Given the nonlinear system
$$ \begin{align} \dot{x}_1 &= -4x_1 + 10x_2 + u \\ \dot{x}_2 &= -x_1 - 2x_2 - \log(1 + x_1^2) \\ y &= x_1 + x_2 \end{align} $$
Assume the system should be forced to stay at the operating point $x^*_1 = 5$, $y^* = 10$. From that follows that $x^*_2 = 5$.
However, by choosing an input $u^*$ for the operating point, I can only make $\dot{x}_1 = 0$, but not $\dot{x}_2 = 0$: Assume $u^* = -30$, then
$$ \begin{align} \dot{x}^*_1 &= 0 \\ \dot{x}^*_2 &= -18.2581 \\ \end{align} $$
So it seems there is no way to make this operating point an equilibrium. However, the linearization at $(x^*_1,x^*_2,u^*) = (5,5,-30)$ is stable, controllable and observable...
Question: I would like to use linearization at the required operating point, but how to deal with non-existing equilibrium at this point?
Edit: To give another example, take the normalized pendulum:
$$ \begin{align} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= \sin(x_1) - \frac{1}{2}x_2 + u \end{align} $$
Usually, the task is to balance the pendulum at the top position s.t. $x_1 = x_2 = 0$.
But what if the task is to rotate the pendulum at a constant rate of change of its angle, such that $\dot{\phi} = \dot{x}_1 = c \neq 0$? To achieve this, $x_2 = c$ must hold, and then I can choose $u$ such that $\dot{x}_2$ gets zero, but what to do whith the non-zero $\dot{x}_1$?