HW2

Problem 1:

A) Convexity Properties

Prove for \(\phi = \frac{1}{4} x^4\), prove for convexity:

Its first derivative is \(\phi' = x^3\) and its second derivative is \(\phi'' = 3x^2\). Since \(\phi'' \geq 0\) for all \(x\), \(\phi\) is convex.

Since for all \(x \neq 0\), \(\phi'' > 0\), therefore, \(\forall x < y, f'(x) < f'(y)\) due to the fact that \(\phi'\) is strictly increasing, so \(\phi\) is strictly convex.

the strongly convex requires that \(\phi'' \geq m > 0\) for some constant \(m\). However, when \(x = 0\), \(\phi'' = 0\), so we cannot find such a constant \(m\) that satisfies the condition. Therefore, \(\phi\) is not strongly convex.

B) show that for any \(c>0\), the origin is UGAS for this system

\[\dot{x} = -c \nabla \phi(x) = -c x^3\]

Construct the Lyapunov function \(V(x) = x^2\). we need to show that

1. \(V(x)\) is positive definite
   • for \(x \neq 0\), \(V(x) = x^2 > 0\)
   • and for \(x = 0\), \(V(0) = 0\)
   • thus, \(V(x)\) is positive definite
2. \(V(x)\) is radially unbounded
   • as \(|x| \to \infty\), \(V(x) = x^2 \to \infty\)
   • thus, \(V(x)\) is radially unbounded
3. \(\dot{V}(x)\) is negative definite
   • \(\dot{V} = 2x \dot{x} = 2 * x * (-c x^3) = -2c x^4 \leq 0\).
   • for \(x \neq 0, c > 0\), \(\dot{V} = -2c x^4 < 0\)
   • thus, \(\dot{V}(x)\) is negative definite

therefore, the function \(V(x)\) is GAS.

since the original dynamic system \(\dot{x} = -c \nabla \phi(x)\) is autonomous, by Lyapunov's theorem, the origin is UGAS for this system.

C) show that the origin is not exponentially stable for this system

Using the previous Lyapunov function \(V(x) = x^2\), we want to show that for some \(M > 0\) and \(\alpha > 0\), \(V(x(t)) \leq M e^{-\alpha t} V(x(0))\) does not hold for all \(t \geq 0\), we want show that its rate of change is less than exponential decay at origin.

from the part B, we have \(\dot{V} = -2c x^4\). We can rewrite this as \(\dot{V} = -2c V^2\) since \(V(x) = x^2\).

\[dV = -2cV^2 dt \\ \frac{dV}{V^2} = -2c dt \]

Then take the definite integral from \(\tau = 0\) to \(\tau = t\) of both sides, we have:

\[\int_{V(0)}^{V(t)} \frac{dV}{V^2} = \int_0^t -2c dt \\ -\frac{1}{V(t)} + \frac{1}{V(0)} = -2ct \\ V(t) = \frac{v(0)}{1 + 2c t v(0)}.\]

Since this function \(V(t) = \frac{v(0)}{1 + 2c t v(0)} = O(\frac{1}{t})\) which decays slower than any exponential function \(M e^{-\alpha t} = O(e^{-\alpha t})\), The origin is not exponentially stable for this system. (for example, when t approach infinity)

D) consider the second order system, use the hybrid invarianc principle and the following lyapunov to establish that the set is UGAS

The set is \(A = \{0\} \times \{0\} \times \{T_{\min}, T_{\max}\}\)

\[V(x) = \frac{1}{4}|x_2 - x_1|^2 + \frac{1}{4} |x_2|^2 + cx_3^2 \phi(x_1)\]

1. check for positive definite:
   • the set A implies that \(x_1 = 0, x_2 = 0, x_3 = [T_\text{min}, T_\text{max}]\)
   • Simply the Lyapunov function as: \(V([0,0,x_3]) = 0 + 0 + cx_3^2 * 0 = 0\).
   • when $ x \neq 0$, \(V(x) > 0\) since all the terms are positive, thus, \(V(x)\) is positive definite with respect to the set A.
2. check for radially unbounded:
   • as \(|x| \to \infty\), \(V(x) \to \infty\) since at least \(|x_2|^2\) will go to infinity, thus, \(V(x)\) is radially unbounded.
3. check for negative definite:

