Relation of parameter-dependent center manifold and parameter-independent center manifold
$begingroup$
Considering a parameter-dependent ODE
$dot{x} = Ax + f_A(x,y,z,mu) \
dot{y} = By + f_B(x,y,z,mu) \
dot{z} = Cz + f_C(x,y,z,mu) $
where A has only eigenvalues with zero real part, B only eigenvalues with negative real part and C has only eigenvalues with positive real part and such that $(x,y,z,mu) in mathbb{R}^{n_0} timesmathbb{R}^{n_-} timesmathbb{R}^{n_+} timesmathbb{R}^{p}$. Suppose the system has an equilibrium in $(x,y,z,mu) = (0,0,0,0)$. From the center manifold theorem I know there exists a neighbourhood $U_{n_0} subset mathbb R^{n_0}$ of the origin and a $n_0$-dimensional center manifold $c: U_{n_0}to mathbb{R}^{n_-}times mathbb{R}^{n_+}$.
The standard trick is to extend the system to
$dot{x} = Ax + f_A(x,y,z,mu) \
dot{mu} = 0 \
dot{y} = By + f_B(x,y,z,mu) \
dot{z} = Cz + f_C(x,y,z,mu) $
and the center manifold theorem implies there exists neighbourhood $mathbb{R}^{n_0} times mathbb{R}^{p}$ of the origin and a $n_0+p$-dimensional center manifold $tilde{c}: U_{n_0} times U_{p}times mathbb{R}^{p} to mathbb{R}^{n_-}times mathbb{R}^{n_+}$.
My question is, how are $c$ and $tilde{c}$ related? Is there a way to express $c$ in terms of $tilde{c}$ or vice versa?
ordinary-differential-equations dynamical-systems
asked Dec 12 '18 at 11:03
JDoeJDoe
320113
$endgroup$
add a comment |
$begingroup$
Considering a parameter-dependent ODE
$dot{x} = Ax + f_A(x,y,z,mu) \
dot{y} = By + f_B(x,y,z,mu) \
dot{z} = Cz + f_C(x,y,z,mu) $
where A has only eigenvalues with zero real part, B only eigenvalues with negative real part and C has only eigenvalues with positive real part and such that $(x,y,z,mu) in mathbb{R}^{n_0} timesmathbb{R}^{n_-} timesmathbb{R}^{n_+} timesmathbb{R}^{p}$. Suppose the system has an equilibrium in $(x,y,z,mu) = (0,0,0,0)$. From the center manifold theorem I know there exists a neighbourhood $U_{n_0} subset mathbb R^{n_0}$ of the origin and a $n_0$-dimensional center manifold $c: U_{n_0}to mathbb{R}^{n_-}times mathbb{R}^{n_+}$.
The standard trick is to extend the system to
$dot{x} = Ax + f_A(x,y,z,mu) \
dot{mu} = 0 \
dot{y} = By + f_B(x,y,z,mu) \
dot{z} = Cz + f_C(x,y,z,mu) $
and the center manifold theorem implies there exists neighbourhood $mathbb{R}^{n_0} times mathbb{R}^{p}$ of the origin and a $n_0+p$-dimensional center manifold $tilde{c}: U_{n_0} times U_{p}times mathbb{R}^{p} to mathbb{R}^{n_-}times mathbb{R}^{n_+}$.
My question is, how are $c$ and $tilde{c}$ related? Is there a way to express $c$ in terms of $tilde{c}$ or vice versa?
ordinary-differential-equations dynamical-systems
asked Dec 12 '18 at 11:03
JDoeJDoe
320113
$endgroup$
$begingroup$
You are missing the main point: when you introduce a parameter depended center manifold you matrices $A,B,C$ usually change. Otherwise this "standard trick" would be a waist of time.
$endgroup$
– Artem
Dec 15 '18 at 2:07
$begingroup$
@Artem I think this is covered by letting $f_A$ and others to have non-zero derivatives at the origin. At least it was done this way in Shilnikov-Shilnikov-Turaev-Chua book, as far as I remember.
$endgroup$
– Evgeny
Dec 17 '18 at 16:32
$begingroup$
@Evgeny What book are you referring to? But there should be a relation between both center manifolds. In the first case you get a family of center manifolds (one for each parameter value) and in the second case you obtain a center manifold without the parameter. It seems natural, that there must be a relation between both center manifolds and I was hoping, somebody can shed some light into this matter.
$endgroup$
– JDoe
Dec 19 '18 at 10:37
1
$begingroup$
I refer to this book. You might be interested in part of book around Theorem 5.2.
