Stability of Mathieu equation: $x''(t)+cos t ,x(t)=0$
The equation
$$
x''(t)+cos t ,x(t)=0 quad (1)
$$
can be transformed to the system:
$$vec{x}'=
begin{pmatrix}
0 & 1\
-cos t & 0
end{pmatrix} vec{x}=A(t) cdot x(t)
$$
with minimum period $T=2pi$. Let $mu_1,mu_2$ be its characteristic values. A theorem gives:
$$mu_1mu_2=expBigg{int_0^{2pi} tr(A(t))dtBigg}=1 quad (2)
$$
Therefore, the Wronskian of any two linearly independent solutions satisfies:
$$
W(t+2pi)=W(t) quad (3)
$$
Does $(3)$ imply that all solutions are bounded and thus we have asymptotic stability? If not, in what way could we use $(2)$ to determine $(1)$'s stability?
differential-equations dynamical-systems periodic-functions stability-in-odes
asked Nov 30 at 20:03
Jevaut
64010
add a comment |
The equation
$$
x''(t)+cos t ,x(t)=0 quad (1)
$$
can be transformed to the system:
$$vec{x}'=
begin{pmatrix}
0 & 1\
-cos t & 0
end{pmatrix} vec{x}=A(t) cdot x(t)
$$
with minimum period $T=2pi$. Let $mu_1,mu_2$ be its characteristic values. A theorem gives:
$$mu_1mu_2=expBigg{int_0^{2pi} tr(A(t))dtBigg}=1 quad (2)
$$
Therefore, the Wronskian of any two linearly independent solutions satisfies:
$$
W(t+2pi)=W(t) quad (3)
$$
Does $(3)$ imply that all solutions are bounded and thus we have asymptotic stability? If not, in what way could we use $(2)$ to determine $(1)$'s stability?
differential-equations dynamical-systems periodic-functions stability-in-odes
asked Nov 30 at 20:03
Jevaut
64010
Determining the numerical solution of $Phi'(t)=A(t)Phi(t)$, $Phi(0)=I_2$ gives $$Phi(2pi)=pmatrix{-8.06518375& 14.50923829\ 4.41423509& -8.06518374}$$ with eigenvalues $-16.06813251$ and $-0.06223499$, so that you have one contracting and one expanding eigenspace.
– LutzL
Dec 1 at 11:25
add a comment |
The equation
$$
x''(t)+cos t ,x(t)=0 quad (1)
$$
can be transformed to the system:
$$vec{x}'=
begin{pmatrix}
0 & 1\
-cos t & 0
end{pmatrix} vec{x}=A(t) cdot x(t)
$$
with minimum period $T=2pi$. Let $mu_1,mu_2$ be its characteristic values. A theorem gives:
$$mu_1mu_2=expBigg{int_0^{2pi} tr(A(t))dtBigg}=1 quad (2)
$$
Therefore, the Wronskian of any two linearly independent solutions satisfies:
$$
W(t+2pi)=W(t) quad (3)
$$
Does $(3)$ imply that all solutions are bounded and thus we have asymptotic stability? If not, in what way could we use $(2)$ to determine $(1)$'s stability?
differential-equations dynamical-systems periodic-functions stability-in-odes
asked Nov 30 at 20:03
Jevaut
64010
The equation
$$
x''(t)+cos t ,x(t)=0 quad (1)
$$
can be transformed to the system:
$$vec{x}'=
begin{pmatrix}
0 & 1\
-cos t & 0
end{pmatrix} vec{x}=A(t) cdot x(t)
$$
with minimum period $T=2pi$. Let $mu_1,mu_2$ be its characteristic values. A theorem gives:
$$mu_1mu_2=expBigg{int_0^{2pi} tr(A(t))dtBigg}=1 quad (2)
$$
Therefore, the Wronskian of any two linearly independent solutions satisfies:
$$
W(t+2pi)=W(t) quad (3)
$$
Does $(3)$ imply that all solutions are bounded and thus we have asymptotic stability? If not, in what way could we use $(2)$ to determine $(1)$'s stability?
