Picard-Lindelöf Theorem application












2












$begingroup$


I'm really lost here guys. I'd appreciate it if you could help.



Consider the differential equation



$$
frac{dy}{dt} = f(t, y) tag{1}
$$



with $f$ satisfying the conditions of the Picard-Lindelöf Theorem. Also, $y_1(t) = 3 , t in mathbb{R}$ is a solution of $(1)$. What can we conclude about the solution $y(t)$ which satisfies the initial condition $y(0) = 1$ ?










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$endgroup$

















    2












    $begingroup$


    I'm really lost here guys. I'd appreciate it if you could help.



    Consider the differential equation



    $$
    frac{dy}{dt} = f(t, y) tag{1}
    $$



    with $f$ satisfying the conditions of the Picard-Lindelöf Theorem. Also, $y_1(t) = 3 , t in mathbb{R}$ is a solution of $(1)$. What can we conclude about the solution $y(t)$ which satisfies the initial condition $y(0) = 1$ ?










    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      I'm really lost here guys. I'd appreciate it if you could help.



      Consider the differential equation



      $$
      frac{dy}{dt} = f(t, y) tag{1}
      $$



      with $f$ satisfying the conditions of the Picard-Lindelöf Theorem. Also, $y_1(t) = 3 , t in mathbb{R}$ is a solution of $(1)$. What can we conclude about the solution $y(t)$ which satisfies the initial condition $y(0) = 1$ ?










      share|cite|improve this question











      $endgroup$




      I'm really lost here guys. I'd appreciate it if you could help.



      Consider the differential equation



      $$
      frac{dy}{dt} = f(t, y) tag{1}
      $$



      with $f$ satisfying the conditions of the Picard-Lindelöf Theorem. Also, $y_1(t) = 3 , t in mathbb{R}$ is a solution of $(1)$. What can we conclude about the solution $y(t)$ which satisfies the initial condition $y(0) = 1$ ?







      ordinary-differential-equations initial-value-problems






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      edited Dec 7 '18 at 19:43







      user618968

















      asked Jul 6 '16 at 21:09









      Da MikeDa Mike

      527




      527






















          1 Answer
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          $begingroup$

          So, for anyone interested, after a couple of days and with some indications, I solved my question.



          Let $y_2(t)$ be the solution with initial condition $y(0) = 1$.



          Now suppose that $y_1$ and $y_2$ have a common point at $(t_0, y_1(t_0)) =(t_0, 1)$. Then the following problem



          $$
          frac{dy}{dt} = f(t, y) , y(t_0) = 1
          $$



          has 2 solutions, $y_1$ and $y_2$. But that can't be true because of the Picard-Lindelof Theorem. So $y_1(t) neq y_2(t)$ for all $t in mathbb{R}$ and because $y_2(0) < y_1(t)$ we conclude that $y_2(t) < y_1(t) implies y_2(t) < 3$ for all $t in mathbb{R}$.






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            $begingroup$

            So, for anyone interested, after a couple of days and with some indications, I solved my question.



            Let $y_2(t)$ be the solution with initial condition $y(0) = 1$.



            Now suppose that $y_1$ and $y_2$ have a common point at $(t_0, y_1(t_0)) =(t_0, 1)$. Then the following problem



            $$
            frac{dy}{dt} = f(t, y) , y(t_0) = 1
            $$



            has 2 solutions, $y_1$ and $y_2$. But that can't be true because of the Picard-Lindelof Theorem. So $y_1(t) neq y_2(t)$ for all $t in mathbb{R}$ and because $y_2(0) < y_1(t)$ we conclude that $y_2(t) < y_1(t) implies y_2(t) < 3$ for all $t in mathbb{R}$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              So, for anyone interested, after a couple of days and with some indications, I solved my question.



              Let $y_2(t)$ be the solution with initial condition $y(0) = 1$.



              Now suppose that $y_1$ and $y_2$ have a common point at $(t_0, y_1(t_0)) =(t_0, 1)$. Then the following problem



              $$
              frac{dy}{dt} = f(t, y) , y(t_0) = 1
              $$



              has 2 solutions, $y_1$ and $y_2$. But that can't be true because of the Picard-Lindelof Theorem. So $y_1(t) neq y_2(t)$ for all $t in mathbb{R}$ and because $y_2(0) < y_1(t)$ we conclude that $y_2(t) < y_1(t) implies y_2(t) < 3$ for all $t in mathbb{R}$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                So, for anyone interested, after a couple of days and with some indications, I solved my question.



                Let $y_2(t)$ be the solution with initial condition $y(0) = 1$.



                Now suppose that $y_1$ and $y_2$ have a common point at $(t_0, y_1(t_0)) =(t_0, 1)$. Then the following problem



                $$
                frac{dy}{dt} = f(t, y) , y(t_0) = 1
                $$



                has 2 solutions, $y_1$ and $y_2$. But that can't be true because of the Picard-Lindelof Theorem. So $y_1(t) neq y_2(t)$ for all $t in mathbb{R}$ and because $y_2(0) < y_1(t)$ we conclude that $y_2(t) < y_1(t) implies y_2(t) < 3$ for all $t in mathbb{R}$.






                share|cite|improve this answer









                $endgroup$



                So, for anyone interested, after a couple of days and with some indications, I solved my question.



                Let $y_2(t)$ be the solution with initial condition $y(0) = 1$.



                Now suppose that $y_1$ and $y_2$ have a common point at $(t_0, y_1(t_0)) =(t_0, 1)$. Then the following problem



                $$
                frac{dy}{dt} = f(t, y) , y(t_0) = 1
                $$



                has 2 solutions, $y_1$ and $y_2$. But that can't be true because of the Picard-Lindelof Theorem. So $y_1(t) neq y_2(t)$ for all $t in mathbb{R}$ and because $y_2(0) < y_1(t)$ we conclude that $y_2(t) < y_1(t) implies y_2(t) < 3$ for all $t in mathbb{R}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jul 9 '16 at 11:54









                Da MikeDa Mike

                527




                527






























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