16th Conference on Applied Mathematics, Univ. of Central Oklahoma, Electronic Journal of Differential Equations, Conf. 07, 2001, pp. 47–59. ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu ftp ejde.math.swt.edu (login: ftp)

A new a priori estimate for multi-point boundary-value problems ^{∗}

Chaitan P. Gupta

Abstract

Let f : [0, 1] × ℝ^{2} → ℝ be a function satisfying Caratheodory’s conditions and e(t) ∈ L^{1}[0, 1]. Let 0 < ξ_{1} < ξ_{2} < ⋯ < ξ_{m−2} < 1 and a_{i} ∈ ℝ for i = 1, 2, …, m − 2 be given. A priori estimates of the form

∥x∥_{∞} ≤ C∥x′′∥_{1}, ∥x′∥_{∞} ≤ C∥x′′∥_{1},

are needed to obtain the existence of a solution for the multi-point boundary-value problem

x′′(t) = f(t, x(t), x′(t)) + e(t), 0 < t < 1,

x(0) = 0, x(1) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}),

using Leray Schauder continuation theorem. The purpose of this paper is to obtain a new a priori estimate of the form ∥x∥_{∞} ≤ C∥x′′∥_{1}. This new estimate then enables us to obtain a new existence theorem. Further, we obtain a new a priori estimate of the form ∥x∥_{∞} ≤ C∥x′′∥_{1} for the three-point boundary-value problem

x′′(t) = f(t, x(t), x′(t)) + e(t), 0 < t < 1,
x′(0) = 0, x(1) = αx(η),

where η ∈ (0, 1) and α ∈ ℝ are given. The estimate obtained for the three-point boundary-value problem turns out to be sharper than the one obtained by particularizing the m-point boundary value estimate to the three-point case.

1 Introduction

Let f : [0, 1] × ℝ^{2} → ℝ be a function satisfying Caratheodory’s conditions and e(t) ∈ L^{1}[0, 1]. Let 0 < ξ_{1} < ξ_{2} < ⋯ < ξ_{m−2} < 1 and a_{i} ∈ ℝ for i = 1, 2, …, m − 2 be given. Let us consider the problem of existence of a solution for the multi-point boundary-value problem

∗Mathematics Subject Classifications: 34B10, 34B15, 34G20.

Key words: Three-point boundary-value problem, m-point boundary-value problem, a-priori estimates, Leray-Schauder Continuation theorem, Caratheodory’s conditions.

\begin{matrix}c \\ ◯ \end{matrix}2001 Southwest Texas State University.

Published July 20, 2001.

\begin{matrix}x′′(t) = f(t, x(t), x′(t)) + e(t), 0 < t < 1, \\ x(0) = 0, x(1) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}). \end{matrix}  (1)

In [2] the author and Sergei Trofimchuk had studied this problem earlier and obtained existence results using the Leray-Schauder continuation theorem. Now, to apply the Leray-Schauder continuation theorem requires a priori estimates of the form

∥x∥_{∞} ≤ C∥x′′∥_{1}, ∥x′∥_{∞} ≤ C∥x′′∥_{1}.

For a function x(t) ∈ W^{2,1}(0, 1) with x(0) = 0, x(1) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}), and ∑_{i=1}^{m−2} a_{i}ξ_{i} ≠ 1, Gupta and Trofimchuk obtained the a priori estimate

∥x′∥_{∞} ≤ \frac{1}{1 − τ} ∥x′′∥_{1},

where, 0 ≤ τ < 1 is suitable constant defined by a_{i}, and ξ_{i}, i = 1, 2, …, m − 2. Using, then the estimate ∥x∥_{∞} ≤ ∥x′∥_{∞}, for functions x(t) ∈ W^{2,1}(0, 1) with x(0) = 0, they obtained the estimate

∥x∥_{∞} ≤ \frac{1}{1 − τ} ∥x′′∥_{1}.

The purpose of this paper is to obtain a new and sharper estimate ∥x∥_{∞} ≤ C∥x′′∥_{1} for x(t) ∈ W^{2,1}(0, 1) with x(0) = 0, x(1) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}), and ∑_{i=1}^{m−2} a_{i}ξ_{i} ≠ 1. This new estimate then enables us to obtain a new existence theorem for the above boundary-value problem. Further, we obtain a new a priori estimate of the form ∥x∥_{∞} ≤ C∥x′′∥_{1} for the three-point boundary-value problem

\begin{matrix}x′′(t) = f(t, x(t), x′(t)) + e(t), 0 < t < 1, \\ x′(0) = 0, x(1) = αx(η), \end{matrix}  (2)

where η ∈ (0, 1) and α ∈ ℝ are given. The estimate obtained for the threepoint boundary-value problem turns out to be sharper than the one obtained by particularizing the m-point boundary-value estimate to the three-point case. These a priori estimates have been motivated by the results of [1].