differential-equations dynamical-systems periodic-functions stability-in-odes
differential-equations dynamical-systems periodic-functions stability-in-odes
asked Nov 30 at 20:03
Jevaut
64010
asked Nov 30 at 20:03
Jevaut
64010
asked Nov 30 at 20:03
Jevaut
64010
asked Nov 30 at 20:03
Jevaut
64010
asked Nov 30 at 20:03
Jevaut
64010
64010
Determining the numerical solution of $Phi'(t)=A(t)Phi(t)$, $Phi(0)=I_2$ gives $$Phi(2pi)=pmatrix{-8.06518375& 14.50923829\ 4.41423509& -8.06518374}$$ with eigenvalues $-16.06813251$ and $-0.06223499$, so that you have one contracting and one expanding eigenspace.
– LutzL
Dec 1 at 11:25
add a comment |
Determining the numerical solution of $Phi'(t)=A(t)Phi(t)$, $Phi(0)=I_2$ gives $$Phi(2pi)=pmatrix{-8.06518375& 14.50923829\ 4.41423509& -8.06518374}$$ with eigenvalues $-16.06813251$ and $-0.06223499$, so that you have one contracting and one expanding eigenspace.
– LutzL
Dec 1 at 11:25
Determining the numerical solution of $Phi'(t)=A(t)Phi(t)$, $Phi(0)=I_2$ gives $$Phi(2pi)=pmatrix{-8.06518375& 14.50923829\ 4.41423509& -8.06518374}$$ with eigenvalues $-16.06813251$ and $-0.06223499$, so that you have one contracting and one expanding eigenspace.
– LutzL
Dec 1 at 11:25
Determining the numerical solution of $Phi'(t)=A(t)Phi(t)$, $Phi(0)=I_2$ gives $$Phi(2pi)=pmatrix{-8.06518375& 14.50923829\ 4.41423509& -8.06518374}$$ with eigenvalues $-16.06813251$ and $-0.06223499$, so that you have one contracting and one expanding eigenspace.
– LutzL
Dec 1 at 11:25
add a comment |
1 Answer
1
active
oldest
votes
Hint :
If you want to make a conclusion about the stability of the given system and thus the initial equation via Floquet Theory (as your initial approach), then there is a theorem, that states that if for a multiplier $mu_j$ for your given system, it is $|mu_j| <1$, then the system is unstable. But, that's true, if you can prove that :
$$mu_1mu_2 = 1 Rightarrow |mu_1mu_2| = 1 Leftrightarrow |mu_1| = frac{1}{|mu_2|} <1, ; text{if} ; mu_1, mu_2 neq 1$$
In order to conclude that, use the case of the characteristic matrix being periodic, thus $Phi(t+T) = Phi(t) Rightarrow Phi(2 pi) = Phi(0)E Rightarrow E= Phi(0)^{-1}Phi(2pi)$.
Now, the characteristic values will be the eigenvalues of the matrix $E$. You can calculate them (or approximate them) and conclude if $mu_1,mu_2 neq 1$.
You can find more information and elaborations (proofs etc) about that theorem (which also states 2 cases about stability and asymptotic stability) and Floquet Theory in general, here.
Graphs :
(A phase portrait for a certain $t$) For a simple case of time $t$ such that $cos t = -1$, the system has the image of the phase portrait :
$qquad qquad qquad quad$
which is a saddle, thus unstable.
Now, a sample solution family by sampling some initial values for the solution of the given equation and its derivative, one can see the unstability :
$qquad qquad qquad qquad$
answered Nov 30 at 21:16
Rebellos
14.4k31245
You cannot use the linearization to infer the stability of a time dependent system.
