# Uniqueness of ODE solution with Dirichlet conditions on the half line

I have this ODE problem $-u''(x)=(1-u^2(x))u(x)$ where $x \in (0,{\infty})$ and $u \in C^2 (0,{\infty}) \cap C [0,{\infty})$ , with the initial condition $u(0)=0$,

I have proved the existence of a solution, but i have trouble proving its uniqueness .

This is a crucial part to complete my thesis,

I would be very thankful if you can help.

• you may want look up gradient equation. these have the total energy $e = \frac12u'^2 + \frac12u^2 - \frac18u^4$ is conserved on the trajectories.
– abel
Feb 16 '15 at 19:22
• That should be $1/4$, not $1/8$. Feb 18 '15 at 18:04

For uniqueness of a second-order differential equation you generally need two initial conditions, typically $u(0) = \ldots$ and $u'(0) = \ldots$. In this case, solutions other than $u = 0$ are $$u(x) = c \sqrt{\dfrac{2}{c^2+1}} \text{sn} \left(\dfrac{x}{\sqrt{c^2+1}},c\right)$$ for arbitrary real $c$, where $\text{sn}$ is a Jacobi elliptic function. In particular, for $c=\pm 1$ this is $\pm \tanh(x/\sqrt{2})$, corresponding to $u'(0) = \pm 1/\sqrt{2}$. If $|u'(0)| > 1/\sqrt{2}$, it seems the solutions blow up at a finite value of $x$, but if $0 < |u'(0)| < 1/\sqrt{2}$ you get a periodic solution with a real value of $c$.
• Thanks a lot, Is there a chance to have uniqueness under these conditions only $u(0)=0$ and $u(x)\to1$ when $x\to+ \infty$? Because i have no conditions on $u'(0)$ and it is stated in a paper that there is a unique solution for the following ODE problem $−u′′(x) = f(u(x)), ∀x ∈ R^{+∗} , u(0) = 0, u′(0) > 0, u′ > 0, 0 < u < 1\,in\,R^{+∗}$ where $f$ be a Lipschitz-continuous function in $[0, 1]$such that $f(0) = f(1) = 0$ and f is nonincreasing in $[1 − δ, 1]$ for some $δ > 0$, $∀ 0 ≤ s < 1$, $\int_{s}^{1} {f(τ )}\,dτ >0$,which is my case taking $f(u)=(1-u^2)u$ Feb 17 '15 at 18:47
• Additional condition on this DE $\lim_{x\to{+\infty}}\,{u(x)}=1$ Feb 17 '15 at 18:53
• Yes, with the additional condition, the only solution is $u(x) = \tanh(x/\sqrt{2})$. Feb 17 '15 at 19:57