Frozen time equilibria are not solutions of nonautonomous system












0














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







share|cite|improve this question


















  • 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
















0














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







share|cite|improve this question


















  • 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














0












0








0







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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















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