Mixing
Frozen time equilibria are not solutions of nonautonomous system
I am reading the following textbook:
Introduction to Applied Nonlinear Dynamical Systems and Chaos by Wiggins
p.6 (top)
Suppose we have $$dot{x} = f(x,t), xin mathbb{R}^n $$
By implicit function theorem, if we can find $(bar{x},bar{t})$ such that $f(bar{x},bar{t})=0$ and $D_xf(bar{x},bar{t})neq 0$, then we can find $bar{x}(t)$, with $bar{x}(bar{t})=bar{x}$ such that $f(bar{x}(t),t)=0$.
If $bar{x}(t)$ is a solution of the nonautonomous vector field, then it must be constant in time, i.e., $dot{bar{x}}(t)=0$.
Why this is the case? I am confused about it.
differential-equations dynamical-systems
asked Nov 20 '18 at 8:19
sleeve chen
3,05941852
add a comment |
I am reading the following textbook:
Introduction to Applied Nonlinear Dynamical Systems and Chaos by Wiggins
p.6 (top)
Suppose we have $$dot{x} = f(x,t), xin mathbb{R}^n $$
By implicit function theorem, if we can find $(bar{x},bar{t})$ such that $f(bar{x},bar{t})=0$ and $D_xf(bar{x},bar{t})neq 0$, then we can find $bar{x}(t)$, with $bar{x}(bar{t})=bar{x}$ such that $f(bar{x}(t),t)=0$.
If $bar{x}(t)$ is a solution of the nonautonomous vector field, then it must be constant in time, i.e., $dot{bar{x}}(t)=0$.
Why this is the case? I am confused about it.
differential-equations dynamical-systems
asked Nov 20 '18 at 8:19
sleeve chen
3,05941852
2
The implicit function theorem guarantees the existence of a $C^1$ function $bar{x}colon(bar{t}-epsilon,bar{t}+epsilon)tomathbb{R}^n$ such that $f(bar{x}(t),t)=0$ for all $tin(bar{t}-epsilon,bar{t}+epsilon)$. But that $bar{x}$ being a solution means that $dot{bar{x}}(t)=f(bar{x}(t),t)$, which is equal to $0$ by the above reasoning. Consequently, $bar{x}$ must be constant, at least on $(bar{t}-epsilon,bar{t}+epsilon)$.
– user539887
Nov 20 '18 at 8:41
@user539887 Thanks! I get it.
– sleeve chen
Nov 20 '18 at 8:44
add a comment |
I am reading the following textbook:
Introduction to Applied Nonlinear Dynamical Systems and Chaos by Wiggins
p.6 (top)
Suppose we have $$dot{x} = f(x,t), xin mathbb{R}^n $$
By implicit function theorem, if we can find $(bar{x},bar{t})$ such that $f(bar{x},bar{t})=0$ and $D_xf(bar{x},bar{t})neq 0$, then we can find $bar{x}(t)$, with $bar{x}(bar{t})=bar{x}$ such that $f(bar{x}(t),t)=0$.
If $bar{x}(t)$ is a solution of the nonautonomous vector field, then it must be constant in time, i.e., $dot{bar{x}}(t)=0$.
Why this is the case? I am confused about it.
differential-equations dynamical-systems
asked Nov 20 '18 at 8:19
sleeve chen
3,05941852
I am reading the following textbook:
Introduction to Applied Nonlinear Dynamical Systems and Chaos by Wiggins
p.6 (top)
Suppose we have $$dot{x} = f(x,t), xin mathbb{R}^n $$
By implicit function theorem, if we can find $(bar{x},bar{t})$ such that $f(bar{x},bar{t})=0$ and $D_xf(bar{x},bar{t})neq 0$, then we can find $bar{x}(t)$, with $bar{x}(bar{t})=bar{x}$ such that $f(bar{x}(t),t)=0$.
If $bar{x}(t)$ is a solution of the nonautonomous vector field, then it must be constant in time, i.e., $dot{bar{x}}(t)=0$.
