POSITIVE RADIAL SOLUTIONS FOR ELLIPTIC ...

COMMUNICATIONS ON PURE AND APPLIED ANALYSIS Volume 13, Number 6, November 2014

doi:10.3934/cpaa.2014.13.2713 pp. 2713–2731

POSITIVE RADIAL SOLUTIONS FOR ELLIPTIC EQUATIONS ON EXTERIOR DOMAINS WITH NONLINEAR BOUNDARY CONDITIONS

Dagny Butler, Eunkyung Ko Department of Mathematics & Statistics, Mississippi State University Mississippi State, MS 39762, USA Department of Mathematics & Statistics, University of North Carolina at Greensboro, Greensboro, NC 27412, USA TIFR Center for Applicable Mathematics, Yelahanka, Bangalore 560065, India,

Eun Kyoung Lee and R. Shivaji Department of Mathematics Education, Pusan National University, Busan, South Korea, Department of Mathematics & Statistics, University of North Carolina at Greensboro, Greensboro, NC 27412, USA,

(Communicated by Wei Feng) Abstract. We study positive radial solutions to the boundary value problem  −∆u = λK(|x|)f (u), x ∈ Ω,     ∂u + c˜(u)u = 0, |x| = r0 ,  ∂η    u(x) → 0, |x| → ∞, where ∆u = div ∇u is the Laplacian of u, λ is a positive parameter, Ω = {x ∈ RN |N > 2, |x| > r0 with r0 > 0}, K : [r0 , ∞) → (0, ∞) is a continuous ∂ function such that lim K(r) = 0, ∂η is the outward normal derivative, and r→∞

c˜ : [0, ∞) → (0, ∞) is a continuous function. We considervarious C 1 classes of f (s) the reaction term f : [0, ∞) → R that are sublinear at ∞ i.e. lim =0 . s→∞ s In particular, we discuss existence and multiplicity results for classes of f with (a) f (0) > 0, (b) f (0) < 0, and (c) f (0) = 0. We establish our existence and multiplicity results via the method of sub-super solutions. We also discuss some uniqueness results.

1. Introduction. In reaction diffusion processes, steady states define the long term dynamics. Here we consider a steady state reaction diffusion equation on an exterior domain with a nonlinear boundary condition on the interior boundary. Namely, we 2000 Mathematics Subject Classification. Primary: 35J66, 34B18; Secondary: 35J60. Key words and phrases. Positive solution, nonlinear boundary conditions, exterior domains. The third author is supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (2012R1A1A1011225).

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DAGNY BUTLER, EUNKYUNG KO, EUN KYOUNG LEE AND R. SHIVAJI

study positive radial solutions to:  x ∈ Ω,   −∆u = λK(|x|)f (u),   ∂u + c˜(u)u = 0, |x| = r0 , (1)  ∂η    u(x) → 0, |x| → ∞, where ∆u = div ∇u is the Laplacian of u, λ is a positive parameter, Ω = {x ∈ RN |N > 2, |x| > r0 with r0 > 0}, K : [r0 , ∞) → (0, ∞) is a continuous function ∂ such that lim K(r) = 0, ∂η is the outward normal derivative, and c˜ : [0, ∞) → r→∞

(0, ∞) is a continuous function. Here the reaction term f : [0, ∞) → R is a C 1 function. There is a rich history in the literature on the study of steady state reaction diffusion equations on bounded domains with Dirichlet boundary conditions when f (0) > 0 (see [2, 4, 8, 10, 11, 15], and [22]), when f (0) < 0 (see [1, 3, 5, 7, 9, 12], and [18]), and when f (0) = 0 (see [17] and [19]). See also [6, 16], and [20] where recently various results have been discussed for exterior domains with Dirichlet boundary conditions on |x| = r0 . Recently, the authors of [13] studied steady state equations on bounded domains arising in combustion theory with nonlinear boundary conditions (as in equation (1)). Here our focus will be to establish several results for exterior domain problems with nonlinear boundary conditions at |x| = r0 as described in (1). See [13] for the motivation of studying reaction diffusion equations with such nonlinear boundary conditions. Instead of working directly with (1), we note that the change of variables r = |x| and s = ( rr0 )2−N transforms (1) into the following boundary value problem (for details, see the Appendix 8.1):  ˜ −u00 (t) = λh(t)f (u(t)), t ∈ (0, 1),     (N − 2) 0 u (1) + c˜(u(1))u(1) = 0,  r0    u(0) = 0, ˜ where h(t) =

−2(N −1) 1 r02 N −2 K(r0 t 2−N (2−N )2 t

). We will only assume K(r) ≤ rN1+µ for ˜ ∈ C((0, 1], (0, ∞)) could be singular r 1 and for some µ ∈ (0, N − 2). Then, h ˜ ˜ = inf h(t) > 0, and there exists a constant d˜ > 0 such that at 0. Note that h t∈(0,1]

˜ ≤ dα˜˜ for all t ∈ (0, 1] where α h(t) ˜ = (NN−2)−µ −2 . t Motivated by the above discussion, in this paper, we will study positive solutions in C 2 (0, 1) ∩ C 1 [0, 1] to the following boundary value problem:  00 t ∈ (0, 1),   −u (t) = λh(t)f (u(t)), 0 u (1) + c(u(1))u(1) = 0, (2)   u(0) = 0, where c : [0, ∞) → (0, ∞) is a continuous function and h ∈ C((0, 1], (0, ∞)) is such that h = inf h(t) > 0 and there exists a constant d > 0 such that h(t) ≤ tdα for t∈(0,1]

all t ∈ (0, 1] where α ∈ (0, 1). We consider various C 1 classes of the reaction term f : [0, ∞) → R that are sublinear at ∞; we state this as our first main hypothesis:

POSITIVE RADIAL SOLUTIONS FOR ELLIPTIC EQUATIONS

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f (s) = 0. s In particular, we discuss existence and multiplicity results for classes of f with (a) f (0) > 0, (b) f (0) < 0, and (c) f (0) = 0. We will establish several existence and multiplicity results via the method of sub-super solutions. We also discuss some uniqueness results. Our first five results concern existence, multiplicity, and uniqueness in the case when f (0) > 0 (see Figure 1). (H1) lim

s→∞

fHuL

u

λ vs. k|uk∞

Graph of f (u)

Figure 1. Graph of f (u) where f (0) > 0 and a possible Bifurcation Diagram We state two additional hypotheses which we use for subsequent results: (H2) The function f (s) is increasing for s ∈ [0, ∞). s (H3) The function f (s) is nondecreasing for s ∈ [0, ∞). Theorem 1.1. Suppose f (s) > 0 for s ∈ [0, ∞) and (H1) hold. Then, the boundary value problem (2) has at least one positive solution for all λ > 0. Theorem 1.2. Suppose f (s) > 0 for s ∈ [0, ∞), (H1) and (H3) hold, and c is a strictly increasing function. Then, the boundary value problem (2) has a unique positive solution for all λ > 0. Next, we introduce e the unique positive solution of  00   −e (t) = h(t), t ∈ (0, 1), e0 (1) = 0,   e(0) = 0,

(3)

which we will use in the statement of our following result: Theorem 1.3. Suppose f (0) > 0, (H1) and (H2) hold. If there exists 0 < a < b a 16kek∞ f (a) such that b > , then (2) has at least three positive solutions for λ∗ < h f (b) λ < λ∗ , where λ∗ =

16b hf (b)

and λ∗ =

a f (a)kek∞ .

