Inequality of ODE












3












$begingroup$


Suppose $u(t)>0$ satisfies following inequality, how do we solve $u$?
$$u^{'}(t)leq a cdot u^{frac{n+2}{n}}(t)$$
for $tgeq 0$. This seems a little different from Gronwall's inequality, how do we find inequality of $u(t)$, you can add some conditions to $u$ if you want. Thanks for any help.






ordinary-differential-equations







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    $frac{du}{u^{frac{n+2}{n}}}le a dtto frac 12 n u^{-frac 2n}le a t + C_0$
    $endgroup$
    – Cesareo
    Jan 3 at 10:18








  • 2




    $begingroup$
    @Cesareo: There is a minus sign missing in $-frac12nu(t)^{-frac2n}le at-frac12nu_0^{-frac2n}$, so that then $$u(t)le frac{u_0}{(1-frac2nu_0^{frac2n}at)^{frac{n}2}}.$$
    $endgroup$
    – LutzL
    Jan 3 at 23:15












  • $begingroup$
    @LutzL Thanks for the hint.
    $endgroup$
    – Cesareo
    Jan 3 at 23:38










  • $begingroup$
    This bound is only valid until its singularity, even if there is a real branch of the root after that, it does no longer relate to $u$.
    $endgroup$
    – LutzL
    Jan 3 at 23:51
















3












$begingroup$


Suppose $u(t)>0$ satisfies following inequality, how do we solve $u$?
$$u^{'}(t)leq a cdot u^{frac{n+2}{n}}(t)$$
for $tgeq 0$. This seems a little different from Gronwall's inequality, how do we find inequality of $u(t)$, you can add some conditions to $u$ if you want. Thanks for any help.






ordinary-differential-equations







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    $frac{du}{u^{frac{n+2}{n}}}le a dtto frac 12 n u^{-frac 2n}le a t + C_0$
    $endgroup$
    – Cesareo
    Jan 3 at 10:18








  • 2




    $begingroup$
    @Cesareo: There is a minus sign missing in $-frac12nu(t)^{-frac2n}le at-frac12nu_0^{-frac2n}$, so that then $$u(t)le frac{u_0}{(1-frac2nu_0^{frac2n}at)^{frac{n}2}}.$$
    $endgroup$
    – LutzL
    Jan 3 at 23:15












  • $begingroup$
    @LutzL Thanks for the hint.
    $endgroup$
    – Cesareo
    Jan 3 at 23:38










  • $begingroup$
    This bound is only valid until its singularity, even if there is a real branch of the root after that, it does no longer relate to $u$.
    $endgroup$
    – LutzL
    Jan 3 at 23:51














3












3








3


2



$begingroup$


Suppose $u(t)>0$ satisfies following inequality, how do we solve $u$?
$$u^{'}(t)leq a cdot u^{frac{n+2}{n}}(t)$$
for $tgeq 0$. This seems a little different from Gronwall's inequality, how do we find inequality of $u(t)$, you can add some conditions to $u$ if you want. Thanks for any help.






ordinary-differential-equations







share|cite|improve this question









$endgroup$




Suppose $u(t)>0$ satisfies following inequality, how do we solve $u$?
$$u^{'}(t)leq a cdot u^{frac{n+2}{n}}(t)$$
for $tgeq 0$. This seems a little different from Gronwall's inequality, how do we find inequality of $u(t)$, you can add some conditions to $u$ if you want. Thanks for any help.






