If $vert F(t,x)vert leq alpha (t)vert xvert + beta(t)$ the maximal solutions are global for an ODE












1












$begingroup$


Let $F: mathbb{R} timesmathbb{R}^2 rightarrow mathbb{R}$ localy lipschitz in its second variable. Let the Cauchy problem be:



$$x'=F(t,x),\ x(0) = x_0 $$
Let $alpha : mathbb{R} rightarrow mathbb{R}$ and $beta : mathbb{R} rightarrow mathbb{R}$. Show that if $forall x in mathbb{R}^2, forall t in mathbb{R}$ $$|F(t,x)| leq alpha(t)|x| + beta (t) $$
then any maximal solution is global.



I am unable to do this. I suppose I have to either show that $F$ is uniformly bounded or that it is uniformly Lipschitz, but I have no clue how to do either. Any help will be appreciated.






ordinary-differential-equations cauchy-problem







share|cite|improve this question











$endgroup$












  • $begingroup$
    (by considering $x=0$, $beta$ maps into $[0,infty)$, and by considering $x$ sufficiently large, the same is true for $alpha$)
    $endgroup$
    – Calvin Khor
    Jan 3 at 17:16
















1












$begingroup$


Let $F: mathbb{R} timesmathbb{R}^2 rightarrow mathbb{R}$ localy lipschitz in its second variable. Let the Cauchy problem be:



$$x'=F(t,x),\ x(0) = x_0 $$
Let $alpha : mathbb{R} rightarrow mathbb{R}$ and $beta : mathbb{R} rightarrow mathbb{R}$. Show that if $forall x in mathbb{R}^2, forall t in mathbb{R}$ $$|F(t,x)| leq alpha(t)|x| + beta (t) $$
then any maximal solution is global.



I am unable to do this. I suppose I have to either show that $F$ is uniformly bounded or that it is uniformly Lipschitz, but I have no clue how to do either. Any help will be appreciated.






ordinary-differential-equations cauchy-problem







share|cite|improve this question











$endgroup$












  • $begingroup$
    (by considering $x=0$, $beta$ maps into $[0,infty)$, and by considering $x$ sufficiently large, the same is true for $alpha$)
    $endgroup$
    – Calvin Khor
    Jan 3 at 17:16














1












1








1





$begingroup$


Let $F: mathbb{R} timesmathbb{R}^2 rightarrow mathbb{R}$ localy lipschitz in its second variable. Let the Cauchy problem be:



$$x'=F(t,x),\ x(0) = x_0 $$
Let $alpha : mathbb{R} rightarrow mathbb{R}$ and $beta : mathbb{R} rightarrow mathbb{R}$. Show that if $forall x in mathbb{R}^2, forall t in mathbb{R}$ $$|F(t,x)| leq alpha(t)|x| + beta (t) $$
then any maximal solution is global.



I am unable to do this. I suppose I have to either show that $F$ is uniformly bounded or that it is uniformly Lipschitz, but I have no clue how to do either. Any help will be appreciated.






ordinary-differential-equations cauchy-problem







share|cite|improve this question











$endgroup$




Let $F: mathbb{R} timesmathbb{R}^2 rightarrow mathbb{R}$ localy lipschitz in its second variable. Let the Cauchy problem be:



$$x'=F(t,x),\ x(0) = x_0 $$
Let $alpha : mathbb{R} rightarrow mathbb{R}$ and $beta : mathbb{R} rightarrow mathbb{R}$. Show that if $forall x in mathbb{R}^2, forall t in mathbb{R}$ $$|F(t,x)| leq alpha(t)|x| + beta (t) $$
then any maximal solution is global.



I am unable to do this. I suppose I have to either show that $F$ is uniformly bounded or that it is uniformly Lipschitz, but I have no clue how to do either. Any help will be appreciated.






