Mixing
Find a solution curve that connects a saddle point.
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
The entire problem is as follows:
Consider $dotx = y+x^2-y^2$ and $doty = -x-2xy$. (a) Find all equilibria for this system and determine their stability. (b) Find the Hamiltonian of the system. (c) Find a solution curve that connects a saddle point. (d) Does the curve obtained in (c) connect any other saddle points?
Below is my answer to parts $(a)$ and $(b)$:
a)
There are four equilibrium points here, all of which can be found using relatively simple algebra. They are $E_1 = (0,0), E_2 = (0,1), E_3,4 = left( pm fracsqrt32, - frac12 right)$. The stability of these is found below:
$$Dfleft( x, y right) = left( beginmatrix 2x & 1-2y \ -1-2y & -2xendmatrixright)$$
$$Dfleft( E_1 right) = left( beginmatrix 0 & 1 \ -1 & 0endmatrixright) implies lambda_1,2 = pm i$$
Thus $E_1$ is a center.
$$Dfleft( E_2 right) = left( beginmatrix 0 & -1 \ -3 & 0endmatrixright) implies lambda_1,2 = pm sqrt 3$$
Thus $E_2$ is a saddle.
$$Dfleft( E_3 right) = left( beginmatrix sqrt 3 & 2 \ 0 & - sqrt 3endmatrixright) implies lambda_1,2 = pm sqrt 3$$
Thus $E_3$ is a saddle.
$$Dfleft( E_4 right) = left( beginmatrix -sqrt3 & 2 \ 0 & sqrt 3endmatrixright) implies lambda_1,2 = pm sqrt 3$$
Thus $E_4$ is a saddle.
b)
$$H(x,y) = int (y + x^2 - y^2)dy = frac12y^2 + x^2 y - frac13y^3 + f(x)$$
Thus
$$-fracddx H(x,y) = -fracddx [x^2 y + f(x)] = dot y = -x-2xy$$
which implies
$$-2xy -f'(x) = -x-2xy implies f'(x) = x$$
thus $f(x) = frac12x^2$. So the Hamiltonian is
$$H(x,y) = frac12y^2 + x^2 y - frac13y^3 + frac12x^2$$.
Now that I have all of this information however, (c) seems like a strangely vague question. Is all I have to do here plug in a saddle point and see what solution it gives me? A walkthrough of this would be helpful, and then would in turn help me with part (d). Thanks in advance for any and all advice!
dynamical-systems mathematical-modeling nonlinear-system
edited Jul 19 at 0:51
Key Flex
4,346425
asked Jul 19 at 0:10
obewanjacobi
312110
add a comment |Â
up vote
1
down vote
favorite
The entire problem is as follows:
Consider $dotx = y+x^2-y^2$ and $doty = -x-2xy$. (a) Find all equilibria for this system and determine their stability. (b) Find the Hamiltonian of the system. (c) Find a solution curve that connects a saddle point. (d) Does the curve obtained in (c) connect any other saddle points?
Below is my answer to parts $(a)$ and $(b)$:
a)
There are four equilibrium points here, all of which can be found using relatively simple algebra. They are $E_1 = (0,0), E_2 = (0,1), E_3,4 = left( pm fracsqrt32, - frac12 right)$. The stability of these is found below:
$$Dfleft( x, y right) = left( beginmatrix 2x & 1-2y \ -1-2y & -2xendmatrixright)$$
$$Dfleft( E_1 right) = left( beginmatrix 0 & 1 \ -1 & 0endmatrixright) implies lambda_1,2 = pm i$$
Thus $E_1$ is a center.
$$Dfleft( E_2 right) = left( beginmatrix 0 & -1 \ -3 & 0endmatrixright) implies lambda_1,2 = pm sqrt 3$$
Thus $E_2$ is a saddle.
$$Dfleft( E_3 right) = left( beginmatrix sqrt 3 & 2 \ 0 & - sqrt 3endmatrixright) implies lambda_1,2 = pm sqrt 3$$
Thus $E_3$ is a saddle.
