# Chain Rule and Multivariable Calculus Question

1. Jan 11, 2012

### 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

2. Jan 12, 2012

### 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

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