Take a as the reference system. Divided into two small problems.

Find the velocity v(x) of point e with distance a as x:

Because the midpoint of two points is always the midpoint, the velocities of two AE points relative to the midpoint must be the same, v(x)-v(x/2) = 0-(-v(x/2)), that is, v (x) = 2v (x/2) = 2 n * v (x/2 n), and x > =0.

Lim(n tends to positive infinity) V (L)/L = Lim V (L/2 N)/(L/2 N) = V/L. If v(x) is derivable at 0 (it seems to be a philosophical problem, but it can't be proved) and makes L/2 n = z, then when N tends to positive infinity, Z tends to 0, Lim V.

V (x) = x * v (x/2 n)/(x/2 n), let x/2 n = p, fix x, and n tends to positive infinity, then p tends to 0, and v (x) = x * limv (p)/p = v/l * x v/l * x.

Find the motion equation of insects;

When the worm moves to point E, where A is far away from X, the speed of E relative to A is V _ EA = V/L * X.

The speed of the worm relative to a is v _ buga = v _ buge+v _ ea = c+v/l * x = dx/dt, and the initial condition x(t=0) = 0. Exp(V/L*t) = 1+V/C/L*x when solving this first-order differential equation.

Let x = L+V*t, and solve it simultaneously with the above formula: exp (vt/l) =1+v/c+v * vt/c/l let Vt/L = T, v/c = a (> 0), the solution exp (t) =1+a+a * t.

This equation has only one positive real number solution, but it is a LambertW. So the symbolic solution is enough here, and then the numerical method (Newton iteration or something) is used when there are specific values.