Solving problems involving numerical integration (Ⅱ): Modified Simpson's methods for equal intervals of odd numbers

William Guo; William Guo

doi:10.3934/steme.2023011

STEM Education

2023, Volume 3, Issue 3: 171-189. doi: 10.3934/steme.2023011

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Technical note

Solving problems involving numerical integration (Ⅱ): Modified Simpson's methods for equal intervals of odd numbers

William Guo ^,

School of Engineering and Technology, Central Queensland University, Bruce Highway, North Rockhampton, QLD 4702, Australia

Academic Editor: Dianshun Hu

Received: 06 July 2023 Revised: 27 July 2023 Accepted: 09 August 2023 Published: 17 August 2023

The trapezium and Simpson's methods are widely used for numerical integration. In most circumstances, Simpson's method is more accurate than the trapezium method but only applicable to cases with equal intervals of even numbers. This technical note reports the formulation of two modified Simpson's methods, the trapezium-corrected Simpson's method (TCSM) and cubic-corrected Simpson's method (CCSM), as general-purpose symmetric formulas to solve numerical integrations with equal intervals of odd numbers (n ≥ 5) with the same level of accuracy as that of Simpson's method applied to the even number near n. Error analysis in terms of the order of error bound and case studies in this note demonstrate and validate the usefulness of the proposed formulas for solving different types of theoretical problems and real-world applications. In terms of accuracy of approximation for cases with equal intervals of odd numbers, CCSM performs better than TCSM by at least one order in error bound whereas TCSM performs better than the trapezium method by at least one order in error bound.

Keywords:

numerical integration,
Simpson's method,
trapezium method,
order of error bound,
trapezium-corrected Simpson's method (TCSM),
cubic-corrected Simpson's method (CCSM)

Citation: William Guo. Solving problems involving numerical integration (Ⅱ): Modified Simpson's methods for equal intervals of odd numbers[J]. STEM Education, 2023, 3(3): 171-189. doi: 10.3934/steme.2023011

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Abstract

1. Introduction

The Sturm-Liouville problem arises within many areas of science, engineering and applied mathematics. It has been studied for more than two decades. Many physical, biological and chemical processes are described using models based on it (see ^[1,2,3], ^[8], ^[9] and ^[11]).

For the homogeneous Sturm-Liouville problem with nonlocal conditions you can see ^[2], ^[9] and ^{[11,12,13,14,15]}. For the nonhomogeneous equation see ^[7]. In ^[7] the authors studied the nonhomogeneous Sturm-Liouville boundary value problem of the differential equation

$x''(t)+ m(t) = -\lambda^2 x(t),\; \; \; t \in (0,\pi),$

with the conditions

$x(0) = 0,\; \; \; \; \; x'(\xi)+\lambda x(\xi) = 0,\; \; \; \; \xi \in(0, \pi].$

Here, we are concerned, firstly, with the nonlocal problem of the nonlinear differential inclusion

$\begin{equation} -x''(t) \in F(t,\lambda x(t)),\; \; \; \; \; a.e.\; \; \; \; t\in (0,\pi), \end{equation}$

(1.1)

with the nonlocal conditions ( $\; \eta > \xi$ )

$\begin{equation} x'(0)-\lambda x(0) = 0\; \; \; \mbox{and} \; \int_\xi^\eta \; x(\tau)d\tau = 0,\; \; \; \xi\in[0,\pi),\; \; \; \eta\in(0,\pi]. \end{equation}$

(1.2)

For

$\; h(t, \lambda)\; +\; \lambda^2 x(t)\; = \; f(t,\lambda x(t)) \in F(t,\lambda x(t)),$

we study the existence of multiple solutions (eignevalues and eignefunctions) of the nonhomogeneous Sturm-Liouville problem of the differential equation

$\begin{equation} x''(t)+ h(t , \lambda) = -\lambda^2 x(t),\; \; \; t \in (0,\pi), \end{equation}$

(1.3)

with the conditions (1.2).

The special case of the nonlocal condition (1.2)

$\begin{equation} x'(0)-\lambda x(0) = 0\; \; \; \mbox{and} \; \int_0^\pi \; x(\tau)d\tau = 0, \end{equation}$

(1.4)

will be considered.

2. Existence of solutions

Consider the nonlocal boundary value problem of the nonlinear differential inclusion (1.1)-(1.2) under the following assumptions.

(ⅰ) The set $F(t, x)$ is nonempty, closed and convex for all $(t, x)\in [0, 1]\times R\times R$ .

(ⅱ) $F(t, x)$ is measurable in $t\in [0, 1]$ for every $x, y\in R$ .

(ⅲ) $F(t, x)$ is upper semicontinuous in $x$ and $y$ for every $t\in [0, 1]$ .

