Conditions when solutions of $dot{x} = f(x)$ exist for all time
I am reading the following textbook:
Introduction to Applied Nonlinear Dynamical Systems and Chaos by Wiggins
p.92 (top)
Consider $$dot{x} = f(x)$$ where $f(x)$ is $C^r$, $rgeq 1$, on some open set $Uin mathbb{R}^n$. Suppose that the solutions exist for all time (Leave it as an exercise to make the necessary modifications when solutions exist only on finite time intervals).
My problem is what are the conditions to make solutions exist for all time.
Consider and example: $$dot{x} = x^2, xin mathbb{R}$$
the solution through $x_0$ at $t = 0$ is $$x(t,0,x_0) = frac{-x_0}{x_0t - 1}$$
this solution (trajectory) does not exist for all time, since it becomes infinite at $t = 1/x_0$.
So what are the conditions to make solution exist for all time
differential-equations dynamical-systems
asked Dec 7 at 8:21
sleeve chen
3,01541852
add a comment |
I am reading the following textbook:
Introduction to Applied Nonlinear Dynamical Systems and Chaos by Wiggins
p.92 (top)
Consider $$dot{x} = f(x)$$ where $f(x)$ is $C^r$, $rgeq 1$, on some open set $Uin mathbb{R}^n$. Suppose that the solutions exist for all time (Leave it as an exercise to make the necessary modifications when solutions exist only on finite time intervals).
My problem is what are the conditions to make solutions exist for all time.
Consider and example: $$dot{x} = x^2, xin mathbb{R}$$
the solution through $x_0$ at $t = 0$ is $$x(t,0,x_0) = frac{-x_0}{x_0t - 1}$$
this solution (trajectory) does not exist for all time, since it becomes infinite at $t = 1/x_0$.
So what are the conditions to make solution exist for all time
differential-equations dynamical-systems
asked Dec 7 at 8:21
sleeve chen
3,01541852
add a comment |
I am reading the following textbook:
Introduction to Applied Nonlinear Dynamical Systems and Chaos by Wiggins
p.92 (top)
Consider $$dot{x} = f(x)$$ where $f(x)$ is $C^r$, $rgeq 1$, on some open set $Uin mathbb{R}^n$. Suppose that the solutions exist for all time (Leave it as an exercise to make the necessary modifications when solutions exist only on finite time intervals).
My problem is what are the conditions to make solutions exist for all time.
Consider and example: $$dot{x} = x^2, xin mathbb{R}$$
the solution through $x_0$ at $t = 0$ is $$x(t,0,x_0) = frac{-x_0}{x_0t - 1}$$
this solution (trajectory) does not exist for all time, since it becomes infinite at $t = 1/x_0$.
So what are the conditions to make solution exist for all time
differential-equations dynamical-systems
asked Dec 7 at 8:21
sleeve chen
3,01541852
I am reading the following textbook:
Introduction to Applied Nonlinear Dynamical Systems and Chaos by Wiggins
p.92 (top)
Consider $$dot{x} = f(x)$$ where $f(x)$ is $C^r$, $rgeq 1$, on some open set $Uin mathbb{R}^n$. Suppose that the solutions exist for all time (Leave it as an exercise to make the necessary modifications when solutions exist only on finite time intervals).
My problem is what are the conditions to make solutions exist for all time.
Consider and example: $$dot{x} = x^2, xin mathbb{R}$$
the solution through $x_0$ at $t = 0$ is $$x(t,0,x_0) = frac{-x_0}{x_0t - 1}$$
this solution (trajectory) does not exist for all time, since it becomes infinite at $t = 1/x_0$.
So what are the conditions to make solution exist for all time
differential-equations dynamical-systems
differential-equations dynamical-systems
asked Dec 7 at 8:21
sleeve chen
3,01541852
asked Dec 7 at 8:21
sleeve chen
3,01541852
asked Dec 7 at 8:21
sleeve chen
3,01541852
asked Dec 7 at 8:21
sleeve chen
3,01541852
asked Dec 7 at 8:21
sleeve chen
3,01541852
3,01541852
add a comment |
add a comment |
1 Answer
1
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oldest
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Usual conditions for the infinite existence of solutions of $dot x=f(x)$ are restrictions to linear growth like
$$
|dot x|=|f(x)|le C(1+|x|)
$$
or
$$
frac12frac{d}{dt}|x|^2=|x^Tdot x|=|x^Tf(x)|le C(1+|x|^2)
$$
which impose an exponential bound on the growth of solutions. The solution of the corresponding differential inequalities is with the Grönwall lemma.
