Solving a Class of Singular Fifth-Order Boundary Value Problems Using Reproducing Kernel Hilbert Space Method

Abstract

We use the reproducing kernel Hilbert space method to solve the fifth-order boundary value problems. The exact solution to the fifth-order boundary value problems is obtained in reproducing kernel space. The approximate solution is given by using an iterative method and the finite section method. The present method reveals to be more effective and convenient compared with the other methods.

1. Introduction

The reproducing kernel Hilbert space method has been shown [1–7] to solve effectively, easily, and accurately a large class of linear and nonlinear, ordinary, partial differential equations. However, in [1–7], it cannot be used directly boundary value problems with mixed boundary conditions, since it is very difficult to obtain a reproducing kernel function satisfying mixed nonlinear boundary conditions. The aim of this work is to fill this gap. In [8], we give a new reproducing kernel Hilbert space for solving singular linear fourth-order boundary value problems with mixed boundary conditions. In this paper, we use the new reproducing kernel Hilbert function space method to solve the nonlinear fifth-order boundary value problems.

Singular fifth-order boundary value problems arise in the fields of gas dynamics, Newtonian fluid mechanics, fluid mechanics, fluid dynamics, elasticity, reaction-diffusion processes, chemical kinetics, and other branches of applied mathematics.

Let us consider the following class of singular fifth-order mixed boundary value problems: 𝑢⁽⁵⁾(𝑥)+𝑝₁(𝑥)𝑥^𝛼1(1−𝑥)^𝛽1𝑢⁽⁴⁾(𝑥)𝑝+⋯+⁴(𝑥)𝑥^𝛼4(1−𝑥)^𝛽4𝑢^′(𝑥)+𝑝₅(𝑥)𝑥^𝛼5(1−𝑥)^𝛽5𝑢(𝑥)(𝑥)(),𝐵=𝑓,𝑥∈0,1_𝑖𝑢=𝑟_𝑖,(𝑖=1,2,…,5), (1) where 𝑝_𝑗(𝑥),𝑓(𝑥)∈𝐿²[0,1] (𝑗=1,…,5) are known functions. 𝐵_𝑖𝑢 (𝑖=1,2,…,5) are linear conditions. We assume that (1) has a unique solution which belongs to 𝑊⁶₂[0,1], where 𝑊⁶₂[0,1] is a reproducing kernel space.

Remark 1. If 𝐵_𝑖𝑢=𝑢^(𝑖)(0) (𝑖=1,…,𝑚), then (1) is an initial value problem. If 𝐵_𝑖𝑢=𝑢(𝑥_𝑖) (𝑖=1,2,…,𝑚), then (1) is a multipoint problem and so on. That is, problem (1) has a rather general form.
Let 𝛼=max_1≤𝑖≤5{𝛼_𝑖} and 𝛽=max_1≤𝑖≤5{𝛽_𝑖}, 𝐹(𝑥)=𝑥^𝛼(1−𝑥)^𝛽𝑓(𝑥).
Consider (𝐿𝑢)(𝑥)=𝑥^𝛼(1−𝑥)^𝛽𝑢⁽⁵⁾(𝑥)+𝑥^𝛼−𝛼1(1−𝑥)^𝛽−𝛽1𝑝₁(𝑥)𝑢⁽⁴⁾(𝑥)+⋯+𝑥^𝛼−𝛼4(1−𝑥)^𝛽−𝛽4𝑝₄(𝑥)𝑢^′(𝑥)+𝑥^𝛼−𝛼5(1−𝑥)^𝛽−𝛽5𝑝₅(𝑥)𝑢(𝑥). (2) It is easy to prove that 𝐿∶𝑊⁶₂[0,1]→𝐿²[0,1] is a bounded linear operator. On the other hand, we suppose that the linear conditions can also always be homogenized; after homogenization of these conditions, we put these conditions into the reproducing kernel space 𝑊⁶₂[0,1] constructed in the following section. Equation (1) can be transformed into the following form in 𝑊⁶₂[0,1]: (𝐿𝑢)(𝑥)=𝐹(𝑥). (3)

2. Reproducing Kernel Hilbert Space

Definition 2. Let 𝐻 be a real Hilbert space of functions 𝑓∶Ω→𝑅. A function 𝐾∶Ω×Ω→𝑅 is called reproducing kernel for 𝐻 if(i)𝐾(𝑥,⋅)∈𝐻 for all 𝑥∈Ω, (ii)𝑓(𝑥)=⟨𝑓,𝐾(⋅,𝑥)⟩_𝐻 for all 𝑓∈𝐻 and all 𝑥∈Ω.

