Integral curves and flows of bounded vector fields
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Let $X=(X_1,dots, X_n)in L^infty$ be a smooth (non Lipschitz) vector field such that
begin{equation}tag{1}
X_n ge c|(X_1, dots, X_{n-1})|,.
end{equation}
Does the cone condition (1) imply that the flow $Phi(t,x)$ is globally defined? I would say so, as it seems that forces integral curves to be confined.
Is this true?
Take $bar x$ and given the maximal interval $(0, t_{bar x})$, I argue by contradiction that $t_{bar x}$ is finite. Take a sequence of times $t_n$ converging to $t_{bar x}$ and let $x_n$ be the points corresponding to the flow $Phi(t_n,bar x)$. If I am able to prove that the sequence $x_n$ is bounded, then it must converge and thus I would be able to extend the flow at $t_{bar x}$ by setting it equal to $lim_n x_n$.
How do I prove that these $x_n$ stay bounded, given (1)?
vector-fields
asked Nov 16 at 15:06
Paolo Intuito
988218
add a comment |
up vote
2
down vote
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Let $X=(X_1,dots, X_n)in L^infty$ be a smooth (non Lipschitz) vector field such that
begin{equation}tag{1}
X_n ge c|(X_1, dots, X_{n-1})|,.
end{equation}
Does the cone condition (1) imply that the flow $Phi(t,x)$ is globally defined? I would say so, as it seems that forces integral curves to be confined.
Is this true?
Take $bar x$ and given the maximal interval $(0, t_{bar x})$, I argue by contradiction that $t_{bar x}$ is finite. Take a sequence of times $t_n$ converging to $t_{bar x}$ and let $x_n$ be the points corresponding to the flow $Phi(t_n,bar x)$. If I am able to prove that the sequence $x_n$ is bounded, then it must converge and thus I would be able to extend the flow at $t_{bar x}$ by setting it equal to $lim_n x_n$.
How do I prove that these $x_n$ stay bounded, given (1)?
vector-fields
asked Nov 16 at 15:06
Paolo Intuito
988218
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $X=(X_1,dots, X_n)in L^infty$ be a smooth (non Lipschitz) vector field such that
begin{equation}tag{1}
X_n ge c|(X_1, dots, X_{n-1})|,.
end{equation}
Does the cone condition (1) imply that the flow $Phi(t,x)$ is globally defined? I would say so, as it seems that forces integral curves to be confined.
Is this true?
Take $bar x$ and given the maximal interval $(0, t_{bar x})$, I argue by contradiction that $t_{bar x}$ is finite. Take a sequence of times $t_n$ converging to $t_{bar x}$ and let $x_n$ be the points corresponding to the flow $Phi(t_n,bar x)$. If I am able to prove that the sequence $x_n$ is bounded, then it must converge and thus I would be able to extend the flow at $t_{bar x}$ by setting it equal to $lim_n x_n$.
How do I prove that these $x_n$ stay bounded, given (1)?
vector-fields
asked Nov 16 at 15:06
Paolo Intuito
988218
Let $X=(X_1,dots, X_n)in L^infty$ be a smooth (non Lipschitz) vector field such that
begin{equation}tag{1}
X_n ge c|(X_1, dots, X_{n-1})|,.
end{equation}
Does the cone condition (1) imply that the flow $Phi(t,x)$ is globally defined? I would say so, as it seems that forces integral curves to be confined.
Is this true?
Take $bar x$ and given the maximal interval $(0, t_{bar x})$, I argue by contradiction that $t_{bar x}$ is finite. Take a sequence of times $t_n$ converging to $t_{bar x}$ and let $x_n$ be the points corresponding to the flow $Phi(t_n,bar x)$. If I am able to prove that the sequence $x_n$ is bounded, then it must converge and thus I would be able to extend the flow at $t_{bar x}$ by setting it equal to $lim_n x_n$.
How do I prove that these $x_n$ stay bounded, given (1)?
vector-fields
vector-fields
asked Nov 16 at 15:06
Paolo Intuito
988218
asked Nov 16 at 15:06
Paolo Intuito
988218
edited Nov 23 at 15:57
asked Nov 16 at 15:06
Paolo Intuito
988218
asked Nov 16 at 15:06
Paolo Intuito
988218
asked Nov 16 at 15:06
Paolo Intuito
988218
988218
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You don't even need the assumption $(1)$.
Local existence and uniqueness is ensured because $X$ is locally Lipschitz.
