Property of ODE $y'=sin(y^2)$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Second order linear homogeneous ODE with constant coefficientsQuestions about the Picard–Lindelöf theorem for an ODEHow to apply Picard–Lindelöf theorem to the whole domain.Solution of ODE is also lip-continuous, if ODE is lip-continuous?Do continuously differentiable functions which are not Lipschitz have uniqueness of solutions of ODEProving $f(x,y)$ is not Lipschitz but still has a unique solution to initial value problemLipschitz continuous ODE solution intersecting a hyperplane infinitely often in finite timeUnique global solution of $x'=-sin(x)$Picard–Lindelöf theorem conditions not met with unique IVP solutionLipschitz continuity of $e^sin$
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Property of ODE $y'=sin(y^2)$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Second order linear homogeneous ODE with constant coefficientsQuestions about the Picard–Lindelöf theorem for an ODEHow to apply Picard–Lindelöf theorem to the whole domain.Solution of ODE is also lip-continuous, if ODE is lip-continuous?Do continuously differentiable functions which are not Lipschitz have uniqueness of solutions of ODEProving $f(x,y)$ is not Lipschitz but still has a unique solution to initial value problemLipschitz continuous ODE solution intersecting a hyperplane infinitely often in finite timeUnique global solution of $x'=-sin(x)$Picard–Lindelöf theorem conditions not met with unique IVP solutionLipschitz continuity of $e^sin$
$begingroup$
Let $y(x)'=sin(y(x)^2), xinmathbbR$ be given.
I want to show without solving that
$$y=const. Leftrightarrow (y(0))^2=kpi, kinmathbbN$$
and that $y$ is monotone.
The direction $Rightarrow$ is clear. Now for the interesting part ($Leftarrow$):
Since $sin(y)$ is countinously differentiable (thus locally Lipschitz continous), I can show that every local solution is unique by Picard-Lindelöf Theorem. It follows that the solution must not intersect the lines $y^2=kpi, kinmathbbN$ and lies between them. Now I am stuck, as I can not show that this already leads to a constant solution.
ordinary-differential-equations
asked Apr 2 at 15:32
EpsilonDeltaEpsilonDelta
7301615
$endgroup$
add a comment |
$begingroup$
Let $y(x)'=sin(y(x)^2), xinmathbbR$ be given.
I want to show without solving that
$$y=const. Leftrightarrow (y(0))^2=kpi, kinmathbbN$$
and that $y$ is monotone.
The direction $Rightarrow$ is clear. Now for the interesting part ($Leftarrow$):
Since $sin(y)$ is countinously differentiable (thus locally Lipschitz continous), I can show that every local solution is unique by Picard-Lindelöf Theorem. It follows that the solution must not intersect the lines $y^2=kpi, kinmathbbN$ and lies between them. Now I am stuck, as I can not show that this already leads to a constant solution.
ordinary-differential-equations
asked Apr 2 at 15:32
EpsilonDeltaEpsilonDelta
7301615
$endgroup$
$begingroup$
This is an autonomous equation, so you can treat it by looking for equilibrium points and thinking about their stability. If $0<y^2(0)<pi$, then $y'$ is positive and the trajectories must be increasing for all $x$, but can't cross the equilibrium point at $y=sqrtpi.$ so you can squeeze the solution at $y(0)=sqrtpi$ between increasing and decreasing trajectories.
$endgroup$
– B. Goddard
Apr 2 at 15:40
$begingroup$
Maybe I thought too complicated, but if the solutions are locally unique and I know that the value $kpi$ is taken at $x=0$ while knowing that $kpi$ is a solution for all $x$, they must coincide, correct? I think I overthought on this one...
$endgroup$
– EpsilonDelta
Apr 2 at 16:09
$begingroup$
You need to separate what exactly you want to show, it seems that your task description mixes two problems. The first you have solved, characterizing the constant solutions. Now it remains to show that non-constant solutions are monotone, which is a trivial consequence of the constant sign of $f(y)=sin(y^2)$ between roots.
$endgroup$
– LutzL
Apr 2 at 16:10
$begingroup$
@LutzL So my argument in my first comment is correct for showing that the solution must be constant if the said value is attained?
$endgroup$
– EpsilonDelta
Apr 2 at 16:15
1
$begingroup$
Yes. Roots of $f$ produce constant solutions, and by uniqueness, as you said, any solution having a root as one value has to be constant all along.
$endgroup$
– LutzL
Apr 2 at 16:34
add a comment |
$begingroup$
Let $y(x)'=sin(y(x)^2), xinmathbbR$ be given.