(ⅳ) There exist a bounded measurable function $m: [0, 1]\longrightarrow R$ and a constant $\lambda$ , such that

$\|F(t,x)\| = \sup\{|f|:f\in F(t,x)\}\; \leq |m(t)|+ \lambda^2 |x|.$

Remark 1. From the assumptions (i)-(iv) we can deduce that (see ^[1], ^[5] and ^[6]) there exists $f\in F(t, x)$ , such that

(v) $f:I\times R\longrightarrow R$ is measurable in $t$ for every $x, y\in R$ and continuous in $x$ for $t\in [0, 1]$ and there exist a bounded measurable function $m:[0, \pi] \rightarrow R$ and a constant $\lambda^2$ such that

$|f(t,x)|\leq |m(t)|+ \lambda^2 |x|,$

and $f$ satisfies the nonlinear differential equation

$\begin{equation} -x''(t)\; = \; f(t,\lambda x(t)),\; \; \; \; \; a.e.\; \; \; \; t\in (0,\pi). \end{equation}$

(2.1)

So, any solution of (2.1) is a solution of (1.1).

(ⅵ) $\lambda \, (\eta-\xi) \neq -2, \; \lambda \in R$ .

(ⅶ)

$\frac{2(1+ |\lambda| \pi)\pi^2+\pi}{|{\mathcal{A}}|}\; \lambda^2 \pi < 1.$

For the integral representation of the solution of (2.1) and (1.2) we have the following lemma.

Lemma 2.1. If the solution of the problem (2.1) and (1.2) exists, then it can be represented by the integral equation

$\begin{eqnarray} x(t)& = & \frac{2(1+ \lambda t)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x (s))ds \bigg] \\ && - \int_0^t (t-s)f (s,\lambda x (s))ds, \end{eqnarray}$

(2.2)

where ${\mathcal{A}} = (\eta-\xi)[2+\lambda(\eta-\xi)]\neq 0$ .

Proof. Integrating both sides of Eq (2.1) twice, we obtain

$\begin{equation} x(t)-x (0)- tx' (0) = - \int_0^t (t-s)f (s,\lambda x (s))ds \end{equation}$

(2.3)

and using the assumption $x' (0)-\lambda x (0) = 0$ , we obtain

$\begin{equation} x (0) = \frac{1}{\lambda}x' (0). \end{equation}$

(2.4)

The assumption $\; \int_\xi^\eta x(\tau)d\tau = 0$ implies that

$\begin{eqnarray*} x (0)\int_\xi^\eta d\tau+x'(0)\int_\xi^\eta \tau d\tau& = &\int_\xi^\eta \int_0^\tau (\tau-s)f (s,\lambda x (s))ds\; d\tau,\\ (\eta-\xi) x(0)+ \frac{(\eta-\xi)^2}{2} \lambda x(0)& = &\int_0^\xi\,\int_\xi^\eta (\tau-s)d \tau\; f (s,\lambda x (s))ds\\ &&+\int_\xi^\eta\,\int_s^\eta (\tau-s)d \tau\; f (s,\lambda x (s))ds,\\ \frac{(\eta-\xi)[2+\lambda(\eta-\xi)]}{2} x(0)& = &\int_0^\xi\bigg[\frac{(\eta-s)^2}{2}-\frac{(\xi-s)^2}{2}\bigg] f (s,\lambda x (s))ds\\ &&+\int_\xi^\eta \frac{(\eta-s)^2}{2}f (s,\lambda x (s))ds\\ & = &\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x (s))ds \end{eqnarray*}$

and we can get

$\begin{eqnarray} x(0)& = &\frac{2}{(\eta-\xi)[2+\lambda(\eta-\xi)]}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x (s))ds \bigg] \\ & = &\frac{2}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x (s))ds \bigg]. \end{eqnarray}$

(2.5)

Substituting (2.5) into (2.4), we obtain

$\begin{equation} x' (0) = \frac{2 \lambda}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x (s))ds \bigg]. \end{equation}$

(2.6)

Now from (2.3), (2.5) and (2.6), we obtain

$\begin{eqnarray*} x(t)& = & \frac{2(1+ \lambda t)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x (s))ds \bigg] \nonumber\\ && - \int_0^t (t-s)f (s,\lambda x (s))ds. \end{eqnarray*}$

To complete the proof, differentiate equation (2.2) twice, we obtain

$\begin{eqnarray*} x'(t)& = & \frac{2 \lambda }{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x (s))ds \bigg] \nonumber\\ && - \int_0^t f (s,\lambda x (s))ds, \end{eqnarray*}$

and

$x'' (t)\; = \; -f (t,\lambda x(t)),\; \; \; a.e. t \in (0,T).$

Now

$x' (0) = \frac{2 \lambda }{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x (s))ds \bigg],$

and

$\lambda x (0) = \frac{2 \lambda }{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x (s))ds \bigg].$

From that, we get $x' (0)-\lambda x (0) = 0$ .

Now, to ensure that $\; \int_\xi^\eta x(\tau) d\tau = 0$ ,

we have

$\int_\xi^\eta\; \frac{2(1+ \lambda t)}{{\mathcal{A}}} = \frac{2(\eta-\xi)+\lambda (\eta^2-\xi^2)}{{\mathcal{A}}} = \frac{(\eta-\xi)[2+\lambda(\eta-\xi)]}{{\mathcal{A}}} = 1,$

from that, we obtain as before

$\begin{eqnarray*} \int_\xi^\eta x(\tau) d\tau& = & \int_\xi^\eta \frac{2(1+ \lambda \tau)}{{\mathcal{A}}}\,d\tau\; \bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x (s))ds \bigg] \nonumber\\ && - \int_\xi^\eta \int_0^\tau (\tau-s)f (s,\lambda x (s))ds \,d \tau,\\\\ & = &\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x (s))ds \nonumber\\ &&-\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x (s))ds+\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x (s))ds = 0. \end{eqnarray*}$

This proves the equivalence between the integral equation (2.2) and the nonlocal boundary value problem (1.1)-(1.2).

