How to show the maximal interval for $dotx(t)=-cx^2(t)+fracx^2(t)1+x^2(t)$ is $[0,infty)$? The Next CEO of Stack OverflowHow to show the solution for $dotx(t)=-cx^2(t)+fracx^2(t)1+x^2(t)$ is bounded?Maximal interval of existence for autonomous systemsDifferential equation maximal interval and solutionExistence of a solution for a nonlinear ODE on $[0,infty)$Dynamical Systems Maximal Interval of ExistenceProve that any maximal interval of existence is openInterval of definition for the solution $ln|y|+y^2=sin(x)+1$Maximal solution and interval of differential equation $y'=yln(1+y)$The upper bound of the definition interval of a maximal solution ($x'(t) = x(t)^3$)On the maximal solutions of an ODE: how to find a Maximal IntervalWhy are the maximal interval's endpoints semicontinuous functions of the initial condition?
How to use ReplaceAll on an expression that contains a rule
Is there such a thing as a proper verb, like a proper noun?
Is Nisuin Biblical or Rabbinic?
Players Circumventing the limitations of Wish
Can Sneak Attack be used when hitting with an improvised weapon?
It is correct to match light sources with the same color temperature?
Calculate the Mean mean of two numbers
What flight has the highest ratio of timezone difference to flight time?
Audio Conversion With ADS1243
Aggressive Under-Indexing and no data for missing index
Getting Stale Gas Out of a Gas Tank w/out Dropping the Tank
Small nick on power cord from an electric alarm clock, and copper wiring exposed but intact
How to get the last not-null value in an ordered column of a huge table?
Is it ok to trim down a tube patch?
Where do students learn to solve polynomial equations these days?
Traveling with my 5 year old daughter (as the father) without the mother from Germany to Mexico
Purpose of level-shifter with same in and out voltages
Can this note be analyzed as a non-chord tone?
How do I fit a non linear curve?
Expressing the idea of having a very busy time
Why is the US ranked as #45 in Press Freedom ratings, despite its extremely permissive free speech laws?
Inductor and Capacitor in Parallel
What would be the main consequences for a country leaving the WTO?
Which Pokemon have a special animation when running with them out of their pokeball?
How to show the maximal interval for $dotx(t)=-cx^2(t)+fracx^2(t)1+x^2(t)$ is $[0,infty)$?
The Next CEO of Stack OverflowHow to show the solution for $dotx(t)=-cx^2(t)+fracx^2(t)1+x^2(t)$ is bounded?Maximal interval of existence for autonomous systemsDifferential equation maximal interval and solutionExistence of a solution for a nonlinear ODE on $[0,infty)$Dynamical Systems Maximal Interval of ExistenceProve that any maximal interval of existence is openInterval of definition for the solution $ln|y|+y^2=sin(x)+1$Maximal solution and interval of differential equation $y'=yln(1+y)$The upper bound of the definition interval of a maximal solution ($x'(t) = x(t)^3$)On the maximal solutions of an ODE: how to find a Maximal IntervalWhy are the maximal interval's endpoints semicontinuous functions of the initial condition?
$begingroup$
The following differential equation has the maximal interval as $[0,infty)$:
$$
dotx(t)=-cx^2(t)+fracx^2(t)1+x^2(t)
$$
where $c>0$.
My try:
We need to show the solution lives in a bounded set, then we are done. Since $fracx^2(t)1+x^2(t) leq 1$ we can write
$$
dotx(t)leq-cx^2(t)+1
$$
I do not know how to find the upper bound for $x(t)$ and how to come up with a lower bound for $x(t)$.
ordinary-differential-equations dynamical-systems
asked Mar 28 at 3:46
SepideSepide
5258
$endgroup$
add a comment |
$begingroup$
The following differential equation has the maximal interval as $[0,infty)$:
$$
dotx(t)=-cx^2(t)+fracx^2(t)1+x^2(t)
$$
where $c>0$.
