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I need help how to mathematically interpret an ODE (Newton's second law). I used to the ODE in this form: $$ m\ddot x(t)=F(t)\tag{1} $$

However, in another book they wrote: $$ m\ddot x=F(x,\dot x) \tag{2} $$ where $F: \mathbb{R}^n \times \mathbb{R}^n\rightarrow \mathbb{R}^n$.

Questions:

  1. I guess $F(x,\dot x)$ is an abbreviation for $F(x(t),\dot x(t))$, is it correct?

  2. What it the difference between writing $F(t)$ or $F(x,\dot x)$?

  3. What is the meaning of the notation $F: \mathbb{R}^n \times \mathbb{R}^n\rightarrow \mathbb{R}^n$?

Thanks!

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F is obviously force.

The force on an object can always be considered a function of time. In some scenarios it may be independent of position or velocity. (E.g. an object falling in a constant gravitational field experiences a constant force.) In such a case it would be reasonable to give an ODE in the first form for you to solve.

In other scenarios the force is a function of position and velocity. In these cases it is only indirectly a function of time (because position and velocity are functions of time). For example, in a non constant gravitational field the force depends on position, and in the air, resistance depends on velocity. In these scenarios the second form makes more sense.

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  • $\begingroup$ This answer would be more directly helpful if you addressed the three numbered questions that the OP has asked. $\endgroup$ – postmortes May 25 '17 at 13:56
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To your first question, yes. The dot notation $\dot{x}$ is used in ODEs and Physics to indicate differentiation with respect to time, so it can be also written as $\dot{x}(t)$ to make the time-dependence explicit.

The difference between $F(t)$ and $F(\dot{x}, \ddot{x})$ can be one of emphasis: the first emphasizes the the time dependence is the main concern, and the second emphasizes that the ODE is the main concern. If you're interested in the level curves of $F$ though, the choice of variable will determine the curves you draw. $F(t)$ would be a graph of $F$ against $t$, while $F(\dot{x}, \ddot{x})$ would be a vector field represenation (1-d vs 2-d in a sense).

$F:{\mathbb R}^n \times {\mathbb R}^n \rightarrow {\mathbb R}^n$ tells you the domain and range of $F$. Here, $F$ is acting on two variables each of which takes values in ${\mathbb R}^n$ and it produces a result which is also a value in ${\mathbb R}^n$. If you had a constraint, say, that the acceleration $\ddot{x}$ were never negative then you would change the domain to be ${\mathbb R}^n \times [0,\infty]^n \rightarrow {\mathbb R}^n$. This can be extended to as many variables as $F$ needs in order to produce its output (e.g. if you added the third derivative of $x$ as well then F might act on ${\mathbb R}^n \times {\mathbb R}^n \times {\mathbb R}^n$.

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