Electronic Journal of Qualitative Theory of Differential Equations 2007, No. 4, 1-17; http://www.math.u-szeged.hu/ejqtde/

Positive Solutions to an N^{th} Order Right Focal Boundary Value Problem

Mariette Maroun Department of Mathematics and Physics University of Louisiana at Monroe Monroe, Louisiana 71209, USA maroun@ulm.edu

Abstract

The existence of a positive solution is obtained for the n^{th} order right focal boundary value problem y^{(n)} = f(x, y), 0 < x ≤ 1, y^{(i)}(0) = y^{(n−2)}(p) = y^{(n−1)}(1) = 0, i = 0, ⋯, n − 3, where \frac{1}{2} < p < 1 is fixed and where f(x, y) is singular at x = 0, y = 0, and possibly at y = ∞. The method applies a fixed-point theorem for mappings that are decreasing with respect to a cone.

Key words: Fixed point theorem, boundary value problem. AMS Subject Classification: 34B15.

1 Introduction

In this paper, we establish the existence of a positive solution for the n^{th} order right focal boundary value problem,

y^{(n)} = f(x, y), for x ∈ (0, 1],  (1)
y^{(i)}(0) = y^{(n−2)}(p) = y^{(n−1)}(1) = 0, i = 0, ⋯, n − 3,  (2)

where \frac{1}{2} < p < 1 is fixed and f(x, y) is singular at x = 0, y = 0, and may be singular at y = ∞.

We assume the following conditions hold for f:

(H1) f(x, y) : (0, 1]×(0, ∞) ⟶ (0, ∞) is continuous, and f(x, y) is decreasing in y for every x.

(H2) lim_{y→0^{+}} f(x, y) = +∞ and lim_{y→∞} f(x, y) = 0 uniformly on compact subsets of (0, 1].

We reduce the problem to a third order integro-differential problem. We establish decreasing operators for which we find fixed points that are solutions to this third order problem. Then, we use Gatica, Oliker, and Waltman methods to find a positive solution to the integro-differential third order problem. We integrate the positive solution n−3 times to obtain the positive solution to the n^{th} order right focal boundary value problem. The role of \frac{1}{2} < p < 1 is fundamental for the positivity of the Green’s function which in turn is fundamental for the positivity of desired solutions. The existence of positive solutions to a similar third order right focal boundary value problem was established in [22].

Singular boundary value problems for ordinary differential equations have arisen in numerous applications, especially when only positive solutions are useful. For example, when n = 2, Taliaferro [28] has given a nice treatment of the general problem, Callegari and Nachman [9] have studied existence questions of this type in boundary layer theory, and Lunning and Perry [21] have established constructive results for generalized Emden-Fowler boundary value problems. Also, Bandle, Sperb, and Stakgold [3] and Bobisud, et al. [6], [7], [8], have obtained results for singular boundary value problems that arise in reaction-diffusion theory, while Callegari and Nachman [10] have considered such boundary conditions in non-Newtonian fluid theory as well as in the study of pseudoplastic fluids. Nachman and Callegari point to applications in glacial advance and transport of coal slurries down conveyor belts. See [10] for references. Other applications for these boundary value problems appear in problems such as in draining flows [1], [5] and semipositone and positone problems [2].

In addition, much attention has been devoted to theoretical questions for singular boundary value problems. In some studies on singular boundary value problems, the underlying technique has been to obtain a priori estimates on solutions to an associated two-parameter family of problems, and then use these bounds along with topological transversality theorems to obtain solutions of the original problem; for example, see Granas, Guenther, and Lee [15] and Dunninger and Kurtz [11]. This method has been fairly exploited in a number of recent papers by O’Regan, [24], [25], [26]. Baxley [4] also used to some degree this latter technique in his work on singular boundary value problems for membrane response of a spherical cap. Wei [30] gave necessary and sufficient conditions for the existence of positive solutions for the singular Emden-Fowler equation satisfying Sturm-Liouville boundary conditions employing upper and lower solutions methods. Guoliang [16] also gave necessary and sufficient conditions for a higher order singular boundary value problem, using superlinear and sublinear conditions to show the existence of a positive solution.

