7 Linear DAEs
This chapter introduces linear differential-algebraic equations, the matrix pencil that governs them, and the canonical form that makes their solution explicit.
We consider linear implicit differential equations of the form
\bm{E}\dot{\bm{x}}(t) = \bm{A}\bm{x}(t) + \bm{f}(t), \tag{7.1}
where \bm{E},\bm{A}\in\mathbb{R}^{n\times n} are constant matrices and \bm{f}:\mathbb{R}\to\mathbb{R}^n is a prescribed forcing term. Such equations arise from linearized nonlinear models, electrical circuits, constrained mechanical systems, and systems with algebraic conservation laws.
The treatment here is theoretical: it answers when a solution exists and how many initial conditions may be prescribed, through the Weierstrass canonical form and the Kronecker index. The computational counterpart — how to obtain the same information without computing a canonical form — is the subject of Chapter 8.
7.1 Linear Differential-Algebraic Equations
Why Differential-Algebraic Equations?
A differential-algebraic equation (DAE) has the implicit form
\bm{F}(\bm{x},\dot{\bm{x}},t)=\bm{0}.
Unlike an explicit ODE, it need not determine every component of \dot{\bm{x}} directly. This is useful when physical constraints are simpler to express than to eliminate. In multibody dynamics, each body has its own equations of motion, while joints impose geometric constraints. In electrical circuits, element constitutive equations are supplemented by Kirchhoff current and voltage laws. The present chapter focuses on constant-coefficient linear DAEs; the nonlinear examples below motivate the subject.
Pendulum: minimal coordinates versus constrained coordinates
For a mass m on a rigid rod of length \ell, with y positive upwards, minimal coordinates give
\left\{ \begin{aligned} \dot\theta &= \omega,\\ \dot\omega &= -\frac{g}{\ell}\sin\theta. \end{aligned} \right.
Cartesian coordinates with a Lagrange multiplier instead give
\left\{ \begin{aligned} \dot x &= u,\\ \dot y &= v,\\ m\dot u &= \lambda x,\\ m\dot v &= -mg+\lambda y,\\ 0 &= x^2+y^2-\ell^2. \end{aligned} \right.
The multiplier \lambda represents the radial constraint force per unit distance. This system is a nonlinear DAE, not an instance of the linear pencil studied below. Differentiating its position constraint produces the hidden velocity constraint xu+yv=0, followed by u^2+v^2+x\dot u+y\dot v=0.
Electrical circuits
For a series RLC circuit driven by a voltage source e(t), eliminating the resistor and inductor voltages gives the minimal-coordinate ODE
\left\{ \begin{aligned} \dot v_C &= \frac{i}{C},\\ \dot i &= \frac{e(t)-Ri-v_C}{L}. \end{aligned} \right.
If the branch voltages v_R and v_L are retained as unknowns, the same circuit is described by the DAE
\left\{ \begin{aligned} C\dot v_C-i &= 0,\\ L\dot i-v_L &= 0,\\ v_R-Ri &= 0,\\ v_R+v_C+v_L-e(t) &= 0. \end{aligned} \right.
The last two equations are algebraic: they encode the resistor constitutive law and Kirchhoff’s voltage law. More generally, Kirchhoff’s current law enforces \sum_j i_j=0 at each node. Retaining branch currents and node voltages usually gives a sparse DAE, whereas eliminating all constraints first can produce a more complicated ODE.
Three fundamental cases
Consider \bm E\dot{\bm x}=\bm A\bm x+\bm f(t) with square constant matrices.
| Structure of \bm E | Interpretation | Initial data |
|---|---|---|
| \det\bm E\ne0 | Explicit ODE after multiplying by \bm E^{-1} | All components of \bm x(0) are free. |
| \bm E=\bm0 | Algebraic system \bm A\bm x=-\bm f | If \bm A is invertible, \bm x(0) is fixed by \bm f(0). |
| \bm E singular and nonzero | Genuine mixed DAE, possibly with hidden constraints | Consistency and uniqueness depend on the pencil. |
Singularity of \bm E does not imply singularity of the pencil \bm A-\lambda\bm E.
A primary consideration is the existence and uniqueness of a solution
\bm{x}(t): \mathbb{R} \rightarrow \mathbb{R}^{n}
The intriguing case arises when \bm{E} is singular (i.e., not invertible). In contrast, if \bm{E} is non-singular, we can simply invert it, allowing us to express the DAEs as ordinary differential equations (ODEs) as follows:
\dot{\bm{x}}(t) = \bm{E}^{-1}\bm{A}\bm{x}(t) + \bm{E}^{-1}\bm{f}(t). \tag{7.2}
For the equation in Equation 7.2, it suffices to assume that \bm{f}(t) is piecewise continuous1, which guarantees the existence and uniqueness of solutions. For a detailed discussion, see, for instance, [1].
