Exponential spline method for singularly perturbed third-order boundary value problems

Yohannis Alemayehu Wakjira; Gemechis File Duressa

doi:10.1515/dema-2020-0024

Article Open Access

Exponential spline method for singularly perturbed third-order boundary value problems

Yohannis Alemayehu Wakjira and Gemechis File Duressa

Published/Copyright: December 31, 2020

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From the journal Demonstratio Mathematica Volume 53 Issue 1

Abstract

The exponential spline function is presented to find the numerical solution of third-order singularly perturbed boundary value problems. Convergence analysis of the method is briefly discussed, and it is shown to be sixth order convergence. To validate the applicability of the method, some model problems are solved for different values of the perturbation parameter, and the numerical results are presented both in tables and graphs. Furthermore, the present method gives more accurate solution than some methods existing in the literature.

Keywords: exponential spline; singular perturbation; boundary layers; convergence

MSC 2010: 65L10; 65L11

1 Introduction

Singularly perturbed problems arise frequently in the mathematical modelling of real-life phenomena in science and engineering areas such as fluid mechanics, elasticity, quantum mechanics, chemical-reactor theory, aerodynamics, plasma dynamics, rarefied-gas dynamics, oceanography, meteorology, modelling of semiconductor devices, geophysics, optimal control theory, diffraction theory and reaction–diffusion processes [1,2]. Solutions of singularly perturbed boundary value problems manifest multi-scale character. Due to the presence of perturbation parameter, ε, the solution varies quickly near thin transition layer and performs regularly and varies slowly away from the layer. Hence, the main concern with such problems is the swift growth or deterioration of their solutions in one or more narrow boundary layer region(s). As a result, not only determining analytical solutions to such problems is difficult but also the convergence analysis.

In recent years, a considerable number of numerical methods that deal with quartic, quintic and septic splines with polynomials and non-polynomials; combination of asymptotic expansion approximations; shooting method and finite difference methods; subdivision collocation methods and B-splines collocation methods have been developed for solving singularly perturbed boundary value problems using various splines [1,2,3,4,5,6,7,8]. Furthermore, associating quadratic spline method with other techniques was introduced for the time fractional sub-diffusion and the Helmholtz equation with the Sommerfeld boundary conditions; for details one can refer to [9,10] and references therein.

However, classical finite difference methods are not reliable to preserve the stability property as they require the introduction of very fine meshes inside the boundary layers, which requires more computational cost. Furthermore, they could not capture the solutions in the layer region of the domain as the solution profile depends on the perturbation parameter [3,11]. Thus, it is crucial to develop more accurate numerical method which works suitably for ε≪h, where most of the numerical methods fail to give smooth solution.

Hence, the purpose of the study is to develop a convergent and more accurate spline method for solving third-order singularly perturbed boundary value problem and that works for the cases where other numerical methods fail to give good results. This method depends on exponential spline function which has exponential and polynomial parts.

We consider singularly perturbed reaction–diffusion boundary value problems of the form:

(1)Ly(x)≡−εy‴(x)+u(x)y=f(x),y(0)=α1, y(1)=β1, y′(0)=γ1

(2)Ly(x)≡−εy‴(x)+u(x)y=f(x),y(0)=α1, y(1)=β1, y″(0)=γ2,

where α1, β1, γ1 and γ2 are constants; ε is the small positive parameter; and u(x) and f(x) are sufficiently smooth functions. The spline function proposed in this paper has the form: Tn=span1,x,x2,x3,ekx,e−kx, where k can be real or imaginary.

2 Formulation of the method

We consider a uniform mesh Δ with nodal points xi on interval [a, b], such that:

Δ: a=x0<x1<x2<⋯<xn−1<xn=b, xi=x0+ih, i=0,1,…,n, where h=b−an.

An exponential spline function SΔ(x) of a class C4[a,b] which interpolates y(x) at mesh points xi, i=0,1,2,…,n, depends on a parameter k. If k→0, then it reduces to ordinary quintic spline in [a,b].

