I am reading in a book and I encountered this step that I do not understand in the proof. The book is proving the following:
$ \frac{dx}{dt} = A(t) x + f(t) $
where $A(t)$ is a continuous $n \times n$ matrix on $\mathbb{R}$, and $A(t + \omega) = A(t)$, $\omega > 0$, $t \in \mathbb{R}$. Let $\phi(t)$ be a solution matrix with $\det \phi(t) \neq 0$. Show that if the differential equation has a unique $\omega$-periodic solution, then $\left[\phi^{-1}(\omega) – \phi^{-1}(0) \right]$ is non-singular.
In the end, they were able to show this
$ \phi(\omega) \left[ \phi^{-1}(\omega) – I \right] x_0 = \phi(\omega) \int_0^\omega \phi^{-1}(\omega) f(s) \,ds$
and then jumped to the conclusion that
$\left| \phi^{-1}(\omega) – \phi^{-1}(0) \right| \neq 0$
or else this is no unique solution. I am confused on why they can say if you want a unique solution, then $\left| \phi^{-1}(\omega) – \phi^{-1}(0) \right| \neq 0$.
Then I am also confused to about why saying $\left| \phi^{-1}(\omega) – \phi^{-1}(0) \right| \neq 0$ should make $\left[ \phi^{-1}(\omega) – \phi^{-1}(0) \right]$ non-singular.
Please show me the reasons, thank you. Let me know if you don't have enough information. I only gave the end of the proof.
Best Answer
I think some of your formulas have typos. The solution of the system with initial value $x(0)=x_0$ is $$ x(t) = \phi(t) \left(x_0 + \int_0^t \phi(s)^{-1} f(s)\; ds\right)$$ In particular, with $t = \omega$, the solution is periodic iff $x(\omega) = x_0$, i.e. $$ (I - \phi(\omega)) \; x_0 = \phi(\omega) \int_{0}^{\omega} \phi(s)^{-1} f(s) ds$$ The periodic solution is unique (i.e. there is exactly one $x_0$ that satisfies this) if and only if $I - \phi(\omega)$ is nonsingular, i.e. $\left| I- \phi(\omega) \right| \ne 0$.
I think the authors are assuming $\phi(0)=I$, and since $I - \phi(\omega) = \phi(\omega) (\phi(\omega)^{-1} - I)$ and the determinant of a product is the product of the determinants this is also equivalent to $\left| \phi(\omega)^{-1} - \phi(0)^{-1}\right| \ne 0$