Boundary Value Problem with Unknown Integration Interval In this paper (pg. 15), the propagation of nerve impulses is modeled as, 
$$
y_1' = 3 (y_1 + y_2 - 1/3 y_1^3 - 1.3)
$$
$$
y_2' = -(y_1 - 0.7 + 0.8 y_2) / 3
$$
The period $T$ however is unknown. With change of parameter the equation is changed to,
$$
y_1' = 3 T (y_1 + y_2 - 1/3 y_1^3 - 1.3)
$$
$$
y_2' = -T (y_1 - 0.7 + 0.8 y_2) / 3
$$
$$
T' = 0
$$
where boundary conditions are,
$$
y_1(0) = y_1(1), \quad y_2(0) = y_2(1)
$$
All of this makes sense. But then the problem says "in order to eliminate degenerate solutions $y'(t) = 0$ and sols with $T=0$, we have to use $dy_2 / d\tau (0) = 1$". As a result a new boundary condition is added,
$$ -T(y_1(0) - 0.7 + 0.8 y_2 (0) / 3 = 1$$
I did not understand where this comes from. Why was 1 chosen, not zero, or any other value? How was the new boundary condition obtained?
If I had to guess a non-zero value was chosen to stop $T=0$ being picked as a solution? But doesn't that artificially push the solution certain set of values?
Thanks,
 A: The system is autonomous, thus invariant under shifts of the initial time. This freedom is also reflected in the three components of the state. So to fix a specific solution, one would need another boundary condition that is not related to the periodicity. One could set $y_1(0)=0$ or better $y_1'(0)=0$ or $y_1'(0)=0$. But that is also satisfied by stationary solutions.
So setting $y_2'(0)$ to some fixed non-zero value will prevent finding stationary solutions. However you are right, it seems to be a lucky guess that this value $1$ will give a solution at all. As a periodic solution, the component $y_2$ will have extrema where the derivative is zero, so setting the derivative to some smaller value would appear to give a greater chance to actually find a solution.
As the solution plot shows,

the $y_2$ range includes $[0.5,2]$. This gives by the mean value theorem derivatives for unit parameter $\tau$ including the range $[-1.5,1.5]$. So setting the derivative to $b=1$ is a safe choice to target this solution. As the first plot shows, these different solutions for different choices of $y_2'(0)=b$ give the same curve in the phase space, just started at a different point.
The paper only gives a demonstration of the method, so that it is sufficient that the given parameters indeed lead to the presented solution.
A: Based on this book by Shampine, pg. 174, 
https://www.mobt3ath.com/uplode/book/book-50091.pdf
I was able to solve the non-seperated version of the problem, 
from scipy.integrate import solve_bvp

def fun(x, y, p):
    T = p[0]
    return np.vstack((
        3.0 * T *  (y[0] + y[1] - (y[0]**3)/3.0 - 1.3),
        -T / 3.0 * (y[0] - 0.7 + 0.8*y[1]) 
    ))

def bc(ya, yb, p):
    return np.array( [ ya[0], yb[0], ya[1]-yb[1] ]   )

x = np.linspace(0, 1, 5)
y[0] = np.sin(2 * np.pi * x)
y[1] = np.cos(2 * np.pi * x)
sol = solve_bvp(fun, bc, x, y, p=[2*np.pi])
print (sol.p)

which reports 10.71, the same value shared in the book.
