Derivative of an autonomous system of ODEs












0















Consider the $N$-dimensional autonomous system of ODEs
$$dot{x}= f(x),$$
where a locally unique solution $x(t)$, starting from the initial condition $x$, is denoted as $x(t)=phi(t,x)$. Show that



$$Big(frac{partial}{partial{x}}phi(t,x)Big)f(x)=f(phi(t,x))$$




Here is my proposed solution.



We are given that $dot{x}= f(x),$ so we know that



$$frac{dx(t)}{dt}=f(x(t))$$



If we integrate both sides with respect to $t$, we will get



$$int_{t_0}^{t}frac{dx(t)}{dt}dt =int_{t_0}^{t} f(x(s)) ds$$



Therefore,



$$x(t) = x_0 + int_{t_0}^{t} f(x(s)) ds$$



where the integration constant is chosen such that $x(t_0)=x_0$. As $x(t)=phi(t,x)$, we have



$$phi(t,x) = x_0 + int_{t_0}^{t} f(x(s)) ds$$



Hence,



begin{equation}
begin{split}
frac{partialphi}{partial{t}}(t,x) & = 0 + frac{partial}{partial{t}}int_{t_0}^{t} f(x(s)) ds \
& = f(x(t))\
& = f(phi(t,x))
end{split}
end{equation}



where the second equality follows from the fundamental theorem of calculus. Next, we are given that





  1. $phi$ depends on $(t,x)$.


  2. $x$ depends on $t$.


Therefore, by the chain rule,



begin{equation}
begin{split}
frac{partialphi}{partial{t}}(t,x) & = frac{partialphi}{partial{x}}(t,x)frac{dx}{dt} \
& = frac{partialphi}{partial{x}}(t,x)f(x)\
& = f(phi(t,x))
end{split}
end{equation}



Thus, $Big(frac{partial}{partial{x}}phi(t,x)Big)f(x)=f(phi(t,x))$.



It does seem rather strange that I concluded that



$$frac{partialphi}{partial{t}}(t,x) = frac{partialphi}{partial{x}}(t,x)frac{dx}{dt}$$



This must be because $frac{partialphi}{partial{t}}(t,x) = frac{partialphi}{partial{x}}(t,x)frac{dx}{dt}$ instead of $frac{partialphi}{partial{t}}(t,x) = frac{partialphi}{partial{t}}(t,x)frac{dx}{dt}$ because of the chain rule.



Is this approach correct? Please let me know if there are any better alternatives.










share|cite|improve this question





























    0















    Consider the $N$-dimensional autonomous system of ODEs
    $$dot{x}= f(x),$$
    where a locally unique solution $x(t)$, starting from the initial condition $x$, is denoted as $x(t)=phi(t,x)$. Show that



    $$Big(frac{partial}{partial{x}}phi(t,x)Big)f(x)=f(phi(t,x))$$




    Here is my proposed solution.



    We are given that $dot{x}= f(x),$ so we know that



    $$frac{dx(t)}{dt}=f(x(t))$$



    If we integrate both sides with respect to $t$, we will get



    $$int_{t_0}^{t}frac{dx(t)}{dt}dt =int_{t_0}^{t} f(x(s)) ds$$



    Therefore,



    $$x(t) = x_0 + int_{t_0}^{t} f(x(s)) ds$$



    where the integration constant is chosen such that $x(t_0)=x_0$. As $x(t)=phi(t,x)$, we have



    $$phi(t,x) = x_0 + int_{t_0}^{t} f(x(s)) ds$$



    Hence,



    begin{equation}
    begin{split}
    frac{partialphi}{partial{t}}(t,x) & = 0 + frac{partial}{partial{t}}int_{t_0}^{t} f(x(s)) ds \
    & = f(x(t))\
    & = f(phi(t,x))
    end{split}
    end{equation}



    where the second equality follows from the fundamental theorem of calculus. Next, we are given that





    1. $phi$ depends on $(t,x)$.


    2. $x$ depends on $t$.


    Therefore, by the chain rule,



    begin{equation}
    begin{split}
    frac{partialphi}{partial{t}}(t,x) & = frac{partialphi}{partial{x}}(t,x)frac{dx}{dt} \
    & = frac{partialphi}{partial{x}}(t,x)f(x)\
    & = f(phi(t,x))
    end{split}
    end{equation}



    Thus, $Big(frac{partial}{partial{x}}phi(t,x)Big)f(x)=f(phi(t,x))$.



