Mixing
Finding incomplete solutions of second order ODE system
$begingroup$
For a smooth, 1-periodic, positive function $c:mathbb{R}rightarrow (0,infty)$ consider the following second order ODE system for the curve $gamma(t)= (x(t),y(t),z(t)):mathbb{R}rightarrow mathbb{R}^3$ (note that I will denote the derivative of $c$ by a prime and the derivatives of the curve components $x,y,z$ by dots).
begin{align}
(i) hspace{15mm} ddot{x}(t)+ frac{c'(z(t))}{c(z(t))}~ dot{x}(t)dot{z}(t)=0 \
(ii) hspace{14mm} ddot{y}(t) +frac{c'(z(t))}{c(z(t)))}~dot{y}(t)dot{z}(t)=0 \
(iii) hspace{16mm} ddot{z}(t)-c'(z(t)) ~ dot{x}(t)dot{y}(t)=0
end{align}
My question: I'm interested in finding a function $c$ as described above such that the system has "incomplete" solutions, i.e. solutions $gamma:mathbb{R}rightarrow mathbb{R}^3$ which are not defined for all times $tinmathbb{R}$. Or alternatively I want to show that for all $c$ as above there are no incomplete solutions.
Background: the background to this problem is a geometric one: I have a family of metrics ${g_c}$ given on $mathbb{R}^3$ (where $c$ is as described above) and I'm asking myself if all those metrics are geodasically complete. Solving the geodesic equation for those metrics gives the above system of ODEs.
ordinary-differential-equations riemannian-geometry nonlinear-system
edited Jan 30 at 9:51
YuiTo Cheng
2,0592837
asked Jan 26 at 15:45
JS.JS.
1709
$endgroup$
add a comment |
$begingroup$
For a smooth, 1-periodic, positive function $c:mathbb{R}rightarrow (0,infty)$ consider the following second order ODE system for the curve $gamma(t)= (x(t),y(t),z(t)):mathbb{R}rightarrow mathbb{R}^3$ (note that I will denote the derivative of $c$ by a prime and the derivatives of the curve components $x,y,z$ by dots).
begin{align}
(i) hspace{15mm} ddot{x}(t)+ frac{c'(z(t))}{c(z(t))}~ dot{x}(t)dot{z}(t)=0 \
(ii) hspace{14mm} ddot{y}(t) +frac{c'(z(t))}{c(z(t)))}~dot{y}(t)dot{z}(t)=0 \
(iii) hspace{16mm} ddot{z}(t)-c'(z(t)) ~ dot{x}(t)dot{y}(t)=0
end{align}
My question: I'm interested in finding a function $c$ as described above such that the system has "incomplete" solutions, i.e. solutions $gamma:mathbb{R}rightarrow mathbb{R}^3$ which are not defined for all times $tinmathbb{R}$. Or alternatively I want to show that for all $c$ as above there are no incomplete solutions.
Background: the background to this problem is a geometric one: I have a family of metrics ${g_c}$ given on $mathbb{R}^3$ (where $c$ is as described above) and I'm asking myself if all those metrics are geodasically complete. Solving the geodesic equation for those metrics gives the above system of ODEs.
ordinary-differential-equations riemannian-geometry nonlinear-system
edited Jan 30 at 9:51
YuiTo Cheng
2,0592837
asked Jan 26 at 15:45
JS.JS.
1709
$endgroup$
add a comment |
$begingroup$
For a smooth, 1-periodic, positive function $c:mathbb{R}rightarrow (0,infty)$ consider the following second order ODE system for the curve $gamma(t)= (x(t),y(t),z(t)):mathbb{R}rightarrow mathbb{R}^3$ (note that I will denote the derivative of $c$ by a prime and the derivatives of the curve components $x,y,z$ by dots).
begin{align}
(i) hspace{15mm} ddot{x}(t)+ frac{c'(z(t))}{c(z(t))}~ dot{x}(t)dot{z}(t)=0 \
(ii) hspace{14mm} ddot{y}(t) +frac{c'(z(t))}{c(z(t)))}~dot{y}(t)dot{z}(t)=0 \
(iii) hspace{16mm} ddot{z}(t)-c'(z(t)) ~ dot{x}(t)dot{y}(t)=0
end{align}
My question: I'm interested in finding a function $c$ as described above such that the system has "incomplete" solutions, i.e. solutions $gamma:mathbb{R}rightarrow mathbb{R}^3$ which are not defined for all times $tinmathbb{R}$. Or alternatively I want to show that for all $c$ as above there are no incomplete solutions.
