Perturbation Methods for 1st order ODE - Find the asymptotic solution

[VinnyCee]

Homework Statement

Apply the regular perturbation method to solve the following ordinary differential equation

[tex]\left(1\,+\,\epsilon\,y\right)\,\frac{dy}{dx}\,+\,y\,=\,0[/tex]

subject to

[tex]x\,=\,1;\,y\,=\,1[/tex]

Show that the asymptotic solution is of the form

[tex]y\,=\,e^{1\,-\,x}\,+\,\epsilon\,\left[e^{1\,-\,x}\,-\,e^{2\,(1\,-\,x)}\right]\,+\,\dots[/tex]

Homework Equations

http://www.sm.luth.se/~johanb/applmath/chap2en/index.html"

The Attempt at a Solution

First, I get the base case by setting epsilon to zero.

[tex]\frac{dy_0}{dx}\,+\,y_0\,=\,0[/tex]

Am I supposed to use the "subject to" conditions now?

[tex]\frac{dy_0}{dx}\,=\,-y_0\,=\,-1[/tex]

I have no idea if that is what I am supposed to do, I hope so otherwise I am completely lost here.

Now, I assume that a solution is of the form

[tex]y\,=\,y_0\,+\,\epsilon\,y_1\,+\,\epsilon^2\,y_2\,+\,\dots[/tex]

Differentiating that approximation

[tex]\frac{dy}{dx}\,=\,\frac{dy_0}{dx}\,+\,\epsilon\,\frac{dy_1}{dx}\,+\,\epsilon^2\,\frac{dy_2}{dx}\,+\,\dots[/tex]

Expanding the original equation given

[tex]\frac{dy}{dx}\,+\,\epsilon\,y\,\frac{dy}{dx}\,+\,y\,=\,0[/tex]

and substituting the approximations into it

[tex]\left[\frac{dy_0}{dx}\,+\,\epsilon\,\frac{dy_1}{dx}\,+\,\epsilon^2\,\frac{dy_2}{dx}\,+\,\dots\right]\,+\,\left[\epsilon\,y_0\,\frac{dy_0}{dx}\,+\,2\,\epsilon^2\,y_0\,\frac{dy_1}{dx}\,+\,\epsilon^3\,y_1\,\frac{dy_1}{dx}\,+\,\dots\right]\,+\,\left[y_0\,+\,\epsilon\,y_1\,+\,\epsilon^2\,y_2\,+\,\dots\right]\,=\,0[/tex]

Please tell me if the above substitution and expansion are correct.


Now I start matching terms according to their order.

O(1): [tex]\frac{dy_0}{dx}\,=\,-y_0[/tex]

O([itex]\epsilon[/itex]): [tex]\frac{dy_1}{dx}\,=\,-y_0\,\frac{dy_0}{dx}\,-\,y_1[/tex]

O([itex]\epsilon^2[/itex]): [tex]\frac{dy_2}{dx}\,=\,-2\,y_0\,\frac{dy_1}{dx}\,-\,y_2[/tex]

Solving for order one

[tex]\int\,\frac{dy_0}{dx}\,=\,\int\,-y_0[/tex]

[tex]y_0\,=\,K_0[/tex]

OR using the base case that I am unsure about

[tex]y_0\,=\,1[/tex]

But which one is it?

Thanks in advance for your help!

 

[Master1022]

Hi,

I know this was posted many years ago, but I just thought I would respond in case anyone else is taking a look at this thread. (I am not sure whether I have done the quoting correctly, but it just follows through your work)

So for this problem (which has no boundary layer), I would:
1. Define your series expansion (which you did)
2. Substitute that into the ODE (which you did, although I disagree with your expansion)
3. Re-group and solve the sub-problems
4. Recombine solutions

Problem Statement:

Apply the regular perturbation method to solve the following ordinary differential equation

[tex]\left(1\,+\,\epsilon\,y\right)\,\frac{dy}{dx}\,+\,y\,=\,0, s.t. y(1) = 1 [/tex]

Attempt:

First, I get the base case by setting epsilon to zero.

[tex]\frac{dy_0}{dx}\,+\,y_0\,=\,0[/tex]

Am I supposed to use the "subject to" conditions now?

I don't think this part is strictly necessary here. I suppose we can find the solution to the unperturbed problem (i.e. ##\epsilon = 0 ##), and then compare to our series solution. Otherwise, it is not needed to find the series solution.

