Differential Equations Bounded

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Q1



What does it mean when it says the solutions become unbounded in finite time? How do solutions that look bounded look like. I don't really understand how this question relates back to the generalized equation r^2 = x^2 + y^2




differential-equations




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  • This means that exists a finite time $t_0$ such that $|x(t)| undersett to t_0^-to infty$
    – nicomezi
    Jul 26 at 12:50















up vote
0
down vote

favorite












Q1



What does it mean when it says the solutions become unbounded in finite time? How do solutions that look bounded look like. I don't really understand how this question relates back to the generalized equation r^2 = x^2 + y^2




differential-equations




share|cite|improve this question



















  • This means that exists a finite time $t_0$ such that $|x(t)| undersett to t_0^-to infty$
    – nicomezi
    Jul 26 at 12:50













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Q1



What does it mean when it says the solutions become unbounded in finite time? How do solutions that look bounded look like. I don't really understand how this question relates back to the generalized equation r^2 = x^2 + y^2




differential-equations




share|cite|improve this question











Q1



What does it mean when it says the solutions become unbounded in finite time? How do solutions that look bounded look like. I don't really understand how this question relates back to the generalized equation r^2 = x^2 + y^2





differential-equations





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 26 at 12:48









KhanMan

387




387











  • This means that exists a finite time $t_0$ such that $|x(t)| undersett to t_0^-to infty$
    – nicomezi
    Jul 26 at 12:50

















  • This means that exists a finite time $t_0$ such that $|x(t)| undersett to t_0^-to infty$
    – nicomezi
    Jul 26 at 12:50
















This means that exists a finite time $t_0$ such that $|x(t)| undersett to t_0^-to infty$
– nicomezi
Jul 26 at 12:50





This means that exists a finite time $t_0$ such that $|x(t)| undersett to t_0^-to infty$
– nicomezi
Jul 26 at 12:50











1 Answer
1






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oldest

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up vote
2
down vote



accepted










Hint.



(a)



The equilibrium points are those such that $x' = y' = 0;$



Considering the linearised approximation about $(0,0)$ we have



$$
x' = y\
y' = -x
$$



which gives the orbits



$$
(r^2)' = 0 Rightarrow r = C_0
$$



because $frac 12(r^2)' = xfracdxdt+yfracdydt ;$ characterizing a center.



(b)



Multiplying respectively by $x, y;$ and adding we have



(1)
$$
xfracdxdt+yfracdydt = (x^2+y^2)^2Rightarrow frac 12 (r^2)' = r^4
$$



(2)
$$
xfracdxdt+yfracdydt = -(x^2+y^2)^2Rightarrow frac 12 (r^2)' = -r^4
$$






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    Hint.



    (a)



    The equilibrium points are those such that $x' = y' = 0;$



    Considering the linearised approximation about $(0,0)$ we have



    $$
    x' = y\
    y' = -x
    $$



    which gives the orbits



    $$
    (r^2)' = 0 Rightarrow r = C_0
    $$



    because $frac 12(r^2)' = xfracdxdt+yfracdydt ;$ characterizing a center.



    (b)



    Multiplying respectively by $x, y;$ and adding we have



    (1)
    $$
    xfracdxdt+yfracdydt = (x^2+y^2)^2Rightarrow frac 12 (r^2)' = r^4
    $$



    (2)
    $$
    xfracdxdt+yfracdydt = -(x^2+y^2)^2Rightarrow frac 12 (r^2)' = -r^4
    $$






    share|cite|improve this answer



























      up vote
      2
      down vote



      accepted










      Hint.



      (a)



      The equilibrium points are those such that $x' = y' = 0;$



      Considering the linearised approximation about $(0,0)$ we have



      $$
      x' = y\
      y' = -x
      $$



      which gives the orbits



      $$
      (r^2)' = 0 Rightarrow r = C_0
      $$



      because $frac 12(r^2)' = xfracdxdt+yfracdydt ;$ characterizing a center.



      (b)



      Multiplying respectively by $x, y;$ and adding we have



      (1)
      $$
      xfracdxdt+yfracdydt = (x^2+y^2)^2Rightarrow frac 12 (r^2)' = r^4
      $$



      (2)
      $$
      xfracdxdt+yfracdydt = -(x^2+y^2)^2Rightarrow frac 12 (r^2)' = -r^4
      $$






      share|cite|improve this answer

























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        Hint.



        (a)



        The equilibrium points are those such that $x' = y' = 0;$



        Considering the linearised approximation about $(0,0)$ we have



        $$
        x' = y\
        y' = -x
        $$



        which gives the orbits



        $$
        (r^2)' = 0 Rightarrow r = C_0
        $$



        because $frac 12(r^2)' = xfracdxdt+yfracdydt ;$ characterizing a center.



        (b)



        Multiplying respectively by $x, y;$ and adding we have



        (1)
        $$
        xfracdxdt+yfracdydt = (x^2+y^2)^2Rightarrow frac 12 (r^2)' = r^4
        $$



        (2)
        $$
        xfracdxdt+yfracdydt = -(x^2+y^2)^2Rightarrow frac 12 (r^2)' = -r^4
        $$






        share|cite|improve this answer















        Hint.



        (a)



        The equilibrium points are those such that $x' = y' = 0;$



        Considering the linearised approximation about $(0,0)$ we have



        $$
        x' = y\
        y' = -x
        $$



        which gives the orbits



        $$
        (r^2)' = 0 Rightarrow r = C_0
        $$



        because $frac 12(r^2)' = xfracdxdt+yfracdydt ;$ characterizing a center.



        (b)



        Multiplying respectively by $x, y;$ and adding we have



        (1)
        $$
        xfracdxdt+yfracdydt = (x^2+y^2)^2Rightarrow frac 12 (r^2)' = r^4
        $$



        (2)
        $$
        xfracdxdt+yfracdydt = -(x^2+y^2)^2Rightarrow frac 12 (r^2)' = -r^4
        $$







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Jul 26 at 13:28


























        answered Jul 26 at 13:17









        Cesareo

        5,5912412




        5,5912412






















             

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