Help finding the critical values of α where the qualitative nature of the phase portrait for the system...
$begingroup$
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 ordinary-differential-equations eigenvalues-eigenvectors matrix-equations eigenfunctions
asked Jun 1 '15 at 5:17
Angel GarciaAngel Garcia
264
$endgroup$
add a comment |
$begingroup$
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 ordinary-differential-equations eigenvalues-eigenvectors matrix-equations eigenfunctions
asked Jun 1 '15 at 5:17
Angel GarciaAngel Garcia
264
$endgroup$
1
$begingroup$
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 $-frac{25}{11}<alpha<-frac{24}{11}$.
$endgroup$
– symplectomorphic
Jun 1 '15 at 5:24
$begingroup$
Thank you very much! you made me realize that i had forgotten to enter the alpha <, and alpha is >
$endgroup$
– Angel Garcia
Jun 1 '15 at 5:50
add a comment |
$begingroup$
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 ordinary-differential-equations eigenvalues-eigenvectors matrix-equations eigenfunctions
asked Jun 1 '15 at 5:17
Angel GarciaAngel Garcia
264
$endgroup$
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 ordinary-differential-equations eigenvalues-eigenvectors matrix-equations eigenfunctions
matrices ordinary-differential-equations eigenvalues-eigenvectors matrix-equations eigenfunctions
asked Jun 1 '15 at 5:17
Angel GarciaAngel Garcia
264
asked Jun 1 '15 at 5:17
Angel GarciaAngel Garcia
264
asked Jun 1 '15 at 5:17
Angel GarciaAngel Garcia
264
asked Jun 1 '15 at 5:17
Angel GarciaAngel Garcia
264
asked Jun 1 '15 at 5:17
Angel GarciaAngel Garcia
264
264
1
$begingroup$
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 $-frac{25}{11}<alpha<-frac{24}{11}$.
$endgroup$
– symplectomorphic
Jun 1 '15 at 5:24
$begingroup$
Thank you very much! you made me realize that i had forgotten to enter the alpha <, and alpha is >
$endgroup$
– Angel Garcia
Jun 1 '15 at 5:50
add a comment |
1
$begingroup$
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 $-frac{25}{11}<alpha<-frac{24}{11}$.
$endgroup$
– symplectomorphic
Jun 1 '15 at 5:24
$begingroup$
Thank you very much! you made me realize that i had forgotten to enter the alpha <, and alpha is >
$endgroup$
– Angel Garcia
Jun 1 '15 at 5:50
1
1
$begingroup$
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 $-frac{25}{11}<alpha<-frac{24}{11}$.
$endgroup$
– symplectomorphic
Jun 1 '15 at 5:24
$begingroup$
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 $-frac{25}{11}<alpha<-frac{24}{11}$.
$endgroup$
– symplectomorphic
Jun 1 '15 at 5:24
$begingroup$
Thank you very much! you made me realize that i had forgotten to enter the alpha <, and alpha is >
$endgroup$
– Angel Garcia
Jun 1 '15 at 5:50
$begingroup$
Thank you very much! you made me realize that i had forgotten to enter the alpha <, and alpha is >
$endgroup$
– Angel Garcia
Jun 1 '15 at 5:50
add a comment |
2 Answers
2
active
oldest
votes
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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 $A = frac{(100 + 44 alpha)^{1/2}}{2}$
answered Feb 4 '18 at 0:19
Dong LeDong Le
717
$endgroup$
$begingroup$
The last is wrong, as $(-1-A)(-1+A)=1-A^2$ by binomial formula.
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– LutzL
Jun 13 '18 at 7:45
add a comment |
$begingroup$
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 $−frac{25}{11}<α<−frac{24}{11}$.
$endgroup$
add a comment |
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2 Answers
2
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2 Answers
2
active
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$begingroup$
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 $A = frac{(100 + 44 alpha)^{1/2}}{2}$
answered Feb 4 '18 at 0:19
Dong LeDong Le
717
$endgroup$
$begingroup$
The last is wrong, as $(-1-A)(-1+A)=1-A^2$ by binomial formula.
$endgroup$
– LutzL
Jun 13 '18 at 7:45
add a comment |
$begingroup$
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 $A = frac{(100 + 44 alpha)^{1/2}}{2}$
answered Feb 4 '18 at 0:19
Dong LeDong Le
717
$endgroup$
$begingroup$
The last is wrong, as $(-1-A)(-1+A)=1-A^2$ by binomial formula.
$endgroup$
– LutzL
Jun 13 '18 at 7:45
add a comment |
$begingroup$
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 $A = frac{(100 + 44 alpha)^{1/2}}{2}$
answered Feb 4 '18 at 0:19
Dong LeDong Le
717
$endgroup$
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 $A = frac{(100 + 44 alpha)^{1/2}}{2}$
answered Feb 4 '18 at 0:19
Dong LeDong Le
717
answered Feb 4 '18 at 0:19
Dong LeDong Le
717
answered Feb 4 '18 at 0:19
Dong LeDong Le
717
answered Feb 4 '18 at 0:19
Dong LeDong Le
717
717
$begingroup$
The last is wrong, as $(-1-A)(-1+A)=1-A^2$ by binomial formula.
$endgroup$
– LutzL
Jun 13 '18 at 7:45
add a comment |
$begingroup$
The last is wrong, as $(-1-A)(-1+A)=1-A^2$ by binomial formula.
$endgroup$
– LutzL
Jun 13 '18 at 7:45
$begingroup$
The last is wrong, as $(-1-A)(-1+A)=1-A^2$ by binomial formula.
$endgroup$
– LutzL
Jun 13 '18 at 7:45
$begingroup$
The last is wrong, as $(-1-A)(-1+A)=1-A^2$ by binomial formula.
$endgroup$
– LutzL
Jun 13 '18 at 7:45
add a comment |
$begingroup$
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 $−frac{25}{11}<α<−frac{24}{11}$.
$endgroup$
add a comment |
$begingroup$
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 $−frac{25}{11}<α<−frac{24}{11}$.
$endgroup$
add a comment |
$begingroup$
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 $−frac{25}{11}<α<−frac{24}{11}$.
$endgroup$
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 $−frac{25}{11}<α<−frac{24}{11}$.
answered Jun 13 '18 at 7:43
community wiki
LutzL
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$begingroup$
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 $-frac{25}{11}<alpha<-frac{24}{11}$.
$endgroup$
– symplectomorphic
Jun 1 '15 at 5:24
$begingroup$
Thank you very much! you made me realize that i had forgotten to enter the alpha <, and alpha is >
$endgroup$
– Angel Garcia
Jun 1 '15 at 5:50