In the following two theorems, we focus on additional uniqueness results when f (0) > 0:

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Theorem 1.4. Suppose f (s) > 0 for s ∈ [0, ∞), (H1) hold, and c is a strictly increasing function. Then, there exists λ > 0 such that for λ ∈ (0, λ), (2) has a unique positive solution. To state the next theorem, we introduce some additional assumptions: p (H4) There exists A, B > 0 such that c(u) ≤ Au + B for some p ∈ (0, 1 − α). (p+1)(1−α) D (H5) There exists γ ∈ p + 1, such that f 0 (u) ≤ (1+u) γ for all u > 0 p and some D > 0.

Theorem 1.5. Suppose f (s) > 0 for s ∈ [0, ∞), (H4), (H5) hold, and c is a strictly increasing function. Then, (2) has a unique positive solution for λ 1. Before we discuss our existence and multiplicity results for the cases when f (0) < 0 and f (0) = 0, we state the following nonexistence result for λ small: Theorem 1.6. Suppose f (0) ≤ 0 and (H1) holds. Then, there does not exist a positive solution of (2) when λ is small. Our next theorem concerns existence in the case that f (0) < 0 (see Figure 2). fHuL

u

Graph of f (u)

λ vs. k|uk∞

Figure 2. Graph of f (u) where f (0) < 0 and a typical Bifurcation Diagram To state our next result, we introduce another hypothesis: (H6) lim f (s) = ∞. s→∞

Theorem 1.7. Suppose f (0) < 0, (H1) and (H6) hold. Then, (2) has at least one positive solution for λ 1. Our remaining two results concern the cases when f (0) = 0 with (i) f 0 (0) = 0 and (ii) f 0 (0) > 0 (see Figures 3 and 4). We first introduce another hypothesis we use to establish our results. f (s) (H7) lim 1+α = 0 for α ∈ (0, 1). s→0 s We establish: Theorem 1.8. Suppose f (0) = 0, f 0 (0) = 0, (H1), (H2), (H6), and (H7) hold. Then, (2) has at least two positive solutions for λ 1. Theorem 1.9. Suppose f (0) = 0, f 0 (0) > 0, (H1), (H3), and (H6) hold, and c is a strictly increasing function. Then (2) has a unique positive solution for λ 1.


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fHuL

u

f 0 (0) = 0

λ vs. k|uk∞

Figure 3. Graph of f (u) where f (0) = 0 and a typical Bifurcation Diagram

fHuL

u

f 0 (0) > 0

λ vs. k|uk∞

Figure 4. Graph of f (u) where f (0) = 0 with a typical Bifurcation Diagram We will use the method of sub-supersolutions to prove our existence and multiplicity results. By a subsolution of (2), we mean a function ψ ∈ C 2 (0, 1) ∩ C 1 [0, 1] that satisfies  00 0 < t < 1,   −ψ (t) ≤ λh(t)f (ψ), 0 ψ (1) + c(ψ(1))ψ(1) ≤ 0, (4)   ψ(0) ≤ 0, and by a supersolution of (2), we mean a function Z ∈ C 2 (0, 1) ∩ C 1 [0, 1] that satisfies  00 0 < t < 1,   −Z (t) ≥ λh(t)f (Z), 0 Z (1) + c(Z(1))Z(1) ≥ 0, (5)   Z(0) ≥ 0. We also define u ≺ v if: (i) u(t) < v(t) on (0, 1] and (ii) u(0) < v(0) or u0 (0) < v 0 (0). We say that a function ψ is a strict subsolution of (2) if: (i) ψ is a subsolution of (2) and (ii) u ψ for all solutions u of (2) such that u(t) ≥ ψ(t) for all t ∈ [0, 1]. Similarly, we say that a function Z is a strict supersolution of (2) if: (i) Z is a supersolution of (2) and (ii) u ≺ Z for all solutions u of (2) such that u(t) ≤ Z(t) for all t ∈ [0, 1].

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We first make an observation on strict sub (super) solutions. Lemma 1.10. If f is increasing, then every sub (super) solution ψ (Z) which is not a solution is a strict sub (super) solution of (2). Proof. We will begin with the proof that ψ is a strict subsolution. Let u be any solution such that u ≥ ψ. Then (u − ψ) satisfies: −(u − ψ)00 (t) ≥ λh(t)[f (u(t)) − f (ψ(t))]

(6)

(u − ψ)(0) ≥ 0

(7)

(u − ψ) (1) ≥ [c(ψ(1))ψ(1) − c(u(1))u(1)].

(8)

0

Note that since ψ is not a solution, at least one of the above inequalities is strict. Further, since f is increasing, from (6) we have (u − ψ)00 ≤ 0. Now if u(1) = ψ(1), then necessarily (u−ψ)0 (1) < 0 while from (8) we get (u−ψ)0 (1) ≥ 0, a contradiction. Hence, u(t) > ψ(t), t ∈ (0, 1]. Now if u(0) > ψ(0), then we are done. If u(0) = ψ(0), then u0 (0) > ψ 0 (0) must hold since if u0 (0) = ψ 0 (0), then again the concavity of u−ψ implies u(1) < ψ(1), a contradiction. Hence, the proof that ψ is a strict subsolution is complete. A similar argument shows that Z is a strict supersolution. Next, we let X = {u ∈ C 1 [0, 1]|u(0) = 0}. We observe (see the Appendix 8.2 for details) that solving problem (2) is equivalent to finding u ∈ X such that Z 1 u(t) = (T u)(t) := λ G(t, s)h(s)f (u(s))ds − c(u(1))u(1)t; t ∈ [0, 1], (9) 0

where G(t, s) =

s,

0 ≤ s ≤ t,

t,

t ≤ s ≤ 1.