Conversely, when the matrix \bm{E} is singular, the existence of a solution becomes uncertain. A singular matrix \bm{E} indicates that ODEs are coupled with linear algebraic constraints. Therefore, it is essential to consider certain regularity conditions for the pair (\bm{E}, \bm{A}), as illustrated in the following example.
To establish the existence and uniqueness of a solution for Equation 7.1, we need to introduce the following definition.
Definition 7.1 (Matrix Pencils) A matrix pair (\bm{E},\bm{A}) is defined as a regular pencil if there exists a scalar \lambda_{0} \in \mathbb{R} such that:
\begin{vmatrix} \bm{A} - \lambda_{0} \bm{E} \end{vmatrix} \neq 0.
The central idea is that if the pair (\bm{E}, \bm{A}) is a regular pencil, then we can simplify the problem through an appropriate change of coordinates. This concept is further clarified by the following theorem.
Theorem 7.1 (Weierstrass Form) A matrix pair \bm{E}, \bm{A} \in \mathbb{R}^{n \times n} is classified as a regular pencil if, and only if, there exist invertible matrices
\bm{P} \in \mathbb{R}^{n \times n} \quad\text{and}\quad \bm{Q} \in \mathbb{R}^{n \times n}
such that: (\bm{PEQ}, \bm{PAQ}) = \left( \begin{bmatrix} \bm{N} & 0 \\ 0 & \bm{I} \end{bmatrix}, \begin{bmatrix} \bm{I} & 0 \\ 0 & \bm{J} \end{bmatrix} \right), \tag{7.3}
where \bm{N} is an upper triangular nilpotent matrix, \bm{J} is a matrix in Jordan canonical form, and \bm{I} is the identity matrix of appropriate dimensions.
To enhance understanding, we will briefly recall the definition of a nilpotent matrix.
Definition 7.2 (Nilpotent Block) A square matrix \bm{N} is termed nilpotent if there exists an integer \nu \in \mathbb{N} such that:
\bm{N}^{\nu} = 0.
Remark 7.1. It is important to note that the smallest integer \nu for which \bm{N}^{\nu} = 0 is referred to as the degree of the nilpotent matrix \bm{N}. Furthermore, any upper triangular matrix that has zeros along its main diagonal qualifies as a nilpotent matrix.
The decoupling form presented in Equation 7.3 is known as the Weierstrass canonical form. For further details, refer to [2]. The matrices \bm{N} and \bm{J} in this form possess the following structures:
\bm{N} = \begin{bmatrix} \bm{N}_{1} & 0 & \dots & 0 \\ 0 & \bm{N}_{2} & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & \bm{N}_{n_{1}} \end{bmatrix}, \quad \bm{J} = \begin{bmatrix} \bm{J}_{1} & 0 & \dots & 0 \\ 0 & \bm{J}_{2} & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & \bm{J}_{n_{2}}. \end{bmatrix}
Here, \bm{N}_{i} for i=1, \ldots, n_{1} and \bm{J}_{i} for i=1, \ldots, n_{2} are Jordan blocks. The blocks \bm{N}_{i} are associated with the zero eigenvalues of the matrix \bm{E}, and are structured as follows:
\bm{N}_{i} = \begin{bmatrix} 0 & 1 & & \\ & 0 & 1 & \\ & & \ddots & \ddots \\ & & & \ddots & 1 \\ & & & & 0 \end{bmatrix}, \quad i = 1, \ldots, n_{1}.
The dimension of each block \bm{N}_{i} corresponds to the difference between the algebraic and geometric multiplicities of the respective eigenvalue. Conversely, the blocks \bm{J}_{i} for i=1, \ldots, n_{2} are classical Jordan blocks, structured as follows:
\bm{J}_{k} = \begin{bmatrix} \lambda_{k} & 1 & & \\ & \lambda_{k} & 1 & \\ & & \ddots & \ddots \\ & & & \ddots & 1 \\ & & & & \lambda_{k} \end{bmatrix}, \quad k = 1, \ldots, n_{2}.
Next, we provide a constructive proof of the Weierstrass canonical form. This proof is based on a sequential multiplication of non-singular matrices. We will demonstrate only the sufficiency part; for the necessity, please refer to [2].
