Initial Value Approach for a Class of Singular Perturbation Problems

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Table 1(b). Numerical Results of Example 1 with ε=10^-4 , h=10^-4

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The numerical results are given in tables 1(a), 1(b) for ε=10^-3 and 10^-4 respectively.

Example 2.: Now consider the following non-homogeneous singular perturbation problem from fluid dynamics for fluid of small viscosity.

with y(0)=0 and y(1)=1.

Table 2(a). Numerical Results of Example 2 with ε=10^-3_, h=10^-3

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Table 2(b). Numerical Results of Example 2 with ε=10^-4 , h=10^-4

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The exact solution is given by

The set of initial value problems corresponding to this example are

The numerical results are given in tables 2(a), 2(b) for ε=10^-3 and 10^-4 respectively.

Example 3.: Now we consider the following variable coefficient singular perturbation problem from Kevorkian and Cole ^[5], page: 33; equations 2.3.26 and 2.3.27 with α=-1/2;

with y(0)=0 and y(1)=1.

We have chosen to use uniformly valid approximation (which is obtained by the method given by Nayfeh ^[7], page: 148; equation 4.2.32) as our ‘exact’ solution; The set of initial value problems related to this example are with y₀(1)=1, z(1)=ε+1/2 with y(0)=0. The numerical results are given in tables 3(a), 3(b) for ε=10^-3 and 10^-4 respectively.

Table 3(a). Numerical Results of Example 3 with ε=10^-3_, h=10^-3

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Table 3(b). Numerical Results of Example 3 with ε=10^-4 , h=10^-4

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4. Non-Linear Problems

We now extend this method for a class of non-linear singularly perturbed two point boundary value problems with left end boundary layer of the underlying interval. For this we consider a class of non-linear singularly perturbed two point boundary value problems of the form:

(13)

with

(14)

where ε is a small positive parameter (0<ε<<1) and α, β are known constants. We assume that a(x), b(x, y) are sufficiently continuously differentiable functions in [0, 1]. Furthermore, we assume that (13)-(14) has a solution which displays a boundary layer of width O(ε) at x=1 for small values of ε.

Step 1. Setup the two first order equations equivalent to equation (12) as follows:

(15)

and

(16)

Step2. Obtain the reduced problem by setting ε=0 in equation (13) with appropriate boundary condition

(17)

with

(18)

Step 3. Setup the initial conditions as follows:

Using , the solution of the reduced problem, in equation (15) we have

(19)

This will be the initial condition for equation (15) and y(0)=α will be the initial condition for equation (16).

Step 4. Get the set of initial value problems as follows:

Replacing y(x) by in (15), we get

(20)

Now the differential equation (20) with the condition (19) and the differential equation (17) with (18) constitute an initial value problem and the differential equation (16) with the condition y(0)=α constitute another initial value problem.

Therefore the set of initial value problems corresponding to equation (12)-(13) are given by

(21)

(22)

Thus in a manner of speaking, we have replaced the original boundary value problem (13)-(14) by a set of initial value problems. The integration of these initial value problems goes in opposite direction, and the second problem is solved only if the solution of the first one is known. The present method does not require the analytical solution of the reduced problem. The initial value problem (21) does not contain the perturbation parameter. It is not a perturbation problem. Hence we solve (21) to get using classical Runge-Kutta method. In fact any standard method can be used. The initial value problem (22) is a singular perturbation problem. To solve the initial value problem (22) we use the trapezoidal formula for the numerical integration of the first order differential equation to obtain a two term relationship which is described in section 2. By using the initial condition and the two term relationship, we obtain the numerical solution of the original boundary value problem.

5. Non Linear Examples

Again to demonstrate the applicability of the method, we have applied it to a non-linear singular perturbation problem with left-end boundary layer.

Example 4.: Consider the following non linear singular perturbation problem from Bender and Orszag ^[2], page: 463; equations: 9.7.1;

with y(0)=0 and y(1)=0.

We have chosen to use Bender and Orszag’s uniformly valid approximation (ref. ^[2], page: 463; equation: 9.7.6) for comparison. y(x)=log_e(2/(1+x))-(log_e2)e^-2x/ε. For this example, we have boundary layer of thickness O(ε) at x=0. (cf. Bender and Orszag ^[2]).

Table 4(a). Numerical Results of Example 4 with ε=10^-3_, h=10^-3

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Table 4(b). Numerical Results of Example 4 with ε=10^-4 , h=10^-4

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The set of initial value problems related to this example are with and with . The numerical results are given in tables 4(a), 4(b) for ε=10^-3 and 10^-4 respectively.