I want to show without solving that
$$y=const. Leftrightarrow (y(0))^2=kpi, kinmathbbN$$
and that $y$ is monotone.
The direction $Rightarrow$ is clear. Now for the interesting part ($Leftarrow$):
Since $sin(y)$ is countinously differentiable (thus locally Lipschitz continous), I can show that every local solution is unique by Picard-Lindelöf Theorem. It follows that the solution must not intersect the lines $y^2=kpi, kinmathbbN$ and lies between them. Now I am stuck, as I can not show that this already leads to a constant solution.
ordinary-differential-equations
asked Apr 2 at 15:32
EpsilonDeltaEpsilonDelta
7301615
$endgroup$
Let $y(x)'=sin(y(x)^2), xinmathbbR$ be given.
I want to show without solving that
$$y=const. Leftrightarrow (y(0))^2=kpi, kinmathbbN$$
and that $y$ is monotone.
The direction $Rightarrow$ is clear. Now for the interesting part ($Leftarrow$):
Since $sin(y)$ is countinously differentiable (thus locally Lipschitz continous), I can show that every local solution is unique by Picard-Lindelöf Theorem. It follows that the solution must not intersect the lines $y^2=kpi, kinmathbbN$ and lies between them. Now I am stuck, as I can not show that this already leads to a constant solution.
ordinary-differential-equations
ordinary-differential-equations
asked Apr 2 at 15:32
EpsilonDeltaEpsilonDelta
7301615
asked Apr 2 at 15:32
EpsilonDeltaEpsilonDelta
7301615
asked Apr 2 at 15:32
EpsilonDeltaEpsilonDelta
7301615
asked Apr 2 at 15:32
EpsilonDeltaEpsilonDelta
7301615
asked Apr 2 at 15:32
EpsilonDeltaEpsilonDelta
7301615
7301615
$begingroup$
This is an autonomous equation, so you can treat it by looking for equilibrium points and thinking about their stability. If $0<y^2(0)<pi$, then $y'$ is positive and the trajectories must be increasing for all $x$, but can't cross the equilibrium point at $y=sqrtpi.$ so you can squeeze the solution at $y(0)=sqrtpi$ between increasing and decreasing trajectories.
$endgroup$
– B. Goddard
Apr 2 at 15:40
$begingroup$
Maybe I thought too complicated, but if the solutions are locally unique and I know that the value $kpi$ is taken at $x=0$ while knowing that $kpi$ is a solution for all $x$, they must coincide, correct? I think I overthought on this one...
$endgroup$
– EpsilonDelta
Apr 2 at 16:09
$begingroup$
You need to separate what exactly you want to show, it seems that your task description mixes two problems. The first you have solved, characterizing the constant solutions. Now it remains to show that non-constant solutions are monotone, which is a trivial consequence of the constant sign of $f(y)=sin(y^2)$ between roots.
$endgroup$
– LutzL
Apr 2 at 16:10
$begingroup$
@LutzL So my argument in my first comment is correct for showing that the solution must be constant if the said value is attained?
$endgroup$
– EpsilonDelta
Apr 2 at 16:15
1
$begingroup$
Yes. Roots of $f$ produce constant solutions, and by uniqueness, as you said, any solution having a root as one value has to be constant all along.
$endgroup$
– LutzL
Apr 2 at 16:34
add a comment |
$begingroup$
This is an autonomous equation, so you can treat it by looking for equilibrium points and thinking about their stability. If $0<y^2(0)<pi$, then $y'$ is positive and the trajectories must be increasing for all $x$, but can't cross the equilibrium point at $y=sqrtpi.$ so you can squeeze the solution at $y(0)=sqrtpi$ between increasing and decreasing trajectories.
$endgroup$
– B. Goddard
Apr 2 at 15:40
$begingroup$
Maybe I thought too complicated, but if the solutions are locally unique and I know that the value $kpi$ is taken at $x=0$ while knowing that $kpi$ is a solution for all $x$, they must coincide, correct? I think I overthought on this one...
$endgroup$
– EpsilonDelta
Apr 2 at 16:09
$begingroup$
You need to separate what exactly you want to show, it seems that your task description mixes two problems. The first you have solved, characterizing the constant solutions. Now it remains to show that non-constant solutions are monotone, which is a trivial consequence of the constant sign of $f(y)=sin(y^2)$ between roots.
$endgroup$
– LutzL
Apr 2 at 16:10
$begingroup$
@LutzL So my argument in my first comment is correct for showing that the solution must be constant if the said value is attained?