\[\frac{d}{dx}\frac{1}{4}|x_2 - x_1|^2 \\ = \frac{1}{2} (x_2 - x_1) (x_2' - x_1') \\ = \frac{1}{2} (x_2 - x_1) (-2cx_3 * x_1^3 - \frac{2}{x_3}(x_2 - x_1)) \\ = -c x_3 (x_2 - x_1) \nabla\phi - \frac{1}{x_3} \|x_2 - x_1\|^2 \tag{A} \]

\[\frac{d}{dx} \frac{1}{4} * |x_2|^2 \\ = \frac{1}{2} x_2 x_2' \\ = \frac{1}{2} x_2 (-2cx_3 * \nabla\phi) \\ = -c x_3 x_2 \nabla\phi \tag{B} \]

\[\frac{d}{dx} cx_3^2\phi(x_1) \\ = cx_3\phi + 2 c x_2 x_3 \nabla \phi = 2c x_1 x_3 \nabla \phi \\ = c x_3 \phi + 2 c x_3 \nabla \phi (x_2 - x_1) \tag{C} \]

Adding (A), (B), and (C) together, we have:

\[\dot{V} = \Bigl[-c x_3 (x_2 - x_1)^T \nabla\phi - \frac{1}{x_3} \|x_2 - x_1\|^2\Bigr] + \Bigl[-c x_3 x_2^T \nabla\phi\Bigr] + \Bigl[c x_3 \phi + 2c x_3 (x_2 - x_1)^T \nabla\phi\Bigr] \\ = \dot{V} = -3cx_3 \phi(x_1) - \frac{1}{x_3}|x_2 - x_1|^2\]

since all \(x_1, x_2, x_3 > 0\), \(\dot{V} \leq 0\). Therefore, it is negative definite

Therefore, by the hybrid invariance principle, the set A is UGAS.

E) compare the soluations of the solution 1 and 2 from the same initial conditions of \(x_1\)

F) does the set A set still UGAS when T -> infinity?

since the definition for the jump set is when \(x_3 \rightarrow T_{\max}\), the system will never reach the jump set when \(T_{\max} \to \infty\). Therefore, the question is equivalent to asking whether the flow system at set \(A = \{0\} \times \{0\} \times \{T_{\min}, \infty\}\) is UGAS.

as we can observe from the part E, when the x_3 approach to infinity, for an arbitary value of \(X = [\delta, 0, x_3]\) the change in the V(x) will be dominated by the term \(cx_3^2 \phi(x_1)\), which will also grow to infinity, but the change distance from \([\delta, 0, x_3]\) to the set A is only \(\delta\), which is a constant.

therefore, we cannot find a \(\delta\) such that \(\Delta V(x) \leq \alpha(\delta)\) for some \(\alpha \in \mathcal{R}\), which is required for the definition of UGAS. Therefore, the set A is not UGAS when \(T_{\max} \to \infty\).

Problem 2:

A) show that \(\phi\) is radically unbounded

\[\frac{1}{2} \mu |x_1 - x^*|^2 <= \phi (x_1) - \phi^* \\ \rightarrow \phi(x_1) \ge \frac{1}{2} \mu |x_1 - x^*|^2 + \phi^* \]

as \(|x_1| \to \infty\), since both \(x^*\) and \(\phi^*\) are constants (global minimum), the RHS approaches to infinity. therefore, \(\phi\) is radically unbounded.

B) Consider the Hybrid system

B.i) find conditions one the parameters \(c, T_{\min}, T_{\max}\) such that the set \(A = \{x^*\} \times \{x^*\} \times \{T_{\min}, T_{\max}\}\) is UGES

The Lyapunov function is given as:

to show that the set A is UGES, we need to show prove the 3 conditions listed above:

5a) the Lyapunov function is quadratically bounded

let \(e_1 = x_1 - x^*, e_2 = x_2 - x^*\), so \(s = e_2 -e_1 = x_2 - x_1\)

\[V(x) = \frac{1}{4} |s|^2 + \frac{1}{4} |e_2|^2 + c x_3^2 (\phi(x) - \phi ^*)\]

from part A, we have that \(\phi(x) - \phi^* \geq \frac{1}{2} \mu |e_1|^2\), therefore,

\[V(x) \geq \frac{1}{4} |s|^2 + \frac{1}{4} |e_2|^2 + c x_3^2 * \frac{1}{2} \mu |e_1|^2\]

let p = 2, in this case, we have \(c_1|x|^2_A \leq V(x) \leq c_2 |x|^2_A\)

