Electronic Journal of Qualitative Theory of Differential Equations 2006, No. 12 1, 1-15; http://www.math.u-szeged.hu/ejqtde/

LOCALIZED SOLUTIONS OF ELLIPTIC

EQUATIONS: LOITERING AT THE HILLTOP

Joseph A. Iaia

Abstract. We find an infinite number of smooth, localized, radial solutions of Δ_{p}u + f(u) = 0 in ℝ^{N} - one with each prescribed number of zeros - where Δ_{p}u is the p-Laplacian of the function u.

1. Introduction

In this paper we will prove the existence of smooth, radial solutions with any prescribed number of zeros to:

Δ_{p}u + f(u) = 0 in ℝ^{N},  (1.1)
u(x) → 0 as |x| → ∞,  (1.2)

where Δ_{p}u = ∇ · (|∇u|^{p−2}∇u) (p > 1) is the p-Laplacian of the function u (note that p = 2 is the usual Laplacian operator), f is the nonlinearity described below, and N ≥ 2,.

Solutions of (1.1)-(1.2) arise as critical points of the functional J : S → ℝ defined by:

J(u) = ∫_{ℝ^{N}} \frac{1}{p}|∇u|^{p} − F (u) dx

where F (u) = ∫_{0}^{u} f(t) dt and S = {u ∈ W^{1,p}(ℝ^{N}) | F (u) ∈ L^{1}(ℝ^{N})}. Setting r = |x| and assuming that u is a radial function so that u(x) = u(|x|) = u(r) then:

Δ_{p}u = |u′|^{p−2}[(p − 1)u′′ + \frac{N − 1}{r}u′] = \frac{1}{r^{N−1}} (r^{N−1}|u′|^{p−2}u′)′

where ′ denotes differentiation with respect to the variable r.

We consider therefore looking for solutions of:

|u′|^{p−2}[(p − 1)u′′ + \frac{N − 1}{r}u′] + f(u) = \frac{1}{r^{N−1}} (r^{N−1}|u′|^{p−2}u′)′ + f(u) = 0  (1.3)
lim_{r→0^{+}} u′(r) = 0,  (1.4)
lim_{r→∞} u(r) = 0.  (1.5)

Remark: The case p = 2 was examined in [2]. There the authors proved the existence of an infinite number of solutions of (1.3)-(1.5) - one with each precribed number of zeros - for nonlinearities f similar to the ones examined in this paper. In this paper we have weaker assumptions than those in [2] and we also have only

Typeset by AMS-TEX

EJQTDE, 2006 No. 12, p. 1 that p > 1. Existence of ground states of (1.3)-(1.5) for quite general nonlinearities f was established in [1]. Our extra assumptions on f allow us to prove the existence of an infinite number of solutions of (1.3)-(1.5).

1991 Mathematics Subject Classification. Primary 34B15: Secondary 35J65.

Key words and phrases. radial, p-Laplacian.

For p ≠ 2, equation (1.3) is degenerate at points where u′ = 0 and we will see later that in some instances this prevents u from being twice differentiable at some points. We see however that by multiplying (1.3) by r^{N−1}, integrating on (0, r), and using (1.4) we obtain:

−r^{N−1}|u′(r)|^{p−2}u′(r) = ∫_{0}^{r} t^{N−1}f(u(t)) dt.  (1.6)

Therefore, instead of seeking solutions of (1.3)-(1.5) in C^{2}[0, ∞) we will attempt to find u ∈ C^{1}[0, ∞) satisfying (1.4)-(1.6).

The type of nonlinearity we are interested in is one for which F (u) ≡ ∫_{0}^{u} f(t) dt has the shape of a “hilltop.” We require that f : [−δ, δ] → ℝ and:

f is odd, there exists K > 0 such that |f(x) − f(y)| ≤ K|x − y| for all x, y ∈ [−δ, δ] and  (1.7)
there exists β, δ such that 0 < β < δ with f < 0 on (0, β), f > 0 on (β, δ), and f(δ) = 0.  (1.8)

We also require:

there exists γ with β < γ < δ such that F < 0 on (0, γ) and F > 0 on (γ, δ).  (1.9)

Finally we assume:

∫_{0} \frac{1}{\sqrt[p]{|F (t)|}} dt = ∞ if p > 2  (1.10)

and:

∫ ^{δ} \frac{1}{\sqrt[p]{F (δ) − F (t)}} dt = ∞ if p > 2.  (1.11)

Main Theorem. Let f be a function satisfying (1.7)-(1.11). Then there exist an infinite number of solutions of (1.4)-(1.6), at least one with each prescribed number of zeros.

Remark: Assumption (1.8) can be weakened to allow f to have a finite number of zeros, 0 < β_{1} < β_{2} < ⋯ < β_{n} < δ where f < 0 on (0, β_{1}), f > 0 on (β_{n−1}, β_{n}) and we still require assumption (1.9). A key fact that we would then need to prove is that the solution of a certain initial value problem is unique. Sufficient conditions to assure this are (1.10)-(1.11) and the following:

∫ ^{β_{l+1}} \frac{1}{\sqrt[p]{F (β_{l+1}) − F (t)}} dt = ∞ if p > 2 and if f > 0 on (β_{l}, β_{l+1})

and

∫_{β_{l}} \frac{1}{\sqrt[p]{F (β_{l}) − F (t)}} dt = ∞ if p > 2 and if f < 0 on (β_{l}, β_{l+1}).

Remark: Let 0 < β < δ and suppose q_{i} ≥ 1 for i = 1, 2, 3. If p > 2 then also suppose q_{1} ≥ p − 1 and q_{3} ≥ p − 1. Let f be an odd function such that f(u) = u^{q_{1}}|u − β|^{q_{2}−1}(u − β)(δ − u)^{q_{3}} for 0 < u < δ and suppose F (δ) > 0. Then (1.7)-(1.11) are satisfied and the Main Theorem applies to all such functions f.

Remark: If 1 < p ≤ 2 then it follows from the fact that f is locally Lipschitz that (1.10) and (1.11) are satisfied. Since f is locally Lipschitz at u = 0, it follows that |F (u)| ≤ Cu^{2} in some neighborhood of u = 0 for some C > 0. Then since 1 < p ≤ 2:

∫_{0} \frac{1}{\sqrt[p]{|F (t)|}} dt ≥ \frac{1}{C^{\frac{1}{p}}} ∫_{0} \frac{1}{t^{\frac{2}{p}}} = ∞.

A similar argument shows that (1.11) also holds for 1 < p ≤ 2.