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 3 at 18:20









amWhy

1




1










asked Jan 3 at 16:41









John MayneJohn Mayne

1,109719




1,109719












  • $begingroup$
    (by considering $x=0$, $beta$ maps into $[0,infty)$, and by considering $x$ sufficiently large, the same is true for $alpha$)
    $endgroup$
    – Calvin Khor
    Jan 3 at 17:16


















  • $begingroup$
    (by considering $x=0$, $beta$ maps into $[0,infty)$, and by considering $x$ sufficiently large, the same is true for $alpha$)
    $endgroup$
    – Calvin Khor
    Jan 3 at 17:16
















$begingroup$
(by considering $x=0$, $beta$ maps into $[0,infty)$, and by considering $x$ sufficiently large, the same is true for $alpha$)
$endgroup$
– Calvin Khor
Jan 3 at 17:16




$begingroup$
(by considering $x=0$, $beta$ maps into $[0,infty)$, and by considering $x$ sufficiently large, the same is true for $alpha$)
$endgroup$
– Calvin Khor
Jan 3 at 17:16










2 Answers
2






active

oldest

votes


















2












$begingroup$

See Grönwall lemma on a bound for solutions, using the solution of $$u'(t)=α(t)u+(α(t)|x_0|+β(t)), ~~ u(0)=0.$$ Then after establishing that $$|x(t)-x_0|le |u(t)|$$ wherever the solution exists, apply theorems on the maximal solution, and why the bound prevents finite boundaries of the maximal domain.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Suppose $x(t)$ is not global, namely $exists t_0>$ such that
    $$ lim_{ttoto t_0^-}|x(t)|=infty. tag{1}$$
    Integrating from $0$ to $t<t_0$ gives
    $$ x(t)=x(0)+int_0^tF(s,x)ds $$
    which implies
    $$ |x(t)|le |x(0)|+int_0^t|F(s,x)|ds. $$
    Using $F(t,x)lealpha(t)|x|+beta(t)$, one has
    $$ |x(t)|le |x(0)|+int_0^t(alpha(s)|x(s)|+beta(s))ds.tag{2}$$
    Define
    $$ g(t)=int_0^t(alpha(s)|x(s)|+beta(s))ds tag{3} $$
    and then
    $$ g'(t)=alpha(t)|x(t)|+beta(t). tag{4} $$
    By (1), it is not hard to see
    $$ lim_{ttoto t_0^-}g(t)=infty. tag{5}$$
    Using (3) and (4) in (2), one has
    $$ g'(t)-beta(t)le|x_0|alpha(t)+alpha(t)g(t) $$
    or
    $$ g'(t)-alpha(t)g(t)lebeta(t)+|x_0|alpha(t). $$
    It is easy to see
    $$ left(e^{-int_0^talpha(s)ds}g(t)right)'le (beta(t)+|x_0|alpha(t))e^{-int_0^talpha(s)ds}. $$
    Integrating from $0$ to $t$, one has
    $$ g(t)le g(0)e^{-int_0^talpha(s)ds}+ e^{-int_0^talpha(s)ds}int_0^t(beta(r)+|x_0|alpha(r))e^{-int_0^ralpha(s)ds}dr. $$
    Using this, one has
    $$ g(t)le C, tin[0,t_0]$$
    for some constant $C$, which is against (5).






    share|cite|improve this answer









    $endgroup$













      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%2f3060755%2fif-vert-ft-x-vert-leq-alpha-t-vert-x-vert-betat-the-maximal-soluti%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      See Grönwall lemma on a bound for solutions, using the solution of $$u'(t)=α(t)u+(α(t)|x_0|+β(t)), ~~ u(0)=0.$$ Then after establishing that $$|x(t)-x_0|le |u(t)|$$ wherever the solution exists, apply theorems on the maximal solution, and why the bound prevents finite boundaries of the maximal domain.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        See Grönwall lemma on a bound for solutions, using the solution of $$u'(t)=α(t)u+(α(t)|x_0|+β(t)), ~~ u(0)=0.$$ Then after establishing that $$|x(t)-x_0|le |u(t)|$$ wherever the solution exists, apply theorems on the maximal solution, and why the bound prevents finite boundaries of the maximal domain.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          See Grönwall lemma on a bound for solutions, using the solution of $$u'(t)=α(t)u+(α(t)|x_0|+β(t)), ~~ u(0)=0.$$ Then after establishing that $$|x(t)-x_0|le |u(t)|$$ wherever the solution exists, apply theorems on the maximal solution, and why the bound prevents finite boundaries of the maximal domain.