2 Definitions and Properties of Cones

In this section, we begin by giving some definitions and some properties of cones in a Banach space.

Let (B, ∥ · ∥) be a real Banach space. A nonempty set K ⊂ B is called a cone if the following conditions are satisfied:

(a) the set K is closed;

(b) if u, v ∈ K then αu + βv ∈ K, for all α, β ≥ 0;

(c) u, −u ∈ K imply u = 0. Given a cone, K, a partial order, ≤, is induced on B by x ≤ y, for x, y ∈ B iff y−x ∈ K. (For clarity we sometimes write x ≤ y (w.r.t. K)). If x, y ∈ B with x ≤ y , let < x, y > denote the closed order interval between x and y given by, < x, y >= {z ∈ K|x ≤ z ≤ y}. A cone K is normal in B provided there exists δ > 0 such that ∥ e_{1} +e_{2} ∥≥ δ, for all e_{1}, e_{2} ∈ K with ∥ e_{1} ∥=∥ e_{2} ∥= 1.

Remark: If K is a normal cone in B, then closed order intervals are norm bounded.

3 Gatica, Oliker, and Waltman Fixed Point Theorem

Now we state the fixed point theorem due to Gatica, Oliker, and Waltman on which most of the results of this paper depend.

Theorem 3.1 Let B be a Banach space, K a normal cone in B, C a subset of K such that if x, y are elements of C, x ≤ y, then < x, y > is contained in C, and let T:C → K be a continuous decreasing mapping which is compact on any closed order interval contained in C . Suppose there exists x_{0} ∈ C such that T^{2}(x_{0}) is defined (where T^{2}(x_{0}) = T (T x_{0})), and furthermore, T x_{0} and T^{2}x_{0} are order comparable to x_{0}. Then T has a fixed point in C provided that either, (I) T x_{0} ≤ x_{0} and T^{2}x_{0} ≤ x_{0}, or T x_{0} ≥ x_{0} and T^{2}x_{0} ≥ x_{0}, or (II) The complete sequence of iterates {T^{n}x_{0}}_{n=0}^{∞} is defined, and there exists y_{0} ∈ C such that y_{0} ≤ T^{n}x_{0}, for every n.

We consider the following Banach space, B, with associated norm, ∥ · ∥:

B = {u : [0, 1] → R | u is continuous},

∥u∥ = sup_{x∈[0,1]} |u(x)|.

We also define a cone, K, in B by, K = {u ∈ B|u(x) ≥ 0, g(x)u(p) ≤ u(x) ≤ u(p) and u(x) is concave on [0, 1]}, where

g(x) = \frac{x(2p − x)}{p^{2}}, for 0 ≤ x ≤ 1.

4 The Integral Operator

In this section, we will define a decreasing operator T that will allow us to use the stated fixed point theorem.

First, we define k(x) by,

k(x) = ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} g(s)ds,

Given g(x) and k(x) above, we define g_{θ}(x) and k_{θ}(x), for θ > 0, by

g_{θ}(x) = θ · g(x),

and

k_{θ}(x) = θ · k(x),

and we will assume hereafter (H3) ∫_{0}^{1} f(x, k_{θ}(x))dx < ∞, for each θ > 0.

We note that the function f(x, y) = \frac{1}{\sqrt[4]{xy}} also satisfies (H3). In particular, for each θ > 0, ∫_{0}^{1} f(x, g_{θ}(x))dx = \sqrt[4]{(\frac{p^{2}}{θ})}[ \frac{2}{\sqrt[4]{2p}} + 4\frac{(2p−1)^{\frac{3}{4}} −(2p)^{\frac{3}{4}}}{3(\sqrt{2p})} < ∞.

