Help finding the critical values of α where the qualitative nature of the phase portrait for the system...












3














I was asked to solved for the eigenvalues in terms of α for 2X2 matrix and so i did and my answer was marked as correct. Then I was asked to solve for this:



The roots are complex when?



There is a saddle point for?



The equilibrium point is a stable node where?



I tried solving the problem by finding values which would give me 1 and zero for the eigenvalues and got the values of -24/11, -25/11 and -26/11 but the were all wrong. How would i solve for these three questions?
can someone walk me through the steps?



The eigenvalues in terms of α are:



$r = -1 + dfrac{sqrt{100 + 44 alpha}}{2}$



$r_2 = -1 - dfrac{sqrt{100 + 44 alpha}}{2}$






matrices differential-equations eigenvalues-eigenvectors matrix-equations eigenfunctions







share|cite|improve this question


















  • 1




    The roots are complex when $100+44alpha<0$, i.e. when $alpha<-frac{25}{11}$. There is a saddle point when the eigenvalues are real and have opposite sign, so you need $r_1>0$ and $r_2<0$, hence $frac{1}{2}sqrt{100+44alpha}>1$, so $sqrt{100+44alpha}>2$, so $100+44alpha>4$, so $alpha>-frac{24}{11}$. A stable node occurs when both roots are negative but real, so $r_1<0$ and $r_2<0$, which happens when $-frac{25}{11}<alpha<-frac{24}{11}$.
    – symplectomorphic
    Jun 1 '15 at 5:24










  • Thank you very much! you made me realize that i had forgotten to enter the alpha <, and alpha is >
    – Angel Garcia
    Jun 1 '15 at 5:50
















3














I was asked to solved for the eigenvalues in terms of α for 2X2 matrix and so i did and my answer was marked as correct. Then I was asked to solve for this:



The roots are complex when?



There is a saddle point for?



The equilibrium point is a stable node where?



I tried solving the problem by finding values which would give me 1 and zero for the eigenvalues and got the values of -24/11, -25/11 and -26/11 but the were all wrong. How would i solve for these three questions?
can someone walk me through the steps?



The eigenvalues in terms of α are:



$r = -1 + dfrac{sqrt{100 + 44 alpha}}{2}$



$r_2 = -1 - dfrac{sqrt{100 + 44 alpha}}{2}$






matrices differential-equations eigenvalues-eigenvectors matrix-equations eigenfunctions







share|cite|improve this question


















  • 1




    The roots are complex when $100+44alpha<0$, i.e. when $alpha<-frac{25}{11}$. There is a saddle point when the eigenvalues are real and have opposite sign, so you need $r_1>0$ and $r_2<0$, hence $frac{1}{2}sqrt{100+44alpha}>1$, so $sqrt{100+44alpha}>2$, so $100+44alpha>4$, so $alpha>-frac{24}{11}$. A stable node occurs when both roots are negative but real, so $r_1<0$ and $r_2<0$, which happens when $-frac{25}{11}<alpha<-frac{24}{11}$.
    – symplectomorphic
    Jun 1 '15 at 5:24










  • Thank you very much! you made me realize that i had forgotten to enter the alpha <, and alpha is >
    – Angel Garcia
    Jun 1 '15 at 5:50














3












3








3







I was asked to solved for the eigenvalues in terms of α for 2X2 matrix and so i did and my answer was marked as correct. Then I was asked to solve for this:



The roots are complex when?



There is a saddle point for?



The equilibrium point is a stable node where?



I tried solving the problem by finding values which would give me 1 and zero for the eigenvalues and got the values of -24/11, -25/11 and -26/11 but the were all wrong. How would i solve for these three questions?
can someone walk me through the steps?



The eigenvalues in terms of α are:



$r = -1 + dfrac{sqrt{100 + 44 alpha}}{2}$



$r_2 = -1 - dfrac{sqrt{100 + 44 alpha}}{2}$






matrices differential-equations eigenvalues-eigenvectors matrix-equations eigenfunctions







share|cite|improve this question













I was asked to solved for the eigenvalues in terms of α for 2X2 matrix and so i did and my answer was marked as correct. Then I was asked to solve for this:



The roots are complex when?



There is a saddle point for?



The equilibrium point is a stable node where?



I tried solving the problem by finding values which would give me 1 and zero for the eigenvalues and got the values of -24/11, -25/11 and -26/11 but the were all wrong. How would i solve for these three questions?
can someone walk me through the steps?



