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Stability of equilibrium points based on characteristic equation
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I am reading a paper, and there is a coupled differential equations:
$$begincases
dotx = ax^2-x^3-y-z\
doty = (a+alpha)x^2-y\
dotz = mu(bx+c-z)
endcases
$$
Let the first two equations be $0$, we have $$z = -x^3-alpha x^2 = f(x)$$
By the above, we can find three equilibrium points: $$N_1(x_1,y_1), O(x_0,y_0), N_2(x_2,y_2)$$
Then the paper use the following characteristic equation to discuss their stability:
$$s^2-sigma(x_i)s-f'(x_i)=0, i = 0,1,2$$
where divergence $sigma = -(1-2ax+3x^2)$ and the slope $f'$.
I just know the stability of equilibrium can be found by the phase plots and characteristic equation. However, I only know the characteristic equation of second or higher order ODEs. I have no idea about the above formula and cannot find some useful information by typing key words.
Can anyone please let me know what is the above formula (some lecture notes are fine). Thanks!
Paper link:
https://aip.scitation.org/doi/abs/10.1063/1.3563581 (from p.3 section II. to p.4, you don't need any background of networks.)
differential-equations
asked Jul 30 at 8:17
sleeve chen
2,77531646
add a comment |Â
up vote
1
down vote
favorite
I am reading a paper, and there is a coupled differential equations:
$$begincases
dotx = ax^2-x^3-y-z\
doty = (a+alpha)x^2-y\
dotz = mu(bx+c-z)
endcases
$$
Let the first two equations be $0$, we have $$z = -x^3-alpha x^2 = f(x)$$
By the above, we can find three equilibrium points: $$N_1(x_1,y_1), O(x_0,y_0), N_2(x_2,y_2)$$
Then the paper use the following characteristic equation to discuss their stability:
$$s^2-sigma(x_i)s-f'(x_i)=0, i = 0,1,2$$
where divergence $sigma = -(1-2ax+3x^2)$ and the slope $f'$.
I just know the stability of equilibrium can be found by the phase plots and characteristic equation. However, I only know the characteristic equation of second or higher order ODEs. I have no idea about the above formula and cannot find some useful information by typing key words.
Can anyone please let me know what is the above formula (some lecture notes are fine). Thanks!
Paper link:
https://aip.scitation.org/doi/abs/10.1063/1.3563581 (from p.3 section II. to p.4, you don't need any background of networks.)
differential-equations
asked Jul 30 at 8:17
sleeve chen
2,77531646
2
Can you provide a link to the paper or its title? From your question it is hard to understand why everything goes down to quadratic equation when eigenvalues of Jacobi matrix must come from cubic equation here.
– Evgeny
Jul 30 at 10:02
@Evgeny Just fix it, thanks for suggestion.
– sleeve chen
Jul 30 at 20:02
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am reading a paper, and there is a coupled differential equations:
$$begincases
dotx = ax^2-x^3-y-z\
doty = (a+alpha)x^2-y\
dotz = mu(bx+c-z)
endcases
$$
Let the first two equations be $0$, we have $$z = -x^3-alpha x^2 = f(x)$$
By the above, we can find three equilibrium points: $$N_1(x_1,y_1), O(x_0,y_0), N_2(x_2,y_2)$$
Then the paper use the following characteristic equation to discuss their stability:
$$s^2-sigma(x_i)s-f'(x_i)=0, i = 0,1,2$$
where divergence $sigma = -(1-2ax+3x^2)$ and the slope $f'$.
I just know the stability of equilibrium can be found by the phase plots and characteristic equation. However, I only know the characteristic equation of second or higher order ODEs. I have no idea about the above formula and cannot find some useful information by typing key words.
Can anyone please let me know what is the above formula (some lecture notes are fine). Thanks!
Paper link:
https://aip.scitation.org/doi/abs/10.1063/1.3563581 (from p.3 section II. to p.4, you don't need any background of networks.)
differential-equations
asked Jul 30 at 8:17
sleeve chen
2,77531646
I am reading a paper, and there is a coupled differential equations:
$$begincases
dotx = ax^2-x^3-y-z\
doty = (a+alpha)x^2-y\
dotz = mu(bx+c-z)
endcases
$$
Let the first two equations be $0$, we have $$z = -x^3-alpha x^2 = f(x)$$
By the above, we can find three equilibrium points: $$N_1(x_1,y_1), O(x_0,y_0), N_2(x_2,y_2)$$
Then the paper use the following characteristic equation to discuss their stability:
$$s^2-sigma(x_i)s-f'(x_i)=0, i = 0,1,2$$
where divergence $sigma = -(1-2ax+3x^2)$ and the slope $f'$.
I just know the stability of equilibrium can be found by the phase plots and characteristic equation. However, I only know the characteristic equation of second or higher order ODEs. I have no idea about the above formula and cannot find some useful information by typing key words.
Can anyone please let me know what is the above formula (some lecture notes are fine). Thanks!
Paper link:
https://aip.scitation.org/doi/abs/10.1063/1.3563581 (from p.3 section II. to p.4, you don't need any background of networks.)
differential-equations
asked Jul 30 at 8:17
sleeve chen
2,77531646
edited Jul 30 at 20:02
asked Jul 30 at 8:17
sleeve chen
2,77531646
asked Jul 30 at 8:17
sleeve chen
2,77531646
asked Jul 30 at 8:17
sleeve chen
2,77531646
2,77531646
2
Can you provide a link to the paper or its title? From your question it is hard to understand why everything goes down to quadratic equation when eigenvalues of Jacobi matrix must come from cubic equation here.
