Eigenvalue of a given operator If $u_0$ is a positive radial symmetric nontrival solution of 
$$
-\frac{1}{2}\frac{d^2u}{dx^2}+\lambda u -u^3=0
$$
Then how to show $-3\lambda$ is a eigenvalue of 
$$
Lu=-\frac{1}{2}\frac{d^2u}{dx^2}+\lambda u -3u_0^2 u
$$
and the corresponding eigenfunction is $u_0^2$ ?
What I try:
\begin{align}
Lu_0^2
&=-\frac{1}{2}\frac{d^2u_0^2}{dx^2}+\lambda u_0^2 -3u_0^4 \\
&=-(\frac{du_0}{dx})^2 -(u_0 \frac{d^2 u_0}{dx^2}) + \lambda u_0^2 -3 u_0^4  \\
&=-(\frac{du_0}{dx})^2-\frac{1}{2}(u_0 \frac{d^2 u_0}{dx^2})-2u_0^4
+ [-\frac{1}{2}(u_0 \frac{d^2 u_0}{dx^2})+ \lambda u_0^2 - u_0^4]  \\
&=-(\frac{du_0}{dx})^2-\frac{1}{2}(u_0 \frac{d^2 u_0}{dx^2})-2u_0^4
\end{align}
then I don't how to deal the $-(\frac{du_0}{dx})^2$, what should I do ?
 A: Here is my try.  Let $v_0=\frac{du_0}{dx}$.  From
$$-\frac12\frac{d^2u_0}{dx^2}+\lambda u_0-u_0^3=0\implies \frac{d^2u_0}{dx^2}=2\lambda u_0-2u_0^4,$$
we have as you computed
$$Lu_0^2=-v_0^2-u_0\frac{d^2u_0}{dx^2}+\lambda u_0^2-3u_0^4=-v_0^2-u_0(2\lambda u_0-2u_0^4)+\lambda u_0^2-3u_0^4.$$
So $Lu_0^2=-v_0^2-\lambda u_0^2+u_0^4$.  Also from
$$-\frac12\frac{d^2u_0}{dx^2}+\lambda u_0-u_0^3=0\implies-\frac{1}{2}v_0\frac{dv_0}{dx}+\lambda u_0\frac{du_0}{dx}-u_0^3\frac{du_0}{dx}=0,$$
we get by integrating wrt $x$ that
$$-\frac{v_0^2}{4}+\frac{\lambda u_0^2}{2}-\frac{u_0^4}{4}=c$$
for some constant $c$.  So $-v_0^2+2\lambda u_0^2-u_0^4=4c$.  That is,
$$Lu_0^2=-v_0^2-\lambda u_0^2+u_0^4=(-v_0^2+2\lambda u_0^2-u_0^4)-3\lambda u_0^2=4c-3\lambda u_0^2.$$
Presumably, you have some conditions to make $c=0$ (is there anything missing?), because if $c=0$, $Lu_0^2=-3\lambda u_0^2$ as required.
A: This answer is a supplement to Snookie's answer above.  All notations are borrowed from there.  Some information still seems to be missing.  Therefore, I assume that you want a non-periodic solution $u_0$.  (Alternatively, you can require that $u_0$ is a nontrivial solution such that $\lim\limits_{x\to\pm\infty}\,u_0(x)=0$ and the proof is essentially unchanged.)
