# Uniqueness of the solution of non-linear ODE of second order

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?

## 1 Answer

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