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Chain Rule and Multivariable Calculus Question

  1. Jan 11, 2012 #1
    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
     
  2. jcsd
  3. Jan 12, 2012 #2
    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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