Uniqueness of $y'=cos(y)$ solutions












1












$begingroup$


Consider the equation
begin{cases}
displaystyle y'(x)=cosleft(y(x)right)\
y(0)=0
end{cases}
I've found explicitely $y$ and I know this equation has a unique solution. However is there a way to prove unicity of such an $y$ without using Cauchy-Lipschitz Theorem ? I've tried to suppose two solutions $y_1$ and $y_2$ to show $y_1-y_2=0$. I only found
$$
(y_1-y_2)'=-2sinleft(frac{y_1+y_2}{2}right)sinleft(frac{y_1-y_2}{2}right)
$$
But it did not help me. Can anyone got an idea to show it this way ?






ordinary-differential-equations







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    If you set $z:=y_1-y_2$ and integrate what you found, using that $z(0)=0$, you should be able to conclude via Gronwall's inequality (integral form).
    $endgroup$
    – Giuseppe Negro
    Jan 24 at 19:38










  • $begingroup$
    @GiuseppeNegro I wonder how you would use it ? Because $cosleft(yright)$ does satisfy $leq y$ ?
    $endgroup$
    – Atmos
    Jan 25 at 17:32
















1












$begingroup$


Consider the equation
begin{cases}
displaystyle y'(x)=cosleft(y(x)right)\
y(0)=0
end{cases}
I've found explicitely $y$ and I know this equation has a unique solution. However is there a way to prove unicity of such an $y$ without using Cauchy-Lipschitz Theorem ? I've tried to suppose two solutions $y_1$ and $y_2$ to show $y_1-y_2=0$. I only found
$$
(y_1-y_2)'=-2sinleft(frac{y_1+y_2}{2}right)sinleft(frac{y_1-y_2}{2}right)
$$
But it did not help me. Can anyone got an idea to show it this way ?






ordinary-differential-equations







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    If you set $z:=y_1-y_2$ and integrate what you found, using that $z(0)=0$, you should be able to conclude via Gronwall's inequality (integral form).
    $endgroup$
    – Giuseppe Negro
    Jan 24 at 19:38










  • $begingroup$
    @GiuseppeNegro I wonder how you would use it ? Because $cosleft(yright)$ does satisfy $leq y$ ?
    $endgroup$
    – Atmos
    Jan 25 at 17:32














1












1








1





$begingroup$


Consider the equation
begin{cases}
displaystyle y'(x)=cosleft(y(x)right)\
y(0)=0
end{cases}
I've found explicitely $y$ and I know this equation has a unique solution. However is there a way to prove unicity of such an $y$ without using Cauchy-Lipschitz Theorem ? I've tried to suppose two solutions $y_1$ and $y_2$ to show $y_1-y_2=0$. I only found
$$
(y_1-y_2)'=-2sinleft(frac{y_1+y_2}{2}right)sinleft(frac{y_1-y_2}{2}right)
$$
But it did not help me. Can anyone got an idea to show it this way ?






ordinary-differential-equations







share|cite|improve this question









$endgroup$




Consider the equation
begin{cases}
displaystyle y'(x)=cosleft(y(x)right)\
y(0)=0
end{cases}
I've found explicitely $y$ and I know this equation has a unique solution. However is there a way to prove unicity of such an $y$ without using Cauchy-Lipschitz Theorem ? I've tried to suppose two solutions $y_1$ and $y_2$ to show $y_1-y_2=0$. I only found
$$
(y_1-y_2)'=-2sinleft(frac{y_1+y_2}{2}right)sinleft(frac{y_1-y_2}{2}right)
$$
But it did not help me. Can anyone got an idea to show it this way ?






