5 DAEs Forms
Many physical systems are governed not by ordinary differential equations (ODEs) but by differential-algebraic equations (DAEs), which combine differential dynamics with algebraic constraints. A general DAE can be written as:
\bm{f}(\bm{x}'(t), \bm{x}(t), t) = \bm{0},
where bm{x}(t) \in \Bbb{R}^n is the state vector, \bm{x}'(t) is its time derivative, and \bm{f} represents a (possibly nonlinear) vector function.
Unlike ODEs, DAEs may not permit direct solution for \bm{x}'(t) due to the potential singularity of the Jacobian:
\dfrac{\partial \bm{f}}{\partial \bm{x}'} = \dfrac{\partial \bm{f}(\bm{x}', \bm{x}, t)}{\partial \bm{x}'}.
When this Jacobian is singular (i.e., not invertible), implicit or indirect techniques are needed. The differential index of a DAE quantifies how many times we must differentiate the equation to isolate \bm{x}'(t).
5.1 Semi-Explicit Form of DAEs
A more structured (and often more tractable) form of DAEs is:
\bm{A}(\bm{x}, t) \bm{x}' = \bm{g}(\bm{x}, t),
where \bm{A}(\bm{x}, t) is singular. When \bm{A}(\bm{x}, t) has a block-diagonal structure:
\bm{A}(\bm{x}, t) = \begin{pmatrix} \bm{I} & \bm{0} \\ \bm{0} & \bm{0} \end{pmatrix},
the DAE is said to be in semi-explicit form.
Why is this useful?
- The semi-explicit form naturally separates the system into differential and algebraic components:
- Differential part: Explicit ODEs (e.g., \bm{y}' = \bm{h}(\bm{y}, \bm{z}, t)).
- Algebraic part: Constraints (e.g., \bm{0} = \bm{k}(\bm{y}, \bm{z}, t)).
- Differential part: Explicit ODEs (e.g., \bm{y}' = \bm{h}(\bm{y}, \bm{z}, t)).
- This separation simplifies index analysis (discussed later).
A generic DAE:
\bm{f}(\bm{x}',\bm{x},t)=\bm{0}
can always be rewritten in semi-explicit form by introducing the auxiliary variable \bm{x}_{\mathrm{dot}}(t)=\bm{x}'(t):
\begin{aligned} \bm{x}'&= \bm{x}_{\mathrm{dot}} \\[1em] \bm{0} &= \bm{f}(\bm{x}_{\mathrm{dot}},\bm{x},t) \end{aligned}
Important:
- This transformation increases the differential index by 1.
- For example, an index-1 DAE becomes index-2 after conversion.
5.2 Index-0 DAEs are ODEs
By definition, index-0 DAEs are simply ODEs of the form:
\bm{x}'(t) = \bm{f}(\bm{x}, t),
where the system is fully explicit with no algebraic constraints.
5.3 Index-1 DAEs in Semi-Explicit Form
An index-1 DAE in semi-explicit form is:
\left\{ \begin{aligned} \bm{x}' &= \bm{f}(\bm{x}, \bm{y}, t) \quad &\text{(Differential part)}, \\[0.5em] \bm{0} &= \bm{g}(\bm{x}, \bm{y}, t) \quad &\text{(Algebraic part)}, \end{aligned} \right.
where:
- \bm{x} \in \Bbb{R}^n are the differential variables,
- \bm{y} \in \Bbb{R}^m are the algebraic variables,
- \bm{f} \in C^1(\Bbb{R}^{n+m+1}, \Bbb{R}^n),
- \bm{g} \in C^1(\Bbb{R}^{n+m+1}, \Bbb{R}^m).
The Jacobian of \bm{g} w.r.t. \bm{y},
\bm{g}_{\bm{y}}(\bm{x}, \bm{y}, t) = \frac{\partial \bm{g}}{\partial \bm{y}},
must be non-singular. This ensures that the Implicit Function Theorem applies, allowing us to express \bm{y} as a function of \bm{x} and t:
\bm{y} = \bm{y}(\bm{x}, t).
Substituting back into the differential part reduces the DAE to an ODE:
\bm{x}'(t) = \bm{g}(\bm{x}(t),t),\qquad\textrm{where}\qquad \bm{g}(\bm{x},t)= \bm{f}(\bm{x},\bm{y}(\bm{x},t),t)
5.4 The Implicit Function Theorem
The Implicit Function Theorem ensures that the algebraic constraints can be locally solved for \bm{y}.
Theorem 5.1 Let A\subset\Bbb{R}^{n+m} an open set and \bm{f}\in C^1(A,\Bbb{R}^m), moreover
(\bm{x}_0,\bm{y}_0)\in A;
\bm{f}(\bm{x}_0,\bm{y}_0)=\bm{0};
\dfrac{\partial\bm{f}(\bm{x}_0,\bm{y}_0)}{\partial\bm{y}} non-singular.
