Dynamic Systems and Applications 15 (2006) 3-20

EXISTENCE PRINCIPLES AND A UPPER AND LOWER SOLUTION THEORY FOR THE ONE-DIMENSION p-LAPLACIAN

SINGULAR BOUNDARY VALUE PROBLEM WITH SIGN CHANGING NONLINEARITIES

R. P. AGARWAL^{1}, H. GAO^{2}, D. JIANG^{3}, D. O’REGAN^{4}, AND X. ZHANG^{2}

1Department of Mathematical Sciences, Florida Institute of Technology, Melbourne, Florida 32901, USA. 2Applied Science College, Changahun University, Changchun 130022, P.R. China. 3Department of Mathematics, Northeast Normal University, Changchun 130024, P.R. China. 4Department of Mathematics, National University of Ireland, Galway, Ireland.

Dedicated to John Baxley on the occasion of his retirement from Wake Forest University.

ABSTRACT. The singular boundary value problem

\{ \begin{matrix}(Φ(u′))′ + q(t)g(t, u) = 0, 0 < t < 1, \\ u(0) = 0, u(1) + φ(u′(1)) = 0 \end{matrix}

is studied in this paper; here Φ(s) = |s|^{p−2}s, p > 1, and φ is a continuous, strictly increasing odd function defined on (−∞, +∞). The singularity may occur at u = 0 or t = 0, and the function g may be superlinear at u = ∞ and may change signs. The existence of solutions is obtained via an upper and lower solutions method and existence principles.

AMS (MOS) Subject Classification. 34B15.

1. INTRODUCTION

In this paper, we study the existence of positive solutions to the nonlinear singular boundary value problem

\{ \begin{matrix}(Φ(u′))′ + q(t)g(t, u(t)) = 0, 0 < t < 1, \\ u(0) = 0, u(1) + φ(u′(1)) = 0 \end{matrix}  (1.1)

where Φ(s) = |s|^{p−2}s, p > 1, and φ is a continuous, strictly increasing odd function defined on (−∞, +∞). Here q(t) may be singular at t = 0, the nonlinearity g may be singular at u = 0, and g may be superlinear at u = ∞ and may change signs.

Received August 20, 2004                1056-2176 $15.00 \begin{matrix}c \\ ◯ \end{matrix}Dynamic Publishers, Inc.

4 AGARWAL, GAO, JIANG, O’REGAN, AND ZHANG

Problem (1.1) has been studied extensively in the literature; we refer the reader to [1-6, 8, 9, 11, 12, 17-20] (p = 2) and [7, 10, 13-16, 22-24]. In [7, 22] the problem is nonsingular and in [10, 15, 19, 23] the problem is only able to possess singularities at t = 0 or t = 1. In [9] (p = 2), the authors established a theorem on upper and lower solutions (Theorem 1) for the singular Dirichlet problem, and using this theorem, they proved the existence and uniqueness of solutions. In [13, 14], using the method in [9], the authors studied the Dirichlet problem in which q(t)g(t, u) may be singular at u = 0, t = 0, or t = 1, and g may be superlinear at u = ∞ and may change signs. Motivated by the results mentioned above, the purpose of this paper is to establish the existence of a positive solution to problem (1.1) by applying the method used in [9, 13, 14]. In Section 2, we establish some existence principles which are extensions of the existence principles in [2, 22]. In Section 3, we present result on upper and lower solutions. Section 4 is devoted to the proof of our main result. Finally, an example is given in Section 5.

2. EXISTENCE PRINCIPLES

In this section, we establish existence principles for the singular boundary value problem

\{ \begin{matrix}(Φ(u′))′ + F (t, u) = 0, 0 < t < 1, \\ u(0) = A, u(1) + φ(u′(1)) = B, \end{matrix}  (2.1)

where Φ(s) = |s|^{p−2}s, p > 1. Suppose the following three conditions are satisfied: (A1) φ is a continuous, strictly increasing odd function defined on (−∞, +∞); (A2) F (t, u) : (0, 1] × R → R is continuous; (A3) for each r > 0 there exists h_{r} ∈ C((0, 1], (0, ∞)) such that |u| ≤ r implies

|F (t, u)| ≤ h_{r}(t) for t ∈ (0, 1]. Moreover, there exist 0 ≤ μ < p − 1 such that

∫_{0}^{1} t^{μ}h_{r}(t)dt < ∞,  (2.2)

with

lim_{t→0^{+}} t^{μ+1}h_{r}(t) = 0, if ∫_{0}^{1/2} h_{r}(t)dt = ∞.  (2.3)

Remark 2.1. For example, the function

h_{r}(t) = t^{−a}, 0 < t ≤ 1, 0 ≤ a < p

satisfies the condition (A3) provided μ ∈ (a − 1, p − 1) ∩ [0, p − 1).

Remark 2.2. It is easy to check that the condition (A3) implies that

∫_{0}^{1/2} Φ^{−1}(Φ ∘ (I + φ)^{−1}(|B − A|) + ∫_{s}^{1/2} h_{r}(x)dx)ds < ∞,

and

∫_{1/2}^{1} Φ^{−1}(Φ ∘ (I + φ)^{−1}(|B − A|) + ∫_{1/2}^{s} h_{r}(x)dx)ds < ∞,

where Φ^{−1}(u) := |u|^{\frac{1}{p−1}} sgnu is the inverse function to Φ(v).

In fact,

∫_{0}^{1/2} Φ^{−1}(Φ ∘ (I + φ)^{−1}(|B − A|) + ∫_{s}^{1/2} h_{r}(x)dx)ds
≤ ∫_{0}^{1/2} Φ^{−1}(Φ ∘ (I + φ)^{−1}(|B − A|) + ∫_{s}^{1/2} \frac{x^{μ}}{s^{μ}} h_{r}(x)dx)ds
≤ ∫_{0}^{1/2} s^{− \frac{μ}{p−1}} Φ^{−1}(Φ ∘ (I + φ)^{−1}(|B − A|) + ∫_{s}^{1/2} x^{μ}h_{r}(x)dx)ds
≤ ∫_{0}^{1/2} s^{− \frac{μ}{p−1}} dsΦ^{−1}(Φ ∘ (I + φ)^{−1}(|B − A|) + ∫_{0}^{1/2} x^{μ}h_{r}(x)dx) < ∞,

and clearly, we have

∫_{1/2}^{1} Φ^{−1}(Φ ∘ (I + φ)^{−1}(|B − A|) + ∫_{1/2}^{s} h_{r}(x)dx)ds < ∞.

