determine the stability of an equilibrium point(x,0).
up vote
1
down vote
favorite
I would like to determine the stability of equilibrium point $(x,0)$ of the differential equation
$dot x = Ax$
$ A=
bigg[
begin{matrix}
0&0\0&a
end{matrix}
bigg]
$ and $ a >0 $
I got
$ x'(t) = 0 $
$ y'(t) = a*y$
So, the equilibrium point is $(x,0)$.
How can I determine the stability of these equilibrium points?
Here is the phase portrait:
Any help will be appreciated!
calculus differential-equations stability-theory
asked Nov 15 at 1:48
HIABCD
83
add a comment |
up vote
1
down vote
favorite
I would like to determine the stability of equilibrium point $(x,0)$ of the differential equation
$dot x = Ax$
$ A=
bigg[
begin{matrix}
0&0\0&a
end{matrix}
bigg]
$ and $ a >0 $
I got
$ x'(t) = 0 $
$ y'(t) = a*y$
So, the equilibrium point is $(x,0)$.
How can I determine the stability of these equilibrium points?
Here is the phase portrait:
Any help will be appreciated!
calculus differential-equations stability-theory
asked Nov 15 at 1:48
HIABCD
83
The entire $x$ axis is an equilibrium point.
– copper.hat
Nov 15 at 1:50
@Moo I draw the phase portrait, and I think the equilibrium point is unstable.
– HIABCD
Nov 15 at 17:33
Thanks! Please add.
– HIABCD
Nov 15 at 17:49
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I would like to determine the stability of equilibrium point $(x,0)$ of the differential equation
$dot x = Ax$
$ A=
bigg[
begin{matrix}
0&0\0&a
end{matrix}
bigg]
$ and $ a >0 $
I got
$ x'(t) = 0 $
$ y'(t) = a*y$
So, the equilibrium point is $(x,0)$.
How can I determine the stability of these equilibrium points?
Here is the phase portrait:
Any help will be appreciated!
calculus differential-equations stability-theory
asked Nov 15 at 1:48
HIABCD
83
I would like to determine the stability of equilibrium point $(x,0)$ of the differential equation
$dot x = Ax$
$ A=
bigg[
begin{matrix}
0&0\0&a
end{matrix}
bigg]
$ and $ a >0 $
I got
$ x'(t) = 0 $
$ y'(t) = a*y$
So, the equilibrium point is $(x,0)$.
How can I determine the stability of these equilibrium points?
Here is the phase portrait:
Any help will be appreciated!
calculus differential-equations stability-theory
calculus differential-equations stability-theory
asked Nov 15 at 1:48
HIABCD
83
asked Nov 15 at 1:48
HIABCD
83
edited Nov 15 at 17:37
asked Nov 15 at 1:48
HIABCD
83
asked Nov 15 at 1:48
HIABCD
83
asked Nov 15 at 1:48
HIABCD
83
83
The entire $x$ axis is an equilibrium point.
– copper.hat
Nov 15 at 1:50
@Moo I draw the phase portrait, and I think the equilibrium point is unstable.
– HIABCD
Nov 15 at 17:33
Thanks! Please add.
– HIABCD
Nov 15 at 17:49
add a comment |
The entire $x$ axis is an equilibrium point.
– copper.hat
Nov 15 at 1:50
@Moo I draw the phase portrait, and I think the equilibrium point is unstable.
– HIABCD
Nov 15 at 17:33
Thanks! Please add.
– HIABCD
Nov 15 at 17:49
The entire $x$ axis is an equilibrium point.
– copper.hat
Nov 15 at 1:50
The entire $x$ axis is an equilibrium point.
– copper.hat
Nov 15 at 1:50
@Moo I draw the phase portrait, and I think the equilibrium point is unstable.
– HIABCD
Nov 15 at 17:33
@Moo I draw the phase portrait, and I think the equilibrium point is unstable.
– HIABCD
Nov 15 at 17:33
Thanks! Please add.
– HIABCD
Nov 15 at 17:49
Thanks! Please add.
