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Uniqueness of positive solutions to a class of semilinear elliptic equations

Chunming Li and Yong Zhou*

Author Affiliations

Department of Mathematics, Zhejiang Normal University, Jinhua 321004, Zhejiang, PR China

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Boundary Value Problems 2011, 2011:38  doi:10.1186/1687-2770-2011-38

The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2011/1/38

Received:2 June 2011
Accepted:24 October 2011
Published:24 October 2011

© 2011 Li and Zhou; licensee Springer.

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


In this article, we consider the uniqueness of positive radial solutions to the Dirichlet boundary value problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M1">View MathML</a>

where Ω denotes an annulus in ℝn (n ≥ 3). The uniqueness criterion is established by applying shooting method.

positive solution; semilinear elliptic equation; uniqueness

1 Introduction

This article is concerned with the positive radial solutions to a class of semilinear elliptic equations

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M2','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M2">View MathML</a>


where Ω: = {x | x ∈ ℝn, a < |x| < b}, a and b are positive real numbers, f C1((0, + ∞) × [0, + ∞)) and g : [0, + ∞) → ℝ is differentiable. Equation 1.1 describes stationary states for many reaction-diffusion equations. The absence of positive solutions to the elliptic equations also means that the existing solutions oscillate, which is also important information in applications.

In recent years, there is a widespread concern over the positive solutions to the Dirichlet boundary value problem (1.1) when g(|x|) = 0, i.e.,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M3','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M3">View MathML</a>


When the nonlinear term just depends on u, the uniqueness of (1.2) has been exhaustively studied (see [1-6]). In 1985, the uniqueness of (1.2) was discussed in different domains by Ni and Nussbaum [7] to the case when f depends on |x| and u, f(|x|,u) > 0 and f(|x|,u) satisfies some growth conditions. Erbe and Tang [8] presented a new uniqueness criterion using a shooting method and Sturm comparison theorem.

So far it seems that nobody considers the uniqueness to problem (1.1). Inspired by the above articles, the aim of the present article is to establish some simple criteria for the uniqueness of positive radial solutions to problem (1.1). Obviously, what we investigate in this article has a more general form than (1.2). Although due to technical reasons, when g(|x|) = 0 it does not hold in this article, there exist many other g(|x|) which satisfy our main result.

We now conclude this introduction by outlining the rest of this article. In Section 2, we will show the existence and uniqueness of positive solutions of the initial problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M4','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M4">View MathML</a>

where α > 0. Our method is the Schauder-Tikhonov fixed point theory. The existence and uniqueness of this initial problem is important to prove our main result. In Section 3, we will give the proof of our main result, i.e., show the uniqueness of positive solutions to Equation 1.1, using a shooting method and Sturm theorem.

2 Preliminaries

To consider the positive radial solutions of Equation 1.1, it is reasonable to investigate the corresponding radial equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M5','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M5">View MathML</a>

where t = |x|. For giving a proof of uniqueness of problem (1.1), let us consider the initial problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M6','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M6">View MathML</a>


where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M7">View MathML</a>. We shall show that problem (2.1) has a unique positive solution. By a solution to problem (1.2), we mean u C2 and u > 0 for all t ∈ (a, b). First of all, we give a well-known lemma.

Lemma 2.1 (The Schauder-Tikhonov fixed point theorem [9]). Let × be a Banach space and K X be a nonempty, closed, bounded and convex set. If the operator T : K X continuously maps K into itself and T(K) is relatively compact in X, then T has a fixed point x K.

Theorem 2.1 If there exist m and M, such that 0 < m u M for u C([a, b], (0, ∞)) and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M8">View MathML</a>


Then, Equation 2.1 has a unique positive solution.

Proof. We assume that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M9">View MathML</a>

endowed with the supremum norm ||u|| = supatb |u(t)|. Let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M10','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M10">View MathML</a>

Define the operator T : K X, by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M11','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M11">View MathML</a>

We shall apply the Schauder-Tikhonov theorem to prove that there exists a fixed point u(t), which is a positive solution of problem (2.1), for the operator T in the non-empty closed convex set K.

We shall do it by several steps as follows:

Step 1: Check that T : K K is well defined. Obviously, by (2.2), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M12','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M12">View MathML</a>

thus T : K K is well defined.

Step 2: Verify that T : K K is continuous. Note that h(t), f(t, u) are continuous, they are integrable on [a, b], there exists a constant M1 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M13','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M13">View MathML</a>


The function f(t,u) is continuous, thus for ∀ ε > 0, there exists δ > 0 such that for any u(t),v(t) ∈ K with ||u-v|| ≤ δ,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M14','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M14">View MathML</a>

From this, it follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M15','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M15">View MathML</a>

Thus, T is continuous on K.

