Unstable fixed pointsClassifying local behavior of fixed points using eigenvalues from linear stability...
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Unstable fixed points
Classifying local behavior of fixed points using eigenvalues from linear stability analysis of 3D systemstable and unstable manifolds of singular points and periodic orbitsUsing the Lyapunov-Perron method to find the local stable/unstable manifoldsFinding Lyapunov function for a given system of differential equationsFixed points of dynamical systemsFinding the stable and unstable manifold of this systemShow that $(x_1,x_2)=(0,0)$ is an unstable fixed point for $dot{x_1} = -x_1 + x_2^6$, $dot{x_2} =x_2^3 + x_1^6$Determining whether a fixed point of differential equation system is stable or notHow to prove a fixed point is stable?Stability of ODE involving trig functions and nonhyperbolic fixed points
$begingroup$
When considering the system
begin{cases}
x' = (A-By)x \
y' = (C-Dx)y, &
end{cases}
($A,B,C,D > 0$)
I am trying to understand how to tell that the fixed points $(0,0)$ and $(C/D,A/B)$ are unstable. If I'm not mistaken, Lyapunov functions can be used to show that a point is stable but not that it is unstable. I was thinking of finding the eigenvalues (if we consider the system as $bf{x'} = Ax$), but the eigenvalues will involve $x$ and $y$ and that hasn't happened before as I've been solving exercises, so I'm not sure if that is the way to go...
ordinary-differential-equations
asked Mar 13 at 23:43
RickSRickS
778
$endgroup$
add a comment |
$begingroup$
When considering the system
begin{cases}
x' = (A-By)x \
y' = (C-Dx)y, &
end{cases}
($A,B,C,D > 0$)
I am trying to understand how to tell that the fixed points $(0,0)$ and $(C/D,A/B)$ are unstable. If I'm not mistaken, Lyapunov functions can be used to show that a point is stable but not that it is unstable. I was thinking of finding the eigenvalues (if we consider the system as $bf{x'} = Ax$), but the eigenvalues will involve $x$ and $y$ and that hasn't happened before as I've been solving exercises, so I'm not sure if that is the way to go...
ordinary-differential-equations
asked Mar 13 at 23:43
RickSRickS
778
$endgroup$
$begingroup$
Incidentally, there are Lyapunov functions (sort of) that can be used to prove that an equilibrium is unstable; see, e.g., Chetaev function.
$endgroup$
– user539887
Mar 14 at 8:07
add a comment |
$begingroup$
When considering the system
begin{cases}
x' = (A-By)x \
y' = (C-Dx)y, &
end{cases}
($A,B,C,D > 0$)
I am trying to understand how to tell that the fixed points $(0,0)$ and $(C/D,A/B)$ are unstable. If I'm not mistaken, Lyapunov functions can be used to show that a point is stable but not that it is unstable. I was thinking of finding the eigenvalues (if we consider the system as $bf{x'} = Ax$), but the eigenvalues will involve $x$ and $y$ and that hasn't happened before as I've been solving exercises, so I'm not sure if that is the way to go...
ordinary-differential-equations
asked Mar 13 at 23:43
RickSRickS
778
$endgroup$
When considering the system
begin{cases}
x' = (A-By)x \
y' = (C-Dx)y, &
end{cases}
($A,B,C,D > 0$)
I am trying to understand how to tell that the fixed points $(0,0)$ and $(C/D,A/B)$ are unstable. If I'm not mistaken, Lyapunov functions can be used to show that a point is stable but not that it is unstable. I was thinking of finding the eigenvalues (if we consider the system as $bf{x'} = Ax$), but the eigenvalues will involve $x$ and $y$ and that hasn't happened before as I've been solving exercises, so I'm not sure if that is the way to go...
ordinary-differential-equations
ordinary-differential-equations
asked Mar 13 at 23:43
RickSRickS
778
asked Mar 13 at 23:43
RickSRickS
778
asked Mar 13 at 23:43
RickSRickS
778
asked Mar 13 at 23:43
RickSRickS
778
asked Mar 13 at 23:43
RickSRickS
778
778
$begingroup$
Incidentally, there are Lyapunov functions (sort of) that can be used to prove that an equilibrium is unstable; see, e.g., Chetaev function.
$endgroup$
– user539887
Mar 14 at 8:07
add a comment |
$begingroup$
Incidentally, there are Lyapunov functions (sort of) that can be used to prove that an equilibrium is unstable; see, e.g., Chetaev function.
$endgroup$
– user539887
Mar 14 at 8:07
$begingroup$
Incidentally, there are Lyapunov functions (sort of) that can be used to prove that an equilibrium is unstable; see, e.g., Chetaev function.
