HW1

Hybrid Systems Dictionary

The limit points p of a set \(X ⊂ R^n\).

\(p\) is a limit point of the set \(X\) if every open neighborhood of \(p\) (i.e., any open interval of the form \((p - ε, p + ε)\) for some \(ε > 0\)) contains at least one point of \(X\) different from \(p\) itself.

the set of limit points of \(X\) is denoted by \(X'\).

The closure of a set \(X ⊂ R^n\).

\(\operatorname{cl}(X)\) is the smallest closed set containing \(X\).

Math Definition:

\[\operatorname{cl}(X) = X \cup \{ \text{boundary/limit points of } X \}\]

Example: the closure of the set \(X = (0, 1)\) is \(\operatorname{cl}(X) = [0, 1]\) because it includes all the points in \((0, 1)\) as well as the boundary points 0 and 1.

A closed set \(X ⊂ R^n\).

A set \(X\) is closed if

1. \(X\) contains all of its limit points, or equivalently

2. the complement of \(X\) is open.

3. \(X\) contains all of its boundary points.

Math Definition:

\[ X \text{ is closed} \iff \forall \text{ limit points } p, p \in X \]

or

\[ X \text{ is closed} \iff X = \operatorname{cl}(X) \]

Example: the set \(X = [0, 1]\) is closed because it contains all of its limit points (in this case, every point of [0,1] is a limit point of [0,1], including the endpoints 0 and 1) and is a subset of \(\mathbb{R}\).

the set \(X = \R\) is also closed because it contains all of its limit points (every point in \(\R\) is a limit point of \(\R\)) and is a subset of \(\mathbb{R}\).

An open set \(X ⊂ R^n\).

Math Def: A set \(X\) is open if for every point \(x \in X\), there exists an \(ε > 0\) such that the open ball \(B(x, ε) = \{ y \in R^n : ||y - x|| < ε \}\) is entirely contained within \(X\).

Example: the set \(X = R\) is open because for any point \(x \in R\), we can choose an \(ε > 0\) such that the open interval \((x - ε, x + ε)\) is entirely contained within \(R\).

the set \(X = (-\infty, 0)\) is also open because for every point \(x \in (-\infty, 0)\), there is an \(ε > 0\) such that the open interval \((x - ε, x + ε)\) is entirely contained within \((-\infty, 0)\).

Neither Open nor Closed sets

\((-1, 0]\) is neither open nor closed because

1. Not open: for the point 0, there is no \(ε > 0\) such that the open interval \((0 - ε, 0 + ε)\) is entirely contained within \((-1, 0]\) (\(0 + ε > 0\)).

2. Not closed: the limit point 0 is not contained in \((-1, 0]\).

The Boundary of a set \(X ⊂ R^n\).

Math Def: A point \(p\) is a boundary point of the set \(X\) if every open neighborhood/ball of \(p\) contains at least one point of \(X\) and at least one point not in \(X\).

the set of boundary points of \(X\) is denoted by \(\partial X\).

\[ \exists M > 0 \text{ such that } ||x|| \le M, \forall x \in X \]

example: the boundary of a disk \(X = \{ (x, y) \in R^2 : x^2 + y^2 \leq 1 \}\) is the circle \(\partial X = \{ (x, y) \in R^2 : x^2 + y^2 = 1 \}\).

The interior of a set \(X ⊂ R^n\).

Math Def: The interior of a set \(X\) is the set of all points in \(X\) that are not boundary points. It is denoted by \(\operatorname{int}(X)\).

Example:

• If \(X = [0, 1]\), then \(\operatorname{int}(X) = (0, 1)\).
• If \(X = (0, 1)\), then \(\operatorname{int}(X) = (0, 1)\) -> the interior of an open set is the set itself.

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A convex set \(X ⊂ R^n\).

math def: A set \(X\) is convex if for any two points \(x, y \in X\), the line segment connecting \(x\) and \(y\) is entirely contained within \(X\). Formally, for all \(x, y \in X\) and for all \(\lambda \in [0, 1]\), we have:

\[ \forall x,y \in X \text{ and } \forall \lambda \in [0, 1], \quad \lambda x + (1 - \lambda) y \in X \]

example:

1. the set \(X = (2,7)\) is convex (note CANNOT take the endpoints 2 and 7 because the set is open).
2. A Ball is a convex set because for any two points inside the ball, the line segment connecting them is also entirely contained within the ball.

The convex hull of a set \(X ⊂ R^n\).

Math def: \(\text{conv} X\) is the smallest convex set containing \(X\). It is the set of all convex combinations of points in \(X\).

Example:

1. for the set \(X = [1,3]\), the convex hull is \(\text{conv} X = [1,3]\)
2. if X is an open ball, then the convex hull is the same open ball.

the empty set \(∅\).

math def: The empty set \(∅\) is the set that contains no elements. It is a subset of every set and is considered both open and closed in any topological space.

example: \(∅ = \{ \}\), or the set of all \(x\) such that \(x^2 = -1\).

A compact set \(X ⊂ R^n\).

math def: A set \(X\) is compact if it is closed and bounded. This means that \(X\) contains all its limit points (closed) and can be contained within some finite region of space (bounded).

example: the set \(X = [0, 1]\) is compact because it is closed (it contains its limit points 0 and 1) and bounded (it lies within the interval from 0 to 1).