– Artem
Nov 30 at 21:25
Once again, you cannot use eigenvalues of time dependent systems to conclude the stability. -1.
– Artem
Nov 30 at 21:28
@Artem Fine, I altered my answer to only use the Floquet Theory approach which is the standard and theoritical way of handling time dependent systems (also given abstracts and conent for further justification and studying).
– Rebellos
Nov 30 at 21:45
@Artem Any issues now for that -1 ?
– Rebellos
Nov 30 at 21:54
1. There is no such thing as a "phase portrait" for a time dependent system. Also, your reasoning about multipliers is incorrect.
– Artem
Nov 30 at 22:19
|
show 7 more comments
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%2f3020568%2fstability-of-mathieu-equation-xt-cos-t-xt-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint :
If you want to make a conclusion about the stability of the given system and thus the initial equation via Floquet Theory (as your initial approach), then there is a theorem, that states that if for a multiplier $mu_j$ for your given system, it is $|mu_j| <1$, then the system is unstable. But, that's true, if you can prove that :
$$mu_1mu_2 = 1 Rightarrow |mu_1mu_2| = 1 Leftrightarrow |mu_1| = frac{1}{|mu_2|} <1, ; text{if} ; mu_1, mu_2 neq 1$$
In order to conclude that, use the case of the characteristic matrix being periodic, thus $Phi(t+T) = Phi(t) Rightarrow Phi(2 pi) = Phi(0)E Rightarrow E= Phi(0)^{-1}Phi(2pi)$.
Now, the characteristic values will be the eigenvalues of the matrix $E$. You can calculate them (or approximate them) and conclude if $mu_1,mu_2 neq 1$.
You can find more information and elaborations (proofs etc) about that theorem (which also states 2 cases about stability and asymptotic stability) and Floquet Theory in general, here.
Graphs :
(A phase portrait for a certain $t$) For a simple case of time $t$ such that $cos t = -1$, the system has the image of the phase portrait :
$qquad qquad qquad quad$
which is a saddle, thus unstable.
Now, a sample solution family by sampling some initial values for the solution of the given equation and its derivative, one can see the unstability :
$qquad qquad qquad qquad$
answered Nov 30 at 21:16
Rebellos
14.4k31245
You cannot use the linearization to infer the stability of a time dependent system.
– Artem
Nov 30 at 21:25
Once again, you cannot use eigenvalues of time dependent systems to conclude the stability. -1.
– Artem
Nov 30 at 21:28
@Artem Fine, I altered my answer to only use the Floquet Theory approach which is the standard and theoritical way of handling time dependent systems (also given abstracts and conent for further justification and studying).
– Rebellos
Nov 30 at 21:45
@Artem Any issues now for that -1 ?
– Rebellos
Nov 30 at 21:54
1. There is no such thing as a "phase portrait" for a time dependent system. Also, your reasoning about multipliers is incorrect.
– Artem
Nov 30 at 22:19
|
show 7 more comments
Hint :
If you want to make a conclusion about the stability of the given system and thus the initial equation via Floquet Theory (as your initial approach), then there is a theorem, that states that if for a multiplier $mu_j$ for your given system, it is $|mu_j| <1$, then the system is unstable. But, that's true, if you can prove that :
$$mu_1mu_2 = 1 Rightarrow |mu_1mu_2| = 1 Leftrightarrow |mu_1| = frac{1}{|mu_2|} <1, ; text{if} ; mu_1, mu_2 neq 1$$
In order to conclude that, use the case of the characteristic matrix being periodic, thus $Phi(t+T) = Phi(t) Rightarrow Phi(2 pi) = Phi(0)E Rightarrow E= Phi(0)^{-1}Phi(2pi)$.
Now, the characteristic values will be the eigenvalues of the matrix $E$. You can calculate them (or approximate them) and conclude if $mu_1,mu_2 neq 1$.