2 A priori estimates

We begin this section by first describing an estimate obtained by Gupta and Trofimchuk. Let a_{i} ∈ ℝ, ξ_{i} ∈ (0, 1), i = 1, 2, …, m − 2, 0 < ξ_{1} < ξ_{2} < ⋯ < ξ_{m−2} < 1, with ∑_{i=1}^{m−2} a_{i}ξ_{i} ≠ 1, be given. Let x(t) ∈ W^{2,1}(0, 1) be such that x(0) = 0, x(1) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}). Let us write the condition x(1) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}) in symmetric form ∑_{i=1}^{m−1} a_{i}x(ξ_{i}) = 0 by setting a_{m−1} = −1 and ξ_{m−1} = 1. Then the assumption ∑_{i=1}^{m−2} a_{i}ξ_{i} ≠ 1 is equivalent to ∑_{i=1}^{m−1} a_{i}ξ_{i} ≠ 0. Let us, define, for i, j = 1, 2, …, m − 1,

σ_{ij} = a_{i}(ξ_{i} − ξ_{j}) for i ≠ j ,
σ_{jj} = ( ∑_{i=1}^{m−1} a_{i})ξ_{j} .

We observe that

∑_{i=1}^{m−1} σ_{ij} = ∑_{i=1}^{m−1} a_{i}ξ_{i} ≠ 0, for j = 1, 2, …, m − 1 .

For a ∈ ℝ, setting a_{+} = max(a, 0) and a_{−} = max(−a, 0) so that a = a_{+} − a_{−}, |a| = a_{+} + a_{−}, we see that

∑_{i=1}^{m−1}(σ_{ij})_{+} ≠ ∑_{i=1}^{m−1}(σ_{ij})_{−}.  (3)

We, next, define

σ_{+}^{j} = ∑_{i=1}^{m−1}(σ_{ij})_{+}, σ_{−}^{j} = ∑_{i=1}^{m−1}(σ_{ij})_{−} for j = 1, 2, …, m − 1 ,

and

τ = min\{\frac{σ_{+}^{j}}{σ_{−}^{j}} , \frac{σ_{−}^{j}}{σ_{+}^{j}} : j = 1, 2, …, m − 1\}.  (4)

We, note, that 0 ≤ τ < 1 in view of (3).

Proposition 1 Let a_{i} ∈ ℝ, ξ_{i} ∈ (0, 1), i = 1, 2, …, m − 2, 0 < ξ_{1} < ξ_{2} < ⋯ < ξ_{m−2} < 1, with ∑_{i=1}^{m−2} a_{i}ξ_{i} ≠ 1, be given. Then for x(t) ∈ W^{2,1}(0, 1) with x(0) = 0, x(1) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}) we have

∥x∥_{∞} ≤ \frac{1}{1 − τ} ∥x′′∥_{1},  (5)

where τ is as given in (4).

We refer the reader to [2] for a proof of this proposition.

Theorem 2 Let a_{i} ∈ ℝ, ξ_{i} ∈ (0, 1), i = 1, 2, …, m − 2, 0 < ξ_{1} < ξ_{2} < ⋯ < ξ_{m−2} < 1, with ∑_{i=1}^{m−2} a_{i}ξ_{i} ≠ 1, ∑_{i=1}^{m−2} a_{i} ≠ 1, be given. Then for x(t) ∈ W^{2,1}(0, 1) with x(0) = 0, x(1) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}) we have

∥x∥_{∞} ≤ C∥x′′∥_{1},  (6)

where

C = min\{\frac{1}{1 − τ}, C_{1}\},

with τ as defined in (4),

C_{1} = max\{C_{2}, \frac{1}{1 − τ} ∑_{i=1}^{m−2} |\frac{a_{i}(1 − ξ_{i})}{1 − ∑_{i=1}^{m−2} a_{i}} |\},

and C_{2} as defined below in (12).

Proof Let ξ_{m−1} = 1, a_{m−1} = −1 so that the condition x(1) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}) may be written in the symmetric form ∑_{i=1}^{m−1} a_{i}x(ξ_{i}) = 0 and ∑_{i=1}^{m−1} a_{i} ≠ 0. Since x(t) ∈ W^{2,1}(0, 1) there exists a c ∈ [0, 1] such that ∥x∥_{∞} = |x(c)|. We may assume that x(c) > 0, by replacing x(t) by −x(t), if necessary. Next, since x(0) = 0, we see that c ∈ (0, 1]. In case, c ∈ (0, 1) we must have x′(c) = 0. Applying, now, the Taylor’s formula with integral remainder after the second term at each ξ_{i}, i = 1, 2, …, m − 1, to get

x(ξ_{i}) = x(c) + r_{i},  (7)

where

r_{i} = ∫_{c}^{ξ_{i}}(ξ_{i} − s)x′′(s)ds ≤ 0,  (8)

i = 1, 2, …, m − 1. Multiplying the equation (7) by a_{i}, i = 1, 2, …, m − 1, and adding the resulting equations we obtain