Now, for the existence of at least one continuous solution for the problem of the integral equation (2.2), we have the following theorem.

Theorem 2.1. Let the assumptions (v)-(vii) be satisfied, then there exists at least one solution $x\in C[0, \pi]$ of the nonlocal boundary value problem (2.1) and (1.2). Moreover, from Remark 1, then there exists at least one solution $x\in C[0, \pi]$ of the nonlocal boundary value problem (1.1)-(1.2).

Proof. Define the set $Q_r\subset C[0, \pi]$ by

$Q_r = \{x\in C:\parallel x\parallel\leq r\},\; \; r\geq \frac{ 2(1+ |\lambda| \pi)\pi^2+\pi\; \|m \|_{L^1} }{|{\mathcal{A}}|-\big[2(1+ |\lambda| \pi)\pi^2+\pi\big]\; \lambda^2 \pi }.$

It is clear that the set $Q_r$ is nonempty, closed and convex.

Define the operator $T$ associated with (2.2) by

$\begin{eqnarray*} T x(t)& = & \frac{2(1+ \lambda t)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x (s))ds \bigg] \nonumber\\ && - \int_0^t (t-s)f (s,\lambda x (s))ds. \end{eqnarray*}$

Let $x\in Q_r$ , we have

$\begin{eqnarray*} | T x(t)|&\leq&\frac{2(1+ |\lambda| t)}{|{\mathcal{A}}|}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} |f (s,\lambda x (s))|ds+\int^\xi_0 \frac{(\xi-s)^2}{2}|f (s,\lambda x (s))ds| \bigg] \nonumber\\ && + \int_0^t (t-s)|f (s,\lambda x (s))|ds,\\ &\leq&\frac{2(1+ |\lambda| \pi)\pi^2}{|{\mathcal{A}}|}\; \int_0^\pi \{|m (s)|+\lambda^2|x (s)|\}ds + \pi\int_0^\pi \{|m (s)|+\lambda^2|x (s)|\}ds, \nonumber\\ &\leq&\bigg[\frac{2(1+ |\lambda| \pi)\pi^2}{|{\mathcal{A}}|}+\pi \bigg]\; \{\|m \|_{L^1}+\lambda^2 \pi \|x\| \},\\ &\leq&\frac{2(1+ |\lambda| \pi)\pi^2+\pi}{|{\mathcal{A}}|}\; \{\|m \|_{L^1}+\lambda^2 \pi r \}\leq r,\\ \end{eqnarray*}$

and we have

$\frac{2(1+ |\lambda| \pi)\pi^2+\pi}{|{\mathcal{A}}|}\; \|m \|_{L^1} \leq r\bigg(1-\frac{2(1+ |\lambda| \pi)\pi^2+\pi}{|{\mathcal{A}}|}\; \lambda^2 \pi \bigg).$

Then $T:Q_r\rightarrow Q_r$ and the class $\{Tx\}\subset Q_r$ is uniformly bounded in $Q_r$ .

In what follows we show that the class $\{Tx\}$ , $x\in Q_r$ is equicontinuous. For $t_1, t_2\in [0, \pi], \; t_1 < t_2$ such that $|t_2-t_1| < \delta$ , we have

$\begin{eqnarray*} Tx(t_2)-Tx (t_1)& = & \frac{2(1+ \lambda t_2)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x (s))ds \bigg] \nonumber\\ && - \int_0^{t_2} (t_2-s)f (s,\lambda x (s))ds-\frac{2(1+ \lambda t_1)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x (s))ds\\ &&-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x (s))ds \bigg] - \int_0^{t_1} (t_2-s)f (s,\lambda x (s))ds,\\ |Tx (t_2)-Tx (t_1)|& = &\bigg|\frac{2(1+ \lambda t_2)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x (s))ds \bigg] \nonumber\\ && - \int_0^{t_2} (t_2-s)f (s,\lambda x (s))ds-\frac{2(1+ \lambda t_1)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x (s))ds\\ &&-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x (s))ds \bigg] + \int_0^{t_1} (t_1-s)f (s,\lambda x (s))ds \bigg|, \nonumber\\ \\ &\leq&\frac{2 |\lambda|( t_2-t_1)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} |f (s,\lambda x (s))|ds+\int^\xi_0 \frac{(\xi-s)^2}{2}|f (s,\lambda x (s))|ds \bigg] \nonumber\\ && + (t_2-t_1)\int_0^{t_1} |f (s,\lambda x (s))|ds+\pi \int_{t_1}^{t_2} |f (s,\lambda x (s))|ds,\\ \\ &\leq&\frac{2 |\lambda|( t_2-t_1)\pi^2}{{\mathcal{A}}}\; \int_0^\pi |f (s,\lambda x (s))|ds \nonumber\\ && + (t_2-t_1)\int_0^{\pi} |f (s,\lambda x (s))|ds+\pi \int_{t_1}^{t_2} |f (s,\lambda x (s))|ds,\\ \\ &\leq&\frac{2 |\lambda|( t_2-t_1)\pi^2}{{\mathcal{A}}}\; \{\|m \|_{L^1}+\lambda^2\|x \|\} + (t_2-t_1)\; \{\|m \|_{L^1}+\lambda^2\|x \|\} \nonumber\\ &&+\pi \int_{t_1}^{t_2} \{|m (s)|+\lambda^2|x(s) |\}ds. \end{eqnarray*}$

Hence the class of function $\{Tx\}$ , $x\in Q_r$ is equicontinuous. By Arzela-Ascolis ^[4] Theorem, we found that the class $\{Tx\}$ is relatively compact.

