Internal Dynamics

🧩 Exponential Matrix

Let

$$M=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$$

1. 🧮

Compute the exponential of $M$.

🗝️ Hint:

$$\cosh x:=\frac{e^x+e^{-x}}{2},\sinh x:=\frac{e^x-e^{-x}}{2}.$$

2. 🐍 💻 🔬

Compute numerically:

• exp(M) (numpy)

• expm(M) (scipy.linalg)

and check the results consistency.

🔓 Exponential Matrix

1. 🔓

We have

$$M=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],M^2=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$$

and hence for any $j\in \mathbb{N}$,

$$M^{2j+1}=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],M^{2j}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I.$$

$$\begin{array}{rl}{e}^M& =\sum _{k=0}^{+\mathrm{\infty }}\frac{M^k}{k!}\\ & =\left(\sum _{j=0}^{+\mathrm{\infty }}\frac{1}{(2j)!}\right)I+\left(\sum _{j=0}^{+\mathrm{\infty }}\frac{1}{(2j+1)!}\right)M\\ & =\left(\sum _{k=0}^{+\mathrm{\infty }}\frac{1^k+(-1{)}^k}{2(k!)}\right)I+\left(\sum _{k=0}^{+\mathrm{\infty }}\frac{1^k-(-1{)}^k}{2(k!)}\right)M\\ & =(\cosh 1)I+(\sinh 1)M\end{array}$$

Thus,

$$e^M=\left[\begin{array}{cc}\cosh 1& \sinh 1\\ \sinh 1& \cosh 1\end{array}\right].$$

2. 🔓

>>> M = [[0.0, 1.0], [1.0, 0.0]]

>>> exp(M)
array([[1.        , 2.71828183],
       [2.71828183, 1.        ]])

>>> expm(M)
array([[1.54308063, 1.17520119],
       [1.17520119, 1.54308063]])

These results are consistent:

>>> array([[exp(0.0), exp(1.0)], 
...        [exp(1.0), exp(0.0)]])
array([[1.        , 2.71828183],
       [2.71828183, 1.        ]])

>>> array([[cosh(1.0), sinh(1.0)], 
...        [sinh(1.0), cosh(1.0)]])
array([[1.54308063, 1.17520119],
       [1.17520119, 1.54308063]])

🧩 G.A.S. $\iff$ L.A.

📝 For any dynamical system, if the origin is a globally asymptotically stable equilibrium, then it is a locally attractive equilbrium.

💎 For linear systems, the converse result also holds.

🚀 Let’s prove this!

1. 🧠

Show that for any linear system $\dot{x}=Ax$, if the origin is locally attractive, then it is also globally attractive.

2. 🧠

Show that linear system $\dot{x}=Ax$, if the origin is globally attractive, then it is also globally asymptotically stable.

🗝️ Hint: Consider the solutions $e_k(t):=e^{At}{e}_k$ associated to $e_k(0)=e_k$ where $(e_1,\dots ,e_n)$ is the canonical basis of the state space.

🔓 G.A.S. $\iff$ L.A.

1. 🔓

If the origin is locally attractive, then there is a $\epsilon >0$ such that for any $x_0\in {\mathbb{R}}^n$ such that $\parallel x_0\parallel \le \epsilon$,

$$\underset{t\to +\mathrm{\infty }}{lim}{e}^{At}{x}_0=0.$$

Now, let any $x_0\in {\mathbb{R}}^n$. Since the norm of $\epsilon x_0/\parallel x_0\parallel$ is $\epsilon$, and by linearity of $e^{At}$, we obtain

$$\begin{array}{rl}\underset{t\to +\mathrm{\infty }}{lim}{e}^{At}{x}_0& =\underset{t\to +\mathrm{\infty }}{lim}{e}^{At}\left(\frac{\parallel x_0\parallel }{\epsilon }\epsilon \frac{x_0}{\parallel x_0\parallel }\right)\\ & =\frac{\parallel x_0\parallel }{\epsilon }\underset{t\to +\mathrm{\infty }}{lim}{e}^{At}\left(\epsilon \frac{x_0}{\parallel x_0\parallel }\right)\\ & =0.\end{array}$$

Thus the origin is globally attractive.

2. 🔓

Let $X_0$ be a bounded set of ${\mathbb{R}}^n$. Since $$x_0=\sum _{k=1}^n{x}_{0k}{e}_k,$$

the solution $x(t)$ of $\dot{x}=Ax$, $x(0)=x_0$ satisfies $$\begin{array}{rl}x(t)& =e^{At}{x}_0=e^{At}\left(\sum _{k=1}^n{x}_{0k}{e}_k\right)=\sum _{k=1}^n{x}_{0k}{e}^{At}{e}_k.\end{array}$$

$$\begin{array}{rl}\parallel x(t)\parallel & =\Vert \sum _{k=1}^n{x}_{0k}{e}^{At}{e}_k\Vert \\ & \le \sum _{k=1}^n|x_{0k}|\Vert e^{At}{e}_k\Vert \\ & =\sum _{k=1}^n|x_{0k}|\Vert e_k(t)\Vert \\ & \le (\sum _{k=1}^n|x_{0k}|)\underset{k=1,\dots ,n}{max}\parallel e_k(t)\parallel \end{array}$$