find \(|x|^2_A\), the distance from x to A

\[|x|^2_A = |x_1 - x^*|^2 + |x_2 - x^*|^2 + \min_{x_3 \in \{T_{\min}, T_{\max}\}} |x_3 - x_3|^2 \\ = |x_1 - x^*|^2 + |x_2 - x^*|^2 = |e_1|^2 + |e_2|^2 \]

the lower bound:

\[c_1 |x|^2_A = c_1 (|x_1 - x^*|^2 + |x_2 - x^*|^2) = c_1 (|e_1|^2 + |e_2|^2) \]

the lower bound reached global min when \(x_3 = T_{\min}\), therefore, we have:

\[V \geq \frac{1}{4} (|s|^2 + |e_2|^2) + cT_{\min}^2 \frac{\mu}{2}|e_1|^2\]

for this to \(\geq c_1 |x|^2_A\), we need to have:

\[c_1 ( |e_1|^2 + |e_2|^2) \leq \frac{1}{4} (|s|^2 + |e_2|^2) + cT_{\min}^2 \frac{\mu}{2}|e_1|^2\]

since |s|^2 is positive, therefore, the \(c_1\) needs to be less than \(min\{\frac{1}{4}, c T_{\min}^2 \frac{\mu}{2}\}\)

similarly for the upper bound, there is a solution for c_1, and c_2

5b) the flow condition

there is such a lambda that satisfies the flow condition.

5c) the jump condition

\(\(V(g) \leq \exp(-\lambda)V(x)\)\), means the post-jump state, \(g\), has a smaller value of the Lyapunov function than the pre-jump state, \(x\), by a factor of \(\exp(-\lambda)\).

therefore, we need to show that \(V(x^+) \leq V(x)\).

calculate for \(V(x)\), before jump: when \(x_3 = T_{\max}\):

\[ V(x) = \frac{1}{4} |x_2 - x_1|^2 + \frac{1}{4} |x_2 - x^*|^2 + c T_{\max}^2 (\phi(x) - \phi^*) \]

calculate for \(V(x^+)\), after jump: when \(x_3 = T_{\min}\):

note that \(x_1^+ = x_1, x_2^+ = x_1, x_3^+ = T_{\min}\)

\[ V(x^+) = \frac{1}{4} |x_1 - x_1|^2 + \frac{1}{4} |x_1 - x^*|^2 + c T_{\min}^2 (\phi(x^+) - \phi^*) \\ = \frac{1}{4} |x_1 - x^*|^2 + c T_{\min}^2 (\phi(x) - \phi^*) \]

AND WE WANT TO MAKE SURE THAT \(V(x^+) \leq V(x)\), therefore, we need to have:

\[\frac{1}{4} |x_1 - x^*|^2 + c T_{\min}^2 (\phi(x) - \phi^*) \leq \frac{1}{4} |x_2 - x_1|^2 + \frac{1}{4} |x_2 - x^*|^2 + c T_{\max}^2 (\phi(x) - \phi^*) \\ \frac{1}{4} (|x_1 - x^*|^2 - |x_2 - x_1| ^ 2 - |x_2 - x^*|^2) \le c(\phi(x) - \phi^*)(T_{\max}^2 - T_{\min}^2) \]

Triangular inequality: \(|x_1 - x^*| \leq |x_1 - x_2| + |x_2 - x^*|\)

\[|x_2 - x_1|^2 - |x_2 - x^*| ^ 2 \geq \frac{1}{2} |x_1 - x^*|^2\]

part A:

\(\phi(x) - \phi^* \geq \frac{1}{2} \mu |x_1 - x^*|^2\)

\[c(\phi(x) - \phi^*)(T_{\max}^2 - T_{\min}^2) \geq c * \frac{1}{2} \mu |x_1 - x^*|^2 (T_{\max}^2 - T_{\min}^2) \\ \ge \frac{1}{4} (\frac{1}{2} |x_1 - x^*|^2)\]

\[\rightarrow c \geq \frac{1}{4 \mu (T_{\max}^2 - T_{\min}^2)} \]

therefore, for the set A to be UGES, we need to have \(c \geq \frac{1}{4 \mu (T_{\max}^2 - T_{\min}^2)}\)

part B.ii) simulate the solutions of the system (TODO: IDK)

c - the higher the c, the faster the convergence, and the movement isfaster

• the friction level after a restart. the smaller the \(T_{\min}\), the more intense is the initial damping.

the larger the \(T_{\max}\), the system resets less frequently, which means the system will have more time to converge before it resets, therefore, the system will converge faster.