    It does seem rather strange that I concluded that



    $$frac{partialphi}{partial{t}}(t,x) = frac{partialphi}{partial{x}}(t,x)frac{dx}{dt}$$



    This must be because $frac{partialphi}{partial{t}}(t,x) = frac{partialphi}{partial{x}}(t,x)frac{dx}{dt}$ instead of $frac{partialphi}{partial{t}}(t,x) = frac{partialphi}{partial{t}}(t,x)frac{dx}{dt}$ because of the chain rule.



    Is this approach correct? Please let me know if there are any better alternatives.










    share|cite|improve this question



























      0












      0








      0








      Consider the $N$-dimensional autonomous system of ODEs
      $$dot{x}= f(x),$$
      where a locally unique solution $x(t)$, starting from the initial condition $x$, is denoted as $x(t)=phi(t,x)$. Show that



      $$Big(frac{partial}{partial{x}}phi(t,x)Big)f(x)=f(phi(t,x))$$




      Here is my proposed solution.



      We are given that $dot{x}= f(x),$ so we know that



      $$frac{dx(t)}{dt}=f(x(t))$$



      If we integrate both sides with respect to $t$, we will get



      $$int_{t_0}^{t}frac{dx(t)}{dt}dt =int_{t_0}^{t} f(x(s)) ds$$



      Therefore,



      $$x(t) = x_0 + int_{t_0}^{t} f(x(s)) ds$$



      where the integration constant is chosen such that $x(t_0)=x_0$. As $x(t)=phi(t,x)$, we have



      $$phi(t,x) = x_0 + int_{t_0}^{t} f(x(s)) ds$$



      Hence,



      begin{equation}
      begin{split}
      frac{partialphi}{partial{t}}(t,x) & = 0 + frac{partial}{partial{t}}int_{t_0}^{t} f(x(s)) ds \
      & = f(x(t))\
      & = f(phi(t,x))
      end{split}
      end{equation}



      where the second equality follows from the fundamental theorem of calculus. Next, we are given that





      1. $phi$ depends on $(t,x)$.


      2. $x$ depends on $t$.


      Therefore, by the chain rule,



      begin{equation}
      begin{split}
      frac{partialphi}{partial{t}}(t,x) & = frac{partialphi}{partial{x}}(t,x)frac{dx}{dt} \
      & = frac{partialphi}{partial{x}}(t,x)f(x)\
      & = f(phi(t,x))
      end{split}
      end{equation}



      Thus, $Big(frac{partial}{partial{x}}phi(t,x)Big)f(x)=f(phi(t,x))$.



      It does seem rather strange that I concluded that



      $$frac{partialphi}{partial{t}}(t,x) = frac{partialphi}{partial{x}}(t,x)frac{dx}{dt}$$



      This must be because $frac{partialphi}{partial{t}}(t,x) = frac{partialphi}{partial{x}}(t,x)frac{dx}{dt}$ instead of $frac{partialphi}{partial{t}}(t,x) = frac{partialphi}{partial{t}}(t,x)frac{dx}{dt}$ because of the chain rule.



      Is this approach correct? Please let me know if there are any better alternatives.










      share|cite|improve this question
















      Consider the $N$-dimensional autonomous system of ODEs
      $$dot{x}= f(x),$$
      where a locally unique solution $x(t)$, starting from the initial condition $x$, is denoted as $x(t)=phi(t,x)$. Show that



      $$Big(frac{partial}{partial{x}}phi(t,x)Big)f(x)=f(phi(t,x))$$




      Here is my proposed solution.



      We are given that $dot{x}= f(x),$ so we know that



      $$frac{dx(t)}{dt}=f(x(t))$$



      If we integrate both sides with respect to $t$, we will get



      $$int_{t_0}^{t}frac{dx(t)}{dt}dt =int_{t_0}^{t} f(x(s)) ds$$



      Therefore,



      $$x(t) = x_0 + int_{t_0}^{t} f(x(s)) ds$$



      where the integration constant is chosen such that $x(t_0)=x_0$. As $x(t)=phi(t,x)$, we have



      $$phi(t,x) = x_0 + int_{t_0}^{t} f(x(s)) ds$$



      Hence,



      begin{equation}
      begin{split}
      frac{partialphi}{partial{t}}(t,x) & = 0 + frac{partial}{partial{t}}int_{t_0}^{t} f(x(s)) ds \
      & = f(x(t))\
      & = f(phi(t,x))
      end{split}
      end{equation}



      where the second equality follows from the fundamental theorem of calculus. Next, we are given that





      1. $phi$ depends on $(t,x)$.


