Electronic Journal of Qualitative Theory of Differential Equations 2008, No. 38, 1-13; http://www.math.u-szeged.hu/ejqtde/

RADIAL SOLUTIONS TO A SUPERLINEAR DIRICHLET PROBLEM USING BESSEL FUNCTIONS

JOSEPH IAIA* AND SRIDEVI PUDIPEDDI**

Abstract. We look for radial solutions of a superlinear problem in a ball. We show that for if n is a sufficiently large nonnegative integer, then there is a solution u which has exactly n interior zeros. In this paper we give an alternate proof to that which was given in [1].

1. Introduction

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

(1.1)  \begin{cases}Δu + f(u) = g(|x|) for x ∈ Ω \\ u = 0 for x ∈ ∂Ω, \end{cases}

for N ≥ 2 and where Ω is the ball of radius T > 0 centered at the origin in ℝ^{N}, Δ is the Laplacian operator, and f : ℝ → ℝ is a continuous function and where g ∈ ℂ^{1}[0, T].

Motivation: A. Castro and A. Kurepa proved existence of solutions of (1.1) for a wide variety of nonlinearities, f. See [1]. In this paper we give an alternate and, in our estimation, a somewhat easier proof of this result by approximating solutions of (1.1) with appropriate linear equations. In a groundbreaking paper in 1979, B. Gidas, W. Ni, and L. Nirenberg [2] proved that if Ω is a ball then all positive solutions of

Δu + f(u) = 0 in Ω
u = 0 on ∂Ω

are spherically symmetric. K. McLeod, W.C. Troy and F.B. Weissler studied the radial solutions of

Δu + f(u) = 0 in Ω

lim_{|x|→∞} u(x) = 0

for Ω ∈ ℝ^{N} in [3].

We assume the following hypotheses: (H1) f is a locally Lipschitz continuous function, f is increasing for large |u| and f(0) = 0.

(H2) lim_{|u|→∞} \frac{f(u)}{u} = ∞ (that is, f is superlinear).

Let F (u) = ∫_{0}^{u} f(s)ds and note that from (H2) it follows that

(1.2)  lim_{|u|→∞} \frac{F (u)}{u^{2}} = ∞.

(H3) There exists a k with 0 < k ≤ 1, such that

lim_{u→∞} (\frac{u}{f(u)})^{\frac{N}{2}} (NF (ku) − \frac{(N − 2)}{2}uf(u) − \frac{N + 2}{2}||g|| |u| − T ||g′|| |u|) = ∞

where || || is the supremum norm on [0, T]. (H3*) There exists a k with 0 < k ≤ 1, such that

lim_{u→−∞} (\frac{u}{f(u)})^{\frac{N}{2}} (NF (ku) − \frac{(N − 2)}{2}uf(u) − \frac{N + 2}{2}||g|| |u| − T ||g′|| |u|) = ∞.

EJQTDE, 2008 No. 38, p. 1 (H4) There exists an M > 0 such that

NF (u) − \frac{N − 2}{2}uf(u) − \frac{N + 2}{2}||g|| |u| − T ||g′|| |u| > −M

for all u.

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

(1.3)  u′′ + \frac{N − 1}{r}u′ + f(u) = g(r) for 0 < r < T
(1.4)  u′(0) = 0, u(T) = 0.

Main Theorem: If (H1)-(H4) are satisfied then (1.1) has infinitely many radially symmetric solutions with u(0) > 0. If in place of (H3) we have (H3*) then (1.1) has infinitely many radially symmetric solutions with u(0) < 0.

2. Preliminaries

The technique used to solve (1.3) - (1.4) is the shooting method. That is, we first look at the initial value problem

(2.1)  u′′ + \frac{N − 1}{r}u′ + f(u) = g(r) for 0 < r < T

(2.2)  u(0) = d > 0, u′(0) = 0.

By varying d appropriately, we attempt to find a d such that u(r, d) has exactly n zeros on [0, T) and u(T) = 0.

Multiplying (2.1) by r^{N−1} and integrating on (0, r) gives

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

Integrating (2.3) and applying the initial conditions we get

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

Let ϕ(u) be equal to the right hand side of (2.4). It is straightforward to show that ϕ(u) is a contraction mapping on C[0, ϵ], the set of continuous functions with supremum norm on [0, ϵ], for some ϵ > 0. Then by the contraction mapping principle there exists a u ∈ C[0, ϵ] such that ϕ(u) = u. Thus, u is continuous solution of (2.4). Then by (H1), (2.2), and (2.3), we see that u′ is continuous on [0, ϵ].

From (H1) and (2.3) it follows that \frac{u′}{r} is bounded, that lim_{r→0^{+}} \frac{u′}{r} exists, and so that \frac{u′}{r} is continuous

on [0, ϵ]. Then it follows from (2.1) that u′′ is continuous on [0, ϵ].

