Solution to Cauchy Problem
$begingroup$
I am trying to solve the following Cauchy Problem:
$y'(t) = A(t)y(t),
A=begin{pmatrix}
t &-1 \
1 &t
end{pmatrix}, y(0)=y_0$
What I did:
I know that $ forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$
So the solution would be:
$y(t)= e^{ begin{pmatrix}
t^2/2 &-t \
t &t^2/2
end{pmatrix}} y_0$
Is that a good reasonning? Please note that I don't have more information regarding the domain of study of this Cauchy Problem so I don't know if that plays any decisive role in the solution found above.
Many thanks!
cauchy-problem
asked Feb 9 '18 at 13:13
PerelManPerelMan
654313
$endgroup$
add a comment |
$begingroup$
I am trying to solve the following Cauchy Problem:
$y'(t) = A(t)y(t),
A=begin{pmatrix}
t &-1 \
1 &t
end{pmatrix}, y(0)=y_0$
What I did:
I know that $ forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$
So the solution would be:
$y(t)= e^{ begin{pmatrix}
t^2/2 &-t \
t &t^2/2
end{pmatrix}} y_0$
Is that a good reasonning? Please note that I don't have more information regarding the domain of study of this Cauchy Problem so I don't know if that plays any decisive role in the solution found above.
Many thanks!
cauchy-problem
asked Feb 9 '18 at 13:13
PerelManPerelMan
654313
$endgroup$
add a comment |
$begingroup$
I am trying to solve the following Cauchy Problem:
$y'(t) = A(t)y(t),
A=begin{pmatrix}
t &-1 \
1 &t
end{pmatrix}, y(0)=y_0$
What I did:
I know that $ forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$
So the solution would be:
$y(t)= e^{ begin{pmatrix}
t^2/2 &-t \
t &t^2/2
end{pmatrix}} y_0$
Is that a good reasonning? Please note that I don't have more information regarding the domain of study of this Cauchy Problem so I don't know if that plays any decisive role in the solution found above.
Many thanks!
cauchy-problem
asked Feb 9 '18 at 13:13
PerelManPerelMan
654313
$endgroup$
I am trying to solve the following Cauchy Problem:
$y'(t) = A(t)y(t),
A=begin{pmatrix}
t &-1 \
1 &t
end{pmatrix}, y(0)=y_0$
What I did:
I know that $ forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$
So the solution would be:
$y(t)= e^{ begin{pmatrix}
t^2/2 &-t \
t &t^2/2
end{pmatrix}} y_0$
Is that a good reasonning? Please note that I don't have more information regarding the domain of study of this Cauchy Problem so I don't know if that plays any decisive role in the solution found above.
Many thanks!
cauchy-problem
cauchy-problem
asked Feb 9 '18 at 13:13
PerelManPerelMan
654313
asked Feb 9 '18 at 13:13
PerelManPerelMan
654313
edited Dec 5 '18 at 15:21
PerelMan
asked Feb 9 '18 at 13:13
PerelManPerelMan
654313
asked Feb 9 '18 at 13:13
PerelManPerelMan
654313
asked Feb 9 '18 at 13:13
PerelManPerelMan
654313
654313
add a comment |
add a comment |
1 Answer
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active
oldest
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You indeed found the general form of a solution. See Wikipedia linear differential equation. However , this is only true because $A$ and $A^prime$ commute.
answered Feb 9 '18 at 13:54
mathcounterexamples.netmathcounterexamples.net
27k22157
$endgroup$
$begingroup$
Thanks for useful link. I guess the condition $forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$ is stronger than the condition $A$ and $A'$ commute, in the sense that the first condition imply the second ?
$endgroup$
– PerelMan
Feb 9 '18 at 14:11
1
$begingroup$
Yes the condition you gave is stronger. You can easily derive the second by differentiating both sides of your equality in $s$and making $s=t$ afterwards.
$endgroup$
– mathcounterexamples.net
Feb 9 '18 at 15:01
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You indeed found the general form of a solution. See Wikipedia linear differential equation. However , this is only true because $A$ and $A^prime$ commute.
answered Feb 9 '18 at 13:54
mathcounterexamples.netmathcounterexamples.net
27k22157
$endgroup$
$begingroup$
Thanks for useful link. I guess the condition $forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$ is stronger than the condition $A$ and $A'$ commute, in the sense that the first condition imply the second ?
