7 Linear DAE examples and properties
A primary consideration is the existence and uniqueness of a solution
\bm{x}(t): \Bbb{R} \rightarrow \Bbb{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{7.1}
For the equation in Equation 7.1, 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 Equation 1, we need to introduce the following definition.
Definition 7.1 (Regular Pencil) A matrix pair (\bm{E},\bm{A}) is defined as a regular pencil if there exists a scalar \lambda_{0} \in \Bbb{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 Canonical Form) A matrix pair \bm{E}, \bm{A} \in \Bbb{R}^{n \times n} is classified as a regular pencil if, and only if, there exist invertible matrices
\bm{P} \in \Bbb{R}^{n \times n} \quad\text{and}\quad \bm{Q} \in \Bbb{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{7.2}
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 Matrix) A square matrix \bm{N} is termed nilpotent if there exists an integer \nu \in \Bbb{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 7.2 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:
\begin{aligned} \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}, \\ \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} \end{aligned}
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{7.3}
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 7.3 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{7.4}
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 7.4 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}.
7.1 Separation of algebraic and differential part
The Weierstrass canonical form is particularly valuable because it allows us to rewrite the system described by Equation 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{7.5}
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{7.6}
Solution of \bm{z}_2'(t) = \bm{J}\bm{z}_2(t)+\bm{g}_2(t)
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].
Solution of \bm{N}\bm{z}_1'(t) = \bm{z}_1(t)+\bm{g}_1(t)
On the other hand, Equation 7.6 remains a DAE. However, since the matrix \bm{N} is nilpotent, we can manipulate it effectively. To facilitate this, we can rewrite Equation 7.6 as:
\bm{z}_{1}(t) = \bm{N}\dot{\bm{z}}_{1}(t) - \bm{g}_{1}(t). \tag{7.7}
Substituting Equation 7.7 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 index \nu is sometimes referred to as the Kronecker index. It is important to note that \nu represents the number of times the function \bm{g}_{1}(t) is differentiated in order to obtain the solution \bm{z}_{1}(t).
With this understanding, we can now express the unique solution to the original problem stated in Equation 1. To do this, we simply invert the change of coordinates described in Equation 7.5. 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{7.8}
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 Equation 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 1:
Theorem 7.2 The problem in Equation 1 with initial condition \bm{x}_{0} = \bm{x}(t=0) has a solution if the following conditions are satisfied:
- The function \bm{f}(t) is piecewise continuous.
- 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} \dfrac{\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 7.8.
Weaker conditions can be applied to \bm{f}(t); for instance, it is sufficient for \bm{f}(t) to be Lebesgue measurable.↩︎