Examples on Picard-Lindelöf’s theorem for odes

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I am trying to find an example of ode, $x^‘=f(t,x)$, where $f$ does not satisfies Picard-Lindelöf theorem, but it still have unique solution.



Is it possible?







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  • You can look at Alexander Eremenko's answer to Solution set of non-unique solutions to First order ODE's on MO.
    – user539887
    Aug 10 at 19:10










  • And if you need a concrete example, see, e.g., Example: f(x,y) is not Lipschitz in y but still has a unique solution to initial value problem.
    – user539887
    Aug 11 at 8:00














up vote
0
down vote

favorite












I am trying to find an example of ode, $x^‘=f(t,x)$, where $f$ does not satisfies Picard-Lindelöf theorem, but it still have unique solution.



Is it possible?







share|cite|improve this question




















  • You can look at Alexander Eremenko's answer to Solution set of non-unique solutions to First order ODE's on MO.
    – user539887
    Aug 10 at 19:10










  • And if you need a concrete example, see, e.g., Example: f(x,y) is not Lipschitz in y but still has a unique solution to initial value problem.
    – user539887
    Aug 11 at 8:00












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am trying to find an example of ode, $x^‘=f(t,x)$, where $f$ does not satisfies Picard-Lindelöf theorem, but it still have unique solution.



Is it possible?







share|cite|improve this question












I am trying to find an example of ode, $x^‘=f(t,x)$, where $f$ does not satisfies Picard-Lindelöf theorem, but it still have unique solution.



Is it possible?









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 10 at 18:25









Charles

533420




533420











  • You can look at Alexander Eremenko's answer to Solution set of non-unique solutions to First order ODE's on MO.
    – user539887
    Aug 10 at 19:10










  • And if you need a concrete example, see, e.g., Example: f(x,y) is not Lipschitz in y but still has a unique solution to initial value problem.
    – user539887
    Aug 11 at 8:00
















  • You can look at Alexander Eremenko's answer to Solution set of non-unique solutions to First order ODE's on MO.
    – user539887
    Aug 10 at 19:10










  • And if you need a concrete example, see, e.g., Example: f(x,y) is not Lipschitz in y but still has a unique solution to initial value problem.
    – user539887
    Aug 11 at 8:00















You can look at Alexander Eremenko's answer to Solution set of non-unique solutions to First order ODE's on MO.
– user539887
Aug 10 at 19:10




You can look at Alexander Eremenko's answer to Solution set of non-unique solutions to First order ODE's on MO.
– user539887
Aug 10 at 19:10












And if you need a concrete example, see, e.g., Example: f(x,y) is not Lipschitz in y but still has a unique solution to initial value problem.
– user539887
Aug 11 at 8:00




And if you need a concrete example, see, e.g., Example: f(x,y) is not Lipschitz in y but still has a unique solution to initial value problem.
– user539887
Aug 11 at 8:00










1 Answer
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oldest

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up vote
1
down vote



accepted










Yes, it is.



  1. Osgood's criterion

  2. $f(t,x)$ is decreasing as a function of $x$ for all $t$.





share|cite|improve this answer




















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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Yes, it is.



    1. Osgood's criterion

    2. $f(t,x)$ is decreasing as a function of $x$ for all $t$.





    share|cite|improve this answer
























      up vote
      1
      down vote



      accepted










      Yes, it is.



      1. Osgood's criterion

      2. $f(t,x)$ is decreasing as a function of $x$ for all $t$.





      share|cite|improve this answer






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Yes, it is.



        1. Osgood's criterion

        2. $f(t,x)$ is decreasing as a function of $x$ for all $t$.





        share|cite|improve this answer












        Yes, it is.



        1. Osgood's criterion

        2. $f(t,x)$ is decreasing as a function of $x$ for all $t$.






        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 10 at 18:43









        Julián Aguirre

        64.9k23894




        64.9k23894






















             

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