Finding the Critical points of a system












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$begingroup$


Find all critical points of the system



$y_1'= y_1(10-y_1-y_2)$



$y_2'= y_2(30-2y_1-y_2)$



then classify them as stable, asymptotically stable, or unstable.



I need help with this particular question, as you may see, the only problem is that I have NO idea how to determine the critical points of the system , nonetheless, I have a thought, if the critical points are the eigenvalues of the system, then can't I turn this into a matrix and determine the eigenvalues from the matrix, or does this not work? How would I go about determining these critical points as eigenvalues. Any suggestion of literature that may help or explanations would be immensely appreciated.






ordinary-differential-equations proof-writing







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    0












    $begingroup$


    Find all critical points of the system



    $y_1'= y_1(10-y_1-y_2)$



    $y_2'= y_2(30-2y_1-y_2)$



    then classify them as stable, asymptotically stable, or unstable.



    I need help with this particular question, as you may see, the only problem is that I have NO idea how to determine the critical points of the system , nonetheless, I have a thought, if the critical points are the eigenvalues of the system, then can't I turn this into a matrix and determine the eigenvalues from the matrix, or does this not work? How would I go about determining these critical points as eigenvalues. Any suggestion of literature that may help or explanations would be immensely appreciated.






    ordinary-differential-equations proof-writing







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Find all critical points of the system



      $y_1'= y_1(10-y_1-y_2)$



      $y_2'= y_2(30-2y_1-y_2)$



      then classify them as stable, asymptotically stable, or unstable.



      I need help with this particular question, as you may see, the only problem is that I have NO idea how to determine the critical points of the system , nonetheless, I have a thought, if the critical points are the eigenvalues of the system, then can't I turn this into a matrix and determine the eigenvalues from the matrix, or does this not work? How would I go about determining these critical points as eigenvalues. Any suggestion of literature that may help or explanations would be immensely appreciated.






      ordinary-differential-equations proof-writing







      share|cite|improve this question









      $endgroup$




      Find all critical points of the system



      $y_1'= y_1(10-y_1-y_2)$



      $y_2'= y_2(30-2y_1-y_2)$



      then classify them as stable, asymptotically stable, or unstable.



      I need help with this particular question, as you may see, the only problem is that I have NO idea how to determine the critical points of the system , nonetheless, I have a thought, if the critical points are the eigenvalues of the system, then can't I turn this into a matrix and determine the eigenvalues from the matrix, or does this not work? How would I go about determining these critical points as eigenvalues. Any suggestion of literature that may help or explanations would be immensely appreciated.






      ordinary-differential-equations proof-writing




      ordinary-differential-equations proof-writing






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      asked Dec 13 '18 at 12:10









      lastgunslingerlastgunslinger

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          The stationary or critical points are the points where the solutions through them are constant. Which means that the derivatives are zero.



          In $y_1'=0=y_2'$ you get trivially $(y_1,y_2)=(0,0)$ as solution, then setting only one component to zero $(y_1,y_2)=(0,30)$ and $(y_1,y_2)=(10,0)$ and then as fourth point the solution of the linear system of the second factors.






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            $begingroup$

            Critical points are points where $(y'_1, y_2') = (0, 0)$, so in your case these are the possible cases




            $y_1 = 0$ and $y_2 = 0$




            Let's call that point ${bf x}_1 = (0, 0)$




            $y_1 = 0$ and $30−2y_1−y_2 = 0$




            This leads to $y_2 = 30$, let's call that point ${bf x}_2 = (0, 30)$




            $10−y_1−y_2 = 0$ and $y_2 = 0$




            This leads to $y_1 = 10$, let's call that point ${bf x}_3 = (10, 0)$




            $10−y_1−y_2 = 0$ and $30−2y_1−y_2 = 0$




            In this case the solution is ${bf x}_4 = (20, -10)$



            To test the stability, you just need to calculate the Jacobian matrix and evaluate its eigenvalues at each of the four locations above.



            $$
            J = pmatrix{partial y_1'/partial y_1 & partial y_1'/partial y_2 \
            partial y_2'/partial y_1 & partial y_2'/partial y_2}
            $$



            The plot below will help you confirm that the numbers you calculate for the eigenvalues make sense



            enter image description here






            share|cite|improve this answer











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              2 Answers
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              active

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              2 Answers
              2






              active

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              active

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              active

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              0












              $begingroup$

              The stationary or critical points are the points where the solutions through them are constant. Which means that the derivatives are zero.



              In $y_1'=0=y_2'$ you get trivially $(y_1,y_2)=(0,0)$ as solution, then setting only one component to zero $(y_1,y_2)=(0,30)$ and $(y_1,y_2)=(10,0)$ and then as fourth point the solution of the linear system of the second factors.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                The stationary or critical points are the points where the solutions through them are constant. Which means that the derivatives are zero.



                In $y_1'=0=y_2'$ you get trivially $(y_1,y_2)=(0,0)$ as solution, then setting only one component to zero $(y_1,y_2)=(0,30)$ and $(y_1,y_2)=(10,0)$ and then as fourth point the solution of the linear system of the second factors.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The stationary or critical points are the points where the solutions through them are constant. Which means that the derivatives are zero.



                  In $y_1'=0=y_2'$ you get trivially $(y_1,y_2)=(0,0)$ as solution, then setting only one component to zero $(y_1,y_2)=(0,30)$ and $(y_1,y_2)=(10,0)$ and then as fourth point the solution of the linear system of the second factors.






                  share|cite|improve this answer









                  $endgroup$



                  The stationary or critical points are the points where the solutions through them are constant. Which means that the derivatives are zero.