For each subinterval [xi,xi+1], i=1,2,…,n−1, we consider the exponential spline, SΔ(x) of the form:

(3)SΔ(x)=aiek(x−xi)+bie−k(x−xi)+ci(x−xi)3+di(x−xi)2+ei(x−xi)+fi,

where i=1,2, …,n−1; ai, bi, ci, di, ei and fi are unknown coefficients, and k is a free parameter which will be used to raise the accuracy of the method.

Let y(x) be the exact solution of Eqs. (1) and (2) and Si be an approximation to yi=y(xi) obtained by the spline function SΔ(x) passing through the points (xi,Si) and xi+1,Si+1.

To develop the consistency relations between the value of spline and its derivatives at knots, let

(4)SΔ(xi)=yi,SΔ(xi+1)=yi+1,SΔ″(xi)=Di,SΔ″(xi+1)=Di+1,SΔ‴(xi)=Ti,SΔ‴(xi+1)=Ti+1,SΔ(4)(xi)=Fi,SΔ(4)(xi+1)=Fi+1,fori=0,1,…,n.

To define spline in terms of yi, yi+1, Di, Di+1, Ti, Ti+1, Fi and Fi+1, the coefficients introduced in Eq. (3) are calculated as:

(5)ai=Fi+1−Fieθ2k4sinh(θ),bi=Fieθ−Fi+12k4sinh(θ)ci=Di+1−Di6h −Fi+1−Fi6hk2,di=12Di+Fik2,ei=yi+1−yih−Fi−Fi+1hk4−h6(Di+1+2Di)−h6k2(Fi+1+2Fi),fi=yi−Fik4,

where θ=kh and i=0,1,2, …,n.

Using the continuity condition of the first derivatives at knots, we have:

(6)SΔ−1′(xi) = SΔ′(xi).

Then from Eq. (6), we do have:

(7)−ai−1ksinh(θ)+bi−1kcosh(θ)+3ci−1h2+2di−1h+ei−1=bik+ei.

By reducing indices of Eq. (5) by one and replacing in Eq. (7), we obtain:

(8)Fi−1sinh(θ)k3+Ficos(θ)k3sin(θ)−Ficos2(θ)k3sin(θ)+3hDi6−3hDi−16+3hFi6k2−3hFi−16k2+hDi−1+hFik2+Fi−1hk4−Fihk4−Dih6−2hDi−16−hFi6k2−2hFi−16k2= Fi+11k3sinh(θ)−h6k2−1hk4−Ficoth(θ)k3−2h6k2+1hk2−2hDi6−hDi+16+yi+1−yih.

On simplification, Eq. (8) yields:

(9)Di+1+4Di+Di+1=6h2yi+1−2yi+yi−1−6h2λ1Fi−1+2ρ1Fi+λ1Fi+1,

where λ1=3h6k2−2h3k2+1k2sinhθ, ρ1=6cothθhk3.

Again, using the continuity condition of the third derivatives at knots, we have:

(10)SΔ−1‴(xi) = SΔ‴(xi).

Then, from Eq. (10), we do have:

(11)ai−1k3sinh(θ)−bi−1k3cosh(θ)+6ci−1=−bik3+6ci.

By reducing indices of Eq. (5) by one and replacing in Eq. (11), we obtain:

(12)sinh(θ)k−Fiksinh(θ)+cos2(θ)ksinh(θ)+Dih−Di−1h+Fihk2−Fi−1hk2=−Fi+1ksinh(θ)+Ficosh(θ)ksinh(θ)+Di+1h−Dih+Fi+1hk2−Fihk2.

On simplifying Eq. (12), we obtain:

(13)Di+1−2Di+Di+1=λ2Fi−1+2ρ2Fi+λ2Fi+1,

where λ2=hk3−1k2 and ρ2=1k2−hcothθk3.

Now, subtracting Eq. (13) from Eq. (9), we obtain:

(14)Di=1h2yi+1−2yi+yi−1−h2λ1+λ26Fi+1+2ρ1+ρ26Fi+λ1+λ26Fi−1.