The eigenvalues in terms of α are:



$r = -1 + dfrac{sqrt{100 + 44 alpha}}{2}$



$r_2 = -1 - dfrac{sqrt{100 + 44 alpha}}{2}$






matrices differential-equations eigenvalues-eigenvectors matrix-equations eigenfunctions




matrices differential-equations eigenvalues-eigenvectors matrix-equations eigenfunctions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jun 1 '15 at 5:17









Angel Garcia

264




264








  • 1




    The roots are complex when $100+44alpha<0$, i.e. when $alpha<-frac{25}{11}$. There is a saddle point when the eigenvalues are real and have opposite sign, so you need $r_1>0$ and $r_2<0$, hence $frac{1}{2}sqrt{100+44alpha}>1$, so $sqrt{100+44alpha}>2$, so $100+44alpha>4$, so $alpha>-frac{24}{11}$. A stable node occurs when both roots are negative but real, so $r_1<0$ and $r_2<0$, which happens when $-frac{25}{11}<alpha<-frac{24}{11}$.
    – symplectomorphic
    Jun 1 '15 at 5:24










  • Thank you very much! you made me realize that i had forgotten to enter the alpha <, and alpha is >
    – Angel Garcia
    Jun 1 '15 at 5:50














  • 1




    The roots are complex when $100+44alpha<0$, i.e. when $alpha<-frac{25}{11}$. There is a saddle point when the eigenvalues are real and have opposite sign, so you need $r_1>0$ and $r_2<0$, hence $frac{1}{2}sqrt{100+44alpha}>1$, so $sqrt{100+44alpha}>2$, so $100+44alpha>4$, so $alpha>-frac{24}{11}$. A stable node occurs when both roots are negative but real, so $r_1<0$ and $r_2<0$, which happens when $-frac{25}{11}<alpha<-frac{24}{11}$.
    – symplectomorphic
    Jun 1 '15 at 5:24










  • Thank you very much! you made me realize that i had forgotten to enter the alpha <, and alpha is >
    – Angel Garcia
    Jun 1 '15 at 5:50








1




1




The roots are complex when $100+44alpha<0$, i.e. when $alpha<-frac{25}{11}$. There is a saddle point when the eigenvalues are real and have opposite sign, so you need $r_1>0$ and $r_2<0$, hence $frac{1}{2}sqrt{100+44alpha}>1$, so $sqrt{100+44alpha}>2$, so $100+44alpha>4$, so $alpha>-frac{24}{11}$. A stable node occurs when both roots are negative but real, so $r_1<0$ and $r_2<0$, which happens when $-frac{25}{11}<alpha<-frac{24}{11}$.
– symplectomorphic
Jun 1 '15 at 5:24




The roots are complex when $100+44alpha<0$, i.e. when $alpha<-frac{25}{11}$. There is a saddle point when the eigenvalues are real and have opposite sign, so you need $r_1>0$ and $r_2<0$, hence $frac{1}{2}sqrt{100+44alpha}>1$, so $sqrt{100+44alpha}>2$, so $100+44alpha>4$, so $alpha>-frac{24}{11}$. A stable node occurs when both roots are negative but real, so $r_1<0$ and $r_2<0$, which happens when $-frac{25}{11}<alpha<-frac{24}{11}$.
– symplectomorphic
Jun 1 '15 at 5:24












Thank you very much! you made me realize that i had forgotten to enter the alpha <, and alpha is >
– Angel Garcia
Jun 1 '15 at 5:50




Thank you very much! you made me realize that i had forgotten to enter the alpha <, and alpha is >
– Angel Garcia
Jun 1 '15 at 5:50










2 Answers
2






active

oldest

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0














You can use Viete's formula to determine the sign of $alpha$:
so you will have:
$r_1 + r_2 = -2$ and $r_1 * r_2 = 1 - A*(2+A)$ where $A = frac{(100 + 44 alpha)^{1/2}}{2}$






share|cite|improve this answer





















  • The last is wrong, as $(-1-A)(-1+A)=1-A^2$ by binomial formula.
    – LutzL
    Jun 13 at 7:45



















0














symplectomorphic:




  • The roots are complex when $100+44α<0$, i.e. when $α<−frac{25}{11}$.

  • There is a saddle point when the eigenvalues are real and have opposite sign, so you need $r_1>0$ and $r_2<0$, hence $frac12sqrt{100+44α}>1$, so $100+44α>4$, so $α>−frac{24}{11}$.