Background: the background to this problem is a geometric one: I have a family of metrics ${g_c}$ given on $mathbb{R}^3$ (where $c$ is as described above) and I'm asking myself if all those metrics are geodasically complete. Solving the geodesic equation for those metrics gives the above system of ODEs.
ordinary-differential-equations riemannian-geometry nonlinear-system
edited Jan 30 at 9:51
YuiTo Cheng
2,0592837
asked Jan 26 at 15:45
JS.JS.
1709
$endgroup$
For a smooth, 1-periodic, positive function $c:mathbb{R}rightarrow (0,infty)$ consider the following second order ODE system for the curve $gamma(t)= (x(t),y(t),z(t)):mathbb{R}rightarrow mathbb{R}^3$ (note that I will denote the derivative of $c$ by a prime and the derivatives of the curve components $x,y,z$ by dots).
begin{align}
(i) hspace{15mm} ddot{x}(t)+ frac{c'(z(t))}{c(z(t))}~ dot{x}(t)dot{z}(t)=0 \
(ii) hspace{14mm} ddot{y}(t) +frac{c'(z(t))}{c(z(t)))}~dot{y}(t)dot{z}(t)=0 \
(iii) hspace{16mm} ddot{z}(t)-c'(z(t)) ~ dot{x}(t)dot{y}(t)=0
end{align}
My question: I'm interested in finding a function $c$ as described above such that the system has "incomplete" solutions, i.e. solutions $gamma:mathbb{R}rightarrow mathbb{R}^3$ which are not defined for all times $tinmathbb{R}$. Or alternatively I want to show that for all $c$ as above there are no incomplete solutions.
Background: the background to this problem is a geometric one: I have a family of metrics ${g_c}$ given on $mathbb{R}^3$ (where $c$ is as described above) and I'm asking myself if all those metrics are geodasically complete. Solving the geodesic equation for those metrics gives the above system of ODEs.
ordinary-differential-equations riemannian-geometry nonlinear-system
ordinary-differential-equations riemannian-geometry nonlinear-system
edited Jan 30 at 9:51
YuiTo Cheng
2,0592837
asked Jan 26 at 15:45
JS.JS.
1709
edited Jan 30 at 9:51
YuiTo Cheng
2,0592837
asked Jan 26 at 15:45
JS.JS.
1709
edited Jan 30 at 9:51
YuiTo Cheng
2,0592837
edited Jan 30 at 9:51
YuiTo Cheng
2,0592837
edited Jan 30 at 9:51
YuiTo Cheng
2,0592837
2,0592837
asked Jan 26 at 15:45
JS.JS.
1709
asked Jan 26 at 15:45
JS.JS.
1709
asked Jan 26 at 15:45
JS.JS.
1709
1709
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The equation system has the following first integrals:
$$c(z)^2left(dot x^2+dot y^2right)=I_1=const$$
$$c(z)left(ydot x-xdot yright)=I_2=const$$
If $c_{MIN}<c(z)<c_{MAX}$, then $dot x$ and $dot y$ are bounded:
$$|dot x|<frac{sqrt{I_1}}{c_{MIN}}$$
$$|dot y|<frac{sqrt{I_1}}{c_{MIN}}$$
Assuming that $c'(z)$ is a bounded smooth function, $ddot z(t)$ is also bounded. This means that there are no singular solutions (such that $x$, $y$ or $z$ reach infinity for finite $t$). Also, existence theorem guarantees existence of solution to the initial value problem in the neighborhood of every $t=t_0$, so there are no incomplete solutions.
answered Jan 30 at 9:36
atarasenkoatarasenko
3565
$endgroup$
$begingroup$
@user450093: to find $I_1$: multiply equation (i) by $dot x$, (ii) by $dot y$, and add them. To find $I_2$: multiply equation (i) by $y$, (ii) by $x$, and subtract the latter from the former. However, such procedure does not work with equation (iii).
