Solving a second-order matrix differential equation
$begingroup$
I have the differential equation $frac{d^2 x}{dt^2}+Ax=0$ where $A$ is a matrix and $$frac{d^2 x}{dt^2}=left(frac{d^2 x_1}{dt^2},frac{d^2 x_2}{dt^2},ldots,frac{d^2 x_n}{dt^2}right)^Ttext{ for }x=(x_1,x_2,ldots,x_n).$$
I know the solution is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0$, but I have no idea why this is.
How would I go about deriving this solution? The literature I have looked at simply states the solution with little or no explanation.
matrices ordinary-differential-equations
edited Apr 26 '13 at 17:02
Michael Hardy
1
asked Apr 26 '13 at 15:41
hello123hello123
211212
$endgroup$
add a comment |
$begingroup$
I have the differential equation $frac{d^2 x}{dt^2}+Ax=0$ where $A$ is a matrix and $$frac{d^2 x}{dt^2}=left(frac{d^2 x_1}{dt^2},frac{d^2 x_2}{dt^2},ldots,frac{d^2 x_n}{dt^2}right)^Ttext{ for }x=(x_1,x_2,ldots,x_n).$$
I know the solution is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0$, but I have no idea why this is.
How would I go about deriving this solution? The literature I have looked at simply states the solution with little or no explanation.
matrices ordinary-differential-equations
edited Apr 26 '13 at 17:02
Michael Hardy
1
asked Apr 26 '13 at 15:41
hello123hello123
211212
$endgroup$
add a comment |
$begingroup$
I have the differential equation $frac{d^2 x}{dt^2}+Ax=0$ where $A$ is a matrix and $$frac{d^2 x}{dt^2}=left(frac{d^2 x_1}{dt^2},frac{d^2 x_2}{dt^2},ldots,frac{d^2 x_n}{dt^2}right)^Ttext{ for }x=(x_1,x_2,ldots,x_n).$$
I know the solution is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0$, but I have no idea why this is.
How would I go about deriving this solution? The literature I have looked at simply states the solution with little or no explanation.
matrices ordinary-differential-equations
edited Apr 26 '13 at 17:02
Michael Hardy
1
asked Apr 26 '13 at 15:41
hello123hello123
211212
$endgroup$
I have the differential equation $frac{d^2 x}{dt^2}+Ax=0$ where $A$ is a matrix and $$frac{d^2 x}{dt^2}=left(frac{d^2 x_1}{dt^2},frac{d^2 x_2}{dt^2},ldots,frac{d^2 x_n}{dt^2}right)^Ttext{ for }x=(x_1,x_2,ldots,x_n).$$
I know the solution is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0$, but I have no idea why this is.
How would I go about deriving this solution? The literature I have looked at simply states the solution with little or no explanation.
matrices ordinary-differential-equations
matrices ordinary-differential-equations
edited Apr 26 '13 at 17:02
Michael Hardy
1
asked Apr 26 '13 at 15:41
hello123hello123
211212
edited Apr 26 '13 at 17:02
Michael Hardy
1
asked Apr 26 '13 at 15:41
hello123hello123
211212
edited Apr 26 '13 at 17:02
Michael Hardy
1
edited Apr 26 '13 at 17:02
Michael Hardy
1
edited Apr 26 '13 at 17:02
Michael Hardy
1
1
asked Apr 26 '13 at 15:41
hello123hello123
211212
asked Apr 26 '13 at 15:41
hello123hello123
211212
asked Apr 26 '13 at 15:41
hello123hello123
211212
211212
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Let $A$ be diagonalisable ($A=P^{-1}A_dP$ with $A_d$ diagonal). Then the equation $vec{x}''+Avec{x}=0$ can be rewritten as $Pvec{x}''+A_dPvec{x}=0$. Denote $vec{y}=Pvec{x}$, then
$$vec{y}''+A_dvec{y}=0.$$
But this is equivalent to $n$ independent scalar ODEs
$$ y_k''+lambda_k y_k=0,$$
where $lambda_k$ denotes the eigenvalue of $A$ standing in the $k$th place of the diagonal of $A_d$. Their solutions are
$$y_k(t)=cossqrt{lambda_k}(t-t_0),y_k(t_0)+frac{1}{sqrt{lambda_k}}sinsqrt{lambda_k}(t-t_0),y'_k(t_0).$$
Or, in matrix form
$$ vec{y}(t)=cos sqrt{A_d}(t-t_0),vec{y}(t_0)+frac{1}{sqrt{A_d}}sin sqrt{A_d}(t-t_0),vec{y},'(t_0).