          share|cite|improve this answer











          $endgroup$



          See Grönwall lemma on a bound for solutions, using the solution of $$u'(t)=α(t)u+(α(t)|x_0|+β(t)), ~~ u(0)=0.$$ Then after establishing that $$|x(t)-x_0|le |u(t)|$$ wherever the solution exists, apply theorems on the maximal solution, and why the bound prevents finite boundaries of the maximal domain.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 3 at 17:41

























          answered Jan 3 at 17:01









          LutzLLutzL

          57.2k42054




          57.2k42054























              0












              $begingroup$

              Suppose $x(t)$ is not global, namely $exists t_0>$ such that
              $$ lim_{ttoto t_0^-}|x(t)|=infty. tag{1}$$
              Integrating from $0$ to $t<t_0$ gives
              $$ x(t)=x(0)+int_0^tF(s,x)ds $$
              which implies
              $$ |x(t)|le |x(0)|+int_0^t|F(s,x)|ds. $$
              Using $F(t,x)lealpha(t)|x|+beta(t)$, one has
              $$ |x(t)|le |x(0)|+int_0^t(alpha(s)|x(s)|+beta(s))ds.tag{2}$$
              Define
              $$ g(t)=int_0^t(alpha(s)|x(s)|+beta(s))ds tag{3} $$
              and then
              $$ g'(t)=alpha(t)|x(t)|+beta(t). tag{4} $$
              By (1), it is not hard to see
              $$ lim_{ttoto t_0^-}g(t)=infty. tag{5}$$
              Using (3) and (4) in (2), one has
              $$ g'(t)-beta(t)le|x_0|alpha(t)+alpha(t)g(t) $$
              or
              $$ g'(t)-alpha(t)g(t)lebeta(t)+|x_0|alpha(t). $$
              It is easy to see
              $$ left(e^{-int_0^talpha(s)ds}g(t)right)'le (beta(t)+|x_0|alpha(t))e^{-int_0^talpha(s)ds}. $$
              Integrating from $0$ to $t$, one has
              $$ g(t)le g(0)e^{-int_0^talpha(s)ds}+ e^{-int_0^talpha(s)ds}int_0^t(beta(r)+|x_0|alpha(r))e^{-int_0^ralpha(s)ds}dr. $$
              Using this, one has
              $$ g(t)le C, tin[0,t_0]$$
              for some constant $C$, which is against (5).






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Suppose $x(t)$ is not global, namely $exists t_0>$ such that
                $$ lim_{ttoto t_0^-}|x(t)|=infty. tag{1}$$
                Integrating from $0$ to $t<t_0$ gives
                $$ x(t)=x(0)+int_0^tF(s,x)ds $$
                which implies
                $$ |x(t)|le |x(0)|+int_0^t|F(s,x)|ds. $$
                Using $F(t,x)lealpha(t)|x|+beta(t)$, one has
                $$ |x(t)|le |x(0)|+int_0^t(alpha(s)|x(s)|+beta(s))ds.tag{2}$$
                Define
                $$ g(t)=int_0^t(alpha(s)|x(s)|+beta(s))ds tag{3} $$
                and then
                $$ g'(t)=alpha(t)|x(t)|+beta(t). tag{4} $$
                By (1), it is not hard to see
                $$ lim_{ttoto t_0^-}g(t)=infty. tag{5}$$
                Using (3) and (4) in (2), one has
                $$ g'(t)-beta(t)le|x_0|alpha(t)+alpha(t)g(t) $$
                or
                $$ g'(t)-alpha(t)g(t)lebeta(t)+|x_0|alpha(t). $$
                It is easy to see
                $$ left(e^{-int_0^talpha(s)ds}g(t)right)'le (beta(t)+|x_0|alpha(t))e^{-int_0^talpha(s)ds}. $$
                Integrating from $0$ to $t$, one has
                $$ g(t)le g(0)e^{-int_0^talpha(s)ds}+ e^{-int_0^talpha(s)ds}int_0^t(beta(r)+|x_0|alpha(r))e^{-int_0^ralpha(s)ds}dr. $$
                Using this, one has
                $$ g(t)le C, tin[0,t_0]$$
                for some constant $C$, which is against (5).