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 24 at 18:51









AtmosAtmos

4,832420




4,832420








  • 2




    $begingroup$
    If you set $z:=y_1-y_2$ and integrate what you found, using that $z(0)=0$, you should be able to conclude via Gronwall's inequality (integral form).
    $endgroup$
    – Giuseppe Negro
    Jan 24 at 19:38










  • $begingroup$
    @GiuseppeNegro I wonder how you would use it ? Because $cosleft(yright)$ does satisfy $leq y$ ?
    $endgroup$
    – Atmos
    Jan 25 at 17:32














  • 2




    $begingroup$
    If you set $z:=y_1-y_2$ and integrate what you found, using that $z(0)=0$, you should be able to conclude via Gronwall's inequality (integral form).
    $endgroup$
    – Giuseppe Negro
    Jan 24 at 19:38










  • $begingroup$
    @GiuseppeNegro I wonder how you would use it ? Because $cosleft(yright)$ does satisfy $leq y$ ?
    $endgroup$
    – Atmos
    Jan 25 at 17:32








2




2




$begingroup$
If you set $z:=y_1-y_2$ and integrate what you found, using that $z(0)=0$, you should be able to conclude via Gronwall's inequality (integral form).
$endgroup$
– Giuseppe Negro
Jan 24 at 19:38




$begingroup$
If you set $z:=y_1-y_2$ and integrate what you found, using that $z(0)=0$, you should be able to conclude via Gronwall's inequality (integral form).
$endgroup$
– Giuseppe Negro
Jan 24 at 19:38












$begingroup$
@GiuseppeNegro I wonder how you would use it ? Because $cosleft(yright)$ does satisfy $leq y$ ?
$endgroup$
– Atmos
Jan 25 at 17:32




$begingroup$
@GiuseppeNegro I wonder how you would use it ? Because $cosleft(yright)$ does satisfy $leq y$ ?
$endgroup$
– Atmos
Jan 25 at 17:32










1 Answer
1






active

oldest

votes


















1












$begingroup$

In order to prove the unicity in this particular case let $U$ be an open interval around $0in{mathbb R}$, where $|y(x)|<{piover2}$, and consider the auxiliary function
$$u(x):=log{1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}qquad(xin U) .tag{1}$$
One computes
$$u'(x)=left({1over1+sinbigl(y(x)bigr)}-{-1over1-sinbigl(y(x)bigr)}right)cosbigl(y(x)bigr)y'(x)equiv2qquad(xin U) .$$
I think you will accept that this together with $u(0)=0$ implies $u(x)=2x$ for all $xin U$. From $(1)$ we then get
$${1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}=e^{2x}qquad(xin U) ,$$
and this leads to
$$y(x)=arcsinbigl(tanh xbigr)qquad(xin U) .tag{2}$$
As $|tanh x|<1$ for all $xin {mathbb R}$ we see that the |RHS| of $(2)$ is $<{piover2}$ for all $xin{mathbb R}$, so that we may choose $U={mathbb R}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I found that $u'(x)=-sin(y(x))$
    $endgroup$
    – Atmos
    Jan 24 at 20:51











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3086233%2funiqueness-of-y-cosy-solutions%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









1












$begingroup$

In order to prove the unicity in this particular case let $U$ be an open interval around $0in{mathbb R}$, where $|y(x)|<{piover2}$, and consider the auxiliary function
$$u(x):=log{1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}qquad(xin U) .tag{1}$$
One computes
$$u'(x)=left({1over1+sinbigl(y(x)bigr)}-{-1over1-sinbigl(y(x)bigr)}right)cosbigl(y(x)bigr)y'(x)equiv2qquad(xin U) .$$
I think you will accept that this together with $u(0)=0$ implies $u(x)=2x$ for all $xin U$. From $(1)$ we then get
$${1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}=e^{2x}qquad(xin U) ,$$
and this leads to
$$y(x)=arcsinbigl(tanh xbigr)qquad(xin U) .tag{2}$$
As $|tanh x|<1$ for all $xin {mathbb R}$ we see that the |RHS| of $(2)$ is $<{piover2}$ for all $xin{mathbb R}$, so that we may choose $U={mathbb R}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I found that $u'(x)=-sin(y(x))$
    $endgroup$
    – Atmos
    Jan 24 at 20:51
















