5. Linear Quadratic Problem (LQ)

5.1 LQ Problem

Considering a linear system: \(\dot x = Ax + Bu\) is A.S. \(\Leftrightarrow\) (is necessary and sufficient to (N&S) ) \(\dot x = Ax\) is A.S. Thus, we can give the theorem:

A N&S condition for the A.S. of a linear system is :

• Given any \(Q = Q^T \succ 0, \exists P = P^T \succ 0\), satisfying:

  \[ A^TP + PA = -Q \]

To prove the sufficiency:

\[ \begin{aligned} V(x) &= x^TPx \\ \dot V(x) &= \dot x^TPx + x^TP\dot x\\ &=x^TA^TPx + x^TPAx \\ &= x^T(A^TP + PA)x \\ &= -x^TQx \end{aligned} \]

\(\bar x = 0\) is G.A.S.

Example

\[ \begin{aligned} \dot x_1 &= -x_2 \\ \dot x_2 &= x_1 + (x_1^2 - 1)x_2 \end{aligned} \]

\(\bar x = \begin{bmatrix} 0\\0 \end{bmatrix}\) is an equilibrium point, find the stability.

\[ \begin{aligned} A &= \begin{bmatrix} 0&-1\\1&-1 \end{bmatrix} \\ & \Rightarrow x_A(\lambda) = \lambda^2 + \lambda + 1\\ & \Rightarrow \lambda = \frac{-1 \pm i\sqrt 3}{2} \end{aligned} \]

The system is A.S. for \(Q = Q^T\succ 0\)

Let \(A^TP+PA=-Q\), we can choose the value \(Q = \mathbf I\), \(P = \begin{bmatrix} P_{11} & P_{12} \\ P_{21} & P_{22}\end{bmatrix} = \begin{bmatrix} \frac32 & -\frac12 \\ -\frac12 & 1\end{bmatrix}\), the eigenvalue of \(P\) is \(\lambda = \frac54 \pm \frac{\sqrt 5}{4}\), \(P \succ 0\). Let \(V(x) = x^TPx\) is globally PD in \(\bar x = 0\) and radially unbounded

\[ \dot V(x) = \dot x^TPx + x^TP\dot x \]

Notes

We consider the nonlinear and higher oder part of the system,

\[ \dot x = Ax + \underbrace{\varphi(x) - Ax}_{\Delta \varphi(x)} \]

Where \(\Delta \varphi(x)\) is the nonlinear and higher order part of the system, it should be \(0\) near the linearization point,

\[ \lim_{||x||\to 0}\frac{\Delta \varphi(x)}{||x||} = 0 \]

Unless we will have,

\[ \begin{aligned} \dot V(x) &= x^T(A^TP + PA)x + \text{higher order terms} \\ &= -x^TQx + \text{higher order terms}\\ \end{aligned} \]

\(-x^TQx\) is ND but due to the effect of higher order terms, \(\dot V(x)\) could be non ND.

Back to our case,

\[ \begin{aligned} \dot V(x) &= -(x_1^2 + x_2^2) - x_1^3x_2 + 2x_1^2x_2^2 \end{aligned} \]

\(- x_1^3x_2 + 2x_1^2x_2^2\) are higher order terms, the system is ND in \(\bar x = 0\) but not globally, \(\bar x\) is A.S.

To find the border of \(V(x)\), we make some approximation:

\[ \begin{aligned} \dot V(x) &= -||x||^2 + x_1 x_1x_2 (-x_1 +2x_2) \\ &\leq -||x||^2 + |x_1| |x_1x_2| |-x_1 +2x_2| \\ \end{aligned} \]

We can further approximate the Lyapunov function by using some approximation techniques,

\[ \begin{aligned} (|x_1|-|x_2|)^2 = x_1^2 +x_2^2 - 2|x_1||x_2| &\geq 0 \\ \frac12 ||x||^2 &\geq |x_1x_2| \\ \hfill \\ |\begin{bmatrix} -1 \\ 2\end{bmatrix}^T x| &\leq \sqrt 5 ||x|| \end{aligned} \]

Now we can get

\[ \dot V(x) \leq -||x||^2 + ||x|| \frac12 ||x||^2 \sqrt 5 ||x|| = -||x||^2 (1- \frac{\sqrt 5}{2}||x||^2) \]

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