$endgroup$
– PerelMan
Feb 9 '18 at 14:11
1
$begingroup$
Yes the condition you gave is stronger. You can easily derive the second by differentiating both sides of your equality in $s$and making $s=t$ afterwards.
$endgroup$
– mathcounterexamples.net
Feb 9 '18 at 15:01
add a comment |
$begingroup$
You indeed found the general form of a solution. See Wikipedia linear differential equation. However , this is only true because $A$ and $A^prime$ commute.
answered Feb 9 '18 at 13:54
mathcounterexamples.netmathcounterexamples.net
27k22157
$endgroup$
$begingroup$
Thanks for useful link. I guess the condition $forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$ is stronger than the condition $A$ and $A'$ commute, in the sense that the first condition imply the second ?
$endgroup$
– PerelMan
Feb 9 '18 at 14:11
1
$begingroup$
Yes the condition you gave is stronger. You can easily derive the second by differentiating both sides of your equality in $s$and making $s=t$ afterwards.
$endgroup$
– mathcounterexamples.net
Feb 9 '18 at 15:01
add a comment |
$begingroup$
You indeed found the general form of a solution. See Wikipedia linear differential equation. However , this is only true because $A$ and $A^prime$ commute.
answered Feb 9 '18 at 13:54
mathcounterexamples.netmathcounterexamples.net
27k22157
$endgroup$
You indeed found the general form of a solution. See Wikipedia linear differential equation. However , this is only true because $A$ and $A^prime$ commute.
answered Feb 9 '18 at 13:54
mathcounterexamples.netmathcounterexamples.net
27k22157
answered Feb 9 '18 at 13:54
mathcounterexamples.netmathcounterexamples.net
27k22157
answered Feb 9 '18 at 13:54
mathcounterexamples.netmathcounterexamples.net
27k22157
answered Feb 9 '18 at 13:54
mathcounterexamples.netmathcounterexamples.net
27k22157
27k22157
$begingroup$
Thanks for useful link. I guess the condition $forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$ is stronger than the condition $A$ and $A'$ commute, in the sense that the first condition imply the second ?
$endgroup$
– PerelMan
Feb 9 '18 at 14:11
1
$begingroup$
Yes the condition you gave is stronger. You can easily derive the second by differentiating both sides of your equality in $s$and making $s=t$ afterwards.
$endgroup$
– mathcounterexamples.net
Feb 9 '18 at 15:01
add a comment |
$begingroup$
Thanks for useful link. I guess the condition $forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$ is stronger than the condition $A$ and $A'$ commute, in the sense that the first condition imply the second ?
$endgroup$
– PerelMan
Feb 9 '18 at 14:11
1
$begingroup$
Yes the condition you gave is stronger. You can easily derive the second by differentiating both sides of your equality in $s$and making $s=t$ afterwards.
$endgroup$
– mathcounterexamples.net
Feb 9 '18 at 15:01
$begingroup$
Thanks for useful link. I guess the condition $forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$ is stronger than the condition $A$ and $A'$ commute, in the sense that the first condition imply the second ?
$endgroup$
– PerelMan
Feb 9 '18 at 14:11
$begingroup$
Thanks for useful link. I guess the condition $forall t, s in mathbb{R}: A(s)A(t)=A(t)A(s)$ is stronger than the condition $A$ and $A'$ commute, in the sense that the first condition imply the second ?
$endgroup$
– PerelMan
Feb 9 '18 at 14:11
1
1
$begingroup$
Yes the condition you gave is stronger. You can easily derive the second by differentiating both sides of your equality in $s$and making $s=t$ afterwards.
$endgroup$
– mathcounterexamples.net
Feb 9 '18 at 15:01
$begingroup$
Yes the condition you gave is stronger. You can easily derive the second by differentiating both sides of your equality in $s$and making $s=t$ afterwards.
$endgroup$
– mathcounterexamples.net
Feb 9 '18 at 15:01
add a comment |
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