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 3 at 7:24









STUDENTSTUDENT

1036




1036








  • 4




    $begingroup$
    $frac{du}{u^{frac{n+2}{n}}}le a dtto frac 12 n u^{-frac 2n}le a t + C_0$
    $endgroup$
    – Cesareo
    Jan 3 at 10:18








  • 2




    $begingroup$
    @Cesareo: There is a minus sign missing in $-frac12nu(t)^{-frac2n}le at-frac12nu_0^{-frac2n}$, so that then $$u(t)le frac{u_0}{(1-frac2nu_0^{frac2n}at)^{frac{n}2}}.$$
    $endgroup$
    – LutzL
    Jan 3 at 23:15












  • $begingroup$
    @LutzL Thanks for the hint.
    $endgroup$
    – Cesareo
    Jan 3 at 23:38










  • $begingroup$
    This bound is only valid until its singularity, even if there is a real branch of the root after that, it does no longer relate to $u$.
    $endgroup$
    – LutzL
    Jan 3 at 23:51














  • 4




    $begingroup$
    $frac{du}{u^{frac{n+2}{n}}}le a dtto frac 12 n u^{-frac 2n}le a t + C_0$
    $endgroup$
    – Cesareo
    Jan 3 at 10:18








  • 2




    $begingroup$
    @Cesareo: There is a minus sign missing in $-frac12nu(t)^{-frac2n}le at-frac12nu_0^{-frac2n}$, so that then $$u(t)le frac{u_0}{(1-frac2nu_0^{frac2n}at)^{frac{n}2}}.$$
    $endgroup$
    – LutzL
    Jan 3 at 23:15












  • $begingroup$
    @LutzL Thanks for the hint.
    $endgroup$
    – Cesareo
    Jan 3 at 23:38










  • $begingroup$
    This bound is only valid until its singularity, even if there is a real branch of the root after that, it does no longer relate to $u$.
    $endgroup$
    – LutzL
    Jan 3 at 23:51








4




4




$begingroup$
$frac{du}{u^{frac{n+2}{n}}}le a dtto frac 12 n u^{-frac 2n}le a t + C_0$
$endgroup$
– Cesareo
Jan 3 at 10:18






$begingroup$
$frac{du}{u^{frac{n+2}{n}}}le a dtto frac 12 n u^{-frac 2n}le a t + C_0$
$endgroup$
– Cesareo
Jan 3 at 10:18






2




2




$begingroup$
@Cesareo: There is a minus sign missing in $-frac12nu(t)^{-frac2n}le at-frac12nu_0^{-frac2n}$, so that then $$u(t)le frac{u_0}{(1-frac2nu_0^{frac2n}at)^{frac{n}2}}.$$
$endgroup$
– LutzL
Jan 3 at 23:15






$begingroup$
@Cesareo: There is a minus sign missing in $-frac12nu(t)^{-frac2n}le at-frac12nu_0^{-frac2n}$, so that then $$u(t)le frac{u_0}{(1-frac2nu_0^{frac2n}at)^{frac{n}2}}.$$
$endgroup$
– LutzL
Jan 3 at 23:15














$begingroup$
@LutzL Thanks for the hint.
$endgroup$
– Cesareo
Jan 3 at 23:38




$begingroup$
@LutzL Thanks for the hint.
$endgroup$
– Cesareo
Jan 3 at 23:38












$begingroup$
This bound is only valid until its singularity, even if there is a real branch of the root after that, it does no longer relate to $u$.
$endgroup$
– LutzL
Jan 3 at 23:51




$begingroup$
This bound is only valid until its singularity, even if there is a real branch of the root after that, it does no longer relate to $u$.
$endgroup$
– LutzL
Jan 3 at 23:51










1 Answer
1






active

oldest

votes


















1












$begingroup$

We first take note of a few simple facts concerning $1 le n in Bbb N$:



$dfrac{n + 2}{n} = 1 + dfrac{2}{n}; tag 1$



set



$k = dfrac{2}{n}; tag 2$



then



$dfrac{n + 2}{n} = 1 + k; tag 3$



$u'(t) le au^{1 + k}; tag 3$



$u^{-k - 1} u'(t) le a; tag 4$



$(-dfrac{1}{k} u^{-k})' le a; tag 5$



we next integrate 'twixt $t_0$ and $t > t_0$:



$-dfrac{1}{k}u^{-k}(t) + dfrac{1}{k}u^{-k}(t_0) = displaystyle int_{t_0}^t (-dfrac{1}{k} u^{-k}(s))' ; ds le a(t - t_0); tag 6$



