Gradient dynamical systems have no nonconstant recurrent solutions
This is given as something we should intuitively understand, but I don't see how this is trivial. We were given that a solution is recurrent if $X(t_n) to X(0)$ for some sequence from $t_n$ to infinity. Why is this true?
differential-equations dynamical-systems
edited Dec 3 '18 at 1:13
caverac
13.8k21130
asked Dec 3 '18 at 1:07
MathGuyForLife
1007
add a comment |
This is given as something we should intuitively understand, but I don't see how this is trivial. We were given that a solution is recurrent if $X(t_n) to X(0)$ for some sequence from $t_n$ to infinity. Why is this true?
differential-equations dynamical-systems
edited Dec 3 '18 at 1:13
caverac
13.8k21130
asked Dec 3 '18 at 1:07
MathGuyForLife
1007
Just to mention why this should be "intuitive": If $f$ is the function whose gradient we are considering, and $X(t)$ follows $nabla f$, then $f(X(t))$ should be increasing for all $t$ since the gradient points in the direction of steepest ascent for $f$.
– user25959
Dec 3 '18 at 1:29
Hint: if $X(t_n) rightarrow X(0)$, what happens to $f(X(t_n))$?
– Evgeny
Dec 3 '18 at 14:20
add a comment |
This is given as something we should intuitively understand, but I don't see how this is trivial. We were given that a solution is recurrent if $X(t_n) to X(0)$ for some sequence from $t_n$ to infinity. Why is this true?
differential-equations dynamical-systems
edited Dec 3 '18 at 1:13
caverac
13.8k21130
asked Dec 3 '18 at 1:07
MathGuyForLife
1007
This is given as something we should intuitively understand, but I don't see how this is trivial. We were given that a solution is recurrent if $X(t_n) to X(0)$ for some sequence from $t_n$ to infinity. Why is this true?
differential-equations dynamical-systems
differential-equations dynamical-systems
edited Dec 3 '18 at 1:13
caverac
13.8k21130
asked Dec 3 '18 at 1:07
MathGuyForLife
1007
edited Dec 3 '18 at 1:13
caverac
13.8k21130
asked Dec 3 '18 at 1:07
MathGuyForLife
1007
edited Dec 3 '18 at 1:13
caverac
13.8k21130
edited Dec 3 '18 at 1:13
caverac
13.8k21130
edited Dec 3 '18 at 1:13
caverac
13.8k21130
13.8k21130
asked Dec 3 '18 at 1:07
MathGuyForLife
1007
asked Dec 3 '18 at 1:07
MathGuyForLife
1007
asked Dec 3 '18 at 1:07
MathGuyForLife
1007
1007
Just to mention why this should be "intuitive": If $f$ is the function whose gradient we are considering, and $X(t)$ follows $nabla f$, then $f(X(t))$ should be increasing for all $t$ since the gradient points in the direction of steepest ascent for $f$.
– user25959
Dec 3 '18 at 1:29
Hint: if $X(t_n) rightarrow X(0)$, what happens to $f(X(t_n))$?
– Evgeny
Dec 3 '18 at 14:20
add a comment |
Just to mention why this should be "intuitive": If $f$ is the function whose gradient we are considering, and $X(t)$ follows $nabla f$, then $f(X(t))$ should be increasing for all $t$ since the gradient points in the direction of steepest ascent for $f$.
– user25959
Dec 3 '18 at 1:29
Hint: if $X(t_n) rightarrow X(0)$, what happens to $f(X(t_n))$?
– Evgeny
Dec 3 '18 at 14:20
Just to mention why this should be "intuitive": If $f$ is the function whose gradient we are considering, and $X(t)$ follows $nabla f$, then $f(X(t))$ should be increasing for all $t$ since the gradient points in the direction of steepest ascent for $f$.
– user25959
Dec 3 '18 at 1:29
Just to mention why this should be "intuitive": If $f$ is the function whose gradient we are considering, and $X(t)$ follows $nabla f$, then $f(X(t))$ should be increasing for all $t$ since the gradient points in the direction of steepest ascent for $f$.
– user25959
Dec 3 '18 at 1:29
Hint: if $X(t_n) rightarrow X(0)$, what happens to $f(X(t_n))$?
– Evgeny
Dec 3 '18 at 14:20
Hint: if $X(t_n) rightarrow X(0)$, what happens to $f(X(t_n))$?
