This question has been put on [MSE](https://math.stackexchange.com/questions/1826265/how-can-i-prove-the-uniqueness-of-an-ode-vx-fracvx1v2x2v) 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](https://en.wikipedia.org/wiki/Quasiconvex_function).


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!



How do I show the following ODE as a unique solution in $(0,1)$?
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. 