Problem 3:

Calculate the convergence time as a function of initial conditions. show the system is UGFTAS at origin, and simulate the system and describe their emerging behavior from different initial conditions.

A) \(\dot{x} = - \text{sign}(x)\)

prove for UGFTAS:

construct the lyapunov function \(V(x) = x^2\), we want to show that it is UGFTAS at origin

1. \(V(x)\) is positive definite
   • for \(x \neq 0\), \(V(x) = x^2 > 0\)
   • and for \(x = 0\), \(V(0) = 0\)
   • thus, \(V(x)\) is positive definite
2. \(V(x)\) is radially unbounded
   • as \(x \to \infty\), \(V(x) = x^2 \to \infty\)
   • thus, \(V(x)\) is radially unbounded
3. \(\dot{V}(x)\) is negative definite
   • \(\dot{V} = 2x \dot{x} = 2 * x * (-\text{sign}(x)) = -2|x| \leq 0\).
   • thus, \(\dot{V}(x)\) is negative definite

therefore, the function \(V(x)\) is GAS. and since the system is autonomous, by Lyapunov's theorem, the origin is UGAS for this system.

to prove finite time stable:

we need to satisfy the condition that \(\dot{V} \leq -c V(x)^p\), for some \(c > 0\) and \(p \in (0,1)\)

LHS: \(\dot{V} = -2|x|\). Since \(V(x) = x^2\), we have \(|x| = \sqrt{V(x)}\), therefore, \(\dot{V} = -2 \sqrt{V(x)}\)

RHS: \(-c V(x)^p = -c x^{2p}\), choose \(p = \frac{1}{2}\), we have \(-c x = -c \sqrt{V(x)}\)

therefore, we can have a \(c = 2\) and \(p = \frac{1}{2}\) such that \(-2\sqrt{V(x)} \leq -2\sqrt{V(x)}\), which means the system is finite time stable.

convergence time:

from previous, we have

\[\dot{V} \leq -2 \sqrt{V(x)} \\ \frac{dV}{dt} \leq -c V^p \\ \frac{dV}{V^p} \leq -c dt \\ \]

when the time t goes from 0 to T, and V goes from \(V(x_0)\) to 0, we have:

\[\int_{V(x_0)}^0 \frac{dV}{V^p} \leq \int_0^T -c dt \]

LHS: \(\int_{V(x_0)}^0 V^{-p}dV = [\frac{v^{1-p}}{1-p}]^0_{V(x_0)} = \frac{0}{1-p} - \frac{V(x_0)^{1-p}}{1-p} \\ = - \frac{V(x_0)^{1-p}}{1-p}\)

RHS: \(\int_0^T -c dt = -cT\)

therefore, we have:

\[- \frac{V(x_0)^{1-p}}{1-p} \leq -cT \\ T \leq \frac{V(x_0)^{1-p}}{c(1-p)}\]

for \(p = \frac{1}{2}\), and \(c = 2\), we have:

\[ T = \frac{V(x_0) ^ \frac{1}{2}}{2 * \frac{1}{2}} = \sqrt{V(x_0) ^ \frac{1}{2}} = |x_0|\]

B) \(\dot{x} = -x^{\frac{1}{3}}\)

Choose the same Lyapunov function \(V(x) = x^2\),

we have shown that V(x) is positive definite and radially unbounded in part A, so we only need to check whether \(\dot{V}\) is negative definite:

\[\dot{V} = 2x \dot{x} = 2 * x * (-x^{\frac{1}{3}}) = -2 x^{\frac{4}{3}} \leq 0\]

therefore, \(\dot{V}\) is negative definite, and the system is UGAS.

similarly, $\dot{V} = -2 x^\frac{4}{3} = -2 V(x)^\frac{2}{3} \leq -c V(x)^p $.

therefore, we only need to choose \(c = 2\) and \(p = \frac{2}{3}\) to satisfy the condition for finite time stability, therefore, the system is finite time stable, therefore the system is UGFTAS at origin.

converge time:

similarly, we have:

\[ T \leq \frac{V(x_0)^{1-p}}{c(1-p)}\]

for \(p = \frac{2}{3}\), and \(c = 2\), we have:

\[ T = \frac{V(x_0) ^ {\frac{1}{3}}}{2 * (\frac{1}{3})} = \frac{3}{2} V(x_0) ^ {\frac{1}{3}} = \frac{3}{2} |x_0|^{\frac{2}{3}}\]