6. Right End Boundary Layer Problems

Finally, we extend this method for singularly perturbed two point boundary value problems with right-end boundary layer of the underlying interval. To be specific, we consider a class of singular perturbation problem of the form:

(23)

with

(24)

where ε is a small positive parameter (0<ε<<1) and α, β are known constants. We assume that a(x), b(x) and f(x) are sufficiently continuously differentiable functions in [0, 1]. Furthermore, we assume that a(x)≤ M<0 throughout the interval [0,1], where M is some negative constant. This assumption merely implies that the boundary layer will be in the neighbourhood of x=1.

Step 1. Setup the two first order equations equivalent to the equation (23) as follows:

(25)

and

(26)

Step 2. Obtain the reduced problem by setting ε=0 in equation (23) with the appropriate boundary condition.

(27)

with

(28)

Step 3. Set up the initial conditions as follows:

Using , the solution of the reduced problem, in equation (26) we have

(29)

From (3.27) we have

Substituting this in (3.29), we get

(30)

This is the initial condition for equation (25). and y(1)=β will be the initial condition for equation (26).

Step 4. Get the set of initial value problems as follows:

Replacing y(x) by in (25), we get

(31)

Now the differential equation (31) with the condition (30) and the differential equation (27) with (28) constitute an initial value problem and the differential equation (26) with the condition y(1)= β constitute another initial value problem.

Therefore the set of initial value problems corresponding to equation (25)-(26) are given by

with

(32)

(33)

Thus in a manner of speaking, we have replaced the original boundary value problem (23)-(24) by a set of initial value problems. The integration of these initial value problems goes in opposite direction, and the second problem is solved only if the solution of the first one is known. We solve these initial value problems to obtain the solution over the interval [0, 1]. The present method does not require the analytical solution of the reduced problem. The initial value problem (32) does not contain the perturbation parameter. It is not a perturbation problem. Hence we solve (32) to get using classical Runge-Kutta method. In fact any standard method can be used.

The initial value problem (33) is a singular perturbation problem. To solve the initial value problem (33) we use the trapezoidal formula for the numerical integration of the first order differential equation to obtain a two term relationship. By using the initial condition and the two term relationship, we obtain the numerical solution of the original boundary value problem. The main feature of this method is that it does not require very fine mesh. Some numerical experiments have been included to demonstrate the applicability of the method.

Now we consider the initial value problem (33) i.e., with y(1)= β.

We now divide the interval [0, 1] into N equal parts with mesh size h. i.e., and for

Integrating by parts the equation (33) in , we get

By making use of the trapezoidal formula for evaluating the integrals approximately, we obtain

We consider

The above relation becomes,

After simple manipulation, we obtain a two term recurrence relationship,

(34)

The initial condition is used in (34) to obtain the numerical solution, in the interval [0, 1].

7. Examples with Right-End Boundary Layer

To illustrate the method for singularly perturbed two point boundary value problems with right-end boundary layer of the underlying interval we considered two examples.

Table 5(a). Numerical Results of Example 5 with ε=10^-3_, h=10^-3

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Table 5(b). Numerical Results of Example 5 with ε=10^-4 , h=10^-4

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Example 5.: Consider the following singular perturbation problem

with y(0)=1 and y(1)=0.

Clearly, this problem has a boundary layer at x=1. i.e.; at the right end of the underlying interval.

The exact solution is given by y(x)=(e^(x-1)/ε-1)/(e^-1/ε-1).

The set of initial value problems for this are:

The numerical results are given in tables 5(a), 5(b) for ε=10^-3 and 10^-4 respectively.

Example 6.: Now we consider the following singular perturbation problem ; x∈[0, 1] with y(0)=1+exp(-(1+ε)/ε); and y(1)=1+1/e.

Clearly this problem has a boundary layer at x=1.

The exact solution is given by y(x)=e⁽¹⁺ε^)(x-1)/ε+e^-x

The set of initial value problems related to this example are with y₀(0)=1, z(0)=-ε-1 and with y(1)=1+1/e.

The numerical results are given in tables 6(a), 6(b) for ε=10^-3 and 10^-4 respectively.

Table 6(a). Numerical Results of Example 6 with ε=10^-3_, h=10^-3

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Table 6(b). Numerical Results of Example 6 with ε=10^-4 , h=10^-4

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