$endgroup$
– atarasenko
Jan 30 at 14:36
$begingroup$
This is already really helpful! But to me it is not clear yet how you conclude that all solutions are complete from the fact that $dot{x}$,$dot{y}$ and $ddot{z}$ are bounded (especially I don't see how the boundedness of $ddot{z}$ is used).
$endgroup$
– JS.
Jan 31 at 13:44
$begingroup$
The idea is that if a solution cannot be extended beyond $t=t_0$, then $x(t_0)$ or $y(t_0)$ or $z(t_0)$ is infinite. This is impossible because $x(t_0)=infty$ means that $dot x(t_0) = infty$, so $|dot x|$ must be unbounded. Similarly, $z(t_0)=infty$ means that $ddot z(t_0)=infty$.
$endgroup$
– atarasenko
Feb 1 at 7:04
$begingroup$
However, I agree that my proof is not very strict, it does not rule out other types of singularities, for example $dot x(t)=sin{(1/(t-t_0))}$.
$endgroup$
– atarasenko
Feb 1 at 7:05
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%2f3088396%2ffinding-incomplete-solutions-of-second-order-ode-system%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
$begingroup$
The equation system has the following first integrals:
$$c(z)^2left(dot x^2+dot y^2right)=I_1=const$$
$$c(z)left(ydot x-xdot yright)=I_2=const$$
If $c_{MIN}<c(z)<c_{MAX}$, then $dot x$ and $dot y$ are bounded:
$$|dot x|<frac{sqrt{I_1}}{c_{MIN}}$$
$$|dot y|<frac{sqrt{I_1}}{c_{MIN}}$$
Assuming that $c'(z)$ is a bounded smooth function, $ddot z(t)$ is also bounded. This means that there are no singular solutions (such that $x$, $y$ or $z$ reach infinity for finite $t$). Also, existence theorem guarantees existence of solution to the initial value problem in the neighborhood of every $t=t_0$, so there are no incomplete solutions.
answered Jan 30 at 9:36
atarasenkoatarasenko
3565
$endgroup$
$begingroup$
@user450093: to find $I_1$: multiply equation (i) by $dot x$, (ii) by $dot y$, and add them. To find $I_2$: multiply equation (i) by $y$, (ii) by $x$, and subtract the latter from the former. However, such procedure does not work with equation (iii).
$endgroup$
– atarasenko
Jan 30 at 14:36
$begingroup$
This is already really helpful! But to me it is not clear yet how you conclude that all solutions are complete from the fact that $dot{x}$,$dot{y}$ and $ddot{z}$ are bounded (especially I don't see how the boundedness of $ddot{z}$ is used).
$endgroup$
– JS.
Jan 31 at 13:44
$begingroup$
The idea is that if a solution cannot be extended beyond $t=t_0$, then $x(t_0)$ or $y(t_0)$ or $z(t_0)$ is infinite. This is impossible because $x(t_0)=infty$ means that $dot x(t_0) = infty$, so $|dot x|$ must be unbounded. Similarly, $z(t_0)=infty$ means that $ddot z(t_0)=infty$.
$endgroup$
– atarasenko
Feb 1 at 7:04
$begingroup$
However, I agree that my proof is not very strict, it does not rule out other types of singularities, for example $dot x(t)=sin{(1/(t-t_0))}$.
$endgroup$
– atarasenko
Feb 1 at 7:05
add a comment |
$begingroup$
The equation system has the following first integrals:
$$c(z)^2left(dot x^2+dot y^2right)=I_1=const$$
$$c(z)left(ydot x-xdot yright)=I_2=const$$
If $c_{MIN}<c(z)<c_{MAX}$, then $dot x$ and $dot y$ are bounded:
$$|dot x|<frac{sqrt{I_1}}{c_{MIN}}$$
$$|dot y|<frac{sqrt{I_1}}{c_{MIN}}$$
Assuming that $c'(z)$ is a bounded smooth function, $ddot z(t)$ is also bounded. This means that there are no singular solutions (such that $x$, $y$ or $z$ reach infinity for finite $t$). Also, existence theorem guarantees existence of solution to the initial value problem in the neighborhood of every $t=t_0$, so there are no incomplete solutions.
answered Jan 30 at 9:36
atarasenkoatarasenko
3565
$endgroup$
$begingroup$
@user450093: to find $I_1$: multiply equation (i) by $dot x$, (ii) by $dot y$, and add them. To find $I_2$: multiply equation (i) by $y$, (ii) by $x$, and subtract the latter from the former. However, such procedure does not work with equation (iii).