In order to show that u ∈ C^{2}[0, T], we define the energy equation of (2.1)-(2.2) as

(2.5)  E = \frac{u^{′2}}{2} + F (u).

Note that from (1.2) there exists a J > 0 such that

(2.6)  F (u) ≥ −J

for all u ∈ ℝ.

From (2.5) and (2.6) we see that

(2.7)  u^{′2} ≤ 2(E + J).

EJQTDE, 2008 No. 38, p. 2 Using (2.1) we see that

E′ = −\frac{N − 1}{r}u^{′2} − g(r)u′
≤ ||g|||u′| (defined in (H3))
≤ ||g||\sqrt{2}\sqrt{E + J} (by (2.7)).

Dividing by \sqrt{E + J} and integrating gives

\frac{1}{\sqrt{2}}|u′| ≤ \sqrt{E(t) + J} ≤ \sqrt{F (d) + J} + ||g||t ≤ \sqrt{F (d) + J} + ||g||T.

Thus, from (2.7) it follows that |u′| is uniformly bounded wherever it is defined and since u(0) = d, thus |u| is uniformly bounded wherever it is defined. It follows from this that u and u′ are defined on all of [0, T] and from (2.1) it then follows that u ∈ C^{2}[0, T].

The next several arguments presented were essentially originally proved in [1] and are included here for completeness.

Since f(u) > 0 for sufficiently large u > 0 (by (H2)), we see from (2.3) that u′ < 0 on (0, r) for small r > 0 if d is sufficiently large. Let k be the number given by (H3). Now for sufficiently large d it follows that u′ < 0 on (0, r_{kd}) where r_{kd} is the smallest positive value of r such that u(r_{kd}) = kd.

Remark 1: First, we want to find a lower bound for r_{kd}. Since f is increasing for large u (by (H1)), we see from (2.3) that

−r^{N−1}u′ ≤ [f(d) + ||g||] ∫_{0}^{r} t^{N−1}dt

= [f(d) + ||g||]\frac{r^{N}}{N} .

Dividing by r^{N−1} and integrating on [0, r_{kd}] we see that

(1 − k)d = ∫_{0}^{r_{kd}} −u′dt ≤ ∫_{0}^{r_{kd}} \frac{t[f(d) + ||g||]}{N}dt = \frac{t[f(d) + ||g||]}{2N}r_{kd}^{2}.

Thus,

r_{kd} ≥ \sqrt{\frac{2N(1 − k)d}{f(d) + ||g||}}.

For sufficiently large d we have ||g|| ≤ f(d) (by (H2)), thus we obtain for sufficiently large d

r_{kd} ≥ \sqrt{\frac{2N(1 − k)d}{2f(d)}}.

So,

(2.8)  r_{kd} ≥ \sqrt{\frac{N(1 − k)d}{f(d)}}

for sufficiently large d.

Remark 2: Because of its appearance in Pohozaev’s identity we will see that it will be important to find a lower bound on

(2.9)  ∫_{0}^{r_{kd}} t^{N−1} (NF (u) − \frac{N − 2}{2}u f(u) − \frac{N + 2}{2}g(t) u − t g′(t) u) dt.

EJQTDE, 2008 No. 38, p. 3 By hypothesis (H2), F′ = f > 0 for large u. Therefore, F is increasing for large u. Since for large d, u is decreasing for 0 ≤ t ≤ r_{kd}, and kd ≤ u(t) ≤ d, this implies F (kd) ≤ F (u) ≤ F (d). So on [0, r_{kd}] we have

(2.10)  ∫_{0}^{r_{kd}} t^{N−1}NF (u) dt ≥ F (kd) r_{kd}^{N} for large d

then by hypothesis (H1), f is increasing for large u and using this we have

∫_{0}^{r_{kd}} t^{N−1} \frac{N − 2}{2}u f(u) dt ≤ \frac{N − 2}{2N}d f(d) r_{kd}^{N} for large d

so,

(2.11)  − ∫_{0}^{r_{kd}} t^{N−1} \frac{N − 2}{2}uf(u)dt ≥ −\frac{N − 2}{2N}df(d)r_{kd}^{N}.

Now using the estimates in (2.8), (2.10), (2.11) and using the fact that g and g′ are bounded, we estimate (2.9) as follows:

(2.12)

∫_{0}^{r_{kd}} t^{N−1} (NF (u) − \frac{N − 2}{2}uf(u) − \frac{N + 2}{2}g(t)u − tg′(t)u) dt ≥ (F (kd) − \frac{N − 2}{2N}df(d) − \frac{N + 2}{2N}||g||d − \frac{1}{N} T ||g′||d) r_{kd}^{N}

≥ (NF (kd) − \frac{N − 2}{2}df(d) − \frac{N + 2}{2}||g||d − T ||g′||d) (\frac{1}{N} (\sqrt{\frac{N(1 − k)d}{f(d)}})^{N} )

= C(N, k) (NF (kd) − \frac{N − 2}{2}df(d) − \frac{N + 2}{2}||g||d − T ||g′||d) (\frac{d}{f(d)})^{\frac{N}{2}}

where C(N, k) = \frac{1}{N} [N(1 − k)]^{\frac{N}{2}} .

