Mixing
Solving a non-homogenous system of differential equations
$begingroup$
If X' = AX + b then can I use the same principle as if this was an first order ODE?
What do I mean by that, to solve a first order ODE we multiply with the so-called integrating factor: $e^{-int a(t)dt}$ can I apply this here and solve it this way?
So what would I do:
$$X'-AX=b$$
multiply both sides with $e^{-At}$
$$(Xe^{-At})'=be^{-At}$$
integrate both sides
$$Xe^{-At}=int be^{-At}dt$$
Giving us:
$$X=e^{At}int be^{-At}dt$$
Where $A in mathcal M_n(mathbb{R})$
Is this proof correct? How do I integrate a Matrix or a vector? Do I integrate on every component? Also how do I raise $e$ to some matrix power? Do I use the fact that:
$$e^{At}=sum_{n=0}^{infty}frac{(At)^n}{n!}$$
And if $A$ is diagonalizable I can:
$$e^{At}=P(sum_{n=0}^{infty}frac{(Dt)^n}{n!})P^{-1}$$
Now:
$$e^{At}=Pe^{Dt}P^{-1}$$
where: $D$ is a diagonal matrix, but still how do I compute $e^{Dt}$?
And what if $A$ is not diagnolaziable and it's root are complex, conjugate?
How can I solve then?
ordinary-differential-equations systems-of-equations
asked Jan 15 at 13:41
C. CristiC. Cristi
1,629218
$endgroup$
add a comment |
$begingroup$
If X' = AX + b then can I use the same principle as if this was an first order ODE?
What do I mean by that, to solve a first order ODE we multiply with the so-called integrating factor: $e^{-int a(t)dt}$ can I apply this here and solve it this way?
So what would I do:
$$X'-AX=b$$
multiply both sides with $e^{-At}$
$$(Xe^{-At})'=be^{-At}$$
integrate both sides
$$Xe^{-At}=int be^{-At}dt$$
Giving us:
$$X=e^{At}int be^{-At}dt$$
Where $A in mathcal M_n(mathbb{R})$
Is this proof correct? How do I integrate a Matrix or a vector? Do I integrate on every component? Also how do I raise $e$ to some matrix power? Do I use the fact that:
$$e^{At}=sum_{n=0}^{infty}frac{(At)^n}{n!}$$
And if $A$ is diagonalizable I can:
$$e^{At}=P(sum_{n=0}^{infty}frac{(Dt)^n}{n!})P^{-1}$$
Now:
$$e^{At}=Pe^{Dt}P^{-1}$$
where: $D$ is a diagonal matrix, but still how do I compute $e^{Dt}$?
And what if $A$ is not diagnolaziable and it's root are complex, conjugate?
How can I solve then?
ordinary-differential-equations systems-of-equations
asked Jan 15 at 13:41
C. CristiC. Cristi
1,629218
$endgroup$
$begingroup$
Did you review something like math.berkeley.edu/~conway/Teaching/old/summer2016-2552bc/… and math.okstate.edu/people/binegar/4233/4233-l03.pdf and ndsu.edu/pubweb/~novozhil/Teaching/266%20Data/lecture_24.pdf?
$endgroup$
– Moo
Jan 15 at 13:53
$begingroup$
No, but can you exaplin me in more details what really happens when you raise $e$ to a matrix?
$endgroup$
– C. Cristi
Jan 15 at 14:31
$begingroup$
Sure, read this - cs.cornell.edu/cv/researchpdf/19ways+.pdf
$endgroup$
– Moo
Jan 15 at 16:07
$begingroup$
Your proof is formally correct although it doesn't make much sense if you don't know about the matrix exponential. For the integral of a vector-valued function, you integrate each entry, yes. For the calculation of the matrix exponential see my answer below.
$endgroup$
– Christoph
Jan 16 at 5:50
add a comment |
$begingroup$
If X' = AX + b then can I use the same principle as if this was an first order ODE?
What do I mean by that, to solve a first order ODE we multiply with the so-called integrating factor: $e^{-int a(t)dt}$ can I apply this here and solve it this way?