$endgroup$
– atarasenko
Jan 30 at 14:36
$begingroup$
This is already really helpful! But to me it is not clear yet how you conclude that all solutions are complete from the fact that $dot{x}$,$dot{y}$ and $ddot{z}$ are bounded (especially I don't see how the boundedness of $ddot{z}$ is used).
$endgroup$
– JS.
Jan 31 at 13:44
$begingroup$
The idea is that if a solution cannot be extended beyond $t=t_0$, then $x(t_0)$ or $y(t_0)$ or $z(t_0)$ is infinite. This is impossible because $x(t_0)=infty$ means that $dot x(t_0) = infty$, so $|dot x|$ must be unbounded. Similarly, $z(t_0)=infty$ means that $ddot z(t_0)=infty$.
$endgroup$
– atarasenko
Feb 1 at 7:04
$begingroup$
However, I agree that my proof is not very strict, it does not rule out other types of singularities, for example $dot x(t)=sin{(1/(t-t_0))}$.
$endgroup$
– atarasenko
Feb 1 at 7:05
add a comment |
$begingroup$
The equation system has the following first integrals:
$$c(z)^2left(dot x^2+dot y^2right)=I_1=const$$
$$c(z)left(ydot x-xdot yright)=I_2=const$$
If $c_{MIN}<c(z)<c_{MAX}$, then $dot x$ and $dot y$ are bounded:
$$|dot x|<frac{sqrt{I_1}}{c_{MIN}}$$
$$|dot y|<frac{sqrt{I_1}}{c_{MIN}}$$
Assuming that $c'(z)$ is a bounded smooth function, $ddot z(t)$ is also bounded. This means that there are no singular solutions (such that $x$, $y$ or $z$ reach infinity for finite $t$). Also, existence theorem guarantees existence of solution to the initial value problem in the neighborhood of every $t=t_0$, so there are no incomplete solutions.
answered Jan 30 at 9:36
atarasenkoatarasenko
3565
$endgroup$
The equation system has the following first integrals:
$$c(z)^2left(dot x^2+dot y^2right)=I_1=const$$
$$c(z)left(ydot x-xdot yright)=I_2=const$$
If $c_{MIN}<c(z)<c_{MAX}$, then $dot x$ and $dot y$ are bounded:
$$|dot x|<frac{sqrt{I_1}}{c_{MIN}}$$
$$|dot y|<frac{sqrt{I_1}}{c_{MIN}}$$
Assuming that $c'(z)$ is a bounded smooth function, $ddot z(t)$ is also bounded. This means that there are no singular solutions (such that $x$, $y$ or $z$ reach infinity for finite $t$). Also, existence theorem guarantees existence of solution to the initial value problem in the neighborhood of every $t=t_0$, so there are no incomplete solutions.
answered Jan 30 at 9:36
atarasenkoatarasenko
3565
answered Jan 30 at 9:36
atarasenkoatarasenko
3565
answered Jan 30 at 9:36
atarasenkoatarasenko
3565
answered Jan 30 at 9:36
atarasenkoatarasenko
3565
3565
$begingroup$
@user450093: to find $I_1$: multiply equation (i) by $dot x$, (ii) by $dot y$, and add them. To find $I_2$: multiply equation (i) by $y$, (ii) by $x$, and subtract the latter from the former. However, such procedure does not work with equation (iii).
$endgroup$
– atarasenko
Jan 30 at 14:36
$begingroup$
This is already really helpful! But to me it is not clear yet how you conclude that all solutions are complete from the fact that $dot{x}$,$dot{y}$ and $ddot{z}$ are bounded (especially I don't see how the boundedness of $ddot{z}$ is used).
$endgroup$
– JS.
Jan 31 at 13:44
$begingroup$
The idea is that if a solution cannot be extended beyond $t=t_0$, then $x(t_0)$ or $y(t_0)$ or $z(t_0)$ is infinite. This is impossible because $x(t_0)=infty$ means that $dot x(t_0) = infty$, so $|dot x|$ must be unbounded. Similarly, $z(t_0)=infty$ means that $ddot z(t_0)=infty$.