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A function \(f: A \rightarrow B\), and its domain and range

math def

• domain: set A
• codomain: set B
• range: the set of all actual outputs of the function, i.e., \(\{ f(x) | x \in A \} \subseteq B\)

example: for the function \(f(x) = 1/x, f: \N \rightarrow \R\), domain is \(\N\), codomain is \(\R\), range is \(\{ 1, 1/2, 1/3, ... \} \subset \R\).

A continuous function \(f: A \rightarrow B\).

math def: a function \(f\) is continuous at a point \(c \in A\) if:

\[\forall \epsilon > 0, \exists \delta > 0 \text{ such that } ||x - c|| < \delta \rightarrow \text{ then } ||f(x) - f(c)|| < \epsilon.\]

example: \(f(x) = 4\) is continous, because for every \(ε > 0\), we can choose any \(δ > 0\) (e.g., \(δ = 1\)), and the condition holds true since \(|f(x) - f(c)| = |4 - 4| = 0 < ε\) for all \(x\) in the domain. same as \(f(x) = min(x, 5)\).

A continously differentiable function \(f: A \rightarrow B\).

math def: A function \(f\) is continuously differentiable on a set \(A\) if it is differentiable at every point in \(A\) and its derivative \(f'\) is a continuous function on \(A\).

example: the function \(f(x) = x^2\) is continuously differentiable on \(\R\) because its derivative \(f'(x) = 2x\) exists for all \(x \in \R\) and is continuous everywhere on \(\R\).

A locally Lipschitz function \(f: A \rightarrow B\).

math def: A function \(f\) is locally Lipschitz on a set \(A\) if for every point \(x_0 \in A\), there exists a neighborhood \(U\) of \(x_0\) and a constant \(L > 0\) such that: $$ \forall x, y \in U, \quad ||f(x) - f(y)|| \leq L ||x - y|| $$

example: the function \(f(x) = x^2\) is locally Lipschitz on \(\R\): just plug in the values:

\[|f(x) - f(y)| = |x^2 - y^2| = |x - y||x + y| \leq L|x - y|\]

for any bounded neighborhood around \(x_0\), we can find a suitable \(L \geq | x + y | = 2(x_0 + \epsilon)\). so \(f\) is locally Lipschitz.

but we cannot find a global Lipschitz constant for all of \(\R\) because as \(x\) and \(y\) grow larger, \(|x + y|\) can become arbitrarily large, making it impossible to find a single constant \(L\) that works for all pairs of points in \(\R\).

An absolutely continous function \(f: A \rightarrow B\).

math def: A function \(f\) is absolutely continuous on an interval \([a, b]\) if for every \(ε > 0\), there exists a \(δ > 0\) such that for any finite collection of non-overlapping sub-intervals \((x_i, y_i) \in [a, b]\), if the \(\sum |y_i - x_i| < δ\), then \(\sum |f(y_i) - f(x_i)| < ε\).

it can be expressed as: $$ f(x) = f(a) + \int_{a}^{x} f'(t) dt, \quad \forall x \in [a, b] $$

example: the function \(f(x) = x^2\) is absolutely continuous on any closed interval \([a, b]\)

A locally bounded function \(f: A \rightarrow B\).

math def: A function \(f\) is locally bounded on a set \(A\) if for every point \(x \in A\), there exists a neighborhood \(U\) around \(x\) and a constant \(M > 0\) such that: $$ \forall y \in U, \quad ||f(y)|| \leq M $$ (i.e., the function's values are bounded within that neighborhood).

example: the function \(f(x) = 1/x\) is locally bounded on the set \(A = (0, \infty)\) because for any point \(x_0 > 0\), we can choose a neighborhood \(U = (x_0 - ε, x_0 + ε)\) for some small \(ε > 0\). Within this neighborhood, the function values are bounded above by \(M = 1/(x_0 - ε)\) and below by \(m = 1/(x_0 + ε)\).

A set-valued mapping \(F: A \doublerightarrow B\)., its domain and range.

math def: A set-valued mapping (or multifunction) \(F\) from a set \(A\) to a set \(B\) assigns to each element \(x \in A\) a subset of \(B\), denoted as \(F(x) \subseteq B\).

• domain: set A
• range: \(\bigcup_{x \in A} F(x) \subseteq B\)

example: consider the set-valued mapping \(F: \R \doublerightarrow \R\) defined by \(F(x) = [x, x + 1]\). in this case, the domain is R, and range is the set of all intervals of the form \([x, x + 1]\) for \(x \in \R\), which is \(\R\).

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the class \(\mathcal{K}\) functions.

Math Def: A function \(f : [0, a) \to [0, \infty)\) is of class \(\mathcal{K}\) if it is strictly increasing and \(f(0) = 0\).

Example: \(f(s) = s^2\), \(f(s) = \frac{s}{1+s}\), \(f(s) = \tanh(s)\) are class \(\mathcal{K}\) functions on \([0, \infty)\).

The class \(\mathcal{K}_\infty\) functions.

