- #1
AxiomOfChoice
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Suppose you start with a function [itex]f(x,y,t)[/itex] which satisfies some partial differential equation in the variables [itex]x,y,t[/itex]. Suppose you make a change of variables [itex]x,y,t \to \xi,z,\tau[/itex], where [itex]\tau = g_\tau(x,y,t)[/itex] and similarly for [itex]\xi[/itex] and [itex]z[/itex]. If you want to know what the differential operators [itex]\partial_t, \partial_x[/itex], and [itex]\partial_y[/itex] look like in these variables, don't you need to do something like
[tex]
\frac{\partial}{\partial t} = \frac{\partial}{\partial \tau}\frac{\partial \tau}{\partial t} + \frac{\partial}{\partial \xi}\frac{\partial \xi}{\partial t} + \frac{\partial}{\partial z}\frac{\partial z}{\partial t} = \frac{\partial}{\partial \tau}\frac{\partial}{\partial t}g_\tau + \frac{\partial}{\partial \xi}\frac{\partial}{\partial t} g_\xi+ \frac{\partial}{\partial z}\frac{\partial}{\partial t}g_z,
[/tex]
and similarly for the other variables?
[tex]
\frac{\partial}{\partial t} = \frac{\partial}{\partial \tau}\frac{\partial \tau}{\partial t} + \frac{\partial}{\partial \xi}\frac{\partial \xi}{\partial t} + \frac{\partial}{\partial z}\frac{\partial z}{\partial t} = \frac{\partial}{\partial \tau}\frac{\partial}{\partial t}g_\tau + \frac{\partial}{\partial \xi}\frac{\partial}{\partial t} g_\xi+ \frac{\partial}{\partial z}\frac{\partial}{\partial t}g_z,
[/tex]
and similarly for the other variables?
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