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I want to solve the problem : $$\frac{\partial u}{\partial t}=\frac{1}{\left\vert \nabla u \right\vert^p} \operatorname{div} \left( \frac{\nabla u}{\left\vert \nabla u\right\vert^p}\right) $$ We search for a self-similarity solution, the general form of which is as follows $$u(x,y,t) = f(\xi), \text{ with } \xi = \frac{(x^2+y^2)^n}{a (t)}$$ from which we obtain

$$\alpha \xi =(1-p) (2n)^{-2p+2}\left( \left( \frac{1}{2n(1-p)} + \frac{2n-1}{2n}\right) \left( \frac{df}{d\xi }\right)^{-2p}+\xi \left( \frac{df}{d\xi } \right)^{-2p-1} \dfrac{d^2 f}{d\xi^2}\right) $$ 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?

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Quite easily, a complicated non-linear ODE like that could have no "explicit" solutions. So you should always keep in mind what properties of the solution you would like to understand, even if it cannot be written down explicitly. Fortunately, this ODE has some simplifying features, making it solvable in quadratures, at least presuming that $\alpha$ is a constant (does not depend on $\xi$).

The first observation is that the equation explicitly depends only on $df/d\xi$, so might as well let $g(\xi) = df/d\xi$ and reduce the order of the equation. Then, changing the variable to $\eta = \log(\xi)$ gets rid of the explicit dependence on $\xi$ in the right-hand side. After that, doing some standard manipulations, the equation becomes separable. As a shortcut to these manipulations, you may notice that a general solution will have the form \begin{align*} g(\xi) &= A_1 \xi^{A_2} \left(\xi^{A_3} + c_1\right)^{A_4} , \\ f(\xi) &= c_0 + \int g(\xi) d\xi , \end{align*} where the constants $A_1, A_2, A_3, A_4$ need to be chosen to satisfy the equation and the constants $c_0$, $c_1$ are free integration constants. For example $A_4 = -1/(2p)$, but the other constants will have a more complicated form.

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