How do I classify the equilibrium points of this dynamical system?
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Consider the dynamical system $$dot{r}=r(3-2r-s)$$
$$dot{s}=s(2-r-s)$$
Find and classify the equilibrium points of the system.
I have found the equilibrium points to be $(0,0), (0,2), (frac{3}{2},0)$ and $(1,1)$ simply by equating the given equations to $0$ and solving.
What we have been taught in my lectures is how to classify equilibrium points of a dynamical system in the form
$$dot{x}=ax+by$$
$$dot{y}=cx+dy$$
by writing the system in matrix form and using the method of finding the eigenvalues to determine the nature of the points (this is quite straightforward but let me know if I should expand on it).
How would I use this method to classify the points in the dynamical system in question?
differential-equations dynamical-systems nonlinear-system
asked Nov 23 at 3:14
user499701
1037
add a comment |
up vote
1
down vote
favorite
Consider the dynamical system $$dot{r}=r(3-2r-s)$$
$$dot{s}=s(2-r-s)$$
Find and classify the equilibrium points of the system.
I have found the equilibrium points to be $(0,0), (0,2), (frac{3}{2},0)$ and $(1,1)$ simply by equating the given equations to $0$ and solving.
What we have been taught in my lectures is how to classify equilibrium points of a dynamical system in the form
$$dot{x}=ax+by$$
$$dot{y}=cx+dy$$
by writing the system in matrix form and using the method of finding the eigenvalues to determine the nature of the points (this is quite straightforward but let me know if I should expand on it).
How would I use this method to classify the points in the dynamical system in question?
differential-equations dynamical-systems nonlinear-system
asked Nov 23 at 3:14
user499701
1037
1
You would need to linearize the system around those points, and once you get a linear system you can use your method.
– user25959
Nov 23 at 3:16
@user25959 What do you mean by linearising the system around the points?
– user499701
Nov 23 at 3:20
1
You are basically approximating your system of equations by a linear system of equations. The coefficients of the linear system will come from evaluating the Jacobian where the rows are partial deriviatives of $f(r,s) = r(3-2r-s)$ and $g(r,s)=s(2-r-s)$.
– user25959
Nov 23 at 3:23
1
examples here: sosmath.com/diffeq/system/nonlinear/linearization/…
– user25959
Nov 23 at 3:23
@Moo Maybe I have been careless about this. Thank you anyways
– user499701
Nov 23 at 6:59
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the dynamical system $$dot{r}=r(3-2r-s)$$
$$dot{s}=s(2-r-s)$$
Find and classify the equilibrium points of the system.
I have found the equilibrium points to be $(0,0), (0,2), (frac{3}{2},0)$ and $(1,1)$ simply by equating the given equations to $0$ and solving.
What we have been taught in my lectures is how to classify equilibrium points of a dynamical system in the form
$$dot{x}=ax+by$$
$$dot{y}=cx+dy$$
by writing the system in matrix form and using the method of finding the eigenvalues to determine the nature of the points (this is quite straightforward but let me know if I should expand on it).
How would I use this method to classify the points in the dynamical system in question?
differential-equations dynamical-systems nonlinear-system
asked Nov 23 at 3:14
user499701
1037
Consider the dynamical system $$dot{r}=r(3-2r-s)$$
$$dot{s}=s(2-r-s)$$
Find and classify the equilibrium points of the system.
I have found the equilibrium points to be $(0,0), (0,2), (frac{3}{2},0)$ and $(1,1)$ simply by equating the given equations to $0$ and solving.
What we have been taught in my lectures is how to classify equilibrium points of a dynamical system in the form
$$dot{x}=ax+by$$
$$dot{y}=cx+dy$$
by writing the system in matrix form and using the method of finding the eigenvalues to determine the nature of the points (this is quite straightforward but let me know if I should expand on it).
How would I use this method to classify the points in the dynamical system in question?
differential-equations dynamical-systems nonlinear-system
differential-equations dynamical-systems nonlinear-system
asked Nov 23 at 3:14
user499701
1037
asked Nov 23 at 3:14
user499701
1037
asked Nov 23 at 3:14
user499701
1037
asked Nov 23 at 3:14
user499701
1037
asked Nov 23 at 3:14
user499701
1037
1037
1
You would need to linearize the system around those points, and once you get a linear system you can use your method.
– user25959
Nov 23 at 3:16
@user25959 What do you mean by linearising the system around the points?
– user499701
Nov 23 at 3:20
1
You are basically approximating your system of equations by a linear system of equations. The coefficients of the linear system will come from evaluating the Jacobian where the rows are partial deriviatives of $f(r,s) = r(3-2r-s)$ and $g(r,s)=s(2-r-s)$.
– user25959
Nov 23 at 3:23
1
examples here: sosmath.com/diffeq/system/nonlinear/linearization/…
– user25959
Nov 23 at 3:23
@Moo Maybe I have been careless about this. Thank you anyways
– user499701
Nov 23 at 6:59
add a comment |
1
You would need to linearize the system around those points, and once you get a linear system you can use your method.
