Electronic Journal of Qualitative Theory of Differential Equations 2008, No. 20, 1-23; http://www.math.u-szeged.hu/ejqtde/

LOCALIZED RADIAL SOLUTIONS FOR A NONLINEAR P-LAPLACIAN EQUATION IN ℝ^{ℕ}

SRIDEVI PUDIPEDDI

Abstract. We establish the existence of radial solutions to the p-Laplacian equation Δ_{p}u + f(u) = 0 in ℝ^{ℕ}, where f behaves like |u|^{q−1}u when u is large and f(u) < 0 for small positive u. We show that for each nonnegative integer n, there is a localized solution u which has exactly n zeros.

1. Introduction

In this paper we look for solutions u : ℝ^{ℕ} → ℝ of the nonlinear partial differential equation

(1.1)  ∇ · (|∇u|^{p−2}∇u) + f(u) = 0,
(1.2)  lim_{|x|→∞} u(x) = 0,

with 1 < p < N. We also assume f(u) behaves like |u|^{q−1}u where u is large and f(u) < 0 for small positive u.

Motivation: When p = 2 then (1.1) is

Δu + f(u) = 0.

McLeod, Troy and Weissler studied the radial solutions of the above mentioned equation in [5]. In this paper they made a remark that their result could be extended to the p-Laplacian. In this paper we show that their conjecture is true. Also, Castro and Kurepa studied

Δu + g(u) = q(x),

subject to Dirichlet boundary conditions on a ball in ℝ^{ℕ}, where g is superlinear and q ∈ 𝕃^{2} in [1]. The p-Laplacian equation has been studied in different settings. Gazzola, Serrin and Tang [9] have proved existence of radial solutions to a p-Laplacian equation with Dirichlet and Neumann boundary conditions. Calzolari, Filippucci and Pucci [8] have proved existence of radial solutions for the p-Laplacian with weights.

We assume that the function f satisfies the following hypotheses:

(H1) f is an odd locally Lipschitz continuous function,

(H2) f(u) < 0 for 0 < u < ϵ_{1} for some ϵ_{1} > 0,

(H3) f(u) = |u|^{q−1}u + g(u) with \frac{g(|u|)}{|u|^{q}} → 0 as |u| → ∞ where 1 < p < q + 1 < \frac{Np}{N−p}.

From (H2) and (H3) we see that f(u) has at least one positive zero.

(H4) Let α be the least positive zero of f and β be the greatest positive zero of f,

(H5) Let F (u) ≡ ∫_{0}^{u} f(s)ds with exactly one positive zero γ, with γ > β,

(H6) If p > 2 we also assume for some ϵ_{2} > 0

∫_{0}^{ϵ_{2}} \frac{1}{\sqrt[p]{|F (u)|}}du = ∞.

We assume that u(x) = u(|x|) and let r = |x|. In this case (1.1)-(1.2) becomes the nonlinear ordinary differential equation

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

EJQTDE, 2008 No. 20, p. 1 for 0 < r < ∞, with

(1.4)  lim_{r→∞} u(r) = 0, lim_{r→0^{+}} u′(r) = 0.

We would like to find C^{2} solutions of (1.3)-(1.4) but we will see later that this is not always possible (see the proof of Lemma 2.1). However multiplying (1.3) by r^{N−1} and integrating gives

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

Instead of looking for solutions of (1.3)-(1.4) in C^{2} we look for solutions of (1.4)-(1.5) in C^{1}.

Our Main Theorem is

Let the nonlinearity f have the properties (H1)-(H6), and let n be a nonnegative integer. Then there is a solution u ∈ C^{1}[0, ∞) of (1.4)-(1.5) such that u has exactly n zeros.

The technique used to solve (1.4)-(1.5) is the shooting method. That is, we first solve the initial value problem

r^{N−1}|u′|^{p−2}u′ = − ∫_{0}^{r} s^{N−1}f(u(s))ds
u(0) = d ≥ 0.

By varying d appropriately, we attempt to find a d such that u(r, d) has exactly n zeros and u satisfies (1.4). In section 2, we establish the existence of solutions of this initial value problem by the contraction mapping principle. In section 3, we see that after a rescaling of u we get a family of functions {u_{λ}}, which converges to the solution of

−r^{N−1}|v′|^{p−2}v′ = ∫_{0}^{r} s^{N−1}|v|^{q−1}v ds,
v(0) = 1, v′(0) = 0,

where 1 < p < q + 1 < \frac{Np}{N−p}. We will then show that v has infinitely many zeros which will imply that there are solutions, u, with any given number of zeros. In section 4, we prove our Main Theorem.

Note: From (H3) and (H5) we see that

(1.6)  F (u) = \frac{1}{q + 1}|u|^{q+1} + G(u),

where G(u) = ∫_{0}^{u} g(s)ds. Dividing both sides by |u|^{q+1} and taking the limit as |u| → ∞ gives

(1.7)  lim_{|u|→∞} \frac{F (u)}{|u|^{q+1}} = lim_{|u|→∞} (\frac{1}{q + 1} + \frac{G(u)}{|u|^{q+1}} ) .

Using L’Hopital’s rule and (H3) we see that

(1.8)  lim_{u→∞} \frac{G(u)}{|u|^{q+1}} = 0.

Thus, we have

lim_{|u|→∞} \frac{F (u)}{|u|^{q+1}} = \frac{1}{q + 1}.

This implies that F (u) ≥ 0 for |u| sufficiently large, so F (u) ≥ 0 for |u| ≥ M. Also since F is continuous on the compact set [−M, M] we see that F is bounded below and there is a −L < 0 such that

(1.9)  F (u) ≥ −L

for all u.

EJQTDE, 2008 No. 20, p. 2

Note: When 1 < p ≤ 2, then assumption (H6) also holds. This follows from (H1). The details of this are as follows: since f is locally Lipschitz and since f(0) = 0 we have

|f(u)| = |f(u) − f(0)| ≤ c|u − 0| = c|u|

for |u| < ϵ_{2} for some ϵ_{2} > 0, and where c > 0 is a Lipschitz constant for f. Integrating on (0, u) where 0 ≤ u ≤ ϵ_{2} gives:

− ∫_{0}^{u} c t dt ≤ ∫_{0}^{u} f(t) dt ≤ ∫_{0}^{u} c t dt.

Thus,

\frac{−cu^{2}}{2} ≤ F (u) ≤ \frac{cu^{2}}{2}

for |u| < ϵ_{2}. So, |F (u)| ≤ \frac{cu^{2}}{2} for |u| ≤ ϵ_{2}. Thus, |F (u)|^{\frac{1}{p}} ≤ (\frac{c}{2})^{\frac{1}{p}} u^{\frac{2}{p}} for |u| < ϵ_{2}. Hence,

∫_{0}^{ϵ_{2}} \frac{1}{|F (u)|^{\frac{1}{p}}} du ≥ (\frac{2}{c})^{\frac{1}{p}} ∫_{0}^{ϵ_{2}} \frac{1}{u^{\frac{2}{p}}} du = ∞, if 1 < p ≤ 2.

2. Existence of solutions of the initial value problem

Now let us consider the initial value problem

(2.1)  r^{N−1}|u′|^{p−2}u′ = − ∫_{0}^{r} s^{N−1}f(u(s))ds,

with

(2.2)  u(0) = d ≥ 0.

The local existence of solutions of (2.1) and (2.2) is well known, see [6] and [7], so u ∈ C^{1}[0, ϵ] for ϵ > 0 and small.

We define Φ_{p}(x) = |x|^{p−2}x for x ∈ ℝ and p > 1. Note that the inverse of Φ_{p}(x) is Φ_{p′}(x) where \frac{1}{p} + \frac{1}{p′} = 1, that is p′ = \frac{p}{p−1}. Note that both Φ_{p} and Φ_{p′} are odd for every p. Now dividing (2.1) by r^{N−1}, gives

(2.3)  |u′|^{p−2}u′ = \frac{−1}{r^{N−1}} ∫_{0}^{r} s^{N−1}f(u(s))ds.

Using the definition of Φ_{p}, we get

Φ_{p}(u′) = \frac{−1}{r^{N−1}} ∫_{0}^{r} s^{N−1}f(u(s))ds.

Now applying Φ_{p′} on both sides, leads to

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

Note that if f(d) = 0, then u ≡ d is a solution of (2.1)-(2.2). So, we now assume that

(2.5)  f(d) ≠ 0.

Now we explain why we aim at solutions of (1.4)-(1.5) instead of solutions of (1.3)-(1.4).

Lemma 2.1.

u ∈ C^{2}[0, ϵ)

if 1 < p ≤ 2 and

u ∈ C^{2}{r ∈ [0, ϵ) | u′(r) ≠ 0}

if p > 2.

EJQTDE, 2008 No. 20, p. 3 Proof. Now let us consider equation (2.4) which is

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

Since Φ_{p′}(x) = |x|^{p′−2}x, so Φ′_{p′} = (p′ − 1)|x|^{p′−2}. Since p′ − 2 = \frac{2 − p}{p − 1}, we see that Φ′_{p′} is continuous

for all x, if 1 < p ≤ 2 and Φ′_{p′} is continuous at all x ≠ 0, if p > 2.