$-dfrac{1}{k}u^{-k}(t) + dfrac{1}{k}u^{-k}(t_0) le a(t - t_0); tag 7$



$-dfrac{1}{k}u^{-k}(t) le a(t - t_0) - dfrac{1}{k}u^{-k}(t_0) ; tag 8$



$u^{-k}(t) ge u^{-k}(t_0) -ka(t - t_0); tag 9$



since



$u(t) > 0, tag{10}$



it follows that the both sides of (9) are positive for $t$ sufficiently close to $t_0$, and always if $t > t_0$ provided $a le 0$; hence for such $t$,



$0 < u^k(t) le (u^{-k}(t_0) -ka(t - t_0))^{-1} = dfrac{1}{u^{-k}(t_0) -ka(t - t_0)} = dfrac{u^k(t_0) }{1 -ku^k(t_0)a(t - t_0)}, tag{11}$



whence



$0 < u(t) le ((u^{-k}(t_0) -ka(t - t_0))^{-1/k} = dfrac{1}{sqrt[k]{u^{-k}(t_0) -ka(t - t_0)}} = dfrac{u(t_0) }{sqrt[k]{1 - ku^k(t_0)a(t - t_0)}}. tag{12}$



Inspection of (12) reveals that



$displaystyle lim_{t to infty} u(t) = 0, ; a < 0, tag{13}$



whereas the limit, if one exists, is not determined from (12) if $a ge 0$. In fact, a singularity is encountered as



$t to t_0 + dfrac{1}{k u^k(t_0)a} tag{14}$



from below. The situation as $t to -infty$ is similar, with the cases $a < 0$, $a ge 0$ interchanged.






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    1 Answer
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    active

    oldest

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    1












    $begingroup$

    We first take note of a few simple facts concerning $1 le n in Bbb N$:



    $dfrac{n + 2}{n} = 1 + dfrac{2}{n}; tag 1$



    set



    $k = dfrac{2}{n}; tag 2$



    then



    $dfrac{n + 2}{n} = 1 + k; tag 3$



    $u'(t) le au^{1 + k}; tag 3$



    $u^{-k - 1} u'(t) le a; tag 4$



    $(-dfrac{1}{k} u^{-k})' le a; tag 5$



    we next integrate 'twixt $t_0$ and $t > t_0$:



    $-dfrac{1}{k}u^{-k}(t) + dfrac{1}{k}u^{-k}(t_0) = displaystyle int_{t_0}^t (-dfrac{1}{k} u^{-k}(s))' ; ds le a(t - t_0); tag 6$



    $-dfrac{1}{k}u^{-k}(t) + dfrac{1}{k}u^{-k}(t_0) le a(t - t_0); tag 7$



    $-dfrac{1}{k}u^{-k}(t) le a(t - t_0) - dfrac{1}{k}u^{-k}(t_0) ; tag 8$



    $u^{-k}(t) ge u^{-k}(t_0) -ka(t - t_0); tag 9$



    since



    $u(t) > 0, tag{10}$



    it follows that the both sides of (9) are positive for $t$ sufficiently close to $t_0$, and always if $t > t_0$ provided $a le 0$; hence for such $t$,



    $0 < u^k(t) le (u^{-k}(t_0) -ka(t - t_0))^{-1} = dfrac{1}{u^{-k}(t_0) -ka(t - t_0)} = dfrac{u^k(t_0) }{1 -ku^k(t_0)a(t - t_0)}, tag{11}$



    whence



    $0 < u(t) le ((u^{-k}(t_0) -ka(t - t_0))^{-1/k} = dfrac{1}{sqrt[k]{u^{-k}(t_0) -ka(t - t_0)}} = dfrac{u(t_0) }{sqrt[k]{1 - ku^k(t_0)a(t - t_0)}}. tag{12}$



    Inspection of (12) reveals that



    $displaystyle lim_{t to infty} u(t) = 0, ; a < 0, tag{13}$



    whereas the limit, if one exists, is not determined from (12) if $a ge 0$. In fact, a singularity is encountered as



