First integral of a vectorial field












2














I have a problem that says:




Let



$$dot x = f(x,y) $$
$$dot y = g(x,y) $$



be a 2-dimensional ODE with vector field $X:Uto mathbb{R}^2 $, $X=(f,g)in mathcal C^1 (U)$, and $U$ an open star domain of $mathbb{R}^2 $. X is integrable if there exists a $mathcal C^1 $ function $H: Uto mathbb R$ which is not constant over an open set, but it is constant over the solutions of the ODE.




  1. Prove that $X$ is integrable if, and only if, there exists a $mathcal C^1 $ function $H$ such that


$$DH(x,y)X(x,y)=frac{partial H}{partial x} (x,y) f(x,y)+frac{partial H}{partial y} (x,y) g(x,y)=0 $$




If $varphi(t)$ is a solution of the ODE, then $dot varphi = (f(varphi),g(varphi)) =X(varphi)$, so if $X$ is integrable there exists a function $H$ such that $DH(varphi)X(varphi)=0$. Also, $DH(varphi)X(x,y)=0, ; forall x,y in U$, but I don't know how to prove it in general. Maybe I should use that $U$ is a star domain, but I'm not sure because the problem has more sections.



Any help is welcome. Thanks!






integration differential-equations vector-fields







share|cite|improve this question




















  • 1




    Note that since $X$ is in $mathcal{C}^1(U)$ an existence and uniqueness theorem works and it is guaranteed that there exists a solution to IVP $varphi(0) = (x_0, y_0)$. Hence you can do a simple substitution in $DH(varphi(t)) X(varphi(t)) equiv 0$.
    – Evgeny
    Nov 26 at 10:00








  • 1




    Also, "only if" part of the question can be proven by plugging solution $phi(t)$ into $H$ and taking a derivative w.r.t. $t$.
    – Evgeny
    Nov 26 at 10:21










  • ¡Thansk Evgeny! Very clear, I've understood it. For the "only if" part, if we suppose that $DH(x,y)X(x,y)equiv 0 $, then $frac{d}{dt}H(varphi(t))= DH(varphi(t))X(varphi(t))=0 $, so $H$ is constant over a solution $varphi$. But I'm not sure why $H$ is not constant (over an open set). Or is this just a hypothesis?
    – Relativo
    Nov 26 at 10:42






  • 1




    You pick a local solution with initial values $φ(0)=(x,y)$ so that at $t=0$ you get then $DH(x,y)X(x,y)=0$.
    – LutzL
    Nov 26 at 10:51






  • 1




    @Relativo I think that since this is a short piece of text, $H$ has all the same properties as it was mentioned before. So, the statement should be read: "Vector field $X$ has a first integral $H$ if this locally non-constant function is constant along solutions. Such function $H$ is also a first integral $iff$ $DH(X) equiv 0$."
    – Evgeny
    Nov 26 at 10:56
















2














I have a problem that says:




Let



$$dot x = f(x,y) $$
$$dot y = g(x,y) $$



be a 2-dimensional ODE with vector field $X:Uto mathbb{R}^2 $, $X=(f,g)in mathcal C^1 (U)$, and $U$ an open star domain of $mathbb{R}^2 $. X is integrable if there exists a $mathcal C^1 $ function $H: Uto mathbb R$ which is not constant over an open set, but it is constant over the solutions of the ODE.




  1. Prove that $X$ is integrable if, and only if, there exists a $mathcal C^1 $ function $H$ such that


$$DH(x,y)X(x,y)=frac{partial H}{partial x} (x,y) f(x,y)+frac{partial H}{partial y} (x,y) g(x,y)=0 $$




If $varphi(t)$ is a solution of the ODE, then $dot varphi = (f(varphi),g(varphi)) =X(varphi)$, so if $X$ is integrable there exists a function $H$ such that $DH(varphi)X(varphi)=0$. Also, $DH(varphi)X(x,y)=0, ; forall x,y in U$, but I don't know how to prove it in general. Maybe I should use that $U$ is a star domain, but I'm not sure because the problem has more sections.



Any help is welcome. Thanks!






integration differential-equations vector-fields







share|cite|improve this question




















  • 1




    Note that since $X$ is in $mathcal{C}^1(U)$ an existence and uniqueness theorem works and it is guaranteed that there exists a solution to IVP $varphi(0) = (x_0, y_0)$. Hence you can do a simple substitution in $DH(varphi(t)) X(varphi(t)) equiv 0$.
    – Evgeny
    Nov 26 at 10:00








  • 1




    Also, "only if" part of the question can be proven by plugging solution $phi(t)$ into $H$ and taking a derivative w.r.t. $t$.
    – Evgeny
    Nov 26 at 10:21