These are not the only conditions, only the relatively easy ones. If the vector field $f(x)$ points inwards on the boundary of a bounded set $U$, all solutions are restricted to $U$ and thus bounded and exist for all positive times. This looks geometrically simple but can be hard to establish
answered Dec 7 at 8:47
LutzL
55.6k42054
How do you relate that to the ODE ?
– Yves Daoust
Dec 7 at 8:50
For your second condition, it looks like the example $dot{x} = x^2$ in my problem satisfies this; however, it does not have solution for all time?
– sleeve chen
Dec 7 at 8:51
1
@sleevechen : No, $x^Tf(x)=x^3$ grows faster than $|x|^2$.
– LutzL
Dec 7 at 8:56
Thanks I see. It looks a bit like the definition of Lipshitz continuity.
– sleeve chen
Dec 7 at 9:02
2
Yes. I do not know of necessary conditions, ODE can have a vast variety of behaviors. You can only select classes of nice ones.
– LutzL
Dec 7 at 9:08
|
show 1 more comment
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Usual conditions for the infinite existence of solutions of $dot x=f(x)$ are restrictions to linear growth like
$$
|dot x|=|f(x)|le C(1+|x|)
$$
or
$$
frac12frac{d}{dt}|x|^2=|x^Tdot x|=|x^Tf(x)|le C(1+|x|^2)
$$
which impose an exponential bound on the growth of solutions. The solution of the corresponding differential inequalities is with the Grönwall lemma.
These are not the only conditions, only the relatively easy ones. If the vector field $f(x)$ points inwards on the boundary of a bounded set $U$, all solutions are restricted to $U$ and thus bounded and exist for all positive times. This looks geometrically simple but can be hard to establish
answered Dec 7 at 8:47
LutzL
55.6k42054
How do you relate that to the ODE ?
– Yves Daoust
Dec 7 at 8:50
For your second condition, it looks like the example $dot{x} = x^2$ in my problem satisfies this; however, it does not have solution for all time?
– sleeve chen
Dec 7 at 8:51
1
@sleevechen : No, $x^Tf(x)=x^3$ grows faster than $|x|^2$.
– LutzL
Dec 7 at 8:56
Thanks I see. It looks a bit like the definition of Lipshitz continuity.
– sleeve chen
Dec 7 at 9:02
2
Yes. I do not know of necessary conditions, ODE can have a vast variety of behaviors. You can only select classes of nice ones.
– LutzL
Dec 7 at 9:08
|
show 1 more comment
Usual conditions for the infinite existence of solutions of $dot x=f(x)$ are restrictions to linear growth like
$$
|dot x|=|f(x)|le C(1+|x|)
$$
or
$$
frac12frac{d}{dt}|x|^2=|x^Tdot x|=|x^Tf(x)|le C(1+|x|^2)
$$
which impose an exponential bound on the growth of solutions. The solution of the corresponding differential inequalities is with the Grönwall lemma.
These are not the only conditions, only the relatively easy ones. If the vector field $f(x)$ points inwards on the boundary of a bounded set $U$, all solutions are restricted to $U$ and thus bounded and exist for all positive times. This looks geometrically simple but can be hard to establish
answered Dec 7 at 8:47
LutzL
55.6k42054
How do you relate that to the ODE ?
– Yves Daoust
Dec 7 at 8:50
For your second condition, it looks like the example $dot{x} = x^2$ in my problem satisfies this; however, it does not have solution for all time?
– sleeve chen
Dec 7 at 8:51
1
@sleevechen : No, $x^Tf(x)=x^3$ grows faster than $|x|^2$.
– LutzL
Dec 7 at 8:56
Thanks I see. It looks a bit like the definition of Lipshitz continuity.
– sleeve chen
Dec 7 at 9:02
2
Yes. I do not know of necessary conditions, ODE can have a vast variety of behaviors. You can only select classes of nice ones.
– LutzL
Dec 7 at 9:08
|
show 1 more comment
Usual conditions for the infinite existence of solutions of $dot x=f(x)$ are restrictions to linear growth like
$$
|dot x|=|f(x)|le C(1+|x|)
$$
or
$$
frac12frac{d}{dt}|x|^2=|x^Tdot x|=|x^Tf(x)|le C(1+|x|^2)
$$
which impose an exponential bound on the growth of solutions. The solution of the corresponding differential inequalities is with the Grönwall lemma.