$endgroup$
– Evgeny
Dec 19 '18 at 16:22
add a comment |
$begingroup$
Considering a parameter-dependent ODE
$dot{x} = Ax + f_A(x,y,z,mu) \
dot{y} = By + f_B(x,y,z,mu) \
dot{z} = Cz + f_C(x,y,z,mu) $
where A has only eigenvalues with zero real part, B only eigenvalues with negative real part and C has only eigenvalues with positive real part and such that $(x,y,z,mu) in mathbb{R}^{n_0} timesmathbb{R}^{n_-} timesmathbb{R}^{n_+} timesmathbb{R}^{p}$. Suppose the system has an equilibrium in $(x,y,z,mu) = (0,0,0,0)$. From the center manifold theorem I know there exists a neighbourhood $U_{n_0} subset mathbb R^{n_0}$ of the origin and a $n_0$-dimensional center manifold $c: U_{n_0}to mathbb{R}^{n_-}times mathbb{R}^{n_+}$.
The standard trick is to extend the system to
$dot{x} = Ax + f_A(x,y,z,mu) \
dot{mu} = 0 \
dot{y} = By + f_B(x,y,z,mu) \
dot{z} = Cz + f_C(x,y,z,mu) $
and the center manifold theorem implies there exists neighbourhood $mathbb{R}^{n_0} times mathbb{R}^{p}$ of the origin and a $n_0+p$-dimensional center manifold $tilde{c}: U_{n_0} times U_{p}times mathbb{R}^{p} to mathbb{R}^{n_-}times mathbb{R}^{n_+}$.
My question is, how are $c$ and $tilde{c}$ related? Is there a way to express $c$ in terms of $tilde{c}$ or vice versa?
ordinary-differential-equations dynamical-systems
asked Dec 12 '18 at 11:03
JDoeJDoe
320113
$endgroup$
Considering a parameter-dependent ODE
$dot{x} = Ax + f_A(x,y,z,mu) \
dot{y} = By + f_B(x,y,z,mu) \
dot{z} = Cz + f_C(x,y,z,mu) $
where A has only eigenvalues with zero real part, B only eigenvalues with negative real part and C has only eigenvalues with positive real part and such that $(x,y,z,mu) in mathbb{R}^{n_0} timesmathbb{R}^{n_-} timesmathbb{R}^{n_+} timesmathbb{R}^{p}$. Suppose the system has an equilibrium in $(x,y,z,mu) = (0,0,0,0)$. From the center manifold theorem I know there exists a neighbourhood $U_{n_0} subset mathbb R^{n_0}$ of the origin and a $n_0$-dimensional center manifold $c: U_{n_0}to mathbb{R}^{n_-}times mathbb{R}^{n_+}$.
The standard trick is to extend the system to
$dot{x} = Ax + f_A(x,y,z,mu) \
dot{mu} = 0 \
dot{y} = By + f_B(x,y,z,mu) \
dot{z} = Cz + f_C(x,y,z,mu) $
and the center manifold theorem implies there exists neighbourhood $mathbb{R}^{n_0} times mathbb{R}^{p}$ of the origin and a $n_0+p$-dimensional center manifold $tilde{c}: U_{n_0} times U_{p}times mathbb{R}^{p} to mathbb{R}^{n_-}times mathbb{R}^{n_+}$.
My question is, how are $c$ and $tilde{c}$ related? Is there a way to express $c$ in terms of $tilde{c}$ or vice versa?
ordinary-differential-equations dynamical-systems
ordinary-differential-equations dynamical-systems
asked Dec 12 '18 at 11:03
JDoeJDoe
320113
asked Dec 12 '18 at 11:03
JDoeJDoe
320113
asked Dec 12 '18 at 11:03
JDoeJDoe
320113
asked Dec 12 '18 at 11:03
JDoeJDoe
320113
asked Dec 12 '18 at 11:03
JDoeJDoe
320113
320113
$begingroup$
You are missing the main point: when you introduce a parameter depended center manifold you matrices $A,B,C$ usually change. Otherwise this "standard trick" would be a waist of time.
$endgroup$
– Artem
Dec 15 '18 at 2:07
$begingroup$
@Artem I think this is covered by letting $f_A$ and others to have non-zero derivatives at the origin. At least it was done this way in Shilnikov-Shilnikov-Turaev-Chua book, as far as I remember.
$endgroup$
– Evgeny
Dec 17 '18 at 16:32
$begingroup$
@Evgeny What book are you referring to? But there should be a relation between both center manifolds. In the first case you get a family of center manifolds (one for each parameter value) and in the second case you obtain a center manifold without the parameter. It seems natural, that there must be a relation between both center manifolds and I was hoping, somebody can shed some light into this matter.
