Second Order Matrix ODE
$begingroup$
Given the equation $frac{d^2X}{dt^2} = MX$ and the appropriate initial values, how would one go about solving this equation?
I've looked at Qualitative dependence of solution to second-order matrix differential equation on eigenvalues, which was very useful but I don't really understand how the change of basis was performed nor how the eigenvalues could be found.
Are there any resources that explain how to tackle second order matrix equations out there?
matrices ordinary-differential-equations
asked Dec 31 '18 at 8:37
Jonathan LowJonathan Low
635
$endgroup$
add a comment |
$begingroup$
Given the equation $frac{d^2X}{dt^2} = MX$ and the appropriate initial values, how would one go about solving this equation?
I've looked at Qualitative dependence of solution to second-order matrix differential equation on eigenvalues, which was very useful but I don't really understand how the change of basis was performed nor how the eigenvalues could be found.
Are there any resources that explain how to tackle second order matrix equations out there?
matrices ordinary-differential-equations
asked Dec 31 '18 at 8:37
Jonathan LowJonathan Low
635
$endgroup$
$begingroup$
If you don't know how to find eigenvalues of a matrix then I suggest you find an introductory text on linear algebra.
$endgroup$
– Eddy
Dec 31 '18 at 18:04
$begingroup$
Sorry, @Eddy. My question was less about the process which I understand, than the reasons for doing so - why do we do those things? Are there any links you can point me to if you can't explain it yourself?
$endgroup$
– Jonathan Low
Jan 2 at 14:17
add a comment |
$begingroup$
Given the equation $frac{d^2X}{dt^2} = MX$ and the appropriate initial values, how would one go about solving this equation?
I've looked at Qualitative dependence of solution to second-order matrix differential equation on eigenvalues, which was very useful but I don't really understand how the change of basis was performed nor how the eigenvalues could be found.
Are there any resources that explain how to tackle second order matrix equations out there?
matrices ordinary-differential-equations
asked Dec 31 '18 at 8:37
Jonathan LowJonathan Low
635
$endgroup$
Given the equation $frac{d^2X}{dt^2} = MX$ and the appropriate initial values, how would one go about solving this equation?
I've looked at Qualitative dependence of solution to second-order matrix differential equation on eigenvalues, which was very useful but I don't really understand how the change of basis was performed nor how the eigenvalues could be found.
Are there any resources that explain how to tackle second order matrix equations out there?
matrices ordinary-differential-equations
matrices ordinary-differential-equations
asked Dec 31 '18 at 8:37
Jonathan LowJonathan Low
635
asked Dec 31 '18 at 8:37
Jonathan LowJonathan Low
635
asked Dec 31 '18 at 8:37
Jonathan LowJonathan Low
635
asked Dec 31 '18 at 8:37
Jonathan LowJonathan Low
635
asked Dec 31 '18 at 8:37
Jonathan LowJonathan Low
635
635
$begingroup$
If you don't know how to find eigenvalues of a matrix then I suggest you find an introductory text on linear algebra.
$endgroup$
– Eddy
Dec 31 '18 at 18:04
$begingroup$
Sorry, @Eddy. My question was less about the process which I understand, than the reasons for doing so - why do we do those things? Are there any links you can point me to if you can't explain it yourself?
$endgroup$
– Jonathan Low
Jan 2 at 14:17
add a comment |
$begingroup$
If you don't know how to find eigenvalues of a matrix then I suggest you find an introductory text on linear algebra.
$endgroup$
– Eddy
Dec 31 '18 at 18:04
$begingroup$
Sorry, @Eddy. My question was less about the process which I understand, than the reasons for doing so - why do we do those things? Are there any links you can point me to if you can't explain it yourself?
$endgroup$
– Jonathan Low
Jan 2 at 14:17
$begingroup$
If you don't know how to find eigenvalues of a matrix then I suggest you find an introductory text on linear algebra.
$endgroup$
– Eddy
Dec 31 '18 at 18:04
$begingroup$
If you don't know how to find eigenvalues of a matrix then I suggest you find an introductory text on linear algebra.