Let

k(r) = \frac{1}{r^{N−1}} ∫_{0}^{r} t^{N−1}f(u) dt.

Then using the fact that f is bounded, it is straight forward to show that k is continuous on [0, ϵ). Now,

k′(r) = [\frac{−(N − 1)}{r^{N}} ∫_{0}^{r} t^{N−1} f(u) dt + f(u)]

so k′ continuous on (0, ϵ).

Claim: k′ is continuous on [0, ϵ). Proof of the Claim: We do this in two steps: Step 1: We show k′(0) = \frac{f(d)}{N}. By definition

k′(0) = lim_{r→0} \frac{k(r) − k(0)}{r − 0}

= lim_{r→0} \frac{\frac{1}{r^{N−1}} ∫_{0}^{r} t^{N−1} f(u) dt − 0}{r − 0}

= lim_{r→0} \frac{∫_{0}^{r} t^{N−1}f(u) dt}{r^{N}}.

Applying L’Hopital’s rule gives k′(0) = \frac{f(d)}{N}.

Step 2: We show lim_{r→0} k′(r) = k′(0) = \frac{f(d)}{N}.

Differentiating k(r) and taking the limit as r → 0 gives

lim_{r→0} k′(r) = lim_{r→0} \frac{−(N − 1)}{r^{N}} ∫_{0}^{r} t^{N−1} f(u) dt + f(u)

= \frac{−(N − 1)}{N}f(d) + f(d)

= \frac{f(d)}{N}.

We get the second equality by using L’Hopital’s rule.

Steps 1 and 2 imply that k′ is continuous on [0, ϵ). EJQTDE, 2008 No. 20, p. 4 Finally, by the chain rule and (2.1) we see u′ is differentiable and that

−u′′ = Φ′_{p′} (\frac{1}{r^{N−1}} ∫_{0}^{r} t^{N−1} f(u) dt) k′(r)

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

= \frac{1}{p − 1} |\frac{1}{r^{N−1}} ∫_{0}^{r} t^{N−1} f(u) dt|^{\frac{2−p}{p−1}} [\frac{−(N − 1)}{r^{N}} ∫_{0}^{r} t^{N−1} f(u) dt + f(u)]

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

By the previous claim, k′ is continuous. Note that |u′|^{2−p} is continuous for 1 < p ≤ 2 and |u′|^{2−p} is continuous at all points where u′ ≠ 0 for p > 2 and hence the lemma follows. □

Remark: If p > 2, u′(r_{0}) = 0, and f(u(r_{0})) ≠ 0, then u′′(r_{0}) is undefined.

To see this, suppose on the contrary that u′′(r_{0}) is defined. Using the fact that u′(r_{0}) = 0, (2.1) becomes

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

Dividing by (r − r_{0}) and taking the limit as r → r_{0} gives

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

Using L’ Hopital’s rule we obtain

0 = −|u′(r_{0})|^{p−2}u′′(r_{0}) = f(u(r_{0})).

Thus, |f(u(r_{0}))| = 0 which is a contradiction to our assumption that f(u(r_{0})) ≠ 0. Thus, u′′(r_{0}) is undefined.

Remark: If p > 2, u′(r_{0}) = 0, and f(u(r_{0})) = 0, then it is not clear whether u is C^{2} in a neighborhood of r_{0} when u′(r_{0}) = 0. However, for the purposes of this paper a more detailed analysis of this situation is not needed.

To prove the following two lemmas, let [0, R) be the maximal interval of existence for which u is a solution for (2.1)-(2.2).

Our goal is to show that u solves (2.1)-(2.2) on [0, ∞). So, we aim at proving R = ∞, and we will do this in two lemmas. In the first lemma we show that if R < ∞ then the limits of u and u′ as r → R^{−} are defined. Once the limits exist then in the second lemma, we establish that R = ∞.

Lemma 2.2. Suppose u solves (2.1)-(2.2) on [0, R) with R < ∞, then there exists u_{0}, u′_{0} ∈ ℝ such that

lim_{r→R^{−}} u(r) = u_{0},

lim_{r→R^{−}} u′(r) = u′_{0}.

Proof. The following is the energy equation for (2.1)-(2.2)

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

Using (2.1) we see that

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

EJQTDE, 2008 No. 20, p. 5 Note that E′(r) ≤ 0, so E is decreasing, and so E(r) ≤ E(0) which is

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

Then by (1.9)

\frac{(p − 1)|u′|^{p}}{p} − L ≤ F (d).

Further simplification gives

|u′|^{p} ≤ \frac{p(F (d) + L)}{p − 1}.

Then |u′| ≤ M where M = [\frac{p(F (d)+L)}{p−1}]^{\frac{1}{p}} . So, by the mean value theorem we have

|u(x) − u(y)| ≤ M|x − y|

for all x, y ∈ [0, R). This implies that u has a limit as x → R^{−}. So, there exists a u_{0} ∈ ℝ such that lim_{r→R^{−}} u(r) = u_{0}. Taking the limit as r → R^{−} on both sides of (2.1), we see that lim_{r→R^{−}} u′(r) exists, and

we call it u′_{0}. □

Lemma 2.3. A solution exists for (2.1)-(2.2) on [0, ∞).

Proof. If R = ∞, we are done. Suppose R < ∞.

Case(i): If u′(R) ≠ 0, then by Lemma 2.1, u ∈ C^{2} in a neighborhood of R, so differentiating (1.5) and then dividing by |u′|^{p−2}, we have

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

Since u′(R) ≠ 0, then by the standard existence theorem for ordinary differential equations there exists a solution for the differential equation on [R, R + ϵ) for some ϵ > 0 with u(R) = u_{0} and u′(R) = u′_{0}. This contradicts the definition of R, hence, R = ∞.

Case(ii): If u′(R) = 0 and f(u(R)) ≠ 0, then we can use the contraction mapping principle and extend our solution u to [R, R + ϵ) for some ϵ > 0. This contradicts the definition of R.

Case(iii): If u′(R) = 0 and f(u(R)) = 0 we can extend u ≡ u(R) for r > R. Again this contradicts the definition of R. □

Lemma 2.4. Let d > β, then |u(r)| < d for 0 < r < ∞ and f(d) ≠ 0.

Proof. From (2.6)-(2.7) it follows that

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

If there exists a r_{0} > 0 such that |u(r_{0})| = d, then

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

This implies |u′| = 0 on [0, r_{0}]. Hence, u(r) ≡ d on [0, r_{0}]. Then by (1.5), f(d) = 0, but this contradicts our assumption that f(d) ≠ 0. □

Lemma 2.5. If z_{1} < z_{2}, with u(z_{1}) = u(z_{2}) = 0, and |u| > 0 on (z_{1}, z_{2}), then there is exactly one extremum, m, between (z_{1}, z_{2}) and also |u(m)| > γ.

EJQTDE, 2008 No. 20, p. 6 Proof. Suppose without loss of generality that u > 0 on (z_{1}, z_{2}). Then there exists an extremum, m, such that u′(m) = 0. And

F (u(m)) = E(m) ≥ E(z_{2}) = \frac{p − 1}{p}|u′(z_{2})|^{p} ≥ 0.

Thus |u(m)| ≥ γ for any extremum. Suppose there exists consecutive extrema m_{1} < m_{2} < m_{3} such that at m_{1} and m_{3} we have local maxima and m_{2} is a local minimum with u′ < 0 on (m_{1}, m_{2}) and u′ > 0 on (m_{2}, m_{3}). We have z_{1} < m_{1} < m_{2} < m_{3} < z_{2} and since the energy is decreasing we obtain E(m_{2}) ≥ E(m_{3}) ≥ E(z_{2}). Since u′(m_{2}) = u′(m_{3}) = 0 and since F (u(z_{2})) = 0 this gives

(2.8)  F (u(m_{2})) ≥ F (u(m_{3})) ≥ \frac{p − 1}{p}|u′(z_{2})|^{p} ≥ 0.

And by (H5) it follows that u(m_{2}) ≥ γ and u(m_{3}) ≥ γ. Also, since m_{2} is a local minimum and m_{3} is a local maximum we have γ ≤ u(m_{2}) < u(m_{3}). But by (H5), F is increasing for u > γ and this implies F (u(m_{2})) < F (u(m_{3})) which is a contradiction to (2.8). □

Lemma 2.6. If u(r_{0}) = u′(r_{0}) = 0 then u ≡ 0.

Proof. Suppose u(r_{0}) = 0 and u′(r_{0}) = 0. First we will do the easy case, and show that u ≡ 0 on (r_{0}, ∞). Since E′ ≤ 0 and E(r_{0}) = 0 then either E < 0 for r > r_{0} or E ≡ 0 on (r_{0}, r_{0} + ϵ) for some ϵ > 0. We will show E ≡ 0 on (r_{0}, r_{0} + ϵ). For suppose E < 0 for r > r_{0}. Then we see that |u| > 0 for r > r_{0}, for if there exists an r_{1} > r_{0} such that u(r_{1}) = 0 then

0 ≤ \frac{p − 1}{p}|u′(r_{1})|^{p} = E(r_{1}) < 0.