Now, I assume that a solution is of the form

[tex]y\,=\,y_0\,+\,\epsilon\,y_1\,+\,\epsilon^2\,y_2\,+\,\dots[/tex]

Differentiating that approximation

[tex]\frac{dy}{dx}\,=\,\frac{dy_0}{dx}\,+\,\epsilon\,\frac{dy_1}{dx}\,+\,\epsilon^2\,\frac{dy_2}{dx}\,+\,\dots[/tex]

Agreed, this is step 1 done.

Expanding the original equation given
[tex]\frac{dy}{dx}\,+\,\epsilon\,y\,\frac{dy}{dx}\,+\,y\,=\,0[/tex]

and substituting the approximations into it
[tex]\left[\frac{dy_0}{dx}\,+\,\epsilon\,\frac{dy_1}{dx}\,+\,\epsilon^2\,\frac{dy_2}{dx}\,+\,\dots\right]\,+\,\left[\epsilon\,y_0\,\frac{dy_0}{dx}\,+\,2\,\epsilon^2\,y_0\,\frac{dy_1}{dx}\,+\,\epsilon^3\,y_1\,\frac{dy_1}{dx}\,+\,\dots\right]\,+\,\left[y_0\,+\,\epsilon\,y_1\,+\,\epsilon^2\,y_2\,+\,\dots\right]\,=\,0[/tex]

Please tell me if the above substitution and expansion are correct.

I don't think your expansions are quite right (slightly off - I think there shouldn't be any coefficients of 2, and there are some missing terms). In fact, I would have just left the equation in the un-expanded form as we can just see, by eye, which terms fall into which orders.
[tex] \left( 1 + \epsilon \cdot (y_0 + \epsilon y_1 + ... ) \right) \left(y_0 ' + \epsilon y_1 ' + ... \right) + (y_0 + \epsilon y_1 + ... ) = 0 [/tex]
where I have used short-hand notation for the ##y_i = y_i (x) ## and ## y_{j}' = \frac{dy_j}{dx} ##

We can also substitute into the boundary conditions to get:
[tex] y(1) = 1 \rightarrow 1 = y_0 (1) + \epsilon y_{1}(1) + ... [/tex]

This is step 2 completed.

Now I start matching terms according to their order.

O(1): [tex]\frac{dy_0}{dx}\,=\,-y_0[/tex]
O([itex]\epsilon[/itex]): [tex]\frac{dy_1}{dx}\,=\,-y_0\,\frac{dy_0}{dx}\,-\,y_1[/tex]

These are correct, but you should also include the boundary conditions.

So for order ## \epsilon ^0##, we have:
[tex] y_0 ' + y_0 = 0, y_0 (1) = 1 [/tex]

and for order ## \epsilon ^1##, we have:
[tex] y_1 ' + y_1 = -y_0 y_0 ', y_1 (1) = 0 [/tex]

To find the boundary conditions, we compare coefficients of different powers of ##\epsilon##.

O([itex]\epsilon^2[/itex]): [tex]\frac{dy_2}{dx}\,=\,-2\,y_0\,\frac{dy_1}{dx}\,-\,y_2[/tex]

You don't really need this one (so I haven't checked it). We know we don't need it because the solution just goes up to the ## \epsilon ^{1} ## term (for this more simple problem, we don't need higher order equations)

Solving for order one

[tex]\int\,\frac{dy_0}{dx}\,=\,\int\,-y_0[/tex]

[tex]y_0\,=\,K_0[/tex]

So you are right that you need to solve this problem, but you haven't done this part correctly. You may want to revisit your notes on first order differential equations.

Given that this post is so old, I'll just write it here that solving this leads to:

Spoiler

## y_0 (x) = e^{1 - x} ##

OR using the base case that I am unsure about

[tex]y_0\,=\,1[/tex]

But which one is it?

You shouldn't use your base/unperturbed solution in this solution.

From here, now we need to solve the problem for order ## \epsilon ^{1} ##. The answer for that is shown in the spoiler below (just basic diff eq's):

Spoiler

## y_1 (x) = e^{1 - x} - e^{2 - 2x} ##

Then we recombine those two solutions to get the first few terms of our series solution:
[tex] y(x) = y_0 (x) + \epsilon y_1 (x) + ... [/tex]

Just substitute in the answers from above to get the required expression. I hope this has been of some help.