If $x$ is a solution to the equation $dot y = f(y) quad f in C^infty (mathbb{R}^d)$, and $x(0) = x(T)$ for...











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In the book of The Elements of Mechanics by Gallovotti, at page 16, it is asked that




If $x$ is a solution to the equation $dot y = f(y) quad f in
C^infty (mathbb{R}^d)$, and $x(0) = x(T)$ for some T, prove that
$x(t) = x(t+T)$. Would this also be true if $f in C^1(mathbb{R}) ?$ [Hint: Use uniqueness.]




However, we do know that



$$x(t) = x(0) + int_{tau = 0}^{tau=t} f(x(tau)),$$



and if we looked at the difference
$$x(t+T) - x(t) = int_{tau=t}^{tau=t+T} f(x(tau)),$$



I cannot see any reason why this integral should be zero.






differential-equations classical-mechanics







share|cite|improve this question


























    up vote
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    In the book of The Elements of Mechanics by Gallovotti, at page 16, it is asked that




    If $x$ is a solution to the equation $dot y = f(y) quad f in
    C^infty (mathbb{R}^d)$    , and $x(0) = x(T)$ for some T, prove that
    $x(t) = x(t+T)$. Would this also be true if $f in C^1(mathbb{R}) ?$ [Hint: Use uniqueness.]




    However, we do know that



    $$x(t) = x(0) + int_{tau = 0}^{tau=t} f(x(tau)),$$



    and if we looked at the difference
    $$x(t+T) - x(t) = int_{tau=t}^{tau=t+T} f(x(tau)),$$



    I cannot see any reason why this integral should be zero.






    differential-equations classical-mechanics







    share|cite|improve this question
























      up vote
      2
      down vote

      favorite
      2









      up vote
      2
      down vote

      favorite
      2






      2





      In the book of The Elements of Mechanics by Gallovotti, at page 16, it is asked that




      If $x$ is a solution to the equation $dot y = f(y) quad f in
      C^infty (mathbb{R}^d)$      , and $x(0) = x(T)$ for some T, prove that
      $x(t) = x(t+T)$. Would this also be true if $f in C^1(mathbb{R}) ?$ [Hint: Use uniqueness.]




      However, we do know that



      $$x(t) = x(0) + int_{tau = 0}^{tau=t} f(x(tau)),$$



      and if we looked at the difference
      $$x(t+T) - x(t) = int_{tau=t}^{tau=t+T} f(x(tau)),$$



      I cannot see any reason why this integral should be zero.






      differential-equations classical-mechanics







      share|cite|improve this question













      In the book of The Elements of Mechanics by Gallovotti, at page 16, it is asked that




      If $x$ is a solution to the equation $dot y = f(y) quad f in
      C^infty (mathbb{R}^d)$      , and $x(0) = x(T)$ for some T, prove that
      $x(t) = x(t+T)$. Would this also be true if $f in C^1(mathbb{R}) ?$ [Hint: Use uniqueness.]




      However, we do know that



      $$x(t) = x(0) + int_{tau = 0}^{tau=t} f(x(tau)),$$



      and if we looked at the difference
      $$x(t+T) - x(t) = int_{tau=t}^{tau=t+T} f(x(tau)),$$



      I cannot see any reason why this integral should be zero.






      differential-equations classical-mechanics




      differential-equations classical-mechanics






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Nov 22 at 3:43









      onurcanbektas

      3,3081936




      3,3081936






















          2 Answers
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          accepted










          Set



          $z(t) = x(t + T); tag 1$



          then



          $z(0) = x(0 + T) = x(T) = x(0); tag 2$



          also



          $dot z(t) = dot x(t + T) = f(x(t + T)) = f(z(t)); tag{2.5}$



          now since



          $f(y) in C^infty(Bbb R^n) tag 3$



          is differentiable, it follows that $f(y)$ is locally Lipschitz continuous at each point $y_0 in Bbb R^n$; and from this it further follows, via the Picard-Lindeloef theorem, that through any $y_0 in Bbb R^n$ there is precisely one solution to



          $dot y = f(y), ; y(0) = y_0; tag 4$



          therefore, in light of (2), which says that both $x(t)$ and $z(t)$ satisfy



          $x(0) = z(0) = y_0 tag 5$



          for some $y_0$, we conclude that



          $x(t) = z(t) = x(t + T). tag 6$



          Since $f(y) in C^1(Bbb R^n)$ is sufficient for $f(y)$ Lipschitz, the result holds in this case as well.






          share|cite|improve this answer




























            up vote
            4
            down vote













            Let $y_1(t) = x(t)$ and $y_2(t) = x(t+T)$. Then both $y_1$ and $y_2$ satisfy
            begin{align}
            y' = f(y)
            end{align}
            with the same initial condition. Hence by uniqueness we are done.






            share|cite|improve this answer





















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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

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              active

              oldest

              votes








              up vote
              2
              down vote



              accepted










              Set



              $z(t) = x(t + T); tag 1$



              then



              $z(0) = x(0 + T) = x(T) = x(0); tag 2$



              also



              $dot z(t) = dot x(t + T) = f(x(t + T)) = f(z(t)); tag{2.5}$



              now since



              $f(y) in C^infty(Bbb R^n) tag 3$



              is differentiable, it follows that $f(y)$ is locally Lipschitz continuous at each point $y_0 in Bbb R^n$; and from this it further follows, via the Picard-Lindeloef theorem, that through any $y_0 in Bbb R^n$ there is precisely one solution to