Definition 3. A real Hilbert space 𝐻 of functions on a set Ω is called a reproducing kernel Hilbert space if there exists a reproducing kernel 𝐾 of 𝐻.

One defines that the inner product space 𝑊₂^𝑚+1[0,1]={𝑢∣𝑢,𝑢^′,…,𝑢^(𝑚) are absolutely continuous  function, 𝑢^(𝑚+1)∈𝐿²[0,1]}.

The inner product in 𝑊₂^𝑚+1[0,1] is given by ⟨𝑢(𝑥)(𝑥=,𝑣)⟩_𝑚𝑖=0𝑢^(𝑖)(0)𝑣^(𝑖)(0)+¹₀𝑢^(𝑚+1)(𝑡)𝑣^(𝑚+1)(𝑡)𝑑𝑡. (4)

Theorem 4 (see [8]). The space 𝑊₂^𝑚+1[0,1] is a reproducing kernel space, and its reproducing kernel is 𝑅^{𝑚+1}𝑥,𝑦=^𝑚_𝑖∑₌₀1(𝑖!)²𝑥^𝑖𝑦^𝑖+1(𝑚!)²∫^𝑥₀(𝑥−𝑡)^𝑚𝑦−𝑡^𝑚𝑑𝑡,𝑥<𝑦,_𝑚𝑖∑₌₀1(𝑖!)²𝑥^𝑖𝑦^𝑖+1(𝑚!)²∫^𝑦₀(𝑥−𝑡)^𝑚𝑦−𝑡^𝑚𝑑𝑡,𝑦<𝑥. (5)

For studying the solution of (1) in the homogenized form, we give space (6) as follows: 𝑊⁶₂[0,1]=𝑢∣𝑢∈𝑊⁶₂[0,1],𝐵_𝑖𝑢=0,𝑖=1,2,…,5. (6)𝑊⁶₂[0,1] is equipped with the same inner product 𝑊⁶₂[0,1]. In the following, we construct a reproducing kernel for the space 𝑊⁶₂[0,1], and we give Lemmas 5 and 6.

Lemma 5. Let 𝐴∶𝐻[𝑎,𝑏]→𝐿²[𝑎,𝑏] be a bounded linear operator; function 𝑅_𝑥(𝑦) is the reproducing kernel of space 𝐻[𝑎,𝑏]. Let 𝑔_𝑧(𝑥)=(𝐴_𝑠𝑅_𝑥(𝑠))(𝑧); then ‖𝑔_𝑧(𝑥)‖²=(𝐴_𝑠(𝐴_𝑡𝑅_𝑠(𝑡))(𝑧))(𝑧), where 𝐻[𝑎,𝑏] denotes any reproducing kernel space of functions over [𝑎,𝑏], the symbol 𝐴_𝑠 indicates that the operator 𝐴 applies to functions of the variable 𝑠, and the symbol (𝐴_𝑠𝑅_𝑥(𝑠))(𝑧) indicates that the operator 𝐴 applies to function 𝑅_𝑥(𝑠) of the variable 𝑠 and 𝑠=𝑧.

Proof. Consider 𝑔_𝑧(𝑥)²=𝐴_𝑠𝑅_𝑥(𝑠)(𝑧),𝐴_𝑡𝑅_𝑥(𝑡)(𝑧=𝐴_𝑠)𝐴_𝑡𝑅_𝑥(𝑠),𝑅_𝑥(𝑡𝑧𝑧)=𝐴)()(_𝑠𝐴_𝑡𝑅_𝑠(𝑡)(𝑧)(𝑧). (7)

Lemma 6. If 𝐴, 𝑔_𝑧(𝑥), and 𝑅_𝑥(𝑦) are defined as in Lemma 5, let 𝐾_𝑥(𝑦)=𝑅_𝑥(𝑦)−𝑔_𝑧(𝑥)𝑔_𝑧(𝑦)/‖𝑔_𝑧(𝑥)‖²; consider the space 𝐻₁={𝑢(𝑦)∣𝑢(𝑦)∈𝐻[𝑎,𝑏],𝑎𝑛𝑑(𝐴_𝑦𝑢(𝑦))(𝑧)=0}, then, 𝐾_𝑥(𝑦) is the reproducing kernel of space 𝐻₁.