0 = ∑_{i=1}^{m−1} a_{i}x(ξ_{i}) = ∑_{i=1}^{m−1} a_{i}x(c) + ∑_{i=1}^{m−1} a_{i}r_{i}.  (9)

Now, equations (8), (9) imply that

0 < x(c) = −\frac{1}{∑_{i=1}^{m−1} a_{i}} ∑_{i=1}^{m−1} a_{i}r_{i} = − ∑_{i=1}^{m−1}(\frac{a_{i}}{∑_{i=1}^{m−1} a_{i}}) ∫_{c}^{ξ_{i}}(ξ_{i} − s)x′′(s)ds

≤ ∑_{i=1}^{m−1}(\frac{a_{i}}{∑_{i=1}^{m−1} a_{i}})_{+}| ∫_{c}^{ξ_{i}}(ξ_{i} − s)x′′(s)ds|.  (10)

We, next, observe that

| ∫_{c}^{ξ_{i}}(ξ_{i} − s)x′′(s)ds| ≤ |ξ_{i} − c| ∫_{c}^{ξ_{i}} |x′′(s)|ds ≤ |ξ_{i} − c| ∫_{0}^{1} |x′′(s)|ds,

for i = 1, 2, …, m − 1. We thus see from (8) that

∥x∥_{∞} = x(c) ≤ ∑_{i=1}^{m−1}(\frac{a_{i}}{∑_{i=1}^{m−1} a_{i}})_{+}| ∫_{c}^{ξ_{i}}(ξ_{i} − s)x′′(s)ds|

≤ ∑_{i=1}^{m−1}(\frac{a_{i}}{∑_{i=1}^{m−1} a_{i}})_{+}|ξ_{i} − c| ∫_{0}^{1} |x′′(s)|ds

≤ max_{u∈[0,1]}( ∑_{i=1}^{m−1}(\frac{a_{i}}{∑_{i=1}^{m−1} a_{i}})_{+}|ξ_{i} − u|) ∫_{0}^{1} |x′′(s)|ds. (11)

Since, now, ∑_{i=1}^{m−1}(\frac{a_{i}}{∑_{i=1}^{m−1} a_{i}})_{+}|ξ_{i} −u| is a piecewise linear function, its maximum value is attained at one of the points, 0, ξ_{j}, j = 1, 2, …, m − 1. Accordingly, we get

max_{u∈[0,1]}( ∑_{i=1}^{m−1}(\frac{a_{i}}{∑_{i=1}^{m−1} a_{i}})_{+}|ξ_{i} − u|)

= max \{ \begin{matrix}∑_{i=1}^{m−1} ξ_{i}(\frac{a_{i}}{∑_{i=1}^{m−1} a_{i}})_{+}, \\ ∑_{i=1,i≠j}^{m−1}(\frac{a_{i}}{∑_{i=1}^{m−1} a_{i}})_{+}|ξ_{i} − ξ_{j}|, j = 1, 2, …, m − 1, \end{matrix} \}  (12)

= max \{ \begin{matrix}∑_{i=1}^{m−2} ξ_{i}(\frac{a_{i}}{1−∑_{i=1}^{m−2} a_{i}})_{−} + (_{\frac{1}{1−∑_{i=1}^{m−2} a_{i}}})_{+}, \\ ∑_{i=1,i≠j}^{m−2}(\frac{a_{i}}{1−∑_{i=1}^{m−2} a_{i}})_{−}|ξ_{i} − ξ_{j}| + (_{\frac{1}{1−∑_{i=1}^{m−2} a_{i}}})_{+}(1 − ξ_{j}), \\ j = 1, 2, …, m − 2, \\ ∑_{i=1}^{m−2}(\frac{a_{i}}{1−∑_{i=1}^{m−2} a_{i}})_{−}(1 − ξ_{i}) \end{matrix} \} ≡ C_{2}.

Accordingly, when x(c) = ∥x∥_{∞} with c ∈ (0, 1) we see that

∥x∥_{∞} ≤ C_{2}∥x′′∥_{1}.  (13)

Let, now, c = 1 so that ∥x∥_{∞} = x(1). We, then, see that there exists a λ_{i}, for each i = 1, 2, …, m − 2, such that

x(1) − x(ξ_{i}) = (1 − ξ_{i})x′(λ_{i}).  (14)

It follows from equations (14) that

( ∑_{i=1}^{m−2} a_{i} − 1)x(1) = ∑_{i=1}^{m−2} a_{i}(x(1) − x(ξ_{i})) = ∑_{i=1}^{m−2} a_{i}(1 − ξ_{i})x′(λ_{i}).

