Homework 2

Due at 11:59pm on Jan 30th. Submit to https://canvas.uw.edu/courses/1862386/assignments/11123228.

Problem 1

Consider the system \(\dot{x}=A x\). Show that the nonnegative orthant (\(\mathbb{R}_+^n = \{x:x_i \geq 0 \; \forall i\}\)) is invariant if and only if \(A_{ij} \geq 0\) for all \(i \neq j\). That is, show that nonegative trajectories stay nonengative if and only if the off diagonal entries of \(A\) are nonnegative.

Solution

Assume that \(A_{ij} \geq 0\) for all \(i \neq j\) and \(x(0)\) is in the nonnegative orthant. We have \(x(t)=e^{tA}x(0)=\lim_{k \rightarrow \infty} (I+(t/k)A)^k x(0)\). For large enough \(k\), all entries of \((I+(t/k)A)\) are nonnegative, and hence \((I+(t/k)A)^k x(0)\) have nonnegative entries.

For the othre direction, suppose one of the elements of \(A_{ij}\) is negative (\(i \neq j\)). Now take \(x(0)=e_j\), where \(e_j\) is the \(j'th\) standard basis. For small enough \(\tau\), \(x(\tau)\approx e_j+\tau A e_j\). The $i$’th component of this vector is approximately \(h A_{ij}<0\), and \(\mathbb{R}_+^n\) isn’t invariant.

Problem 2

The system \(\dot{x}=Ax\) is called constant norm if for every trajectory, \(\Vert x(t) \Vert\) is constant, that is, it doesn’t depend on \(t\). The system is called constant speed if for every trajectory, \(\Vert \dot{x}(t)\Vert\) is constant.

a) Find the conditions on \(A\) under which the system is constant norm.

b) Find the conditions on \(A\) under which the system is constant speed.

c) Is every constant norm system a constant speed system?

d) Is every constant speed system a constant norm system?

Solution

a) We want \(\Vert x(t) \Vert^2\) to be constant, or

\[\frac{d}{dt} \Vert x(t) \Vert^2 = x^T (A^T+A) x=0\]

For this to hold for all \(x\), \(A\) need to be skew-symmetric.

b) Using the fact that \(\dot{x}=A x\), we want

\[\frac{d}{dt} \Vert \dot x(t) \Vert^2 = x^T A^T (A^T+A) A x=0\]

Or \(A^T (A^T+A) A=0\).

c) Yes, since if \(A^T +A=0\), then \(A^T (A^T+A) A=0\).

d) Not necessarily. For example, \(A=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\).

Problem 3

Consider the nonlinear system

\(\ddot{x} = -g(x)\),

where \(g\) is some increasing function with \(g(0)=0\). Find a conserved quantity for this system. Hint: think back to the case of \(g(x)=x\), and see how the conserved quantity was constructed (think about integrals).

Solution

Following what we have done in class, the conserved quantity (or energy) is

\[V(x,\dot{x})=\frac{1}{2} \dot{x}+\int_0^x g(z) dz.\]

Problem 4

Compute the graident of the following functions:

a) \(f(x_1,x_2)= x_1 x_2 + x_1 \cos(x_2)\)

b) \(f(x)=\frac{1}{\Vert x \Vert} a^Tx\), where \(x\) is a vector in \(\mathbb{R}^n\) and \(a\) is a constant vector. To be tehnical, we can define \(f(0)=0\), although it doesn’t really matter in the gradident calculation.

solution

a) We have

\[\frac{\partial f}{\partial x_1} = x_2 + \cos(x_2)\] \[\frac{\partial f}{\partial x_2} = x_1 - x_1\sin(x_2)\]

Therefore:

\[\nabla f = \begin{pmatrix} x_2 + \cos(x_2) \\ x_1(1 - \sin(x_2)) \end{pmatrix}\]

b) Write \(f(x) = \Vert x\Vert^{-1} a^T x\). We use the product rule, the first part is easy:

\[\nabla(a^T x) = a\]

Since \(\Vert x\Vert = (x^Tx)^{1/2}\),

\[\nabla\Vert x\Vert = \frac{x}{\Vert x\Vert} \implies \nabla\left(\Vert x\Vert^{-1}\right) = -\frac{1}{\Vert x\Vert^2}\cdot\frac{x}{\Vert x\Vert} = -\frac{x}{\Vert x\Vert^3}\]

We have

\[\nabla f = \nabla\!\left(\Vert x\Vert^{-1}\right) a^Tx \;+\; \Vert x\Vert^{-1}\nabla(a^Tx)\] \[= -\frac{x}{\Vert x\Vert^3}(a^T x) + \frac{a}{\Vert x\Vert}\]

Factoring out \(\Vert x\Vert^{-1}\):

\[\nabla f = \frac{1}{\Vert x\Vert}\left(a - \frac{a^T x}{\Vert x\Vert^2}x\right)\]

Problem 5

Consider the single bus swing equation:

\[\begin{align} \dot{\theta} &=\omega \\ M \dot{\omega} &= -D \omega - B \theta \end{align}\]

Take \(M\), \(D\) and \(B\) to all be \(1\). We know the system is stable. But let’s say we have a more strigent requirement: the frequency should stay within \(\pm 0.1\). Find all initial points such that all the frequency is within this bound for all \(t\geq 0\).

Solution

This is where the Lyapunov method tend to struggle, since the set we want the trajectories to be in isn’t an ellipse. The easiest way is to compute \(e^{tA}\) and simulate with different \(x(0)\) to get the set. Here,

\[e^{tA} = e^{-t/2}\begin{pmatrix} \cos\!\left(\frac{\sqrt{3}}{2}t\right) + \frac{1}{\sqrt{3}}\sin\!\left(\frac{\sqrt{3}}{2}t\right) & \frac{2}{\sqrt{3}}\sin\!\left(\frac{\sqrt{3}}{2}t\right) \\[6pt] -\frac{2}{\sqrt{3}}\sin\!\left(\frac{\sqrt{3}}{2}t\right) & \cos\!\left(\frac{\sqrt{3}}{2}t\right) - \frac{1}{\sqrt{3}}\sin\!\left(\frac{\sqrt{3}}{2}t\right) \end{pmatrix}\]

and the feasible initial points are

But this approach obviously won’t scale to larger systems.