$endgroup$
– EpsilonDelta
Apr 2 at 16:15
1
$begingroup$
Yes. Roots of $f$ produce constant solutions, and by uniqueness, as you said, any solution having a root as one value has to be constant all along.
$endgroup$
– LutzL
Apr 2 at 16:34
$begingroup$
This is an autonomous equation, so you can treat it by looking for equilibrium points and thinking about their stability. If $0<y^2(0)<pi$, then $y'$ is positive and the trajectories must be increasing for all $x$, but can't cross the equilibrium point at $y=sqrtpi.$ so you can squeeze the solution at $y(0)=sqrtpi$ between increasing and decreasing trajectories.
$endgroup$
– B. Goddard
Apr 2 at 15:40
$begingroup$
This is an autonomous equation, so you can treat it by looking for equilibrium points and thinking about their stability. If $0<y^2(0)<pi$, then $y'$ is positive and the trajectories must be increasing for all $x$, but can't cross the equilibrium point at $y=sqrtpi.$ so you can squeeze the solution at $y(0)=sqrtpi$ between increasing and decreasing trajectories.
$endgroup$
– B. Goddard
Apr 2 at 15:40
$begingroup$
Maybe I thought too complicated, but if the solutions are locally unique and I know that the value $kpi$ is taken at $x=0$ while knowing that $kpi$ is a solution for all $x$, they must coincide, correct? I think I overthought on this one...
$endgroup$
– EpsilonDelta
Apr 2 at 16:09
$begingroup$
Maybe I thought too complicated, but if the solutions are locally unique and I know that the value $kpi$ is taken at $x=0$ while knowing that $kpi$ is a solution for all $x$, they must coincide, correct? I think I overthought on this one...
$endgroup$
– EpsilonDelta
Apr 2 at 16:09
$begingroup$
You need to separate what exactly you want to show, it seems that your task description mixes two problems. The first you have solved, characterizing the constant solutions. Now it remains to show that non-constant solutions are monotone, which is a trivial consequence of the constant sign of $f(y)=sin(y^2)$ between roots.
$endgroup$
– LutzL
Apr 2 at 16:10
$begingroup$
You need to separate what exactly you want to show, it seems that your task description mixes two problems. The first you have solved, characterizing the constant solutions. Now it remains to show that non-constant solutions are monotone, which is a trivial consequence of the constant sign of $f(y)=sin(y^2)$ between roots.
$endgroup$
– LutzL
Apr 2 at 16:10
$begingroup$
@LutzL So my argument in my first comment is correct for showing that the solution must be constant if the said value is attained?
$endgroup$
– EpsilonDelta
Apr 2 at 16:15
$begingroup$
@LutzL So my argument in my first comment is correct for showing that the solution must be constant if the said value is attained?
$endgroup$
– EpsilonDelta
Apr 2 at 16:15
1
1
$begingroup$
Yes. Roots of $f$ produce constant solutions, and by uniqueness, as you said, any solution having a root as one value has to be constant all along.
$endgroup$
– LutzL
Apr 2 at 16:34
$begingroup$
Yes. Roots of $f$ produce constant solutions, and by uniqueness, as you said, any solution having a root as one value has to be constant all along.
$endgroup$
– LutzL
Apr 2 at 16:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, roots $f(y^*)$ of the right side of $y'=f(y)$ produce constant solutions $y(t)=y^*$.
Yes, by uniqueness any solution having a root $y^*$ of $f$ as one value has to be constant all along.
As this scalar function $f$ has infinitely many roots converging to infinity, any initial value (that is not a root itself) is between two roots, the corresponding solution is bounded above and below by the respective constant solutions and exists thus for all times.
As this is a scalar equation and $f$ is continuous, the values of $f$ between two roots have a constant sign, so solutions are either monotonically increasing or monotonically falling.
The left and right limits $xtopminfty$ of any such solutions are the bounding roots of $f$, as the limits exist and $y'$ converges to zero.
answered Apr 2 at 16:07
LutzLLutzL
60.9k42157
$endgroup$
add a comment |
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$begingroup$
Yes, roots $f(y^*)$ of the right side of $y'=f(y)$ produce constant solutions $y(t)=y^*$.
Yes, by uniqueness any solution having a root $y^*$ of $f$ as one value has to be constant all along.
As this scalar function $f$ has infinitely many roots converging to infinity, any initial value (that is not a root itself) is between two roots, the corresponding solution is bounded above and below by the respective constant solutions and exists thus for all times.
