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How would one analyse the solutions to this non linear ODE system?
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I've come upon the following system of ODEs, where $f_1, f_2, f_3 : I subset mathbbRto mathbbR$:
$$f_1^2 + f_2^2 = 1$$
$$(f_1')^2 + (f_2')^2 + (f_3')^2 = 1$$
$$-f_2f_1'' + f_1f_2'' = (f_1'')^2 + (f_2'')^2 + (f_3'')^2 $$
I'm trying to find solutions (even trivial ones, really) but it hasn't been easy. What would be some useful restrictions on $f_1, f_2$ and $f_3$? If it's too much of a task to find particular solutions, how can one best understand this system?
Oh, and if anyone knows whether there's a way to tackle this using numerical methods I'd appreciate it too.
calculus real-analysis differential-equations
asked Aug 2 at 21:24
Matheus Andrade
587214
add a comment |Â
up vote
1
down vote
favorite
I've come upon the following system of ODEs, where $f_1, f_2, f_3 : I subset mathbbRto mathbbR$:
$$f_1^2 + f_2^2 = 1$$
$$(f_1')^2 + (f_2')^2 + (f_3')^2 = 1$$
$$-f_2f_1'' + f_1f_2'' = (f_1'')^2 + (f_2'')^2 + (f_3'')^2 $$
I'm trying to find solutions (even trivial ones, really) but it hasn't been easy. What would be some useful restrictions on $f_1, f_2$ and $f_3$? If it's too much of a task to find particular solutions, how can one best understand this system?
Oh, and if anyone knows whether there's a way to tackle this using numerical methods I'd appreciate it too.
calculus real-analysis differential-equations
asked Aug 2 at 21:24
Matheus Andrade
587214
Have you tried for example polar-coordinates (on the first two coordinates)? I do not have a solution yet, but it seems promising.
– WalterJ
Aug 2 at 21:53
@WalterJ I'm not sure I understand your suggestion. How can polar coordinates help me here?
– Matheus Andrade
Aug 2 at 23:18
I meant using the problem structure directly, i.e. $(f_1,f_2)$ living on the circle and $(dotf_1,dotf_2,dotf_3)$ living on the sphere. This might give you a lot more intuition, but it requires some work. For example the first equation can become $dotr(t)=0$, $r(t_0)=1$, for $r$ some radius.
– WalterJ
Aug 2 at 23:34
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I've come upon the following system of ODEs, where $f_1, f_2, f_3 : I subset mathbbRto mathbbR$:
$$f_1^2 + f_2^2 = 1$$
$$(f_1')^2 + (f_2')^2 + (f_3')^2 = 1$$
$$-f_2f_1'' + f_1f_2'' = (f_1'')^2 + (f_2'')^2 + (f_3'')^2 $$
I'm trying to find solutions (even trivial ones, really) but it hasn't been easy. What would be some useful restrictions on $f_1, f_2$ and $f_3$? If it's too much of a task to find particular solutions, how can one best understand this system?
Oh, and if anyone knows whether there's a way to tackle this using numerical methods I'd appreciate it too.
calculus real-analysis differential-equations
asked Aug 2 at 21:24
Matheus Andrade
587214
I've come upon the following system of ODEs, where $f_1, f_2, f_3 : I subset mathbbRto mathbbR$:
$$f_1^2 + f_2^2 = 1$$
$$(f_1')^2 + (f_2')^2 + (f_3')^2 = 1$$
$$-f_2f_1'' + f_1f_2'' = (f_1'')^2 + (f_2'')^2 + (f_3'')^2 $$
I'm trying to find solutions (even trivial ones, really) but it hasn't been easy. What would be some useful restrictions on $f_1, f_2$ and $f_3$? If it's too much of a task to find particular solutions, how can one best understand this system?
Oh, and if anyone knows whether there's a way to tackle this using numerical methods I'd appreciate it too.
calculus real-analysis differential-equations
asked Aug 2 at 21:24
Matheus Andrade
587214
edited Aug 2 at 21:30
asked Aug 2 at 21:24
Matheus Andrade
587214
asked Aug 2 at 21:24
Matheus Andrade
587214
asked Aug 2 at 21:24
Matheus Andrade
587214
587214
Have you tried for example polar-coordinates (on the first two coordinates)? I do not have a solution yet, but it seems promising.
– WalterJ
Aug 2 at 21:53
@WalterJ I'm not sure I understand your suggestion. How can polar coordinates help me here?