This is a contradiction. So suppose without loss of generality that u > 0 for r > r_{0}. Then for r > r_{0} and r close to r_{0} and by (H2), f(u) < 0 so

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

Thus u is increasing on (r_{0}, r_{0} + ϵ) for some ϵ > 0. Now since E(r) < 0 on (r_{0}, r_{0} + ϵ) therefore

\frac{p − 1}{p}|u′|^{p} + F (u) < 0,

and so

|u′| < (\frac{p}{p − 1})^{\frac{1}{p}} |F (u)|^{\frac{1}{p}} .

Therefore,

∞ = ∫_{0}^{u(r_{0}+ϵ)} \frac{ds}{\sqrt[p]{|F (s)|}} = ∫_{r_{0}}^{r_{0}+ϵ} \frac{|u′|}{|F (u)|^{\frac{1}{p}}} dt < ∫_{r_{0}}^{r_{0}+ϵ} (\frac{p}{p − 1})^{\frac{1}{p}} dt < ∞.

This is a contradiction to (H6) and to the note at the end of the introduction. Then E ≡ 0 on [r_{0}, r_{0}+ϵ) and so

\frac{−(N − 1)}{r}|u′|^{p} = E′ ≡ 0

on [r_{0}, r_{0} + ϵ) and thus u ≡ 0 on [r_{0}, r_{0} + ϵ). Denote [r_{0}, r_{1}) as the maximal half open interval for which u ≡ 0. If r_{1} < ∞, again we can show that u ≡ 0 on [r_{1}, r_{1} + ϵ), but this will contradict the definition of r_{1}. Thus, E ≡ 0 on (r_{0}, ∞). Hence u ≡ 0 on [r_{0}, ∞).

Now we will prove that u ≡ 0 on (0, r_{0}). To prove this we use the idea from [2] and do the required modifications to fit our case. We will use hypothesis (H6). Let

r_{1} = inf_{r>0}{r | u(r) = 0, u′(r) = 0}.

EJQTDE, 2008 No. 20, p. 7 If r_{1} = 0 then u ≡ 0 on (0, ∞) and then by continuity u ≡ 0 on [0, ∞) and we are done. So suppose by the way of contradiction that r_{1} > 0. Let \frac{r_{1}}{2} < r < r_{1}, so \frac{2}{r_{1}} > \frac{1}{r}. Now consider the derivative of the energy function given in equation (2.7) and then integrate it between r and r_{1} to obtain

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

Since u(r_{1}) = 0, so F (u(r_{1})) = 0 and u′(r_{1}) = 0, we get

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

Now let

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

Differentiating we get

w′ = −\frac{(N − 1)|u′(r)|^{p}}{r} .

Solving this for |u′|^{p}, gives

|u′(r)|^{p} = \frac{−rw′}{N − 1}.

Substituting this in (2.9) gives

(2.10)  \frac{−(p − 1)rw′}{p(N − 1)} + F (u) = w

and rearranging terms, we get

\frac{(p − 1)rw′}{p(N − 1)} + w = F (u).

Letting η = \frac{(N−1)p}{p−1} then we have

w′ + \frac{ηw}{r} = \frac{ηF (u)}{r}.

Multiplying both sides by r^{η}, gives

(r^{η}w)′ = ηr^{η−1}F (u).

Integrating between r and r_{1} for r sufficiently close to r_{1}, gives

r_{1}^{η}w(r_{1}) − r^{η}w = ∫_{r}^{r_{1}} ηt^{η−1}F (u)dt.

Since w(r_{1}) = 0, and by (H2), F (u(t)) ≤ 0 for t sufficiently close to r_{1} we obtain

w = \frac{−η}{r^{η}} ∫_{r}^{r_{1}} t^{η−1}F (u(t))dt = \frac{η}{r^{η}} ∫_{r}^{r_{1}} t^{η−1}|F (u(t))|dt.

Now plugging w and w′ in (2.10) we have

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

Solving this for |u′|^{p} gives (for r close to r_{1})

(2.11)  |u′|^{p} = \frac{p}{p − 1} [\frac{η}{r^{η}} ∫_{r}^{r_{1}} t^{η−1}|F (u(t))|dt + |F (u(r))|] .

Observe next that for r < r_{1} and r sufficiently close to r_{1} that u′(r) ≠ 0; for if there exists r_{2} < r_{1} such that u′(r_{2}) = 0 then from (2.11), u ≡ 0 on (r_{2}, r_{1}), this contradicts the definition of r_{1}. Hence without loss of generality assume that u′(r) < 0 for r < r_{1} and r sufficiently close to r_{1}. Now for r < t < r_{1}, u

EJQTDE, 2008 No. 20, p. 8 is decreasing so u(r) > u(t) > 0 which implies F (u(r)) < F (u(t)) < 0 and so |F (u(r))| > |F (u(t))| > 0, which leads to the following

|u′|^{p} ≤ \frac{p}{p − 1} [|F (u(r))| + \frac{η}{r^{η}} |F (u(r))| ∫_{r}^{r_{1}} t^{η−1}dt]

= \frac{p}{p − 1} [|F (u(r))| + \frac{η}{r^{η}} \frac{|F (u(r))|}{η}(r_{1}^{η} − r^{η})]

= \frac{p|F (u(r))|r_{1}^{η}}{(p − 1)r^{η}}

≤ \frac{p2^{η}|F (u(r))|}{p − 1}.

The last inequality follows as \frac{2}{r_{1}} > \frac{1}{r}, so

|u′|^{p} ≤ \frac{p2^{η}|F (u(r))|}{p − 1}.

Solving this for |u′|, we get

|u′| ≤ \sqrt[p]{\frac{p2^{η}}{p − 1}} \sqrt[p]{|F (u(r))|}.

Dividing by \sqrt[p]{|F (u(r))|}, integrating on (r, r_{1}) and using (H6) and the remark following (H6) we obtain

∞ = ∫_{0}^{u(r)} \frac{1}{\sqrt[p]{|F (s)|}}ds = ∫_{r}^{r_{1}} \frac{|u′|}{\sqrt[p]{|F (u)|}}dt

≤ \sqrt[p]{\frac{p2^{η}}{p − 1}} ∫_{r}^{r_{1}} dt

= \sqrt[p]{\frac{p2^{η}}{p − 1}}(r_{1} − r)

< ∞.

Thus we get a contradiction and so r_{1} = 0 and hence u ≡ 0. □

3. Solutions with a prescribed number of zeros

In this section we show that there are solutions for (2.1)-(2.2) with a large number of zeros. For this we study the behavior of solutions as d grows large. We consider the idea from [5], page 371 and we do the required modifications to fit our case. Given λ > 0, let u(r) be the solution of (2.1)-(2.2) with d = λ^{\frac{p}{q−p+1}} . Define

(3.1)  u_{λ} = λ^{\frac{−p}{q−p+1}} u(\frac{r}{λ}).

Then u_{λ} satisfies

(3.2)  r^{N−1}|u′_{λ}|^{p−2}u′_{λ} = − ∫_{0}^{r} s^{N−1}λ^{\frac{−pq}{q−p+1}} f(λ^{\frac{p}{q−p+1}} u_{λ}(s))ds,

and

(3.3)  u_{λ}(0) = 1.

Lemma 3.1. As λ → ∞, u_{λ} → v, uniformly on compact subsets of [0, ∞), where v is a solution of

(3.4)  r^{N−1}|v′|^{p−2}v′ = − ∫_{0}^{r} s^{N−1}|v(s)|^{q−1}v(s)ds,
(3.5)  v(0) = 1.

Proof. Let

E(r, λ) = \frac{(p − 1)|u′_{λ}|^{p}}{p} + λ^{\frac{−pq}{q−p+1}} F (λ^{\frac{p}{q−p+1}} u_{λ})

then

\frac{∂}{∂r} E(r, λ) ≡ E′(r, λ) = −\frac{(N − 1)|u′_{λ}|^{p}}{r} .

This implies E(r, λ) is decreasing in r. So for λ > 0

E(r, λ) ≤ E(0, λ)
= λ^{\frac{−pq}{q−p+1}} F (λ^{\frac{p}{q−p+1}} )

Using (1.6) to simplify the right hand side, gives the following:

λ^{\frac{−p(q+1)}{q−p+1}} F (λ^{\frac{p}{q−p+1}} ) = λ^{\frac{−p(q+1)}{q−p+1}} \frac{λ^{\frac{p(q+1)}{q−p+1}}}{q + 1} + λ^{\frac{−p(q+1)}{q−p+1}} G(λ^{\frac{p}{q−p+1}} )

(3.6)  λ^{\frac{−p(q+1)}{q−p+1}} F (λ^{\frac{p}{q−p+1}} ) = \frac{1}{q + 1} + λ^{\frac{−p(q+1)}{q−p+1}} G(λ^{\frac{p}{q−p+1}} ).

