Which way of solving from nonlinear control to choose?

I have a nonlinear system:

$$\begin{cases} x'=f(x)+u \\ y=f(x) \end{cases}$$

where $$f(x)$$ - gradient of some one-extremal function (for example $$f=e^{-(x)^2}$$), i.e. $$\frac{df}{dx}$$.

Task: I want construct a continuous control $$u$$ that ensures the following condition:

$$y(t)=y(0) e^{-\beta t}, \beta>0$$

Which method to use for the solution: MRAC, asymptotic output tracking, feedback linearization, or something else?

I am not an expert, please do not pass by and give advice.

• $\dot{y}=\frac{d f(x)}{dx} \dot{x}=\frac{d f(x)}{dx} (f(x)+u)$ so provided that $\frac{d f(x)}{dx} \neq 0$ take $u=-f(x)-\frac{\beta y}{\frac{d f(x)}{dx}}$ Feb 7 at 9:18
• @abc1455 thank you for your feedback, could you please fulfill my request: explain how to get rid of the constraint you indicated if this is the purpose of management? and if you are ready, please fill out your comment in the form of an answer
– dtn
Feb 7 at 18:43
• @abc1455 And one more thing. I see that you need a Hessian to solve the problem? (The second derivative of the function with respect to the argument).
– dtn
Feb 8 at 4:33

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 be invertible assume that $$y,x$$ are scalars then if hessian is not invertible this means that $$\frac{\partial f}{\partial x}=0 \implies \dot{y}=0$$ This means that your system is not output controllable even though it is state controllable.
• And in any small neighborhood of this point, where the Hessian $\frac{df}{dx}≠0$ , such control is applicable, I understand correctly?
• This is applicable anywhere the hessian is invertible. However, you have to take care because it might be that the hessian is not invertible at your extremum e.g.$x^4$, even though it is invertible in the neighborhood of $x=0$ Feb 8 at 10:18