Accordingly, we get

∥x∥_{∞} = x(1) = ∑_{i=1}^{m−2} \frac{a_{i}(1 − ξ_{i})}{∑_{i=1}^{m−2} a_{i} − 1}x′(λ_{i})

≤ ∑_{i=1}^{m−2} |\frac{a_{i}(1 − ξ_{i})}{∑_{i=1}^{m−2} a_{i} − 1}|∥x′∥_{∞}

≤ (\frac{1}{1 − τ} ∑_{i=1}^{m−2} |\frac{a_{i}(1 − ξ_{i})}{∑_{i=1}^{m−2} a_{i} − 1}|)∥x′′∥_{1}.  (15)

Thus from estimates (13), (15) we obtain

∥x∥_{∞} ≤ max\{C_{2}, \frac{1}{1 − τ} ∑_{i=1}^{m−2} |\frac{a_{i}(1 − ξ_{i})}{∑_{i=1}^{m−2} a_{i} − 1}|\}∥x′′∥_{1} ≡ C_{1}∥x′′∥_{1}. (16)

The estimate (6) is now immediate since ∥x∥_{∞} ≤ \frac{1}{1−τ} ∥x′′∥_{1}, from Proposition 1. This completes the proof of Theorem 2. □

Remark 3 Let η ∈ (0, 1), α ∈ ℝ with αη ≠ 1 be given. It was proved earlier by Gupta and Trofimchuk for x(t) ∈ W^{2,1}(0, 1) with x(0) = 0, x(1) = αx(η) that

∥x∥_{∞} ≤ ∥x′′∥_{1} if α ≤ 1,

∥x∥_{∞} ≤ \frac{1 − η}{1 − αη} ∥x′′∥_{1} if αη < 1 and α > 1,

∥x∥_{∞} ≤ \frac{α − 1}{αη − 1}∥x′′∥_{1} if α > 1 and αη > 1,

so that

τ = 0 if α ≤ 1,

\frac{1}{1 − τ} = \frac{1 − η}{1 − αη} if α > 1 and αη < 1,

\frac{1}{1 − τ} = \frac{α − 1}{αη − 1} if α > 1 and αη > 1 .

Remark 4 Let us note that for x(t) ∈ W^{2,1}(0, 1) with x(0) = 0, x(1) = αx(η) the constant C_{2} defined in (12) is given by

C_{2} = max\{η(\frac{α}{1 − α})_{−} + (\frac{1}{1 − α})_{+}, (\frac{1}{1 − α})_{+}(1 − η), (\frac{α}{1 − α})_{−}(1 − η)\}.

It follows that

C_{2} = \{ \begin{matrix}max\{\frac{1+|α|η}{1+|α|} , \frac{|α|(1−η)}{1+|α|}\} & for α ≤ 0, \\ \frac{1}{1−α} & for 0 ≤ α < 1, \\ max\{\frac{αη}{α−1}, \frac{α(1−η)}{α−1}\} & for α > 1 . \end{matrix}

Next, we see from the definition of C_{1} in (16) and (3) that

C_{1} = \{ \begin{matrix}max\{\frac{1+|α|η}{1+|α|} , \frac{|α|(1−η)}{1+|α|}\} & for α ≤ 0, \\ \frac{1}{1−α} & for 0 ≤ α < 1, \\ max\{\frac{αη}{α−1}, \frac{α(1−η)^{2}}{(α−1)(1−αη)}\} & for αη < 1 and α > 1, \\ max\{\frac{αη}{α−1}, \frac{α(1−η)}{(αη−1)}\} & for αη > 1 and α > 1 . \end{matrix}

Finally, we see that for x(t) ∈ W^{2,1}(0, 1) with x(0) = 0, x(1) = αx(η) we have

∥x∥_{∞} ≤ C∥x′′∥_{1},  (17)

where C = min\{\frac{1}{1−τ}, C_{1}\} is given by

C = \{ \begin{matrix}max\{\frac{1+|α|η}{1+|α|} , \frac{|α|(1−η)}{1+|α|}\} & for α ≤ 0, \\ 1 & for 0 ≤ α < 1, \\ min\{\frac{1−η}{1−αη} , max\{\frac{αη}{α−1}, \frac{α(1−η)^{2}}{(α−1)(1−αη)}\}\} & for αη < 1 and α > 1, \\ min\{\frac{α−1}{αη−1}, max\{\frac{αη}{α−1}, \frac{α(1−η)}{(αη−1)}\}\} & for αη > 1 and α > 1. \end{matrix}

The following theorem gives a better estimate than (17) for an x(t) ∈ W^{2,1}(0, 1) with x(0) = 0, x(1) = αx(η).