Math Def: A function \(f : [0, \infty) \to [0, \infty)\) is of class \(\mathcal{K}_\infty\) if it is of class \(\mathcal{K}\) and \(\lim_{s \to \infty} f(s) = \infty\).

Example: \(f(s) = s\), \(f(s) = s^3\), \(f(s) = e^s - 1\) are class \(\mathcal{K}_\infty\) functions on \([0, \infty)\).

The class \(\mathcal{KL}\) functions.

Math Def: A function \(f : [0, \infty) \times [0, \infty) \to [0, \infty)\) is of class \(\mathcal{KL}\) if for each fixed \(t \geq 0\), the function \(f(\cdot, t)\) is of class \(\mathcal{K}\), and for each fixed \(s \geq 0\), the function \(f(s, \cdot)\) is decreasing and \(\lim_{t \to \infty} f(s, t) = 0\).

• s: distance from origin / size
• t: time

example: \(f(s,t) = se^{-t}\).

suppose the t is fixed, then increase s will increase f(s,t), and \(\lim_{s \to \infty} f(s,t) \to \infty\) for any fixed t.

suppose the s is fixed, then increase t will decrease f(s,t), and when \(\lim_{t \to \infty} f(s,t) \to 0\) for any fixed s.

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A positive definite function \(f: A \rightarrow B\) with respect to a set \(C \subset A\).

Math Def: A function \(f: A \rightarrow B\) is positive definite with respect to a set \(C \subset A\) if:

1. \(f(x) > 0\) for all \(x \notin C\)
2. \(f(x) = 0\) for all \(x \in C\)

example: consider that \(C = \{0\} \subset \R\), then the function \(f(x) = x^2\) is positive definite with respect to the set \(C\) because:

1. For all \(x \neq 0\), \(f(x) = x^2 > 0\).
2. For \(x = 0\), \(f(0) = 0^2 = 0\).

A radically unbounded function

Math Def: A function \(V: \mathbb{R}^n \to \mathbb{R}\) is radially unbounded if \(V(x) \to \infty\) as \(\|x\| \to \infty\).

example: The function \(V(x) = x^2\) is radially unbounded because as \(x\) increases without bound, \(V(x)\) also increases without bound.

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The outer semicontinuous hall of a set

math def: The outer semicontinuous (osc) hull of a set is the smallest set-valued mapping that is outer semicontinuous and contains the original set-valued mapping.

for a set X, \(\text{osc}(X) = cl(X)\)

example:

1. if \(X = (0, 1)\), then \(\text{osc}(X) = cl(X) = [0, 1]\)
2. if \(X = Q \cap (0, 1)\), then \(\text{osc}(X) = cl(X) = [0, 1]\)

the outer limit of a sequence of sets

(\(\limsup_{i \to \infty} X_i\))

math def: the set of all points \(x\) such that every neighborhood of \(x\) intersects infinitely many of the sets \(X_i\). Formally,

\[\limsup_{n \to \infty} S_n = \bigcap_{N=1}^\infty \left( \bigcup_{n \geq N} S_n \right)\]

example: 1. if \(S_n = [0, 1 - 1/n]\), then \(\limsup_{n \to \infty} S_n = [0, 1)\)

1. if $S_n = \(S_n =\begin{cases} [0,1] & \text{if } n \text{ is odd} \\ [3,4] & \text{if } n \text{ is even} \end{cases}\). Then the outer limit is \(\limsup_{n \to \infty} S_n = [0,1] \cup [3,4]\) because every neighborhood of any point in \([0,1]\) intersects infinitely many odd-indexed sets, and every neighborhood of any point in \([3,4]\) intersects infinitely many even-indexed sets.

the inner limit of a sequence of sets

(\(\liminf_{i \to \infty} X_i\))

math def: the set of all points \(x\) such that every neighborhood of \(x\) intersects all but finitely many of the sets \(X_i\). Formally,

\[\liminf_{n \to \infty} S_n = \bigcup_{N=1}^\infty \left( \bigcap_{n \geq N} S_n \right)\]

example:

1. if \(S_n = [0, 1 - 1/n]\), then \(\liminf_{n \to \infty} S_n = [0, 1)\)

2. if $S_n = \(S_n =\begin{cases} [0,1] & \text{if } n \text{ is odd} \\ [3,4] & \text{if } n \text{ is even} \end{cases}\). Then the inner limit is \(\liminf_{n \to \infty} S_n = ∅\) because there are no points that belong to all but finitely many of the sets.

A locally eventually bounded sequence

math def: A sequence of sets \(\{X_n\}\) is locally eventually bounded if for every \(x \in \mathbb{R}^n\), there exists a neighborhood \(U\) of \(x\) and an integer \(N > 0\) such that the set

\[\bigcup_{n \geq N} (X_n \cap U)\]

is bounded.

example: for a set \(X_n = \[n, n+1\]\), for a neighborhood around any point x = 4, we can choose U = (3, 5), and N = 5, then the union of the sets from n = 5 to infinity intersected with U is empty, which is bounded. Thus, the sequence {X_n} is locally eventually bounded.

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Pointwise Convergence

\(f_n \rightarrow f\) pointwise on a set \(A\) if for every \(x \in A\),

\[\lim_{n \to \infty} f_n(x) = f(x).\]

example:

\[f_n(x) = \frac{x}{n}, x \in [0, 1]\]

pointwise converges to \(f(x) = 0\) as \(n \to \infty, \forall x \in [0, 1]\).