$endgroup$
– Eddy
Dec 31 '18 at 18:04
$begingroup$
Sorry, @Eddy. My question was less about the process which I understand, than the reasons for doing so - why do we do those things? Are there any links you can point me to if you can't explain it yourself?
$endgroup$
– Jonathan Low
Jan 2 at 14:17
$begingroup$
Sorry, @Eddy. My question was less about the process which I understand, than the reasons for doing so - why do we do those things? Are there any links you can point me to if you can't explain it yourself?
$endgroup$
– Jonathan Low
Jan 2 at 14:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The idea is to simplify the equation by transforming it to the form
$$
frac{dY} {dt} = D Y
$$
where $D$ is a constant diagonal matrix. This matrix equation is can be easily rewritten as a system of equations where each equation only features a single element of $Y$:
$$
frac{dY_{ij}} {dt} = D_{ii} Y_{ij} quad forall i, j
$$
The problem is, what systems of the form
$$
frac{dX} {dt} = M X
$$
where $M$ is a constant matrix, can transform variables so that we get an equation of the simplified firm above? Well, if $M$ is a matrix with a complete set of right eigenvectors which form a matrix $R$, then the corresponding matrix of left eigenvectors is $R^{-1}$ and the diagonal matrix of eigenvalues is $Lambda=R^{-1}MR$. So, how do we get a nice diagonal matrix in our equation? If we premultiply by $R^{-1} $ and insert a factor of $I=RR^{-1}$ between $M$ and $X$ then
$$
R^{-1} frac{dX} {dt} = R^{-1} M R R^{-1} X
$$
$$
frac{dR^{-1}X} {dt} = Lambda R^{-1} X
$$
Using a letter to the repeated combination, $eta=R^{-1} X$, yields
$$
frac{deta} {dt} = Lambda eta
$$
which is in the simplified form desired.
answered Jan 2 at 17:22
EddyEddy
959612
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The idea is to simplify the equation by transforming it to the form
$$
frac{dY} {dt} = D Y
$$
where $D$ is a constant diagonal matrix. This matrix equation is can be easily rewritten as a system of equations where each equation only features a single element of $Y$:
$$
frac{dY_{ij}} {dt} = D_{ii} Y_{ij} quad forall i, j
$$
The problem is, what systems of the form
$$
frac{dX} {dt} = M X
$$
where $M$ is a constant matrix, can transform variables so that we get an equation of the simplified firm above? Well, if $M$ is a matrix with a complete set of right eigenvectors which form a matrix $R$, then the corresponding matrix of left eigenvectors is $R^{-1}$ and the diagonal matrix of eigenvalues is $Lambda=R^{-1}MR$. So, how do we get a nice diagonal matrix in our equation? If we premultiply by $R^{-1} $ and insert a factor of $I=RR^{-1}$ between $M$ and $X$ then
$$
R^{-1} frac{dX} {dt} = R^{-1} M R R^{-1} X
$$
$$
frac{dR^{-1}X} {dt} = Lambda R^{-1} X
$$
Using a letter to the repeated combination, $eta=R^{-1} X$, yields
$$
frac{deta} {dt} = Lambda eta
$$
which is in the simplified form desired.
answered Jan 2 at 17:22
EddyEddy
959612
$endgroup$
add a comment |
$begingroup$
The idea is to simplify the equation by transforming it to the form
$$
frac{dY} {dt} = D Y
$$
where $D$ is a constant diagonal matrix. This matrix equation is can be easily rewritten as a system of equations where each equation only features a single element of $Y$:
$$
frac{dY_{ij}} {dt} = D_{ii} Y_{ij} quad forall i, j
$$
The problem is, what systems of the form
$$
frac{dX} {dt} = M X
$$
where $M$ is a constant matrix, can transform variables so that we get an equation of the simplified firm above? Well, if $M$ is a matrix with a complete set of right eigenvectors which form a matrix $R$, then the corresponding matrix of left eigenvectors is $R^{-1}$ and the diagonal matrix of eigenvalues is $Lambda=R^{-1}MR$. So, how do we get a nice diagonal matrix in our equation? If we premultiply by $R^{-1} $ and insert a factor of $I=RR^{-1}$ between $M$ and $X$ then
$$
R^{-1} frac{dX} {dt} = R^{-1} M R R^{-1} X
$$
$$
frac{dR^{-1}X} {dt} = Lambda R^{-1} X
$$
Using a letter to the repeated combination, $eta=R^{-1} X$, yields
$$
frac{deta} {dt} = Lambda eta
$$
which is in the simplified form desired.
