Is speed a function of position?

Question

Let $x$ be a smooth function from $[0,\infty)$ to $\mathbb{R}^n$ satisfying the following differential equation $x''(t) = f(x(t))$, where $f$ is a smooth function from $\mathbb{R}^n$ to itself. Then is it true that $||x'(t)||$ is a function of $x(t)$? (For example, if we suppose that $x(0) = x(1)$, will $||x'(0)|| = ||x'(1)||$?)

In the case $n=1$, the answer to my question is yes. Here is an easy proof:

$$ \int_{x(0)}^{x(T)} f(x)dx = \int_{0}^{T}f(x(t))x'(t)dt = \int_{0}^{T} x''(t)x'(t)dt = \int_0^T x'(t)d(x'(t)) = \frac{\lVert x'(T) \rVert^2 - \lVert x'(0) \rVert^2}{2} $$

Thus, knowing $x(T)$, we can find explicitly $\lVert x'(T) \rVert$, and therefore $||x'(t)||$ is a function of x(t). If we let $T = 1$ and assume that $x(0) = x(1)$, then we get immediately that $\lVert x'(0) \rVert$ = $\lVert x'(1) \rVert$.

However, I do not know how to process the case $n>1$, or find a counter example.

Thank you.

Extra information: Stated in the language of physics, my question is: if the force acting on an object is a function of its position, will the object's speed (not velocity!) also be a function of its position? I think of this question while I am studying the Work - Kinetic Energy theorem in physics. In fact, the proof I just gave for $n=1$ case is just the mathematical form of the Work - Kinetic Energy theorem.

Answer 1

you will need that $f$ is a conservative force field. For $n=1$ the statement $$0= \int_{x(0)}^{x(1)} f(x)dx$$ is always true because $x(1)=x(0)$ means your integration is empty. However this doesn't have to be the case in higher dimensions. If $f$ is conservativ the proof is quite similiar to what you've done (and only then the statement is true). Let $\gamma(t)$ be a smooth path with $\gamma(0)=x(0)$ and $\gamma(1)=x(1)$ then $$0=\int_\gamma f(x)~ dx=\int_0^1\langle f(\gamma(t)),\dot{\gamma}(t)\rangle ~ dt$$ \begin{align} 0=\int_\gamma f(x)~ dx&=\int_0^1\langle f(\gamma(t)),\dot{\gamma}(t)\rangle ~ dt\\ &=\int_0^1\langle \ddot{\gamma}(t),\dot{\gamma}(t)\rangle ~ dt\\ &=\int_0^1\frac{d}{dt}\left[\frac{1}{2}\|\dot{\gamma}(t)\|^2\right] ~ dt\\ &=\frac{1}{2}\|\dot{\gamma}(1)\|^2-\frac{1}{2}\|\dot{\gamma}(0)\|^2\quad \end{align} this leads to the same conclusion that you made. To answer the question, wheater the velocity is a function of time, you can do the following: assume that there exists an potential $V$ ($-\nabla V= m\ddot{x}$), that does not explicitly depend on time. \begin{align} V(x)+T&= E\\ \implies V(x)+\frac{1}{2}mv^2&= E\\ \implies \|v(x)\|&=\sqrt{\frac{2}{m}\bigg(E-T(x)\bigg)} \end{align} so $\|v(x)\|$ is a function of $x$.

Answer 2

I am going to try to give a physical point of view to this question.

You are considering a differential equation of the form $x'' (t) = f (x(t))$. It corresponds to the Newton's equation of motion, where the acceleration $x''$ is equal to a force field $f (x(t))$. It is crucial to note that $f$ is only a function of the position, and do not depend on its speed. A damping force of the form $f \sim - \lambda v$ do not enter in this situation.

If we assume that the force field $f(x(t))$ is a conservative force, then it derives from a potential energy $\phi$ so that

$$ f (x) = - \nabla_{x} \, \phi $$.

