Numerical methods for singularly perturbed differential ... methods for DEs.pdfsingularly perturbed...

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Numerical methods for singularly perturbeddifferential equations with turning points

J. Christy Roja

Assistant ProfessorDepartment of Mathematics

Bharathidasan UniversityTiruchirappalli-620 024

June 5, 2018

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 1 / 74

Plan of talk

1 Introduction

2 Mathematical model for a turning point problem

3 Applications

4 Problems studied in the thesis

5 References

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 2 / 74

Introduction to Singular Perturbation Problems (SPPs)

The birth of the SPPs was introduced by Prandtl at the ThirdInternational Congress of Mathematicians in Heidelberg in 1904and it was reported in the proceedings of the conference.Many practical problems, such as the mathematical boundarylayer theory or approximation of solutions of various problems aredescribed by differential equations involving large or smallparameters.The solutions of SPPs have non-uniform behavior. That is, thereare thin layer(s) (boundary layer region) where the solution variesrapidly while away from the layer(s) (outer region) the solutionbehaves regularly and varies slowly.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 3 / 74

Let Pε denote the original problem and uε be its solution.Let P0 denote the reduced problem of Pε (setting ε = 0 in Pε) and u0be its solution.Then the problem Pε is called a Singular Perturbation Problem(SPP)if and only if uε does not converge uniformly to u0 in the entire domainof the definition of the problem. Otherwise the problem is calledRegular Perturbation Problem (RPP).

Example 1 (Regular Perturbation Problem)

Pε :

u′ε(x) = −εuε(x), x ∈ (0,1],

uε(0) = 1, 0 < ε 1.

P0 : u′0(x) = 0, x ∈ (0,1],u0(0) = 1,

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 4 / 74

Example 2 (Singular Perturbation Problem)

Pε :

εu′ε(x) = −uε(x), x ∈ (0,1],

uε(0) = 1, 0 < ε 1.

P0 : u0(x) = 0, x ∈ [0,1],

The exact solution is given by

uε(x) = exp(−x/ε).

Note that,

limε→0

limx→0

uε(x) = 1,

limx→0

limε→0

uε(x) = 0.

That is, uε(x) does not converge uniformly to the reduced problemsolution on [0,1].

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 5 / 74

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x − axis

− a

xis

(x)=e−x/ε

Figure: Graph of the Solution for ε = 0.01

The solution changes very rapidly near the neighborhood of x = 0.This neighborhood is called a boundary layer.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 6 / 74

Numerical Methods for SPPs

Often these mathematical problems are extremely difficult (or evenimpossible) to solve exactly and in these circumstances approximatesolutions are necessary. One can obtain an approximate solutionthrough the use of perturbation methods.In general, regular numerical methods like Euler method, Runge Kuttamethods, finite difference methods, etc cannot be applied to these SPPs.

For Example

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

−1

−0.5

0

0.5

1

x − axis

y −

axis

Numerical Solutio for N=50 and ε=0.01

Exact Solution for ε=0.01

Figure: Euler Method for Example (2)

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 7 / 74

The disadvantage in the classical numerical methods (finitedifference/ finite element) is due to the nature of the coefficients.That is, inaccurate solution due to perturbation parameter.Classical numerical methods on equidistant grids yieldsatisfactory numerical solution for singularly perturbed boundaryvalue problems only if one uses an unacceptably large number ofgrid points.In order to overcome this difficulty we apply numerical methods onappropriate meshes like Shishkin mesh, Bakhvalov mesh,Bakhvalov Shishkin mesh etc.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 8 / 74

Introduction to Singularly Perturbed Turning Point Problems(SPTPPs)

The main difference between singular perturbation problem andsingularly perturbed turning point problem is the coefficient of theconvection term vanishes inside the domain of the differentialequation.If the turning point occur at the interior of the domain, then theproblem is called as an interior turning point problem, otherwise itis a boundary turning point problem.If the velocity distribution is linear, then the problem is known as asimple turning point problem, otherwise it is a multiple turningpoint problem.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 9 / 74

Mathematical model for a turning point problem

Consider the one dimensional equation [3] which describes a quantummechanical particle in a potential V (x)(

−ε2 d2

dx2 + V (x)− E)

y(x) = 0,

where V (x) is the potential energy of the particle andE is the total energy of the particle

For this equation , Q(x) = V (x)− E , so Q(x) vanishes at pointswhere V (x) = E and these are called turning points.The classical orbit of a particle in the potential V (x) is confined tothe regions where V (x) ≤ E .The particle moves until it reaches a point where V = E and thenit stops, turns around and moves of in the opposite direction.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 10 / 74

Applications

SPTPPs occur in the modelling of following problems.

Modeling of steady and unsteady viscous flow problems with largeReynolds number

Navier Stokes flows with large Reynolds numbers

Magneto-hydrodynamic duct problems at high Hartman numbers

Heat transport problem with large Peclet numbers

One dimensional version of stationary convection-diffusion problemswith a dominant convective term

Speed field that changes its sign in the catch basin

Geophysics and modeling thermal boundary layers in laminar flow.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 11 / 74

A typical linear turning point problem in one dimension[12] is given by

Example 3

−εu′′

(x) + xb(x)u′(x) + c(x)u(x) = f (x), x ∈ (−1,1), u(−1) = u(1) = 0

under the following assumptions:(i) b(x) 6= 0 on [−1,1] (ii) c(x) ≥ 0, c(0) > 0.

The location of any boundary layer(s) depends on the sign of theconvection term.

From our experience, we expect a boundary layer at x = −1 if thecoefficient of the convection term xb(x) is negative at x = −1, and aboundary layer at x = 1 if the same coefficient is positive at x = 1.

If b(x) is positive on [−1,1], we have xb(x)|x=−1 < 0 and xb(x)|x=1 > 0.

