> restart;

Asymptotic Expansion of the Roots of [Maple Math] and the Solution to a Boundary Value Problem

Prepared by Douglas B. Meade

25 September 1998

>

Quadratic Equation

Root #1: x=O( 1 )

Iterations: [Maple Math]

> G := x -> sort( expand( (-epsilon*x^2-1)/3 ) );

[Maple Math]

Initial approximation:

> x0 := -1/3;

[Maple Math]

Successive approximations: note that each iteration doubles the number of terms in the approximation but only one additional term converges to its final value.

> for k from 0 to 5 do

> x.(k+1) := G(x.k);

> od;

[Maple Math]

[Maple Math]

[Maple Math]

[Maple Math]
[Maple Math]
[Maple Math]

[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]

[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]
[Maple Math]

>

Root #2: x=O( 1/ [Maple Math] )

Iterations: [Maple Math]

> H := x -> normal( (-1/x - 3 )/epsilon );

[Maple Math]

> x.0 := -3/epsilon;

[Maple Math]

> for k from 0 to 5 do

> x.(k+1) := H(x.k);

> series( x.(k+1), epsilon, k+2 ); # these show how the iterates converge

> od;

[Maple Math]

[Maple Math]

[Maple Math]

[Maple Math]

[Maple Math]

[Maple Math]

[Maple Math]

[Maple Math]

[Maple Math]

[Maple Math]

[Maple Math]

[Maple Math]

>

Boundary Value Problem

The differential equation and boundary conditions are defined as follows:

> ODE := epsilon*diff(x(t),t$2) + 3*diff(x(t),t) + x(t) = 0;

[Maple Math]

> BC1 := x(0)=0;

[Maple Math]

> BC2 := x(1)=1;

[Maple Math]

>

Exact Solution

> SOLN := dsolve( { ODE, BC1, BC2 }, x(t) );

[Maple Math]

>

Graphical Solution

3D plot (note that Maple is able to handle epsilon=0 in a reasonable way

> plot3d( rhs(SOLN), t=0..1, epsilon=0..0.1 );

>

2D plots for selected values of epsilon. Note the rapid initial growth that occurs as [Maple Math] -> 0; the solution is nearly linear closer to 1.

> plot( [ seq( eval( rhs(SOLN), epsilon=e ), e=[0.001,0.01,0.1,1] )], t=0..1 );

>