I want to solve the problem : $$\dfrac{\partial u}{\partial t}=\frac{1}{\left\vert \nabla u\right\vert ^{p}}{div}\left( \dfrac{\nabla u}{\left\vert \nabla u\right\vert ^{p}}\right) $$$$\frac{\partial u}{\partial t}=\frac{1}{\left\vert \nabla u \right\vert^p} \operatorname{div} \left( \frac{\nabla u}{\left\vert \nabla u\right\vert^p}\right) $$ We search for a self-similarity solution, the general form of which is as follows $$u(x,y,t) = f \left( \xi \right), \text{ with } \xi = \dfrac{(x^{2}+y^{2})^{n}}{a (t)}$$$$u(x,y,t) = f(\xi), \text{ with } \xi = \frac{(x^2+y^2)^n}{a (t)}$$ from which we obtain
$$\alpha \xi =\left( 1-p\right) \left( 2n\right) ^{-2p+2}\left( \left( \frac{1}{2n\left( 1-p\right) }+\frac{2n-1}{2n}\right) \left( \dfrac{df}{d\xi }\right) ^{-2p}+\xi \left( \dfrac{df}{d\xi }\right) ^{-2p-1}\dfrac{d^{2}f}{d\xi ^{2}}\right) $$$$\alpha \xi =(1-p) (2n)^{-2p+2}\left( \left( \frac{1}{2n(1-p)} + \frac{2n-1}{2n}\right) \left( \frac{df}{d\xi }\right)^{-2p}+\xi \left( \frac{df}{d\xi } \right)^{-2p-1} \dfrac{d^2 f}{d\xi^2}\right) $$ Now, I am very confused on how to solve the above equation and find the exact solution of $f(\xi)$, thereby finding the exact solution of $u(x,y,t)$. So my question is, does anyone know how to solve this ordinary differential equation?