Lemma 2.1. There exists an unique solution V_{r} ∈ C([0, 1], R) ∩ C^{1}((0, 1], R), with t^{\frac{μ}{p−1}} V′_{r}(t) ∈ C([0, 1], R), to the problem

\{ \begin{matrix}(Φ(V′_{r}))′ + h_{r}(t) = 0, 0 < t < 1, \\ V_{r}(0) = A, V_{r}(1) + φ(V′_{r}(1)) = B; \end{matrix}  (2.4)

here h_{r} is as described in (A3).

Proof. First we prove uniqueness. Suppose that V_{r}, V̄_{r} are two solutions to (2.4) and let z(t) = V_{r}(t) − V̄_{r}(t). If z(t) ≢ 0 on [0,1], since z(0) = 0, then |z(σ)| = max_{t∈[0,1]} |z(t)| > 0, and σ ∈ (0, 1]. Without loss of generality, we assume that z(σ) > 0. Since z(0) = 0, then there exists an interval (a, σ] ⊂ [0, 1] such that

z(a) = 0, z(t) > 0, t ∈ (a, σ].

If σ = 1, then (V_{r} − V̄_{r})′(1) ≥ 0, and so we have φ(V_{r}′(1)) ≥ φ(V̄_{r}′(1)). It follows from (2.4) that z(1) = −φ(V_{r}′(1)) + φ(V̄_{r}′(1)) ≤ 0, which is a contradiction. Thus σ ∈ (a, 1) and z′(σ) = 0.

Notice that

(Φ(V_{r}′(x)))′ = −h_{r}(x) in (0, 1).

Integrating both sides of the above equality over [s, σ], a < s ≤ σ, we have

V_{r}′(s) = Φ^{−1}(Φ(V_{r}′(σ)) + ∫_{s}^{σ} h_{r}(x)dx)ds, a < s ≤ σ.

Integrate both sides of the above equality from a to σ to obtain

V_{r}(σ) − V_{r}(a) = ∫_{a}^{σ} Φ^{−1}(Φ(V_{r}′(σ)) + ∫_{s}^{σ} h_{r}(x)dx)ds.

Similarly, we have

V̄_{r}(σ) − V̄_{r}(a) = ∫_{a}^{σ} Φ^{−1}(Φ(V̄_{r}′(σ)) + ∫_{s}^{σ} h_{r}(x)dx)ds.

6 AGARWAL, GAO, JIANG, O’REGAN, AND ZHANG

Consequently, z(σ) = V_{r}(σ) − V̄_{r}(σ) = 0, which is a contradiction.

To prove the existence of solutions, we set, for 0 < t < 1,

y(t) := A + ∫_{0}^{t} Φ^{−1}( Φ ∘ (I + φ)^{−1}(B − A) + ∫_{s}^{t} h_{r}(x) dx) ds
−B − φ ∘ Φ^{−1}[Φ ∘ (I + φ)^{−1}(A − B) + ∫_{t}^{1} h_{r}(x) dx]
− ∫_{t}^{1} Φ^{−1}( Φ ∘ (I + φ)^{−1}(A − B) + ∫_{t}^{s} h_{r}(x) dx) ds.

Clearly, by Remark 2.2, y(t) is continuous in (0,1] and y(0^{+}) < 0 < y(1) . Thus, y(t) has zeros in (0,1). Let σ be a zero of y(t) in (0,1). Then

A + ∫_{0}^{σ} Φ^{−1}( Φ ∘ (I + φ)^{−1}(B − A) + ∫_{s}^{σ} h_{r}(x) dx) ds

= B + φ ∘ Φ^{−1}[Φ ∘ (I + φ)^{−1}(A − B) + ∫_{σ}^{1} h_{r}(x) dx]
+ ∫_{σ}^{1} Φ^{−1}( Φ ∘ (I + φ)^{−1}(A − B) + ∫_{σ}^{s} h_{r}(x) dx) ds.

Put

V_{r}(t) = \{ \begin{matrix}A + ∫_{0}^{t} Φ^{−1}( Φ ∘ (I + φ)^{−1}(B − A) + ∫_{s}^{σ} h_{r}(x) dx) ds ,  0 < t ≤ σ, \\ B + φ ∘ Φ^{−1}[Φ ∘ (I + φ)^{−1}(A − B) + ∫_{σ}^{1} h_{r}(x) dx] \\ + ∫_{t}^{1} Φ^{−1}( Φ ∘ (I + φ)^{−1}(A − B) + ∫_{σ}^{s} h_{r}(x) dx) ds ,  σ ≤ t ≤ 1. \end{matrix}
(2.5)

Then, V_{r} is a well defined function on (0,1]. Moreover

V′_{r}(t) = \{ \begin{matrix}Φ^{−1}(Φ ∘ (I + φ)^{−1}(B − A) + ∫_{t}^{σ} h_{r}(x)dx), 0 < t ≤ σ, \\ −Φ^{−1}(Φ ∘ (I + φ)^{−1}(A − B) + ∫_{σ}^{t} h_{r}(x)dx), σ ≤ t ≤ 1. \end{matrix}  (2.6)
For 0 < t ≤ σ,
|V_{r}(t) − A| ≤ ∫_{0}^{t} Φ^{−1}(Φ ∘ (I + φ)^{−1}(|B − A|) + ∫_{s}^{σ} h_{r}(x)dx)ds
≤ ∫_{0}^{t} Φ^{−1}(Φ ∘ (I + φ)^{−1}(|B − A|) + ∫_{s}^{σ} \frac{x^{μ}}{s^{μ}} h_{r}(x)dx)ds
≤ ∫_{0}^{t} s^{− \frac{μ}{p−1}} dsΦ^{−1}(Φ ∘ (I + φ)^{−1}(|B − A|) + ∫_{0}^{σ} x^{μ}h_{r}(x)dx)ds
= \frac{p−1}{p−1−μ}t^{\frac{p−1−μ}{p−1}} Φ^{−1}(Φ ∘ (I + φ)^{−1}(|B − A|) + ∫_{0}^{σ} x^{μ}h_{r}(x)dx),

and we have V_{r}(0) = A.