  • ¡Thansk Evgeny! Very clear, I've understood it. For the "only if" part, if we suppose that $DH(x,y)X(x,y)equiv 0 $, then $frac{d}{dt}H(varphi(t))= DH(varphi(t))X(varphi(t))=0 $, so $H$ is constant over a solution $varphi$. But I'm not sure why $H$ is not constant (over an open set). Or is this just a hypothesis?
    – Relativo
    Nov 26 at 10:42






  • 1




    You pick a local solution with initial values $φ(0)=(x,y)$ so that at $t=0$ you get then $DH(x,y)X(x,y)=0$.
    – LutzL
    Nov 26 at 10:51






  • 1




    @Relativo I think that since this is a short piece of text, $H$ has all the same properties as it was mentioned before. So, the statement should be read: "Vector field $X$ has a first integral $H$ if this locally non-constant function is constant along solutions. Such function $H$ is also a first integral $iff$ $DH(X) equiv 0$."
    – Evgeny
    Nov 26 at 10:56














2












2








2







I have a problem that says:




Let



$$dot x = f(x,y) $$
$$dot y = g(x,y) $$



be a 2-dimensional ODE with vector field $X:Uto mathbb{R}^2 $, $X=(f,g)in mathcal C^1 (U)$, and $U$ an open star domain of $mathbb{R}^2 $. X is integrable if there exists a $mathcal C^1 $ function $H: Uto mathbb R$ which is not constant over an open set, but it is constant over the solutions of the ODE.




  1. Prove that $X$ is integrable if, and only if, there exists a $mathcal C^1 $ function $H$ such that


$$DH(x,y)X(x,y)=frac{partial H}{partial x} (x,y) f(x,y)+frac{partial H}{partial y} (x,y) g(x,y)=0 $$




If $varphi(t)$ is a solution of the ODE, then $dot varphi = (f(varphi),g(varphi)) =X(varphi)$, so if $X$ is integrable there exists a function $H$ such that $DH(varphi)X(varphi)=0$. Also, $DH(varphi)X(x,y)=0, ; forall x,y in U$, but I don't know how to prove it in general. Maybe I should use that $U$ is a star domain, but I'm not sure because the problem has more sections.



Any help is welcome. Thanks!






integration differential-equations vector-fields







share|cite|improve this question















I have a problem that says:




Let



$$dot x = f(x,y) $$
$$dot y = g(x,y) $$



be a 2-dimensional ODE with vector field $X:Uto mathbb{R}^2 $, $X=(f,g)in mathcal C^1 (U)$, and $U$ an open star domain of $mathbb{R}^2 $. X is integrable if there exists a $mathcal C^1 $ function $H: Uto mathbb R$ which is not constant over an open set, but it is constant over the solutions of the ODE.




  1. Prove that $X$ is integrable if, and only if, there exists a $mathcal C^1 $ function $H$ such that


$$DH(x,y)X(x,y)=frac{partial H}{partial x} (x,y) f(x,y)+frac{partial H}{partial y} (x,y) g(x,y)=0 $$




If $varphi(t)$ is a solution of the ODE, then $dot varphi = (f(varphi),g(varphi)) =X(varphi)$, so if $X$ is integrable there exists a function $H$ such that $DH(varphi)X(varphi)=0$. Also, $DH(varphi)X(x,y)=0, ; forall x,y in U$, but I don't know how to prove it in general. Maybe I should use that $U$ is a star domain, but I'm not sure because the problem has more sections.



Any help is welcome. Thanks!






integration differential-equations vector-fields




integration differential-equations vector-fields






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 at 10:48

























asked Nov 26 at 9:54









Relativo

27019




27019








  • 1




    Note that since $X$ is in $mathcal{C}^1(U)$ an existence and uniqueness theorem works and it is guaranteed that there exists a solution to IVP $varphi(0) = (x_0, y_0)$. Hence you can do a simple substitution in $DH(varphi(t)) X(varphi(t)) equiv 0$.
    – Evgeny
    Nov 26 at 10:00








  • 1




    Also, "only if" part of the question can be proven by plugging solution $phi(t)$ into $H$ and taking a derivative w.r.t. $t$.
    – Evgeny
    Nov 26 at 10:21










  • ¡Thansk Evgeny! Very clear, I've understood it. For the "only if" part, if we suppose that $DH(x,y)X(x,y)equiv 0 $, then $frac{d}{dt}H(varphi(t))= DH(varphi(t))X(varphi(t))=0 $, so $H$ is constant over a solution $varphi$. But I'm not sure why $H$ is not constant (over an open set). Or is this just a hypothesis?
    – Relativo
    Nov 26 at 10:42






  • 1




    You pick a local solution with initial values $φ(0)=(x,y)$ so that at $t=0$ you get then $DH(x,y)X(x,y)=0$.
    – LutzL
    Nov 26 at 10:51






