Lipschitz continuity of $e^{sin}$
We want to use the Picard-Lindelöf-Theorem to show that the ODE
$$y'=mathrm{e}^{sin(ty)}$$
has a unique solution on $mathbb{R}$ with the initial value $y(0)=0$. As far as I know, we have to show that the right hand side is Lipschitz-continuous with respect to $y$. So we need to proof
$$vert mathrm{e}^{sin(tx)}-mathrm{e}^{sin(ty)}vert leq L vert x-yvert$$
for all $t,x,y in mathbb{R}$. I don't get it, to show this. I heard that because the right hand side is differentiable, that is eventually possible to show this with the mean-value theorem or gradient. But I don't know whether this works because I think the derivative is not bounded. What can one do?
real-analysis differential-equations multivariable-calculus lipschitz-functions
edited Jan 5 at 22:52
Gnumbertester
1455
asked Jan 5 at 21:43
JanJan
116113
add a comment |
We want to use the Picard-Lindelöf-Theorem to show that the ODE
$$y'=mathrm{e}^{sin(ty)}$$
has a unique solution on $mathbb{R}$ with the initial value $y(0)=0$. As far as I know, we have to show that the right hand side is Lipschitz-continuous with respect to $y$. So we need to proof
$$vert mathrm{e}^{sin(tx)}-mathrm{e}^{sin(ty)}vert leq L vert x-yvert$$
for all $t,x,y in mathbb{R}$. I don't get it, to show this. I heard that because the right hand side is differentiable, that is eventually possible to show this with the mean-value theorem or gradient. But I don't know whether this works because I think the derivative is not bounded. What can one do?
real-analysis differential-equations multivariable-calculus lipschitz-functions
edited Jan 5 at 22:52
Gnumbertester
1455
asked Jan 5 at 21:43
JanJan
116113
What makes you think that the derivative is not bounded?
– A.Γ.
Jan 5 at 21:54
@A.Γ. We have $vert partial_y f(t,y)vert = vert t cos(ty)mathrm{e}^{sin(ty)}vert$ and then we can make $t$ arbitrarily large? Or is this not allowed?
– Jan
Jan 5 at 22:01
But it is bounded on $[0,T]$ for every $T>0$.
– A.Γ.
Jan 5 at 22:48
@A.Γ. So $t$ is fixed, as I understand. Ok, it is bounded. How does the rest of the argumentation work? The boundary constant is $tmathrm{e}$, right?
– Jan
Jan 5 at 22:51
Two possibilities: 1. Say that the function is uniformly Lipschitz on $[0,T]$ and use the global version of the theorem. Then say that $T$ is arbitrary. or 2. Prove existence of a solution on $[0,+infty)$ using $|y'(t)|le e$ (no blow-up) and then use the local version of the theorem to prove uniqueness.
– A.Γ.
Jan 5 at 22:58
add a comment |
We want to use the Picard-Lindelöf-Theorem to show that the ODE
$$y'=mathrm{e}^{sin(ty)}$$
has a unique solution on $mathbb{R}$ with the initial value $y(0)=0$. As far as I know, we have to show that the right hand side is Lipschitz-continuous with respect to $y$. So we need to proof
$$vert mathrm{e}^{sin(tx)}-mathrm{e}^{sin(ty)}vert leq L vert x-yvert$$
for all $t,x,y in mathbb{R}$. I don't get it, to show this. I heard that because the right hand side is differentiable, that is eventually possible to show this with the mean-value theorem or gradient. But I don't know whether this works because I think the derivative is not bounded. What can one do?
real-analysis differential-equations multivariable-calculus lipschitz-functions
edited Jan 5 at 22:52
Gnumbertester
1455
asked Jan 5 at 21:43
JanJan
116113
We want to use the Picard-Lindelöf-Theorem to show that the ODE
$$y'=mathrm{e}^{sin(ty)}$$
has a unique solution on $mathbb{R}$ with the initial value $y(0)=0$. As far as I know, we have to show that the right hand side is Lipschitz-continuous with respect to $y$. So we need to proof
$$vert mathrm{e}^{sin(tx)}-mathrm{e}^{sin(ty)}vert leq L vert x-yvert$$
for all $t,x,y in mathbb{R}$. I don't get it, to show this. I heard that because the right hand side is differentiable, that is eventually possible to show this with the mean-value theorem or gradient. But I don't know whether this works because I think the derivative is not bounded. What can one do?