(10)

Then, the following lemma holds. Lemma 1.11. If ψ is a subsolution of (2) and Z is a supersolution of (2) such that ψ ≤ Z then (2) has a solution u ∈ [ψ, Z]. Moreover, if ψ and Z are strict sub and super solutions respectively such that ψ ≺ Z, then there exists C 1 such that deg(I − T, D, 0) = 1 where D := {v ∈ X|ψ ≺ v ≺ Z, ||v||1 < C}. To establish our multiplicity results, we use the following lemma. Lemma 1.12. Suppose there exists a subsolution ψ1 , a strict supersolution Z1 , a strict subsolution ψ2 , and a supersolution Z2 for (2) such that ψ1 ≤ Z1 ≤ Z2 , ψ1 ≤ ψ2 ≤ Z2 , and ψ2 6≤ Z1 . Then, (2) has at least three distinct solutions u1 , u2 , and u3 such that ψ1 ≤ u1 ≺ Z1 , ψ2 ≺ u2 ≤ Z2 , and u3 ∈ [ψ1 , Z2 ] \ ([ψ1 , Z1 ] ∪ [ψ2 , Z2 ]). Remark 1. For results similar to Lemmas 1.11 and 1.12 in the case when the reaction term has no singularities and with Dirichlet boundary conditions, see [2] and [21], and in the case when the reaction term has no singularities and with nonlinear boundary conditions, see [14]. In Section 2, we discuss some necessary preliminaries, including establishing Lemmas 1.11 and 1.12. Then, in Section 3, we prove Theorems 1.1 - 1.5 for the case f (0) > 0. We discuss a non-existence result when f (0) ≤ 0 (Theorem 1.6) in Section 4. Section 5 is concerned with proving an existence result for the case f (0) < 0 (Theorem 1.7). In Section 6, we prove results when f (0) = 0 (Theorems 1.8 - 1.9). In Section 7, we discuss examples that satisfy the hypotheses of our theorems. Finally, in the Appendix in Section 8.1, we show the details of the transformation used


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to obtain the boundary value problem (2) from (1), and in Section 8.2, we show that solving (2) is equivalent to finding a solution u ∈ X to the integral equation (9). 2. Preliminaries. Here we restate and prove our main results related to sub-super solutions, namely Lemma 1.11 and Lemma 1.12, discussed in the introduction. 2.1. Lemma 2.1. Lemma 2.1. (see also Lemma 1.11) If ψ is a subsolution of (2) and Z is a supersolution of (2) such that ψ ≤ Z then (2) has a solution u ∈ [ψ, Z]. Moreover, if ψ and Z are strict sub and super solutions respectively such that ψ ≺ Z, then there exists C 1 such that deg(I − T, D, 0) = 1 where D := {v ∈ X|ψ ≺ v ≺ Z, ||v||1 < C}. Proof of Lemma 2.1. Consider  00   u (t) + λh(t)f (γ(t, u(t))) = 0, u0 (1) = −¯ c(u(1)),   u(0) = 0, where γ : (0, 1) × R → R is defined by    Z(t), u, γ(t, u) =   ψ(t), and

   c(Z(1))Z(1), c(a)a, c¯(a) =   c(ψ(1))ψ(1),

t ∈ (0, 1), (M)

u > Z(t), ψ(t) ≤ u ≤ Z(t),

(11)

u < ψ(t), a > Z(1), ψ(1) ≤ a ≤ Z(1),

(12)

a < ψ(1).

Define T¯ : X → X by T¯(u)(t) = λ

Z

1

G(t, s)h(s)f (γ(s, u(s)))ds − c¯(u(1))t 0

We claim that T¯ is completely continuous in X. To prove this, we need to show that T : X → X is completely continuous, because that will imply that T¯ = T ◦ γ : X → X is completely continuous. That is, we need to show that T is continuous and that T takes bounded sets of X into relatively compact sets of X. To show the latter, we let B be a bounded set in X. If {un } ⊂ B, then it suffices to show that there exist {unk } ⊂ {un } such that T unk converge in X. Since ||u0n ||∞ ≤ M , {un } is equicontinuous in C[0, 1]. By Arzela-Ascoli Theorem, there exists {unk } ⊂ {un } such that unk → u in C[0, 1]. Then, T unk → T u in C[0, 1], since T is continuous in C[0, 1], and |T 0 (unk )(t) − T 0 (u)(t)| Z 1 ≤ λh(s)|f (unk (s)) − f (u(s))|ds + |c(unk (1))unk (1) − c(u(1))u(1)| t

Z ≤

1

λh(s)||f (unk ) − f (u)||∞ ds + |c(unk (1))unk (1) − c(u(1))u(1)| 0

→0 uniformly as k → ∞. Hence, T unk converges to T u in X which implies T (B) is relatively compact in X.

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We also need to show that T : X → X is continuous. If un → u on X, then similar to the above calculations, T un → T u in C[0, 1] and (T un )0 → (T u)0 in C[0, 1]. Hence, T is continuous in X. Therefore, we have shown that T is completely continuous in X, and it follows that T¯ is also completely continuous in X. Furthermore, we claim that T¯ is bounded in X; that is, there exists C 1 such that ||T¯u||1 < C for all u ∈ X. To prove this claim, for all u ∈ X, we have ||T¯u||1 = sup |(T¯u)(x)| + sup |(T¯u)0 (x)| x∈[0,1]

x∈[0,1] 1

λG(x, s)h(s)f (γ(s, u(s)))ds¯ c(u(1))x x∈[0,1] 0 Z 1 λh(s)f (γ(s, u(s)))ds + c¯(u(1)) + sup x∈[0,1] x Z 1 h(s)ds + 2M2 ≤2M1 λ

Z = sup

0

u(t) in (a, b), ψ(a) = u(a), and ψ(b) = u(b) or (ii) there exists (a, 1) ⊂ (0, 1) such that ψ(t) > u(t) in (a, 1] and ψ(a) = u(a). If (i) holds, then define w(t) := ψ(t) − u(t) which implies w > 0 on (a, b) and w(a) = 0 = w(b). Then, there exists t˜ ∈ (a, b) such that w0 (t˜) = ψ 0 (t˜) − u0 (t˜) = 0 R t˜ R t˜ and w0 (t) > 0 for t ∈ (a, t˜). Thus, 0 > −w0 (t) = t w00 (s)ds = t ψ 00 (s) − u00 (s)ds ≥ R t˜ (−λh(s)f (ψ(s)) + λh(s)f (γ(s, u(s))))ds = 0, a contradiction. t If (ii) holds, then define w(t) := ψ(t) − u(t) which implies w > 0 on (a, 1] and w(a) = 0. Since ψ(1) > u(1), w0 (1) = ψ 0 (1) − u0 (1) ≤ −c(ψ(1))ψ(1) + c¯(u(1)) = 0. This implies that there exists t˜ ∈ (a, 1] such that w0 (t˜) = 0 and w0 (t) > 0 for t ∈ (a, t˜), which gives a similar contradiction as in (i). Furthermore, u ≤ Z follows by a similar argument. Hence, u ∈ [ψ, Z], and consequently, u is a solution of (2). Next, if ψ and Z are strict sub and super solutions respectively, then ψ ≺ u ≺ Z. Also, all solutions v of (2) such that v ∈ [ψ, Z] satisfy ψ ≺ v ≺ Z and ||v||1 < C. Let D := {v ∈ X|ψ ≺ v ≺ Z, ||v||1 < C}. Clearly, D is bounded. Also, since v ∈ D satisfy ψ ≺ v ≺ Z, there exists > 0 such that B (v) := {z ∈ X|kz − vk1 < } ⊂ D, and hence, D is open.