Step 3: We check that T(K) is relatively compact in X.

Since TK K, TK is uniformly bounded. Now, verify that TK is equicontinuous. Let u K, then we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M16','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M16">View MathML</a>


Similar to (2.3), there exists a constant M2 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M17">View MathML</a>

Take a sequence {un} ⊂ K, by the mean value theorem, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M18','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M18">View MathML</a>

Thus, TK is equicontinuous. Arzela-Ascoli theorem [9] implies TK is relatively compact. Now, we have verified that T : K K satisfies all assumptions of the Schauder-Tikhonov theorem. Thus, there exists a fixed point u which is a positive solution of problem (2.1).

Now, we are in a position to prove the uniqueness of problem (2.1). The proof of the uniqueness of solution is based on the work of [10]. Suppose that u and v are two different solutions of problem (2.1), then the function

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M19','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M19">View MathML</a>

is a solution of Cauchy problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M20','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M20">View MathML</a>

where ψ = f(t,v) - f(t,u). It follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M21','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M21">View MathML</a>

Hence, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M22','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M22">View MathML</a>

where M3 is a constant, such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M23','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M23">View MathML</a>

On the other hand, since the function f(t, u) is Hölder continuous with respect to the second variable on (0, + ∞), we obtain, for appropriate values t0, L > 0,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M24','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M24">View MathML</a>

From this, we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M25','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M25">View MathML</a> for t t0. It now follows from Gronwall's inequality that ω ≡ 0 for a < t t0, consequently u' ≡ v' for t t0. We find u(t) ≡ v(t) for all t ∈ (a,t0]. With the initial point t0 replace by ρ > t0, for an appropriate value ρ, the same proof can be reapplied as often as necessary to give uniqueness of any continuation of the solution whose values lie in (a, b). The proof is complete.

3 Uniqueness

Theorem 3.1 Assume that h(t) and f(t,u) for a < t < b, u(t) > 0, satisfy inequality (2.2) and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M26','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M26">View MathML</a>

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M27','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M27">View MathML</a>, then problem (1.1) has at most one positive radial solution.

Example 3.1 For the equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M28','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M28">View MathML</a>

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M29','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M29">View MathML</a> Let t = |x|, then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M30','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M30">View MathML</a>

A straightforward computation yields

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M31','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M31">View MathML</a>


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M32','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M32">View MathML</a>

Therefore, Theorem 3.1 ensures that there exists at most one positive radial solution.

Before proving our main result, we will do some preliminaries and give some useful lemmas.

Let u(t,α) denote the unique solution of Equation 2.1. If α > 0, then the solution u(t,α) is positive for t slightly larger than a. When it vanishes in (a, b), we define b(α) to be the first zero of u(t, α). More precisely, b(α) is a function of α which has the property that u(t, α) > 0 for t ∈ (a, b(α)) and u(b(α), α) = 0. Let N denote the set of α > 0 for which the solution u(t, α) has a finite zero b(α). The variation of u(t, α) is defined by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M33','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M33">View MathML</a>

and satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M34','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M34">View MathML</a>


Let L be the linear operator given by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M35','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M35">View MathML</a>


By (2.4), it is easy to show that u(t, α) has a unique critical point c(α) in (a, b(α)), and at this point, u(t, α) obtains a local maximum value.

Lemma 3.1 Assume that (F2) holds, then ϕ(t, α) > 0 for all t ∈ (a, c(α)).

Proof. We introduce a function

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M36','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M36">View MathML</a>


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M37','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M37">View MathML</a>

and accordingly

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M38','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M38">View MathML</a>

It is easy to see that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M39','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M39">View MathML</a>

Differentiating Q(t, α) with respect to t, we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M40','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M40">View MathML</a>


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M41','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M41">View MathML</a>

Hence, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M42','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M42">View MathML</a>

From hypotheses (F2), we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M43','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M43">View MathML</a>


Since Q(t,α) > 0 in t ∈ (a,c(α)) and inequality (3.3) holds, by the Sturm comparison principle (see [2]), we see that Q(t,α) oscillates faster that ϕ(t,α). Hence, ϕ(t,α) has no zero in t ∈ (a,c(α)). From ϕ(a,α) = 0 and ϕ'(a,α) = 1, it follows that ϕ(t, α) > 0 for all t ∈ (a, c(α)). The proof is complete.

Remark 3.1 Lemma 3.1 was already proved in [11]. Here we give a simpler proof, directly using Sturm comparison principle.

Now, we present a lemma which has been given to the case g(|x|) = 0 (see [8]). To make the article as self-contained as possible, we will give a simple proof with a slight modification to [8].

