solving coupled equation












1












$begingroup$


$$begin{aligned}dot{x}&=xleft ( 3-x-2y right )\
dot{y}&=yleft ( 2-x-y right )end{aligned}$$



The above is a coupled equation.



The fixed point condition requires all $;x,y;$ for which $x^{ast} =0;$ and $; y^{ast}=0.$

Solving, I arrive at $left ( x^*=0,y^*=0 right );$ and $;left ( x^*=1,y^*=1 right ).$

But there are $2$ other solutions: $left ( x^*=0,y^*=2 right );$ and $;left ( x^*=3,y^*=0 right ).$



I am unable to determine the last $2$ ordered pairs of fixed points. Any help would be good.






linear-algebra dynamical-systems







share|cite|improve this question











$endgroup$












  • $begingroup$
    Concentrate on the RHS. WHAT IF $x=0$ ? Take cases. As far as I can see you need only to find the equilibrium points.
    $endgroup$
    – dmtri
    Dec 21 '18 at 11:49


















1












$begingroup$


$$begin{aligned}dot{x}&=xleft ( 3-x-2y right )\
dot{y}&=yleft ( 2-x-y right )end{aligned}$$



The above is a coupled equation.



The fixed point condition requires all $;x,y;$ for which $x^{ast} =0;$ and $; y^{ast}=0.$

Solving, I arrive at $left ( x^*=0,y^*=0 right );$ and $;left ( x^*=1,y^*=1 right ).$

But there are $2$ other solutions: $left ( x^*=0,y^*=2 right );$ and $;left ( x^*=3,y^*=0 right ).$



I am unable to determine the last $2$ ordered pairs of fixed points. Any help would be good.






linear-algebra dynamical-systems







share|cite|improve this question











$endgroup$












  • $begingroup$
    Concentrate on the RHS. WHAT IF $x=0$ ? Take cases. As far as I can see you need only to find the equilibrium points.
    $endgroup$
    – dmtri
    Dec 21 '18 at 11:49
















1












1








1





$begingroup$


$$begin{aligned}dot{x}&=xleft ( 3-x-2y right )\
dot{y}&=yleft ( 2-x-y right )end{aligned}$$



The above is a coupled equation.



The fixed point condition requires all $;x,y;$ for which $x^{ast} =0;$ and $; y^{ast}=0.$

Solving, I arrive at $left ( x^*=0,y^*=0 right );$ and $;left ( x^*=1,y^*=1 right ).$

But there are $2$ other solutions: $left ( x^*=0,y^*=2 right );$ and $;left ( x^*=3,y^*=0 right ).$



I am unable to determine the last $2$ ordered pairs of fixed points. Any help would be good.






linear-algebra dynamical-systems







share|cite|improve this question











$endgroup$




$$begin{aligned}dot{x}&=xleft ( 3-x-2y right )\
dot{y}&=yleft ( 2-x-y right )end{aligned}$$



The above is a coupled equation.



The fixed point condition requires all $;x,y;$ for which $x^{ast} =0;$ and $; y^{ast}=0.$

Solving, I arrive at $left ( x^*=0,y^*=0 right );$ and $;left ( x^*=1,y^*=1 right ).$

But there are $2$ other solutions: $left ( x^*=0,y^*=2 right );$ and $;left ( x^*=3,y^*=0 right ).$



I am unable to determine the last $2$ ordered pairs of fixed points. Any help would be good.






linear-algebra dynamical-systems




linear-algebra dynamical-systems






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 11:21









user376343

3,9584829




3,9584829










asked Apr 3 '16 at 10:53









MathematicingMathematicing

2,46521858




2,46521858












  • $begingroup$
    Concentrate on the RHS. WHAT IF $x=0$ ? Take cases. As far as I can see you need only to find the equilibrium points.
    $endgroup$
    – dmtri
    Dec 21 '18 at 11:49




















  • $begingroup$
    Concentrate on the RHS. WHAT IF $x=0$ ? Take cases. As far as I can see you need only to find the equilibrium points.
    $endgroup$
    – dmtri
    Dec 21 '18 at 11:49


















$begingroup$
Concentrate on the RHS. WHAT IF $x=0$ ? Take cases. As far as I can see you need only to find the equilibrium points.
$endgroup$
– dmtri
Dec 21 '18 at 11:49






$begingroup$
Concentrate on the RHS. WHAT IF $x=0$ ? Take cases. As far as I can see you need only to find the equilibrium points.
$endgroup$
– dmtri
Dec 21 '18 at 11:49












3 Answers
3






active

oldest

votes


















1












$begingroup$

Consider first solving the $dot{x}$ equation. You should conclude that either $x=0$ or $x=3-2y$.