So what would I do:
$$X'-AX=b$$
multiply both sides with $e^{-At}$
$$(Xe^{-At})'=be^{-At}$$
integrate both sides
$$Xe^{-At}=int be^{-At}dt$$
Giving us:
$$X=e^{At}int be^{-At}dt$$
Where $A in mathcal M_n(mathbb{R})$
Is this proof correct? How do I integrate a Matrix or a vector? Do I integrate on every component? Also how do I raise $e$ to some matrix power? Do I use the fact that:
$$e^{At}=sum_{n=0}^{infty}frac{(At)^n}{n!}$$
And if $A$ is diagonalizable I can:
$$e^{At}=P(sum_{n=0}^{infty}frac{(Dt)^n}{n!})P^{-1}$$
Now:
$$e^{At}=Pe^{Dt}P^{-1}$$
where: $D$ is a diagonal matrix, but still how do I compute $e^{Dt}$?
And what if $A$ is not diagnolaziable and it's root are complex, conjugate?
How can I solve then?
ordinary-differential-equations systems-of-equations
asked Jan 15 at 13:41
C. CristiC. Cristi
1,629218
$endgroup$
If X' = AX + b then can I use the same principle as if this was an first order ODE?
What do I mean by that, to solve a first order ODE we multiply with the so-called integrating factor: $e^{-int a(t)dt}$ can I apply this here and solve it this way?
So what would I do:
$$X'-AX=b$$
multiply both sides with $e^{-At}$
$$(Xe^{-At})'=be^{-At}$$
integrate both sides
$$Xe^{-At}=int be^{-At}dt$$
Giving us:
$$X=e^{At}int be^{-At}dt$$
Where $A in mathcal M_n(mathbb{R})$
Is this proof correct? How do I integrate a Matrix or a vector? Do I integrate on every component? Also how do I raise $e$ to some matrix power? Do I use the fact that:
$$e^{At}=sum_{n=0}^{infty}frac{(At)^n}{n!}$$
And if $A$ is diagonalizable I can:
$$e^{At}=P(sum_{n=0}^{infty}frac{(Dt)^n}{n!})P^{-1}$$
Now:
$$e^{At}=Pe^{Dt}P^{-1}$$
where: $D$ is a diagonal matrix, but still how do I compute $e^{Dt}$?
And what if $A$ is not diagnolaziable and it's root are complex, conjugate?
How can I solve then?
ordinary-differential-equations systems-of-equations
ordinary-differential-equations systems-of-equations
asked Jan 15 at 13:41
C. CristiC. Cristi
1,629218
asked Jan 15 at 13:41
C. CristiC. Cristi
1,629218
asked Jan 15 at 13:41
C. CristiC. Cristi
1,629218
asked Jan 15 at 13:41
C. CristiC. Cristi
1,629218
asked Jan 15 at 13:41
C. CristiC. Cristi
1,629218
1,629218
$begingroup$
Did you review something like math.berkeley.edu/~conway/Teaching/old/summer2016-2552bc/… and math.okstate.edu/people/binegar/4233/4233-l03.pdf and ndsu.edu/pubweb/~novozhil/Teaching/266%20Data/lecture_24.pdf?
$endgroup$
– Moo
Jan 15 at 13:53
$begingroup$
No, but can you exaplin me in more details what really happens when you raise $e$ to a matrix?
$endgroup$
– C. Cristi
Jan 15 at 14:31
$begingroup$
Sure, read this - cs.cornell.edu/cv/researchpdf/19ways+.pdf
$endgroup$
– Moo
Jan 15 at 16:07
$begingroup$
Your proof is formally correct although it doesn't make much sense if you don't know about the matrix exponential. For the integral of a vector-valued function, you integrate each entry, yes. For the calculation of the matrix exponential see my answer below.
$endgroup$
– Christoph
Jan 16 at 5:50
add a comment |
$begingroup$
Did you review something like math.berkeley.edu/~conway/Teaching/old/summer2016-2552bc/… and math.okstate.edu/people/binegar/4233/4233-l03.pdf and ndsu.edu/pubweb/~novozhil/Teaching/266%20Data/lecture_24.pdf?
$endgroup$
– Moo
Jan 15 at 13:53
$begingroup$
No, but can you exaplin me in more details what really happens when you raise $e$ to a matrix?
$endgroup$
– C. Cristi
Jan 15 at 14:31
$begingroup$
Sure, read this - cs.cornell.edu/cv/researchpdf/19ways+.pdf
$endgroup$
– Moo
Jan 15 at 16:07
$begingroup$
Your proof is formally correct although it doesn't make much sense if you don't know about the matrix exponential. For the integral of a vector-valued function, you integrate each entry, yes. For the calculation of the matrix exponential see my answer below.