If y is a solution of (1)-(2), then

u(x) = y^{(n−3)}(x),

is positive and concave. Hence, if in addition u ∈ K, then ||u|| = u(p).

Also, we get,

u′′′ = f(x, ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} u(s)ds),  (3)
u(0) = u′(p) = u′′(1) = 0.  (4)

Since g(x) is concave with g(0) = 0 and ||g(x)|| = g(p), then we observe, that for each positive solution, u(x), of (3)-(4), there is some θ > 0, such that g_{θ}(x) ≤ u(x), for 0 ≤ x ≤ 1.

Next, we let D ⊆ K be defined by

D = {u ∈ K| there exists θ(u) > 0 so that g_{θ}(x) ≤ u(x), 0 ≤ x ≤ 1}.

We note that for each u ∈ K,

∥u∥ = sup_{x∈[0,1]} |u(x)| = u(p).

Next, we define an integral operator T : D → K by

(T u)(x) = ∫_{0}^{1} G(x, t)f(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} u(s)ds)dt,

where G(x, t) is the Green’s function for y′′′ = 0 satisfying (4), and given by

G(x, t) = \begin{cases}\frac{x(2t−x)}{2}, x ≤ t ≤ p , \\ \frac{t^{2}}{2} , t ≤ x, t ≤ p , \\ \frac{x(2p−x)}{2}, x ≤ t, t ≥ p , \\ \frac{x(2p−x)}{2} + \frac{(x−t)^{2}}{2} , x ≥ t ≥ p ; \end{cases}

see [14] .

First, we show T is a decreasing operator. Let u ∈ D be given. Then there exists θ > 0 such that g_{θ}(x) ≤ u(x). Then, by condition (H1), f(x, u(x)) ≤ f(x, g_{θ}(x)). Now, let u(x) ≤ v(x) for u(x), v(x) ∈ D. Then,

∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} u(s)ds ≤ ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} v(s)ds.

Then by condition (H1),

f(x, ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} v(s)ds) ≤ f(x, ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} u(s)ds).

And since G(x, t) > 0, we have by (H1) and (H3),

∫_{0}^{1} G(x, t)f(t, ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} v(s)ds)dt < ∫_{0}^{1} G(x, t)f(t, ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} u(s)ds)dt
≤ ∫_{0}^{1} G(x, t)f(t, ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} g_{θ}(s)ds)dt
= ∫_{0}^{1} G(x, t)f(x, k_{θ}(x))
< ∞,

where g_{θ}(x) ≤ u(x).

Therefore, T is well-defined on D and T is a decreasing operator.

Remark: We claim that T : D → D. To see this, suppose u ∈ D and let

w(x) = (T u)(x) = ∫_{0}^{1} G(x, t)f(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} u(s)ds)dt ≥ 0.

Thus, for 0 ≤ x ≤ 1, w(x) ≥ 0. Also by properties of G,

w′′′(x) = f(x, ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} u(s)ds) > 0, for 0 < x ≤ 1,

and w(x) satisfies (5.4). As we argued previously, ∥w∥ = w(p). Since we have that w′′(1) = 0 and w′′′(x) > 0, then w is concave. Also, with w(p) = ||w(x)||, then w(x) ≥ w(p)g(x) = g_{w(p)}(x). Therefore, w ∈ D, and T : D → D. Remark: It is well-known that T u = u iff u is a solution of (3)-(4). Hence, we seek solutions of (3)-(4) that belong to D.

5 A Priori Bounds on Norms of Solutions

In this section, we will show that solutions of (3)-(4) have positive a priori upper and lower bounds on their norms. The proofs will be done by contradiction.

Lemma 5.1 If f satisfies (H1)-(H3), then there exists S > 0 such that ||u|| ≤ S for any solution u of (3)-(4) in D.