Now solve the $dot{y}$ equation in each of these cases i.e. first suppose $x=0$, then substitute this into the $dot{y}$ equation to get $0=y(2-y)$, so this is satisfied when $y=0$ and when $y=2$.



So you will have found the points $(0,0)$ and $(0,2)$. Repeat this process for the other relation found on $x$ (i.e. $x=3-2y$) to get the other points.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    make the equations quadratic:



    $y^2+(x-2)y=0$



    solving for this we get $y=0,-(x-2)$ so when you substitute $y=0$ in the above equation you will get $x=3~x=0$, now when you substitute $x=0$ in the second equation you get $y=2,0$ when you solve for the equation on the whole, you get $(1,1),~(0,0)$ as the fixed points. Now from the above obtained fixed point pairs you can see that the fixed points are $(0,2),~(3,0)$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      You appear to have taken something in factorised form, expanded it, then re-factorised it into the same form as before in order to solve it. One should simply read off the solutions from the original factorised form.
      $endgroup$
      – Mar5bar
      Apr 3 '16 at 11:54



















    -1












    $begingroup$

    Analytical solution for the third and fourth fixed points will be a bit tricky (messy and cumbersome). However, you can use substitution in any one of the equations (replace $y$ in $dot x$ or $x$ in $dot y$) and solve the quadratic equation to get your other two fixed points.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      I dont know why you have downvoted my answer, so here you go for the complete answer:
      $endgroup$
      – Raaja
      Apr 3 '16 at 11:24











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Consider first solving the $dot{x}$ equation. You should conclude that either $x=0$ or $x=3-2y$.



    Now solve the $dot{y}$ equation in each of these cases i.e. first suppose $x=0$, then substitute this into the $dot{y}$ equation to get $0=y(2-y)$, so this is satisfied when $y=0$ and when $y=2$.



    So you will have found the points $(0,0)$ and $(0,2)$. Repeat this process for the other relation found on $x$ (i.e. $x=3-2y$) to get the other points.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Consider first solving the $dot{x}$ equation. You should conclude that either $x=0$ or $x=3-2y$.



      Now solve the $dot{y}$ equation in each of these cases i.e. first suppose $x=0$, then substitute this into the $dot{y}$ equation to get $0=y(2-y)$, so this is satisfied when $y=0$ and when $y=2$.



      So you will have found the points $(0,0)$ and $(0,2)$. Repeat this process for the other relation found on $x$ (i.e. $x=3-2y$) to get the other points.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Consider first solving the $dot{x}$ equation. You should conclude that either $x=0$ or $x=3-2y$.



        Now solve the $dot{y}$ equation in each of these cases i.e. first suppose $x=0$, then substitute this into the $dot{y}$ equation to get $0=y(2-y)$, so this is satisfied when $y=0$ and when $y=2$.



        So you will have found the points $(0,0)$ and $(0,2)$. Repeat this process for the other relation found on $x$ (i.e. $x=3-2y$) to get the other points.






        share|cite|improve this answer









        $endgroup$



        Consider first solving the $dot{x}$ equation. You should conclude that either $x=0$ or $x=3-2y$.



        Now solve the $dot{y}$ equation in each of these cases i.e. first suppose $x=0$, then substitute this into the $dot{y}$ equation to get $0=y(2-y)$, so this is satisfied when $y=0$ and when $y=2$.



        So you will have found the points $(0,0)$ and $(0,2)$. Repeat this process for the other relation found on $x$ (i.e. $x=3-2y$) to get the other points.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 3 '16 at 11:22









        Mar5barMar5bar

        36717




        36717























            0












            $begingroup$

            make the equations quadratic:



            $y^2+(x-2)y=0$



            solving for this we get $y=0,-(x-2)$ so when you substitute $y=0$ in the above equation you will get $x=3~x=0$, now when you substitute $x=0$ in the second equation you get $y=2,0$ when you solve for the equation on the whole, you get $(1,1),~(0,0)$ as the fixed points. Now from the above obtained fixed point pairs you can see that the fixed points are $(0,2),~(3,0)$.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              You appear to have taken something in factorised form, expanded it, then re-factorised it into the same form as before in order to solve it. One should simply read off the solutions from the original factorised form.
              $endgroup$
              – Mar5bar
              Apr 3 '16 at 11:54
















            0












            $begingroup$

            make the equations quadratic:



            $y^2+(x-2)y=0$



            solving for this we get $y=0,-(x-2)$ so when you substitute $y=0$ in the above equation you will get $x=3~x=0$, now when you substitute $x=0$ in the second equation you get $y=2,0$ when you solve for the equation on the whole, you get $(1,1),~(0,0)$ as the fixed points. Now from the above obtained fixed point pairs you can see that the fixed points are $(0,2),~(3,0)$.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              You appear to have taken something in factorised form, expanded it, then re-factorised it into the same form as before in order to solve it. One should simply read off the solutions from the original factorised form.
              $endgroup$
              – Mar5bar
              Apr 3 '16 at 11:54