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By the Excision Property, for C˜ 1 such that D ⊂ BC˜ (0), it follows that deg(I − T¯, D, 0) = deg(I − T¯, BC˜ (0), 0) = 1. Since T = T¯ on D, we have deg(I − T, D, 0) = deg(I − T¯, D, 0) = 1. 2.2. Lemma 2.2. Lemma 2.2. (see also Lemma 1.12) Suppose there exists a subsolution ψ1 , a strict supersolution Z1 , a strict subsolution ψ2 , and a supersolution Z2 for (2) such that ψ1 ≤ Z1 ≤ Z2 , ψ1 ≤ ψ2 ≤ Z2 , and ψ2 6≤ Z1 . Then, (2) has at least three distinct solutions u1 , u2 , and u3 such that ψ1 ≤ u1 ≺ Z1 , ψ2 ≺ u2 ≤ Z2 , and u3 ∈ [ψ1 , Z2 ] \ ([ψ1 , Z1 ] ∪ [ψ2 , Z2 ]). Proof of Lemma 2.2. Consider the modified problem  00 t ∈ (0, 1),   u (t) + λh(t)f (γ1,2 (t, u(t))) = 0, u(1) = −¯ c1,2 (u(1)),   u(0) = 0,

(M1,2 )

where γ1,2 : (0, 1) × R → R is defined by  u > Z2 (t),   Z2 (t), u, ψ1 (t) ≤ u ≤ Z2 (t), (13) γ1,2 (t, u) =   ψ1 (t), u < ψ1 (t), and  a > Z2 (1),   c(Z2 (1))Z2 (1), ψ1 (1) ≤ a ≤ Z2 (1), (14) c¯1,2 (a) = c(a)a,   c(ψ1 (1))ψ1 (1), a < ψ1 (1). For any > 0, we see that ψ1 − and Z2 + are strict sub and super solutions of (M1,2 ) (because if u is a solution of (M1,2 ), then this implies that ψ1 (t) − < ψ1 (t) ≤ u(t) ≤ Z2 (t) < Z2 (t) + .) By Lemma 2.1, there exists C 1 such that deg(I − T1,2 , D1,1 , 0) = 1, deg(I − T1,2 , D2,2 , 0) = 1 and deg(I − T1,2 , D1,2 , 0) = 1 where T1,2 : X → X is defined by T1,2 (u)(t) = T (γ1,2 (t, u(t))) and D1,1 = {u ∈ X|ψ1 − ≺ u ≺ Z1 , ||u||1 < C}, D2,2 = {u ∈ X|ψ2 ≺ u ≺ Z2 + , ||u||1 < C}, and D1,2 = {u ∈ X|ψ1 − ≺ u ≺ Z2 + , ||u||1 < C}. ¯ 1,1 ∪ D ¯ 2,2 ), 0) = −1. By the Excision and Additive Property, deg(I −T1,2 , D1,2 \(D Hence, there exists at least three distinct solutions u1 , u2 , and u3 of (2) such that ψ ≤ u1 ≺ Z1 , ψ2 ≺ u2 ≤ Z2 and u3 ∈ [ψ1 , Z2 ] \ ([ψ1 , Z1 ] ∪ [ψ2 , Z2 ]). 3. Proofs of Theorems 1.1 - 1.5. 3.1. Proof of Theorem 1.1. First, we construct a subsolution ψ1 . Clearly, ψ1 ≡ 0 is a subsolution of (2), because −ψ100 < λh(t)f (ψ1 ). Also, ψ 0 (1) + c(ψ1 (1))ψ1 (1) = 0 and ψ1 (0) = 0. Next, we will construct a positive supersolution Z2 . Let fˆ(x) := max[0,x] f (t). ˆ

Note that since limu→∞ f (u) = 0, there exists Mλ 1 such that u 1 . Let Z = M e. Then, 2 λ λkek∞

fˆ(Mλ kek∞ ) Mλ kek∞

−Z200 = Mλ h(t) ≥ λh(t)fˆ(Mλ kek∞ ) ≥ λh(t)fˆ(Mλ e) ≥ λh(t)f (Z2 ).

≤

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Also, Z20 (1) + c(Z2 (1))Z2 (1) = Mλ e0 (1) + c(Mλ e(1))Mλ e(1) = c(Mλ e(1))Mλ e(1) ≥ 0. Finally, Z2 (0) = Mλ e(0) = 0. Hence, Z2 = Mλ e is a supersolution of (2). Thus, by Lemma 2.1, (2) has at least one positive solution for all λ > 0. 3.2. Proof of Theorem 1.2. We already know that (2) has at least one positive solution for all λ > 0. Since ψ = 0 is a strict subsolution of (2), then there exists a minimal positive solution, u. If v is any other positive solution of (2) such R 1 00 R1 that v 6≡ u, then u < v on (0, 1]. Then, [vu − iuv 00 ]dx = 0 [v(−λh(x)f (u)) − 0 h R1 R1 v u dx. Thus, 0 [vu00 − uv 00 ]dx ≤ − f (v) u(−λh(x)f (v))]dx = λ 0 h(x)f (u)f (v) f (u) u 0, since f (u) is nondecreasing. R1 On the other hand, 0 [vu00 −uv 00 ]dx = v(1)u0 (1)−u(1)v 0 (1) = −v(1)c(u(1))u(1)+ R1 u(1)c(v(1))v(1) = u(1)v(1)[c(v(1)) − c(u(1))]. Since c is strictly increasing, 0 [vu00 − uv 00 ]dx > 0, which is a contradiction. Therefore, u = v, so (2) has a unique positive solution for λ > 0. 3.3. Proof of Theorem 1.3. We already know that (2) has at least one positive solution for λ > 0. We let ψ1 and Z2 be as in the proof of Theorem 1.1. We need to construct a strict subsolution and a strict supersolution. First, we construct a a . Let Z1 := a ||e||e ∞ . Then, ||Z1 ||∞ = a. strict supersolution Z1 when λ < f (a)||e|| ∞ a e It follows that −Z100 (t) = ||e|| h(t) > λh(t)f (a) ≥ λh(t)f a = λh(t)f (Z1 ). ||e||∞ ∞ a a a Also, Z10 (1) + c(Z1 (1))Z1 (1) = ||e|| e0 (1) + c ||e|| e(1) ||e|| e(1) = ∞ ∞ ∞ a a c ||e||∞ e(1) ||e||∞ e(1) ≥ 0, and Z1 (0) = 0. Hence, Z1 is a strict supersolution of a . (2) if λ < f (a)||e|| ∞ Next, we will construct a strict subsolution ψ2 with ||ψ2 ||∞ > b when λ > hf16b (b) . Define 1 − (1 − ( r )µ )δ , 0 ≤ r ≤ , ρ(r) = (15) 1, ≤ r ≤ 21 ,

where δ, µ > 1 and ≈ 0, and ρ(r) = ρ(1 − r). Let v(r) = bρ(r) (see Figure 5). Note that |v 0 (r)| ≤ b δµ .