$endgroup$
– JDoe
Dec 19 '18 at 10:37
1
$begingroup$
I refer to this book. You might be interested in part of book around Theorem 5.2.
$endgroup$
– Evgeny
Dec 19 '18 at 16:22
add a comment |
$begingroup$
You are missing the main point: when you introduce a parameter depended center manifold you matrices $A,B,C$ usually change. Otherwise this "standard trick" would be a waist of time.
$endgroup$
– Artem
Dec 15 '18 at 2:07
$begingroup$
@Artem I think this is covered by letting $f_A$ and others to have non-zero derivatives at the origin. At least it was done this way in Shilnikov-Shilnikov-Turaev-Chua book, as far as I remember.
$endgroup$
– Evgeny
Dec 17 '18 at 16:32
$begingroup$
@Evgeny What book are you referring to? But there should be a relation between both center manifolds. In the first case you get a family of center manifolds (one for each parameter value) and in the second case you obtain a center manifold without the parameter. It seems natural, that there must be a relation between both center manifolds and I was hoping, somebody can shed some light into this matter.
$endgroup$
– JDoe
Dec 19 '18 at 10:37
1
$begingroup$
I refer to this book. You might be interested in part of book around Theorem 5.2.
$endgroup$
– Evgeny
Dec 19 '18 at 16:22
$begingroup$
You are missing the main point: when you introduce a parameter depended center manifold you matrices $A,B,C$ usually change. Otherwise this "standard trick" would be a waist of time.
$endgroup$
– Artem
Dec 15 '18 at 2:07
$begingroup$
You are missing the main point: when you introduce a parameter depended center manifold you matrices $A,B,C$ usually change. Otherwise this "standard trick" would be a waist of time.
$endgroup$
– Artem
Dec 15 '18 at 2:07
$begingroup$
@Artem I think this is covered by letting $f_A$ and others to have non-zero derivatives at the origin. At least it was done this way in Shilnikov-Shilnikov-Turaev-Chua book, as far as I remember.
$endgroup$
– Evgeny
Dec 17 '18 at 16:32
$begingroup$
@Artem I think this is covered by letting $f_A$ and others to have non-zero derivatives at the origin. At least it was done this way in Shilnikov-Shilnikov-Turaev-Chua book, as far as I remember.
$endgroup$
– Evgeny
Dec 17 '18 at 16:32
$begingroup$
@Evgeny What book are you referring to? But there should be a relation between both center manifolds. In the first case you get a family of center manifolds (one for each parameter value) and in the second case you obtain a center manifold without the parameter. It seems natural, that there must be a relation between both center manifolds and I was hoping, somebody can shed some light into this matter.
$endgroup$
– JDoe
Dec 19 '18 at 10:37
$begingroup$
@Evgeny What book are you referring to? But there should be a relation between both center manifolds. In the first case you get a family of center manifolds (one for each parameter value) and in the second case you obtain a center manifold without the parameter. It seems natural, that there must be a relation between both center manifolds and I was hoping, somebody can shed some light into this matter.
$endgroup$
– JDoe
Dec 19 '18 at 10:37
1
1
$begingroup$
I refer to this book. You might be interested in part of book around Theorem 5.2.
$endgroup$
– Evgeny
Dec 19 '18 at 16:22
$begingroup$
I refer to this book. You might be interested in part of book around Theorem 5.2.
$endgroup$
– Evgeny
Dec 19 '18 at 16:22
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2f3036549%2frelation-of-parameter-dependent-center-manifold-and-parameter-independent-center%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
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.
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%2f3036549%2frelation-of-parameter-dependent-center-manifold-and-parameter-independent-center%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
$begingroup$
You are missing the main point: when you introduce a parameter depended center manifold you matrices $A,B,C$ usually change. Otherwise this "standard trick" would be a waist of time.
$endgroup$
– Artem
Dec 15 '18 at 2:07
$begingroup$
@Artem I think this is covered by letting $f_A$ and others to have non-zero derivatives at the origin. At least it was done this way in Shilnikov-Shilnikov-Turaev-Chua book, as far as I remember.
$endgroup$
– Evgeny
Dec 17 '18 at 16:32
$begingroup$
@Evgeny What book are you referring to? But there should be a relation between both center manifolds. In the first case you get a family of center manifolds (one for each parameter value) and in the second case you obtain a center manifold without the parameter. It seems natural, that there must be a relation between both center manifolds and I was hoping, somebody can shed some light into this matter.
$endgroup$
– JDoe
Dec 19 '18 at 10:37
1
$begingroup$
I refer to this book. You might be interested in part of book around Theorem 5.2.
$endgroup$
– Evgeny
Dec 19 '18 at 16:22