Bounding a solution for an ODE












0














Given
$$dot{x}=tcdot sin^2(xt), quad x(0)=0 quad tin[0,1]$$
I need to bound the solution to that ODE.



I know that $lvert dot{x} rvert = lvert f(x,y) rvert le 1$, but how can I continue from here and bound $x$ itself? I don't know how to approach that.






differential-equations







share|cite|improve this question



























    0














    Given
    $$dot{x}=tcdot sin^2(xt), quad x(0)=0 quad tin[0,1]$$
    I need to bound the solution to that ODE.



    I know that $lvert dot{x} rvert = lvert f(x,y) rvert le 1$, but how can I continue from here and bound $x$ itself? I don't know how to approach that.






    differential-equations







    share|cite|improve this question

























      0












      0








      0







      Given
      $$dot{x}=tcdot sin^2(xt), quad x(0)=0 quad tin[0,1]$$
      I need to bound the solution to that ODE.



      I know that $lvert dot{x} rvert = lvert f(x,y) rvert le 1$, but how can I continue from here and bound $x$ itself? I don't know how to approach that.






      differential-equations







      share|cite|improve this question













      Given
      $$dot{x}=tcdot sin^2(xt), quad x(0)=0 quad tin[0,1]$$
      I need to bound the solution to that ODE.



      I know that $lvert dot{x} rvert = lvert f(x,y) rvert le 1$, but how can I continue from here and bound $x$ itself? I don't know how to approach that.






      differential-equations




      differential-equations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 27 '18 at 15:55









      galah92galah92

      25418




      25418






















          2 Answers
          2






          active

          oldest

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          3














          Your equation tells you that $dot{x}(t)geq 0$ for all $t$, so let's forget about the absolute value. Moreover, since $sin^2leq 1$, we have
          $$dot{x}(t)leq t$$ for all $t$, and integrating we have
          $$int_0^sdot{x}(t)dtleq int_0^stdt Rightarrow x(s)leq {1over 2}s^2,$$ so $1/2$ is a bound, since you are saying $sin [0,1]$.



          Anyway, notice that this procedure can be iterated: since $sin theta<theta$ for positive $theta$, you can conclude that $dot{x}(t)leq {1over 2}t$, which now gives $1/4$ as a bound. Continuing like that, you have what @juliàn was saying: $0$ is actually the only solution.






          share|cite|improve this answer























          • Thanks. I can't detect the use of the IV though. Why is it needed?
            – galah92
            Nov 27 '18 at 16:19










          • You need it to say that $int_0^sdot{x}(t)dt=[x(s)-x(0)]=x(s)$
            – Leo163
            Nov 27 '18 at 16:22










          • Oh, of course..
            – galah92
            Nov 27 '18 at 16:23



















          5














          Well, the unique solution is $x(t)=0$ for all $t$.






          share|cite|improve this answer





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3














            Your equation tells you that $dot{x}(t)geq 0$ for all $t$, so let's forget about the absolute value. Moreover, since $sin^2leq 1$, we have
            $$dot{x}(t)leq t$$ for all $t$, and integrating we have
            $$int_0^sdot{x}(t)dtleq int_0^stdt Rightarrow x(s)leq {1over 2}s^2,$$ so $1/2$ is a bound, since you are saying $sin [0,1]$.



            Anyway, notice that this procedure can be iterated: since $sin theta<theta$ for positive $theta$, you can conclude that $dot{x}(t)leq {1over 2}t$, which now gives $1/4$ as a bound. Continuing like that, you have what @juliàn was saying: $0$ is actually the only solution.






            share|cite|improve this answer























            • Thanks. I can't detect the use of the IV though. Why is it needed?
              – galah92
              Nov 27 '18 at 16:19










            • You need it to say that $int_0^sdot{x}(t)dt=[x(s)-x(0)]=x(s)$
              – Leo163
              Nov 27 '18 at 16:22










            • Oh, of course..
              – galah92
              Nov 27 '18 at 16:23
















            3














            Your equation tells you that $dot{x}(t)geq 0$ for all $t$, so let's forget about the absolute value. Moreover, since $sin^2leq 1$, we have
            $$dot{x}(t)leq t$$ for all $t$, and integrating we have
            $$int_0^sdot{x}(t)dtleq int_0^stdt Rightarrow x(s)leq {1over 2}s^2,$$ so $1/2$ is a bound, since you are saying $sin [0,1]$.