EJQTDE, 2006 No. 12, p. 2

2. Existence, Uniqueness, and Continuity

We denote C(S) = {f : S → ℝ | f is continuous on S.} Let f be locally Lipschitz and let d ∈ ℝ with |d| ≤ δ. Denote u(r, d) as a solution of the initial value problem:

−r^{N−1}|u′(r)|^{p−2}u′(r) = ∫_{0}^{r} t^{N−1}f(u(t)) dt.  (2.1)
u(0) = d.  (2.2)

We will show using the contraction mapping principle that a solution of (2.1)-(2.2) exists. For p > 1 we denote Φ_{p}(x) = |x|^{p−2}x. Note that Φ_{p} is continuous for p > 1 and Φ_{p}^{−1} = Φ_{p′} where \frac{1}{p} + \frac{1}{p′} = 1. For future reference we note that Φ′_{p}(x) = (p − 1)|x|^{p−2} and |Φ_{p}(x)| = |x|^{p−1}. We rewrite (2.1) as:

−u′ = \frac{1}{r^{\frac{N−1}{p−1}}} Φ_{p′}[∫_{0}^{r} t^{N−1}f(u(t)) dt].  (2.3)

Integrating on (0, r) and using (2.2) gives:

u = d − ∫_{0}^{r} \frac{1}{t^{\frac{N−1}{p−1}}} Φ_{p′}[∫_{0}^{t} s^{N−1}f(u(s)) ds] dt.  (2.4)

Thus we see that solutions of (2.1)-(2.2) are fixed points of the mapping:

T u = d − ∫_{0}^{r} \frac{1}{t^{\frac{N−1}{p−1}}} Φ_{p′}[∫_{0}^{t} s^{N−1}f(u(s)) ds] dt.  (2.5)

Lemma 2.1. Let f be locally Lipschitz and let d be a real number such that |d| ≤ δ. Then there exists a solution u ∈ C^{1}[0, ϵ) of (2.1)-(2.2) for some ϵ > 0. In addition, u′(0) = 0. Proof. First, if f(d) = 0 then u ≡ d is a solution of (2.1)-(2.2) and u′(0) = 0. So we now assume that f(d) ≠ 0. Denote B_{R}^{ϵ}(d) = {u ∈ C[0, ϵ) such that ∥u − d∥ < R} where ∥ · ∥ is the supremum norm. We will now show that if ϵ > 0 and R > 0 are small enough then T : B_{R}^{ϵ}(d) → B_{R}^{ϵ}(d) and that T is a contraction mapping. Since f is bounded on [\frac{|d|}{2}, \frac{|d|+δ}{2} ], say by M, it follows from (2.5) that:

|T u − d| ≤ ∫_{0}^{r} \frac{1}{t^{\frac{N−1}{p−1}}} (\frac{Mt^{N}}{N})^{\frac{1}{p−1}} = (\frac{p − 1}{p})(\frac{M}{N})^{\frac{1}{p−1}} r^{\frac{p}{p−1}} ≤ (\frac{p − 1}{p})(\frac{M}{N})^{\frac{1}{p−1}} ϵ^{\frac{p}{p−1}} .

Therefore we see that ∥T u − d∥ < R if ϵ is chosen small enough and hence T : B_{R}^{ϵ}(d) → B_{R}^{ϵ}(d) for ϵ small enough. Next by the mean value theorem we see that for some h with 0 < h < 1 we have:

|Φ_{p′}[∫_{0}^{t} s^{N−1}f(u(s)) ds] − Φ_{p′}[∫_{0}^{t} s^{N−1}f(v(s)) ds]| =
\frac{1}{p − 1}| ∫_{0}^{t} s^{N−1}[hf(u) + (1 − h)f(v)] ds|^{\frac{2−p}{p−1}} | ∫_{0}^{t} s^{N−1}[f(u) − f(v)] ds|.  (2.6)

Case 1: 1 < p ≤ 2 EJQTDE, 2006 No. 12, p. 3 Using again that f is bounded on [d − 1, d + 1] by M and that the local Lipschitz constant is K (i.e. for u, v ∈ B_{1}^{ϵ}(d) we have |f(u) − f(v)| ≤ K|u − v|) we obtain by (2.5)-(2.6):

∥T u − T v∥ ≤ \frac{K}{p − 1}∥u − v∥ ∫_{0}^{r} \frac{1}{t^{\frac{N−1}{p−1}}} M^{\frac{2−p}{p−1}} (\frac{t^{N}}{N})^{\frac{2−p}{p−1}} \frac{t^{N}}{N}

= C_{1}∥u − v∥ ∫_{0}^{r} t^{\frac{1}{p−1}} dt ≤ C_{2}ϵ^{\frac{p}{p−1}} ∥u − v∥

where C_{1}, C_{2} are constants depending only on p, N, K, and M.

Case 2: p > 2

Since f(d) ≠ 0 and f is continuous we may choose R small enough so that:

L ≡ min_{[d−R,d+R]} |f| > 0.

Therefore,

| ∫_{0}^{t} s^{N−1}[hf(u) + (1 − h)f(v)] ds| ≥ \frac{Lt^{N}}{N} .  (2.7)

Thus, by (2.5)-(2.7) we have

∥T u − T v∥ ≤ \frac{K}{p − 1}(\frac{L}{N})^{\frac{2−p}{p−1}} ∥u − v∥ ∫_{0}^{r} \frac{1}{t^{\frac{N−1}{p−1}}} t^{\frac{N(2−p)}{p−1}} \frac{t^{N}}{N} dt

= \frac{K}{(p − 1)} \frac{1}{N^{\frac{1}{p−1}} L^{\frac{p−2}{p−1}}} ∥u − v∥ ∫_{0}^{r} t^{\frac{1}{p−1}} dt ≤ C_{3}ϵ^{\frac{p}{p−1}} ∥u − v∥

where C_{3} depends only on p, N, K, and M.

Therefore in both cases we see that T is a contraction for R and ϵ small enough. Thus by the contraction mapping principle, there is a unique u ∈ C[0, ϵ_{1}) such that T u = u. That is, there is a continuous function u such that u satisfies (2.4) on [0, ϵ_{1}) for some ϵ_{1} > 0. In addition, since f(d) ≠ 0 we see that the right hand side of (2.4) is continuously differentiable on (0, ϵ) for some ϵ with 0 < ϵ ≤ ϵ_{1} and therefore u ∈ C^{1}(0, ϵ). Also, subtracting d from (2.4), dividing by r, and taking the limit as r → 0^{+} gives u′(0) = 0. Finally, dividing (2.1) by r^{N−1} and taking the limit as r → 0^{+} we see that lim_{r→0^{+}} u′(r) = 0. Therefore, u ∈ C^{1}[0, ϵ). □

Note we see from (2.3) that u ∈ C^{2} at all points where u′ ≠ 0.

If u′(r_{0}) = 0 then using (2.1) we obtain:

−|u′(r)|^{p−2}u′(r) = \frac{1}{r^{N−1}} ∫_{r_{0}}^{r} t^{N−1}f(u(t)) dt.

It then follows that:

lim_{r→r_{0}} \frac{|u′(r)|^{p−2}u′(r)}{r − r_{0}} = \{ \begin{matrix}−\frac{f(u(r_{0}))}{N} if r_{0} = 0 \\ −f(u(r_{0})) if r_{0} > 0. \end{matrix}  (2.8)

Remark: If 1 < p ≤ 2 then we see from (2.8) that u′′(r_{0}) exists and rewriting (1.3) as:

(p − 1)u′′ + \frac{N − 1}{r}u′ + |u′|^{2−p}f(u) = 0,

we see that u ∈ C^{2}[0, ϵ).

Remark: If p > 2 then u might not be twice differentiable at points where u′ = 0. In fact if u′(r_{0}) = 0 and f(u(r_{0})) ≠ 0 then by (2.8) we see that lim_{r→r_{0}} |\frac{u′(r)}{r−r_{0}}| = ∞ and so u is not twice differentiable at r_{0}.

Lemma 2.2. Let f satisfy (1.7)-(1.9). If u is a solution of the initial value problem (2.1)-(2.2) with |d| ≤ δ on some interval (0, R) with R ≤ ∞, then:

F (u) ≤ F (d) on (0, R)  (2.9)

and

\frac{p − 1}{p}|u′|^{p} ≤ F (d) + |F (β)| ≤ F (δ) + |F (β)| on (0, R).  (2.10)

Proof.