Consequently, if b is positive on [−1,1], then the solution u has twoboundary layers at x = 1 and x = −1 otherwise the solution has interiorlayer at x = 0.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 12 / 74

Example 4 (Exhibiting layers at the boundary)

Consider the BVP

εu′′(x)− 2(2x − 1)u′(x)− 4u(x) = 0 ∀ x ∈ (0,1)

u(0) = 1, u(1) = 1

The exact solution is given by

u(x) = e−2x(1−x)/ε

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 13 / 74

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

u

Figure: Exact solution of example 4 for ε = 2−2 to ε = 2−10 and N = 1024

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 14 / 74

Example 5 (Exhibiting layers at the interior)

Consider the BVP

εu′′(x) + 2xu′(x) = 0 ∀ x ∈ (−1,1)

u(−1) = −1, u(1) = 1

The exact solution is given by

u(x) = erf (x/√ε)

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 15 / 74

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

x

u

Figure: Exact solution of example 5 for ε = 2−2 to ε = 2−10 and N = 1024

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 16 / 74

Numerical methods studied in the thesis

In the present thesis, motivated by the works of[1, 4, 6, 9, 11, 16, 17, 19, 20, 21], two methods are given namely

Parameter Uniform Finite DifferenceMethod(PUFDM)Variable Mesh Spline ApproximationMethod(VMSAM)

for various singularly perturbed turning point problems.

The PUFDM and VMSAM are discussed for Problem class I,where as the PUFDM is applied for Problem classes II to V.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 17 / 74

Problem class I: Second order SPTPPs with Robin boundaryconditions

Find u ∈ C1(Ω = [−1,1]) ∩ C2(Ω = (−1,1)) such that

Lu ≡ εu′′(x) + a(x)u′(x)− b(x)u(x) = f (x), ∀ x ∈ Ω (1)

with Robin boundary conditionsB1u(−1) = β1u(−1)− εβ2u′(−1) = A (2)B2u(1) = γ1u(1) + εγ2u

′(1) = B

and the asssumptionsa(0) = 0, a

′(0) < 0, |a(x)| ≤ α0 > 0,0 < β0 ≤ b(x),

α0 < β0, |a′(x)| ≥ |a

′(0)|2∀x ∈ Ω, β1, β2 ≥ 0,

β1 − εβ2 > 0, γ2 ≥ 0 & γ1 > 0.

(3)

where ε(0 < ε 1) is a small positive parameter, a(x),b(x) and f (x)are sufficiently smooth functions on Ω.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 18 / 74

Theorem 6 (Minimum Principle)

Let L be the differential operator defined in (1) and v ∈ C1(Ω) ∩ C2(Ω).If B1v(−1) ≥ 0, B2v(1) ≥ 0 and Lv ≤ 0 ∀ x ∈ Ω, then v(x) ≥ 0 ∀x ∈ Ω.

Lemma 7 (Stability Result)

If u ∈ C0(Ω) ∩ C2(Ω), then

|u(x)| ≤ C max max|B1u|, |B2u|, ||Lu||x∈Ω , ∀x ∈ Ω.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 19 / 74

From now on we shall denote the subdomains of Ω = [−1,1] asΩ1 = [−1,−δ], Ω2 = [−δ, δ] and Ω3 = [δ, 1], 0 < δ ≤ 1/2. The choice ofδ = 1/2 can be found in [4].

Lemma 8

Let u be the solution of (1)-(3). Then

||u(k)|| ≤ Cε−(k) max||f ||, ||u||, k = 1,2||u(3)|| ≤ Cε−(3) max||f ||, ||f ′||, ||u||,

∀x ∈ Ω1 ∪ Ω3, where C depends on ||a||, ||a′||, ||b|| and ||b′||.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 20 / 74

The following lemma gives estimates for u and its derivatives in theinterval Ω2 which includes the turning point x = 0.

Lemma 9

Let u be the solution of (1)-(3). Then

||u(k)(x)|| ≤ C, ∀x ∈ Ω2,

where C depends on ||a||, ||a′ ||, ||b||, ||b′ ||, ||f ||, ||f ′ || and β.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 21 / 74

To derive ε - uniform error estimates we require sharper bounds of thesolution and its derivatives. For this we use Shishkin decomposition of thesolution u as

u = v + w .

Here v is the solution of the problem

Lv = f (4)

β1v(−1)− εβ2v′(−1) = β1v0(−1)− εβ2v

0(−1) + ε(β1v1(−1)− εβ2v′

1(−1)),

γ1v(1) + εγ2v′(1) = γ1v0(1) + εγ2v

0(1) + ε(γ1v1(1) + εγ2v′

1(1))

where v = v0 + εv1 + ε2v2.Also v0 and v1 are defined respectively, to be the solutions of the reducedproblem:

av′

0 − bv0 = f and av′

1 − bv1 = −v′′

0 (5)

and v2 is the solution of the problem similar to that defining u

Lv2 = −v′′

1 , (6)

β1v2(−1)− εβ2v′

2(−1) = 0, γ1v2(1) + εγ2v′

2(1) = 0.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 22 / 74

The singular component w is the solution of the homogeneous problem

Lw = 0, (7)

β1w(−1)− εβ2w′(−1) = (β1u(−1)− εβ2u

′(−1))− (β1v(−1)− εβ2v

′(−1)),

γ1w(1) + εγ2w′(1) = (γ1u(1) + εγ2u

′(1))− (γ1v(1) + εγ2v

′(1)).

