Cauchy problem with specific value of starting point, find limit












0












$begingroup$


I have the following Cauchy problem:



$$ y' = vert y vert - arctan{e^x} $$
$$ y(0) = y_0 $$



I want to prove that there exists a specific value of $y_0$ such that:



$$ lim_{xtoinfty} y(x) = frac{pi}{2}.$$



I am not too sure how to even start on this, beside saying that f(x,y) is Lipschizt and therefore I have local uniqueness and existence of solution which can be extended globally to $mathbb{R}$. Any help?






ordinary-differential-equations cauchy-problem







share|cite|improve this question











$endgroup$












  • $begingroup$
    As long as $y geq 0$ you can study the linear ODE $y' = y - arctan(e^x)$.
    $endgroup$
    – Christoph
    Jan 29 at 12:10
















0












$begingroup$


I have the following Cauchy problem:



$$ y' = vert y vert - arctan{e^x} $$
$$ y(0) = y_0 $$



I want to prove that there exists a specific value of $y_0$ such that:



$$ lim_{xtoinfty} y(x) = frac{pi}{2}.$$



I am not too sure how to even start on this, beside saying that f(x,y) is Lipschizt and therefore I have local uniqueness and existence of solution which can be extended globally to $mathbb{R}$. Any help?






ordinary-differential-equations cauchy-problem







share|cite|improve this question











$endgroup$












  • $begingroup$
    As long as $y geq 0$ you can study the linear ODE $y' = y - arctan(e^x)$.
    $endgroup$
    – Christoph
    Jan 29 at 12:10














0












0








0





$begingroup$


I have the following Cauchy problem:



$$ y' = vert y vert - arctan{e^x} $$
$$ y(0) = y_0 $$



I want to prove that there exists a specific value of $y_0$ such that:



$$ lim_{xtoinfty} y(x) = frac{pi}{2}.$$



I am not too sure how to even start on this, beside saying that f(x,y) is Lipschizt and therefore I have local uniqueness and existence of solution which can be extended globally to $mathbb{R}$. Any help?






ordinary-differential-equations cauchy-problem







share|cite|improve this question











$endgroup$




I have the following Cauchy problem:



$$ y' = vert y vert - arctan{e^x} $$
$$ y(0) = y_0 $$



I want to prove that there exists a specific value of $y_0$ such that:



$$ lim_{xtoinfty} y(x) = frac{pi}{2}.$$



I am not too sure how to even start on this, beside saying that f(x,y) is Lipschizt and therefore I have local uniqueness and existence of solution which can be extended globally to $mathbb{R}$. Any help?






ordinary-differential-equations cauchy-problem




ordinary-differential-equations cauchy-problem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 29 at 11:44









Julián Aguirre

69.6k24297




69.6k24297










asked Jan 29 at 10:14









qcc101qcc101

629213




629213












  • $begingroup$
    As long as $y geq 0$ you can study the linear ODE $y' = y - arctan(e^x)$.
    $endgroup$
    – Christoph
    Jan 29 at 12:10


















  • $begingroup$
    As long as $y geq 0$ you can study the linear ODE $y' = y - arctan(e^x)$.
    $endgroup$
    – Christoph
    Jan 29 at 12:10
















$begingroup$
As long as $y geq 0$ you can study the linear ODE $y' = y - arctan(e^x)$.
$endgroup$
– Christoph
Jan 29 at 12:10




$begingroup$
As long as $y geq 0$ you can study the linear ODE $y' = y - arctan(e^x)$.
$endgroup$
– Christoph
Jan 29 at 12:10










