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I am reading the following book:

Introduction to applied nonlinear dynamical systems and chaos, Stephen Wiggins

On p. 77, for a general vector field $$\dot{x} = f(x), \ \ \ x\in \mathbb{R}^n.$$

A scalar valued function $I(x)$ is said to be an integral if it is contant on trajectories: $$\dot{I}(x) = \nabla I(x)\cdot \dot{x}= \nabla I(x)\cdot f(x)=0.$$

It says, from the above the level sets of $I(x)$ are invariant sets.

How to see the level sets of $I(x)$ are invariant?

sleeve chen
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    A set $M$ is invariant if for any point $p$ its trajectory $\lbrace \phi^t(p) \rbrace_{t \in \mathbb{R}}$ belongs to $M$, where $\phi^t$ is a phase flow. Pick any point $y$ in level set $\lbrace I(x) = C \rbrace$: since $I(x)$ is a first integral, then $I(\phi^t(y)) \equiv C$, hence $\phi^t(y)$ belongs to $\lbrace I(x) = C \rbrace$ for all $t$ and $\lbrace \phi^t(y) \rbrace_{t \in \mathbb{R}} \subset \lbrace I(x) = C \rbrace$ . – Evgeny Apr 24 '19 at 05:55

1 Answers1

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The level sets of a regular functions $f$ which is a first integral of a vector field (i.e. for a first order ODE) are invariant in the sense that $f$ does not vary along the integral curves.

This means that $\nabla f(x)\cdot X(x) = 0 $ for any $x$ in the space domain. Hence if $f$ is not a constant function, this means that the variation of $f$ happens along directions which live in the orthogonal space to the one generated by the vectors of the vector field $X$ associated to the dynamical system $\dot{x}=f(x)$ you defined in the question.

Hence the level sets $\{x\in\mathbb{R}^n : f(x)=c\}=f^{-1}(\{c\})$ are mapped by the flow of the vector field in a subset of the level set itself. In formulas, let $\Phi^t:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be the flow of the system, then we have that $\forall c\in f(\mathbb{R}^n)$ and $\forall t\in\mathbb{R}$ it follows $\phi^t(f^{-1}(\{c\}))\subset f^{-1}(\{c\})$.

The reason behind this invariancy is simply because the gradient $\nabla f$ is orthogonal to the level sets of $f$, hence since the gradient vector is orthogonal again to the vector field $X$ by definition of first integral, then it follows that the level sets are parallel to the vector field $X$, or more precisely to its integral curves. Hence the level sets are invariant with respect to the flow of $X$.

Dadeslam
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