Conditions when solutions of $dot{x} = f(x)$ exist for all time












4














I am reading the following textbook:



Introduction to Applied Nonlinear Dynamical Systems and Chaos by Wiggins
p.92 (top)



Consider $$dot{x} = f(x)$$ where $f(x)$ is $C^r$, $rgeq 1$, on some open set $Uin mathbb{R}^n$. Suppose that the solutions exist for all time (Leave it as an exercise to make the necessary modifications when solutions exist only on finite time intervals).




My problem is what are the conditions to make solutions exist for all time.




Consider and example: $$dot{x} = x^2, xin mathbb{R}$$



the solution through $x_0$ at $t = 0$ is $$x(t,0,x_0) = frac{-x_0}{x_0t - 1}$$



this solution (trajectory) does not exist for all time, since it becomes infinite at $t = 1/x_0$.



So what are the conditions to make solution exist for all time






differential-equations dynamical-systems







share|cite|improve this question



























    4














    I am reading the following textbook:



    Introduction to Applied Nonlinear Dynamical Systems and Chaos by Wiggins
    p.92 (top)



    Consider $$dot{x} = f(x)$$ where $f(x)$ is $C^r$, $rgeq 1$, on some open set $Uin mathbb{R}^n$. Suppose that the solutions exist for all time (Leave it as an exercise to make the necessary modifications when solutions exist only on finite time intervals).




    My problem is what are the conditions to make solutions exist for all time.




    Consider and example: $$dot{x} = x^2, xin mathbb{R}$$



    the solution through $x_0$ at $t = 0$ is $$x(t,0,x_0) = frac{-x_0}{x_0t - 1}$$



    this solution (trajectory) does not exist for all time, since it becomes infinite at $t = 1/x_0$.



    So what are the conditions to make solution exist for all time






    differential-equations dynamical-systems







    share|cite|improve this question

























      4












      4








      4







      I am reading the following textbook:



      Introduction to Applied Nonlinear Dynamical Systems and Chaos by Wiggins
      p.92 (top)



      Consider $$dot{x} = f(x)$$ where $f(x)$ is $C^r$, $rgeq 1$, on some open set $Uin mathbb{R}^n$. Suppose that the solutions exist for all time (Leave it as an exercise to make the necessary modifications when solutions exist only on finite time intervals).




      My problem is what are the conditions to make solutions exist for all time.




      Consider and example: $$dot{x} = x^2, xin mathbb{R}$$



      the solution through $x_0$ at $t = 0$ is $$x(t,0,x_0) = frac{-x_0}{x_0t - 1}$$



      this solution (trajectory) does not exist for all time, since it becomes infinite at $t = 1/x_0$.



      So what are the conditions to make solution exist for all time






      differential-equations dynamical-systems







      share|cite|improve this question













      I am reading the following textbook:



      Introduction to Applied Nonlinear Dynamical Systems and Chaos by Wiggins
      p.92 (top)



      Consider $$dot{x} = f(x)$$ where $f(x)$ is $C^r$, $rgeq 1$, on some open set $Uin mathbb{R}^n$. Suppose that the solutions exist for all time (Leave it as an exercise to make the necessary modifications when solutions exist only on finite time intervals).




      My problem is what are the conditions to make solutions exist for all time.




      Consider and example: $$dot{x} = x^2, xin mathbb{R}$$



      the solution through $x_0$ at $t = 0$ is $$x(t,0,x_0) = frac{-x_0}{x_0t - 1}$$



      this solution (trajectory) does not exist for all time, since it becomes infinite at $t = 1/x_0$.



      So what are the conditions to make solution exist for all time






      differential-equations dynamical-systems




      differential-equations dynamical-systems






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 7 at 8:21









      sleeve chen

      3,01541852




      3,01541852






















          1 Answer
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          3














          Usual conditions for the infinite existence of solutions of $dot x=f(x)$ are restrictions to linear growth like
          $$
          |dot x|=|f(x)|le C(1+|x|)
          $$
          or
          $$
          frac12frac{d}{dt}|x|^2=|x^Tdot x|=|x^Tf(x)|le C(1+|x|^2)
          $$
          which impose an exponential bound on the growth of solutions. The solution of the corresponding differential inequalities is with the Grönwall lemma.




