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The Hill's equation $(H)$ is defined as $y'' + p(t)y=0$ where $p(t+T)=p(t) \forall t$. Let $x_1=y$ and $x_2=y'$, and let $x=\begin{bmatrix}{x_1}\\{x_2}\end{bmatrix}$, so $x'=\begin{bmatrix}{0}&{1}\\{-p(t)}&{0}\end{bmatrix}x$.

Now we consider the solutions of $(H)$ with the following initial conditions: $\varphi_1(0) = 1$, $\varphi_2(0) = 0$, $\varphi_1'(0) = 0$ and $\varphi_2'(0) = 1$.

My Floquet's Theory notes says that $\phi(t) = \begin{bmatrix}{\varphi_1}&{\varphi_2}\\{\varphi_1'}&{\varphi_2'}\end{bmatrix}$ is a fundamental matrix of $(H)$, but I don't know why. Could anybody explain that? Thanks

southernKid33
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1 Answers1

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The condition that $p(t)$ is periodic is not necessary. You also have an error in the definition of $\varphi_1, \varphi_2$.

Consider a linear system of ODEs, $x'(t)=A(t)x(t)$, where $x(t)$ is an $\mathbb{R}^n$-valued unknown function and $A(t)$ is a given $n\times n$ matrix-valued function. This system defines for each $t\in\mathbb{R}$ a linear transformation $F(t):\mathbb{R}^n\to\mathbb{R}^n$, sending $v\in\mathbb{R}^n$ to the value at $t$ of the solution to the system whose value at $t=0$ is $v$. The matrix of $F(t)$ with respect to the standard basis of $\mathbb{R}^n$, as a function of $t$, is called the fundamental matrix of solutions of the system. Its columns form a basis of the space of solutions of the system. The $j$-th column is the solution whose value at $t=0$ is the $j$-th standard vector (1 at the $j$-th place, the rest 0). If you apply this to the Hill's equation you get that, in your notation, $\varphi_1(0)=1,\varphi'_1(0)=0, \varphi_2(0)=0,\varphi'_2(0)=1$.

Gil Bor
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  • It's true, the initial conditions were wrong, sorry for the mistake and than you for the explanation. Anyway, would you mind explainning why that $\phi(t)$ satisfies the equation?I will really appreciate it – southernKid33 Jan 02 '19 at 20:58
  • Just chasing definitions. $\phi(t)$, by definition, is the matrix of $F(t)$ with respect to the standard basis of $\mathbb{R}^n$. That is, the $j$-th column of $\phi(t)$ is the image of the $j$-th standard basis vector under $F(t)$. By definition of $F(t)$, it is the solution to $x'=Ax$ such that $x(0)$ is the standard $j$-th vector. This is the same as $\phi'=A\phi$, with $\phi(0)=I_n$ (the $n$ by $n$ identity matrix). – Gil Bor Jan 02 '19 at 21:17
  • I mean the Hill's equation, not $x'=Ax$ – southernKid33 Jan 03 '19 at 08:23
  • Can you define ''the fundamental matrix of of (H)''? As far as I know, the fundamental matrix of a 2nd order ODE is defined as the fundamental matrix of the associated 1st order linear system. In other words, the fundamental matrix $\phi(t)$ of $y''+py=0$ is not supposed to satisfy $\phi''+p\phi=0$, but rather $\phi'=A\phi$, with $A$ the 2 by 2 matrix in your question. – Gil Bor Jan 05 '19 at 07:54