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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.

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

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  • $\begingroup$ 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 $f(u)=(1-u^2)u$ $\endgroup$
    – user215931
    Feb 17 '15 at 18:47
  • $\begingroup$ Additional condition on this DE $\lim_{x\to{+\infty}}\,{u(x)}=1$ $\endgroup$
    – user215931
    Feb 17 '15 at 18:53
  • $\begingroup$ Yes, with the additional condition, the only solution is $u(x) = \tanh(x/\sqrt{2})$. $\endgroup$ Feb 17 '15 at 19:57
  • $\begingroup$ Thanks I really appreciate your help, $\endgroup$
    – user215931
    Feb 18 '15 at 17:37
  • $\begingroup$ Is there any references for the uniqueness proof,because ive been asked to give the proof details,its obligatory, i tried hard to find it but i couldnt till now $\endgroup$
    – user215931
    Feb 18 '15 at 17:42

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