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?
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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 |
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1 Answer
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1 Answer
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$begingroup$
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 |
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6oLE9 matV9Mv8FTDD4c6yt7r9CoAjS,Gcu8D99VHIerHPJAi5YAje6 3Di734MWt7L6,GY,PLJrE
$begingroup$
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