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Uniqueness of positive solutions to a class of semilinear elliptic equations
Boundary Value Problems volume 2011, Article number: 38 (2011)
Abstract
In this article, we consider the uniqueness of positive radial solutions to the Dirichlet boundary value problem
where Ω denotes an annulus in ℝn(n ≥ 3). The uniqueness criterion is established by applying shooting method.
1 Introduction
This article is concerned with the positive radial solutions to a class of semilinear elliptic equations
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.,
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
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
where t = |x|. For giving a proof of uniqueness of problem (1.1), let us consider the initial problem
where . We shall show that problem (2.1) has a unique positive solution. By a solution to problem (1.2), we mean u ∈ C2and 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
Then, Equation 2.1 has a unique positive solution.
Proof. We assume that
endowed with the supremum norm ||u|| = supa≤t≤b|u(t)|. Let
Define the operator T : K → X, by
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
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 M 1 such that
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,
From this, it follows that
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
Similar to (2.3), there exists a constant M2 such that
Take a sequence {u n } ⊂ K, by the mean value theorem, we have
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
is a solution of Cauchy problem
where ψ = f(t,v) - f(t,u). It follows that
Hence, we have
where M3 is a constant, such that
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,
From this, we have 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
where , then problem (1.1) has at most one positive radial solution.
Example 3.1 For the equation
where Let t = |x|, then
A straightforward computation yields
and
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
and satisfies
Let L be the linear operator given by
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
where
and accordingly
It is easy to see that
Differentiating Q(t, α) with respect to t, we get
and
Hence, we have
From hypotheses (F 2), we obtain
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 (F 1), then
(H 1) 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
Using the definition of L, we have
Similar to (3.4), we have
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
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
Thus, . 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 (F 3) 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
Differentiating G(t, α) with respect to t, we get
and
Hence,
Hence, we have
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
Note that ϕ(b(α), α) = 0, thus integrating both sides of (3.8) from τ(α) to b(α), we obtain
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 (F 3) 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 (F 3) 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
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
From above Lemma 3.3, we have ϕ(b(α),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
However, it seems that these assumptions are too strict.
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Acknowledgements
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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CL and YZ both carried out all studies in the article and approved the final version.
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Li, C., Zhou, Y. Uniqueness of positive solutions to a class of semilinear elliptic equations. Bound Value Probl 2011, 38 (2011). https://doi.org/10.1186/1687-2770-2011-38
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DOI: https://doi.org/10.1186/1687-2770-2011-38