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Taking the time derivative of $y$ yields: $$\dot{y}=\frac{\partial f}{\partial x}(f(x)+u)$$ We need $y(t)=y(0) e^{-\beta t}$ but this is possible if and only if $\dot{y}=-\beta y$ , if the hessian is invertible then this is possible if and only if $u=-f(x)-\left(\frac{\partial f}{\partial x}\right)^{-1}\beta y$. To get an intuition on why the hessian must ...
The problem is complicated by the fact that the function f(t) and the point at which its minimum/maximum x∗ is located, generally speaking, are not known to us. So if I understand it correctly: $f(t)$ is a unknown function, of which your controlled system should approach either the maximum or the minimum value $f(t)$ has over $t$. If I look at your system ...