Since $X_0$ is bounded, there is a $\alpha >0$ such that for any $x_0=(x_{01},\dots ,x_{0n})$ in $X_0$,

$$\parallel x_0{\parallel }_1:=\sum _{k=1}^n|x_{0k}|\le \alpha .$$

Since for every $k=1,\dots ,n$, $\underset{t\to +\mathrm{\infty }}{lim}\parallel e_k(t)\parallel =0$,

$$\underset{t\to +\mathrm{\infty }}{lim}\underset{k=1,\dots ,n}{max}\parallel e_k(t)\parallel =0.$$

Finally

$$\begin{array}{rl}\parallel x(t,x_0)\parallel & \le (\sum _{k=1}^n|x_{0k}|)\underset{k=1,\dots ,n}{max}\parallel e_k(t)\parallel \\ & \le \alpha \underset{k=1,\dots ,n}{max}\parallel e_k(t)\parallel \end{array}$$

Thus $\parallel x(t,x_0)\parallel \to 0$ when $t\to \mathrm{\infty }$, uniformly w.r.t. $x_0\in X_0$. In other words, the origin is globally asymptotically stable.

🧩 Spring-Mass System

Consider the scalar ODE

$$\ddot{x}+kx=0,\text{ with }k>0$$

1. 🧮

Represent this system as a first-order ODE.

2. 🧠 🧮

Is this system asymptotically stable?

3. 🧠 🧮

Do the solutions have oscillatory components?

Find the set of associated rotational frequencies.

4. 🧠 🧮

Same set of questions (1., 2., 3.) for

$$\ddot{x}+b\dot{x}+kx=0$$

when $b>0$.

🔓 Spring-Mass System

1. 🔓

$$\frac{d}{dt}\left[\begin{array}{c}x\\ \dot{x}\end{array}\right]=\left[\begin{array}{rr}0& 1\\ -k& 0\end{array}\right]\left[\begin{array}{c}x\\ \dot{x}\end{array}\right]=A\left[\begin{array}{c}x\\ \dot{x}\end{array}\right]$$

2. 🔓

We have

$$max\{\mathrm{\Re }s|s\in \sigma (A)\}=0,$$

hence the system is not globally asymptotically stable.

3. 🔓

Since

$$det(sI-A)=det\left(\begin{array}{rr}s& -1\\ k& s\end{array}\right)=s^2+k,$$

the spectrum of $A$ is

$$\sigma (A)=\{s\in \mathbb{C}|det(sI-A)=0\}=\{i\sqrt{k},-i\sqrt{k}\}.$$

The system poles are $\pm i\sqrt{k}$.

The general solution $x(t)$ can be decomposed as $$x(t)=x_{+}{e}^{i\sqrt{k}t}+x_{-}{e}^{-i\sqrt{k}t}.$$

Thus the components of $x(t)$ oscillate at the rotational frequency

$$\omega =\sqrt{k}.$$

4. 🔓

$$\frac{d}{dt}\left[\begin{array}{c}x\\ \dot{x}\end{array}\right]=\left[\begin{array}{rr}0& 1\\ -k& -b\end{array}\right]\left[\begin{array}{c}x\\ \dot{x}\end{array}\right]=A\left[\begin{array}{c}x\\ \dot{x}\end{array}\right]$$

$$det(sI-A)=det\left(\begin{array}{rr}s& -1\\ k& s+b\end{array}\right)=s^2+bs+k,$$

Let $\mathrm{\Delta }:=b^2-4k$. If $b\ge 2\sqrt{k}$, then $\mathrm{\Delta }\ge 0$ and

$$\sigma (A)=\{\frac{-b+\sqrt{\mathrm{\Delta }}}{2},\frac{-b-\sqrt{\mathrm{\Delta }}}{2}\}.$$

Otherwise,

$$\sigma (A)=\{\frac{-b+i\sqrt{-\mathrm{\Delta }}}{2},\frac{-b-i\sqrt{-\mathrm{\Delta }}}{2}\}.$$

Thus, if $b\ge 2\sqrt{k}$,

$$max\{\mathrm{\Re }s|s\in \sigma (A)\}=\frac{-b+\sqrt{b^2-4k}}{2}<0$$

and otherwise

$$max\{\mathrm{\Re }s|s\in \sigma (A)\}=-\frac{b}{2}<0.$$

In each case, the system is globally asymptotically stable.

If $b\ge 2\sqrt{k}$, the poles are real-valued; the components of the solution do not oscillate.