C) for strongly convex function \(\phi(x)\) with \(x^* = \arg \min \phi (x) = 0\)

The dynamics is \(\dot{x} = - \frac{\nabla \phi (x)}{|\nabla \phi (x)| ^ \frac{1}{2}}\)

using the Lyapunov function \(V(x) = \frac{1}{2} |\nabla \phi(x)| ^ 2\).

prove for the UGAFTS:

1. \(V(x)\) is positive definite
   • for \(x \neq 0\), since \(\phi\) is strongly convex, \(\nabla \phi(x) \neq 0\), therefore, \(V(x) = \frac{1}{2} |\nabla \phi(x)| ^ 2 > 0\)
   • and for \(x = 0\), since \(x^* = 0\) is the global minimum, \(\nabla \phi(0) = 0\), therefore, \(V(0) = 0\)
   • thus, \(V(x)\) is positive definite
2. \(V(x)\) is radially unbounded
   • as \(|x| \to \infty\), since \(\phi\) is strongly convex, \(|\nabla \phi(x)| \to \infty\), therefore, \(V(x) = \frac{1}{2} |\nabla \phi(x)| ^ 2 \to \infty\)
   • thus, \(V(x)\) is radially unbounded
3. \(\dot{V}(x)\) is negative definite

let $ y(x) = \nabla \phi(x), \frac{dy}{dx} = \nabla^2 \phi(x), V(y) = \frac{1}{2} y(x)^T y(x), dV = \frac{1}{2} (dy^Ty + y^T dy) = y^T dy = y^T \nabla^2 \phi(x)$

\[\dot{V} = \nabla V(x) \dot{x} = (dV)^T \dot{x} \\ = (y^T \nabla^2 \phi(x))^T \dot{x} = -(\nabla \phi(x)) (\nabla ^ 2 \phi (x))^T \frac{\nabla \phi(x)}{|\nabla \phi (x)| ^ \frac{1}{2}}\]

because the strong convexity, we know that \(y^T \nabla^2 \phi(x) y \geq m |y|^2\) for some constant \(m > 0\), therefore, we have:

\[\dot{V} \leq -m |\nabla \phi(x)|^2 / |\nabla \phi(x)|^{\frac{1}{2}} = -m |\nabla \phi(x)|^{\frac{3}{2}} = -c V(x)^p\]

in terms of V, we have $ c = m * 2^{\frac{3}{4}}$ and \(p = \frac{3}{4}\), therefore, the system is finite time stable, therefore the system is UGFTAS at origin.

convergence time

from the previous, we have:

\[ T \leq \frac{V(x_0)^{1-p}}{c(1-p)}\]

for \(p = \frac{3}{4}\), and \(c = m * 2^{\frac{3}{4}}\), we have

\(V(x_0) = \frac{1}{2} |\nabla \phi(x_0)|^2\), therefore, we have:

\[T \geq \frac{2}{m} |\nabla \phi (x_0)|^{\frac{1}{2}}\]

Problem 4:

to show a UGAS is fixed time stable in premise of UGAS, we need to show

\[\dot{V} (X) \leq - (\alpha_1 V^p(x) + \alpha_2 V^q(x))^k\]

A) the dynamics \(\dot{x} = -x^\frac{1}{3} - x^3\)

Prove for UGAFxTS

Construct the Lyapunov function \(V(x) = x^2\), we want to show that it is UGAS at origin

the \(V(x)\) is positive definite and radially unbounded as proved in all the previous quesitons, we only want to show that \(\dot{V}\) is negative definite:

\[\dot{V} = 2x \dot{x} = 2 * x * (-x^\frac{1}{3} - x^3) = -2 x^{\frac{4}{3}} - 2 x^4 \leq 0\]

therefore, \(\dot{V}\) is negative definite, and the system is UGAS.

to show that the system is fixed time stable, rearrange the \(\dot{V}\) using \(V(x) = x^2\):

\[\dot{V} = -2 V(x) ^ \frac{2}{3} - 2 V(x)^2\]

therefore, we can choose \(\alpha_1 = 2, \alpha_2 = 2, p = \frac{2}{3}, q = 2, k = 1\) to satisfy the condition for fixed time stability, therefore, the system is fixed time stable at origin.

