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Multivariable Calculus Chain Rule Problem: Wave equation

  1. Nov 12, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that any function of the form

    ##z = f(x + at) + g(x - at)##

    is a solution to the wave equation

    ##\frac {\partial^2 z} {\partial t^2} = a^2 \frac {\partial^2 z} {\partial x^2}##

    [Hint: Let u = x + at, v = x - at]

    2. The attempt at a solution

    My problem with this is not that I haven't been able to solve it. The book's solution is right here:
    hLVjQdS.png

    I began to have trouble when I decided not to use f'(u) and g'(v) instead I used ##\frac {\partial z} {\partial u}## and ##\frac {\partial z} {\partial v}##

    When I did this I got

    ##\frac {\partial^2 z} {\partial x^2} = \frac {\partial^2 z} {\partial u^2} + \frac {2 \partial^2 z } {\partial u \partial v} + \frac {\partial^2 z} {\partial v^2}##

    and

    ##\frac {\partial^2 z} {\partial t^2} = a^2 (\frac {\partial^2 z} {\partial u^2} - \frac {2 \partial^2 z } {\partial u \partial v} + \frac {\partial^2 z} {\partial v^2})##

    And that does not fulfill the condition stated in the beginning.

    So I have a two questions in the end:

    1)What is the difference between writing f'(u) and ##\frac {\partial z} {\partial u}##?

    2)Do ##\frac {2 \partial^2 z } {\partial u \partial v}## and ##- \frac {2 \partial^2 z } {\partial u \partial v}## both evaluate to 0? (That's the only way ##\frac {\partial^2 z} {\partial t^2} = a^2 \frac {\partial^2 z} {\partial x^2}##) or did I do something wrong in my calculations?
     
  2. jcsd
  3. Nov 12, 2013 #2
    One ordinarily would use the notation f'(u) when f is deemed to be a function of a single variable, and ∂f/∂u when f is a function of several variables including u. However, the notation shouldn't make any difference.

    Your problem is that when you went for ∂##^2##z/∂x##^2## you regarded ∂f/∂u as a function of both u and v, which it is not -- it is a function of u alone. Same with v. So you are not going to get any cross terms.

    This may be a good reason why ∂f/∂u isn't a good way to notate f'(u).
     
  4. Nov 12, 2013 #3
    Thank you so much. This has been bothering me for a while. So basically I interpreted this as ##f(u,v)## instead of ##f(u)## and ##g(v)## separately. Right?
     
  5. Nov 12, 2013 #4
    Exactly right.
     
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