I'm reading Solitons: An introduction [1], and in the chapter two (§2.2 p. 21), the author make a qualitative analysis of KDV: $u_t -6uu_x + u_{xxx} = 0$, then he try find solutions on this form $u(x,t)=f(x-ct)$, and then make the following substitution $\xi =x-ct$ and then we have: $-cf'-6ff'+f'''=0$, integrating twice we have:
$$\frac{1}{2}(f')^2 = f^3 + \frac{1}{2}c f^2 + Af +B \equiv F(f)$$
where $A$ and $B$ are constants.
So I don't understand this step (cf. book [1] p. 23). He wrote the Taylor series for $F(f)$ around $f_1$, where $f(\xi_1) = f_1$ and $F(f_1) = 0$
$$(f')^2 = 2(f-f_1)F'(f_1) + O((f-f_1)^2)) \tag{1}$$
And then:
$$f = f_1 + \frac{1}{2}(\xi - \xi_1)^2F'(f_1) + O((\xi - \xi_1)^4) \tag{2}$$
I did not understand what he did in p. 23 from $(\text{1})$ to $(\text{2})$, someone can explain to me?
[1] P.G. Drazin, R.S. Johnson, Solitons: An Introduction, Cambridge University Press, 1989
The form of your equation can be thought of as an energy conservation law, with $K = (f')^2$ as kinetic energy, $U = -F(f)$ as potential energy and $$K+U = E$$ forms the energy of the system.
Since your $F(f)$ is a cubic polynomial, the zeros of the polynomial correspond to $0$ energy. i.e. $f' =0 $. Perhaps Newton-Raphson method is used to "iteratively" converge to their solution.