Phase portrait of a gradient system
$begingroup$
Let $T$ be the torus defined as the square $0leq theta_1, theta_2leq 2pi$ with opposite sides identified. Let $F(theta_1,theta_2)=costheta_1+costheta_2$. Sketch the phase portrait for the system $-text{grad}, F$ in $T$. Sketch a three-dimensional representation of this phase portrait with $T$ represented as the surface of a doughnut.
I understand that the $-text{grad}(F) = (sin(theta_1), sin(theta_2))$, but I'm very confused on how the torus is defined here. Why is a doughnut defined as a square?
I'm guessing in essence that we can find $theta'_1$ and $theta'_2$ and apply the domain to it, but the confusion on $T$ leaves me stuck?
ordinary-differential-equations
edited Dec 5 '18 at 12:52
Harry49
6,13931132
asked Dec 4 '18 at 19:14
MathGuyForLifeMathGuyForLife
1007
$endgroup$
add a comment |
$begingroup$
Let $T$ be the torus defined as the square $0leq theta_1, theta_2leq 2pi$ with opposite sides identified. Let $F(theta_1,theta_2)=costheta_1+costheta_2$. Sketch the phase portrait for the system $-text{grad}, F$ in $T$. Sketch a three-dimensional representation of this phase portrait with $T$ represented as the surface of a doughnut.
I understand that the $-text{grad}(F) = (sin(theta_1), sin(theta_2))$, but I'm very confused on how the torus is defined here. Why is a doughnut defined as a square?
I'm guessing in essence that we can find $theta'_1$ and $theta'_2$ and apply the domain to it, but the confusion on $T$ leaves me stuck?
ordinary-differential-equations
edited Dec 5 '18 at 12:52
Harry49
6,13931132
asked Dec 4 '18 at 19:14
MathGuyForLifeMathGuyForLife
1007
$endgroup$
add a comment |
$begingroup$
Let $T$ be the torus defined as the square $0leq theta_1, theta_2leq 2pi$ with opposite sides identified. Let $F(theta_1,theta_2)=costheta_1+costheta_2$. Sketch the phase portrait for the system $-text{grad}, F$ in $T$. Sketch a three-dimensional representation of this phase portrait with $T$ represented as the surface of a doughnut.
I understand that the $-text{grad}(F) = (sin(theta_1), sin(theta_2))$, but I'm very confused on how the torus is defined here. Why is a doughnut defined as a square?
I'm guessing in essence that we can find $theta'_1$ and $theta'_2$ and apply the domain to it, but the confusion on $T$ leaves me stuck?
ordinary-differential-equations
edited Dec 5 '18 at 12:52
Harry49
6,13931132
asked Dec 4 '18 at 19:14
MathGuyForLifeMathGuyForLife
1007
$endgroup$
Let $T$ be the torus defined as the square $0leq theta_1, theta_2leq 2pi$ with opposite sides identified. Let $F(theta_1,theta_2)=costheta_1+costheta_2$. Sketch the phase portrait for the system $-text{grad}, F$ in $T$. Sketch a three-dimensional representation of this phase portrait with $T$ represented as the surface of a doughnut.
I understand that the $-text{grad}(F) = (sin(theta_1), sin(theta_2))$, but I'm very confused on how the torus is defined here. Why is a doughnut defined as a square?
I'm guessing in essence that we can find $theta'_1$ and $theta'_2$ and apply the domain to it, but the confusion on $T$ leaves me stuck?
ordinary-differential-equations
ordinary-differential-equations
edited Dec 5 '18 at 12:52
Harry49
6,13931132
asked Dec 4 '18 at 19:14
MathGuyForLifeMathGuyForLife
1007
edited Dec 5 '18 at 12:52
Harry49
6,13931132
asked Dec 4 '18 at 19:14
MathGuyForLifeMathGuyForLife
1007
edited Dec 5 '18 at 12:52
Harry49
6,13931132
edited Dec 5 '18 at 12:52
Harry49
6,13931132
edited Dec 5 '18 at 12:52
Harry49
6,13931132
6,13931132
asked Dec 4 '18 at 19:14
MathGuyForLifeMathGuyForLife
1007
asked Dec 4 '18 at 19:14
MathGuyForLifeMathGuyForLife
1007
asked Dec 4 '18 at 19:14
MathGuyForLifeMathGuyForLife
1007
1007
add a comment |
add a comment |
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$begingroup$
It's called flat torus. Here is the animation from Wikipedia page about torus, that shows how you can map points of a square with identical sides to a torus surface in 3D.
answered Dec 4 '18 at 19:25
Vasily MitchVasily Mitch
1,69638
$endgroup$
add a comment |
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$begingroup$
It's called flat torus. Here is the animation from Wikipedia page about torus, that shows how you can map points of a square with identical sides to a torus surface in 3D.
answered Dec 4 '18 at 19:25
Vasily MitchVasily Mitch
1,69638
$endgroup$
add a comment |
$begingroup$
It's called flat torus. Here is the animation from Wikipedia page about torus, that shows how you can map points of a square with identical sides to a torus surface in 3D.
answered Dec 4 '18 at 19:25
Vasily MitchVasily Mitch
1,69638
$endgroup$
add a comment |
$begingroup$
It's called flat torus. Here is the animation from Wikipedia page about torus, that shows how you can map points of a square with identical sides to a torus surface in 3D.
answered Dec 4 '18 at 19:25
Vasily MitchVasily Mitch
1,69638
$endgroup$
It's called flat torus. Here is the animation from Wikipedia page about torus, that shows how you can map points of a square with identical sides to a torus surface in 3D.
answered Dec 4 '18 at 19:25
Vasily MitchVasily Mitch
1,69638
answered Dec 4 '18 at 19:25
Vasily MitchVasily Mitch
1,69638
answered Dec 4 '18 at 19:25
Vasily MitchVasily Mitch
1,69638
answered Dec 4 '18 at 19:25
Vasily MitchVasily Mitch
1,69638
1,69638
add a comment |
add a comment |
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