Proof: We assume that the conclusion of the lemma is false. Then there exists a sequence, {u_{m}}_{m=1}^{∞}, of solutions of (3)-(4) in D such that u_{m}(x) > 0, for x ∈ (0, 1], and

∥u_{m}∥ ≤ ∥u_{m+1}∥ and lim_{m→∞} ∥u_{m}∥ = ∞.

For a solution u of (3)-(4), we have

u′′′ = f(x, ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} u(s)ds) > 0, for 0 < x ≤ 1,

or u′′ < 0, for 0 < x ≤ 1.

This says that u is concave. In particular, the graphs of the sequence of solutions, u_{m}, are concave. Furthermore, for each m, the boundary conditions (4) and the concavity of u_{m} give us,

u_{m}(x) ≥ u_{m}(p)g(x) = ∥u_{m}∥g(x) = g_{∥u_{m}∥}(x) for all x,

and so for every 0 < c < 1,

lim_{m→∞} u_{m}(x) = ∞ uniformly on [c, 1].

Now, let us define

M := max{G(x, t) : (x, t) ∈ [0, 1] × [0, 1]}.

Then, from condition (H2), there exists m_{0} such that, for all m ≥ m_{0} and x ∈ [p, 1],

f(x, ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} u_{m}(s)ds) ≤ \frac{1}{M(1 − p)}.

Let

θ = ∥u_{m_{0}}∥ = u_{m_{0}}(p).

Then, for all m ≥ m_{0},

u_{m}(x) ≥ g_{θ}(x) = ∥u_{m_{0}}∥g(x), for 0 ≤ x ≤ 1.

So, for m ≥ m_{0}, and for 0 ≤ x ≤ 1, we have

u_{m}(x) = (T u_{m})(x)
= ∫_{0}^{1} G(x, t)f(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} u_{m}(s)ds)dt
= ∫_{0}^{p} G(x, t)f(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} u_{m}(s)ds)dt

+ ∫_{p}^{1} G(x, t)f(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} u_{m}(s)ds)dt

≤ ∫_{0}^{p} G(x, t)f(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} g_{θ}(s)ds)dt + ∫_{p}^{1} M \frac{1}{M(1 − p)}dt
≤ ∫_{0}^{p} G(x, t)f(t, k_{θ}(t))dt + 1
≤ ∫_{0}^{1} G(x, t)f(t, k_{θ}(t))dt + 1
≤ M ∫_{0}^{1} f(t, k_{θ}(t))dt + 1.

This is a contradiction to lim_{m→∞} ∥u_{m}∥ = ∞. Hence, there exists an S > 0 such that ∥u∥ < S for any solution u ∈ D of (3)-(4). □

Now we deal with positive a priori lower bounds on the solution norms.

Lemma 5.2 If f satisfies (H1)-(H3), then there exists R > 0 such that ||u|| ≥ R for any solution u of (5.3)-(5.4) in D.

Proof: We assume the conclusion of the lemma is false. Then, there exists a sequence {u_{m}}_{m=1}^{∞} of solutions of (3)-(4) in D such that u_{m}(x) > 0, for x ∈ (0, 1], and

∥u_{m}∥ ≥ ∥u_{m+1}∥

and

lim_{m→∞} ∥u_{m}∥ = 0.

Now we define

m̄ := min{G(x, t) : (x, t) ∈ [p, 1] × [p, 1]} > 0.

From condition (H2), lim_{y→0^{+}} f(x, y) = ∞ uniformly on compact subsets of (0, 1]. Thus, there exists δ > 0 such that, for x ∈ [p, 1] and 0 < y < δ,

f(x, y) > \frac{1}{m̄(1 − p)}.

In addition, there exists m_{0} such that, for all m ≥ m_{0} and x ∈ (0, 1]

0 < u_{m}(x) < \frac{δ}{2},

0 < ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} u_{m}(s)ds < \frac{δ}{2}.

