HW3

Question 1

we want to show that the completeness of the solution. To do that, we need to show that

1. there is no finite escape time: every state variable is bounded for all time.
2. No deadlock: the solutions can always evolve, either by flowing or by jumping.
3. No Zeno behavior: the solution does not have infinitely many jumps in a finite amount of time. There is a minimal time between jumps.

Proving the state variables are bounded

The subsystem \(\mathbb{S}^1\) is a circle. we have:

\[\mathbb{S}^1 = V = x_2^2 + x_3^2 = 1 \\ \frac{dV}{dt} = 2x_2\frac{dx_2}{dt} + 2x_3\frac{dx_3}{dt} \\= 2x_2(\frac{2 \pi x_3}{\epsilon}) + 2x_3(\frac{-2 \pi x_2}{\epsilon}) \\ = \frac{4 \pi x_2 x_3}{\epsilon} - \frac{4 \pi x_2 x_3}{\epsilon} = 0\]

Therefore, \(\mathbb{S}^1\) is invariant. since the jump map is identity, it is invarient under jumps as well.

therefore, the state variables (\(x_2\) and \(x_3\)) are bounded for all time.

since the \(x_2, x_3\) are eventually bounded, it just oscillating every fast, in this case, the result of the rate of change of \(x_1\) can be approximated by the average value of \(x_2\) over time, which is zero (since it is oscillating, proved above). Therefore, we have:

\[\dot{x_1} = -x_1 + x_2 = -x_1\]

to show that \(x_1\) is bounded, we need to show that for any interval of \(\delta\), there is a time \(T\) such that for all \(t > T\), \(|x_1(t)| < \delta\).

Looking at the \(x_4\), its rate of change is 1 and it is reset to 0 after reaching 1. Therefore, the flow period is strictly 1, and there is a minimal time between jumps, which is 1. Therefore, there is no Zeno behavior.

therefore:

\[\frac{dx_1}{dt} = - x_1 \\ \int_0^1 \frac{dx_1}{d\tau} d\tau = \int_0^1 -x_1 d\tau \\ x_1(1) = x_1(0)e^{-1}\]

Then at the next jump, \(x_1\) is reset to \(0.5x_1(1) = 0.5x_1(0)e^{-1}\).

therfore, \(\frac{x_1(1)}{x_1(0)} = 0.5e^{-1} < 1\), the change in the next jump is smaller than the change in the previous jump, which means that \(x_1\) is getting smaller and smaller after each jump, therefore, there is no finite escape time. In another words, if there are \(k\) jumps, we have:

\[x_1(t_k) = (0.5e^{-1})^k x_1(0)\]

This is bounded and globally asymptotically stable at \(x_1 = 0\).

prove \(x_1\) is GAS

construct the lyapunov function \(V(x_1) = x_1^2\). we have:

$$V(x_1) = x_1^2 $$, this is positive definite. and radically unbounded.

$$ \dot{V} = 2x_1\dot{x_1} = 2x_1(-x_1) = -2x_1^2 = -2V$$, this is negative definite

therefore, all the state variables are bounded for all time, there is no finite escape time.

Proving there is no deadlock

since \(x_4\) is strictly defined, it always flows when \(x_4 \in [0, 1)\) and it always jumps when \(x_4 = 1\). Therefore, there is no deadlock.

and \(C \cup D = \mathbb{R}^ \times \mathbb{S}^1 \times [0, 1]\), which is the entire state space. Therefore, there is no deadlock.

Proving there is no Zeno behavior

as shown above, the flow period is strictly 1. Therefore, there is no Zeno behavior

Therefore the solutin is complete

Prove it is SPGAS at A:

since the solution of \(x_1\) without the \(x_2\) term is GAS as shown above, adding the ocillating term \(x_2\) will make it converge into a small neighborhood of size \(\epsilon\) around \(x_1 = 0\). since the oscillation is a cycle, value of \(\epsilon\) is 1.

therefore, for any compact set K, for every $\delta >0 $, there exists a \(\epsilon > 0\) such that the solution eventually enters the \(\epsilon\)-neighborhood of \(A\) and stays there for all future time. Therefore, it is SPGAS at \(A\).

Question 2

Since \(\phi\) is strictly convex, therefore, we have:

\[\phi(\lambda x + (1-\lambda)y) < \lambda \phi(x) + (1-\lambda)\phi(y)\]

since it is strict, therefore, there is no flat region in the graph where the LHS = RHS, therefore, there is at most one minimizer for the set of minizers of \(\phi(x)\).

Since \(\phi\) is radially unbounded, therefore, there is at least one minimizer. Let \(x^* = \arg \min \phi(x)\), such that \(\nabla \phi(x_1^*) = 0\).

the slow variables are \(x_1, x_3\), and the fast variable is \(x_2\).