– Evgeny
Dec 3 '18 at 14:20
add a comment |
1 Answer
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Consider the gradient function $Phi = Phi({bf x})$,
$$
frac{{rm d}{bf x}}{{rm d}t} = -nabla Phi tag{1}
$$
Assume the system is recurrent
$$
Phi({bf x}(t_n)) = Phi({bf x}(0)) tag{2}
$$
But on the other hand
$$
0 stackrel{(2)}{=}int {rm d}Phi = int_0^{t_n} nabla Phi cdot frac{{rm d}{bf x}}{{rm d}t} {rm d}t stackrel{(1)}{=} -int_0^{t_n} left| frac{{rm d}{bf x}}{{rm d }t}right|^2{rm d}t < 0
$$
So you get to a contradiction. A gradient dynamical system cannot have recurrent orbits!
answered Dec 3 '18 at 1:29
caverac
13.8k21130
4
Just a note: OP's question mentions "recurrent" trajectories which are more general than periodic orbits
– user25959
Dec 3 '18 at 1:31
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Consider the gradient function $Phi = Phi({bf x})$,
$$
frac{{rm d}{bf x}}{{rm d}t} = -nabla Phi tag{1}
$$
Assume the system is recurrent
$$
Phi({bf x}(t_n)) = Phi({bf x}(0)) tag{2}
$$
But on the other hand
$$
0 stackrel{(2)}{=}int {rm d}Phi = int_0^{t_n} nabla Phi cdot frac{{rm d}{bf x}}{{rm d}t} {rm d}t stackrel{(1)}{=} -int_0^{t_n} left| frac{{rm d}{bf x}}{{rm d }t}right|^2{rm d}t < 0
$$
So you get to a contradiction. A gradient dynamical system cannot have recurrent orbits!
answered Dec 3 '18 at 1:29
caverac
13.8k21130
4
Just a note: OP's question mentions "recurrent" trajectories which are more general than periodic orbits
– user25959
Dec 3 '18 at 1:31
add a comment |
Consider the gradient function $Phi = Phi({bf x})$,
$$
frac{{rm d}{bf x}}{{rm d}t} = -nabla Phi tag{1}
$$
Assume the system is recurrent
$$
Phi({bf x}(t_n)) = Phi({bf x}(0)) tag{2}
$$
But on the other hand
$$
0 stackrel{(2)}{=}int {rm d}Phi = int_0^{t_n} nabla Phi cdot frac{{rm d}{bf x}}{{rm d}t} {rm d}t stackrel{(1)}{=} -int_0^{t_n} left| frac{{rm d}{bf x}}{{rm d }t}right|^2{rm d}t < 0
$$
So you get to a contradiction. A gradient dynamical system cannot have recurrent orbits!
answered Dec 3 '18 at 1:29
caverac
13.8k21130
4
Just a note: OP's question mentions "recurrent" trajectories which are more general than periodic orbits
– user25959
Dec 3 '18 at 1:31
add a comment |
Consider the gradient function $Phi = Phi({bf x})$,
$$
frac{{rm d}{bf x}}{{rm d}t} = -nabla Phi tag{1}
$$
Assume the system is recurrent
$$
Phi({bf x}(t_n)) = Phi({bf x}(0)) tag{2}
$$
But on the other hand
$$
0 stackrel{(2)}{=}int {rm d}Phi = int_0^{t_n} nabla Phi cdot frac{{rm d}{bf x}}{{rm d}t} {rm d}t stackrel{(1)}{=} -int_0^{t_n} left| frac{{rm d}{bf x}}{{rm d }t}right|^2{rm d}t < 0
$$
So you get to a contradiction. A gradient dynamical system cannot have recurrent orbits!
answered Dec 3 '18 at 1:29
caverac
13.8k21130
Consider the gradient function $Phi = Phi({bf x})$,
$$
frac{{rm d}{bf x}}{{rm d}t} = -nabla Phi tag{1}
$$
Assume the system is recurrent
$$
Phi({bf x}(t_n)) = Phi({bf x}(0)) tag{2}
$$
But on the other hand
$$
0 stackrel{(2)}{=}int {rm d}Phi = int_0^{t_n} nabla Phi cdot frac{{rm d}{bf x}}{{rm d}t} {rm d}t stackrel{(1)}{=} -int_0^{t_n} left| frac{{rm d}{bf x}}{{rm d }t}right|^2{rm d}t < 0
$$
So you get to a contradiction. A gradient dynamical system cannot have recurrent orbits!
answered Dec 3 '18 at 1:29
caverac
13.8k21130
edited Dec 3 '18 at 1:32
answered Dec 3 '18 at 1:29
caverac
13.8k21130
answered Dec 3 '18 at 1:29
caverac
13.8k21130
answered Dec 3 '18 at 1:29
caverac
13.8k21130
13.8k21130
4
Just a note: OP's question mentions "recurrent" trajectories which are more general than periodic orbits
– user25959
Dec 3 '18 at 1:31
add a comment |
4
Just a note: OP's question mentions "recurrent" trajectories which are more general than periodic orbits
– user25959
Dec 3 '18 at 1:31
4
4
Just a note: OP's question mentions "recurrent" trajectories which are more general than periodic orbits
– user25959
Dec 3 '18 at 1:31
Just a note: OP's question mentions "recurrent" trajectories which are more general than periodic orbits
– user25959
Dec 3 '18 at 1:31
add a comment |
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Just to mention why this should be "intuitive": If $f$ is the function whose gradient we are considering, and $X(t)$ follows $nabla f$, then $f(X(t))$ should be increasing for all $t$ since the gradient points in the direction of steepest ascent for $f$.
– user25959
Dec 3 '18 at 1:29
Hint: if $X(t_n) rightarrow X(0)$, what happens to $f(X(t_n))$?
– Evgeny
Dec 3 '18 at 14:20