For 0 < t ≤ σ, it follows from (2.2)-(2.4) and (2.6) that, if ∫_{0}^{1/2} h_{r}(x)dx < ∞ and μ = 0, then V′_{r}(t) is continuous at t = 0; if 0 < μ < p − 1 and ∫_{0}^{1/2} h_{r}(x)dx = ∞, then

lim_{t→0^{+}} t^{μ} ∫_{t}^{σ} h_{r}(x)dx = lim_{t→0^{+}} \frac{∫_{t}^{σ} h_{r}(x)dx}{t^{−μ}} = lim_{t→0^{+}} \frac{h_{r}(t)}{μt^{−μ−1}} = lim_{t→0^{+}} \frac{1}{μ}t^{μ+1}h_{r}(t) = 0,

and so lim_{t→0^{+}} t^{\frac{μ}{p−1}} V′_{r}(t) = 0.

Therefore, V_{r}(t) is continuous on [0,1], t^{\frac{μ}{p−1}} V′_{r}(t) ∈ C([0, 1], R) and

V_{r}(0) = A, V_{r}(1) + φ(V′_{r}(1)) = B,
[Φ(V′_{r}(t))]′ = −h_{r}(t), for t ∈ (0, 1).

This complete the proof.

Remark 2.3. Here σ in Lemma 2.1 is the unique point in (0,1) with V′_{r}(σ) = (I + φ)^{−1}(B − A), since [Φ(V′_{r})]′ < 0 on (0, 1] and V_{r} is concave on [0,1].

EXISTENCE PRINCIPLES 7

Similarly we have the following result.

Lemma 2.2. There exists an unique solution v_{r} ∈ C([0, 1], R) ∩ C^{1}((0, 1], R), with t^{\frac{μ}{p−1}} v′_{r}(t) ∈ C([0, 1], R) to the problem

\{ \begin{matrix}(Φ(v′_{r}))′ − h_{r}(t) = 0, 0 < t < 1, \\ v_{r}(0) = A, v_{r}(1) + φ(v′_{r}(1)) = B; \end{matrix}

here h_{r} is as in (A3).

Suppose that D ⊂ C([0, 1], R) is a bounded set, and there exists a constant r > 0 such that ||u|| ≤ r for u ∈ D̄, where

||u|| = max_{t∈[0,1]} |u(t)|.

Thus |F (t, u(t))| ≤ h_{r}(t) for u ∈ D̄.

For each fixed u ∈ D, we consider the boundary value problem

\{ \begin{matrix}(Φ(w′(t)))′ + F (t, u(t)) = 0, 0 < t < 1, \\ w(0) = A, w(1) + φ(w′(1)) = B. \end{matrix}  (2.7)

Now, we consider the modified boundary value problem

\{ \begin{matrix}(Φ(w′(t)))′ + η_{n}(t)F (t, u(t)) = 0, 0 < t < 1, \\ w(0) = A, w(1) + φ(w′(1)) = B, \end{matrix}  (2.7)^{n}

where n ≥ 4 is a natural number, η_{n}(t) is a continuous function such that 0 ≤ η_{n}(t) ≤ 1 on [0,1] and

η_{n}(t) = \{ \begin{matrix}0 ,  0 ≤ t < \frac{1}{2n}, \\ 1 ,  \frac{1}{n} ≤ t ≤ 1. \end{matrix}

From Lemma 2.1 for each fixed u ∈ D, the boundary value problem (2.7)^{n} has a unique solution w_{n} ∈ C^{1}[0, 1]. We also see that (2.7)^{n} is equivalent to

w_{n}(t) = A + ∫_{0}^{t} Φ^{−1}(τ_{n} + ∫_{x}^{1} η_{n}(s)F (s, u(s))ds)dx =: (T_{n}u)(t), 0 ≤ t ≤ 1, (2.8)^{n}

where τ_{n} = Φ(w′_{n}(1)) is the solution of the equation

∫_{0}^{1} Φ^{−1}(τ_{n} + ∫_{x}^{1} η_{n}(s)F (s, u(s))ds)dx + φ(Φ^{−1}(τ_{n})) = B − A.  (2.9)^{n}

In fact, τ_{n} is the unique solution of (2.9)^{n}, since Φ^{−1} and φ are continuous, strictly increasing function on R with Φ^{−1}(R) = R and φ(R) = R (cf. [22, Theorem 2.1]).

Lemma 2.3. Let w_{n}(t) be a solution to problem (2.7)^{n}. Then

v_{r}(t) ≤ w_{n}(t) = (T_{n}u)(t) ≤ V_{r}(t), 0 ≤ t ≤ 1.

Proof. We shall only prove w_{n}(t) ≤ V_{r}(t) on [0,1] since the arguments are essentially the same for the case w_{n}(t) ≥ v_{r}(t) on [0,1].

8 AGARWAL, GAO, JIANG, O’REGAN, AND ZHANG

Assume that w_{n}(t) > V_{r}(t) for some t ∈ [0, 1]. Since w_{n}(0) = V_{r}(0), w_{n}(1) + φ(w′_{n}(1)) = V_{r}(1) + φ(V′_{r}(1)), we have as in Lemma 2.1 that there exists an interval (a, σ] ⊂ (0, 1) such that the function y(t) := w_{n}(t) − V_{r}(t) > 0 in (a, σ], and

y(a) = 0, y(σ) = max_{t∈[0,1]} y(t) > 0 with y′(σ) = 0.

Notice that

[Φ( w′_{n}(t))]′ − [Φ( V′_{r}(t))]′ = −η_{n}(t)F (t, u(t)) + h_{r}(t) ≥ 0, for all t ∈ (0, 1). Integrate both sides of the above inequality over [t, σ], a < t < σ, and we get

[Φ( w′_{n}(σ))] − [Φ( w′_{n}(t))] ≥ [Φ( V′_{r}(σ))] − [Φ( V′_{r}(t))] for all t ∈ (a, σ], i.e.,

y′(t) = w′_{n}(t) − V′_{r}(t) ≤ 0, in (a, σ],

so we have y(σ) ≤ y(a) = 0, a contradiction.

Lemma 2.4. Let w_{n}(t) = (T_{n}u)(t) be a solution to (2.7)^{n}. Then there is a nonnegative function y_{r} ∈ C([0, 1], [0, ∞)) such that

|t^{\frac{μ}{p−1}} w′_{n}(t)| ≤ y_{r}(t).