Why this is the case? I am confused about it.
differential-equations dynamical-systems
differential-equations dynamical-systems
asked Nov 20 '18 at 8:19
sleeve chen
3,05941852
asked Nov 20 '18 at 8:19
sleeve chen
3,05941852
asked Nov 20 '18 at 8:19
sleeve chen
3,05941852
asked Nov 20 '18 at 8:19
sleeve chen
3,05941852
asked Nov 20 '18 at 8:19
sleeve chen
3,05941852
3,05941852
2
The implicit function theorem guarantees the existence of a $C^1$ function $bar{x}colon(bar{t}-epsilon,bar{t}+epsilon)tomathbb{R}^n$ such that $f(bar{x}(t),t)=0$ for all $tin(bar{t}-epsilon,bar{t}+epsilon)$. But that $bar{x}$ being a solution means that $dot{bar{x}}(t)=f(bar{x}(t),t)$, which is equal to $0$ by the above reasoning. Consequently, $bar{x}$ must be constant, at least on $(bar{t}-epsilon,bar{t}+epsilon)$.
– user539887
Nov 20 '18 at 8:41
@user539887 Thanks! I get it.
– sleeve chen
Nov 20 '18 at 8:44
add a comment |
2
The implicit function theorem guarantees the existence of a $C^1$ function $bar{x}colon(bar{t}-epsilon,bar{t}+epsilon)tomathbb{R}^n$ such that $f(bar{x}(t),t)=0$ for all $tin(bar{t}-epsilon,bar{t}+epsilon)$. But that $bar{x}$ being a solution means that $dot{bar{x}}(t)=f(bar{x}(t),t)$, which is equal to $0$ by the above reasoning. Consequently, $bar{x}$ must be constant, at least on $(bar{t}-epsilon,bar{t}+epsilon)$.
– user539887
Nov 20 '18 at 8:41
@user539887 Thanks! I get it.
– sleeve chen
Nov 20 '18 at 8:44
2
2
The implicit function theorem guarantees the existence of a $C^1$ function $bar{x}colon(bar{t}-epsilon,bar{t}+epsilon)tomathbb{R}^n$ such that $f(bar{x}(t),t)=0$ for all $tin(bar{t}-epsilon,bar{t}+epsilon)$. But that $bar{x}$ being a solution means that $dot{bar{x}}(t)=f(bar{x}(t),t)$, which is equal to $0$ by the above reasoning. Consequently, $bar{x}$ must be constant, at least on $(bar{t}-epsilon,bar{t}+epsilon)$.
– user539887
Nov 20 '18 at 8:41
The implicit function theorem guarantees the existence of a $C^1$ function $bar{x}colon(bar{t}-epsilon,bar{t}+epsilon)tomathbb{R}^n$ such that $f(bar{x}(t),t)=0$ for all $tin(bar{t}-epsilon,bar{t}+epsilon)$. But that $bar{x}$ being a solution means that $dot{bar{x}}(t)=f(bar{x}(t),t)$, which is equal to $0$ by the above reasoning. Consequently, $bar{x}$ must be constant, at least on $(bar{t}-epsilon,bar{t}+epsilon)$.
– user539887
Nov 20 '18 at 8:41
@user539887 Thanks! I get it.
– sleeve chen
Nov 20 '18 at 8:44
@user539887 Thanks! I get it.
– sleeve chen
Nov 20 '18 at 8:44
add a comment |
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2
The implicit function theorem guarantees the existence of a $C^1$ function $bar{x}colon(bar{t}-epsilon,bar{t}+epsilon)tomathbb{R}^n$ such that $f(bar{x}(t),t)=0$ for all $tin(bar{t}-epsilon,bar{t}+epsilon)$. But that $bar{x}$ being a solution means that $dot{bar{x}}(t)=f(bar{x}(t),t)$, which is equal to $0$ by the above reasoning. Consequently, $bar{x}$ must be constant, at least on $(bar{t}-epsilon,bar{t}+epsilon)$.
– user539887
Nov 20 '18 at 8:41
@user539887 Thanks! I get it.
– sleeve chen
Nov 20 '18 at 8:44