Integral of matrix exponential

Question

Let $A$ be an $n \times n$ matrix. Then the solution of the initial value problem \begin{align*} \dot{x}(t) = A x(t), \quad x(0) = x_0 \end{align*} is given by $x(t) = \mathrm{e}^{At} x_0$.

I am interested in the following matrix \begin{align*} \int_{0}^T \mathrm{e}^{At}\, dt \end{align*} for some $T>0$. Can one write down a general solution to this without distinguishing cases (e.g. $A$ nonsingular)?

Is this matrix always invertible?

Answer 1

Case I. If $A$ is nonsingular, then $$ \int_0^T\mathrm{e}^{tA}\,dt=\big(\mathrm{e}^{TA}-I\big)A^{-1}, $$ where $I$ is the identity matrix.

Case II. If $A$ is singular, then using the Jordan form we can write $A$ as $$ A=U^{-1}\left(\begin{matrix}B&0\\0&C\end{matrix}\right)U, $$ where $C$ is nonsingular, and $B$ is strictly upper triangular. Then $$ \mathrm{e}^{tA}=U^{-1}\left(\begin{matrix}\mathrm{e}^{tB}&0\\0&\mathrm{e}^{tC} \end{matrix}\right)U, $$ and $$ \int_0^T\mathrm{e}^{tA}\,dt=U^{-1}\left(\begin{matrix}\int_0^T\mathrm{e}^{tB}dt&0\\0&C^{-1}\big(\mathrm{e}^{TC}-I\big) \end{matrix}\right)U $$ But $\int_0^T\mathrm{e}^{tB}dt$ may have different expressions. For example if $$ B_1=\left(\begin{matrix}0&0\\0&0\end{matrix}\right), \quad B_2=\left(\begin{matrix}0&1\\0&0\end{matrix}\right), $$ then $$ \int_0^T\mathrm{e}^{tB_1}dt=\left(\begin{matrix}T&0\\0&T\end{matrix}\right), \quad \int_0^T\mathrm{e}^{tB_2}dt=\left(\begin{matrix}T&T^2/2\\0&T\end{matrix}\right). $$

Answer 2

The general formula is the power series

$$ \int_0^T e^{At} dt = T \left( I + \frac{AT}{2!} + \frac{(AT)^2}{3!} + \dots + \frac{(AT)^{n-1}}{n!} + \dots \right) $$

Note that also

$$ \left(\int_0^T e^{At} dt \right) A + I = e^{AT} $$

is always satisfied.

A sufficient condition for this matrix to be non-singular is the so-called Kalman-Ho-Narendra Theorem, which states that the matrix $\int_0^T e^{At} dt$ is invertible if

$$ T(\mu - \lambda) \neq 2k \pi i $$

for any nonzero integer $k$, where $\lambda$ and $\mu$ are any pair of eigenvalues of $A$.

Note to the interested: This matrix also comes from the discretization of a continuous linear time invariant system. It can also be said that controllability is preserved under discretization if and only if this matrix has an inverse.

Answer 3

The way I like to do it is based on the following observation: let $$ \bar{A} := \begin{bmatrix} A & B \\ 0 & 0 \end{bmatrix}, $$

where $0$ is the zero matrix (dimensions s.t. $\bar{A}$ is square). Then, $$ \mathrm{e}^{\bar{A}t} = \begin{bmatrix} \mathrm{e}^{At} & \int_0^t\mathrm{e}^{A\tau}\mathrm{d}\tau B \\ 0 & I \end{bmatrix}. $$

Hence, for the integral you can just build this block matrix with $B=I$, compute the matrix exponential of it, then extract the top right block. In a more "closed" form: $$ \int_0^t\mathrm{e}^{A\tau}\mathrm{d}\tau B = \begin{bmatrix}I & 0\end{bmatrix}\mathrm{e}^{\bar{A}t}\begin{bmatrix}0 \\ I\end{bmatrix}. $$

The advantage of this method with respect to using matrix inversion and/or Jordan form is that this method is numerically stable even when $A$ is singular (or close to singular). The disadvantage, of course, is that it takes a 4x bigger matrix as an input.

Why it works

It follows from this observation: if you have the non-homogeneous ODE $$ \dot{X}(t) = AX(t) + B, $$ its solution is $$ X(t) = \mathrm{e}^{At}X(0) + \int_0^t\mathrm{e}^{A\tau}\mathrm{d}\tau B. $$

Define the auxiliary variable $U(t)$, which is constant (i.e., $U(t) = U(0)$ for all positive $t$). Then $\dot{U}(t) = 0$ and we have the system of ODEs \begin{align*} \dot{X}(t) &= AX(t) + BU(t), \\ \dot{U}(t) &= 0, \end{align*}

which is a homogeneous ODE on the augmented variable $\begin{bmatrix} X(t) \\ U(t) \end{bmatrix}$. Therefore, we have

$$\begin{bmatrix} \dot{X}(t) \\ \dot{U}(t) \end{bmatrix} = \begin{bmatrix} A & B \\ 0 & 0 \end{bmatrix}\begin{bmatrix} {X}(t) \\ {U}(t) \end{bmatrix} = \bar{A}\begin{bmatrix} {X}(t) \\ {U}(t) \end{bmatrix}$$ whose solution is $$\begin{bmatrix} {X}(t) \\ {U}(t) \end{bmatrix} = \mathrm{e}^{\bar{A}t}\begin{bmatrix} {X}(0) \\ {U}(0) \end{bmatrix},$$

but also,

$$\begin{bmatrix} {X}(t) \\ {U}(t) \end{bmatrix} = \begin{bmatrix} \mathrm{e}^{At}X(0) + \int_0^t\mathrm{e}^{A\tau}\mathrm{d}\tau BU(0) \\ U(0) \end{bmatrix} = \begin{bmatrix} \mathrm{e}^{At} & \int_0^t\mathrm{e}^{A\tau}\mathrm{d}\tau B \\ 0 & I \end{bmatrix}\begin{bmatrix} X(0) \\ U(0) \end{bmatrix}.$$