Using continuity of third derivative and Eq. (14), we obtain the relation:

(15)Ti=1h3yi+2−3yi+1+3yi−yi−1−hpFi+2+(p0−p+α)Fi+1+(p−p0+β)Fi−pFi−1,

where p=λ1+λ26, p0=2ρ1+ρ26, β=1θ21−θcothθ and α=1θ2θcschθ−1.

Defining the operator L by LTi≡pTi+2+Ti−2+sTi+qTi+1+Ti−1, for any function T evaluated at the mesh points [12], we have:

(16)LT≡pTi+2+Ti−2+sTi+qTi+1+Ti−1.

Using Eqs. (15) and (16), we obtain the relation:

(17)α+βh3yi+2−2yi+1+2yi−1−yi−2=pTi+2+Ti−2+sTi+qTi+1+Ti−1,

for i=2,3,…,n−2.

At nodal point xi, the proposed singularly perturbed Eq. (1) can be evaluated as:

(18)−εy‴(xi)+u(xi)y(xi)=f(xi).

From Eq. (18) we obtain:

y‴(xi)=u(xi)y(xi)−f(xi)ε,

where fi=f(xi) and ui=u(xi).

Using spline’s third derivative, we have:

(19)Ti=uiyi−fiε,Ti−1=ui−1yi−1−fi−1ε,Ti−2=ui−2yi−2−fi−2ε,Ti+1=ui+1yi+1−fi+1ε,Ti+2=ui+2yi+2−fi+2ε.

Substituting Eq. (19) into Eq. (17) and simplifying, we obtain:

(20)((α+β)ε+ui−2ph3)yi−2−((α+β)ε−qui−1h3)yi−1+suih3yi+((α+β)ε+ui+1qh3)yi+1−((α+β)ε+ui+2ph3)yi+2=−h3(p(fi+2+fi−2)+sfi+q(fi−1+fi+1))+ti,

for i=2,…, n−2 and ti is the local truncation error with:

(21)ti=4h(2α−β)yi′+13h3(6p+6q+3s−10α−4β)yi(3)+130h5(−30(4p+q)+17α+14β)yi(5)+11260h7(−105(16p+q)+65α+62β)yi7+176048h9(α+β)−8p45−q360yi9 + O(h11).

Again, by truncating terms in Eq. (20) that contains h9 and above, for arbitrary α and β provided that α+β=12, and evaluated for free parameters p, q and s, we obtain:

2p+2q+s=1,120p+30q=7.5,16080p+105q=31.5.

On solving we obtain: (α, β, p , q, s)=16,13,1240,730,2140, and the local truncation error for Eq. (20) is as follows:

ti=−16048εh9yi9+O(h11),i=2,3,…,n−2.

3 End conditions

The relation given in Eq. (20) has (n−3) linear algebraic equations in the (n−1) unknown yi, for i=1,2,…,n−1. So we need two more equations at each end. Following the procedure given in [13], the required end condition of Eq. (1) can be written as:

(22)∑l=03blyl+c1hy0′+h3∑l=03dlyl‴+t1=0,i=0,

(23)∑l=n−3nmlyl+h3∑l=n−4nklyl‴+tn−1=0,i=n−1,

where bl,c1,dl,ml and kl,l=0,1,2,3,n−4,n−3,n−2,n−1,n are arbitrary parameters which can be calculated using the method of undetermined coefficients.

Thus, the end condition of Eq. (1) can be calculated as:

(24)b0, b1, b2, b3, c1, d0, d1, d2, d3=−8009,−1600,2720,−92809,−22403,4489,22003,14723,569,

(25)mn−3, mn−2, mn−1, mn, kn−4, kn−3, kn−2, kn−1, kn=(480,−1440,1440,−480,2,−8,252,232,2).

By using Eqs. (18) and (24), we obtain the first boundary equation as:

(26)−1600ε+22003u1h3y1+2720ε+14723u2h3y2+−92809+569u3h3y3=22403εhγ1+8009ε+4489u0h3α1+h34489f0−22003f1+14723f2+569f3+t1.