Then, there exists two open set
U\subset\Bbb{R}^n\quad\text{and}\quad V\subset\Bbb{R}^m
and a function \bm{\phi}:U\mapsto V such that
\bm{y}_0=\bm{\phi}(\bm{x}_0);
\bm{f}(\bm{x},\bm{y})=\bm{0} iff \bm{y}=\bm{\phi}(\bm{x}) (for \bm{x}\in U and \bm{y}\in V);
\bm{\phi}\in C^1(U,V)
and
\dfrac{\partial\bm{\phi}(\bm{x})}{\partial\bm{x}}=- \left(\dfrac{\partial\bm{f}(\bm{x},\bm{\phi}(\bm{x}))}{\partial\bm{y}}\right)^{-1} \dfrac{\partial\bm{f}(\bm{x},\bm{\phi}(\bm{x}))}{\partial\bm{x}}
Illustrative Example: A Simple Index-1 DAE
Consider the semi-explicit DAE:
\begin{aligned} x' &= -y, \\ 0 &= x + y - \sin(t). \end{aligned}
Step 1: Apply the Implicit Function Theorem
The algebraic equation is:
y = \sin(t) - x.
Step 2: Substitute Back into the ODE
The system reduces to:
x' = -\sin(t) + x,
which is an ODE solvable with standard methods.
DAE of index 2
DAE of index 1 are in practice ODE with implicit parameters. Index 2 DAE are more complex
\begin{aligned} \bm{x}' &= \bm{f}(\bm{x},\bm{y},t)\\[1em] \bm{0} &= \bm{g}(\bm{x},t) \end{aligned}
where \bm{x}(t)\in\Bbb{R}^n, \bm{y}(t)\in\Bbb{R}^m, \bm{f}\in C^1(\Bbb{R}^{n+m+1},\Bbb{R}^n) and \bm{g}\in C^1(\Bbb{R}^{n+1},\Bbb{R}^m). Moreover
\dfrac{\partial\bm{g}(\bm{x},t)}{\partial\bm{x}} \times\dfrac{\partial\bm{f}(\bm{x},\bm{y},t)}{\partial\bm{y}}
ia a non singular matrix. In this case implicit function Theorem cannot be used to determine \bm{y} as a function of \bm{x}.
To check the index perform one derivation of the constraints
\dfrac{\mathrm{d}\bm{g}(\bm{x}(t),t)}{\mathrm{d}t}= \dfrac{\partial\bm{g}(\bm{x}(t),t)}{\partial\bm{x}}\bm{x}'(t)+ \dfrac{\partial\bm{g}(\bm{x}(t),t)}{\partial t}=\bm{0}
obtaining the new (hidden) constraints
\bm{h}(\bm{x},\bm{y},t)=\bm{0}
where
\bm{h}(\bm{x},\bm{y},t)= \dfrac{\partial\bm{g}(\bm{x},t)}{\partial\bm{x}} \bm{f}(\bm{x},\bm{y},t)+ \dfrac{\partial\bm{g}(\bm{x},t)}{\partial t}
Thus, a new DAE is obtained
\begin{aligned} \bm{x}' &= \bm{f}(\bm{x},\bm{y},t)\\[1em] \bm{0} &= \bm{h}(\bm{x},\bm{y},t) \end{aligned}
The derived DAE is now of index 1:
\begin{aligned} \bm{x}' &= \bm{f}(\bm{x},\bm{y},t)\\[1em] \bm{0} &= \bm{h}(\bm{x},\bm{y},t) \end{aligned}
to check it must be
\dfrac{\partial\bm{h}(\bm{x},\bm{y},t)}{\partial\bm{y}}= \dfrac{\partial\bm{g}(\bm{x},t)}{\partial\bm{x}}\times \dfrac{\partial\bm{f}(\bm{x},\bm{y},t)}{\partial\bm{y}}
non singular.
