New Numerical Method for Solving Tenth Order Boundary Value Problems

Ali Akgül; Esra Karatas Akgül; Dumitru Baleanu; Mustafa Inc

doi:10.3390/math6110245

,

and

¹

Department of Mathematics, Art and Science Faculty, Siirt University, TR-56100 Siirt, Turkey

²

Department of Mathematics, Faculty of Education, Siirt University, TR-56100 Siirt, Turkey

³

Department of Mathematics and Computer Sciences, Art and Science Faculty, Çankaya University, TR-06300 Ankara, Turkey

⁴

Department of Mathematics, Institute of Space Sciences, 077125 Bucharest, Romania

Mathematics2018, 6(11), 245;https://doi.org/10.3390/math6110245

This article belongs to the Special Issue Advances in Differential and Difference Equations with Applications 2019

Version Notes

Order Reprints

Abstract

In this paper, we implement reproducing kernel Hilbert space method to tenth order boundary value problems. These problems are important for mathematicians. Different techniques were applied to get approximate solutions of such problems. We obtain some useful reproducing kernel functions to get approximate solutions. We obtain very efficient results by this method. We show our numerical results by tables.

Keywords:

reproducing kernel Hilbert space method; tenth-order boundary value problems; reproducing kernel functions

AMS:

35C10; 46E22; 30E25

1. Introduction

A numerical approximation of tenth-order boundary value problems has been given in [1]. Usmani [2] has solved fourth order boundary value problems by using the quartic spline method. Twizell and Boutayeb [3] have improved the numerical approximations for higher order eigenvalue value problems. The approximation of second order boundary value problems has been presented by Alberg and Ito [4]. Siraj-ul-Islam et al. [5] has used a non-polynomial spline method to approximate the sixth-order boundary value problems. Papamichael and Worsey [6] have applied the cubic spline algorithm to solve linear fourth-order boundary value problems. Siddiqi and Twizell [7,8] have enhanced numerical approximations of tenth and twelfth-order boundary value problems.

We consider the following problem in the reproducing kernel Hilbert space:

\begin{matrix} q^{(x)} (τ) + a_{1} (τ) q^{(i x)} (τ) + a_{2} (τ) q^{(v i i i)} (τ) \\ + a_{3} (τ) q^{(v i i)} (τ) + a_{4} (τ) q^{(v i)} (τ) + a_{5} (τ) q^{(v)} (τ) \\ + a_{6} (τ) q^{(i v)} (τ) + a_{7} (τ) q^{‴} (τ) + a_{8} (τ) q^{″} (τ) \\ + a_{9} (τ) q^{'} (τ) + a_{10} (τ) q (τ) = H (τ), τ \in [a, b], \end{matrix}

with boundary conditions:

\begin{matrix} q (a) = α_{0}, q (b) = β_{0}, q^{″} (a) = α_{1}, q^{″} (b) = β_{1}, \\ q^{(i v)} (a) = α_{2}, q^{(i v)} (b) = β_{2}, q^{(v i)} (a) = α_{3}, \\ q^{(v i)} (b) = β_{3}, q^{(v i i i)} (a) = α_{4}, q^{(v i i i)} (b) = β_{4}, \end{matrix}

where

α_{j}, β_{j}, j = 0, 1, 2, 3, 4

are arbitrary fixed real constants, and

a_{j} (τ), j = 1, 2, \dots, 10

and

H (τ)

are continuous functions given on

[a, b] .

The notion of reproducing kernel has been presented by Zaremba [9]. Aronszajn has given a systematic reproducing kernel theory containing the Bergman kernel function [10]. For more details, see [11,12,13,14,15,16,17,18,19,20,21,22,23,24,25].

This paper is constructed as: Section 2 shows the reproducing kernel functions. The exact and approximate solutions of the problem are acquired in Section 3. Numerical results are given in Section 4. Conclusions are presented in the last section.

