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Critical Points Clarification
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I'm uncertain if I have all the critical points required of the system
I do $fracdxdt = x(x-2) + y^2$...and likewise..I do $fracdydt= y(1-x)$
I then set both equations to zero
If we set $x = 0$ in the first equation, we solve the second equation $y = 0$
If we set $y = 0$ in the second equation, we solve the first equation for $x = 0$ and $x = 2$.
Likewise, I solve for y - xy = 0. Then xy = y so x = 1. We plug this solution into the first equation so that y^2 = 1 and hence, have (1,1),(1,-1) as our results as well.
Hence, my critical points are then $(0,0);(2,0)$;(1,1);(1,-1)
Is this the correct approach for this particular question?
Likewise, what does it mean to determine the stability of the critical point and the type of critical point? Is it referring to stable, unstable, and semi-stable? Likewise, is the type of critical point choosing amongst center, node, saddle point, and spiral?
Thank you
differential-equations nonlinear-system homogeneous-equation
asked Jul 22 at 19:44
KhanMan
447
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up vote
2
down vote
favorite
I'm uncertain if I have all the critical points required of the system
I do $fracdxdt = x(x-2) + y^2$...and likewise..I do $fracdydt= y(1-x)$
I then set both equations to zero
If we set $x = 0$ in the first equation, we solve the second equation $y = 0$
If we set $y = 0$ in the second equation, we solve the first equation for $x = 0$ and $x = 2$.
Likewise, I solve for y - xy = 0. Then xy = y so x = 1. We plug this solution into the first equation so that y^2 = 1 and hence, have (1,1),(1,-1) as our results as well.
Hence, my critical points are then $(0,0);(2,0)$;(1,1);(1,-1)
Is this the correct approach for this particular question?
Likewise, what does it mean to determine the stability of the critical point and the type of critical point? Is it referring to stable, unstable, and semi-stable? Likewise, is the type of critical point choosing amongst center, node, saddle point, and spiral?
Thank you
differential-equations nonlinear-system homogeneous-equation
asked Jul 22 at 19:44
KhanMan
447
1
Your critical points are correct. Regarding stability, without your giving any context (that is, what do you know?) it is not possible to respond. Perhaps Principle of Linearized Stability? Incidentally, "semi-stable" (understood in the standard way) does not make much sense for critical points in multidimensional systems of ODEs.
– user539887
Jul 22 at 20:38
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm uncertain if I have all the critical points required of the system
I do $fracdxdt = x(x-2) + y^2$...and likewise..I do $fracdydt= y(1-x)$
I then set both equations to zero
If we set $x = 0$ in the first equation, we solve the second equation $y = 0$
If we set $y = 0$ in the second equation, we solve the first equation for $x = 0$ and $x = 2$.
Likewise, I solve for y - xy = 0. Then xy = y so x = 1. We plug this solution into the first equation so that y^2 = 1 and hence, have (1,1),(1,-1) as our results as well.
Hence, my critical points are then $(0,0);(2,0)$;(1,1);(1,-1)
Is this the correct approach for this particular question?
Likewise, what does it mean to determine the stability of the critical point and the type of critical point? Is it referring to stable, unstable, and semi-stable? Likewise, is the type of critical point choosing amongst center, node, saddle point, and spiral?
Thank you
differential-equations nonlinear-system homogeneous-equation
asked Jul 22 at 19:44
KhanMan
447
I'm uncertain if I have all the critical points required of the system
I do $fracdxdt = x(x-2) + y^2$...and likewise..I do $fracdydt= y(1-x)$
I then set both equations to zero
If we set $x = 0$ in the first equation, we solve the second equation $y = 0$
If we set $y = 0$ in the second equation, we solve the first equation for $x = 0$ and $x = 2$.
Likewise, I solve for y - xy = 0. Then xy = y so x = 1. We plug this solution into the first equation so that y^2 = 1 and hence, have (1,1),(1,-1) as our results as well.
Hence, my critical points are then $(0,0);(2,0)$;(1,1);(1,-1)
Is this the correct approach for this particular question?