Lemma 2.1. If (H1) - (H4) are satisfied, then

(2.13)  lim_{d→∞} inf_{[0,T]} E(r, d) = ∞.

Proof. Let us suppose 0 ≤ r ≤ T. Consider Pohozaev’s identity which states

[r^{N}E − r^{N}g(r)u + \frac{N − 2}{2}r^{N−1}uu′]′ = r^{N−1} [NF (u) − \frac{N − 2}{2}uf(u) − \frac{N + 2}{2}g(r)u − rg′(r)u] .

This can be verified by simply differentiating and then using (2.1).

Integrating Pohozaev’s identity on [0, r], and using (H4) and (2.12) gives

r^{N}E(r, d) − r^{N}g(r)u + \frac{N − 2}{2}r^{N−1}uu′ = ∫_{0}^{r} t^{N−1} [NF (u) − \frac{N − 2}{2}uf(u) − \frac{N + 2}{2}g(t)u − tg′(t)u] dt

= ∫_{0}^{r_{kd}} t^{N−1} [NF (u) − \frac{N − 2}{2}uf(u) − \frac{N + 2}{2}g(t)u − tg′(t)u] dt

+ ∫_{r_{kd}}^{r} t^{N−1} [NF (u) − \frac{N − 2}{2}uf(u) − \frac{N + 2}{2}g(t)u − tg′(t)u] dt

≥ C(N, k) (\frac{d}{f(d)})^{\frac{N}{2}} [NF (kd) − \frac{N − 2}{2}df(d) − \frac{N + 2}{2}||g||d − T ||g′||d] − M (\frac{r^{N} − r_{kd}^{N}}{N} ) .

Ignoring the last term on the right hand side we get (2.14)

r^{N}E(r, d)−r^{N}g(r)u+\frac{N − 2}{2}r^{N−1}uu′ ≥ C(N, k) (\frac{d}{f(d)})^{\frac{N}{2}} [NF (kd) − \frac{N − 2}{2}df(d) − \frac{N + 2}{2}||g||d − T ||g′||d]−\frac{MT^{N}}{N}

EJQTDE, 2008 No. 38, p. 4 Now let us estimate uu′.

First note from (1.2) that there exists a B such that if |u| ≥ B then \frac{u^{2}}{F (u)} ≤ 1. That is if |u| ≥ B

then u^{2} ≤ F (u) ≤ F (u) + J. On other hand if |u| ≤ B then u^{2} ≤ B^{2}. And since F (u) + J ≥ 0 (by (2.6)) we see that for all u we have

(2.15)  u^{2} ≤ F (u) + J + B^{2}.

Using Young’s inequality, (2.5), and (2.15) gives us the following:

uu′ ≤ \frac{1}{2}u^{2} + \frac{1}{2}u^{′2}

≤ (F (u) + J + B^{2}) + \frac{1}{2}u^{′2}

= (\frac{1}{2}u^{′2} + F (u)) + J + B^{2}

= E(r, d) + J + B^{2}.

Substituting this into the left hand side of (2.14), rewriting, and estimating we see that

r^{N}E − r^{N}g(r)u + \frac{N − 2}{2}r^{N−1}uu′ ≤ T^{N}E + T^{N}||g|| |u| + \frac{N − 2}{2}T^{N−1}|uu′|

≤ T^{N}E + T^{N}||g||^{2} + T^{N}u^{2} + \frac{N − 2}{2}T^{N−1}[E + J + B^{2}]

≤ T^{N}E + T^{N}||g||^{2} + T^{N}[E + J + B^{2}] + \frac{N − 2}{2}T^{N−1}[E + J + B^{2}]

= (2T^{N} + \frac{N − 2}{2}T^{N−1}) E + T^{N−1} ((T + \frac{N − 2}{2}) (J + B^{2}) + ||g||^{2})

= C_{1}E + C_{2}

where C_{1} > 0 and C_{2} > 0 depend only on T, N, J, B and ||g||.

Thus, combining the above with (2.14) gives:

C(N, k) (\frac{d}{f(d)})^{\frac{N}{2}} [NF (kd) − \frac{N − 2}{2}df(d) − \frac{N + 2}{2}||g||d − T ||g′||d] − \frac{MT^{N}}{N}

≤ C_{1}E + C_{2}.