2. Reproducing Kernel Spaces and Their Reproducing Kernel Functions

We give some important reproducing kernel functions in this section. We define

V_{2}^{11} [0, 1]

by:

V_{2}^{11} [0, 1] = \{\begin{matrix} h, h^{^{'}}, h^{^{″}}, h^{^{‴}}, h^{(4)}, h^{(5)}, h^{(6)}, h^{(7)}, h^{(8)}, h^{(9)}, h^{(10)} \\ are absolutely continuous functions, h^{(11)} \in L^{2} [0, 1], \\ h (0) = h^{″} (0) = h^{(4)} (0) = h^{(6)} (0) = h^{(8)} (0) = 0 \\ h (1) = h^{″} (1) = h^{(4)} (1) = h^{(6)} (1) = h^{(8)} (1) = 0 . \end{matrix}\}

We define the inner product as:

\begin{matrix} {⟨ h, P_{y} ⟩}_{V_{2}^{11} [0, 1]} = \sum_{i = 0}^{10} h^{i} (0) P_{y}^{i} (0) + + \int_{0}^{1} h^{(11)} (τ) P_{y}^{(11)} (τ) d (τ) . \end{matrix}

We can obtain:

\begin{matrix} {⟨ h, P_{y} ⟩}_{V_{2}^{11} [0, 1]} & = & h (0) P_{y} (0) + h^{^{'}} (0) P_{y}^{^{'}} (0) + h^{^{″}} (0) P_{y}^{^{″}} (0) \\ + & h^{^{‴}} (0) P_{y}^{^{‴}} (0) + h^{(4)} (0) P_{y}^{(4)} (0) \\ + & h^{(5)} (0) P_{y}^{(5)} (0) + h^{(6)} (0) P_{y}^{(6)} (0) \\ + & h^{(7)} (0) P_{y}^{(7)} (0) + h^{(8)} (0) P_{y}^{(8)} (0) \\ + & h^{(9)} (0) P_{y}^{(9)} (0) + h^{(10)} (0) P_{y}^{(10)} (0) \\ + & \int_{0}^{1} h^{(11)} (τ) P_{y}^{(11)} (τ) d (τ) . \end{matrix}

P_{y} (τ)

is the reproducing kernel function of

V_{2}^{11} [0, 1] .

Therefore, we have

\begin{matrix} P_{y} (0) = P_{y}^{″} (0) = P_{y}^{(4)} (0) = P_{y}^{(6)} (0) = P_{y}^{(8)} (0) = 0, \\ P_{y} (1) = P_{y}^{″} (1) = P_{y}^{(4)} (1) = P_{y}^{(6)} (1) = P_{y}^{(8)} (1) = 0 . \end{matrix}

Thus, we get

\begin{matrix} {⟨ h, P_{y} ⟩}_{V_{2}^{11} [0, 1]} & = & h^{^{'}} (0) P_{y}^{^{'}} (0) + h^{^{‴}} (0) P_{y}^{^{‴}} (0) \\ + & h^{(5)} (0) P_{y}^{(5)} (0) + h^{(7)} (0) P_{y}^{(7)} (0) \\ + & h^{(9)} (0) P_{y}^{(9)} (0) + h^{(10)} (0) P_{y}^{(10)} (0) \\ + & \int_{0}^{1} h^{(11)} (τ) P_{y}^{(11)} (τ) d (τ) . \end{matrix}

