Electronic Journal of Qualitative Theory of Differential Equations 2010, No. 16, 1-8; http://www.math.u-szeged.hu/ejqtde/

On existence and uniqueness of positive solutions for integral boundary boundary value problems ^{∗}

Jinxiu Mao, Zengqin Zhao School of Mathematical Sciences, Qufu Normal University, Qufu, Shandong, 273165, People’s Republic of China maojinxiu1982@163.com

Abstract: By applying the monotone iterative technique, we obtain the existence and uniqueness of C^{1}[0, 1] positive solutions in some set for singular boundary value problems of second order ordinary differential equations with integral boundary conditions.

2000 MSC: 34B10; 34B15.

Keywords: Boundary value problem; Positive solution.

1 Introduction and the main result

In this paper, we consider the existence of positive solutions for the following nonlinear singular boundary value problem:

\{ \begin{matrix}−u′′ + k^{2}u = f(t, u), t ∈ (0, 1), \\ u(0) = 0, u(1) = ∫_{0}^{1} u(t)dA(t), \end{matrix}  (1.1)

where A is right continuous on [0, 1), left continuous at t = 1, and nondecreasing on [0, 1), with A(0) = 0. ∫_{0}^{1} u(t)dA(t) denotes the Riemann-Stieltjes integral of u with respect to A. k is a constant. Problems involving Riemann-Stieltjes integral boundary condition have been studied in [3,7–9,13]. These boundary conditions includes multipoint and integral boundary conditions, and sums of these, in a single framework. By changing variables t ↦ 1 − t, studying (1.1) also covers the case

u(0) = ∫_{0}^{1} u(t)dA(t), u(1) = 0.

For a comprehensive study of the case when there is a Riemann-Stieltjes integral boundary condition at both ends, see [7].

∗Research supported by the National Natural Science Foundation of China (10871116) and the Doctoral Program Foundation of Education Ministry of China(200804460001).

In recent years, there are many papers investigating nonlocal boundary value problems of the second order ordinary differential equation u′′ +f(t, u) = 0. For example, we refer the reader to [1,3–5,7–9,11,12] for some work on problems with integral type boundary conditions. However, there are fewer papers investigating boundary value problems of the equation −u′′+k^{2}u = f(t, u). In [6], Du and Zhao investigated the following multi-point boundary value problem

\{ \begin{matrix}− u′′ = f(t, u), t ∈ (0, 1), \\ u(0) = ∑_{i=1}^{m−2} α_{i}u(η_{i}), u(1) = 0. \end{matrix}

They assumed f is decreasing in u and get existence of C[0, 1] positive solutions ω with the property that ω(t) ≥ m(1−t) for some m > 0. In a recent paper [5], Webb and Zima studied the problem (1.1) (and others) when dA is allowed to be a signed measure, and obtained existence of multiple positive solutions under suitable conditions on f(t, u). Here we only study the positive measure case. We impose stronger restrictions on f. We suppose f is increasing in u, satisfies a strong sublinear property and may be singular at t = 0, 1. By applying the monotone iterative technique, we obtain the existence and uniqueness of C^{1}[0, 1] positive solutions in some set D. Also, we use iterative methods, we establish uniqueness, obtain error estimates and the convergence rate of C^{1}[0, 1] positive solutions with the property that there exists M > m > 0 such that mt ≤ u(t) ≤ Mt.

In this paper, we first introduce some preliminaries and lemmas in Section 2, and then we state our main results in Section 3.

2 Preliminaries and lemmas

We make the following assumptions: (H_{1}) There exists k > 0 such that sinh(k) > ∫_{0}^{1} sinh(k(1 − t))dA(t); (H_{2}) f ∈ C((0, 1) × [0, +∞), [0, +∞)), f(t, u) is increasing in u and there exists a constant b ∈ (0, 1) such that

f(t, ru) ≥ r^{b}f(t, u), for all r ∈ (0, 1) and (t, u) ∈ (0, 1) × [0, +∞).  (1.2)

Remark 2.1. If M > 1, condition (1.2) is equivalent to

f(t, Mu) ≤ M^{b}f(t, u), for all (t, u) ∈ (0, 1) × [0, +∞).  (1.3)

Our discussion is in the space E = C[0, 1] of continuous functions endowed with the usual supremum norm. Let P = {u ∈ C[0, 1] : u ≥ 0} be the standard cone of nonnegative continuous functions. Definition 2.1. A function u ∈ C[0, 1] ⋂ C^{2}(0, 1) is called a C[0, 1] solution if it satisfies (1.1). A C[0, 1] solution u is called a C^{1}[0, 1] solution if both u′(0+) and u′(1−) exist. A solution u is called a positive solution if u(t) > 0, t ∈ (0, 1).