real-analysis differential-equations multivariable-calculus lipschitz-functions
real-analysis differential-equations multivariable-calculus lipschitz-functions
edited Jan 5 at 22:52
Gnumbertester
1455
asked Jan 5 at 21:43
JanJan
116113
edited Jan 5 at 22:52
Gnumbertester
1455
asked Jan 5 at 21:43
JanJan
116113
edited Jan 5 at 22:52
Gnumbertester
1455
edited Jan 5 at 22:52
Gnumbertester
1455
edited Jan 5 at 22:52
Gnumbertester
1455
1455
asked Jan 5 at 21:43
JanJan
116113
asked Jan 5 at 21:43
JanJan
116113
asked Jan 5 at 21:43
JanJan
116113
116113
What makes you think that the derivative is not bounded?
– A.Γ.
Jan 5 at 21:54
@A.Γ. We have $vert partial_y f(t,y)vert = vert t cos(ty)mathrm{e}^{sin(ty)}vert$ and then we can make $t$ arbitrarily large? Or is this not allowed?
– Jan
Jan 5 at 22:01
But it is bounded on $[0,T]$ for every $T>0$.
– A.Γ.
Jan 5 at 22:48
@A.Γ. So $t$ is fixed, as I understand. Ok, it is bounded. How does the rest of the argumentation work? The boundary constant is $tmathrm{e}$, right?
– Jan
Jan 5 at 22:51
Two possibilities: 1. Say that the function is uniformly Lipschitz on $[0,T]$ and use the global version of the theorem. Then say that $T$ is arbitrary. or 2. Prove existence of a solution on $[0,+infty)$ using $|y'(t)|le e$ (no blow-up) and then use the local version of the theorem to prove uniqueness.
– A.Γ.
Jan 5 at 22:58
add a comment |
What makes you think that the derivative is not bounded?
– A.Γ.
Jan 5 at 21:54
@A.Γ. We have $vert partial_y f(t,y)vert = vert t cos(ty)mathrm{e}^{sin(ty)}vert$ and then we can make $t$ arbitrarily large? Or is this not allowed?
– Jan
Jan 5 at 22:01
But it is bounded on $[0,T]$ for every $T>0$.
– A.Γ.
Jan 5 at 22:48
@A.Γ. So $t$ is fixed, as I understand. Ok, it is bounded. How does the rest of the argumentation work? The boundary constant is $tmathrm{e}$, right?
– Jan
Jan 5 at 22:51
Two possibilities: 1. Say that the function is uniformly Lipschitz on $[0,T]$ and use the global version of the theorem. Then say that $T$ is arbitrary. or 2. Prove existence of a solution on $[0,+infty)$ using $|y'(t)|le e$ (no blow-up) and then use the local version of the theorem to prove uniqueness.
– A.Γ.
Jan 5 at 22:58
What makes you think that the derivative is not bounded?
– A.Γ.
Jan 5 at 21:54
What makes you think that the derivative is not bounded?
– A.Γ.
Jan 5 at 21:54
@A.Γ. We have $vert partial_y f(t,y)vert = vert t cos(ty)mathrm{e}^{sin(ty)}vert$ and then we can make $t$ arbitrarily large? Or is this not allowed?
– Jan
Jan 5 at 22:01
@A.Γ. We have $vert partial_y f(t,y)vert = vert t cos(ty)mathrm{e}^{sin(ty)}vert$ and then we can make $t$ arbitrarily large? Or is this not allowed?
– Jan
Jan 5 at 22:01
But it is bounded on $[0,T]$ for every $T>0$.
– A.Γ.
Jan 5 at 22:48
But it is bounded on $[0,T]$ for every $T>0$.
– A.Γ.