1 Answer
1






active

oldest

votes


















0












$begingroup$

Let $y($ be a solution and suppose that $y(x)=0$ for some $xge0$. Then $y'(x)<0$. This implies that once a solution takes a negative value, it remains negative, Thus, the only way to solve the problem is to look for positive solutions. This means solving the linear equation $y'=y-arctan(e^x)$. Its solution is
$$
y(x)=e^xBigl(y_0-int_0^xe^{-t}arctan(e^t),dtBigr).
$$
This will be positive if and only if
$$
y_0ge int_0^infty e^{-t}arctan(e^t),dt=a.
$$
If $y_0>a$, then $lim_{xtoinfty}y(x)=infty$. If $y_0=a$, then
$$
lim_{xtoinfty}y(x)=lim_{xtoinfty}e^xint_x^infty e^{-t}arctan(e^t),dt=fracpi2.
$$
Note. Although it is not needed for the argument, the actual value of $a$ is
$$
frac{pi+log4}{4}.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If $y_0 = a $ wouldn't y(x) be always 0?
    $endgroup$
    – qcc101
    Jan 31 at 10:04












  • $begingroup$
    No. If $x<infty$, then $y_0>int_0^xe^{-t}arctan(e^t),dt$.
    $endgroup$
    – Julián Aguirre
    Jan 31 at 12:30












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%2f3092003%2fcauchy-problem-with-specific-value-of-starting-point-find-limit%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









0












$begingroup$

Let $y($ be a solution and suppose that $y(x)=0$ for some $xge0$. Then $y'(x)<0$. This implies that once a solution takes a negative value, it remains negative, Thus, the only way to solve the problem is to look for positive solutions. This means solving the linear equation $y'=y-arctan(e^x)$. Its solution is
$$
y(x)=e^xBigl(y_0-int_0^xe^{-t}arctan(e^t),dtBigr).
$$
This will be positive if and only if
$$
y_0ge int_0^infty e^{-t}arctan(e^t),dt=a.
$$
If $y_0>a$, then $lim_{xtoinfty}y(x)=infty$. If $y_0=a$, then
$$
lim_{xtoinfty}y(x)=lim_{xtoinfty}e^xint_x^infty e^{-t}arctan(e^t),dt=fracpi2.
$$
Note. Although it is not needed for the argument, the actual value of $a$ is
$$
frac{pi+log4}{4}.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If $y_0 = a $ wouldn't y(x) be always 0?
    $endgroup$
    – qcc101
    Jan 31 at 10:04












  • $begingroup$
    No. If $x<infty$, then $y_0>int_0^xe^{-t}arctan(e^t),dt$.
    $endgroup$
    – Julián Aguirre
    Jan 31 at 12:30
















0












$begingroup$

Let $y($ be a solution and suppose that $y(x)=0$ for some $xge0$. Then $y'(x)<0$. This implies that once a solution takes a negative value, it remains negative, Thus, the only way to solve the problem is to look for positive solutions. This means solving the linear equation $y'=y-arctan(e^x)$. Its solution is
$$
y(x)=e^xBigl(y_0-int_0^xe^{-t}arctan(e^t),dtBigr).
$$
This will be positive if and only if
$$
y_0ge int_0^infty e^{-t}arctan(e^t),dt=a.
$$
If $y_0>a$, then $lim_{xtoinfty}y(x)=infty$. If $y_0=a$, then
$$
lim_{xtoinfty}y(x)=lim_{xtoinfty}e^xint_x^infty e^{-t}arctan(e^t),dt=fracpi2.
$$
Note. Although it is not needed for the argument, the actual value of $a$ is
$$
frac{pi+log4}{4}.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If $y_0 = a $ wouldn't y(x) be always 0?
    $endgroup$
    – qcc101
    Jan 31 at 10:04












  • $begingroup$
    No. If $x<infty$, then $y_0>int_0^xe^{-t}arctan(e^t),dt$.
    $endgroup$
    – Julián Aguirre
    Jan 31 at 12:30