Uniform convergence in \(\mathcal{L}_2\) ?????????????

\(L_2\): The space of square-integrable functions, i.e., functions \(f\) such that

\[\int |f(x)|^2 \, dx < \infty.\]

and the distance between two functions \(f\) and \(g\) in \(L_2\) is defined by the \(L_2\) norm:

\[\|f - g\|_{L^2} = \left( \int |f(x) - g(x)|^2 \, dx \right)^{1/2}\]

and the sequence of functions \(F_n(x)\) converges to \(f(x)\) in \(L_2\) means, as \(n\) approaches infinity, the averaged squared error goes to 0:

\[\lim_{n \to \infty} \int_a^b |F_n(x) - F(x)|^2 \, dx = 0\]

math def: let S be a subset of real number system and let uniform convergence means that for every \(ε > 0\), there exists an integer \(N\) such that for all \(n \geq N\),

\[n > N \implies \|F_n(x) - f(x)\|_{L^2} < ε\]

plug in the \(L_2\) norm definition, the uniform convergence in \(L_2\) can be expressed as:

\[n > N \implies \left( \int_a^b |F_n(x) - f(x)|^2 \, dx \right)^{1/2} < ε\]

example:

\[ f_n(t) = t^n, t \in [0, 1] \]

as \(n \to \infty\), \(f_n(t)\) converges to the function \(f(t) = 0\) for \(t \in [0, 1)\) and \(f(1) = 1\), because

\[ \lim_{n \to \infty} \int_0^1 |t^n - 0|^2 \, dt = \lim_{n \to \infty} \int_0^1 t^{2n} \, dt = \lim_{n \to \infty} \left[ \frac{t^{2n+1}}{2n+1} \right]_0^1 = \lim_{n \to \infty} \frac{1}{2n + 1} = 0 \]

Graphical Convergence

math def:

Let Fₙ : ℝᵐ ⇉ ℝⁿ and F : ℝᵐ ⇉ ℝⁿ be set-valued mappings.

The sequence {Fₙ} converges graphically to F (written Fₙ →^G F or Fₙ \xrightarrow{G} F) if

\[\limsup_{n\to\infty} \mathrm{gph}(F_n) \subset \mathrm{gph}(F) \subset \liminf_{n\to\infty} \mathrm{gph}(F_n)\]

where gph(F) = { (x,y) ∈ ℝᵐ × ℝⁿ | y ∈ F(x) } is the graph of the set-valued mapping F, or:

\[\mathrm{gph}(F) = \lim_{n \to \infty} \mathrm{gph}(F_n)\]

In other words:

The graphs of the mappings Fₙ converge in the Kuratowski–Painlevé sense (i.e., both outer and inner limits of the graphs coincide) to the graph of F.

Example:

for the sequence of continuous functions \(f_n(x) = \arctan(nx)\), the corresponding set-valued mappings \(F_n(x) = \{ f_n(x) \}\) converge graphically to the set-valued mapping \(F(x)\) defined by:

\[ F(x) = \begin{cases} \left\{ \frac{\pi}{2} \right\} & \text{if } x > 0 \\ \left\{ -\frac{\pi}{2} \right\} & \text{if } x < 0 \\ [-\frac{\pi}{2}, \frac{\pi}{2}] & \text{if } x = 0 \end{cases} \]

first, we want to show that the inner limit of a sequence of sets of the graphs of \(F_n\) contains the graph of \(F\), i.e., $$ \mathrm{gph}(F) \subseteq \liminf_{n \to \infty} \mathrm{gph}(F_n) $$:

• case A: if x > 0 , as n → ∞, f_n(x) = arctan(nx) → π/2
• case B: if x < 0 , as n → ∞, f_n(x) = arctan(nx) → -π/2
• case C: if x = 0 , as n → ∞, f_n(0) = arctan(0) = 0. so
  1. subcase 1: to prove that \(y \in (-π/2, π/2)\): then we set \(y_n = \arctan(n x_n)\), it is clear that $x_n = \tan(y_n)/n $. Since \(\tan(y)\) is a finite function, as n → ∞, \(x_n\) → 0. Thus the sequence \((x_n, y_n) \in \mathrm{gph}(F_n)\) converges to (0, y).
  2. subcase 2: to prove \(y = π/2\): lets give a very small value to the \(x_n\), lets say make \(x_n = \frac{1}{\sqrt{n}}\). it is clear that as n → ∞, \(x_n\) → 0. so it satisfy the condition that x = 0 when n → ∞. now when we plug in \(x_n\) into \(f_n(x)\), we have \(y_n = f_n(x_n) = \arctan(n \cdot \frac{1}{\sqrt{n}}) = \arctan(\sqrt{n})\). as n → ∞, \(y_n\) → π/2. thus the sequence \((x_n, y_n) \in \mathrm{gph}(F_n)\) converges to (0, π/2).
  3. subcase 3: to prove that \(y = -π/2\): lets give a very small negative value to the \(x_n\), lets say make \(x_n = -\frac{1}{\sqrt{n}}\). it is clear that as n → ∞, \(x_n\) → 0. so it satisfy the condition that x = 0 when n → ∞. now when we plug in \(x_n\) into \(f_n(x)\), we have \(y_n = f_n(x_n) = \arctan(n \cdot -\frac{1}{\sqrt{n}}) = \arctan(-\sqrt{n})\). as n → ∞, \(y_n\) → -π/2. thus the sequence \((x_n, y_n) \in \mathrm{gph}(F_n)\) converges to (0, -π/2).