The Green’s function for (1.1) is given in the following Lemma which was proved in [5] for the general case when dA is a signed measure. Lemma 2.1 [5] Suppose that g ∈ C(0, 1) and (H_{1}) holds. Then the following linear boundary value problem

\{ \begin{matrix}−u′′ + k^{2}u = g(t), t ∈ (0, 1), \\ u(0) = 0, u(1) = ∫_{0}^{1} u(t)dA(t) \end{matrix}  (2.1)

has a unique positive solution u and u can be expressed in the form

u(t) = ∫_{0}^{1} F (t, s)g(s)ds,

where

F (t, s) = G(t, s) + \frac{sinh(kt)}{sinh(k) − ∫_{0}^{1} sinh(kτ)dA(τ)} ∫_{0}^{1} G(τ, s)dA(τ), s, t ∈ [0, 1],  (2.2)

G(t, s) = \{ \begin{matrix}\frac{sinh (ks) sinh (k (1 − t))}{sinh (k) k}, & 0 ≤ s ≤ t, \\ \frac{sinh (kt) sinh (k (1 − s))}{sinh (k) k}, & t ≤ s ≤ 1. \end{matrix}  (2.3)

Remark 2.2. We call F (t, s) the Green’s function of problem (1.1). Suppose that (H_{1}), (H_{2}) hold. Then solutions of (1.1) are equivalent to continuous solutions of the integral equation

u(t) = ∫_{0}^{1} F (t, s)f(s, u(s))ds,

where F (t, s) is mentioned in (2.2). Lemma 2.2 For any t, s ∈ [0, 1], there exist constants c_{1}, c_{2} > 0 such that

c_{2}e(t)e(s) ≤ F (t, s) ≤ c_{1}e(s), s, t ∈ [0, 1],  (2.4)

where e(s) = s(1 − s). Proof. Suppose that

I(t) = sinh(k)t − sinh(kt), t ∈ [0, 1].

EJQTDE, 2010 No. 16, p. 3 Then I(0) = I(1) = 0 and I′′(t) = −k^{2} sinh(kt) ≤ 0, t ∈ [0, 1]. So I(t) ≥ 0, i.e.

sinh(kt) ≤ sinh(k)t, t ∈ [0, 1].  (2.5)

Similarly we have

kt ≤ sinh(kt), t ∈ [0, 1].  (2.6)

From (2.3) we know

\frac{k}{sinh(k)}G(t, t)G(s, s) ≤ G(t, s) ≤ G(t, t).  (2.7)

By using (2.3), (2.5) and (2.6) we obtain

G(t, t) ≥ \frac{(kt)(k(1 − t))}{sinh(k)k} = \frac{ke(t)}{sinh(k)},  (2.8)

and

G(t, t) ≤ \frac{(sinh(k)t)(sinh(k)(1 − t))}{sinh(k)k} = \frac{sinh(k)e(t)}{k}.  (2.9)

From (2.2), (2.7), (2.8) and (2.9) we have

F (t, s) ≥ G(t, s) ≥ \frac{k}{sinh(k)}G(t, t)G(s, s) ≥ (\frac{k}{sinh(k)})^{3}e(t)e(s)  (2.10)

and

F (t, s) ≤ G(s, s) + G(s, s)\frac{sinh(k)}{sinh(k) − ∫_{0}^{1} sinh(kτ)dA(τ)} ∫_{0}^{1} dA(τ)

≤ \frac{sinh(k)}{k}e(s)[1 + \frac{sinh(k)}{sinh(k) − ∫_{0}^{1} sinh(kτ)dA(τ)} ∫_{0}^{1} dA(τ)].

(2.11)

Letting c_{1} = \frac{sinh(k)}{k}[1 + \frac{sinh(k)}{sinh(k) − ∫_{0}^{1} sinh(kτ)dA(τ)} ∫_{0}^{1} dA(τ)] and c_{2} = (\frac{k}{sinh(k)})^{3}, we have c_{2}e(t)e(s) ≤ F (t, s) ≤ c_{1}e(s).

Thus, (2.4) holds.

