Find the largest interval where the Initial Value Problem $y'(t)=t+sin(y(t))$ with $y(2)=1$ has a unique...












1














Consider the initial value problem (IVP)
begin{cases}
y'(t)= t + sin(y(t)), \
y(2) = 1. \
end{cases}
Find the largest interval $mathcal{I}subset mathbb{R}$ containing $t_0=2$ so that the problem has a unique solution $y$ in $mathcal{I}$.




My proof attempt:



Let $L>frac{pi}{2}>0$. Define
$$ R := {(t,y)in mathbb{R}^2:|t-2|leq L, |y-1|leq L } $$
Then
$$ 2-Lleq t leq 2+L text{ and } 1-Lleq y leq 1+L $$
Since $L>frac{pi}{2} Rightarrow exists y_0in (1-L, 1+L)$ so that $sin(y_0)=1$. Hence,
$$ M = underset{R}{sup}|F(t,y)|=3+L $$
Then
$$ |partial_y F(t,y)|=|cos(y)|leq 1 $$
Put $c = 1$, then by the Mean Value theorem, for $(t,y),(t,u)in R Rightarrow$
$$ |F(t,y)-F(t,u)|leq c|y-u| $$
Let $a_{*}=min left(L, frac{L}{M}right)=min left(L, frac{L}{3+L}right)=frac{L}{3+L}$. Using Picard-Lindelof's theorem, there should exist a unique solution on the interval $mathcal{I}=left[2-frac{L}{3+L}, 2+frac{L}{3+L}right]$.



Since $underset{Lrightarrow infty}{lim}frac{L}{3+L}=1$, our largest interval should be $mathcal{I}=left[1+varepsilon,3-varepsilonright]$ where $varepsilon in (0,1)$.



Am I on the right track here?






real-analysis differential-equations proof-verification initial-value-problems







share|cite|improve this question




















  • 1




    I have also read somewhere that there is a global solution (i.e. $mathcal{I}=mathbb{R}$) if F(t,y) is globally lipschitz. Which I believe it is since $nabla F=(1, cos(y))$. Which should be bounded under the operator norm. So how do I reconcile this fact with what I did up above? Is what I did wrong?
    – Joe Man Analysis
    Nov 19 '18 at 4:49


















1














Consider the initial value problem (IVP)
begin{cases}
y'(t)= t + sin(y(t)), \
y(2) = 1. \
end{cases}
Find the largest interval $mathcal{I}subset mathbb{R}$ containing $t_0=2$ so that the problem has a unique solution $y$ in $mathcal{I}$.




My proof attempt:



Let $L>frac{pi}{2}>0$. Define
$$ R := {(t,y)in mathbb{R}^2:|t-2|leq L, |y-1|leq L } $$
Then
$$ 2-Lleq t leq 2+L text{ and } 1-Lleq y leq 1+L $$
Since $L>frac{pi}{2} Rightarrow exists y_0in (1-L, 1+L)$ so that $sin(y_0)=1$. Hence,
$$ M = underset{R}{sup}|F(t,y)|=3+L $$
Then
$$ |partial_y F(t,y)|=|cos(y)|leq 1 $$
Put $c = 1$, then by the Mean Value theorem, for $(t,y),(t,u)in R Rightarrow$
$$ |F(t,y)-F(t,u)|leq c|y-u| $$
Let $a_{*}=min left(L, frac{L}{M}right)=min left(L, frac{L}{3+L}right)=frac{L}{3+L}$. Using Picard-Lindelof's theorem, there should exist a unique solution on the interval $mathcal{I}=left[2-frac{L}{3+L}, 2+frac{L}{3+L}right]$.



Since $underset{Lrightarrow infty}{lim}frac{L}{3+L}=1$, our largest interval should be $mathcal{I}=left[1+varepsilon,3-varepsilonright]$ where $varepsilon in (0,1)$.



Am I on the right track here?






real-analysis differential-equations proof-verification initial-value-problems







share|cite|improve this question




















  • 1




    I have also read somewhere that there is a global solution (i.e. $mathcal{I}=mathbb{R}$) if F(t,y) is globally lipschitz. Which I believe it is since $nabla F=(1, cos(y))$. Which should be bounded under the operator norm. So how do I reconcile this fact with what I did up above? Is what I did wrong?
    – Joe Man Analysis
    Nov 19 '18 at 4:49
















1












1








1







Consider the initial value problem (IVP)
begin{cases}
y'(t)= t + sin(y(t)), \
y(2) = 1. \
end{cases}
Find the largest interval $mathcal{I}subset mathbb{R}$ containing $t_0=2$ so that the problem has a unique solution $y$ in $mathcal{I}$.




