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Let $n$ be an integer with $n>3$ and $f \colon [0,\infty ) \to \mathbb{R}$ be a solution of $t^{1-n}(t^{n-1}f'(t))'=f(t)|f(t)|^{\frac{4}{n-2}}$ with initial values $f(0)=a$ and $f'(0)=0$. Then, is the solution is unique?

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Hint:

$t^{1-n}(t^{n-1}f'(t))'=f(t)|f(t)|^\frac{4}{n-2}$

$\dfrac{d^2f}{dt^2}-\dfrac{1-n}{t}\dfrac{df}{dt}=f|f|^\frac{4}{n-2}$

This belongs to a modified Emden–Fowler equation according to http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=378.

Let $r=t^{2-n}$ ,

Then $\dfrac{df}{dt}=\dfrac{df}{dr}\dfrac{dr}{dt}=(2-n)t^{1-n}\dfrac{df}{dr}$

$\dfrac{d^2f}{dt^2}=\dfrac{d}{dt}\left((2-n)t^{1-n}\dfrac{df}{dr}\right)=(2-n)t^{1-n}\dfrac{d}{dt}\left(\dfrac{df}{dr}\right)+(2-n)(1-n)t^{-n}\dfrac{df}{dr}=(2-n)t^{1-n}\dfrac{d}{dr}\left(\dfrac{df}{dr}\right)\dfrac{dr}{dt}+(2-n)(1-n)t^{-n}\dfrac{df}{dr}=(2-n)t^{1-n}\dfrac{d^2f}{dr^2}(2-n)t^{1-n}+(2-n)(1-n)t^{-n}\dfrac{df}{dr}=(2-n)^2t^{2-2n}\dfrac{d^2f}{dr^2}+(2-n)(1-n)t^{-n}\dfrac{df}{dr}$

$\therefore(2-n)^2t^{2-2n}\dfrac{d^2f}{dr^2}+(2-n)(1-n)t^{-n}\dfrac{df}{dr}-(1-n)(2-n)t^{-n}\dfrac{df}{dr}=f|f|^\frac{4}{n-2}$

$(n-2)^2t^{2-2n}\dfrac{d^2f}{dr^2}=f|f|^\frac{4}{n-2}$

$\dfrac{d^2f}{dr^2}=\dfrac{t^{2n-2}f|f|^\frac{4}{n-2}}{(n-2)^2}$

$\dfrac{d^2f}{dr^2}=\dfrac{r^{-\frac{2n-2}{n-2}}f|f|^\frac{4}{n-2}}{(n-2)^2}$

Which reduces to an Emden–Fowler equation.

Let $\begin{cases}f=\dfrac{g}{s}\\r=\dfrac{1}{s}\end{cases}$ ,

Then $\dfrac{d^2g}{ds^2}=\dfrac{s^{-\frac{2}{n-2}-1}g|g|^\frac{4}{n-2}}{(n-2)^2}$

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