$ -\frac{1}{3}(1-x^2)G''=2(1-\frac{1}{3}G')G, x\in(-1,1)$ only admits zero solution. 
Suppose there is a smooth function $u$ on $[-1,1]$ satisfying
$$
-\frac{1}{3}((1-x^2)u')'+1=e^{2u}.
$$
Prove that $u\equiv 0$.

If we define $G(x)=(1-x^2)u'$, simple conputation gives
$$
-\frac{1}{3}(1-x^2)G''=2(1-\frac{1}{3}G')G, x\in(-1,1),
$$
and resonable to assume $G(1)=G(-1)=0$. Thus it suffices to prove $G\equiv 0$.
I have tried to give some expressions of $\int_{-1}^1G^2$, for example, by the ODE with respect to $G$ and the boundary value it is easy to show
$$
\int_{-1}^1(1-x^2)(G')^2=5\int_{-1}^1G^2,
$$
but I have no idea how to continue.
This problem is actually from the post Axially symmetric solution of a PDE on $S^2$.
Appreciate any help!
 A: Solutions to similar equations:


*

*$ f\left(x\right)\left(1-x^2\right)u'+1=e^{2u}$
Since $u$ is continuous on the compact set $\left[-1,1\right]$ it attains its maximum and minimum at the endpoints of this interval or where its derivative is $0$.
In both cases ($x=\pm1$ and $u’=0$) the differential equation reduces to $1=e^{2u}$ so that $u=0$.
Therefore both minimum and maximum of $u$ are $0$ that implies $u\equiv0$.


*

*$ f\left(x\right)\left(\left(1-x^2\right)u'\right)'+1=e^{2u}$ with $f>0$
Suppose that $u>0$. Then $\left(\left(1-x^2\right)u'\right)'>0$ so that $\left(1-x^2\right)u'$ is increasing. If $u'=0$ at $x_0$ then in a neighborhood of $x_0$ we have that $u'\left(x\right)<0$ for $x<x_0$ and $u'\left(x\right)>0$ for $x>x_0$ (since $\left(1-x^2\right)$ is alway positive for $x\in \left(-1,1\right)$). Then $x_0$ is a local minima. This means that all local maxima are not positive. Similarly, when $u<0$, all local minima are not negative.
Then if $M$ is the maximum and $m$ is the minimum we have $M\leq 0 \leq m$ that is possible only when $m=M=0$. Since $u$ is $0$ at the endpoints of the interval $\left[-1,1\right]$ we have $u\equiv0$.
