I am looking how to apply a change the variables to the heat equation. It may be quite simple but I want to make sure that I did it correctly (or find my error if I made one). The equation is given as
$$\frac{\partial T} {\partial t} = \alpha \big(\frac{\partial^2T}{\partial r^2} + \frac{2}{r} \frac{\partial T}{\partial r}\big) $$
The change of variable is given by $ u = T r$.
Therefore I started by deriving the first and second derivative $ \frac{\partial u}{\partial r} = T$ and $\frac{\partial^2 u}{\partial r^2} = 0$.
Than I substituted it into the heat equation by
$\frac{\partial} {\partial t} \frac{\partial u}{\partial r}= \alpha \bigg(\frac{\partial^2 }{\partial r^2} \frac{\partial u}{\partial r} + \frac{2}{r} \frac{\partial }{\partial r} \frac{\partial u}{\partial r}\bigg) $
$\frac{\partial^2 u}{\partial r \partial t} = \alpha \big(\frac{\partial^3 u}{\partial r^3} + \frac{2}{r} \frac{\partial^2 u}{\partial r^2}\big) $
For the next steps I am not sure whether I did it correctly. I use the definition of the second derivative and also simplified the expression by dividing with $\frac{\partial u}{\partial r}$
$\frac{\partial^2 u}{\partial r \partial t} = \alpha \big(\frac{\partial^3 u}{\partial r^3} + \frac{2}{r} 0\big) $
$\frac{\partial u}{\partial t} = \alpha \frac{\partial^2 u}{\partial r^2}$
The solution should be correct, but evidently the last two steps are not.
I think you are meant to assume $T$ also depends on $r$, so using product rule the partial derivatives should be $$ \frac{\partial u}{\partial r} = r\frac{\partial T}{\partial r} + T, \quad \frac{\partial^2 u}{\partial r^2} = r\frac{\partial^2 T}{\partial r^2} + 2\frac{\partial T}{\partial r}, \quad \frac{\partial u}{\partial t} = r\frac{\partial T}{\partial t} $$ $$ \implies \frac{\partial T}{\partial r} = \frac{1}{r}\left[\frac{\partial u}{\partial r} - T \right], \quad \frac{\partial^2 T}{\partial r^2} = \frac{1}{r}\left[\frac{\partial^2 u}{\partial r^2} - 2\frac{\partial T}{\partial r} \right], \quad \frac{\partial T}{\partial t} = \frac{1}{r}\frac{\partial u}{\partial t}. $$ Now plugging these in you get
\begin{align*} \frac{1}{r}\frac{\partial u}{\partial t} &= \alpha\left( \frac{1}{r}\left[\frac{\partial^2 u}{\partial r^2} - 2\frac{\partial T}{\partial r} \right] + \frac{2}{r^2}\left[\frac{\partial u}{\partial r} - T \right] \right)\\ &= \alpha\left( \frac{1}{r}\left[\frac{\partial^2 u}{\partial r^2} - \frac{2}{r}\left[\frac{\partial u}{\partial r} - T \right] \right] + \frac{2}{r^2}\left[\frac{\partial u}{\partial r} - T \right] \right)\\ &= \frac{\alpha}{r} \frac{\partial^2 u}{\partial r^2}\\ \implies \frac{\partial u}{\partial t} &= \alpha \frac{\partial^2 u}{\partial r^2}. \end{align*}
Also, for future reference, usually we avoid dividing by $\frac{\partial u}{\partial r}$ because $u$ or its derivatives may be zero somewhere in the domain.