To show that the set is SGPAS at \(A_\phi \times M\mathbb{B} \times [0,1]\) when \(\epsilon \rightarrow 0\), we need to show that:

showing the fast system variable \(x_2\) system is GAS while treating \(x_1\) as a constant

Settings the fast dynamics system as \(\tau = \frac{t}{\epsilon}\) from the flow of \(x_2\), we have:

\[\frac{dx_2}{d\tau} = -x_2 + \nabla \phi(x_1)\]

therefore, the fast system is stable when \(x_2 = \nabla \phi(x_1)\), when \(\frac{dx_2}{d\tau} = 0\).

As \(\epsilon \rightarrow 0\), in fact when \(\epsilon = 0\), we have

\[ \epsilon \dot(x_2) = -x_2 + \nabla \phi(x_1) \\ 0 = -x_2 + \nabla \phi(x_1) \\ x_2 = \nabla \phi(x_1)\]

as the slow manifold

Let $c = x_2 - \nabla \phi(x_1), \frac{dc}{d\tau} = \frac{dx_2}{d\tau} $, given we treat \(x_1\) as a constant. Since \(x_2\) changes depend on the time period \(\tau\), we have \(c(\tau) = x_2(\tau) - \nabla \phi(x_1)\). we have:

\[\frac{dx_2}{d\tau} = -c \\ \frac{dc}{d\tau} = -c \\ \frac{dc}{c} = -d\tau \\ \int \frac{dc}{c} = \int -d\tau \\ \ln c = -\tau + C \\ c(\tau) = Ae^{-\tau} \]

when taking \(\tau = 0\), we have \(c(0) = A\), therefore, \(c(\tau) = c(0)e^{-\tau}\). As \(\tau \rightarrow \infty\), \(\x_2(\tau) \rightarrow \nabla \phi(x_1)\). Therefore, the fast system is GAS at \(x_2 = \nabla \phi(x_1)\) when flow. and since the jump map is identity, it is GAS at jumps as well. therefore, the fast system is GAS at \(x_2 = \nabla \phi(x_1)\).

prove that the reduced system is GAS

construct the lyapunov function \(V(x_1) = \phi(x_1) - \phi(x_1^*)\). we have:

1. positive definite: since \(\phi\) is strictly convex, and \(x_1^*\) is the unique minimizer, therefore, \(\phi(x_1) > \phi(x_1^*)\) for all \(x_1 \neq x_1^*\). therefore, \(V(x_1) > 0\) for all \(x_1 \neq x_1^*\), and \(V(x_1^*) = 0\). therefore, it is positive definite.
2. radially unbounded: since \(\phi\) is radially unbounded, therefore, \(V(x_1) = \phi(x_1) - \phi(x_1^*)\) is also radially unbounded.
3. negative definite: we have:

\[\dot{V} = \nabla \phi(x_1) \dot{x_1} = \nabla \phi(x_1) (-\nabla \phi(x_1)) = -||\nabla \phi(x_1)||^2 \leq 0\]

since \(x_1^*\) is the unique minimizer, therefore, \(\nabla \phi(x_1) = 0\) if and only if \(x_1 = x_1^*\). therefore, \(\dot{V} = 0\) if and only if \(x_1 = x_1^*\). therefore, it is negative definite.

therefore, \(x_1\) is GAS at flows

in addition, since \(x_1^+ = x_1\), it does not change during jumps, therefore, \(x_1\) is GAS at \(x_1^*\).

therefore, the reduced system is GAS at \(A_\phi \times M\mathbb{B} \times [0,1]\).

Since the boundary layer is GAS and the reduced system is GAS, therefore, the system is SGPAS at \(A_\phi \times M\mathbb{B} \times [0,1]\) when \(\epsilon \rightarrow 0\).

completeness of the solution

since the \(x_2\) = \(\nabla \phi(x_1)\) in the reduced system, it is dependent on the gradient descent of \(\phi\). But the system range is \(M\mathbb{B}\), it is a constant ball with range \(x_2 \in [-M, M]\). There must exists some \(\nabla \phi(x_1)\) that is outside of the range. Therefore, the solution is not complete.

For example, when the \(\phi(x_1) = x_1^2\), then \(\nabla \phi(x_1) = 2x_1\). Let $ M = 1$, and let the start condition be some x_1 such that \(2x_1 > 1\) (like when \(x_1 = 10\)), then even when the initial condition is bounded at \(t = 0\), the \(x_2\) for \(t = 0^+\) will be \(2x_1 = 20 > 1\), which is outside of the range. Therefore, the solution is not complete.