– Matheus Andrade
Aug 2 at 23:18
I meant using the problem structure directly, i.e. $(f_1,f_2)$ living on the circle and $(dotf_1,dotf_2,dotf_3)$ living on the sphere. This might give you a lot more intuition, but it requires some work. For example the first equation can become $dotr(t)=0$, $r(t_0)=1$, for $r$ some radius.
– WalterJ
Aug 2 at 23:34
add a comment |Â
Have you tried for example polar-coordinates (on the first two coordinates)? I do not have a solution yet, but it seems promising.
– WalterJ
Aug 2 at 21:53
@WalterJ I'm not sure I understand your suggestion. How can polar coordinates help me here?
– Matheus Andrade
Aug 2 at 23:18
I meant using the problem structure directly, i.e. $(f_1,f_2)$ living on the circle and $(dotf_1,dotf_2,dotf_3)$ living on the sphere. This might give you a lot more intuition, but it requires some work. For example the first equation can become $dotr(t)=0$, $r(t_0)=1$, for $r$ some radius.
– WalterJ
Aug 2 at 23:34
Have you tried for example polar-coordinates (on the first two coordinates)? I do not have a solution yet, but it seems promising.
– WalterJ
Aug 2 at 21:53
Have you tried for example polar-coordinates (on the first two coordinates)? I do not have a solution yet, but it seems promising.
– WalterJ
Aug 2 at 21:53
@WalterJ I'm not sure I understand your suggestion. How can polar coordinates help me here?
– Matheus Andrade
Aug 2 at 23:18
@WalterJ I'm not sure I understand your suggestion. How can polar coordinates help me here?
– Matheus Andrade
Aug 2 at 23:18
I meant using the problem structure directly, i.e. $(f_1,f_2)$ living on the circle and $(dotf_1,dotf_2,dotf_3)$ living on the sphere. This might give you a lot more intuition, but it requires some work. For example the first equation can become $dotr(t)=0$, $r(t_0)=1$, for $r$ some radius.
– WalterJ
Aug 2 at 23:34
I meant using the problem structure directly, i.e. $(f_1,f_2)$ living on the circle and $(dotf_1,dotf_2,dotf_3)$ living on the sphere. This might give you a lot more intuition, but it requires some work. For example the first equation can become $dotr(t)=0$, $r(t_0)=1$, for $r$ some radius.
– WalterJ
Aug 2 at 23:34
add a comment |Â
1 Answer
1
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oldest
votes
up vote
4
down vote
accepted
The first two equations say this is a motion on the cylinder with speed $1$.
One family of trivial solutions is where the motion is in the $f_3$ direction only:
$$ f_1(t) = cos(alpha), f_2(t) = sin(alpha), f_3(t) = t + c$$
EDIT: Following WalterJ's suggestion, if $f_1(t) = cos(g(t))$ and $f2(t)=sin(g(t))$, the first equation is satisfied automatically and the last two become
$$ eqaligndotg^2 + dotf_3^2 &= 1 cr
ddotg = dotg^4 + ddotg^2 + ddotf_3^2cr$$
We can then eliminate $f_3$ from this: note that
$$ dotf_3 ddotf_3 = - dotg ddotg $$
so $$ddotf_3^2 = dfracdotg^2 ddotg^21-dotg^2 $$
and thus we get one equation for $g$:
$$ ddotg = dotg^4 + ddotg^2 + dfracdotg^2 ddotg^21-dotg^2 $$
Note this is a separable (but rather nasty) first-order equation in $dotg$.
answered Aug 2 at 23:07
Robert Israel
303k22200440
Do you mean that the equation is separable if we solve for $ddotg$?
– Evgeny
Aug 3 at 0:14
I just arrived at something almost equivalent to your edit, I'm glad I saw it! Thanks for the help
– Matheus Andrade
Aug 3 at 1:44
This is also what I got, but I desperately tried to simplify the last expression
– WalterJ
Aug 3 at 11:02
@Evgeny I mean that if you write $dotg = v$, the equation $dotv = v^4 + dotv^2 + dfracv^2 dotv^21-v^2$ is separable. You have to first solve a quadratic to get it in the form $dotv = f(v)$.
– Robert Israel
Aug 3 at 19:17
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The first two equations say this is a motion on the cylinder with speed $1$.
