Question about equilibrium points of autonomous ODEs
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My professor stated that if $x' = f(x)$ (so it is autonomous), then the equilibrium points are the set of $x_{eq}$ such that $f(x_{eq}) = 0$. Although this makes intuitive sense, I believe that this statement is not perfectly true. For example, consider $x ' = x^{frac{1}{3}}$. There are infinitely many solutions, one of which is $x(t) = left(frac{2}{3}tright)^{frac{3}{2}}$. We have there that $f(0) = 0$, yet $0$ is not an equilibrium point of $x(t)$ since $x(0)neq x(1)$. Is there a condition I am missing for this condition to classify equilibrium points, or am I misunderstanding equilibrium points?
ordinary-differential-equations
asked Jan 12 at 20:58
J. PistachioJ. Pistachio
488212
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add a comment |
$begingroup$
My professor stated that if $x' = f(x)$ (so it is autonomous), then the equilibrium points are the set of $x_{eq}$ such that $f(x_{eq}) = 0$. Although this makes intuitive sense, I believe that this statement is not perfectly true. For example, consider $x ' = x^{frac{1}{3}}$. There are infinitely many solutions, one of which is $x(t) = left(frac{2}{3}tright)^{frac{3}{2}}$. We have there that $f(0) = 0$, yet $0$ is not an equilibrium point of $x(t)$ since $x(0)neq x(1)$. Is there a condition I am missing for this condition to classify equilibrium points, or am I misunderstanding equilibrium points?
ordinary-differential-equations
asked Jan 12 at 20:58
J. PistachioJ. Pistachio
488212
$endgroup$
add a comment |
$begingroup$
My professor stated that if $x' = f(x)$ (so it is autonomous), then the equilibrium points are the set of $x_{eq}$ such that $f(x_{eq}) = 0$. Although this makes intuitive sense, I believe that this statement is not perfectly true. For example, consider $x ' = x^{frac{1}{3}}$. There are infinitely many solutions, one of which is $x(t) = left(frac{2}{3}tright)^{frac{3}{2}}$. We have there that $f(0) = 0$, yet $0$ is not an equilibrium point of $x(t)$ since $x(0)neq x(1)$. Is there a condition I am missing for this condition to classify equilibrium points, or am I misunderstanding equilibrium points?
ordinary-differential-equations
asked Jan 12 at 20:58
J. PistachioJ. Pistachio
488212
$endgroup$
My professor stated that if $x' = f(x)$ (so it is autonomous), then the equilibrium points are the set of $x_{eq}$ such that $f(x_{eq}) = 0$. Although this makes intuitive sense, I believe that this statement is not perfectly true. For example, consider $x ' = x^{frac{1}{3}}$. There are infinitely many solutions, one of which is $x(t) = left(frac{2}{3}tright)^{frac{3}{2}}$. We have there that $f(0) = 0$, yet $0$ is not an equilibrium point of $x(t)$ since $x(0)neq x(1)$. Is there a condition I am missing for this condition to classify equilibrium points, or am I misunderstanding equilibrium points?
ordinary-differential-equations
ordinary-differential-equations
asked Jan 12 at 20:58
J. PistachioJ. Pistachio
488212
asked Jan 12 at 20:58
J. PistachioJ. Pistachio
488212
asked Jan 12 at 20:58
J. PistachioJ. Pistachio
488212
asked Jan 12 at 20:58
J. PistachioJ. Pistachio
488212
asked Jan 12 at 20:58
J. PistachioJ. Pistachio
488212
488212
add a comment |
add a comment |
1 Answer
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This is just the definition of a stationary or equilibrium point, to be a root of the right side. The constant function with that point as value is a solution. This does not include stability or uniqueness properties, these are additional.
answered Jan 12 at 21:21
LutzLLutzL
57.5k42054
$endgroup$
$begingroup$
Correct me if I'm wrong, but stability seems to be focused on the behavior of solutions with initial conditions near the equilibrium point. My issue is that I don't see $0$ as an equilibrium point for the solution $x(t) = left(frac{2}{3}tright)^{frac{3}{2}}$. Does this mean that equilibrium points only exist for a particular solution of an ODE? If we just check that $f'(x_{eq}) = 0$, how do we know which solutions of the ODE this makes sense for?
$endgroup$
– J. Pistachio
Jan 12 at 21:28
1
$begingroup$
But $x(t)=0$ is also a solution, and that is all that is required to characterize this as stationary point. You seem to be concentrated on solutions, but equilibrium points are a property of the phase space, not of particular solutions.
$endgroup$
– LutzL
Jan 12 at 21:31
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
This is just the definition of a stationary or equilibrium point, to be a root of the right side. The constant function with that point as value is a solution. This does not include stability or uniqueness properties, these are additional.
answered Jan 12 at 21:21
LutzLLutzL
57.5k42054
$endgroup$
$begingroup$
Correct me if I'm wrong, but stability seems to be focused on the behavior of solutions with initial conditions near the equilibrium point. My issue is that I don't see $0$ as an equilibrium point for the solution $x(t) = left(frac{2}{3}tright)^{frac{3}{2}}$. Does this mean that equilibrium points only exist for a particular solution of an ODE? If we just check that $f'(x_{eq}) = 0$, how do we know which solutions of the ODE this makes sense for?
