In my textbook (end of page 282) it goes over the calculation of the Asymptotic Solution and Stability of Wavefront Solutions of the Fisher Equation.
In this it transforms the Fisher equation $$U''+cU'+U(1-U)=0$$ into the equation $$\epsilon\frac{d^2g}{d\xi^2}+\frac{dg}{d\xi}+g(1-g)=0 \;\;\;\;\oplus$$ with the boundary conditions $$g(-\infty)=1, \;g(\infty)=0, \;0<\epsilon\leq \frac{1}{c^2},\; g(0)=1/2$$
The text then states to look for solutions of the above equation as a regular perturbation series (a standard power series) in $\epsilon$, that is let
$$g(\xi;\epsilon)=g_0(\xi)+ \epsilon g_1(\xi)+...$$
With the conditions
$$U(0)=1/2 \Rightarrow g(0;\epsilon)=1/2, \;\;\forall\epsilon$$ $$g_0(-\infty)=1, \;g_0(\infty)=0, \;g_0(0)=1/2,$$ $$g_i(\pm\infty)=0, \;g_i(0)=0, \;\;i=1,2,3... $$
The textbook then states that it takes the regular perturbation series and substitutes it into equation $\oplus$, then equating powers of $\epsilon$ gets
$$O(1):\frac{dg_0}{d\xi}=-g_0(1-g_0) \Rightarrow g_0(\xi)=(1+e^\xi)^{-1}$$ $$O(\epsilon):\frac{dg_1}{d\xi}+(1-2g_0)g_1=-\frac{d^2g_0}{d\xi^2}$$
My problem is I don't understand how they got to the last part. Can someone show me what happens when I substitute my power series into $\oplus$ and how they took $O(1)$ and calculated $g_0(\xi)$? (I will provide more information if needed)
Ok, take $g=g_0+\epsilon g_1+\ldots$ and plug it in: $$ \epsilon(g_0+\epsilon g_1+\ldots)''+(g_0+\epsilon g_1+\ldots)'+(g_0+\epsilon g_1+\ldots)(1-g_0-\epsilon g_1-\ldots)=0. $$ The derivatives are linear so we get $$ \epsilon g_0''+\epsilon^2 g_1''+\ldots+g_0'+\epsilon g_1'+\ldots+(g_0+\epsilon g_1+\ldots)(1-g_0-\epsilon g_1-\ldots)=0. $$ Now collect all the terms on the left without $\epsilon$: $$ g_0'+g_0(1-g_0)=0, $$ you get the first equation. Collect all the terms with $\epsilon$ and so on...