      2. $x$ depends on $t$.


      Therefore, by the chain rule,



      begin{equation}
      begin{split}
      frac{partialphi}{partial{t}}(t,x) & = frac{partialphi}{partial{x}}(t,x)frac{dx}{dt} \
      & = frac{partialphi}{partial{x}}(t,x)f(x)\
      & = f(phi(t,x))
      end{split}
      end{equation}



      Thus, $Big(frac{partial}{partial{x}}phi(t,x)Big)f(x)=f(phi(t,x))$.



      It does seem rather strange that I concluded that



      $$frac{partialphi}{partial{t}}(t,x) = frac{partialphi}{partial{x}}(t,x)frac{dx}{dt}$$



      This must be because $frac{partialphi}{partial{t}}(t,x) = frac{partialphi}{partial{x}}(t,x)frac{dx}{dt}$ instead of $frac{partialphi}{partial{t}}(t,x) = frac{partialphi}{partial{t}}(t,x)frac{dx}{dt}$ because of the chain rule.



      Is this approach correct? Please let me know if there are any better alternatives.







      differential-equations proof-verification






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 1:53







      Axion004

















      asked Jan 1 at 23:22









      Axion004Axion004

      295212




      295212






















          1 Answer
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          active

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          2














          If you make the flow definition more specific, such as not re-using the simple $x$ in two different meanings, you could define the flow as
          $$x(t)=ϕ(t;x_0) ~~ text{ where } ~~ x(0)=ϕ(0;x_0)=x_0.$$





          You need to invoke the autonomous nature of the ODE which has as consequence that $tmapsto ϕ_t$ is a group action/representation of the additive group $(Bbb R,+)$. This means that $ϕ_tcirc ϕ_s=ϕ_{t+s}$, or
          $$
          ϕ(t+s;x_0)=ϕ(t;ϕ(s;x_0))iff x(t+s)=ϕ(t;x(s)).
          $$

          From your point-of-view the closest variant might be $x(t)= ϕ(t-s;x(s))$. Anyway, to get the $x$-derivative of $ϕ$ involved you need first some kind of explicit variability in the $x$-argument of $ϕ$ which is easiest done by moving the initial point along the solution curve.



          Taking, in the first variant, the derivative for $s$ at $s=0$ then gives indeed by the chain rule
          $$
          f(ϕ(t+s;x_0))=frac{∂}{∂s}ϕ(t;ϕ(s;x_0))=frac{∂ϕ}{∂x}(t;ϕ(s;x_0))cdot f(ϕ(s;x_0))\
          overset{s=0}implies
          f(ϕ(t;x_0))=frac{∂ϕ}{∂x}(t;x_0)f(x_0)\
          $$






          share|cite|improve this answer























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            1 Answer
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            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2














            If you make the flow definition more specific, such as not re-using the simple $x$ in two different meanings, you could define the flow as
            $$x(t)=ϕ(t;x_0) ~~ text{ where } ~~ x(0)=ϕ(0;x_0)=x_0.$$





            You need to invoke the autonomous nature of the ODE which has as consequence that $tmapsto ϕ_t$ is a group action/representation of the additive group $(Bbb R,+)$. This means that $ϕ_tcirc ϕ_s=ϕ_{t+s}$, or
            $$
            ϕ(t+s;x_0)=ϕ(t;ϕ(s;x_0))iff x(t+s)=ϕ(t;x(s)).
            $$

            From your point-of-view the closest variant might be $x(t)= ϕ(t-s;x(s))$. Anyway, to get the $x$-derivative of $ϕ$ involved you need first some kind of explicit variability in the $x$-argument of $ϕ$ which is easiest done by moving the initial point along the solution curve.