$endgroup$
– user539887
Mar 14 at 8:07
$begingroup$
Incidentally, there are Lyapunov functions (sort of) that can be used to prove that an equilibrium is unstable; see, e.g., Chetaev function.
$endgroup$
– user539887
Mar 14 at 8:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The eigenvalues will stop involving $x, y$ after we evaluate the linearization at the fixed points. Here is how.
Let's compute the linearization of the right-hand side:
$$
L(x,y) =
left[
begin{array}{ll}
A - By & -Bx\
Dy & C - Dx\
end{array}
right]
$$
Evaluating it at $(x,y) = (0,0)$, we get:
$$
L(0,0) =
left[
begin{array}{ll}
A & 0\
0 & C\
end{array}
right],
$$
so the eigenvalues are $A, C$. Since they each have a positive real part (I know, you said they are real and positive, but I am using language suitable for a more general setting), the fixed point $(x, y) = (0,0)$ is unstable.
Similarly inspect the eigenvalues of $L(C/D, A/B)$.
answered Mar 14 at 0:06
avsavs
3,714514
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The eigenvalues will stop involving $x, y$ after we evaluate the linearization at the fixed points. Here is how.
Let's compute the linearization of the right-hand side:
$$
L(x,y) =
left[
begin{array}{ll}
A - By & -Bx\
Dy & C - Dx\
end{array}
right]
$$
Evaluating it at $(x,y) = (0,0)$, we get:
$$
L(0,0) =
left[
begin{array}{ll}
A & 0\
0 & C\
end{array}
right],
$$
so the eigenvalues are $A, C$. Since they each have a positive real part (I know, you said they are real and positive, but I am using language suitable for a more general setting), the fixed point $(x, y) = (0,0)$ is unstable.
Similarly inspect the eigenvalues of $L(C/D, A/B)$.
answered Mar 14 at 0:06
avsavs
3,714514
$endgroup$
add a comment |
$begingroup$
The eigenvalues will stop involving $x, y$ after we evaluate the linearization at the fixed points. Here is how.
Let's compute the linearization of the right-hand side:
$$
L(x,y) =
left[
begin{array}{ll}
A - By & -Bx\
Dy & C - Dx\
end{array}
right]
$$
Evaluating it at $(x,y) = (0,0)$, we get:
$$
L(0,0) =
left[
begin{array}{ll}
A & 0\
0 & C\
end{array}
right],
$$
so the eigenvalues are $A, C$. Since they each have a positive real part (I know, you said they are real and positive, but I am using language suitable for a more general setting), the fixed point $(x, y) = (0,0)$ is unstable.
Similarly inspect the eigenvalues of $L(C/D, A/B)$.
answered Mar 14 at 0:06
avsavs
3,714514
$endgroup$
add a comment |
$begingroup$
The eigenvalues will stop involving $x, y$ after we evaluate the linearization at the fixed points. Here is how.
Let's compute the linearization of the right-hand side:
$$
L(x,y) =
left[
begin{array}{ll}
A - By & -Bx\
Dy & C - Dx\
end{array}
right]
$$
Evaluating it at $(x,y) = (0,0)$, we get:
$$
L(0,0) =
left[
begin{array}{ll}
A & 0\
0 & C\
end{array}
right],
$$
so the eigenvalues are $A, C$. Since they each have a positive real part (I know, you said they are real and positive, but I am using language suitable for a more general setting), the fixed point $(x, y) = (0,0)$ is unstable.
Similarly inspect the eigenvalues of $L(C/D, A/B)$.
answered Mar 14 at 0:06
avsavs
3,714514
$endgroup$
The eigenvalues will stop involving $x, y$ after we evaluate the linearization at the fixed points. Here is how.
Let's compute the linearization of the right-hand side:
$$
L(x,y) =
left[
begin{array}{ll}
A - By & -Bx\
Dy & C - Dx\
end{array}
right]
$$
Evaluating it at $(x,y) = (0,0)$, we get:
$$
L(0,0) =
left[
begin{array}{ll}
A & 0\
0 & C\
end{array}
right],
$$
so the eigenvalues are $A, C$. Since they each have a positive real part (I know, you said they are real and positive, but I am using language suitable for a more general setting), the fixed point $(x, y) = (0,0)$ is unstable.
Similarly inspect the eigenvalues of $L(C/D, A/B)$.
answered Mar 14 at 0:06
avsavs
3,714514
answered Mar 14 at 0:06
avsavs
3,714514
answered Mar 14 at 0:06
avsavs
3,714514
answered Mar 14 at 0:06
avsavs
3,714514
3,714514
add a comment |
add a comment |
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$begingroup$
Incidentally, there are Lyapunov functions (sort of) that can be used to prove that an equilibrium is unstable; see, e.g., Chetaev function.
$endgroup$
– user539887
Mar 14 at 8:07