convergence time upper bound

from the lecture, we know that the upper bound for convergence time is given by:

\[T(x_0) \leq \frac{1}{\alpha_1^k(1-pk)} + \frac{1}{\alpha_2^k(qk-1)}\]

for \(\alpha_1 = 2, \alpha_2 = 2, p = \frac{2}{3}, q = 2, k = 1\), we have:

\[T(x_0) \leq \frac{1}{2^1(1-\frac{2}{3}*1)} + \frac{1}{2^1(2*1-1)} = \frac{3}{2} + \frac{1}{2} = 2\]

B)

From the conclusion in the question 3, we know that

\[\dot{V} = \nabla \phi^T \nabla^2 \phi \dot{x} \\ = \nabla \phi^T \nabla^2 \phi (- c_1 \frac{\nabla \phi}{|\nabla \phi|^{\frac{p_1 - 2}{p_1 - 1}}} - c_2 \frac{\nabla \phi}{|\nabla \phi|^{\frac{p_2 - 2}{p_2 - 1}}})\]

from the strong convexity, we have \(\nabla \phi^T \nabla^2 \phi \nabla \phi \geq m |\nabla \phi|^2\), therefore, we have:

\[\dot{V} \leq - c_1m\frac{|\nabla \phi|^2}{|\nabla \phi|^{\frac{p_1 - 2}{p_1 - 1}}} - c_2 m \frac{|\nabla \phi|^2}{|\nabla \phi|^{\frac{p_2 - 2}{p_2 - 1}}} \\ \leq -c_1 m |\nabla \phi| ^ \frac{p_1}{p_1 - 1} - c_2 m |\nabla \phi| ^ \frac{p_2}{p_2 - 1}\]

since \(V(x) = \frac{1}{2} |\nabla \phi|^2\), we have \(|\nabla \phi| = \sqrt{2V}\), therefore, we have:

\[\dot{V} \leq - (c_1 m 2^{\frac{p_1}{2(p_1 - 1)}} V^{\frac{p_1}{2(p_1 - 1)}} + c_2 m 2^{\frac{p_2}{2(p_2 - 1)}} V^{\frac{p_2}{2(p_2 - 1)}})\]

therefore, we can choose \(\alpha_1 = c_1 m 2^{\frac{p_1}{2(p_1 - 1)}}, \alpha_2 = c_2 m 2^{\frac{p_2}{2(p_2 - 1)}}, p = \frac{p_1}{2(p_1 - 1)}, q = \frac{p_2}{2(p_2 - 1)}, k = 1\) to satisfy the condition for fixed time stability, therefore, the system is fixed time stable at origin.

find the upper bound for convergence

from the lecture, we know that the upper bound for convergence time is given by:

\[T(x_0) \leq \frac{1}{\alpha_1^k(1-pk)} + \frac{1}{\alpha_2^k(qk-1)}\]

plug in the values for \(\alpha_1, \alpha_2, p, q, k\), we have:

\[T(x_0) \leq \frac{1}{(c_1 m 2^{\frac{p_1}{2(p_1 - 1)}})^1(1-\frac{p_1}{2(p_1 - 1)}*1)} + \frac{1}{(c_2 m 2^{\frac{p_2}{2(p_2 - 1)}})^1(\frac{p_2}{2(p_2 - 1)}*1-1)}\]

Problem 5:

A) show that every solution of the systen satisfies \(|x(t)| \rightarrow \infty\) as \(t \to \infty\)

when \(x(0) \in \R \setminus \{0\}, x_2 \in {0}\).

since \(\dot{x_2} = 0\), therefore, \(x_2(t) = C\). And from the given condition, C = 0, therefore, in the initial condition, we always have \(x_2(0) = 0\).

for \(\dot{x_1}\), since in the initial condition, \(x_2(0) = 0\), therefore, \(\dot{x_1} = x_1\). Therefore:

\[\int \frac{dx_1}{x_1} = \int dt \\ \ln |x_1(t)| = t + C \\ |x_1(t)| = e^t * e^C = x_1(0) e^t\]

therefore, as \(t \to \infty\), \(|x_1(t)| \to \infty\), and \(|x(t)| \geq |x_1(t)|\). Therefore, every solution of the system satisfies \(|x(t)| \to \infty\) as \(t \to \infty\).