therefore, the inner limit is proved, every point in the graph of S can be approximated by a sequence of points from the graphs of \(F_n\).

similarly, we can show that the outer limit of the graphs of \(F_n\) is contained in the graph of \(F\), i.e., $$ \limsup_{n \to \infty} \mathrm{gph}(F_n) \subseteq \mathrm{gph}(F) $$:

• case A: if x > 0 , as n → ∞, f_n(x) = arctan(nx) → π/2
• case B: if x < 0 , as n → ∞, f_n(x) = arctan(nx) → -π/2
• case C: if x = 0 , as n → ∞, we want to find out the possible limit points of the sequence.

• since the range of arctan function is \((-π/2, π/2)\), thus the possible limit points are in the interval \([-π/2, π/2]\). therefore, any converging subsequence of \((0, y_n) \in \mathrm{gph}(F_n)\) must converge to a point in the graph of F.

• therefore, the outer limit is proved, every limit point of sequences from the graphs of \(F_n\) belongs to the graph of F.

Therefore, we have shown that both conditions for graphical convergence are satisfied, and thus \(F_n\) converges graphically to \(F\).

an outer semicontinuous set-valued mapping (osc)

math def: A set-valued mapping \(F: \mathbb{R}^n \rightrightarrows \mathbb{R}^m\) is outer semicontinuous at \(x\) if its graph is closed.Specifically, for every sequence of points \(x_i\) converging to \(x\) and every sequence \(y_i \in F(x_i)\) converging to \(y\), the limit \(y\) must be in \(F(x)\).

\[\{ x_i \to x, \; y_i \to y, \; y_i \in F(x_i) \} \implies y \in F(x)\]

example:

\[F(x) = \begin{cases} \{-1\} & x < 0 \\ [-1, 1] & x = 0 \\ \{1\} & x > 0 \end{cases}\]

is outer semicontinuous at \(x = 0\).

Prove: we want to show that that graph is closed. all other x ≠ 0 are continuous functions since they are constants, so they are osc. now we only need to prove at x = 0.

let \(x_i \to 0\), and \(y_i \in F(x_i)\)

• case A: if \(x_i < 0\), then \(y_i = -1\). so as \(x_i \to 0\), \(y_i \to -1\). and \(-1 \in F(0)\).
• case B: if \(x_i > 0\), then \(y_i = 1\). so as \(x_i \to 0\), \(y_i \to 1\). and \(1 \in F(0)\).
• case C: if \(x_i = 0\), then \(y_i \in [-1, 1]\). so as \(x_i \to 0\), \(y_i \to y \in [-1, 1]\). and \(y \in F(0)\).

therefore, in all cases, the limit \(y\) belongs to \(F(0)\), proving that the graph of \(F\) is closed at \(x = 0\). thus F is outer semicontinuous at x = 0.

an upper semicontinuous set-valued mapping (usc)

math def: A set-valued mapping \(F\) is upper semicontinuous at \(x\) if, for every open set \(V\) containing \(F(x)\), there exists a neighborhood \(U\) of \(x\) such that for all \(z \in U\), \(F(z) \subset V\).

example: the example from osc is also usc.

Proof: we want to show that for every open set V containing F(0) = [-1, 1], there exists a neighborhood U of 0 such that for all z in U, F(z) ⊆ V.

since F(0) = [-1, 1], let V be any open set containing [-1, 1]. we can choose a neighborhood U = (-δ, δ) for some small δ > 0.

• case A: if z < 0, then F(z) = {-1}. since -1 ∈ V, we have F(z) ⊆ V.
• case B: if z > 0, then F(z) = {1}. since 1 ∈ V, we have F(z) ⊆ V.

therefore, no matter how small the U is, for all z in U, F(z) is always a subset of V. thus F is upper semicontinuous at x = 0.

a lower semicontinuous set-valued mapping (lsc)

math def: A set-valued mapping \(F\) is lower semicontinuous at \(x\) if for every open set \(V\) that intersects \(F(x)\) (i.e., \(F(x) \cap V \neq \emptyset\)), there exists a neighborhood \(U\) of \(x\) such that for all \(z \in U\), \(F(z)\) also intersects \(V\).Intuitively, \(F\) does not "shrink" or "lose values" abruptly.

example:

the above example is NOT lsc.

Disprove by counterexample: we want to show that there exists an open set V that intersects F(0) = [-1, 1], but for any neighborhood U of 0, there exists a z in U such that F(z) does not intersect V.

let V = (-0.5, 0.5), which intersects F(0) = [-1, 1]. the intersection with V is (-0.5, 0.5).

now, lets say for any z > 0, F(z) = {1}, which does not intersect V.

An example of LSC is \(F(x) = [-1, 1]\) for all \(x\).

A locally Lipschitz set-valued mapping.