You can find more information and elaborations (proofs etc) about that theorem (which also states 2 cases about stability and asymptotic stability) and Floquet Theory in general, here.
Graphs :
(A phase portrait for a certain $t$) For a simple case of time $t$ such that $cos t = -1$, the system has the image of the phase portrait :
$qquad qquad qquad quad$
which is a saddle, thus unstable.
Now, a sample solution family by sampling some initial values for the solution of the given equation and its derivative, one can see the unstability :
$qquad qquad qquad qquad$
answered Nov 30 at 21:16
Rebellos
14.4k31245
You cannot use the linearization to infer the stability of a time dependent system.
– Artem
Nov 30 at 21:25
Once again, you cannot use eigenvalues of time dependent systems to conclude the stability. -1.
– Artem
Nov 30 at 21:28
@Artem Fine, I altered my answer to only use the Floquet Theory approach which is the standard and theoritical way of handling time dependent systems (also given abstracts and conent for further justification and studying).
– Rebellos
Nov 30 at 21:45
@Artem Any issues now for that -1 ?
– Rebellos
Nov 30 at 21:54
1. There is no such thing as a "phase portrait" for a time dependent system. Also, your reasoning about multipliers is incorrect.
– Artem
Nov 30 at 22:19
|
show 7 more comments
Hint :
If you want to make a conclusion about the stability of the given system and thus the initial equation via Floquet Theory (as your initial approach), then there is a theorem, that states that if for a multiplier $mu_j$ for your given system, it is $|mu_j| <1$, then the system is unstable. But, that's true, if you can prove that :
$$mu_1mu_2 = 1 Rightarrow |mu_1mu_2| = 1 Leftrightarrow |mu_1| = frac{1}{|mu_2|} <1, ; text{if} ; mu_1, mu_2 neq 1$$
In order to conclude that, use the case of the characteristic matrix being periodic, thus $Phi(t+T) = Phi(t) Rightarrow Phi(2 pi) = Phi(0)E Rightarrow E= Phi(0)^{-1}Phi(2pi)$.
Now, the characteristic values will be the eigenvalues of the matrix $E$. You can calculate them (or approximate them) and conclude if $mu_1,mu_2 neq 1$.
You can find more information and elaborations (proofs etc) about that theorem (which also states 2 cases about stability and asymptotic stability) and Floquet Theory in general, here.
Graphs :
(A phase portrait for a certain $t$) For a simple case of time $t$ such that $cos t = -1$, the system has the image of the phase portrait :
$qquad qquad qquad quad$
which is a saddle, thus unstable.
Now, a sample solution family by sampling some initial values for the solution of the given equation and its derivative, one can see the unstability :
$qquad qquad qquad qquad$
answered Nov 30 at 21:16
Rebellos
14.4k31245
Hint :
If you want to make a conclusion about the stability of the given system and thus the initial equation via Floquet Theory (as your initial approach), then there is a theorem, that states that if for a multiplier $mu_j$ for your given system, it is $|mu_j| <1$, then the system is unstable. But, that's true, if you can prove that :
$$mu_1mu_2 = 1 Rightarrow |mu_1mu_2| = 1 Leftrightarrow |mu_1| = frac{1}{|mu_2|} <1, ; text{if} ; mu_1, mu_2 neq 1$$
In order to conclude that, use the case of the characteristic matrix being periodic, thus $Phi(t+T) = Phi(t) Rightarrow Phi(2 pi) = Phi(0)E Rightarrow E= Phi(0)^{-1}Phi(2pi)$.
Now, the characteristic values will be the eigenvalues of the matrix $E$. You can calculate them (or approximate them) and conclude if $mu_1,mu_2 neq 1$.
You can find more information and elaborations (proofs etc) about that theorem (which also states 2 cases about stability and asymptotic stability) and Floquet Theory in general, here.
Graphs :
(A phase portrait for a certain $t$) For a simple case of time $t$ such that $cos t = -1$, the system has the image of the phase portrait :
$qquad qquad qquad quad$
which is a saddle, thus unstable.