My proof attempt:



Let $L>frac{pi}{2}>0$. Define
$$ R := {(t,y)in mathbb{R}^2:|t-2|leq L, |y-1|leq L } $$
Then
$$ 2-Lleq t leq 2+L text{ and } 1-Lleq y leq 1+L $$
Since $L>frac{pi}{2} Rightarrow exists y_0in (1-L, 1+L)$ so that $sin(y_0)=1$. Hence,
$$ M = underset{R}{sup}|F(t,y)|=3+L $$
Then
$$ |partial_y F(t,y)|=|cos(y)|leq 1 $$
Put $c = 1$, then by the Mean Value theorem, for $(t,y),(t,u)in R Rightarrow$
$$ |F(t,y)-F(t,u)|leq c|y-u| $$
Let $a_{*}=min left(L, frac{L}{M}right)=min left(L, frac{L}{3+L}right)=frac{L}{3+L}$. Using Picard-Lindelof's theorem, there should exist a unique solution on the interval $mathcal{I}=left[2-frac{L}{3+L}, 2+frac{L}{3+L}right]$.



Since $underset{Lrightarrow infty}{lim}frac{L}{3+L}=1$, our largest interval should be $mathcal{I}=left[1+varepsilon,3-varepsilonright]$ where $varepsilon in (0,1)$.



Am I on the right track here?






real-analysis differential-equations proof-verification initial-value-problems







share|cite|improve this question















Consider the initial value problem (IVP)
begin{cases}
y'(t)= t + sin(y(t)), \
y(2) = 1. \
end{cases}
Find the largest interval $mathcal{I}subset mathbb{R}$ containing $t_0=2$ so that the problem has a unique solution $y$ in $mathcal{I}$.




My proof attempt:



Let $L>frac{pi}{2}>0$. Define
$$ R := {(t,y)in mathbb{R}^2:|t-2|leq L, |y-1|leq L } $$
Then
$$ 2-Lleq t leq 2+L text{ and } 1-Lleq y leq 1+L $$
Since $L>frac{pi}{2} Rightarrow exists y_0in (1-L, 1+L)$ so that $sin(y_0)=1$. Hence,
$$ M = underset{R}{sup}|F(t,y)|=3+L $$
Then
$$ |partial_y F(t,y)|=|cos(y)|leq 1 $$
Put $c = 1$, then by the Mean Value theorem, for $(t,y),(t,u)in R Rightarrow$
$$ |F(t,y)-F(t,u)|leq c|y-u| $$
Let $a_{*}=min left(L, frac{L}{M}right)=min left(L, frac{L}{3+L}right)=frac{L}{3+L}$. Using Picard-Lindelof's theorem, there should exist a unique solution on the interval $mathcal{I}=left[2-frac{L}{3+L}, 2+frac{L}{3+L}right]$.



Since $underset{Lrightarrow infty}{lim}frac{L}{3+L}=1$, our largest interval should be $mathcal{I}=left[1+varepsilon,3-varepsilonright]$ where $varepsilon in (0,1)$.



Am I on the right track here?






real-analysis differential-equations proof-verification initial-value-problems




real-analysis differential-equations proof-verification initial-value-problems






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edited Nov 19 '18 at 10:52

























asked Nov 19 '18 at 4:47









Joe Man Analysis

33419




33419








  • 1




    I have also read somewhere that there is a global solution (i.e. $mathcal{I}=mathbb{R}$) if F(t,y) is globally lipschitz. Which I believe it is since $nabla F=(1, cos(y))$. Which should be bounded under the operator norm. So how do I reconcile this fact with what I did up above? Is what I did wrong?
    – Joe Man Analysis
    Nov 19 '18 at 4:49
















  • 1




    I have also read somewhere that there is a global solution (i.e. $mathcal{I}=mathbb{R}$) if F(t,y) is globally lipschitz. Which I believe it is since $nabla F=(1, cos(y))$. Which should be bounded under the operator norm. So how do I reconcile this fact with what I did up above? Is what I did wrong?
    – Joe Man Analysis
    Nov 19 '18 at 4:49










1




1




I have also read somewhere that there is a global solution (i.e. $mathcal{I}=mathbb{R}$) if F(t,y) is globally lipschitz. Which I believe it is since $nabla F=(1, cos(y))$. Which should be bounded under the operator norm. So how do I reconcile this fact with what I did up above? Is what I did wrong?
– Joe Man Analysis
Nov 19 '18 at 4:49