Theorem 5 Let α ∈ ℝ and η ∈ (0, 1) with α ≠ 1, αη ≠ 1, be given. Then for x(t) ∈ W^{2,1}(0, 1) with x(0) = 0, x(1) = αx(η) we have

∥x∥_{∞} ≤ M∥x′′∥_{1}

where

M = \{ \begin{matrix}max\{\frac{1+|α|η}{1+|α|} , \frac{|α|(1−η)}{1+|α|}\} & if α ≤ −1. \\ \frac{1−αη}{1−α} & if − 1 ≤ α < 0, \\ 1 & if 0 ≤ α < 1, \\ max\{\frac{η}{2} , \frac{α(1−η)}{α−1}, \frac{αη(1−η)}{1−αη}\} & if α > 1 and αη < 1, \\ max\{\frac{η}{2} , \frac{αη−1}{α−1}, \frac{αη(1−η)}{αη−1}\} & if α > 1 and αη > 1 . \end{matrix}

Proof For α ≤ 0 we see from Theorem 2 and remark 4 that

M = max\{\frac{1 + |α|η}{1 + |α|} , \frac{|α|(1 − η)}{1 + |α|}\}.

This implies, in particular, for α ≤ −1 that M = max\{\frac{1+|α|η}{1+|α|}, \frac{|α|(1−η)}{1+|α|}\}. Note that for −1 ≤ α < 0,

\frac{1 − αη}{1 − α} = \frac{1 + η|α|}{1 + |α|} ≥ \frac{|α|(1 + η)}{1 + |α|} > \frac{|α|(1 − η)}{1 + |α|}

and so we again see from Theorem 2 and Remark 4 that

M = \{ \begin{matrix}\frac{1−αη}{1−α} & if − 1 ≤ α < 0 \\ 1 & if 0 ≤ α < 1 . \end{matrix}

Finally, we consider the case α > 1. Let x(η) = z so that x(1) = αz. We may assume without loss of generality that z ≥ 0, replacing x(t) by −x(t) if necessary. Suppose, now, ∥x∥_{∞} = 1 so that there exists a c ∈ [0, 1] such that either x(c) = 1 or x(c) = −1. We consider all possible cases of the location for c.

(i) Suppose that c ∈ (0, η] and x(c) = 1. Then x′(c) = 0, c ≠ η. Now, by mean value theorem there exist ν_{1} ∈ [c, η], ν_{2} ∈ [η, 1] such that

x′(ν_{1}) = \frac{x(η) − x(c)}{η − c} = −\frac{1 − z}{η − c} , x′(ν_{2}) = \frac{x(1) − x(η)}{1 − η} = \frac{αz − z}{1 − η} .

We note that x′(ν_{1}) ≤ 0, x′(ν_{2}) ≥ 0 since 0 ≤ z ≤ 1 and α > 1. It follows that

∫_{0}^{1} |x′′(s)|ds ≥| ∫_{c}^{ν_{1}} x′′(s)ds| + | ∫_{ν_{1}}^{ν_{2}} x′′(s)ds|
=2|x′(ν_{1})| + x′(ν_{2}) = 2\frac{1 − z}{η − c} + \frac{αz − z}{1 − η}
≥ min_{c∈[0,η),z∈[0, \frac{1}{α}]}\{2\frac{1 − z}{η − c} + \frac{αz − z}{1 − η} \}
≥ min_{c∈[0,η)}\{\frac{2}{η − c}, \frac{2(α − 1)}{α(η − c)} + \frac{α − 1}{α(1 − η)}\}
≥ min\{\frac{2}{η} , \frac{α − 1}{α(1 − η)}\}.

(ii) Let, now, c ∈ (0, η], x(c) = −1. Then since x′(c) = 0, c ≠ η, we again see from mean value theorem that there exist ν_{3} ∈ [c, η], ν_{4} ∈ [η, 1] such that

x′(ν_{3}) = \frac{x(η) − x(c)}{η − c} = \frac{z + 1}{η − c}, x′(ν_{4}) = \frac{x(1) − x(η)}{1 − η} = \frac{αz − z}{1 − η} .

Again we note that x′(ν_{3}) > 0, x′(ν_{4}) ≥ 0 since 0 ≤ z ≤ 1 and α > 1 and we have

\begin{matrix}∫_{0}^{1} |x′′(s)|ds ≥ & | ∫_{c}^{ν_{3}} x′′(s)ds| + | ∫_{ν_{3}}^{ν_{4}} x′′(s)ds| \\ = & x′(ν_{3}) + |x′(ν_{4}) − x′(ν_{3})| = \frac{1 + z}{η − c} + |\frac{αz − z}{1 − η} − \frac{1 + z}{η − c} |. \end{matrix} (18)

Let F (z, c) = \frac{1+z}{η−c} +|\frac{αz−z}{1−η} − \frac{1+z}{η−c}|. We need to estimate min_{c∈[0,η),z∈[0, \frac{1}{α}]} F (z, c). We note that

F (0, c) = \frac{2}{η − c} ≥ \frac{2}{η} for c ∈ [0, η),

F (\frac{1}{α}, c) = \frac{α + 1}{α(η − c)} + |\frac{α − 1}{α(1 − η)} − \frac{α + 1}{α(η − c)}| ≥ \frac{α − 1}{α(1 − η)} for c ∈ [0, η).