Math Def: A set-valued mapping \(F\) is locally Lipschitz near \(x\) if there exist positive constants \(\lambda\) (Lipschitz constant) and \(\delta\) such that for all \(x_1, x_2\) within distance \(\delta\) of \(x\):

\[F(x_1) \subset F(x_2) + \lambda \|x_1 - x_2\| \mathbb{B}\]

where \(\mathbb{B}\) is the unit ball.

This implies the "distance" between the sets (Hausdorff distance) is bounded linearly by the distance between the input points.

example:

\(F(x) = \{x\}\) for all \(x\).

Proof: based on the definition, the distance between the sets (which is \(d_H(F(x_1), F(x_2)) = \|x_1 - x_2\|\)) needs to be bounded linearly by the distance between the input points (which is \(\lambda \|x_1 - x_2\| \mathbb{B} = \lambda \|x_1 - x_2\|\)).

Thus, we just need to satisfy the requirement \(\|x_1 - x_2\| \leq \lambda \|x_1 - x_2\|\), which is true when we choose \(\lambda \geq 1\).

therefore, the mapping \(F(x) = \{x\}\) is locally Lipschitz near any point \(x\) with Lipschitz constant \(\lambda \geq 1\).

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Bouncing Ball (2)

A) simulate:

assumptions

assume that \(\gamma = 0.5\), and \(\lambda = 0.8\)

so here, \(x_1(0,0) = 10\) means at time 0, and when there is no jumps yet, the start height is 10.

\(x_2(0,0) = 1\) means at time 0, and when there is no jumps yet, the start velocity is 1.

the definition of x is therefore \(x = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}\)

and as the problem states, the derivative of x ( flow map ) which is \(\dot{x} = \begin{pmatrix} x_2 \\ -\gamma \end{pmatrix}\). which means the derivative of height is velocity, and the derivative of velocity is -γ (gravity pulling down) = \(-0.5\).

and the Jump matrix, what is the value of \(x\) after each bounce, is defined as:

\[ x^+ = \begin{pmatrix} x_1 \\ -\lambda x_2 \end{pmatrix} \]

so we can calculate some local max/min points:

the local max height when #jumps = 0: - should occur at \(t = \frac{x_2(0,0)}{\gamma} = \frac{1}{0.5} = 2s\), - and the max height is \(x_1(0,0) + x_2 = 11\)

the local min height when #jumps = 0: - should occur at

\[v_f ^ 2 = 2 * \gamma * h \Rightarrow v_f = \sqrt{2 * 0.5 * 11} = \sqrt{11} \approx 3.3166 m/s\]

• and the time to reach the ground is \(\(t = \frac{v_f - v_i}{\gamma} + 2 = \frac{\sqrt{11}}{0.5} + 2 \approx 8.63s\)\)

therefore, the first bounce should occur at around 8.63s, when the height is 0, and \(x_1(8.63, 0) = 0\), \(x_2(8.63, 0) = -\sqrt{11} \approx -3.3166 m/s\).

after the bounce, looking at the jump map, we have:

\[ x_1(0, 1) = x_1(8.63, 0) = 0 \]

\[ x_2(0,1) = -\lambda x_2(8.63, 0) = -0.8 * -\sqrt{11} = 0.8\sqrt{11} \approx 2.6533 m/s\]

Here is the simulation result using the Simulator.m file provided in the class notes:

B) what is the the limit point for \(x_1\) and \(x_2\)? is the limiting point part of the sets C and D?

as $ t \to \infty $, the limit point for \(x_1\) (height) is approaching 0, and the limit point for \(x_2\) (velocity) is also approaching 0.

the limiting point (0, 0) is neither part of the set C nor D. because both \(x_1\) and \(x_2\) need to be exactly 0 to be in the set.

C) now, consider the closure set of C and D, i.e., \(\text{cl}(C)\) and \(\text{cl}(D)\), denoted as \(\bar{C}\) and \(\bar{D}\). is the limiting point of part A contain in \(\bar{C} \cup \bar{D}\)?

the closure of the set C is:

• for the case \(x_1 > 0\), since \(x_1 = 0\) is a boundary point (because for any open neighborhood around (0, x2), there are points both in C and not in C), thus we include the boundary point, so \(\bar{C}\) includes the points where \(x_1 \geq 0\), which means \(\bar{C} = \{x \in \mathbb{R}^2 : x_1 \geq 0\}\).

• the second case is included in the first case, so we don't need to consider it separately.

the closure of the set D is:

• since \(x_2 = 0\) is a boundary point (because for any open neighborhood around (0, 0), there are points both in D and not in D), thus we include the boundary point, so \(\bar{D}\) includes the points where \(x_2 \leq 0\), which means \(\bar{D} = \{x \in \mathbb{R}^2 : x_1 = 0, x_2 \leq 0\}\).

therefore, the \(\bar{C} \cup \bar{D} = \{x \in \mathbb{R}^2 : x_1 \geq 0 \text{ or } (x_1 = 0, x_2 \leq 0)\}\). the point (0, 0) is included in this set, because it satisfies either conditions.