Again using Eqs (18) and (25) we obtain the other end condition as:

(27)2un−4h3yn−4+480ε+8un−3h3yn−3+(−1440ε+252un−2h3)yn−2+(1440ε+232un−1h3)yn−1=−480ε+2unh3β1+h32fn−4−8fn−3+252fn−2+232fn−1+2fn+tn−1.

Similarly, besides Eq. (23), it requires additional equation to determine the end condition of Eq. (2) which can be written as:

(28)∑l=03blyl+c1hy0″+h3∑l=03dlyl‴+t1=0,i=1.

After solving coefficients of Eq. (28) using Eq. (18), we obtain end condition of Eq. (2) as:

(29)−936011ε+24811u1h3y1+1008011ε+234611u2h3y2+−360011+170411u3h3y3=72011εhγ2−288011ε−44811u0h3α1+h324811f0+234611f1+170411f2+2f3+t1.

Using values given on the Eqs. (24) and (25), we get the local truncation error for Eqs. (26) and (27) as:

ti=122105εh9yi9+O(h11),i=1,251126εh9yi9+O(h11),i=n−1.

4 Convergence analysis

The main purpose here is to drive a bound on ∥E∥∞. Considering Eqs. (20), (26) and (27), we obtain linear system of order (n−1)×(n−1) and that can be rewritten in the matrix form:

(30)AY = G+T,

where A=N+h3BU, U=diag(ui),

N=−1600ε2720ε−92809ε…−ε0ε−0.5ε…0.5ε−ε0ε−0.5ε…⋮⋮⋮⋱⋮…0.5ε−ε0ε−0.5ε……0.5ε−ε0ε……0480ε−1440ε1440ε,

B=d1d2d3…qsqp…pqsqp…⋱⋱⋱⋱⋱…pqsqp…pqsq…kn−4kn−3kn−2kn−1,

Y=(y1,y2,…,yn−2,yn−1)t

and

G=g1,g2,…,gn−2,gn−1t,

where

gi=22403εhγ1−8009ε−4489u0h3α1+h3∑j=03djfj,i=1,(0.5ε−u0ph3)α1+h3(p(f0+f4)+sf2+q(f1+f3)),i=2,h3(p(fi+2+fi−2)+sfi+q(fi+1+fi−1)),i=3,4,…,n−3,(0.5ε−unph3)β1+h3(p(fn+fn−4)+sf(i)+q(fn−1+fn−3)),i=n−2,(−480ε+2unh3)β1+h3∑j=0nkn−jfn−j,i=n−1.

Assuming that Y¯=(y(x1), y(x2), …, y(xn−1))t is the exact solution of Eq. (1) at the points: xi,fori=1,2,…,n−1, we have:

(31)AY¯=G

with truncation error: T=(t1, t2, …, tn−1)t and

(32)Ti=122105εh9yi9+O(h11),i=1,−16048εh9yi9+O(h11),i=2,3, …,n−2,251126εh9yi9+O(h11),i=n−1.

Now, subtracting Eq. (30) from Eq. (31), we obtain:

(33)A(Y¯−Y)=AE=T,

where E is the discretization error, and E=Y¯−Y=(e1, e2, …, en−1)t.

In order to derive the bound on ∥E∥∞ the following two lemmas are important.

Lemma 1

IfHis the square matrix of order n and∥H∥∞<1, then(I+H)−1exists and∥(I+H)−1∥∞<11−∥H∥∞.

Proof

For the details of the proof, one can refer to [14,15].□

Lemma 2

The matrix A is non-singular, if∥u∥<162w, wherew=480(b−a)3(2+3h2(b−a)2).

Proof

According to [16,17], the matrix N is non-singular and its inverse satisfies the inequality:□

∥N−1∥∞≤281(b−a)31+3h22(b−a)2.

Since N is the non-singular matrix, we have:

A=N+h3BU=(I+N−1h3BU)N.

So, to prove the non-singularity of A, it is sufficient to show that I+N−1h3BU is non-singular.