            0












            0








            0





            $begingroup$

            make the equations quadratic:



            $y^2+(x-2)y=0$



            solving for this we get $y=0,-(x-2)$ so when you substitute $y=0$ in the above equation you will get $x=3~x=0$, now when you substitute $x=0$ in the second equation you get $y=2,0$ when you solve for the equation on the whole, you get $(1,1),~(0,0)$ as the fixed points. Now from the above obtained fixed point pairs you can see that the fixed points are $(0,2),~(3,0)$.






            share|cite|improve this answer









            $endgroup$



            make the equations quadratic:



            $y^2+(x-2)y=0$



            solving for this we get $y=0,-(x-2)$ so when you substitute $y=0$ in the above equation you will get $x=3~x=0$, now when you substitute $x=0$ in the second equation you get $y=2,0$ when you solve for the equation on the whole, you get $(1,1),~(0,0)$ as the fixed points. Now from the above obtained fixed point pairs you can see that the fixed points are $(0,2),~(3,0)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Apr 3 '16 at 11:32









            RaajaRaaja

            208312




            208312








            • 1




              $begingroup$
              You appear to have taken something in factorised form, expanded it, then re-factorised it into the same form as before in order to solve it. One should simply read off the solutions from the original factorised form.
              $endgroup$
              – Mar5bar
              Apr 3 '16 at 11:54














            • 1




              $begingroup$
              You appear to have taken something in factorised form, expanded it, then re-factorised it into the same form as before in order to solve it. One should simply read off the solutions from the original factorised form.
              $endgroup$
              – Mar5bar
              Apr 3 '16 at 11:54








            1




            1




            $begingroup$
            You appear to have taken something in factorised form, expanded it, then re-factorised it into the same form as before in order to solve it. One should simply read off the solutions from the original factorised form.
            $endgroup$
            – Mar5bar
            Apr 3 '16 at 11:54




            $begingroup$
            You appear to have taken something in factorised form, expanded it, then re-factorised it into the same form as before in order to solve it. One should simply read off the solutions from the original factorised form.
            $endgroup$
            – Mar5bar
            Apr 3 '16 at 11:54











            -1












            $begingroup$

            Analytical solution for the third and fourth fixed points will be a bit tricky (messy and cumbersome). However, you can use substitution in any one of the equations (replace $y$ in $dot x$ or $x$ in $dot y$) and solve the quadratic equation to get your other two fixed points.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              I dont know why you have downvoted my answer, so here you go for the complete answer:
              $endgroup$
              – Raaja
              Apr 3 '16 at 11:24
















            -1












            $begingroup$

            Analytical solution for the third and fourth fixed points will be a bit tricky (messy and cumbersome). However, you can use substitution in any one of the equations (replace $y$ in $dot x$ or $x$ in $dot y$) and solve the quadratic equation to get your other two fixed points.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              I dont know why you have downvoted my answer, so here you go for the complete answer:
              $endgroup$
              – Raaja
              Apr 3 '16 at 11:24














            -1












            -1








            -1





            $begingroup$

            Analytical solution for the third and fourth fixed points will be a bit tricky (messy and cumbersome). However, you can use substitution in any one of the equations (replace $y$ in $dot x$ or $x$ in $dot y$) and solve the quadratic equation to get your other two fixed points.






            share|cite|improve this answer









            $endgroup$



            Analytical solution for the third and fourth fixed points will be a bit tricky (messy and cumbersome). However, you can use substitution in any one of the equations (replace $y$ in $dot x$ or $x$ in $dot y$) and solve the quadratic equation to get your other two fixed points.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Apr 3 '16 at 11:15









            RaajaRaaja

            208312




            208312








            • 1




              $begingroup$
              I dont know why you have downvoted my answer, so here you go for the complete answer:
              $endgroup$
              – Raaja
              Apr 3 '16 at 11:24














            • 1




              $begingroup$
              I dont know why you have downvoted my answer, so here you go for the complete answer:
              $endgroup$
              – Raaja
              Apr 3 '16 at 11:24








            1




            1




            $begingroup$
            I dont know why you have downvoted my answer, so here you go for the complete answer:
            $endgroup$
            – Raaja
            Apr 3 '16 at 11:24




            $begingroup$
            I dont know why you have downvoted my answer, so here you go for the complete answer:
            $endgroup$
            – Raaja
            Apr 3 '16 at 11:24


















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