$endgroup$
– Christoph
Jan 16 at 5:50
$begingroup$
Did you review something like math.berkeley.edu/~conway/Teaching/old/summer2016-2552bc/… and math.okstate.edu/people/binegar/4233/4233-l03.pdf and ndsu.edu/pubweb/~novozhil/Teaching/266%20Data/lecture_24.pdf?
$endgroup$
– Moo
Jan 15 at 13:53
$begingroup$
Did you review something like math.berkeley.edu/~conway/Teaching/old/summer2016-2552bc/… and math.okstate.edu/people/binegar/4233/4233-l03.pdf and ndsu.edu/pubweb/~novozhil/Teaching/266%20Data/lecture_24.pdf?
$endgroup$
– Moo
Jan 15 at 13:53
$begingroup$
No, but can you exaplin me in more details what really happens when you raise $e$ to a matrix?
$endgroup$
– C. Cristi
Jan 15 at 14:31
$begingroup$
No, but can you exaplin me in more details what really happens when you raise $e$ to a matrix?
$endgroup$
– C. Cristi
Jan 15 at 14:31
$begingroup$
Sure, read this - cs.cornell.edu/cv/researchpdf/19ways+.pdf
$endgroup$
– Moo
Jan 15 at 16:07
$begingroup$
Sure, read this - cs.cornell.edu/cv/researchpdf/19ways+.pdf
$endgroup$
– Moo
Jan 15 at 16:07
$begingroup$
Your proof is formally correct although it doesn't make much sense if you don't know about the matrix exponential. For the integral of a vector-valued function, you integrate each entry, yes. For the calculation of the matrix exponential see my answer below.
$endgroup$
– Christoph
Jan 16 at 5:50
$begingroup$
Your proof is formally correct although it doesn't make much sense if you don't know about the matrix exponential. For the integral of a vector-valued function, you integrate each entry, yes. For the calculation of the matrix exponential see my answer below.
$endgroup$
– Christoph
Jan 16 at 5:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In general you use the Jordan normal form $J$ of $A$, $A = P J P^{-1}$. The matrix $J$ can be written as $J = D + N$ with a diagonal matrix $D$ (containing the eigenvalues of $A$) and with a nilpotent matrix $N$ ($N=0$ if and only if $A$ is diagonalizable). From the power series definition of the exponential function which you already wrote down we obtain
begin{equation}
e^{tA} = P e^{tJ} P^{-1} = P e^{t(D+N)} P^{-1} = P e^{tD} e^{tN} P^{-1}.
end{equation}
Now the matrix exponentials $e^{tD}$ and $e^{tN}$ are easy to evaluate:
1.
begin{equation}
e^{tD} = sum_{k=0}^{infty} frac{(tD)^k}{k!} = sum_{k=0}^{infty} frac{t^k}{k!} D^k = sum_{k=0}^{infty} frac{t^k}{k!} left( begin{array}{cccc}
d_1^k\
& d_2^k\
& & ddots\
& & & d_n^k
end{array}
right) = left( begin{array}{cccc}
sum_{k=0}^{infty} frac{(td_1)^k}{k!}\
& sum_{k=0}^{infty} frac{(td_2)^k}{k!}\
& & ddots\
& & & sum_{k=0}^{infty} frac{(td_n)^k}{k!}
end{array}
right) = left( begin{array}{cccc}
e^{td_1}\
& e^{td_2}\
& & ddots\
& & & e^{td_n}
end{array}
right).
end{equation}
2. For the matrix exponential of the nilpotent matrix $N$, which by definition satisfies $N^k = 0$, $k geq m$, for some $m in mathbb{N}$, you simply obtain a finite sum of powers of $tN$,
begin{equation}
e^{tN} = sum_{k=0}^{m-1} frac{(tN)^k}{k!}.
end{equation}
Finally you compute the product $e^{tA} = P e^{tD} e^{tN} P^{-1}$.
By the way all of this also works for complex-valued entries, there is no difference.
answered Jan 16 at 5:46
ChristophChristoph
58116
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
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oldest
votes
$begingroup$
In general you use the Jordan normal form $J$ of $A$, $A = P J P^{-1}$. The matrix $J$ can be written as $J = D + N$ with a diagonal matrix $D$ (containing the eigenvalues of $A$) and with a nilpotent matrix $N$ ($N=0$ if and only if $A$ is diagonalizable). From the power series definition of the exponential function which you already wrote down we obtain
begin{equation}
e^{tA} = P e^{tJ} P^{-1} = P e^{t(D+N)} P^{-1} = P e^{tD} e^{tN} P^{-1}.