We then take the scalar product of Newton's equation of motion with $x' (t)$ and we notice that it takes the form

$$ \frac{d^{2} x}{d t^{2}} \cdot \frac{d x}{d t} = - \frac{d \phi}{d x} \cdot \frac{\partial x}{\partial t} $$

This can be rewritten as

$$ \frac{d }{d t} \left[ \frac{1}{2} \left| \frac{d x}{d t} \right|^{2} + \phi (x(t)) \right] = 0 $$

We find back that the total energy (kinetic+potential) is conserved along the motion. This immediately shows that at any admissible position $x(t)$, one has

$$ \left| \frac{d x}{d t} \right|^{2} = \frac{E - \phi (x (t))}{2} $$

so that the module of the velocity is only a function of the position, as long as the force is conservative. (This result does not depend on the dimension considered)

A few additional remarks :

• The $1D$ case is degenerate since to any unidimensionnal force field $f(x)$ can be associated a potential, thanks to the fundamental theorem of calculus. For higher dimensions, most of the physical cases considered derive from a potential.

• If you consider a force field of the form $f (x,v)$, for example a damped motion, then you can't express the velocity as a function the position, since energy is no more conserved.

Answer 3

For $F=Ay^N$ where A is a drag coefficient that balances the units in the equation (presumably playing a role akin to that of a higher-dimensional Reynolds number), we can find a special solution using group theory. $$m\ddot{y}=Ay^N$$ is invariant to the stretching group $G(t,y)=(\lambda t, \lambda^\beta y)\lambda_o =1$. Since $dt'=\lambda dt$ and $dy'=\lambda^\beta dy$, $\dot{y}'=\frac{dy'}{dt'}=\frac{\lambda^\beta dy}{\lambda dt}=\lambda^{\beta-1}\dot{y}$. Likewise, $\ddot{y}'=\lambda^{\beta -2}\ddot{y}$. Substituting these values into the DEQ yields $$m\lambda^{\beta -2}\ddot{y}=A\lambda^{N\beta} y^N$$ This results in $\beta -2=N\beta$ or $\beta=\frac{2}{1-N}$. Non-trivial stabilizers for this group include $\mu=\frac{y}{t^\beta}$, $\nu=\frac{\dot{y}}{t^{\beta -1}}$, and $\eta=\frac{\ddot{y}}{t^{\beta -2}}$. By dividing our DEQ by $t^{\beta -2}$ and remembering that $\beta -2=N\beta$, we have the DEQ in stabilizer form: $$m\eta=A\mu^N$$ The Lie algebra between the stabilizers at singularities and saddle points is $\eta=\beta(\beta -1)\mu$. Substituting this into the DEQ results in $$m(\beta^2 -\beta)\mu=A\mu^N$$ Solving this for $\mu$ and substituting $\mu=\frac{y}{t^\beta}$ yields $$y=\bigg(\frac{A}{m(\beta^2 -\beta)}\bigg)^{\frac{\beta}{2}}t^\beta$$ with $\beta=\frac{2}{1-N}$. This solution is easily checked. $$m\ddot{y}=m(\beta^2 -\beta)\bigg(\frac{A}{m(\beta^2 -\beta)}\bigg)^{\frac{\beta}{2}}t^{\beta -2}=A\bigg(\frac{A}{m(\beta^2 -\beta)}\bigg)^{-1}\bigg(\frac{A}{m(\beta^2 -\beta)}\bigg)^{\frac{\beta}{2}}t^{\beta -2}$$ Since $\beta -2=N\beta$, we have $$m\ddot{y}=A\bigg(\frac{A}{m(\beta^2 -\beta)}\bigg)^{\frac{N\beta}{2}}t^{N\beta}=Ay^N$$ Now that we've established that the solution is valid (except for $N=1$, but you've already got that result), we have $$\dot{y}=C_N t^{\beta -1}$$ where $$C_N=\frac{2}{1-N}\bigg(\frac{A(1-N)^2}{2m(1+N)}\bigg)^{\frac{1}{1-N}}$$ Since $\beta -1=\frac{2}{1-N}-1=\frac{1+N}{1-N}$, we have the result that as N becomes very large, $$vt\approx C_N$$ Don't be confused by this result. In cases where there is no acceleration, distance is velocity times total time, so this result does not indicate that an object goes a certain distance and stops, which would be a nonsensical result.

Now to answer your question. Since the dot product of force and distance is work, which decreases the kinetic energy of the particle, we expect the velocity to decrease with distance in every case. What this result shows is that the expected result holds true for all values of N, as least insofar as the chosen DEQ represents a physical reality. For very large N, as time and distance increase, velocity decreases in an almost linear fashion.