Proof. For any 𝑢(𝑦)∈𝐻₁, next, we will prove ⟨𝑢(𝑦),𝐾_𝑥(𝑦)⟩=𝑢(𝑥).
Consider 𝑢𝑦,𝐾_𝑥𝑦=𝑢𝑦,𝑅_𝑥𝑦−𝑔_𝑧(𝑥)𝑔_𝑧𝑦𝑔_𝑧(𝑥)²=𝑢𝑦,𝑅_𝑥𝑦−𝑢𝑦,𝑔_𝑧(𝑥)𝑔_𝑧𝑦‖𝑔_𝑧(𝑥)‖²=𝑢(𝑥)−𝑔_𝑧(𝑥)𝑢𝑦,𝐴_𝑠𝑅_𝑦(𝑠)(𝑧)𝑔_𝑧(𝑥)²=𝑢(𝑥)−𝑔_𝑧(𝑥)𝐴_𝑠𝑢𝑦,𝑅_𝑦(𝑠)(𝑧)𝑔_𝑧(𝑥)²(𝑥)−𝑔=𝑢^𝑧(𝑥)𝐴_𝑠𝑢(𝑠)(𝑧)𝑔_𝑧(𝑥)²=𝑢(𝑥). (8)

Let ℎ₁(𝑥)=𝐵_1𝑦𝑅^{6}(𝑥,𝑦), ℎ₂(𝑥)=𝐵_2𝑦(𝑅^{6}(𝑥,𝑦)−ℎ₁(𝑥)ℎ₁(𝑦)/‖ℎ₁(𝑥)‖²), ℎ₃(𝑥)=𝐵_3𝑦(𝑅^{6}(𝑥,𝑦)−ℎ₁(𝑥)ℎ₁(𝑦)/‖ℎ₁(𝑥)‖²−ℎ₂(𝑥)ℎ₂(𝑦)/‖ℎ₂(𝑥)‖²), ℎ₄(𝑥)=𝐵_4𝑦(𝑅^{6}(𝑥,𝑦)−ℎ₁(𝑥)ℎ₁(𝑦)/‖ℎ₁(𝑥)‖²−ℎ₂(𝑥)ℎ₂(𝑦)/‖ℎ₂(𝑥)‖²−ℎ₃(𝑥)ℎ₃(𝑦)/‖ℎ₃(𝑥)‖²), and ℎ₅(𝑥)=𝐵_5𝑦(𝑅^{6}(𝑥,𝑦)−ℎ₁(𝑥)ℎ₁(𝑦)/‖ℎ₁(𝑥)‖²−ℎ₂(𝑥)ℎ₂(𝑦)/‖ℎ₂(𝑥)‖²−ℎ₃(𝑥)ℎ₃(𝑦)/‖ℎ₃(𝑥)‖²−ℎ₄(𝑥)ℎ₄(𝑦)/‖ℎ₄(𝑥)‖²), where the symbol 𝐵_𝑖𝑦 (𝑖=1,2,3,4,5) indicates that the operator 𝐵_𝑖 (𝑖=1,2,3,4,5) applies to functions of the variable 𝑦. Using Lemma 6, we get Theorem 7.

Theorem 7. The space 𝑊⁶₂[0,1] is a reproducing kernel space, and its reproducing kernel is 𝐾𝑥,𝑦=𝑅^{6}−ℎ𝑥,𝑦¹(𝑥)ℎ₁𝑦ℎ₁(𝑥)²−ℎ₂(𝑥)ℎ₂𝑦ℎ₂(𝑥)²−ℎ₃(𝑥)ℎ₃𝑦ℎ₃(𝑥)²−ℎ₄(𝑥)ℎ₄𝑦ℎ₄(𝑥)²−ℎ₅(𝑥)ℎ₅𝑦ℎ₅(𝑥)². (9)

3. Analytical Solution

Let 𝜓_𝑖(𝑥)=(𝐿_𝑦𝐾(𝑥,𝑦))(𝑥_𝑖), 𝑖=1,2,…. Via Gram-Schmidt orthonormalization for {𝜓_𝑖(𝑥)}^∞_𝑖=1, we get 𝜓_𝑖(𝑥)=^𝑖_𝑘=1𝛽_𝑖𝑘𝜓_𝑘(𝑥), (10) where the 𝛽_𝑖𝑘 are the coefficients resulting from Gram-Schmid orthonormalization.

Lemma 8. If {𝑥_𝑖}^∞_𝑖=1 are distinct points dense in [0,1] and 𝐿⁻¹ is existent, then {𝜓_𝑖(𝑥)}^∞_𝑖=1 is a complete function system in 𝑊^m₂⁺¹[0,1].

