Matched Asymptotic ODE Solution for ##\epsilon d_x(xd_xf)-xf=0##

[member 428835]

Homework Statement

Solve to order $\epsilon$ $$\epsilon d_x(xd_xf)-xf=0$$ subject to $|f(0)|<\infty$ and $f(1)=1$ via matched asymptotic expansions.

Homework Equations

Nothing comes to mind.

The Attempt at a Solution

Perform a matched asymptotic analysis. In this case when I take a series expansion $$f = \sum \epsilon^nf_n$$ the governing ODE yields the following two weighted equations $$xf_0=0\\
d_xf_0 - xf_1 + xd^2_x f_0=0.
$$
Notice the first equation implies $f_0=0$. This is where I am stuck. Any help?

For the inner part I believe an appropriate substitution is $x=(1-y)/g(\epsilon)$, but I thought this change of coordinates was typically introduced after first making an expansion and solving for the outer part.

 

[DrDu]

Should this really read $d_f$ in your formula?

 

[member 428835]

Thanks, just fixed that!

 

[DrDu]

One problem is that you never know whether a simple power series in epsilon is appropriate. E.g. you could transform $y=x/\epsilon$ or $y=x/\epsilon^{1/3}$.

 

[member 428835]

One problem is that you never know whether a simple power series in epsilon is appropriate. E.g. you could transform $y=x/\epsilon$ or $y=x/\epsilon^{1/3}$.

It's a little unclear to me, but what are you suggesting? Typically I would guess $x=(1-y)/g(\epsilon)$ and weight $g$ to balance any two terms I'd like, so something like $g=\epsilon^n$. However, I would only do this after computing the outer part, which as I showed in post 1, isn't giving good results $f_0=f_1=0$.

Any ideas?

 

[DrDu]

So what is the inner and what the outer part, and what distinguishes the two that you would only transform one of them?

 

[member 428835]

So what is the inner and what the outer part, and what distinguishes the two that you would only transform one of them?

That's interesting. I suppose I'm not sure why I don't transform both. Have anything in mind?