Likewise, what does it mean to determine the stability of the critical point and the type of critical point? Is it referring to stable, unstable, and semi-stable? Likewise, is the type of critical point choosing amongst center, node, saddle point, and spiral?
Thank you
differential-equations nonlinear-system homogeneous-equation
asked Jul 22 at 19:44
KhanMan
447
edited Jul 22 at 20:06
asked Jul 22 at 19:44
KhanMan
447
asked Jul 22 at 19:44
KhanMan
447
asked Jul 22 at 19:44
KhanMan
447
447
1
Your critical points are correct. Regarding stability, without your giving any context (that is, what do you know?) it is not possible to respond. Perhaps Principle of Linearized Stability? Incidentally, "semi-stable" (understood in the standard way) does not make much sense for critical points in multidimensional systems of ODEs.
– user539887
Jul 22 at 20:38
add a comment |Â
1
Your critical points are correct. Regarding stability, without your giving any context (that is, what do you know?) it is not possible to respond. Perhaps Principle of Linearized Stability? Incidentally, "semi-stable" (understood in the standard way) does not make much sense for critical points in multidimensional systems of ODEs.
– user539887
Jul 22 at 20:38
1
1
Your critical points are correct. Regarding stability, without your giving any context (that is, what do you know?) it is not possible to respond. Perhaps Principle of Linearized Stability? Incidentally, "semi-stable" (understood in the standard way) does not make much sense for critical points in multidimensional systems of ODEs.
– user539887
Jul 22 at 20:38
Your critical points are correct. Regarding stability, without your giving any context (that is, what do you know?) it is not possible to respond. Perhaps Principle of Linearized Stability? Incidentally, "semi-stable" (understood in the standard way) does not make much sense for critical points in multidimensional systems of ODEs.
– user539887
Jul 22 at 20:38
add a comment |Â
1 Answer
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up vote
1
down vote
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I would say a good start has been made on the right track.
Given the system
$dot x = X(x, y) = -2x + x^2 + y^2, tag 1$
$dot y = Y(x, y) = y - xy, tag 2$
I would find the critical points by first noting that
$y - xy = dot y = 0 tag 3$
implies
$y = xy; tag 4$
thus if $y ne 0, tag 5$
we find
$x = 1; tag 6$
now from (1) with $dot x = 0$,
$-2(1) + 1^2 + y^2 = 0, tag 7$
that is,
$y^2 - 1 = 0 Longrightarrow y = pm 1; tag 8$
thus the critical points with $y ne 0$ are $(1, -1)$ and $(1, 1)$. If $y = 0$ then (1) becomes
$x^2 - 2x = 0, tag 9$
yielding
$x = 0, 2; tag10$
therefore the critical points with $y = 0$ are $(0, 0)$ and $(2, 0)$.
So far, so good; our OP Khan Man and myself agree on the locations of the zeroes of the system (1)-(2).
I guess it's easiest to explain what it means to determine the stability and type of the critical points by proceeding with this example by way of illustration. We may find the Jacobean, or matrix of derivatives, of the system (1)-(2):
$J(x, y) = beginbmatrix dfracpartial Xpartial x & dfracpartial Xpartial y \ dfracpartial Ypartial x & dfracpartial Ypartial y endbmatrix = beginbmatrix 2x - 2 & 2y \ -y & 1 - x endbmatrix; tag11$
at the critical point $(0, 0)$ this becomes
$J(0, 0) = beginbmatrix -2 & 0 \ 0 & 1 endbmatrix; tag12$
since the eigenvalues of $J(0, 0)$, $-2$ and $1$, are real and of opposite signs, $(0, 0)$ is an unstable saddle point; similarly, we may compute the Jacobean at the other zeroes of our system:
$J(2, 0) = beginbmatrix 2 & 0 \ 0 & -1 endbmatrix; tag13$
now the eigenvalues are $2$, $-1$, so we again find an unstable saddle; we proceed:
$J(1, -1) = beginbmatrix 0 & -2 \ 1 & 0 endbmatrix; tag14$
since $J(1, -1)$ is not diagonal we must resort to its characterisic polynomial to find the eigenvalues:
$det (J(1, -1) - mu I) = det left (beginbmatrix -mu & -2 \ 1 & -mu endbmatrix right ) = mu^2 + 2 = 0; tag15$
here the eigenvalues are $mu = pm sqrt 2 i$; the point $(1, -1)$ is a center; since the eigenvalues have $0$ real part, we cannot determine the stability from the Jacobean alone; in such cases, a more detaied analysis is required; likewise,
$J(1, 1) = beginbmatrix 0 & 2 \ -1 & 0 endbmatrix, tag16$
and again we find that
$mu^2 + 2 = 0, ; mu = pm sqrt 2 i, tag17$
so again we conclude that $(1, 1)$ is a center and the stability is indeterminate from this linearized analysis alone.