Now we prove that $T:Q_r\rightarrow Q_r$ is continuous.

Let $\{x_n\} \subset Q_r$ , such that $x_n\rightarrow x_0\in Q_r$ , then

$\begin{eqnarray*} Tx_n(t)& = & \frac{2(1+ \lambda t)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x_n (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x_n (s))ds \bigg] \nonumber\\ && - \int_0^t (t-s)f (s,\lambda x_n (s))ds, \end{eqnarray*}$

and

$\begin{eqnarray*} \lim\limits_{n\to\infty} Tx_n (t)& = &\lim\limits_{n\to\infty}\bigg\{\frac{2(1+ \lambda t)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x_n (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x_n (s))ds \bigg] \nonumber\\ && - \int_0^t (t-s)f (s,\lambda x_n (s))ds\bigg\}. \end{eqnarray*}$

Now, we have

$f (s,x_n (s))\rightarrow f (s,x_0 (s))\; \; \mbox{as}\; n\to\infty,$

and

$|f (s,\lambda x_n (s))|\leq m (s)+\lambda^2|x_n|\in L^1[0, \pi],$

then applying Lebesgue Dominated convergence theorem ^[4], we obtain

$\begin{eqnarray*} \lim\limits_{n\to\infty} Tx_n (t)& = &\frac{2(1+ \lambda t)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} \lim\limits_{n\to\infty} f (s,\lambda x_n (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}\lim\limits_{n\to\infty}f (s,\lambda x_n (s))ds \bigg] \nonumber\\ && - \int_0^t (t-s)\lim\limits_{n\to\infty}f (s,\lambda x_n (s))ds,\\ \\ & = &\frac{2(1+ \lambda t)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x_0 (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2} f (s,\lambda x_0 (s))ds \bigg] \nonumber\\ && - \int_0^t (t-s)f (s,\lambda x_0 (s))ds = F (x_0). \end{eqnarray*}$

Then $Tx_n (t)\rightarrow Tx_0 (t)$ . Which means that the operator $T$ is continuous.

Since all conditions of Schauder theorem ^[4] are hold, then $T$ has a fixed point in $Q_r$ , then the integral equation (2.2) has at least one solution $x\in C[0, \pi]$ .

Consequently the nonlocal boundary value problem (2.1)-(1.2) has at least one solution $x\in C[0, \pi].$ Moreover, from Remark 1, then there exists at least one solution $x\in C[0, \pi]$ of the nonlocal boundary value problem (1.1)-(1.2).

Now, we have the following corollaries

Corollary 1. Let $\; \lambda^2 x(t)\; = \; f(t, \lambda x(t))\; \in F(t, \lambda x(t)).\;$ Let the assumptions of Theorem 2.1 be satisfied. Then there exists at lease one solution $x \in C[0, \pi]$ of

$- x'' (t)\; = \; \lambda^2 x(t),\; \; \; \; \; t \in (0,T).$

with the nonlocal condition (1.2). Moreover, from Remark 1, there exists at lease one solution $x \in C[0, \pi]$ of the problem (1.1)-(1.2).

Corollary 2. Let the assumptions of Theorem 2.1 be satisfied. Then there exists a solution $x \in C[0, \pi]$ of the problem (2.1) and (1.4).

Proof. Putting $\xi = 0$ and $\eta = \pi$ and applying Theorem 2.1 we get the result.

3. Maximal and minimal solutions

Taking $\; J = (0, \pi).\;$ Here, we study the existence of maximal and minimal solutions of the problem (2.1) and (1.2) which is equivalent to the integral equation (2.2).

Definition 3.1. ^[10] Let $\; q(t)\;$ be a solution $\; x(t)\;$ of (2.2) Then $\; q(t)\;$ is said to be a maximal solution of (2.2) if every solution of (2.2) on $\; J\;$ satisfies the inequality $\; x(t)\; \leq \; q(t), \; \; t\in J\;$ . A minimal solution $\; s(t)\;$ can be defined in a similar way by reversing the above inequality i.e. $\; x(t)\; \geq\; s(t), \; t\in J.\;$

We need the following lemma to prove the existence of maximal and minimal solutions of (2.2).

Lemma 3.2. Let $\; f(t, x)\;$ satisfies the assumptionsin Theorem 2.1 and let $\; x(t), \; y(t)\;$ be continuous functions on $J$ satisfying

$\begin{eqnarray*} x(t)&\leq& \frac{2(1+ \lambda t)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x (s))ds \bigg] \nonumber\\ && - \int_0^t (t-s)f (s,\lambda x (s))ds, \\ y(t)&\geq& \frac{2(1+ \lambda t)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda y (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda y (s))ds \bigg] \nonumber\\ && - \int_0^t (t-s)f (s,\lambda y (s))ds \end{eqnarray*}$

where one of them is strict.