Moreover, ∥U∥∞≤∥u∥=maxa≤x≤b|u(x)| [18].

By Cauchy–Schwarz and triangle inequalities [19], we get:

(34)∥N−1h3BU∥∞≤∥N−1∥∞∥h3BU∥∞≤∥N−1∥∞h3∥B∥∞∥U∥∞≤h3∥N−1∥∞∥B∥∞∥u∥,

where ∥B∥∞= kn−1+kn−2+kn−3+kn−4 = 480.

Therefore, substituting ∥N∥∞, ∥B∥∞ and ∥u∥ in Eq. (34), we get the following relation:

(35)∥N−1h3BU∥∞≤1,

Using Lemma 1 and Eq. (35), we deduce that the matrix A is non-singular.

Since A is the non-singular matrix, Eq. (33) can be written as:

E=A−1T=(N+h3BU)−1T.

Since N is non-singular, we can re-write the matrix E in the form:

E=(I+N−1h3BU)−1N−1T,

and using the Cauchy–Schwarz inequality we obtain:

(36)∥E∥∞≤∥(I+N−1h3BU)−1∥∞∥N−1∥∞∥T∥∞.

Hence, from Eq. (36) and Lemma 1, it follows that:

∥E∥∞≤∥N−1∥∞∥T∥∞1−∥N−1h3BU∥∞.

Furthermore, from Eq. (32), we have:

∥T∥∞ = −160480εh9M9, where M9=maxa≤xi≤b|yi9(xi)| and hence

(37)∥E∥∞≤∥N−1∥∞∥T∥∞1−h3∥N−1∥∞∥B∥∞∥u∥≅O(h6),

∥E∥∞≤Mh6, where M is a constant independent of h.

Therefore, from Eq. (37) it follows that ∥E∥∞≤O(h6).

5 Numerical examples and results

To demonstrate the applicability of the method, four singularly perturbed model problems were considered. These examples were chosen because they have been widely discussed in the literature, and their exact solutions were available for comparison.

Example 1

Consider the following singularly perturbed problem:

−εy‴(x)+y(x)=f(x),0≤x≤1,

y(0)=0,y(1)=0,y′(0)=9ε,

where f(x)=6ε(1−x)5x3−6ε2(6(1−x)5−90(1−x)4x+180(1−x)3x2−60(1−x)2x3) and the analytic solution is

y(x)=6x3ε(1−x)5.

Example 2

Consider the following singularly perturbed problem:

−εy‴(x)+y(x)=f(x),0≤x≤1,

y(0)=0,y(1)=0,y″(0)=0,

where f(x)=6ε(1−x)5x3−6ε2(6(1−x)5−90(1−x)4x+180(1−x)3x2−60(1−x)2x3) and the analytic solution is

y(x)=6x3ε(1−x)5.

Example 3

Consider the following singularly perturbed problem:

−εy‴(x)+y(x)=f(x),0≤x≤1,

y(0)=0,y(1)=3εsin(3),y′(0)=9ε,

where f(x)=81ε2cos3x+3εsin3x, and the analytic solution is

y(x)=3εsin(3x).

Example 4

Consider the following singularly perturbed problem:

−εy‴(x)+y(x)=f(x),0≤x≤1,

y(0)=0,y(1)=3εsin(3),y″(0)=0,

where f(x)=81ε2cos3x+3εsin3x, and the analytic solution is

y(x)=3εsin(3x).

The numerical solutions in terms of maximum absolute errors and comparison with other findings existing in the literature are given in Tables 1–4 along with its graph in Figures 1–4 for different values of the perturbation parameters ε and mesh points N. The numerical rate of convergence for all the examples has been calculated by the formula:

rN=logEN2−log(EN)log(2),

where r is the rate of convergence (Tables 5 and 6).