Lemma 10

The smooth component v and singular component w and their derivativessatisfy the bounds for k=0,1,2,3

||v (k)(x)|| ≤ C(1 + ε2−k ), ∀ x ∈ Ω1 ∪ Ω3 and

|w (k)(x)| ≤

Cε−k e−α(1+x)/ε, ∀ x ∈ Ω1

Cε−k e−α(1−x)/ε, ∀ x ∈ Ω3

where |a(x)| ≥ α > 0, ∀x ∈ Ω1 ∪ Ω3.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 23 / 74

Theorem 11

The smooth component v and singular component w and theirderivatives satisfy the bounds for k=0,1,2,3

||v (k)(x)|| ≤ C(1 + ε2−k ), and|w (k)(x)| ≤ Cε−k (e−α(1+x)/ε + e−α(1−x)/ε), ∀ x ∈ Ω.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 24 / 74

Finite Difference SchemeThe problem (1)-(2) is discretized using classical finite differencescheme on piecewise uniform meshes (Shiskin mesh).The domain Ω is divide into three subintervals ΩL = [−1,−1 + τ ],ΩC = [−1 + τ,1− τ ] and ΩR = [1− τ,1] such thatΩ = ΩL ∪ ΩC ∪ ΩR.

The transition parameter τ is chosen to be min

12,2εlnNα

.

The domain ΩN is obtained by putting a uniform mesh with N/4mesh elements in both ΩL and ΩR and a uniform mesh with N/2elements in ΩC .

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 25 / 74

The resulting fitted finite difference scheme is to find U(xi) fori = 0,1,2, · · ·N such that for xi ∈ ΩN ,

LNU(xi) := εδ2U(xi) + a(xi)D∗U(xi)− b(xi)U(xi), (8)BN

1 U(x0) = β1U(x0)− εβ2D+U(x0)

BN2 U(xN) = γ1U(xN) + εγ2D−U(xN), (9)

where D+U(xi) =U(xi+1)− U(xi)

xi+1 − xi, D−U(xi) =

U(xi)− U(xi−1)

xi − xi−1,

δ2U(xi) =D+U(xi)− D−U(xi)

(xi+1 − xi−1)/2and

D∗U(xi) =

D+U(xi) if a(xi) > 0D−U(xi) if a(xi) < 0

.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 26 / 74

Theorem 12 (Discrete minimum principle)

Let LN be the finite difference operator defined in (8)- (9) and let ΩN bean arbitrary mesh of N + 1 mesh points. If ψ is any mesh functiondefined on this mesh such thatBN

1 ψ(x0) ≥ 0, BN2 ψ(xN) ≥ 0 and LNψ(xi) ≤ 0, for i = 1(1)N − 1 then

ψ(xi) ≥ 0, ∀ xi ∈ ΩN .

Lemma 13 (Discrete stability result)

Consider the scheme (8)- (9) to problem (1)-(3). If ψ(xi) is any meshfunction then, for all xi ∈ ΩN

|ψ(xi)| ≤ Cmax|BN1 ψ(x0)|, |BN

2 ψ(xN)|, max1≤i≤N−1

|LNψ(xi)|.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 27 / 74

Analogous to the continuous case, the discrete solution U can bedecomposed as

U = V + W ,

where V and W are respectively the solutions of the problems

LNV = f (xi), xi ∈ ΩN , (10)β1V (−1)− εβ2D+V (−1) = β1v(−1)− εβ2v

′(−1),

γ1V (1) + εγ2D−V (1) = γ1v(1) + εγ2v′(1)

and

LNW = 0, xi ∈ ΩN , (11)β1W (−1)− εβ2D+W (−1) = β1w(−1)− εβ2w

′(−1),

γ1W (1) + εγ2D−W (1) = γ1w(1) + εγ2w′(1).

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 28 / 74

Lemma 14

The error in the smooth component of the numerical solution is bounded as

|(V − v)(xi )| ≤ CN−1, for all xi ∈ ΩN ,

where v is the solution of (4) and V is the solution of (10).

Lemma 15

The error in the singular component of the numerical solution is bounded as

|(W − w)(xi )| ≤ CN−1lnN,∀xi ∈ ΩN ,

where w is the solution of (7) and W is the solution of (11).

Theorem 16If u is the solution of the problem (1)− (3) and U is the correspondingnumerical solution using the method outlined in (8)-(9), then we have

sup0<ε≤1

||U − u||ΩN ≤ CN−1lnN ∀ N ≥ 4,

where the constant C is independent of ε and N.J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 29 / 74

The following example is given to illustrate the numerical method. Weuse the double mesh principle given as in [5] to estimate the error andcompute the experimental rate of convergence of the numericalmethod.Define the double mesh differences to be

DNε =

maxxi∈ΩN

|UN(xi)− U2N(xi)|, and DN = max

εDNε

where UN(xi) and U2N(xi) respectively, denote the numerical solutionobtained using N and 2N mesh intervals. Further, we calculate theRobust order of convergence as

pN = log2

(DN

D2N

).

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 30 / 74

The following example has a turning point at x = 1/2.

Example 17

εu′′

(x)− 2(2x − 1)u′(x)− 4u(x) = 4(4x − 15), x ∈ (0,1)

u(0)− εu′(0) = 1, u(1) + εu

′(1) = 1

Table: Values of DN , pN for the solution u for Example (17)

Number of mesh points N64 128 256 512 1024

DN 9.2291e-1 5.6755e-1 3.7305e-1 2.2919e-1 1.3447e-1pN 7.0144e-1 6.0539e-1 7.0282e-1 7.6921e-1 -

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 31 / 74

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 16

7

8

9

10

11

12

13

14

x

u

eps=2−2

eps=2−3

eps=2−3

eps=2−4

Figure: Solution graph of Example 17 for various values of ε(eps) andN = 64

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 32 / 74

Variable mesh spline approximation method

Let the positive constants h and K be known.We construct a non uniform mesh on ΩL as follows:

hj = hj−1 + K

[hj−1

ε

]min(h2

j−1, ε), j = 2(1)N/4.

Let q =

N/4∑j=1

hj , q =τ

q, hj = qhj , j = 1(1)N/4

An uniform mesh on Ωc , is defined as

hj =4(1− τ)

N, j = N/4 + 1 (1) 3N/4.

