0.6 A. CASTRO, J. KWON, C. M. TAN EJDE-2007/1114. Second zeroLet τ0 > τ1 be such that u(s) ≤ 0.5u(τ1) for all s ∈ [τ1, τ0]. Imitating thearguments leading to (2.2) we see thatτ1 + |u(τ1)|(1−q)/2 ≤ τ0 ≤ τ1 +√2qN |u(τ1)|(1−q)/2. (4.1)Henceu′(τ0) = τ−n0∫ τ0τ1sn|u(s)|qds≥ |u(τ1)|q(τN0 − τN1 )N2qτn0≥ |u(τ1)|q(τ0 − τ1)2qN≥ |u(τ1)|(1+q)/22qN,(4.2)andτn0 ≥ .9sn for any s ∈ (τ0, τ0 + 2q+2N |u(τ1)|(1−q)/2], (4.3)for d > 0 sufficiently large.Suppose now that for all s ∈ [τ0, r ≡ τ0+2q+1N |u(τ1)|(1−q)/2] we have u(s) ≤ 0.Thenu′(s) ≥ 0.9u′(τ0) for all s ∈ [τ0, r]. (4.4)This and and the definition of r give0 ≥ u(r) ≥ u(τ1)2+ .9(r − τ0)u′(r)≥ u(τ1)2+ .9(2q+1)N |u(τ1)|(1−q)/2 |u(τ1)|(1+q)/22qN= 1.3|u(τ1)|,which is a contradiction. From (3.3), |u(τ0)| ≥ 0.2(2/(q + 1))q/(q+1)|u′(t1)|2/(1+q).Since also τ0 ≤ t1+(κ1+0.2√2qN(2/(q+1))q/(q+1))|u′(t1)|(1−q)/(1+q) (see Lemma3.1 and (4.2)). Thusτ0 + 2q+1N |u(τ1)|(1−q)/2≤ t1 + (κ1 + .2(2q+2N)(2/(q + 1))q/(q+1))|u′(t1)|(1−q)/(1+q)≡ t1 + k2|u′(t1)|(1−q)/(1+q).Thus we have proven the following lemma.Lemma 4.1. There exists t2 ∈ [t1, t1 + k2|u′(t1)|(1−q)/(1+q)] such that u(t2) = 0and u(s) < 0 in (t1, t2).EJDE-2007/111 A SUB-SUPER CRITICAL DIRICHLET PROBLEM 75. First positive maximumLet t > t2 be such that u′(s) > 0 on [t2, t]. Thus u′′ ≤ 0 in [t2, t]. Henceu(s) ≤ u′(t2)(s− t2) for all s ∈ [t2, t]. Integrating (1.5) on [t2, s], we havesnu′(s) = tn2u′(t2)−∫ st2rn|u(r)|p−1u(r)dr≥ tn2u′(t2)− sn|u′(t2)|p(s− t2)p+1p+ 1≥ u′(t2)(tn2 −snp+ 1),(5.1)for s ≤ t2 + u′(t2)(1−p)/(1+p). Since tN2 |u′(t2)|2 ≥ 2c1dξ (see (2.4)) and (u′(t2))2 ≤2dp+1/(p+ 1), we havetN2 ≥ 2c1(p+ 12)ξ/(p+1)|u′(t2)|N(1−p)/(1+p). (5.2)Now fors ∈ [t2,min{21/n, 1 + (2c1)−1/N( 2p+ 1) ξN(p+1)}t2 ≡ αt2],from (5.1) and (5.2), we haveu′(s) ≥ u′(t2)( tn2sn− 1p+ 1) ≥ u′(t2)p− 1p+ 1. (5.3)Integration on [t2, αt2] yieldsu(αt2) ≥ p− 1p+ 1αt2u′(t2). (5.4)Therefore, assuming again that u′ > 0 on [t2, t], we havetnu′(t) ≤ tn2u′(t2)−∫ tαt2rn|u(r)|p−1u(r)dr≤ tn2u′(t2)− tn2 (t− αt2)(p− 1p+ 1αt2u′(t2))p.(5.5)This and (5.2) implyt− αt2 ≤(p− 1p+ 1α)−pt−p2 |u′(t2)|1−p≤ (p− 1p+ 1α)−p(2c1)−p/N(p+ 12 )−pξ/(N(p+1))|u′(t2)|(1−p)/(p+1)≡ κ2|u′(t2)|(1−p)/(p+1).(5.6)This proves the following result.Lemma 5.1. There exists τ2 ∈ [t2, αt2 + κ2|u′(t2)|(1−p)/(p+1)] such that u′(τ2) = 0and u′(s) > 0 on [t2, τ2).6. Energy on the interval [t0, τ2]Now we estimate the energy on [t0, τ2].Lemma 6.1. For t ∈ [t0, τ2],tnH(t) ≥ c1dξ. (6.1)8 A. CASTRO, J. KWON, C. M. TAN EJDE-2007/111Proof. Let us prove first that∫ t1t0tnγup+1(t)dt ≥∫ t2t1tnγ1|u(t)q+1|dt, (6.2)where γ1 = ((q+1)(N−2)−2N)/(2(q+1)). Let tˆ0 ∈ [t0, t1] be such that u(tˆ0) = d/4.Then, for t ∈ [t0, tˆ0], we have−u′(t) = t−n∫ t0snup(s) ds ≤ tdpN. (6.3)Integrating on [t0, tˆ0] we have (d/4) ≤ (tˆ20 − t20)dp/(2N). This and (2.2) yieldtˆ0 ≥√Nd1−p2+ t20 = t0√1 +Nd1−p2t20≥ t0√1 +12p+1, (6.4)which combined with (2.2) gives∫ t1t0tnγup+1(t)dt ≥∫ tˆ0t0tnγup+1(t)dt≥ γ(d/4)p+1 tˆN0 − tN0N≥ γ4p+1NtN0((1 +12p+1)N/2 − 1)dp+1≥ γ4p+1N((1 +12p+1)N/2 − 1)NN/2dξ.