We use integration by parts and we get

\begin{matrix} {⟨ h, P_{y} ⟩}_{V_{2}^{11} [0, 1]} & = & h^{^{'}} (0) P_{y}^{^{'}} (0) + h^{^{‴}} (0) P_{y}^{^{‴}} (0) \\ + & h^{(5)} (0) P_{y}^{(5)} (0) + h^{(7)} (0) P_{y}^{(7)} (0) \\ + & h^{(9)} (0) P_{y}^{(9)} (0) + h^{(10)} (0) P_{y}^{(10)} (0) \\ + & h^{(10)} (1) P_{y}^{(11)} (1) - h^{(10)} (0) P_{y}^{(11)} (0) \\ - & h^{(9)} (1) P_{y}^{(12)} (1) + h^{(9)} (0) P_{y}^{(12)} (0) \\ + & h^{(8)} (1) P_{y}^{(13)} (1) - h^{(8)} (0) P_{y}^{(13)} (0) \\ - & h^{(7)} (1) P_{y}^{(14)} (1) + h^{(4)} (1) P_{y}^{(17)} (1) \\ + & h^{^{″}} (1) P_{y}^{(19)} (1) + h^{(7)} (0) P_{y}^{(14)} (0) \\ + & h^{(6)} (1) P_{y}^{(15)} (1) + h (1) P_{y}^{(21)} (1) \\ - & h^{(6)} (0) P_{y}^{(15)} (0) - h^{(5)} (1) P_{y}^{(16)} (1) \\ + & h^{(5)} (0) P_{y}^{(16)} (0) - h^{(4)} (0) P_{y}^{(17)} (0) \\ - & h^{^{‴}} (1) P_{y}^{(18)} (1) + h^{^{‴}} (0) P_{y}^{(18)} (0) \\ - & h^{^{″}} (0) P_{y}^{(19)} (0) - h^{^{'}} (1) P_{y}^{(20)} (1) \\ + & h^{^{'}} (0) P_{y}^{(20)} (0) - h (0) P_{y}^{(21)} (0) \\ - & \int_{0}^{1} h (τ) P_{y}^{(22)} (τ) d (τ) . \end{matrix}

Therefore, we achieve

\begin{matrix} {⟨ h, P_{y} ⟩}_{V_{2}^{11} [0, 1]} & = & h^{^{'}} (0) P_{y}^{^{'}} (0) + h^{^{‴}} (0) P_{y}^{^{‴}} (0) \\ + & h^{(5)} (0) P_{y}^{(5)} (0) + h^{(7)} (0) P_{y}^{(7)} (0) \\ + & h^{(9)} (0) P_{y}^{(9)} (0) + h^{(10)} (0) P_{y}^{(10)} (0) \\ + & h^{(10)} (1) P_{y}^{(11)} (1) - h^{(10)} (0) P_{y}^{(11)} (0) \\ - & h^{(9)} (1) P_{y}^{(12)} (1) + h^{(9)} (0) P_{y}^{(12)} (0) \\ - & h^{(7)} (1) P_{y}^{(14)} (1) + h^{(7)} (0) P_{y}^{(14)} (0) \\ - & h^{(5)} (1) P_{y}^{(16)} (1) + h^{(5)} (0) P_{y}^{(16)} (0) \\ - & h^{^{‴}} (1) P_{y}^{(18)} (1) + h^{^{‴}} (0) P_{y}^{(18)} (0) \\ - & h^{^{'}} (1) P_{y}^{(20)} (1) + h^{^{'}} (0) P_{y}^{(20)} (0) \\ - & \int_{0}^{1} h (τ) P_{y}^{(22)} (τ) d (τ) . \end{matrix}

We choose

\begin{matrix} P_{y}^{^{'}} (0) + P_{y}^{(20)} (0) = 0, P_{y}^{‴} (0) + P_{y}^{(18)} (0) = 0, \\ P_{y}^{(5)} (0) + P_{y}^{(16)} (0) = 0, P_{y}^{(7)} (0) + P_{y}^{(14)} (0) = 0, \\ P_{y}^{(9)} (0) + P_{y}^{(12)} (0) = 0, P_{y}^{(10)} (0) - P_{y}^{(11)} (0) = 0, \\ P_{y}^{(11)} (1) = 0, P_{y}^{(12)} (1) = 0, P_{y}^{(14)} (1) = 0, \\ P_{y}^{(16)} (1) = 0, P_{y}^{(18)} (1) = 0, P_{y}^{(20)} (1) = 0 . \end{matrix}