$endgroup$
– J. Pistachio
Jan 12 at 21:28
1
$begingroup$
But $x(t)=0$ is also a solution, and that is all that is required to characterize this as stationary point. You seem to be concentrated on solutions, but equilibrium points are a property of the phase space, not of particular solutions.
$endgroup$
– LutzL
Jan 12 at 21:31
add a comment |
$begingroup$
This is just the definition of a stationary or equilibrium point, to be a root of the right side. The constant function with that point as value is a solution. This does not include stability or uniqueness properties, these are additional.
answered Jan 12 at 21:21
LutzLLutzL
57.5k42054
$endgroup$
$begingroup$
Correct me if I'm wrong, but stability seems to be focused on the behavior of solutions with initial conditions near the equilibrium point. My issue is that I don't see $0$ as an equilibrium point for the solution $x(t) = left(frac{2}{3}tright)^{frac{3}{2}}$. Does this mean that equilibrium points only exist for a particular solution of an ODE? If we just check that $f'(x_{eq}) = 0$, how do we know which solutions of the ODE this makes sense for?
$endgroup$
– J. Pistachio
Jan 12 at 21:28
1
$begingroup$
But $x(t)=0$ is also a solution, and that is all that is required to characterize this as stationary point. You seem to be concentrated on solutions, but equilibrium points are a property of the phase space, not of particular solutions.
$endgroup$
– LutzL
Jan 12 at 21:31
add a comment |
$begingroup$
This is just the definition of a stationary or equilibrium point, to be a root of the right side. The constant function with that point as value is a solution. This does not include stability or uniqueness properties, these are additional.
answered Jan 12 at 21:21
LutzLLutzL
57.5k42054
$endgroup$
This is just the definition of a stationary or equilibrium point, to be a root of the right side. The constant function with that point as value is a solution. This does not include stability or uniqueness properties, these are additional.
answered Jan 12 at 21:21
LutzLLutzL
57.5k42054
answered Jan 12 at 21:21
LutzLLutzL
57.5k42054
answered Jan 12 at 21:21
LutzLLutzL
57.5k42054
answered Jan 12 at 21:21
LutzLLutzL
57.5k42054
57.5k42054
$begingroup$
Correct me if I'm wrong, but stability seems to be focused on the behavior of solutions with initial conditions near the equilibrium point. My issue is that I don't see $0$ as an equilibrium point for the solution $x(t) = left(frac{2}{3}tright)^{frac{3}{2}}$. Does this mean that equilibrium points only exist for a particular solution of an ODE? If we just check that $f'(x_{eq}) = 0$, how do we know which solutions of the ODE this makes sense for?
$endgroup$
– J. Pistachio
Jan 12 at 21:28
1
$begingroup$
But $x(t)=0$ is also a solution, and that is all that is required to characterize this as stationary point. You seem to be concentrated on solutions, but equilibrium points are a property of the phase space, not of particular solutions.
$endgroup$
– LutzL
Jan 12 at 21:31
add a comment |
$begingroup$
Correct me if I'm wrong, but stability seems to be focused on the behavior of solutions with initial conditions near the equilibrium point. My issue is that I don't see $0$ as an equilibrium point for the solution $x(t) = left(frac{2}{3}tright)^{frac{3}{2}}$. Does this mean that equilibrium points only exist for a particular solution of an ODE? If we just check that $f'(x_{eq}) = 0$, how do we know which solutions of the ODE this makes sense for?
$endgroup$
– J. Pistachio
Jan 12 at 21:28
1
$begingroup$
But $x(t)=0$ is also a solution, and that is all that is required to characterize this as stationary point. You seem to be concentrated on solutions, but equilibrium points are a property of the phase space, not of particular solutions.
$endgroup$
– LutzL
Jan 12 at 21:31
$begingroup$
Correct me if I'm wrong, but stability seems to be focused on the behavior of solutions with initial conditions near the equilibrium point. My issue is that I don't see $0$ as an equilibrium point for the solution $x(t) = left(frac{2}{3}tright)^{frac{3}{2}}$. Does this mean that equilibrium points only exist for a particular solution of an ODE? If we just check that $f'(x_{eq}) = 0$, how do we know which solutions of the ODE this makes sense for?
$endgroup$
– J. Pistachio
Jan 12 at 21:28
$begingroup$
Correct me if I'm wrong, but stability seems to be focused on the behavior of solutions with initial conditions near the equilibrium point. My issue is that I don't see $0$ as an equilibrium point for the solution $x(t) = left(frac{2}{3}tright)^{frac{3}{2}}$. Does this mean that equilibrium points only exist for a particular solution of an ODE? If we just check that $f'(x_{eq}) = 0$, how do we know which solutions of the ODE this makes sense for?
$endgroup$
– J. Pistachio
Jan 12 at 21:28
1
1
$begingroup$
But $x(t)=0$ is also a solution, and that is all that is required to characterize this as stationary point. You seem to be concentrated on solutions, but equilibrium points are a property of the phase space, not of particular solutions.
$endgroup$
– LutzL
Jan 12 at 21:31
$begingroup$
But $x(t)=0$ is also a solution, and that is all that is required to characterize this as stationary point. You seem to be concentrated on solutions, but equilibrium points are a property of the phase space, not of particular solutions.
$endgroup$
– LutzL
Jan 12 at 21:31
add a comment |
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