Proof. Let [a, 1] ⊂ (0, 1] be a compact interval. Notice that

w′_{n}(t) = Φ^{−1}(C_{n} + ∫_{t}^{1} η_{n}(s)F (s, u(s))ds), a ≤ t ≤ 1,  (2.10)

where C_{n} is a solution of the equation

∫_{a}^{1} Φ^{−1}(C_{n} + ∫_{r}^{1} η_{n}(s)F (s, u(s))ds)dr = w_{n}(1) − w_{n}(a).  (2.11)

From the first mean value theorem, there exists a ξ_{n} ∈ [a, 1] such that

Φ^{−1}(C_{n} + ∫_{ξ_{n}}^{1} η_{n}(s)F (s, u(s))ds) = \frac{w_{n}(1) − w_{n}(a)}{1 − a},

i.e.,

C_{n} = − ∫_{ξ_{n}}^{1} η_{n}(s)F (s, u(s))ds + Φ(\frac{w_{n}(1) − w_{n}(a)}{1 − a}).

From Lemma 2.3, v_{r}(t) ≤ w_{n}(t) ≤ V_{r}(t), so there is a M = M(r, a) > 0 such that

|C_{n}| ≤ ∫_{a}^{1} h_{r}(s)ds + Φ(\frac{|w_{n}(1) − w_{n}(a)|}{1 − a}) ≤ M,  (2.12)

and

|C_{n} + ∫_{t}^{1} η_{n}(s)F (s, u(s))ds| ≤ M.  (2.13)

From (2.10), (2.12) and (2.13), we have |w′_{n}(t)| ≤ C(r, a), a ≤ t ≤ 1, where C(r, a) is a positive constant independent of n, u(t), and t ∈ [a, 1]. Thus, there is a positive constant C_{r} such that

|w′_{n}(1)| ≤ C_{r}.

Now since

w′_{n}(t) = Φ^{−1}(Φ[w′_{n}(1)] + ∫_{t}^{1} η_{n}(s)F (s, u(s))ds), 0 < t < 1,

then there is a nonnegative function R_{r} ∈ C(0, 1) such that

|w′_{n}(t)| ≤ Φ^{−1}(Φ(C_{r}) + | ∫_{t}^{1} h_{r}(s)ds|) =: R_{r}(t), 0 < t < 1.

If we put y_{r}(t) := t^{\frac{μ}{p−1}} R_{r}(t), as in the proof of Lemma 2.1, we see that y_{r} ∈ C([0, 1], [0, ∞)).

Lemma 2.5. For each fixed n ≥ 4, the mapping T_{n} : D̄ → C([0, 1], R) is continuous and compact, where D is defined as before.

Proof We only prove the continuity, since the boundedness and equicontinuity of T_{n} follows from Lemma 2.4. Assume that u_{0}, u_{j} ∈ D̄and ||u_{j} − u_{0}|| → 0 as j → ∞. Then we have

(T_{n}u_{j})(t) = ∫_{0}^{t} Φ^{−1}[τ_{nj} + ∫_{r}^{1} η_{n}(s)F (s, u_{j}(s))ds]dr + A,

where τ_{nj} = Φ((T_{n}u_{j})′(1)) is the unique solution of the following equation

∫_{0}^{1} Φ^{−1}(τ_{nj} + ∫_{r}^{1} η_{n}(s)F (s, u_{j}(s))ds)dr + φ[Φ^{−1}(τ_{nj})] + A − B = 0. (2.14)_{nj}

From Lemma 2.4, we know that |(T_{n}u_{j})′(1)| ≤ C_{r}, thus, |τ_{nj}| ≤ Φ(C_{r}), j = 1, 2, · · ·. Suppose that τ_{n∗} ∈ [−Φ(C_{r}), Φ(C_{r})] is an accumulation point of {τ_{nj}}. Then there is a subsequence of {τ_{nj}}, {τ_{nj(m)}}, which converges to τ_{n∗}. From Lemma 2.4 and the dominated convergence theorem, it follows from (2.14)_{nj(m)} that

∫_{0}^{1} Φ^{−1}(τ_{n∗} + ∫_{r}^{1} η_{n}(s)F (s, u_{0}(s))ds)dr + φ[Φ^{−1}(τ_{n∗})] + A − B = 0.

This shows that τ_{n∗} = τ_{n0}, and hence τ_{nj} → τ_{n0}. Thus, we use the dominated convergence theorem to obtain

lim_{j→∞}(T_{n}u_{j})(t) = lim_{j→∞} ∫_{0}^{t} Φ^{−1}[τ_{nj} + ∫_{r}^{1} η_{n}(s)F (s, u_{j}(s))ds]dr + A
= ∫_{0}^{t} Φ^{−1}[τ_{n0} + ∫_{r}^{1} η_{n}(s)F (s, u_{0}(s))ds]dr + A
= (T_{n}u_{0})(t), uniformly for t ∈ [0, 1].

This shows that T_{n} is continuous from D̄ to C([0, 1], R).

10 AGARWAL, GAO, JIANG, O’REGAN, AND ZHANG

Since {w_{n}(t) = (T_{n}u)(t)}_{n=4}^{∞} is equicontinuous on [0,1], applying the Ascoli-Arzela theorem, we know that there exist an uniformly convergent subsequence in {w_{n}(t) = (T_{n}u)(t)}_{n=4}^{∞} denoted again by {w_{n}(t) = (T_{n}u)(t)}_{n=4}^{∞} which converges to w(t) := (T u)(t) uniformly on [0,1]. On the other hand, since |τ_{n}| = Φ(|w′_{n}(1)|) ≤ Φ(C_{r}), there exists a subsequence in {τ_{n}}_{n=4}^{∞} denoted again by {τ_{n}}_{n=4}^{∞} which converges to τ. Thus, we have ||T_{n} − T || → 0 when n → ∞, and T : D̄ → C[0, 1] is continuous and compact. Moreover,

w(t) = A + ∫_{0}^{t} Φ^{−1}(τ + ∫_{x}^{1} F (s, u(s))ds)dx =: (T u)(t), 0 ≤ t ≤ 1,  (2.15)

where τ = Φ(w′(1)) is the unique solution of the equation

∫_{0}^{1} Φ^{−1}(τ + ∫_{x}^{1} F (s, u(s))ds)dx + φ(Φ^{−1}(τ)) = B − A.  (2.16)

This implies that w = (T u)(t) is a solution of (2.7).

Since D is an arbitrary bounded subset in C([0, 1], R), we have the following.

Lemma 2.6. T : C([0, 1], R) → C([0, 1], R) is continuous and completely continuous.