We define the “energy” of a solution as:

E = \frac{p − 1}{p}|u′|^{p} + F (u).  (2.11)

Differentiating E and using (2.1) gives:

E′ = −\frac{N − 1}{r}|u′|^{p} ≤ 0.  (2.12)

Integrating this on (0, r) and using (1.8) gives:

\frac{p − 1}{p}|u′|^{p} + F (u) = E ≤ E(0) = F (d) ≤ F (δ) for r > 0.  (2.13)

Inequalities (2.9)-(2.10) follow from (1.8)-(1.9) and (2.13).

Now by (1.9) we know that F is negative on (0, γ) and by (1.8) we know that F is increasing on (β, δ). Therefore if |d| < δ then F (d) < F (δ). On the other hand if |u(r_{0})| = δ for some r_{0} > 0 then by (2.9) F (δ) ≤ F (d) - a contradiction. Hence if |d| < δ then |u| < δ. □

Lemma 2.3. Let f satisfy (1.7)-(1.9). Let d be a real number such that |d| ≤ δ. Then a solution of (2.1)- (2.2) exists on [0, ∞).

Proof.

If |d| = δ then u ≡ d is a solution on [0, ∞) and so we now suppose that |d| < δ.

Let [0, R) be the maximal interval of existence for a solution of (2.1)-(2.2). From lemma 2.1 we know that R > 0. Now suppose that R < ∞. By lemma 2.2, it follows that u and u′ are uniformly bounded by M = δ + F (δ) + |F (β)| on [0, R). Therefore by the mean value theorem |u(x) − u(y)| ≤ M|x − y| for all x, y ∈ [0, R). Thus, there exists b_{1} ∈ ℝ such that:

lim_{r→R^{−}} u(r) = b_{1}.

By (2.3) there exists b_{2} ∈ ℝ such that:

lim_{r→R^{−}} u′(r) = b_{2}.

If b_{2} ≠ 0 we can apply the standard existence theorem for ordinary differential equations and extend our solution of (2.1)-(2.2) to [0, R + ϵ) for some ϵ > 0 contradicting the maximality of [0, R).

If b_{2} = 0 and f(b_{1}) ≠ 0 we can again apply the contraction mapping principle as we did in lemma 2.1 to extend our solution of (2.1)-(2.2) to [0, R + ϵ) for some ϵ > 0 contradicting the maximality of [0, R).

Finally, if b_{2} = 0 and f(b_{1}) = 0, we can extend our solution by defining u(r) ≡ b_{1} for r > R contradicting the maximality of [0, R).

Thus in each of these cases we see that R cannot be finite and so a solution of (2.1)-(2.2) exists on [0, ∞). □

Lemma 2.4. Let f satisfy (1.7)-(1.10). Let d be a real number such that |d| < δ. Then there is a unique solution of (2.1)-(2.2) on [0, ∞).

Proof.

Case 1: d = ±β

In this case we have E(0) = F (β) (recall that F is even) and since E′ ≤ 0 (by (2.12)) we have E(r) ≤ E(0) = F (β) for r ≥ 0. On the other hand, F has a minimum at u = ±β and so we see that E(r) = \frac{p−1}{p}|u′|^{p} + F (u) ≥ F (β). Thus E ≡ F (β). Thus, −\frac{N−1}{r}|u′|^{p} = E′ ≡ 0 and hence u(r) ≡ ±β.

Case 2: d = 0.

Here we have E(0) = 0 and since E′ ≤ 0 we have E(r) ≤ 0 for r ≥ 0.

Let r_{1} = sup{ r ≥ 0 | E(r) = 0}. If r_{1} = ∞ then u(r) ≡ 0.

So suppose r_{1} < ∞. If r_{1} = 0 then we have u(r_{1}) = 0 and u′(r_{1}) = 0.

If r_{1} > 0 then since E′ ≤ 0 we have E(r) ≡ 0 on [0, r_{1}] hence −\frac{N−1}{r}|u′|^{p} = E′ ≡ 0 and so u ≡ 0 on [0, r_{1}]. Therefore we also have u(r_{1}) = 0 and u′(r_{1}) = 0.

Now using (2.1) we obtain:

−r^{N−1}|u′|^{p−2}u′ = ∫_{r_{1}}^{r} t^{N−1}f(u) dt.  (2.15)

Since:

\frac{p − 1}{p}|u′|^{p} + F (u) = E(r) < E(0) = 0 for r > r_{1},  (2.16)

it follows that |u(r)| > 0 for r > r_{1}. Combining this with the fact that u(r_{1}) = 0, we see that there exists an ϵ > 0 such that 0 < |u(r)| < β for r_{1} < r < r_{1} + ϵ. By (1.8) it follows that |f(u)| > 0 for r_{1} < r < r_{1} + ϵ. Therefore, by (2.15) we see that |u′| > 0 for r_{1} < r < r_{1} + ϵ. Using this fact and rewriting (2.16) we see that:

\frac{|u′|}{\sqrt[p]{|F (u)|}} < (\frac{p}{p − 1})^{\frac{1}{p}} for r_{1} < r < r_{1} + ϵ.  (2.17)

Integrating (2.17) on (r_{1}, r_{1} + ϵ), using (1.10), and that F is even gives:

∞ = ∫_{0}^{|u(r_{1}+ϵ)|} \frac{1}{\sqrt[p]{|F (t)|}} dt = ∫_{r_{1}}^{r_{1}+ϵ} \frac{|u′|}{\sqrt[p]{|F (u)|}} ≤ (\frac{p}{p − 1})^{\frac{1}{p}} ϵ,

a contradiction. Thus we see that r_{1} = ∞ and hence u ≡ 0.

Case 3: f(d) ≠ 0.

We saw that the mapping T defined in lemma 2.1 is a contraction mapping. Therefore, T has a unique fixed point so that if u_{1} and u_{2} are solutions of (2.1)-(2.2) then there exists an ϵ > 0 such that u_{1}(r) ≡ u_{2}(r) on [0, ϵ). Let [0, R) be the maximal half-open interval such that u_{1}(r) ≡ u_{2}(r) on [0, R). By continuity, u_{1}(r) ≡ u_{2}(r) on [0, R] and u′_{1}(r) ≡ u′_{2}(r) on [0, R].

As in the proof of lemma 2.3, if u′_{1}(R) ≠ 0 then it follows from the standard existence-uniqueness theorem of ordinary differential equations that u_{1}(r) ≡ u_{2}(r) on [0, R + ϵ) for some ϵ > 0 contradicting the maximality of [0, R).

If u′_{1}(R) = 0 and f(u_{1}(R)) ≠ 0 then we can again apply the contraction mapping principle as in lemma 2.1 and show that u_{1}(r) ≡ u_{2}(r) on [0, R + ϵ) for some ϵ > 0 contradicting the maximality of [0, R).

If u′_{1}(R) = 0 and u_{1}(R) = β then as in Case 1 above we can show that u_{1}(r) ≡ β for r > R and u_{2}(r) ≡ β for r > R. This contradicts the definition of R. A similar argument applies if u′_{1}(R) = 0 and u_{1}(R) = −β.

