Finding the Critical points of a system
$begingroup$
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
asked Dec 13 '18 at 12:10
lastgunslingerlastgunslinger
728
$endgroup$
add a comment |
$begingroup$
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
asked Dec 13 '18 at 12:10
lastgunslingerlastgunslinger
728
$endgroup$
add a comment |
$begingroup$
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
asked Dec 13 '18 at 12:10
lastgunslingerlastgunslinger
728
$endgroup$
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
asked Dec 13 '18 at 12:10
lastgunslingerlastgunslinger
728
asked Dec 13 '18 at 12:10
lastgunslingerlastgunslinger
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asked Dec 13 '18 at 12:10
lastgunslingerlastgunslinger
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asked Dec 13 '18 at 12:10
lastgunslingerlastgunslinger
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asked Dec 13 '18 at 12:10
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2 Answers
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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.
answered Dec 13 '18 at 12:21
LutzLLutzL
59.2k42057
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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
answered Dec 13 '18 at 12:24
caveraccaverac
14.6k31130
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2 Answers
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2 Answers
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$begingroup$
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.
answered Dec 13 '18 at 12:21
LutzLLutzL
59.2k42057
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 13 '18 at 12:21
LutzLLutzL
59.2k42057
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 13 '18 at 12:21
LutzLLutzL
59.2k42057
$endgroup$
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.
answered Dec 13 '18 at 12:21
LutzLLutzL
59.2k42057
answered Dec 13 '18 at 12:21
LutzLLutzL
59.2k42057
answered Dec 13 '18 at 12:21
LutzLLutzL
59.2k42057
answered Dec 13 '18 at 12:21
LutzLLutzL
59.2k42057
59.2k42057
add a comment |
add a comment |
$begingroup$
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
answered Dec 13 '18 at 12:24
caveraccaverac
14.6k31130
$endgroup$
add a comment |
$begingroup$
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
answered Dec 13 '18 at 12:24
caveraccaverac
14.6k31130
$endgroup$
add a comment |
$begingroup$
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
answered Dec 13 '18 at 12:24
caveraccaverac
14.6k31130
$endgroup$
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
answered Dec 13 '18 at 12:24
caveraccaverac
14.6k31130
edited Dec 13 '18 at 12:29
answered Dec 13 '18 at 12:24
caveraccaverac
14.6k31130
answered Dec 13 '18 at 12:24
caveraccaverac
14.6k31130
answered Dec 13 '18 at 12:24
caveraccaverac
14.6k31130
14.6k31130
add a comment |
add a comment |
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