Suppose $\; f(t, x)\;$ is nondecreasing function in $\; x.\;$ Then

$\begin{equation} x(t)\; < \; y(t), \; t\in J. \end{equation}$

(3.1)

Proof. Let the conclusion (3.1) be false; then there exists $\; t_1\;$ such that

$x(t_1)\; = \; y(t_1),\; \; \; \; \; \; \; t_1\; > \; 0$

and

$x(t)\; < \; y(t),\; \; \; \; \; \; \; \; 0\; < \; t\; < \; t_1.$

From the monotonicity of the function $\; f\;$ in $\; x, \;$ we get

$\begin{eqnarray*} x(t_1)&\leq& \frac{2(1+ \lambda t_1)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda x (s))ds \bigg] \nonumber\\ && - \int_0^{t_1} (t-s)f (s,\lambda x (s))ds, \\ \; & < & \frac{2(1+ \lambda t_1)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda y (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda y (s))ds \bigg] \nonumber\\ && - \int_0^{t_1} (t-s)f (s,\lambda y (s))ds\\ \; & < &\; y(t_1). \end{eqnarray*}$

This contradicts the fact that $\; x(t_1)\; = \; y(t_1);\;$ then

$x(t)\; < \; y(t), \; t\in J.$

Theorem 3.2. Let the assumptions of Theorem 2.1 besatisfied. Furthermore, if $\; f(t, x)\;$ is nondecreasing function in $\; x, \;$ then there exist maximal and minimal solutions of (2.2).

Proof. Firstly, we shall prove the existence of maximal solution of (2.2). Let $\epsilon > 0$ be given. Now consider the integral equation

(3.2)

where

$\; f_{\epsilon} (t,x_\epsilon (t)) \; = \; f(t,x_{\epsilon}(t))\; +\; \epsilon.$

Clearly the function $\; f_{\epsilon} (t, x_\epsilon)\;$ satisfies assumption (v) and

$|\; f_{\epsilon} (t,x_\epsilon)\; |\; \leq\; |m(t)|+ \lambda^2 |x|+\; \epsilon \leq |m_1(t)|+ \lambda^2 |x|,\; |m_1(t)| = |m(t)|+\epsilon.$

Therefore, Equation (3.2) has a continuous solution $\; x_\epsilon(t)\;$ according to Theorem 2.1.

Let $\epsilon_1$ and $\epsilon_2$ be such that $0 < \epsilon_2 < \epsilon_1 < \epsilon.$ Then

$\begin{eqnarray} x_{\epsilon_1}(t)& = & \frac{2(1+ \lambda t)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f_{\epsilon_1} (s,\lambda x_{\epsilon_1} (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f_{\epsilon_1} (s,\lambda x_{\epsilon_1} (s))ds \bigg] \\ && - \int_0^t (t-s)f_{\epsilon_1} (s,\lambda x_{\epsilon_1} (s))ds,\\ & = & \frac{2(1+ \lambda t)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} ( f (s,\lambda x_{\epsilon_1} (s))+\epsilon_1)ds-\int^\xi_0 \frac{(\xi-s)^2}{2}( f (s,\lambda x_{\epsilon_1} (s))+\epsilon_1)ds \bigg] \\ && - \int_0^t (t-s)( f (s,\lambda x_{\epsilon_1} (s))+\epsilon_1)ds,\\ &\; > \; &\frac{2(1+ \lambda t)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} ( f (s,\lambda x_{\epsilon_1} (s))+\epsilon_2)ds-\int^\xi_0 \frac{(\xi-s)^2}{2}( f (s,\lambda x_{\epsilon_1} (s))+\epsilon_2)ds \bigg] \\ && - \int_0^t (t-s)( f (s,\lambda x_{\epsilon_1} (s))+\epsilon_2)ds, \end{eqnarray}$

(3.3)

$\begin{eqnarray} x_{\epsilon_2}(t)& = & \frac{2(1+ \lambda t)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} ( f (s,\lambda x_{\epsilon_2} (s))+\epsilon_2)ds-\int^\xi_0 \frac{(\xi-s)^2}{2}( f (s,\lambda x_{\epsilon_2} (s))+\epsilon_2)ds \bigg] \\ && - \int_0^t (t-s)( f (s,\lambda x_{\epsilon_2} (s))+\epsilon_2)ds. \end{eqnarray}$

(3.4)

Applying Lemma 3.2, then (3.3) and (3.4) imply that

$x_{\epsilon_2}(t) < x_{\epsilon_1}(t)\; \; \; \; \; \; \; for\; t \in\; J.$

As shown before in the proof of Theorem 2.1, the family of functions $\; x_{\epsilon}(t)\;$ defined by Eq (3.2) is uniformly bounded and of equi-continuous functions. Hence by the Arzela-Ascoli Theorem, there exists a decreasing sequence $\epsilon _n$ such that $\; \epsilon_n \rightarrow 0$ as $n \rightarrow \infty,$ and $\; \lim\limits_{n \rightarrow \infty} x_{\epsilon_n}(t)$ exists uniformly in $\; I.\;$ We denote this limit by $\; q(t).\;$ From the continuity of the function $f_{\epsilon_n}$ in the second argument, we get

$\begin{eqnarray*} x(t)& = & \lim\limits_{n \rightarrow \infty} x_{\epsilon_n}(t)\; = \frac{2(1+ \lambda t)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda q (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f (s,\lambda q (s))ds \bigg] \nonumber\\ && - \int_0^t (t-s)f (s,\lambda q (s))ds, \end{eqnarray*}$

which proves that $\; q(t)\;$ is a solution of (2.2).

