LaSalle for time varying systems
I am looking for an explanation, why LaSalles theorem is in general not applicable to time varying systems. Can someone provide an example system with
$$
dot{x}(t) = A(t)x(t) tag{1}
$$
I.e., why can't LaSalles theorem be used if I have a Lyapunov function $V(t, x)$ for the system $(1)$ with $dot{V}(t, x) leq 0$?
dynamical-systems control-theory stability-in-odes
asked Nov 28 '18 at 20:38
SampleTime
52539
add a comment |
I am looking for an explanation, why LaSalles theorem is in general not applicable to time varying systems. Can someone provide an example system with
$$
dot{x}(t) = A(t)x(t) tag{1}
$$
I.e., why can't LaSalles theorem be used if I have a Lyapunov function $V(t, x)$ for the system $(1)$ with $dot{V}(t, x) leq 0$?
dynamical-systems control-theory stability-in-odes
asked Nov 28 '18 at 20:38
SampleTime
52539
add a comment |
I am looking for an explanation, why LaSalles theorem is in general not applicable to time varying systems. Can someone provide an example system with
$$
dot{x}(t) = A(t)x(t) tag{1}
$$
I.e., why can't LaSalles theorem be used if I have a Lyapunov function $V(t, x)$ for the system $(1)$ with $dot{V}(t, x) leq 0$?
dynamical-systems control-theory stability-in-odes
asked Nov 28 '18 at 20:38
SampleTime
52539
I am looking for an explanation, why LaSalles theorem is in general not applicable to time varying systems. Can someone provide an example system with
$$
dot{x}(t) = A(t)x(t) tag{1}
$$
I.e., why can't LaSalles theorem be used if I have a Lyapunov function $V(t, x)$ for the system $(1)$ with $dot{V}(t, x) leq 0$?
dynamical-systems control-theory stability-in-odes
dynamical-systems control-theory stability-in-odes
asked Nov 28 '18 at 20:38
SampleTime
52539
asked Nov 28 '18 at 20:38
SampleTime
52539
asked Nov 28 '18 at 20:38
SampleTime
52539
asked Nov 28 '18 at 20:38
SampleTime
52539
asked Nov 28 '18 at 20:38
SampleTime
52539
52539
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Consider the system
$$
left{begin{array}{lll}tag{1}
dot x&=&0\
dot y&=&0\
end{array}right.
$$
and the function $V(t,x,y)=e^{-t}(x^2+y^2)$. The directional derivative is negative definite:
$$
dot V= -e^{-t}(x^2+y^2)+e^{-t}(2xdot x+2ydot y)=-e^{-t}(x^2+y^2),
$$
but the solutions of (1) do not approach the set
$$
E={ (x,y):; dot V= 0}={(0,0)}.
$$
This example is possible because $V(t,x)$ may decrease due to the explicit dependence on $t$, regardless of the approaching of the solution to the set $E$.
answered Nov 29 '18 at 19:06
AVK
2,0961517
add a comment |
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%2f3017681%2flasalle-for-time-varying-systems%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
Consider the system
$$
left{begin{array}{lll}tag{1}
dot x&=&0\
dot y&=&0\
end{array}right.
$$
and the function $V(t,x,y)=e^{-t}(x^2+y^2)$. The directional derivative is negative definite:
$$
dot V= -e^{-t}(x^2+y^2)+e^{-t}(2xdot x+2ydot y)=-e^{-t}(x^2+y^2),
$$
but the solutions of (1) do not approach the set
$$
E={ (x,y):; dot V= 0}={(0,0)}.
$$
This example is possible because $V(t,x)$ may decrease due to the explicit dependence on $t$, regardless of the approaching of the solution to the set $E$.
answered Nov 29 '18 at 19:06
AVK
2,0961517
add a comment |
Consider the system
$$
left{begin{array}{lll}tag{1}
dot x&=&0\
dot y&=&0\
end{array}right.
$$
and the function $V(t,x,y)=e^{-t}(x^2+y^2)$. The directional derivative is negative definite:
$$
dot V= -e^{-t}(x^2+y^2)+e^{-t}(2xdot x+2ydot y)=-e^{-t}(x^2+y^2),
$$
but the solutions of (1) do not approach the set
$$
E={ (x,y):; dot V= 0}={(0,0)}.
$$
This example is possible because $V(t,x)$ may decrease due to the explicit dependence on $t$, regardless of the approaching of the solution to the set $E$.
answered Nov 29 '18 at 19:06
AVK
2,0961517
add a comment |
Consider the system
$$
left{begin{array}{lll}tag{1}
dot x&=&0\
dot y&=&0\
end{array}right.
$$
and the function $V(t,x,y)=e^{-t}(x^2+y^2)$. The directional derivative is negative definite:
$$
dot V= -e^{-t}(x^2+y^2)+e^{-t}(2xdot x+2ydot y)=-e^{-t}(x^2+y^2),
$$
but the solutions of (1) do not approach the set
$$
E={ (x,y):; dot V= 0}={(0,0)}.
$$
This example is possible because $V(t,x)$ may decrease due to the explicit dependence on $t$, regardless of the approaching of the solution to the set $E$.
answered Nov 29 '18 at 19:06
AVK
2,0961517
Consider the system
$$
left{begin{array}{lll}tag{1}
dot x&=&0\
dot y&=&0\
end{array}right.
$$
and the function $V(t,x,y)=e^{-t}(x^2+y^2)$. The directional derivative is negative definite:
$$
dot V= -e^{-t}(x^2+y^2)+e^{-t}(2xdot x+2ydot y)=-e^{-t}(x^2+y^2),
$$
but the solutions of (1) do not approach the set
$$
E={ (x,y):; dot V= 0}={(0,0)}.
$$
This example is possible because $V(t,x)$ may decrease due to the explicit dependence on $t$, regardless of the approaching of the solution to the set $E$.
answered Nov 29 '18 at 19:06
AVK
2,0961517
answered Nov 29 '18 at 19:06
AVK
2,0961517
answered Nov 29 '18 at 19:06
AVK
2,0961517
answered Nov 29 '18 at 19:06
AVK
2,0961517
2,0961517
add a comment |
add a comment |
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%2f3017681%2flasalle-for-time-varying-systems%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
