I am considering the PDE
$$u_t = \nabla \cdot (\nabla u^m) + \nabla \cdot(u V)$$
where $V$ is some vector field in $\mathbb{R}^n$, and $x \in \mathbb{R}^n$. I am having a trouble showing that the rescaled solution
$$ \phi = k u(ax,a^2 k^{m-1}t)=: ku(\tilde{x},\tilde{t}) $$
where $k,a,m>0$ are constants, solves the following PDE
$$\phi_t = \nabla \cdot (\nabla \phi^m) + \nabla \cdot(\phi V_{\phi}) \ \ \ \ \ \ \ \star$$
with
$V_{\phi} = a k^{m-1} V(ax)$
Noting that $\partial x/ \partial \tilde{x} = 1/a$ and $\partial t/ \partial \tilde{t} = 1/a^2k^{m-1}$, we have
$$\phi_{\tilde{t}} = \phi_{t} \cdot \dfrac{\partial t}{ \partial \tilde{t}} = \phi_{t} \cdot \dfrac{1}{a^2k^{m-1}} $$
$$\nabla_{\tilde{x}}\phi^m = \nabla_{x}\phi^m \cdot \dfrac{\partial x}{ \partial \tilde{x}} = \nabla_{x}\phi^m \cdot \dfrac{1}{a} $$
$$\nabla_{\tilde{x}} \cdot(\nabla_{\tilde{x}}\phi^m) = \nabla_{\tilde{x}} \cdot\left(\nabla_{x}\phi^m \cdot \dfrac{1}{a}\right) = \dfrac{1}{a^2} \sum_{i=1}^n \dfrac{\partial}{\partial x_i} [\nabla_{x}\phi^m]_i = \dfrac{1}{a^2} \nabla_{x} \cdot(\nabla_{x}\phi^m) $$
$$\nabla_{\tilde{x}} \cdot(\phi V) = \nabla_{\tilde{x}}\phi \cdot V + \phi \nabla_{\tilde{x}} \cdot V = \dfrac{1}{a}\nabla_{x}\phi \cdot V + \dfrac{1}{a} \phi (\nabla_{x} \cdot V) = \dfrac{1}{a}\left[ \nabla_{x} \cdot(\phi V) \right]$$
When I substitute into the original PDE I get:
$$\phi_t - \nabla \cdot (\nabla \phi^m)- \nabla \cdot(\phi V) = a^2k^{m-1} \phi_{\tilde{t}} - a^2 \nabla_{\tilde{x}} \cdot(\nabla_{\tilde{x}}\phi^m) - a \nabla_{\tilde{x}} \cdot(\phi V) = 0$$
which I don't see how it is related to $\star$. Does anyones see if I made a mistake or another thing I have to do?
Thanks
Note that the expression requires you to compute $\phi_{t}$, not $\phi_{\tilde{t}}$. Similarly, the gradient terms of $\phi$ are in terms of $x$ not $\tilde{x}$. Using $(X, T)$ rather than $(\tilde{x}, \tilde{t})$, computing derivatives we find
\begin{align} \phi_{t} &\to (k) (a^{2} k^{m-1}) u_{T} \\ &= k^{m} a^{2} u_{T} \\ \nabla \cdot (\nabla \phi^{m}) &= m (m-1) \phi^{m-2} \nabla \phi \cdot \nabla \phi + m \phi^{m-1} \nabla^{2} \phi \\ &\to m (m-1) (k^{m-2} u^{m-2}) (a^{2} k^{2} \nabla_{X} u \cdot \nabla_{X} u) + m (k^{m-1} u^{m-1}) (a^{2} \nabla^{2}_{X} u) \\ &= k^{m} a^{2} \nabla_{X} \cdot \nabla_{X} u^{m} \\ \nabla \cdot (\phi V_{\phi}) &= \phi \nabla \cdot V_{\phi} + V_{\phi} \cdot \nabla \phi \\ &\to (k u) (k^{m-1} a^{2} \nabla_{X} \cdot V) + (k^{m-1} a V) \cdot (a k \nabla_{X} u) \\ &= k^{m} a^{2} \nabla_{X} \cdot (u V) \end{align}
and hence
$$0 = \phi_{t} - \nabla \cdot (\nabla \phi^{m}) - \nabla \cdot (\phi V_{\phi}) = k^{m} a^{2} [u_{T} - \nabla_{X} \cdot (\nabla_{X} u^{m}) - \nabla_{X} \cdot (u V) ]$$
which holds as the RHS is just the PDE that $u$ solves but in the new coordinates $(X, T)$.
However, it might be easier to derive the result here rather than show it using a given result. Working with $n = 1$ first for simplicity, let
\begin{align} u(x, t) &\to \lambda_{u} u(\lambda_{x} x, \lambda_{t} t) := \lambda_{u} u(X, T), & V(x, t) \to \lambda_{v} V(\lambda_{x} x, \lambda_{t} t) := \lambda_{v} V(X, T) \end{align}
then
\begin{align} u_{t} &\to \lambda_{t} \lambda_{u} u_{T}, & \partial_{x}^{2} (u^{m}) &\to \lambda_{u}^{m} \lambda_{x}^{2} \partial_{X}^{2} (u^{m}), & \partial_{x} (u V) &\to \lambda_{u} \lambda_{v} \lambda_{x} \partial_{X} (u V) \end{align}
and so for invariance we require
$$\lambda_{t} = \lambda_{u}^{m-1} \lambda_{x}^{2} = \lambda_{x} \lambda_{v}$$
Setting $\lambda_{u} = k, \lambda_{x} = a$ then gives the results for $\lambda_{t}, \lambda_{v}$.
When $n \ge 2$ we just apply the same approach to each coordinate $x_{i}$ and vector field component $V_{i}$, using Einstein summation notation where repeated indices are summed over
\begin{align} u_{t} &\to \lambda_{t} \lambda_{u} u_{T}, & \nabla \cdot \nabla (u^{m}) &\to \lambda_{u}^{m} \lambda_{x_{i}}^{2} \partial_{X_{i}}^{2} (u^{m}), & \nabla \cdot (u V) &\to \lambda_{u} \lambda_{v_{i}} \lambda_{x_{i}} \partial_{X_{i}} (u V_{i}) \end{align}
and so, for each $i = 1, \dots, n$, the following relations must hold
$$\lambda_{t} = \lambda_{u}^{m-1} \lambda_{x_{i}}^{2} = \lambda_{x_{i}} \lambda_{v_{i}}$$
from which we get the same result.