Finally, we shall show that $\; q(t)\;$ is maximal solution of (2.2). To do this, let $\; x(t)\;$ be any solution of (2.2). Then

$\begin{eqnarray} x_\epsilon(t)& = & \frac{2(1+ \lambda t)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f_\epsilon (s,\lambda x_\epsilon (s))ds-\int^\xi_0 \frac{(\xi-s)^2}{2}f_\epsilon (s,\lambda x_\epsilon (s))ds \bigg] \\ && - \int_0^t (t-s)f_\epsilon (s,\lambda x_\epsilon (s))ds,\\ & = &\frac{2(1+ \lambda t)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} ( f (s,\lambda x_\epsilon (s))+\epsilon) ds-\int^\xi_0 \frac{(\xi-s)^2}{2}( f (s,\lambda x_\epsilon (s))+\epsilon)ds \bigg] \\ && - \int_0^t (t-s)( f (s,\lambda x_\epsilon (s))+\epsilon)ds,\\ & > & \frac{2(1+ \lambda t)}{{\mathcal{A}}}\bigg[\int_0^\eta\frac{(\eta-s)^2}{2} f (s,\lambda x_\epsilon (s)) ds-\int^\xi_0 \frac{(\xi-s)^2}{2} f (s,\lambda x_\epsilon (s))ds \bigg] \\ && - \int_0^t (t-s) f (s,\lambda x_\epsilon (s))ds. \end{eqnarray}$

(3.5)

Applying Lemma 3.2, then (2.2) and (3.5 imply that

$x_\epsilon (t)\; > \; x(t) \; \; \; \; \; for\; t \; \in \; J.$

From the uniqueness of the maximal solution (see ^[10]), it is clear that $x_\epsilon (t)$ tends to $\; q(t)\;$ uniformly in $t\; \in\; J\; \; as\; \; \epsilon \rightarrow 0.$

In a similar way we can prove that there exists a minimal solution of (2.2).

4. The homogeneous problem

Here, we study the existence and some general properties of the eigenvalues and eigenfunctions of the problem of the homogeneous equation

$\begin{equation} x''(t)\; = -\lambda^2 x(t),\; \; \; t \in (0,\pi), \end{equation}$

(4.1)

with the nonlocal condition (1.2).

Lemma 4.3. The eigenfunctions of the nonlocal boundary value problem (4.1) and (1.2) are in the form of

$\begin{equation} x_n (t) = c_n \bigg(\sin \frac{(-\pi+4\pi n)\, t}{2(\eta + \xi)}\; +\; \cos \frac{(-\pi+4\pi n)\, t}{2(\eta + \xi)}\bigg),\; \; \; n = 1,2,\cdots . \end{equation}$

(4.2)

Proof. Firstly, we prove that the eigenvalues are

$\begin{equation} \lambda_n = \frac{-\pi+4\pi n}{2(\eta + \xi)},\; \; \; n = 1,2,\cdots. \end{equation}$

(4.3)

The general solution of the problem (4.1) and (1.2) is given by

$\begin{equation} x(t) = c_1 \sin\lambda t + c_2 \cos\lambda t. \end{equation}$

(4.4)

Differentiating equation (4.4), we obtain

$x' (t) = \lambda c_1 \cos \lambda t - \lambda c_2 \sin \lambda t.$

Using the first condition, when $t = 0$ , we obtain

$\begin{equation} c_1 = c_2. \end{equation}$

(4.5)

Integrating both sides of (4.4) from $\xi$ to $\eta$ , we obtain

$\frac{c_1}{\lambda}\cos\lambda \xi - \frac{c_1}{\lambda}\cos\lambda \eta + \frac{c_2}{\lambda}\sin\lambda \eta - \frac{c_2}{\lambda}\sin\lambda \xi = 0.$

Substituting $c_1 = c_2$ , we obtain

$\begin{equation} \frac{c_1}{\lambda}\cos\lambda \xi - \frac{c_1}{\lambda}\cos\lambda \eta + \frac{c_1}{\lambda}\sin\lambda \eta - \frac{c_1}{\lambda}\sin\lambda \xi = 0. \end{equation}$

(4.6)