We have the following general existence principles from the Schauder fixed point theorem and the Leray–Schauder alternative.

Theorem 2.1. Suppose (A1)−(A3) hold. In addition, suppose there is a constant M > |A| + |B|, independent of λ, with

||u|| = max_{t∈[0,1]} |u(t)| ≠ M  (2.17)

for any solution u ∈ C([0, 1], R) ∩ C^{1}((0, 1], R) to

\{ \begin{matrix}(Φ(u′))′ + λ^{p−1}F (t, u) = 0, 0 < t < 1, \\ u(0) = λA, u(1) + λφ(\frac{u′(1)}{λ}) = λB \end{matrix}  (2.18)_{λ}

for each λ ∈ (0, 1). Then (2.1) has a solution u with ||u|| ≤ M.

Proof. Notice (2.18)_{λ} is equivalent to the fixed point problem

u = λT u, u ∈ C([0, 1], R),  (2.19)_{λ}

where T is as in (2.15). Set

U = {u ∈ C([0, 1], R) : ||u|| < M}.

Since T : C([0, 1], R) → C([0, 1], R) is continuous and completely continuous the Leray–Schauder alterative [19] guarantees that T has a fixed point i.e., (2.19)_{1} has a solution in Ū.

EXISTENCE PRINCIPLES 11

Theorem 2.2. Suppose that (A1) holds, and there exists h ∈ C((0, 1], (0, ∞)) with |F (t, u)| ≤ h(t) for t ∈ (0, 1] and u ∈ R. Moreover assume there exist 0 ≤ μ < p − 1 such that

∫_{0}^{1} t^{μ}h(t)dt < ∞,

with

lim_{t→0^{+}} t^{μ+1}h(t) = 0, if ∫_{0}^{1/2} h(t)dt = ∞.

Then (2.1) has a solution u.

Proof. Solving (2.1) is equivalent to the fixed point problem u = T u. Since T : C([0, 1], R) → C([0, 1], R) is continuous and compact the result follows from Schauder’s fixed point theorem.

3. UPPER AND LOWER SOLUTIONS

Consider the two-point boundary value problem

\{ \begin{matrix}[Φ(u′)]′ + f(t, u) = 0, t ∈ (0, 1), \\ u(0) = A, u(1) + φ(u′(1)) = B, \end{matrix}  (3.1)

where Φ(s) = |s|^{p−2}s , p > 1 , f : D → R is a continuous function and D ⊂ (0, 1] × R.

By a solution u(•) of (3.1) we mean a function u ∈ C([0, 1], R) ∩ C^{1}((0, 1], R), Φ(u′) ∈ C^{1}((0, 1], R) such that (t, u(t)) ∈ D, for all t ∈ (0, 1] and [Φ(u′)]′+f(t, u(t)) = 0, for t ∈ (0, 1), with u(0) = A and u(1) + φ(u′(1)) = B.

Let α ∈ C([0, 1], R) ∩ C^{1}((0, 1], R), Φ(α′) ∈ C^{1}((0, 1], R) satisfy the following condition: (t, α(t)) ∈ D, for t ∈ (0, 1] and

\{ \begin{matrix}[Φ(α′)]′ + f(t, α) ≥ 0, t ∈ (0, 1), \\ α(0) ≤ A, α(1) + φ(α′(1)) ≤ B. \end{matrix}

In this case, we say that α(•) is a lower solution for problem (3.1). The definition of an upper solution β(•) for problem (3.1) is given in a completely similar way, by just reversing the above inequalities. Also, if α, β ∈ C([0, 1], R) are such that α(t) ≤ β(t), for all t ∈ [0, 1], we define the set

D_{α}^{β} := {(t, x) ∈ (0, 1] × R : α(t) ≤ x ≤ β(t)}.

Theorem 3.1. Let α, β be respectively a lower solution and an upper solution for problem (3.1) such that

(a_{1}) φ is a continuous, strictly increasing odd function defined on (−∞, +∞), (a_{2}) α(t) ≤ β(t) for all t ∈ [0, 1],

and suppose that

(a_{3}) D_{α}^{β} ⊂ D. Assume also that there is a function h ∈ C( (0, 1], (0, ∞)) such that

12 AGARWAL, GAO, JIANG, O’REGAN, AND ZHANG

(a_{4})  |f(t, x)| ≤ h(t)  for all (t, x) ∈ D_{α}^{β}.

Moreover, assume there exist 0 ≤ μ < p − 1 such that

(a_{5}) ∫_{0}^{1} t^{μ}h(t)dt < +∞, with

lim_{t→0^{+}} t^{μ+1}h(t) = 0, if ∫_{0}^{1/2} h(t)dt = ∞.

Then problem (3.1) has at least one solution u(t) such that

α(t) ≤ u(t) ≤ β(t) for all t ∈ [0, 1].

Consider now the boundary value problem

\{ \begin{matrix}[Φ(u′)]′ + f^{∗}(t, u) = 0 , t ∈ (0, 1), \\ u(0) = A, u(1) + φ(u′(1)) = B, \end{matrix}  (3.2)^{∗}

where

f^{∗}(t, x) = \{ \begin{matrix}f(t, α(t)),  if x < α(t), \\ f(t, x),  if α(t) ≤ x ≤ β(t), \\ f(t, β(t)),  if x > β(t). \end{matrix}

From (a_{3}) and the definition of f^{∗} we see that f^{∗} : (0, 1) × R → R is continuous and by (a_{4}), it satisfies

|f^{∗}(t, x)| ≤ h(t)  for all (t, x) ∈ (0, 1) × R.  (3.3)

By Theorem 2.2, (3.2)^{∗} has a solution u ∈ C([0, 1], R) ∩ C^{1}((0, 1], R).

Lemma 3.1. Let u(t) be a solution to (3.2)^{∗}. Then

α(t) ≤ u(t) ≤ β(t) for all t ∈ [0, 1],

i.e., u(t) is a solution to (3.1).

Proof. We prove that u(t) ≤ β(t) on [0,1]. Let x(t) := u(t) − β(t). Assume that u(t) > β(t) for some t ∈ [0, 1]. Since x(0) ≤ 0, as in the proof of Lemma 2.3, there exists an interval (a, σ] ⊂ (0, 1) such that x(t) > 0 in (a, σ], and

x(a) = 0, x(σ) = max_{t∈[0,1]} x(t) > 0, x′(σ) = 0.

