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.(adsbygoogle = window.adsbygoogle || []).push({});

[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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Chain Rule and Multivariable Calculus Question

**Physics Forums | Science Articles, Homework Help, Discussion**