Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Provided some initial point x(0), how do I convert the function for velocity vs. position, v(x), into a function for position vs. time, x(t), with time derivative v(x(t))? Constant acceleration is not guaranteed. Surely this must always be possible?

share|improve this question
Short answer: integration / solving a differential equation. Long answer is first a bunch of questions: Is your "position" x or p(t) just a real number — that is, do you have a particle moving along a line? Also, is $v(x)$ guaranteed to correspond to some actual motion? Do you know the position $p(t)$ for at least one time $t$? – ShreevatsaR Jul 30 '11 at 12:47
@ShreevatsaR, the particle is moving along a line, so the position should be a real number. Also we know the initial position of the particle, p(0). And it is safe to assume that v(x) is a smooth function. – B.M. Jul 30 '11 at 12:53
@ShreevatsaR, actually, I would also be interested in the case where the particle is moving in more than one dimension and we are provided a velocity vector. – B.M. Jul 30 '11 at 12:55
I'm worried that you're using both $x$ and $p$ to mean position. I'd phrase the question this way (at least in one dimension): there's an unknown function $x(t)$, whose time-derivative is called $v(t)$, and we know $v(x)$; how do we recover $x(t)$? – Gerry Myerson Jul 30 '11 at 13:27
@Gerry, thanks, I rewrote the question! – B.M. Jul 30 '11 at 13:41

3 Answers

up vote 6 down vote accepted

In one dimension and assuming that the velocity is never zero, the velocity at time $t$ is $v(x(t))$ and also $\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}x(t)$, hence $\mathrm{d}t=\displaystyle\frac{\mathrm{d}x}{v(x)}$, which is solved by $$ t=\int_{x(0)}^{x(t)}\frac{\mathrm{d}z}{v(z)}. $$ This can be rewritten as follows: for every $q$, let $$ U(q)=\int_{0}^{q}\frac{\mathrm{d}z}{v(z)}, $$ then, for every nonnegative $t$, $t=U(x(t))-u_0$ with $u_0=U(x(0))$, hence $$ x(t)=U^{-1}(t+u_0). $$ Example If $v(x)=\mathrm{e}^{−x}$, then $U(q)=\mathrm{e}^{q}−1$ hence $U^{−1}(s)=\log(1+s)$ and $u_0=\mathrm{e}^{x(0)}−1$, which gives $x(t)=\log(\mathrm{e}^{x(0)}+t)$ for every nonnegative $t$.

share|improve this answer
If p(t) is part of the definite integral for the function U(q), how are we able to ultimately find p(t)? – B.M. Jul 30 '11 at 14:04
As I wrote: by inverting the function $U$. For example, if $v(x)=e^{-x}$, then $U(q)=e^q-1$ hence $U^{-1}(s)=\log(1+s)$ and $u_0=e^{p(0)}-1$, which gives $p(t)=\log(e^{p(0)}+t)$. (By the way, $p(t)$ is not part of the definition of $U$.) – Did Jul 30 '11 at 14:36
Note that the two comments above refer to a previous version of the notations in the question, where roughly speaking, one used p for x. – Did Jul 30 '11 at 15:12
very clear now, thank you! – B.M. Jul 30 '11 at 15:33

This is an attempt to solve the 2-D case. Let $p = (x,y)$ be a general position. Then we are given constraints of the form: $$ \begin{eqnarray*} \frac{\mathrm{d}x}{\mathrm{d}t}=f(x,y) \\ \frac{\mathrm{d}y}{\mathrm{d}t}=g(x,y) \end{eqnarray*} $$ Then applying the chain rule to eliminate the variable $t$, $$ \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{g(x,y)}{f(x,y)}. $$ This is a quite general differential equation and I think there is no neat way to write out the solution. But assuming we could solve this, we get $y$ as a function of $x$. Now, plugging this in the first constraint $\frac{\mathrm{d}x}{\mathrm{d}t} = f(x, y(x))$, we get the 1-D version of OP's problem. Now, we can use the solution of @DidierPiau to write down $x(t)$. The general solution is then $p = (x(t), y(x(t)))$.

The above explanation was very handwavy and I do not know what details are needed to make this rigorous. (E.g.:, should I worry about the existence of solutions to the differential equation? Could there be multiple solutions? and so on).

share|improve this answer

$[x+a]/e=[x+b]/f=[z+c]/g=[w+d]/h$; $\{x,y,z,w\}=$ space-time or space-velocity?

share|improve this answer
What do these variables mean and how do these formulas relate to the question? – robjohn Dec 26 '12 at 13:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.