I want to find a two term asymptotic expansion, for small $\epsilon$, of the solution of the following problem:
$$
y'' – \epsilon y' – y = 1\tag{1}, \, y(0) = 0, \,y(1) = 1
$$
My approach: I assume that the solution has an asymptotic expansion that looks like this:
$$
y\sim y_0 + \epsilon^\alpha y_1 + \epsilon^\beta y_2 + \ldots \tag{2}
$$
with $0<\alpha <\beta<\ldots$.
If I substitute $(2)$ into $(1)$ I get the following equation:
$$
(y_0 + \epsilon y_1 + \ldots)'' – \epsilon(y_0 + \epsilon y_1 + \ldots)' – (y_0 + \epsilon y_1 + \ldots) = 1
$$
I will now try to find $y_0, y_1$ by inspecting different order terms:
$\mathcal{O}(1):$ In order to have balance in the equation, $y_0$ must satisfy
$$
y_0'' – y_0 = 1
$$
Solving this equation for $y_0$ gives
$$y_0 = c_1\exp((\frac{1}{2} + \sqrt{5}/2)t) + c_2\exp((\frac{1}{2}
) – \sqrt{5}/2)t)$$
This solution needs to satisfy $y(0) = 0$. Substituting $t = 0$ and setting $y_0(0) = 0$ leads to the solution:
$$
y_0 = -\dfrac{1}{e^{1/2 – \sqrt{5}/2}}\exp((1/2 + \sqrt{5}/2)t) + \dfrac{1}{e^{1/2 – \sqrt{5}/2}}\exp((1/2 – \sqrt{5}/2)t)
$$
Furthermore, to have balance we need $\alpha = 1$.
$\mathcal{O}(\epsilon):$ In order to have balance in the equation $y_1$ must satisfy
$$
y_1'' – y_1 = -y_0'
$$
Or
$$
y_1'' – y_1 = \dfrac{d}{dt}(-\dfrac{1}{e^{1/2 – \sqrt{5}/2}}\exp((1/2 + \sqrt{5}/2)t) + \dfrac{1}{e^{1/2 – \sqrt{5}/2}}\exp((1/2 – \sqrt{5}/2)t))
$$
While I could plug this into wolfram to find a solution for $y_1$, I'm not sure whether this is the correct way to find a two term asymptotic expansion for the initial problem.
Question: Am I on the right track? It feels like there must be an easier way to do this. Could someone point me in the right direction?
Best Answer
My calculation is different from yours. Let $$ y=y_0 + \epsilon y_1 +O(\epsilon^2) $$ in the equation to get $$ (y''_0 + \epsilon y''_1 +O(\epsilon^2))-\epsilon(y'_0 + \epsilon y'_1 +O(\epsilon^2))-(y_0 + \epsilon y_1 +O(\epsilon^2))=1. \tag{1} $$ Then the boundary conditions $y(0)=0,y(1)=1$ become $$ y_0(0)=y_1(0)=0, y_0(1)=1,y_1(0)=0.$$ $O(1)$: $$ y''_0-y_0=1, y_0(0)=0,y_0(1)=1 $$ which has the solution $$ y_0= \frac{e^{-x} \left(e^x-e^{2 x}-e^{x+2}+2 e^{2 x+1}-2 e+e^2\right)}{e^2-1}. $$ $O(\epsilon)$: $$ y''_1-y_1=y_0', y_1(0)=0,y_1(1)=0 $$ which has the solution \begin{eqnarray*} y_1&=& \frac1{2 \left(e^2-1\right)^2}\bigg[e^{-x} (-e^{2 x+2} (x-4)+2 e^x-4 e^{x+2}+2 e^{x+4}+e^{2 x} (x-2)+2 e^{2 x+3} (x-2)\\ &&-2 e^{2 x+1} x+e^2 x-2 e x-e^4 (x+2)+2 e^3 (x+2))\bigg]. \end{eqnarray*} Thus \begin{eqnarray*} y&=&y_0+\epsilon y_1+O(\epsilon^2)\\ &=&\frac{e^{-x} \left(e^x-e^{2 x}-e^{x+2}+2 e^{2 x+1}-2 e+e^2\right)}{e^2-1}\\ && +\frac{\epsilon}{2 \left(e^2-1\right)^2}\bigg[e^{-x} (-e^{2 x+2} (x-4)+2 e^x-4 e^{x+2}+2 e^{x+4}+e^{2 x} (x-2)+2 e^{2 x+3} (x-2)\\ &&-2 e^{2 x+1} x+e^2 x-2 e x-e^4 (x+2)+2 e^3 (x+2))\bigg]. \end{eqnarray*}