Thus, we acquire

\begin{matrix} {⟨ h, P_{y} ⟩}_{V_{2}^{11} [0, 1]} = - \int_{0}^{1} h (τ) P_{y}^{(22)} (τ) d (τ) = h (y) . \end{matrix}

The reproducing kernel function

P_{y}

is obtained as

P_{y} (τ) = \{\begin{matrix} \sum_{i = 1}^{22} c_{i} τ^{i - 1}, τ \leq y, \\ \sum_{i = 1}^{22} d_{i} τ^{i - 1}, τ > y, \end{matrix}

by

\begin{matrix} P_{y}^{(22)} (τ) = - δ (τ - y) . \end{matrix}

Definition 1.

V_{2}^{1} [0, 1]

is given as:

V_{2}^{1} [0, 1] = {m \in A C [0, 1] : m^{'} \in L^{2} [0, 1]},

where

A C

shows the space of absolutely continuous functions.

{⟨m, n⟩}_{V_{2}^{1}} = \int_{0}^{1} (m (τ) n (τ) + m^{'} (τ) n^{'} (τ)) d τ, m, n \in V_{2}^{1} [0, 1]

(1)

and

{∥m∥}_{V_{2}^{1}} = \sqrt{{⟨m, m⟩}_{V_{2}^{1}}}, m \in V_{2}^{1} [0, 1],

(2)

are the inner product and the norm in

V_{2}^{1} [0, 1] .

T_{τ} (ς)

of

V_{2}^{1} [0, 1]

is obtained as [26]:

T_{τ} (ς) = \frac{1}{2 sinh (1)} [cosh (τ + ς - 1) + cosh (|τ - ς| - 1)] .

(3)

3. Solutions in $V_{2}^{11} [0, 1]$

The solution of the problem is considered in the

V_{2}^{11} [0, 1] .

We define

Y : V_{2}^{11} [0, 1] \to V_{2}^{1} [0, 1]

as

\begin{matrix} Y h (τ) = h^{(x)} (τ) + a_{1} (τ) h^{(i x)} (τ) + a_{2} (τ) h^{(v i i i)} (τ) + a_{3} (τ) h^{(v i i)} (τ) \\ + a_{4} (τ) h^{(v i)} (τ) + a_{5} (τ) h^{(v)} (τ) + a_{6} (τ) h^{(i v)} (τ) + a_{7} (τ) h^{‴} (τ) \\ + a_{8} (τ) h^{″} (τ) + a_{9} (τ) h^{'} (τ) + a_{10} (τ) h (τ) . \end{matrix}

The model problem changes to the following problem:

Y h = M (τ), τ \in [0, 1],

(4)

with the following boundary conditions:

\begin{matrix} h (a) = 0, h (b) = 0, h^{″} (a) = 0, h^{″} (b) = 0, \\ h^{(i v)} (a) = 0, h^{(i v)} (b) = 0, h^{(v i)} (a) = 0, \\ h^{(v i)} (b) = 0, h^{(v i i i)} (a) = 0, h^{(v i i i)} (b) = 0 . \end{matrix}

Theorem 1.

Y is a bounded linear operator.

Proof.

We need to show

{∥Y h∥}_{V_{2}^{1}}^{2} \leq M {∥h∥}_{V_{2}^{11}}^{2}

, where

M > 0

. We acquire

{∥Y h∥}_{V_{2}^{1}}^{2} = {⟨Y h, Y h⟩}_{V_{2}^{1}} = \int_{0}^{1} (Y h {(τ)}^{2} + Y h^{'} {(τ)}^{2}) d τ

by Equations (1) and (2). We obtain

h (τ) = {⟨h (\cdot), P_{τ} (\cdot)⟩}_{V_{2}^{11}}

and

Y h (τ) = {⟨h (\cdot), Y P_{τ} (\cdot)⟩}_{V_{2}^{11}},

by reproducing property. Thus, we reach

|Y h (τ)| \leq {∥h∥}_{V_{2}^{11}} {∥Y P_{τ}∥}_{V_{2}^{11}} = M_{1} {∥h∥}_{V_{2}^{11}},

where

M_{1} > 0 .