$$
Now using that $vec{x}=P^{-1}vec{y}$ and that $P^{-1}f(A_d)P=f(A)$ for any function $f$, we obtain the quoted result.
edited Jan 11 at 4:50
Rócherz
3,0263823
answered Apr 26 '13 at 16:05
Start wearing purpleStart wearing purple
47.4k12135192
$endgroup$
$begingroup$
Thanks. What about in the inhomogeneous case? I believe the answer is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0 +(sqrt{A})^{-1}int_{t_0}^tsinleft(sqrt{A}(t-s)right)f(s)ds$, but this can't be derived in the same way since we would have a $Pf(t)$ on the RHS.
$endgroup$
– hello123
Apr 27 '13 at 15:00
1
$begingroup$
It will be fine for the very same reason. You will have the non-homogeneity $vec{F}=Pvec{f}$ for $vec{y}$, then solving for $vec{y}$, coming back to $vec{x}$ and using $P^{-1}mu(A_d)P=mu(A)$, you will get something like that.
$endgroup$
– Start wearing purple
Apr 27 '13 at 15:18
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Let $A$ be diagonalisable ($A=P^{-1}A_dP$ with $A_d$ diagonal). Then the equation $vec{x}''+Avec{x}=0$ can be rewritten as $Pvec{x}''+A_dPvec{x}=0$. Denote $vec{y}=Pvec{x}$, then
$$vec{y}''+A_dvec{y}=0.$$
But this is equivalent to $n$ independent scalar ODEs
$$ y_k''+lambda_k y_k=0,$$
where $lambda_k$ denotes the eigenvalue of $A$ standing in the $k$th place of the diagonal of $A_d$. Their solutions are
$$y_k(t)=cossqrt{lambda_k}(t-t_0),y_k(t_0)+frac{1}{sqrt{lambda_k}}sinsqrt{lambda_k}(t-t_0),y'_k(t_0).$$
Or, in matrix form
$$ vec{y}(t)=cos sqrt{A_d}(t-t_0),vec{y}(t_0)+frac{1}{sqrt{A_d}}sin sqrt{A_d}(t-t_0),vec{y},'(t_0).
$$
Now using that $vec{x}=P^{-1}vec{y}$ and that $P^{-1}f(A_d)P=f(A)$ for any function $f$, we obtain the quoted result.
edited Jan 11 at 4:50
Rócherz
3,0263823
answered Apr 26 '13 at 16:05
Start wearing purpleStart wearing purple
47.4k12135192
$endgroup$
$begingroup$
Thanks. What about in the inhomogeneous case? I believe the answer is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0 +(sqrt{A})^{-1}int_{t_0}^tsinleft(sqrt{A}(t-s)right)f(s)ds$, but this can't be derived in the same way since we would have a $Pf(t)$ on the RHS.
$endgroup$
– hello123
Apr 27 '13 at 15:00
1
$begingroup$
It will be fine for the very same reason. You will have the non-homogeneity $vec{F}=Pvec{f}$ for $vec{y}$, then solving for $vec{y}$, coming back to $vec{x}$ and using $P^{-1}mu(A_d)P=mu(A)$, you will get something like that.