Proof. For each fixed 𝑢(𝑥)∈𝑊^m₂⁺¹[0,1], if ⟨𝑢(𝑥),𝜓_𝑖(𝑥)⟩=0, then 𝑢(𝑥),𝜓_𝑖(𝑥=𝐿_𝑦)𝑢(𝑥),𝐾𝑥,𝑦𝑥_𝑖=𝐿_𝑦𝑢𝑦𝑥_𝑖=0. (11) Taking into account the density of {𝑥_𝑖}^∞_𝑖=1, it results in (𝐿_𝑦𝑢(𝑦))(𝑥)=0. It follows that 𝑢(𝑥)≡0 from the existence of 𝐿⁻¹.

Theorem 9. If {𝑥_𝑖}^∞_𝑖=1 are distinct points dense in [0,1] and 𝐿⁻¹ is existent, then 𝑢(𝑥)=^∞𝑖_𝑖=1𝑘=1𝛽_𝑖𝑘𝐹𝑥_𝑘𝜓_𝑖(𝑥) (12) is an analytical solution of (3).

Proof. 𝑢(𝑥) can be expanded to Fourier series in terms of normal orthogonal basis {𝜓_𝑖(𝑥)}^∞_𝑖=1 in 𝑊₂^𝑚+1[0,1] as follows: 𝑢(𝑥)=^∞_𝑖=1𝑢(𝑥),𝜓_𝑖(𝑥)𝜓_𝑖(𝑥)=^∞𝑖_𝑖=1𝑘=1𝛽_𝑖𝑘𝑢(𝑥),𝜓_𝑘(𝑥)𝜓_𝑖(𝑥)=^∞𝑖_𝑖=1𝑘=1𝛽_𝑖𝑘𝑢(𝑥),𝐿_𝑠𝐾_𝑥(𝑠)𝑥_𝑘𝜓_𝑖(𝑥)=^∞𝑖_𝑖=1𝑘=1𝛽_𝑖𝑘𝐿_𝑠𝑢(𝑥),𝐾_𝑥(𝑠)𝑥_𝑘𝜓_𝑖(𝑥)=^∞𝑖_𝑖=1𝑘=1𝛽_𝑖𝑘𝐿_𝑠𝑢(𝑠)𝑥_𝑘𝜓_𝑖(𝑥)=^∞𝑖_𝑖=1𝑘=1𝛽_𝑖𝑘𝐹𝑥_𝑘𝜓_𝑖(𝑥). (13)

4. Numerical Solution

We define an approximate solution 𝑢_𝑛(𝑥) by 𝑢_𝑛(𝑥)=^𝑛𝑖_𝑖=1𝑘=1𝛽_𝑖𝑘𝐹𝑥_𝑘𝜓_𝑖(𝑥). (14)

Theorem 10. Let 𝜀²_𝑛=‖𝑢(𝑥)−𝑢_𝑛(𝑥)‖², where 𝑢(𝑥) and 𝑢_𝑛(𝑥) are given by (12) and (14); then the sequence of real numbers 𝜀_𝑛(𝑥) is monotonously decreasing and 𝜀_𝑛(𝑥)→0.

Proof. We have 𝜀²_𝑛=𝑢(𝑥)−𝑢_𝑛(𝑥)²=^∞_𝑖=𝑛+1𝑢(𝑥),𝜓_𝑖(𝑥)𝜓_𝑖(𝑥)²=^∞_𝑖=𝑛+1𝑢(𝑥),𝜓_𝑖(𝑥)². (15) Clearly, 𝜀_𝑛−1≥𝜀_𝑛 and consequently {𝜀_𝑛} is monotonously decreasing in the sense of ‖⋅‖. By Theorem 9, we know that ∑^∞_𝑖=1⟨𝑢(𝑥),𝜓_𝑖(𝑥)⟩𝜓_𝑖(𝑥) is convergent in the norm of ‖⋅‖; then we have 𝜀²_𝑛=𝑢(𝑥)−𝑢_𝑛(𝑥)²⟶0. (16) Hence, 𝜀_𝑛→0.

Theorem 11 (convergence analysis). 𝑢_𝑛(𝑥) and 𝑢_𝑛^(𝑘)(𝑥) are uniformly convergent to 𝑢(𝑥) and 𝑢^(𝑘)(𝑥), 𝑘=0,1,2,…,𝑚, where 𝑢(𝑥) and 𝑢_𝑛(𝑥) are given by (12) and (14).

