Let $A$ be a (finite) Hurwitz matrix.

In this related question of mine, (see also https://en.wikipedia.org/wiki/Lyapunov_equation) it is shown that $$ \int_0^\infty \sum_{j,k} (e^{At})_{ij} Q_{jk} (e^{At})_{\ell,k}dt = W_{i,\ell}, $$ where the matrix $W$ is the solution of some matrix equation $AW+WA^T+Q = 0$. Numerically, the solution of such an equation can be computed in $O(d^3)$, where $d$ is the dimension of my system.

I am wondering if this result can be easily extended to multi-linear maps? In particular, is it possible to compute (numerically) the quantity $$ \int_0^\infty \sum_{j_1,j_2,j_3}Q_{j_1,j_2,j_3}(e^{At})_{i_1,j_1} (e^{At})_{i_2,j_2}(e^{At})_{i_3,j_3}dt $$ by an efficient procedure?

Also, if we go to higher order, like $$ \int_0^\infty \sum_{j_1,j_2,\dots,j_k}Q_{j_1,\dots,j_k}(e^{At})_{i_1,j_1} \dots (e^{At})_{i_k,j_k}dt, $$ can we compute the solution of such an equation with a (relatively efficient) numerical procedure? If yes, how does its complexity grow with $k$?

[EDIT] : after my answer below, there still remains a question whether a more efficient algorithm exists (like the Lyapunov equation for $k=2$).


Inspired by Will's answer of Integral of the entrywise square of the exponential of a matrix I realized that the above equation can also be computed by using the Kronecker product.

Let $$M= A \oplus A \oplus A = A \otimes I_n \otimes I_n + I_n \otimes A \otimes I_n + I_n \otimes I_n \otimes A.$$ In terms of indices, this means: $$M_{i_1i_2i_3,j_1j_2j_3} = A_{i_1j_1}\delta_{i_2j_2}\delta_{i_3j_3} + \delta_{i_1j_1}A_{i_2j_2}\delta_{i_3j_3}+ \delta_{i_1j_1}\delta_{i_2j_2}A_{i_3j_3}$$ We have: $$\exp(Mt) = \exp(At)\otimes \exp(At) \otimes \exp(At).$$ In terms of indices this means that: $$\exp(Mt)_{i_1i_2i_3,j_1j_2j_3} = \exp(At)_{i_1,j_1}\exp(At)_{i_2,j_2}\exp(At)_{i_3,j_3}$$ In particular, $$ \int_0^\infty \exp(Mt)dt = -M^{-1},$$ which gives the solution to the above integral in $O(n^9)$.

This can be easily generalized to higher order, with a complexity in $O(n^{3k})$. For $k=2$, it is less efficient than solving the Lyapunov equation (which can be done in $O(n^3)$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.