              $dot y = f(y), ; y(0) = y_0; tag 4$



              therefore, in light of (2), which says that both $x(t)$ and $z(t)$ satisfy



              $x(0) = z(0) = y_0 tag 5$



              for some $y_0$, we conclude that



              $x(t) = z(t) = x(t + T). tag 6$



              Since $f(y) in C^1(Bbb R^n)$ is sufficient for $f(y)$ Lipschitz, the result holds in this case as well.






              share|cite|improve this answer

























                up vote
                2
                down vote



                accepted










                Set



                $z(t) = x(t + T); tag 1$



                then



                $z(0) = x(0 + T) = x(T) = x(0); tag 2$



                also



                $dot z(t) = dot x(t + T) = f(x(t + T)) = f(z(t)); tag{2.5}$



                now since



                $f(y) in C^infty(Bbb R^n) tag 3$



                is differentiable, it follows that $f(y)$ is locally Lipschitz continuous at each point $y_0 in Bbb R^n$; and from this it further follows, via the Picard-Lindeloef theorem, that through any $y_0 in Bbb R^n$ there is precisely one solution to



                $dot y = f(y), ; y(0) = y_0; tag 4$



                therefore, in light of (2), which says that both $x(t)$ and $z(t)$ satisfy



                $x(0) = z(0) = y_0 tag 5$



                for some $y_0$, we conclude that



                $x(t) = z(t) = x(t + T). tag 6$



                Since $f(y) in C^1(Bbb R^n)$ is sufficient for $f(y)$ Lipschitz, the result holds in this case as well.






                share|cite|improve this answer























                  up vote
                  2
                  down vote



                  accepted







                  up vote
                  2
                  down vote



                  accepted






                  Set



                  $z(t) = x(t + T); tag 1$



                  then



                  $z(0) = x(0 + T) = x(T) = x(0); tag 2$



                  also



                  $dot z(t) = dot x(t + T) = f(x(t + T)) = f(z(t)); tag{2.5}$



                  now since



                  $f(y) in C^infty(Bbb R^n) tag 3$



                  is differentiable, it follows that $f(y)$ is locally Lipschitz continuous at each point $y_0 in Bbb R^n$; and from this it further follows, via the Picard-Lindeloef theorem, that through any $y_0 in Bbb R^n$ there is precisely one solution to



                  $dot y = f(y), ; y(0) = y_0; tag 4$



                  therefore, in light of (2), which says that both $x(t)$ and $z(t)$ satisfy



                  $x(0) = z(0) = y_0 tag 5$



                  for some $y_0$, we conclude that



                  $x(t) = z(t) = x(t + T). tag 6$



                  Since $f(y) in C^1(Bbb R^n)$ is sufficient for $f(y)$ Lipschitz, the result holds in this case as well.






                  share|cite|improve this answer












                  Set



                  $z(t) = x(t + T); tag 1$



                  then



                  $z(0) = x(0 + T) = x(T) = x(0); tag 2$



                  also



                  $dot z(t) = dot x(t + T) = f(x(t + T)) = f(z(t)); tag{2.5}$



                  now since



                  $f(y) in C^infty(Bbb R^n) tag 3$



                  is differentiable, it follows that $f(y)$ is locally Lipschitz continuous at each point $y_0 in Bbb R^n$; and from this it further follows, via the Picard-Lindeloef theorem, that through any $y_0 in Bbb R^n$ there is precisely one solution to



                  $dot y = f(y), ; y(0) = y_0; tag 4$



                  therefore, in light of (2), which says that both $x(t)$ and $z(t)$ satisfy



                  $x(0) = z(0) = y_0 tag 5$



                  for some $y_0$, we conclude that



                  $x(t) = z(t) = x(t + T). tag 6$



                  Since $f(y) in C^1(Bbb R^n)$ is sufficient for $f(y)$ Lipschitz, the result holds in this case as well.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 at 7:28









                  Robert Lewis

                  42.8k22862




                  42.8k22862






















                      up vote
                      4
                      down vote













                      Let $y_1(t) = x(t)$ and $y_2(t) = x(t+T)$. Then both $y_1$ and $y_2$ satisfy
                      begin{align}
                      y' = f(y)
                      end{align}
                      with the same initial condition. Hence by uniqueness we are done.






                      share|cite|improve this answer

























                        up vote
                        4
                        down vote













                        Let $y_1(t) = x(t)$ and $y_2(t) = x(t+T)$. Then both $y_1$ and $y_2$ satisfy
                        begin{align}
                        y' = f(y)
                        end{align}
                        with the same initial condition. Hence by uniqueness we are done.






                        share|cite|improve this answer























                          up vote
                          4
                          down vote










                          up vote
                          4
                          down vote









                          Let $y_1(t) = x(t)$ and $y_2(t) = x(t+T)$. Then both $y_1$ and $y_2$ satisfy
                          begin{align}
                          y' = f(y)
                          end{align}
                          with the same initial condition. Hence by uniqueness we are done.






                          share|cite|improve this answer












                          Let $y_1(t) = x(t)$ and $y_2(t) = x(t+T)$. Then both $y_1$ and $y_2$ satisfy
                          begin{align}
                          y' = f(y)
                          end{align}
                          with the same initial condition. Hence by uniqueness we are done.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 22 at 5:17









                          Jacky Chong

                          17.4k21128




                          17.4k21128






























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