EJQTDE, 2006 No. 12, p. 6 Finally, if u′_{1}(R) = 0 and u_{1}(R) = 0, then as in Case 2 above we can show that u_{1}(r) ≡ 0 for r > R and u_{2}(r) ≡ 0 for r > R. This contradicts the definition of R.

Thus we see that in all cases we have R = ∞. This completes the proof. □

Remark: Without assumptions (1.10) and (1.11), solutions of the initial value problem (2.1)-(2.2) are not necessarily unique! For example, let f(u) = −|u|^{q−1}u where 1 ≤ q < p − 1. In addition to u ≡ 0,

u = C(p, q, N)r^{\frac{p}{p−1−q}}

where C(p, q, N) = [\frac{(p−1−q)^{p}}{p^{p−1}[pq+N(p−1−q)]}]^{\frac{1}{p−1−q}} is also a solution of (2.1)-(2.2) with u(0) = 0 and u′(0) = 0. Note

however that ∫_{0} \frac{1}{\sqrt[p]{|F (t)|}} dt = ∫_{0} \frac{(q+1)^{\frac{1}{p}}}{t^{\frac{q+1}{p}}} dt < ∞ since 1 ≤ q < p − 1. Similarly, if f(u) = −|δ − u|^{q−1}(δ − u)

and 1 ≤ q < p − 1 then u ≡ δ and

u = δ − C(p, q, N)r^{\frac{p}{p−1−q}}

(with the same C(p, q, N) as earlier ) are both solutions of (2.1)-(2.2) but (1.11) is not satisfied.

Lemma 2.5. Let u be a solution of (2.1)-(2.2) with γ < d < δ and suppose there exists an r_{1} > 0 such that u(r_{1}) = 0. If (1.10) holds then u′(r_{1}) ≠ 0.

Proof.

This proof is from [1].

Suppose by way of contradiction that u(r_{1}) = 0 and u′(r_{1}) = 0. It follows that E(r_{1}) = 0. (In fact, it follows from lemma 2.4 that u ≡ 0 on [r_{1}, ∞)). Now let r_{0} = inf{r ≤ r_{1} | E(r) = 0}. Since E is continuous, decreasing, and E(0) = F (d) > 0 we see that r_{0} > 0 and that E(r) > 0 for 0 ≤ r < r_{0}. If r_{0} < r_{1} then E(r) ≡ 0 on (r_{0}, r_{1}) and thus −\frac{N−1}{r}|u′|^{p} = E′(r) ≡ 0 on (r_{0}, r_{1}). Therefore u ≡ 0 on (r_{0}, r_{1}) and thus u(r_{0}) = u′(r_{0}) = 0.

Integrating (2.12) on (r, r_{0}) and using that E(r_{0}) = 0 gives:

\frac{p − 1}{p}|u′|^{p} + F (u) = ∫_{r}^{r_{0}} \frac{N − 1}{r}|u′|^{p} dt.  (2.18)

Letting w = ∫_{r}^{r_{0}} \frac{N−1}{r}|u′|^{p} dt, we see that w′ = −\frac{N−1}{r}|u′|^{p}. Thus (2.18) becomes:

w′ + \frac{α}{r}w = \frac{α}{r}F (u) where α = \frac{p(N − 1)}{p − 1}.  (2.19)

By (1.9) it follows that there is an ϵ with 0 < ϵ < \frac{1}{2}r_{0} such that F (u(r)) ≤ 0 on (r_{0} − ϵ, r_{0}). and so solving the first order linear equation (2.19) gives:

w = \frac{α}{r^{α}} ∫_{r}^{r_{0}} t^{α−1}|F (u)| dt for r_{0} − ϵ < r < r_{0}.

Rewriting (2.18) we obtain:

|u′|^{p} = \frac{p}{p − 1}[|F (u)| + \frac{α}{r^{α}} ∫_{r}^{r_{0}} t^{α−1}|F (u(t))| dt] for r_{0} − ϵ < r < r_{0}.  (2.20)

In addition, since E(r) > 0 for r < r_{0}, we see that:

|u′| > (\frac{p}{p − 1})^{\frac{1}{p}} \sqrt[p]{|F (u)|} ≥ 0 for r_{0} − ϵ < r < r_{0}.

EJQTDE, 2006 No. 12, p. 7 Thus u is monotone on (r_{0} − ϵ, r_{0}). Since F′ = f < 0 on (0, β) (by (1.8)) we see that:

|F (u(t))| < |F (u(r))| for r_{0} − ϵ < r < t < r_{0}.  (2.21)

Substituting (2.21) into (2.20) gives:

|u′|^{p} ≤ (\frac{p}{p − 1})\frac{r_{0}^{α}}{r^{α}} |F (u)| ≤ (\frac{p}{p − 1})(\frac{r_{0}}{r_{0} − ϵ})^{α}|F (u)| ≤ 2^{α}(\frac{p}{p − 1})|F (u)| for r_{0} − ϵ < r < r_{0}.

Finally, dividing by |F (u)|, taking roots, integrating on (r, r_{0}), and using (1.10) we obtain:

∞ = ∫_{0}^{|u(r)|} \frac{1}{\sqrt[p]{|F (t)|}} dt ≤ 2^{\frac{N−1}{p−1}} (\frac{p}{p − 1})^{\frac{1}{p}} (r_{0} − r)

a contradiction. Thus u′(r_{1}) ≠ 0 and this completes the proof. □ Lemma 2.6. Let u be a solution of (2.1)-(2.2) where γ < d < δ. Then u′ < 0 on a maximal nonempty open interval (0, M_{d,1}), where either:

(a) M_{d,1} = ∞, lim_{r→∞} u′(r) = 0, lim_{r→∞} u(r) = L where |L| < d and f(L) = 0, or

(b) M_{d,1} is finite, u′(M_{d,1}) = 0, and f(u(M_{d,1})) ≤ 0. In either case, it follows that there exists a unique (finite) number τ_{d} ∈ (0, M_{d,1}) such that u(τ_{d}) = γ and u′ < 0 on (0, τ_{d}].

Proof. From (2.8) we have:

lim_{r→0^{+}} \frac{|u′(r)|^{p−2}u′(r)}{r} = −\frac{f(d)}{N}.

For γ < d < δ the right hand side of the above equation is negative by (1.8). Hence for small values of r > 0 we see that u(r, d) is decreasing. If u is not everywhere decreasing, then there is a first critical point, r = M_{d,1} > 0, with u′(M_{d,1}) = 0 and u′ < 0 on (0, M_{d,1}). From (2.1) we have:

r^{N−1}|u′(r)|^{p−2}u′(r) = ∫_{r}^{M_{d,1}} t^{N−1}f(u(t)) dt.

If f(u(M_{d,1})) > 0 then the above equation implies u′ > 0 for r < M_{d,1} and r sufficiently close to M_{d,1} which contradicts that u′ < 0 on (0, M_{d,1}). Therefore f(u(M_{d,1})) ≤ 0 and so u(M_{d,1}) ≤ β < γ. Thus, there exists τ_{d} ∈ (0, M_{d,1}) with the stated properties. On the other hand, suppose that u(r) is decreasing for all r > 0. We showed in lemma 2.2 that |u(r)| < d < δ for r > 0. Thus lim_{r→∞} u(r) = L with |L| ≤ d < δ. Dividing (2.1) by r^{N} and taking limits as r → ∞ we see that:

lim_{r→∞} \frac{|u′|^{p−2}u′}{r} = −\frac{f(L)}{N}.  (2.22)

EJQTDE, 2006 No. 12, p. 8 We know from (2.10) that u′ is bounded for all r ≥ 0 and so the limit of the left hand side of (2.22) is 0. Thus f(L) = 0 and since |L| ≤ d < δ we see that L = −β, 0, or β. Thus there exists a (finite) τ_{d} with the stated properties.