To summarize what has been illustrated here, the stability of a critical point is the property that trajectories initiated sufficiently close to it will remain so; generally this is associated with every eigenvalue of $J$ at the point in question having negative real part; if some eigenvalue has positive real part, then the point is unstable, and there will generally be a trajectory which moves away from the point, at least for awhile; our saddles exhibit this behavior, which is explained rigorously and in detail in the fine reference cited by user539887 in his comment on the question itself. There are no nodes or spiral points here, since they require all eiigenvalues to have non-zero real part of the same sign. A detailed explanation of all this may be found here.
answered Jul 22 at 21:07
Robert Lewis
36.9k22255
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I would say a good start has been made on the right track.
Given the system
$dot x = X(x, y) = -2x + x^2 + y^2, tag 1$
$dot y = Y(x, y) = y - xy, tag 2$
I would find the critical points by first noting that
$y - xy = dot y = 0 tag 3$
implies
$y = xy; tag 4$
thus if $y ne 0, tag 5$
we find
$x = 1; tag 6$
now from (1) with $dot x = 0$,
$-2(1) + 1^2 + y^2 = 0, tag 7$
that is,
$y^2 - 1 = 0 Longrightarrow y = pm 1; tag 8$
thus the critical points with $y ne 0$ are $(1, -1)$ and $(1, 1)$. If $y = 0$ then (1) becomes
$x^2 - 2x = 0, tag 9$
yielding
$x = 0, 2; tag10$
therefore the critical points with $y = 0$ are $(0, 0)$ and $(2, 0)$.
So far, so good; our OP Khan Man and myself agree on the locations of the zeroes of the system (1)-(2).
I guess it's easiest to explain what it means to determine the stability and type of the critical points by proceeding with this example by way of illustration. We may find the Jacobean, or matrix of derivatives, of the system (1)-(2):
$J(x, y) = beginbmatrix dfracpartial Xpartial x & dfracpartial Xpartial y \ dfracpartial Ypartial x & dfracpartial Ypartial y endbmatrix = beginbmatrix 2x - 2 & 2y \ -y & 1 - x endbmatrix; tag11$
at the critical point $(0, 0)$ this becomes
$J(0, 0) = beginbmatrix -2 & 0 \ 0 & 1 endbmatrix; tag12$
since the eigenvalues of $J(0, 0)$, $-2$ and $1$, are real and of opposite signs, $(0, 0)$ is an unstable saddle point; similarly, we may compute the Jacobean at the other zeroes of our system:
$J(2, 0) = beginbmatrix 2 & 0 \ 0 & -1 endbmatrix; tag13$
now the eigenvalues are $2$, $-1$, so we again find an unstable saddle; we proceed:
$J(1, -1) = beginbmatrix 0 & -2 \ 1 & 0 endbmatrix; tag14$
since $J(1, -1)$ is not diagonal we must resort to its characterisic polynomial to find the eigenvalues:
$det (J(1, -1) - mu I) = det left (beginbmatrix -mu & -2 \ 1 & -mu endbmatrix right ) = mu^2 + 2 = 0; tag15$
here the eigenvalues are $mu = pm sqrt 2 i$; the point $(1, -1)$ is a center; since the eigenvalues have $0$ real part, we cannot determine the stability from the Jacobean alone; in such cases, a more detaied analysis is required; likewise,
$J(1, 1) = beginbmatrix 0 & 2 \ -1 & 0 endbmatrix, tag16$
and again we find that
$mu^2 + 2 = 0, ; mu = pm sqrt 2 i, tag17$
so again we conclude that $(1, 1)$ is a center and the stability is indeterminate from this linearized analysis alone.