    $t to t_0 + dfrac{1}{k u^k(t_0)a} tag{14}$



    from below. The situation as $t to -infty$ is similar, with the cases $a < 0$, $a ge 0$ interchanged.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      We first take note of a few simple facts concerning $1 le n in Bbb N$:



      $dfrac{n + 2}{n} = 1 + dfrac{2}{n}; tag 1$



      set



      $k = dfrac{2}{n}; tag 2$



      then



      $dfrac{n + 2}{n} = 1 + k; tag 3$



      $u'(t) le au^{1 + k}; tag 3$



      $u^{-k - 1} u'(t) le a; tag 4$



      $(-dfrac{1}{k} u^{-k})' le a; tag 5$



      we next integrate 'twixt $t_0$ and $t > t_0$:



      $-dfrac{1}{k}u^{-k}(t) + dfrac{1}{k}u^{-k}(t_0) = displaystyle int_{t_0}^t (-dfrac{1}{k} u^{-k}(s))' ; ds le a(t - t_0); tag 6$



      $-dfrac{1}{k}u^{-k}(t) + dfrac{1}{k}u^{-k}(t_0) le a(t - t_0); tag 7$



      $-dfrac{1}{k}u^{-k}(t) le a(t - t_0) - dfrac{1}{k}u^{-k}(t_0) ; tag 8$



      $u^{-k}(t) ge u^{-k}(t_0) -ka(t - t_0); tag 9$



      since



      $u(t) > 0, tag{10}$



      it follows that the both sides of (9) are positive for $t$ sufficiently close to $t_0$, and always if $t > t_0$ provided $a le 0$; hence for such $t$,



      $0 < u^k(t) le (u^{-k}(t_0) -ka(t - t_0))^{-1} = dfrac{1}{u^{-k}(t_0) -ka(t - t_0)} = dfrac{u^k(t_0) }{1 -ku^k(t_0)a(t - t_0)}, tag{11}$



      whence



      $0 < u(t) le ((u^{-k}(t_0) -ka(t - t_0))^{-1/k} = dfrac{1}{sqrt[k]{u^{-k}(t_0) -ka(t - t_0)}} = dfrac{u(t_0) }{sqrt[k]{1 - ku^k(t_0)a(t - t_0)}}. tag{12}$



      Inspection of (12) reveals that



      $displaystyle lim_{t to infty} u(t) = 0, ; a < 0, tag{13}$



      whereas the limit, if one exists, is not determined from (12) if $a ge 0$. In fact, a singularity is encountered as



      $t to t_0 + dfrac{1}{k u^k(t_0)a} tag{14}$



      from below. The situation as $t to -infty$ is similar, with the cases $a < 0$, $a ge 0$ interchanged.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        We first take note of a few simple facts concerning $1 le n in Bbb N$:



        $dfrac{n + 2}{n} = 1 + dfrac{2}{n}; tag 1$



        set



        $k = dfrac{2}{n}; tag 2$



        then



        $dfrac{n + 2}{n} = 1 + k; tag 3$



        $u'(t) le au^{1 + k}; tag 3$



        $u^{-k - 1} u'(t) le a; tag 4$



        $(-dfrac{1}{k} u^{-k})' le a; tag 5$



        we next integrate 'twixt $t_0$ and $t > t_0$:



        $-dfrac{1}{k}u^{-k}(t) + dfrac{1}{k}u^{-k}(t_0) = displaystyle int_{t_0}^t (-dfrac{1}{k} u^{-k}(s))' ; ds le a(t - t_0); tag 6$



        $-dfrac{1}{k}u^{-k}(t) + dfrac{1}{k}u^{-k}(t_0) le a(t - t_0); tag 7$



        $-dfrac{1}{k}u^{-k}(t) le a(t - t_0) - dfrac{1}{k}u^{-k}(t_0) ; tag 8$



        $u^{-k}(t) ge u^{-k}(t_0) -ka(t - t_0); tag 9$



        since



        $u(t) > 0, tag{10}$



        it follows that the both sides of (9) are positive for $t$ sufficiently close to $t_0$, and always if $t > t_0$ provided $a le 0$; hence for such $t$,



