Find path length from ODEs I am trying to integrate a system of differential equations. Let $p(t)$ be a path in the unit disk in the complex plane with $p(0)=0$. Write $p(t)=r\exp(i\theta)$ for some $\theta\in[0,2\pi)$ and $r\geq0$. Suppose
$$\frac{d}{dt}p(t)=\exp(i[\theta+\phi(r)])$$
where $r=|p(t)|$ and $\phi(r)=\phi_0(1-r)$ for some constant $\phi_0$. Also define $\frac{d}{dt}p(0)=1$. In other words, $p(t)$ is always rotating according to its distance from the origin. I have two questions:


*

*Is there an explicit formula for $p(t)$?

*What is path length of $p(t)$ from $t=1$ to $T$?
Below is a plot of the vector field and $p(t)$ for $\phi_0=-1.197$.

 A: $$\frac{d}{dt}p(t)=\exp(i[\theta+\phi(r) ])=\cos(\theta+\phi(r))+i\sin(\theta+\phi(r))$$
$$p=r\cos(\theta)+ir\sin(\theta)$$
$$\frac{d}{dt}(r\cos(\theta))+i\frac{d}{dt}(r\sin(\theta))=\cos(\theta+\phi(r))+i\sin(\theta+\phi(r))$$
$$\begin{cases}
\frac{d}{dt}(r\cos(\theta))=\frac{dr}{dt}\cos(\theta)-r\sin(\theta)\frac{d\theta}{dt}=\cos(\theta+\phi(r))\\
\frac{d}{dt}(r\sin(\theta))=\frac{dr}{dt}\sin(\theta)+r\cos(\theta)\frac{d\theta}{dt}=\sin(\theta+\phi(r))
\end{cases}$$
$$\begin{cases}
\cos(\theta)dr-r\sin(\theta)d\theta=\cos(\theta+\phi(r))dt\\
\sin(\theta)dr+r\cos(\theta)d\theta=\sin(\theta+\phi(r))dt
\end{cases}$$
$$\frac{\sin(\theta)\frac{dr}{d\theta}+r\cos(\theta)}{\cos(\theta)\frac{dr}{d\theta}-r\sin(\theta)}=\frac{\sin(\theta+\phi(r))}{\cos(\theta+\phi(r)) }$$
$$\left(-\sin(\theta)\cos(\theta+\phi(r))+\cos(\theta)\sin(\theta+\phi(r))\right)\frac{dr}{d\theta}=r\left(\cos(\theta)\cos(\theta+\phi(r))+\sin(\theta)\sin(\theta+\phi(r))\right)$$
$$\sin(\phi(r))\frac{dr}{d\theta}=r\cos(\phi(r))$$
$$\theta=\int \frac{1}{r}\tan(\phi(r))dr$$
With $\phi(r)=\phi_0(1-r)$ :
$$\theta(r)=\int \frac{1}{r}\tan(\phi_0(1-r))dr$$
This is the equation of the trajectory (in polar coordinates) expressed on the form of a function defined by an integral.
Theoretically in is possible to express $\theta(r)$ on closed form. But this is complicated and the closed form includes a special function (Polylogarithm).
Practically one have to use numerical calculus to continue. A-fortiori for the inverse function $r(\theta)$. 
A: Using the chain rule, we have $ p'(t) = (r'(t) + ir(t)\theta'(t))e^{i\theta(t)} $, so the equation becomes
$$ r'(t) + ir(t)\theta'(t) = e^{i\phi_0(1-r)} $$
where $r(0)=r_0$ and $\theta(0)=\theta_0$. Matching real and imaginary parts, we get
\begin{align}
r' &= \cos(\phi_0(1-r)) \\
\theta' &= \frac{1}{r}\sin(\phi_0(1-r)) 
\end{align}
Then $r(t)$ is given (implicitly) by
$$ t = \int_{r_0}^r \frac{1}{\cos(\phi_0(1-u))}du $$
Since $r'(0)=\operatorname{Re}(p'(0)) = 1$, $r_0$ is given by solving $1 = \frac{1}{\cos(\phi_0(1-r_0))}$
As for the path length, perform the integration
$$ \int_{t_1}^{t_2} \left\vert p'(t) \right\vert dt = \int_{t_1}^{t_2} 1\ dt = t_2 - t_1 $$ 
Since $p'(t)$ has constant magnitude.
