Fundamental matrix of Hill's equation
$begingroup$
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
asked Dec 31 '18 at 13:09
vicase98vicase98
214
$endgroup$
add a comment |
$begingroup$
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
asked Dec 31 '18 at 13:09
vicase98vicase98
214
$endgroup$
add a comment |
$begingroup$
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
asked Dec 31 '18 at 13:09
vicase98vicase98
214
$endgroup$
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
asked Dec 31 '18 at 13:09
vicase98vicase98
214
asked Dec 31 '18 at 13:09
vicase98vicase98
214
edited Jan 2 at 22:11
vicase98
asked Dec 31 '18 at 13:09
vicase98vicase98
214
asked Dec 31 '18 at 13:09
vicase98vicase98
214
asked Dec 31 '18 at 13:09
vicase98vicase98
214
214
add a comment |
add a comment |
1 Answer
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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, $varphi_1(0)=1,varphi'_1(0)=0, varphi_2(0)=0,varphi'_2(0)=1$.
answered Jan 2 at 19:48
Gil BorGil Bor
2,985812
$endgroup$
$begingroup$
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
add a comment |
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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, $varphi_1(0)=1,varphi'_1(0)=0, varphi_2(0)=0,varphi'_2(0)=1$.
answered Jan 2 at 19:48
Gil BorGil Bor
2,985812
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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, $varphi_1(0)=1,varphi'_1(0)=0, varphi_2(0)=0,varphi'_2(0)=1$.
answered Jan 2 at 19:48
Gil BorGil Bor
2,985812
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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, $varphi_1(0)=1,varphi'_1(0)=0, varphi_2(0)=0,varphi'_2(0)=1$.
answered Jan 2 at 19:48
Gil BorGil Bor
2,985812
$endgroup$
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, $varphi_1(0)=1,varphi'_1(0)=0, varphi_2(0)=0,varphi'_2(0)=1$.
answered Jan 2 at 19:48
Gil BorGil Bor
2,985812
answered Jan 2 at 19:48
Gil BorGil Bor
2,985812
answered Jan 2 at 19:48
Gil BorGil Bor
2,985812
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
add a comment |
$begingroup$
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
add a comment |
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