How do I show the following ODE as a unique solution in $(0,1)$? This question has been put on MSE for a while and I move it here for a better luck.
Given the ODE
$$
-v''(x)+\frac{v(x)}{(1+v^2(x))^2}+v(x)=1
$$
satisfies the condition $x\in(0,1)$, $v(0)=v(1)=1$, and $v(\cdot)$ is symmetric with respect to $1/2$.

I am wondering that can we determine the solution $v$ uniquely such that the above condition satisfies? It looks to me that the solution of this ODE has a "wall" shape, but this ODE is not linear nor any standard form so I am not quiet sure about it...although the two boundary condition is given. Moreover, I know $v$ is non-negative and quasi-convex.
Typically, by quasi-convexity and symmetricity, we have $v$ is monotone decreasing in $(0,1/2)$ and monotone increasing in $(1/2,1)$.
However, I am still not so sure about the uniqueness...
Thank you!
 A: Write the quasilinear, second-order problem as:
$$
-v''(x) + f(v(x)) = 0 \;, \quad v(0) = v(1) = 1
$$ 
Since $f$ is smooth and $ f'(s) > 1/2 $ for all real inputs $s$,
you can use Nagumo's theory of upper and lower solutions to get the existence of a solution which is bounded and unique.  To read more about this, see


*

*R. E. O’Malley. Singular perturbation methods for ordinary
differential equations. Springer-Verlag, Berlin, 1991.

*K. W. Chang and F. A. Howes. Nonlinear singular perturbation phenomena: theory and application. Springer-Verlag, Berlin, 1984.
A: By conservation of energy for your equation, the quantity $\dot v^2-P(v)$ is constant, with
$$ P(y):= {(1-y+y^2)^2\over 1+y^2} = -{1\over 1+y^2}-2y+y^2+2$$
(just derive to check). Moreover, for solutions $v(t)$ to your equation,  symmetry w.r.to $t=1/2$ is equivalent to the condition $\dot v(1/2)=0$  (one implication is obvious, for the other observe that $v(1-t)$ is a solution too, and use the uniqueness of the Cauchy problem). Therefore, putting $v(1/2)=\lambda$, one has $\dot v(t)^2=P(v(t))-P(\lambda)\le 0$ for all $0\le t\le 1/2$, hence $P(\lambda)$ is the minimum of $P(y)$ in the interval $[\lambda, 1]$, which means that $\lambda$ lies between the minimum point of $P(y)$ on $[0,1]$, $\lambda_*=0.6823..$, and $1$. 
Since you want the solution to be decreasing on $[0,1/2]$, the first order equation
$$\dot v(t)= -\sqrt{P(v(t))-P(\lambda) },\qquad 0<t<1/2 $$
characterizes all symmetric solution which are decreasing on $[0,1/2]$, with $v(1/2)=\lambda$. Solving the first order equation, the condition $v(0)=1$ thus translates into an equation for $\lambda$:
$$\int_{\lambda}^{1} {dy \over\sqrt{P(y)-P(\lambda)  } }={1\over 2}.$$
The integral on the LHS is strictly decreasing from $+\infty$ to $0$ w.r.to $\lambda$ in the  interval $(\lambda_*, 1]$, whence the value for the solution $\lambda=0.971..$, that identifies the unique solution of the initial problem. 
$$*$$
Rmk. Just existence and uniqueness are immediate from the variational nature of the equation, which is the Euler-Lagrange equation for the functional $J(u):=\int_0^1\big( \dot u^2+P(u)\big)dt$. Since $f(x):=x+{x\over 1+x^2}-1$ is an increasing function with linear growth, its antiderivative  ${1\over 2}P(x)$  is a  convex function with quadratic growth. So $J$ is a  strictly convex, coercive functional on the affine space $H^1_0(0,1)+1$, and  has a unique critical point $v $ which is symmetric by uniqueness, as $v(1-t)$ is also a minimizer. 