As in ΩL, a nonuniform mesh is constructed on ΩR ashj = hN+1−j , j = 3N/4 + 1(1)Nand define x0 = −1, xj = xj−1 + hj , j = 1(1)N.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 33 / 74

The cubic spline interpolating polynomial will have the followingproperties:

(i) Sj(x) coincides with the polynomial of degree three on eachinterval [xj−1, xj ], j = 1,2, · · · ,N

(ii) Sj(x) ∈ C2[0,1],(iii) Sj(xj) = u(xj), j = 0(1)N.Then we have the cubic spline functions,

Sj(x) =(xj − x)3

6hjMj−1 +

(x − xj−1)3

6hjMj +

(uj−1 −

h2j Mj−1

6

)(xj − x

hj

)+(

uj −h2

j Mj

6

)(x − xj−1

hj

),

where, x ∈ [xj−1, xj ], hj = xj − xj−1, j = 1(1)Nand Mj = S

′′

j (xj), j = 0(1)N.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 34 / 74

We obtain the difference scheme as

LNuj = Qfj , j = 1(1)N − 1. (12)

where, LNuj = r−j uj−1 + r cj uj + r+

j uj+1, Qfj = q−j fj−1 + qcj fj + q+

j fj+1

r−j =2hj + hj+1

6(hj + hj+1)aj−1 +

hj+1

3hjaj −

h2j+1

6hj (hj + hj+1)aj+1 +

hj

6bj−1 −

ε

hj,

r+j =

h2j

6hj+1(hj + hj+1)aj−1 −

hj

3hj+1aj −

2hj+1 + hj

6(hj + hj+1)aj+1 +

hj+1

6bj+1 −

ε

hj+1,

rcj = −

hj + hj+1

6hj+1aj−1 −

h2j+1 − h2

j

3hj hj+1aj +

hj+1 + hj

6hjaj+1 +

hj+1 + hj

3bj +

ε

hj+

ε

hj+1,

q−j = −hj

6, q+

j = −hj+1

6, qc

j = −hj + hj+1

3

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 35 / 74

We approximate the first derivative by centered finite differenceoperator:B1u(x0) ≡ β1u(x0)− εβ2D0u(x0) = A andB2u(xN) ≡ γ1u(xN) + εγ2D0u(xN) = BThat is,

β1u0 −εβ2

2h0[u1 − u−1] = A (or) u−1 =

−2β1h0

εβ2u0 + u1 +

2h0Aεβ2

(13)

and

γ1uN +εγ2

2hN[uN+1−uN−1] = B (or) uN+1 =

−2γ1hN

εγ2uN + uN−1 +

2hNBεγ2(14)

where u(x−1) and u(xN+1) are the functional values at x−1 and xN+1.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 36 / 74

The nodes x−1 and xN+1 lie outside the interval [0,1] and are calledfictitious nodes. The values u(x−1) and u(xN+1) may be eliminated byassuming that the difference equation (12) holds also for i = 0 andi = N, that is at the boundary points x0 and xN . Substituting the valuesof u−1 and uN+1 from (13) and (14) into the equations (12) for i = 0and i = N, we get respectively,

BN1 ≡

[r c0 −

2β1h0r−0εβ2

]u0 + [r+

0 + r−0 ]u1 = q−0 f−1 + qc

0 f0 + q+0 f1 −

2h0Ar−0εβ2

(15)

and

BN2 ≡

[r cN −

2γ1hN r+N

εγ2

]uN + [r−N + r+

N ]uN−1 = q−N fN−1 + qc

N fN + q+N

fN+1 −2hNBr+

Nεγ2

. (16)

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 37 / 74

Theorem 18Let uj, j = 0(1)N, be a set of values of the approximate solution tou(x) of (1)-(3), obtained by using (12), (15) and (16). Then there arepositive constants C and α (independent of h and ε) such that thefollowing estimate holds:

maxj|uj − u(xj)| ≤ Ch2

c

[exp

−α(1 + xj)

ε

+ exp

−α(1− xj)

ε

]where hc = max

jhj = a constant .

Table: Values of DN , pN for the solution components u for the above Example(17)

Number of mesh points N64 128 256 512 1024

DN 1.4089 3.0437e-1 7.3039e-2 1.7835e-2 5.2069e-3pN 2.2107 2.0591 2.0339 1.7762 -

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 38 / 74

Problem class II: Weakly coupled system of second order SPTPPs

Find u1,u2 ∈ Y = C0(Ω) ∩ C2(Ω) such that

Lu(x) =

L1u(x) = εu

′′

1 (x) + a1(x)u′

1(x) + b11(x)u1(x)

+b12(x)u2(x) = f1(x), x ∈ Ω,

L2u(x) = εu′′

2 (x) + a2(x)u′

2(x) + b21(x)u1(x)

+b22(x)u2(x) = f2(x), x ∈ Ω,

(17)

u1(−1) = l1, u2(−1) = l2, u1(1) = l3, u2(1) = l4 (18)b12 ≥ 0, b21 ≥ 0, b11 + b12 ≤ 0, b22 + b21 ≤ 0|ak (x)| ≤ αk > 0, for 0 < |x | ≤ 1, ak (0) = 0, a

k (0) < 0,and |a′k (x)| ≥ |a′k (0)|/2 ∀x ∈ Ω, for k = 1,2

(19)

where the functions a1(x),a2(x),b11(x),b12(x),b21(x),b22(x), f1(x)and f2(x) are sufficiently smooth on Ω.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 39 / 74

Finite difference scheme for the problem (17)-(18)

The fitted finite difference scheme is to find U(xi) = (U1(xi),U2(xi))T

for i = 0,1,2, · · ·N such that for xi ∈ ΩN ,

LN1 U(xi) := εδ2U1(xi) + a1(xi)D∗U1(xi) + b11(xi)U1(xi) (20)