$endgroup$
– atarasenko
Feb 1 at 7:04
$begingroup$
However, I agree that my proof is not very strict, it does not rule out other types of singularities, for example $dot x(t)=sin{(1/(t-t_0))}$.
$endgroup$
– atarasenko
Feb 1 at 7:05
add a comment |
$begingroup$
@user450093: to find $I_1$: multiply equation (i) by $dot x$, (ii) by $dot y$, and add them. To find $I_2$: multiply equation (i) by $y$, (ii) by $x$, and subtract the latter from the former. However, such procedure does not work with equation (iii).
$endgroup$
– atarasenko
Jan 30 at 14:36
$begingroup$
This is already really helpful! But to me it is not clear yet how you conclude that all solutions are complete from the fact that $dot{x}$,$dot{y}$ and $ddot{z}$ are bounded (especially I don't see how the boundedness of $ddot{z}$ is used).
$endgroup$
– JS.
Jan 31 at 13:44
$begingroup$
The idea is that if a solution cannot be extended beyond $t=t_0$, then $x(t_0)$ or $y(t_0)$ or $z(t_0)$ is infinite. This is impossible because $x(t_0)=infty$ means that $dot x(t_0) = infty$, so $|dot x|$ must be unbounded. Similarly, $z(t_0)=infty$ means that $ddot z(t_0)=infty$.
$endgroup$
– atarasenko
Feb 1 at 7:04
$begingroup$
However, I agree that my proof is not very strict, it does not rule out other types of singularities, for example $dot x(t)=sin{(1/(t-t_0))}$.
$endgroup$
– atarasenko
Feb 1 at 7:05
$begingroup$
@user450093: to find $I_1$: multiply equation (i) by $dot x$, (ii) by $dot y$, and add them. To find $I_2$: multiply equation (i) by $y$, (ii) by $x$, and subtract the latter from the former. However, such procedure does not work with equation (iii).
$endgroup$
– atarasenko
Jan 30 at 14:36
$begingroup$
@user450093: to find $I_1$: multiply equation (i) by $dot x$, (ii) by $dot y$, and add them. To find $I_2$: multiply equation (i) by $y$, (ii) by $x$, and subtract the latter from the former. However, such procedure does not work with equation (iii).
$endgroup$
– atarasenko
Jan 30 at 14:36
$begingroup$
This is already really helpful! But to me it is not clear yet how you conclude that all solutions are complete from the fact that $dot{x}$,$dot{y}$ and $ddot{z}$ are bounded (especially I don't see how the boundedness of $ddot{z}$ is used).
$endgroup$
– JS.
Jan 31 at 13:44
$begingroup$
This is already really helpful! But to me it is not clear yet how you conclude that all solutions are complete from the fact that $dot{x}$,$dot{y}$ and $ddot{z}$ are bounded (especially I don't see how the boundedness of $ddot{z}$ is used).
$endgroup$
– JS.
Jan 31 at 13:44
$begingroup$
The idea is that if a solution cannot be extended beyond $t=t_0$, then $x(t_0)$ or $y(t_0)$ or $z(t_0)$ is infinite. This is impossible because $x(t_0)=infty$ means that $dot x(t_0) = infty$, so $|dot x|$ must be unbounded. Similarly, $z(t_0)=infty$ means that $ddot z(t_0)=infty$.
$endgroup$
– atarasenko
Feb 1 at 7:04
$begingroup$
The idea is that if a solution cannot be extended beyond $t=t_0$, then $x(t_0)$ or $y(t_0)$ or $z(t_0)$ is infinite. This is impossible because $x(t_0)=infty$ means that $dot x(t_0) = infty$, so $|dot x|$ must be unbounded. Similarly, $z(t_0)=infty$ means that $ddot z(t_0)=infty$.
$endgroup$
– atarasenko
Feb 1 at 7:04
$begingroup$
However, I agree that my proof is not very strict, it does not rule out other types of singularities, for example $dot x(t)=sin{(1/(t-t_0))}$.
$endgroup$
– atarasenko
Feb 1 at 7:05
$begingroup$
However, I agree that my proof is not very strict, it does not rule out other types of singularities, for example $dot x(t)=sin{(1/(t-t_0))}$.
$endgroup$
– atarasenko
Feb 1 at 7:05
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.
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%2f3088396%2ffinding-incomplete-solutions-of-second-order-ode-system%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