Finally, the fact that lim_{r→∞} u′(r) = 0 can be seen as follows. In lemma 2.2 we saw that the energy E(r) =

\frac{p−1}{p}|u′(r)|^{p}+F (u(r)) is decreasing and bounded below by F (β), therefore lim_{r→∞} E(r) exists. Since lim_{r→∞} u(r) =

L, we see that lim_{r→∞} F (u(r)) = F (L). Also, since \frac{p−1}{p}|u′(r)|^{p} = E(r) − F (u(r)) and both E(r) and F (u(r)) have a limit as r → ∞, it follows that |u′| has a limit as r → ∞. This limit must be zero since u is bounded. This completes the proof. □

Lemma 2.7. Suppose γ < d^{∗} < δ. Then lim_{d→d^{∗}} u(r, d) = u(r, d^{∗}) uniformly on compact subsets of ℝ and

lim_{d→d^{∗}} u′(r, d) = u′(r, d^{∗}) uniformly on compact subsets of ℝ. Further, if (1.11) holds then lim_{d→δ^{−}} u(r, d) = δ

uniformly on compact subsets of ℝ.

Proof.

If not, then there exists an ϵ_{0} > 0, a compact set K, and sequences r_{j} ∈ K, d_{j} with γ < d_{j} < δ and lim_{j→∞} d_{j} = d^{∗} such that

|u(r_{j}, d_{j}) − u(r_{j}, d^{∗})| ≥ ϵ_{0} > 0 for every j.  (2.23)

However, by lemma 2.2 we know that |u(r, d_{j})| < δ and |u′(r, d_{j})| ≤ (\frac{p}{p−1})^{\frac{1}{p}} [F (δ) + |F (β)|]^{\frac{1}{p}} for all j so that by the Arzela-Ascoli theorem there is a subsequence of the d_{j} (still denote d_{j}) such that u(r, d_{j}) converges uniformly on K to a function u(r) and |u(r)| ≤ δ. From (2.3) we see that u′(r, d_{j}) converges uniformly on K a function v(r) where −v = _{\frac{1}{r^{\frac{N−1}{p−1}}}} Φ_{p′}[∫_{0}^{r} t^{N−1}f(u(t)) dt].

Taking limits in the equation u(r, d_{j}) = d_{j} + ∫_{0}^{r} u′(s, d_{j}) ds, we see that u(r) = d + ∫_{0}^{r} v(s) ds. Hence u′(r) = v(r), that is −u′ = _{\frac{1}{r^{\frac{N−1}{p−1}}}} Φ_{p′}[∫_{0}^{r} t^{N−1}f(u(t)) dt], and thus u is a solution of (2.1)-(2.2) with d = d^{∗}.

So by lemma 2.4, u(r) = u(r, d^{∗}). Therefore, given ϵ = ϵ_{0} > 0 and the compact set K we see that for all r ∈ K we have:

|u(r, d_{j}) − u(r, d^{∗})| < ϵ_{0}

which contradicts (2.23). This completes the proof of the first part of the theorem.

An identical argument shows that lim_{d→δ^{−}} u(r, d) = u(r) uniformly on compact sets where |u(r)| ≤ δ and u

solves (2.1)-(2.2) with d = δ. To complete the proof we need to show u(r) ≡ δ. Let r_{1} = sup{ r ≥ 0 | E(r) = E(0) = F (δ)}. Since E is decreasing we see that if r_{1} = ∞ then E is constant and hence u ≡ δ and we are done.

Therefore we suppose r_{1} < ∞.

By the definition of r_{1} we have:

\frac{p − 1}{p}|u′|^{p} + F (u) = E(r) < E(0) = F (δ) for r > r_{1}.  (2.24)

Thus, it follows that u(r) < δ for r > r_{1}. Also by (1.8) it follows that f(u) > 0 for r_{1} < r < r_{1} + ϵ for some ϵ > 0. Therefore, by (2.15) we see that u′ < 0 for r_{1} < r < r_{1} + ϵ. Using this fact and rewriting (2.24) we see that:

\frac{−u′}{\sqrt[p]{F (δ) − F (u)}} < (\frac{p}{p − 1})^{\frac{1}{p}} for r_{1} < r < r_{1} + ϵ.  (2.25)

Integrating (2.25) on (r_{1}, r_{1} + ϵ) and using (1.11) gives:

∞ = ∫_{u(r_{1}+ϵ)}^{δ} \frac{1}{\sqrt[p]{F (δ) − F (t)}} dt = ∫_{r_{1}}^{r_{1}+ϵ} \frac{−u′}{\sqrt[p]{F (δ) − F (u)}} ≤ (\frac{p}{p − 1})^{\frac{1}{p}} ϵ,

a contradiction. Hence r_{1} = ∞ and u ≡ δ. □

3. Energy Estimates

From lemma 2.6 we saw for γ < d < δ that u(r, d) is decreasing on [0, τ_{d}]. Therefore u^{−1}(y, d) exists for γ ≤ y ≤ d. Lemma 3.1. For γ ≤ y < d < δ we have:

lim_{d→δ^{−}} u^{−1}(y, d) = ∞.

Note: In particular this implies that τ_{d} → ∞ as d → δ^{−} since u^{−1}(γ, d) = τ_{d}. Proof. We fix y_{0} with γ ≤ y_{0} < d and suppose by way of contradiction that there exists d_{k} with d_{k} < δ and d_{k} → δ, u^{−1}(y_{0}, d_{k}) = b_{k}, and that the b_{k} are bounded. Then there is a subsequence of the b_{k} (still denote b_{k}) such that b_{k} → b_{0} for some real number b. By lemma 2.2 we have that |u(r, d_{k})| and |u′(r, d_{k})| are uniformly bounded on say [0, b + 1]. Thus by lemma 2.7, lim_{k→∞} u(r, d_{k}) = δ uniformly on [0, b + 1]. On the other hand, y_{0} = lim_{k→∞} u(b_{k}, d_{k}) = δ - a contradiction since y_{0} < d < δ. □ Lemma 3.2.

(\frac{p − 1}{p})^{\frac{1}{p}} \frac{d − y}{[F (d) − F (y)]^{\frac{1}{p}}} ≤ (\frac{p − 1}{p})^{\frac{1}{p}} ∫_{y}^{d} \frac{dt}{[F (d) − F (t)]^{\frac{1}{p}}} ≤ u^{−1}(y, d) for γ < y < d.

Proof. Rewriting (2.13) gives:

(\frac{p − 1}{p})^{\frac{1}{p}} \frac{|u′(r, d)|}{[F (d) − F (u(r, d)]^{\frac{1}{p}}} ≤ 1.  (3.1)

Since u′(r, d) < 0 on (0, τ_{d}), integrating (3.1) on (0, r) where 0 < r ≤ τ_{d} we obtain:

(\frac{p − 1}{p})^{\frac{1}{p}} ∫_{u(r,d)}^{d} \frac{dt}{[F (d) − F (t)]^{\frac{1}{p}}} ≤ r.