(6.5)Using (1.7), we have |u(t)|q+1 ≤ (q+1)dp+1/(p+1). Also from (2.3) and (2.4), wehave tN1 |u′(t1)|2/2 = t1H(t1) ≥ c1dξ. This implies that k2|u′(t1)|(1−q)/(1+q) < t1for d > 0 large. These inequalities and Lemma 4.1 imply∫ t2t1tn|u(t)|q+1dt ≤ (q + 1p+ 1dp+1) tN2 − tN1N≤ (q + 1p+ 1dp+1) (t1 + k2|u′(t1)|(1−q)/(1+q))N − tN1N≤ (q + 1p+ 1dp+1)tn1(2N − 1)k2|u′(t1)|(1−q)/(1+q)N≤ (q + 1p+ 1dp+1) (2N − 1)k2Ntn1(dξ/2t−N/21)(1−q)/(1+q)=(q + 1p+ 1) (2N − 1)k2Ndp+1+(ξ(1−q)/(2(1+q))tN−1−N(1−q)/(2(1+q))1≤ (q + 1p+ 1) (2N − 1)k2NlnM/γ(d) dη,(6.6)whereη = p+ 1 +ξ(1− q) + (1− p)(2(N − 1)(1 + q)−N(1− q))2(1 + q),M = (2(N − 1)(1 + q)−N(1− q))/(2(1 + q)).EJDE-2007/111 A SUB-SUPER CRITICAL DIRICHLET PROBLEM 9An elementary calculation shows that ξ > η. Thus from (6.5) and (6.6), (6.2)follows. Thus for t ∈ [t1, τ2],tnH(t) = tn0H(t0) +∫ tt0sn(NG(u(s))− N − 22u(s)g(u(s)))ds≥ tn0H(t0) +∫ t2t0sn(NG(u(s))− N − 22u(s)g(u(s)))ds≥ tn0H(t0)≥ c1dξ,(6.7)which proves the lemma. 7. Proof of Theorem 1.1Arguing as in Lemmas 2.1 and 4.1, we see that for d > 0 sufficiently large thereexist numberst3 < · · · < tk ≤ 1 (7.1)such thatu(t) < 0 in (t2i−1, t2i), and u(t) > 0 in (t2i, t2i+1), i = 1, . . .min{k2 ,k + 12}.(7.2)Imitating the arguments leading to (6.2), one sees that∫ t2i+1t2itnγup+1(t)dt ≥∫ t2i+2t2i+1tnγ1|u(t)q+1|dt. (7.3)This in turn (see (6.7)) leads totnH(t) ≥ c1dξ for all t ∈ [t0, 1]. (7.4)This, together with Lemma 2.1, proves Theorem 1.2. From (7.4) we see thatρ2(t) ≡ u2(t) + (u′(t))2 →∞ as d→ +∞, (7.5)uniformly for t ∈ [0, 1]. Therefore, there exists a continuous argument functionθ(t, d) ≡ θ(t) such thatu(t) = ρ(t) cos(θ(t)) and u′(t) = −ρ(t) sin(θ(t)). (7.6)From this we see that θ′(t) = {((n/t)u′(t) + g(u(t)))u(t) + (u′(t))2}/ρ2(t). Thusθ′(t) > 0 for θ(t) = jpi/2 with j = 1, . . . , which implies that if θ(t) = jpi/2 thenθ(s) > jpi/2 for all s ∈ (t, 1].Imitating the arguments of Lemmas 2.1 and 4.1, we see that t2i − t2(i−1) ≤c3 ln1/γ(d)d(1−p)/2. Thus k ≥ c4 ln−1/γ(d)d(p−1)/2 (see (7.1)), which implies thatlimd→+∞θ(1, d) = +∞. (7.7)By the continuity of θ and the intermediate value theorem we see that there existsa sequence d1 < · · · < dj < · · · → ∞ such that θ(1, dj) = jpi+(pi/2). Hence u(t, dj)is a solution to (1.1) having exactly j zeroes in (0, 1), which proves Theorem 1.1.10 A. CASTRO, J. KWON, C. M. TAN EJDE-2007/111References[1] F. Atkinson, H. Brezis and L. Peletier; Solutions d’e´quations elliptiques avec exposant deSobolev critique que changent de signe, C. R. Acad. Sci. Paris Serie I 306 (1988), pp. 711-714.