For t ∈ (a, σ] we have that f^{∗}(t, u(t)) = f(t, β(t)) and therefore

[Φ(u′(t))]′ + f(t, β(t)) = 0 for all t ∈ (a, σ].

On the other hand, as β is an upper solution for (3.1), we also have

[Φ(β′(t))]′ + f(t, β(t)) ≤ 0 for all t ∈ (a, σ].

Thus, we obtain

[Φ(u′(t))]′ ≥ [Φ(β′(t))]′ for all t ∈ (a, σ].

EXISTENCE PRINCIPLES 13

Integrating both sides of this inequality over [t, σ] ⊂ (a, σ], we obtain

Φ(u′(σ)) − Φ(u′(t)) ≥ Φ(β′(σ)) − Φ(β′(t)),

i.e.,

x′(t) ≤ 0 in (a, σ].

Consequently, 0 = x(a) ≥ x(σ) > 0, which is a contradiction. Similarly u(t) ≥ α(t) on [0,1]. The Lemma is proved.

Thus the proof of Theorem 3.1 is complete.

4. MAIN RESULTS

Let g : [0, 1] × (0, ∞) → R be a continuous function and consider the two-point boundary value problem

\{ \begin{matrix}(Φ(u′))′ + q(t)g(t, u) = 0,  t ∈ (0, 1), \\ u(0) = 0, u(1) + φ(u′(1)) = 0. \end{matrix}  (4.1)

Motivated by the example q(t)g(t, u) = σt^{−m}(u^{−a} + u^{b} − sin(8πt)), 0 ≤ m < p, a > 0, σ > 0, b ≥ 0, we have the following result.

Theorem 4.1. Assume that

(H_{1}) φ is a continuous, strictly increasing odd function defined on (−∞, +∞), and \frac{φ(x)}{x} is nondecreasing on (−∞, 0) and nonincreasing on (0, ∞); (H_{2}) |g(t, x)| ≤ F (x) + Q(x) on [0, 1] × (0, ∞) with F > 0 continuous and nonin-

creasing on (0, ∞), Q ≥ 0 continuous on [0, ∞), and Q/F nondecreasing on (0, ∞);

(H_{3}) there exist constants L > 0 and ε > 0 such that g(t, x) > L for all (t, x) ∈ [0, 1] × (0, ε], and F (x) > L, x ∈ (0, ε];

(H_{4}) there exist 0 ≤ μ < p − 1 such that

∫_{0}^{1} t^{μ}q(t)dt < +∞,  (4.2)

with

lim_{t→0^{+}} t^{μ+1}q(t) = 0, if ∫_{0}^{1/2} q(t)dt = ∞,

and

sup_{c∈(0,∞)} (\frac{1}{Φ^{−1}(1 + \frac{Q(c)}{F (c)})} ∫_{0}^{c} \frac{du}{Φ^{−1}(F (u))}) > b_{0},  (4.3)

where

b_{0} = ∫_{0}^{1} Φ^{−1}(∫_{s}^{1} q(r)dr)ds.

Then problem (4.1) has at least one solution u ∈ C([0, 1], [0, ∞)) ∩ C^{1}((0, 1], R).

Note that as in Lemma 2.1, we have the following lemma.

Lemma 4.1. There exists an unique solution W ∈ C([0, 1], [0, ∞))∩C^{1}((0, 1], R) with W (t) > 0 on (0, 1] to the problem

\{ \begin{matrix}(Φ(W′))′ + q(t) = 0, 0 < t < 1, \\ W (0) = 0, W (1) + φ(W′(1)) = 0. \end{matrix}  (4.4)

Choose M > 0, δ > 0 (δ < M) such that

\frac{1}{Φ^{−1}(1 + \frac{Q(M)}{F (M)})} ∫_{δ}^{M} \frac{du}{Φ^{−1}(F (u))} > b_{0}.  (4.5)

Let n_{0} ∈ \{1, 2, ⋯ \} be chosen so that \frac{1}{n_{0}} < min\{ε − m||W ||, δ\}, where W is the solution of (4.4), and 0 < m < min\{L^{\frac{1}{p−1}} , ε/||W ||, 1\} is chosen and fixed. Let N^{+} = {n_{0}, n_{0} + 1, ⋯ }.

We first to show that the following boundary value problem

\{ \begin{matrix}(Φ(u′))′ + q(t)g(t, u) = 0, 0 < t < 1, \\ u(0) = \frac{1}{n}, u(1) + φ(u′(1)) = \frac{1}{n}, n ∈ N^{+} \end{matrix}  (4.6)^{n}

has a solution u_{n} for each n ∈ N^{+} with u_{n}(t) ≥ \frac{1}{n} for t ∈ [0, 1] and ||u_{n}|| < M.

Claim 1 Let α_{n}(t) = mW (t) + \frac{1}{n}, t ∈ [0, 1], so α_{n}(t) is a (strict) lower solution for problem (4.6)^{n}.

Proof. For the choice of m and n, we have mW (t) + \frac{1}{n} ≤ m||W || + \frac{1}{n_{0}} < ε, so from (H_{3})

g(t, mW (t) + \frac{1}{n}) > L > m^{p−1} for all t ∈ [0, 1].

Then we obtain

(Φ(α′_{n}(t)))′ + q(t)g(t, α_{n}(t))
= m^{p−1}(Φ(W′(t)))′ + q(t)g(t, mW (t) + \frac{1}{n})
= q(t)(g(t, mW (t) + \frac{1}{n}) − m^{p−1}) > 0, 0 < t < 1.

And from (H_{1}), we have φ(mW′(1)) ≤ mφ(W′(1)), since W′(1) < 0, so we obtain α_{n}(0) = mW (0) + \frac{1}{n} = \frac{1}{n}, and

\begin{aligned}α_{n}(1) + φ(α′_{n}(1)) & = mW (1) + \frac{1}{n} + φ(mW′(1)) \\ & ≤ mW (1) + \frac{1}{n} + mφ(W′(1)) = \frac{1}{n}. \end{aligned}

Thus, the proof of Claim 1 is complete.