My try:
We need to show the solution lives in a bounded set, then we are done. Since $fracx^2(t)1+x^2(t) leq 1$ we can write
$$
dotx(t)leq-cx^2(t)+1
$$
I do not know how to find the upper bound for $x(t)$ and how to come up with a lower bound for $x(t)$.
ordinary-differential-equations dynamical-systems
asked Mar 28 at 3:46
SepideSepide
5258
$endgroup$
$begingroup$
Is there some condition like $x(0)>0$? If the initial value is negative and large enough, then the slope is negative and the dynamic essentially $dot x=-cx^2$, which has a pole at some finite time. Are there any conditions on $c$, like $c>1$?
$endgroup$
– LutzL
Mar 28 at 7:40
$begingroup$
@LutzL: Sorry $c>0$ but no condition on initial condition.
$endgroup$
– Sepide
Mar 28 at 14:37
$begingroup$
if $x^2(t)$ gets too big, then $dot x$
$endgroup$
– Jane
Mar 28 at 15:19
$begingroup$
Can you check again that the first exponent is correct? The claim would be correct for an odd integer power, $-cx$ or $-cx^3$.
$endgroup$
– LutzL
Mar 28 at 15:31
add a comment |
$begingroup$
The following differential equation has the maximal interval as $[0,infty)$:
$$
dotx(t)=-cx^2(t)+fracx^2(t)1+x^2(t)
$$
where $c>0$.
My try:
We need to show the solution lives in a bounded set, then we are done. Since $fracx^2(t)1+x^2(t) leq 1$ we can write
$$
dotx(t)leq-cx^2(t)+1
$$
I do not know how to find the upper bound for $x(t)$ and how to come up with a lower bound for $x(t)$.
ordinary-differential-equations dynamical-systems
asked Mar 28 at 3:46
SepideSepide
5258
$endgroup$
The following differential equation has the maximal interval as $[0,infty)$:
$$
dotx(t)=-cx^2(t)+fracx^2(t)1+x^2(t)
$$
where $c>0$.
My try:
We need to show the solution lives in a bounded set, then we are done. Since $fracx^2(t)1+x^2(t) leq 1$ we can write
$$
dotx(t)leq-cx^2(t)+1
$$
I do not know how to find the upper bound for $x(t)$ and how to come up with a lower bound for $x(t)$.
ordinary-differential-equations dynamical-systems
ordinary-differential-equations dynamical-systems
asked Mar 28 at 3:46
SepideSepide
5258
asked Mar 28 at 3:46
SepideSepide
5258
edited Mar 28 at 14:36
Sepide
asked Mar 28 at 3:46
SepideSepide
5258
asked Mar 28 at 3:46
SepideSepide
5258
asked Mar 28 at 3:46
SepideSepide
5258
5258
$begingroup$
Is there some condition like $x(0)>0$? If the initial value is negative and large enough, then the slope is negative and the dynamic essentially $dot x=-cx^2$, which has a pole at some finite time. Are there any conditions on $c$, like $c>1$?
$endgroup$
– LutzL
Mar 28 at 7:40
$begingroup$
@LutzL: Sorry $c>0$ but no condition on initial condition.
$endgroup$
– Sepide
Mar 28 at 14:37
$begingroup$
if $x^2(t)$ gets too big, then $dot x$
$endgroup$
– Jane
Mar 28 at 15:19
$begingroup$
Can you check again that the first exponent is correct? The claim would be correct for an odd integer power, $-cx$ or $-cx^3$.
$endgroup$
– LutzL
Mar 28 at 15:31
add a comment |
$begingroup$
Is there some condition like $x(0)>0$? If the initial value is negative and large enough, then the slope is negative and the dynamic essentially $dot x=-cx^2$, which has a pole at some finite time. Are there any conditions on $c$, like $c>1$?
