Solution to SPDDE using Exponentially Fitted Spline Method

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ISSN: 2005-4238 IJAST 452

Solution to SPDDE using Exponentially Fitted Spline Method

V.Vidyasagar^1*, MadhuLatha K², B.Ravindra Reddy³

1,2Department of Mathematics, Kamala Institute of Technology & Science, Huzurabad- 505468, India

3Department of Mathematics, JNTUH College of Engineering Hyderabad, Kukatpally- 500085, India

Abstract

Within this paper, We have answered of SPDDE along with delay. Also, advanced parameters utilizing an exponentially fitted spline approach. At first, the given second order differential-difference equation is replaced by an asymptotically proportionate second order singular perturbation problem. At that point, a fitting factor is brought into the exponentially fitted spline Method. The value of fitting factor is obtained by the singular perturbations theory. The Thomas algorithm is used to solve the tridiagonal system obtained by the method. We have charted maximum absolute errors for the instances selected from the literature. Numerical results, along with comparison, along with the various other strategies, are revealed to expatiate the effectiveness of the technique for arbitrary λ1, λ2 such that λ1 +λ2 =1/2.

Index Terms: Central differences, Differential-difference equations, Boundary layer, Fitting Factor, Cubic Spline.

1. INTRODUCTION

Any ordinary differential equation in which the highest derivative is multiplied by a small positive parameter and also have at least one shift term( negative shift or positive shift) is called singularly perturbed differential-difference equation(SPDDE). In such problems, usually there are thin transition layers, where the solution varies rapidly or jumps abruptly, while away from the layers, the solution behaves regularly and varies slowly. Lange and also Miura [1] - [5] developed a series of papers to get an approximate solution of SPDDEs. Kadalbajoo started the numerical analysis of SPDDE turning point problems and also Sharma.

Kadalbajoo as well as Sharma [6] - [10], they offered numerous robust, precise strategies for the solution of such kind of problems. Kadalbajoo and Sharma [11] illuminate a numerical technique to solve boundary value problems for SPDDEs. Kadalbajoo and also Sharma [12] recommended a mathematical procedure to solve an SPDDE of a mixed kind with the situation in which the solution of the issue exhibits quick oscillations. Kadalbajoo and Sharma [13] defined a numerical approach based on ﬁnite diﬀerence approach to solve a mathematical model arising from a design of neuronal variability. Patidar as well as Sharma [14]

incorporated fitted-operator methods with Micken's non-standard finite difference strategies for the numerical estimates of singularly perturbed linear delay differential equations. Kadalbajoo et al. [15]

derived ε-uniformly convergent ﬁtted approaches for the solution of SPDDE. Kumar and also Sharma [16]

presented a mathematical plan based on B-spline collocation to approximate the solution of boundary value problems for SPDDEs with delay in addition to advance. Amiraliyev and Cimen [17] derived a numerical method for singularly perturbed boundary value problem for a linear second order delay diﬀerential equation with a large delay in the reaction term.

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ISSN: 2005-4238 IJAST 453

2. CONTINUOUS EQUATION

Think about SPDDE along with little delay and additionally advance parameters of the kind:

𝜀𝑦^′′(𝑥) + 𝑝(𝑥)𝑦^′(𝑥) + 𝑞(𝑥)𝑦(𝑥 − 𝛿) + 𝑟(𝑥)𝑦(𝑥)+𝑠(𝑥)𝑦(𝑥 + 𝜂) = 𝑓(𝑥) (1)

∀𝑥 ∈ (0,1) and under the boundary conditions

𝑦(𝑥) = 𝜑(𝑥), 𝑜𝑛 − 𝛿 ≤ 𝑥 ≤ 0 (2)

𝑦(𝑥) = 𝛾(𝑥), 𝑜𝑛 1 ≤ 𝑥 ≤ 1 + 𝜂 (3)

Here 𝑝(𝑥), 𝑞(𝑥), 𝑟(𝑥), 𝑠(𝑥), 𝜑(𝑥) 𝑎𝑛𝑑 𝛾(𝑥) are sufficiently0smooth functions on (0, 1), the singular0perturbation parameter



is small0positive parameter (0 <



<<1), and 0 < 𝛿 = 𝑜(𝜀) and 0 <

𝜂 = 𝑜(𝜀) are0the delay(negative shift)0and the advance(positive shift) parameters0respectively.Typically, the solution for Eq.(1)-Eq.(3) reveals layer behaviour at one end of the interval [0,1] depending upon the sign of 𝑝(𝑥) + 𝑠(𝑥)𝜂 − 𝑞(𝑥)𝛿.