So, for x ∈ [p, 1] and m ≥ m_{0},

u_{m}(x) = (T u_{m})(x)
= ∫_{0}^{1} G(x, t)f(t, ∫_{0}^{t} \frac{(x − s)^{n−4}}{(n − 4)!} u_{m}(s)ds)dt
≥ ∫_{p}^{1} G(x, t)f(t, ∫_{0}^{t} \frac{(x − s)^{n−4}}{(n − 4)!} u_{m}(s)ds)dt

≥ m̄ ∫_{p}^{1} f(t, ∫_{0}^{t} \frac{(x − s)^{n−4}}{(n − 4)!} u_{m}(s)ds)dt
> m̄ ∫_{p}^{1} f(t, \frac{δ}{2})dt
> m̄ ∫_{p}^{1} \frac{1}{m̄(1 − p)}dt
= 1.

Which is a contradiction to lim_{m→∞} ∥u_{m}(x)∥ = 0 uniformly on [0, 1]. Thus, there exists R > 0 such that R ≤ ∥u∥ for any solution u in D of (5.3)-(5.4). □

In summary, there exist 0 < R < S such that, for u ∈ D, a solution of (3)-(4), Lemma 5.1 and Lemma 5.2 give us

R ≤ ∥u∥ ≤ S.

The next section gives the main result, an existence theorem, for this problem.

6 Existence Result

In this section, we will construct a sequence of operators, {T_{m}}_{m=1}^{∞}, each of which is defined on all of K. We will then show, by applications of Theorem 3.1, that each T_{m} has a fixed point, ϕ_{m}, for every m, in K. Then, we will show that some subsequence of the {ϕ_{m}}_{m=1}^{∞} converges to a fixed point of T .

Theorem 6.1 If f satisfies (H1)-(H3), then (3)-(4) has at least one positive solution u in D, such that y(x) = ∫_{0}^{x} \frac{(x−s)^{n−4}}{(n−4)!} u(s)ds is a positive solution of (1)-(2).

Proof: For all m, let u_{m}(x) := T (m), where m is the constant function of that value on [0, 1]. In particular,

u_{m}(x) = ∫_{0}^{1} G(x, t)f(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} mds)dt
= ∫_{0}^{1} G(x, t)f(t, \frac{m(−s)^{n−3}}{(n − 3)!} )dt, for 0 ≤ x ≤ 1.

But f is decreasing in its second component, giving us,

0 < u_{m+1}(x) ≤ u_{m}(x), for 0 ≤ x ≤ 1.

By condition (H2), lim_{m→∞} u_{m}(x) = 0, uniformly on [0, 1].

Now, we define f_{m}(x, y) : (0, 1] × [0, ∞) → (0, ∞) by

f_{m}(x, y) = f(x, max \{y, ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} u_{m}(s)ds\}).

Then f_{m} is continuous and f_{m} does not possess the singularities as found in f at y = 0. Moreover, for (x, y) ∈ (0, 1] × (0, ∞) we have that,

f_{m}(x, y) ≤ f(x, y)

and, moreover,

f_{m}(x, y) = f(x, max \{y, ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} u_{m}(s)ds\}) ≤ f(x, ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} u_{m}(s)ds).

Next, we define a sequence of operators, T_{m} : K → K, for ϕ ∈ K and x ∈ [0, 1], by

T_{m}ϕ(x) := ∫_{0}^{1} G(x, t)f_{m}(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ(s)ds)dt.

It is standard that each T_{m} is a compact mapping on K. Moreover,

T_{m}(0) = ∫_{0}^{1} G(x, t)f_{m}(t, 0)dt
= ∫_{0}^{1} G(x, t)f(t, max\{0, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} u_{m}(s)ds\})dt
= ∫_{0}^{1} G(x, t)f(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} u_{m}(s)ds)dt
> 0.

Also,

T_{m}^{2}(0) = T_{m}( ∫_{0}^{1} G(x, t)f_{m}(t, 0)dt) ≥ 0.