To summarize what has been illustrated here, the stability of a critical point is the property that trajectories initiated sufficiently close to it will remain so; generally this is associated with every eigenvalue of $J$ at the point in question having negative real part; if some eigenvalue has positive real part, then the point is unstable, and there will generally be a trajectory which moves away from the point, at least for awhile; our saddles exhibit this behavior, which is explained rigorously and in detail in the fine reference cited by user539887 in his comment on the question itself. There are no nodes or spiral points here, since they require all eiigenvalues to have non-zero real part of the same sign. A detailed explanation of all this may be found here.
answered Jul 22 at 21:07
Robert Lewis
36.9k22255
add a comment |Â
up vote
1
down vote
accepted
I would say a good start has been made on the right track.
Given the system
$dot x = X(x, y) = -2x + x^2 + y^2, tag 1$
$dot y = Y(x, y) = y - xy, tag 2$
I would find the critical points by first noting that
$y - xy = dot y = 0 tag 3$
implies
$y = xy; tag 4$
thus if $y ne 0, tag 5$
we find
$x = 1; tag 6$
now from (1) with $dot x = 0$,
$-2(1) + 1^2 + y^2 = 0, tag 7$
that is,
$y^2 - 1 = 0 Longrightarrow y = pm 1; tag 8$
thus the critical points with $y ne 0$ are $(1, -1)$ and $(1, 1)$. If $y = 0$ then (1) becomes
$x^2 - 2x = 0, tag 9$
yielding
$x = 0, 2; tag10$
therefore the critical points with $y = 0$ are $(0, 0)$ and $(2, 0)$.
So far, so good; our OP Khan Man and myself agree on the locations of the zeroes of the system (1)-(2).
I guess it's easiest to explain what it means to determine the stability and type of the critical points by proceeding with this example by way of illustration. We may find the Jacobean, or matrix of derivatives, of the system (1)-(2):
$J(x, y) = beginbmatrix dfracpartial Xpartial x & dfracpartial Xpartial y \ dfracpartial Ypartial x & dfracpartial Ypartial y endbmatrix = beginbmatrix 2x - 2 & 2y \ -y & 1 - x endbmatrix; tag11$
at the critical point $(0, 0)$ this becomes
$J(0, 0) = beginbmatrix -2 & 0 \ 0 & 1 endbmatrix; tag12$
since the eigenvalues of $J(0, 0)$, $-2$ and $1$, are real and of opposite signs, $(0, 0)$ is an unstable saddle point; similarly, we may compute the Jacobean at the other zeroes of our system:
$J(2, 0) = beginbmatrix 2 & 0 \ 0 & -1 endbmatrix; tag13$
now the eigenvalues are $2$, $-1$, so we again find an unstable saddle; we proceed:
$J(1, -1) = beginbmatrix 0 & -2 \ 1 & 0 endbmatrix; tag14$
since $J(1, -1)$ is not diagonal we must resort to its characterisic polynomial to find the eigenvalues:
$det (J(1, -1) - mu I) = det left (beginbmatrix -mu & -2 \ 1 & -mu endbmatrix right ) = mu^2 + 2 = 0; tag15$
here the eigenvalues are $mu = pm sqrt 2 i$; the point $(1, -1)$ is a center; since the eigenvalues have $0$ real part, we cannot determine the stability from the Jacobean alone; in such cases, a more detaied analysis is required; likewise,
$J(1, 1) = beginbmatrix 0 & 2 \ -1 & 0 endbmatrix, tag16$
and again we find that
$mu^2 + 2 = 0, ; mu = pm sqrt 2 i, tag17$
so again we conclude that $(1, 1)$ is a center and the stability is indeterminate from this linearized analysis alone.