If $0<b<2\sqrt{k}$, the imaginary part of the poles is

$$\pm \frac{\sqrt{4k-b^2}}{2}=\pm \sqrt{k-(b/2{)}^2},$$

thus the solution components oscillate at the rotational frequency

$$\omega =\sqrt{k-(b/2{)}^2}.$$

🧩 Integrator Chain

Consider the system

$$\dot{x}=Jx\text{ with }J=\left[\begin{array}{ccccc}0& 1& 0& \cdots & 0\\ 0& 0& 1& \cdots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \cdots & 1\\ 0& 0& 0& \cdots & 0\end{array}\right]$$

1. 🧮

Compute the solution $x(t)$ when

$$x(0)=\left[\begin{array}{c}0\\ ⋮\\ 0\\ 1\end{array}\right].$$

2. 🦊 🧮

Compute the solution for an arbitrary $x(0)$

$$x(0)=\left[\begin{array}{c}{x}_1(0)\\ ⋮\\ ⋮\\ x_n(0)\end{array}\right].$$

3. 🧮

Same questions for the system

$$\dot{x}=(\lambda I+J)x$$

for some $\lambda \in \mathbb{C}$.

🗝️ Hint: Find the ODE satisfied by $y(t):=x(t)e^{-\lambda t}$.

4. 🧮

Is the system asymptotically stable ?

5. 🧠

Why does the stability analysis of this system matter ?

🔓 Integrator Chain

1. 🔓

Let $x=(x_1,\dots ,x_n)$.

The ODE $\dot{x}=Jx$ is equivalent to: $$\begin{array}{lcl}{\dot{x}}_1& =& x_2\\ {\dot{x}}_2& =& x_3\\ ⋮& ⋮& ⋮\\ {\dot{x}}_{n-1}& =& x_n\\ {\dot{x}}_n& =& 0.\end{array}$$

When $x(0)=(0,\dots ,0,1)$,

• ${\dot{x}}_n=0$ yields $x_n(t)=1$, then

• ${\dot{x}}_{n-1}=x_n$ yields $x_{n-1}(t)=t$,

• …

• ${\dot{x}}_k=x_{k+1}$ yields $$x_k(t)=\frac{t^{n-k}}{(n-k)!}.$$

To summarize: $$x(t)=\left[\begin{array}{c}{t}^{n-1}/(n-1)!\\ ⋮\\ t^{n-1-k}/(n-1-k)!\\ ⋮\\ t\\ 1\end{array}\right]$$

2. 🔓

We note that

$$x(0)=x_1(0)\left[\begin{array}{c}1\\ ⋮\\ 0\\ 0\end{array}\right]+\cdots +x_{n-1}(0)\left[\begin{array}{c}0\\ ⋮\\ 1\\ 0\end{array}\right]+x_n(0)\left[\begin{array}{c}0\\ ⋮\\ 0\\ 1\end{array}\right].$$

Similarly to the previous question, we find that:

$$x(0)=\left[\begin{array}{c}0\\ ⋮\\ 0\\ 1\\ 0\end{array}\right]\to x(t)=\left[\begin{array}{c}{t}^{n-2}/(n-2)!\\ ⋮\\ t\\ 1\\ 0\end{array}\right]$$

$$x(0)=\left[\begin{array}{c}0\\ ⋮\\ 1\\ 0\\ 0\end{array}\right]\to x(t)=\left[\begin{array}{c}{t}^{n-3}/(n-3)!\\ ⋮\\ 1\\ 0\\ 0\end{array}\right]$$

And more generally, by linearity:

$$x(t)=\left[\begin{array}{c}{\displaystyle x_1(0)+\cdots +x_{n-1}(0)\frac{t^{n-2}}{(n-2)!}+x_n(0)\frac{t^{n-1}}{(n-1)!}}\\ ⋮\\ {\displaystyle x_{n-2}(0)+x_{n-1}(0)t+x_n(0)\frac{t^2}{2}}\\ x_{n-1}(0)+x_n(0)t\\ x_n(0)\end{array}\right].$$

3. 🔓

If $\dot{x}(t)=(\lambda I+J)x(t)$ and $y(t)=x(t)e^{-\lambda t}$, then

$$\begin{array}{rl}\dot{y}(t)& =\dot{x}(t)e^{-\lambda t}+x(t)(-\lambda e^{-\lambda t})\\ & =(\lambda I+J)x(t)e^{-\lambda t}-\lambda Ix(t)e^{-\lambda t}\\ & =Jx(t)e^{-\lambda t}\\ & =Jy(t).\end{array}$$

Since $y(0)=x(0)e^{-\lambda 0}=x(0)$ we get

$$x(t)=\left[\begin{array}{c}{\displaystyle x_1(0)+\cdots +x_n(0)\frac{t^{n-1}}{(n-1)!}}\\ ⋮\\ x_{n-1}(0)+x_n(0)t\\ x_n(0)\end{array}\right]e^{\lambda t}.$$

4. 🔓

The structure of $x(t)$ shows that

• If $\mathrm{\Re }\lambda <0$, then the system is asymptotically stable.

• If $\mathrm{\Re }\lambda \ge 0$, then the system is not.

  For example when $x(0)=(1,0,\dots ,0)$, we have $$x(t)=(1,0,\dots ,0).$$

5. 🔓

Every square complex matrix $A$, even if it is not diagonalizable, can be decomposed into a block-diagonal matrix where each block has the structure $\lambda I+J$.

Thus, the result of the previous question allows to prove the 💎 Stability Criteria in the general case.