By utilizing Taylor series almost the aspect x, the deviating argument conditions may be taken as

𝑦(𝑥 − 𝛿) ≈ 𝑦(𝑥) − 𝛿𝑦^′(𝑥) (4)

𝑦(𝑥 + 𝜂) ≈ 𝑦(𝑥) + 𝜂𝑦^′(𝑥) (5)

Using Eq.(4) and Eq.(5) in Eq.(1) we receive an asymptotically equal singularly perturbed boundary value problem of the type:

𝜀𝑦^′′(𝑥) + 𝛼(𝑥)𝑦^′(𝑥) + 𝛽(𝑥)𝑦(𝑥) = 𝑓(𝑥) (6)

𝑦(0) = 𝜑(0) = 𝜑0 (7)

𝑦(1) = 𝛾(1) = 𝛾₁ (8)

where 𝛼(𝑥) = 𝑝(𝑥) + 𝑠(𝑥)𝜂 − 𝑞(𝑥)𝛿 (9)

𝛽(𝑥) = 𝑞(𝑥) + 𝑟(𝑥) + 𝑠(𝑥) (10)

Since0 < 𝛿 << 1 and 0 < 𝜂 << 1, The transition from Eq.(1) to Eq.(6) is admissible. Further details on the validity of this transition is found in El’sgol’ts and Norkin[18].

3. DESCRIPTION OF METHOD Let x0 = a, xN = b, xi = a+ih, h= (b−a)/N.

S(x,τ) function of class C²[a,b] which will interpolates y(x) at xi which is depending on specification τ, and minimizes to cubic spline within [a, b] as τ → 0 is called as parametric cubic-spline function.

The spline function S(x,τ)= S(x) satisfies in[xi,xi+1], the differential equation

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ISSN: 2005-4238 IJAST 454

𝑆^′′(𝑥) + τS(𝑥) = [𝑆^′′(𝑥_𝑖) + τS(𝑥_𝑖)]^(𝑥^𝑖+1^−𝑥)

ℎ + [𝑆^′′(𝑥_𝑖+1) + 𝜏𝑆(𝑥_𝑖+1)]^(𝑥−𝑥^𝑖⁾

ℎ

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where 𝑆(𝑥_𝑖) = 𝑦_𝑖and τ > 0 is termed as cubic spline in compression.

By Solving the linear second-order differential equation (11) and also influencing arbitrary constants from interpolatory conditions 𝑆(𝑥_𝑖+1) = 𝑦_𝑖+1, 𝑆(𝑥_𝑖) = 𝑦_𝑖 we achieve, after writing as 𝜆 = ℎ𝜏^1/2 S(x) = − ^ℎ²

𝜆²𝑠𝑖𝑛𝜆[𝑀_𝑖+1𝑠𝑖𝑛^{𝜆(𝑥−𝑥}^𝑖⁾

ℎ + 𝑀_𝑖𝑠𝑖𝑛^𝜆(𝑥^𝑖+1^−𝑥)

ℎ ] +^ℎ²

𝜆²[^(𝑥−𝑥^𝑖⁾

ℎ (𝑀_𝑖+1+^𝜆²

ℎ²𝑦_𝑖+1) +^(𝑥^𝑖+1^−𝑥)

ℎ (𝑀_𝑖+

𝜆²

ℎ²𝑦_𝑖)] (12)

By differentiating Eq.(12) and saying x tend to xi, we get

𝑆^′(𝑥_𝑖+) =𝑦𝑖+1− 𝑦𝑖

ℎ + ℎ

𝜆²[(1 − ℎ

𝑠𝑖𝑛𝜆) 𝑀_𝑖+1− (1 − 𝜆𝑐𝑜𝑡𝜆)𝑀_𝑖]

By considering the interval (xi-1,xi) and also proceeding similarly, we achieve

𝑆^′(𝑥_𝑖−) =𝑦_𝑖− 𝑦_𝑖−1

ℎ + ℎ

𝜆²[(1 − 𝜆𝑐𝑜𝑡𝜆)𝑀_𝑖− (1 − 𝜆

𝑠𝑖𝑛𝜆) 𝑀_𝑖−1] Equating right and also left hand derivatives at xi,we can have

𝑦_𝑖− 𝑦_𝑖−1

ℎ + ℎ

𝜆²[(1 − 𝜆𝑐𝑜𝑡𝜆)𝑀_𝑖− (1 − 𝜆

𝑠𝑖𝑛𝜆) 𝑀_𝑖−1]

= ^𝑦^𝑖+1^−𝑦^𝑖

ℎ + ^ℎ

𝜆²[(1 − ^𝜆

𝑠𝑖𝑛𝜆) 𝑀_𝑖+1− (1 − 𝜆𝑐𝑜𝑡𝜆)𝑀_𝑖] (13)

This tends to tridiagonal system

ℎ²(𝜆1𝑀𝑖−1+ 2𝜆2𝑀𝑖+ 𝜆1𝑀𝑖+1) = 𝑦𝑖+1− 2𝑦𝑖+ 𝑦𝑖−1

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where 𝜆₁ = ¹

𝜆²( ^𝜆

𝑠𝑖𝑛𝜆− 1) , 𝜆₂ = ¹

𝜆²(1 − 𝜆𝑐𝑜𝑡𝜆), 𝑀_𝑖 = 𝑆^′′(𝑥_𝑖) , i= 1(1)N-1.