          These are not the only conditions, only the relatively easy ones. If the vector field $f(x)$ points inwards on the boundary of a bounded set $U$, all solutions are restricted to $U$ and thus bounded and exist for all positive times. This looks geometrically simple but can be hard to establish






          share|cite|improve this answer























          • How do you relate that to the ODE ?
            – Yves Daoust
            Dec 7 at 8:50










          • For your second condition, it looks like the example $dot{x} = x^2$ in my problem satisfies this; however, it does not have solution for all time?
            – sleeve chen
            Dec 7 at 8:51






          • 1




            @sleevechen : No, $x^Tf(x)=x^3$ grows faster than $|x|^2$.
            – LutzL
            Dec 7 at 8:56










          • Thanks I see. It looks a bit like the definition of Lipshitz continuity.
            – sleeve chen
            Dec 7 at 9:02






          • 2




            Yes. I do not know of necessary conditions, ODE can have a vast variety of behaviors. You can only select classes of nice ones.
            – LutzL
            Dec 7 at 9:08











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3














          Usual conditions for the infinite existence of solutions of $dot x=f(x)$ are restrictions to linear growth like
          $$
          |dot x|=|f(x)|le C(1+|x|)
          $$
          or
          $$
          frac12frac{d}{dt}|x|^2=|x^Tdot x|=|x^Tf(x)|le C(1+|x|^2)
          $$
          which impose an exponential bound on the growth of solutions. The solution of the corresponding differential inequalities is with the Grönwall lemma.




          These are not the only conditions, only the relatively easy ones. If the vector field $f(x)$ points inwards on the boundary of a bounded set $U$, all solutions are restricted to $U$ and thus bounded and exist for all positive times. This looks geometrically simple but can be hard to establish






          share|cite|improve this answer























          • How do you relate that to the ODE ?
            – Yves Daoust
            Dec 7 at 8:50










          • For your second condition, it looks like the example $dot{x} = x^2$ in my problem satisfies this; however, it does not have solution for all time?
            – sleeve chen
            Dec 7 at 8:51






          • 1




            @sleevechen : No, $x^Tf(x)=x^3$ grows faster than $|x|^2$.
            – LutzL
            Dec 7 at 8:56










          • Thanks I see. It looks a bit like the definition of Lipshitz continuity.
            – sleeve chen
            Dec 7 at 9:02






          • 2




            Yes. I do not know of necessary conditions, ODE can have a vast variety of behaviors. You can only select classes of nice ones.
            – LutzL
            Dec 7 at 9:08
















          3














          Usual conditions for the infinite existence of solutions of $dot x=f(x)$ are restrictions to linear growth like
          $$
          |dot x|=|f(x)|le C(1+|x|)
          $$
          or
          $$
          frac12frac{d}{dt}|x|^2=|x^Tdot x|=|x^Tf(x)|le C(1+|x|^2)
          $$
          which impose an exponential bound on the growth of solutions. The solution of the corresponding differential inequalities is with the Grönwall lemma.




          These are not the only conditions, only the relatively easy ones. If the vector field $f(x)$ points inwards on the boundary of a bounded set $U$, all solutions are restricted to $U$ and thus bounded and exist for all positive times. This looks geometrically simple but can be hard to establish






          share|cite|improve this answer























          • How do you relate that to the ODE ?
            – Yves Daoust
            Dec 7 at 8:50










          • For your second condition, it looks like the example $dot{x} = x^2$ in my problem satisfies this; however, it does not have solution for all time?
            – sleeve chen
            Dec 7 at 8:51






          • 1




            @sleevechen : No, $x^Tf(x)=x^3$ grows faster than $|x|^2$.
            – LutzL
            Dec 7 at 8:56










          • Thanks I see. It looks a bit like the definition of Lipshitz continuity.
            – sleeve chen
            Dec 7 at 9:02






          • 2




            Yes. I do not know of necessary conditions, ODE can have a vast variety of behaviors. You can only select classes of nice ones.
            – LutzL
            Dec 7 at 9:08














          3












          3








          3






          Usual conditions for the infinite existence of solutions of $dot x=f(x)$ are restrictions to linear growth like
          $$
          |dot x|=|f(x)|le C(1+|x|)
          $$
          or
          $$
          frac12frac{d}{dt}|x|^2=|x^Tdot x|=|x^Tf(x)|le C(1+|x|^2)
          $$
          which impose an exponential bound on the growth of solutions. The solution of the corresponding differential inequalities is with the Grönwall lemma.