D) let \(x(0,0) = [0,0]^T\), and consider the hybrid system with flow set \(\bar{C}\) and jump set \(\bar{D}\). what are the hybrid time domains of the solution of the system from this initial condition?

if x(0,0) = [0,0]^T, that means the ball start with height 0 and velocity 0. it has no energy to begin with. ideally, it should just stay there forever, without any jumps or flows.

but if we consider the hybrid system with flow set \(\bar{C}\) and jump set \(\bar{D}\), then:

• the flow set \(\bar{C}\) allows the system to flow when \(x_1 \geq 0\). since \(x(0,0) = [0,0]^T\) satisfies this condition. Although the point (0,0) is in the set C, but since the flow is in a continuous time domain, we need to check if it can stay in the set C, so we need to check the \(\dot{x_2}\), the vertical acceleration, and find that it is \(-0.5\) and therefore it cannot stay in the flow set (flow is impossible)

• then, we check if it can jump. Since jump is in a discrete time domain, we only need to check if the point (0,0) is in the jump set D. since \(x(0,0) = [0,0]^T\) satisfies the condition of set D, it can jump.

• so after the jump, we have: the point \((0,-\lambda * 0) = (0,0)\) again. In this case, we satisfy to jump again.

Therefore, in this case, the hybrid time domain of the solution from this initial condition is: $ \text{dom } x = {(t, j) : t = 0, j \in \mathbb{N} } $, meaning that the system can only jump at discrete time steps j, but cannot flow in continuous time t.

therefore, the hybrid time domain of the system is:

\[\text{dom } x = \{0\} \times \{0, 1, 2, \dots \} = \bigcup_{j=0}^{\infty} ([0, 0] \times \{j\})\]

4) Hybrid system

A) draw the hybrid time domain of the solutions:

so in this case, we want to find the value of z at different time

1): \(x(0,0) = [5,1,0]\):

• at time 0 and 0 switch of modes, z(0,0) = \([5,1,0]\), means the value is 5, mode is 1, and the internal timer \(\tau\) is 0.
• from the matrix \(\dot(x)\), we know that the change rate of \(z\) when q = 1 is \(f_1(z) = -z\); the rate of change of the internal timer \(\tau\) is always \(\delta = 1\); and the mode q remains the same during flow.
• therefore, the interval of each mode change is exactly \(1 / 1 = 1s\). and during the initial state (no switch of modes yet), the value of z will decrease exponentially from 5 to \(5 + \int_{0}^{1} -z dt = 5 - \int_{0}^{1} 5e^{-t} dt = 5 - 5(1 - e^{-1}) = 5e^{-1} \approx 1.839\).

Find the intergral of \(z(t)\) during flow

since \(\dot{z} = -z\), we have

\[ \frac{dz}{dt} = -z \]

\[ \frac{dz}{z} = -dt \]

\[ \int \frac{1}{z} dz = \int -1 dt \]

\[ \ln|z| = -t + C \]

\[ z = e^{-t + C} = e^C e^{-t} = z(0) e^{-t} \]

given the initial condition \(z(0) = 5\), we have \(z(t) = 5e^{-t}\).

therefore, \(\int_{0}^{1} -z dt = \int_{0}^{1} -5e^{-t} dt = -5 \int_{0}^{1} e^{-t} dt = -5[-e^{-t}]_{0}^{1} = -5(-e^{-1} + 1) = 5(1 - e^{-1})\)

here is the graph of hybrid time domain (t) versus number of jumps (j):

here is the graph of z value over time:

2): \(x(0,0) = [5,1,0.5]\):

similar to the first case, but the initial internal timer \(\tau\) is 0.5 instead of 0. that means the first mode switch will occur at time \(t = (1 - 0.5) / 1 = 0.5s\).

after the mode switch, the value of z will grow at a rate of \(f_2(z) = 0.5z\) for 1s, until the next mode switch at \(t = 1.5s\).

at t = 0, #switch = 0, z = 5

at t = 0.5, #switch = 1, z = \(5 + \int_{0}^{0.5} -z dt = 5 - \int_{0}^{0.5} 5e^{-t} dt = 5 - 5(1 - e^{-0.5}) = 5e^{-0.5} \approx 3.0327\)

at t = 1.5, #switch = 2, z = \(3.0327 + \int_{0.5}^{1.5} 0.5z dt = 3.0327 + 2.5 (e^{0.75} - e^{0.25}) = \approx 5.1151\)

here is the graph of hybrid time domain (t) versus number of jumps (j):

here is the graph of z value over time:

3): \(x(0,0) = [5,1,1]\):

similar to the first case, but the initial internal timer \(\tau\) is 1 instead of 0. that means the first mode switch will occur at time \(t = (1 - 1) / 1 = 0s\). so the system will immediately switch to mode 2 at time 0.

here is the graph of hybrid time domain (t) versus number of jumps (j):

here is the graph of z value over time:

B) can you find a \(\delta > 0\) such that the system becomes unstable?