Then, by theorem (3.1) with x_{0} = 0, T_{m} has a fixed point in K for every m. Thus, for every m, there exists a ϕ_{m} ∈ K so that

T_{m}ϕ_{m}(x) = ϕ_{m}(x), 0 ≤ x ≤ 1.

Hence, for m ≥ 1, ϕ_{m} satisfies the boundary conditions (4) of the problem. Also,

T_{m}ϕ_{m}(x) = ∫_{0}^{1} G(x, t)f_{m}(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds)dt
= ∫_{0}^{1} G(x, t)f(t, max\{ ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} u_{m}(s)ds\})dt
≤ ∫_{0}^{t} G(x, t)f(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} u_{m}(s)ds)dt
= T u_{m}(x).

That is, ϕ_{m}(x) = T_{m}ϕ_{m}(x) ≤ T u_{m}(x), for 0 ≤ x ≤ 1, and for every m.

Proceeding as in lemmas 5.1 and 5.2, there exists S > 0 and R > 0 such that

R < ∥ϕ_{m}∥ < S

for every m.

Now, let θ = R. Since ϕ_{m} ∈ K, then for x ∈ [0, 1] and every m,

ϕ_{m}(x) ≥ ϕ_{m}(p)g(x) = ∥ϕ_{m}∥ · g(x) > R · g(x) = θ · g(x) = g_{θ}(x).

Thus, with θ = R, g_{θ}(x) ≤ ϕ_{m}(x) for x ∈ [0, 1], for every m. Thus, {ϕ_{m}}_{m=1}^{∞} is contained in the closed order interval < g_{θ}, S >. Therefore, the sequence {ϕ_{m}}_{m=1}^{∞} is contained in D. Since T is a compact mapping, we may assume lim_{m→∞} T ϕ_{m} exist; say the limit is ϕ^{∗}.

To conclude the proof of this theorem, we still need to show that

lim_{m→∞} (T ϕ_{m}(x) − ϕ_{m}(x)) = 0

uniformly on [0, 1]. This will give us that ϕ^{∗} ∈< g_{θ}, S >. Still with θ = R, then k_{θ}(x) = ∫_{0}^{1} \frac{(x−s)^{n−4}}{(n−4)!} g_{θ}(s)ds ≤ ∫_{0}^{1} \frac{(x−s)^{n−4}}{(n−4)!} ϕ_{m}(s)ds for every m and 0 ≤ x ≤ 1. Let ϵ > 0 be given and choose δ, 0 < δ < 1, such that

∫_{0}^{δ} f(t, k_{θ}(t))dt < \frac{ϵ}{2M} ,

where again M := max{G(x, t) : (x, t) ∈ [0, 1] × [0, 1]}. Then, there exists m_{0} such that, for m ≥ m_{0} and for x ∈ [δ, 1],

∫_{0}^{1} \frac{(x − s)^{n−4}}{(n − 4)!} u_{m}(x) ≤ k_{θ}(x) ≤ ∫_{0}^{1} \frac{(x − s)^{n−4}}{(n − 4)!} ϕ_{m}(x).

So, for x ∈ [δ, 1],

f_{m}( x , ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds) =

f(x, max \{ ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds, ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} u_{m}(s)ds\}) =

f(x, ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds).

Then, for 0 ≤ x ≤ 1,

T ϕ_{m}(x) − ϕ_{m}(x) = T ϕ_{m}(x) − T_{m}ϕ_{m}(x)
= ∫_{0}^{1} G(x, t)f(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds)dt

− ∫_{0}^{1} G(x, t)f_{m}(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds)dt

= ∫_{0}^{δ} G(x, t)f(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds)dt

+ ∫_{δ}^{1} G(x, t)f(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds)dt

− ∫_{0}^{δ} G(x, t)f_{m}(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds)dt

− ∫_{δ}^{1} G(x, t)f_{m}(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds)dt

= ∫_{0}^{δ} G(x, t)f(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds)dt

− ∫_{0}^{δ} G(x, t)f_{m}(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds)dt.