            Anyway, notice that this procedure can be iterated: since $sin theta<theta$ for positive $theta$, you can conclude that $dot{x}(t)leq {1over 2}t$, which now gives $1/4$ as a bound. Continuing like that, you have what @juliàn was saying: $0$ is actually the only solution.






            share|cite|improve this answer























            • Thanks. I can't detect the use of the IV though. Why is it needed?
              – galah92
              Nov 27 '18 at 16:19










            • You need it to say that $int_0^sdot{x}(t)dt=[x(s)-x(0)]=x(s)$
              – Leo163
              Nov 27 '18 at 16:22










            • Oh, of course..
              – galah92
              Nov 27 '18 at 16:23














            3












            3








            3






            Your equation tells you that $dot{x}(t)geq 0$ for all $t$, so let's forget about the absolute value. Moreover, since $sin^2leq 1$, we have
            $$dot{x}(t)leq t$$ for all $t$, and integrating we have
            $$int_0^sdot{x}(t)dtleq int_0^stdt Rightarrow x(s)leq {1over 2}s^2,$$ so $1/2$ is a bound, since you are saying $sin [0,1]$.



            Anyway, notice that this procedure can be iterated: since $sin theta<theta$ for positive $theta$, you can conclude that $dot{x}(t)leq {1over 2}t$, which now gives $1/4$ as a bound. Continuing like that, you have what @juliàn was saying: $0$ is actually the only solution.






            share|cite|improve this answer














            Your equation tells you that $dot{x}(t)geq 0$ for all $t$, so let's forget about the absolute value. Moreover, since $sin^2leq 1$, we have
            $$dot{x}(t)leq t$$ for all $t$, and integrating we have
            $$int_0^sdot{x}(t)dtleq int_0^stdt Rightarrow x(s)leq {1over 2}s^2,$$ so $1/2$ is a bound, since you are saying $sin [0,1]$.



            Anyway, notice that this procedure can be iterated: since $sin theta<theta$ for positive $theta$, you can conclude that $dot{x}(t)leq {1over 2}t$, which now gives $1/4$ as a bound. Continuing like that, you have what @juliàn was saying: $0$ is actually the only solution.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 27 '18 at 16:25

























            answered Nov 27 '18 at 16:06









            Leo163Leo163

            1,600512




            1,600512












            • Thanks. I can't detect the use of the IV though. Why is it needed?
              – galah92
              Nov 27 '18 at 16:19










            • You need it to say that $int_0^sdot{x}(t)dt=[x(s)-x(0)]=x(s)$
              – Leo163
              Nov 27 '18 at 16:22










            • Oh, of course..
              – galah92
              Nov 27 '18 at 16:23


















            • Thanks. I can't detect the use of the IV though. Why is it needed?
              – galah92
              Nov 27 '18 at 16:19










            • You need it to say that $int_0^sdot{x}(t)dt=[x(s)-x(0)]=x(s)$
              – Leo163
              Nov 27 '18 at 16:22










            • Oh, of course..
              – galah92
              Nov 27 '18 at 16:23
















            Thanks. I can't detect the use of the IV though. Why is it needed?
            – galah92
            Nov 27 '18 at 16:19




            Thanks. I can't detect the use of the IV though. Why is it needed?
            – galah92
            Nov 27 '18 at 16:19












            You need it to say that $int_0^sdot{x}(t)dt=[x(s)-x(0)]=x(s)$
            – Leo163
            Nov 27 '18 at 16:22




            You need it to say that $int_0^sdot{x}(t)dt=[x(s)-x(0)]=x(s)$
            – Leo163
            Nov 27 '18 at 16:22












            Oh, of course..
            – galah92
            Nov 27 '18 at 16:23




            Oh, of course..
            – galah92
            Nov 27 '18 at 16:23











            5














            Well, the unique solution is $x(t)=0$ for all $t$.






            share|cite|improve this answer


























              5














              Well, the unique solution is $x(t)=0$ for all $t$.






              share|cite|improve this answer
























                5












                5








                5






                Well, the unique solution is $x(t)=0$ for all $t$.






                share|cite|improve this answer












                Well, the unique solution is $x(t)=0$ for all $t$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 27 '18 at 16:01









                Julián AguirreJulián Aguirre

                67.7k24094




                67.7k24094






























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