Boundary Value Problem: $u'' + u-u^3 = 0$ Consider the differential equation
$$\frac{d^2u}{dx^2} + u − u^3 = 0$$
where $u'(0) = u(L) = 0$ .
If there is a solution which isn’t identically $0$ which satisfies $u(0) = u_0$, then find a relationship between $L$ and $u_0$.
I derived,
$$L =\int_0^1 \frac{dz}{\sqrt{(1-z^2)-(2/4)u_{0}(1-z^4)}}$$
where $z=u/u_{0}$ and $dz = du/u_{0}$
I was wondering if I was able to/how to approximate the minimal value of $L$ for which a non-zero solution exists.
 A: Multiply by $2u'$ and integrate to get $$u'(t)^2+u(t)^2-\frac12u(t)^4=u_0^2-\frac12u_0^4.$$ The solution curves $(x,x')$ follow the level curves. As $x\to x-\frac12x^2$ is monotonically increasing for $0\le x\le 1$ one concludes $|u(t)|\le|u_0|$, as states with higher potential are impossible to reach. Because of the symmetries each quarter of the oscillation $u(0)=u_0\to0\to-u_0\to0\to u_0$ takes the same time so that the period $T$ can be computed via
\begin{align}
L=\frac{T}4=\int_0^{u_0}\frac{du}{u'}
&=\int_0^1\frac{u_0dz}{\sqrt{u_0^2(1-z^2)-\frac12u_0^4(1-z^4)}}\\
&=\int_0^1\frac{dz}{\sqrt{(1-z^2)-\frac12u_0^2(1-z^4)}}
\end{align}

The smallest oscillation periods $4L\approx 2\pi$ occur for very small $u_0$ where the cubic term is negligible and the ODE in first order the oscillation equation $u''+u=0$. Then for $u_0\approx 0$, and using the substitution $z=\sin(\phi)$, the integral can be approximated as
\begin{align}
L&=\int_0^1\frac{dz}{\sqrt{1-z^2}\sqrt{1-\frac12u_0^2(1+z^2)}}
\\
&=\int_0^{\pi/2}\left[1+\frac14u_0^2(1+\sin^2ϕ)+O(u_0^4)\right]dϕ
\\
&=\frac\pi2+\frac{3\pi}{16}u_0^2+O(u_0^4)
\end{align}
A: There is a closed form expression for the solution of PDE in terms of Jacobi elliptic functions.
Using following relations among elliptic functions ${\rm sn}(z;m)$, ${\rm cn}(z;m)$ and ${\rm dn}(z;m)$ for a given modulus $m$:
$$
\begin{cases}
{\rm cn}^2 &= 1 - {\rm sn}^2\\
{\rm dn}^2 &= 1 - m\, {\rm sn}^2
\end{cases}
\quad\text{and}\quad
\begin{cases}
{\rm sn}' &= {\rm cn}\,{\rm dn}\\
{\rm cn}' &= -{\rm sn}\,{\rm dn},\\
{\rm dn}' &= -m\,{\rm sn}\,{\rm cn}
\end{cases}
$$
It is easy to verify ${\rm sn}(z;m)$ is a solution of the PDE
$$\phi(z)'' + (1+m)\phi(z) - 2m\phi^3(z) = 0$$
With a little bit of algebra and the fact ${\rm sn}(z;m)$ is an odd function in $z$, one find the function
$$u(x) = \pm\sqrt{\frac{2m}{m+1}}{\rm sn}\left(\frac{L-x}{\sqrt{m+1}} ; m\right)$$
is a solution to the PDE $u'' + u - u^3 = 0$ satisfying the initial condition $u(L) = 0$.
As a function in $z$, ${\rm sn}(z;m)$ is doubly periodic function with period $4\,K(m)$ and $2i\,K(1-m)$  where
$$K(m) = \int_0^1 \frac{dt}{\sqrt{(1-t^2)(1-mt^2)}}$$
is the complete elliptic integral of the first kind. 
Along the real axis, ${\rm sn}(x;m)$ looks qualitatively like the sine function. 
It is real for all $x$. It starts off with at $0$ at $x = 0$, increases to its maximum value $1$ at $x = K(m)$, decreases to its minimum $-1$ at $x = 3K(m)$ and then back to $0$ at $x = 4K(m)$. In general, its extrema are located at $(2k+1)K(m)$ for integer $k$. In order for $u'(x)$ to vanish at and only at $x = 0$, we need
$$\frac{L}{\sqrt{m+1}} = K(m) \implies L = K(m)\sqrt{m+1}$$
Together with
$$|u_0| = \sqrt{\frac{2m}{m+1}} \quad\implies\quad m = \frac{u_0^2}{2-u_0^2}$$
We obtain a closed form expression of $L$ for $|u_0| < \sqrt{2}$.
$$
L = K\left(\frac{u_0^2}{2-u_0^2}\right)\sqrt{\frac{2}{2-u_0^2}}$$
With help of a CAS, $L$ has following approximation for small $u_0$.
$$\frac{L}{\pi} \sim \frac12 +\frac{3}{16}u_0^2 + \frac{57}{512}u_0^4 + \frac{315}{4096} u_0^6 + \frac{30345}{524288} u_0^8 + \cdots
$$