3.1 Left end boundary layer problem

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ISSN: 2005-4238 IJAST 455

Under the conditions 𝛽(x) ≤ 0, 𝛼(x) ≥ M >0 through the interval [0, 1], where M is positive constant, Eq.(6) possesses an unique solution y(x) which as a whole, presents a boundary layer of width O(𝜀) at x = 0 for little values of 𝜀.

From singular perturbation theory, the solution of Eqs.(6)-(8) is of the type

𝑦(𝑥) = 𝑦₀(𝑥) +^𝛼(0)

𝛼(𝑥)(



₀ − 𝑦₀(0)) 𝑒^{− ∫ (}⁰^𝑥 ^𝛼(𝑥)^𝜀 ^)𝑑𝑥+ 𝑂(𝜀) (15) wherey₀(x) is the solution of reduced problem of Eq. (6) given by

𝛼(𝑥)𝑦₀^′(𝑥) + 𝛽(𝑥)𝑦₀(𝑥) = 𝑓(𝑥), 𝑦₀(1) = 𝛾₁ (16)

Right now dividing the domain [0, 1] into N equal subintervals with constant mesh length h to make sure that 0=x x₁, ₂,...,x_N=1 be mesh points, then we have x_i ih:i0, 1, 2, … ,N.

From Eq. (15), we have

ℎ→0lim𝑦(𝑖ℎ) = 𝑦₀(0) + (𝜑₀− 𝑦₀(0))𝑒^{−𝛼(0)𝑖𝜌}+ 𝑂(𝜀) (17)

where 𝜌 =^ℎ

𝜀

Now to manage the small perturbation parameter, introduce an exponentially fitting factor 𝜎(𝜌) in Eq.(6), we get

𝜎(𝜌)𝜀𝑦^′′(𝑥) + 𝛼(𝑥)𝑦^′(𝑥) + 𝛽(𝑥)𝑦(𝑥) = 𝑓(𝑥) (18)

𝑦(0) = 𝜑(0) = 𝜑₀ , 𝑦(1) = 𝛾(1) = 𝛾₁

The fitting factor 𝜎(𝜌)is to be found out as if the solution of Eq.( 18) uniformly converges to the solution of Eq.(6).

By the Eq. (6), we have

𝜎(𝜌) 𝜀𝑀_𝑖 = 𝑓(𝑥_𝑖) − 𝛼(𝑥_𝑖)𝑦_𝑖^′− 𝛽(𝑥_𝑖)𝑦_𝑖 for i = j-1, j, j+1.

Utilizing the equation in Eq. (14) and also using the following to approximations for first derivative of y : 𝑦_𝑖^′ ≅^𝑦^𝑖+1^−𝑦^𝑖−1

2ℎ , 𝑦_𝑖+1^′ ≅^3𝑦^𝑖+1^−4𝑦^𝑖^+𝑦^𝑖−1

2ℎ ,𝑦_𝑖−1^′ ≅^−𝑦^𝑖+1^+4𝑦^𝑖^−3𝑦^𝑖−1

2ℎ

(19)

We get the following difference equation

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ISSN: 2005-4238 IJAST 456

𝐸_𝑖𝑦_𝑖−1− 𝐹_𝑖𝑦_𝑖+ 𝐺_𝑖𝑦_𝑖+1= 𝐻_𝑖, 𝑖 = 1(1)𝑁 − 1 (20)

where E_i =^σ(ρ)

hρ −^3λ¹^𝛼ⁱ⁻¹

2h −^λ²^𝛼ⁱ

h +^λ¹^𝛼ⁱ⁺¹

2h + λ₁𝛽_i−1

F_i=2σ(ρ)

hρ −2λ₁𝛼_i−1

h +2λ₁𝛼_i+1

h − 2λ₂𝛽_i

G_i=σ(ρ)

hρ −λ₁𝛼_i−1

2h +λ₂𝛼_i

h +3λ₁𝛼_i+1

2h + λ₁𝛽_i+1 𝐻_𝑖 = 𝜆₁𝑓_𝑖−1+ 2𝜆₂𝑓_𝑖+ 𝜆₁𝑓_𝑖+1

We solve the tridiagonal system of Eq.(20) by using the Thomas algorithm.