Then by (1.8)

(3.7)  \frac{G(λ^{\frac{p}{q−p+1}} )}{(λ^{\frac{p}{q−p+1}} )^{q+1}} → 0,

as λ → ∞. Thus, E(r, λ) < \frac{2}{q+1} for large λ. Moreover E(r, λ) is bounded above independently of r and for large λ.

The usual trick to show the convergence of u_{λ} is to use Arzela-Ascoli’s Theorem. For this it suffices to show u_{λ} and u′_{λ} are bounded.

Claim: u_{λ}(r) and u′_{λ}(r) are bounded.

Proof of Claim: By Lemma 2.4, |u(r)| ≤ d = λ^{\frac{p}{q−p+1}} . Thus, by (3.1), |u_{λ}(r)| ≤ 1. Also, since E(r, λ) ≤ E(0, λ) we have

\frac{(p − 1)|u′_{λ}|^{p}}{p} + λ^{\frac{−pq}{q−p+1}} F (λ^{\frac{p}{q−p+1}} u_{λ}) ≤ λ^{\frac{−pq}{q−p+1}} F (λ^{\frac{p}{q−p+1}} ).

Since F (u) ≥ −L (proved in introduction) then we get

\frac{(p − 1)|u′_{λ}|^{p}}{p} ≤ \frac{1}{q + 1} + λ^{\frac{−p(q+1)}{q−p+1}} G(λ^{\frac{p}{q−p+1}} ) + Lλ^{\frac{−pq}{q−p+1}} .

By (3.7) we see that

(3.8)  \frac{(p − 1)|u′_{λ}|^{p}}{p} ≤ \frac{2}{q + 1}

for large λ. Hence, |u′_{λ}| is bounded independent of r and for large λ. By Arzela-Ascoli’s theorem and by a standard diagonal argument there is a subsequence of u_{λ}(r), denoted by u_{λ_{k}}(r), such that

lim_{k→∞} u_{λ_{k}}(r) = v(r)

EJQTDE, 2008 No. 20, p. 10 uniformly on compact subsets of ℝ and v is continuous. End of proof of Claim.

We have

(3.9)  r^{N−1}|u′_{λ}|^{p−2}u′_{λ} = − ∫_{0}^{r} s^{N−1}λ^{\frac{−pq}{q−p+1}} f(λ^{\frac{p}{q−p+1}} u_{λ}(s))ds

r^{N−1}|u′_{λ}|^{p−2}u′_{λ} = − ∫_{0}^{r} s^{N−1} [|u_{λ}|^{q−1}u_{λ} + λ^{\frac{−pq}{q−p+1}} g(λ^{\frac{p}{q−p+1}} u_{λ}(s))] ds

also,

(3.10)  u′_{λ_{k}} = −Φ_{p′} (\frac{1}{r^{N−1}} ∫_{0}^{r} s^{N−1} [|u_{λ_{k}}|^{q−1}u_{λ_{k}} + λk^{\frac{−pq}{q−p+1}} g(λk^{\frac{p}{q−p+1}} )u_{λ_{k}}] ds) .

Since u_{λ_{k}}(r) → v(r) uniformly on compact subsets of ℝ and using (H3), gives

lim_{k→∞} u′_{λ_{k}} = −Φ_{p′} (\frac{1}{r^{N−1}} ∫_{0}^{r} s^{N−1}|v|^{q−1}vds)

≡ ϕ.

Hence, u′_{λ_{k}} → ϕ (pointwise) and since v is continuous it follows that ϕ is continuous. We also have

u_{λ_{k}} = 1 + ∫_{0}^{r} u′_{λ_{k}}ds.

Since u_{λ_{k}} → v uniformly, and u′_{λ_{k}} → ϕ pointwise, and by (3.8), u′_{λ_{k}} is uniformly bounded say by, M, applying dominated convergence theorem we get

v(r) = 1 + ∫_{0}^{r} ϕ(s) ds.

So,

v′ = ϕ.

Thus, from (3.10) we see that

v′ = −Φ_{p′} (\frac{1}{r^{N−1}} ∫_{0}^{r} s^{N−1} |v|^{q−1} v ds) .

Hence,

−r^{N−1}|v′|^{p−2}v′ = ∫_{0}^{r} s^{N−1}|v|^{q−1}v ds.

Note that v(0) = 1, v′(0) = 0. Hence, v ∈ C^{1}[0, ∞) and v satisfies (3.4)-(3.5) for 1 < p < q + 1 < \frac{Np}{N−p}. □

As u_{λ_{k}} converges to v uniformly on compact subsets of ℝ, so now we look for zeros of v. This is done in two steps. In step one we show v has a zero and in step two we show v has infinitely many zeros. The following lemma is technical and we use the result in the subsequent lemma.

Lemma 3.2. Let v solve (3.4)-(3.5). If 1 < p < q + 1 < \frac{Np}{N−p} and if v > 0, then

∫_{0}^{∞} s^{N−1}v^{q+1}ds < ∞.

Proof. By Lemma 3.1, we know that v is continuous and hence bounded on any compact set so to prove this lemma it is sufficient to show ∫_{1}^{∞} s^{N−1}v^{q+1}ds < ∞. We have

−r^{N−1}|v′|^{p−2}v′ = ∫_{0}^{r} s^{N−1}|v|^{q−1}v ds

and v > 0. So v′ < 0 and so v is decreasing. Therefore,

r^{N−1}|v′|^{p−1} = ∫_{0}^{r} s^{N−1}v^{q}ds

≥ v(r)^{q} ∫_{0}^{r} s^{N−1}ds

= \frac{v^{q}r^{N}}{N} .

Thus,

|v′|^{p−1} ≥ \frac{v^{q}r}{N}

−v′ = |v′| ≥ \frac{r^{\frac{1}{p−1}} v^{\frac{q}{p−1}}}{N^{\frac{1}{p−1}}}

\frac{−v′}{v^{\frac{q}{p−1}}} ≥ \frac{r^{\frac{1}{p−1}}}{N^{\frac{1}{p−1}}} .

Integrating this on (0, r), gives

[\frac{−v^{\frac{−q}{p−1} +1}}{\frac{−q}{p−1} + 1} ]_{0}^{r} ≥ ∫_{0}^{r} \frac{s^{\frac{1}{p−1}}}{N^{\frac{1}{p−1}}} ds

further simplification gives

[\frac{(p − 1)v^{\frac{p−q−1}{p−1}}}{q − p + 1} ]_{0}^{r} ≥ \frac{(p − 1)r^{\frac{p}{p−1}}}{pN^{\frac{1}{p−1}}} .

Since by assumption \frac{q−p+1}{p−1} > 0, multiplying both sides with \frac{q−p+1}{p−1} leads to

[v^{\frac{p−1−q}{p−1}} ]_{0}^{r} ≥ \frac{(q − p + 1)r^{\frac{p}{p−1}}}{pN^{\frac{1}{p−1}}} .

Thus,

v(r)^{\frac{p−1−q}{p−1}} − 1 ≥ C r^{\frac{p}{p−1}} ,

where C = \frac{q − p + 1}{pN^{\frac{1}{p−1}}}. So,

\frac{1}{v^{\frac{q+1−p}{p−1}}} = v^{\frac{p−1−q}{p−1}} ≥ 1 + C r^{\frac{p}{p−1}} ≥ C r^{\frac{p}{p−1}} .

Thus,

v^{\frac{q+1−p}{p−1}} ≤ C_{1} r^{\frac{−p}{p−1}} .

So,

v ≤ C_{1} r^{\frac{−p}{q+1−p}} .

Thus we see that

∫_{1}^{∞} s^{N−1}v^{q+1}ds ≤ C_{1}^{q+1} ∫_{1}^{∞} s^{N−1− \frac{p(q+1)}{q+1−p}} ds < ∞.

The last inequality is due to our assumption that 1 < p < q + 1 < \frac{Np}{N−p}. □

Lemma 3.3. Let v be a solution of (3.4)-(3.5). Then v has a zero.

EJQTDE, 2008 No. 20, p. 12 Proof. To prove this lemma, we use an idea of paper [3]. Suppose v > 0 for all r, and consider integrating

(r^{N−1}vv′|v′|^{p−2})′ = r^{N−1}|v′|^{p} − r^{N−1}v^{q+1}

on (0, r), which leads to

r^{N−1}vv′|v′|^{p−2} = ∫_{0}^{r} s^{N−1}|v′|^{p}ds − ∫_{0}^{r} s^{N−1} v^{q+1} ds.

After rearranging terms, we have

(3.11)  −r^{N−1}vv′|v′|^{p−2} + ∫_{0}^{r} s^{N−1}|v′|^{p} ds = ∫_{0}^{r} s^{N−1} v^{q+1} ds.

