Chain Rule and Multivariable Calculus Question

[Stalker_VT]

I think i found the solution to my problem but i was hoping to have someone check to make sure i did not make a mistake.

\xi = x - ct...... (1)

u(t,x) = v(t,\xi)......(2)

Taking the derivative

d[u(t,x) = v(t,\xi)]

\frac{\partial u}{\partial t}dt + \frac{\partial u}{\partial x}dx = \frac{\partial v}{\partial t}dt + \frac{\partial v}{\partial \xi} d\xi.....(3)

\frac{\partial u}{\partial t}dt + \frac{\partial u}{\partial x}dx = \frac{\partial v}{\partial t}dt + \frac{\partial v}{\partial \xi}[ \frac{\partial \xi}{\partial x}dx + \frac{\partial \xi}{\partial t}dt]......(4)

\frac{\partial u}{\partial t}dt + \frac{\partial u}{\partial x}dx = \frac{\partial v}{\partial t}dt + \frac{\partial v}{\partial \xi} \frac{\partial \xi}{\partial x}dx + \frac{\partial v}{\partial \xi}\frac{\partial \xi}{\partial t}dt.....(5)

Question 1
Is it legal to cancel out partial fractions as such

\frac{\partial v}{\partial t}dt + \frac{\partial v}{\partial \xi} \frac{\partial \xi}{\partial x}dx + \frac{\partial v}{\partial \xi}\frac{\partial \xi}{\partial t}dt = \frac{\partial v}{\partial x}dx + \frac{\partial v}{\partial t} dt

Question 2
Is it legal to group like terms to get two separate equations as such:

Using (1) to get \frac{\partial \xi}{\partial x} = 1 and \frac{\partial \xi}{\partial t} = -c we obtain

\frac{\partial u}{\partial t}dt + \frac{\partial u}{\partial x}dx = \frac{\partial v}{\partial t}dt + \frac{\partial v}{\partial\xi}dx - c\frac{\partial v}{\partial \xi} dt

simplifying slightly

\frac{\partial u}{\partial t}dt + \frac{\partial u}{\partial x}dx = [\frac{\partial v}{\partial t} - c\frac{\partial v}{\partial \xi}]dt + \frac{\partial v}{\partial\xi}dx

From this can we conclude the following:

A) \frac{\partial u}{\partial t} = \frac{\partial v}{\partial t} - c\frac{\partial v}{\partial \xi}

B) \frac{\partial u}{\partial x} = \frac{\partial v}{\partial\xi}

by matching parameters in front of dt and dx?

Thank you for any insight

 

[conquest]

I seems like what you're doing is not taking the derivative but calculating the differential of the maps v and u.
In that case if you have coordinates (t,x) and (t,ξ) which are both valid coordinates for the plane. than since u(t,x)=v(t,ξ) all that has happened there is dat you´ve changed coordinates to get from a map u to a map v.

then most of the things you did are correct the grouping i scertainly allowed. since you're basically writing everything in two local frames and equating coefficients.

but bear in mind that ∂/∂ξ = ∂x/∂ξ*∂/∂x + ∂t/∂ξ*∂/∂t