$$Dfleft( E_4 right) = left( beginmatrix -sqrt3 & 2 \ 0 & sqrt 3endmatrixright) implies lambda_1,2 = pm sqrt 3$$
Thus $E_4$ is a saddle.
b)
$$H(x,y) = int (y + x^2 - y^2)dy = frac12y^2 + x^2 y - frac13y^3 + f(x)$$
Thus
$$-fracddx H(x,y) = -fracddx [x^2 y + f(x)] = dot y = -x-2xy$$
which implies
$$-2xy -f'(x) = -x-2xy implies f'(x) = x$$
thus $f(x) = frac12x^2$. So the Hamiltonian is
$$H(x,y) = frac12y^2 + x^2 y - frac13y^3 + frac12x^2$$.
Now that I have all of this information however, (c) seems like a strangely vague question. Is all I have to do here plug in a saddle point and see what solution it gives me? A walkthrough of this would be helpful, and then would in turn help me with part (d). Thanks in advance for any and all advice!
dynamical-systems mathematical-modeling nonlinear-system
edited Jul 19 at 0:51
Key Flex
4,346425
asked Jul 19 at 0:10
obewanjacobi
312110
For a Hamiltonian system, solutions are level curves of the Hamiltonian (sort of like "conservation of energy"). Note that $H$ gives the same value at $E_2$, $E_3$, and $E_4$; hence there exist solution curves between each pair. The wording on part (c) is rather strange to...."connects a saddle point" to what?
– erfink
Jul 19 at 2:16
There lies part of my problem. I'm not quite sure how to handle it, and this was all I was given.
– obewanjacobi
Jul 19 at 4:01
It should help to notice that $y+frac12$ is a factor of $H(x,y)-frac16$: $$H(x,y)-frac16=-frac13left(y+frac12right) (1 - 3 x^2 - 2 y + y^2).$$ From there it is straightforward to draw the phase portrait and so to interpret what c) could be: seems likely that it means "Find a solution curve that is asymptotic to a saddle point".
– John B
Jul 19 at 8:54
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The entire problem is as follows:
Consider $dotx = y+x^2-y^2$ and $doty = -x-2xy$. (a) Find all equilibria for this system and determine their stability. (b) Find the Hamiltonian of the system. (c) Find a solution curve that connects a saddle point. (d) Does the curve obtained in (c) connect any other saddle points?
Below is my answer to parts $(a)$ and $(b)$:
a)
There are four equilibrium points here, all of which can be found using relatively simple algebra. They are $E_1 = (0,0), E_2 = (0,1), E_3,4 = left( pm fracsqrt32, - frac12 right)$. The stability of these is found below:
$$Dfleft( x, y right) = left( beginmatrix 2x & 1-2y \ -1-2y & -2xendmatrixright)$$
$$Dfleft( E_1 right) = left( beginmatrix 0 & 1 \ -1 & 0endmatrixright) implies lambda_1,2 = pm i$$
Thus $E_1$ is a center.
$$Dfleft( E_2 right) = left( beginmatrix 0 & -1 \ -3 & 0endmatrixright) implies lambda_1,2 = pm sqrt 3$$
Thus $E_2$ is a saddle.
$$Dfleft( E_3 right) = left( beginmatrix sqrt 3 & 2 \ 0 & - sqrt 3endmatrixright) implies lambda_1,2 = pm sqrt 3$$
Thus $E_3$ is a saddle.
$$Dfleft( E_4 right) = left( beginmatrix -sqrt3 & 2 \ 0 & sqrt 3endmatrixright) implies lambda_1,2 = pm sqrt 3$$
Thus $E_4$ is a saddle.
b)
$$H(x,y) = int (y + x^2 - y^2)dy = frac12y^2 + x^2 y - frac13y^3 + f(x)$$
Thus
$$-fracddx H(x,y) = -fracddx [x^2 y + f(x)] = dot y = -x-2xy$$
which implies
$$-2xy -f'(x) = -x-2xy implies f'(x) = x$$
thus $f(x) = frac12x^2$. So the Hamiltonian is
$$H(x,y) = frac12y^2 + x^2 y - frac13y^3 + frac12x^2$$.
