I need a check on the following exercise, especially a help in the second point, which I am not able to solve
Consider the ode $$-u'' +u = u^3$$
Given that for all $t \in \mathbb{R}$: $$\frac{1}{2}|u'(t)|^2 - \frac{1}{2}|u(t)|^2 + \frac{1}{4}|u(t)|^4=C$$ with $C \in \mathbb{R}$, show that
- if $\lim_{|t| \rightarrow \infty} u(t)=0$ $(\star)$, then $C=0$.
- Every even solution $u$ for which $(\star)$ holds is such that $|u(0)|^2=2$ or $u(0)=0$
- Show that there are exactly three even solutions to the ODE for which $(\star)$ holds.
- From the asymptot theorem, if the limit is equal to a constant, then $$\lim_{|t| \rightarrow \infty} |u'(t)|=0$$
From $$\frac{1}{2}|u'(t)|^2 - \frac{1}{2}|u(t)|^2 + \frac{1}{4}|u(t)|^4=C$$ taking the limit it follows that $$\lim_{|t| \rightarrow \infty} 0.5 |u'(t)|=C=0$$ hence $C=0$.
- Here's the most difficult point for me. I know that $u(t)=u(-t)$ and $\frac{d}{dt} u(-t) = - u'(-t)$. I'd like to use the identity above and the fact that $C=0$, since the solution satisfies ($\star$), to obtain $$\frac{1}{2} |u(t)|^2 \Bigl( \frac{1}{2}|u(t)|^2 - 1 \Bigr)=0$$ and hence, plugging in $t=0$, I find $$|u(0)|^2 = 2$$ or $$|u(0)|=0$$
but I don't know how to do that.
- I think this point is trivial:
Recasting the system as a first order ODE, I have the vector function $$F(y,y')=[y',y-y^3]^T$$ which is clearly $C^{\infty}$. Therefore, I have at least local existence and uniqueness. So I have just shown above that if I have an even solution, for which $\star$ holds, then I can have $u(0)= \pm \sqrt{2}$ or $u(0)=0$, therefore three possibile initial data, hence three possibile solutions.
Any hint or help in the second point is highly appreciated
Note that since $u'(t) = -u'(-t)$, plugging $t=0$ gives $$u'(0) = - u'(0) \Rightarrow 2u'(0) = 0\Rightarrow u'(0) = 0.$$
From the first part we have $C=0$. Thus $$\frac{1}{2}|u'(t)|^2 - \frac{1}{2}|u(t)|^2 + \frac{1}{4}|u(t)|^4=0.$$ Putting $t=0$ and use $u'(0) = 0$ gives $$ - \frac{1}{2}|u(0)|^2 + \frac{1}{4}|u(0)|^4=0.$$ Solving the above gives $u(0) = 0$ or $|u(0)|^2 = 2$.
For your last part, we have the uniqueness theorem, but you have to be a bit careful on the initial conditions: since you transform it into a first order system, an initial condition is a pair $(u(0), u'(0))$ instead of just $u(0)$. So the three initials conditions are
$$\begin{pmatrix} u(0) \\ u'(0)\end{pmatrix} = \begin{pmatrix} 0 \\ 0\end{pmatrix}, \begin{pmatrix}-\sqrt 2 \\ 0\end{pmatrix}, \begin{pmatrix} \sqrt 2 \\ 0\end{pmatrix}$$ instead of just $\pm \sqrt 2, 0$.