Since v > 0, v′ < 0, and since p < q + 1, it follows from (3.11) and Lemma 3.2 that

(3.12)  ∫_{0}^{∞} s^{N−1}|v′|^{p} ≤ ∫_{0}^{∞} s^{N−1}v^{q+1}ds < ∞.

Then using (3.12) in (3.11) and taking the limit as r → ∞, gives

(3.13)  − lim_{r→∞} r^{N−1}vv′|v′|^{p−2} exists and is finite.

Now integrating the following identity

(\frac{(p − 1) r^{N} |v′|^{p}}{p} + \frac{r^{N}v^{q+1}}{q + 1} )′ = \frac{−(N − p)|v′|^{p} r^{N−1}}{p} + \frac{Nr^{N−1} v^{q+1}}{q + 1}

on (0, r), gives

(3.14)  (\frac{(p − 1) r^{N} |v′|^{p}}{p} + \frac{r^{N}v^{q+1}}{q + 1} ) = ∫_{0}^{r} \frac{−(N − p)|v′|^{p}s^{N−1}}{p} ds + ∫_{0}^{r} \frac{Ns^{N−1}v^{q+1}}{q + 1} ds.

Then by (3.12), both the integrals on the right hand side of (3.14) converge, hence

lim_{r→∞} \frac{(p − 1) r^{N} |v′|^{p}}{p} + \frac{r^{N}v^{q+1}}{q + 1}

exists. Denote

h(r) = \frac{(p − 1) r^{N} |v′|^{p}}{p} + \frac{r^{N}v^{q+1}}{q + 1} .

We have shown that lim_{r→∞} h(r) = l for some l ≥ 0. Then by (3.12),

∫_{0}^{∞} \frac{h(s)}{s}ds < ∞.

Thus, it follows that l = 0, so that

lim_{r→∞} \frac{(p − 1) r^{N} |v′|^{p}}{p} + \frac{r^{N}v^{q+1}}{q + 1} = 0.

Then taking the limit as r → ∞ in (3.14) gives

0 = ∫_{0}^{∞} \frac{−(N − p)|v′|^{p}s^{N−1}}{p} ds + ∫_{0}^{∞} \frac{Ns^{N−1}v^{q+1}}{q + 1} ds.

So,

∫_{0}^{∞} s^{N−1}|v′|^{p}ds = \frac{Np}{(N − p)(q + 1)} ∫_{0}^{∞} s^{N−1}v^{q+1}ds.

But by (3.12) we have

∫_{0}^{∞} s^{N−1}|v′|^{p} ≤ ∫_{0}^{∞} s^{N−1}v^{q+1}ds.

EJQTDE, 2008 No. 20, p. 13 So it follows that

\frac{Np}{(N − p)(q + 1)} ≤ 1.

This contradicts our assumption that q + 1 < \frac{Np}{N−p}. So, v is not positive for all r. Hence, v has a zero. □

Lemma 3.4. Let v be the solution of (3.4)-(3.5). Then v has infinitely many zeros.

Proof. We have from the above lemma that there exists a z_{1} such that v > 0 on [0, z_{1}) and v(z_{1}) = 0. So after z_{1} we have two cases, Case(i): v has a first local minimum call it m_{1} > z_{1}, or Case (ii): v′ < 0 for all r > z_{1}. We want to show that the Case(ii) is not possible. Suppose v′ < 0 for all r > 0. Then

E ≡ \frac{(p − 1)|v′|^{p}}{p} + \frac{1}{q + 1}|v|^{q+1} ≥ 0

and E′ ≤ 0 so

\frac{1}{q + 1}|v|^{q+1} ≤ E(r, d) ≤ E(0, d) = \frac{1}{q + 1}.

Thus |v| ≤ 1. So v is bounded and v′ < 0 and thus lim_{r→∞} v = J. Also since E is bounded and since

E′ ≤ 0, so lim_{r→∞} E(r, d) exists and thus lim_{r→∞} v′(r) exists.

Claim: lim_{r→∞} v′(r) = 0.

Proof of Claim: Suppose not, which means −v′(r) > m > 0 for large r. Then integrating from (0, r), gives

−v(r) + v(0) > mr.

Taking the limit as r → ∞, we see that −v is unbounded, which contradicts our assumption that v is bounded. So, we have the claim. End of proof of Claim.

Consider dividing (3.4) by r^{N} and then taking the limit as r → ∞ and using the above claim, gives

0 = lim_{r→∞} \frac{− ∫_{0}^{r} t^{N−1}|v|^{q−1}v dt}{r^{N}}.

Applying L’Hopital’s rule on right hand side and using lim_{r→∞} v(r) = J < 0 gives

0 = \frac{−|J|^{q−1}J}{N} .

This contradicts our assumption that J < 0. So Case(ii) is not possible.

Hence, v has a first local minimum call it m_{1}, where m_{1} > z_{1}, and let v_{1} = v(m_{1}) < 0. Now v satisfies

r^{N−1}|v′|^{p−2}v′ = − ∫_{m_{1}}^{r} t^{N−1}|v|^{q−1}v dt

and

v(m_{1}) = v_{1}.

We may now use the same argument as in Lemma 3.3 to show that v has a second zero at z_{2} > z_{1}. Proceeding inductively, we can show that v has infinitely many zeros. □

As u_{λ} → v on any fixed compact set when λ is large, this means that the graph of u_{λ} is uniformly close to the graph of v. Since v has infinitely many zeros, suppose the first ρ zeros of v are on [0, K] for K > 0. By uniform convergence on compact subsets u_{λ} will have at least ρ zeros on [0, K + 1] for large λ. By (3.1), u_{λ}(r) = λ^{\frac{−p}{q−p+1}} u (\frac{r}{λ}) , so u will have at least ρ zeros on [0, ∞). So now we are ready to shift gears from v to u.

EJQTDE, 2008 No. 20, p. 14

The following lemma is technical and we mimic the idea from [4] and we do necessary changes to fit our case.

Lemma 3.5. Let u(r, d) be the solution of (2.1)-(2.2). Let us suppose that u(r, d^{∗}) has exactly k zeros and u(r, d^{∗}) → 0 as r → ∞. If |d − d^{∗}| is sufficiently small, then u(r, d) has at most k + 1 zeros on [0, ∞).

Proof. Our goal is to show that for d close to d^{∗}, u(., d) has at most (k + 1) zeros in [0, ∞). So we suppose there is a sequence of values d_{j} converging to d^{∗} and such that u(., d_{j}) has at least (k + 1) zeros on [0, ∞) (if there is no such sequence, we are done). We write u_{j}(r) = u(r, d_{j}) and we denote by z_{j} the (k + 1)st zero of u_{j}, counting from the smallest. We will show that if u_{j} has a (k + 2)nd zero, then u(r, d^{∗}) is going to have a (k + 1)st zero, which is a contradiction.

First we show that u(r, d_{j}) → u(r, d^{∗}) and u′(r, d_{j}) → u′(r, d^{∗}) on compact subsets of [0, ∞) as d_{j} → d^{∗} and j → ∞. We prove this in two claims.

Claim 1: If lim_{j→∞} d_{j} = d^{∗}, then |u(r, d_{j})| ≤ M_{1} and |u′(r, d_{j})| ≤ M_{2} for some M_{1}, M_{2} > 0 for all j.

Proof of Claim 1: We use the fact from (2.6) and (2.7) that energy is decreasing and hence E is bounded by E(0, d_{j}) = F (d_{j}), we can write the energy at r as the following

E(r, d_{j}) = \frac{(p − 1)|u′(r, d_{j})|^{p}}{p} + F (u(r, d_{j})) ≤ F (d_{j}) ≤ F (d^{∗}) + 1

for large j. Also, by (1.9), F (u) ≥ −L thus

\frac{(p − 1)|u′(r, d_{j})|^{p}}{p} ≤ F (d^{∗}) + 1 + L ≤ C

for large j and for some C > 0. Thus, for j large, |u′(r, d_{j})| ≤ M_{2} for some M_{2} > 0. Also, note that since lim_{r→∞} E(r, d^{∗}) exists and since lim_{r→∞} u(r, d^{∗}) = 0 it follows that

F (d^{∗}) = E(0, d^{∗}) > lim_{r→∞} E(r, d^{∗}) ≥ 0.

Thus, F (d^{∗}) > 0. Hence by (H4) and (H5), d^{∗} > γ. By lemma 2.4, |u(r, d_{j})| ≤ d_{j} and since lim_{j→∞} d_{j} = d^{∗}

we have |u(r, d_{j})| ≤ d^{∗} + 1 = M_{1} for large j. End of proof of Claim 1.

Claim 2: u(r, d_{j}) → u(r, d^{∗}) and u′(r, d_{j}) → u′(r, d^{∗}) uniformly on compact subsets of [0, ∞) as j → ∞.