Proof. Let us consider the regular pencil pair (\bm{E}, \bm{A}). We can express this as:
(\bm{E}, \bm{A} + \lambda_{0}\bm{E} - \lambda_{0}\bm{E}).
Now, define the matrix \bm{T}_{0} = (\bm{A} - \lambda_{0}\bm{E}), which is non-singular due to the regular pencil assumption. This leads us to consider the expression:
\bm{T}_{0}^{-1}(\bm{E}, \bm{A} + \lambda_{0}\bm{E} - \lambda_{0}\bm{E}) = \left(\bm{T}_{0}^{-1}\bm{E}, \bm{I} + \lambda_{0}\bm{T}_{0}^{-1}\bm{E}\right).
Next, let \bm{T}_{1} be the non-singular transformation matrix such that \bm{T}_{1}^{-1}\bm{T}_{0}^{-1}\bm{ET}_{1} is in Jordan canonical form. Thus, the pair
\bm{T}_{1}^{-1}\left(\bm{T}_{0}^{-1}\bm{E}, \bm{I} + \lambda_{0}\bm{T}_{0}^{-1}\bm{E}\right)\bm{T}_{1}
takes the following structure:
\left( \begin{bmatrix} \tilde{\bm{N}} & 0 \\ 0 & \tilde{\bm{J}} \end{bmatrix}, \begin{bmatrix} \bm{I} + \lambda_{0}\tilde{\bm{N}} & 0 \\ 0 & \bm{I} + \lambda_{0}\tilde{\bm{J}} \end{bmatrix} \right). \tag{7.4}
Here, the block \tilde{\bm{N}} corresponds to the zero eigenvalues of the singular matrix \bm{T}_{0}^{-1}\bm{E}. Therefore, \tilde{\bm{N}} is upper triangular, with zeros on the main diagonal. As a result, the matrix (\bm{I} + \lambda_{0}\tilde{\bm{N}}) is invertible.
Next, we define the non-singular transformation matrix \bm{T}_{2} as follows:
\bm{T}_{2} = \begin{bmatrix} (\bm{I} + \lambda_{0}\tilde{\bm{N}}) & 0 \\ 0 & \tilde{\bm{J}} \end{bmatrix}.
Multiplying the expression in Equation 7.4 by \bm{T}_{2}^{-1} from the left, we obtain the following structure:
\bm{\mathcal{S}} = \left( \begin{bmatrix} (\bm{I} + \lambda_{0}\tilde{\bm{N}})^{-1}\tilde{\bm{N}} & 0 \\ 0 & \bm{I} \end{bmatrix}, \begin{bmatrix} \bm{I} & 0 \\ 0 & \tilde{\bm{J}}^{-1}(\bm{I} + \lambda_{0}\tilde{\bm{J}}) \end{bmatrix} \right). \tag{7.5}
Finally, we need to consider matrices \bm{T}_{31} and \bm{T}_{32} such that:
- \bm{T}_{31}^{-1}(\bm{I} + \lambda_{0}\tilde{\bm{N}})^{-1}\tilde{\bm{N}}\bm{T}_{31} is in Jordan canonical form.
- \bm{T}_{32}^{-1}\tilde{\bm{J}}^{-1}(\bm{I} + \lambda_{0}\tilde{\bm{J}})\bm{T}_{32} is also in Jordan canonical form.
We construct the transformation matrix \bm{T}_{3} as follows:
\bm{T}_{3} = \begin{bmatrix} \bm{T}_{31} & 0 \\ 0 & \bm{T}_{32} \end{bmatrix}.
Thus, from Equation 7.5 the final change of coordinates yields:
\bm{T}_{3}^{-1}\bm{\mathcal{S}} \bm{T}_{3} = \left( \begin{bmatrix} \bm{N} & 0 \\ 0 & \bm{I} \end{bmatrix}, \begin{bmatrix} \bm{I} & 0 \\ 0 & \bm{J} \end{bmatrix} \right).
It remains to show that the matrix \bm{N} is nilpotent. To do this, we need to demonstrate that the matrix
(\bm{I} + \lambda_{0}\tilde{\bm{N}})^{-1}\tilde{\bm{N}}
is nilpotent. First, observe that the matrices (\bm{I} + \lambda_{0}\tilde{\bm{N}}) and \tilde{\bm{N}} commute since:
(\bm{I} + \lambda_{0}\tilde{\bm{N}})\tilde{\bm{N}} = \tilde{\bm{N}} + \lambda_{0}\tilde{\bm{N}}^{2} = \tilde{\bm{N}}(\bm{I} + \lambda_{0}\tilde{\bm{N}}).
