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I am given with the following:

$$\epsilon y'' + e^x (xy' -y ) = x^2,\quad -1<x<1,\quad 0< \epsilon \ll 1,\text{ and }y(-1)=1,\ y(1)=-1 $$

We learned that in such a case we can expect an interior layer around $x=0$ .

I need to find the leading term of the asymptotic expansion.

Outer solutions are easily found: $$y_0 ^L =(e-1)x -xe^{-x},\quad y_0 ^R = \left(\frac{1}{e} -1 \right) x -xe^{-x}.$$

The inner solution is: $$c_1 \frac{x}{\sqrt{\epsilon}}\int _{0}^{\frac{x}{\sqrt{\epsilon}}} e^{-\frac{s^2}{2} } ds - c_1 e^{- \frac{s^2}{2} } + c_2 \frac{x}{\sqrt{\epsilon}}.$$

I have no idea how to match the inner and outer solutions now.

Will someone please help me figure this out?

Thanks in advance !

Michael Hardy
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Alexandra
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  • You don't need to write $\varepsilon<<1$; you can write $\varepsilon\ll 1$. I changed that and did some other math-notation copy-editing. – Michael Hardy May 25 '14 at 19:50

0 Answers0