One family of trivial solutions is where the motion is in the $f_3$ direction only:
$$ f_1(t) = cos(alpha), f_2(t) = sin(alpha), f_3(t) = t + c$$
EDIT: Following WalterJ's suggestion, if $f_1(t) = cos(g(t))$ and $f2(t)=sin(g(t))$, the first equation is satisfied automatically and the last two become
$$ eqaligndotg^2 + dotf_3^2 &= 1 cr
ddotg = dotg^4 + ddotg^2 + ddotf_3^2cr$$
We can then eliminate $f_3$ from this: note that
$$ dotf_3 ddotf_3 = - dotg ddotg $$
so $$ddotf_3^2 = dfracdotg^2 ddotg^21-dotg^2 $$
and thus we get one equation for $g$:
$$ ddotg = dotg^4 + ddotg^2 + dfracdotg^2 ddotg^21-dotg^2 $$
Note this is a separable (but rather nasty) first-order equation in $dotg$.
answered Aug 2 at 23:07
Robert Israel
303k22200440
Do you mean that the equation is separable if we solve for $ddotg$?
– Evgeny
Aug 3 at 0:14
I just arrived at something almost equivalent to your edit, I'm glad I saw it! Thanks for the help
– Matheus Andrade
Aug 3 at 1:44
This is also what I got, but I desperately tried to simplify the last expression
– WalterJ
Aug 3 at 11:02
@Evgeny I mean that if you write $dotg = v$, the equation $dotv = v^4 + dotv^2 + dfracv^2 dotv^21-v^2$ is separable. You have to first solve a quadratic to get it in the form $dotv = f(v)$.
– Robert Israel
Aug 3 at 19:17
add a comment |Â
up vote
4
down vote
accepted
The first two equations say this is a motion on the cylinder with speed $1$.
One family of trivial solutions is where the motion is in the $f_3$ direction only:
$$ f_1(t) = cos(alpha), f_2(t) = sin(alpha), f_3(t) = t + c$$
EDIT: Following WalterJ's suggestion, if $f_1(t) = cos(g(t))$ and $f2(t)=sin(g(t))$, the first equation is satisfied automatically and the last two become
$$ eqaligndotg^2 + dotf_3^2 &= 1 cr
ddotg = dotg^4 + ddotg^2 + ddotf_3^2cr$$
We can then eliminate $f_3$ from this: note that
$$ dotf_3 ddotf_3 = - dotg ddotg $$
so $$ddotf_3^2 = dfracdotg^2 ddotg^21-dotg^2 $$
and thus we get one equation for $g$:
$$ ddotg = dotg^4 + ddotg^2 + dfracdotg^2 ddotg^21-dotg^2 $$
Note this is a separable (but rather nasty) first-order equation in $dotg$.
answered Aug 2 at 23:07
Robert Israel
303k22200440
Do you mean that the equation is separable if we solve for $ddotg$?
– Evgeny
Aug 3 at 0:14
I just arrived at something almost equivalent to your edit, I'm glad I saw it! Thanks for the help
– Matheus Andrade
Aug 3 at 1:44
This is also what I got, but I desperately tried to simplify the last expression
– WalterJ
Aug 3 at 11:02
@Evgeny I mean that if you write $dotg = v$, the equation $dotv = v^4 + dotv^2 + dfracv^2 dotv^21-v^2$ is separable. You have to first solve a quadratic to get it in the form $dotv = f(v)$.
– Robert Israel
Aug 3 at 19:17
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The first two equations say this is a motion on the cylinder with speed $1$.
One family of trivial solutions is where the motion is in the $f_3$ direction only:
$$ f_1(t) = cos(alpha), f_2(t) = sin(alpha), f_3(t) = t + c$$
EDIT: Following WalterJ's suggestion, if $f_1(t) = cos(g(t))$ and $f2(t)=sin(g(t))$, the first equation is satisfied automatically and the last two become
$$ eqaligndotg^2 + dotf_3^2 &= 1 cr
ddotg = dotg^4 + ddotg^2 + ddotf_3^2cr$$
We can then eliminate $f_3$ from this: note that
$$ dotf_3 ddotf_3 = - dotg ddotg $$
so $$ddotf_3^2 = dfracdotg^2 ddotg^21-dotg^2 $$
and thus we get one equation for $g$:
$$ ddotg = dotg^4 + ddotg^2 + dfracdotg^2 ddotg^21-dotg^2 $$
Note this is a separable (but rather nasty) first-order equation in $dotg$.
answered Aug 2 at 23:07
Robert Israel
303k22200440
The first two equations say this is a motion on the cylinder with speed $1$.
