Suppose I have some nonlinear PDE in two dimensions $x$ and $t$ for the scalar field $u(x,t)$, $$ R(x,t,u,u_x,u_t,u_{xx},u_{xt},u_{tt},\ldots) = 0, \quad t \in [0,\infty), \quad x \in [0, at+b], $$ with $a$ and $b$ positive, some boundary conditions at $x=0$ and $x=at+b$, and some initial condition for $u(x,t)$ at $t=0$.
To address the growing domain I apply the variable transform $$ s = \frac{x}{at+b}, \quad s \in [0,1], $$ and using the chain rule replace derivatives w.r.t. $x$ with those w.r.t. $s$: $$ \frac{\partial}{\partial x}=\frac{1}{at+b}\frac{\partial}{\partial s}. $$ But I can do the same with the $t$-derivatives: $$ \frac{\partial}{\partial t}=\frac{-as}{at+b}\frac{\partial}{\partial s}. $$ Then I can rewrite my residual as an ODE in $s$, with $t$ serving only as a parameter: $$ \hat{R}_{(t)}(s,u,u_s,u_{ss},\ldots)=0. $$
So to numerically solve this PDE, I don't have to do any time integration: merely solve an ODE BVP for each $t$ of interest (probably using each previous timestep solution as a starting guess for the next). But this strikes me as too good to be true.
Am I making a mistake here? If not, is there a name for this type of transformation? Thanks.
You are not using the chain rule correctly. To clarify what is going on, let's use $x, t$ as you originally did and define the new variables $s := \frac{x}{at + b}$ and $u := t$. Then $$\frac{\partial}{\partial x} = \frac{\partial s}{\partial x}\frac{\partial}{\partial s} + \frac{\partial u}{\partial x}\frac{\partial}{\partial u} = \frac{1}{at + b}\frac{\partial}{\partial s} = \frac{1}{au + b}\frac{\partial}{\partial s} $$ as you said but $$\frac{\partial}{\partial t} = \frac{\partial s}{\partial t}\frac{\partial}{\partial s} + \frac{\partial u}{\partial t}\frac{\partial}{\partial u} = \frac{-ax}{(at + b)^2}\frac{\partial}{\partial s} + \frac{\partial}{\partial u} = \frac{-as}{au+b}\frac{\partial}{\partial s} + \frac{\partial}{\partial u}. $$ So you still end up with two partial derivatives to deal with here.