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I am interested in computing the following integral of a matrix exponential.

\begin{equation} \int_0^t e^{A(t-t')} e^{A^T (t-t')} dt' \end{equation}

The only assumption is that $A_{n\times n}$ is real. This is simple (albeit cumbersome) to compute, given a particular $A$. I was wondering, however, if there were any other steps I could take to progress the problem further a little bit further.

For instance (assuming I did not make any mistakes), if $A$ is normal then $A = U \Lambda U^H$, where $\Lambda=\text{diag}(\lambda1,\lambda2,\ldots,\lambda_n)$ contains the eigenvalues and $U$ is unitary. Then

\begin{equation} \int_0^t e^{A(t-t')} e^{A^T (t-t')} dt' = U \left[\int_0^t e^{2\Lambda(t-t')} dt'\right]U^H = U \left[(2\Lambda)^{-1} (e^{2\Lambda t} - I)\right]U^H \end{equation}

However, in general $A$ is not normal.

I appreciate the support in advance.

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A trivial remark: we can do the substitution $s=t-t'$. The trace of the integral looks like a inner product. –  Davide Giraudo Jul 19 '12 at 9:46

1 Answer 1

When the sum of any two complex eigenvalues of $A$ -- counting multiplicity -- is nonzero, some simplifications are possible.

(I) This is similar to solving the Lyapunov equation. Let $f(s)=e^{-sA}e^{-sA^T},\ C=\int_0^t f(s)ds$ and $B = \int_0^t e^{(t-s)A}e^{(t-s)A^T}ds = e^{tA} C e^{tA^T}$. Then \begin{align*} f'(s) &= - Af(s) - f(s)A^T,\\ f(t)-f(0) &= - AC - CA^T,\\ e^{-tA}e^{-tA^T} - I &= - AC - CA^T,\\ I - e^{tA}e^{tA^T} &= -AB-BA^T,\\ AB+BA^T &= e^{tA}e^{tA^T} - I,\\ (I\otimes A+A\otimes I)\operatorname{vec}(B)&=\operatorname{vec}(e^{tA}e^{tA^T} - I).\tag{1} \end{align*} Since no two eigenvalues of $A$ sum up to zero, $I\otimes A+A\otimes I$ is invertible. Therefore the unique solution of $(1)$ corresponds to our desired $B$.

(II) When $A$ is also $\mathbb{C}$-diagonalisable. Let $A=P\Lambda P^{-1}$ be a (not necessarily orthogonal) diagonalisation, where $\Lambda=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$ and $P$ may be nonreal. Then \begin{align*} f(s)&=Pe^{-s\Lambda}(P^{-1}P^{-T})e^{-s\Lambda}P^T\\ &=P\left((e^{-s(\lambda_i+\lambda_j)})\circ(P^{-1}P^{-T})\right)P^T\\ \end{align*} where "$\circ$" denotes a Hadamard product. Therefore $$ B=e^{tA}P\left(\left(\frac{1-e^{-t(\lambda_i+\lambda_j)}}{\lambda_i+\lambda_j}\right)\circ(P^{-1}P^{-T})\right)P^Te^{tA^T}. $$

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