        $0 < u^k(t) le (u^{-k}(t_0) -ka(t - t_0))^{-1} = dfrac{1}{u^{-k}(t_0) -ka(t - t_0)} = dfrac{u^k(t_0) }{1 -ku^k(t_0)a(t - t_0)}, tag{11}$



        whence



        $0 < u(t) le ((u^{-k}(t_0) -ka(t - t_0))^{-1/k} = dfrac{1}{sqrt[k]{u^{-k}(t_0) -ka(t - t_0)}} = dfrac{u(t_0) }{sqrt[k]{1 - ku^k(t_0)a(t - t_0)}}. tag{12}$



        Inspection of (12) reveals that



        $displaystyle lim_{t to infty} u(t) = 0, ; a < 0, tag{13}$



        whereas the limit, if one exists, is not determined from (12) if $a ge 0$. In fact, a singularity is encountered as



        $t to t_0 + dfrac{1}{k u^k(t_0)a} tag{14}$



        from below. The situation as $t to -infty$ is similar, with the cases $a < 0$, $a ge 0$ interchanged.






        share|cite|improve this answer











        $endgroup$



        We first take note of a few simple facts concerning $1 le n in Bbb N$:



        $dfrac{n + 2}{n} = 1 + dfrac{2}{n}; tag 1$



        set



        $k = dfrac{2}{n}; tag 2$



        then



        $dfrac{n + 2}{n} = 1 + k; tag 3$



        $u'(t) le au^{1 + k}; tag 3$



        $u^{-k - 1} u'(t) le a; tag 4$



        $(-dfrac{1}{k} u^{-k})' le a; tag 5$



        we next integrate 'twixt $t_0$ and $t > t_0$:



        $-dfrac{1}{k}u^{-k}(t) + dfrac{1}{k}u^{-k}(t_0) = displaystyle int_{t_0}^t (-dfrac{1}{k} u^{-k}(s))' ; ds le a(t - t_0); tag 6$



        $-dfrac{1}{k}u^{-k}(t) + dfrac{1}{k}u^{-k}(t_0) le a(t - t_0); tag 7$



        $-dfrac{1}{k}u^{-k}(t) le a(t - t_0) - dfrac{1}{k}u^{-k}(t_0) ; tag 8$



        $u^{-k}(t) ge u^{-k}(t_0) -ka(t - t_0); tag 9$



        since



        $u(t) > 0, tag{10}$



        it follows that the both sides of (9) are positive for $t$ sufficiently close to $t_0$, and always if $t > t_0$ provided $a le 0$; hence for such $t$,



        $0 < u^k(t) le (u^{-k}(t_0) -ka(t - t_0))^{-1} = dfrac{1}{u^{-k}(t_0) -ka(t - t_0)} = dfrac{u^k(t_0) }{1 -ku^k(t_0)a(t - t_0)}, tag{11}$



        whence



        $0 < u(t) le ((u^{-k}(t_0) -ka(t - t_0))^{-1/k} = dfrac{1}{sqrt[k]{u^{-k}(t_0) -ka(t - t_0)}} = dfrac{u(t_0) }{sqrt[k]{1 - ku^k(t_0)a(t - t_0)}}. tag{12}$



        Inspection of (12) reveals that



        $displaystyle lim_{t to infty} u(t) = 0, ; a < 0, tag{13}$



        whereas the limit, if one exists, is not determined from (12) if $a ge 0$. In fact, a singularity is encountered as



        $t to t_0 + dfrac{1}{k u^k(t_0)a} tag{14}$



        from below. The situation as $t to -infty$ is similar, with the cases $a < 0$, $a ge 0$ interchanged.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 4 at 22:41

























        answered Jan 3 at 22:41









        Robert LewisRobert Lewis

        48.6k23167




        48.6k23167






























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            Dieringhausen

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            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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