I have a silly question about making a change of variable in a PDE. I will explain it using a toy problem.
Suppose that we have the following PDE: $$\frac{\partial u}{\partial t} = k\frac{\partial^2 u}{\partial x^2}$$
where $u(x,t)$ and $k$ is a positive constant. Now, for some reason, I would like to introduce the following change of variable: $$x = \eta \cos t,$$ so the first order partial derivative with respect to $x$ would be: $$\frac{\partial}{\partial x} = \frac{\partial\eta}{\partial x}\frac{\partial}{\partial \eta} + \frac{\partial t}{\partial x}\frac{\partial}{\partial t},$$ however I have doubts about how the partial derivative with respect to $t$ should be: $$\frac{\partial}{\partial t} = \frac{\partial\eta}{\partial t}\frac{\partial}{\partial \eta} + \frac{\partial t}{\partial t}\frac{\partial}{\partial t},$$ or just stays the same $$\frac{\partial}{\partial t} = \frac{\partial}{\partial t} ?$$
I guess that is the first option (using the chain rule) but I am not sure.
PD: One more question, are $t$ and $\eta$ independent? If they are I do not understand why because of $x = \eta(\cos t)$?
Thanks in advance.
The expression $x(\eta,\, t) = \eta (\cos t)$, implies that $x$ is a function of both $\eta$ and $t$. Since I am not sure if your expression means $\eta$ times $\cos t$ or $\eta$ as a function of $t$, I will answer the simpler version.
Suppose $x(\eta,\, t) = \eta\cdot\cos t$ for the PDE, $$ \frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2}, $$ then (as you have already written) the first derivative becomes, $$ \frac{\partial}{\partial x} = \frac{\partial\eta}{\partial x}\frac{\partial}{\partial \eta} + \frac{\partial t}{\partial x}\frac{\partial}{\partial t} = \sec t \frac{\partial}{\partial\eta} - \frac{1}{\eta}\csc t \frac{\partial}{\partial t}. $$ Consequently, one has, $$ \frac{\partial^2}{\partial x^2} = \sec^2t \frac{\partial^2}{\partial\eta^2} + 0\cdot \frac{\partial}{\partial\eta} + \frac{1}{\eta^2}\csc^2t\frac{\partial^2}{\partial t^2} - \frac{1}{\eta^2}\cot t\csc^2t\frac{\partial}{\partial t}. $$ This expression assumes $\eta$, and $t$ are independent, and thus leads to a new (worse looking) PDE in terms of $\eta$, and $t$, i.e., $$ \frac{\partial u}{\partial t} = k \left(\sec^2t \frac{\partial^2 u}{\partial\eta^2} + \frac{1}{\eta^2}\csc^2t\frac{\partial^2 u}{\partial t^2} - \frac{1}{\eta^2}\cot t\csc^2t\frac{\partial u}{\partial t}\right). $$ So in the simplest most reasonable interpretation answering your question directly, the variables $t$, and $\eta$ are independent. Hence, you are correct that the partial derivative in $t$ remains unchanged.
The purpose of the change of variables really mattes. In the above, the only reason I can think of off the top of my head as to why someone would want to make such a change of variables would be to convert the parabolic PDE into a pseudo-elliptic one. On the other hand, $\eta$ may in fact be a constant and the purpose is to convert the parabolic PDE into an ODE, which occurs when $u$ does not depend on $\eta$, i.e. iff $\frac{\partial u}{\partial\eta}=0$.