Bifurcation in 1D with two parameters
up vote
0
down vote
favorite
I'm trying to do bifurcation analysis of the system
$x'=-x((x-2)^{2} - mu^{2} -9)((x+2)^2 - mu^2 -9)+ epsilon$
When $epsilon = 0$ this is easy because we have some nice circles. But I'm stuck as to what to do when $epsilon > 0 $. I've been advised not to find analytic expressions for equilibria. I'm not looking for an explicit solution spelled out for me, just a shove in the right direction to be able to get there myself. How can one determine location/type of bifurcation for a system like this without known equilibria? Can I use my $epsilon = 0$ diagram and infer something from it for $epsilon >0$? What are my options here, methodology-wise?
differential-equations dynamical-systems bifurcation
asked Nov 17 at 16:23
Jo Fisher
566
add a comment |
up vote
0
down vote
favorite
I'm trying to do bifurcation analysis of the system
$x'=-x((x-2)^{2} - mu^{2} -9)((x+2)^2 - mu^2 -9)+ epsilon$
When $epsilon = 0$ this is easy because we have some nice circles. But I'm stuck as to what to do when $epsilon > 0 $. I've been advised not to find analytic expressions for equilibria. I'm not looking for an explicit solution spelled out for me, just a shove in the right direction to be able to get there myself. How can one determine location/type of bifurcation for a system like this without known equilibria? Can I use my $epsilon = 0$ diagram and infer something from it for $epsilon >0$? What are my options here, methodology-wise?
differential-equations dynamical-systems bifurcation
asked Nov 17 at 16:23
Jo Fisher
566
It is easy to find the equilibria for $epsilon=0$, as you mention yourself. Then you just need to translate the graph of the right-hand side by $epsilon$ and you can proceed from there. PS: I wonder if you want $x''$ instead of $x'$.
– John B
Nov 17 at 20:57
Thank you for the suggestion. It is only one derivative, although it should really be a dot since it's w.r.t. time.
– Jo Fisher
Nov 18 at 9:51
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to do bifurcation analysis of the system
$x'=-x((x-2)^{2} - mu^{2} -9)((x+2)^2 - mu^2 -9)+ epsilon$
When $epsilon = 0$ this is easy because we have some nice circles. But I'm stuck as to what to do when $epsilon > 0 $. I've been advised not to find analytic expressions for equilibria. I'm not looking for an explicit solution spelled out for me, just a shove in the right direction to be able to get there myself. How can one determine location/type of bifurcation for a system like this without known equilibria? Can I use my $epsilon = 0$ diagram and infer something from it for $epsilon >0$? What are my options here, methodology-wise?
differential-equations dynamical-systems bifurcation
asked Nov 17 at 16:23
Jo Fisher
566
I'm trying to do bifurcation analysis of the system
$x'=-x((x-2)^{2} - mu^{2} -9)((x+2)^2 - mu^2 -9)+ epsilon$
When $epsilon = 0$ this is easy because we have some nice circles. But I'm stuck as to what to do when $epsilon > 0 $. I've been advised not to find analytic expressions for equilibria. I'm not looking for an explicit solution spelled out for me, just a shove in the right direction to be able to get there myself. How can one determine location/type of bifurcation for a system like this without known equilibria? Can I use my $epsilon = 0$ diagram and infer something from it for $epsilon >0$? What are my options here, methodology-wise?
differential-equations dynamical-systems bifurcation
differential-equations dynamical-systems bifurcation
asked Nov 17 at 16:23
Jo Fisher
566
asked Nov 17 at 16:23
Jo Fisher
566
edited Nov 17 at 19:19
asked Nov 17 at 16:23
Jo Fisher
566
asked Nov 17 at 16:23
Jo Fisher
566
asked Nov 17 at 16:23
Jo Fisher
566
566
It is easy to find the equilibria for $epsilon=0$, as you mention yourself. Then you just need to translate the graph of the right-hand side by $epsilon$ and you can proceed from there. PS: I wonder if you want $x''$ instead of $x'$.
– John B
Nov 17 at 20:57
Thank you for the suggestion. It is only one derivative, although it should really be a dot since it's w.r.t. time.
– Jo Fisher
Nov 18 at 9:51
add a comment |
It is easy to find the equilibria for $epsilon=0$, as you mention yourself. Then you just need to translate the graph of the right-hand side by $epsilon$ and you can proceed from there. PS: I wonder if you want $x''$ instead of $x'$.
– John B
Nov 17 at 20:57
Thank you for the suggestion. It is only one derivative, although it should really be a dot since it's w.r.t. time.
– Jo Fisher
Nov 18 at 9:51
It is easy to find the equilibria for $epsilon=0$, as you mention yourself. Then you just need to translate the graph of the right-hand side by $epsilon$ and you can proceed from there. PS: I wonder if you want $x''$ instead of $x'$.
– John B
Nov 17 at 20:57
It is easy to find the equilibria for $epsilon=0$, as you mention yourself. Then you just need to translate the graph of the right-hand side by $epsilon$ and you can proceed from there. PS: I wonder if you want $x''$ instead of $x'$.
– John B
Nov 17 at 20:57
Thank you for the suggestion. It is only one derivative, although it should really be a dot since it's w.r.t. time.
– Jo Fisher
Nov 18 at 9:51
Thank you for the suggestion. It is only one derivative, although it should really be a dot since it's w.r.t. time.
– Jo Fisher
Nov 18 at 9:51
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3002537%2fbifurcation-in-1d-with-two-parameters%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
It is easy to find the equilibria for $epsilon=0$, as you mention yourself. Then you just need to translate the graph of the right-hand side by $epsilon$ and you can proceed from there. PS: I wonder if you want $x''$ instead of $x'$.
– John B
Nov 17 at 20:57
Thank you for the suggestion. It is only one derivative, although it should really be a dot since it's w.r.t. time.
– Jo Fisher
Nov 18 at 9:51