Let z_{0} be such that \frac{αz_{0}−z_{0}}{1−η} − \frac{1+z_{0}}{η−c} = 0 so that z_{0} = \frac{1−η}{αη−1−c(α−1)}. It is easy to see that z_{0} ∈ [0, \frac{1}{α}] if η > \frac{α+1}{2α} and c ∈ (0, \frac{2αη−α−1}{α−1}). In this case we get F (z_{0}, c) = \frac{α−1}{αη−1−c(α−1)} ≥ \frac{α−1}{αη−1}. Accordingly we see that F (z, c) ≥ min\{\frac{2}{η} , \frac{α−1}{α(1−η)}\} if αη ≤ 1 and F (z, c) ≥ min\{\frac{2}{η} , \frac{α−1}{α(1−η)}, \frac{α−1}{αη−1}\} if αη > 1. We thus have from (18) that

∫_{0}^{1} |x′′(s)|ds ≥| ∫_{c}^{ν_{3}} x′′(s)ds| + | ∫_{ν_{3}}^{ν_{4}} x′′(s)ds| = x′(ν_{3}) + |x′(ν_{4}) − x′(ν_{3})|
=\frac{1 + z}{η − c} + |\frac{αz − z}{1 − η} − \frac{1 + z}{η − c} |

≥ \{ \begin{matrix}min\{\frac{2}{η} , \frac{α−1}{α(1−η)}\}, & if αη ≤ 1, \\ min\{\frac{2}{η} , \frac{α−1}{α(1−η)}, \frac{α−1}{αη−1}\}, & if αη > 1. \end{matrix}

(iii) Next, suppose that c ∈ (η, 1), x(c) = 1. Again, x′(c) = 0 and we have from mean value theorem that there exist ν_{5} ∈ [η, c], ν_{6} ∈ [c, 1] such that

x′(ν_{5}) = \frac{x(c) − x(η)}{c − η} = \frac{1 − z}{c − η} , x′(ν_{6}) = \frac{x(1) − x(c)}{1 − c} = \frac{αz − 1}{1 − c}.

Note that x′(ν_{5}) ≥ 0, x′(ν_{6}) ≤ 0 since x(1) = αz ≤ 1. Accordingly, we obtain

\begin{matrix}∫_{0}^{1} |x′′(s)|ds ≥ & | ∫_{0}^{ν_{5}} x′′(s)ds| + | ∫_{ν_{5}}^{ν_{6}} x′′(s)ds| \\ = & x′(ν_{5}) + |x′(ν_{6}) − x′(ν_{5})| = 2x′(ν_{5}) + |x′(ν_{6})| \\ = & 2\frac{1 − z}{c − η} + \frac{1 − αz}{1 − c} ≥ \frac{2(α − 1)}{α(1 − η)}, since 0 ≤ z ≤ \frac{1}{α}. \end{matrix}  (19)

(iv) Next, suppose that c ∈ (η, 1), x(c) = −1. Again, x′(c) = 0 and we have from mean value theorem that there exist ν_{7} ∈ [η, c], ν_{8} ∈ [c, 1] such that

x′(ν_{7}) = \frac{x(c) − x(η)}{c − η} = \frac{−1 − z}{c − η} , x′(ν_{8}) = \frac{x(1) − x(c)}{1 − c} = \frac{αz + 1}{1 − c}.

Note that x′(ν_{7}) ≤ 0, x′(ν_{8}) ≥ 0. Accordingly, we obtain

∫_{0}^{1} |x′′(s)|ds ≥| ∫_{0}^{ν_{7}} x′′(s)ds| + | ∫_{ν_{7}}^{ν_{8}} x′′(s)ds|
=|x′(ν_{7})| + |x′(ν_{8}) − x′(ν_{7})| = 2|x′(ν_{7})| + x′(ν_{8})
=2\frac{1 + z}{c − η} + \frac{1 + αz}{1 − c} ≥ \frac{2}{c − η} + \frac{1}{1 − c}
≥\frac{2}{1 − η} ≥ \frac{2(α − 1)}{α(1 − η)}.

(v) Finally suppose that c = 1, so that x(1) = 1 = αz. We then have that there exists a ν_{9} ∈ (η, 1) such that

x′(ν_{9}) = \frac{x(1) − x(η)}{1 − η} = \frac{1 − \frac{1}{α}}{1 − η} = \frac{α − 1}{α(1 − η)}.

Also, there exists a ν_{10} ∈ (0, η) such that

x′(ν_{10}) = \frac{x(η) − x(0)}{η − 0} = \frac{1}{αη} .

Thus

∫_{0}^{1} |x′′(s)|ds ≥ | ∫_{υ_{10}}^{ν_{9}} x′′(s)ds| = |x′(ν_{9}) − x′(ν_{10})|
= |\frac{1 − \frac{1}{α}}{1 − η} − \frac{1}{αη} | = |\frac{αη − 1}{αη(1 − η)}|.