            Taking, in the first variant, the derivative for $s$ at $s=0$ then gives indeed by the chain rule
            $$
            f(ϕ(t+s;x_0))=frac{∂}{∂s}ϕ(t;ϕ(s;x_0))=frac{∂ϕ}{∂x}(t;ϕ(s;x_0))cdot f(ϕ(s;x_0))\
            overset{s=0}implies
            f(ϕ(t;x_0))=frac{∂ϕ}{∂x}(t;x_0)f(x_0)\
            $$






            share|cite|improve this answer




























              2














              If you make the flow definition more specific, such as not re-using the simple $x$ in two different meanings, you could define the flow as
              $$x(t)=ϕ(t;x_0) ~~ text{ where } ~~ x(0)=ϕ(0;x_0)=x_0.$$





              You need to invoke the autonomous nature of the ODE which has as consequence that $tmapsto ϕ_t$ is a group action/representation of the additive group $(Bbb R,+)$. This means that $ϕ_tcirc ϕ_s=ϕ_{t+s}$, or
              $$
              ϕ(t+s;x_0)=ϕ(t;ϕ(s;x_0))iff x(t+s)=ϕ(t;x(s)).
              $$

              From your point-of-view the closest variant might be $x(t)= ϕ(t-s;x(s))$. Anyway, to get the $x$-derivative of $ϕ$ involved you need first some kind of explicit variability in the $x$-argument of $ϕ$ which is easiest done by moving the initial point along the solution curve.



              Taking, in the first variant, the derivative for $s$ at $s=0$ then gives indeed by the chain rule
              $$
              f(ϕ(t+s;x_0))=frac{∂}{∂s}ϕ(t;ϕ(s;x_0))=frac{∂ϕ}{∂x}(t;ϕ(s;x_0))cdot f(ϕ(s;x_0))\
              overset{s=0}implies
              f(ϕ(t;x_0))=frac{∂ϕ}{∂x}(t;x_0)f(x_0)\
              $$






              share|cite|improve this answer


























                2












                2








                2






                If you make the flow definition more specific, such as not re-using the simple $x$ in two different meanings, you could define the flow as
                $$x(t)=ϕ(t;x_0) ~~ text{ where } ~~ x(0)=ϕ(0;x_0)=x_0.$$





                You need to invoke the autonomous nature of the ODE which has as consequence that $tmapsto ϕ_t$ is a group action/representation of the additive group $(Bbb R,+)$. This means that $ϕ_tcirc ϕ_s=ϕ_{t+s}$, or
                $$
                ϕ(t+s;x_0)=ϕ(t;ϕ(s;x_0))iff x(t+s)=ϕ(t;x(s)).
                $$

                From your point-of-view the closest variant might be $x(t)= ϕ(t-s;x(s))$. Anyway, to get the $x$-derivative of $ϕ$ involved you need first some kind of explicit variability in the $x$-argument of $ϕ$ which is easiest done by moving the initial point along the solution curve.



                Taking, in the first variant, the derivative for $s$ at $s=0$ then gives indeed by the chain rule
                $$
                f(ϕ(t+s;x_0))=frac{∂}{∂s}ϕ(t;ϕ(s;x_0))=frac{∂ϕ}{∂x}(t;ϕ(s;x_0))cdot f(ϕ(s;x_0))\
                overset{s=0}implies
                f(ϕ(t;x_0))=frac{∂ϕ}{∂x}(t;x_0)f(x_0)\
                $$






                share|cite|improve this answer














                If you make the flow definition more specific, such as not re-using the simple $x$ in two different meanings, you could define the flow as
                $$x(t)=ϕ(t;x_0) ~~ text{ where } ~~ x(0)=ϕ(0;x_0)=x_0.$$





                You need to invoke the autonomous nature of the ODE which has as consequence that $tmapsto ϕ_t$ is a group action/representation of the additive group $(Bbb R,+)$. This means that $ϕ_tcirc ϕ_s=ϕ_{t+s}$, or
                $$
                ϕ(t+s;x_0)=ϕ(t;ϕ(s;x_0))iff x(t+s)=ϕ(t;x(s)).
                $$

                From your point-of-view the closest variant might be $x(t)= ϕ(t-s;x(s))$. Anyway, to get the $x$-derivative of $ϕ$ involved you need first some kind of explicit variability in the $x$-argument of $ϕ$ which is easiest done by moving the initial point along the solution curve.



                Taking, in the first variant, the derivative for $s$ at $s=0$ then gives indeed by the chain rule
                $$
                f(ϕ(t+s;x_0))=frac{∂}{∂s}ϕ(t;ϕ(s;x_0))=frac{∂ϕ}{∂x}(t;ϕ(s;x_0))cdot f(ϕ(s;x_0))\
                overset{s=0}implies
                f(ϕ(t;x_0))=frac{∂ϕ}{∂x}(t;x_0)f(x_0)\
                $$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 4 at 11:02

























                answered Jan 1 at 23:58









                LutzLLutzL

                56.5k42054




                56.5k42054






























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