In order to find an upper solution of (4.6)^{n}, we consider the problem

\{ \begin{matrix}(Φ(u′))′ + q(t)F (u)(1 + \frac{Q(M)}{F (M)}) = 0, 0 < t < 1, \\ u(0) = \frac{1}{n}, u(1) + φ(u′(1)) = \frac{1}{n}. \end{matrix}  (4.7)^{n}

To show (4.7)^{n} has a solution, we examine

\{ \begin{matrix}(Φ(u′))′ + q(t)F^{∗}(u)(1 + \frac{Q(M)}{F (M)}) = 0, 0 < t < 1, \\ u(0) = \frac{1}{n}, u(1) + φ(u′(1)) = \frac{1}{n}, \end{matrix}  (4.8)^{n}

where

F^{∗}(u) = \{ \begin{matrix}F (u), & u ≥ \frac{1}{n}, \\ F (\frac{1}{n}), & u ≤ \frac{1}{n}. \end{matrix}

Note F^{∗}(u) ≤ F (u) for u > 0.

In the same way as Claim 1, we can easily prove α_{n}(t) = \frac{1}{n} + mW (t) is also a (strict) lower solution of (4.8)^{n}.

From Lemma 2.1, let β_{n}^{0} ∈ C([0, 1], R) ∩ C^{1}((0, 1], R) be the unique solution of the boundary value problem

\{ \begin{matrix}(Φ(u′))′ + q(t)F (α_{n}(t))(1 + \frac{Q(M)}{F (M)}) = 0, 0 < t < 1, \\ u(0) = \frac{1}{n}, u(1) + φ(u′(1)) = \frac{1}{n}. \end{matrix}  (4.9)^{n}

A similar argument as in the proof of Lemma 3.1 yields the following claim. Claim 2. Let y_{1}, y_{2} ∈ C([0, 1], R) ∩ C^{1}((0, 1], R) satisfy

\{ \begin{matrix}−(Φ(y′_{2}(t)))′ ≥ −(Φ(y′_{1}(t)))′, 0 < t < 1, \\ y_{2}(0) ≥ y_{1}(0), \\ y_{2}(1) + φ(y′_{2}(1)) ≥ y_{1}(1) + φ(y′_{1}(1)). \end{matrix}

Then y_{2}(t) ≥ y_{1}(t) for t ∈ [0, 1].

Since β_{n}^{0} is a solution of (4.9)^{n}, then

\{ \begin{matrix}(Φ(β_{n}^{0}′))′ + q(t)F (α_{n}(t))(1 + \frac{Q(M)}{F (M)}) = 0, 0 < t < 1, \\ β_{n}^{0}(0) = \frac{1}{n}, β_{n}^{0}(1) + φ(β_{n}^{0}′(1)) = \frac{1}{n}. \end{matrix}

On the other hand, as α_{n} is a lower solution of (4.8)^{n}, and α_{n} ≥ \frac{1}{n}, we also have

\{ \begin{matrix}(Φ(α_{n}′))′ + q(t)F (α_{n}(t))(1 + \frac{Q(M)}{F (M)}) ≥ 0, 0 < t < 1, \\ α_{n}(0) = \frac{1}{n}, α_{n}(1) + φ(α′_{n}(1)) ≤ \frac{1}{n}. \end{matrix}

From Claim 2, we obtain α_{n}(t) ≤ β_{n}^{0}(t) for t ∈ [0, 1]. Thus, we have

(Φ(β_{n}^{0}′))′ + q(t)F^{∗}(β_{n}^{0})(1 + \frac{Q(M)}{F (M)})

= −q(t)F (α_{n})(1 + \frac{Q(M)}{F (M)}) + q(t)F (β_{n}^{0})(1 + \frac{Q(M)}{F (M)})
= q(t)(1 + \frac{Q(M)}{F (M)})(F (β_{n}^{0}) − F (α_{n})) ≤ 0,

so that β_{n}^{0} is an upper solution for problem (4.8)^{n}.

If we now take α_{n}^{0} ≡ α_{n}, we have that α_{n}^{0} and β_{n}^{0} are, respectively, a lower and an upper solution of (4.8)^{n} with α_{n}^{0}(t) ≤ β_{n}^{0}(t), for all t ∈ [0, 1]. From Theorem 3.1, we know that there exists a solution β_{n} ∈ C([0, 1], R) ∩ C^{1}((0, 1], R) of (4.8)^{n} such that

α_{n}(t) = α_{n}^{0}(t) ≤ β_{n}(t) ≤ β_{n}^{0}(t), ∀t ∈ [0, 1].

16 AGARWAL, GAO, JIANG, O’REGAN, AND ZHANG

Now we claim that ||β_{n}|| < M. Suppose this is false, i.e., suppose ||β_{n}|| ≥ M. Since (Φ(β′_{n}(t)))′ ≤ 0 on (0,1) and β_{n} ≥ \frac{1}{n} on [0,1], there exists σ ∈ (0, 1) with β′_{n}(t) ≥ 0 on (0, σ), β′_{n}(t) ≤ 0 on (σ, 1) and β_{n}(σ) = ||β_{n}||.

Then for z ∈ (0, 1), we have

−(Φ(β′_{n}(z)))′ ≤ F (β_{n}(z))(1 + \frac{Q(M)}{F (M)})q(z).  (4.10)

Integrate from t (0 < t ≤ σ) to σ to obtain

β′_{n}(t) ≤ Φ^{−1}((1 + \frac{Q(M)}{F (M)}) ∫_{t}^{σ} F (β_{n}(z))q(z)dz),

so we have

\frac{β′_{n}(t)}{Φ^{−1}(F (β_{n}(t)))} ≤ Φ^{−1}(1 + \frac{Q(M)}{F (M)})Φ^{−1}(∫_{t}^{σ} q(z)dz),

and then integrate from 0 to σ to obtain

∫ β_{n}(σ)_{\frac{1}{n}} \frac{dy}{Φ^{−1}(F (y)))} ≤ Φ^{−1}(1 + \frac{Q(M)}{F (M)}) ∫_{0}^{σ} Φ^{−1}(∫_{t}^{σ} q(z)dz)dt.

Consequently,

∫_{δ}^{M} \frac{dy}{Φ^{−1}(F (y))} ≤ Φ^{−1}(1 + \frac{Q(M)}{F (M)}) ∫_{0}^{1} Φ^{−1}(∫_{t}^{1} q(z)dz)dt.  (4.11)

This contradicts (4.5) and so ||β_{n}|| < M.

It follows from the fact β_{n} ≥ \frac{1}{n} that β_{n} is a solution of (4.7)^{n} also. Since F is nonincreasing on (0, ∞), as in the proof of Lemma 3.1, we obtain the uniqueness of solutions to (4.7)^{n}.