$endgroup$
– LutzL
Mar 28 at 7:40
$begingroup$
@LutzL: Sorry $c>0$ but no condition on initial condition.
$endgroup$
– Sepide
Mar 28 at 14:37
$begingroup$
if $x^2(t)$ gets too big, then $dot x$
$endgroup$
– Jane
Mar 28 at 15:19
$begingroup$
Can you check again that the first exponent is correct? The claim would be correct for an odd integer power, $-cx$ or $-cx^3$.
$endgroup$
– LutzL
Mar 28 at 15:31
$begingroup$
Is there some condition like $x(0)>0$? If the initial value is negative and large enough, then the slope is negative and the dynamic essentially $dot x=-cx^2$, which has a pole at some finite time. Are there any conditions on $c$, like $c>1$?
$endgroup$
– LutzL
Mar 28 at 7:40
$begingroup$
Is there some condition like $x(0)>0$? If the initial value is negative and large enough, then the slope is negative and the dynamic essentially $dot x=-cx^2$, which has a pole at some finite time. Are there any conditions on $c$, like $c>1$?
$endgroup$
– LutzL
Mar 28 at 7:40
$begingroup$
@LutzL: Sorry $c>0$ but no condition on initial condition.
$endgroup$
– Sepide
Mar 28 at 14:37
$begingroup$
@LutzL: Sorry $c>0$ but no condition on initial condition.
$endgroup$
– Sepide
Mar 28 at 14:37
$begingroup$
if $x^2(t)$ gets too big, then $dot x$
$endgroup$
– Jane
Mar 28 at 15:19
$begingroup$
if $x^2(t)$ gets too big, then $dot x$
$endgroup$
– Jane
Mar 28 at 15:19
$begingroup$
Can you check again that the first exponent is correct? The claim would be correct for an odd integer power, $-cx$ or $-cx^3$.
$endgroup$
– LutzL
Mar 28 at 15:31
$begingroup$
Can you check again that the first exponent is correct? The claim would be correct for an odd integer power, $-cx$ or $-cx^3$.
$endgroup$
– LutzL
Mar 28 at 15:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As with any one-dimensional autonomous dynamical system with differentiable right side, it is sufficient to consider the roots and signs between roots of the right side function. The roots are at $x_0=0$ and $c(1+x^2)=1iff x_pm=pmsqrtfrac1c-1$ which are only real for $cin(0,1]$.
As $x=0$ is a constant solution, any other solution has an invariant sign.
For $c>1$, the right side is negative everywhere for $xne 0$, so that all solutions are falling. This means that positive solutions are bounded and thus defined on all of $[0,infty)$. Negative solutions are unbounded and will reach a point where the quadratic term is dominant over the second bounded term. From that point on there will be a blow-up to $-infty$.
For $cle 1$ the global image is the same, only that there will be a strip of bounded solutions between the stationary solutions $x=x_pm$.
answered Mar 28 at 15:29
LutzLLutzL
60.1k42057
$endgroup$
add a comment |
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%2f3165447%2fhow-to-show-the-maximal-interval-for-dotxt-cx2t-fracx2t1x2t%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
$begingroup$
As with any one-dimensional autonomous dynamical system with differentiable right side, it is sufficient to consider the roots and signs between roots of the right side function. The roots are at $x_0=0$ and $c(1+x^2)=1iff x_pm=pmsqrtfrac1c-1$ which are only real for $cin(0,1]$.
As $x=0$ is a constant solution, any other solution has an invariant sign.
For $c>1$, the right side is negative everywhere for $xne 0$, so that all solutions are falling. This means that positive solutions are bounded and thus defined on all of $[0,infty)$. Negative solutions are unbounded and will reach a point where the quadratic term is dominant over the second bounded term. From that point on there will be a blow-up to $-infty$.