You have that
$$
|Phi(t,x)-x| = left|int_0^t X(Phi(s,x))dsright| leq int_0^t |X(Phi(s,x))|ds
leq |X|_infty t.
$$
This means that solutions stay in a compact set in finite time, which implies that they can be extended to $[0,infty)$.
answered Nov 23 at 17:25
Federico
2,556510
Great thanks. I somehow got distracted by the assumtpion (1) which apparently was not needed. Maybe I can infer something more from it, such that the integral curves will pass through every hyperplane of fixed $n$-th coordinate, $H_h:={x_n = h}$?
– Paolo Intuito
Nov 26 at 9:17
1
Nope: try $Xequiv 0$.
– Federico
Nov 26 at 17:16
Besides the trivial case, i.e. if $|X|_infty>0$?
– Paolo Intuito
Nov 27 at 11:10
1
For that you would $X_n>0$.
– Federico
Nov 27 at 13:36
1
Just consider a trajectory $x(t)$. Assume $M=sup_t x_n(t)=lim_{ttoinfty} x_n(t)<infty$. Then by $(1)$ you get that $x^*=lim_{ttoinfty} x(t)$ exists. But then $x^*$ needs to be a stationary point, giving $X_n(x^*)=0$, contradiction.
– Federico
Nov 28 at 16:13
|
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You don't even need the assumption $(1)$.
Local existence and uniqueness is ensured because $X$ is locally Lipschitz.
You have that
$$
|Phi(t,x)-x| = left|int_0^t X(Phi(s,x))dsright| leq int_0^t |X(Phi(s,x))|ds
leq |X|_infty t.
$$
This means that solutions stay in a compact set in finite time, which implies that they can be extended to $[0,infty)$.
answered Nov 23 at 17:25
Federico
2,556510
Great thanks. I somehow got distracted by the assumtpion (1) which apparently was not needed. Maybe I can infer something more from it, such that the integral curves will pass through every hyperplane of fixed $n$-th coordinate, $H_h:={x_n = h}$?
– Paolo Intuito
Nov 26 at 9:17
1
Nope: try $Xequiv 0$.
– Federico
Nov 26 at 17:16
Besides the trivial case, i.e. if $|X|_infty>0$?
– Paolo Intuito
Nov 27 at 11:10
1
For that you would $X_n>0$.
– Federico
Nov 27 at 13:36
1
Just consider a trajectory $x(t)$. Assume $M=sup_t x_n(t)=lim_{ttoinfty} x_n(t)<infty$. Then by $(1)$ you get that $x^*=lim_{ttoinfty} x(t)$ exists. But then $x^*$ needs to be a stationary point, giving $X_n(x^*)=0$, contradiction.
– Federico
Nov 28 at 16:13
|
show 4 more comments
up vote
1
down vote
accepted
You don't even need the assumption $(1)$.
Local existence and uniqueness is ensured because $X$ is locally Lipschitz.
You have that
$$
|Phi(t,x)-x| = left|int_0^t X(Phi(s,x))dsright| leq int_0^t |X(Phi(s,x))|ds
leq |X|_infty t.
$$
This means that solutions stay in a compact set in finite time, which implies that they can be extended to $[0,infty)$.
answered Nov 23 at 17:25
Federico
2,556510
Great thanks. I somehow got distracted by the assumtpion (1) which apparently was not needed. Maybe I can infer something more from it, such that the integral curves will pass through every hyperplane of fixed $n$-th coordinate, $H_h:={x_n = h}$?
– Paolo Intuito
Nov 26 at 9:17
1
Nope: try $Xequiv 0$.
– Federico
Nov 26 at 17:16
Besides the trivial case, i.e. if $|X|_infty>0$?
– Paolo Intuito
Nov 27 at 11:10
1
For that you would $X_n>0$.
– Federico
Nov 27 at 13:36
1
Just consider a trajectory $x(t)$. Assume $M=sup_t x_n(t)=lim_{ttoinfty} x_n(t)<infty$. Then by $(1)$ you get that $x^*=lim_{ttoinfty} x(t)$ exists. But then $x^*$ needs to be a stationary point, giving $X_n(x^*)=0$, contradiction.
– Federico
Nov 28 at 16:13
|
show 4 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You don't even need the assumption $(1)$.
Local existence and uniqueness is ensured because $X$ is locally Lipschitz.
You have that
$$
|Phi(t,x)-x| = left|int_0^t X(Phi(s,x))dsright| leq int_0^t |X(Phi(s,x))|ds
leq |X|_infty t.
$$
This means that solutions stay in a compact set in finite time, which implies that they can be extended to $[0,infty)$.
answered Nov 23 at 17:25
Federico
2,556510
You don't even need the assumption $(1)$.