Lemma 3.2 Assume α N and f(t,u) satisfies (F1), then

(H1) ϕ(t,α) vanishes at least once and at most finitely many times in (a,b(α)),

(H2) if 0 < α1 < α2, and at least one of u(t,α1) and u(t,α2) has a finite zero, then they intersect in (a,min{b(α1),b(α2)}).

Proof. We shall prove this by contradiction. Suppose to the contrary that ϕ(t, α) does not vanish in (a, b(α)), then ϕ(t, α) > 0, t ∈ (a, b(α)). Note that L(ϕ(t, α)) = 0, so we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M44','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M44">View MathML</a>


Using the definition of L, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M45','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M45">View MathML</a>

Similar to (3.4), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M46','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M46">View MathML</a>


Multiply both sides of (3.4) by u(t, α) and (3.5) by ϕ(t, α), then subtract the resulting identities and we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M47','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M47">View MathML</a>


By (F1), we have the right side of (3.6) is positive in (a, b(α)). The left side of (3.6) is then a strictly increasing function of t in (3.6). We get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M48','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M48">View MathML</a>

Thus, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M49','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M49">View MathML</a>. However, it contradicts u'(b(α),α) < 0 and ϕ(b(α),α) ≥ 0.

Since the rest of proof can be completed by the same argument as [8], we omit them.

Lemma 3.3 If (F1) and (F3) hold, then ϕ(b(α), α) ≠ 0.

Proof. We shall prove this by contradiction. Suppose to the contrary that ϕ(b(α), α) = 0. Now, we may as well define τ(α) to be the last zero of ϕ(t, α) in (a, b(α)). By Lemma 3.1, it is easy to get c(α) ≤ τ(α), thus u'(τ(α), α) ≤ 0 and u'(t, α) < 0 for all t ∈ (τ(α), b(α)]. We introduce a function

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M50','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M50">View MathML</a>

Differentiating G(t, α) with respect to t, we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M51','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M51">View MathML</a>


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M52','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M52">View MathML</a>


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M53','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M53">View MathML</a>

Hence, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M54','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M54">View MathML</a>


Similar to the argument of Lemma 3.2, multiply both sides of (3.4) by G(t, α), and (3.7) by ϕ(t, α) then we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M55','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M55">View MathML</a>


Note that ϕ(b(α), α) = 0, thus integrating both sides of (3.8) from τ(α) to b(α), we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M56','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M56">View MathML</a>


Since τ(α) to be the last zero of ϕ(t, α) in (a, b(α)), the behavior of ϕ(t, α) in (τ(α), b(α)) can be classified into two cases as follows:

(i) ϕ (t, α) > 0 in (τ(α), b(α)), then the left side of (3.9) is negative, but by (F3) the right side is positive.

It is impossible.

(ii) ϕ (t, α) < 0 in (τ(α), b(α)), then the left side of (3.9) is positive, but by (F3) the right side is negative.

It is also impossible. The proof is complete.

The proof of Theorem 3.1 We will prove it as a standard process. Assume that N is a nonempty set, otherwise we have nothing to prove. Let α N, then u(b(α), α) = 0. It is easy to see that u'(b(α), α) ≤ 0. If u'(b(α), α) = 0, then the assumption f(t, 0) ≡ 0 for all t ≥ 0, and the uniqueness of solution of initial value problems for ordinary differential equations imply that u(t, α) ≡ 0 for all t ∈ [a, b(α)], which contradicts the initial condition of u(t, α). Hence, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M57','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M57">View MathML</a>


and the implicit function theorem implies that b(α) is well-defined as a function of α in N and b(α) ∈ C1(N). Furthermore, it follows from (3.10) that N is an open set. By Lemma 3.2, we have N is an open interval (see [8]).

Differentiate both sides of the identity u(b(α), α) = 0 with respect to α, we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M58','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/38/mathml/M58">View MathML</a>

From above Lemma 3.3, we have ϕ(b(α)) ≠ 0. Thus, b'(α) ≠ 0, α N. It means that b'(α) does not change sign, i.e., b(α) is monotone. The proof is complete.

Remark 3.2 Actually, if the functions f(|x|,u) and g(|x|) satisfy some suitable conditions, it is not difficult to get the existence of positive radial solutions to the Dirichlet boundary value problem (1.1). We just need that for Equation 2.1, the functions f(|x|,u) and g(|x|) satisfy inequality (2.2) and

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However, it seems that these assumptions are too strict.

Competing interests

The authors declare that they have no competing interests.

Authors' contributions

CL and YZ both carried out all studies in the article and approved the final version.


Li thanks Zhou for enthusiastic guidance and constant encouragement. The authors were very grateful to the anonymous referees for careful reading and valuable comments. This study was partially supported by the Zhejiang Innovation Project (Grant No. T200905), ZJNSF (Grant No. R6090109) and NSFC (Grant No. 10971197).


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