For $cle 1$ the global image is the same, only that there will be a strip of bounded solutions between the stationary solutions $x=x_pm$.
answered Mar 28 at 15:29
LutzLLutzL
60.1k42057
$endgroup$
add a comment |
$begingroup$
As with any one-dimensional autonomous dynamical system with differentiable right side, it is sufficient to consider the roots and signs between roots of the right side function. The roots are at $x_0=0$ and $c(1+x^2)=1iff x_pm=pmsqrtfrac1c-1$ which are only real for $cin(0,1]$.
As $x=0$ is a constant solution, any other solution has an invariant sign.
For $c>1$, the right side is negative everywhere for $xne 0$, so that all solutions are falling. This means that positive solutions are bounded and thus defined on all of $[0,infty)$. Negative solutions are unbounded and will reach a point where the quadratic term is dominant over the second bounded term. From that point on there will be a blow-up to $-infty$.
For $cle 1$ the global image is the same, only that there will be a strip of bounded solutions between the stationary solutions $x=x_pm$.
answered Mar 28 at 15:29
LutzLLutzL
60.1k42057
$endgroup$
add a comment |
$begingroup$
As with any one-dimensional autonomous dynamical system with differentiable right side, it is sufficient to consider the roots and signs between roots of the right side function. The roots are at $x_0=0$ and $c(1+x^2)=1iff x_pm=pmsqrtfrac1c-1$ which are only real for $cin(0,1]$.
As $x=0$ is a constant solution, any other solution has an invariant sign.
For $c>1$, the right side is negative everywhere for $xne 0$, so that all solutions are falling. This means that positive solutions are bounded and thus defined on all of $[0,infty)$. Negative solutions are unbounded and will reach a point where the quadratic term is dominant over the second bounded term. From that point on there will be a blow-up to $-infty$.
For $cle 1$ the global image is the same, only that there will be a strip of bounded solutions between the stationary solutions $x=x_pm$.
answered Mar 28 at 15:29
LutzLLutzL
60.1k42057
$endgroup$
As with any one-dimensional autonomous dynamical system with differentiable right side, it is sufficient to consider the roots and signs between roots of the right side function. The roots are at $x_0=0$ and $c(1+x^2)=1iff x_pm=pmsqrtfrac1c-1$ which are only real for $cin(0,1]$.
As $x=0$ is a constant solution, any other solution has an invariant sign.
For $c>1$, the right side is negative everywhere for $xne 0$, so that all solutions are falling. This means that positive solutions are bounded and thus defined on all of $[0,infty)$. Negative solutions are unbounded and will reach a point where the quadratic term is dominant over the second bounded term. From that point on there will be a blow-up to $-infty$.
For $cle 1$ the global image is the same, only that there will be a strip of bounded solutions between the stationary solutions $x=x_pm$.
answered Mar 28 at 15:29
LutzLLutzL
60.1k42057
answered Mar 28 at 15:29
LutzLLutzL
60.1k42057
answered Mar 28 at 15:29
LutzLLutzL
60.1k42057
answered Mar 28 at 15:29
LutzLLutzL
60.1k42057
60.1k42057
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.
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%2f3165447%2fhow-to-show-the-maximal-interval-for-dotxt-cx2t-fracx2t1x2t%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
$begingroup$
Is there some condition like $x(0)>0$? If the initial value is negative and large enough, then the slope is negative and the dynamic essentially $dot x=-cx^2$, which has a pole at some finite time. Are there any conditions on $c$, like $c>1$?
$endgroup$
– LutzL
Mar 28 at 7:40
$begingroup$
@LutzL: Sorry $c>0$ but no condition on initial condition.
$endgroup$
– Sepide
Mar 28 at 14:37
$begingroup$
if $x^2(t)$ gets too big, then $dot x$
$endgroup$
– Jane
Mar 28 at 15:19
$begingroup$
Can you check again that the first exponent is correct? The claim would be correct for an odd integer power, $-cx$ or $-cx^3$.
$endgroup$
– LutzL
Mar 28 at 15:31