Therefore, we find

\int_{0}^{1} {[(Y h) (τ)]}^{2} d τ \leq M_{1}^{2} {∥h∥}_{V_{2}^{11}}^{2} .

Since

{(Y h)}^{'} (τ) = {⟨h (\cdot), {(Y P_{η})}^{'} (\cdot)⟩}_{V_{2}^{11}},

then

|{(Y h)}^{'} (η)| \leq {∥h∥}_{V_{2}^{11}} {∥{(Y P_{τ})}^{'}∥}_{V_{2}^{11}} = M_{2} {∥f∥}_{V_{2}^{11}},

where

M_{2} > 0 .

Therefore, we reach

{[{(Y h)}^{'} (τ)]}^{2} \leq M_{2}^{2} {∥h∥}_{V_{2}^{11}}^{2}

and

\int_{0}^{1} {[{(Y h)}^{'} (τ)]}^{2} d τ \leq M_{2}^{2} {∥h∥}_{V_{2}^{11}}^{2} .

Finally, we obtain

\begin{matrix} {∥Y h∥}_{V_{2}^{1}}^{2} & \leq & \int_{0}^{1} ({[(Y h) (τ)]}^{2} + {[{(Y h)}^{'} (τ)]}^{2}) d τ \\ \leq & (M_{1}^{2} + M_{2}^{2}) {∥h∥}_{V_{2}^{11}}^{2} = M {∥h∥}_{V_{2}^{11}}^{2}, \end{matrix}

where

M = M_{1}^{2} + M_{2}^{2} > 0 .

☐

The Main Results

Let

φ_{i} (τ) = T_{τ_{i}} (τ)

and

ψ_{i} (τ) = Y^{*} φ_{i} (τ)

,

Y^{*}

is an adjoint operator of Y. The orthonormal system

{\{{\bar{Ψ}}_{i} (τ)\}}_{i = 1}^{\infty}

of

V_{2}^{11} [0, 1]

can be obtained as:

{\bar{ψ}}_{i} (τ) = \sum_{k = 1}^{i} β_{i k} ψ_{k} (τ), (β_{i i} > 0, i = 1, 2, \dots) .

(5)

Theorem 2.

Let

{\{τ_{i}\}}_{i = 1}^{\infty}

be dense in

[0, 1]

and

ψ_{i} (τ) = {Y_{ς} T_{τ} (ς)|}_{ς = τ_{i}}

. Then, the sequence

{\{ψ_{i} (τ)\}}_{i = 1}^{\infty}

is a complete system in

V_{2}^{11} [0, 1]

.

Proof.

We get

\begin{matrix} ψ_{i} (τ) = (Y^{*} φ_{i}) (τ) = ⟨(Y^{*} φ_{i}) (ς), T_{τ} (ς)⟩ \\ = ⟨φ_{i}) (ς), Y ς T_{τ} (ς)⟩ = {Y_{ς} T_{τ} (ς)|}_{ς = τ_{i}} . \end{matrix}

We know that

ψ_{i} (τ) \in V_{2}^{11} [0, 1]

. For each fixed

h (τ) \in V_{2}^{11} [0, 1]

, let

⟨h (τ), ψ_{i} (η)⟩ = 0, (i = 1, 2, \dots),

⟨h (τ), (Y^{*} φ_{i}) (τ)⟩ = ⟨Y h (\cdot), φ_{i} (\cdot)⟩ = (Y h) (τ_{i}) = 0 .

{\{τ_{i}\}}_{i = 1}^{\infty}

is dense in

[0, 1]

. Thus,

(Y h) (τ) = 0

.

h \equiv 0

by the

Y^{- 1}

. ☐

Theorem 3.

