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I don't see why the following three forms of the LQR optimal control problem are equivalent:

For $\begin{cases} x'=Ax+Bu \\ x(t_0)=x_0\end{cases}$ find

  1. $$\min_{u\in L^2(t_0,T; \mathbb{R}^m)} J(u)=\frac{1}{2}\left\{\int_{t_0}^T ||Cx||^2+||u||^2 dt + ||Rx(T)||^2\right \}$$ where || || is the euclidean norm, and $C,R$ are arbitrary matrices.

  2. $$\min_{u\in L^2(t_0,T; \mathbb{R}^m)} J(u)=\frac{1}{2}\left\{\int_{t_0}^T x^TQx+u^TRu+2x^TNu\ dt + x(T)^TPx(T)\right\}$$

where $Q,P$ are positive semi-definite matrices and $R$ is positive definite.

  1. $$\min_{u\in L^2(t_0,T; \mathbb{R}^m)} J(u)=\frac{1}{2}\left\{\int_{t_0}^T x^TQx+u^TRu\ dt + x(T)^TPx(T)\right\}$$

where $Q,P$ are positive semi-definite matrices and $R$ is positive definite.

I found them posed in all 3 versions but I don't see how to pass from one form to another.

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  • $\begingroup$ Are you sure about 1? The last term doesn't look "quadratic" at all. As far as I understand, 2. is the most general version, and 3 is just a common special case (N=0) that people often consider. $\endgroup$ Jan 27, 2019 at 13:25
  • $\begingroup$ If you're simply missing a square in 1, then it's easy to see that 1 is also a special case of 3 by using $\|Cx\|^2 = x^TC^TCx$. $\endgroup$ Jan 27, 2019 at 13:27
  • $\begingroup$ I have edited my post. I see now 1 and 3 are equivalent, but how about 2 and 3? $\endgroup$
    – Bogdan
    Jan 27, 2019 at 13:47

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