– user25959
Nov 23 at 3:16
@user25959 What do you mean by linearising the system around the points?
– user499701
Nov 23 at 3:20
1
You are basically approximating your system of equations by a linear system of equations. The coefficients of the linear system will come from evaluating the Jacobian where the rows are partial deriviatives of $f(r,s) = r(3-2r-s)$ and $g(r,s)=s(2-r-s)$.
– user25959
Nov 23 at 3:23
1
examples here: sosmath.com/diffeq/system/nonlinear/linearization/…
– user25959
Nov 23 at 3:23
@Moo Maybe I have been careless about this. Thank you anyways
– user499701
Nov 23 at 6:59
1
1
You would need to linearize the system around those points, and once you get a linear system you can use your method.
– user25959
Nov 23 at 3:16
You would need to linearize the system around those points, and once you get a linear system you can use your method.
– user25959
Nov 23 at 3:16
@user25959 What do you mean by linearising the system around the points?
– user499701
Nov 23 at 3:20
@user25959 What do you mean by linearising the system around the points?
– user499701
Nov 23 at 3:20
1
1
You are basically approximating your system of equations by a linear system of equations. The coefficients of the linear system will come from evaluating the Jacobian where the rows are partial deriviatives of $f(r,s) = r(3-2r-s)$ and $g(r,s)=s(2-r-s)$.
– user25959
Nov 23 at 3:23
You are basically approximating your system of equations by a linear system of equations. The coefficients of the linear system will come from evaluating the Jacobian where the rows are partial deriviatives of $f(r,s) = r(3-2r-s)$ and $g(r,s)=s(2-r-s)$.
– user25959
Nov 23 at 3:23
1
1
examples here: sosmath.com/diffeq/system/nonlinear/linearization/…
– user25959
Nov 23 at 3:23
examples here: sosmath.com/diffeq/system/nonlinear/linearization/…
– user25959
Nov 23 at 3:23
@Moo Maybe I have been careless about this. Thank you anyways
– user499701
Nov 23 at 6:59
@Moo Maybe I have been careless about this. Thank you anyways
– user499701
Nov 23 at 6:59
add a comment |
1 Answer
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We find the Jacobian of the system as
$$J(r, s) = begin{bmatrix} dfrac{partial r'}{partial r} & dfrac{partial r'}{partial s} \ dfrac{partial s'}{partial r} & dfrac{partial s'}{partial s} end{bmatrix} = begin{bmatrix} -4 r-s+3 & -r \ -s & -r-2 s+2 \end{bmatrix}$$
Your critical points are correct. Next, we evaluate the eigenvalues of each critical point and have
$$J(0,0) implies lambda_{1,2} = 2, 3 \ J(0,2) implies lambda_{1,2} = -2,1 \ J(3/2,2) implies lambda_{1,2} = -3,dfrac{1}{2} \ J(1,1) implies lambda_{1,2} = dfrac{1}{2} left(-sqrt{5}-3right),dfrac{1}{2} left(sqrt{5}-3right)$$
It is worth mentioning to be careful for marginal points when using linearization. Marginal cases are those that have at least one eigenvalue with zero real part. These are centers, degenerate nodes, and non-isolated fixed points. The smallest perturbation will ”knock” them into a different type of behavior. We do not have any of these in this particular problem.
Next, you can classify those knowing the approach you mention.
We can compare that analysis to the phase portrait for the nonlinear system
answered Nov 23 at 6:46
Moo
5,51631020
add a comment |
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1 Answer
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active
oldest
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up vote
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accepted
We find the Jacobian of the system as
$$J(r, s) = begin{bmatrix} dfrac{partial r'}{partial r} & dfrac{partial r'}{partial s} \ dfrac{partial s'}{partial r} & dfrac{partial s'}{partial s} end{bmatrix} = begin{bmatrix} -4 r-s+3 & -r \ -s & -r-2 s+2 \end{bmatrix}$$
Your critical points are correct. Next, we evaluate the eigenvalues of each critical point and have
$$J(0,0) implies lambda_{1,2} = 2, 3 \ J(0,2) implies lambda_{1,2} = -2,1 \ J(3/2,2) implies lambda_{1,2} = -3,dfrac{1}{2} \ J(1,1) implies lambda_{1,2} = dfrac{1}{2} left(-sqrt{5}-3right),dfrac{1}{2} left(sqrt{5}-3right)$$
It is worth mentioning to be careful for marginal points when using linearization. Marginal cases are those that have at least one eigenvalue with zero real part. These are centers, degenerate nodes, and non-isolated fixed points. The smallest perturbation will ”knock” them into a different type of behavior. We do not have any of these in this particular problem.
Next, you can classify those knowing the approach you mention.