Furthermore, (\bm{I} + \lambda_{0}\tilde{\bm{N}})^{-1} also commutes with \tilde{\bm{N}}:
\begin{aligned} (\bm{I} + \lambda_{0}\tilde{\bm{N}})^{-1}\tilde{\bm{N}} & = (\bm{I} + \lambda_{0}\tilde{\bm{N}})^{-1}\tilde{\bm{N}}(\bm{I} + \lambda_{0}\tilde{\bm{N}})(\bm{I} + \lambda_{0}\tilde{\bm{N}})^{-1} \\[0.5em] &= \tilde{\bm{N}}(\bm{I} + \lambda_{0}\tilde{\bm{N}})^{-1}, \end{aligned}
where we have used the fact that
\tilde{\bm{N}}(\bm{I} + \lambda_{0}\tilde{\bm{N}}) = (\bm{I} + \lambda_{0}\tilde{\bm{N}})\tilde{\bm{N}}
Given that these matrices commute, let \nu be the index such that \tilde{\bm{N}}^{\nu} = 0. The following holds:
\left[(\bm{I} + \lambda_{0}\tilde{\bm{N}})\tilde{\bm{N}}\right]^{\nu} = (\bm{I} + \lambda_{0}\tilde{\bm{N}})^{\nu}\tilde{\bm{N}}^{\nu} = 0.
This completes the proof.
In summary, the overall transformation matrices (\bm{P}, \bm{Q}) are given by:
\bm{P} = (\bm{T}_{0}\bm{T}_{1}\bm{T}_{2}\bm{T}_{3})^{-1}, \quad \bm{Q} = \bm{T}_{1}\bm{T}_{3}.
Differential–Algebraic Decomposition
The Weierstrass canonical form is particularly valuable because it allows us to rewrite the system described by Equation 7.1 using the transformation matrices (\bm{P}, \bm{Q}) defined earlier. This results in the following expression:
\bm{PEQQ^{-1}}\dot{\bm{x}}(t) = \bm{PAQQ^{-1}}\bm{x}(t) + \bm{Pf}(t).
Next, we apply the following change of coordinates:
\begin{aligned} \bm{z}(t) &= \bm{Q}^{-1}\bm{x}(t), \\ \bm{g}(t) &= \bm{Pf}(t), \end{aligned} \tag{7.6}
leading to the equivalent differential-algebraic equations (DAEs):
\begin{bmatrix} \bm{N} & 0 \\ 0 & \bm{I} \end{bmatrix} \begin{bmatrix} \dot{\bm{z}}_{1}(t) \\ \dot{\bm{z}}_{2}(t) \end{bmatrix} = \begin{bmatrix} \bm{I} & 0 \\ 0 & \bm{J} \end{bmatrix} \begin{bmatrix} \bm{z}_{1}(t) \\ \bm{z}_{2}(t) \end{bmatrix} + \begin{bmatrix} \bm{g}_{1}(t) \\ \bm{g}_{2}(t) \end{bmatrix}.
In this formulation, the vectors \bm{z}(t) and \bm{g}(t) are partitioned according to the dimensions of matrices \bm{J} and \bm{N}. Specifically, we have:
\bm{z}(t) = \begin{pmatrix} \bm{z}_{1}(t) \\ \bm{z}_{2}(t) \end{pmatrix}, \quad \bm{g}(t) = \begin{pmatrix} \bm{g}_{1}(t) \\ \bm{g}_{2}(t) \end{pmatrix}.
Notably, the two sets of equations corresponding to \bm{z}_{1}(t) and \bm{z}_{2}(t) are decoupled, yielding:
\begin{aligned} \bm{N}\dot{\bm{z}}_{1} &= \bm{z}_{1} + \bm{g}_{1}(t), \\ \dot{\bm{z}}_{2} &= \bm{J}\bm{z}_{2} + \bm{g}_{2}(t). \end{aligned} \tag{7.7}
Differential Component
The equation associated with \bm{z}_{2}(t) is straightforward; it represents an ordinary differential equation (ODE). Therefore, the unique solution given the initial condition \bm{z}_{2}(0) is:
\bm{z}_{2}(t) = \mathrm{e}^{\bm{J}t}\bm{z}_{2}(0) + \int_{0}^{t} \mathrm{e}^{\bm{J}(t-\tau)}\bm{g}_{2}(\tau)\,\mathrm{d}\tau, \quad \forall t \geq 0.
where matrix exponential is
\exp\bm{A} = \bm{I} + \bm{A} + \frac{1}{2} \bm{A}^2 + \frac{1}{6}\bm{A}^3+\cdots+ \frac{1}{k!}\bm{A}^k+\cdots
For further details, refer to [3].