Thus,

C_{1}E ≥ C(N, k) (\frac{d}{f(d)})^{\frac{N}{2}} [NF (kd) − \frac{N − 2}{2}df(d) − \frac{N + 2}{2}||g||d − T ||g′||d] − C_{3}

where C_{3} depends on T, N, J, B, ||g|| and M.

By assumption the right hand side of the above inequality goes to infinity as d → ∞. Therefore,

lim_{d→∞} inf_{[0,T]} E(r, d) = ∞.

□

Lemma 2.2. If d is sufficiently large and u(r_{0}) = 0, then u′(r_{0}) ≠ 0.

Proof. By Lemma 2.1, if d is sufficiently large then inf_{[0,T]} E(r, d) > 0. So if u(r_{0}) = 0 then we have

\frac{1}{2}u′(r_{0})^{2} = E(r_{0}) ≥ inf_{[0,T]} E(r, d) > 0. □

Lemma 2.3. For d sufficiently large u has a finite number of zeros on [0, T].

Proof. Suppose there exists 0 < z_{1} < z_{2} < ... < z_{n} < ... < T and u(z_{i}) = 0. Then by the mean value theorem there exists m_{1} < m_{2} < ... such that u′(m_{k}) = 0 and where z_{k} < m_{k} < z_{k+1} < T. So there exists z = lim_{n→∞} z_{n} and by continuity u(z) = 0. Also, lim_{k→∞} m_{k} = z and u′(z) = 0 but by the above Lemma 2.2, this cannot happen for sufficiently large d. □

3. Finding zeros

Now we want to show that if d is sufficiently large then u(r, d) will have lots of zeros on [0, T]. From (1.2) we know that F (u) → ∞ as |u| → ∞. Therefore, since lim_{d→∞} inf_{[0,T]} E(r, d) = ∞ (by Lemma

2.1), and since F (u) is increasing for large u and decreasing when u is a large negative number, then

for sufficiently large d there are exactly two solutions of F (u) = \frac{1}{2} inf_{[0,T]} E(r, d) which we denote as

h_{2}(d) < 0 < h_{1}(d). For d > 0 sufficiently large we see from (H2) that u′′(0) = \frac{−f(d) + g(0)}{N} < 0 and u′(0) = 0 so u is initially decreasing on (0, r). Note that h_{1}(d) → ∞ as d → ∞. From (2.3) we see that u will be decreasing as long as f(u) ≥ ||g||. So we see that there is a smallest r > 0, r_{1}(d), such that u(r_{1}(d)) = h_{1}(d) and d ≥ u > h_{1}(d) on [0, r_{1}(d)).

Let

(3.1)  C(d) = \frac{1}{2} min_{r∈[0,r_{1}(d)]} \frac{f(u)}{u} = \frac{1}{2} min_{u∈[h_{1}(d),d]} \frac{f(u)}{u}.

Then by (H2) we see that C(d) → ∞ as d → ∞.

Lemma 3.1. r_{1}(d) → 0 as d → ∞.

Proof. To show this we compare

(3.2)  u′′ + \frac{N − 1}{r}u′ + \frac{f(u)}{u}u = g(r)

with initial conditions u(0) = d > 0 and u′(0) = 0 with

(3.3)  v′′ + \frac{N − 1}{r}v′ + C(d)v = 0

with initial conditions v(0) = d and v′(0) = 0. Note from (3.1) that

(3.4)  \frac{f(u)}{u} ≥ 2C(d) > C(d) on [0, r_{1}(d)].

Claim: u < v on (0, r_{1}(d)] for sufficiently large d.

Proof of the Claim: Since

u(0) = d = v(0)
u′(0) = 0 = v′(0)

then for large d we see from (3.4) that

u′′(0) = \frac{−f(d)}{N} + \frac{g(0)}{N} < −\frac{C(d)}{N}d = v′′(0).

Thus, u < v on (0, ϵ) for some ϵ > 0.

Multiplying (3.2) by r^{N−1}v, (3.3) by r^{N−1}u, and then taking the difference of the resultant equations gives

(r^{N−1}(u′v − uv′))′ + r^{N−1}uv (\frac{f(u)}{u} − \frac{g(r)}{u} − C(d)) = 0.

EJQTDE, 2008 No. 38, p. 6 Since g is bounded, for sufficiently large d we see from (3.4) that

\frac{f(u)}{u} − \frac{g(r)}{u} − C(d) ≥ 2C(d) − \frac{||g||}{u} − C(d) on [0, r_{1}(d)]

= C(d) − \frac{||g||}{u}

≥ C(d) − \frac{||g||}{h_{1}(d)}

> 0 (since C(d) → ∞ as d → ∞ and h_{1}(d) → ∞ as d → ∞).