We can compare that analysis to the phase portrait for the nonlinear system
answered Nov 23 at 6:46
Moo
5,51631020
add a comment |
up vote
1
down vote
accepted
We find the Jacobian of the system as
$$J(r, s) = begin{bmatrix} dfrac{partial r'}{partial r} & dfrac{partial r'}{partial s} \ dfrac{partial s'}{partial r} & dfrac{partial s'}{partial s} end{bmatrix} = begin{bmatrix} -4 r-s+3 & -r \ -s & -r-2 s+2 \end{bmatrix}$$
Your critical points are correct. Next, we evaluate the eigenvalues of each critical point and have
$$J(0,0) implies lambda_{1,2} = 2, 3 \ J(0,2) implies lambda_{1,2} = -2,1 \ J(3/2,2) implies lambda_{1,2} = -3,dfrac{1}{2} \ J(1,1) implies lambda_{1,2} = dfrac{1}{2} left(-sqrt{5}-3right),dfrac{1}{2} left(sqrt{5}-3right)$$
It is worth mentioning to be careful for marginal points when using linearization. Marginal cases are those that have at least one eigenvalue with zero real part. These are centers, degenerate nodes, and non-isolated fixed points. The smallest perturbation will ”knock” them into a different type of behavior. We do not have any of these in this particular problem.
Next, you can classify those knowing the approach you mention.
We can compare that analysis to the phase portrait for the nonlinear system
answered Nov 23 at 6:46
Moo
5,51631020
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We find the Jacobian of the system as
$$J(r, s) = begin{bmatrix} dfrac{partial r'}{partial r} & dfrac{partial r'}{partial s} \ dfrac{partial s'}{partial r} & dfrac{partial s'}{partial s} end{bmatrix} = begin{bmatrix} -4 r-s+3 & -r \ -s & -r-2 s+2 \end{bmatrix}$$
Your critical points are correct. Next, we evaluate the eigenvalues of each critical point and have
$$J(0,0) implies lambda_{1,2} = 2, 3 \ J(0,2) implies lambda_{1,2} = -2,1 \ J(3/2,2) implies lambda_{1,2} = -3,dfrac{1}{2} \ J(1,1) implies lambda_{1,2} = dfrac{1}{2} left(-sqrt{5}-3right),dfrac{1}{2} left(sqrt{5}-3right)$$
It is worth mentioning to be careful for marginal points when using linearization. Marginal cases are those that have at least one eigenvalue with zero real part. These are centers, degenerate nodes, and non-isolated fixed points. The smallest perturbation will ”knock” them into a different type of behavior. We do not have any of these in this particular problem.
Next, you can classify those knowing the approach you mention.
We can compare that analysis to the phase portrait for the nonlinear system
answered Nov 23 at 6:46
Moo
5,51631020
We find the Jacobian of the system as
$$J(r, s) = begin{bmatrix} dfrac{partial r'}{partial r} & dfrac{partial r'}{partial s} \ dfrac{partial s'}{partial r} & dfrac{partial s'}{partial s} end{bmatrix} = begin{bmatrix} -4 r-s+3 & -r \ -s & -r-2 s+2 \end{bmatrix}$$
Your critical points are correct. Next, we evaluate the eigenvalues of each critical point and have
$$J(0,0) implies lambda_{1,2} = 2, 3 \ J(0,2) implies lambda_{1,2} = -2,1 \ J(3/2,2) implies lambda_{1,2} = -3,dfrac{1}{2} \ J(1,1) implies lambda_{1,2} = dfrac{1}{2} left(-sqrt{5}-3right),dfrac{1}{2} left(sqrt{5}-3right)$$
It is worth mentioning to be careful for marginal points when using linearization. Marginal cases are those that have at least one eigenvalue with zero real part. These are centers, degenerate nodes, and non-isolated fixed points. The smallest perturbation will ”knock” them into a different type of behavior. We do not have any of these in this particular problem.
Next, you can classify those knowing the approach you mention.
We can compare that analysis to the phase portrait for the nonlinear system
answered Nov 23 at 6:46
Moo
5,51631020
edited Nov 23 at 6:54
answered Nov 23 at 6:46
Moo
5,51631020
answered Nov 23 at 6:46
Moo
5,51631020
answered Nov 23 at 6:46
Moo
5,51631020
5,51631020
add a comment |
add a comment |
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1
You would need to linearize the system around those points, and once you get a linear system you can use your method.
– user25959
Nov 23 at 3:16
@user25959 What do you mean by linearising the system around the points?
– user499701
Nov 23 at 3:20
1
You are basically approximating your system of equations by a linear system of equations. The coefficients of the linear system will come from evaluating the Jacobian where the rows are partial deriviatives of $f(r,s) = r(3-2r-s)$ and $g(r,s)=s(2-r-s)$.
– user25959
Nov 23 at 3:23
1
examples here: sosmath.com/diffeq/system/nonlinear/linearization/…
– user25959
Nov 23 at 3:23
@Moo Maybe I have been careless about this. Thank you anyways
– user499701
Nov 23 at 6:59