Local existence and uniqueness is ensured because $X$ is locally Lipschitz.
You have that
$$
|Phi(t,x)-x| = left|int_0^t X(Phi(s,x))dsright| leq int_0^t |X(Phi(s,x))|ds
leq |X|_infty t.
$$
This means that solutions stay in a compact set in finite time, which implies that they can be extended to $[0,infty)$.
answered Nov 23 at 17:25
Federico
2,556510
answered Nov 23 at 17:25
Federico
2,556510
answered Nov 23 at 17:25
Federico
2,556510
answered Nov 23 at 17:25
Federico
2,556510
2,556510
Great thanks. I somehow got distracted by the assumtpion (1) which apparently was not needed. Maybe I can infer something more from it, such that the integral curves will pass through every hyperplane of fixed $n$-th coordinate, $H_h:={x_n = h}$?
– Paolo Intuito
Nov 26 at 9:17
1
Nope: try $Xequiv 0$.
– Federico
Nov 26 at 17:16
Besides the trivial case, i.e. if $|X|_infty>0$?
– Paolo Intuito
Nov 27 at 11:10
1
For that you would $X_n>0$.
– Federico
Nov 27 at 13:36
1
Just consider a trajectory $x(t)$. Assume $M=sup_t x_n(t)=lim_{ttoinfty} x_n(t)<infty$. Then by $(1)$ you get that $x^*=lim_{ttoinfty} x(t)$ exists. But then $x^*$ needs to be a stationary point, giving $X_n(x^*)=0$, contradiction.
– Federico
Nov 28 at 16:13
|
show 4 more comments
Great thanks. I somehow got distracted by the assumtpion (1) which apparently was not needed. Maybe I can infer something more from it, such that the integral curves will pass through every hyperplane of fixed $n$-th coordinate, $H_h:={x_n = h}$?
– Paolo Intuito
Nov 26 at 9:17
1
Nope: try $Xequiv 0$.
– Federico
Nov 26 at 17:16
Besides the trivial case, i.e. if $|X|_infty>0$?
– Paolo Intuito
Nov 27 at 11:10
1
For that you would $X_n>0$.
– Federico
Nov 27 at 13:36
1
Just consider a trajectory $x(t)$. Assume $M=sup_t x_n(t)=lim_{ttoinfty} x_n(t)<infty$. Then by $(1)$ you get that $x^*=lim_{ttoinfty} x(t)$ exists. But then $x^*$ needs to be a stationary point, giving $X_n(x^*)=0$, contradiction.
– Federico
Nov 28 at 16:13
Great thanks. I somehow got distracted by the assumtpion (1) which apparently was not needed. Maybe I can infer something more from it, such that the integral curves will pass through every hyperplane of fixed $n$-th coordinate, $H_h:={x_n = h}$?
– Paolo Intuito
Nov 26 at 9:17
Great thanks. I somehow got distracted by the assumtpion (1) which apparently was not needed. Maybe I can infer something more from it, such that the integral curves will pass through every hyperplane of fixed $n$-th coordinate, $H_h:={x_n = h}$?
– Paolo Intuito
Nov 26 at 9:17
1
1
Nope: try $Xequiv 0$.
– Federico
Nov 26 at 17:16
Nope: try $Xequiv 0$.
– Federico
Nov 26 at 17:16
Besides the trivial case, i.e. if $|X|_infty>0$?
– Paolo Intuito
Nov 27 at 11:10
Besides the trivial case, i.e. if $|X|_infty>0$?
– Paolo Intuito
Nov 27 at 11:10
1
1
For that you would $X_n>0$.
– Federico
Nov 27 at 13:36
For that you would $X_n>0$.
– Federico
Nov 27 at 13:36
1
1
Just consider a trajectory $x(t)$. Assume $M=sup_t x_n(t)=lim_{ttoinfty} x_n(t)<infty$. Then by $(1)$ you get that $x^*=lim_{ttoinfty} x(t)$ exists. But then $x^*$ needs to be a stationary point, giving $X_n(x^*)=0$, contradiction.
– Federico
Nov 28 at 16:13
Just consider a trajectory $x(t)$. Assume $M=sup_t x_n(t)=lim_{ttoinfty} x_n(t)<infty$. Then by $(1)$ you get that $x^*=lim_{ttoinfty} x(t)$ exists. But then $x^*$ needs to be a stationary point, giving $X_n(x^*)=0$, contradiction.
– Federico
Nov 28 at 16:13
|
show 4 more comments
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