Now integrating this from 0 to r where 0 < r ≤ r_{1}(d) and using u(0) = v(0) = d and u′(0) = v′(0) = 0 gives

u′(r)v(r) − v′(r)u(r) < 0 on (0, r_{1}(d)].

Suppose now there is a first r_{0} with 0 < r_{0} ≤ r_{1}(d) such that 0 < u(r_{0}) = v(r_{0}) and u < v on (0, r_{0}). Then we see from the above inequality that u′(r_{0}) < v′(r_{0}). On other hand, u(r) < v(r) on (0, r_{0}) and u(r_{0}) = v(r_{0}). So

u(r) − u(r_{0}) < v(r) − v(r_{0}) on (0, r_{1}(d)].

Thus, for r < r_{0} we have

lim_{r→r_{0}^{−}} \frac{u(r) − u(r_{0})}{r − r_{0}} ≥ lim_{r→r_{0}^{−}} \frac{v(r) − v(r_{0})}{r − r_{0}}

which gives

u′(r_{0}) ≥ v′(r_{0}).

This is a contradiction since u′(r_{0}) < v′(r_{0}). Hence this proves the claim.

Now let z(r) = (r/\sqrt{C(d)})^{\frac{N−2}{2}} v (r/\sqrt{C(d)}) . Then

(3.5)  z′′ + \frac{z′}{r} + (1 − \frac{(\frac{N−2}{2})2}{r^{2}} ) z = 0.

The above equation is Bessel’s equation of order \frac{N − 2}{2}. Thus, z(r) = A_{1}J_{\frac{N−2}{2}} (r) + A_{2}Y_{\frac{N−2}{2}} (r) for

constants A_{1} and A_{2} and where J_{\frac{N−2}{2}} is the Bessel function of order \frac{N−2}{2} which is bounded at r = 0 and Y_{\frac{N−2}{2}} is unbounded at r = 0. Since z is bounded at r = 0 and Y_{\frac{N−2}{2}} is not, it must be that z(r) = A_{1}J_{\frac{N−2}{2}} (r), and A_{1} is a positive constant.

Denoting β_{\frac{N−2}{2} ,1} as the first positive zero of J_{\frac{N−2}{2}} (r), we see that the first positive zero of v is \frac{β_{\frac{N−2}{2} ,1}}{\sqrt{C(d)}}

and since u < v on [0, r_{1}(d)] (by the Claim) we see that

r_{1}(d) < \frac{β_{\frac{N−2}{2} ,1}}{\sqrt{C(d)}}.

Since C(d) → ∞ as d → ∞ (as mentioned after (3.1)) it then follows that lim_{d→∞} r_{1}(d) = 0. □

Lemma 3.2. For large d, u has a first positive zero, z_{1}(d), and z_{1}(d) → 0 as d → ∞.

Proof. First we show that u has a zero. We prove this by contradiction. Suppose u > 0 on [0, T] and consider r > r_{1}(d). Then 0 < u < u(r_{1}(d)) = h_{1}(d) so F (u) < F (h_{1}(d)). Also since F (h_{1}(d)) =

\frac{1}{2} inf_{[0,T]} E(r, d) we obtain

\frac{u^{′2}}{2} + F (h_{1}(d)) > \frac{u^{′2}}{2} + F (u) ≥ inf_{[0,T]} E(r, d) = 2F (h_{1}(d))

EJQTDE, 2008 No. 38, p. 7 for r > r_{1}(d).

Thus,

u^{′2} ≥ 2F (h_{1}(d)) for r > r_{1}(d)

and thus

− ∫_{r_{1}(d)}^{r} u′(t)dt ≥ ∫_{r_{1}(d)}^{r} \sqrt{2F (h_{1}(d))}dt

and since u is decreasing and u(r_{1}(d)) = h_{1}(d) this gives

(3.6)  h_{1}(d) − u(r) = u(r_{1}(d)) − u(r) ≥ \sqrt{2F (h_{1}(d))}(r − r_{1}(d))

so,

h_{1}(d) − \sqrt{2F (h_{1}(d))}(r − r_{1}(d)) ≥ u(r) > 0.

Thus,

(3.7)  \frac{h_{1}(d)}{\sqrt{2F (h_{1}(d))}} ≥ r − r_{1}(d).

Evaluating at r = T gives

T − r_{1}(d) ≤ \frac{h_{1}(d)}{\sqrt{2F (h_{1}(d))}}

for large d.

Since h_{1}(d) → ∞ as d → ∞, taking the limit of the above, using Lemma 3.1 and (1.2) we see that

0 < T = lim_{d→∞}[T − r_{1}(d)] ≤ lim_{d→∞} \frac{h_{1}(d)}{\sqrt{2F (h_{1}(d))}} = 0.

