$ R(t) = R_0 - \int_0^t v(T) dT , v(t) = \tanh(\int_0^t \frac{C dT}{R^2(T)})$ ??

Question

Let $R_0,C$ be positive real constants and consider the following differential equation :

$$ R(t) = R_0 - \int_0^t v(T) dT $$ $$ v(t) = \tanh(\int_0^t \frac{C dT}{R^2(T)})$$

How to solve this ? How to simplify it or change it in a simpler differential equation ? How to solve it numerically ?

I considered the special functions :

$$ \frac{d B(x)}{d x} = - \tanh(\frac{C}{B(x)^2}) $$ $$ \frac{d C(x)}{d x} = - \frac{1}{\tanh(\frac{C}{x^2})} $$

( constants of integration may be picked as desired )

As possible helping functions but not sure if they are related.

The functional inverses of $R(t),v(t)$ are also of interest to me.

In particular how do we find $y$ such that

$$ R(y) = 0 $$

??

I assume iterative methods exists ?

Answer 1

What you presented are integral equations. You can transform them into differential equations, \begin{align} R'(t)&= -v(t),& R(0)&=R_0,\\ v'(t)&=(1-v(t)^2)\frac{C}{R(t)^2},&v(0)&=0. \end{align} This now can be solved numerically like any other first order system. One has to observe for the singularity at $R(t)=0$.