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I am studying dynamical systems and I have some trouble in understanding the notation used for differential equations.

For example when I read

$$\overset{..}{x}=F(x),$$

how should I interpret it? Is it asking to find a function $x$ (defined on an open set) such that $x''(t)=F(x(t))$ for all $t$? If so, why don't they write $x''=F\circ x$, or $\overset{..}x =F\circ x$?

Then I see:

Let's call $y$ the velocity, i.e., the function $\overset.x$. Suppose that exists a function $V$ such that $F=\dfrac{-dV}{dx}$. Then the quantity $E(x,y)=\frac12y^2+V(x)$ is constant when $t$ changes.

Does it mean that the function $t\mapsto E(x(t),y(t))=\frac12y(t)^2+V(x(t))$ is constant? What does $\frac{-dV}{dx}$ mean? Simply $V'$? In the definition of derivative of a function I have never seen something like "derivative with respect to x" or "derivative with respect to t". My guess is that when they write $\frac{dV}{dx}$ they mean the derivative of the function that maps every point to its potential (in this case simply $V'$); and when they write $\frac{dV}{dt}$ they mean the derivative of the function that maps the time to the potential at that time (which in this case is the derivative of another function, the function $V\circ x$. So it is $(V\circ x)'=(V'\circ x)\cdot x'$). Is it all right?

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    $\begingroup$ All your interpretations are correct. Your puzzlement is understandable if you are a "pure" mathematician entering the field. After a while, and if one takes care to always know precisely what one is talking about, one gets used to these notations and, even, one can find them useful... $\endgroup$
    – Did
    Aug 9, 2014 at 8:43
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    $\begingroup$ The reason for writing $F(x)$ is that one thinks of it as a given vector field: at each point $x$ of your configuration space there is attached a vector $F(x)$. Given this setup, one then looks for the trajectory of a moving particle whose acceleration agrees with this vector field at each point, and it is here that one abuses notation by writing $x=x(t)$ instead of, say, $x=\phi(t)$ (where $x$ is the position of the particle and $\phi$ is the function describing how this position depends on time; it satisfies $\phi'' = F \circ \phi$). $\endgroup$ Aug 9, 2014 at 9:37
  • $\begingroup$ @HansLundmark I just CWed our two comments. If you disagree, please say so. $\endgroup$
    – Did
    Aug 9, 2014 at 12:04
  • $\begingroup$ @Did: Good idea! $\endgroup$ Aug 9, 2014 at 15:03

1 Answer 1

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These two comments might answer the question.


All your interpretations are correct. Your puzzlement is understandable if you are a "pure" mathematician entering the field. After a while, and if one takes care to always know precisely what one is talking about, one gets used to these notations and, even, one can find them useful... – Did


The reason for writing $F(x)$ is that one thinks of it as a given vector field: at each point $x$ of your configuration space there is attached a vector $F(x)$. Given this setup, one then looks for the trajectory of a moving particle whose acceleration agrees with this vector field at each point, and it is here that one abuses notation by writing $x=x(t)$ instead of, say, $x=ϕ(t)$ (where $x$ is the position of the particle and $ϕ$ is the function describing how this position depends on time; it satisfies $ϕ″=F∘ϕ$). – Hans Lundmark

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