Denoting y = u(r, d) and using the fact that F′ = f > 0 on (γ, δ) we obtain:

(\frac{p − 1}{p})^{\frac{1}{p}} \frac{d − y}{[F (d) − F (y)]^{\frac{1}{p}}} ≤ (\frac{p − 1}{p})^{\frac{1}{p}} ∫_{y}^{d} \frac{dt}{[F (d) − F (t)]^{\frac{1}{p}}} ≤ u^{−1}(y, d).  (3.2)

This completes the proof. □ Lemma 3.3.

lim_{d→δ^{−}}[E(0) − E(τ_{d})] = 0.

Integrating (2.12) on (0, τ_{d}) gives:

E(0) − E(τ_{d}) = ∫_{0}^{τ_{d}} \frac{N − 1}{t}|u′(t, d)|^{p} dt.

Using (2.13) we obtain:

E(0) − E(τ_{d}) ≤ (\frac{p}{p − 1})^{\frac{p−1}{p}} (N − 1) ∫_{0}^{τ_{d}} \frac{1}{t}[F (d) − F (u(t, d)]^{\frac{p−1}{p}} |u′(t, d)| dt.

EJQTDE, 2006 No. 12, p. 10 Now changing variables with y = u(t, d) we obtain:

E(0) − E(τ_{d}) ≤ (\frac{p}{p − 1})^{\frac{p−1}{p}} (N − 1) ∫_{γ}^{d} \frac{[F (d) − F (y)]^{\frac{p−1}{p}}}{u^{−1}(y, d)} dy.  (3.3)

Since [F (d) − F (y)]^{\frac{p−1}{p}} ≤ F (δ)^{\frac{p−1}{p}} for γ ≤ y ≤ d we see by lemma 3.1 that:

lim_{d→δ^{−}} \frac{[F (d) − F (y)]^{\frac{p−1}{p}}}{u^{−1}(y, d)} = 0 for γ ≤ y < d.  (3.4)

Also, by (3.2) and the mean value theorem we see that:

∫_{γ}^{d} \frac{[F (d) − F (y)]^{\frac{p−1}{p}}}{u^{−1}(y, d)} dy ≤ (\frac{p}{p − 1})^{\frac{1}{p}} ∫_{γ}^{d} \frac{F (d) − F (y)}{d − y} dy ≤ (\frac{p}{p − 1})^{\frac{1}{p}} (δ − γ) max_{[γ,δ]} f.

Therefore by (3.4) and the dominated convergence theorem it follows that:

lim_{d→δ^{−}} ∫_{γ}^{d} \frac{[F (d) − F (y)]^{\frac{p−1}{p}}}{u^{−1}(y, d)} dy = 0.

Therefore by (3.3):

lim_{d→δ^{−}}[E(0) − E(τ_{d})] = 0.

This completes the proof. □ Lemma 3.4. Suppose u is monotonic on (τ_{d}, t). Then

E(τ_{d}) − E(t) ≤ \frac{C}{τ_{d}}

where C = 2δ(N − 1)(\frac{p}{p−1})^{\frac{p−1}{p}} [F (δ) + |F (β)|]^{\frac{p−1}{p}}. (Note that C is independent of d). Proof. Integrating (2.12) on (τ_{d}, t), estimating, and using (2.13) gives:

E(τ_{d}) − E(t) = ∫_{τ_{d}}^{t} \frac{N − 1}{s}|u′|^{p} ds ≤ \frac{N − 1}{τ_{d}} ∫_{τ_{d}}^{t} |u′|^{p−1}|u′| ds

≤ \frac{N − 1}{τ_{d}}(\frac{p}{p − 1})^{\frac{p−1}{p}} ∫_{τ_{d}}^{t}[F (δ) − F (u)]^{\frac{p−1}{p}} |u′| ds = \frac{N − 1}{τ_{d}}(\frac{p}{p − 1})^{\frac{p−1}{p}} ∫_{u(t)}^{γ}[F (δ) − F (t)]^{\frac{p−1}{p}} dt

≤ \frac{2δ(N − 1)}{τ_{d}}(\frac{p}{p − 1})^{\frac{p−1}{p}} [F (δ) + |F (β)|]^{\frac{p−1}{p}} = \frac{C}{τ_{d}}

where C = 2δ(N − 1)(\frac{p}{p−1})^{\frac{p−1}{p}} [F (δ) + |F (β)|]^{\frac{p−1}{p}} . This completes the proof. □

Lemma 3.5. Suppose γ < d^{∗} < δ. Let u(r, d^{∗}) be a solution of (2.1)-(2.2) with k zeros and suppose lim_{r→∞} u(r, d^{∗}) = 0. Then for d sufficiently close to d^{∗}, u(r, d) has at most k + 1 zeros.

EJQTDE, 2006 No. 12, p. 11 Proof.

From (2.12) we know that E′(r, d^{∗}) ≤ 0 and since E is bounded from below by F (β), we see that lim_{r→∞} E(r, d^{∗}) exists. Also by assumption lim_{r→∞} u(r, d^{∗}) = 0 and since F is continuous we have lim_{r→∞} F (u(r, d^{∗})) = 0. Since

\frac{p−1}{p}|u′(r, d^{∗})|^{p} = E(r, d^{∗}) − F (u(r, d^{∗})) and the limits of both terms on the right hand side of this equation exist as r → ∞ we see that lim_{r→∞} |u′(r, d^{∗})| exists and since by assumption lim_{r→∞} u(r, d^{∗}) = 0 (so that u(r, d^{∗}) is bounded) we therefore must have:

lim_{r→∞} u′(r, d^{∗}) = 0.  (3.5)

Combining (3.5) with the assumption that lim_{r→∞} u(r, d^{∗}) = 0, we see by (2.11) that:

lim_{r→∞} E(r, d^{∗}) = 0.  (3.6)

Combining (3.6) with the fact that E′(r, d^{∗}) ≤ 0, we see that E(r, d^{∗}) ≥ 0 for all r ≥ 0.

Claim.

E(r, d^{∗}) > 0 for all r ≥ 0.  (3.7)

Proof of claim. First note that E(0, d^{∗}) = F (d^{∗}) > 0. Now suppose E(r_{0}, d^{∗}) = 0 for some r_{0} > 0. Then from (3.6) and the fact that E is decreasing it then follows that E ≡ 0 on [r_{0}, ∞). Thus, −\frac{N−1}{r}|u′|^{p−1} = E′ ≡ 0 on [r_{0}, ∞). Therefore u(r, d^{∗}) ≡ u(r_{0}, d^{∗}) for r ≥ r_{0} and since lim_{r→∞} u(r, d^{∗}) = 0 we see that u(r, d^{∗}) ≡ 0 for r ≥ r_{0}. This implies u′(r_{0}, d^{∗}) = 0. However, by lemma 2.5 u′(r_{0}, d^{∗}) ≠ 0 - a contradiction. This completes the proof of the claim.

By assumption u(r, d^{∗}) has k zeros. Let us denote the kth zero of u(r, d^{∗}) as y^{∗}. Henceforth we assume without loss of generality that u(r, d^{∗}) > 0 for r > y^{∗}. By (3.7) we see that \frac{p−1}{p}|u′(y^{∗}, d^{∗})|^{p} = E(y^{∗}, d^{∗}) > 0. Also since lim_{r→∞} u(r, d^{∗}) = 0 it follows that there exists an M^{∗} > y^{∗} such that u′(M^{∗}, d^{∗}) = 0. Again by (3.7) we see that F (u(M^{∗}, d^{∗})) = E(M^{∗}, d^{∗}) > 0 which implies u(M^{∗}, d^{∗}) > γ. Now by (2.1) we obtain:

−r^{N−1}|u′(r, d^{∗})|^{p−1}u′(r, d^{∗}) = ∫_{M^{∗}}^{r} s^{N−1}f(u(s, d^{∗})) ds.

