1. Lagrangian mechanics

1.1. Euler-Lagrange’s equation + generalized force

Define the Lagrangian as:

Euler–Lagrange equations with genralized force is:

where $Q_j$ is the generalized force which is defined as:

There is a velocity formulation of the generalized force which leverage the following relation:

1. Norm

Definition: For vector $x\in\mathbb{R}^n$, $p\in(0,\infty]$

\begin{gather*}
||x||_p=\left(
x_1^p+\cdots+x_n^p
\right)^{\frac{1}{p}}.
\end{gather*}

When $p\in[1,\infty]$, $||x||_p$ is a norm.

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Theorem: .

Proof of the Theorem

Consider the auxiliary vectors . Note that each entry of $y$ is less than 1:

The following inequality holds for each entry:

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2. Cone

Definition: A set $\mathcal{K}$ is called cone if for all $x\in \mathcal{K},\theta\geq 0$, it holds that
\begin{gather*}
\theta x\in \mathcal{K}
\end{gather*}

Definition: Let $\mathcal{K}$ be a cone, the dual cone $\mathcal{K}^*$ is
\begin{gather*}
\mathcal{K}^*:= \left\{y\mid \langle x,y\rangle \geq 0, \forall x\in\mathcal{K}\right\}
\end{gather*}

Definition: Let $\mathcal{K}$ be a cone, the polar cone $\mathcal{K}^o$ is
\begin{gather*}
\mathcal{K}^o:= \left\{y\mid \langle x,y\rangle \leq 0, \forall x\in\mathcal{K}\right\}
\end{gather*}

Apparently, by the definition, $\mathcal{K}^o=-\mathcal{K}^*$.

Definition: Let $C$ be a set, for any $x\in C$ the normal cone $\mathcal{N}_C(x)$ at this point is:
\begin{gather*}
\mathcal{N}_C(x):= \left\{y\mid \langle y,c-x\rangle\leq 0, \forall c\in\mathcal{K}\right\}
\end{gather*}

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Lemma: If $\mathcal{K}$ is convex and closed, then $\mathcal{K}^{**}=\mathcal{K}$.

Proof

Just use the definition, it is easy to verify that $\mathcal{K}\subseteq \mathcal{K}^{**}$. To show that they are actually the same set. Suppose there exists $x\in\mathcal{K}^{**},x\notin \mathcal{K}$. By Hyperplane separation theorem, there exists $c$ such that \begin{gather*} \mathcal{K}\subseteq H^+=\left\{z\mid c^Tz\geq 0\right\},\\ c^Tx < 0. \end{gather*} From the first relation, it holds that $c\in \mathcal{K}^*$. However, the second relation indicates that such $x\notin \mathcal{K}^{**}$ which contradicts the assumption. Therefore $\mathcal{K}^{**}=\mathcal{K}$.

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Lemma: The normal cone for a cone $\mathcal{K}$ is: \begin{gather*} \mathcal{N}_{\mathcal{K}} (x)=\left\{ \begin{array}{lc} \left\{y\mid\langle y,x \rangle=0\right\}&x\in \mathcal{K}\\ \phi&x\notin \mathcal{K} \end{array} \right. \end{gather*}

Proof

We only need to show that there does not exists such $y\in\mathcal{N}_{\mathcal{K}}$ that $\langle y,x\rangle < 0$. Suppose there exists, then for $x\in \mathcal{K}$, $$ \langle y, \frac{1}{2}x - x\rangle=\langle y, -\frac{1}{2}x \rangle > 0 $$ which contradicts the definition of the normal cone, as $\frac{1}{2}x\in \mathcal{K}$.

3. Positive Definite Matrix

Lemma: Given $A\succeq 0,B\succeq 0$, $AB=0$ iff. $tr(AB)=0$.

