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Assume two parallel systems of the form $\dot{x}=f(x_t)+u_t$ and $\dot{y}=f(y_t)+u_t$. The input that is added to these two systems is the same. Also assume that $f$ is globally unstable, meaning starting at any initial condition $x_0$ except the equilibrium point ($x=0$), the zero input system' states; i.e., $\dot{x}_t=f(x_t)$ diverges to infinity. Does there exist any $u_t$ that can stabilize both parallel systems with different initial conditions $x_0\neq y_0$?

For linear systems obviously the answer is no because by subtracting both equations we get $\dot{y}-\dot{x}=A(y_t-x_t)$ and with unstable $A$ and $y_0-x_0\neq 0$ that is not orthogonal to unstable eigenvector, the difference diverges to infinity. Therefore, at least one of the systems will diverge to infinity. What can we say for nonlinear systems?

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    If it fails with the linear case, then nonlinear system will be worse. – CroCo Nov 16 '21 at 13:04
  • Your linear system can be written as the nonlinear system described, so naturally, as CroCo points out, you will have the same result. It is the very same counterexample. – Rollen S. D'Souza Nov 20 '21 at 01:47
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    @CroCo I don't think it is this easy. The important difference is that the proof sketch in the question covers all linear systems. Can you prove that there is no nonlinear system that actually is stable in the described setting? Because that's what the question is asking as far as I understand it. – SampleTime Nov 20 '21 at 19:48
  • @Rollen That qualifies as a (trivial) counterexample but the more interesting question is whether this is true for any nonlinear system. I am pretty sure it is but I couldn't find a proof yet. – SampleTime Nov 20 '21 at 19:55

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