Jan 5 at 22:48
@A.Γ. So $t$ is fixed, as I understand. Ok, it is bounded. How does the rest of the argumentation work? The boundary constant is $tmathrm{e}$, right?
– Jan
Jan 5 at 22:51
@A.Γ. So $t$ is fixed, as I understand. Ok, it is bounded. How does the rest of the argumentation work? The boundary constant is $tmathrm{e}$, right?
– Jan
Jan 5 at 22:51
Two possibilities: 1. Say that the function is uniformly Lipschitz on $[0,T]$ and use the global version of the theorem. Then say that $T$ is arbitrary. or 2. Prove existence of a solution on $[0,+infty)$ using $|y'(t)|le e$ (no blow-up) and then use the local version of the theorem to prove uniqueness.
– A.Γ.
Jan 5 at 22:58
Two possibilities: 1. Say that the function is uniformly Lipschitz on $[0,T]$ and use the global version of the theorem. Then say that $T$ is arbitrary. or 2. Prove existence of a solution on $[0,+infty)$ using $|y'(t)|le e$ (no blow-up) and then use the local version of the theorem to prove uniqueness.
– A.Γ.
Jan 5 at 22:58
add a comment |
1 Answer
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The actual hypothesis required by the Picard-Linddelöf theorem, if I am not mistaken, is: for every pair $(t,y)$, there exists some open intervals $t in I$, $y in J$ and some constant $K > 0$, such that if $t’ in I$ and $y’,y’’ in J$, $|e^{sin(ty’)}-e^{sin(ty’’)}| leq K|y’-y’’|$.
(ie the function is locally Lipschitz continuous wrt y).
So here, you take $I=]t-1,t+1[$, $J=mathbb{R}$, and recall that $u longmapsto e^{sin(su)}$ is $e|s|$-Lipschitz continuous.
answered Jan 5 at 22:27
MindlackMindlack
2,16717
How do you show the Lipschitz-continuity?
– Jan
Jan 5 at 22:57
The $sin$ function has its range in $[-1,1]$ and is $1$-Lipschitz continuous, and $x longmapsto e^x$ is $e$-Lipschitz continuous on $(-infty,1]$.
– Mindlack
Jan 5 at 23:00
add a comment |
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1 Answer
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1 Answer
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oldest
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The actual hypothesis required by the Picard-Linddelöf theorem, if I am not mistaken, is: for every pair $(t,y)$, there exists some open intervals $t in I$, $y in J$ and some constant $K > 0$, such that if $t’ in I$ and $y’,y’’ in J$, $|e^{sin(ty’)}-e^{sin(ty’’)}| leq K|y’-y’’|$.
(ie the function is locally Lipschitz continuous wrt y).
So here, you take $I=]t-1,t+1[$, $J=mathbb{R}$, and recall that $u longmapsto e^{sin(su)}$ is $e|s|$-Lipschitz continuous.
answered Jan 5 at 22:27
MindlackMindlack
2,16717
How do you show the Lipschitz-continuity?
– Jan
Jan 5 at 22:57
The $sin$ function has its range in $[-1,1]$ and is $1$-Lipschitz continuous, and $x longmapsto e^x$ is $e$-Lipschitz continuous on $(-infty,1]$.
– Mindlack
Jan 5 at 23:00
add a comment |
The actual hypothesis required by the Picard-Linddelöf theorem, if I am not mistaken, is: for every pair $(t,y)$, there exists some open intervals $t in I$, $y in J$ and some constant $K > 0$, such that if $t’ in I$ and $y’,y’’ in J$, $|e^{sin(ty’)}-e^{sin(ty’’)}| leq K|y’-y’’|$.
(ie the function is locally Lipschitz continuous wrt y).
So here, you take $I=]t-1,t+1[$, $J=mathbb{R}$, and recall that $u longmapsto e^{sin(su)}$ is $e|s|$-Lipschitz continuous.
answered Jan 5 at 22:27
MindlackMindlack
2,16717
How do you show the Lipschitz-continuity?
– Jan
Jan 5 at 22:57
The $sin$ function has its range in $[-1,1]$ and is $1$-Lipschitz continuous, and $x longmapsto e^x$ is $e$-Lipschitz continuous on $(-infty,1]$.