There is no such δ > 0 that can make the system unstable. Since the \(\delta\) only affects the rate of change of the internal timer τ, but does not affect the stability of the value z itself. A change in the rate of mode switching does not inherently lead to instability in the system.

when q = 1, the solution for z is decreasing at a rate of \(z(t) = z(0)e^{-t}\)

when q = 2, the solution for z is increasing at a rate of \(z(t) = z(0)e^{0.5t}\)

The period of time spent in each mode is determined by δ, so the period \(T = 1 / \delta\).

therefore, the value of z after one complete cycle (one switch to q=2 and back to q=1) can be expressed as:

\[ z(T) = z(0)e^{-T}e^{0.5T} = z(0)e^{-0.5T} = z(0)e^{-0.5 / \delta} \]

for any δ > 0, \(e^{-0.5 / \delta} < 1\), which means that after each complete cycle, the value of z decrease and stablize.

C) now consider the case when \(f_1(z) = -\frac{z}{t+1}, t \geq t_0 > 0\) and \(\delta = 1\). write the system as an autonomous hybrid system and simulate the system for 50s. Describe the emerging behavior of the system.

The system will eventually diverges and becomes unstable, because the decay rate in mode 1 decreases over time as t increases, leading to less effective reduction of z during mode 1.

here is the simulation result:

D) now, suppose that \(f_2\) is given by \(f_2 = -z^2\). Consider the initial condition \(x(0, 0) = [-1,2,0]\) Draw the hybrid time domain solution.

calculate the integral of z during flow when q = 2:

since \(\frac{dz}{dt} = -z^2\), we have \(-z^{-2}dz = dt\), take integral on both sides, we have

$ \int -z^{-2} dz = \int 1 dt $, then $ z^{-1} = t + C $. take the value \(z(0) = -1\), we have \(C = -1\). therefore, \(z(t) = \frac{1}{t - 1}\).

therefore, when t -> 1 from the left, z(t) -> -∞. so the value of z diverges to negative infinity as t approaches 1 during mode 2.

however that is not the case in the matlab simulation, because the min step size is still to large to show the limit behavior. here is the simulation result when the step size is set to 1e-3:

here is the simulation result when the step size is set to 1e-5:

we can see that as the step size decreases, the simulation gets closer to the theoretical behavior where z diverges to negative infinity as t approaches 1.

17) Hybrid Lyapunov functions for Linear/Affine systems

Lyapunov function \(V(x): \mathbb{R}^n \to \mathbb{R}\) is a scalar function that helps to analyze the stability of a dynamical system by checking if the energy of an object will decrease to 0 over time. it must satisfy those: 1. \(V(0) = 0\) (if the system is at equilibrium (state x=0, velocity=0), the energy is 0) 2. \(V(x) > 0\) for all \(x \neq 0\) (if there is any energy in the system, then the state (x) is not at equilibrium) 3. \(V(x) \rightarrow \infty\) as \(|x| \rightarrow \infty\) (radially unbounded) (if the state x is very far from equilibrium, the energy is very high, meaning it can never stabilize at infinity) 4. \(\dot{V}(x) < 0\) along the flow of the system (the energy decreases constant during continuous evolution)

The defition of the energy of a classic Kinetic Energy is given by:

$$ V(x) = \frac{1}{2} mv^2$$, where m is mass, and v is velocity.

for vector systems, we can generalize the Lyapunov function as:

$$ V(x) = x^T P x $$,

in this case, x is the state vector that has 2 dimensions (position and velocity), therefore:

\[ V(x) = 1/2 x^T x = \frac{1}{2} (x_1^2 + x_2^2) \]

to prove the UGAS (Uniformly Globally Asymptotically Stable) of the hybrid system, we need to find a Lyapunov function \(V(x): \mathbb{R}^n \to \mathbb{R}\) that satisfies the following conditions:

1. V(x) decreases along the flow of the system.
2. V(x) decreases across jumps of the system.

Flow condition

for the V(x) to converge and stablize during flow, need $\dot{V}(x)\leq 0, \quad \forall x \in C $.

from the definition of \(\dot{V}(x)\), we have:

\[ \dot{V}(x) = \frac{\partial V}{\partial x} \cdot \dot{x} = x_1(\beta x_1 + 1) + x_2(\beta x_2 + 2) = \beta(x_1^2 + x_2^2) + (x_1 + 2x_2) \]

\(\dot{V}(x) = \beta |x| ^ 2 + P^Tx\), so the \(P = [1; 2], \quad |P| = \sqrt{1^2 + 2^2} = \sqrt{5}\).

therefore, to make sure that \(\dot{V}(x) < 0\), we need to choose \(\beta\) such that in the worst case when \(x = 1\):

\[ \beta |x| ^ 2 + |P||x| < 0 \]

\[ \beta * 1^2 + \sqrt{5} * 1 < 0 \]

\[ \beta < -\sqrt{5} \approx -2.236 \]

Jump condition

we need to make sure that $V(x^+) - V(x) \leq 0, \quad \forall x \in D $.

\[V(x^+) = \frac{1}{2} * (\lambda x)^ T (\lambda x) = \frac{1}{2} \lambda^2 x^T x = \frac{1}{2} \lambda^2 V(x)\]

and we need \(\frac{V(x^+)}{V(x)} = \lambda^2 < 1\), therefore, we have \(|\lambda| < 1\).

Conclusion

therefore, to make the hybrid system UGAS at origin (\(\mathcal{A} = \{0\} \times \{0\}\)), we need to choose \(\beta < -\sqrt{5} \approx -2.236\) and \(|\lambda| < 1\) .