+b12(xi)U2(xi) = f1(xi) i = 1(1)N − 1,LN

2 U(xi) := εδ2U2(xi) + a2(xi)D∗U2(xi) + b21(xi)U1(xi) (21)+b22(xi)U2(xi) = f2(xi), i = 1(1)N − 1,U1(x0) = u1(−1),U1(xN) = u1(1),

U2(x0) = u2(−1),U2(xN) = u2(1).

where D+Uj(xi) =Uj(xi+1)− Uj(xi)

xi+1 − xi, D−Uj(xi) =

Uj(xi)− Uj(xi−1)

xi − xi−1,

δ2Uj(xi) =D+Uj(xi)− D−Uj(xi)

(xi+1 − xi−1)/2and

D∗Uj(xi) =

D+Uj(xi) if aj(xi) > 0D−Uj(xi) if aj(xi) < 0

.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 40 / 74

Theorem 19

Let u(x) = (u1(x),u2(x))T , for all x ∈ Ω be the solution of (17)-(19)andlet U(xi) = (U1(xi),U2(xi))T , for all xi ∈ ΩN be the numerical solutionof problem (20)-(21). Then we have

sup0<ε≤1

||U1−u1||ΩNε≤ CN−1(lnN)2 and sup

0<ε≤1||U2−u2||ΩN ≤ CN−1(lnN)2.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 41 / 74

Example 20

Consider the following system of singularly perturbed turning pointproblem

εu′′

1 (x)− 2(2x − 1)u′

1(x)− 9u′

1(x) + 2u2(x) = 0, x ∈ (0,1)

εu′′

2 (x)− 4(2x − 1)u′

2(x)− 6u′

2(x) + u1(x) = 0, x ∈ (0,1)

u1(0) = 1, u2(0) = 1, u1(1) = 1, u2(1) = 1.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 42 / 74

Table: Values of DN1 , pN

1 and DN2 , pN

2 for the solution components U1 and U2respectively for Example 20

Number of mesh points N64 128 256 512 1024

DN1 2.3351e-2 1.4544e-2 8.9904e-3 5.2098e-3 2.9596e-3

pN1 6.8307e-1 6.9394e-1 7.8715e-1 8.1584e-1 -

DN2 8.0236e-2 4.1792e-2 2.1304e-2 1.1884e-2 6.1696e-3

pN2 9.4103e-1 9.7207e-1 8.4213e-1 9.4575e-1 -

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 43 / 74

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

u

u1

u2

Figure: Solution graph of Example 20 for ε = 2−4 and N = 27

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 44 / 74

101

102

103

104

10−4

10−3

10−2

10−1

100

N

Err

or(

u1)

ε = 2−1

ε = 2−4

ε = 2−7

ε = 2 −10

N−1

(lnN)2

101

102

103

104

10−4

10−3

10−2

10−1

100

N

Err

or(

u2)

ε = 2−1

ε = 2−4

ε = 2−7

ε = 2 −10

N−1

(lnN)2

Figure: Maximum pointwise errors as a function of N and ε for the solution U1and U2 for Example 20

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 45 / 74

Problem class III: Weakly coupled system of second order SPTPPswith Robin boundary conditions

Find u = (u1,u2)T ∈ Y = C1(Ω) ∩ C2(Ω) such that

Lu(x) =

L1u(x) = εu

′′

1 (x) + a1(x)u′

1(x) + b11(x)u1(x)

+b12(x)u2(x) = f1(x), x ∈ Ω,

L2u(x) = εu′′

2 (x) + a2(x)u′

2(x) + b21(x)u1(x)

+b22(x)u2(x) = f2(x), x ∈ Ω,

(22)

with the boundary conditionsB10u1(−1) ≡ β10u1(−1)− εβ11u

1(−1) = A1,

B11u1(1) ≡ γ11u1(1) + εγ12u′

1(1) = B1,

B20u2(−1) ≡ β20u2(−1)− εβ21u′

2(−1) = A2,

B21u2(1) ≡ γ21u2(1) + εγ22u′

2(1) = B2,

(23)

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 46 / 74

and the assumptionsb12 ≥ 0, b21 ≥ 0, b11 + b12 ≤ b1 < 0, b22 + b21 ≤ b2 < 0|ak (x)| ≤ αk > 0, for 0 < |x | ≤ 1, ak (0) = 0, a

k (0) < 0,αk + bk < 0 and |a′k (x)| ≥ |a′k (0)|/2 ∀x ∈ Ω, for k = 1,2βj0, βj1 ≥ 0, βj0 − εβj1 ≥ 0, γj1, γj2 ≥ 0, j = 1,2.

(24)

where the functions a1(x),a2(x),b11(x),b12(x),b21(x),b22(x),f1(x) and f2(x) are sufficiently smooth on Ω,

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 47 / 74

Finite difference scheme for the problem (22)-(23)

The fitted finite difference scheme is to find U(xi) = (U1(xi),U2(xi))T

for i = 0,1,2, · · ·N such that for xi ∈ ΩN ,

LNU(xi) =

LN

1 U(xi) := εδ2U1(xi) + a1(xi)D∗U1(xi) + b11(xi)U1(xi)

+b12(xi)U2(xi) = f1(xi), i = 1(1)N − 1,LN

2 U(xi) := εδ2U2(xi) + a2(xi)D∗U2(xi) + b21(xi)U1(xi)

+b22(xi)U2(xi) = f2(xi), i = 1(1)N − 1,

(25)

BN

10U1(x0) ≡ β10U1(x0)− εβ11D+U1(x0) = A1,

BN11U1(xN) ≡ γ11U1(xN) + εγ12D−U1(xN) = B1,

BN20U2(x0) ≡ β20U2(x0)− εβ21D+U2(x0) = A2,

BN21U2(xN) ≡ γ21U2(xN) + εγ22D−U2(xN) = B2,

(26)