3.2 Fitting factor Determination

Following the procedure given in Doolan et al. [19], we have

ℎ→0lim(^𝜎(𝜌)

𝜌 (𝑦(𝑖ℎ + ℎ) − 2𝑦(𝑖ℎ) + 𝑦(𝑖ℎ − ℎ))) + lim

ℎ→0((𝜆₁+ 𝜆₂)𝑝(𝑖ℎ)(𝑦(𝑖ℎ + ℎ) − 𝑦(𝑖ℎ − ℎ))) = 0 (21)

Here,we assume that f(xi)- 𝛽(xi)y(xi) is bounded. Substituting Equation (17) in Equation (21) and then after simplifying,

we achieve 𝜎(𝜌) = 𝛼(0)𝜌(𝜆₁+ 𝜆₂)coth (^𝛼(0)𝜌

2 ) (22)

Eq. (22) gives required fitting factor of left end boundary layer.

3.3 Right end boundary layer

Under the assumptions 𝛽(x) ≤ 0, 𝛼(x) ≤ M < 0 throughout the interval [0, 1], where M is some negative constant. Under these assumptions, Eq. (6) has a unique solution y(x) which in general, shows a boundary layer of width O(𝜀) at x = 1 towards small values of 𝜀.

By the theory of singular perturbation, asymptotic solution for right end layer problem is specified as

𝑦(𝑥) = 𝑦₀(𝑥) + (𝛾₁− 𝑦₀(1))𝑒⁽⁻^{𝛼(1)(1−𝑥)}^𝜀 ⁾+ 𝑂(𝜀) (23)

As h0 Eq.(22) becomes

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ℎ→0lim𝑦(𝑖ℎ) = 𝑦₀(0) + (𝛾₁− 𝑦₀(1))𝑒^(−𝛼(1)(¹^𝜀^{−𝑖𝜌))}+ 𝑂(𝜀) (24)

where 𝜌 =^ℎ

𝜀 , introducing 𝜎(𝜌) in Eq.(6), and applying the technique as in left end boundary layer and on simplifying,

we get 𝜎(𝜌) = 𝛼(1)𝜌(𝜆₁+ 𝜆₂)coth (^𝛼(1)𝜌

2 ) (25)

Now we have the three term recurrence equation as:

𝐸𝑖𝑦𝑖−1− 𝐹𝑖𝑦𝑖+ 𝐺𝑖𝑦𝑖+1= 𝐻𝑖, 𝑖 = 1(1)𝑁 − 1 (26)

where

E_i =σ(ρ)

hρ −3λ₁𝛼_i−1 2h −λ₂𝛼_i

h +λ₁𝛼_i+1

2h + λ₁𝛽_i−1

F_i=2σ(ρ)

hρ −2λ₁𝛼_i−1

h +2λ₁𝛼_i+1

h − 2λ₂𝛽_i

G_i=σ(ρ)

hρ −λ₁𝛼_i−1

2h +λ₂𝛼_i

h +3λ₁𝛼_i+1

2h + λ₁𝛽_i+1 𝐻_𝑖 = 𝜆₁𝑓_𝑖−1+ 2𝜆₂𝑓_𝑖+ 𝜆₁𝑓_𝑖+1

We then solve the tridiagonal system of Eq.(26) by utilizing Thomas algorithm.

4. NUMERICAL EXAMPLES

In order to show the recommended strategy, we have used it to problems with left-end boundary layer as well as with the right-end boundary layer. These model instances have been widely discussed in literary works.

The exact solution of Eq.(1) to Eq.(3) with constant coefficients(i.e,

𝑝(𝑥) = 𝑝, 𝑞(𝑥) = 𝑞, 𝑟(𝑥) = 𝑟, 𝑠(𝑥) = 𝑠, 𝑓(𝑥) = 𝑓, 𝜑(𝑥) = 𝜑 𝑎𝑛𝑑 𝛾(𝑥) = 𝛾)

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is given by 𝑦(𝑥) = 𝑐₁𝑒^𝑚¹^𝑥+ 𝑐₂𝑒^𝑚²^𝑥+ 𝑓/𝑐₃ where

𝑚₁=−(𝑝+𝑠𝜂−𝑞𝛿)+√(𝑝+𝑠𝜂−𝑞𝛿)²−4𝜀𝑐₃

2𝜀 , 𝑚₂=−(𝑝+𝑠𝜂−𝑞𝛿)−√(𝑝+𝑠𝜂−𝑞𝛿)²−4𝜀𝑐₃ 2𝜀

𝑐₁=^{−𝑓+𝛾𝑐}³^+𝑒^𝑚2^{(𝑓−𝜑𝑐}³⁾

(𝑒^𝑚1−𝑒^𝑚2)𝑐₃ , 𝑐2=^{𝑓−𝛾𝑐}³^+𝑒^𝑚1^{(−𝑓+𝜑𝑐}³⁾

(𝑒^𝑚1−𝑒^𝑚2)𝑐₃ , 𝑐3= 𝑞 + 𝑟 + 𝑠

Using double mesh principle [19] the maximum absolute errors are calculated when the exact solutions are unknown,

𝐸^𝑁= max

0≤𝑖≤𝑁|𝑦_𝑖^𝑁− 𝑦_2𝑖^2𝑁| Example 1. Consider the SPDDE with left end boundary layer:

𝜀𝑦^′′(𝑥) + 𝑦^′(𝑥) + 2𝑦(𝑥 − 𝛿) − 3𝑦(𝑥) = 0 , 𝜑(𝑥) = 1, 𝛾(𝑥) = 1 In Table 1 and Figure 1, numerical results are tabulated.