3 Main results

Now we state the main results as follows. Theorem 3.1 Suppose that (H_{1}), (H_{2}) hold. Let D = {u(t) ∈ C[0, 1] | ∃L_{u} ≥ l_{u} > 0, l_{u}t ≤ u(t) ≤ L_{u}t, t ∈ [0, 1]}. If

0 < ∫_{0}^{1} f(t, t)dt < +∞  (3.1)

holds. Then problem (1.1) has a unique C^{1}[0, 1] positive solution u^{∗} in D. Moreover, for any initial x_{0} ∈ D, the sequence of functions defined by

x_{n} = ∫_{0}^{1} F (t, s)f(s, x_{n−1}(s))ds, n = 1, 2, … converges uniformly to the unique solution u^{∗}(t) on [0, 1] as n → ∞. Furthermore, we have the error estimation

∥x_{n}(t) − u^{∗}(t)∥ ≤ 2(1 − (t_{0}^{2})^{b^{n}})∥v_{0}∥,  (3.2)

where t_{0}, v_{0} are defined below, and F (t, s) is mentioned in (2.2). Proof. From u(t) ∈ D we know there exists L_{u} > 1 > l_{u} > 0 such that

l_{u}s ≤ u(s) ≤ L_{u}s, s ∈ [0, 1].

This, together with (H_{2}), (1.2) and (1.3), implies that

(l_{u})^{b}f(s, s) ≤ f(s, u(s)) ≤ f(s, L_{u}s) ≤ (L_{u})^{b}f(s, s), s ∈ (0, 1).  (3.3)

Let us define an operator T by

T u = ∫_{0}^{1} F (t, s)f(s, u(s))ds, u ∈ D.  (3.4)

From (3.1) and (3.3) and Lemma 2.2 we can have

∫_{0}^{1} F (t, s)f(s, u(s))ds ≤ c_{1}(L_{u})^{b} ∫_{0}^{1} s(1 − s)f(s, s)ds < +∞.

So the integral operator T makes sense. By (2.2), (2.3), (2.5), (2.6) and (2.7), we have that

F (t, s) ≥ sinh(kt)\frac{∫_{0}^{1} G(τ, s)dA(τ)}{sinh(k) − ∫_{0}^{1} sinh(kτ)dA(τ)}

≥ kt\frac{∫_{0}^{1} G(τ, s)dA(τ)}{sinh(k) − ∫_{0}^{1} sinh(kτ)dA(τ)},

(3.5)

F (t, s) ≤ G(t, t) + \frac{sinh(kt)}{sinh(k) − ∫_{0}^{1} sinh(kτ)dA(τ)} ∫_{0}^{1} G(τ, s)dA(τ)

= sinh(kt) (\frac{sinh(k(1 − t))}{sinh(k)k} + \frac{∫_{0}^{1} G(τ, s)dA(τ)}{sinh(k) − ∫_{0}^{1} sinh(kτ)dA(τ)})

≤ t sinh(k) (\frac{1}{k} + \frac{∫_{0}^{1} G(τ, s)dA(τ)}{sinh(k) − ∫_{0}^{1} sinh(kτ)dA(τ)}) .

(3.6)

Thus

T u(t) ≥ t\frac{k (l_{u})^{b} ∫_{0}^{1} (∫_{0}^{1} G(τ, s)f(s, s)ds) dA(τ)}{sinh(k) − ∫_{0}^{1} sinh(kτ)dA(τ)}, t ∈ [0, 1],  (3.7)
T u(t) ≤ t (L_{u})^{b} sinh(k)×

∫_{0}^{1} (\frac{1}{k} + \frac{∫_{0}^{1} G(τ, s)dA(τ)}{sinh(k) − ∫_{0}^{1} sinh(kτ)dA(τ)}) f(s, s)ds, t ∈ [0, 1]. (3.8)

Thus, from (3.1), (3.7) and (3.8), we obtain

T : D → D.

It is known from Remark 2.2 that a fixed point of the operator T is a solution of BVP (1.1). From condition (1.2) we obtain

T (ru) = ∫_{0}^{1} F (t, s)f(s, ru(s))ds ≥ r^{b} ∫_{0}^{1} F (t, s)f(s, u(s))ds = r^{b}T u,  (3.9)

Obviously T is an increasing operator and from (1.3) we have

T (Mu) ≤ M^{b}T u.  (3.10)
Let x_{0} ∈ D be given. Choose t_{0} ∈ (0, 1) such that

t_{0}^{1−b}x_{0} ≤ T x_{0} ≤ (\frac{1}{t_{0}} )^{1−b}x_{0}.