Proof of Claim 2: By Claim 1, |u(r, d_{j})| ≤ M_{1} and |u′(r, d_{j})| ≤ M_{2}. So the u(r, d_{j}) are bounded and equicontinuous. Then by Arzela-Ascoli’s theorem we have a subsequence (still denoted by d_{j}) such that u(r, d_{j}) → u(r, d^{∗}) uniformly on compact subsets of [0, ∞) as j → ∞. Then by (2.1) and since u_{j} → u uniformly on compact subsets of [0, ∞) we have

lim_{j→∞} |u′_{j}|^{p−2}u′_{j} = lim_{j→∞} \frac{−1}{r^{N−1}} ∫_{0}^{r} s^{N−1}f(u_{j})ds

= \frac{−1}{r^{N−1}} ∫_{0}^{r} s^{N−1}f(u)ds.

Therefore, |u′_{j}|^{p−2}u′_{j} converges uniformly on compact subsets of [0, ∞). Thus, u′_{j}(r) converges uniformly say to g(r). We now show that g(r) = u′(r, d^{∗}). Integrating on (0, r), gives

lim_{j→∞} ∫_{0}^{r} u′_{j}(t)dt = ∫_{0}^{r} g(t)dt

lim_{j→∞}(u_{j}(r) − u_{j}(0)) = ∫_{0}^{r} g(t)dt.

EJQTDE, 2008 No. 20, p. 15 Since u_{j}(r, d_{j}) → u(r, d^{∗}), we get

u(r, d^{∗}) − u(0, d^{∗}) = ∫_{0}^{r} g(t)dt.

Differentiating this we get u′(r, d^{∗}) = g(r) = lim_{j→∞} u′_{j}. End of proof of Claim 2.

Let t_{j} be the (k + 2)nd zero of u_{j}. Then there exists an l_{j} such that z_{j} < l_{j} < t_{j} and l_{j} is a local extremum. So by Lemma 2.6

F (u(l_{j})) = E(l_{j}, d_{j}) ≥ E(t_{j}, d_{j}) = \frac{p − 1}{p}|u′(t_{j})|^{p} > 0.

Then by (H5), |u(l_{j})| > γ. Now let b_{j} be the smallest number greater than z_{j} such that |u_{j}(b_{j})| = α. Let a_{j} be the smallest number greater than z_{j} such that |u_{j}(a_{j})| = \frac{α}{2}. Let m_{j} be the local extrema between the kth and (k + 1)st zeros of u_{j}. So we have m_{j} < z_{j} < a_{j} < b_{j}. Since the energy is decreasing we have E(z_{j}, d_{j}) ≤ E(m_{j}, d_{j}). Since u′_{j}(m_{j}) = 0, u_{j}(z_{j}) = 0, F (u_{j}(z_{j})) = 0, and by Lemma 2.6, we have

0 < \frac{(p − 1)|u′_{j}(z_{j})|^{p}}{p} ≤ F (u_{j}(m_{j})).

Thus, |u_{j}(m_{j})| > γ. So there exists a largest number q_{j} less than z_{j} such that |u_{j}(q_{j})| = γ. Note m_{j} < q_{j} < z_{j} < a_{j} < b_{j} < l_{j} < t_{j}.

Claim 3: b_{j}−a_{j} ≥ K_{1} > 0, where K_{1} is independent of j for sufficiently large j. Also, ξ_{2}−ξ_{1} ≥ K_{2} > 0 where ξ_{1} and ξ_{2} are two consecutive zeros of u_{j}.

Proof of Claim 3: Since the energy is decreasing and since d_{j} → d^{∗} for j large we have

\frac{p − 1}{p}|u′_{j}|^{p} + F (u_{j}) ≤ F (d_{j}) ≤ F (d^{∗}) + 1

for large j. Rewriting this inequality gives

(3.15)  \frac{|u′_{j}|}{(F (d^{∗}) + 1 − F (u_{j}))^{\frac{1}{p}}} ≤ (\frac{p}{p − 1})^{\frac{1}{p}} .

So integrate (3.15) on (a_{j}, b_{j}) to get

∫ α_{\frac{α}{2}} \frac{dt}{(F (d^{∗}) + 1 − F (t))^{\frac{1}{p}}} = ∫_{a_{j}}^{b_{j}} \frac{|u′_{j}|}{(F (d^{∗}) + 1 − F (u_{j}))^{\frac{1}{p}}} ds ≤ (\frac{p}{p − 1})^{\frac{1}{p}} (b_{j} − a_{j}).

So letting

(3.16)  K_{1} ≡ (\frac{p − 1}{p})^{\frac{1}{p}} ∫ α_{\frac{α}{2}} \frac{dt}{(F (d^{∗}) + 1 − F (t))^{\frac{1}{p}}}

we see that K_{1} ≤ b_{j} − a_{j} for all j.

Turning to the second part of the claim, using Lemma 2.5, let m be the extremum between ξ_{1} and ξ_{2}. Let us integrate (3.15) on (ξ_{1}, m) and using (3.15) and that |u(m)| > γ (by Lemma 2.5) gives

(\frac{p − 1}{p})^{\frac{1}{p}} ∫_{0}^{γ} \frac{dt}{(F (d^{∗}) + 1 − F (t))^{\frac{1}{p}}} ≤ (\frac{p − 1}{p})^{\frac{1}{p}} ∫_{0}^{|u(m)|} \frac{dt}{(F (d^{∗}) + 1 − F (t))^{\frac{1}{p}}}

= (\frac{p − 1}{p})^{\frac{1}{p}} ∫_{ξ_{1}}^{m} \frac{|u′_{j}|}{(F (d^{∗}) + 1 − F (t))^{\frac{1}{p}}} ds

≤ m − ξ_{1}.

EJQTDE, 2008 No. 20, p. 16 Similarly on [m, ξ_{2}] we have,

(\frac{p − 1}{p})^{\frac{1}{p}} ∫_{0}^{γ} \frac{dt}{(F (d^{∗}) + 1 − F (t))^{\frac{1}{p}}} ≤ ξ_{2} − m.

So,

(3.17)  K_{2} ≡ 2 (\frac{p − 1}{p})^{\frac{1}{p}} ∫_{0}^{γ} \frac{dt}{(F (d^{∗}) + 1 − F (t))^{\frac{1}{p}}} ≤ ξ_{2} − ξ_{1}.

End of proof of Claim 3.

In particular u_{j} → u uniformly on [0, y^{∗} + \frac{K_{2}}{2} ] where y^{∗} is the kth zero of u(r, d^{∗}). Along with

Lemma 2.6 and the previous claim, it follows that for large j, u_{j} has exactly k zeros on [0, y^{∗} + \frac{K_{2}}{2} ] .

Let y_{j} be the kth zero of u_{j}, then by Claim 2, u_{j}(r, d_{j}) → u(r, d^{∗}) as j → ∞ on [0, y^{∗} + \frac{K_{2}}{2} ] , so

y_{j} → y^{∗} as j → ∞.

Claim 4: z_{j} → ∞ as j → ∞.

Proof of Claim 4: Suppose not, that is if |z_{j}| ≤ A then there exists a subsequence j_{k} such that z_{j_{k}} → z and u(r, d_{j_{k}}) → u(r, d^{∗}) on [0, A] which in turn implies

0 = u(z_{j_{k}}, d_{j_{k}}) → u(z, d^{∗}).

Since z_{j_{k}} > y_{j_{k}} and y_{j_{k}} → y^{∗} as j → ∞, then z ≥ y^{∗}. On the other hand, u(r, d^{∗}) has exactly k zeros, therefore z = y^{∗}. Thus u_{j}(y_{j}) = 0 = u_{j}(z_{j}). By the mean value theorem, u′_{j}(w_{j}) = 0 for some w_{j} with y_{j} ≤ w_{j} ≤ z_{j}. Since u_{j} → u uniformly on [0, A] and y_{j} → y^{∗} ← z_{j}, so taking the limit gives u′(y^{∗}) = 0, but by Lemma 2.6, this implies u ≡ 0. This is a contradiction to our assumption that u has exactly k zeros. End of Claim 4.

Now let us show that the q_{j} are bounded as j → ∞. Since u_{j} → u and u′_{j} → u′ uniformly on compact subsets of [y^{∗}, m^{∗} + 1], where m^{∗} is the local extremum of u(r, d^{∗}) that occurs after y^{∗}, we see that u′_{j} must be zero on [y^{∗}, m^{∗} + 1] for j large. Thus there exists an m_{j} with y_{j} < m_{j} < m^{∗} + 1 such that u′_{j}(m_{j}) = 0.

Next, we estimate q_{j} − m_{j} on [m_{j}, q_{j}], since u ≥ γ > β on [m_{j}, q_{j}] so we have f(u) ≥ C > 0. So

−r^{N−1}|u′_{j}|^{p−2}u′_{j} = − ∫_{m_{j}}^{r}(r^{n−1}|u′_{j}|^{p−2}u′_{j})′ dt = ∫_{m_{j}}^{r} r^{N−1}f(u_{j}) dt ≥ \frac{C(r^{N} − m_{j}^{N})}{N} ≥ \frac{C}{N} (r − m_{j})r^{N−1}.