– Mindlack
Jan 5 at 23:00
add a comment |
The actual hypothesis required by the Picard-Linddelöf theorem, if I am not mistaken, is: for every pair $(t,y)$, there exists some open intervals $t in I$, $y in J$ and some constant $K > 0$, such that if $t’ in I$ and $y’,y’’ in J$, $|e^{sin(ty’)}-e^{sin(ty’’)}| leq K|y’-y’’|$.
(ie the function is locally Lipschitz continuous wrt y).
So here, you take $I=]t-1,t+1[$, $J=mathbb{R}$, and recall that $u longmapsto e^{sin(su)}$ is $e|s|$-Lipschitz continuous.
answered Jan 5 at 22:27
MindlackMindlack
2,16717
The actual hypothesis required by the Picard-Linddelöf theorem, if I am not mistaken, is: for every pair $(t,y)$, there exists some open intervals $t in I$, $y in J$ and some constant $K > 0$, such that if $t’ in I$ and $y’,y’’ in J$, $|e^{sin(ty’)}-e^{sin(ty’’)}| leq K|y’-y’’|$.
(ie the function is locally Lipschitz continuous wrt y).
So here, you take $I=]t-1,t+1[$, $J=mathbb{R}$, and recall that $u longmapsto e^{sin(su)}$ is $e|s|$-Lipschitz continuous.
answered Jan 5 at 22:27
MindlackMindlack
2,16717
answered Jan 5 at 22:27
MindlackMindlack
2,16717
answered Jan 5 at 22:27
MindlackMindlack
2,16717
answered Jan 5 at 22:27
MindlackMindlack
2,16717
2,16717
How do you show the Lipschitz-continuity?
– Jan
Jan 5 at 22:57
The $sin$ function has its range in $[-1,1]$ and is $1$-Lipschitz continuous, and $x longmapsto e^x$ is $e$-Lipschitz continuous on $(-infty,1]$.
– Mindlack
Jan 5 at 23:00
add a comment |
How do you show the Lipschitz-continuity?
– Jan
Jan 5 at 22:57
The $sin$ function has its range in $[-1,1]$ and is $1$-Lipschitz continuous, and $x longmapsto e^x$ is $e$-Lipschitz continuous on $(-infty,1]$.
– Mindlack
Jan 5 at 23:00
How do you show the Lipschitz-continuity?
– Jan
Jan 5 at 22:57
How do you show the Lipschitz-continuity?
– Jan
Jan 5 at 22:57
The $sin$ function has its range in $[-1,1]$ and is $1$-Lipschitz continuous, and $x longmapsto e^x$ is $e$-Lipschitz continuous on $(-infty,1]$.
– Mindlack
Jan 5 at 23:00
The $sin$ function has its range in $[-1,1]$ and is $1$-Lipschitz continuous, and $x longmapsto e^x$ is $e$-Lipschitz continuous on $(-infty,1]$.
– Mindlack
Jan 5 at 23:00
add a comment |
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What makes you think that the derivative is not bounded?
– A.Γ.
Jan 5 at 21:54
@A.Γ. We have $vert partial_y f(t,y)vert = vert t cos(ty)mathrm{e}^{sin(ty)}vert$ and then we can make $t$ arbitrarily large? Or is this not allowed?
– Jan
Jan 5 at 22:01
But it is bounded on $[0,T]$ for every $T>0$.
– A.Γ.
Jan 5 at 22:48
@A.Γ. So $t$ is fixed, as I understand. Ok, it is bounded. How does the rest of the argumentation work? The boundary constant is $tmathrm{e}$, right?
– Jan
Jan 5 at 22:51
Two possibilities: 1. Say that the function is uniformly Lipschitz on $[0,T]$ and use the global version of the theorem. Then say that $T$ is arbitrary. or 2. Prove existence of a solution on $[0,+infty)$ using $|y'(t)|le e$ (no blow-up) and then use the local version of the theorem to prove uniqueness.
– A.Γ.
Jan 5 at 22:58