Let us define u_{0} = t_{0}x_{0}, v_{0} = \frac{1}{t_{0}} x_{0}, t_{0} ∈ (0, 1). Then u_{0} ≤ v_{0} and from (3.9) and (3.10) we have

T u_{0} ≥ t_{0}^{b}T x_{0} ≥ t_{0}x_{0} = u_{0}, T v_{0} ≤ (\frac{1}{t_{0}} )^{b}T x_{0} ≤ \frac{1}{t_{0}} x_{0} = v_{0}.  (3.11)

Now we define

u_{n} = T u_{n−1}, v_{n} = T v_{n−1}, (n = 1, 2, 3, …).

It is easy to verify from (3.11) that

u_{0} ≤ u_{1} ≤ … ≤ u_{n} ≤ … ≤ v_{n} ≤ … ≤ v_{1} ≤ v_{0}.  (3.12)

Clearly, u_{0} = t_{0}^{2}v_{0}. By induction, we see that

u_{n} ≥ (t_{0}^{2})^{b^{n}}v_{n}, (n = 0, 1, 2, …).  (3.13)

Since P is a normal cone with normality constant 1, it follows that

∥v_{n} − u_{n}∥ ≤ ∥u_{n+p} − u_{n}∥ ≤ (1 − (t_{0}^{2})^{b^{n}})∥v_{0}∥.  (3.14)

So {u_{n}} is a cauchy sequence, therefore u_{n} converges to some u^{∗} ∈ D. From this inequality it also follows that v_{n} → u^{∗}.

We see that u^{∗} is a fixed point of T. Thus, u^{∗} ∈ D from u_{0}, v_{0} ∈ D and u^{∗} ∈ [u_{0}, v_{0}]. It follows from u_{0} ≤ x_{0} ≤ v_{0} that u_{n} ≤ x_{n} ≤ v_{n}, (n = 1, 2, 3, …). So

\begin{aligned}∥x_{n} − u^{∗}∥ & ≤ ∥x_{n} − u_{n}∥ + ∥u_{n} − u^{∗}∥ ≤ 2∥v_{n} − u_{n}∥ \\ & ≤ 2(1 − (t_{0}^{2})^{b^{n}})∥v_{0}∥. \end{aligned}  (3.15)

Next we prove the uniqueness of fixed points of T. Let \overline{x} ∈ D be any fixed points of T. From u^{∗}, \overline{x} ∈ D and the definition of D, we can put t_{1} = sup\{t > 0 | \overline{x} ≥ tu^{∗}\}. Evidently 0 < t_{1} < ∞. We now prove t_{1} ≥ 1. In fact, if 0 < t_{1} < 1, then

\overline{x} = T \overline{x} ≥ T (t_{1}u^{∗}) ≥ (t_{1})^{b}T u^{∗} = (t_{1})^{b}u^{∗},

which contradicts the definition of t_{1} since (t_{1})^{b} > t_{1}. Thus t_{1} ≥ 1 and \overline{x} ≥ u^{∗}. In the same way, we can prove \overline{x} ≤ u^{∗} and hence \overline{x} = u^{∗}. The uniqueness of fixed points of A in D is proved. For any initial z_{0} ∈ D, z_{n} = T^{n}z_{0} → u^{∗} with rate of convergence

∥z_{n} − u^{∗}∥ = o(1 − (t_{0}^{2})^{b^{n}})  (3.16)

from the results above. Choosing z_{0} = x_{0}, we obtain

∥x_{n} − u^{∗}∥ = o(1 − (t_{0}^{2})^{b^{n}}).  (3.17)

This completes the proof of Theorem 3.1. Remark Suppose that β_{i}(t)(i = 0, 1, 2, … m) are nonnegative continuous functions on (0, 1), which may be unbounded at the end points of (0, 1). Ω is the set of functions f(t, u) which satisfy the condition (H_{2}). Then we have the following conclusions:

(1) β_{i}(t) ∈ Ω, u^{b} ∈ Ω, where 0 < b < 1;

(2) If 0 < b_{i} < +∞(i = 1, 2, … m) and b > max_{1≤i≤m}\{b_{i}\}, then [β_{0}(t) + ∑_{i=1}^{m} β_{i}(t)u^{b_{i}}]^{\frac{1}{b}} ∈ Ω;

(3) If f(t, u) ∈ Ω, then β_{i}(t)f(t, u) ∈ Ω;

(4) If f_{i}(t, u) ∈ Ω(i = 1, 2, … m), then max_{1≤i≤m}{f_{i}(t, u)} ∈ Ω, min_{1≤i≤m}{f_{i}(t, u)} ∈ Ω.

The above four facts can be verified directly. This indicates that there are many kinds of functions which satisfy the condition (H_{2}).

EJQTDE, 2010 No. 16, p. 7

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(Received January 1, 2010)