So we have

−|u′_{j}|^{p−2}u′_{j} ≥ \frac{C}{N} (r − m_{j}).

Further simplification and integrating on [m_{j}, q_{j}] gives

d_{j} − γ ≥ u(m_{j}) − γ = ∫_{m_{j}}^{q_{j}} u′_{j} dt ≥ (\frac{C}{N} )^{\frac{1}{p−1}} ∫_{m_{j}}^{q_{j}}(r − m_{j})^{\frac{1}{p−1}} dt.

Now using Lemma 2.4 and the fact that j is large gives

d^{∗} + 1 ≥ d_{j} − γ ≥ (\frac{C}{N} )^{\frac{1}{p−1}} (\frac{p − 1}{p}) (q_{j} − m_{j})^{\frac{p}{p−1}}

for large j. As we saw in a previous paragraph that m_{j} are bounded by m^{∗} + 1, it follows that q_{j} are bounded.

Claim 5: For sufficiently large j, |u_{j}(r)| < γ for all r > z_{j}.

EJQTDE, 2008 No. 20, p. 17

Proof of Claim 5: Suppose on the contrary that there is a smallest c_{j} > z_{j} such that |u_{j}(c_{j})| = γ. Thus, on (z_{j}, c_{j}) we have 0 < |u_{j}| < γ. Hence, F (u_{j}) ≤ 0 on (z_{j}, c_{j}). So there exists an a_{j} and a b_{j} such that z_{j} < a_{j} < b_{j} < c_{j} with |u_{j}(a_{j})| = \frac{α}{2}, |u_{j}(b_{j})| = α. Also, F is decreasing on [\frac{α}{2}, α], so that F (\frac{α}{2}) ≥ F (u_{j}) for all \frac{α}{2} ≤ u_{j} ≤ α.

Now integrating the following identity on (q_{j}, c_{j})

(r^{\frac{p(N−1)}{p−1}} E)′ = \frac{p(N − 1)F (u_{j}(r))r^{\frac{Np−2p+1}{p−1}}}{p − 1}

and since |u_{j}(q_{j})| = |u_{j}(c_{j})| = γ and F (u_{j}(q_{j})) = 0 = F (u_{j}(c_{j})), gives

(3.18) 0 ≤ \frac{cj^{\frac{p(N−1)}{p−1}} |u′_{j}(c_{j})|^{p}(p − 1)}{p} = \frac{qj^{\frac{p(N−1)}{p−1}} |u′_{j}(q_{j})|^{p}(p − 1)}{p} + ∫_{q_{j}}^{c_{j}} \frac{p(N − 1)F (u_{j}(r))t^{\frac{Np−2p+1}{p−1}}}{p − 1} dt.

Since q_{j} is bounded, for an appropriate subsequence q_{j} → q^{∗} where u(q^{∗}, d^{∗}) = γ and since u′_{j} → u′ uniformly on [0, q^{∗} + 1], then u′_{j}(q_{j}) → u′(q^{∗}, d^{∗}). Hence

(3.19)  lim_{j→∞} inf ∫_{q_{j}}^{c_{j}} t^{\frac{(N−2)p+1}{p−1}} F (u_{j}(r))dt ≥ − \frac{(q^{∗})^{\frac{p(N−1)}{p−1}} |u′(q^{∗}, d^{∗})|^{p}(p − 1)}{p} > −∞.

Also, since z_{j} → ∞ and since z_{j} < a_{j} < b_{j}, so a_{j} → ∞.

On other hand, by Claim 3

∫_{q_{j}}^{c_{j}} t^{\frac{(N−2)p+1}{p−1}} F (u_{j}(t))dt ≤ ∫_{a_{j}}^{b_{j}} t^{\frac{(N−2)p+1}{p−1}} F (u_{j}(t))dt

≤ F (\frac{α}{2}) (bj^{\frac{(N−1)p}{p−1}} − aj^{\frac{(N−1)p}{p−1}} ) (\frac{p − 1}{(N − 1)p})

≤ F (\frac{α}{2}) a_{j}^{\frac{(N−1)p}{p−1} −1}(b_{j} − a_{j}) (\frac{p − 1}{(N − 1)p})

≤ K_{1}F (\frac{α}{2}) a_{j}^{\frac{(N−1)p}{p−1} −1} (\frac{p − 1}{(N − 1)p}) → −∞

as j → ∞. (We obtain the last inequality by using K_{1} ≤ (b_{j} − a_{j}) from Claim 3 and also F (\frac{α}{2}) < 0.) Thus,

∫_{q_{j}}^{c_{j}} t^{\frac{(N−2)p+1}{p−1}} F (u_{j}(t))dt → −∞,

but this is a contradiction to (3.19). Hence, |u_{j}(r)| < γ for large j and for r > z_{j}. End of proof of Claim 5.

Now suppose u_{j} has another zero, call it t_{j} > z_{j}. Then there is a local extrema for u_{j} at a value s_{j} such that z_{j} < s_{j} < t_{j}. Since the energy is decreasing, we have E(t_{j}) ≤ E(s_{j}). By Lemmas 2.5 and 2.6 we have

0 < \frac{(p − 1)|u′_{j}(t_{j})|^{p}}{p} ≤ F (u_{j}(s_{j})).

Thus |u_{j}(s_{j})| > γ (by (H5)). By Claim 5, for sufficiently large j and for all r > z_{j} we have |u_{j}(r)| < γ. In particular |u_{j}(s_{j})| < γ, a contradiction. Hence, there is no zero of u_{j} larger than z_{j}. □

4. Proof of Main Theorem

To prove the Main Theorem we construct the following sets such that u has any prescribed number of zeros.

Let S_{k} = {d ≥ γ | u(r, d) has exactly k zeros for r ≥ 0 } and let d_{k} = sup S_{k}.

We will then show that S_{k} for k ≥ 0 is nonempty and bounded above.

Let S_{0} = { d ≥ γ | u(r, d) > 0 for all r ≥ 0 }.

Claim: γ ∈ S_{0}.

Proof of Claim: If d = γ, then u(0) = γ > 0. So the energy at r = 0 is

E(0, γ) = \frac{p − 1}{p}|u′(0)|^{p} + F (u(0)) = 0.

So E < 0 for r > 0; for if there is an r_{1} > 0 such that E(r_{1}, γ) = 0 then E ≡ 0 on [0, r_{1}], this implies u ≡ 0 on [0, r_{1}], but u(0) = γ > 0. Thus E < 0 for r > 0. If there exists an r_{2} such that u(r_{2}) = 0 then

E(r_{2}, d) = \frac{p − 1}{p}|u′(r_{2})|^{p} ≥ 0

contradicting E < 0 for all r > 0. Therefore, u(r, γ) > 0 for all r ≥ 0. Hence γ ∈ S_{0}. End of proof of Claim.

Lemma 4.1. S_{0} ≠ ∅ and S_{0} is bounded above.

Proof. S_{0} is nonempty by the above Claim and S_{0} is bounded above by the Lemmas 3.1 and 3.3. □

Let d_{0} = sup S_{0}. Since d > γ for all d ∈ S_{0} so d_{0} > γ > 0.

Now our goal is to show that u(r, d_{0}) > 0 and that u(r, d_{0}) satisfies (1.4). As d_{0} is the supremum of S_{0} we expect u(r, d_{0}) > 0. We prove this in two lemmas. In the first lemma we show u(r, d_{0}) ≥ 0 and in the second lemma we show u(r, d_{0}) > 0.

Lemma 4.2. u(r, d_{0}) ≥ 0 for all r ≥ 0.

Proof. If u(r_{0}, d_{0}) < 0 for some r_{0}, then by continuity with respect to initial conditions on compact sets for d close to d_{0} and d < d_{0} , we have u(r_{0}, d) < 0. This contradicts the definition of S_{0}. □

Lemma 4.3. u(r, d_{0}) > 0.

Proof. Suppose there exists an r_{1} such that u(r_{1}, d_{0}) = 0. By Lemma 4.2, we know u(r, d_{0}) ≥ 0. So u(r, d_{0}) has a minimum at r_{1} and also since u ∈ C^{1}[0, ∞), this implies u′(r_{1}) = 0. Then by Lemma 2.6, u ≡ 0 which is a contradiction to u(0) = d_{0} ≠ 0.

□

Let d > d_{0}. Then u(r, d) has at least one zero, otherwise d would be in S_{0} which it is not. Moreover, as d approaches d_{0} from above, we expect that the first zero of u, z_{1}(d), should go to infinity. This is shown in the following lemma.

Lemma 4.4. lim_{d→d_{0}^{+}} z_{1}(d) = ∞.

