How to read double integrator equations?

Question

From what I understand, double integrator is a model where some entity can move according to some speed, which depends on the acceleration force exerted on the entity.

If someone would ask me to represent such a model, with $x$ the position of the entity, $v$ its velocity, and $a$ its acceleration, I would write it down as simply :

$$ \dot{x} = v $$ $$ \ddot{x} = a $$

However these do not seem to be the same relations as given by the Wikipedia page (see https://en.wikipedia.org/wiki/Double_integrator). Maybe I'm having trouble with the notation. What do the following equations, meant to represent a double integrator system in a single dimension, mean exactly?

$$ \ddot{q} = u(t) $$ $$ y = q(t) $$

$u$ is described as the control input, which I assume is the acceleration, and $q$ the output, which is the entity's position? So, what is $y$? It seems to be equal to $q$, so what is its use here?

Answer 1

The wikipedia article is inconsistent in notation and in form.

Also, the article contains constraint equations between the degrees of freedom which further complicates the notation. You don't have such considerations for a 1 DOF problem.

So take your example, with $n=1$ DOF and consider the following quantities

• The generalized coordinates are vector of $n$ values

  $$\boldsymbol{q} = \pmatrix{x} \tag{1}$$

• The differential equation is given in terms of the degrees of freedom as a system of $n$ equations

  $$ \ddot{\boldsymbol{q}} = \boldsymbol{\rm f}(t, \boldsymbol{q}, \dot{\boldsymbol{q}}) \tag{2}$$

  $$ \ddot{x} = {\rm f}(t,x,\dot{x}) $$

• As an ODE the above is of second order and it is setup as the integration of two variables (double integrator) in order to be solved. As a system of two first order ODEs the above is expressed with $2n$ equations.

  $$ \tfrac{\rm d}{{\rm d}t} \pmatrix{ \boldsymbol{q} \\ \boldsymbol{\dot{q}} } = \pmatrix{ \boldsymbol{\dot q} \\ \boldsymbol{\rm f}(t,\boldsymbol{q},\boldsymbol{\dot{q}}) } \tag{3}$$ $$ \tfrac{\rm d}{{\rm d}t} \pmatrix{x \\ v} = \pmatrix{v \\ {\rm f}(t,x,v)} $$

• More formally with a state vector $\boldsymbol{x} = \pmatrix{ \boldsymbol{q} \\ \boldsymbol{\dot q}}$ the above system of equations is brought into a more canonical form which is what the article should have shown

  $$ \tfrac{\rm d}{{\rm d}t} \boldsymbol{x} = \boldsymbol{\rm u}(t, \boldsymbol{x}) \tag{4}$$

  $$ \begin{aligned} \dot{x} & = v \\ \dot{v} & = {\rm f}(t,x,v) \end{aligned}$$

  _{Note that the state vector isn't a real vector in terms of physics, but rather a mathematical construct.}

• The system can also describe constraints which tie together different degrees of freedom and their derivatives

  $$ \boldsymbol{y} = \boldsymbol{\rm g}(t, \boldsymbol{x}) \tag{5} $$

  but this does not apply in your case.

• All of the above become a little more formal when expressed in terms of linear algebra as a DAE system (differential and algebraic equations)

  $$ \tfrac{\rm d}{{\rm d}t} \boldsymbol{x} = \mathbf{A} \boldsymbol{x} + \boldsymbol{b} \tag{6} $$ $$ \boldsymbol{0} = \mathbf{G} \boldsymbol{x} + \boldsymbol{c} \tag{7} $$