We thus see from (i), (ii), (iii), (iv) and (v) that for α > 1, ∥x∥_{∞} ≤ M∥x′′∥_{1} with

M = \{ \begin{matrix}max\{\frac{η}{2} , \frac{α(1−η)}{α−1}, \frac{αη(1−η)}{1−αη}\} & if αη ≤ 1, \\ max\{\frac{η}{2} , \frac{αη−1}{α−1}, \frac{αη(1−η)}{αη−1}\} & if αη > 1, \end{matrix}

since for α > 1, αη > 1, \frac{αη(1−η)}{αη−1} > \frac{α(1−η)}{α−1}. This completes the present proof. □

Remark 6 Let α = 4 and η = \frac{1}{2}. Let us consider the estimate

∥x∥_{∞} ≤ C∥x′′∥_{1},  (20)

for x(t) ∈ W^{2,1}(0, 1) with x(0) = 0, x(1) = 4x(\frac{1}{2}). Now, the function

φ(t) = \{ \begin{matrix}2t^{3}, & for t ∈ [0, \frac{1}{2}], \\ \frac{3t−1}{2}, & for t ∈ [\frac{1}{2}, 1], \end{matrix}  (21)

is such that φ(t) ∈ W^{2,1}(0, 1) with φ(0) = 0 and φ(1) = 4φ(\frac{1}{2}). Moreover, ∥φ∥_{∞} = 1 and ∥φ′′∥_{1} = \frac{3}{2}. It follows that C ≥ \frac{2}{3} in (20). Now, Proposition 1 and Remark 3 give C = 3 in (20); while Theorem 2 and Remark 4 give C = 2 in (20); and Theorem 5 gives C = 1 in (20). This shows that Theorem 5 gives the best estimate ∥x∥_{∞} ≤ ∥x′′∥_{1} for x(t) ∈ W^{2,1}(0, 1) with x(0) = 0, x(1) = 4x(\frac{1}{2}). However, the function φ(t) defined in (21) indicates that it may be possible to improve C in (20). This question remains open at this time.

To explore this further we introduce the notion of approximate best constant in the following.

Definition B ∈ ℝ is called “approximate best constant” if for every ε > 0 there exists an α ∈ ℝ and an η ∈ (0, 1) such that (i) for every x(t) ∈ W^{2,1}(0, 1) with x(0) = 0, x(1) = αx(η), ∥x∥_{∞} ≤ (B + ε)∥x′′∥_{1}; (ii) there exists a function ϕ(t) ∈ W^{2,1}(0, 1) with ϕ(0) = 0, ϕ(1) = αϕ(η), and ∥ϕ∥_{∞} > B∥ϕ′′∥_{1}.

Theorem 7 For every k > 1, 1 − \frac{1}{k} is an approximate best constant.

Proof For each integer n > 2, consider the function ϕ_{kn}(t) ∈ W^{2,1}(0, 1) defined by

ϕ_{kn}(t) = \{ \begin{matrix}t^{n}, & for t ∈ [0, \frac{1}{k}], \\ \frac{nt}{k^{n−1}} − \frac{n−1}{k^{n}}, & for t ∈ [\frac{1}{k} , 1]. \end{matrix}

It is easy to see that ϕ_{kn}(t) ∈ W^{2,1}(0, 1), with ϕ_{kn}(0) = 0, ϕ_{kn}(1) = α_{kn}ϕ_{kn}(\frac{1}{k}), where α_{kn} = n(k − 1) + 1, and

∥ϕ′′_{kn}∥_{1} = \frac{n}{k^{n−1}} , ∥ϕ_{kn}∥_{∞} = ϕ_{kn}(1) = \frac{n(k − 1) + 1}{k^{n}},

so that

∥ϕ_{kn}∥_{∞} = \frac{n(k − 1) + 1}{nk}∥ϕ′′_{kn}∥_{1}.  (22)

Now, since α_{kn} · \frac{1}{k} = \frac{n(k−1)+1}{k} = n − \frac{n−1}{k} > 1 for n > 2, we obtain using Theorem 5 the estimate

\begin{matrix}∥x∥_{∞} ≤ \frac{n(k − 1) + 1}{k(n − 1)}∥x′′∥_{1} for x(t) ∈ W^{2,1}(0, 1) \\ x(0) = 0, x(1) = α_{kn}x(\frac{1}{k}). \end{matrix}  (23)

Let us set B_{kn} = \frac{n(k−1)+1}{nk} = 1 − \frac{1}{k} + \frac{1}{nk}, M_{kn} = \frac{n(k−1)+1}{k(n−1)} = 1 − \frac{1}{k} + \frac{1}{n−1}. We notice that

M_{kn} − B_{kn} = \frac{1}{n − 1} − \frac{1}{nk} = \frac{n(k − 1) + 1}{n(n − 1)k} > 0,

so that M_{kn} − B_{kn} > 0. Also, we note that

lim_{n→∞} B_{kn} = lim_{n→∞} M_{kn} = 1 − \frac{1}{k} .