These are not the only conditions, only the relatively easy ones. If the vector field $f(x)$ points inwards on the boundary of a bounded set $U$, all solutions are restricted to $U$ and thus bounded and exist for all positive times. This looks geometrically simple but can be hard to establish
answered Dec 7 at 8:47
LutzL
55.6k42054
Usual conditions for the infinite existence of solutions of $dot x=f(x)$ are restrictions to linear growth like
$$
|dot x|=|f(x)|le C(1+|x|)
$$
or
$$
frac12frac{d}{dt}|x|^2=|x^Tdot x|=|x^Tf(x)|le C(1+|x|^2)
$$
which impose an exponential bound on the growth of solutions. The solution of the corresponding differential inequalities is with the Grönwall lemma.
These are not the only conditions, only the relatively easy ones. If the vector field $f(x)$ points inwards on the boundary of a bounded set $U$, all solutions are restricted to $U$ and thus bounded and exist for all positive times. This looks geometrically simple but can be hard to establish
answered Dec 7 at 8:47
LutzL
55.6k42054
edited Dec 7 at 9:01
answered Dec 7 at 8:47
LutzL
55.6k42054
answered Dec 7 at 8:47
LutzL
55.6k42054
answered Dec 7 at 8:47
LutzL
55.6k42054
55.6k42054
How do you relate that to the ODE ?
– Yves Daoust
Dec 7 at 8:50
For your second condition, it looks like the example $dot{x} = x^2$ in my problem satisfies this; however, it does not have solution for all time?
– sleeve chen
Dec 7 at 8:51
1
@sleevechen : No, $x^Tf(x)=x^3$ grows faster than $|x|^2$.
– LutzL
Dec 7 at 8:56
Thanks I see. It looks a bit like the definition of Lipshitz continuity.
– sleeve chen
Dec 7 at 9:02
2
Yes. I do not know of necessary conditions, ODE can have a vast variety of behaviors. You can only select classes of nice ones.
– LutzL
Dec 7 at 9:08
|
show 1 more comment
How do you relate that to the ODE ?
– Yves Daoust
Dec 7 at 8:50
For your second condition, it looks like the example $dot{x} = x^2$ in my problem satisfies this; however, it does not have solution for all time?
– sleeve chen
Dec 7 at 8:51
1
@sleevechen : No, $x^Tf(x)=x^3$ grows faster than $|x|^2$.
– LutzL
Dec 7 at 8:56
Thanks I see. It looks a bit like the definition of Lipshitz continuity.
– sleeve chen
Dec 7 at 9:02
2
Yes. I do not know of necessary conditions, ODE can have a vast variety of behaviors. You can only select classes of nice ones.
– LutzL
Dec 7 at 9:08
How do you relate that to the ODE ?
– Yves Daoust
Dec 7 at 8:50
How do you relate that to the ODE ?
– Yves Daoust
Dec 7 at 8:50
For your second condition, it looks like the example $dot{x} = x^2$ in my problem satisfies this; however, it does not have solution for all time?
– sleeve chen
Dec 7 at 8:51
For your second condition, it looks like the example $dot{x} = x^2$ in my problem satisfies this; however, it does not have solution for all time?
– sleeve chen
Dec 7 at 8:51
1
1
@sleevechen : No, $x^Tf(x)=x^3$ grows faster than $|x|^2$.
– LutzL
Dec 7 at 8:56
@sleevechen : No, $x^Tf(x)=x^3$ grows faster than $|x|^2$.
– LutzL
Dec 7 at 8:56
Thanks I see. It looks a bit like the definition of Lipshitz continuity.
– sleeve chen
Dec 7 at 9:02
Thanks I see. It looks a bit like the definition of Lipshitz continuity.
– sleeve chen
Dec 7 at 9:02
2
2
Yes. I do not know of necessary conditions, ODE can have a vast variety of behaviors. You can only select classes of nice ones.
– LutzL
Dec 7 at 9:08
Yes. I do not know of necessary conditions, ODE can have a vast variety of behaviors. You can only select classes of nice ones.
– LutzL
Dec 7 at 9:08
|
show 1 more comment
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