end{equation}
Now the matrix exponentials $e^{tD}$ and $e^{tN}$ are easy to evaluate:
1.
begin{equation}
e^{tD} = sum_{k=0}^{infty} frac{(tD)^k}{k!} = sum_{k=0}^{infty} frac{t^k}{k!} D^k = sum_{k=0}^{infty} frac{t^k}{k!} left( begin{array}{cccc}
d_1^k\
& d_2^k\
& & ddots\
& & & d_n^k
end{array}
right) = left( begin{array}{cccc}
sum_{k=0}^{infty} frac{(td_1)^k}{k!}\
& sum_{k=0}^{infty} frac{(td_2)^k}{k!}\
& & ddots\
& & & sum_{k=0}^{infty} frac{(td_n)^k}{k!}
end{array}
right) = left( begin{array}{cccc}
e^{td_1}\
& e^{td_2}\
& & ddots\
& & & e^{td_n}
end{array}
right).
end{equation}
2. For the matrix exponential of the nilpotent matrix $N$, which by definition satisfies $N^k = 0$, $k geq m$, for some $m in mathbb{N}$, you simply obtain a finite sum of powers of $tN$,
begin{equation}
e^{tN} = sum_{k=0}^{m-1} frac{(tN)^k}{k!}.
end{equation}
Finally you compute the product $e^{tA} = P e^{tD} e^{tN} P^{-1}$.
By the way all of this also works for complex-valued entries, there is no difference.
answered Jan 16 at 5:46
ChristophChristoph
58116
$endgroup$
add a comment |
$begingroup$
In general you use the Jordan normal form $J$ of $A$, $A = P J P^{-1}$. The matrix $J$ can be written as $J = D + N$ with a diagonal matrix $D$ (containing the eigenvalues of $A$) and with a nilpotent matrix $N$ ($N=0$ if and only if $A$ is diagonalizable). From the power series definition of the exponential function which you already wrote down we obtain
begin{equation}
e^{tA} = P e^{tJ} P^{-1} = P e^{t(D+N)} P^{-1} = P e^{tD} e^{tN} P^{-1}.
end{equation}
Now the matrix exponentials $e^{tD}$ and $e^{tN}$ are easy to evaluate:
1.
begin{equation}
e^{tD} = sum_{k=0}^{infty} frac{(tD)^k}{k!} = sum_{k=0}^{infty} frac{t^k}{k!} D^k = sum_{k=0}^{infty} frac{t^k}{k!} left( begin{array}{cccc}
d_1^k\
& d_2^k\
& & ddots\
& & & d_n^k
end{array}
right) = left( begin{array}{cccc}
sum_{k=0}^{infty} frac{(td_1)^k}{k!}\
& sum_{k=0}^{infty} frac{(td_2)^k}{k!}\
& & ddots\
& & & sum_{k=0}^{infty} frac{(td_n)^k}{k!}
end{array}
right) = left( begin{array}{cccc}
e^{td_1}\
& e^{td_2}\
& & ddots\
& & & e^{td_n}
end{array}
right).
end{equation}
2. For the matrix exponential of the nilpotent matrix $N$, which by definition satisfies $N^k = 0$, $k geq m$, for some $m in mathbb{N}$, you simply obtain a finite sum of powers of $tN$,
begin{equation}
e^{tN} = sum_{k=0}^{m-1} frac{(tN)^k}{k!}.
end{equation}
Finally you compute the product $e^{tA} = P e^{tD} e^{tN} P^{-1}$.
By the way all of this also works for complex-valued entries, there is no difference.
answered Jan 16 at 5:46
ChristophChristoph
58116
$endgroup$
add a comment |
$begingroup$
In general you use the Jordan normal form $J$ of $A$, $A = P J P^{-1}$. The matrix $J$ can be written as $J = D + N$ with a diagonal matrix $D$ (containing the eigenvalues of $A$) and with a nilpotent matrix $N$ ($N=0$ if and only if $A$ is diagonalizable). From the power series definition of the exponential function which you already wrote down we obtain
begin{equation}
e^{tA} = P e^{tJ} P^{-1} = P e^{t(D+N)} P^{-1} = P e^{tD} e^{tN} P^{-1}.
end{equation}
Now the matrix exponentials $e^{tD}$ and $e^{tN}$ are easy to evaluate:
1.