Table 1

Maximum absolute errors and numerical rate of convergence for Example 1

ε↓	N=10	N=20	N=40
New method
1/16	1.0028 × 10⁻⁶	7.7262 × 10⁻⁹	5.9445 × 10⁻¹¹
r→	1.3824	3.2636	3.2736
1/32	4.2762 × 10⁻⁷	3.2959 × 10⁻⁹	2.5316 × 10⁻¹¹
r→	3.2610	3.2660	3.2743
1/64	1.7927 × 10⁻⁷	1.3211 × 10⁻⁹	1.0236 × 10⁻¹¹
r→	3.3258	3.2535	3.2756
Reference [4]
1/16	6.8572 × 10⁻⁶	1.1698 × 10⁻⁷	1.8578 × 10⁻⁹
1/32	2.9156 × 10⁻⁶	4.9916 × 10⁻⁸	7.9252 × 10⁻¹⁰
1/64	1.2223 × 10⁻⁶	2.0000 × 10⁻⁸	3.2111 × 10⁻¹⁰
Reference [5]
1/16	4.8700 × 10⁻⁴	1.8600 × 10⁻⁵	1.9500 × 10⁻⁵
1/32	1.9500 × 10⁻⁴	8.7600 × 10⁻⁶	8.6300 × 10⁻⁶
1/64	7.9700 × 10⁻⁵	4.0000 × 10⁻⁶	3.6100 × 10⁻⁶

Table 2

Maximum absolute errors and numerical rate of convergence for Example 2

ε↓	N=10	N=20	N=40
New method
1/16	1.5925 × 10⁻⁵	2.5004 × 10⁻⁷	3.9129 × 10⁻⁹
r→	2.2345	2.2393	2.2416
1/32	5.6471 × 10⁻⁶	8.8838 × 10⁻⁸	1.3917 × 10⁻⁹
r→	2.2317	2.2470	2.2517
1/64	1.8565 × 10⁻⁶	2.8912 × 10⁻⁸	4.5267 × 10⁻¹⁰
r→	2.2463	2.2386	2.3977
Reference [2]
1/16	2.8930 × 10⁻⁴	5.3006 × 10⁻⁶	2.6033 × 10⁻⁸
1/32	1.0962 × 10⁻⁴	1.9394 × 10⁻⁶	1.3221 × 10⁻⁸
1/64	3.8007 × 10⁻⁵	6.8026 × 10⁻⁷	6.2298 × 10⁻⁹
Reference [3]
1/16	6.2854 × 10⁻³	—	—
1/32	1.9707 × 10⁻³	—	—
1/64	3.9065 × 10⁻⁴	—	—

Table 3

Maximum absolute errors and numerical rate of convergence for Example 3

ε↓	N=10	N=20	N=40
New method
1/16	4.4336 × 10⁻⁸	2.0866 × 10⁻¹⁰	1.0750 × 10⁻¹²
r→	3.9727	3.8422	3.9422
1/32	1.8916 × 10⁻⁸	8.8814 × 10⁻¹¹	4.5211 × 10⁻¹³
r→	2.0969	3.8595	4.9501
1/64	7.9396 × 10⁻⁹	3.5668 × 10⁻¹¹	1.8448 × 10⁻¹³
r→	4.0398	3.8366	4.9522
Reference [4]
1/16	3.1247 × 10⁻⁷	4.9269 × 10⁻⁹	7.4543 × 10⁻¹¹
1/32	1.3421 × 10⁻⁷	2.1095 × 10⁻⁹	3.1741 × 10⁻¹¹
1/64	5.6587 × 10⁻⁸	8.4937 × 10⁻¹⁰	1.2904 × 10⁻¹¹
Reference [5]
1/16	2.3200 × 10⁻⁴	6.1200 × 10⁻⁵	1.5200 × 10⁻⁶
1/32	9.7700 × 10⁻⁵	2.5900 × 10⁻⁵	6.4500 × 10⁻⁶
1/64	3.7800 × 10⁻⁵	1.0000 × 10⁻⁶	2.5000 × 10⁻⁶

