Little of the following may make sense to you, but I hope it does. I am starting out with PDEs, and there are particular 'tricks' in my notes that are used to reduce the dimensions of a PDE. This is done through variable substitution.
My question concerns the finding of new 'derivative' terms? Not sure what they are called. Here is an example.
Suppose I have the following term in my PDE: $\frac{\partial U}{\partial t}$ ($t$ stands for time), and I am making a following substitution: $\tau = T-t$, where $T$ is just a constant. Clearly: $\frac{\partial}{\partial t}=\frac{\partial \tau}{\partial t} \frac{\partial}{\partial \tau}$ which leads to the following: $\frac{\partial}{\partial t} = - \frac{\partial}{\partial \tau}$, because $\frac{\partial \tau}{\partial t} = -1$. But then I struggle to process the following.
Now suppose there is translation of the co-ordinate system such that we are transforming from $(\eta,\tau)$ to $(x,\tau^{'})$. Now, in my notes I have that:
$$\frac{\partial}{\partial \tau} = \frac{\partial x}{\partial \tau} \frac{\partial}{\partial x} + \frac{\partial \tau^{'}}{\partial \tau} \frac{\partial}{\partial \tau^{'}}$$
where $x=\eta+(r-\frac{1}{2} \sigma^2)\tau$ and $\tau=\tau^{'}$
And I do not follow why we suddenly have additional terms in that partial derivative expression. Is that because both changes of variable contain $\tau$?
Then why is this the case, there is no $\eta$ in both expressions:
$$\frac{\partial}{\partial \eta} = \frac{\partial x}{\partial \eta} \frac{\partial}{\partial x} + \frac{\partial \tau^{'}}{\partial \eta} \frac{\partial}{\partial \tau^{'}}$$