Now that I have all of this information however, (c) seems like a strangely vague question. Is all I have to do here plug in a saddle point and see what solution it gives me? A walkthrough of this would be helpful, and then would in turn help me with part (d). Thanks in advance for any and all advice!
dynamical-systems mathematical-modeling nonlinear-system
edited Jul 19 at 0:51
Key Flex
4,346425
asked Jul 19 at 0:10
obewanjacobi
312110
The entire problem is as follows:
Consider $dotx = y+x^2-y^2$ and $doty = -x-2xy$. (a) Find all equilibria for this system and determine their stability. (b) Find the Hamiltonian of the system. (c) Find a solution curve that connects a saddle point. (d) Does the curve obtained in (c) connect any other saddle points?
Below is my answer to parts $(a)$ and $(b)$:
a)
There are four equilibrium points here, all of which can be found using relatively simple algebra. They are $E_1 = (0,0), E_2 = (0,1), E_3,4 = left( pm fracsqrt32, - frac12 right)$. The stability of these is found below:
$$Dfleft( x, y right) = left( beginmatrix 2x & 1-2y \ -1-2y & -2xendmatrixright)$$
$$Dfleft( E_1 right) = left( beginmatrix 0 & 1 \ -1 & 0endmatrixright) implies lambda_1,2 = pm i$$
Thus $E_1$ is a center.
$$Dfleft( E_2 right) = left( beginmatrix 0 & -1 \ -3 & 0endmatrixright) implies lambda_1,2 = pm sqrt 3$$
Thus $E_2$ is a saddle.
$$Dfleft( E_3 right) = left( beginmatrix sqrt 3 & 2 \ 0 & - sqrt 3endmatrixright) implies lambda_1,2 = pm sqrt 3$$
Thus $E_3$ is a saddle.
$$Dfleft( E_4 right) = left( beginmatrix -sqrt3 & 2 \ 0 & sqrt 3endmatrixright) implies lambda_1,2 = pm sqrt 3$$
Thus $E_4$ is a saddle.
b)
$$H(x,y) = int (y + x^2 - y^2)dy = frac12y^2 + x^2 y - frac13y^3 + f(x)$$
Thus
$$-fracddx H(x,y) = -fracddx [x^2 y + f(x)] = dot y = -x-2xy$$
which implies
$$-2xy -f'(x) = -x-2xy implies f'(x) = x$$
thus $f(x) = frac12x^2$. So the Hamiltonian is
$$H(x,y) = frac12y^2 + x^2 y - frac13y^3 + frac12x^2$$.
Now that I have all of this information however, (c) seems like a strangely vague question. Is all I have to do here plug in a saddle point and see what solution it gives me? A walkthrough of this would be helpful, and then would in turn help me with part (d). Thanks in advance for any and all advice!
dynamical-systems mathematical-modeling nonlinear-system
edited Jul 19 at 0:51
Key Flex
4,346425
asked Jul 19 at 0:10
obewanjacobi
312110
edited Jul 19 at 0:51
Key Flex
4,346425
edited Jul 19 at 0:51
Key Flex
4,346425
edited Jul 19 at 0:51
Key Flex
4,346425
4,346425
asked Jul 19 at 0:10
obewanjacobi
312110
asked Jul 19 at 0:10
obewanjacobi
312110
asked Jul 19 at 0:10
obewanjacobi
312110
312110
For a Hamiltonian system, solutions are level curves of the Hamiltonian (sort of like "conservation of energy"). Note that $H$ gives the same value at $E_2$, $E_3$, and $E_4$; hence there exist solution curves between each pair. The wording on part (c) is rather strange to...."connects a saddle point" to what?
– erfink
Jul 19 at 2:16
There lies part of my problem. I'm not quite sure how to handle it, and this was all I was given.