b

1

Figure 5. Graph of v


Define ψ2 on [0, 21 ] to be the solution of −ψ200 (r) = λhf (v(r)), ψ2 (0) = 0 = ψ20 ( 12 ),

r ∈ (0, 12 ),

2723

(16)

and extend ψ2 on [ 21 , 1] such that ψ2 (r) = ψ2 (1 − r). Note that ψ20 (0) > 0 implies that ψ20 (1) < 0. We claim that ψ2 (r) > v(r) on (0, 1). If this claim is true, then −ψ200 = λhf (v(r)) < λh(r)f (ψ2 (r)) for r ∈ (0, 1). Also, ψ20 (1) + c(ψ2 (1))ψ2 (1) = ψ20 (1) < 0 and ψ2 (0) = 0. Therefore, ψ2 is a strict subsolution of (2). To prove the claim that ψ2 (r) > v(r) on (0, 1), it is enough to show that ψ20 (r) > R1 R1 v 0 (r) for r ∈ [0, ). From (16), we have t2 −ψ200 (r)dr = λh t2 f (v(r))dr, t ∈ [0, ]. R1 R1 Thus, ψ20 (t) = λh t2 f (v(r))dr ≥ λh 2 f (v(r))dr = λhf (b) 12 − . δµ b 0 Hence, if λ > f (b) , then ψ20 (r) > b δµ ≥ v (t), t ∈ [0, ). Now choosing ( 12 −)h 1 = 41 and δ ≈ 1, µ ≈ 1, this can be achieved since λ > hf16b (b) . Note that ψ2 ( 2 ) > v( 12 ) = b > a = ||Z1 ||∞ , so ψ2 6≤ Z1 . Moreover, Mλ can be chosen large enough so that Z2 > Z1 and Z2 > ψ2 . Thus, by Lemma 2.2, (2) has at least three positive a solutions when hf16b (b) < λ < f (a)kek∞ . 3.4. Proof of Theorem 1.4. We already know that (2) has at least one positive solution for all λ > 0. Also, recall that if u is a solution to (2) then u(t) = R1 λG(t, s)h(s)f (u(s))ds − c(u(1))u(1)t. Since ψ = 0 is a strict subsolution of (2), 0 then there exists a minimal positive solution, u. If v is any other positive solution of (2) such that v 6≡ u, then u < v on (0, 1]. Then, Z 1 v(t) − u(t) =λ G(t, s)h(s)[f (v(s)) − f (u(s))]ds + t[c(u(1))u(1) − c(v(1))v(1)] 0

Z

1

G(t, s)h(s)[f (v(s)) − f (u(s))]ds (since c is strictly increasing)

0 such that f 0 (s) ≤ M for all s ≥ 0. Thus, R1 R1 it follows that w(t0 ) ≤ λ 0 G(t0 , s)h(s)f 0 (θ)w(t0 )ds ≤ λw(t0 )M 0 G(t0 , s)h(s)ds. R1 Hence, 1 ≤ λM 0 G(t0 , s)h(s)ds, which is a contradiction for λ ≈ 0. Therefore, u = v, so (2) has a unique positive solution for λ small. 3.5. Proof of Theorem 1.5. We first state and prove two lemmas that we need to establish Theorem 1.5. Lemma 3.1. If uλ is a positive solution of (2), then uλ ≥ eλ ; (0, 1), where eλ is the unique solution of:  00   −eλ (t) = λf0 h(t), t ∈ (0, 1), e0λ (1) + c(eλ (1))eλ (1) = 0, (17)   eλ (0) = 0, where f0 = inf [0,∞) f (u) > 0.

2724


Proof. Consider (uλ − eλ )00 (t) = −λh(t)f (u(t)) + λf0 h(t) = λh(t)[f0 − f (u(t))] ≤ 0, which implies uλ − eλ is concave down. Also, (uλ − eλ )(0) = 0. Suppose (uλ − eλ )(1) < 0. Then, (uλ − eλ )0 (1) < 0. This contradicts [u0λ (1) + c(uλ (1))uλ (1)]−[e0λ (1)+c(eλ (1))eλ (1)] = (u0λ −e0λ )(1)+c(uλ (1))uλ (1)−c(eλ (1))eλ (1) = 0, since c is strictly increasing. Thus, uλ ≥ eλ . Remark 2. We see that the boundary value problem (17) has a positive solution, because ψ = 0 is a strict subsolution of (17), and there exists M (λ) > 0 such that Z = M (λ)e is a supersolution of (17). Since ψ < Z, there exists a positive solution u u ∈ (ψ, Z]. Also, if we define f˜ := f0 h(t), then f˜(u) is nondecreasing. By a similar argument as in the proof of Theorem 1.2, we can prove that this solution must be unique. Note that eλ is a subsolution of (2). Lemma 3.2. Let eλ be the solution of (17), and let Mλ := eλ (1). Then there exists 1 K > 0 such that Mλ ≥ Kλ 1+p for λ 1. Proof. If wλ is a solution of  00   −w (t) = λf0 h, t ∈ (0, 1), w0 (1) + c(Mλ )w(1) = 0,   w(0) = 0,

(18)

then we first claim that eλ ≥ wλ . To prove this claim, consider (eλ − wλ )00 = −λf0 h(t) + λf0 h = λf0 [h − h(t)] ≤ 0, which implies eλ − wλ is concave down. Also, (eλ − wλ )(0) = 0. Suppose (eλ − wλ )(1) < 0. Then, (eλ − wλ )0 (1) < 0, which contradicts (e0λ − wλ0 )(1) + c(Mλ )(eλ − wλ )(1) = 0. Thus, eλ ≥ wλ . Now integrating (18) we obtain −wλ0 (t) = λf0 ht + C1 , and integrating again 2 −wλ (t) = λf0 h t2 + C1 t + C2 for some constants C1 and C2 . Since wλ (0) = 0, C2 = 0. Also, from the boundary condition at t = 1, we see that λf0 h 0 = 0. wλ (1) + c(Mλ )wλ (1) = −C1 − λf0 h + c(Mλ ) −C1 − 2 Thus, C1 =

i h c(M ) −λf0 h 1+ 2 λ

. Plugging back in we obtain " # λ) 1 + c(M λf0 h λf0 h 2 wλ (1) = − + λf0 h = . 2 1 + c(Mλ ) 2 + 2c(Mλ ) 1+c(Mλ )

λf0 h Thus, Mλ = eλ (1) ≥ wλ (1) = 2+2c(M . Rearranging, f0 hλ ≤ Mλ [2 + 2c(Mλ )] ≤ λ) p Mλ [2 + 2(AMλ + B)] by (H4). This implies that there exists K such that Mλ ≥ 1 Kλ p+1 for λ 1.