By (1.8) we have f(u(M^{∗}, d^{∗})) > 0, so from the above equation we see that u(r, d^{∗}) is decreasing for r > M^{∗} as long as u(r, d^{∗}) remains greater than β. In particular, since lim_{r→∞} u(r, d^{∗}) = 0, we see that there exists s^{∗},

t^{∗} with M^{∗} < s^{∗} < t^{∗} such that u(s^{∗}) = \frac{u(M^{∗})+γ}{2} and u(t^{∗}) = γ.

Now let d_{n} be any sequence such that lim_{n→∞} d_{n} = d^{∗}. Then by lemmas 2.4 and 2.7, for some subsequence of d_{n} (still denoted d_{n}) we see that u(r, d_{n}) converges uniformly on compact sets to u(r, d^{∗}) and that u′(r, d_{n}) converges uniformly on compact sets to u′(r, d^{∗}).

In particular we see that u(r, d_{n}) converges uniformly to u(r, d^{∗}) on [0, t^{∗} + 1]. Since γ < d < δ, we see by lemma 2.5 that if u(r_{0}, d^{∗}) = 0 and r_{0} > 0 then u′(r_{0}, d^{∗}) ≠ 0 and so by lemma 2.7 for sufficiently large n we see that u(r, d_{n}) has exactly k zeros on [0, t^{∗} +1]. Further for sufficiently large n there exists a t_{n} ∈ [s^{∗}, t^{∗} +1] such that u(t_{n}, d_{n}) = γ and lim_{n→∞} t_{n} = t^{∗}.

We now assume by way of contradiction that u(r, d_{n}) has at least (k + 2) interior zeros. We denote z_{n} as the (k + 1)st zero of u(r, d_{n}) and w_{n} as the (k + 2)nd zero of u(r, d_{n}). Since u(r, d_{n}) converges uniformly to u(r, d^{∗}) on [0, t^{∗} + 1], we see that for large n we have z_{n} > t^{∗} + 1 and in fact:

lim_{d→∞} z_{n} = ∞,  (3.8)

EJQTDE, 2006 No. 12, p. 12 for if some subsequence of z_{n} (still denoted z_{n}) were uniformly bounded by some B < ∞ then a further subsequence (still denoted z_{n}) would converge to some z^{∗} with y^{∗} < t^{∗}+1 ≤ z^{∗} ≤ B. Since u(r, d_{n}) converges uniformly to u(r, d^{∗}) on [0, z^{∗} + 1], we would then have that u(z^{∗}, d^{∗}) = 0 and since z^{∗} ≥ t^{∗} + 1 > y^{∗}, z^{∗} would then be a (k + 1)st zero of u(r, d^{∗}). However by assumption u(r, d^{∗}) has only k zeros - a contradiction. Thus (3.8) holds.

By assumption γ < d^{∗} < δ so that for sufficiently large n we have that γ < d_{n} < δ so by lemma 2.5 we have that u′(w_{n}, d_{n}) ≠ 0. Thus \frac{p−1}{p}|u′(z_{n})|^{p} = E(z_{n}) ≥ E(w_{n}) = \frac{p−1}{p}|u′(w_{n})|^{p} > 0 so we see that there exists m_{n} with z_{n} < m_{n} < w_{n}, u′(r, d_{n}) < 0 on [z_{n}, m_{n}), and u′(m_{n}, d_{n}) = 0. Also |u(m_{n}, d_{n})| > γ since F (u(m_{n})) = E(m_{n}) ≥ E(w_{n}) > 0. Hence there exists a_{n}, b_{n}, c_{n} with z_{n} < a_{n} < b_{n} < c_{n} < m_{n} such that u(a_{n}) = −β, u(b_{n}) = −\frac{β+γ}{2} ≡ τ, and u(c_{n}) = −γ. Now as in the proof of lemma 2.7 with α = \frac{p(N−1)}{p−1} we have (r^{α}E)′ = αr^{α−1}F (u). Integrating this on [t_{n}, c_{n}], using the fact that F (u) ≤ 0 on [t_{n}, c_{n}], and that F (u(r, d_{n})) ≤ F (τ) < 0 on [a_{n}, b_{n}] we obtain:

0 ≤ \frac{p − 1}{p}c_{n}|u′(c_{n})|^{p} = c_{n}^{α}E(c_{n}) = t_{n}^{α}E(t_{n}) + ∫_{t_{n}}^{c_{n}} αr^{α−1}F (u) dr ≤ t_{n}^{α}E(t_{n}) + ∫_{a_{n}}^{b_{n}} αr^{α−1}F (u) dr

≤ t_{n}^{α}E(t_{n}) + F (τ)[b_{n}^{α} − a_{n}^{α}] ≤ t_{n}^{α}E(t_{n}) + F (τ)b_{n}^{α−1}[b_{n} − a_{n}].  (3.9)

From lemma 2.2 we know that |u′| ≤ (\frac{p}{p−1})^{\frac{1}{p}} [F (δ) + |F (β)|]^{\frac{1}{p}} . Integrating this on [a_{n}, b_{n}] gives:

b_{n} − a_{n} ≥ c > 0  (3.10)

where c = (\frac{γ−β}{2} )(\frac{p}{p−1})^{\frac{−1}{p}} [F (δ) + |F (β)|]^{\frac{−1}{p}}. Substituting (3.10) into (3.9) and using the fact that F (τ) < 0 we see that we obtain:

0 ≤ t_{n}^{α}E(t_{n}) + cF (τ)b_{n}^{α−1}.  (3.11)

In addition, since b_{n} ≥ z_{n} we see from (3.8) that:

lim_{n→∞} b_{n} = ∞.  (3.12)

Finally, by lemma 2.7 we know that u(r, d_{n}) converges uniformly to u(r, d^{∗}) on [0, t^{∗} + 1] and u′(r, d_{n}) converges uniformly to u′(r, d^{∗}) on [0, t^{∗} + 1] and t_{n} → t^{∗}. Therefore, we see that:

lim_{n→∞} t_{n}^{α}E(t_{n}, d_{n}) = (t^{∗})^{α}E(t^{∗}, d^{∗})  (3.13)

Substituting (3.12)-(3.13) into (3.11) and recalling that F (τ) < 0, and α = \frac{p(N−1)}{p−1} > 1 (since N ≥ 2), we see that the right hand side of (3.11) goes to −∞ as n → ∞ which contradicts the inequality in (3.11). This completes the proof. □

4. Proof of the Main Theorem

Proof.

For k ∈ ℕ ∪ {0}, define

A_{k} = {d ∈ (β, δ)|u(r, d) has exactly k zeros on [0, ∞)}.

Observe first that (β, γ) ⊂ A_{0} because for any d ∈ (β, γ) we have E(0, d) = F (d) < 0 so that by (2.12) E(r, d) < 0 for all r > 0. Thus u(r, d) > 0 for if u(r_{0}, d) = 0 then E(r_{0}, d) = \frac{p−1}{p}|u′(r_{0}, d)|^{p} ≥ 0 - a contradiction. Thus we see that A_{0} is nonempty.

