Matched asymptotic expansion

Resume

Remerciement

Introduction

Exercise 1 :

We consider the following Cauchy problem for nonlinear ordinary differential equation :

[pic 1]

With  is a non-negative small parameter.[pic 2]

1.  We want to find such as  is a solution of (1).[pic 3][pic 4]

 is solution of [pic 5][pic 6]

[pic 7]

Finally, we find the following result:

[pic 8]

1. We want to find the exact solution of the equation (1) :

If we use the characteristic equation, we can say that the exact solution would be of the following form :

[pic 9]

With C1 and C2 which are representing real constants.

After that, we are using the initial conditions, and we can obtain the exact solution :

[pic 10]

[pic 11]

1. We are looking for a second order asymptotic expansion of in the form :[pic 12]

[pic 13]

We want to show that :

[pic 14]

We first obtain the following equations :

[pic 15]

We can identify the following problems:

[pic 16]

[pic 17]

[pic 18]

We have these systems :

[pic 19]

[pic 20]

Now we use the initial conditions to obtain:

[pic 21]

[pic 22]

Using the initial conditions, we have:

[pic 23]

[pic 24]

Then:

[pic 25]

Finally, we have:

[pic 26]

We have the following initial conditions :

[pic 27]

So finally, we obtain :

[pic 28]

We can write the following equation to find the solution using Matlab :

[pic 29]

Chapter 1: Regulier

[pic 30]

We start by taking in account the equation (1):

[pic 31]

We write all the equation in function of :[pic 32]

[pic 33]

By rewriting we get:

[pic 34]

Then we integer this equation

[pic 35]

An other time:

[pic 36]

We assume the  characteristic equation:

[pic 37]

So:

[pic 38]

Now we are going to determine a particular solution of (1) on the form:

[pic 39]

[pic 40]

[pic 41]

Then we introduce this new equations (3) and (4) on the first one, and we get:

[pic 42]

so

[pic 43]

And finnaly we obtain:

[pic 44]

By identification:

[pic 45]

So we have:

[pic 46]

Then we obtain the exact solution using the general solution (2) and the particular one (5) :

[pic 47]

Then by using the boundary conditions we have:

[pic 48]

[pic 49]

[pic 50]

Then we do (7)-(6):

[pic 51]

and:

[pic 52]

Finally

[pic 53]

Outer-limit

[pic 54]

So:

[pic 55]

And

[pic 56]

We put into the equation (1), the equations (8) and (9):

[pic 57]

And we rearrange:

[pic 58]

Then we can easily identify:

[pic 59]

And by using the boundary conditions[pic 60]

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