If

h (τ)

is the exact solution of Equation (4), then

h = \sum_{i = 1}^{\infty} \sum_{k = 1}^{i} β_{i k} M (τ_{k}) {\bar{ψ}}_{i} (τ),

(6)

where

{(τ_{i})}_{i = 1}^{\infty}

is dense in

[0, 1]

.

Proof.

We obtain

\begin{matrix} h (τ) & = & \sum_{i = 1}^{\infty} {⟨h (τ), {\bar{ψ}}_{i} (τ)⟩}_{V_{2}^{11}} {\bar{ψ}}_{i} (τ) \\ = & \sum_{i = 1}^{\infty} \sum_{k = 1}^{i} β_{i k} {⟨h (τ), Ψ_{k} (τ)⟩}_{V_{2}^{11}} {\bar{ψ}}_{i} (τ) \\ = & \sum_{i = 1}^{\infty} \sum_{k = 1}^{i} β_{i k} {⟨h (τ), Y^{*} φ_{k} (τ)⟩}_{V_{2}^{11}} {\bar{ψ}}_{i} (τ) \\ = & \sum_{i = 1}^{\infty} \sum_{k = 1}^{i} β_{i k} {⟨Y h (τ), φ_{k} (η)⟩}_{V_{2}^{1}} {\bar{ψ}}_{i} (τ) \\ = & \sum_{i = 1}^{\infty} \sum_{k = 1}^{i} β_{i k} Y h (τ_{k}) {\bar{ψ}}_{i} (τ) \\ = & \sum_{i = 1}^{\infty} \sum_{k = 1}^{i} β_{i k} M (τ_{k}) {\bar{ψ}}_{i} (η), \end{matrix}

by Equation (5) and the uniqueness of solution of Equation (4). ☐

The approximate solution

h_{n}

can be obtained by:

h_{n} = \sum_{i = 1}^{n} \sum_{k = 1}^{i} β_{i k} M (τ_{k}) {\bar{ψ}}_{i} (τ) .

(7)

4. Numerical Results

Example 1.

We consider the following tenth-order equation as:

\begin{matrix} q^{(10)} (τ) = - (80 + 19 τ + τ^{2}) exp (τ), 0 \leq τ \leq 1, \end{matrix}

with boundary conditions:

\begin{matrix} q (0) = 0, & q (1) = 0, \\ q^{″} (0) = 0, & q^{″} (1) = - 4 exp (1), \\ q^{(i v)} (0) = - 8, & q^{(i v)} (1) = - 16 exp (1), \\ q^{(v i)} (0) = - 24, & q^{(v i)} (1) = - 36 exp (1), \\ q^{(v i i i)} (0) = - 48, & q^{(v i i i)} (1) = - 64 exp (1) . \end{matrix}

The exact solution to the above boundary value problem is given by [1]:

\begin{matrix} q (τ) = τ (1 - τ) exp (τ) . \end{matrix}

We need to homogenize the boundary conditions to apply the reproducing kernel Hilbert space method:

\begin{matrix} r_{1} (x) = c_{1} + c_{2} x + c_{3} x^{2} + c_{4} x^{3} + c_{5} x^{4} \\ + c_{6} x^{5} + c_{7} x^{6} + c_{8} x^{7} + c_{9} x^{8} + c_{10} x^{9} \\ s_{1} = c_{1} = 0, \\ s_{2} = 2 c_{3} = 0, \\ s_{3} = 24 c_{5} = - 8, \\ s_{4} = 720 c_{7} = - 24, \\ s_{5} = 40, 320 c_{9} = - 48, \\ s_{6} = c_{1} + c_{2} + c_{3} + c_{4} + c_{5} + c_{6} + c_{7} + c_{8} + c_{9} + c_{10} = 0, \\ s_{7} = 72 c_{10} + 56 c_{9} + 42 c_{8} + 30 c_{7} + 20 c_{6} + 12 c_{5} + 6 c_{4} + 2 c_{3} = - 4 e, \\ s_{8} = 3024 c_{10} + 1680 c_{9} + 840 c_{8} + 360 c_{7} + 120 c_{6} + 24 c_{5} = - 16 e, \\ s_{9} = 60, 480 c_{10} + 20, 160 c_{9} + 5040 c_{8} + 720 c_{7} = - 36 e, \\ s_{10} = 362, 880 c_{10} + 40, 320 c_{9} = - 64 e, \end{matrix}