0












0








0





$begingroup$

Let $y($ be a solution and suppose that $y(x)=0$ for some $xge0$. Then $y'(x)<0$. This implies that once a solution takes a negative value, it remains negative, Thus, the only way to solve the problem is to look for positive solutions. This means solving the linear equation $y'=y-arctan(e^x)$. Its solution is
$$
y(x)=e^xBigl(y_0-int_0^xe^{-t}arctan(e^t),dtBigr).
$$
This will be positive if and only if
$$
y_0ge int_0^infty e^{-t}arctan(e^t),dt=a.
$$
If $y_0>a$, then $lim_{xtoinfty}y(x)=infty$. If $y_0=a$, then
$$
lim_{xtoinfty}y(x)=lim_{xtoinfty}e^xint_x^infty e^{-t}arctan(e^t),dt=fracpi2.
$$
Note. Although it is not needed for the argument, the actual value of $a$ is
$$
frac{pi+log4}{4}.
$$






share|cite|improve this answer









$endgroup$



Let $y($ be a solution and suppose that $y(x)=0$ for some $xge0$. Then $y'(x)<0$. This implies that once a solution takes a negative value, it remains negative, Thus, the only way to solve the problem is to look for positive solutions. This means solving the linear equation $y'=y-arctan(e^x)$. Its solution is
$$
y(x)=e^xBigl(y_0-int_0^xe^{-t}arctan(e^t),dtBigr).
$$
This will be positive if and only if
$$
y_0ge int_0^infty e^{-t}arctan(e^t),dt=a.
$$
If $y_0>a$, then $lim_{xtoinfty}y(x)=infty$. If $y_0=a$, then
$$
lim_{xtoinfty}y(x)=lim_{xtoinfty}e^xint_x^infty e^{-t}arctan(e^t),dt=fracpi2.
$$
Note. Although it is not needed for the argument, the actual value of $a$ is
$$
frac{pi+log4}{4}.
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 29 at 21:01









Julián AguirreJulián Aguirre

69.6k24297




69.6k24297












  • $begingroup$
    If $y_0 = a $ wouldn't y(x) be always 0?
    $endgroup$
    – qcc101
    Jan 31 at 10:04












  • $begingroup$
    No. If $x<infty$, then $y_0>int_0^xe^{-t}arctan(e^t),dt$.
    $endgroup$
    – Julián Aguirre
    Jan 31 at 12:30


















  • $begingroup$
    If $y_0 = a $ wouldn't y(x) be always 0?
    $endgroup$
    – qcc101
    Jan 31 at 10:04












  • $begingroup$
    No. If $x<infty$, then $y_0>int_0^xe^{-t}arctan(e^t),dt$.
    $endgroup$
    – Julián Aguirre
    Jan 31 at 12:30
















$begingroup$
If $y_0 = a $ wouldn't y(x) be always 0?
$endgroup$
– qcc101
Jan 31 at 10:04






$begingroup$
If $y_0 = a $ wouldn't y(x) be always 0?
$endgroup$
– qcc101
Jan 31 at 10:04














$begingroup$
No. If $x<infty$, then $y_0>int_0^xe^{-t}arctan(e^t),dt$.
$endgroup$
– Julián Aguirre
Jan 31 at 12:30




$begingroup$
No. If $x<infty$, then $y_0>int_0^xe^{-t}arctan(e^t),dt$.
$endgroup$
– Julián Aguirre
Jan 31 at 12:30


















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%2f3092003%2fcauchy-problem-with-specific-value-of-starting-point-find-limit%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

MongoDB - Not Authorized To Execute Command

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
Read more

in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

Image
1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
Read more

Npm cannot find a required file even through it is in the searched directory

Image
up vote 0 down vote favorite I got that error: users-Air:pdfuploader user$ npm run start client@0.1.0 start /Users/user/Documents/pdfuploader react-scripts start Could not find a required file. Name: index.html Searched in: /Users/user/Documents/pdfuploader/public npm ERR! code ELIFECYCLE npm ERR! errno 1 npm ERR! client@0.1.0 start: `react-scripts start` npm ERR! Exit status 1 npm ERR! npm ERR! Failed at the client@0.1.0 start script. npm ERR! This is probably not a problem with npm. There is likely additional logging output above. npm ERR! A complete log of this run can be found in: npm ERR! /Users/user/.npm/_logs/2018-11-19T10_19_14_555Z-debug.log users-Air:pdfuploader user$ And the file is in /Users/user/Documents/pdfuploader/public as you can see in my github repository: https://github.c...
Read more