The Student Room Group

Asymptotic expansion

Can some one check if this correct, please
Reply 1
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You're intermediate steps are very confusing.

if y(x)=y0(x)+εy1(x)+O(ε2) y(x) = y_0(x)+\varepsilon y_1(x) + O(\varepsilon ^2), then y(0)=1y0(0)+εy1(0)=1 y(0) = 1 \Rightarrow y_0(0) + \varepsilon y_1(0) = 1

equate powers of epsilon for the BCs

y1(0)=0 y_1(0)= 0 it looks like you set it equal to 1?
(edited 3 years ago)
Reply 3
I wrote yo(0) = 1 not y1(0) = 1
Original post by meme12
I wrote yo(0) = 1 not y1(0) = 1


Yeah, but you applied it when you said c4 = 1 and why have you y=... in stead of y_1 = ... ?
Reply 5
I redid it again so my final answer should be 1 + epsilon (x^2)/2 + O(.....
Thanks , is this correct
Original post by meme12
I redid it again so my final answer should be 1 + epsilon (x^2)/2 + O(.....
Thanks , is this correct

yes
Reply 7
If you don't mind can you check this too , I wrote that it's uniformly valid as the second term is much less than the first term and it becomes non uniform when x^2 << O( epsilon^-1)
Original post by meme12
If you don't mind can you check this too , I wrote that it's uniformly valid as the second term is much less than the first term and it becomes non uniform when x^2 << O( epsilon^-1)


You're not answering this part correctly, though you have the somewhat basic understanding of what you need to mention here.

When x2x^2 is of the same magnitude of as ϵ1\epsilon^{-1}, that's when the term ϵx22\epsilon \dfrac{x^2}{2} starts behaving like the leading order order term. This is bad, since it means there is a substantial contribution coming from here to the leading order, which pollutes the solution. This term is known in the biz as a secular term because it grows with the independent variable.

Hence, the expansion is nonuniform for all xx, and becomes bad when x2=O(ϵ1)x^2 = O(\epsilon^{-1}). Otherwise, the expansion is good enough when x2O(ϵ1)x^2 \ll O(\epsilon^{-1}).

This is why in practice when we are interested in long-time asymototics, this regular perturbation is somewhat rubbish since it does not hold when x is sufficiently large ('sufficiently large' here differs from context to context). One way to ensure our solution is valid for as much of the domain as possible is to decrease ϵ\epsilon further and further. But this can be upgraded a whole lot using the method of multiple-scales asymptotic analysis (which I hope you are going to learn if you're already on the regular perturbation expansion!) whereby we do not need to decrease epsilon as much to obtain an accurate solution over a sufficiently large domain.
(edited 3 years ago)
Reply 9
Thank you

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