a question on visualizing a state of DAE and a question from a continous time nonlinear dynamics 
*

*Could someone explain me what is the fundamental difference between the dynamical system of the kind $\dot x = f(x)$ and $E \dot x= f(x)$ where $E$ is a singular matrix with real entries. For the first system, a state is $x(t)$ at time $t$; is it the same for the second system? I wish to visualise trajectories of such type of dynamical system to be more precise.


*Consider a continuous time dynamical system
\begin{align}
    \dot x= f\left(x\right)
\end{align}
on a state-space $\mathcal{X}$, where $x$ is a coordinate vector of the state, $f$ is a non-linear vector valued smooth function of the same dimension as its argument $x$.
Could anyone tell me why assume $f$ to be smooth and what it means by ''function of the same dimension as its argument $x$'', what is a dimension of function, by the way, and why its dimension needs to be equal as $x$?
Thank you for your clarification.
 A: *

*The first kind is the classic ordinary differential equation (ODE), the 2nd is a generalized version of it. The matrix $E$ is sometimes called the mass matrix (in analogy of Newtonian Mechanism). In the most general form, it could be state and/or time dependent.

a. If the matrix $E$ is invertible/nonsingular, then we can easily convert the 2nd form into the 1st.
$$
\dot x = E^{-1}f(x)
$$
So mathematically they are the exact same system, but writing it in the 2nd forms could make the numerical solver's life easier. Why?
Intuitively if $E$ has a large condition number, i.e. some variables change very fast, some change very slow, then the numerical solver could have a hard time determine step size without this knowledge. Check out Stiff Equation
b. If the matrix $E$ is singular, this is no longer a classic ODE. Consider a vector $v\neq 0$ s.t. $v^TE=0$ and a vector $u\neq 0,s.t. Eu^T=0$, then we actually get a constraint equation along the projection $v$.
$$
0=v^TE\dot x=v^T f(x)
$$
With a $d$ dimensional null space of $E$ we get $d$ constraint equations, which defines the $n-d$ dimension manifold which the solution trajectory shall live on.
One simple example
$$
E = \begin{bmatrix}1& 0\\0&0\end{bmatrix}\\
E[\dot x,\dot y]^T= [-y, x^2+y^2-1]^T
$$
Then the dynamics of $y$ is not explicitly defined, but we got a constraint function of $x,y$ s.t. the trajectory shall live on the unit circle $x^2+y^2-1=0$. Then the dynamics of $y$ could be figured out from the constraint via implicit function theorem.
Also check this out, from computational perspective.
How to handle mass matrix



*"function of the same dimension as its argument x'', what is a dimension of function, by the way, and why its dimension needs to be equal as x?"

This simply means if $x\in \mathbb R^n$, then $f:\mathbb R^n\to \mathbb R^n$. The simple reason if $x\in \mathbb R^n$, then its time derivative shall also be a vector in $n$ dimension vector space. This is like if your position in in 3d space, then your velocity vector is also a 3d vector. or you have no way adding them up.
Geometrically, each dynamic system/ODE describes a vector field on the space $\mathcal{X}$, these vectors live in the tangent space of the manifold $\mathcal{X}$. In this case if $x$ live in $n$ dim Euclidean space, its tangent space is also $n$ dim.
For smoothness, I feel it's a requirement from technical convenience to prove something down the line.... In real life we don't necessarily use smooth $f$