B) Filippov with initial solution \(x_0 = x(0) \in (\R \setminus \{0\} \times \{0\})\) that satisfies \(|x(t)| \to 0\) as \(t \to \infty\)

The Filippov solution is given by:

\[F(x) = \bigcap_{\delta > 0} \bigcap_{\mu(N)=0} \overline{\text{co}} \left( f(B(x,\delta) \setminus N) \right)\]

the filippov set therefore exlucluding all the points that has measure 0, which means the filippov solution will exclude the points where \(x_2 = 0\).

in this case, for any point near x with \(x_2 = 0\), the vector field is \((-x_1, 0)\), therefore, the Filippov set is \(F(x) = {(-x_1, 0)^T}\).

therefore, \(\dot(x_1) = -x_1, \dot{x_2} = 0\), and as \(t \to \infty\), \(|x_1(t)| \to 0\), and \(|x_2(t)| = 0\). Therefore, the filippov solution with initial condition \(x(0) \in (\R \setminus \{0\} \times \{0\})\) satisfies \(|x(t)| \to 0\) as \(t \to \infty\).

C) for the initial condition \(x(0) \in \R \setminus \{0\} \times \{0\}\), there exists a Krasovskii solution that satisfies \(x(t) = x_0\) for all \(t \geq 0\)

the definition of Krasovskii solution is given by:

\[K(x) = \bigcap_{\delta > 0} \overline{\text{co}} \left( f(B(x,\delta)) \right)\]

at the point where \(x = (x_1, 0)\), we can find the points where \(x_2 = 0\), whose vector field is \((x_1, 0)\), and the points where \(x_2 \neq 0\), whose vector field is \((-x_1, 0)\).

therefore, the Krasovskii set is given by \(K(x) = \{(x_1, 0)^T, (-x_1, 0)^T\}\), which means the K(x) is a line segment between \((x_1, 0)^T\) and \((-x_1, 0)^T\). And this contains the point (0, 0), therefore, we can find a Krasovskii solution that satisfies \(x(t) = x_0\) for all \(t \geq 0\).

For each initial condition \(x(0) \in \R \setminus \{0\} \times \{0\}\), find an arbitrary small state perturbation \(e : \R_{\geq 0} \to \R^2\) that

From Part C, the Krasovskii set is given by \(K(x) = \{(x_1, 0)^T, (-x_1, 0)^T\}\), therefore, the average velocity of the system is given by

\[v_\text{avg} = \lambda v_1 + (1-\lambda) v_2\]

from part C, we have \(v_1 = (x_1, 0)^T\) and \(v_2 = (-x_1, 0)^T\), therefore, we have:

\[v_\text{avg} = \lambda (x_1, 0)^T + (1-\lambda) (-x_1, 0)^T \]

since \(x_2\) is always 0, the distance of the point to the origin is determined by the change in \(x_1\), and we can find that the change in the distance to the origin is given by:

\[\dot{x_1} = \lambda x_1 + (1-\lambda) (-x_1) = (2\lambda - 1) x_1\]

D) ... keeps the trajectory of the system in an arbitrarily small neighborhood of the \(x_0\)

if the change in \(x_1\) is 0, then the change of the x is also 0, therefore, the trajectory of the system will be in an arbitrarily small neighborhood of \(x_0\).

therefore, when \(\lambda = \frac{1}{2}\), or for a small cycle from \([0, T]\), the perturbation function is given by:

\[e(t) = \begin{cases} 0 & t \in [0, \frac{T}{2}) \\ \epsilon & t \in [\frac{T}{2}, T) \end{cases}\]

for any \(\epsilon > 0\)

E) ... yielding convergence to 0 arbitrarily slow

similarly, if the change in \(x_1\) is negative, then the change of the x is also negative, therefore, the trajectory of the system will converge to 0.

therefore, when \(\lambda < \frac{1}{2}\), or for a small \(\delta > 0, \lambda = \frac{1 - \delta}{2}\). In a small cycle from \([0, T]\), the perturbation function is given by:

\[e(t) = \begin{cases} 0 & t \in [0, T\frac{1 - \delta}{2}) \\ \epsilon & t \in [T\frac{1 - \delta}{2}, T) \end{cases}\]

F) ... yielding divergence to infinity arbitrarily slow

similarly, if the change in \(x_1\) is positive, then the change of the x is also positive, therefore, the trajectory of the system will diverge to infinity.

\[ e(t) = \begin{cases} 0 & t \in [0, T\frac{1 + \delta}{2}) \\ \epsilon & t \in [T\frac{1 + \delta}{2}, T) \end{cases}\]