Proof. For any 𝑥∈[0,1], 𝑘=0,1,2,…,5, 𝑢_𝑛^(𝑘)(𝑥)−𝑢^(𝑘)(𝑥)=𝑢_𝑛(𝑡)(𝑡),𝜕−𝑢^𝑘𝐾(𝑥,𝑡)𝜕𝑥^𝑘≤𝑢_𝑛(𝑡)(𝑡⋅𝜕−𝑢)^𝑘𝐾(𝑥,𝑡)𝜕𝑥^𝑘. (17) Then there exists 𝐶_𝑘>0 such that 𝑢_𝑛^(𝑘)(𝑥)−𝑢^(𝑘)(𝑥)≤𝐶_𝑘𝑢_𝑛(𝑡)−𝑢(𝑡)=𝐶_𝑘𝜀_𝑛⟶0. (18)

The numerical solution to (3) can be obtained using the following method: 𝑢_𝑛(𝑥)=^𝑛_𝑖=1𝑑_𝑖𝜓_𝑖(𝑥), (19) where the coefficients 𝑑_𝑖, 𝑖=1,…,𝑚, are determined by the equation ^𝑛_𝑖=1𝑑_𝑖𝐿𝜓_𝑖(𝑥)|_{𝑥=𝑥𝑗}=𝐹𝑥_𝑗,𝑗=1,2,…,𝑛. (20) Using (19) and (20), we have (𝐿𝑢_𝑚)(𝑥_𝑗)=𝐹(𝑥_𝑗), 𝑗=1,2,…,𝑛. So, 𝑢_𝑛(𝑥) is the approximation solution of (3).

5. Numerical Experiment

In this section, two numerical examples are studied to demonstrate the accuracy of the present method.

Example 12. Consider the following fifth-order boundary value problem with nonclassical side condition (the right-hand side of this problem has a singularity at 𝑥=0, 𝑥=1): 𝑢⁽⁵⁾(𝑥)−𝑒^𝑥𝑥𝑢1−𝑥^′′(𝑥)+𝑒^𝑥𝑥𝑢(𝑥)(𝑥),𝑢(0)1−𝑥=𝑓0<𝑥<1,=𝑢^′(0)1=𝑢4=0,5𝑢^′′(0)+42¹₀𝑒^−𝑥𝑢(𝑥)𝑑𝑥=0,4𝑢^′′′(1)+𝑢^′(1)=10𝑢^′′(1), (21) where 𝑓(𝑥)=𝑒^𝑥(−45+195𝑥−750𝑥²+320𝑥^5/2−600𝑥³−32(5+𝑒^𝑥)𝑥^7/2+40(20+3𝑒^𝑥)𝑥⁴−16(9+4𝑒^𝑥)𝑥^9/2+16(23+10𝑒^𝑥)𝑥⁵−16𝑥^11/2+32𝑥⁶)/32(−1+𝑥)𝑥^5/2. The exact solution is 𝑢_𝑇(𝑥)=𝑥²(√𝑥−1/2)𝑒^𝑥. The numerical results are presented in Table 1.

Table 1 

The numerical results of Example 12.

Example 13 (see [9]). Consider the following fifth-order boundary value problem (the right-hand side of this problem has a singularity at 𝑥=0): 𝑢⁽⁵⁾(𝑥)−𝑒^−𝑥𝑢(𝑥)=−𝑒^−𝑥𝑥92()+()1−𝑥9451−11𝑥32√𝑥𝑢(0),0<𝑥<1,=𝑢^′(0)=𝑢^′′(0)=𝑢(1)=0,𝑢^′(1)=−1, (22) where the exact solution is 𝑢_𝑇(𝑥)=𝑥^9/2(1−𝑥). By the homogeneous process of the boundary condition, letting 𝑣(𝑥)=𝑢(𝑥)−𝑥³(1−𝑥), the problem can be transformed into the equivalent form 𝑣⁽⁵⁾(𝑥)−𝑒^−𝑥𝑣(𝑥)=−𝑒^−𝑥𝑥92−𝑥³()+()1−𝑥9451−11𝑥32√𝑥𝑣(0),0≤𝑥≤1,=𝑣^′(0)=𝑣^′′(0)=𝑣(1)=𝑣^′(1)=0. (23) The numerical results are presented in Tables 2, 3, and 4.

Table 2 

Comparison of the absolute error of Example 13.

Table 3 

The numerical results of Example 13.

Table 4 

The numerical results of Example 13.

6. Conclusions and Remarks

In this paper, a new reproducing kernel space satisfying mixed boundary value conditions is constructed skillfully. This makes it easy to solve such kind of problems. Furthermore, the exact solution of the problem can be expressed in series form. The numerical results demonstrate that the new method is quite accurate and efficient for singular problems of fifth-order ordinary differential equations. All computations have been performed using the Mathematica 7.0 software package.

Acknowledgments

The authors would like to express their thanks to the unknown referees for their careful reading and helpful comments. This paper is supported by the Natural Science Foundation of Inner Mongolia (2013MS0109) and Project Application Technology Research and Development Foundation of Inner Mongolia.