Linear independence of solutions to Floquet equation

Question

I have a question about solutions to the Floquet equation, i.e.: $$ \dot{x} = A(t) x $$ where $x$ is an $N$-dimensional time-dependent column vector and $A(t)$ is a $T$-periodic $N \times N$ matrix.

To solve the equation I first find the eigenvalues $q_\beta$ and eigenvectors $\phi_\beta(t)$ of the equation $$ ( A(t) - \frac{d}{dt} ) \phi_\beta = q_\beta \phi_\beta $$ where the $\phi_\beta$ live in the space of $T$-periodic column vectors. With the $q_\beta$ and $\phi_\beta$ we form the Floquet equation solutions $\psi_\beta(t)$ by $$ \psi_\beta(t) = \phi_\beta(t) e^{q_\beta t} $$ As I understand it, there can only be $N$ linearly independent solutions $\psi_\beta$ out of the infinite number of $q_\beta, \phi_\beta$ pairs. This is because the Floquet equation is an ordinary linear differential equation and so possesses an $N$-dimensional space of solutions. The redundancy of solutions (again, as I understand it) is explained by the fact that for every periodic eigenvector $\phi$ with eigenvalue $q$ we can generate another $\phi'(t) = \phi(t) e^{i 2 \pi n \frac{t}{T}}$ with eigenvalue $q' = q - i \frac{2 \pi n}{T}$.

My question is: if $N$ eigenvectors $\phi_\beta$ are selected so that their associated Floquet solutions $\psi_\beta$ are linearly independent, under what conditions are the solutions orthogonal? For instance, the matrix $A(t)$ that I currently working on is real, normal ($AA^T=A^TA$), and has the property that $A + A^T$ is constant.

Answer 1

I assume that $\|A(t)\|\le C$ for all $t\in [0,T]$, that $A(\cdot)$ is continuously differentiable, and that $A(0) = A(T)$.

Let $M_A$ be the operator of multiplication in $L^2(0,T;\mathbb R^N)$ with the matrix $A(\cdot)$, that is $M_Ax(t) = A(t)x(t)$. It's adjoint is given by $M_A^*=M_{A^T}$. Since $AA^T = A^TA$, it follows that $M_A$ is a bounded normal operator.

Now, let $Dx(t) = x'(t)$, defined on $\mathcal D = \{x\in H^1 : x(0) = x(T)\}$, where $H^1 := H^1(0,T;\mathbb R^n)$. It is not hard to see that its adjoint is then given by $D^* = -D$. Now, consider the operator $T := M_A-D$, defined on $\mathcal D$. Its adjoint is $T^* = M_{A^T}+D$.

Ok, consider the operators $TT^*$ and $T^*T$. We have $$ dom(TT^*) = \{x\in dom(T^*) : T^*x\in dom(T)\} = \{x\in\mathcal D : M_{A^T}x+x'\in\mathcal D\} $$ By the assumptions on $A$, $\mathcal D$ is invariant under both $M_A$ and $M_{A^T}$. Thus, $dom(TT^*) = \{x\in\mathcal D : x'\in\mathcal D\}$, that is, $$ dom(TT^*) = \{x\in H^2 : x(0) = x(T),\;x'(0) = x'(T)\}. $$ By the same reasoning, we get $\mathcal D' := dom(T^*T) = dom(TT^*)$. Now, for $x\in\mathcal D'$ we have \begin{align*} TT^*x &= (M_A-D)(M_{A^T}+D) = M_AM_{A^T}x - DM_{A^T}x + M_Ax' - x''\\ TT^*x &= (M_{A^T}+D)(M_A-D) = M_{A^T}M_Ax + DM_Ax - M_{A^T}x' - x''. \end{align*} Furthermore, \begin{align*} -DM_{A^T}x + M_Ax' &= -\tfrac d{dt}A^Tx + Ax' = -(A^T)'x-A^Tx' + Ax'\\ DM_Ax - M_{A^T}x'&= \tfrac d{dt}Ax -A^Tx' = A'x + Ax' - A^Tx', \end{align*} For these to be the same we need that $A' + (A^T)' = 0$, which means that $A+A^T$ is constant, which you say is the case. So, $T$ is normal.

Now, let us see that the eigenvectors corresponding to different eigenvalues are orthogonal in $L^2$. If $S$ is a normal operator and $x,u\in dom (SS^*)$, then $$ \langle Sx,Su\rangle = \langle S^*Sx,u\rangle = \langle SS^*x,u\rangle = \langle S^*x,S^*u\rangle. $$ In particular, if we set $u=x$, then $\|Sx\| = \|S^*x\|$. Now, since for any $\lambda\in\mathbb C$ the operator $T-\lambda I$ is normal, this implies $\ker(T-\lambda I) = \ker(T^* - \overline\lambda I)$. So, let $x,u\in\mathcal D$ such that $Tx = \lambda x$ and $Tu = \mu u$ with $\lambda\neq \mu$. Then $Tx = \lambda x\in\mathcal D$ and so $x,u\in\mathcal D'$. We obtain $$ \lambda\langle x,u\rangle = \langle\lambda x,u\rangle = \langle Tx,u\rangle = \langle x,T^*u\rangle = \langle x,\overline\mu u\rangle = \mu\langle x,u\rangle, $$ thus $\langle x,u\rangle=0$.

This is of course the $L^2$-inner product. So, $\langle x,u\rangle = 0$ means that $$ \int_0^T\langle x(t),u(t)\rangle\,dt = 0. $$ It does not say anything about the orthogonality between $x(t)$ and $u(t)$ at particular times $t$.

Answer 2

Ok, I've been looking a little more thoroughly at what you're doing. ;-) To keep things simple, let's assume the matrix $A$ is constant. Then what you're looking at are the ODEs $\dot x = (A-q)x$ and you want to find solutions $x$ that are $T$-periodic. Now, the solution to this equation is $x(t) = e^{t(A-q)}a = e^{-tq}e^{tA}a$, where $a$ is some vector. Periodicity means that $x(0) = x(T)$, i.e., $a = e^{-Tq}e^{TA}a$ or $e^{TA}a = e^{Tq}a$. This is an eigenvalue equation for $A$ which can only be satisfied if $e^{Tq}$ is an eigenvalue for $A$. There are $N$ of these pairs $(q_k,a_k)$. In fact, you may choose the eigenvalues $q_k$ and eigenvectors $a_k$ for $A$ itself because $Aa_k = q_ka_k$ implies $e^{TA}a_k = e^{Tq_k}a_k$.

Now, if $(q_1,a_1)$ and $(q_2,a_2)$ are such that $q_1\neq q_2$, then $a_1$ and $a_2$ are indeed orthogonal because $A$ is normal. So, for the solutions $x_1$ and $x_2$ of the Floquet equation you get $x_j(t) = e^{-tq_j}e^{tA}a_j = e^{-tq_j}e^{tq_j}a_j = a_j$, which is constant. So, $$ \langle x_1(t),x_2(t)\rangle = \langle a_1,a_2\rangle = 0. $$ This of course has to be modified for non-constant $A$, where the $x_j(t)$ are surely not constant.