I have also read somewhere that there is a global solution (i.e. $mathcal{I}=mathbb{R}$) if F(t,y) is globally lipschitz. Which I believe it is since $nabla F=(1, cos(y))$. Which should be bounded under the operator norm. So how do I reconcile this fact with what I did up above? Is what I did wrong?
– Joe Man Analysis
Nov 19 '18 at 4:49












1 Answer
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From the Picard theorem one can infer the following: Given an ODE $y'=f(t,y)$ with $f$ defined in an open set $Omegasubset{mathbb R}^2$ and fulfilling the assumptions of the theorem in the neighborhood of each point $(t,y)inOmega$, any solution of an IVP $y(t_0)=y_0$ can be extended to a maximal solution $$tilde y:quad Jmapsto{mathbb R},quad xmapsto tilde y(x)$$
in an unique way. Here $J$ is an open interval which may depend on the given initial point $(t_0,y_0)$. Furthermore the graph ${cal G}(tilde y)subsetOmega$ of this maximal solution will "ultimately" leave any compact set $KsubsetOmega$ given in advance. (For example, the solution cannot develop a $xmapstosin{1over x}$ singularity in the interior of $Omega$.)



In the case at hand we have $f(t,y)=t+sin y$ and $Omega={mathbb R}^2$. It follows that for any solution $tmapsto y(t)$ one has $|y'(t)|leq |t|+1$. This allows to conclude that $|tilde y(t)|leq C(1+t^2)$ for a suitable $C>0$; hence $tilde y$ cannot drift away to $pm infty$ in finite time. Since ${cal G}(tilde y)$ will ultimately leave any compact rectangle $K:[-M,M]times[-C(2+M^2),C(2+M^2]$, and cannot do so across the bottom and top edges, it follows that $tilde y$ is defined on all of ${mathbb R}$.






share|cite|improve this answer



















  • 1




    Could you please explain how the bound on the maximal solution was obtained?
    – Joe Man Analysis
    Nov 19 '18 at 10:49











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From the Picard theorem one can infer the following: Given an ODE $y'=f(t,y)$ with $f$ defined in an open set $Omegasubset{mathbb R}^2$ and fulfilling the assumptions of the theorem in the neighborhood of each point $(t,y)inOmega$, any solution of an IVP $y(t_0)=y_0$ can be extended to a maximal solution $$tilde y:quad Jmapsto{mathbb R},quad xmapsto tilde y(x)$$
in an unique way. Here $J$ is an open interval which may depend on the given initial point $(t_0,y_0)$. Furthermore the graph ${cal G}(tilde y)subsetOmega$ of this maximal solution will "ultimately" leave any compact set $KsubsetOmega$ given in advance. (For example, the solution cannot develop a $xmapstosin{1over x}$ singularity in the interior of $Omega$.)



In the case at hand we have $f(t,y)=t+sin y$ and $Omega={mathbb R}^2$. It follows that for any solution $tmapsto y(t)$ one has $|y'(t)|leq |t|+1$. This allows to conclude that $|tilde y(t)|leq C(1+t^2)$ for a suitable $C>0$; hence $tilde y$ cannot drift away to $pm infty$ in finite time. Since ${cal G}(tilde y)$ will ultimately leave any compact rectangle $K:[-M,M]times[-C(2+M^2),C(2+M^2]$, and cannot do so across the bottom and top edges, it follows that $tilde y$ is defined on all of ${mathbb R}$.






share|cite|improve this answer



















  • 1




    Could you please explain how the bound on the maximal solution was obtained?
    – Joe Man Analysis
    Nov 19 '18 at 10:49
















2














From the Picard theorem one can infer the following: Given an ODE $y'=f(t,y)$ with $f$ defined in an open set $Omegasubset{mathbb R}^2$ and fulfilling the assumptions of the theorem in the neighborhood of each point $(t,y)inOmega$, any solution of an IVP $y(t_0)=y_0$ can be extended to a maximal solution $$tilde y:quad Jmapsto{mathbb R},quad xmapsto tilde y(x)$$
in an unique way. Here $J$ is an open interval which may depend on the given initial point $(t_0,y_0)$. Furthermore the graph ${cal G}(tilde y)subsetOmega$ of this maximal solution will "ultimately" leave any compact set $KsubsetOmega$ given in advance. (For example, the solution cannot develop a $xmapstosin{1over x}$ singularity in the interior of $Omega$.)