Proof of the Lemma

$\Rightarrow$: This is trivial by imposing the trace operation to $AB$.
$\Leftarrow$: Let $A=PP^T,B=QQ^T$, where $P,Q$ can be chosen such as Cholesky decomposition or square root of the corresponding matrice. It holds that \begin{gather*} tr(AB)=tr(PP^TQQ^T)=tr(Q^TPP^TQ)=tr(P^TQ)^2=\|P^TQ\|_F^2=0\\ \Rightarrow AB=PP^TQQ^T=P 0 Q^T=0. \end{gather*}

3.1. Linear Matrix Inequality(LMI) and SemiDefinite Programming(SDP)

1. Stability

1.1. Continuous-time Stability

Definition: A square matrix $A\in\mathbb{R}^{n\times n}$ is Hurwitz-stable iff.
\begin{gather*}
Re[\lambda_i(A)]<0,\ \forall i=1,\cdots,n
\end{gather*}

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Theorem(Lyapunov Stability): $A\in\mathbb{R}^{n\times n}$ is Hurwitz-stable iff. there exists $P\succ 0$ such that \begin{gather} PA+A^TP\prec 0. \label{eq:continuous-time Lyapunov inequality} \end{gather}

This Theorem was originated from control theory. I would like to present an algebra view-point.

Proof of the Theorem

Necessary condition '$\Leftarrow$'

This is a common result in control theory in which one can prove using contradiction. Suppose there exists $P\succ 0,1\leq i\leq n$ such that \eqref{eq:continuous-time Lyapunov inequality} holds and $Re[\lambda_i(A)]\geq 0$. Consider the following dynamics: \begin{gather*} \dot{x}(t)=Ax(t), x(0)=x_0. \end{gather*} For any $x_0\neq 0,\displaystyle\lim_{t\to\infty}x(t)\neq 0$. However, consider the Lyapunov function $V(x)=x^TPx$. From \eqref{eq:continuous-time Lyapunov inequality}, it holds that: \begin{gather*} \frac{d}{dt}{V}(x)=x^T\left(PA+A^TP\right)x<0,\forall x\neq 0\\ v(x)> 0,\forall x(t)\neq 0 \end{gather*} By LaSalle's invariance principle, it holds that $x(t)\to 0$, which contradicts the condition that $\displaystyle\lim_{t\to\infty}x(t)\neq 0$.

Sufficient condition '$\Rightarrow$'

Since $A$ is real, from linear algebra, there always exists invertible $Q\in\mathbb{R}^{n\times n}$ such that $A=Q^{-1}JQ$, where \begin{gather} \label{eq: similar transfrom for J} \begin{array}{c} J=diag\left\{ J_1,\cdots, J_m \right\}, J_i=\begin{bmatrix} C_i&I&0&0\\ 0&C_i&I&0\\ 0&0&\ddots&\vdots\\ 0&0&\cdots&C_i \end{bmatrix},\\ C_i=\lambda_i\in\mathbb{R}, \text{ or } C_i= \begin{bmatrix} a_i&b_i\\ -b_i&a_i \end{bmatrix}. \end{array} \end{gather} These $J_i$ are called real Jordan block. Apply the congruent transformation to \eqref{eq:continuous-time Lyapunov inequality} with $Q^{-1}$, one can find: \begin{gather*} Q^{-T}PQ^{-1}J+J^TQ^{-T}PQ^{-1}\prec0. \end{gather*} Denote $\hat{P}:=Q^{-T}PQ^{-1}$, it leads to \begin{gather} \hat{P}J+J^T\hat{P}\prec0. \label{eq: transoformed continuous-time Lyapunov equation} \end{gather} If we construct $\hat{P}$ to be also diagonally arranged with corresponding dimension: $ diag\{\hat{P}_1,\cdots,\hat{P}_m\}$, the resulting \eqref{eq: transoformed continuous-time Lyapunov equation} will be block diagonal. Therefore, it suffices to prove the case that $J$ is a single Jordan block, i.e., $m=1$.