This is impossible. Thus u has a first zero, z_{1}(d). Then repeating the above argument on [0, z_{1}(d)] and letting r = z_{1}(d) in (3.7) we get

0 ≤ z_{1}(d) − r_{1}(d) ≤ \frac{h_{1}(d)}{\sqrt{2F (h_{1}(d))}} → 0

as d → ∞. Also, since r_{1}(d) → 0 as d → ∞ (by Lemma 3.1) we see that z_{1}(d) → 0 as d → ∞. □

We next show for sufficiently large d that u attains the value h_{2}(d) at some r_{2}(d) where z_{1}(d) < r_{2}(d) < T. So we suppose u′ < 0 on a maximal interval (z_{1}(d), r). Here h_{2}(d) < u < 0 and this implies F (u) ≤ F (h_{2}(d)) for sufficiently large d. Then as in the beginning of the proof of Lemma 3.2

\frac{1}{2}u^{′2} + F (h_{2}(d)) ≥ \frac{1}{2}u^{′2} + F (u) ≥ inf_{[0,T]} E(r, d) = 2F (h_{2}(d))

so,

u^{′2} ≥ 2F (h_{2}(d)) on (z_{1}(d), r).

Then

∫_{z_{1}(d)}^{r} −u′dt = ∫_{z_{1}(d)}^{r} |u′|dt ≥ ∫_{z_{1}(d)}^{r} \sqrt{2F (h_{2}(d))}dt

and since u(z_{1}(d)) = 0 this leads to

−u(r) ≥ \sqrt{2F (h_{2}(d))}(r − z_{1}(d))

and therefore

(3.8)  u(r) ≤ −\sqrt{2}\sqrt{F (h_{2}(d))}(r − z_{1}(d)).

Now suppose by the way of contradiction that u > h_{2}(d) on (z_{1}(d), T). Then from (3.8) we see that

h_{2}(d) ≤ u(r) ≤ −\sqrt{2}\sqrt{F (h_{2}(d))}(r − z_{1}(d))

EJQTDE, 2008 No. 38, p. 8

−h_{2}(d) ≥ \sqrt{2}\sqrt{F (h_{2}(d))}(r − z_{1}(d)).

Evaluating this at r = T gives

T − z_{1}(d) ≤ \frac{−h_{2}(d)}{\sqrt{2}\sqrt{F (h_{2}(d))}}

and now taking the limit, using Lemma 3.2, and (1.2) we see that

0 < T = lim_{d→∞}[T − z_{1}(d)] ≤ lim_{d→∞} \frac{−h_{2}(d)}{\sqrt{2}\sqrt{F (h_{2}(d))}} = 0.

And again this is impossible. Therefore, there exists a smallest value of r, r_{2}(d), such that z_{1}(d) < r_{2}(d) < T with u(r_{2}(d)) = h_{2}(d) and u > h_{2}(d) on [0, r_{2}(d)). Now evaluating (3.8) at r = r_{2}(d) and using that u(r_{2}(d)) = h_{2}(d) we obtain

h_{2}(d) = u(r_{2}(d)) ≤ −\sqrt{2}\sqrt{F (h_{2}(d))}(r_{2}(d) − z_{1}(d))

now taking the limit as d → ∞ and (1.2) gives

lim_{d→∞} \sqrt{2}[r_{2}(d) − z_{1}(d)] ≤ lim_{d→∞} \frac{−h_{2}(d)}{\sqrt{F (h_{2}(d))}} = 0.

Hence r_{2}(d) − z_{1}(d) → 0 as d → ∞ and since z_{1}(d) → 0 as d → ∞ (from Lemma 3.2) it follows that

(3.9)  r_{2}(d) → 0 as d → ∞.

We next want to show that u has a minimum on (r_{2}(d), T). Suppose again by contradiction that u is decreasing on (r_{2}(d), T). We want to show that there exists an extremum of u at r where r > r_{2}(d).

Let C(d) = \frac{1}{2} min_{(−∞,h_{2}(d)]} \frac{f(u)}{u}. Note that C(d) → ∞ as d → ∞ by (H2). Now as in the proof of

Lemma 3.1 we compare

(3.10)  u′′ + \frac{N − 1}{r}u′ + \frac{f(u)}{u}u = g(r)

with

(3.11)  v′′ + \frac{N − 1}{r}v′ + C(d)v = 0

with initial conditions v(r_{2}(d)) = u(r_{2}(d)) and v′(r_{2}(d)) = u′(r_{2}(d)). With an argument similar to the Claim in Lemma 3.1 we can show that u > v on (r_{2}(d), T) for sufficiently large d. Let z(r) =

(r/\sqrt{C(d)})^{\frac{N−2}{2}} v (r/\sqrt{C(d)}) . Then again as earlier z solves Bessel’s equation

(3.12)  z′′ + \frac{z′}{r} + (1 − \frac{(\frac{N−2}{2})2}{r^{2}} ) z = 0

of order \frac{N − 2}{2}.

