I have been attempting to perform a change of variables on the radial component of a PDE. We assume (by ansatz) that $f(r)$ is a function that admits at least two derivatives.
Taking a change of variables $r = e^{t}$ with $r_t = e^t = r$ and $t_r = r^{-1}$, we may write the multivariate chain rule for the first derivative as $f_t = f_r r_t = r f_r$. My confusion stems from taking the second derivative to determine a formula for $f_{tt}$. I my first method, I write that
$$(r f_r)_r = f_{tr}$$ $$r f_{rr} + f_r = f_{tt} t_r = f_{tt} r^{-1}$$ yielding $$f_{tt} = r^2 f_{rr} + r f_r$$
I believe that this approach is correct as it directly quotes definitions and does not employ any special properties. For the second method, I try to commute the partial derivatives of $f$ to provide what I had hoped would be a more intuitive derivation.
$$ \begin{align} f_{tt} &= (f_r r)_{t} = f_{rt} r + f_r r_t = f_{rt} r + f_r r \\ &= f_{tr} r + f_r r = (f_r r)_r r + f_r r = r^2 f_{rr} + 2 r f_r \end{align} $$
I believe that the second $r f_r$ term arises from an unjustifiable equivalency $f_{rt} = f_{tr}$. I might also have been careless with my notation and that could have concealed a mistake.
Indeed. The indices commute when they are independent variables, but $r$ is dependent on $t$.
So don't do that. Instead, just apply the chain rule again. $(f_r)_t=(f_r)_r~r_t$
$\qquad\begin{align}f_{tt} &= f_{rt}r+f_r r\\&=f_{rr}r_t~r+f_rr\\&=f_{rr}r^2+f_rr\end{align}$