I was reading https://www.ucl.ac.uk/~ucahhwi/LTCC/sectionB-similarity.pdf. The introduce new variable as for example $s=\varepsilon^b t$ and say that $\partial t = \varepsilon^b \partial s$. I think this is not true. If I interpret $s$ as a function of $t$, $s(t)=\varepsilon^b t$ then \begin{align} \frac{\partial s}{\partial t}= \varepsilon^b \end{align} and so $\partial s=\partial t \varepsilon^b$. Then it should be $\partial t= \varepsilon^{-b} \partial s$.
Am I mistaking?
They are actually right but are using a potentially sloppy notation. When they transform the heat equation $$ \partial_t u=\kappa\,\partial_{xx}u $$ to a new time variable $s=e^bt$ they introduce a new function $v(s,x)=u(t,x)=u(e^{-b}s,x)\,.$ By the chain rule $\partial_s v=\partial_t u\,e^{-b}\,.$ Therefore, $v$ satisfies the equivalent heat equation $$ e^{b}\partial_s v=\kappa\,\partial_{xx}v\,. $$