  • A stable node occurs when both roots are negative but real, so $r_1<0$ and $r_2<0$, which happens when $−frac{25}{11}<α<−frac{24}{11}$.






share|cite|improve this answer























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    2 Answers
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    2 Answers
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    0














    You can use Viete's formula to determine the sign of $alpha$:
    so you will have:
    $r_1 + r_2 = -2$ and $r_1 * r_2 = 1 - A*(2+A)$ where $A = frac{(100 + 44 alpha)^{1/2}}{2}$






    share|cite|improve this answer





















    • The last is wrong, as $(-1-A)(-1+A)=1-A^2$ by binomial formula.
      – LutzL
      Jun 13 at 7:45
















    0














    You can use Viete's formula to determine the sign of $alpha$:
    so you will have:
    $r_1 + r_2 = -2$ and $r_1 * r_2 = 1 - A*(2+A)$ where $A = frac{(100 + 44 alpha)^{1/2}}{2}$






    share|cite|improve this answer





















    • The last is wrong, as $(-1-A)(-1+A)=1-A^2$ by binomial formula.
      – LutzL
      Jun 13 at 7:45














    0












    0








    0






    You can use Viete's formula to determine the sign of $alpha$:
    so you will have:
    $r_1 + r_2 = -2$ and $r_1 * r_2 = 1 - A*(2+A)$ where $A = frac{(100 + 44 alpha)^{1/2}}{2}$






    share|cite|improve this answer












    You can use Viete's formula to determine the sign of $alpha$:
    so you will have:
    $r_1 + r_2 = -2$ and $r_1 * r_2 = 1 - A*(2+A)$ where $A = frac{(100 + 44 alpha)^{1/2}}{2}$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Feb 4 at 0:19









    Dong Le

    517




    517












    • The last is wrong, as $(-1-A)(-1+A)=1-A^2$ by binomial formula.
      – LutzL
      Jun 13 at 7:45


















    • The last is wrong, as $(-1-A)(-1+A)=1-A^2$ by binomial formula.
      – LutzL
      Jun 13 at 7:45
















    The last is wrong, as $(-1-A)(-1+A)=1-A^2$ by binomial formula.
    – LutzL
    Jun 13 at 7:45




    The last is wrong, as $(-1-A)(-1+A)=1-A^2$ by binomial formula.
    – LutzL
    Jun 13 at 7:45











    0














    symplectomorphic:




    • The roots are complex when $100+44α<0$, i.e. when $α<−frac{25}{11}$.

    • There is a saddle point when the eigenvalues are real and have opposite sign, so you need $r_1>0$ and $r_2<0$, hence $frac12sqrt{100+44α}>1$, so $100+44α>4$, so $α>−frac{24}{11}$.

    • A stable node occurs when both roots are negative but real, so $r_1<0$ and $r_2<0$, which happens when $−frac{25}{11}<α<−frac{24}{11}$.






    share|cite|improve this answer




























      0














      symplectomorphic:




      • The roots are complex when $100+44α<0$, i.e. when $α<−frac{25}{11}$.

      • There is a saddle point when the eigenvalues are real and have opposite sign, so you need $r_1>0$ and $r_2<0$, hence $frac12sqrt{100+44α}>1$, so $100+44α>4$, so $α>−frac{24}{11}$.

      • A stable node occurs when both roots are negative but real, so $r_1<0$ and $r_2<0$, which happens when $−frac{25}{11}<α<−frac{24}{11}$.






      share|cite|improve this answer


























        0












        0








        0






        symplectomorphic:




        • The roots are complex when $100+44α<0$, i.e. when $α<−frac{25}{11}$.

        • There is a saddle point when the eigenvalues are real and have opposite sign, so you need $r_1>0$ and $r_2<0$, hence $frac12sqrt{100+44α}>1$, so $100+44α>4$, so $α>−frac{24}{11}$.

        • A stable node occurs when both roots are negative but real, so $r_1<0$ and $r_2<0$, which happens when $−frac{25}{11}<α<−frac{24}{11}$.






        share|cite|improve this answer














        symplectomorphic:




        • The roots are complex when $100+44α<0$, i.e. when $α<−frac{25}{11}$.

        • There is a saddle point when the eigenvalues are real and have opposite sign, so you need $r_1>0$ and $r_2<0$, hence $frac12sqrt{100+44α}>1$, so $100+44α>4$, so $α>−frac{24}{11}$.

        • A stable node occurs when both roots are negative but real, so $r_1<0$ and $r_2<0$, which happens when $−frac{25}{11}<α<−frac{24}{11}$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        answered Jun 13 at 7:43


























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