Proof. Suppose lim_{d→d_{0}^{+}} z_{1}(d) = z_{d_{0}} < ∞. Since u(r, d) → u(r, d_{0}) uniformly on compact subsets as d → d_{0},

this implies u(z_{d_{0}}) = lim_{d→d_{0}} u(z_{1}(d), d), and which in turn implies u(z_{d_{0}}, d_{0}) = 0. However, by Lemma

4.3, u(r, d_{0}) > 0, which is a contradiction. □ EJQTDE, 2008 No. 20, p. 19

Next we want to show the energy E(r, d_{0}) ≥ 0. This is crucial, as if E(r, d_{0}) < 0 at some point, say n_{1}, then u will not have any zeros after n_{1}, and also u will not decay as r → ∞. So we have the following lemma.

Lemma 4.5. E(r, d_{0}) ≥ 0 for all r ≥ 0.

Proof. If E(r_{0}, d_{0}) < 0 then by continuity E(r_{0}, d) < 0 for d > d_{0} and d close to d_{0}. On other hand, since d > d_{0} then u(r, d) has a first zero, z_{1}(d), (F (u(z_{1}(d))) = 0) so the energy is

E(z_{1}(d), d) = \frac{p − 1}{p}|u′|^{p} ≥ 0.

But since E′ ≤ 0, we must have that z_{1}(d) ≤ r_{0}. This contradicts Lemma 4.4. Hence the result follows. □

Lemma 4.6. u′(r, d_{0}) < 0 on (0, ∞).

Proof. Since u(0) = d and u′(0) = 0, first we want to show that u is decreasing on (0, ϵ) for some ϵ > 0. Dividing both sides of (2.1) by r^{N} and then taking the limit as r → 0, and applying L’Hopital’s rule, gives

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

The last inequality is true since by the definition of S_{0}, we have that d_{0} > γ and then by (H5), γ > β where β is the largest zero of f. Thus, f(d_{0}) > 0. So, u′ < 0 on (0, ϵ) for some ϵ > 0.

Let [0, R_{d_{0}}] be the maximal interval so that u′ < 0 on (0, R_{d_{0}}). If R_{d_{0}} = ∞, then u′ < 0 on (0, ∞) and we are done. Otherwise R_{d_{0}} < ∞ and u′(R_{d_{0}}) = 0.

Claim: 0 < u(R_{d_{0}}) ≤ β.

Proof of Claim: Suppose f(u(R_{d_{0}})) > 0 and let us look at the following identity:

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

Using u′(R_{d_{0}}) = 0, this gives

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

for r < R_{d_{0}} and r close to R_{d_{0}}. We get the last inequality since f(u(R_{d_{0}})) > 0, and by continuity, f(u) > 0 for r near R_{d_{0}}. This implies u′ > 0 on (r, R_{d_{0}}) for r close to R_{d_{0}}. But by assumption u′ < 0 on (0, R_{d_{0}}). Hence, f(u(R_{d_{0}})) ≤ 0 and since we also know u(R_{d_{0}}) > 0, this implies 0 < u(R_{d_{0}}) ≤ β. End of proof of Claim.

The previous claim implies F (u(R_{d_{0}})) < 0. Since u′(R_{d_{0}}) = 0 we obtain

E(R_{d_{0}}, d_{0}) = F (u(R_{d_{0}})) < 0,

which is a contradiction to Lemma 4.5. Hence, u′(r, d_{0}) < 0 for all r > 0. □

Since we now know that u(r, d_{0}) > 0 and u′(r, d_{0}) < 0, it follows that lim_{r→∞} u(r, d_{0}) exists.

Lemma 4.7. lim_{r→∞} u(r, d_{0}) = U ≥ 0 where U is some nonnegative zero of f.

Proof. Since E is decreasing and bounded below we see that lim_{r→∞} E(r, d_{0}) = E. Rewriting (2.6) we obtain:

\frac{p − 1}{p}|u′|^{p} = E(r, d_{0}) − F (u(r, d_{0})).

The limit of both terms on the right exists and so we have

lim_{r→∞} \frac{(p − 1)|u′|^{p}}{p} = E − F (U).

Thus, lim_{r→∞} |u′| exists.

Claim 1: lim_{r→∞} |u′| = 0.

Proof of Claim 1: Suppose not, then lim_{r→∞} |u′| = L > 0. So |u′(r)| > \frac{L}{2} > 0 if r ≥ R . Suppose

|u′(r)| = −u′(r), now integrating |u′(r)| > \frac{L}{2} > 0, gives

∫_{R}^{r} −u′(r)dt > ∫_{R}^{r} \frac{L}{2}dt

for r ≥ R, this implies

d_{0} ≥ u(R) ≥ u(R) − u(r) ≥ \frac{L}{2}(r − R) → ∞,

which is a contradiction. Hence, lim_{r→∞} |u′| = 0 and so lim_{r→∞} u′ = 0. End of proof of Claim 1.

Dividing both sides of (2.1) by r^{N} gives

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

Taking the limit as r → ∞, and then doing simplification on right hand side by L’Hopital’s rule, gives

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

So, f(U) = 0. □

Lemma 4.8. U = 0.

Proof. Taking the limit as r → ∞ in (2.1), gives E = F (U). By Lemma 4.4, E ≥ 0. Hence F (U) ≥ 0. Also by Lemma 4.7, f(U) = 0. Thus by (H5) and (H6), U ≡ 0. □

Let S_{1} = { d > d_{0} | u(r, d) has exactly one zero for all r ≥ 0 }.

Lemma 4.9. S_{1} ≠ ∅ and S_{1} is bounded above.

Proof. By Lemma 3.5, if d > d_{0} and d close to d_{0} then u(r, d) has at most one zero. Also, if d > d_{0} then d ∉ S_{0} so u(r, d) has at least one zero. Therefore, for d > d_{0} and d close to d_{0}, u(r, d) has exactly one zero. Hence S_{1} is nonempty. Also by Lemmas 3.1 - 3.4, S_{1} is bounded above. □

Define d_{1} = sup S_{1}.

As above we can show that u(r, d_{1}) has exactly one zero and u(r, d_{1}) → 0 as r → ∞.

Proceeding inductively, we can find solutions that tend to zero at infinity and with any prescribed number of zeros. Hence, we complete the proof of the main theorem.

Here is an example of a u that satisfies the hypotheses (H1)-(H6):

(4.1)  u′′ + \frac{2}{r}u′ + u^{3} − u = 0

where p = 2, N = 3, and f(u) = u^{3} − u. The graph of f(u) is EJQTDE, 2008 No. 20, p. 21 As F is the anti derivative of f, the graph of F is The only positive zero of F occurs at γ = \sqrt{2}.

Here are some graphs of solutions of (4.1) for different values of d, all graphs are generated numerically using Mathematica:

(a) Solution that remains positive when d = 1.4 < γ = \sqrt{2} :

(b) Solution with exactly one zero when d = 4.7 > γ = \sqrt{2} :

(c) Solution with exactly two zeros when d = 15.1 > γ = \sqrt{2} :

□ Acknowledgments

I thank my advisor, J. Iaia who suggested this problem to me and guided me in writing up this paper.

References

[1] A. CASTRO and A. KUREPA, Infinitely Many Radially Symmetric Solutions to a Superlinear Dirichlet Problem in a Ball, Proceedings of American Mathematical Society, Volume 101, Number 1, Sept 1987.

[2] B. FRANCHI, E. LANCONELLI, and J. SERRIN, Existence and Uniqueness of Non-negative Solutions of Quasilinear Equations in ℝ^{n}, Advances in Mathematics 118,177-243(1996), Article No. 0021.

[3] A. HARAUX and F.B. WEISSLER, Non-uniqueness for a Semilinear Initial Value Problem, Indiana University Mathematics Journal, Vol. 31, No. 2(1982).

[4] J. IAIA, and H. WARCHALL, Nonradial Solutions of a Semilinear Elliptic Equation in Two Dimensions, Journal of Differential Equations 119, No.2, 533-558(1995).

[5] K. McLEOD, W.C.TROY, AND F.B. WEISSLER, Radial solutions of Δu + f(u) = 0 with Prescribed Numbers of Zeros, Journal of Differential Equations 83,368-378(1990).

[6] W.M. NI and J. SERRIN, Nonexistence Theorems for Quasilinear Partial Differential Equations, Rend. Circolo Mat. Palermo, Series II, No. 8, 171-185(1985).

[7] W.M. NI and J. SERRIN, Existence and Nonexistence Theorems for Ground States for Quasilinear Partial Differential Equations, The anomalous case, Rome, Accad. Naz. dei Lincei, Atti dei Convegni, No. 77, 181-197(1986).

[8] E. CALZOLARI, R. FILIPPUCCI, and P. PUCCI, Existence of Radial Solutions for the p-Laplacian Elliptic Equations With Weights, Discrete and Continuous Dynamical Systems, Volume 15, 447-479(2006)

[9] F. GAZZOLA and J. SERRIN and M. TANG, Existence of Ground States And Free Boundary Problems For Quasilinear Elliptic Operators, Advances in Differential Equations, Volume 5, 1-30(2000)

(Received August 10, 2007)

Department of Mathematics, University of North Texas, Denton, Texas-76203-1430

   E-mail address: sridevi@unt.edu, sridevi.pudipeddi@gmail.com