Example 2.Consider the SPDDE with left end boundary layer:

𝜀𝑦^′′(𝑥) + 𝑦^′(𝑥) − 3𝑦(𝑥)+2𝑦(𝑥 + 𝜂) = 0 , 𝜑(𝑥) = 1, 𝛾(𝑥) = 1 In Tables 2, 3 and Figure 2, numerical results are tabulated.

Example 3. Consider the SPDDE with left end boundary layer:

𝜀𝑦^′′(𝑥) + 𝑦^′(𝑥) − 2𝑦(𝑥 − 𝛿) − 5𝑦(𝑥)+𝑦(𝑥 + 𝜂) = 0 , 𝜑(𝑥) = 1, 𝛾(𝑥) = 1 In Tables 4,5, 6 and in Figures 3, 4, the numerical results are tabulated.

Example 4. Consider SPDDE with right end boundary layer:

𝜀𝑦^′′(𝑥) − 𝑦^′(𝑥) − 2𝑦(𝑥 − 𝛿) + 𝑦(𝑥) = 0 , 𝜑(𝑥) = 1, 𝛾(𝑥) = −1 In Table 7 and Figure 5, numerical results are tabulated.

Example 5. Consider the SPDDE with right end boundary layer:

𝜀𝑦^′′(𝑥) − 𝑦^′(𝑥) + 𝑦(𝑥) − 2𝑦(𝑥 + 𝜂) = 0 , 𝜑(𝑥) = 1, 𝛾(𝑥) = −1 In Table 8 and Figure 6, numerical results are tabulated.

𝜀𝑦^′′(𝑥) − (1 + 𝑒^𝑥²)𝑦^′(𝑥) − 𝑥𝑦(𝑥 − 𝛿) + 𝑥²𝑦(𝑥) − (1 − 𝑒^−𝑥)𝑦(𝑥 + 𝜂) = 1 , 𝜑(𝑥) = 1, 𝛾(𝑥) = −1 Where the exact solution is unknown. In Table 9 and Figure 7, the numerical results are given.

𝜀𝑦^′′(𝑥) − 𝑦^′(𝑥) − 2𝑦(𝑥 − 𝛿) + 𝑦(𝑥) − 2𝑦(𝑥 + 𝜂) = 0 , 𝜑(𝑥) = 1, 𝛾(𝑥) = −1 In Table 10, 11 and Figure 8, 9 the numerical results are tabulated.

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Example 8. Consider the SPDDE with left end boundary layer:

𝜀𝑦^′′(𝑥) + 𝑦^′(𝑥) − 2𝑦(𝑥 − 𝛿) + 𝑦(𝑥) − 𝑦(𝑥 + 𝜂) = −1 , 𝜑(𝑥) = 1, 𝛾(𝑥) = 1 In Table 12 and Figure 10, numerical results are tabulated.

Figure 1. In Example 1, numerical solution Figure 2. In Example 2, numerical solution for 𝜺 = 𝟎. 𝟏 and 𝜼 = 𝟎 for 𝜺 = 𝟎. 𝟏 and 𝜹 = 𝟎

Figure 3. In Example 3, numerical solution Figure 4. In Example 3, numerical solution for 𝜺 = 𝟎. 𝟏 and 𝜼 = 𝟎. 𝟎𝟓for 𝜺 = 𝟎. 𝟏 and 𝜹 = 𝟎. 𝟎𝟓

Figure 5. In Example 4, numerical solution Figure 6. In Example 5, numerical solution

for 𝜺 = 𝟎. 𝟏 and 𝜼 = 𝟎 for 𝜺 = 𝟎. 𝟏 and 𝜹 = 𝟎

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for 𝜺 = 𝟎. 𝟏 and 𝜹 = 𝟎 for 𝜺 = 𝟎. 𝟏 and 𝜼 = 𝟎. 𝟎𝟓

for 𝜺 = 𝟎. 𝟏 and 𝜹 = 𝟎. 𝟎𝟓 for 𝜺 = 𝟎. 𝟏 and 𝜹 = 𝟎. 𝟎𝟓

Table 1. In solution of Example 1, the maximum absolute errors for 𝜺 = 𝟎. 𝟏 and 𝜼 = 𝟎 𝛿\ N 8 32 128