Now, we are ready to prove Theorem 1.5. From Lemma 3.1 , we know that there exists uλ , a minimal positive solution of (2). If vλ is any other positive solution of (2), then uλ ≤ vλ and uλ 6≡ vλ . Define wλ := vλ − uλ . Then, Z 1 wλ =λ G(t, s)h(s)[f (vλ (s)) − f (uλ (s))]ds − [c(vλ (1))vλ (1) − c(uλ (1))uλ (1)]t 0

Z ≤λ

1

G(t, s)h(s)f 0 (θλ (s))wλ (s)ds,

0

for some θλ ∈ [uλ , vλ ] (by the Mean Value Theorem). Let xλ be such that wλ (xλ ) = R1 ||wλ ||∞ . Then, wλ (xλ ) ≤ λ 0 G(xλ , s)h(s)f 0 (θλ (s))wλ (xλ )ds. This implies 1 ≤


λ

R1

G(xλ , s)h(s)f 0 D (1+θλ )γ , and θλ (s)

0

(θλ (s))ds. Now using, G(xλ , s) ≤ s, h(s) ≤ 1 ≥ eλ (s) ≥ Mλ s ≥ Kλ 1+p s we obtain 1

Z

2725

d sα ,

f 0 (θλ (s)) ≤

λs1−α

1 ≤Dd

ds 1 (1 + Kλ 1+p s)γ "Z 1 # Z 1 λ λs1−α 1−α ≤Dd λs ds + 1 1 1+p s)γ 0 λ (Kλ 0

γ

˜ 1 λα−1 + D ˜ 2 λ1− 1+p − D ˜ 2λ =D

γp−(1+p)(1−α) 1+p

,

˜ 1 and D ˜ 2 . But, since α ∈ (0, 1) and γ ∈ p + 1, (p+1)(1−α) , for some constants D p γ

γp−(1+p)(1−α) 1+p

˜ 1 λα−1 + D ˜ 2 λ1− 1+p − D ˜ 2λ D Hence, Theorem 1.5 is proven.

→ 0 as λ → ∞, which is a contradiction.

4. Proof of Theorem 1.6. Here since f (0) ≤ 0, there exists N > 0 such that f (u) ≤ N u for all u > 0. Consider the eigenvalue problem:  00   −φ (t) = λh(t)φ(t), t ∈ (0, 1), φ0 (1) = 0, (19)   φ(0) = 0. Let λ0 (> 0) be the principal eigenvalue and φ0 ∈ C 2 (0, 1) ∩ C 1 [0, 1] with φ0 (t) > 0 for t ∈ (0, 1] be a corresponding eigenfunction of (19). Suppose u is a positive solution. Multiplying (2) by φ0 and integrating by parts we obtain Z 1 0 =c(u(1))u(1)φ0 (1) + φ0 (s)h(s)[u(s)λ0 − λf (u(s))]ds 0

Z ≥u(1)φ0 (1)c(u(1)) +

1

φ0 (s)h(s)u(s)[λ0 − λN ]ds, 0

which is a contradiction if λ
0) be the principal eigenvalue and φ1 ∈ C 2 [0, 1] with φ1 > 0; t ∈ (0, 1) be a corresponding eigenfunction of (20). Then, there exists d1 > 0 such that 0 < φ1 (t) ≤ d1 t(1 − t) for t ∈ (0, 1). Let β ∈ (1, 2 − α), > 0, m > 0, and µ > 0 such that −m > [λ1 βφ21 − β(β − 1)|φ01 |2 ] in (0, ] ∪ [1 − , 1) and φ1 > µ in (, 1 − ). This is possible since φ1 = 0 and |φ01 | = 6 0 at t = 0, 1. Now, we construct a positive subsolution of (2). Define ψ = λk0 φβ1 where −k0
0 and e is as before (see equation (3)). Recall fˆ(x) = maxu∈[0,x] f (u); then fˆ satisfies ˆ lims→∞ fˆ(s) = ∞ and lims→∞ f (s) = 0, and fˆ is nondecreasing. Choose M (λ) ˆ

s

(λ)||e||∞ ) 00 1 such that ||e||1∞ λ ≥ f (M = −M (λ)e00 (t) = M (λ)h(t) ≥ M (λ)||e||∞ . Then, −Z λfˆ(M (λ)||e||∞ )h(t) ≥ λfˆ(M (λ)e(t))h(t) ≥ λh(t)f (Z). Also, Z(0) = M (λ)e(0) = 0, and Z 0 (1) + c(Z(1))Z(1) = M (λ)e0 (1) + c(M (λ)e(1))M (λ)e(1) = c(M (λ)e(1)) M (λ)e(1) ≥ 0. Thus, Z is a supersolution of (2). Choose M (λ) 1 such that ψ ≤ Z. Hence, (2) has a positive solution u ∈ [ψ, Z].

6. Proofs of Theorems 1.8-1.9. 6.1. Proof of Theorem 1.8. Clearly, ψ1 = 0 is a solution to (2). Consider the boundary value problem  00   −φ (t) = λφ(t), t ∈ (0, 1), φ0 (1) + c0 φ(1) = 0, (21)   φ(0) = 0. ˜ (> 0) be the principal eigenvalue and φ˜ ∈ C 2 [0, 1] with φ˜ > 0; t ∈ (0, 1] be a Let λ corresponding eigenfunction of (21) and c0 = c(0). ˜ φ. ˜ Consider the function Define Z1 = φ˜ where > 0. Then, −Z100 = −φ˜00 = λ ˜ ˜ ˜ ˜ ˜ > 0 for H(t, s) := λs−λh(t)f (s). We need to show H(t, φ(t)) = λφ(t)−λh(t)f (φ) ˜ t ∈ (0, 1). We know there exists d1 , d2 > 0 such that d2 t(1 − t) ≤ φ(t) ≤ d1 t(1 − t). Also, recall that h(t) ≤ tdα for all t ∈ (0, 1] and α ∈ (0, 1). Then, for t ∈ (0, 1) and ≈ 0, we have ˜ ˜ φ(t) ˜ − λh(t)f (φ) ˜ ≥ λd ˜ 2 t(1 − t) − λ d (φ(t)) ˜ 1+α H(t, φ(t)) = λ tα