– Evgeny
Jul 30 at 10:02
@Evgeny Just fix it, thanks for suggestion.
– sleeve chen
Jul 30 at 20:02
add a comment |Â
2
Can you provide a link to the paper or its title? From your question it is hard to understand why everything goes down to quadratic equation when eigenvalues of Jacobi matrix must come from cubic equation here.
– Evgeny
Jul 30 at 10:02
@Evgeny Just fix it, thanks for suggestion.
– sleeve chen
Jul 30 at 20:02
2
2
Can you provide a link to the paper or its title? From your question it is hard to understand why everything goes down to quadratic equation when eigenvalues of Jacobi matrix must come from cubic equation here.
– Evgeny
Jul 30 at 10:02
Can you provide a link to the paper or its title? From your question it is hard to understand why everything goes down to quadratic equation when eigenvalues of Jacobi matrix must come from cubic equation here.
– Evgeny
Jul 30 at 10:02
@Evgeny Just fix it, thanks for suggestion.
– sleeve chen
Jul 30 at 20:02
@Evgeny Just fix it, thanks for suggestion.
– sleeve chen
Jul 30 at 20:02
add a comment |Â
1 Answer
1
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oldest
votes
up vote
1
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accepted
They study this system as slow-fast system: they "freeze" $z$-variable, thinking about it as a parameter for $(x, y)$-system. After that they study stability of equilibria of two-dimensional system regular way. Namely, if you have a system $dotx = P(x, y), ; doty = Q(x, y)$, then eigenvalues of Jacobi matrix $ J = left ( beginarraycc P'_x & P'_y \ Q'_x & Q'_y endarray right )$ computed at equilibrium determine its stability. The characteristic equation for $2 times 2$ system is always $lambda^2 - rm tr, J cdotlambda + rm det, J = 0$ and $rm tr, J = P'_x + Q'_y = rm div, (P, Q)$. The final formula from their paper just follows from the used notation.
answered Aug 1 at 9:45
Evgeny
4,4622921
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
They study this system as slow-fast system: they "freeze" $z$-variable, thinking about it as a parameter for $(x, y)$-system. After that they study stability of equilibria of two-dimensional system regular way. Namely, if you have a system $dotx = P(x, y), ; doty = Q(x, y)$, then eigenvalues of Jacobi matrix $ J = left ( beginarraycc P'_x & P'_y \ Q'_x & Q'_y endarray right )$ computed at equilibrium determine its stability. The characteristic equation for $2 times 2$ system is always $lambda^2 - rm tr, J cdotlambda + rm det, J = 0$ and $rm tr, J = P'_x + Q'_y = rm div, (P, Q)$. The final formula from their paper just follows from the used notation.
answered Aug 1 at 9:45
Evgeny
4,4622921
add a comment |Â
up vote
1
down vote
accepted
They study this system as slow-fast system: they "freeze" $z$-variable, thinking about it as a parameter for $(x, y)$-system. After that they study stability of equilibria of two-dimensional system regular way. Namely, if you have a system $dotx = P(x, y), ; doty = Q(x, y)$, then eigenvalues of Jacobi matrix $ J = left ( beginarraycc P'_x & P'_y \ Q'_x & Q'_y endarray right )$ computed at equilibrium determine its stability. The characteristic equation for $2 times 2$ system is always $lambda^2 - rm tr, J cdotlambda + rm det, J = 0$ and $rm tr, J = P'_x + Q'_y = rm div, (P, Q)$. The final formula from their paper just follows from the used notation.
answered Aug 1 at 9:45
Evgeny
4,4622921
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
They study this system as slow-fast system: they "freeze" $z$-variable, thinking about it as a parameter for $(x, y)$-system. After that they study stability of equilibria of two-dimensional system regular way. Namely, if you have a system $dotx = P(x, y), ; doty = Q(x, y)$, then eigenvalues of Jacobi matrix $ J = left ( beginarraycc P'_x & P'_y \ Q'_x & Q'_y endarray right )$ computed at equilibrium determine its stability. The characteristic equation for $2 times 2$ system is always $lambda^2 - rm tr, J cdotlambda + rm det, J = 0$ and $rm tr, J = P'_x + Q'_y = rm div, (P, Q)$. The final formula from their paper just follows from the used notation.
answered Aug 1 at 9:45
Evgeny
4,4622921
They study this system as slow-fast system: they "freeze" $z$-variable, thinking about it as a parameter for $(x, y)$-system. After that they study stability of equilibria of two-dimensional system regular way. Namely, if you have a system $dotx = P(x, y), ; doty = Q(x, y)$, then eigenvalues of Jacobi matrix $ J = left ( beginarraycc P'_x & P'_y \ Q'_x & Q'_y endarray right )$ computed at equilibrium determine its stability. The characteristic equation for $2 times 2$ system is always $lambda^2 - rm tr, J cdotlambda + rm det, J = 0$ and $rm tr, J = P'_x + Q'_y = rm div, (P, Q)$. The final formula from their paper just follows from the used notation.
answered Aug 1 at 9:45
Evgeny
4,4622921
answered Aug 1 at 9:45
Evgeny
4,4622921
answered Aug 1 at 9:45
Evgeny
4,4622921
answered Aug 1 at 9:45
Evgeny
4,4622921
4,4622921
add a comment |Â
add a comment |Â
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2
Can you provide a link to the paper or its title? From your question it is hard to understand why everything goes down to quadratic equation when eigenvalues of Jacobi matrix must come from cubic equation here.
– Evgeny
Jul 30 at 10:02
@Evgeny Just fix it, thanks for suggestion.
– sleeve chen
Jul 30 at 20:02