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Consider the following matrix function $$ f(t) = \cos(\omega_1t) A_1 + \cos(\omega_2t) A_2, \quad t\ge 0, $$ where $A_1$, $A_2$ are real square matrices and $\omega_1$, $\omega_2$ positive numbers.

Now, consider the following series of nested integrals $$ S(t)=\sum_{n=1}^\infty \int_0^{t} f(t_1)\left(\int_0^{t_1} f(t_2)\cdots \left(\int_0^{t_{n-1}} f(t_n)\, \mathrm{d}t_n\right)\cdots\mathrm{d}t_2\right) \mathrm{d}t_1. $$

If $\omega_1=\omega_2=\bar \omega$ then the above series converges to $$ \exp\left({(A_1+A_2) \int_0^t \cos(\bar \omega \tau )\, \mathrm{d}\tau \,}\right)-I, $$ so that, in this case, a (rather loose) upper bound to the 2-norm of $S(t)$ is $$ \|S(t)\|\le \exp\left({\frac{2\alpha}{\bar \omega}}\right)-1, $$ where $\alpha:=\max\{\|A_1\|,\|A_2\|\}$.

My question. With reference to the case $\omega_1\ne \omega_2$ does the following upper bound hold true $$ \|S(t)\|\le \exp\left({\frac{2\alpha}{\bar \omega}}\right)-1 $$ where $\bar \omega:=\min\{\omega_1,\omega_2\}$? [If this is not true, is it possible to find an upper bound on $\|S(t)\|$ depending on $\bar \omega$?]

A (perhaps) simplifying assumption. From the comments of some of you, I've realized that finding such a bound is a rather difficult task for general matrices $A_1$ and $A_2$. So let us suppose that $A_1$ and $A_2$ are such that $[A_1,A_2]=V$ where $V$ is a skew-symmetric matrix and $[\cdot,\cdot]$ denotes the matrix commutator. Does this assumption lead to a simplification of the problem?

Any comment is very welcome.

Consider the following matrix function $$ f(t) = \cos(\omega_1t) A_1 + \cos(\omega_2t) A_2, \quad t\ge 0, $$ where $A_1$, $A_2$ are real square matrices and $\omega_1$, $\omega_2$ positive numbers.

Now, consider the following series of nested integrals $$ S(t)=\sum_{n=1}^\infty \int_0^{t} f(t_1)\left(\int_0^{t_1} f(t_2)\cdots \left(\int_0^{t_{n-1}} f(t_n)\, \mathrm{d}t_n\right)\cdots\mathrm{d}t_2\right) \mathrm{d}t_1. $$

If $\omega_1=\omega_2=\bar \omega$ then the above series converges to $$ \exp\left({(A_1+A_2) \int_0^t \cos(\bar \omega \tau )\, \mathrm{d}\tau \,}\right)-I, $$ so that, in this case, a (rather loose) upper bound to the 2-norm of $S(t)$ is $$ \|S(t)\|\le \exp\left({\frac{2\alpha}{\bar \omega}}\right)-1, $$ where $\alpha:=\max\{\|A_1\|,\|A_2\|\}$.

My question. With reference to the case $\omega_1\ne \omega_2$ does the following upper bound hold true $$ \|S(t)\|\le \exp\left({\frac{2\alpha}{\bar \omega}}\right)-1 $$ where $\bar \omega:=\min\{\omega_1,\omega_2\}$? [If this is not true, is it possible to find an upper bound on $\|S(t)\|$ depending on $\bar \omega$?]

$ $

A (perhaps) simplifying assumption. From the comments of some of you, I've realized that finding such a bound is a rather difficult task for general matrices $A_1$ and $A_2$. So let us suppose that $A_1$ and $A_2$ are such that $[A_1,A_2]=V$ where $V$ is a skew-symmetric matrix and $[\cdot,\cdot]$ denotes the matrix commutator. Does this assumption lead to a simplification of the problem?

Any comment is very welcome.

Consider the following matrix function $$ f(t) = \cos(\omega_1t) A_1 + \cos(\omega_2t) A_2, \quad t\ge 0, $$ where $A_1$, $A_2$ are real square matrices and $\omega_1$, $\omega_2$ positive numbers.

Now, consider the following series of nested integrals $$ S(t)=\sum_{n=1}^\infty \int_0^{t} f(t_1)\left(\int_0^{t_1} f(t_2)\cdots \left(\int_0^{t_{n-1}} f(t_n)\, \mathrm{d}t_n\right)\cdots\mathrm{d}t_2\right) \mathrm{d}t_1. $$

If $\omega_1=\omega_2=\bar \omega$ then the above series converges to $$ \exp\left({(A_1+A_2) \int_0^t \cos(\bar \omega \tau )\, \mathrm{d}\tau \,}\right)-I, $$ so that, in this case, a (rather loose) upper bound to the 2-norm of $S(t)$ is $$ \|S(t)\|\le \exp\left({\frac{2\alpha}{\bar \omega}}\right)-1, $$ where $\alpha:=\max\{\|A_1\|,\|A_2\|\}$.

