Fundamental matrix of Hill's equation












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$begingroup$


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






differential







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$endgroup$

















    0












    $begingroup$


    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






    differential







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      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






      differential







      share|cite|improve this question











      $endgroup$




      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






      differential




      differential






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      share|cite|improve this question













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      share|cite|improve this question








      edited Jan 2 at 22:11







      vicase98

















      asked Dec 31 '18 at 13:09









      vicase98vicase98

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          $begingroup$

          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 $ntimes n$ matrix-valued function. This system defines for each $tinmathbb{R}$ a linear transformation $F(t):mathbb{R}^ntomathbb{R}^n$, sending $vinmathbb{R}^n$ 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, $varphi_1(0)=1,varphi'_1(0)=0, varphi_2(0)=0,varphi'_2(0)=1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
            $endgroup$
            – vicase98
            Jan 2 at 20:58










          • $begingroup$
            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 $phi'=Aphi$, with $phi(0)=I_n$ (the $n$ by $n$ identity matrix).
            $endgroup$
            – Gil Bor
            Jan 2 at 21:17










          • $begingroup$
            I mean the Hill's equation, not $x'=Ax$
            $endgroup$
            – vicase98
            Jan 3 at 8:23










          • $begingroup$
            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 $phi''+pphi=0$, but rather $phi'=Aphi$, with $A$ the 2 by 2 matrix in your question.
            $endgroup$
            – Gil Bor
            Jan 5 at 7:54













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          $begingroup$

          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 $ntimes n$ matrix-valued function. This system defines for each $tinmathbb{R}$ a linear transformation $F(t):mathbb{R}^ntomathbb{R}^n$, sending $vinmathbb{R}^n$ 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, $varphi_1(0)=1,varphi'_1(0)=0, varphi_2(0)=0,varphi'_2(0)=1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
            $endgroup$
            – vicase98
            Jan 2 at 20:58










          • $begingroup$
            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 $phi'=Aphi$, with $phi(0)=I_n$ (the $n$ by $n$ identity matrix).
            $endgroup$
            – Gil Bor
            Jan 2 at 21:17










          • $begingroup$
            I mean the Hill's equation, not $x'=Ax$
            $endgroup$
            – vicase98
            Jan 3 at 8:23










          • $begingroup$
            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 $phi''+pphi=0$, but rather $phi'=Aphi$, with $A$ the 2 by 2 matrix in your question.
            $endgroup$
            – Gil Bor
            Jan 5 at 7:54


















          1












          $begingroup$

          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 $ntimes n$ matrix-valued function. This system defines for each $tinmathbb{R}$ a linear transformation $F(t):mathbb{R}^ntomathbb{R}^n$, sending $vinmathbb{R}^n$ 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, $varphi_1(0)=1,varphi'_1(0)=0, varphi_2(0)=0,varphi'_2(0)=1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
            $endgroup$
            – vicase98
            Jan 2 at 20:58










          • $begingroup$
            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 $phi'=Aphi$, with $phi(0)=I_n$ (the $n$ by $n$ identity matrix).
            $endgroup$
            – Gil Bor
            Jan 2 at 21:17










          • $begingroup$
            I mean the Hill's equation, not $x'=Ax$
            $endgroup$
            – vicase98
            Jan 3 at 8:23










          • $begingroup$
            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 $phi''+pphi=0$, but rather $phi'=Aphi$, with $A$ the 2 by 2 matrix in your question.
            $endgroup$
            – Gil Bor
            Jan 5 at 7:54
















          1












          1








          1





          $begingroup$

          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 $ntimes n$ matrix-valued function. This system defines for each $tinmathbb{R}$ a linear transformation $F(t):mathbb{R}^ntomathbb{R}^n$, sending $vinmathbb{R}^n$ 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, $varphi_1(0)=1,varphi'_1(0)=0, varphi_2(0)=0,varphi'_2(0)=1$.






          share|cite|improve this answer









          $endgroup$



          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 $ntimes n$ matrix-valued function. This system defines for each $tinmathbb{R}$ a linear transformation $F(t):mathbb{R}^ntomathbb{R}^n$, sending $vinmathbb{R}^n$ 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, $varphi_1(0)=1,varphi'_1(0)=0, varphi_2(0)=0,varphi'_2(0)=1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 2 at 19:48









          Gil BorGil Bor

          2,985812




          2,985812












          • $begingroup$
            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
            $endgroup$
            – vicase98
            Jan 2 at 20:58










          • $begingroup$
            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 $phi'=Aphi$, with $phi(0)=I_n$ (the $n$ by $n$ identity matrix).
            $endgroup$
            – Gil Bor
            Jan 2 at 21:17










          • $begingroup$
            I mean the Hill's equation, not $x'=Ax$
            $endgroup$
            – vicase98
            Jan 3 at 8:23










          • $begingroup$
            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 $phi''+pphi=0$, but rather $phi'=Aphi$, with $A$ the 2 by 2 matrix in your question.
            $endgroup$
            – Gil Bor
            Jan 5 at 7:54




















          • $begingroup$
            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
            $endgroup$
            – vicase98
            Jan 2 at 20:58










          • $begingroup$
            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 $phi'=Aphi$, with $phi(0)=I_n$ (the $n$ by $n$ identity matrix).
            $endgroup$
            – Gil Bor
            Jan 2 at 21:17










          • $begingroup$
            I mean the Hill's equation, not $x'=Ax$
            $endgroup$
            – vicase98
            Jan 3 at 8:23










          • $begingroup$
            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 $phi''+pphi=0$, but rather $phi'=Aphi$, with $A$ the 2 by 2 matrix in your question.
            $endgroup$
            – Gil Bor
            Jan 5 at 7:54


















          $begingroup$
          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
          $endgroup$
          – vicase98
          Jan 2 at 20:58




          $begingroup$
          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
          $endgroup$
          – vicase98
          Jan 2 at 20:58












          $begingroup$
          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 $phi'=Aphi$, with $phi(0)=I_n$ (the $n$ by $n$ identity matrix).
          $endgroup$
          – Gil Bor
          Jan 2 at 21:17




          $begingroup$
          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 $phi'=Aphi$, with $phi(0)=I_n$ (the $n$ by $n$ identity matrix).
          $endgroup$
          – Gil Bor
          Jan 2 at 21:17












          $begingroup$
          I mean the Hill's equation, not $x'=Ax$
          $endgroup$
          – vicase98
          Jan 3 at 8:23




          $begingroup$
          I mean the Hill's equation, not $x'=Ax$
          $endgroup$
          – vicase98
          Jan 3 at 8:23












          $begingroup$
          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 $phi''+pphi=0$, but rather $phi'=Aphi$, with $A$ the 2 by 2 matrix in your question.
          $endgroup$
          – Gil Bor
          Jan 5 at 7:54






          $begingroup$
          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 $phi''+pphi=0$, but rather $phi'=Aphi$, with $A$ the 2 by 2 matrix in your question.
          $endgroup$
          – Gil Bor
          Jan 5 at 7:54




















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