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 48 / 74

Theorem 21

Let u(x) = (u1(x),u2(x))T , for all x ∈ Ω be the solution of (22)-(24)andlet U(xi) = (U1(xi),U2(xi))T , for all xi ∈ ΩN be the numerical solutionof problem (25)-(26). Then we have

sup0<ε≤1

||U1 − u1||ΩNε≤ CN−1lnN and sup

0<ε≤1||U2 − u2||ΩN ≤ CN−1lnN.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 49 / 74

Example 22

Consider the following system of singularly perturbed turning pointproblem

εu′′

1 (x)− 7(2x − 1)u′

1(x)− 10u1(x) + 2u2(x) = −ex , x ∈ (0,1)

εu′′

2 (x)− 3(2x − 1)u′

2(x)− 7u2(x) + 3u1(x) = x + 5, x ∈ (0,1)

u1(0)− εu′1(0) = 2, u2(0)− εu′2(0) = 2,

u1(1) + εu′

1(1) = 2, u2(1) + εu′

2(1) = 2.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 50 / 74

Table: Values of DN1 , pN

1 and DN2 , pN

2 for the solution components U1 and U2respectively for Example 22

Number of mesh points N64 128 256 512 1024

DN1 5.3386e-2 3.3316e-2 2.0917e-2 1.2528e-2 6.3892e-3

pN1 6.8027e-1 6.7153e-1 7.3946e-1 9.7149e-1 -

DN2 5.7580e-2 3.4862e-2 2.0375e-2 1.1785e-2 6.5330e-3

pN2 7.2390e-1 7.7486e-1 7.8991e-1 8.5108e-1 -

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 51 / 74

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

x

u

u1

u2

Figure: Solution graph of Example 22 for ε = 2−4 and N = 27

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 52 / 74

101

102

103

104

10−3

10−2

10−1

N

Err

or(

u1)

ε = 2−4

ε = 2−7

ε = 2−10

ε = 2 −13

N−1

(lnN)

101

102

103

104

10−3

10−2

10−1

N

Err

or(

u2)

ε = 2−1

ε = 2−4

ε = 2−7

ε = 2 −10

N−1

(lnN)

Figure: Maximum pointwise errors as a function of N and ε for the solution u1and u2 for Example 22

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 53 / 74

Problem class IV: Third order SPTPPs

Find u ∈ C1(Ω) ∩ C3(Ω) such thatLu = εu

′′′(x) + a(x)u

′′(x)− b(x)u

′(x) + c(x)u(x) = f (x), x ∈ Ω,

u(−1) = l1, u′(−1) = l2, u

′(1) = l3,

(27)

|a(x)| ≤ α > 0, for 0 < |x | ≤ 1, a(0) = 0, a

′(0) < 0,

β0 ≥ b(x) ≥ β0 > 0, γ0 ≥ c(x) ≥ γ0 > 0, α < β0 − γ0,

and |a′(x)| ≥ |a′(0)|/2 ∀x ∈ Ω,

(28)

where a(x),b(x), c(x) and f (x) are smooth functions on Ω.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 54 / 74

The above problem is equivalent to the following problem:

Find u = (u1,u2)T , u1,u2 ∈ C0(Ω) ∩ C2(Ω) such that

Lu =

L1u := u2(x)− u

1(x) = 0, x ∈ Ω,

L2u := εu′′

2 (x) + a(x)u′2(x)− b(x)u2(x)

+c(x)u1(x) = f (x), x ∈ Ω,

(29)

u1(−1) = l1, u2(−1) = l2, u2(1) = l3. (30)

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 55 / 74

Finite difference scheme for the problem (29)-(30)

The fitted finite difference scheme is to find U(xi) = (U1(xi),U2(xi))T

for i = 0,1,2, · · ·N such that for xi ∈ ΩN ,

LN1 U(xi) := U2(xi)− D−U1(xi) = 0, i = 1(1)N,

LN2 U(xi) := εδ2U2(xi) + a(xi)D∗U2(xi)− b(xi)U2(xi) (31)

+c(xi)U1(xi) = f (xi), i = 1(1)N − 1,U1(x0) = u1(x0), U2(x0) = u2(x0), U2(xN) = u2(xN).

where D+Uj(xi) =Uj(xi+1)− Uj(xi)

xi+1 − xi, D−Uj(xi) =

Uj(xi)− Uj(xi−1)

xi − xi−1,

δ2Uj(xi) =D+Uj(xi)− D−Uj(xi)

(xi+1 − xi−1)/2and

D∗Uj(xi) =

D+Uj(xi) if a(xi) > 0D−Uj(xi) if a(xi) < 0

.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 56 / 74

Theorem 23

Let u(x) = (u1(x),u2(x))T , for all x ∈ Ω be the solution of (29)-(30)andlet U(xi) = (U1(xi),U2(xi))T , for all xi ∈ ΩN

ε be the numerical solutionof problem (31). Then we have

sup0<ε≤1

||U1 − u1||ΩNε≤ CN−1lnN and sup

0<ε≤1||U2 − u2||ΩN

ε≤ CN−1lnN.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 57 / 74

Example 24

Consider the following singularly perturbed turning point problem

εu′′′

(x)− 5xu′′

(x)− (x + 4)u′(x) + (2 + x)u(x) = −ex , x ∈ (−1,1)

u(−1) = 1, u′(−1) = 1, u

′(1) = 1.

