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

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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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            Marschland

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            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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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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