I've been stuck on this problem for the past few days, and was wondering if someone could clarify part ii) of the following question. In order to understand the question, I will also include my solution for part i).
The question is as follows:
Part i) For the dynamical system below, derive a linear stability analysis:
$$\qquad \qquad \qquad \qquad \qquad \frac{dx}{dt}=f(x) \qquad \qquad \qquad \qquad \qquad \qquad (1)$$
Additional information: The time-independent variable $\bar{x}$ denotes a fixed point for the dynamical system and thus $f(\bar{x})=0$.
Part ii) Repeat the derivation in part i) and continue the Taylor expansion to 4th order, for the case of $f'(\bar{x})=f''(\bar{x})=f'''(\bar{x})=0$. What is the stability condition of $f$ in this case?
Hint: look for real solutions in the ODE for the small perturbation function, $\epsilon(t)$, and consider the sign of both the initial condition, $\epsilon(0)$, and the fourth derivative, $f^{(4)}(\bar{x})$.
My solution to part i):
If we consider a small perturbation $\epsilon(t)$. Then let: $$\qquad \qquad \qquad \qquad \qquad x(t)=\bar{x}+\epsilon(t) \qquad \qquad \qquad \qquad \qquad \qquad (2)$$
Taking the derivative of equation $(2)$, we find:
$$\qquad \qquad \qquad \qquad \qquad \frac{dx}{dt}=\frac{d\epsilon}{dt} \qquad \text{(since $\bar{x}$ is time independent)} \qquad (3)$$
Thus substituting equation $(2)$ & $(3)$ into equation $(1)$, we have:
$$\frac{dx}{dt}=f(\bar{x}+\epsilon(t))$$ $$\quad \Rightarrow \frac{dx}{dt}=\frac{d\epsilon}{dt}= f(\bar{x}+\epsilon(t))$$ $$\qquad\qquad\qquad\qquad\quad\quad\Rightarrow \frac{d\epsilon}{dt}=f(\bar{x}+\epsilon(t)) \qquad\qquad\qquad\qquad\qquad (4) $$
Now taking the Taylor series expansion of RHS of equation $(4)$ about $x=\bar{x}$ we have:
$$f(x)=f(\bar{x}+\epsilon(t))=f(\bar{x})+f'(\bar{x})((\epsilon+\bar{x})-\bar{x})+\frac{f''(\bar{x})((\epsilon+\bar{x})-\bar{x})^2}{2!}+\frac{f'''(\bar{x})((\epsilon+\bar{x})-\bar{x})^3}{3!}+... \qquad \qquad (5)$$ $$\Rightarrow f(x)=f(\bar{x})+f'(\bar{x})(\epsilon)+O(\epsilon^2)$$ $$\qquad\qquad\qquad\qquad\qquad \Rightarrow f(x)=f'(\bar{x})(\epsilon) \qquad \text{(since $f(\bar{x})=0$ and $O(\epsilon^2)$ becomes negligible)}$$ $$\Rightarrow\frac{d\epsilon}{dt}=f'(\bar{x})(\epsilon)$$
Thus we are left with a separable equation, solving for $\epsilon(t)$ as follows:
$$\frac{d\epsilon}{dt}=f'(\bar{x})(\epsilon)$$ $$\Rightarrow\frac{d\epsilon}{\epsilon}=f'(\bar{x})dt$$ $$\Rightarrow log(\epsilon)=f'(\bar{x})t+c$$ $$\Rightarrow \epsilon(t)=e^{f'(\bar{x})t+c}$$ $$\qquad\qquad\qquad\qquad\qquad\Rightarrow \epsilon(t)=\epsilon_0e^{f'(\bar{x})t}\qquad\qquad\qquad\qquad\qquad (6)$$
Thus from this result, we can find the stability of our fixed points using the following properties:
If $f'(\bar{x})<0$ then $\epsilon(t)$ decays to zero, and thus we approach our fixed point $$\Rightarrow \text{Our fixed point $f(\bar{x})$ is stable} $$
If $f'(\bar{x})>0$ then $\epsilon(t)$ grows exponentially, and thus we move away from our fixed point $$\Rightarrow \text{Our fixed point $f(\bar{x})$ is unstable} $$
- If $f'(\bar{x})=0$ linear stability analysis cannot determine the stability of the fixed point, requiring other techniques (i.e. graphical approach)
My solution to part ii) so far:
Now if we expand the Taylor expansion to fourth order as in equation $(5)$ then we have:
$$f(x)=f(\bar{x})+f'(\bar{x})(\epsilon)+\frac{f''(\bar{x})(\epsilon^2)}{2!}+\frac{f'''(\bar{x})(\epsilon^3)}{3!}+\frac{f^{(4)}(\bar{x})(\epsilon^4)}{4!}+O(\epsilon^5)$$ $$\Rightarrow \frac{d\epsilon}{dt}=f(\bar{x})+f'(\bar{x})(\epsilon)+\frac{f''(\bar{x})(\epsilon^2)}{2!}+\frac{f'''(\bar{x})(\epsilon^3)}{3!}+\frac{f^{(4)}(\bar{x})(\epsilon^4)}{4!}+O(\epsilon^5)$$ $$\qquad\qquad\Rightarrow \frac{d\epsilon}{dt}=\frac{f^{(4)}(\bar{x})(\epsilon^4)}{4!} \qquad\qquad \text{(since $f(\bar{x})=f'(\bar{x})=f''(\bar{x})=f'''(\bar{x})=0$ and $O(\epsilon^5)$ becomes negligible)} $$
Thus we are left with a nonlinear equation:
$$\qquad\qquad\qquad\qquad\qquad\frac{d\epsilon}{dt}-\frac{f^{(4)}(\bar{x})(\epsilon^4)}{24}=0\qquad\qquad\qquad\qquad\qquad (7)$$
This is where I'm stuck on how to solve this.
My question/clarification for part ii)
I was wondering if someone could point me in the right direction as to how to solve the above nonlinear equation. Do I use my solution for $\epsilon(t)$ in equation $(6)$ somewhere? Or am I allowed to substitute $\frac{dx}{dt}$ instead of $\frac{d\epsilon}{dt}$, making equation $(7)$ a linear ODE?
Additionally, in the hint they say use the real solutions of the ODE for the perturbation function $\epsilon(t)$. Do they mean the solution I found for $\epsilon(t)$ in equation $(6)$?
Finally, could someone explain why constants of integration can be ignored or subsumed into $\epsilon_0$ such as in equation $(6)$? If we aren't specified an initial condition for $\epsilon(t)$, do we just define the initial condition as $\epsilon(t)=\epsilon_0$?
I'd appreciate any clarifications/suggestions on where to go from here.
Thanks