As this is a scalar equation and $f$ is continuous, the values of $f$ between two roots have a constant sign, so solutions are either monotonically increasing or monotonically falling.
The left and right limits $xtopminfty$ of any such solutions are the bounding roots of $f$, as the limits exist and $y'$ converges to zero.
answered Apr 2 at 16:07
LutzLLutzL
60.9k42157
$endgroup$
add a comment |
$begingroup$
Yes, roots $f(y^*)$ of the right side of $y'=f(y)$ produce constant solutions $y(t)=y^*$.
Yes, by uniqueness any solution having a root $y^*$ of $f$ as one value has to be constant all along.
As this scalar function $f$ has infinitely many roots converging to infinity, any initial value (that is not a root itself) is between two roots, the corresponding solution is bounded above and below by the respective constant solutions and exists thus for all times.
As this is a scalar equation and $f$ is continuous, the values of $f$ between two roots have a constant sign, so solutions are either monotonically increasing or monotonically falling.
The left and right limits $xtopminfty$ of any such solutions are the bounding roots of $f$, as the limits exist and $y'$ converges to zero.
answered Apr 2 at 16:07
LutzLLutzL
60.9k42157
$endgroup$
add a comment |
$begingroup$
Yes, roots $f(y^*)$ of the right side of $y'=f(y)$ produce constant solutions $y(t)=y^*$.
Yes, by uniqueness any solution having a root $y^*$ of $f$ as one value has to be constant all along.
As this scalar function $f$ has infinitely many roots converging to infinity, any initial value (that is not a root itself) is between two roots, the corresponding solution is bounded above and below by the respective constant solutions and exists thus for all times.
As this is a scalar equation and $f$ is continuous, the values of $f$ between two roots have a constant sign, so solutions are either monotonically increasing or monotonically falling.
The left and right limits $xtopminfty$ of any such solutions are the bounding roots of $f$, as the limits exist and $y'$ converges to zero.
answered Apr 2 at 16:07
LutzLLutzL
60.9k42157
$endgroup$
Yes, roots $f(y^*)$ of the right side of $y'=f(y)$ produce constant solutions $y(t)=y^*$.
Yes, by uniqueness any solution having a root $y^*$ of $f$ as one value has to be constant all along.
As this scalar function $f$ has infinitely many roots converging to infinity, any initial value (that is not a root itself) is between two roots, the corresponding solution is bounded above and below by the respective constant solutions and exists thus for all times.
As this is a scalar equation and $f$ is continuous, the values of $f$ between two roots have a constant sign, so solutions are either monotonically increasing or monotonically falling.
The left and right limits $xtopminfty$ of any such solutions are the bounding roots of $f$, as the limits exist and $y'$ converges to zero.
answered Apr 2 at 16:07
LutzLLutzL
60.9k42157
edited Apr 2 at 16:55
answered Apr 2 at 16:07
LutzLLutzL
60.9k42157
answered Apr 2 at 16:07
LutzLLutzL
60.9k42157
answered Apr 2 at 16:07
LutzLLutzL
60.9k42157
60.9k42157
add a comment |
add a comment |
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$begingroup$
This is an autonomous equation, so you can treat it by looking for equilibrium points and thinking about their stability. If $0<y^2(0)<pi$, then $y'$ is positive and the trajectories must be increasing for all $x$, but can't cross the equilibrium point at $y=sqrtpi.$ so you can squeeze the solution at $y(0)=sqrtpi$ between increasing and decreasing trajectories.
$endgroup$
– B. Goddard
Apr 2 at 15:40
$begingroup$
Maybe I thought too complicated, but if the solutions are locally unique and I know that the value $kpi$ is taken at $x=0$ while knowing that $kpi$ is a solution for all $x$, they must coincide, correct? I think I overthought on this one...
$endgroup$
– EpsilonDelta
Apr 2 at 16:09
$begingroup$
You need to separate what exactly you want to show, it seems that your task description mixes two problems. The first you have solved, characterizing the constant solutions. Now it remains to show that non-constant solutions are monotone, which is a trivial consequence of the constant sign of $f(y)=sin(y^2)$ between roots.
$endgroup$
– LutzL
Apr 2 at 16:10
$begingroup$
@LutzL So my argument in my first comment is correct for showing that the solution must be constant if the said value is attained?
$endgroup$
– EpsilonDelta
Apr 2 at 16:15
1
$begingroup$
Yes. Roots of $f$ produce constant solutions, and by uniqueness, as you said, any solution having a root as one value has to be constant all along.
$endgroup$
– LutzL
Apr 2 at 16:34