Algebraic Component
On the other hand, Equation 7.7 remains a DAE. However, since the matrix \bm{N} is nilpotent, we can manipulate it effectively. To facilitate this, we can rewrite Equation 7.7 as:
\bm{z}_{1}(t) = \bm{N}\dot{\bm{z}}_{1}(t) - \bm{g}_{1}(t). \tag{7.8}
Substituting Equation 7.8 into itself and differentiating gives us:
\begin{aligned} \bm{z}_{1}(t) &= \bm{N}\frac{\mathrm{d}}{\mathrm{d} t}\left[\bm{N}\dot{\bm{z}}_{1}(t) - \bm{g}_{1}(t)\right] - \bm{g}_{1}(t) \\ &= \bm{N}^{2}\frac{\mathrm{d}^{2}\bm{z}_{1}(t)}{\mathrm{d} t^{2}} - \bm{N}\frac{\mathrm{d}\bm{g}_{1}(t)}{\mathrm{d} t} - \bm{g}_{1}(t) \\ &= \bm{N}^{3}\frac{\mathrm{d}^{3}\bm{z}_{1}(t)}{\mathrm{d} t^{3}} - \bm{N}^{2}\frac{\mathrm{d}^{2}\bm{g}_{1}(t)}{\mathrm{d} t^{2}} - \cdots \\ &\vdots \\ &= \bm{N}^{\nu}\frac{\mathrm{d}^{\nu}\bm{z}_{1}(t)}{\mathrm{d} t^{\nu}} - \bm{N}^{\nu-1}\frac{\mathrm{d}^{\nu-1}\bm{g}_{1}(t)}{\mathrm{d} t^{\nu-1}} - \cdots - \bm{N}\frac{\mathrm{d}\bm{g}_{1}(t)}{\mathrm{d} t} - \bm{g}_{1}(t). \end{aligned}
This process establishes differential properties of the solutions. Note that the procedure terminates when we reach the index \nu of matrix \bm{N}, since \bm{N}^{\nu} = 0. Consequently, we can express a unique solution as a finite summation of increasing derivatives of \bm{g}(t):
\bm{z}_{1}(t) = -\sum_{k=0}^{\nu-1} \bm{N}^{k} \frac{\mathrm{d}^{k} \bm{g}_{1}(t)}{\mathrm{d} t^{k}}.
Remark 7.2. The nilpotency degree \nu is often called the Kronecker (differentiation) index for a regular constant-coefficient DAE. The highest derivative of \bm g_1 appearing in the algebraic solution is of order \nu-1, not \nu. An invertible \bm E corresponds to index zero (no nilpotent block).
With this understanding, we can now express the unique solution to the original problem stated in Equation 7.1. To do this, we simply invert the change of coordinates described in Equation 7.6. The solution is given by:
\bm{x}(t) = \bm{Q}\bm{z}(t) = \bm{Q} \begin{bmatrix} -\displaystyle\sum_{k=0}^{\nu-1}\bm{N}^{k}\frac{\mathrm{d}^{k} \bm{g}_{1}(t)}{\mathrm{d} t^{k}} \\ \mathrm{e}^{\bm{J}t}\bm{z}_{2}(0) + \displaystyle\int_{0}^{t} \mathrm{e}^{\bm{J}(t-\tau)}\bm{g}_{2}(\tau)\,\mathrm{d}\tau \end{bmatrix} \tag{7.9}
Remark 7.3. It is crucial to observe that the initial condition \bm{z}_{1}(0) cannot be chosen arbitrarily; it must satisfy the following relationship:
\bm{z}_{1}(0) = -\sum_{k=0}^{\nu-1}\bm{N}^{k}\frac{\mathrm{d}^{k}\bm{g}_{1}(0)}{\mathrm{d} t^{k}}.
As a result, the initial condition \bm{x}_{0} = \bm{x}(t=0) for Equation 7.1 is also not freely chosen. This reflects the requirement that the initial condition must be consistent with the constraints imposed by the matrix \bm{E}.