Now it is a well known fact about Bessel functions (see [4], Page 165, Theorem C) that there exists a constant K such that every interval of length K has at least one zero of z(r). This implies that every

interval of length \frac{K}{\sqrt{C(d)}} has a zero of v. Thus for large d, we see that v must have a zero on (r_{2}(d), T).

And since u > v on (r_{2}(d), T) we see that u gets positive which contradicts that u is decreasing on (r_{2}(d), T). Thus we see that there exists an m_{1}(d) with r_{2}(d) < m_{1}(d) < T such that u decreases on (r_{2}(d), m_{1}(d)) and m_{1}(d) is a local minimum of u. Also we see that

m_{1}(d) − r_{2}(d) ≤ \frac{K}{\sqrt{C(d)}} → 0

as d → ∞. And since r_{2}(d) → 0 as d → ∞ (by (3.9)) we see that m_{1}(d) → 0 as d → ∞. Also, F (u(m_{1})) = E(m_{1}(d)) ≥ inf_{[0,T]} E(r, d) → ∞ as d → ∞ (by Lemma 2.1). In a similar way we can show

that for large d, u has a second zero, z_{2}(d), with m_{1}(d) < z_{2}(d) < T and z_{2}(d) → 0 as d → ∞ and u has a second extremum, m_{2}(d), with z_{2}(d) < m_{2}(d) < T and m_{2}(d) → 0 as d → ∞. Continuing in this way we can get as many zeros of u(r, d) as desired on (0, T) for large enough d.

4. Proof of the Main Theorem

To prove the Main Theorem we construct the following sets.

Let S_{k} = { d | u(r, d) has exactly k zeros for all r ∈ [0, T) and inf_{[0,T]} E > 0 }.

Let us denote k_{0} ≥ 0 as the smallest value of k such that S_{k} ≠ ∅. Also, as we saw at the end of section 3, u(r, d) has more and more zeros on (0, T) provided d is chosen large enough. And also inf_{[0,T]} E > 0 if

d is chosen large enough (by Lemma 2.1). Hence it follows that S_{k_{0}} is bounded above and nonempty.

Let d_{k_{0}} = sup S_{k_{0}}.

Lemma 4.1. u(r, d_{k_{0}}) has exactly k_{0} zeros on [0, T).

Proof. By definition of k_{0}, u(r, d_{k_{0}}) has at least k_{0} zeros on [0, T). Suppose u(r, d_{k_{0}}) has more than k_{0} zeros on [0, T). Then for d close to d_{k_{0}} and d < d_{k_{0}}, by continuity with respect to initial conditions and by Lemma 2.2, u(r, d) also has more than k_{0} zeros on [0, T). However, if d ∈ S_{k_{0}}, then u(r, d) has exactly k_{0} zeros on [0, T). This is a contradiction to the definition of d_{k_{0}}. Thus, u(r, d_{k_{0}}) has exactly k_{0} zeros on [0, T). □

Lemma 4.2. u(T, d_{k_{0}}) = 0.

Proof. If u(T, d_{k_{0}}) ≠ 0 then by continuity with respect to initial conditions and Lemma 2.2, u(r, d) has the same number of zeros as u(r, d_{k_{0}}) for d close to d_{k_{0}}. But if d > d_{k_{0}} then d ∉ S_{k_{0}} so u(r, d) cannot have the same number of zeros as u(r, d_{k_{0}}). This is a contradiction. Thus, u(T, d_{k_{0}}) = 0. □

Let S_{k_{0}+1} = { d > d_{k_{0}} | u(r, d) has exactly k_{0} + 1 zeros on [0, T) and inf_{[0,T]} E > 0 }.

Lemma 4.3. S_{k_{0}+1} ≠ ∅ and S_{k_{0}+1} is bounded above.

Proof. By continuity with respect to initial conditions and Lemma 2.2, if d > d_{k_{0}} and d close to d_{k_{0}} then u(r, d) has at most k_{0} + 1 zeros on [0, T). Also, if d > d_{k_{0}} then d ∉ S_{k_{0}} so u(r, d) does not have exactly k_{0} zeros on [0, T). Now u(r, d) cannot have less than k_{0} zeros because this would imply that S_{k_{0}} = ∅ for some value of k smaller than k_{0} which contradicts the definition of k_{0}. Thus, u(r, d) has at least k_{0} + 1 zeros on [0, T). Since we already showed that u(r, d) for d > d_{k_{0}} and d close to d_{k_{0}} has at most k_{0} + 1 zeros on [0, T) therefore, for d > d_{k_{0}} and d close to d_{k_{0}}, u(r, d) has exactly k_{0} + 1 zeros on [0, T). Hence S_{k_{0}+1} is nonempty. Then by remarks at the end of section 3, S_{k_{0}+1} is bounded above. □

Define d_{k_{0}+1} = sup S_{k_{0}+1}.