          These are not the only conditions, only the relatively easy ones. If the vector field $f(x)$ points inwards on the boundary of a bounded set $U$, all solutions are restricted to $U$ and thus bounded and exist for all positive times. This looks geometrically simple but can be hard to establish






          share|cite|improve this answer














          Usual conditions for the infinite existence of solutions of $dot x=f(x)$ are restrictions to linear growth like
          $$
          |dot x|=|f(x)|le C(1+|x|)
          $$
          or
          $$
          frac12frac{d}{dt}|x|^2=|x^Tdot x|=|x^Tf(x)|le C(1+|x|^2)
          $$
          which impose an exponential bound on the growth of solutions. The solution of the corresponding differential inequalities is with the Grönwall lemma.




          These are not the only conditions, only the relatively easy ones. If the vector field $f(x)$ points inwards on the boundary of a bounded set $U$, all solutions are restricted to $U$ and thus bounded and exist for all positive times. This looks geometrically simple but can be hard to establish







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 7 at 9:01

























          answered Dec 7 at 8:47









          LutzL

          55.6k42054




          55.6k42054












          • How do you relate that to the ODE ?
            – Yves Daoust
            Dec 7 at 8:50










          • For your second condition, it looks like the example $dot{x} = x^2$ in my problem satisfies this; however, it does not have solution for all time?
            – sleeve chen
            Dec 7 at 8:51






          • 1




            @sleevechen : No, $x^Tf(x)=x^3$ grows faster than $|x|^2$.
            – LutzL
            Dec 7 at 8:56










          • Thanks I see. It looks a bit like the definition of Lipshitz continuity.
            – sleeve chen
            Dec 7 at 9:02






          • 2




            Yes. I do not know of necessary conditions, ODE can have a vast variety of behaviors. You can only select classes of nice ones.
            – LutzL
            Dec 7 at 9:08


















          • How do you relate that to the ODE ?
            – Yves Daoust
            Dec 7 at 8:50










          • For your second condition, it looks like the example $dot{x} = x^2$ in my problem satisfies this; however, it does not have solution for all time?
            – sleeve chen
            Dec 7 at 8:51






          • 1




            @sleevechen : No, $x^Tf(x)=x^3$ grows faster than $|x|^2$.
            – LutzL
            Dec 7 at 8:56










          • Thanks I see. It looks a bit like the definition of Lipshitz continuity.
            – sleeve chen
            Dec 7 at 9:02






          • 2




            Yes. I do not know of necessary conditions, ODE can have a vast variety of behaviors. You can only select classes of nice ones.
            – LutzL
            Dec 7 at 9:08
















          How do you relate that to the ODE ?
          – Yves Daoust
          Dec 7 at 8:50




          How do you relate that to the ODE ?
          – Yves Daoust
          Dec 7 at 8:50












          For your second condition, it looks like the example $dot{x} = x^2$ in my problem satisfies this; however, it does not have solution for all time?
          – sleeve chen
          Dec 7 at 8:51




          For your second condition, it looks like the example $dot{x} = x^2$ in my problem satisfies this; however, it does not have solution for all time?
          – sleeve chen
          Dec 7 at 8:51




          1




          1




          @sleevechen : No, $x^Tf(x)=x^3$ grows faster than $|x|^2$.
          – LutzL
          Dec 7 at 8:56




          @sleevechen : No, $x^Tf(x)=x^3$ grows faster than $|x|^2$.
          – LutzL
          Dec 7 at 8:56












          Thanks I see. It looks a bit like the definition of Lipshitz continuity.
          – sleeve chen
          Dec 7 at 9:02




          Thanks I see. It looks a bit like the definition of Lipshitz continuity.
          – sleeve chen
          Dec 7 at 9:02




          2




          2




          Yes. I do not know of necessary conditions, ODE can have a vast variety of behaviors. You can only select classes of nice ones.
          – LutzL
          Dec 7 at 9:08




          Yes. I do not know of necessary conditions, ODE can have a vast variety of behaviors. You can only select classes of nice ones.
          – LutzL
          Dec 7 at 9:08


















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