Equilibrium point stability given the derivative

Question

Given a system described by the following equation:

$$y'' + y'^4 + y'^2*u + y^3 = 0 $$ where $y(0) = 0$ and $ y'(0) \neq 0 $ , what is the stability of the equilibrium point? The eq. point is $0$ and $y$ is the output while $u$ is the input.

Answer 1

First of all, as this system is time-invariant, the stability of an equilibrium point does depend on time, and hence on the initial condition. So, what I think you mean with $y(0)=0, \dot{y}(0)\neq 0$ is that for which initial condition $(y^* ,\dot{y}^* )=(y(0),\dot{y}(0)) $ am I converging to a stable equilibrium point...?

You can approach this question in multiple manners. For me the most systematic way is to transform the system into a state space system, with $x_1=y$ and $x_2=\dot{y} $. This yields

$$\begin{align} \dot{x}_1 &= x_2 \\\ \dot{x}_2 &= -x_1^3 - x_2^4 - x_2^2u \end{align} \qquad \to \ \dot{x}=f(x,u)$$

Locally linearize at the point $(x_1^* , x_2^* ) = ( y^* , \dot{y}^* )=(0 , x_2^* )$ gives

$$\left.\frac{\partial f}{\partial x}\right|_{x^*} = \begin{pmatrix} 0 & 1 \\\ -3x_1^2 & -4x_2^3-2x_2u \end{pmatrix}$$

The eigenvalues of this matrix are for the equilibrium point equal to $\lambda_1=0$ and $\lambda_2=-4(x_2^* )^3-2x_2^* u^* $. So, with linearisation you cannot say anything about the stability. If you also analyse the phase plane of the DE, you can conclude that no point is stable, as any small pertubation away from $x_1=0$, gives that the velocity vectors point away from the equilibrium point.