As above we can show that u(r, d_{k_{0}+1}) has exactly k_{0} + 1 zeros on [0, T) and u(T, d_{k_{0}+1}) = 0. Proceeding inductively, we can find solutions that tend to zero at infinity and with any prescribed number, n, of zeros on [0, T) where n ≥ k_{0}. Hence, this completes the proof of the Main Theorem if (H3) holds.

If (H3*) holds instead of (H3) let v(r) = −u(r). Then v satisfies

(4.1)  v′′ + \frac{N − 1}{r}v′ + f_{2}(v) = g_{2}(r)
(4.2)  v(0) = −d

EJQTDE, 2008 No. 38, p. 10

(4.3)  v′(0) = 0

where

f_{2}(v) = −f(−v)
g_{2}(r) = −g(r)
F_{2}(v) = ∫_{0}^{v} f_{2}(u)du = ∫_{0}^{v} −f(−u)du = F (−v).

And, now we look for solutions of (4.1)-(4.3) with −d > 0 (that is d < 0) along with v(T) = 0. It is straightforward to show that (H1), (H2) and (H4) are satisfied by f_{2} (and F_{2}).

Then by (H3*)

∞ = lim_{u→−∞} (\frac{u}{f(u)})^{\frac{N}{2}} (NF (ku) − \frac{(N − 2)}{2}uf(u) − \frac{N + 2}{2}||g|| |u| − T ||g′|| |u|)

= lim_{u→∞} (\frac{−u}{f(−u)})^{\frac{N}{2}} (NF (−ku) − \frac{(N − 2)}{2}(−u)f(−u) − \frac{N + 2}{2}||g|| |u| − T ||g′|| |u|)

= lim_{u→∞} (\frac{u}{f_{2}(u)})^{\frac{N}{2}} (NF_{2}(ku) − \frac{(N − 2)}{2}uf_{2}(u) − \frac{N + 2}{2}||g_{2}|| |u| − T ||g′_{2}|| |u|) .

Thus (H3) is satisfied by g_{2} and f_{2} (and F_{2}).

Also defining

E_{2}(r, d) = \frac{1}{2}v^{′2} + F_{2}(v)

we see that

E_{2}(r, d) = \frac{1}{2}u^{′2} + F_{2}(−u)

= \frac{1}{2}u^{′2} + F (u)

= E(r, d).

Therefore, (H1)-(H4) are satisfied by f_{2} (and F_{2}) and so by the first part of the theorem we see that there are an infinite number of solutions of (4.1)-(4.3) with v(0) = −d > 0 and v(T) = 0. Thus, u(r) = −v(r) satisfies (1.3)-(1.4) with u(0) = −v(0) = d < 0. This completes the proof of the Main Theorem.

Here is an example of a u that satisfies the hypotheses (H1)-(H4):

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

where N = 3, f(u) = u^{3} − u and g(r) = 0.

Here are some graphs of solutions of (4.4) for different values of d, all graphs are generated numerically using Mathematica:

(a) Solution that remains positive when d = 4 (b) Solution with exactly one zero when d = 4.5

(c) Solution with exactly two zeros when d = 15

(d) Solution with exactly three zeros when d = 35 Now let us consider another example, here u satisfies the hypotheses (H1)-(H4):

(4.5)  u′′ + \frac{2}{r}u′ + u^{3} − u = \frac{1}{r^{2} + 1}

where N = 3, f(u) = u^{3} − u and g(r) = \frac{1}{r^{2} + 1}.

Here are some graphs of solutions of (4.5) for different values of d, as above all graphs are generated numerically using Mathematica:

(a) Solution that remains positive when d = 5 EJQTDE, 2008 No. 38, p. 12 (b) Solution with exactly one zero when d = 6

(c) Solution with exactly three zeros when d = 50

□

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. Gidas and W.M. Ni and L. Nirenberg, Symmetry and Related Properties via the Maximum Principle, Comm. Math. Phys., Volume 68, 209-243(1979)

[3] K. McLeod and W.C. Troy and F.B. Weissler, Radial solutions of Δu + f(u) = 0 with Prescribed Numbers of Zeros, Journal of Differential Equations, Volume 83, 368-378(1990)

[4] G.F. Simmons, Differential Equations with Applications and Historical Notes, 2nd edition, McGraw-Hill Science/Engineering/Math(1991)

(Received August 8, 2008)

*Department of Mathematics, University of North Texas, Denton, Texas, **Department of Mathematics, Augsburg College, Minneapolis, MN

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