Table: Values of DN1 , pN

1 and DN2 , pN

2 for the solution components U1 and U2respectively for Example 24

Number of mesh points N64 128 256 512 1024

DN1 2.7087e-2 1.5076e-2 8.0262e-3 4.0350e-3 2.0241e-3

pN1 8.4539e-1 9.0943e-1 9.9214e-1 9.9530e-1 -

DN2 6.7643e-2 4.5055e-2 2.5059e-2 1.2807e-2 7.7298e-3

pN2 5.8625e-1 8.4638e-1 9.6833e-1 7.2848e-1 -

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 58 / 74

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

1.2

x

U

U1

U2

Figure: Solution graph of Example 24 for ε = 2−4 and N = 27

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 59 / 74

101

102

103

104

10−5

10−4

10−3

10−2

10−1

N

Err

or(

u1)

ε = 2−4

ε = 2−8

ε = 2−12

ε = 2 −16

N−1

(lnN)

101

102

103

104

10−3

10−2

10−1

N

Err

or(

u2)

ε = 2−4

ε = 2−8

ε = 2−12

ε = 2 −16

N−1

(lnN)

Figure: Maximum pointwise errors as a function of N and ε for the solution u1and u2 for Example 24

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 60 / 74

Problem class V: Fourth order SPTPPs

Find u ∈ C2(Ω) ∩ C4(Ω) such thatLu = −εuiv (x)− a(x)u

′′′(x) + b(x)u

′′(x)

+c(x)u(x) = f (x), x ∈ Ω,

u(−1) = l1, u(1) = l2, u′′

(−1) = l3, u′′

(1) = l4,(32)

with the assumptions|a(x)| ≤ α > 0, for 0 < |x | ≤ 1, a(0) = 0,a′(0) < 0, β0 ≥ b(x) ≥ β0 > 0, γ0 ≥ c(x) ≥ γ0 > 0,

α < β0 − γ0, and |a′(x)| ≥ |a′(0)|/2 ∀x ∈ Ω,

(33)

where a(x),b(x), c(x) and f (x) are smooth functions on Ω.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 61 / 74

The above problem is equivalent to the following problem:

Find u = (u1,u2)T , u1,u2 ∈ C0(Ω) ∩ C2(Ω) such that

Lu =

L1u := u

′′

1 (x) + u2(x) = 0, x ∈ Ω,

L2u := εu′′

2 (x) + a(x)u′2(x)− b(x)u2(x)

+c(x)u1(x) = f (x), x ∈ Ω,

(34)

u1(−1) = l1, u1(1) = l2, u2(−1) = l3, u2(1) = l4. (35)

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 62 / 74

Finite difference scheme for the problem (34)-(35)

The fitted finite difference scheme is to find U(xi) = (U1(xi),U2(xi))T

for i = 0,1,2, · · ·N such that for xi ∈ ΩNε , is LN = (LN

1 ,LN2 ) where

LN1 U(xi) := δ2U1(xi) + U2(xi) = 0, i = 1(1)N − 1,

LN2 U(xi) := εδ2U2(xi) + a(xi)D∗U2(xi)− b(xi)U2(xi) (36)

+c(xi)U1(xi) = f (xi), i = 1(1)N − 1,U1(x0) = u1(x0), U1(xN) = u1(xN),

U2(x0) = u2(x0), U2(xN) = u2(xN),

where D+Uj(xi) =Uj(xi+1)− Uj(xi)

xi+1 − xi, D−Uj(xi) =

Uj(xi)− Uj(xi−1)

xi − xi−1,

δ2Uj(xi) =D+Uj(xi)− D−Uj(xi)

(xi+1 − xi−1)/2and

D∗Uj(xi) =

D+Uj(xi) if a(xi) > 0D−Uj(xi) if a(xi) < 0

.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 63 / 74

Theorem 25

Let u(x) = (u1(x),u2(x))T , for all x ∈ Ω be the solution of (34)-(35)andlet U(xi) = (U1(xi),U2(xi))T , for all xi ∈ ΩN be the numerical solutionof problem (36). Then we have

sup0<ε≤1

||U1 − u1||ΩNε≤ CN−1lnN and sup

0<ε≤1||U2 − u2||ΩN ≤ CN−1lnN

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 64 / 74

Example 26

Consider the following singularly perturbed turning point problem

−εuiv (x)+5xu′′′

(x)+(4+x)u′′

(x)+(2+x)u(x) = −exp(x), x ∈ (−1,1)

u(−1) = 1,u(1) = 1,u′′

(−1) = 1, u′′

(1) = 1.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 65 / 74

Table: Values of DN1 , pN

1 and DN2 , pN

2 for the solution components U1 and U2respectively for Example 26

Number of mesh points N64 128 256 512 1024

DN1 3.9254e-2 1.9943e-2 1.0050e-2 5.0450e-3 2.5274e-3

pN1 9.7697e-1 9.8861e-1 9.9433e-1 9.9717e-1 -

DN2 3.5852e-2 2.5491e-2 1.4071e-2 8.0757e-3 4.5981e-3

pN2 4.9207e-1 8.5727e-1 8.0104e-1 8.1256e-1 -

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 66 / 74

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10.8

0.9

1

1.1

1.2

1.3

1.4

1.5

x

u

u1

u2

Figure: Solution graph of Example 26 for ε = 2−4 and N = 27

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 67 / 74

101

102

103

104

10−3

10−2

10−1

N

Err

or(

u1)

e−4

e−7

e−10

e−13

N−1

(lnN)

101

102

103

104

10−3

10−2

10−1

N

Err

or(

u2)

e−4

e−7

e−10

e−13

N−1

(lnN)

Figure: Maximum pointwise errors as a function of N and ε for the solution u1and u2 for Example 26

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 68 / 74

Conclusion and Scope

Conclusion and Scope

Finite Difference method and Variable mesh spline approximationmethod are applied for problem class I. Finite Difference method isused solve the remaining problems.One can apply these method for other class of problems likemultiple turning point problems, turning point problem with interiorlayers, two parameter turning point problems, turning pointproblem with discontinuous source term, etc.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 69 / 74

List of Publications

List of Publications

(1) N. Geetha, A. Tamilselvan, V. Subburayan, Parameter uniformnumerical method for third order singularly perturbed turning pointproblems exhibiting boundary layers, Int. J. Appl. Comput. Math.,DOI 10.1007/s40819-015-0064-4. (2015) (Springer)