  • 1




    @Relativo I think that since this is a short piece of text, $H$ has all the same properties as it was mentioned before. So, the statement should be read: "Vector field $X$ has a first integral $H$ if this locally non-constant function is constant along solutions. Such function $H$ is also a first integral $iff$ $DH(X) equiv 0$."
    – Evgeny
    Nov 26 at 10:56














  • 1




    Note that since $X$ is in $mathcal{C}^1(U)$ an existence and uniqueness theorem works and it is guaranteed that there exists a solution to IVP $varphi(0) = (x_0, y_0)$. Hence you can do a simple substitution in $DH(varphi(t)) X(varphi(t)) equiv 0$.
    – Evgeny
    Nov 26 at 10:00








  • 1




    Also, "only if" part of the question can be proven by plugging solution $phi(t)$ into $H$ and taking a derivative w.r.t. $t$.
    – Evgeny
    Nov 26 at 10:21










  • ¡Thansk Evgeny! Very clear, I've understood it. For the "only if" part, if we suppose that $DH(x,y)X(x,y)equiv 0 $, then $frac{d}{dt}H(varphi(t))= DH(varphi(t))X(varphi(t))=0 $, so $H$ is constant over a solution $varphi$. But I'm not sure why $H$ is not constant (over an open set). Or is this just a hypothesis?
    – Relativo
    Nov 26 at 10:42






  • 1




    You pick a local solution with initial values $φ(0)=(x,y)$ so that at $t=0$ you get then $DH(x,y)X(x,y)=0$.
    – LutzL
    Nov 26 at 10:51






  • 1




    @Relativo I think that since this is a short piece of text, $H$ has all the same properties as it was mentioned before. So, the statement should be read: "Vector field $X$ has a first integral $H$ if this locally non-constant function is constant along solutions. Such function $H$ is also a first integral $iff$ $DH(X) equiv 0$."
    – Evgeny
    Nov 26 at 10:56








1




1




Note that since $X$ is in $mathcal{C}^1(U)$ an existence and uniqueness theorem works and it is guaranteed that there exists a solution to IVP $varphi(0) = (x_0, y_0)$. Hence you can do a simple substitution in $DH(varphi(t)) X(varphi(t)) equiv 0$.
– Evgeny
Nov 26 at 10:00






Note that since $X$ is in $mathcal{C}^1(U)$ an existence and uniqueness theorem works and it is guaranteed that there exists a solution to IVP $varphi(0) = (x_0, y_0)$. Hence you can do a simple substitution in $DH(varphi(t)) X(varphi(t)) equiv 0$.
– Evgeny
Nov 26 at 10:00






1




1




Also, "only if" part of the question can be proven by plugging solution $phi(t)$ into $H$ and taking a derivative w.r.t. $t$.
– Evgeny
Nov 26 at 10:21




Also, "only if" part of the question can be proven by plugging solution $phi(t)$ into $H$ and taking a derivative w.r.t. $t$.
– Evgeny
Nov 26 at 10:21












¡Thansk Evgeny! Very clear, I've understood it. For the "only if" part, if we suppose that $DH(x,y)X(x,y)equiv 0 $, then $frac{d}{dt}H(varphi(t))= DH(varphi(t))X(varphi(t))=0 $, so $H$ is constant over a solution $varphi$. But I'm not sure why $H$ is not constant (over an open set). Or is this just a hypothesis?
– Relativo
Nov 26 at 10:42




¡Thansk Evgeny! Very clear, I've understood it. For the "only if" part, if we suppose that $DH(x,y)X(x,y)equiv 0 $, then $frac{d}{dt}H(varphi(t))= DH(varphi(t))X(varphi(t))=0 $, so $H$ is constant over a solution $varphi$. But I'm not sure why $H$ is not constant (over an open set). Or is this just a hypothesis?
– Relativo
Nov 26 at 10:42




1




1




You pick a local solution with initial values $φ(0)=(x,y)$ so that at $t=0$ you get then $DH(x,y)X(x,y)=0$.
– LutzL
Nov 26 at 10:51




You pick a local solution with initial values $φ(0)=(x,y)$ so that at $t=0$ you get then $DH(x,y)X(x,y)=0$.
– LutzL
Nov 26 at 10:51




1




1




@Relativo I think that since this is a short piece of text, $H$ has all the same properties as it was mentioned before. So, the statement should be read: "Vector field $X$ has a first integral $H$ if this locally non-constant function is constant along solutions. Such function $H$ is also a first integral $iff$ $DH(X) equiv 0$."
– Evgeny
Nov 26 at 10:56




@Relativo I think that since this is a short piece of text, $H$ has all the same properties as it was mentioned before. So, the statement should be read: "Vector field $X$ has a first integral $H$ if this locally non-constant function is constant along solutions. Such function $H$ is also a first integral $iff$ $DH(X) equiv 0$."
– Evgeny
Nov 26 at 10:56















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