Multiplying (4.6) by $\frac{\lambda}{c_1}$ , we obtain

$\begin{eqnarray} &&\cos\lambda \xi -\cos\lambda \eta + \sin\lambda \eta - \sin\lambda \xi = 0, \\ &&2\sin\frac{\lambda(\xi+\eta)}{2}\; \sin\frac{\lambda(\eta-\xi)}{2}\; +\; 2\sin\frac{\lambda(\eta-\xi)}{2}\; \cos\frac{\lambda(\eta+\xi)}{2} = 0, \\ &&\sin\frac{\lambda(\xi+\eta)}{2}\; +\; \cos\frac{\lambda(\eta+\xi)}{2} = 0, \\ &&\tan \frac{\lambda(\xi+\eta)}{2} = -1, \\ &&\frac{\lambda(\xi+\eta)}{2} = -\frac{\pi}{4}+n \pi. \end{eqnarray}$

(4.7)

From (4.7), we deduce that

$\lambda_n = \frac{-\pi+4\pi n}{2(\eta + \xi)},\; \; \; n = 1,2,....\; .$

Therefore, from (4.4) we can get

$x_n (t) = c_n \bigg(\sin \frac{(-\pi+4\pi n)\, t}{2(\eta + \xi)}\; +\; \cos \frac{(-\pi+4\pi n)\, t}{2(\eta + \xi)}\bigg),\; \; \; n = 1,2,...\; .$

Corollary 3. The eigenfunctions of the nonlocal boundary value problem (4.1) and (1.4) are in the form of

$\begin{equation} x_n (t) = c_n \bigg(\sin \frac{(-1+4 n)\, t}{2}\; +\; \cos \frac{(-1+4 n)\, t}{2}\bigg),\; \; \; n = 1,2,....\; . \end{equation}$

(4.8)

Proof. Putting $\xi = 0$ and $\eta = \pi$ and applying Lemma 4.3 we obtain the result.

5. The nonhomogeneous problem

Now, we study the existence of multiple solutions of the nonhomogeneous problem (1.3) and (1.2). Let $x_1, \; x_2$ be two solutions of the problem (1.3) and (1.2). Let $u(t) = x_1(t)-x_2(t)$ , then the function $u$ satisfy the Sturm-Liouville problem

$u''(t) = -\lambda^2 u(t)$

with the nonlocal conditions

$u'(0)-\lambda u(0) = 0\; \; \; \mbox{and}\; \; \; \; \int_\xi^\eta u(\tau) d\tau = 0,\; \; \; \xi\in[0,\pi),\; \; \; \eta\in(0,\pi].$

So, the values of (eigenvalues) $\lambda_n$ for the non zero solution of (4.1) and (1.2) is the same values (eigenvalues) of $\lambda_n$ for the multiple solutions (eigenfunctions) of (1.3) and (1.2), i.e.

$\lambda_n = \frac{-\pi+4\pi n}{2(\eta + \xi)},\; \; \; n = 1,2,....\; .$

Theorem 5.3. The multiple solutions (eigenfunctions) $x_n(t)$ of the problem (1.3) and (1.2) are given by

$\begin{equation} x_n (t) = A_n \bigg(\sin \frac{(-\pi+4\pi n)\, t}{2(\eta + \xi)}+ \cos \frac{(-\pi+4\pi n)\, t}{2(\eta + \xi)}\bigg) - \int_0^t \frac{\sin\frac{(-\pi+4\pi n)(t-s)}{2(\eta + \xi)} }{\frac{-\pi+4\pi n}{2(\eta + \xi)}} h(s, \lambda) ds. \end{equation}$

(5.1)

Proof. Here we use the variation of parameter method to get the solution of (1.3) and (1.2). Assume that the solutions of (1.3) and (1.2) are given by

$\begin{equation} x_n (t) = A_1 \cos\lambda t + A_2 \sin\lambda t + x_p(t). \end{equation}$

(5.2)

So, we have

$x_1 (t) = \cos\lambda t,\; \; \; x_2 (t) = \sin\lambda t.$

Now, we can get $W (x_1, x_2) = \lambda$ . Hence

$x_p(t) = - \cos\lambda t \int_0^t \frac{\sin\lambda s}{\lambda} h(s, \lambda) ds + \sin\lambda t \int_0^t \frac{\cos\lambda s}{\lambda} h(s, \lambda) ds,$

thus

$\begin{equation} x_p(t) = -\int_0^t \frac{\sin\lambda (t-s)}{\lambda} h(s, \lambda) ds. \end{equation}$

(5.3)

From (5.3) and (5.2), we obtain

$\begin{equation} x_n (t) = A_1 \sin \frac{(-\pi+4\pi n)\, t}{2(\eta + \xi)}+ A_2 \cos \frac{(-\pi+4\pi n)\, t}{2(\eta + \xi)} - \int_0^t \frac{\sin\frac{(-\pi+4\pi n)(t-s)}{2(\eta + \xi)} }{\frac{-\pi+4\pi n}{2(\eta + \xi)}} h(s, \lambda) ds. \end{equation}$

(5.4)

By using the first condition $x' (0)- \lambda x (0) = 0$ , we get

$A_1 = A_2,$

therefore the multiple solutions of the nonlocal problem (1.3) and (1.2) are given by

$x_n (t) = A_n \bigg( sin \frac{(-\pi+4\pi n)\, t}{2(\eta + \xi)}+ cos \frac{(-\pi+4\pi n)\, t}{2(\eta + \xi)}\bigg) - \int_0^t \frac{\sin\frac{(-\pi+4\pi n)\, (t-s)}{2(\eta + \xi)}}{\frac{(-\pi+4\pi n)}{2(\eta + \xi)}} h(s,\lambda) ds,\; \; \; n = 1,2,....\; .$