– HIABCD
Nov 15 at 17:49
add a comment |
1 Answer
1
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oldest
votes
up vote
0
down vote
accepted
We want to determine the stability of the equilibrium points of the system $dot x = Ax$, where
$$ A=
bigg[
begin{matrix}
0&0\0&a
end{matrix}
bigg], ~text{with}~a >0 $$
The critical points are where we simultaneously have $x' = y' = 0$ and we get the entire $x-$axis as
$$(x, y) = (x, 0)$$
Since this system is decoupled, we can write
$$begin{align} x'(t) &= 0 implies x(t) = c \ y'(t) &= ay implies y(t) = c e^{a t} end{align}$$
The phase portrait is
Using all of the information above, we determine that the critical point is unstable.
answered Nov 15 at 18:30
Moo
5,2883920
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
We want to determine the stability of the equilibrium points of the system $dot x = Ax$, where
$$ A=
bigg[
begin{matrix}
0&0\0&a
end{matrix}
bigg], ~text{with}~a >0 $$
The critical points are where we simultaneously have $x' = y' = 0$ and we get the entire $x-$axis as
$$(x, y) = (x, 0)$$
Since this system is decoupled, we can write
$$begin{align} x'(t) &= 0 implies x(t) = c \ y'(t) &= ay implies y(t) = c e^{a t} end{align}$$
The phase portrait is
Using all of the information above, we determine that the critical point is unstable.
answered Nov 15 at 18:30
Moo
5,2883920
add a comment |
up vote
0
down vote
accepted
We want to determine the stability of the equilibrium points of the system $dot x = Ax$, where
$$ A=
bigg[
begin{matrix}
0&0\0&a
end{matrix}
bigg], ~text{with}~a >0 $$
The critical points are where we simultaneously have $x' = y' = 0$ and we get the entire $x-$axis as
$$(x, y) = (x, 0)$$
Since this system is decoupled, we can write
$$begin{align} x'(t) &= 0 implies x(t) = c \ y'(t) &= ay implies y(t) = c e^{a t} end{align}$$
The phase portrait is
Using all of the information above, we determine that the critical point is unstable.
answered Nov 15 at 18:30
Moo
5,2883920
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
We want to determine the stability of the equilibrium points of the system $dot x = Ax$, where
$$ A=
bigg[
begin{matrix}
0&0\0&a
end{matrix}
bigg], ~text{with}~a >0 $$
The critical points are where we simultaneously have $x' = y' = 0$ and we get the entire $x-$axis as
$$(x, y) = (x, 0)$$
Since this system is decoupled, we can write
$$begin{align} x'(t) &= 0 implies x(t) = c \ y'(t) &= ay implies y(t) = c e^{a t} end{align}$$
The phase portrait is
Using all of the information above, we determine that the critical point is unstable.
answered Nov 15 at 18:30
Moo
5,2883920
We want to determine the stability of the equilibrium points of the system $dot x = Ax$, where
$$ A=
bigg[
begin{matrix}
0&0\0&a
end{matrix}
bigg], ~text{with}~a >0 $$
The critical points are where we simultaneously have $x' = y' = 0$ and we get the entire $x-$axis as
$$(x, y) = (x, 0)$$
Since this system is decoupled, we can write
$$begin{align} x'(t) &= 0 implies x(t) = c \ y'(t) &= ay implies y(t) = c e^{a t} end{align}$$
The phase portrait is
Using all of the information above, we determine that the critical point is unstable.
answered Nov 15 at 18:30
Moo
5,2883920
edited Nov 15 at 18:45
answered Nov 15 at 18:30
Moo
5,2883920
answered Nov 15 at 18:30
Moo
5,2883920
answered Nov 15 at 18:30
Moo
5,2883920
5,2883920
add a comment |
add a comment |
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The entire $x$ axis is an equilibrium point.
– copper.hat
Nov 15 at 1:50
@Moo I draw the phase portrait, and I think the equilibrium point is unstable.
– HIABCD
Nov 15 at 17:33
Thanks! Please add.
– HIABCD
Nov 15 at 17:49