One family of trivial solutions is where the motion is in the $f_3$ direction only:
$$ f_1(t) = cos(alpha), f_2(t) = sin(alpha), f_3(t) = t + c$$
EDIT: Following WalterJ's suggestion, if $f_1(t) = cos(g(t))$ and $f2(t)=sin(g(t))$, the first equation is satisfied automatically and the last two become
$$ eqaligndotg^2 + dotf_3^2 &= 1 cr
ddotg = dotg^4 + ddotg^2 + ddotf_3^2cr$$
We can then eliminate $f_3$ from this: note that
$$ dotf_3 ddotf_3 = - dotg ddotg $$
so $$ddotf_3^2 = dfracdotg^2 ddotg^21-dotg^2 $$
and thus we get one equation for $g$:
$$ ddotg = dotg^4 + ddotg^2 + dfracdotg^2 ddotg^21-dotg^2 $$
Note this is a separable (but rather nasty) first-order equation in $dotg$.
answered Aug 2 at 23:07
Robert Israel
303k22200440
edited Aug 2 at 23:50
answered Aug 2 at 23:07
Robert Israel
303k22200440
answered Aug 2 at 23:07
Robert Israel
303k22200440
answered Aug 2 at 23:07
Robert Israel
303k22200440
303k22200440
Do you mean that the equation is separable if we solve for $ddotg$?
– Evgeny
Aug 3 at 0:14
I just arrived at something almost equivalent to your edit, I'm glad I saw it! Thanks for the help
– Matheus Andrade
Aug 3 at 1:44
This is also what I got, but I desperately tried to simplify the last expression
– WalterJ
Aug 3 at 11:02
@Evgeny I mean that if you write $dotg = v$, the equation $dotv = v^4 + dotv^2 + dfracv^2 dotv^21-v^2$ is separable. You have to first solve a quadratic to get it in the form $dotv = f(v)$.
– Robert Israel
Aug 3 at 19:17
add a comment |Â
Do you mean that the equation is separable if we solve for $ddotg$?
– Evgeny
Aug 3 at 0:14
I just arrived at something almost equivalent to your edit, I'm glad I saw it! Thanks for the help
– Matheus Andrade
Aug 3 at 1:44
This is also what I got, but I desperately tried to simplify the last expression
– WalterJ
Aug 3 at 11:02
@Evgeny I mean that if you write $dotg = v$, the equation $dotv = v^4 + dotv^2 + dfracv^2 dotv^21-v^2$ is separable. You have to first solve a quadratic to get it in the form $dotv = f(v)$.
– Robert Israel
Aug 3 at 19:17
Do you mean that the equation is separable if we solve for $ddotg$?
– Evgeny
Aug 3 at 0:14
Do you mean that the equation is separable if we solve for $ddotg$?
– Evgeny
Aug 3 at 0:14
I just arrived at something almost equivalent to your edit, I'm glad I saw it! Thanks for the help
– Matheus Andrade
Aug 3 at 1:44
I just arrived at something almost equivalent to your edit, I'm glad I saw it! Thanks for the help
– Matheus Andrade
Aug 3 at 1:44
This is also what I got, but I desperately tried to simplify the last expression
– WalterJ
Aug 3 at 11:02
This is also what I got, but I desperately tried to simplify the last expression
– WalterJ
Aug 3 at 11:02
@Evgeny I mean that if you write $dotg = v$, the equation $dotv = v^4 + dotv^2 + dfracv^2 dotv^21-v^2$ is separable. You have to first solve a quadratic to get it in the form $dotv = f(v)$.
– Robert Israel
Aug 3 at 19:17
@Evgeny I mean that if you write $dotg = v$, the equation $dotv = v^4 + dotv^2 + dfracv^2 dotv^21-v^2$ is separable. You have to first solve a quadratic to get it in the form $dotv = f(v)$.
– Robert Israel
Aug 3 at 19:17
add a comment |Â
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Have you tried for example polar-coordinates (on the first two coordinates)? I do not have a solution yet, but it seems promising.
– WalterJ
Aug 2 at 21:53
@WalterJ I'm not sure I understand your suggestion. How can polar coordinates help me here?
– Matheus Andrade
Aug 2 at 23:18
I meant using the problem structure directly, i.e. $(f_1,f_2)$ living on the circle and $(dotf_1,dotf_2,dotf_3)$ living on the sphere. This might give you a lot more intuition, but it requires some work. For example the first equation can become $dotr(t)=0$, $r(t_0)=1$, for $r$ some radius.
– WalterJ
Aug 2 at 23:34