$endgroup$
– Start wearing purple
Apr 27 '13 at 15:18
add a comment |
$begingroup$
Let $A$ be diagonalisable ($A=P^{-1}A_dP$ with $A_d$ diagonal). Then the equation $vec{x}''+Avec{x}=0$ can be rewritten as $Pvec{x}''+A_dPvec{x}=0$. Denote $vec{y}=Pvec{x}$, then
$$vec{y}''+A_dvec{y}=0.$$
But this is equivalent to $n$ independent scalar ODEs
$$ y_k''+lambda_k y_k=0,$$
where $lambda_k$ denotes the eigenvalue of $A$ standing in the $k$th place of the diagonal of $A_d$. Their solutions are
$$y_k(t)=cossqrt{lambda_k}(t-t_0),y_k(t_0)+frac{1}{sqrt{lambda_k}}sinsqrt{lambda_k}(t-t_0),y'_k(t_0).$$
Or, in matrix form
$$ vec{y}(t)=cos sqrt{A_d}(t-t_0),vec{y}(t_0)+frac{1}{sqrt{A_d}}sin sqrt{A_d}(t-t_0),vec{y},'(t_0).
$$
Now using that $vec{x}=P^{-1}vec{y}$ and that $P^{-1}f(A_d)P=f(A)$ for any function $f$, we obtain the quoted result.
edited Jan 11 at 4:50
Rócherz
3,0263823
answered Apr 26 '13 at 16:05
Start wearing purpleStart wearing purple
47.4k12135192
$endgroup$
$begingroup$
Thanks. What about in the inhomogeneous case? I believe the answer is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0 +(sqrt{A})^{-1}int_{t_0}^tsinleft(sqrt{A}(t-s)right)f(s)ds$, but this can't be derived in the same way since we would have a $Pf(t)$ on the RHS.
$endgroup$
– hello123
Apr 27 '13 at 15:00
1
$begingroup$
It will be fine for the very same reason. You will have the non-homogeneity $vec{F}=Pvec{f}$ for $vec{y}$, then solving for $vec{y}$, coming back to $vec{x}$ and using $P^{-1}mu(A_d)P=mu(A)$, you will get something like that.
$endgroup$
– Start wearing purple
Apr 27 '13 at 15:18
add a comment |
$begingroup$
Let $A$ be diagonalisable ($A=P^{-1}A_dP$ with $A_d$ diagonal). Then the equation $vec{x}''+Avec{x}=0$ can be rewritten as $Pvec{x}''+A_dPvec{x}=0$. Denote $vec{y}=Pvec{x}$, then
$$vec{y}''+A_dvec{y}=0.$$
But this is equivalent to $n$ independent scalar ODEs
$$ y_k''+lambda_k y_k=0,$$
where $lambda_k$ denotes the eigenvalue of $A$ standing in the $k$th place of the diagonal of $A_d$. Their solutions are
$$y_k(t)=cossqrt{lambda_k}(t-t_0),y_k(t_0)+frac{1}{sqrt{lambda_k}}sinsqrt{lambda_k}(t-t_0),y'_k(t_0).$$
Or, in matrix form
$$ vec{y}(t)=cos sqrt{A_d}(t-t_0),vec{y}(t_0)+frac{1}{sqrt{A_d}}sin sqrt{A_d}(t-t_0),vec{y},'(t_0).
$$
Now using that $vec{x}=P^{-1}vec{y}$ and that $P^{-1}f(A_d)P=f(A)$ for any function $f$, we obtain the quoted result.
edited Jan 11 at 4:50
Rócherz
3,0263823
answered Apr 26 '13 at 16:05
Start wearing purpleStart wearing purple
47.4k12135192
$endgroup$
Let $A$ be diagonalisable ($A=P^{-1}A_dP$ with $A_d$ diagonal). Then the equation $vec{x}''+Avec{x}=0$ can be rewritten as $Pvec{x}''+A_dPvec{x}=0$. Denote $vec{y}=Pvec{x}$, then
$$vec{y}''+A_dvec{y}=0.$$
But this is equivalent to $n$ independent scalar ODEs
$$ y_k''+lambda_k y_k=0,$$
where $lambda_k$ denotes the eigenvalue of $A$ standing in the $k$th place of the diagonal of $A_d$. Their solutions are
$$y_k(t)=cossqrt{lambda_k}(t-t_0),y_k(t_0)+frac{1}{sqrt{lambda_k}}sinsqrt{lambda_k}(t-t_0),y'_k(t_0).$$
Or, in matrix form
$$ vec{y}(t)=cos sqrt{A_d}(t-t_0),vec{y}(t_0)+frac{1}{sqrt{A_d}}sin sqrt{A_d}(t-t_0),vec{y},'(t_0).