To summarize what has been illustrated here, the stability of a critical point is the property that trajectories initiated sufficiently close to it will remain so; generally this is associated with every eigenvalue of $J$ at the point in question having negative real part; if some eigenvalue has positive real part, then the point is unstable, and there will generally be a trajectory which moves away from the point, at least for awhile; our saddles exhibit this behavior, which is explained rigorously and in detail in the fine reference cited by user539887 in his comment on the question itself. There are no nodes or spiral points here, since they require all eiigenvalues to have non-zero real part of the same sign. A detailed explanation of all this may be found here.
answered Jul 22 at 21:07
Robert Lewis
36.9k22255
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I would say a good start has been made on the right track.
Given the system
$dot x = X(x, y) = -2x + x^2 + y^2, tag 1$
$dot y = Y(x, y) = y - xy, tag 2$
I would find the critical points by first noting that
$y - xy = dot y = 0 tag 3$
implies
$y = xy; tag 4$
thus if $y ne 0, tag 5$
we find
$x = 1; tag 6$
now from (1) with $dot x = 0$,
$-2(1) + 1^2 + y^2 = 0, tag 7$
that is,
$y^2 - 1 = 0 Longrightarrow y = pm 1; tag 8$
thus the critical points with $y ne 0$ are $(1, -1)$ and $(1, 1)$. If $y = 0$ then (1) becomes
$x^2 - 2x = 0, tag 9$
yielding
$x = 0, 2; tag10$
therefore the critical points with $y = 0$ are $(0, 0)$ and $(2, 0)$.
So far, so good; our OP Khan Man and myself agree on the locations of the zeroes of the system (1)-(2).
I guess it's easiest to explain what it means to determine the stability and type of the critical points by proceeding with this example by way of illustration. We may find the Jacobean, or matrix of derivatives, of the system (1)-(2):
$J(x, y) = beginbmatrix dfracpartial Xpartial x & dfracpartial Xpartial y \ dfracpartial Ypartial x & dfracpartial Ypartial y endbmatrix = beginbmatrix 2x - 2 & 2y \ -y & 1 - x endbmatrix; tag11$
at the critical point $(0, 0)$ this becomes
$J(0, 0) = beginbmatrix -2 & 0 \ 0 & 1 endbmatrix; tag12$
since the eigenvalues of $J(0, 0)$, $-2$ and $1$, are real and of opposite signs, $(0, 0)$ is an unstable saddle point; similarly, we may compute the Jacobean at the other zeroes of our system:
$J(2, 0) = beginbmatrix 2 & 0 \ 0 & -1 endbmatrix; tag13$
now the eigenvalues are $2$, $-1$, so we again find an unstable saddle; we proceed:
$J(1, -1) = beginbmatrix 0 & -2 \ 1 & 0 endbmatrix; tag14$
since $J(1, -1)$ is not diagonal we must resort to its characterisic polynomial to find the eigenvalues:
$det (J(1, -1) - mu I) = det left (beginbmatrix -mu & -2 \ 1 & -mu endbmatrix right ) = mu^2 + 2 = 0; tag15$
here the eigenvalues are $mu = pm sqrt 2 i$; the point $(1, -1)$ is a center; since the eigenvalues have $0$ real part, we cannot determine the stability from the Jacobean alone; in such cases, a more detaied analysis is required; likewise,
$J(1, 1) = beginbmatrix 0 & 2 \ -1 & 0 endbmatrix, tag16$
and again we find that
$mu^2 + 2 = 0, ; mu = pm sqrt 2 i, tag17$
so again we conclude that $(1, 1)$ is a center and the stability is indeterminate from this linearized analysis alone.
To summarize what has been illustrated here, the stability of a critical point is the property that trajectories initiated sufficiently close to it will remain so; generally this is associated with every eigenvalue of $J$ at the point in question having negative real part; if some eigenvalue has positive real part, then the point is unstable, and there will generally be a trajectory which moves away from the point, at least for awhile; our saddles exhibit this behavior, which is explained rigorously and in detail in the fine reference cited by user539887 in his comment on the question itself. There are no nodes or spiral points here, since they require all eiigenvalues to have non-zero real part of the same sign. A detailed explanation of all this may be found here.
answered Jul 22 at 21:07
Robert Lewis
36.9k22255
I would say a good start has been made on the right track.