In the case at hand we have $f(t,y)=t+sin y$ and $Omega={mathbb R}^2$. It follows that for any solution $tmapsto y(t)$ one has $|y'(t)|leq |t|+1$. This allows to conclude that $|tilde y(t)|leq C(1+t^2)$ for a suitable $C>0$; hence $tilde y$ cannot drift away to $pm infty$ in finite time. Since ${cal G}(tilde y)$ will ultimately leave any compact rectangle $K:[-M,M]times[-C(2+M^2),C(2+M^2]$, and cannot do so across the bottom and top edges, it follows that $tilde y$ is defined on all of ${mathbb R}$.






share|cite|improve this answer



















  • 1




    Could you please explain how the bound on the maximal solution was obtained?
    – Joe Man Analysis
    Nov 19 '18 at 10:49














2












2








2






From the Picard theorem one can infer the following: Given an ODE $y'=f(t,y)$ with $f$ defined in an open set $Omegasubset{mathbb R}^2$ and fulfilling the assumptions of the theorem in the neighborhood of each point $(t,y)inOmega$, any solution of an IVP $y(t_0)=y_0$ can be extended to a maximal solution $$tilde y:quad Jmapsto{mathbb R},quad xmapsto tilde y(x)$$
in an unique way. Here $J$ is an open interval which may depend on the given initial point $(t_0,y_0)$. Furthermore the graph ${cal G}(tilde y)subsetOmega$ of this maximal solution will "ultimately" leave any compact set $KsubsetOmega$ given in advance. (For example, the solution cannot develop a $xmapstosin{1over x}$ singularity in the interior of $Omega$.)



In the case at hand we have $f(t,y)=t+sin y$ and $Omega={mathbb R}^2$. It follows that for any solution $tmapsto y(t)$ one has $|y'(t)|leq |t|+1$. This allows to conclude that $|tilde y(t)|leq C(1+t^2)$ for a suitable $C>0$; hence $tilde y$ cannot drift away to $pm infty$ in finite time. Since ${cal G}(tilde y)$ will ultimately leave any compact rectangle $K:[-M,M]times[-C(2+M^2),C(2+M^2]$, and cannot do so across the bottom and top edges, it follows that $tilde y$ is defined on all of ${mathbb R}$.






share|cite|improve this answer














From the Picard theorem one can infer the following: Given an ODE $y'=f(t,y)$ with $f$ defined in an open set $Omegasubset{mathbb R}^2$ and fulfilling the assumptions of the theorem in the neighborhood of each point $(t,y)inOmega$, any solution of an IVP $y(t_0)=y_0$ can be extended to a maximal solution $$tilde y:quad Jmapsto{mathbb R},quad xmapsto tilde y(x)$$
in an unique way. Here $J$ is an open interval which may depend on the given initial point $(t_0,y_0)$. Furthermore the graph ${cal G}(tilde y)subsetOmega$ of this maximal solution will "ultimately" leave any compact set $KsubsetOmega$ given in advance. (For example, the solution cannot develop a $xmapstosin{1over x}$ singularity in the interior of $Omega$.)



In the case at hand we have $f(t,y)=t+sin y$ and $Omega={mathbb R}^2$. It follows that for any solution $tmapsto y(t)$ one has $|y'(t)|leq |t|+1$. This allows to conclude that $|tilde y(t)|leq C(1+t^2)$ for a suitable $C>0$; hence $tilde y$ cannot drift away to $pm infty$ in finite time. Since ${cal G}(tilde y)$ will ultimately leave any compact rectangle $K:[-M,M]times[-C(2+M^2),C(2+M^2]$, and cannot do so across the bottom and top edges, it follows that $tilde y$ is defined on all of ${mathbb R}$.







share|cite|improve this answer














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edited Nov 19 '18 at 10:56

























answered Nov 19 '18 at 10:07









Christian Blatter

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  • 1




    Could you please explain how the bound on the maximal solution was obtained?
    – Joe Man Analysis
    Nov 19 '18 at 10:49














  • 1




    Could you please explain how the bound on the maximal solution was obtained?
    – Joe Man Analysis
    Nov 19 '18 at 10:49








1




1




Could you please explain how the bound on the maximal solution was obtained?
– Joe Man Analysis
Nov 19 '18 at 10:49




Could you please explain how the bound on the maximal solution was obtained?
– Joe Man Analysis
Nov 19 '18 at 10:49


















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