Lipschitz continuity of $e^{sin}$
$begingroup$
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 ordinary-differential-equations multivariable-calculus lipschitz-functions
$endgroup$
add a comment |
$begingroup$
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 ordinary-differential-equations multivariable-calculus lipschitz-functions
$endgroup$
$begingroup$
What makes you think that the derivative is not bounded?
$endgroup$
– A.Γ.
Jan 5 at 21:54
$begingroup$
@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?
$endgroup$
– Jan
Jan 5 at 22:01
$begingroup$
But it is bounded on $[0,T]$ for every $T>0$.
$endgroup$
– A.Γ.
Jan 5 at 22:48
$begingroup$
@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?
$endgroup$
– Jan
Jan 5 at 22:51
$begingroup$
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.
$endgroup$
– A.Γ.
Jan 5 at 22:58
add a comment |
$begingroup$
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 ordinary-differential-equations multivariable-calculus lipschitz-functions
$endgroup$
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 ordinary-differential-equations multivariable-calculus lipschitz-functions
real-analysis ordinary-differential-equations multivariable-calculus lipschitz-functions
edited Jan 5 at 22:52
Gnumbertester
6841114
6841114
asked Jan 5 at 21:43
JanJan
658315
658315
$begingroup$
What makes you think that the derivative is not bounded?
$endgroup$
– A.Γ.
Jan 5 at 21:54
$begingroup$
@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?
$endgroup$
– Jan
Jan 5 at 22:01
$begingroup$
But it is bounded on $[0,T]$ for every $T>0$.
$endgroup$
– A.Γ.
Jan 5 at 22:48
$begingroup$
@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?
$endgroup$
– Jan
Jan 5 at 22:51
$begingroup$
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.
$endgroup$
– A.Γ.
Jan 5 at 22:58
add a comment |
$begingroup$
What makes you think that the derivative is not bounded?
$endgroup$
– A.Γ.
Jan 5 at 21:54
$begingroup$
@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?
$endgroup$
– Jan
Jan 5 at 22:01
$begingroup$
But it is bounded on $[0,T]$ for every $T>0$.
$endgroup$
– A.Γ.
Jan 5 at 22:48
$begingroup$
@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?
$endgroup$
– Jan
Jan 5 at 22:51
$begingroup$
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.
$endgroup$
– A.Γ.
Jan 5 at 22:58
$begingroup$
What makes you think that the derivative is not bounded?
$endgroup$
– A.Γ.
Jan 5 at 21:54
$begingroup$
What makes you think that the derivative is not bounded?
$endgroup$
– A.Γ.
Jan 5 at 21:54
$begingroup$
@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?
$endgroup$
– Jan
Jan 5 at 22:01
$begingroup$
@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?
$endgroup$
– Jan
Jan 5 at 22:01
$begingroup$
But it is bounded on $[0,T]$ for every $T>0$.
$endgroup$
– A.Γ.
Jan 5 at 22:48
$begingroup$
But it is bounded on $[0,T]$ for every $T>0$.
$endgroup$
– A.Γ.
Jan 5 at 22:48
$begingroup$
@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?
$endgroup$
– Jan
Jan 5 at 22:51
$begingroup$
@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?
$endgroup$
– Jan
Jan 5 at 22:51
$begingroup$
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.
$endgroup$
– A.Γ.
Jan 5 at 22:58
$begingroup$
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.
$endgroup$
– A.Γ.
Jan 5 at 22:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The actual hypothesis required by the Picard-Lindelöf 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.
$endgroup$
$begingroup$
How do you show the Lipschitz-continuity?
$endgroup$
– Jan
Jan 5 at 22:57
1
$begingroup$
The $sin$ function has its range in $[-1,1]$ and is $1$-Lipschitz continuous, and $x longmapsto e^x$ is $e$-Lipschitz continuous on $(-infty,1]$.
$endgroup$
– Mindlack
Jan 5 at 23:00
add a comment |
Your Answer
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%2f3063223%2flipschitz-continuity-of-e-sin%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
$begingroup$
The actual hypothesis required by the Picard-Lindelöf 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.
$endgroup$
$begingroup$
How do you show the Lipschitz-continuity?
$endgroup$
– Jan
Jan 5 at 22:57
1
$begingroup$
The $sin$ function has its range in $[-1,1]$ and is $1$-Lipschitz continuous, and $x longmapsto e^x$ is $e$-Lipschitz continuous on $(-infty,1]$.
$endgroup$
– Mindlack
Jan 5 at 23:00
add a comment |
$begingroup$
The actual hypothesis required by the Picard-Lindelöf 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.
$endgroup$
$begingroup$
How do you show the Lipschitz-continuity?
$endgroup$
– Jan
Jan 5 at 22:57
1
$begingroup$
The $sin$ function has its range in $[-1,1]$ and is $1$-Lipschitz continuous, and $x longmapsto e^x$ is $e$-Lipschitz continuous on $(-infty,1]$.
$endgroup$
– Mindlack
Jan 5 at 23:00
add a comment |
$begingroup$
The actual hypothesis required by the Picard-Lindelöf 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.
$endgroup$
The actual hypothesis required by the Picard-Lindelöf 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.
edited Feb 3 at 17:38
answered Jan 5 at 22:27
MindlackMindlack
4,900211
4,900211
$begingroup$
How do you show the Lipschitz-continuity?
$endgroup$
– Jan
Jan 5 at 22:57
1
$begingroup$
The $sin$ function has its range in $[-1,1]$ and is $1$-Lipschitz continuous, and $x longmapsto e^x$ is $e$-Lipschitz continuous on $(-infty,1]$.
$endgroup$
– Mindlack
Jan 5 at 23:00
add a comment |
$begingroup$
How do you show the Lipschitz-continuity?
$endgroup$
– Jan
Jan 5 at 22:57
1
$begingroup$
The $sin$ function has its range in $[-1,1]$ and is $1$-Lipschitz continuous, and $x longmapsto e^x$ is $e$-Lipschitz continuous on $(-infty,1]$.
$endgroup$
– Mindlack
Jan 5 at 23:00
$begingroup$
How do you show the Lipschitz-continuity?
$endgroup$
– Jan
Jan 5 at 22:57
$begingroup$
How do you show the Lipschitz-continuity?
$endgroup$
– Jan
Jan 5 at 22:57
1
1
$begingroup$
The $sin$ function has its range in $[-1,1]$ and is $1$-Lipschitz continuous, and $x longmapsto e^x$ is $e$-Lipschitz continuous on $(-infty,1]$.
$endgroup$
– Mindlack
Jan 5 at 23:00
$begingroup$
The $sin$ function has its range in $[-1,1]$ and is $1$-Lipschitz continuous, and $x longmapsto e^x$ is $e$-Lipschitz continuous on $(-infty,1]$.
$endgroup$
– Mindlack
Jan 5 at 23:00
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%2f3063223%2flipschitz-continuity-of-e-sin%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
$begingroup$
What makes you think that the derivative is not bounded?
$endgroup$
– A.Γ.
Jan 5 at 21:54
$begingroup$
@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?
$endgroup$
– Jan
Jan 5 at 22:01
$begingroup$
But it is bounded on $[0,T]$ for every $T>0$.
$endgroup$
– A.Γ.
Jan 5 at 22:48
$begingroup$
@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?
$endgroup$
– Jan
Jan 5 at 22:51
$begingroup$
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.
$endgroup$
– A.Γ.
Jan 5 at 22:58