EJQTDE, 2006 No. 12, p. 13 We now assume that d > γ and we apply lemma 3.4 at t = M_{d,1} where M_{d,1} is defined in lemma 2.6 and we combine this with lemma 3.3 to obtain:

lim_{d→δ^{−}} F (u(M_{d,1})) = F (δ) > 0.

Thus

|u(M_{d,1})| > γ for d sufficiently close to δ.  (4.1)

This implies that M_{d,1} < ∞ for if M_{d,1} = ∞, then from lemma 2.6 we see that u(M_{d,1}) = lim_{r→∞} u(r), |u(M_{d,1})| < d < δ, and f(u(M_{d,1})) = 0 which implies |u(M_{d,1})| ≤ β - contradicting (4.1). Thus M_{d,1} < ∞ and by lemma 2.6 we see that f(u(M_{d,1})) ≤ 0 so by (1.8) we have u(M_{d,1}) ≤ β. Combining this with (4.1) we see that we must have u(M_{d,1}) < −γ < 0. Therefore for d < δ and d sufficiently close to δ, we see that u(r, d) must have a first zero, z_{d,1}.

Thus we see that A_{0} is bounded above by a quantity that is strictly less than δ. We now define:

d_{0} = sup A_{0}

and we note that d_{0} < δ.

Lemma 4.1.

u(r, d_{0}) > 0 for r ≥ 0.

Proof.

Suppose there exists a smallest value of r, r_{0}, such that u(r_{0}, d_{0}) = 0. By Lemma 2.5, u′(r_{0}, d_{0}) ≠ 0 thus u(r, d_{0}) becomes negative for r slightly larger than r_{0}. By lemma 2.7 it follows that if d < d_{0} is sufficiently close to d_{0} then u(r, d) must also have a zero close to r_{0}. However by the definition of d_{0} if d < d_{0} then u(r, d) > 0 - a contradiction. This completes the proof. □

Lemma 4.2

u′(r, d_{0}) < 0 for r > 0.

Proof

We will show that M_{d_{0},1} = ∞ where M_{d_{0},1} is defined in lemma 2.6. If M_{d_{0},1} < ∞ then by lemma 2.7 for d slightly larger than d_{0} we also have M_{d,1} < ∞. Also, since u(r, d_{0}) > 0 then u(M_{d_{0},1}, d_{0}) > 0 and again by lemma 2.7 we also have u(M_{d,1}, d) > 0 for d sufficiently close to d_{0}. By lemma 2.6 it follows that f(u(M_{d,1}, d)) ≤ 0 so that 0 ≤ u(M_{d,1}, d) ≤ β thus E(M_{d,1}, d) < 0. Since E is decreasing we see that E(r, d) < 0 for r ≥ M_{d,1}. For d slightly larger than d_{0}, u(r, d) must have a first zero, z_{d,1}, (by definition of d_{0}) and z_{d,1} > M_{d,1} since u(r, d) > 0 on [0, M_{d,1}]. Thus, we have 0 ≤ E(z_{1}, d) ≤ E(M_{d,1}, d) < 0 - a contradiction. This completes the proof. □

From lemmas 2.6, 4.1, and 4.2 we see that lim_{r→∞} u(r, d_{0}) = L where f(L) = 0 where L < d_{0} < δ and since u(r, d_{0}) > 0 we have that L = 0 or L = β. We also see that lim_{r→∞} E(r, d_{0}) = F (L).

Lemma 4.3.

lim_{d→d_{0}^{+}} z_{d,1} = ∞

Proof.

If z_{d,1} ≤ C for d > d_{0} then as in the proof of (3.8) there would be a subsequence d_{n} with d_{n} → d_{0} and z_{d_{n},1} → z. By lemma 2.7 it then would follow that u(z, d_{0}) = 0 which contradicts that u(r, d_{0}) > 0. This completes the proof. □

EJQTDE, 2006 No. 12, p. 14 Lemma 4.4. L = 0

Proof.

We know that L = 0 or L = β so suppose L = β. Then lim_{r→∞} E(r, d_{0}) = F (L) = F (β) < 0 so there exists an r_{0} such that E(r_{0}, d_{0}) < 0. Thus for d > d_{0} and d sufficiently close to d_{0} we have by lemma 2.7 E(r_{0}, d) < 0. Since E(z_{d,1}, d) ≥ 0 we see that z_{d,1} < r_{0} which contradicts lemma 4.3. Thus lim_{r→∞} u(r, d_{0}) = 0 and this completes the proof. □

By definition of d_{0}, if d > d_{0} then u(r, d) has at least one zero. By lemma 3.4, if d is close to d_{0} then u(r, d) has at most one zero. Therefore for d > d_{0} and d sufficiently close to d_{0}, u(r, d) has exactly one zero. Thus the set A_{1} is nonempty and d_{0} < sup A_{1}.

As we saw in the first part of the proof of the main theorem, M_{d,1} < ∞ and u(M_{d,1}) < −γ for d sufficiently close to δ. By a similar argument as in lemma 2.6, it can be shown that there exists an M_{d,2} with M_{d,1} < M_{d,2} ≤ ∞ such that u′(r, d) > 0 on (M_{d,1}, M_{d,2}). Also, by lemma 3.4 we see that

0 ≤ E(0) − E(M_{d,2}) = [E(0) − E(τ_{d})] + [E(τ_{d}) − E(M_{d,1})] + [E(M_{d,1}) − E(M_{d,2}]
≤ [E(0) − E(τ_{d})] + \frac{C}{τ_{d}} + \frac{C}{M_{d,1}} where C is independent of d.

By lemmas 3.1, 3.3 and the fact that τ_{d} < M_{d,1} we see:

lim_{d→δ^{−}} F (u(M_{d,2})) = E(0) = F (δ) > 0.

As at the beginning of the proof of the main theorem we may also show that M_{d,2} < ∞ and u(M_{d,2}) > γ for d sufficiently close to δ. Therefore, there exists z_{d,2} such that M_{d,1} < z_{d,2} < M_{d,2} and u(z_{d,2}, d) = 0. Therefore A_{1} is bounded above by a quantity strictly less than δ.

Let:

d_{1} = sup A_{1}

and note that d_{0} < d_{1} < δ.

In a similar way in which we proved that u(r, d_{0}) > 0 and lim_{r→∞} u(r, d_{0}) = 0 we can show that u(r, d_{1}) has exactly one zero and that lim_{r→∞} u(r, d_{1}) = 0.

In a similar way we may show by induction that A_{k} is nonempty and bounded above by a quantity strictly less than δ. Let

d_{k} = sup A_{k}.

It can be shown that u(r, d_{k}) has exactly k zeros and that lim_{r→∞} u(r, d_{k}) = 0.

This completes the proof of the main theorem. □

References

1. B. Franchi, E. Lanconelli, and J. Serrin, Existence and Uniqueness of Nonnegative Solutions of Quasilinear Equations in ℝ^{N}, Advances in Mathematics 118 (1996), 177–243.

2. J. Iaia, H. Warchall, and F.B. Weissler, Localized Solutions of Sublinear Elliptic Equations: Loitering at the Hilltop, Rocky Mountain Journal of Mathematics 27(4) (1997), 1131–1157.

(Received June 6, 2006)

Dept. of Mathematics, University of North Texas, Denton, TX 76203-1430 E-mail address: iaia@unt.edu

EJQTDE, 2006 No. 12, p. 15