Assuming that $x$ is the radius, then radial symmetry of $u_0$ implies that $v_0(0)=u_0'(x)=0$.  Therefore, if $s:=u_0(0)$, then from Snookie's answer, we have $$2\lambda s^2-s^4=4c\,.$$
That is,
$$-v_0^2+2\lambda u_0^2-u_0^4=2\lambda s^2-s^4\,.$$
Hence,
$$v_0=\pm\sqrt{(u_0^2-s^2)\,\left(2\lambda-u_0^2-s^2\right)}\,.$$
That is,
$$\frac{1}{\sqrt{(u_0^2-s^2)\,\left(2\lambda-s^2-u_0^2\right)}}\,\left(\frac{\text{d}u_0}{\text{d}x}\right)=\pm1\tag{*}\,.$$
Note from $-v_0^2+2\lambda\,u_0-u_0^4=4c$, we have
$$(u_0^2-\lambda)^2+v_0^2=\lambda^2-4c\,.$$
Thus, if we look at the $(u_0,v_0)$-diagram, then we see that any solution is either trivial or lies in a closed loop.  A nonperiodic solution can only arise if the period of the trajectory in the corresponding closed loop in the $(u_0,v_0)$-diagram is infinite.

Without loss of generality, suppose that $s\geq 0$ (otherwise, note that swapping the sign of $u_0$ also yields a solution).  If $s\neq 0$ or $s^2\neq 2\lambda$, then there are three cases: $s^2<2\lambda -s^2$, $s^2=2\lambda-s^2$, and $s^2>2\lambda-s^2$.  If $s^2<2\lambda-s^2$, then note from (*) that
$$\pm x=\int_{s}^{u_0(x)}\,\frac{1}{\sqrt{(t^2-s^2)\,\left(2\lambda-s^2-t^2\right)}}\,\text{d}t\,.$$
Therefore, $u_0(x)$ lies between $s$ and $\sqrt{2\lambda-s^2}$, and so
$$|x|\leq \int_{s}^{\sqrt{2\lambda-s^2}}\,\frac{1}{\sqrt{(t^2-s^2)\,\left(2\lambda-s^2-t^2\right)}}\,\text{d}t<\infty\,,\tag{#}$$
whence $u_0$ is periodic.  (I omit the proof of (#) here, but it is due to the fact that $\displaystyle\int_0^y\,\frac{1}{\sqrt{t}}\,\text{d}t$ is finite for every $y>0$.)
Similarly, if $s^2>2\lambda-s^2$, then according to (*), we have
$$\pm x=\int_s^{u_0(x)}\,\frac{1}{\sqrt{(t^2-2\lambda+s^2)\,(s^2-t^2)}}\,\text{d}t\,.$$
This means $u_0(x)$ lies between $\sqrt{\max\{0,2\lambda-s^2\}}$ and $s$, making
$$|x|\leq \int_{\sqrt{\max\{0,2\lambda-s^2\}}}^{s}\,\frac{1}{\sqrt{(t^2-2\lambda+s^2)\,(s^2-t^2)}}\,\text{d}t<\infty\,.$$  Ergo, $u_0$ is periodic.  Consequently, $s^2=\lambda$ is the only possibility.
However, there exists a unique solution $u:=u_0$ to
$$-\frac{1}{2}\,\frac{\text{d}^2u}{\text{d}x^2}+\lambda\,u-u^3=0$$
provided that $u_0(0)$ and $u_0'(0)$ is known (this is due to the Picard-Lindelöf Theorem).  In the case that $s^2=\lambda$, we have $s=\sqrt{\lambda}$, and we already have one solution $u_0(x)=\sqrt{\lambda}$ for all $x$.  Therefore, this is the only solution, which is a constant (whence periodic) solution.  Therefore, the assumption that $s\neq 0$ and $s^2\neq 2\lambda$ is false.  However, if $s=0$, then we have another constant (whence periodic) solution $u_0\equiv 0$.  Hence, $s^2=2\lambda$, which gives $c=0$.  This is the only case that yields a non-periodic solution $u_0$.
For $\lambda>0$ and $s=\sqrt{2\lambda}$, we have
$$u_0(x)=\sqrt{2\lambda}\,\text{sech}(\sqrt{2\lambda}\,x)\,.$$
This solution is indeed non-periodic (and $\lim\limits_{x\to\pm\infty}\,u_0(x)=0$).  For $\lambda \leq 0$, there does not exist a nontrivial, non-periodic, radially symmetric solution $u_0$.