answered Jan 2 at 17:22
EddyEddy
959612
$endgroup$
add a comment |
$begingroup$
The idea is to simplify the equation by transforming it to the form
$$
frac{dY} {dt} = D Y
$$
where $D$ is a constant diagonal matrix. This matrix equation is can be easily rewritten as a system of equations where each equation only features a single element of $Y$:
$$
frac{dY_{ij}} {dt} = D_{ii} Y_{ij} quad forall i, j
$$
The problem is, what systems of the form
$$
frac{dX} {dt} = M X
$$
where $M$ is a constant matrix, can transform variables so that we get an equation of the simplified firm above? Well, if $M$ is a matrix with a complete set of right eigenvectors which form a matrix $R$, then the corresponding matrix of left eigenvectors is $R^{-1}$ and the diagonal matrix of eigenvalues is $Lambda=R^{-1}MR$. So, how do we get a nice diagonal matrix in our equation? If we premultiply by $R^{-1} $ and insert a factor of $I=RR^{-1}$ between $M$ and $X$ then
$$
R^{-1} frac{dX} {dt} = R^{-1} M R R^{-1} X
$$
$$
frac{dR^{-1}X} {dt} = Lambda R^{-1} X
$$
Using a letter to the repeated combination, $eta=R^{-1} X$, yields
$$
frac{deta} {dt} = Lambda eta
$$
which is in the simplified form desired.
answered Jan 2 at 17:22
EddyEddy
959612
$endgroup$
The idea is to simplify the equation by transforming it to the form
$$
frac{dY} {dt} = D Y
$$
where $D$ is a constant diagonal matrix. This matrix equation is can be easily rewritten as a system of equations where each equation only features a single element of $Y$:
$$
frac{dY_{ij}} {dt} = D_{ii} Y_{ij} quad forall i, j
$$
The problem is, what systems of the form
$$
frac{dX} {dt} = M X
$$
where $M$ is a constant matrix, can transform variables so that we get an equation of the simplified firm above? Well, if $M$ is a matrix with a complete set of right eigenvectors which form a matrix $R$, then the corresponding matrix of left eigenvectors is $R^{-1}$ and the diagonal matrix of eigenvalues is $Lambda=R^{-1}MR$. So, how do we get a nice diagonal matrix in our equation? If we premultiply by $R^{-1} $ and insert a factor of $I=RR^{-1}$ between $M$ and $X$ then
$$
R^{-1} frac{dX} {dt} = R^{-1} M R R^{-1} X
$$
$$
frac{dR^{-1}X} {dt} = Lambda R^{-1} X
$$
Using a letter to the repeated combination, $eta=R^{-1} X$, yields
$$
frac{deta} {dt} = Lambda eta
$$
which is in the simplified form desired.
answered Jan 2 at 17:22
EddyEddy
959612
answered Jan 2 at 17:22
EddyEddy
959612
answered Jan 2 at 17:22
EddyEddy
959612
answered Jan 2 at 17:22
EddyEddy
959612
959612
add a comment |
add a comment |
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$begingroup$
If you don't know how to find eigenvalues of a matrix then I suggest you find an introductory text on linear algebra.
$endgroup$
– Eddy
Dec 31 '18 at 18:04
$begingroup$
Sorry, @Eddy. My question was less about the process which I understand, than the reasons for doing so - why do we do those things? Are there any links you can point me to if you can't explain it yourself?
$endgroup$
– Jonathan Low
Jan 2 at 14:17