Table 4

Maximum absolute errors and numerical rate of convergence for Example 4

ε↓	N=10	N=20	N=40
New method
1/16	7.1881 × 10⁻⁷	6.8962 × 10⁻⁹	6.9667 × 10⁻¹¹
r→	2.9452	2.8707	3.4707
1/32	2.5476 × 10⁻⁷	2.4500 × 10⁻⁹	2.4726 × 10⁻¹¹
r→	2.9417	2.8722	3.5260
1/64	8.3850 × 10⁻⁸	7.9666 × 10⁻¹⁰	8.0634 × 10⁻¹²
r→	2.9592	2.8680	3.8707
Reference [2]
1/16	9.4405 × 10⁻⁶	5.4886 × 10⁻⁷	2.5658 × 10⁻⁸
1/32	3.1645 × 10⁻⁶	1.9215 × 10⁻⁷	9.1282 × 10⁻⁹
1/64	9.9920 × 10⁻⁷	6.1969 × 10⁻⁸	2.9364 × 10⁻⁹

$Figure 1 The graph of exact and numerical solutions of Example 1 for N=40N=40 and ε=164\varepsilon =\tfrac{1}{64}.$

Figure 1

The graph of exact and numerical solutions of Example 1 for N=40 and ε=164.

$Figure 2 The graph of exact and numerical solutions of Example 2 for N=40N=40 and ε=132\varepsilon =\tfrac{1}{32}.$

Figure 2

The graph of exact and numerical solutions of Example 2 for N=40 and ε=132.

$Figure 3 The graph of exact and numerical solutions of Example 3 for N=20N=20 and ε=164\varepsilon =\tfrac{1}{64}.$

Figure 3

The graph of exact and numerical solutions of Example 3 for N=20 and ε=164.

$Figure 4 The graph of exact and numerical solutions of Example 4 for N=20N=20 and ε=132\varepsilon =\tfrac{1}{32}.$

Figure 4

The graph of exact and numerical solutions of Example 4 for N=20 and ε=132.

Table 5

Maximum absolute errors for Example 2 when ε≪h

ε↓	N=10	N=50	N=100	N=150	N=200	N=250
New method
10−1	3.0119 × 10⁻⁵	1.9396 × 10⁻⁹	3.0307 × 10⁻¹¹	2.6779 × 10⁻¹²	4.3687 × 10⁻¹³	8.3924 × 10⁻¹⁴
10−2	8.6774 × 10⁻⁷	5.6414 × 10⁻¹¹	8.8095 × 10⁻¹³	7.6186 × 10⁻¹⁴	1.0994 × 10⁻¹⁴	9.0289 × 10⁻¹⁵
10−3	1.7611 × 10⁻⁸	1.2017 × 10⁻¹²	1.8835 × 10⁻¹⁴	1.6632 × 10⁻¹⁵	2.7283 × 10⁻¹⁶	8.8663 × 10⁻¹⁷
10−4	1.7081 × 10⁻¹⁰	2.5046 × 10⁻¹⁴	4.0023 × 10⁻¹⁶	3.6209 × 10⁻¹⁷	7.5647 × 10⁻¹⁸	4.1056 × 10⁻¹⁸
Reference [2]
10−1	5.3363 × 10⁻⁴	5.9813 × 10⁻⁸	9.1841 × 10⁻⁹	2.1617 × 10⁻⁹	7.3808 × 10⁻¹⁰	—
10−2	1.8773 × 10⁻⁵	2.2337 × 10⁻⁹	2.5972 × 10⁻¹⁰	6.0042 × 10⁻¹¹	2.0403 × 10⁻¹¹	—
10−3	1.5441 × 10⁻⁶	5.5630 × 10⁻¹¹	3.8697 × 10⁻¹²	8.5417 × 10⁻¹³	2.8902 × 10⁻¹³	—
10−4	1.8248 × 10⁻⁸	3.2291 × 10⁻¹²	5.6471 × 10⁻¹⁴	1.0585 × 10⁻¹⁴	3.3903 × 10⁻¹⁵	—
Reference [3]
10−1	1.6190 × 10⁻²	7.3371 × 10⁻⁴	6.4463 × 10⁻⁴	6.3671 × 10⁻⁴	6.3496 × 10⁻⁴	7.0694 × 10⁻⁴
10−2	5.4777 × 10⁻⁴	3.5302 × 10⁻⁵	3.2708 × 10⁻⁵	3.3005 × 10⁻⁵	3.3331 × 10⁻⁵	4.8167 × 10⁻⁵
10−3	4.3814 × 10⁻⁵	2.4150 × 10⁻⁶	1.3966 × 10⁻⁶	1.1544 × 10⁻⁶	1.2348 × 10⁻⁶	1.5754 × 10⁻⁶
10−4	7.5623 × 10⁻⁶	2.4329 × 10⁻⁷	1.1223 × 10⁻⁷	7.6323 × 10⁻⁸	6.1521 × 10⁻⁶	3.3448 × 10⁻⁸