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Suppose $x(t)$ is not global, namely $exists t_0>$ such that
                  $$ lim_{ttoto t_0^-}|x(t)|=infty. tag{1}$$
                  Integrating from $0$ to $t<t_0$ gives
                  $$ x(t)=x(0)+int_0^tF(s,x)ds $$
                  which implies
                  $$ |x(t)|le |x(0)|+int_0^t|F(s,x)|ds. $$
                  Using $F(t,x)lealpha(t)|x|+beta(t)$, one has
                  $$ |x(t)|le |x(0)|+int_0^t(alpha(s)|x(s)|+beta(s))ds.tag{2}$$
                  Define
                  $$ g(t)=int_0^t(alpha(s)|x(s)|+beta(s))ds tag{3} $$
                  and then
                  $$ g'(t)=alpha(t)|x(t)|+beta(t). tag{4} $$
                  By (1), it is not hard to see
                  $$ lim_{ttoto t_0^-}g(t)=infty. tag{5}$$
                  Using (3) and (4) in (2), one has
                  $$ g'(t)-beta(t)le|x_0|alpha(t)+alpha(t)g(t) $$
                  or
                  $$ g'(t)-alpha(t)g(t)lebeta(t)+|x_0|alpha(t). $$
                  It is easy to see
                  $$ left(e^{-int_0^talpha(s)ds}g(t)right)'le (beta(t)+|x_0|alpha(t))e^{-int_0^talpha(s)ds}. $$
                  Integrating from $0$ to $t$, one has
                  $$ g(t)le g(0)e^{-int_0^talpha(s)ds}+ e^{-int_0^talpha(s)ds}int_0^t(beta(r)+|x_0|alpha(r))e^{-int_0^ralpha(s)ds}dr. $$
                  Using this, one has
                  $$ g(t)le C, tin[0,t_0]$$
                  for some constant $C$, which is against (5).






                  share|cite|improve this answer









                  $endgroup$



                  Suppose $x(t)$ is not global, namely $exists t_0>$ such that
                  $$ lim_{ttoto t_0^-}|x(t)|=infty. tag{1}$$
                  Integrating from $0$ to $t<t_0$ gives
                  $$ x(t)=x(0)+int_0^tF(s,x)ds $$
                  which implies
                  $$ |x(t)|le |x(0)|+int_0^t|F(s,x)|ds. $$
                  Using $F(t,x)lealpha(t)|x|+beta(t)$, one has
                  $$ |x(t)|le |x(0)|+int_0^t(alpha(s)|x(s)|+beta(s))ds.tag{2}$$
                  Define
                  $$ g(t)=int_0^t(alpha(s)|x(s)|+beta(s))ds tag{3} $$
                  and then
                  $$ g'(t)=alpha(t)|x(t)|+beta(t). tag{4} $$
                  By (1), it is not hard to see
                  $$ lim_{ttoto t_0^-}g(t)=infty. tag{5}$$
                  Using (3) and (4) in (2), one has
                  $$ g'(t)-beta(t)le|x_0|alpha(t)+alpha(t)g(t) $$
                  or
                  $$ g'(t)-alpha(t)g(t)lebeta(t)+|x_0|alpha(t). $$
                  It is easy to see
                  $$ left(e^{-int_0^talpha(s)ds}g(t)right)'le (beta(t)+|x_0|alpha(t))e^{-int_0^talpha(s)ds}. $$
                  Integrating from $0$ to $t$, one has
                  $$ g(t)le g(0)e^{-int_0^talpha(s)ds}+ e^{-int_0^talpha(s)ds}int_0^t(beta(r)+|x_0|alpha(r))e^{-int_0^ralpha(s)ds}dr. $$
                  Using this, one has
                  $$ g(t)le C, tin[0,t_0]$$
                  for some constant $C$, which is against (5).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 4 at 16:50









                  xpaulxpaul

                  22.6k24455




                  22.6k24455






























                      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%2f3060755%2fif-vert-ft-x-vert-leq-alpha-t-vert-x-vert-betat-the-maximal-soluti%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