To complete the proof and to ensure that $x_n (t)$ is the solution of (1.3) and (1.2), we firstly prove that

$x''_n (t) + h (t, \lambda) = - \lambda^2 x_n (t).$

Differentiating (5.4) twice, we get

$\begin{eqnarray*} x'_n (t)& = &A_n \frac{-\pi+4\pi n}{2(\eta + \xi)} \bigg( \cos\frac{(-\pi+4\pi n)\, t}{2(\eta + \xi)} - \sin \frac{(-\pi+4\pi n)\, t}{2(\eta + \xi)}\bigg)\\ &&- \int_0^t \cos \frac{(-\pi+4\pi n)\, (t-s)}{2(\eta + \xi)} h(s, \lambda)ds \end{eqnarray*}$

and

$\begin{eqnarray*} x''_n (t)& = &A_n \bigg(\frac{-\pi+4\pi n}{2(\eta + \xi)}\bigg)^2 \bigg(- \sin \frac{(-\pi+4\pi n)\, t}{2(\eta + \xi)} - \cos \frac{(-\pi+4\pi n)\, t}{2(\eta + \xi)}\bigg)\\ &&-g(t)+ \frac{-\pi+4\pi n}{2(\eta + \xi)}\int_0^t \sin \frac{(-\pi+4\pi n)\, (t-s)}{2(\eta + \xi)} h(s,\lambda)ds \end{eqnarray*}$

and

$\begin{eqnarray*} x''_n (t) + h(t, \lambda)& = &A_n\bigg(\frac{-\pi+4\pi n}{2(\eta + \xi)}\bigg)^2 \bigg(- \sin \frac{(-\pi+4\pi n)\, t}{2(\eta + \xi)} - \cos \frac{(-\pi+4\pi n)\, t}{2(\eta + \xi)}\bigg)\\ &&-h(t, \lambda)+ \frac{-\pi+4\pi n}{2(\eta + \xi)}\int_0^t \sin \frac{(-\pi+4\pi n)\, (t-s)}{2(\eta + \xi)} h(s, \lambda)ds + h(t, \lambda)\\ & = & - \lambda^2 x_n (t). \end{eqnarray*}$

Also we have $x' (0)- \lambda x (0) = 0$ .

Example 1. Let $\; h(t, \lambda)\; = \; \lambda^2.\;$ Then we find that

$x_p(t)\; = \; -\int_0^t \frac{\sin\lambda (t-s)}{\lambda} \lambda^2 ds\; = \; cos \lambda t - 1$

and the multiple solutions of the nonlocal problem (1.3) and (1.2) are given by

$x_n (t) = A_1 \sin \frac{(-\pi+4\pi n)\, t}{2(\eta + \xi)}+ A_2 \cos \frac{(-\pi+4\pi n)\, t}{2(\eta + \xi)} + cos \lambda t - 1.$

Now consider the Riemann integral boundary condition (1.4).

Corollary 4. The multiple solutions (eigenfunctions) $x_n(t)$ of the problem (1.3)-(1.4) are given by

$x_n (t) = A_n \bigg(\sin \frac{(-1+4 n)\, t}{2}+ \cos \frac{(-1+4 n)\, t}{2}\bigg) - \int_0^t \frac{\sin\frac{(-1+4 n)(t-s)}{2} }{\frac{-1+4 n}{2}} h(s, \lambda) ds.$

Proof. In this special case, we put $\xi = 0$ and $\eta = \pi$ and applying Theorem 5.3 we get the result.

Example 2. Let $\; h(t, \lambda)\; = \; \lambda^2.\;$ Then we find that

$x_p(t)\; = \; -\int_0^t \frac{\sin\lambda (t-s)}{\lambda} \lambda^2 ds\; = \; cos \lambda t - 1,$

and the solution $\; x_n(t)\;$ of the problem (1.3)-(1.4) are given by

$x_n (t)\; = \; A_n \bigg(\sin \frac{(-1+4 n)\, t}{2}+ \cos \frac{(-1+4 n)\, t}{2}\bigg) + cos \lambda t - 1.$

6. Conclusions

Here, we proved the existence of solutions $\; x \in C[0, \pi]\;$ of the nonlocal boundary value problem of the differential inclusion (1.1) with the nonlocal condition (1.2).

The maximal and minimal solutions of the problem (1.1)-(1.2) have been proved. The eigenvalues and eigenfunctions of the homogeneous and nonhomogeneous equations (4.1) and (1.3) with the nonlocal condition (1.2) have been obtained. Two examples have been studied to illustrate our results.

Acknowledgments

We thank the referees for their constructive remarks and comments on our work which reasonably improved the presentation and the structure of the manuscript.

Conflict of interest

The authors declare no conflict of interest.

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Author's biography Dr. William Guo is a professor in mathematics education with Central Queensland University, Australia. He is specialized in teaching applied mathematics for both engineering and education students. His research interests include mathematics education, applied computing, data analysis and numerical modeling. He is a member of IEEE and Australian Mathematical Society

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