Chain Rule and Multivariable Calculus Question

  • Thread starter Stalker_VT
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  • #1
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Main Question or Discussion Point

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.

[itex]\xi[/itex] = x - ct........................ (1)

u(t,x) = v(t,[itex]\xi[/itex])...............................(2)

Taking the derivative

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

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

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

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

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

[itex]\frac{\partial v}{\partial t}[/itex]dt + [itex]\frac{\partial v}{\partial \xi}[/itex] [itex]\frac{\partial \xi}{\partial x}[/itex]dx + [itex]\frac{\partial v}{\partial \xi}[/itex][itex]\frac{\partial \xi}{\partial t}[/itex]dt = [itex]\frac{\partial v}{\partial x}[/itex]dx + [itex]\frac{\partial v}{\partial t}[/itex] dt

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

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

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

simplifying slightly

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

From this can we conclude the following:

A) [itex]\frac{\partial u}{\partial t}[/itex] = [itex]\frac{\partial v}{\partial t}[/itex] - c[itex]\frac{\partial v}{\partial \xi}[/itex]

B) [itex]\frac{\partial u}{\partial x}[/itex] = [itex]\frac{\partial v}{\partial\xi}[/itex]

by matching parameters in front of dt and dx?

Thank you for any insight
 

Answers and Replies

  • #2
133
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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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