Table 6

Maximum absolute errors for Example 3 when ε≪h

ε↓	N=10	N=50	N=100	N=150	N=200	N=250
New method
10−1	2.1007 × 10⁻¹⁰	8.4524 × 10⁻¹⁶	5.6292 × 10⁻¹⁶	1.7495 × 10⁻¹⁶	4.9739 × 10⁻¹⁷	2.5405 × 10⁻¹⁸
10−2	4.1057 × 10⁻¹²	4.0874 × 10⁻¹⁷	3.8625 × 10⁻¹⁸	1.4474 × 10⁻¹⁸	1.1709 × 10⁻¹⁸	5.4454 × 10⁻¹⁹
10−3	1.3838 × 10⁻¹³	1.9331 × 10⁻¹⁸	5.8445 × 10⁻²⁰	1.5543 × 10⁻²⁰	1.2955 × 10⁻²⁰	4.8789 × 10⁻²¹
Reference [3]
10−1	7.4049 × 10⁻⁴	1.1109 × 10⁻⁴	4.4739 × 10⁻⁵	2.3062 × 10⁻⁵	1.2307 × 10⁻⁵	6.0161 × 10⁻⁶
10−2	3.0209 × 10⁻²	1.8618 × 10⁻⁵	9.0607 × 10⁻⁶	5.9279 × 10⁻⁶	4.3685 × 10⁻⁶	3.4355 × 10⁻⁶
10−3	—	2.1405 × 10⁻⁶	1.0275 × 10⁻⁶	6.7970 × 10⁻⁷	5.0807 × 10⁻⁷	4.0537 × 10⁻⁷

6 Conclusion

The exponential spline method is developed to approximate solution of a third-order singularly perturbed two point boundary value problems. The convergence analysis is investigated and revealed that the present method is of sixth order convergence. Moreover, the study analysed by taking different mesh size h and sufficiently small perturbation parameter ε. From the results in Tables 1–4, one can see that maximum absolute error decreases as both the perturbation parameter and mesh size decrease, which in turn shows the convergence of the computed solution. Furthermore, the result of the present method is compared with the existing findings and observed that it gives more accurate solution than some existing numerical methods reported in the literature. The present method approximated the exact solution very well. Generally, the present method is convergent and more accurate for solving third-order singularly perturbed two point boundary value problems and the scheme can also work properly for the case of variable coefficients but it needs the linearization technique, a supporting technique to be applicable for nonlinear problems.

Acknowledgments

The authors would like to express their gratitude to the authors of literature for the provision of initial idea for this work. The authors also thank Jimma University and Dambi Dollo University for necessary supports.

Funding: Not applicable.
Conflict of interest: The authors declare that they have no competing interest.
Author contributions: YAW proposed the main idea of this paper. YAW and GFD prepared the manuscript and performed all the steps of the proofs in this research. Both authors contributed equally and significantly in writing this paper. Both authors read and approved the final manuscript.

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Received: 2020-06-08

Revised: 2020-09-15

Accepted: 2020-10-10

Published Online: 2020-12-31

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/dema-2020-0024

Keywords for this article

exponential spline; singular perturbation; boundary layers; convergence

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