[2] R. Benguria, J. Dolbeault, and M. Esteban; Classification of the solutions of semilinearelliptic problems in a ball, J. Differential Equations 167 (2000), no. 2, pp. 438-466.[3] H. Brezis and L. Nirenberg; Positive solutions of nonlinear elliptic equations involving criticalSobolev exponents, Comm. Pure Appl. Math. 36 (1983), no. 4, pp. 437-477.[4] A. Castro and A. Kurepa; Infinitely many solutions to a superlinear Dirichlet problem in aball, Proc. Amer. Math. Soc., 101 (1987), No. 1, pp. 57-64.[5] A. Castro and A. Kurepa; Radially symmetric solutions to a superlinear Dirichlet problemwith jumping nonlinearities, Trans. Amer. Math. Soc. 315 (1989), pp. 353-372[6] A. Castro and A. Kurepa, Radially symmetric solutions to a Dirichlet problem involvingcritical exponents, Trans. Amer. Math. Soc., 348 (1996), no. 2, pp. 781-798.[7] G. Cerami, S. Solimini, M. Struwe, Some existence results for superlinear elliptic boundaryvalue problems involving critical exponents, J. Funct. Anal. 69 (1986), 289-306.[8] A. El Hachimi and F. de Thelin; Infinitely many radially symmetric solutions for a quasilinearelliptic problem in a ball, J. Differential Equations 128 (1996), pp. 78-102.[9] L. Erbe and M. Tang; Structure of positive radial solutions of semilinear elliptic equations,J. Differential Equations 133 (1997), pp. 179-202.[10] M. Garc´ıa-Huidobro, R. Mansevich, and F. Zanolin; Infinitely many solutions for a Dirich-let problem with a nonhomogeneous p-Laplacian-like operator in a ball. Adv. DifferentialEquations (1997), no. 2, pp. 203-230.[11] J. Jacobsen, and K. Schmitt; Radial solutions of quasilinear elliptic differential equations.Handbook of differential equations, pp. 359-435, Elsevier/North-Holland, Amsterdam, (2004).[12] S. I. Pohozaev; On the eigenfunctions of the equation ∆u + λf(u) = 0. Dokl. Akad. NaukSSSR 165 (1965), pp. 36-39.Departament of Mathematics, Harvey Mudd College, Claremont, CA 91711, USAE-mail address, A. Castro: castro@math.hmc.eduE-mail address, J. Kwon: kwonjy@math.uci.eduE-mail address, C. M. Tan: ctan@hmc.edu

Infinitely Many Radial Solutions for a Sub-Super Critical Dirichlet Boundary Value Problem in a Ball

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Infinitely Many Radial Solutions for a Sub-Super Critical Dirichlet Boundary Value Problem in a Ball

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