$$
Now using that $vec{x}=P^{-1}vec{y}$ and that $P^{-1}f(A_d)P=f(A)$ for any function $f$, we obtain the quoted result.
edited Jan 11 at 4:50
Rócherz
3,0263823
answered Apr 26 '13 at 16:05
Start wearing purpleStart wearing purple
47.4k12135192
edited Jan 11 at 4:50
Rócherz
3,0263823
edited Jan 11 at 4:50
Rócherz
3,0263823
edited Jan 11 at 4:50
Rócherz
3,0263823
3,0263823
answered Apr 26 '13 at 16:05
Start wearing purpleStart wearing purple
47.4k12135192
answered Apr 26 '13 at 16:05
Start wearing purpleStart wearing purple
47.4k12135192
answered Apr 26 '13 at 16:05
Start wearing purpleStart wearing purple
47.4k12135192
47.4k12135192
$begingroup$
Thanks. What about in the inhomogeneous case? I believe the answer is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0 +(sqrt{A})^{-1}int_{t_0}^tsinleft(sqrt{A}(t-s)right)f(s)ds$, but this can't be derived in the same way since we would have a $Pf(t)$ on the RHS.
$endgroup$
– hello123
Apr 27 '13 at 15:00
1
$begingroup$
It will be fine for the very same reason. You will have the non-homogeneity $vec{F}=Pvec{f}$ for $vec{y}$, then solving for $vec{y}$, coming back to $vec{x}$ and using $P^{-1}mu(A_d)P=mu(A)$, you will get something like that.
$endgroup$
– Start wearing purple
Apr 27 '13 at 15:18
add a comment |
$begingroup$
Thanks. What about in the inhomogeneous case? I believe the answer is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0 +(sqrt{A})^{-1}int_{t_0}^tsinleft(sqrt{A}(t-s)right)f(s)ds$, but this can't be derived in the same way since we would have a $Pf(t)$ on the RHS.
$endgroup$
– hello123
Apr 27 '13 at 15:00
1
$begingroup$
It will be fine for the very same reason. You will have the non-homogeneity $vec{F}=Pvec{f}$ for $vec{y}$, then solving for $vec{y}$, coming back to $vec{x}$ and using $P^{-1}mu(A_d)P=mu(A)$, you will get something like that.
$endgroup$
– Start wearing purple
Apr 27 '13 at 15:18
$begingroup$
Thanks. What about in the inhomogeneous case? I believe the answer is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0 +(sqrt{A})^{-1}int_{t_0}^tsinleft(sqrt{A}(t-s)right)f(s)ds$, but this can't be derived in the same way since we would have a $Pf(t)$ on the RHS.
$endgroup$
– hello123
Apr 27 '13 at 15:00
$begingroup$
Thanks. What about in the inhomogeneous case? I believe the answer is $x=cos(sqrt{A}(t-t_0))x_0+(sqrt{A})^{-1}sin(sqrt{A}(t-t_0))dot{x}_0 +(sqrt{A})^{-1}int_{t_0}^tsinleft(sqrt{A}(t-s)right)f(s)ds$, but this can't be derived in the same way since we would have a $Pf(t)$ on the RHS.
$endgroup$
– hello123
Apr 27 '13 at 15:00
1
1
$begingroup$
It will be fine for the very same reason. You will have the non-homogeneity $vec{F}=Pvec{f}$ for $vec{y}$, then solving for $vec{y}$, coming back to $vec{x}$ and using $P^{-1}mu(A_d)P=mu(A)$, you will get something like that.
$endgroup$
– Start wearing purple
Apr 27 '13 at 15:18
$begingroup$
It will be fine for the very same reason. You will have the non-homogeneity $vec{F}=Pvec{f}$ for $vec{y}$, then solving for $vec{y}$, coming back to $vec{x}$ and using $P^{-1}mu(A_d)P=mu(A)$, you will get something like that.
$endgroup$
– Start wearing purple
Apr 27 '13 at 15:18
add a comment |
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