Given the system
$dot x = X(x, y) = -2x + x^2 + y^2, tag 1$
$dot y = Y(x, y) = y - xy, tag 2$
I would find the critical points by first noting that
$y - xy = dot y = 0 tag 3$
implies
$y = xy; tag 4$
thus if $y ne 0, tag 5$
we find
$x = 1; tag 6$
now from (1) with $dot x = 0$,
$-2(1) + 1^2 + y^2 = 0, tag 7$
that is,
$y^2 - 1 = 0 Longrightarrow y = pm 1; tag 8$
thus the critical points with $y ne 0$ are $(1, -1)$ and $(1, 1)$. If $y = 0$ then (1) becomes
$x^2 - 2x = 0, tag 9$
yielding
$x = 0, 2; tag10$
therefore the critical points with $y = 0$ are $(0, 0)$ and $(2, 0)$.
So far, so good; our OP Khan Man and myself agree on the locations of the zeroes of the system (1)-(2).
I guess it's easiest to explain what it means to determine the stability and type of the critical points by proceeding with this example by way of illustration. We may find the Jacobean, or matrix of derivatives, of the system (1)-(2):
$J(x, y) = beginbmatrix dfracpartial Xpartial x & dfracpartial Xpartial y \ dfracpartial Ypartial x & dfracpartial Ypartial y endbmatrix = beginbmatrix 2x - 2 & 2y \ -y & 1 - x endbmatrix; tag11$
at the critical point $(0, 0)$ this becomes
$J(0, 0) = beginbmatrix -2 & 0 \ 0 & 1 endbmatrix; tag12$
since the eigenvalues of $J(0, 0)$, $-2$ and $1$, are real and of opposite signs, $(0, 0)$ is an unstable saddle point; similarly, we may compute the Jacobean at the other zeroes of our system:
$J(2, 0) = beginbmatrix 2 & 0 \ 0 & -1 endbmatrix; tag13$
now the eigenvalues are $2$, $-1$, so we again find an unstable saddle; we proceed:
$J(1, -1) = beginbmatrix 0 & -2 \ 1 & 0 endbmatrix; tag14$
since $J(1, -1)$ is not diagonal we must resort to its characterisic polynomial to find the eigenvalues:
$det (J(1, -1) - mu I) = det left (beginbmatrix -mu & -2 \ 1 & -mu endbmatrix right ) = mu^2 + 2 = 0; tag15$
here the eigenvalues are $mu = pm sqrt 2 i$; the point $(1, -1)$ is a center; since the eigenvalues have $0$ real part, we cannot determine the stability from the Jacobean alone; in such cases, a more detaied analysis is required; likewise,
$J(1, 1) = beginbmatrix 0 & 2 \ -1 & 0 endbmatrix, tag16$
and again we find that
$mu^2 + 2 = 0, ; mu = pm sqrt 2 i, tag17$
so again we conclude that $(1, 1)$ is a center and the stability is indeterminate from this linearized analysis alone.
To summarize what has been illustrated here, the stability of a critical point is the property that trajectories initiated sufficiently close to it will remain so; generally this is associated with every eigenvalue of $J$ at the point in question having negative real part; if some eigenvalue has positive real part, then the point is unstable, and there will generally be a trajectory which moves away from the point, at least for awhile; our saddles exhibit this behavior, which is explained rigorously and in detail in the fine reference cited by user539887 in his comment on the question itself. There are no nodes or spiral points here, since they require all eiigenvalues to have non-zero real part of the same sign. A detailed explanation of all this may be found here.
answered Jul 22 at 21:07
Robert Lewis
36.9k22255
edited Jul 22 at 22:31
answered Jul 22 at 21:07
Robert Lewis
36.9k22255
answered Jul 22 at 21:07
Robert Lewis
36.9k22255
answered Jul 22 at 21:07
Robert Lewis
36.9k22255
36.9k22255
add a comment |Â
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1
Your critical points are correct. Regarding stability, without your giving any context (that is, what do you know?) it is not possible to respond. Perhaps Principle of Linearized Stability? Incidentally, "semi-stable" (understood in the standard way) does not make much sense for critical points in multidimensional systems of ODEs.
– user539887
Jul 22 at 20:38