512

8 32 128 512

𝜆₁=1

6 , 𝜆₂=1

3 𝜆₁= 1

12 , 𝜆₂= 5 12

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0.00 0.05 0.09

3.0138e-03 1.8911e-04 1.1678e-05 7.2965e-07

3.3814e-03 1.9902e-04 1.2535e-05 7.8318e-07

3.6674e-03 2.1269e-04 1.3255e-05 8.2832e-07

2.8681e-03 1.9231e-04 1.2064e-05 7.5418e-07

2.8400e-03 1.8703e-04 1.1730e-05 7.3326e-07

2.7729e-03 1.8094e-04 1.1351e-05 7.0958e-07

0.00 0.05 0.09

Results in [9]

0.09907804 0.03700736 0.00954678 0.00214501 0.09659609 0.03640566 0.00924661 0.00202998 0.09277401 0.03556652 0.00895172 0.00192488

Table 2. In solution of Example 2, the maximum absolute errors for 𝜺 = 𝟎. 𝟏 and 𝜹 = 𝟎 𝜂\N 8 32 128

512

8 32 128 512

𝜆₁=1

6 , 𝜆₂=1

3 𝜆₁= 1

12 , 𝜆₂= 5 12 0.00

0.05 0.09

3.0138e-03 1.8911e-04 1.1678e-05 7.2965e-07

2.6508e-03 1.7770e-04 1.0893e-05 6.8004e-07

2.5307e-03 1.7128e-04 1.0754e-05 6.7236e-07

2.8681e-03 1.9231e-04 1.2064e-05 7.5418e-07

2.9150e-03 1.9563e-04 1.2274e-05 7.6734e-07

2.9436e-03 1.9702e-04 1.2371e-05 7.7341e-07

0.00 0.05 0.09

Results in [9]

0.09907804 0.03700736 0.00954678 0.00214501 0.09977501 0.03727087 0.00979659 0.00224472 0.10031348 0.03723863 0.00996284 0.00458698

Table 3. In solution of Example 2, the maximum absolute errors for 𝜹 = 𝟎. 𝟓𝜺 and 𝜼 = 𝟎. 𝟓𝜺 𝜀 \𝑁 8 32 128

512

8 32 128 512

𝜆1=1

6 , 𝜆2=1

3 𝜆₁= 1

12 , 𝜆₂= 5 12

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10^-1 10^-2 10^-3 10^-4 10^-5 10^-6

2.6508e-03 1.7770e-04 1.0893e-05 6.8004e-07

1.6707e-02 2.4621e-03 1.7817e-04 1.1252e-05

1.9981e-02 5.2476e-03 1.0620e-03 1.0972e-04

2.0320e-02 5.5820e-03 1.3926e-03 3.2198e-04

2.0354e-02 5.6156e-03 1.4258e-03 3.5511e-04

2.0358e-02 5.6189e-03 1.4291e-03 3.5843e-04

2.9150e-03 1.9563e-04 1.2274e-05 7.6734e-07

1.6920e-02 2.4947e-03 1.8010e-04 1.1373e-05

2.0198e-02 5.2639e-03 1.0634e-03 1.0983e-04

2.0538e-02 5.5984e-03 1.3936e-03 3.2205e-04

2.0572e-02 5.6319e-03 1.4269e-03 3.5518e-04

2.0575e-02 5.6353e-03 1.4302e-03 3.5849e-04

Table 4. In solution of Example 3, the maximum absolute errors for 𝜹 = 𝟎. 𝟓𝜺 and 𝜼 = 𝟎. 𝟓𝜺 𝜀 \𝑁 8 32 128

512

8 32 128 512

𝜆₁=1

6 , 𝜆₂=1

3 𝜆₁= 1

12 , 𝜆₂= 5 12 10^-1

10^-2 10^-3 10^-4 10^-5 10^-6

3.7071e-02 2.3757e-03 1.4566e-04 9.0928e-06

6.8761e-02 2.3421e-02 2.1293e-03 1.2739e-04

8.5568e-02 2.8080e-02 7.2036e-03 1.3931e-03

8.7402e-02 3.0061e-02 8.1092e-03 1.9160e-03

8.7587e-02 3.0261e-02 8.3086e-03 2.1147e-03

8.7606e-02 3.0282e-02 8.3285e-03 2.1346e-03

1.5547e-02 1.0631e-03 6.6867e-05 4.1808e-06

7.6701e-02 1.3338e-02 1.0174e-03 6.3136e-05

9.3655e-02 2.8920e-02 6.2077e-03 6.8250e-04

9.5503e-02 3.0903e-02 8.1721e-03 1.9202e-03

9.5690e-02 3.1103e-02 8.3715e-03 2.1188e-03

9.5709e-02 3.1123e-02 8.3915e-03 2.1387e-03

Table 5. In solution of Example 3, the maximum absolute errors for 𝜺 = 𝟎. 𝟏 and 𝜼 = 𝟎. 𝟎𝟓 𝛿\N 8 32 128