– obewanjacobi
Jul 19 at 4:01
It should help to notice that $y+frac12$ is a factor of $H(x,y)-frac16$: $$H(x,y)-frac16=-frac13left(y+frac12right) (1 - 3 x^2 - 2 y + y^2).$$ From there it is straightforward to draw the phase portrait and so to interpret what c) could be: seems likely that it means "Find a solution curve that is asymptotic to a saddle point".
– John B
Jul 19 at 8:54
add a comment |Â
For a Hamiltonian system, solutions are level curves of the Hamiltonian (sort of like "conservation of energy"). Note that $H$ gives the same value at $E_2$, $E_3$, and $E_4$; hence there exist solution curves between each pair. The wording on part (c) is rather strange to...."connects a saddle point" to what?
– erfink
Jul 19 at 2:16
There lies part of my problem. I'm not quite sure how to handle it, and this was all I was given.
– obewanjacobi
Jul 19 at 4:01
It should help to notice that $y+frac12$ is a factor of $H(x,y)-frac16$: $$H(x,y)-frac16=-frac13left(y+frac12right) (1 - 3 x^2 - 2 y + y^2).$$ From there it is straightforward to draw the phase portrait and so to interpret what c) could be: seems likely that it means "Find a solution curve that is asymptotic to a saddle point".
– John B
Jul 19 at 8:54
For a Hamiltonian system, solutions are level curves of the Hamiltonian (sort of like "conservation of energy"). Note that $H$ gives the same value at $E_2$, $E_3$, and $E_4$; hence there exist solution curves between each pair. The wording on part (c) is rather strange to...."connects a saddle point" to what?
– erfink
Jul 19 at 2:16
For a Hamiltonian system, solutions are level curves of the Hamiltonian (sort of like "conservation of energy"). Note that $H$ gives the same value at $E_2$, $E_3$, and $E_4$; hence there exist solution curves between each pair. The wording on part (c) is rather strange to...."connects a saddle point" to what?
– erfink
Jul 19 at 2:16
There lies part of my problem. I'm not quite sure how to handle it, and this was all I was given.
– obewanjacobi
Jul 19 at 4:01
There lies part of my problem. I'm not quite sure how to handle it, and this was all I was given.
– obewanjacobi
Jul 19 at 4:01
It should help to notice that $y+frac12$ is a factor of $H(x,y)-frac16$: $$H(x,y)-frac16=-frac13left(y+frac12right) (1 - 3 x^2 - 2 y + y^2).$$ From there it is straightforward to draw the phase portrait and so to interpret what c) could be: seems likely that it means "Find a solution curve that is asymptotic to a saddle point".
– John B
Jul 19 at 8:54
It should help to notice that $y+frac12$ is a factor of $H(x,y)-frac16$: $$H(x,y)-frac16=-frac13left(y+frac12right) (1 - 3 x^2 - 2 y + y^2).$$ From there it is straightforward to draw the phase portrait and so to interpret what c) could be: seems likely that it means "Find a solution curve that is asymptotic to a saddle point".
– John B
Jul 19 at 8:54
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2856135%2ffind-a-solution-curve-that-connects-a-saddle-point%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
For a Hamiltonian system, solutions are level curves of the Hamiltonian (sort of like "conservation of energy"). Note that $H$ gives the same value at $E_2$, $E_3$, and $E_4$; hence there exist solution curves between each pair. The wording on part (c) is rather strange to...."connects a saddle point" to what?
– erfink
Jul 19 at 2:16
There lies part of my problem. I'm not quite sure how to handle it, and this was all I was given.
– obewanjacobi
Jul 19 at 4:01
It should help to notice that $y+frac12$ is a factor of $H(x,y)-frac16$: $$H(x,y)-frac16=-frac13left(y+frac12right) (1 - 3 x^2 - 2 y + y^2).$$ From there it is straightforward to draw the phase portrait and so to interpret what c) could be: seems likely that it means "Find a solution curve that is asymptotic to a saddle point".
– John B
Jul 19 at 8:54