My question. With reference to the case $\omega_1\ne \omega_2$ does the following upper bound hold true $$ \|S(t)\|\le \exp\left({\frac{2\alpha}{\bar \omega}}\right)-1 $$ where $\bar \omega:=\min\{\omega_1,\omega_2\}$? [If this is not true, is it possible to find an upper bound on $\|S(t)\|$ depending on $\bar \omega$?]

Consider the following matrix function $$ f(t) = \cos(\omega_1t) A_1 + \cos(\omega_2t) A_2, \quad t\ge 0, $$ where $A_1$, $A_2$ are real square matrices and $\omega_1$, $\omega_2$ positive numbers.

Now, consider the following series of nested integrals $$ S(t)=\sum_{n=1}^\infty \int_0^{t} f(t_1)\left(\int_0^{t_1} f(t_2)\cdots \left(\int_0^{t_{n-1}} f(t_n)\, \mathrm{d}t_n\right)\cdots\mathrm{d}t_2\right) \mathrm{d}t_1. $$

If $\omega_1=\omega_2=\bar \omega$ then the above series converges to $$ \exp\left({(A_1+A_2) \int_0^t \cos(\bar \omega \tau )\, \mathrm{d}\tau \,}\right)-I, $$ so that, in this case, a (rather loose) upper bound to the 2-norm of $S(t)$ is $$ \|S(t)\|\le \exp\left({\frac{2\alpha}{\bar \omega}}\right)-1, $$ where $\alpha:=\max\{\|A_1\|,\|A_2\|\}$.

My question. With reference to the case $\omega_1\ne \omega_2$ does the following upper bound hold true $$ \|S(t)\|\le \exp\left({\frac{2\alpha}{\bar \omega}}\right)-1 $$ where $\bar \omega:=\min\{\omega_1,\omega_2\}$? [If this is not true, is it possible to find an upper bound on $\|S(t)\|$ depending on $\bar \omega$?]

A (perhaps) simplifying assumption. From the comments of some of you, I've realized that finding such a bound is a rather difficult task for general matrices $A_1$ and $A_2$. So let us suppose that $A_1$ and $A_2$ are such that $[A_1,A_2]=V$ where $V$ is a skew-symmetric matrix and $[\cdot,\cdot]$ denotes the matrix commutator. Does this assumption lead to a simplification of the problem?

Any comment is very welcome.

Consider the following matrix function $$ f(t) = \cos(\omega_1t) A_1 + \cos(\omega_2t) A_2, \quad t\ge 0, $$ where $A_1$, $A_2$ are real square matrices and $\omega_1$, $\omega_2$ positive numbers.

Now, consider the following series of nested integrals $$ S(t)=\sum_{n=1}^\infty \int_0^{t} f(t_1)\left(\int_0^{t_1} f(t_2)\cdots \left(\int_0^{t_{n-1}} f(t_n)\, \mathrm{d}t_n\right)\cdots\mathrm{d}t_2\right) \mathrm{d}t_1. $$

If $\omega_1=\omega_2=\bar \omega$ then the above series converges to $$ \exp\left({(A_1+A_2) \int_0^t \cos(\bar \omega \tau )\, \mathrm{d}\tau \,}\right)-I, $$ so that, in this case, a (rather loose) upper bound to the norm of $S(t)$ is $$ \|S(t)\|\le \exp\left({\frac{2\alpha}{\bar \omega}}\right)-1, $$ where $\alpha:=\max\{\|A_1\|,\|A_2\|\}$.

My question. With reference to the case $\omega_1\ne \omega_2$ does the following upper bound hold true $$ \|S(t)\|\le \exp\left({\frac{2\alpha}{\bar \omega}}\right)-1 $$ where $\bar \omega:=\min\{\omega_1,\omega_2\}$? [If this is not true, is it possible to find an upper bound on $\|S(t)\|$ depending on $\bar \omega$?]

Consider the following matrix function $$ f(t) = \cos(\omega_1t) A_1 + \cos(\omega_2t) A_2, \quad t\ge 0, $$ where $A_1$, $A_2$ are real square matrices and $\omega_1$, $\omega_2$ positive numbers.

Now, consider the following series of nested integrals $$ S(t)=\sum_{n=1}^\infty \int_0^{t} f(t_1)\left(\int_0^{t_1} f(t_2)\cdots \left(\int_0^{t_{n-1}} f(t_n)\, \mathrm{d}t_n\right)\cdots\mathrm{d}t_2\right) \mathrm{d}t_1. $$

If $\omega_1=\omega_2=\bar \omega$ then the above series converges to $$ \exp\left({(A_1+A_2) \int_0^t \cos(\bar \omega \tau )\, \mathrm{d}\tau \,}\right)-I, $$ so that, in this case, a (rather loose) upper bound to the 2-norm of $S(t)$ is $$ \|S(t)\|\le \exp\left({\frac{2\alpha}{\bar \omega}}\right)-1, $$ where $\alpha:=\max\{\|A_1\|,\|A_2\|\}$.

My question. With reference to the case $\omega_1\ne \omega_2$ does the following upper bound hold true $$ \|S(t)\|\le \exp\left({\frac{2\alpha}{\bar \omega}}\right)-1 $$ where $\bar \omega:=\min\{\omega_1,\omega_2\}$? [If this is not true, is it possible to find an upper bound on $\|S(t)\|$ depending on $\bar \omega$?]