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I have a brief question in regards to defining outer and inner expansion in a boundary layer problem. We have given:

$\epsilon \frac{d^{2}f}{dx^2} + (2+x)\frac{df}{dx} + f = 1, \quad 0 < x < 1$

$0 < \epsilon \ll 1, \quad f(0) = 2, \quad f(1) = 0$

I am able to define the outer expansion, but I have a question regarding the inner expansion. Here I am supposed to let $s = x/\epsilon$. This will then give me:

$\frac{1}{\epsilon}\frac{d^{2}f}{ds^2} + (2 + \epsilon s)\frac{1}{\epsilon}\frac{df}{ds} + f = 1$

Multiplying every term with $\epsilon$ yields:

$\frac{d^{2}f}{ds^2} + 2\frac{df}{ds} + \epsilon s \frac{df}{ds} + \epsilon f = \epsilon$

If we then define $f = f_0 + \epsilon f_1 + . . .$

Should the $\epsilon$ on the right side of this equality be included in the equation for $f_0$ or $f_1$? In other words should I let:

$O(1): \frac{d^{2}f_0}{ds^2} + 2\frac{df_0}{ds} = 0$


$O(1): \frac{d^{2}f_0}{ds^2} + 2\frac{df_0}{ds} = \epsilon$

And likewise, should I let:

$O(\epsilon): \frac{d^{2}f_1}{ds^2} + 2\frac{df_1}{ds} = -s\frac{df_0}{ds} - f_0$


$O(\epsilon): \frac{d^{2}f_1}{ds^2} + 2\frac{df_1}{ds} = -s\frac{df_0}{ds} - f_0 + 1$

If anyone can explain this to me I will be very grateful!

share|improve this question
Include the $\epsilon$ on the right hand side with all the other terms of order $\epsilon$, i.e. $f''_{0} + 2f'_{0} = 0$ and $f''_{1} + 2f'_{1} = -sf_{0} - f_{0} + 1$. Let me know if you want more details and I'll write up an answer. –  in_wolfram_we_trust May 7 '12 at 13:35
The $\TeX$ code for $\ll$ is \ll. Also, you can use double dollar signs to get displayed equations, which look nicer and are easier to read. –  joriki May 7 '12 at 14:28
Thanks a lot to the both of you! I really appreciate this! I know the procedure for solving these problems, but this is the first time I encountered the situation above where we have a constant $\epsilon$ on the right side of the equality. –  Kristian May 7 '12 at 14:42
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