1. When $C$ is real. Note here if $J$ is a scalar, then we are done. Because any $\hat{P}>0$ satisfies \eqref{eq: transoformed continuous-time Lyapunov equation}. What we are really dealing with, is the case where $\ell=dim(J)\geq 2$. To show there exists $\hat{P}$ such that \eqref{eq: transoformed continuous-time Lyapunov equation} holds, constructing $\hat{P}=diag\{\hat{p}_1,\cdots,\hat{p}_\ell\}$. With the proposed $\hat{P}$, \eqref{eq: transoformed continuous-time Lyapunov equation} becomes: \begin{gather} \label{eq: expended continuous-time Lyapunov function} \begin{bmatrix} 2\hat{p}_1\lambda&\hat{p}_1&&&\\ \hat{p}_1&2\hat{p}_2\lambda&\hat{p}_2&&\\ &\hat{p}_2&\ddots&\ddots&\\ &&\ddots&\ddots&\hat{p}_{\ell-1}\\ &&&\hat{p}_{\ell-1}&2\hat{p}_\ell\lambda \end{bmatrix}\prec0, \lambda<0. \end{gather} finite induction is used to prove such $\hat{p}$ exists. pick any $\hat{p}_\ell>0$
   • For the case $\ell =2$, \eqref{eq: transoformed continuous-time Lyapunov equation} becomes: \begin{gather*} \begin{bmatrix} 2\hat{p}_1\lambda&\hat{p}_1\\ \hat{p}_1&2\hat{p}_2\lambda \end{bmatrix}\prec0\Leftrightarrow 2\hat{p}_2\lambda-\frac{1}{2}\hat{p}_1\lambda^{-1}<0\leftrightarrow 4\hat{p}_2\lambda^2>\hat{p}_1, \end{gather*} where such $\hat{p}_1$ always exists.
   • For the case that $\ell>2$, suppose $\hat{p}_\ell,\cdots,\hat{p}_{2}$ have been chosen, \eqref{eq: transoformed continuous-time Lyapunov equation} looks like: \begin{gather*} \begin{bmatrix} \begin{array}{c|ccccc} 2\hat{p_1}\lambda&\hat{p}_1&&\\ \hline \hat{p}_1& 2\hat{p}_2\lambda&\hat{p}_2&&\\ &\hat{p}_2&\ddots&\ddots&\\ &&\ddots&\ddots&\hat{p}_{\ell-1}\\ &&&\hat{p}_{\ell-1}&2\hat{p}_\ell\lambda \end{array} \end{bmatrix}\prec0 \end{gather*} By Schur complement, it is equivalent to \begin{gather*} \begin{bmatrix} 2\hat{p_1}\lambda&\hat{p}_1\\ \hat{p}_1&\alpha \end{bmatrix}\prec 0, \end{gather*} for some $\alpha<0$, in which such $\hat{p}_1$ always exists.
     
2. When $C= \begin{bmatrix} a&b\\ -b&a \end{bmatrix}$, I would like to transfer it back to case 1. Consider $\hat{P}=diag\{\hat{p}_1,\hat{p}_1,\hat{p}_2,\hat{p}_2,\cdots,\hat{p}_\ell,\hat{p}_\ell\}$. Then \eqref{eq: transoformed continuous-time Lyapunov equation} becomes: \begin{gather*} \begin{bmatrix} 2\hat{p}_1\lambda&0&\hat{p}_1&0 &&&&\\ 0&2\hat{p}_1\lambda&0&\hat{p}_1 &&&&\\ \hat{p}_1&0&2\hat{p}_2\lambda&0&\hat{p}_2&0\\ 0&\hat{p}_1&0&2\hat{p}_2\lambda&0&\hat{p}_2 &&\\ && \hat{p}_2&0&\ddots&\ddots&&\\ &&0&\hat{p}_2&\ddots&\ddots&\ddots&\\ &&&&\ddots&\ddots&\ddots&\ddots&\\ &&&&&&2\hat{p}_{\ell-1}\lambda&0&\hat{p}_{\ell-1}&0\\ &&&&&&0&2\hat{p}_{\ell-1}\lambda&0&\hat{p}_{\ell-1}\\ &&&&&&\hat{p}_{\ell-1}&0&2\hat{p}_\ell\lambda&0\\ &&&&&&0&\hat{p}_{\ell-1}&0&2\hat{p}_\ell\lambda \end{bmatrix}\prec 0. \end{gather*} By selecting odd rows and columns and even rows and columns, one can find it gets identical expressions as in \eqref{eq: expended continuous-time Lyapunov function}.
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