2727

by (H7). Thus, d 1+α 1+α 1+α d1 t (1 − t)1+α tα ˜ 2 − λdα d1+α tα (1 − t)α ] > 0 =t(1 − t)[λd 1

˜ ˜ 2 t(1 − t) − λ H(t, φ(t)) ≥λd

˜ φ(t) ˜ > λh(t)f (φ) ˜ = λh(t)f (Z1 (t)). Hence, Z1 is a for ≈ 0. Thus, −Z100 (t) = λ strict supersolution of (2). Next, consider a C 1 function f˜ satisfying the hypotheses in Theorem 1.7 such that f˜(u) < f (u) for all u ≥ 0. This is possible by our assumptions on f (u). Let ψ2 = ψ(t) be a positive solution for λ 1 as described in Theorem 1.7. Then, −ψ200 (t) = λh(t)f˜(ψ2 ) < λh(t)f (ψ2 ), and hence, ψ2 is a strict subsolution of (2) for λ 1. ˜ e where M ˜ > 0. Then, Finally, we construct a large supersolution; let Z2 = M 00 00 ˜ ˜ ˜ is large −Z2 (t) = −M e (t) = M h(t) ≥ λh(t)f (Z2 (t)) for t ∈ (0, 1) provided M ˆ ˜ ≥ λfˆ(M ˜ ||e||∞ ), which is possible since lims→∞ f (s) = 0. Also, enough so that M s ˜ large enough so that Z2 (t) > ψ2 (t) and Z2 (t) > Z1 (t) for t ∈ (0, 1). choose M Furthermore, choose > 0 small enough so that S = {t ∈ (0, 1)|ψ2 (t) − Z1 (t) > 0} is nonempty. By Lemma 2.2, there exists at least two positive solutions u1 ∈ [ψ2 , Z2 ] and u2 ∈ [0, Z2 ] \ ([0, Z1 ] ∪ [ψ2 , Z2 ]) 6.2. Proof of Theorem 1.9. First, we will prove the existence of a positive solution for λ 1. Consider a C 1 function f˜ satisfying the hypotheses in Theorem 1.7 such that f˜(u) < f (u) for all u ≥ 0. Let ψ2 = ψ(t) be a positive solution for λ 1 as described in Theorem 1.7. Hence, ψ2 is a subsolution of (2) for λ 1 as ˜ e with M ˜ 1 so discussed in the proof of Theorem 1.8. Similarly, choose Z2 = M that Z2 will be a supersolution of (2) such that Z2 (t) ≥ ψ2 (t) for t ∈ (0, 1). Hence, by Lemma 2.1, (2) has at least one positive solution u ∈ (ψ2 , Z2 ] for λ 1. Next to prove the uniqueness result, first note that since lims→∞ f (s) s = 0, given R z ¯ ¯ where A = 1 h(s)ds. Let u be λ there exists K(λ) such that f (z) ≤ 2λ(A+1) +K 0 any solution of (2), and let t0 ∈ (0, 1) be such that ||u||∞ = u(t i 0) = h 0 ). Thus, u(t R1 R1 ¯ ds < λ G(t0 , s)h(s)f (u(s))ds − c(u(1))u(1)t0 ≤ λ G(t0 , s)h(s) u(t0 ) + K 0 1 u(t 0) 2

0

2λ(A+1)

¯ ¯ := M ¯ (λ). Let e be as before (see equation + λKA. Hence, ||u||∞ ≤ 2λKA ˆ ¯ (3)), and let J > 0 such that J > λf (M (λ)) and J ≥ λfˆ(J||e||∞ ). Then for any ¯ (λ)) > 0. nonnegative solution v, −(Je − v)00 = Jh(t) − λh(t)f (v) ≥ h(t)(J − λfˆ(M Also, (Je − v)(0) = 0, and (Je − v)0 (1) + c0 (Je − v)(1) = Jc0 e(1) − [v 0 (1) + c0 v(1)] ≥ −[v 0 (1) + c(v(1))v(1)] = 0. Hence, by the Maximum Principle, Je − v ≥ 0. Furthermore, −(Je)00 = Jh(t) ≥ λfˆ(J||e||∞ )h(t) ≥ λf (Je(t))h(t), and (Je)(0) = 0, and (Je)0 (1)+c((Je)(1))(Je)(1) ≥ (Je)0 (1) = 0. Thus, Je is a supersolution. Hence, there exists a maximal positive solution of (2). Suppose v is the maximal solution of (2) and u is any other positive solution R1 00 where u 6≡ v. Then, u < v on (0, 1], and it follows thath 0 [vu00 − uv i ]dx = R1 R1 u v [v(−λh(x)f (u)) − u(−λh(x)f (v))]dx = λ 0 h(x)f (u)f (v) f (u) − f (v) dx ≤ 0, 0 u since f (u) is nondecreasing. (Note that since (H3) is also satisfied, f (s) > 0 for s ∈ (0, ∞).) R1 On the other hand, 0 [vu00 −uv 00 ]dx = v(1)u0 (1)−u(1)v 0 (1) = −v(1)c(u(1))u(1)+ u(1)c(v(1))v(1) = u(1)v(1)[c(v(1)) − c(u(1))] > 0, (since c is strictly increasing)

2728


which is a contradiction. Hence, u = v, so (2) has a unique positive solution for λ 1. 7. Examples. Here, we will discuss some examples that satisfy our theorems. First, we provide examples of K(r), h(t), and c(u) that satisfy the hypotheses 7 for all of our theorems. For instance, choose N = 3. Then let K(r) = r− 2 , which fulfills the conditions for K. Using the formula for h and taking the case r0 = 1, 1 we find that the corresponding function is h(t) = t− 2 , which satisfies the necessary 1 assumptions on h. Furthermore, choose c(u) = u 4 + 1, which fulfills the hypotheses on c. τu For Theorems 1.1 - 1.5, consider the example f (u) = e τ +u for τ > 0, which is a model that arises in combustion theory and has been recently studied in [13] (see Figure 6). Clearly, f satisfies the hypotheses of Theorems 1.1 and 1.4. Also, if 0 < fHuL

u

τu

Figure 6. Graph of e τ +u u τ < 4, then f (u) is nondecreasing, and hence for this range of τ , f satisfies Theorem b 1.2. Furthermore, for τ 1, we choose a = 1 and b = τ . Then, limτ →∞ f (b) = τ a 1 limτ →∞ τ2 = 0 (by L’Hopital’s Rule), and limτ →∞ f (a) = limτ →∞ τ +1 = 1e . τ e