begin{equation}
e^{tD} = sum_{k=0}^{infty} frac{(tD)^k}{k!} = sum_{k=0}^{infty} frac{t^k}{k!} D^k = sum_{k=0}^{infty} frac{t^k}{k!} left( begin{array}{cccc}
d_1^k\
& d_2^k\
& & ddots\
& & & d_n^k
end{array}
right) = left( begin{array}{cccc}
sum_{k=0}^{infty} frac{(td_1)^k}{k!}\
& sum_{k=0}^{infty} frac{(td_2)^k}{k!}\
& & ddots\
& & & sum_{k=0}^{infty} frac{(td_n)^k}{k!}
end{array}
right) = left( begin{array}{cccc}
e^{td_1}\
& e^{td_2}\
& & ddots\
& & & e^{td_n}
end{array}
right).
end{equation}
2. For the matrix exponential of the nilpotent matrix $N$, which by definition satisfies $N^k = 0$, $k geq m$, for some $m in mathbb{N}$, you simply obtain a finite sum of powers of $tN$,
begin{equation}
e^{tN} = sum_{k=0}^{m-1} frac{(tN)^k}{k!}.
end{equation}
Finally you compute the product $e^{tA} = P e^{tD} e^{tN} P^{-1}$.
By the way all of this also works for complex-valued entries, there is no difference.
answered Jan 16 at 5:46
ChristophChristoph
58116
$endgroup$
In general you use the Jordan normal form $J$ of $A$, $A = P J P^{-1}$. The matrix $J$ can be written as $J = D + N$ with a diagonal matrix $D$ (containing the eigenvalues of $A$) and with a nilpotent matrix $N$ ($N=0$ if and only if $A$ is diagonalizable). From the power series definition of the exponential function which you already wrote down we obtain
begin{equation}
e^{tA} = P e^{tJ} P^{-1} = P e^{t(D+N)} P^{-1} = P e^{tD} e^{tN} P^{-1}.
end{equation}
Now the matrix exponentials $e^{tD}$ and $e^{tN}$ are easy to evaluate:
1.
begin{equation}
e^{tD} = sum_{k=0}^{infty} frac{(tD)^k}{k!} = sum_{k=0}^{infty} frac{t^k}{k!} D^k = sum_{k=0}^{infty} frac{t^k}{k!} left( begin{array}{cccc}
d_1^k\
& d_2^k\
& & ddots\
& & & d_n^k
end{array}
right) = left( begin{array}{cccc}
sum_{k=0}^{infty} frac{(td_1)^k}{k!}\
& sum_{k=0}^{infty} frac{(td_2)^k}{k!}\
& & ddots\
& & & sum_{k=0}^{infty} frac{(td_n)^k}{k!}
end{array}
right) = left( begin{array}{cccc}
e^{td_1}\
& e^{td_2}\
& & ddots\
& & & e^{td_n}
end{array}
right).
end{equation}
2. For the matrix exponential of the nilpotent matrix $N$, which by definition satisfies $N^k = 0$, $k geq m$, for some $m in mathbb{N}$, you simply obtain a finite sum of powers of $tN$,
begin{equation}
e^{tN} = sum_{k=0}^{m-1} frac{(tN)^k}{k!}.
end{equation}
Finally you compute the product $e^{tA} = P e^{tD} e^{tN} P^{-1}$.
By the way all of this also works for complex-valued entries, there is no difference.
answered Jan 16 at 5:46
ChristophChristoph
58116
edited Jan 16 at 6:09
answered Jan 16 at 5:46
ChristophChristoph
58116
answered Jan 16 at 5:46
ChristophChristoph
58116
answered Jan 16 at 5:46
ChristophChristoph
58116
58116
add a comment |
add a comment |
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$begingroup$
Did you review something like math.berkeley.edu/~conway/Teaching/old/summer2016-2552bc/… and math.okstate.edu/people/binegar/4233/4233-l03.pdf and ndsu.edu/pubweb/~novozhil/Teaching/266%20Data/lecture_24.pdf?
$endgroup$
– Moo
Jan 15 at 13:53
$begingroup$
No, but can you exaplin me in more details what really happens when you raise $e$ to a matrix?
$endgroup$
– C. Cristi
Jan 15 at 14:31
$begingroup$
Sure, read this - cs.cornell.edu/cv/researchpdf/19ways+.pdf
$endgroup$
– Moo
Jan 15 at 16:07
$begingroup$
Your proof is formally correct although it doesn't make much sense if you don't know about the matrix exponential. For the integral of a vector-valued function, you integrate each entry, yes. For the calculation of the matrix exponential see my answer below.
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– Christoph
Jan 16 at 5:50