Next, we show that β_{n} is an upper solution of (4.6)^{n}. Observe that

|g(t, x)| ≤ F (x) + Q(x) on [0, 1] × (0, ∞).

Then, we have

(Φ(β′_{n}(t)))′ + q(t)g(t, β_{n}(t)) ≤ −q(t)F (β_{n}(t))(1 + \frac{Q(M)}{F (M)}) + q(t)|g(t, β_{n}(t))|
≤ q(t)F (β_{n}(t))(\frac{Q(β_{n}(t))}{F (β_{n}(t))} − \frac{Q(M)}{F (M)})
≤ 0, t ∈ (0, 1).

Thus, β_{n} is an upper solution for problem (4.6)^{n}.

This together with Claim 1 implies that α_{n} and β_{n} are, respectively, a lower and an upper solution for problem (4.6)^{n} with α_{n} ≤ β_{n} for all t ∈ [0, 1]. So we conclude (4.6)^{n} has a solution u_{n} ∈ C([0, 1], R) ∩ C^{1}((0, 1], R) such that

mW (t) + \frac{1}{n} = α_{n}(t) ≤ u_{n}(t) ≤ β_{n}(t) ≤ M, ∀ t ∈ [0, 1].

Let e = [a, 1] ⊂ (0, 1] so
u_{n}(t) = ∫_{a}^{t} Φ^{−1}[Φ(u_{n}′(1)) + ∫_{r}^{1} q(s)g(s, u_{n}(s))ds]dr + u_{n}(a), t ∈ e.

Note that |u′_{n}(1)| = −u′_{n}(1) = φ^{−1}(u_{n}(1) − \frac{1}{n}) ≤ φ^{−1}(M). Thus, \{u′_{n}(1)\}_{n∈N^{+}} has a convergent subsequence; for convenience let {u′_{n}(1)}_{n∈N^{+}} denote this subsequence also, and let r_{0} ∈ R be its limit. Let

M_{e} = sup{|g(t, x)| : t ∈ e, mW (t) ≤ x ≤ M} < +∞.  (4.12)

From (4.12 ) and 0 < mW (t) ≤ u_{n}(t) ≤ M for t ∈ e, as in the proof of Lemma 2.5, we obtain

|u′_{n}(t)| ≤ C_{e}, t ∈ e = [a, 1],

where C_{e} is a positive constant dependent on e and independent of u_{n}, and t ∈ e.

Then the Ascoli-Arzela theorem guarantees the existence of a subsequence N_{1} of N^{+} and a function z ∈ C([0, 1], R) ∩ C^{1}((0, 1], R) with u_{n} → z in C([0, 1], R) ∩ C^{1}((0, 1], R) as n → ∞ through N_{1}. Thus, we have

z(t) = ∫_{a}^{t} Φ^{−1}[Φ(r_{0}) + ∫_{r}^{1} q(s)g(s, z(s))ds]dr + z(a), t ∈ e,

and

z′(t) = Φ^{−1}[Φ(r_{0}) + ∫_{t}^{1} q(s)g(s, z(s))ds], t ∈ e,

so r_{0} = z′(1), and

(Φ(z′(t)))′ + q(t)g(t, z(t)) = 0 for t ∈ e.

Since e is an arbitrary compact subset in [0,1], we find that

z ∈ C([0, 1], [0, ∞)) ∩ C^{1}((0, 1], R), and (Φ(z′(t)))′ + q(t)g(t, z(t)) = 0 for t ∈ (0, 1).

We also have

z(0) = z(1) + φ(z′(1)) = lim_{n→+∞} \frac{1}{n} = 0.

The proof of Theorem 4.1 is complete.

Essentially the same argument as in Theorem 4.1 and [14, Theorem 2] yields the following result.

Theorem 4.2. Suppose (H_{1}) in Theorem 4.1 hold. In addition assume that (H_{2}^{∗}) for any r > 0 there is h_{r} ∈ C((0, 1], (0, ∞)):

|q(t)g(t, x)| ≤ h_{r}(t), for all (t, x) ∈ (0, 1] × [r, ∞),

and there exist 0 ≤ μ < p − 1 such that

∫_{0}^{1} t^{μ}h_{r}(t)dt < +∞,

with

lim_{t→0^{+}} t^{μ+1}h_{r}(t) = 0, if ∫_{0}^{1/2} h_{r}(t)dt = ∞;

(H_{3}^{∗}) there exist constants L > 0 and ε > 0 such that

g(t, x) > L for all (t, x) ∈ [0, 1] × (0, ε].

Then problem (4.1) has at least one solution u ∈ C([0, 1], [0, ∞) ∩ C^{1}((0, 1], R). If, moreover, g(t, x) is nonincreasing in x > 0, then the solution is unique.

5. AN EXAMPLE

Example 5.1. Consider the singular boundary value problem

\{ \begin{matrix}(Φ(u′))′ + σt^{−m}(u^{−α} + u^{β} − sin(8πt)) = 0, 0 < t < 1, \\ u(0) = 0, u(1) + (u′(1))^{\frac{1}{3}} = 0, \end{matrix}  (5.1)

with 0 ≤ m < p, σ > 0, α > 0, β ≥ 0.

Set

φ(x) = x^{\frac{1}{3}} , F (u) = u^{−α}, Q(u) = u^{β} + 1, q_{1}(t) = t^{−m}, q(t) = σt^{−m}
b_{1} := ∫_{0}^{1} Φ^{−1}( ∫_{t}^{1} q_{1}(x)dx)dt.

Then b_{0} = σ^{\frac{1}{p−1}} b_{1}.

Applying Theorem 4.1, we can find that (5.1) has a positive solution if

σ < [\frac{p − 1}{b_{1}(α + p − 1)}]^{p−1} sup_{x∈(0,∞)} \frac{x^{α+p−1}}{1 + x^{α} + x^{α+β}} .  (5.2)

Obviously, (H_{1}) − (H_{4}) in Theorem 4.1 are satisfied. Thus, (5.1) has a solution u ∈ C([0, 1], [0, ∞) ∩ C^{1}((0, 1], R) with u > 0 on (0,1].

Remark 5.1. If 0 ≤ β < p − 1, then BVP (5.1) has at least one positive solution for all σ > 0, since the right-hand side of (5.2) is infinity.

ACKNOWLEDGMENTS

The research by D. Jiang was supported by NNSF of China.

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