Inverse of $r sin(\omega t) + v t$?

Question

I am wondering if there is an inverse for this function, $x(t)=r sin(\omega t) + v t$. The inverse function theorem suggests that an inverse for this function does exist, although it may have to be piecewise defined around wherever $x'(t)=0$, that is, wherever $t=\frac{1}{\omega}arccos(\frac{v}{r\omega})$.

Here is my progress so far:

Differentiate implicitly considering $t$ as a function of $x$, giving $1=vt'-r\omega sin(\omega t)t'$.

Making the substitutions $u=\omega t$, $\beta=\frac{\omega}{v}$, $\rho=\frac{r\omega}{v}$ gives the nicer form $u'=\beta+\rho u'sin(u)$.

After some derivatives with respect to $u$ and backwards substitution I obtained the polynomial (?) differential equation $y''y=3(y')^2+y^4-\alpha y^5$ where $y=u'$ and $\alpha=\frac{1}{\beta}$. I don't know any methods for solving this.

If there is a particular method for solving this, I would be more interested in that method than in the actual solution. For demonstration purposes it might be easier to find the inverse of something like $sin(t)+t$ instead.

A solution to this problem in general could give a mathematically precise time of impact calculation for a rotating body with constant velocity.

Answer 1

By inspecting the derivative, we get a condition on the existence: $$ x'(t) = r \omega \mathrm{cos}(\omega t) + v \geq 0 \Leftrightarrow v \geq |r \omega| $$ or if $v$ is negative: $$ x'(t) \leq 0 \Leftrightarrow v \leq -|r \omega| $$ So under those conditions the inverse of $x(t)$ exists. However the inverse function is not "nice" in the sense that it can't be expressed as a combination of the elementary functions. That means that you can not find an explicit formula for the inverse function.

EDIT:

To actually prove that a function is not elementary is not a trivial task and I'm not sure if a proof exists for the function in question.

To answer the question in the comment: The inverse function can be expressed as a Taylor series, which is an infinite polynomial (locally if $|v|=|r \omega|$, globally if $|v|>|r \omega|$). I will omit the proof but it is a result of complex analysis (both the function and it's inverse are holomorphic).

As for the differential equation, I don't know any methods $-$ save for numerical $-$ to solve such equations. And since we know that the answer is not going to be very nice, the best you could hope is for some kind of integral representation for the answer. I personally think trying to solve the differential equation will not be worthwhile.

Answer 2

Note that there is a big difference between the questions "is there an inverse function" and "find an inverse function". Starting with a different example, if $$f(t)=\sin t+2t$$ then $$f'(t)=\cos t+2\ ,$$ and this is always positive, so the Inverse Function Theorem says that $f$ has an inverse. However to find the inverse means in effect to solve $$y=\sin t+2t$$ for $t$ in terms of $y$, and it is going to be difficult (to say the least!) to find an explicit formula. For your later example $$f(t)=\sin t+t$$ you can do something similar: in this case it is possible for the derivative to be zero, but only at isolated points, so with a bit of care you can show that the inverse exists. However, once again, to find a formula for the inverse is a very different problem. For your original question $$x(t)=r \sin(\omega t) + v t$$ the derivative $$x'(t)=r\omega\cos(\omega t)+v$$ could be always positive, or always negative, or sometimes one and sometimes the other. It depends on the values of $r,v$ and $\omega$. So unless you have some further information about these, I don't think it is possible to answer your question.