\begin{matrix} \{\begin{matrix} c_{1} = 0, c_{2} = - \frac{24}{175} + \frac{1123}{2700} e, c_{3} = 0, \\ c_{4} = \frac{352}{945} - \frac{719}{2268}, c_{5} = - \frac{1}{3}, c_{6} = \frac{28}{225} - \frac{253}{2700} e, \\ c_{7} = - \frac{1}{30}, c_{8} = \frac{1}{126} - \frac{19}{3780} e, \\ c_{9} = - \frac{1}{840}, c_{10} = \frac{1}{7560} - \frac{1}{5670} e . \end{matrix}\} \end{matrix}

Therefore, we obtain

\begin{matrix} r_{1} (x) = - \frac{1}{113, 400} x (20 x^{8} e - 15 x^{8}, \\ + 570 x^{6} e + 135 x^{7} - 900 x^{6} + 10, 626 x^{4} e, \\ + 3780 x^{5} - 14, 112 x^{4} + 35, 950 x^{2} e, \\ + 37, 800 x^{3} - 42, 240 x^{2} - 47, 166 e + 15, 552) . \end{matrix}

We obtain Table 1 by our accurate technique.

Table 1. Exact solution (ES) and absolute errors (AE) of Example 1.

5. Conclusions

In this paper, we used an accurate technique for investigating tenth order boundary value problems. An example was chosen to prove the computational accuracy. As shown in Table 1, our method is very accurate. We acquired some significant reproducing kernel functions to get approximate solutions and absolute errors.

Author Contributions

These authors contributed equally to this work.

Funding

This research was supported by 2017-SİÜFED-39, 2017-SİÜFEB-40 and 2018-SİÜFEB-012.

Conflicts of Interest

The authors declare no conflict of interest.

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Table 1. Exact solution (ES) and absolute errors (AE) of Example 1.

$τ$	ES	AE (NPCSM) [1]	AE (PCSM) [1]	AE (RKHSM)
$0.2$	$0.195424441$	$2.433 \times 10^{- 7}$	$3.982 \times 10^{- 4}$	$3.330 \times 10^{- 8}$
$0.4$	$0.358037927$	$3.986 \times 10^{- 7}$	$6.663 \times 10^{- 4}$	$7.031 \times 10^{- 8}$
$0.6$	$0.358037927$	$4.428 \times 10^{- 7}$	$7.598 \times 10^{- 4}$	$6.076 \times 10^{- 8}$
$0.8$	$0.356086549$	$3.328 \times 10^{- 7}$	$5.885 \times 10^{- 4}$	$2.682 \times 10^{- 8}$

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New Numerical Method for Solving Tenth Order Boundary Value Problems

Abstract

1. Introduction

2. Reproducing Kernel Spaces and Their Reproducing Kernel Functions

3. Solutions in $V_{2}^{11} [0, 1]$

The Main Results

4. Numerical Results

5. Conclusions

Author Contributions

Funding

Conflicts of Interest

References

Article Metrics

Citations

Article Access Statistics

New Numerical Method for Solving Tenth Order Boundary Value Problems

Abstract

1. Introduction

2. Reproducing Kernel Spaces and Their Reproducing Kernel Functions

3. Solutions in V 2 11 [ 0 , 1 ]

The Main Results

4. Numerical Results

5. Conclusions

Author Contributions

Funding

Conflicts of Interest

References

Article Metrics

Citations

Article Access Statistics

3. Solutions in $V_{2}^{11} [0, 1]$