To summarize our discussion, we can outline the conditions for the existence of a solution to the problem stated in Equation 7.1:
Theorem 7.2 The problem in Equation 7.1 with initial condition \bm{x}_{0} = \bm{x}(t=0) has a solution if the following conditions are satisfied:
- The differential forcing is locally integrable (piecewise continuity is sufficient), and the transformed algebraic forcing \bm g_1 has the derivatives through order \nu-1 required by Equation 7.9 (with additional smoothness when a classical continuously differentiable solution is sought).
- The pair (\bm{E}, \bm{A}) forms a regular pencil.
- The initial condition \bm{x}_{0} meets the requirement: \bm{x}_{0} = \bm{Q}\begin{bmatrix} -\displaystyle\sum_{k=0}^{\nu-1}\bm{N}^{k} \frac{\mathrm{d}^{k} \bm{g}_{1}(0)}{\mathrm{d} t^{k}} \\ \bm{z}_{2}(0) \end{bmatrix}
Furthermore, the corresponding solution is given by Equation 7.9.
Jordan form and the infinite eigenvalues of a pencil
Over \mathbb C, every square matrix \bm A is similar to a block-diagonal Jordan matrix \bm J=\bm T^{-1}\bm A\bm T. Its blocks are \bm J_k(\lambda)=\lambda\bm I_k+\bm N_k, where \bm N_k has ones on the superdiagonal and \bm N_k^k=\bm0. The number of blocks associated with \lambda is its geometric multiplicity; a matrix is diagonalizable exactly when all its Jordan blocks have size one.
For a regular pencil, Weierstrass equivalence uses two matrices \bm P,\bm Q, rather than a similarity transformation. The Jordan block \bm J describes finite generalized eigenvalues; the nilpotent block \bm N describes eigenvalues at infinity. The largest nilpotent block size is the differentiation index. A regular pencil can therefore have singular \bm E without loss of uniqueness for consistent initial data.
Worked examples: indices one, two, and three
Use the sign convention \bm N\dot{\bm z}+\bm z=\bm g(t) in the following examples. This convention is different from the \bm N\dot{\bm z}-\bm z=\bm g_1 convention in Equation 7.8, so the alternating signs below are intentional:
\bm z=\sum_{k=0}^{\nu-1}(-1)^k\bm N^k\bm g^{(k)}.
Index 1. For \bm N=[0], the constraint is z=g; no differentiation is required.
Index 2. For \bm N=\begin{bmatrix}0&1\\0&0\end{bmatrix} and \bm g=(t,t^2)^T,
\bm z=\bm g-\bm N\bm g'=(t-2t,t^2)^T=(-t,t^2)^T.
Indeed \bm N\dot{\bm z}+\bm z=(2t-t,t^2)^T=\bm g.
Index 3. For \bm N=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix} and \bm g=(0,0,t^2)^T,
\bm z=\bm g-\bm N\bm g'+\bm N^2\bm g''=(2,-2t,t^2)^T.
Direct substitution verifies \bm N\dot{\bm z}+\bm z=\bm g. In each case the initial algebraic state is fixed by \bm g(0),\ldots,\bm g^{(\nu-1)}(0), whereas the initial state of any accompanying ODE block is free. Inconsistent algebraic initial values do not produce a classical solution satisfying both the equation and the prescribed initial condition.
| Index | Nilpotency | Highest forcing derivative |
|---|---|---|
| 0 | No algebraic block | Not applicable |
| 1 | \bm N=\bm0 | 0 |
| 2 | \bm N^2=\bm0 | 1 |
| 3 | \bm N^3=\bm0 | 2 |
| \nu | \bm N^\nu=\bm0 | \nu-1 |
Further regular and singular pencils
For \bm E=\operatorname{diag}(1,0) and \bm A=\operatorname{diag}(2,1), \det(\bm A-\lambda\bm E)=2-\lambda: the pencil is regular although \bm E is singular. The DAE \bm E\dot{\bm x}=\bm A\bm x gives \dot x_1=2x_1, x_2=0, so x_2(0)=0 is mandatory. In contrast, \bm E=\operatorname{diag}(1,0) and \bm A=\begin{bmatrix}0&1\\0&0\end{bmatrix} have identically zero determinant and permit the arbitrary-function solutions derived above. Regularity is therefore the key structural hypothesis, not invertibility of \bm E.
Weaker conditions can be applied to \bm{f}(t); for instance, it is sufficient for \bm{f}(t) to be Lebesgue measurable.↩︎