512

8 32 128 512

𝜆1=1

6 , 𝜆2=1

3 𝜆₁= 1

12 , 𝜆₂ = 5 12 0.00

0.05 0.09

3.7240e-02 2.3116e-03 1.4203e-04 8.8821e-06

3.7071e-02 2.3757e-03 1.4566e-04 9.0928e-06

1.4743e-02 1.0010e-03 6.2934e-05 3.9348e-06

1.5547e-02 1.0631e-03 6.6867e-05 4.1808e-06

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3.6691e-02 2.4121e-03 1.4765e-04 9.2249e-06

1.6086e-02 1.1077e-03 6.9761e-05 4.3619e-06

0.00 0.05 0.09

Results in [9]

0.09190267 0.03453494 0.01164358 0.00300463 0.10233615 0.03823132 0.01295871 0.00335137 0.11018870 0.04110846 0.01400144 0.00362925

Table 6. In solution of Example 3, the maximum absolute errors for 𝜺 = 𝟎. 𝟏 and 𝜹 = 𝟎. 𝟎𝟓 𝜂\N 8 32 128

512

8 32 128 512

𝜆₁=1

6 , 𝜆₂=1

3 𝜆₁= 1

12 , 𝜆₂= 5 12 0.00

0.05 0.09

3.7240e-02 2.3116e-03 1.4203e-04 8.8821e-06

3.7071e-02 2.3757e-03 1.4566e-04 9.0928e-06

3.6691e-02 2.4121e-03 1.4765e-04 9.2249e-06

1.5164e-02 1.0332e-03 6.4947e-05 4.0606e-06

1.5547e-02 1.0631e-03 6.6867e-05 4.1808e-06

1.5828e-02 1.0855e-03 6.8335e-05 4.2731e-06

0.00 0.05 0.09

Results in [9]

0.09720029 0.03640446 0.01229476 0.00317786 0.10233615 0.03823132 0.01295871 0.00335137 0.10632014 0.03965833 0.01348348 0.00349050

Table 7. In solution of Example 4, the maximum absolute errors for 𝜺 = 𝟎. 𝟏 and 𝜼 = 𝟎 𝛿\N 8 32 128

512

8 32 128 512

𝜆₁=1

6 , 𝜆₂ =1

3 𝜆₁= 1

12 , 𝜆₂= 5 12 0.00

0.05 0.09

1.5647e-02 8.9130e-04 5.5464e-05 3.4650e-06

1.4912e-02 8.4701e-04 5.2996e-05 3.3113e-06

9.0740e-03 5.2137e-04 3.2532e-05 2.0332e-06

8.6107e-03 5.0104e-04 3.1181e-05 1.9483e-06

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1.4174e-02 8.1460e-04 5.0696e-05 3.1680e-06

8.1459e-03 4.7837e-04 2.9870e-05 1.8670e-06

0.00 0.05 0.09

Results in [9]

0.07847490 0.04678972 0.01727912 0.00443086 0.09222560 0.03828329 0.01487799 0.00380679 0.10509460 0.03149275 0.01299340 0.00331935

Table 8. In solution of Example 5, the maximum absolute errors for 𝜺 = 𝟎. 𝟏 and 𝜹 = 𝟎 𝜂\N 8 32 128

512

8 32 128 512

𝜆₁=1

6 , 𝜆₂=1

3 𝜆₁ = 1

12 , 𝜆₂= 5 12 0.00

0.05 0.09

1.5647e-02 8.9130e-04 5.5464e-05 3.4650e-06

1.6193e-02 9.2853e-04 5.7550e-05 3.5950e-06

1.6511e-02 9.5072e-04 5.8983e-05 3.6849e-06

9.0740e-03 5.2137e-04 3.2532e-05 2.0332e-06

9.4199e-03 5.3953e-04 3.3648e-05 2.1027e-06

9.6224e-03 5.5419e-04 3.4383e-05 2.1493e-06

0.00 0.05 0.09

Results in [9]

0.07847490 0.04678972 0.01727912 0.00443086 0.06834579 0.05516436 0.01972508 0.00506769 0.08328237 0.06168267 0.02169662 0.00558451

Table 9. In solution of Example 6, by double mesh principle the maximum absolute errors for 𝜺 = 𝟎. 𝟏 and 𝜹 = 𝟎

𝜂\N 8 32 128 512

8 32 128 512

𝜆₁ =1

6 , 𝜆₂=1

3 𝜆₁= 1

12 , 𝜆₂= 5 12 0.00

0.05 0.09

2.5508e-01 5.2265e-02 2.7442e-02 8.0024e-03

2.5649e-01 5.3356e-02 2.7967e-02 8.1504e-03

2.5648e-01 5.2531e-02 2.7464e-02 8.0038e-03

2.5788e-01 5.3621e-02 2.7989e-02 8.1519e-03

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2.5760e-01 5.4224e-02 2.8388e-02 8.2689e-03

2.5900e-01 5.4490e-02 2.8410e-02 8.2704e-03

Table 10. In solution of Example 7, the maximum absolute errors for 𝜺 = 𝟎. 𝟏 and 𝜼 = 𝟎. 𝟎𝟓 𝛿\N 8 32 128

