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












1














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.






differential-equations cauchy-problem







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  • (by considering $x=0$, $beta$ maps into $[0,infty)$, and by considering $x$ sufficiently large, the same is true for $alpha$)
    – Calvin Khor
    2 days ago
















1














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.






differential-equations cauchy-problem







share|cite|improve this question
























  • (by considering $x=0$, $beta$ maps into $[0,infty)$, and by considering $x$ sufficiently large, the same is true for $alpha$)
    – Calvin Khor
    2 days ago














1












1








1







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.






differential-equations cauchy-problem







share|cite|improve this question















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.






differential-equations cauchy-problem




differential-equations cauchy-problem






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edited 2 days ago









amWhy

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192k28224439










asked 2 days ago









John Mayne

1,109719




1,109719












  • (by considering $x=0$, $beta$ maps into $[0,infty)$, and by considering $x$ sufficiently large, the same is true for $alpha$)
    – Calvin Khor
    2 days ago


















  • (by considering $x=0$, $beta$ maps into $[0,infty)$, and by considering $x$ sufficiently large, the same is true for $alpha$)
    – Calvin Khor
    2 days ago
















(by considering $x=0$, $beta$ maps into $[0,infty)$, and by considering $x$ sufficiently large, the same is true for $alpha$)
– Calvin Khor
2 days ago




(by considering $x=0$, $beta$ maps into $[0,infty)$, and by considering $x$ sufficiently large, the same is true for $alpha$)
– Calvin Khor
2 days ago










2 Answers
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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.






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    0














    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





















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      2 Answers
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      2 Answers
      2






      active

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      active

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      active

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      2














      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




























        2














        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


























          2












          2








          2






          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














          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 2 days ago

























          answered 2 days ago









          LutzL

          56.3k42054




          56.3k42054























              0














              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


























                0














                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
























                  0












                  0








                  0






                  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












                  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 yesterday









                  xpaul

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