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)=0U”+cU’+U(1-U)=0 into the equation \epsilon\frac{d^2g}{d\xi^2}+\frac{dg}{d\xi}+g(1-g)=0 \;\;\;\;\oplus\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)=1/2g(-\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\epsilon, that is let g(\xi;\epsilon)=g_0(\xi)+ \epsilon g_1(\xi)+...g(\xi;\epsilon)=g_0(\xi)+ \epsilon g_1(\xi)+... With the conditions U(0)=1/2 \Rightarrow g(0;\epsilon)=1/2, \;\;\forall\epsilonU(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_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... 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\oplus, then equating powers of \epsilon\epsilon gets O(1):\frac{dg_0}{d\xi}=-g_0(1-g_0) \Rightarrow g_0(\xi)=(1+e^\xi)^{-1}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}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\oplus and how they took O(1)O(1) and calculated g_0(\xi)g_0(\xi)? (I will provide more information if needed) ================= Why not to take this power series and plug it in? – Artem Oct 21 at 2:59 Well that's my problem right now. Like I'm having a mental blank on how to plug it in... It's such a stupid thing... – Bradley Gray Oct 21 at 3:00 ================= 1 Answer 1 ================= Ok, take g=g_0+\epsilon g_1+\ldotsg=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. \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. \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\epsilon: g_0'+g_0(1-g_0)=0, g_0'+g_0(1-g_0)=0, you get the first equation. Collect all the terms with \epsilon\epsilon and so on...