                  In $y_1'=0=y_2'$ you get trivially $(y_1,y_2)=(0,0)$ as solution, then setting only one component to zero $(y_1,y_2)=(0,30)$ and $(y_1,y_2)=(10,0)$ and then as fourth point the solution of the linear system of the second factors.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 13 '18 at 12:21









                  LutzLLutzL

                  58k42054




                  58k42054























                      0












                      $begingroup$

                      Critical points are points where $(y'_1, y_2') = (0, 0)$, so in your case these are the possible cases




                      $y_1 = 0$ and $y_2 = 0$




                      Let's call that point ${bf x}_1 = (0, 0)$




                      $y_1 = 0$ and $30−2y_1−y_2 = 0$




                      This leads to $y_2 = 30$, let's call that point ${bf x}_2 = (0, 30)$




                      $10−y_1−y_2 = 0$ and $y_2 = 0$




                      This leads to $y_1 = 10$, let's call that point ${bf x}_3 = (10, 0)$




                      $10−y_1−y_2 = 0$ and $30−2y_1−y_2 = 0$




                      In this case the solution is ${bf x}_4 = (20, -10)$



                      To test the stability, you just need to calculate the Jacobian matrix and evaluate its eigenvalues at each of the four locations above.



                      $$
                      J = pmatrix{partial y_1'/partial y_1 & partial y_1'/partial y_2 \
                      partial y_2'/partial y_1 & partial y_2'/partial y_2}
                      $$



                      The plot below will help you confirm that the numbers you calculate for the eigenvalues make sense



                      enter image description here






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        Critical points are points where $(y'_1, y_2') = (0, 0)$, so in your case these are the possible cases




                        $y_1 = 0$ and $y_2 = 0$




                        Let's call that point ${bf x}_1 = (0, 0)$




                        $y_1 = 0$ and $30−2y_1−y_2 = 0$




                        This leads to $y_2 = 30$, let's call that point ${bf x}_2 = (0, 30)$




                        $10−y_1−y_2 = 0$ and $y_2 = 0$




                        This leads to $y_1 = 10$, let's call that point ${bf x}_3 = (10, 0)$




                        $10−y_1−y_2 = 0$ and $30−2y_1−y_2 = 0$




                        In this case the solution is ${bf x}_4 = (20, -10)$



                        To test the stability, you just need to calculate the Jacobian matrix and evaluate its eigenvalues at each of the four locations above.



                        $$
                        J = pmatrix{partial y_1'/partial y_1 & partial y_1'/partial y_2 \
                        partial y_2'/partial y_1 & partial y_2'/partial y_2}
                        $$



                        The plot below will help you confirm that the numbers you calculate for the eigenvalues make sense



                        enter image description here






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Critical points are points where $(y'_1, y_2') = (0, 0)$, so in your case these are the possible cases




                          $y_1 = 0$ and $y_2 = 0$




                          Let's call that point ${bf x}_1 = (0, 0)$




                          $y_1 = 0$ and $30−2y_1−y_2 = 0$




                          This leads to $y_2 = 30$, let's call that point ${bf x}_2 = (0, 30)$




                          $10−y_1−y_2 = 0$ and $y_2 = 0$




                          This leads to $y_1 = 10$, let's call that point ${bf x}_3 = (10, 0)$




                          $10−y_1−y_2 = 0$ and $30−2y_1−y_2 = 0$




                          In this case the solution is ${bf x}_4 = (20, -10)$



                          To test the stability, you just need to calculate the Jacobian matrix and evaluate its eigenvalues at each of the four locations above.



                          $$
                          J = pmatrix{partial y_1'/partial y_1 & partial y_1'/partial y_2 \
                          partial y_2'/partial y_1 & partial y_2'/partial y_2}
                          $$



                          The plot below will help you confirm that the numbers you calculate for the eigenvalues make sense



                          enter image description here






                          share|cite|improve this answer











                          $endgroup$



                          Critical points are points where $(y'_1, y_2') = (0, 0)$, so in your case these are the possible cases




                          $y_1 = 0$ and $y_2 = 0$




                          Let's call that point ${bf x}_1 = (0, 0)$




                          $y_1 = 0$ and $30−2y_1−y_2 = 0$




                          This leads to $y_2 = 30$, let's call that point ${bf x}_2 = (0, 30)$




                          $10−y_1−y_2 = 0$ and $y_2 = 0$




                          This leads to $y_1 = 10$, let's call that point ${bf x}_3 = (10, 0)$




                          $10−y_1−y_2 = 0$ and $30−2y_1−y_2 = 0$




                          In this case the solution is ${bf x}_4 = (20, -10)$



                          To test the stability, you just need to calculate the Jacobian matrix and evaluate its eigenvalues at each of the four locations above.



                          $$
                          J = pmatrix{partial y_1'/partial y_1 & partial y_1'/partial y_2 \
                          partial y_2'/partial y_1 & partial y_2'/partial y_2}
                          $$



                          The plot below will help you confirm that the numbers you calculate for the eigenvalues make sense



                          enter image description here







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Dec 13 '18 at 12:29

























                          answered Dec 13 '18 at 12:24









                          caveraccaverac

                          14.6k31130




                          14.6k31130






























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