Thus, for 0 ≤ x ≤ 1, we have,

|T ϕ_{m}(x) − ϕ_{m}(x)| = | ∫_{0}^{δ} G(x, t)f(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds)dt

− ∫_{0}^{δ} G(x, t)f_{m}(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds)dt|

≤ M[| ∫_{0}^{δ} f(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds)dt|

+| ∫_{0}^{δ} f_{m}(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds)d|]

= M[ ∫_{0}^{δ} f(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds)dt

+ ∫_{0}^{δ} f(t, max \{ ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} u_{m}(s)ds,

∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds\})dt]

= M[ ∫_{0}^{δ} f(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds)dt

+ ∫_{0}^{δ} f(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds)dt]

= 2M ∫_{0}^{δ} f(t, ∫_{0}^{t} \frac{(t − s)^{n−4}}{(n − 4)!} ϕ_{m}(s)ds)dt
≤ 2M ∫_{0}^{δ} f(t, ∫_{0}^{1} \frac{(x − s)^{n−4}}{(n − 4)!} g_{θ}(s)ds)ds
= 2M ∫_{0}^{δ} f(t, k_{θ}(t))dt
= 2M \frac{ϵ}{2M} = ϵ.

Thus, for m ≥ m_{0},

∥T ϕ_{m} − ϕ_{m}∥ < ϵ.

In particular, lim_{m→∞} (T ϕ_{m}(x) − ϕ_{m}(x)) = 0 uniformly on [0, 1], and for 0 ≤ x ≤ 1

T ϕ^{∗}(x) = T ( lim_{m→∞} T ϕ_{m}(x))
= T ( lim_{m→∞} ϕ_{m}(x))
= lim_{m→∞} (T ϕ_{m}(x))
= ϕ^{∗}(x).

Thus,

T ϕ^{∗} = ϕ^{∗},

and ϕ^{∗} is a desired solution of (3)-(4).

Now, if ϕ^{∗}(x) is the solution of (3)-(4), let y(x) = ∫_{0}^{x} \frac{(x−s)^{n−4}}{(n−4)!} ϕ^{∗}(s)ds. Then we have

y(0) = ∫_{0}^{0} \frac{(x − s)^{n−4}}{(n − 4)!} ϕ^{∗}(s)ds = 0,

and by the Fundamental Theorem of Calculus,

y^{(n−3)}(x) = ϕ^{∗}(x).

Thus,

y^{(n−3)}(0) = ϕ^{∗}(0) = 0.

Also,

y^{(n−2)}(x) = (ϕ^{∗})′(x),

thus,

y^{(n−2)}(p) = (ϕ^{∗})′(p) = 0.

And,

y^{(n−1)}(x) = (ϕ^{∗})′′(x)
y^{(n−1)}(1) = (ϕ^{∗})′′(1) = 0.

Moreover,

y^{(n)}(x) = (ϕ^{∗})′′′(x) = (T ϕ^{∗})′′′(x) = f(x, ∫_{0}^{x} \frac{(x − s)^{n−4}}{(n − 4)!} ϕ^{∗}(s)ds) = f(x, y).

Thus, y(x) = ∫_{0}^{x} \frac{(x−s)^{n−4}}{(n−4)!} ϕ^{∗}(s)ds > 0, 0 ≤ x ≤ 1 solves (1)-(2).

This completes the proof. □

Remark: The results of this paper extend to Boundary Value Problems for y^{(n)} = f(x, y, y′, ·, y^{(n−3)}) under the same boundary conditions.

Acknowledgment: The author is indebted to the referee’s suggestions. These have greatly improved this paper.

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(Received June 22, 2006)