(2) N. Geetha, A. Tamilselvan, Numerical method for system ofsecond order singularly perturbed turning point problems withRobin boundary conditions, Procedia Engineering, 127 (2015),670 â 677. (Elsevier)

(3) N. Geetha, A. Tamilselvan, Variable mesh spline approximationmethod for solving second order singularly perturbed turning pointproblems with Robin boundary conditions, International Journal ofApplied and Computational Mathematics, DOI10.1007/s40819-015-0064-4. (2016). (Springer)

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 70 / 74

List of Publications

(4) N. Geetha, A. Tamilselvan, Parameter uniform numerical methodfor fourth order singularly perturbed turning point problemsexhibiting boundary layers, Ain Shams Engineering Journal(2016), http://dx.doi.org/10.1016/j.asej.2016.04.018. (Elsevier)

(5) N. Geetha, A. Tamilselvan, Numerical method for system ofsecond order singularly perturbed turning point problemsexhibiting boundary layers, Journal of Mathematical Modeling, Vol.6, No. 2, 2016, pp. 211-232.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 71 / 74

List of articles communicated for publication

List of articles communicated for publication

1 N. Geetha, A. Tamilselvan, Parameter uniform numerical methodfor second order singularly perturbed turning point problems withRobin boundary conditions.

2 J. Christy Roja, A. Tamilselvan, N. Geetha, A parameter- uniformsecond order Schwarz method for a weakly coupled system ofsingularly perturbed convection diffusion equations.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 72 / 74

Problems working on

N. Geetha, A. Tamilselvan, Numerical method for a two parametersingularly perturbed turning point problems.N. Geetha, A. Tamilselvan, Parameter uniform numerical methodfor singularly perturbed turning point problems with discontinuoussource term.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 73 / 74

References

[1] L. R. Abrahamsson, A priori estimates for solutions of singularperturbations with a turning point, Studies in applied mathematics56, 51-69 (1977).

[2] R. C. Ackerberg and R. E. O’Malley ,Boundary layer problemsexibiting resonance, Stdies in aplied mathematics XLIX (30),277-295 (1970).

[3] C. M. Bender, S. A. Orzag, Advanced Mathematical methods forScientist and Engineers, McGrawhill. New York, (1978).

[4] A. E. Berger, H. Han and R. B. Kellogg, A priori estimates andanalysis of a numerical method for a turning point problem,Mathematics of computation 42, 465-492,(1984).

[5] E. P. Doolan, J. J. H. Miller, W. H. A. Schilders, Uniform numericalmethods for problems with initial and boundary layers, Boole press,Dublin(1980).

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 73 / 74

[6] F. A. Howes, An asymptotic-numerical method for a model systemwith turning points, Journal of Computational and AppliedMathematics, 29, 343-356, (1990).

[7] T. C. Hanks, Model relating heat flow values near, and verticlevelocities of mass transport beneath, ocean rices, Geohys. Res.76, 537-544 (1971).

[8] H. Schlichting, Boundary-layer theory, McGRaw-hill, New York,(1979.)

[9] Kapil K. sharma, Pratima rai, Kailash C. Patidar, A review onsingularly perturbed differential equations with turning points andinteriror layers, Applied Mathematics and computation, 219,10575-10609, (2013).

[10] R. E. O’ Malley, Introduction to singular perturbations, AcademicPress, New York, (1974).

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 73 / 74

[11] Anthony Leung, A third order linear differential equation on thereal line with two turning points, Journal of differential equations29, 304-328, (1978).

[12] H. G. Roos, M. Stynes, L. Tobiska, Numerical methods forsingularly perturbed differential equations-convection-diffusion andflow problems, Springer, (2006).

[13] W. Wasow, Linear turning point theory, Springer Verlag, NewYork, (1984).

[14] A. M. Watts, A Singular perturbation problem with a turning point,Bull. Austral. Math. Soc, Vol 5, 61-73, (1971).

[15] P. A. Ferrell, Sufficient conditions for the uniform convergence of adifference scheme for a singularly perturbed turning point problem,SIAM J. on Numerical Analysis 25(3), 618-643 (1988).

[16] M. K. Kadalbajoo, K. C. Patidar, Variable mesh splineapproximation method for solving singularly perturbed turning point

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 73 / 74

problems having boundary layer(s), Computer Mathematics withApplications, Vol 42, 1439- 1453, (2001).

[17] A. Tamilselvan, N. Ramanujam, R. Mythili Priyadharshini and T.Valanarasu Parameter uniform numerical method a system ofcoupled singularly perturbed convection diffusion equations withmixed type boundary conditions, J. Appl. Math. & Informatics, Vol28, 109-130,(2010)

[18] S. Natesan, N. Ramanujam, A computational method for solvingsingularly perturbed turning point problems exhibiting twinboundary layers, Applied Mathematics and computation, Vol 93,259-275, (1998).

[19] S. Natesan, N. Ramanujam, Initial-Value Technique for SingularlyPerturbed Turning Point Problems Exhibiting Twin BoundaryLayers, Journal of Optimization Theory and Applications, 99(1),37-52, (1998).

[20] S. Natesan, J. Jayakumar, J. Vigo-Aguiar, Parameter uniformnumerical method for singularly perturbed turning point problems

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 73 / 74

exhibiting boundary layers, Journal computational and appliedMathematics, vol 158, 121-134, (2003)

[21] Jia-qi Mo, Zhao-hui Wen, A class of boundary value problems forthird order differential equation with a turning point, AppliedMathematics and Mechnics 31(8), 1027-1032, (2010)

[22] F.Z. Geng, S.P. Qian, S. Li., A numerical method for singularlyperturbed turning point problems with an interior layer, Journal ofComputational and Applied Mathematics 255, 97-105, 2014.

J. Christy Roja (Department of Mathematics) Numerical methods ... June 5, 2018 74 / 74

Thank You

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