1












$begingroup$

In order to prove the unicity in this particular case let $U$ be an open interval around $0in{mathbb R}$, where $|y(x)|<{piover2}$, and consider the auxiliary function
$$u(x):=log{1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}qquad(xin U) .tag{1}$$
One computes
$$u'(x)=left({1over1+sinbigl(y(x)bigr)}-{-1over1-sinbigl(y(x)bigr)}right)cosbigl(y(x)bigr)y'(x)equiv2qquad(xin U) .$$
I think you will accept that this together with $u(0)=0$ implies $u(x)=2x$ for all $xin U$. From $(1)$ we then get
$${1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}=e^{2x}qquad(xin U) ,$$
and this leads to
$$y(x)=arcsinbigl(tanh xbigr)qquad(xin U) .tag{2}$$
As $|tanh x|<1$ for all $xin {mathbb R}$ we see that the |RHS| of $(2)$ is $<{piover2}$ for all $xin{mathbb R}$, so that we may choose $U={mathbb R}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I found that $u'(x)=-sin(y(x))$
    $endgroup$
    – Atmos
    Jan 24 at 20:51














1












1








1





$begingroup$

In order to prove the unicity in this particular case let $U$ be an open interval around $0in{mathbb R}$, where $|y(x)|<{piover2}$, and consider the auxiliary function
$$u(x):=log{1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}qquad(xin U) .tag{1}$$
One computes
$$u'(x)=left({1over1+sinbigl(y(x)bigr)}-{-1over1-sinbigl(y(x)bigr)}right)cosbigl(y(x)bigr)y'(x)equiv2qquad(xin U) .$$
I think you will accept that this together with $u(0)=0$ implies $u(x)=2x$ for all $xin U$. From $(1)$ we then get
$${1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}=e^{2x}qquad(xin U) ,$$
and this leads to
$$y(x)=arcsinbigl(tanh xbigr)qquad(xin U) .tag{2}$$
As $|tanh x|<1$ for all $xin {mathbb R}$ we see that the |RHS| of $(2)$ is $<{piover2}$ for all $xin{mathbb R}$, so that we may choose $U={mathbb R}$.






share|cite|improve this answer











$endgroup$



In order to prove the unicity in this particular case let $U$ be an open interval around $0in{mathbb R}$, where $|y(x)|<{piover2}$, and consider the auxiliary function
$$u(x):=log{1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}qquad(xin U) .tag{1}$$
One computes
$$u'(x)=left({1over1+sinbigl(y(x)bigr)}-{-1over1-sinbigl(y(x)bigr)}right)cosbigl(y(x)bigr)y'(x)equiv2qquad(xin U) .$$
I think you will accept that this together with $u(0)=0$ implies $u(x)=2x$ for all $xin U$. From $(1)$ we then get
$${1+sinbigl(y(x)bigr)over1-sinbigl(y(x)bigr)}=e^{2x}qquad(xin U) ,$$
and this leads to
$$y(x)=arcsinbigl(tanh xbigr)qquad(xin U) .tag{2}$$
As $|tanh x|<1$ for all $xin {mathbb R}$ we see that the |RHS| of $(2)$ is $<{piover2}$ for all $xin{mathbb R}$, so that we may choose $U={mathbb R}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 25 at 15:07

























answered Jan 24 at 19:03









Christian BlatterChristian Blatter

175k8115327




175k8115327












  • $begingroup$
    I found that $u'(x)=-sin(y(x))$
    $endgroup$
    – Atmos
    Jan 24 at 20:51


















  • $begingroup$
    I found that $u'(x)=-sin(y(x))$
    $endgroup$
    – Atmos
    Jan 24 at 20:51
















$begingroup$
I found that $u'(x)=-sin(y(x))$
$endgroup$
– Atmos
Jan 24 at 20:51




$begingroup$
I found that $u'(x)=-sin(y(x))$
$endgroup$
– Atmos
Jan 24 at 20:51


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3086233%2funiqueness-of-y-cosy-solutions%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

Image
28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
Read more

Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

Image
up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
Read more

A Topological Invariant for $pi_3(U(n))$

Image
3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
Read more