5.5 DAE of index 3
Index 3 DAE semi-esplicit take the form
\begin{aligned} \bm{x}' &= \bm{f}(\bm{x},\bm{y},\bm{z},t)\\[1em] \bm{y}' &= \bm{g}(\bm{x},\bm{y},t)\\[1em] \bm{0} &= \bm{h}(\bm{y},t) \end{aligned}
where
\bm{x}(t)\in\Bbb{R}^n\qquad \bm{y}(t)\in\Bbb{R}^m\qquad \bm{z}(t)\in\Bbb{R}^p \\[1em] \bm{f}\in C^1(\Bbb{R}^{n+m+p+1},\Bbb{R}^n)\quad \bm{g}\in C^1(\Bbb{R}^{n+m+1},\Bbb{R}^m)\quad \bm{h}\in C^1(\Bbb{R}^{m+1},\Bbb{R}^n)
To check the index derive the constraints
\dfrac{\mathrm{d}\bm{h}(\bm{y}(t),t)}{\mathrm{d}t}= \dfrac{\partial\bm{h}(\bm{y}(t),t)}{\partial\bm{y}}\bm{y}'(t)+ \dfrac{\partial\bm{h}(\bm{y}(t),t)}{\partial t}=\bm{0}
obtaining the new (hidden) constraints
\bm{a}(\bm{x},\bm{y},t)=\bm{0}
where
\bm{a}(\bm{x},\bm{y},t)= \dfrac{\partial\bm{h}(\bm{y},t)}{\partial\bm{y}}\bm{g}(\bm{x},\bm{y},t)+ \dfrac{\partial\bm{h}(\bm{y},t)}{\partial t}
Thus, a new DAE is obtained
\begin{aligned} \bm{x}' &= \bm{f}(\bm{x},\bm{y},\bm{z},t)\\[1em] \bm{y}' &= \bm{g}(\bm{x},\bm{y},t)\\[1em] \bm{0} &= \bm{a}(\bm{x},\bm{y},t) \end{aligned}
The DAE
\begin{aligned} \begin{pmatrix} \bm{x}' \\ \bm{y}' \end{pmatrix} &= \begin{pmatrix} \bm{f}(\bm{x},\bm{y},\bm{z},t)\\ \bm{g}(\bm{x},\bm{y},t) \end{pmatrix}\\[1em] \bm{0} &= \bm{a}(\bm{x},\bm{y},t) \end{aligned}
is index 2 iff
\dfrac{\partial\bm{a}(\bm{x},\bm{y},t)}{\partial\bm{x}}\times \dfrac{\partial\bm{f}(\bm{x},\bm{y},\bm{z},t)}{\partial\bm{z}}+ \dfrac{\partial\bm{a}(\bm{x},\bm{y},t)}{\partial\bm{y}}\times \dfrac{\partial\bm{g}(\bm{x},\bm{y},t)}{\partial\bm{z}}
is non singular, i.e.
\dfrac{\partial\bm{h}(\bm{y},t)}{\partial\bm{y}}\times \dfrac{\partial\bm{g}(\bm{x},\bm{y},t)}{\partial\bm{x}}\times \dfrac{\partial\bm{f}(\bm{x},\bm{y},\bm{z},t)}{\partial\bm{z}}
is non singular.
5.6 DAE in semi-esplicit form
DAE of index 1
\left\{\begin{aligned} \bm{x}' &= \bm{f}(\bm{x},\bm{y},t)\\[1em] \bm{0} &= \bm{g}(\bm{x},\bm{y},t) \end{aligned}\right. \qquad \qquad \dfrac{\partial\bm{g}(\bm{x},\bm{y},t)}{\partial\bm{y}}\quad \textbf{non singular}
DAE of index 2
\left\{\begin{aligned} \bm{x}' &= \bm{f}(\bm{x},\bm{y},t)\\[1em] \bm{0} &= \bm{g}(\bm{x},t) \end{aligned}\right. \qquad \underbrace{ \dfrac{\partial\bm{g}(\bm{x},t)}{\partial\bm{x}} \times\dfrac{\partial\bm{f}(\bm{x},\bm{y},t)}{\partial\bm{y}} }_{\textbf{non singular}}
DAE of index 3
\left\{\begin{aligned} \bm{x}' &= \bm{f}(\bm{x},\bm{y},\bm{z},t)\\[1em] \bm{y}' &= \bm{g}(\bm{x},\bm{y},t)\\[1em] \bm{0} &= \bm{h}(\bm{y},t) \end{aligned}\right. \quad \underbrace{ \dfrac{\partial\bm{h}(\bm{y},t)}{\partial\bm{y}}\times \dfrac{\partial\bm{g}(\bm{x},\bm{y},t)}{\partial\bm{x}}\times \dfrac{\partial\bm{f}(\bm{x},\bm{y},\bm{z},t)}{\partial\bm{z}} }_{\textbf{non singular}}
DAE of index \mu (in the Hessemberg form)
The DAE
\begin{aligned} x_1' &= f_1(x_1,x_2,\ldots,x_{\mu},t)\\[1em] x_2' &= f_2(x_1,x_2,\ldots,x_{\mu-1},t)\\[1em] &\vdots&\\ x_k' &= f_k(x_{k-1},x_k,\ldots,x_{\mu-1},t)\qquad k=3,\ldots,\mu-1\\[1em] &\vdots\\ 0 &= f_{\mu}(x_{\mu-1},t) \end{aligned}
with
\dfrac{\partial f_{\mu}}{\partial x_{\mu-1} } \times \dfrac{\partial f_{\mu-1}}{\partial x_{\mu-2} } \times\cdots\times \dfrac{\partial f_{2}}{\partial x_1} \times \dfrac{\partial f_{1}}{\partial x_{\mu} }
non singular is a DAE of index \mu.