τu

e

2

a τ 0 τ +u Hence, for τ 1, ( hf16b (b) , f (a)||e||∞ ) is nonempty. Also, f (u) = e (τ +u)2 > 0 for u ≥ 0, so f satisfies the hypotheses of Theorem 1.3 when τ 1. Finally, D 1 1 since f 0 (u) ≤ (1+u) 2 for some D > 0, choosing α = 2 , p = 4 , and γ = 2 ∈

(1 + p, (1+p)(1−α) ) = ( 54 , 25 ), we see that the hypotheses in Theorem 1.5 are satisfied p u+1 ˜>1 for any τ > 0. Similarly, it is easy to verify that the function f (u) = u+˜ a for a will satisfy the hypotheses for Theorems 1.1, 1.2, 1.4, and 1.5. For Theorems 1.6 and 1.7, the function f (u) = (u + 1)η − σ for η ∈ (0, 1) and 2.5 σ > 1 will fulfill the necessary assumptions. Furthermore, the function f (u) = uu2 +1 will satisfy the hypotheses for Theorem 1.8. Finally, the function f (u) = (u+1)η −1 for η ∈ (0, 1) will fulfill the assumptions for Theorem 1.9. 8. Appendix. 8.1. Consider the p radial positive solutions u of (2): Let r = |x| = x21 + x22 + ... + x2N and v(r) = u(x). Then: ∂r 2xi xi 1. = p 2 = 2 2 ∂xi r 2 x1 + x2 + ... + xN ∂u ∂r ∂u xi 2. = v 0 (r) ⇒ = v 0 (r) ∂xi ∂xi ∂xi r


PN

3. ∆u = div(∇u) =

i=1

∂ ∂xi

∂u ∂xi

=

2729

N N X ∂ 0 xi X 00 xi xi v (r) = (v (r) + ∂xi r r r i=1 i=1

N N N v 00 (r) X 2 X v 0 (r) v 0 (r) v 0 (r) X 2 xi ) + v 0 (r)xi (− 3 )) = x + (− x = v 00 (r) + + r r r2 i=1 i i=1 r r3 i=1 i N −1 0 v 0 (r) v 0 (r) − = v 00 (r) + v (r). N r r r Thus, (2) implies

N −1 0 v (r) =λK(r)f (v(r)) r −(v 00 (r)rN −1 + v 0 (r)(N − 1)rN −2 ) =λrN −1 K(r)f (v(r)) −v 00 (r) −

−(rN −1 v 0 (r))0 =λrN −1 K(r)f (v(r)); r ∈ (r0 , ∞). ∂u 1 ∂u ∂u ∂η = η · ∇u = r0 h−x1 , −x2 , ..., −xN i · h ∂x1 , ..., ∂xN i = 0 0 P N v (r0 ) v (r0 ) v (r0 ) 1 2 0 i=1 xi = −v (r0 ), and r0 h−x1 , −x2 , ..., −xN i · h r0 x1 , ..., r0 xN i = − r02 hence, the boundary condition at |x| = r0 reduces to −v 0 (r0 ) + c(v(r0 ))v(r0 ) = 0.

Also, on |x| = r0 , we have 0

Hence, (2) reduces to   −(rN −1 v 0 (r))0 = λrN −1 K(r)f (v(r)), −v 0 (r0 ) + c(v(r0 ))v(r0 ) = 0,  v(r) → 0, Now let t = 1.

∂t ∂r 0

r r0

2−N

= (2 − N )

r r0

∂t 2. v (r) = z 0 (t) ∂r =

3. r = r0 t

1 2−N

r ∈ (r0 , ∞), (22) r → ∞,

and z(t) = v(r). Then:

1−N

1 r0 z 0 (t) r2−N 2−N 0

r

2−N r02−N 1−N

=

r1−N

.

∂ ∂ 1−N Furthermore, −(rN −1 v 0 (r))0 = − ∂r (rN −1 v 0 (r)) = − ∂r (rN −1 z 0 (t) r2−N ) = 2−N r 2−N r02−N

∂t (−z 00 (t)) ∂r =

0

2−N r02−N

00

−z (t) =λ =λ

1−N (−z 00 (t)) r2−N . Hence from (22), we obtain 2−N r 0

r02−N 2−N r04−2N (2 − N )

!

r02−N 2−N

!

1

1 N −1 2−N r K r t f (z(t)) 0 1−N

r 1 2(N −1) 2−N r K r t f (z(t)) 0 2

−1) 1 r04−2N 2(N −1) 2(N r0 t 2−N K r0 t 2−N f (z(t)) 2 (2 − N ) 2(N −1) 1 r02 =λ t− N −2 K r0 t 2−N f (z(t)). 2 (2 − N )

=λ

2(N −1) 1 r02 − N −2 K r0 t 2−N . Also, (2−N )2 t (N −2) 0 r0 z (1), and from (22) we obtain

Thus, −z 00 (t) = λh(t)f (z); (0, 1) where h(t) = 1−N = −z 0 (1) 2−N −v 0 (r0 ) = −z 0 (1) r2−N 2−N r0 r0 = 0

(N −2) 0 r0 z (1)

+ c(z(1))z(1) = 0. Finally, as r → ∞, t → 0 (since N > 2), and hence,

2730


we obtain z(0) = 0. Combining the above calculations, we have:  00 t ∈ (0, 1),  −z (t) = λh(t)f (z(t)), (N −2) 0 z (1) + c(z(1))z(1) = 0,  r0 z(0) = 0, 2(N −1) 1 r02 − N −2 2−N where h(t) = (2−N K r t . t 2 0 )

(23)

8.2. Suppose u ∈ C 2 (0, 1) ∩ C 1 [0, 1] is a solution of (2). Integrating (2) from t to 1, we obtain: Z 1 Z 1 λh(s)f (u(s))ds = λh(s)f (u(s))ds − c(u(1))u(1). u0 (t) = u0 (1) + t

t

Integrating again from 0 to x and using u(0) = 0, we obtain: Z xZ 1 u(x) = λh(s)f (u(s))dsdt − c(u(1))u(1)x 0

Z

t x

s

Z

=

Z

1

Z

0

Z =

0

x

x

Z

0

1

λh(s)f (u(s))xds − c(u(1))u(1)x.

λh(s)f (u(s))sds + 0

x

λh(s)f (u(s))dtds − c(u(1))u(1)x

λh(s)f (u(s))dtds +

x

Thus, Z

1

G(x, s)h(s)f (u(s))ds − c(u(1))u(1)x

u(x) = λ 0

where

s; 0 ≤ s ≤ x (24) x; x ≤ s ≤ 1. R1 Conversely, if u ∈ X satisfies u(x) = λ 0 G(x, s)h(s)f (u(s))ds − c(u(1))u(1)x where G(x, s) is given by (24), it is straightforward to show that u ∈ C 2 (0, 1) ∩ C 1 [0, 1] and satisfies (2). G(x, s) =

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