Let, now, ε > 0 be given. Choose, n_{0} such that M_{kn_{0}} < 1 − \frac{1}{k} + ε. It, now, follows from (23) and (22) that

\begin{matrix}∥x∥_{∞} ≤ (1 − \frac{1}{k} + ε)∥x′′∥_{1} for x(t) ∈ W^{2,1}(0, 1) \\ x(0) = 0, x(1) = α_{kn_{0}}x(\frac{1}{k}), \end{matrix}

and

∥ϕ_{kn_{0}}∥_{∞} = (1 − \frac{1}{k} + \frac{1}{n_{0}k})∥ϕ′′_{kn}∥_{1} > (1 − \frac{1}{k})∥ϕ′′_{kn}∥_{1}.

This completes the proof of the Theorem. □

Remark 8 We note that lim_{k→∞}(1 − \frac{1}{k}) = 1. In view of this, it may be conjectured that 1 may be a best constant in the sense that there exists an α ∈ ℝ and an η ∈ (0, 1) such that for x(t) ∈ W^{2,1}(0, 1) with x(0) = 0, x(1) = αx(η) one has the estimate

∥x∥_{∞} ≤ ∥x′′∥_{1}.

However, since lim_{k→∞} α_{kn} = ∞ and lim_{k→∞} \frac{1}{k} = 0, it is not clear if such α ∈ ℝ and an η ∈ (0, 1) exist.

3 Existence theroems

We state below the existence theorems one obtains using the a priori estimates obtained above. We omit the proof of these theorems as they are similar to the corresponding theorems in [2].

Theorem 9 Let f : [0, 1] × ℝ^{2} → ℝ be a function satisfying Caratheodory’s conditions. Assume that there exist functions p(t), q(t), r(t) in L^{1}(0, 1) such that

|f(t, x_{1}, x_{2})| ≤ p(t) |x_{1}| + q(t) |x_{2}| + r(t)

for a.e. t ∈ [0, 1] and all (x_{1}, x_{2}) ∈ ℝ^{2}. Let a_{i} ∈ ℝ, ξ_{i} ∈ (0, 1), i = 1, 2, …, m − 2, 0 < ξ_{1} < ξ_{2} < ⋯ < ξ_{m−2} < 1 with ∑_{i=1}^{m−2} a_{i}ξ_{i} ≠ 1 and ∑_{i=1}^{m−2} a_{i} ≠ 1, be given. Then the multi-point boundary-value problem

\begin{matrix}x′′(t) = f(t, x(t), x′(t)) + e(t), 0 < t < 1, \\ x(0) = 0, x(1) = ∑_{i=1}^{m−2} a_{i}x(ξ_{i}). \end{matrix}

has at least one solution in C^{1}[0, 1] provided

C ∥p(t)∥_{1} + \frac{1}{1 − τ} ∥q(t)∥_{1} < 1,

where C is as given in Theorem 2 and τ as given in Proposition 1.

Theorem 10 Let f : [0, 1] × ℝ^{2} ↦ ℝ be a function satisfying Caratheodory’s conditions. Assume that there exist functions p(t), q(t), r(t) such that the functions p(t), q(t), r(t) are in L^{1}(0, 1) and

|f(t, x_{1}, x_{2})| ≤ p(t) |x_{1}| + q(t) |x_{2}| + r(t)

for a.e. t ∈ [0, 1] and all (x_{1}, x_{2}) ∈ ℝ^{2}. Let α ∈ ℝ, η ∈ (0, 1), α ≠ 1, and αη ≠ 1 be given. Then, the three-point boundary-value problem

\begin{matrix}x′′(t) = f(t, x(t), x′(t)) + e(t), 0 < t < 1, \\ x(0) = 0, x(1) = αx(η). \end{matrix}

has at least one solution in C^{1}[0, 1] provided

M ∥p(t)∥_{1} + \frac{1}{1 − τ} ∥q(t)∥_{1} < 1.

where M is as given in Theorem 5 and τ as given in Proposition 1.

References

[1] C. P. Gupta, S. I. Trofimchuk, a priori Estimates for the Existence of a Solution for a Multi-Point Boundary Value Problem, J. of Inequalities and Applications, 5(2000) p. 351-365.

[2] C. P. Gupta, S. I. Trofimchuk, Solvability of a multi-point boundary-value problem and related a priori estimates, Canadian Applied Mathematics Quarterly, Vol. 6 (1998) pp. 45-60.

Chaitan P. Gupta Department of Mathematics University of Nevada, Reno Reno, NV 89557 USA email: gupta@unr.edu chaitang@hotmail.com