Now, a sample solution family by sampling some initial values for the solution of the given equation and its derivative, one can see the unstability :
$qquad qquad qquad qquad$
answered Nov 30 at 21:16
Rebellos
14.4k31245
edited Dec 1 at 11:00
answered Nov 30 at 21:16
Rebellos
14.4k31245
answered Nov 30 at 21:16
Rebellos
14.4k31245
answered Nov 30 at 21:16
Rebellos
14.4k31245
14.4k31245
You cannot use the linearization to infer the stability of a time dependent system.
– Artem
Nov 30 at 21:25
Once again, you cannot use eigenvalues of time dependent systems to conclude the stability. -1.
– Artem
Nov 30 at 21:28
@Artem Fine, I altered my answer to only use the Floquet Theory approach which is the standard and theoritical way of handling time dependent systems (also given abstracts and conent for further justification and studying).
– Rebellos
Nov 30 at 21:45
@Artem Any issues now for that -1 ?
– Rebellos
Nov 30 at 21:54
1. There is no such thing as a "phase portrait" for a time dependent system. Also, your reasoning about multipliers is incorrect.
– Artem
Nov 30 at 22:19
|
show 7 more comments
You cannot use the linearization to infer the stability of a time dependent system.
– Artem
Nov 30 at 21:25
Once again, you cannot use eigenvalues of time dependent systems to conclude the stability. -1.
– Artem
Nov 30 at 21:28
@Artem Fine, I altered my answer to only use the Floquet Theory approach which is the standard and theoritical way of handling time dependent systems (also given abstracts and conent for further justification and studying).
– Rebellos
Nov 30 at 21:45
@Artem Any issues now for that -1 ?
– Rebellos
Nov 30 at 21:54
1. There is no such thing as a "phase portrait" for a time dependent system. Also, your reasoning about multipliers is incorrect.
– Artem
Nov 30 at 22:19
You cannot use the linearization to infer the stability of a time dependent system.
– Artem
Nov 30 at 21:25
You cannot use the linearization to infer the stability of a time dependent system.
– Artem
Nov 30 at 21:25
Once again, you cannot use eigenvalues of time dependent systems to conclude the stability. -1.
– Artem
Nov 30 at 21:28
Once again, you cannot use eigenvalues of time dependent systems to conclude the stability. -1.
– Artem
Nov 30 at 21:28
@Artem Fine, I altered my answer to only use the Floquet Theory approach which is the standard and theoritical way of handling time dependent systems (also given abstracts and conent for further justification and studying).
– Rebellos
Nov 30 at 21:45
@Artem Fine, I altered my answer to only use the Floquet Theory approach which is the standard and theoritical way of handling time dependent systems (also given abstracts and conent for further justification and studying).
– Rebellos
Nov 30 at 21:45
@Artem Any issues now for that -1 ?
– Rebellos
Nov 30 at 21:54
@Artem Any issues now for that -1 ?
– Rebellos
Nov 30 at 21:54
1. There is no such thing as a "phase portrait" for a time dependent system. Also, your reasoning about multipliers is incorrect.
– Artem
Nov 30 at 22:19
1. There is no such thing as a "phase portrait" for a time dependent system. Also, your reasoning about multipliers is incorrect.
– Artem
Nov 30 at 22:19
|
show 7 more comments
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%2f3020568%2fstability-of-mathieu-equation-xt-cos-t-xt-0%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
Determining the numerical solution of $Phi'(t)=A(t)Phi(t)$, $Phi(0)=I_2$ gives $$Phi(2pi)=pmatrix{-8.06518375& 14.50923829\ 4.41423509& -8.06518374}$$ with eigenvalues $-16.06813251$ and $-0.06223499$, so that you have one contracting and one expanding eigenspace.
– LutzL
Dec 1 at 11:25