512

8 32 128 512

𝜆₁=1

6 , 𝜆₂=1

3 𝜆₁= 1

12 , 𝜆₂= 5 12 0.00

0.05 0.09

3.1378e-02 1.8002e-03 1.1207e-04 7.0036e-06

2.9748e-02 1.7000e-03 1.0542e-04 6.5861e-06

2.8294e-02 1.6111e-03 9.9794e-05 6.2345e-06

1.8115e-02 1.0328e-03 6.4031e-05 4.0020e-06

1.6810e-02 9.5187e-04 5.9260e-05 3.7026e-06

1.5665e-02 8.8187e-04 5.5198e-05 3.4484e-06

0.00 0.05 0.09

Results in [9]

0.09930002 0.03685072 0.01331683 0.00342882 0.09997296 0.03218424 0.01167102 0.00299572 0.10044578 0.02850398 0.01038902 0.00266379

Table 11. In solution of Example 7, the maximum absolute errors for 𝜺 = 𝟎. 𝟏 and 𝜹 = 𝟎. 𝟎𝟓 𝜂\N 8 32 128

512

8 32 128 512

𝜆₁=1

6 , 𝜆₂=1

3 𝜆₁= 1

12 , 𝜆₂= 5 12 0.00

0.05 0.09

2.7911e-02 1.5877e-03 9.8361e-05 6.1442e-06

2.9748e-02 1.7000e-03 1.0542e-04 6.5861e-06

3.1069e-02 1.7812e-03 1.1080e-04 6.9223e-06

1.5365e-02 8.6624e-04 5.4148e-05 3.3829e-06

1.6810e-02 9.5187e-04 5.9260e-05 3.7026e-06

1.7866e-02 1.0172e-03 6.3105e-05 3.9437e-06

0.00 0.05 0.09

Results in [9]

0.10055269 0.02759534 0.01007834 0.00258299 0.09997296 0.03218424 0.01167102 0.00299572 0.09944067 0.03591410 0.01297367 0.00334044

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Table 12. In solution of Example 8, the maximum absolute errors for 𝜹 = 𝟎. 𝟓𝜺 and 𝜼 = 𝟎. 𝟓𝜺 𝜀\𝑁 8 32 128

512

8 32 128 512

𝜆₁=1

6 , 𝜆₂=1

3 𝜆1= 1

12 , 𝜆2= 5 12 10^-1

10^-2 10^-3 10^-4 10^-5 10^-6

4.7078e-03 2.9344e-04 1.8125e-05 1.1323e-06

1.5855e-02 2.4000e-03 2.4806e-04 1.4485e-05

1.9105e-02 5.1245e-03 1.0544e-03 1.3676e-04

1.9443e-02 5.4565e-03 1.3836e-03 3.2142e-04

1.9477e-02 5.4898e-03 1.4167e-03 3.5453e-04

1.9480e-02 5.4932e-03 1.4201e-03 3.5784e-04

3.2577e-03 2.1561e-04 1.3525e-05 8.4554e-07

1.6542e-02 2.4538e-03 1.7806e-04 1.1249e-05

1.9798e-02 5.1791e-03 1.0581e-03 1.0960e-04

2.0136e-02 5.5112e-03 1.3872e-03 3.2165e-04

2.0170e-02 5.5446e-03 1.4204e-03 3.5476e-04

2.0174e-02 5.5479e-03 1.4237e-03 3.5807e-04

5. DISCUSSIONS AND CONCLUSION

We have implemented a numerical scheme using an exponentially fitted spline scheme for the solution of SPDDE along with the delay and also advance parameters whose answers show layer behaviour on the left or right end interval. This technique manages the perturbation parameter and also the fast changes that take place in the boundary layer region. This method gives good results in both cases i.e., ℎ ≤ 𝜀 𝑎𝑛𝑑 𝜀 <<

ℎ. To validate the method, we have solved model examples with left –end and right-end boundary layer by taking various values of 𝑁, 𝜀, 𝛿 𝑎𝑛𝑑 𝜂. To support and strengthen the method, numerical results are actually compared to the outcomes of Kadalbajoo as well as Sharma [9]. It is monitored from the results that, this strategy estimates the exact solution exceptionally well and also produce a precise treatment. Coming from the graphs, we noted that, when the solution of the boundary value problem exhibits layer behaviour on the left wing side, the affect of delay or advance on the solution in the boundary layer region is very small, while in the exterior region it is actually considerable, i.e., the rise in the delay increases the width of exterior region while the rise in the advance lowers the width of exterior region (Fig: 1-4). When the solution of the boundary value problem shows layer behaviour on the right side, the solution in the boundary layer region in addition to the outer region is affected by the changes in delay and also advance. The thickness of the layer rises as the size of the delay increases while it lowers as the size of the advance increases (Fig: 5-9).

REFERENCES

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