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Partial derivatives

  1. Mar 1, 2015 #1
    1. The problem statement, all variables and given/known data


    w(u,v) = f(u) + g(v)

    u(x,t) = x - at
    v(x,t) = x + at

    show that:

    [tex]\frac{\partial ^{2}w}{\partial t^{2}} = a^{2}\frac{\partial ^{2}w}{\partial x^{2}}[/tex]

    3. The attempt at a solution

    w(x-at, x+at) = f(x-at) + g(x+at)

    [itex]\frac{\partial }{\partial t}(\frac{\partial w}{\partial t})[/itex]

    [itex](\frac{\partial w}{\partial t}) = \frac{\partial w}{\partial u}\frac{\partial u}{\partial t}[/itex]

    [itex]\frac{\partial w}{\partial u}=f'(u) = f'(x-at)[/itex]

    [itex]\frac{\partial u}{\partial t}=-a[/itex]

    [itex]\therefore \frac{\partial w}{\partial t} = -af'(x-at)[/itex]

    I don't know where to go from here. I'm not even sure if what I've done is correct, I don't know how to properly approach this question.

    thanks for any guidance.
    Last edited: Mar 1, 2015
  2. jcsd
  3. Mar 1, 2015 #2
    Take it step by step.

    First take the left hand side of what you want to prove, that is, the second derivative of w with respect to t.

    Next, do the same with the other side, that is find the second derivative of w with respect to x.

    (This step like the step before this will need you do some differentiation using the chain rule)

    Once here, you can equate both these equations which will be what you set out to prove.

    NOTE: you were on the right track with taking the partial derivative of w w.r.t t but I think you forgot g(v) = g(x+at)
    Last edited: Mar 1, 2015
  4. Mar 1, 2015 #3

    Stephen Tashi

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    I think you meant: w(u,v) = f(u) + g(v)

    You mean:

    [itex] \frac{\partial w}{\partial t} = \frac{\partial w}{\partial u} \frac{\partial u}{\partial t} + \frac{\partial w}{\partial v}\frac{\partial v}{\partial t} [/itex].
  5. Mar 1, 2015 #4
    I tried caltulating the individual components and I did what titas.b said to do. It didn't workout when I tried to show each side is equal.

    for example, how would you do partial w/ partial u? is that equal to f'(u)? maybe my individual compenents are incorrect.
  6. Mar 1, 2015 #5

    Stephen Tashi

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    Yes, or you can denote it [itex] \frac{df}{du} [/itex].

    We can't comment on that unless you show what you did.
  7. Mar 1, 2015 #6
    [itex]\frac{\partial w}{\partial t}=\frac{\partial w}{\partial u}\frac{\partial u}{\partial t}+\frac{\partial w}{\partial v}\frac{\partial v}{\partial t}[/itex].

    [itex]\frac{\partial w}{\partial u} = f'(u)[/itex].

    [itex]\frac{\partial w}{\partial v} = g'(v)[/itex].

    [itex]\frac{\partial u}{\partial t}=-a[/itex].

    [itex]\frac{\partial v}{\partial t}=a[/itex].

    [itex]\therefore \frac{\partial w}{\partial t} = a(g'(x+at) - f'(x-at))[/itex].

    [itex]\frac{\partial w}{\partial x}=\frac{\partial w}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial w}{\partial v}\frac{\partial v}{\partial x}[/itex].

    [itex]\frac{\partial u}{\partial x}=\frac{\partial v}{\partial x}=1[/itex].

    [itex]\therefore \frac{\partial w}{\partial x}=f'(x-at)+g'(x+at)[/itex].

    [itex]\frac{\partial^{2}w}{\partial t^{2}}=a(g''(x+at)-f''(x-at))[/itex].

    [itex]\frac{\partial^{2}w}{\partial x^{2}}=g''(x+at)+f''(x-at)[/itex].

    we want to show that:

    [itex]\frac{\partial^{2}w}{\partial t^{2}}= a^{2}\frac{\partial^{2}w}{\partial x^{2}}[/itex].

    plug in and manipulate and you'll find that it doesn't work.
  8. Mar 1, 2015 #7
    Check your differentiation for w with respect to t for the second order. Everything else seems to be fine
  9. Mar 1, 2015 #8
    I realized my mistake. I rewrote f'(u) as partial f/partial u, and I saw my mistake when taking the 2nd derivative (it was more clear).

    However, it STILL doesn't work.
  10. Mar 1, 2015 #9
    The following equation is where you made your mistake.
    You make a couple of mistakes in arriving at your final result. Please try again, and try to be more careful with the calculus and algebra.

  11. Mar 1, 2015 #10
    yes I know. I fixed it. I rewrote f'(u) as partial f/partial u and I treated it as multiplication with the partial derivatives. You should end up with two different partials at the bottom.

    if you plug it in I don't know how to show LHS = RHS.
  12. Mar 1, 2015 #11
    I get the following:

    [itex]\frac{\partial^{2}w}{\partial t^{2}}=a^2(g''(x+at)+f''(x-at))[/itex]

    Compare this with what you got to see where you went wrong.

  13. Mar 1, 2015 #12

    Stephen Tashi

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    You seem to have computed the derivative of [itex] g'(x + at) [/itex] with respect to [itex] t [/itex] as [itex]g''(x + at) [/itex]. That derivative requires the chain rule. It would be [itex] a\ g''(x + at) [/itex]. The derivative of [itex] f'(x - at) [/itex] also requires using the chain rule: [itex] D_t ( f'(x - at)) = f'(v) \frac{dv}{dt} [/itex]
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